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BY TIWARI SIR, ________________________________, , ___________________________________________________________________________________________, 12. A metallic sphere having no net charge is placed near a, 1. One metallic sphere A is given positive charge, finite metal plate carrying a positive charge. The, whereas another indentical metallic sphere B of, electric force on the sphere will be, exactly same mass as of A is given equal, (a) towards the plate, amount of negative charge. Then., (b) Away from the palte, (a) mass of A and mass of B still remain equal (b), (c) Parallel to the palte, mass of A increases, (d) zero, (c) mass of B decreases, 13. Two charged spheres separated at a distance d exert a, (d) mass of B increases, force F on each other. If they are immersed in a liquid, 2. When 1014 electrons are removed from a, of dielectric constant. then what is the force (if all, neutral metal sphere, then the charge on the, conditions are same.), (a) F/2, (b)F, (c) 2F, (d) 4F, sphere becomes., 14., Two, charges, placed, in, air, repel, each, other, by a force, (a)16C, (b) -16C, -4, of, 10, N, ., when, oil, is, introduced, between, the, charges, (c) 32C, (d) -32C, then the force becomes 2.5×10-9NThe dielectric constant, -19, 3. A conductor has 14.4×10 C positive charge ., of oil is –, The conductor has (charge on electron = 1.6×10 (a) 2.5, (b) 0.25, (c) 2.0, (d) 4.0, 19, C), 15. Fg and Fe represent gravitational and electrostatic force, (a) 9 electrons in excess, respectively between electrons situated at a distance, (b) 27 electrons in short, 10cm The ratio of Fg/Fe is of the order of –, (a) 1042, (b) 10-21, (c) 27 electrons in excess, 24, (c), 10, (d) 10-43, (d) 9 electrons in short, 16. Two protons are a distance of 1×10-10cm from each, 4. Charge on is - particle is –, other the forces acting on them are –, (a) 4.8 × 10-19C, (b) 1.6× 10-19C, (a) nuclear force and gravitational force, -19, -19, (c) 3.5× 10 C, (d) 6.4×10 C, (b) nuclear force and coulomb force, 5. A body has -80 micro coulomb of charge, (c) coulomb force and gravitational force, number of additional electrons in it will be –, (d), nuclear, coulomb and gravitational force, (a) 8× 10-5, (b) 80× 10-17, 17. Charge Q is divided into two parts which are then kept, (c) 5× 1014, (d) 1.28×10-17, some distance apart . The force between them will be, 6. When a body is earth connected electrons from, maximum , if the two parts are having the charge., the earth flow into the body. This means the, (a) each, (b) and, body is(c) and, (d) None of these, (a) unchanged, (b) Charged positively, 19, 18. When 10 electrons are removed from a neutral metal, (c) Charged negatively (d) an insulator, 7., , There are two charges +1c and +5C The ratio of, the forces acting on them will be(a) 1:5, (b) 1:1, (c) 5:1, (d) 1:25, 8. The value of electric permittivity of free space is –, (a) 9× 109NC2/m2, (b) 8.85× 10-12Nm2/C2, (c) 8.85× 10-12C2/N-m2, (d) 9×109C2/N-m2, 9. Two charges each equal to 2C are 0.5m apart. If, both of them exist inside vacuum. then the force, between them is –, (a) 1.89 N, (b) 2.44 N, (c) 0.144 N, (d) 3.144 N, 10. The charges on two spheres are +7C and, -5C respectively. They experience a force F. If each, of them is given an additional charge of -2C then, the new force attraction will be(a) F`, (b) F/2, (c) F/√3, (d) 2F, 11. Two charges are of equal of magnitudes are, separated by a distance r experiences a forces F . if, magnitudes become half of two charges and distance, between them becomes two times then new force, between charges will be, (a) F, (b) F/16, (c) F/√3, (d) none of these, , plate through some process, then the charge on it., becomes(a) +1.6 C, (b) -1.6 C, (c) 1019 C, (d) 10-19 C, 19. If a change on the body is 1nC, then how many, electrons are present on the body?, (a) 1.6 ×1019, (b) 6.25×109, (c) 6.25× 1027, (d) 6.25×1028, 20. Dielectric constant of metal is –, (a) 1, (b) 0, (c) 8, (d) ∞, 21. How many electrons are there in one coulomb of, negative eharge., (a) 6.25× 1017, (b) 62.5×1018, 18, (c) 6.25×10, (d) 6.25×1019, 22. Which of the following relation is correct ?, ∈, (a) ∈=, (b) ∈= 𝐾 ∈ (c) ∈= 𝐾 ∈ (d) All are correct, 23. Coulomb's Law in veetorial form is –, (a) → =, 𝑟̂, (b) )→ =, →, (c) )→ =, , 𝑟̂, , (d) a&c both, , 24. Farad/m is equivalent to(a) coluom/N-m2, (b) coluom/N-m, (c) culom2/N-m2, (d) none, 25. 1 Faraday is equivalent to –, (a) 96500 C (b) 96485 C, (c) both, (d) none
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26. Two fixed charges +4q and +q are at a distance 3m, apart . At what point between the charges a third, charge +q must be placed to keep it in equilibrium.?, (a)2m from +q, (b) 2m from +4q, (c) 1m from +4q, (d) none of these, 27. Four Charges Q, q, Q and q are kept at the four, corners of a square . What is the relation ; between, Q and q, so that the net force on a charge 'q' is zero?, ¼a½ q=2√2Q, ¼b½ Q=-2√2q, ¼c½ q=-2√2Q, ¼d½ q=-Q, 28. If two charges +4e and +e are at a distance 'x' apart, then at what distance charge q must be placed from, +e , so that it is in equilibrium?, ¼a½ x/2, ¼b½x/3, ¼c½ x/6, ¼d½ 2x/3, 29. A charge'q' is lying at mid point of the line joining, the two similar charges Q. The system will be in, equilibrium , if the value of 'q' is:¼a½ Q/2, ¼b½ &Q/2, ¼c½ Q/4, ¼d½ &Q/4, 30. Two point charges +2C and +6C repel each other, with a force of 12 N . If a charge of -4C is given to, each of these charges , the force now is –, (a)4N (repulsive), (b) 4N (attractive), (c) 12N (attractive), (d) 8 N (repulsive), 31. Ratio of the permittivity of medium to the, permittivity of free space is known as(a) Permittivity, (b) Absolute permittivity, (b) Relative permittivity, (d) All of above, 32. Four charges q,2q , 3q, 4q are placed at corners, A,B,C and D of a square as shown below in the, figure direction of net force at is –, , 2q B, , Aq, (a), (b), (c), (d), , Along AB, Along CB, Along AC, Along DB, , 3q, D, , 4q, C, , 33. Four charges each equal to –Q are placed at the, corners of a square and a charge +q is placed at its, centre If the system is in equilibrium the value of 'q', is –, ¼a½, 1 + 2√2, ¼b½ − (1 + 2√2), ¼c½, , 1 + 2√2, , ¼d½, , (1 + 2√2), , 34. Four equal charges 'q' are lying at the corners, A,B,C and D of square of side 'a' . The, resultant force on charge at D will be¼a½ zero, , ¼b½, , (, , ), , ¼c½, ¼d½, 35. The dielectric constant of water is 81. Its, absolute pumittivity will be., ¼a½ 9-12×10&10 Nm2@C2, ¼b½ 7-12× 10-15Nm2@C2, -10, 2, 2, ¼c½ 7-12×10 Nm @C, ¼d½ 1-02×10-13 Nm2@C2, 36. Dielectric constant is –, (a) Dimensionless quantity (b) universal constant, (c) Conversion Factor, (d) None of above, , 37., (a), (b), (c), (d), 38., , If a body is charged by rubbing , its weight, decreases slightly, Increases slightly, Remains constant, May increase or decrease slightly, Four charges are arranged at the corners of a, square ABCD as shown in the figure . The force, on the charge kept at the centre 'O' is –, (a)zero, (b)Along the diagonal, (c)Along the diagonal, (d) perpendicular to, , A +q, , -2q, C, , B, +2q, +q, D, , 39. Two charges each of 1C are at a distance 1km, apart the force between them is¼a½ 9×103N, ¼b½ 9×10-3 N, -4, ¼c½ 1-1×10 N, ¼d½ 104 N, 40. Two charges each equal to 2C are placed in, Vacuum at a distance 0.5 cm , then, electrostatics force between then is ¼a½ 1-89N, ¼b½ 2-44 N ¼c½ 0-144 N ¼d½ 3-144 N, 41. The unit of intensity of electric field is, (a) newton/coulomb, (b) Joule/coulomb, (c) Volt-metre, (d) newton/metre, 42. Which of the following is deflected by electric field?, (a) x-rays, (b) - rays, (c)Neutrons, (d) -particles, 43. Electric field strength due to a point charge of 5C, at a distance 80cm from the charge is, ¼a½ 8×104 NC-1, ¼b½ 7×104 NC-1, 4, -1, ¼c½5×10 NC, ¼d½ 4×104 NC-1, 44. The electric field due to a charge at a distance of 3 m, from it is 500 NC-1 . The magnitude of the charge is –, ¼a½2-5 C, ¼b½ 2-0C, ¼c½ 1-0C, ¼d½ 0-5C, 45. Two charges +5C and +10C are placed 20cm, apart . The net electric field at the mid-point, between the two charges is –, ¼a½ 4-5×106 NC-1 , towards +5C, ¼b½ 4-5×106 NC-1 , towards +10 C, ¼c½ 13-5×106 NC-1 ,towards +5C, ¼d½ 13-5×106 NC-1 , towards +10C, 46. 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 –, ¼a½ 8L, ¼b½ 4L, ¼c½ 2L, ¼d½ L/4, 47. Two point charges of 2C and 80C are 10cm apart, where will the electric field strength be zero on the, line joining the charges from 20 C charge., (a) 0.1m, (b) 0.04m (c)0.033m, (d) 0.33m, 48. q,2q,3q and 4q charges are placed at the four, corners A,B,C and D of a square. The field at the, centre P of the square has the direction along., ¼a½ AB, ¼b½ CB, ¼c½ AC, ¼d½ BD
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49. The distance between the two charges 25C and, 36C is 11 cm. At what point on the line joining the, two, the intensity will be zero ?, (a) At a distance of 5 cm from 25C, (b) At a distance of 5 cm from 36C, (c) At a distance of 4 cm from 25C, (d) At a distance of 4 cm from 36C, 50. ABC is an equilateral triangle, Charges -2q are, placed at each corner . The electric intensity at O, will be-2q, A, ¼a½, a, , ¼b½, ¼c½ zero, ¼d½, , -2q, , a, , O, , a, C -2q, , B, , 51. Volt/m is – equivalent to –, , ¼a½ N/m, , ¼b½ N-m, , ¼c½ N-C, , ¼d½ N/C, , 61. Charges 20nC are separated by 5 mm. Calculate, the magnitude of dipole moment., ¼a½ 10&12 m, ¼b½ 10&10 m, &11, ¼c½ 10 m, ¼d½ None of these, 62. The electric dipole moment of an electron and a, proton 4.3 mm apart, is¼a½ 6.8×10-28 C-m, ¼b½ 2-56×10-28 C2-m, -14, ¼c½ 3-72×10 C-m, ¼d½ 11×10-28 C-m, 63. If Ea be the electric field strength of a short dipole at, a point on the axial line and Ee that on the, equatorial line at the same distance, then¼a½ Eb=2Ea, ¼b½ Ea =2Eb, ¼c½ Ea =Eb, ¼d½ None of these, 64. Electric field due to an electric dipole at a distance r, from its centre in axial position is E. If the dipole is, rotated through an angle of 900 about its, perpendicular axis. then the magnitude of electric, field at the same point will be –, ¼a½ E, ¼b½ E/4, ¼c½ E/2, ¼d½ 2E, 65. The torque acting on a dipole of moment → in an, , 52. Dimensional formula of electric field is, , ¼a½[MLT-2A-1], ¼c½ [ML2T-2A-2], , ¼b½ [ML2TA-2], ¼d½ [MLT-3A-1], , 53. Two positive point charges q1=16C and q2= 4 C, are separated in vacuum by a distance of 3.0m , find, the point on the line between the charges, where the, net electric field is zero., , ¼a½ 2m from q1, ¼c½ 1m from q2, , ¼a½An electric field only ¼b½ A magnetic field only, ¼c½ Both magnetic and electric field, ¼d½ Meither electric nor magnetic field, 55. Electric lines of force about negative point charge, are-, , ¼a½, ¼b½, ¼c½, ¼d½, , Circular, anticlockwise, Circular , clockwise, Radial inward, radial outward, , 56. An electron and a proton are lying ina uniform, electric field . The ratio of their acceleration will be :, , ¼a½ One, , ¼b½ zero ¼c½ mp@me, , ¼a½ Expands, ¼b½ Contraets, ¼c½ neithes expands nor contraets, ¼d½ None of above, 58. Two like charges are placed at some distance apart., The resultant electric field at their qutre will be-, , ¼b½ infinity, ¼d½ 2kq/r2, , 59. are placed on comers of a square as shown in figure, having side of 5 cm. I Q is one micro coulomb. , then, electric field intensity at centre will be -, , ¼a½, ¼b½, ¼c½, ¼d½, , 1-02×107N/C Upwards, 2-04×107N/C downwards, 2-04×107N/C Upwards, 1-05×107N/C downwards, , 60. The figure shows some of the electric field lines, corresponding to an electric field. The figure, suggests., , ¼a½ EA>EB>E, ¼c½ EA=EB>EC, , ¼b½ EA=EB=EC, ¼d½ EA=EB<EC, , 𝑬, , ¼b½ ⎯ →, ×, , ¼c½ 𝑧𝑒𝑟𝑜, , ¼d½ ⎯ →, ×, , 66. When an electric dipole → is placed in a uniform, 𝑷, , electric field→ . Then at what angle between → and, 𝑬, , → the value of torque. will be maximum., , 𝑷, , 𝑬, , 67., (a), (b), (c), (d), 68., , 69., , ¼d½ me @mp, , 57. On being negatively charged , a soap bubble, , ¼a½ zero, ¼c½ Kq/r2, , ¼a½ → .→, , ¼b½ 2m from q2, ¼d½0.5m from q2, , 54. The electric charge in uniform motion produces-, , 𝑷, , electric field → is –, , 70., , 71., , ¼a½ 900, ¼b½ 00, ¼c½ 1800, ¼d½ 450, An electric dipole is kept in non-uniform electric, field it., A force and a torque, a force but not a torque, a torque but not a force, Neither a force nor a torque, An electric dipole of moment p is placed normal to, the lines of force of electric intensity E. then the, work done in deflecting it through an angle of 1800 is, ¼a½ PE, ¼b½ 2PE, ¼c½& 2PE ¼d½ zero, For a dipole q=2×10-6C and d=0.01m calculate the, maximum torque for this dipole, if E= 5×105NC-1, ¼a½ 1×10-3 Nm-1, ¼b½ 10×10-3N-m-1, -3, ¼c½ 10×10 N-m, ¼d½ 1×102N-m2, A charge of -2Q coulomb is revolving round another, , stationary charge of +2Q coulomb in a circular, path of radius 0.05 m. The work done in revolving, the charge will be:, ¼a½, 𝐽, ¼b½ zero ¼c½ 𝐽, ¼d½ 𝐽, The value of 'V' and 'E' at the mid point of the line, joining an electron and a proton will be¼a½ V0 & E=o, ¼b½ V=0 & E=o, ¼c½ V0 & Eo, ¼d½ V=0 &, , 72. When an electric dipole is placed in a uniform, electric field then(a) Only torque acts on it., (b) Only force acts on it., (c) both force and torque act on it., (d) neither force nor torque on its., 73. Electric potential due to an electric dipole on its, axis at distance r will be –, ¼a½, ¼b½, ¼c½ zero, ¼d½ KP
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74. Equal charges 'q' are situated at the three corners of an, equilateral triangle . The value of 'V' and 'E' at the, centroid will be –, ¼a½ V=0 & Eo, ¼b½ V0 & E=o, ¼c½ V0 & Eo, ¼d½ V=0 & E=o, 75. The electric potential while moving along the lines of, force:, (a) decrease, (b) increases, (b) remains constant, (d) becomes infinite, 76. An electric dipole when placed in a non-uniform electric, field , will execute(a) Only linear motion, (b) only rotator motion, (c) both linear and rotator motions, (d) neither linear motion nor rotator, 77. Electric field due to electric dipole varries as –, ¼a½ 𝐸 = ¼b½ 𝐸 = ¼c½ 𝐸 =, ¼d½ 𝐸 = 𝑟, 78. Electric potential due to electric dipole varries as –, ¼a½ 𝑉 ∝ 𝑟 ¼b½ 𝑉 ∝ 𝑟 ¼c½ 𝑉 ∝ 𝑟, ¼d½ 𝑉 ∝ 𝑟, 79. At the mid point of electric dipole¼a½ E=o, V0, ¼b½ Eo, V=0, ¼c½ E=o, V=0, ¼d½ Eo, V0, 80. Potential energy of electric dipole in uniform electric field, is –, ¼a½ -PE sin, ¼b½ PEcos, ¼c½ - PEcos, ¼d½ 0, 81. Find the work done by some external force in moving a, charge q=4C from infinity to a point. Where electric, potential is 104V., ¼a½ 4×10&2 J, ¼b½ 2×10&2 J, &2, ¼c½ 8×10 J, ¼d½ 1×10&2 J, 82. Equal charges are given to two spheres of different radii., The potential will –, (a) be more one the smaller sphere, (b) be more on the bigger sphere, (c) be equal on both the spheres, (d) depend on the nature of the materials of the spheres, 83. Three charges 2q,-q,-q are located at the vertices of an, equilateral, triangle At the centre of the triangle., (a) the field is zero but potential is non-zero, (b) the field is non-zero but potential is zero, (c) Both field and potential are zero, (d) Both field and potential are non-zero, 84. Electric field due to point charge varies¼a½ E r0, ¼b½ 𝐸 ∝, 85., (a), (b), 86., , ¼c½ 𝐸 ∝, ¼d½ 𝐸 ∝, S.I. unit electric potential is :, Joule/C, (b) Volt, a and b both, (d) none of these, Variation of electric potential with distance is –, ¼a½ V r0, ¼b½ 𝑉 ∝, , ¼c½ 𝑉 ∝, ¼d½ 𝑉 ∝, 87. The work done in bringing a 20C charge from point A to, point B for distance 0.2 m is 2 J. The potential difference, between the two points will be (in volt)., ¼a½ 0-2, ¼b½ 8, ¼c½ 0-1, ¼d½ 0-4, 88. As shown in the figure, charges +q and –q are placed at, the vertices B and C isosceles triangle. The potential at the, vertex A is –, A, ¼a½, √, , ¼b½zero, ¼c½, ¼d½, , a, √, (, , ), , b, , √, , B +q, , b, , -q C, , 89. Two plates are 2cm apart a potential difference of 10V is, applied between them, the electric field between the plates, is –, ¼a½ 20NC-1, ¼b½ 500 NC-1, -1, ¼c½ 5 NC, ¼d½ 250 NC-1h, 90. At a certain distance from a point charge the electric field, is 500Vm-1 and the potential is 3000V . What is this, distance ?, ¼a½ 6 m, ¼b½ 12 m, ¼c½ 36 m, ¼d½ 144 m, 91. Two charges of 4C each are placed at the corners A and, B of an equilateral triangle of side length 0.2 m in air . The, 𝟏, electric potential at C, [𝟒𝝅𝜺 𝟗 × 𝟏𝟎𝟗 N-m2C-2], 𝟎, , ¼a½ 9×104V, ¼b½ 18×104V, ¼c½ 36×104V, ¼d½ 36×10-4V, 92. The electric potential V at any point O (x,y,z all in meters ), in space is given by V=4x2 Volt . The electric field at the, point (1m, 0, 2m) in volt, metre-1 is (a)8 along negative X-axis, (b) 8 along positive x – axis, (c) 16 along negative X-axis, (d) 16 along positive Z -axis, 93. The variation of potential 'V' with distance 'r' in a region, with E=0 , will be¼a½ V r0, ¼b½ 𝑉 ∝, , ¼c½ 𝑉 ∝, ¼d½ 𝑉 ∝, 94. A particle of mass 'm' and charge 'q' is accelerated, through a potential difference of V volt . Its energy will be¼a½ q.v, ¼b½ mqv, ¼c½, .𝑉, ¼d½, 95. The electric potential V is given as a function of distance x, (metre by V= (5x2+10x-9) Volt. value of electric field at, x=1 is ., ¼a½ -20Vm-1, ¼b½ 6 Vm-1, ¼c½ 11 Vm-1, ¼d½ -23 Vm-1, 96. 1MeV is equivalent –, ¼a½ 1-6×10&12 Joule, ¼b½1-6×10&13 Joule, ¼c½ 1-6×10&10 Joule, ¼d½ 1-6×10&19 Joule, 97. An electron is accelerated through a potential difference of, 'V' volt , then its momentum will be:, ¼a½ √2𝑚𝑒𝑉, , ¼b½, , ¼c½ √𝑚𝑒𝑉, ¼d½𝑚𝑒𝑣, 98. Work done on equipotential surfaces is :(a) finite and positive, (b) Infinite, (c) finite ane negative, (d) zero, 99. Unit of dipole moment is –, (a) Amp-m, (b) Coulomb-m, (c) Amp-m2, (d) Coulmb-m2, 100. Direction of dipole moment is –, (a) –Q to +Q is Charge, (b) +Q to –Q Charge, (c) a and b, (d) None
