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Page 1 of 16, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, 2.1 Fill in the blanks, (a) The volume of a cube of side 1 cm is equal to ..m3, (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to...(mm)?, (c) A vehicle moving with a speed of 18 km h-1 covers....m in 1 s, (d) The relative density of lead is 11.3. Its density is ...g cm-3 or ..kg m., Solution:, (a) Volume of cube, V = (1 cm)3 = (102 m)3 = 10-6 m3., (b) Surface area = curved area + area on top /base = 2Trrh + 2Tr2 = 2Tr (h + r), r = 2 cm = 20 mm, h = 10 cm = 100 mm, Surface area = 2TTT (h + r) = 2 x 3.14 x 20 (100 + 20) = 15072 mm2, Hence, answer is 15072 mm?, (c) Speed of vehicle = 18 km/h, 1 km = 1000m, 1 hr = 60 x 60 = 3600 s, 1 km/hr = 1000 m/3600 s = 5/18 m/s, 18 km/h = = (18 x 1000)/3600, = 5 m/s, Distance travelled by the vehicle in 1, (d) The Relative density of lead is 11.3 g cm3, => 11.3 x 103 kg m3 [1 kilogra, 10°g, 1 meter = 102 cm], => 11.3 x 103 kg m-4, 2.2 Fill in the bla, by suitable conversion of units, (a) 1 kg m?, ..g cm? s-2, (c) 3.0 m s = ... km h-2, (d) G = 6.67 x 10-11 N m? (kg)-2 = .... (cm)3 s-2 g1, Solution:, (a) 1 kg m? s-2 = ...g cm? s-2, 1 kg m2 s2 = 1kg x 1m2 x 1s -2, We know that,, 1kg = 103, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734), NDAMENTAL OF PHYSICS

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Page 2 of 16, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, 1m = 100cm = 102cm, When the values are put together, we get:, 1kg x 1m? x 1s2 = 10°g x (10?cm)? x 1s2 = 10°g x 104 cm2 x 1s2 = 107 gcm?s2, =>1kg m? s? = 107 gcm?s?, (b) 1 m = ..... ly, Using the formula,, Distance = speed x time, Speed of light = 3 x 108 m/s, Time = 1 yr = 365 days = 365 x 24 hr = 365 x 24 x 60 x 60 sec, Put these values in the formula mentioned above, we get:, One light year distance = (3 x 108 m/s) x (365 x 24 x 60 x 60) = 9.46x1015m, %3D, 9.46 x 1015 m = 1ly, So that, 1m = 1/9.46 x 1015ly, => 1.06 x 10-16ly, =>1 meter = 1.06 x 1016ly, (c) 3.0 m s-2 = .. km h-2, 1 km = 1000m so that 1m = 1/1000 km, 3.0 m s2 = 3.0 (1/1000 km), /3600 hour) 2 = 3.0 x 103 km x ((1/3600)-2h-2), = 3 x 10-3km x (3600)2, 3.88 x 104 km h-2, => 3.0 m s2 = 3.88, km h2, (d) G = 6.67, 10-" N m2 (kg)-2 = .... (cm)3s-2 g-1, 101 N m2 (kg)2, RUNDAMENTAL OF PHYSICS, 1N = 1kg m s2, 1 kg = 103 g, 1m = 100cm= 102 cm, Put the values together, we get:, => 6.67 x 1011 Nm2 kg2 = 6.67 x 1011 x (1kg m s 2) (1m2) (1kg²), Solve the following and cancelling out the units, we get:, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 3 of 16, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, => 6.67 x 1011 x (1kg -1 x 1m³ x 1s2), Put the above values together to convert kg to g and m to cm, => 6.67 x 10:11 x (10°g)1 x (102 cm)3 x (1s2), => 6.67 x 10-8 cm3 s-2 g -1, =>G = 6.67 x 10-11 Nm?(kg)2 = 6.67 x 108 (cm) s? g1, 2.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J =1 kg m2s., Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length, equals B m, the unit of time is y s. Show that a calorie has a magnitude of 4.2 a1 B-2 y? in terms, of the new units., CS, Solution:, 1 calorie = 4.2 J = 4.2 kg m? s-2, The standard formula for the conversion is, .(눈), (국).(₩), Given unit, new unit, Mg, Dimensional formula for energyD, [M'L'T], Here, x 1, y = 2 and z =-2, M, = 1 kg, L, = 1m, T; = 1s, and M2 = a kg, L2 = B m, T2 = YS, Calorie, new unit, = 4.2 ()(, Calorie = 4.2 arB2, 2.4 Explain this statement clearly:, "To call a dimensional quantity 'large' or 'small' is meaningless without specifying a standard for, comparison". In view of this, reframe the following statements wherever necessary:, (a) atoms are very small objects, (b) a jet plane moves with great speed, (c) the mass of Jupiter is very large, (d) the air inside this room contains a large number of molecules, (e) a proton is much more massive than an electron, (f) the speed of sound is much smaller than the speed of light., NDAMENTAL OF PHY, Solution:, (a) In comparison with a soccer ball, atoms are very small, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 4 of 16, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, (b) When compared with a bicycle, jet plane travels at high speed., (c) When compared with the mass of a cricket ball, the mass of Jupiter is very large., (d) As compared with the air inside a lunch box, the air inside the room has a large number of, molecules., (e) A proton is massive when compared with an electron., (f) Like comparing the speed of a bicycle and a jet plane, the speed of light is more than the speed of, sound., 2.5 A new unit of length is chosen such that the speed of light in vacuum, distance between the Sun and the Earth in terms of the new unit if lig, unity. What is the, takes 8 min and 20 s to, cover this distance?, Solution:, Distance between them = Speed of light x Time taken by light to cover the distance, Speed of light = 1 unit, Time taken = 8 x 60 + 20 = 480 + 20 = 500s, The distance between Sun and Earth = 1 x 500 = 500 units., 2.6 Which of the following is the most precise device for measuring length:, (a) a vernier callipers with 20 divisions on the sliding scale, (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale, (c) an optical instrument that can measure length to within a wavelength of light?, Solution:, (a) Least count = 1-, (b) Least count, Amberof divisions, (c) least count = wavelength of light = 10-5 cm, = 0.00001 cm, We can come to the conclusion that the optical instrument is the most precise device used to measure length., 2.7. A student measures the thickness of a human hair by looking at it through a, microscope of magnification 100. He makes 20 observations and finds that the average width of, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 5 of 16, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness, of the hair?, Solution:, Magnification of the microscope = 100, Average width of the hair in the field of view of the microscope 3.5 mm, %3D, Actual thickness of hair =3.5 mm/100 = 0.035 mm, 2. 8. Answer the following:, (a)You are given a thread and a metre scale. How will you estimate the diameter of, the thread?, (b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do, you think it is possible to increase the accuracy of the screw gauge arbitrarily by, increasing the number of divisions on the circular scale?, (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why, is a set of 100 measurements of the diameter expected to yield a, estimate than a set of 5 measurements only?, more reliable, Solution:, (a) The thread should be wrapped around a pencil a number of times so as to form a coil having its, turns touching each other closely. Measure the length of this coil with a metre scale. If L be the length, of the coil and n be the number of turns of the coil then the diameter of the thread is given by the, relation, Diameter = L/n., (b) Least count of the screw gauge = Pitch/number of divisions on the circular scale, So, theoretically when the number of divisions on the circular scale is increased the least count of the, screw gauge will decrease. Hence, the accuracy of the screw gauge will increase. However, this is only, a theoretical idea. Practically, there will be many other difficulties when the number of turns is, increased., (c) The probability of making random errors can be reduced to a larger extent in 100 observations than, in the case of 5 observations., 2.9. The photograph of a house occupies an area of 1.75 cm? on a 35 mm slide. The slide is, projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear, magnification of the projector-screen arrangement?, uDMSTAOF SICS, Solution:, Arial Magnification = Area of the image/Area of the object, = 1.55/1.75 x 104, = 8.857x 103, Linear Magnification = VArial magnification, = v8.857x 103, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)