Page 3 : 45, 2nd Floor, Maharishi Dayanand Marg,, Corner Market, Malviya Nagar, New Delhi - 110017, Tel : 49842349 / 49842350, , No part of this publication may be reproduced in, any form without prior permission of the publisher., The author and the publisher do not take any legal, responsibility for any errors or misrepresentations that, might have crept in. We have tried and made our best, efforts to provide accurate up-to-date information in, this book., , All Right Reserved, , © Copyright, Disha, , Corporate, Office, , DISHA PUBLICATION, , Disha Experts, Sanjeev Kumar Jha, Govind Thakur, Typeset by Disha DTP Team, , www.dishapublication.com, , www.mylearninggraph.com, , Books &, ebooks for, School &, Competitive, Exams, , Etests, for, Competitive, Exams, , Write to us at
[email protected]
Page 4 :
Contents, Physical World, Units and Measurements, , P-1 – P-13, , , , 1., , Topic 1 : Unit of Physical Quantities, Topic 2 : Dimensions of Physical Quantities, Topic 3 : Errors in Measurements, Motion in a Straight Line, , P-14 – P-25, , , , 2., , Topic 1 : Distance, Displacement & Uniform Motion, Topic 2 : Non-uniform Motion, Topic 3 : Relative Velocity, Topic 4 : Motion Under Gravity, Motion in a Plane, , P-26 – P-35, , , , 3., , Topic 1 : Vectors, Topic 2 : Motion in a Plane with Constant Acceleration, Topic 3 : Projectile Motion, Topic 4 : Relative Velocity in Two Dimensions & Uniform Circular Motion, Laws of Motion, , P-36 – P-53, , , , 4., , Topic 1 : Ist, IInd & IIIrd Laws of Motion, Topic 2 : Motion of Connected Bodies, Pulley & Equilibrium of Forces, Topic 3 : Friction, Topic 4 : Circular Motion, Banking of Road, Work, Energy and Power, , P-54 – P-75, , , , 5., , Topic 1 : Work, Topic 2 : Energy, Topic 3 : Power, 6., , System of Particles and Rotational Motion, , , , Topic 4 : Collisions, Topic 1 : Centre of Mass, Centre of Gravity & Principle of Moments, Topic 2 : Angular Displacement, Velocity and Acceleration, Topic 3 : Torque, Couple and Angular Momentum, Topic 4 : Moment of Inertia and Rotational K.E., Topic 5 : Rolling Motion, , P-76 – P-112
Page 5 :
Gravitation, , P-113 – P-130, , , , 7., , Topic 1 : Kepler’s Laws of Planetary Motion, Topic 2 : Newton’s Universal Law of Gravitation, Topic 3 : Acceleration due to Gravity, Topic 4 : Gravitational Field and Potential Energy, Topic 5 : Motion of Satellites, Escape Speed and Orbital Velocity, Mechanical Properties of Solids, , , , 8., , P-131 – P-138, , Topic 1 : Hooke’s Law & Young’s Modulus, Topic 2 : Bulk and Rigidity Modulus and Work Done in Stretching a Wire, Mechanical Properties of Fluids, , , , 9., , P-139 – P-154, , Topic 1 : Pressure, Density, Pascal’s Law and Archimedes’ Principle, Topic 2 : Fluid Flow, Reynold’s Number and Bernoulli’s Principle, Topic 3 : Viscosity and Terminal Velocity, Topic 4 : Surface Tension, Surface Energy and Capillarity, , , 10. Termal Properties of Matter, , P-155 – P-168, , Topic 1 : Termometer & Termal Expansion, Topic 2 : Calorimetry and Heat Transfer, Topic 3 : Newton’s Law of Cooling, , , 11. Termodynamics, , P-169 – P-185, , Topic 1 : First Law of Termodynamics, Topic 2 : Specifc Heat Capacity and Termodynamical Processes, Topic 3 : Carnot Engine, Refrigerators and Second Law of Termodynamics, , , 12. Kinetic Teory, , P-186 – P-198, , Topic 1 : Kinetic Teory of an Ideal Gas and Gas Laws, Topic 2 : Speed of Gas, Pressure and Kinetic Energy, Topic 3 : Degree of Freedom, Specifc Heat Capacity, and Mean Free Path, , , 13. Oscillations, , P-199 – P-218, , Topic 1 : Displacement, Phase, Velocity and Acceleration in S.H.M., Topic 2 : Energy in Simple Harmonic Motion, Topic 3 : Time Period, Frequency, Simple Pendulum and Spring Pendulum, 14. Waves, , , , Topic 4 : Damped, Forced Oscillations and Resonance, Topic 1 : Basic of Mechanical Waves, Progressive and Stationary Waves, Topic 2 : Vibration of String and Organ Pipe, , P-219 – P-234
Page 6 :
Topic 3 : Beats, Interference and Superposition of Waves, Topic 4 : Musical Sound and Doppler’s Efect, P-235 – P-253, , , , 15. Electric Charges and Fields, , Topic 1 : Electric Charges and Coulomb’s Law, Topic 2 : Electric Field and Electric Field Lines, 16. Electrostatic Potential and Capacitance, , , , Topic 3 : Electric Dipole, Electric Flux and Gauss’s Law, P-254 – P-280, , Topic 1 : Electrostatic Potential and Equipotential Surfaces, Topic 2 : Electric Potential Energy and Work Done in Carrying a Charge, Topic 3 : Capacitors, Grouping of Capacitor and Energy Stored in a Capacitor, P-281 – P-311, , , , 17. Current Electricity, , Topic 1 : Electric Current, Drif of Electrons, Ohm’s Law, Resistance and Resistivity, Topic 2 : Combination of Resistances, Topic 3 : Kirchhof ’s Laws, Cells, Termo e.m.f. Electrolysis, Topic 4 : Heating Efect of Current, 18. Moving Charges and Magnetism, , , , Topic 5 : Wheatstone Bridge and Diferent Measuring Instruments, P-312 – P-339, , Topic 1 : Motion of Charged Particle in Magnetic Field, Topic 2 : Magnetic Field Lines, Biot-Savart’s Law and Ampere’s Circuital Law, Topic 3 : Force and Torque on Current Carrying Conductor, Topic 4 : Galvanometer and its Conversion into Ammeter and Voltmeter, P-340 – P-347, , , , 19. Magnetism and Matter, , Topic 1 : Magnetism, Gauss’s Law, Magnetic Moment, Properties of Magnet, Topic 2 : Te Earth Magnetism, Magnetic Materials and their Properties, 20. Electromagnetic Induction, , , , Topic 3 : Magnetic Equipment, P-348 – P-360, , Topic 1 : Magnetic Flux, Faraday’s and Lenz’s Law, 21. Alternating Current, , , , Topic 2 : Motional and Static EMI and Application of EMI, Topic 1 : Alternating Current, Voltage and Power, Topic 2 : AC Circuit, LCR Circuit, Quality and Power Factor, Topic 3 : Transformers and LC Oscillations, , P-361 – P-376
Page 7 :
P-377 – P-388, , , , 22. Electromagnetic Waves, , Topic 1 : Electromagnetic Waves, Conduction and Displacement Current, Topic 2 : Electromagnetic Spectrum, P-389 – P-414, , , , 23. Ray Optics and Optical Instruments, , Topic 1 : Plane Mirror, Spherical Mirror and Refection of Light, Topic 2 : Refraction of Light at Plane Surface and Total Internal Refection, Topic 3 : Refraction at Curved Surface Lenses and Power of Lens, Topic 4 : Prism and Dispersion of Light, Topic 5 : Optical Instruments, P-415 – P-432, , , , 24. Wave Optics, , Topic 1 : Wavefront, Interference of Light, Coherent and Incoherent Sources, Topic 2 : Young’s Double Slit Experiment, Topic 3 : Difraction, Polarisation of Light and Resolving Power, P-433 – P-448, , , , 25. Dual Nature of Radiation and Matter, , Topic 1 : Matter Waves, Cathode and Positive Rays, Topic 2 : Photon, Photoelectric Efect X-rays and Davisson-Germer Experiment, P-449 – P-460, , , , 26. Atoms, , Topic 1 : Atomic Structure and Rutherford’s Nuclear Model, Topic 2 : Bohr’s Model and the Spectra of the Hydrogen Atom, P-461 – P-472, , , , 27. Nuclei, , Topic 1 : Composition and Size of the Nuclei, Topic 2 : Mass-Energy Equivalence and Nuclear Reactions, Topic 3 : Radioactivity, , , 28. Semiconductor Electronics : Materials, Devices and Simple Circuits, , P-473 – P-493, , Topic 1 : Solids, Semiconductors and P-N Junction Diode, Topic 2 : Junction Transistor, 29. Communication Systems, , , , Topic 3 : Digital Electronics and Logic Gates, P-494 – P-500, , Topic 1 : Communication Systems, MT-1 – MT-8, , Mock Test 2 with Solutions, , MT-9 – MT-16, , , , , , Mock Test 1 with Solutions
Page 8 :
Opening/ Closing Rank for TOP NITs & List of NITs in India, Opening/ Closing Rank for Top NITS, , College Name, NIT Trichy, NIT Rourkela, NIT Surathkal, NIT Warangal, NIT Calicut, NIT Kurukshetra, NIT Durgapur, MNIT Allahabad, NIT Silchar, MNIT Jaipur, , OR/CR, ORF, CR, OR, CR, OR, CR, OR, CR, OR, CR, OR, , CSE, 2060, 5317, 2253, 9420, 960, 3181, 978, 2341, 2201, 10222, 2268, , ECE, 5325, 8011, 8571, 12009, 3378, 5608, 2919, 2919, 8023, 14769, 8320, , ME, 4154, 12970, 11662, 20304, 6315, 11788, 4340, 10209, 10629, 20480, 11195, , EE/EEE, 5708, 10353, 4084, 19168, 3456, 6801, 5270, 8152, 9703, 18966, 9454, , CR, , 6170, , 12067, , 18115, , 16273, , OR, CR, OR, CR, OR, CR, OR, , 5611, 12095, 1449, 4051, 8699, 23882, 1148, , 12509, 16098, 3600, 7128, 17899, 40841, 3881, , 14511, 22753, 5884, 11145, 21851, 49215, 9277, , 13595, 19325, 5879, 8790, 32579, 56958, 4119, , CR, , 3831, , 7868, , 11426, , 9179, , List of NITs in India, , After IITs, NITs form the second layer of topmost engineering Institutes in India, Rank Of (Amongst NITs), , Name, , State, , NIRF Score, , NIRF Ranking, , 1, , NIT Trichy, , Tamil Nadu, , 61.62, , 10, , 2, , NIT Rourkela, , Odisha, , 57.75, , 16, , 3, , NIT Karnataka, , Karnataka, , 55.25, , 21, , 4, , NIT Warangal, , Telangana, , 53.21, , 26, , 5, , NIT Calicut, , Kerala, , 52.69, , 28, , 6, , V-NIT, , Maharashtra, , 51.27, , 31, , 7, , NIT Kurukshetra, , Haryana, , 47.58, , 41, , 7, , MN-NIT, , Uttar Pradesh, , 47.49, , 42, , 8, , NIT Durgapur, , West Bengal, , 46.47, , 46, , 9, , NIT Silchar, , Assam, , 45.61, , 51, , 10, , M-NIT, , Rajasthan, , 45.20, , 53, , 11, , SV-NIT, , Gujarat, , 41.88, , 58, , 12, , NIT Hamirpur, , Himachal Pradesh, , 41.48, , 60, , 13, , MA-NIT, , Madhya Pradesh, , 40.98, , 62, , 14, , NITIE, , Maharashtra, , 40.48, , 66, , 15, , NIT Meghalaya, , Meghalaya, , 40.32, , 67, , 16, , NIT Agartala, , Tripura, , 39.53, , 70, , 17, , NIT Raipur, , Chattisgarh, , 39.09, , 74, , 18, , NIT Goa, , Goa, , 37.06, , 87
Page 9 :
FAQs - Frequently Asked Questions, QUESTION: Which Colleges other than IITs accept JEE Advanced scores?, , at the end of each paper of the examination, given to them at the start, of the paper., , ANSWER:, , QUESTION: During examination can I change my answers?, , 1., , Institute of Science (IISc), Bangalore, , 2., , Indian Institute of Petroleum and Energy (IIPE), Visakhapatnam, , ANSWER: Candidate will have the option to change previously, saved answer of any question, anytime during the entire duration of, the test., , 3. Indian Institutes of Science Education and Research (IISER),, Bhopal, 4. Indian Institutes of Science Education and Research (IISER),, Mohali, 5. Indian Institutes of Science Education and Research (IISER),, Kolkata, 6. Indian Institutes of Science Education and Research (IISER),, Pune, 7. Indian Institutes of Science Education and Research (IISER),, Thiruvananthapuram, 8. Indian Institute of Space Science and Technology (IIST), Thiruvananthapuram, , QUESTION: How can I change a previously saved answer during the, CBT of JEE (Advanced)-2018?, ANSWER: To change the answer of a question that has already been, answered and saved, first select the corresponding question from, the Question Palette, then click on “Clear Response” to clear the, previously entered answer and subsequently follow the procedure for, answering that type of question., QUESTION: Will I be given a printout/hard copy of the questions, papers along with my responses to questions in Paper-I and Paper-II, after the completion of the respective papers?, ANSWER: No., , 9. Rajiv Gandhi Institute of Petroleum Technology (RGIPT), Rae, Bareli, , QUESTION: How will I be getting a copy of the questions papers, and my responses to questions in Paper-I and Paper-II?, , QUESTION: If I am absent in one of the papers (Paper 1, Paper 2),, will my result be declared?, , ANSWER: The responses of all the candidates who have appeared, for both Paper 1 and Paper 2, recorded during the exam, along with, the questions of each paper, will be electronically mailed to their, registered email ids, by Friday, May 25, 2018, 10:00 IST., , ANSWER: NO. You will be considered absent in JEE, (Advanced)-2018 and the result will not be prepared/declared. It is, compulsory to appear in both the papers for result preparation., QUESTION: Do I have to choose my question paper language at the, time of JEE (Advanced)-2018 registration?, ANSWER: NO. There is no need to indicate question paper language, at the time of JEE (Advanced)-2018 registration. Candidates will, have the option to choose their preferred language (English or Hindi),, as the default language for viewing the questions, at the start of the, Computer Based Test (CBT) examination of JEE (Advanced)-2018., QUESTION: Can I change the language (from English to Hindi, and vice versa) of viewing the questions during the CBT of JEE, (Advanced)-2018?, ANSWER: Questions will be displayed on the screen of the Candidate, in the chosen default language (English or Hindi). Further, the, candidate can also switch/toggle between English or Hindi languages,, as the viewing language of any question, anytime during the entire, period of the examination. The candidate will also be having the, option of changing default question viewing language anytime during, the examination., QUESTION: Will I be given rough sheets for my calculations during, the CBT of JEE (Advanced)-2018?, ANSWER: Yes, you will be given “Scribble Pad” (containing, blank sheets, for rough work) at the start of every paper of JEE, (Advanced)-2018. You can do all your calculations inside this, “Scribble Pad”. Candidates MUST submit their signed Scribble Pads, , QUESTION: Suppose two candidates have same JEE, (Advanced)-2018 aggregate marks. Will the two candidates be given, the same rank?, ANSWER: If the aggregate marks scored by two or more candidates, are the same, then the following tie-break policy will be used for, awarding ranks: Step 1: Candidates having higher positive marks, will be awarded higher rank. If the tie breaking criterion at Step 1, fails to break the tie, then the following criterion at Step 2 will be, followed. Step 2: Higher rank will be assigned to the candidate who, has obtained higher marks in Mathematics. If this does not break the, tie, higher rank will be assigned to the candidate who has obtained, higher marks in Physics. If there is a tie even after this, candidates will, be assigned the same rank., QUESTION: I have read in newspapers that for the academic year, 2018-2019, supernumerary seats for female candidates would be there, in IITs. Does this mean that the non-females will get reduced number, of seats in IITs in 2018?, ANSWER: A decision has been taken at the level of the IIT Council to,, inter alia, improve the gender balance in the undergraduate programs, at the IITs from the current (approximately) 8% to 14% in 2018-19, by creating supernumerary seats specifically for female candidates,, without any reduction in the number of seats that was made available, to non-female candidates in the previous academic year (i.e. academic, year 2017-2018).
Page 10 :
1, , P-1, , Physical World, Units and Measurements, , Physical World, Units, and Measurements, 6., , TOPIC 1 Unit of Physical Quantities, 1., , 2., , The density of a material in SI unit is 128 kg m–3. In, certain units in which the unit of length is 25 cm and, the unit of mass is 50 g, the numerical value of density, of the material is:, [10 Jan. 2019 I], (a) 40, (b) 16, (c) 640, (d) 410, A metal sample carrying a current along X-axis with density Jx is, subjected to a magnetic field Bz (along z-axis). The electric field, Ey developed along Y-axis is directly proportional to Jx as well, as Bz. The constant of proportionality has SI unit, [Online April 25, 2013], 3, 2, As, m, m2, m, (a), (b), (c), (d), As, As, m3, A, , Dimensions of Physical, TOPIC 2, Quantities, 3., , 4., , 5., , 1, E, 1, , y=, and z =, are, B, CR, m0 e0, defined where C-capacitance, R-Resistance, l-length,, E-Electric field, B-magnetic field and e 0 , m 0 , - free, space permittivity and permeability respectively. Then :, [Sep. 05, 2020 (II)], (a) x, y and z have the same dimension., (b) Only x and z have the same dimension., (c) Only x and y have the same dimension., (d) Only y and z have the same dimension., Dimensional formula for thermal conductivity is (here K, denotes the temperature :, [Sep. 04, 2020 (I)], –2, –2, (a) MLT K, (b) MLT K–2, –3, (c) MLT K, (d) MLT–3 K–1, A quantity x is given by (IFv2/WL4) in terms of moment of, inertia I, force F, velocity v, work W and Length L. The, dimensional formula for x is same as that of :, [Sep. 04, 2020 (II)], (a) planck’s constant, (b) force constant, (c) energy density, (d) coefficient of viscosity, The quantities x =, , 7., , 8., , Amount of solar energy received on the earth's surface, per unit area per unit time is defined a solar constant., Dimension of solar constant is : [Sep. 03, 2020 (II)], (a) ML2T–2, (b) ML0T–3, (c) M2L0T–1, (d) MLT–2, If speed V, area A and force F are chosen as fundamental, units, then the dimension of Young's modulus will be :, [Sep. 02, 2020 (I)], 2, –1, 2, –3, (a) FA V, (b) FA V, (c) FA2V–2, (d) FA–1V0, If momentum (P), area (A) and time (T) are taken to be the, fundamental quantities then the dimensional formula for, energy is :, [Sep. 02, 2020 (II)], (a) [P2AT –2], (b) [PA–1T–2], (c) [PA1/ 2T-1 ], , 9., , (d) [P1/ 2AT-1], , Which of the following combinations has the dimension, of electrical resistance (Î0 is the permittivity of vacuum, and m0 is the permeability of vacuum)?, [12 April 2019 I], (a), , m0, e0, , (b), , m0, e0, , (c), , e0, m0, , e0, (d) m, 0, , 10. In the formula X = 5YZ2, X and Z have dimensions of, capacitance and magnetic field, respectively. What are, the dimensions of Y in SI units ?, [10 April 2019 II], –3 –2 8 4, –1 –2 4 2, (a) [M L T A ], (b) [M L T A ], (c) [M–2 L0 T–4 A–2], (d) [M–2 L–2 T6 A3], 11. In SI units, the dimensions of, , Î0, is: [8 April 2019 I], m0, , (a) A–1TML3, (b) AT2 M–1L–1, (c) AT–3ML3/2, (d) A2T3 M–1L–2, 12. Let l, r, c and v represent inductance, resistance,, capacitance and voltage, respectively. The dimension of, l, in SI units will be :, [12 Jan. 2019 II], rcv, –, 2, –1, (a) [LA ], (b) [A ], (c) [LTA], (d) [LT2]
Page 11 :
P-2, , Physics, , 13. The force of interaction between two atoms is given by, æ x2 ö, F = ab exp ç ç akT ÷÷ ; where x is the distance, k is the, è, ø, Boltzmann constant and T is temperature and a and b are, two constants. The dimensions of b is: [11 Jan. 2019 I], (a) M0L2T–4, (b) M2LT–4, (c) MLT–2, (d) M2 L2 T–2, , 14. If speed (V), acceleration (A) and force (F) are considered, as fundamental units, the dimension of Young’s modulus, will be :, , [11 Jan. 2019 II], , (a), , V–2A2F–2, , V–2A2F2, , (b), , (c), , V–4 A–2 F, , (d) V–4A2F, , hc5, where c is speed of, G, light, G universal gravitational constant and h is the Planck’s, constant. Dimension of f is that of :, [9 Jan. 2019 I], (a) area, (b) energy, (c) momentum, (d) volume, 16. Expression for time in terms of G (universal gravitational, constant), h (Planck's constant) and c (speed of light) is, proportional to:, [9 Jan. 2019 II], , 15. A quantity f is given by f =, , (a), (c), , hc5, G, , (b), , Gh, , (d), , 5, , c3, Gh, , Gh, , c, c3, 17. The dimensions of stopping potential V0 in photoelectric, effect in units of Planck’s constant ‘h’, speed of light ‘c’, and Gravitational constant ‘G’ and ampere A is:, [8 Jan. 2019 I], (a) hl/3G2/3cl/3 A –1, (b) h2/3c5/3G1/3A –1, (c) h–2/3 e–1/3 G4/3 A–1, (d) h2G3/2C1/3 A–1, 2, B, , where B is magnetic field and m0, 18. The dimensions of, 2m 0, is the magnetic permeability of vacuum, is:, [8 Jan. 2019 II], (a) MLT–2, (b) ML2T–1, (c) ML2T–2, (d) ML–1T–2, 19. The characteristic distance at which quantum gravitational, effects are significant, the Planck length, can be determined, from a suitable combination of the fundamental physical, constants G, h and c. Which of the following correctly, gives the Planck length?, [Online April 15, 2018], 1, ö2, , 1, æ Gh, 2 3, (a) G2hc (b) ç 3 ÷, (c), 2 h 2 c (d) Gh c, G, èc ø, 20. Time (T), velocity (C) and angular momentum (h) are, chosen as fundamental quantities instead of mass, length, and time. In terms of these, the dimensions of mass would, be :, [Online April 8, 2017], , (a) [ M ]=[ T–1 C–2 h ], (b) [ M ]=[ T–1 C2 h ], –1, –2, –1, (c) [ M ]=[ T C h ] (d) [ M ]=[ T C–2 h ], 21. A, B, C and D are four different physical quantities having, different dimensions. None of them is dimensionless. But, we know that the equation AD = C ln (BD) holds true., Then which of the combination is not a meaningful quantity ?, [Online April 10, 2016], C AD 2, (b) A2 –B2C2, BD, C, A, (A - C), -C, (c), (d), B, D, 22. In the following 'I' refers to current and other symbols, have their usual meaning, Choose the option that, corresponds to the dimensions of electrical conductivity :, [Online April 9, 2016], (a) M–1 L–3T 3 I, (b) M–1 L–3 T3 I2, (c) M–1L3T3 I, (d) ML–3 T–3 I2, 23. If electronic charge e, electron mass m, speed of light in, vacuum c and Planck’s constant h are taken as fundamental, quantities, the permeability of vacuum m0 can be expressed, in units of :, [Online April 11, 2015], , (a), , æ hc ö, (b) ç 2 ÷, è me ø, , æ h ö, (a) ç 2 ÷, è me ø, , æ mc 2 ö, (d) çç 2 ÷÷, è he ø, 24. If the capacitance of a nanocapacitor is measured in terms, of a unit ‘u’ made by combining the electric charge ‘e’,, Bohr radius ‘a0’, Planck’s constant ‘h’ and speed of light, ‘c’ then:, [Online April 10, 2015], , æ h ö, (c) ç 2 ÷, è ce ø, , (a) u =, , e2 h, a0, , (b) u =, , (c) u =, , e2c, ha 0, , (d) u =, , hc, 2, , e a0, e2 a 0, hc, , 25. From the following combinations of physical constants, (expressed through their usual symbols) the only, combination, that would have the same value in different, systems of units, is:, [Online April 12, 2014], (a), , (b), , (c), , (d), , ch, 2peo2, , e2, 2pe o Gme2, , (me = mass of electron), , m o eo G, , c2 he 2, , 2p m o eo h, G, ce2
Page 12 :
P-3, , Physical World, Units and Measurements, , 26. In terms of resistance R and time T, the dimensions of ratio, m, of the permeability m and permittivity e is:, e, [Online April 11, 2014], (a) [RT–2] (b) [R2T–1] (c) [R2], (d) [R2T2], 27. Let [ Î0 ] denote the dimensional formula of the permittivity, of vacuum. If M = mass, L = length, T = time and A =, electric current, then:, [2013], –1, –3, 2, –1, –3, 4, (a) Î0 = [M L T A] (b) Î0 = [M L T A2], (c) Î0 = [M1 L2 T1 A2] (d) Î0 = [M1 L2 T1 A], 28. If the time period t of the oscillation of a drop of liquid of, density d, radius r, vibrating under surface tension s is given, by the formula t = r 2b s c d a / 2 . It is observed that the, d, . The value of b, s, [Online April 23, 2013], , time period is directly proportional to, should therefore be :, (a), , 3, 4, , (b), , 3, , 2, 3, (d), 3, 2, 29. The dimensions of angular momentum, latent heat and, capacitance are, respectively. [Online April 22, 2013], , (c), , (a) ML2 T1A 2 , L2 T -2 , M -1L-2 T 2, (b) ML2 T -2 , L2 T 2 , M -1L-2 T 4 A 2, (c) ML2 T -1, L2 T -2 , ML2 TA 2, (d) ML2 T -1 , L2 T -2 , M -1L-2 T 4 A 2, 30. Given that K = energy, V = velocity, T = time. If they are, chosen as the fundamental units, then what is dimensional, formula for surface tension?, [Online May 7, 2012], (a) [KV–2T –2 ], (b) [K2 V2T–2 ], (c) [K2V–2 T–2 ], (d) [KV2 T2 ], 31. The dimensions of magnetic field in M, L, T and C, (coulomb) is given as, [2008], (a) [MLT–1 C–1], (b) [MT2 C–2], (c) [MT–1 C–1], (d) [MT–2 C–1], 32. Which of the following units denotes the dimension, ML2, , , where Q denotes the electric charge?, [2006], Q2, 2, (a) Wb/m, (b) Henry (H), (c) H/m2, (d) Weber (Wb), 33. Out of the following pair, which one does NOT have, identical dimensions ?, [2005], (a) Impulse and momentum, (b) Angular momentum and planck’s constant, (c) Work and torque, (d) Moment of inertia and moment of a force, 34. Which one of the following represents the correct, dimensions of the coefficient of viscosity?, [2004], , -1 -1, (a) éë ML T ùû, , -1, (b) éë MLT ùû, , -1 -2, (c) éë ML T ùû, , -2 -2, (d) éë ML T ùû, , 35. Dimensions of, meaning, are, , 1 , where symbols have their usual, mo eo, [2003], , (b) [L-2 T 2 ], , (a) [L-1T], , (c) [L2 T -2 ], (d) [LT -1 ], 36. The physical quantities not having same dimensions are, (a) torque and work, [2003], (b) momentum and planck’s constant, (c) stress and young’s modulus, (d) speed and (m o e o ) -1 / 2, 37. Identify the pair whose dimensions are equal, [2002], (a) torque and work, (b) stress and energy, (c) force and stress, (d) force and work, , TOPIC 3 Errors in Measurements, 38. A screw gauge has 50 divisions on its circular scale. The, circular scale is 4 units ahead of the pitch scale marking,, prior to use. Upon one complete rotation of the circular, scale, a displacement of 0.5 mm is noticed on the pitch, scale. The nature of zero error involved, and the least, count of the screw gauge, are respectively :, [Sep. 06, 2020 (I)], (a) Negative, 2 mm, (b) Positive, 10 mm, (c) Positive, 0.1 mm, (d) Positive, 0.1 mm, 39. The density of a solid metal sphere is determined by, measuring its mass and its diameter. The maximum error in, æ x ö, the density of the sphere is çè, ÷ %. If the relative errors, 100 ø, in measuring the mass and the diameter are 6.0% and 1.5%, respectively, the value of x is ______., [NA Sep. 06, 2020 (I)], 40. A student measuring the diameter of a pencil of circular, cross-section with the help of a vernier scale records the, following four readings 5.50 mm, 5.55 mm, 5.45 mm,, 5.65 mm, The average of these four reading is 5.5375 mm, and the standard deviation of the data is 0.07395 mm. The, average diameter of the pencil should therefore be recorded as :, [Sep. 06, 2020 (II)], (a) (5.5375 ± 0.0739) mm (b) (5.5375 ± 0.0740) mm, (c) (5.538 ± 0.074) mm, (d) (5.54 ± 0.07) mm, 41. A physical quantity z depends on four observables a, b, c, and d, as z =, , 2, 2 3, a b, , . The percentages of error in the meacd 3, surement of a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in z is :, [Sep. 05, 2020 (I)]
Page 13 :
P-4, , 42., , 43., , 44., , 45., , 46., , 47., , 48., , Physics, , (a) 12.25%, (b) 16.5%, (c) 13.5%, (d) 14.5%, Using screw gauge of pitch 0.1 cm and 50 divisions on its, circular scale, the thickness of an object is measured. It, should correctly be recorded as : [Sep. 03, 2020 (I)], (a) 2.121 cm, (b) 2.124 cm, (c) 2.125 cm, (d) 2.123 cm, The least count of the main scale of a vernier callipers is, 1 mm. Its vernier scale is divided into 10 divisions and, coincide with 9 divisions of the main scale. When jaws are, touching each other, the 7th division of vernier scale, coincides with a division of main scale and the zero of, vernier scale is lying right side of the zero of main scale., When this vernier is used to measure length of a cylinder, the zero of the vernier scale betwen 3.1 cm and 3.2 cm and, 4th VSD coincides with a main scale division. The length, of the cylinder is : (VSD is vernier scale division), [Sep. 02, 2020 (I)], (a) 3.2 cm, (b) 3.21 cm, (c) 3.07 cm, (d) 2.99 cm, If the screw on a screw-gauge is given six rotations, it moves, by 3 mm on the main scale. If there are 50 divisions on the, circular scale the least count of the screw gauge is:, [9 Jan. 2020 I], (a) 0.001 cm, (b) 0.02 mm, (c) 0.01 cm, (d) 0.001 mm, For the four sets of three measured physical quantities as, given below. Which of the following options is correct?, [9 Jan. 2020 II], (A) A1 = 24.36, B1 = 0.0724, C1 = 256.2, (B) A2 = 24.44, B2 = 16.082, C2 = 240.2, (C) A3 = 25.2, B3 = 19.2812, C3 = 236.183, (D) A4 = 25, B4 = 236.191, C4 = 19.5, (a) A4 + B4 + C4 < A1 + B1 + C1 < A3 + B3 + C3 < A2 + B2 + C2, (b) A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3 = A4 + B4 + C4, (c) A4 + B4 + C4 < A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3, (d) A1 + B1 + C1 < A3 + B3 + C3 < A2 + B2 + C2 < A4 + B4 + C4, A simple pendulum is being used to determine the value, of gravitational acceleration g at a certain place. The length, of the pendulum is 25.0 cm and a stop watch with 1 s, resolution measures the time taken for 40 oscillations to, be 50 s. The accuracy in g is:, [8 Jan. 2020 II], (a) 5.40%, (b) 3.40%, (c) 4.40%, (d) 2.40%, In the density measurement of a cube, the mass and edge, length are measured as (10.00 ± 0.10) kg and (0.10 ± 0.01), m, respectively. The error in the measurement of density, is:, [9 April 2019 I], 3, (a) 0.01 kg/m, (b) 0.10 kg/m3, 3, (c) 0.013 kg/m, (d) 0.07 kg/m3, The area of a square is 5.29 cm 2. The area of 7 such, squares taking into account the significant figures is:, [9 April 2019 II], , (a) 37cm2, 49., , 50., , 51., , 52., , 53., , 54., , (b) 37.030 cm2, , (c) 37.03 cm2, (d) 37.0 cm2, In a simple pendulum experiment for determination of, acceleration due to gravity (g), time taken for 20, oscillations is measured by using a watch of 1 second, least count. The mean value of time taken comes out to be, 30 s. The length of pendulum is measured by using a, meter scale of least count 1 mm and the value obtained is, 55.0 cm. The percentage error in the determination of g is, close to :, [8 April 2019 II], (a) 0.7% (b) 0.2%, (c) 3.5%, (d) 6.8%, The least count of the main scale of a screw gauge is 1 mm., The minimum number of divisions on its circular scale, required to measure 5 µm diameter of a wire is:, [12 Jan. 2019 I], (a) 50, (b) 200, (c) 100, (d) 500, The diameter and height of a cylinder are measured by, a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm,, respectively. What will be the value of its volume in, appropriate significant figures?, [10 Jan. 2019 II], 3, (a) 4264 ± 81 cm, (b) 4264.4 ± 81.0 cm3, 3, (c) 4260 ± 80 cm, (d) 4300 ± 80 cm3, The pitch and the number of divisions, on the circular, scale for a given screw gauge are 0.5 mm and 100, respectively. When the screw gauge is fully tightened, without any object, the zero of its circular scale lies 3, division below the mean line., The readings of the main scale and the circular scale, for, a thin sheet, are 5.5 mm and 48 respectively, the, thickness of the sheet is:, [9 Jan. 2019 II], (a) 5.755 mm, (b) 5.950 mm, (c) 5.725 mm, (d) 5.740 mm, The density of a material in the shape of a cube is, determined by measuring three sides of the cube and its, mass. If the relative errors in measuring the mass and, length are respectively 1.5% and 1%, the maximum error, in determining the density is:, [2018], (a) 2.5% (b) 3.5%, (c) 4.5%, (d) 6%, The percentage errors in quantities P, Q, R and S are 0.5%,, 1%, 3% and 1.5% respectively in the measurement of a, physical quantity A =, , P 3Q 2, ., RS, , The maximum percentage error in the value of A will be, [Online April 16, 2018], (a) 8.5%, (b) 6.0%, (c) 7.5%, (d) 6.5%, 55. The relative uncertainty in the period of a satellite orbiting, around the earth is 10–2. If the relative uncertainty in the, radius of the orbit is negligible, the relative uncertainty in, the mass of the earth is, [Online April 16, 2018], (a) 3×10–2, (b) 10–2, (c) 2 × 10–2, (d) 6 × 10–2
Page 14 :
P-5, , Physical World, Units and Measurements, , 56. The relative error in the determination of the surface area, of a sphere is a. Then the relative error in the determination, of its volume is, [Online April 15, 2018], 2, 2, 3, a (b), a, a, (c), (d) a, 3, 3, 2, In a screw gauge, 5 complete rotations of the screw cause, it to move a linear distance of 0.25 cm. There are 100 circular, scale divisions. The thickness of a wire measured by this, screw gauge gives a reading of 4 main scale divisions and, 30 circular scale divisions. Assuming negligible zero error,, the thickness of the wire is:, [Online April 15, 2018], (a) 0.0430 cm, (b) 0.3150 cm, (c) 0.4300 cm, (d) 0.2150 cm, The following observations were taken for determining, surface tensiton T of water by capillary method :, Diameter of capilary, D = 1.25 × 10–2 m, rise of water, h = 1.45 × 10–2 m, Using g = 9.80 m/s2 and the simplified relation, rhg, T=, ´ 10 3 N/m, the possible error in surface tension, 2, is closest to :, [2017], (a) 2. 4 % (b) 10 %, (c) 0.15%, (d) 1.5%, A physical quantity P is described by the relation, P = a1/2 b2 c3 d –4, If the relative errors in the measurement of a, b, c and d, respectively, are 2%, 1%, 3% and 5%, then the relative, error in P will be :, [Online April 9, 2017], (a) 8%, (b) 12%, (c) 32%, (d) 25%, A screw gauge with a pitch of 0.5 mm and a circular scale, with 50 divisions is used to measure the thickness of a, thin sheet of Aluminium. Before starting the measurement,, it is found that wen the two jaws of the screw gauge are, brought in contact, the 45th division coincides with the, main scale line and the zero of the main scale is barely, visible. What is the thickness of the sheet if the main scale, reading is 0.5 mm and the 25th division coincides with the, main scale line?, [2016], (a) 0.70 mm, (b) 0.50 mm, (c) 0.75 mm, (d) 0.80 mm, A student measures the time period of 100 oscillations of, a simple pendulum four times. The data set is 90 s, 91 s, 95, s, and 92 s. If the minimum division in the measuring clock, is 1 s, then the reported mean time should be:, [2016], (a) 92 ± 1.8 s, (b) 92 ± 3s, (c) 92 ± 1.5 s, (d) 92 ± 5.0 s, , (a), , 57., , 58., , 59., , 60., , 61., , 62. The period of oscillation of a simple pendulum is, L, . Measured value of L is 20.0 cm known to 1 mm, g, accuracy and time for 100 oscillations of the pendulum is, found to be 90 s using a wrist watch of 1s resolution. The, accuracy in the determination of g is :, [2015], (a) 1%, (b) 5%, (c) 2%, (d) 3%, , T = 2p, , 63. Diameter of a steel ball is measured using a Vernier callipers, which has divisions of 0.1 cm on its main scale (MS) and, 10 divisions of its vernier scale (VS) match 9 divisions on, the main scale. Three such measurements for a ball are, given as:, [Online April 10, 2015], S.No. MS(cm) VS divisions, 1., , 0.5, , 8, , 2., 3., , 0.5, 0.5, , 4, 6, , If the zero error is – 0.03 cm, then mean corrected diameter, is:, (a) 0.52 cm, (b) 0.59 cm, (c) 0.56 cm, (d) 0.53 cm, 64. The current voltage relation of a diode is given by, I = ( e1000V T - 1) mA, where the applied voltage V is in, , volts and the temperature T is in degree kelvin. If a student, makes an error measuring ±0.01 V while measuring the, current of 5 mA at 300 K, what will be the error in the, value of current in mA?, [2014], (a) 0.2 mA (b) 0.02 mA (c) 0.5 mA (d) 0.05 mA, 65. A student measured the length of a rod and wrote it as 3.50, cm. Which instrument did he use to measure it?, [2014], (a) A meter scale., (b) A vernier calliper where the 10 divisions in vernier, scale matches with 9 division in main scale and main, scale has 10 divisions in 1 cm., (c) A screw gauge having 100 divisions in the circular, scale and pitch as 1 mm., (d) A screw gauge having 50 divisions in the circular scale, and pitch as 1 mm., 66. Match List - I (Event) with List-II (Order of the time interval, for happening of the event) and select the correct option, from the options given below the lists:, [Online April 19, 2014], List - I, (1) Rotation, period of earth, , List - II, (i) 10 5 s, , (2) Revolution, (ii) 10 7 s, period of earth, (3) Period of light (iii) 10 –15 s, wave, (4) Period of, (iv) 10 –3 s, sound wave, (a) (1)-(i), (2)-(ii), (3)-(iii), (4)-(iv), , (b) (1)-(ii), (2)-(i), (3)-(iv), (4)-(iii), (c) (1)-(i), (2)-(ii), (3)-(iv), (4)-(iii), (d) (1)-(ii), (2)-(i), (3)-(iii), (4)-(iv)
Page 15 :
P-6, , Physics, , 67. In the experiment of calibration of voltmeter, a standard cell, of e.m.f. 1.1 volt is balanced against 440 cm of potential wire., The potential difference across the ends of resistance is, found to balance against 220 cm of the wire. The, corresponding reading of voltmeter is 0.5 volt. The error in, the reading of volmeter will be: [Online April 12, 2014], (a) – 0. 15 volt, (b) 0.15 volt, (c) 0.5 volt, (d) – 0.05 volt, 68. An experiment is performed to obtain the value of, acceleration due to gravity g by using a simple pendulum of, length L. In this experiment time for 100 oscillations is, measured by using a watch of 1 second least count and the, value is 90.0 seconds. The length L is measured by using a, meter scale of least count 1 mm and the value is 20.0 cm. The, error in the determination of g would be:, [Online April 9, 2014], (a) 1.7% (b) 2.7%, (c) 4.4%, (d) 2.27%, 69. Resistance of a given wire is obtained by measuring the, current flowing in it and the voltage difference applied across, it. If the percentage errors in the measurement of the current, and the voltage difference are 3% each, then error in the, value of resistance of the wire is, [2012], (a) 6%, (b) zero, (c) 1%, (d) 3%, 70. A spectrometer gives the following reading when used to, measure the angle of a prism., Main scale reading : 58.5 degree, Vernier scale reading : 09 divisions, Given that 1 division on main scale corresponds to 0.5, degree. Total divisions on the Vernier scale is 30 and match, with 29 divisions of the main scale. The angle of the prism, from the above data is, [2012], (a) 58.59 degree, (b) 58.77 degree, (c) 58.65 degree, (d) 59 degree, 71. N divisions on the main scale of a vernier calliper coincide, with (N + 1) divisions of the vernier scale. If each division of, main scale is ‘a’ units, then the least count of the instrument, is, [Online May 19, 2012], (a) a, (c), , N, ´a, N +1, , (b), , a, N, , (d), , a, N +1, , 72. A student measured the diameter of a wire using a screw, gauge with the least count 0.001 cm and listed the, measurements. The measured value should be recorded, as, [Online May 12, 2012], (a) 5.3200 cm, (b) 5.3 cm, (c) 5.32 cm, (d) 5.320 cm, 73. A screw gauge gives the following reading when used to, measure the diameter of a wire., Main scale reading : 0 mm, Circular scale reading : 52 divisions, Given that 1mm on main scale corresponds to 100 divisions, of the circular scale. The diameter of wire from the above, data is, [2011], (a) 0.052 cm, (b) 0.026 cm, (c) 0.005 cm, (d) 0.52 cm, 74. The respective number of significant figures for the, numbers 23.023, 0.0003 and 2.1 × 10–3 are, [2010], (a) 5, 1, 2, (b) 5, 1, 5, (c) 5, 5, 2, (d) 4, 4, 2, 75. In an experiment the angles are required to be measured, using an instrument, 29 divisions of the main scale exactly, coincide with the 30 divisions of the vernier scale. If the, smallest division of the main scale is half- a degree, (= 0.5°), then the least count of the instrument is: [2009], (a) half minute, (b) one degree, (c) half degree, (d) one minute, 76. A body of mass m = 3.513 kg is moving along the x-axis, with a speed of 5.00 ms–1. The magnitude of its momentum, is recorded as, [2008], (a) 17.6 kg ms–1, (b) 17.565 kg ms–1, (c) 17.56 kg ms–1, (d) 17.57 kg ms–1, 77. Two full turns of the circular scale of a screw gauge cover a, distance of 1mm on its main scale. The total number of, divisions on the circular scale is 50. Further, it is found that, the screw gauge has a zero error of – 0.03 mm. While, measuring the diameter of a thin wire, a student notes the, main scale reading of 3 mm and the number of circular scale, divisions in line with the main scale as 35. The diameter of, the wire is, [2008], (a) 3.32 mm, (b) 3.73 mm, (c) 3.67 mm, (d) 3.38 mm
Page 19 :
P-10, , Physics, , 32. (b) Mutual inductance =, , f BA, =, I, I, , [ MT -1Q -1L2 ], , = ML2Q -2, [QT -1 ], 33. (d) Moment of Inertia, I = MR2, [I] = [ML2], r, r r, Moment of force, t = r ´ F, r, t = [ L][ MLT -2 ] = [ML2T -2 ], 34. (a) According to, Stokes law,, F, F = 6phrv Þ h =, 6pr v, [Henry] =, , h=, , [ MLT –2 ], –1, , Þ h = [ ML-1T -1 ], , [ L][ LT ], 35. (c) As we know, the velocity of light in free space is, given by, , c=, , 1, 1, \, = e 2 = Z12T 2, m 0 e0, mo eo, , 1, 2, 2, mo eo = C [m/s], , = [LT–1]2, = [M0L2T–2], 36. (b) Momentum, = mv = [MLT–1], Planck’s constant,, 2 –2, E = [ ML T ], = [ML2T–1], [T –1 ], v, r r, 37. (a) Work W = F × s = Fs cos q, r r, Q A × B = AB cos q, , h=, , = [ MLT -2 ][ L] = [ML2T -2 ] ;, r r r, Torque, t = r ´ F Þ t = rF sin q, r r, Q A ´ B = AB sin q, , = [ L ] [MLT -2 ] = [ ML2T -2 ], 38. (b) Given : No. of division on circular scale of screw gauge = 50, Pitch = 0.5 mm, Least count of screw gauge, , =, , Pitch, No. of division on circular scale, , 0.5, mm = 1 ´ 105 m = 10 mm, 50, And nature of zero error is positive., 39. (1050), =, , Density, r =, , 6, M, M, Þ r = MD -3, =, 3, p, V, 4 æ Dö, pç ÷, 3 è 2ø, , æ Dr ö Dm, æ DD ö, \%ç ÷ =, + 3ç, = 6 + 3 ´ 1.5 = 10.5%, è D ÷ø, è rø, m, æ Dr ö 1050, æ x ö, %ç ÷ =, %=ç, %, è 100 ÷ø, è r ø 100, , \ x = 1050.00, 40. (d) Average diameter, dav = 5.5375 mm, Deviation of data, Dd = 0.07395 mm, As the measured data are upto two digits after decimal,, therefore answer should be in two digits after decimal., , \ d = (5.54 ± 0.07) mm, 41. (d) Given : Z =, , a 2b 2/ 3, cd 3, , Percentage error in Z,, =, , DZ 2 Da 2 Db 1 Dc 3Dd, =, +, +, +, Z, a, 3 b 2 c, d, , 2, 1, ´ 1.5 + ´ 4 + 3 ´ 2.5 = 14.5%., 3, 2, 42. (a) Thickness = M.S. Reading + Circular Scale Reading, (L.C.), = 2´ 2+, , Here LC =, , Pitch, 0.1, =, = 0.002 cm per, Circular scale division 50, , division, So, correct measurement is measurement of integral, multiple of L.C., 43. (c) L.C. of vernier callipers = 1 MSD – 1 VSD, , 9ö, æ, = ç1 - ÷ ´ 1 = 0.1 mm = 0.01 cm, 10, è, ø, Here 7th division of vernier scale coincides with a division, of main scale and the zero of vernier scale is lying right, side of the zero of main scale., Zero error = 7 × 0.1 = 0.7 mm = 0.07 cm., Length of the cylinder = measured value – zero error, = (3.1 + 4 × 0.01) – 0.07 = 3.07 cm., 44. (d) When screw on a screw-gauge is given six rotations,, it moves by 3mm on the main scale, 3, = 0.5mm, 6, Pitch 0.5 mm, =, \ Least count L.C. =, CSD, 50, 1, mm = 0.01 mm = 0.001cm, =, 100, 45. (None), D1 = A1 + B1 + C1 = 24.36 + 0.0724 + 256.2 = 280.6, D2 = A2 + B2 + C2 = 24.44 + 16.082 + 240.2 = 280.7, D3 = A3 + B3 + C3 = 25.2 + 19.2812 + 236.183= 280.7, , \, , Pitch =
Page 20 :
P-11, , Physical World, Units and Measurements, , D4 = A4 + B4 + C4 = 25 + 236.191 + 19.5 = 281, None of the option matches., 46. (c) Given, Length of simple pendulum, l = 25.0 cm, Time of 40 oscillation, T = 50s, Time period of pendulum, , T = 2p, , l, g, , 4p2 l, 4p 2l, Þg= 2, g, T, Dg Dl 2DT, =, +, Þ Fractional error in g =, g, l, T, Þ T2 =, , Dg æ 0.1 ö, æ 1ö, =ç, + 2 ç ÷ = 0.044, ÷, è 50 ø, g è 25.0 ø, Dg, \ Percentage error in g =, ´ 100 = 4.4%, g, 47. (Bonus) d = M = M = Ml -3, V, l3, Þ, , 0.10, æ 0.01ö, Dd DM, Dl, 3, =, + 3ç, =, +3, è 0.10 ÷ø = 0.31 kg/m ,, 10.00, d, M, l, 48. (d) A = 7 × 5.29 = 37.03 cm2, The result should have three significant figures, so, A = 37.0 cm2, 49. (d) We have, , T = 2p, , l, 2 l, or g = 4p 2, g, T, , Dg, DR, DT, ´ 100 =, ´ 100 + 2, ´ 100, g, Q, T, 0.1, æ 1ö, ´ 100 + 2 ç ÷ ´ 100, è 30 ø, 55, = 0.18 + 6.67 = 6.8%, 50. (b) Least count of main scale of screw gauge = 1 mm, Least count of screw gauge, =, , =, , Pitch, Number of division on circular scale, , 10-3, N, Þ N = 200, 51. (c), 52. (c) Least count of screw gauge,, 5 ´ 10 -6 =, , Pitch, LC = No. of division, = 0.5 × 10–3 = 0.5 × 10–2 mm + ve error = 3 × 0.5 × 10–2 mm, = 1.5 × 10–2 mm = 0.015 mm, Reading = MSR + CSR – (+ve error), = 5.5 mm + (48 × 0.5 × 10–2) – 0.015, = 5.5 + 0.24 – 0.015 = 5.725 mm, , 53. (c) = 1.5 % + 3 (1%) = 4.5%, 54. (d) Maximum percentage error in A, = 3(% error in P) + 2(% error in Q), 1, + (% error in R) + 1(% error in S), 2, 1, = 3 ´ 0.5 + 2 ´ 1 + ´ 3 + 1 ´ 1.5, 2, = 1.5 + 2 + 1.5 + 1.5 = 6.5%, 55. (c) From Kepler's law, time period of a satellite,, r3, 4p 2 3, T2 =, r, Gm, GM, Relative uncertainty in the mass of the earth, T = 2p, , DM, DT, =2, = 2 ´ 10-2, M, T, , (Q 4p & G constant and, , Dr, negligible), r, Ds, Dr, 56. (c) Relative error in Surface area,, = 2 ´ = a and, s, r, Dv, Dr, relative error in volume,, = 3´, v, r, \ Relative error in volume w.r.t. relative error in area,, Dv 3, = a, v, 2, Value of 1 part on main scale, 57. (d) Least count =, Number of parts on vernier scale, 0.25, cm = 5 × 10–4 cm, =, 5×100, Reading = 4 × 0.05 cm + 30 × 5 × 10–4 cm, = (0.2 + 0.0150) cm = 0.2150 cm (Thickness of wire), , relative uncertainty in radius, , rhg, ´ 103, 2, Relative error in surface tension,, , 58. (d) Surface tension, T =, , DT Dr Dh, =, +, + 0 (Q g, 2 & 103 are constant), T, r, h, Percentage error, DT æ 10 –2 ´ 0.01 10 –2 ´ 0.01ö, = ç, +, ÷ 100, T, è 1.25 ´ 10 –2 1.45 ´ 10 –2 ø, = (0.8 + 0.689), = (1.489) = 1.489% @ 1.5%, 59. (c) Given, P = a1/2 b2 c2 d–4,, Maximum relative error,, DP 1 Da, Db, Dc, Dd, =, +2, +3 +4, P, 2 a, b, c, d, 1, = ´ 2 + 2 ´ 1 + 3 ´ 3 + 4 ´ 5 = 32%, 2, 0.5, 60. (d) L.C. =, = 0.01 mm, 50, Zero error = 5 × 0.01 = 0.05 mm (Negative), Reading = (0.5 + 25 × 0.01) + 0.05 = 0.80 mm, 100 ´
Page 21 :
P-12, , Physics, , | DT1 | + | DT2 | + | DT3 | + | DT4 |, 4, 2 +1+ 3 + 0, =, = 1.5, 4, As the resolution of measuring clock is 1.5 therefore, the mean time should be 92 ± 1.5, L, 62. (d) As, g = 4p2, T2, Dg, DL, DT, So,, ´100 =, ´100 + 2, ´100, g, L, T, , 61. (c) DT =, , 0.1, 1, ´100 + 2 ´ ´100 = 2.72 ; 3%, =, 20, 90, 0.1, 63. (b) Least count =, = 0.01 cm, 10, d1 = 0.5 + 8 × 0.01 + 0.03 = 0.61 cm, d2 = 0.5 + 4 × 0.01 + 0.03 = 0.57 cm, d3 = 0.5 + 6 × 0.01 + 0.03 = 0.59 cm, 0.61 + 0.57 + 0.59, Mean diameter =, 3, , 68. (b) According to the question., t = (90 ± 1) or,, , l = (20 ± 0.1) or,, , t = 2p, , 69., , When, I = 5mA, e1000 V /T = 6mA, 1000, T, Error = ± 0.01 (By exponential function), )´, , 1000, ´ (0.01) = 0.2 mA, 300, 65. (b) Measured length of rod = 3.50 cm, For Vernier Scale with 1 Main Scale Division = 1 mm, 9 Main Scale Division = 10 Vernier Scale Division,, Least count = 1 MSD –1 VSD = 0.1 mm, 66. (a) Rotation period of earth is about 24 hrs ; 105 s, Revolution period of earth is about 365 days ; 107 s, Speed of light wave C = 3 × 108 m/s, Wavelength of visible light of spectrum, l = 4000 – 7800 Å, , 70., , = (6 mA) ´, , 1, C = f l æç and T = ö÷, è, fø, Therefore period of light wave is 10–15 s (approx), 67. (d) In a voltmeter, V µl, V = kl, Now, it is given E = 1.1 volt for l1 = 440 cm, and V = 0.5 volt for l2 = 220 cm, Let the error in reading of voltmeter be DV then,, 1.1 = 400 K and (0.5 – DV) = 220 K., Þ, , 1.1 0.5 - DV, =, 440, 220, , \, , DV = -0.05 volt, , l, 4p 2l, Þ g= 2, g, t, , 1 ö, æ 0.1, Dg, Dt ö, æ Dl, + 2 ´ ÷ = 0.027, = ± ç +2 ÷ = ç, 90 ø, è l, è 20, g, t ø, Dg, % = 2.7%, \, g, (a) According to ohm’s law, V = IR, V, R=, I, Absolute error, ´102, \ Percentage error =, Measurement, DV, DI, ´100 =, ´ 100 = 3%, where,, V, I, DR, DV, DI, ´ 100 =, ´ 102 +, ´ 102, then,, R, V, I, = 3% + 3% = 6%, (c) Q Reading of Vernier = Main scale reading, + Vernier scale reading × least count., Main scale reading = 58.5, Vernier scale reading = 09 division, least count of Vernier = 0.5°/30, 0.5°, Thus, R = 58.5° + 9 ×, 30, R = 58.65°, (d) No. of divisions on main scale = N, No. of divisions on vernier scale = N + 1, size of main scale division = a, Let size of vernier scale division be b, then we have, aN, aN = b (N + 1) Þ b =, N +1, aN, Least count is a – b = a –, N +1, a, é N +1 - N ù, = aê, ú = N +1, ë N +1 û, (d) The least count (L.C.) of a screw guage is the smallest, length which can be measured accurately with it., or,, , I = (e1000 V /T - 1) mA (given), , V /T, , Dl 0.1, =, l, 20, , Dg, %=?, g, As we know,, , = 0.59 cm, 64. (a) The current voltage relation of diode is, , Also, dI = (e1000, , Dt 1, =, t, 90, , 71., , 72., , 1, cm, 1000, Hence measured value should be recorded upto 3 decimal, places i.e., 5.320 cm, , As least count is 0.001 cm =
Page 22 :
P-13, , Physical World, Units and Measurements, , 73. (a) Least count, L.C. =, , 1, mm, 100, , Diameter of wire = MSR + CSR × L.C., Q 1 mm = 0.1 cm, = 0+, , 1, × 52 = 0.52 mm = 0.052 cm, 100, , 74. (a) Number of significant figures in 23.023 = 5, Number of significant figures in 0.0003 = 1, Number of significant figures in 2.1 × 10–3 = 2, So, the radiation belongs to X-rays part of the spectrum., 75. (d) 30 Divisions of V.S. coincide with 29 divisions of M.S., 29, \ 1 V.S.D =, MSD, 30, L.C. = 1 MSD – 1VSD, 29, = 1 MSD MSD, 30, , 1, , MSD, 30, 1, ´ 0.5° = 1 minute., =, 30, 76. (a) Momentum, p = m × v, , =, , Given, mass of a body = 3.513 kg speed of body, = (3.513) × (5.00) = 17.565 kg m/s, = 17.6 (Rounding off to get three significant figures), 77. (d) Least count of screw gauge = 0.01 mm, Q, , 0.5, mm, 50, , Reading = [M.S.R. + C.S.R. × L.C.] – (zero error), = [3 + 35 × 0.01] – (–0.03) = 3.38 mm
Page 23 :
14, , 2, , Physics, , Motion in a Straight, Line, Distance, Displacement &, TOPIC 1, Uniform Motion, 1., , TOPIC 2 Non-uniform Motion, , A particle is moving with speed v = b x along positive, x-axis. Calculate the speed of the particle at time t = t (assume, that the particle is at origin at t = 0)., [12 Apr. 2019 II], , 6., , The velocity (v) and time (t) graph of a body in a straight, line motion is shown in the figure. The point S is at 4.333, seconds. The total distance covered by the body in 6 s is:, [05 Sep. 2020 (II)], , b2 t, b2 t, b2 t, (b), (c) b2 t, (d), 2, 4, 2, All the graphs below are intended to represent the same, motion. One of them does it incorrectly. Pick it up., [2018], , v (m/s) 4, 2, 0, –2, , (a), 2., , distance, , velocity, , (a), , 3., , 4., , 5., , B, S, , D, , t (in s), , 1 2 3 4 5 6, C, , 37, 49, m (b) 12 m, (c) 11 m, (d), m, 3, 4, The speed verses time graph for a particle is shown in the, figure. The distance travelled (in m) by the particle during, the time interval t = 0 to t = 5 s will be __________., [NA 4 Sep. 2020 (II)], , (a), position, , (b), , time, , 7., , velocity, , position, , (c), , A, , time, , (d), , 10, 8, u, –1 6, (ms ), 4, 2, , time, , A car covers the first half of the distance between two, places at 40 km/h and other half at 60 km/h. The average, speed of the car is, [Online May 7, 2012], (a) 40 km/h, (b) 45 km/h, (c) 48 km/h, (d) 60 km/h, The velocity of a particle is v = v0 + gt + ft2. If its position, is x = 0 at t = 0, then its displacement after unit time (t =, 1) is, [2007], (a) v0 + g /2 + f, (b) v0 + 2g + 3f, (c) v0 + g /2 + f/3, (d) v0 + g + f, A particle located at x = 0 at time t = 0, starts moving, along with the positive x-direction with a velocity 'v' that, varies as v = a x . The displacement of the particle, varies with time as, [2006], 2, 1/2, 3, (a) t, (b) t, (c) t, (d) t, , 1 2 3 4 5, time, (s), , 8., , 9., , The distance x covered by a particle in one dimensional, motion varies with time t as x2 = at2 + 2bt + c. If the, acceleration of the particle depends on x as x–n, where n, is an integer, the value of n is ______. [NA 9 Jan 2020 I], A bullet of mass 20g has an initial speed of 1 ms–1, just, before it starts penetrating a mud wall of thickness 20 cm., If the wall offers a mean resistance of 2.5×10–2 N, the speed, of the bullet after emerging from the other side of the wall, is close to :, [10 Apr. 2019 II], (a) 0.1 ms–1, (b) 0.7 ms–1, (c) 0.3 ms–1, (d) 0.4 ms–1
Page 24 :
P-15, , Motion in a Straight Line, , (a) a +, , b2, 4c, , (b) a +, , b2, 3c, , b2, b2, (d) a +, c, 2c, 11. A particle starts from origin O from rest and moves with a, uniform acceleration along the positive x-axis. Identify all, figures that correctly represents the motion qualitatively, (a = acceleration, v = velocity, x = displacement, t = time), [8 Apr. 2019 II], , (c) a +, , 14. An automobile, travelling at 40 km/h, can be stopped at a, distance of 40 m by applying brakes. If the same automobile, is travelling at 80 km/h, the minimum stopping distance, in, metres, is (assume no skidding) [Online April 15, 2018], (a) 75 m, (b) 160 m (c) 100 m (d) 150 m, 15. The velocity-time graphs of a car and a scooter are shown, in the figure. (i) the difference between the distance, travelled by the car and the scooter in 15 s and (ii) the time, at which the car will catch up with the scooter are,, respectively, [Online April 15, 2018], (a) 337.5m and 25s, , (b) 225.5m and 10s, (c) 112.5m and 22.5s, , A Car B, , 45, Velocity (ms –1) ®, , 10. The position of a particle as a function of time t, is given, by, x(t) = at + bt2 – ct3, where, a, b and c are constants. When the particle attains, zero acceleration, then its velocity will be:, [9 Apr. 2019 II], , 30, , (B), , (C), , (D), , G, , 15, O, 0, , (A), , F, Scooter, , E, , 5, , C, D, 10 15 20 25, Time in (s) ®, , (d) 11.2.5m and 15s, 16. A man in a car at location Q on a straight highway is moving, with speed v. He decides to reach a point P in a field at a, distance d from highway (point M) as shown in the figure., Speed of the car in the field is half to that on the highway., What should be the distance RM, so that the time taken to, reach P is minimum?, [Online April 15, 2018], P, d, , (a) (B), (C), (b) (A), (c) (A), (B), (C), (d) (A), (B), (D), 12. A particle starts from the origin at time t = 0 and moves, along the positive x-axis. The graph of velocity with, respect to time is shown in figure. What is the position, of the particle at time t = 5s?, [10 Jan. 2019 II], v, (m/s), 3, , Q, , M, , d, d, (c), (d) d, 2, 2, 17. Which graph corresponds to an object moving with a, constant negative acceleration and a positive velocity ?, [Online April 8, 2017], , (a), , d, 3, , R, , (b), , (a), , 2, , (b), Velocity, , Velocity, , 1, 0, , 13., , 1 2 3 4 5 6 7 8 9 10, (a) 10 m, (b) 6 m, (c) 3 m, (d) 9 m, In a car race on straight road, car A takes a time t less, than car B at the finish and passes finishing point with, a speed 'v' more than of car B. Both the cars start from, rest and travel with constant acceleration a1 and a2, respectively. Then 'v' is equal to:, [9 Jan. 2019 II], 2a1 a 2, (a), (b), t, 2a1 a 2 t, a1 + a 2, a1 + a 2, (c), a1 a 2 t, (d), t, 2, , Time, , (c), , Time, , (d), Velocity, , Velocity, , Distance, , Distance, , 18. The distance travelled by a body moving along a line in, time t is proportional to t3., The acceleration-time (a, t) graph for the motion of the, body will be, [Online May 12, 2012]
Page 25 :
P-16, , Physics, (x1 – x2), , (x1 – x2), , a, , a, , (a), , (b), , (c), , t, , t, , a, (d), , t, , The graph of an object’s motion (along the x-axis) is shown, in the figure. The instantaneous velocity of the object at, points A and B are vA and vB respectively. Then, [Online May 7, 2012], , 23., , x(m), 15, , 10, B, 5, Dt = 8, , 0, , 20., , 24., , A, Dx = 4 m, 10, , 25., , 20 t (s), , (a) vA = vB = 0.5 m/s, (b) vA = 0.5 m/s < vB, (c) vA = 0.5 m/s > vB, (d) vA = vB = 2 m/s, An object, moving with a speed of 6.25 m/s, is decelerated, at a rate given by, , 26., , dv, = -2.5 v where v is the instantaneous speed. The time, dt, , 21., , taken by the object, to come to rest, would be: [2011], (a) 2 s, (b) 4 s, (c) 8 s, (d) 1 s, A body is at rest at x = 0. At t = 0, it starts moving in the, positive x-direction with a constant acceleration. At the, same instant another body passes through x = 0 moving, in the positive x-direction with a constant speed. The, position of the first body is given by x1(t) after time ‘t’;, and that of the second body by x2(t) after the same time, interval. Which of the following graphs correctly, describes (x1 – x2) as a function of time ‘t’?, [2008], (x1 – x2), , (x1 – x2), , (a), , O, , t, , (b), , (d), , O, , t, , f, to come to rest. If the, 2, total distance traversed is 15 S , then, [2005], 1, (a) S = ft 2, (b) S = f t, 6, 1 2, 1 2, ft, (c) S = ft, (d) S =, 4, 72, A particle is moving eastwards with a velocity of 5 ms–1., In 10 seconds the velocity changes to 5 ms–1 northwards., The average acceleration in this time is, [2005], 1 -2, (a), ms towards north, 2, 1, (b), ms - 2 towards north - east, 2, 1, (c), ms - 2 towards north - west, 2, (d) zero, The relation between time t and distance x is t = ax2 + bx, where a and b are constants. The acceleration is [2005], (a) 2bv3, (b) –2abv 2 (c) 2av2, (d) –2av 3, An automobile travelling with a speed of 60 km/h, can, brake to stop within a distance of 20m. If the car is going, twice as fast i.e., 120 km/h, the stopping distance will be, [2004], (a) 60 m, (b) 40 m, (c) 20 m, (d) 80 m, A car, moving with a speed of 50 km/hr, can be stopped by, brakes after at least 6 m. If the same car is moving at a, speed of 100 km/hr, the minimum stopping distance is, [2003], (a) 12 m, (b) 18 m, (c) 24 m, (d) 6 m, If a body looses half of its velocity on penetrating 3 cm in, a wooden block, then how much will it penetrate more, before coming to rest?, [2002], (a) 1 cm, (b) 2 cm, (c) 3 cm, (d) 4 cm., Speeds of two identical cars are u and 4u at the specific, instant. The ratio of the respective distances in which the, two cars are stopped from that instant is, [2002], (a) 1 : 1, (b) 1 : 4, (c) 1 : 8, (d) 1 : 16, , and then decelerates at the rate, , t, 19., , t, , 22. A car, starting from rest, accelerates at the rate f through a, distance S, then continues at constant speed for time t, , a, , (c), , O, , O, , 27., , 28., , TOPIC 3 Relative Velocity, t, , 29. Train A and train B are running on parallel tracks in the, opposite directions with speeds of 36 km/hour and 72, km/hour, respectively. A person is walking in train A in, the direction opposite to its motion with a speed of 1.8
Page 26 :
P-17, , Motion in a Straight Line, , km/hour. Speed (in ms–1) of this person as observed from, train B will be close to : (take the distance between the, tracks as negligible), [2 Sep. 2020 (I)], (a) 29.5 ms–1, (b) 28.5 ms–1, (c) 31.5 ms–1q, (d) 30.5 ms–1, 30. A passenger train of length 60 m travels at a speed of 80, km/hr. Another freight train of length 120 m travels at a, speed of 30 km/h. The ratio of times taken by the, passenger train to completely cross the freight train when:, (i) they are moving in same direction, and (ii) in the, opposite directions is:, [12 Jan. 2019 II], 11, 5, 3, 25, (a), (b), (c), (d), 5, 2, 2, 11, 31. A person standing on an open ground hears the sound of, a jet aeroplane, coming from north at an angle 60º with, ground level. But he finds the aeroplane right vertically, above his position. If v is the speed of sound, speed of the, plane is:, [12 Jan. 2019 II], , 32., , 2v, v, 3, v, (a), (b), (c) v, (d), 3, 2, 2, A car is standing 200 m behind a bus, which is also at rest., The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2, and the car has acceleration 4 m/s2. The car will catch up, with the bus after a time of :, [Online April 9, 2017], , (a), 33., , 34., , 110 s, , (b), , (c), , u+v, 2, , uv, , (b), , 1 2, u + v2, 2, , (d), , æ u 2 + v2 ö, ç, ÷, 2 ø, è, , TOPIC 4 Motion Under Gravity, 35., , 2 æhö, ç ÷, 3 ègø, , h, g, , (b) t = 1.8, , æhö, (c) t = 3.4 ç ÷, ègø, , (d) t =, , 2h, 3g, , 36. A Tennis ball is released from a height h and after freely, falling on a wooden floor it rebounds and reaches height, h, . The velocity versus height of the ball during its motion, 2, may be represented graphically by :, (graph are drawn schematically and on not to scale), [4 Sep. 2020 (I)], , v, , v, h/2, , (a), , h/2, h, , h(v) (b), , v, , v, (c), , h(v), , h, , h, h/2, , h(v) (d), , h, h/2, , h(v), , 120 s, , (c) 10 2 s, (d) 15 s, A person climbs up a stalled escalator in 60 s. If standing, on the same but escalator running with constant velocity, he takes 40 s. How much time is taken by the person to, walk up the moving escalator? [Online April 12, 2014], (a) 37 s, (b) 27 s, (c) 24 s, (d) 45 s, A goods train accelerating uniformly on a straight railway, track, approaches an electric pole standing on the side of, track. Its engine passes the pole with velocity u and the, guard’s room passes with velocity v. The middle wagon of, the train passes the pole with a velocity., [Online May 19, 2012], (a), , (a) t =, , A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is, dropped from the helicopter when it is at a height h. The, time taken by the packet to reach the ground is close to, [g is the accelertion due to gravity] : [5 Sep. 2020 (I)], , 37. A ball is dropped from the top of a 100 m high tower on a, 1, planet. In the last s before hitting the ground, it covers a, 2, distance of 19 m. Acceleration due to gravity (in ms–2) near, the surface on that planet is _______., [NA 8 Jan. 2020 II], 38. A body is thrown vertically upwards. Which one of the, following graphs correctly represent the velocity vs time?, [2017], , (a), , (c), , (b), , (d), , 39. Two stones are thrown up simultaneously from the edge, of a cliff 240 m high with initial speed of 10 m/s and 40, m/s respectively. Which of the following graph best, represents the time variation of relative position of the, second stone with respect to the first ?
Page 27 :
P-18, , Physics, , (Assume stones do not rebound after hitting the ground, and neglect air resistance, take g = 10 m/ s2), [2015], , y, h, , (The figures are schematic and not drawn to scale), (a), 240, , (b), , (y2 – y1) m, , 240, , (c), , (c), , 240, , 41., , 240, , 12, , t(s), , t(s), , From a tower of height H, a particle is thrown vertically, upwards with a speed u. The time taken by the particle, to, hit the ground, is n times that taken by it to reach the, highest point of its path. The relation between H, u and n, is:, [2014], 2, 2, 2, 2, (a) 2gH = n u, (b) gH = (n – 2) u d, (c) 2gH = nu2 (n – 2), (d) gH = (n – 2)u2, Consider a rubber ball freely falling from a height h = 4.9 m, onto a horizontal elastic plate. Assume that the duration, of collision is negligible and the collision with the plate is, totally elastic., Then the velocity as a function of time and the height as, a function of time will be :, [2009], v, , (a), , y, h, , t, , O, –v1, , t, , v, +v1, , (b), , O, –v1, , y, , t1, , 2t1, , 4t1, , t, , t, , O, , t, , t, , 42., , (y2 – y1) m, , 12, , +v1, , y, h, , t(s), , (d), 12, , t, , 2t1, , v, v1, , 8, , (y2 – y1 ) m, , t® 8, , 40., , t(s), , 12, , t1, , t, , (y2 – y1) m, , (d), 8, , O, , h, , t, , A parachutist after bailing out falls 50 m without friction. When, parachute opens, it decelerates at 2 m/s2 . He reaches the ground, with a speed of 3 m/s. At what height, did he bail out ? [2005], (a) 182 m, (b) 91 m, (c) 111m, (d) 293m, 43. A ball is released from the top of a tower of height h meters., It takes T seconds to reach the ground. What is the position, T, of the ball at, second, [2004], 3, 8h, (a), meters from the ground, 9, 7h, meters from the ground, (b), 9, (c), , h, meters from the ground, 9, , 17 h, meters from the ground, 18, 44. From a building two balls A and B are thrown such that A is, thrown upwards and B downwards (both vertically). If vA, and vB are their respective velocities on reaching the, ground, then, [2002], (a) vB > vA, (b) vA = vB, (c) vA > vB, (d) their velocities depend on their masses., , (d)
Page 32 :
P-23, , Motion in a Straight Line, , Þ 16 u2 = 2as2, 8u 2, Þ s2 =, a, Dividing (i) and (ii),, , 29., , ...(ii), , s1 u 2 a, 1, =, × 2 =, s2 2a 8u, 16, (a) According to question, train A and B are running on, parallel tracks in the opposite direction., , 1.8 km/h, , 36 km/h, A, , B, , v, , R (Observer), Distance, PQ = vp × t (Distance = speed × time), Distance, QR = V.t, , PQ, QR, , 120, = 24 second., 5, 34 (d) Let 'S' be the distance between two ends 'a' be the, constant acceleration, As we know v2 – u2 = 2aS, , vc2 = u 2 + aS, vc2 = u 2 +, , v2 - u 2, 2, , u 2 + v2, 2, 35. (c) For upward motion of helicopter,, , vc =, , v2 = u 2 + 2 gh Þ v 2 = 0 + 2 gh Þ v = 2gh, Now, packet will start moving under gravity., Let 't' be the time taken by the food packet to reach the, ground., 1, s = ut + at 2, 2, 1, 1, Þ -h = 2 gh t - gt 2 Þ gt 2 - 2 gh t - h = 0, 2, 2, , or, t =, , Bus, , 2´, , 2 m/sec2, , 200 m, Given, uC = uB = 0, aC = 4 m/s2, aB = 2 m/s2, hence relative acceleration, aCB = 2 m/sec2, 1, Now, we know, s = ut + at 2, 2, 1, 200 = ´ 2t 2 Q u = 0, 2, Hence, the car will catch up with the bus after time, , t = 10 2 second, , 1, 1, 15 "escalator", +, =, 60 40 120, second, , 2 gh ± 2 gh + 4 ´, , 1 vp ´ t, v, =, Þ vp =, 2, V.t, 2, 4 m/sec2, Car, , So, the person’s speed is, , v2 - u2, 2, Let v be velocity at mid point., S, 2, 2, Therefore, vc - u = 2a, 2, , VMA = –1.8 km/h = –0.5 m/s, Vman, B = Vman, A + VA, B, = Vman, A + VA – VB = –0.5 + 10 – (–20), = – 0.5 + 30 = 29.5 m/s., 30. (a), vP, P, 31. (d) Q, o, 60, , 32. (c), , 1 "escalator", 40 second, Walking with the escalator going, the speed add., , or, aS =, , VB = -72 km/h = –20 m/s, , cos 60° =, , Person’s speed walking only is, , So, the time to go up the escalator t =, , VA = 36 km/h = 10 m/s, , 72 km/h, , 1 "escalator", 60 second, Standing the escalator without walking the speed is, , 33. (c), , or, t =, , g, ´h, 2, , g, 2, , 2 gh, (1 + 2) Þ t =, g, , or, t = 3.4, , 2h, (1 + 2), g, , h, g, , 36. (c) For uniformly accelerated/ deaccelerated motion :, , v 2 = u 2 ± 2 gh, As equation is quadratic, so, v-h graph will be a parabola
Page 33 :
P-24, , Physics, , v, at t= 0, h = d, 2, , 1 ® 2 : V increases downwards, h 2 ® velocity changes its direction, 2 ® 3 : V decreases upwards, , d, 1, , 3, collision, takes 2, place, , v=, , Using, S = ut +, Þ S = 0´t +, , 1 2, gt, 2, , 1 2, gt, 2, , Þ 200 = gt2, , Þt=, In last, , 200, g, , [Q 2S = 100m], …(i), , 1, s, body travels a distance of 19 m, so in, 2, , æ 1ö, çt – ÷, è 2ø, , distance travelled = 81, 2, , 1 æ 1ö, g ç t – ÷ = 81, 2 è 2ø, 2, , æ 1ö, \ g ç t – ÷ = 81´ 2, è 2ø, 81´ 2, æ 1ö, Þ çt – ÷ =, 2, g, è, ø, , Þ, , u + 2 gh, H, , 37. (08.00) Let the ball takes time t to reach the ground, , \, , u, , 2, , Now, v = u + at, , Initially velocity is downwards (–ve) and then after, collision it reverses its direction with lesser magnitude, i.e., velocity is upwards (+ve)., Note that time t = 0 corresponds to the point on the graph, where h = d., Next time collision takes place at 3., , Now,, , y1 = 10t – 5t2 ; y2 = 40t – 5t2, y1 = – 240m, t = 8s, y2 – y1 = 30t for t < 8s., t > 8s,, 1, y2 – y1 = 240 – 40t – gt2, 2, 40. (c) Speed on reaching ground, , 39. (b), for, \, for, , 1, 1, =, ( 200 – 81 ´ 2), 2, g, , using (i), , g = 2(10 2 – 9 2), , Þ g =2 2, \ g = 8 m/s2, 38. (a) For a body thrown vertically upwards acceleration, remains constant (a = – g) and velocity at anytime t is, given by V = u – gt, During rise velocity decreases linearly and during fall, velocity increases linearly and direction is opposite to, each other., Hence graph (a) correctly depicts velocity versus time., , Þ, , u 2 + 2 gh = -u + gt, , Time taken to reach highest point is t =, Þt =, , u + u 2 + 2 gH, , g, (from question), , =, , u, ,, g, , nu, g, , Þ 2gH = n(n –2)u2, 41. (b) For downward motion v = –gt, The velocity of the rubber ball increases in downward, direction and we get a straight line between v and t with a, negative slope., 1 2, Also applying y - y0 = ut + at, 2, 1 2, 1 2, We get y - h = - gt Þ y = h - gt, 2, 2, The graph between y and t is a parabola with y = h at t = 0., As time increases y decreases., For upward motion., The ball suffer elastic collision with the horizontal elastic, plate therefore the direction of velocity is reversed and the, magnitude remains the same., Here v = u – gt where u is the velocity just after collision., As t increases, v decreases. We get a straight line between v, and t with negative slope., 1 2, Also y = ut - gt, 2, All these characteristics are represented by graph (b)., 42. (d) Initial velocity of parachute, after bailing out,, u=, , 2gh, , u = 2 ´ 9.8 ´ 50 = 14 5, The velocity at ground,, v = 3m/s, v2 - u2, 32 - 980, » 243 m, S=, =, 2´2, 4, Initially he has fallen 50 m., \ Total height from where, he bailed out = 243 + 50 = 293 m, , 50 m, v, a = - 2 m / s2, , 3m / s
Page 34 :
P-25, , Motion in a Straight Line, , 43. (a), , We have s = ut +, , 44. (b), , 1 2, gt ,, 2, , velocity u, , 1, Þ h = 0 × T + gT2, 2, 1 2, Þ h = gT, 2, , Vertical distance moved in time, , Ball A is thrown upwards with, , from the building. During its, downward journey when it comes, back to the point of throw, its, speed is equal to the speed of, throw, (u). So, for the journey of both, the balls from point A to B., , T, is, 3, , 2, , 1 æTö, 1 gT 2 h, h' = g ç ÷ Þ h' = ´, =, 2 è 3ø, 2, 9, 9, \ Position of ball from ground = h -, , h, 8h, =, 9, 9, , We can apply v2 – u2 = 2gh., As u, g, h are same for both the, balls, vA = vB, , A, , h, B, , u, u
Page 35 :
P-26, , 3, , Physics, , Motion in a Plane, 5., , TOPIC 1 Vectors, 1., , 2., , 3., , ®, , A force F = (i$ + 2 $j + 3k$ ) N acts at a point (4$i + 3 $j - k$ ) m., Then the magnitude of torque about the point ($i + 2 $j + k$ ) m, will be x N-m. The value of x is ______., , (, , [NA 7 Jan. 2020 II], uuur, uur uuur, A 2 = 5 and A1 + A 2 = 5. The value of, uur uuur, 3A1 - 2A 2 is :, [8 April 2020 II], , )(, , (a) – 106.5, 4., , (c) 90°, 6., , [NA Sep. 05, 2020 (I)], r, r, r, r, The sum of two forces P and Q is R such that | R | =, r, r, | P | . The angle q (in degrees) that the resultant of 2 P, r, r, and Q will make with Q is _______., uur, Let A1 = 3,, uur uuur, 2A1 + 3A 2 ·, , ), , Two forces P and Q, of magnitude 2F and 3F, respectively,, are at an angle q with each other. If the force Q is, doubled, then their resultant also gets doubled. Then, the, angle q is:, [10 Jan. 2019 II], (a) 120°, (b) 60°, (d) 30°, ur, ur, Two vectors A and B have equal magnitudes. The, ur ur, ur ur, magnitude of A + B is ‘n’ times the magnitude of A - B ., ur, ur, The angle between A and B is:, [10 Jan. 2019 II], , (, , é n 2 - 1ù, (a) cos -1 ê 2 ú, ë n + 1û, , 7., , (b) – 99.5, , (c) – 112.5, (d), – 118.5, In the cube of side ‘a’ shown in the figure, the vector, from the central point of the face ABOD to the central, point of the face BEFO will be:, [10 Jan. 2019 I], 8., , (, , ), , 1 ˆ ˆ, a i -k, (b), 2, , (c), , (, , ), , (d), , 1, a ˆj - iˆ, 2, , (, , ), , (, , ), , 1, a ˆj - kˆ, 2, , 9., , (, , ), , é n - 1ù, (b) cos -1 ê, ë n + 1úû, , é n2 -1ù, é n -1ù, (c) sin -1 ê 2 ú, (d) sin -1 ê, n, +, 1, ë n + 1úû, ë, û, r, r, Let A = (iˆ + ˆj) and B = (iˆ - ˆj) . The magnitude of a, r, r r r r r r, coplanar vector C such that A.C = B.C = A.B is given, by, [Online April 16, 2018], , (a), , 5, 9, , (b), , 10, 9, , (c), , 20, 9, , (d), , 9, 12, , ur, A vector A is rotated by a small angle Dq radian (Dq << 1), ur, ur ur, to get a new vector B . In that case B - A is :, [Online April 11, 2015], ur, ur, ur, (a) A Dq, (b) B Dq - A, , ur æ Dq2 ö, A çç 1 (d) 0, ÷, 2 ÷ø, è, r r r r, If A ´ B = B ´ A, then the angle between A and B is [2004], p, p, (a), (b), 2, 3, p, (c) p, (d), 4, , (c), 1, a kˆ - iˆ, (a), 2, , )
Page 36 :
P-27, , Motion in a Plane, , 15. A particle starts from the origin at t = 0 with an initial, velocity of 3.0 iˆ m/s and moves in the x-y plane with a, , Motion in a Plane with, TOPIC 2, Constant Acceleration, 10., , A balloon is moving up in air vertically above a point A on the, ground. When it is at a height h1, a girl standing at a distance, d (point B) from A (see figure) sees it at an angle 45º with, respect to the vertical. When the balloon climbs up a further, height h2, it is seen at an angle 60º with respect to the vertical, if the girl moves further by a distance 2.464 d (point C). Then the, height h2 is (given tan 30º = 0.5774):, [Sep. 05, 2020 (I)], , is given as v . Then v (in m/s) is____ [NA 8 Jan. 2020 I], r, 17. A particle moves such that its position vector r (t) = cos, , h2, h1, A, , 60°, , 45°, d, , (a) 1.464 d, 11., , B 2.464d C, (b) 0.732 d, , (c) 0.464 d, (d) d, Starting from the origin at time t = 0, with initial velocity, , 5 ˆj ms–1, a particle moves in the x–y plane with a constant, acceleration of (10iˆ + 4 ˆj) ms–2. At time t, its coordiantes, are (20 m, y0 m). The values of t and y0 are, respectively :, [Sep. 04, 2020 (I)], (a) 2 s and 18 m, (b) 4 s and 52 m, (c) 2 s and 24 m, (d) 5 s and 25 m, 12., , The position vector of a particle changes with time, r, according to the relation r (t) = 15 t 2 $i + (4 - 20 t 2 ) $j., What is the magnitude of the acceleration at t = 1?, (a) 40, , 13., , (b) 25, , [9 April 2019 II], (d) 50, , (c) 100, , (, , ), , A particle moves from the point 2.0iˆ + 4.0 ˆj m , at t = 0,, , (, , ), , with an initial velocity 5.0iˆ + 4.0 ˆj ms -1 . It is acted upon, by a constant force which produces a constant acceleration, 4.0iˆ + 4.0 ˆj ms -2 . What is the distance of the particle, , (, , 14., , constant acceleration (6.0 iˆ + 4.0 ˆj) m/ s 2. The xcoordinate of the particle at the instant when its ycoordinate is 32 m is D meters. The value of D is:, [9 Jan. 2020 II], (a) 32, (b) 50, (c) 60, (d) 40, 16. A particle is moving along the x-axis with its coordinate, with time ‘t’ given by x(t) = 10 + 8t – 3t2. Another particle is, moving along the y-axis with its coordinate as a function of, time given by y(t) = 5 – 8t3. At t = 1 s, the speed of the, second particle as measured in the frame of the first particle, , ), , from the origin at time 2s?, [11 Jan. 2019 II], (a) 15 m, (b) 20 2m, (c) 5 m, (d) 10 2m, A particle is moving with a velocity vr = K (y iˆ + x ĵ ),, where K is a constant. The general equation for its path, is:, [9 Jan. 2019 I], 2, 2, (a) y = x + constant, (b) y = x + constant, (c) y2 = x2 + constant, , (d) xy = constant, , wt iˆ + sin wt ĵ where w is a constant and t is time. Then, r, which of the following statements is true for the velocity v, r, (t) and acceleration a (t) of the particle: [8 Jan. 2020 II], r, r, r, (a) v is perpendicular to r and a is directed away from, the origin, r, r, r, (b) v and a both are perpendicular to r, r, r, r, (c) v and a both are parallel to r, r, r, r, (d) v is perpendicular to r and a is directed towards, the origin, r, 18. A particle is moving with velocity n = k ( yiˆ + xjˆ) , where k, is a constant. The general equation for its path is [2010], (a) y = x2 + constant, (b) y2 = x + constant, (c) xy = constant, (d) y2 = x2 + constant, 19. A particle has an initial velocity of 3iˆ + 4 ˆj and an, acceleration of 0.4iˆ + 0.3 ˆj . Its speed after 10 s is : [2009], (b) 7 units, (a) 7 2 units, (c) 8.5 units, (d) 10 units, 20. The co-ordinates of a moving particle at any time ‘t’are, given by x = a t 3 and y = b t 3 . The speed of the particle, at time ‘t’ is given by, [2003], (a) 3t a 2 + b2, , (b) 3t 2 a 2 + b2, , (c) t 2 a 2 + b2, , (d), , a 2 + b2, , TOPIC 3 Projectile Motion, 21. A particle of mass m is projected with a speed u from the, p, w.r.t. horizontal (x-axis). When, 3, it has reached its maximum height, it collides completely, , ground at an angle q =, , inelastically with another particle of the same mass and, velocity uiˆ. The horizontal distance covered by the combined, mass before reaching the ground is:, , [9 Jan. 2020 II]
Page 37 :
P-28, , Physics, , (a), , 3 3 u2, 8 g, , (b), , 3 2 u2, 4 g, , 5 u2, u2, (d) 2 2, g, 8 g, The trajectory of a projectile near the surface of the earth, is given as y = 2x – 9x2. If it were launched at an angle q0, with speed v0 then (g = 10 ms–2):, [12 April 2019 I], , (c), , 22., , (a) q0 = sin –1, , 1, 5, , and v0 =, , 5, ms–1, 3, , 24., , 25., , 26., , (a) 1.0 m (b) 4.2 m, , (c) 6.1 m, , (d) 9.8 m, , æ 2 ö, 3, (b) q0 = cos–1 çè 5 ÷ø and v0 = ms–1, 5, , 29. The position of a projectile launched from the origin at t =, r, 0 is given by r = 40iˆ + 50 ˆj m at t = 2s. If the projectile, , æ 1 ö, 9, (c) q0 = cos–1 çè ÷ø and v0 = ms–1, 5, 3, , was launched at an angle q from the horizontal, then q is, (take g = 10 ms–2), [Online April 9, 2014], 2, -1, -1 3, (a) tan, (b) tan, 3, 2, 4, 7, -1, -1, (c) tan, (d) tan, 5, 4, 30. A projectile is given an initial velocity of (iˆ + 2 ˆj ) m/s,, , æ 2 ö, 3, (d) q0 = sin –1 çè ÷ø and v0 = ms–1, 5, 5, , 23., , 27. Two guns A and B can fire bullets at speeds 1 km/s and, 2 km/s respectively. From a point on a horizontal, ground, they are fired in all possible directions. The, ratio of maximum areas covered by the bullets fired by, the two guns, on the ground is:, [10 Jan. 2019 I], (a) 1 : 16 (b) 1 : 2, (c) 1 : 4, (d) 1 : 8, 28. The initial speed of a bullet fired from a rifle is 630 m/s. The, rifle is fired at the centre of a target 700 m away at the same, level as the target. How far above the centre of the target ?, [Online April 11, 2014], , A shell is fired from a fixed artillery gun with an initial, speed u such that it hits the target on the ground at a, distance R from it. If t1 and t2 are the values of the time, taken by it to hit the target in two possible ways, the, product t1t2 is :, [12 April 2019 I], (a) R/4g (b) R/g, (c) R/2g, (d) 2R/g, Two particles are projected from the same point with the, same speed u such that they have the same range R, but, different maximum heights, h1 and h2. Which of the, following is correct ?, [12 April 2019 II], 2, 2, (a) R = 4 h1h2, (b) R =16 h1h2, 2, (c) R = 2 h1h2, (d) R2 = h1h2, A plane is inclined at an angle a = 30o with respect to the, horizontal. A particle is projected with a speed u = 2 ms–1,, from the base of the plane, as shown in figure. The distance, from the base, at which the particle hits the plane is close to, : (Take g=10 ms–2), [10 April 2019 II], , (a) 20 cm (b) 18 cm, (c) 26 cm, (d) 14 cm, A body is projected at t = 0 with a velocity 10 ms–1 at an, angle of 60° with the horizontal. The radius of curvature, of its trajectory at t = 1s is R. Neglecting air resistance, and taking acceleration due to gravity g = 10 ms–2, the, value of R is:, [11 Jan. 2019 I], (a) 10.3 m, (b) 2.8 m, (c) 2.5 m, (d) 5.1 m, , (, , ), , where iˆ is along the ground and ĵ is along the vertical., If g = 10 m/s2 , the equation of its trajectory is : [2013], (a) y = x - 5 x 2, , (b) y = 2 x - 5 x 2, , (c) 4 y = 2 x - 5 x 2, , (d) 4 y = 2 x - 25 x 2, , 31. The maximum range of a bullet fired from a toy pistol, mounted on a car at rest is R0= 40 m. What will be the acute, angle of inclination of the pistol for maximum range when, the car is moving in the direction of firing with uniform, velocity v = 20 m/s, on a horizontal surface ? (g = 10 m/s 2), [Online April 25, 2013], (a) 30°, , (b) 60°, , (c) 75°, , (d) 45°, , 32. A ball projected from ground at an angle of 45° just clears, a wall in front. If point of projection is 4 m from the foot of, wall and ball strikes the ground at a distance of 6 m on the, other side of the wall, the height of the wall is :, [Online April 22, 2013], (a) 4.4 m (b) 2.4 m, (c) 3.6 m, (d) 1.6 m, 33. A boy can throw a stone up to a maximum height of 10 m., The maximum horizontal distance that the boy can throw, the same stone up to will be, [2012], (a) 20 2 m, (b) 10 m, (c) 10 2 m, (d) 20 m, 34. A water fountain on the ground sprinkles water all around, it. If the speed of water coming out of the fountain is v, the, total area around the fountain that gets wet is:, [2011], (a) p, , v4, g, , 2, , p v4, (b) 2 2, g, , (c) p, , v2, g, , 2, , (d) p, , v2, g
Page 38 :
P-29, , Motion in a Plane, , 35. A projectile can have the same range ‘R’ for two angles, of projection. If ‘T1’ and ‘T2’ to be time of flights in the, two cases, then the product of the two time of flights is, directly proportional to., [2004], 1, 1, (a) R, (b), (c) 2, (d) R2, R, R, 36. A ball is thrown from a point with a speed ' v0 ' at an, elevation angle of q. From the same point and at the same, ' v0 ', 2, to catch the ball. Will the person be able to catch the ball? If, yes, what should be the angle of projection q?, [2004], (a) No, (b) Yes, 30°, (c) Yes, 60°, (d) Yes, 45°, A boy playing on the roof of a 10 m high building throws, a ball with a speed of 10m/s at an angle of 30º with the, horizontal. How far from the throwing point will the ball be, at the height of 10 m from the ground ?, [2003], , instant, a person starts running with a constant speed, , 37., , 1, 3, [ g = 10m/s , sin 30 = , cos 30o =, ], 2, 2, (a) 5.20m (b) 4.33m, (c) 2.60m (d) 8.66m, 2, , o, , Relative Velocity in Two, TOPIC 4 Dimensions & Uniform, Circular Motion, 38. A clock has a continuously moving second's hand of 0.1, m length. The average acceleration of the tip of the hand, (in units of ms–2) is of the order of: [Sep. 06, 2020 (I)], (a) 10 –3, , (b) 10 –4, , (c) 10 –2, (d) 10 –1, 39. When a carsit at rest, its driver sees raindrops falling on, it vertically. When driving the car with speed v, he sees, that raindrops are coming at an angle 60º from the horizontal. On furter increasing the speed of the car to (1 +, b)v, this angle changes to 45º. The value of b is close to:, [Sep. 06, 2020 (II)], (a) 0.50, (b) 0.41, (c) 0.37, (d) 0.73, 40. The stream of a river is flowing with a speed of 2 km/h., A swimmer can swim at a speed of 4 km/h. What should, be the direction of the swimmer with respect to the flow, of the river to cross the river straight? [9 April 2019 I], (a) 90°, (b) 150°, (c) 120°, (d) 60°, , 41. Ship A is sailing towards north-east with velocity km/hr, where points east and , north. Ship B is at a distance of 80, km east and 150 km north of Ship A and is sailing towards, west at 10 km/hr. A will be at minimum distance from B in:, [8 April 2019 I], (a) 4.2 hrs., (b) 2.6 hrs., (c) 3.2 hrs., (d) 2.2 hrs., 42. Two particles A, B are moving on two concentric circles, of radii R1 and R2 with equal angular speed w. At t = 0,, their positions and direction of motion are shown in the, figure :, [12 Jan. 2019 II], Y, , A, X, , R1, B, , R2, , (a) w(R1 + R2) iˆ, , p, is given by:, 2w, (b) –w(R1 + R2) iˆ, , (c) w(R2 – R1) iˆ, , (d) w(R1 – R2) iˆ, , ®, ® and t =, The relative velocity vA, - vB, , 43. A particle is moving along a circular path with a constant, speed of 10 ms–1. What is the magnitude of the change in, velocity of the particle, when it moves through an angle of, 60° around the centre of the circle?, [Online April 10, 2015], (a), , (b) zero, , 10 3m/s, , (c) 10 2m/s, (d) 10 m/s, 44. If a body moving in circular path maintains constant speed, of 10 ms–1, then which of the following correctly describes, relation between acceleration and radius?, [Online April 10, 2015], a, , a, , (a), , (b), r, , r, , a, , a, , (c), , (d), r, , r
Page 42 :
P-33, , Motion in a Plane, , = (2cos15°), , 2, 4 sin15°, æ1, ö 16sin 15°, - ç ´ 10sin 30°÷, ø 100cos 2 30°, 10 10cos 30° è 2, , 16 3 - 16, ; 0.1952m ; 20cm, 60, (b), =, , 26., , 10 m/s, , q, , g, o, , ux =, , v, , gcosq, , 60, , 5, , g, (10 - 5 3), , Horizontal component of velocity, vx = 10cos 60° = 5 m/s, vertical component of velocity, vy = 10cos 30° = 5 3 m/s, After t = 1 sec., Horizontal component of velocity vx = 5 m/s, Vertical component of velocity, , (, , ), , vy = | 5 3 –10 | m / s = 10 – 5 3, Centripetal, acceleration an =, v +v, 2, x, , v2, R, , 2, y, , 25 +100 + 75 –100 3, ...(i), =, an, 10cos q, From figure (using (i)), , ÞR=, , tan q=, R=, , 27., , (, , 100 2 – 3, 10cos15, , ) = 2.8m, , (a) As we know, range R =, , Vertical velocity (initial), 50 = uy t +, , u 2 sin 2q, g, , 4, , A1 u14 é 1 ù, 1, =, =, =, A 2 u 42 êë 2 úû 16, (c) Let ‘t’ be the time taken by the bullet to hit the target., \ 700 m = 630 ms–1 t, \, , 700m, 630ms -1, , =, , 10, sec, 9, , 1, (–10) ×4, 2, or, 50 = 2uy – 20, 70, = 35m / s, or, uy =, 2, u y 35 7, =, =, \ tan q =, u x 20 4, , 7, 4, r ˆ, 30. (b) From equation, v = i + 2 ˆj, Þ x=t, , Þ Angle q = tan–1, , 1, y = 2t - (10t 2 ), 2, From (i) and (ii), y = 2x – 5x2, , … (ii), , P, wall, 45°, O, 4m A, 6m, As ball is projected at an angle 45° to the horizontal, therefore Range = 4H, 10, = 2.5 m, or 10 = 4H Þ H =, 4, (Q Range = 4 m + 6 m = 10m), Maximum height, H =, \ u2 =, , H ´ 2g, 2, , sin q, , =, , u 2 sin 2 q, 2g, , 2.5 ´ 2 ´10, æ 1 ö, ç, ÷, è 2ø, , or, u = 100 = 10 ms -1, Height of wall PA, , 1 2, \ h = gt, 2, , = OA tan q 2, , … (i), , 31. (b), , For vertical motion,, Here, u = 0, , 1, æ 10 ö, = ´ 10 ´ ç ÷, è 9ø, 2, , 1 2, gt, 2, , Þ uy × 2 +, , 32. (b), , \ A µ R2 or, A µ u4, , Þ t=, , 40, = 20m/s, 2, , 10 – 5 3, = 2 – 3 Þq= 15°, 5, , and, area A = p R2, , 28., , 500, m = 6.1 m, 81, Therefore, the rifle must be aimed 6.1 m above the centre, of the target to hit the target., 29. (c) From question,, Horizontal velocity (initial),, , =, , 2, , = 100, , 1 g(OA) 2, 2 u 2 cos2 q, 1, 10 ´ 16, = 4- ´, = 2.4 m, 1, 1, 2, 10 ´ 10 ´, ´, 2, 2
Page 43 :
P-34, , 33., , Physics, , (d) R =, , u 2 sin2 q, u 2 sin 2 q, ,H=, g, 2g, , vr, , Hmax at 2q = 90°, , vr, , u2, Hmax =, 2g, , u 2 sin 2q, u2, Þ Rmax =, g, g, , tan 60° =, , v 2 sin 2q v 2 sin 90° v 2, =, =, Where Rmax =, g, g, g, , tan 45° =, ...(i), , 35., , v4, g2, , (a) A projectile have same range for two angle, Let one angle be q, then other is 90° – q, , T1 =, , vr, v, , ...(i), , 2u sin q, 2u cos q, , T2 =, g, g, , vr, (b + 1)v, , 3v = (b + 1)v Þ b = 3 - 1 = 0.732., 40. (c), , sin q =, , u, 2 1, = =, v, 4 2, , 37., , 38., , 36., , q, , with respect to flow,, , ĵ (North), , 39., , B, , 41. (b), , rBA, , iˆ (East), , A, , r, vA = 30iˆ + 50 ˆj km/hr, r, vB = (-10iˆ) km/hr, rBA = (80iˆ + 150 ˆj ) km, r, r, r, vBA = vB - v A = -10iˆ - 30iˆ - 50iˆ = 40iˆ - 50 ˆj, , u 2 sin 2q (10)2 sin(2 ´ 30°), =, = 5 3 = 8.66 m, g, 10, (a) Here, R = 0.1 m, , tminimum =, , 2p 2 p, =, = 0.105 rad /s, T, 60, Acceleration of the tip of the clock second's hand,, , =, , a = w 2 R = (0.105)2 (0.1) = 0.0011 = 1.1 ´ 10 -3 m/s2, Hence, average acceleration is of the order of 10–3., (d) The given situation is shown in the diagram. Here vr, be the velocity of rain drop., , u, , = 90° + 30° = 120°, , R=, , w=, , v, , or q = 30°, , 4u 2 sin q cos q, then, T1T2 =, = 2R, g, , u 2 sin 2 q, ), (Q R =, g, Thus, it is proportional to R. (Range), (c) Yes, Man will catch the ball, if the horizontal, component of velocity becomes equal to the constant, speed of man., vo, = vo cos q, 2, or q = 60°, (d) Horizontal range is required, , ...(ii), , Dividing (i) by (ii) we get,, , ...(ii), , From equation (i) and (ii), A=p, , –vcar = (b + 1)v, , When car is moving with speed (1 + b)v ,, , 10 ´ g ´ 2, = 20 metre, g, (a) Let, total area around fountain, 2, A = pRmax, , 45°, , When car is moving with speed v,, , Rmax =, , 34., , 60°, –vcar = v, , v, , u2, = 10 Þ u2 = 10 g ´ 2, 2g, , R=, , vr, , ( rrBA )(· vrBA ), 2, ( vrBA ), , (80iˆ + 150 ˆj )( -40iˆ - 50 ˆj ), , \t=, , (10 41) 2, , 10700, 10 41 ´ 10 41, , =, , 107, = 2.6 hrs., 41
Page 44 :
P-35, , Motion in a Plane, , 42., , Change in velocity,, , p p, =, 2w 2, So, both have completed quater circle, , (c) From, q = wt = w, , | Dv | = v12 + v 22 + 2v1 v 2 cos ( p – q ), , = 2vsin, , wR1 A, , r, , r, , (Q| v1 | = | v 2 |) = v, , = (2 × 10) × sin(30°) = 2 × 10 ×, , wR2 B, , V2, r, ra = constant, Hence graph (c) correctly describes relation between, acceleration and radius., a=, , Relative velocity,, , ( ), , v A – v B =wR1 –iˆ - wR 2 ( –i ) =w ( R 2 – R1 ) i, , v2, , (d), , q, , 1, 2, , = 10 m/s, 44. (c) Speed, V = constant (from question), Centripetal acceleration,, , X, , 43., , q, 2, , v1, v2, , v1, , (p - q), - v1
Page 45 :
P-36, , 4, , Physics, , Laws of Motion, TOPIC 1, 1., , 2., , Ist, Ind & IIIrd Laws of, Motion, , A particle moving in the xy plane experiences a velocity, r, dependent force F = k ( v y i$ + v x $j ) , where vx and vy are x, r, r, and y components of its velocity v . if a is the acceleration of the particle, then which of the following statements, is true for the particle?, [Sep. 06, 2020 (II)], r r, (a) quantity v ´ a is constant in time, r, (b) F arises due to a magnetic field, (c) kinetic energy of particle is constant in time, r r, (d) quantity v × a is constant in time, A spaceship in space sweeps stationary interplanetary dust., dM (t ), = bv 2 (t ),, As a result, its mass increases at a rate, dt, where v (t) is its instantaneous velocity. The instantaneous, acceleration of the satellite is :, [Sep. 05, 2020 (II)], (a) -bv3 (t ), , (b) -, , 5., , 6., , bv 3, M (t ), , 2bv 3, bv 3, (d) M (t ), 2 M (t ), A small ball of mass m is thrown upward with velocity u, from the ground. The ball experiences a resistive force, mkv2 where v is its speed. The maximum height attained, by the ball is :, [Sep. 04, 2020 (II)], , (c) -, , 3., , (a), , (b), , 1 æ ku 2 ö, ln 1 +, 2 g ÷ø, k çè, , 1 æ ku, ln 1 +, 2 k çè, g ÷ø, A ball is thrown upward with an initial velocity V0 from the, surface of the earth. The motion of the ball is affected by a, drag force equal to mgv2 (where m is mass of the ball, v is, its instantaneous velocity and g is a constant). Time taken, by the ball to rise to its zenith is :, [10 April 2019 I], , (c), , 4., , ku 2, 1, tan -1, 2k, g, 1, ku 2, tan -1, k, 2g, , (d), , æ g ö, æ g ö, 1, tan -1 ç, V0 ÷, sin -1 ç, V, (b), ç, ÷, ç g 0 ÷÷, gg, gg, è g ø, è, ø, æ, ö, æ, 1, g, 1, 2g ö, l n ç1 +, V0 ÷ (d), tan -1 ç, V, (c), ç g 0 ÷÷, g ÷ø, g g çè, 2g g, è, ø, A ball is thrown vertically up (taken as + z-axis) from the, ground. The correct momentum-height (p-h) diagram is:, [9 April 2019 I], , (a), , 2ö, , 1, , (a), , (b), , (c), , (d), , A particle of mass m is moving in a straight line with, momentum p. Starting at time t = 0, a force F = kt acts in the, same direction on the moving particle during time interval, T so that its momentum changes from p to 3p. Here k is a, constant. The value of T is :, [11 Jan. 2019 II], (a), , 2, , k, p, , (b) 2, , p, k, , 2k, 2p, (d), p, k, A particle of mass m is acted upon by a force F given by, , (c), , 7., , R, the empirical law F = 2 v(t). If this law is to be tested, t, experimentally by observing the motion starting from rest,, the best way is to plot :, [Online April 10, 2016], 1, (a) log v(t) against, (b) v(t) against t2, t, 1, (d) log v(t) against t, (c) log v(t) against 2, t
Page 46 :
P-37, , Laws of Motion, , 8., , A large number (n) of identical beads, each of mass m, and radius r are strung on a thin smooth rigid horizontal, rod of length L (L >> r) and are at rest at random, positions. The rod is mounted between two rigid, supports (see figure). If one of the beads is now given, a speed v, the average force experienced by each support, after a long time is (assume all collisions are elastic):, [Online April 11, 2015], , (a) Statement 1 is true, Statement 2 is true, Statement 2 is, the correct explanation of Statement 1., (b) Statement 1 is false, Statement 2 is true., (c) Statement 1 is true, Statement 2 is false., (d) Statement 1 is true, Statement 2 is true, Statement 2 is, not the correct explanation of Statement 1., 12. Two fixed frictionless inclined planes making an angle 30°, and 60° with the vertical are shown in the figure. Two, blocks A and B are placed on the two planes. What is the, relative vertical acceleration of A with respect to B ? [2010], A, , L, , B, , (a), , mv 2, 2(L - nr), , (b), , mv 2, L - 2nr, , mv 2, (d) zero, L - nr, A body of mass 5 kg under the action of constant force, r, r, F = Fxˆi + Fy ˆj has velocity at t = 0 s as v = 6iˆ - 2ˆj m/s, , (c), 9., , (, , ), , r, r, and at t = 10s as v = +6ˆj m / s . The force F is:, [Online April 11, 2014], , æ 3 ˆ 4 ˆö, (a), (b) ç - i + j ÷ N, è 5 5 ø, æ 3ˆ 4 ˆö, (d) ç i - j ÷ N, (c) 3iˆ - 4ˆj N, è5 5 ø, 10. A particle of mass m is at rest at the origin at time, t = 0. It is subjected to a force F(t) = F0e–bt in the x direction., Its speed v(t) is depicted by which of the following, curves?, [2012], , (, , ), , F0, mb, , F0, mb, v (t ), , 15., , 16., , (a), (b), (c) Zero, (d) 4.9 ms–2 in vertical direction, A ball of mass 0.2 kg is thrown vertically upwards by applying, a force by hand. If the hand moves 0.2 m while applying the, force and the ball goes upto 2 m height further, find the, magnitude of the force. (Consider g = 10 m/s2)., [2006], (a) 4 N, (b) 16 N, (c) 20 N, (d) 22 N, A player caught a cricket ball of mass 150 g moving at a, rate of 20 m/s. If the catching process is completed in 0.1s,, the force of the blow exerted by the ball on the hand of the, player is equal to, [2006], (a) 150 N (b) 3 N, (c) 30 N, (d) 300 N, A particle of mass 0.3 kg subject to a force F = – kx with, k = 15 N/m . What will be its initial acceleration if it is, released from a point 20 cm away from the origin ?[2005], (a) 15 m/s2 (b) 3 m/s2 (c) 10 m/s2 (d) 5 m/s2, A block is kept on a frictionless inclined surface with angle, of inclination ‘a’. The incline is given an acceleration ‘a’, to keep the block stationary. Then a is equal to [2005], , (b) v (t ), , v (t ), , t, , t, , (c), , 14., , ), , F0, mb, , (a), , 13., , -3jˆ + 4ˆj N, , (, , 30°, , 60°, , 4.9 ms–2 in horizontal direction, 9.8 ms–2 in vertical direction, , F0, mb, , (d) v (t ), , t, t, 11. This question has Statement 1 and Statement 2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., Statement 1: If you push on a cart being pulled by a horse, so that it does not move, the cart pushes you back with an, equal and opposite force., Statement 2: The cart does not move because the force, described in statement 1 cancel each other., [Online May 26, 2012], , a, , a, , (a) g cosec a, (b) g / tan a, (c) g tan a, (d) g, 17. A rocket with a lift-off mass 3.5 × 104 kg is blasted upwards, with an initial acceleration of 10m/s2. Then the initial thrust, of the blast is, [2003], (a) 3.5 ´ 10 5 N, (b) 7.0 ´ 10 5 N, (c) 14.0 ´ 10 5 N, (d) 1.75 ´ 10 5 N, 18. Three forces start acting simultaneously on a particle, r, moving with velocity, v . These forces are represented, in magnitude and direction by the three sides of a triangle, ABC. The particle will now move with velocity [2003]
Page 47 :
P-38, , Physics, , C, , r, (a) less than v, , A, , (a), , 2g, 3, , (b), , g, 2, , (c), , 5g, 6, , B, , r, (b) greater than v, r, (c) v in the direction of the largest force BC, (d) vr , remaining unchanged, , 19. A solid sphere, a hollow sphere and a ring are released, from top of an inclined plane (frictionless) so that they, slide down the plane. Then maximum acceleration down, the plane is for (no rolling), [2002], (a) solid sphere, (b) hollow sphere, (c) ring, (d) all same, , Motion of Connected Bodies,, TOPIC 2 Pulley & Equilibrium of, Forces, 20. A mass of 10 kg is suspended by a rope of length 4 m, from, the ceiling. A force F is applied horizontally at the midpoint of the rope such that the top half of the rope makes, an angle of 45° with the vertical. Then F equals:, (Take g = 10 ms–2 and the rope to be massless), [7 Jan. 2020 II], (a) 100 N, (b) 90 N, (c) 70 N, (d) 75 N, 21. An elevator in a building can carry a maximum of 10, persons, with the average mass of each person being 68, kg. The mass of the elevator itself is 920 kg and it moves, with a constant speed of 3 m/s. The frictional force, opposing the motion is 6000 N. If the elevator is moving, up with its full capacity, the power delivered by the motor, to the elevator (g =10 m/s2) must be at least:, [7 Jan. 2020 II], (a) 56300 W, (b) 62360 W, (c) 48000 W, (d) 66000 W, 22. A mass of 10 kg is suspended vertically by a rope from, the roof. When a horizontal force is applied on the rope, at some point, the rope deviated at an angle of 45°at the, roof point. If the suspended mass is at equilibrium, the, magnitude of the force applied is (g = 10 ms–2), [9 Jan. 2019 II], (a) 200 N, (b) 140 N, (c) 70 N, (d) 100 N, 23. A mass ‘m’ is supported by a massless string wound around, a uniform hollow cylinder of mass m and radius R. If the, str ing does not slip on the cylinder, with what, acceleration will the mass fall or release?, [2014], , R, m, , m, (d) g, 24. Two blocks of mass M1 = 20 kg and M2 = 12 kg are, connected by a metal rod of mass 8 kg. The system is, pulled vertically up by applying a force of 480 N as shown., The tension at the mid-point of the rod is :, [Online April 22, 2013], 480 N, , (a) 144 N, , M1, , (b) 96 N, (c) 240 N, (d) 192 N, , M2, , 25. A uniform sphere of weight W and radius 5 cm is being, held by a string as shown in the figure. The tension in the, string will be :, [Online April 9, 2013], , 8 cm, , W, W, W, W, (b) 5, (c) 13, (d) 13, 5, 12, 5, 12, 26. A spring is compressed between two blocks of masses m1, and m2 placed on a horizontal frictionless surface as shown, in the figure. When the blocks are released, they have, initial velocity of v1 and v2 as shown. The blocks travel, distances x1 and x2 respectively before coming to rest., , (a), , 12, , æx ö, The ratio ç 1 ÷ is, è x2 ø, , [Online May 12, 2012], m2, , m1, , v2, , v1, , (a), , m2, m1, , (b), , m1, m2, , (c), , m2, m1, , (d), , m1, m2
Page 48 :
P-39, , Laws of Motion, , 27. A block of mass m is connected to another block of mass, M by a spring (massless) of spring constant k. The block, are kept on a smooth horizontal plane. Initially the blocks, are at rest and the spring is unstretched. Then a constant, force F starts acting on the block of mass M to pull it., Find the force of the block of mass m., [2007], mF, MF, (a), (b), M, (m + M ), mF, (c) ( M + m) F, (d), (m + M ), m, 28. Two masses m1 = 5g and m2 = 4.8 kg tied to a string, are hanging over a light frictionless pulley. What is the, acceleration of the masses when left free to move ?, [2004], ( g = 9.8m / s 2 ), , 33. When forces F1, F2, F3 are acting on a particle of mass m, such that F2 and F3 are mutually perpendicular, then the, particle remains stationary. If the force F1 is now removed, then the acceleration of the particle is, [2002], (a) F1/m, (b) F2F3 /mF1, (c) (F2 - F3)/m, (d) F2 /m., 34. Two forces are such that the sum of their magnitudes is, 18 N and their resultant is 12 N which is perpendicular, to the smaller force. Then the magnitudes of the forces, are, [2002], (a) 12 N, 6 N, (b) 13 N, 5 N, (c) 10 N, 8 N, (d) 16N, 2N., 35. A light string passing over a smooth light pulley connects, two blocks of masses m1 and m2 (vertically). If the, acceleration of the system is g/8, then the ratio of the, masses is, [2002], (a) 8 : 1, (b) 9 : 7, (c) 4 : 3, (d) 5 : 3, 36. Three identical blocks of masses m = 2 kg are drawn by a, force F = 10. 2 N with an acceleration of 0. 6 ms-2 on a, frictionless surface, then what is the tension (in N) in, the string between the blocks B and C?, [2002], C, , (a) 5 m/s2, (b) 9.8 m/s2, (c) 0.2 m/s2, (d) 4.8 m/s2, 29. A spring balance is attached to the ceiling of a lift. A man, hangs his bag on the spring and the spring reads 49 N,, when the lift is stationary. If the lift moves downward with, an acceleration of 5 m/s2, the reading of the spring balance, will be, [2003], (a) 24 N, (b) 74 N, (c) 15 N, (d) 49 N, 30. A block of mass M is pulled along a horizontal frictionless, surface by a rope of mass m. If a force P is applied at the, free end of the rope, the force exerted by the rope on the, block is, [2003], Pm, Pm, PM, (a), (b), (c) P, (d), M +m, M -m, M +m, 31. A light spring balance hangs from the hook of the other, light spring balance and a block of mass M kg hangs from, the former one. Then the true statement about the scale, reading is, [2003], (a) both the scales read M kg each, (b) the scale of the lower one reads M kg and of the, upper one zero, (c) the reading of the two scales can be anything but the, sum of the reading will be M kg, (d) both the scales read M/2 kg each, 32. A lift is moving down with acceleration a. A man in the lift, drops a ball inside the lift. The acceleration of the ball as, observed by the man in the lift and a man standing, stationary on the ground are respectively, [2002], (a) g, g, (b) g – a, g – a, (c) g – a, g, (d) a, g, , B, , A, , F, , (a) 9.2, (b) 3.4, (c) 4, (d) 9.8, 37. One end of a massless rope, which passes over a massless, and frictionless pulley P is tied to a hook C while the other, end is free. Maximum tension that the rope can bear is 360, N. With what value of maximum safe acceleration (in ms-2), can a man of 60 kg climb on the rope?, [2002], P, C, , (a) 16, , (b) 6, , (c) 4, , (d) 8, , TOPIC 3 Friction, 38. An insect is at the bottom of a hemispherical ditch of, radius 1 m. It crawls up the ditch but starts slipping, after it is at height h from the bottom. If the coefficient, of friction between the ground and the insect is 0.75,, then h is : (g = 10 ms–2), [Sep. 06, 2020 (I)], (a) 0.20 m, , (b) 0.45 m, , (c) 0.60 m, (d) 0.80 m, 39. A block starts moving up an inclined plane of inclination, 30° with an initial velocity of v0. It comes back to its, v, initial position with velocity 0 . The value of the, 2, coefficient of kinetic friction between the block and the, I, . The nearest integer to I, inclined plane is close to, 1000, is _________., [NA Sep. 03, 2020 (II)]
Page 49 :
P-40, , Physics, , 40. A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled, in case (B), by a force F=20 N, making an angle of 30 o with, the horizontal, as shown in the figures. The coefficient of, friction between the block and floor is m = 0.2. The, difference between the accelerations of the block, in case, (B) and case (A) will be : (g =10 ms–2), [12 April 2019 II], , 44. Two masses m1 = 5 kg and m2 = 10 kg, connected by an, inextensible string over a frictionless pulley, are moving, as shown in the figure. The coefficient of friction of, horizontal surface is 0.15. The minimum weight m that, should be put on top of m2 to stop the motion is: [2018], T, m, m2, (a) 18.3 kg, m, 2, (b) 27.3 kg, T, , (c) 43.3 kg, , m1, , (d) 10.3 kg, , (a) 0.4 ms–2, (b) 3.2 ms–2, –2, (c) 0.8 ms, (d) 0 ms–2, 41. Two blocks A and B masses mA=1 kg and mB = 3 kg are kept, on the table as shown in figure. The coefficient of friction, between A and B is 0.2 and between B and the surface of, the table is also 0.2. The maximum force F that can be, applied on B horizontally, so that the block A does not, slide over the block B is : [Take g = 10 m/s2], [10 April 2019 II], , (a) 8 N, (b) 16 N (c) 40 N, (d) 12 N, 42. A block kept on a rough inclined plane, as shown in the, figure, remains at rest upto a maximum force 2 N down, the inclined plane. The maximum external force up the, inclined plane that does not move the block is 10 N. The, coefficient of static friction between the block and the, plane is : [Take g = 10 m/s2], [12 Jan. 2019 II], 10, , N, , 2N, 30°, , 3, 3, (b), 2, 4, 2, 1, (c), (d), 3, 2, 43. A block of mass 10 kg is kept on a rough inclined plane as, shown in the figure. A force of 3 N is applied on the block., The coefficient of static friction between the plane and, the block is 0.6. What should be the minimum value of, force P, such that the block doesnot move downward?, (take g = 10 ms–2), [9 Jan. 2019 I], (a), , m1g, , 45. A given object takes n times more time to slide down a 45°, rough inclined plane as it takes to slide down a perfectly, smooth 45° incline. The coefficient of kinetic friction, between the object and the incline is :, , (a), (c), , 1-, , n2, , [Online April 15, 2018], 1, (b) 1 - 2, n, , 2, , (d), , 1, , 1, 2-n, , 1, 1 - n2, , 46. A body of mass 2kg slides down with an acceleration of, 3m/s2 on a rough inclined plane having a slope of 30°., The external force required to take the same body up the, plane with the same acceleration will be: (g = 10m/s2), [Online April 15, 2018], (a) 4N, (b) 14N, (c) 6N, (d) 20N, 47. A rocket is fired vertically from the earth with an acceleration, of 2g, where g is the gravitational acceleration. On an, inclined plane inside the rocket, making an angle q with, the horizontal, a point object of mass m is kept. The, minimum coefficient of friction mmin between the mass and, the inclined surface such that the mass does not move is :, [Online April 9, 2016], (a) tan2q, (b) tanq, (c) 3 tanq, (d) 2 tan q, 48. Given in the figure are two blocks A and B of weight 20 N, and 100 N, respectively. These are being pressed against a, wall by a force F as shown. If the coefficient of friction, between the blocks is 0.1 and between block B and the, wall is 0.15, the frictional force applied by the wall on, block B is:, [2015], F, , A, , B, , P, , 3, , N, , k, 10, , (a) 32 N, , g, , 45°, , (b) 18 N, , (c) 23 N, , (d), , 25 N, , (a) 120 N, (c) 100 N, , (b) 150 N, (d) 80 N
Page 50 :
P-41, , Laws of Motion, , 49. A block of mass m = 10 kg rests on a horizontal table. The, coefficient of friction between the block and the table is, 0.05. When hit by a bullet of mass 50 g moving with speed, n, that gets embedded in it, the block moves and comes to, stop after moving a distance of 2 m on the table. If a freely, n, falling object were to acquire speed, after being dropped, 10, from height H, then neglecting energy losses and taking g, = 10 ms–2, the value of H is close to:, [Online April 10, 2015], (a) 0.05 km, (b) 0.02 km, (c) 0.03 km, (d) 0.04 km, 50. A block of mass m is placed on a surface with a vertical, x3, . If the coefficient of friction, 6, is 0.5, the maximum height above the ground at which the, block can be placed without slipping is:, [2014], , cross section given by y =, , 1, 2, 1, 1, m, m, m, m, (b), (c), (d), 6, 3, 3, 2, 51. Consider a cylinder of mass M resting on a rough horizontal, rug that is pulled out from under it with acceleration ‘a’, perpendicular to the axis of the cylinder. What is Ffriction, at point P? It is assumed that the cylinder does not slip., [Online April 19, 2014], w, v, O, , (a), , A, , h, , vo2, , L, , C, 2, vo, , (a), , 2h, +, m 2mg, , (b), , h, +, m 2mg, , (c), , h v2o, +, 2m mg, , (d), , v2, h, + o, 2m 2mg, , 54. A block A of mass 4 kg is placed on another block B of, mass 5 kg, and the block B rests on a smooth horizontal, table. If the minimum force that can be applied on A so, that both the blocks move together is 12 N, the maximum, force that can be applied to B for the blocks to move, together will be:, [Online April 9, 2014], (a) 30 N, (b) 25 N, (c) 27 N, (d) 48 N, 55. A block is placed on a rough horizontal plane. A time, dependent horizontal force F = kt acts on the block, where, k is a positive constant. The acceleration - time graph of, the block is :, [Online April 25, 2013], a, a, , (a), , P, , B, , (b), , O, a, , a, , O, a, , t, , t, , F friction, Ma, Ma, (d), 3, 2, 52. A heavy box is to dragged along a rough horizontal floor., To do so, person A pushes it at an angle 30° from the, horizontal and requires a minimum force FA, while person, B pulls the box at an angle 60° from the horizontal and, needs minimum force FB. If the coefficient of friction, , (a) Mg, , (b) Ma, , (c), , between the box and the floor is, , (a), , 3, , (b), , FA, 3, , the ratio, is, FB, 5, [Online April 19, 2014], , 5, 3, , 2, 3, (d), 3, 2, 53. A small ball of mass m starts at a point A with speed vo, and moves along a frictionless track AB as shown. The, track BC has coefficient of friction m. The ball comes to, stop at C after travelling a distance L which is:, [Online April 11, 2014], (c), , (c), , (d), , O, O, t, t, 56. A body starts from rest on a long inclined plane of slope, 45°. The coefficient of friction between the body and, the plane varies as m = 0.3 x, where x is distance travelled, down the plane. The body will have maximum speed, (for g = 10 m/s2) when x =, [Online April 22, 2013], (a) 9.8 m, (b) 27 m, (c) 12 m, (d) 3.33 m, 57. A block of weight W rests on a horizontal floor with, coefficient of static friction m. It is desired to make the, block move by applying minimum amount of force. The, angle q from the horizontal at which the force should be, applied and magnitude of the force F are respectively., [Online May 19, 2012], mW, , (a), , q = tan -1 ( m) , F =, , (b), , æ 1ö, mW, q = tan -1 ç ÷ , F =, è mø, 1 + m2, , 1 + m2
Page 51 :
P-42, , Physics, , (c), , q = 0, F = mW, , æ m ö, mW, ,F =, (d) q = tan -1 ç, ÷, 1+ m, è 1 + mø, 58. An insect crawls up a hemispherical surface very slowly., The coefficient of friction between the insect and the, surface is 1/3. If the line joining the centre of the, hemispherical surface to the insect makes an angle a with, the vertical, the maximum possible value of a so that the, insect does not slip is given by [Online May 12, 2012], , 64. A block rests on a rough inclined plane making an angle of, 30° with the horizontal. The coefficient of static friction, between the block and the plane is 0.8. If the frictional, force on the block is 10 N, the mass of the block (in kg) is, 2, (take g = 10 m / s ), [2004], (a) 1.6, (b) 4.0, (c) 2.0, (d) 2.5, 65. A horizontal force of 10 N is necessary to just hold a block, stationary against a wall. The coefficient of friction between, the block and the wall is 0.2. The weight of the block is, [2003], , a, , 10N, (a) cot a = 3, (b) sec a = 3, (c) cosec a = 3, (d) cos a = 3, 59. The minimum force required to start pushing a body up, rough (frictional coefficient m) inclined plane is F1 while, the minimum force needed to prevent it from sliding down, is F2. If the inclined plane makes an angle q from the, horizontal such that tan q = 2m then the ratio, , F1, is, F2, [2011 RS], , (a) 1, (b) 2, (c) 3, (d) 4, 60. If a spring of stiffness ‘k’ is cut into parts ‘A’ and ‘B’ of, length l A : l B = 2 : 3, then the stiffness of spring ‘A’ is, given by, (a), , [2011 RS], , 3k, 5, , (b), , 2k, 5, , 5k, 2, 61. A smooth block is released at rest on a 45° incline and then, slides a distance ‘d’. The time taken to slide is ‘n’ times as, much to slide on rough incline than on a smooth incline., The coefficient of friction is, [2005], , (c) k, , (a), , (c), , (d), , 1, mk = 1 – 2, n, ms = 1 -, , 1, n, , 2, , (b), , (d), , mk = 1ms = 1-, , (c) 400 m, , TOPIC 4, , Circular Motion, Banking of, Road, , 67. A disc rotates about its axis of symmetry in a hoizontal, plane at a steady rate of 3.5 revolutions per second. A coin, placed at a distance of 1.25cm from the axis of rotation, remains at rest on the disc. The coefficient of friction, between the coin and the disc is (g = 10m/s2), [Online April 15, 2018], (a) 0.5, (b) 0.7, (c) 0.3, (d) 0.6, 68. A conical pendulum of length 1 m makes an angle q = 45°, w.r.t. Z-axis and moves in a circle in the XY plane.The, radius of the circle is 0.4 m and its centre is vertically below O. The speed of the pendulum, in its circular path, will, be :, (Take g = 10 ms–2), [Online April 9, 2017], Z, , 1, , (a) 0.4 m/s, , n2, , (b) 4 m/s, , 1, , (c) 0.2 m/s, , n2, , 62. The upper half of an inclined plane with inclination f is, perfectly smooth while the lower half is rough. A body, starting from rest at the top will again come to rest at the, bottom if the coefficient of friction for the lower half is, given by, [2005], (a) 2 cos f (b) 2 sin f (c) tan f (d) 2 tan f, 63. Consider a car moving on a straight road with a speed of, 100 m/s . The distance at which car can be stopped is, [ m k = 0.5 ], (a) 1000 m (b) 800 m, , (a) 20 N, (b) 50 N, (c) 100 N (d) 2 N, 66. A marble block of mass 2 kg lying on ice when given a, velocity of 6 m/s is stopped by friction in 10 s. Then the, coefficient of friction is, [2003], (a) 0.02, (b) 0.03, (c) 0.04, (d) 0.06, , O, q, , C, , (d) 2 m/s, 69. A particle is released on a vertical smooth semicircular, track from point X so that OX makes angle q from the, vertical (see figure). The normal reaction of the track on, the particle vanishes at point Y where OY makes angle f, with the horizontal. Then:, [Online April 19, 2014], X, Y, , q, , [2005], (d) 100 m, , f, O
Page 52 :
P-43, , Laws of Motion, , 1, cos q, 2, 2, 3, (c) sin f = cos q, (d) sin f = cos q, 3, 4, 70. A body of mass ‘m’ is tied to one end of a spring and, whirled round in a horizontal plane with a constant angular, velocity. The elongation in the spring is 1 cm. If the, angular velocity is doubled, the elongation in the spring, is 5 cm. The original length of the spring is :, [Online April 23, 2013], (a) 15 cm, (b) 12 cm, (c) 16 cm, (d) 10 cm, 71. A point P moves in counter-clockwise direction on a, circular path as shown in the figure. The movement of 'P' is, such that it sweeps out a length s = t3 + 5, where s is in, metres and t is in seconds. The radius of the path is 20 m., The acceleration of 'P' when t = 2 s is nearly., [2010], y, , (a) sin f = cos f, , (b), , sin f =, , B, P(x,y), m, 20, , O, , A, , x, , (a) 13m/s2, (b) 12 m/s2, (c) 7.2 ms2, (d) 14m/s2, 72. For a particle in uniform circular motion, the acceleration, r, a at a point P(R,q) on the circle of radius R is (Here q is, measured from the x-axis), [2010], , (a), , -, , n2, n2, cos q iˆ +, sin q ˆj, R, R, , (b), , -, , n2, n2, sin q iˆ +, cos q ˆj, R, R, , (c) -, , n2, n2, cos q iˆ sin q ˆj, R, R, , n2 ˆ n2 ˆ, i+, j, R, R, 73. An annular ring with inner and outer radii R1 and R2 is, rolling without slipping with a uniform angular speed. The, ratio of the forces experienced by the two particles situated, F1, on the inner and outer parts of the ring , F is [2005], 2, , (d), , (a), , æ R1 ö, çè R ÷ø, 2, , 2, , (b), , R2, R1, , R1, (d) 1, R2, 74. Which of the following statements is FALSE for a particle, moving in a circle with a constant angular speed ? [2004], (a) The acceleration vector points to the centre of the, circle, (b) The acceleration vector is tangent to the circle, (c) The velocity vector is tangent to the circle, (d) The velocity and acceleration vectors are, perpendicular to each other., 75. The minimum velocity (in ms-1) with which a car driver must, traverse a flat curve of radius 150 m and coefficient of friction, 0.6 to avoid skidding is, [2002], (a) 60, (b) 30, (c) 15, (d) 25, , (c)
Page 54 :
P-45, , Laws of Motion, , Þ 40 - 0 = 2 (a) 0.2 Þ a = 100 m/s2, \ F = ma = 0.2 ´ 100 = 20 N, Þ N - mg = 20 Þ N = 20 + 2 = 22 N, Note :, Whand + Wgravity = DK, , mv, , L–2nr, , mv, , 9., , (a) From question,, Mass of body, m = 5 kg, Velocity at t = 0,, , Þ F (0.2) + (0.2)(10)(2.2) = 0 Þ F = 22 N, 14. (c) Given, mass of cricket ball, m = 150 g = 0.15 kg, Initial velocity, u = 20 m/s, Force,, , u = (6iˆ - 2 ˆj) m/s, Velocity at t = 10s,, , m(v - u ) 0.15(0 - 20), =, = 30 N, t, 0.1, 15. (c) Mass (m) = 0.3 kg, Force, F = m.a = –kx, Þ ma = –15x, Þ 0.3a = –15x, 15, -150, x=, x = - 50 x, Þ a= –, 0.3, 3, a = –50 × 0.2 = 10m/s2, 16. (c) When the incline is given an acceleration a towards, the right, the block receives a reaction ma towards left., F=, , v = + 6 ĵ m/s, Force, F = ?, Acceleration, a =, , v -u, t, -3iˆ + 4 ˆj, , 6 ˆj - (6iˆ - 2 ˆj ), =, m/s2, 10, 5, Force, F = ma, ( -3iˆ + 4 ˆj ), = 5´, = ( -3iˆ + 4 ˆj ) N, 5, 10. (c) Given that F(t) = F0e–bt, dv, Þ m, = F0e–bt, dt, dv, F0 -bt, e, =, dt, m, =, , v, , ò dv =, 0, , ma, g cos, , a, mg cosa, + ma sina mg, , F0 -bt, e dt, mò, 0, , F0 é, F é e - bt ù, - e - bt - e -0 ù, v= 0ê, ú =, û, mb ë, m ëê -b ûú, , (, , ), , 0, , F0 é, 1 - e -bt ù, û, mb ë, 11. (a) According to newton third law of motion i.e. every, action is associated with equal and opposite reaction., 12. (d) mg sin q = ma, \ a = g sin q, \ Vertical component of acceleration, = g sin2 q, \ Relative vertical acceleration of A with respect to B is, , Þ v=, , g (sin 2 60 - sin 2 30], æ3 1ö g, = g ç – ÷ = = 4.9 m/s2, è4 4ø 2, in vertical direction, 13. (d) For the motion of ball, just after the throwing, v = 0, s = 2m, a = –g = –10ms–2, v2 – u2 = 2as for upward journey, , Þ -u 2 = 2( -10) ´ 2 Þ u 2 = 40, When the ball is in the hands of the thrower, , u = 0, v = 40 ms–1, s = 0.2 m, v2 – u2 = 2as, , a, , a, , t, , t, , N, a, , mg sin a, , For block to remain stationary, Net force along the incline, should be zero., mg sin a = ma cos a Þ a = g tan a, 17. (b) In the absence of air resistance, if, Thrust (F), the rocket moves up with an acceleration, a, then thrust, F = mg + ma, a, , \ F = m ( g + a) = 3.5 × 104 ( 10 + 10), = 7 × 105 N, , mg, , 18. (d) Resultant force is zero, as three forces are represented, by the sides of a triangle taken in the same order. From, r, r, Newton’s second law, Fnet = ma., Therefore, acceleration is also zero i.e., velocity remains, unchanged., 19. (d) This is a case of sliding (if plane is friction less) and, therefore the acceleration of all the bodies is same., 20. (a) From the free body diagram
Page 55 :
P-46, , Physics, , T cos45° = 100 N, T sin45° = F, On dividing (i) by (ii) we get, , ...(i), ...(ii), , 480, = 12 ms -2, 20 + 12 + 8, Tension at the mid point, =, , T cos 45° 100, =, T sin 45°, F, Þ F = 100 N, , Mass of rod ö, æ, T = ç M2 +, ÷a, 2, è, ø, = (12 + 4) × 12 = 192 N, 25. (d) P, , 21. (d) Net force on the elevator = force on elevator, + frictional force, , q, , Þ F = (10 m + M)g + f, where, m = mass of person, M = mass of elevator,, , 8 cm, T, , f = frictional force, Þ F = (10 × 68 + 920) × 9.8 + 600, , Q 5 cm O, , Þ F = 22000 N, Þ P = FV = 22000 × 3 = 66000 W, 22. (d), , w wcosq, PQ = OP 2 + OQ2, , o, , 45, , = 132 + 52 = 12, Tension in the string, o, , 45, , T = w cos q =, , F, , 26. (a), 27. (d) Writing free body-diagrams for m & M,, , 100 N, , At equilibrium,, mg 100, =, tan 45° =, F, F, \ F = 100 N, 23. (b) From figure,, , m, , M, K, , F, N, , N, m, , R, , mg, , a, , T, T, m a, mg, , Acceleration a = Ra, and mg – T = ma, From equation (i) and (ii), æ aö, T × R = mR2a = mR2 çè ÷ø, R, or T = ma, Þ mg – ma = ma, g, Þ a=, 2, 24. (d) Acceleration produced in upward direction, a=, , 13, W, 12, , F, M1 + M 2 + Mass of metal rod, , …(i), …(ii), , a, T T, , M, , F, , Mg, , we get T = ma and F – T = Ma, where T is force due to spring, Þ F – ma = Ma or,, F = Ma + ma, \ Acceleration of the system, F, a=, ., M +m, Now, force acting on the block of mass m is, æ F ö = mF, ma = m ç, ., è M + m ÷ø m + M, If a is the acceleration along the inclined plane, then, 28. (c) Here, m1 = 5kg and m2 = 4.8 kg., If a is the acceleration of the masses,, m1a = m1g – T ...(i), m2a = T – m2g ...(ii), Solving (i) and (ii) we get, æ m - m2 ö, a=ç 1, ÷g, è m1 + m2 ø, , Þa=, , (5 - 4.8) ´ 9.8, m / s2 = 0.2 m/s2, (5 + 4.8)
Page 56 :
P-47, , Laws of Motion, , 29. (a) When lift is stationary, W1 = mg ...(i), When the lift descends with acceleration, a, W2 = m(g – a), 49, (10 – 5) = 24.5 N, W2 =, 10, , 34. (b) Let the two forces be F1 and F2 and let F2 < F1. R is, the resultant force., Given F1 + F2 = 18, ...(i), From the figure F22 + R 2 = F12, F12 - F22 = R 2, , \ F12 - F22 = 144, Only option (b) follows equation (i) and (ii)., F1, , T, a, mg, , R, , F2, , 30. (d) Taking the rope and the block as a system, a, M, T, , m, , P, , we get P = (m + M)a, P, \ Acceleration produced, a =, m+M, Taking the block as a system,, Force on the block, F = Ma, MP, m+M, 31. (a) The Earth exerts a pulling force Mg. The block in turn, exerts a reaction force Mg on the spring of spring balance, S1 which therefore shows a reading of M kgf., As both the springs are massless. Therefore, it exerts a, force of Mg on the spring of spring balance S2 which, shows the reading of M kgf., , 35. (b) For mass m1, m1g – T = m1a, For mass m2, T–m2g = m2a, , \ F=, , s2, , Mkgf, , Mkgf, , s1, , M, Mg, , 32. (c) Case - I: For the man standing in the lift, the, acceleration of the ball, r, r r, abm = ab - am Þ abm = g – a, Case - II: The man standing on the ground, the acceleration, of the ball, r, r r, abm = ab - am Þ abm = g – 0 = g, 33. (a) When forces F1, F2 and F3 are acting on the particle,, it remains in equilibrium. Force F2 and F3 are perpendicular, to each other,, F1 = F2 + F3, F22 + F32, The force F1 is now removed, so, resultant of F2 and F3, will now make the particle move with force equal to F1., F, Then, acceleration, a = 1, m, , \ F1 =, , F1, , ...(i), ...(ii), , T, , a m2, , T, m1, , m2g, , a, , m1g, Adding the equations we get, , a=, , (m1 - m2 ) g, m1 + m2, , g, 8, m1, -1, m, m, m, 9, 1 m2, Þ 1 +1 = 8 1 - 8 Þ 1 =, \, =, m, m, m, m, 7, 8, 1 +1, 2, 2, 2, m2, 36. (b) Force = mass × acceleration, \ F = (m + m + m) × a, F = 3m × a, F, a=, 3m, 10.2, m / s2, \a =, 6, 10.2, \ T2 = ma = 2 ´, = 3.4N, 6, , Here a =, , 2 kg, C, , T2, T2, , 2 kg, , 2 kg, , B, , A, , T1, , T1, , 37. (c) Tension, T = 360 N, Mass of a man m = 60 kg, mg – T = ma, , F, , ...(ii)
Page 58 :
P-49, , Laws of Motion, , 45. (b) The coefficients of kinetic friction between the object, and the incline, , But, amax = mg = 0.2 × 10 = 2, F -8, =2, 4, Þ F = 16 N, \, , 1, 1ö, æ, m = tan q ç1 - 2 ÷ Þ m = 1 - 2, è n ø, n, , 42. (a) From figure, 2 + mg sin 30° = mmg cos 30° and, 10 = mg sin 30° + m mg cos 30°, = 2mmg cos 30° – 2, Þ 6 = mmg cos 30° and, 4 = mg cos 30°, By dividing above two, , 3, 2, , 2, , 10, k, , g, , P, , µ = 0.6, , Ma, C, , kg, , q 30°, , B, , A, , k, , m3mgcosq, , µmgcos?, , 3 + mgsing?, , N, , 3, , 2, , A, , Þ F – 10 – 4 = 6, F 30°, q, F = 20 N, C, B, 47. (b) Let m be the minimum coefficient of friction, , P, , 43. (a), , 46. (d) Equation of motion when the mass slides down, Mg sin q – f = Ma, Þ 10 – f = 6 (M = 2 kg, a = 3 m/s2, q = 30° given), \ f = 4N, f, Equation of motion when the block is, pushed up, Let the external force required to take, the block up the plane with same, Ma, acceleration be F, f, F – Mg sin q – f = Ma, g, , 3, Þ =m´ 3, 2, , \ Coefficient of friction, m =, , (Q q = 45°), , 45°, , mg sin 45° =, , 3mgsinq, , 100, 2, , = 50 2, , [Qm = 10kg, g = 9.8 m s, , mmg cosq = 0.6 × mg ×, , -2, , At equilibrium, mass does not move so,, 3mg sinq = m3mg cosq, , ], , 1, , 2, 3 + mg sinq = P + mmg cosq, , 3g, , [ μ min < tan θ, , = 0.6 ´ 50 2, , 3 + 50 2 = P + 30 2, \ P = 31.28 = 32 N, 44. (b) Given : m1 = 5kg; m2 = 10kg; m = 0.15, FBD for m1, m1g – T = m1a, Þ 50 – T = 5 × a, and, T – 0.15 (m + 10)g = (10 + m)a, For rest a = 0, or, 50 = 0.15 (m + 10) 10, N, m, T, m2, m(m+m2)g (m+m2)g, , f2, , f1, , 48. (a), , F, , A, , 20N, , B, f1, , N, , 100N, , Assuming both the blocks are stationary, N= F, f1 = 20 N, f2 = 100 + 20 = 120N, f, , T, m1, m1g = 50N, , Þ 5=, , 3, (m + 10), 20, , 100, = m + 10 \ m = 23.3kg; close to option (b), 3, , 120N, , Considering the two blocks as one system and due to, equilibrium f = 120N
Page 61 :
P-52, , Physics, , 60. (d) It is given lA : lB = 2 : 3, lA, , 2l, æ 3l ö, = , lB = ç ÷, è 5ø, 5, , 1, l, If initial spring constant is k, then, , \ We know that k µ, k l = k Al A = k B l B, , æ 2l ö, kl = kA ç ÷, è 5ø, 5k, kA =, 2, 61. (b), , a = g sin q - mg cos q, , d q, sin d, g, =, a 45°, 45°, smooth, rough, On smooth inclined plane, acceleration of the body = g, sin q. Let d be the distance travelled, 1, \ d = ( g sin q)t12 ,, 2, 2d, ,, t1 =, g sin q, On rough inclined plane,, mg sin q – mR, a=, m, mg sin q – mmg cos q, Þ a=, m, Þ a = g sin q – mkg cos q, 1, ˆ cos q) t 2, \ d = ( g sin q - mkg, 2, 2, 2d, ˆ cos q, g sin q - mkg, According to question, t2 = nt1, , g sin f = -( g sin f - mg cos f), Þ m = 2 tan f, NOTE, According to work-energy theorem, W = DK = 0, (Since initial and final speeds are zero), \ Workdone by friction + Work done by gravity = 0, l, i.e., -( µ mg cos f ) + mg l sin f = 0, 2, µ, cos f = sin f or µ = 2 tan f, or, 2, 63. (a) Given, initial velocity, u = 100m/s., Final velocity, v = 0., Acceleration, a = mkg = 0.5 × 10, v2 – u2 = 2as or, Þ 02 – u2 = 2(–mkg)s, 1, Þ -1002 = 2 ´ - ´ 10 ´ s, 2, Þ s = 1000 m, 64. (c), , fs, N, , m, , in, gs, , 2d, ˆ cos q, g sin q - mkg, , 2d, =, g sin q, , Here, m is coefficient of kinetic friction as the block, moves over the inclined plane., \ sin q = (sin q – mk̂ cos q)n2, 1, 1, 2, Þ n =, Þ n=, 1 - mk, 1 - mk, Þ mk = 1 -, , 1, , n2, 62. (d) For first half, acceleration = g sin f;, For second half, , acceleration = – ( g sin f - mg cos f), For the block to come to rest at the bottom, acceleration, in I half = retardation in II half., , sq, co, , 30°, , Since the body is at rest on the inclined plane,, mg sin 30° = Force of friction, Þ m ´ 10 ´ sin 30° = 10, Þ m ´ 5 = 10 Þ m = 2.0 kg, 65. (d) Horizontal force, N = 10 N., Coefficient of friction m = 0.2., f = mN, , 10N, , t2 =, , n, , °, 30, , mg, mg, , 10N, , 10N, , W, , The block will be stationary so long as, Force of friction = weight of block, \ mN = W, Þ 0.2 × 10 = W, Þ W = 2N, 66. (d) u = 6 m/s, v = 0, t = 10s, a = ?, Acceleration a =, , v–u, t, , 0–6, 10, -6, = -0.6m / s2, Þ a=, 10, , Þ a=, , mg, , f = mN N, The retardation is due to the frictional force, \ f = ma Þ mN = ma
Page 63 :
P-54, , 5, , Physics, , Work, Energy and, Power, 1., , 2., , 3., , A person pushes a box on a rough horizontal platform, surface. He applies a force of 200 N over a distance of, 15 m. Thereafter, he gets progressively tired and his applied, force reduces linearly with distance to 100 N. The total, distance through which the box has been moved is 30 m., What is the work done by the person during the total, movement of the box ?, [4 Sep. 2020 (II)], (a) 3280 J, (b) 2780 J, (c) 5690 J, (d) 5250 J, , C, , 1, 3, J, (c) 1J, (d), J, 2, 2, A block of mass m is kept on a platform which starts, from rest with constant acceleration g/2 upward, as, shown in fig. work done by normal reaction on block in, time t is:, [10 Jan. 2019 I], , (a) 2J, , TOPIC 1 Work, 4., , (a) -, , B, , q, A, A small block starts slipping down from a point B on an, inclined plane AB, which is making an angle q with the, horizontal section BC is smooth and the remaining section, CA is rough with a coefficient of friction m. It is found that, the block comes to rest as it reaches the bottom (point A), of the inclined plane. If BC = 2AC, the coefficient of friction, is given by m = k tanq. The value of k is _________., , 5., , [NA 2 Sep. 2020 (I)], r, Consider a force F = - xiˆ + yjˆ . The work done by this, , 6., , force in moving a particle from point A(1, 0) to B(0, 1), along the line segment is: (all quantities are in SI units), [9 Jan. 2020 I], , m g2 t 2, 8, , (b), , m g2 t2, 8, , 3m g 2 t 2, 8, A body of mass starts moving from rest along x-axis so, that its velocity varies as v = a s where a is a constant s, and is the distance covered by the body. The total work, done by all the forces acting on the body in the first second, after the start of the motion is: [Online April 16, 2018], 1 4 2, ma t, (a), (b) 4ma 4 t 2, 8, 1, ma 4 t 2, (c) 8ma 4 t 2, (d), 4, When a rubber-band is stretched by a distance x, it exerts, restoring force of magnitude F = ax + bx2 where a and b are, constants. The work done in stretching the unstretched, rubber-band by L is:, [2014], (c) 0, , (d), , (a) aL2 + bL3, , (b), , (c), 7., , (b), , aL2 bL3, +, 2, 3, , (, , 1, aL2 + bL3, 2, , ), , 1 æ aL2 bL3 ö, (d) 2 çç 2 + 3 ÷÷, è, ø, , A uniform chain of length 2 m is kept on a table such that, a length of 60 cm hangs freely from the edge of the table., The total mass of the chain is 4 kg. What is the work done, in pulling the entire chain on the table ?, [2004], (a) 12 J, (b) 3.6 J, (c) 7.2 J, (d) 1200 J
Page 64 :
P-55, , Work, Energy & Power, , 8., , r, r, r, r, A force F = (5i + 3 j + 2k ) N is applied over a particle, which displaces it from its origin to the point, r r, r, r = (2i - j )m. The work done on the particle in joules is, , [2004], (a) +10, (b) +7, (c) –7, (d) +13, 9. A spring of spring constant 5 × 103 N/m is stretched initially, by 5cm from the unstretched position. Then the work, required to stretch it further by another 5 cm is [2003], (a) 12.50 N-m, (b) 18.75 N-m, (c) 25.00 N-m, (d) 6.25 N-m, 10. A spring of force constant 800 N/m has an extension of 5 cm., The work done in extending it from 5 cm to 15 cm is, [2002], (a) 16 J, (b) 8 J, (c) 32 J, (d) 24 J, , th, , TOPIC 2 Energy, 11. A cricket ball of mass 0.15 kg is thrown vertically up by a, bowling machine so that it rises to a maximum height of 20, m after leaving the machine. If the part pushing the ball, applies a constant force F on the ball and moves, horizontally a distance of 0.2 m while launching the ball,, the value of F (in N) is (g = 10 ms–2) __________., [NA 3 Sep. 2020 (I)], 12. A particle (m = l kg) slides down a frictionless track, (AOC) starting from rest at a point A (height 2 m). After, reaching C, the particle continues to move freely in air, as a projectile. When it reaching its highest point P, (height 1 m), the kinetic energy of the particle (in J) is:, (Figure drawn is schematic and not to scale; take g = 10, ms–2) ¾¾¾ ., [NA 7 Jan. 2020 I], Height, , A, , 14. A spring whose unstretched length is l has a force, constant k. The spring is cut into two pieces of, unstretched lengths 11 and l2 where, l1 = nl2 and n is an, integer. The ratio k 1/k 2 of the corresponding force, constants, k1 and k2 will be:, [12 April 2019 II], 1, 1, (c), (d) n 2, (a) n, (b) 2, n, n, 15. A body of mass 1 kg falls freely from a height of 100m, on, a platform of mass 3 kg which is mounted on a spring, having spring constant k = 1.25 × 106 N/m. The body sticks, to the platform and the spring’s maximum compression is, found to be x. Given that g = 10 ms–2, the value of x will be, close to :, [11 April 2019 I], (a) 40 cm (b) 4 cm, (c) 80 cm, (d) 8 cm, 16. A uniform cable of mass ‘M’ and length ‘L’ is placed on a, , P, C, , 2m, , æ 1ö, horizontal surface such that its ç ÷ part is hanging, è nø, below the edge of the surface. To lift the hanging part of, the cable upto the surface, the work done should be:, [9 April 2019 I], 2MgL, MgL, MgL, (a), (b), (c), (d) nMgL, 2, 2, 2n, n, n2, 17. A wedge of mass M = 4m lies on a frictionless plane. A, particle of mass m approaches the wedge with speed v., There is no friction between the particle and the plane, or between the particle and the wedge. The maximum, height climbed by the particle on the wedge is given by:, [9 April 2019 II], , (a), , v2, g, , (b), , 2v 2, 7g, , 2v 2, v2, (d), 5g, 2g, 18. A particle moves in one dimension from rest under the, influence of a force that varies with the distance travelled, by the particle as shown in the figure. The kinetic energy, of the particle after it has travelled 3 m is :, [8 April 2019 I], , (c), , O, , 13. A particle moves in one dimension from rest under the, influence of a force that varies with the distance travelled, by the particle as shown in the figure. The kinetic energy, of the particle after it has travelled 3 m is :, [7 Jan. 2020 II], , (a) 4 J, (c) 6.5 J, , (b) 2.5 J, (d) 5 J, , (a) 4 J, (b) 2.5 J, (c) 6.5 J, (d) 5 J, 19. A particle which is experiencing a force, given by, r, r, r, r, r, F = 3i - 12 j, undergoes a displacement of d = 4i. If, the particle had a kinetic energy of 3 J at the beginning, of the displacement, what is its kinetic energy at the end, of the displacement?, [10 Jan. 2019 II], (a) 9 J, (b) 12 J, (c) 10 J, (d) 15 J
Page 65 :
P-56, , Physics, , 20. A block of mass m, lying on a smooth horizontal surface,, is attached to a spring (of negligible mass) of spring, constant k. The other end of the spring is fixed, as shown, in the figure. The block is initally at rest in its equilibrium, position. If now the block is pulled with a constant force, F, the maximum speed of the block is: [9 Jan. 2019 I], , v (m/s), -1, , 50 ms, , (0,0), , m, , (a), , 2F, mk, , (b), , F, p mk, , (c), , F, , pF, mk, , F, , (d), , mk, , 21. A force acts on a 2 kg object so that its position is given, as a function of time as x = 3t2 + 5. What is the work, done by this force in first 5 seconds?, [9 Jan. 2019 II], (a) 850 J (b) 950 J, (c) 875 J, (d) 900 J, 22. A particle is moving in a circular path of radius a under the, action of an attractive potential U = (a) -, , k, , (b), , 4a 2, , k, 2r 2, , . Its total energy is:, [2018], , (a) 3h, , (b), , ¥, , (c), , 5, h, 3, , (d), , 8, h, 3, , 26. A time dependent force F = 6t acts on a particle of mass, 1 kg. If the particle starts from rest, the work done by the, force during the first 1 second will be, [2017], (a) 9 J, (b) 18 J, (c) 4.5 J, (d) 22 J, 27. Velocity–time graph for a body of mass 10 kg is shown in, figure. Work–done on the body in first two seconds of, the motion is :, [Online April 10, 2016], , t(s), , P, , k, , 2a 2, 3 k, (c) zero, (d) - 2, 2a, 23. Two particles of the same mass m are moving in circular, -16 3, -r, orbits because of force, given by F(r) =, r, The first particle is at a distance r = 1, and the second, at, r = 4. The best estimate for the ratio of kinetic energies, of the first and the second particle is closest to, [Online April 16, 2018], (a) 10–1, (b) 6 × 10–2 (c) 6 × 102 (d) 3 × 10–3, 24. A body of mass m = 10–2 kg is moving in a medium and, experiences a frictional force F = –kv2. Its intial speed is v0 =, 1 2, 10 ms–1. If, after 10 s, its energy is mv0 , the value of k will, 8, be:, [2017], (a) 10–4 kg m–1, (b) 10–1 kg m–1 s–1, (c) 10–3 kg m–1, (d) 10–3 kg s–1, 25. An object is dropped from a height h from the ground., Every time it hits the ground it looses 50% of its kinetic, energy. The total distance covered as t ® ¥ is, [Online April 8, 2017], , 10s, , (a) – 9300 J, (b) 12000 J, (c) –4500 J, (d) –12000 J, 28. A point particle of mass m, moves long the uniformly, rough track PQR as shown in the figure. The coefficient, of friction, between the particle and the rough track, equals m. The particle is released, from rest from the, point P and it comes to rest at a point R. The energies,, lost by the ball, over the parts, PQ and QR, of the track,, are equal to each other, and no energy is lost when particle, changes direction from PQ to QR., The value of the coefficient of friction m and the distance, x (= QR), are, respectively close to :, [2016], h=2m, 30°, Horizontal, Surface, , R, Q, , (a) 0.29 and 3.5 m, (b) 0.29 and 6.5 m, (c) 0.2 and 6.5 m, (d) 0.2 and 3.5 m, 29. A person trying to lose weight by burning fat lifts a mass, of 10 kg upto a height of 1 m 1000 times. Assume that the, potential energy lost each time he lowers the mass is, dissipated. How much fat will he use up considering the, work done only when the weight is lifted up? Fat supplies, 3.8 × 107 J of energy per kg which is converted to, mechanical energy with a 20% efficiency rate. Take g = 9.8, ms–2 :, [2016], (a) 9.89 × 10–3 kg, (b) 12.89 × 10–3 kg, (c) 2.45 × 10–3 kg, (d) 6.45 × 10–3 kg, 30. A particle is moving in a circle of radius r under the action, of a force F = ar2 which is directed towards centre of the, circle. Total mechanical energy (kinetic energy + potential, energy) of the particle is (take potential energy = 0 for r = 0) :, [Online April 11, 2015], 5 3, 4 3, 1 3, αr, ar (b) ar, (c), (d) ar3, 6, 3, 2, 31. A block of mass m = 0.1 kg is connected to a spring of, unknown spring constant k. It is compressed to a distance, x from its equilibrium position and released from rest. After, approaching half the distance æç x ö÷ from equilibrium, è 2ø, position, it hits another block and comes to rest, momentarily, while the other block moves with a velocity, 3 ms–1., The total initial energy of the spring is :, [Online April 10, 2015], , (a)
Page 66 :
P-57, , Work, Energy & Power, , (a) 0.3 J, (c) 0.8 J, , (a) constant, , (b) 0.6 J, (d) 1.5 J, , (c), , th, , æ1ö, 32. A bullet looses ç ÷ of its velocity passing through, ènø, one plank. The number of such planks that are required, to stop the bullet can be:, [Online April 19, 2014], , (c), , 1 2, mv, 3, , (d), , 1, mv 2, 6, , 34. Two springs of force constants 300 N/m, (Spring A) and 400 N/m (Spring B) are joined together in, series. The combination is compressed by 8.75 cm. The, E, E, ratio of energy stored in A and B is A . Then A is, EB, EB, equal, to :, [Online April 9, 2013], (a), , 4, 3, , (b), , 16, 9, , (c), , 3, 4, , (d), , 3, , F (in N), , 2, 1, x (in m), , 0, –1, –2, 0 1 2 3 4 5 6 78, , (a) 34 J, (b) 34.5 J, (c) 4.5 J, (d) 29.4 J, 36. A particle gets displaced by, D r = 2iˆ + 3 ˆj + 4kˆ m under the action of a force, r, F = 7iˆ + 4 ˆj + 3kˆ . The change in its kinetic energy is, , (, , (, , ), , ), , [Online May 7, 2012], (a) 38 J, (b) 70 J, (c) 52.5 J, (d) 126 J, 37. At time t = 0 a particle starts moving along the x-axis. If, its kinetic energy increases uniformly with time ‘t’, the, net force acting on it must be proportional to [2011 RS], , t, , D = éëU ( x = ¥) - U at equilibrium ùû , D is, , [2010], , b2, b2, b2, b2, (b), (c), (d), 2a, 12a, 4a, 6a, An athlete in the olympic games covers a distance of 100, m in 10 s. His kinetic energy can be estimated to be in the, range, [2008], 5, 5, (a) 200 J - 500 J, (b) 2 × 10 J - 3 × 10 J, (c) 20, 000 J - 50,000 J, (d) 2,000 J - 5, 000 J, A 2 kg block slides on a horizontal floor with a speed of 4m/, s. It strikes a uncompressed spring, and compresses it till, the block is motionless. The kinetic friction force is 15N and, spring constant is 10,000 N/m. The spring compresses by, [2007], (a) 8.5 cm, (b) 5.5 cm, (c) 2.5 cm, (d) 11.0 cm, A particle is projected at 60o to the horizontal with a kinetic, energy K. The kinetic energy at the highest point is, (a) K /2, (b) K, [2007], (c) Zero, (d) K /4, A particle of mass 100g is thrown vertically upwards with, a speed of 5 m/s. The work done by the force of gravity, during the time the particle goes up is, [2006], , (a), , 39., , 40., , 9, 16, , r, 35. The force F = Fiˆ on a particle of mass 2 kg, moving along, the x-axis is given in the figure as a function of its position, x. The particle is moving with a velocity of 5 m/s along the, x-axis at x = 0. What is the kinetic energy of the particle at, x = 8 m?, [Online May 26, 2012], , (d), , t, , atoms in a diatomic molecule is approximately given by, a, b, U(x) = 12 - 6 , where a and b are constants and x is, x, x, the distance between the atoms. If the dissociation energy, of the molecule is, , n2, 2n 2, (b), (c) infinite (d) n, 2n - 1, n -1, 33. A spring of unstretched length l has a mass m with one, end fixed to a rigid support. Assuming spring to be made, of a uniform wire, the kinetic energy possessed by it if its, free end is pulled with uniform velocity v is:, [Online April 12, 2014], 1, mv 2 (b) mv2, 2, , 1, , 38. The potential energy function for the force between two, , (a), , (a), , (b) t, , 41., , 42., , (a) –0.5 J, (b) –1.25 J, (c) 1.25 J, (d) 0.5 J, 43. The potential energy of a 1 kg particle free to move along, æ x 4 x2 ö, the x-axis is given by V ( x) = ç, - ÷ J., 2ø, è 4, The total mechanical energy of the particle is 2 J. Then,, the maximum speed (in m/s) is, [2006], (a), , 3, , (b), , 1, , (d) 2, 2, 2, 44. A mass of M kg is suspended by a weightless string. The, horizontal force that is required to displace it until the, string makes an angle of 45° with the initial vertical, direction is, [2006], (a) Mg ( 2 + 1), Mg, (c), 2, , 2, , (c), , (b) Mg 2, (d) Mg ( 2 - 1)
Page 67 :
P-58, , Physics, , 45. A spherical ball of mass 20 kg is stationary at the top of, a hill of height 100 m. It rolls down a smooth surface to, the ground, then climbs up another hill of height 30 m and, finally rolls down to a horizontal base at a height of 20 m, above the ground. The velocity attained by the ball is, [2005], (a) 20 m/s, (b) 40 m/s, 46., , 47., , 48., , 49., , (c) 10 30 m/s, (d) 10 m/s, A particle moves in a straight line with retardation, proportional to its displacement. Its loss of kinetic energy, for any displacement x is proportional to, [2004], (a) x, (b) e x, (c) x2, (d) loge x, A particle is acted upon by a force of constant magnitude, which is always perpendicular to the velocity of the particle,, the motion of the particles takes place in a plane. It follows, that, [2004], (a) its kinetic energy is constant, (b) its acceleration is constant, (c) its velocity is constant, (d) it moves in a straight line, A wire suspended vertically from one of its ends is, stretched by attaching a weight of 200N to the lower end., The weight stretches the wire by 1 mm. Then the elastic, energy stored in the wire is, [2003], (a) 0.2 J, (b) 10 J, (c) 20 J, (d) 0.1 J, A ball whose kinetic energy is E, is projected at an angle of, 45° to the horizontal. The kinetic energy of the ball at the, highest point of its flight will be, [2002], (a) E, , (b) E / 2, , (c) E/2, , (d) zero, , TOPIC 3 Power, 50. A body of mass 2 kg is driven by an engine delivering a, constant power of 1 J/s. The body starts from rest and, moves in a straight line. After 9 seconds, the body has, moved a distance (in m) ________. [5 Sep. 2020 (II)], 51. A particle is moving unidirectionally on a horizontal plane, under the action of a constant power supplying energy, source. The displacement (s) - time (t) graph that describes, the motion of the particle is (graphs are drawn schematically, and are not to scale) :, [3 Sep. 2020 (II)], S, , S, , (a), , (b), t, , t, , S, , S, , (c), , (d), t, , t, , 52. A 60 HP electric motor lifts an elevator having a, maximum total load capacity of 2000 kg. If the frictional, force on the elevator is 4000 N, the speed of the elevator, at full load is close to: (1 HP = 746 W, g = 10 ms–2), [7 Jan. 2020 I], (a) 1.7 ms–1, (b) 1.9 ms–1, (c) 1.5 ms–1, (d) 2.0 ms–1, 53. A particle of mass M is moving in a circle of fixed radius, R in such a way that its centripetal acceleration at time t, is given by n2 R t2 where n is a constant. The power, delivered to the particle by the force acting on it, is :, [Online April 10, 2016], 1, M n2 R2t2, (b) M n2R2t, 2, (c) M n R2t2, (d) M n R2t, 54. A car of weight W is on an inclined road that rises by 100, m over a distance of 1 Km and applies a constant frictional, , (a), , W, on the car. While moving uphill on the road at, 20, a speed of 10 ms–1, the car needs power P. If it needs, , force, , P, while moving downhill at speed v then value of, 2, v is:, [Online April 9, 2016], (a) 20 ms–1 (b) 5 ms–1 (c) 15 ms–1 (d) 10 ms–1, 55. A wind-powered generator converts wind energy into electrical, energy. Assume that the generator converts a fixed fraction, of the wind energy intercepted by its blades into electrical, energy. For wind speed n, the electrical power output will be, most likely proportional to, [Online April 25, 2013], 4, 2, (a) n, (b) n, (c) n, (d) n, 56. A 70 kg man leaps vertically into the air from a crouching, position. To take the leap the man pushes the ground with, a constant force F to raise himself. The center of gravity, rises by 0.5 m before he leaps. After the leap the c.g. rises, by another 1 m. The maximum power delivered by the, muscles is : (Take g = 10 ms–2) [Online April 23, 2013], (a) 6.26 × 103 Watts at the start, (b) 6.26 × 103 Watts at take off, (c) 6.26 × 104 Watts at the start, (d) 6.26 × 104 Watts at take off, 57. A body of mass ‘m’, accelerates uniformly from rest to ‘v1’, in time ‘t1’. The instantaneous power delivered to the body, as a function, of time ‘t’ is, [2004], mv12t, mv1t 2, (a), (b), t1, t12, mv1t, mv12t, (c) t, (d), t1, 1, , power, , 58. A body is moved along a straight line by a machine, delivering a constant power. The distance moved by the, body in time ‘t’ is proportional to, [2003], (a) t3/4, (b) t3/2, (c) t1/4, (d) t1/2
Page 68 :
P-59, , Work, Energy & Power, , TOPIC 4 Collisions, 59. Two bodies of the same mass are moving with the same, speed, but in different directions in a plane. They have a, completely inelastic collision and move together, thereafter with a final speed which is half of their initial, speed. The angle between the initial velocities of the, two bodies (in degree) is ______.[NA 6 Sep. 2020 (I)], 60. Particle A of mass m moving with velocity ( 3$i + $j ) ms-1, 1, , collides with another particle B of mass m2 which is at rest, r, r, initially. Let V1 and V2 be the velocities of particles A and, B after collision respectively. If m1 = 2m2 and after collision, r, r, r, V1 = ($i + 3 $j ) ms -1 , the angle between V1 and V2 is :, , 64. A particle of mass m is moving along the x-axis with initial, velocity uiˆ. It collides elastically with a particle of mass, 10 m at rest and then moves with half its initial kinetic, energy (see figure). If sin q1 = n sin q2 , then value of n, is ___________., [NA 2 Sep. 2020 (II)], m, q1, q2, m, 10, m, ui$, 10 m, 65. Two particles of equal mass m have respective initial, æ iˆ + ˆj ö, velocities uiˆ and u ç 2 ÷ . They collide completely, è, ø, inelastically. The energy lost in the process is: [9 Jan. 2020 I], (a), , 1, mu2, 3, , (b), , (c), , 3, mu2, 4, , (d), , [6 Sep. 2020 (II)], (a) 15º, , (b) 60º, , (c) – 45º, (d) 105º, 61. Blocks of masses m, 2m, 4m and 8m are arranged in a line on, a frictionless floor. Another block of mass m, moving with, speed v along the same line (see figure) collides with mass, m in perfectly inelastic manner. All the subsequent collisions, are also perfectly inelastic. By the time the last block of, mass 8m starts moving the total energy loss is p% of the, original energy. Value of ‘p’ is close to :, [4 Sep. 2020 (I)], v, m, , m, , 2m, , 4m, , 8m, , (a) 77, (b) 94, (c) 37, (d) 87, 62. A block of mass 1.9 kg is at rest at the edge of a table, of, height 1 m. A bullet of mass 0.1 kg collides with the block, and sticks to it. If the velocity of the bullet is 20 m/s in, the horizontal direction just before the collision then the, kinetic energy just before the combined system strikes, the floor, is [Take g = 10 m/s2 . Assume there is no, rotational motion and losss of energy after the collision, is negligiable.], [3 Sep. 2020 (II)], (a) 20 J, (b) 21 J, (c) 19 J, (d) 23 J, 63. A particle of mass m with an initial velocity u iˆ collides, perfectly elastically with a mass 3 m at rest. It moves, with a velocity v ˆj after collision, then, v is given by :, [2 Sep. 2020 (I)], (a) v =, (c) v =, , 2, u, 3, u, , 2, , (b) v =, (d) v =, , u, 3, 1, 6, , u, , 1, mu2, 8, , 2, mu2, 3, , 66. A body A, of mass m = 0.1 kg has an initial velocity of 3 iˆ ms–1., It collides elastically with another body, B of the same, mass which has an initial velocity of 5 ĵ ms–1. After, r, collision, A moves with a velocity v = 4 iˆ + ˆj . The, , (, , ), , x, J. The value of, 10, x is _______., [NA 8 Jan. 2020 I], 67. A particle of mass m is dropped from a height h above the, ground. At the same time another particle of the same, mass is thrown vertically upwards from the ground with a, , energy of B after collision is written as, , speed of, , 2 gh . If they collide head-on completely, inelastically, the time taken for the combined mass to reach, the ground, in units of, (a), , 1, 2, , h, is:, g, (b), , [8 Jan. 2020 II], 3, 4, , 1, 3, (d), 2, 2, 68. A man (mass = 50 kg) and his son (mass = 20 kg) are, standing on a frictionless surface facing each other. The, man pushes his son so that he starts moving at a speed of, 0.70 ms–1 with respect to the man. The speed of the man, with respect to the surface is :, [12 April 2019 I], (a) 0.28 ms–1, (b) 0.20 ms–1, (c) 0.47 ms–1, (d) 0.14 ms–1, , (c), , 69. Two particles, of masses M and 2M, moving, as shown,, with speeds of 10 m/s and 5 m/s, collide elastically at the, origin. After the collision, they move along the indicated, directions with speeds v1 and v2, respectively. The values, of v1 and v2 are nearly :, [10 April 2019 I]
Page 69 :
P-60, , Physics, , 5, th of the initial kinetic, 6, energy is lost in whole process. What is value of M/m?, [9 Jan. 2019 I], , C, also perfectly inelastically, , A, m, (a) 6.5 m/s and 6.3 m/s (b) 3.2 m/s and 6.3 m/s, (c) 6.5 m/s and 3.2 m/s (d) 3.2 m/s and 12.6 m/s, 70. A body of mass 2 kg makes an elastic collision with a, second body at rest and continues to move in the original, direction but with one fourth of its original speed. What, is the mass of the second body?, [9 April 2019 I], (a) 1.0 kg (b) 1.5 kg, (c) 1.8 kg, (d), 1.2 kg, 71. A particle of mass ‘m’ is moving with speed ‘2v’ and, collides with a mass ‘2m’ moving with speed ‘v’ in the, same direction. After collision, the first mass is stopped, completely while the second one splits into two particles, each of mass ‘m’, which move at angle 45° with respect to, the original direction., [9 April 2019 II], The speed of each of the moving particle will be:, (a), , 2 v, , (b) 2 2 v, , (c) v / (2 2), (d) v/ 2, 72. A body of mass m1 moving with an unknown velocity of, v1 iˆ , undergoes a collinear collision with a body of mass, m2 moving with a velocity v2 iˆ . After collision, m1 and m2, move with velocities of v3 iˆ and v4 iˆ , respectively.., If m2 = 0.5 m1 and v3 = 0.5 v1, then v1 is : [8 April 2019 II], v, v, (a) v4 – 2 (b) v4 – v2 (c) v4 – 2 (d) v4 + v2, 2, 4, 73. An alpha-particle of mass m suffers 1-dimensional elastic, collision with a nucleus at rest of unknown mass. It is, scattered directly backwards losing, 64% of its initial, kinetic energy. The mass of the nucleus is :, [12 Jan. 2019 II], (a) 2 m, (b) 3.5 m, (c) 1.5 m, (d) 4 m, 74. A piece of wood of mass 0.03 kg is dropped from the, top of a 100 m height building. At the same time, a bullet, of mass 0.02 kg is fired vertically upward, with a velocity, 100 ms–1, from the ground. The bullet gets embedded in, the wood. Then the maximum height to which the, combined system reaches above the top of the building, before falling below is: (g = 10 ms–2) [10 Jan. 2019 I], (a) 20 m, (b) 30 m, (c) 40 m, (d), 10 m, 75. There block A, B and C are lying on a smooth horizontal, surface, as shown in the figure. A and B have equal, masses, m while C has mass M. Block A is given an inital, speed v towards B due to which it collides with B, perfectly inelastically. The combined mass collides with, , (a) 5, , (b) 2, , B, m, , C, M, (c) 4, , (d), , 3, , 76. In a collinear collision, a particle with an initial speed n0, strikes a stationary particle of the same mass. If the final, total kinetic energy is 50% greater than the original, kinetic energy, the magnitude of the relative velocity, between the two particles, after collision, is:, [2018], n0, n0, n0, (a), (b) 2n0, (c), (d), 2, 4, 2, 77. The mass of a hydrogen molecule is 3.32×10–27 kg. If 1023, hydrogen molecules strike, per second, a fixed wall of area, 2 cm2 at an angle of 45° to the normal, and rebound, elastically with a speed of 103 m/s, then the pressure on, the wall is nearly:, [2018], (a) 2.35 × 103 N/m2, (b) 4.70 × 103 N/m2, (c) 2.35 × 102 N/m2, (d) 4.70 × 102 N/m2, 78. It is found that if a neutron suffers an elastic collinear, collision with deuterium at rest, fractional loss of its energy, is pd; while for its similar collision with carbon nucleus at, rest, fractional loss of energy is Pc. The values of Pd and, Pc are respectively:, [2018], (a), , ( ×89, ×28) (b) ( ×28, ×89 ) (c), , (0, 0), , (d) (0, 1), , 79. A proton of mass m collides elastically with a particle of, unknown mass at rest. After the collision, the proton and, the unknown particle are seen moving at an angle of 90°, with respect to each other. The mass of unknown particle, is:, [Online April 15, 2018], m, 3, , m, (c) 2m, (d) m, 2, 80. Two particles A and B of equal mass M are moving with, the same speed v as shown in the figure. They collide, completely inelastically and move as a single particle C., The angle q that the path of C makes with the X-axis is, given by:, [Online April 9, 2017], (a), , (a) tanq =, (b) tanq =, (c) tanq =, , (d) tanq =, , (b), , 3+ 2, , Y, , 1- 2, C, , 3- 2, 1- 2, , 1- 2, 2(1 + 3), 1- 3, 1+ 2, , q, A, , 30°, , X, , 45°, B
Page 70 :
P-61, , Work, Energy & Power, , 81. A neutron moving with a speed 'v' makes a head on, collision with a stationary hydrogen atom in ground state., The minimum kinetic energy of the neutron for which, inelastic collision will take place is :, [Online April 10, 2016], (a) 20.4 eV (b) 10.2 eV (c) 12.1 eV (d) 16.8 eV, 82. A particle of mass m moving in the x direction with speed, 2v is hit by another particle of mass 2m moving in the y, direction with speed v. If the collision is perfectly inelastic,, the percentage loss in the energy during the collision is, close to :, [2015], (a) 56%, (b) 62%, (c) 44%, (d) 50%, 83. A bullet of mass 4g is fired horizontally with a speed of, 300 m/s into 0.8 kg block of wood at rest on a table. If the, coefficient of friction between the block and the table is, 0.3, how far will the block slide approximately?, [Online April 12, 2014], (a) 0.19 m (b) 0.379 m (c) 0.569 m (d) 0.758 m, 84. Three masses m, 2m and 3m are moving in x-y plane with, speed 3u, 2u and u respectively as shown in figure. The, three masses collide at the same point at P and stick, together. The velocity of resulting mass will be:, [Online April 12, 2014], y, , 2m, 2u, 60°, m, 3u, , P, , 60°, , (, , ), , (, , x, , ), , u ˆ, u ˆ, i + 3jˆ, i - 3jˆ, (b), 12, 12, u ˆ, u ˆ, -i + 3jˆ, -i - 3jˆ, (c), (d), 12, 12, 85. This question has statement I and statement II. Of the four, choices given after the statements, choose the one that, best describes the two statements., Statement - I: Apoint particle of mass m moving with, speed u collides with stationary point particle of mass, M. If the maximum energy loss possible is given as, æ1, ö, f ç mv2 ÷ then f = æç m ö÷ ., è2, ø, èM + mø, , (, , ), , (, , (a) m, , v02, x02, , v0, (b) m 2 x, 0, 2, , 2 æ v0 ö, m, (d), 3 çè x0 ÷ø, 89. A projectile moving vertically upwards with a velocity of, 200 ms–1 breaks into two equal parts at a height of 490 m., One part starts moving vertically upwards with a velocity, of 400 ms–1. How much time it will take, after the break, up with the other part to hit the ground?, [Online May 12, 2012], v0, (c) 2m x, 0, , 3m, u, (a), , 86. A projectile of mass M is fired so that the horizontal, range is 4 km. At the highest point the projectile explodes, in two parts of masses M/4 and 3M/4 respectively and, the heavier part starts falling down vertically with zero, initial speed. The horizontal range (distance from point, of firing) of the lighter part is :, [Online April 23, 2013], (a) 16 km (b) 1 km, (c) 10 km (d) 2 km, 87. A moving particle of mass m, makes a head on elastic, collision with another particle of mass 2m, which is initially, at rest. The percentage loss in energy of the colliding, particle on collision, is close to, [Online May 19, 2012], (a) 33%, (b) 67%, (c) 90%, (d) 10%, 88. Two bodies A and B of mass m and 2m respectively are, placed on a smooth floor. They are connected by a spring, of negligible mass. A third body C of mass m is placed, on the floor. The body C moves with a velocity v0 along, the line joining A and B and collides elastically with A., At a certain time after the collision it is found that the, instantaneous velocities of A and B are same and the, compression of the spring is x0. The spring constant k, will be, [Online May 12, 2012], , ), , Statement - II: Maximum energy loss occurs when the, particles get stuck together as a result of the collision., [2013], (a) Statement - I is true, Statment - II is true, Statement, - II is the correct explanation of Statement - I., (b) Statement-I is true, Statment - II is true, Statement II is not the correct explanation of Statement - II., (c) Statement - I is true, Statment - II is false., (d) Statement - I is false, Statment - II is true., , (a) 2 10 s, , (b) 5 s, , (c) 10 s, , (d), , 10 s, , 90. Statement -1: Two particles moving in the same direction, do not lose all their energy in a completely inelastic, collision., Statement -2 : Principle of conservation of momentum, holds true for all kinds of collisions., [2010], (a) Statement -1 is true, Statement -2 is true ; Statement, -2 is the correct explanation of Statement -1., (b) Statement -1 is true, Statement -2 is true; Statement -2, is not the correct explanation of Statement -1, (c) Statement -1 is false, Statement -2 is true., (d) Statement -1 is true, Statement -2 is false., 91. A block of mass 0.50 kg is moving with a speed of 2.00, ms–1 on a smooth surface. It strikes another mass of 1.00, kg and then they move together as a single body. The, energy loss during the collision is, [2008], (a) 0.16 J (b) 1.00 J (c) 0.67 J (d) 0.34 J
Page 71 :
P-62, , Physics, , (a) 0.16 J (b) 1.00 J (c) 0.67 J (d) 0.34 J, 92. A bomb of mass 16kg at rest explodes into two pieces of, masses 4 kg and 12 kg. The velolcity of the 12 kg mass is, 4 ms–1. The kinetic energy of the other mass is [2006], (a) 144 J (b) 288 J, (c) 192 J, (d) 96 J, 93. The block of mass M moving on the frictionless horizontal, surface collides with the spring of spring constant k and, compresses it by length L. The maximum momentum of, the block after collision is, [2005], , M, , kL2, ML2, (b) Mk L (c), (d) zero, 2M, k, 94. A mass ‘m’ moves with a velocity ‘v’ and collides, inelastically with another identical mass. After collision, , (a), , the l st mass moves with velocity, , v, , in a direction, 3, perpendicular to the initial direction of motion. Find the, speed of the 2 nd mass after collision., m, A before, collision, , m, , v, , [2005], , 3, Aafter, collision, v, , 2, , v, 3, 3, 95. Consider the following two statements :, [2003], A. Linear momentum of a system of particles is zero, B. Kinetic energy of a system of particles is zero., Then, (a) A does not imply B and B does not imply A, (b) A implies B but B does not imply A, (c) A does not imply B but B implies A, (d) A implies B and B implies A, (a), , 3v, , (b) v, , (c), , (d)
Page 76 :
P-67, , Work, Energy & Power, , Assuming the mass of athelet to 40 kg his average K.E, would be, 1, K.E = ´ 40 ´ (10)2 = 2000J, 2, Assuming mass to 100 kg average kinetic energy, 1, K.E. = ´ 100 ´ (10) 2 = 5000 J, 2, 40. (b) Suppose the spring gets compressed by x before, stopping., kinetic energy of the block = P.E. stored in the spring +, work done against friction., 1, 1, 2, 2, Þ ´ 2 ´ (4) = ´ 10,000 ´ x + 15 ´ x, 2, 2, Þ 10,000 x2 + 30x – 32 = 0, Þ 5000 x 2 + 15 x - 16 = 0, , -15 ± (15)2 - 4 ´ (5000)(-16), 2 ´ 5000, = 0.055m = 5.5cm., 41. (d) Let u be the velocity with which the particle is thrown, and m be the mass of the particle. Then, \ x=, , 1, K = mu 2 ., ... (1), 2, At the highest point the velocity is u cos 60° (only the, horizontal component remains, the vertical component, being zero at the top-most point). Therefore kinetic energy, at the highest point., 1, K, 1, K ' = m(u cos 60°) 2 = mu 2 cos2 60° =, [From 1], 2, 2, 4, 42. (b) Given, Mass of the particle, m = 100g, Initial speed of the particle, m = 5 m/s, Final speed of the particle, v = 0, Work done by the force of gravity, = Loss in kinetic energy of the body., 1, 1 100, = m (v2 – u2) = ´, (02 – 52), 2, 2 1000, = –1.25 J, 43. (a) Potential energy, x4 – x2, joule, 4 2, For maxima of minima, dV, = 0 Þ x 3 - x = 0 Þ x = ±1, dx, 1 1, 1, Þ Min. P.E. = - = - J, 4 2, 4, K.E.(max.) + P.E.(min.) = 2 (Given), 1 9, \ K.E.(max.) = 2 + =, 4 4, 1 2, K.E.max . = mvmax ., 2, 1, 9, 3, 2, Þ ´ 1 ´ vmax, . = Þ vmax. =, 2, 4, 2, , V(x) =, , 44. (d) Work done by tension + Work done by force (applied), + Work done by gravitational force = change in kinetic, energy, Work done by tension is zero, O, 45°, , l, B, , A, , F, , C, , F, Þ 0 + F ´ AB - Mg ´ AC = 0, 1, é, ê1 æ AC ö, 2, Þ F = Mgç, ÷ = Mg ê, ê 1, è AB ø, êë 2, l, [Q AB = l sin 45° =, and, 2, , ù, ú, ú, ú, úû, , 1 ö, æ, AC = OC - OA = l - l cos 45° = l ç1 ÷, è, 2ø, , where l = length of the string.], , Þ F = Mg ( 2 - 1), , 45. (b), , 100, , 30, , mgH, Using conservation of energy,, Total energy at 100 m height, = Total energy at 20m height, , 20, 1, mv 2 + mgh, 2, , æ1 2, ö, m (10 × 100) = m çè v + 10 ´ 20÷ø, 2, 1 2, or, v = 800 or v = 1600 = 40 m/s, 2, Note :, Loss in potential energy = gain in kinetic energy, 1, m ´ g ´ 80 = mv 2, 2, 1 2, 10 ´ 80 = v, 2, v2 = 1600 or v = 40 m/s, 46. (c) Given : retardation µ displacement, i.e., a = –kx [Here, k = constant], dv, But a = v, dx, vdv, \, = - kx Þ, dx, , v2, , x, , v1, , 0, , ò v dv = - ò kxdx
Page 82 :
P-73, , Work, Energy & Power, 1, , 1, mv 2 < 2´10.2 < 20.4 eV, 2, Y, pf = 3 m V, , ..... (ii), , 90°, , M, u2 = 0, , ,v2, M, , m, u1 = u, Proton, , m, ,v, , 1, o = (mv1 – Mv2), 2, , Unknown mass, , Before collision, , After collision, , v2 - v1 cos90, u cos 45, (Q Collision is elastic), , m, , 82. (a), , 2v, , pi, , Coefficient of restution e = 1 =, , v2, =1, u, 2, ..... (iii), Þ u = 2v2, Solving eqs (i), (ii), & (iii), we get mass of unknown, particle, M = m, 80. (a) For particle C,, According to law of conservation of linear momentum,, verticle component,, 2 mv' sin q = mv sin 60° + mv sin 45°, Þ, , mv mv 3 ...... (i), +, 2, 2, Horizontal component,, 2 mv' cos q = mv sin 60° – mv cos 45°, mv mv, ...... (ii), 2mv'cos q =, +, 2, 2, Y, A, B, v sin 60°, , 2mv 'sin q =, , Y', v sin 45°, , 30°, 60°, , 45°, X', – v cos 45°, For particle B, , X, v cos 60°, For particle A, , 1, v, <, v<, (1 ∗ 1), 2, n ↑ v(H), , (n)(H) ↑, , v, 2, , (m1 ∗ m 2 ), , v, , Before, After, , æ v ö2 1, 1, 1, mv 2 , (2m) çç ÷÷÷ < mv2, èç 2 ø, 2, 2, 4, K.E. lost is used to jump from 1st orbit to 2nd orbit, DK.E. = 10.2ev, Minimum K.E. of neutron for inelastic collision, , Loss in K.E. =, , v, , 2m, Initial momentum of the system, , pi =, , [m(2V) 2 ´ 2m(2V) 2 ], , = 2m ´ 2V, Final momentum of the system = 3mV, By the law of conservation of momentum, 2 2mv = 3mV Þ, , 2 2v, = Vcombined, 3, , Loss in energy, 1, 1, 1, 2, DE = m1V12 + m2V22 - (m1 + m2 )Vcombined, 2, 2, 2, 4, 5, DE = 3mv2 - mv 2 = mv 2 = 55.55%, 3, 3, Percentage loss in energy during the collision ; 56%, 83. (b) Given, m1 = 4g, u1 = 300 m/s, m2 = 0.8 kg = 800 g, u2 = 0 m/s, From law of conservation of momentum,, m1u1 + m2u2 = m1v1 + m2v2, Let the velocity of combined system = v m/s, then,, 4 × 300 + 800 × 0 = (800 + 4) × v, 1200, = 1.49 m / s, 804, Now, m = 0.3 (given), a = mg, a = 0.3 × 10, = 3 m/s2, then, from v2 = u2 + 2as, (1.49)2 = 0 + 2 × 3 × s, , 1, 3, +, 2+ 3, 2, tan q = 2, =, 1 1, 1, - 2, 2, 2, m1, , X, , v=, , Dividing eqn (i) by eqn (ii),, , 81. (a) For inelastic collision v ' <, , 45°, , s=, , (take g = 10 m/s2), , (1.49 )2, , 6, 2.22, s=, 6, = 0.379 m, 84. (d) From the law of conservation of momentum we know, that,, m1u1 + m2u2 + .... = m1v1 + m2v2 + ...., Given m1 = m, m2 = 2m and m3 = 3m, and u1 = 3u, u2 = 2u and u3 = u, ®, , Let the velocity when they stick = v
Page 84 :
P-75, , Work, Energy & Power, , Momentum of the block, = M × v, , 1, ´ 0.5 ´ 2 ´ 2 + 0 = 1J, 2, Momentum before collision, = Momentum after collision, m1u1 + m2u2 = (m + M) × v, =, , k, × L = kM × L, M, 94. (d) Considering conservation of momentum along x-direction,, mv = mv1 cos q, ...(1), where v1 is the velocity of second mass, In y-direction,, mv, 0=, - mv1 sin q, 3, mv, or m1v1 sin q =, ...(2), 3, , =M×, , \ 0.5 × 2 + 1 × 0 = (0.5 + 1) × v Þ v =, , 2, m/s, 3, , Final kinetic energy of the system is, 1, K.E f = (m + M )v 2, 2, 1, 2 2 1, = (0.5 + 1) ´ ´ = J, 2, 3 3 3, \ Energy loss during collision, æ 1ö, = ç1 - ÷ J = 0.67J, è 3ø, 92. (b) Let the velocity and mass of 4 kg piece be v1 and m1, and that of 12 kg piece be v2 and m2., , v/ 3, , m, , v, v, , 16 kg, , Situation 1, 4 kg = m1, v1, , Initial momentum, =0, , m2 = 12 kg Final momentum, v2, = m2v2 – m1v1, , Situation 2, , Applying conservation of linear momentum, 16 × 0 = 4 × v1 + 12 × 4, 12 ´ 4, = –12 ms -1, Þ v1 = –, 4, Kinetic energy of 4 kg mass, 1, 1, \ K . E. = m1v12 = ´ 4 ´144 = 288 J, 2, 2, 93. (b) When the spring gets compressed by length L., K.E. lost by mass M = P.E. stored in the compressed spring., 1, 1, Mv 2 = k L2, 2, 2, k, ×L, Þ v=, M, , M, , v1 cosq, , q, v1, , v1 sinq, Squaring and adding eqns. (1) and (2) we get, , v12 = v2 +, , v2, , Þ v1 =, , 2, , v, 3, 3, 95. (c) Kinetic energy of a system of particle is zero only, when the speed of each particles is zero. This implies, momentum of each particle is zero, thus linear momentum, of the system of particle has to be zero., Also if linear momentum of the system is zero it does not, mean linear momentum of each particle is zero. This is because, linear momentum is a vector quantity. In this case the kinetic, energy of the system of particles will not be zero., \ A does not imply B but B implies A., Given, force, F = 200 N extension of wire, x = 1mm.
Page 85 :
P-76, , 6, , Physics, , System of Particles and, Rotational Motion, Centre of Mass, Centre of, TOPIC 1 Gravity & Principle of, Moments, 1., , 2., , The centre of mass of a solid hemisphere of radius 8 cm is, x cm from the centre of the flat surface. Then value of x is, ______., [NA Sep. 06, 2020 (II)], a, A square shaped hole of side l =, is carved out at a, 2, a, distance d = from the centre ‘O’ of a uniform circular, 2, disk of radius a. If the distance of the centre of mass of the, a, remaining portion from O is - , value of X (to the nearest, X, integer) is ___________., [NA Sep. 02, 2020 (II)], , 5., , (a) (1.25 m, 1.50 m), (b) (0.75 m, 1.75 m), (c) (0.75 m, 0.75 m), (d) (1 m, 1.75 m), As shown in fig. when a spherical cavity (centred at O) of, radius 1 is cut out of a uniform sphere of radius R (centred, at C), the centre of mass of remaining (shaded) part of, sphere is at G, i.e on the surface of the cavity. R can be, determined by the equation:, [8 Jan. 2020 II], (a) (R2 + R + 1) (2 – R) = 1, (b) (R2 – R – l) (2 – R) = 1, , a, O, , (c) (R2 – R + l) (2 – R) = l, d, , (d) (R2 + R – 1) (2 – R) = 1, , l = a/2, , 6., 3., , A rod of length L has non-uniform linear mass density, 2, , æ xö, given by r(x) = a + b ç ÷ , where a and b are constants, èLø, and 0 £ x £ L. The value of x for the centre of mass of the, rod is at:, [9 Jan. 2020 II], , (a), , (b), , 3 æ 2a + b ö, ç, ÷L, 4 è 3a + b ø, , 4æ a +b ö, 3 æ 2a + b ö, (d) ç, ç, ÷L, ÷L, 3 è 2a + 3b ø, 2 è 3a + b ø, The coordinates of centre of mass of a uniform flag shaped, lamina (thin flat plale) of mass 4 kg. (The coordinates of, the same are shown in figure) are:, [8 Jan. 2020 I], , (c), 4., , 3æ a +b ö, ç, ÷L, 2 è 2a + b ø, , Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg, are placed at three corners of a right angle triangle of, sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure., The center of mass of the system is at a point:, [7 Jan. 2020 I], , (a), (b), (c), (d), , 0.6 cm right and 2.0 cm above 1 kg mass, 1.5 cm right and 1.2 cm above 1 kg mass, 2.0 cm right and 0.9 cm above 1 kg mass, 0.9 cm right and 2.0 cm above 1 kg mass
Page 86 :
P-77, , System of Particles and Rotational Motion, , 7., , Three particles of masses 50 g, 100 g and 150 g are placed, at the vertices of an equilateral triangle of side 1 m (as, shown in the figure). The (x, y) coordinates of the centre of, mass will be :, [12 Apr. 2019 II], , 10. The position vector of the centre of mass r cm of an, asymmetric uniform bar of negligible area of crosssection as shown in figure is:, [12 Jan. 2019 I], , L, L, , 8., , æ 3, 5 ö, (a) çç 4 m, 12 m ÷÷, è, ø, , æ 7, 3 ö, (b) çç 12 m, 8 m ÷÷, è, ø, , (a), , æ 7, 3 ö, (c) çç 12 m, 4 m ÷÷, è, ø, , æ 3, 7 ö, (d) çç 8 m, 12 m ÷÷, è, ø, , (c), , Four particles A, B, C and D with masses mA = m, mB =, 2m, mC = 3m and mD = 4m are at the corners of a square., They have accelerations of equal magnitude with, directions as shown. The acceleration of the centre of, mass of the particles is :, [8 April 2019 I], , 2L, , 3L, , r, 13, 5, rcm = L xˆ + L yˆ, 8, 8, , (b), , r, 5, 13, rcm = L xˆ + L yˆ, 8, 8, , r, 3, 11, rcm = L xˆ + L yˆ, 8, 8, , (d), , r, 11, 3, rcm = L xˆ + L yˆ, 8, 8, , 11. A force of 40 N acts on a point B at the end of an L-shaped, object, as shown in the figure. The angle q that will produce, maximum moment of the force about point A is given by:, [Online April 15, 2018], A, 1, (a) tan q =, 4, (b) tan q = 2, (c), , 4m, , 1, tan q =, 2, , ®, , F, , (d) tan q = 4, , (a), , ( ), , a $ $, i– j, 5, , ( ), , a $ $, i+ j, 5, A uniform rectangular thin sheet ABCD of mass M has, length a and breadth b, as shown in the figure. If the shaded, portion HBGO is cut-off, the coordinates of the centre of, mass of the remaining portion will be : [8 Apr. 2019 II], , (c) Zero, , 9., , (b) a, (d), , 2m, , q, , B, , 12. In a physical balance working on the principle of moments,, when 5 mg weight is placed on the left pan, the beam, becomes horizontal. Both the empty pans of the balance, are of equal mass. Which of the following statements is, correct ?, [Online April 8, 2017], (a) Left arm is longer than the right arm, (b) Both the arms are of same length, (c) Left arm is shorter than the right arm, (d) Every object that is weighed using this balance, appears lighter than its actual weight., 13. In the figure shown ABC is a uniform wire. If centre of, BC, is, AB, [Online April 10, 2016], , mass of wire lies vertically below point A, then, close to :, A, , æ 3a 3b ö, (a) ç , ÷, è 4 4 ø, , æ 5a 5b ö, (b) ç , ÷, è 3 3 ø, , æ 2a 2b ö, (c) ç , ÷, è 3 3 ø, , 5a 5b, (d) æç , ö÷, 12, 12 ø, è, , 60°, (a), (c), , B, 1.85, 1.37, , C, (b) 1.5, (d) 3
Page 87 :
P-78, , Physics, , 14. Distance of the centre of mass of a solid uniform cone, from its vertex is z0. If the radius of its base is R and its, height is h then z0 is equal to :, [2015], , 5h, 3h, 3h 2, h2, (b), (c), (d), 8, 4, 8R, 4R, 15. A uniform thin rod AB of length L has linear mass, bx, density m (x) = a +, , where x is measured from A. If, L, æ 7ö, the CM of the rod lies at a distance of ç ÷ L from A,, è 12 ø, then a and b are related as : [Online April 11, 2015], (a) a = 2b, (b) 2a = b, (c) a = b, (d) 3a = 2b, 16. A thin bar of length L has a mass per unit length l, that, increases linearly with distance from one end. If its total, mass is M and its mass per unit length at the lighter end is, lO, then the distance of the centre of mass from the lighter, end is:, [Online April 11, 2014], , (a), , 2, L l o L2, L loL, +, (b), 3 8M, 2 4M, L l o L2, 2L l o L2, +, (c), (d), 3, 4M, 3, 6M, 17. A boy of mass 20 kg is standing on a 80 kg free to move, long cart. There is negligible friction between cart and, ground. Initially, the boy is standing 25 m from a wall. If he, walks 10 m on the cart towards the wall, then the final, distance of the boy from the wall will be, [Online April 23, 2013], (a) 15 m, (b) 12.5 m (c) 15.5 m (d) 17 m, 18. A thin rod of length ‘L’ is lying along the x-axis with its, ends at x = 0 and x = L. Its linear density (mass/length), n, æ xö, varies with x as k ç ÷ , where n can be zero or any, è Lø, positive number. If the position xCM of the centre of mass, of the rod is plotted against ‘n’, which of the following, graphs best approximates the dependence of xCM on n?, [2008], xCM, xCM, , (a), , (a), , (c), , L, L, 2, O, xCM, L, L, 2, O, , (b), n, , (d), n, , L, 2, O, xCM, L, L, 2, O, , n, , 20. Consider a two particle system with particles having masses, m1 and m2. If the first particle is pushed towards the centre, of mass through a distance d, by what distance should the, second particle is moved, so as to keep the centre of mass, at the same position?, [2006], m2, m1, (a) m d, (b) m + m d, 1, 1, 2, (c), , m1, d, m2, , (d) d, , 21. A body A of mass M while falling vertically downwards, 1, under gravity breaks into two parts; a body B of mass, 3, 2, M and a body C of mass, M. The centre of mass of, 3, bodies B and C taken together shifts compared to that of, body A towards, [2005], (a) does not shift, (b) depends on height of breaking, (c) body B, (d) body C, 22. A ‘T’ shaped object with dimensions shown in the figure,, r, is lying on a smooth floor. A force ‘ F ’ is applied at the, point P parallel to AB, such that the object has only the, translational motion without rotation. Find the location of, P with respect to C., [2005], l, B, , A, P, , 2l, , F, , C, , (a), , 3, l, 2, , TOPIC 2, , (b), , 2, l, 3, , (c) l, , 4, l, 3, , Angular Displacement,, Velocity and Aceleration, , 23. A bead of mass m stays at point P(a, b) on a wire bent in, the shape of a parabola y = 4Cx2 and rotating with angular, speed w (see figure). The value of w is (neglect friction) :, [Sep. 02, 2020 (I)], y, w, P (a, b), , n, , 19. A circular disc of radius R is removed from a bigger circular, disc of radius 2R such that the circumferences of the discs, coincide. The centre of mass of the new disc is, a/R form the centre of the bigger disc. The value of a is, [2007], (a) 1/4, (b) 1/3, (c) 1/2, (d) 1/6, , (d), , x, , O, , (a) 2 2gC, (c), , 2gC, ab, , (b) 2 gC, (d), , 2g, C
Page 88 :
P-79, , System of Particles and Rotational Motion, , 24. A cylindrical vessel containing a liquid is rotated about, its axis so that the liquid rises at its sides as shown in the, figure. The radius of vessel is 5 cm and the angular speed, of rotation is w rad s –1. The difference in the height,, h (in cm) of liquid at the centre of vessel and at the side, will be :, [Sep. 02, 2020 (I)], , (a), , w, , (b), h, , 10 cm, , (a), , 2w2, 25g, , (b), , 25w2, (c), 2g, , 5w2, 2g, , 2w2, (d), 5g, , 25. A spring mass system (mass m, spring constant k and, natural length l) rests in equilibrium on a horizontal disc., The free end of the spring is fixed at the centre of the disc., If the disc together with spring mass system, rotates about, it’s axis with an angular velocity w, (k >> mw2) the relative, change in the length of the spring is best given by the, option:, [9 Jan. 2020 II], (a), , 2 æ mw2 ö, ç, ÷, 3 çè k ÷ø, , (b), , 2mw2, k, , mw2, mw2, (d), k, 3k, 26. A particle of mass m is fixed to one end of a light spring, having force constant k and unstretched length l. The other, end is fixed. The system is given an angular speed w about, the fixed end of the spring such that it rotates in a circle in, gravity free space. Then the stretch in the spring is:, [8 Jan. 2020 I], , (c), , (a), , (c), , ml w2, k - wm, , (b), , ml w 2, k + mw, , (c), , 2, , (d), , ml w 2, k - mw 2, , ml w 2, k + mw, , 27. A uniform rod of length l is being rotated in a horizontal, plane with a constant angular speed about an axis passing, through one of its ends. If the tension generated in the rod, due to rotation is T(x) at a distance x from the axis, then, which of the following graphs depicts it most closely?, [12 Apr. 2019 II], , (d), , 28. A smooth wire of length 2pr is bent into a circle and kept, in a vertical plane. A bead can slide smoothly on the wire., When the circle is rotating with angular speed w about, the vertical diameter AB, as shown in figure, the bead is, at rest with respect to the circular ring at position P as, shown. Then the value of w2 is equal to :, [12 Apr. 2019 II], , 3g, (b) 2 g / (r 3), 2r, (d) 2g/r, (c) ( g 3) / r, 29. A long cylindrical vessel is half filled with a liquid. When, the vessel is rotated about its own vertical axis, the liquid, rises up near the wall. If the radius of vessel is 5 cm and its, rotational speed is 2 rotations per second, then the, difference in the heights between the centre and the sides, in, cm, will be :, [12 Jan. 2019 II], (a) 2.0, (b) 0.1, (c) 0.4, (d) 1.2, (a)
Page 89 :
P-80, , Physics, , 30. A particle is moving with a uniform speed in a circular, orbit of radius R in a central force inversely proportional, to the nth power of R. If the period of rotation of the, particle is T, then:, [2018], (b) T µ R n /2+1, (c) T µ R (n +1)/2, (d) T µ R n /2, 31. The machine as shown has 2 rods of length 1 m connected, by a pivot at the top. The end of one rod is connected to, the floor by a stationary pivot and the end of the other rod, has a roller that rolls along the floor in a slot., As the roller goes back and forth, a 2 kg weight moves up, and down. If the roller is moving towards right at a constant, speed, the weight moves up with a : [Online April 9, 2017], (a) T µ R3/2 for any n., , 2 kg, , F, , Fixed pivot, x, , 34. A cubical block of side 30 cm is moving with velocity, 2 ms–1 on a smooth horizontal surface. The surface has a, bump at a point O as shown in figure. The angular velocity, (in rad/s) of the block immediately after it hits the bump,, is :, [Online April 9, 2016], a = 30 cm, , O, (a) 13.3, (b) 5.0, (c) 9.4, (d) 6.7, 35. Two point masses of mass m1 = fM and m2 = (1 – f) M (f, < 1) are in outer space (far from gravitational influence of, other objects) at a distance R from each other. They move, in circular orbits about their centre of mass with angular, velocities w1 for m1 and w2 for m2. In that case, [Online May 19, 2012], (a) (1 – f) w1 = fw, (b) w1 = w2 and independent of f, (c) fw1 = (1 – f)w2, (d) w1 = w2 and depend on f, , Movable roller, , TOPIC 3, , 36. Four point masses, each of mass m, are fixed at the corners, of a square of side l. The square is rotating with angular, frequency w, about an axis passing through one of the, corners of the square and parallel to its diagonal, as, shown in the figure. The angular momentum of the square, about this axis is :, [Sep. 06, 2020 (I)], , (a) ml 2w, , (a), , 3g, cos q, 2l, , (b), , 2g, cos q, 3l, , 3g, 2g, sin q, sin q, (d), 2l, 2l, 33. Concrete mixture is made by mixing cement, stone and, sand in a rotating cylindrical drum. If the drum rotates too, fast, the ingredients remain stuck to the wall of the drum, and proper mixing of ingredients does not take place. The, maximum rotational speed of the drum in revolutions per, minute (rpm) to ensure proper mixing is close to :, (Take the radius of the drum to be 1.25 m and its axle to, be horizontal):, [Online April 10, 2016], (a) 27.0, (b) 0.4, (c) 1.3, (d) 8.0, , (c), , Torque, Couple and, Angular Momentum, , ax, is, , (a) constant speed, (b) decreasing speed, (c) increasing speed, 3, (d) speed which is th of that of the roller when the, 4, weight is 0.4 m above the ground, 32. A slender uniform rod of mass M and length l is pivoted, at one end so that it can rotate in a vertical plane (see, figure). There is negligible friction at the pivot. The free, end is held vertically above the pivot and then released., The angular acceleration of the rod when it makes an angle, [2017], q with the vertical is, , (b) 4 ml2w, , (c) 3 ml2w, (d) 2 ml2w, 37. A thin rod of mass 0.9 kg and length 1 m is suspended, at, rest, from one end so that it can freely oscillate in the, vertical plane. A particle of move 0.1 kg moving in a straight, line with velocity 80 m/s hits the rod at its bottom most, point and sticks to it (see figure). The angular speed, (in rad/s) of the rod immediately after the collision will be, ______________., [NA Sep. 05, 2020 (II)], 38. A person of 80 kg mass is standing on the rim of a circular, platform of mass 200 kg rotating about its axis at 5, revolutions per minute (rpm). The person now starts, moving towards the centre of the platform. What will be, the rotational speed (in rpm) of the platform when the, person reaches its centre __________., [NA Sep. 03, 2020 (I)]
Page 90 :
P-81, , System of Particles and Rotational Motion, , 39. A block of mass m = 1 kg slides with velocity v = 6 m/s, on a frictionless horizontal surface and collides with a, uniform vertical rod and sticks to it as shown. The rod is, pivoted about O and swings as a result of the collision, making angle q before momentarily coming to rest. If, the rod has mass M = 2 kg, and length l = 1 m, the value, of q is approximately: (take g = 10 m/s2), , 42. A uniform cylinder of mass M and radius R is to be pulled, over a step of height a (a < R) by applying a force F at its, centre 'O' perpendicular to the plane through the axes of, the cylinder on the edge of the step (see figure). The, minimum value of F required is :, [Sep. 02, 2020 (I)], F, , [Sep. 03, 2020 (I)], O, , M, l, , q, m v, , (a) 63°, (c) 69°, 40., , æ R-aö, (a) Mg 1 - ç, ÷, è R ø, , m, , m, , (b) 55°, (d) 49°, w, , l, q, , A uniform rod of length 'l' is pivoted at one of its ends on, a vertical shaft of negligible radius. When the shaft rotates, at angular speed w the rod makes an angle q with it (see, figure). To find q equate the rate of change of angular, momentum (direction going into the paper), , ml 2 2, w sin q cos q about the centre of mass (CM) to the, 12, torque provided by the horizontal and vertical forces FH, and FV about the CM. The value of q is then such that :, [Sep. 03, 2020 (II)], (a) cos q =, (c) cos q =, , 2g, , (b) cos q =, , 3l w2, g, lw, , (d) cos q =, , 2, , g, 2l w2, 3g, 2l w2, , 41., , A, , 25, , 50, , 75, 2m, , a, 2, , æ R ö, (b) Mg ç, ÷ -1, è R-aø, , a, a2, (d) Mg 1 - 2, R, R, 43. Consider a uniform rod of mass M = 4m and length l pivoted, about its centre. A mass m moving with velocity v making, p, angle q = to the rod’s long axis collides with one end of, 4, the rod and sticks to it. The angular speed of the rod-mass, system just after the collision is:, [8 Jan. 2020 I], 3 v, 3v, (a), (b), 7 2l, 7l, 4v, 3 2v, (d), 7l, 7 l, 44. A particle of mass m is moving along a trajectory given by, x = x0 + a cos w1t, y = y0 + b cos w2 t, The torque, acting on the particle about the origin, at t = 0, is:, [10 Apr. 2019 I], 2$, 2$, +, my, a, (a) m ( - x0 b + y0 a ) w1 k, (b), 0 w1 k, 2, 2 $, (c) zero, (d) -m( x0 bw2 - y0 aw1 )k, 45. The time dependence of the position of a particle of mass, r, m = 2 is given by r (t ) = 2 t i$ - 3t 2 $j . Its angular momentum,, with respect to the origin, at time t = 2 is :, [10 Apr. 2019 II], $, $, (a) 48 i + j, (b) 36k$, , (c), , (, , (, , ), , (c) -34 k$ - $i, , 0, , 2, , R, , (c) Mg, , FV, FH, , O, , 100, B, , Shown in the figure is rigid and uniform one meter long, rod AB held in horizontal position by two strings tied to, its ends and attached to the ceiling. The rod is of mass 'm', and has another weight of mass 2 m hung at a distance of, 75 cm from A. The tension in the string at A is :, [Sep. 02, 2020 (I)], (a) 0.5 mg, (b) 2 mg, (c) 0.75 mg, (d) 1 mg, , ), , (d) -48k$, 46. A metal coin of mass 5 g and radius 1 cm is fixed to a thin, stick AB of negligible mass as shown in the figure The, system is initially at rest. The constant torque, that will, make the system rotate about AB at 25 rotations per second, in 5s, is close to :, [10 Apr. 2019 II]
Page 91 :
P-82, , Physics, , (a) 4.0×10–6 Nm, (b) 1.6×10–5 Nm, (c) 7.9×10–6 Nm, (d) 2.0×10–5 Nm, 47. A rectangular solid box of length 0.3 m is held, horizontally, with one of its sides on the edge of a, platform of height 5m. When released, it slips off the, table in a very short time t = 0.01 s, remaining essentially, horizontal. The angle by which it would rotate when it, hits the ground will be (in radians) close to :, [8 Apr. 2019 II], , (a) 0.5, (b) 0.3, (c) 0.02, (d) 0.28, 48. A particle of mass 20 g is released with an initial velocity, 5 m/s along the curve from the point A, as shown in the, figure. The point A is at height h from point B. The particle, slides along the frictionless surface. When the particle, reaches point B, its angular momentum about O will be :, (Take g = 10 m/s2), [12 Jan. 2019 II], O, a = 10 m, A, , h = 10 m, B, , (a) 2 kg-m2/s, , (b) 8 kg-m2/s, , 50. The magnitude of torque on a particle of mass 1 kg is 2.5, Nm about the origin. If the force acting on it is 1 N, and the, distance of the particle from the origin is 5m, the angle, between the force and the position vector is (in radians):, [11 Jan. 2019 II], p, p, p, p, (b), (c), (d), 6, 3, 8, 4, 51. To mop-clean a floor, a cleaning machine presses a, circular mop of radius R vertically down with a total, force F and rotates it with a constant angular speed, about its axis. If the force F is distributed uniformly, over the mop and if coefficient of friction between the, mop and the floor is m, the torque, applied by the, machine on the mop is:, [10 Jan. 2019 I], (a) m FR/3, (b) m FR/6, 2, μ mFR, (c) m FR/2, (d), 3, 52. A rigid massless rod of length 3l has two masses attached, at each end as shown in the figure. The rod is pivoted at, point P on the horizontal axis (see figure). When released, from initial horizontal position, its instantaneous angular, acceleration will be:, [10 Jan. 2019 II], 2l, l, , (a), , P, , 5 Mo, , 2 Mo, , g, g, (b), 13l, 3l, g, 7g, (c), (d), 2l, 3l, 53. An L-shaped object, made of thin rods of uniform mass, density, is suspended with a string as shown in figure. If, AB = BC, and the angle made by AB with downward, vertical is q, then:, [9 Jan. 2019 I], (a), , (d) 3 kg-m2/s, uur, uur, 49. A slab is subjected to two forces F1 and F2 of same, uur, magnitude F as shown in the figure. Force F2 is in XY(c) 6, , kg-m2/s, , plane while force F1 acts along z-axis at the point, r r, 2i + 3 j . The moment of these forces about point O will, , (, , ), , be :, , [11 Jan. 2019 I], Z, , F1, F2, , O, , 30º, , 6m, , (a), (c), , ( 3$i - 2 $j + 3k$ ) F, ( 3$i + 2 $j – 3k$ ) F, , 54., (b), (d), , tan q =, , (c), , tan q =, , y, , 4m, , x, , (a), , ( 3$i - 2 $j – 3k$ ) F, ( 3$i + 2 $j + 3k$ ) F, , A, , 1, , (b) tan q =, , 2 3, 2, , (d) tan q =, , 3, B, , x, , 1, 2, , 1, 3
Page 92 :
P-83, , System of Particles and Rotational Motion, , A uniform rod AB is suspended from a point X, at a variable, distance from x from A, as shown. To make the rod, horizontal, a mass m is suspended from its end A. A set of, (m, x) values is recorded. The appropriate variable that, give a straight line, when plotted, are:, [Online April 15, 2018], 1, 1, (a) m,, (b) m, 2 (c) m, x, (d) m, x2, x, x, 55. A thin uniform bar of length L and mass 8m lies on a smooth, horizontal table. Two point masses m and 2m moving in, the same horizontal plane from opposite sides of the bar, with speeds 2v and v respectively. The masses stick to the, L, L, bar after collision at a distance, and, respectively, 3, 6, from the centre of the bar. If the bar starts rotating about, its center of mass as a result of collision, the angular speed, of the bar will be:, [Online April 15, 2018], L/6, , v, , L/3, , 2v, , O, , 58. A bob of mass m attached to an inextensible string of, length l is suspended from a vertical support. The bob, rotates in a horizontal circle with an angular speed w rad/s, about the vertical. About the point of suspension: [2014], (a) angular momentum is conserved., (b) angular momentum changes in magnitude but not in, direction., (c) angular momentum changes in direction but not in, magnitude., (d) angular momentum changes both in direction and, magnitude., 59. A ball of mass 160 g is th rown up at an an gle, of 60° to the horizontal at a speed of 10 ms–1. The angular, momentum of the ball at the highest point of the trajectory, with respect to the point from which the ball is thrown is, nearly (g = 10 ms–2), [Online April 19, 2014], (a) 1.73 kg m2/s, (b) 3.0 kg m2/s, (c) 3.46 kg m2/s, (d) 6.0 kg m2/s, 60. A particle is moving in a circular path of radius a, with a, constant velocity v as shown in the figure. The centre of, circle is marked by ‘C’. The angular momentum from the, origin O can be written as:, [Online April 12, 2014], y, , v, 6v, 3v, v, (b), (c), (d), 6L, 5L, 5L, 5L, 56. A particle of mass m is moving along the side of a square, of side 'a', with a uniform speed v in the x-y plane as shown, in the figure :, [2016], , (a), , O, , C, , y, , D, , O, , a, V, , C, , a V, V a, V, A, B, a, V, 45° R, , 61., , a, , Which of the following statements is false for the angular, ur, momentum L about the origin?, ur, éR, ù, + a ú k$ when the particle is moving from, (a) L = mv ê, ë 2, û, B to C., ur mv, Rk$ when the particle is moving from D to A., (b) L =, 2, ur, mv $, Rk when the particle is moving from A to B., (c) L = –, 2, ur, éR, ù, - a ú k$ when the particle is moving from, (d) L = mv ê, ë 2, û, C to D., 57. A particle of mass 2 kg is on a smooth horizontal table, and moves in a circular path of radius 0.6 m. The height, of the table from the ground is 0.8 m. If the angular speed, of the particle is 12 rad s–1, the magnitude of its angular, momentum about a point on the ground right under the, centre of the circle is :, [Online April 11, 2015], (a) 14.4 kg m2s–1, (b) 8.64 kg m2s–1, (c) 20.16 kg m2s–1, (d) 11.52 kg m2s–1, , a, , q, , 62., , 63., , 64., , x, , (a) va (1 + cos 2q), (b) va (1 + cos q), (c) va cos 2q, (d) va, A particle of mass 2 kg is moving such that at time t, its, r, position, in meter, is given by r (t ) = 5iˆ - 2t 2 ˆj . The angular, momentum of the particle at t = 2s about the origin in, kg m–2 s–1 is :, [Online April 23, 2013], (a) - 80 k$, (b) (10iˆ - 16 ˆj ), (c) -40k$, (d) 40k$, A bullet of mass 10 g and speed 500 m/s is fired into a door, and gets embedded exactly at the centre of the door. The, door is 1.0 m wide and weighs 12 kg. It is hinged at one end, and rotates about a vertical axis practically without friction., The angular speed of the door just after the bullet embeds, into it will be :, [Online April 9, 2013], (a) 6.25 rad/sec, (b) 0.625 rad/sec, (c) 3.35 rad/sec, (d) 0.335 rad/sec, A stone of mass m, tied to the end of a string, is whirled, around in a circle on a horizontal frictionless table. The, length of the string is reduced gradually keeping the, angular momentum of the stone about the centre of the, circle constant. Then, the tension in the string is given by, T = Arn, where A is a constant, r is the instantaneous, radius of the circle. The value of n is equal to, [Online May 26, 2012], (a) – 1, (b) – 2, (c) – 4, (d) – 3, A thin horizontal circular disc is rotating about a vertical, axis passing through its centre. An insect is at rest at a
Page 93 :
P-84, , Physics, , point near the rim of the disc. The insect now moves along, a diameter of the disc to reach its other end. During the, journey of the insect, the angular speed of the disc.[2011], (a) continuously decreases, (b) continuously increases, (c) first increases and then decreases, (d) remains unchanged, 65. A small particle of mass m is projected at an angle q with, the x-axis with an initial velocity n0 in the x-y plane as, , n sin q, shown in the figure. At a time t < 0, , the angular, g, , momentum of the particle is, , [2010], , y, v0, , (a) Angular velocity, (b) Angular momentum, (c) Moment of inertia, (d) Rotational, kinetic energy, r, 70. Let F be the force, r acting on a particle having position, r, vector r , and T be the torque of this force about the, origin. Then, [2003], r r, r r, (a) r .Tr = 0 and Fr .Tr ¹ 0, r, (b) r .T ¹ 0 and F .T = 0, r r, r r, (c) r .T ¹ 0 and F .T ¹ 0, r, r r, r, (d) r .T = 0 and F .T = 0, 71. A particle of mass m moves along line PC with velocity v, as shown. What is the angular momentum of the particle, about P?, [2002], C, L, , q, , P, , x, , (a), , -mg n0t 2 cos q ˆj, , 1, (c) - mg n 0 t 2 cos q kˆ, 2, , (b), , mg n 0t cos q kˆ, , (d), , 1, mgn 0 t 2 cos q iˆ, 2, , where iˆ, ˆj and k̂ are unit vectors along x, y and z-axis, respectively., 66. Angular momentum of the particle rotating with a central, force is constant due to, [2007], (a) constant torque, (b) constant force, (c) constant linear momentum, (d) zero torque, 67. A thin circular ring of mass m and radius R is rotating, about its axis with a constant angular velocity w. Two, objects each of mass M are attached gently to the, opposite ends of a diameter of the ring. The ring now, rotates with an angular velocity w' =, [2006], w(m - 2M ), w (m + 2 M ), (b), (a), (m + 2M ), m, wm, wm, (c), (d), (m + M ), (m + 2M ), 68. A force of – Fkˆ acts on O, the origin of the coordinate, system. The torque about the point (1, –1) is, [2006], Z, , O, , Y, , X, , (a) F (iˆ - ˆj ), (b) - F (iˆ + ˆj ), ˆ, ˆ, (c) F (i + j ), (d) - F (iˆ - ˆj ), 69. A solid sphere is rotating in free space. If the radius of, the sphere is increased keeping mass same, which one, of the following will not be affected ?, [2004], , l, , (a) mvL, , TOPIC 4, , O, , (b) mvl, , (c) mvr, , (d) zero., , Moment of Inertia and, Rotational K.E., , 72. Shown in the figure is a hollow icecream cone (it is open at, the top). If its mass is M, radius of its top, R and height, H,, then its moment of inertia about its axis is :, [Sep. 06, 2020 (I)], R, , H, , MR 2, M (R2 + H 2 ), (b), 2, 4, MH 2, MR 2, (c), (d), 3, 3, 73. The linear mass density of a thin rod AB of length L varies, xö, æ, from A to B as l ( x ) = l 0 ç1 + ÷ , where x is the distance, è Lø, from A. If M is the mass of the rod then its moment of inertia, about an axis passing through A and perpendicular to the, rod is :, [Sep. 06, 2020 (II)], , (a), , 5, 7, ML2, (b), ML2, 12, 18, 2, 3, (c) ML2, (d) ML2, 5, 7, 74. A wheel is rotating freely with an angular speed w on a, shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel, , (a)
Page 94 :
P-85, , System of Particles and Rotational Motion, , of moment of inertia 3I initially at rest is suddenly coupled, to the same shaft. The resultant fractional loss in the, kinetic energy of the system is :, [Sep. 05, 2020 (I)], (a), , 5, 6, , (b), , 1, 4, , 3, 4, 75. ABC is a plane lamina of the shape of an equilateral, triangle. D, E are mid points of AB, AC and G is the, centroid of the lamina. Moment of inertia of the lamina, about an axis passing through G and perpendicular to the, plane ABC is I0. If part ADE is removed, the moment of, , (c) 0, , (d), , inertia of the remaining part about the same axis is, , NI0, 16, , where N is an integer. Value of N is _______., [NA Sep. 04, 2020 (I)], A, , 79. Moment of inertia of a cylinder of mass M, length L and, radius R about an axis passing through its centre and, perpendicular to the axis of the cylinder is, æ R 2 L2 ö, + ÷÷ . If such a cylinder is to be made for a, I = M çç, è 4 12 ø, given mass of a material, the ratio L/R for it to have minimum, possible I is :, [Sep. 03, 2020 (I)], , (a), , 2, 3, , (b), , 3, 2, (d), 2, 3, 80. An massless equilateral triangle EFG of side 'a' (As shown, in figure) has three particles of mass m situated at its, vertices. The moment of inertia of the system about the, N, ma 2, line EX perpendicular to EG in the plane of EFG is, 20, where N is an integer. The value of N is _________., [Sep. 03, 2020 (II)], , (c), , X, E, , D, , 3, 2, , F, , G, B, , C, , 76. A circular disc of mass M and radius R is rotating about, its axis with angular speed w1. If another stationary disc, R, having radius, and same mass M is dropped co-axially, 2, on to the rotating disc. Gradually both discs attain, constant angular speed w2. The energy lost in the process, is p% of the initial energy. Value of p is ___________., [NA Sep. 04, 2020 (I)], 77. Consider two uniform discs of the same thickness and, different radii R1 = R and R2 = aR made of the same material., If the ratio of their moments of inertia I1 and I2, respectively,, about their axes is I1 : I2 = 1 : 16 then the value of a is :, [Sep. 04, 2020 (II)], (a) 2 2, (c) 2, 78. y, , O, , (b) 2, (d) 4, O', , 80 cm, , x, 60 cm, For a uniform rectangular sheet shown in the figure, the, ratio of moments of inertia about the axes perpendicular to, the sheet and passing through O (the centre of mass) and, O' (corner point) is :, [Sep. 04, 2020 (II)], (a) 2/3, (b) 1/4, (c) 1/8, (d) 1/2, , E, , a, , G, , (b), , 20, J, 3, , 81. Two uniform circular discs are rotating independently in, the same direction around their common axis passing, through their centres. The moment of inertia and angular, velocity of the first disc are 0.1 kg-m2 and 10 rad s–1, respectively while those for the second one are 0.2 kg-m 2, and 5 rad s–1 respectively. At some instant they get stuck, together and start rotating as a single system about their, common axis with some angular speed. The kinetic energy, of the combined system is :, [Sep. 02, 2020 (II)], (a), , 10, J, 3, , 5, 2, J, J, (d), 3, 3, 82. Three solid spheres each of mass m and diameter d are, stuck together such that the lines connecting the centres, form an equilateral triangle of side of length d. The ratio, I0, of moment of inertia I0 of the system about an axis, IA, passing the centroid and about center of any of the spheres, IA and perpendicular to the plane of the triangle is:, [9 Jan. 2020 I], 13, (a), 23, 15, (b), 13, 23, (c), 13, 13, (d), 15, , (c)
Page 95 :
P-86, , Physics, , 83. One end of a straight uniform 1 m long bar is pivoted on, horizontal table. It is released from rest when it makes, an angle 30° from the horizontal (see figure). Its angular, speed when it hits the table is given as ns -1 , where n is, an integer. The value of n is ________ . [9 Jan. 2020 I], , 84. A uniformly thick wheel with moment of inertia I and, radius R is free to rotate about its centre of mass (see, fig). A massless string is wrapped over its rim and two, blocks of masses m1 and m2 (m1 > m2) are attached to the, ends of the string. The system is released from rest. The, angular speed of the wheel when m1 descents by a distance, h is:, [9 Jan. 2020 II], , 87. Mass per unit area of a circular disc of radius a depends, on the distance r from its centre as s(r) = A + Br. The, moment of inertia of the disc about the axis, perpendicular, to the plane and passing through its centre is:, [7 Jan. 2020 II], , aB ö, 4æ A, (a) 2pa çè + ÷ø, 4 5, , Bö, 4 æ A aB ö, 4æA, (c) pa çè + ÷ø, (d) 2pa çè + ÷ø, 4 5, 4 5, 88. A circular disc of radius b has a hole of radius a at its centre, (see figure). If the mass per unit area of the disc varies as, æ s0 ö, ç, ÷ , then the radius of gyration of the disc about its axis, è r ø, passing through the centre is :, [12 Apr. 2019 I], , 1/ 2, , (a), , a2 + b2 + ab, 2, , (b), , a+b, 2, , 1/ 2, , (c), , a2 + b2 + ab, 3, , (d), , a+b, 3, , é 2(m1 - m2 ) gh ù, ú, (a) ê, 2, ëê (m1 + m2 )R + 1 ûú, é 2(m1 + m2 ) gh ù, ú, (b) ê, 2, êë (m1 + m2 )R + 1 úû, , 89. Two coaxial discs, having moments of inertia I1 and, , 1/2, , é (m1 - m2 ) ù, ú, (c) ê, 2, ëê (m1 + m2 )R + 1 ûú, , gh, , 1/2, , é, ù, m1 + m2, ú gh, (d) ê, 2, êë (m1 + m2 )R + 1 úû, 85. As shown in the figure, a bob of mass m is tied by a, massless string whose other end portion is wound on a, fly wheel (disc) of radius r and mass m. When released, from rest the bob starts falling vertically. When it has, covered a distance of h, the angular speed of the wheel, will be:, [7 Jan. 2020 I], , 1 4 gh, r, 3, , (b) r, , 3, 2 gh, , w1, ,, 2, about their common axis. They are brought in contact with, each other and thereafter they rotate with a common angular, velocity. If Ef and Ei are the final and initial total energies,, then (Ef – Ei) is :, [10 Apr. 2019 I], 3 2, I w2, I1w12, I w2, (b) 1 1 (c) I1w1, (d) - 1 1, 8, 6, 12, 24, 90. A thin disc of mass M and radius R has mass per unit area, s(r) = kr2 where r is the distance from its centre. Its moment, of inertia about an axis going through its centre of mass, and perpendicular to its plane is :, [10 Apr. 2019 I], , (a) -, , 3, 1 2 gh, r, (d), 4 gh, r, 3, 86. The radius of gyration of a uniform rod of length l, about, l, an axis passing through a point, away from the centre, 4, of the rod, and perpendicular to it, is: [7 Jan. 2020 I], , (a), , 1, l, 4, , (b), , 1, l, 8, , (c), , 7, l, 48, , (d), , 3, l, 8, , MR 2, 3, , (b), , MR 2, 6, , (d), , 2MR 2, 3, , MR 2, 2, 91. A solid sphere of mass M and radius R is divided into two, (c), , (c), , I1, , are, 2, , rotating with respective angular velocities w1 and, , (a), (a), , Bö, 4 æ aA, + ÷, (b) 2pa çè, 4 5ø, , 7M, and is, 8, converted into a uniform disc of radius 2R. The second, part is converted into a uniform solid sphere. Let I1 be, the moment of inertia of the new sphere about its axis., The ratio I1/I2 is given by :, [10 Apr. 2019 II], (a) 185, (b) 140, (c) 285, (d) 65, , unequal parts. The first part has a mass of
Page 96 :
P-87, , System of Particles and Rotational Motion, , 92. A stationary horizontal disc is free to rotate about its axis., When a torque is applied on it, its kinetic energy as a, function of q, where q is the angle by which it has rotated,, is given as kq2. If its moment of inertia is I then the angular, acceleration of the disc is:, [9 April 2019 I], k, k, k, 2k, q, q (c) 2 I q, q, (a), (b), (d), 4I, I, I, 93. Moment of inertia of a body about a given axis is 1.5 kg, m2. Initially the body is at rest. In order to produce a, rotational kinetic energy of 1200 J, the angular, acceleration of 20 rad/s2 must be applied about the axis, for a duration of:, [9 Apr. 2019 II], (a) 2.5s, (b) 2s, (c) 5s, (d) 3s, 94. A thin smooth rod of length L and mass M is rotating, freely with angular speed w0 about an axis perpendicular, to the rod and passing through its center. Two beads of, mass m and negligible size are at the center of the rod, initially. The beads are free to slide along the rod. The, angular speed of the system, when the beads reach the, opposite ends of the rod, will be:, [9 Apr. 2019 II], M w0, M w0, (a), (b), M +m, M + 3m, M w0, M w0, (d), (c), M + 6m, M + 2m, 95. A thin circular plate of mass M and radius R has its density, varying as r(r) = r0 r with r0 as constant and r is the, distance from its center. The moment of Inertia of the, circular plate about an axis perpendicular to the plate and, passing through its edge is I = a MR2. The value of the, coefficient a is:, [8 April 2019 I], 3, (a) 1 2, (b) 3 5, (c) 8 5, (d), 2, 96. Let the moment of inertia of a hollow cylinder of length 30, cm (inner radius 10 cm and outer radius 20 cm), about its, axis be 1. The radius of a thin cylinder of the same mass, such that its moment of inertia about its axis is also I, is:, [12 Jan. 2019 I], (a) 12 cm, (b) 16 cm, (c) 14 cm, (d) 18 cm, 97. The moment of inertia of a solid sphere, about an axis, parallel to its diameter and at a distance of x from it, is ‘I(x)’., Which one of the graphs represents the variation of I(x), with x correctly?, [12 Jan. 2019 II], , I(x), , I(x), , (a), , (b), , x, , O, , x, , O, , I(x), , I(x), , (c), , (d), O, , x, , O, , x, , 98. An equilateral triangle ABC is cut from a thin solid sheet, of wood. (See figure) D, E and F are the mid-points of its, sides as shown and G is the centre of the triangle. The, moment of inertia of the triangle about an axis passing, through G and perpendicular to the plane of the triangle is, I0. If the smaller triangle DEF is removed from ABC, the, moment of inertia of the remaining figure about the same, axis is I. Then :, [11 Jan. 2019 I], A, , D, , E, G, , B, , C, , F, , 15, 3, I0, (b) I = I0, 16, 4, 9, I0, (c) I = I 0, (d) I =, 16, 4, 99. a string is wound around a hollow cylinder of mass 5 kg, and radius 0.5 m. If the string is now pulled with a, horizontal force of 40 N, and the cylinder is rolling, without slipping on a horizontal surface (see figure), then, the angular acceleration of the cylinder will be (Neglect, the mass and thickness of the string) [11 Jan. 2019 II], 40 N, , (a), , I=, , (a) 20 rad/s2, (b) 16 rad/s2, 2, (c) 12 rad/s, (d) 10 rad/s2, 100. A circular disc D1 of mass M and radius R has two, identical discs D2 and D3 of the same mass M and radius, R attached rigidly at its opposite ends (see figure). The, moment of inertia of the system about the axis OO’,, passing through the centre of D1, as shown in the figure,, will be :, [11 Jan. 2019 II], O', , D2, , O, , D3, , D1, , (a) MR2, (b) 3MR2, 4, 2, MR 2, MR 2, (d), (c), 5, 3, 101. Two identical spherical balls of mass M and radius R, each are stuck on two ends of a rod of length 2R and, mass M (see figure). The moment of inertia of the, system about the axis passing perpendicularly through, the centre of the rod is:, [10 Jan. 2019 II]
Page 97 :
P-88, , Physics, , (a), (c), , 137, MR 2, 15, , (b), , 209, MR 2, 15, , (d), , 17, MR 2, 15, 152, MR 2, 15, , m, are connected at the two ends of, 2, a massless rigid rod of length l. The rod is suspended by, a thin wire of torsional constant k at the centre of mass, of the rod-mass system (see figure). Because of, torsional constant k, the restoring toruque is t= kq for, angular displacement q. If the rod is rotated by q0 and, released, the tension in it when it passes through its mean, position will be:, [9 Jan. 2019 I], , 102. Two masses m and, , 105. From a uniform circular disc of radius R and mass 9 M, a, R, is removed as shown in the figure., 3, The moment of inertia of the remaining disc about an, axis perpendicular to the plane of the disc and passing,, through centre of disc is :, [2018], , small disc of radius, , 2R, 3, R, , 40, 37, MR2 (c) 10 MR2 (d), MR2, 9, 9, 106. A thin circular disk is in the xy plane as shown in the, figure. The ratio of its moment of inertia about z and z¢ axes, will be, [Online April 16, 2018], , (a) 4 MR2 (b), , z, , z¢, , O, , 3k q0 2, 2k q0 2, k q0 2, k q0 2, (b), (c), (d), l, l, l, 2l, 103. A rod of length 50 cm is pivoted at one end. It is raised, such that if makes an angle of 30° from the horizontal as, shown and released from rest. Its angular speed when, it passes through the horizontal (in rads–1) will be, (g = 10 ms–2), [9 Jan. 2019 II], , x, , (a), , (a) 1 : 2, (b) 1 : 4, (c) 1 : 3, (d) 1 : 5, 107. A thin rod MN, free to rotate in the vertical plane about the, fixed end N, is held horizontal. When the end M is released, the speed of this end, when the rod makes an angle a with, the horizontal, will be proportional to: (see figure), [Online April 15, 2018], M, , 30°, , (a), , 30, 7, , (b), , 30, , 20, 30, (d), 3, 2, 104. Seven identical circular planar disks, each of mass M and, radius R are welded symmetrically as shown. The moment, of inertia of the arrangement about the axis normal to the, plane and passing through the point P is:, [2018], (c), , P, O, , (a), , 19, MR 2, 2, , (b), , 55, MR 2 (c), 2, , 73, 181, MR 2 (d), MR2, 2, 2, , y, , a, , N, , (a), (b) cosa, cos a, (c) sin a, (d), sin a, 108. The moment of inertia of a uniform cylinder of length l and, radius R about its perpendicular bisector is I. What is the, ratio l/R such that the moment of inertia is minimum ?, [2017], 3, 3, 3, (a) 1, (b), (c), (d), 2, 2, 2, 109. Moment of inertia of an equilateral triangular lamina ABC,, about the axis passing thr ough its centre O and, perpendicular to its plane is Io as shown in the figure. A, cavity DEF is cut out from the lamina, where D, E, F are, the mid points of the sides. Moment of inertia of the, remaining part of lamina about the same axis is :, [Online April 8, 2017], (a), , 7, Io, 8, , (b), , 15, Io, 16, , (c), , 3Io, 4, , (d), , 31I o, 32, , C, , E, , F, O, A, , D, , B
Page 98 :
P-89, , System of Particles and Rotational Motion, , R, is made in a thin uniform, 4, disc having mass M and radius R, as shown in figure. The, moment of inertia of the remaining portion of the disc about, an axis passing through the point O and perpendicular to, the plane of the disc is :, [Online April 9, 2017], 219 MR 2, (a), 256, , 110. A circular hole of radius, , R, 237 MR 2, R/4, 512, o', O, 19 MR 2, (c), 3R/4, 512, 197 MR 2, (d), 256, 111. From a solid sphere of mass M and radius R a cube of, maximum possible volume is cut. Moment of inertia of cube, about an axis passing through its center and perpendicular, to one of its faces is :, [2015], 4MR 2, 4MR 2, (a), (b), 9 3p, 3 3p, 2, MR, MR 2, (c), (d), 32 2p, 16 2p, 112. Consider a thin uniform square sheet made of a rigid, material. If its side is ‘a’ mass m and moment of inertia I, about one of its diagonals, then :[Online April 10, 2015], , (b), , (a), , I>, , ma 2, 12, , (b), , ma 2, ma 2, <I<, 24, 12, , ma 2, ma 2, (d) I =, 24, 12, 113. A ring of mass M and radius R is rotating about its axis, with angular velocity w. Two identical bodies each of mass, m are now gently attached at the two ends of a diameter of, the ring. Because of this, the kinetic energy loss will be:, [Online April 25, 2013], , (c), , I=, , Mm, m( M + 2m) 2 2, w2 R2, w R, (b), ( M + m), M, Mm, ( M + m) M 2 2, w R, w2 R2, (c), (d), ( M + 2 m), (M + 2m), 114. This question has Statement 1and Statement 2. Of the four, choices given after the Statements, choose the one that, best describes the two Statements., Statement 1: When moment of inertia I of a body rotating, about an axis with angular speed w increases, its angular, momentum L is unchanged but the kinetic energy K, increases if there is no torque applied on it., Statement 2: L = Iw, kinetic en ergy of rotation, 1 2, = Iw, [Online May 12, 2012], 2, (a) Statement 1 is true, Statement 2 is true, Statement, 2 is not the correct explanation of Statement 1., (b) Statement 1 is false, Statement 2 is true., (a), , (c) Statement 1 is true, Statement 2 is true, Statement, 2 is correct explanation of the Statement 1., (d) Statement 1 is true, Statement 2 is false., 115. A solid sphere having mass m and radius r rolls down an, inclined plane. Then its kinetic energy is, [Online May 7, 2012], 5, 2, rotational and translational, (a), 7, 7, 2, 5, (b), rotational and translational, 7, 7, 2, 3, (c), rotational and translational, 5, 5, 1, 1, (d), rotational and translational, 2, 2, 116. A circular hole of diameter R is cut from a disc of mass M, and radius R; the circumference of the cut passes through, the centre of the disc. The moment of inertia of the, remaining portion of the disc about an axis perpendicular, to the disc and passing through its centre is, [Online May 7, 2012], (a), , æ 15 ö, 2, çè ÷ø MR, 32, , (b), , æ 1ö, 2, çè ÷ø MR, 8, , æ 3ö, æ 13 ö, 2, 2, (d) çè ÷ø MR, çè ÷ø MR, 8, 32, 117. A mass m hangs with the help of a string wrapped around, a pulley on a frictionless bearing. The pulley has mass m, and radius R. Assuming pulley to be a perfect uniform, circular disc, the acceleration of the mass m, if the string, does not slip on the pulley, is:, [2011], (c), , (a) g, , (b), , 2, g, 3, , (c), , g, 3, , (d), , 3, g, 2, , 118. A pulley of radius 2 m is rotated about its axis by a force, F = (20t – 5t2) newton (where t is measured in seconds), applied tangentially. If the moment of inertia of the pulley, about its axis of rotation is 10 kg-m2 the number of rotations, made by the pulley before its direction of motion is, reversed, is:, [2011], (a) more than 3 but less than 6, (b) more than 6 but less than 9, (c) more than 9, (d) less than 3, 119. A thin uniform rod of length l and mass m is swinging, freely about a horizontal axis passing through its end. Its, maximum angular speed is w. Its centre of mass rises to, a maximum height of, [2009], 2, 2, 1 lw, 1l w, (a), (b), 6 g, 2 g, 2 2, 1l w, 1 l 2w2, (c), (d), 6 g, 3 g, 120. Consider a uniform square plate of side ‘a’ and mass ‘M’., The moment of inertia of this plate about an axis, perpendicular to its plane and passing through one of its, corners is, [2008]
Page 99 :
P-90, , Physics, , (a), , 5, Ma 2, 6, , (b), , 1, Ma 2, 12, , 7, 2, Ma 2, Ma 2, (d), 12, 3, 121. For the given uniform square lamina ABCD, whose centre, is O,, [2007], , (c), , D, , F, , C, , O, A, , (a), , I AC = 2 I EF, , B, , E, , (b), , 2 I AC = I EF, , (c) I AD = 3I EF, (d) I AC = I EF, 122. Four point masses, each of value m, are placed at the, corners of a square ABCD of side l. The moment of inertia, of this system about an axis passing through A and parallel, to BD is, [2006], (a), , 2ml 2, , (b), , 3ml2, , (c) 3ml 2, (d) ml 2, 123. The moment of inertia of a uniform semicircular disc of, mass M and radius r about a line perpendicular to the, plane of the disc through the centre is, [2005], 1, 2, Mr, (a), (b), Mr 2 `, 4, 5, (c), , 1, Mr 2, 2, , (d), , (d), , IA dA, =, I B dB, , where dA and dB are their densities., 125. A circular disc X of radius R is made from an iron plate of, thickness t, and another disc Y of radius 4R is made from an, t, . Then the relation between the, 4, moment of inertia IX and II is, [2003], , iron plate of thickness, , (a), , ΙY = 32 Ι X, , (c) Ι Y = Ι X, , (b), , (c), , æ M ö, çè, ÷w, M + 4m ø 1, , (d), , æ M ö, çè, ÷ w1, M + 2mø, , TOPIC 5 Rolling Motion, 129.A uniform sphere of mass 500 g rolls without slipping, on a plane horizontal surface with its centre moving at a, speed of 5.00 cm/s. Its kinetic energy is:, [8 Jan. 2020 II], (a) 8.75 × 10–4 J, (b) 8.75 × 10–3 J, (c) 6.25 × 10–4 J, (d) 1.13 × 10–3 J, 130., , Mr 2, , 124. One solid sphere A and another hollow sphere B are of, same mass and same outer radii. Their moment of inertia, about their diameters are respectively IA and IB Such that, [2004], (a) IA < IB, (b) IA > IB, (c) IA = IB, , 126. A particle performing uniform circular motion has angular, frequency is doubled & its kinetic energy halved, then the, new angular momentum is, [2003], L, (a), (b) 2 L, 4, L, (c) 4 L, (d), 2, 127. Moment of inertia of a circular wire of mass M and radius, R about its diameter is, [2002], (a) MR2/2 (b) MR 2, (c) 2MR 2 (d) MR2/4, 128. Initial angular velocity of a circular disc of mass M is, w 1. Then two small spheres of mass m are attached gently, to diametrically opposite points on the edge of the disc., What is the final angular velocity of the disc? [2002], æ M + mö, æ M + mö, (a) çè, (b) çè, ÷w, ÷w, M ø 1, m ø 1, , Consider a uniform cubical box of side a on a rough floor, that is to be moved by applying minimum possible force F, at a point b above its centre of mass (see figure). If the, coefficient of friction is m = 0.4, the maximum possible value, b, for box not to topple before moving is, a, ________., [NA 7 Jan. 2020 II], 131.A solid sphere and solid cylinder of identical radii approach, an incline with the same linear velocity (see figure). Both, roll without slipping all throughout. The two climb, maximum heights hsph and hcyl on the incline. The ratio, , of 100 ×, , hsph, hcyl, , is given by :, , [8 Apr. 2019 II], , ΙY = 16 Ι X, , (d) ΙY = 64 Ι X, (a), , 2, 5, , (b) 1, , (c), , 14, 15, , (d), , 4, 5
Page 100 :
P-91, , System of Particles and Rotational Motion, , 132.The following bodies are made to roll up (without, slipping) the same inclined plane from a horizontal plane:, (i) a ring of radius R, (ii) a solid cylinder of radius, , R, 2, , R, and (iii) a solid sphere of radius . If, in each case, the, 4, speed of the center of mass at the bottom of the incline, is same, the ratio of the maximum heights they climb is:, [9 April 2019 I], (a) 4 : 3 : 2, (b) 10 : 15 : 7, (c) 14 : 15 : 20, (d) 2 : 3 : 4, 133. A homogeneous solid cylindrical roller of radius R and, mass M is pulled on a cricket pitch by a horizontal, force. Assuming rolling without slipping, angular, acceleration of the cylinder is:, [10 Jan. 2019 I], , (a), , 3F, 2mR, , (b), , F, 3m R, , (c), , F, 2mR, , (d), , 2F, 3m R, , 134. A roller is made by joining together two cones at their, vertices O. It is kept on two rails AB and CD, which are, placed asymmetrically (see figure), with its axis, perpendicular to CD and its centre O at the centre of line, joining AB and Cd (see figure). It is given a light push so, that it starts rolling with its centre O moving parallel to, CD in the direction shown. As it moves, the roller will, tend to:, [2016], B, D, , O, , Figure). If they roll on the incline without slipping such, sin qc, that their accelerations are the same, then the ratio, sin q s, is:, [Online April 9, 2014], MC, , A, , M, , S, , B, , qC, qS, , C, , (a), , 8, 7, , D, , 15, 14, , (b), , 8, 15, (d), 7, 14, 137. A loop of radius r and mass m rotating with an angular, velocity w0 is placed on a rough horizontal surface., The initial velocity of the centre of the hoop is zero.What, will be the velocity of the centre of the hoop when it ceases, to slip ?, [2013], , (c), , (a), , rw0, 4, , (b), , rw0, 3, , rw0, (d) rw0, 2, 138. A tennis ball (treated as hollow spherical shell) starting, from O rolls down a hill. At point A the ball becomes air, borne leaving at an angle of 30° with the horizontal. The, ball strikes the ground at B. What is the value of the, distance AB ?, (Moment of inertia of a spherical shell of mass m and radius, 2, R about its diameter = mR 2 ), 3, [Online April 22, 2013], , (c), , O, , A, , C, , (a) go straight., (b) turn left and right alternately., (c) turn left., (d) turn right., 135. A uniform solid cylindrical roller of mass ‘m’ is being, pulled on a horizontal surface with force F parallel to the, surface and applied at its centre. If the acceleration of the, cylinder is ‘a’ and it is rolling without slipping then the, value of ‘F’ is:, [Online April 10, 2015], 5, ma, (a) ma, (b), 3, 3, ma, (c), (d) 2 ma, 2, 136. A cylinder of mass Mc and sphere of mass Ms are placed, at points A and B of two inclines, respectively (See, , 2.0 m, 30°, 0.2 m, , A, , B, , (a) 1.87 m, (b) 2.08 m, (c) 1.57 m, (d) 1.77 m, 139. A thick-walled hollow sphere has outside radius R0. It rolls, down an incline without slipping and its speed at the bottom, is v0. Now the incline is waxed, so that it is practically, frictionless and the sphere is observed to slide down, (without any rolling). Its speed at the bottom is observed, to be 5v0/4. The radius of gyration of the hollow sphere, about an axis through its centre is [Online May 26, 2012], (a) 3R0/2, (b) 3R0/4, (c) 9R0 /16, (d) 3R0
Page 101 :
P-92, , Physics, , 140. A solid sphere is rolling on a surface as shown in figure,, with a translational velocity v ms–1. If it is to climb the, inclined surface continuing to roll without slipping, then, minimum velocity for this to happen is, [Online May 12, 2012], , h, , v, , (a), , (b), , 2gh, , 7, gh, 5, , 7, 10, gh, gh, (d), 2, 7, 141. A round uniform body of radius R, mass M and moment of, inertia I rolls down (without slipping) an inclined plane, making an angle q with the horizontal. Then its, acceleration is, [2007], , (c), , (a), (c), , g sin q, 2, , 1 - MR / I, g sin q, , 1 + MR 2 / I, , (b), (d), , g sin q, 1 + I / MR 2, g sin q, 1 - I / MR 2
Page 116 :
P-107, , System of Particles and Rotational Motion, , 106. (c) As we know, moment of inertia of a disc about an axis, passing through C.G. and perpendicular to its plane,, 2, , mR, 2, Moment of inertia of a disc about a tangential axis, perpendicular to its own plane,, Iz =, , 3, Iz' = mR 2, 2, , q, , q, , r, mR 2 3mR 2, Iz Iz' =, =1 3, \, 2, 2, 107. (a) When the rod makes an angle a, l, Displacement of centre of mass = cos a, 2, l, l, mg cos a = I w 2, 2, 2, l, ml 2 2, mg cos a =, w (Q M.I. of thin uniform rod, 2, 6, about an axis passing through its centre of mass and, , perpendicular to the rod I =, Þ, , w=, , Speed of end = w ´ l = 3g cos al, i.e., Speed of end, w µ cos a, 108. (c) As we know, moment of inertia of a solid cylinder, about an axis which is perpendicular bisector, , I=, , l, , m é V l2 ù, = ê + ú, 4 ë pl 3 û, , V, 2l, =, 2, 3, pl, , Þ, , ÞV =, 2, , 3, , dl m é -V 2l ù, =, +, =0, dl 4 êë pl 2 3 úû, , 2 pl 3, 3, , l, 3, 3, l, 2pl, Þ 2 = or, =, 2, 2, R, 3, R, 109. (b) According to theorem of perpendicular axes, moment, of inertia of triangle (ABC), I0 = kml2, ..... (i), BC = 1, Moment of inertia of a cavity DEF, , pR 2 l =, , IDEF = K, =, , mæ lö, ç ÷, 4 è 2ø, , k, ml 2, 16, , I0, 16, Moment of inertia of remaining part, IDEF =, , I0 15I 0, =, 16, 16, 110. (b) Moment of Inertia of complete disc about 'O' point, MR 2, Itotal =, 2, Radius of removed disc = R/4, \ Mass of removed disc = M/16, [As M µ R2], M.I of removed disc about its own axis (O'), Iremain = I0 -, , 2, , 1 Mæ Rö, MR 2, çè ÷ø =, 2 16 4, 512, M.I of removed disc about O, Iremoved disc = Icm + mx2, =, , 2, , MR 2 M æ 3R ö, 19 MR 2, + ç ÷ =, 512, 512 16 è 4 ø, M.I of remaining disc, =, , 2, 237, Iremaining = MR - 19 MR 2 =, MR 2, 512, 2, 512, 2, R, 111. (a) Here a =, 3, 4 3, pR, M, Now,, =3 3, M¢, a, , ml 2, ), 12, , 3 g cos a, l, , mR 2 ml 2, +, 4, 12, é, m, l2 ù, I = ê R2 + ú, 4ë, 3û, , From equation (i),, , 4 3, pR, 3, = 3, =, p., 3, 2, æ 2 ö, R÷, ç, è 3 ø, , M¢=, , a, , 2M, 3p, , Moment of inertia of the cube about the given axis,, I=, , M ¢a 2, 6, 2, , 2M æ 2 ö, ´ç, R÷, 2, 3p è 3 ø = 4 MR, =, 9 3p, 6, 112. (d) For a thin uniform square sheet, , I1 = I2 = I3 =, , ma 2, 12, , 2, , I1, I2, , I3
Page 118 :
P-109, , System of Particles and Rotational Motion, , The moment of inertia of the rod about O is, , 1 2, ml . The, 3, , 1 2, I w where I is, 2, the moment of inertia of the rod about O. When the rod is in, position B, its angular velocity is zero. In this case, the, energy of the rod is mgh where h is the maximum height to, which the centre of mass (C.M) rises, Gain in potential energy = Loss in kinetic energy, 1 2 1 æ 1 2ö 2, \ mgh = I w = 2 çè 3 ml ÷ø w, 2, kinetic energy of the rod at position A =, , Þ h=, , l 2 w2, 6g, , 120. (d) Inn' =, , 1, Ma 2, M (a 2 + a 2 ) =, 12, 6, n, , Again, by the same theorem I = IAC + IBD = 2 IAC, (\ IAC = IBD by symmetry of the figure), I, 2, From (i) and (ii), we get, IEF = IAC., , ...(ii), , \ I AC =, , 122. (c), , l, , D, , n, , A, , 2, l/, , C, , O, B, n', , Inn' = M.I due to the point mass at B +, M.I due to the point mass at D +, M.I due to the point mass at C., m, , A, D, , æ l ö, Inn' = m ç, è 2 ÷ø, , 2, , æ l ö, +m ç, è 2 ÷ø, , 2, , +m, , O, , ( 2l), , 2, 2, , B, , C, n, , m, , 1, , DB, 2a, a, =, =, 2, 2, 2, By parallel axes the orem, moment of inertia of plate about, an axis through one of its corners., , Also, DO =, , 2, , Ma 2 Ma 2, æ a ö, Imm ' = I nn ' + M ç, =, +, ÷, è 2ø, 6, 2, 2, , 2, , Ma + 3Ma, 2, = Ma 2, 6, 3, 121. (d) By the theorem of perpendicular axes,, I = IEF + IGH, Here, I is the moment of inertia of square lamina about an, axis through O and perpendicular to its plane., \ IEF = IGH (By Symmetry of Figure), =, , Z, , Y, F, , D, , æ l ö, + m( 2l) 2, Þ I nn ' = 2 ´ m ç, è 2 ÷ø, = ml2 + 2ml2 = 3ml2, 123. (c) The disc may be assumed as combination of two semi, circular parts. Therefore, circular disc will have twice, the mass of semicircular disc., 1, (2m)r2 = Mr2, 2, Let I be the moment of inertia of the uniform semicircular, disc, , Moment of inertia of disc =, , Mr 2, 2, 124. (a) The moment of inertia of solid sphere A about its, , Þ 2 I = 2Mr 2 Þ I =, , 2, MR 2 ., 5, The moment of inertia of a hollow sphere B about its, 2, 2, diameter I B = MR ., 3, , diameter I A =, , \ I A < IB, , C, , 125. (d) We know that density (d ) =, X, , O, , mass( M ), volume(V ), , \ M = d ´ V = d ´ ( pR 2 ´ t ) ., , The moment of inertia of a disc is given by I =, A, , \ I EF =, , I, 2, , E, , B, , \ Ix =, , ...(i), , 1, 1, MxRx2 = (d ´ pR 2 ´ t ) R 2, 2, 2, , 1, MR 2, 2
Page 122 :
7, , P-113, , Gravitation, , Gravitation, TOPIC 1 Kepler's Laws of Planetary, Motion, 1., , 4., , If the angular momentum of a planet of mass m, moving, around the Sun in a circular orbit is L, about the center of, the Sun, its areal velocity is:, [9 Jan. 2019 I], , L, 4L, L, 2L, (b), (c), (d), m, m, 2m, m, Figure shows elliptical path abcd of a planet around the, (a), , 2., , 1, sun S such that the area of triangle csa is the area of the, 4, ellipse. (See figure) With db as the semimajor axis, and ca, as the semiminor axis. If t1 is the time taken for planet to go, over path abc and t2 for path taken over cda then:, [Online April 9, 2016], , TOPIC 2 Newton's Universal Law of, Gravitation, 5., , c, , d, , (a) 500 days, (b) 320 days, (c) 260 days, (d) 220 days, The time period of a satellite of earth is 5 hours. If the, separation between the earth and the satellite is increased, to 4 times the previous value, the new time period will, become, [2003], (a) 10 hours, (b) 80 hours, (c) 40 hours, (d) 20 hours, , A straight rod of length L extends from x = a to x = L + a., The gravitational force it exerts on point mass ‘m’ at x = 0,, if the mass per unit length of the rod is A + Bx2, is given, by:, [12 Jan. 2019 I], (a), , é æ 1, 1ö, ù, Gm ê A ç, - ÷ - BL ú, ë èa+L aø, û, , (b), , é æ1, 1 ö, ù, Gm ê A ç ÷ - BL ú, a, a, +, L, ø, ë è, û, , (c), , é æ 1, 1ö, ù, - ÷ + BL ú, Gm ê A ç, ë èa+L aø, û, , (d), , é æ1, 1 ö, ù, Gm ê A ç ÷ + BL ú, a, a, +, L, è, ø, ë, û, , b, S, a, , 3., , (a) t1 = 4t2, (b) t1 = 2t2, (c) t1 = 3t2, (d) t1 = t2, India’s Mangalyan was sent to the Mars by launching it, into a transfer orbit EOM around the sun. It leaves the, earth at E and meets Mars at M. If the semi-major axis of, Earth’s orbit is ae = 1.5 × 1011 m, that of Mars orbit am =, 2.28 × 1011 m, taken Kepler’s laws give the estimate of time, for Mangalyan to reach Mars from Earth to be close to:, [Online April 9, 2014], Mars orbit, O, , M, , am, , ae, Sun, , E, , 6., , Take the mean distance of the moon and the sun from the, earth to be 0.4 × 106 km and 150 × 106 km respectively., Their masses are 8 × 1022 kg and 2 × 1030 kg respectively., The radius of the earth is 6400 km. Let DF1 be the difference, in the forces exerted by the moon at the nearest and farthest, points on the earth and DF2 be the difference in the force, exerted by the sun at the nearest and farthest points on, the earth. Then, the number closest to, , Earth’s orbit, , (a) 2, , (b) 6, , DF1, is:, DF2, , [Online April 15, 2018], (c) 10–2, (d) 0.6
Page 123 :
P-114, , 7., , Physics, , Four particles, each of mass M and equidistant from each, other, move along a circle of radius R under the action of, their mutual gravitational attraction. The speed of each, particle is:, [2014], GM, R, , (a), , (, , GM, R, , ), , (, , (a), , ), , GM, 1 GM, 1+ 2 2, 1+ 2 2, (d), R, 2 R, From a sphere of mass M and radius R, a smaller sphere of, R, radius, is carved out such that the cavity made in the, 2, original sphere is between its centre and the periphery, (See figure). For the configuration in the figure where the, distance between the centre of the original sphere and the, removed sphere is 3R, the gravitational force between the, two sphere is:, [Online April 11, 2014], , (c), 8., , 2 2, , (b), , 12. The acceleration due to gravity on the earth’s surface at, the poles is g and angular velocity of the earth about the, axis passing through the pole is w. An object is weighed at, the equator and at a height h above the poles by using a, spring balance. If the weights are found to be same, then h, is : (h<<R, where R is the radius of the earth), , 13., , R 2w 2, 2g, , (b), , R 2w 2, [5 Sep. 2020 (II)], g, , R 2w 2, R 2w 2, (c), (d), 4g, 8g, The height 'h' at which the weight of a body will be the, same as that at the same depth 'h' from the surface of the, earth is (Radius of the earth is R and effect of the rotation, of the earth is neglected) :, [2 Sep. 2020 (II)], , (a), , 5, R-R, 2, , (b), , R, 2, , 5R - R, 3R - R, (d), 2, 2, 14. A box weighs 196 N on a spring balance at the north pole., Its weight recorded on the same balance if it is shifted to, the equator is close to (Take g = 10 ms –2 at the north pole, and the radius of the earth = 6400 km): [7 Jan. 2020 II], (a) 195.66 N, (b) 194.32 N, (c) 194.66 N, (d) 195.32 N, 15. The ratio of the weights of a body on the Earth’s surface to, that on the surface of a planet is 9:4. The mass of the, (c), , 3R, (a), , 9., , 10., , 41 GM 2, , (c), (d), 450 R 2, 225 R 2, 450 R 2, Two particles of equal mass ‘m’ go around a circle of radius, R under the action of their mutual gravitational attraction., The speed of each particle with respect to their centre of, mass is, [2011 RS], Gm, Gm, Gm, Gm, (a), (b), (c), (d), 4R, 3R, R, 2R, Two spherical bodies of mass M and 5M & radii R & 2R, respectively are released in free space with initial separation, between their centres equal to 12 R. If they attract each, other due to gravitational force only, then the distance, covered by the smaller body just before collision is [2003], (a) 2.5 R, (b) 4.5 R, (c) 7.5 R, (d) 1.5 R, 3600 R 2, , (b), , GM 2, , 59 GM 2, , 41 GM 2, , TOPIC 3 Acceleration due to Gravity, 11., , The value of acceleration due to gravity is g1 at a height, R, (R = radius of the earth) from the surface of the, 2, earth. It is again equal to g1 and a depth d below the sur-, , h=, , ædö, face of the earth. The ratio ç ÷ equals : [5 Sep. 2020 (I)], è Rø, (a), , 4, 9, , (b), , 5, 9, , (c), , 1, 3, , (d), , 7, 9, , 1, th of that of the Earth. If ‘R’ is the radius of the, 9, Earth, what is the radius of the planet ? (Take the planets, to have the same mass density)., [12 April 2019 II], , planet is, , R, R, R, R, (b), (c), (d), 3, 4, 9, 2, 16. The value of acceleration due to gravity at Earth’s surface, is 9.8 ms– 2. The altitude above its surface at which the, acceleration due to gravity decreases to 4.9 ms– 2, is close, to : (Radius of earth = 6.4 × 106 m), [10 April 2019 I], 6, 6, (a) 2.6×10 m, (b) 6.4×10 m, (c) 9.0×106 m, (d) 1.6×106 m, 17. Suppose that the angular velocity of rotation of earth is, increased. Then, as a consequence., [Online April 16, 2018], (a) There will be no change in weight anywhere on the, earth, (b) Weight of the object, everywhere on the earth, wild, decrease, (c) Weight of the object, everywhere on the earth, will, increase, (d) Except at poles, weight of the object on the earth will, decrease, , (a)
Page 124 :
P-115, , Gravitation, , 18. The variation of acceleration due to gravity g with distance, d from centre of the earth is best represented by (R =, Earth's radius):, [2017, Online May 7, 2012], g, g, (a), , (b), , d, R, , O, , R, , g, , g, (c), , (d), , d, , d, O, O, R, 19. The mass density of a spherical body is given by r (r) =, k, for r < R and r (r) = 0 for r > R,, r, where r is the distance from the centre., The correct graph that describes qualitatively the acceleration, a, of a test particle as a function of r is :, [Online April 9, 2017], a, , a, (a), , (b), , r, , R, , R, , (c), , (d), r, , r, R, If the Earth has no rotational motion, the weight of a person, on the equator is W. Determine the speed with which the, earth would have to rotate about its axis so that the person, R, , 20., , r, , a, , a, , 3, at the equator will weight W . Radius of the Earth is, 4, 2, 6400 km and g =10 m/s ., [Online April 8, 2017], , (a) 1.1×10–3 rad/s, (c), , 0.63 × 10–3 rad/s, , (b) 0.83×10–3 rad/s, (d) 0.28×10–3 rad/s, , 21. The change in the value of acceleration of earth towards, sun, when the moon comes from the position of solar, eclipse to the position on the other side of earth in line, with sun is:, = 7.36 × 1022 kg, radius of the moon’s, , (mass of the moon, orbit = 3.8 × 108 m)., (a), , 6.73 × 10–5 m/s2, , (c), , 6.73 × 10–2 m/s2, , [Online April 22, 2013], (b), , 6.73 × 10–3 m/s2, , (d), , 6.73 × 10–4 m/s2, , g, (where g = the acceleration due to gravity on, 9, the surface of the earth) in terms of R, the radius of the, earth, is, [2009], R, (a), (b) R / 2, (c), (d) 2 R, 2R, 2, 24. The change in the value of ‘g’ at a height ‘h’ above the, surface of the earth is the same as at a depth ‘d’ below the, surface of earth. When both ‘d’ and ‘h’ are much smaller, than the radius of earth, then which one of the following is, correct?, [2005], h, 3h, (a) d =, (b) d =, 2, 2, (c) d = h, (d) d =2 h, 25. Average density of the earth, [2005], (a) is a complex function of g, (b) does not depend on g, (c) is inversely proportional to g, (d) is directly proportional to g, becomes, , d, O, , 22. Assuming the earth to be a sphere of uniform density, the, acceleration due to gravity inside the earth at a distance of, r from the centre is proportional to[Online May 12, 2012], (a) r, (b) r–1, (c) r2, (d) r–2, 23. The height at which the acceleration due to gravity, , Gravitational Field and, Potential Energy, , TOPIC 4, , 26. Two planets have masses M and 16 M and their radii are a, and 2a, respectively. The separation between the centres, of the planets is 10a. A body of mass m is fired from the, surface of the larger planet towards the smaller planet along, the line joining their centres. For the body to be able to, reach the surface of smaller planet, the minimum firing, speed needed is :, [6 Sep. 2020 (II)], (a) 2, , GM, a, , (b) 4, , GM, a, , 3 5GM, GM 2, (d), 2, a, ma, 27. On the x-axis and at a distance x from the origin, the, gravitational field due to a mass distribution is given by, , (c), , Ax, 2, , ( x + a 2 )3/2, , in the x-direction. The magnitude of, , gravitational potential on the x-axis at a distance x, taking, its value to be zero at infinity, is :, [4 Sep. 2020 (I)], (a), , A, ( x2 + a2 ), , 1, , (b), 2, , (c) A( x 2 + a 2 ), , 1, , 2, , A, ( x2 + a2 ), , 3, , 2, , 2, 2, (d) A( x + a ), , 3, , 2
Page 125 :
P-116, , 28., , Physics, , The mass density of a planet of radius R varies with the, æ, r2 ö, distance r from its centre as r(r ) = r0 çç1 - 2 ÷÷ . Then the, è R ø, gravitational field is maximum at :, [3 Sep. 2020 (II)], , (a) r = 3 R, 4, , 1, , R, , Gravitational field E, , (b) 1.16, , GM, a, , (c) 1.21, , GM, a, , (d) 1.41, , GM, a, , K, . Identify the, r2, correct relation between the radius R of the particle’s orbit, and its period T:, [8 April 2019 II], 2, 3, (a) T/R is a constant, (b) T /R is a constant, , field produced by a mass density r (r ) =, , (d) r =, , 4, , (c) T/R2 is a constant, (d) TR is a constant, 34. A body of mass m is moving in a circular orbit of radius R, about a planet of mass M. At some instant, it splits into, two equal masses. The first mass moves in a circular orbit, of radius, , 3, , radius, , 1, , 1, , 1, , 2, , 3, , 4, , 5, radius R, , 2, 1, 1, 1, (b), (c), (d), 3, 6, 2, 3, 30. An asteroid is moving directly towards the centre of the, earth. When at a distance of 10 R (R is the radius of the, earth) from the earths centre, it has a speed of 12 km/s., Neglecting the effect of earths atmosphere, what will be, the speed of the asteroid when it hits the surface of the, earth (escape velocity from the earth is 11.2 km/ s)? Give, your answer to the nearest integer in kilometer/s _____., [NA 8 Jan. 2020 II], 31. A solid sphere of mass ‘M’ and radius ‘a’ is surrounded by, a uniform concentric spherical shell of thickness 2a and, mass 2M. The gravitational field at distance ‘3a’ from the, centre will be:, [9 April 2019 I], 2, , (b), , GM, 2, , (c), , GMm, GMm, GMm, GMm, (b) +, (c) (d), 6, R, 6, R, 2R, 2R, 35. From a solid sphere of mass M and radius R, a spherical, portion of radius R/2 is removed, as shown in the figure., Taking gravitational potential V = 0 at r = ¥, the potential at, the centre of the cavity thus formed is :, (G = gravitational constant), [2015], , (a), , (a), , 2GM, , R, , and the other mass, in a circular orbit of, 2, , 3R, . The difference between the final and initial, 2, total energies is:, [Online April 15, 2018], , 2, , 2, , 0, , (a), , GM, a, , 33. A test particle is moving in circular orbit in the gravitational, , (b) r = R, , 5, R, 3, 9, 29. Consider two solid spheres of radii R1 = 1m, R2=2m and, masses M1 and M2, respectively. The gravitational field, m1, due to sphere 1 and 2 are shown. The value of m is:, 2, [8 Jan. 2020 I], , (c) r =, , (a) 1.35, , GM, 2, , (d), , 2GM, , 9a, 9a, 3a, 3a 2, 32. Four identical particles of mass M are located at the corners, of a square of side ‘a’. What should be their speed if each, of them revolves under the influence of others’, gravitational field in a circular orbit circumscribing the, square ?, [8 April 2019 I], , -, , -2GM, -2GM, -GM, -GM, (b), (c), (d), 3R, R, 2R, R, Which of the following most closely depicts the correct, variation of the gravitational potential V(r) due to a large, planet of radius R and uniform mass density ? (figures, are not drawn to scale), [Online April 11, 2015], , (a), 36., , V(r), V(r), , (a), , V(r), (c), , (b), , r, , O, , O, , r, , (d), , r, , O, V(r), O, , r
Page 126 :
P-117, , Gravitation, , 37. The gravitational field in a region is given by, , (, , ), , ®, , (a), , 2Gm, r, , potential energy of a particle of mass 1 kg when it is taken, from the origin to a point (7 m, – 3 m) is:, [Online April 19, 2014], , (c), , 2Gm æ, 1 ö, 1÷, r çè, 2ø, , g = 5N / kgiˆ + 12N / kgjˆ . The change in the gravitational, , (a) 71 J, m1, , 38., , (b) 13 58J (c) – 71 J, m2, v1 v2, , (d) 1 J, , Two hypothetical planets of masses m1 and m2 are at rest, when they are infinite distance apart. Because of the, gravitational force they move towards each other along, the line joining their centres. What is their speed when, their separation is ‘d’?, [Online April 12, 2014], (Speed of m1 is v1 and that of m2 is v2), (a) v1 = v2, (b), , v1 = m 2, , 2G, 2G, v 2 = m1, d ( m1 + m 2 ), d ( m1 + m 2 ), , (c), , v1 = m1, , 2G, 2G, v 2 = m2, d ( m1 + m 2 ), d ( m1 + m 2 ), , (d), , 2G, 2G, v1 = m 2, v2 = m 2, m1, m2, , 39. The gravitational field, due to the 'left over part' of a uniform, sphere (from which a part as shown, has been 'removed, out'), at a very far off point, P, located as shown, would be, (nearly) :, [Online April 9, 2013], Mass of complete, sphere = M, , Removed, Part, R, , P, , R, , (b), , Gm, r, , (d), , 2Gm, r, , 42. Two bodies of masses m and 4 m are placed at a distance r., The gravitational potential at a point on the line joining, them where the gravitational field is zero is:, [2011], (a), , d, , 2 -1, , -, , 4Gm, r, , (b), , -, , 6Gm, r, , (c), , -, , 9Gm, r, , (d) zero, , 43. This question contains Statement-1 and Statement-2. Of, the four choices given after the statements, choose the, one that best describes the two statements., [2008], Statement-1 : For a mass M kept at the centre of a cube of, side ‘a’, the flux of gravitational field passing through its, sides 4 p GM. and, Statement-2: If the direction of a field due to a point source, is radial and its dependence on the distance ‘r’ from the, source is given as, , 1, , , its flux through a closed surface, r2, depends only on the strength of the source enclosed by, the surface and not on the size or shape of the surface., (a) Statement -1 is false, Statement-2 is true, (b) Statement -1 is true, Statement-2 is true; Statement -2, is a correct explanation for Statement-1, (c) Statement -1 is true, Statement-2 is true; Statement 2 is not a correct explanation for Statement-1, (d) Statement -1 is true, Statement-2 is false, , 44. A particle of mass 10 g is kept on the surface of a uniform, sphere of mass 100 kg and radius 10 cm. Find the work to, be done against the gravitational force between them to, take the particle far away from the sphere, (you may take G = 6.67× 10 -11 Nm 2 / kg 2 ), , [2005], , x, , (a), , 5 GM, 6 x2, , (b), , 8 GM, 9 x2, , (c), , 7 GM, 8 x2, , (d), , 6 GM, 7 x2, , 40. The mass of a spaceship is 1000 kg. It is to be launched, from the earth's surface out into free space. The value of g, and R (radius of earth) are 10 m/s2 and 6400 km respectively., The required energy for this work will be, [2012], 11, 8, (a) 6.4 × 10 Joules, (b) 6.4 × 10 Joules, (c) 6.4 × 109 Joules, (d) 6.4 × 1010 Joules, 41. A point particle is held on the axis of a ring of mass m and, radius r at a distance r from its centre C. When released, it, reaches C under the gravitational attraction of the ring. Its, speed at C will be, [Online May 26, 2012], , (a) 3.33 × 10 -10 J, , (b) 13.34 × 10 -10 J, , (c) 6.67 × 10 -10 J, , (d) 6.67 × 10 -9 J, , 45. If ‘g’ is the acceleration due to gravity on the earth’s, surface, the gain in the potential energy of an object of, mass ‘m’ raised from the surface of the earth to a height, equal to the radius ‘R' of the earth is, [2004], (a), , 1, mgR (b), 4, , 1, mgR (c) 2 mgR, 2, , (d) mgR, , 46. Energy required to move a body of mass m from an orbit of, radius 2R to 3R is, [2002], (a) GMm/12R2, , (b) GMm/3R2, , (c) GMm/8R, , (d) GMm/6R.
Page 127 :
P-118, , Physics, , Motion of Satellites, Escape, TOPIC 5, Speed and Orbital Velocity, 47., , 48., , 49., , A satellite is in an elliptical orbit around a planet P. It is, observed that the velocity of the satellite when it is farthest, from the planet is 6 times less than that when it is closest, to the planet. The ratio of distances between the satellite, and the planet at closest and farthest points is :, [NA 6 Sep. 2020 (I)], (a) 1 : 6, (b) 1 : 3, (c) 1 : 2, (d) 3 : 4, A body is moving in a low circular orbit about a planet of, mass M and radius R. The radius of the orbit can be taken, to be R itself. Then the ratio of the speed of this body in, the orbit to the escape velocity from the planet is :, 1, (a), (b) 2, [4 Sep. 2020 (II)], 2, (c) 1, (d) 2, A satellite is moving in a low nearly circular orbit around, the earth. Its radius is roughly equal to that of the earth’s, radius Re. By firing rockets attached to it, its speed is, instantaneously increased in the direction of its motion so, 3, times larger. Due to this the farthest, 2, distance from the centre of the earth that the satellite, reaches is R. Value of R is :, [3 Sep. 2020 (I)], (a) 4Re, (b) 2.5Re, (c) 3Re, (d) 2Re, , a circular orbit at this height is E2. The value of h for, which E1 and E2 are equal, is:, [9 Jan. 2019 II], 3, (a) 1.6 × 10 km, (b) 3.2 × 103 km, 3, (c) 6.4 × 10 km, (d) 28 × 104 km, 53. Planet A has mass M and radius R. Planet B has half the, mass and half the radius of Planet A. If the escape velocities, from the Planets A and B are vA and vB, respectively, then, vA n, = ., vB 4 The value of n is :, , (a) 4, , (b) 1, , (c) 2, , K, over, r, a large distance 'r' from its centre. In that region, a small, star is in a circular orbit of radius R. Then the period of, revolution, T depends on R as :, [2 Sep. 2020 (I)], , The mass density of a spherical galaxy varies as, , (a), , T 2 µ R (b) T 2 µ R3 (c), , T2 µ, , 1, R3, , (d) T µ R, , 51. A body A of mass m is moving in a circular orbit of radius, R about a planet. Another body B of mass, , m, collides with, 2, , r, æ vö, A with a velocity which is half çè ÷ø the instantaneous, 2, r, velocity v or A. The collision is completely inelastic., Then, the combined body:, [9 Jan. 2020 I], (a) continues to move in a circular orbit, (b) Escapes from the Planet’s Gravitational field, (c) Falls vertically downwards towards the planet, (d) starts moving in an elliptical orbit around the planet, 52. The energy required to take a satellite to a height 'h', above Earth surface (radius of Eareth = 6.4 × 10 3 km) is, E1 and kinetic energy required for the satellite to be in, , (d) 3, , 54. A satellite of mass m is launched vertically upwards, with an initial speed u from the surface of the earth., After it reaches height R (R = radius of the earth), it, m, so that subsequently the, 10, satellite moves in a circular orbit. The kinetic energy of, the rocket is (G is the gravitational constant; M is the, mass of the earth):, [7 Jan. 2020 I], , ejects a rocket of mass, , (a), , m æ 2 113 GM ö, çu +, ÷, 20 è, 200 R ø, , that it become, , 50., , [9 Jan. 2020 II], , 3m æ, 5GM ö, u+, (c), ç, 8 è, 6 R ÷ø, , 2, , æ 2 119 GM ö, (b) 5m çè u ÷, 200 R ø, mæ, 2GM ö, (d) 20 çè u - 3 R ÷ø, , 2, , 55. A spaceship orbits around a planet at a height of 20 km, from its surface. Assuming that only gravitational field of, the planet acts on the spaceship, what will be the number, of complete revolutions made by the spaceship in 24 hours, around the planet ? [Given : Mass of Planet = 8×10 22 kg,, Radius of planet = 2×10 6 m, Gravitational constant, G = 6.67×10–11Nm2/kg2], [10 April 2019 II], (a) 9, , (b) 17, , (c) 13, , (d) 11, , 56. A rocket has to be launched from earth in such a way that, it never returns. If E is the minimum energy delivered by, the rocket launcher, what should be the minimum energy, that the launcher should have if the same rocket is to be, launched from the surface of the moon? Assume that the, density of the earth and the moon are equal and that the, earth’s volume is 64 times the volume of the moon., [8 April 2019 II], (a), , E, 64, , (b), , E, 32, , (c), , E, 4, , (d), , E, 16, , 57. A satellite of mass M is in a circular orbit of radius R about, the centre of the earth. A meteorite of the same mass, falling, towards the earth collides with the satellite completely in, elastically. The speeds of the satellite and the meteorite are, the same, Just before the collision. The subsequent motion, of the combined body will be, [12 Jan. 2019 I]
Page 128 :
P-119, , Gravitation, , (a) such that it escape to infinity, (b) In an elliptical orbit, (c) in the same circular orbit of radius R, (d) in a circular orbit of a different radius, 58. Two satellites, A and B, have masses m and 2m respectively., A is in a circular orbit of radius R, and B is in a circular orbit, of radius 2R around the earth. The ratio of their kinetic, energies, TA/TB, is :, [12 Jan. 2019 II], (a), , 1, 2, , (b) 1, , 1, 2, 59. A satellite is revolving in a circular orbit at a height h from, the earth surface, such that h << R where R is the radius of, the earth. Assuming that the effect of earth’s atmosphere, can be neglected the minimum increase in the speed, required so that the satellite could escape from the, gravitational field of earth is:, [11 Jan. 2019 I], , (c) 2, , (d), , (a), , 2gR, , (b), , gR, , (c), , gR, 2, , (d), , gR, , (, , ), , 2 -1, , 60. A satellite is moving with a constant speed v in circular, orbit around the earth. An object of mass ‘m’ is ejected, from the satellite such that it just escapes from the, gravitational pull of the earth. At the time of ejection,, the kinetic energy of the object is:, [10 Jan. 2019 I], 2, 2, (a) 2 m v, (b) m v, (c), , 1, m v2, 2, , (d), , 3, m v2, 2, , 61. Two stars of masses 3 × 1031 kg each, and at distance, 2 × 1011 m rotate in a plane about their common centre, of mass O. A meteorite passes through O moving, perpen-dicular to the star’s rotation plane. In order to, escape from the gravitational field of this double star,, the minimum speed that meteorite should have at O is:, (Take Gravitational constant G = 66 × 10–11 Nm2 kg–2), [10 Jan. 2019 II], 4, (a) 2.4 × 10 m/s, (b) 1.4 × 105 m/s, (c) 3.8 × 104 m/s, (d) 2.8 × 105 m/s, 62. A satellite is revolving in a circular orbit at a height 'h' from, the earth's surface (radius of earth R; h < < R). The minimum, increase in its orbital velocity required, so that the satellite, could escape from the earth's gravitational field, is close, to : (Neglect the effect of atmosphere.), [2016], (a), , gR / 2, , (b), , gR, , (c), , 2gR, , (d), , gR, , (, , ), , 2-1, , 63. An astronaut of mass m is working on a satellite orbiting, the earth at a distance h from the earth's surface. The radius, of the earth is R, while its mass is M. The gravitational pull, FG on the astronaut is :, [Online April 10, 2016], (a) Zero since astronaut feels weightless, GMm, , (b), , (R + h), , (c), , FG =, , 2, , < FG <, , GMm, R2, , GMm, (R + h) 2, , (d) 0 < FG <, , GMm, R2, , 64. A very long (length L) cylindrical galaxy is made of, uniformly distributed mass and has radius R(R < < L). A, star outside the galaxy is orbiting the galaxy in a plane, perpendicular to the galaxy and passing through its centre., If the time period of star is T and its distance from the, galaxy’s axis is r, then :, [Online April 10, 2015], (a) T µ r, , (b), , (c) T µ r2, , (d) T2 µ r3, , Tµ r, , 65. What is the minimum energy required to launch a satellite, of mass m from the surface of a planet of mass M and, radius R in a circular orbit at an altitude of 2R? [2013], 5GmM, (b), 6R, , 2GmM, (c), 3R, , GmM, GmM, (d), 2R, 2R, 66. A planet in a distant solar system is 10 times more massive, than the earth and its radius is 10 times smaller. Given that, the escape velocity from the earth is 11 km s–1, the escape, velocity from the surface of the planet would be [2008], (a) 1.1 km s–1, (b) 11 km s–1, (c) 110 km s–1, (d) 0.11 km s–1, 67. Suppose the gravitational force varies inversely as the nth, power of distance. Then the time period of a planet in circular, orbit of radius ‘R’ around the sun will be proportional to, (a), , (a), , R, , n, , æ n +1ö, çè, ÷ø, , (b), , R, , æ n -1ö, çè, ÷, 2 ø, æ n- 2ö, ç, ÷, , [2004], , (c) R 2, (d) Rè 2 ø, 68. The time period of an earth satellite in circular orbit is, independent of, [2004], (a) both the mass and radius of the orbit, (b) radius of its orbit, (c) the mass of the satellite, (d) neither the mass of the satellite nor the radius of its, orbit.
Page 129 :
P-120, , 69., , A satellite of mass m revolves around the earth of radius R, at a height x from its surface. If g is the acceleration due to, gravity on the surface of the earth, the orbital speed of the, satellite is, [2004], (a), , 70., , Physics, , gR 2, R+ x, , (b), , gR, R-x, , (c) gx, , æ gR2 ö 1/ 2, (d) ç, ÷, è R + xø, , The escape velocity for a body projected vertically, upwards from the surface of earth is 11 km/s. If the body is, projected at an angle of 45°with the vertical, the escape, velocity will be, [2003], (a) 11 2 km / s, , (b) 22 km/s, , (c) 11 km/s, , (d), , 11, 2, , km / s, , 71. The kinetic energy needed to project a body of mass m from, the earth surface (radius R) to infinity is, [2002], (a) mgR/2 (b) 2mgR (c) mgR, (d) mgR/4., 72. If suddenly the gravitational force of attraction between, Earth and a satellite revolving around it becomes zero,, then the satellite will, [2002], (a) continue to move in its orbit with same velocity, (b) move tangentially to the original orbit in the same, velocity, (c) become stationary in its orbit, (d) move towards the earth, 73. The escape velocity of a body depends upon mass as, [2002], (a) m0, (b) m1, (c) m2, (d) m3
Page 140 :
8, , P-131, , Mechanical Properties of Solids, , Mechanical Properties, of Solids, 5., , TOPIC 1 Hooke's Law & Young's Modulus, 1., , If the potential energy between two molecules is given by, A, B, U = - 6 + 12 , then at equilibrium, separation between, r, r, molecules, and the potential energy are: [Sep. 06, 2020 (I)], 1, ö6, , B, A2, (a) æç ÷ , è 2 Aø, 2B, 1, , A2, æ 2B ö 6, (c) ç ÷ , è Aø, 4B, , 2., , 3., , 4., , 1, Bö 6, , (b) æç ÷ , 0, è Aø, , [10 April 2019 II], , 6., , 1, , A2, æ 2B ö 6, (d) ç ÷ , è Aø, 2B, , A body of mass m = 10 kg is attached to one end of a wire of, length 0.3 m. The maximum angular speed (in rad s–1) with, which it can be rotated about its other end in space station is, (Breaking stress of wire = 4.8 × 107 Nm–2 and area of crosssection of the wire = 10–2 cm2) is _______ ., [9 Jan 2020 I], A uniform cylindrical rod of length L and radius r, is made, from a material whose Young’s modulus of Elasticity, equals Y. When this rod is heated by temperature T and, simultaneously subjected to a net longitudinal, compressional force F, its length remains unchanged. The, coefficient of volume expansion, of the material of the, rod, is (nearly) equal to :, [12 April 2019 II], (a) 9F/(pr 2YT), (b) 6F/(pr 2YT, (c) 3F/(pr 2YT), (d) F/(3pr 2YT), In an environment, brass and steel wires of length 1 m, each with areas of cross section 1 mm2 are used. The, wires are connected in series and one end of the combined, wire is connected to a rigid support and other end is, subjected to elongation. The stress required to produce a, net elongation of 0.2 mm is,, [Given, the Young’s modulus for steel and brass are,, respectively, 120×109N/m2 and 60×109N/m2], [10 April 2019 II], (a) 1.2×106 N/m 2, (b) 4.0×106 N/m 2, (c) 1.8×106 N/m2, (d) 0.2×106 N/m2, , The elastic limit of brass is 379 MPa. What should be the, minimum diameter of a brass rod if it is to support a 400, N load without exceeding its elastic limit?, , 7., , 8., , 9., , (a) 1.00 mm, , (b) 1.16 mm, , (c) 0.90 mm, , (d) 1.36 mm, , A steel wire having a radius of 2.0 mm, carrying a load of, 4kg, is hanging from a ceiling. Given that g = 3.1 À ms–2,, what will be the tensile stress that would be developed in, the wire?, [9 April 2019 I], (a) 6.2 × 106 Nm–2, (b) 5.2 × 106 Nm–2, (c) 3.1 × 106 Nm–2, (d) 4.8 × 106 Nm–2, A steel wire having a radius of 2.0 mm, carrying a load of, 4kg, is hanging from a ceiling. Given that g = 3.1 À ms–2,, what will be the tensile stress that would be developed in, the wire?, [8 April 2019 I], 6, –2, 6, (a) 6.2 × 10 Nm, (b) 5.2 × 10 Nm–2, 6, –2, (c) 3.1 × 10 Nm, (d) 4.8 × 106 Nm–2, Young’s moduli of two wires A and B are in the ratio 7 : 4., Wire A is 2 m long and has radius R. Wire B is 1.5 m long, and has radius 2 mm. If the two wires stretch by the same, length for a given load, then the value of R is close to :, [8 April 2019 II], (a) 1.5 mm (b) 1.9 mm (c) 1.7 mm, (d) 1.3 mm, As shown in the figure, forces of 105N each are applied in, opposite directions, on the upper and lower faces of a, cube of side 10cm, shifting the upper face parallel to itself, by 0.5cm. If the side of another cube of the same material, is, 20cm, then under similar conditions as above, the, displacement will be:, [Online April 15, 2018], F, , F, (a) 1.00cm, (c) 0.37cm, , (b) 0.25cm, (d) 0.75cm
Page 141 :
P-132, , 10., , 11., , 12., , Physics, , A thin 1 m long rod has a radius of 5 mm. A force of 50 pkN, is applied at one end to determine its Young's modulus., Assume that the force is exactly known. If the least count, in the measurement of all lengths is 0.01 mm, which of the, following statements is false ? [Online April 10, 2016], (a) The maximum value of Y that can be determined is, 2 × 1014N/m2., DY, (b), gets minimum contribution from the uncertainty, Y, in the length, DY, (c), gets its maximum contribution from the, Y, uncertainty in strain, (d) The figure of merit is the largest for the length of the, rod., A uniformly tapering conical wire is made from a material, of Young's modulus Y and has a normal, unextended length, L. The radii, at the upper and lower ends of this conical, wire, have values R and 3R, respectively. The upper end of, the wire is fixed to a rigid support and a mass M is, suspended from its lower end. The equilibrium extended, length, of this wire, would equal : [Online April 9, 2016], (a), , æ 2 Mg ö, L ç1 +, è 9 pYR 2 ÷ø, , (b), , æ 1 Mg ö, L ç1 +, è 9 pYR 2 ÷ø, , (c), , æ 1 Mg ö, L ç1 +, è 3 pYR 2 ÷ø, , (d), , æ 2 Mg ö, L ç1 +, è 3 pYR 2 ÷ø, , The pressure that has to be applied to the ends of a steel, wire of length 10 cm to keep its length constant when its, temperature is raised by 100ºC is:, (For steel Young’s modulus is 2 ´ 1011 Nm -2 and, coefficient of thermal expansion is 1.1 ´ 10-5 K -1 ) [2014], (a), , (b), , 2.2 ´ 10 9 Pa, , (d) 2.2 ´ 106 Pa, Two blocks of masses m and M are connected by means, of a metal wire of cross-sectional area A passing over a, frictionless fixed pulley as shown in the figure. The system, is then released. If M = 2 m, then the stress produced in, the wire is :, [Online April 25, 2013], (c), , 13., , 2.2 ´ 108 Pa, 2.2 ´ 10 7 Pa, , 14. A copper wire of length 1.0 m and a steel wire of length, 0.5 m having equal cross-sectional areas are joined end to, end. The composite wire is stretched by a certain load, which stretches the copper wire by 1 mm. If the Young’s, modulii of copper and steel are respectively 1.0 × 1011 Nm–, 2 and 2.0 × 1011 Nm–2, the total extension of the composite, wire is :, [Online April 23, 2013], (a) 1.75 mm (b) 2.0 mm (c) 1.50 mm (d) 1.25 mm, 15. A uniform wire (Young’s modulus 2 × 1011 Nm–2) is, subjected to longitudinal tensile stress of 5 × 107 Nm–2. If, the overall volume change in the wire is 0.02%, the, fractional decrease in the radius of the wire is close to :, [Online April 22, 2013], (a) 1.0 × 10–4, (b) 1.5 × 10–4, (c) 0.25 × 10–4, (d) 5 × 10–4, 16. If the ratio of lengths, radii and Young's moduli of steel, and brass wires in the figure are a, b and c respectively,, then the corresponding ratio of increase in their lengths is :, [Online April 9, 2013], Steel, M, Brass, 2M, , 3a, 2ac, 2a 2 c, (c), (d), 2, 2b c, b2, b, 2ab2, 17. A steel wire can sustain 100 kg weight without breaking. If, the wire is cut into two equal parts, each part can sustain, a weight of, [Online May 19, 2012], (a) 50 kg, (b) 400 kg (c) 100 kg (d) 200 kg, 18. A structural steel rod has a radius of 10 mm and length of, 1.0 m. A 100 kN force stretches it along its length. Young’s, modulus of structural steel is 2 × 1011 Nm–2. The percentage, strain is about, [Online May 7, 2012], (a) 0.16% (b) 0.32% (c) 0.08% (d) 0.24%, 19. The load versus elongation graphs for four wires of same, length and made of the same material are shown in the, figure. The thinnest wire is represented by the line, [Online May 7, 2012], Load, D, (a), , 3c, , (b), , C, B, A, T, , O, , m, T, , M, , (a), , 2mg, 3A, , (b), , 4mg, 3A, , (c), , mg, A, , (d), , 3mg, 4A, , Elongation, , (a) OA, (b) OC, (c) OD, (d) OB, 20. Two wires are made of the same material and have the, same volume. However wire 1 has cross-sectional area A, and wire 2 has cross-sectional area 3A. If the length of wire, 1 increases by Dx on applying force F, how much force is, needed to stretch wire 2 by the same amount?, [2009], (a) 4 F, (b) 6 F, (c) 9 F, (d) F
Page 142 :
P-133, , Mechanical Properties of Solids, , 21., , a, , A wire elongates by l mm when a load W is hanged from it. If, the wire goes over a pulley and two weights W each are hung, at the two ends, the elongation of the wire will be (in mm), [2006], (a) l, (b) 2l, (c) zero, (d) l/2, , TOPIC 2, , Bulk and Rigidity Modulus and, Work Done in Stretching a Wire, , 22. Two steel wires having same length are suspended from a, ceiling under the same load. If the ratio of their energy, stored per unit volume is 1 : 4, the ratio of their diameters, is:, [9 Jan 2020 II], (a), , b, , 2 :1, , 26., , (b) 1 : 2, , (c) 2 : 1, (d) 1 : 2, 23. A boy’s catapult is made of rubber cord which is 42 cm, long, with 6 mm diameter of cross-section and of negligible, mass. The boy keeps a stone weighing 0.02 kg on it and, stretches the cord by 20 cm by applying a constant force., When released, the stone flies off with a velocity of 20 ms–, 1. Neglect the change in the area of cross-section of the, cord while stretched. The Young’s modulus of rubber is, closest to :, [8 April 2019 I], (a) 106 N/m–2, (b) 104 N/m–2, (c) 108 N/m–2, (d) 103 N/m–2, 24. A solid sphere of radius r made of a soft material of bulk, modulus K is surrounded by a liquid in a cylindrical, container. A massless piston of area a floats on the surface, of the liquid, covering entire cross-section of cylindrical, container. When a mass m is placed on the surface of the, piston to compress the liquid, the fractional decrement in, æ dr ö, the radius of the sphere çè ÷ø ,is :, [2018], r, , 27., , Ka, Ka, mg, mg, (b), (c), (d), mg, 3mg, 3Ka, Ka, A bottle has an opening of radius a and length b. A cork of, length b and radius (a + Da) where (Da < < a) is compressed, to fit into the opening completely (see figure). If the bulk, modulus of cork is B and frictional coefficient between the, bottle and cork is m then the force needed to push the cork, into the bottle is :, [Online April 10, 2016], , 29., , (a), 25., , 28., , (a) (pmB b) a, (b) (2pmBb) Da, (c) (pmB b) Da, (d) (4 pmB b) Da, Steel ruptures when a shear of 3.5 × 108 N m–2 is applied., The force needed to punch a 1 cm diameter hole in a steel, sheet 0.3 cm thick is nearly:, [Online April 12, 2014], (a) 1.4 × 104 N, (b) 2.7 × 104 N, (c) 3.3 × 104 N, (d) 1.1 × 104 N, The bulk moduli of ethanol, mercury and water are given, as 0.9, 25 and 2.2 respectively in units of 109 Nm–2. For a, given value of pressure, the fractional compression in, DV, volume is, . Which of the following statements about, V, DV, for these three liquids is correct ?[Online April 11, 2014], V, (a) Ethanol > Water > Mercury, (b) Water > Ethanol > Mercury, (c) Mercury > Ethanol > Water, (d) Ethanol > Mercury > Water, In materials like aluminium and copper, the correct order of, magnitude of various elastic modului is:, [Online April 9, 2014], (a) Young’s modulus < shear modulus < bulk modulus., (b) Bulk modulus < shear modulus < Young’s modulus, (c) Shear modulus < Young’s modulus < bulk modulus., (d) Bulk modulus < Young’s modulus < shear modulus., If ‘S’ is stress and ‘Y’ is young’s modulus of material of a, wire, the energy stored in the wire per unit volume is [2005], , 2, 2Y, S, S, (b) 2S 2Y (c), (d), 2Y, S2, 2Y, 30. A wire fixed at the upper end stretches by length l by, applying a force F. The work done in stretching is [2004], , (a), , (a) 2Fl, , (b) Fl, , (c), , F, 2l, , (d), , Fl, 2
Page 145 :
P-136, , 15., , Physics, , 19. (a) From the graph, it is clear that for the same value of, load, elongation is maximum for wire OA. Hence OA is the, thinnest wire among the four wires., , (c) Given , y = 2 ´ 1011 Nm -2, , æ Fö, Stress ç ÷ = 5 ´ 107 Nm -2, è Aø, , l, , DV = 0.02% = 2 × 10–4 m3, , 20. (c), , A, , Y, , Dr, =?, r, , g=, , Wire (1), , æ Dl ö, g, stress, Þ strain ç ÷ =, strain, è l 0 ø stress, , … (i), , DV = 2prl 0 Dr - pr 2 Dl, and solving we get, Dr, = 0.25 ´10-4, r, (c) According to questions,, , Dl b =, , prs2 .ys, , 2Mgl b, prb2 .y b, , Y=, , [Q Fs = (M + 2M)g], [Q Fb = 2Mg], , Þ, , Stress, Strain, , 105, pr 2Y, , Therefore % strain =, , =, , For wire 2 ,, ...(ii), , 21. (a) Case (i), , T, , T, , T, , W, , W, , Stress, F, Strain =, =, Y, AY, , =, , ...(i), , F l, F', l, ´, =, ´, Þ F¢ = 9F, A Dx 3 A 3Dx, , (c) Breaking force a area of cross section of wire, Load hold by wire is independent of length of the wire., (a) Given: F = 100 kN = 105 N, Y = 2 × 1011 Nm–2, l0 = 1.0 m, radius r = 10 mm = 10– 2 m, From formula, Y =, , F/A, D x/l, , F '/ 3 A, Dx /( l / 3), From (i) and (ii) we get,, , 3Mgl s, , 18., , l, 3, , Y=, , Dl s, pr 2 .y, 3a, = s s = 2, \ Dl b 2Mg.l b 2b C, prb2 .y b, , 17., , For wire 1, Length, L1 = l, Area, A1 = A, For wire 2, , Area, A2 = 3A, As the wires are made of same material, so they will have, same young’s modulus., For wire 1,, , Fl, Fl, As, y = ADl Þ Dl = Ay, Dl s =, , l/3, Wire (2), , Length, L2 =, , ls, r, y, Dls, = a, s = b, s = c,, =?, lb, rb, yb, Dl b, , 3mgl s, , Y, , … (ii), , From eqns (i) and (ii) putting the value of Dl , l 0 and DV, , 16., , 3A, , At equilibrium, T = W, , 105, , 1, -4, 11 = 628, 3.14 ´ 10 ´ 2 ´ 10, , 1, ´100 = 0.16%, 628, , Young’s modules, Y =, Elongation, l =, , W L, ´, A Y, , W/A, .....(1), l/L, , W
Page 146 :
P-137, , Mechanical Properties of Solids, , Case (ii) At equilibrium T = W, \ Young’s moduls, Y =, , Fractional change in volume, , W/A, l/2, L/2, , Using eq. (i) & (ii), , W L, W/A, ´, Þ l=, A Y, l/ L, Þ Elongation is the same., , Þ, , Y=, , \, , 22. (a) If force F acts along the length L of the wire of crosssection A, then energy stored in unit volume of wire, is given by, 1, Energy density = stress × strain, 2, , F, X ö, 1 F, F æ, = ´ ´, çèQ stress = and strain =, ÷, A, AY ø, 2 A AY, , 3dr mg, =, r, Ka, , dr mg, =, (fractional decrement in radius), r 3Ka, , Stress <, , N, N, Normal force, < <, A, (2, p, a)b, Area, , Stress = B×strain, N, 2paΧa ≥ b, <B, (2pa)b, pa 2 b, ÞN<B, , (2pa)2 Χab 2, , pa 2 b, Force needed to push the cork., , D, , 26. (c), , If u1 and u2 are the densities of two wires, then, 4, , d1, d, u1 æ d 2 ö, 14, = ( 4) Þ 1 = 2 :1, =ç ÷ Þ, d2, d2, u2 è d1 ø, , F, , 23. (a) When a catapult is stretched up to length l, then the, stored energy in it = Dk. E Þ, , 1 æ YA ö, 1, . ç ÷ ( DI )2 = mv 2, 2 è Lø, 2, , Þy=, , 0.02 ´ 400 ´ 0.42 ´ 4, p ´ 36 ´ 10 –6 ´ 0.04, , D (Dl ) 2, , Shearing strain is created along the side surface of the, punched disk. Note that the forces exerted on the disk are, exerted along the circumference of the disk, and the total, force exerted on its center only., Let us assume that the shearing stress along the side, surface of the disk is uniform, then, , F=, , = 2.3 × 106 N/m2, , (c), , Bulk modulus, K =, , K=, , Þ, , dFmax =, , ò, , s max dA = s max, , surface, , ò, , dA, , surface, , æDö, = ò s max .A = smax .2p ç ÷ h, è2ø, , volumetric stress, volumetric strain, , 8 æ1, -2 ö, -2, = 3.5 ´10 ´ ç ´ 10 ÷ ´ 0.3 ´10 ´ 2p, è2, ø, , = 3.297 ´ 104 ; 3.3 ´ 10 4 N, , mg, æ dV ö, aç, ÷, èV ø, , 1, Bulk modulus, As bulk modulus is least for ethanol (0.9) and maximum for, mercury (25) among ehtanol, mercury and water. Hence, , 27. (a) Compressibility =, , dV mg, =, V, Ka, , volume of sphere, V =, , ò, , surface, , So, order is 106., 24., , h, , mv 2 L, , m = 0.02 kg, v = 20 ms–1, L = 0.42 m, A = (p d2)/(4), d = 6 × 10–3 m, Dl = 0.2 m, y=, , ...(ii), , f < m N < m 4pbΧaB = (4pmBb)Da, , 1 F2, 1 F 2 ´ 16 1 F 2 ´ 16, =, =, 2 A2Y 2 (pd 2 )2 Y 2 pd 4Y, , =, , 25. (d), , dV 3dr, =, V, r, , ....(i), 4 3, pR, 3, , DV, V, Ethanol > Water > Mercury, , compression in volume
Page 147 :
P-138, , 28. (c), , Physics, , Poisson’s ratio, s =, , lateral strain ( b ), , longitudinal strain ( a ), , For material like copper, s = 0.33, And, Y = 3k (1 – 2 s), , 9 1 3, Also,, = +, Y k h, , 1, ´ stress ´ strain, 2, We know that,, Y=, , stress, strain, , 1, stress 1 S 2, ´ stress ´, = ×, 2, Y, 2 Y, 30. (d) Let A and L be the area and length of the wire., Work done by constant force in displacing the wire by a, distance l., = change in potential energy, , E=, , Y = 2h (1+ s), Hence, h < Y < k, 29. (a) Energy stored in the wire per unit volume,, E=, , stress, Y, On substituting the expression of strain in equation (i) we, get, , Þ strain =, , ...(i), , 1, = × stress × strain × volume, 2, =, , 1 F l, Fl, ´ ´ ´ A´ L =, 2 A L, 2
Page 148 :
9, , P-139, , Mechanical Properties of Fluids, , Mechanical Properties, of Fluids, Pressure, Density, Pascal's Law, TOPIC 1 and Archimedes' Principle, 1., , 2., , 3., , A hollow spherical shell at outer radius R floats just, submerged under the water surface. The inner radius of, the shell is r. If the specific gravity of the shell material is, 27, w.r.t water, the value of r is :, [5 Sep. 2020 (I)], 8, 8, 4, 2, 1, R, R, R, R, (a), (b), (c), (d), 9, 9, 3, 3, An air bubble of radius 1 cm in water has an upward, acceleration 9.8 cm s–2. The density of water is 1 gm, cm–3 and water offers negligible drag force on the bubble. The, mass of the bubble is (g = 980 cm/s2), [4 Sep. 2020 (I)], (a) 4.51 gm (b) 3.15 gm (c) 4.15 gm, (d) 1.52 gm, Two identical cylindrical vessels are kept on the ground, and each contain the same liquid of density d. The area of, the base of both vessels is S but the height of liquid in one, vessel is x1 and in the other, x2. When both cylinders are, connected through a pipe of negligible volume very close, to the bottom, the liquid flows from one vessel to the other, until it comes to equilibrium at a new height. The change in, energy of the system in the process is : [4 Sep. 2020 (II)], (a) gdS ( x22 + x12 ), , (b) gdS ( x2 + x1 )2, , 3, 1, gdS ( x2 - x1 )2, gdS ( x2 - x1 )2, (d), 4, 4, A leak proof cylinder of length 1 m, made of a metal which, has very low coefficient of expansion is floating vertically, in water at 0°C such that its height above the water surface, is 20 cm. When the temperature of water is increased to, 4°C, the height of the cylinder above the water surface, becomes 21 cm. The density of water at T = 4°C, relative to, the density at T = 0°C is close to:, [8 Jan 2020 (I)], (a) 1.26, (b) 1.04, (c) 1.01, (d) 1.03, Consider a solid sphere of radius R and mass density, , (c), , 4., , 5., , æ, r2 ö, r(r) = r0 ç 1 - 2 ÷ , 0 < r £ R. The minimum density of a, è R ø, liquid in which it will float is:, [8 Jan 2020 (I)], r0, r0, 2r0, 2r0, (a), (b), (c), (d), 3, 5, 5, 3, , 6., , M, , 5m, N, 5m, O, , Two liquids of densities r1 and r2(r2 = 2r1) are filled up, behind a square wall of side 10 m as shown in figure. Each, liquid has a height of 5 m. The ratio of the forces due to these, liquids exerted on upper part MN to that at the lower part NO, is (Assume that the liquids are not mixing): [8 Jan 2020 (II)], (a) 1/3, (b) 2/3, (c) 1/2, (d) 1/4, 7. A cubical block of side 0.5 m floats on water with 30% of, its volume under water. What is the maximum weight that, can be put on the block without fully submerging it under, water? [Take, density of water = 103 kg/m3], [10 April 2019 (II)], (a) 46.3 kg (b) 87.5 kg (c) 65.4 kg, (d) 30.1 kg, 8. A submarine experiences a pressure of 5.05×106 Pa at depth, of d1 in a sea. When it goes further to a depth of d2, it, experiences a pressure of 8.08×106 Pa. Then d1– d1 is, approximately (density of water=103 kg/m3 and acceleration, due to gravity = 10 ms–2):, [10 April 2019 (II)], (a) 300 m, (b) 400 m (c) 600 m, (d) 500 m, 4, 9. A wooden block floating in a bucket of water has of its, 5, volume submerged. When certain amount of an oil poured, into the bucket, it is found that the block is just under the, oil surface with half of its volume under water and half in, oil. The density of oil relative to that of water is:, [9 April 2019 (II)], (a) 0.5, (b) 0.8, (c) 0.6, (c) 0.7, 10. A load of mass M kg is suspended from a steel wire of, length 2 m and radius 1.0 mm in Searle’s apparatus, experiment. The increase in length produced in the wire is, 4.0 mm. Now the load is fully immersed in a liquid of relative, density 2. The relative density of the material of load is 8., The new value of increase in length of the steel wire is :, [12 Jan. 2019 (II)], (a) 3.0 mm (b) 4.0 mm (c) 5.0 mm, (d) Zero
Page 149 :
P-140, , Physics, , 11. A soap bubble, blown by a mechanical pump at the mouth, of a tube, increases in volume, with time, at a constant rate., The graph that correctly depicts the time dependence of, pressure inside the bubble is given by: [12 Jan. 2019 (II)], P, , P, , (a), , (b), 1, t, , log(t), , P, , P, (c), , (d), , 13, t, t, 12. A liquid of density r is coming out of a hose pipe of radius, a with horizontal speed v and hits a mesh. 50% of the liquid, passes through the mesh unaffected. 25% looses all of its, momentum and 25% comes back with the same speed. The, resultant pressure on the mesh will be: [11 Jan. 2019 (I)], 1 2, 3 2, 1 2, ρv, ρv (c), ρv, (b), (d) ρv 2, 4, 4, 2, 13. A thin uniform tube is bent into a circle of radius r in the, vertical plane. Equal volumes of two immiscible liquids, whose, densities are r1 and r1 (r1 > r2) fill half the circle. The, angle q between the radius vector passing through the common, interface and the vertical is, [Online April 15, 2018], -1 é p æ r - r ö ù, (a) q = tan ê ç 1 2 ÷ ú, ë 2 è r1 + r2 ø û, -1 p æ r1 - r2 ö, (b) q = tan, ç, ÷, 2 è r1 + r2 ø, , (a), , ær ö, (c) q = tan -1 p ç 1 ÷, è r2 ø, (d) None of above, 14. There is a circular tube in a vertical plane. Two liquids, which do not mix and of densities d1 and d2 are filled in the, tube. Each liquid subtends 90º angle at centre. Radius, joining their interface makes an angle a with vertical. Ratio, d1, is:, d2, , [2014], , a, , d1, , d2, , 1 + sin a, 1 + cos a, (b), 1 - sin a, 1 - cos a, 1 + tan a, 1 + sin a, (c), (d), 1 - tan a, 1 - cos a, 15. A uniform cylinder of length L and mass M having crosssectional area A is suspended, with its length vertical, from, a fixed point by a massless spring such that it is half, submerged in a liquid of density s at equilibrium position., The extension x0 of the spring when it is in equilibrium, is:, [2013], , (a), , (a), , Mg, k, , (b), , Mg æ, LAs ö, ç1 –, ÷, k è, M ø, , (c), , Mg æ, LAs ö, ç1 –, ÷, k è, 2M ø, , (d), , Mg æ, LAs ö, ç1 +, ÷, k è, M ø, , 16. A ball is made of a material of density r where roil < r < rwater, with roil and rwater represe-nting the densities of oil and, water, respectively. The oil and water are immiscible. If the, above ball is in equilibrium in a mixture of this oil and water,, which of the following pictures represents its equilibrium, position ?, [2010], , water, (a), , oil, , oil, (b), , water, , water, (c), , oil, (d), , oil, , water, , 17. Two identical charged spheres are suspended by strings, of equal lengths. The strings make an angle of 30°, with each other. When suspended in a liquid of density, 0.8g cm–3, the angle remains the same. If density of the, material of the sphere is 1.6 g cm–3 , the dielectric constant, of the liquid is, [2010], (a) 4, (b) 3, (c) 2, (d) 1, 18. A jar is filled with two non-mixing liquids 1 and 2 having, densities r1 and, r2 respectively. A solid ball, made of a, material of density r3 , is dropped in the jar. It comes to, equilibrium in the position shown in the figure.Which of, the following is true for r1, r2and r3?, [2008]
Page 150 :
P-141, , Mechanical Properties of Fluids, r1, , r3, , (a) r3 < r1 < r2, (c) r1 < r2 < r3, , (b) r1 > r3 > r2, (d) r1 < r3 < r2, , Fluid Flow, Reynold's Number, TOPIC 2 and Bernoulli's Principle, 19. A fluid is flowing through a horizontal pipe of varying crosssection, with speed v ms–1 at a point where the pressure is, P, P Pascal. At another point where pressure is, Pascal its, 2, –1, speed is V ms . If the density of the fluid is r kg m–3 and the, flow is streamline, then V is equal to : [6 Sep. 2020 (II)], P, 2P 2, (a), (b), +v, +v, r, r, P, P 2, (c), (d), + v2, +v, 2r, r, 20. Water flows in a horizontal tube (see figure). The pressure, of water changes by 700 Nm–2 between A and B where the, area of cross section are 40 cm2 and 20 cm2, respectively., Find the rate of flow of water through the tube., (density of water = 1000 kgm–3), [9 Jan. 2020 (I)], A, B, , (a) 3020 cm3/s, (b) 2720 cm3/s, (c) 2420 cm3/s, (d) 1810 cm3/s, 21. An ideal fluid flows (laminar flow) through a pipe of nonuniform diameter. The maximum and minimum diameters of, the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of, the minimum and the maximum velocities of fluid in this, pipe is:, [7 Jan. 2020 (II)], 81, 9, 3, 3, (a), (b), (c), (d) 256, 16, 4, 2, 22. Water from a tap emerges vertically downwards with an, initial speed of 1.0 ms–1. The cross-sectional area of the, tap is 10–4 m 2. Assume that the pressure is constant, throughout the stream of water and that the flow is streamlined., The cross-sectional area of the stream, 0.15 m below the tap, would be : [Take g = 10 ms–2), [10 April 2019 (II)], (a) 2×10–5 m2, (b) 5×10–5 m2, (c) 5×10–4 m2, (d) 1×10–5 m2, 23. Water from a pipe is coming at a rate of 100 liters per, minute. If the radius of the pipe is 5 cm, the Reynolds, number for the flow is of the order of : (density of water =, 1000 kg/m3, coefficient of viscosity of water = 1 mPa s), [8 April 2019 I], (a) 10 3, (b) 10 4, (c) 10 2, (d), 10 6, , 24. Water flows into a large tank with flat bottom at the rate, of 10–4 m3 s–(1) Water is also leaking out of a hole of area, 1 cm2 at its bottom. If the height of the water in the tank, remains steady, then this height is:, [10 Jan. 2019 I], (a) 5.1 cm, (b) 7 cm (c) 4 cm, (d), 9 cm, 25. The top of a water tank is open to air and its water lavel, is maintained. It is giving out 0.74m3 water per minute, through a circular opening of 2 cm radius in its wall. The, depth of the centre of the opening from the level of water, in the tank is close to:, [9 Jan. 2019 (II)], (a) 6.0 m, (b) 4.8 m (c) 9.6 m, (d) 2.9 m, 26. When an air bubble of radius r rises from the bottom to the, 5r, . Taking the, 4, atmospheric pressure to be equal to 10m height of water, column, the depth of the lake would approximately be, (ignore the surface tension and the effect of temperature):, [Online April 15, 2018], (a) 10.5m, (b) 8.7m (c) 11.2m, (d) 9.5m, 27. Two tubes of radii r1 and r2, and lengths l1 and l2, respectively, are connected in series and a liquid flows through, each of them in streamline conditions. P1 and P2 are pressure, , surface of a lake, its radius becomes, , differences across the two tubes. If P2 is 4P1 and l2 is l1 ,, 4, then the radius r2 will be equal to :, [Online April 9, 2017], r1, 2, 28. Consider a water jar of radius R that has water filled up to, height H and is kept on a stand of height h (see figure)., Through a hole of radius r (r << R) at its bottom, the water, leaks out and the stream of water coming down towards, the ground has a shape like a funnel as shown in the figure., If the radius of the cross–section of water stream when it, hits the ground is x. Then :, [Online April 9, 2016], , (a) r1, , (b) 2r1, , (c) 4r1, , (d), , R, H, 2r, , h, 2x, 1, ö4, , æ H, (a) x = r ç, è H + h ÷ø, , 2, , æ H ö, (b) x = r çè, ÷, H + hø, 1, , æ H ö, æ H ö2, (c) x = r ç, (d) x = r ç, è H + h ÷ø, è H + h ÷ø, 29. If it takes 5 minutes to fill a 15 litre bucket from a water tap, , of diameter, , 2, , cm then the Reynolds number for the, p, flow is (density of water = 103 kg/m3) and viscosity of, water = 10–3 Pa.s) close to :, [Online April 10, 2015], (a) 1100, (b) 11,000 (c) 550, (d) 5500
Page 151 :
P-142, , Physics, , 30. An open glass tube is immersed in mercury in such a way, that a length of 8 cm extends above the mercury level. The, open end of the tube is then closed and sealed and the tube, is raised vertically up by additional 46 cm. What will be, length of the air column above mercury in the tube now?, (Atmospheric pressure = 76 cm of Hg), [2014], (a) 16 cm, (b) 22 cm (c) 38 cm, (d) 6 cm, 31. In the diagram shown, the difference in the two tubes of, the manometer is 5 cm, the cross section of the tube at A, and B is 6 mm2 and 10 mm2 respectively. The rate at which, water flows through the tube is (g = 10 ms–2), [Online April 19, 2014], , 5 cm, , (a) Depends on H, (b) 1 : 1, (c) 2 : 2, (d) 1 : 2, 36. Water is flowing through a horizontal tube having crosssectional areas of its two ends being A and A¢ such that the, ratio A/A¢ is 5. If the pressure difference of water between, the two ends is 3 × 105 N m–2, the velocity of water with, which it enters the tube will be (neglect gravity effects), [Online May 12, 2012], (a) 5 m s–1, (b) 10 m s–1, (c) 25 m s–1, (d) 50 10 m s–1, 37. A square hole of side length l is made at a depth of h and, a circular hole of radius r is made at a depth of 4h from the, surface of water in a water tank kept on a horizontal surface., If l << h, r << h and the rate of water flow from the holes is, the same, then r is equal to, [May 7, 2012], , B, , A, , h, 4h, , (a) 7.5 cc/s (b) 8.0 cc/s (c) 10.0 cc/s (d) 12.5 cc/s, 32. A cylindrical vessel of cross-section A contains water to a, height h. There is a hole in the bottom of radius ‘a’. The, time in which it will be emptied is: [Online April 12, 2014], 2A h, 2A h, (a), (b), 2 g, pa, pa 2 g, , 2 2A h, A, h, (d), 2, 2 g, g, pa, 2pa, 33. Water is flowing at a speed of 1.5 ms–1 through horizontal, tube of cross-sectional area 10–2 m2 and you are trying to, stop the flow by your palm. Assuming that the water stops, immediately after hitting the palm, the minimum force, that you must exert should be, (density of water = 103 kgm–3) [Online April 9, 2014], (a) 15 N, (b) 22.5 N (c) 33.7 N, (d) 45 N, 34. Air of density 1.2 kg m–3 is blowing across the horizontal, wings of an aeroplane in such a way that its speeds above, and below the wings are 150 ms –1 and 100 ms –1 ,, respectively. The pressure difference between the upper, and lower sides of the wings, is : [Online April 22, 2013], (a) 60 Nm–2, (b) 180 Nm–2, (c) 7500 Nm–2, (d) 12500 Nm–2, 35. In a cylindrical water tank, there are two small holes A and, B on the wall at a depth of h1, from the surface of water and at, a height of h2 from the bottom of water tank. Surface of water, is at heigh H from the bottom of water tank. Water coming out, from both holes strikes the ground at the same point S. Find, the ratio of h1 and h2, [Online May 26, 2012], (c), , h1, A, H, B, h2, S, , A, , v1, , B, , v2, , l, l, l, l, (b), (c), (d), 2p, 3p, 3p, 2p, 38. Water is flowing continuously from a tap having an internal, diameter 8 × 10–3 m. The water velocity as it leaves the tap, is 0.4 ms–1. The diameter of the water stream at a distance 2, × 10–1 m below the tap is close to:, [2011], (a) 7.5 × 10–3 m, (b) 9.6 × 10–3 m, (c) 3.6 × 10–3 m, (d) 5.0 × 10–3 m, 39. A cylinder of height 20 m is completely filled with water., The velocity of efflux of water (in ms-1) through a small hole, on the side wall of the cylinder near its bottom is [2002], (a) 10, (b) 20, (c) 25.5, (d) 5, (a), , TOPIC 3 Viscosity and Terminal Velocity, 40. In an experiment to verify Stokes law, a small spherical ball of, radius r and density r falls under gravity through a distance h, in air before entering a tank of water. If the terminal velocity of, the ball inside water is same as its velocity just before entering, the water surface, then the value of h is proportional to :, (ignore viscosity of air), [5 Sep. 2020 (II)], 4, 3, (a) r, (b) r, (c) r (d), r2, 41. A solid sphere, of radius R acquires a terminal velocity, v1 when falling (due to gravity) through a viscous fluid, having a coefficient of viscosity h. The sphere is broken, into 27 identical solid spheres. If each of these spheres, acquires a terminal velocity, v2, when falling through the, same fluid, the ratio (v1/v2) equals: [12 April 2019 (II)], (a) 9, (b) 1/27 (c) 1/9, (d) 27
Page 152 :
P-143, , Mechanical Properties of Fluids, , 42. Which of the following option correctly describes the, variation of the speed v and acceleration 'a' of a point mass, falling vertically in a viscous medium that applies a force, F = –kv, where 'k' is a constant, on the body? (Graphs are, schematic and not drawn to scale), [Online April 9, 2016], a, , (a), v, t, v, , (b), , a, , 47. The terminal velocity of a small sphere of radius a in a, viscous liquid is proportional to [Online May 26, 2012], (a) a 2, (b) a 3, (c) a, (d) a–1, –3, 48. If a ball of steel (density r = 7.8 g cm ) attains a terminal, velocity of 10 cm s–1 when falling in water (Coefficient, of viscosity hwater = 8.5 × 10–4 Pa.s), then, its terminal, velocity in glycerine (r = 1.2 g cm–3, h = 13.2 Pa.s) would, be, nearly, [2011 RS], (a) 6.25 × 10–4 cm s–1, (b) 6.45 × 10–4 cm s–1, (c) 1.5 × 10–5 cm s–1, (d) 1.6 × 10–5 cm s–1, 49. A spherical solid ball of volume V is made of a material, of density r1. It is falling through a liquid of density, r1 (r2< r1). Assume that the liquid applies a viscous force, on the ball that is proportional to the square of its speed, v, i.e., Fviscous = –kv2 (k > 0). The terminal speed of the ball, is, [2008], (a), , t, , v, , Vg (r1 – r2 ), k, , (b), , Vg r1, k, , Vgr1, Vg (r1 – r2 ), (d), k, k, 50. If the terminal speed of a sphere of gold (density, = 19.5 kg/m3) is 0.2 m/s in a viscous liquid (density = 1.5, kg/m 3), find the terminal speed of a sphere of silver, (density = 10.5 kg/m3) of the same size in the same liquid, [2006], (a) 0.4 m/s, (b) 0.133 m/s, (c) 0.1 m/s, (d) 0.2 m/s, 51. Spherical balls of radius ‘R’ are falling in a viscous fluid, of viscosity ‘h’ with a velocity ‘v’. The retarding viscous, force acting on the spherical ball is, [2004], (a) inversely proportional to both radius ‘R’ and velocity ‘v’, (b) directly proportional to both radius ‘R’ and velocity ‘v’, (c) directly proportional to ‘R’ but inversely proportional, to ‘v’, (d) inversely proportional to ‘R’ but directly proportional, to velocity ‘v’, , (c), , (c), , a, t, v, , (d), , a, t, , 43. The velocity of water in a river is 18 km/hr near the surface., If the river is 5 m deep, find the shearing stress between the, horizontal layers of water. The co-efficient of viscosity of, water = 10–2 poise., [Online April 19, 2014], (a) 10–1 N/m2, (b) 10–2 N/m2, (c) 10–3 N/m2, (d) 10–4 N/m2, 44. The average mass of rain drops is 3.0 × 10–5 kg and their, avarage terminal velocity is 9 m/s. Calculate the energy, transferred by rain to each square metre of the surface at a, place which receives 100 cm of rain in a year., [Online April 11, 2014], (a) 3.5 × 105 J, (b) 4.05 × 104 J, (c) 3.0 × 105 J, (d) 9.0 × 104 J, 45. A tank with a small hole at the bottom has been filled with, water and kerosene (specific gravity 0.8). The height of, water is 3m and that of kerosene 2m. When the hole is, opened the velocity of fluid coming out from it is nearly:, (take g = 10 ms–2 and density of water = 103 kg m–3), [Online April 11, 2014], (a) 10.7 ms–1, (b) 9.6 ms–1, (c) 8.5 ms–1, (d) 7.6 ms–1, 46. In an experiment, a small steel ball falls through a liquid at, a constant speed of 10 cm/s. If the steel ball is pulled upward, with a force equal to twice its effective weight, how fast, will it move upward ?, [Online April 25, 2013], (a) 5 cm/s (b) Zero (c) 10 cm/s, (d) 20 cm/s, , TOPIC 4, , Surface Tension, Surface, Energy and Capillarity, , 52. When a long glass capillary tube of radius 0.015 cm is, dipped in a liquid, the liquid rises to a height of 15 cm, within it. If the contact angle between the liquid and glass, to close to 0°, the surface tension of the liquid, in, milliNewton m–1, is [r(liquid) = 900 kgm–3, g = 10 ms–2], (Give answer in closest integer) __________., [NA 3 Sep. 2020 (I)], 53. Pressure inside two soap bubbles are 1.01 and 1.02, atmosphere, respectively. The ratio of their volumes is :, [3 Sep. 2020 (I)], (a) 4 : 1, (b) 0.8 : 1 (c) 8 : 1, (d) 2 : 1, 54. A capillary tube made of glass of radius 0.15 mm is dipped, vertically in a beaker filled with methylene iodide (surface, tension = 0.05 Nm–1, density = 667 kg m–3) which rises
Page 153 :
P-144, , Physics, , to height h in the tube. It is observed that the two tangents, drawn from liquid-glass interfaces (from opp. sides of, the capillary) make an angle of 60° with one another. Then, h is close to (g = 10 ms–2)., [2 Sep. 2020 (II)], (a) 0.049 m, (b) 0.087 m, (c) 0.137 m, (d) 0.172 m, 55. A small spherical droplet of density d is floating exactly, half immersed in a liquid of density r and surface tension, T. The radius of the droplet is (take note that the surface, tension applies an upward force on the droplet):, [9 Jan. 2020 (II)], (a) r =, , 2T, 3(d + r) g, , (c) r =, , T, (d + r) g, , (b) r =, , (d) r =, , T, (d - r) g, 3T, (2d - r) g, , 56. The ratio of surface tensions of mercury and water is, given to be 7.5 while the ratio of their densities is 13.6., Their contact angles, with glass, are close to 135o and 0o,, respectively. It is observed that mercury gets depressed, by an amount h in a capillary tube of radius r1, while water, rises by the same amount h in a capillary tube of radius, r2. The ratio, (r1/r2), is then close to :, [10 April 2019 (I)], (a) 4/5, (b) 2/5, (c) 3/5, (d) 2/3, 57. If ‘M’ is the mass of water that rises in a capillary tube of, radius ‘r’, then mass of water which will rise in a capillary, tube of radius ‘2r’ is :, [9 April 2019 I], M, (a) M, (b), (c) 4 M, (d), 2M, 2, 58. If two glass plates have water between them and are, separated by very small distance (see figure), it is very, difficult to pull them apart. It is because the water in, between forms cylindrical surface on the side that gives, rise to lower pressure in the water in comparison to, atmosphere. If the radius of the cylindrical surface is R, and surface tension of water is T then the pressure in water, between the plates is lower by : [Online April 10, 2015], Cylindrical surface, of water, , 2T, 4T, T, T, (a), (b), (c), (d), R, R, 4R, R, 59. On heating water, bubbles being formed at the bottom of, the vessel detach and rise. Take the bubbles to be spheres, of radius R and making a circular contact of radius r with, the bottom of the vessel. If r << R and the surface tension, of water is T, value of r just before bubbles detach is:, (density of water is rw), [2014], , R, 2r, 2, (a) R, , 2 rwg, 3T, , 2, (b) R, , rw g, 6T, , rw g, 2 3r w g, (d) R, T, T, 60. A large number of liquid drops each of radius r coalesce, to from a single drop of radius R. The energy released in, the process is converted into kinetic energy of the big, drop so formed. The speed of the big drop is (given,, surface tension of liquid T, density r), [Online April 19, 2014, 2012], 2, (c) R, , (a), , T æ1 1 ö, r çè r R ÷ø, , (b), , 2T æ 1 1 ö, r çè r R ÷ø, , 4T æ 1 1 ö, 6T æ 1 1 ö, - ÷, (d), ç, r èr Rø, r çè r R ÷ø, 61. Two soap bubbles coalesce to form a single bubble. If V, is the subsequent change in volume of contained air and S, change in total surface area, T is the surface tension and P, atmospheric pressure, then which of the following relation, is correct?, [Online April 12, 2014], (a) 4PV + 3ST = 0, (b) 3PV + 4ST = 0, (c) 2PV + 3ST = 0, (d) 3PV + 2ST = 0, 62. An air bubble of radius 0.1 cm is in a liquid having surface, tension 0.06 N/m and density 103 kg/m3. The pressure inside, the bubble is 1100 Nm–2 greater than the atmospheric, pressure. At what depth is the bubble below the surface of, the liquid? (g = 9.8 ms–2), [Online April 11, 2014], (a) 0.1 m, (b) 0.15 m (c) 0.20 m, (d) 0.25 m, 63. A capillary tube is immersed vertically in water and the height, of the water column is x. When this arrangement is taken into, a mine of depth d, the height of the water column is y. If R is, (c), , x, the radius of earth, the ratio y is:, , [Online April 9, 2014], , dö, æ, æ 2d ö, (a) ç 1 - ÷, (b) ç 1 - ÷, R, Rø, è, ø, è, æ R -d ö, æR+dö, (c) ç, (d) ç, ÷, ÷, èR+dø, è R -d ø, 64. Wax is coated on the inner wall of a capillary tube and the, tube is then dipped in water. Then, compared to the unwaxed, capillary, the angle of contact q and the height h upto which, water rises change. These changes are :, [Online April 23, 2013]
Page 154 :
P-145, , Mechanical Properties of Fluids, , (a) q increases and h also increases, (b) q decreases and h also decreases, (c) q increases and h decreases, (d) q decreases and h increases, 65. A thin tube sealed at both ends is 100 cm long. It lies, horizontally, the middle 20 cm containing mercury and two, equal ends containing air at standard atmospheric pressure. If, the tube is now turned to a vertical position, by what amount, will the mercury be displaced ?, [Online April 23, 2013], , l0, , 69. Two mercury drops (each of radius ‘r’) merge to form, bigger drop. The surface energy of the bigger drop, if T is, the surface tension, is :, [2011 RS], (a) 4pr 2T, , (c) 28 / 3 pr 2T, (d) 25/ 3 pr 2T, 70. A capillary tube (A) is dipped in water. Another identical, tube (B) is dipped in a soap-water solution. Which of the, following shows the relative nature of the liquid columns, in the two tubes?, [2008], , l0, , 20 cm, , 100 cm, (Given : cross-section of the tube can be assumed to be, uniform), (a) 2.95 cm (b) 5.18 cm (c) 8.65 cm, (d) 0.0 cm, 66. This question has Statement-1 and Statement-2. Of the four, choices given after the Statements, choose the one that, best describes the two Statetnents., Statement-1: A capillary is dipped in a liquid and liquid, rises to a height h in it. As the temperature of the liquid is, raised, the height h increases (if the density of the liquid, and the angle of contact remain the same)., Statement-2: Surface tension of a liquid decreases with, the rise in its temperature., [Online April 9, 2013], (a) Statement-1 is true, Statement-2 is true; Statement-2 is, not the correct explanation for Statement-1., (b) Statement-1 is false, Statement-2 is true., (c) Statement-1 is true, Statement-2 is false., (d) Statement-1 is true, Statement-2 is true; Statement-2 is, the correct explanation for Statement-1., 67. A thin liquid film formed between a U-shaped wire and a, light slider supports a weight of 1.5 × 10–2 N (see figure)., The length of the slider is 30 cm and its weight is negligible., The surface tension of the liquid film is, [2012], , FILM, , z, W, (a) 0.0125 Nm–1, (b) 0.1 Nm–1, (c) 0.05 Nm–1, (d) 0.025 Nm–1, 68. Work done in increasing the size of a soap bubble from a, radius of 3 cm to 5 cm is nearly (Surface tension of soap, solution = 0.03 Nm–1), [2011], (a) 0.2 p mJ (b) 2p mJ (c) 0.4p mJ, (d) 4p mJ, , (b) 2pr 2T, , A, , B, , A, , B, , A, , B, , A, , B, , (a), , (b), , (c), , (d), , 71. A 20 cm long capillary tube is dipped in water. The water, rises up to 8 cm. If the entire arrangement is put in a freely, falling elevator the length of water column in the capillary, tube will be, [2005], (a) 10 cm, (b) 8 cm (c) 20 cm, (d) 4 cm, 72. If two soap bubbles of different radii are connected by a, tube, [2004], (a) air flows from the smaller bubble to the bigger, (b) air flows from bigger bubble to the smaller bubble till, the sizes are interchanged, (c) air flows from the bigger bubble to the smaller bubble, till the sizes become equal, (d) there is no flow of air.
Page 161 :
P-152, , Physics, , v1, v2 = 9., 42. (c) When a point mass is falling vertically in a viscous, medium, the medium or viscous fluid exerts drag force on, the body to oppose its motion and at one stage body, falling with constant terminal velocity., 43. (b) h = 10–2 poise, , W–T –F=0, , or, , v = 18 km/h =, , 47., , 18000, = 5 m/s, 3600, , l =5m, v, Strain rate =, l, shearing stress, strain rate, \ Shearing stress = h × strain rate, , Coefficient of viscosity, h =, , \ VT µ a 2, 48. (a) When the ball attains terminal velocity, Weight of the ball = viscous force + buoyant force, \V rg = 6phrv + V rl g, Þ Vg ( r - rl ) = 6phrv, , (, , ', \ v ' h ' = (r - r l ) ´ v h, (r - r l ), , Þ v' =, , 2, = 1m ´, , E=, =, , ), , ', Also Vg r- ri = 6ph ' rv ', , 5, = 10 ´ = 10–2 Nm–2, 5, (b) Total volume of rain drops, received 100 cm in a year, by area 1 m2, , 100, m = 1 m3, 100, As we know, density of water,, d = 103 kg/m3, Therefore, mass of this volume of water, M = d × v = 103 × 1 = 103 kg, Average terminal velocity of rain drop, v = 9 m/s (given), Therefore, energy transferred by rain,, , 2 a 2 (r - s ) g, 9h, , VT =, , -2, , 44., , 2 r 2 (r - s)g, 9, h, As in case of upward motion upward force is twice its, effective weight, therefore, it will move with same speed, 10 cm/s, (a) Terminal velocity in a viscous medium is given by:, , And terminal velocity v =, , =, 49., , (r - rl' ) vh, ´, (r - rl ) h ', , (7.8 - 1.2) 10 ´ 8.5 ´ 10-4, ´, (7.8 - 1), 13.2, , \ v ' = 6.25 ´ 10-4 cm/s, (a) When the ball attains terminal velocity, Weight of the ball = Buoyant force + Viscous force, , 1, mv2, 2, , 1, × 103 × (9)2, 2, , 1, × 103 × 81 = 4.05 × 104 J, 2, (b) According to Toricelli’s theorem,, Velocity of efflex,, , =, , 45., , W=V r1g, , \ V r1 g, , Veff = 2 gh = 2 ´ 9.8 ´ 5 @ 9.8 ms -1, 46., , B=Vr2 g, , Fv, , (c), , T (upthrust), F, (viscous, force), , r, , W(weight), Weight of the body, 4 3, W = mg = pr rg, 3, 4 3, T = pr sg, 3, and F = 6phvr, When the body attains terminal velocity net force acting, on the body is zero. i.e.,, , = V r2 g + kvt2, , Þ Vg ( r1 – r2 ) g = kvt2, , Vg (r1 - r 2 ), k, (c) Given,, Density of gold, rG = 19.5 kg/m3, Density of silver, r5 = 10.5kg/m3, Density of liquid, s = 1.5kg/m3, Þ vt =, , 50., , Terminal velocity, vT =, vT, , 2 r 2 (r - s ) g, 9h, , (10.5 - 1.5), 9, Þ vT = 0.2 ´, 2, 0.2 (19.5 - 1.5), 18, \ vT = 0.1 m/s, , \, , 2, , =, , 2, , 51. (b) From Stoke's law, force of viscosity acting on a, spherical body is, F = 6phrv, hence F is directly proportional to radius & velocity.
Page 163 :
P-154, , Physics, , 1 é4 3 ù 2, é1 1 ù, ´ ê pR rú v = 4pR 3T ê - ú, 2 ë3, û, ër R û, v2 =, , 6T é 1 1 ù, r êë r R úû, , v=, , 6T é 1 1 ù, r êë r R úû, , 61. (b), 62. (a) Given: Radius of air bubble,, r = 0.1 cm = 10–3 m, Surface tension of liquid,, S = 0.06 N/m = 6 × 10–2 N/m, Density of liquid, r = 103 kg/m3, Excess pressure inside the bubble,, rexe = 1100 Nm–2, Depth of bubble below the liquid surface,, h=?, As we know,, 2s, rExcess = hrg +, r, 2 ´ 6 ´ 10-2, Þ 1100 = h × 103 × 9.8 +, 10-3, Þ 1100 = 9800 h + 120, Þ 9800h = 1100 – 120, 980, Þ h=, = 0.1 m, 9800, 63. (a) Acceleration due to gravity changes with the depth,, , dö, æ, g¢ = g ç1 - ÷, R, è, ø, Pressure, P = rgh, Hence ratio,, 64., , x æ dö, is ç1 - ÷, y è Rø, , (c) Angle of contact q, , TSA - TSL, TLA, when water is on a waxy or oily surface, TSA < TSL cos q is negative i.e.,, 90° < q < 180°, i.e., angle of contact q increases, And for q > 90° liquid level in capillary tube fall., i.e., h decreases, 65. (b), cos q =, , 66. (b) Surface tension of a liquid decreases with the rise in, temperture. At the boiling point of liquid, surface tension, is zero., Capillary rise h =, , 2T cos q, rdg, , As surface tension T decreases with rise in temperature, hence capillary rise also decreases., 67. (d) Let T is the force due to surface tension per unit length,, then, F = 2lT, l = length of the slider., At equilibrium, F = W, \ 2Tl = mg, Þ T=, , mg, 1.5 ´ 10 -2, 1.5, =, =, = 0.025 Nm–1, 2l 2 ´ 30 ´ 10-2 60, , 68. (c) Work done = increase in surface area × surface tension, Þ W = 2T4p[(52) – (3)2] × 10–4, = 2 × 0.03 × 4p [25 – 9] × 10–4 J, = 0.4p × 10–3 J = 0.4p mJ, 69. (c) As volume remains constant, \Sum of volumes of 2 smaller drops, = Volume of the bigger drop, 4, 4, 2. pr 3 = pR 3 Þ R = 21/ 3 r, 3, 3, 2, Surface energy = Surface tension × Surface area = T .4 pR, , = T 4p 22 / 3 r 2 = T .28 / 3 pr 2 ., 70. (c) In case of water, the meniscus shape is concave, upwards. From ascent formula h =, , 2s cos q, r rg, , The surface tension (s) of soap solution is less than water., Therefore height of capillary rise for soap solution should, be less as compared to water. As in the case of water, the, meniscus shape of soap solution is also concave upwards., 71. (c) Water fills the tube entirely in gravityless condition, i.e., 20 cm., 72. (a) Let pressure outside be P0 and r and R be the radius, of smaller bubble and bigger bubble respectively., 2T, \ Pressure P1 For smaller bubble = P0 +, r, 2T, P2 For bigger bubble = P0 +, ( R > r), R, \ P1 > P2, hence air moves from smaller bubble to bigger bubble.
Page 164 :
10, Thermal Properties, of Matter, TOPIC 1 Thermometer & Thermal, Expansion, 1., , Two different wires having lengths L1 and L2 , and, respective temperature coefficient of linear expansion a1, and a 2 , are joined end-to-end. Then the effective, temperature coefficient of linear expansion is :, [Sep. 05, 2020 (II)], (a), , a1 L1 + a 2 L2, L1 + L2, , a + a2, (c) 1, 2, , 2., , 3., , 4., , 5., , 6., , 7., , (b) 2 a1a 2, aa, L2 L1, (d) 4 1 2, a1 + a 2 ( L2 + L1 ) 2, , A bakelite beaker has volume capacity of 500 cc at 30°C., When it is partially filled with Vm volume (at 30°C) of, mercury, it is found that the unfilled volume of the beaker, remains constant as temperature is varied. If g(beaker) = 6 ×, 10–6 °C–1 and g(mercury) = 1.5 × 10–4 °C–1, where g is the, coefficient of volume expansion, then Vm (in cc) is close to, __________., [NA Sep. 03, 2020 (I)], When the temperature of a metal wire is increased from, 0°C to 10°C, its length increased by 0.02%. The percentage, change in its mass density will be closest to :, [Sep. 02, 2020 (II)], (a) 0.06, (b) 2.3, (c) 0.008, (d) 0.8, A non-isotropic solid metal cube has coefficients of linear, expansion as: 5 ´ l0–5/°C along the x-axis and 5 ´ 10–6/°C, along the y and the z-axis. If the coefficient of volume, expansion of the solid is C ´ 10–6/°C then the value of C is, ¾¾¾ ., [NA 7 Jan. 2020 I], o, At 40 C, a brass wire of 1 mm radius is hung from the, ceiling. A small mass, M is hung from the free end of the, wire. When the wire is cooled down from 40oC to 20oC it, regains its original length of 0.2 m. The value of M is close, to:, [12 April 2019 I], (Coefficient of linear expansion and Young’s modulus of, brass are 10–5/oC and 1011 N/m2, respectively; g = 10 ms–2), (a) 9 kg, (b) 0.5 kg (c) 1.5 kg, (d) 0.9 kg, Two rods A and B of identical dimensions are at, temperature 30°C. If A is heated upto 180°C and B upto, , 8., , 9., , T°C, then the new lengths are the same. If the ratio of the, coefficients of linear expansion of A and B is 4 : 3, then, the value of T is :, [11 Jan. 2019 II], (a) 230°C, (b) 270°C, (c) 200°C, (d) 250°C, A thermometer graduated according to a linear scale reads, a value x0 when in contact with boiling water, and x0/3, when in contact with ice. What is the temperature of an, object in °C, if this thermometer in the contact with the, object reads x0/2?, [11 Jan. 2019 II], (a) 25, (b) 60, (c) 40, (d) 35, A rod, of length L at room temperature and uniform area, of cross section A, is made of a metal having coefficient, of linear expansion a/°C. It is observed that an external, compressive force F, is applied on each of its ends,, prevents any change in the length of the rod, when its, temperature rises by DT K. Young’s modulus, Y, for this, metal is:, [9 Jan. 2019 I], F, F, (a), (b), AaDT, A a ( D T - 273), F, 2F, (c), (d), 2A a D T, AaDT, An external pressure P is applied on a cube at 0oC so that, it is equally compressed from all sides. K is the bulk, modulus of the material of the cube and a is its, coefficient of linear expansion. Suppose we want to bring, the cube to its original size by heating. The temperature, should be raised by :, [2017], , 3a, (b) 3PKa, PK, P, P, (c), (d), 3aK, aK, 10. A steel rail of length 5 m and area of cross-section 40, cm2 is prevented from expanding along its length while, the temperature rises by 10°C. If coefficient of linear, expansion and Young’s modulus of steel are 1.2 × 10–5 K–1, and 2 × 1011 Nm–2 respectively, the force developed in, the rail is approximately:, [Online April 9, 2017], (a) 2 × 107 N, (b) 1 × 105 N, (c) 2 × 109 N, (d) 3 × 10–5 N, , (a)
Page 165 :
P-156, , 11., , Physics, , A compressive force, F is applied at the two ends of a, long thin steel rod. It is heated, simultaneously, such that, its temperature increases by DT. The net change in its, length is zero. Let l be the length of the rod, A its area of, cross–section, Y its Youn g’s modulus, and a its, coefficient of linear expansion. Then, F is equal to :, [Online April 8, 2017], (a) l2 Ya DT, (b) lA Ya DT, AY, aDT, The ratio of the coefficient of volume expansion of a glass, container to that of a viscous liquid kept inside the, container is 1 : 4. What fraction of the inner volume of the, container should the liquid occupy so that the volume of, the remaining vacant space will be same at all temperatures ?, [Online April 23, 2013], (a) 2 : 5, (b) 1 : 4, (c) 1 : 64, (d) 1 : 8, On a linear temperature scale Y, water freezes at – 160° Y, and boils at – 50° Y. On this Y scale, a temperature of 340 K, would be read as : (water freezes at 273 K and boils at 373 K), [Online April 9, 2013], (a) – 73.7° Y, (b) – 233.7° Y, (c) – 86.3° Y, (d) – 106.3° Y, A wooden wheel of radius R is made of two semicircular, part (see figure). The two parts are held together by a ring, made of a metal strip of cross sectional area S and length, L. L is slightly less than 2pR. To fit the ring on the wheel, it, is heated so that its temperature rises by DT and it just, steps over the wheel. As it cools down to surrounding, temperature, it presses the semicircular parts together. If, the coefficient of linear expansion of the metal is a, and its, Young's modulus is Y, the force that one part of the wheel, applies on the other part is :, [2012], , (c) A Ya DT, 12., , 13., , 14., , (a), , 2pSY aDT, , (b), , SY aDT, , (c), , pSY aDT, , (d), , 2SY aDT, , (d), , R, , TOPIC 2 Calorimetry and Heat, Transfer, 15., , Three rods of identical cross-section and lengths are made, of three different materials of thermal conductivity K1, K2, and K3, respectively. They are joined together at their ends, to make a long rod (see figure). One end of the long rod is, maintained at 100ºC and the other at 0ºC (see figure). If the, joints of the rod are at 70ºC and 20ºC in steady state and, there is no loss of energy from the surface of the rod,, the correct relationship between K1, K2 and K3 is :, [Sep. 06, 2020 (II)], , 100°C, , K1, , K2, , K3, , 0°C, , 70°C 20°C, , 16., , 17., , 18., , 19., , (a) K1 : K3 = 2 : 3, K1 < K3 = 2 : 5, (b) K1 < K2 < K3, (c) K1 : K2 = 5 : 2, K1 : K3 = 3 : 5, (d) K1 > K2 > K3, A bullet of mass 5 g, travelling with a speed of 210 m/s,, strikes a fixed wooden target. One half of its kinetics energy is converted into heat in the bullet while the other, half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is, 0.030 cal/(g – ºC) (1 cal = 4.2 × 107 ergs) close to :, [Sep. 05, 2020 (I)], (a) 87.5ºC, (b) 83.3ºC, (c) 119.2ºC, (d) 38.4ºC, The specific heat of water = 4200 J kg–1 K–1 and the latent, heat of ice = 3.4 × 105 J kg–1. 100 grams of ice at 0°C is, placed in 200 g of water at 25°C. The amount of ice that will, melt as the temperature of water reaches 0°C is close to (in, grams) :, [Sep. 04, 2020 (I)], (a) 61.7, (b) 63.8, (c) 69.3, (d) 64.6, A calorimter of water equivalent 20 g contains 180 g of, water at 25°C. 'm' grams of steam at 100°C is mixed in it till, the temperature of the mixture is 31°C. The value of 'm' is, close to (Latent heat of water = 540 cal g–1, specific heat of, water = 1 cal g–1 °C–1), [Sep. 03, 2020 (II)], (a) 2, (b) 4, (c) 3.2, (d) 2.6, Three containers C1, C2and C3 have water at different, temperatures. The table below shows the final temperature, T when different amounts of water (given in liters) are taken, from each container and mixed (assume no loss of heat, during the process), [8 Jan. 2020 II], C1, , C2, , C3, , T, , 1l, , 2l, , —, , 60°C, , –, , 1l, , 2l, , 30°C, , 2l, , —, , 1l, , 60°C, , 1l, , 1l, , 1l, , q, , The value of q (in °C to the nearest integer) is______., 20. M grams of steam at 100°C is mixed with 200 g of ice at its, melting point in a thermally insulated container. If it, produces liquid water at 40°C [heat of vaporization of water, is 540 cal/ g and heat of fusion of ice is 80 cal/g], the value, of M is ________, [NA 7 Jan. 2020 II], o, 21. When M1 gram of ice at –10 C (Specific heat = 0.5 cal g–1 oC–1), is added to M2 gram of water at 50oC, finally no ice is, left and the water is at 0oC. The value of latent heat of, ice, in cal g–1 is:, [12 April 2019 I]
Page 166 :
P-157, , Thermal Properties of Matter, , (a), , 50 M 2, -5, M1, , 5M1, (b) M - 50, 2, , (c), , 50M 2, M1, , (d), , 5 M2, -5, M1, , 22. A massless spring (K = 800 N/m), attached with a mass, (500 g) is completely immersed in 1kg of water. The spring, is stretched by 2cm and released so that it starts vibrating., What would be the order of magnitude of the change in, the temperature of water when the vibrations stop, completely? (Assume that the water container and spring, receive negligible heat and specific heat of mass = 400 J/kg K,, specific heat of water = 4184 J/kg K), [9 April 2019 II], (a) 10–4 K, (b) 10–5 K (c) 10–1 K, (d) 10–3 K, 23. Two materials having coefficients of thermal conductivity, ‘3K’ and ‘K’ and thickness ‘d’ and ‘3d’, respectively, are, joined to form a slab as shown in the figure. The, temperatures of the outer surfaces are ‘q2’ and ‘q1’, respectively, (q2 > q1). The temperature at the interface is:, [9 April 2019 II], , q1 9q 2, q 2 + q1, +, (b), 10 10, 2, q1 5q2, q 2q, +, (d) 1 + 2, (c), 6, 6, 3, 3, 24. A cylinder of radius R is surrounded by a cylindrical shell, of inner radius R and outer radius 2R. The thermal, conductivity of the material of the inner cylinder is K1 and, that of the outer cylinder is K2. Assuming no loss of, heat, the effective thermal conductivity of the system for, heat flowing along the length of the cylinder is:, [12 Jan. 2019 I], K1 + K 2, (a), (b) K1 + K 2, 2, 2K1 + 3K 2, K1 + 3K 2, (c), (d), 5, 4, 25. Ice at –20°C is added to 50 g of water at 40°C, When the, temperature of the mixture reaches 0°C, it is found that 20, g of ice is still unmelted. The amount of ice added to the, water was close to, [11 Jan. 2019 I], (Specific heat of water = 4.2J/g/°C, Specific heat of Ice = 2.1 J/g/°C, Heat of fusion of water at 0°C = 334J/g), (a) 50g, (b) 100 g, (c) 60 g, (d) 40 g, 26. When 100 g of a liquid A at 100°C is added to 50 g of a, , (a), , liquid B at temperature 75°C, the temperature of the mixture, becomes 90°C. The temperature of the mixture, if 100 g of, liquid A at 100°C is added to 50 g of liquid B at 50°C, will, be :, [11 Jan. 2019 II], (a) 85°C, (b) 60°C, (c) 80°C, (d) 70°C, , 27. A metal ball of mass 0.1 kg is heated upto 500°C and, dropped into a vessel of heat capacity 800 JK–1 and, containing 0.5 kg water. The initial temperature of water, and vessel is 30°C. What is the approximate percentage, increment in the temperature of the water? [Specific Heat, Capacities of water and metal are, respectively, 4200 Jkg–, 1K–1 and 400 Jkg–1 K–1 ], [11 Jan. 2019 II], (a) 15%, (b) 30%, (c) 25%, (d) 20%, 28. A heat source at T = 103 K is connected to another heat, reservoir at T = 102 K by a copper slab which is 1 m, thick. Given that the thermal conductivity of copper is, 0.1 WK–1 m–1, the energy flux through it in the steady, state is:, [10 Jan. 2019 I], (a) 90 Wm–2, (b) 120 Wm–2, (c) 65 Wm–2, (d) 200 Wm–2, 29. An unknown metal of mass 192 g heated to a temperature, of 100°C was immersed into a brass calorimeter of mass, 128 g containing 240 g of water at a temperature of 8.4°C., Calculate the specific heat of the unknown metal if water, temperature stablizes at 21.5°C. (Specific heat of brass is, 394 J kg–1 K–1), [10 Jan. 2019 II], –1, –1, (a) 458 J kg K, (b) 1232 J kg–1 K–1, (c) 916 J kg–1 K–1, (d) 654 J kg–1 K–1, 30. Temperature difference of 120°C is maintained between, two ends of a uniform rod AB of length 2L. Another bent, , 3L, , is, 2, connected across AB (See figure). In steady state,, temperature difference between P and Q will be close to:, [9 Jan. 2019 I], , rod PQ, of same cross-section as AB and length, , L, 4, A, L, 2, , P, , L, , Q, , B, , (a) 45°C, (b) 75°C, (c) 60°C, (d) 35°C, 31. A copper ball of mass 100 gm is at a temperature T. It is, dropped in a copper calorimeter of mass 100 gm, filled with, 170 gm of water at room temperature. Subsequently, the, temperature of the system is found to be 75°C. T is given, by (Given : room temperature = 30° C, specific heat of, copper = 0.1 cal/gm°C, [2017], (a) 1250°C (b) 825°C (c) 800°C, (d) 885° C, 32. In an experiment a sphere of aluminium of mass 0.20 kg, is heated upto 150°C. Immediately, it is put into water of, volume 150 cc at 27°C kept in a calorimeter of water, equivalent to 0.025 kg. Final temperature of the system, is 40°C. The specific heat of aluminium is :, (take 4.2 Joule=1 calorie), [Online April 8, 2017], (a) 378 J/kg – °C, (b) 315 J/kg – °C, (c) 476 J/kg – °C, (d) 434 J/kg – °C
Page 167 :
P-158, , Physics, , 33. An experiment takes 10 minutes to raise the temperature, of water in a container from 0ºC to 100ºC and another 55, minutes to convert it totally into steam by a heater, supplying heat at a uniform rate. Neglecting the specific, heat of the container and taking specific heat of water, to be 1 cal / g ºC, the heat of vapourization according, to this experiment will come out to be :, [Online April 11, 2015], (a) 560 cal/ g, (b) 550 cal/ g, (c) 540 cal/ g, (d) 530 cal/ g, 34. Three rods of Copper, Brass and Steel are welded together, to form a Y shaped structure. Area of cross - section of, each rod = 4 cm2. End of copper rod is maintained at 100ºC, where as ends of brass and steel are kept at 0ºC. Lengths, of the copper, brass and steel rods are 46, 13 and 12 cms, respectively. The rods are thermally insulated from, surroundings excepts at ends. Thermal conductivities of, copper, brass and steel are 0.92, 0.26 and 0.12 CGS units, respectively. Rate of heat flow through copper rod is:[2014], (a) 1.2 cal/s, (b) 2.4 cal/s, (c) 4.8 cal/s, (d) 6.0 cal/s, 35. A black coloured solid sphere of radius R and mass M is, inside a cavity with vacuum inside. The walls of the cavity, are maintained at temperature T0. The initial temperature, of the sphere is 3T0. If the specific heat of the material of, the sphere varies as aT3 per unit mass with the temperature, T of the sphere, where a is a constant, then the time taken, for the sphere to cool down to temperature 2T0 will be (s is, Stefan Boltzmann constant), [Online April 19, 2014], (a), , (c), 36., , 37., , Ma, , æ3ö, In ç ÷, 4pR s è 2 ø, , (b), , 2, , Ma, , æ 16 ö, In ç ÷, 16pR s è 3 ø, 2, , (d), , Ma, , æ 16 ö, In ç ÷, 4pR s è 3 ø, 2, , Ma, , æ3ö, In ç ÷, 16pR s è 2 ø, 2, , Water of volume 2 L in a closed container is heated with a, coil of 1 kW. While water is heated, the container loses, energy at a rate of 160 J/s. In how much time will the, temperature of water rise from 27°C to 77°C? (Specific heat, of water is 4.2 kJ/kg and that of the container is negligible)., [Online April 9, 2014], (a) 8 min 20 s, (b) 6 min 2 s, (c) 7 min, (d) 14 min, Assume that a drop of liquid evaporates by decrease in its, surface energy, so that its temperature remains, unchanged.What should be the minimum radius of the, drop for this to be possible? The surface tension is T,, density of liquid is r and L is its latent heat of, vaporization., [2013], (a) rL/T, , (b), , T / rL (c) T/rL, , (d) 2T/rL, , 38. A mass of 50g of water in a closed vessel, with, surroundings at a constant temperature takes 2 minutes to, cool from 30°C to 25°C. A mass of 100g of another liquid in, an identical vessel with identical surroundings takes the, same time to cool from 30° C to 25° C. The specific heat of, the liquid is :, (The water equivalent of the vessel is 30g.), [Online April 25, 2013], (a) 2.0 kcal/kg, (b) 7 kcal/kg, (c) 3 kcal/kg, (d), 0.5 kcal/kg, 39. 500 g of water and 100 g of ice at 0°C are in a calorimeter, whose water equivalent is 40 g. 10 g of steam at 100°C is, added to it. Then water in the calorimeter is : (Latent heat, of ice = 80 cal/g, Latent heat of steam = 540 cal/ g), [Online April 23, 2013], (a) 580 g, (b) 590 g, (c) 600 g, (d) 610 g, 40. Given that 1 g of water in liquid phase has volume 1 cm 3, and in vapour phase 1671 cm3 at atmospheric pressure, and the latent heat of vaporization of water is 2256 J/g; the, change in the internal energy in joules for 1 g of water at, 373 K when it changes from liquid phase to vapour phase, at the same temperature is :, [Online April 22, 2013], (a) 2256, (b) 167, (c) 2089, (d) 1, 41. A large cylindrical rod of length L is made by joining two, , L, identical rods of copper and steel of length æç ö÷ each., è 2ø, The rods are completely insulated from the surroundings., If the free end of copper rod is maintained at 100°C and, that of steel at 0°C then the temperature of junction is, (Thermal conductivity of copper is 9 times that of steel), [Online May 19, 2012], (a) 90°C, (b) 50°C, (c) 10°C, (d) 67°C, 42. The heat radiated per unit area in 1 hour by a furnace, whose temperature is 3000 K is (s = 5.7 × 10–8 W m–2 K–4), [Online May 7, 2012], (a) 1.7 × 1010 J, (b) 1.1 × 1012 J, (c) 2.8 × 108 J, (d) 4.6 × 106 J, 43. 100g of water is heated from 30°C to 50°C. Ignoring the, slight expansion of the water, the change in its internal, energy is (specific heat of water is 4184 J/kg/K): [2011], (a) 8.4 kJ (b) 84 kJ, (c) 2.1 kJ (d) 4.2 kJ, 44. The specific heat capacity of a metal at low temperature, (T) is given as, 3, , æ T ö, C p (kJK -1kg -1 ) = 32 ç, è 400 ÷ø, A 100 gram vessel of this metal is to be cooled from 20ºK, to 4ºK by a special refrigerator operating at room, temperature (27°C). The amount of work required to, cool the vessel is, [2011 RS], (a) greater than 0.148 kJ, (b) between 0.148 kJ and 0.028 kJ, (c) less than 0.028 kJ, (d) equal to 0.002 kJ
Page 168 :
P-159, , Thermal Properties of Matter, , 45. A long metallic bar is carrying heat from one of its ends, to the other end under steady–state. The variation of, temperature q along the length x of the bar from its hot end, is best described by which of the following figures?[2009], , q, , q, , (a), , r1, , (b), , x, , x, , q, , r2, , q, , (c), , One end of a thermally insulated rod is kept at a temperature, T1 and the other at T2. The rod is composed of two sections, of length l1 and l2 and thermal conductivities K1 and K2, respectively. The temperature at the interface of the two, section is, [2007], T1 l1, , K1, , 47., , x, , l2, , T2, , K2, , (a), , ( K1l1T1 + K 2l2T2 ), ( K1l1 + K 2l2 ), , (b), , ( K 2l2T1 + K1l1T2 ), ( K1l1 + K 2 l2 ), , (c), , ( K 2l1T1 + K1l2T2 ), ( K 2 l1 + K1l2 ), , (d), , ( K1l2T1 + K 2l1T2 ), ( K1l2 + K 2 l1 ), , (a), (c), , r02 R2 s, , T, , T4, r, , 2, , (b), , 4, , (b), , (r2 - r1 ), (r1 r2 ), , (c), , ( r2 - r1 ), , (d), , r1 r2, (r2 - r1 ), , 50. If the temperature of the sun were to increase from T to, 2T and its radius from R to 2R, then the ratio of the radiant, energy received on earth to what it was previously will, be, [2004], (a) 32, (b) 16, (c) 4, (d) 64, 51. The temperature of the two outer surfaces of a composite, slab, consisting of two materials having coefficients of, thermal conductivity K and 2K and thickness x and 4x,, respectively, are T2 and T1 (T2 > T1 ) . The rate of heat, transfer through the slab, in a steady state is, æ A(T2 - T1 ) K ö, çè, ÷ø f , with f equal to, x, , (d), , R2 s, , T, , r, , 7, T0, 3, 5, (d) T f = T0, 2, , (b) T f =, , [2004], , 4x, , 2K, , T1, , 4, , temperature T0, while Box contains one mole of helium at, æ 7ö, temperature çè ÷ø T0 . The boxes are then put into thermal, 3, contact with each other, and heat flows between them until, the gases reach a common final temperature (ignore the, heat capacity of boxes). Then, the final temperature of, the gases, Tf in terms of T0 is, [2006], 3, T0, 7, 3, (c) T f = T0, 2, , K, , 2, , Two rigid boxes containing different ideal gases are placed, on a table. Box A contains one mole of nitrogen at, , (a) T f =, , x, , T4, , 4pr, r2, where r0 is the radius of the Earth and s is Stefan's constant., 48., , 2, , pr02 R2 s, , T2, , ær ö, ln ç 2 ÷, è r1 ø, , Assuming the Sun to be a spherical body of radius R at a, temperature of TK, evaluate the total radiant powerd, incident of Earth at a distance r from the Sun, [2006], , 4pr02 R 2s, , T1, , (a), (d), , x, 46., , 49. The figure shows a system of two concentric spheres of, radii r1 and r2 are kept at temperatures T1 and T2,, respectively. The radial rate of flow of heat in a substance, between the two concentric spheres is proportional to, [2005], , 2, 1, 1, (b), (c) 1, (d), 3, 2, 3, 52. The earth radiates in the infra-red region of the spectrum., The spectrum is correctly given by, [2003], (a) Rayleigh Jeans law, (b) Planck’s law of radiation, (c) Stefan’s law of radiation, (d) Wien’s law, 53. Heat given to a body which raises its temperature by 1°C, is, [2002], (a) water equivalent, (b) thermal capacity, (c) specific heat, (d) temperature gradient, , (a)
Page 169 :
P-160, , Physics, , 62. A hot body, obeying Newton’s law of cooling is cooling, down from its peak value 80°C to an ambient temperature, of 30°C. It takes 5 minutes in cooling down from 80°C to, 40°C. How much time will it take to cool down from 62°C to, 32°C?, (Given In 2 = 0.693, In 5 = 1.609) [Online April 11, 2014], (a) 3.75 minutes, (b) 8.6 minutes, (c) 9.6 minutes, (d) 6.5 minutes, 63. If a piece of metal is heated to temperature q and then, allowed to cool in a room which is at temperature q0, the, graph between the temperature T of the metal and time t, will be closest to, [2013], , T, (a), , (c), , T, q0, , (a), , 60., , (d), , T, q0, , (b), , 0, , t, , t, , (d), , A body takes 10 minutes to cool from 60°C to 50°C. The, temperature of surroundings is constant at 25°C. Then,, the temperature of the body after next 10 minutes will, be approximately, [Online April 15, 2018], (a) 43°C, (b) 47°C, (c) 41°C, (d) 45°C, 61. Hot water cools from 60°C to 50°C in the first 10 minutes, and to 42°C in the next 10 minutes. The temperature of, the surroundings is:, [Online April 12, 2014], (a) 25°C, (b) 10°C, (c) 15°C, (d) 20°C, , 0, , (d), , t, , loge (q – q0), , loge (q – q0), , (b), , (c), (c), , t, , O, , t, t, O, O, 64. A liquid in a beaker has temperature q(t) at time t and q0, is temperature of surroundings, then according to, Newton's law of cooling the correct graph between, loge(q – q0) and t is :, [2012], , 0, , (a), , T, q0, , loge (q – q0), , A metallic sphere cools from 50°C to 40°C in 300 s. If, atmospheric temperature around is 20°C, then the sphere's, temperature after the next 5 minutes will be close to :, [Sep. 03, 2020 (II)], (a) 31°C, (b) 33°C, (c) 28°C, (d) 35°C, 59. Two identical beakers A and B contain equal volumes of, two different liquids at 60°C each and left to cool down., Liquid in A has density of 8 × 102 kg/m3 and specific heat of, 2000 J kg–1 K–1 while liquid in B has density of 103 kg m–3, and specific heat of 4000 J kg–1 K–1. Which of the following, best describes their temperature versus time graph, schematically ? (assume the emissivity of both the beakers, to be the same), [8 April 2019 I], , t, , O, , TOPIC 3 Newton's Law of Cooling, 58., , (b), , loge (q – q0), , 54. Infrared radiation is detected by, [2002], (a) spectrometer, (b) pyrometer, (c) nanometer, (d) photometer, 55. Which of the following is more close to a black body?, [2002], (a) black board paint, (b) green leaves, (c) black holes, (d) red roses, 56. If mass-energy equivalence is taken into account, when, water is cooled to form ice, the mass of water should[2002], (a) increase, (b) remain unchanged, (c) decrease, (d) first increase then decrease, 57. Two spheres of the same material have radii 1 m and 4 m, and temperatures 4000 K and 2000 K respectively. The, ratio of the energy radiated per second by the first sphere, to that by the second is, [2002], (a) 1 : 1, (b) 16 : 1, (c) 4 : 1, (d) 1 : 9., , 0, , t, , 65. According to Newton’s law of cooling, the rate of cooling, of a body is proportional to (Dq)n , where Dq is the, difference of the temperature of the body and the, surroundings, and n is equal to, [2003], (a) two, (b) three, (c) four, (d) one
Page 173 :
P-164, , Physics, , and P × 55 × 60 = mL, ...(ii), Dividing equation (i) by (ii) we get, 10 C ´100, =, 55, L, \ L = 550 cal./g., 34. (c) Rate of heat flow is given by,, , DT, 6.4, ´100 =, ´100 = 21%,, T, 30, so the closest answer is 20%., Temp. of, Temp. of, heat source heat reservoir, 1m, 28. (a), 3, 2, 10 K, 10 K, , Q=, , æ dQ ö kADT, çè ÷ø =, dt, l, , Where, K = coefficient of thermal conductivity, l = length of rod and A = area of cross-section of rod, , 1 æ dQ ö, kD T, Energy flux, çè ÷ø =, A dt, l, , 29., , 100°C, , ( 0.1)( 900), , = 90 W/m 2, 1, (c) Let specific heat of unknown metal be ‘s’ According, to principle of calorimetry, Heat lost, = Heat gain m × sDq = m1sbrass (Dq1 + m2 swater + Dq2), Þ 192 × S × (100 – 21.5), = 128 × 394 × (21.5 – 8.4), Solving we get,+ 240 × 4200 × (21.5 – 8.4), S = 916 Jkg–1k–1, =, , 30. (a), , Copper, B, , 120, , R/2, , R/4, , 32., , 33., , Brass, 0°C, , 0.92 ´ 4(100 - T ), 46, 0.26 ´ 4 ´ (T - 0), 0.12 ´ 4 ´ (T - 0), +, 13, 12, Þ 200 – 2T = 2T + T, Þ T = 40°C, , =, , R, , P, , R/4, , L/4, Q, , R/2, , B, , 120 ´ 5 3, 360, DTPQ =, = 45°C, × R=, 8R, 5, 8, (d) According to principle of calorimetry,, Heat lost = Heat gain, 100 × 0.1(T – 75) = 100 × 0.1 × 45 + 170 × 1 × 45, 10 T– 750 = 450 + 7650 = 8100, Þ T – 75 = 810, T = 885°C, (d) According to principle of calorimetry,, Qgiven = Qused, 0.2 × S × (150 – 40) = 150 × 1 × (40 – 27) + 25 × (40 – 27), 0.2 × S × 110 = 150 × 13 + 25 × 13, Specific heat of aluminium, 13 ´ 25 ´ 7, S=, = 434 J/kg-°C, 0.2 ´ 110, (b) As Pt = mCDT, So, P × 10 × 60 = mC 100 ...(i), , 0.92 ´ 4 ´ 60, = 4.8 cal/s, 46, (c) In the given problem, fall in temperature of sphere,, , \, , O, , In steady state temperature difference between P and, Q,, , 31., , Steel, , If the junction temperature is T, then, QCopper = QBrass + QSteel, , L, L/4, , T, , 0°C, , DTAB 120 120 ´ 5, =, =, 8, R AB, 8R, R, 5, , A, , KA(q1 - q 2 ), l, , 35., , QCopper =, , dT = ( 3T0 - 2T0 ) = T0, Temperature of surrounding, Tsurr = T0, Initial temperature of sphere, Tinitial = 3T0, Specific heat of the material of the sphere varies as,, c = aT 3 per unit mass (a = a constant), Applying formula,, , (, , dT sA 4, 4, =, T - Tsurr, dt McJ, , Þ, , ), , T0, s 4pR 2 é, =, ( 3T )4 - ( T0 )4 ùûú, dt Ma ( 3T )3 J ëê 0, 0, , Þ dt =, , Ma 27T04 J, , s 4 pR 2 ´ 80T04, , Solving we get,, Time taken for the sphere to cool down temperature 2T 0,, , t=, , Ma, , æ 16 ö, ln ç ÷, 16pR s è 3 ø, 2
Page 174 :
P-165, , Thermal Properties of Matter, , 36. (a) From question,, In 1 sec heat gained by water, = 1 KW – 160 J/s, = 1000 J/s – 160 J/s, = 840 J/s, Total heat required to raise the temperature of water (volume 2L) from 27°c to 77°c, = mwater ×sp. ht × Dq, = 2 × 103 × 4.2 × 50 [Q mass = density × volume], And, 840 × t = 2 × 103 × 4.2 × 50, 2 ´103 ´ 4.2 ´ 50, 840, = 500 s = 8 min 20s, 37. (d) When radius is decrease by DR,, , or, t =, , 4pR 2 DRrL = 4pT[R 2 - (R - DR) 2 ], Þ rR2 DRL = T[R 2 - R 2 + 2RDR - DR 2 ], Þ rR 2 DRL = T2RDR, , [ DR is very small], , 2T, rL, 38. (d) As the surrounding is identical, vessel is identical, time taken to cool both water and liquid (from 30°C to 25°C), is same 2 minutes, therefore, ÞR=, , æ dQ ö, æ dQ ö, =ç ÷, çè ÷ø, dt water è dt ø liquid, (mw C w + W)DT (ml Cl + W) D T, =, t, t, (W = water equivalent of the vessel), or , m w C w = m l C l, , =, , L, L, + k ´ 0´, 2, 2, L, L, 9k ´ + k ´, 2, 2, , 9k ´ 100 ´, , =, , 900, kL, = 2, = 90°C, 10kL, 2, 42. (a) According to Stefan’s law, E = s T4, Heat radiated per unit area in 1 hour (3600s) is, = 5.7× 10–8 × (300)4 × 3600 = 1.7 × 1010 J, 43. (a) DU = DQ = mcDT, 100, × 4184 (50 – 30) » 8.4 kJ, 1000, 44. (d) Required work = energy released, , =, , ò, , Here, Q = mc dT, 4, , æ T3 ö, = ò 0.1 ´ 32 ´ ç, dT =, 3÷, è, 400, ø, 20, 4, , òT, , 3, , 4, , 3.2, , ò 64 ´ 106 T, , 3, , dT, , 20, , dT = 0.002kJ, , Therefore, required work = 0.002 kJ, 45. (a) Let Q be the temperature at a distance x from hot end, of bar. Let Q is the temperature of hot end., The heat flow rate is given by, , mWCW, ml, , dQ kA(q1 - q), =, dt, x, , 39. (b) As 1g of steam at 100°C melts 8g of ice at 0°C., 10 g of steam will melt 8× 10 g of ice at 0°C, Water in calorimeter = 500 + 80 + 10g = 590g, 40. (c), 41. (a), L, Copper, , K copper lsteel + Ksteel lcopper, , 20, , 50 ´ 1, = 0.5 kcal / kg, 100, , 100°C, , K copper qcopper lsteel + K steel qsteel lcopper, , q=, , = 5 ´ 10 –8, , or,, , \ Specific heat of liquid , Cl =, , From formula temperature of junction;, , Steel, , 0°C, , L/2, L/2, Let conductivity of steel Ksteel = k then from question, Conductivity of copper Kcopper = 9k, qcopper = 100°C, qsteel = 0°C, L, lsteel = lcopper =, 2, , x dQ, kA dt, Thus, the graph of Q versus x is a straight line with a positive, intercept and a negative slope., The above equation can be graphically represented by, option (a)., 46. (d) Let T be the temperature of the interface., In the steady state, Q1 = Q2, Þ q - q = x dQ, 1, kA dt, , T1, , \, , Þ q = q1 -, , 1, , 2, , K1, , K2, , T2, , K1 A(T1 - T ) K 2 A(T - T2 ), ,, =, l1, l2
Page 177 :
P-168, , Physics, , 62. (b) From Newton’s law of cooling,, t=, , 63., , æ q - q0 ö, 1, log e ç 2, ÷, k, è q1 - q0 ø, , 64., , 1, (40 - 30), log e, k, (80 - 30), , And, t =, , 1, (32 - 30), log e, (62 - 30), k, , ...(1), , Þ, , ...(2), , Þ, , Dividing equation (2) by (1),, 1, log e, t k, =, 5 1, log e, k, , (a) According to newton's law of cooling, dq, = - k(q - q0 ), dt, , From question and above equation,, 5=, , (c) According to Newton’s law of cooling, the, temperature goes on decreasing with time non-linearly., , (32 - 30), (62 - 30), (40 - 30), (80 - 30), , On solving we get, time taken to cool down from 62°C, to 32°C, t = 8.6 minutes., , dq, = - kdt, (q - q0 ), q, , ò, , q0, , t, , dq, = - k ò dt, (q - q 0 ), q, , Þ, , log(q - q0 ) = -kt + c, Which represents an equation of straight line., Thus the option (a) is correct., 65. (d) From Newton’s law of cooling -, , dQ, µ (Dq), dt
Page 178 :
11, , P-169, , Thermodynamics, , Thermodynamics, TOPIC 1 First Law of Thermodynamics, 1., , A gas can be taken from A to B via two different processes, ACB and ADB., , (c), 5., , 6., , 2., , 3., , 4., , When path ACB is used 60 J of heat flows into the system, and 30J of work is done by the system. If path ADB is, used work done by the system is 10 J. The heat Flow into, the system in path ADB is :, [9 Jan. 2019 I], (a) 40 J, (b) 80 J (c) 100 J, (d), 20 J, 200g water is heated from 40°C to 60°C. Ignoring the slight, expansion of water, the change in its internal energy is, close to (Given specific heat of water = 4184 J/kgK):, [Online April 9, 2016], (a) 167.4 kJ (b) 8.4 kJ (c) 4.2 kJ (d) 16.7 kJ, A gas is compressed from a volume of 2m3 to a volume of, 1m3 at a constant pressure of 100 N/m2. Then it is heated at, constant volume by supplying 150 J of energy. As a result,, the internal energy of the gas:, [Online April 19, 2014], (a) increases by 250 J, (b) decreases by 250 J, (c) increases by 50 J, (d) decreases by 50 J, An insulated container of gas has two chambers separated, by an insulating partition. One of the chambers has volume, V1 and contains ideal gas at pressure P1 and temperature, T1. The other chamber has volume V2 and contains, ideal gas at pressure P2 and temperature T2. If the partition, is removed without doing any work on the gas, the, final equilibrium temperature of the gas in the container, will be, [2008], (a), , T1T2 ( PV, 1 1 + P2V2 ), PV, 1 1T2 + P2V2T1, , (b), , PV, 1 1T1 + P2V2T2, PV, 1 1 + P2V2, , PV, 1 1T2 + P2V2T1, PV, 1 1 + P2V2, , (d), , T1T2 ( PV, 1 1 + P2V2 ), PV, T, 1 1 1 + P2V2T2, , When a system is taken from state i to state f along the, path iaf, it is found that Q =50 cal and W = 20 cal. Along the, path ibf Q = 36 cal. W along the path ibf is, [2007], a, f, , i, b, (a) 14 cal (b) 6 cal, (c) 16 cal (d) 66 cal, A system goes from A to B via two processes I and II as, shown in figure. If DU1 and DU2 are the changes in internal, energies in the processes I and II respectively, then [2005], p, II, A, , B, , I, v, , (a) relation between DU1 and DU 2 can not be determined, , 7., , (b), , DU1 = DU 2, , (c), , DU 2 < DU1, , (d) DU 2 > DU1, Which of the following is incorrect regarding the first law, of thermodynamics?, [2005], (a) It is a restatement of the principle of conservation of, energy, (b) It is not applicable to any cyclic process, (c) It does not introduces the concept of the entropy, (d) It introduces the concept of the internal energy, , TOPIC 2, 8., , Specific Heat Capacity and, Thermodynamical Processes, , Three different processes that can occur in an ideal, monoatomic gas are shown in the P vs V diagram. The, paths are lebelled as A ® B, A ® C and A ® D. The change
Page 179 :
P-170, , Physics, , in internal energies during these process are taken as EAB,, EAC and EAD and the workdone as WAB, WAC and WAD., The correct relation between these parameters are :, [5 Sep. 2020 (I)], D T1>T2, C, P, B, T1, A, T2, V, , 9., , (a) EAB = EAC < EAD, WAB > 0, WAC = 0, WAD < 0, (b) EAB = EAC = EAD, WAB > 0, WAC = 0, WAD > 0, (c) EAB < EAC < EAD, WAB > 0, WAC > WAD, (d) EAB > EAC > EAD, WAB < WAC < WAD, In an adiabatic process, the density of a diatomic gas, becomes 32 times its initial value. The final pressure of the, gas is found to be n times the initial pressure. The value of, n is :, [5 Sep. 2020 (II)], , 1, 32, 10. Match the thermodynamic processes taking place in a, system with the correct conditions. In the table : DQ is the, heat supplied, DW is the work done and DU is change in, internal energy of the system., [4 Sep. 2020 (II)], Process, Condition, (I) Adiabatic, (A) DW = 0, (II) Isothermal, (B) DQ = 0, , (a) 32, , (III) Isochoric, , (b) 326, , (c) 128, , (d), , (a), , (b), , (c), , (d), , 14. Starting at temperature 300 K, one mole of an ideal diatomic, gas (g = 1.4) is first compressed adiabatically from volume, V1, . It is then allowed to expand isobarically to, 16, volume 2V2. If all the processes are the quasi-static then, the final temperature of the gas (in °K) is (to the nearest, integer) ______., [9 Jan. 2020 II], 15. A thermodynamic cycle xyzx is shown on a V-T diagram., , V1 to V2 =, , (C) DU ¹ 0, DW ¹ 0,, , DQ ¹ 0, (IV) Isobaric, (D) DU = 0, (a) (I)-(A), (II)-(B), (III)-(D), (IV)-(D), (b) (I)-(B), (II)-(A), (III)-(D), (IV)-(C), (c) (I)-(A), (II)-(A), (III)-(B), (IV)-(C), (d) (I)-(B), (II)-(D), (III)-(A), (IV)-(C), 11. A balloon filled with helium (32°C and 1.7 atm.) bursts., Immediately afterwards the expansion of helium can be, considered as :, [3 Sep. 2020 (I)], (a) irreversible isothermal (b) irreversible adiabatic, (c) reversible adiabatic, (d) reversible isotherm7al, 12. An engine takes in 5 mole of air at 20°C and 1 atm, and, compresses it adiabaticaly to 1/10th of the original, volume. Assuming air to be a diatomic ideal gas made up, of rigid molecules, the change in its internal energy during, this process comes out to be X kJ. The value of X to the, nearest integer is ________., [NA 2 Sep. 2020 (I)], 13. Which of the following is an equivalent cyclic process, corresponding to the thermodynamic cyclic given in the, figure?, where, 1 ® 2 is adiabatic., (Graphs are schematic and are not to scale) [9 Jan. 2020 I], , The P-V diagram that best describes this cycle is:, (Diagrams are schematic and not to scale), [8 Jan. 2020 I], , (a), , (b), , (c), , (d), , 16. A litre of dry air at STP expands adiabatically to a volume, of 3 litres. If g = 1.40, the work done by air is:, (31.4 = 4.6555) [Take air to be an ideal gas], [7 Jan. 2020 I], (a) 60.7 J (b) 90.5 J (c) 100.8 J (d) 48 J
Page 180 :
P-171, , Thermodynamics, , 17. Under an adiabatic process, the volume of an ideal gas gets, doubled. Consequently the mean collision time between, the gas molecule changes from t1 to t2. If, , Cp, Cv, , = g for this, , t2, gas then a good estimate for t is given by:, 1, [7 Jan. 2020 I], , (a) 2, , (b), , 1, 2, , (c), , æ 1ö, çè ÷ø, 2, , g, , (d), , æ 1ö, çè ÷ø, 2, , g +1, 2, , 18. A sample of an ideal gas is taken through the cyclic process, abca as shown in the figure. The change in the internal, energy of the gas along the path ca is – 180 J, The gas, absorbs 250 J of heat along the path ab and 60 J along the, path bc. The work down by the gas along the path abc is:, [12 Apr. 2019 I], , (a) 120 J, (b) 130 J (c) 100 J, (d) 140 J, 19. A cylinder with fixed capacity of 67.2 lit contains helium, gas at STP. The amount of heat needed to raise the, temperature of the gas by 20oC is : [Given that R = 8.31 J, mol – 1 K – 1], [10 Apr. 2019 I], (a) 350 J, (b) 374 J (c) 748 J, (d) 700 J, , (a) DQA < DQB, DUA < DUB, (b) DQA > DQB, DUA > DUB, (c) DQA > DQB, DUA = DUB, (d) DQA = DQB; DUA = DUB, 23. A thermally insulted vessel contains 150 g of water at 0°C., Then the air from the vessel is pumped out adiabatically. A, fraction of water turns into ice and the rest evaporates at, 0°C itself. The mass of evaporated water will be closed to:, (Latent heat of vaporization of water = 2.10 × 106 J kg–1 and, Latent heat of Fusion of water = 3.36 × 10 5 J kg–1), [8 April 2019 I], (a) 150 g, (b) 20 g, (c) 130 g, (d), 35 g, 24. The given diagram shows four processes i.e., isochoric,, isobaric, isothermal and adiabatic. The correct assignment, of the processes, in the same order is given by :, [8 Apr. 2019 II], , (a) a d b c, (b) d a c b (c) a d c b, (d) d a b c, 25. For the given cyclic process CAB as shown for gas, the, work done is:, [12 Jan. 2019 I], 6.0, 5, , 20. n moles of an ideal gas with constant volume heat capacity, CV undergo an isobaric expansion by certain volume. The, ratio of the work done in the process, to the heat supplied, is:, [10 Apr. 2019 I], nR, CV + nR, , nR, (b) C - nR, V, , 4nR, (c), CV - nR, , 4nR, (d) C + nR, V, , (a), , 21. One mole of an ideal gas passes through a process where, é 1 æ V ö2 ù, P, =, P, ê1 - ç 0 ÷ ú ., 0, pressure and volume obey the relation, êë 2 è V ø úû, , Here Po and Vo are constants. Calculate the charge in the, temperature of the gas if its volume changes from Vo to 2Vo., [10 Apr. 2019 II], 1 Po Vo, 5 Po Vo, 3 Po Vo, 1 Po Vo, (b), (c), (d), 2 R, 4 R, 4 R, 4 R, 22. Following figure shows two processes A and B for a gas., If DQA and DQB are the amount of heat absorbed by the, system in two cases, and DUA and DUB are changes in, internal energies, respectively, then: [9 April 2019 I], , (a), , A, , C, , 4, p(Pa) 3, 2, B, , 1, 1, , 2, , 3, , 4, , 5, V(m3), , (a) 30 J, (b) 10 J, (c) 1 J, (d) 5 J, 26. A rigid diatomic ideal gas undergoes an adiabatic process, at room temperature. The relation between temperature and, volume for this process is TVx = constant, then x is:, [11 Jan. 2019 I], 3, 2, 2, 5, (b), (c), (d), 5, 5, 3, 3, 27. Half mole of an ideal monoatomic gas is heated at constant, pressure of 1 atm from 20°C to 90°C. Work done by gas is, close to: (Gas constant R = 8.31 J/mol-K) [10 Jan. 2019 II], (a) 581 J, (b) 291 J (c) 146 J, (d) 73 J, 28. One mole of an ideal monoatomic gas is taken along the, path ABCA as shown in the PV diagram. The maximum, temperature attained by the gas along the path BC is given, by, [Online April 16, 2018], , (a)
Page 181 :
P-172, , Physics, V, , P, 3P0, , b, C, , P0 A, V0, , 2V0, , P, , P, d, , (a), , 9P0 V0, 9P0 V0, 9P0 V0, 3P0 V0, (b), (c), (d), 2nR, nR, 4nR, 2nR, 31. The ratio of work done by an ideal monoatomic gas to the, heat supplied to it in an isobaric process is :, [Online April 9, 2016], , (a), , 2, 3, 3, 2, (a), (b), (c), (d), 5, 2, 5, 3, 32. Consider an ideal gas confined in an isolated closed, chamber. As the gas undergoes an adiabatic expansion,, the average time of collision between molecules increases, as Vq, where V is the volume of the gas. The value of q, , Cp ö, æ, is : ç g =, [2015], ÷, Cv ø, è, 3g + 5, 3g - 5, g +1, g -1, (a), (b), (c), (d), 6, 6, 2, 2, 33. Consider a spherical shell of radius R at temperature T. The, black body radiation inside it can be considered as an ideal, gas of photons with internal energy per unit volume u =, U, 1æUö, µ T 4 and pressure p = ç ÷ . If the shell now, V, 3è V ø, undergoes an adiabatic expansion the relation between T, and R is :, [2015], , Tµ, , 1, , (c) T µ e–R (d) T µ e–3R, R3, 34. An ideal gas goes through a reversible cycle a®b®c®d, has the V - T diagram shown below. Process d®a and, b®c are adiabatic., , a, , c, a, , (b), , b, , b, c, , d, , V, , V, P, , P, d, , c, , a, , b, , B, V0 2V0 V, , (b), , T, , (c), , P0, , a, , The corresponding P - V diagram for the process is (all, figures are schematic and not drawn to scale) :, [Online April 10, 2015], , A, , 2P0, , 1, R, , d, , V, , (a) 25 P0 V0 (b) 25 P0 V0 (c) 25 P0 V0 (d) 5 P0V0, 8 R, 4 R, 16 R, 8 R, 29. One mole of an ideal monoatomic gas is compressed, isothermally in a rigid vessel to double its pressure at room, temperature, 27°C. The work done on the gas will be:, [Online April 15, 2018], (a) 300R ln 6, (b) 300R, (c) 300R ln 7, (d) 300R ln 2, 30. 'n' moles of an ideal gas undergoes a process A ® B as, shown in the figure. The maximum temperature of the gas, during the process will be :, [2016], P, , (a) T µ, , c, , B, , (d), V, , a, , b, d, , c, V, , 35. One mole of a diatomic ideal gas undergoes a cyclic process, ABC as shown in figure. The process BC is adiabatic. The, temperatures at A, B and C are 400 K, 800 K and 600 K, respectively. Choose the correct statement:, [2014], , B, , 800 K, , P, , A, , 400 K, , 600 k, C, , V, (a) The change in internal energy in whole cyclic process, is 250 R., (b) The change in internal energy in the process CA is 700 R., (c) The change in internal energy in the process AB is 350 R., (d) The change in internal energy in the process BC is –, 500 R., 36. An ideal monoatomic gas is confined in a cylinder by a spring, loaded piston of cross section 8.0 × 10–3 m2. Initially the gas, is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the, spring is in its relaxed state as shown in figure. The gas is, heated by a small heater until the piston moves out slowly, by 0.1 m. The force constant of the spring is 8000 N/m and, the atmospheric pressure is 1.0 × 105 N/m2. The cylinder and
Page 182 :
P-173, , Thermodynamics, , the piston are thermally insulated. The piston and the spring, are massless and there is no friction between the piston and, the cylinder. The final temperature of the gas will be:, (Neglect the heat loss through the lead wires of the heater., The heat capacity of the heater coil is also negligible)., [Online April 11, 2014], , 41. Helium gas goes through a cycle ABCDA (consisting of, two isochoric and isobaric lines) as shown in figure. The, efficiency of this cycle is nearly : (Assume the gas to be, close to ideal gas), [2012], B, , 2P0, , (a) 15.4 %, (b) 9.1 %, , P0, , Po To R, (a), Po - a, , Po To R, (b), Po + a, , (c) PoToRIn 2, (d) PoToR, 39. A certain amount of gas is taken through a cyclic process, (A B C D A) that has two isobars, one isochore and one, isothermal. The cycle can be represented on a P-V indicator, diagram as :, [Online April 22, 2013], P, , (a), , A, , B, , C, , B, , P, , D, , B, , A, , A, , ), , (, , B, , C, , (d), A, , (b), , ), , D, , V, V, , 40. An ideal gas at atmospheric pressure is adiabatically, compressed so that its density becomes 32 times of its, initial value. If the final pressure of gas is 128 atmospheres,, the value of ‘g’of the gas is :, [Online April 22, 2013], (a) 1.5, (b) 1.4, (c) 1.3, (d) 1.6, , ), , (, , a 2V 2 2, m -1, 2, , (, , ), , a 2, aV 2 2, m -1, (d), m -1, 2, 2, 44. n moles of an ideal gas undergo a process A ® B as shown, in the figure. Maximum temperature of the gas during the, process is, [Online May 12, 2012], , (c), , A, , 2P0, , D, , C, , D, , (, , aV 2, m -1, 2, , B, , P0, , V, , P, , (c), , (a), , C, , V, , P, , (d) 12.5 %, 2V0, V0, 42. An ideal monatomic gas with pressure P, volume V and, temperature T is expanded isothermally to a volume 2V, and a final pressure Pi. If the same gas is expanded, adiabatically to a volume 2V, the final pressure is Pa. The, P, ratio a is, [Online May 26, 2012], Pi, (a) 2–1/3, (b) 21/3, (c) 22/3, (d) 2–2/3, 43. The pressure of an ideal gas varies with volume as P = aV,, where a is a constant. One mole of the gas is allowed to, undergo expansion such that its volume becomes ‘m’ times, its initial volume. The work done by the gas in the process, is, [Online May 19, 2012], , P, , (b), , D, , A, , (c) 10.5%, , (a) 300 K, (b) 800 K (c) 500 K, (d) 1000 K, 37. During an adiabatic compression, 830 J of work is done on, 2 moles of a diatomic ideal gas to reduce its volume by, 50%. The change in its temperature is nearly:, (R = 8.3 JK–1 mol–1), [Online April 11, 2014], (a) 40 K, (b) 33 K (c) 20 K, (d) 14 K, 38. The equation of state for a gas is given by PV = nRT + aV,, where n is the number of moles and a is a positive constant., The initial temperature and pressure of one mole of the gas, contained in a cylinder are To and Po respectively. The, work done by the gas when its temperature doubles, isobarically will be:, [Online April 9, 2014], , C, , V0 2V, 0, , 3P0V0, 9 P0V0, 9 P0V0, (b), (c), (d), nR, 2 nR, 2 nR, 4 nR, 45. This question has Statement 1 and Statement 2. Of the four, choices given after the Statements, choose the one that, best describes the two Statements., Statement 1: In an adiabatic process, change in internal, energy of a gas is equal to work done on/by the gas in the, process., Statement 2: The temperature of a gas remains constant, in an adiabatic process., [Online May 7, 2012], (a) Statement 1 is true, Statement 2 is true, Statement 2 is a, correct explanation of Statement 1., (b) Statement 1 is true, Statement 2 is false., , (a), , 9P0V0, , V
Page 183 :
P-174, , Physics, , (c) Statement 1 is false, Statement 2 is true., (d) Statement 1 is false, Statement 2 is true, Statement 2 is, not a correct explanation of Statement 1., 46. A container with insulating walls is divided into equal parts, by a partition fitted with a valve. One part is filled, with an ideal gas at a pressure P and temperature T, whereas, the other part is completly evacuated. If the valve is, suddenly opened, the pressure and temperature of the gas, will be :, [2011 RS], T, P, P T, ,T, (b) P, T, (c) P,, (d), ,, 2, 2, 2 2, Directions for questions 47 to 49: Questions are based on, the following paragraph., Two moles of helium gas are taken over the cycle ABCDA, as, shown in the P-T diagram., [2009], , (a), , 5, , 2 × 10, , A, , B, , D, , C, , 300K, , 500K, , P (Pa), 1 × 10, , 5, , T, , 47. Assuming the gas to be ideal the work done on the gas in, taking it from A to B is, (a) 300 R, (b) 400 R (c) 500 R, (d) 200 R, 48. The work done on the gas in taking it from D to A is, (a) + 414 R (b) – 690 R (c) + 690 R, (d) – 414 R, 49. The net work done on the gas in the cycle ABCDA is, (a) 279 R, (b) 1076 R (c) 1904 R, (d) zero, 50. The work of 146 kJ is performed in order to compress one, kilo mole of gas adiabatically and in this process the, temperature of the gas increases by 7°C. The gas is [2006], (R = 8.3 J mol–1 K–1), (a) diatomic, (b) triatomic, (c) a mixture of monoatomic and diatomic, (d) monoatomic, 51. Which of the following parameters does not characterize, the thermodynamic state of matter?, [2003], (a) Temperature, (b) Pressure, (c) Work, (d) Volume, , Carnot Engine, Refrigerators, TOPIC 3 and Second Law of, Thermodynamics, 52. An engine operates by taking a monatomic ideal gas through, the cycle shown in the figure. The percentage efficiency of, the engine is close is ______., [NA 6 Sep. 2020 (II)], 3 PO, , B, , C, , A, , D, , 2 PO, PO, , VO, , 2VO, , 53. If minimum possible work is done by a refrigerator in, converting 100 grams of water at 0°C to ice, how much heat, (in calories) is released to the surroundings at temperature, 27°C (Latent heat of ice = 80 Cal/gram) to the nearest integer?, [NA 3 Sep. 2020 (II)], 54. A heat engine is involved with exchange of heat of 1915 J,, – 40 J, +125 J and – Q J, during one cycle achieving an, efficiency of 50.0%. The value of Q is :, [2 Sep. 2020 (II)], (a) 640 J, (b) 40 J, (c) 980 J, (d) 400 J, 1, 55. A Carnot engine having an efficiency of, is being used, 10, as a refrigerator. If the work done on the refrigerator is 10, J, the amount of heat absorbed from the reservoir at lower, temperature is:, [8 Jan. 2020 II], (a) 99 J, (b) 100 J (c) 1 J, (d) 90 J, 56. A Carnot engine operates between two reservoirs of, temperatures 900 K and 300 K. The engine performs 1200 J, of work per cycle. The heat energy (in J) delivered by the, engine to the low temperature reservoir, in a cycle, is, _______., [NA 7 Jan. 2020 I], 57. Two ideal Carnot engines operate in cascade (all heat, given up by one engine is used by the other engine to, produce work) between temperatures, T1 and T2. The, temperature of the hot reservoir of the first engine is T1, and the temperature of the cold reservoir of the second, engine is T2. T is temperature of the sink of first engine, which is also the source for the second engine. How is, T related to T1 and T2, if both the engines perform equal, amount of work ?, [7 Jan. 2020 II], 2T1T2, T1 + T2, (a) T = T + T, (b) T =, 2, 1, 2, (c) T = T1T2, , (d) T = 0, , 58. A Carnot engine has an efficiency of 1/6. When the, temperature of the sink is reduced by 62oC, its efficiency, is doubled. The temperatures of the source and the sink, are, respectively., [12 Apr. 2019 II], o, o, o, (a) 62 C, 124 C, (b) 99 C, 37oC, o, o, (c) 124 C, 62 C, (d) 37oC, 99oC, 59. Three Carnot engines operate in series between a heat, source at a temperature T1 and a heat sink at temperature, T4 (see figure). There are two other reservoirs at temperature, T2 and T3, as shown, with T1 > T2 > T3 > T(4) The three, engines are equally efficient if:, [10 Jan. 2019 I]
Page 184 :
P-175, , Thermodynamics, , ( ), = (T T ) ; T = (T T ), = (T T ) ; T = (T T ), = ( T T ) ; T = (T T ), , (a) T2 = ( T1 T4 ), , 1/2, , (b) T2, (c) T2, (d) T2, , 2, 1 4, , 2, 1 4, , 3, 1 4, , ; T3 = T12 T4, , 1/3, , 3, , 2, 1 4, , 3, , 2, 1, , 4, , 3, , 1, , 3, 4, , 1/3, , 1/3, , (ii) Sequentially keeping in contact with 8 reservoirs such, that each reservoir supplies same amount of heat., , 1/3, , In both the cases body is brought from initial temperature, 100°C to final temperature 200°C. Entropy change of the, body in the two cases respectively is :, , 1/3, , 1/4, , 1/4, , 60. Two Carnot engines A and B are operated in series. The, first one, A receives heat at T1 (= 600 K) and rejects to a, reservoir at temperature T2. The second engine B receives, heat rejected by the first engine and in turn, rejects to a, heat reservoir at T3 (= 400 K). Calculate the temperature, T2 if the work outputs of the two engines are equal:, [9 Jan. 2019 II], (a) 600 K, (b) 400 K (c) 300 K, (d) 500 K, 61. A Carnot's engine works as a refrigerator between 250 K, and 300 K. It receives 500 cal heat from the reservoir at the, lower temperature.The amount of work done in each cycle, to operate the refrigerator is: [Online April 15, 2018], (a) 420 J, (b) 2100 J (c) 772 J, (d) 2520 J, 62. Two Carnot engines A and B are operated in series. Engine, A receives heat from a reservoir at 600K and rejects heat to, a reservoir at temperature T. Engine B receives heat rejected, by engine A and in turn rejects it to a reservoir at 100K. If, the efficiencies of the two engines A and B are represented, h, by hA and hB respectively, then what is the value of A, hB, [Online April 15, 2018], 12, 12, 5, 7, (b), (c), (d), 7, 5, 12, 12, 63. An engine operates by taking n moles of an ideal gas, through the cycle ABCDA shown in figure. The thermal, efficiency of the engine is : (Take Cv =1.5 R, where R is gas, constant), [Online April 8, 2017], , (a), , 2P0, , (a) 0.24, (b) 0.15, , B, , C, , P, P0, , A, , D, , (c) 0.32, V0, , 2V0, , V, (d) 0.08, 64. A Carnot freezer takes heat from water at 0°C inside it and, rejects it to the room at a temperature of 27°C. The latent, heat of ice is 336 × 103 J kg–1. If 5 kg of water at 0°C is, converted into ice at 0°C by the freezer, then the energy, consumed by the freezer is close to :, [Online April 10, 2016], (a) 1.51 × 105 J, (b) 1.68 × 106 J, (c) 1.71 × 107 J, (d) 1.67 × 105 J, 65. A solid body of constant heat capacity 1 J/°C is being heated, by keeping it in contact with reservoirs in two ways : [2015], (i) Sequentially keeping in contact with 2 reservoirs such, that each reservoir supplies same amount of heat., , (a) ln2, 2ln2, (b) 2ln2, 8ln2, (c) ln2, 4ln2, (d) ln2, ln2, 66. A Carnot engine absorbs 1000 J of heat energy from a, reservoir at 127°C and rejects 600 J of heat energy during, each cycle. The efficiency of engine and temperature of, sink will be:, [Online April 12, 2014], (a) 20% and – 43°C, (b) 40% and – 33°C, (c) 50% and – 20°C, (d) 70% and – 10°C, 67., , p, 2p0, p0, , A, D, v0, , B, C, 2v0 v, , The above p-v diagramrepresents the thermodynamic cycle, of an engine, operating with an ideal monatomic gas. The, amount of heat, extracted from the source in a single cycle, is, [2013], (a) p 0 v 0, , æ 13 ö, (b) ç ÷ p0 v0, è2ø, , æ 11 ö, (c) ç ÷ p0 v0, è2ø, , (d) 4p0v0, , 68. A Carnot engine, whose efficiency is 40%, takes in heat, from a source maintained at a temperature of 500K. It is, desired to have an engine of efficiency 60%. Then, the, intake temperature for the same exhaust (sink) temperature, must be :, [2012], (a) efficiency of Carnot engine cannot be made larger than 50%, (b) 1200 K, (c) 750 K, (d) 600 K, 69. The door of a working refrigerator is left open in a, well insulated room. The temperature of air in the room, will, [Online May 26, 2012], (a) decrease, (b) increase in winters and decrease in summers, (c) remain the same, (d) increase, 70. This question has Statement 1 and Statement 2. Of the four, choices given after the Statements, choose the one that, best describes the two Statements., Statement 1: An inventor claims to have constructed an, engine that has an efficiency of 30% when operated, between the boiling and freezing points of water. This is, not possible.
Page 185 :
P-176, , Physics, , Statement 2: The efficiency of a real engine is always, less than the efficiency of a Carnot engine operating, between the same two temperatures., [Online May 19, 2012], (a) Statement 1 is true, Statement 2 is true, Statement 2 is, not the correct explanation of Statement 1., (b) Statement 1 is true, Statement 2 is false., (c) Statement 1 is false, Statement 2 is true., (d) Statement 1 is true, Statement 2 is true, Statement 2 is, the correct explanation of Statement 1., 71. A Carnot engine operating between temperatures T1 and T2, has efficiency, increases to, , 75., , 76., , 1, . When T2 is lowered by 62 K its efficiency, 6, , 1, . Then T1 and T2 are, respectively:, 3, , [2011], , (a) 372 K and 310 K, (b) 330 K and 268 K, (c) 310 K and 248 K, (d) 372 K and 310 K, 72. A diatomic ideal gas is used in a Carnot engine as the, working substance. If during the adiabatic expansion part, of the cycle the volume of the gas increases from V to 32, V, the efficiency of the engine is, [2010], (a) 0.5, (b) 0.75, (c) 0.99, (d) 0.25, 73. A Carnot engine, having an efficiency of h = 1/10 as heat, engine, is used as a refrigerator. If the work done on the, system is 10 J, the amount of energy absorbed from the, reservoir at lower temperature is, [2007], (a) 100 J, (b) 99 J, (c) 90 J, (d) 1 J, 74. The temperature-entropy diagram of a reversible engine, cycle is given in the figure. Its efficiency is, [2005], T, , 2T0, T0, S0, , 2S0, , S, , 1, 2, 1, 1, (b), (c), (d), 3, 2, 4, 3, Which of the following statements is correct for any, thermodynamic system ?, [2004], (a) The change in entropy can never be zero, (b) Internal energy and entropy are state functions, (c) The internal energy changes in all processes, (d) The work done in an adiabatic process is always zero., “Heat cannot by itself flow from a body at lower temperature, to a body at higher temperature” is a statement or, consequence of, [2003], (a) second law of thermodynamics, (b) conservation of momentum, (c) conservation of mass, (d) first law of thermodynamics, A Carnot engine takes 3 × 106 cal of heat from a reservoir at, 627°C, and gives it to a sink at 27°C. The work done by the, engine is, [2003], (a) 4.2 × 106 J, (b) 8.4 × 106 J, (c) 16.8 × 106 J, (d) zero, Which statement is incorrect?, [2002], (a) All reversible cycles have same efficiency, (b) Reversible cycle has more efficiency than an, irreversible one, (c) Carnot cycle is a reversible one, (d) Carnot cycle has the maximum efficiency in all cycles, Even Carnot engine cannot give 100% efficiency because, we cannot, [2002], (a) prevent radiation, (b) find ideal sources, (c) reach absolute zero temperature, (d) eliminate friction, , (a), , 77., , 78., , 79.
Page 194 :
P-185, , Thermodynamics, , Q2, TS, 1, = 1- 0 0 =, 3, Q1, 3, T S, 2 0 0, 75. (b) Internal energy and entropy are state function, they are, independent of path taken., 76. (a) This is a consequence of second law of, thermodynamics, 77. (b) Here, T1 = 627 + 273 = 900 K, T2 = 27 + 273 = 300 K, , = 1-, , Efficiency, h = 1 = 1-, , T2, T1, , 300, 1 2, = 1- =, 900, 3 3, , But h =, , W, Q, , 2, 2, W 2, = Þ W = ´ Q = ´ 3 ´ 106, Q 3, 3, 3, = 2 × 106 cal, = 2 × 106 × 4.2 J = 8.4 × 106 J, 78. (a) All reversible engines have same efficiencies if they, are working for the same temperature of source and sink., If the temperatures are different, the efficiency is, different., 79. (c) In Carnot’s cycle we assume frictionless piston,, absolute insulation and ideal source and sink (reservoirs)., \, , The efficiency of carnot’s cycle h = 1 -, , T2, T1, , The efficiency of carnot engine will be 100% when its, sink (T2) is at 0 K., The temperature of 0 K (absolute zero) cannot be realised, in practice so, efficiency is never 100%.
Page 195 :
12, , P-186, , Physics, , Kinetic Theory, TOPIC 1, 1., , 2., , 3., , 4., , Kinetic Theory of an Ideal, Gas and Gas Laws, , Initially a gas of diatomic molecules is contained in a, cylinder of volume V1 at a pressure P1 and temperature, 250 K. Assuming that 25% of the molecules get, dissociated causing a change in number of moles. The, pressure of the resulting gas at temperature 2000 K,, when contained in a volume 2V1 is given by P2. The ratio, P2/P1 is ______., [NA Sep. 06, 2020 (I)], The change in the magnitude of the volume of an ideal gas, when a small additional pressure DP is applied at a constant, temperature, is the same as the change when the, temperature is reduced by a small quantity DT at constant, pressure. The initial temperature and pressure of the gas, were 300 K and 2 atm. respectively. If | DT |= C | DP | ,, then value of C in (K/atm.) is __________., [NA Sep. 04, 2020 (II)], The number density of molecules of a gas depends on, their distance r from the origin as , n(r) = n0e–ar4. Then, the total number of molecules is proportional to :, [12 April 2019 II], –3/4, (a) n0a, (b) n0 a1/2, (c) n0 a1/4, (d) n0 a–3, A vertical closed cylinder is separated into two parts by a, frictionless piston of mass m and of negligible thickness., The piston is free to move along the length of the cylinder., The length of the cylinder above the piston is l1, and that, below the piston is l2, such that l1 > l2. Each part of the, cylinder contains n moles of an ideal gas at equal temperature, T. If the piston is stationary, its mass, m, will be given by:, (R is universal gas constant and g is the acceleration due to, gravity), [12 Jan. 2019 II], (a), , RT é l1 - 3l2 ù, ê, ú, ng ë l1 I 2 û, , RT é 2l1 + l2 ù, (b) g ê l I ú, ë 1 2 û, , nRT é l1 - l2 ù, nRT é 1 1 ù, ê, ú, (d), ê + ú, g ë l1 l2 û, g ë l2 l1 û, The temperature of an open room of volume 30 m3 increases, from 17°C to 27°C due to sunshine. The atmospheric pressure, in the room remains 1 × 105 Pa. If ni and nf are the number of, molecules in the room before and after heating, then nf – ni, will be :, [2017], (a) 2.5 × 1025, (b) –2.5 × 1025, (c) –1.61 × 1023, (d) 1.38 × 1023, For the P-V diagram given for an ideal gas,, , (c), , 5., , 6., , 1, P, P=, , Constant, V, , 2, V, , out of the following which one correctly represents the, T-P diagram ?, [Online April 9, 2017], 2, 2, T, (a) T, (b), , 1, , 1, P, , P, , T, , T, 2, (c), 7., , 1, , 1, , 2, , (d), , P, Chamber I, ideal, gas, 1, , P, Chamber II, real, gas, 2, , 3, , 4
Page 196 :
P-187, , Kinetic Theory, , 8., , There are two identical chambers, completely thermally, insulated from surroundings. Both chambers have a, partition wall dividing the chambers in two compartments., Compartment 1 is filled with an ideal gas and, Compartment 3 is filled with a real gas. Compartments 2, and 4 are vacuum. A small hole (orifice) is made in the, partition walls and the gases are allowed to expand in, vacuum., Statement-1: No change in the temperature of the gas, takes place when ideal gas expands in vacuum. However,, the temperature of real gas goes down (cooling) when it, expands in vacuum., Statement-2: The internal energy of an ideal gas is only, kinetic. The internal energy of a real gas is kinetic as, well as potential., [Online April 9, 2013], (a) Statement-1 is false and Statement-2 is true., (b) Statement-1 and Statement-2 both are true., Statement-2 is the correct explanation of Statement-1., (c) Statement-1 is true and Statement-2 is false., (d) Statement-1 and Statement-2 both are true., Statement-2 is not correct explanation of Statement-1., Cooking gas containers are kept in a lorry moving with, uniform speed. The temperature of the gas molecules, inside will, [2002], (a) increase, (b) decrease, (c) remain same, (d) decrease for some, while increase for others, , (a) 104 N/m2, (b) 108 N/m2, 3, 2, (c) 10 N/m, (d) 1016 N/m2, 13. The temperature, at which the root mean square velocity, of hydrogen molecules equals their escape velocity from, the earth, is closest to :, [8 April 2019 II], –23, [Boltzmann Constant kB = 1.38 × 10 J/K, Avogadro Number NA = 6.02 × 1026 /kg, Radius of Earth : 6.4 × 106 m, Gravitational acceleration on Earth = 10 ms–2], (a) 800 K, (b) 3 × 105 K, 4, (c) 10 K, (d) 650 K, 14. A mixture of 2 moles of helium gas (atomic mass = 4u), and, 1 mole of argon gas (atomic mass = 40u) is kept at 300 K in, a container. The ratio of their rms speeds, , é Vrms ( helium ) ù, ê, ú is close to :, [9 Jan. 2019 I], ë Vrms ( argon ) û, (a) 3.16, (b) 0.32, (c) 0.45, (d) 2.24, 15. N moles of a diatomic gas in a cylinder are at a temperature, T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get, converted into monoatomic gas. What is the change in, the total kinetic energy of the gas ?, [Online April 9, 2017], 1, nRT, (b) 0, 2, 3, 5, nRT, nRT, (c), (d), 2, 2, 16. In an ideal gas at temperature T, the average force that a, molecule applies on the walls of a closed container, depends on T as Tq. A good estimate for q is:, [Online April 10, 2015], 1, (a), (b) 2, 2, 1, (c) 1, (d), 4, 17. A gas molecule of mass M at the surface of the Earth has, kinetic energy equivalent to 0°C. If it were to go up, straight without colliding with any other molecules, how, high it would rise? Assume that the height attained is much, less than radius of the earth. (kB is Boltzmann constant)., [Online April 19, 2014], 273k B, (a) 0, (b), 2Mg, , (a), , TOPIC 2, , Speed of Gas, Pressure, and Kinetic Energy, , Number of molecules in a volume of 4 cm 3 of a perfect, monoatomic gas at some temperature T and at a pressure, of 2 cm of mercury is close to? (Given, mean kinetic energy of a molecule (at T) is 4 × 10–14 erg, g = 980 cm/s2,, density of mercury = 13.6 g/cm3) [Sep. 05, 2020 (I)], (a) 4.0 × 1018, (b) 4.0 × 1016, 16, (c) 5.8 × 10, (d) 5.8 × 1018, 10. Nitrogen gas is at 300°C temperature. The temperature, (in K) at which the rms speed of a H2 molecule would be, equal to the rms speed of a nitrogen molecule, is, _____________. (Molar mass of N2 gas 28 g);, [NA Sep. 05, 2020 (II)], 11. For a given gas at 1 atm pressure, rms speed of the, molecules is 200 m/s at 127°C. At 2 atm pressure and at, 227°C, the rms speed of the molecules will be:, [9 April 2019 I], (a) 100 m/s, (b) 80 5 m/s, (c) 100 5 m/s, (d) 80 m/s, 12. If 1022 gas molecules each of mass 10–26 kg collide with a, surface (perpendicular to it) elastically per second over, an area 1 m2 with a speed 104 m/s, the pressure exerted, by the gas molecules will be of the order of :, [8 April 2019 I], 9., , 819k B, 546k B, (d), 2Mg, 3Mg, 18. At room temperature a diatomic gas is found to have an, r.m.s. speed of 1930 ms–1. The gas is:, [Online April 12, 2014], (a) H2, (b) Cl2, (c) O2, (d) F2, (c)
Page 197 :
P-188, , Physics, , 19. In the isothermal expansion of 10g of gas from volume, V to 2V the work done by the gas is 575J. What is the, root mean square speed of the molecules of the gas at, that temperature?, [Online April 25, 2013], (a) 398m/s, (b) 520m/s, (c) 499m/s, (d) 532m/s, 20. A perfect gas at 27°C is heated at constant pressure so as, to double its volume. The final temperature of the gas will, be, close to, [Online May 7, 2012], (a) 327°C, (b) 200°C, (c) 54°C, (d) 300°C, 21. A thermally insulated vessel contains an ideal gas of, molecular mass M and ratio of specific heats g. It is, moving with speed v and it's suddenly brought to rest., Assuming no heat is lost to the surroundings, its, temperature increases by:, [2011], (a), , 22., , 23., , 24., , 25., , ( g - 1) Mv 2 K, 2 gR, , The total internal energy, U of a mole of this gas, and the, æ Cp ö, value of g ç =, ÷ are given, respectively, by:, è Cv ø, [Sep. 06, 2020 (I)], (a) U =, , ( g - 1), , ( g - 1), Mv 2 K, 2R, , (a), , n1T1 + n2T2 + n3T3, n1 + n2 + n3, , (b), , (c), , n12T12 + n22T22 + n32T32, n1T1 + n2T2 + n3T3, , (d), , (b) U = 5RT and g =, , 7, 5, , 5, 7, 6, RT and g =, (d) U = 5RT and g =, 2, 5, 5, 27. In a dilute gas at pressure P and temperature T, the mean, time between successive collisions of a molecule varies, with T is :, [Sep. 06, 2020 (II)], , (c) U =, , (a) T, , gM 2v, (b), K, 2R, , (c), , 5, 6, RT and g =, 2, 5, , (b), , 1, T, , 1, (d) T, T, Match the Cp/Cv ratio for ideal gases with different type, of molecules :, [Sep. 04, 2020 (I)], Column-I, Column-II, Molecule Type, Cp/Cv, (A) Monatomic, (I) 7/5, (B) Diatomic rigid molecules (II) 9/7, (C) Diatomic non-rigid molecules(III) 4/3, (D) Triatomic rigid molecules (IV) 5/3, (a) (A)-(IV), (B)-(II), (C)-(I), (D)-(III), (b) (A)-(III), (B)-(IV), (C)-(II), (D)-(I), (c) (A)-(IV), (B)-(I), (C)-(II), (D)-(III), (d) (A)-(II), (B)-(III), (C)-(I), (D)-(IV), A closed vessel contains 0.1 mole of a monatomic ideal, gas at 200 K. If 0.05 mole of the same gas at 400 K is, added to it, the final equilibrium temperature (in K) of, the gas in the vessel will be close to _________., [NA Sep. 04, 2020 (I)], , (c), , 2, , (d) 2( g + 1) R Mv K, Three perfect gases at absolute temperatures T1, T2 and T3, are mixed. The masses of molecules are m1, m2 and m3 and, the number of molecules are n1, n2 and n3 respectively., Assuming no loss of energy, the final temperature of the, mixture is :, [2011], , 28., , n1T12 + n2T22 + n3T32, n1T1 + n2T2 + n3T3, , (T1 + T2 + T3 ), 3, , One kg of a diatomic gas is at a pressure of, 8 × 104N/m2. The density of the gas is 4kg/m3. What is, the energy of the gas due to its thermal motion?[2009], (a) 5 × 104 J, (b) 6 × 104 J, 4, (c) 7 × 10 J, (d) 3 × 104 J, The speed of sound in oxygen (O2) at a certain temperature, is 460 ms–1. The speed of sound in helium (He) at the same, temperature will be (assume both gases to be ideal), [2008], –1, –1, (a) 1421 ms, (b) 500 ms, (c) 650 ms–1, (d) 330 ms–1, At what temperature is the r.m.s velocity of a hydrogen, molecule equal to that of an oxygen molecule at 47°C?, [2002], (a) 80 K, (b) –73 K, (c) 3 K, (d) 20 K, , 29., , 30., , Consider a gas of triatomic molecules. The molecules are, assumed to be triangular and made of massless rigid rods, whose vertices are occupied by atoms. The internal energy, of a mole of the gas at temperature T is :, [Sep. 03, 2020 (I)], , Degree of Freedom, Specific, TOPIC 3 Heat Capacity, and Mean, Free Path, 26., , Molecules of an ideal gas are known to have three, translational degrees of freedom and two rotational, degrees of freedom. The gas is maintained at a, temperature of T., , (a), , 5, RT, 2, , (b), , 3, RT, 2, , 9, RT, (d) 3RT, 2, 31. To raise the temperature of a certain mass of gas by 50°C, at a constant pressure, 160 calories of heat is required., When the same mass of gas is cooled by 100°C at constant, , (c)
Page 198 :
P-189, , Kinetic Theory, , 32., , 33., , 34., , 35., , 36., , volume, 240 calories of heat is released. How many, degrees of freedom does each molecule of this gas have, (assume gas to be ideal)?, [Sep. 03, 2020 (II)], (a) 5, (b) 6, (c) 3, (d) 7, A gas mixture consists of 3 moles of oxygen and 5 moles of, argon at temperature T. Assuming the gases to be ideal, and the oxygen bond to be rigid, the total internal energy, (in units of RT) of the mixture is : [Sep. 02, 2020 (I)], (a) 15, (b) 13, (c) 20, (d) 11, An ideal gas in a closed container is slowly heated. As its, temperature increases, which of the following statements, are true?, [Sep. 02, 2020 (II)], (1) The mean free path of the molecules decreases, (2) The mean collision time between the molecules, decreases, (3) The mean free path remains unchanged, (4) The mean collision time remains unchanged, (a) (2) and (3), (b) (1) and (2), (c) (3) and (4), (d) (1) and (4), Consider two ideal diatomic gases A and B at some, temperature T. Molecules of the gas A are rigid, and have, a mass m. Molecules of the gas B have an additional, m, vibrational mode, and have a mass, . The ratio of the, 4, B, specific heats (CA, V and CV ) of gas A and B, respectively, is:, [9 Jan 2020 I], (a) 7 : 9, (b) 5 : 9, (c) 3 : 5, (d) 5 : 7, Two gases-argon (atomic radius 0.07 nm, atomic weight, 40) and xenon (atomic radius 0.1 nm, atomic weight 140), have the same number density and are at the same, temperature. The ratio of their respective mean free, times is closest to:, [9 Jan 2020 II], (a) 3.67, (b) 1.83, (c) 2.3, (d) 4.67, The plot that depicts the behavior of the mean free time t, (time between two successive collisions) for the molecules, of an ideal gas, as a function of temperature (T),, qualitatively, is: (Graphs are schematic and not drawn to, scale), [8 Jan. 2020 I], , 37. Consider a mixture of n moles of helium gas and 2n, moles of oxygen gas (molecules taken to be rigid) as an, ideal gas. Its CP/CV value will be:, [8 Jan. 2020 II], (a) 19/13, (b) 67/45, (c) 40/27, (d) 23/15, Cp 5, 38. Two moles of an ideal gas with C = are mixed with 3, 3, V, , Cp 4, moles of another ideal gas with C = . The value of, 3, V, Cp, , 39., , 40., , 41., , 42., , for the mixture is:, [7 Jan. 2020 I], CV, (a) 1. 45, (b) 1.50, (c) 1.47, (d) 1.42, Two moles of helium gas is mixed with three moles of, hydrogen molecules (taken to be rigid). What is the molar, specific heat of mixture at constant volume?, (R = 8.3 J/mol K), [12 April 2019 I], (a) 19.7 J/mol L, (b) 15.7 J/mol K, (c) 17.4 J/mol K, (d) 21.6 J/mol K, A diatomic gas with rigid molecules does 10 J of work, when expanded at constant pressure. What would be the, heat energy absorbed by the gas, in this process ?, [12 April 2019 II], (a) 25 J, (b) 35 J, (c) 30 J, (d) 40 J, A 25×10 – 3 m3 volume cylinder is filled with 1 mol of O2, gas at room temperature (300 K) . The molecular diameter, of O2, and its root mean square speed, are found to be 0.3, nm and 200 m/s, respectively. What is the average collision, rate (per second) for an O2 molecule?, [10 April 2019 I], (a) ~1012, (b) ~1011, (c) ~1010, (d) ~1013, When heat Q is supplied to a diatomic gas of rigid, molecules, at constant volume its temperature increases, by DT. The heat required to produce the same change in, temperature, at a constant pressure is :, [10 April 2019 II], (a), , 2, Q, 3, , (b), , 5, Q, 3, , 7, 3, Q, Q, (d), 5, 2, An HCl molecule has rotational, translational and, vibrational motions. If the rms velocity of HCl molecules, in its gaseous phase is v , m is its mass and k B is, Boltzmann constant, then its temperature will be:, [9 April 2019 I], , (c), , (a), , 43., , t, , t, , (b), 1, T, , T, , t, , (a), , t, , (c), , (d), 1, T, , T, , mv 2, 6kB, , mv 2, (c), 7k B, , (b), , mv 2, 3k B, , mv 2, (d), 5k B
Page 199 :
P-190, , 44., , 45., , 46., , 47., , 48., , 49., , 50., , Physics, , The specific heats, C p and Cv of a gas of diatomic, molecules, A, are given (in units of J mol–1 k–1) by 29 and, 22, respectively. Another gas of diatomic molecules, B,, has the corresponding values 30 and 21. If they are treated, as ideal gases, then:, [9 April 2019 II], (a) A is rigid but B has a vibrational mode., (b) A has a vibrational mode but B has none., (c) A has one vibrational mode and B has two., (d) Both A and B have a vibrational mode each., An ideal gas occupies a volume of 2 m 3 at a pressure of 3, × 106 Pa. The energy of the gas:, [12 Jan. 2019 I], (a) 9 × 106 J, (b) 6 × 104 J, (c) 108 J, (d) 3 × 102 J, An ideal gas is enclosed in a cylinder at pressure of 2 atm, and temperature, 300 K. The mean time between two, successive collisions is 6 × 10–8 s. If the pressure is doubled, and temperature is increased to 500 K, the mean time, between two successive collisions wiil be close to:, [12 Jan. 2019 II], –7, –8, (a) 2 × 10 s, (b) 4 × 10 s, (c) 0.5 × 10–8 s, (d) 3 × 10–6 s, , 51. Two moles of an ideal monoatomic gas occupies a volume, V at 27°C. The gas expands adiabatically to a volume 2 V., Calculate (1) the final temperature of the gas and (2) change, in its internal energy., [2018], (a) (1) 189 K, (2) 2.7 kJ, (b) (1) 195 K, (2) –2.7 kJ, (c) (1) 189 K, (2) –2.7 kJ, (d) (1) 195 K, (2) 2.7 kJ, 52. Two moles of helium are mixed with n with moles of, 3, C, hydrogen. If P = for the mixture, then the value of n, CV 2, is, [Online April 16, 2018], (a) 3/2, (b) 2, (c) 1, (d) 3, 53. Cp and Cv are specific heats at constant pressure and, constant volume respectively. It is observed that, Cp – Cv = a for hydrogen gas, Cp – Cv = b for nitrogen gas, The correct relation between a and b is :, [2017], (a) a = 14 b, (b) a = 28 b, , A gas mixture consists of 3 moles of oxygen and 5 moles, of argon at temperature T. Considering only translational, and rotational modes, the total internal energy of the, system is :, [11 Jan. 2019 I], (a) 15 RT, (b) 12 RT, (c) 4 RT, (d) 20 RT, In a process, temperature and volume of one mole of an, ideal monoatomic gas are varied according to the relation, VT = K, where K is a constant. In this process the, temperature of the gas is increased by DT. The amount of, heat absorbed by gas is (R is gas constant) :, [11 Jan. 2019 II], 1, 1, KRΔT, (a) RΔT, (b), 2, 2, 2K, 3, ΔT, RΔT, (c), (d), 3, 2, Two kg of a monoatomic gas is at a pressure of 4 × 104, N/m2. The density of the gas is 8 kg/m3. What is the, order of energy of the gas due to its thermal motion?, [10 Jan 2019 II], 3, 5, (a) 10 J, (b) 10 J, (c) 104 J, (d) 106 J, A 15 g mass of nitrogen gas is enclosed in a vessel at, a temperature 27°C. Amount of heat transferred to the, gas, so that rms velocity of molecules is doubled, is, about: [Take R = 8.3 J/K mole], [9 Jan. 2019 II], (a) 0.9 kJ, (b) 6 kJ, (c) 10 kJ, (d) 14 kJ, , 54., , 1, b, (d) a = b, 14, An ideal gas has molecules with 5 degrees of freedom., The ratio of specific heats at constant pressure (C p) and, at constant volume (Cv) is : [Online April 8, 2017], 7, (a) 6, (b), 2, 7, 5, (c), (d), 5, 2, An ideal gas undergoes a quasi static, reversible process, in which its molar heat capacity C remains constant. If, during this process the relation of pressure P and volume, V is given by PVn = constant, then n is given by (Here CP, and CV are molar specific heat at constant pressure and, constant volume, respectively) :, [2016], CP - C, C - CV, (a) n =, (b) n =, C - CV, C - CP, C – CP, CP, (c) n =, (d) n =, C – CV, CV, Using equipartition of energy, the specific heat, (in J kg–1 K–1) of aluminium at room temperature can, be estimated to be (atomic weight of aluminium = 27), [Online April 11, 2015], (a) 410, (b) 25, (c) 1850, (d) 925, Modern vacuum pumps can evacuate a vessel down to a, pressure of 4.0 × 10–15 atm. at room temperature (300, K). Taking R = 8.0 JK–1 mole–1, 1 atm = 105 Pa and, NAvogadro = 6 × 1023 mole–1, the mean distance between, molecules of gas in an evacuated vessel will be of the, order of:, [Online April 9, 2014], (a) 0.2 mm, (b) 0.2 mm, (c) 0.2 cm, (d) 0.2 nm, , (c) a =, , 55., , 56., , 57.
Page 200 :
P-191, , Kinetic Theory, , 58., , Figure shows the variation in temperature (DT) with the, amount of heat supplied (Q) in an isobaric process, corresponding to a monoatomic (M), diatomic (D) and a, polyatomic (P) gas. The initial state of all the gases are the, same and the scales for the two axes coincide. Ignoring, vibrational degrees of freedom, the lines a, b and c, respectively correspond to :, [Online April 9, 2013], , oxygen. The ratio, , a, b, , Q, , c, DT, , 59., , (a) P, M and D, , (b) M, D and P, , (c) P, D and M, , (d) D, M and P, , A given ideal gas with g =, , Cp, , = 1.5 at a temperature T. If, Cv, the gas is compressed adiabatically to one-fourth of its, initial volume, the final temperature will be, [Online May 12, 2012], , (a) 2 2T, (c) 2 T, , 60. If CP and CV denote the specific heats of nitrogen per unit, mass at constant pressure and constant volume, respectively, then, [2007], (a) CP – CV = 28R, (b) CP – CV = R/28, (c) CP – CV = R/14, (d) CP – CV = R, 61. A gaseous mixture consists of 16 g of helium and 16 g of, , (b) 4 T, (d) 8 T, , Cp, Cv, , of the mixture is, , [2005], , (a) 1.62, (b) 1.59, (c) 1.54, (d) 1.4, 62. One mole of ideal monatomic gas (g = 5/3) is mixed with, one mole of diatomic gas (g = 7/5). What is g for the, mixture? g Denotes the ratio of specific heat at constant, pressure, to that at constant volume, [2004], (a) 35/23, (b) 23/15, (c) 3/2, (d) 4/3, 63. During an adiabatic process, the pressure of a gas is found, to be proportional to the cube of its absolute temperature., The ratio CP/CV for the gas is, [2003], (a), , 4, 3, , (b) 2, , (c), , 5, 3, , (d), , 3, 2
Page 202 :
P-193, , Kinetic Theory, , Here, M = Molar mass of gas molecule, T = temperature of the gas molecule, We have given vN2 = vH2, 3RTN2, , \, , M N2, , Þ, 11., , TH2, 2, , =, , 3RTH 2, M H2, , 573, Þ TH2 = 41 K, 28, , (c) Vrms =, v1, =, v2, , =, , 3RT, M, , T1 (273 + 127), =, =, T2 (273 + 237), , 400, =, 500, , 4, 2, =, 5, 5, , 5, 5, v1 =, ´ 200 = 100 5 m/s., 2, 2, (Bouns) Rate of change of momentum during collision, \ v2 =, , 12., , =, , mv – (– mv ), 2mv, =, N, Dt, Dt, , so pressure P = N ´ (2mv), Dt ´ A, 1022 ´ 2 ´ 10–26 ´ 104, = 2 N / m2, 1´ 1, (c) vrms = ve, =, , 13., , 3RT, M, 3 ´ 8.314 ´ 300, 2, (1930 ) =, M, 3 ´ 8.314 ´ 300, M=, » 2 ´10 -3 kg, 1930 ´1930, The gas is H2., , 18. (a) Q, , 3RT, = 11.2 ´ 103, M, , or, , or, , 14., , 3kT, = 11.2 ´ 103, m, , 3 ´ 1.38 ´ 10-23 T, 2 ´ 10-3, , V1rms, (a) Using V, =, 2rms, , = 11.2 ´ 10, , 3, , \ v = 104 K, , M2, M1, , Vrms ( He ), , 15., , M Ar, 40, =, M He = 4 = 3.16, Vrms ( Ar ), (a) Energy associated with N moles of diatomic gas,, 5, Ui = N RT, 2, Energy associated with n moles of monoatomic gas, 3, = n RT, 2, Total energy when n moles of diatomic gas converted into, 3, 5, monoatomic (Uf) = 2n RT + (N - n) RT, 2, 2, , =, , 1, 5, nRT + NRT, 2, 2, , Now, change in total kinetic energy of the gas, 1, DU = Q = nRT, 2, 1 mN 2, V rm s, 16 . (c) Pressure, P =, 3 V, (mN )T, or, P =, V, If the gas mass and temperature are constant then, P µ (Vrms)2 µ T, So, force µ (Vrms)2 µ T, i.e., Value of q = 1, 17. (d) Kinetic energy of each molecule,, 3, K.E. = K B T, 2, In the given problem,, Temperature, T = 0°C = 273 K, Height attained by the gas molecule, h = ?, 819K B, 3, K.E. = K B ( 273) =, 2, 2, K.E. = P.E., 819K B, = Mgh, Þ, 2, 819K B, or h =, 2Mg, C=, , 3rv, mass of the gas, 20. (a) Given, V1 = V, V2 = 2V, , 19. (c) v rms =, , T1 = 27° + 273 = 300 K, T2 = ?, From charle’s law, V1 V2, =, Q Pressure is constant ), T1 T2 (, V, 2V, or, 300 = T, 2, \ T2 = 600 K = 600 – 273 = 327°C, 21. (c) As, work done is zero., So, loss in kinetic energy = heat gain by the gas, 1 2, R, mv = nCv DT = n, DT, 2, g -1, 1 2 m R, mv =, DT, 2, M g -1
Page 203 :
P-194, , Physics, , \ DT =, 22., , Mv 2 ( g - 1), K, 2R, , And g =, n1, , (a) Number of moles of first gas = N, A, , But vrms =, , n1 + n2 + n3, RTmix, NA, , (a) Given, mass = 1 kg, Density = 4 kg m–3, , g=, , \, , vO2, vHe, , =, , M O2, , ´, , \ g = 1+, , 2 9, =, 7 7, (D) Triatomic rigid molecules, f = 6, \ g = 1+, , gRT, M, , M He, g He, , 1.4, 4, ´, = 0.3237, 32 1.67, vO 2, 460, \ vHe =, =, = 1421 m / s, 0.3237 0.3237, (d) RMS velocity of a gas molecule is given by, 3RT, M, Let T be the temperature at which the velocity of hydrogen, molecule is equal to the velocity of oxygen molecule., , 2 4, =, 6 3, 29. (266.67) Here work done on gas and heat supplied to the, gas are zero., Let T be the final equilibrium temperature of the gas in the, vessel., Total internal energy of gases remain same., \ g = 1+, , i.e., u1 + u2 = u '1 + u '2, or, n1Cv DT1 + n2Cv DT2 = (n1 + n2 )CvT, Þ (0.1)Cv (200) + (0.05)Cv (400) = (0.15)CvT, , Vrms =, , 3RT, 3R ´ (273 + 47), =, 2, 32, Þ T = 20K, (c) Total degree of freedom f = 3 + 2 = 5, , \, , 26., , Total energy, U =, , nfRT 5RT, =, 2, 2, , 2, , where f = degree of freedom, f, , 2 7, =, 5 5, (C) Diatomic non-rigid molecules, f = 7, , [Q PM = dRT ], , =, , 25., , Cv, , = 1+, , 2 5, =, 3 3, (B) Diatomic rigid molecules, f = 5, , [As R and T is constant], g O2, , Cp, , \ g = 1+, , (a) The speed of sound in a gas is given by v =, g, M, , 3kT, R, , (A) Monatomic, f = 3, , =, , \ vµ, , vrms, , l, 1, Þtµ, 3kT, T, m, 28. (c) As we know,, , mass, 1, = m3, Volume =, density 4, Internal energy of the diatomic gas, , 24., , l, , \t =, , n1T1 + n2T2 + n3T3, n1 + n2 + n3, , 5, 5, 1, PV = ´ 8 ´ 104 ´ = 5 ´ 10 4 J, 2, 2, 4, Alternatively:, 5, 5m, 5 m PM, RT =, ´, K.E = nRT =, d, 2, 2M, 2M, 4, 5 mP 5 1 ´ 8 ´ 10, =, = ´, = 5 ´ 104 J, 2 d, 2, 4, , 1, , But, mean time of collision, t =, , n1, n, n, RT1 + 2 RT2 + 3 RT3, NA, NA, NA, , 23., , 2, 2 7, = 1+ =, f, 5 5, , 2pnd 2, where, d = diameter of the molecule, n = number of molecules per unit volume, , If there is no loss of energy then, P1V1 + P2V2 + P3V3 = PV, , Tmix =, , = 1+, , Cv, , 27. (b) Mean free path, l =, , n2, Number of moles of second gas = N, A, n3, Number of moles of third gas = N, A, , =, , Cp, , 800, = 266.67 K, 3, (d) Here degree of freedom, f = 3 + 3 = 6 for triatomic nonlinear molecule., Internal energy of a mole of the gas at temperature T,, \T =, , 30., , f, 6, nRT = RT = 3RT, 2, 2, 31. (b) Let Cp and Cv be the specific heat capacity of the gas, at constant pressure and volume., U=
Page 204 :
P-195, , Kinetic Theory, , At constant pressure, heat required, , DQ1 = nC p DT, Þ 160 = nC p × 50, , ...(i), , At constant volume, heat required, DQ2 = nCv DT, Þ 240 = nCv ×100, , ...(ii), , Dividing (i) by (ii), we get, Cp 4, 160 C p 50, =, ×, Þ, =, Cv 3, 240 Cv 100, g=, , Cp, Cv, , =, , 4, 2, = 1+, 3, f, , (Here, f = degree of freedom), , 32., , Þ f = 6., (a) Total energy of the gas mixture,, , 33., , f1n1 RT1 f 2 n2 RT2, +, Emix =, 2, 2, 5, 5, = 3 ´ RT + ´ 3RT = 15RT, 2, 2, (a) As we know mean free path, 1, æ Nö, 2 ç ÷ pd 2, èV ø, , l=, , Here,, , N = no. of molecule, V = volume of container, d = diameter of molecule, , But PV = nRT = nNKT, N, P, Þ =, =n, V KT, 1 KT, l=, 2 pd 2 P, For constant volume and hence constant number density, P, is constant., T, So mean free path remains same., As temperature increases no. of collision increases so, relaxation time decreases., (d) Specific heat of gas at constant volume, , n of gas molecules, , 34., , 5, R, 5, 7, = 2 =, B, B, C, =, R, \, Hence C, 7, v, 7, v, 2, R, 2, CvA, , 35. (Bonus) Mean free path of a gas molecule is given by, 1, l=, 2pd 2 n, Here, n = number of collisions per unit volume, d = diameter of the molecule, If average speed of molecule is v then, l, Mean free time, t =, v, M, 1, 1, Þ t=, =, 2, 2 3RT, 2pnd v, 2pnd, æ, 3RT ö, çQ v = M ÷, è, ø, \ tµ, , 2, M \ t1 = M1 ´ d2, t2, M2, d12, d2, , 2, , =, , 40 æ 0.1 ö, ´ç, ÷ = 1.09, 140 è 0.07 ø, , 36. (c) Relaxation time (t ) µ, , mean free path, 1, Þ tµ, speed, v, , and, v µ T, 1, \tµ, T, Hence graph between t v/s, , 1, , is a straight line which, T, is correctly depicted by graph shown in option (c)., 37. (a) Helium is a monoatomic gas and Oxygen is a diatomic, gas., For helium, CV1 =, , 3, 5, R and CP1 = R, 2, 2, , For oxygen, CV2 =, g=, , 5, 7, R and CP = R, 2, 2, 2, , N1CP1 + N 2 CP2, N1CV1 + N 2 CV2, , Cv =, , 1, fR; f = degree of freedom, 2, For gas A (diatomic), f = 5 (3 translational + 2 rotational), , 5, 7, n. R + 2n. R, 2 = 19nR ´ 2, g= 2, Þ, 3, 5, 2(13nR ), n. R + 2n. R, 2, 2, , 5, R, 2, For gas B (diatomic) in addition to (3 translational + 2, rotational) 2 vibrational degree of freedom., , æC ö, 19, \ç P ÷, =, C, è V ø mixture 13, , \C A =, v
Page 207 :
P-198, , Physics, , No. of moles of helium,, mHe, 16, n1 = M, =4, =, 4, He, Number of moles of oxygen,, 16 1, =, n2 =, 32 2, 3, 1 5, 5, 4 ´ R + ´ R 6R + R, 2, 2, 2, 4, =, =, 9, 1ö, æ, \ Cv, çè 4 + ÷ø, 2, 2, , 29 R ´ 2 29 R, =, and, 9´ 4, 18, Specific heat at constant pressure, =, , Cp =, , n1C p1 + n2C p2, (n1 + n2 ), , 7, 10 R + R, 4 = 47 R, =, 9, 18, 2, , 5R 1 7 R, + ´, 4´, 2 2 2, =, 1ö, æ, çè 4 + ÷ø, 2, , Þ, \, , Cp, Cv, , =, , 47 R 18, ´, = 1.62, 18 29 R, , 5, 7, g =, 3 2 5, n1 = 1, n2 = 1, , 62. (c) g 1 =, , n1 + n2, n, n, = 1 + 2, g -1, g1 - 1 g 2 - 1, 1 +1, 1, 1, 3 5, =, +, = + =4, g - 1 5 - 1 7 -1 2 2, 3, 5, 2, 3, \, =4 Þ g=, g -1, 2, Þ, , 63. (d) P µ T 3 Þ PT -3 = constant, ....(i), But for an adiabatic process, the pressure temperature, relationship is given by, P1-g T g = constant, , Þ PT, , g, 1-g, , = constt., , From (i) and (ii), , ....(ii), , g, 3, = -3 Þ g = -3 + 3g Þ g =, 1- g, 2
Page 208 :
13, , P-199, , Oscillations, , Oscillations, TOPIC 1, 1., , 2., , 3., , Displacement, Phase, Velocity, and Acceleration in S.H.M., , 5., , The position co-ordinates of a particle moving in a 3-D, coordinate system is given by, [9 Jan 2019, II], x = a coswt, y = a sinwt, and z = awt, The speed of the particle is:, (a), (c), 2 aw (b) aw, 3 aw (d) 2aw, Two simple harmonic motions, as shown, are at right angles., They are combined to form Lissajous figures., x(t) = A sin (at + d), y(t) = B sin (bt), Identify the correct match below, [Online April 15, 2018], p, (a) Parameters: A = B, a = 2b; d = ; Curve: Circle, 2, p, (b) Parameters: A = B, a = b; d = ; Curve: Line, 2, p, (c) Parameters: A ¹ B, a = b; d = ; Curve: Ellipse, 2, (d) Parameters: A ¹ B, a = b; d = 0; Curve: Parabola, The ratio of maximum acceleration to maximum velocity in, a simple harmonic motion is 10 s–1. At, t = 0 the displacement, is 5 m. What is the maximum acceleration ? The initial phase, is p, [Online April 8, 2017], 4, (a), , 500 m/s2, , (b), , 500, , 2, , m/s2, , Two particles are performing simple harmonic motion in a, straight line about the same equilibrium point. The, amplitude and time period for both particles are same, and equal to A and T, respectively. At time t = 0 one particle, has displacement A while the other one has displacement, -A, and they are moving towards each other. If they cross, 2, each other at time t, then t is:, [Online April 9, 2016], , (a), 6., , 7., , 5T, 6, , (b), , T, 3, , (c), , T, 4, , (d), , T, 6, , A simple harmonic oscillator of angular frequency 2 rad, s–1 is acted upon by an external force F = sin t N. If the, oscillator is at rest in its equilibrium position at t = 0, its, position at later times is proportional to :, [Online April 10, 2015], 1, 1, (a) sin t + cos 2t, (b) cos t - sin 2t, 2, 2, 1, 1, (c) sin t - sin 2t, (d) sin t + sin 2t, 2, 2, x and y displacements of a particle are given as x(t) = a sin, wt and y (t) = a sin 2wt. Its trajectory will look like :, [Online April 10, 2015], y, , y, , x, , (a), , 750 m/s2, , 4., , (c), (d), 750 2 m/s2, A particle performs simple harmonic mition with amplitude, A. Its speed is trebled at the instant that it is at a distance, , x, , (b), , 2A, from equilibrium position. The new amplitude of the, 3, motion is :, [2016], 7A, A, 41 (d) 3A, (a) A 3, (b), (c), 3, 3, , y, , y, , (c), , x, , (d), , x
Page 209 :
P-200, , 8., , 9., , 10., , Physics, , A body is in simple harmonic motion with time period half, second (T = 0.5 s) and amplitude one cm (A = 1 cm). Find, the average velocity in the interval in which it moves form, equilibrium position to half of its amplitude., [Online April 19, 2014], (a) 4 cm/s (b) 6 cm/s (c) 12 cm/s (d) 16 cm/s, Which of the following expressions corresponds to simple, harmonic motion along a straight line, where x is the, displacement and a, b, c are positive constants?, [Online April 12, 2014], 2, (a) a + bx – cx, (b) bx2, (c) a – bx + cx2, (d) – bx, A particle which is simultaneously subjected to two, perpendicular simple harmonic motions represented by;, x = a1 cos wt and y = a2 cos 2 wt traces a curve given by:, [Online April 9, 2014], , (a), , y, , y, , a2, , a2, , a1, , x, , O, , (b), , 12. Two particles are executing simple harmonic motion of, the same amplitude A and frequency w along the x-axis., Their mean position is separated by distance X0(X0 > A)., If the maximum separation between them is (X0 + A), the, phase difference between their motion is:, [2011], (a), , (c), , 11., , x, , O, , (d), , a1, , y, , (c), , (d), , [2011], 1, , (a), , M +m, M, , (c), , æ M + mö 2, çè, ÷, M ø, , (b), , æ M ö2, çè, ÷, M + mø, , (d), , M, M +m, , 1, , is written as a = A cos(wt + d ) ,then, , a2, x, , t, , y, , t, , p, 2, , [2007], , A = x0 w 2 , d = 3p / 4 (b) A = x0, d = -p / 4, 2, , O, , (b), , (d), , æ A1 ö, , (c) A = x0 w 2 , d = p / 4 (d) A = x0 w , d = -p / 4, 15. A coin is placed on a horizontal platform which undergoes, vertical simple harmonic motion of angular frequency w., The amplitude of oscillation is gradually increased. The, coin will leave contact with the platform for the first time, (a) at the mean position of the platform, [2006], g, (b) for an amplitude of 2, w, (c) for an amplitude of, 16., , t, , p, 6, , law x = x0 cos(wt - p / 4) . If the acceleration of the particle, , The displacement y(t) = A sin (wt + f) of a pendulum for, 2p, is correctly represented by, f=, 3, [Online May 19, 2012], y, y, , (a), , (c), , of ç A ÷ is:, è 2ø, , x, , y, , a2, , p, 4, , 13. A mass M, attached to a horizontal spring, executes S.H.M., with amplitude A1. When the mass M passes through its, mean position then a smaller mass m is placed over it and, both of them move together with amplitude A2. The ratio, , (a), , a1, , (b), , 14. A point mass oscillates along the x-axis according to the, , O a1, , y, , p, 3, , t, , g2, , w2, (d) at the highest position of the platform, The maximum velocity of a particle, executing simple, harmonic motion with an amplitude 7 mm, is 4.4 m/s. The, period of oscillation is, [2006], (a) 0.01 s (b) 10 s, (c) 0.1 s, (d) 100 s, , 17. The function sin 2 (wt ) represents, [2005], (a) a periodic, but not simple harmonic motion with a, p, period, w, (b) a periodic, but not simple harmonic motion with a, 2p, period, w, (c) a simple harmonic motion with a period, , p, w, , (d) a simple harmonic motion with a period, , 2p, w
Page 210 :
P-201, , Oscillations, , 18., , Two simple harmonic motions are represented by the, pö, æ, equations y1 = 0.1 sin ç100pt + ÷ and y 2 = 0.1 cos pt ., 3ø, è, , 19., , The phase difference of the velocity of particle 1 with, respect to the velocity of particle 2 is, [2005], p, -p, p, -p, (a), (b), (c), (d), 6, 6, 3, 3, Two particles A and B of equal masses are suspended from, two massless springs of spring constants k 1 and k 2,, respectively. If the maximum velocities, during, oscillation, are equal, the ratio of amplitude of A and B, is, [2003], k2, k1, k1, k2, (b) k, (c), (d) k, k2, k1, 1, 2, The displacement of a particle varies according to the, , (a), 20., , relation x = 4(cos pt + sin pt ). The amplitude of the, particle is, [2003], (a) – 4, , TOPIC 2, , (d) 8, , Energy in Simple Harmonic, Motion, , The displacement time graph of a particle executing S.H.M., is given in figure : (sketch is schematic and not to scale), , displacement, , 21., , (c) 4 2, , (b) 4, , O, , 2T, 4, T, 4, , 3T T, 4, , time (s), , 5T, 4, , Which of the following statements is/are true for this, motion?, [Sep. 02, 2020 (II)], 3T, 4, (2) The acceleration is maximum at t = T, , (1) The force is zero at t =, , (3) The speed is maximum at t =, , T, 4, , (4) The P.E. is equal to K.E. of the oscillation at t =, , 22., , T, 2, , (a) (1), (2) and (4), (b) (2), (3) and (4), (c) (1), (2) and (3), (d) (1) and (4), A particle undergoing simple harmonic motion has time, , pt, . The, 90, ratio of kinetic to potential energy of this particle at, t = 210s will be :, [11 Jan 2019, I], , 23. A pendulum is executing simple harmonic motion and its, maximum kinetic energy is K1. If the length of the, pendulum is doubled and it performs simple harmonic, motion with the same amplitude as in the first case, its, maximum kinetic energy is K2., [11 Jan 2019, II], (a) K2 = 2K1, , 24., , (a), , (b) 1, , (c) 2, , (d), , 1, 3, , K2 =, , K1, 2, , K1, (d) K2 = K1, 4, A particle is executing simple harmonic motion (SHM) of, amplitude A, along the x-axis, about x = 0. When its, potential Energy (PE) equals kinetic energy (KE), the, position of the particle will be:, [9 Jan 2019, II], , (c), , K2 =, , (a), , A, 2, , (b), , A, 2 2, , A, (d) A, 2, 25. A particle is executing simple harmonic motion with a time, period T. At time t = 0, it is at its position of equilibrium., The kinetic energy-time graph of the particle will look like:, [2017], (c), , (a), , (b), , (c), , (d), , 26. A block of mass 0.1 kg is connected to an elastic spring of, spring constant 640 Nm–1 and oscillates in a medium of, constant 10–2 kg s–1. The system dissipates its energy, gradually. The time taken for its mechanical energy of, vibration to drop to half of its initial value, is closest to :, [Online April 9, 2017], (a) 2 s, (b) 3.5 s, (c) 5 s, (d) 7 s, 27. For a simple pendulum, a graph is plotted between its, kinetic energy (KE) and potential energy (PE) against its, displacement d. Which one of the following represents these, correctly? (graphs are schematic and not drawn to scale), [2015], E, , KE, , E, PE, , (a), , d, , dependent displacement given by x ( t ) = A sin, , 1, 9, , (b), , (b), , KE, , PE, , E, , (c), , E, , KE, PE, , (d), d, , PE, KE, , d
Page 211 :
P-202, , 28., , 29., , 30., , A pendulum with time period of 1s is losing energy. At, certain time its energy is 45 J. If after completing 15, oscillations, its energy has become 15 J, its damping, constant (in s–1) is :, [Online April 11, 2015], 1, 1, 1, ln3 (c) 2, ln3, (a), (b), (d), 30, 15, 2, This question has Statement 1 and Statement 2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., If two springs S1 and S2 of force constants k1 and k 2, respectively, are stretched by the same force, it is found, that more work is done on spring S1 than on spring S2., Statement 1 : If stretched by the same amount work done, on S1, Statement 2 : k1 < k2, [2012], (a) Statement 1 is false, Statement 2 is true., (b) Statement 1 is true, Statement 2 is false., (c) Statement 1 is true, Statement 2 is true, Statement 2 is, the correct explanation for Statement 1, (d) Statement 1 is true, Statement 2 is true, Statement 2 is, not the correct explanation for Statement 1, A particle of mass m executes simple harmonic motion with, amplitude a and frequency n. The average kinetic energy, during its motion from the position of equilibrium to the, end is, [2007], 2, 2 2, 2, 2, 2, (a) 2p ma n, (b) p ma n, 1, 2 2, ma n, (d) 4p 2 ma 2 n2, 4, Starting from the origin a body oscillates simple, harmonically with a period of 2 s. After what time will its, kinetic energy be 75% of the total energy?, [2006], , (c), , 31., , 1, 1, 1, 1, s, s, s, s, (b), (d), (c), 6, 3, 4, 12, The total energy of a particle, executing simple harmonic, motion is, [2004], (a) independent of x, (b) µ x2, (c) µ x, (d) µ x1/2, where x is the displacement from the mean position, hence, total energy is independent of x., A body executes simple harmonic motion. The potential, energy (P.E), the kinetic energy (K.E) and total energy (T.E), are measured as a function of displacement x. Which of, the following statements is true ?, [2003], (a) K.E. is maximum when x = 0, (b) T.E is zero when x = 0, (c) K.E is maximum when x is maximum, (d) P.E is maximum when x = 0, In a simple harmonic oscillator, at the mean position, [2002], (a) kinetic energy is minimum, potential energy is maximum, (b) both kinetic and potential energies are maximum, (c) kinetic energy is maximum, potential energy is minimum, (d) both kinetic and potential energies are minimum, , (a), , 32., , 33., , 34., , Physics, , Time Period, Frequency,, TOPIC 3 Simple Pendulum and Spring, Pendulum, 35. An object of mass m is suspended at the end of a massless, wire of length L and area of cross-section, A. Young, modulus of the material of the wire is Y. If the mass is, pulled down slightly its frequency of oscillation along the, vertical direction is:, [Sep. 06, 2020 (I)], (a), , f =, , 1 mL, 2p YA, , (b) f =, , 1 YA, 2p mL, , 1 mA, 1 YL, (d) f =, 2p YL, 2p mA, 36. When a particle of mass m is attached to a vertical spring, of spring constant k and released, its motion is described, by y (t) = y0 sin2wt, where ‘y’ is measured from the lower, , (c), , f =, , end of unstretched spring. Then w is :, [Sep. 06, 2020 (II)], (a), , 1, 2, , g, y0, , (b), , g, 2 y0, , (d), , g, y0, , 2g, y0, 37. A block of mass m attached to a massless spring is, performing oscillatory motion of amplitude 'A' on a, frictionless horizontal plane. If half of the mass of the block, breaks off when it is passing through its equilibrium point,, the amplitude of oscillation for the remaining system, become fA. The value of f is :, [Sep. 03, 2020 (II)], 1, 1, (a), (b) 1, (c), (d), 2, 2, 2, 38. A person of mass M is, sitting on a swing of length L and, swinging with an angular amplitude q0. If the person stands, up when the swing passes through its lowest point, the, work done by him, assuming that his centre of mass, moves by a distance l (l<<L), is close to :, [12 April 2019, II], (b) mgl (1+q02), (a) mgl (1– q02), , (c), , æ q0 2, (d) Mgl çç 1 + 2, è, , (c) mgl, , ö, ÷, ÷, ø, , 39. A simple pendulum oscillating in air has period T. The bob, of the pendulum is completely immersed in a non-viscous, liquid. The density of the liquid is, , 1, th of the material of, 16, , the bob. If the bob is inside liquid all the time, its period of, oscillation in this liquid is :, [9 April 2019 I], (a) 2T, , 1, 10, , (b) 2T, , 1, 1, (c) 4T, 14, 15, , (d) 4T, , 1, 14
Page 212 :
P-203, , Oscillations, , 40., , Two light identical springs of spring constant k are attached, horizontally at the two ends of a uniform horizontal rod AB of, length l and mass m. The rod is pivoted at its centre ‘O’ and, can rotate frreely in horizontal plane. The other ends of two, springs are fixed to rigid supports as shown in figure. The rod, is gently pushed through a small angle and released. The, frequency of resulting oscillation is:, [12 Jan 2019, I], A, , 2, 3, s, s, (b), 3, 2, 3, s, (c), (d) 2 3s, 2, 45. A particle executes simple harmonic motion with an, amplitude of 5 cm. When the particle is at 4 cm from the, mean position, the magnitude of its velocity in SI units, is equal to that of its acceleration. Then, its periodic time, in seconds is:, [10 Jan 2019, II], (a), , y, , (a), , O, , 4p, 3, , (b), , 3, p, 8, , 8p, 7, (d), p, 3, 3, A cylindrical plastic bottle of negligible mass is filled, with 310 ml of water and left floating in a pond with still, water. If pressed downward slightly and released, it, starts performing simple harmonic motion at angular, frequency w. If the radius of the bottle is 2.5 cm then w, is close to: (density of water = 103 kg/m3), [10 Jan 2019, II], –1, (a) 3.75 rad s, (b) 1.25 rad s–1, –1, (c) 2.50 rad s, (d) 5.00 rad s–1, A rod of mass 'M' and length '2L' is suspended at its middle, by a wire. It exhibits torsional oscillations; If two masses, each of 'm' are attached at distance 'L/2' from its centre on, both sides, it reduces the oscillation frequency by 20%., The value of ratio m/M is close to :, [9 Jan 2019, II], (a) 0.77, (b) 0.57, (c) 0.37, (d) 0.17, A silver atom in a solid oscillates in simple harmonic motion, in some direction with a frequency of 1012/sec. What is the, force constant of the bonds connecting one atom with the, other? (Mole wt. of silver = 108 and Avagadro number, = 6.02 ×1023 gm mole –1), [2018], (a) 6.4 N/m (b) 7.1 N/m (c) 2.2 N/m (d) 5.5 N/m, A particle executes simple harmonic motion and is located, at x = a, b and c at times t0, 2t0 and 3t0 respectively. The, frequency of the oscillation is [Online April 16, 2018], (c), , x, , 46., , B, , 1 3k, 1 2k, 1 6k, 1 k, (b), (c), (d), 2π m, 2π m, 2π m, 2π m, A simple pendulum, made of a string of length l and a bob, of mass m, is released from a small angle q0. It strikes a, block of mass M, kept on a horizontal surface at its lowest, point of oscillations, elastically. It bounces back and goes, up to an angle q1. The M is given by : [12 Jan 2019, I], , (a), , 41., , (a), , m æ θ0 + θ1 ö, ç, ÷, 2 è θ0 - θ1 ø, , (b), , 47., , æ θ -θ ö, mç 0 1 ÷, è θ0 + θ1 ø, , æθ +θ ö, m æ θ0 - θ1 ö, mç 0 1 ÷, ç, ÷, (d), 2 è θ0 + θ1 ø, è θ0 - θ1 ø, A simple harmonic motion is represented by :, , (c), 42., , 48., , y = 5(sin3pt + 3 cos3pt) cm, The amplitude and time period of the motion are :, [12 Jan 2019, II], 3, s, 2, 3, 2, (c) 5 cm, s, (d) 5 cm, s, 2, 3, A simple pendulum of length 1 m is oscillating with an, angular frequency 10 rad/s. The support of the pendulum, starts oscillating up and down with a small angular, frequency of 1 rad/s and an amplitude of 10–2 m. The, relative change in the angular frequency of the pendulum, is best given by :, [11 Jan 2019, II], (a) 10–3 rad/s, (b) 1 rad/s, (c) 10–1 rad/s, (d) 10–5 rad/s, The mass and the diameter of a planet are three times the, respective values for the Earth. The period of oscillation, of a simple pendulum on the Earth is 2s. The period of, oscillation of the same pendulum on the planet would be:, [11 Jan 2019, II], , (a) 10 cm,, , 43., , 44., , 2, s, 3, , (b) 10 cm,, , 49., , (a), , 1, æa+bö, cos -1 ç, ÷, 2pt 0, è 2c ø, , (b), , 1, æa+bö, cos -1 ç, ÷, 2pt 0, è 3c ø, , (c), , 1, æ 2a + 3c ö, cos -1 ç, ÷ (d), 2 pt 0, è b ø, , 1, æa +cö, cos-1 ç, ÷, 2 pt 0, è 2b ø, , 50. In an experiment to determine the period of a simple pendulum of length 1 m, it is attached to different spherical, bobs of radii r1 and r2. The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be 5 × 10–4 s, the difference in radii, |r1– r2|, is best given by:, [Online April 9, 2017], (a) 1 cm, (b) 0.1 cm (c) 0.5 cm (d) 0.01 cm
Page 213 :
P-204, , 52., , 53., , 54., , 1, 1, 1, Hz (c), Hz, Hz, (a), (b), (d) 2 Hz, 2 2, 4, 2, A pendulum clock loses 12 s a day if the temperature is, 40°C and gains 4 s a day if the temperature is 20° C. The, temperature at which the clock will show correct time, and, the co-efficient of linear expansion (a) of the metal of the, pendulum shaft are respectively :, [2016], (a) 30°C; a = 1.85 × 10–3/°C, (b) 55°C; a = 1.85 × 10–2/°C, (c) 25°C; a = 1.85 × 10–5/°C, (d) 60°C; a = 1.85 × 10–4/°C, In an engine the piston undergoes vertical simple harmonic, motion with amplitude 7 cm. A washer rests on top of the, piston and moves with it. The motor speed is slowly, increased. The frequency of the piston at which the washer, no longer stays in contact with the piston, is close to :, [Online April 10, 2016], (a) 0.7 Hz (b) 1.9 Hz (c) 1.2 Hz (d) 0.1 Hz, A pendulum made of a uniform wire of cross sectional area, A has time period T. When an additional mass M is added, to its bob, the time period changes to TM. If the Young's, 1, modulus of the material of the wire is Y then, is equal, Y, , 2, 2, , A 1 kg block attached to a spring vibrates with a frequency, of 1 Hz on a frictionless horizontal table. Two springs, identical to the original spring are attached in parallel to an, 8 kg block placed on the same table. So, the frequency of, vibration of the 8 kg block is :, [Online April 8, 2017], , T (in s ), , 51., , Physics, , 8.0, 6.0, 4.0, 2.0, 0.5 1.0 1.5 2.0 L (in m), , What is the value of g at the place?, (a) 9.81 m/s2, (b) 9.87 m/s2, 2, (c) 9.91 m/s, (d) 10.0 m/s2, 57. The amplitude of a simple pendulum, oscillating in air with, a small spherical bob, decreases from 10 cm to 8 cm in 40, seconds. Assuming that Stokes law is valid, and ratio of, the coefficient of viscosity of air to that of carbon dioxide, is 1.3. The time in which amplitude of this pendulum will, reduce from 10 cm to 5 cm in carbon dioxide will be close to, (In 5 = 1.601, In 2 = 0.693)., [Online April 9, 2014], (a) 231 s, (b) 208 s, (c) 161 s, (d) 142 s, 58. Two bodies of masses 1 kg and 4 kg are connected to a, vertical spring, as shown in the figure. The smaller mass, executes simple harmonic motion of angular frequency, 25 rad/s, and amplitude 1.6 cm while the bigger mass, remains stationary on the ground. The maximum force, exerted by the system on the floor is (take g = 10 ms–2), [Online April 9, 2014], , to:, , 1 kg, , (g = gravitational acceleration), (a), , é æ T ö2 ù A, ê1 - ç M ÷ ú, ë è T ø û Mg, éæ T ö2 ù A, êç M ÷ - 1ú, ëè T ø, û Mg, , 55., , 56., , [2015], , é æ T ö2 ù A, (b) ê1 - ç T ÷ ú Mg, ëê è M ø úû, , éæ T, êç M, ëè T, , 2, ù Mg, ö, (d), (c), ÷ - 1ú, ø, û A, A particle moves with simple harmonic motion in a straight, line. In first t s, after starting from rest it travels a distance, a, and in next t s it travels 2a, in same direction, then:, (a) amplitude of motion is 3a, [2014], (b) time period of oscillations is 8t, (c) amplitude of motion is 4a, (d) time period of oscillations is 6t, In an experiment for determining the gravitational, acceleration g of a place with the help of a simple, pendulum, the measured time period square is plotted, against the string length of the pendulum in the figure., [Online April 19, 2014], , 4 kg, , (a) 20 N, (b) 10 N, (c) 60 N, (d) 40 N, 59. An ideal gas enclosed in a vertical cylindrical container, supports a freely moving piston of mass M. The piston, and the cylinder have equal cross sectional area A. When, the piston is in equilibrium, the volume of the gas is V0, and its pressure is P0. The piston is slightly displaced, from the equilibrium position and released. Assuming that, the system is completely isolated from its surrounding,, the piston executes a simple harmonic motion with, frequency, [2013], V, MP, 1 0 0, 1 AgP0, (a), (b), 2p A 2 g, 2p V0 M, (c), , 1, 2p, , A 2 gP0, MV0, , (d), , 1, 2p, , MV0, AgP0
Page 214 :
P-205, , Oscillations, , 60., , A mass m = 1.0 kg is put on a flat pan attached to a vertical, spring fixed on the ground. The mass of the spring and the, pan is negligible. When pressed slightly and released, the, mass executes simple harmonic motion. The spring, constant is 500 N/m. What is the amplitude A of the, motion, so that the mass m tends to get detached from, the pan ?, (Take g = 10 m/s2)., The spring is stiff enough so that it does not get distorted, during the motion., [Online April 22, 2013], m, , 61., , 62., , 63., , (b) b, , (c), , 1, b, , (d), , 2, b, , A ring is suspended from a point S on its rim as shown in, the figure. When displaced from equilibrium, it oscillates, with time period of 1 second. The radius of the ring is, (take g =, , p2), , [Online May 19, 2012], S, , (a) 0.15 m, 64., , (b) 1.5 m, , (b), , ld, (r - d ) g, , (d) 2p, , 2p, , lr, dg, lr, (r - d ) g, , 65. If x, v and a denote the displacement, the velocity and, the acceleration of a particle executing simple harmonic, motion of time period T, then, which of the following, does not change with time?, [2009], (a) aT/x, (b) aT + 2pv, (c) aT/v, (d) a2T2 + 4p2v2, , (a) 2 f, , (d) 0.5 m, , A wooden cube (density of wood ‘d’) of side ‘l’ floats in a, liquid of density ‘r’ with its upper and lower surfaces, horizontal. If the cube is pushed slightly down and, released, it performs simple harmonic motion of period, ‘T’, [2011 RS], , (b) f /2, , k2, (c) f /4, , (d) 4 f, , 67. The displacement of an object attached to a spring and, executing simple harmonic motion is given by x = 2 × 10–2, cos pt metre.The time at which the maximum speed first, occurs is, [2007], (a) 0.25 s, , (b) 0.5 s, , (c) 0.75 s, , (d) 0.125 s, , 68. The bob of a simple pendulum is a spherical hollow ball, filled with water. A plugged hole near the bottom of the, oscillating bob gets suddenly unplugged. During, observation, till water is coming out, the time period of, oscillation would, [2005], (a) first decrease and then increase to the original value, (b) first increase and then decrease to the original value, (c) increase towards a saturation value, (d) remain unchanged, 69. If a simple harmonic motion is represented by, 2, , d x, dt 2, (a), , (c) 1.0 m, , m, , k1, , If a simple pendulum has significant amplitude (up to a, factor of 1/e of original) only in the period between t =, 0s to t = t s, then t may be called the average life of the, pendulum. When the spherical bob of the pendulum, suffers a retardation (due to viscous drag) proportional, to its velocity with b as the constant of proportionality,, the average life time of the pendulum in second is, (assuming damping is small), [2012], 0.693, b, , (c) 2p, , ld, rg, , (b) A = 2.0 cm, , (c) A < 2.0 cm, (d) A = 1.5 cm, Two simple pendulums of length 1 m and 4 m respectively, are both given small displacement in the same direction, at the same instant. They will be again in phase after the, shorter pendulum has completed number of oscillations, equal to :, [Online April 9, 2013], (a) 2, (b) 7, (c) 5, (d) 3, , (a), , 2p, , 66. Two springs, of force constants k1 and k2 are connected, to a mass m as shown. The frequency of oscillation of, the mass is f. If both k1 and k2 are made four times their, original values, the frequency of oscillation becomes, [2007], , k, , (a) A > 2.0 cm, , (a), , + ax = 0 , its time period is, 2p, a, , (b), , 2p, a, , (c), , [2005], , 2p a, , (d), , 2pa, , 70. The bob of a simple pendulum executes simple harmonic, motion in water with a period t, while the period of oscillation, of the bob is t0 in air. Neglecting frictional force of water and, given that the density of the bob is (4 / 3) ´ 1000 kg/m 3 ., Which relationship between t and t0 is true?, [2004], (a) t = 2t 0, , (b) t = t 0 / 2, , (c) t = t0, , (d) t = 4t 0
Page 215 :
P-206, , 71., , 72., , 73., , Physics, , A particle at the end of a spring executes S.H.M with a, period t1. while the corresponding period for another spring, is t2. If the period of oscillation with the two springs in, series is T then, [2004], -1, , -1, , -1, , = t1 + t 2, , (a), , T, , (c), , T = t1 + t2, , (b) T 2 = t12 + t 22, (d) T -2 = t1-2 + t 2-2, , A mass M is suspended from a spring of negligible mass., The spring is pulled a little and then released so that the, mass executes SHM of time period T. If the mass is, 5T, increased by m, the time period becomes, . Then the, 3, m, ratio of, is, [2003], M, 3, 25, 16, 5, (b), (c), (d), (a), 5, 9, 9, 3, The length of a simple pendulum executing simple harmonic, motion is increased by 21%. The percentage increase in, the time period of the pendulum of increased length is, [2003], (a) 11%, , 74., , (c) 42%, , (d) 10%, , If a spring has time period T, and is cut into n equal parts,, then the time period of each part will be, [2002], T n, , (b) T / n (c) nT, (d) T, A child swinging on a swing in sitting position, stands up,, then the time period if the swing will, [2002], (a), , 75., , (b) 21%, , (a) increase, (b) decrease, (c) remains same, (d) increases if the child is long and decreases if the, child is short, , TOPIC 4, , Damped, Forced, Oscillations and Resonance, , 76. A damped harmonic oscillator has a frequency of 5, oscillations per second. The amplitude drops to half its, value for every 10 oscillations. The time it will take to drop, to, , 1, of the original amplitude is close to :, 1000, , (a) 50s, , (b) 100s, , (c) 20s, , [8 April 2019, II], (d) 10s, , 77. The displacement of a damped harmonic oscillator is given, by x(t) = e–0.1t. cos(10pt + j). Here t is in seconds., The time taken for its amplitude of vibration to drop to half, of its initial value is close to :, [9 Jan 2019, II], (a) 4s, (b) 7s, (c) 13s, (d) 27s, , 78. An oscillator of mass M is at rest in its equilibrium position, 1, 2, in a potential V = k(x - X) . A particle of mass m comes, 2, from right with speed u and collides completely inelastically, with M and sticks to it. This process repeats every time, the oscillator crosses its equilibrium position. The, amplitude of oscillations after 13 collisions is:, (M = 10, m = 5, u = 1, k = 1)., [Online April 16, 2018], , (a), , 1, 2, , (b), , 1, 3, , (c), , 2, 3, , (d), , 3, 5, , 79. The angular frequency of the damped oscillator is given, æk, r2 ö, ÷ where k is the spring constant, m, by, w = çç m 4m 2 ÷ø, è, , is the mass of the oscillator and r is the damping constant., , r2, is 8%, the change in time period, mk, compared to the undamped oscillator is approximately, as follows:, [Online April 11, 2014], , If the ratio, , (a) increases by 1%, , (b) increases by 8%, , (c) decreases by 1%, , (d) decreases by 8%, , 80. The amplitude of a damped oscillator decreases to 0.9 times, its original magnitude in 5s. In another 10s it will decrease, to a times its original magnitude, where a equals [2013], (a) 0.7, , (b) 0.81, , (c) 0.729, , (d) 0.6, , 81. A uniform cylinder of length L and mass M having crosssectional area A is suspended, with its length vertical, from, a fixed point by a massless spring, such that it is half, submerged in a liquid of density s at equilibrium position., When the cylinder is given a downward push and released,, it starts oscillating vertically with a small amplitude. The, time period T of the oscillations of the cylinder will be :, [Online April 25, 2013], é, ù, M, (a) Smaller than 2p ê, ú, ë (k + Asg ) û, , (b) 2p, , M, k, , é, ù, M, (c) Larger than 2p ê, ú, ë (k + Asg ) û, , (d), , 1, , é, ù, M, 2p ê, ú, (, k, +, A, s, g, ), ë, û, , 1, , 2, , 1, , 2, , 2
Page 216 :
P-207, , Oscillations, , 82. Bob of a simple pendulum of length l is made of iron., The pendulum is oscillating over a horizontal coil, carrying direct current. If the time period of the pendulum, is T then :, [Online April 23, 2013], (a) T < 2p, , l, and damping is smaller than in air alone., g, , (b) T = 2p, , l, and damping is larger than in air alone., g, , (c) T > 2p, , l, and damping is smaller than in air alone., g, , (d) T < 2p, , l, and damping is larger than in air alone., g, , 83. In forced oscillation of a particle the amplitude is, maximum for a frequency w1 of the force while the energy, is maximum for a frequency w2 of the force; then, [2004], (a) w1 < w2 when damping is small and w1 > w2 when, damping is large, (b) w1 > w2, (c) w1 = w2, (d) w1 < w2, 84. A particle of mass m is attached to a spring (of spring, constant k) and has a natural angular frequency w0. An, external force F(t) proportional to cos wt (w ¹ w 0 ) is, applied to the oscillator. The displacement of the oscillator, will be proportional to, [2004], 1, 1, (a), (b) m (w 2 - w 2 ), 2, 2, m (w 0 + w ), 0, m, m, (c), 2, 2, (d), 2, 2, w0 - w, (w 0 + w )
Page 217 :
P-208, , 1., , Physics, , (a) Here, vx = – a w sin wt, vy = a w cos wt and vz = aw, Þ v = v 2x + v 2y + v z2, Þv=, , 2., , ( –aw sin wt ) + ( aw cos wt ), 2, , 2, , + ( aw ), , 5., , 12 0°, , y2, , 2 xy, + 2cos d = sin 2 d, 2, AB, A, B, x = A sin (at + d), y = B sin (bt + r), 3., , 60°, , Clearly A ¹ B hence ellipse., (b) Maximum velocity in SHM, vmax = aw, Maximum acceleration in SHM, Amax = aw2, where a and w are maximum amplitude and angular, frequency., A, Given that, max = 10, v max, i.e., w = 10 s–1, Displacement is given by, x = a sin (wt + p/4), at t = 0, x = 5, 5 = a sin p/4, , p, rad., 3, If they cross each other at time t then, , Angle covered to meet q = 60° =, , t<, 6., , 2, , 2 æ 2A ö, Finally 3V = w A' - ç, ÷, è 3 ø, , 7., 2, , trebled), On dividing we get, , æ 2A ö, A2 - ç, ÷, è 3 ø, , 0, , 0, , x, , t, , 0, , 0, , 1, sin 2t, 2, (c) At t = 0, x (t) = 0 ; y (t) = 0, x (t) is a sinusoidal function, , x = sin t –, , Where A'= final amplitude (Given at x =, , 3, =, 1, , t, , ò dV µ ò sin t dt, , ò dx = ò (- cos t + 1) dt, , 2, 2, (b) We know that V = w A - x, , æ 2A ö, A' 2 - ç, ÷, è 3 ø, , 0, , V µ – cos t + 1, , Maximum acceleration Amax = aw2 = 500 2 m/s2, 2 æ 2A ö, Initially V = w A - ç, ÷, è 3 ø, , q, p, T, <, T<, 2p 3≥2p, 6, , (c) As we know,, F = ma Þ a µ F, or, a µ sin t, dv, Þ, µ sin t, dt, Þ, , 5 = a sin 45° Þ a = 5 2, , 4., , A, , O, , A, 2, (at time t = 0), , 2, , v = 2aw, (c) From the two mutually perpendicular S.H.M.’s, the, general equation of Lissajous figure,, x2, , Equilibrium point, , (d), , 2A, , velocity to, 3, , 2, , p, ; x (t) = a and y (t) = 0, 2w, Hence trajectory of particle will look like as (c)., (c) Given: Time period, T = 0.5 sec, Amplitude, A = 1 cm, Average velocity in the interval in which body moves, from equilibrium to half of its amplitude, v = ?, S, , At t =, , 8., , 2, , é 2 4A 2 ù, 4A 2, 9 ê A - 9 ú = A'2 –, 9, ëê, ûú, , \ A' =, , 7A, 3, , O
Page 219 :
P-210, , 17., , Physics, , (a) Clearly sin 2wt is a periodic function with period, , p, w, , x = 4 2 sin(p t + 45°), On comparing it with standard equation x = A sin(wt + f), we get A = 4 2, 21. (c) From graph equation of SHM, , 0, 2, , For SHM, , d y, dt, , 2, , y = sin2 wt =, =, , p/w, , 2p/w, , X = A cos wt, , 3p/w, , 3T, particle is at mean position., 4, \ Acceleration = 0, Force = 0, (2) At T particle again at extreme position so acceleration, is maximum., , (1) At, , µ -y, , 1 – cos 2wt, 2, , 1 1, – cos 2wt, 2 2, , (3) At t =, , dy 1, = ´ 2w sin 2wt = 2w sin wt cos wt, dt 2, = w sin 2wt, , maximum., Acceleration = 0, (4) When KE = PE, , v=, , 2, , d y, , 2, , = 2w cos 2wt which is not, dt 2, proportional to –y. Hence, it is not in SHM., 18. (b) Velocity of particle 1,, , Acceleration, a =, , dy1, pö, æ, = 0.1 ´ 100p cos ç100pt + ÷, è, dt, 3ø, Velocity of particle 2,, v1 =, , p p 2 p - 3p, p, =–, = f1 - f 2 = - =, 3 2, 6, 6, (c) Maximum velocity during SHM, Vmax = Aw, But k = mw2, , \ w=, , k, m, , \ Maximum velocity of the body in SHM, =A, , k, m, , 20., , (c), , A, , 1, 22. (d) Kinetic energy, k = mw 2 A 2 cos 2 wt, 2, 1, Potential energy, U = mw 2 A 2 sin 2 wt, 2, k, 1, 2, 2 p, = cot wt = cot, (210) =, U, 90, 3, 1, 2 2, 23. (a) K = mw x, 2, , w=, k2, , Þ, , 2, , 1, mw 2 A 2, 2, , A = Lq, , k1, k, = A2 2, m, m, , Þ A1 k 1 = A 2, , +A, , T, = A cos wt Þ t =, 2, 2, \ x = – A which is not possible, \ 1, 2 and 3 are correct., , Þ, , Þ K max =, , As maximum velocities are equal, \ A1, , 1, 1, k ( A2 - x 2 ) = kx 2, 2, 2, Here, A = amplitude of SHM, x = displacement from mean position, Þ, , Þ A2 = 2 x 2 Þ x =, , dy, pö, æ, v2 = 2 = - 0.1p sin pt = 0.1p cos ç pt + ÷, è, dt, 2ø, \ Phase difference of velocity of particle 1 with respect to, the velocity of particle 2 is, , 19., , T, , particle is at mean position so velocity is, 4, , A1, =, A2, , Displacement, x = 4(cos pt + sin pt ), , æ sin pt cos pt ö, = 2 ´ 4ç, +, ÷, è 2, 2 ø, , = 4 2(sin p t cos 45° + cos p t sin 45°), , k2, k1, , g, L, , Þ K=, , 1 g 2 2, m. .L q, 2 L, , 1, mgLq 2, 2, K, L 1, \ 1 =, =, Þ K 2 = 2K1, K 2 2L 2, =
Page 220 :
P-211, , Oscillations, , \, , K1, L 1, =, =, Þ K 2 = 2K1, K 2 2L 2, 1 2, kx, 2, , 24. (c) Potential energy (U) =, , 1 2 1 2, kA - kx, 2, 2, According to the question, U = k, , 25., , Kinetic energy (K) =, , Work done by spring S1, w1 =, , 1, k1x 2, 2, , 1, 1, 1, \ kx 2 = kA2 - kx 2, 2, 2, 2, A, Þ x2 = A2 or, x =±, 2, (b) For a particle executing SHM, At mean position; t = 0, wt = 0, y = 0, V = Vmax = aw, 1, \ K.E. = KEmax = mw2a2, 2, T, p, At extreme position : t = , wt = , y = A, V = Vmin = 0, 4, 2, \ K.E. = KEmin = 0, , Work done by spring S2, w2 =, , 1, k2 x 2, 2, , 1, Kinetic energy in SHM, KE = mw2(a2 – y2), 2, 1, = mw2a2cos2wt, 2, Hence graph (b) correctly depicts kinetic energy time graph., 26. (b) Since system dissipates its energy gradually, and, hence amplitude will also decreases with time according to, a = a0 e–bt/m, ....... (i), Q Energy of vibration drop to half of its initial value, , (E0), as E µ a2 Þ a µ E, -2, , a, a = 0 Þ bt = 10 t = t, 2, m, 0.1, 10, From eqn (i),, a0, = a 0e - t 10, 2, t, , 27., , 28., , Taking log on both sides, b, 1, =, ln3, m 15, 1 2, 29. (b) Work done, w = kx, 2, , 1, = e - t 10 or 2 = e10, 2, t, ln 2 =, \ t = 3.5 seconds, 10, 1, 2, 2, (d) K.E = k ( A - d ), 2, 1 2, and P.E. = kd, 2, At mean position d = 0. At extreme positions d = A, , (d) As we know, E = E0 e, –, , –, , bt, m, , b15, , 15 = 45e m, [As no. of oscillations = 15 so t = 15sec], –, 1, =e, 3, , b15, m, , Since w1 > w2 Thus (k1 > k2), 30. (b) The kinetic energy of a particle executing S.H.M. at, any instant t is given by, 1 2 2 2, ma w sin wt, 2, where, m = mass of particle, a = amplitude, w = angular frequency, t = time, , K=, , The average value of sin 2wt over a cycle is, , 1, ., 2, , 1ö, æ 1ö æ, 1, 2, \KE = mw2a2 çè ÷ø çè Q < sin q > = ÷ø, 2, 2, 2, =, , 1, 1, 2 2, mw a = ma2 (2pn)2, 4, 4, 2, , (Q w = 2pn), , 2 2, , or, < K > = p ma n, 31. (a) K.E. of a body undergoing SHM is given by,, 1, 2 2, 2, ma w cos wt, 2, Here, a = amplitude of SHM, w = angular velocity of SHM, K .E. =, , 1 2 2, Total energy in S.H.M = ma w, 2, Given K.E. = 75% T.E., 1, 75 1, ma 2 w 2 cos 2 w t =, ´ ma 2 w 2, 2, 100 2, p, 2, Þ 0.75 = cos wt Þ wt =, 6, p, p´2, 1, Þt=, Þt=, Þt= s, 6´w, 6 ´ 2p, 6, 32. (a) At any instant the total energy in SHM is, 1 2, kA = constant,, 2 0, where A0 = amplitude, k = spring constant, hence total energy is independent of x., 33. (a) K.E. of simple harmonic motion, 1, = mw2 (a 2 - x 2 ), 2
Page 225 :
P-216, , 62., , Physics, , (d) The equation of motion for the pendulum, for, damped harmonic motion, F = – kx - bv, Þ ma + kx + bv = 0, d 2x, , dx, m 2 + kx + b, =0, dt, dt, , Þ, , k, b dx, + x+, =0, m dt, dt 2 m, , d2 x, , Þ, , dt, , 2, , +, , b dx k, + x =0, m dt m, , … (1), , Let x = elt is the solution of the equation (1), dx, d2 x, = lelt Þ, = l 2 elt, dt, dt 2, Substituting in the equation (1), b, k, l 2 elt + l elt + elt = 0, m, m, b, k, l2 + l + = 0, m, m, , b, b2, k, ±, -4, 2, m, m, -b ±, m, l=, =, 2, Solving the equation (1) for x,, -, , x=, , b2 - 4km, 2m, we have, , -b, t, 2, e m, , k, +b, w = w0 2 - l 2 where w0 = , l =, m, 2, 1 2, The average life = =, l b, 63. (a), 64. (a) Let the cube be at a depth x from the equilibrium, position., Force acting on the cube = up thrust on the portion of, length x., , F = – rl 2 xg [\ mass density X volume ] ....(i), Clearly F µ – x, Hence it is a SHM., Equation of SHM is F = –kx, ....(ii), Comparing equation (i) and (ii) we have, k = rl2g, Now, Time period, T = 2p, Þ T = 2p, , = 2p, , a = -w2 x where w 2 is a constant., , a=, , d2 x, , Þ, , rg, ld, Þ T = 2p, dl, rg, 65. (a) For an SHM, the acceleration, , m, k, , w=, , \, , –4p 2 x, T2, , Þ, , aT –4p2, =, x, T, , aT, The time period T is also constant. Therefore,, is a, x, constant., 66. (a) The two springs are in parallel., \ Effective spring constant,, k = k1 + k2, Initial frequency of oscillation is given by, 1 k1 + k 2, ....(i), m, 2p, When both k1 and k2 are made four times their original, values, the new frequency is given by, , v =, , v' =, =, , 1, 2p, , 1, 2p, , 4 k1 + 4 k 2, m, , æ 1, 4(k1 + 4k 2 ), = 2ç, ç 2p, m, è, , dx, = 2 × 10–2 p sin p t, dt, For the first time, the when velocity becomes maximum,, sin p t = 1, Þ sin p t = sin p, 2, v=, , 1, p, or,, t = = 0.5 sec., 2, 2, 68. (b) When plugged hole near the bottom of the oscillating, bob gets suddenly unplugged, centre of mass of, combination of liquid and hollow portion (at position l ),, Þ pt =, , first goes down ( to l + D l) and when total water is drained, , out, centre of mass regain its original position (to l ),, Time period, T = 2p, , l, g, , \ ‘T ’ first increases and then decreases to original value., , l d, , rl 2 g, , Comparing the above equation with, a = –w2x, we get, , ö, ÷÷ = 2v, ø, , 67. (b) Here, Displacement, x = 2 × 10–2 cos p t, Velocity is given by, , 3, , ld, rg, , k1 + k2, m, , c
Page 226 :
P-217, , Oscillations, , 70., , (a) Time period, t = 2p, , l, , g eff, , Þ 1+, , ;, , l, g, , In air, t 0 = 2p, , Þ, , Buoyant, force, , 4, ´ 1000 Vg, 3, , 1000, æ4 ö, Vg, Net force = ç - 1 ÷ ´ 1000 Vg =, 3, è3 ø, 1000 Vg, g, geff =, =, 4, 3 ´ ´1000V 4, 3, , \ t = 2p, , l, l, = 2 ´ 2p, g /4, g, , m, (b) Time period for first spring, t1 = 2p, ,, k1, m, Time period for second spring, t 2 = 2p, k2, , k1k 2, kl + k 2, \ Time period of oscillation for series combination, , Force constant of the series combination keff =, , m( k l + k 2 ), k1k 2, , T = 2p, , \ T = 2p, , 2, t1, , m m, +, = 2p, +, 2, 2, k 2 k1, (2p), (2p), , Þ T 2 = t12 + t 22, 72., , where x is the displacement from the mean position, (c) With mass M, the time period of the spring., M, k, With mass M + m, the time period becomes,, , T = 2p, , T ' = 2p, \ 2p, , Þ, , l' = l + 0.21 l, Þ l' = 1.21 l, , T ' = 2p, , M + m 5T, =, k, 3, , 1.21l, g, , % increase in length =, 1.21l - l, l, , ´ 100 =, , T '- T, ´100, T, , (, , ), , 1.21 - 1 ´ 100, , = (1.1 - 1) ´ 100 = 10%, 74. (b) Let k be the spring constant of the original spring., Time period T = 2p, , m, where m is the mas s of oscillating, k, , body., When the spring is cut into n equal parts, the spring, constant of one part becomes nk. Therefore the new time, period,, T ' = 2p, , 2, t2, , l, g, , New length, l ' = l + 2 1 % o f l, , =, , t = 2t0, 71., , 16, m 25, =, -1 =, M, 9, 9, , 73. (d) Time period, T = 2p, , 1000 Vg, , Weight, , m 25, =, M, 9, , m, T, =, nk, n, , l, where l = distance, g, between the point of suspension and the centre of mass of, the child., As the child stands up, her centre of mass is raised. The, distance between point of suspension and centre of mass, decreases ie length l decreases., , 75. (b) The time period T = 2p, , \ l¢ < l, \ T ¢ < T i.e., the period decreases., point of suspension, l', , l, , M +m 5, M, = ´ 2p, 3, k, k, , M +m =, , 25, ´M, 9, , Case (ii) child standing, , Case (i) child sitting
Page 227 :
P-218, , Physics, , Putting value of M, m, u and K we get amplitude, , point of suspension, , 1 75, 1, =, 15 1, 3, 79. (b) The change in time period compared to the undamped, oscillator increases by 8%., A=, , l', , l, , Case (ii) child standing, , 80., , Case (i) child sitting, , 0.9A 0 = A 0 e, , ...(i), , A0, = Ae e- kt, and, 1000, , ...(ii), , Dividing (i) by (ii) and solving, we get, t = 20 s, 77. (b) Amplitude of vibration at time t = 0 is given by, A = A0e –0.1× 0 = 1 × A0 = A0, also at t = t, if A =, , Þ, , 78., , A0, 2, , -, , b(5), 2m, , … (i), , After 10 more second,, A = A0 e, , -, , b(15), 2m, , …(ii), , From eqns (i) and (ii), A = 0.729 A0, \ a = 0.729, 81. (a), 82. (d) When the pendulum is oscillating over a current, carrying coil, and when the direction of oscillating, pendulum bob is opposite to the direction of current. Its, instantaneous acceleration increases., , l, g, and damping is larger than in air alone due energy dissipation., 83. (c) As energy µ ( Amplitude)2, the maximum for both, of them occurs at the same frequency and this is only, possible in case of resonance., In resonance state w1 = w 2, 84. (b) Equation of displacement in forced oscillation is given, by, Hence time period T < 2p, , 1, = e –0.1t, 2, , t = 10 ln 2 ;7s, (b) In first collision mu momentum will be imparted to, system, in second collision when momentum of (M + m) is in, opposite direction mu momentum of particle will make its, momentum zero., On 13th collision, m ® M + 12 ; M + 13m ® V, mu = (M + 13m)v Þ v =, , bt, 2m, , (where, A0 = maximum amplitude), According to the questions, after 5 second,, , 76. (c) Time of half the amplitude is = 2s, Using, A = A0e–kt, A0, = Ae e - k ´2, 2, , (c) Q A = A 0e, , -, , mu, u, =, M + 13m 15, , u, K, ´A, v = wA Þ =, 15, M - 13m, , y=, , F0, 2, , 2 2, , m (w 0 - w ), , =, , F0, , m (w 0 2 - w 2 ), Here damping effect is considered to be zero, , \x µ, , 1, 2, , 2, , m (w 0 - w )
Page 228 :
14, Waves, Basic of Mechanical Waves,, TOPIC 1 Progressive and Stationary, Waves, 1., , 2., , Assume that the displacement (s) of air is proportional, to the pressure difference (Dp) created by a sound wave., Displacement (s) further depends on the speed of sound, (v), density of air (r) and the frequency ( f ). If Dp ~, 10Pa, v ~ 300 m/s, r ~ 1 kg/m3 and f ~ 1000 Hz, then s, will be of the order of (take the multiplicative constant, to be 1), [Sep. 05, 2020 (I)], 3, (a), mm, (b) 10 mm, 100, 1, (c), mm, (d) 1 mm, 10, For a transverse wave travelling along a straight line, the, distance between two peaks (crests) is 5 m, while the, distance between one crest and one trough is 1.5 m. The, possible wavelengths (in m) of the waves are :, [Sep. 04, 2020 (I)], 1 1 1, , , , ......., 1 3 5, 1 1 1, , , , ......., (c) 1, 2, 3, ....., (d), 2 4 6, A progressive wave travelling along the positive x-direction, is represented by y(x,t) = Asin (kx – wt + f). Its snapshot at, t = 0 is given in the figure., [12 April 2019 I], , (a) 1, 3, 5, ....., , 3., , (b), , 4., , 5., , 6., , 7., , 8., , For this wave, the phase f is :, p, (a) (b) p, (c) 0, 2, , (d), , p, 2, , A small speaker delivers 2 W of audio output. At what, distance from the speaker will one detect 120 dB intensity, sound ? [Given reference intensity of sound as 10–12W/m2], [12 April 2019 II], (a) 40 cm (b) 20 cm, (c) 10 cm (d) 30 cm, The pressure wave,, P = 0.01 sin[1000t – 3x] Nm–2, corresponds to the sound, produced by a vibrating blade on a day when atmospheric, temperature is 0°C. On some other day when temperature, is T, the speed of sound produced by the same blade and, at the same frequency is found to be 336 ms–1. Approximate, value of T is :, [9 April 2019 I], (a) 4°C, (b) 11°C, (c) 12°C (d), 15°C, A travelling harmonic wave is represented by the equation y(x, t)=10–3 sin(50t+2x), where x and y are in meter and, t is in seconds. Which of the following is a correct statement about the wave?, [12 Jan. 2019 I], (a) The wave is propagating along the negative x-axis, with speed 25 ms–1., (b) The wave is propagating along the positive x-axis with, speed 100 ms–1., (c) The wave is propagating along the positive x-axis with, speed 25 ms–1., (d) The wave is propagating along the negative x-axis, with speed 100 ms–1., A transverse wave is represented by, 10 æ 2p 2p ö, y = sin ç t x÷, p, l ø, è T, For what value of the wavelength the wave velocity is, twice the maximum particle velocity?, [Online April 9, 2014], (a) 40 cm (b) 20 cm, (c) 10 cm (d) 60 cm, In a transverse wave the distance between a crest and, neighbouring trough at the same instant is 4.0 cm and the, distance between a crest and trough at the same place is, 1.0 cm. The next crest appears at the same place after a, time interval of 0.4s. The maximum speed of the vibrating, particles in the medium is :, [Online April 25, 2013], 3p, 5p, cm/s, cm/s, (a), (b), 2, 2, p, cm/s, (c), (d) 2p cm/s, 2
Page 229 :
P-220, , Physics, , 9., , When two sound waves travel in the same direction in a, medium, the displacements of a particle located at 'x' at, time ‘t’ is given by :, y1 = 0.05 cos (0.50 px – 100 pt), y2 = 0.05 cos (0.46 px – 92 pt), where y1, y2 and x are in meters and t in seconds. The, speed of sound in the medium is : [Online April 9, 2013], (a) 92 m/s (b) 200 m/s (c) 100 m/s (d) 332 m/s, 10. The disturbance y (x, t) of a wave propagating in the, 1, positive x-direction is given by y =, at time t = 0, 1 + x2, 1, and by y =, at t = 2 s, where x and y are in, é1 + ( x - 1) 2 ù, ë, û, meters. The shape of the wave disturbance does not change, during the propagation. The velocity of wave in m/s is, [Online May 26, 2012], (a) 2.0, (b) 4.0, (c) 0.5, (d) 1.0, 11. The transverse displacement y (x, t) of a wave is given by, y( x, t ) = e, , (, , - ax 2 + bt 2 + 2 ab ) xt, , This represents a:, , )., , [2011], , (a) wave moving in – x direction with speed, , b, a, , (b) standing wave of frequency, , b, 1, (c) standing wave of frequency, b, , (d) wave moving in + x direction speed, , a, b, , 12., , A wave travelling along the x-axis is described by the, equation y(x, t) = 0.005 cos (a x – bt). If the wavelength, and the time period of the wave are 0.08 m and 2.0s,, respectively, then a and b in appropriate units are [2008], 0.08, 2.0, ,b =, (a) a = 25.00 p , b = p (b) a =, p, p, 0.04, 1.0, p, ,b =, (c) a =, (d) a = 12.50p, b =, p, p, 2.0, 13. A sound absorber attenuates the sound level by 20 dB., The intensity decreases by a factor of, [2007], (a) 100, (b) 1000, (c) 10000 (d) 10, 14. The displacement y of a particle in a medium can be, , pö, æ, -6, expressed as, y = 10 sin ç100t + 20 x + ÷ m where t is, è, 4ø, in second and x in meter. The speed of the wave is [2004], (a) 20 m/s, (b) 5 m/s, (c) 2000 m/s, (d) 5p m/s, 15. The displacement y of a wave travelling in the x -direction, pö, æ, is given by y = 10 - 4 sin ç 600 t - 2 x + ÷ metres, 3ø, è, where x is expressed in metres and t in seconds. The speed, of the wave - motion, in ms–1 , is, [2003], (a) 300, (b) 600, (c) 1200 (d) 200, , 16. When temperature increases, the frequency of a tuning, fork, [2002], (a) increases, (b) decreases, (c) remains same, (d) increases or decreases depending on the material, , TOPIC 2 Vibration of String and Organ, Pipe, 17. In a resonance tube experiment when the tube is filled, with water up to a height of 17.0 cm from bottom, it resonates with a given tuning fork. When the water level is, raised the next resonance with the same tuning fork occurs at a height of 24.5 cm. If the velocity of sound in air, is 330 m/s, the tuning fork frequency is :, [Sep. 05, 2020 (I)], (a) 2200 Hz, (b) 550 Hz, (c) 1100 Hz, (d) 3300 Hz, 18. A uniform thin rope of length 12 m and mass 6 kg hangs, vertically from a rigid support and a block of mass 2 kg, is attached to its free end. A transverse short wave-train, of wavelength 6 cm is produced at the lower end of the, rope. What is the wavelength of the wavetrain (in cm), when it reaches the top of the rope ?[Sep. 03, 2020 (I)], (a) 3, (b) 6, (c) 12, (d) 9, 19. Two identical strings X and Z made of same material have, tension T X an d T Z in them. If their fundamental, frequencies are 450 Hz and 300 Hz, respectively, then, the ratio TX/TZ is:, [Sep. 02, 2020 (I)], (a) 2.25, (b) 0.44, (c) 1.25, (d) 1.5, 20. A wire of density 9 × 10–3 kg cm–3 is stretched between, two clamps 1 m apart. The resulting strain in the wire is, 4.9 × 10–4. The lowest frequency of the transverse, vibrations in the wire is, (Young’s modulus of wire Y = 9 × 1010 Nm–2), (to the nearest, integer), ___________., [Sep. 02, 2020 (II)], 21. A one metre long (both ends open) organ pipe is kept in a, gas that has double the density of air at STP. Assuming, the speed of sound in air at STP is 300 m/s, the frequency, difference between the fundamental and second harmonic, of this pipe is ______ Hz., [NA 8 Jan. 2020 (I)], 22. A transverse wave travels on a taut steel wire with a, velocity of v when tension in it is 2.06 ´ l04 N. When, the tension is changed to T, the velocity changed to v/2., The value of T is close to:, [8 Jan. 2020 (II)], (a) 2.50 ´ l04 N, (b) 5.15 ´ l03 N, (c) 30.5 ´ l04 N, (d) 10.2 ´ l02 N, 23. Speed of a transverse wave on a straight wire (mass 6.0, g, length 60 cm and area of cross-section 1.0 mm2) is 90, ms–1. If the Young’s modulus of wire is 16 ´ l011 Nm–2 the, extension of wire over its natural length is:, [7 Jan. 2020 (I)]
Page 230 :
P-221, , Waves, , (a) 0.03 mm, (c) 0.04 mm, , (b) 0.02 mm, (d) 0.01 mm, , 24. Equation of travelling wave on a stretched string of linear, density 5 g/m is y = 0.03 sin (450 t – 9x) where distance and, time are measured in SI units. The tension in the string is:, [11 Jan 2019 (I)], (a) 10 N, (b) 7.5 N, (c) 12.5 N (d) 5 N, 25. A heavy ball of mass M is suspended from the ceiling of, a car by a light string of mass m (m<<M). When the car is, at rest, the speeed of transverse waves in the string is 60, ms–1. when the car has acceleration a, the wave-speed, increases to 60.5 ms–1. The value of a, in terms of, gravitational acceleration g, is closest to: [9 Jan. 2019 (I)], , g, g, g, g, (b), (c), (d), 30, 5, 10, 20, A wire of length L and mass per unit length 6.0 × 10–3 kgm–1, is put under tension of 540 N. Two consecutive, frequencies that it resonates at are: 420 Hz and 490 Hz., Then L in meters is:, [9 Jan. 2020 (II)], (a) 2.1 m, (b) 1.1 m, (c) 8.1 m, (d) 5.1 m, A tuning fork of frequency 480 Hz is used in an experiment, for measuring speed of sound (v) in air by resonance, tube method. Resonance is observed to occur at two, successive lengths of the air column, l1 = 30 cm and l2 = 70, cm. Then, v is equal to :, [12 April 2019 (II)], (a) 332 ms–1, (b) 384 ms–1, (c) 338 ms–1, (d) 379 ms–1, A string 2.0 m long and fixed at its ends is driven by a 240, Hz vibrator. The string vibrates in its third harmonic mode., The speed of the wave and its fundamental frequency is:, [9 April 2019 (II)], (a) 180 m/s, 80 Hz, (b) 320 m/s, 80 Hz, (c) 320 m/s, 120 Hz, (d) 180 m/s, 120 Hz, A string is clamped at both the ends and it is vibrating in, its 4th harmonic. The equation of the stationary wave is Y, = 0.3 sin(0.157x) cos(200At). The length of the string is:, (All quantities are in SI units.), [9 April 2019 (I)], (a) 20 m, (b) 80 m, (c) 40 m (d), 60 m, A wire of length 2L, is made by joining two wires A and B, of same length but different radii r and 2r and made of the, same material. It is vibrating at a frequency such that the, joint of the two wires forms a node. If the number of, antinodes in wire A is p and that in B is q then the ratio, p : q is :, [8 April 2019 (I)], (a), , 26., , 27., , 28., , 29., , 30., , (a) 3 : 5, (b) 4 : 9, (c) 1 : 2, (d), 1: 4, 31. A closed organ pipe has a fundamental frequency of 1.5, kHz. The number of overtones that can be distinctly, heard by a person with this organ pipe will be: (Assume, that the highest frequency a person can hear is 20,000 Hz), [10 Jan. 2019 (I)], (a) 6, (b) 4, (c) 7, (d) 5, , 32. A string of length 1 m and mass 5 g is fixed at both, ends. The tension in the string is 8.0 N. The string is, set into vibration using an external vibrator of frequency, 100 Hz. The separation between successive nodes on, the string is close to:, [10 Jan. 2019 (I)], (a) 10.0 cm (b) 33.3 cm (c) 16.6 cm (d) 20.0 cm, 33. A granite rod of 60 cm length is clamped at its middle point, and is set into longitudinal vibrations. The density of, granite is 2.7 × 103 kg/m3 and its Young's modulus is, 9.27×1010 Pa., What will be the fundamental frequency of the longitudinal, vibrations?, [2018], (a) 5 kHz (b) 2.5 kHz (c) 10 kHz (d) 7.5 kHz, 34. The end correction of a resonance column is 1cm. If the, shortest length resonating with the tuning fork is 10cm,, the next resonating length should be, [Online April 16, 2018], (a) 32cm, (b) 40cm, (c) 28cm (d) 36cm, 35. Two wires W1 and W2 have the same radius r and, respective densities r1 and r2 such that r2 = 4r1. They, are joined together at the point O, as shown in the figure., The combination is used as a sonometer wire and kept, under tension T. The point O is midway between the two, bridges. When a stationary waves is set up in the, composite wire, the joint is found to be a node. The ratio, of the number of antinodes formed in W1 to W2 is :, [Online April 8, 2017], r1, r2, O, W, W1, 2, (a) 1 : 1, (b) 1 : 2, (c) 1 : 3, (d) 4 : 1, 36. A uniform string of length 20 m is suspended from a rigid, support. A short wave pulse is introduced at its lowest, end. It starts moving up the string. The time taken to reach, the supports is :, [2016], (take g = 10 ms–2), (a) 2 2s (b) 2 s, (d)2 s, (c) 2p 2 s, 37. A pipe open at both ends has a fundamental frequency f, in air. The pipe is dipped vertically in water so that half of, it is in water. The fundamental frequency of the air column, is now :, [2016], f, 3f, (a) 2f, (b) f, (c), (d), 2, 4, 38. A pipe of length 85 cm is closed from one end. Find the, number of possible natural oscillations of air column in, the pipe whose frequencies lie below 1250 Hz. The velocity, of sound in air is 340 m/s., [2014], (a) 12, (b) 8, (c) 6, (d) 4, 39. The total length of a sonometer wire between fixed ends is, 110 cm. Two bridges are placed to divide the length of wire, in ratio 6 : 3 : 2. The tension in the wire is 400 N and the, mass per unit length is 0.01 kg/m. What is the minimum, common frequency with which three parts can vibrate?, [Online April 19, 2014]
Page 231 :
P-222, , 40., , 41., , 42., , 43., , 44., , 45., , 46., , 47., , (a) 1100 Hz, (b) 1000 Hz, (c) 166 Hz, (d) 100 Hz, A sonometer wire of length 1.5 m is made of steel. The, tension in it produces an elastic strain of 1%. What is the, fundamental frequency of steel if density and elasticity of, steel are 7.7 × 103 kg/m3 and 2.2 × 1011 N/m2 respectively?, (a) 188.5 Hz, (b) 178.2 Hz, [2013], (c) 200.5 Hz, (d) 770 Hz, A sonometer wire of length 114 cm is fixed at both the ends., Where should the two bridges be placed so as to divide the, wire into three segments whose fundamental frequencies, are in the ratio 1 : 3 : 4 ?, [Online April 23, 2013], (a) At 36 cm and 84 cm from one end, (b) At 24 cm and 72 cm from one end, (c) At 48 cm and 96 cm from one end, (d) At 72 cm and 96 cm from one end, A cylindrical tube, open at both ends, has a fundamental, frequency f in air. The tube is dipped vertically in water so, that half of it is in water. The fundamental frequency of the, air-column is now :, [2012], (a) f, (b) f / 2, (c) 3 f /4 (d) 2 f, An air column in a pipe, which is closed at one end, will, be in resonance wtih a vibrating tuning fork of frequency, 264 Hz if the length of the column in cm is (velocity of, sound = 330 m/s), [Online May 26, 2012], (a) 125.00 (b) 93.75, (c) 62.50 (d) 187.50, A uniform tube of length 60.5 cm is held vertically with, its lower end dipped in water. A sound source of frequency, 500 Hz sends sound waves into the tube. When the length, of tube above water is 16 cm and again when it is 50 cm,, the tube resonates with the source of sound. Two lowest, frequencies (in Hz), to which tube will resonate when it, is taken out of water, are (approximately)., [Online May 19, 2012], (a) 281, 562, (b), 281, 843, (c) 276, 552, (d), 272, 544, The equation of a wave on a string of linear mass density, 0.04 kg m–1 is given by, é æ t, x öù, y = 0.02(m) sin ê2p ç, ÷ú ., ë è 0.04(s ) 0.50(m) ø û, The tension in the string is, [2010], (a) 4.0 N (b) 12.5 N (c) 0.5 N (d) 6.25 N, While measuring the speed of sound by performing a, resonance column experiment, a student gets the first, resonance condition at a column length of 18 cm during, winter. Repeating the same experiment during summer,, she measures the column length to be x cm for the second, resonance. Then, [2008], (a) 18 > x, (b) x > 54, (c) 54 > x > 36, (d) 36 > x > 18, A string is stretched between fixed points separated by, 75.0 cm. It is observed to have resonant frequencies of, 420 Hz and 315 Hz. There are no other resonant, frequencies between these two. Then, the lowest resonant, frequency for this string is, [2006], , Physics, , (a) 105 Hz, (b) 1.05 Hz, (c) 1050 Hz, (d) 10.5 Hz, 48. Tube A has both ends open while tube B has one end, closed, otherwise they are identical. The ratio of, fundamental frequency of tube A and B is, [2002], (a) 1 : 2, (b) 1 : 4, (c) 2 : 1, (d) 4 : 1, 49. A wave y = a sin(wt–kx) on a string meets with another, wave producing a node at x = 0. Then the equation of the, unknown wave is, [2002], (a) y = a sin( w t + kx), (b) y = –a sin( w t + kx), (c) y = a sin( w t – kx), (d) y = –a sin( w t – kx), , TOPIC 3, , Beats, Interference and, Superposition of Waves, , 50. There harmonic waves having equal frequency n and same, p, p, and - respectively.., 4, 4, When they are superimposed the intensity of the resultant, wave is close to:, [9 Jan. 2020 I], (a) 5.8 I0, (b) 0.2 I0, (c) 3 I0, (d) I0, 51. The correct figure that shows, schematically, the wave, pattern produced by superposition of two waves of, frequencies 9 Hz and 11 Hz is :, [10 April 2019 II], , intensity I0, have phase angles 0,, , (a), , (b), , (c), , (d), 52. A resonance tube is old and has jagged end. It is still used, in the laboratory to determine velocity of sound in air. A, tuning fork of frequency 512 Hz produces first resonance, when the tube is filled with water to a mark 11 cm below, a reference mark, near the open end of the tube. The, experiment is repeated with another fork of frequency, 256 Hz which produces first resonance when water
Page 232 :
P-223, , Waves, , reaches a mark 27 cm below the reference mark. The, velocity of sound in air, obtained in the experiment, is, close to:, [12 Jan. 2019 II], (a) 322 ms–1, (b) 341 ms–1, (c) 335 ms–1, (d) 328 ms–1, 53. A tuning fork vibrates with frequency 256 Hz and gives, one beat per second with the third normal mode of, vibration of an open pipe. What is the length of the pipe?, (Speed of sound of air is 340 ms–1), [Online April 15, 2018], (a) 190 cm, (b) 180 cm, (c) 220 cm, (d) 200 cm, 54. 5 beats/ second are heard when a turning fork is sounded, with a sonometer wire under tension, when the length of, the sonometer wire is either 0.95m or 1m . The frequency, of the fork will be:, [Online April 15, 2018], (a) 195Hz (b) 251Hz (c) 150Hz (d) 300Hz, 55. A standing wave is formed by the superposition of two, waves travelling in opposite directions. The transverse, displacement is given by, , æ 5p ö, y(x, t) = 0.5 sin ç x ÷ cos(200 pt)., è 4 ø, What is the speed of the travelling wave moving in the, positive x direction ?, (x and t are in meter and second, respectively.), [Online April 9, 2017], (a) 160 m/s (b) 90 m/s (c) 180 m/s (d) 120 m/s, 56. A wave represented by the equation y1 = acos (kx – wt) is, superimposed with another wave to form a stationary wave, such that the point x – 0 is node. The equation for the, other wave is, [Online May 12, 2012], (a) a cos (kx – wt + p), (b) a cos (kx + wt + p), pö, pö, æ, æ, a cos ç kx + wt + ÷, a cos ç kx - wt + ÷, (d), è, è, 2ø, 2ø, 57. Following are expressions for four plane simple harmonic, waves, [Online May 7, 2012], æ, ö, x, (i) y1 = A cos 2p ç n1t + ÷, l1 ø, è, , 58. A travelling wave represented by, y = A sin (wt – kx) is superimposed on another wave, represented by y = A sin (wt + kx). The resultant is, (a) A wave travelling along + x direction [2011 RS], (b) A wave travelling along – x direction, (c) A standing wave having nodes at, , 59., , 60., , 61., , 62., , (c), , æ, ö, x, (ii) y2 = A cos 2p ç n1t + + p÷, l1, è, ø, æ, xö, (iii) y3 = A cos 2p ç n2t + ÷, l2 ø, è, , æ, xö, (iv) y4 = A cos 2p ç n2t - ÷, l2 ø, è, The pairs of waves which will produce destructive, interference and stationary waves respectively in a, medium, are, (a) (iii, iv), (i, ii), (b) (i, iii), (ii, iv), (c) (i, iv), (ii, iii), (d) (i, ii), (iii, iv), , nl, , n = 0,1, 2...., 2, (d) A standing wave having nodes at, 1ö l, æ, x = ç n + ÷ ; n = 0,1, 2...., è, 2ø 2, Statement - 1 : Two longitudinal waves given by, equations : y1(x, t) = 2a sin (wt – kx) and y2(x, t) = a, sin (2wt - 2kx) will have equal intensity.., [2011 RS], Statement - 2 : Intensity of waves of given frequency in, same medium is proportional to square of amplitude only., (a) Statement-1 is true, statement-2 is false., (b) Statement-1 is true, statement-2 is true, statement2 is the correct explanation of statement-1, (c) Statement-1 is true, statement-2 is true, statement2 is not the correct explanation of statement-1, (d) Statement-1 is false, statement-2 is true., Three sound waves of equal amplitudes have frequencies, (n –1), n, (n + 1). They superpose to give beats. The number, of beats produced per second will be :, [2009], (a) 3, (b) 2, (c) 1, (d) 4, When two tuning forks (fork 1 and fork 2) are sounded, simultaneously, 4 beats per second are heard. Now, some, tape is attached on the prong of the fork 2. When the, tuning forks are sounded again, 6 beats per second are, heard. If the frequency of fork 1 is 200 Hz, then what was, the original frequency of fork 2?, [2005], (a) 202 Hz (b) 200 Hz (c) 204 Hz (d) 196 Hz, A tuning fork of known frequency 256 Hz makes 5 beats, per second with the vibrating string of a piano. The beat, frequency decreases to 2 beats per second when the, tension in the piano string is slightly increased. The, frequency of the piano string before increasing the tension, was, [2003], (a) (256 + 2) Hz, (b) (256 – 2) Hz, (c) (256 – 5) Hz, (d) (256 + 5) Hz, A tuning fork arrangement (pair) produces 4 beats/sec with, one fork of frequency 288 cps. A little wax is placed on the, unknown fork and it then produces 2 beats/sec. The, frequency of the unknown fork is, [2002], (a) 286 cps (b) 292 cps (c) 294 cps (d) 288 cps, x=, , 63., , TOPIC 4, , Musical Sound and Doppler's, Effect, , 64. A sound source S is moving along a straight track with, speed v, and is emitting sound of frequency vo (see, figure). An observer is standing at a finite distance, at the, point O, from the track. The time variation of frequency, heard by the observer is best represented by:, [Sep. 06, 2020 (I)]
Page 233 :
P-224, , Physics, , (t0 represents the instant when the distance between the, source and observer is minimum), v, , v, , (a) vo, , v, , (b), t0 t, , vo, , 71., t0, , v, , t, , vo, , vo, , (c), , 72., , (d), t0, , t, , t0, , t, , 65. A driver in a car, approaching a vertical wall notices that, the frequency of his car horn, has changed from 440 Hz to, 480 Hz, when it gets reflected from the wall. If the speed of, sound in air is 345 m/s, then the speed of the car is :, [Sep. 05, 2020 (II)], (a) 54 km/hr, (b) 36 km/hr, (c) 18 km/hr, (d) 24 km/hr, 66. The driver of a bus approaching a big wall notices that the, frequency of his bus’s horn changes from 420 Hz to 490 Hz, when he hears it after it gets reflected from the wall. Find, the speed of the bus if speed of the sound is 330 ms–1., [Sep. 04, 2020 (II)], (a) 91 kmh–1, (b) 81 kmh–1, (c) 61 kmh–1, (d) 71 kmh–1, 67. Magnetic materials used for making permanent magnets, (P) and magnets in a transformer (T) have different, properties of the following, which property best matches, for the type of magnet required?, [Sep. 02, 2020 (I)], (a) T : Large retentivity, small coercivity, (b) P : Small retentivity, large coercivity, (c) T : Large retentivity, large coercivity, (d) P : Large retentivity, large coercivity, 68. A stationary observer receives sound from two identical, tuning forks, one of which approaches and the other one, recedes with the same speed (much less than the speed, of sound). The observer hears 2 beats/sec. The oscillation, frequency of each tuning fork is v0 = 1400 Hz and the, velocity of sound in air is 350 m/s. The speed of each, tuning fork is close to:, [7 Jan. 2020 I], 1, 1, 1, m/s (b) 1m/s, m/s, (a), (c), (d) 8 m/s, 2, 4, 69. A submarine (A) travelling at 18 km/hr is being chased, along the line of its velocity by another submarine (B), travelling at 27 km/hr. B sends a sonar signal of 500 Hz to, detect A and receives a reflected sound of frequency v., The value of v is close to :, [12 April 2019 I], (Speed of sound in water = 1500 ms–1), (a) 504 Hz, (b) 507 Hz, (c) 499 Hz, (d) 502 Hz, 70. Two sources of sound S1 and S2 produce sound waves of, same frequency 660 Hz. A listener is moving from source, S1 towards S2 with a constant speed u m/s and he hears, , 73., , 74., , 75., , 76., , 77., , 78., , 10 beats/s. The velocity of sound is 330 m/s. Then u, equals:, [12 April 2019 II], (a) 5.5 m/s, (b) 15.0 m/s, (c) 2.5 m/s, (d) 10.0 m/s, A stationary source emits sounds waves of frequency, 500 Hz. Two observers moving along a line passing, through the source detect sound to be of frequencies, 4801 Hz and 530 Hz. Their respective speeds are, in, ms–1,, (Given speed of sound = 300 m/s), [10 April 2019 I], (a) 12, 16 (b) 12, 18 (c) 16, 14 (d) 8, 18, A source of sound S is moving with a velocity of 50 m/s, towards a stationary observer. The observer measures, the frequency of the source as 1000 Hz. What will be the, apparent frequency of the source when it is moving away, from the observer after crossing him? (Take velocity of, sound in air 350 m/s), [10 April 2019 II], (a) 750 Hz (b) 857 Hz (c) 1143 Hz (d) 807 Hz, Two cars A and B are moving away from each other in, opposite directions. Both the cars are moving with a speed, of 20 ms–1 with respect to the ground. If an observer in car A, detects a frequency 2000 Hz of the sound coming from car B,, what is the natural frequency of the sound source in car B?, (speed of sound in air = 340 ms–1), [9 April 2019 II], (a) 2250 Hz, (b) 2060 Hz, (c) 2300 Hz, (d) 2150 Hz, A train moves towards a stationary observer with speed, 34 m/s. The train sounds a whistle and its frequency, registered by the observer is f1 If the speed of the train, is reduced to 17 m/s, the frequency registered is f2 If, speed of sound is 340 m/s, then the ratio f1/f2 is:, [10 Jan. 2019 I], (a) 18/17, (b) 19/18, (c) 20/19 (d) 21/20, A musician using an open flute of length 50 cm, produces second harmonic sound waves. A person runs, towards the musician from another end of a hall at a, speed of 10 km/h. If the wave speed is 330 m/s, the, frequency heard by the running person shall be close to:, [9 Jan. 2019 II], (a) 666 Hz, (b) 753 Hz, (c) 500 Hz, (d) 333 Hz, Two sitar strings, A and B, playing the note 'Dha' are slightly, out of tune and produce beats and frequency 5 Hz. The, tension of the string B is slightly increased and the beat, frequency is found to decrease by 3 Hz . If the frequency, of A is 425 Hz, the original frequency of B is, [Online April 16, 2018], (a) 430 Hz (b) 428 Hz (c) 422 Hz (d) 420 Hz, A toy–car, blowing its horm, is moving with a steady speed, of 5 m/s, away from a wall. An observer, towards whom the, toy car is moving, is able to hear 5 beats per second. If the, velocity of sound in air is 340 m/s, the frequency of the, horn of the toy car is close to : [Online April 10, 2016], (a) 680 Hz (b) 510 Hz (c) 340 Hz (d) 170 Hz, Two engines pass each other moving in opposite, directions with uniform speed of 30 m/s. One of them is, blowing a whistle of frequency 540 Hz. Calculate the
Page 234 :
P-225, , Waves, , frequency heard by driver of second engine before they, pass each other. Speed of sound is 330 m/sec:, [Online April 9, 2016], (a) 450 Hz (b) 540 Hz (c) 270 Hz (d) 648 Hz, 79. A train is moving on a straight track with speed 20 ms–1., It is blowing its whistle at the frequency of 1000 Hz. The, percentage change in the frequency heard by a person, standing near the track as the train passes him is (speed, of sound = 320 ms–1) close to :, [2015], (a) 18%, (b) 24%, (c) 6%, (d) 12%, 80. A source of sound emits sound waves at frequency f0. It, is moving towards an observer with fixed speed, vs (vs < v, where v is the speed of sound in air). If the, observer were to move towards the source with speed, v0, one of the following two graphs (A and B) will given, the correct variation of the frequency f heard by the, observer as v0 is changed., , (B), , (A), f, , f, , v0, 1/v0, The variation of f with v0 is given correctly by :, [Online April 11, 2015], f0, (a) graph A with slope =, (v + vs ), , (b) graph B with slope =, , f0, (v – vs ), , (c) graph A with slope =, , f0, (v – vs ), , f0, (v + vs ), 81. A bat moving at 10 ms–1 towards a wall sends a sound, signal of 8000 Hz towards it. On reflection it hears a, sound of frequency f. The value of f in Hz is close to, (speed of sound = 320 ms–1), [Online April 10, 2015], (a) 8516, (b) 8258, (c) 8424 (d) 8000, 82. A source of sound A emitting waves of frequency 1800, Hz is falling towards ground with a terminal speed v. The, observer B on the ground directly beneath the source, receives waves of frequency 2150 Hz. The source A, receives waves, reflected from ground of frequency, nearly: (Speed of sound = 343 m/s), [Online April 12, 2014], (a) 2150 Hz, (b) 2500 Hz, (c) 1800 Hz, (d) 2400 Hz, 83. Two factories are sounding their sirens at 800 Hz. A man, goes from one factory to other at a speed of 2m/s. The, velocity of sound is 320 m/s. The number of beats heard, by the person in one second will be:, [Online April 11, 2014], (d) graph B with slope =, , (a) 2, (b) 4, (c) 8, (d) 10, 84. A and B are two sources generating sound waves. A, listener is situated at C. The frequency of the source at A, is 500 Hz. A, now, moves towards C with a speed 4 m/s., The number of beats heard at C is 6. When A moves away, from C with speed 4 m/s, the number of beats heard at C, is 18. The speed of sound is 340 m/s. The frequency of, the source at B is :, [Online April 22, 2013], A, C, B, (a) 500 Hz (b) 506 Hz (c) 512 Hz (d) 494 Hz, 85. An engine approaches a hill with a constant speed. When, it is at a distance of 0.9 km, it blows a whistle whose echo, is heard by the driver after 5 seconds. If the speed of, sound in air is 330 m/s, then the speed of the engine is :, [Online April 9, 2013], (a) 32 m/s (b) 27.5 m/s (c) 60 m/s (d) 30 m/s, 86. This question has Statement 1 and Statement 2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., Statement 1: Bats emitting ultrasonic waves can detect, the location of a prey by hearing the waves reflected from it., Statement 2: When the source and the detector are, moving, the frequency of reflected waves is changed., [Online May 12, 2012], (a) Statement 1 is false, Statement 2 is true., (b) Statement 1 is true, Statement 2 is false., (c) Statement 1 is true, Statement 2 is true, Statement, 2 is not the correct explanation of Statement 1., (d) Statement 1 is true, Statement 2 is true, Statement, 2 is the correct explanation of Statement 1., 87. A motor cycle starts from rest and accelerates along a, straight path at 2m/s2. At the starting point of the motor, cycle there is a stationary electric siren. How far has the, motor cycle gone when the driver hears the frequency of, the siren at 94% of its value when the motor cycle was at, rest? (Speed of sound = 330 ms–1), [2009], (a) 98 m, (b) 147 m (c) 196 m (d) 49 m, 88. A whistle producing sound waves of frequencies 9500, HZ and above is approaching a stationary person with, speed v ms–1. The velocity of sound in air is 300 ms–1. If, the person can hear frequencies upto a maximum of, 10,000 HZ, the maximum value of v upto which he can, hear whistle is, [2006], (a) 15 2 ms -1, , (b), , 15, , ms -1, , 2, (c) 15, (d) 30 ms–1, 89. An observer moves towards a stationary source of sound,, with a velocity one-fifth of the velocity of sound. What is, the percentage increase in the apparent frequency ?[2005], (a) 0.5%, (b) zero, (c) 20 % (d) 5 %, , ms–1
Page 235 :
P-226, , 1., , Physics, , 6., , (a) As we know,, w, ´ rV 2, V, é, w, Bù, êQ K = , V =, ú, V, rû, ë, , Pressure amplitude, DP0 = aKB = S 0 KB = S 0 ´, , DP0, 10, 1, 3, », m=, mm », mm, rV w 1 ´ 300 ´ 1000, 30, 100, (b) Given : Distance between one crest and one trough, = 1.5 m, Þ S0 =, , 2., , = (2n1 + 1), , l, 2, , Distance between two crests = 5 m = n2 l, 1.5 (2n1 + 1), =, Þ 3n2 = 10n1 + 5, 5, 2n2, , Here n1 and n2 are integer., If, , 3., 4., , n1 = 1, n2 = 5, , \l = 1, , n1 = 4, n2 = 15, , \l = 1/ 3, , n1 = 7, n2 = 25, , \l = 1/ 5, , 1 1 1, Hence possible wavelengths , , metre., 1 3 5, (b) At t = 0, x = 0, y = 0, f = p rad, (a) Using, b = 10, æ I ö, or 120 = 10 log10 çè -12 ÷ø, 10, , Also I =, , 5., , P, , =, , 2, , ...(i), , ...(ii), 4pr, 4pr 2, On solving above equations, we get, r = 40 cm., (a) On comparing with P = P0 sin (wt – kx), we have, w = 1000 rad/s, K = 3 m–1, \ v0 =, , 2, , w 1000, =, = 333.3m/s, k, 3, , v1, T, = 1, v2, T2, 333.3, =, 336, \ t = 4°C, or, , 273 + 0, 273 + t, , (a) Comparing the given equation, y = 10–3sin(50t + 2x) with standard equation,, y = a sin(wt – kx), Þ wave is moving along –ve x-axis with speed, w, 50, v= Þv=, = 25 m/sec., K, 2, 7. (a) Given, amplitude a = 10 cm, wave velocity = 2 × maximum particle velocity, i.e, wl = 2 aw, 2p, p, or, l = 4a = 4 × 10 = 40 cm, 8. (b), 9. (b) Standard equation, æw, ö, y(x, t) = A cos ç x - wt ÷, èV, ø, From any of the displacement equation, Say y1, w, = 0.50 p and w = 100 p, V, 100p, = 0.5p, \, V, 100p, \ V=, = 200 m/s, 0.5p, 10. (c) The equation of wave at any time is obtained by, putting X = x – vt, 1, 1, y=, ...(i), 2 = 1 + ( x - vt ) 2, 1+ x, We know at t = 2 sec,, 1, ...(ii), y=, 1 + ( x - 1) 2, On comparing (i) and (ii) we get, vt = 1, 1, V=, t, As t = 2 sec, 1, \V=, =0.5 m/s., 2, 11. (a) Given, y (x, t) = e, , (-ax2 +bt2 +2, , ab xt, , ), , -[( ax )2 +( b t )2 + 2 a x . b t ], = e, - ( a x + bt )2, = e, æ, ö, b ÷, -ç x +, t, è, a ø, e, , 2, , =, It is a function of type y = f (x + vt), \ y (x, t) represents wave travelling along –ve x direction, w, b, =, Þ Speed of wave =, a, k
Page 237 :
P-228, , Physics, , Fundamental frequency in the string, , f =, , =, , 1 T, 1 T, =, 2l m 2l sA, 1, 9 ´ 109 ´ 4.9 ´ 10 -4, 2 ´1, 9 ´ 103, , DL =, , 1, 1, 49 ´ 109 - 4 -3 = ´ 70 = 35 Hz, 2, 2, 21. (106) Given : Vair = 300 m/s, rgas = 2 r air, =, , B, r, , Using, V =, , Vgas, Vair, , =, , Þ Vgas =, , B, 2rair, B, rair, , Vair, 2, , =, , 300, 2, , And fnth harmonic =, , =, , Tl mv 2 ´ l, =, YA l (YA), , 6 ´ 10 –3 ´ 902, 11, , –6, , = 3 ´ 10 –4 m, , 16 ´ 10 ´ 10, = 0.03 mm, 24. (c) We have given,, y = 0.03 sin(450 t – 9x), Comparing it with standard equation of wave, we get w =, 450 k = 9, w 450, = 50m/s, \ v= =, k, 9, Velocity of travelling wave on a stretched string is given, by, T, T, Þ = 2500, m, m, m = linear mass density, Þ T = 2500 × 5 × 10–3, Þ 12.5 N, v=, , = 150 2m/s, nv, (in open organ pipe), 2L, , (L = 1 metre given), \ f2nd harmonic – ffundamental =, , 2v, v, v, –, =, 2 ´1 2 ´ 1 2, , \ f2n harmonic – ffundamental = 150 2 = 150 » 106 Hz, 2, 2, 22. (b) The velocity of a transverse wave in a stretched wire, is given by, v=, , T, mv 2, ´l ÞT =, m, I, T, Again from, Y = DL / L0, A, , Using, v =, , T, m, , 25. (b) Wave speed V =, , T, m, , when car is at rest a = 0, , \ 60 =, , Mg, m, , Similarly when the car is moving with acceleration a,, , (, , M g2 + a 2, , 60.5 =, , ), , 12, , m, , Where,, T = Tension in the wire, m = linear density of wire, , 60.5, =, 60, , (Q V µ T ), , g2 + a2, 2, æ 0.5 ö, =1+, çè1+, ÷ø =, 2, 60, 60, g, 2, Þ g2 + a 2 = g 2 + g 2 ×, 60, , g2, 4, , v, T, \ 1 = 1, v2, T2, 2.06 ´ 10 4, v, Þ ´2=, v, T2, 4, , 2.06 ´ 10, = 0.515 ´ 104 N, 4, Þ T2 = 5.15 × 103 N, 23. (a) Given, l = 60 cm, m = 6 g, A = 1 mm2, v = 90 m/s and Y, = 16 × 1011 Nm–2, Þ T2 =, , g2 + a2, , a =g, , 2, g, =, 60, 30, , [which is closest to g/5], , 26. (a) Fundamental frequency, f = 70 Hz., The fundamental frequency of wire vibrating under, tension T is given by, , f =, , 1 T, 2L m
Page 238 :
P-229, , Waves, , Here, µ = mass per unit length of the wire, L = length of wire, , Since rod is clamped at middle fundamental wave shape, is as follow, , 1, 540, 2 L 6 ´ 10-3, Þ L » 2.14 m, (b) V = f l = f × 2 (l 2 - l1 ), = 480 × 2(0.70 – 0.30), = 384 m/s, 3l, 4, = 2 or l = m, (b), 2, 3, 4, Velocity, v = f l = 240 ´ = 320 m/sec, 3, 240, = 80 Hz, Also f1 =, 3, (b) Given, y = 0.3 sin (0.157 x) cos (200 pt), So k = 0.157 and w = 200p, w 200p, = 4000m/s, or f = 100 Hz, v = =, k 0.157, nv 4v 2v, =, =, Now, using f =, 2l 2l, l, 2v 2 ´ 4000, \l =, =, = 80m, f, 100, (c) As there must be node at both ends and at the joint, of the wire A and B so, , l, A, N, = L Þ l = 2L, 2, l/2, l = 1.2m (Q L = 60 cm = 0.6m (given), Using v = fl, , 70 =, , 27., , 28., , 29., , 30., , v 5.85 ´103, =, l, 1.2, = 4.88 × 103 Hz ; 5 KHz, Þ, , l, = l1 + e = 11 cm, 4, (Q end correction e = 1 cm given), 3l, For second resonance,, = l2 + e, 4, Þ l 2 = 3 ´ 11 - 1 = 32 cm, 35. (b) n1 = n2, T ® Same, r ® Same, l ® Same, Frequency of vibration, , T, 8, =, ´ 1000 = 40m/s, m, 5, Here, T = tension and µ = mass/length, v 40, m, Wavelength of wave l = =, n 100, Separation b/w successive nodes,, l, 40, 20, =, =, m = 20 cm, 2 2 ´ 100 100, 33. (a) In solids, Velocity of wave, , n=, , p, 2l, , p1, , =, , T, , pr 2 r, As T, r, and l are same for both the wires, n1 = n2, p2, , r1, r2, p, 1, Þ 1 =, p2 2, , Þ l A = 2l B, , P 1, =, q 2, 31. (a) If a closed pipe vibration in Nth mode then frequency, ( 2N - 1) v = 2N - 1 n, of vibration n =, (, ) 1, 4l, (where n1 = fundamental frequency of vibration), Hence 20,000 = (2N – 1) × 1500, Þ N = 7.1 » 7, \ Number of over tones = (No. of mode of vibration) – 1, =7–1=6, 32. (d) Velocity of wave on string, , f=, , 34. (a) For first resonance,, , VA, uB rB, l, =, =, =2= A, VB, u A rA, lB, , Þ, , 36., , Y, 9.27 ´ 1010, =, r, 2.7 ´ 103, 3, v = 5.85 × 10 m/sec, , Q r2 = 4 r1, , (a ) We know that velocity in string is given by, , v=, , T, m, , where m =, , ...(i), , m mass of string, =, l length of string, , The tension T =, From (1) and (2), , m, ´ x´g, l, , ..(ii), , l, , V=, , V=, , A, , T, x, dx, = gx, dt, , x -1/2 dx = g dt, l, , l, , 0, , 0, , \ ò x -1/2 dx - g ò dt, Þ2 l
Page 239 :
P-230, , Physics, , = g´t \ t = 2, 37., , 41. (d) Total length of the wire, L = 114 cm, n1 : n2 : n3 = 1 : 3 : 4, Let L1, L2 and L3 be the lengths of the three parts, 1, As n µ, L, 1 1 1, \ L1 : L2 : L3 = : : = 12: 4 : 3, 1 3 4, æ 12, ö, \ L1 = ç, ´ 114÷ = 72cm, è 12 + 4 + 3, ø, , 20, l, =2, =2 2, g, 10, , f, , (b), l, , l, (a), , (b), , The fundamental frequency in case (a) is f =, The fundamental frequency in case (b) is, , v, 2l, , v, v, =, =f, 4(l / 2) 2l, 38. (c) Length of pipe = 85 cm = 0.85m, Frequency of oscillations of air column in closed organ, pipe is given by,, (2n - 1)u, f =, 4L, (2n - 1)u, f =, £ 1250, 4L, (2n - 1) ´ 340, £ 1250, Þ, 0.85 ´ 4, Þ 2n – 1 < 12.5 » 6, 39. (b) Total length of sonometer wire, l = 110 cm = 1.1 m, Length of wire is in ratio, 6 : 3 : 2 i.e. 60 cm, 30 cm, 20 cm., Tension in the wire, T = 400 N, Mass per unit length, m = 0.01 kg, Minimum common frequency = ?, As we know,, f'=, , 1 T 1000, Hz, =, 2l m, 11, 1000, Hz, Similarly, n1 =, 6, 1000, n2 =, Hz, 3, 1000, n3 =, Hz, 2, Hence common frequency = 1000 Hz, 40. (b) Fundamental frequency,, é, T, mù, v, 1 T, 1 T, and m = ú, f =, =, =, êQ v =, m, lû, 2l 2l m 2l Ar ë, Tl, T Y Dl, Also, Y =, Þ, =, ADl, A, l, , Frequency, n =, , Þ f =, , 1, 2l, , y Dl, lr, , ....(i), , Dl, = 0.01,, l, r = 7.7 × 103 kg/m3 (given), y = 2.2 × 1011 N/m2 (given), Dl, Putting the value of l,, , r and y in eqn. (i) we get,, l, 2 103, f =, ´, 7 3 or f » 178.2 Hz, , l = 1.5 m,, , æ 4, ö, L2 = ç ´ 114÷ = 24 cm, è 19, ø, æ 3, ö, and L3 = ç ´ 114÷ = 18 cm, è 19, ø, Hence the bridges should be placed at 72 cm and 72 + 24, = 96 cm from one end., 42. (a) Initially for open organ pipe, fundamental frequency, v, n0 =, … (i), 2l0, where l0 is the length of the tube, v = speed of sound, But when it is half dipped in water, it becomes closed organ, l, pipe of length 0 ., 2, Fundamental frequency of closed organ pipe, v, nc =, … (ii), 4lc, l0, New length, lc =, 2, v, v, Þ nc =, Thus n c =, … (iii), 4l0 / 2, 2l, From equations (i) and (iii), n 0 = nc, Thus, nc = f ( Q n0 = f is given), 43. (b) Given : Frequency of tuning fork, n = 264 Hz, Length of column L = ?, For closed organ pipe, v, n=, 4l, v, 330, Þl=, =, = 0.3125, 4n, 4 ´ 264, or, l = 0.3125 × 100 = 31.25 cm, In case of closed organ pipe only odd harmonics are, possible., Therefore value of l will be (2n – 1) l, Hence option (b) i.e. 3 × 31.25 = 93.75 cm is correct., 44. (d) Two lowest frequencies to which tube will resonates, are 272 Hz and 544 Hz., é æ t ö, x ù, 45. (d) y = 0.02(m)sin ê2p ç, ú, ÷0.04(, s, ), 0.50(, m) û, ø, ë è, Comparing it with the standard wave equation, y = a sin(wt - kx ), we get, 2p, w=, rad s–1, 0.04
Page 240 :
P-231, , Waves, , and k =, , 2p, 0.50, , Wave velocity, v =, , w, k, , 2p / 0.04, = 12.5 m / s, 2p / 0.5, Velocity on a string is given by, , 49., , Þv=, , v=, , T, m, , \ T = v2 ´ m = (12.5)2 × 0.04 = 6.25 N, 46. (b) Fundamental frequency for first resonant length, v, v, n=, =, (in winter), 4l1 4 ´ 18, Fundamental frequency for second resonant length, 3v ' 3v ', (in summer), n' =, =, 4l 2 4x, According to questions,, v, 3v', \, =, 4 ×18 4× x, v', \ x = 3 ´ 18 ´, v, v', \ x = 54 ´ cm, v, v' > v because velocity of light is greater in summer as, , compared to winter (v µ T ), \ x > 54cm, 47. (a) It is given that 315 Hz and 420 Hz are two resonant, frequencies, let these be n th and (n + 1)th harmonies, then, we have nv = 315, 2l, v, = 420, and (n + 1), 2l, n + 1 420, Þ, =, n, 315, Þn=3, v, v, = 105 Hz, Hence 3 ´, = 315 Þ, 2, l, 2l, The lowest resonant frequency is when, n=1, Therefore lowest resonant frequency, = 105 Hz., 48. (c) The fundamental frequency for tube B closed at one, end is given by, lù, v, é, Ql = ú, uB =, ê, 4û, 4l, ë, Where l = length of the tube and v is the velocity of, sound in air., The fundamental frequency for tube A open with both ends, is given by, lù, é, v, Ql = ú, uA =, ê, 2û, 2l, ë, , 50., 51., 52., 53., , 54., , 55., , uA, v 4l 2, \ u = 2l ´ v = 1, B, (b) To form a node there should be superposition of this, wave with the reflected wave. The reflected wave should, travel in opposite direction with a phase change of p. The, equation of the reflected wave will be, y = a sin (wt + kx + p), Þ y = – a sin (wt + kx), (a), (c) Beat frequency, = difference in frequencies of two waves, = 11 – 9 = 2 Hz, (d), (d) According to question, tuning fork gives 1 beat/second, with (N) 3rd normal mode. Therefore, organ pipe will have, frequency (256 ± 1) Hz. In open organ pipe, frequency, NV, n=, 2l, 3 ´ 340, Þ l = 2 m = 200 cm, or, 255 =, 2´ l, (a) Probable frequencies of tuning fork be n ± 5, 1, Frequency of sonometer wire, n µ, l, n + 5 100, \, =, Þ 95(n + 5) = 100( n - 5), n - 5 95, or, 95 n + 475 = 100 n – 500, or, 5 n = 975, 975, or, n =, = 195 Hz, 5, æ 5p ö, (a) Given, y (x, t) = 0.5 sin ç x ÷ cos (200 pt),, è 4 ø, , comparing with equation – y (x, t) = 2 a sin kx cos wt, 5p, w = 200 p, k =, 4, w 200p, speed of travelling wave v = =, = 160 m/s, k 5p 4, 56. (b) Since the point x = 0 is a node and reflection is taking, place from point x = 0. This means that reflection must be, taking place from the fixed end and hence the reflected ray, must suffer an additional phase change of p or a path, l, change of ., 2, So, if yincident = a cos ( kx – wt), Þ yincident = a cos (– kx – wt + p), = – a cos (wt + kx), Hence equation for the other wave, y = a cos(kx + wt + p), 57. (d) In case of destructive interference, Phase difference f = 180° or p, So wave pair (i) and (ii) will produce destructive, interference., Stationary or standing waves will produce by equations, (iii) & (iv) as two waves travelling along the same line but, in opposite direction., n¢ = n + x, 58. (d) y = A sin (wt – kx) + A sin (wt + kx), y = 2A sin wt cos kx
Page 241 :
P-232, , Physics, , This is an equation of standing wave. For position of, nodes, cos kx = 0, 2p, p, .x = (2n + 1), Þ, l, 2, 2 n + 1) l, (, Þ x=, , n = 0,1, 2,3,..........., 4, 59. (a) Intensity of a wave, 1, I = pw2 A2 v, 2, Since, I µ A2w 2, \ I1 µ (2a)2 w 2, , 60., 61., , 62., , 63., , and I 2 µ a 2 (2w ) 2, I1 = I 2, In the same medium, p and v are same., Intensity depends on amplitude and frequency., (b) Maximum number of beats, = Maximum frequency – Minimum frequency, = ( n + 1) – ( n – 1) = 2 Beats per second, (d) Frequency of fork 1, no = 200 Hz, No. of beats heard when fork 2 is sounded with fork 1 = Dn =, 4, Now on loading (attaching tape) on unknown fork, the, mass of tuning fork increases, So the beat frequency, increases (from 4 to 6 in this case) then the frequency, of the unknown fork 2 is given by,, n = n 0 – Dn = 200 – 4 = 196 Hz, (c) It is given that tuning fork of frequency 256 Hz makes, 5 beats/second with the vibrating string of a piano., Therefore, possible frequency of the piano are (256 ± 5), Hz. i.e., either 261Hz or 251 Hz. When the tension in the, piano string increases, its frequency will increases. As, the original frequency was 261Hz, the beat frequency, should decreases, we can conclude that the frequency of, piano string is 251Hz, (b) Frequency of unknown fork = known frequency ± Beat, frequency = 288 + 4 cps or 288 – 4 cps i.e. 292 cps or 284, cps. When a little wax is placed on the unknown fork, it, produces 2 beats/sec. When a little wax is placed on the, unknown fork, its frequency decreases and simultaneously, the beat frequency decreases confirming that the frequency, of the unknown fork is 292 cps., Note : Had the frequency of unknown fork been 284 cps,, then on placing wax its frequency would have decreased, thereby increasing the gap between its frequency and the, frequency of known fork. This would produce high beat, frequency., , 64. (b) Frequency heard by the observer, æ, ö, vsound, vobserved = ç, ÷ v0, è vsound - v cos q ø, Observer, O, D, q, Source V, , Initially q will be less so cos q more., \ vobserved more, then it will decrease., , 65. (a) Let f 1 be the frequency heard by wall,, æ v ö, f1 = ç, f0, è v - vc ÷ø, Here, v = Velocity of sound,, vc = Velocity of Car,, f0 = actual frequency of car horn, Let f2 be the frequency heard by driver after reflection, from wall., æ v + vc ö, æ v + vc ö, f2 = ç, f =, f, è v ÷ø 1 çè v - v ÷ø 0, c, , é 345 + vc ù, 12 345 + vc, Þ 480 = ê, =, ú 440 Þ, 11 345 - vc, ë 345 - vc û, Þ vc = 54 km/hr, 66. (a) From the Doppler's effect of sound, frequency, appeared at wall, 330, fw =, ×f, ...(i), 330 - v, Here, v = speed of bus,, f = actual frequency of source, Frequency heard after reflection from wall (f') is, 330 + v, 330 + v, f '=, × fw =, ×f, 330, 330 - v, 330 + v, Þ 490 =, × 420, 330 - v, 330 ´ 7, Þv=, » 25.38 m/s = 91 km/s, 91, 67. (d) Permanent magnets (P) are made of materials with large, retentivity and large coercivity. Transformer cores (T) are, made of materials with low retentivity and low coercivity., 68. (c) From Doppler’s effect, frequency of sound heard (f1), when source is approaching, c, f1 = f 0, c–v, Here, c = velocity of sound, v = velocity of source, Frequency of sound heard (f2) when source is receding, , c, c+v, Beat frequency = f1 – f2, 1 ù, é 1, Þ 2 = f1 – f 2 = f 0c ê, –, –, c, v, c, +, v úû, ë, 2v, = f0c, é v2 ù, c 2 ê1 – 2 ú, êë c úû, For c>> v, c, 2c, 350 1, Þ v=, =, =, = m/s, f 0 1400 4, 2 f0, f2 = f0, , æ v - vo ö, æ 1500 - 5 ö, 69. (d) f1 = f ç v - v ÷ = f çè 1500 - 7.5 ÷ø, è, ø, s, No reflected signal,
Page 244 : 15, , P-235, , Electric Charges and Fields, , Electric Charges, and Fields, TOPIC 1, 1., , 2., , Electric Charges and Coulomb's, Law, , +Q, , Three charges + Q, q, + Q are placed respectively, at, distance, d/2 and d from the origin, on the x-axis. If the net, force experienced by + Q, placed at x = 0, is zero, then, value of q is:, [9 Jan. 2019 I], (a) – Q/4, (b) + Q/2 (c) + Q/4, (d), – Q/2, Charge is distributed within a sphere of radius R with a, , A -2r a, where A and a, e, r2, are constants. If Q is the total charge of this charge, distribution, the radius R is:, [9 Jan. 2019, II], , –Q, , volume charge density p(r) =, , Q ö, æ, (a) a log ç 1 ÷, 2paA ø, è, , 3., , 4., , æ, ö, ç, ÷, a, 1, log ç, (b), ÷, 2, çç 1 - Q ÷÷, 2paA ø, è, , æ, ö, ç, ÷, 1, a, Q ö, æ, (c) a log ç, log ç 1 ÷ (d), ÷, 2, 2paA ø, è, çç 1 - Q ÷÷, 2paA ø, è, Two identical conducting spheres A and B, carry equal, charge. They are separated by a distance much larger than, their diameter, and the force between them is F. A third, identical conducting sphere, C, is uncharged. Sphere C is, first touched to A, then to B, and then removed. As a, result, the force between A and B would be equal to, [Online April 16, 2018], 3F, F, 3F, (b), (c) F, (d), 4, 2, 8, Shown in the figure are two point charges +Q and –Q, inside the cavity of a spherical shell. The charges are kept, near the surface of the cavity on opposite sides of the, centre of the shell. If s1 is the surface charge on the inner, surface and Q1 net charge on it and s2 the surface charge, on the outer surface and Q2 net charge on it then :, [Online April 10, 2015], (a), , 5., , (a) s1 ¹ 0, Q1 = 0, (b) s1 ¹ 0, Q1 = 0, s2 = 0, Q2 = 0, s2 ¹ 0, Q2 = 0, (c) s1 = 0, Q1 = 0, (d) s1 ¹ 0, Q1 ¹ 0, s2 = 0, Q2 = 0, s2 ¹ 0, Q2 ¹ 0, Two charges, each equal to q, are kept at x = – a and x = a, q, on the x-axis. A particle of mass m and charge q 0 =, is, 2, placed at the origin. If charge q0 is given a small, displacement (y <<a) along the y-axis, the net force acting, on the particle is proportional to, [2013], 1, 1, (d) –, y, y, Two balls of same mass and carrying equal charge are, hung from a fixed support of length l. At electrostatic, equilibrium, assuming that angles made by each thread is, small, the separation, x between the balls is proportional, to :, [Online April 9, 2013], (a) l, (b) l 2, (c) l 2/3, (d) l 1/3, Two identical charged spheres suspended from a common, point by two massless strings of length l are initially a, distance d(d << l) apart because of their mutual repulsion., The charge begins to leak from both the spheres at a, constant rate. As a result charges approach each other, with a velocity v. Then as a function of distance x between, them,, [2011], (a) v µ x–1 (b) v µ x½ (c) v µ x, (d) v µ x–½, A charge Q is placed at each of the opposite corners of a, square. A charge q is placed at each of the other two, corners. If the net electrical force on Q is zero, then Q/q, equals:, [2009], 1, (a) –1, (b) 1, (c) (d) -2 2, 2, If gE and gM are the accelerations due to gravity on the, surfaces of the earth and the moon respectively and if, Millikan’s oil drop experiment could be performed on the, , (a) y, , 6., , 7., , 8., , 9., , (b) –y, , (c)
Page 245 :
P-236, , Physics, , [2007], , two surfaces, one will find the ratio, electronic charge on the moon, to be, electronic charge on the earth, 10., , 11., , (a) gM / g E (b) 1, (c) 0, (d) g E / g M, Two spherical conductors B and C having equal radii and, carrying equal charges on them repel each other with a, force F when kept apart at some distance. A third spherical, conductor having same radius as that B but uncharged is, brought in contact with B, then brought in contact with C, and finally removed away from both. The new force of, repulsion between B and C is, [2004], (a) F/8, (b) 3 F/4 (c) F/4, (d) 3 F/8, Three charges –q1 , +q2 and –q3 are place as shown in the, figure. The x - component of the force on –q 1 is, proportional to, [2003], Y, q3, a, , b, , +q 2 X, q2 q3, q 2 q3, (a) 2 - 2 cos q, (b) 2 +, sin q, b, b, a, a2, q, q, q, q, (c) 2 + 3 cos q, (d) 2 - 3 sin q, 2, 2, 2, b, b, a, a2, If a charge q is placed at the centre of the line joining two, equal charges Q such that the system is in equilibrium, then the value of q is, [2002], (a) Q/2, (b) –Q/2 (c) Q/4, (d) –Q/4, q1, , 12., , TOPIC 2, 13., , x1, , 14., , x13, x23, , (b), , Q2, B, , x2, , O, x2, x1, , (c), , x1, x2, , (d), , x2 2, x12, , Consider the force F on a charge ‘q’ due to a uniformly, charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if ‘q’ is placed at distance r from the, centre of the shell?, [Sep. 06, 2020 (II)], , 1 Qq, 1 Qq, for r < R (b), > F > 0 for r < R, 4pe 0 R 2, 4pe0 R 2, , 1 Qq, 1 Qq, for r > R (d) F =, for all r, 4pe0 R 2, 4pe0 R 2, 15. Two charged thin infinite plane sheets of uniform surface, , (c) F =, , charge density s + and s – , where | s + | > | s – |, intersect, at right angle. Which of the following best represents the, electric field lines for this system ? [Sep. 04, 2020 (I)], s–, , s+, , (a), , s–, , s+, , (b), , s–, , s+, , (c), , s–, , Electric Field and Electric Field, Lines, , Charges Q1 and Q2 are at points A and B of a right angle, triangle OAB (see figure). The resultant electric field at, point O is perpendicular to the hypotenuse, then Q1/Q2 is, proportional to :, [Sep. 06, 2020 (I)], A, Q1, , (a), , (a) F =, , s+, , (d), , 16. A particle of charge q and mass m is subjected to an electric, field E = E0 (1 – ax2) in the x-direction, where a and E0 are, constants. Initially the particle was at rest at x = 0. Other, than the initial position the kinetic energy of the particle, becomes zero when the distance of the particle from the, origin is :, [Sep. 04, 2020 (II)], 3, 1, 2, (c), (d), a, a, a, 17. A charged particle (mass m and charge q) moves along X, axis with velocity V0. When it passes through the origin it, r, enters a region having uniform electric field E = - Ejˆ which, , (a) a, , (b), , extends upto x = d. Equation of path of electron in the, region x > d is :, [Sep. 02, 2020 (I)]
Page 246 :
P-237, ur, 20. An electric dipole of moment p = (iˆ - 3 ˆj + 2kˆ) ´ 10-29 C.m, is at the origin (0, 0, 0). The electric field due to this dipole, r, at r = +iˆ + 3 ˆj + 5kˆ, r ur, (note that r . p = 0) is parallel to:, [9 Jan. 2020, I], , Electric Charges and Fields, , Y, E, O, , (a) y =, (c) y =, 18., , qEd, mV02, qEd, , V0, , X, d, , (x - d ), , (b) y =, , x, , (d) y =, , qEd æ d, ö, - x÷, 2 ç, mV0 è 2, ø, , qEd 2, , x, mV02, A small point mass carrying some positive charge on it, is, released from the edge of a table. There is a uniform electric, field in this region in the horizontal direction. Which of the, following options then correctly describe the trajectory of, the mass ? (Curves are drawn schematically and are not to, scale)., [Sep. 02, 2020 (II)], E, x, mV02, , (a) (+iˆ - 3 ˆj - 2kˆ), , (c) (+iˆ + 3 ˆj - 2kˆ), (d) (-iˆ - 3 ˆj + 2kˆ), 21. A charged particle of mass ‘m’ and charge ‘q’ moving under, the influence of uniform electric field Eiˆ and a uniform, r, magnetic field Bk follows a trajectory from point P to Q as, shown in figure. The velocities at P and Q are respectively,, r, r, vi and -2vj . Then which of the following statements, (A, B, C, D) are the correct? (Trajectory shown is, schematic and not to scale), [9 Jan. 2020, I], Y, , E, P, , y, y, , a, , y, , (a), , O, , (b), , x, , x, , y, , y, , (c), 19., , (d), , x, x, Consider a sphere of radius R which carries a uniform, R, charge density r. If a sphere of radius, is carved out of, 2, ur, EA, it, as shown, the ratio ur, of magnitude of electric, EB, ur, ur, field E A and E B , respectively, at points A and B due to, the remaining portion is:, [9 Jan. 2020, I], , (b) (-iˆ + 3 ˆj - 2kˆ), , B, , v, , Q, 2v, , 2a, , 3 æ mv 2 ö, (A) E = 4 çç qa ÷÷, è, ø, , X, , 3 æ mv 2 ö, (B) Rate of work done by the electric field at P is 4 çç a ÷÷, è, ø, (C) Rate of work done by both the fields at Q is zero, (D) The difference between the magnitude of angular, momentum of the particle at P and Q is 2 mav., (a) (A), (C), (D), (b) (B), (C), (D), (c) (A), (B), (C), (d) (A), (B), (C) , (D), 22. Three charged particles, y, , 2q, B d, , 150°, O, , d, 30°, 30°, d, , –4q, A, x, C, –2q, , A, B and C with charges – 4q, 2q and –2q are present on, the circumference of a circle of radius d. The charged, particles A, C and centre O of the circle formed an, equilateral triangle as shown in figure. Electric field at O, along x-direction is:, [8 Jan. 2020, I], (a), , 21, 34, , (b), , 18, 34, , (c), , 17, 54, , (d), , 18, 54, , (a), , 3q, p Î0 d, , 2, , (b), , 2 3q, p Î0 d, , 2, , (c), , 3q, 4p Î0 d, , 2, , (d), , 3 3q, 4p Î0 d 2
Page 247 :
P-238, , 23., , Physics, , A particle of mass m and charge q is released from rest in, a uniform electric field. If there is no other force on the, particle, the dependence of its speed v on the distance x, travelled by it is correctly given by (graphs are schematic, and not drawn to scale), [8 Jan. 2020, II], v, , v, , (a), , x, , v, , v, , (c), , (d), x, , x, , Two infinite planes each with uniform surface charge, density +s are kept in such a way that the angle between, them is 30°. The electric field in the region shown between, them is given by:, [7 Jan. 2020, I], , 30°, , x, é, æ, s, 3ö, xˆ ù, s é, xˆ ù, yˆ + ú, (a), (1 + 3) yˆ - ú (b), êç1 +, ÷, ê, Î0 ëêè, 2 ø, 2 ûú, 2 Î0 ë, 2û, s éæ, 3ö, xˆ ù, xˆ ù, s é, yˆ - ú, 1 + 3 yˆ + ú (d), êç1 ÷, ê, 2 Î0 ëêè, 2 ø, 2 ûú, 2 Î0 ë, 2û, A particle of mass m and charge q has an initial velocity, r, r, r, v = v0 $j . If an electric field E = E0 i and magnetic field, r, B = B0iˆ act on the particle, its speed will double after a, , (, , 25., , ), , time:, , 26., , [7 Jan 2020, II], 2mv0, 2mv0, 3mv0, 3mv0, (a) qE, (b) qE, (c), (d) qE0, qE0, 0, 0, A simple pendulum of length L is placed between the, plates of a parallel plate capacitor having electric field E,, as shown in figure. Its bob has mass m and charge q. The, time period of the pendulum is given by :, [10 April 2019, II], , L, 2p, qE ö, æ, (a), çg+, ÷, m ø, è, 2p, , (c), , L, qE ö, æ, çg÷, m ø, è, , 2p, , (b), , (d), , 1, , R, , (b) E µ, , R, (c) R, (d), R 2, 5, 2, 30. Two point charges q1 ( 10 mC) and q2 (– 25 mC) are, placed on the x-axis at x = 1 m and x = 4 m respectively., The electric field (in V/m) at a point y = 3 m on y-axis, is,, [9 Jan 2019, II], é, ù, 1, = 9 ´ 109 Nm 2 C -2 ú, ê take, 4p Î0, ë, û, (b), , (a) (63 î – 27 ĵ ) × 102, , (b) (– 63 î + 27 ĵ ) × 102, , (c) (81 î – 81 ĵ ) × 102 (d) (–81 î + 81 ĵ ) × 102, 31. A body of mass M and charge q is connected to a spring, of spring constant k. It is oscillating along x-direction about, its equilibrium position, taken to be at x = 0, with an, amplitude A. An electric field E is applied along the, x-direction. Which of the following statements is correct?, [Online April 15, 2018], , 1, 1 q2 E2, mw2 A2 +, 2, 2 k, 2qE, (b) The new equilibrium position is at a distance:, k, from x = 0, qE, (c) The new equilibrium position is at a distance:, 2k, from x = 0, 2 2, 1, 2 2 1q E, (d) The total energy of the system is mw A –, 2, 2 k, 32. A solid ball of radius R has a charge density r given by, (a) The total energy of the system is, , q2 E2, m2, , the ball is:, , L, , (a), , æ qE ö, g2 + ç, ÷, è m ø, , 3, , rö, æ, r = r0 ç1 - ÷ for 0 £ r £ R. The electric field outside, è Rø, , L, g2 -, , 2p, , 29., , (a), , y, , (c), , 1, 1, 1, (c) E µ 4 (d) E µ 2, D, D, D, D, The bob of a simple pendulum has mass 2 g and a charge, of 5.0 ¼C. It is at rest in a uniform horizontal electric field, of intensity 2000 V/m. At equilibrium, the angle that the, pendulum makes with the vertical is :, [8 April 2019 I], (take g = 10 m/s2), (a) tan–1 (2.0) (b) tan –1 (0.2), (c) tan–1 (5.0) (d) tan –1 (0.5), For a uniformly charged ring of radius R, the electric field, on its axis has the largest magnitude at a distance h from, its centre. Then value of h is:, [9 Jan. 2019 I], , (a) E µ, 28., , (b), x, , 24., , 27. Four point charges –q, +q, + q and –q are placed on y-axis, at y = –2d, y = –d, y = +d and y = +2d, respectively. The, magnitude of the electric field E at a point on the x-axis at, x = D, with D>> d, will behave as:, [9 April 2019, II], , 2, , r0 R 3, e0 r 2, , [Online April 15, 2018], (b), , 4r0 R 3, 3e 0 r 2, , (c), , 3r 0 R 3, 4e 0 r 2, , (d), , r0 R 3, 12e0 r 2
Page 248 :
P-239, , Electric Charges and Fields, , 33., , 34., , A long cylindrical shell carries positive surface charge s in, the upper half and negative surface charge - s in the lower, half. The electric field lines around the cylinder will look, like figure given in : (figures are schematic and not drawn, to scale), [2015], (a), , (b), , (c), , (d), , C ield Lines, , A wire of length L (=20 cm), is bent into a semicircular, arc. If the two equal halves of the arc were each to be, uniformly charged with charges ± Q, [|Q| = 103e0, Coulomb where e0 is the permittivity (in SI units) of free, space] the net electric field at the centre O of the, semicircular arc would be : [Online April 11, 2015], Y, , X, , O, , (a) (50 × 10 N/C) $j, 3, , 35., , 36., , O, , (b) (50 × 103 N/C) $i, , (c) (25 × 103 N/C) $j, (d) (25 × 103 N/C) $i, A thin disc of radius b = 2a has a concentric hole of radius, ‘a’ in it (see figure). It carries uniform surface charge ‘s’, on it. If the electric field on its axis at height ‘h’ (h << a), from its centre is given as ‘Ch’ then value of ‘C’ is :, [Online April 10, 2015], s, (a), 4aÎ0, s, (b), 8aÎ0, s, (c), aÎ0, s, (d), 2aÎ0, A spherically symmetric charge distribution is characterised, by a charge density having the following variations:, rö, æ, r ( r ) = ro ç1 - ÷ for r < R, è Rø, r(r) = 0 for r ³ R, Where r is the distance from the centre of the charge, distribution ro is a constant. The electric field at an internal, point (r < R) is:, [Online April 12, 2014], (a), , ro æ r r 2 ö, ç ÷, 4eo çè 3 4R ÷ø, , ro æ r r 2 ö, (c), ç ÷, 3eo çè 3 4R ÷ø, , (b), , 37. The magnitude of the average electric field normally, present in the atmosphere just above the surface of the, Earth is about 150 N/C, directed inward towards the center, of the Earth. This gives the total net surface charge carried, by the Earth to be:, [Online April 9, 2014], [Given eo = 8.85 × 10–12 C2/N-m2, RE = 6.37 × 106 m], (a) + 670 kC, (b) – 670 kC, (c) – 680 kC, (d) + 680 kC, 38. The surface charge density of a thin charged disc of radius, R is s. The value of the electric field at the centre of the, disc is, , s, . With respect to the field at the centre, the, 2 Î0, , electric field along the axis at a distance R from the centre, of the disc :, [Online April 25, 2013], (a) reduces by 70.7%, (b) reduces by 29.3%, (c) reduces by 9.7%, (d) reduces by 14.6%, 39. A liquid drop having 6 excess electrons is kept stationary, under a uniform electric field of 25.5 kVm–1. The density of, liquid is 1.26 × 103 kg m–3. The radius of the drop is (neglect, buoyancy)., [Online April 23, 2013], (a) 4.3 × 10–7 m, (b) 7.8 × 10–7 m, (c) 0.078 × 10–7 m, (d) 3.4 × 10–7 m, 40. In a uniformly charged sphere of total charge Q and radius, R, the electric field E is plotted as function of distance, from the centre, The graph which would correspond to the, above will be:, [2012], E(r), , E(r), , (a), , (b), r, , r, , E(r), , E(r), , (c), , (d), r, , r, , 41. Three positive charges of equal value q are placed at, vertices of an equilateral triangle. The resulting lines of, force should be sketched as in [Online May 26, 2012], , (a), , (b), , (c), , (d), , ro æ r r 2 ö, ç ÷, eo çè 3 4R ÷ø, , ro æ r r 2 ö, (d), ç ÷, 12eo çè 3 4R ÷ø
Page 249 :
P-240, , 42., , Physics, , A thin semi-circular ring of radius r has a positive charge q, ur, distributed uniformly over it. The net field E at the centre, O is, [2010], (a), , j, , (b), O, , 43., , q, , ˆj, , 2, , 2, , (b) -, , 2, , æ5 rö, with charge density varying as r(r ) = r0 çè - ÷ø upto r, 4 R, = R , and r(r ) = 0 for r > R , where r is the distance from, the origin. The electric field at a distance r(r < R) from the, origin is given by, [2010], r0 r æ 5 r ö, 4pr0 r æ 5 r ö, (a), ç - ÷, (b) 3e çè 3 - R ÷ø, 4e 0 è 3 R ø, 0, , r0 r æ 5 r ö, r0 r æ 5 r ö, ç - ÷, (d) 3ε çè 4 - ÷ø, R, 4ε0 è 4 R ø, 0, This question contains Statement-1 and Statement-2. Of, the four choices given after the statements, choose the, one that best describes the two statements., Statement-1 : For a charged particle moving from point P, to point Q, the net work done by an electrostatic field on, the particle is independent of the path connecting point P, to point Q., Statement-2 : The net work done by a conservative force, on an object moving along a closed loop is zero. [2009], (a) Statement-1 is true, Statement-2 is true; Statement-2, is the correct explanation of Statement-1., (b) Statement-1 is true, Statement-2 is true; Statement-2, is not the correct explanation of Statement-1., (c) Statement-1 is false, Statement-2 is true., (d) Statement-1 is true, Statement-2 is false., Q, Let r (r ) =, r be the charge density distribution for, p R4, a solid sphere of radius R and total charge Q. For a point, ‘P’ inside the sphere at distance r1 from the centre of the, sphere, the magnitude of electric field is :, [2009], 2, Q, Qr1, (a), (b), 2, 4p Î0 r1, 4p Î0 R4, , (c), , 45., , (c), , 46., , Qr12, , r, , E(r), , (d) 0, 3p Î0 R4, A thin spherical shell of radus R has charge Q spread, uniformly over its surface. Which of the following graphs, most closely represents the electric field E(r) produced by, the shell in the range 0 £ r < ¥, where r is the distance from, the centre of the shell?, [2008], , O, , r, , R, , (d), O, , R, , r, , 47. Two spherical conductors A and B of radii 1 mm and 2 mm, are separated by a distance of 5 cm and are uniformly, charged. If the spheres are connected by a conducting, wire then in equilibrium condition, the ratio of the, magnitude of the electric fields at the surfaces of spheres, A and B is, [2006], (a) 4 : 1, (b) 1 : 2, (c) 2 : 1, (d) 1 : 4, 48. Two point charges + 8q and – 2q are located at, x = 0 and x = L respectively. The location of a point on the, x axis at which the net electric field due to these two point, charges is zero is, [2005], , (c), , 44., , R, , R, , q, , ˆj, 4p e 0 r, 4p e 0 r 2, q, q, ˆj, ˆ, (c) - 2 2 j, (d), 2, 2p e 0 r, 2p e 0 r 2, Let there be a spherically symmetric charge distribution, (a), , O, , r, , E(r), , i, , O, , E(r), , E(r), , L, (b) 2 L, (c) 4 L, (d) 8 L, 4, A charged ball B hangs from a silk thread S, which makes, an angle q with a large charged conducting sheet P, as, shown in the figure. The surface charge density s of the, sheet is proportional to, [2005], , (a), 49., , P, , q, S, , B, (a) cot q, (b) cos q (c) tan q, (d) sin q, 50. Four charges equal to -Q are placed at the four corners of, a square and a charge q is at its centre. If the system is in, equilibrium the value of q is, [2004], (a) -, , (b), , Q, (1 + 2 2), 4, , Q, Q, (1 + 2 2), (1 + 2 2), (d), 4, 2, A charged oil drop is suspended in a uniform field of 3×104, v/m so that it neither falls nor rises. The charge on the, drop will be (Take the mass of the charge = 9.9×10–15 kg, and g = 10 m/s2), [2004], (a) 1.6×10–18 C, (b) 3.2×10–18 C, (c) 3.3×10–18 C, (d) 4.8×10–18 C, , (c) -, , 51., , Q, (1 + 2 2), 2
Page 250 :
P-241, , Electric Charges and Fields, , (a) surface change density on the inner surface is uniform, , Electric Dipole, Electric Flux, TOPIC 3 and Gauss's Law, 52., , Two identical electric point dipoles have dipole moments, ®, , ®, , P1 = P$i and P2 = - P$i and are held on the x axis at distance, ‘a’ from each other. When released, they move along xaxis with the direction of their dipole moments remaining, unchanged. If the mass of each dipole is ‘m’, their speed, when they are infinitely far apart is : [Sep. 06, 2020 (II)], , P, 1, (a), a pe 0 ma, , (c), 53., , P, 1, (b), a 2pe 0 ma, , P, 2, a pe 0 ma, , (d), , P, 2, a 2 pe 0 ma, , ur, An electric field E = 4 xiˆ - ( y 2 + 1) ˆj N/C passes through, the box shown in figure. The flux of the electric field, through surfaces ABCD and BCGF are marked as f1 and, f11 respectively. The difference between (f1 – f11) is (in, Nm2/C) _______., [9 Jan 2020, II], z, A (0, 0, 2), (0, 2, 2), , 54., , y, , D, , B, , (3, 0, 2), , C, , (3, 2, 2), E, F, x, (0, 0, 0), (3, 0, 0), H, G, (0, 2, 0), (3, 2, 0), , In finding the electric field using Gauss law the formula, r, q, | E | = enc is applicable. In the formula Î is, 0, Î | A|, 0, , 55., , and equal to, , permittivity of free space, A is the area of Gaussian surface, and qenc is charge enclosed by the Gaussian surface. This, equation can be used in which of the following situation?, [8 Jan 2020, I], (a) Only when the Gaussian surface is an equipotential, surface., Only when the Gaussian surface is an, r, (b) equipotential surface and | E | is constant on the surface., r, (c) Only when | E | = constant on the surface., (d) For any choice of Gaussian surface., Shown in the figure is a shell made of a conductor. It has, inner radius a and outer radius b, and carries charge Q. At, ur, its centre is a dipole p as shown. In this case :, [12 April 2019, I], , 56., , Q/2, , 4 pa 2, (b) electric field outside the shell is the same as that of a, point charge at the centre of the shell., (c) surface charge density on the outer surface depends, r, on P, (d) surface charge density on the inner surface of the, shell is zero everywhere.Let a total charge 2 Q be distributed in a sphere of radius, R, with the charge density given by r(r) = kr, where r is, the distance from the centre. Two charges A and B, of – Q, each, are placed on diametrically opposite points, at equal, distance, a, from the centre. If A and B do not experience, any force, then., [12 April 2019, II], , (a) a = 8–1/4 R, , (b) a =, , (c) a = 2–1/4 R, , (d), , 3R, 21/ 4, , a = R/ 3, 57. An electric dipole is formed by two equal and opposite, charges q with separation d. The charges have same mass, m. It is kept in a uniform electric field E. If it is slightly, rotated from its equilibrium orientation, then its angular, frequency w is :, [8 April 2019, II], qE, qE, qE, 2qE, (b), (c) 2, (d), md, md, 2md, md, 58. An electric field of 1000 V/m is applied to an electric dipole, at angle of 45°. The value of electric dipole moment is, 10–29 C.m. What is the potential energy of the electric, dipole?, [11 Jan 2019, II], (a) –20 × 10–18 J, (b) –7 × 10 –27 J, (c) –10 × 10–29 J, (d) – 9 × 10–20 J, 59. Charges – q and + q located at A and B, respectively,, constitute an electric dipole. Distance AB = 2a, O is the, mid point of the dipole and OP is perpendicular to AB., A charge Q is placed at P where OP = y and y >> 2a. The, charge Q experiences an electrostatic force F. If Q is now, moved along the equatorial line to P¢ such that OP¢, , (a), , æ yö, æy, ö, = ç ÷ , the force on Q will be close to: ç >> 2a ÷, 3, 3, è ø, è, ø, [10 Jan 2019, II], P, , Q P¢, A, , (a) 3 F, , O, –q, (b), , +q, , F, 3, , (c) 9 F, , B, , (d) 27 F
Page 251 :
P-242, , 60., , Physics, , A charge Q is placed at a distance a/2 above the centre of the, square surface of edge a as shown in the figure. The electric, flux through the square surface is:, [Online April 15, 2018], , Q, (a) 3e, 0, , P, a/2, , Q, (b) 6e, 0, Q, (c) 2e, 0, , 61., , 62., , a, Q, (d) e, 0, ur, An electric dipole has a fixed dipole moment p , which, makes angle q with respect to x-axis. When subjected to, uur, ur, an electric field E1 = Eiˆ , it experiences a torque T1 = t iˆ ., uur, When subjected to another electric field E2 = 3E1 ˆj it, uur, ur, experiences torque T2 = -T1 . The angle q is :, [2017], (a) 60°, (b) 90°, (c) 30°, (d) 45°, Four closed surfaces and corresponding charge distributions are shown below., [Online April 9, 2017], , 5q, , q, , 2q, , q, , S1, , 8q, –2q, –4q, , –q, q, , q, , q, , S3, , S2, , Q, b, , (c), 64., , (, , 2, , p a -b, , Q, 2 pa 2, , 2, , ), , (b), , (d), , (a) pa2 E, , S4, , a, , 2Q, , 45°, , 2Q, pa 2, , (, , ®, E, , 3q, , Let the respective electric fluxes through the surfaces be, F 1, F 2, F 3, and F 4. Then :, (a) F 1< F2 = F 3 > F4, (b) F 1> F2 > F 3 > F4, (c) F 1= F2 = F 3 = F4, (d) F 1> F 3 ; F 2 < F4, 63. The region between two concentric spheres of radii 'a' and, 'b', respectively (see figure), have volume charge density, A, r= , where A is a constant and r is the distance from, r, the centre. At the centre of the spheres is a point charge, Q. The value of A such that the electric field in the region, between the spheres will be constant, is:, [2016], , (a), , through a circular surface of radius 0.02 m parallel to the YZ plane is nearly:, [Online April 19, 2014], (a) 0.125 Nm2/C, (b) 0.02 Nm2/C, 2, (c) 0.005 Nm /C, (d) 3.14 Nm2/C, ur, ur, 65. Two point dipoles of dipole moment p1 and p 2 are at a, ur ur, distance x from each other and p1 || p 2 . The force between, the dipoles is :, [Online April 9, 2013], 1 4 p1 p2, 1 3 p1 p2, (a), (b), 4pe0 x 4, 4pe0 x3, 1 8 p1 p2, 1 6 p1 p2, (c), (d), 4pe0 x 4, 4pe0 x 4, 66. The flat base of a hemisphere of radius a with no charge, inside it lies in a horizontal plane. A uniform electric field, ®, p, E is applied at an angle, with the vertical direction. The, 4, electric flux through the curved surface of the hemisphere, is, [Online May 19, 2012], , (, , ), , The electric field in a region of space is given by,, r, E = Eoˆi + 2Eoˆj where Eo = 100 N/C. The flux of the field, , ), , 2, (d), 2 2, 2 2, 67. An electric dipole is placed at an angle of 30° to a nonuniform electric field. The dipole will experience [2006], (a) a translational force only in the direction of the field, (b) a translational force only in a direction normal to, the direction of the field, (c) a torque as well as a translational force, (d) a torque only, 68. If the electric flux entering and leaving an enclosed surface, respectively is f1 and f2, the electric charge inside the surface, will be, [2003], (a) (f2 – f1)eo, (b) (f1 – f2)/eo, (c) (f2 – f1)/eo, (d) (f1 – f2)eo, 69. A charged particle q is placed at the centre O of cube of, length L (A B C D E F G H). Another same charge q is, placed at a distance L from O. Then the electric flux, through ABCD is, [2002], E, F, , (c), , c, , O, , 2 p b2 - a 2, , 2, , ( p + 2) pa 2 E, , pa2 E, , D, , Q, , pa2 E, , (b), , H, A, , (a) q /4 p Î0 L, (c) q/2 p Î0 L, , q, , q, G, B, , L, , (b) zero, (d) q/3 p Î0 L
Page 263 :
16, , P-254, , Physics, , Electrostatic Potential, and Capacitance, 4., , Electrostatic Potential and, TOPIC 1, Equipotential Surfaces, 1., , Ten charges are placed on the circumference of a circle of, radius R with constant angular separation between, successive charges. Alternate charges 1, 3, 5, 7, 9 have, charge (+q) each, while 2, 4, 6, 8, 10 have charge (–q), each. The potential V and the electric field E at the centre, of the circle are respectively :, (Take V = 0 at infinity), [Sep. 05, 2020 (II)], (a) V =, , r, R, , (a), , 5., , 10 q, , 4pe 0 R 2, (c) V = 0; E = 0, , (d) V =, 2., , 10 q, 10 q, ;E=, 4 pe0 R, 4 pe 0 R 2, , Two isolated conducting spheres S1 and S2 of radius, , 2, R, 3, , 1, R have 12 mC and –3 mC charges, respectively, and, 3, are at a large distance from each other. They are now, connected by a conducting wire. A long time after this is, done the charges on S1 and S2 are respectively :, , and, , 3., , 1, (R + r ), Q, 4pe 0 2( R 2 + r 2 ), , (b), , 1 (2 R + r ), Q, 4 pe0 ( R 2 + r 2 ), , 1, (R + r ), 1 ( R + 2 r )Q, Q, (d), 2, 2, 4pe0 ( R 2 + r 2 ), 4pe 0 2( R + r ), ur, A point dipole = p – po $x kept at the origin. The potential, and electric field due to this dipole on the y-axis at a, distance d are, respectively : (Take V = 0 at infinity), [12 April 2019 I], ur, ur, ur, p, -p, p, ,, (a), (b) 0,, 4pe 0 d 3, 4pe 0 d 2 4pe 0 d 3, ur, ur, ur, p, p, -p, ,, (c) 0,, (d), 4pe 0 d 3, 4pe 0 d 2 4pe 0 d 3, A uniformly charged ring of radius 3a and total charge q, is placed in xy-plane centred at origin. A point charge q is, moving towards the ring along the z-axis and has speed v, at z = 4a. The minimum value of v such that it crosses the, origin is :, [10 April 2019 I], , (c), , 10q, ;E=0, 4pe 0 R, , (b) V = 0; E =, , A charge Q is distributed over two concentric conducting, thin spherical shells radii r and R (R > r). If the surface, charge densities on the two shells are equal, the electric, potential at the common centre is : [Sep. 02, 2020 (II)], , [Sep. 03, 2020 (I)], (a) 4.5 mC on both, (b) +4.5 mC and –4.5 mC, (c) 3 mC and 6 mC, (d) 6 mC and 3 mC, Concentric metallic hollow spheres of radii R and 4R hold, charges Q1 and Q2 respectively. Given that surface charge, densities of the concentric spheres are equal, the potential, difference V(R) – V(4R) is :, [Sep. 03, 2020 (II)], (a), , 3Q1, 16pe 0 R, , (b), , 3Q2, 4pe0 R, , (c), , Q2, 4pe0 R, , (d), , 3Q1, 4pe0 R, , 6., , 1/2, , (a), , 2 æ 4 q2 ö, ç, ÷, m è 15 4pe0a ø, , 1/2, , 7., , 1/2, , (b), , 2 æ 1 q2 ö, ç, ÷, m è 5 4pe0a ø, , 1/2, , 2 æ 2 q2 ö, 2 æ 1 q2 ö, (c), (d), ç, ÷, ç, ÷, m è 15 4pe0a ø, m è 15 4pe0a ø, A solid conducting sphere, having a charge Q, is, surrounded by an uncharged conducting hollow spherical, shell. Let the potential difference between the surface of, the solid sphere and that of the outer surface of the hollow, shell be V. If the shell is now given a charge of – 4 Q, the, new potential difference between the same two surfaces, is :, [8 April 2019 I], (a) – 2V, (b) 2 V, (c) 4 V, (d), V
Page 264 :
P-255, , Electrostatic Potential and Capacitance, , 8., , 9., , r, The electric field in a region is given by E = ( Ax + B ) iˆ ,, where E is in NC–1 and x is in metres. The values of, constants are A = 20 SI unit and B = 10 SI unit. If the, potential at x = 1 is V1 and that at x = –5 is V2, then, V1 – V2 is :, [8 Jan. 2019 II], (a) 320 V, (b) – 48V (c) 180 V, (d) – 520 V, The given graph shows variation (with distance r from centre ), of :, [11 Jan. 2019 I], , rO, , r, rO, (a) Electric field of a uniformly charged sphere, (b) Potential of a uniformly charged spherical shell, (c) Potential of a uniformly charged sphere, (d) Electric field of a uniformly charged spherical shell, 10. A charge Q is distributed over three concentric spherical, shells of radii a, b, c (a < b < c) such that their surface, charge densities are equal to one another., The total potential at a point at distance r from their, common centre, where r < a, would be:, [10 Jan. 2019 I], Q ab + bc + ca, (a) 12pÎ, abc, 0, , (b), , Q (a 2 + b 2 + c 2 ), 4pÎ0 (a 3 + b 3 + c3 ), , Q (a + b + c), Q, (c) 4pÎ (a + b + c), (d) 4pÎ (a 2 + b 2 + c2 ), 0, 0, 11. Two electric dipoles, A, B with respective dipole, r, r, moments d A = – 4 qa iˆ and d B = – 2 qa iˆ are placed on, the x-axis with a separation R, as shown in the figure, , The distance from A at which both of them produce the, same potential is:, [10 Jan. 2019 I], R, 2R, (a) 2 + 1, (b), 2 +1, (c), , R, , (d), , 2R, , 2 -1, 2 -1, 12. Consider two charged metallic spheres S1 and S2 of radii, R1 and R2, respectively. The electric fields E1 (on S1) and, E2 (on S2) on their surfaces are such that E1/E2 = R1/R2., Then the ratio V1(on S1)/V2(on S2) of the electrostatic, potentials on each sphere is:, [8 Jan. 2019 II], (a) R1/R 2, (b) (R1/R2)2, 3, , æR ö, (c) (R2/R1), (d) ç 1 ÷, è R2 ø, 13. Three concentric metal shells A, B and C of respective, radii a, b and c (a < b < c) have surface charge densities, +s, –s and +s respectively. The potential of shell B is:, [2018], , s é a2 - b2 ù, (a) Î ê a +c ú, 0 êë, ûú, , s é a 2 - b2 ù, (b) Î ê b +cú, 0 êë, ûú, , s é b2 - c2 ù, (c) Î ê b +a ú, 0 ëê, ûú, , s é b 2 - c2 ù, (d) Î ê c +a ú, 0 ëê, ûú, , 14. There is a uniform electrostatic field in a region. The, potential at various points on a small sphere centred at P,, in the region, is found to vary between in the limits 589.0 V, to 589.8 V. What is the potential at a point on the sphere, whose radius vector makes an angle of 60° with the direction, of the field ?, [Online April 8, 2017], (a) 589.5 V (b) 589.2 V (c) 589.4 V (d) 589.6 V, 15. Within a spherical charge distribution of charge density, r(r), N equipotential surfaces of potential V0, V0 + DV, V0, + 2DV, .........V0 + NDV (DV > 0), are drawn and have, increasing radii r0, r1, r2,......... rN, respectively. If the, difference in the radii of the surfaces is constant for all, values of V0 and DV then :, [Online April 10, 2016], 1, (a) r(r) = constant, (b) r(r) µ 2, r, 1, (c) r(r) µ, (d) r(r) µ r, r, 16. The potential (in volts) of a charge distribution is given by, V(z) = 30 – 5z2 for |z| £ 1m, V(z) = 35 – 10 |z| for |z| ³ 1 m., V(z) does not depend on x and y. If this potential is, generated by a constant charge per unit volume r0 (in, units of e0) which is spread over a certain region, then, choose the correct statement., [Online April 9, 2016], (a) r0 = 20 e0 in the entire region, (b) r0 = 10 e0 for |z| £ 1 m and p0 = 0 elsewhere, (c) r0 = 20 e0 for |z| £ 1 m and p0 = 0 elsewhere, (d) r0 = 40 e0 in the entire region, 17. A uniformly charged solid sphere of radius R has potential, V0 (measured with respect to ¥) on its surface. For this, sphere the equipotential surfaces with potentials, 3V0 5V0 3V0, V0, ,, ,, and, have radius R1, R2, R3 and R4, 2, 4, 4, 4, respectively. Then, [2015], (a) R1 = 0 and R2 < (R4 – R3), (b) 2R = R4, (c) R1 = 0 and R2 > (R4 – R3), (d) R1 ¹ 0 and (R2 – R1) > (R4 – R3), r, -1 exists in a region of, $, 18. An electric field E = (25i$ + 30j)NC, space. If the potential at the origin is taken to be zero, then the potential at x = 2 m, y = 2 m is :, [Online April 11, 2015], (a) –110 J (b) –140 J (c) –120 J (d) –130 J, r, 19. Assume that an electric field E = 30x 2 ˆi exists in space., Then the potential difference VA - VO , where VO is the, potential at the origin and VA the potential at x = 2 m is:, (a) 120 J/C, (b) –120 J/C, [2014], (c) –80 J/C, (d) 80 J/C
Page 265 :
P-256, , Physics, , 20. Consider a finite insulated, uncharged conductor placed, near a finite positively charged conductor. The uncharged, body must have a potential : [Online April 23, 2013], (a) less than the charged conductor and more than at, infinity., (b) more than the charged conductor and less than at, infinity., (c) more than the charged conductor and more than at, infinity., (d) less than the charged conductor and less than at, infinity., 21. Two small equal point charges of magnitude q are, suspended from a common point on the ceiling by, insulating mass less strings of equal lengths. They come, to equilibrium with each string making angle q from the, vertical. If the mass of each charge is m, then the, electrostatic potential at the centre of line joining them will, æ 1, ö, be ç, = k÷., è 4p Î0, ø, , [Online April 22, 2013], , (a) 2 k mg tan q, , (b), , k mg tan q, , (c) 4 k mg / tan q, , (d), , k mg / tan q, , (, , ), , 1 ( R + r) Q, (c) 4pe, 0 R2 + r 2, , (, , ), , ( R + r) Q, 1, (b), 4pe 0 2 R 2 + r 2, , (, , ( R - r) Q, 1, (d), 4pe 0 2 R 2 + r 2, , (, , q, A, , q, B, , D, -q, , C, -q, , ur, E, ur changes, V remains unchanged, unchanged, V changes, E remains, ur, both E and V change, ur, (d) E and V remain unchanged, 28. The potential at a point x (measured in m m) due to some, charges situated on the x-axis is given by V(x) = 20/(x2 – 4), volt. The electric field E at x = 4 m m is given by [2007], (a) (10/9) volt/ m m and in the +ve x direction, (b) (5/3) volt/ m m and in the –ve x direction, (c) (5/3) volt/ m m and in the +ve x direction, (d) (10/9) volt/ m m and in the –ve x direction, 29. Two thin wire rings each having a radius R are placed at a, distance d apart with their axes coinciding. The charges, on the two rings are +q and -q. The potential difference, between the centres of the two rings is, [2005], (a), (b), (c), , 22. A point charge of magnitude + 1 mC is fixed at (0, 0, 0). An, isolated uncharged spherical conductor, is fixed with its, center at (4, 0, 0). The potential and the induced electric, field at the centre of the sphere is :[Online April 22, 2013], (a) 1.8 × 105 V and – 5.625 × 106 V/m, (b) 0 V and 0 V/m, (c) 2.25 × 105 V and – 5.625 × 106 V/m, (d) 2.25 × 105 V and 0 V/m, 23. A charge of total amount Q is distributed over two, concentric hollow spheres of radii r and R (R > r) such that, the surface charge densities on the two spheres are equal., The electric potential at the common centre is, [Online May 19, 2012], , 1 ( R - r) Q, (a), 4pe 0 R2 + r 2, , 26. An electric charge 10–3 m C is placed at the origin (0, 0) of, X – Y co-ordinate system. Two points A and B are situated, at ( 2, 2) and (2, 0) respectively. The potential, difference between the points A and B will be, [2007], (a) 4.5 volts, (b) 9 volts, (c) Zero, (d) 2 volt, 27. Charges are placed on the vertices of a square as shown., r, Let E be the electric field and V the potential at the, centre. If the charges on A and B are interchanged with, those on D and C respectively, then, [2007], , 24. The electric potential V(x) in a region around the origin is, given by V(x) = 4x2 volts. The electric charge enclosed in, a cube of 1 m side with its centre at the origin is (in coulomb), [Online May 7, 2012], (a) 8e0, (b) – 4e0 (c) 0, (d) – 8e0, 25. The electrostatic potential inside a charged spherical ball is, given by f = ar2 + b where r is the distance from the centre, and a, b are constants. Then the charge density inside the, ball is:, [2011], (a) –6ae0r, (b) –24pae0, (c) –6ae0, (d) –24pae0r, , ù, ú (b), R 2 + d 2 úû, , q, 2 p Î0, , (c), , q é1, 1, ê 2, 4 p Î0 ê R, R + d2, ë, , ), ), , é1, ê êë R, , (a), , 1, , ù, ú (d), úû, , qR, 4p Î0 d 2, , zero, , 30. A thin spherical conducting shell of radius R has a charge, q. Another charge Q is placed at the centre of the shell., R, 2, [2003], , The electrostatic potential at a point P , a distance, from the centre of the shell is, 2Q, (a) 4pe R, o, , 2Q, 2q, (b) 4pe R - 4pe R, o, o, , 2Q, q, (c) 4pe R + 4pe R, o, o, , (q + Q)2, (d) 4pe R, o
Page 266 :
P-257, , Electrostatic Potential and Capacitance, , Electric Potential Energy and, TOPIC 2 Work Done in Carrying a, Charge, 31. A solid sphere of radius R carries a charge Q + q distributed, uniformaly over its volume. A very small point like piece of, it of mass m gets detached from the bottom of the sphere, and falls down vertically under gravity. This piece carries, charge q. If it acquires a speed v when it has fallen through, a vertical height y (see figure), then : (assume the remaining, portion to be spherical)., [Sep. 05, 2020 (I)], , Q, , R, , 33. Hydrogen ion and singly ionized helium atom are, accelerated, from rest, through the same potential, difference. The ratio of final speeds of hydrogen and, helium ions is close to :, [Sep. 03, 2020 (II)], (a) 1 : 2, (b) 10 : 7, (c) 2 : 1, (d) 5 : 7, 34. In free space, a particle A of charge 1 mC is held fixed at a, point P. Another particle B of the same charge and mass 4, mg is kept at a distance of 1 mm from P. If B is released,, then its velocity at a distance of 9 mm from P is :, é, ù, 1, = 9 ´ 109 Nm 2C -2 ú, êTake, [10 April 2019 II], 4pe 0, ë, û, (a) 1.0m/s, (b) 3.0×104 m/s, 3, (c) 2.0×10 m/s, (d) 1.5×102 m/s, 35. A system of three charges are placed as shown in the, figure:, , q, y, v, , é, ù, qQ, (a) v 2 = y ê, +, g, ú, 2, êë 4pe 0 R ym, úû, é, ù, qQ, + gú, (b) v 2 = y ê, ë 4pe 0 R( R + y )m, û, é, ù, Qq R, + gú, (c) v 2 = 2 y ê, 3, êë 4pe0 ( R + y ) m, úû, é, ù, qQ, 2, + gú, (d) v = 2 y ê, 4, (, ), pe, R, R, +, y, m, ë, 0, û, 32. A two point charges 4q and –q are fixed on the x-axis at, , If D >> d, the potential energy of the system is best given, by, [9 April 2019 I], (a), , 1 é - q 2 - qQd ù, 1 é - q 2 2qQd ù, +, ê, ú, ê, 2 ú (b), 4p Î0 ë d 2 D û, 4p Î0 ë d, D2 û, , (c), , 1 é q 2 qQd ù, 1 é q 2 qQd ù, + 2 ú (d), - 2 ú, ê+, ê4p Î0 ë d, 4p Î0 ë d, D û, D û, , 36. A positive point charge is released from rest at a distance, r0 from a positive line charge with uniform density. The, speed (v) of the point charge, as a function of, instantaneous distance r from line charge, is proportional, to :, [8 April 2019 II], , d, d, and x = , respectively. If a third point charge, ge, 2, 2, ‘q’ is taken from the origin to x = d along the semicircle as, shown in the figure, the energy of the charge will :, [Sep. 04, 2020 (I)], x=-, , 4q, , (a) increase by, , 3q 2, 4pe0 d, , (b) increase by, , 2q 2, 3pe0 d, , (c) decrease by, , q2, 4pe0 d, , (d) decrease by, , 4q 2, 3pe0 d, , –q, , +r /r, (a) v µ e 0, , (b) v µ, , ærö, ln ç ÷, è r0 ø, , ærö, ærö, (c) v µ ln ç ÷, (d) v µ ç ÷, r, è 0ø, è r0 ø, 37. There is a uniform spherically symmetric surface charge, density at a distance Ro from the origin. The charge, distribution is initially at rest and starts expanding because, of mutual repulsion. The figure that represents best the, speed V (R(t)) of the distribution as a function of its, instantaneous radius R(t) is:, [12 Jan. 2019 I]
Page 267 :
P-258, , Physics, V(R(t)), , V(R(t)), , (a), , (b), R (t), , Ro, , Ro, , R (t), , V(R(t)), , V(R(t)), Vo, , (c), , (d), Ro, , R (t), , Ro, , R (t), , 38. Three charges Q, + q and + q are placed at the vertices of, a right-angle isosceles triangle as shown below. The net, electrostatic energy of the configuration is zero, if the value, of Q is :, [11 Jan. 2019 I], Q, , +q, , +q, , - 2q, (b), 2 +1, , (a) + q, , -q, (d) –2q, (c), 1+ 2, 39. Four equal point charges Q each are placed in the xy, plane at (0, 2), (4, 2), (4, – 2) and (0, – 2). The work, required to put a fifth charge Q at the origin of the, coordinate system will be:, [10 Jan. 2019 II], (a), , Q2, 4 pe 0, , (c), , Q2, 2 2 pe0, , 1 ö, æ, ç1 +, ÷, 3ø, è, , (b), , Q2, 4 pe 0, , (d), , Q2, 4pe 0, , 1 ö, æ, ç1 +, ÷, 5ø, è, , 40. Statement 1 : No work is required to be done to move a, test charge between any two points on an equipotential, surface., Statement 2 : Electric lines of force at the equipotential, surfaces are mutually perpendicular to each other., [Online April 25, 2013], (a) Statement 1 is true, Statement 2 is true, Statement 2 is, the correct explanation of Statement 1., (b) Statement 1 is true, Statement 2 is true, Statement 2 is, not the correct explanation of Statement 1., (c) Statement 1 is true, Statement 2 is false., (d) Statement 1 is false, Statement 2 is true., , 41. An insulating solid sphere of radius R has a uniformly, positive charge density r. As a result of this uniform charge, distribution there is a finite value of electric potential at, the centre of the sphere, at the surface of the sphere and, also at a point outside the sphere. The electric potential, at infinite is zero., [2012], Statement -1 When a charge q is taken from the centre, to the surface of the sphere its potential energy changes, qr, ., by, 3e0, Statement -2 The electric field at a distance r (r <R) from, rr, the centre of the sphere is, ., 3e0, (a) Statement 1 is true, Statement 2 is true; Statement 2 is, not the correct explanation of statement 1., (b) Statement 1 is true Statement 2 is false., (c) Statement 1 is false Statement 2 is true., (d) Statement 1 is true, Statement 2 is true, Statement 2 is, the correct explanation of Statement 1, 42. Two positive charges of magnitude ‘q’ are placed, at the, ends of a side (side 1) of a square of side ‘2a’. Two negative, charges of the same magnitude are kept at the other corners., Starting from rest, if a charge Q moves from the middle of, side 1 to the centre of square, its kinetic energy at the, centre of square is, [2011 RS], (a) zero, (c), , 1 2qQ æ, 2 ö, 1÷, 4pe 0 a çè, 5ø, , (b), , 1 2qQ æ, 1 ö, 1+, ÷, 4pe 0 a çè, 5ø, , (d), , 1 2qQ æ, 1 ö, 1÷, 4pe 0 a çè, 5ø, , 43. Two points P and Q are maintained at the potentials of 10, V and – 4 V, respectively. The work done in moving 100, electrons from P to Q is:, [2009], (a) 9.60 × 10–17J, (b) –2.24 × 10–16 J, (c) 2.24 × 10–16 J, (d) –9.60× 10–17 J, 44. Two insulating plates are both uniformly charged in such, a way that the potential difference between them is V2 –, V1 = 20 V. (i.e., plate 2 is at a higher potential). The plates, are separated by d = 0.1 m and can be treated as infinitely, large. An electron is released from rest on the inner surface, of plate 1. What is its speed when it hits plate 2? (e = 1.6 ×, 10–19 C, me = 9.11 × 10–31 kg), [2006], Y, 0.1 m, X, , 1, 2.65 × 106 m/s, , (a), (c) 1.87 × 106 m/s, , 2, (b) 7.02 × 1012 m/s, (d) 32 × 10–19 m/s
Page 268 :
P-259, , Electrostatic Potential and Capacitance, , 45. A charged particle ‘q’ is shot towards another charged, particle ‘Q’ which is fixed, with a speed ‘v’. It approaches, ‘Q’ upto a closest distance r and then returns. If q were, given a speed of ‘2v’ the closest distances of approach, would be, [2004], (a) r/2, (b) 2 r, (c) r, (d) r/4, 46. On moving a charge of 20 coulomb by 2 cm, 2 J of work, is done, then the potential difference between the points, is, [2002], (a) 0.1 V, (b) 8 V, (c) 2 V, (d) 0.5 V, , Capacitors, Grouping of, TOPIC 3 Capacitors and Energy Stored, in a Capacitor, 47. Two capacitors of capacitances C and 2C are charged to, potential differences V and 2V, respectively. These are then, connected in parallel in such a manner that the positive, terminal of one is connected to the negative terminal of the, other. The final energy of this configuration is :, [Sep. 05, 2020 (I)], 25 2, 3 2, (a), CV, (b), CV, 6, 2, 9 2, (c) zero, (d), CV, 2, 48. In the circuit shown, charge on the 5 mF capacitor is :, [Sep. 05, 2020 (II)], 2 mF, , 4 mF, , (a) 450 mC, (b) 590 mC, (c) 160 mC, (d) 650 mC, 51. A 5 mF capacitor is charged fully by a 220 V supply. It is, then disconnected from the supply and is connected in, series to another uncharged 2.5 mF capacitor. If the energy, change during the charge redistribution is, , X, J then value, 100, , of X to the nearest integer is ________., [NA Sep. 02, 2020 (I)], 52. A 10 mF capacitor is fully charged to a potential difference, of 50 V. After removing the source voltage it is connected, to an uncharged capacitor in parallel. Now the potential, difference across them becomes 20 V. The capacitance of, the second capacitor is :, [Sep. 02, 2020 (II)], (a) 15 mF, (b) 30 mF, (c) 20 mF, (d) 10 mF, 53. Effective capacitance of parallel combination of two, capacitors C1 and C2 is 10 mF. When these capacitors are, individually connected to a voltage source of 1 V, the, energy stored in the capacitor C2 is 4 times that of C1. If, these capacitors are connected in series, their effective, capacitance will be:, [8 Jan. 2020 I], (a) 4.2 mF, (b) 3.2 mF (c) 1.6 mF, (d) 8.4 mF, 54. A capacitor is made of two square plates each of side ‘a’, making a very small angle a between them, as shown in, figure. The capacitance will be close to: [8 Jan. 2020 II], V1, a, , 5 mF, d, , O, , 6V, , (a) 18.00 mC, (b) 10.90 mC, (c) 16.36 mC, (d) 5.45 mC, 49. A capacitor C is fully charged with voltage V0 . After, disconnecting the voltage source, it is connected in parallel, C, with another uncharged capacitor of capacitance . The, 2, energy loss in the process after the charge is distributed, between the two capacitors is :, [Sep. 04, 2020 (II)], 1, 1, (b) CV02, CV02, 3, 2, 1, 1, CV02, (c), (d), CV02, 4, 6, 50. In the circuit shown in the figure, the total charge is 750 mC, and the voltage across capacitor C2 is 20 V. Then the, charge on capacitor C2 is :, [Sep. 03, 2020 (I)], , (a), , C2, , C1 = 15 mF, , C3 = 8 m F, +, , V, , a, , 6V, , –, , Î0 a 2, (a), d, , æ aa ö, çè1 - ÷ø, 2d, , V2, , Î0 a 2, (b), d, , æ aa ö, çè1 - ÷ø, 4d, , Î0 a 2 æ aa ö, Î0 a 2 æ 3aa ö, (d), çè1 + ÷ø, ç1 ÷, d, d, d è, 2d ø, 55. A parallel plate capacitor has plates of area A separated, by distance ‘d’ between them. It is filled with a dielectric, which has a dielectric constant that varies as k(x) = K(1 +, ax) where ‘x’ is the distance measured from one of the, plates. If (ad) << l, the total capacitance of the system is, best given by the expression:, [7 Jan. 2020 I], AK Î0 æ ad ö, (a), ç1 +, ÷, d è, 2 ø, 2, A Î0 K æ æ ad ö ö, ç1 + ç, ÷, (b), ÷, d ç è 2 ø ÷, è, ø, æ, 2 2ö, A Î0 K ç a d ÷, 1+, (c), 2 ÷, d ç, è, ø, AK Î0, (1 + ad ), (d), d, , (c)
Page 269 :
P-260, , Physics, , 56. A 60 pF capacitor is fully charged by a 20 V supply. It is, then disconnected from the supply and is connected, to another uncharged 60 pF capacitor in parallel. The, electrostatic energy that is lost in this process by the, time the charge is redistributed between them is (in nJ), [NA 7 Jan. 2020 II], 57. The parallel combination of two air filled parallel plate, capacitors of capacitance C and nC is connected to a, battery of voltage, V. When the capacitors are fully, charged, the battery is removed and after that a dielectric, material of dielectric constant K is placed between the, two plates of the first capacitor. The new potential, difference of the combined system is: [9 April 2020 II], nV, (a), K+n, , (b) V, , 60. Figure shows charge (q) versus voltage (V) graph for, series and parallel combination of two given capacitors., The capacitances are :, [10 April 2019 I], , 61., , (n + 1) V, V, (d), (K + n), K+n, 58. Two identical parallel plate capacitors, of capacitance C, each, have plates of area A, separated by a distance d., The space between the plates of the two capacitors, is, filled with three dielectrics, of equal thickness and dielectric, constants K1, K2 and K3. The first capacitors is filled as, shown in Fig. I, and the second one is filled as shown in, Fig. II., If these two modified capacitors are charged by the same, potential V, the ratio of the energy stored in the two, would, be (E1 refers to capacitors (I) and E2 to capacitors (II) :, [12 April 2019 I], (c), , 62., , 63., , 64., K1 K 2 K 3, E1, (a) E = ( K + K + K )( K K + K K + K K, 2, 1, 2, 3, 2 3, 3 1, 1 2, , (a) 40 mF and 10 mF, (b) 60 mF and 40 mF, (c) 50 mF and 30 mF, (d) 20 mF and 30 mF, A capacitor with capacitance 5mF is charged to 5 mC. If, the plates are pulled apart to reduce the capacitance to 2, ¼F, how much work is done?, [9 April 2019 I], (a) 6.25 × 10–6 J, (b) 3.75 × 10–6 J, (c) 2.16 × 10–6 J, (d) 2.55 × 10–6 J, Voltage rating of a parallel plate capacitor is 500 V. Its, dielectric can withstand a maximum electric field of 106 V/, m. The plate area is 10–4 m2. What is the dielectric constant, if the capacitance is 15 pF ?, [8 April 2019 I], –12, 2, 2, (given “0 = 8.86 × 10 C /Nm ), (a) 3.8, (b) 8.5, (c) 4.5, (d), 6.2, A parallel plate capacitor has 1mF capacitance. One of its, two plates is given + 2mC charge and the other plate,, +4mC charge. The potential difference developed across, the capacitor is :, [8 April 2019 II], (a) 3 V, (b) 1 V, (c) 5 V, (d) 2 V, In the figure shown, after the switch ‘S’ is turned from, position ‘A’ to position ‘B’, the energy dissipated in the, circuit in terms of capacitance ‘C’ and total charge ‘Q’ is:, [12 Jan. 2019 I], A, , E1 ( K1 + K 2 + K3 )( K 2 K3 + K3 K1 + K1 K 2, (b) E =, K1 K 2 K 3, 2, 9 K1 K 2 K3, E1, (c) E = ( K + K + K )( K K + K K + K K ), 2, 1, 2, 3, 2 3, 3 1, 1 2, E1 ( K1 + K 2 + K3 )(K 2 K 3 + K 3 K1 + K1 K 2, (d) E =, 9 K1 K 2 K3, 2, , 59. In the given circuit, the charge on 4 mF capacitor will be :, [12 April 2019 II], , S, e, , (b) 9.6 mC (c) 13.4 mC, , (d) 24 mC, , C, , 3C, , 1 Q2, 3 Q2, 5 Q2, 3 Q2, (b), (c), (d), 8 C, 8 C, 8 C, 4 C, 65. A parallel plate capacitor with plates of area 1 m2 each, are, at a separation of 0.1 m. If the electric field between the, plates is 100 N/C, the magnitude of charge on each plate is :, (a), , (Take Î0 = 8.85, (a) 5.4 mC, , B, , × 10–12, , C2, ), N – M2, , [12 Jan. 2019 II], , (a) 7.85 × 10–10 C, , (b) 6.85 × 10–10 C, , (c) 8.85 × 10–10 C, , (d) 9.85 × 10–10 C
Page 270 : P-261, , Electrostatic Potential and Capacitance, , 66. In the circuit shown, find C if the effective capacitance of, the whole circuit is to be 0.5 µF. All values in the circuit are, in µF., [12 Jan. 2019 II], A, , C, , 2, 2, , 2, , 1, 2, , 2, , 2, , B, , 7, µF, 11, , 6, 7, µF (c) 4 µF, (d), µF, 5, 10, 67. In the figure shown below, the charge on the left plate of, the 10 mF capacitor is –30mC. The charge on the right plate, of the 6mF capacitor is :, [11 Jan. 2019 I], (a), , (b), , 6 mF, 10 m F, 4 mF, , 2 mF, , (a) –12 m C, (b) +12 m C, (c) –18 m C, (d) +18 m C, 68. Seven capacitors, each of capacitance 2 µF, are to be, connected in a configuration to obtain an effective, , 69. A parallel plate capacitor having capacitance 12 pF is, charged by a battery to a potential difference of 10 V, between its plates. The charging battery is now, disconnected and a porcelain slab of dielectric constant, 6.5 is slipped between the plates. The work done by the, capacitor on the slab is:, [10 Jan. 2019 II], (a) 692 pJ, (b) 508 pJ, (c) 560 pJ, (d) 600 pJ, 70. A parallel plate capacitor is of area 6 cm2 and a, separation 3 mm. The gap is filled with three dielectric, materials of equal thickness (see figure) with dielectric, constants K1 = 10, K2 = 12 and K3 = 1(4) The dielectric, constant of a material which when fully inserted in, above capacitor, gives same capacitance would be:, [10 Jan. 2019 I], , (a) 4, (b) 14, (c) 12, (d) 36, 71. A parallel plate capacitor is made of two square plates of, side ‘a’, separated by a distance d (d<<a). The lower, triangular portion is filled with a dielectric of dielectric, constant K, as shown in the figure. Capacitance of this, capacitor is:, [9 Jan. 2019 I], , d, , æ 6ö, capacitance of ç ÷ µF. Which of the combinations,, è 13 ø, shown in figures below, will achieve the desired value?, [11 Jan. 2019 II], , K, a, , (a), , (a), , K Î0 a 2, 2d (K + 1), , K Î0 a 2, 1 K Î0 a 2, In K, (d), 2, d, d, 72. A parallel plate capacitor with square plates is filled with, four dielectrics of dielectric constants K1, K2, K3, K4, arranged as shown in the figure. The effective dielectric, constant K will be:, [9 Jan. 2019 II], , (a) K =, , (b) K =, (d), , K Î0 a 2, In K, d (K – 1), , (c), , (b), , (c), , (b), , K1, , K2, , L/2, , K3, , K4, , L/2, , d/2, , d/2, , ( K1 + K3 ) ( K 2 + K 4 ), K1 + K 2 + K 3 + K 4, , ( K1 + K 2 ) ( K3 + K 4 ), 2(K1 + K 2 + K 3 + K 4 )
Page 271 :
P-262, , Physics, , ( K1 + K 2 ) ( K 3 + K 4 ), , (c) K =, , K1 + K 2 + K 3 + K 4, , ( K1 + K 4 ) ( K 2 + K3 ), , (d) K =, , 2(K1 + K 2 + K 3 + K 4 ), , 73. A parallel plate capacitor of capacitance 90 pF is connected, to a battery of emf 20V. If a dielectric material of dielectric, 5, constant k = is inserted between the plates, the, 3, magnitude of the induced charge will be:, [2018], (a) 1.2 n C (b) 0.3 n C (c) 2.4 n C (d) 0.9 n C, 74. In the following circuit, the switch S is closed at t = 0. The, charge on the capacitor C1 as a function of time will be, æ, CC ö, given by ç Ceq = 1 2 ÷ ., C1 + C2 ø, è, , [Online April 16, 2018], , C1, , (a) CeqE[1 – exp(–t/RCeq)], , C2, , (b) C1E[1 – exp(–tR/C1)], S, (c) C2E[1 – exp(–t/RC2)], R, E, (d) CeqE exp(–t/RCeq), 75. The equivalent capacitance between A and B in the circuit, given below is:, 6 µF, , 2 µF, , A, 5 µF, , 5 µF, , 4 µF, , 2 µF, B, , [Online April 15, 2018], (a) 4.9 µF, (b) 3.6 µF (c) 5.4 µF, (d) 2.4 µF, 76. A parallel plate capacitor with area 200cm2 and separation, between the plates 1.5cm, is connected across a battery of, emf V. If the force of attraction between the plates is 25 × 10–, 6N, the value of V is approximately: [Online April 15, 2018], , 78. A capacitance of 2m F is required in an electrical circuit, across a potential difference of 1.0 kV. A large number of, 1m F capacitors are available which can withstand a, potential difference of not more than 300 V. The minimum, number of capacitors required to achieve this is [2017], (a) 24, (b) 32, (c) 2, (d) 16, 79. A combination of parallel plate capacitors is maintained at, a certain potential difference., , C1, A, , D, , When a 3 mm thick slab is introduced between all the, plates, in order to maintain the same potential difference,, the distance between the plates is increased by 2.4 mm., Find the dielectric constant of the slab., [Online April 9, 2017], (a) 3, (b) 4, (c) 5, (d) 6, 80. The energy stored in the electric field produced by a metal, sphere is 4.5 J. If the sphere contains 4 mC charge, its, radius will be : [Take : 1 = 9 ´ 109 N - m 2 / C 2 ], 4 pe0, [Online April 8, 2017], (a) 20mm (b) 32mm (c) 28mm (d) 16mm, 81. A combination of capacitors is set up as shown in the, figure. The magnitude of the electric field, due to a point, charge Q (having a charge equal to the sum of the charges, on the 4 mF and 9 mF capacitors), at a point distance 30 m, from it, would equal :, [2016], 3m F, 4m F, 9m F, 2m F, + –, 8V, , (a) 150V, (b) 100V (c) 250V, (d) 300V, 77. A capacitor C1 is charged up to a voltage V = 60V by, connecting it to battery B through switch (1), Now C1 is, disconnected from battery and connected to a circuit, consisting of two uncharged capacitors C2 = 3.0mF and C3 =, 6.0mF through a switch (2) as shown in the figure. The sum, of final charges on C2 and C3 is: [Online April 15, 2018], , (a) 36mC, , (b) 20mC, , B, , E, , 2 ö, æ, -12 C, ç e 0 = 8.85 ´ 10, ÷, N.m 2 ø, è, , (1), B, 60 V, , C3, , C2, , C, , (2), , (c) 54mC, , C, , 1, , A, 8, , C2, C1, , (a) 420 N/C, (b) 480 N/C, (c) 240 N/C, (d) 360 N/C, 82. Figure shows a network of capacitors where the numbers, indicates capacitances in micro Farad. The value of, capacitance C if the equivalent capacitance between point, A and B is to be 1 mF is :, [Online April 10, 2016], , C3, , 2, , (d) 40mC, , 6, 2, , 4, 12, B
Page 272 :
P-263, , Electrostatic Potential and Capacitance, , 32, 31, 33, 34, mF (b), mF (c), mF (d), mF, 23, 23, 23, 23, 83. Three capacitors each of 4 mF are to be connected in such, a way that the effective capacitance is 6mF. This can be, done by connecting them :, [Online April 9, 2016], (a) all in series, (b) all in parallel, (c) two in parallel and one in series, (d) two in series and one in parallel, 84. In the given circuit, charge Q2 on the 2µF capacitor, changes as C is varied from 1µF to 3µF. Q2 as a function of, 'C' is given properly by: (figures are drawn schematically, and are not to scale), [2015], , (a), , 1µF, , C, , 2µF, , E, Charge, , dielectric whose permittivity varies linearly from Î1 at one, plate to Î2 at the other. The capacitance of capacitor is:, [Online April 19, 2014], (a) Î0 (Î1 + Î2 ) A / d, (b) Î0 ( Î2 + Î1 ) A / 2d, (c) Î0 A / éëd ln ( Î2 / Î1 ) ùû, (d) Î0 ( Î2 - Î1 ) A / ëé d ln ( Î2 / Î1 ) ûù, 88. The space between the plates of a parallel plate capacitor, is filled with a ‘dielectric’ whose ‘dielectric constant’ varies, with distance as per the relation:, K(x) = Ko + lx (l = a constant), The capacitance C, of the capacitor, would be related to its, vacuum capacitance Co for the relation :, [Online April 12, 2014], (a) C =, , ld, Co, ln (1 + K o ld ), , (b) C =, , l, C, d.ln (1 + K o ld ) o, , (c) C =, , ld, C, ln (1 + ld / K o ) o, , (d) C =, , l, Co, d.ln (1 + Ko / ld ), , Charge, , Q2, , Q2, , (a), , (b), 1µF, , 3µF, , C, , 1µF, , 3µF, , 89. A parallel plate capacitor is made of two plates of length l,, width w and separated by distance d. A dielectric slab, (dielectric constant K) that fits exactly between the plates, is held near the edge of the plates. It is pulled into the, , C, , Charge, , Charge, , Q2, , Q2, , (c), 3µF, , C, , 1µF, , 3µF, , C, , 85. In figure a system of four capacitors connected across, a 10 V battery is shown. Charge that will flow from, switch S when it is closed is : [Online April 11, 2015], , 2mF a, , ¶U, where U is the energy of, ¶x, the capacitor when dielectric is inside the capacitor up to, distance x (See figure). If the charge on the capacitor is Q, then the force on the dielectric when it is near the edge is:, [Online April 11, 2014], , capacitor by a force F = -, , (d), 1µF, , 87. The gap between the plates of a parallel plate capacitor of, area A and distance between plates d, is filled with a, , 3mF, x, l, , S, , d, , 3mF b 2mF, , 2, , (a), , 10 V, (a) 5 µC from b to a, (b) 20 µC from a to b, (c) zero, (d) 5 µC from a to b, 86. A parallel plate capacitor is made of two circular plates, separated by a distance 5 mm and with a dielectric of, dialectric constant 2.2 between them. When the electric, field in the dielectric is 3 ´ 104 V m the charge density of, the positive plate will be close to:, [2014], (a) 6 ´ 10-7 C m 2, (c) 3 ´ 104 C m 2, , (b) 3 ´10-7 C m2, (d) 6 ´104 C m 2, , Q d, 2wl 2 e o, Q 2d, , K, , (b), , ( K - 1), , (d), , Q2 w, 2dl 2 e0, Q2w, , ( K - 1), , K, 2wl 2 e o, 2dl 2 e o, 90. Three capacitors, each of 3 mF, are provided. These cannot, be combined to provide the resultant capacitance of:, [Online April 9, 2014], (a) 1 mF, (b) 2 mF, (c) 4.5 mF, (d) 6 mF, 91. A parallel plate capacitor having a separation between the, plates d, plate area A and material with dielectric constant, K has capacitance C0. Now one-third of the material is, replaced by another material with dielectric constant 2K,, so that effectively there are two capacitors one with area, , (c)
Page 273 :
P-264, , Physics, , 1, 2, A, dielectric constant 2K and another with area, A, 3, 3, and dielectric constant K. If the capacitance of this new, C, capacitor is C then, is, [Online April 25, 2013], C0, 4, 2, 1, (c), (d), 3, 3, 3, 92. To establish an instantaneous current of 2 A through a 1, mF capacitor ; the potential difference across the capacitor, plates should be changed at the rate of :, [Online April 22, 2013], (a) 2 × 104 V/s, (b) 4 × 106 V/s, (c) 2 × 106 V/s, (d) 4 × 104 V/s, ur, 93. A uniform electric field E exists between the plates of a, charged condenser. A charged particle enters the space, ur, between the plates and perpendicular to E . The path of, the particle between the plates is a :, [Online April 9, 2013], (a) straight line, (b) hyperbola, (c) parabola, (d) circle, 94. The figure shows an experimental plot discharging of a, capacitor in an RC circuit. The time constant t of this circuit, lies between :, [2012], , (b), , Potential difference, V in volts, , (a) 1, , 25, 20, 15, 10, 5, 0, , 50 100 150, , 200, , 250, , Time in seconds, , 300, , (a) 150 sec and 200 sec (b) 0 sec and 50 sec, (c) 50 sec and 100 sec, (d) 100 sec and 150 sec, 95. The capacitor of an oscillatory circuit is enclosed in a, container. When the container is evacuated, the resonance, frequency of the circuit is 10 kHz. When the container is, filled with a gas, the resonance frequency changes by 50, Hz. The dielectric constant of the gas is, [Online May 26, 2012], (a) 1.001, (b) 2.001 (c) 1.01, (d) 3.01, 96. Statement 1: It is not possible to make a sphere of, capacity 1 farad using a conducting material., Statement 2: It is possible for earth as its radius is, 6.4 × 106 m., [Online May 26, 2012], (a) Statement 1 is true, Statement 2 is true, Statement 2 is, the correct explanation of Statement 1., (b) Statement 1 is false, Statement 2 is true., (c) Statement 1 is true, Statement 2 is true, Statement 2 is, not the correct explanation of Statement 1., (d) Statement 1 is true, Statement 2 is false., , 97. A series combination of n1 capacitors, each of capacity, C1 is charged by source of potential difference 4 V. When, another parallel combination of n2 capacitors each of, capacity C2 is charged by a source of potential difference, V, it has the same total energy stored in it as the first, combination has. The value of C2 in terms of C1 is then, [Online May 12, 2012], n2, 2 C1, (a) 16 n C1, (b) n n, 1, 1 2, n2, 16 C1, (c) 2 n C1, (d) n n, 1, 1 2, 98. Two circuits (a) and (b) have charged capacitors of, capacitance C, 2C and 3C with open switches. Charges on, each of the capacitor are as shown in the figures. On closing, the switches, [Online May 7, 2012], S, S, Q, C, , 2Q, 3C, L, , 2Q, 2C, , Q, 2C, , L, , R, Circuit (a), , R, Circuit (b), , (a) No charge flows in (a) but charge flows from R to L in (b), (b) Charges flow from L to R in both (a) and (b), (c) Charges flow from R to L in (a) and from L to R in (b), (d) No charge flows in (a) but charge flows from L to R in (b), 99. Let C be the capacitance of a capacitor discharging through, a resistor R. Suppose t1 is the time taken for the energy, stored in the capacitor to reduce to half its initial value and, t2 is the time taken for the charge to reduce to one-fourth, its initial value. Then the ratio t1/ t2 will be, [2010], (a) 1, , (b), , 1, 2, , (c), , 1, 4, , (d) 2, , 100. A parallel plate capacitor with air between the plates has, capacitance of 9 pF. The separation between its plates is, ‘d’. The space between the plates is now filled with two, dielectrics. One of the dielectrics has dielectric constant, d, k1 = 3 and thickness, while the other one has dielectric, 3, 2d, . Capacitance of the, constant k2 = 6 and thickness, 3, capacitor is now, [2008], (a) 1.8 pF, (b) 45 pF (c) 40.5 pF (d) 20.25 pF, 101. A parallel plate condenser with a dielectric of dielectric, constant K between the plates has a capacity C and is, charged to a potential V volt. The dielectric slab is slowly, removed from between the plates and then reinserted. The, net work done by the system in this process is [2007]
Page 274 :
P-265, , Electrostatic Potential and Capacitance, , 1, ( K - 1) CV 2, 2, , (a) zero, , (b), , 2, (c) CV ( K - 1), K, , (d) ( K - 1) CV 2, , 102. A parallel plate capacitor is made by stacking n equally, spaced plates connected alternatively. If the capacitance, between any two adjacent plates is ‘C’ then the resultant, capacitance is, [2005], (a) (n + 1) C, (b) (n – 1) C, (c) nC, , (d) C, , 103. A fully charged capacitor has a capacitance ‘C’. It, is discharged through a small coil of resistance wire, embedded in a thermally insulated block of specific heat, capacity ‘s’ and mass ‘m’. If the temperature of the block is, raised by ‘DT’, the potential difference ‘V’ across the, capacitance is, [2005], (a), (c), , mC DT, s, , 2ms DT, C, , (b), (d), , 2mC DT, s, ms DT, C, , 104. A sheet of aluminium foil of negligible thickness is, introduced between the plates of a capacitor. The, capacitance of the capacitor, [2003], (a) decreases, (b) remains unchanged, (c) becomes infinite, (d) increases, -18, 105. The work done in placing a charge of 8 ´ 10 coulomb, on a condenser of capacity 100 micro-farad is [2003], , (a) 16 ´ 10 -32 joule, , (b) 3.1´10 -26 joule, , (c) 4 ´10-10 joule, , (d) 32 ´ 10-32 joule, , 106. If there are n capacitors in parallel connected to V volt, source, then the energy stored is equal to, [2002], (a) CV, , (b), , 1, nCV2 (c) CV2, 2, , (d), , 1, CV2, 2n, , 107. Capacitance (in F) of a spherical conductor with radius 1 m, is, [2002], (a) 1.1´ 10 -10, , (b) 10 -6, , (c) 9 ´ 10 -9, , (d) 10 -3
Page 278 :
P-269, , Electrostatic Potential and Capacitance, , 21. (c), , By Gauss's theorem, , O, , E=, , q, , Tcosq, q, C Tsinq q, x, Fe, , q, , Charge density, r =, , In equilibrium, Fe = T sin q, mg = T cos q, , Y, , Fe, q2, =, mg 4p Î0 x 2 ´ mg, , \ x=, , 26. (c), , q2, 4p Î0 tan q mg, , kq, kq, V=, +, = 4 kmg / tan q, x/2 x/2, 22. (c) q = 1µC = 1 × 10–6C, r = 4 cm = 4 ×10–2 m, , kq 9 ´ 109 ´10 -6, =, Potential V =, = 2.25 × 105 V., r, 4 ´ 10-2, , =, , 9 ´109 ´1´10-6, 16 ´10-4, , \, , r pr 2, , So, q1 =, , =, , 4pR 2, , Qr 2, R2 + r 2, , and q2 =, , Now, potential, V =, =, =, , QR 2, R2 + r 2, , 1 é q1 q 2 ù, +, ê, ú, 4 pe 0 ë r, R û, , 1 é Qr, QR ù, +, 4pe 0 êë R 2 + r 2 R 2 + r 2 úû, , Q( R + r ), 2, , 2, , 1, 4pe0, , A(Ö2,Ö2), , X, , (0,0) ®, rB B (2,0), , rB =, , (, , ), , 2, 2 from the origin,, 4 = 2 units., , (2) 2 + (0) 2 = 2 units., , Now, potential at A, due to charge q = 10 –3 mC, Q, 1, ×, 4 p Î0 ( rA ), , Potential at B, due to charge Q = 10–3 QC VB =, , Q, 1, ×, 4 p Î0 ( rB ), , \ Potential difference between the points A and B is given, by, , VA – VB =, , 1 10 –3, 1 10 –3, ×, ×, 4p Î0 rA, 4p Î0 rB, , 10–3 æ 1 1 ö 10 –3 æ 1 1 ö, ç - ÷=, 4p Î0 è rA rB ø 4 p Î0 çè 2 2 ÷ø, , Q, ´ 0 = 0., 4p Î0, 27. (a) As shown in the figure, the resultant electric fields, before and after interchanging the charges will have the, same magnitude, but opposite directions., As potential is a scalar quantity, So the potential will be, same in both cases., q, q, A, B, =, , R +r, 24. (c) Charges reside only on the outer surface of a, conductor with cavity., 25. (c) Electric field, df, E== – 2ar, dr, , = – 6e0a, , The distance of point B(2, 0) from the origin,, , =, , q2, , 4pr 2dr, , rA = ( 2)2 + ( 2 ) 2 =, , r2, , = –5.625 × 106 V/m, , dq, , The distance of point A, , VA =, , kq, , 23. (c) Let q1 and q2 be charge on two spheres of radius, 'r' and 'R' respectively, As, q1 + q2 = Q, and s1 = s2 [Surface charge density are equal], q1, , ®, rA, O, , Electric potential at the centre of the line, , Induced electric field E = –, , ....(ii), , From (i) and (ii),, Q = –8 pe0ar3, Þdq = – 24pe0ar2 dr, , mg, , tan q =, , 1 q, 4 pe 0 r 2, , ®, E, , ....(i), , D, -q, , C, -q
Page 289 :
P-280, , Physics, , 103. (c) Applying conservation of energy,, Electric potential energy of capacitor = heat absorbed, 100. (c), , The capacitance with air between the plates, e0 A, = 9pF, d, On introducing two dielectric between the plates, the given, capacitance is equal to two capacitances connected in, series where, k Î A 3Î A, C1 = 1 0 = 0, d /3, d /3, 3 ´ 3Î0 A, 9 Î0 A, =, =, d, d, and, 3k2Î0 A, kÎ A, C2 = 2 0 =, 2d / 3, 2d, 3 ´6 Î0 A, 9 Î0 A, =, =, 2d, d, The equivalent capacitance Ceq is, C=, , 1, 1, 1, =, +, Ceq C1 C2, =, , d, d, 2d, +, =, 9 Î0 A 9 Î0 A 9 Î0 A, , 9 Î0 A 9, = ´ 9 pF = 40.5 pF, 2 d, 2, 101. (a) The potential energy of a charged capacitor is given, \ Ceq =, , (, , ), , -18 2, , 8 ´ 10, = 32 × 10–32 J, \ U = 1´, 2 100 ´ 10 -6, 106. (b) In parallel, equivalent capacitance of n capacitor of, capacitance C, C¢ = nC, Energy stored in this capacitor, 1 1 2, E= C V, 2, 1, 1, 2, 2, Þ E = (nC )V = nCV, 2, 2, C, C, n times, , C, , 2, , Q, ., 2C, When a dielectric slab is introduced between the plates, , by U =, , 2m. s. Dt, 1, CV 2 = m. s Dt ; V =, C, 2, 104. (b) The capacitance without aluminium foil is, e A, C= 0, d, Here, d is distance between the plates of a capacitor, A = Area of plates of capacitor, When an aluminium foil of thickness t is introduced, between the plates., e A, Capacitance, C¢ = 0, d –t, If thickness of foil is negligible 50 d – t ~ d. Hence, C = C¢., 105. (d) The work done is stored in the form of potential energy, which is given by, 1 Q2, U=, 2 C, , the energy is given by, , Q2, , where K is the dielectric, 2KC, , constant., Again, when the dielectric slab is removed slowly its, energy increases to initial potential energy. Thus, work, done is zero., 102. (b) As n plates are joined alternately positive plate of all, (n – 1) capacitor are connected to one point and negative, plate of all (n – 1) capacitors are connected to other point., It means (n – 1) capacitors joined in parallel., \ Resultant capacitance = (n – 1)C, , V, V, Alternatively, Each capacitor has a potential difference of V between the, plates., So, energy stored in each capacitor, 1, CV 2 ., 2, \ Energy stored in n capacitor, , =, , é1, 2ù, = ê CV ú ´ n, ë2, û, 107. (a) Capacitance of spherical conductor = 4pE0R, Here, R is radius of conductor, 1, ´ 1 = 1.1 ´ 10-10 F, \ C = 4p Î0 R =, 9 ´109
Page 290 :
17, , P-281, , Current Electricity, , Current Electricity, Electric Current, Drift of, TOPIC 1 Electrons, Ohm's Law,, Resistance and Resistivity, 1., , 2., , A circuit to verify Ohm’s law uses ammeter and voltmeter, in series or parallel connected correctly to the resistor., In the circuit :, [Sep. 06, 2020 (II)], (a) ammeter is always used in parallel and voltmeter is, series, (b) Both ammeter and voltmeter must be connected in, parallel, (c) ammeter is always connected in series and voltmeter, in parallel, (d) Both, ammeter and voltmeter must be connected in, series, Consider four conducting materials copper, tungsten,, mercury and aluminium with resistivity rC, rT, rM and rA, respectively. Then :, [Sep. 02, 2020 (I)], (a) rC > r A > rT, (b) r M > r A > rC, (c), , 3., , r A > rT > rC, , 4., , 5., , (d) r A > rM > rC, , (a) The emf of the battery is 1.5 V and its internal, resistance is 1.5 W, (b) The value of the resistance R is 1.5 W, (c) The potential difference across the battery is 1.5 V when, it sends a current of 1000 mA, (d) The emf of the battery is 1.5 V and the value of R is 1.5 W, A current of 5 A passes through a copper conductor, (resistivity) = 1.7×10 – 8Wm) of radius of cross-section, 5 mm. Find the mobility of the charges if their drift, velocity is 1.1×10 – 3 m/s., [10 Apr. 2019 I], (a) 1.8m2/Vs, (b) 1.5 m2/Vs, (c) 1.3 m2/Vs, (d), 1.0m2/Vs, In an experiment, the resistance of a material is plotted as, a function of temperature (in some range). As shown in, the figure, it is a straight line., [10 Apr. 2019 I], , One may canclude that:, , To verify Ohm’s law, a student connects the voltmeter, across the battery as, shown in the figure. The measured, voltage is plotted as a function of the current, and the, following graph is obtained :, [12 Apr. 2019 I], , (a), , 6., , If Vo is almost zero, identify the correct statement:, , R0, , 2, , (b) R(T) = R 0 e - T0 /T, , T2, - T 2 /T02, , 2, , 2, , 2, , (d) R(T) = R 0 eT /T0, Space between two concentric conducting spheres of radii, a and b (b > a) is filled with a medium of resistivity r. The, resistance between the two spheres will be :, [10 Apr. 2019 II], (c), , R(T) = R 0 e, , (a), , r æ1 1ö, ç - ÷, 4p è a b ø, , (b), , r æ1 1ö, ç - ÷, 2p è a b ø, , r æ1 1ö, r æ1 1ö, (d), ç + ÷, ç + ÷, 4p è a b ø, 2p è a b ø, In a conductor, if the number of conduction electrons, per unit volume is 8.5 × 1028 m –3 and mean free time is, 25 fs (femto second), it’s approximate resistivity is:, (m e = 9.1 × 10–31 kg), [9 Apr. 2019 II], (a) 10–6 W m, (b) 10–7 Wm, (c) 10–8 Wm, (d) 10–5 Wm, , (c), , 7., , R(T) =
Page 291 :
P-282, , Physics, , A 200 W resistor has a certain color code. If one replaces, the red color by green in the code, the new resistance will, be :, [8 April 2019 I], (a) 100 W (b) 400 W (c) 300 W, (d) 500 W, 9. The charge on a capacitor plate in a circuit, as a function of, time, is shown in the figure:, [12 Jan. 2019 II], 6, 5, 4, q(µc) 3, 2, 0, 6, 8, 2, 4, t(s), What is the value of current at t = 4 s ?, (a) Zero, (b) 3 µA, (c) 2 µA, (d) 1.5 µA, 10. A resistance is shown in the figure. Its value and tolerance, are given respectively by:, [9 Jan. 2019 I], 8., , 16. A uniform wire of length l and radius r has a resistance of, r, 100 W. It is recast into a wire of radius . The resistance, 2, of new wire will be :, [Online April 9, 2017], (a) 1600 W (b) 400 W (c) 200 W, (d) 100 W, 17. When 5V potential difference is applied across a wire of, length 0.1 m, the drift speed of electrons is 2.5 × 10–4 ms–1., If the electron density in the wire is 8 × 1028 m–3, the, resistivity of the material is close to :, [2015], (a) 1.6 × 10–6 Wm, (b) 1.6 × 10–5 Wm, (c) 1.6 × 10–8 Wm (d), 1.6 × 10–7 Wm, 18. Suppose the drift velocity nd in a material varied with the, applied electric field E as nd µ E . Then V – I graph for, a wire made of such a material is best given by :, [Online April 10, 2015], V, , V, , (a), , (b), I, , (a) 270 W, 10%, (b) 27 kW, 10%, (c) 27 kW, 20%, (d) 270 W, 5%, 11. Drift speed of electrons, when 1.5 A of current flows in a, copper wire of cross section 5 mm2, is v. If the electron, density in copper is 9 × 1028/m3 the value of v in mm/s, close to (Take charge of electron to be = 1.6 × 10–19C), [9 Jan. 2019 I], (a) 0.02, (b) 3, (c) 2, (d), 0.2, 12. A copper wire is stretched to make it 0.5% longer. The, percentage change in its electrical resistance if its volume, remains unchanged is:, [9 Jan. 2019 I], (a) 2.0%, (b) 2.5%, (c) 1.0%, (d), 0.5%, 13. A carbon resistance has following colour code. What is, the value of the resistance?, [9 Jan. 2019 II], , GOY Golden, (a) 530 kW ± 5% (b), 5.3 kW ± 5%, (c) 6.4 MW ± 5%(d), 64 MW ± 10%, 14. A heating element has a resistance of 100W at room, temperature. When it is connected to a supply of 220 V,, a steady current of 2 A passes in it and temperature is, 500°C more than room temperature. What is the, temperature coefficient of resistance of the heating, element?, [Online April 16, 2018], (a) 1 × 10–4°C–1, (b) 5 × 10–4°C–1, (c) 2 × 10–4°C–1, (d) 0.5 × 10–4°C–1, 15. A copper rod of cross-sectional area A carries a uniform, current I through it. At temperature T, if the volume charge, density of the rod is r, how long will the charges take to, travel a distance d ?, [Online April 15, 2018], (a), , 2rdA, IT, , (b), , 2rdA, I, , (c), , rdA, I, , (d), , rdA, IT, , I, , V, , V, , (c), , (d), I, , I, , 19. Correct set up to verify Ohm’s law is :, [Online April 23, 2013], A, V, , (a), , (b), , A, , V, , A, , (c), V, , V, , (d), A
Page 292 :
P-283, , Current Electricity, , 20. The resistance of a wire is R. It is bent at the middle by 180°, and both the ends are twisted together to make a shorter wire., The resistance of the new wire is [Online May 26, 2012], (a) 2 R, (b) R/2, (c) R/4, (d) R/8, 21. If a wire is stretched to make it 0.1% longer, its resistance, will :, [2011], (a) increase by 0.2%, (b) decrease by 0.2%, (c) decrease by 0.05%, (d) increase by 0.05%, DIRECTIONS : Question No. 22 and 23 are based on the, following paragraph., Consider a block of conducting material of resistivity ‘r’ shown in, the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply, superposition principle to find voltage ‘DV’ developed between ‘B’, and ‘C’. The calculation is done in the following steps:, (i) Take current ‘I’ entering from ‘A’ and assume it to spread, over a hemispherical surface in the block., (ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s, law E = r j, where j is the current per unit area at ‘r’., (iii) From the ‘r’ dependence of E(r), obtain the potential V(r) at r., (iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and, superpose results for ‘A’ and ‘D’., I, , a, A, , I, , DV, , a, , b, B, , C, , 27. The length of a given cylindrical wire is increased by 100%., Due to the consequent decrease in diameter the change in, the resistance of the wire will be, [2003], (a) 200%, (b) 100%, (c) 50%, (d) 300%, , TOPIC 2 Combination of Resistances, 28. In the given circuit diagram, a wire is joining points B and, D. The current in this wire is:, [9 Jan. 2020 I], , (a) 0.4A, (b) 2A, (c) 4A, (d) zero, 29. The series combination of two batteries, both of the same, emf 10 V, but different internal resistance of 20 W and 5, W, is connected to the parallel combination of two, resistors 30 W and R W. The voltage difference across, the battery of internal resistance 20 W is zero, the value, of R (in W) is _________., [NA. 8 Jan. 2020 II], 30. The current I1 (in A) flowing through 1 W resistor in the, following circuit is:, [7 Jan. 2020 I], , D, , 22. DV measured between B and C is, [2008], rI, rI, rI, rI, –, –, (a), (b), pa p(a + b), a (a + b), rI, rI, rI, –, (c), (d), 2pa 2p(a + b), 2p(a - b), 23. For current entering at A, the electric field at a distance ‘r’, from A is, [2008], rI, rI, rI, rI, (a), (b), (c), (d), 2, 2, 2, 8pr, r, 2pr, 4pr 2, 24. The resistance of a wire is 5 ohm at 50°C and 6 ohm at, 100°C. The resistance of the wire at 0°C will be, [2007], (a) 3 ohm (b) 2 ohm (c) 1 ohm, (d) 4 ohm, 25. A material 'B' has twice the specific resistance of 'A'. A, circular wire made of 'B' has twice the diameter of a wire, made of 'A'. then for the two wires to have the same, resistance, the ratio lB/lA of their respective lengths must, be, [2006], 1, 1, (a) 1, (b), (c), (d) 2, 2, 4, 26. An electric current is passed through a circuit containing, two wires of the same material, connected in parallel. If the, 4, 2, lengths and radii are in the ratio of and , then the ratio, 3, 3, of the current passing through the wires will be [2004], (a) 8/9, (b) 1/3, (c) 3, (d) 2, , (a) 0.4, (b) 0.5, (c) 0.2, (d) 0.25, 31. A wire of resistance R is bent to form a square ABCD as, shown in the figure. The effective resistance between E, and C is: (E is mid-point of arm CD) [9 April 2019 I], A, B, , D, , (a) R, , (b), , C, , E, 7, R, 64, , (c), , 3, R, 4, , (d), , 1, R, 16, , 32. A metal wire of resistance 3 W is elongated to make a uniform, wire of double its previous length. This new wire is now, bent and the ends joined to make a circle. If two points on, the circle make an angle 60° at the centre, the equivalent, resistance between these two points will be: [9 Apr. 2019 II], (a), , 12, W, 5, , (b), , 5, W, 2, , (c), , 5, W, 3, , (d), , 7, W, 2
Page 293 :
P-284, , Physics, , 33. In the figure shown, what is the current (in Ampere) drawn, from the battery? You are given :, [8 Apr. 2019 II], R1 = 15 W, R2 = 10 W, R3 = 20 W, R4 = 5 W, R5 = 25 W,, R6 = 30 W, E = 15 V, , A, , B, , 5R, 5R, (c), (d) 3 R, 2, 3, 38. In the given circuit diagram when the current reaches steady, state in the circuit, the charge on the capacitor of, capacitance C will be :, [2017], r2, (a) CE +, (r r2 ), r1, (b) CE, (r1 + r), (c) CE, r, (d) CE 1, (r2 + r), (a) 2 R, , (a) 13/24, (b) 7/18, (c) 9/32, (d) 20/3, 34. A uniform metallic wire has a resistance of 18 W and is, bent into an equilateral triangle. Then, the resistance, between any two vertices of the triangle is:, [10 Jan. 2019 I], (a) 4 W, (b) 8 W, (c) 12 W, (d) 2 W, 35. The actual value of resistance R, shown in the figure is 30, W. This is measured in an experiment as shown using the, V, standard formula R = , where V and I are the reading of, I, the voltmeter and ammeter, respectively. If the measured, value of R is 5% less, then the internal resistance of the, voltmeter is:, [10 Jan. 2019 II], V, A, , 39., , In the above circuit the current in each resistance is, [2017], (a) 0.5A, (b) 0 A, (c) 1 A, (d) 0.25 A, 1W, , A1, , 40., , 1W, , V, , R3, , R4, , R1, , A3, , 4W, 1W, , (a) 600 W (b) 570 W (c) 35 W, (d) 350 W, 36. In the given circuit the internal resistance of the 18 V cell, is negligible. If R1 = 400W, R3 = 100 W and R4 = 500 W, and the reading of an ideal voltmeter across R4 is 5 V,, then the value of R2 will be:, [9 Jan. 2019 II], , 1W, , A2, B, 9V, 0.5W, , R, , (b), , 4W, , 4W, 1W, , 1W, , A 9 V battery with internal resistance of 0.5 W is connected, across an infinite network as shown in the figure. All, ammeters A1, A2, A3 and voltmeter V are ideal., Choose correct statement., [Online April 8, 2017], (a) Reading of A1 is 2 A (b) Reading of A1 is 18 A, (c) Reading of V is 9 V, (d) Reading of V is 7 V, 41. Six equal resistances are connected between points P, Q, and R as shown in figure. Then net resistance will be, maximum between :, [Online April 25, 2013], P, , R2, , (a) 300 W, (c) 550 W, , 18 V, , r, r, , (b) 450 W, , r, , r, , (d) 230 W, , 37. In the given circuit all resistances are of value R ohm each., The equivalent resistance between A and B is :, [Online April 15, 2018], , ¥, , r, Q, , (a) P and R, (c) Q and R, , r, , R, , (b) P and Q, (d) Any two points
Page 294 :
P-285, , Current Electricity, , 42. A letter 'A' is constructed of a uniform wire with resistance, 1.0 W per cm, The sides of the letter are 20 cm and the cross, piece in the middle is 10 cm long. The apex angle is 60 . The, resistance between the ends of the legs is close to:, [Online April 9, 2013], (a) 50.0 W (b) 10 W, (c) 36.7 W (d) 26.7 W, 43. Two conductors have the same resistance at 0°C but their, temperature coefficients of resistance are a1 and a2. The, respective temperature coefficients of their series and, parallel combinations are nearly, [2010], a1 + a 2, a1 + a 2, , a1 + a 2, (a), (b) a1 + a 2 ,, 2, 2, a1a 2, a, +, a, a, +, 2 , 1 a2, (c) a1 + a 2 ,, (d) 1, a1 + a 2, 2, 2, 44. The current I drawn from the 5 volt source will be [2006], , 10W, 5W, , 10W, , I, , 20W, , 10W, +–, , 48., , 5 volt, , 5W, , 6W, , (a) 0.71 A from positive to negative terminal, (b) 0.42 A from positive to negative terminal, (c) 0.21 A from positive to negative terminal, (d) 0.36 A from negative to positive terminal, 49. In the circuit, given in the figure currents in different branches, and value of one resistor are shown. Then potential at point, B with respect to the point A is :, [Sep. 05, 2020 (II)], 2V, D, E, B, 1A, 2W, F, C, 2A, 2A, 1V, (a) + 2 V, (b) – 2 V, (c) – 1 V, (d) + 1 V, 50. The value of current i1 flowing from A to C in the circuit, diagram is :, [Sep. 04, 2020 (II)], , 8V, , 3W, , i, , (a) 4 A, (b) 2 A, (c) 1 A, (d) 6 A, 46. The resistance of the series combination of two resistances, is S. when they are joined in parallel the total resistance is, P. If S = nP then the minimum possible value of n is, [2004], (a) 2, (b) 3, (c) 4, (d) 1, 47. A 3 volt battery with negligible internal resistance is, connected in a circuit as shown in the figure. The current, I, in the circuit will be, [2003], , 3W, , 3W, , B, 2W, , 2W, , 5W, 4W, , 4W i1, A, , C, , 2W, , (a) 2 A, 51., , i, , 2W, , (b) 4 A, B, , D, , (c) 1 A, , (d) 5 A, , 60W, , 40W, A, , C, , 90W, , 110W, D, , 3W, , (b) 1.5 A, , 10V, , 2W, 4W, In the figure shown, the current in the 10 V battery is, close to :, [Sep. 06, 2020 (II)], , 1.5W, , (a) 1 A, , 10W, , 20V, , 2W, , 3V, , Kirchhoff 's Laws, Cells,, Thermo e.m.f. & Electrolysis, , A, , (a) 0.33 A (b) 0.5 A, (c) 0.67 A, (d) 0.17 A, 45. The total current supplied to the circuit by the battery is, [2004], 6V, , TOPIC 3, , (c) 2 A, , (d) 1/3 A, , 40 V, Four resistances 40 W, 60 W, 90 W and 110 W make the, arms of a quadrilateral ABCD. Across AC is a battery of, emf 40 V and internal resistance negligible. The potential, difference across BD in V is __________., [NA. Sep. 04, 2020 (II)]
Page 295 :
P-286, , Physics, , 52. An ideal cell of emf 10 V is connected in circuit shown in, figure. Each resistance is 2 W. The potential difference (in, V) across the capacitor when it is fully charged is, __________., [Sep. 02, 2020 (II)], R1, R3, , R2, , C, R5, , 57. In the circuit shown, the potential difference between A, and B is :, [11 Jan. 2019 II], 1V, , 1W, M, 5W, A, , 1W, , 2V, , 1W, , 3V, , D, , C, , B, , N, , R4, 10 V, 53. In the given circuit, an ideal voltmeter connected across, the 10 W resistance reads 2V. The internal resistance r, of, each cell is :, [10 Apr. 2019 I], , (a) 1 W, (b) 0.5 W, (c) 1.5 W, (d) 0 W, 54. For the circuit shown, with R1 = 1.0 W, R2 = 2.0 W, E1 = 2 V, and E2 = E3 = 4 V, the potential difference between the, points ‘a’ and ‘b’ is approximately (in V) : [8 April 2019 I], , (a) 2.7, (b) 2.3, (c) 3.7, (d), 3.3, 55. A cell of internal resistance r drives current through an, external resistance R. The power delivered by the cell to, the external resistance will be maximum when :, [8 Apr. 2019 II], (a) R = 0.001 r, (b) R = 1000 r, (c) R = 2r, (d) R = r, 56. In the given circuit diagram, the currents, I1 = – 0.3 A, I4 = 0.8, A and I5 = 0.4 A, are flowing as shown. The currents I 2, I3, and I6, respectively, are :, [12 Jan. 2019 II], I, Q, P, 6, I3, , (a) 1 V, (b) 2 V, (c) 3 V, (d) 6 V, 58. In the given circuit the cells have zero internal resistance., The currents (in Amperes) passing through resistance, R1 and R2 respectively, are:, [10 Jan. 2019 I], , (a) 1, 2, (b) 2, 2, (c) 0.5, 0, (d) 0, 1, 59. When the switch S, in the circuit shown, is closed then, the valued of current i will be:, [9 Jan. 2019 I], , (a) 3A, (b) 5A, (c) 4A, (d), 2A, 60. Two batteries with e.m.f. 12 V and 13 V are connected in, parallel across a load resistor of 10W. The internal, resistances of the two batteries are 1W and 2W respectively., The voltage across the load lies between:, [2018], (a) 11.6 V and 11.7 V, (b) 11.5 V and 11.6 V, (c) 11.4 V and 11.5 V, , (d) 11.7 V and 11.8 V, , 61. In the circuit shown, the current in the 1W resistor is:, [2015], , 6V, , P 2W, 1W, , I5, , 10W, , I2, , I1, , S, I4, (a) 1.1 A, – 0.4 A, 0.4 A, , R, (b) 1.1 A, 0.4 A, 0.4 A, , (c) 0.4 A, 1.1 A, 0.4 A, , (d) –0.4 A, 0.4 A, 1.1 A, , 3W, (a) 0.13 A, from Q to P, (c) 1.3A from P to Q, , 9V, , W 3W, (b) 0.13 A, from P to Q, (d) 0A
Page 296 :
P-287, , Current Electricity, , 62. In the electric network shown, when no current flows, through the 4W resistor in the arm EB, the potential, difference between the points A and D will be :, [Online April 11, 2015], , 2W, , F, , D, , E, , r = 0.5 W, , E1, , E2, , 2V, 4W, , 2W, , R, R, , 4V, A, , B, , 9V, , 3V, , C, , (a) 6 V, (b) 3 V, (c) 5 V, (d) 4 V, 63. The circuit shown here has two batteries of 8.0 V and 16.0, V and three resistors 3 W, 9 W and 9 W and a capacitor of, 5.0 mF., [Online April 12, 2014], I, , 3W, , 5 mF, , P2, , 5V, 2W, , 9W, 9W, , 8.0 V, , (a) 5.5 W, (b) 3.5 W, (c) 4.5 W, (d) 2.5 W, 67. A 5V battery with internal resistance 2W and a 2V battery, with internal resistance 1W are connected to a 10W resistor, as shown in the figure., [2008], , 68., , 5W, , 69., 20 W 10 W, , 2V, 1W, , 16.0 V, , I2, I1, How much is the current I in the circuit in steady state?, (a) 1.6 A, (b) 0.67 A, (c) 2.5 A, (d) 0.25 A, 64. In the circuit shown, current (in A) through 50 V and 30 V, batteries are, respectively., [Online April 11, 2014], , 50 V, , 10W, , 30 V, , 5W, , (a) 2.5 and 3, (b) 3.5 and 2, (c) 4.5 and 1, (d) 3 and 2.5, 65. A d.c. main supply of e.m.f. 220 V is connected across a, storage battery of e.m.f. 200 V through a resistance of 1W., The battery terminals are connected to an external resistance, ‘R’. The minimum value of ‘R’, so that a current passes, through the battery to charge it is: [Online April 9, 2014], (a) 7 W, (b) 9 W, (c) 11 W, (d) Zero, 66. A dc source of emf E1 = 100 V and internal resistance r = 0.5, W, a storage battery of emf E2 = 90 V and an external, resistance R are connected as shown in figure. For what, value of R no current will pass through the battery ?, [Online April 22, 2013], , 70., , 71., , The current in the 10W resistor is, (a) 0.27 A P2 to P1, (b) 0.03 A P1 to P2, (c) 0.03 A P2 to P1, (d) 0.27 A P1 to P2, A battery is used to charge a parallel plate capacitor till the, potential difference between the plates becomes equal to, the electromotive force of the battery. The ratio of the, energy stored in the capacitor and the work done by the, battery will be, [2007], (a) 1/2, (b) 1, (c) 2, (d) 1/4, The Kirchhoff's first law (Si = 0) and second law (SiR = SE),, where the symbols have their usual meanings, are, respectively based on, [2006], (a) conservation of charge, conservation of momentum, (b) conservation of energy, conservation of charge, (c) conservation of momentum, conservation of charge, (d) conservation of charge, conservatrion of energy, A thermocouple is made from two metals, Antimony and, Bismuth. If one junction of the couple is kept hot and the, other is kept cold, then, an electric current will, [2006], (a) flow from Antimony to Bismuth at the hot junction, (b) flow from Bismuth to Antimony at the cold junction, (c) now flow through the thermocouple, (d) flow from Antimony to Bismuth at the cold junction, Two sources of equal emf are connected to an external, resistance R. The internal resistance of the two sources are, R1and R2 (R1 > R1). If the potential difference across the, source having internal resistance R2 is zero, then, [2005]
Page 297 :
P-288, , (a) R = R2 - R1, (b) R = R2 ´ ( R1 + R2 ) /( R2 - R1 ), (c) R = R1R2 /( R2 - R1 ), (d) R = R1R2 /( R1 - R2 ), 72. Two voltameters, one of copper and another of silver, are, joined in parallel. When a total charge q flows through the, voltameters, equal amount of metals are deposited. If the, electrochemical equivalents of copper and silver are Z1, and Z2 respectively the charge which flows through the, silver voltameter is, [2005], Z, q, q, Z2, (a), (d) q 1, (c) q, `(b), Z, Z, Z2, Z1, 1+ 1, 1+ 2, Z2, Z1, 73. An energy source will supply a constant current into the, load if its internal resistance is, [2005], (a) very large as compared to the load resistance, (b) equal to the resistance of the load, (c) non-zero but less than the resistance of the load, (d) zero, 74. The thermo emf of a thermocouple varies with the, temperature q of the hot junction as E = aq + bq2 in volts, where the ratio a/b is 700°C. If the cold junction is kept at, 0°C, then the neutral temperature is, [2004], (a) 1400°C, (b) 350°C, (c) 700°C, (d) No neutral temperature is possible for this termocouple., 75. The electrochemical equivalent of a metal is 3.35 × 10–7 kg, per Coulomb. The mass of the metal liberated at the cathode, when a 3A current is passed for 2 seconds will be, [2004], (a) 6.6×1057kg, (b) 9.9×10–7 kg, (c) 19.8×10–7 kg, (d) 1.1×10–7 kg, 76. The thermo e.m.f. of a thermo-couple is 25 mV/°C at room, temperature. A galvanometer of 40 ohm resistance, capable, of detecting current as low as 10–5 A, is connected with, the thermo couple. The smallest temperature difference that, can be detected by this system is [2003], (a) 16°C, (b) 12°C, (c) 8°C, (d) 20°C, 77. The negative Zn pole of a Daniell cell, sending a constant, current through a circuit, decreases in mass by 0.13g in 30, minutes. If the electeochemical equivalent of Zn and Cu, are 32.5 and 31.5 respectively, the increase in the mass of, the positive Cu pole in this time is, [2003], (a) 0.180 g (b) 0.141g (c) 0.126 g (d) 0.242 g, 78. The mass of product liberated on anode in an, electrochemical cell depends on, [2002], (a) (It)1/2, (b) It, (c) I/t, (d) I2 t, (where t is the time period for which the current is passed)., , Physics, , TOPIC 4 Heating Effect of Current, 79. An electrical power line, having a total resistance of 2 W,, delivers 1 kW at 220 V. The efficiency of the transmission, line is approximately :, [Sep. 05, 2020 (I)], (a) 72%, (b) 91%, (c) 85%, (d) 96%, 80. Model a torch battery of length l to be made up of a thin, cylindrical bar of radius ‘a’ and a concentric thin cylindrical, shell of radius ‘b’ filled in between with an electrolyte of resistivity, r (see figure). If the battery is connected to a resistance of, value R, the maximum Joule heating in R will take place for :, [Sep. 03, 2020 (I)], , r, , l, , a, b, , r, æbö, ln ç ÷, 2pl è a ø, 2r æ b ö, r, b, (c) R = l n æç ö÷, (d) R =, ln ç ÷, pl è a ø, pl è a ø, 81. In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W,, 15 small fans of 10 W and 2 heaters of 1 kW. The voltage, of electric main is 220 V. The minimum fuse capacity (rated, value) of the building will be:, [7 Jan. 2020 II], (a) 10 A, (b) 25 A, (c) 15 A, (d) 20 A, 82. The resistive network shown below is connected to a D.C., source of 16 V. The power consumed by the network is 4, Watt. The value of R is :, [12 Apr. 2019 I], (a), , r æbö, R=, ç ÷, 2pl è a ø, , (b) R =, , (a) 6W, (b) 8W, (c) 1W, (d) 16W, o, 83. One kg of water, at 20 C, is heated in an electric kettle, whose heating element has a mean (temperature averaged), resistance of 20 W. The rms voltage in the mains is 200 V., Ignoring heat loss from the kettle, time taken for water to, evaporate fully, is close to :, [Specific heat of water = 4200 J/(kgoC), Latent heat of water, = 2260 kJ/kg], [12 Apr. 2019 II], (a) 16 minutes, (b) 22 minutes, (c) 3 minutes, (d) 3 minutes
Page 298 :
Current Electricity, , 84. Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V),, are connected in series across a 220 V voltage source. If, the 25 W and 100 W bulbs draw powers P1 and P2, respectively, then:, [12 Jan. 2019 I], (a) P1=16 W, P2=4 W, (b) P1=16 W, P2=9 W, (c) P1=9 W, P2=16 W, (d) P1=4 W, P2=16 W, 85. Two equal resistances when connected in series to a battery,, consume electric power of 60 W. If these resistance are now, connected in parallel combination to the same battery, the, electric power consumed will be :, [11 Jan. 2019 I], (a) 60 W, (b) 240 W (c) 120 W, (d) 30 W, 86. A 2 W carbon resistor is color coded with green, black,, red and brown respectively. The maximum current which, can be passed through this resistor is: [10 Jan. 2019 I], (a) 20 mA (b) 100 mA (c) 0.4 mA (d) 63 mA, 87. A current of 2 mA was passed through an unknown resistor, which dissipated a power of 4.4 W. Dissipated power when, an ideal power supply of 11 V is connected across it is:, [10 Jan. 2019 II], (a) 11 × 10–5 W, (b) 11 × 10–3 W, –4, (c) 11 × 10 W, (d) 11 × 105 W, 88. A constant voltage is applied between two ends of a metallic, wire. If the length is halved and the radius of the wire is, doubled, the rate of heat developed in the wire will be:, [Online April 15, 2018], (a) Increased 8 times, (b) Doubled, (c) Halved, (d) Unchanged, 89. The figure shows three circuits I, II and III which are, connected to a 3V battery. If the powers dissipated by the, configurations I, II and III are P1, P2 and P3 respectively,, then :, [Online April 9, 2017], 1W, , 1W, , 1W, , 1W, 1W, , 1W, 1W, 1W, (I), , 3V, , 3V 1W, , 1W, 1W, , 1W, , 1W, , 1W, , 1W, , 1W, , (II), , 3V, , (a), , 5, 400ln J, 6, , (c) 300 J, , 2, (b) 200ln J, 3, 1.5, J, (d) 400 ln, 1.3, , 91., , R, , r, , In the circuit shown, the resistance r is a variable resistance., If for r = fR, the heat generation in r is maximum then the, value of f is :, [Online April 9, 2016], 1, 1, 3, (b) 1, (c), (d), 2, 4, 4, 92. In a large building, there are 15 bulbs of 40 W, 5 bulbs of, 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage, of electric mains is 220 V. The minimum capacity of the, main fuse of the building will be:, [2014], (a) 8 A, (b) 10 A, (c) 12 A, (d) 14 A, 93. Four bulbs B1, B2, B3 and B4 of 100 W each are connected, to 220 V main as shown in the figure., [Online April 19, 2014], , (a), , 220 V, , B1, , B2, , B3, , B4, , Ammeter, , The reading in an ideal ammeter will be:, (a) 0.45 A (b) 0.90 A (c) 1.35 A, (d) 1.80 A, 94. The supply voltage to room is 120V. The resistance of the, lead wires is 6W. A 60 W bulb is already switched on., What is the decrease of voltage across the bulb, when a, 240 W heater is switched on in parallel to the bulb?[2013], (a) zero, (b) 2.9 Volt, (c) 13.3 Volt, (d) 10.04Volt, 95. Which of the four resistances P, Q, R and S generate the, greatest amount of heat when a current flows from A to, B?, [Online April 23, 2013], P = 2W, , (III), , (a) P1 > P2 > P3, (b) P1 > P3 > P2, (c) P2 > P1 > P3, (d) P3 > P2 > P1, 90. The resistance of an electrical toaster has a temperature, dependence given by R(T) = R0 [1 + a(T – T0)] in its range, of operation. At T0 = 300K, R = 100 W and at T = 500 K, R =, 120 W. The toaster is connected to a voltage source at 200, V and its temperature is raised at a constant rate from 300, to 500 K in 30 s. The total work done in raising the, temperature is :, [Online April 10, 2016], , P-289, , R, , Q = 4W, , A, , B, R = 1W, , S = 2W, , (a) Q, (b) S, (c) P, (d) R, 96. Two electric bulbs rated 25W – 220 V and 100W – 220V are, connected in series to a 440 V supply. Which of the bulbs, will fuse?, [2012], (a) Both, (b) 100 W (c) 25 W, (d) Neither, 97. A 6.0 volt battery is connected to two light bulbs as shown in, figure. Light bulb 1 has resistance 3 ohm while light bulb 2, has resistance 6 ohm. Battery has negligible internal, resistance. Which bulb will glow brighter?, [Online May 19, 2012]
Page 299 :
P-290, , Physics, , Bulb 1, , Bulb 2, , +, , –, , 6.0 V, (a) Bulb 1 will glow more first and then its brightness will, become less than bulb 2, (b) Bulb 1, (c) Bulb 2, (d) Both glow equally, 98. Three resistors of 4 W, 6 W and 12 W are connected in, parallel and the combination is connected in series with a, 1.5 V battery of 1 W internal resistance. The rate of Joule, heating in the 4 W resistor is [Online May 12, 2012], (a) 0.55 W (b) 0.33 W (c) 0.25 W (d) 0.86 W, 99. This question has Statement 1 and Statement 2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., Statement 1: The possibility of an electric bulb fusing is, higher at the time of switching ON., Statement 2: Resistance of an electric bulb when it is not lit, up is much smaller than when it is lit up., [Online May 7, 2012], (a) Statement 1 is true, Statement 2 is false, (b) Statement 1 is false, Statement 2 is true, Statement, 2 is not a correct explanation of Statement 1., (c) Statement 1 is true, Statement 2 is true, Statement 2, is a correct explanation of Statement 1., (d) Statement 1 is false, Statement 2 is true., 100. The resistance of a bulb filmanet is 100W at a temperature, of 100°C. If its temperature coefficient of resistance be, 0.005 per °C, its resistance will become 200 W at a, temperature of, [2006], (a) 300°C, (b) 400°C (c) 500°C, (d) 200°C, 101. An electric bulb is rated 220 volt - 100 watt. The power, consumed by it when operated on 110 volt will be [2006], (a) 75 watt (b) 40 watt (c) 25 watt (d) 50 watt, 102. A heater coil is cut into two equal parts and only one part, is now used in the heater. The heat generated will now, be, [2005], (a) four times, (b) doubled, (c) halved, (d) one fourth, 103. The resistance of hot tungsten filament is about 10 times, the cold resistance. What will be the resistance of 100 W, and 200 V lamp when not in use ?, [2005], (a) 20 W, (b) 40 W, (d) 400W, (c) 200 W, 104. The thermistors are usually made of, [2004], (a) metal oxides with high temperature coefficient of, resistivity, (b) metals with high temperature coefficient of, resistivity, , (c) metals with low temperature coefficient of resistivity, (d) semiconducting materials having low temperature, coefficient of resistivity, 105. Time taken by a 836 W heater to heat one litre of water, from 10°C to 40°C is, [2004], (a) 150 s, (b) 100 s, (c) 50 s, (d) 200 s, 106. A 220 volt, 1000 watt bulb is connected across a 110 volt, mains supply. The power consumed will be, [2003], (a) 750 watt, (b) 500 watt, (c) 250 watt, (d) 1000 watt, 107. A wire when connected to 220 V mains supply has power, dissipation P1. Now the wire is cut into two equal pieces, which are connected in parallel to the same supply. Power, dissipation in this case is P2. Then P2 : P1 is, [2002], (a) 1, (b) 4, (c) 2, (d) 3, 108. If in the circuit, power dissipation is 150 W, then R is, [2002], R, , 2W, (a) 2W, , (b) 6W, , 15 V, (c) 5W, , (d) 4W, , Wheatstone Bridge and, TOPIC 5 Different Measuring, Instruments, 109. Two resistors 400W and 800W are connected in series across, a 6 V battery. The potential difference measured by a, voltmeter of 10 kW across 400W resistor is close to:, [Sep. 03, 2020 (II)], (a) 2 V, (b) 1.8 V, (c) 2.05 V, (d) 1.95 V, 110. Which of the following will NOT be observed when a, multimeter (operating in resistance measuring mode), probes connected across a component, are just reversed?, [Sep. 03, 2020 (II)], (a) Multimeter shows an equal deflection in both cases, i.e. before and after reversing the probes if the chosen, component is resistor., (b) Multimeter shows NO deflection in both cases i.e., before and after reversing the probes if the chosen, component is capacitor., (c) Multimeter shows a deflection, accompanied by a, splash of light out of connected and NO deflection, on reversing the probes if the chosen component is, LED., (d) Multimeter shows NO deflection in both cases i.e., before and after reversing the probes if the chosen, component is metal wire., 111. A potentiometer wire PQ of 1 m length is connected to a, standard cell E 1 . Another cell E 2 of emf 1.02 V is, connected with a resistance 'r' and switch S (as shown in, figure). With switch S open, the null position is obtained, at a distance of 49 cm from Q. The potential gradient in, the potentiometer wire is :, [Sep. 02, 2020 (II)]
Page 300 :
P-291, , Current Electricity, , E1, J, , P, r, , Q, G, , E2, , S, (a) 0.02 V/cm, (b), 0.01 V/cm, (c) 0.03 V/cm, (d), 0.04 V/cm, 112. In a meter bridge experiment S is a standard resistance. R, is a resistance wire. It is found that balancing length is, l = 25 cm. If R is replaced by a wire of half length and half, diameter that of R of same material, then the balancing, distance l¢ (in cm) will now be ____. [NA. 9 Jan. 2020 II], , 113. The length of a potentiometer wire is 1200 cm and it carries, a current of 60 mA. For a cell of emf 5 V and internal, resistance of 20 W, the null point on it is found to be at, 1000 cm. The resistance of whole wire is:, [8 Jan. 2020 I], (a) 80 W, (b) 120 W (c) 60 W, (d) 100 W, 114. Four resistances of 15 W, 12 W, 4 W and 10 W respectively, in cyclic order to form Wheatstone’s network. The, resistance that is to be connected in parallel with the, resistance of 10 W to balance the network is _____ W., [NA. 8 Jan. 2020 I], 115. The balancing length for a cell is 560 cm in a potentiometer, experiment. When an external resistance of 10 W is, connected in parallel to the cell, the balancing length, N, changes by 60 cm. If the internal resistance of the cell is, 10, W, where N is an integer then value of N is ________., [NA. 7 Jan. 2020 II], 116. In a meter bridge experiment, the circuit diagram and the, corresponding observation table are shown in figure., [10 Apr. 2019 I], , Sl.No., 1., 2., 3., , RW, 1000, 100, 10, , l (cm), 60, 13, 1.5, , 4., , 1, , 1.0, , Which of the reading is consistent ?, (a) 3, (b) 2, (c) 4, (d) 1, 117. In the circuit shown, a four-wire potentiometer is made of, a 400 cm long wire, which extends between A and B. The, resistance per unit length of the potentiometer wire is, r = 0.01 W/cm. If an ideal voltmeter is connected as shown, with jockey J at 50 cm from end A, the expected reading of, the voltmeter will be :, [8 Apr. 2019 II], , (a) 0.50 V (b) 0.75 V (c) 0.25 V, (d) 0.20 V, 118. In a meter bridge, the wire of length 1 m has a non-uniform, dR, cross-section such that, the variation, of its resistance, dl, dR, 1, R with length l is, . Two equal resistances are, µ, dl, l, connected as shown in the figure. The galvanometer has, zero deflection when the jockey is at point P. What is the, length AP?, [12 Jan. 2019 I], R', , R', , G, P, l, , 1 l, (a) 0.2 m, (b) 0.3 m, (c) 0.25 m, (d) 0.35 m, 119. An ideal battery of 4 V and resistance R are connected in, series in the primary circuit of a potentionmeter of length, 1 m and resistance 5W. The value of R, to give a potential, difference of 5 mV across 10 cm of potentiometer wire is:, [12 Jan. 2019 I], (a) 490W, (b) 480W, (c) 395W, (d) 495W, 120. The resistance of the meter bridge AB in given figure is 4W., With a cell of emf e= 0.5 V and rheostat resistance Rh = 2W, the null point is obtained at some point J. When the cell is, replaced by another one of emf e =e2 the same null point, J is found for Rh = 6W. The emf e2 is:, [11 Jan. 2019 I]
Page 301 :
P-292, , Physics, , e, , A, , B, , J, , Rh, 6V, (a) 0.4 V, (b) 0.3 V, (c) 0.6 V, (d) 0.5 V, 121. In a Wheatstone bridge (see fig.), Resistances P and Q are, approximately equal. When R = 400 W, the bridge is balanced., On interchanging P and Q, the value of R, for balance, is, 405W. The value of X is close to :, [11 Jan. 2019 I], B, Q, , P, A, , G, , K2, , 11, 11, 13, 5, L, L, L, L, (b), (c), (d), 12, 24, 24, 12, 124. The Wheatstone bridge shown in Fig. here, gets balanced, when the carbon resistor used as R1 has the colour code, (Orange, Red, Brown). The resistors R2 and R4 are 80 W, and 40 W, respectively., Assuming that the colour code for the carbon resistors, gives their accurate values, the colour code for the carbon, resistor, used as R3, would be:, [10 Jan. 2019 II], , (a), , C, , R1, G, , X, , R, , R3, , D, , K1, (a) 401.5 ohm, (b) 404.5 ohm, (c) 403.5 ohm, (d) 402.5 ohm, 122. In the experimental set up of metre bridge shown in the, figure, the null point is obtaine data distance of 40 cm from, A. If a 10 W resistor is connected in series with R1, the null, point shifts by 10 cm. The resistance that should be, connected in parallel with (R1 + 10) W such that the null, point shifts back to its initial position is :, [11 Jan. 2019 II], R1, , R2, , +, , B, , –, (b) Brown, Blue, Black, (d) Grey, Black, Brown, , 125. In a potentiometer experiment, it is found that no current, passes through the galvanometer when the terminals of, the cell are connected across 52 cm of the potentiometer, wire. If the cell is shunted by a resistance of 5 W, a balance, is found when the cell is connected across 40 cm of the, wire. Find the internal resistance of the cell. [2018], (b) 1.5 W, , (c) 2 W, , (d) 2.5 W, , 126. On interchanging the resistances, the balance point of a, meter bridge shifts to the left by 10 cm. The resistance of, their series combination is 1kW. How much was the resistance, on the left slot before interchanging the resistances?, [2018], (a) 990 W, , (a) 20 W, (b) 40 W, (c) 60 W, (d) 30 W, 123. A potentiometer wire AB having length L and resistance, 12 r is joined to a cell D of emf e and internal resistance, r. A cell C having emf e/2 and internal resistance 3r is, connected. The length AJ at which the galvanometer as, shown in fig. shows no deflection is: [10 Jan. 2019 I], , R4, , (a) Brown, Blue, Brown, (c) Red, Green, Brown, , (a) 1 W, , G, A, , R2, , (b) 505 W, , (c) 550 W, , (d) 910 W, , 127. In a meter bridge, as shown in the figure, it is given that, resistance Y=12.5 W and that the balance is obtained at a, distance 39.5 cm from end A (by jockey J). After, interchanging the resistances X and Y, a new balance point, is found at a distance l2 from end A. What are the values of, X and l2 ?, [Online April 15, 2018]
Page 302 :
P-293, , Current Electricity, X, , Y, , 131. A 10V battery with internal resistance 1W and a 15V battery, with internal resistance 0.6 W are connected in parallel to, a voltmeter (see figure). The reading in the voltmeter will, be close to :, [Online April 10, 2015], , B, G, l1, , A, , (100 – l1), , C, , 39.5 J wire, , 10V, , METER SCALE, , 1W, 15V, Key, , Battery, , (a) 19.15 W and 39.5 cm, (b) 8.16 W and 60.5 cm, (c) 19.15 W and 60.5 cm, (d) 8.16 W and 39.5 cm, 128. Which of the following statements is false ?, [2017], (a) A rheostat can be used as a potential divider, (b) Kirchhoff's second law represents energy conservation, (c) Wheatstone bridge is the most sensitive when all the, four resistances are of the same order of magnitude, (d) In a balanced wheatstone bridge if the cell and the, galvanometer are exchanged, the null point is disturbed., 129. In a meter bridge experiment resistances are connected as, shown in the figure. Initially resistance P = 4 W and the, neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of, the neutral point is at 80 cm from A. The value of unknown, resistance R is :, [Online April 9, 2017], Q, , P, , G, , A, , B, , N, , E, , Rh, , ( ), K, , 33, 20, W, (b) 6 W, (c) 7 W, (d), W, 5, 3, 130. A potentiometer PQ is set up to compare two resistances as, shown in the figure. The ammeter A in the circuit reads 1.0 A, when two way key K3 is open. The balance point is at a length, l1 cm from P when two way key K3 is plugged in between 2, and 1, while the balance point is at a length l2 cm from P when, key K3 is plugged in between 3 and 1. The ratio of two, , (a), , resistances R1 , is found to be :, R2, , (a), , l1, l1 + l2, , (b), , l2, l2 - l1, , (c), , [Online April 8, 2017], , l1, l1 - l2, , (d), , l1, l2 - l1, , 0.6W, V, , (a) 12.5 V (b) 24.5 V (c) 13.1 V, (d) 11.9 V, 132. In an experiment of potentiometer for measuring the internal, resistance of primary cell a balancing length l is obtained on, the potentiometer wire when the cell is open circuit. Now the, cell is short circuited by a resistance R. If R is to be equal to, the internal resistance of the cell the balancing length on the, potentiometer wire will be, [Online May 26, 2012], (a) l, (b) 2l, (c) l/2, (d) l/4, 133. It is preferable to measure the e.m.f. of a cell by, potentiometer than by a voltmeter because of the following, possible reasons., [Online May 12, 2012], (i) In case of potentiometer, no current flows through, the cell., (ii) The length of the potentiometer allows greater precision., (iii) Measurement by the potentiometer is quicker., (iv) The sensitivity of the galvanometer, when using a, potentiometer is not relevant., Which of these reasons are correct?, (a) (i), (iii), (iv), (b) (i), (iii), (iv), (c) (i), (ii), (d) (i), (ii), (iii), (iv), 134. In a sensitive meter bridge apparatus the bridge wire should, possess, [Online May 12, 2012], (a) high resistivity and low temperature coefficient., (b) low resistivity and high temperature coefficient., (c) low resistivity and low temperature coefficient., (d) high resistivity and high temperature coefficient., 135. In a metre bridge experiment null point is obtained at 40 cm, from one end of the wire when resistance X is balanced, against another resistance Y. If X < Y, then the new position, of the null point from the same end, if one decides to, balance a resistance of 3X against Y, will be close to :, [Online April 9, 2013], (a) 80 cm, (b) 75 cm, (c) 67 cm, (d) 50 cm, 136. The current in the primary circuit of a potentiometer is 0.2, A. The specific resistance and cross-section of the, potentiometer wire are 4 × 10–7 ohm metre and 8 × 10–7 m2,, respectively. The potential gradient will be equal to, [2011 RS], (a) 1 V /m (b) 0.5 V/m (c) 0.1 V/m (d) 0.2 V/m, 137. Shown in the figure below is a meter-bridge set up with null, deflection in the galvanometer.
Page 303 :
P-294, , Physics, , R, , 55W, , G, 20 cm, , The value of the unknown resistor R is, [2008], (a) 13.75 W (b) 220 W (c) 110 W, (d) 55 W, 138. In a Wheatstone's bridge, three resistances P, Q and R, connected in the three arms and the fourth arm is formed, by two resistances S1 and S2 connected in parallel. The, condition for the bridge to be balanced will be [2006], (a), , P, 2R, =, Q S1 + S2, , (b), , P R ( S1 + S2 ), =, Q, S1 S2, , (c), , P R ( S1 + S2 ), =, 2S1S2, Q, , (d), , P, R, =, Q S1 + S2, , 139. In a potentiometer experiment the balancing with a cell is at, length 240 cm. On shunting the cell with a resistance of 2W,, the balancing length becomes 120 cm. The internal, resistance of the cell is, [2005], (a) 0.5W, (b) 1W, (c) 2W, (d) 4W, , 140. In a meter bridge experiment null point is obtained at 20 cm., from one end of the wire when resistance X is balanced, against another resistance Y. If X < Y, then where will be the, new position of the null point from the same end, if one, decides to balance a resistance of 4 X against Y [2004], (a) 40 cm, (b) 80 cm, (c) 50 cm, (d) 70 cm, 141. The length of a wire of a potentiometer is 100 cm, and the, e. m.f. of its standard cell is E volt. It is employed to measure, the e.m.f. of a battery whose internal resistance is 0.5W. If, the balance point is obtained at l = 30 cm from the positive, end, the e.m.f. of the battery is, [2003], (a), , 30 E, 100.5, 30 ( E - 0.5i ), , (b), , 30 E, (100 - 0.5), , 30 E, 100, 100, where i is the current in the potentiometer wire., 142. An ammeter reads upto 1 ampere. Its internal resistance is, 0.81ohm. To increase the range to 10 A the value of the, required shunt is, [2003], (a) 0.03 W (b) 0.3 W, (c) 0.9 W, (d) 0.09 W, 143. If an ammeter is to be used in place of a voltmeter, then we, must connect with the ammeter a, [2002], (a) low resistance in parallel, (b) high resistance in parallel, (c) high resistance in series, (d) low resistance in series., , (c), , (d)
Page 304 :
P-295, , Current Electricity, , 1., , 2., , (c) Ammeter : In series connection, the same current, flows through all the components. It aims at measuring, the current flowing through the circuit and hence, it is, connected in series., Voltmeter : A voltmeter measures voltage change between, two points in a circuit. So we have to place the voltmeter in, parallel with the cicuit component., (b) rM = 98 ´ 10-8, , r A = 2.65 ´ 10-8, rT = 5.65 ´ 10-8, \rM > rT > r A > rC, (a) When i = 0, V = e = 1.5 volt, (d) Charge mobility, ( m) =, , Vd, E, , [ Where Vd = drift velocity ], , and resistivity ( r ) =, , Þ m=, , =, , 5., , E EA, I(r), =, Þ E=, j, I, A, , Vd Vd A, =, E, Ir, , (, , 1.1 ´10 –3 ´ p ´ 5 ´10 –3, , ), , 2, , 5 ´ 1.7 ´10 –8, , m = 1.0, , 2, , m, Vs, , é æ 1 ö ù, ê- mç 2 ÷+ cú, = eë è T ø û, , -T02, , R(T) = R 0e, (a), , dR =, , T2, , (r)(dx), 4px 2, , R = ò dR, b, , ò dR = rò, a, , 7., , (c), , r=, , m, ne 2 t, 9.1 ´ 10-31, , 8.5 ´1028 ´ (1.6 ´10 -19 ) 2 ´ 25 ´ 10 -15, = 10–8 W-m, 8., (d) Number 2 is associated with the red colour. This, colour is replaced by green., Q Colour code figure for green is 5, \ New resistance = 500 W, 9., (a) Clearly, from graph, dq, = 0at t = 4s [Since q is constant], Current, I =, dt, 10. (b) Color code : Bl, Br, R, O, Y, G, B, V, Gr, W, 0, 1, 2, 3, 4, 5, 6, 7, 8 9, R = AB × C ± D% where D = tolerance, Dgold = ±5%, Dsilver = ±10%; Dno colour = ±20%, Red voilet orange silver, R = 27 × 103 W ± 10% = 27 kW ± 10%, 11. (a) Using, I = neAvd, , \Drift speed v d =, , 1, neA, , = 0.02 mms–1, ´1.6 ´ 10-19 ´ 5 ´ 10-6, rl, 12. (c) Resistance, R =, A, 2, l l rl, R=r ´ =, [?Volume (V) = A?.], A l, V, Since resistivity and volume remains constant therefore, % change in resistance, DR 2Dl, =, = 2 ´ (0.5) = 1%, R, l, 13. (a) Colour code for carbon resistor, Bl, Br, R, O, Y, G, Blue, V, Gr, W, 0 1 2 3 4 5 6 7 8 9, Resistance, R = AB × C ± D, Bands A and B are the first two significant figures of, resistance, B and C indicates the decimal multiplier or the number of, zeros that follow A and B, B and D is tolerance: Gold = ± 5%,, Silver = ± 10 % No colour = ± 20%, 9 ´ 10, , æ 1 ö, Þ l nR = - m ç, ÷+c, è T2 ø, here, m & c are constants, , 6., , æ r ö æ1 1ö, R = ç ÷ .ç - ÷, è 4p ø è a b ø, , 1.5, , (b) Equation of straight line from graph, y = – mx + c, , R, , r é -1 ù, 4p êë x úû a, , =, , rC = 1.724 ´ 10-8, , 3., 4., , b, , ÞR=, , dx, 4 px 2, , æ 1 ö, - mç, 2÷, = e è T ø ´ ec, , 28, , R = 53 ´ 104 ± 5% = 530kW ± 5%
Page 305 :
P-296, , 14., , 15., , Physics, , (c) Resistance after temperature increases by 500°C i.e.,, V 220, Rt = =, = 110W, I, 2, R0 = 100 (given) temperature coefficient of resistance, a = ?, using Rt = R0 (1 + at), 110 = 100 (1 + a500), a=, , 10, 100 ´ 500, , or,, , a = 2 ´ 10 -4°C -1, , (c) Charge density r =, , charge, q, =, Þ q = rAd, volume Ad, , q rAd, =, I, I, (a) Given, R1 = 100 W, r' = r/2, R2 = ?, rl, Resistivity of wire, R =, Q Area × length = volume, A, rV, Hence, R = 2, A, Since, r ® constant, V ® constant, 1, Rµ 2, A, 1, Q A = pr2, or R µ, r4, R2, = 16 Þ R2 = 16 ´ 100 = 1600 W, Resistance of new wire., R1, l, (b) V = IR = (neAvd )r, A, V, \ r = V lne, d, Here V = potential difference, l = length of wire, n = no. of electrons per unit volume of conductor., e = no. of electrons, Placing the value of above parameters we get resistivity, , Also, q = IT Þ T =, , 16., , 17., , r=, , 5, , 8 ´ 1028 ´ 1.6 ´ 10-19 ´ 2.5 ´ 10-4 ´ 0.1, = 1.6 × 10–5Wm, 18., , 19., , (c) i = neAVd and Vd µ E (Given), or, i µ E, i2 µ E, i2 µ V, Hence graph (c) correctly dipicts the V-I graph for a wire, made of such type of material., (a) In ohm's law, we check V = IR where I is the corrent, flowing through a resistor and V is the potential difference, across that resistor. Only option (a) fits the above criteria., Remember that ammeter is connected in series with, resistance and voltmeter parallel with the resistance., , l, A, If wire is bent in the middle then, l, l¢ = , A¢ = 2 A, 2, l¢, \ New resistance, R¢ = r, A¢, l, r, rl R, 2, = ., =, =, 2A, 4A 4, (a) Resistance of wire, , 20. (c) Resistance of wire (R) = r, , 21., , R=, , rl rl 2, =, (Q V = Al), A V, , l2, = constant × l2, V, \ Fractional change in resistance, , Hence, R = r, DR, Dl, =2, R, l, , DR, æ dl ö, = 200 ´ ç ÷, R, è l ø, Q dl/l = 0.1%, 100 ´, , { }, , 0.1 ù, é, \ % change in R = ê 200 ´, = 0.2%, 100 úû, ë, \ Resistance will increase by 0.2%., 22. (a) Let j be the current density., I, Then j ´ 2pr 2 = I Þ j =, 2 pr 2, rI, \ E = rj =, 2 pr 2, Now, VB – VC, a, , =-, , ò, , r uur, E × dr = -, , a +b, a, , a, , ò, , rI, , 2 pr 2, a+b, , dr, , rI é 1 ù, rI, rI, - ú, =, ê, 2 p ë r û a + b 2 pa 2 p ( a + b ), On applying superposition as mentioned we get, rI, rI, ', DVBC = 2 ´ DVBC, =, pa p ( a + b ), rI, 23. (c) As shown in Answer (a) E =, 2pr 2, 24. (d) Resistance of a metal conductor at temperature t°C, is given by, Rt = R0 (1 + at),, R0 is the resistance of the wire at 0ºC, and a is the temperature coefficient of resistance., Resistance at 50°C, R50 = R0 (1 + 50a), .. (i), Resistance at 100°C, R100 = R0 (1 + 100a) ... (ii), From (i), R50 – R0 = 50aR0, ... (iii), From (ii), R100 – R0 = 100aR0, ... (iv), =-
Page 312 :
P-303, , Current Electricity, B, , i, , P2, , i–i1, , m = Zq, , C, , i1, 10W, , 5V, , 1W, , Þ Zµ, , 2V, , 2W, , 69., 70., , 71., , D, , P1, , Again applying Kirchoff's second law in P2 CDP1P2 we, get,, 10 i1 + 2 – i + i1= 0, 2i – 22i1 = 4, ....(ii), From (i) and (ii), 32i1 = 1, 1, Þ i1 =, A from P2 to P1, 32, 1, 2, (a) Energy in capacitor = CV, 2, Work done by battery = QV = CV2, where C = Capacitance of capacitor, V = Potential difference,, e = emf of battery, 1, CV 2, 1, 2, = (Q V = e), Required ratio =, 2, 2, CV, (d) Note : Kirchhoff's first law is based on conservation, of charge and Kirchhoff's second law is based on, conservation of energy., (d) At cold junction, current flows from Antimony to, Bismuth because current flows from metal occurring later, in the series to metal occurring earlier in the thermoelectric, series. In thermoelectric series, Bismuth comes earlier than, Antimony so at cold junction, current. Flow from Antimony, to Bismuth., (a), , R1, , R2, , E, , E, , I, , R, , Let E be the emf of each source of current, 2E, Current in the circuit I =, R + R1 + R2, Potential difference across cell having internal resistance R2, V = E – iR2 = 0, 2E, × R2 = 0, E–, R + R1 + R2, Þ R + R1 + R 2 - 2R 2 = 0, , 72., , .... (i), , Also q = q1 + q2, , A, , 68., , 1 Þ Z1 = q2, Z 2 q1, q, , q, q, = 1 +1, q2 q2, , Þ, , .... (ii), (Dividing (ii) by q2), , q, q, 1+ 1, q2, From equation (i) and (iii),, , Þ q2 =, , .... (iii), , q, Z, 1+ 2, Z1, (d) Current is given by, , q2 =, , 73., , E, ,, R+r, If internal resistance (r) is zero,, E, I=, = constant., R, Thus, energy source will supply a constant current if its, internal resistance is zero., 74. (d) Given E = aq + bq2, dE, Þ, = a + 2bq, dq, At neutral temperature, dE, q = qn :, =0, dq, , I=, , d2E, -a, = -350 Þ, = 2b, 2b, d q2, hence no q is possible for E to be maximum no neutral, temperature is possible., 75. (c) From the Faraday’s first law of electrolysis,, m = Zit, Þ m = 3.3 × 10–7 × 3 × 2, = 19.8 × 10–7 kg, 76. (a) Let the smallest temperature difference be q°C that, can be detected by the thermocouple, then, Thermo emf = (25 × 10–6) q, Let I is the smallest current which can be detected by the, galvanometer of resistance R., Potential difference across galvanometer, IR = 10–5 × 40, \ 10–5 × 40 = 25 × 10–6 × q, Þ q = 16°C., 77. (c) According to Faraday’s first law of electrolysis, m= Z×I×t, When I and t is same, m µ Z, Þ qn =, , mCu ZCu Þ m = Z Cu ´ m, =, Cu, Zn, Z Zn, mZn Z Zn, , Þ R + R1 - R 2 = 0, , \, , Þ R = R 2 - R1, (a) From Faraday’s first law of electrolysis, mass, deposited, , Þ mCu =, , 31.5, ´ 0.13 = 0.126 g, 32.5
Page 313 :
P-304, , 78., 79., , Physics, , (b) From the Faraday’s first law of electrolysis, m = ZIt Þ m µ It, (b) Given : Power, P = 1 kW = 1000 W, R = 2W, V = 220 V, Current, I =, , 84. (a) As R =, , Current flown i =, , P 1000, =, V, 220, , P1 = i 2 R1 =, , 2, , æ 1000 ö, ´2, Ploss = I 2 R = ç, è 220 ÷ø, 1000, \ Efficiency = 1000 + P ´ 100 = 96%., loss, , 80., , (b) Maximum power in external resistance is generated, when it is equal to internal resistance of battery i.e., PR, maximum when r = R, The maximum Joule heating in R will take place for, the, resistance of small element, , r, , R, r, b, ln, 2 pl a, (d) Net Power, P, = 15 × 45 + 15 × 100 + 15 × 10 + 2 × 1000, = 15 × 155 + 2000 W, Power, P = VI, P, V, , 15 ´155 + 2000, \ I main =, = 19.66 A » 20 A, 220, (b) Equivalent resistance,, 4 R ´ 4R, 6 R ´ 12 R, + R+, +R, 4R + 4R, 6 R + 12 R, = 2R + R + 4R + R, = 8R., , V2, 162, Using, P = R Þ 4 =, eq, 8R, , \, 83., , (b), , R=, , 162, =8W, 4´8, , e2, (R/2), or P' = 4P = 240 W(Q P = 60 W), (a) Colour code for carbon resistor, Bl, Br, R, O, Y, G,, Blue, V,, 0 1, 2 3, 4 5, 6, 7, Resistance, R = AB × C ± D, \ Resistance, R = 50 × 102 W, Now using formula, Power, P = i 2R, , Gr, W, 8 9, , P, 2, = 20mA, =, R, 50 ´102, 87. (a) Power, P = I2R, 4.4 = 4 × 10–6 × R, Þ R = 1.1 × 106W, When supply of 11 v is connected, , \i=, , or, R =, , Req =, , 2202, = 16 W, 25, , e2, e2, =, R eq 2R, In parallel condition, Req = R/2., , 86., , b, , b, , 82., , æ 2202 2202 ö, ç 25 + 100 ÷, è, ø, , ´, , New power, P¢ =, , l, , Þ I=, , 2202, , Power consumed, P =, , b, , 81., , 220, R1 + R 2, , Similarly, P2 = i2R2 = 4 W, 85. (b) When two resistances are connected in series,, Req = 2R, , rdr, r dr, DR =, ÞR=, 2prl, 2pl òa r, a, , V2, 2202, 2202, , so R1 =, and R 2 =, P, 25, 100, , v 2 112 112, ´, ´10 –6, =, R, 1.1 1.1, =11´10 –5 W, (a) Rate of heat i.e., Power developed in the wire =, , Power, P’=, , 88., , P=, , V2, R, , Resistance of the wire of length, L R1 =, , rL rL, =, A pr 2, , V2, R1, Resistance of the wire when length is halved i.e., L/2, , \, , Power, P1 =, , L, 2 = rL = R1, R2 =, 8, p(2r )2 p8r 2, r, , V, 8V, =, R1 R1, 8, or, P2 = 8P1 i.e., power, or original wire., , \, , Power, P2 =, , increased 8 times of previous
Page 315 :
P-306, , I=, , Physics, , 101. (c) The resistance of the electric bulb is, , 440, R eff, , I=, , V 2 (220) 2, =, P, 100, The power consumed when operated at 110 V is, R=, , 440, 2, , (220), (220) 2, +, 25, 100, 440, =, ;, 1 ù, é1, (220) 2 ê +, ë 25 100 úû, , I=, , 40, Am p, 220, , æ 25 ö, æ 40 ö < I æ = 100 A ö, A÷ < I ç =, A÷, Q I1 ç =, 2ç, ÷, è 200 ø, è 220 ø, è 220 ø, Thus the bulb rated 25 W–220 will fuse., , 97., , (b) Total resistance =, , 6´3, = 2W, 6+3, , 6, = 3A, 2, Therefore current through bulb 1 is 2A and bulb 2 is 1A., So bulb 1 will glow more, (c) Resistors 4 W, 6 W and 12 W are connected in parallel,, its equivalent resistance (R) is given by, , Current in circuit =, , 98., , 1 1 1 1, 12, = + +, Þ R=, = 2W, R 4 6 12, 6, Again R is connected to 1.5 V battery whose internal, resistance r = 1 W., Equivalent resistance now,, R¢ = 2W + 1W = 3W, V 1.5 1, =, = A, Current, Itotal =, R', 3, 2, 1, Itotal = = 3x + 2x + x = 6x, 2, , 1, 12, \ Current through 4W resistor = 3x, , Þx=, , 1, 1, = A, 12 4, Therefore, rate of Joule heating in the 4W resistor, , = 3×, , 2, , 1, æ1ö, = I2R = ç ÷ ´ 4 = = 0.25W, 4, è4ø, 99. (c), 100. (b) Let resistance of bulb filament be R0 at 0°C using R =, R0 (1 + a Dt) we have, R1 = R0 [1 + a × 100] = 100 ....(1), R2 = R0 [1 + a × T] = 200, ....(2), On dividing we get, 200, 1 + aT, 1 + 0.005 T, =, Þ2=, 100 1 + 100a, 1 + 100 ´ 0.005, , Þ T = 400°C, Note : We may use this expression as an approximation, because the difference in the answers is appreciable. For, accurate results one should use R = R0eaDT, , P¢ =, , V2, R, , Þ P=, , (110) 2, 2, , =, , (220) /100, 102. (b) Heat generated,, , 100, = 25 W, 4, , V 2t, R, After cutting equal length of heater coil will become half., As R µ l, R, Resistance of half the coil =, 2, , H=, , V 2t, = 2H, R, 2, \ As R reduces to half, ‘H’ will be doubled., , H¢ =, , V2, R, \ Resistance of tungsten filament when in use, , 103. (b) Power, P = Vi =, , V 2 200 ´ 200, =, = 400 W, P, 100, Resistance when not in use i.e., cold resistance, Rhot =, , 400, = 40 W, 10, 104. (a) Thermistors are usually made of metaloxides with, high temperature coefficient of resistivity., 105. (a) Heat supplied in time t for heating 1L water from, 10°C to 40°C, DQ = mCp × DT, = 1 × 4180 × (40 – 10) = 4180 × 30, But DQ = P × t = 836 × t, Rcold =, , 4180 ´ 30, = 150s, 836, 106. (c) We know that resistance,, Þt=, , 2, Vrated, (220) 2, =, = 48.4 W, Prated, 1000, When this bulb is connected to 110 volt mains supply we, get, , R=, , V 2 (110)2, =, = 250W, R, 48.4, 107. (b) Case 1 Initial power dissipation,, P=, , R, P1 =, , V2, R, , V
Page 316 :
P-307, , Current Electricity, , Case 2, When wire is cut into two equal pieces, the resistance of, R, each piece is . When they are connected in parallel, 2, , Equivalent resistance, Req =, , R/2, , R/4, , V, , æV ö, V, = 4ç, = 4 P1, ç R ÷÷, R/4, è, ø, 108. (b) The equivalent resistance of parallel combination of, 2W and R is, P2 =, , Req =, , R 25 1, =, =, S 75 3, New resistance,, l, r, l´2, R' = 2 = r, A, A, 4, Þ R¢ = 2R, l2, R', =, Þ, S 100 – l 2, , Þ, , =, , 2, , 1.02, = 0.02 volt/cm, 100 - 49, 112. (40) For the given meter bridge, \x =, , l1, R, =, Where, l1 = balancing length, S 100 – l1, , R/2 R, =, 2, 4, , R/2, , V, Power dissipated,, , Balancing length from P = 100 – 49, , 2, , 2´R, 2+ R, , Þ, , \ Power dissipation P =, , V2, Re q, , \ 15 0 =, , (15)2, Req, , Þ, , ...(i), , lö, æ, çQ R = r ÷, Aø, è, , l2, 1, 2´ =, 3 100 – l 2, , l2, 2R, =, S 100 – l 2, , Using (i), , l!, 2 = 40 cm, , 113. (d), , 225 ´ ( R + 2), 2R, 3, Þ, =, 2R, 2+ R 2, Þ 4 R = 6 + 3R Þ R = 6W, 109. (d) The voltmeter of resistance 10kW is parallel to the, resistance of 400W. So, their equivalent resistance is, , Þ 150 =, , 1, 1, 1, 1, 1, =, +, =, +, R ' 10 k W 400W 10000 400, Þ, , Let R be the resistance of the whole wire, Potential gradient for the potentiometer wire, , 1 1 + 25, 26, =, =, R ' 10000 10000, , ' AB ' = -, , 10000, Þ R' =, W, 26, Using Ohm's law, current in the circuit, , æ dV ö, 60 ´ R, V AP = ç, lAP =, ´ 1000mV, 1200, è d l AB ÷ø, Þ VAP = 50 R mV, Also, VAP = 5 V (for balance point at P), , Voltage, 6, =, Net Resistance 10000 + 800, 26, Potential difference measured by voltmeter, I=, , V = IR ' =, , 6, 10000, ´, 10000, 26, + 800, 26, , 150, = 1.95 volt, 77, 110. (b) Multimeter shows deflection in both cases i.e. before, and after reversing the probes if the chosen component is, capacitor., Potential drop, 111. (a) Potential gradient, x =, length, Here, Potential drop = 1.02, , dV, I ´ R é 60 ´ R ù, =, =ê, ú mv / m, dl, l, ë l AB û, , \R =, , V AP, 50 ´ 10, , –3, , =, , 5, = 100W, 50 ´ 10 –3, , 114. (10), , ÞV =, , As per Wheatstone bridge balance condition, , P S, =, Q R
Page 317 :
P-308, , Physics, , Let resistance R’ is connected in parallel with resistance S, of 10W, 10 R ', 15, 10 R ', Þ 5=, \ =, 10 + R ', 12 10 + R ', 4, Þ 50 + 5R’ = 10R’, 50, = 10W, 5, 115. (12) We know that, E µ l where l is the balancing length, \ E = k (560), ....(i), When the balancing length changes by 60 cm, , ...(ii), , dl, l, Let R1 and R2 be the resistance of AP and PB respectively., Using wheatstone bridge principle, \, , R ' R1, or R1 = R 2, =, R ' R2, , Now, ò dR = k ò, l, , r + 10 56, =, Þ 50 r + 500 = 560, 10, 50, , 6, N, Þ r = W = W Þ N = 12, 5, 10, , 116. (c) For a balanced bridge, R1 l 2, =, R 2 l1, , \ R1 = k ò l -1 2dl = k.2. l, , R, l, =, X 100 – l, Using the above expression, , 1, , R 2 = k ò l -1 2dl = k.(2 - 2 l ), l, , Putting R1 = R2, , k2 l = k(2 - 2 l ), \ 2 l =1, l=, , R(100 – l ), l, , 100 ´ 40 2000, =, W, for observation (1) X =, 60, 3, , for observation (2) X =, , 100 ´ 87 8700, =, W, 13, 13, , for observation (3) X =, , 10 ´ 98.5 1970, =, W, 1.5, 3, , 1 ´ 99, = 99W, 1, Clearly we can see that the value of x calculated in, observation (4) is inconsistent than other., 117. (c) The resistance of potentiometer wire, R = 0.01 × 400 = 4 W, Current in the wire, , for observation (4) X =, , i=, , 3, 1, V, =, = A, RT 4 + 0.5 + 0.57 + 1 2, , Now V = iRAJ =, , 1, × (0.01 × 50) = 0.25 V.., 2, , 1, 2, , 1, i.e., l = m Þ 0.25 m, 4, R, 4v, , So, , X=, , dl, l, , 0, , Dividing (i) by (ii) we get, Þ, , dR, 1, dR, 1, (where k is constant), µ, Þ, = k´, dl, dl, l, l, , dR = k, , \ R' =, , E, 10 = k (500), r + 10, , 118. (c) We have given, , 119. (c), , i, , 5W, , i, , 1m, Current flowing through the circuit (I) is given by, , æ 4 ö, I=ç, A, è R + 5 ÷ø, Resistance of length 10 cm of wire, 10, = 0.5W, 100, According to question,, æ 4 ö, 5 ´ 10 -3 = ç, .(0.5), è R + 5 ÷ø, 4, \, = 10 -2 or R + 5 = 400 W, R +5, \ R = 395W, 120. (b) Given, Emf of cell, e = 0.5 v, Rheostat resistance, Rh = 2W, Potential gradient is, =5´, , dv æ 6 ö 4, =, ´, dL çè 2 + 4 ÷ø L, Let null point be at l cm when cell of emfe = 0.5 v is used., æ 6 ö 4, thus e1 = 0.5V = ç, ´ ´l, è 2 + 4 ÷ø L, , ... (i)
Page 319 :
P-310, , Physics, , P, l, =, Q (100 - l ), Initially neutral position is 60 cm. from A, so, , 129. (d) In balance position of bridge,, , 4, Q, 16 8, =, ÞQ= = W, 60 40, 6 3, Now, when unknown resistance R is connected in series, to P, neutral point is 80 cm from A then,, 4+ R Q, =, 80, 20, 4+ R 8, =, 80, 60, 64, 64 - 24 40, R=, -4=, =, W, 6, 6, 6, 20, Hence, the value of unknown resistance R is =, W, 3, 130. (d) When key is at point (1), V1 = iR1 = xl1, When key is at (3), V2 = i (R1 + R2) = xl2, , R1, l, R, l, = 1 Þ 1= 1, R1 + R 2 l2, R 2 l2 - l1, 131. (c) As the two cells oppose each other hence, the, effective emf in closed circuit is 15 – 10 = 5 V and net, resistance is 1 + 0.6 = 1.6 W (because in the closed, circuit the internal resistance of two cells are in series., Current in the circuit,, I=, , effective emf, 5, =, A, total resistance 1.6, , The potential difference across voltmeter will be same as, the terminal voltage of either cell., Since the current is drawn from the cell of 15 V, \ V1 = E1 – Ir1, 5, ´ 0.6 = 13.1 V, = 15 –, 1.6, , 132. (c) Balancing length l will give emf of cell, \ E = Kl, Here K is potential gradient., If the cell is short circuited by resistance 'R', Let balancing length obtained be l¢ then, V = kl¢, æ E -V ö, R, r= ç, è V ÷ø, Þ V = E – V [Q r = R given], Þ 2V = E, or, 2Kl¢ = Kl, , \, , l¢ =, , l, 2, , 133. (c) To measure the emf of a cell we prefer potentiometer, rather than voltmeter because, (i) the length of potentiometer which allows greater, precision., (ii) in case of potentiometer, no current flows through the, cell., (iii) of high sensitivity., 134. (a) Bridge wire in a sensitive meter bridge wire should be, of high resistivity and low temperature coefficient., x, 40, 2, 135. (c) From question, y = 100 - 40 = 3, Þx=, , 2, y, 3, , Again,, , 3x, Z, =, y 100 - Z, , or, , 2y, Z, 3 =, y, 100 - Z, , 3´, , Solving we get Z = 67 cm, Therefore new position of null point @ 67 cm, 136. (c) Potential gradient, Þk=, , k=, , V IR I æ rl ö I r, =, = ç ÷=, l, l lè A ø A, , 0.2 ´ 4 ´ 10 -7 0.8, =, = 0.1 V/m, 8, 8 ´ 10-7, , 137. (b) Given,, Balance point from one end, l1 = 20 cm, From the condition for balance of metre bridge, we have, 55, l1, =, R 100 – l1, , 55 20, =, R 80, , Þ R = 220W, 138. (b) From balanced wheat stone bridge, S=, , P R, where, =, Q S, , S1S 2, S1 + S 2, , 139. (c) Initial balancing length, l1 = 240 cm New balancing, length, l2 = 120 cm., The internal resistance of the cell,, æl -l ö, 240 - 120, ´ 2 = 2W, r= ç 1 2 ÷´R =, l, 120, è, 2 ø
Page 320 :
P-311, , Current Electricity, , 140. (c) From the balanced wheat stone bridge, E, , R1 l1, =, R2 l 2, , i, , i, , where l2 = 100 – l1, In the first case X = 20, Y 80, Y = 4X, In the second case, 4X, l, =, Y, 100 - l, , Þ, , 4X, l, =, 4 X 100 – l, , Þ l = 50, 141. (d) From the principle of potentiometer, V µ l, If a cell of emF E is employed in the circuit between the, ends of potentiometer wire of length L, then, V, l, = ;, E L, , Þ V=, , El 30E, =, L 100, , r, E', , Note : In this arrangement, the internal resistance of the, battery E does not play any role as current is not passing, through the battery., 142. (d) ig × G = (i – ig) S, \ S=, , ig ´ G, i - ig, , =, , 1 ´ 0.81, = 0.09W, 10 - 1, , 143. (c) To use an ammeter in place of voltmeter, we must, connect a high resistance in series with the ammeter., Connecting high resistance in series makes its resistance, much higher.
Page 321 :
18, , P-312, , Physics, , Moving Charges, and Magnetism, Motion of Charged Particle in, TOPIC 1, Magnetic Field, 1., , An electron is moving along + x direction with a velocity, of 6 × 106 ms–1. It enters a region of uniform electric field of, 300 V/cm pointing along + y direction. The magnitude and, direction of the magnetic field set up in this region such, that the electron keeps moving along the x direction will, be :, [Sep. 06, 2020 (I)], , 4., , 5., , (a) 3 × 10–4 T, along + z direction, (b) 5 × 10–3 T, along – z direction, (c) 5 × 10–3 T, along + z direction, 2., , (b) 0.88 m, , (d) 3 × 10–4 T, along – z direction, A particle of charge q and mass m is moving with a, , (c) 0.44 m, , velocity – v $i (v ¹ 0) towards a large screen placed in the, Y-Z plane at a distance d. If there is a magnetic field, ur, B = B0 k$ , the minimum value of v for which the particle, will not hit the screen is:, [Sep. 06, 2020 (I)], , 6., , qdB0, 2qdB0, (b), 3m, m, qdB0, qdB0, (c), (d), m, 2m, A charged particle carrying charge 1 mC is moving with, velocity (2iˆ + 3 ˆj + 4kˆ) ms–1. If an external magnetic field, , 7., , (a), , 3., , of (5iˆ + 3 ˆj - 6kˆ) ´10-3 T exists in the region where the, particle is moving then the force on the particle is, ur, ur, [Sep. 03, 2020 (I)], F ´10-9 N. The vector F is :, (a), , - 0.30iˆ + 0.32 ˆj - 0.09kˆ, , (b), , - 30iˆ + 32 ˆj - 9kˆ, , (c), , - 300iˆ + 320 ˆj - 90kˆ, , (d), , - 3.0iˆ + 3.2 ˆj - 0.9kˆ, , A beam of protons with speed 4 × 105 ms–1 enters a uniform, magnetic field of 0.3 T at an angle of 60° to the magnetic, field. The pitch of the resulting helical path of protons is, close to : (Mass of the proton = 1.67 × 10–27 kg, charge of, the proton = 1.69 × 10–19 C), [Sep. 02, 2020 (I)], (a) 2 cm, (b) 5 cm, (c) 12 cm (d) 4 cm, The figure shows a region of length ‘l’ with a uniform, magnetic field of 0.3 T in it and a proton entering the region, with velocity 4 × 105 ms–1 making an angle 60° with the, field. If the proton completes 10 revolution by the time it, cross the region shown, ‘l’ is close to (mass of proton, = 1.67 × 10–27 kg, charge of the proton = 1.6 × 10–19 C), [Sep. 02, 2020 (II)], B, (a) 0.11 m, 60°, , l, (d) 0.22 m, Proton with kinetic energy of 1 MeV moves from south to, north. It gets an acceleration of 1012 m/s2 by an applied, magnetic field (west to east). The value of magnetic field:, (Rest mass of proton is 1.6 ´ 10–27 kg) [8 Jan 2020, I], (a) 0.71 mT, (b) 7.1 mT, (c) 0.071 mT, (d) 71 mT, A particle having the same charge as of electron moves, in a circular path of radius 0.5 cm under the influence of, a magnetic field of 0.5T. If an electric field of 100V/m, makes it to move in a straight path then the mass of the, particle is (Given charge of electron = 1.6 × 10–19C), [12 April 2019, I], kg, , (b) 1.6 × 10–27 kg, , (c) 1.6 × 10–19 kg, , (d) 2.0 × 10–24 kg, , (a) 9.1 ×, 8., , 10–31, , An electron, moving along the x-axis with an initial energy, ur, of 100 eV, enters a region of magnetic field B = (1.5×10–3T) k$, at S (see figure). The field extends between x = 0 and x =, 2 cm. The electron is detected at the point Q on a screen
Page 322 :
P-313, , Moving Charges and Magnetism, , placed 8 cm away from the point S. The distance d between, P and Q (on the screen) is :, (Electron’s charge = 1.6 × 10–19 C, mass of electron, = 9.1 × 10–31 kg), [12 April 2019, II], , (a) 11.65 cm, (c) 1.22 cm, , (b) 12.87 cm, (d) 2.25 cm, , 9., , A proton, an electron, and a Helium nucleus, have the, same energy. They are in circular orbits in a plane due to, magnetic field perpendicular to the plane. Let rp, re and rHe, be their respective radii, then,, [10 April 2019, I], (a) re > rp = rHe, (b) re < rp = rHe, (c) re < rp < rHe, (d) re > rp > rHe, 10. A proton and an α -particle (with their masses in the ratio, of 1 : 4 and charges in the ratio 1 : 2) are accelerated from, rest through a potential difference V. If a uniform magnetic, field (B) is set up perpendicular to their velocities, the ratio, of the radii rp : ra of the circular paths descrfibed by them, will be:, [12 Jan 2019, I], 11., , (a) 1: 2, (b) 1: 2, (c) 1: 3, (d) 1: 3, In an experiment, electrons are accelerated, from rest, by, applying a voltage of 500 V. Calculate the radius of the, path if a magnetic field 100 mT is then applied., [Charge of the electron = 1.6 × 10–19 C, Mass of the electron = 9.1 × 10–31 kg] [11 Jan 2019, I], (a) 7.5 × 10–3 m, , (b) 7.5 × 10–2 m, , (c) 7.5 m, (d) 7.5 ×10–4 m, 12. The region between y = 0 and y = d contains a magnetic, r, field B = Bzˆ . A particle of mass m and charge q enters, mv, the region with a velocity vr = viˆ . if d =, , the, 2qB, acceleration of the charged particle at the point of its, emergence at the other side is :, [11 Jan 2019, II], (a), , qv B æ 1 ˆ, 3 ˆö, j÷, ç im è2, 2 ø, , (b), , qv B æ 3 ˆ 1 ˆ ö, i + j÷, ç, m è 2, 2 ø, , (c), , qvB æ - ˆj + iˆ ö, ç, ÷, m è 2 ø, , (d), , qvB æ iˆ + ˆj ö, ç, ÷, m è 2 ø, , 13. An electron, a proton and an alpha particle having the same, kinetic energy are moving in circular orbits of radii re,, rp, ra respectively in a uniform magnetic field B. The, relation between re, rp, ra is :, [2018], (a) re > rp = ra, (b) re < rp = ra, (c) re < rp < ra, (d) re < ra < rp, 14. A negative test charge is moving near a long straight wire, carrying a current. The force acting on the test charge is, parallel to the direction of the current. The motion of the, charge is :, [Online April 9, 2017], (a) away from the wire, (b) towards the wire, (c) parallel to the wire along the current, (d) parallel to the wire opposite to the current, 15. In a certain region static electric and magnetic fields exist., r, ˆ . If a, The magnetic field is given by B = B0 (iˆ + 2ˆj - 4k), r, ˆ, test charge moving with a velocity v = v 0 (3iˆ - ˆj + 2k), experiences no force in that region, then the electric field, in the region, in SI units, is :, [Online April 8, 2017], , r, r, ˆ (b) E = -v B (iˆ + ˆj + 7k), ˆ, E = -v0B0(3iˆ - 2jˆ - 4k), 0 0, r, r, ˆ, ˆ, (c) E = v0 B0 (14jˆ + 7k), (d) E = -v0 B0 (14jˆ + 7k), 16. Consider a thin metallic sheet perpendicular to the plane, of the paper moving with speed 'v' in a uniform magnetic, field B going into the plane of the paper (See figure). If, charge densities s1 and s2 are induced on the left and, right surfaces, respectively, of the sheet then (ignore fringe, effects):, [Online April 10, 2016], (a), , v, , - Î0 vB, Î vB, , s2 = 0, 2, 2, , (a), , s1 =, , (b), , s1 = Î0 vB, s 2 = - Î0 vB, , (c), , Î vB, - Î0 vB, s1 = 0, , s2 =, 2, 2, , B, , (d) s1 = s 2 = Î0 vB, s1 s2, 17. A proton (mass m) accelerated by a potential difference V, flies through a uniform transverse magnetic field B. The, field occupies a region of space by width ‘d’. If a be the, angle of deviation of proton from initial direction of motion, (see figure), the value of sin a will be :, [Online April 10, 2015], , B, a, d, Bd, 2m, , (a), , qV, , (c), , B, q, d 2mV, , (b), , B qd, 2 mV, , (d) Bd, , q, 2mV
Page 323 :
P-314, , Physics, , 18. A positive charge ‘q’ of mass ‘m’ is moving along the + x, axis. We wish to apply a uniform magnetic field B for, time Dt so that the charge reverses its direction crossing, the y axis at a distance d. Then: [Online April 12, 2014], , 19., , 20., , (a) B =, , pd, mv, and Dt =, v, qd, , (c) B =, , 2mv, pd, and Dt =, (d), qd, 2v, , 22., , pd, 2mv, and Dt =, v, qd, , ra = rp = rd, , (b), , ra = rp < rd, , rd > rp, (c) ra > rd > rp, (d) ra =, This question has Statement 1 and Statement 2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., Statement 1: A charged particle is moving at right angle, to a static magnetic field. During the motion the kinetic, energy of the charge remains unchanged., Statement 2: Static magnetic field exert force on a moving, charge in the direction perpendicular to the magnetic field., [Online May 26, 2012], (a) Statement 1 is false, Statement 2 is true., (b) Statement 1 is true, Statement 2 is true, Statement 2 is, not the correct explanation of Statement 1., (c) Statement 1 is true, Statement 2 is false., (d) Statement 1 is true, Statement 2 is true, Statement 2 is, the correct explanation of Statement 1., A proton and a deuteron are both accelerated through the, same potential difference and enter in a magnetic field, perpendicular to the direction of the field. If the deuteron, follows a path of radius R, assuming the neutron and proton, masses are nearly equal, the radius of the proton’s path, will be, [Online May 19, 2012], R, , R, (c), (d) R, 2, 2, The magnetic force acting on charged particle of charge 2, mC in magnetic field of 2 T acting in y-direction, when the, , (a), 23., , B=, , pd, mv, and Dt =, 2v, 2qd, , A particle of charge 16 × 10–16 C moving with velocity, 10 ms–1 along x-axis enters a region where magnetic field, ur, of induction B is along the y-axis and an electric field, of magnitude 104 Vm–1 is along the negative z-axis. If, the charged particle continues moving along x-axis, the, ur, magnitude of B is :, [Online April 23, 2013], (a) 16 × 103 Wb m–2, (b) 2 × 103 Wb m–2, (c) 1 × 103 Wb m–2, (d) 4 × 103 Wb m–2, Proton, deuteron and alpha particle of same kinetic energy, are moving in circular trajectories in a constant magnetic, field. The radii of proton, deuteron and alpha particle are, respectively rp, rd and ra. Which one of the following, relation is correct?, [2012], (a), , 21., , (b) B =, , 2R, , (b), , (, , ), , 6, -1, particle velocity is 2iˆ + 3 ˆj ´ 10 ms is, , [Online May 12, 2012], , (a) 8 N in z-direction, (b) 8 N in y-direction, (c) 4 N in y-direction, (d) 4 N in z-direction, 24. The velocity of certain ions that pass undeflected through, crossed electric field E = 7.7 k V/m and magnetic field, B = 0.14 T is, [Online May 7, 2012], (a) 18 km/s, (b) 77 km/s, (c) 55 km/s, (d) 1078 km/s, r, 25. An electric charge +q moves with velocity v = 3iˆ + 4 ˆj + kˆ, ur, in an electromagnetic field given by E = 3i$ + $j + 2k$ and, ur, B = iˆ + ˆj - 3kˆ The y - component of the force experienced, by + q is :, [2011 RS], (a) 11 q, (b) 5 q, (c) 3 q, (d) 2 q, 26. A charged particle with charge q enters a region of, ur, constant, uniform and mutually orthogonal fields E and, ur, ur, ur, r, B with a velocity v perpendicular to both E and B ,, and comes out without any change in magnitude or, r, direction of v . Then, [2007], r ur ur 2, r ur ur 2, (a) v = B ´ E / E, (b) v = E ´ B / B, r ur ur 2, r ur ur, (c) v = B ´ E / B, (d) v = E ´ B / E 2, 27. A charged particle moves through a magnetic field, perpendicular to its direction. Then, [2007], (a) kinetic energy changes but the momentum is, constant, (b) the momentum changes but the kinetic energy is, constant, (c) both momentum and kinetic energy of the particle, are not constant, (d) both momentum and kinetic energy of the particle, are constant, 28. In a region, steady and uniform electric and magnetic fields, are present. These two fields are parallel to each other. A, charged particle is released from rest in this region. The, path of the particle will be a, [2006], (a) helix, (b) straight line, (c) ellipse, (d) circle, 29. A charged particle of mass m and charge q travels on a, circular path of radius r that is perpendicular to a magnetic, field B. The time taken by the particle to complete one, revolution is, [2005], 2 pmq, 2pm, 2pqB, 2pq 2 B, (b), (d), (c), B, qB, m, m, 30. A uniform electric field and a uniform magnetic field are, acting along the same direction in a certain region. If an, electron is projected along the direction of the fields with, a certain velocity then, [2005], (a) its velocity will increase, (b) Its velocity will decrease, (c) it will turn towards left of direction of motion, (d) it will turn towards right of direction of motion, , (a)
Page 324 :
P-315, , Moving Charges and Magnetism, , 31. A particle of mass M and charge Q moving with velocity, r, v describe a circular path of radius R when subjected to, a uniform transverse magnetic field of induction B. The, work done by the field when the particle completes one, full circle is, [2003], , 32., , 33., , æ Mv 2 ö, (a) ç, (b) zero, ÷ 2 pR, è R ø, (d) B Qv 2p R, (c) B Q 2 p R, If an electron and a proton having same momenta enter, perpendicular to a magnetic field, then, [2002], (a) curved path of electron and proton will be same, (ignoring the sense of revolution), (b) they will move undeflected, (c) curved path of electron is more curved than that of, the proton, (d) path of proton is more curved., The time period of a charged particle undergoing a circular, motion in a uniform magnetic field is independent of its, (a) speed, (b) mass, [2002], (c) charge, (d) magnetic induction, , 36. Magnitude of magnetic field (in SI units) at the centre of, a hexagonal shape coil of side 10 cm, 50 turns and, m0 I, is :, p, [Sep. 03, 2020 (I)], , carrying current I (Ampere) in units of, , (a), , 250 3 (b) 50 3 (c) 500 3 (d) 5 3, 37. A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the, a, magnetic fields due to the wire at distance, and 2a,, 3, respectively from the axis of the wire is: [9 Jan 2020, I], 2, 1, 3, (b) 2, (c), (d), 3, 2, 2, An electron gun is placed inside a long solenoid of radius, R on its axis. The solenoid has n turns/length and carries, a current I. The electron gun shoots an electron along the, radius of the solenoid with speed v. If the electron does, not hit the surface of the solenoid, maximum possible value, of v is (all symbols have their standard meaning):, , (a), , 38., , [9 Jan 2020, II], , Magnetic Field Lines,, TOPIC 2 Biot-Savart's, Law and Ampere's Circuital, 34. A charged particle going around in a circle can be considered to be a current loop. A particle of mass m carrying charge q is moving in a plane wit speed v under the, ®, , influence of magnetic field B . The magnetic moment, of this moving particle :, [Sep. 06, 2020 (II)], ®, , (a), , mv 2 B, 2 B2, , ®, , (b) -, , 35., , 2 pB 2, mv 2 B, , mv 2 B, , (d) 2 B2, B2, A wire A, bent in the shape of an arc of a circle, carrying a, current of 2 A and having radius 2 cm and another wire B,, also bent in the shape of arc of a circle, carrying a current, of 3 A and having radius of 4 cm, are placed as shown in, the figure. The ratio of the magnetic fields due to the wires, A and B at the common centre O is :, [Sep. 04, 2020 (I)], A, O, , (a) 4 : 6, (c) 2 : 5, , 39., , (b), , em0 nIR, 2m, , em0 nIR, 2em 0 nIR, (d), 4m, m, A very long wire ABDMNDC is shown in figure carrying, current I. AB and BC parts are straight, long and at, right angle. At D wire forms a circular turn DMND of, radius R., AB, BC parts are tangential to circular turn at N and D., Magnetic field at the centre of circle is:, [8 Jan 2020, II], , (a), , m0 I æ, 1 ö, p+, ÷, 2 pR çè, 2ø, , (b), , m0 I æ, 1 ö, p÷, 2 pR çè, 2ø, , (c), , m0 I, (p + 1), 2pR, , (d), , m0 I, 2R, , B, , 90°, 60°, (b) 6 : 4, (d) 6 : 5, , em0 nIR, m, , (c), , mv 2 B, , ®, , ®, , (c) -, , (a)
Page 325 :
P-316, , 45. As shown in the figure, two infinitely long, identical wires, are bent by 90º and placed in such a way that the segments, LP and QM are along the x-axis, while segments PS and, QN are parallel to the y-axis. If OP = OQ = 4 cm, and the, magnitude of the magneticf field at O is 10–4 T, and the, two wires carry equal currents (see figure), the magnitude, of the current in each wire and the direction of the magnetic, field at O will be (µ0 = 4p × 10–7 NA–2): [12 Jan 2019, I], , S, These wires carry currents of equal magnitude I, whose, directions are shown in the figure. The net magnetic field, at point P will be :, [12 April 2019, I], (a) Zero, , (b), , –, , m0 I, ( xˆ + yˆ ), 2 pd, , +m 0 I, m0 I, ( zˆ ), ( xˆ + yˆ ), (d), pd, 2 pd, A thin ring of 10 cm radius carries a uniformly distributed, charge. The ring rotates at a constant angular speed of 40, À rad s–1 about its axis, perpendicular to its plane. If the, magnetic field at its centre is 3.8 × 10–9 T, then the charge, carried by the ring is close to (µ0 = 4p × 10–7 N/A2)., [12 April 2019, I], –6, –5, (a) 2×10 C (b) 3×10 C (c) 4×10–5C (d) 7×10–6C, Find the magnetic field at point P due to a straight line, segment AB of length 6 cm carrying a current of 5 A. (See, figure) (mo=4p×10–7 N-A–2), [12 April 2019, II], , (c), , 41., , OQ, P, , L, , m, 3c, , °, , Q, , 2 cm, , 2 cm, , R, , P, , 43., , 44., , (a) 2.0×10–5T, (b) 1.5×10–5T, (c) 3.0×10–5T, (d) 2.5×10–5T, The magnitude of the magnetic field at the center of an, equilateral triangular loop of side 1 m which is carrying a, current of 10 A is :, [10 April 2019, II], [Take mo = 4p×10–7 NA–2], (a) 18 mT (b) 9 mT, (c) 3 mT, (d) 1 mT, A square loop is carrying a steady current I and the, magnitude of its magnetic dipole moment is m. If this, square loop is changed to a circular loop and it carries the, same current, the magnitude of the magnetic dipole moment, of circular loop will be :, [10 April 2019, II], (a), , m, p, , (b), , 3m, p, , (c), , 2m, p, , (d), , 4m, p, , x, , M, , N, (a) 20 A, perpendicular out of the page, (b) 40 A, perpendicular out of the page, (c) 20 A, perpendicular into the page, (d) 40 A, perpendicular into the page, 46. A current loop, having two circular arcs joined by two, radial lines is shown in the figure. It carries a current of 10, A. The magnetic field at point O will be close to:, [9 Jan. 2019 I], O, 3 cm, , 42., , y, , 5, , Two very long, straight, and insulated wires are kept at, 90° angle from each other in xy-plane as shown in the, figure., , q =4, , 40., , Physics, , S, , i = 10A, (a) 1.0 × 10–7 T, (b) 1.5 × 10–7 T, –5, (c) 1.5 × 10 T, (d) 1.0 × 10–5 T, 47. One of the two identical conducting wires of length L is, bent in the form of a circular loop and the other one into, a circular coil of N identical turns. If the same current is, passed in both, the ratio of the magnetic field at the, central of the loop (B1) to that at the centre of the coil, B, (BC), i.e., L will be:, [9 Jan 2019, II], BC, , (a) N, , (b), , (c) N2, , (d), , 1, N, 1, , N2, 48. The dipole moment of a circular loop carrying a current I,, is m and the magnetic field at the centre of the loop is B1., When the dipole moment is doubled by keeping the current
Page 326 :
P-317, , Moving Charges and Magnetism, , current constant, the magnetic field at the centre of the, B, loop is B2. The ratio 1 is:, [2018], B2, 1, 2, A Helmholtz coil has pair of loops, each with N turns and, radius R. They are placed coaxially at distance R and the, same current I flows through the loops in the same, direction. The magnitude of magnetic field at P, midway, between the centres A and C, is given by (Refer to figure):, , (a) 2, 49., , (b), , (c), , 3, , (d), , 2, , [Online April 15, 2018], , 2R, , A, , C, , P, , R, , (a), , 50., , 4 N m0 I, , 8 N m0 I, , 8 N m0 I, , (b), (c), (d), 53/ 2 R, 53/ 2 R, 51/ 2 R, 51/ 2 R, A current of 1A is flowing on the sides of an equilateral, triangle of side 4.5 × 10–2m . The magnetic field at the, centre of the triangle will be:, [Online April 15, 2018], (a) 4 × 10–5Wb/m2, , 51., , 4 N m0 I, , (b) Zero, , (c) 2 × 10–5Wb/m2, (d) 8 × 10–5Wb/m2, Two identical wires A and B, each of length 'l', carry the, same current I. Wire A is bent into a circle of radius R, and wire B is bent to form a square of side 'a'. If BA and, BB are the values of magnetic field at the centres of the, B, circle and square respectively, then the ratio A is:, BB, [2016], p2, p2, p2, p2, (b), (c), (d), 16, 8, 8 2, 16 2, Two long current carrying thin wires, both with current I,, are held by insulating threads of length L and are in, equilibrium as shown in the figure, with threads making, an angle 'q' with the vertical. If wires have mass l per unit, length then the value of I is :, (g = gravitational acceleration), [2015], , (a), , 52., , (a), (b), (c), , 2, , pgL, tan q, µ0, , plgL, tan q, µ0, , sin q, , (d) 2sin q, , L, q, , plgL, µ0 cos q, plgL, µ0 cos q, , I, , I, , 53. Consider two thin identical conducting wires covered, with very thin insulating material. One of the wires is, bent into a loop and produces magnetic field B1, at its, centre when a current I passes through it. The ratio B1 :, B2 is:, [Online April 12, 2014], (a) 1 : 1, (b) 1 : 3, (c) 1 : 9, (d) 9 : 1, 54. A parallel plate capacitor of area 60 cm2 and separation 3, mm is charged initially to 90 mC. If the medium between the, plate gets slightly conducting and the plate loses the charge, initially at the rate of 2.5 × 10–8 C/s, then what is the magnetic, field between the plates ?, [Online April 23, 2013], –8, (a) 2.5 × 10 T, (b) 2.0 × 10–7 T, (c) 1.63 × 10–11 T (d), Zero, 55. A current i is flowing in a straight conductor of length L., The magnetic induction at a point on its axis at a distance, L, from its centre will be :, 4, , [Online April 22, 2013], , (a) Zero, , (b), , m 0i, , (d), , (c), , m0i, 2pL, , 4m0i, , 2L, 5pL, 56. Choose the correct sketch of the magnetic field lines of a, circular current loop shown by the dot e and the cross, [Online April 22, 2013], Ä., , (a), , (b), , (c), , (d), , 57. An electric current is flowing through a circular coil of, radius R. The ratio of the magnetic field at the centre of, the coil and that at a distance 2 2R from the centre of, the coil and on its axis is :, [Online April 9, 2013], (a) 2 2, (b) 27, (c) 36, (d) 8, 58. A charge Q is uniformly distributed over the surface of, non-conducting disc of radius R. The disc rotates about, an axis perpendicular to its plane and passing through its, centre with an angular velocity w. As a result of this, rotation a magnetic field of induction B is obtained at, the centre of the disc. If we keep both the amount of charge, placed on the disc and its angular velocity to be constant, and vary the radius of the disc then the variation of the, magnetic induction at the centre of the disc will be, represented by the figure :, [2012]
Page 327 :
P-318, , Physics, , B, , B, , (a), , (b), R, , R, , B, , B, , (c), 59., , R, , R, , A current I flows in an infinitely long wire with cross section, in the form of a semi-circular ring of radius R. The magnitude, of the magnetic induction along its axis is:, [2011], (a), , 60., , (d), , m0 I, , 2p2 R, , m0 I, 2pR, , (b), , (c), , m0 I, 4pR, , (d), , m0 I, , p2R, , Two long parallel wires are at a distance 2d apart. They, carry steady equal currents flowing out of the plane of, the paper as shown. The variation of the magnetic field B, along the line XX' is given by, [2010], B, , X¢, , (a) X, d, , X¢, d, , d, , X¢, , X, , d, , d, , X¢, , X, , d, , 61., , i, . The value of the magnetic field at its centre is, 3, [2006], (a) 1.05 × 10–2 Weber/m2 (b) 1.05 × 10–5 Weber/m2, (c) 1.05 × 10–3 Weber/m2 (d) 1.05 × 10–4 Weber/m2, 66. Two concentric coils each of radius equal to 2 p cm are, placed at right angles to each other. 3 ampere and 4 ampere are the currents flowing in each coil respectively., The magnetic induction in Weber/m2 at the centre of the, , coils will be (m0 = 4p ´10-7 Wb / A.m), , B, , (d), , ), , 1, , a current, , B, , (c), , (, , (b), , m 0 æ I1 + I 2 ö 2, ç, ÷, 2p è d ø, , m0, m0, (d), I 2 + I 22 2, ( I1 + I 2 ), 2 pd, 2pd 1, 65. A long solenoid has 200 turns per cm and carries a current, i. The magnetic field at its centre is 6.28 × 10–2 Weber/m2., Another long solenoid has 100 turns per cm and it carries, , d, , X, , (a), , 1, , m0, ( I12 + I 2 2 ), 2pd, , (c), , B, , (b), , (a) 2.5 × 10–7 T southward, (b) 5 × 10–6 T northward, (c) 5 × 10–6 T southward, (d) 2.5 × 10–7 T northward, 62. A long straight wire of radius a carries a steady current i., The current is uniformly distributed across its cross, section. The ratio of the magnetic field at a/2 and 2a is, [2007], (a) 1/2, (b) 1/4, (c) 4, (d) 1, 63. A current I flows along the length of an infinitely long,, straight, thin walled pipe. Then, [2007], (a) the magnetic field at all points inside the pipe is the, same, but not zero, (b) the magnetic field is zero only on the axis of the, pipe, (c) the magnetic field is different at different points, inside the pipe, (d) the magnetic field at any point inside the pipe is zero, 64. Two identical conducting wires AOB and COD are placed, at right angles to each other. The wire AOB carries an, electric current I1 and COD carries a current I2. The, magnetic field on a point lying at a distance d from O, in, a direction perpendicular to the plane of the wires AOB, and COD, will be given by, [2007], , d, , A horizontal overhead powerline is at height of 4m from, the ground and carries a current of 100A from east to west., The magnetic field directly below it on the ground is, (m0 = 4p×10 –7 Tm A–1), [2008], , (a) 10 -5, , [2005], , (b) 12 ´ 10 -5, , (c) 7 ´ 10 -5, (d) 5 ´ 10 -5, 67. A current i ampere flows along an infinitely long straight, thin walled tube, then the magnetic induction at any point, inside the tube is, [2004], m 0 2i, ., tesla, (a), (b) zero, 4p r, 2i, tesla, (c) infinite, (d), r
Page 328 :
P-319, , Moving Charges and Magnetism, , 68. A long wire carries a steady current. It is bent into a circle, of one turn and the magnetic field at the centre of the coil, is B. It is then bent into a circular loop of n turns. The, magnetic field at the centre of the coil will be, [2004], (a) 2n B, (b) n2 B, (c) nB, (d) 2 n2 B, 69. The magnetic field due to a current carrying circular loop, of radius 3 cm at a point on the axis at a distance of 4 cm, from the centre is 54 mT. What will be its value at the, centre of loop?, [2004], (a) 125 mT (b) 150 mT (c) 250 mT (d) 75 mT, 70. If in a circular coil A of radius R, current I is flowing and in, another coil B of radius 2R a current 2I is flowing, then the, ratio of the magnetic fields BA and BB, produced by them, will be, [2002], (a) 1, (b) 2, (c) 1/2, (d) 4, , TOPIC 3, 71., , A square loop of side 2a and carrying current I is kept is xz, plane with its centre at origin. A long wire carrying the, same current I is placed parallel to z-axis and passing, through point (0, b, 0), (b >> a). The magnitude of torque, on the loop about z-axis will be : [Sep. 06, 2020 (II)], (a), , (a), , 73., , 2m 0 I 2 a 2, pb, , (b), , m 0 I2 a 2b, , 2m 0 I 2 a 2 b, p(a 2 + b2 ), , m 0 I2 a 2, 2 p (a + b ), 2 pb, A square loop of side 2a, and carrying current I, is kept in, XZ plane with its centre at origin. A long wire carrying the, same current I is placed parallel to the z-axis and passing, through the point (0, b, 0), (b >> a). The magnitude of the, torque on the loop about z-axis is given by :, [Sep. 05, 2020 (I)], , (c), , 72., , Force and Torque on Current, Carrying Conductor, , 2, , m0 I 2a2, 2 pb, , 2, , æ ˆj, kˆ ö, abI , along ç, +, ÷, ç 2, 2 ÷ø, è, æ ˆj, kˆ ö, (b), 2abI , along ç, +, ÷, ç 2, 2 ÷ø, è, æ ˆj, 2kˆ ö, (c), 2abI , along ç, +, ÷, ç 5, 5 ÷ø, è, æ ˆj, 2kˆ ö, (d) abI , along çç, +, ÷, 5 ÷ø, è 5, 74. A small circular loop of conducting wire has radius a and, carries current I. It is placed in a uniform magnetic field B, perpendicular to its plane such that when rotated slightly, about its diameter and released, it starts performing simple, harmonic motion of time period T. If the mass of the loop is m, then :, [9 Jan 2020, II], , (a), , (a) T =, , 2m, IB, , (b) T =, , pm, 2 IB, , 2pm, pm, (c) T =, IB, IB, 75. Two wires A & B are carrying currents I1 and I2 as shown, in the figure. The separation between them is d. A third, wire C carrying a current I is to be kept parallel to them, at a distance x from A such that the net force acting on it, is zero. The possible values of x are : [10 April 2019, I], , (c) T =, , (d), , (b), , (a), , æ I ö, I2, x = ç 1 ÷ d and x =, d, (I1 + I2 ), è I1 - I2 ø, , (b), , æ I2 ö, æ I2 ö, x=ç, d and x = ç, d, ÷, è (I1 + I2 ) ø, è (I1 - I2 ) ÷ø, , m 0 I 2 a3, 2pb2, , 2m 0 I 2 a 3, 2m 0 I 2 a 2, (c), (d), pb, pb2, A wire carrying current I is bent in the shape ABCDEFA as, shown, where rectangle ABCDA and ADEFA are, perpendicular to each other. If the sides of the rectangles, are of lengths a and b, then the magnitude and direction of, magnetic moment of the loop ABCDEFA is :, [Sep. 02, 2020 (II)], Z, E, I, I, F, C, D, Y, O, b, A a, B, X, , æ I1 ö, æ I2 ö, (c) x = çè (I + I ) ÷ø d and x = çè (I - I ) ÷ø d, 1, 2, 1, 2, , (d), , x=±, , I1d, (I1 - I2 ), , 76. A rectangular coil (Dimension 5 cm × 2.5 cm) with 100, turns, carrying a current of 3 A in the clock-wise, direction, is kept centered at the origin and in the X-Z, plane. A magnetic field of 1 T is applied along X-axis. If, the coil is tilted through 45° about Z-axis, then the torque, on the coil is:, [9 April 2019 I], (a) 0.38 Nm, (b) 0.55 Nm, (c) 0.42 Nm, (d) 0.27 Nm
Page 329 :
P-320, , 77., , Physics, , A rigid square of loop of side ‘a’ and carrying current I2 is, lying on a horizontal surface near a long current I1 carrying, wire in the same plane as shown in figure. The net force, on the loop due to the wire will be:, [9 April 2019 I], I2, , I1, , cm carries a current 1 of 12 A. Out of the following different orientations which one corresponds to stable equilibrium ?, [Online April 9, 2017], Z, Z, , (a), a, , X, , a, , 78., , m II, (a) Repulsive and equal to o 1 2, 2p, mo I1I 2, (b) Attractive and equal to, 3p, mo I1I 2, (c) Repulsive and equal to, 4p, (d) Zero, A circular coil having N turns and radius r carries a current, I. It is held in the XZ plane in a magnetic field B. The, torque on the coil due to the magnetic field is :, [8 April 2019 I], , (a), , Br 2 I, pN, , (b) Bpr2I N, , Bpr 2 I, (d) Zero, N, An infinitely long current carrying wire and a small current, carrying loop are in the plane of the paper as shown. The, redius of the loop is a and distance of its centre from the, wire is d (d>>a). If the loop applies a force F on the wire, then:, [9 Jan. 2019 I], , (c), 79., , B, , Bd c, I, I, a b, , c, I, , Y (b), , Z, B, , B, (c), , b, , a, , I d, I c, , Y (d), , b, , I, I, , I, , y, , I, , x, , d, , z, B, , 2, æ a2 ö, æ aö, (c) F µ ç 3 ÷, (d) F µ ç ÷, è dø, èd ø, 80. A charge q is spread uniformly over an insulated loop of, radius r . If it is rotated with an angular velocity w with, respect to normal axis then the magnetic moment of the, loop is, [Online April 16, 2018], , 81., , 1, 4, 3, qwr 2 (b), qwr 2 (c), qwr 2 (d) qwr 2, (a), 2, 3, 2, A uniform magnetic field B of 0.3 T is along the positive Zdirection. A rectangular loop (abcd) of sides 10 cm × 5, , I d, I c, , Y, , 83. A rectangular loop of sides 10 cm and 5 cm carrying a, current 1 of 12 A is placed in different orientations as shown, in the figures below :, z, , (A), , æ aö, (b) F µ çè ÷ø, d, , a, , X, X, 82. Two coaxial solenoids, of, different, radius, carry current I in, uur, force, on the inner, the same direction. F1 be the magnetic, uur, solenoid due to the outer one and F2 be the magnetic, force on the outer solenoid due to the inner one. Then :, [2015], uur, uur, (a) F1 is radially inwards and F2 = 0, uur, uur, (b) F1 is radially outwards and F2 = 0, uur uur, (c) F1 = F2 = 0, uur, uur, (d) F1 is radially inwards and F2 is radially outwards, , B, , (a) F = 0, , Y, , b, , X, , Z, , d, I, a, , (B), , I, , I, I, z, , x, , y, , B, , I, (C), , I, , I, , I, , y, , I, , x, , z, B, , (D), , I, , I, x, , I, , I, , y
Page 330 :
P-321, , Moving Charges and Magnetism, , 84., , If there is a uniform magnetic field of 0.3 T in the positive, z direction, in which orientations the loop would be in (i), stable equilibrium and (ii) unstable equilibrium ?, [2015], (a) (B) and (D), respectively, (b) (B) and (C), respectively, (c) (A) and (B), respectively, (d) (A) and (C), respectively, Two long straight parallel wires, carrying (adjustable), current I1 and I2, are kept at a distance d apart. If the force, ‘F’ between the two wires is taken as ‘positive’ when the, wires repel each other and ‘negative’ when the wires, attract each other, the graph showing the dependence of, ‘F’, on the product I1 I2, would be :, [Online April 11, 2015], F, F, (a), , O, , I1I2, , (b), , O, , 85., , O, I1I2, , (b), , (c) IBR, (d), 86., , (d), , O, , q0, , y, , B, , x, –1.5, , (a) 1.57 W (b) 2.97 W (c) 14.85 W (d) 29.7 W, 87. Three straight parallel current carrying conductors are, shown in the figure. The force experienced by the middle, conductor of length 25 cm is: [Online April 11, 2014], I1 = 30 A, , I2 = 20 A, , 5 cm, , I1I2, , Q, , P, , I, , 2.0, , 3 cm, , A wire carrying current I is tied between points P and, Q and is in the shape of a circular arc of radius R due, to a uniform magnetic field B (perpendicular to the plane, of the paper, shown by xxx) in the vicinity of the wire., If the wire subtends an angle 2q0 at the centre of the, circle (of which it forms an arc) then the tension in the, wire is :, [Online April 11, 2015], IBR, (a), B, 2sin q0, , IBRq0, sin q0, , 1.5, , F, , F, (c), , I1I2, , z, , I = 10 A, , 10–4, , (a) 3 ×, N toward right, –4, (b) 6 × 10 N toward right, (c) 9 × 10–4 N toward right, (d) Zero, 88. A rectangular loop of wire, supporting a mass m, hangs, r, with one end in a uniform magnetic field B pointing out, of the plane of the paper. A clockwise current is set up, such that i > mg/Ba, where a is the width of the loop., Then :, [Online April 23, 2013], , R, , IBR, sin q0, , A conductor lies along the z-axis at -1.5 £ z < 1.5 m and, carries a fixed current of 10.0 A in -â z direction (see figure)., r, For a field B = 3.0 ´10-4 e-0.2x aˆ y T, find the power required, to move the conductor at constant speed to x = 2.0 m,, y = 0 m in 5 ´10-3 s. Assume parallel motion along the, x-axis., [2014], , Q, , P, , y, , Ii, S, , x, , a, mg, , R, , (a) The weight rises due to a vertical force caused by, the magnetic field and work is done on the system., (b) The weight do not rise due to vertical for caused by, the magnetic field and work is done on the system.
Page 331 :
P-322, , 89., , 90., , Physics, , (c) The weight rises due to a vertical force caused by the, magnetic field but no work is done on the system., (d) The weight rises due to a vertical force caused by, the magnetic field and work is extracted from the, magnetic field., Currents of a 10 ampere and 2 ampere are passed through, two parallel thin wires A and B respectively in opposite, directions. Wire A is infinitely long and the length of the, wire B is 2 m. The force acting on the conductor B, which, is situated at 10 cm distance from A will be, [Online May 26, 2012], (a) 8 × 10–5 N, (b) 5 × 10–5 N, (c) 8p × 10–7 N, (d) 4p × 10–7 N, The circuit in figure consists of wires at the top and bottom, and identical springs as the left and right sides. The wire, at the bottom has a mass of 10 g and is 5 cm long. The wire, is hanging as shown in the figure. The springs stretch 0.5, cm under the weight of the wire and the circuit has a total, resistance of 12 W. When the lower wire is subjected to a, static magnetic field, the springs, stretch an additional, 0.3 cm. The magnetic field is, [Online May 12, 2012], , mo I é b - a ù, 4p êë ab úû, mo I, [2(b - a ) + p / 3(a + b)], (c), 4p, (d) zero, 92. Due to the presence of the current I1 at the origin:, (a) The forces on AD and BC are zero., (b) The magnitude of the net force on the loop is given, (b), , I1 I, mo [2(b - a ) + p / 3(a + b] ., 4p, (c) The magnitude of the net force on the loop is given, , by, , m o II1, (b - a)., 24 ab, (d) The forces on AB and DC are zero., Two long conductors, separated by a distance d carry, current I1 and I2 in the same direction. They exert a force F, on each other. Now the current in one of them is increased, to two times and its direction is reversed. The distance is, also increased to 3d. The new value of the force between, them is, [2004], , by, , 93., , 24 V, , 2F, F, F, (b), (c) –2 F, (d) 3, 3, 3, If a current is passed through a spring then the spring, will, [2002], (a) expand, (b) compress, (c) remains same, (d) none of these, Wires 1 and 2 carrying currents i1 and i2 respectively are, inclined at an angle θ to each other. What is the force on, a small element dl of wire 2 at a distance of r from wire 1 (as, shown in figure) due to the magnetic field of wire 1? [2002], (a) m 0 i1i2 dl tan q, 1, 2, 2 pr, m0, i1, i1i2 dl sin q, (b), i2, r, 2pr, m0, dl, q, (c), i1i2 dl cos q, 2 pr, m0, (d), i1i2 dl sin q, 4 pr, , (a), Magnetic, field region, , 94., , 5 cm, , 95., (a) 0.6 T and directed out of page, (b) 1.2 T and directed into the plane of page, (c) 0.6 T and directed into the plane of page, (d) 1.2 T and directed out of page, Directions : Question numbers 91 and 92 are based on the, following paragraph., A current loop ABCD is held fixed on the plane of the paper as, shown in the figure. The arcs BC (radius = b) and DA (radius =, a) of the loop are joined by two straight wires AB and CD. A, steady current I is flowing in the loop. Angle made by AB and, CD at the origin O is 30°. Another straight thin wire with steady, current I1 flowing out of the plane of the paper is kept at the, origin., [2009], B, a, I1, , A, 30°, , O, , I, , D, b, , 91., , C, , The magnitude of the magnetic field (B) due to the loop, ABCD at the origin (O) is :, (a), , m o I (b - a ), 24ab, , -, , Galvanometer and its, TOPIC 4 Conversion into Ammeter and, Voltmeter, 96. A galvanometer of resistance G is converted into a voltmeter of ragne 0 – 1V by connecting a resistance R1 in, series with it. The additional resistance R1 in series with, it. The additional resistance that should be connected in, series with R1 to increase the range of the voltmeter to 0, – 2V will be :, [Sep. 05, 2020 (I)], (a) G, (b) R1, (c) R1 – G, (d) R1 + G
Page 332 :
P-323, , Moving Charges and Magnetism, , 97. A galvanometer is used in laboratory for detecting the, null point in electrical experiments. If, on passing a, current of 6 mA it produces a deflection of 2°, its figure, of merit is close to :, [Sep. 05, 2020 (II)], (a) 333° A/div. (b), 6 × 10–3 A/div., (c) 666° A/div. (d), 3 × 10–3 A/div., 98. A galvanometer coil has 500 turns and each turn has an, average area of 3 × 10–4 m2. If a torque of 1.5 Nm is required, to keep this coil parallel to a magnetic field when a current, of 0.5 A is flowing through it, the strength of the field (in T), is __________., [NA Sep. 03, 2020 (II)], 99. A galvanometer of resistance 100 W has 50 divisions on, its scale and has sensitivity of 20 µA/division. It is to be, converted to a voltmeter with three ranges, of 0–2V, 0–10, V and 0–20 V. The appropriate circuit to do so is :, [12 April 2019, I], , 102., , 103., , 104., (a), , (b), 105., (c), , 0 – 5 V. Therefore the value of shunt resistance required, to convert the above galvanometer into an ammeter of range, 0 – 10 mA is :, [10 April 2019, I], (a) 500 W (b) 100 W (c) 200 W (d) 10 W, A moving coil galvanometer has resistance 50 W and it, indicates full deflection at 4 mA current. A voltmeter is, made using this galvanometer and a 5 k W resistance. The, maximum voltage, that can be measured using this, voltmeter, will be close to:, [9 April 2019 I], (a) 40 V, (b) 15 V, (c) 20 V, (d), 10 V, A moving coil galvanometer has a coil with 175 turns and, area 1 cm2. It uses a torsion band of torsion constant 10–, 6 N-m/rad. The coil is placed in a magnetic field B parallel, to its plane. The coil deflects by 1° for a current of 1mA., The value of B (in Tesla) is approximately:, [9 April 2019, II], (a) 10–4, (b) 10–2, (c) 10–1, (c) 10–3, The resistance of a galvanometer is 50 ohm and the, maximum current which can be passed through it is 0.002, A. What resistance must be connected to it order to, convert it into an ammeter of range 0 – 0.5 A?, [9 April 2019, II], (a) 0.5 ohm, (b) 0.002 ohm, (c) 0.02 ohm, (d) 0.2 ohm, The galvanometer deflection, when key K1 is closed but, K2 is open, equals θ 0 (see figure). On closing K2 also, and adjusting R2 to 5W, the deflection in galvanometer, θ0, . The resistance of the galvanometer is, then,, 5, given by [Neglect the internal resistance of battery]:, [12 Jan 2019, I], , becomes, , (d), 100. A moving coil galvanometer, having a resistance G,, produces full scale deflection when a current Ig flows, through it. This galvanometer can be converted into (i), an ammeter of range 0 to I0 (I0 > Ig) by connecting a shunt, resistance RA to it and (ii) into a voltmeter of range 0 to V, (V=GI0) by connecting a series resistance Rv to it. Then,, [12 April 2019, II], ö, RA æ I g, =ç, ÷÷ and, RV çè I 0 - I g, ø, , (a), , æ I0 - I g, R A RV = G ç, ç Ig, è, , (b), , æ Ig, R, RA RV = G and A = çç, RV è I 0 - I g, , (c), , æ Ig, R A RV = G ç, ç I0 - I g, è, , (d), , 2, , 2, , R1 = 220 W, , G, , 2, , ö, ÷÷, ø, , 2, , K1, , ö, ÷÷, ø, , ö, RA æ I 0 - I g, =ç, ÷÷ and, RV çè I g, ø, Ig, RA, =, RA RV = G 2 and, RV ( I0 - I g ), 2, , R2, , K2, , 2, , ö, ÷÷, ø, , 101. A moving coil galvanometer allows a full scale current, of 10– 4 A. A series resistance of 2 MW is required to, convert the above galvanometer into a voltmeter of range, , (a) 5W, (b) 22W, (c) 25W, (d) 12W, 106. A galvanometer, whose resistance is 50 ohm, has 25, divisions in it. When a current of 4 × 10–4 A passes through, it, its needle (pointer) deflects by one division. To use this, galvanometer as a voltmeter of range 2.5 V, it should be, connected to a resistance of :, [12 Jan 2019, II], (a) 250 ohm, (b) 200 ohm, (c) 6200 ohm, (d) 6250 ohm
Page 333 :
P-324, , Physics, , 107. A galvanometer having a resistance of 20 W and 30, division on both sides has figure of merit 0.005 ampere/, division. The resistance that should be connected in series, such that it can be used as a voltmeter upto 15 volt, is:, [11 Jan 2019, II], (a) 100 W (b) 120 W (c) 80 W, (d) 125 W, 108. A galvanometer having a coil resistance 100 W gives a full, scale deflection when a current of 1 mA is passed through, it. What is the value of the resistance which can convert, this galvanometer into a voltmeter giving full scale, deflection for a potential difference of 10 V?, [8 Jan 2019, II], (a) 10 kW, (b) 8.9 kW, (c) 7.9 kW, (d) 9.9 kW, 109. In a circuit for finding the resistance of a galvanometer by, half deflection method, a 6 V battery and a high resistance, of 11kW are used. The figure of merit of the galvanometer, 60mA/division. In the absence of shunt resistance, the, galvanometer produces a deflection of q = 9 divisions when, current flows in the circuit. The value of the shunt, resistance that can cause the deflection of q/2 , is closest, to, [Online April 16, 2018], (a) 55W, (b) 110W, (c) 220W (d) 550W, 110. A galvanometer with its coil resistance 25W requires a, current of 1mA for its full deflection. In order to construct, an ammeter to read up to a current of 2A, the approximate, value of the shunt resistance should be, [Online April 16, 2018], (a) 2.5 × 10–2W, (b) 1.25 × 10–3W, (c) 2.5 × 10–3W, (d) 1.25 × 10–2W, 111. When a current of 5 mA is passed through a galvanometer, having a coil of resistance 15 W , it shows full scale, deflection. The value of the resistance to be put in series, with the galvanometer to convert it into to voltmeter of, range 0 - 10 V is, [2017], (a), , 2.535 ´10 W, 3, , (b), , 114. To know the resistance G of a galvanometer by half, deflection method, a battery of emf VE and resistance R, is used to deflect the galvanometer by angle q. If a shunt, of resistance S is needed to get half deflection then G, R, and S related by the equation:, [Online April 9, 2016], (a) S (R + G) = RG, (b) 2S (R + G) = RG, (c) 2G = S, (d) 2S = G, 115. The AC voltage across a resistance can be measured, using a :, [Online April 11, 2015], (a) hot wire voltmeter, (b) moving coil galvanometer, (c) potential coil galvanometer, (d) moving magnet galvanometer, 116. In the circuit diagrams (A, B, C and D) shown below, R is, a high resistance and S is a resistance of the order of, galvanometer r esistance G. The correct circuit,, corresponding to the half deflection method for finding, the resistance and figure of merit of the galvanometer, is, the circuit labelled as:, [Online April 11, 2014], K2, , R, S, , G, , (A), K1, , K2, , R, G, S, , (B), , 4.005 ´10 W, 3, , (c) 1.985 ´ 10 W, (d) 2.045 ´ 103 W, 112. A galvanometer having a coil resistance of 100 W gives a, full scale deflection, when a currect of 1 mA is passed, through it. The value of the resistance, which can convert, this galvanometer into ammeter giving a full scale deflection, for a current of 10 A, is :, [2016], (a) 0.1 W (b) 3W, (c) 0.01W (d) 2W, 113. A 50 W resistance is connected to a battery of 5V. A, galvanometer of resistance 100 W is to be used as an, ammeter to measure current through the resistance, for, this a resistance rs is connected to the galvanometer. Which, of the following connections should be employed if the, measured current is within 1% of the current without the, ammeter in the circuit ?, [Online April 9, 2016], (a) rs = 0.5 W in series with the galvanometer, (b) rs = 1 W in series with galvanometer, (c) rs = 1W in parallel with galvanometer, (d) rs = 0.5 W in parallel with the galvanometer., 3, , K1, , K2, , S, G, R, , (C), K1, S, R, , G, , (D), , K1, , K2
Page 334 :
P-325, , Moving Charges and Magnetism, , (a) Circuit A with G =, , RS, , Ii, , ( R - S), , R1, , (b) Circuit B with G = S, (c) Circuit C with G = S, (d) Circuit D with G =, , R2, , G, , RS, ( R - S), , 117. This questions has Statement I and Statement II. Of the, four choices given after the Statements, choose the one, that best describes into two Statements., Statement-I : Higher the range, greater is the resistance of, ammeter., Statement-II : To increase the range of ammeter, additional, shunt needs to be used across it., [2013], (a) Statement-I is true, Statement-II is true, Statement-II, is the correct explanation of Statement-I., (b) Statement-I is true, Statement-II is true, Statement-II, is not the correct explanation of Statement-I., , R3, , (a) 107 W (b) 137 W (c) 107/2 W (d) 77 W, 119. A shunt of resistance 1 W is connected across a, galvanometer of 120 W resistance. A current of 5.5 ampere, gives full scale deflection in the galvanometer. The current, that will give full scale deflection in the absence of the, shunt is nearly :, [Online April 9, 2013], (a) 5.5 ampere, (b) 0.5 ampere, (c) 0.004 ampere, (d) 0.045 ampere, 120. In the circuit , the galvanometer G shows zero deflection., If the batteries A and B have negligible internal resistance,, the value of the resistor R will be [2005], 500 W, G, , (c) Statement-I is true, Statement-II is false., (d) Statement-I is false, Statement-II is true., 118. To find the resistance of a galvanometer by the half, deflection method the following circuit is used with, resistances R1= 9970 W, R2 = 30 W and R3 = 0. The, deflection in the galvanometer is d. With R3 = 107 W the, deflection changed to, , d, . The galvanometer resistance is, 2, , approximately :, [Online April 22, 2013], , 2V, 12V, , B, , R, , A, , (a) 100 W (b) 200W (c) 1000 W (d) 500 W, 121. A moving coil galvanometer has 150 equal divisions. Its, current sensitivity is 10-divisions per milliampere and, voltage sensitivity is 2 divisions per millivolt. In order that, each division reads 1 volt, the resistance in ohms needed, to be connected in series with the coil will be [2005], (a) 105, (b) 103, (c) 9995, (d) 99995
Page 335 :
P-326, , 1., , Physics, , r, (c) E = 300 ˆj V/cm = 3 ´ 104 V/m, r, V = 6 ´ 106 iˆ, E, , 4., , \ Pitch = (V cos q), , y, E = 300 j, z, 5., , r, B must be in +z axis., r, r r, qE + qV ´ B = 0, , 2pm, qB, , 2pm, qB, , = (4 ´ 105 cos 60°), , x, , V/cm = 3– ´ 104 V/m, e, V, V = 6 ´ 106 i$, , 2p æ 1.67 ´10-27, ç, 0.3 çè 1.69 ´10-19, , E 3 ´ 104, =, = 5 ´ 10-3 T, V 6 ´ 106, Hence, magnetic field B = 5 × 10–3 T along +z direction., (c) In uniform magnetic field particle moves in a circular, path, if the radius of the circular path is 'r', particle will not, hit the screen., \ B=, , (c) Time period of one revolution of proton, T =, , \ Length of region, l = 10 ´ (v cos q)T, Þ l = 10 ´ v cos 60° ´, , Þl=, , 2pm, qB, , 20pmv 20 ´ 3.14 ´ 1.67 ´ 10 -27 ´ 4 ´ 105, =, qB, 1.6 ´ 10 -19 ´ 0.3, , Þ l = 0.44 m, d, , 6., r=, , mv, qB0, , é mv, ù, = qvB0 ú, êQ, r, ë, û, , (a), , 2, , Hence, minimum value of v for which the particle will not, hit the screen., qB0 d, m, (a) [Given: q = 1mC = 1 ´ 10-6 C;, r, V = (2iˆ + 3 ˆj + 4kˆ) m/s and, r, B = (5iˆ + 3 ˆj - 6kˆ) ×10 –3 T ], , As we know, magnetic force F = qvB = ma, , r æ qvB ö, \ a =ç, è m ÷ø perpendicular to velocity.., , v=, , 3., , iˆ ˆj, r, r r, -6, -3, F = q(V ´ B) = 10 ´ 10 2 3, , \ Also v =, , 5 3 -6, , = (-30iˆ + 32 ˆj - 9kˆ) ´ 10-9 N, r, \ F = ( -30iˆ + 32 ˆj - 9kˆ), , 2KE, 2 ´ e ´ 106, =, m, m, , \ a=, , qvB eB 2 ´ e ´ 106, =, m, m, m, , \ 1012, , æ 1.6 ´ 10 –19 ö 2, . 2 ´ 10 3 B, =ç, –27 ÷, ´, 1.67, 10, è, ø, , kˆ, 4, , ö, ÷÷ = 4 cm, ø, , Here, m = mass of proton, q = charge of proton, B = magnetic field., Linear distance travelled in one revolution,, p = T(v cos q) (Here, v = velocity of proton), , E = VB, , 2., , (d) Pitch = (v cos q)T and T =, , 3, , \B;, , 1, 2, , ´ 10–3 T = 0.71 mT (approx), , 2pm, qB
Page 338 :
P-329, , Moving Charges and Magnetism, , 24., 25., , 26., , E 7.7 ´ 103, (c) As velocity v = =, = 55 km/s, B, 0.14, (a) The charge experiences both electric and magnetic, force., Electric force, Fe = qE, r r, Magnetic force, Fm = q ( v ´ B ), ur, ur r ur, \ Net force, F = q é E + v ´ B ù, ë, û, é, iˆ ˆj kˆ ù, ê, ú, = q ê 3$i + $j + 2k$ + 3 4 1 ú, ê, 1 1 -3 ú, ë, û, $, $, $, ˆ, = q é3i + j + 2k + i ( -12 - 1) - $j ( -9 - 1) + k ( 3 - 4 )ù, ë, û, $, $, $, $, $, ˆ, é, ù, = q ë3i + j + 2k - 13i + 10 j - k û, = q éë -10i$ + 11$j + k$ ùû, Fy = 11qjˆ, , Thus, the y component of the force., (b) As velocity is not changing, charge particle must go, undeflected, then, qE = qvB, E, Þ v=, B, Also,, r r, E´B, E B sin q, =, 2, B, B2, , E, r, = |v| =v, B, B, (b) When a charged particle enters a magnetic field at a, direction perpendicular to the direction of motion, the path, of the motion is circular. In circular motion the direction of, velocity changes at every point (the magnitude remains, constant)., Therefore, the tangential momentum will change at every, point. But kinetic energy will remain constant as it is given, 1, by mv 2 and v2 is the square of the magnitude of velocity, 2, which does not change., (b) The charged particle will move along the lines of, electric field (and magnetic field). Magnetic field will exert, no force. The force by electric field will be along the lines, of uniform electric field. Hence the particle will move in a, straight line., (c) Equating magnetic force to centripetal force,, =, , 27., , 28., , 29., , E B sin 90°, 2, , 30. (b) Due to electric field, it experiences force and, accelerates i.e. its velocity decreases., 31. (b) The workdone, dW = Fds cosq, The angle between force and displacement is 90°., Therefore work done is zero., ×, , ×, , ×, , ×, , ×, , F, ×, , ×, , S, , 32. (a) When a moving charged particle is subjected to a, perpendicular magnetic field, then it describes a circular, path of radius., p, r=, qB, where q = Charge of the particle, p = Momentum of the particle, B = Magnetic field, Here p, q and B are constant for electron and proton, therefore, the radius will be same., 33. (a) The time period of a charged particle of charge q and, 2 pm, mass m moving in a magnetic field (B) is T =, qB, Clearly time period is independent of speed of the particle., 34. (d), , v, , =, , mv 2, = qvB sin 90º, r, Bqr, mv, Þ, = Bq Þ v =, r, m, Time to complete one revolution,, 2pr 2pm, =, T=, v, qB, , ×, , +, , +q, , Length of the circular path, l = 2pr, q, qv, =, T 2 pr, Magnetic moment M = Current × Area, , Current, i =, , = i ´ pr 2 =, M =, , qv, ´ pr 2, 2 pr, , 1, q×v×r, 2, , Radius of circular path in magnetic field, r =, , mv, qB, , 1, mv, mv 2, qv ´, ÞM =, 2, qB, 2B, r, r, Direction of M is opposite of B therefore, r, r - mv 2 B, M =, 2B 2, (By multiplying both numerator and denominator by B)., \M =
Page 345 :
P-336, , Physics, , or, Total force, Ftotal, , =, , m 0 Ia 2, 2d 2, , =, , m0 I a 2, 2(d 2 + a 2 ), , [Q d >> a ], , a2, Clearly Ftotal µ 2, d, 80., , (a) Magnetic moment, m = IA =, , qv, ( pr 2 ), 2 pr, , qrw 2, 1, (pr ) = qr 2 w, 2 pr, 2, (c) Magnetic moment of current carrying rectangular loop, of area A is given by M = NIA, magnetic moment of current carrying coil is a vector and, its direction is given by right hand thumb rule, for, r, rectangular loop, B at centre due to current in loop and, r, M are always parallel., , or, m =, , 81., , B, M, , B, M, Ä, Inwards, , e, Outwards, , 82., 83., 84., , 85., , Hence, (c) corresponds to stable equilibrium., r r, (c) F1 = F2 = 0, because of action and reaction pair, r r, (a) For stable equilibrium M || B, r, r, For unstable equilibrium M || (–B), (a) I1 I2 = Positive, (attract) F = Negative, I1 I2= Negative, (repell) F = Positive, Hence, option (a) is the correct answer., (c) For small arc length, 2T sin q = BIR 2 q, (As F = BIL and L = RZq), T = BIR, Q, , P, T, , 86., , 2, , 2, , 0, , 0, , 2, , -3, -0.2 x, dx, = 9 ´ 10 ò e, , l=3m, , 0, , 9 ´ 10, 0.2, , P, Q, Also given; length of wire Q, = 25 cm = 0.25 m, Force on wire Q due to wire R, FQR = 10–7 ´, , [- e-0.2´ 2 + 1], , R, , 2 ´ 20 ´ 10, ´ 0.25, 0.05, , = 20 × 10–5 N (Towards left), Force on wire Q due to wire P, FQP = 10–7 ×, , 2 ´ 30 ´ 10, ´ 0.25, 0.03, , = 50 × 10–5 N (Towards right), Hence, Fnet = FQP – FQR, = 50 × 10–5 N – 20 × 10–5 N, = 3 × 10–4 N towards right, 88. (c), 89. (a) Force acting on conductor B due to conductor A is, given by relation, F=, , m 0 I1 I 2 l, 2 pr, , \F=, , W = ò Fdx = ò 3.0 ´ 10 -4 e -0.2 x ´ 10 ´ 3dx, , =, , 5 cm, , Tsinq, , q, , (b) Work done in moving the conductor is,, , -3, , 3 cm, , l-length of conductor B, r-distance between two conductors, , T, , R, Tsinq, , 9 ´ 10-3, ´ [1 - e -0.4 ], 0.2, 2.97×10 –3, 9×10 –3 ´ (0.33), =, =, 2, 2, Power required to move the conductor is,, W, P=, t, 2.97 ´ 10-3, P=, = 2.97 W, (0.2) ´ 5 ´ 10-3, 87. (a) I1 = 30 A I = 10 A, I2 = 20 A, , =, , I = 10 A, , z, x, , 4p´ 10-7 ´ 10 ´ 2 ´ 2, = 8 × 10–5 N, 2 ´ p´ 0.1, , 90. (a), 91. (a) The magnetic field at O due to current in DA is, m I p, B1 = o ´ (directed vertically upwards), 4p a 6, The magnetic field at O due to current in BC is, m I p, B2 = o ´, (directed vertically downwards), 4p b 6, The magnetic field due to current AB and CD at O is zero., Therefore the net magnetic field is
Page 346 :
P-337, , Moving Charges and Magnetism, , 92., , 93., , 94., , 95., , B = B1 – B2 (directed vertically upwards), m I p mo I p, = o, ´, 4 p a 6 4p b 6, m I æ 1 1ö m I, = o ç - ÷ = o (b - a), 24 è a b ø 24ab, r, r r, (d) F = I ( l ´ B), The force on AD and BC due to current I1 is zero. This is, uur, because the directions of current element I d l and, r, magnetic field B are parallel., (a) Force acting between two long conductor carrying, current,, m 2I I, F = 0 1 2 ´l, ...(i), 4p d, Where d = distance between the conductors, l = length of conductor, m 2(2 I1 ) I 2, In second case, F ¢ = - 0, ..(ii), l, 4 p 3d, From equation (i) and (ii), we have, F ¢ -2, =, \, F, 3, (b) When current is passed through a spring then current, flows parallel in the adjacent turns in the same direction., As a result the various turn attract each other and spring, get compress., (c) Magnetic field due to current in wire 1 at point P distant, r from the wire is, m i, q, B = 0 1 [ cos q + cos q ], i2, 4p r, i, 1, , B=, , m 0 i1 cos q, 2p r, , r, , P, dl, , q, , This magnetic field is directed perpendicular to the plane of, paper, inwards., The force exerted due to this magnetic field on current, element i2 dl is, dF = i2 dl B sin 90°, \ dF = i2 dlB, æ m i cos q ö, Þ dF = i2 dl ç 0 1, ÷, è 4p r ø, m0, i1 i2 dl cos q, 2 pr, 96. (d) Galvanometer of resistance (G) converted into a, voltmeter of range 0-1 V., ig, , R1, , V = 1 = ig (G + R1 ), , ...(i), , To increase the range of voltmeter 0-2 V, G, , R1, , R2, , ...(ii), , Dividing eq. (i) by (ii),, Þ, , G + R1, 1, =, 2 G + R1 + R2, , Þ G + R1 + R2 = 2G + 2 R1, \ R2 = G + R1, 97. (d) Given,, Current passing through galvanometer, I = 6 mA, Deflection, q = 2°, Figure of merit of galvanometer, I 6 ´10-3, =, = 3 ´10-3 A/div, 2, q, 98. (20), Given,, Area of galvanometer coil, A = 3 × 10–4 m2, Number of turns in the coil, N = 500, Current in the coil, I = 0.5 A, r r, Torque t =| M ´ B |= NiAB sin(90°) = NiAB, =, , ÞB=, , t, 1.5, =, = 20 T, NiA 500 ´ 0.5 ´ 3 ´10 -4, , 99. (c) ig = 20 × 50 = 1000 µA = 1 mA, Using, V = ig (G + R), we have, 2 = 10–3 (100 + R1), R1 = 1900 W, when, V = 10 volt, 10 = 10–3 (100 + R2 + R1), 10000 = (100 + R2 + 1900), \ R2 = 8000 W, 100. (b) In an ammeter,, ig = i0, , =, , G, , 2 = ig ( R1 + R2 + G), , RA, RA + G, , and for voltmeter,, V = ig (G + RV) = Gi0, On solving above equations, we get, RARV = G2, R A æ ig ö, =ç, and, ÷, RV è i0 - ig ø, , 2, , 101. (Bonus) v = ig (R + G), Þ 5 = 10–4 (2 × 106 + x), x = – 195 × 104W
Page 347 :
P-338, , Physics, , 102. (c) V = ig (G + R) = 4 × 10–3 (50 + 5000) = 20V, 103. (d) Cq = NBiA sin 90°, , æ p ö, or 10-6 ç, = 175B(10-3 ) ´ 10-4, è 180 ÷ø, \ B = 10–3 T, 104. (d) Using, ig = i, , S, S+G, , S, S + 50, On solving, we get, 0.002 = 0.5, , 100, ; 0.2 W, 498, 105. (b) When key K1 is closed and key K2 is open, S=, , ig =, , E, = Cq 0, 220 + R g, , ... (i), , When both the keys are closed, Cq0, E, 5, æ, ö, ´, =, ig =, ç, ÷, 5R g, (R g + 5), 5, ç 220 +, ÷, 5 + Rg ø, è, Þ, , S, , ... (ii), , E, = Cq0, 220 + R g, , ... (i), , G, , G, , I, R, , Þ 5500 + 25Rg = 225Rg + 1100, 200Rg = 4400, Rg = 22W, 106. (b) Galvanometer has 25 divisions Ig = 4 × 10–4 × 25 = 10–2 A, ig, G, R, , V = Ig Rnet, , 2.5V, , v = Ig (G + R), 2.5 = (50 + R) 10–2 \ R = 200W, 107. (c) Deflection current, = Igmax = nxk =0.005 × 30, Where, n = Number of divisions = 30 and k = 0.005 amp/, division, = 15 × 10–2 = 0.15, v = Ig[20 + R], 15 = 0.15 [20 + R], 100 = 20 + R, R = 80 W, , I/2, , R, E, , Dividing (i) by (ii), we get, 225R g + 1100, Þ, =5, 1100 + 5R g, , v, , 1, 2, S=, (R + G)I, e2, RG ´, , Cq 0, 5E, =, 225R g + 1100, 5, , 50W, , 108. (d) Given,, Resistance of galvanometer, G = 100W, Current, ig = 1 mA, A galvanometer can be converted into voltmeter by, connecting a large resistance R in series with it., Total resistance of the combination = G + R, According to Ohm’s law, V = ig (G + R), \ 10 = 1 × 10–3 (100 + R0), Þ 10000 – 100 = 9900 W = R0, Þ R0 = 9.9 kW, 109. (b) Figure of merit of a galvanometer is the correct required, to produce a deflection of one division in the galvanometer, I, i.e., figure of merit =, q, e, 1, I=, G = KW, R+G, 9, 1, S, 1, e, eS, =, ´, Þ =, 2 R + GS, S+ G, 2 R(S + G) + GS, G+S, , E, , 1, ´ 102 ´ 270 ´ 10 -6, 2, S=, = 110 W, æ 6ö, 6-ç ÷, è 2ø, 110. (d) According to question, current through galvanometer,, Ig = 1 mA, Current through shunt (I – Ig) = 2 A, Galvanometer resistance Rg = 25W, Resistance of shunt, S = ?, Ig, G, I0R0 = (I – Ig)S, 11 ´ 103 ´, , 10-3 ´ 25, I – Ig, S, 2, S ; 1.25 × 10–2W, 111. (c) Given : Current through the galvanometer,, ÞS=, , ig = 5 × 10–3 A, Galvanometer resistance, G = 15W, Let resistance R to be put in series with the galvanometer, to convert it into a voltmeter., V = ig (R + G), 10 = 5 × 10–3 (R + 15), \ R = 2000 – 15 = 1985 = 1.985 × 103 W, 112. (c) Ig G = ( I – Ig)s, \ 10–3 × 100 = (10 – 10–3) × S, \ S » 0.01W
Page 348 :
P-339, , Moving Charges and Magnetism, , 113. (d) As we know, I =, , V 5, =, = 0.1, R 50, , I' = 0.099, When Galvanometer is connected, 100S, V, R eq = 50 +, =, 100 + S I, 100S, 5, Þ, =, - 50, 100 + S 0.099, 100S, 100S, = 50.50 - 50 Þ, = 0.5, 100 + S, 100 + S, Þ 100S = 50 + 0.55 Þ 99.5S = 50, 50, S=, = 0.5 W, 99.05, So, shunt of resistance = 0.5W is connected in parallel with, the galvanometer., V, 114. (a) According to Ohm's Law, I =, R, V, Ig =, R +G, where, Ig-Galvanometer current, G-Galvonometer resistance, IG, R, G, Þ, , Þ R(G + S) + GS = 2S(R + G), Þ RG + RS + GS = 2S(R + G), Þ RG = 2S(R + G) - S(R + G), \ RG = S(R + G), 115. (b) To measure AC voltage across a resistance a, moving coil galvanometer is used., 116. (d) The correct circuit diagram is D with galvanometer, resistance, RS, R-S, 117. (d) Statements I is false and Statement II is true, , G=, , IgG, I – Ig, Therefore for I to increase, S should decrease, So additional, S can be connected across it., 118. (d), 119. (d) The current that will given full scale deflection in the, absence of the shunt is nearly equal to the current through, the galvanometer when shunt is connected i.e. Ig, For ammeter, shunt resistance, S =, , =, , V, When shunt of resistance S is connected parallel to the, GS, Galvanometer then G =, G +S, V, \ I=, GS, R+, G +S, R, , IG, 2, G, , Equal potential difference is given by, I'g G = (I - I'g )S, , I 'g (G + S) = IS, Ig, 2, , =, , IS, G +S, , V, V, S, =, ´, 2(R + G ) R + GS, G+S, G +S, 1, S, Þ, =, 2(R + G) R(G + S) + GS, Þ, , 5.5 ´ 1, = 0.045 ampere., 120 + 1, , 500 W, , 120. (a), , A, , i, 12V, , Again, i =, , 2V, , R, , 12 – 2 = (500W)i Þ i =, S, , Þ, , IS, G+S, , As Ig =, , 10, 1, =, 500 50, , 12, 1, =, 500 + R 50, , Þ 500 + R = 600, Þ R = 100 W, 121. (c) Resistance of Galvanometer,, Current sensitivity, 10, = 5W, ÞG=, G=, Voltage sensitivity, 2, Here ig = Full scale deflection current =, V = voltage to be measured = 150 volts, (such that each division reads 1 volt), 150, ÞR=, - 5 = 9995W, 15 ´ 10 -3, , 150, = 15 mA, 10
Page 349 :
19, , P-340, , Physics, , Magnetism and, Matter, Magnetism, Gauss’s Law,, TOPIC 1 Magnetic Moment, Properties, of Magnet, 1., , 2., , 3., , A small bar magnet placed with its axis at 30° with an external, field of 0.06 T experiences a torque of 0.018 Nm. The, minimum work required to rotate it from its stable to, unstable equilibrium position is :, [Sep. 04, 2020 (I)], (a) 6.4 × 10–2 J, (b) 9.2 × 10–3 J, (c) 7.2 × 10–2 J, (d) 11.7 × 10–3 J, A circular coil has moment of inertia 0.8 kg m2 around any, diameter and is carrying current to produce a magnetic, moment of 20 Am2. The coil is kept initially in a vertical, position and it can rotate freely around a horizontal, diameter. When a uniform magnetic field of 4 T is applied, along the vertical, it starts rotating around its horizontal, diameter. The angular speed the coil acquires after rotating, by 60° will be :, [Sep. 04, 2020 (II)], (a) 10 rad s–1, (b) 10p rad s–1, (c) 20p rad s–1, (d) 20 rad s–1, Two magnetic dipoles X and Y are placed at a separation, d, with their axes perpendicular to each other. The dipole, moment of Y is twice that of X. A particle of charge q is, passing through their midpoint P, at angle q = 45° with the, horizontal line, as shown in figure. What would be the, magnitude of force on the particle at that instant? (d is, much larger than the dimensions of the dipole), [8 April 2019 II], , (c), 4., , 5., , 6., , 7., , 8., , 9., æ m0 ö M, ç ÷, (a) è 4p ø d, 2, , ( ), , 3, , ´ qv, , (b) 0, , æm ö M, 2ç 0÷, è 4p ø d, 2, , ( ), , 3, , ´ qv, , æ m0 ö 2M, ´ qv, ç ÷, (d) è 4p ø d 3, 2, , ( ), , A magnet of total magnetic moment 10 -2 iˆ A-m2 is, placed in a time varying magnetic field,, B iˆ (coswt)where B = 1 Tesla and w = 0.125 rad/s. The, work done for reversing the direction of the magnetic, moment at t = 1 second, is:, [10 Jan. 2019 I], (a) 0.01 J, (b) 0.007 J, (c) 0.028 J, (d) 0.014 J, A magnetic dipole in a constant magnetic field has :, [Online April 8, 2017], (a) maximum potential energy when the torque is maximum, (b) zero potential energy when the torque is minimum., (c) zero potential energy when the torque is maximum., (d) minimum potential energy when the torque is maximum., A magnetic dipole is acted upon by two magnetic fields, which are inclined to each other at an angle of 75°. One of, the fields has a magnitude of 15 mT. The dipole attains, stable equilibrium at an angle of 30° with this field. The, magntidue of the other field (in mT) is close to :, [Online April 9, 2016], (a) 1, (b) 11, (c) 36, (d) 1060, A 25 cm long solenoid has radius 2 cm and 500 total number, of turns. It carries a current of 15 A. If it is equivalent to a, r, magnet of the same size and magnetization M (magnetic, uur, moment/volume), then M is : [Online April 10, 2015], (a) 30000p Am–1, (b) 3pAm–1, –1, (c) 30000 Am, (d) 300 Am–1, A bar magnet of length 6 cm has a magnetic moment of, 4 J T–1. Find the strength of magnetic field at a distance of, 200 cm from the centre of the magnet along its equatorial, line., [Online May 7, 2012], (a) 4 × 10–8 tesla, (b) 3.5 × 10–8 tesla, (c) 5 × 10–8 tesla, (d) 3 × 10–8 tesla, A thin circular disc of radius R is uniformly charged with, density s > 0 per unit area. The disc rotates about its axis, with a uniform angular speed w.The magnetic moment of, the disc is, [2011 RS]
Page 350 :
P-341, , Magnetism and Matter, , pR 4, (b), sw, 2, , 4, , (a) pR sw, 4, , pR, sw, (d) 2pR 4 sw, 4, 10. A magnetic needle is kept in a non-uniform magnetic field., It experiences, [2005], (a) neither a force nor a torque, (b) a torque but not a force, (c) a force but not a torque, (d) a force and a torque, 11. The length of a magnet is large compared to its width and, breadth. The time period of its oscillation in a vibration, magnetometer is 2s. The magnet is cut along its length, into three equal parts and these parts are then placed on, each other with their like poles together. The time period, of this combination will be, [2004], , (c), , r, (a) B, (b) zero, , r, r, (c) much large than | B | and parallel to B, r, r, (d) much large than | B | but opposite to B, 17. Magnetic materials used for making permanent magnets, (P) and magnets in a transformer (T) have different, properties of the following, which property best matches, for the type of magnet required? [Sep. 02, 2020 (I)], (a) T : Large retentivity, small coercivity, (b) P : Small retentivity, large coercivity, (c) T : Large retentivity, large coercivity, (d) P : Large retentivity, large coercivity, 18., , 2, 2, s, s, (c) 2 s, (d), 3, 3, 12. A magnetic needle lying parallel to a magnetic field requiers, (a) 2 3 s, , (b), , W units of work to turn it through 600 . The torque needed, to maintain the needle in this position will be, [2003], 3, (d) 2 W, W, 2, 13. The magnetic lines of force inside a bar magnet [2003], (a) are from north-pole to south-pole of the magnet, (b) do not exist, (c) depend upon the area of cross-section of the bar, magnet, (d) are from south-pole to north-pole of the Magnet, , (a), , 3W, , (b) W, , (c), , The Earth Magnetism, Magnetic, TOPIC 2, Materials and their properties, 14. An iron rod of volume 10–3 m3 and relative permeability, 1000 is placed as core in a solenoid with 10 turns/cm. If a, current of 0.5 A is passed through the solenoid, then the, magnetic moment of the rod will be : [Sep. 05, 2020 (II)], (a) 50 × 102 Am2, (b) 5 × 102 Am2, 2, 2, (c) 500 × 10 Am, (d) 0.5 × 102 Am2, 15. A paramagnetic sample shows a net magnetisation of 6 A/m, when it is placed in an external magnetic field of 0.4 T at a, temperature of 4 K. When the sample is placed in an external, magnetic field of 0.3 T at a temperature of 24 K, then the, magnetisation will be :, [Sep. 04, 2020 (II)], (a) 1 A/m, (b) 4 A/m, (c) 2.25 A/m, (d) 0.75 A/m, 16. A perfectly diamagnetic sphere has a small spherical cavity, at its centre, which is filled with a paramagnetic substance., r, The whole system is placed in a uniform magnetic field B., Then the field inside the paramagnetic substance is :, [Sep. 03, 2020 (II)], , P, , 19., , 20., , 21., , 22., , The figure gives experimentally measured B vs. H variation, in a ferromagnetic material. The retentivity, co-ercivity, and saturation, respectively, of the material are:, [7 Jan. 2020 II], (a) 1.5 T, 50 A/m and 1.0 T, (b) 1.5 T, 50 A/m and 1.0 T, (c) 150 A/m, 1.0 T and 1.5 T, (d) 1.0 T, 50 A/m and 1.5 T, A paramagnetic material has 1028 atoms/m3. Its magnetic, susceptibility at temperature 350 K is 2.8 × 10–4 . Its, susceptibility at 300 K is:, [12 Jan. 2019 II], (a) 3.267 × 10–4, (b) 3.672 × 10–4, (c) 3.726 × 10–4, (d) 2.672 × 10 –4, A paramagnetic substance in the form of a cube with sides, 1 cm has a magnetic dipole moment of 20 × 10–6 J/T when, a magnetic intensity of 60 × 103 A/m is applied. Its magnetic, susceptibility is:, [11 Jan. 2019 II], (a) 3.3 × 10–2, (b) 4.3 × 10 –2, (c) 2.3 × 10–2, (d) 3.3 × 10–4, At some location on earth the horizontal component of, earth’s magnetic field is 18 × 10–6 T. At this location,, magnetic needle of length 0.12 m and pole strength 1.8, Am is suspended from its mid-point using a thread, it, makes 45° angle with horizontal in equilibrium. To keep, this needle horizontal, the vertical force that should be, applied at one of its ends is:, [10 Jan. 2019 II], (a) 3.6 × 10–5 N, (b) 1.8 × 10–5 N, (c) 1.3 × 10–5 N, (d) 6.5 × 10–5 N, A bar magnet is demagnetized by inserthing it inside a, solenoid of length 0.2 m, 100 turns, and carrying a current, of 5.2 A. The coercivity of the bar magnet is:, [9 Jan. 2019 I], (a) 285 A/m, (b) 2600 A/m, (c) 520 A/m, (d) 1200 A/m
Page 351 :
P-342, , Physics, , 23. The B-H curve for a ferromagnet is shown in the figure., The ferromagnet is placed inside a long solenoid with, 1000 turns/cm.. The current that should be passed in the, solenoid to demagnetise the ferromagnet completely is:, [Online April 15, 2018], B(T), 2, 0, 1, 0, , H(A/m), , –200 –100 100 200, –1, 0, –2, 0, , (a) 2 mA, (b) 1 mA (c) 40 µA, (d) 20 µA, 24. Hysteresis loops for two magnetic materials A and B are, given below :, D, , B, , H, (A), , H, (B), , These materials are used to make magnets for elecric, generators, transformer core and electromagnet core., Then it is proper to use :, [2016], (a) A for transformers and B for electric generators., (b) B for electromagnets and transformers., (c) A for electric generators and trasformers., (d) A for electromagnets and B for electric generators., 25. A fighter plane of length 20 m, wing span (distance from, tip of one wing to the tip of the other wing) of 15m and, height 5m is lying towards east over Delhi. Its speed is, 240 ms–1. The earth's magnetic field over Delhi is 5 ×, 10–5T with the declination angle ~0° and dip of q such, 2, . If the voltage developed is VB between, 3, the lower and upper side of the plane and VW between the, tips of the wings then VB and VW are close to :, [Online April 10, 2016], (a) VB = 40 mV; VW = 135 mV with left side of pilot at, higher voltage, (b) VB = 45 mV; VW = 120 mV with right side of pilot, at higher voltage, (c) VB = 40 mV; VW = 135 mV with right side of pilot, at higher voltage, (d) VB = 45 mV; VW = 120 mV with left side of pilot at, higher voltage, 26. A short bar magnet is placed in the magnetic meridian, of the earth with north pole pointing north. Neutral, points are found at a distance of 30 cm from the magnet, on the East – West line, drawn through the middle point, of the magnet. The magnetic moment of the magnet in, , that sin q =, , Am2 is close to: (Given, , =Horizontal component of earth’s magnetic field = 3.6, × 10–5 tesla), [Online April 11, 2015], (a) 14.6, (b) 19.4, (c) 9.7, (d) 4.9, 27. The coercivity of a small magnet where the ferromagnet, gets demagnetized is 3 ´ 103 Am -1. The current required, to be passed in a solenoid of length 10 cm and number of, turns 100, so that the magnet gets demagnetized when, inside the solenoid, is:, [2014], (a) 30 mA (b) 60 mA (c) 3 A, (d) 6 A, 28. An example of a perfect diamagnet is a superconductor., This implies that when a superconductor is put in a, magnetic field of intensity B, the magnetic field Bs inside, the superconductor will be such that:, [Online April 19, 2014], (a) Bs = – B, (b) Bs = 0, (c) Bs = B, (d) Bs < B but Bs ¹ 0, 29. Three identical bars A, B and C are made of different, magnetic materials. When kept in a uniform magnetic field,, the field lines around them look as follows:, , m0, = 10–7 in SI units and BH, 4p, , A, , C, , B, , Make the correspondence of these bars with their, material being diamagnetic (D), ferromagnetic (F) and, paramagnetic (P):, [Online April 11, 2014], (a) A « D, B « P, C « F, (b) A « F, B « D, C « P, (c) A « P, B « F, C « D, (d) A « F, B « P, C « D, 30. The magnetic field of earth at the equator is approximately, 4 × 10–5 T. The radius of earth is 6.4 × 106 m. Then the, dipole moment of the earth will be nearly of the order of:, [Online April 9, 2014], (a) 1023 A m2, (b) 1020 A m2, (c) 1016 A m2, (d) 1010 A m2, 31. The mid points of two small magnetic dipoles of length d, in end-on positions, are separated by a distance x, (x > >, d). The force between them is proportional to x–n where n, is:, [Online April 9, 2014], N, , S, , N, x, , S, , (a) 1, (b) 2, (c) 3, (d) 4, 32. Two short bar magnets of length 1 cm each have magnetic, moments 1.20 Am2 and 1.00 Am2 respectively. They are, placed on a horizontal table parallel to each other with, their N poles pointing towards the South. They have a, common magnetic equator and are separated by a distance, of 20.0 cm. The value of the resultand horizontal magnetic, induction at the mid-point O of the line joining their, centres is close to (Horizontal component of earth.s, magnetic induction is 3.6× 10.5Wb/m2), [2013], (a) 3.6 × 10.5 Wb/m2, (b) 2.56 × 10.4 Wb/m2, (c) 3.50 × 10.4 Wb/m2, (d) 5.80 × 10.4 Wb/m2
Page 352 :
P-343, , Magnetism and Matter, , 33. The earth’s magnetic field lines resemble, that of a dipole at the centre of the earth. If the magnetic, moment of this dipole is close to 8 × 1022 Am2, the, value of earth’s magnetic field near the equator is close, to (radius of the earth = 6.4 × 106 m), [Online April 25, 2013], (a) 0.6 Gauss, (b) 1.2 Gauss, (c) 1.8 Gauss, (d) 0.32 Gauss, 34. Relative permittivity and permeability of a material e r and, m r , respectively. Which of the following values of these, quantities are allowed for a diamagnetic material? [2008], (a) e r = 0.5, m r = 1.5, (b) e r = 1.5, mr = 0.5, (c) e r = 0.5, m r = 0.5, (d) e r = 1.5, mr = 1.5, 35. Needles N1, N2 and N3 are made of a ferromagnetic, a, paramagnetic and a diamagnetic substance respectively., A magnet when brought close to them will, [2006], (a) attract N1 and N2 strongly but repel N3, (b) attract N1 strongly, N2 weakly and repel N3 weakly, (c) attract N1 strongly, but repel N2 and N3 weakly, (d) attract all three of them, 36. The materials suitable for making electromagnets should, have, [2004], (a) high retentivity and low coercivity, (b) low retentivity and low coercivity, (c) high retentivity and high coercivity, (d) low retentivity and high coercivity, 37. A thin rectangular magnet suspended freely has a period, of oscillation equal to T. Now it is broken into two equal, halves (each having half of the original length) and one, piece is made to oscillate freely in the same field. If its, period of oscillation is T¢, the ratio, (a), , 1, 2 2, , (c) 2, , (b), , 1, 2, , (d), , 1, 4, , T', is, T, , [2003], , 38. Curie temperature is the temperature above which, [2003], (a) a ferromagnetic material becomes paramagnetic, (b) a paramagnetic material becomes diamagnetic, (c) a ferromagnetic material becomes diamagnetic, (d) a paramagnetic material becomes ferromagnetic, , TOPIC 3 Magnetic Equipment, 39. A ring is hung on a nail. It can oscillate, without slipping, or sliding (i) in its plane with a time period T1 and, (ii) back, and forth in a direction perpendicular to its plane, with a, period T2. The ratio, (a), , 2, 3, 3, , T1, will be :, T2, , (b), , [Sep. 05, 2020 (II)], 2, 3, , 2, 3, 2, 40. A magnetic compass needle oscillates 30 times per minute, at a place where the dip is 45o, and 40 times per minute, where the dip is 30o. If B1 and B2 are respectively the total, magnetic field due to the earth and the two places, then, the ratio B1/B2 is best given by :, (c), , (d), , [12 April 2019 I], (a) 1.8, (b) 0.7, (c) 3.6, (d) 2.2, 41. A hoop and a solid cylinder of same mass and radius are, made of a permanent magnetic material with their, magnetic moment parallel to their respective axes. But, the magnetic moment of hoop is twice of solid cylinder., They are placed in a uniform magnetic field in such a, manner that their magnetic moments make a small angle, with the field. If the oscillation periods of hoop and, cylinder are Th and Tc respectively, then:, [10 Jan. 2019 II], (a) Th = Tc, (b) Th = 2 Tc, (c) Th = 1.5 Tc, (d) Th = 0.5 Tc, 42. A magnetic needle of magnetic moment 6.7 × 10–2 Am2, and moment of inertia 7.5 × 10–6 kg m2 is performing, simple harmonic oscillations in a magnetic field of 0.01, T. Time taken for 10 complete oscillations is : [2017], (a) 6.98 s, (b) 8.76 s, (c) 6.65 s, (d) 8. 89 s
Page 353 :
P-344, , 1., , Physics, , (c) Here, q = 30°, t = 0.018 N-m, B = 0.06 T, Torque on a bar magnet :, , and B2 =, , t = MB sin q, , µ0 2M, 4p ( d / 2 ) 3, µ0 2M, 4p ( d / 2) 3, , 0.018 = M ´ 0.06 ´ sin 30°, , 1, Þ M = 0.6 A-m 2, 2, Position of stable equilibrium (q = 0°), Position of unstable equilibrium (q = 180°), Minimum work required to rotate bar magnet from stable, to unstable equilibrium, Þ 0.018 = M ´ 0.06 ´, , DU = U f - U i = -MB cos180° - (- MB cos0°), W = 2MB = 2 ´ 0.6 ´ 0.06, , 2., , \ W = 7.2 ´ 10-2 J, (a) Given,, Moment of inertia of circular coil, I = 0.8 kg m2, Magnetic moment of circular coil, M = 20 Am2, Rotational kinetic energy of circular coil,, 1 2, Iw, 2, Here, w = angular speed of coil, Potential energy of bar magnet = – MB cos f, From energy conservation, , KE =, , \, , or q = 45º, , 4., , 5., , 6., , 1 2, I w = U in - U f = - MB cos 60° - (- MB ), 2, Þ, , MB 1 2, = Iw, 2, 2, , Þ, , 20 ´ 4 1, = (0.8)w 2, 2, 2, , Þ 100 = w 2 Þ w = 10 rad, 3., , (b), , µ, 2K, B1 = 0, 4 p ( d / 2 )3, , B, tan q = 2 =, B1 µ0 2M = 1, 4p ( d / 2) 3, , 7., , 8., , uur, The resultant field is 45º from B1. The angle between B, uur, and v zero, so force on the particle is zero., (c), Work done, W = 2 m·B, = 2 × 10–2 × 1 cos (0.125), = 0.02 J, (c) Potential energy of dipole,, U = – pE cos q, Torque experienced by dipole, t = pE sin q, Torque will be maximum (tmax) when q = 90° then potential, energy U = 0, (b) We know that, magnetic dipole moment, M = NiA cosθ i.e., M µ cosθ, When two magnetic fields are inclined at an angle of 75°, the equilibrium will be at 30°, so, 1, cos θ < cos(75°, 30°) < cos 45° <, 2, x, 15, < [ x » 11, 2, 2, r, NiA, (c) M (mag. moment/volume) =, Al, Ni, (500)15, =, =, = 30000 Am–1, l, 25 ´ 10 –2, (c) Along the equatorial line, magnetic field strength, m, M, B= 0, 3/2, 4p 2, r + l2, , (, , ), , Given: M = 4JT–1, , r = 200 cm = 2 m, 6cm, l=, = 3 cm = 3 × 10–2 m, 2, \B=, , 4p ´ 10 -7, ´, 4p, , 4, , (, , ), , 3, , é 2, -2 2 ù 2, ê 2 + 3 ´ 10, ú, ë, û, Solving we get, B = 5 × 10–8 tesla
Page 354 :
P-345, , Magnetism and Matter, , 9., , (c), , 14. (b) Given,, Volume of iron rod, V = 10-3 m3, Relative permeability, m r = 1000, Number of turns per unit length, n = 10, Magnetic moment of an iron core solenoid,, M = (m r - 1) ´ NiA, , q, Magnetic dipole moment, =, 2m, Angular momentum, , a, , Þ M = (mr -1) ´ Ni, \ Magnetic dipole moment (M), q æ mR 2 ö, 1, 4, .ç, M =, ÷ .w = s.pR w., 2m è 2 ø, 4, , 10, ´ 0.5 ´ 10 -3 = 499.5 » 500., 10 -2, 15. (d) For paramagnetic material. According to curies law, Þ M = 999 ´, , 10. (d) A magnetic needle kept in non uniform magnetic field, experience a force and torque due to unequal forces acting, on poles., 11. (b) Initially, time period of magnet, 1, 2, I, = 25 where I = ml, 12, MB, When the magnet is cut into three pieces the pole strength, will remain the same and, Moment of inertia of each part,, 1 æ mö æ l ö, I, (I¢) =, çè ÷ø çè ÷ø ´ 3 =, 12 3 3, 9, We have, Magnetic moment (M), = Pole strength (m) × l, \ New magnetic moment,, æ lö, M ' = m ´ ç ÷ ´ 3 = ml = M, è 3ø, , 1, T, For two temperatures T1 and T2, cµ, , c1T1 = c2T2, , T = 2p, , New time period, T¢ = 2p, , V, N, Þ M = (mr - 1) ´ iV, l, l, , I¢, M ¢B, , I, 2, T, Þ T¢ =, = s., 9 MB, 9 3, 12. (a) Workdone to turn a magnetic needle from angle q1 to, q2 is given by, W = MB (cos q1 - cos q2 ), , = 2p, , \ W = MB (cos 0° - cos 60°), , But c =, \, , I, B, , I1, I, T1 = 2 T2, B1, B2, , I, 6, 0.3, ´ 4 = 2 ´ 24 Þ I 2 =, = 0.75 A/m, 0.4, 0.3, 0.4, 16. (b) When magnetic field is applied to a diamagnetic, substance, it produces magnetic field in opposite direction, so net magnetic field inside the cavity of sphere will be, zero. So, field inside the paramagnetic substance kept, inside the cavity is zero., 17. (d) Permanent magnets (P) are made of materials with large, retentivity and large coercivity. Transformer cores (T) are, made of materials with low retentivity and low coercivity., Þ, , 18. (d), , æ 1 ö MB, = MB ç 1 - ÷ =, è 2ø, 2, MB, = 3W, 2, 13. (d) The magnetif field lines of bar magnet form closed, lines. As shown in the figure, the magnetic lines of force, are directed from south to north inside a bar magnet., Outside the bar magnet magnetic field lines directed from, north to south pole., , \ Torque, t = MB sin q = MB sin 60° = 3, , 19. (a) According to Curie law for paramagnetic substance,, 1, , c, , TC2, , cµ T Þ 1 = T, c2, C1, C, N, , S, , 2.8 ´ 10 –4, 300, =, c2, 350, , c2 =, , 2.8 ´ 350 ´10 –4, = 3.266 × 10–4, 300
Page 355 :
P-346, , Physics, , 20. (d) Magnetic susceptibility,, c=, , I, H, , where, I =, , Now, c =, , Magnetic moment 20 ´10 –6, = 20 N/m2, =, Volume, 10 –6, 20, , 1, = ´10 –3 = 3.3 ´10–4, 60´10 3, 3, , 21. (d) using, MB sinq = F l Sinq (t), , 28. (b) Magnetic field inside the superconductor is zero., Diamagnetic substances are repelled in external magnetic, field., 29. (b) Diamagnetic materials are repelled in an external, magnetic field., Bar B represents diamagnetic materials., 30. (a) Given, B = 4 × 10–5 T, RE = 6.4 × 106 m, Dipole moment of the earth M = ?, m M, B= 0 3, 4p d, 4 ´10-5 =, , F, B, 45°, m, , l, MB sin 45° = F sin 45°, 2, , F = 2MB = 2 × 1.8 × 18 × 10–6 = 6.5 × 10–5N, , 4p´10-7 ´ M, , 24., 25., 26., , 3.6 ´10–5, , (0.3)3, 10 –7, Hence, M = 9.7 Am2, 27. (c) Magnetic field in solenoid B = m0ni, Þ M =, , Þ, , B, = ni, m0, , (Where n = number of turns per unit length), B Ni, 100i, =, Þ 3 ´ 103 =, m0 L, 10 ´ 10-2, Þ i = 3A, , Þ, , 3, , B1, B2, O, , S, N, , or, H =, , 23., , ), , \ M @ 1023 Am2, 31. (d) In magnetic dipole, 1, Force µ, r4, In the given question,, Force µ x– n, Hence, n = 4, 32. (b) Given : M1 = 1.20 Am2, N, BH, , B, Nö, æ, 22. (b) Corecivity, H = m and B = m 0 ni çè n = ÷ø, l, 0, , N, 100, i=, × 5.2 = 2600 A/m, l, 0.2, (b) Given Number of turns,, n = 1000 turns/cm = 1000 × 100 turns/m, Coercivity of ferromagnet, H = 100 A/m, Current to demagnetise the ferromagnet, I = ?, Using,, H = nI, or, 100 = 105 × I, 100, \ I = 5 = 1 mA, 10, (b) Graph [A] is for material used for making permanent, magnets (high coercivity), Graph [B] is for making electromagnets and transformers., (d) VB = VBHl = 240 × 5 × 10–5 cos(q) × 5 = 44.7 mv, By right hand rule, the charge moves to the left of pilot., (c) Here, r = 30cm = 0.3cm, m0 M, = BH = 3.6 ´ 10 –5, we know, 4 pr 3, , (, , 4p´ 6.4 ´ 106, , r, , S, , r, , N, , S, 20, cm = 0.1m, M2 = 1.00 Am2 ; r =, 2, Bnet = B1 + B2 + BH, , m ( M1 + M 2 ), Bnet = 0, + BH, 4p, r3, =, , 10 -7 (1.2 + 1), (0.1)3, , + 3.6 ´ 10 -5 = 2.56 ´ 10 -4 wb/m2, , 33. (a) Given M = 8 × 1022 Am2, d = Re = 6.4 × 106m, m0 2M, ., Earth’s magnetic field, B =, 4p d3, 4 p ´ 10 -7 2 ´ 8 ´ 1022, =, ´, @ 0.6 Gauss, 4p, (6.4 ´ 106 )3, 34. (b) For a diamagnetic material, the value of µr is slightly, less than one. For any material, the value of Îr is always, greater than 1., 35. (b) Ferromagnetic substance has magnetic domains, whereas paramagnetic substances have magnetic dipoles, which get attracted to a magnetic field. Ferromagnetic, material magnetised strongly in the direction of magnetism, field, Hence, N1 will be attracted paramagnetic substance, attract weekly in the direction of field. Hence, N2 will weakly, attracted. Diamagnetic substances do not have magnetic, dipole but in the presence of external magnetic field due to, their orbital motion of electrons these substances are, repelled. Hence, N3 will be repelled.
Page 356 :
P-347, , Magnetism and Matter, , 36. (b) Electromagnet should be amenable to magnetisation, & demagnetization., \ Materials suitable for making electromagnets should, have low retentivity and low coercivity should be low., 37. (b) The time period of a rectangular magnet oscillating in, earth’s magnetic field is given by T = 2p, , I, MBH, , where I = Moment of inertia of the rectangular magnet, M = Magnetic moment, BH = Horizontal component of the earth’s magnetic field, Initially, the time period of the magnet, T = 2p, , I, 1, where I = M l2, MBH, 12, , I¢, M ¢BH, , \, , T12, T22, , I, We have, T = 2p MB, x, , =, , Bx 2, Bx1, , B2 cos 45°, æ 2ö, =, çè ÷ø =, 1.5, B1 cos 30°, , 2, , 2, , B2, 2, æ 4ö, ´, çè ÷ø =, 3, B1, 6, B1, 9, =, B2 8 6, , T = 2p, , T¢, I¢ M, I /8 M, 1 1, ´, =, ´, =, =, \ =, M /2 I, T, M I, 4 2, 38. (a) The temperature above which a ferromagnetic, substance becomes paramagnetic is called Curie’s, temperature., 39. (a) Let I1 and I2 be the moment of inertia in first and second, case respectively., , I1 = 2 MR 2, , I, MB, , Where, M = magnetic moment, I moment of inertia and B =, magnetic field, mR 2, Th = 2p, ( 2MB), , Tc = 2p, , 1 / 2mR 2, MB, , X, Using, T = 2p, = 2p, , TµI, , T2, , I, mgd, , = 0.46, , Clearly, Th = Tc, 42. (c) Given : Magnetic moment, M = 6.7 × 10–2 Am2, Magnetic field, B = 0.01 T, Moment of inertia, I = 7.5 × 10–6 Kgm2, , MR 2 3, I 2 = MR +, = MR 2, 2, 2, Axis of rotation, 2, , Time period, T = 2p, , B2 ´ 2, 2 ´ B1 ´ 3, , 41. (a) Using, time /oscillation period,, , I, M, 1 æ M öæ l ö, ç ÷ ç ÷ = and M ¢ =, 12 è 2 ø è 2 ø, 8, 2, , T1, , 2MR 2, 2, =, 3, 2, 3, MR, 2, , I1, =, I2, , 2, , \, , Moment of inertia of each part, I¢ =, , T1, =, T2, , 40. (B onus), , or, , Case 2, Magnet is cut into two identical pieces such that each, piece has half the original length., Then T ¢ = 2p, , \, , I, MB, , 7.5 ´ 10-6, -2, , =, , 2p, ´ 1.06 s, 10, , 6.7 ´ 10 ´ 0.01, Time taken for 10 complete oscillations, t = 10T = 2p × 1.06, = 6.6568 » 6.65 s
Page 357 :
20, , P-348, , Physics, , Electromagnetic, Induction, Magnetic Flux, Faraday's, TOPIC 1, and Lenz's Law, 1., , 3., , Two concentric circular coils, C1 and C2, are placed in the, XY plane. C1 has 500 turns, and a radius of 1 cm. C2 has 200, turns and radius current 20 cm. C2 carries a time dependent, current I(t) = (5t2 – 2t + 3) A Where t is in s. The emf induced, , An elliptical loop having resistance R, of semi major axis a,, and semi minor axis b is placed in a magnetic field as shown, in the figure. If the loop is rotated about the x-axis with, angular frequency w, the average power loss in the loop, due to Joule heating is :, [Sep. 03, 2020 (I)], B, z, , 2., , y, (a), , a, , magnet, c, b, Three positions shown describe : (1) the magnet’s entry, (2) magnet is completely inside and (3) magnet’s exit., (1), (2), (3), ®, (1), , (2), ®, , (b), , (1), , (2), , (1), (d), , (3), ®, , ®, , (c), , (3), , 6., , ®, (2), , ®, , 5., , ®, , (3), ®, , p 2 a 2 b 2 B 2 w2, 2R, , a, , x, , y, (b) zero, , pabBw, p 2 a 2 b 2 B 2 w2, (d), R, R, A uniform magnetic field B exists in a direction, perpendicular to the plane of a square loop made of a, metal wire. The wire has a diameter of 4 mm and a total, length of 30 cm. The magnetic field changes with time at a, steady rate dB/dt = 0.032 Ts–1. The induced current in, the loop is close to (Resistivity of the metal wire is, 1.23 × 10–8 W m), [Sep. 03, 2020 (II)], , (c), , 4., , (a), , b, , x, , 4, . The value of x is, x, ______., [NA Sep. 05, 2020 (I)], A small bar magnet is moved through a coil at constant, speed from one end to the other. Which of the following, series of observations will be seen on the galvanometer G, attached across the coil ?, [Sep. 04, 2020 (I)], G, , in C1 (in mV), at the instant t = 1 s is, , 7., , (a) 0.43 A, (b) 0.61 A (c) 0.34 A, (d) 0.53 A, A circular coil of radius 10 cm is placed in a uniform magnetic, field of 3.0 × 10–5 T with its plane perpendicular to the field, initially. It is rotated at constant angular speed about an, axis along the diameter of coil and perpendicular to, magnetic field so that it undergoes half of rotation in 0.2 s., The maximum value of EMF induced (in mV) in the coil will, be close to the integer _____. [NA Sep. 02, 2020 (I)], In a fluorescent lamp choke (a small transformer) 100 V of, reverse voltage is produced when the choke current, changes uniformly from 0.25 A to 0 in a duration of 0.025, ms. The self-inductance of the choke (in mH) is estimated, to be ______., [NA 9 Jan. 2020 I], At time t = 0 magnetic field of 1000 Gauss is passing, perpendicularly through the area defined by the closed, loop shown in the figure. If the magnetic field reduces, linearly to 500 Gauss, in the next 5 s, then induced EMF, in the loop is:, [NA 8 Jan. 2020 I]
Page 358 :
P-349, , Electromagnetic Induction, , (c), , (d), (a) 56 mV (b) 28 mV (c) 48 mV (d) 36 mV, Consider a circular coil of wire carrying constant current, I, forming a magnetic dipole. The magnetic flux through, an infinite plane that contains the circular coil and, excluding the circular coil area is given by fi .The, magnetic flux through the area of the circular coil area is, given by f0. Which of the following option is correct?, [7 Jan. 2020 I], (a) fi = f0 (b) fi > f0 (c) fi < f0 (d) fi = – f0, 9. A long solenoid of radius R carries a time (t) - dependent, current I(t) = I0t(l – t). A ring of radius 2R is placed coaxially, near its middle. During the time interval 0 £ t £ 1, the, induced current (IR) and the induced EMF(VR) in the ring, change as:, [7 Jan. 2020 I], (a) Direction of IR remains unchanged and VR is maximum, at t = 0.5, (b) At t = 0.25 direction of IR reverses and VR is maximum, (c) Direction of IR remains unchanged and VR is zero at t = 0.25, (d) At t = 0.5 direction of IR reverses and VR is zero, 10. A loop ABCDEFA of straight edges has six corner points, A(0, 0, 0), B{5, 0, 0), C(5, 5, 0), D(0, 5, 0), E(0, 5, 5) and, F(0, 0, 5). The magnetic field in this region is, r, B = ( 3iˆ + 4kˆ )T. The quantity of flux through the loop, ABCDEFA (in Wb) is _________ . [NA 7 Jan. 2020 I], 11. A planar loop of wire rotates in a uniform magnetic field., Initially, at t = 0, the plane of the loop is perpendicular to, the magnetic field. If it rotates with a period of 10 s about, an axis in its plane then the magnitude of induced emf will, be maximum and minimum, respectively at:[7 Jan. 2020 II], (a) 2.5 s and 7.5 s, (b) 2.5 s and 5.0 s, (c) 5.0 s and 7.5 s, (d) 5.0 s and 10.0 s, 12. A very long solenoid of radius R is carrying current, I(t) = kte–at (k >0), as a function of time (t >0). Counter, clockwise current is taken to be positive. A circular, conducting coil of radius 2R is placed in the equatorial, plane of the solenoid and concentric with the solenoid., The current induced in the outer coil is correctly depicted,, as a function of time, by:, [9 Apr. 2019 II], 8., , (a), , (b), , 13. Two coils ‘P’ and ‘Q’ are separated by some distance. When, a current of 3A flows through coil ‘P’, a magnetic flux of, 10–3 Wb passes through ‘Q’. No current is passed through, ‘Q’. When no current passes through ‘P’ and a current of, 2A passes through ‘Q’, the flux through ‘P’ is:, [9 Apr. 2019 II], (a) 6.67 × 10–4 Wb, (b) 3.67 × 10–3 Wb, (c) 6.67 × 10–3 Wb, (d) 3.67 × 10–4 Wb, 14. The self induced emf of a coil is 25 volts. When the, current in it is changed at uniiform rate from 10 A to 25, A in 1s, the change in the energy of the inductance is:, [9 Jan. 2019 II], (a) 740 J, (b) 437.5 J, (c) 540 J, (d) 637.5 J, 15. A conducting circular loop made of a thin wire, has area, 3.5 × 10 –3m2 and resistance 10W. It is placed perpendicular, to a time dependent magnetic field B (t) = (0.4T) sin (50pt)., The the net charge flowing through the loop during t = 0, s and t = 10 ms is close to:, [9 Jan. 2019 I], (a) 14 mC, (b) 7 mC (c) 21 mC, (d) 6 mC, 16. In a coil of resistance 100 W , a current is induced by, changing the magnetic flux through it as shown in the, figure. The magnitude of change in flux through the coil is, [2017], (a), , 250 Wb, , (b) 275 Wb, (c) 200 Wb, (d) 225 Wb, 17. A conducting metal circular–wire–loop of radius r is placed, perpendicular to a magnetic field which varies with time as, -t, , B = B0 e t , where B0 and t are constants, at time t = 0. If, the resistance of the loop is R then the heat generated in, the loop after a long time (t ® ¥) is ;, [Online April 10, 2016], (a), , p 2 r 4 B04, 2tR, , (b), , p 2 r 4 B02, 2tR, , (c), , p 2 r 4 B02 R, t, , (d), , p 2 r 4 B02, tR
Page 359 :
P-350, , Physics, , 18. When current in a coil changes from 5 A to 2 A in 0.1 s,, average voltage of 50 V is produced. The self - inductance, of the coil is :, [Online April 10, 2015], (a) 6 H, (b) 0.67 H, (c) 3 H, (d) 1.67 H, 19. Figure shows a circular area of radius, , V (t ), t, , ®, , R where a uniform magnetic field B is, going into the plane of paper and, increasing in magnitude at a constant, rate., , R, , (A), , (B), , I (t ), , I (t ), t, , t, , In that case, which of the following graphs, drawn, schematically, correctly shows the variation of the induced, electric field E(r)?, [Online April 19, 2014], (C), , (D), , I (t ), , I (t ), , t, , E, , t, , E, , (a), , (b), R, , r, , R, , r, , R, , r, , E, , E, , (c), , (d), R, , r, , 20. A coil of circular cross-section having 1000 turns and 4, cm2 face area is placed with its axis parallel to a magnetic, field which decreases by 10–2 Wb m–2 in 0.01 s. The e.m.f., induced in the coil is:, [Online April 11, 2014], (a) 400 mV, (b) 200 mV, (c) 4 mV, (d) 0.4 mV, 21. A circular loop of radius 0.3 cm lies parallel to amuch bigger, circular loop of radius 20 cm. The centre of the small loop, is on the axis of the bigger loop. The distance between, their centres is 15 cm. If a current of 2.0 A flows through, the smaller loop, then the flux linked with bigger loop is, [2013], (a) 9.1 × 10–11 weber, , (b) 6 × 10–11 weber, , (c) 3.3 × 10–11 weber, , (d) 6.6 × 10–9 weber, , 22. Two coils, X and Y, are kept in close vicinity of each other., When a varying current, I(t), flows through coil X, the, induced emf (V(t)) in coil Y, varies in the manner shown, here. The variation of I(t), with time, can then be, represented by the graph labelled as graph :, [Online April 9, 2013], , (a) A, (b) C, (c) B, (d) D, 23. A coil is suspended in a uniform magnetic field, with the, plane of the coil parallel to the magnetic lines of force., When a current is passed through the coil it starts, oscillating; It is very difficult to stop. But if an aluminium, plate is placed near to the coil, it stops. This is due to :, [2012], (a) developement of air current when the plate is placed, (b) induction of electrical charge on the plate, (c) shielding of magnetic lines of force as aluminium is a, paramagnetic material., (d) electromagnetic induction in the aluminium plate, giving rise to electromagnetic damping., 24. Magnetic flux through a coil of resistance 10 W is changed, by Df in 0.1 s. The resulting current in the coil varies with, time as shown in the figure. Then |Df| is equal to (in weber), [Online May 12, 2012], i(A), , 4, , 0.1, , t(s), , (a) 6, (b) 4, (c) 2, (d) 8, 25. The flux linked with a coil at any instant 't' is given by, f = 10t2 – 50t + 250. The induced emf at t = 3s is [2006], (a) –190 V, (b) –10 V, (c) 10 V, (d) 190 V
Page 360 :
P-351, , Electromagnetic Induction, , Motional and Static EMI and, TOPIC 2 Application of EMI, 26. An infinitely long straight wire carrying current I, one side, opened rectangular loop and a conductor C with a sliding, connector are located in the same plane, as shown in the, figure. The connector has length l and resistance R. It, slides to the right with a velocity v. The resistance of the, conductor and the self inductance of the loop are, negligible. The induced current in the loop, as a function, of separation r, between the connector and the straight, wire is :, [Sep. 05, 2020 (II)], , one side opened long, conducting wire loop, , C, I, , R, , v, , l, , r, , (a), , m0 Ivl, 4p Rr, , (b), , m 0 Ivl, p Rr, , m0 Ivl, 2m 0 Ivl, (d), p Rr, 2p Rr, 27. The figure shows a square loop L of side 5 cm which is, connected to a network of resistances. The whole setup is, moving towards right with a constant speed of 1 cm s–1. At, some instant, a part of L is in a uniform magnetic field of 1, T, perpendicular to the plane of the loop. If the resistance, of L is 1.7 &!, the current in the loop at that instant will be, close to :, [12 Apr. 2019 I], , (c), , (a) 60µA, (b) 170 µA, (c) 150 µA, (d) 115 µA, 28. The total number of turns and cross-section area in a, solenoid is fixed. However, its length L is varied by, adjusting the separation between windings. The, inductance of solenoid will be proportional to:, [9 April 2019 I], (a) L, (b) L2, (c) 1/ L2, (d), 1/L, , 29. A thin strip 10 cm long is on a U shaped wire of negligible, resistance and it is connected to a spring of spring constant, 0.5 Nm–1 (see figure). The assembly is kept in a uniform, magnetic field of 0.1 T. If the strip is pulled from its, equilibrium position and released, the number of, oscillations it performs before its amplitude decreases by, a factor of e is N. If the mass of strip is 50 grams, its, resistance 10W and air drag negligible, N will be close to :, [8 April 2019 I], , (a) 1000, (b) 50000 (c) 5000, (d) 10000, 30. A 10 m long horizontal wire extends from North East to, South West. It is falling with a speed of 5.0 ms–1, at right, angles to the horizontal component of the earth’s magnetic, field, of 0.3 × 10–4 Wb/m2. The value of the induced emf in, wire is :, [12 Jan. 2019 II], (a) 1.5 × 10–3 V, (b) 1.1 × 10–3 V, (c) 2.5 × 10–3V, (d) 0.3 × 10–3 V, 31. There are two long co-axial solenoids of same length l., The inner and outer coils have radii r1 and r2 and number, of turns per unit length n1 and n2, respectively. The ratio of, mutual inductance to the self-inductance of the inner-coil, is :, [11 Jan. 2019 I], (a), , n1, n2, , (b), , n2 r1, ×, n1 r2, , (c), , n2 r22, ×, n1 r12, , (d), , n2, n1, , 32. A copper wire is wound on a wooden frame, whose shape, is that of an equilateral triangle. If the linear dimension of, each side of the frame is increased by a factor of 3, keeping, the number of turns of the coil per unit length of the frame, the same, then the self inductance of the coil:, [11 Jan. 2019 II], (a) decreases by a factor of 9, (b) increases by a factor of 27, (c) increases by a factor of 3, (d) decreases by a factor of 9 3, 33. A solid metal cube of edge length 2 cm is moving in a, positive y-direction at a constant speed of 6 m/s. There, is a uniform magnetic field of 0.1 T in the positive, z-direction. The potential difference between the two, faces of the cube perpendicular to the x-axis, is:, [10 Jan. 2019 I], (a) 12 mV, (b) 6 mV, (c) 1 mV, (d) 2 mV
Page 361 :
P-352, , Physics, , 34. An insulating thin rod of length l has a linear charge, x, on it. The rod is rotated about an, l, axis passing through the origin (x = 0) and perpendicular, to the rod. If the rod makes n rotations per second, then, the time averaged magnetic moment of the rod is:, [10 Jan. 2019 I], p, n r l3, (a) p n r l3, (b), 3, p, n r l3, (c), (d) n r l3, 4, 35. A coil of cross-sectional area A having n turns is placed in, a uniform magnetic field B. When it is rotated with an, angular velocity w, the maximum e.m.f. induced in the coil, will be, [Online April 16, 2018], 3, nBAw, (a) nBAw, (b), 2, 1, nBAw, (c) 3nBAw, (d), 2, 36. An ideal capacitor of capacitance 0.2 mF is charged to a, potential difference of 10V. The charging battery is then, disconnected. The capacitor is then connected to an ideal, inductor of self inductance 0.5mH. The current at a time, when the potential difference across the capacitor is 5V, is:, [Online April 15, 2018], (a) 0.17A (b) 0.15A (c) 0.34A (d) 0.25A, 37. A copper rod of mass m slides under gravity on two smooth, parallel rails, with separation 1 and set at an angle of q, with the horizontal. At the bottom, rails are joined by a, resistance R.There is a uniform magnetic field B normal to, the plane of the rails, as shown in the figure. The terminal, speed of the copper rod is:, [Online April 15, 2018], , density r(x) = r0, , 39. A square frame of side 10 cm and a long straight wire, carrying current 1 A are in the plate of the paper. Starting, from close to the wire, the frame moves towards the right, with a constant speed of 10 ms–1 (see figure)., I = 1A, , x, v, , 10 cm, , The e.m.f induced at the time the left arm of the frame is at, x = 10 cm from the wire is:, [Online April 19, 2014], (a) 2 mV, (b) 1 mV, (c) 0.75 mV, (d) 0.5 mV, 40. A metallic rod of length ‘l’ is tied to a string of length 2l, and made to rotate with angular speed w on a horizontal, table with one end of the string fixed. If there is a vertical, magnetic field ‘B’ in the region, the e.m.f. induced across, the ends of the rod is, [2013], , ®, , B, , l, R, , (a), (c), , mgR cos q, B2l 2, mgR tan q, , (a), , q, , (b), , mgR sin q, B2l 2, mgR cot q, , (d), B2l 2, B2l 2, 38. At the centre of a fixed large circular coil of radius R, a much, smaller circular coil of radius r is placed. The two coils are, concentric and are in the same plane. The larger coil carries, a current I. The smaller coil is set to rotate with a constant, angular velocity w about an axis along their common, diameter. Calculate the emf induced in the smaller coil after a, time t of its start of rotation., [Online April 15, 2018], (a), , m0 I 2, wr sin wt, 2R, , (c), , m0 I, wpr 2 sin wt, 2R, , (b), , m0 I, wpr 2 sin wt, 4R, , (d), , m0 I 2, wr sin wt, 4R, , 2 Bwl2, 2, , (b), , 3Bwl 2, 2, , 4 Bwl2, 5Bwl 2, (d), 2, 2, 41. A coil of self inductance L is connected at one end of two, rails as shown in figure. A connector of length l, mass m, can slide freely over the two parallel rails. The entire set up, is placed in a magnetic field of induction B going into the, page. At an instant t = 0 an initial velocity v0 is imparted to, it and as a result of that it starts moving along x-axis. The, displacement of the connector is represented by the figure., [Online May 19, 2012], (c), , B, L, , x-axis
Page 362 :
P-353, , (b), , Time, , Displacement, , (a), , Displacement, , Electromagnetic Induction, , RW, , Time, , Displacement, , Displacement, , (d), , m0 NI, in the middle of the, L, solenoid but becomes less as we move towards its ends., [Online May 19, 2012], (a) Statement 1 is true, Statement 2 is false., (b) Statement 1 is true, Statement 2 is true, Statement 2 is, the correct explanation of Statement 1., (c) Statement 1 is false, Statement 2 is true., (d) Statement 1 is true, Statement 2 is true, Statement 2 is, not the correct explanation of Statement 1., 43. A boat is moving due east in a region where the earth's, magnetic field is 5.0 × 10–5 NA–1 m–1 due north and, horizontal. The boat carries a vertical aerial 2 m long. If the, speed of the boat is 1.50 ms–1, the magnitude of the induced, emf in the wire of aerial is:, [2011], (a) 0.75 mV, (b) 0.50 mV, (c) 0.15 mV, (d) 1mV, 44. A horizontal straight wire 20 m long extending from east to, west falling with a speed of 5.0 m/s, at right angles, to the horizontal component of the earth’s magnetic field, 0.30 × 10–4 Wb/m2. The instantaneous value of the e.m.f., induced in the wire will be, [2011 RS], (a) 3 mV, (b) 4.5 mV (c) 1.5 mV (d) 6.0 mV, 45. A rectangular loop has a sliding connector PQ of length l, and resistance R W and it is moving with a speed v as, shown. The set-up is placed in a uniform magnetic field, going into the plane of the paper. The three currents I1, I2, and I are, [2010], , I2, , Q, , Blv, 2 Blv, , I=, 6R, 6R, , (a), , I1 = - I 2 =, , (b), , I1 = I 2 =, , (c), , I1 = I 2 = I =, , Time, , pm0 N 2 r 2, ., L, Statement 2: The magnetic induction in the solenoid in, , RW, , I, , Time, , 42. This question has Statement 1 and Statement 2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., Statement 1: Self inductance of a long solenoid of length, L, total number of turns N and radius r is less than, , Statement 1 carrying current I is, , v, , RW, , I1, , (c), , l, , P, , Blv, 2 Blv, ,I =, 3R, 3R, Blv, R, , Bl n, Bl n, , I=, 6R, 3R, 46. Two coaxial solenoids are made by winding thin insulated, wire over a pipe of cross-sectional area A = 10 cm2 and, length = 20 cm. If one of the solenoid has 300 turns and the, other 400 turns, their mutual inductance is, [2008], (m0 = 4p × 10 –7 Tm A–1), (a) 2.4p × 10–5 H, (b) 4.8p × 10–4 H, (c) 4.8p × 10–5 H, (d) 2.4p × 10–4 H, 47. One conducting U tube can slide inside another as shown, in figure, maintaining electrical contacts between the tubes., The magnetic field B is perpendicular to the plane of the, figure . If each tube moves towards the other at a constant, speed v, then the emf induced in the circuit in terms of B, l, and v where l is the width of each tube, will be [2005], , (d), , I1 = I 2 =, , A, , B, v, , v, , X, C, , (a) – Blv, (b) Blv, (c) 2 Blv, (d) zero, 48. A metal conductor of length 1 m rotates vertically about one, of its ends at angular velocity 5 radians per second. If the, horizontal component of earth’s magnetic field is 0.2×10–4T,, then the e.m.f. developed between the two ends of the, conductor is, [2004], (a) 5 mV, (b) 50 mV, (c) 5 mV, (d) 50mV, 49. A coil having n turns and resistance RW is connected with, a galvanometer of resistance 4RW. This combination is, moved in time t seconds from a magnetic field W1 weber to, W2 weber. The induced current in the circuit is [2004]
Page 363 :
P-354, , (a), , Physics, , -, , (W2 - W1 ), Rnt, , (b), , (W2 - W1 ), 5 Rnt, , (d), , -, , n (W 2 - W1 ), 5 Rt, , n(W2 - W1 ), Rt, 50. Two coils are placed close to each other. The mutual, inductance of the pair of coils depends upon, [2003], (a) the rates at which currents are changing in the two, coils, (b) relative position and orientation of the two coils, (c) the materials of the wires of the coils, (d) the currents in the two coils, 51. When the current changes from +2 A to –2A in 0.05 second,, an e.m.f. of 8 V is induced in a coil. The coefficient of self, -induction of the coil is, [2003], (a) 0.2 H, (b) 0.4 H, (c) 0.8 H, (d) 0.1 H, , (c), , -, , -, , 52. A conducting square loop of side L and resistance R moves, in its plane with a uniform velocity v perpendicular to one, of its sides. A magnetic induction B constant in time and, space, pointing perpendicular and into the plane at the, loop exists everywhere with half the loop outside the field,, as shown in figure. The induced emf is, [2002], , L, , (a) zero, , (b) RvB, , v, , (c) vBL/R, , (d) vBL
Page 364 : P-355, , Electromagnetic Induction, , 1., , (5), For coil C1, No. of turns N1 = 500 and radius, r = 1 cm., For coil C2, No. of turns N2 = 200 and radius, R = 20 cm, dI, I = (5t - 2t + 3) Þ, = (10t - 2), dt, æ m IN ö, fsmall = BA = ç 0 2 ÷ (pr 2 ), è 2R ø, Induced emf in small coil,, , Current, i =, , But, resistance of wire, R = r, , 2, , e=, , \i =, , 5., , d f æ m 0 N 2 ö 2 di æ m 0 N1 N 2 pr 2 ö, =, pr N1 = ç, ÷ (10t - 2), dt çè 2r ÷ø, dt è, 2R, ø, , -d f, = ABw sin(wt ), dt, Max. value of Emf = ABw = pR2Bw, , 4(4p)10-7 ´ 200, 10 -4, ´ 500 ´ -2 p, 20, 10, , = 3.14 ´ 0.1 ´ 0.1 ´ 3 ´ 10 -5 ´, , 4, Þ x = 5., x, (b) Case (a) : When bar magnet is entering with constant, speed, flux (f) will change and an e.m.f. is induced, so, galvanometer will deflect in positive direction., Case (b) : When magnet is completely inside, flux (f) will, not change, so galvanometer will show null deflection., Case (c) : When bar magnet is making on exit, again flux, (f) will change and an e.m.f. is induced in opposite direction, so galvanometer will deflect in negative direction i.e., reverse direction., (a) As we know, emf e = NABw cos wt , Here N = 1, Average power,, A2 B 2 w 2 æ 1 ö, e, A B w cos wt, <P> = <, >=<, >=, çè ÷ø, R, 2, R, R, Therefore average power loss in the loop due to Joule, heating, 2, , 2, , 2, , p, 0.2, , = 15 ´ 10 -6 V = 15 mV, , = 8 ´ 10-4 volt = 0.8 mV=, , 2, , 2p, p, =, 2T 0.2, , Emf induced, e =, , = 80 ´ p 2 ´ 10-7 ´ 10 ´ 102 ´ 10-2, , 3., , dB ( A)2 0.032 ´ {p´ 2 ´ 10-3 }2, =, = 0.61 A., dt rl, 1.23 ´10-8 ´ 0.3, , r r, Flux as a function of time f = B × A = AB cos(wt ), , æ m N N pr 2 ö, m N N pr 2, e=ç 0 1 2, 8=4 0 1 2, ÷, 2R, R, è, ø, , 2., , l, A, , (15), Here, B = 3.0 × 10–5 T, R = 10 cm = 0.1 m, w=, , At t = 1 s, , =, , e, R, , 2, , p 2 a 2 b2 B 2 2, (w ), 2R, (b) Given,, Length of wire, l = 30 cm, Radius of wire, r = 2 mm = 2 × 10–3 m, , 6., , (10) Given dI = 0.25 – 0 = 0.25 A, dt = 0.025 ms, Induced voltage, Eind = 100 v, Self-inductance, L = ?, , L (0.25 – 0), Df, Þ 100 =, Dt, .025 ´10-3, –3, Þ L = 10 H = 10 mH, (a) According to question, dB = 1000 – 500 = 500 gauss, = 500 × 10–4T, Time dt = 5 s, Using faraday law, Using, Eind =, , 7., , Induced EMF , e = –, , df, dB, = A, dt, dt, , dB 1000 – 500, =, ´ 10 –4 = 10 –2 T/sec, dt, 5, , <P>=, , 4., , Resistivity of metal wire, r = 1.23 ´ 10 - 8 W m, d f dB, Emf generated, | e | =, =, ( A), dt, dt, , (Q f = B.A.), , Area, A = ar of X –2 ar of D = (16 × 4 – 2 × Area of triangle) cm2, , 1, æ, ö, = ç 64 – 2 ´ ´ 2 ´ 4 ÷ cm 2, 2, è, ø, = 56 × 10–4 m2
Page 366 :
P-357, , Electromagnetic Induction, , Magnitude of change in flux (df) = R × area under current, vs time graph, 1 1, ´ ´ 10 = 250 Wb, 2 2, 17. (b) Electric flux is given by, f = B.A, , or, df = 100 ´, , (Q B = B0e-t/ t ), , f = B0 pr 2 e - t / t, , Induced E.m.f. e =, Heat =, , df B0 pr 2 - t / t, e, =, dt, t2, , ¥ 2, , p 2 r 4 B02, e, =, ò R 2tR, , 18. (d) According to Faraday’s law of electromagnetic, induction,, Ldi, dt, , æ 5–2 ö, 50 = L ç, ÷, è 0.1sec ø, 50 ´ 0.1 5, = = 1.67 H, Þ L=, 3, 3, 19. (a) Inside the sphere field varies linearly i.e., E µ r with, 1, distance and outside varies according to E µ, r2, Hence the variation is shown by curve (a), 20. (a) Given: No. of turns N = 1000, Face area, A = 4 cm2 = 4 × 10–4 m2, Change in magnetic field,, DB = 10–2 wbm–2, Time taken, t = 0.01s = 10–2 sec, Emf induced in the coil e = ?, Applying formula,, Induced emf, e =, =, , -d f, DB ö, = N æç, ÷ A cos q, dt, è Dt ø, , 1000 ´ 10-2 ´ 4 ´ 10 -4, 10-2, , 2, , 2, , f = B × A = Bldr, Þf=, , ´ p(0.3 ´ 10 -2 ) 2, , 2[(0.2) + (0.15) ], On solving, = 9.216 × 10–11 = 9.2 × 10–11 weber, 22. (a) Induced emf, , - di, eµ, dt, 23. (d) Because of the Lenz’s law of conservation of energy., Length of straignt wire, l = 20m Earth’s Magneti field,, B = 0.30 × 10–4 Wb/m2., , [Q A = l dr and B.A = BA cos 0°], , m0 I, l dr, 2pr, , V, , I, r, Emf, e =, , dr, , m Ivl, d f m 0 Il dr, =, × Þe= 0 ×, dt, 2 pr dt, 2p r, , Induce current in the loop, i =, , e m 0 Ivl, =, ×, R 2p Rr, , 27. (b) Induced emf,, e = Bvl= 1 × 10–2 × 0.05 = 5 × 10–4 V, Equivalent resistance,, R=, , 4´ 2, 4, + 1.7 = + 1.7 ; 3 W, 3, 4+2, , Current, i =, , = 400 mV, , 21. (a) As we know, Magnetic flux, f = B. A, m0 (2)(20 ´ 10 -2 ) 2, , m0 I, 2 pr, Magnetic flux for small displacement dr,, B=, , 0, , Induced emf, e =, , Df, or Ri = Df, (Q e = Ri ), Dt, Dt, Þ Df = R(i.Dt), = R × area under i – t graph, 1, = 10 × × 4 × 0.1 = 2 weber, 2, 25. (b) Electric flux, f = 10t2 – 50t + 250, df, = - (20t - 50), Induced emf, e = dt, et = 3 = –10 V, 26. (d) Magnetic field at a distance r from the wire, 24. (c) As e =, , e 5 ´10-4, =, ; 170 m A, R, 3, , 28. (d) Inductance =, , m0 N 2 A, L, , 29. (c) Force on the strip when it is at stretched position x, from mean position is, F = -kx - iIB = -kx -, , BIv, ´ IB, R, , B2 I 2, ´v, R, Above expression shows that it is case of damped, oscillation, so its amplitude can be given by, F = -kx -
Page 368 :
P-359, , Electromagnetic Induction, , 38. (c) According to Faraday’s law of electromagnetic, induction,, df, e= and f = BA cos wt = Bpr 2 cos wt, dt, d, 2, 2, Þ e = - (pr B cos wt ) = pr B sin wt (w ), dt, m0 I, m0 I ö, æ, e, pwr 2 sin wt çQ=, B, \=, ÷, è, 2R, 2R ø, 39. (b) In the given question,, Current flowing through the wire, I = 1A, Speed of the frame, v = 10 ms–1, Side of square loop, l = 10 cm, Distance of square frame from current carrying wires, x = 10 cm., We have to find, e.m.f induced e = ?, According to Biot-Savart’s law, m Idlsin q, B= 0, 4p x 2, =, , 4p´10-7 1´ 10-1, ´, 2, 4p, 10-1, , ( ), , = 10–6, Induced e.m.f. e = Blv, = 10–6 × 10–1 × 10 = 1 mv, 40. (d) Here, induced e.m.f., w, l, 2l, x, , e=, , 3l, , dx, , ò (wx) Bdx = Bw, , 2l, , 45. (b) Due to the movement of resistor R, an emf equal to, Blv will be induced in it as shown in figure clearly,, P, l Blv, RW, , 5 Bl 2 w, =, 2, 41. (d), 42. (b) Self inductance of a long solenoid is given by, m0 N 2 A, l, Magnetic field at the centre of solenoid, m0 NI, B=, l, So both the statements are correct and statement 2 is, correct explanation of statement 1, 43. (d) As magnetic field lines form close loop,, hence every magnetic field line creating magnetic flux, through the inner region (fi) must be passing through, the outer region. Since flux in two regions are in opposite, region., , L=, , \ fi = –f0, 44. (a) Induced, emF, e = Bvl, = 0.3 × 10–4 × 5 × 20, = 3 × 10–3 V = 3 mV., , RW, , I, I2, , I1, , I = I1 + I 2, , Q, , Also, I1 = I2, , Solving the circuit,, we get I1 = I 2 =, and I = 2 I1 =, , Blv, 3R, , 2 Blv, 3R, , 46. (d) Given, Area of cross-section of pipe,, A = 10 cm2, Length of pipe, l = 20 cm, M=, =, , m0 N1 N2 A, l, , 4p ´ 10-7 ´ 300 ´ 400 ´ 100 ´ 10 -4, 0.2, , M=, , [(3l)2 – (2l)2 ], 2, , v, , RW, , m0 N1 N2 A, l, , = 2.4p × 10–4 H, 47. (c) Relative velocity of the tube of width l,, = v – (–v) v = 2v, \ Induced emf. = B.l (2v), 48. (b) Given, length of conductor l = 1m,, Angular speed, w = 5 rad/s,, Magnetic field, B = 0.2 × 10–4 T, EmF generated between two ends of conductor, e=, , 49. (b), , Bwl 2 0.2 ´ 10-4 ´ 5 ´ 1, =, = 50mV, 2, 2, , Df (W2 - W1 ), =, Dt, t, , Rtot = ( R + 4 R )W = 5R W, i=, , - n(W2 - W1 ), nd f, =, Rtot dt, 5Rt, , (Q W2 & W1 are magnetic flux), 50. (b) Mutual inductance depends on the relative position, and orientation of the two coils.
Page 369 :
P-360, , 51. (d) Induced emf,, Df -D ( LI ), DI, =, = -L, Dt, Dt, Dt, DI, \ | e |= L, Dt, [2 – (–2)], Þ 8= L ´, 0.05, 8 ´ 0.05, Þ L=, = 0.1H, 4, 52. (d) As the side BC is outside the field, no emf is induced, across BC. Further, sides AB and CD are not cutting any, flux. So, they will not centribute in flux., Only side AD is cutting the flux 50 emf will be induced due, to AD only., The induced emf is, e=-, , Physics, , r r, -d f, d ( B × A) -d ( BA cos0º ), e=, ==, dt, dt, dt, ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , × l, , ×, , ×, , ×, , ×, , ×, , A, , D, , ×, , ×, , C, , ×, , dA, d (l ´ x ), = -B, dt, dt, dx, \ e = - Bl = - Blv, dt, , \ e = –B, , V, , ×, X, , ×, , B
Page 370 :
21, Alternating Current, Alternating Current,, TOPIC 1, Voltage and Power, 1., , 2., , An alternating voltage v(t) = 220 sin 100Àt volt is applied, to a purely resistive load of 50W. The time taken for the, current to rise from half of the peak value to the peak, value is :, [8 April 2019 I], (a) 5 ms, (b) 2.2 ms (c) 7.2 ms (d) 3.3 ms, A small circular loop of wire of radius a is located at the, centre of a much larger circular wire loop of radius b. The, two loops are in the same plane. The outer loop of radius, b carries an alternating current I = Io cos (wt). The emf, induced in the smaller inner loop is nearly :, [Online April 8, 2017], (a), , 2, pm0 Io a 2, . w sin (wt) (b) pm0 Io . a w cos (wt), 2, b, 2, b, , a2, pm0 I o b 2, w sin (wt) (d), w cos ( wt), b, a, A sinusoidal voltage V(t) = 100 sin (500t) is applied across, a pure inductance of L = 0.02 H. The current through the, coil is:, [Online April 12, 2014], (a) 10 cos (500 t), (b) – 10 cos (500t), (c) 10 sin (500t), (d) – 10 sin (500t), In an a.c. circuit the voltage applied is E = E0 sin wt. The, , (a), , 4., , pö, æ, resulting current in the circuit is I = I 0 sin ç wt - ÷ . The, è, 2ø, power consumption in the circuit is given by, [2007], E I, (a) P = 2 E0 I0, (b) P = 0 0, 2, , E0 I 0, (d) P =, 2, In a uniform magnetic field of induction B a wire in the, form of a semicircle of radius r rotates about the diameter, of the circle with an angular frequency w. The axis of, rotation is perpendicular to the field. If the total resistance, of the circuit is R, the mean power generated per period of, rotation is, [2004], , (c) P = zero, 5., , (b), , ( B pr 2 w ) 2, 8R, , B pr 2 w, ( B pr w 2 ) 2, (d), 2R, 8R, Alternating current can not be measured by D.C. ammeter, because, [2004], (a) Average value of current for complete cycle is zero, (b) A.C. Changes direction, (c) A.C. can not pass through D.C. Ammeter, (d) D.C. Ammeter will get damaged., , (c), , 6., , AC Circuit, LCR Circuit,, TOPIC 2, Quality and Power Factor, 7., , (c) pm0 Io, 3., , ( B pr w )2, 2R, , A part of a complete circuit is shown in the figure. At some, instant, the value of current I is 1 A and it is decreasing at a, rate of 102A s–1. The value of the potential difference VP –VQ,, (in volts) at that instant, is ______., [NA Sep. 06, 2020 (I)], L=50 mH I, , R=2W, , P, , 8., , 9., , Q, 30 V, An AC circuit has R = 100 W, C = 2 mF and L = 80 mH,, connected in series. The quality factor of the circuit is :, [Sep. 06, 2020 (I)], (a) 2, (b) 0.5, (c) 20, (d) 400, , In a series LR circuit, power of 400 W is dissipated from a, source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In, order to bring the power factor to unity, a capacitor of value C, is added in series to the L and R. Taking the value C as, , æ nö, çè ÷ø mF , then value of n is ______. [NA Sep. 06, 2020 (II)], 3p, 10. A series L-R circuit is connected to a battery of emf V. If, the circuit is switched on at t = 0, then the time at which the, , æ 1ö, energy stored in the inductor reaches ç ÷ times of its, è nø, maximum value, is :, [Sep. 04, 2020 (II)]
Page 371 :
P-362, , L æ n ö, ÷, (a) R ln ç, è n -1 ø, , (b), , L æ n ö, ln ç, ÷, R è n +1ø, , (d), , L æ n +1ö, ln ç, ÷, R è n -1 ø, , L æ n -1 ö, ln ç, ÷, R è n ø, A 750 Hz, 20 V (rms) source is connected to a resistance of, 100 W, an inductance of 0.1803 H and a capacitance of 10, mF all in series. The time in which the resistance (heat, capacity 2 J/°C) will get heated by 10°C. (assume no loss, of heat to the surroundings) is close to :, , (c), 11., , Physics, , 2, 1, (a) ln 2, (b), ln 2, 2, (c) 2 ln 2, (d) ln 2, 16. A circuit connected to an ac source of emf e = e0sin(100t), , p, between, 4, the emf e and current i. Which of the following circuits, will exhibit this ?, [8 April 2019 II], , with t in seconds, gives a phase difference of, , [Sep. 03, 2020 (I)], , 12., , (a) 418 s, , (b) 245 s, , (c) 365 s, , (d) 348 s, , (a) 1.1 × 10–2 H, , (b) 1.1 × 10–1 H, , 5.5 × 10–5 H, , 6.7 × 10–7 H, , (c), 13., , (a) RL circuit with R = 1 kW and L = 10 mH, , An inductance coil has a reactance of 100 W. When an AC, signal of frequency 1000 Hz is applied to the coil, the applied, voltage leads the current by 45°. The self-inductance of, the coil is :, [Sep. 02, 2020 (II)], (d), , Consider the LR circuit shown in the figure. If the switch, S is closed at t = 0 then the amount of charge that passes, through the battery between t = 0 and t =, , L, is :, R, , (b) RL circuit with R = 1 kW and L = 1 mH, (d) RC circuit with R = 1 kW and C = 1 mF, (d) RC circuit with R = 1 kW and C = 10 mF., 17. In the figure shown, a circuit contains two identical, resistors with resistance R = 5 W and an inductance with L, = 2 mH. An ideal battery of 15 V is connected in the circuit., What will be the current through the battery long after the, switch is closed?, [12 Jan. 2019 I], , [12 April 2019 II], , S, L, R, , R, , (a), , (c), 14., , 2.7EL, R, , 2, , 7.3EL, R2, , (d), , EL, , (a) 5.5 A, (c) 3 A, , 2.7 R 2, EL, , 18., , 7.3R 2, , A coil of self inductance 10 mH and resistance 0.1 W is, connected through a switch to a battery of internal, resistance 0.9 W. After the switch is closed, the time taken, for the current to attain 80% of the saturation value is, [take ln 5 = 1.6], , 15., , (b), , (b) 7.5 A, (d) 6 A, I2, R2, R1, , C, L, , I1, , ~, , [10 April 2019 II], , (a) 0.324 s, , (b) 0.103 s, , (c) 0.002 s, , (d) 0.016 s, , A 20 Henry inductor coil is connected to a 10 ohm, resistance in series as shown in figure. The time at which, rate of dissipation of energy (Joule’s heat) across, resistance is equal to the rate at which magnetic energy is, stored in the inductor, is :, [8 April 2019 I], , 3, 3, µF, R2 = 20 W, L =, H and, 10, 2, R1 = 10 W. Current in L-R1 path is I1 and in C-R2 path it is, , In the above circuit, C =, , I2. The voltage of A.C source is given by, V = 200 2 sin, (100 t) volts. The phase difference between I1 and I2 is :, [12 Jan. 2019 II], (a) 60°, (b) 30°, (c) 90°, (d) 0
Page 372 :
P-363, , Alternating Current, , 19., , In the circuit shown,, R, , (a), , L, , (c), , 23., , S2, S1, e, the switch S1 is closed at time t = 0 and the switch S2 is, kept open. At some later time (t0), the switch S1 is opened, and S2 is closed. the behaviour of the current I as a function, of time ‘t’ is given by:, [11 Jan. 2019 II], I, , I, , (a), , (b), tO, , tO, , t, , tO, , 21., , t, , t, , tO, , t, , A series AC circuit containing an inductor (20 mH), a, capacitor (120 mF) and a resistor (60 W) is driven by an, AC source of 24 V/50 Hz. The energy dissipated in the, circuit in 60 s is:, [9 Jan. 2019 I], (a) 5.65 × 102 J, , (b) 2.26 × 103 J, , (c) 5.17 × 102 J, , (d) 3.39 × 103 J, , In LC circuit the inductance L = 40 mH and capacitance C, = 100 mF. If a voltage V(t) = 10 sin(314 t) is applied to the, circuit, the current in the circuit is given as:, [9 Jan. 2019 II], (a) 0.52 cos 314 t, (c) 5.2 cos 314 t, , (b), , 2, , (d), , eL, ÎL, , Î L æ 1ö, ç1 - ÷, R2 è e ø, ÎR, , R, eL2, A LCR circuit behaves like a damped harmonic oscillator., Comparing it with a physical spring-mass damped, oscillator having damping constant ‘b’, the correct, equivalence would be:, [7 Jan. 2020 I], (a) L « m, C « k, R « b, 1, 1, 1, (b) L « , C « , R «, b, m, k, (c) L « k, C « b, R « m, 1, (d) L « m, C « , R « b, k, An emf of 20 V is applied at time t = 0 to a circuit containing, in series 10 mH inductor and 5 W resistor. The ratio of the, currents at time t = ¥ and at t = 40 s is close to:, , [7 Jan. 2020 II], (b) 1.15, , (c) 1.46, (d) 0.84, 25. In an a.c. circuit, the instantaneous e.m.f. and current are, given by, e = 100 sin 30 t, , (d), , (c), , 2, , (Take e2 = 7.389), (a) 1.06, , I, , I, , 20., , 24., , ÎR, , (b) 10 cos 314 t, (d) 0.52 sin 314 t, , 22., , As shown in the figure, a battery of emf Î is connected to an, inductor L and resistance R in series. The switch is closed at, t = 0. The total charge that flows from the battery, between t, = 0 and t = tc (tc is the time constant of the circuit) is:, [8 Jan. 2020 II], , pö, æ, i = 20 sin ç 30 t - ÷, 4ø, è, In one cycle of a.c., the average power consumed by the, circuit and the wattless current are, respectively: [2018], 1000, (a) 50W, 10A, (b), W, 10A, 2, 50, W, 0, (c), (d) 50W, 0, 2, 26. For an RLC circuit driven with voltage of amplitude vm and, frequency w0 =, , 1, , the current exhibits resonance. The, LC, quality factor, Q is given by:, [2018], , CR, w0 L, w0 R, R, (b), (c), (d), w0, R, L, (w0C), 27. A sinusoidal voltage of peak value 283 V and angular, frequency 320/s is applied to a series LCR circuit. Given, that R = 5 W, L= 25 mH and C = 1000 mF. The total impedance,, and phase difference between the voltage across the, source and the current will respectively be :, [Online April 9, 2017], (a), , (a) 10 W and tan–1 æç 5 ö÷ (b) 7 W and 45°, è 3ø, , æ8ö, 5, (c) 10 W and tan –1 ç ÷ (d) 7 W and tan–1 æç ö÷, è3ø, è 3ø
Page 373 :
P-364, , 28., , 29., , Physics, , An arc lamp requires a direct current of 10 A at 80 V to, function. If it is connected to a 220 V (rms), 50 Hz AC, supply, the series inductor needed for it to work is close to :, [2016], (a) 0.044 H, (b) 0.065 H, (c) 80 H, (d) 0.08 H, A series LR circuit is connected to a voltage source with, V(t) = V0 sinwt. After very large time, current l(t) behaves, , Lö, æ, as çè t 0 >> ÷ø :, R, , (a) 6.7 mA, (b) 0.67 mA, (c) 100 mA, (d) 67 mA, 31. An LCR circuit is equivalent to a damped pendulum. In an, LCR circuit the capacitor is charged to Q0 and then, connected to the L and R as shown below :, , [Online April 9, 2016], , C, If a student plots graphs of the square of maximum charge, Q 2Max on the capacitor with time(t) for two different, values L1 and L2 (L1 > L2) of L then which of the following, represents this graph correctly ? (plots are schematic and, not drawn to scale), [2015], , I(t), , (, , (a), , ), , t, , t = t0, , L1, , 2, , I(t), , (a), , QMax, , L2, , 2, , (b), , QMax, , Q0 (For both L1 and L2), , t, , t, , t, , (b) t = t, 0, , 2, , (c), , QMax, , L1, , 2, , Q, (d) Max L1, , L2, , L2, , t, , I(t), , t, , t = t0, , t, , 32. For the LCR circuit, shown here, the current is observed, to lead the applied voltage. An additional capacitor C’,, when joined with the capacitor C present in the circuit,, makes the power factor of the circuit unity. The, capacitor C’, must have been connected in :, [Online April 11, 2015], , (c), , L, , I(t), , (d), , L, , R, , C, , R, , t, , t0, , ~, , 30., , An inductor (L = 0.03 H) and a resistor (R = 0.15 kW) are, connected in series to a battery of 15V emf in a circuit, shown below. The key K1 has been kept closed for a long, time. Then at t = 0, K1 is opened and key K2 is closed, simultaneously. At t = l ms, the current in the circuit will, be : ( e5 @ 150 ), , [2015], , 0.03 H, , 0.15 kW, K2, , 15V, , K1, , V = V0sintw, , (a) series with C and has a magnitude, , (b) series with C and has a magnitude, , 2, , C, , (w LC –1), , 1 - w2 LC, w2 L, , (c) parallel with C and has a magnitude, , 1 - w2 LC, w2 L, , (d) parallel with C and has a magnitude, , C, (w LC - 1), 2
Page 374 :
P-365, , Alternating Current, , 33., , V, , In the circuits (a) and (b) switches S1 and S2 are closed, at t = 0 and are kept closed for a long time. The variation of, current in the two circuits for t ³ 0 are roughly shown by, figure (figures are schematic and not drawn to scale) :, [Online April 10, 2015], C, , L, , R, , R, S1, C, , R, , L, , S1, E, E, R, , E, , (a), (b), , (a), , (a) i, , E, R, , (b), (a), , (b) i, (b), t, , t, E, R, , (a), , (b), , (c) i, , E, R, , (b), , (d) i, , (a) Work done by the battery is half of the energy, dissipated in the resistor, (b) At, t = t, q = CV/2, (c) At, t = 2t, q = CV (1 – e–2), (d) At, t = 2 t, q = CV (1 – e–1), 37. A series LR circuit is connected to an ac source of, frequency w and the inductive reactance is equal to 2R. A, capacitance of capacitive reactance equal to R is added in, series with L and R. The ratio of the new power factor to, the old one is :, [Online April 25, 2013], , (a), , (a), In the circuit shown here, the point ‘C’ is kept connected, to point ‘A’ till the current flowing through the circuit, becomes constant. Afterward, suddenly, point ‘C’ is, disconnected from point ‘A’ and connected to point ‘B’ at, time t = 0. Ratio of the voltage across resistance and the, inductor at t = L/R will be equal to:, [2014], A, , C, , R, L, , B, , (a), 35., , 36., , e, 1- e, , (b) 1, , (c) –1, , (d), , 1- e, e, , When the rms voltages VL, VC and VR are measured, respectively across the inductor L, the capacitor C and the, resistor R in a series LCR circuit connected to an AC source,, it is found that the ratio VL : VC : VR = 1 : 2 : 3. If the rms, voltage of the AC sources is 100 V, the VR is close to:, [Online April 9, 2014], (a) 50 V, (b) 70 V (c) 90 V, (d) 100 V, In an LCR circuit as shown below both switches are open, initially. Now switch S1 is closed, S2 kept open. (q is charge, on the capacitor and t = RC is Capacitive time constant)., Which of the following statement is correct ?, [2013], , 2, 3, , 2, 5, , (b), , (c), , 3, 2, , (d), , 5, 2, , 38. When resonance is produced in a series LCR circuit, then, which of the following is not correct ?, [Online April 25, 2013], (a) Current in the circuit is in phase with the applied, voltage., (b) Inductive and capacitive reactances are equal., (c) If R is reduced, the voltage across capacitor will, increase., (d) Impedance of the circuit is maximum., 39. The plot given below is of the average power delivered to, an LRC circuit versus frequency. The quality factor of the, circuit is :, [Online April 23, 2013], average power (microwatts), , t, , t, , 34., , S2, , 1.0, 0.5, 0.0, 3, , 4, 5, 6, 7, frequency (kHz), , (a) 5.0, (b) 2.0, (c) 2.5, (d) 0.4, –11, 40. In a series L-C-R circuit, C = 10 Farad, L = 10–5 Henry, and R = 100 Ohm, when a constant D.C. voltage E is applied, to the circuit, the capacitor acquires a charge 10 –9 C. The, D.C. source is replaced by a sinusoidal voltage source in
Page 375 :
P-366, , 41., , Physics, , which the peak voltage E0 is equal to the constant D.C., voltage E. At resonance the peak value of the charge, acquired by the capacitor will be : [Online April 22, 2013], (a) 10–15 C (b) 10–6 C (c) 10–10 C (d) 10–8 C, An LCR circuit as shown in the figure is connected to a, voltage source Vac whose frequency can be varied., V, , 24 H, , 2 µF, , ~, , 42., , 15 W, , Vac = V0 sin wt, , The frequency, at which the voltage across the resistor is, maximum, is :, [Online April 22, 2013], (a) 902 Hz (b) 143 Hz (c) 23 Hz, (d) 345 Hz, In the circuit shown here, the voltage across E and C are, respectively 300 V and 400 V. The voltage E of the ac source, is :, [Online April 9, 2013], L, , C, , ~, , E, , (a) 400 Volt (b) 500 Volt(c) 100 Volt (d) 700 Volt, 43., , (a), , p, LC, 4, , (a) 1.7 × 105 W, , (b) 2.7 × 106 W, , (c) 3.3 × 107 W, , (d) 1.3 × 104 W, , 47. Combination of two identical capacitors, a resistor R and, a dc voltage source of voltage 6V is used in an experiment, on a (C-R) circuit. It is found that for a parallel combination, of the capacitor the time in which the voltage of the fully, charged combination reduces to half its original voltage is, 10 second. For series combination the time for needed for, reducing the voltage of the fully charged series, combination by half is, [2011 RS], (a) 10 second, , (b) 5 second, , (c) 2.5 second, , (d) 20 second, , 48. In the circuit shown below, the key K is closed at t = 0. The, current through the battery is, [2010], , (c) 2 RC ln 7, , C, , R1, , R2, , In an LCR circuit shown in the following figure, what will, be the readings of the voltmeter across the resistor and, ammeter if an a.c. source of 220V and 100 Hz is connected, to it as shown?, [Online May 7, 2012], , L, , L, , 3 RC ln 7, , (d), , K, , V, , 2 RC ln 2, , (b), , (b) 2p LC, , (c) LC, (d) p LC, 46. A resistor ‘R’ and 2µF capacitor in series is connected, through a switch to 200 V direct supply. Across the, capacitor is a neon bulb that lights up at 120 V. Calculate, the value of R to make the bulb light up 5 s after the switch, has been closed. (log10 2.5 = 0.4), [2011], , A resistance R and a capacitance C are connected in series, to a battery of negligible internal resistance through a key., The key is closed at t = 0. If after t sec the voltage across, the capacitance was seven times the voltage across R, the, value of t is, [Online May 12, 2012], (a) 3 RC ln 2, , 44., , 45. A fully charged capacitor C with initial charge q0 is, connected to a coil of self inductance L at t = 0. The time at, which the energy is stored equally between the electric, and the magnetic fields is:, [2011], , 100 W, , (a), , VR1R2, R12 + R22, , V, at t = 0 and R at t = ¥, 2, , V ( R1 + R2 ), V, at t = 0 and, at t = ¥, R1 R2, R2, VR1R2, V, (c), at t = 0 and, at t = ¥, R2, R 2 + R2, , (b), , 1, , V, , V, , V, , 300V 300 V, , A, , VR, , 220 V, 100 Hz, (a) 800 V, 8 A, (c) 300 V, 3 A, , (b) 110 V, 1.1 A, (d) 220V, 2.2 A, , (d), , 2, , V ( R1 + R2 ), V, at t = 0 and, at t = ¥, R1 R2, R2, , 49. In a series LCR circuit R = 200W and the voltage and the, frequency of the main supply is 220V and 50 Hz, respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. On taking out, the inductor from the circuit the current leads the voltage, by 30°. The power dissipated in the LCR circuit is [2010], (a) 305 W (b) 210 W (c) Zero W (d) 242 W
Page 376 :
P-367, , Alternating Current, , 50., , E, L, R1, R2, S, , An inductor of inductance L = 400 mH and resistors of, resistance R1 = 2W and R2 = 2W are connected to a battery, of emf 12 V as shown in the figure. The internal resistance, of the battery is negligible. The switch S is closed at t = 0., The potential drop across L as a function of time is [2009], (a), , 51., , 52., , (, , 12 -3t, e V, t, , ), , -t / 0.2, V, (b) 6 1 - e, , (c) 12e–5t V, (d) 6e–5t V, In a series resonant LCR circuit, the voltage across R is, 100 volts and R = 1 kW with C = 2mF. The resonant, frequency w is 200 rad/s. At resonance the voltage across, L is, [2006], –2, (a) 2.5 × 10 V, (b) 40 V, (c) 250 V, (d) 4 × 10–3 V, An inductor (L = 100 mH), a resistor (R = 100 W) and a, battery (E = 100 V) are initially connected in series as, shown in the figure. After a long time the battery is, disconnected after short circuiting the points A and B., The current in the circuit 1 ms after the short circuit is, [2006], L, , B, E, , 53., , (a) 1/eA, (b) eA, (c) 0.1 A, (d) 1 A, In an AC generator, a coil with N turns, all of the same area, A and total resistance R, rotates with frequency w in a, magnetic field B. The maximum value of emf generated in, the coil is, [2006], (a) N.A.B.R.w, (b) N.A.B, (c) N.A.B.R., (d) N.A.B.w, The phase difference between the alternating current and, p, emf is . Which of the following cannot be the constituent, 2, of the circuit?, [2005], (a) R, L, (b) C alone(c) L alone (d) L, C, A circuit has a resistance of 12 ohm and an impedance of, 15 ohm. The power factor of the circuit will be, [2005], (a) 0.4, (b) 0.8, (c) 0.125, (d) 1.25, , (c) 2 mF, , (d) 1 mF, , 58. In an LCR series a.c. circuit, the voltage across each of the, components, L, C and R is 50V. The voltage across the LC, combination will be, [2004], (b) 50 2 V, , (a) 100 V, , (c) 50 V, (d) 0 V (zero), 59. In a LCR circuit capacitance is changed from C to 2 C. For, the resonant frequency to remain unchanged, the, inductance should be changed from L to, [2004], (a) L/2, (b) 2 L, (c) 4 L, (d) L/4, 60. The power factor of an AC circuit having resistance (R), and inductance (L) connected in series and an angular, velocity w is, [2002], (a) R/ w L, (c), , w 2L2)1/2, (d) R/(R2 – w 2L2)1/2, (b) R/(R2 +, , w L/R, , 61. The inductance between A and D is, , (a) 3.66 H, (c) 0.66 H, , TOPIC 3, , 3H, , 3H, , 3H, , [2002], , D, , (b) 9 H, (d) 1 H, , Transformers and LC, Oscillations, , 62. For the given input voltage waveform Vin(t), the output, voltage waveform Vo(t), across the capacitor is correctly, depicted by :, [Sep. 06, 2020 (I)], 1kW, +5V 5m s, 0V, , 10nF, t, , m, , VO(t), , s, , 55., , (b) 4 mF, , 5, , 54., , (a) 8 mF, , A, , R, A, , 56. A coil of inductance 300 mH and resistance 2 W is, connected to a source of voltage 2V. The current reaches, half of its steady state value in, [2005], (a) 0.1 s, (b) 0.05 s (c) 0.3 s, (d) 0.15 s, 57. The self inductance of the motor of an electric fan is 10 H., In order to impart maximum power at 50 Hz, it should be, connected to a capacitance of, [2005], , 0, Vo(t), 3V, , (a) 2V, 5ms, , 10ms 15ms, , t
Page 377 :
P-368, , Physics, , Vo(t), (b), , having 4000 turns. The output power is delivered at 230, V by the transformer. If the current in the primary of the, transformer is 5A and its efficiency is 90%, the output, current would be:, [9 Jan. 2019 II], , 2V, , 5ms, , 10ms 15ms, , t, , Vo(t), (c), , 2V, 5ms, Vo(t), , (d), , 63., , t, , (b) 45 A, , (c) 35 A, , (d) 25 A, , 65. A power transmission line feeds input power at 2300 V to a, step down transformer with its primary windings having, 4000 turns, giving the output power at 230 V. If the current in, the primary of the transformer is 5 A, and its efficiency is, 90%, the output current would be: [Online April 16, 2018], (a) 20 A, (b) 40 A (c) 45 A, (d) 25 A, 66. In an oscillating LC circuit the maximum charge on the, capacitor is Q. The charge on the capacitor when the, energy is stored equally between the electric and magnetic, field is, [2003], , 2V, , t, 5ms 10ms 15ms, A transformer consisting of 300 turns in the primary and, 150 turns in the secondary gives output power of 2.2kW., If the current in the secondary coil is 10 A, then the input, voltage and current in the primary coil are :, (a) 220 V and 20 A, (c) 440 V and 5 A, , 64., , 10ms 15ms, , (a) 50 A, , [10 April 2019 I], (b) 440 V and 20 A, (d) 220 V and 10 A, , A power transmission line feeds input power at 2300 V, to a step down transformer with its primary windings, , (a), , Q, 2, , (b), , Q, 3, , (c), , Q, 2, , (d) Q, , 67. The core of any transformer is laminated so as to [2003], (a) reduce the energy loss due to eddy currents, (b) make it light weight, (c) make it robust and strong, (d) increase the secondary voltage, 68. In a transformer, number of turns in the primary coil are 140, and that in the secondary coil are 280. If current in primary, coil is 4 A, then that in the secondary coil is, [2002], (a) 4 A, (b) 2 A, (c) 6 A, (d) 10 A.
Page 383 :
P-374, , 40., 41., 42., 43., 44., , Physics, , (d), (c) Frequency f =, , 1, 2p LC, , =, , V = V0 e, , 1, 2 ´ 3.4 24 ´ 2 ´ 10 -6, , ; 23Hz, (c) Voltage E of the ac source, E = VC – VL = 400 V – 300 V = 100 V, (a) t = 3 RC ln 2, (d) In case of series RLC circuit,, Equation of voltage is given by, , 48., , 2, , \ VR = V 2 = 220V, , 45., , V 220, =, = 2.2A, R 100, , (a) Energy stored in magnetic field =, , 1 2, Li, 2, , 49., , 1 q2, Energy stored in electric field =, 2 C, , Energy will be equal when, \, , 1 2 1 q2, Li =, 2, 2 C, , tan wt = 1, q = q0 cos wt, 1, ( q cos wt )2, Þ L(wq0 sin wt)2 = 0, 2, 2C, p, 1, Þw=, Þ wt =, 4, LC, Þ t=, 46., , ....(2), , tan f =, , 1, w CR, , Þ, , (b) We have, V = V0(1 – e–t/RC), , æ 1, ö, Z = R2 + ç, - wL ÷, w, C, è, ø, , 200 – 120, 80, =, 200, 200, t = loge(2.5), , e–t/r =, , Þ t = RC in (2.5), , [Q r = RC], , =, , = Vrms ×, , =, , ...(1), , In series grouping, equivalent capacitance =, , æ 200 200 ö, (200) 2 + ç, ÷ = 200 W, 3ø, è 3, , Power dissipated in the circuit = VrmsIrms cos f, , ), , V, V = V0, = 0, 2, In second case :, , 2, , 2, , Þ R = 2.71 × 106 W, (c) Time constant for parallel combination, = 2RC, RC, Time constant for series combination =, 2, In first case :, t, –, t, V0, –, CR, = V0 – V0e, V = V0, Þ, 1 – e CR, 2, t, - 1, RC, 2, e, , V0, 2, , 1, 200, 1, =, = R tan f = 200 ´, 3, 3, wC, Impedance of the circuit,, , p, LC, 4, , (, , =, , t 10, Þ t2 = 1 = = 2.5 sec., 4 4, (c) At t = 0, no current will flow through L and R1 as, inductor will offer infinite resistance., V, \ Current through battery, i =, R2, At t = ¥, inductor behave as conducting wire, RR, Effective resistance, Reff = 1 2, R1 + R2, V, V ( R1 + R2 ), \ Current through battery = R =, R1R2, eff, (d) When only the capacitance is removed phase, difference between current and voltage is, XL, tan f =, R, wL, Þ tan f =, R, 1, 200, =, Þ wL = R tan f = 200 ´, 3, 3, When only inductor is removed, phase difference between, current and voltage is, , \, , Þ 120 = 200(1 – e–t/RC), , 47., , t2, ( RC / 2), , From (1) and (2), t1, t2, =, 2 RC ( RC / 2 ), , V 2 = VR2 + (VL - VC ), Here, V = 220 V; VL = VC = 300 V, , Current i =, , -, , C, 2, , Vrms R, ×, Z Z, , (220)2 ´ 200, (200)2, , R ö V 2rms R, æ, çQ cos f = ÷ =, Zø, è, Z2, , =, , 220 ´ 220, = 242 W, 200, , 50. (c) Growth in current in branch containing L and R2 when, switch is closed is given by, E, [1 - e - R2t / L ], i=, R2, Þ, , di E R2 - R2t / L, E =, × ×e, = e, dt R2 L, L, , R2t, L
Page 384 :
P-375, , Alternating Current, , Hence, potential drop across L, Ldi æ E - R2t / L ö, =ç e, VL =, ÷L, dt è L, ø, = Ee, , 51., , - R2t / L, , = 12e, , -, , 2t, , 400 ´10 -3, , (c) Across resistor, I =, At resonance,, , = 12e–5tV, , V, 100, =, = 0.1 A, R 1000, , 1, 1, =, = 2500, wC 200 ´ 2 ´ 10-6, Voltage across L is, I X L = 0.1 ´ 2500 = 250 V, (a) Initially, when steady state is achieved,, X L = XC =, , 52., , E, i=, R, Let E is short circuited at t = 0. Then, At t = 0, E 100, = 1A, =, R 100, Let during decay of current at any time the current flowing, di, is - L - iR = 0, dt, di, R, Þ = - dt, i, L, , Maximum current, i0 =, , i, , t, , i0, , 0, , i, R, =- t, i0, L, , Þ i = i0, , 54., , 55., , 56., , 1, 2, , (2 pn ) L, , i = i0 1 - e, , -, , 4p ´ 50 ´ 50 ´ 10, , \ Net voltage difference across, LC = 50 – 50 = 0, , 1, 2p LC, For resonant frequency to remain same, LC = constant, \ LC = L' C', Þ LC = L¢ × 2C, , 59. (a) Resonant frequency, Fr =, , L, 2, 60. (b) Resistance of the inductor, XL = wL, The impedance triangle for resistance (R) and inductor (L), connected in series is shown in the figure., R, , -100´10 -3, , ), , Rt, , 2, , 58. (d) In a series LCR circuit voltage across the inductor, and capacitor are in opposite phase, , 2, , Rt, L, , 1, , =, , \ C = 0.1 ´ 10-5 F = 1mF, , 2, , -3, 1, E - t, Þ i = e L = 1 ´ e 100´10 =, R, e, ur ur, df, d ( N B. A), =(d) e = dt, dt, d, = - N ( BA cos wt ) = NBAw sin wt, dt, Þ e max = NBAw, (a) Phase difference for R–L circuit lies between, æ pö, ç 0, ÷ but 0 or p/2, è 2ø, (b) Given, Resistance of circuit, R = 12 W, Imedance of circuit, Z = 15 W, R, 12 4, =, = = 0.8, Power factor = cos f =, Z, 15 5, (a) Current in inductor circuit is given by,, , (, , C=, , R, - t, e L, , R, , 53., , L, 300 ´ 10 -3, ´ 0.69, log 2 =, R, 2, Þ t = 0.1 sec., 57. (d) For maximum power, XL = XC, which yields, , Þ t=, , Þ L' =, , di, R, Þ ò = ò - dt, i, L, Þ log e, , Taking log on both the sides,, Rt, = log1 - log 2, L, , Rt, , i0, 1, = i0 (1 - e L ) Þ e L =, 2, 2, , +w, , 2, , L, , XL= w L, , f, R, , Net impedance of circuit Z =, Power factor, cos f =, Þ cos f =, , X L2 + R 2, , R, Z, , R, 2, , R + w 2 L2, 61. (d) All three inductors are connected in parallel. The, equivalent inductance Lp is given by, 1, 1, 1, 1, 1 1 1 3, = +, +, = + + = =1, L p L1 L2 L3 3 3 3 3, , \ Lp = 1, 62. (a) When first pulse is applied, the potential across, capacitor, 1 ö, æ, V0 (t ) = Vin ç1 - e RC ÷, è, ø, , At t = 5ms = 5 ´ 10-6 s
Page 385 :
P-376, , Physics, , Þ Is = 0.9 ´ 50 = 45A, , 10m F, , Vin, , V0(t), , Efficiency n = 0.9 =, , 5´10-6, æ, ö, 3, -9, 10, V0 (t ) = 5 ç 1 - e ´10 ´10 ÷ = 5(1 - e -0.5 ) = 2V, ç, ÷, è, ø, When no pulse is applied, capacitor will discharge., Now, Vin = 0 means discharging., 1, , V0 (t ) = 2e RC = 2e -0.5 = 1.21 V, , Now for next 5 ms, 1, , V0 (t ) = 5 - 3.79e RC, , After 5 ms again, V0 (t ) = 2.79 Volt » 3 V, Hence, graph (a) correctly depicts., 63., , (c) Power output (V2I2) = 2.2 kW, 2.2kW, = 220 volts, (10A ), , \ V2 =, , \ Input voltage for step-down transformer, V1 N1, =, =2, V2 N 2, , Vinput = 2 × Voutput = 2 × 220, = 440 V, I1 N 2, Also I = N, 2, 1, , \, , 64., , I1 =, , 1, ´ 10 = 5A, 2, , P, VI, (b) Efficiency, h= out = s s, Pin VpIp, Þ 0.9 =, , Output current = 45A, 65. (c) Given : VP = 2300 V, Vs = 230 V, IP = 5 A, n = 90% = 0.9, , 230 ´ Is, 2300 ´ 5, , Ps, Þ Ps = 0.9 Pp, PP, , VsIs = 0.9 ×VPIP (Q P = VI), Is =, , 0.9 ´ 2300 ´ 5, = 45A, 230, , 66. (c) When the capacitor is completely charged, the total, energy in the LC circuit is with the capacitor and that, energy is given by, Umax =, , 1 Q2, 2 C, , When half energy is with the capacitor in the form of electric, field between the plates of the capacitor we get, U max 1 q¢2, =, 2, 2 C, , Here q' is the charge on the plate of capacitor when energy, is shared equally., \, , Q, 1 1 Q 2 1 q¢2, Þ q¢ =, ´, =, 2 2 C, 2 C, 2, , 67. (a) Laminated core provide less area of cross-section for, the current to flow. Because of this, resistance of the core, increases and current decreases there by decreasing the, energy loss due to eddy current., 68. (b) Number of turns in primary, Np = 140, Number of turns in secondary Ns = 280, Ip = 4A, Is = ?, Using transformation ratio for a transformer, I s 140, =, 4 280, Þ Is = 2 A, , Þ, , Is N p, =, Ip, Ns
Page 386 :
22, Electromagnetic, Waves, Electromagnetic Waves,, TOPIC 1 Conduction, and Displacement Current, 1., , For a plane electromagnetic wave, the magnetic field at a, point x and time t is, , 4., , ®, , B( x, t ) = [1.2 ´ 10-7 sin(0.5 ´ 103®x + 1.5 ´ 1011t )k$ ]T ®, The instantaneous electric field E corresponding to B, is: (speed of light c = 3 × 108 ms–1) [Sep. 06, 2020 (II)], , where c = 3 × 108 ms–1 is the speed of light., The corresponding electric field is : [Sep. 03, 2020 (I)], ur, (a) E = 9sin[200p ( y + ct )]kˆ V/m, ur, (b) E = -10-6 sin[200p ( y + ct )]kˆ V/m, ur, (c) E = 3 ´ 10-8 sin[200p ( y + ct )]kˆ V/m, ur, (d) E = -9sin[200p ( y + ct )]kˆ V/m, , ®, , V, 3, 11, (a) E ( x, t ) = [ -36 sin(0.5 ´ 10 x + 1.5 ´ 10 t ) $j ], m, ®, , V, (b) E( x, t ) = [36 sin(1 ´ 103 x + 0.5 ´ 1011 t ) $j ], m, ®, , V, (c) E ( x, t ) = [36 sin(0.5 ´ 103 x + 1.5 ´ 1011 t ) k$ ], m, , 5., , ®, , 2., , V, (d) E ( x, t ) = [36 sin(1 ´ 103 x + 1.5 ´ 1011 t )$i], m, An electron is constrained to move along the y-axis with a, speed of 0.1 c (c is the speed of light) in the presence, of, ®, electromagnetic wave, whose electric field is E = 30j$, sin(1.5 × 107t – 5 × 10–2x) V/m. The maximum magnetic force, experienced by the electron will be :, , (given c = 3 × 108 ms–1 & electron charge = 1.6 × 10–19C), [Sep. 05, 2020 (I)], (a), 3., , 3.2 × 10–18 N, , (b), , E0, (- xˆ + yˆ ) sin(kz - wt ), c, E, (b) 0 ( xˆ + yˆ )sin(kz - wt ), c, , (a), , The electric field of a plane electromagnetic wave, propagating along the x direction in vacuum is, r, r, E = E0 ˆj cos(wt - kx ) . The magnetic field B, at the, moment t = 0 is :, [Sep. 03, 2020 (II)], r, E0, (a) B =, cos(kx)kˆ, m0 e 0, r, (b) B = E0 m0 e0 cos(kx) ˆj, r, (c) B = E0 m0 e0 cos(kx)kˆ, , r, (d) B =, , 2.4 × 10–18 N, , (c) 4.8 × 10–19 N, (d) 1.6 × 10–19 N, The electric field of a plane electromagnetic wave is given, r, by E = E0 ( xˆ + yˆ ) sin(kz - wt ), Its magnetic field will be given by : [Sep. 04, 2020 (II)], , E0, ( xˆ - yˆ ) sin(kz - wt ), c, E, (d) 0 ( xˆ - yˆ ) cos(kz - wt ), c, The magnetic field of a plane electromagnetic wave is, ur, ˆ, B = 3 ´ 10-8 sin[200p ( y + ct )]iT, , (c), , 6., , E0, m0 e 0, , cos(kx) ˆj, , A plane electromagnetic wave, has frequency of 2.0 × 1010, Hz and its energy density is 1.02 × 10–8 J/m3 in vacuum., The amplitude of the magnetic field of the wave is close to, , Nm 2, 1, = 9 ´109 2 and speed of light = 3 × 108 ms–1) :, 4pe0, C, [Sep. 02, 2020 (I)], (a) 150 nT, (b) 160 nT, (c) 180 nT, (d) 190 nT, , (
Page 387 :
P-378, , 7., , In a plane electromagnetic wave, the directions of electric, field and magnetic field are represented by k̂ and 2iˆ - 2 ˆj,, respectively. What is the unit vector along direction of, propagation of the wave., [Sep. 02, 2020 (II)], (a), , 8., , Physics, , 1 ˆ ˆ, (i + j ), 2, , (b), , 1, 2, , ( ˆj + kˆ), , 1 ˆ ˆ, 1 ˆ, (2i + j ), (c), (i + 2 ˆj ), (d), 5, 5, The electric fields of two plane electromagnetic plane, waves in vacuum are given by, ur, ur, E1 = E0 ˆj cos(wt - kx ) and E2 = E 0 kˆ cos(wt - ky ) ., , At t = 0, a particle of charge q is at origin with a velocity, r, v = 0.8 cjˆ (c is the speed of light in vacuum). The, instantaneous force experienced by the particle is:, [9 Jan 2020, I], ˆ, (a) E0 q(0.8iˆ - ˆj + 0.4k ) (b) E0 q(0.4iˆ - 3 ˆj + 0.8kˆ), 9., , (c) E0 q(- 0.8iˆ + ˆj + kˆ) (d) E0 q( 0.8iˆ + ˆj + 0.2kˆ), A plane electromagnetic wave is propagating along the, iˆ + ˆj, , with its polarization along the direction, direction, 2, kˆ. The correct form of the magnetic field of the wave, would be (here B0 is an appropriate constant):, [9 Jan 2020, II], æ, iˆ - ˆj, iˆ + ˆj ö, cos ç wt - k, (a) B0, è, 2, 2 ÷ø, (b) B0, , ˆj - iˆ, , æ, iˆ + ˆj ö, cos ç wt + k, è, 2, 2 ÷ø, , æ, iˆ + ˆj ö, ˆ, (c) B0 k cos ç wt - k, è, 2 ÷ø, æ, iˆ + ˆj ö, cos ç wt - k, è, 2, 2 ÷ø, A plane electromagnetic wave of frequency 25 GHz is, propagating in vacuum along the z-direction. At a, particular point in space and time, the magnetic field is, r, given by B = 5 ´ 10-8 ˆj T . The corresponding electric, r, field E is (speed of light c = 3 ´ l08 ms–l), [8 Jan 2020, II], (a) 1.66 ´ 10–16 iˆ V/m, (b) – 1.66 ´ 10–16 iˆ V/m, , (d) B0, , 10., , 11., , iˆ + ˆj, , (c) –15 iˆ V/m, (d) 15 iˆ V/m, If the magnetic field in a plane electromagnetic wave is, ur, given by B = 3 ´ 10–8 sin (l.6 ´ 103x + 48 ´ 1010t) ĵ T, then, what will be expression for electric field?, [7 Jan 2020, I], ur, (a) E = (60 sin (1.6 ´ l03x + 48 ´ l010t) k̂ v/m), ur, (b) E = (9 sin (1.6 ´ l03x + 48 ´ l010t) k̂ v/m), , ur, (c) E = (3 ´ l0–8 sin (l.6 ´ l03x + 48 ´ l010t) k̂ v/m), ur, (d) E = (3 ´ l0–8sin (l.6 ´ l03x + 48 ´ l010t) k̂ v/m), 12. The electric field of a plane electromagnetic wave is, given by, r E iˆ + ˆj cos(kz + wt ), E = 0 2, At t = 0, a positively charged particle is at the point, , pö, æ, (x, y, z) = çè 0, 0, ÷ø . If its instantaneous velocity at (t = 0), k, is v0 kˆ , the force acting on it due to the wave is:, [7 Jan 2020, II], iˆ + ˆj, (b) zero, (a) parallel to, 2, iˆ + ˆj, (c) antiparallel to, (d) parallel to k̂, 2, 13. An electromagnetic wave is represented by the electric, ur, field E = E n$ sin[w t + (6 y - 8z)] . Taking unit vectors in, 0, , x, y and z directions to be $i , $j , k$ , the direction of, propogation $s is :, 3$i - 4 $j, (a) s$ =, 5, , [12 April 2019, I], , -4k$ + 3 $j, (b) s$ =, 5, 3 $j - 3k$, (d) s$ =, 5, , æ -3$j + 4k$ ö, (c) s$ = ç, ÷÷, ç, 5, è, ø, 14. A plane electromagnetic wave having a frequency, v = 23.9 GHz propagates along the positive z-direction in, free space. The peak value of the Electric Field is 60 V/m., Which among the following is the acceptable magnetic, field component in the electromagnetic wave ?, [12 April 2019, II], ur, 7, 3, (a) B = 2 ´ 10 sin(0.5 ´ 10 z + 1.5 ´ 1011 t )$i, ur, (b) B = 2 ´ 10-7 sin(0.5 ´ 103 z - 1.5 ´ 1011 t )$i, ur, (c) B = 60sin(0.5 ´ 103 x + 1.5 ´ 1011 t )k$, ur, -7, 2, 11 $, (d) B = 2 ´ 10 sin(1.5 ´ 10 x + 0.5 ´ 10 t ) j, 15. The electric field of a plane electromagnetic wave is given, ur, by E = E $i cos(kz) cos(w t), 0, , The corresponding magnetic field is then given by :, [10 April 2019, I], ur E 0, $j sin (kz) sin (w t), (a) B =, C, ur E 0, $j sin(kz) cos ( w t), (b) B =, C, ur E 0, $j cos (kz) sin (w t), (c) B =, C, ur E 0, k$ sin (kz) cos( w t), (d) B =, C
Page 388 :
P-379, , Electromagnetic Waves, , 16., , 17., , Light is incident normally on a completely absorbing, surface with an energy flux of 25 Wcm–2. If the surface has, an area of 25 cm2, the momentum transferred to the surface, in 40 min time duration will be:, [10 April 2019, II], (a) 6.3×10–4 Ns, (b) 1.4×10–6 Ns, (c) 5.0×10–3 Ns, (d) 3.5×10–6 Ns, The magnetic field of a plane electromagnetic wave is, given by:, r, B = B $i [ cos ( kz – w t ) ] + B $j cos ( kz + wt ), 0, , 18., , 1, , Where B0 = 3 × 10–5 T and B1 = 2 × 10–6 T., The rms value of the force experienced by a stationary, charge Q = 10–4 C at z = 0 is closest to: [9 April 2019 I], (a) 0.6 N, (b) 0.1 N, (c) 0.9 N, (d) 3 × 10–2 N, A plane electromagnetic wave of frequency 50 MHz travels, in free space along the positive x-direction. At a particular, , r, , point in space and time, E = 6.3 ˆj V / m. The, , r, , corresponding magnetic field B , at that point will be:, [9 April 2019 I], (a) 18.9 × 10–8 k̂T, (b) 2.1 × 10–8 k̂T, 19., , 20., , 21., , (c) 6.3 × 10–8 k̂T, (d) 18.9 × 108 k̂T, 50 W/m2 energy density of sunlight is normally incident, on the surface of a solar panel. Some part of incident, energy (25%) is reflected from the surface and the rest is, absorbed. The force exerted on 1m2 surface area will be, close to (c = 3 × 108 m/s):, [9 April 2019, II], (a) 15 × 10–8 N, (b) 20 × 10–8 N, (c) 10 × 10–8 N, (d) 35 × 10–8 N, A plane electromagnetic wave travels in free space along, the x-direction. The electric field component of the wave, at a particular point of space and time is E = 6 Vm –1 along, y-direction. Its corresponding magnetic field component,, B would be:, [8 April 2019 I], (a) 2 × 10–8 T along z-direction, (b) 6 × 10–8 T along x-direction, (c) 6 × 10–8 T along z-direction, (d) 2 × 10–8 T along y-direction, The magnetic field of an electromagnetic wave is given by:, ur, Wb, B = 1.6 ´ 10 –6 cos 2 ´ 10 7 z + 6 ´ 1015 t 2iˆ + ˆj 2, m, The associated electric field will be : [8 April 2019, II], , (, , )(, , ), , V, ur, (a) E = 4.8 × 102 cos(2 × 107 z – 6 × 1015 t) 2iˆ + ˆj, m, ur, V, (b) E = 4.8 ´ 10 2 cos(2 ´ 10 7 z - 6 ´ 1015 t)( -2 $j + $i), m, , (, , ), , V, ur, (c) E = 4.8 × 102 cos(2 × 107 z + 6 × 1015 t) –iˆ + 2 ˆj, m, , (, , V, ur, (d) E = 4.8 × 102 cos(2 × 107 z + 6 × 1015 t) iˆ – 2 ˆj, m, The mean intensity of radiation on the surface of the Sun, is about 108 W/m2. The rms value of the corresponding, magnetic field is closest to :, [12 Jan 2019, II], (a) 1 T, (b)102 T, (c) 10–2 T (d) 10–4 T, , (, , 22., , ), , ), , 23. An electromagnetic wave of intensity 50 Wm–2 enters in a, medium of refractive index ‘n’ without any loss. The ratio, of the magnitudes of electric fields, and the ratio of the, magnitudes of magnetic fields of the wave before and after, entering into the medium are respectively, given by :, [11 Jan 2019, I], æ 1 1 ö, ,, n, n, (a) ç, (b), ÷, è n nø, 1 ö, æ, æ 1, ö, , n÷, (c) ç n ,, (d) ç, ÷, nø, è, è n, ø, 24. A 27 mW laser beam has a cross-sectional area of 10 mm 2., The magnitude of the maximum electric field in this, electromagnetic wave is given by :, [Given permittivity of space Î0 = 9 × 10 –12 SI units, Speed, of light c = 3 × 108 m/s], [11 Jan 2019, II], (a) 2 kV/m, (c) 0.7 kV/m, (b) 1 kV/m, (d) 1.4 kV/m, 25. If the magnetic field of a plane electromagnetic wave is, given by (The speed of light = 3 × 108 m/s), , (, , ), , é, xö ù, 15 æ, B = 100 × 10–6 sin ê 2 p ´ 2 ´ 10 çè t - ÷ø ú, c û, ë, then the maximum electric field associated with it is:, [10 Jan. 2019 I], 4, 4, (a) 6 × 10 N/C, (b) 3 × 10 N/C, (c) 4 × 104 N/C, (d) 4.5 104 N/C, 26. The electric field of a plane polarized electromagnetic, wave in free space at time t = 0 is given by an expression, ur, E ( x, y ) = 10 ˆj cos [(6 x + 8 z)], ur, The magnetic field B ( x, z, t ) is given by: (c is the, velocity of light), [10 Jan 2019, II], 1 ˆ, ˆ cos [ (6 x - 8 z + 10ct ) ], (a), (6k + 8i), c, 1 ˆ, ˆ cos [ (6 x + 8 z - 10ct ) ], (b), (6k - 8i), c, 1 ˆ, ˆ cos [ (6 x + 8 z - 10ct ) ], (c), (6k + 8i), c, 1 ˆ, ˆ cos [(6 x + 8 z + 10ct ) ], (d), (6k - 8i), c, 27. An EM wave from air enters a medium. The electric fields, , r, é, æ z öù, are E1 = E01 xˆ cos ê 2pv ç - t ÷ú in air and, è c øû, ë, r, E2 = E02 xˆ cos [ k (2 z - ct )] in medium, where the wave, number k and frequency v refer to their values in air. The, medium is nonmagnetic. If Îr1 and Îr refer to relative, 2, permittivities of air and medium respectively, which of the, following options is correct?, [9 Jan 2019, I], Î, Îr, r1, 1 =4, =2, (a), (b), Îr, Îr, 2, , (c), , Îr, Îr, , 1, , 2, , 2, , =, , 1, 4, , (d), , Îr, Îr, , 1, 2, , =, , 1, 2
Page 390 :
P-381, , Electromagnetic Waves, , 36., , 37., , 38., , 39., , 40., , 41., , 42., , 43., , During the propagation of electromagnetic waves in a, medium:, [2014], (a) Electric energy density is double of the magnetic, energy density., (b) Electric energy density is half of the magnetic energy, density., (c) Electric energy density is equal to the magnetic energy, density., (d) Both electric and magnetic energy densities are zero., A lamp emits monochromatic green light uniformly in all, directions. The lamp is 3% efficient in converting electrical, power to electromagnetic waves and consumes 100 W of, power. The amplitude of the electric field associated with, the electromagnetic radiation at a distance of 5 m from the, lamp will be nearly:, [Online April 12, 2014], (a) 1.34 V/m, (b) 2.68 V/m, (c) 4.02 V/m, (d) 5.36 V/m, An electromagnetic wave of frequency 1 × 1014 hertz is, propagating along z-axis. The amplitude of electric field is, 4 V/m. If e0 = 8.8 × 10–12 C2/N-m2, then average energy, density of electric field will be: [Online April 11, 2014], (a) 35.2 × 10–10 J/m3, (b) 35.2 × 10–11 J/m3, –12, 3, (c) 35.2 × 10 J/m, (d) 35.2 × 10–13 J/m3, The magnetic field in a travelling electromagnetic wave, has a peak value of 20 nT. The peak value of electric field, strength is :, [2013], (a) 3 V/m, (b)6 V/m, (c) 9 V/m, (d) 12 V/m, A plane electromagnetic wave in a non-magnetic dielectric, ur ur, medium is given by E = E 0 (4 ´ 10 -7 x - 50t ) with, distance being in meter and time in seconds. The dielectric, constant of the medium is :, [Online April 22, 2013], (a) 2.4, (b)5.8, (c) 8.2, (d) 4.8, Select the correct statement from the following :, [Online April 9, 2013], (a) Electromagnetic waves cannot travel in vacuum., (b) Electromagnetic waves are longitudinal waves., (c) Electromagnetic waves are produced by charges, moving with uniform velocity., (d) Electromagnetic waves carry both energy and, momentum as they propagate through space., An electromagnetic wave in vacuum has the electric and, r, r, magnetic field E and B , which are always perpendicular, r, to each other. The direction of polarization is given by X, r, and that of wave propagation by k . Then, [2012], r, r r r, r, (a) X || B and k || B ´ E, r r r, r r, (b) X || E and k || E ´ B, r r r, r r, (c) X || B and k || E ´ B, r, r r, r r, (d) X || E and k || B ´ E, An electromagnetic wave with frequency w and, wavelength l travels in the + y direction. Its magnetic field, is along + x-axis. The vector equation for the associated, electric field (of amplitude E0) is [Online May 19, 2012], ®, 2p, æ, (a) E = - E0 cos ç wt +, è, l, , ö, y ÷ xˆ, ø, , ®, 2p, æ, (b) E = E0 cos çè wt l, , ö, y ÷ xˆ, ø, , ®, 2p, æ, (c) E = E0 cos ç wt è, l, , ö, y÷ zˆ, ø, , ®, 2p, æ, (d) E = - E0 cos ç wt +, è, l, , ö, y÷ zˆ, ø, , 44. An electromagnetic wave of frequency v = 3.0 MHz, passes from vacuum into a dielectric medium with, permittivity Î = 4.0. Then, [2004], (a) wave length is halved and frequency remains, unchanged, (b) wave length is doubled and frequency becomes half, (c) wave length is doubled and the frequency remains, unchanged, (d) wave length and frequency both remain unchanged., 45. Electromagnetic waves are transverse in nature is evident, by, [2002], (a) polarization, (b) interference, (c) reflection, (d) diffraction, , TOPIC 2 Electromagnetic Spectrum, 46. The correct match between the entries in column I and, column II are :, [Sep. 05, 2020 (II)], I, II, Radiation, Wavelength, (A) Microwave, (i) 100 m, (B) Gamma rays, (ii) 10–15 m, (C) A.M. radio waves, (iii) 10–10 m, (D) X-rays, (iv) 10–3 m, (a) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii), (b) (A)-(i), (B)-(iii), (C)-(iv), (D)-(ii), (c) (A)-(iii), (B)-(ii), (C)-(i), (D)-(iv), (d) (A)-(iv), (B)-(ii), (C)-(i), (D)-(iii), 47. Chosse the correct option relating wavelengths of different, parts of electromagnetic wave spectrum :, [Sep. 04, 2020 (I)], (a) l visible < l micro waves < l radio waves < l X - rays, (b) l radio waves > l micro waves > l visible > l x-rays, (c) l x- rays < l micro waves < l radio waves < l visible, (d) l visible > l x-rays > l radio waves > l micro waves, 48. Given below in the left column are different modes of, communication using the kinds of waves given in the, right column., [10 April 2019, I], A. Optical Fibre, P. Ultrasound, Communication, B. Radar, Q. Infrared Light, C. Sonar, R. Microwaves, D. Mobile Phones, S. Radio Waves
Page 391 :
P-382, , 49., , 50., , 51., , From the options given below, find the most appropriate, match between entries in the left and the right column., (a) A – Q, B – S, C – R, D – P, (b) A – S, B – Q, C – R, D – P, (c) A – Q, B – S, C – P, D – R, (d) A – R, B – P, C – S, D – Q, Arrange the following electromagnetic radiations per, quantum in the order of increasing energy :, [2016], A : Blue light B : Yellow light, C : X-ray, D : Radiowave., (a) C, A, B, D, (b) B, A, D, C, (c) D, B, A, C, (d) A, B, D, C, Microwave oven acts on the principle of :, [Online April 9, 2016], (a) giving rotational energy to water molecules, (b) giving translational energy to water molecules, (c) giving vibrational energy to water molecules, (d) transferring electrons from lower to higher energy, levels in water molecule, Match List - I (Electromagnetic wave type) with List - II, (Its association/application) and select the correct option, from the choices given below the lists:, [2014], List 2, (i) To treat muscular, strain, 2. Radio waves, (ii) For broadcasting, 3. X-rays, (iii) To detect fracture of, bones, 4. Ultraviolet rays, (iv) Absorbed by the, ozone layer of the, atmosphere, 1, 2, 3, 4, (a) (iv), (iii), (ii), (i), (b) (i), (ii), (iv), (iii), (c) (iii), (ii), (i), (iv), (d) (i), (ii), (iii), (iv), If microwaves, X rays, infrared, gamma rays, ultra-violet,, radio waves and visible parts of the electromagnetic, spectrum are denoted by M, X, I, G, U, R and V then which, of the following is the arrangement in ascending order of, wavelength ?, [Online April 19, 2014], (a) R, M, I, V, U, X and G, (b) M, R, V, X, U, G and I, (c) G, X, U, V, I, M and R, (d) I, M, R, U, V, X and G, Match the List-I (Phenomenon associated with, electromagnetic radiation) with List-II (Part of, electromagnetic spectrum) and select the correct code from, the choices given below this lists:[Online April 11, 2014], 1., , 52., , 53., , Physics, , List 1, Infrared waves, , I, , List I, Doublet of sodium, , List II, (A) Visible radiation, , II Wavelength, corresponding to, temperature associated, with the isotropic, radiation filling all space, , (B) Microwave, , III Wavelength emitted by, atomic hydrogen in, interstellar space, , (C) Short radio wave, , IV Wavelength of radiation (D) X-rays, arising from two close, energy levels in hydrogen, , (a) (I)-(A), (II)-(B), (III)-(B), (IV)-(C), (b) (I)-(A), (II)-(B), (III)-(C), (IV)-(C), (c) (I)-(D), (II)-(C), (III)-(A), (IV)-(B), (d) (I)-(B), (II)-(A), (III)-(D), (IV)-(A), 54. Match List I (Wavelength range of electromagnetic, spectrum) with List II (Method of production of these, waves) and select the correct option from the options given, below the lists., [Online April 9, 2014], List I, , Lis t II, , (1) 700 nm to, 1 mm, , (i) Vibration of atoms, and molecules ., , (2) 1 nm to, 400 nm, , (ii) Inner s hell electrons, in atoms moving, from one energy, level to a lower level., , (3) < 10–3 nm (iii) Radioactive decay of, the nucleus ., (4) 1 mm to, 0.1 m, , (iv) Magnetron valve., , (a) (1)-(iv), (2)-(iii), (3)-(ii), (4)-(i), (b) (1)-(iii), (2)-(iv), (3)-(i), (4)-(ii), (c) (1)-(ii), (2)-(iii), (3)-(iv), (4)-(i), (d) (1)-(i), (2)-(ii), (3)-(iii), (4)-(iv), 55. Photons of an electromagnetic radiation has an energy, 11 keV each. To which region of electromagnetic spectrum, does it belong ?, [Online April 9, 2013], (a) X-ray region, (b) Ultra violet region, (c) Infrared region, (d) Visible region, 56. The frequency of X-rays; g-rays and ultraviolet rays are, respectively a, b and c then, [Online May 26, 2012], (a) a < b; b > c, (b) a > b ; b > c, (c) a < b < c, (d) a = b = c
Page 396 :
P-387, , Electromagnetic Waves, , Oscillation of B can be only along ĵ or k̂ direction., w = 2pf = 2p × 2 × 1014 Hz, ur, \ B ( x, t ) = (9 ´ 10 –8 T )kˆ sin[2p(1.5 ´ 10 –6 ´ –2 ´104 t )], 36., , (c) E0 = CB0 and C =, Electric energy density =, , 1, m 0e 0, 1, e0 E02 = m E, 2, , Magnetic energy density =, , 37., , 1 Bo 2, = mB, 2 m0, , Thus, mE = mB, Energy is equally divided between electric and magnetic, field., (b) Wavelength of monochromatic green light, = 5.5 × 10–5 cm, Power, Area, , Intensity I =, =, , 100 ´ ( 3 /100 ), , =, , 3, Wm -2, 100p, , 4p ( 5 ), Now, half of this intensity (I) belongs to electric field and, half of that to magnetic field, therefore,, I 1, = e0 E 02 C, 2 4, , or E 0 =, , =, , =, , 38., , 39., , 2, , 2I, e0 C, , æ 3 ö, 2´ç, p÷, è 100 ø, 1, æ, ö, ´ 3 ´108, ç, 9÷, è 4p ´ 9 ´10 ø, , (, , 6, ´ 30 =, 25, , ), , 7.2, , \ E 0 = 2.68 V / m, (c) Given: Amplitude of electric field,, E0 = 4 v/m, Absolute permitivity,, e0 = 8.8 × 10–12 c2/N-m2, Average energy density uE = ?, Applying formula,, 1, 2, Average energy density uE = e0 E, 4, 1, -12, 2, Þ uE = ´ 8.8 ´ 10 ´ (4), 4, = 35.2 × 10–12 J/m3, (b) From question,, B0 = 20 nT = 20 × 10–9T, (Q velocity of light in vacuum C = 3 × 108 ms–1), r, r, r, E0 = B0 ´ C, r, r r, | E0 |=| B | ×| C |= 20 ´ 10 -9 ´ 3 ´ 108, = 6 V/m., , 40. (b), 41. (d) Electromagnetic waves do not required any medium, to propagate. They can travel in vacuum. They are, transverse in nature like light. They carry both energy and, momentum., A changing electric field produces a changing magnetic, field and vice-versa. Which gives rise to a transverse wave, known as electromagnetic wave., 42. (b) Q The E.M. wave are transverse in nature i.e.,, r r, k ´E r, =H, =, …(i), m, r, r B, where H =, m, r r, r, k ´H, and, …(ii), = -E, we, r, r, r, r, k is ^ H and k is also ^ to E, r r, The direction of wave propagation is parallel to E ´ B., The direction of polarization is parallel to electric field., 43. (c) In an electromagnetic wave electric field and, magnetic field are perpendicular to the direction of, propagation of wave. The vector equation for the electric, field is, r, E = E0 cos æç wt - 2p y ö÷ zˆ, è, l ø, 44. (a) Frequency remains unchanged during refraction, Velocity of EM wave in vacuum, 1, =C, Vvacuum =, m0 Î0, vmed =, l med, , l vacuum, , 1, µ0 Î0 ´4, =, , vmed, v vacuum, , =, , c, 2, , =, , c/2 1, =, 2, c, , \ Wavelength is halved and frequency remains, unchanged, 45. (a) The phenomenon of polarisation is shown only by, transverse waves. The vibration of electromagnetic wave, are restricted through polarization in a direction, perpendicular to wave propagation., 46. (d) Energy sequence of radiations is, Eg -Rays > EX-Rays > Emicrowave > EAM Radiowaves, , \ l g -Rays < l X-Rays < l microwave < l AM Radiowaves, From the above sequence, we have, (a) Microwave ® 10 -3 m (iv), (b) Gamma Rays ® 10-15 m (ii), (c) AM Radio wave ® 100 m (i), (d) X-Rays ® 10-10 m (iii)
Page 397 :
P-388, , 47., , Physics, , (b) The orderly arrangement of different parts of EM wave, in decreasing order of wavelength is as follows:, , l radiowaves > l microwaves > l visible > l X-rays, 48., , 49., , (c) Optical Fibre Communication – Infrared Light, Radar – Radio Waves, Sonar – Ultrasound, Mobile Phones – Microwaves, (c), , E, Decreases, g-rays X-rays uv-rays Visible rays IR rays, Radio, VIBGYOR Microwaves waves, , Radio wave < yellow light < blue light < X-rays, (Increasing order of energy), 50., 51., , 52., , (c) Microwave oven acts on the principle of giving, vibrational energy to water molecules., (d), (1) Infrared rays are used to treat muscular strain because, these are heat rays., (2) Radio waves are used for broadcasting because these, waves have very long wavelength ranging from few, centimeters to few hundred kilometers., (3) X-rays are used to detect fracture of bones because, they have high penetrating power but they can't penetrate, through denser medium like dones., (4) Ultraviolet rays are absorbed by ozone of the, atmosphere., (c) Gamma rays < X-rays < Ultra violet < Visible rays, < Infrared rays < Microwaves < Radio waves., , 53. (d) Wavelength emitted by atomic hydrogen in interstellar, space - Part of short radio wave of electromagnetic, spectrum., Doublet of sodium - visible radiation., 54. (d) Vibration of atoms and molecules 700 nm to 1 mm, Radioactive decay of the nucleus < 10–3 nm, Magnetron valve 1 mm to 0.1 m, 55. (a) E =, Þ l=, , hc, hc, Þ l=, l, E, , 6.6 ´ 10-34 ´ 3 ´108, 11´1000 ´ 1.6 ´10 -19, , = 12.4 Å, Increasing order of frequency, x-rays u-v rays visible Infrared, , wavelength range of visible region is 4000Å to 7800Å., 56. (a) Frequency range of g-ray,, b = 1018 – 1023 Hz, Frequency range of X-ray,, a = 1016 – 1020 Hz, Frequency range of ultraviolet ray,, c = 1015 – 1017 Hz, \ a < b; b > c
Page 398 :
23, Ray Optics and, Optical Instruments, below. The distance over which the man can see the image, of the light source in the mirror is:, [12 Jan. 2019 I], , Plane Mirror, Spherical Mirror, TOPIC 1, and Reflection of Light, 1., , When an object is kept at a distance of 30 cm from a concave, mirror, the image is formed at a distance of 10 cm from the, mirror. If the object is moved with a speed of 9, cms–1, the speed (in cms–1) with which image moves at, that instant is ________., [NA Sep. 03, 2020 (II)], , d, L, 2L, , 2., (a) d, , Object, , 20, (cm), , 16, , 12, , 8, , 4, , A spherical mirror is obtained as shown in the figure from, a hollow glass sphere. If an object is positioned in front of, the mirror, what will be the nature and magnification of the, image of the object? (Figure drawn as schematic and not, to scale), [Sep. 02, 2020 (I)], (a) Inverted, real and magnified, (b) Erect, virtual and magnified, (c) Erect, virtual and unmagnified, (d) Inverted, real and unmagnified, , 4., , A concave mirror for face viewing has focal length of 0.4 m., The distance at which you hold the mirror from your face, in order to see your image upright with a magnification of, 5 is:, [9 April 2019 I], (a) 0.24 m, (b) 1.60 m (c) 0.32 m, (d) 0.16 m, A point source of light, S is placed at a distance L in front, of the centre of plane mirror of width d which is hanging, vertically on a wall. A man walks in front of the mirror, along a line parallel to the mirror, at a distance 2L as shown, , 5., , 6., , (b) 2d, , d, 2, Two plane mirrors are inclined to each other such that a, ray of light incident on the first mirror (M1) and parallel to, the second mirror (M2) is finally reflected from the second, mirror (M2) parallel to the first mirror (M1). The angle, between the two mirrors will be:, [9 Jan. 2019 II], (a) 45°, (b) 60°, (c) 75°, (d) 90°, , (c) 3d, , 3., , S, , (d), , An object is gradually moving away from the focal point, of a concave mirror along the axis of the mirror. The, graphical representation of the magnitude of linear, magnification (m) versus distance of the object from the, mirror (x) is correctly given by, (Graphs are drawn schematically and are not to scale), [8 Jan. 2020 II], (a)
Page 399 :
P-390, , Physics, , (b), , 10. To get three images of a single object, one should have, two plane mirrors at an angle of, [2003], (a) 60º, (b) 90º, (c) 120º, (d) 30º, 11. If two plane mirrors are kept at 60° to each other, then the, number of images formed by them is, [2002], (a) 5, (b) 6, (c) 7, (d) 8, , Refraction of Light at Plane, TOPIC 2 Surface and Total Internal, Reflection, , (c), , 12. An observer can see through a small hole on the side of a, jar (radius 15 cm) at a point at height of 15 cm from the, bottom (see figure). The hole is at a height of 45 cm. When, the jar is filled with a liquid up to a height of 30 cm the same, observer can see the edge at the bottom of the jar. If the, refractive index of the liquid is N/100, where N is an integer,, the value of N is ___________. [NA Sep. 03, 2020 (I)], , (d), , 45 cm, , 7., , A particle is oscillating on the X-axis with an amplitude, 2 cm about the point x0 = 10 cm with a frequency w. A concave, mirror of focal length 5 cm is placed at the origin (see figure), Identify the correct statements: [Online April 15, 2018], (A) The image executes periodic motion, (B) The image executes non-periodic motion, (C) The turning points of the image are asymmetric w.r.t, the image of the point at x = 10 cm, (D) The distance between the turning points of the, oscillation of the image is, , 8., , 9., , 100, 21, , x0 = 10 cm, , x=0, , (a) (B), (D), , (b) (B), (C), , (c) (A), (C), (D), , (d) (A), (D), , 60 cm, , (b) –24 cm, , (c), , – 60 cm, , (d) 24 cm, , A car is fitted with a convex side-view mirror of focal length, 20 cm. A second car 2.8 m behind the first car is overtaking, the first car at a relative speed of 15 m/s. The speed of the, image of the second car as seen in the mirror of the first, one is :, [2011], (a), , 1, m/s, 15, , (b) 10 m/s, , (c) 15 m/s, , 15 cm, , 13. A light ray enters a solid glass sphere of refractive index, m = 3 at an angle of incidence 60°. The ray is both, reflected and refracted at the farther surface of the sphere., The angle (in degrees) between the reflected and refracted, rays at this surface is ___________., [NA Sep. 02, 2020 (II)], 14. A vessel of depth 2h is half filled with a liquid of refractive, index 2 2 and the upper half with another liquid of, , You are asked to design a shaving mirror assuming that a, person keeps it 10 cm from his face and views the magnified, image of the face at the closest comfortable distance of 25, cm. The radius of curvature of the mirror would then be :, [Online April 10, 2015], (a), , 15 cm, , (d), , 1, m/s, 10, , refractive index 2. The liquids are immiscible. The, apparent depth of the inner surface of the bottom of vessel, will be:, [9 Jan. 2020 I], h, h, (a), (b), 2( 2 + 1), 2, (c), , h, 3 2, , (d), , 3, h 2, 4, , 15. There is a small source of light at some depth below the, 4, surface of water (refractive index = ) in a tank of large, 3, cross sectional surface area. Neglecting any reflection from, the bottom and absorption by water, percentage of light, that emerges out of surface is (nearly):, [Use the fact that surface area of a spherical cap of height, h and radius of curvature r is 2prh], [9 Jan. 2020 II], (a) 21%, (b) 34%, (c) 17%, (d) 50%
Page 400 :
P-391, , Ray Optics and Optical Instruments, , 16. The critical angle of a medium for a specific wavelength, if, the medium has relative permittivity 3 and relative, , (a), , 2 3, + 2b, a, , (b) 2a +, , 2b, 3, , 4, for this wavelength, will be: [8 Jan. 2020 I], 3, (a) 15°, (b) 30°, (c) 45°, (d) 60°, 17. A concave mirror has radius of curvature of 40 cm. It is at, the bottom of a glass that has water filled up to 5 cm (see, figure). If a small partricle is floating on the surface of, water, its image as seen, from directly above the glass, is at, a distance d from the surface of water. The value of d is, close to :, [12 Apr. 2019 I], (Refractive index of water = 1.33), , 20. In figure, the optical fiber is l = 2 m long and has a diameter, of d = 20 mm. If a ray of light is incident on one end of the, fiber at angle q1 = 40°, the number of reflections it makes, before emerging from the other end is close to :, , (a) 6.7 cm, (b) 13.4 cm (c) 8.8 cm, (d) 11.7 cm, 18. A transparent cube of side d, made of a material of refractive, index m2, is immersed in a liquid of refractive index, m1(m1< m2). A ray is incident on the face AB at an angle q, (shown in the figure). Total internal reflection takes place, at point E on the face BC., , (a) 55000, (b) 66000 (c) 45000, (d) 57000, 21. A light wave is incident normally on a glass slab of, refractive index 1.5. If 4% of light gets reflected and the, amplitude of the electric field of the incident light is 30 V/, m, then the amplitude of the electric field for the wave, propogating in the glass medium will be:[12 Jan. 2019 I], , permeability, , Then q must satisfy :, -1, (a) q < sin, , (c) q < sin -1, , m1, m2, m 22, m12, , [12 Apr. 2019 II], (b) q > sin -1, , -1, , (d) q > sin -1, , m 22, m12, , (c) 2a +, , 2b, , (refractive index of fiber is 1.31 and sin 40° = 0.64), [8 April 2019 I], , (a) 30 V/m, , (b) 10 V/m, , (c) 24 V/ m, , (d) 6 V/m, , 22. Let the refractive index of a denser medium with respect to, a rarer medium be n 12 and its critical angle be qC. At an, angle of incidence A when light is travelling from denser, medium to rarer medium, a part of the light is reflected and, the rest is refracted and the angle between reflected and, refracted rays is 90°. Angle A is given by :, [Online April 8, 2017], , -1, , m1, m2, , 19. A ray of light AO in vacuum is incident on a glass slab at, angle 60o and refracted at angle 30o along OB as shown in, the figure. The optical path length of light ray from A to B, is :, [10 Apr. 2019 I], , (d) 2a + 2b, , 3, , (a), , 1, -1, , cos (sin qC ), , (c) cos–1 (sin qC), , (b), , 1, -1, , tan (sin qC ), , (d) tan–1 (sin qC), , 23. A diver looking up through the water sees the outside, world contained in a circular horizon. The refractive index, 4, of water is , and the diver’s eyes are 15 cm below the, 3, surface of water. Then the radius of the circle is:, [Online April 9, 2014], (a), , 15 ´ 3 ´ 5 cm, , (b) 15 ´ 3 7 cm, , (c), , 15 ´ 7, cm, 3, , (d), , 15 ´ 3, 7, , cm
Page 401 :
P-392, , 24., , 25., , Physics, , A printed page is pressed by a glass of water. The refractive, index of the glass and water is 1.5 and 1.33, respectively. If, the thickness of the bottom of glass is 1 cm and depth of, water is 5 cm, how much the page will appear to be shifted, if viewed from the top ?, [Online April 25, 2013], (a) 1.033 cm, , (b) 3.581 cm, , (c) 1.3533 cm, , (d) 1.90 cm, , A light ray falls on a square glass slab as shown in the, diagram. The index of refraction of the glass, if total internal, reflection is to occur at the vertical face, is equal to :, [Online April 23, 2013], , (c), , æ, æ, 1ö, 1 ö, ç1 + ÷ h2 - ç 1 + ÷ h1, è m1 ø, è m2 ø, , æ, æ, 1 ö, 1 ö, ç 1 - ÷ h2 + ç 1 - ÷ h1, è m1 ø, è m2 ø, 29. A transparent solid cylindrical rod has a refractive index of, 2, . It is surrounded by air. A light ray is incident at the, 3, mid-point of one end of the rod as shown in the figure., , (d), , q, , 45° Incident ray, , The incident angle q for which the light ray grazes along, the wall of the rod is :, [2009], æ, ö, 2, 1, -1, 3/2, ÷, (b) sin ç, (a) sin, è 3ø, , (, , ), , æ 1 ö, sin -1 ç, ÷, (d) sin -1 (1/ 2 ), è 3ø, A fish looking up through the water sees the outside world, , (c), , 30., 5, 2, , 26., , 27., , 28., , 3, 3, (a) ( 2 + 1) (b), (c), (d), 2, 2, 2, Light is incident from a medium into air at two possible, angles of incidence (A) 20° and (B) 40°. In the medium, light travels 3.0 cm in 0.2 ns. The ray will :, [Online April 9, 2013], (a) suffer total internal reflection in both cases (A) and, (B), (b) suffer total internal reflection in case (B) only, (c) have partial reflection and partial transmission in case, (B), (d) have 100% transmission in case (A), Let the x-z plane be the boundary between two transparent, media. Medium 1 in z ³ 0 has a refractive index of 2 and, medium 2 with z < 0 has a refractive index of 3 . A ray of, r, light in medium 1 given by the vector A = 6 3iˆ + 8 3 ˆj - 10kˆ, is incident on the plane of separation. The angle of, refraction in medium 2 is:, [2011], (a) 45°, (b) 60°, (c) 75°, (d) 30°, A beaker contains water up to a height h1 and kerosene, of height h2 above water so that the total height of, (water + kerosene) is (h1 + h2). Refractive index of water is, m1and that of kerosene is m2. The apparent shift in the, position of the bottom of the beaker when viewed from, above is, [2011 RS], , æ, æ, 1ö, 1 ö, (a) ç1 + m ÷ h1 - ç1 + m ÷ h2, è, 1ø, è, 2ø, æ, æ, 1 ö, 1 ö, (b) ç 1 - m ÷ h1 + ç 1 - m ÷ h2, è, 1ø, è, 2ø, , contained in a circular horizon. If the refractive index of, 4, water is and the fish is 12 cm below the surface, the, 3, radius of this circle in cm is, [2005], (a), , 36, 7, , (b), , 36 7, , (c), , 4 5, , (d) 36 5, , 31. Consider telecommunication through optical fibres. Which, of the following statements is not true?, [2003], (a) Optical fibres can be of graded refractive index, (b) Optical fibres are subject to electromagnetic, interference from outside, (c) Optical fibres have extremely low transmission loss, (d) Optical fibres may have homogeneous core with a, suitable cladding., 32. Which of the following is used in optical fibres? [2002], (a) total internal reflection (b) scattering, (c) diffraction, (d) refraction., , TOPIC 3, , Refraction at Curved Surface, Lenses and Power of Lens, , 33. A point like object is placed at a distance of 1 m in front of, a convex lens of focal length 0.5 m. A plane mirror is placed, at a distance of 2 m behind the lens. The position and, nature of the final image formed by the system is :, [Sep. 06, 2020 (I)], (a) 2.6 m from the mirror, real, (b) 1 m from the mirror, virtual, (c) 1 m from the mirror, real, (d) 2.6 m from the mirror, virtual
Page 402 :
P-393, , Ray Optics and Optical Instruments, , 34., , A double convex lens has power P and same radii of curvature R of both the surfaces. The radius of curvature of a, surface of a plano-convex lens made of the same material, with power 1.5 P is :, [Sep. 06, 2020 (II)], , R, 3R, R, (c), (d), 2, 2, 3, 35. For a concave lens of focal length f, the relation between, object and image distances u and v, respectively, from its pole, can best be represented by (u = v is the reference line) :, , (a) 2R, , (b), , R, m1 - m2, , (b), , 2R, m1 - m2, , (c), , 2R, 2(m1 - m2 ), , (d), , R, 2 - (m1 - m2 ), , =, u, , (a), , 40. The graph shows how the magnification m produced by a, thin lens varies with image distance v. What is the focal, length of the lens used ?, [10 Apr. 2019 II], , u, , =, , v, , f, u, , v, f, , (a), , v, , [Sep. 05, 2020 (I)], , v, f, , 39. One plano-convex and one plano-concave lens of same, radius of curvature ‘R’ but of different materials are joined, side by side as shown in the figure. If the refractive index, of the material of 1 is m1 and that of 2 is m2, then the focal, length of the combination is :, [10 Apr. 2019 I], , (b), , u, , =, , v, , f, , u, , v, f, (c), , u, , (b), , b2c, a, , (c), , a, c, , (d), , b, c, , u, , =, , 41. A convex lens of focal length 20 cm produces images of, the same magnification 2 when an object is kept at two, distances x1 and x2 (x1 > x2) from the lens. The ratio of x1, and x2 is:, [9 Apr. 2019 II], , (d), 36., , b2, ac, , v, , f, v, f, , (a), , u, f, The distance between an object and a screen is 100 cm. A, lens can produce real image of the object on the screen for, two different positions between the screen and the object., The distance between these two positions is 40 cm. If the, æ N ö, power of the lens is close to ç, D where N is an, è 100 ÷ø, integer, the value of N is ___________., [NA Sep. 04, 2020 (II)], , (a) 2 : 1, , (b) 3 : 1, , (c) 5 : 3, , (d) 4 : 3, , 42. A thin convex lens L (refractive index = 1.5) is placed on a, plane mirror M. When a pin is placed at A, such that OA =, 18 cm, its real inverted image is formed at A itself, as shown, in figure. When a liquid of refractive index µi is put between, the lens and the mirror, the pin has to be moved to A’, such, that OA’ = 27 cm, to get its inverted real image at A’ itself., The value of µi will be:, [9 Apr. 2019 II], , 37. A point object in air is in front of the curved surface of a, plano-convex lens. The radius of curvature of the curved, surface is 30 cm and the refractive index of the lens material, is 1.5, then the focal length of the lens (in cm), is__________., [NA 8 Jan. 2020 I], 38. A thin lens made of glass (refractive index = 1.5) of focal, length f = 16 cm is immersed in a liquid of refractive index, 1.42. If its focal length in liquid is fl ,then the ratio fl /f is, closest to the integer:, [7 Jan. 2020 II], (a) 1, , (b) 9, , (c) 5, , (d) 17, , (a), , 4, 3, , (b), , 3, 2, , (c), , 3, , (d), , 2
Page 403 :
P-394, , Physics, , 43. An upright object is placed at a distance of 40 cm in front, of a convergent lens of focal length 20 cm. A convergent, mirror of focal length 10 cm is placed at a distance of 60 cm, on the other side of the lens. The position and size of the, final image will be :, [8 April 2019 I], (a) 20 cm from the convergent mirror, same size as the object, (b) 40 cm from the convergent mirror, same size as the object, (c) 40 cm from the convergent lens, twice the size of the, object, (d) 20 cm from the convergent mirror, twice the size of the, object, 44. A convex lens (of focal length 20 cm) and a concave mirror,, having their principal axes along the same lines, are kept, 80 cm apart from each other. The concave mirror is to the, right of the convex lens. When an object is kept at a, distance of 30 cm to the left of the convex lens, its image, remains at the same position even if the concave mirror is, removed. The maximum distance of the object for which, this concave mirror, by itself would produce a virtual image, would be :, [8 Apr. 2019 II], (a) 30 cm, (b) 25 cm (c) 10 cm, (d) 20 cm, 45. What is the position and nature of image formed by lens, combination shown in figure? (f1, f2 are focal lengths), [12 Jan. 2019 I], , 2 cm, A, , 20 cm, , B, , f1 = + 5 cm, , f2 = –5 cm, , (a) 70 cm from point B at left; virtual, (b) 40 cm from point B at right; real, , 20, cm from point B at right, real, 3, (d) 70 cm from point B at right; real, Formation of real image using a biconvex lens is shown, below :, [12 Jan. 2019 II], , (c), 46., , 2f, 2f, , 47., , f, , screen, , f, , If the whole set up is immersed in water without disturbing, the object and the screen positions, what will one observe, on the screen ?, (a) Image disappears, (b) Magnified image, (c) Erect real image, (d) No change, A plano-convex lens (focal length f2, refractive index µ2,, radius of curvature R) fits exactly into a plano-concave, lens (focal length f 1 , refractive index µ1 , radius of, curvature R). Their plane surfaces are parallel to each other., Then, the focal length of the combination will be :, [12 Jan. 2019 II], , (a) f1 – f2, , (b), , R, µ2 - µ1, , 2 f1 f 2, (d) f1 + f2, f1 + f 2, 48. An object is at a distance of 20 m from a convex lens of, focal length 0.3 m. The lens forms an image of the object. If, the object moves away from the lens at a speed of 5m/s,, the speed and direction of the image will be :, [11 Jan. 2019 I], (a) 2.26 × 10–3 m/s away from the lens, (b) 0.92 × 10–3 m/s away from the lens, (c) 3.22 × 10–3 m/s towards the lens, (d) 1.16 × 10–3 m/s towards the lens, 49. A plano convex lens of refractive index m1 and focal, length f1 is kept in contact with another plano concave, lens of refractive index m2 and focal length f2 If the, radius of curvature of their spherical faces is R each, and f1 = 2f2, then m1 and m2 are related as:, [10 Jan. 2019 I], (a) m1 + m2 = 3, (b) 2m1 – m2 = 1, , (c), , (c) 3m2 – 2m1 = 1, (d) 2m2 – m1 = 1, 50. The eye can be regarded as a single refracting surface., The radius of curvature of this surface is equal to that, of cornea (7.8 mm). This surface separates two media of, refractive indices 1 and 1.34. Calculate the distance from, the refracting surface at which a parallel beam of light, will come to focus., [10 Jan. 2019 II], (a) 1 cm, (b) 2 cm, (c) 4.0 cm, (d) 3.1 cm, 51. A convex lens is put 10 cm from a light source and it, makes a sharp image on a screen, kept 10 cm from the lens., Now a glass block (refractive index 1.5) of 1.5 cm thickness, is placed in contact with the light source. To get the sharp, image again, the screem is shifted by a distance d. Then d, is:, [9 Jan. 2019 I], (a) 1.1 cm away from the lens, (b) 0, (c) 0.55 cm towards the lens, (d) 0.55 cm away from the lens, 52. A planoconvex lens becomes an optical system of 28 cm, focal length when its plane surface is silvered and, illuminated from left to right as shown in Fig-A. If the same, lens is instead silvered on the curved surface and, illuminated from other side as in Fig. B, it acts like an optical, system of focal length 10 cm. The refractive index of the, material of lens if:, [Online April 15, 2018], , Fig. A, , (a) 1.50, , (b) 1.55, , Fig. B, , (c) 1.75, , (d) 1.51
Page 404 :
P-395, , Ray Optics and Optical Instruments, , 53., , 54., , 55., , A convergent doublet of separated lenses, corrected for, spherical aberration, has resultant focal length of 10cm., The separation between the two lenses is 2cm. The focal, lengths of the component lenses [Online April 15, 2018], (a) 18cm, 20cm, (b) 10cm, 12cm, (c) 12cm, 14cm, (d) 16cm, 18cm, In an experiment a convex lens of focal length 15 cm is, placed coaxially on an optical bench in front of a convex, mirror at a distance of 5 cm from it. It is found that an, object and its image coincide, if the object is placed at a, distance of 20 cm from the lens. The focal length of the, convex mirror is :, [Online April 9, 2017], (a) 27.5 cm (b) 20.0 cm (c) 25.0 cm (d) 30.5 cm, A hemispherical glass body of radius 10 cm and refractive, index 1.5 is silvered on its curved surface. A small air bubble, is 6 cm below the flat surface inside it along the axis. The, position of the iamge of the air bubble made by the mirror, is seen :, [Online April 10, 2016], 10 cm, , f1 and f2 are close to :, (a) f1 = 7.8 cm, (b) f1 = 12.7 cm, (c) f1 = 15.6 cm, (d) f1 = 7.8 cm, , f2 = 12.7 cm, f2 = 7.8 cm, f2 = 25.4 cm, f2 = 25.4 cm, , 58. A thin convex lens of focal length ‘f’ is put on a plane, mirror as shown in the figure. When an object is kept at, a distance ‘a’ from the lens - mirror combination, its, a, image is formed at a distance, in front of the, 3, combination. The value of ‘a’ is :, [Online April 11, 2015], , 6cm, O, Silvered, , 56., , (a) 14 cm below flat surface, (b) 20 cm below flat surface, (c) 16 cm below flat surface, (d) 30 cm below flat surface, A convex lens, of focal length 30 cm, a concave lens of, focal length 120 cm, and a plane mirror are arranged as, shown. For an object kept at a distance of 60 cm from the, convex lens, the final image, formed by the combination, is, a real image, at a distance of :, [Online April 9, 2016], , (a) 3f, , (b), , 3, f, 2, , (c) f, , (d) 2f, , 3ö, æ, 59. A thin convex lens made from crown glass ç m = ÷ has, 2ø, è, focal length f. When it is measured in two different, 4, 5, and , it has the focal, 3, 3, lengths f1 and f2 respectively. The correct relation between, the focal lengths is:, [2014], , liquids having refractive indices, , (a) f1 = f2 < f, (b) f1 > f and f2 becomes negative, |Focal length| |Focal length|, = 30 cm, = 120 cm, 60cm, 20cm, , 57., , 70 cm, (a) 60 cm from the convex lens, (b) 60 cm from the concave lens, (c) 70 cm from the convex lens, (d) 70 cm from the concave lens, To find the focal length of a convex mirror, a student records, the following data :, [Online April 9, 2016], Object Pin Convex Lens Convex Mirror Image Pin, 22.2cm, 32.2cm, 45.8cm, 71.2cm, The focal length of the convex lens is f1 and that of mirror, is f2. Then taking index correction to be negligibly small,, , (c) f2 > f and f1 becomes negative, (d) f1 and f2 both become negative, 60. The refractive index of the material of a concave lens is m., It is immersed in a medium of refractive index m1. A parallel, beam of light is incident on the lens. The path of the, emergent rays when m1 > m is:, [Online April 12, 2014], m, m1, (a), , m1
Page 405 :
m, , P-396, , m1, , Physics, , m1, , (b), , m, m1, , m1, , (c), , m, m1, , (c) remain same, (d) does not depend on colour of light, 66. In an optics experiment, with the position of the object, fixed, a student varies the position of a convex lens and, for each position, the screen is adjusted to get a clear, image of the object. A graph between the object distance u, and the image distance v, from the lens, is plotted using, the same scale for the two axes. A straight line passing, through the origin and making an angle of 45° with the xaxis meets the experimental curve at P. The coordinates of, P will be, [2009], æ f fö, (a) ç , ÷, (b) ( f, f ), è 2 2ø, (c) ( 4 f, 4 f ), (d) ( 2 f, 2 f ), 67. A student measures the focal length of a convex lens by, putting an object pin at a distance ‘u’ from the lens and, measuring the distance ‘v’ of the image pin. The graph, between ‘u’ and ‘v’ plotted by the student should look, like, [2008], , m1, , (d), , v(cm), , (a), , (b), O, , 61., , 62., , 63., , 64., , 65., , An object is located in a fixed position in front of a screen., Sharp image is obtained on the screen for two positions of, a thin lens separated by 10 cm. The size of the images in, two situations are in the ratio 3 : 3. What is the distance, between the screen and the object?, [Online April 11, 2014], (a) 124.5 cm, (b) 144.5 cm, (c) 65.0 cm, (d) 99.0 cm, Diameter of a plano-convex lens is 6 cm and thickness at, the centre is 3 mm. If speed of light in material of lens is, 2× 108 m/s, the focal length of the lens is, [2013], (a) 15 cm (b) 20 cm (c) 30 cm (d) 10 cm, The image of an illuminated square is obtained on a screen, with the help of a converging lens. The distance of the, square from the lens is 40 cm. The area of the image is 9, times that of the square. The focal length of the lens is :, [Online April 22, 2013], (a) 36 cm (b) 27 cm (c) 60 cm (d) 30 cm, An object at 2.4 m in front of a lens forms a sharp image on, a film 12 cm behind the lens. A glass plate 1 cm thick, of, refractive index 1.50 is interposed between lens and film, with its plane faces parallel to film. At what distance, (from lens) should object shifted to be in sharp focus of, film?, [2012], (a) 7.2 m, (b) 2.4 m, (c) 3.2 m, (d) 5.6 m, When monochromatic red light is used instead of blue, light in a convex lens, its focal length will, [2011 RS], (a) increase, (b) decrease, , v(cm), , O, , u(cm), , u(cm), , v(cm), v(cm), , (c), , O, , u(cm), , (d), , O, , u(cm), , 68. Two lenses of power –15 D and +5 D are in contact with, each other. The focal length of the combination is [2007], (a) + 10 cm, (b) – 20 cm, (c) – 10 cm, (d) + 20 cm, 69. A thin glass (refractive index 1.5) lens has optical power of, – 5 D in air. Its optical power in a liquid medium with, refractive index 1.6 will be, [2005], (a) – 1D, (b) 1 D, (c) – 25 D (d) 25 D, 70. A plano convex lens of refractive index 1.5 and radius of, curvature 30 cm. Is silvered at the curved surface. Now this, lens has been used to form the image of an object. At what, distance from this lens an object be placed in order to have, a real image of size of the object, [2004], (a) 60 cm (b) 30 cm (c) 20 cm (d) 80 cm, , TOPIC 4 Prism and Dispersion of Light, 71. The surface of a metal is illuminated alternately with, photons of energies E1 = 4 eV and E2 = 2.5 eV respectively., The ratio of maximum speeds of the photoelectrons emitted, in the two cases is 2. The work function of the metal in (eV), is _______________., [NA Sep. 05, 2020 (II)]
Page 406 :
P-397, , Ray Optics and Optical Instruments, , 72., , The variation of refractive index of a crown glass thin prism, with wavelength of the incident light is shown. Which of, the following graphs is the correct one, if Dm is the angle, of minimum deviation ?, [11 Jan. 2019, I], 1.535, 1.530, , n2, , 1.525, 1.520, 1.515, 1.510, , 400, , 500, , 600, , 700, , l (nm), , Dm, , (a), , Dm, , 400 500 600 700, , l (nm), , 73. A monochromatic light is incident at a certain angle on an, equilateral triangular prism and suffers minimum deviation., 3,, [11 Jan. 2019 II], , If the refractive index of the material of the prism is, , then the angle of incidence is :, (a) 90°, (b) 30°, (c) 60°, (d) 45°, 74. A ray of light is incident at an angle of 60° on one face of, a prism of angle 30°. The emergent ray of light makes an, angle of 30° with incident ray. The angle made by the, emergent ray with second face of prism will be:, [Online April 16, 2018], (a) 30°, (b) 90°, (c) 0°, (d) 45°, 75. In an experiment for determination of refractive index of, glass of a prism by i – d, plot it was found thata ray incident, at angle 35°, suffers a deviation of 40° and that it emerges, at angle 79°. In that case which of the following is closest, to the maximum possible value of the refractive index?, [2016], (a) 1.7, (b) 1.8, (c) 1.5, (d) 1.6, 76. Monochromatic light is incident on a glass prism of angle, A. If the refractive index of the material of the prism is µ,, a ray, incident at an angle q, on the face AB would get, transmitted through the face AC of the prism provided :, [2015], A, q, , (b), , B, 400 500 600 700, , l (nm), , Dm, , C, , é, æ, æ 1 öù, (a) q > cos -1 êµsin ç A + sin -1 ç ÷ ú, è µ ø ûú, è, ëê, é, æ, æ 1 öù, (b) q < cos-1 êµsin ç A + sin -1 ç ÷ ú, è µ ø ûú, è, ëê, æ, -1 é, -1 æ 1 ö ù, (c) q > sin êµsin ç A - sin ç ÷ ú, è µ ø ûú, è, ëê, , (c), 400 500 600 700, , l (nm), , Dm, , æ, -1 é, -1 æ 1 ö ù, (d) q < sin êµsin ç A - sin ç ÷ ú, è µ ø ûú, è, ëê, 77. The graph between angle of deviation (d) and angle of, incidence (i) for a triangular prism is represented by[2013], d, , (d), , (a), 400 500 600 700, , l (nm), , o, , i
Page 407 :
P-398, , Physics, , d, , (b), o, , i, , d, , (c), o, , i, , d, , (d), o, , 78., , i, A beam of light consisting of red, green and blue colours, is incident on a right-angled prism on face AB. The, refractive indices of the material for the above red, green, and blue colours are 1.39, 1.44 and 1.47 respectively. A, person looking on surface AC of the prism will see, [Online May 26, 2012], A, , 80. Which of the following processes play a part in the, formation of a rainbow?, [Online May 7, 2012], (i) Refraction, (ii) Total internal reflection, (iii) Dispersion, (iv) Interference, (a) (i), (ii) and (iii), (b) (i) and (ii), (c) (i), (ii) and (iv), (d) (iii) and (iv), 81. The refractive index of a glass is 1.520 for red light and, 1.525 for blue light. Let D1 and D2 be angles of minimum, deviation for red and blue light respectively in a prism of, this glass. Then,, [2006], (a) D1 < D2, (b) D1 = D2, (c) D1 can be less than or greater than D2 depending, upon the angle of prism, (d) D1 > D2, 82. A light ray is incident perpendicularly to one face of a 90°, prism and is totally internally reflected at the glass-air, interface. If the angle of reflection is 45°, we conclude that, the refractive index n, [2004], , 45°, 45°, , 45°, , 45°, B, , 79., , C, , (a) no light, (b) green and blue colours, (c) red and green colours, (d) red colour only, A glass prism of refractive index 1.5 is immersed in water, 4, (refractive index ) as shown in figure. A light beam, 3, incident normally on the face AB is totally reflected to, reach the face BC, if, [Online May 19, 2012], B, , A, q, , C, , (a), (c), , 5, 9, 8, sin q >, 9, sin q >, , (b), (d), , 2, 3, 1, sin q >, 3, sin q >, , (a), , n>, , (c), , n<, , 1, 2, 1, 2, , (b), , n> 2, , (d), , n< 2, , TOPIC 5 Optical Instruments, 83. A compound microscope consists of an objective lens of, focal length 1 cm and an eye piece of focal length 5 cm with, a separation of 10 cm., The distance between an object and the objective lens, at, n, which the strain on the eye is minimum is, cm., 40, The value of n is ______., [NA Sep. 05, 2020 (I)], 84. In a compound microscope, the magnified virtual image is, formed at a distance of 25 cm from the eye-piece. The focal, length of its objective lens is 1 cm. If the magnification is, 100 and the tube length of the microscope is 20 cm, then the, focal length of the eye-piece lens (in cm) is __________., [NA Sep. 04, 2020 (I)], 85. The magnifying power of a telescope with tube length 60, cm is 5. What is the focal length of its eye piece?, [8 Jan. 2020 I], (a) 20 cm, (b) 40 cm, (c) 30 cm, (d) 10 cm
Page 408 :
P-399, , Ray Optics and Optical Instruments, , 86. If we need a magnification of 375 from a compound, microscope of tube length 150 mm and an objective of, focal length 5 mm, the focal length of the eye-piece, should, be close to:, [7 Jan. 2020 I], (a) 22mm, (b) 12mm, (c) 2 mm, (d) 33mm, 87. An observer looks at a distant tree of height 10 m with a, telescope of magnifying power of 20. To the observer the, tree appears :, [2016], (a) 20 times taller, (b) 20 times nearer, (c) 10 times taller, (d) 10 times nearer, 88. To determine refractive index of glass slab using a, travelling microscope, minimum number of readings, required are :, [Online April 10, 2016], (a) Two, (b) Four, (c) Three (d) Five, 89. A telescope has an objective lens of focal length 150 cm, and an eyepiece of focal length 5 cm. If a 50 m tall tower at, a distance of 1 km is observed through this telescope in, normal setting, the angle formed by the image of the tower, is q, then q is close to :, [Online April 10, 2015], (a) 30°, (b) 15°, (c) 60°, (d) 1°, 90. In a compound microscope, the focal length of objective, lens is 1.2 cm and focal length of eye piece is 3.0 cm. When, object is kept at 1.25 cm in front of objective, final image is, formed at infinity. Magnifying power of the compound, microscope should be:, [Online April 11, 2014], (a) 200, (b) 100, (c) 400, (d) 150, 91. The focal lengths of objective lens and eye lens of a, Galilean telescope are respectively 30 cm and 3.0 cm., telescope produces virtual, erect image of an object, situated far away from it at least distance of distinct vision, from the eye lens. In this condition, the magnifying power, of the Galilean telescope should be:, [Online April 9, 2014], (a) + 11.2 (b) – 11.2 (c) – 8.8, (d) + 8.8, 92. This question has Statement-1 and Statement-2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., Statement 1: Very large size telescopes are reflecting, telescopes instead of refracting telescopes., Statement 2: It is easier to provide mechanical support to, large size mirrors than large size lenses., [Online April 23, 2013], (a) Statement-1 is true and Statement-2 is false., (b) Statement-1 is false and Statement-2 is true., , 93., , 94., , 95., , 96., , 97., , 98., , (c) Statement-1 and statement-2 are true and Statement2 is correct explanation for statement-1., (d) Statements-1 and statement-2 are true and Statement2 is not the correct explanation for statement-1., The focal length of the objective and the eyepiece of a, telescope are 50 cm and 5 cm respectively. If the telescope, is focussed for distinct vision on a scale distant 2 m from, its objective, then its magnifying power will be:, [Online April 22, 2013], (a) – 4, (b) – 8, (c) + 8, (d) – 2, –2, A telescope of aperture 3 × 10 m diameter is focused on, a window at 80 m distance fitted with a wire mesh of spacing, 2 × 10–3 m. Given: l = 5.5 × 10–7 m, which of the following, is true for observing the mesh through the telescope?, [Online May 26, 2012], (a) Yes, it is possible with the same aperture size., (b) Possible also with an aperture half the present, diameter., (c) No, it is not possible., (d) Given data is not sufficient., We wish to make a microscope with the help of two positive, lenses both with a focal length of 20 mm each and the, object is positioned 25 mm from the objective lens. How, far apart the lenses should be so that the final image is, formed at infinity?, [Online May 12, 2012], (a) 20mm, (b) 100 mm, (c) 120 mm, (d) 80mm, An experment is performed to find the refractive index of, glass using a travelling microscope. In this experiment, distances are measured by, [2008], (a) a vernier scale provided on the microscope, (b) a standard laboratory scale, (c) a meter scale provided on the microscope, (d) a screw gauge provided on the microscope, The image formed by an objective of a compound, microscope is, [2003], (a) virtual and diminished, (b) real and diminished, (c) real and enlarged, (d) virtual and enlarged, An astronomical telescope has a large aperture to [2002], (a) reduce spherical aberration, (b) have high resolution, (c) increase span of observation, (d) have low dispersion
Page 409 :
P-400, , 1., , Physics, , Magnification, m =, , 2., , Let angle between the two mirrors be q., Ray PQ P mirror M1 and Rs P mirror M2, \ M1Rs = ÐORQ = Ð M1OM2= q, Similarly, ÐM2QP = ÐOQR = Ð M2OM1= q, , (1), Distance of object, u = – 30 cm, Distance of image, v = 10 cm, -v (-10) 1, =, =, u, -30, 3, , \, , 1, Speed of image = m2 × speed of object = ´ 9 = 1 cm s–1, 9, (d) Object is placed beyond radius of curvature (R) of, concave mirror hence image formed is real, inverted and, diminished or unmagnified., , 6., , C hi, , 3., , F, Image, , (c) Using mirror formula, magnification is given by, f, –1, =, u – f 1– u, f, At focus magnification is ¥, And at u = 2f, magnification is 1., Hence graph (d) correctly depicts ‘m’ versus distance of, object ‘x’ graph., , P, hi < h0, , v, Þ v = -5u, u, 1 1 1, Using + =, v u f, , (c), , +5 = -, , 7., , (c) When object is at 8 cm, f ´ u 5´8, 40, cm, =, =u -f 8-5, 3, When object is at 12 cm, , 1, 1, 1, + =, -5u u 0.4, \ u = 0.32 m, , Image V1 =, , or, , Image V2 =, 3d, 2, , 4., , 180°, = 60°, 3, , m=, , h0, O, , In DORQ, 3q = 180° Þ q =, , S, , L, , (c), , L, , f ´ u 5 ´ 12, 60, =, =cm, u - f 12 - 5, 7, , 40 60 100, cm, =, 3, 7, 21, So A, C and D are correct statements., (c) Convex morror is used as a shaving mirror., , Separation = V1 - V2 =, , 8., 3d, 2, , L, , O, , Total distance =, 5., , (b), , 10 cm, , 3d 3d, +, = 3d, 2, 2, , 15 cm, , M1, q, , R, , From question : v = 15 cm, u = – 10 cm, Radius of curvature, R = 2f = ?, , P, , q, , O, , q, , Using mirror formula,, , q, , q, Q, , M2, , 1 1 1, + =, v u f, , 1, 1, 1, +, =, Þ f = – 30 cm, 15 (-10) f, Therefore radius of curvature,, R = 2f = – 60 cm
Page 410 :
P-401, , Ray Optics and Optical Instruments, , 9., , (a) From mirror formula, 1 1 1, + =, v u f, , Differentiating the above equation, we get, , 13. (90.00), In the figure, QR is the reflected ray and QS is refracted, ray. CQ is normal., , dv, v 2 æ du ö, =- 2ç ÷, dt, u è dt ø, , Q, , Also,, v, f, =, u u– f, , r', P, , 2, , æ f ö du, dv, = -ç, Þ, è u - f ø÷ dt, dt, , r, , 60°, , 2, , 10., , 11., , 12., , m= 3, , dv 1, =, m/s, dt 15, , (b) The number of images formed is given by, 360, n=, -1, q, 360, -1 = 3, Þ, q, 360°, Þ q=, = 90°, 4, (a) When two plane mirrors are inclined at each other at, an angle q then the number of the images (n) of a point, object kept between the plane mirrors is, 360°, - 1,, n=, q, 360°, (if, is even integer), q, 360°, -1 = 5, \Number of images formed =, 60º, (158), 15, From figure, sin i =, and sin r = sin 45°, 2, 15 + 30 2, From Snell's law, m ´ sin i = 1 ´ sin r, Þm´, , 15, 2, , 15 + 30, , 2, , = 1 ´ sin 45° =, , r, , 30 cm, , 1, , Apply Snell's law at P, 1sin 60° = 3 sin r, 1, Þ r = 30°, 2, From geometry, CP = CQ, Þ sin r =, , \ r ' = 30°, Again apply snell's law at Q,, , 3 sin r ' = 1sin e, , 3, = sin e Þ e = 60°, 2, From geometry, (As angles lies on a straight line), r '+ q + e = 180°, Þ, , Þ 30° + q + 60° = 180° Þ q = 90°., , 14. (d) Apparent depth,, , h, , m1 = 2, m2 = 2 2, , t1 t2, h, h, 3h, 3h 2, +, =, +, =, =, m1 m2, 4, 2 2 2 2 2, 15. (c) Given,, Refractive index, m =, , 45° 15 cm, P, 15 cm, , 45 cm, , 4, 3, , 4, sin q = 1sin 90°, 3, 3, Þ sinq =, 4, , 30 cm, , 1, 2 = 158 ´ 10 -2 = N, \m =, 15, 100, 1125, Hence, value of N ; 158 ., , h, , D=, , 2, , 15 cm, , i, , q, S, r' R, reflected ray, , C, , dv æ 0.2 ö, =ç, Þ, ÷ ´ 15, dt è 2.8 - 0.2 ø, , Þ, , e, , cos q =, , 7, 4, , Solid angle, W = 2p(1– cosq) = 2p(1– 7 / 4), Fraction of energy transmitted
Page 412 :
P-403, , Ray Optics and Optical Instruments, , 22., , (d), , Incident ray, denser, A A, (µD), 90°, rarer, (µR), , r, , m R sin i, ..... (i), =, m D sin r, Q Ði = A and Ðr = (90° - A), m, We also know that, sin qC = R, mD, sin A, n, From eq (i), sin qC =, sin(90° - A), sin A, sin q C =, cos A, sin qC = tan A, or, A = tan–1 (sin qC), 4, 23. (d) Given, µ =, 3, h = 15 cm, R= ?, From Snell's law,, , Þ, Þ, or,, or,, 24., , (c), , 1, sin C = =, m, , h, , 3, =, 2, 2, 4, R +h, 2, 2, 2, O, 16 R = 9 R + 9 h, 7R2 = 9 h2, 3, 3, R=, h=, ´ 15cm, 7, 7, Real depth = 5 cm + 1cm = 6 cm, , Water, m = 1.33, , R, , \ (sin 90° - r) =, Þ cos r =, , 5cm, 1cm, , 5, 1, =, +, 1.33 1.5, ; 3.8 + 0.7 ; 4.5 cm, \ Shift = 6 cm – 4.5 cm @ 1.5 cm, (d) At point A by Snell’s law, sin 45°, 1, m=, Þ sin r =, … (i), sin r, m 2, At point B, for total internal reflection,, , sin i1 =, , 1, m, , 1, m, , 1, m, , … (ii), , Now cos r = 1 - sin 2 r = 1 R, , =, , 1, 2m 2, , 2m 2 - 1, , … (iii), , 2m2, , From eqs (ii) and (iii), 1, 2m 2 - 1, =, m, 2m 2, , Squaring both sides and then solving, we get, 3, 2, 26. (b) Velocity of light in medium, m=, , 3cm, 3 ´ 10-2 m, =, = 1.5 m/s, 0.2ns 0.2 ´ 10-9 s, Refractive index of the medium, , Vmed =, , d1 d2, +, + ...., Apparent depth =, m1 m2, , 25., , A, i1, 90° m, , From figure, i1 = 90° - r, , m=, m = 1.5, Glass, , i2, , B, , C, , sin 90°, =m, sin C, , 45°, , Air, , Vair, 3 ´108, =, = 2 m/s, Vmed, 1.5, , As µ =, , 1, sin C, , \ sin C =, , 1 1, = = 30°, m 2, , Condition of TIR is angle of incidence i must be greater, than critical angle. Hence ray will suffer TIR in case of (B), (i = 40° > 30°) only., 27. (a) As refractive index for z > 0 and z £ 0 is different xy, plane should be the boundry between two media., Angle of incidence is given by, cos (p–i) =, , (6, , ), , 3iˆ + 8 3 ˆj - 10 kˆ .kˆ, 20
Page 415 :
P-406, , =, , Physics, , when liquid is put between, then, , m1 – m2, R, , 1, 2 2, = +, f2 fl f, , R, m1 – m2, (d) From the equation of line, m = k1v + k2 (Q y = mx + c), Þ f eq =, , 40., , 1, 2 2, or (27 / 2) = 18 + f, or f = – 54 cm, , Þ, , v, = k1v + k2, u, , vö, æ, çQ m = ÷, uø, è, , Now -, , Þ, , k, 1, = k1 + 2, u, v, , (Dividing both sides by v), , æ 1 ö, = (m1 - 1) ´ ç, ÷, è -18 ø, , k, 1, Þ 2 - – k1, v u, , \m1 =, , 1 1 1, Comparing with lens formula v - u = f , we get, k1 =, , 41., , (b) Using, M <, or, , ,2 <, , v, u, , 1, 4, +1 =, 3, 3, , 40 ´ 20, 43. (Bouns) v1 = ( 40 – 20 ) = 40 cm, , u2 = 60 – 40 = 20 cm, , 1, and k2 = 1, –f, , 1, b, \ f=, = –, slope of m - v graph, c, , 1, 1, = (m1 - 1) ´, 54, R, , \v2 =, , 20 ´10, , ( 20 –10 ), , = 20 cm, , \ Image traces back to object itself as image formed by, lens is a centre of curvature of mirror., 44. (c) For lens, , v1, Þ v1 < ,2 x1, x1, , 1 1 1, We have v , u < f, 1, 1, 1, , <, ,2 x1 x1 20, x1 = 30 cm, , or, , 1 1 1, - =, v u f, , 1, 1, 1, And 2 x , x < 20, 2, 2, or x2 = – 10 cm, x1 30, <, <3, x2 10, 1, 2, (a) f = f, 1, l, , So,, , 42., , Here 2f1 = 18 cm or f1 = 9 cm, , So,, , 1 2, =, or fl = 18 cm, 9 fl, , 1, æ2ö, = (m - 1) ç ÷, Using,, fl, èRø, , 1, æ2ö, = (1.5 - 1) ç ÷, 18, èRø, \ R = 18 cm, or, , 1, 1, 1, =, v -30 20, \ v = + 60 cm, According to the condition, image formed by lens, should be the centre of curvature of the mirror, and so, 2f’ = 20 or f’ = 10 cm, 1 1 1, (d) By lens’s formula, - =, V u f, For first lens, [u1 = –20], , or, , 45., , 1, 1, 1, 20, = Þ VI =, V1 -20 5, 3, , Image formed by first lense will behave as an object for, second lens, so, u 2 =, , 20, 14, -2=, 3, 3, , 1, 1, 1, =, Þ V2 = 70 cm, V2 14 -5, 3
Page 417 :
P-408, , Þ, , Physics, , 1 1 2, 9, = - =, v 5 19 95, , 95, = 10.55cm, 9, Thus, screen is shifted by a distance, d = 10.55 – 10 = 0.55 cm away from the lens., , or, v =, , 52., , (b) Case-1, , +, 1 æ m –1ö, =ç, ÷, f1 è R ø, , P = 2P1 + P2 Þ, , f = –28, , 1, æ m –1 ö, = 2ç, ÷, 28, è R ø, (Q Power, P =, , 1, & fplane mirror = ¥), f, , 1 1 1, 1, 1, 1, + = Þ +, =, Þ v = 20 cm., v u f, v -4 -5, , +, f2 = –, , R, 2, , f = –10 cm, , 1, æ m –1ö 2, = 2ç, ÷+, 10, è 2 ø R, 1, 1 2, =, + Þ 2 = 1 - 2 = 18, 10 28 R, R 10 28 280, 280, R=, cm, 9, , P = 2P1 + P2 Þ, or,, or,, or,, , 5, 9, , 1 1, 1, d, = +, and solving, we get f1, f2 18, F f1 f 2 f1 f 2, , cm and 20 cm respectively., (a) Given, focal length of lens (f) = 15 cm, object is placed at a distance (u) = – 20 cm, By lens formula,, f = 15 cm, 1 1 1, = f v u, 1 1, 1 1 1, = + = v f u 15 20, 1 4 -3, =, v, 60, v = 60 cm, , Apparent height, ha = hr, , m1, 1, = 30 ´, = 20 cm below, m2, 1.5, , flat surface., 56. (a) Len's formula is given by, 1 1 1, = f v u, For convex lens,, 1, 1, 1, 1, 1, = +, Þ, =, 30 v 60, 60 v, , Similarly for concave lens, , m = 1+, , Using, 54., , Þ m–1=, , 5 14, =, = 1.55, 9 9, (a) For minimum spherical aberration separation,, d = f1 – f2 = 2 cm, Resultant focal length = F = 10 cm, , \, , 53., , 1, æ m –1 ö, = 2ç, ÷9, 28, è 280 ø, , 10, = –5 cm, 2, u = (10 – 6) = –4 cm., By using mirror formula,, , \ Focal length f =, , Case-2, , 1 æ m –1 ö, =ç, ÷, f1 è R ø, , The image I gets formed at 60 cm to the right of the lens, and it will be inverted., The rays from the image (I) formed further falls on the, convex mirror forms another image. This image should, formed in such a way that it coincide with object at the, same point due to reflection takes place by convex mirror., Distance between lens and mirror will be, d = image distance (v) – radius of curvature of convex, mirror., 5 = 60 – 2f, 2f = 60 – 5, 55, f=, = 27.5 cm (convex mirror), 2, 55. (b) Given, radius of hemispherical glass R = 10 cm, , d = 5 cm, , 1, 1 1, 1, 1, = Þ =, - 120 v 40, v 60, , Virtual object 10 cm behind plane mirror., Hence real image 10 cm infront of mirror or, 60 cm from, convex lens., 57. (a) Taking f2 = 12.07, Using Mirror's formula, 1 1 1, = +, f, v u, 1, 1, 1, 1, 1, 1, Þ, =, + Þ, =, 12.7 25.4 u, 12.7 25.4 u, u = 25.4 = v', Now using Len's formula, 1 1 1, 1, 1, 1, = - Þ, =, +, f v u, f1 25.4 + 13.6 10, , 1, 1, 1, 390, =, +, Þ f1 =, = 7.96, f1 39 10, 49, The closest answers is (a) as option (c) and (d) are not, possible., Þ
Page 418 :
P-409, , Ray Optics and Optical Instruments, , 58., , (d) When object is kept at a distance ‘a’ from thin, covex lens, , æ 1, 1, 1 ö, = ( a m l - 1) ç, ÷, fa, è R1 R 2 ø, , O, , Þ, , I1 v, , If m1 > m, then fm and fa have opposite signs and the, nature of lens changes i.e. a convex lens diverges the light, rays and concave lens converges the light rays. Thus given, option (a) is correct., 61. (d) Given: Separation of lens for two of its position,, d = 10 cm, Ratio of size of the images in two positions, , a, 1 1 1, By lens formula : v – u = f, 1, 1, 1, –, =, V (– a) f, , I1 3, =, I2 2, , 1 1 1, or, = –, ...(i), v f a, Mirror forms image at equal distance from mirror, , v, , I1, , v, , Distance of object from the screen, D = ?, Applying formula,, I1 ( D + d )2, =, I 2 ( D - d )2, , I2, , Þ, , Now, again from lens formula, , I3, , 3, D 2 + 100 + 20 D, = 2, 2, D + 100 - 20 D, Þ 3D2 + 300 – 60D = 2D2 + 200 + 40D, Þ D2 – 100D + 100 = 0, On solving, we get D = 99 cm, Hence the distance between the screen and the object is 99 cm., , v, 3 1 1, – =, a V f, , 59., , 62. (c) \ n =, [From eqn. (i)], , 3, 2, 32 + (R – 3mm)2 = R2, Þ 32 + R2 – 2R(3mm) + (3mm)2 = R2, Þ R » 15 cm, , öæ 2 ö, 1 æ m, =ç, - 1÷ ç ÷, f è mL, øè R ø, , 60., , 4, , f1 = 4 R, 3, , 5, for m L 2 = , f 2 = -5 R, 3, Þ f2 = (–) ve, (a) If a lens of refractive index m is immersed in a medium, of refractive index m1, then its focal length in medium is, given by, æ 1, 1 ö, - 1) ç, ÷, è R1 R 2 ø, If fa is the focal length of lens in air, then, 1, =, fm, , (, , m ml, , Velocity of light in vacuum, Velocity of light in medium, , \ n=, , Hence, a = 2f, (b) By Lens maker's formula for convex lens, , for, m L1 =, , 3 ( D + 10) 2, =, 2 ( D - 10) 2, , Þ, , I2, , a/3, , 3 1 1 1, – + =, a f a f, , f m ( a ml - 1), =, fa ( m ml - 1), , 63., , 1 æ 3 öæ 1 ö, = ç –1 ÷ç ÷ Þ f = 30 cm, f è 2 øè 15 ø, (d) If side of object square = l, and side of image square = l¢, , From question,, , l '2, =9, l, , l', =3, l, i.e., magnification m = 3, u = – 40 cm, v = 3 × 40 = 120 cm, f= ?, , or, , R = 3cm, 3mm
Page 419 :
P-410, , Physics, , From formula,, , 1 1 1, - =, v u f, , Experimental, curve, , 66. (d) |v|, , 1, 1, 1, =, 120 -40 f, , or,, 64., , Straight, line, , 1, 1, 1 1+ 3, =, +, =, \ f = 30 cm, f 120 40 120, , (d) The focal length of the lens, 1 1 1, = f v u, , 1, 1, +, 12 240, 20 + 1 21, =, =, 240, 240, , 1 1 1, - =, v u f, , 240, cm, 21, When glass plate is interposed between lens and film, so, shift produced will be, , æ, 1ö, Shift = t ç 1 - ÷, è mø, 1 ö, 1, æ, 1ç 1 = 1´, è 3 / 2 ÷ø, 3, Now image should be form at, , v' = 12 - 1 = 35 cm, 3 3, Now the object distance u., Using lens formula again, 1, 1 1, < ,, f, v' u, 1 1 1, Þ < ,, u v' f, 1 3, 21 1 é 3 21 ù, =, Þ =, u 35 240 5 êë 7 48 úû, , 1 1 é 48 - 49 ù, Þ = ê, u 5 ë 7 ´ 16 úû, Þ u = –7 ×16 × 5 = – 560 cm = – 5.6 m, (a) From the Cauchy, B C, Formula, µ = A + 2 + 1, l, l, 1, \µµ, l, As, lblue < lred, \ lblue > µred, From lens maker's formula, , Þ, , æ1, 1, 1ö, = (m - 1) ç - ÷, f, è R1 R2 ø, 1, 1, >, fB, fR, , 45°, , From lens formula, , f =, , and, , (2f, 2f), , |u|, For the graph to intersect y = x line. The value of | v | and |, u | must be equal., , =, , 65., , P, , Þ fR > fB., , When u = -2 f , v = 2 f, f, Also v =, f, 1+, u, As |u| increases, v decreases for |u| > f. The graph between, |v| and |u| is shown in the figure. A straight line passing, through the origin and making an angle of 45°with the, x-axis meets the experimental curve at P (2f, 2f )., , 1 1 1, = f v u, This graph suggest that when, , 67. (c) From the lens formula, u = – f, v = + µ, When u is at – µ , v = f., , v (cm), , f, –f, , u (cm), , When the object is moved further away from the lens, v, decreases but remains positive., 68. (c) When two thin lenses are in contact coaxially, power, of combination is given by, P = P1 + P2, = (– 15 + 5) D, = – 10 D., Also, P =, Þ f =, , 1, f, , 1, 1, metre, =, P -10, , æ1, ö, \ f = - ç ´ 100÷ cm = -10 cm., è 10, ø
Page 421 :
P-412, , Physics, , 5, 5, . That is m max is less than = 1.67, 3, 3, But dm will be less than 40° so, , m max can be, , m<, , 76., , 5, 5, sin 57° < sin 60° Þ m = 1.5, 3, 3, , (c) When r2 = C, ÐN2RC = 90°, Where C = critical angle, As sin C =, , or, , m<, , Þ, , m < 2 Þ m < 1.414, 79. (c) For total internal reflection on face AC, q > critical angle (C), and sinq ³ sinC, sin q ³, , 1, = sin r2, m, , N1, N2, , Q r1, , r2, , P, , 1, wm g, , 4, mw, sin q ³, Þ sin q ³ 3, 3, mg, 2, 8, \ sin q ³ ., 9, 80. (a) Rainbow is formed due to the dispersion of light, suffering refraction and total internal reflection (TIR) in, the droplets present in the atmosphere., , A, , q, , 1, 1, Þm<, sin q, sin 45°, , R, , 81. (a) When angle of prism is small,, B, , Angle of deviation, D = (m – 1) A, , C, , Applying snell's law at ‘R’, µ sin r2 = 1 sin90°, ...(i), Applying snell's law at ‘Q’, 1 × sin q = µ sin r 1, ...(ii), But r1 = A – r2, So, sin q = µ sin (A – r 2), sin q = µ sin A cos r 2 – cos A, From (1), 2, cos r2 = 1 – sin r2 = 1–, , Since lb < lr, Þ mr < mb, , 82. (b) For total internal reflection, Incident angle (i) > critical angle (ic),, ...(iii), , 1, µ2, , [using (i)], , 1, µ2, , ...(iv), , - cos A, , é, –1 æ 1 ö ù, q = sin–1 êm sin(A – sin ç µ ÷ ú, è øû, ë, So, for transmission through face AC, , 78., , é, –1 æ 1 ö ù, q > sin–1 êm sin(A – sin ç µ ÷ ú, è øû, ë, (c) For the prism as the angle of incidence (i) increases,, the angle of deviation (d) first decreases goes to minimum, value and then increases., (d) For light to come out through face 'AC', total, internal reflection must not take place., i.e., q < c Þ sin q < sin c, , 1, Þ sin q <, m, , \ sin 45° >, , Þ, , on further solving we can show for ray not to transmitted, through face AC, , 77., , \ sin i > sin ic, Þ sin 45° > sin ic Þ sin ic =, , By eq. (iii) and (iv), sin q = µsin A 1 -, , Þ D1 < D2, , 1, 2, , >, , 1, n, , 1, n, , 1, n, , Þn> 2, , 83. (50), Given : Length of compound microscope, L = 10 cm, Focal length of objective f0 = 1 cm and of eye-piece,, fe = 5 cm, u0 = fe = 5 cm, Final image formed at infinity ( ¥ ), ve = ¥, v0 = 10 – 5 = 5, Using lens formula,, , 1 1 1, - =, v u f, , 1 1, 1, 1 1 1, 5, =, Þ = Þ u0 = - cm, v0 u0, f0, 5 u0 1, 4, , or,, , 5 N, =, 4 40, , \N =, , 200, = 50 cm., 4
Page 423 :
P-414, , 91., , Physics, , (Q D = 25 cm least distance of distinct vision) = 200, Hence the magnifying power of the compound microscope, is 200, (d) Given, Focal length of objective, f0 = 30 cm, focal length of eye lens, fe = 3.0 cm, Magnifying power, M = ?, Magnifying power of the Galilean telescope,, f æ f ö, M D = 0 ç1 - e ÷, fe è, Dø, , Limit of resolution, Dq =, 1.22 ´ 5.5 ´ 10-7, , = 2.23 × 10–5 rad., 3 ´10-2, At a distance of 80 m , the telescope is able to resolve, between two points which are separated by 2.23 × 10–5, × 80 m, = 1.78 × 10–3 m, , =, , 30 æ, 3 ö, ç1 - ÷ [Q D = 25 cm], 3 è 25 ø, , =, , 22, = 8.8 cm, 25, (c) One side of mirror is opaque and another side is, reflecting this is not in case of lens hence, it is easier to, provide mechanical support to large size mirrors than large, size lenses. Reflecting telescopes are based on the same, principle except that the formation of images takes place, by reflection instead of refraction., , 1.22l, d, , Eye-Piece, Objective, , = 10 ´, , 92., , 93., , (d) Given : f0 = 50 cm, f e = 5cm, d = 25 cm, u0 = –200 cm, Magnification M = ?, As, , 1, 1, 1, =, v0 u 0 f 0, , Þ, , 1, 1 1, 1, 1, 4 -1, 3, = +, =, =, =, v0 f 0 u 0 50 200 200 200, , v0 =, , or, , 200, cm, 3, , Now ve = d = –25cm, From,, , 1, 1, 1, =, ve u e fe, , –, , 1, 1 1, = u e f e ve, , =, , 1 1, 6, +, =, 5 25 25, , -25, cm, 6, Magnification M = M0 × Me, , or,, , ve =, , v v, -200 / 3 -25, = 0´ e =, ´, u0 ue, 200, -25 / 6, , 94., , 1, = - ´ 6 = -2, 3, (a) Given : d = 3 × 10–2 m, l = 5.5 × 10–7 m, , 95. (c), , O, , F0, 20 mm, 25 mm, , I, V0, , fe, , To obtain final image at infinity, object which is the image, formed by objective should be at focal distance of eyepiece., By lens formula (for objective), 1, 1, 1, =, v0 u0, f0, , or,, , 1, 1, 1, =, v0 -25 20, , 1, 1, 1, 5- 4, 1, =, =, mm, v0 20 25 = 100 100, \ v0 = 100 mm, Therefore the distance between the lenses, = v0 + fe = 100 mm + 20 mm = 120 mm, 96. (a) To find the refractive index of glass using a travelling, microscope, a vernier scale is provided on the microscope, 97. (c) A real, inverted and enlarged image of the object is, formed by the objective lens of a compound microscope., 98. (b) The resolving power of a telescope is, , Þ, , R.P =, , D, 1.22 l, , where D = diameter of the objective lens, l = wavelength of light., D, l, Resolving power of telescope resolution will be high if its, objective is of large aperture., , Clearly, R.P µ
Page 424 :
24, Wave Optics, Wavefront, Interference of, TOPIC 1 Light, Coherent and, Incoherent Sources, 1., , In the figure below, P and Q are two equally intense, coherent sources emitting radiation of wavelength 20 m., The separation between P and Q is 5 m and the phase of P, is ahead of that of Q by 90º. A, B and C are three distinct, points of observation, each equidistant from the midpoint, of PQ. The intensities of radiation at A, B, C will be in the, ratio :, [Sep. 06, 2020 (I)], B, , (a), , A, (a) 0 : 1 : 4, (b) 2 : 1 : 0, (c) 0 : 1 : 2, (d) 4 : 1 : 0, Two coherent sources of sound, S1 and S2, produce sound, waves of the same wavelength, l = 1 m, in phase. S1 and S2, are placed 1.5 m apart (see fig.). A listener, located at L,, directly in front of S2 finds that the intensity is at a minimum, when he is 2 m away from S2. The listener moves away, from S1, keeping his distance from S2 fixed. The adjacent, maximum of intensity is observed when the listener is at a, distance d from S1. Then, d is :, [Sep. 05, 2020 (II)], , 4., , 5., , 6., , 2m, , (a) 12 m, , S1, , (b) 5 m, , (b), , 2p æ L1 L2 ö, ç - ÷, l è n1 n2 ø, , 2p, 2p, (n2 L1 - n1 L2 ), ( n1 L1 - n2 L2 ), (d), l, l, In an interference experiment the ratio of amplitudes of, , minimum intensities of fringes will be : [8 April 2019 I], (a) 2, (b) 18, (c) 4, (d) 9, , C, , 1.5 m, , 2 p æ L2 L1 ö, ç - ÷, l è n1 n2 ø, , (c), , Q, , S2, , Two light waves having the same wavelength l in vacuum, are in phase initially. Then the first wave travels a path L1, through a medium of refractive index n1 while the second, wave travels a path of length L2 through a medium of, refractive index n2. After this the phase difference between, the two waves is :, [Sep. 03, 2020 (II)], , a1 1, coherent waves is a = 3 . The ratio of maximum and, 2, , P, , 2., , 3., , 2m, , L, , d, , (c) 2 m, , (d) 3 m, , 7., , Two coherent sources produce waves of different intensities, which interfere. After interference, the ratio of the maximum, intensity to the minimum intensity is 16. The intensity of the, waves are in the ratio:, [9 Jan. 2019 I], (a) 16 : 9, (b) 25 : 9 (c) 4 : 1, (d) 5 : 3, On a hot summer night, the refractive index of air is smallest, near the ground and increases with height from the ground., When a light beam is directed horizontally, the Huygens', principle leads us to conclude that as it travels, the light, beam :, [2015], (a) bends downwards, (b) bends upwards, (c) becomes narrower, (d) goes horizontally without any deflection, Interference pattern is observed at ‘P’ due to, superimposition of two rays coming out from a source ‘S’, as shown in the figure. The value of ‘l’ for which maxima is, obtained at ‘P’ is:, (R is perfect reflecting surface) [Online April 12, 2014]
Page 425 :
P-416, , Physics, , 21, , S, , P, , 30°, , (a) l =, , (c), 8., , 9., , l=, , 2nl, , (b), , 3 -1, , ( 2n - 1) l, , (, , 4 2- 3, , 3, , ), , l=, , (d) l =, , ( 2n - 1) l, 2, , (, , ), , 3 -1, , ( 2n - 1) l, 3 -1, , Two monochromatic light beams of intensity 16 and 9 units, are interfering. The ratio of intensities of bright and dark, parts of the resultant pattern is: [Online April 11, 2014], 16, 4, 7, 49, (a), (b), (c), (d), 3, 1, 1, 9, n identical waves each of intensity I0 interfere with each, other. The ratio of maximum intensities if the interference, is (i) coherent and (ii) incoherent is :, [Online April 23, 2013], 1, 1, (c), (d) n, n, n2, A ray of light of intensity I is incident on a parallel glass, slab at point A as shown in diagram. It undergoes partial, reflection and refraction. At each reflection, 25% of incident, energy is reflected. The rays AB and A'B' undergo, interference. The ratio of Imax and Imin is :, [Online April 9, 2013], B, B', , (a) n2, 10., , R, , (b), , A', A, , C, , 11., , 12., , C', , (a) 49 : 1, (b) 7 : 1, (c) 4 : 1, (d) 8 : 1, Two coherent plane light waves of equal amplitude makes, a small angle a (< < 1) with each other. They fall almost, normally on a screen. If l is the wavelength of light waves,, the fringe width Dx of interference patterns of the two sets, of waves on the screen is, [Online May 19, 2012], l, l, 2l, l, (a), (b), (c) 2a, (d), ( ), a, a, a, This question has a paragraph followed by two statements,, Statement – 1 and Statement – 2. Of the given four, alternatives after the statements, choose the one that, describes the statements., A thin air film is formed by putting the convex surface of a, plane-convex lens over a plane glass plate. With, , monochromatic light, this film gives an interference pattern, due to light reflected from the top (convex) surface and, the bottom (glass plate) surface of the film., Statement – 1 : When light reflects from the air-glass plate, interface, the reflected wave suffers a phase change of p., Statement – 2 : The centre of the interference pattern is, dark., [2011], (a) Statement – 1 is true, Statement – 2 is true, Statement, – 2 is the correct explanation of Statement – 1., (b) Statement – 1 is true, Statement – 2 is true, Statement, – 2 is not the correct explanation of Statement – 1., (c) Statement – 1 is false, Statement – 2 is true., (d) Statement – 1 is true, Statement – 2 is false., Directions : Questions number 13-15 are based on the following, paragraph., An initially parallel cylindrical beam travels in a medium of, refractive index m (I) =m0 + m2 I, where m0 and m2 are positive, constants and I is the intensity of the light beam. The intensity, of the beam is decreasing with increasing radius, 13. As the beam enters the medium , it will, [2010], (a) diverge, (b) converge, (c) diverge near the axis and converge near the periphery, (d) travel as a cylindrical beam, 14. The initial shape of the wavefront of the beam is [2010], (a) convex, (b) concave, (c) convex near the axis and concave near the periphery, (d) planar, 15. The speed of light in the medium is, [2010], (a) minimum on the axis of the beam, (b) the same everywhere in the beam, (c) directly proportional to the intensity I, (d) maximum on the axis of the beam, 16. To demonstrate the phenomenon of interference, we, require two sources which emit radiation, [2003], (a) of nearly the same frequency, (b) of the same frequency, (c) of different wavelengths, (d) of the same frequency and having a definite phase, relationship, , TOPIC 2, , Young's Double Slit, Experiment, , 17. A young’s double-slit experiment is performed using, monocromatic light of wavelength l. The inntensity of, light at a point on the screen, where the path difference is, l, is K units. The intensity of light at a point where the, l, nK, is given by, , where n is an inte6, 12, ger. The value of n is ______. [NA Sep. 06, 2020 (II)], , path difference is
Page 426 :
P-417, , Wave Optics, , 18., , 19., , 20., , In a Young’s double slit experiment, light of 500 nm is used, to produce an interference pattern. When the distance, between the slits is 0.05 mm, the angular width (in degree), of the fringes formed on the distance screen is close to :, [Sep. 03, 2020 (I)], (a) 0.17°, (b) 0.57°, (c) 1.7°, (d) 0.07°, Interference fringes are observed on a screen by, illuminating two thin slits 1 mm apart with a light source (l, = 632.8 nm). The distance between the screen and the slits, is 100 cm. If a bright fringe is observed on a screen at a, distance of 1.27 mm from the central bright fringe, then the, path difference between the waves, which are reaching, this point from the slits is close to : [Sep. 02, 2020 (I)], (a) 1.27 mm, (b) 2.87 nm, (c) 2 nm, (d) 2.05 mm, In a Young's double slit experiment, 16 fringes are observed, in a certain segment of the screen when light of wavelength, 700 nm is used. If the wavelength of light is changed to, 400 nm, the number of fringes observed in the same, segment of the screen would be : [Sep. 02, 2020 (II)], (a) 24, (b) 30, (c) 18, , 21., , 22., , 23., , 24., , 25. The figure shows a Young’s double slit experimental setup., It is observed that when a thin transparent sheet of, thickness t and refractive index ¼ is put in front of one of, the slits, the central maximum gets shifted by a distance, equal to n fringe widths. If the wavelength of light used is, », t will be:, [9 April 2019 I], , (a), , 2nDl, a (m _ 1), , (b), , nDl, a (m _ 1), , (c), , Dl, a ( m _ 1), , (d), , 2 Dl, a (m _ 1), , 26. In a Young’s double slit experiment, the path difference, at, a certain point on the screen, betwen two interfering waves, 1, th of wavelength. The ratio of the intensity at this, 8, point to that at the centre of a bright fringe is close to:, , is, , (d) 28, , In a Young’s double slit experiment 15 fringes are observed, on a small portion of the screen when light of wavelength, 500 nm is used. Ten fringes are observed on the same, section of the screen when another light source of, wavelength l is used. Then the value of l is (in nm), ______., [NA 9 Jan 2020 II], In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is, 1, th of a wavelength. The ratio of the intensity of light at, 8, that point to that at the centre of a bright fringe is:, [8 Jan 2020 II], (a) 0.853, (b) 0.672 (c) 0.568, (d) 0.760, In a Young’s double slit experiment, the separation, between the slits is 0.15 mm. In the experiment, a source, of light of wavelength 589 nm is used and the interference, pattern is observed on a screen kept 1.5 m away. The, separation between the successive bright fringes on the, screen is:, [7 Jan 2020 II], (a) 6.9 mm (b) 3.9 mm (c) 5.9 mm (d) 4.9 mm, In a double slit experiment, when a thin film of thickness t, having refractive index µ is introduced in front of one of, the slits, the maximum at the centre of the fringe pattern, shifts by one fringe width. The value of t is (» is the, wavelength of the light used) :, [12 April 2020 I], , [11 Jan 2019 I], (a) 0.74, (b) 0.85, (c) 0.94, (d) 0.80, 27. In a Young’s double slit experiment with slit separation, 1, rad, 40, by using light of wavelength l1 When the light of, wavelength l2 is used a bright fringe is seen at the, same angle in the same set up. Given that l1 and l2 are, in visible range (380 nm to 740 nm), their values are:, [10 Jan. 2019 I], (a) 625 nm, 500 nm, (b) 380 nm, 525 nm, (c) 380 nm, 500 nm, (d) 400 nm, 500 nm, , 0.1 mm, one observes a bright fringe at angle, , 28. Consider a Young’s double slit experiment as shown in, figure. What should be the slit separation d in terms of, wavelength l such that the first minima occurs directly, in front of the slit (S1)?, [10 Jan 2019 II], S1, Source, , P, , st, , 1 minima, , d, S2, Screen, , 2d, , (a), , 2l, (m - 1), , (b), , l, 2(m - 1), , (a), , (c), , l, (m - 1), , (d), , l, (2m - 1), , (c), , 2, , (, (, , l, 5-2, l, , 2 5- 2, , ), , (b), , ), , (d), , (, , l, 5-2, l, , ), , (5 - 2 )
Page 427 :
P-418, , 29., , 30., , 31., , Physics, , In a Young's double slit experiment, the slits are placed, 0.320 mm apart. Light of wavelength l = 500 nm is, incident on the slits. The total number of bright fringes, that are observed in the angular range – 30° £ q £ 30°, is, [9 Jan 2019 II], (a) 640, (b) 320, (c) 321, (d) 641, In a Young's double slit experiment, slits are separated by, 0.5 mm, and the screen is placed 150 cm away. A beam of, light consisting of two wavelengths, 650 nm and 520 nm,, is used to obtain interference fringes on the screen. The, least distance from the common central maximum to the, point where the bright fringes due to both the wavelengths, coincide is :, [2017], (a) 9.75 mm (b) 15.6 mm (c) 1.56 mm (d) 7.8 mm, In a Young’s double slit experiment with light of, wavelength l the separation of slits is d and distance of, screen is D such that D >> d >> l. If the fringe width, is b, the distance from point of maximum intensity to the, point where intensity falls to half of maximum intensity, on either side is:, [Online April 11, 2015], b, b, b, b, (b), (c), (d), 6, 3, 4, 2, In a Young’s double slit experiment, the distance between, the two identical slits is 6.1 times larger than the slit width., Then the number of intensity maxima observed within the, central maximum of the single slit diffraction pattern is:, [Online April 19, 2014], (a) 3, (b) 6, (c) 12, (d) 24, Using monochromatic light of wavelength l, an, experimentalist sets up the Young’s double slit experiment, in three ways as shown., If she observes that y = b’, the wavelength of light used, is:, [Online April 9, 2014], , s1, , (Fringe width = b¢), d, s2, 2D, , (a) 520 nm (b) 540 nm (c) 560 nm (d) 580 nm, 34. Two coherent point sources S1 and S2 are separated by a, small distance 'd' as shown. The fringes obtained on the, screen will be, [2013], , d, S1 S2, , (a), 32., , 33., , s1, , Central fringe, O, , d, , (Fringe width b), , s2, D, , (Screen), , Central fringe, , Screen, D, , (a) points, , (b) straight lines, , (c) semi-circles, , (d) concentric circles, , 35. The source that illuminates the double - slit in 'double - slit, interference experiment' emits two distinct monochromatic, waves of wavelength 500 nm and 600 nm, each of them, producing its own pattern on the screen. At the central, point of the pattern when path difference is zero, maxima, of both the patterns coincide and the resulting interference, pattern is most distinct at the region of zero path difference., But as one moves out of this central region, the two fringe, systems are gradually out of step such that maximum due, to on wavelength coincides with the minimum due to the, other and the combined fringe system becomes completely, indistinct. This may happen when path difference in nm, is:, [Online April 25, 2013], (a) 2000, (b) 3000, (c) 1000, (d) 1500, 2500, l (l is wavelength, 3, of light used) and refractive index m = 1.5 is inserted, between one of the slits and the screen in Young’s double, slit experiment. At a point on the screen equidistant from, the slits, the ratio of the intensities before and after the, introduction of the glass plate is :, [Online April 25, 2013], (a) 2 : 1, (b) 1 : 4, (c) 4 : 1, (d) 4 : 3, This question has Statement-1 and Statement-2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., , 36. A thin glass plate of thickness is, , s1, , Transparent, mica sheet of, thickness, t = 1.8 mm, m = 1.6, , Central fringe, y (Fringe, width b ), O, , d, , 37., , s2, D, , (Screen)
Page 428 :
P-419, , Wave Optics, , 38., , Statement-1: In Young’s double slit experiment, the number, of fringes observed in the field of view is small with longer, wavelength of light and is large with shorter wavelength, of light., Statement-2: In the double slit experiment the fringe width, depends directly on the wavelength of light., [Online April 22, 2013], (a) Statement-1 is true, Statement-2 is true and the Statement2 is correct explanation of the Statement-1., (b) Statement-1 is false and the Statement-2 is true., (c) Statement-1 is true Statement-2 is true and the, Statement-2 is not correct explanation of the, Statement-1., (d) Statement-1 is true and the Statement-2 is false., In Young's double slit experiment, one of the slit is wider, than other, so that amplitude of the light from one slit is, double of that other slit. If Im be the maximum intensity,, the resultant intensity I when they interfere at phase, difference f is given by :, [2012], , Im æ, 2, (b), ç 1 + 2cos, 3 è, , Im, (4 + 5 cos f), (a), 9, , (c), 39., , 40., , 41., , Im, 5, , æ, 2 fö, ç 1 + 4 cos ÷, 2ø, è, , (d), , fö, ÷, 2ø, , Im æ, 2 fö, ç 1 + 8 cos ÷, 9 è, 2ø, , In Young’s double slit interference experiment, the slit, widths are in the ratio 1 : 25. Then the ratio of intensity at, the maxima and minima in the interference pattern is, [Online May 26, 2012], (a) 3 : 2, (b) 1 : 25 (c) 9 : 4, (d) 1 : 5, The maximum number of possible interference maxima for, slit separation equal to 1.8l, where l is the wavelength of, light used, in a Young’s double slit experiment is, [Online May 12, 2012], (a) zero, (b) 3, (c) infinite (d) 5, In a Young’s double slit experiment with light of wavelength, l, fringe pattern on the screen has fringe width b. When, two thin transparent glass (refractive index m) plates of, thickness t1 and t2 (t1 > t2) are placed in the path of the, two beams respectively, the fringe pattern will shift by a, distance, [Online May 7, 2012], mb t1, b ( m - 1) æ t1 ö, (b) l t, ç ÷, l, 2, è t2 ø, l, b ( m - 1), t1 - t2 ), (c), (d) ( m - 1) ( t1 + t2 ), (, b, l, At two points P and Q on screen in Young’s double slit, experiment, waves from slits S1 and S2 have a path, l, difference of 0 and , respectively. The ratio of intensities, 4, at P and Q will be :, [2011 RS], , (a), , 42., , (a) 2 : 1, , (b), , 2 :1 (c) 4 : 1, , (d) 3 : 2, , 43. In a Young’s double slit experiment, the two slits act as, coherent sources of wave of equal amplitude A and, wavelength l. In another experiment with the same, arrangement the two slits are made to act as incoherent, sources of waves of same amplitude and wavelength. If, the intensity at the middle point of the screen in the first, case is I1 and in the second case is I2, then the ratio, (a) 2, , (b) 1, , (c) 0.5, , I1, is, I2, , [2011 RS], (d) 4, , 44. A mixture of light, consisting of wavelength 590 nm and an, unknown wavelength, illuminates Young’s double slit and, gives rise to two overlapping interference patterns on the, screen. The central maximum of both lights coincide., Further, it is observed that the third bright fringe of known, light coincides with the 4th bright fringe of the unknown, light. From this data, the wavelength of the unknown light, is:, [2009], (a) 885.0 nm, , (b) 442.5 nm, , (c) 776.8 nm, (d) 393.4 nm, 45. In a Young’s double slit experiment the intensity at a point, l, where the path difference is (l being the wavelength of, 6, I, light used) is I. If I0 denotes the maximum intensity,, is, I0, equal to, [2007], 1, 1, 3, (a) 3, (b), (c), (d), 2, 2, 4, 2, 46. A Young’s double slit experiment uses a monochromatic, source. The shape of the interference fringes formed on a, screen is, [2005], , (a) circle, , (b) hyperbola, , (c) parabola, , (d) straight line, , 47. The maximum number of possible interference maxima for, slit-separation equal to twice the wavelength in Young’s, double-slit experiment is, [2004], (a) three, (b) five, (c) infinite (d) zero, , Polarisation of, TOPIC 3 Diffraction,, Light and Resolving Power, 48. A beam of plane polarised light of large cross-sectional, area and uniform intensity of 3.3 Wm–2 falls normally on a, polariser (cross sectional area 3 × 10–4 m2) which rotates, about its axis with an angular speed of 31.4 rad/s. The, energy of light passing through the polariser per, revolution, is close to :, [Sep. 04, 2020 (I)], –5, (a) 1.0 × 10 J, (b) 1.0 × 10–4 J, (c) 1.5 × 10–4 J, (d) 5.0 × 10–4 J
Page 429 :
P-420, , 49., , 50., , 51., , 52., , 53., , 54., , Physics, , Orange light of wavelength 6000 × 10–10 m illuminates a, single slit of width 0.6 × 10–4 m. The maximum possible, number of diffraction minima produced on both sides of, the central maximum is _____________., [NA Sep. 04, 2020 (II)], The aperture diameter of telescope is 5m. The separation, between the moon and the earth is 4 × 105 km. With light, of wavelength of 5500 Å, the minimum separation between, objects on the surface of moon, so that they are just, resolved, is close to:, [9 Jan. 2020 I], (a) 60 m, (b) 20 m, (c) 200 m, (d) 600 m, A polarizer - analyser set is adjusted such that the, intensity of light coming out of the analyser is just 10% of, the original intensity. Assuming that the polarizer analyser set does not absorb any light, the angle by which, the analyser need to be rotated further to reduce the, output intensity to be zero, is:, [7 Jan. 2020 I], (a) 71.6°, (b) 18.4° (c) 90°, (d) 45°, The value of numerical aperature of the objective lens of, a microscope is 1.25. If light of wavelength 5000 Å is used,, the minimum separation between two points, to be seen, as distinct, will be :, [12 April 2019 I], (a) 0.24 µm (b) 0.38 µm (c) 0.12 µm (d) 0.48 µm, A system of three polarizers P1, P2, P3 is set up such that the, pass axis of P3 is crossed with respect to that of P1. The, pass axis of P2 is inclined at 60o to the pass axis of P3. When, a beam of unpolarized light of intensity Io is incident on P1,, the intensity of light transmitted by the three polarizers is I., The ratio (Io/I) equals (nearly) :, [12 April 2019 II], (a) 5.33, (b) 16.00 (c) 10.67, (d) 1.80, Diameter of the objective lens of a telescope is 250 cm., For light of wavelength 600 nm. Coming from a distant, object, the limit of resolution of the telescope is close to:, [9 April 2019 II], (a) 1.5 ´ 10, , 55., , 56., , 57., , -7, , rad(b), , 2.0 ´ 10, , -7, , rad, , (c) 3.0 ´ 10 -7 rad, (d) 4.5 ´ 10-7 rad, Calculate the limit of resolution of a telescope objective, having a diameter of 200 cm, if it has to detect light of, wavelength 500 nm coming from a star. [8 April 2019 II], (a) 305 × 10–9 radian, (b) 610 × 10–9 radian, –9, (c) 152.5 × 10 radian, (d) 457.5 × 10–9 radian, In a double-slit experiment, green light (5303Å) falls on a, double slit having a separation of 19.44 µm and a width of, 4.05 µm. The number of bright fringes between the first, and the second diffraction minima is : [11 Jan 2019 II], (a) 10, (b) 05, (c) 04, (d) 09, Consider a tank made of glass (refractive index 1.5) with a, thick bottom. It is filled with a liquid of refractive index m., A student finds that, irrespective of what the incident, angle i (see figure) is for a beam of light entering the, liquid, the light reflected from the liquid glass interface is, never completely polarized. For this to happen, the, minimum value of m is:, [9 Jan. 2019 I], , 3, , 5, , 4, 3, 5, 3, 58. The angular width of the central maximum in a single slit, diffraction pattern is 60°. The width of the slit is 1 mm. The, slit is illuminated by monochromatic plane waves. If, another slit of same width is made near it, Young's fringes, can be observed on a screen placed at a distance 50 cm, from the slits. If the observed fringe width is 1 cm, what is, slit separation distance?, (i.e. distance between the centres of each slit.) [2018], (a) 25 mm, (b) 50 mm (c) 75 mm, (d) 100 mm, 59. Unpolarized light of intensity I passes through an ideal, polarizer A. Another indentical polarizer B is placed behind, I, A. The intensity of light beyond B is found to be . Now, 2, another identical polarizer C is placed between A and B., (a), , 5, 3, , (b), , (c), , (d), , The intensity beyond B is now found to be I . The angle, 8, between polarizer A and C is:, [2018], (a) 0°, (b) 30°, (c) 45°, (d) 60°, 60. Light of wavelength 550 nm falls normally on a slit of width, 22.0 × 10–5cm. The angular position of the second minima, from the central maximum will be (in radians), [Online April 15, 2018], p, p, p, p, (a), (b), (c), (d), 8, 12, 4, 6, 61. Unpolarized light of intensity I is incident on a system of, two polarizers, A followed by B. The intensity of emergent, light is I/2. If a third polarizer C is placed between A and B,, the intensity of emergent light is reduced to I/3. The angle, between the polarizers A and C is q. Then, [Online April 16, 2018], 1/ 4, , 1/ 4, , æ2ö, (a) cos q = ç ÷, è3ø, , æ1ö, (b) cos q = ç ÷, è3ø, , 1/ 2, , 62., , 1/ 2, , æ1ö, æ2ö, (c) cos q = ç ÷, (d) cos q = ç ÷, è3ø, è3ø, A plane polarized light is incident on a polariser with its pass, axis making angle q with x-axis, as shown in the figure. At four, different values of q, q = 8°, 38°, 188° and 218°, the observed, intensities are same. What is the angle between the direction, of polarization and x-axis, [Online April 15, 2018], y, , q, , (a) 203°, , Pass axis, , (b) 45°, , x, z, , (c) 98°, , (d) 128°
Page 430 :
P-421, , Wave Optics, , 63., , 64., , 65., , 66., , An observer is moving with half the speed of light towards, a stationary microwave source emitting waves at frequency, 10 GHz. What is the frequency of the microwave measured, by the observer? (speed of light = 3 × 108 ms–1) [2017], (a) 17.3 GHz, (b) 15.3 GHz, (c) 10.1 GHz, (d) 12.1 GHz, A single slit of width 0.1 mm is illuminated by a parallel, beam of light of wavelength 6000 Å and diffraction, bands are observed on a screen 0.5 m from the slit. The, distance of the third dark band from the central bright, band is :, [Online April 9, 2017], (a) 3 mm, (b) 9mm (c) 4.5 mm (d) 1.5 mm, A single slit of width b is illuminated by a coherent, monochromatic light of wavelength l. If the second and, fourth minima in the diffraction pattern at a distance 1 m, from the slit are at 3 cm and 6 cm respectively from the, central maximum, what is the width of the central maximum?, (i.e. distance between first minimum on either side of the, central maximum), [Online April 8, 2017], (a) 1.5 cm, (b) 3.0 cm (c) 4.5 cm, (d) 6.0 cm, The box of a pin hole camera, of length L, has a hole of, radius a. It is assumed that when the hole is illuminated by, a parallel beam of light of wavelength l the spread of the, spot (obtained on the opposite wall of the camera) is the, sum of its geometrical spread and the spread due to, diffraction. The spot would then have its minimum size, (say bmin) when :, [2016], (a) a = lL and b min = 4lL, l2, and b min = 4lL, L, æ 2l2 ö, l2, and bmin = ç, (c) a =, ÷, ç L ÷, L, è, ø, æ 2l 2 ö, ÷, (d) a = l l and b min = çç, ÷, è L ø, Two stars are 10 light years away from the earth. They are, seen through a telescope of objective diameter 30 cm. The, wavelength of light is 600 nm. To see the stars just resolved, by the telescope, the minimum distance between them, should be (1 light year = 9.46 × 1015 m) of the order of :, [Online April 10, 2016], 8, 10, (a) 10 km (b) 10 km (c) 1011 km (d) 106 km, In Young's double slit experiment, the distance between, slits and the screen is 1.0 m and monochromatic light of, 600 nm is being used. A person standing near the slits is, looking at the fringe pattern. When the separation between, the slits is varied, the interference pattern disappears for a, particular distance d0 between the slits. If the angular, 1°, resolution of the eye is, , the value of d0 is close to :, 60, [Online April 9, 2016], (a) 1 mm, (b) 3mm (c) 2 mm, (d) 4mm, , (b) a =, , 67., , 68., , 69. Assuming human pupil to have a radius of 0.25 cm and a, comfortable viewing distance of 25 cm, the minimum, separation between two objects that human eye can resolve, at 500 nm wavelength is :, [2015], (a) 100 µm (b) 300 µm (c) 1 µm, (d) 30 µm, 70. Unpolarized light of intensity I0 is incident on surface of, a block of glass at Brewster’s angle. In that case, which, one of the following statements is true ?, [Online April 11, 2015], (a) reflected light is completely polarized with intensity, less than, , I0, 2, , (b) transmitted light is completely polarized with intensity, less than, , I0, 2, , (c) transmitted light is partially polarized with intensity, I0, 2, , (d) reflected light is partially polarized with intensity, I0, 2, , 71. Two beams, A and B, of plane polarized light with mutually, perpendicular planes of polarization are seen through a, polaroid. From the position when the beam A has maximum, intensity (and beam B has zero intensity), a rotation of, polaroid through 30° makes the two beams appear equally, bright. If the initial intensities of the two beams are IA and, IB respectively, then, , [2014], , 1, 3, (c) 1, (d), 3, 2, The diameter of the objective lens of microscope makes an, angle b at the focus of the microscope. Further, the medium, between the object and the lens is an oil of refractive index, n. Then the resolving power of the microscope, [Online April 19, 2014], (a) increases with decreasing value of n, (b) increases with decreasing value of b, (c) increases with increasing value of n sin 2b, , (a) 3, , 72., , IA, equals:, IB, , (b), , 1, n sin 2b, 73. A ray of light is incident from a denser to a rarer medium., The critical angle for total internal reflection is qiC, and Brewster’s angle of incidence is qiB, such that sinqiC/, sinqiB = h = 1.28. The relative refractive index of the two, media is:, [Online April 19, 2014], (a) 0.2, (b) 0.4, (c) 0.8, (d) 0.9, (d) increases with increasing value of
Page 431 :
P-422, , 74., , 75., , 76., , 77., , Physics, , In an experiment of single slit diffraction pattern, first, minimum for red light coincides with first maximum of some, other wavelength. If wavelength of red light is 6600 Å,, then wavelength of first maximum will be:, [Online April 12, 2014], (a) 3300Å (b) 4400Å (c) 5500Å (d) 6600Å, Abeam of unpolarised light of intensity I0 is passed, through a polaroidAand then through another polaroid B, which is oriented so that its principal plane makes an angle, of 45° relative to that of A. The intensity of the emergent, light is, [2013], (a) I0, (b) I0/2, (c) I0/4, (d) I0/8, This question has Statement-1 and Statements-2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., Statement-1 : Out of radio waves and microwaves, the, radio waves undergo more diffraction., Statement-2 : Radio waves have greater frequency, compared to microwaves., [Online April 25, 2013], (a) Statement-l is true, Statement-2 is true and, Statement-2 is the correct explanation of Statement-1, (b) Statement-1 is false , Statement-2 is true., (c) Statement-1 is true, Statement-2 is false., (d) Statement-1 is true, Statement-2 is true but Statement2 is not the correct explanation of Statement-1, A person lives in a high-rise building on the bank of a river, 50 m wide. Across the river is a well lit tower of height 40 m., When the person, who is at a height of 10 m, looks through, a polarizer at an appropriate angle at light of the tower, reflecting from the river surface, he notes that intensity of, light coming from distance X from his building is the least, and this corresponds to the light coming from light bulbs, at height 'Y' on the tower. The values of X and Y are, respectively close to (refractive index of water ;, , 4, ), 3, , [Online April 9, 2013], , 78. The first diffraction minimum due to the single slit, diffraction is seen at q = 30° for a light of wavelength 5000, Å falling perpendicularly on the slit. The width of the slit, is, [Online May 12, 2012], (a) 2.5 × 10– 5 cm, , (b) 1.25 × 10–5 cm, , (c) 10 × 10–5 cm, , (d) 5 × 10–5 cm, , 79. Two polaroids have their polarizing directions parallel so, that the intensity of a transmitted light is maximum. The, angle through which either polaroid must be turned if the, intensity is to drop by one-half is [Online May 7, 2012], (a) 135°, , 10 m, , 50 m, , (a) 25 m, 10 m, (c) 22 m, 13 m, , (b) 13 m, 27 m, (d) 17 m, 20 m, , (d) 180°, , Statement - 2: The light coming from the sky is polarized, due to scattering of sun light by particles in the atmosphere. The scattering is largest for blue light. [2011 RS], (a) Statement -1 is true, statement-2 is false., (b) Statement-1 is true, statement-2 is true, statement-2 is, the correct explanation of statement-1, (c) Statement-1 is true, statement-2 is true, statement-2 is, not the correct explanation of statement-1, (d) Statement-1 is false, statement-2 is true., 81. In an experiment, electrons are made to pass through a narrow, slit of width ‘d’ comparable to their de Broglie wavelength., They are detected on a screen at a distance ‘D’ from the slit, (see figure)., , d, , y=0, D, , Which of the following graphs can be expected to, represent the number of electrons ‘N’ detected as a, function of the detector position ‘y’(y = 0 corresponds to, the middle of the slit), [2008], y, , y, , (a) N, , d, , (b) N, , (c) N, , d, , y, , y, , Y, , X, , (c) 120°, , 80. Statement - 1: On viewing the clear blue portion of the, sky through a Calcite Crystal, the intensity of transmitted light varies as the crystal is rotated., , L, , 40 m, , (b) 90°, , d, , (d) N, , d
Page 432 :
P-423, , Wave Optics, , 82., , If I 0 is the intensity of the principal maximum in the single, slit diffraction pattern, then what will be its intensity when, the slit width is doubled?, [2005], (a) 4 I 0, , 83., , (b) 2 I 0, , (c), , I0, 2, , (d) I 0, , When an unpolarized light of intensity I 0 is incident on a, polarizing sheet, the intensity of the light which does not, get transmitted is, [2005], 1, 1, I0, I0, (b), (c) I 0, (d) zero, 2, 4, Two point white dots are 1 mm apart on a black paper., They are viewed by eye of pupil diameter 3 mm., Approximately, what is the maximum distance at which, these dots can be resolved by the eye?, , (a), 84., , [Take wavelength of light = 500 nm], [2005], (a) 1 m, (b) 5 m, (c) 3 m, (d) 6 m, 85. The angle of incidence at which reflected light is totally, polarized for reflection from air to glass (refractive index, n), is, [2004], (a) tan -1 (1/ n), , (b) sin -1 (1/ n), , (c) sin -1 (n), , (d) tan -1 (n), , 86. Wavelength of light used in an optical instrument are, l1 = 4000 Å and l 2 = 5000 Å , then ratio of their, , respective resolving powers (corresponding to, , l1 and, , l 2 ) is, , [2002], , (a) 16 : 25, , (b) 9 : 1, , (c) 4 : 5, , (d) 5 : 4
Page 433 :
P-424, , 1., , Physics, , (b) For (A), , 2., , xP, P, , (d) Initially, S2 L = 2 m, 2, , Q, , xQ, d, , 5, æ3ö, S1 L = 22 + ç ÷ = = 2.5 m, 2, è2ø, , A, , Path difference, Dx = S1 L - S2 L = 0.5 m =, , xP - xQ = (d + 2.5) - (d - 2.5) = 5 m, Phase difference Df due to path difference, 2p, 2p, p, ( Dx) =, (5) = ., l, 20, 2, At A, Q is ahead of P by path, as wave emitted by Q, reaches before wave emitted by P., , S2, , p p, - =0, 2 2, (due to P being ahead of Q by 90°), , When the listner move from L, first maxima will appear if, path difference is integral multiple of wavelength., For example, (n = 1 for first maxima), Dx = nl = 1l, , = I + I + 2 I I cos(0) = 4 I, , \ Dx = l = S1 L '- S 2 L, , For C,, , Þ1= d - 2 Þ d = 3 m, , Path difference, xQ - xP = 5 m, 2p, 2p, p, ( Dx ) =, (5) =, l, 20, 2, , Total phase difference at C =, , p p, + =p, 2 2, , 3., , (c) The distance traversed by light in a medium of, refractive index m in time t is given by, d = vt, ...(i), where v is velocity of light in the medium. The distance, traversed by light in a vacuum in this time,, D = ct = c ´, , I net = I1 + I 2 + 2 I1 I 2 cos(Df), , = I + I + 2 I I cos(p) = 0, , =d, , For B,, Path difference, xP - xQ = 0, p, 2, (due to P being ahead of Q by 90°), , Phase difference, Df =, , p, I B = I + I + 2 I I cos = 2 I, 2, Therefore intensities of radiation at A, B and C will be in, the ratio, , I A : I B : IC = 4 I : 2 I : 0 = 2 :1: 0., , L, , S1, , I A = I1 + I 2 + 2 I1 I 2 cos Df, , =, , 2 N/c, d, , \ Total phase difference at A, , Phase difference Df due to path difference, , L', , /c, 2N, , =, , l, 2, , d, v, , c, = md, v, , [from equation (i)], ...(ii), , c, (Q m = ), v, , This distance is the equvalent distance in vacuum and is, called optical path., Optical path for first ray which travels a path L1 through a, medium of refractive index n1 = n1 L1, Optical path for second ray which travels a path L2 through, a medium of refractive index n2 = n2 L2, Path difference = n1 L1 - n2 L2, Now, phase difference, =, , 2p, 2p, ´ path difference =, ´ (n1 L1 - n2 L2 ), l, l
Page 437 :
P-428, , 39., , Physics, , (c) We know that,, æ w1, ö, ç w + 1÷, è, ø, 2, , 43., , 2, , I max, =, 2, I min æ w, ö, 1 -1, ç w, ÷, è, ø, 2, , Imax and Imin are maximum and minium intensity, w1 and w2 are widths of two slits, , (a) For coherent sources, intensity at mid point, I1 µ (a + a)2, Þ I1 µ (2a)2, For incoherent sources, intensity of mid point is, I2 µ 2a2, I, 2, \ 1 =, I2 1, 44. (b) Let l be the wavelength of unknown light. Third bright, fringe of known light coincides with the 4th bright fringe of, the unknown light., , \, , I max, I min, , æ 1, ö, + 1÷÷, çç, 25, ø, =è, 2, æ 1, ö, 1, çç, ÷÷, è 25 ø, , æ w1, ö, 1, çè w = 25 given÷ø, 2, , On solving we get,, I max, I min, , 40., , 45., , 36, 9, = 25 =, =9:4, 16 4, 25, , (b) As sin q =, , nl, and sinq cannot be > 1, d, , nl, 1.8l, or n = 1.8, Hence maximum number of possible interference maximas,, 0, ±1 i.e. 3, \1=, , 41., , (c) Shift =, , b ( m - 1), l, , 42., , t1 -, , b ( m - 1), l, , Imax =, , (, , I1 + I 2, , ), , 2, , Maximum resultant intensity,, , t2, , (, , I¢ + I¢, I, 3, \, =, I max 4, =, , b ( m - 1), , ( t1 - t2 ), l, (a) Path difference at P Dx1 = 0, \ Phase difference at P will be, 2p, Df1 =, Dx1, l, 2p, ´0, =, l, = 0°, Resultant Intensity at P, I1 = I 0 + I 0 + 2 I 0 cos 0° = 4 I 0, Path difference at Q, l, Dx2 =, 4, \ Phase difference at Q, 2p l æ p ö, × =ç ÷, D=, l 4 è2ø, Resultant intensity at Q., p, I 2 = I 0 + I 0 + 2 I0 cos = 2 I 0, 2, I1 4 I 0 2, Thus,, =, =, I 2 2I0 1, =, , 4lD, 3l1D, =, d, d, 3(590) D 4lD, \, =, d, d, 3, Þ l = ´ 590 = 442.5 nm, 4, (a) For path difference of l, the phase difference is 2p, l, For path difference of , the phase difference is, 6, 2p ´ l / 6 p, =, l, 3, Resultant intensity, p, I = I1 + I 2 + 2 I1 I 2 cos, 3, \ I = I1 + I 2 + I1 I 2, For two identical source, I1= I2 = I¢ (say), then I = 3I¢, , \, , 2, , ) = (2 I ¢ ), 2, , 2, , = 4I ¢, , The intensity of light at any point of the screen where the, phase difference due to light coming from the two slits is f is, given by, æ fö, I = I o cos 2 ç ÷, è 2ø, Where I0 is the maximum intensity., NOTE This formula is applicable when I1= I2., , Phase difference f =, , 2p l, ´ = p/3, l 6, 2, , p æ 3ö, I, 3, = cos2 = ç, ÷ =, I0, 6 è 2 ø, 4, (d) The light passing through the slits interfere and, produce dark and bright band one screen. The shape of, interference fringes formed on a screen in case of a, monochromatic source is a straight line., (b) For constructive interference path difference (As sin, q £ 1), d sin q = nl, , \, , 46., , 47.
Page 438 :
P-429, , Wave Optics, , Given d = 2l, , 48., , n, \ 2l sin q = nl Þ sin q =, 2, n = 0, 1, – 1, 2, – 2 hence five maxima are possible., (d) Given :, Intensity, I0 = 3.3 Wm–2, Area,, A = 3 × 10–4 m2, Angular speed, w = 31.4 rad/s, , Average energy = I0 A < cos2 q >, Q < cos2 q > =, , 1, per revolution, 2, , \ Average energy =, 49., , (3.3)(3 ´ 10-4 ), ; 5 ´ 10- 4 J, 2, , (198), For obtaining secondary minima at a point path difference, should be integral multiple of wavelength, , =, , = 5368 × 10–2 m = 53.68 m » 60 m, 51. (b) According to question, the intensity of light coming, out of the analyser is just 10% of the original intensity (I0), Using, I = I0cos2q, Þ, , I0, 1, = I 0 cos2 q Þ, = cos 2 q, 10, 10, , Þ cos q =, , For n to be maximum sin q = 1, , 1, 10, , = 0.316 Þ q » 71.6°, , Therefore, the angle by which the analyser need to be, rotated further to reduced the output intensity to be zero, f = 90° – q = 90° – 71.6° = 18.4°, 52. (a) x =, , \ d sin q = nl, nl, \ sin q =, d, , (1.22)(5500 ´ 10 -10 ) ´ 4 ´ 105 ´ 103, 5, , =, , 1.22l, 2µsinq, , 1.22 ´ 5000, = 0.24 µm, 2 ´ 1.25, , æ I0 ö, 53. (c) I = çè 2 ÷ø cos2 30° cos2 60°, , d 6 ´ 10-5, =, = 100, l 6 ´ 10-7, Total number of minima on one side = 99, Total number of minima = 198., n=, , =, 50. (a), \, , I0 3 1, ´ ´, 2 4 4, I 0 32, =, = 10.67, I, 3, , 1.22l 1.22 ´ 600 ´ 10-9, =, d, 250 ´ 10-2, = 3.0 × 10–7 rad, , 54. (c) q =, , 55. (a) q =, Smallest angular separation between two distant objects, here moon and earth,, l, q = 1.22, a, a = aperture diameter of telescope, , Distance O1O2 = (q)d, Minimum separation between objects on the surface of moon,, , lö, æ, = ç 1.22 ÷ d, è, aø, , 1.22l 1.22 ´ 500 ´ 10 -9, =, = 305 × 10–9 rad., d, 2, , 56. (b), 57. (b) According to Brewster’s law, refractive index of material (m) is equal to tangent of polarising angle, Qtan i b = μ =, , 1.5, μ, , 1, 1.5, <, (Qsin ic < sin ib ), 2, 2, μ, μ + (1.5), \sin i b =, , 1.5, μ + (1.5), 2, , 2, , or, μ 2 + (1.5 ) <1.5´ μ, 2
Page 439 :
P-430, , Physics, , Þ μ 2 + (1.5) < ( μ×1.5), 2, , 2, , 3, Þμ <, i.e. minimum value of m should, 5, 3, 5, , be, 58., , (a) Angular width of central maxima =, , c, 2 = 10 3 = 17.3 GHz, f = 10, c, c2, , 2l, d, , d 50 ´ 10-2 10 -6 ´ 50 ´10 -2, 10-2 = ´, =, 2, d', 2´d', Therefore, slit separation distance, d’ = 25mm, (c) Axis of transmission of A & B are parallel., Polariser A Polariser B, , I, , I/2, , I/2, , After introducing polariser C between A and B,, Polariser A, , I, , 60., , Polariser C, , I/2, , Polariser B, , I/2Cos q, 2, , I/2Cos4 q, , I, I, 1, cos4 q =, Þ cos4 q =, 2, 8, 4, 1, or, q = 45°, Þ cos q =, 2, (a) If angular position of 2nd maxima from central maxima, is q then, , (2n - 1) l 3l 3 ´ 550 ´ 10 -9, =, =, 2a, 20 2 ´ 22 ´ 10 -7, \ q ; p rad, 8, (a) Polariser A and B have same alignment of transmission, axis., Lets assume polariser c is introduced at q angle, sin q =, , 61., , 1, 1, cos 2 q ´ cos 2 q =, 2, 3, , or,, 62., 63., , cos 4 q =, , V = relative speed of approach, , c+, , l´D, d, or, l = ; Fringe width, b =, d', 2, , 59., , c+v, ;, c+v, f0 = 10 GHz, f = f0, , 2, æ 2ö, Þ cos q = ç ÷, è 3ø, 3, , 14, , (a), (a) Use relativistic doppler's effect as velocity of, observer is not small as compared to light, , 64. (b) a = 0.1 mm = 10–4 cm,, l = 6000 × 10–10 cm = 6 × 10–7 cm, D = 0.5 m, for 3rd dark band, a sin q = 3 l, 3l x, =, or sin q =, a, D, The distance of the third dark band from the central bright, band, x=, , -7, 3 lD 3 ´ 6 ´ 10 ´ 0.5, =, = 9 mm, a, 10-4, , 65. (b) For secondary minima,, nl, b sin q = nl Þ sin q =, b, Distance of n th secondary minima x = D sin q, x, or sin q1 = 1, D, 2l, sin q1 =, b, n=4, 4l x 2, =, sin q2 =, b, D, 4l 2 l 2 l, x 2 - x1 =, =, b, b, b, 2l, 2l, 3=, Þb=, ..... (i), b, 3, 2l, Width of central maxima =, b, 2l, =, = 3 cm., 2l, 3, .. from eq. (i), 66. (a) Given geometrical spread = a, l, lL, Diffraction spread = ´ L =, a, a, lL, The sum b = a +, a, For b to be minimum, lL ö, db, d æ, =0, ça +, ÷ =0, da, da è, a ø, a = lL, b min = lL + lL = 2 lL = 4lL
Page 440 :
P-431, , Wave Optics, , 67., , where, qiB is Brewster’s angle of incidence,, , 0.61l l, =, 4, R, The minimum distance between them, , (a) We know that Dq =, , And, m =, , R, 9.46´1015 ´10´ 0.61´ 600´10,9, 0.61´l <, 9, 0.3, 11, = 1.15 × 10 m, Þ 1.115 × 108 km., (c) Given D = 1.0m, wavelength of monochromatic light, l = 600 nm., l<, , 68., , d : Dθ < 1´, , On solving we get, relative refractive index of the two media., 74. (b) In a single slit experiment,, For diffraction maxima,, l, 2, and for diffraction minima,, a sin q = ( 2n + 1), , p, 1, ´, 180 60, , a sin q = nl, According to question,, , d0 < 2´10,3 < 2 mm, , 69., , ( 2 ´1 + 1), , 0.25, 1, =, (d) sin q =, 25 100, , 25cm, , 70., , 71., , 1.22l, Resolving power =, = 30 mm., 2m sin q, (a) When unpolarised light is incident at Brewster’s, angle then reflected light is completely polarized and the, intensity of the reflected light is less than half of the, incident light., (d) According to malus law, intensity of emerging beam, is given by,, , I = I0cos2q, Now, IA¢ = IA cos230º, IB¢ = IB cos260º, As IA¢ = IB¢, Þ, 72. (c), , 3, 1 IA 1, =, = IB ´ ;, 4, 4 IB 3, Resolving power of microscope,, IA ´, , 2n sin q, l, l = Wavelength of light used to illuminate the object, n = Refractive index of the medium between object and, objective, q = Angle, R.P. =, , 73. (c) Here, sin qic / sin qiB = 1.28, As we know,, sin qiB, m=, æp, ö, sin ç - qiB ÷, è2, ø, , l, = 1´ 6600, 2, , (Q l R = 6600Å ), l=, , q, , 0.25cm, , 1, sin qic, , 75., , 6600 ´ 2, 3, , l = 4400Å, (c) Relation between intensities, I0, , (I0/2), , (unpolarised), , 45° B, IR, , A, I, 1 I, æI ö, I r = ç 0 ÷ cos2 (45°) = 0 ´ = 0, è 2ø, 2 2 4, , 76. (c) Wavelength of radio waves is greater than, microwaves hence frequency of radio waves is less than, microwaves., The degree of diffraction is greater whose wavelength is, greater., 77. (b), 78. (c) For first minimum,, dsinq = l, l, 5000 ´ 10 -8 cm, Þ d=, =, sin q, sin 30°, 5000 ´10 -8 cm, = 10 ´ 10 -5 cm, =, 1, 2, I0, I0, 2, 79. (a) For I =, and I = I0 cos q =, 2, 2, \ q = 45°, Therefore the angle through which either polaroids turned, is 135° (= 180° – 45°), 80. (b) When viewed through a polaroid which is rotated, then the light from a clear blue portion of the sky shows a, rise and fall of intensity. The light coming from the sky is, polarised due to scattering of sunlight by particles in the, atmosphere.
Page 441 :
P-432, , Physics, Incident sunlight, (unpolarised), , Sun, Scattered light, (polarised), , 81., , to observer, , (d) The electron beam will be diffracted and the maxima, is obtained at y = 0., Also, the diffraction pattern, should be wider than the slit, width., 2, , 82., , 83., , p, æ sin f ö, ÷÷ and f = ( b sin q ), I = I 0 çç, l, f, ø, è, When the slit width is doubled, the amplitude of the wave, at the centre of the screen is doubled, so the intensity at, the centre is increased by a factor 4., (b) From the law of Malus, I = I0cos2 q, When an unpolarised light is converted into plane, polarised light by passing through polariod, its intensity, become hafl., , (a), , I0, 2, Þ Intensity of untransmitted light, I, I, = I0 - 0 = 0, 2, 2, 84. (b) y ³ 1.22 l, D, d, , \ Intensity of polarized light =, , Þ D£, , yd, 10 - 3 ´ 3 ´ 10 - 3 = 30, » 5m, =, 6.1, (1.22) l (1.22) ´ 5 ´ 10 - 7, , \ Dmax = 5m, , 85. (d) From the Brewster’s law, angle of incidence for total, polarization is given by tan q = n, Þ q = tan–1 n, Where n is the refractive index of the glass., 86. (d) The resolving power of an optical instrument is, inversely proportional to the wavelength of light used., ( R.P )1 l 2 5, =, =, ( R.P )2 l1 4
Page 442 :
25, , P-433, , Dual Nature of Radiation and Matter, , Dual Nature of, Radiation and Matter, 5., , Matter Waves, Cathode, TOPIC 1, and Positive Rays, 1., , 2., , 3., , (a) le > lHe+ + > lp, , (b) le < lHe+ + = lp, , (c) le > lp > lHe+ +, , (d) le < lp < lHe+ +, , l, . Value of DE, is: [9 Jan. 2020 I], 2, (a) E, (b) 4E, (c) 3E, (d) 2E, An electron of mass m and magnitude of charge |e| initially, at rest gets accelerated by a constant electric field E. The, rate of change of de-Broglie wavelength of this electron at, time t ignoring relativistic effects is:, [9 Jan. 2020 II], , wavelength become, , (He+ +), , An electron, a doubly ionized helium ion, and a, proton are having the same kinetic energy. The relation, between their respective de-Broglie wavelengths le, lHe++, and lp is :, [Sep. 06, 2020 (I)], , A particle moving with kinetic energy E has de Broglie, wavelength l. If energy DE is added to its energy, the, , 6., , Assuming the nitrogen molecule is moving with, r.m.s.velocity at 400 K, the de-Broglie wavelength of, nitrongen molecule is close to :, , (a) -, , (Given : nitrogen molecule weight : 4.64 × 10–26 kg, Boltzman, constant : 1.38 × 10–23 J/K, Planck constant : 6.63 × 10–34 J.s), [Sep. 06, 2020 (II)], , (c) -, , (a) 0.24 Å, , (b) 0.20 Å, , (c) 0.34 Å, , (d) 0.44 Å, , 7., , m, moving along the x-axis with, 2, velocity v0 collides elastically with another particle B at, , Particle A of mass mA =, , 3, l0, 2, , (c) Dl = 2l 0, 4., , (b) Dl =, , 5, l0, 2, , (d) Dl = 4l 0, , A particle is moving 5 times as fast as an electron. The, ratio of the de-Broglie wavelength of the particle to that of, the electron is 1.878 × 10–4. The mass of the particle is, close to :, [Sep. 02, 2020 (II)], (a) 4.8 × 10–27 kg, (b) 9.1 × 10–31 kg, (c) 1.2 × 10–28 kg, (d) 9.7 × 10–28 kg, , h, | e | Et, , 2, , 2 2, , e E t, m2 v02, l0, , (c), 1+, , 8., , (b), , (d), , l0 2, 1+, , rest having mass mB =, , (a) Dl =, , |e|E t, , | e | Et, h, , -h, | e | Et 2, , r, An electron (mass m) with initial velocity v = v0iˆ + v0 ˆj is, r, in an electric field E = - E0 kˆ. If l 0 is initial de-Broglie, wavelength of electron, its de-Broglie wave length at time, t is given by:, [8 Jan. 2020 II], (a), , m, . If both particles move along, 3, the x-axis after the collision, the change Dl in de-Broglie, wavelength of particle A, in terms of its de-Broglie, wavelength (l0) before collision is : [Sep. 04, 2020 (I)], , h, , e2 E 2 t 2, 2 m2 v02, , l0, , (b), 1+, , e2 E02 t 2, m2 v02, l0, , (d), 2+, , e2 E 2t 2, m 2 v02, , A particle ‘P’ is formed due to a completely inelastic collision, of particles ‘x’ and ‘y’ having de-Broglie wavelengths ‘gx’, and ‘gy’ respectively. If x and y were moving in opposite, directions, then the de-Broglie wavelength of ‘P’ is:, [9 Apr. 2019 II], (a), , g xg y, gx +g y, , (c) g x - g y, , g xg y, (b) | g - g |, x, y, (d) g x + g y
Page 443 :
P-434, , 9., , Physics, , Two particles move at right angle to each other. Their de, Broglie wavelengths are l1 and l2 respectively. The particles, suffer perfectly inelastic collision. The de Broglie, wavelength l, of the final particle, is given by :, [8 April 2019 I], (a), , 1, l, , 2, , =, , 1, l12, , +, , 1, l 22, , (b) l =, , l1l2, , 2 1, 1, l2 + l2, (d) l = l + l, 2, 1, 2, A particle A of mass ‘m’ and charge ‘q’ is accelerated by a, potential difference of 50v Another particle B of mass ‘4m’, and charge‘q’ is accelerated by a potential differnce of, , (c) l =, 10., , 2500V. The ratio of de-Broglie wavelength, , (a), , lA 2, =, lB 3, , (b), , lA 1, =, lB 2, , lA, lA 1, =, (d) l = 2, lB 3, B, 16. A parallel beam of electrons travelling in, x-direction falls on a slit of width d (see figure). If after, passing the slit, an electron acquires momentum py in the, y-direction then for a majority of electrons passing through, the slit (h is Planck’s constant) : [Online April 10, 2015], , (c), , Y, , lA, is, lB, , X, , [12 Jan. 2019 I], (a) 10.00, (b) 0.07, (c) 14.14, (d) 4.47, 11. If the deBroglie wavelength of an electron is equal to 10–, 3 times the wavelength of a photon of frequency 6 × 1014, Hz, then the speed of electron is equal to :, (Speed of light = 3 × 108 m/s), Planck’s constant = 6.63 × 10–34J.s, Mass of electron = 9.1 × 10–31 kg), [11 Jan. 2019 I], 6, (a) 1.1 ×10 m/s, (b) 1.7 ×106 m/s, 6, (c) 1.8 ×10 m/s, (d) 1.45 ×106 m/s, 12. In an electron microscope, the resolution that can be, achieved is of the order of the wavelength of electrons, used. To resolve a width of 7.5 × 10–12 m, the minimum, electron energy required is close to:[10 Jan. 2019 I], (a) 500 keV (b) 100 keV (c) 1 keV (d) 25 keV, 13. Two electrons are moving with non-relativistic speeds, perpendicular to each other. If corresponding de Broglie, wavelengths are l1 and l2, their de Broglie wavelength in, the frame of reference attached to their centre of mass is:, [Online April 15, 2018], (a) lCM = l1 = l2, (c), 14., , 15., , lCM =, , 2l1l 2, l12 + l 22, , (b), , (d), , 1, 1, 1, = +, l1 l1 l 2, æ l + l2 ö, lCM = ç 1, ÷, è 2 ø, , If the de Broglie wavelengths associated with a proton and, an a-particle are equal, then the ratio of velocities of the, proton and the a-particle will be:[Online April 15, 2018], (a) 1 : 4, (b) 1 : 2, (c) 4 : 1, (d) 2 : 1, A particle A of mass m and initial velocity v collides with, m, which is at rest. The collision is, 2, head on, and elastic. The ratio of the de-Broglie, , a particle B of mass, , wavelengths lA to lB after the collision is, , [2017], , (a) |Py| d > h, (b) |Py| d < h, (c) |Py| d ; h, (d) |Py| d > >h, 17. de-Broglie wavelength of an electron accelerated by a voltage, of 50 V is close to (|e| = 1.6 × 10–19 C, me = 9.1 × 10–31 kg,, h = 6.6 × 10–34 Js) :, [Online April 10, 2015], (a) 2.4 Å (b) 0.5 Å, (c) 1.7 Å, (d) 1.2 Å, 18. For which of the following particles will it be most difficult, to experimentally verify the de-Broglie relationship?, [Online April 9, 2014], (a) an electron, (b) a proton, (c) an a-particle, (d) a dust particle, 19. Electrons are accelerated through a potential difference V, and protons are accelerated through a potential difference, 4 V. The de-Broglie wavelengths are le and lp for electrons, le, is given by :, lp, (given me is mass of electron and mp is mass of proton)., , and protons respectively. The ratio of, , [Online April 23, 2013], mp, , (a), , le, =, lp, , (c), , le 1 me, =, l p 2 mp, , me, , (b), , le, =, lp, , (d), , mp, le, =2, lp, me, , me, mp, , 20. If the kinetic energy of a free electron doubles, it’s deBroglie wavelength changes by the factor, [2005], 1, 1, (a) 2, (b), (c), (d), 2, 2, 2, 21. Formation of covalent bonds in compounds exhibits [2002], (a) wave nature of electron, (b) particle nature of electron, (c) both wave and particle nature of electron, (d) none of these
Page 444 :
P-435, , Dual Nature of Radiation and Matter, , Photon, Photoelectric Effect, TOPIC 2 X-rays and Davisson-Germer, Experiment, 22., , A beam of electrons of energy E scatters from a target, having atomic spacing of 1Å. The first maximum intensity, occurs at q = 60º. Then E (in eV) is ______., , (Plank constant h = 6.64 × 10–34 Js, 1 eV = 1.6 × 10–19 J,, electron mass m = 9.1 × 10–31 kg) [NA Sep. 05, 2020 (I)], 23. The surface of a metal is illuminated alternately with, photons of energies E1 = 4 eV and E2 = 2.5 eV respectively., The ratio of maximum speeds of the photoelectrons emitted, in the two cases is 2. The work function of the metal in (eV), is _______________., [NA Sep. 05, 2020 (II)], 24. Given figure shows few data points in a photo electric, effect experiment for a certain metal. The minimum energy, for ejection of electron from its surface is : (Plancks constant, h = 6.62 × 10–34 J.s), [Sep. 04, 2020 (I)], , Vstop (V), , Y, , C, , (6, V), , B (5.5, 0), A, 5, , 14, , X, , f(10 Hz), , (a) 2.27 eV, (b) 2.59 eV, (c) 1.93 eV, (d) 2.10 eV, 25. In a photoelectric effect experiment, the graph of stopping, potential V versus reciprocal of wavelength obtained is, shown in the figure. As the intensity of incident radiation, is increased :, [Sep. 04, 2020 (II)], , V, , q, , 26., , 1/l, , (a) Straight line shifts to right, (b) Slope of the straight line get more steep, (c) Straight line shifts to left, (d) Graph does not change, When the wavelength of radiation falling on a metal is, changed from 500 nm to 200 nm, the maximum kinetic energy, of the photoelectrons becomes three times larger. The work, function of the metal is close to :, [Sep. 03, 2020 (I)], (a) 0.81 eV, (b) 1.02 eV, (c) 0.52 eV, (d) 0.61 eV, , 27. Two sources of light emit X-rays of wavelength 1 nm and, visible light of wavelength 500 nm, respectively. Both the, sources emit light of the same power 200 W. The ratio of, the number density of photons of X-rays to the number, density of photons of the visible light of the given, wavelengths is :, [Sep. 03, 2020 (II)], 1, 1, (b) 250, (c), (d) 500, 500, 250, When radiation of wavelength l is used to illuminate a, metallic surface, the stopping potential is V. When the, same surface is illuminated with radiation of wavelength, , (a), 28., , V, . If the theshold, 4, wavelength for the metallic surface is nl then value of n, will be _________., [NA Sep. 02, 2020 (I)], , 3l, the stopping potential is, , 29. Radiation, with wavelength 6561 Å falls on a metal surface, to produce photoelectrons. The electrons are made to enter, a uniform magnetic field of 3 × 10–4 T. If the radius of the, largest circular path followed by the electrons is 10 mm,, the work function of the metal is close to: [9 Jan. 2020 I], (a) 1.1 ev, (b) 0.8 ev, (c) 1.6 ev, (d) 1.8 ev, 30. When photon of energy 4.0 eV strikes the surface of a, metal A, the ejected photoelectrons have maximum kinetic, energy TA eV and de-Broglie wavelength lA. The maximum, kinetic energy of photoelectrons liberated from another, metal B by photon of energy 4.50 eV is TB=(TA–1.5)eV. If, the de-Broglie wavelength of these photoelectrons lB =, 2lA, then the work function of metal B is:[8 Jan. 2020 I], (a) 4 eV, (b) 2 eV, (c) 1.5 eV, (d) 3 eV, 31. A beam of electromagnetic radiation of intensity, 6.4 × 10–5 W/cm 2 is comprised of wavelength, l = 310, nm. It falls normally on a metal (work function j = 2eV), of surface area of 1 cm 2. If one in 10 3 photons ejects an, electron, total number of electrons ejected in 1 s is 10x ., (hc = 1240 eVnm, l eV = 1.6 × 10–19 J), then x is _______., [NA 7 Jan. 2020 I], 32. The stopping potential V0 (in volt) as a function of frequency, (v) for a sodium emitter, is shown in the figure. The work, function of sodium, from the data plotted in the figure, will be :, (Given : Planck’s constant (h) = 6.63 × 10–34 Js, electron, charge e = 1.6 × 10–19 C), [12 Apr. 2019 I], , (a) 1.82 eV, , (b) 1.66 eV (c) 1.95 eV, , (d) 2.12 eV
Page 445 :
P-436, , Physics, , 33. In a photoelectric effect experiment the threshold, wavelength of light is 380 nm. If the wavelength of incident, light is 260 nm, the maximum kinetic energy of emitted, electrons will be:, Given E (in eV) =, [10 Apr. 2019 I], (a) 1.5 eV, (b) 3.0 eV (c) 4.5 eV, (d) 15.1 eV, 34. A 2 mW laser operates at a wavelength of 500 nm. The, number of photons that will be emitted per second is :, [Given Planck’s constant h= 6.6×10–34 Js, speed of light, c = 3.0×108 m/s], [10 Apr. 2019 II], (a) 5×1015, (b) 1.5×1016, (c) 2×1016, (d) 1×1016, 35. The electric field of light wave is given as, , r, æ 2p x, ö N, E = 103 cos ç, - 2p ´ 6 ´ 1014 t ÷ x$, -7, è 5 ´10, ø C, This light falls on a metal plate of work function 2eV. The, stopping potential of the photo-electrons is:, 12375, [9 April 2019 I], l ( in Å ), (a) 2.0 V, (b) 0.72 V (c) 0.48 V, (d) 2.48 V, 36. When a certain photosensistive surface is illuminated with, monochromatic light of frequency v, the stopping potential, for the photo current is –V 0/2. When the surface is, illuminated by monochromatic light of frequency v/2, the, stoppoing potential is –V0. The threshold frequency for, photoelectric emission is :, [12 Jan. 2019 II], , Given, E (in eV) =, , 37., , 38., , 39., , 5v, 3, , 4, v, 3, , 3v, 2, In a Frank-Hertz experiment, an electron of energy 5.6 eV, passes through mercury vapour and emerges with an, energy 0.7 eV. The minimum wavelength of photons emitted, by mercury atoms is close to :, [12 Jan. 2019 II], (a) 1700 nm, (b) 2020 nm, (c) 220 nm, (d) 250 nm, In a photoelectric experiment, the wavelength of the light, incident on a metal is changed from 300 nm to 400 nm. The, decrease in the stopping potential is close to :, [11 Jan. 2019 II], hc, æ, ö, çè = 1240 nm-V÷ø, e, , (a), , (b), , (c) 2 v, , (d), , (a) 0.5 V, (b) 1.5 V, (c) 1.0 V, (d) 2.0 V, A metal plate of area 1 × 10–4 m2 is illuminated by a, radiation of intensity 16 mW/m2. The work function of the, metal is 5 eV. The energy of the incident photons is 10 eV, and only 10% of it produces photo electrons. The number, of emitted photo electrons per second and their maximum, energy, respectively, will be:, [1 eV = 1.6 × 10–19 J], [10 Jan. 2019 II], (a) 1014 and 10 eV, (b) 1012 and 5 eV, (c) 1011 and 5 eV, (d) 1010 and 5 eV, , 40. Surface of certain metal is first illuminated with light of, wavelength l1 = 350 nm and then, by light of wavelength, l2 = 540 nm. It is found that the maximum speed of the, photo electrons in the two cases differ by a factor of (2), The work function of the metal (in eV) is close to:, , 1240, (Energy of photon = l in nm eV ), (, ), , [9 Jan. 2019 I], , (a) 1.8, (b) 2.5, (c) 5.6, (d), 1.4, 41. The magnetic field associated with a light wave is given, at the origin by, B = B0 [sin(3.14 × 107)ct + sin(6.28 × 107)ct]., If this light falls on a silver plate having a work function, of 4.7 eV, what will be the maximum kinetic energy of the, photoelectrons?, [9 Jan. 2019 II], (c = 3 × 108 ms–1, h = 6.6 × 10–34J-s), (a) 6.82 eV, (b) 12.5 eV, (c) 8.52 eV, (d) 7.72 eV, 42. An electron beam is accelerated by a potential difference V to, hit a metallic target to produce X-rays. It produces continuous, as well as characteristic X-rays.If lmin is the smallest possible, wavelength of X-ray in the spectrum, the variation of log lmin, with log V is correctly represented in :, [2017], (a), , (b), , (c), , (d), , 43. A Laser light of wavelength 660 nm is used to weld Retina, detachment. If a Laser pulse of width 60 ms and power 0.5, kW is used the approximate number of photons in the, pulse are :, [Take Planck’s constant h = 6.62 × 10–34 Js], [Online April 9, 2017], 20, 18, (a) 10, (b) 10, (c) 1022, (d) 1019, 44. The maximum velocity of the photoelectrons emitted from, the surface is v when light of frequency n falls on a metal, surface. If the incident frequency is increased to 3n, the, maximum velocity of the ejected photoelectrons will be :, [Online April 8, 2017], (a) less than 3v, (b) v, (c) more than 3v, (d) equal to 3v, 45. Radiation of wavelength l, is incident on a photocell. The, fastest emitted electron has speed v. If the wavelength is, changed to, will be:, , 3l, , the speed of the fastest emitted electron, 4, [2016], 1, , (a), , æ 4 ö2, = vç ÷, è3ø, , (c), , æ 4 ö2, > vç ÷, è3ø, , 1, , (b), , æ 3 ö2, = vç ÷, è4ø, , (d), , æ 4 ö2, < vç ÷, è3ø, , 1, , 1
Page 446 :
P-437, , Dual Nature of Radiation and Matter, , 46., , A photoelectric surface is illuminated successively by, l, . If the, 2, maximum kinetic energy of the emitted photoelectrons in, the second case is 3 times that in the first case, the work, function of the surface is :, [Online April 10, 2016], , monochromatic light of wavelengths l and, , hc, hc, (b), 2l, l, hc, 3hc, (c), (d), 3l, l, When photons of wavelength l1 are incident on an isolated, sphere, the corresponding stopping potential is found to, be V. When photons of wavelength l2 are used, the, corresponding stopping potential was thrice that of the, above value. If light of wavelength l3 is used then find the, stopping potential for this case : [Online April 9, 2016], , (a), , 47., , (a), , hc é 1, 1, 1ù, - ú, ê +, e ë l 3 l 2 l1 û, , (b), , hc é 1, 1, 1ù, - ú, ê +, e ë l3 2l 2 l1 û, , (c), , hc é 1, 1, 1ù, - ú, ê e ë l 3 l 2 l1 û, , I, , (a), , (b), , O, , l, , O, , Lis t-II, (i) Particle n atu re of, ligh t, (ii) Dis crete en erg y, levels of atom, (iii) W av e natu re o f, electro n, Structu re o f ato m, , (a) (A)-(ii); (B)-(i); (C)-(iii), (b) (A)-(iv); (B)-(iii); (C)-(ii), (c) (A)-(i); (B)-(iv); (C)-(iii), (d) (A)-(ii); (B)-(iv); (C)-(iii), 49. A beam of light has two wavelengths of 4972Å and 6216Å, with a total intensity of 3.6 × 10 –3 Wm –2 equally, distributed among the two wavelengths. The beam falls, normally on an area of 1 cm2 of a clean metallic surface, of work function 2.3 eV. Assume that there is no loss of, light by reflection and that each capable photon ejects, one electron. The number of photoelectrons liberated in, 2s is approximately:, [Online April 12, 2014], (a) 6 × 1011, (b) 9 × 1011, (c) 11 × 1011, (d) 15 × 1011, , l, , I, , (c), , hc é 1, 1, 3 ù, ê +, ú, e ë l3 2l 2 2l1 û, 48. Match List - I (Fundamental Experiment) with List - II (its, conclusion) and select the correct option from the choices, given below the list:, [2015], , (iv ), , I, , I, , (d), , Lis t-I, A . Fran ck-Hertz, Exp eriment, B. Pho to -electric, experimen t, C. Davis on -Germer, experimen t, , 50. A photon of wavelength l is scattered from an electron,, which was at rest. The wavelength shift Dl is three times, of l and the angle of scattering q is 60°. The angle at which, the electron recoiled is f. The value of tan f is : (electron, speed is much smaller than the speed of light), [Online April 11, 2014], (a) 0.16, (b) 0.22, (c) 0.25, (d) 0.28, 51. The anode voltage of a photocell is kept fixed. The, wavelength l of the light falling on the cathode is gradually, changed. The plate current I of the photocell varies as, follows :, [2013], , (d), , O, O, l, l, 52. In an experiment on photoelectric effect, a student plots, stopping potential V0 against reciprocal of the wavelength, l of the incident light for two different metals A and B., These are shown in the figure. [Online April 25, 2013], V0, , Metal A, , Metal B, , 1/l, , Looking at the graphs, you can most appropriately say, that:, (a) Work function of metal B is greater than that of metal A, (b) For light of certain wavelength falling on both metal,, maximum kinetic energy of electrons emitted from A, will be greater than those emitted from B., (c) Work function of metal A is greater than that of metal B, (d) Students data is not correct, 53. A copper ball of radius 1 cm and work function 4.47eV is, irradiated with ultraviolet radiation of wavelength 2500 Å., The effect of irradiation results in the emission of electrons, from the ball. Further the ball will acquire charge and due, to this there will be a finite value of the potential on the ball., The charge acquired by the ball is :, [Online April 25, 2013], –13, (a) 5.5×10 C, (b) 7.5 × 10–13C, –12, (c) 4.5 × 10 C, (d) 2.5 × 10–11C
Page 447 :
P-438, , 54., , 55., , 56., , 57., , This equation has statement 1 and statement 2. Of the four, choices given after the statements, choose the one that, describes the two statements., Statement 1: Davisson-Germer experiment established the, wave nature of electrons., Statement 2 : If electrons have wave nature, they can, interfere and show diffraction., [2012], (a) Statement 1 is false, Statement 2 is true., (b) Statement 1 is true, Statement 2 is false, (c) Statement 1 is true, Statement 2 is true, Statement, 2 is the correct explanation of statement 1, (d) Statement 1 is true, Statement 2 is true, Statement 2 is, not the correct explanation of Statement 1, Photoelectrons are ejected from a metal when light of, frequency u falls on it. Pick out the wrong statement from, the following., [Online May 26, 2012], (a) No electrons are emitted if u is less than, W/h, where W is the work function of the metal, (b) The ejection of the photoelectrons is instantaneous., (c) The maximum energy of the photoelectrons is hu., (d) The maximum energy of the photoelectrons is, independent of the intensity of the light., This question has Statement 1 and Statement 2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., Statement 1: A metallic surface is irradiated by a, monochromatic light of frequency u > u0 (the threshold, frequency). If the incident frequency is now doubled, the, photocurrent and the maximum kinetic energy are also, doubled., Statement 2: The maximum kinetic energy of, photoelectrons emitted from a surface is linearly, dependent on the frequency of the incident light. The, photocurrent depends only on the intensity of the incident, light., [Online May 19, 2012], (a) Statement 1 is true, Statement 2 is true, Statement, 2 is the correct explanation of Statement 1., (b) Statement 1 is false, Statement 2 is true., (c) Statement 1 is true, Statement 2 is false., (d) Statement 1 is true, Statement 2 is true, Statement, 2 is not the correct explanation of Statement 1., This question has Statement – 1 and Statement – 2. Of the, four choices given after the statements, choose the one, that best describes the two statements., [2011], Statement – 1: A metallic surface is irradiated by a, monochromatic light of frequency v > v0 (the threshold, frequency). The maximum kinetic energy and the stopping, potential are Kmax and V0 respectively. If the frequency, incident on the surface is doubled, both the Kmax and V0, are also doubled., Statement – 2 : The maximum kinetic energy and the, stopping potential of photoelectrons emitted from a, surface are linearly dependent on the frequency of incident, light., , Physics, , (a) Statement–1 is true, Statement–2 is true, Statement –, 2 is the correct explanation of Statement – 1., (b) Statement–1 is true, Statement–2 is true, Statement, – 2 is not the correct explanation of Statement – 1., (c) Statement – 1 is false, Statement – 2 is true., (d) Statement – 1 is true, Statement – 2 is false., 58. Statement -1 : When ultraviolet light is incident on a, photocell, its stopping potential is V0 and the maximum, kinetic energy of the photoelectrons is Kmax .When the, ultraviolet light is replaced by X-rays, both V0 and Kmax, increase., Statement -2 : Photoelectrons are emitted with speeds, ranging from zero to a maximum value because of the, range of frequencies present in the incident light. [2010], (a) Statement -1 is true, Statement -2 is true ; Statement, -2 is the correct explanation of Statement -1., (b) Statement -1 is true, Statement -2 is true; Statement, -2 is not the correct explanation of Statement -1, (c) Statement -1 is false, Statement -2 is true., (d) Statement -1 is true, Statement -2 is false., 59. The surface of a metal is illuminted with the light of 400, nm. The kinetic energy of the ejected photoelectrons was, found to be 1.68 eV. The work function of the metal is :, (hc = 1240 eV.nm), [2009], (a) 1.41 eV, (b) 1.51 eV, (c) 1.68 eV, (d) 3.09 eV, Directions: Question No. 60 and 61 are based on the following, paragraph., Wave property of electrons implies that they will show diffraction, effects. Davisson and Germer demonstrated this by diffracting, electrons from crystals. The law governing the diffraction from, a crystal is obtained by requiring that electron waves reflected, from the planes of atoms in a crystal interfere constructively, (see figure)., , Inco m, Electr ing, ons, , i, , g, Outgoin s, n, o, tr, c, Ele, , d, Crystal plane, , 60. Electrons accelerated by potential V are diffracted from a, crystal. If d = 1Å and i = 30°, V should be about [2008], (h = 6.6 × 10–34 Js, me = 9.1 × 10–31 kg, e = 1.6 × 10–19 C), (a) 2000 V (b) 50 V, (c) 500 V (d) 1000 V, 61. If a strong diffraction peak is observed when electrons are, incident at an angle ‘i’ from the normal to the crystal planes, with distance ‘d’ between them (see figure), de Broglie, wavelength ldB of electrons can be calculated by the, relationship ( n is an integer), [2008], (a) d sin i = nldB, , (b) 2d cos i = nldB, , (c) 2d sin i = nldB, , (d) d cos i = nldB
Page 448 :
P-439, , Dual Nature of Radiation and Matter, , 62., 63., , 64., 65., , Photon of frequency n has a momentum associated with, it. If c is the velocity of light, the momentum is [2007], (a) hn / c (b) n /c, (c) h n c, (d) hn / c2, The threshold frequency for a metallic surface corresponds, to an energy of 6.2 eV and the stopping potential for a, radiation incident on this surface is 5 V. The incident, radiation lies in, [2006], (a) ultra-violet region, (b) infra-red region, (c) visible region, (d) X-ray region, The time taken by a photoelectron to come out after the, photon strikes is approximately, [2006], (a) 10–4 s (b) 10–10 s (c) 10–16 s (d) 10–1 s, The anode voltage of a photocell is kept fixed. The, wavelength l of the light falling on the cathode is gradually, changed. The plate current I of the photocell varies as, follows, [2006], , 67., , 68., , 69., (a), , (b), , I, , I, , 70., O, , (c), , I, , (d), O, , 66., , O, , l, , l, , l, , (c) increase by a factor of 2, (d) decrease by a factor of 2, A radiation of energy E falls normally on a perfectly, reflecting surface. The momentum transferred to the surface, is, [2004], (a) Ec, (b) 2 E / c (c) E / c, (d) E / c 2, According to Einstein’s photoelectric equation, the plot, of the kinetic energy of the emitted photo electrons from a, metal Versus the frequency, of the incident radiation gives, a straight line whose slope, [2004], (a) depends both on the intensity of the radiation and, the metal used, (b) depends on the intensity of the radiation, (c) depends on the nature of the metal used, (d) is the same for the all metals and independent of the, intensity of the radiation, The work function of a substance is 4.0 eV. The longest, wavelength of light that can cause photoelectron emission, from this substance is approximately, [2004], (a) 310 nm (b) 400 nm (c) 540 nm (d) 220 nm, Two identical photocathodes receive light of frequencies, f1 and f2. If the velocites of the photo electrons (of mass, m) coming out are respectively v1 and v2, then [2003], 2h, (a) v12 - v2 2 =, ( f - f2 ), m 1, 1/ 2, , I, , é 2h, ù, (b) v1 + v2 = ê ( f1 + f 2 )ú, ëm, û, O, , l, , A photocell is illuminated by a small bright source placed, 1, 1 m away. When the same source of light is placed m, 2, away, the number of electrons emitted by photocathode, would, [2005], (a) increase by a factor of 4, (b) decrease by a factor of 4, , (c) v 2 + v 2 = 2 h ( f + f ), 1, 2, 2, m 1, 1/ 2, é 2h, ù, (d) v1 - v2 = ê ( f1 - f 2 )ú, ëm, û, 71. Sodium and copper have work functions 2.3 eV and 4.5 eV, respectively. Then the ratio of the wavelengths is nearest, to, [2002], (a) 1 : 2, (b) 4 : 1, (c) 2 : 1, (d) 1 : 4
Page 452 :
P-443, , Dual Nature of Radiation and Matter, , 21. (a) Covalent bonds are formed by sharing of electrons, with different compounds. Formation of covalent bond is, best explained by molecular orbital theory., 22. (50), From Bragg's equation 2d sin q = l and de-Broglie, wavelength, l =, , h, h, =, P, 2mE, , 25. (d) According to Einstein's photoelectric equation, K max = hv - f0, Þ eVs =, , hc, - f0, l, , hc f 0, le e, where l = wavelength of incident light, Þ Vs =, , f0 = work function, Vs = stopping potential, Comparing the above equation with y = mx + c, we get, , d, q, 2d sin q = l =, Þ 2 ´ 10 -10 ´, , q, , hc, e, Increasing the frequency of incident radiation has no effect, on work function and frequency. So, graph will not change., , slope =, , h, 2mE, 3 6.6 ´ 10 -34, =, 2, 2mE, , 26. (d) Using equation, =, , [Q q = 60° and d = 1Å = 1 ´ 10 -10 m ], 2, , 1, 6.64 ´ 10, \E = ´, ; 50 eV, 2 9.1 ´ 10 -31 ´ 3 ´ 1.6 ´ 10 -19, , 23., , 2, From the Einstein's photoelectric equation, Energy of photon, = Kinetic energy of photoelectrons + Work function, Þ Kinetic energy = Energy of Photon – Work Function, Let f0 be the work function of metal and v1 and v2 be the, velocity of photoelectrons. Using Einstein's photoelectric, equation we have, 1 2, mv1 = 4 - f 0, ...(i), 2, 1 2, mv2 = 2.5 - f0, ...(ii), 2, 1 2, mv1, 4 - f0, Þ 2, =, 1 2 2.5 - f0, mv2, 2, 4 - f0, Þ (2)2 =, Þ 10 - 4f 0 = 4 - f0, 2.5 - f 0, , f0 = 2eV, 24., , (a) Graph of Vs and f given at B (5.5, 0), Minimum energy for ejection of electron, = Work function (f)., f = hV joule or f =, \f =, , hV, eV (for V = 0), e, , 6.62 ´ 10 -34 ´ 5.5 ´ 1014, 1.6 ´ 10, , -19, , KEmax =, , -48, , eV = 2.27 eV, , hc, -f, l, , hc, hc, -f=, -f, l, 500, , ...(1), , hc, -f, 200, Dividing equation (2) by (1),, , Again, 3 KEmax =, , ...(2), , hc, 3KEmax 3 200 - f, = =, hc, KEmax, 1, -f, 500, , Putting the value of hc = 1237.5 and solving we get, work, function, f = 0.61 eV., 27. (a) Given,, Wavelength of X-rays, l1 = 1 nm = 1 × 10 –9 m, Wavelength of visible light, l 2 = 500 × 10 –9 m, The number of photons emitted per second from a source, of monochromatic radiation of wavelength l and power P, is given as, , n=, , P P Pl, =, =, E hn hc, , ( QE = hn and n =, , c, ), l, , Þ Clearly n µ l, , Þ, , n1 l1, 1, =, =, n2 l 2 500, , 28. (9), When radiation of wavelength A, lA is used to illuminate,, stopping potential VA = V, hc, = f + eV, l, , ...(i)
Page 455 :
P-446, , Physics, , 3.6 ´ 10-3, 2, = 1.8 × 10–3 Wm–2, work function f = hv, , 1, c, 2, – hn0 = mv, 2, l, 4 hc, 1, - hn0 = mv¢ 2, \, 3 l, 2, , 45. (c), , h, , =, , 4, 4, n - n0, n - n0, 3, \, \ v¢ = v 3, =, n - n0, n - n0, v2, , =, , v¢2, , 6.62 ´ 10-34 )(3 ´108 ), (, =, , 4, 3, 46. (a) From Einstein's photoelectric equation, hc, K . E .l = - f, ...(i), l, (for monochromatic light of wavelength l), where f is work function, hc, K . E .l /2 =, -f, ...(ii), l/2, (for monochromatic light of wavelength l/2), From question,, , l, , \ v¢ > v, , 3, , 12.4 ´ 10, ev, l, for different wavelengths, , =, , 2hc, hc, - f = 3 - 3f, l, l, , 47., , hc hc, =, + eV, l 2 l0, , ...(2), , hc hc, =, + 3eV ', ...(3), l3 l 0, From equation (1) & (2), 3, 2, 1, =, 2l1 2l 2 l 0, , and l 2 = 6216Å, and I = 3.6 ´ 10 -3 Wm -2, Intensity associated with each wavelength, , f2 =, , 12.4 ´ 103 12.4 ´ 103, =, = 1.994 eV = 3.184 × 10–19, l2, 6216, , (Q1 cm, , 2, , -4, , = 10 m, , 2, , 1.8 ´ 10-3, 3.984 ´ 10, , -19, , ´ 10-4, , ) = 0.45 × 10, , 12, , So, the number of photo electrons liberated in 2 sec., = 0.45 × 1012 × 2, = 9 × 1011, 50. (b), 51. (d) As l is increased, there will be a value of l above, which photoelectrons will be cease to come out so, photocurrent will become zero. Hence (d) is correct answer., 52. (d), , hc, - f = eV0, l, , hc f, el e, For metal A, , For metal B, , fA 1, =, hc l, , fB 1, =, hc l, , v0 =, , 3, 1 ù, hc é 1, +, ê ú=V', e ë l3 2l1 2l 2 û, , 49. (b) Given, l1 = 4972Å, , 12.4 ´ 103 12.4 ´ 103, =, = 2.493 eV = 3.984 × 10–19 J, l1, 4972, , no. of electrons per m2-s =, , é 3, hc, 1 ù, - hc ê, ú = eV ', l1, ë 2l1 2l 2 û, , 48. (a) Frank-Hertz experiment - Discrete energy levels of, atom, Photoelectric effect - Particle nature of light., Davison - Germer experiment - wave nature of electron., , f1 =, , J, Work function for metallic surface f = 2.3 eV (given), f2 < f, Therefore, f2 will not contribute in this process., Now, no. of electrons per m2-s = no. of photons per m2-s, , hc, æ hc, ö, K . E.l /2 = 3( K . E .l ) Þ, - f = 3ç - f÷, l/2, èl, ø, , hc, hc, \ f=, Þ 2f =, l, 2l, (None), From Einstein's photoelectric equation, we have, hc hc, =, + eV, ...(1), l1 l 0, , hc, l, , 1, (increasing and decreasing) is not, l, specified hence we cannot say that which metal has, comparatively greater or lesser work function (f)., (a), (a) Davisson Germer experiment showed that electron, beams can undergo diffraction when passed through, atomic crystal. This established wave nature of electron, as waves can exhibit interference and diffraction., , As the value of, , 53., 54.
Page 456 :
P-447, , Dual Nature of Radiation and Matter, , 55. (c) According to photo-electric equation :, K.Emax = hv – hv0 (Work function), Some sort of energy is used in ejecting the, photoelectrons., 56. (b) The maximum kinetic energy of photoelectrons, depends upon frequency on incident light and photo, current depends upon intensity of incident light., 57. (c) By Einstein photoelectric equation,, Kmax = eV0 = hv – hv0, When v is doubled, Kmax and V0 become more than double., 58. (d) We know that, eV0 = Kmax = hn – f, where, f is the work function., X-rays have higher frequency (v) than ultraviolet rays., Therefore as v increases K.E and V0 both increases., The kinetic energy ranges from zero to maximum because, of loss of energy due to subsequent collisions before, getting ejected., 59. (a) Wavelength of incident light, l = 400 nm hc = 1240, eV.nm, K.E = 1.68 eV, Using Einstein’s photoelectric equation, hc, - W = K .E, l, , ÞW=, , hc, - K .E, l, , 1240, - 1.68, 400, = 3.1 – 1.68, , ÞW=, , = 1.41 eV, 60. (b) The path difference between the rays APB and CQD, is, Dx = MQ + QN = d cos i + d cos i, Dx = 2d cos i, , A, , B, D, , C, i, , P, N, M i, Q, For constructive interference the path difference is integral, multiple of wavelength, \ nl = 2d cos i, From de-broglie concept, Wavelength,, d, , l=, \, , h, h, h, =, =, p, 2mK.E, 2meV, nh, , = 2d cos i, 2meV, Squaring both side, n 2h 2, = 4d 2 cos2 i, 2meV, For first order interference n = 1, , \V =, =, , h, , 2, , 8med 2 cos 2 i, (6.6 ´ 10 -34 )2, , 8 ´ 9.1´ 10, , -31, , ´1.6 ´10 -19 ´ (10 -10 ) 2 ´ cos 2 30, , = 50 V, 61. (b) For constructive interference,, 2d cos i = nldB, 62. (a) Energy of a photon of frequency n is given by, E = hn ., Also, E = mc2, mc2 = hn, hn, hn, Þ p=, c, c, 63. (a) Work function, f = 6.2 eV = 6.2 × 1.6 × 10–19 J, Stopping potential, V = 5 volt, From the Einstein’s photoelectric equation, hc, - f = eV0, l, hc, Þl=, f + eV0, Þ mc =, , =, , 6.6 ´ 10, , -34, , ´ 3 ´ 108, , » 10, , -7, , m, 1.6 ´ 10 (6.2 + 5), This range lies in ultra violet range., 64. (b) The photoelectric emission is an instantaneous, process without any apparent time lag. It is known that, emission starts in the time of the order of 10–9 second. So,, the approximate time taken by a photoelectron to come, out after the photon strikes is 10–10 second., 65. (b) As l decreases, y increases and hence the speed of, photoelectron increases. The chances of photo electron, to meet the anode increases and hence photo electric, current increases., 66. (a), , -19, , I µ, , I, r, , ;, 2, , 2, , ær ö, 1, =ç 2÷ =, I 2 è r1 ø, 4, I1, , I 2 ® 4 times I1, , When intensity becomes 4 times, no. of photoelectrons, emitted would increase by 4 times, since number of electrons, emitted per second is directly proportional to intensity.
Page 457 :
P-448, , Physics, , E, c, When a photon hits a perfectly reflecting surface, it reflects, black in opposite direction with same energy and, momentum., , 67. (b) Momentum of photon of energy E is =, , E æ –E ö, 2E, –ç, ÷ =, C è C ø, C, This is equal to momentum transferred to the surface., 68. (d) From the Einstein photoelectric equation K.E. = hn – f, Here, f = work function of metal, h = Plank's constant, slope of graph of K.E. & n is h (Plank’s constant) which is, same for all metals., 69. (a) Work function of metal (f) is given by, , \ Change in momentum =, , f=, , hc, l, , Þ l=, Þ l=, , hc, f, , 6.63 ´ 10-34 ´ 3 ´108, , = 310 nm, 4 ´ 1.6 ´ 10–19, 70. (a) Let work function be W and v1 and v2 be the velocity, of electrons for frequencies f1 and f 2., Using Einstein’s photo electric equation for one, photodiode, we get, , 1 2, mv, ....(i), 2 1, Using Einstein’s photo electric equation for another, photodiode we get,, 1 2, hf 2 - W = mv2, ....(ii), 2, Subtracting (ii) from (i) we get, 1 2 1 2, (hf1 – W) – (hf2 – W) = mv1 - mv2, 2, 2, m 2 2, \ h ( f1 - f 2 ) = ( v1 - v2 ), 2, 2h, 2, 2, ( f1 - f 2 ), \ v1 - v 2 =, m, 71. (c) We know that work function,, hC, E = hu =, l, where, h = Planck’s constant, C = velocity of light, l = wavelength of light, E, l, \ Na = Cu, ECu l Na, hf1 - W =, , Þ, , l Na E Cu 4.5 2, =, =, », l Cu E Na 2.3 1
Page 458 :
26, , P-449, , Atoms, , Atoms, TOPIC 1, 1., , 4., , Atomic Structure and, Rutherford's Nuclear Model, , The graph which depicts the results of Rutherford gold, foil experiment with, [8 Jan. 2020 I], a-particles is:, q: Scattering angle, Y: Number of scattered a-particles detected, (Plots are schematic and not to scale), , Model and the Spectra, TOPIC 2 Bohr's, of the Hydrogen Atom, 5., , (b), , (a), , An a-particle of energy 5 MeV is scattered through 180º by a, fixed uranium nucleus. The distance of closest approach is of, the order of, [2004], (a) 10–12 cm, (b) 10–10 cm, (c) 10–14 cm, (d) 10–15 cm, , A particle of mass 200 MeV/c2 collides with a hydrogen, atom at rest. Soon after the collision the particle comes, to rest, and the atom recoils and goes to its first excited, state. The initial kinetic energy of the particle (in eV), is, , 6., (c), 2., , (d), , In the Rutherford experiment, a-particles are scattered from, a nucleus as shown. Out of the four paths, which path is, not possible?, [Online May 7, 2012], , 7., , A, , N, . The value of N is :, 4, , (Given the mass of the hydrogen atom to be 1 GeV/c2), In the line spectra of hydrogen atom, difference between, the largest and the shortest wavelengths of the Lyman, series is 304 Å. The corresponding difference for the, Paschan series in Å is : __________., [NA Sep. 04, 2020 (I)], In a hydrogen atom the electron makes a transition from (n, + 1)th level to the nth level. If n >> 1, the frequency of, radiation emitted is proportional to : [Sep. 02, 2020 (II)], (a), , B, , (c), , C, , 8., , D, , (a) D, 3., , (b) B, , (c) C, , (d) A, , 1 2, mv bombards a heavy, 2, nuclear target of charge Ze. Then the distance of closest, approach for the alpha nucleus will be proportional to, [2006], , An alpha nucleus of energy, , (a) v 2, , (b), , 1, m, , (c), , 1, v, , 2, , (d), , 1, Ze, , 9., , [NA Sep. 05, 2020 (I)], , 1, n, 1, , (b), , 1, n3, 1, , (d) 4, n2, n, The energy required to ionise a hydrogen like ion in its, ground state is 9 Rydbergs. What is the wavelength of, the radiation emitted when the electron in this ion jumps, from the second excited state to the ground state?, [9 Jan. 2020 II], (a) 24.2 nm, (b) 11.4 nm, (c) 35.8 nm, (d) 8.6 nm, The first member of the Balmer series of hydrogen atom, has a wavelength of 6561 Å. The wavelength of the, second member of the Balmer series (in nm) is ______., [NA 8 Jan. 2020 II]
Page 459 :
P-450, , Physics, , 10. The time period of revolution of electron in its ground, state orbit in a hydrogen atom is 1.6 ´ 10–16 s. The, frequency of revolution of the electron in its first excited, state (in s–1) is:, [7 Jan. 2020 I], (a) 1.6 ´ 1014, (b), 7.8 ´ 1014, (c) 6.2 ´ 1015, (d), 5.6 ´ 1012, 11. An excited He+ ion emits two photons in succession, with, wavelengths 108.5 nm and 30.4 nm, in making a transition, to ground state. The quantum number n, corresponding, to its initial excited state is (for photon of wavelength »,, energy E =, 12., , 13., , 14., , 15., , 16., , 17., , 18., , 1240eV, l (innm), , [12 April 2019 II], , (a) n = 4, (b) n = 5 (c) n = 7, (d) n = 6, The electron in a hydrogen atom first jumps from the third, excited state to the second excited state and subsequently, to the first excited state. The ratio of the respective, wavelengths, l1/ l2, of the photons emitted in this process, is :, [12 April 2019 II], (a) 20/7, (b) 27/5, (c) 7/5, (d) 9/7, Consider an electron in a hydrogen atom, revolving in its, second excited state (having radius 4.65 Å). The de-Broglie, wavelength of this electron is :, [12 April 2019 II], (a) 3.5 Å, (b) 6.6 Å (c) 12.9 Å (d) 9.7 Å, In Li++, electron in first Bohr orbit is excited to a level by a, radiation of wavelength l. When the ion gets deexcited, to the ground state in all possible ways (including, intermediate emissions), a total of six spectral lines are, observed. What is the value of l ? [10 April 2019 II], (Given : h = 6.63×10–34 Js; c = 3 × 108 ms–1), (a) 11.4 nm (b) 9.4 nm (c) 12.3 nm (d) 10.8 nm, Taking the wavelength of first Balmer line in hydrogen, spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the, 2nd Balmer line (n = 4 to n = 2) will be; [9 April 2019 I], (a) 889.2 nm, (b) 488.9 nm, (c) 642.7 nm, (d) 388.9 nm, A He+ ion is in its first excited state. Its ionization energy, is:, [9 April 2019 II], (a) 48.36 eV, (b) 54.40 eV, (c) 13.60 eV, (d) 6.04 eV, Radiation coming from transitions n = 2 to n = 1 of, hydrogen atoms fall on He+ ions in n = 1 and n = 2 states., The possible transition of helium ions as they absorb, energy from the radiation is :, [8 April 2019 I], (a) n = 2 ® n = 3, (b) n = 1 ® n = 4, (c) n = 2 ® n = 5, (d) n = 2 ® n = 4, A hydrogen atom, initially in the ground state is excited, by absorbing a photon of wavelength 980Å. The radius of, the atom in the excited state, in terms of Bohr radius a0,, will be:, [11 Jan 2019 I], (a) 25a0, (b) 9a0, (c) 16a0, (d) 4a0, , 19. In a hydrogen like atom, when an electron jumps from the, M-shell to the L-shell, the wavelength of emitted radiation, is l. If an electron jumps from N-shell to the L-shell, the, wavelength of emitted radiation will be: [11 Jan 2019 II], 27, 16, 25, 20, l, l (c), l, l, (b), (d), 20, 25, 16, 27, An electron from various excited states of hydrogen atom, emit radiation to come to the ground state. Let ln, lg be, the de Broglie wavelength of the electron in the nth state, and the ground state respectively. Let L n be the, wavelength of the emitted photon in the transition from, the nth state to the ground state. For large n, (A, B are, constants), [2018], , (a), 20., , B, (a) L n » A+ 2, ln, , (b) Ln » A + Bln, , (c) L 2n » A + B l 2n, , (d) L 2n » l, , 21. If the series limit frequency of the Lyman series is v1, then, the series limit frequency of the P-fund series is :, [2018], (a) 25 nL, (b) 16 nL (c) nL/16, (d) nL/25, 22. The de-Broglie wavelength (lB) associated with the, electron orbiting in the second excited state of hydrogen, atom is related to that in the ground state (lG) by, [Online April 16, 2018], (a) lB = lG/3, (b) lB = lG/2, (c) lB = 2lG, (d) lB = 3lG, 23. The energy required to remove the electron from a singly, ionized Helium atom is 2.2 times the energy required to, remove an electron from Helium atom. The total energy, required to ionize the Helium atom completely is:, [Online April 15, 2018], (a) 20 eV, (b) 79 eV (c) 109 eV (d) 34 eV, 24. Muon (m–1) is negatively charged ( q = e ) with a mass, mm = 200me, where me is the mass of the electron and e is, the electronic charge. If m–1 is bound to a proton to form a, hydrogen like atom, identify the correct statements, [Online April 15, 2018], (A) Radius of the muonic orbit is 200 times smaller than, that of the electron, (B) the speed of the m–1 in the nth orbit is, , 1, times, 200, , that of the election in the nth orbit, (C) The lonization energy of muonic atom is 200 times, more than that of an hydrogen atom, (D) The momentum of the muon in the nth orbit is 200, times more than that of the electron, (a) (A), (B), (D), (b) (B), (D), (c) (C), (D), (d) (A), (C), (D)
Page 460 :
P-451, , Atoms, , 25., , Some energy levels of a molecule are shown in the figure., The ratio of the wavelengths r = l1/l2, is given by, [2017], , 1, 4, 2, 3, (b) r =, (c) r =, (d) r =, 3, 3, 3, 4, The acceleration of an electron in the first orbit of the, hydrogen atom (z = 1) is :, [Online April 9, 2017], , (a) r =, 26., , (a), (c), 27., , 28., , 29., , 30., , h2, p 2 m2 r 3, h, 2, , (b), , 2, 2 3, , 4p m r, , (d), , h2, 8p 2 m 2 r 3, h, , 2, 2 3, , 4 pm r, , According to Bohr’s theory, the time averaged magnetic, field at the centre (i.e. nucleus) of a hydrogen atom due to, the motion of electrons in the nth orbit is proportional to :, (n = principal quantum number) [Online April 8, 2017], (a) n –4, (b) n –5, (c) n –3, (d) n –2, A hydrogen atom makes a transition from n = 2 to n = 1 and, emits a photon. This photon strikes a doubly ionized lithium, atom (z = 3) in excited state and completely removes the, orbiting electron. The least quantum number for the excited, state of the ion for the process is :, [Online April 9, 2016], (a) 2, (b) 4, (c) 5, (d) 3, As an electron makes a transition from an excited state to, the ground state of a hydrogen - like atom/ion :, [2015], (a) kinetic energy decreases, potential energy increases, but total energy remains same, (b) kinetic energy and total energy decrease but potential, energy increases, (c) its kinetic energy increases but potential energy and, total energy decrease, (d) kinetic energy, potential energy and total energy decrease, The de–Broglie wavelength associated with the electron, in the n = 4 level is :, [Online April 11, 2015], 1, th of the de–Broglie wavelength of the electron, 4, in the ground state., (b) four times the de–Broglie wavelength of the electron, in the ground state, (c) two times the de–Broglie wavelength of the electron, in the ground state, (d) half of the de–Broglie wavelength of the electron in, the ground state, , (a), , 31. If one were to apply Bohr model to a particle of mass ‘m’, and charge ‘q’ moving in a plane under the influence of a, magnetic field ‘B’, the energy of the charged particle in, the nth level will be :, [Online April 10, 2015], , æ hqB ö, (a) n çè, ÷, 2pm ø, , æ hqB ö, (b) n çè, ÷, 8pm ø, , hqB ö, (c) n æç, è 4pm ÷ø, , æ hqB ö, (d) n ç, è pm ÷ø, , 32. The radiation corresponding to 3 ® 2 transition of, hydrogen atom falls on a metal surface to produce, photoelectrons. These electrons are made to enter a, magnetic field of 3 × 10–4 T. If the radius of the largest, circular path followed by these electrons is 10.0 mm, the, work function of the metal is close to:, [2014], (a) 1.8 eV, , (b) 1.1 eV, , (c) 0.8 eV, , (d) 1.6 eV, , 33. Hydrogen (1, , (, , H1), Deuterium (, , 2, 1H ), singly ionised Helium, , He4 ) and doubly ionised lithium, +, , 2, , (, , 3, , Li 6 ), , ++, , all have, , one electron around the nucleus. Consider an electron, transition from n = 2 to n = 1. If the wavelengths of emitted, radiation are l1, l 2 , l3 and l 4 respectively then, approximately which one of the following is correct?, [2014], (a) 4l1 = 2l 2 = 2l 3 = l 4, (b) l1 = 2l 2 = 2l 3 = l 4, (c) l1 = l 2 = 4l3 = 9l4, (d) l1 = 2l 2 = 3l3 = 4l4, 34. Match List - I (Experiment performed) with List-II, (Phenomena discovered/associated) and select the correct, option from the options given below the lists:, [Online April 19, 2014], List - I, , List - II, , (1) Davisson and Germer (i) Wave nature of, experiment, electrons, (2) Millikan's oil drop, experiment, , (ii) Charge of an electron, , (3) Rutherford, experiment, , (iii) Quantisation of, energy levels, , (4) Franck-Hertz, experiment, , (iv) Existence of nucleus, , (a), (b), (c), (d), , (1)-(i), (2)-(ii), (3)-(iii), (4)-(iv), (1)-(i), (2)-(ii), (3)-(iv), (4)-(iii), (1)-(iii), (2)-(iv), (3)-(i), (4)-(ii), (1)-(iv), (2)-(iii), (3)-(ii), (4)-(i)
Page 461 :
P-452, , 35., , 36., , The binding energy of the electron in a hydrogen atom is, 13.6 eV, the energy required to remove the electron from, the first excited state of Li++ is: [Online April 9, 2014], (a) 122.4 eV, (b) 30.6 eV, (c) 13.6 eV, (d) 3.4 eV, Ina hydrogen like atom electron make transition from an, energy level with quantum number n to another with, quantum number (n – 1). If n>>1, the frequency of, radiation emitted is proportional to :, [2013], (a), , 37., , 38., , Physics, , 1, n, , (b), , 1, n2, , (c), , 1, n3, , (d), , 1, n3, , 2, A 12.5 eV electron beam is used to bombard gaseous, hydrogen at room temperature. It will emit :, [Online April 25, 2013], (a) 2 lines in the Lyman series and 1 line in the Balmar, series, (b) 3 lines in the Lyman series, (c) 1 line in the Lyman series and 2 lines in the Balmar series, (d) 3 lines in the Balmer series, In the Bohr’s model of hydrogen-like atom the force, between the nucleus and the electron is modified as, , F=, , e2 æ 1 b ö, ç + ÷ , where b is a constant. For this atom,, 4pe0 è r 2 r 3 ø, , the radius of the nth orbit in terms of the Bohr radius, æ, e h2, ç a0 = 0, ç, mp e 2, è, , 39., , 40., , ö, ÷ is :, ÷, ø, , (a) rn = a0n – b, (b) rn = a0n2 + b, 2, (c) rn = a0n – b, (d) rn = a0n + b, Orbits of a particle moving in a circle are such that the, perimeter of the orbit equals an integer number of deBroglie wavelengths of the particle. For a charged, particle moving in a plane perpendicular to a magnetic, field, the radius of the nth orbital will therefore be, proportional to :, [Online April 22, 2013], 2, (a) n, (b) n, (c) n 1/2, (d) n 1/4, In the Bohr model an electron moves in a circular orbit around, the proton. Considering the orbiting electron to be a circular, current loop, the magnetic moment of the hydrogen atom,, when the electron is in nth excited state, is :, [Online April 9, 2013], æ e n2 h ö, (a) ç, ÷, ç 2m 2p ÷, è, ø, , 41., , [Online April 23, 2013], , æ e ö nh, (b) ç ÷, è m ø 2p, , 2, æ e ö nh, æ e ön h, (c) ç, (d) ç ÷, ÷, è 2m ø 2p, è m ø 2p, Hydrogen atom is excited from ground state to another, state with principal quantum number equal to 4. Then the, number of spectral lines in the emission spectra will be :, [2012], (a) 2, (b) 3, (c) 5, (d) 6, , 42. A diatomic molecule is made of two masses m1 and m2, which are separated by a distance r. If we calculate its, rotational energy by applying Bohr's rule of angular, momentum quantization, its energy will be given by : (n is, an integer), [2012], (a), , (c), , (m1 + m2 )2 n2 h 2, 2m12 m22 r 2, 2n 2 h 2, (m1 + m2 )r 2, , (b), , (d), , n2 h 2, 2(m1 + m2 )r 2, (m1 + m2 )n2 h2, 2m1m2 r 2, , 43. Which of the plots shown in the figure represents speed, (vn) of the electron in a hydrogen atom as a function of the, principal quantum number (n)? [Online May 26, 2012], , (a) B, (b) D, (c) C, (d) A, 44. A doubly ionised Li atom is excited from its ground state, (n = 1) to n = 3 state. The wavelengths of the spectral lines, are given by l32, l31 and l21. The ratio l32/l31 and, l21/l31 are, respectively, [Online May 12, 2012], (a) 8.1, 0.67, (b) 8.1, 1.2, (c) 6.4, 1.2, (d) 6.4, 0.67, 45. A hypothetical atom has only three energy levels. The, ground level has energy, E1 = – 8 eV. The two excited, states have energies, E2 = – 6 eV and E3 = – 2 eV. Then, which of the following wavelengths will not be present in, the emission spectrum of this atom?, [Online May 12, 2012], (a) 207 nm, (b) 465 nm, (c) 310 nm, (d) 620 nm, 46. The electron of a hydrogen atom makes a transition from, the (n + 1) th orbit to the nth orbit. For large n the, wavelength of the emitted radiation is proportional to, [Online May 7, 2012], (a) n, (b) n 3, (c) n 4, (d) n 2, 47. Energy required for the electron excitation in Li++ from the, first to the third Bohr orbit is :, [2011], (a) 36.3 eV, (b) 108.8 eV, (c) 122.4 eV, (d) 12.1 eV, 48. The transition from the state n = 4 to n = 3 in a hydrogen, like atom results in ultraviolet radiation. Infrared, radiation will be obtained in the transition from : [2009], (a) 3 ® 2, (b) 4 ® 2 (c) 5 ® 4, (d) 2 ® 1
Page 462 :
P-453, , Atoms, , 49., , Suppose an electron is attracted towards the origin by a, k, where ‘k’ is a constant and ‘r’ is the distance of, r, the electron from the origin. By applying Bohr model to, this system, the radius of the nth orbital of the electron, is found to be ‘rn’ and the kinetic energy of the electron, to be ‘Tn’. Then which of the following is true? [2008], 1, 2, (a) Tn µ 2 , rn µ n, n, (b) Tn independent of n, rn µ n, 1, (c) Tn µ , rn µ n, n, , n=4, n =3, , force, , 50., , 51., , 1, 2, (d) Tn µ 3 , rn µ n, n, Which of the following transitions in hydrogen atoms emit, photons of highest frequency?, [2007], (a) n = 1 to n = 2 (b), n = 2 to n = 6, (c) n = 6 to n = 2 (d), n = 2 to n = 1, The diagram shows the energy levels for an electron in a, certain atom. Which transition shown represents the, emission of a photon with the most energy?, [2005], , n=2, , I, , II, , III, , IV, , n =1, , (a) IV, (b) III, (c) II, (d) I, 52. The wavelengths involved in the spectrum of deuterium, , ( D), 2, 1, , are slightly different from that of hydrogen, , spectrum, because, [2003], (a) the size of the two nuclei are different, (b) the nuclear forces are different in the two cases, (c) the masses of the two nuclei are different, (d) the attraction between the electron and the nucleus, is differernt in the two cases, 53. If 13.6 eV energy is required to ionize the hydrogen atom, then, the energy required to remove an electron from n = 2 is, [2002], (a) 10.2 eV, (b) 0 eV, (c) 3.4 eV, (d) 6.8 eV
Page 470 :
27, , P-461, , Nuclei, , Nuclei, TOPIC 1, 1., , Composition and Size of the, Nuclei, , 7., , The radius R of a nucleus of mass number A can be, estimated by the formula R = (1.3 × 10–15)A1/3 m. It follows, that the mass density of a nucleus is of the order of :, ( M prot. @ M neut. ; 1.67 ´ 10 -27 kg) [Sep. 03, 2020 (II)], , (a) 103 kg m–3, (c) 1024 kg m–3, 2., , 3., , 4., , 5., , (b) 1010 kg m–3, (d) 1017 kg m–3, , 27, 13 Al, , then the radius of, (a) 8 fermi, (c) 5 fermi, , nucleus is estimated to be 3.6 fermi, , 125, 52 Te, , nucleus be nearly, , [2005], , (b) 6 fermi, (d) 4 fermi, , TOPIC 2 Mass-Energy Equivalence and, Nuclear Reactions, 6., , You are given that mass of 73 Li = 7.0160 u,, 4, 2 He = 4.0026 u, and Mass of 11 H = 1.0079 u., When 20 g of 73 Li is converted, , (b) 8 × 106, , (c) 6.82 × 105, , (d) 1.33 × 106, , Given the masses of various atomic particles mp = 1.0072 u,, mn = 1.0087 u, me = 0.000548 u, mv = 0, md = 2.0141 u, where, p º proton, n º neutron, e º electron,, v º antineutrino and d º deuteron. Which of the following process is allowed by momentum and energy conservation?, [Sep. 06, 2020 (II)], (a) n + n ® deuterium atom (electron bound to the nucleus), , The ratio of the mass densities of nuclei of 40Ca and 16O is, close to :, [8 April 2019 II], (a) 1, (b) 0.1, (c) 5, (d) 2, An unstable heavy nucleus at rest breaks into two nuclei, which move away with velocities in the ratio of 8:27. The, ratio of the radii of the nuclei (assumed to be spherical ) is:, [Online April 15, 2018], (a) 8 : 27, (b) 2 : 3, (c) 3 : 2, (d) 4 : 9, Which of the following are the constituents of the nucleus?, [2007], (a) Electrons and protons (b) Neutrons and protons, (c) Electrons and neutrons (d) Neutrons and positrons, If radius of the, , (a) 4.5 × 105, , Mass of, , into 42 He by proton, capture, the energy liberated, (in kWh), is :, [Mass of nucleon = 1 GeV/c2], [Sep. 06, 2020 (I)], , (b) p ® n + e+ + v, (c) n + p ® d + g, (d) e+ + e– ® g, 8., , Find the Binding energy per neucleon for 120, 50 Sn. Mass of, proton mp = 1.00783 U, mass of neutron mn = 1.00867 U and, mass of tin nucleus mSn = 119.902199 U., (take 1U = 931 MeV), [Sep. 04, 2020 (II)], (a) 7.5 MeV, (b) 9.0 MeV, (c) 8.0 MeV, (d) 8.5 MeV, , In a reactor, 2 kg of 92U235 fuel is fully used up in 30 days., The energy released per fission is 200 MeV. Given that the, Avogadro number, N = 6.023 × 1026 per kilo mole and 1 eV, = 1.6 × 10–19 J. The power output of the reactor is close to:, [Sep. 02, 2020 (I)], (a) 35 MW, (b) 60 MW, (c) 125 MW, (d) 54 MW, 10. Consider the nuclear fission, Ne20 ® 2He4 + C12, Given that the binding energy/nucleon of Ne20, He4 and, C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV,, identify the correct statement:, [10 Jan. 2019 II], (a) energy of 12.4 MeV will be supplied, (b) 8.3 MeV energy will be released, (c) energy of 3.6 MeV will be released, (d) energy of 11.9 MeV has to be supplied, 9.
Page 471 :
P-462, , Physics, , 11., , Imagine that a reactor converts all given mass into energy, and that it operates at a power level of 109 watt. The mass, of the fuel consumed per hour in the reactor will be :, (velocity of light, c is 3 × 108 m/s) [Online April 9, 2017], (a) 0.96 gm, (b) 0.8 gm, (c) 4 × 10–2 gm, (d) 6.6 × 10–5 gm, 12. Two deuterons undergo nuclear fusion to form a Helium, nucleus. Energy released in this process is : (given binding, energy per nucleon for deuteron=1.1 MeV and for, helium=7.0 MeV), [Online April 8, 2017], (a) 30.2 MeV, (b) 32.4 MeV, (c) 23.6 MeV, (d) 25.8 MeV, 13. When Uranium is bombarded with neutrons, it undergoes, fission. The fission reaction can be written as :, + 0 n1 ® 56 Ba141 + 36 Kr 92 + 3 x + Q(energy), where three particles named x are produced and energy Q, is released. What is the name of the particle x ?, [Online April 9, 2013], (a) electron, (b) a-particle, (c) neutron, (d) neutrino, 14. Assume that a neutron breaks into a proton and an, electron. The energy released during this process is : (mass, of neutron = 1.6725 × 10–27 kg, mass of proton = 1.6725 ×, 10–27 kg, mass of electron = 9 × 10–31 kg)., [2012], (a) 0.51 MeV, (b) 7.10 MeV, (c) 6.30 MeV, (d) 5.4 MeV, 15. Ionisation energy of Li (Lithium) atom in ground state is, 5.4 eV. Binding energy of an electron in Li+ ion in ground, state is 75.6 eV. Energy required to remove all three, electrons of Lithium (Li) atom is [Online May 19, 2012], (a) 81.0 eV, (b) 135.4 eV, (c) 203.4 eV, (d) 156.6 eV, 16. After absorbing a slowly moving neutron of mass mN, (momentum » 0) a nucleus of mass M breaks into two, nuclei of masses m1 and 5m1 (6m1 = M + mN) respectively., If the de Broglie wavelength of the nucleus with mass m1, is l, the de Broglie wavelength of the nucleus will be [2011], (a) 5l, (b) l / 5, (c) l, (d) 25l, DIRECTIONS: Questions number 17-18 are based on the, following paragraph., A nucleus of mass M + Dm is at rest and decays into two daughter, 92 U, , 235, , M, each. Speed of light is c., [2010], 2, 17. The binding energy per nucleon for the parent nucleus is, E1 and that for the daughter nuclei is E2. Then, (a) E2 = 2E1, (b) E1 > E2, , nuclei of equal mass, , (c) E2 > E1, (d) E1 = 2 E2, 18. The speed of daughter nuclei is, (a) c, (c) c, , Dm, M + Dm, , Dm, M, , (b) c, , 2Dm, M, , (d) c, , Dm, M + Dm, , 19. Statement-1: Energy is released when heavy nuclei, undergo fission or light nuclei undergo fusion and, Statement-2 : For heavy nuclei, binding energy per, nucleon increases with increasing Z while for light nuclei, it decreases with increasing Z., [2008], (a) Statement-1 is false, Statement-2 is true, (b) Statement-1 is true, Statement-2 is true; Statement-2, is a correct explanation for Statement-1, (c) Statement-1 is true, Statement-2 is true; Statement-2, is not a correct explanation for Statement-1, (d) Statement-1 is true, Statement-2 is false, 20. If MO is the mass of an oxygen isotope 8 O17 ,MP and MN, are the masses of a proton and a neutron respectively, the, nuclear binding energy of the isotope is, [2007], (a) (MO –17MN)c2, (b) (MO – 8MP)c2, (c) (MO– 8MP –9MN)c2 (d) MO c 2, 21. When 3Li7 nuclei are bombarded by protons, and the, resultant nuclei are 4Be8, the emitted particles will be, [2006], (a) alpha particles, (b) beta particles, (c) gamma photons, (d) neutrons, 22. If the binding energy per nucleon in 73 Li and 42 He nuclei, are 5.60 MeV and 7.06 MeV respectively, then in the, reaction, p + 73 Li ¾¾, ® 2 42 He, , [2006], , energy of proton must be, (a) 28.24 MeV, (b) 17.28 MeV, (c) 1.46 MeV, (d) 39.2 MeV, , 23. A nuclear transformation is denoted by X (n, a ) 73 Li ., Which of the following is the nucleus of element X ?, [2005], (a) 10, (b) 12 C 6 (c) 11, (d) 95 B, 5 B, 4 Be, 24. A nucleus disintegrated into two nuclear parts which have, their velocities in the ratio of 2 : 1. The ratio of their nuclear, sizes will be, [2004], (a) 3½ : 1, (b) 1:21/3 (c) 21/3:1, (d) 1:3½, , ( ), 2, , 25. The binding energy per nucleon of deuteron 1 H and, , ( ), 4, , helium nucleus 2 He is 1.1 MeV and 7 MeV respectively.., If two deuteron nuclei react to form a single helium nucleus,, then the energy released is, [2004], (a) 23.6 MeV, (b) 26.9 MeV, (c) 13.9 MeV, (d) 19.2 MeV, 26. When a U 238 nucleus originally at rest, decays by emitting, an alpha particle having a speed ‘u’, the recoil speed of, the residual nucleus is, [2003], (a), , 4u, 238, , (b) - 4u, 234, , (c), , 4u, 234, , (d) - 4u, 238
Page 472 :
P-463, , Nuclei, , 33. Two radioactive substances A and B have decay, constants 5l and l respectively. At t = 0, a sample has the, same number of the two nuclei. The time taken for the, , 27. In the nuclear fusion reaction, 2, 3, 4, 1 H + 1 H ® 2 He + n, , given that the repulsive potential energy between the two, nuclei is ~ 7.7 ´ 10 -14 J , the temperature at which the, gases must be heated to initiate the reaction is nearly, [Boltzmann’s Constant k = 1.38 ´ 10 -23 J/K ], , (d) 10 9 K, , (c) 10 3 K, , (b) 10 5 K, , (a) 10 7 K, , [2003], , 2, , æ 1ö, ratio of the number of nuclei to become ç ÷ will be :, è eø, (a) 1/2l, , (b) 1/4l, , (c) 1/l, , [10 April 2019, II], (d) 2/l, , 34. In a radioactive decay chain, the initial nucleus is 232, 90 Th ., At the end there are 6 a-particles and 4 b-particles which, , TOPIC 3 Radioactivity, , are emitted. If the end nucleus is, , 28. Acitvities of three radioactive substances A, B and C are, represented by the curves A, B and C, in the figure. Then, their half-lives T1 (A) : T1 (B) : T1 (C) are in the ratio :, 2, 2, 2 [Sep. 05, 2020 (I)], In R, 6, 4, 2, , B, , C, , 0, , A, , 5, , (a) 2 : 1 : 1, , 10, , t, (yrs), , (b) 3 : 2 : 1, , (c) 2 : 1 : 3, (d) 4 : 3 : 1, 29. A radioactive nucleus decays by two different processes., The half life for the first process is 10 s and that for the, second is 100 s. The effective half life of the nucleus is, close to :, [Sep. 05, 2020 (II)], (a) 9 sec., (b) 6 sec., (c) 55 sec., (d) 12 sec., 30. In a radioactive material, fraction of active material, remaining after time t is 9/16. The fraction that was, remaining after t/2 is :, [Sep. 03, 2020 (I)], (a), , 4, 5, , (b), , 3, 5, , (c), , 3, 4, , (d), , 7, 8, , 31. The activity of a radioactive sample falls from 700 s –1 to 500, s –1 in 30 minutes. Its half life is close to: [7 Jan. 2020, II], (a) 72 min, (b) 62 min, (c) 66 min, (d) 52 min, 32. Two radioactive materials A and B have decay constants, 10 l and l, respectively. If initially they have the same, number of nuclei, then the ratio of the number of nuclei of, A to that of B will be 1/e after a time : [10 April 2019, I], (a), , 1, 9l, , (b), , 1, 11l, , (c), , 11, 10l, , (d), , 1, 10l, , A , A and Z are given by :, ZX, , [12 Jan. 2019, II], (a) A = 208 ; Z = 80, (b) A = 202 ; Z = 80, (c) A = 208 ; Z = 82, (d) A = 200; Z = 81, 35. Using a nuclear counter the count rate of emitted, particles from a radioactive source is measured. At t =, 0 it was 1600 counts per second and t = 8 seconds it, was 100 counts per second. The count rate observed,, as counts per second, at t = 6 seconds is close to:, [10 Jan. 2019 I], (a) 200, (b) 150, (c) 400, (d), 360, 36. A sample of radioactive material A, that has an activity of, 10 mCi (1 Ci = 3.7 × 1010 decays/s), has twice the number of, nuclei as another sample of a different radioactive material B, which has an acitvity of 20 mCi. The correct choices for halflives of A and B would then be respectively: [9 Jan. 2019 I], (a) 5 days and 10 days (b) 10 days and 40 days, (c) 20 days and 5 days (d) 20 days and 10 days, 37. At a given instant, say t = 0, two radioactive substances, R, A and B have equal activities. The ratio B of their, RA, activities after time t itself decays with time t as e–3t. If, the half-life of A is ln2, the half-life of B is:, [9 Jan. 2019, II], , ln2, ln2, (c), (d) 2ln2, 4, 2, 38. At some instant, a radioactive sample S1 having an activity, 5mCi has twice the number of nuclei as another sample S2, which has an activity of 10mCi. The half lives of S1 and S2, are, [Online April 16, 2018], (a) 10 years and 20 years, respectively, (b) 5 years and 20 years, respectively, (c) 20 years and 10 years, respectively, (d) 20 years and 5 years, respectively, 60Co having activity, 39. A solution containing active cobalt 27, of 0.8 µCi and decay constant l is injected in an animal's, body. If 1cm3 of blood is drawn from the animal's body, after 10 hrs of injection, the activity found was 300 decays, per minute. What is the volume of blood that is flowing in, the body? ( 1Ci = 3.7 × 1010 decay per second and at t, = 10 hrs e–lt = 0.84), [Online April 15, 2018], (a) 6 litres (b) 7 litres (c) 4 litres (d) 5 litres, (a) 4ln2, , (b)
Page 473 :
P-464, , 40., , A radioactive nucleus A with a half life T, decays into a, nucleus B. At t = 0, there is no nucleus B. At sometime t, the, ratio of the number of B to that of A is 0.3. Then, t is given, by, [2017], T, (a) t = T log (1.3), (b) t = log(1.3), , log 2, log1.3, (d) t =, log1.3, log 2, 41. Half-lives of two radioactive elements A and B are 20, minutes and 40 minutes, respectively. Initially, the samples, have equal number of nuclei. After 80 minutes, the ratio of, decayed number of A and B nuclei will be :, [2016], (a) 1 : 4, (b) 5 : 4, (c) 1 : 16, (d) 4 : 1, 42. Let Nb be the number of b particles emitted by 1 gram, , (c) t = T, , of Na24 radioactive nuclei, , (half life = 15 hrs) in 7.5 hours,, Nb is close to (Avogadro number = 6.023 × 1023/g. mole):, [Online April 11, 2015], (a) 6.2 × 1021, (b) 7.5 × 1021, (c) 1.25 × 1022, (d) 1.75 × 1022, 43. A piece of wood from a recently cut tree shows 20 decays, per minute. A wooden piece of same size placed in a, museum (obtained from a tree cut many years back) shows, 2 decays per minute. If half life of C14 is 5730 years, then, age of the wooden piece placed in the museum is, approximately:, [Online April 19, 2014], (a) 10439 years, (b) 13094 years, (c) 19039 years, (d) 39049 years, 44. A piece of bone of an animal from a ruin is found to have, 14C activity of 12 disintegrations per minute per gm of its, carbon content. The 14C activity of a living animal is 16, disintegrations per minute per gm. How long ago nearly, did the animal die? (Given half life of 14C is t1/2 = 5760, years), [Online April 12, 2014], (a) 1672 years, (b) 2391 years, (c) 3291 years, (d) 4453 years, 45. A radioactive nuclei with decay constant 0.5/s is being, produced at a constant rate of 100 nuclei/s. If at t = 0 there, were no nuclei, the time when there are 50 nuclei is:, [Online April 11, 2014], 4, (a) 1s, (b) 2ln æç ö÷ s, è 3ø, , æ 4ö, (d) ln ç ÷ s, è 3ø, 46. The half-life of a radioactive element A is the same as the, mean-life of another radioactive element B. Initially both, substances have the same number of atoms, then :, [Online April 22, 2013], (a) A and B decay at the same rate always., (b) A and B decay at the same rate initially., (c) A will decay at a faster rate than B., (d) B will decay at a faster rate than A., (c) ln 2 s, , Physics, , 47. The counting rate observed from a radioactive source at, t = 0 was 1600 counts s–1, and t = 8 s, it was 100 counts, s–1. The counting rate observed as counts s–1 at t = 6 s, will be, [Online May 26, 2012], (a) 250, (b) 400, (c) 300, (d) 200, 48. The decay constants of a radioactive substance for a and, b emission are la and lb respectively. If the substance, emits a and b simultaneously, then the average half life of, the material will be, [Online May 19, 2012], 2Ta Tb, (a) T + T, a, b, , (b) Ta + Tb, , Ta Tb, (c) T + T, a, b, , (d), , (, , ), , 1, Ta + Tb, 2, 49. Which of the following Statements is correct?, [Online May 12, 2012], (a) The rate of radioactive decay cannot be controlled, but that of nuclear fission can be controlled., (b) Nuclear forces are short range, attractive and charge, dependent., (c) Nuclei of atoms having same number of neutrons are, known as isobars., (d) Wavelength of matter waves is given by de Broglie, formula but that of photons is not given by the same, formula, 50. A sample originally contained 1020 radioactive atoms,, which emit a-particles. The ratio of a-particles emitted in, the third year to that emitted during the second year is 0.3., How many a-particles were emitted in the first year?, [Online May 7, 2012], (a) 3 × 1018, (b) 3 × 1019, (c) 5 × 1018, (d) 7 × 1019, 51. The half life of a radioactive substance is 20 minutes. The, , approximate time interval (t2 – t1) between the time t2 when, 2, 1, of it had decayed and time t1 when of it had decayed, 3, 3, , is :, , [2011], , (a) 14 min (b) 20 min (c) 28 min (d) 7 min, 52. Statement - 1 : A nucleus having energy E1 decays by, b– emission to daughter nucleus having energy E2, but, the b– rays are emitted with a continuous energy spectrum, having end point energy E1 – E2., Statement - 2 : To conserve energy and momentum in, b– decay at least three particles must take part in the, transformation., [2011 RS], (a) Statement-1 is correct but statement-2 is not correct., (b) Statement-1 and statement-2 both are correct and, statement-2 is the correct explanation of statement-1., (c) Statement-1 is correct, statement-2 is correct and, statement-2 is not the correct explanation of, Statement-1, (d) Statement-1 is incorrect, statement-2 is correct.
Page 474 :
P-465, , Nuclei, , 53., , A radioactive nucleus (initial mass number A and atomic, number Z emits 3 a - particles and 2 positrons. The ratio of, number of neutrons to that of protons in the final nucleus, will be, [2010], (a), , A- Z -8, Z -4, , (b), , A- Z -4, Z -8, , A - Z - 12, A- Z -4, (d), Z -4, Z -2, 54. The half-life period of a radio-active element X is same as, the mean life time of another radio-active element Y. Initially, they have the same number of atoms. Then, [2007], (a) X and Y decay at same rate always, (b) X will decay faster than Y, (c) Y will decay faster than X, (d) X and Y have same decay rate initially, 55. The energy spectrum of b-particles [number N(E) as, a function of b-energy E] emitted from a radioactive source, is, [2006], , (c), , (a), , (b), , N(E), E, , E0, , (c), , N(E), , E0, , (d), , N(E), , E0, , E, , N(E), , E, , E0, , E, , 7, of it decays into, 8, Zn in 15 minutes. The corresponding half life is [2005], (a) 15 minutes, (b) 10 minutes, 1, (c) 7 minutes, (d) 5 minutes, 2, , 56. Starting with a sample of pure, , 66, , Cu ,, , 57. The intensity of gamma radiation from a given source is I., I, On passing through 36 mm of lead, it is reduced to . The, 8, I, thickness of lead which will reduce the intensity to will, 2, be, [2005], (a) 9 mm, (b) 6mm, (c) 12mm, (d) 18mm, 58. Which of the following cannot be emitted by radioactive, substances during their decay?, [2003], (a) Protons, (b) Neutrinoes, (c) Helium nuclei, (d) Electrons, 59. A nucleus with Z= 92 emits the following in a sequence:, a, b - , b - a, a, a, a, a, b - , b - , a, b + , b + , a, Then Z of the resulting nucleus is, [2003], (a) 76, (b) 78, (c) 82, (d) 74, 60. A radioactive sample at any instant has its disintegration, rate 5000 disintegrations per minute. After 5 minutes, the, rate is 1250 disintegrations per minute. Then, the decay, constant (per minute) is, [2003], (a) 0.4 ln 2, (b) 0.2 ln 2, (c) 0.1 ln 2, (d) 0.8 ln 2, 61. At a specific instant emission of radioactive compound is, deflected in a magnetic field. The compound can emit, (i) electrons, (ii) protons, (iii) He2+, (iv) neutrons, The emission at instant can be, [2002], (a) i, ii, iii, (b) i, ii, iii, iv, (c) iv, (d) ii, iii, 62. If N0 is the original mass of the substance of half-life period, t1/2 = 5 years, then the amount of substance left after 15, years is, [2002], (a) N0/8, (b) N0/16 (c) N0/2, (d) N0/4
Page 475 :
P-466, , 1., , Physics, , (d) Density of nucleus, r =, , Mass, mA, =, Volume, 4 3, pR, 3, , In use of 1 g Li energy released =, In use of 20 g energy released =, , mA, (Q R = R0 A1/ 3 ), 4, 1/ 3 3, p( R0 A ), 3, Here m = mass of a nucleon, Þr=, , 2., 3., , [(7.016 + 1.0079) - 2 ´ 4.0026]u ´ c 2, 7.016 ´ 1.6 ´ 10-24, , ´ 20 g, , æ 0.0187 ´ 1.6 ´ 10 -19 ´ 109, ö, =ç, ´ 20÷ = 480 ´ 1010 J, -24, 7.016 ´ 1.6 ´ 10, è, ø, , Þ r = 2.38 ´1017 kg/m3, , Q1 J =2.778×10–7 kWh, , (a) Nuclear density is independent of atomic number., (c) Let heavy nucleus breaks into two nuclei of mass m1, and m 2 and move away with velocities V 1 and V 2, respectively., , \ Energy released = 480 × 1010 × 2.778 × 10–7, = 1.33 × 106 kWh, (c) For the momentum and energy conservation, mass, defect (Dm) should be positive. Since some energy is lost, in every process., , V1, 8, =, V2 27, , 7., , (m p + mn ) > md, , m1V1 = m2V2 (Law of momentum conservation), , 8., , 4, r ´ pR13, 3, 4, r ´ pR23, 3, , = (50×1.00783 + 70 × 1.008) – (119.902199), = 1.096, , mass ö, æ, çèQ density r =, ÷, volume ø, 1, 27 ö 3, , 1, 3´, æ 3ö 3, , æ R1 ö æ, çè R ÷ø = çè 8 ÷ø = çè 2 ÷ø, 2, , Þ, , (b), (b) Radius of a nucleus,, , æA ö, \, = 1, R2 çè A2 ÷ø, , 1/ 3, , æ 27 ö, =ç, è 125 ÷ø, , 7, 1, 3 Li + 1H, , ¾¾, ®2, , (, , 1/ 3, , 4, 2 He, , ), , Dm ® [mLi + mH ] - 2[ M He ], Energy released = Dmc 2, , Binding energy 1020.5631, =, = 8.5 MeV, Nucleon, 120, , 9., , (b) Power output of the reactor,, , =, , 5, Þ R2 = ´ 3.6 = 6 fermi, 3, , (d), , R1 3, \ R =2, 2, , Binding energy = (Dm)C 2 = ( Dm) ´ 931 = 1020.56, , P=, , R = R0 ( A)1/ 3, Here, R0 is a constant, A = atomic mass number, R1, , (d) Mass defect,, , Dm = (50m p + 70mn ) - (msn ), , m1 V2 27, =, =, 8, m2 V1, , Þ, , 6., , Dmc 2, ´ 20 g, mLi, , 3 ´1.67 ´ 10-27, (Given, R0 = 1.3 × 10–15), \r =, 4 ´ 3.14 ´ (1.3 ´10-15 )3, , According to question,, , 4., 5., , =, , Dmc 2, mLi, , energy, time, , 2, 6.023 ´ 1026 ´ 200 ´ 1.6 ´ 10 -19, ´, ; 60 MW, 235, 30 ´ 24 ´ 60 ´ 60, , 10. (d), 3, =, 5, , E Dmc 2, =, Dt, Dt, mass of the fuel consumed per hour in the reactor,, Dm P, 109, = 2=, = 4 ´ 10 -2 gm, 8, 2, Dt c, (3 ´ 10 ), 12. (c) 1H2 + 1H2 ® 2He4, Total binding energy of two deuterium nuclei = 1.1 × 4 =, 4.4 MeV, Binding energy of a (2He4) nuclei = 4 × 7 = 28 MeV, Energy released in this process = 28 – 4.4 = 23.6 MeV, , 11. (c) Power level of reactor, P =
Page 481 :
P-472, , Physics, , æ I/8ö, - m ´ 36 = logç, ÷ ..........(i), è I ø, For intensity I/2, thickness = d, æ I / 2ö, ...........(ii), -m ´ d = log ç, è I ÷ø, Dividing (i) by (ii),, 36, =, d, , æ 1ö, æ 1ö, log ç ÷ 3log ç ÷, 36, è 8ø, è 2ø, =, = 3 or d =, = 12 mm, 3, æ 1ö, æ 1ö, log ç ÷, log ç ÷, è 2ø, è 2ø, , 58. (a) The radioactive substances emit a -particles (Helium, nucleus), b-particles (electrons) and neutrinoes. Protons, cannot be emitted by radioactive substances during their, decay., 59. (b) The number of a-particles released = 8, Decrease in atomic number = 8 × 2 = 16, The number of b–-particles released = 4, Increase in atomic number = 4 × 1 = 4, Also the number of b+ particles released is 2, which should, decrease the atomic number by 2., , Therefore the final atomic number of resulting nucleus, = Z –16 + 4 – 2 = Z –14, = 92 – 14 = 78, 60. (a) Initial activity, Ao = 5000 disintegration per minute, Activity after 5 min, A = 1250 disintegration per minute, A = Aoe–lt, A, Þ e–lt = o, A, A, 1, 5000, 1, Þ l = log e o = log e, 5, 1250, t, A, , 2, log e 2 = 0.4 log e 2, 5, 61. (a) Charged particles are deflected in magnetic field., Electrons, protons and He2+ all are charged species. Hence,, correct option is (a)., 62. (a) After every half-life, the mass of the substance, reduces to half its initial value., N, N /2, 5 years, 5 years, N 0 ¾¾¾¾, ® 0 ¾¾¾¾, ® 0, 2, 2, N 0 5years N 0 / 4 N 0, =, ¾¾¾¾, ®, =, 4, 2, 8, =
Page 482 :
28, , P-473, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Semiconductor Electronics :, Materials, Devices and, Simple Circuits, 1., , With increasing biasing voltage of a photodiode, the photocurrent magnitude :, [Sep. 05, 2020 (I)], (a) remains constant, , (d), 3., , (b) increases initially and after attaining certain value, it, decreases, (c) Increases linearly, 2., , (d) increases initially and saturates finally, Two Zener diodes (A and B) having breakdown voltages, of 6 V and 4 V respectively, are connected as shown in the, circuit below. The output voltage V0 variation with input, voltage linearly increasing with time, is given by :, (Vinput = 0 V at t = 0), (figures are qualitative), [Sep. 05, 2020 (II)], , Vin, , A, , B, 6V, , 4V, , RL = 400W, , 6V, , V0, , Solids, Semiconductors and, TOPIC 1, P-N Junction Diode, , time, Take the breakdown voltage of the zener diode used in the, given circuit as 6V. For the input voltage shown in figure, below, the time variation of the output voltage is :, (Graphs drawn are schematic and not to scale), [Sep. 04, 2020 (I)], R1, 10 V, Vin, V0, V=0, – 10 V, , V, (a), , t, V, , V0, , (b), , t, , 100W, , V, , V0, , (c), , 4V, , t, , (a), V, , time, 6V, , V0, , 4V, , (b), time, V0, , 6V, 4V, , (c), time, , (d), 4., , t, , When a diode is forward biased, it has a voltage drop of, 0.5 V. The safe limit of current through the diode is 10 mA., If a battery of emf 1.5 V is used in the circuit, the value of, minimum resistance to be connected in series with the, diode so that the current does not exceed the safe limit is :, [Sep. 03, 2020 (I)], (a) 300 W, (b) 50 W, (c) 100 W, (d) 200 W
Page 483 :
P-474, , Physics, , 5., , If a semiconductor photodiode can detect a photon with a, maximum wavelength of 400 nm, then its band gap energy is:, Planck's constant, h = 6.63 × 10–34 J.s., Speed of light, c = 3 × 108 m/s, [Sep. 03, 2020 (II)], (a) 1.1 eV, (b) 2.0 eV, (c) 1.5 eV, (d) 3.1 eV, , 6., , Both the diodes used in the circuit shown are assumed to, be ideal and have negligible resistance when these are, forward biased. Built in potential in each diode is 0.7 V. For, the input voltages shown in the figure, the voltage (in, Volts) at point A is ________., [NA 9 Jan. 2020 I], , 7., , 8., , The current i in the network is:, , VC, CV, , QB =, e, 2, , 9., , (a) 2.5 mA (b) 1.5 mA (c) 7.5 mA, (d) 3.5 mA, 12. The figure represents a voltage regulator circuit using a, Zener diode. The breakdown voltage of the Zener diode is, 6 V and the load resistance is RL = 4 k. The series resistance, of the circuit is Ri=1k. If the battery voltage VB varies from, 8 V to 16 V, what are the minimum and maximum values of, the current through Zener diode ?, [10 Apr. 2019 II], , (b) QA = VC, QB = CV, , VC, VC, CV, (d) QA =, , QB =, e, e, 2, The circuit shown below is working as a 8 V dc regulated, voltage source. When 12 V is used as input, the power, dissipated (in mW) in each diode is; (considering both, zener diodes are identical) _____. [NA 9 Jan. 2020 II], , (c) QA = VC, QB =, , (a) 10 V, (b) 5 V, (c) 15 V, (d) zero, 11. Figure shows a DC voltage regulator circuit, with a Zener, diode of breakdown voltage = 6V. If the unregulated input, voltage varies between 10 V to 16 V, then what is the, maximum Zener current ?, [12 Apr. 2019 II], , [9 Jan. 2020 II], , (a) 0.2 A, (b) 0.6 A, (c) 0.3 A, (d) 0 A, Two identical capacitors A and B, charged to the same, potential 5V are connected in two different circuits as, shown below at time t = 0. If the charge on capacitors A, and B at time t = CR is QA and QB respectively, then (Here, e is the base of natural logarithm), [9 Jan. 2020 II], , (a) QA =, , 10. In the figure, potential difference between A and B is:, [7 Jan. 2020 II], , (a) 0.5 mA; 6 mA, (b) 1 mA; 8.5 mA, (c) 0.5 mA; 8.5 mA, (d) 1.5 mA; 8.5 mA, 13. The reverse breakdown voltage of a Zener diode is 5.6 V in, the given circuit., , The current Iz through the Zener is :, (a) 10 mA, (b) 17 mA, (c) 15 mA, (d) 7 mA, , [8 April 2019 I]
Page 484 :
P-475, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 14., , In the given circuit the current through, Zener Diode is close to :, [11 Jan. 2019 I], 12V, , R1 500 W, , 15., , R2 1500 W, V2 =10 V R2, (a) 0.0 mA, (b) 6.7 mA, (c) 4.0 mA, (d) 6.0 mA, The circuit shown below contains two ideal diodes, each, with a forward resistance of 50 W. If the battery voltage is, 6V, the current through the 100 W resistance (in Amperes), is :, [11 Jan. 2019 II], , 150 W, , D1, , (a) 0.8 V, (b) 0.6 V, (c) 0.2 V, (d) 0.4 V, 19. The reading of the ammeter for a silicon diode in the given, circuit is :, [2018], , 200W, , 3V, (a) 0, (b) 15 mA, (c) 11.5 mA, (d) 13.5 mA, 20. In the given circuit, the current through zener diode is:, [Online April 16, 2018], , 75 W, , R1 500 W, D2, , 15 V, 100 W, , 16., , 6V, (a) 0.036, (b) 0.020, (c) 0.027, (d) 0.030, For the circuit shown below, the current through the, Zener diode is:, [10 Jan. 2019 II], 5 kW, , 50 V, , 120 V, , 10 kW, , (a) 9 mA, (b) 5 mA, (c) Zero, (d) 14 mA, 17. Mobility of electrons in a semiconductor is defined as the, ratio of their drift velocity to the applied electric field. If, for, an n-type semiconductor, the density of electrons is 1019 m –, 3, and their mobility is 1.6m2/(V.s) then the resistivity of the, semiconductor (since it is an n-type semiconductor, contribution of holes is ignored) is close to:[9 Jan. 2019 I], (a) 2 Wm, (b) 4 Wm (c) 0.4 Wm, (d) 0.2 Wm, 18. Ge and Si diodes start conducting at 0.3 V and 0.7 V, respectively. In the following figure if Ge diode, connection are reversed, the value of Vo changes by :, (assume that the Ge diode has large breakdown voltage), [9 Jan. 2019 II], Ge, Vo, 12 V, , Si, , 5K, , Vz = 10 V, , 1500 W R2, , (a) 2.5mA, (b) 3.3mA, (c) 5.5mA, (d) 6.7mA, 21. What is the conductivity of a semiconductor sample having, electron concentration of 5 × 1018 m–3, hole concentration, of 5 × 1019 m–3, electron mobility of 2.0 m2 V–1 s–1 and, hole mobility of 0.01 m2 V–1 s–1 ?, [Online April 8, 2017], (Take charge of electron as 1.6 × 10–19 C), (a) 1.68 (W – m)–1, (b) 1.83 (W – m)–1, –1, (c) 0.59 (W – m), (d) 1.20 (W – m)–1, 22. The V–I characteristic of a diode is shown in the figure., The ratio of forward to reverse bias resistance is :, [Online April 8, 2017], I(mA), 20, 15, 10, –10, 1m A, , (a) 10, (c) 106, , .7, , .8 V (Volt), , (b) 10–6, (d) 100
Page 485 :
P-476, , 23., , Physics, , Identify the semiconductor devices whose characteristics, are given below, in the order (i), (ii), (iii), (iv) :, [2016], I, I, , V, , (i), , Resistance, , V, Illuminated, , 24., , 25., , 26., , 27., , 5W, , D2, , 10 W, , V, , (ii), , I dark, , D1, , Intensity, of light, , (iii), (iv), (a) Solar cell, Light dependent resistance, Zener diode,, simple diode, (b) Zener diode, Solar cell, simple diode, Light dependent, resistance, (c) Simple diode, Zener diode, Solar cell, Light dependent, resistance, (d) Zener diode, Simple diode, Light dependent, resistance, Solar cell, The temperature dependence of resistances of Cu and, undoped Si in the temperature range 300-400 K, is best, described by :, [2016], (a) Linear increase for Cu, exponential decrease of Si., (b) Linear decrease for Cu, linear decrease for Si., (c) Linear increase for Cu, linear increase for Si., (d) Linear increase for Cu, exponential increase for Si., An experiment is performed to determine the 1–V, characteristics of a Zener diode, which has a protective, resistance of R = 100 W, and a maximum power of dissipation, rating of 1W. The minimum voltage range of the DC source, in the circuit is :, [Online April 9, 2016], (a) 0 – 5V, (b) 0 – 24 V, (c) 0 – 12 V, (d) 0 – 8V, A red LED emits light at 0.1 watt uniformly around it. The, amplitude of the electric field of the light at a distance of 1 m, from the diode is :, [2015], (a) 5.48 V/m, (b) 7.75 V/m, (c) 1.73 V/m, (d) 2.45 V/m, A 2V battery is connected across AB as shown in the, figure. The value of the current supplied by the battery, when in one case battery’s positive terminal is connected, to A and in other case when positive terminal of battery, is connected to B will respectively be:, [Online April 11, 2015], , A B, (a) 0.4 A and 0.2 A, (b) 0.2 A and 0.4 A, (c) 0.1 A and 0.2 A, (d) 0.2 A and 0.1 A, 28. In an unbiased n-p junction electrons diffuse from n-region, to p-region because :, [Online April 10, 2015], (a) holes in p-region attract them, (b) electrons travel across the junction due to potential, difference, (c) only electrons move from n to p region and not the, vice-versa, (d) electron concentration in n-region is more compared, to that in p-region, 29. The forward biased diode connection is:, [2014], (a), , +2V, , –2V, , (b), , –3V, , –3V, , (c), , 2V, , 4V, , –2V, , +2V, , (d), 30. For LED’s to emit light in visible region of electromagnetic, light, it should have energy band gap in the range of:, [Online April 12, 2014], (a) 0.1 eV to 0.4 eV, (b) 0.5 eV to 0.8 eV, (c) 0.9 eV to 1.6 eV, (d) 1.7 eV to 3.0 eV, 31. A Zener diode is connected to a battery and a load as, show below:, [Online April 11, 2014], , 4 kW, , IL, , A, IZ, , I, , RL = 2kW, , 10 V = VZ, , 60 V, B, , The currents, I, IZ and IL are respectively., (a) 15 mA, 5 mA, 10 mA, (b) 15 mA, 7.5 mA, 7.5 mA, (c) 12.5 mA, 5 mA, 7.5 mA, (d) 12.5 mA, 7.5 mA, 5 mA, 32. The I-V characteristic of an LED is, , (a), , I, O, , (b), V, , [2013], , B, G, Y, R, O, , V
Page 486 :
P-477, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , V, I, , O, , 33., , V, , D2, , 35., , I, t, , Figure shows a circuit in which three identical diodes are, used. Each diode has forward resistance of 20 W and infinite, backward resistance. Resistors R1 = R2 = R3 = 50 W. Battery, voltage is 6 V. The current through R3 is :, [Online April 22, 2013], , D1, , 34., , (b), , R, Y, (d) G, B, , (c), , I, , O, , I, , (d), , R1, , t, , 36. If in a p-n junction diode, a square input signal of 10 V is, applied as shown, [2007], 5V, , R2, , D3, , + –, R3, 6V, (a) 50 mA, (b) 100 mA, (c) 60 mA, (d) 25 mA, This question has Statement 1 and Statement 2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., Statement 1: A pure semiconductor has negative, temperature coefficient of resistance., Statement 2: On raising the temperature, more charge, carriers are released into the conduction band., [Online May 12, 2012], (a) Statement 1 is false, Statement 2 is true., (b) Statement 1 is true, Statement 2 is false., (c) Statement 1 is true, Statement 2 is true, Statement 2 is, not a correct explanation of Statement 1., (d) Statement 1 is true, Statement 2 is true, Statement 2 is, the correct explanation of Statement 1., A p-n junction (D) shown in the figure can act as a rectifier., An alternating current source (V) is connected in the circuit., D, R, , V, , (c), , ~, , RL, -5V, , Then the output signal across RL will be, 10 V, +5V, (a), , (b), , (c), , (d), -5V, -10 V, 37. Carbon, silicon and germanium have four valence electrons, each. At room temperature which one of the following, statements is most appropriate ?, [2007], (a) The number of free electrons for conduction is, significant only in Si and Ge but small in C., (b) The number of free conduction electrons is significant, in C but small in Si and Ge., (c) The number of free conduction electrons is negligibly, small in all the three., (d) The number of free electrons for conduction is, significant in all the three., 38. If the lattice constant of this semiconductor is decreased,, then which of the following is correct?, [2006], conduction, band width, , The current (I) in the resistor (R) can be shown by :[2009], , band gap, , Ec, Eg, , valence, band width, , (a), , (a), (b), (c), (d), , All Ec, Eg, Ev increase, Ec and Ev increase, but Eg decreases, Ec and Ev decrease, but Eg increases, All Ec, Eg, Ev decrease, , Ev
Page 487 :
P-478, , Physics, , 39., , A solid which is not transparent to visible light and whose, conductivity increases with temperature is formed by, [2006], (a) Ionic bonding, (b) Covalent bonding, (c) Vander Waals bonding, (d) Metallic bonding, , 40., , If the ratio of the concentration of electrons to that of, 7, holes in a semiconductor is and the ratio of currents is, 5, 7, , then what is the ratio of their drift velocities? [2006], 4, 5, 4, 4, 5, (a), (b), (c), (d), 8, 5, 7, 4, The circuit has two oppositively connected ideal diodes, in parallel. What is the current flowing in the circuit?[2006], 4W, , 41., , D1, 12V, , 42., , 3W, , D2, 2W, , (a) 1.71 A (b) 2.00 A (c) 2.31 A (d) 1.33 A, In the following, which one of the diodes reverse, biased?, [2006], +10 V, , R, , (a), +5 V, , (b), , –10 V, R, –5 V, , (c), R, , 44. When p-n junction diode is forward biased then, [2004], (a) both the depletion region and barrier height are reduced, (b) the depletion region is widened and barrier height is, reduced, (c) the depletion region is reduced and barrier height is, increased, (d) Both the depletion region and barrier height are, increased, 45. A strip of copper and another of germanium are cooled, from room temperature to 80K. The resistance of [2003], (a) each of these decreases, (b) copper strip increases and that of germanium, decreases, (c) copper strip decreases and that of germanium increases, (d) each of these increases, 46. The difference in the variation of resistance with, temeperature in a metal and a semiconductor arises, essentially due to the difference in the, [2003], (a) crystal sturcture, (b) variation of the number of charge carriers with, temperature, (c) type of bonding, (d) variation of scattering mechanism with temperature, 47. In the middle of the depletion layer of a reverse- biased, p-n junction, the, [2003], (a) electric field is zero, (b) potential is maximum, (c) electric field is maximum, (d) potential is zero, 48. At absolute zero, Si acts as, [2002], (a) non-metal, (b) metal, (c) insulator, (d) none of these, 49. By increasing the temperature, the specific resistance of a, conductor and a semiconductor, [2002], (a) increases for both, (b) decreases for both, (c) increases, decreases (d) decreases, increases, 50. The energy band gap is maximum in, [2002], (a) metals, (b) superconductors, (c) insulators, (d) semiconductors., , TOPIC 2 Junction Transistor, , –10 V, , +5 V, R, , (IB ), 60 mA, 50 mA, 40 mA, 30 mA, 20 mA, 10 mA, , R, , 8, , (IC) in mA, , (d), , 51. The output characteristics of a transistor is shown in the, figure. When VCE is 10 V and IC = 4.0 mA, then value of bac, is ______., [NA Sep. 06, 2020 (II)], , 6, , 43., , The electrical conductivity of a semiconductor increases, when electromagnetic radiation of wavelength shorter than, 2480 nm is incident on it. The band gap in (eV) for the, semiconductor is, [2005], (a) 2.5 eV (b) 1.1 eV (c) 0.7 eV (d) 0.5 eV, , 4, 2, 2 4, , 6 8 10 12 14, (VCE) in volts
Page 488 :
P-479, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 52. The transfer characteristic curve of a transistor, having, input and output resistance 100 W and 100 k W, respectively, is shown in the figure. The Voltage and Power, gain, are respectively :, [12 Apr. 2019 I], , IC, C, , RB, B, , 56., , (a) 2.5×104, 2.5×106, (b) 5×104, 5×106, (c) 5×104, 5×105, (d) 5×104, 2.5×106, 53. An npn transistor operates as a common emitter amplifier,, with a power gain of 60 dB. The input circuit resistance is, 100W and the output load resistance is 10 kW. The common, emitter current gain b is :, [10 Apr. 2019 I], (a) 102, (b) 60, (c) 6×102, (d) 104, 54. An NPN transistor is used in common emitterconfiguration, as an amplifier with 1 k &! load resistance. Signal voltage, of 10 mV is applied across the base-emitter. This produces a, 3 mA change in the collector current and 15 ¼A change in, the base current of the amplifier. The input resistance and, voltage gain are:, [9 April 2019 I], (a) 0.33 k W 1.5, (b) 0.67 k W 300, (c) 0.67 k W 200, (d) 0.33 k W 300, 55. A common emitter amplifier circuit, built using an npn, transistor, is shown in the figure. Its dc current gain is 250,, RC = 1 k &! and VCC = 10V. What is the minimum base current, for VCE to reach saturation ?, [8 Apr. 2019 II], , v1 ~, , VBB, , (b) 100 mA, (d) 10 mA, , VCC, , IE, , ~ v0, , In the figure, given that VBB supply can vary from 0 to 5.0, V, VCC = 5 V, bdc = 200, RB = 100 kW, RC = 1 KW and, VBE = 1.0 V. The minimum base current and the input, voltage at which the transistor will go to saturation, will, be, respectively :, [12 Jan. 2019 II], (a) 25 µA and 3.5 V, (b) 20 µA and 3.5 V, (c) 25 µA and 2.8 V, (d) 20 µA and 2.8 V, 57. In a common emitter configuration with suitable bias, it is, given than RL is the load resistance and RBE is small signal, dynamic resistance (input side). Then, voltage gain, current, gain and power gain are given, respectively, by:, (b is current gain, IB, IC, IE are respectively base, collector, and emitter currents:), [Online April 15, 2018], RL DI E 2 RL, ,, ,b, RBE DI B, RBE, , (a), , b, , (b), , b2, , RL DI C, R, ,, ,b L, RBE DI B RBE, , (c), , b2, , RL DIC 2 RL, ,, ,b, RBE DI E, RBE, , (d), , b, , RL DIC 2 RL, ,, ,b, RBE DI B, RBE, , 58. The current gain of a common emitter amplifier is 69. If the, emitter current is 7.0 mA, collector current is :, [Online April 9, 2017], (a) 9.6 mA, (b) 6.9 mA, (c) 0.69 mA, (d) 69 mA, 59. In a common emitter amplifier circuit using an n-p-n, transistor, the phase difference between the input and the, output voltages will be :, [Online April 2, 2017], (a) 135°, (b) 180°, (c) 45°, (d) 90°, 60. For a common emitter configuration, if a and b have their, usual meanings, the incorrect relationship between a and, b is :, [2016], (a) a =, , (a) 40 mA, (c) 7 mA, , E, , RC, , (c), , b, 1 +b, , 1 1, = +1, a b, , (b) a =, , b2, 1 +b 2, , (d) None of these
Page 489 :
P-480, , 61., , Physics, , A realistic graph depicting the variation of the reciprocal, of input resistance in an input characteristics measurement, in a common emitter transistor configuration is :, [Online April 10, 2016], (a) 1, ri, , (b), , (c), , (W - 1 ), , 0.01, , 0, , 0.6, VBE (V), , 1, , 66., , 0, , 0.6, VBE (V), , 1, , 67., , 0, , 0.6, VBE (V), , 1, , 1 -1, (W ), ri, 0.01, , 1 -1, (W ), ri, 10, , (d), , 1 -1, (W ), ri, 0.1, , 69., 0, , 62., , 63., , 64., , 65., , 68., , 0.6, VBE (V), , 1, , The ratio (R) of output resistance r 0 , and the input, resistance r i in measurements of input and output, characteristics of a transistor is typically in the range :, [Online April 10, 2016], (a) R ~ 102 – 103, (b) R ~ 1 – 10, (c) R ~ 0.1 – 1.0, (d) R ~0.1 – 0.01, An unknown transistor needs to be identified as a npn or, pnp type. A multimeter, with +ve and –ve terminals, is used, to measure resistance between different terminals of, transistor. If terminal 2 is the base of the transistor then, which of the following is correct for a pnp transistor?, [Online April 9, 2016], (a) +ve terminal 2, –ve terminal 3,resistance low, (b) +ve terminal 2, –ve terminal 1, resistane high, (c) +ve terminal 1, –ve terminal 2, resistance high, (d) +ve terminal 3, –ve terminal 2, resistance high, An n-p-n transistor has three leads A, B and C. Connecting, B and C by moist fingers, A to the positive lead of an, ammeter, and C to the negative lead of the ammeter, one, finds large deflection. Then, A, B and C refer respectively, to:, [Online April 9, 2014], (a) Emitter, base and collector, (b) Base, emitter and collector, (c) Base, collector and emitter, (d) Collector, emitter and base., A working transistor with its three legs marked P, Q and R, is tested using a multimeter. No conduction is found, , 70., , between P and Q. By connecting the common (negative), terminal of the multimeter to R and the other (positive), terminal to P or Q, some resistance is seen on the multimeter., Which of the following is true for the transistor? [2008], (a) It is an npn transistor with R as base, (b) It is a pnp transistor with R as base, (c) It is a pnp transistor with R as emitter, (d) It is an npn transistor with R as collector, In a common base mode of a transistor, the collector current, is 5.488 mA for an emitter current of 5.60 mA. The value of, the base current amplification factor (b) will be [2006], (a) 49, (b) 50, (c) 51, (d) 48, In a common base amplifier, the phase difference between, the input signal voltage and output voltage is, [2005], p, (a) p, (b), 4, p, (d) 0, (c), 2, When npn transistor is used as an amplifier, [2004], (a) electrons move from collector to base, (b) holes move from emitter to base, (c) electrons move from base to collector, (d) holes move from base to emitter, For a transistor amplifier in common emitter configuration, for load impedance of 1k W (h fe = 50 and hoe = 25) the, current gain is, [2004], (a) – 24.8, (b) – 15.7, (c) – 5.2, (d) – 48.78, The part of a transistor which is most heavily doped to, produce large number of majority carriers is, [2002], (a) emmiter, (b) base, (c) collector, (d) can be any of the above three., , TOPIC 3 Digital Electronics and Logic, Gates, 71. Identify the correct output signal Y in the given, combination of gates (as shown) for the given inputs A, and B., [Sep. 06, 2020 (I)], B, Y, A, , A, B, 5, , 10, , 15, , 20, , t
Page 491 :
P-482, , Physics, , 78. The logic gate equivalent to the given logic circuit is:, [9 Apr. 2019 II], , b, a, x, , (a) NAND (b) OR, (c) NOR, (d) AND, 79. The ouput of the given logic cirfcuit is: [12 Jan. 2019 I], A, , (a) OR, (b) NAND, (c) NOT, (d) AND, 83. To get an output of 1 from the circuit shown in figure the, input must be :, [Online April 10, 2016], a, b, , Y, B, (a) AB + AB, (b) AB + AB, (c) AB, (d) AB, 80. To get output ‘1’ at R, for the given logic gate circuit, the input values must be:, [10 Jan. 2019 I], , Y, c, , (a) a = 0, b = 0, c = 1, (b) a = 1, b = 0, c = 0, (c) a = 1, b = 0, c = 1, (d) a = 0, b = 1, c = 0, 84. The truth table given in fig. represents :, [Online April 9, 2016], A B Y, , (a) X = 0, Y = 1, (b) X = 1, Y = 1, (c) X = 1, Y = 0, (d) X = 0, Y = 0, 81. Truth table for the given circuit will be, [Online April 15, 2018], x, z, , (a), , (c), , y z, , 0 0 1, 0 1 1, 1 0 1, , (b), , y z, , 1 0, , 1 1 1, , x, , y z, , x, , 0 0 1, 0 1 1, 1 0 1, 1 1, , (d), , y z, , 0 0 0, 0 1 1, 1 0 1, 1 1 1, , If a, b, c, d are inputs to a gate and x is its output, then, as, per the following time graph, the gate is :, [2016], d, c, , B, , 0 0 0, 0 1 0, 1 0 0, , 1, , 1, , 82., , x, , 0, , 0, , 0, 1, , 1, 0, , 1, 1, , 1, , 1, , 1, , OR – Gate, AND– Gate, , (b) NAND– Gate, (d) NOR– Gate, , 50 W, , 85., A, , y, x, , (a), (c), , 0, , VCC – 6V, , 50 W, Vout, R = 10 kW, , Given: A and B are input terminals., Logic 1 = > 5 V, Logic 0 = < 1 V, Which logic gate operation, the above circuit does?, [Online April 19, 2014], (a) AND Gate, (b) OR Gate, (c) XOR Gate, (d) NOR Gate, 86. Identify the gate and match A, B, Y in bracket to check., [Online April 9, 2014], A, B, , (a), (b), (c), (d), , A.B, Y, , AND (A = 1, B = 1, Y = 1), OR (A = 1, B = 1, Y = 0), NOT (A = 1, B = 1, Y = 1), XOR (A = 0, B = 0, Y = 0)
Page 492 :
P-483, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 87., , Which of the following circuits correctly represents the, following truth table ?, [Online April 25, 2013], , 5V, , A B C, C, , 0 0 0, 0 1 0, , 1, A, , 1 0 1, 1 1 0, , 2, B, , (a), , C, , A, B, , A, (b) B, , C, , A, B, , C, , (c), , A, Y, , (d) A, B, 88., , (a) A NAND B, (b) A OR B, (c) A AND B, (d) A NOR B, 90. Truth table for system of four NAND gates as shown in, figure is :, [2012], , C, , A system of four gates is set up as shown. The ‘truth, table’ corresponding to this system is :, [Online April 23, 2013], , B, , A, , (a), Y, , B, , (a), , (c), , 89., , (c), A B Y, 0 0 1, 0 1 0, 1 0 0, 1 1 1, , (b), , A B Y, 0 0 0, 0 1 0, 1 0 1, 1 1 0, , (d), A B Y, A B Y, 0 0 1, 0 0 1, 0 1 1, 0 1 0, 1 0 0, 1 0 1, 1 1 0, 1 1 0, Consider two npn transistors as shown in figure. If 0 Volts, corresponds to false and 5 Volts correspond to true then, the output at C corresponds to : [Online April 9, 2013], , A, 0, 0, 1, 1, , B, 0, 1, 0, 1, , Y, 0, 1, 1, 0, , A, 0, 0, 1, 1, , B, 0, 1, 0, 1, , Y, 1, 1, 0, 0, , (b), , (d), , A, 0, 0, 1, 1, , B, 0, 1, 0, 1, , Y, 0, 0, 1, 1, , A, 0, 0, 1, 1, , B, 0, 1, 0, 1, , Y, 1, 0, 1, 1, , 91. The figure shows a combination of two NOT gates and a, NOR gate., [Online May 26, 2012], A, Y, B, , The combination is equivalent to a, (a) NAND gate, (b) NOR gate, (c) AND gate, (d) OR gate
Page 493 :
P-484, , 92., , 93., , Physics, , Which one of the following is the Boolean expression for, NOR gate?, [Online May 19, 2012], , A, , (a), , B, , (b) Y = A.B, , Y = A+ B, , (c) Y = A.B, (d) Y = A, Which logic gate with inputs A and B performs the same, operation as that performed by the following circuit?, A [Online May 7, 2012], B, , (a) NAND gate, (b) OR gate, , 94., , 95., , Lamp, , A, , (b) NOT gate, (c) XOR gate, 96., , Input B, , (a), , (d) AND gate, The output of an OR gate is connected to both the inputs, of a NAND gate. The combination will serve as a:, [2011 RS], (a) NOT gate, (b) NOR gate, (c) AND gate, (d) OR gate, The combination of gates shown below yields [2010], (a) OR gate, , Input A, , Output is, , V, , (c) NOR gate, , (b), (c), (d), 97. In the circuit below, A and B represent two inputs and C, represents the output., [2008], A, C, , X, B, , Y, , (d) NAND gate, The logic circuit shown below has the input waveforms, ‘ A’ and ‘B’ as shown. Pick out the correct output, waveform., [2009], , B, The circuit represents, (a) NOR gate, (c) NAND gate, , (b) AND gate, (d) OR gate
Page 494 :
P-485, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 1., , (d) I-V characteristic of a photodiode is as follows :, , VR = 1.5 - 0.5 = 1 volt, iR = 1 (=VR), , mA, , 1, 1, \ Rmin = = -2 = 100 W, i 10, , V, , 5., , Reverse bias, , mA, On increasing the biasing voltage of a photodiode, the, magnitude of photocurrent first increases and then attains, a saturation., 2., , (c) Till input voltage reaches 4 V. No zener is in breakdown, region so V0 = Vi. Then now when Vi changes between 4 V, to 6 V one zener with 4 V will breakdown and P.D. across, this zener will become constant and remaining potential, will dorp across resistance in series with 4 V zener., Now current in circuit increases abruptly and source must, have an internal resistance due to which some potential, will get drop across the source also so correct graph, between V0 and t will be, V0, 6V, , \ Band gap E g =, , (12) Right hand diode is reversed biased and left hand, diode is forward biased., Hence Voltage at ‘A’, VA = 12.7 – 0.7 = 12 volt, , 7., , (c) Both the diodes are reverse biased, so, there is no, flow of current through 5W and 20W resistances., Now, two resistors of 10W and two resistors of 5W are in, series., Hence current I through the network = 0.3 A, , 8., , (c) In case I diode is reverse biased, so no current flows, \ QA = CV, In case II, current will flow as diode is forward biased. So,, it offers negligible resistance to the flow of current and, thus be replaced by short circuit. Now, the charge of, capacitor will leak through the resistance and decay, exponentially with time., During discharging of capacitor, Potential difference across the capacitor at any instant, , t, , 4., , (c) Here two zener diodes are in reverse polarity so if one, is in forward bias the other will be in reverse bias and, above 6V the reverse bias will too be in conduction mode., Hence when V > 6V the output will be constant. And when, V < 6V it will follow the inut voltage., (c) According to question, when diode is forward biased,, Vdiode = 0.5 V, Safe limit of current, I = 10 mA = 10–2 A, Rmin = ?, –2, , 10 A, 0.5 volt, 1.5 V, Voltage through resistance, , hc 1237.5, =, = 3.09 eV, l, 400, , 6., , 4V, , 3., , (d) Given,, Wavelength of photon, l = 400 nm, A photodiode can detect a wavelength corresponding to, the energy of band gap. If the signal is having wavelength, greater than this value, photodiode cannot detect it., , -, , t, CR, , V ' = Ve, But t = CR, , V ' = Ve -1 =, , \, , R, , 9., , V, e, , Charge QB = CV’ =, , CV, e, , 12 – 8, = 10 –2 A, 400, Power dissipited in each diode, P = VI, Þ P = 4 × 10–2 = 40 mW, , (40) Current in the circuit, I =
Page 495 :
P-486, , Physics, , 10. (a) The given circuit has two 10kW resistances in parallel,, so we can reduce this parallel combination to a single, equivalent resistance of 5kW., , 14. (a) Since voltage across zener diode does not reach to, breakdown voltage therefore its resistance will be infinite, & current through it is 0., 15. (b) As D2 is reversed biased, so no current through 75W, resistor., now Req= 150 + 50 + 100, = 300 W, , 11., , Diode is in forward bias. So it will behave like a conducting, wire., 30, ´ 5 = 10 V, VA – VB =, 5 + 10, (d) Current in load resistance,, , 6, , i1 =, , 3, , = 1.5 × 10–3 A = 1.5 mA, , So, required current I =, I=, , is =, \, , 2 ´ 103, , 6, = 0.02, 300, , 16. (a) The voltage across zener diode is constant, , 4 ´10, For V = 16 volt,, (16 - 6), , BatteryVoltage, 300, , i i –i1, R2, 5kW(R1), , 120V, , 10kW, , = 5 mA, , i2 = is – i1 = 5 – 1.5 = 3.5 mA, , 12. (c), , i( R 2 ) =, , V, 50, =, = 5´10 –3 A, R 10 ´103, , i ( R1 ) =, , V 120 – 50, 70, 14 ´ 10–3 A, =, =, 3, 3, R 5 ´10, 5´10, , \ izenerdiode= 14 × 10–3 – 5 × 10–3 = 9 × 10–3 A = 9 mA, 17. (c) As we know, current density,, j = sE = nevd, s = ne, , For voltage, V = 8V, , 3ö 1, æ, Current, I1 = ç 8 - 6 - ÷ = = 0.5mA, 2ø 2, è, For voltage, V = 16V, 3ö, æ, Current, I 2 = ç16 - 6 - ÷ = 8.5 mA, 2ø, è, 13. (a), , vd, = nem, E, , 1, 1, =r=, =Resistivity, s, n e eme, , =, or, , 1, 19, , 10 ´ 1.6 ´ 1019 - 19 ´ 1.6, P = 0.4 Wm, , 18. (d) Initially Ge and Si are both forward biased so current will, effectivily pass through Ge diode \ V° = 12 – 0.3 =11.7 V, And if "Ge" is revesed then current will flow through "Si", diode, \ V° = 12 – 0.7 =11.3 V, Clearly, V° changes by 11.7 – 11.3 = 0.4 V, , P.D. across 800W resistors = 5.6 V, 5.6, A = 7 mA, 800, Now, P.D. across 200W resistors = 9 – 5.6 V = 3.4 V, , so, I 800 W =, , 9 - 5.6, = 17 mA, 200, so, current through zener diode = I 2 = 17 – 7 = 10 mA, , so, I 200 W =, , 19. (c) Clearly from fig. given in question, Silicon diode is, in forward bias., \ Potential barrier across diode, DV = 0.7 volts, Current, I =, , V - DV 3 - 0.7 2.3, =, =, = 11.5mA, R, 200, 200
Page 496 :
P-487, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 20. (b), , 25. (c) The minimum voltage range of DC source is given by, V2 = PR, Q P = 1 watt, R = 100W, = 1×100, \ V = 10 volt., , I, R1, , 500W, , IR2, , 15 V, , IZ, VZ = 10 V, , 1500W R2, , The voltage drop across R2 is VR = VZ = 10V, 2, The current through R2 is, IR 2 =, , VR 2, R2, , 26. (d) Using Uav =, But U av =, P, , \, , 10V, =, = 0.667 ´ 10-2 A, 1500W, , 4p r, , = 6.67 × 10–3 A = 6.67 mA, The voltage drop across R1 is, , E 20 =, , VR1 = 15V - VR 2 = 15V - 10V = 5V, , \, , VR1, R1, , =, , P, 4p r 2 ´ c, =, , 1, e0 E02 ´ c, 2, , 2P, 4 pr 2 e 0 c, , =, , 2 ´ 0.1 ´ 9 ´ 109, 1 ´ 3 ´ 108, , E0 = 6 = 2.45V/m, , 27. (a) When positive terminal connected to A then diode, , The current through R1 is, IR1 =, , 2, , 1, e 0 E02, 2, , 5V, = 10-2 A = 10 ´ 10 -3 A = 10 mA, 500W, , D1 is forward biased, current, I =, , 2, = 0.4 A, 5, , The current through the zener diode is, , When positive terminal connected to B then diode D2, , IZ = IR1 - IR 2 = (10 - 6.67)mA = 3.3mA, , is forward biased, current, I =, , 21. (a) The conductivity of semiconductor, s = e (heµe + hhµh), , E 20 =, , = 1.6 × 10 –19(5 × 1018 ×2 + 5 × 1019 × 0.01), = 1.6 × 1.05 = 1.68, , DV, 0.1, 22. (b) Forward bias resistance =, =, = 10 W, DI 10 ´ 10-3, 10, Reverse bias resistance =, = 107 W, -6, 10, Forward bias resistance, Ratio of resistances =, = 10–6, Reverse bias resistance, 23. (c) Graph (p) is for a simple diode., Graph (q) is showing the V Break down used for zener, diode., Graph (r) is for solar cell which shows cut-off voltage, and open circuit current., Graph (s) shows the variation of resistance h and hence, current with intensity of light., 24. (a), , r, , \, , 2 ´ 0.1 ´ 9 ´ 109, 1 ´ 3 ´ 108, , 28. (d) Electrons in an unbiased p-n junction, diffuse from, n -region i.e. higher electron concentration to p-region, i.e. low electron concentration region., n, 29. (a) P, For forward bias, p-side must be at higher potential than, n-side. DV = (+ )Ve, 30. (d) Energy band gap range is given by,, Eg =, , hc, l, , (, , ), , -7, -7, For visible region l = 4 ´ 10 : 7 ´ 10 m, , Eg =, , =, , B, , 4 pr 2 e 0 c, , =, , E0 = 6 = 2.45V/m, , P, , T, T, Semiconductor, Metal (for limited, –Eg, KT, range of temperature), , 2P, , 2, = 0.2 A, 10, , =, , 6.6 ´ 10 -34 ´ 3 ´ 108, 7 ´ 10 -7, 19.8 ´ 10-26, 7 ´ 10 -7, 2.8 ´ 10-19, 1.6 ´ 10-19, , Eg = 1.75 eV
Page 497 :
P-488, , Physics, , 31. (d) Here, R = 4 kW = 4 × 103 W, Vi = 60 V, Zener voltage Vz = 10 V, RL = 2 kW = 2 × 103 W, Load current, IL =, , VZ, 10, =, = 5 mA, RL 2 ´ 103, , Current through R, I =, =, , 60 - 10, 4 ´ 103, , =, , 50, 4 ´ 103, , Vi - VZ, R, , = 12.5 mA, , Fom circuit diagram,, I = IZ + IL, Þ 12.5 = IZ + 5, Þ IZ = 12.5 – 5 = 7.5 mA, 32. (a) For same value of current higher value of voltage is, required for higher frequency hence (a) is correct answer., 33. (a) Here, diodes D1 and D2 are forward biased and D3 is, reverse biased., Therefore current through R3, i=, , V, 6, 1, =, =, A = 50 mA, R ' 120 20, , 34. (d) Temperature coefficient of resistance is negative for, pure semiconductor. And no. of charge carriers in, conduction band increases with increase in temperature., 35. (b) The given circuit will work as half wave rectifier as it, conducts during the positive half cycle of input AC., Forward biased in one half cycle and reverse biased in, the other half cycle]., 36. (a) The current will flow through RL when the diode is, forward biased., 37. (a) Si and Ge are semiconductors but C is an insulator. In, Si and Ge at room temperature, the energy band gap is low, due to which electrons in the covalent bonds gains kinetic, energy and break the bond and move to conduction band., As a result, hole is created in valence band. So, the number, of free electrons is significant in Si and Ge., 38. (c) A crystal structure is made up of a unit cell, arranged in a particular way; which is periodically, repeated in three dimensions on a lattice. The spacing, between unit cells in various directions is called its, lattice constants. As lattice constants increases the, band-gap (E g ), also increases which means more, energy would be required by electrons to reach the, conduction band from the valence band. Automatically, E c and E v decreases., 39. (b) Van der Waal's bonding is attributed to the attractive, forces between molecules of a liquid. The conductivity of, , semiconductors (covalent bonding) and insulators (ionic, bonding) increases with increase in temperature., Solid which is formed by covalent bond is not transparent, to visible light and its conductivity increase with, temperature., 40. (c) Relation between drift velocity and current is, I = nAeVd, Ie, Ih, Þ, , =, , ne eAve, nh eAvh, , 7 7 ve, = ´, 4 5 vh, , ve 5, =, vh 4, 41. (b) D2 is forward biased., D1 is reversed biased. So, it will act like an open circuit., So effective resistance of the circuit, Þ, , R = 4 + 2 = 6W, , \i =, , E 12, =, =2A, R 6, , 42. (d) p-side connected to low potential and n-side is, connected to high potential., 43. (d) Band gap = energy of photon of wavelength 2480, nm. So,, Band gap, E g =, , hc, l, , æ 6.63 ´ 10 - 34 ´ 3 ´ 108 ö, 1, eV, ÷´, -9, -19, ÷, 2480 ´ 10, è, ø 1.6 ´ 10, , = çç, , = 0.5 eV, 44. (a) In forward biasing, the p type is connected to positive, terminal and n type is connected with negative terminal., So holes from p region and electron from n region are, pushed towards the Junction which reduces the width of, depletion layer. Also, distance between diffused holes, and electrons decrease, which decrease electric field, hence barrier potential., 45. (c) Copper is a conductor and in conductor resistance, decreases with decrease in temperature. Germanium is a, semicon ductor. In semi-conductor resistance increases, with decrease in temperature., 46. (b) When the temperature increases, certain bounded, electrons become free which tend to promote, conductivity. Simultaneously number of collisions, between electrons and positive kernels increases which, decrease the relaxation time., 47. (a) In reverse biasing the width of depletion region, increases, and current flowing through diode is zero. Thus,, electric field is zero at middle of depletion region.
Page 499 :
P-490, , Physics, , 58. (b) Given, current gain of CE amplifier b = 69, IE = 7 mA, IC, or, = 69, IB, b, 69 I C, We know that, a =, =, =, 1 + b 70 I E, 69 69, =, ´7, 70 70, Collector current, IC = 6.9 mA, IC = I E ´, , 59. (b) In common emitter configuration for n-p-n transistor, input and output signals are 180° out of phase i.e., phase, difference between output and input voltage is 180°., 60. (b, d)We know that a =, , Ic, Ic, and b =, Ib, Ie, , Also Ie = Ib + Ic, Ic, I, Ic, b, \ a=, = b =, Ic 1 + b, Ib + Ic, 1+, Ib, , Option (b) and (d) are therefore incorrect., I, slope = 1/r, , 61. (c), , P, , Þ, VBE, , ri, 62. (c) For C.B. configuration r ? 0.1ς, o, , 2, , +ve terminal 1, –ve terminal 2, resistance high., 64. (c) In the given question, A, B and C refer base, collector, and emitter respectively., 65. (b) It is a p-n-p transistor with R as base., 66. (a) Collector current, IC = 5.488 mA,, Emitter current Ie = 5.6 mA, a=, , Ic 5.488, ,, =, Ie, 5.6, , b=, , a, = 49, 1- a, , 67. (d) In common base amplifier circuit, input and output, voltage are in the same phase. So, the phase difference, between input voltage signal and output voltage signal is, zero., 68. (c) In npn transistor, electrons moves from emitter to base., 69. (d) In common emitter configuration for transistor, amplifier current gain, , -h f e, 1 + boe RL, , Where hfe and hoe are hybrid parameters., , 63. (c) Connecting circuit according to question, it is clear, , -50, , 1 + 25 ´ 10-6 ´ 1 ´ 103, = – 48.78, 70. (a) Emitter main function is to supply the majority charge, carrriers towards the collector. Therefore emitter is most, heavily doped., 71. (a) Boolean expression,, y = A×B = A + B = A+ B, B, , For CE and CC-configuration, , ri, » 1ς ., r0, , C, , B, , \ Ai =, , 1/r, , n, , 1, , Ai =, V, , P, , E, , A, , B, , A, , Truth table :, A B Y, 0, , 1, , 1, , 1, 0, , 0, 0, , 1, 0, , 1, , 1, , 1
Page 500 :
P-491, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 1, A 0, , 0, , B 1, , 1, , 1, 0, , 1, , 0, , 0, , 1, , 1, , 1, , 1, , 0, y 1, , 1, , 1, , 0, , 1, , 74. (d), , 0, 1, , 0, 0, , 0, 1, , 1, , y=A+B, , 72., , (c) When two inputs of NAND gate is shorted, it behaves, like a NOT gate so boolen equation will be, , OR + NOT ® NOR Gate, Hence Boolean relation at the output stage – Y for the, circuit,, Y = A + B = A.B, , y = A+ B +C, , 75. (a), , y = A× B × C, A B C, 0, 1, , 0, 0, , 0, 0, , 0, 0, , 0, , 1, , 0, , 0, , 0, 1, , 0, 1, , 1, 0, , 0, 0, , 1, , 0, , 1, , 0, , 0, 1, , 1, 1, , 1, 1, , 0, 1, , Y = AB. A = AB + A = AB + A, For A = 1, B = 0, Y = (1) × 0 + 0, ÞY=0+0=0, 76. (d) A logic gate is reversible if we can recover input data, from the output. Hence NOT gate., 77. (c), , Thus, whole arrangement behaves like a AND gate., (a), , 73., , A, , A, , A× B, , A, B, , x = W + A× B, , A× B, , B, , W = ( A × B ) × ( A + B), , A+ B, , B, , A, , B, , (A + B), , (A + B). A, , 0, , 0, , 0, , 0, , (A + B).A, 1, , 0, 1, 1, , 1, 0, 1, , 1, 1, 1, , 0, 1, 1, , 1, 0, 0, , 78. (b) Truth table ®, The output is of OR-gate, , A, , B, , A× B, , A+ B, , 1, , 0, , 1, , 0, , 0, , 1, , 0, , 0, , 1, , 1, , 0, , 0, , 1, , 0, , 1, , 1, , 0, , 0, , 0, , 0, , 1, , 0, , 0, , 1, , 1, , 1, , 0, , 0, , W = ( A × B) × ( A + B) Q = W + A × B, , Q=x, , A B, , A B, , A.B, , 0, 0, , 0, 1, , 1, 1, , 1, 0, , 0, 1, , 1, , 0, , 0, , 1, , 1, , 1, , 1, , 0, , 0, , 1
Page 501 :
P-492, , Physics, , 86. (a), A, B, , A, , 79. (c) A, , A.B, , Y = (A. AB). (AB + B), , AB, , A.B, B, , B, , (, , A. AB, , ), , Y = AB.AB = AB + AB = AB + AB = AB, , A.B = Y+A = C, , = A × AB + AB × B, , A, , Y+A = C, 0 0 0 1 0, 0 1 0 1 0, 1 0 0 0 1, 1 1 1 0 0, 88. (a) In the given system all four gate is NOR gate, Truth Table, , = A × ( A + B) + AB × B, , = AB, 80. (c) From the given logic circuit,, p= x+y, Q = y.x = y + x, Output, R = P + Q, To make output 1, P + Q must be ‘0’, So, x = 1, y = 0, 81. (c) Truth table of the circuit is as follows, x y, 0 0, , x a = x. y b = x . y, 1, 0, 0, , 0 1, 1 0, , 1, 0, , 0, 0, , 1, 0, , 1, 1, , 1 1, , 0, , 1, , 0, , 1, , A, 0, 0, 1, 1, , B, , B (y ' = A + B) y '' = (A + y ') y ''' = (A + y '') y = y ''+ y ''', 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, , i.e.,, , z = a.b, 1, , a, , b, , (a + b ), , c, , Y = (a + b) .c, , 0, , 0, , 0, , 0, , 0, , 0, , 1, , 1, , 1, , 1, , 1, , 0, , 1, , 1, , 1, , 1, , 1, , 1, , 0, , 0, , Output of OR gate must be 1 and c = 1, So, a = 1, b = 0 or a = 0, b = 1., 84. (a) It represents OR-Gate., A B A + B =Y, 0 0, 0, 1, Y 0 1, 1 0, 1, 1 1, 1, , A, , B y, , 0, 0, , 0, 1, , 1, 0, , 1, 1, , 0, 1, , 0, 1, , 89. (a) The output at C corresponds to A NAND B or, A×B = C, 90. (a), A, , Y = A.AB B.AB, , 82. (a) In case of an 'OR' gate the input is zero when all inputs, are zero. If any one input is ' 1', then the output is '1'., 83. (c) Truth table for given logical circuit, , 85. (a) AND Gate, , Y, , In this case output Y is equivalent to AND gate., 87. (a) For circuit 1, , (AB + B), , Y = A × AB + ( AB + B ), , A, B, , A. B, , Y2 = A.AB, Y1 = AB, B, Y3 = B.AB, , By expanding this Boolen expression, , Y = A.B + B.A, Thus the truth table for this expression should be (a)., 91. (c) Truth table is as shown :, A, 0, , B A, 0 1, , B A+B A+B, 1, 1, 0, , 0, , 1, , 1, , 0, , 1, , 0, , 1, , 0, , 0, , 1, , 1, , 0, , 1, , 1, , 0, , 0, , 0, , 1
Page 502 :
P-493, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Thus the combination of two NOT gates and one NOR, gate is equivalent to a AND gate., 92. (a) NOR gate is the combination of NOT and OR gate., , 95. (a) The final boolean expression of these gates is,, X = ( A . B ) = A + B = A + B Þ OR gate, , It means OR gate is formed., 96. (d) The final boolean expression, , Boolean expression for NOR gate is, Y = A+ B, 93. (b) When either of A or B is 1 i.e. closed then lamp will, glow., In this case, Truth table, Inputs, , Output, , A, , B, , Y, , 0, , 0, , 0, , 0, , 1, , 1, , 1, , 0, , 1, , 1, , 1, , 1, , 0, 1, 1, , 1, 0, 1, , 0, 0, 1, , Y = ( A + B ) = A.B = A × B ., , Thus, it is an AND gate for which truth, table is, 97. (d) A, C, , This represents OR gate., 94. (b) When both inputs of NAND gate are jointed to form, a single input, it behaves as NOT gate, OR + NOT = NOR., , ( A + B) = NOR gate, A, B, , A B Y, 0 0 0, , (A+B), , (A+B), , B, , The truth table for the above circuit is :, A, 1, 1, 0, 0, , B, 1, 0, 1, 0, , C, 1, 1, 1, 0, , when either A or B conducts, the gate conducts. It means, C = A + B which is for OR gate.
Page 503 :
29, , P-494, , Physics, , Communication, Systems, 6., , TOPIC 1 Communication Systems, 1., , An amplitude modulated wave is represented by the, expression vm = 5(1 + 0.6 cos 6280t ) sin(211´104 t ) volts, The minimum and maximum amplitudes of the amplitude, modulated wave are, respectively : [Sep. 02, 2020 (I)], 5, 3, V, 5 V, (b), V, 8 V, 2, 2, (c) 5 V, 8 V, (d) 3 V, 5 V, In an amplitude modulator circuit, the carrier wave is given, by, C(t) = 4 sin (20000 pt) while modulating signal is given, by, m(t) = 2 sin (2000 pt). The values of modulation index, and lower side band frequency are : [12 April 2019 II], (a) 0.5 and 10 kHz, (b) 0.4 and 10 kHz, (c) 0.3 and 9 kHz, (d) 0.5 and 9 kHz, A message signal of frequency 100 MHz and peak, voltage100 V is used to execute amplitude modulation on, a carrier wave of frequency 300 GHz and peak voltage 400, V. The modulation index and difference between the two, side band frequencies are :, [10 April 2019 II], (a) 4 ; 1×108 Hz, (b) 4 ; 2×108 Hz, (c) 0.25 ; 2×108 Hz, (d) 0.25 ; 1 × 10–8T, A signal Acoswt is transmitted using v0 sin w0t as carrier, wave. The correct amplitude modulated (AM) signal is:, [9 April 2019 I], , (a), , 2., , 3., , 4., , (a) v0 sinw0 t + A sin ( w 0 w ) t + A sin ( w 0 + w ) t, ¯, 2, 2, (b) v0 sin[w0(1 + 0.01 Asinwt)t], , The wavelength of the carrier waves in a modern optical, fiber communication network is close to :, [8 April 2019 I], (a) 2400 nm (b) 1500 nm (c) 600 nm (d) 900 nm, 7. In a line of sight ratio communication, a distance of about, 50 km is kept between the transmitting and receiving, antennas. If the height of the receiving antenna is 70m,, then the minimum height of the transmitting antenna, should be :, [8 April 2019 II], (Radius of the Earth = 6.4 × 106 m)., (a) 20 m, (b) 51 m, (c) 32 m, (d) 40 m, 8. A 100 V carrier wave is made to vary between 160 V and, 40 V by a modulating signal. What is the modulation index?, [12 Jan. 2019 I], (a) 0.3, (b) 0.5, (c) 0.6, (d) 0.4, 9. To double the covering range of a TV transmittion tower,, its height should be multiplied by:, [12 Jan 2019 II], 1, (a), (b) 2, (c) 4, (d) 2, 2, 10. An amplitude modulated signal is given by V(t) = 10[1 +, 0.3 cos (2.2 × 104t)]sin(5.5 × 105t). Here t is in seconds., The sideband frequencies (in kHz) are, [Given p = 22/7], [11 Jan 2019 II], (a) 1785 and 1715, (b) 178.5 and 171.5, (c) 89.25 and 85.75, , (d) 892.5 and 857.5, , 11. An amplitude modulated signal is plotted below :, 10 V, 8V, , V (t), t, , (c) v0 sinw0t + Acoswt, 5., , (d) (v0 + A) coswt sinw0t, The physical sizes of the transmitter and receiver antenna, in a communication system are:, [9 April 2019 II], (a) independent of both carrier and modulation frequency, (b) inversely proportional to carrier frequency, (c) inversely proportional to modulation frequency, (d) proportional to carrier frequency, , 8 ms, 100 ms, Which one of the following best describes the above, signal?, [11 Jan. 2019 II], (a) (9 + sin (2.5 p × 105 t)) sin (2p × 104t)V, (b) (1 + 9 sin (2p × 104t)) sin (2.5 p × 105t) V
Page 504 :
P-495, , Communication Systems, , (c) (9 + sin (2p× 104t)) sin (2.5 p × 105t)V, (d) (9 + sin (4p × 104t)) sin (5p × 105t) V, 12. A TV transmission tower has a height of 140 m and the, height of the receiving antenna is 40 m. What is the, maximum distance upto which signals can be broadcasted, from this tower in LOS (Line of Sight) mode? (Given :, radius of earth = 6.4 × 106 m)., [10 Jan. 2019 I], (a) 65 km, , (b) 48 km, , (c) 80 km, , (d) 40 km, , 13. The modulation frequency of an AM radio station is 250, kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency, will you allot?, [10 Jan. 2019 I], (a) 2750 kHz, (b) 2900 kHz, (c) 2250 kHz, (d) 2000 kHz, 14. In a communication system operating at wavelength 800, nm, only one percent of source frequency is available as, signal bandwidth. The number of channels accomodated, for transmitting TV signals of band width 6 MHz are (Take, velocity of light c = 3 × 108m/s, h = 6.6 × 10–34J-s), [9 Jan. 2019 II], (a) 3.75 × 106, (b) 3.86 × 106, (c) 6.25 × 105, (d) 4.87 × 105, 15. A telephonic communication service is working at carrier, frequency of 10 GHz. Only10% of it is utilized for, transmission. How many telephonic channels can be, transmitted simultaneously if each channel requires a, bandwidth of 5 kHz?, [2018], (a) 2 × 103 (b) 2 × 104 (c) 2 × 105 (d) 2 × 106, 16. A carrier wave of peak voltage 14 V is used for transmitting, a message signal. The peak voltage of modulating signal, given to achieve a modulation index of 80% will be: [2018], (a) 11.2 V, (b) 7 V, (c) 22.4 V, (d) 28 V, 17. The number of amplitude modulated broadcast stations, that can be accommodated in a 300 kHz band width for the, highest modulating frequency 15 kHz will be:, [Online April 15, 2018], (a) 20, (b) 10, (c) 8, (d) 15, 18. The carrier frequency of a transmitter is provided by a, tank circuit of a coil of inductance 49mH and a capactiance, of 2.5nF . It is modulated by an audio signal of 12kHz. The, frequency range occupied by the side bands is:, [Online April 15, 2018], (a) 18kHz – 30kHz, (b) 63kHz – 75kHz, (c) 442kHz – 466kHz, (d) 13482kHz – 13494kHz, 19. In amplitude modulation, sinusoidal carrier frequency used, is denoted by ωc and the signal frequency is denoted by, ω m . The bandwidth ( Dωm ) of the signal is such that Dωm, < ωc . Which of the following frequencies is not contained, in the modulated wave ?, [2017], (a) ω m + ω c, (b) ω c - ω m, (c) ω m, (d) ωc, , 20. A signal is to be transmitted through a wave of wavelength l, using a linear antenna. The length 1 of the antenna and effective power radiated Peff will be given respectively as :, (K is a constant of proportionality), [Online April 9, 2017], 2, , æ1ö, (a) l, Peff = K ç ÷, èlø, , (b), , l, æ1ö, , Peff = K ç ÷, 8, èlø, 1, , 3, l, l, æ 1 ö2, 1, (c), , Peff = K æç ö÷, (d), , Peff = K ç ÷, 16, 5, èlø, èlø, 21. A signal of frequency 20 kHz and peak voltage of 5 Volt is, used to modulate a carrier wave of frequency 1.2 MHz and, peak voltage 25 Volts. Choose the correct statement., [Online April 8, 2017], , (a) Modulation index = 5, side frequency bands are at, 1400 kHz and 1000 kHz, (b) Modulation index = 5, side frequency bands are at, 21.2 kHz and 18.8 kHz, (c) Modulation index=0.8, side frequency bands are at, 1180 kHz and 1220 kHz, (d) Modulation index=0.2, side frequency bands are at, 1220 kHz and 1180 kHz, 22. Choose the correct statement :, , [2016], , (a) In frequency modulation the amplitude of the high, frequency carrier wave is made to vary in proportion, to the amplitude of the audio signal., (b) In frequency modulation the amplitude of the high, frequency carrier wave is made to vary in proportion, to the frequency of the audio signal., (c) In amplitude modulation the amplitude of the high, frequency carrier wave is made to vary in proportion, to the amplitude of the audio signal., (d) In amplitude modulation the frequency of the high, frequency carrier wave is made to vary in proportion, to the amplitude of the audio signal., 23. A modulated signal Cm(t) has the form Cm(t) = 30 sin 300pt, + 10 (cos 200pt – cos 400pt). The carrier frequency fc, the, modulating frequency (message frequency) fw and the, modulation indix m are respectively given by :, [Online April 10, 2016], 1, (a) fc = 200 Hz; fw = 50 Hz; m =, 2, 2, (b) fc = 150 Hz; fw = 50 Hz; m =, 3, 1, 3, 1, (d) fc = 200 Hz; fw = 30 Hz; m =, 2, , (c) fc = 150 Hz; fw = 30 Hz; m =
Page 505 :
P-496, , Physics, , 24. An audio signal consists of two distinct sounds: one a, human speech signal in the frequency band of 200 Hz to, 2700 Hz, while the other is a high frequency music signal, in the frequency band of 10200 Hz to 15200 Hz. The ratio of, the AM signal bandwidth required to send both the signals, together to the AM signal bandwidth requried to send, just the human speech is :, [Online April 9, 2016], (a) 2, (b) 5, (c) 6, (d) 3, 25. A signal of 5 kHz frequency is amplitude modulated on a, carrier wave of frequency 2 MHz. The frequencies of the, resultant signal is/are :, [2015], (a) 2005 kHz, 2000 kHz and 1995 kHz, (b) 2000 kHz and 1995 kHz, (c) 2 MHz only, (d) 2005 kHz and 1995 kHz, 26. Long range radio transmission is possible when the radio, waves are reflected from the ionosphere. For this to happen, the frequency of the radio waves must be in the range:, [Online April 19, 2014], (a) 80 - 150 MHz, (b) 8 - 25 MHz, (c) 1 - 3 MHz, (d) 150 - 1500 kHz, 27. For sky wave propagation, the radio waves must have a, frequency range in between:, [Online April 12, 2014], (a) 1 MHz to 2 MHz, (b) 5 MHz to 25 MHz, (c) 35 MHz to 40 MHz, (d) 45 MHz to 50 MHz, 28. A transmitting antenna at the top of a tower has height 32, m and height of the receiving antenna is 50 m. What is the, maximum distance between them for satisfactory, communication in line of sight (LOS) mode?, [Online April 9, 2014], (a) 55.4 km (b) 45.5 km (c) 54.5 km (d) 455 km, 29. A diode detector is used to detect an amplitudemodulated, wave of 60% modulation by using a condenser of capacity, 250 picofarad in parallel with a load resistance 100 kilo, ohm. Find the maximum modulated frequency which could, be detected by it., [2013], , 33., , 34., , 35., , D, Signal, , 32., , C, , R, , (a) 10.62 MHz, (b) 10.62 kHz, (c) 5.31 MHz, (d), 5.31 kHz, 30. Which of the following modulated signal has the best, noise-tolerance ?, [Online April 25, 2013], (a) Long-wave, (b) Short-wave, (c) Medium-wave, (d) Amplitude-modulated, 31. Which of the following statement is NOT correct?, [Online April 23, 2013], (a) Ground wave signals are more stable than the sky, wave signals., , 36., , (b) The critical frequency of an ionospheric layer is the, highest frequency that will be reflected back by the, layer when it is vertically incident., (c) Electromagnetic waves of frequencies higher than, about 30 MHz cannot penetrate the ionosphere., (d) Sky wave signals in the broadcast frequency range, are stronger at night than in the day time., This question has Statement-1 and Statement-2. Of the, four choices given after the Statements, choose the one, that best describes the two Statements., Statement-1: Short wave transmission is achieved due to, the total internal reflection of the e-m wave from an, appropriate height in the ionosphere., Statement-2: Refractive index of a plasma is independent, of the frequency of e-m waves. [Online April 22, 2013], (a) Statement-1 is true, Statement-2 is false., (b) Statement-1 is false, Statement-2 is true., (c) Statement-1 is true, Statement-2 is true but, Statement -2 is not the correct explanation of, statement-1., (d) Statement-1 is true, Statement-2 is true and, Statement -2 is the correct explanation of Statement-1., If a carrier wave c(t) = A sin wct is amplitude modulated by, a modulator signal m(t) = A sin wmt then the equation of, modulated signal [Cm(t)] and its modulation index are, respectively, [Online April 9, 2013], (a) Cm (t) = A (1 + sin wm t) sin wc t and 2, (b) Cm (t) = A (1 + sin wm t) sin wm t and 1, (c) Cm (t) = A (1 + sin wm t) sin wc t and 1, (d) Cm (t) = A (1 + sin wc t) sin wm t and 2, A radar has a power of 1kW and is operating at a frequency, of 10 GHz. It is located on a mountain top of height 500 m., The maximum distance upto which it can detect object, located on the surface of the earth, (Radius of earth = 6.4 × 106m) is :, [2012], (a) 80 km, (b) 16 km (c) 40 km (d) 64 km, A radio transmitter transmits at 830 kHz. At a certain, distance from the transmitter magnetic field has amplitude, 4.82 × 10–11T. The electric field and the wavelength are, respectively, [Online May 26, 2012], (a) 0.014 N/C, 36 m, (b) 0.14 N/C, 36 m, (c) 0.14 N/C, 360 m, (d) 0.014 N/C, 360 m, Given the electric field of a complete amplitude modulated, wave as, ®, æ, ö, ˆ c 1 + Em cos w m t cos w c t ., E = iE, çè, ÷ø, Ec, Where the subscript c stands for the carrier wave and m, for the modulating signal. The frequencies present in the, modulated wave are, [Online May 19, 2012], , (a) w c and wc2 + w 2m, (b) wc , wc + wm and wc - wm
Page 506 :
Communication Systems, , (c) wc and w m, (d) wc and, , wc wm, , 37. A 10 kW transmitter emits radio waves of wavelength 500, m. The number of photons emitted per second by the, transmitter is of the order of, [Online May 12, 2012], (a) 1037, (b) 1031, (c) 1025, (d) 1043, 38. This question has Statement – 1 and Statement – 2. Of the, four choices given after the statements, choose the one, that best describes the two statements., [2011], Statement – 1 : Sky wave signals are used for long, distance radio communication. These signals are in general,, less stable than ground wave signals., Statement – 2 : The state of ionosphere varies from, hour to hour, day to day and season to season., , P-497, , (a) Statement–1 is true, Statement–2 is true, Statement–2, is the correct explanation of Statement–1., (b) Statement–1 is true, Statement–2 is true, Statement–2, is not the correct explanation of Statement – 1., (c) Statement – 1 is false, Statement – 2 is true., (d) Statement – 1 is true, Statement – 2 is false., 39. Which of the following four alternatives is not correct ?, We need modulation :, [2011 RS], (a) to reduce the time lag between transmission and, reception of the information signal, (b) to reduce the size of antenna, (c) to reduce the fractional band width, that is the ratio of, the signal band width to the centre frequency, (d) to increase the selectivity
Page 508 :
P-499, , Communication Systems, , Putting values R, hT and hR, d max = 2×6.4×106 éë 140 + 40 ùû, , or, d max ; 65 km, 13. (d) According to question, modulation frequency, 250, Hz is 10% of carrier wave, f carrier =, , 250, = 2500KHZ, 0.1, , \ Range of signal 2500 ± 250 KHz = 2250 Hz to 2750 Hz For, 2000 KHZ, f mod = 200 Hz, , 1, Power P = m æç ö÷, è lø, , V 3 ´ 108 30, =, ×1014 Hz, =, l 8 ´ 10-7 8, , = 3.75 × 1014 Hz, 1% of f = 0.0375 × 1014 Hz, = 3.75 × 1012 Hz = 3.75 × 106 MHz, As we know, number of channels accomodated for, transmission =, , here µ = K, , 15. (c) If n = no. of channels, 10% of 10 GHz = n × 5 KHz or,, Þ n = 2 × 105, 16. (a) Given : modulation index m = 80% = 0.8, Ec = 14 V, Em = ?, Em, Þ E m = m ´ E c = 0.8 ´ 14 = 11.2V, Ec, , 17. (b) Given, modulating frequency fm = 15 KHz, \ Bandwidth of one channel = 2fm = 30 kHz, \, , No of channels accommodate =, , 300kHz, = 10, 30kHz, , 18. (c) Given : Inductance, L = 49 µH = 49 × 10–6 H,, capacitance C = 2.5 nF = 2.5 × 10–9 F, Using w =, , 1, LC, 1, , =, , 49 ´ 10 -6 ´, , 2.5, ´ 10 -9, 10, , =, , 1, 7 ´ 5 ´ 10-8, , or,, , 108, 22, = 2p ´ f = 2 ´, ´f, 7 ´5, 7, , or,, , f =, , 108, =, 7´5, , (Q w = 2pf ), , 107 104, =, kHz = 454.54kHz, 22, 22, , Vm, 5, =, = 0.2, V0 25, , Given, frequency of carrier wave (fc) = 1.2 × 106 Hz, = 1200 kHz., Frequency of signal (f0) = 20 kHz., Side frequency bands = fc ± f0, f1 = 1200 – 20 = 1180 kHz, f2 = 1200 + 20 = 1220 kHz, 22. (c) In amplitude modulation, the amplitude of the high, frequency carrier wave made to vary in proportional to the, amplitude of audio signal., Audio signal, , 3.75 ´106, total bandwidth of Channel, =, = 6.25 × 105, bandwidth needed per channel, 6, , using, m =, , 2, , 21. (d) Modulation index (m) =, , \ Range = 1800 KHZ to 2200 KHZ, 14. (c) Frequency, f =, , Therefore frequency range 454.54 ± 12kHz, i.e., 442 kHz – 466 kHz, 19. (c) Modulated carrier wave contains frequency wc and, wc ± w m, 20. (a) Length of antenna = comparable to l, Power radiated by linear antenna inversely depends on, the square of wavelength and directly on the length of the, antenna. Hence,, , Carrier wave, , Amplitude modulated wave, 23. (b) Comparing the given equation with standard, mA c, modulated signal wave equation, m = Ac sin wc t +, 2, mA c, cos (wc + ws) t, cos (wc – ws) t –, 2, Ac, 2, < 10 Þ m < (modulation index), 2, 3, Ac = 30, wc – ws = 200p, wc + ws = 400p, m, , Þ f c = 150, fs = 50 Hz ., 24. (c) Ratio of AM signal Bandwidths, <, , 15200 , 200 15000, <, < 6., 2700 , 200, 2500, , 25. (a) Amplitude modulated wave consists of three, frequencies are wc + wm, w,wc – wm, i.e. 2005 kHz, 2000kHz, 1995 kHz, 26. (b) Frequency of radio waves for sky wave propagation, is 2 MHZ to 30 MHZ.
Page 509 :
P-500, , Physics, , 27., , (b) Sky wave propagation is suitable for frequency range, 5 MHz to 25 MHz., 28. (b) Given : hR = 32 m, hT = 50 m, Maximum distance, dM = ?, Applying, dM =, , 2Rh T + 2Rh R, , = 2 ´ 6.4 ´106 ´ 50 + 2 ´ 6.4 ´106 ´ 32 = 45.5 km, 29. (b) Given : Resistance R = 100 kilo ohm, = 100 × 103 W, Capacitance C = 250 picofarad, = 250 × 10–12F, t = RC = 100 × 103 × 250 × 10–12 sec, , = 2.5 × 107 × 10–12 sec, = 2.5 × 10–5 sec, The higher frequency whcih can be detected with tolerable, distortion is, f =, , 1, 1, Hz, =, 2pma RC 2 p ´ 0.6 ´ 2.5 ´ 10 -5, 4, , 4, 100 ´ 10, ´ 10 4 Hz, Hz =, 1.2 p, 25 ´ 1.2p, = 10.61 KHz, , =, , This condition is obtained by applying the condition that, rate of decay of capacitor voltage must be equal or less, than the rate of decay modulated singnal voltage for proper, detection of mdoulated signal., 30. (b) Short-wave has the best noise tolerance., 31. (c) Above critical frequency (fc), an electromagnetic, wave penetrates the ionosphere and is not reflected by it., 32. (a) Effective refractive index of the ionosphere, 1/ 2, , é 80.5N ù, n eff = n 0 ê1 f 2 úû, ë, where f is the frequency of em waves, 33. (c) Modulation index, , ma =, , Em A, = =1, Ec A, , Equation of modulated signal [Cm(t)], = E(C) + maE(C) sin wmt, = A (1+ sin wCt) sin wmt, (As E(C) = A sinwCt), , 34. (a) Let d is the maximum distance, upto which it can, detect the objects, C, From DAOC, 2, , 2, , OC = AC + AO, , d, , 2, A, , (h + R)2 = d 2 + R 2, , Þ, , 2, , 2, , d = (h + R) - R, , q, , h, B, R, , R, O, , 2, , d = (h + R)2 - R2 ; d = h2 + 2hR, d = 5002 + 2 ´ 6.4 ´ 10 6 = 80 km, 35. (d) Frequency of EM wave u = 830 KHz, = 830 × 103 Hz., Magnetic field, B = 4.82 × 10–11 T, c, As we know, frequency, u =, l, , or l =, , c, 3 ´ 108, =, v 830 ´ 103, , l ; 360 m, And, E = BC = 4.82 × 10–11 × 3 × 108, = 0.014 N/C, 36. (b) The frequencies present in amplitude modulated, wave are :, Carrier frequency = wc, Upper side band frequency = wc + wm, Lower side band frequency = wc – wm., nhc, l, (where, n = no. of photons per second), , 37. (b) Power =, , Þ, , n=, , 10 ´ 103 ´ 500, -34, , 8, , ; 1031, , ´ 3 ´ 10, 6.6 ´ 10, 38. (b) For long distance communication, sky wave signals, are used., Also, the state of ionosphere varies every time., So, both statements are correct., 39. (a) Low frequencies cannot be transmitted to long, distances. Therefore, they are super imposed on a high, frequency carrier signal by a process known as modulation., Speed of electro-magnetic waves will not change due to, modulation. So there will be time lag between transmission, and reception of the information signal.
Page 510 :
Mock Test-1, , MT-1, , 1, (c) Resistance of both decrease, , PART-I (Multiple Choice Questions), 1., , 2., , 3., , A d.c. supply of 120V is connected to a large resistance X., A voltmeter of resistance 10kW placed in series in the circuit, reads 4V. What is the value of X?, (a) 190 kW, , (b) 90 kW, , (c) 290 kW, , (d) 390 kW, , (d) Resistnance of both increase, 4., , (a) Velocity of light for violet is greater than the velocity, of light for red colour., , The period of the satellite of the earth orbiting very near to, the surface of the earth is T0. What is the period of the, geostationary satellite in terms of T0, T0, (a), (b) 7T0, 7, (c) 7T0, (d) 7 7T0, In “Al” and “Si”. If temperature is changed from normal, temperature to 70 K then, (a) The resistance of Al will increase and that of Si will, decrease, (b) The resistance of Al will decrease and that of Si will, increase, , When white light passes through a dispersive medium, it, breaks up into various colours. Which of the following, statements is true?, , (b) Velocity of light for violet is less than the velocity of, light for red., (c) Velocity of light is the same for all colours, (d) Velocity of light for differnet colours has nothing to, do with the phenomenon of dispersion, 5., , An engine has an efficiency of 1/6. When the temperature, of sink is reduced by 62°C, its efficiency is doubled., Temperatures of source and sink are, (a) 99°C, 37°C, , (b) 124°C, 62°C, , (c) 37°C, 99°C, , (d) 62°C, 124°C
Page 511 :
MT-2, 6., , Mock Test-1, , x(m), , Two isothermals are shown in figure at temperatures T1, and T2. Which of the following relations is correct?, , 3, , P, , 0, T1, T2, , 10., , 7., , 8., , 9., , (a) T1 > T2, , V, (b) T1 < T2, , (c) T1 = T2, , (d) T1 =, , 1, T, 2 2, , During the experiment to determine the resistivity of a wire, by a metre bridge, the jockey is moved gently along the, wire from left to right to, (a) find a deflection in the galvanometer towards left, (b) find a deflection in the galvanometer towards right, (c) find no deflection in the galvanometer, (d) find a maximum deflection in the galvanometer, The circular head of a screw gauge is divided into 200, divisions and move 1 mm ahead in one revolution. If the, same instrument has a zero error of –0.05 mm and the reading, on the main scale in measuring diameter of a wire is 6 mm, and that on circular scale is 45. The diameter of the wire is, (a) 6.275 mm (b) 6.375 mm (c) 5.75 mm (d) 5.50 mm, The adjacent figure shows the position graph of one, dimensional motion of a particle of mass 4 kg. The impulse, at t =0 s and t = 4 is given respectively as:, , 11., , 12., , 13., , 4, , t(s), , (a) 0,0, , (b) 0, – 3 kg ms – 1, , (c) + 3 kg ms–1, 0, , (d) + 3 kg ms– 1, – 3 kg. ms–1, , In case of a p-n junction diode at high value of reverse, bias, the current rises sharply. The value of reverse bias is, known as, (a) cut off voltage, , (b) zener voltage, , (c) inverse voltage, , (d) critical voltage, , A photon materializes into an electron-positron pair. The, kinetic energy of the electron is found to be 0.19 MeV., What was the energy of the photon?, (a) 0.38 MeV, , (b) 0.70 MeV, , (c) 1.40 MeV, , (d) None, , If 10% of a radioactive material decays in 5 days, then the, amount of the original material left after 20 days is, approximately, (a) 60%, , (b) 65%, , (c) 70%, , (d) 75%, , Lights of two different frequencies, whose photons have, energies 1 eV and 2.5 eV respectively, successively, illuminate a metal whose work function is 0.5 eV. The ratio, of the maximum speeds of the emitted electrons will be, (a) 1 : 5, , (b) 1 : 4, , (c) 1 : 2, , (d) 1 : 1
Page 512 :
Mock Test-1, 14., , MT-3, , A block of mass 15 kg is placed over a frictionless horizontal, surface. Another block of mass 10 kg is placed over it, that, is connected with a light string passing over two pulleys, fastened to the 15 kg block. A force F = 80N is applied, horizontally to the free end of the string. Friction coefficient, between two blocks is 0.6. The portion of the string, between 10 kg block and the upper pulley is horizontal., Pulley, string and connecting rods are massless. (Take g =, 10 m/s²), , 16., , 10kg, , (a), , ( -3jˆ + 4ˆj) N, , æ 3 ˆ 4 ˆö, (b) ç - i + j ÷ N, è 5 5 ø, , (c), , ( 3iˆ - 4ˆj) N, , æ 3ˆ 4 ˆö, (d) ç i - j ÷ N, è5 5 ø, , A plate of mass (M) is placed on a horizontal frictionless, surface and a body of mass (m) is placed on this plate. The, coefficient of dynamic friction between this body and the, plate is m. If a force 3mmg is applied to the body of mass (m), along the horizontal, the acceleration of the plate will be, (a), , 17., , µ=0.6, 15 kg, F = 80N, , (c), 18., , (a) 3.2 m/s² (b) 2.0 m/s² (c) 1.6 m/s² (d) 0.8 m/s², 15., , A body of mass 5 kg under the action of constant force, ®, , ®, , (, , ), , F = Fx ˆi + Fy ˆj has velocity at t = 0 s as v = 6iˆ - 2ˆj m/s, , d1K1 + d 2 K 2, d1 + d 2, , mmg, M+m, , (c), , 3mmg, M, , (d), , 2mmg, M+m, , d1 + d 2, (d) (d / K + d / K ), 1, 1, 2, 2, , A particle starts S.H.M. from the mean position. Its, amplitude is a and total energy E. At one instant5 its kinetic, energy is 3 E/4, its displacement at this instant is, (a) y =, , ®, , and at t = 10s as v = +6jˆ m / s . The force ®, F is:, , (b), , Two rods of length d1 and d2 and coefficients of thermal, conductivities K1 and K2 are kept touching each other., Both have the same area of cross-section the equivalent, of thermal conductivity is, (a) K1 + K2, (b) K1d1 + K2d2, , Smooth, , The magnitude of acceleration of the 10 kg block is, , mm, g, M, , (c) y =, , a, 2, a, 3/ 2, , (b) y =, , a, 2, , (d) y = a
Page 513 :
MT-4, 19., , Two long parallel wires P and Q are both perpendicular to, the plane of the paper with distance of 5 m between them., If P and Q carry current of 2.5 amp and 5 amp respectively, in the same direction, then the magnetic field at a point, half-way between the wires is, (a), , (c), 20., , Mock Test-1, , 3m 0, 2p, , 3m 0, 2p, , (b), , m0, p, , (d), , m0, 2p, , 22., , 23., , In a circuit L, C and R are connected in series with an, alternating voltage source of frequency f. The current leads, the voltage by 45°. The value of C is, 1, , (a) pf (2 pfL - R), 1, , (c) pf (2 pfL + R ), , C, A, , 1, , (b) 2 pf (2 pfL - R), 1, , (d) 2 pf (2 pfL + R), , B, 24., , PART-II (Numerical Answer Questions), 21., , The radius of curvature of a thin plano-convex lens is 10, cm (of curved surface) and the refractive index is 1.5. If the, plane surface is silvered, then it behaves like a concave, mirror of focal length (in cm), Two masses A and B of 10 kg and 5 kg respectively are, connected with a string passing over a frictionless pulley, fixed at the corner of a table (as shown in figure). The, coefficient of friction between the table and the block is, 0.2. The minimum mass (in kg) of C that may be placed on, A to prevent it from moving is equal to, , An insect trapped in a circular groove of radius 12 cm., moves along the groove steadily and completes 7, revolutions in 100 sec. What is the linear speed (in m/s) of, the motion ?, , 25., , Ammeter and voltmeter readings were recorded as 0.25 A, and 0.5 V during the experiment to determine the resistance, of a given wire using Ohm's law. What is the correct value, of the resistance (in ohm)?, A wheel is rotating at 900 r.p.m. about its axis. When power, is cut off it comes to rest in 1 minute. The angular retardation, in (rad/s2) is :
Page 514 :
Mock Test-1, , MT-5, , RESPONSE SHEET, 1., , a, , b c d, , 2., , a, , b c d, , 3., , a, , b c d, , 4., , a, , b c d, , 5., , a, , b c d, , 6., , a, , b c d, , 7., , a, , b c d, , 8., , a, , b c d, , 9., , a, , b c d, , 10., , a, , b c d, , 11., , a, , b c d, , 12., , a, , b c d, , 13., , a, , b c d, , 14., , a, , b c d, , 15., , a, , b c d, , 16., , a, , b c d, , 17., , a, , b c d, , 18., , a, , b c d, , 19., , a, , b c d, , 20., , a, , b c d, , 21., , 22., , 23., , 24., , 25.
Page 515 :
MT-6, , Solutions Mock Test -1, , HINTS & SOLUTIONS, MOCK TEST-1, 1., , (c) Given (10 kW) I = 4V OR 10 kW ., , 120V, = 4 V OR, X + 10kW, , 8., , Number of divisions on circular scale = 200, , 300 kW = X + 10 kW;, X = 290 kW., , L.C =, , 120V, , V, , 2., , (d), , 3., , (b), , X, , =, , 9., , (a) From h = 1 -, , In 2nd case :, , T2 T2, 1 5, ,, = 1 - h = 1 - = ...(i), 6 6, T1 T1, , T2 - 62, 2 2, = 1 - h' = 1 - = .....(ii), T1, 6 3, , 3ö, æ, = m(vf - vi ) = 4 ç 0 - ÷ = -3kg ms -1, è, 4ø, 10., , 11., , 1, T2 = 62, T2 = 310 K = 310 – 273 = 37°C, 5, , T1 =, , 6, 6, T2 = ´ 310 = 372 K = 372 - 273 = 99°C, 5, 5, , 6., , (a) As isothermal at T1 is farther from the origin than the, isothermal at T2, therefore, T1 > T2, , 7., , (c) The jockey is moved so as to get the null point (i.e.,, no deflection in the galvanometer)., , (b) In reverse bias on p-n junction when high voltage is, applied, electric break down of junction takes place,, resulting large increase in reverse current. This high, voltage applied is called zener voltage., (c) Conservation of Energy,, , Photon ® e + + e - + energy, EP = E, , 2, 2 6, 4, Using (i), T2 – 62 = T1 = ´ T2 = T2, 3, 3 5, 5, , or, , 3, m/s, 4, , 3, m / s, little after 4, 4, second, velocity becomes zero, Therefore, Impulse, , (Si), , 5., , (d) t < 0, vi = 0 and t > 0, vf =, , Little before 4 second v =, , T ¯ R metal ¯ R semi - conductor, , (b) As mv > mr therefore, vv < vr., , 1 mm, = 0.005 mm = 0.0005 cm, 200, , = 3 kg ms –1, , T ¯ (300K to 70K ), , 4., , Number of divisions on circular scale, , æ3 ö, \ Impulse = m (vf –vi) = 4 ç - 0÷, è4 ø, , 2, , ö, ÷ = 7 3 ; Ts = 7 7 T0, ÷, ø, , (Al), , Pitch, , Diameter of the wire = (Main scale reading + Circular, scale reading × L.C.) – zero error, = 6 mm + 45 × 0.005 – (– 0.05), = 6 mm + 0.225 mm + 0.05 mm = 6.275 mm, , 10kW, , 2, 3, 2, 3, T 2 µ r 3 law; T0 µ R & Ts µ (> 7R) ;, , æ Ts, ç, çT, è 0, , (a) Pitch = 1 mm, , re +, , +E, , re -, , + K.E. of both e - + e +, , We know that rest mass energy of e + + e - is .5 MeV, Hence E P = .5 + .5 + .19 ´ 2 = 1.40MeV, 12., , (b) Let initial amount be 100 gm., disintegrated, 5 days, , 100 gm ¾¾ ¾®, , 100 ´ 10, 100, , Next 5 days, , 90, , ¾¾ ¾ ¾¾®, , 81, , ¾¾ ¾¾¾®, , 73, , 90 ´ 10, 100, , 81´10, 100, 73, ´ 10, Next 5 days, ¾¾ ¾ ¾¾®, 100, Next 5 days, , Left, , 10, , 90, , 9, , 81, , 8.1, , » 73, , 7.3, , » 65
Page 516 :
Solutions Mock Test -1, 13., , MT-7, , (c) Einstein equation KEmax = E – Work function;, , Force, F = ?, , 1, mv2 = E - W, 2, , Acceleration, a =, , Using this concept,, 1, mV12 max, 1 - .5, 1, V max 1, 2, =, = or 1, =, 1 2, 2.5 - .5 4, V2 max 2, V2 max, 2, , 14., , =, , = 5´, , 16., , f, , 6 ˆj - (6iˆ - 2 ˆj ) -3iˆ + 4 ˆj, =, m/s2, 10, 5, , Force, F = ma, , (a) First, let us check upto what value of F, both blocks, move together. Till friction becomes limiting, they will, be moving together. Using the FBDs, , F, F, , ( -3iˆ + 4 ˆj ), = ( -3iˆ + 4 ˆj ) N, 5, , mmg, M, , (a) a =, , a1, , F, , v -u, t, , f, Body, m, , 15 kg, F, , 3mmg, , fr, , M plate, , a2, , fr, , fr = mmg = Ma, , 10 kg block will not slip over the 15 kg block till, acceleration of 15 kg block becomes maximum as it is, created only by friction force exerted by 10 kg block on, it., , 17., , (d) When two rods are connected in series, Q=, , a1 > a 2(max), , F-f, f, =, for limiting condition as f maximum is 60N., 10, 15, , \, , F = 100 N, Therefore, for F = 80N, both will move together., Their combined acceleration, by applying NLM using, both as system, F = 25a, a=, , 15., , 18., , Mass of body, m = 5 kg, Velocity at t = 0,, u = (6iˆ - 2 ˆj) m/s, Velocity at t = 10s,, v = + 6 ĵ m/s, , d1 + d 2 d1 d 2, ;, =, +, K, K1 K 2, , (b) Total energy, E =, K.E. =, , 80, = 3.2 m / s 2, 25, , (a) From question,, , A(T1 - T2 )t A(T1 - T2 ) t, =, d1 d 2, (d 1 + d 2 ) / K, +, K1 K1, , So,, 19., , \K =, , 1, mw 2 a 2 ;, 2, , (d1 + d 2 ), d1 d 2, +, K1 K 2, , 3E 1, = mw 2 (a 2 - y 2 ) ., 4, 2, , 3 a2 - y2, a, a2, 2, =, y, =, or, or y = ., 2, 4, 2, 4, a, , (a) When current flow in both wire in same direction then, magnetic field at half way due to P wire., uur, m I, m I, m, BP = 0 1 = 0 1 = 0, 5, p .5, p, 2p, 2, , (where I1= 5Amp), , The direction of Bp is downward
Page 517 :
MT-8, , Solutions Mock Test -1, 22., , P, , Q, ×, , 5 Amp, , C, , 2.5 Amp, , (20) The silvered plano convex lens behaves as a concave, mirror; whose focal length is given by, 1 2, 1, = +, F f1 f m, , ×, , 5m, Magnetic field at half way due to Q wire, BQ =, , m0I 2 m0, =, [upward, 5 2p, 2p, 2, , If plane surface is silvered, ×, , ], fm =, , [where I 2 = 2.5Amp. ], Net magnetic field at half way, m, 3m, m, B = B P + B Q = 0 + 0 = 0 (downward, p 2p, 2p, , Hence net magnetic field at midpoint =, 20., , ), , 3m 0, 2p, , R2 ¥, = =¥, 2, 2, , \, , æ 1, 1, 1 ö, æ 1 1 ö m –1, = (m –1) ç, –, ÷ = (m – 1) ç – ÷ =, R, f1, R, R, èR ¥ø, 2ø, è 1, , \, , 1 2(m –1) 1 2(m –1), R, =, + =, ÞF =, F, R, ¥, R, 2(m – 1), , Here R = 20 cm, m = 1.5, , (d) From figure,, , \ F=, , 23., , 20, = 20cm, 2(1.5 –1), , (15) Let T be the tension in the string; f = frictional force, between block A and table; m’ = minimum mass of C., For the just motion of block A on table, T = f = mR = m(m + m ' )g = 0.2(10 + m' )g ....(i), For the just motion of block B, T = 5g, , ....(ii), , From (i) and (ii), 5g = 0.2 (10 + m’)g, 1, - wL, w, tan 45º = C, R, , 1, 1, = R + wL, Þ, - wL = R Þ, wC, wC, , C=, 21., , 1, 1, =, w(R + w), 2 pf ( R + 2 pfL), , (5.3)This of example of uniform circular motion., w=, , 2p, 7, = 2 pn = 2p ´, = 0.44 rad / sec . ;, T, 100, , V = Rw = 0.44 × 12 = 5.3 cm/sec., , or 5 = 2 + 0.2m’ or m’ =, V 0.5, =, = 2 W., I 0.25, , 24., , (2), , 25., , (p/2)Angular retardation,, , R=, , 5-2, = 15kg, 0.2, , w - w1 2p(n 2 - n1 ) 2p(0 - 900 / 60) p, a= 2, =, =, = rad / s 2 ., t, t, 60, 2
Page 518 :
Mock Test-2, , MT-9, , 2, PART-I (Multiple Choice Questions), 1., , The horizontal component of impulse provided by string, tension in moving from extreme point to lower most, point –, ///////////////, , 3., , 4., , A current of 4A produces a deflection of 30° in the, galvanometer. The figure of merit is, (a) 6.5 A/rad, (b) 7.6 A/rad, (c) 7.5 A/rad, (d) 8.0 A/rad, A uniform disc of radius R and mass m is connected to a, wall by string of length 2R. The normal reaction of wall is –, , q, h, , (a) m 2gh, (b) m 2gh (1 - cos q), (c) m 2gh (cos q), , 2., , (d) can’t be found without knowing tension as a function, of time, N divisions on the main scale of a vernier callipers coincide, with N + 1 divisions on the vernier scale. If each division, on the main scale is of a units, determine the least count of, the instrument., a, a, 2a, 2a, (a), (b), (c), (d), N +1, N -1, N -1, N +1, , (a) mg, 5., , (b) mg/2, , (c) mg/ 3, , (d) 2mg, , For the determination of the focal length of a convex mirror,, a convex lens is required because, (a) it is not possible to obtain the image produced by a, convex mirror on the screen, (b) a convex lens has high resolving power so it helps to, measure the focal length correctly, (c) a convex mirror always forms a real image which is, diminished by the convex lens, (d) none of these
Page 519 :
MT-10, 6., , Mock Test-2, , A particle moves along x-axis with initial position x = 0. Its, velocity varies with x-coordinate as shown in graph. The, acceleration ‘a’ of this particle varies with x as –, v, , 8., , A ring of mass M and radius R lies in x-y plane with its, centre at origin as shown. The mass distribution of ring is, nun-uniform such that at any point P on the ring, the mass, per unit length is given by l = l0 cos2q (where l0 is a, positive constant). Then the moment of inertia of the ring, about z-axis is –, y, , x, , P, , M, a, , a, x (b), , (a), , a, (c), , x, , a, x (d), , (a) MR2, x, 9., , 7., , R, q, , Choose the wrong statement regarding the experiment to, identify a diode, an LED, a resistor and a capacitor, (a) when pointer moves in one way when voltage is, applied and doesn't move when reversed and there is, light emission, the item is LED, (b) when pointer moves in one way when voltage is, applied and doesn't move when reversed and there is, no emission of light, the item is a diode, (c) when pointer moves in one way when voltage is, applied and not when reversed, the item is a resistor, (d) none of these, , 10., , (b), , x, , 1M, 1 M, 1, R (d), R, MR 2 (c), 2, l, p, l0, 2, 0, , What happens when we do not stir the mixture, continuously in the experiment to determine the specific, heat capacity of a given solid ?, (a) Temperatue of mixture is increased, (b) Temperature of mixture is not constant, (c) Temperature of mixture is constant, (d) None of these, An ideal monatomic gas with pressure P, volume V and, temperature T is expanded isothermally to a volume 2V, and a final pressure Pi. If the same gas is expanded, adiabatically to a volume 2V, the final pressure is Pa. The, , P, ratio a is, Pi, (a) 2–1/3, , (b) 21/3, , (c) 22/3, , (d) 2–2/3
Page 520 :
Mock Test-2, 11., , 12., , MT-11, , In a Young’s double slit experiment with light of wavelength, l, fringe pattern on the screen has fringe width b. When, two thin transparent glass (refractive index m) plates of, thickness t1 and t2 (t1 > t2) are placed in the path of the, two beams respectively, the fringe pattern will shift by a, distance, (a), , b ( m - 1) æ t1 ö, l çè t ÷ø, , (b), , mb t1, l t2, , (c), , b ( m - 1), ( t1 - t2 ), l, , (d), , ( m - 1) ( t1 + t2 ), , 2, , Consider Gauss’s law, following is true ?, , l, b, , q, , Ñò E.dA = Î0, , . Which of the, , +, , r, (a) E must be the electric field due to the enclosed charge, r, (b) If net charge inside the Gaussian surface = 0, then E, must be zero everywhere over the Gaussian surface., (c) If the only charge inside the Gaussian surface is an, electric dipole, then the integral is zero., uuur, r, (d) E is parallel to dA everywhere over the Gaussian surface., A dipole consisting of two charges +q and –q joined by a, massless rod of length l, is seen oscillating with a small, amplitude in a uniform electric field of magnitude E. The, period of oscillation, (a) is proportional to E, (b) is proportional to 1/E, (c) is p, , ml, 3qE, , 14., , 15., + –, , 13., , r uuur, , (d) is proportional to, , 16., , 17., , 1, E, , but ¹ p, , ml, 3qE, , A sinusoidal wave is propagating in negative x-direction, in a string stretched along x-axis. A particle of string at, x = 2m is found at its mean position and it is moving in, positive y-direction at t = 1 sec. The amplitude of the wave,, the wavelength and the angular frequency of the wave are, 0.1 meter, p/2 meter and 2p rad/sec respectively., The equation of the wave is, (a) y = 0.1 sin (4p (t – 1) + 8 (x – 2)), (b) y = 0.1 sin ((t – 1) – (x – 2)), (c) y = 0.1 sin (2p (t – 1) + 4 (x – 2)), (d) None of these, In an experiment, a small steel ball falls through a liquid at, a constant speed of 10 cm/s. If the steel ball is pulled, upward with a force equal to twice its effective weight,, how fast will it move upward ?, (a) 5 cm/s (b) Zero, (c) 10 cm/s (d) 20 cm/s, A ball of mass 160 g is th rown up at an an gle, of 60° to the horizontal at a speed of 10 ms–1. The angular, momentum of the ball at the highest point of the trajectory, with respect to the point from which the ball is thrown is, nearly (g = 10 ms–2), (a) 1.73 kg m2/s, (b) 3.0 kg m2/s, 2, (c) 3.46 kg m /s, (d) 6.0 kg m2/s, A block of mass m is attached with massless spring of, force constant k. The block is placed over a fixed rough, inclined surface for which the coefficient of friction is, µ = 3/4. The block of mass m is initially at rest. The block of, mass M is released from rest with spring in unstretched, state. The minimum value of M required to move the block, up the plane is (neglect mass of string and pulley and, friction in pulley.)
Page 521 :
MT-12, , Mock Test-2, (a) –10 cm. (b) –10/4 cm. (c) –10/3 cm. (d) None, PART-II (Numerical Answer Questions), 21., M, , m, , 18., , 19., , 37°, /////////////////////////////////////////, 3, 4, 6, 3, m, m, m, m, (a), (b), (c), (d), 5, 5, 5, 2, A charged particle of mass m and having a charge Q is, placed in an electric field E which varies with time as E = E0, sin wt. What is the amplitude of the S.H.M. executed by, the particle?, QE 0, 1 QE 0, (a), (b), 2, 2 mw 2, mw, 2QE 0, (c), (d) None of these, mw 2, In a photoelectric experiment, with light of wavelength l,, the fastest electron has speed v. If the exciting wavelength, is changed to 3l/4, the speed of the fastest emitted electron, will become –, , (a) v, , 3, 4, , (b) v, , 22., , The activity of a radioactive substance drops to 1/32 of its, initial value in 7.5 h. Find the half life (in hours)., Six resistors of 10W each are connected as shown. The, equivalent resistance (in ohms) between points X and Y is –, X, , Y, , 23., , 24., , The electric potential V is given as a function of distance x, by V = (5x2 + 10x – 9) volt. Value of electric field (in v/m) at, x = 1m is, Three moles of an ideal monoatomic gas perform a cycle, shown in figure. The gas temperatures in different states, are T1 = 200K, T2 = 400K, T3 = 1600 K, and T4 = 800K. The, work done (in kJ) by the gas during the cycle is, (Take R = 25/3 J/mol-K), P, 2, , 4, 3, , 4, 4, (d) greater than v, 3, 3, A thin convex lens of focal length 10cm and refractive, index 1.5 is cut vertically into two equal pieces. They are, placed as shown with a liquid of refractive index 3 between, them. What is the focal length of the combination ?, , 3, , (c) less than v, 20., , 1, , 25., , 4, , T, A body is executing simple harmonic motion. At a, displacement x from mean position, its potential energy is, , E1 = 2J and at a displacement y from mean position, its, potential energy is E2 = 8J. The potential energy E (in, Joule) at a displacement (x + y) from mean position is
Page 522 :
MT-14, , Solutions Mock Test -2, , HINTS & SOLUTIONS, MOCK TEST-2, Dp = mv = m 2gh, , 1., , (a), , 2., , (a) Least count of vernier callipers, , a, x, , Hence the graph relating a to x is, , = value of one division of main scale – value of one, division of vernier scale, Now, N × a = (N + 1) a', (a' = value of one division of vernier scale), a¢ =, , N, a, N +1, , \ Least count = a - a ¢ =, 3., , 7., , (c) For a resistor the pointer moves in both ways when, voltage is applied., , 8., , (a) Divide the ring into infinitely small lengths of mass, dm1. Even though mass distribution is non-uniform,, each mass dm1 is at same distance R from origin., , a, N +1, , \ MI of ring about z-axis is, = dm1R2 + dm2 R2 + ..............+ dmnR2 = MR2, , (b) Here I = 4A, c, , pc, æ 30 ´ p ö, q =30° = ç, ÷ =, 6, è 180 ø, Now, k =, , I, 4, 4´6´7 2´6´7, =, =, =, p, 22, 11, q æ ö, ç ÷, è6ø, , 9., , (b) If temp of mixture (T) is not constant then according, to calculation we find wrong value of specific heat. If, we stir continuously then temperature of whole mixture, becomes same., , 10., , (d) For isothermal process :, PV = Pi .2V, P = 2Pi, For adiabatic process, , 84, =, = 7.6 A / rad, 11, , 4., , ...(i), , PVg = Pa (2V)g, (Q for monatomic gas g= 5 3 ), , (c), mg = T cos 30°, , 30°, , T, R, , R, N, , N = T sin 30°, Þ N=, , or,, , Þ, , mg, , 5., , (a) A convex mirror always forms a virtual image which, can't be cast on a screen. Hence, a convex lens is, used to find the focal lenght of a convex mirror., , 6., , (a) The linear relationship between v and x is, , 11., , \ a = m2x – mC, , [From (i)], , dv, = -m ( -mx + C), dx, , P, Pa, 2, = 5 Þ a =2 3, Pi, Pi, 23, , (c) Shift =, , =, , v = – mx + C where m and C are positive constants., \ Acceleration, a = v, , = Pa, , 5, (2V) 3, , -2, , mg, 3, , 5, 2Pi V 3, , 12., , (c), , f=, , b ( m - 1), l, , t1 -, , b ( m - 1), l, , b ( m - 1), l, , t2, , ( t1 - t2 ), , Q net, For dipole Qnet = 0 Þ f = 0, Î0
Page 523 :
Solutions Mock Test -2, , 13., , (c), , I=, , MT-15, , ml 2, ;, 12, , Restoring troque = qElsin q » qEl q =, , \w =, , \T = p, , 14., , qEl, qEl ´12 2p, =, =, I, T, ml 2, ml, 3qE, , 2Mg, (before coming it rest momentarily again)., K, Thus the maximum extension in the spring is, , by, , x=, , 2Mg, k, , .......... (1), , For block of mass m to just move up the incline, kx = mg sin q + µ mg cos q, , Shifting the origin of position to left by 2m, that is, to, x = 0. Also shifting the origin of time backwards by 1, sec, that is to t = 0 sec., y = 0.1 sin (2p (t – 1) + 4 (x – 2)), (c) Weight of the body, , T (upthrust), , F, 4 3, W = mg = pr rg (viscous, 3, force), , r, , 3 3, 4, 2Mg = mg ´ + mg ´, 5 4, 5, , \ maximum acceleration =, , 4 3, pr sg, 3, , W(weight), , When the body attains terminal velocity net force, acting on the body is zero. i.e.,, , \ A=, 19. (d), , 2 r 2 (r - s)g, 9, h, , As in case of upward motion upward force is twice its, effective weight, therefore, it will move with same, speed 10 cm/s, , 20. (c), , æ 1, 1, 1 ö, = (m - 1) ç, f, è R1 R 2 ÷ø, , R = 10 cm., , 1ö, 10, æ 1, f ¢ = (3 - 1) ç, =è -10 10 ÷ø, 4, , v = 10 m/s, , = H mv cos q, , 3, m, 5, , QE 0, = w2A, m, , 1, hc, -f, mv2 =, 2, l, , q = 60°, ®, , M=, , QE 0, mw 2, , (c) Given : m = 0.160 kg, , ®, , Angular momentum L = r ´ m v, , or, , 1, hc, 4hc, -f=, - f . Clearly, v ¢ >, mn '2 =, 2, (3l / 4), 3l, , W–T –F=0, And terminal velocity v =, , .......... (2), , 18. (a) For a particle undergoing SHM with an amplitude A, and angular frequency w, the maximum acceleration =, w2 A, Here the maximum force on the particle = QE0, , and F = 6phvr, , 16., , 102 ´ sin 2 60°´ cos 60°, 2 ´10, , 17. (a) As long as the block of mass m remains stationary,, the block of mass M released from rest comes down, , (c) The equation of wave moving in negative x-direction,, assuming origin of position at x = 2 and origin of time, (i.e. initial time) at t = 1 sec., , T=, , =, , = 3.46 kg m2/s, , y = 0.1 sin (2pt + 4x), , 15., , v2 sin 2 q, cos q, 2g, , d 2q, dt 2, , é, v2 sin 2 q ù, êH =, ú, 2g úû, êë, , =, , f=20, , f', , f=20, , 4, v, 3
