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https://t.me/NEET_StudyMaterial, Chapter, , 16, , Electric Charges and Fields, Solutions, SECTION - A, Objective Type Questions, 1., , If a body has positive charge on it, then it means it has, (1) Gained some protons, , (2) Lost some protons, , (3) Gained some electrons, , (4) Lost some electrons, , Sol. Answer (4), Due to lack of electron body get positive charge., 2., , Sure check for presence of electric charge is, (1) Process of induction, , (2) Repulsion between bodies, , (3) Attraction between bodies, , (4) Frictional force between bodies, , Sol. Answer (2), Due to similar (like charge), repulsion force is possible but attraction force may be due to uncharged body., 3., , If a solid and a hollow conducting sphere have same radius then, (1) Hollow sphere will hold more maximum charge, (2) Solid sphere will hold more maximum charge, (3) Both the spheres will hold same maximum charge, (4) Both the sphere can’t hold charge, , Sol. Answer (3), Excess charge spread on outer surface only from their property., 4., , When a conducting soap bubble is negatively charged then, (1) Its size starts varying arbitrarily, , (2) It expands, , (3) It contracts, , (4) No change in its size takes place, , Sol. Answer (2), Due to repulsion force between diametrically opposite wall, it expands., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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2, 5., , Electric Charges and Fields, , Solution of Assignment, , Five balls marked a to e are suspended using separate threads. Pairs (b, c) and (d, e) show electrostatic, repulsion while pairs (a, b), (c, e) and (a, e) show electrostatic attraction. The ball marked a must be, (1) Negatively charged, , (2) Positively charged, , (3) Uncharged, , (4) Any of the above is possible, , Sol. Answer (3), 6., , When a plastic rod rubbed with wool is brought near the knob of a negatively charged gold leaf electroscope,, the gold leaves, (1) Contract, , (2) Dilate, , (3) Start oscillating, , (4) Collapse completely, , Sol. Answer (2), 7., , Coulomb’s law is analogous to, (1) Charge conservation law, , (2) Newton’s second law of motion, , (3) Law of conservation of energy, , (4) Newton’s law of gravitation, , Sol. Answer (4), Coulomb’s law and Newton’s law of gravitation are inverse square law., 8., , Two point charges Q1 and Q2 exert a force F on each other when kept certain distance apart. If the charge, on each particle is halved and the distance between the two particles is doubled, then the new force between, the two particles would be, (1), , F, 2, , (2), , F, 4, , (3), , F, 8, , (4), , F, 16, , Sol. Answer (4), Given, , F, , KQ1Q2, r, , if, Q1 , , Q2 , , …(i), , Q1, 2, , Q2, 2, , & r = 2r, Then = F1 , 9., , F, 16, , Two equally charged identical small balls kept some fixed distance apart exert a repulsive force F on each, other. A similar uncharged ball, after touching one of them is placed at the mid-point of line joining the two, balls. Force experienced by the third ball is, (1) 4F, , (2) 2F, , (3) F, , (4), , F, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 3, , Sol. Answer (3), First case : Q, , F1  F , , Q, , r, , KQ 2, r2, , …(i), , Q, Second case :, 2, , F1, , Q, 2, , r, 2, , F2, r, 2, , Q, , FNet  F2  F1, , , , K Q2 4 K Q2 4, , 4.r 2, 2r 2, , KQ 2, F, r2, , Force remain’s constant, 10. Two equal point charges A and B are R distance apart. A third point charge placed on the perpendicular, bisector at a distance ‘d’ from the centre will experience maximum electrostatic force when, (1) d , , R, 2 2, , (2) d , , R, , (3) d  R 2, , 2, , Sol. Answer (1), F1  F2 , , KQ, , F, , F2, , 2, , ⎡ 2 R2 ⎤, ⎢d  4 ⎥, ⎣, ⎦, , d2 , , (+)Q, , FN = F1 cos + F2 cos, , F1, ,  , R2, 4, , , R, 2 d, R, , (4) d  2 2R, , a, , (+)Q, , = 2F1 cos =, FN  2., , KQ 2, d, 1, ., ⎛ 2 R2 ⎞ ⎡ 2 R2 ⎤ 2, ⎜⎝ d  4 ⎟⎠ ⎢d  4 ⎥, ⎣, ⎦, , If F = Maximum than, , So we get  , , dF, 0, d ”d ', , R, 2 2, , 11. A charged gold leaf electroscope has its leaves apart by certain amount having enclosed air. When the, electroscope is subjected to X-rays, then the leaves, (1) Further dilate, , (2) Start oscillating, , (3) Collapse, , (4) Remain unaltered, , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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4, , Electric Charges and Fields, , Solution of Assignment, , 12. Two equal positive charges Q are fixed at points (a, 0) and (–a, 0) on the x-axis. An opposite charge –q at, rest is released from point (0, a) on the y-axis. The charge –q will, (1) Move to infinity, , (2) Move to origin and rest there, , (3) Undergo SHM about the origin, , (4) Execute oscillatory periodic motion but not SHM, , Sol. Answer (4), , –q, , In question, , F2, , Net force on q is not proportional to (x), as F  (–x) [For SHM], , (+)Q, , but Net force on q is, , F, , 2, ,  a2, , , , (+)Q, x = –a, , K q. Q x, , x, , F1, , , x, , x = +a, , 1, 2, , This is condition for periodic motion, 13. Four charges each equal to Q are placed at the four corners of a square and a charge q is placed at the, centre of the square. If the system is in equilibrium then the value of q is, (1), , Q, (1  2 2), 2, , (2), , –Q, (1  2 2), 4, , (3), , Q, (1  2 2), 4, , (4), , –Q, (1  2 2), 2, , Sol. Answer (2), Net force on Q due to other corner charge is, , Q, , Q, , F123 = F3  F12  F22, , 2l, , = F3  2 F1, 2, , =, , KQ, 2KQ, , 2l 2, l2, , 2, , Q, , F4, F1, F2, , Q, , F3, , Force on Q1 due to centre charge –q, , F4 , , KQq, .2, l2, , If net force on corner charge Q is zero, Then, F123 + F4 = 0, So q  , , Q⎡, 1  2 2 ⎤⎦, 4⎣, , 14. Which of the following is not true about electric charge?, (1) Charge on a body is always integral multiple of certain charge known as charge of electron, (2) Charge is a scalar quantity, (3) Net charge on an isolated system is always conserved, (4) Charge can be converted into energy and energy can be converted into charge, Sol. Answer (4), A rest charge cannot be converted into energy., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 5, , 15. What is the amount of charge possessed by 1 kg of electrons?, (1) 1.76 × 1011 C, , (2) 1.76 × 10–9 C, , (3) 1.76 × 10–7 C, , (4) 1.76 × 10–5 C, , Sol. Answer (1), ∵ me  9.1  10 31kg, qe  1.6  10 19 C, , So charge due to 1 kg electron, Q, , 1.6  10 19,  1.76  1011C, 9.1  10 31, , 16. According to Coulomb’s Law, which is correct relation for the following diagram?, q1, , (1) q1 q2 < 0, , (2) q1 q2 > 0, , F21 q2, , F12, , (3) q1 q2 = 0, , (4) q1 q2 >> 100 C, , Sol. Answer (1), Both charge should be unlike charge, q1 = +Q, So q1 q2 =, , ,, , q2 = –Q, , –Q2, , So q1 q2 = Negative, So q1 q2 < 0, 17. A charge q is to be distributed on two conducting spheres. What should be the value of the charges on the, spheres so that the repulsive force between them is maximum when they are placed at a fixed distance from, each other in air?, q, q, and, 2, 2, Sol. Answer (1), , (2), , (1), , q, 3q, and, 4, 4, , (3), , q, 2q, and, 3, 3, , (4), , q, 4q, and, 5, 5, , Force between both is, , K Q  q  Q, r2, If F = maximum then, F, , dF, 0, dQ, , So Q , , q, 2, , So both charge be, , q q, ,, 2 2, , 18. A point charge q1 exerts an electric force on a second point charge q2. If third charge q3 is brought near, the, electric force of q1 exerted on q2, (1) Decreases, (2) Increases, (3) Remains unchanged, (4) Increases if q3 is of same sign as q1 and decreases if q3 is of opposite sign, Sol. Answer (3), Electric force between ‘2’ charge do not depend on the ‘3’rd charge., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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6, , Electric Charges and Fields, , Solution of Assignment, , , and  distance away from, 2, one end respectively. What should be Q in order to make the net force on q to be zero?, , 19. Three charges +4q, Q and q are placed in a straight line of length l at points 0,, , (1) –q, , (3) –, , (2) 4q, , q, 2, , (4) –2q, , Sol. Answer (1), FNet on q is, F, , x=0, , K .4q 3 KQq.4, , l2, l2, , l, 2, , x, , +4q, , x=l, , Q, , q, , If F = 0 then, Q = –q, 20. A particle of mass m and carrying charge –q1 is moving around a charge +q2 along a circular path of radius, r. Find period of revolution of the charge –q1, (1), , 163 0 mr 3, q1 q2, , (2), , 83 0 mr 3, q1 q2, , (3), , q1 q2, 163 0 mr 3, , (4) Zero, , Sol. Answer (1), , , , mv 2, 1 q1 q2, , ., 4 0 r 2, r, , ⎡ 1 q1 q 2 ⎤, v ⎢, ⎥, ⎣ 4  0 rm ⎦, , q2, r, , 1, 2, , m, (–q), , For 1 trip, , T , T , , 1, 2 r,  2r  4  0 mr   q1 q 2  2, v, , 16 3  0 mr 3, q1 q2, , 21. Consider three point objects P, Q and R. P and Q repel each other, while P and R attract. What is the nature, of force between Q and R?, (1) Repulsive force, , (2) Attractive force, , (3) No force, , (4) None of these, , Sol. Answer (2), 22. Which of the following processes involves the principle of electrostatic induction?, (1) Pollination, , (2) Chocolate making, , (3) Xerox copying, , (4) All of these, , Sol. Answer (4), These are properties of electrostatic induction., 23. The electric field intensity at a point in vacuum is equal to, (1) Zero, (2) Force a proton would experience there, (3) Force an electron would experience there, (4) Force a unit positive charge would experience there, Sol. Answer (4), This is defination of electric field., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 7, , 24. A sphere of radius r has electric charge uniformly distributed in its entire volume. At a distance d from the, centre inside the sphere (d < r) the electric field intensity is directly proportional to, (1), , 1, d, , (2), , 1, d2, , (4) d 2, , (3) d, , Sol. Answer (3), , q, , Electric field inside volume charge is given by, , d, , 1, qd, E , 4  0 r 3, , r, , Ed, 25. The electric field at 2R from the centre of a uniformly charged non-conducting sphere of radius R is E. The, electric field at a distance, , R, from the centre will be, 2, , (1) Zero, , (2) 2E, , (3) 4E, , (4) 16E, , Sol. Answer (2), , Kq, Given E  2R 2,  , , Then E ' , , Kq., , …(i), , R, 2, , …(ii), , R3, , Find E' = 2E, 26. In a uniform electric field if a charge is fired in a direction different from the line of electric field then the, trajectory of the charge will be a, (1) Straight line, , (2) Circle, , (3) Parabola, , (4) Ellipse, , Sol. Answer (3), , y, , F = qE = m ax, , (+q), , ⎡ qE ⎤, ax  ⎢ ⎥, ⎣m⎦, , Then, x  0 , , y, 1, ax t 2, 2, , But y = uxt then t , So, x , , (E), …(i), , x, , x, , y, ux, , 1 qE y 2, ., 2 m u2, , so x  y2 for parabola, 27. A positively charged pendulum is oscillating in a uniform electric field pointing upwards. Its time period as, compared to that when it oscillates without electric field, (1) Is less, , (2) Is more, , (3) Remains unchanged (4) Starts fluctuating, , Sol. Answer (2), Effective g decreases., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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8, , Electric Charges and Fields, , Solution of Assignment, , 28. How many electrons should be removed from a coin of mas 1.6 g, so that it may float in an electric field of, intensity 109 N/C directed upward?, (1) 9.8 × 107, , (2) 9.8 × 105, , (3) 9.8 × 103, , (4) 9.8 × 101, , Sol. Answer (1), qE = mg, neE = mg, Use n , , mq, eE, , 29. ABC is an equilateral triangle. Charges +q are placed at each corner. The electric field intensity at the centroid, of triangle will be, A +q, , O, , +q, B, (1), , 1, q,  2, 40 r, , (2), , 1, 3q,  2, 40 r, , r, , +q, C, (3), , 1, q, , 40 r, , (4) Zero, , Sol. Answer (4), FN = 0, 30. A charge Q is placed at the centre of a square. If electric field intensity due to the charge at the corners of, E, the square is E1 and the intensity at the mid point of the side of square is E2, then the ratio of 1 will be, E2, (1), , 1, , 2 2, Sol. Answer (3), E1 , , 1 Q2, 4  0 l 2, , E2 , , 1, Q4, 4  0 l 2, , (2), , 2, , (3), , 1, 2, , (4) 2, , …(i), , E2 = 2E1, , E1 1, , E2 2, 31. Point charges each of magnitude Q are placed at three corners of a square as shown in the diagram. What, is the direction of the resultant electric field at the fourth corner?, C, , B, A, , O, , +Q, , (1) OC, , (2) OE, , D, , E, , +Q, , –Q, , (3) OD, , (4) OB, , Sol. Answer (4), Resultant force act along OB, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 9, , 32. Two charges e and 3e are placed at a distance r. The distance of the point where the electric field intensity, will be zero is, r, from 3e charge, (1  3 ), , (1), , r, , (3), , (1– 3 ), , from 3e charge, , (2), , (4), , r, from e charge, (1  3 ), r, 1, 1, 3, , from e charge, , Sol. Answer (2), Net electric field at P is zero then, , e, , P, x, , O = E1 – E2, , 3e, r, , E1 = E2, , ke, k 3e, , 2, x,  r  x 2, , 1, 3, , x r x, , so,, , r  x  3x, , r  x ⎡⎣1  3 ⎤⎦, x, , , , r, , 1 3, , , , 33. If electric lines of force in a region are represented as shown in the figure, then one can conclude that, electric, field is, , (1) Non-uniform, , (2) Uniform, , (3) Both uniform and non-uniform, , (4) Zero everywhere, , Sol. Answer (1), Diverging electric line of force denote non-uniform electric field., 34. An uncharged sphere of metal is placed in a uniform electric field produced by two oppositely charged plates., The lines of force will appear as, , +, (1), , (2), , (3), , (4), , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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10, , Electric Charges and Fields, , Solution of Assignment, , 35. An electron released on the axis of a positively charged ring at a large distance from the centre will, (1) Not move, , (2) Do oscillatory motion, , (3) Do SHM, , (4) Do non periodic motion, , Sol. Answer (2), , FN  q., , k Q x, , x, , 2, ,  r2, , , , 3, 2, , For SHM, F  (–x), 36. Electric charge Q, Q and –2Q respectively are placed at the three corners of an equilateral triangle of side, a. Magnitude of the electric dipole moment of the system is, (1), , 2Qa, , (2), , 3Qa, , (3) Qa, , (4) 2Qa, , Q, , Sol. Answer (2), P1 = Q.a, P2 = Q.a, , q, P1, , P  P12  P22  2PP, 1 2 cos , , P2 , Q, , PN  3 Q.a, , –2Q, , q, , 37. An electric dipole placed in a uniform electric field experiences maximum moment of couple when the dipole, is placed, (1) Against the direction of the field, , (2) Towards the electric field, , (3) Perpendicular to the direction of the field, , (4) At 135° to the direction of the field, , Sol. Answer (3),  = PE sin, For  = 90°,  = Max., 38. Force of interaction between two co-axial short electric dipoles whose centres are R distance apart varies, as, (1), , 1, R, , (2), , 1, R2, , (3), , 1, R3, , (4), , 1, R4, , Sol. Answer (4), F, , K 6P1 P2, r4, , –Q, , –Q, , +Q, , +Q, , r, , 39. Two charges of +25 × 10–9 coulomb and –25 × 10–9 coulomb are placed 6 m apart. Find the electric field, intensity ratio at points 4 m from the centre of the electric dipole (i) on axial line (ii) on equatorial line, (1), , 1000, 49, , (2), , 49, 1000, , (3), , 500, 49, , (4), , 49, 500, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 11, , Sol. Answer (1), , Eaxial , , Eeq. , , k.2Pr, , r, , 2, ,  l2, , , , 2, , …(i), , k.p, , r, , 2, ,  l2, , , , 3, 2, , …(ii), , Eaxial 1000, Find E  49, eq, , 40. The electric force on a point charge situated on the axis of a short dipole is F. If the charge is shifted along, the axis to double the distance, the electric force acting will be, (1) 4F, , (2), , F, 2, , (3), , F, 4, , (4), , F, 8, , Sol. Answer (4), F, , 1⎡, 1⎤, F  3⎥, r 3 ⎢⎣, r ⎦, , If r1 = 2r, Then force become, , F, 8, , 41. An electric dipole is placed at an angle 60° with an electric field of strength 4 × 105 N/C. It experiences a, torque equal to 8 3 Nm . Calculate the charge on the dipole, if dipole is of length 4 cm, (1) 10–1 C, , (2) 10–2 C, , (3) 10–3 C, , (4) 10–4 C, , Sol. Answer (3),  = PE sin, 8 3  ⎡⎣q  4  102 ⎤⎦ . 4  105 sin60, , find q = 10–3, 42. Figure shows electric lines of forces due to charges Q1 and Q2. Hence, , Q1, , Q2, , (1) Q1 and Q2 both are negative, , (2) Q1 and Q2 both are positive, , (3) Q1 > Q2, , (4) Both (2) & (3), , Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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12, , Electric Charges and Fields, , Solution of Assignment, , 43. Figure shows electric lines of force. If Ex and Ey are the magnitudes of electric field at points x and y, respectively, then, , x, , (1) Ex > Ey, , y, , (2) Ex = Ey, , (3) Ex < Ey, , (4) Any of these, , Sol. Answer (1), 44. A charge q is situated at the centre of a cube. Electric flux through one of the faces of the cube is, , q, (2) 3, 0, , q, (1) , 0, , q, (3) 6, 0, , (4) Zero, , Sol. Answer (3), Total  of 6 surface  ,  One surface , , q, 0, , q, 60, , 45. A charge Q is placed at the centre of the open end of a cylindrical vessel. Electric flux through the surface, of the vessel is, , q, (1) 2, 0, , q, (2) , 0, , (3), , 2q, 0, , (4) Zero, , Sol. Answer (1), If charge Q is surrounded with two cylinder then flux of ‘2’ cylinder is, ,  , , q, 0, , q, Flux of one cylinder = 2 , 0, , 46. A hemispherical surface of radius R is kept in a uniform electric field E as shown in figure. The flux through, the curved surface is, , E, , R, (1) E2R2, , (2) ER2, , (3) E4R2, , (4) Zero, , Sol. Answer (2), qNet = 0, , So Net = 0, , in +out = 0, out = –in = –[E.R2cos180°], = E.R2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 13, , 47. Total electric flux associated with unit positive charge in vacuum is, (1) 40, , 1, (2) 4, 0, , 1, (3) , 0, , (4) 0, , Sol. Answer (3), , , q,  0 for q = 1, , , , 1, 0, , 48. A charged body has an electric flux F associated with it. Now if the body is placed inside a conducting shell, then the electric flux outside the shell is, (1) Zero, , (2) Greater than F, , (3) Less than F, , (4) Equal to F, , Sol. Answer (4), Charge remains constant so flux remains constant., 49. A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The, outward flux over the surface of the cylinder is given by, (1) 2R2E, , (2), , R 2E, 2, , (3) 2RLE, , (4) R2E, , Sol. Answer (4), Not net flux only outward flux, φ = ER2, 50. A rectangular surface of sides 10 cm and 15 cm is placed inside a uniform electric field of 25 V/m, such, that the surface makes an angle of 30° with the direction of electric field. Find the flux of the electric field, through the rectangular surface, (1) 0.1675 N/m2C, , (2) 0.1875 Nm2/C, , Sol. Answer (2), , (3) Zero, , (4) 0.1075 Nm2/C, , 25 V/m, , φ = EAcos30°, = 0.1875 Nm2/C, , 30°, , 15 cm, , A, , 10 cm, 51. If an electric field is given by 10iˆ  3 jˆ  4kˆ , calculate the electric flux through a surface of area 10 units lying, in yz plane, (1) 100 units, , (2) 10 units, , (3) 30 units, , (4) 40 units, , Sol. Answer (1), , E  10lˆ  3 jˆ  4kˆ, , , A  10lˆ, ,  , So,   E.A  100 unit, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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14, , Electric Charges and Fields, , Solution of Assignment, , 52. The electric field in a region is radially outward and at a point is given by E = 250 r V/m (where r is the, distance of the point from origin). Calculate the charge contained in a sphere of radius 20 cm centred at the, origin, (1) 2.22 × 10–6 C, , (2) 2.22 × 10–8 C, , (3) 2.22 × 10–10 C, , (4) Zero, , Sol. Answer (3), E , , 1 Q, ., 4  0 r 2, , 9  109.Q, r2, Use r = 20 × 10–2 m, 250r , , Find Q = ?, Q = 2.22 × 10–10 C, 53. There is uniform electric field of 8  103 iˆ N/C. What is the net flux (in SI units) of the uniform electric field, through a cube of side 0.3 m oriented so that its faces are parallel to the coordinate plane?, (1) 2 × 8 × 103, , (2) 0.3 × 8 × 103, , (4) 8 × 106 × 6, , (3) Zero, , Sol. Answer (3), , E  8  103 iˆ N/C (Uniform), here q = 0, So = 0, 54. A charge Q is kept at the corner of a cube. Electric flux passing through one of those faces not touching, that charge is, , Q, (1) 24, 0, , Q, (2) 3, 0, , Q, (3) 8, 0, , Q, (4) 6, 0, , Sol. Answer (1),  Net , , q, 8  0 (Of 3 surface), ,  One surface , , q, 24  0, , 55. An isolated solid metal sphere of radius R is given an electric charge. Which of the graphs below best shows, the way in which the electric field E varies with distance x from the centre of the sphere?, E, , E, , (1), , E, , (2), O, , R, , x, , E, , (3), O, , R, , x, , (4), O, , R, , x, , O, , x, , Sol. Answer (3), E outside , , kq, r2, , Einside = 0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 15, , 56. The electric field intensity at P and Q, in the shown arrangement, are in the ratio, q, a, , r, , P, , b, , 3q, 2r, , Q, , Fig.: Hollow concentric shell, , (1) 1 : 2, , (2) 2 : 1, , (3) 1 : 1, , (4) 4 : 3, , Sol. Answer (3), kq, r2, , EP , , …(i), , kq, , EQ , ,  2r , , 2, , , , k.3q, ,  2r 2, , , , k q k .3q, , 4 r 2 4r 2, , , , kq, r2, , …(ii), , EP : EQ = 1 : 1, 57. Consider an atom with atomic number Z as consisting of a positive point charge at the centre and surrounded, by a distribution of negative electricity uniformly distributed within a sphere of radius R. The electric field at, a point inside the atom at a distance r from the centre is, (1), , Ze ⎡ 1, r ⎤, –, 40 ⎢⎣ r 2 R 3 ⎥⎦, , (2), , Ze ⎡ 1, 1 ⎤, , 40 ⎢⎣ r 2 R 3 ⎥⎦, , (3), , 2Ze, 4 0 r 2, , (4) Zero, , Sol. Answer (1), E = E1 – E2, E=, , k.ze k zer, , r2, R3, , 58. An electron is rotating around an infinite positive linear charge in a circle of radius 0.1 m, if the linear charge, density is 1 C/m, then the velocity of electron in m/s will be, (1) 0.562 × 107, , (2) 5.62 × 107, , (3) 562 × 107, , (4) 0.0562 × 107, , Sol. Answer (2), mv 2, qE, r, , , mv 2,  e., r, 2 0 r, v  5.62  107 m/s, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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16, , Electric Charges and Fields, , Solution of Assignment, , , 59. A dipole with an electric moment p is located at a distance r from a long thread charged uniformly with a, , linear charge density . Find the force F acting on the dipole if the vector p is oriented along the thread, , (1), , p, 20 r 2, , So, FN = 0, , (3), , p, 20 r , , +, , +, , +, , F1, , F2, , + +, , –, , 60. For two infinitely long charged parallel sheets, the electric field at P will be, (1), , (4) Zero, , + +, , Sol. Answer (4), F1 = F2, , p, 20 r, , (2), , , , –, 2 x 2(r – x ), , (2), , , (3) , 0, , , , , 20 x 2(r – x )0, , (4) Zero, , , +, +, +, +, +, +, +, +, +, +, , x, , P, , , +, +, +, +, +, +, +, +, +, +, , r, , Sol. Answer (4), EN = E1 – E2 , , , , , =0, 2 0 2 0, , SECTION - B, Objective Type Questions, 1., , Select the correct statement about electric charge, (1) Charge can be converted into energy and energy can be converted into charge, (2) Charge of a particle increases with increase in its velocity, (3) Charge on a body is always integral multiple of a certain charge called charge of electron, (4) Charge on a body is always positive or zero, , Sol. Answer (3), Quantization of charge., 2., , Figure shows electric field lines due to a charge configuration, from this we conclude that, , q1, , q2, , (1) q1 and q2 are positive and q2 > q1, , (2) q1 and q2 are positive and q1 > q2, , (3) q1 and q2 are negative and |q1| > |q2|, , (4) q1 and q2 are negative and |q2| > |q1|, , Sol. Answer (2), (i) Electric field lines originates from positive charge., (ii) Higher the number of field lines originating from positive charge, greater is magnitude of charge., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 3., , Electric Charges and Fields, , 17, , Figure shows three concentric metallic spherical shells. The outermost shell has charge q2, the inner most shell, has charge q1, and the middle shell is uncharged. The charge appearing on the inner surface of outermost shell is, , r1, , r2, r3, , (1) q1 + q2, , (2), , q2, 2, , (3) –q1, , (4) Zero, , Sol. Answer (3), , –q', , Suppose a guassian surface passes through, conducting shell with radius (r3), , q', , q2, , q1, , Flux through it well be zero. So, net charge, enclosed must be zero., , r1, ,  q1 + q' = 0, , r2, , r3, , q' = –q1, 4., , Six point charges are placed at the vertices of a hexagon of side 1m as shown in figure. Net electric field at the, centre of the hexagon is, –q, –q, , q, , –q, , O, q, , q, , 6q, (2) 4, 0, , (1) Zero, , q, (3) , 0, , q, (4) 4, 0, , Sol. Answer (3), , 1, q, Electric field at O due to each charge is E , 4  0 1 2, So, net electric field (Enet) is,, , ⇒ E net  E 2  E 2  2E 2 cos120  2E, ⇒ E net  4E , , 5., , –q, , –q, –q, , q, q, , q, , q,  0, , A proton and an -particle having equal kinetic energy are projected in a uniform transverse electric field as shown, in figure, (1) Proton trajectory is more curved, (2) -particle trajectory is more curved, (3) Both trajectories are equally curved but in opposite direction, , + + + + + + + + +, , (4) Both trajectories are equally curved and in same direction, Sol. Answer (2), -particle has more charge than proton,  Strong electric force on -particle and more curved path., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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18, 6., , Electric Charges and Fields, , Solution of Assignment, , , Electric field in a region is uniform and is given by E  aiˆ  bˆj  ckˆ . Electric flux associated with a surface of area, , A  R 2 iˆ is, (1) aR2, , (2) 3aR2, , (3) 2abR, , (4) acR, , Sol. Answer (1), ,  ,   E.A  aR 2, 7., , Which of the following is not true about electric charge?, (1) Charge is a scalar quantity, (2) Charge on an isolated system is always conserved, (3) A particle having nonzero rest mass can have zero charge, (4) A particle having zero rest mass can have non zero charge, , Sol. Answer (4), Charge is always associated with mass,  particle with zero rest mass can never have a charge., 8., , If 0 is permittivity of free space, e is charge of proton, G is universal gravitational constant and mp is mass of a, proton then the dimensional formula for, (1) [M1L1T–3A–1], , e2, 4 0Gmp 2, , (2) [M0L0T0A0], , is, (3) [M1L3T–3A–1], , (4) [M–1L–3T4A2], , Sol. Answer (2), Gravitational force F1 , , G M P2, r2, , 1 e2, Electrostatic force F2  4  r 2, 0, F2, F, , 1, , , , e2, 4  0G M P2, ,  Dimension less [ M° L° T° A°], 9., , Two positive point charges of unequal magnitude are placed at a certain distance apart. A small positive test, charge is placed at null point, then, (1) The test charge is in unstable equilibrium, , (2) The test charge is in stable equilibrium, , (3) The test charge is in neutral equilibrium, , (4) The test charge is not in equilibrium, , Sol. Answer (1), N, Q1, , q, , Q2, , When charge is displaced above, it gets repelled and move away from null point., Hence, unstable equilibrium., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 19, , 10. Three particles are projected in a uniform electric field with same velocity perpendicular to the field as shown., Which particle has highest charge to mass ratio?, C, B, , –––––––––––, , A, , +++++++++++, (1) A, , (2) B, , (3) C, , (4) All have same charge to mass ratio, , Sol. Answer (3), Charge with maximum curved path has highest charge to mass ratio., 11. An infinite line charge is at the axis of a cylinder of length 1 m and radius 7 cm. If electric field at any point on the, curved surface of cylinder is 250 NC–1, then net electric flux through the cylinder is, (1) 1.1 × 102 Nm2 C–1, , (2) 9.74 × 10–6 Nm2 C–1, , (3) 5.5 × 106 Nm2 C–1, , (4) 2.5 × 102 Nm2 C–1, , Sol. Answer (1), Charge enclosed is (q) = (1), , 7cm, , , E ,  250, 2  0  0.07 , , +, +, +, +, +, +, +, +, , So,  = 500(0.07), q, Electric flux through cylinder =  = 500 (0.07), 0, , +, +, +, +, +, +, +, +, , 1 cm, ,  1.1 × 102 Nm2 C–1, 12. A small conducting sphere is hanged by an insulating thread between the plates of a parallel plate capacitor as, shown in figure. The net force on the sphere is, , A, , +, +, +, +, +, +, +, +, +, +, +, +, +, , (1) Towards plate A, , (2) Towards plate B, , B, , –, –, –, –, –, –, –, –, –, –, –, –, –, , (3) Upwards, , (4) Zero, , Sol. Answer (4), , A, , +, , –, , +, , – +, –, +, –, + – ++, – +, , –, , +, , –, , +, , –, , B, , –, , Net force on sphere will be zero., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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20, , Electric Charges and Fields, , Solution of Assignment, , 13. Which of the following is not the unit of charge?, (1) Farad, , (2) Coulomb, , (3) Stat coulomb, , (4) Faraday, , Sol. Answer (1), Farad is not a unit of charge, 14. If two charges of 1 coulomb each are placed 1 km apart, then the force between them will be, (1) 9 × 103 N, , (2) 9 × 10–3 N, , (3) 9 × 10–4 N, , (4) 10–6 N, , Sol. Answer (1), F, , 9  109 11, , 1000 , , 2, ,  9  103 N, , 15. The magnitude of electric field strength E such that an electron placed in it would experience an electrical force, equal to its weight is given by, (1) mge, , (2), , mg, e, , (3), , e, mg, , (4), , e2 g, 2m, , Sol. Answer (2), mg = eE, , E, , mg, e, , 16. The figure shown is a plot of electric field lines due to two charges Q1 and Q2. The sign of charges is, , Q1, Q2, , (1) Both negative, , (2) Q1 positive and Q2 negative, , (3) Both positive, , (4) Q1 negative and Q2 positive, , Sol. Answer (1), Electric field lines terminates at negative charge, 17. The figure shows electric field lines. If EA and EB are electric fields at A and B and distance AB is r, then, A, , (1) EA > EB, , (2) EA = EB/r, , B, (3) EA < EB, , (4) EA = EB/r2, , Sol. Answer (1), EA > EB,  Closes the electric field lines, stronger is electric field., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 21, , 18. Electric charge q, q and –2q are placed at the corners of an equilateral triangle ABC of side L. The magnitude, of electric dipole moment of the system is, (1) qL, , (2) 2qL, , (3), , (4) 4qL, , 3qL, +q, , Sol. Answer (3), P = qL, Pnet  P 2  P 2  2P 2 cos 60, , P, 60°, , Pnet  3P  3qL, , –q, , P, , +q, , 19. A given charge situated at a certain distance from a short electric dipole in the end on position experience, a force F. If the distance of the charge is doubled, the force acting on the charge will be, (1) 2F, , (2), , F, 2, , (3), , F, 4, , (4), , F, 8, , Sol. Answer (4), , 1, r3, on doubling the distance, F = qE and E , , E' , , E, 8, , So, F ' , , F, 8, , , , 20. The torque  acting on an electric dipole of dipole moment p in an electric field E is, ,  ,   , ,   , (1)   p ·E, (2)   p  E, (3)   p E, (4)   pE, Sol. Answer (2),   ,   P E, 21. An electric dipole consists of two opposite charges each of magnitude 1C separated by a distance of 2 cm., The dipole is placed in an external field of 105 N/C. The maximum torque on the dipole is, (1) 2 × 10–4 N m, , (2) 2 × 10–3 N m, , (3) 4 × 10–3 N m, , (4) 10–3 N m, , Sol. Answer (2), Max. torque max = pE sin90°, = (1 × 10–6) (2 × 10–2) (105), = 2 × 10–3 Nm, 22. A charge Q is situated at the centre of a cube. The electric flux through one of the faces of the cube is, , Q, (1) , 0, , Q, (2) 2, 0, , Q, (3) 4, 0, , Q, (4) 6, 0, , Sol. Answer (4), , Q, Total flux through cube      (Six surfaces), 0, Q, Flux through each face = 6 , 0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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22, , Electric Charges and Fields, , Solution of Assignment, , 23. A charge q is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the, surface of the vessel is, (1) Zero, , (2), , q, 0, , (3), , q, 20, , (4), , 2q, 0, , Sol. Answer (3), q, Total flux through the cylindrical gaussian surface = , 0, , .q, , 1, Flux through open cylinder = (Total flux), 2, , q, = 2, 0, , 24. A charged body has an electric flux  associated with it. The body is now placed inside a metallic container. The, flux , outside the container will be, (2) Equal to , , (1) Zero, , (3) Greater than , , (4) Less than , , Sol. Answer (2), , Gaussian surface, + +++++++++++, +, +, + – – +–+– – –– +, ++ – +, +, +, +, +, + –, –, +, –, +, +, +, +, –, +, +, + – – – – – – +, +, +, + +++++++++++, Metallic container, Charged body, As same charge is enclosed,  Same flux outside the container, 25. The given figure shows, two parallel plates A and B of charge densities + and – respectively. Electric intensity, will be zero in region, , I, , A, (1) I only, , (2) II only, , III, , II, , B, (3) III only, , (4) Both (1) & (3), , Sol. Answer (4), , +, (i), , (ii), , –, , (iii), , P, , 0, , M, , 0, , A, , B, , , 0, , , 0, , At points P and M is zero., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 23, , 26. If the electric field intensity in a fair weather atmosphere is 100 V/m, then the total charge on the earth’s surface, is (radius of the earth is 6400 km), (1) 4.55 × 107 C, , (2) 4.55 × 108 C, , (3) 4.55 × 105 C, , (4) 4.55 × 106 C, , Sol. Answer (3), E  100, , V, m, , R = 6400 km, , By Gauss law, q, 0, , EA , , ⇒ q  EA 0, , , , , , ⇒ q  200 4 6400  103, ,    8.85  10, 2, , 12, ,  q = 4.55 × 105 C, 27. A sphere of radius R has a uniform distribution of electric charge in its volume. At a distance x from its centre for, x < R, the electric field is directly proportional to, (1), , 1, x2, , (2), , 1, x, , (3) x, , (4) x2, , Sol. Answer (3), In non-conducting sphere,, If x < R (radius), then E , , kQx, R3, , Or E  x, 28. The electric field at 20 cm from the centre of a uniformly charged non-conducting sphere of radius 10 cm is E., Then at a distance 5 cm from the centre it will be, (1) 16 E, , (2) 4 E, , (3) 2 E, , (4) Zero, , Sol. Answer (3), R = 10 cm, r = 20 cm, , E, , kQ, ,  0.22, , Now at r = 5 cm, E' , , Now,, , kQ  0.05, ,  0.13, , E '  0.05, 2, , 0.2  2, 3 , E,  0.1, , E' = 2E, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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24, , Electric Charges and Fields, , Solution of Assignment, , 29. When two particles having charges q1 and q2 are kept at a certain distance, they exert a force F on each other., If the distance between the two particles is reduced to half and the charge on each particle, is doubled then the force between the particles would be, (1) 2F, , (2) 4F, , (3) 8F, , (4) 16F, , Sol. Answer (4), F, , kq1q2, r2, , k  2q1  2q2 , , Now, F ' , , ⎛r⎞, ⎜⎝ ⎟⎠, 2, , 2, , ⎡ kq q ⎤, F '  8 ⎢ 12 2 ⎥  8 F, ⎣ r, ⎦, , 30. Charge 2Q and –Q are placed as shown in figure. The point at which electric field intensity is zero will be, somewhere, , –Q, , +2Q, , (1) Between –Q and 2Q, (2) On the left of –Q, (3) On the right of 2Q, (4) On the perpendicular bisector of line joining the charges, Sol. Answer (2), , –Q, , +2Q, , In case of two charges of opposite polarity, neutral point always lies outside the line joining charges and, closes to smaller magnitude charge., 31. If a small sphere of mass m and charge q is hung from a silk thread at an angle  with the surface of a, vertical charged conducting plate, then for equilibrium of sphere, the surface charge density of the plate is, ⎛ mg ⎞, (1) 0 ⎜, ⎟ tan , ⎝ q ⎠, , ⎛ 2mg ⎞, (2) 0 ⎜, ⎟ tan , ⎝ q ⎠, , (3) 0(mgq)tan , , ⎛ mg ⎞, (4) 0 ⎜, ⎟ tan , ⎝ 3q ⎠, , Sol. Answer (1), , q, , 0, T sin , , T cos , mg, tan  , , , , q,  0 mg, , +, +, +, +, +, +, , +, T, +, +, +, T sin, +, +, , , , T cos, , mg, , q, 0, ,  0 mg tan , q, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 25, , 32. Two long thin charged rods with charge density  each are placed parallel to each other at a distance d apart., ⎛, 1 ⎞, The force per unit length exerted on one rod by the other will be ⎜ where k , ⎟, 40 ⎠, ⎝, , (1), , k 2, d, , (2), , k 2 2, d, , (3), , k 2, d2, , (4), , , , Sol. Answer (2), , Electric field due to rod (1) at distance ‘d’ = 2   d, 0, , ⎡  ⎤, F qE, So, force per unit length l  l   ⎢ 2  d ⎥, 0 ⎦, ⎣, , =, , +, +, +, (1), +, +, +, , d, , k 2 2, d2, , , +, +, +, (2), +, +, +, , k 2 2, d, , 33. The dimensional formula of linear charge densityis, (1) [M–1L–1T+1A], , (2) [M0L–1T+1A], , (3) [M–1L–1T+1A–1], , (4) [M0L–1T+1A–1], , Sol. Answer (2), Linear charge density () =, , , , Q, L, ,  AT   ⎡M0L1 T1A1 ⎤, ⎦, L ⎣, , 34. Two isolated metallic spheres of radii 2 cm and 4 cm are given equal charge, then the ratio of charge density, on the surfaces of the spheres will be, (1) 1 : 2, , (2) 4 : 1, , (3) 8 : 1, , (4) 1 : 4, , Sol. Answer (2), Surface charge density () =, , , , Q, 4 r 2, , 1, r2, , 1 r22 42 4,    r 2  22  1, 2, 1, , 35. If the number of electric lines of force emerging out of a closed surface is 1000, then the charge enclosed, by the surface is, (1) 8.854 × 10–9 C, , (2) 8.854 × 10–4 C, , (3) 8.854 × 10–1 C, , (4) 8.854 C, , Sol. Answer (1),   1000 , , q, 0, , q = 8.854 × 10–12 ×1000, q = 8.854 × 10–9C, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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26, , Electric Charges and Fields, , Solution of Assignment, , 36. A charge of 1 coulomb is located at the centre of a sphere of radius 10 cm and a cube of side 20 cm. The, ratio of outgoing flux from the sphere and cube will be, (1) More than one, , (2) Less than one, , (3) One, , (4) Nothing certain can be said, , Sol. Answer (3), If charge inclosed same, electric flux will be same., 37. Gauss’s law can help in easy calculation of electric field due to, (1) Moving charge only, , (2) Any charge configuration, , (3) Any symmetrical charge configuration, , (4) Some special symmetric charge configuration, , Sol. Answer (4), For easy calculation of electric field using Gauss' law, gaussian surfaces having some special symmetry with, respect to charge configuration is used., 38. An electric dipole when placed in a uniform electric field E will have minimum potential energy, when the angle, made by dipole moment with field E is, (1) , , (2), , 3, 2, , (3) Zero, , (4), , , 2, , Sol. Answer (3), U = –pE cos, For Umin  = 0° So, Umin = –pE, 39. An electric dipole is placed in non-uniform electric field. It may experience, (1) Resultant force and couple, , (2) Only resultant force, , (3) Only couple, , (4) All of these, , Sol. Answer (4), , , Electric dipole in non uniform E may experience force, or couple., 40. Each of two large conducting parallel plates has one sided surface area A. If one of the plates is given a, charge Q whereas the other is neutral, then the electric field at a point in between the plates is given by, Q, (1) A, 0, , Sol. Answer (2), , E net, , Q, , 2, 4 A 0, Q, = 4 A, 0, , Q, (2) 2 A, 0, , Q, (3) 4 A, 0, , (4) Zero, , +Q, , Q Q,  Q Q, 2 2, 2 +2, –, Q, +, –, + 4A 0 –, +, –, Q, +, –, 4 A 0, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 27, , 41. If atmospheric electric field is approximately 150 volt/m and radius of the earth is 6400 km, then the total, charge on the earth’s surface is, (1) 6.8 × 105 coulomb, , (2) 6.8 × 106 coulomb, , (3) 6.8 × 104 coulomb, , (4) 6.8 × 109 coulomb, , Sol. Answer (1), E = 150 V/m, R = 6400 km, Using Gauss' Law, ⇒ EA , , q, 0, , , , ⇒ 150 4 6400  103, , , ,    q, , q = 150 × 4 ×(6400, , ×103)2, , 2, , 0, , × 8.854 × 10–12, , q = 6.8 × 105 C, SECTION - C, Previous Years Questions, 1., , The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in, a sphere of radius 'a' centred at the origin of the field, will be given by, [AIPMT-2015], (1)  0 Aa 3, , (2) 4 0 Aa 2, , (3) A 0a 2, , (4) 4 0 Aa 3, , Sol. Answer (4), Let charge enclosed in the sphere of radius is q. According to Gauss theorem,,  , , q, , E.4r 2 , , q, 0, , 4Ar 3 , , q, 0, , ∫ E.ds   0, a, , r=a, then, q  4  0 Aa 3, , 2., , Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the, equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium, separation between the balls now become, [NEET-2013], , y, r, ⎛ r ⎞, (1) ⎜ 3 ⎟, ⎝ 2⎠, , ⎛ 2r ⎞, ⎟, (2) ⎜, ⎝ 3⎠, , r, , y/2, , ⎛ 2r ⎞, (3) ⎜ ⎟, ⎝ 3 ⎠, , ⎛ 1 ⎞, (4) ⎜, ⎟, ⎝ 2⎠, , 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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28, , Electric Charges and Fields, , Solution of Assignment, , Sol. Answer (1), K .Q 2, r2, , T sin  , , …(i), , T cos, , T cos = mg, , i, ii, , …(ii), , tan  , , T sin, , T, , K .Q 2, mg.r 2, , , , mg, ⇒ r .tan 1  r .tan  2, 2, 1, , ⇒ r 2., , ⇒, , 2, 2, , r, r 2 .2,  r 22 ., 2y, 2.y, , r3, r3,  2, 2y, y, , ⇒ r2 , , r, 1, , 23, , 3., , What is the flux through a cube of side a if a point charge of q is at one of its corner?, [AIPMT (Prelims)-2012], q, (1) , 0, , q, 2, (2) 2 6a, 0, , (3), , 2q, 0, , q, (4) 8, 0, , Sol. Answer (4), net , , Qinc. Q, , 0,  0 (through eight cubes), , a, Flux through one cube = 8 , 0, , 4., , A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward, electric flux will, [AIPMT (Prelims)-2011], (1) Be doubled, , (2) Increase four times, , (3) Be reduced to half, , (4) Remain the same, , Sol. Answer (4), Flux does not depend upon size of surface., 5., , Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between, the ions, the number of electrons missing from each ion will be (e being the charge on an electron), [AIPMT (Prelims)-2010], , 40 Fd 2, (1), e2, , (2), , 40 Fe 2, d2, , (3), , 40 Fd 2, e2, , 40 Fd 2, (4), q2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 29, , Sol. Answer (3), Force between two ions, , K .Q1.Q2, F, d2, , 1  ne , ., F, 4  0 d 2, 2, , ⇒, , ⇒ n2.e2 = Fd2.40, , ⇒ n  Fd 2 .4 0, , 6., , e2, , The electric field at a distance, , 3R, from the centre of a charged conducting spherical shell of radius R is E., 2, , The electric field at a distance, , R, from the centre of the sphere is, 2, , (1) Zero, , (2) E, , (3), , E, 2, , [AIPMT (Mains)-2010], , (4), , E, 3, , Sol. Answer (1), Electric field inside the shell is zero., 7., , A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the, charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of, the ring is, [AIPMT (Prelims)-2008], A, , K, C, , O, , B, , D, (1) 3E along OK, , (2) 3E along KO, , (3) E along OK, , (4) E along KO, , Sol. Answer (3), E along OK, Since E at the centre must be zero., 8., , Three point charges +q, –2q and +q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x =, a, y = 0, z = 0) respectively. The magnitude and direction of the electric dipole moment vector of this charge, assembly are, [AIPMT (Prelims)-2007], (1), , 2 qa along + x direction, , (2), , 2 qa along + y direction, , (3), , 2 qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0), , (4) qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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30, , Electric Charges and Fields, , Solution of Assignment, , Sol. Answer (3), , y, , p, q, , –2q, , o, , q, , x, , p, , z, , pnet  2p  2qa, 9., , A hollow cylinder has a charge q coulomb within it. If the electric flux in units of volt × metre associated with the, curved surface B, the flux linked with the plane surface A in units of volt × metre will be [AIPMT (Prelims)-2007], , B, C, , q, (1)   , 0, , ⎞, 1⎛ q, (2) 2 ⎜    ⎟, ⎝ 0, ⎠, , A, , q, (3) 2, 0, , (4), , , 3, , Sol. Answer (2), Q, , B, , Net flux through the all surface = , 0, , C, , Flux through curved surface = , Hence, flux through plane surface , , A, , ⎞, 1⎛ Q,  ⎟, ⎜, 2 ⎝ 0, ⎠, , , , 10. An electric dipole moment P is lying along a uniform electric field E . The work done in rotating the dipole by, 90° is, [AIPMT (Prelims)-2006], (1), , 2 pE, , (2), , pE, 2, , (3) 2pE, , (3) pE, , Sol. Answer (4), 11. A square surface of side L metre is in the plane of the paper. A uniform electric field E (volt/m), also in the, plane of the paper, is limited only to the lower half of the square surface (see figure). The electric flux in SI units, associated with the surface is, [AIPMT (Prelims)-2006], , E, , (1), , EL2, 2 0, , (2), , EL2, 2, , (3) Zero, , (4) EL2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 31, , Sol. Answer (3), ,  , net  E.A  EA cos , , | ∵   90, , =0, 12. A charged cloud system produces an electric field in the air near the earth’s surface. A particle of charge –2 × 10–9, C is acted on by a downward electrostatic force of 3 × 10–6 N when placed in this field. The gravitational and, electrostatic force, respectively, exerted on a proton placed in this field are, (1) 1.64 × 10–26 N, 2.4 × 10–16 N, , (2) 1.64 × 10–26 N, 1.5 × 103 N, , (3) 1.56 × 10–18 N, 2.4 × 10–16 N, , (4) 1.5 × 103 N, 2.4 × 10–16 N, , Sol. Answer (1), F = Q.E, ⇒E , , 3  10 6, F, ,  1.5  10 3, 2  10 9, q, , Hence force on proton = FP = QP.E, = (1.6 ×10–19) × (1.5 ×103 ), = 2.4 × 10–16 N, Gravity force on proton = FG = mg, = 1.6 × 10–27 × 10, = 1.64 × 10–26 N, 13. The frequency of oscillation of an electric dipole moment having dipole moment p and rotational inertia I,, oscillating in a uniform electric field E is given, (1) (1/2) I /pE, , (2) (1/2) pE /I, , (3) (2) pE /I, , (4) (2) I /pE, , Sol. Answer (2), , n, , 1 w, , , T 2, , PE, , I  1 . P.E, I, 2, 2, , 14. What is the net charge on a conducting sphere of radius 10 cm? Given that the electric field 15 cm from the, center of the sphere is equal to 3 × 103 N/C and is directed inward, (1) –7.5 × 10–5 C, , (2) –7.5 × 10–9 C, , (3) 7.5 × 10–5 C, , (4) 7.5 × 10–9 C, , Sol. Answer (2), E.A. , , Qinc., Q, ⇒ 3  103  4r 2  inc., 0, 0, , ⇒ Qinc. , , , , , , 75  10, 3  103  152,  7.5  10 9 C, =, 9, 109, 9  10  100, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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32, , Electric Charges and Fields, , Solution of Assignment, , 15. The given figure gives electric lines of force due to two charges q1 and q2. What are the signs of the two, charges?, , (1) q1 is positive but q2 is negative, , (2) q1 is negative but q2 is positive, , (3) Both are negative, , (4) Both are positive, , Sol. Answer (3), Electric field is directed from positive to negative charge., 16. A charge q is placed at the centre of the line joining two exactly equal positive charges Q. The system of, three charges will be in equilibrium, if q is equal to, (1) –Q, , (2), , Q, 2, , (3) , , Q, 4, , (4) +Q, , Sol. Answer (3), Net force on Q due to other charges, , +Q, , q, , +Q, , K .Q.q K .Q.Q, , r2, 4r 2, ⇒Q , , Q, 4, , 17. A point charge +q is placed at the mid-point of a cube of side l. The electric flux emerging from the cube is, 6ql 2, 0, , (1), , q, (2) 6l 2 , 0, , q, (4) , 0, , (3) Zero, , Sol. Answer (4),   E .A. , ⇒  net , , Qinc ., 0, , Q, 0, , 18. A point Q lies on the perpendicular bisector of an electrical dipole of dipole moment p. If the distance of Q, from the dipole is r (much larger than the size of the dipole), then the electric field at Q is proportional to, (1) p2 & r –3, , (2) p & r –2, , (3) p–1 & r –2, , (4) p & r –3, , Sol. Answer (4), , E , , Q, , p, , , , 4  0 . r 2  a 2, p, , 4  0 .r 3, , , , 3, 2, , ∵ r >> a, , –q, , r, a, , +q, , ⇒ E  p, E  r 3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 33, , 19. A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic, energy attained by the particle after moving a distance y is, (2) qE2y, , (1) qEy, , (3) qEy2, , (4) q2Ey, , Sol. Answer (1), F = QE, , a, , F qE, , m, m, , Hence, K .E. , , =, , 1, mv 2, 2, , 1, ⎛ 2qEy ⎞, m⎜, ⎝ m ⎟⎠, 2, , = qEy, 20. A hollow insulated conducting sphere is given a positive charge of 10 C. What will be the electric field at the, centre of the sphere if its radius is 2 metre?, (1) 20 C m–2, , (2) 5 C m–2, , (4) 8 C m–2, , (3) Zero, , Sol. Answer (3), Electric field at the centre is zero., 21. When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two, charges separated by a distance, (1) Increases K times, , (3) Decreases K times (4) Increases K–2 times, , (2) Remains unchanged, , Sol. Answer (3), , Fair , , 1 Q1.Q2, ., 4 0 r 2, , Fmedium , , Fair, ⇒ decreases by k times, k, , 22. Electric field at centre O of semicircle of radius a having linear charge density  is given as, , O, , a, , , 2, (1)  a, 0, , , (2)  a, 0, , , (3) 2 a, 0, , , (4)  a, 0, , Sol. Answer (3), 1 ⎛ ⎞, k.. a.d, k. .dl , k.d , k . 2 , .⎜ ⎟, dE  2 =, , .∫ d  =, , 2, 2, ⎝ a⎠, 0, 2, , a, a, a, a, 0, , d, , a, , dl, , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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34, , Electric Charges and Fields, , Solution of Assignment, , , , 23. A dipole of dipole moment P is placed in uniform electric field E , then torque acting on it is given by,  ,  ,  ,   , (2)   P  E, (3)   P  E, (4)   P  E, (1)   P  E, Sol. Answer (2),  , ,  P E, from formula, 24., , A charge q is located at the centre of a cube. The electric flux through any face of the cube is, , 2q, (1) 6( 4 ), 0, , 4q, (2) 6( 4 ), 0, , q, (3) 6( 4 ), 0, , q, (4) 6( 4 ), 0, , Sol. Answer (2), , , qinc. Q, , 0, 0, , 1⎛ Q⎞, Flux through any one surface of the cube = 6 ⎜  ⎟, ⎝ 0⎠, , 25. The unit of permittivity of free space, 0, is, (1) coulomb/newton-metre, , (2) newton-metre2/coulomb2, , (3) coulomb2/(newton-metre2), , (4) coulomb2/(newton-metre)2, , Sol. Answer (3), From formula, F , , 1 Q2, ., 4  0 r 2, , ⇒ 0 , ⇒, , Q2, Fr 2, , C2, Nm2, , 26. A square surface of side L meter in the plane of the paper is placed in a uniform electric field E (volt/m) acting, along the same plane at an angle  with the horizontal side of the square as shown in figure. The electric flux, linked to the surface, in units of volt-m, is, , E, , , (1) Zero, , (2) EL2, , (3) EL2 cos, , (4) EL2 sin, , Sol. Answer (1), ,  , Flux    E.A  EA cos   0, , | ∵   0, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electric Charges and Fields, , 35, , SECTION - D, Assertion-Reason Type Questions, 1., , A : A negatively charged body means that the body has gained electrons while a positively charged body, means the body has lost some of its electrons., R : Charging process involves transfer of electrons., , Sol. Answer (1), 2., , A : Particles such as photon or neutrino which have no rest mass are uncharged., R : Charge cannot exist without mass., , Sol. Answer (1), 3., , A : When a body is charged, its mass changes., R : Charge is quantized., , Sol. Answer (2), 4., , A : Though quark particles have fractional electronic charges, the quantum of charge is still electronic charge, (e)., R : Quark particles do not exist in free state., , Sol. Answer (1), 5., , A : An electron has negative charge by definition., R : Charge of a body depends on its velocity., , Sol. Answer (3), 6., , A : A point charge cannot exert force on itself., R : Coulomb force is a central force., , Sol. Answer (2), 7., , A : Since matter cannot be concentrated at a point, therefore point charge is not possible., R : An electron is a point charge., , Sol. Answer (3), 8., , A : A finite size charged body may behave like a point charge if it produces an inverse square electric field., R : Two charged bodies may be considered as point charges if their distance of separation is very large, compared to their dimensions., , Sol. Answer (2), 9., , A : The path traced by a positive charge is a field line., R : A field line can intersect itself., , Sol. Answer (4), 10. A : If electric flux over a closed surface is negative then the surface encloses net negative charge., R : Electric flux is independent of the charge distribution inside the surface., Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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36, , Electric Charges and Fields, , Solution of Assignment, , 11. A : We may have a Gaussian surface in which less number of field lines enter and more field lines come, out., R : The electric field E in the Gauss’s law is only due to the enclosed charges., Sol. Answer (3), 12. A : The equilibrium of a charged particle under the action of electrostatic force alone can never be stable., R : Coulombian force is an action-reaction pair., Sol. Answer (2), 13. A : The field in a cavity inside a conductor is zero which causes electrostatic shielding., R : Dielectric constant of conductors in electrostatics is infinite., Sol. Answer (2), 14. A : If dipole moment of water molecules were zero, then microwave cooking would not be possible., R : In a microwave oven the water molecules vibrate due to oscillating electric field in microwave and heat, the food., Sol. Answer (1), 15. A : Electric field lines are continuous curves in free space., R : Electric field lines start from negative charge and terminate at positive charge., Sol. Answer (3), 16. A : When an electric dipole is placed in uniform electric field, net force on it will be zero., R : Force on the constituent charges of the dipole will be equal and opposite when it is in uniform electric, field., Sol. Answer (1), 17. A : Gauss’ theorem is applicable on any closed surface., R : In order to find the value of electric field due to a charge distribution, Gauss’ theorem should be applied, on a symmetrical closed surface., Sol. Answer (2), , , , , , , , https://t.me/NEET_StudyMaterial, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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https://t.me/NEET_StudyMaterial, Chapter, , 17, , Electrostatic Potential and Capacitance, Solutions, SECTION - A, Objective Type Questions, 1., , Three isolated equal charges are placed at the three corners of an equilateral triangle as shown in figure. The, statement which is true for net electric potential V and net electric field intensity E at the centre of the triangle, is, q, , O, q, , (1) E = 0, V = 0, , q, , (2) V = 0, E  0, , (3) V  0, E = 0, q, , Sol. Answer (3), l, , Enet = 0, , 3, , ⎛k q 3⎞, kq, Vnet  3 ⎜, 3 3, ⎟, l, ⎝ l ⎠, , 2., , (4) V 0, E  0, , E O, , q, , E, , E, , q, , The potential at a point 0.1 m from an isolated point charge is + 100 volt. The nature of the point charge is, (1) Positive, , (2) Negative, , (3) Zero, , (4) Either positive or zero, , Sol. Answer (1), As potential is +ve, so point charge is also +ve., 3., , A charge of 10C is placed at the origin of x-y coordinate system. The potential difference between two points, (0, a) and (a, 0) in volt will be, (1), , 9  10 4, a, , (2), , 9  10 4, , (3), , a 2, , 9  10 4, 2a, , (4) Zero, , Sol. Answer (4), kq, VA , a, , VB , , kq, a, , A, a, , a B, 10  C, , V = VA – VB = 0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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38, 4., , Electrostatic Potential and Capacitance, , Solution of Assignment, , Four charges of same magnitude q are placed at four corners of a square of side a. The value of electric potential, 1, at the centre of the square will be (Where k  4 ), 0, (1), , 4kq, a, , (2) 4 2, , kq, a, , (3), , 4kq, , (4), , 2a, , kq, a 2, , Sol. Answer (2), , a, , q, , q, , P, , a, 2, , q, , q, , ⎛ kq 2 ⎞, kq, VP  4 ⎜, 4 2, ⎟, a, ⎝ a ⎠, , 5., , Two identical positive charges are placed on the y-axis at y = – a and y = + a. The variation of V (electric, potential) along x-axis is shown by graph, , V, , V, , V, (2), , (1), , V, , (3), , (4), , x, , x, , x, , x, , Sol. Answer (1), , q, a, a, , q, , 2kq, , V , , 6., , x, , a2  x 2, , Which graph best represents the variation of electric potential as a function of distance from the centre of a uniformly, charged solid sphere of charge of radius R?, , V, , V, , V, , (2), , (1), , R, , r, , V, , (3), , R, , r, , (4), , R, , r, , R, , r, , Sol. Answer (3), , ⎡, ⎤, Q, 1, . 3 3R 2  r 2 ⎥, ⎢V , 4, , 2, R, 0, ⎣, ⎦, , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 7., , Electrostatic Potential and Capacitance, , 39, , A hollow charged metal sphere has radius r. If the potential difference between its surface and a point at a distance, 3r from the centre is V, then the electric field intensity at distance 3r from the centre is, (1), , V, 3r, , (2), , V, 4r, , (3), , V, 6r, , (4), , V, 2r, , Sol. Answer (3), VA , , kq, r, , VP , , kq, 3r, , q, , r, , A, 3r, , 2kq, V = VA – VP =, 3r, 3Vr, 2, , kq , , E, , 8., , P, , kq, 3Vr, V, , , 2, 2, 9r, 6r, 2 9r, ,  , , Two concentric hollow conducting spheres of radius r and R are shown. The charge on outer shell is Q. What, charge should be given to inner sphere so that the potential at any point P outside the outer sphere is zero?, , Q, x, , r, , P, , R, (1) , , Qr, R, , (2) , , QR, r, , (3) –Q, , Sol. Answer (3), VP , , x, , r, , P, , R, , q = –Q, 9., , 2QR, r, , Q, , q, , kq KQ, , 0, x, x, , (4) , , The potential gradient is a, (1) Vector quantity, , (2) Scalar quantity, , (3) Conversion factor, , (4) Constant, , Sol. Answer(1), Potential gradient =, , dV,  E (Vector), dr, , 10. The electric potential V at a point P(x, y, z) in space is given by V = 4 x2 volt. Electric field at a point (1m,, 0, 2m) in V/m is, (1) 8 along –ve x-axis, , (2) 8 along +ve x-axis, , (3) 16 along –ve x-axis (4) 16 along +ve x-axis, , Sol. Answer (1), V = 4x2, Ex , , dV,  8 x, dx, , Ex = –8, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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40, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 11. Figure shows the variation of electric field intensity E versus distance x. What is the potential difference between, the points at x = 2 m and at x = 6 m from O?, , E, (N/C), 10, 5, O, (1) 30 V, , 2, , 4, , (2) 60 V, , 6 x (m), (3) 40 V, , (4) 80 V, , Sol. Answer(1), , V2  V6  ∫ Edr, , E 10, 5, , V2  V6  10 2 , , 1, 10 2  30, 2, , 2, , 4, , 6x, , 12. Figure shows a set of equipotential surfaces. The magnitude and direction of electric field that exists in the, region is, y, , 135º, O, , 1, , 2, , 3, , 10 V, 20 V, 30 V, 40 V, , x(m), , (1) 10 2 V/m at 45º with x-axis, , (2) 10 2 V/m at –45º with x-axis, , (3) 5 2 V/m at 45º with x-axis, , (4) 5 2 V/m at –45º with x-axis, , E, , Sol. Answer (1), , E, , 10 2, 1, , 1, 2, , E  10 2 at 45° with x-axis, , 90° 135°, 45°, 1, 2, 3 10, 0, 20, 30, 40, , 13. Determine the electric field strength vector if the potential of this field depends on x, y coordinates as V = 10, axy, (1) 10 a ( yiˆ  xˆj ), , (2)  10 a [ yiˆ  xˆj ], , (3)  a [ yiˆ  xˆj ], , (4)  10a [ xiˆ  ykˆ ], , Sol. Answer (2), V = 10axy, , Ex , , dV, dV,  10ay , Ey ,  10ax, dx, dy, , , E  10a yiˆ  xjˆ, , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 41, , 14. If on the x-axis electric potential decreases uniformly from 60 V to 20 V between x = –2 m to x = +2 m, then the, magnitude of electric field at the origin, (1) Must be 10 V/m, , (2) May be greater than 10 V/m, , (3) Is zero, , (4) Is 5 V/m, , Sol. Answer (2), , Ey, , 40, V, Ex ,  10, 4, m, , 60, , Ey  0, , –2 –1, , 20, 0, , 1, , 2, Ex, , E  E x2  E y2  10, , 15. An infinite conducting sheet has surface charge density . The distance between two equipotential surfaces, is r. The potential difference between these two surfaces is, r, (1) 2, 0, , r, (2) , 0, , , (3)  r, 0, , , (4) 2 r, 0, , Sol. Answer (2), V  Ed , V , , r, 0, , r, 0, , 16. Two small spheres each carrying a charge q are placed, distance r apart. If one of the spheres is taken around, the other in a circular path, the work done will be equal to, (1) Force between them × r, , (2) Force between them, 2r, , (3) Force between them × 2r, , (4) Zero, , Sol. Answer (4), , r, q, , q, , W = 0 as whole path is equipotential., 17. Work done in moving a charge q coulomb on the surface of a given charged conductor of potential V is, , (1), , V, joule, q, , (2) Vq joule, , (3), , q, joule, V, , (4) Zero, , Sol. Answer (4), As the surface of a conductor is equipotential, So w = 0., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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42, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 18. If an -particle and a proton are accelerated from rest by a potential difference of 1 megavolt then the ratio of, their kinetic energy will be, (1), , 1, 2, , (2) 1, , (3) 2, , (4) 4, , Sol. Answer (3), KE = qV, KE  qV q, , , 2, KE P q pV q p, , 19. When a test charge is brought in from infinity along the perpendicular bisector of an electric dipole, the work, done is, (1) Positive, , (2) Zero, , (3) Negative, , (4) None of these, , Sol. Answer (2), W = q(Vf – Vi ) = q(0 – 0) = 0, 20. The work done in moving an electric charge q in an electric field does not depend upon, (1) Mass of the particle, , (2) Potential difference between two points, , (3) Magnitude of charge, , (4) All of these, , Sol. Answer (1), Work done does not depend on mass of the particle., 21. A particle A has charge +q and particle B has charge +4q with each of them having the same mass m. When, VA, allowed to fall from rest through the same electric potential difference, the ratio of their speeds V will become, B, (1) 1 : 2, , (2) 2 : 1, , (3) 1 : 4, , (4) 4 : 1, , Sol. Answer (1), qV , , 1, mVA2, 2, , VA , , 2qV, m, , VB , , 8qV, m, , VA 1, , VB 2, , 22. If 50 joule of work must be done to move an electric charge of 2 C from a point, where potential is –10 volt, to another point, where potential is V volt, the value of V is, (1) 5 V, , (2) – 15 V, , (3) + 15 V, , (4) + 10 V, , Sol. Answer (3), 50 = 2(V – (–10)), 25 = V + 10, V = 15 V, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 43, , 23. A proton has a mass 1.67 × 10–27 kg and charge +1.6 × 10–19 C. If the proton is accelerated through a, potential difference of million volts, then the kinetic energy is, (1) 1.6 × 10–15 J, , (2) 1.6 × 10–13 J, , (3) 1.6 × 10–21 J, , (4) 3.2 × 10–13 J, , Sol. Answer (2), (1.6 × 10–19) (106) =, 1.6  10 13 , , 1, (1.67 × 10–27) v2, 2, , 1.67,  10 27 v 2  KE, 2, , KE = 1.6 × 10–13 J, 24. Calculate the work done in taking a charge –2 × 10–9 C from A to B via C (in diagram), , z(cm), , A(0,0,3), q = 8 mC, , C(0,6,9), , B, O (0,4,0), , y(cm), , x(cm), (1) 0.2 joule, , (2) 1.2 joule, , (3) 2.2 joule, , (4) Zero, , Sol. Answer (2), z, , 9  109  8  10 3, VA ,  24  108 V, 3  10 2, , VB , , 9  109  8  103,  18  108 V, 4  102, , W = –2 × 10–9 (–6 × 108 ), W = 12 × 10–1 J, , C(0,6,9), A(0,0,3),  MC B(0,4,0), , y, , x, , W = 1.2 J, 25. The electric potential at a distance of 3 m on the axis of a short dipole of dipole moment, 4 × 10–12 coulomb-meter is, (1) 1.33 × 10–3 V, , (2) 4 mV, , (3) 12 mV, , (4) 27 mV, , Sol. Answer (2), , V, , 9  109  4  1012,  4  103 V  4 mV, 9, , 26. The electric potential in volts due to an electric dipole of dipole moment 2 × 10–8 coulomb-metre at a distance, of 3m on a line making an angle of 60° with the axis of the dipole is, (1) Zero, , (2) 10, , (3) 20, , (4) 40, , Sol. Answer (2), , V, , 9  109  2  10 8 , 9, , 1, 2  10 V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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44, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 27. An electric dipole of length 2 cm is placed with its axis making an angle of 30° to a uniform electric field 105, N/C. If it experiences a torque of 10 3 Nm, then potential energy of the dipole, (1) – 10 J, , (2) – 20 J, , (3) – 30 J, , (4) – 40 J, , Sol. Answer (3), , 10 3  P105, , 1, 2, , 2 3  10 4  P, , U  2 3  104  105 , , 3, 2, , U = –3 × 10, U = –30 J, 28. Two electrons are moving towards each other, each with a velocity of 106 m/s. What will be closest distance, of approach between them?, (1) 1.53 × 10–8 m, , (2) 2.53 × 10–10 m, , (3) 2.53 × 10–6 m, , (4) Zero, , Sol. Answer (2), , 1, 2. 9.1  10 31 106, 2, , , , 9.1  10 19 , ,  , , 2, , , , , , 9  109  1.6  10 19, , , , 2, , r, , 9  10 9  2.56  10 38, r, , r = 2.56 × 10–10 m, 29. Three charges –q, Q and –q are placed respectively at equal distances on a straight line. If the potential energy, of the system of three charges is zero, then what is the ratio of Q : q ?, (1) 1 : 1, , (2) 1 : 2, , (3) 1 : 3, , Sol. Answer (4), U, , x, ,  kqQ kqQ kq, , , 0, x, x, 2x, 2, , –q, , x, 2x, , Q, , (4) 1 : 4, , –q, , kq 2 2kqQ, , 2x, x, , q = 4Q,, , Q 1, , q 4, , 30. A point charge q is surrounded by eight identical charges at distance r as shown in figure. How much work, is done by the forces of electrostatic repulsion when the point charge at the centre is removed to infinity?, q, q, q, , q, q, , O, q, , (1) Zero, , (2), , 8q 2, 40 r, , q, , q, q, 8q, (3) 4 r, 0, , (4), , 64q 2, 40 r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 45, , Sol. Answer (2), W = –q (Vf – Vi) = –q(V – Vi) = +qVi, V i  8., W , , kq 2, 8q 2, , r, 4  0 r, ,  8q 2, 4  0 r, , 31. 1000 small water drops each of capacitance C join together to form one large spherical drop. The capacitance, of bigger sphere is, (1) C, , (2) 10 C, , (3) 100 C, , (4) 1000 C, , Sol. Answer (2), C = 40r, 4, 4, 1000. r 3  R 3, 3, 3, , R = 10r, C' = 40R = 10 (40r), C' = 10C, 32. Two parallel plate capacitors have their plate areas 100 cm2 and 500 cm2 respectively. If they have the same, charge and potential and the distance between the plates of the first capacitor is 0.5 mm, then the, distance between the plates of the second capacitor is, (1) 0.10 cm, , (2) 0.15 cm, , Sol. Answer (4), , (3) 0.20 cm, , 0.5, , (4) 0.25 cm, , d, , 100 0 500 0, , 0.5, d, q, , d = 2.5 cm = 0.25 cm, , q, , –q, 2, , A1 = 100 cm, , –q, 2, , A2 = 500 cm, , 33. A dielectric slab of dielectric constant K is placed between the plates of a parallel plate capacitor carrying, charge q. The induced charge q' on the surface of slab is given by, (1) q   q , , q, K, , (2) q   q , , q, K, , ⎡1, ⎤, (3) q   q ⎢  1⎥, K, ⎣, ⎦, , 1⎞, ⎛, (4) q   q ⎜1  ⎟, K, ⎝, ⎠, , Sol. Answer (2), , 1⎞, ⎛, q ⎜ 1 ⎟, ⎝ K⎠, , 1⎞, ⎛, q ⎜ 1 ⎟, ⎝ K⎠, , q, , –q, , 1⎞, ⎛, q '  q ⎜ 1  ⎟, ⎝ K⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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46, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 34. Two charged capacitors have their outer plates fixed and inner plates connected by a spring of force constant, ‘k’. The charge on each capacitor is q. Find the extension in the spring at equilibrium, , +, , –, , +, , k, , –, , smooth, (1), , q2, 2A 0 k, , (2), , q2, 4 A 0 k, , (3), , q2, A 0 k, , (4) Zero, , Sol. Answer (1), q2, 2 A 0, , F  kx , , x , , q2, 2 A0k, , 35. A battery does 200 J of work in charging a capacitor. The energy stored in the capacitor is, (1) 200 J, , (2) 100 J, , (3) 50 J, , (4) 400 J, , Sol. Answer (2), , U, , 1 2, Cv, 2, , W = Cv2, U=, , W, = 100 J, 2, , (half of work is lost in heat), , 36. The following arrangement consists of five identical metal plates parallel to each other. Area of each plate is, A and separation between the successive plates is d. The capacitance between P and Q is, P, Q, , (1), , 5 0 A, d, , (2), , 7 A, 0, 3 d, , (3), , 4 0 A, 3 d, , (4), , 5 0 A, 3 d, , Sol. Answer (4), , 1, 2, 3, 4, 5, P, , C, , R 2C, C, , P, Q, , Q, , Cnet , , P Q, Q, 1, 2, 23, 3, 4, 4, 5, R, R, Q, Q, P, R, , Q, , R, , 5C, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 47, , 37. Two similar conducting balls having charges +q and –q are placed at a separation d from each other in air., The radius of each ball is r and the separation between their centres is d(d >>r). Calculate the capacitance, of the two ball system, +q + + + +, +, +, r, +, +, +, +, +, +, ++, + +, , (1) 40r, , VA , , kq k  q , , r, d r, , kq, kq, VB , , r, d r, V  VA  VB , , V , , d, , (3) 4 loge, , (2) 20r, , Sol. Answer (2), , q, +, ++, +, +, r, +, + d, +, +, +, +A, +, +, + + +, , 2kq 2kq, 2q, , , r, d  r 4  0, , –q, , r, , 0r, d, , (4) 4 loge, , r, d, , –q, , B, , 1 ⎤, ⎡1, ⎢r  d  r ⎥, ⎣, ⎦, , q ⎡ d  2r ⎤, ⎢, ⎥, 2 0 ⎣  r  d  r  ⎦, , d >> r, V , , q, 2  0, , q,  2 0 r  C, V, 38. A capacitor is half filled with a dielectric (K = 2) as shown in figure A. If the same capacitor is to be filled, with same dielectric as shown, what would be the thickness of dielectric so that capacitor still has same, capacity?, +, +, , d, , d, t, , –, Fig. A, (1) 2d/3, , (2) 3d/2, , –, Fig. B, (3) 3d/4, , (4) 4d/3, , Sol. Answer (1), , Cnet, , k  1 A 0, , 2, d, , Cnet , , Cnet , , A 0, A 0, , t, 1, ⎛, ⎞, d  t ⎜1  ⎟ d , ⎝, 2, 2⎠, , 3A0, 2d, , 3 A 0, A 0, , t, 2d, d, 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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48, , Electrostatic Potential and Capacitance, , 3d , , Solution of Assignment, , 3t,  2d, 2, , 3t, d, 2, t, , 2d, 3, , 39. Capacitors C1(10 F) and C2(20 F) are connected in series across a 3 kV supply, as shown. What is the, charge on the capacitor C1?, , 3 kV, C2, , C1, 10 F, (1) 45000 C, , 20 F, , (2) 20000 C, , (3) 15000 C, , (4) 10000 C, , Sol. Answer (2), 1, 1, 1, 3, , , , C 10 20 20, , C, , 20, F, 3, , q, , 20, .3000  20000  10 6, 3, , q = 2 × 10–4 C, 40. The charge on the 6 F capacitor in the circuit shown is, , A, 90 V, , 9 F 6 F, , B, , (1) 540 C, Sol. Answer (3), , 1, 1 1 3, ,  , C 18 9 18, C = 6 F, q = 6 × 90 = 540 C, , (2) 270 C, , (3) 180 C, , (4) 90 C, , A, , A, 90 V 9  F 6  F, , B, , 12  F, , X, X, , q1 A, , 12  F, , X, , 540  C, , A, X, , 540  C, , 18 X 9, , B, , q1, 6, , 540  q1 12, , 2q1 = 540 – q1, 3q1 = 540, q1 = 180 C, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 49, , 41. In the circuit below C1  20 F , C2  40 F and C3  50 F. If no capacitor can sustain more than 50 V, then, maximum potential difference between X and Y is, X, C1, , (1) 95 V, , C2, , (2) 75 V, , C3, , Y, , (3) 150 V, , Sol. Answer (1), , (4) 65 V, X, , qmax = 1000 C, 2000 C, 2500 C, , Y, 20  F, 40  F, 50  F, , q = 1000 C, 1000 C, 1000 C, V = 50 V, 25 V, 20 V, Vnet = 50 + 20 + 25, Vnet = 95 V, , 42. In the circuit shown below C1 = 10 F, C2 = C3=20 F, and C4 = 40 F. If the charge on C1 is 20 C then potential, difference between X and Y is, C1, , C3, , X, , Y, C2, , (1) 2 V, Sol. Answer (2), q3  20 C, q4  q4  40 C, V , , 20 20, , 3V, 10 20, , (2) 3 V, , C4, , (3) 6 V, , 10  F, , 20 F, , 20  F, 20c, , qq33, , 20 F, q2, , (4) 3.5 V, , 40 F, q3, , 43. A parallel plate capacitor after charging is kept connected to a battery and the plates are pulled apart with the help, of insulating handles. Now which of the following quantities will decrease ?, (1) Charge, , (2) Capacitance, , (3) Energy stored, , (4) All of these, , Sol. Answer (4), V remains constant, C, , A 0, ⇒ d increases, d, , C decreases, q = CV, 1, 2, q decreases U  CV, 2, , U decreases, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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50, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 44. In the circuit below, if a dielectric is inserted into C2 then the charge on C1 will, C1, , C2, , V, , (1) Increase, , (2) Decrease, , (3) Remain same, , Sol. Answer (1), C1C2V, C1  C2, , q1 , , C2, , C1, q1' , , (4) Be halved, , KC1C2V, C1  KC2, , q1'  q1 , so charge increases., , V, , 45. A capacitor with plate separation d is charged to V volts. The battery is disconnected and a dielectric slab of, d, thickness, and dielectric constant ‘2’ is inserted between the plates. The potential difference across its terminals, 2, becomes, (1) V, , (2) 2V, , (3), , 4V, 3, , (4), , 3V, 4, , Sol. Answer (4), q = CV, C , , A 0, 4 A 0 4C, , , d⎛, 1⎞, 3d, 3, d  ⎜1  ⎟, 2⎝, 2⎠, , 4CV , 3, , q, , CV , , 4CV , 3, , 3V, 4, , V , , 46. An uncharged parallel plate capacitor having a dielectric of constant K is connected to a similar air cored parallel, capacitor charged to a potential V. The two capacitors share charges and the common potential is V. The, dielectric constant K is, (1), , V  V, V V, , Sol. Answer (4), V , , V , ,  KC  0  C V , , (2), , V V, V, , (3), , V V, V, , (4), , V V , V, , KC, 0, , KC  C, , CV, KC  C, , C,V, , KV + V = V, KV = V – V, , K, , V V , V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 51, , 47. Two identical capacitors are connected in parallel across a potenial difference V. After they are fully charged, the, positive plate of first capacitor is connected to negative plate of second and negative plate of first is connected to, positive plate of other. The loss of energy will be, (1), , 1, CV 2, 2, , (2) CV2, , (3), , 1, CV 2, 4, , CV, , –CV, , –CV, , CV, , (4) Zero, , Sol. Answer (2), 1, 1, Ui  CV 2  CV 2  CV 2, 2, 2, , Uf = 0, U = CV2, , –V, , +, , ⇒, –, , +, , V, , SECTION - B, Objective Type Questions, 1., , As in the figure, if a capacitor of capacitance ‘C’ is charged by connecting it with resistance R, then energy given, by the battery will be, C, , R, V, 1, CV 2, (1), 2, , 1, 2, (2) Less than CV, 2, , Sol. Answer (3), , (4) More than CV2, , C, , t, ⎛, ⎞, q  CV ⎜ 1  e RC ⎟, ⎝, ⎠, , t=, , (3) CV2, , q = CV, , V, , Battery gives CV charge, , R, , W = CV2, 2., , Figure shows a solid conducting sphere of radius 1 m, enclosed by a metallic shell of radius 3 m such that their, centres coincide. If outer shell is given a charge of 6 C and inner sphere is earthed, find magnitude charge on the, surface of inner shell, , 3m, 1m, , (1) 1 C, Sol. Answer (2), , V , , kq k 6, , 0, 1, 3, , q = –2 C, , (2) 2 C, , (3) 4 C, , –q, o, 1m, , (4) 6 C, , 6+q, q, 6 C, 3m, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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52, 3., , Electrostatic Potential and Capacitance, , Solution of Assignment, , A positively charged ring is in y – z plane with its centre at origin. A positive test charge q0, held at origin is released, along x-axis, then its speed, (1) Increases continuously, , (2) Decreases continuously, , (3) First increases then decreases, , (4) First decreases then increases, , Sol. Answer (1), , kq, , V, , x  a2, 2, , x, , V decreases, So, U decreases, So, K increases, 4., , Three point charges q, q and –2q are placed at the corners of an equilateral triangle of side ‘L’. Calculate work done, by external force in moving all the charges far apart without acceleration, (1), , 1 3q 2, 40 L, , (2) , , 1 3q 2, 40 L, , (3), , 1 5q 2, 40 L, , Sol. Answer (1), , L, , 3kq 2, L, , U  0, Wext  U   U , , 5., , q, 3kq, L, , 1 5q 2, 40 L, , q, , k  q  2q  k  q  q  k  q  2q , U, , , L, L, L, U, , (4) , , 2, , L, , L, , –2q, , There is a uniformly charged non conducting solid sphere made of material of dielectric constant one. If electric, potential at infinity be zero, then the potential at its surface is V. If we take electric potential at its surface to be, zero, then the potential at the centre will be, (1), , 3V, 2, , (2), , V, 2, , (3) V, , (4) Zero, , Sol. Answer (2), Vsurface = V + V, If V = 0, Vsurface = V, If V = –V, Vsurface = 0, , Vcen , , 6., , 3V, 3V, V,  V , V , 2, 2, 2, , 3x 2 y 2, , Electric potential in a region is varying according to the relation V , , where x and y are in metre and V, 2, 4, is in volt. Electric field intensity (in N/C) at a point (1 m, 2 m) is, (1) 3iˆ  ˆj, , (2)  3iˆ  ˆj, , (3) 6iˆ  2 ˆj, , (4)  6iˆ  2 ˆj, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 53, , Sol. Answer (2), V , , 3x 2 y 2, , 2, 4, , Ex , , dV,  3 x  3, dx, , Ey , , y, dV,   1, 2, dy, , , E  3iˆ  jˆ, 7., , There exists a uniform electric field E = 4 × 105 Vm–1 directed along negative x-axis such that electric potential at, origin is zero. A charge of –200 C is placed at origin, and a charge of +200 C is placed at (3 m, 0). The electrostatic, potential energy of the system is, (1) 120 J, , (2) – 120 J, , (3) – 240 J, , (4) Zero, , Sol. Answer (1), , E, –200 C, , V3  4  105  3  12  105, , kq q, U  1 2  q1V2  q2V2, r, , V0 = 0, , 200 C, V3 = 12 × 10, , 5, , 3, , 9  109  200  200  1012, U, + (–200)(0) + 200 (12 ×105) ×10–6, 3, U = –120 + 240, U = 120 J, 8., , If the electric potential on the axis of an electric dipole at a distance ‘r’ from it is V, then the potential at a point on, its equatorial line at the same distance away from it will be, (1) 2 V, , (2), , V, 2, , (3) 0, , (4) – V, , Sol. Answer (3), Potential at any point on the equatorial line due to an electric dipole is 0., 9., , Three identical charged capacitors each of capacitance 5 F are connected as shown in figure. Potential difference, across capacitor (3), long time after the switches K1 and K2 are is closed, is, 100C, K2, , 1, , +, +, +, +, +, +, +, +, , 100C, , (1) 20 V, , 3, , (2) 10 V, , 100C, +, +, +, +, , 2, , K1, , (3) 5 V, , (4) Zero, , Sol. Answer (4), In the wires, the charge is 0, Thus final charge = 0, Thus V = 0, , 100 –100 100 –100, , –100 100, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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54, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 10. A positive point charge q is placed at a distance 2R from the surface of a metallic shell of radius R. The electric, field at centre of shell due to induced charge has magnitude, , 2R, , q, 1, (2) 4, 2, 0 9R, , (1) Zero, Sol. Answer (2), , , Eint  E ext  0, , 1, q, (3) 4, 2, 0 4R, , ++, , R, +, +, + Eext Eind, ++, , , , E int  E ext, , kq, E int , 9R 2, , q, , 2R, , 1 q, (4) 4, 2, 0 R, , q, , 11. Six point charges are placed at the vertices of a regular hexagon of side a as shown. If E represents electric field, and V represents electric potential at O, then, –q, q, , q, –q, (1) E  0 but V  0, , V , , –q, , 3kq 3kq, , 0, r, r, , from the figure, , a, , (2) E  0 but V  0, , Sol. Answer (3), , –q, , O, q, , (3) E  0 and V  0, , (4) E  0 and V  0, , q, , q, , –q, , 2E, , q, , –q, 2E, 2E, Enet = 0, , 12. A point charge q is held at the centre of a circle of radius r. B, C are two points on the circumference of the circle, and A is a point outside the circle. If WAB represents work done by electric field in taking a charge q0 from A to B, and WAC represents the workdone from A to C, then, , B, r, A, (1) WAB > WAC, , (2) WAB < WAC, , q, C, (3) WAB = WAC  0, , (4) WAB = WAC  0, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 55, , Sol. Answer (3), , B, , WAB  UB  U A  q0 VB  VA , WAC  UC  U A  q0 VC  VA , , A, , C, , As VB – VC, WAB = WAC, , 13. Three charged particles having charges q, –2q and q are placed in a line at points (–a, 0), (0,0) and (a, 0), respectively. The expression for electric potential at P(r, 0) for r >> a is, , 1 qa 2, (1), 40 r 4, , 1 2qa 2, (2), 40 r 3, , Sol. Answer (2), E1 , , r, , 2kp, , E2  E1 , , E2 , , 3, , a⎞, ⎛, ⎜⎝ r  ⎟⎠, 2, , 2kp, a⎞, ⎛, ⎜⎝ r  ⎟⎠, 2, , 3, , , , 2kp, a⎞, ⎛, ⎜⎝ r  ⎟⎠, 2, , 3, , 1 8qa 2, (4), 40 r, , 1 4qa 2, (3), 40 r 2, , P1, q, , P, , P2, –2q q, , E1, , E2, , 2kp, a⎞, ⎛, ⎜⎝ r  ⎟⎠, 2, , 3, , ⎡, ⎤, ⎢, ⎥, 2kp ⎢, 1, 1, ⎥, , 3, 3, r 3 ⎢⎛, a⎞, a⎞ ⎥, ⎛, ⎢ ⎜1  ⎟, ⎜⎝ 1  ⎟⎠ ⎥, 2r ⎠, 2r ⎦, ⎣⎝, 2kp ⎡ 3a ⎛, 3a ⎞ ⎤, 1,  ⎜1  ⎟ ⎥, r 3 ⎢⎣ 2r ⎝, 2r ⎠ ⎦, , 2kp ⎡ 3a ⎤, r 3 ⎢⎣ r ⎥⎦, , E , V , , 6 kpa, , r4,  kp, a⎞, ⎛, ⎜⎝ r  ⎟⎠, 2, , 2, , 3, , 6 pa, 3 pa, , 4, 4, 2, , r,  0 r 4, 0, 2, , , kp, a⎞, ⎛, ⎜⎝ r  ⎟⎠, 2, , 2, , , , 2a ⎤ 2kpa, kp ⎡ 2a, 1,  1,  3, 2r ⎥⎦, r 2 ⎢⎣ 2r, r, , 14. Two metal spheres A and B of radii a & b (a < b) respectively are at a large distance apart. Each sphere carries a, charge of 100 C. The spheres are connected by a conducting wire, then, (1) Charge will flow from A to B, , (2) Charge will flow from B to A, , (3) No charge flows in the wire, , (4) All charges will reside on the connecting wire, , Sol. Answer (1), k 100 , a, a<b, VA , , VB , , k 100, b, , Va > VB, Charge flows from A to B., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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56, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 15. Three different dielectrics are filled in a parallel plate capacitor as shown. What should be the dielectric constant, of a material, which when fully filled between the plates produces same capacitance?, , A, 2, , (1) 4, , (2) 6, , K2=3, , K3=6, , K1=6, , d, , d, , 2, , A, 2, , 2, (3) 5, , (4) 9, , Sol. Answer (3), C1 , , C2 , , C2 , , C2 , , 6 A 0 3 A 0, , 2d, d, , ⎛, 2⎜d, ⎝, , A, 2, , 6, , A 0, d⎛, 1⎞ d ⎛, 1⎞ ⎞,  ⎜1  ⎟  ⎜1  ⎟ ⎟, 2⎝, 3⎠ 2 ⎝, 6⎠ ⎠, , 3, , 6, , A, 2, , A 0, ⎛d d ⎞, 2⎜  ⎟, ⎝ 6 12 ⎠, A 0, ⎛d⎞, 2⎜ ⎟, ⎝ 4⎠, , 2A 0, d, , C2 , , C  C1  C2 , , 5A 0, d, , 16. Consider a sphere of radius R having charge q uniformly distributed inside it. At what minimum distance from its, surface the electric potential is half of the electric potential at its centre?, (1) R, , (2), , R, 2, , (3), , 4R, 3, , (4), , R, 3, , Sol. Answer (4), , VC , , 3kQ, 2R, , VS , , kQ, 3kQ, , Rx, 4R, , q, R, , 4R = 3R + 3x, 3x = R, , x, , R, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 57, , 17. There are two identical capacitors, the first one is uncharged and filled with a dielectric of constant K while the, other one is charged to potential V having air between its plates. If two capacitors are joined end to end, the, common potential will be, (1), , V, K 1, , (2), , KV, K 1, , Sol. Answer (4), , CV, Cl  K, , V , , V, l K, , KV, K 1, , (4), , V, K 1, , KC, , KC  0   CV, V , KC  C, , V , , (3), , C, V, , 18. Seven identical plates each of area A and successive separation d are arranged as shown in figures., The effective capacitance of the system between P & Q is, , Q, , P, , (1), , 7 0 A, d, , (2), , 6 0 A, d, , (3), , 5 0 A, d, , (4), , 3 0 A, d, , Sol. Answer (2), , Q, 1, 2, 3, 4, 5, 6, 7, , P, , 12 23 34 45 56, , 67, , Q, , P, All are in parallel, C net  6C , , 6 A 0, d, , 19. In a certain region of space, variation of potential with distance from origin as we move along x-axis is given, by V = 8x2 + 2, where x is the x-coordinate of a point in space. The magnitude of electric field at a point, (– 4, 0) is, (1) – 16 V/m, , (2) 16 V/m, , (3) – 64 V/m, , (4) 64 V/m, , Sol. Answer (4), V = 8x2 + 2, Ex , , dV,  16 x, dx, , x = –4, Ex = 64 V/m, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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58, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 20. Four charges +q, –q, +q, –q are placed in order on the four consecutive corners of a square of side L. The, work done in inter changing the position of any two neighbouring charges of the opposite sign is, (1), , , , q2, –4  2, 40L, , , , (2), , Sol. Answer (3), , 2kq 2, a 2, , U  U2  U1  , , , , (3), , +q, , 4kq 2 2kq 2, U1 , , a, a 2, U2 , , , , q2, 42 2, 40L, , –q, , –q, 4kq, , 2, , a 2, , , , 4kq, a, , , , q2, 4–2 2, 40L, , +q, , , , (4), , q, , –q, , q, , –q, , , , q2, 4 2, 40L, , , , 2, , kq 2 ⎡⎣ 4  2 2 ⎤⎦, W, a, , 21. The charge q is fired 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 2v, the closest distance of approach would be, , q, (1) r, , v, , r, , (2) 2r, , (3), , Q, , r, 2, , (4), , r, 4, , Sol. Answer (4), 1, kqQ, mv 2 , 2, r, , v, q, , Q, , 1, kqQ, 2, m  2v  , 2, r, , 1 r, , 4 r, r , , r, 4, , 22. Three charges are placed along x-axis at x = – a, x = 0 and x = a as shown in the figure. The potential energy, of the system is, (q ), x = –a, , ⎛ 1 ⎞ q2, (1) – ⎜, ⎝ 4 0 ⎟⎠ a, , –q, , ( q), , x=0, , x=a, , ⎛ 1 ⎞ 3q 2, (2) – ⎜, ⎝ 4 0 ⎟⎠ 2a, , ⎛ 1 ⎞ q2, (3)  ⎜, ⎝ 4 0 ⎟⎠ a, , ⎛ 1 ⎞ 3q 2, (4)  ⎜, ⎝ 4 0 ⎟⎠ 2a, , Sol. Answer (2), U, ,  kq 2 kq 2 ka 2, , , 2a, a, a, , U, , 3kq 2, 2a, , q, x = –a, , –q, , q, , x=0, , x=a, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 59, , 23. Find the charge on capacitor C3, , C1, , C2, , C3, , C4, V, , Given, that C1 = C2 = C and C3 = C4 = 3C., (1), , 3, CV, 2, , (2), , CV, 2, , Sol. Answer (1), Cnet  2C, , q1 C1, , C2, , C, 3C, , C, 3C, , C3, , C4, , q = 2CV, , q2, , q1 1, , q2 3, , (3) 3CV, , (4) 2CV, , V, , 3q1 = q2, q1 + q2 = 2CV, q1 , , CV, 2, , q2 , , 3CV, 2, , 24. If initial charge on all the capacitors were zero, work done by the battery in the circuit shown is, , 4 F, , 4 F, 2 F, , 10 V, (1) 0.2 mJ, , (2) 200 mJ, , (3) 0.4 mJ, , (4) 400 mJ, , Sol. Answer (3), Cnet = 4F, , 4 F, , q = 40 F, W = qV = 40 ×, , 4 F, 2 F, , 10–6, , × 10, , 10 V, , W = 400 J, , 25. While working on a physics project at school physics lab, you require a 4 F capacitor in a circuit across a, potential difference of 1 kV. Unfortunately, 4 F capacitors are out of stock in your lab but 2 F capacitors, which can withstand a potential difference of 400 V are available in plenty. If you decide to use the 2 F, capacitors in place of 4 F capacitor, minimum number of capacitors required are, (1) 16, , (2) 18, , (3) 20, , (4) 12, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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60, , Electrostatic Potential and Capacitance, , Solution of Assignment, , Sol. Answer (2), CAB , , 2, 3, , Total number of benches , , A, 2, , B, , 4 3, 6, 2, , Total number of capacitors = 6 × 3 = 18, 26. A parallel plate capacitor with air between the plates has a capacitance C. If the distance between the plates, is doubled and the space between the plates is filled with a dielectric of dielectric constant 6, then the, capacitance will become, C, C, (3) 12C, (4), (1) 3C, (2), 3, 6, Sol. Answer (1), , C, , A 0, d, , C , , 6 A 0 2 A 0, , 2d, d, , C = 3C, 27. A capacitor of capacitance C is charged with the help of a 200 V battery. It is then discharged through a small, coil of resistance wire embedded in a thermally insulated block of specific heat capacity 2.5 × 102 J/kg and, mass 0.1 kg. If the temperature of the block rises by 0.4 K, the value of C is, (1) 500 F, , (2) 500 F, , (4) 50 F, , (3) 50 F, , Sol. Answer (2), 1, 2,  C   200  2.5  102  0.1  0.4, 2, , 2  104 C  1  10, C, , 1,  500 F, 2  103, , 28. Electric charges having same magnitude of electric charge 'q ' coulombs are placed at x = 1 m, 2 m, 4 m,, 8m ....... so on. If any two consecutive charges have opposite sign but the first charge is necessarily positive,, what will be the potential at x = 0?, (1) Infinity, , (2) Zero, , (3), , 1 ⎛ 2q ⎞, ⎜ ⎟, 4 0 ⎝ 3 ⎠, , 1, (4) 4  2q , 0, , Sol. Answer (3), , kq kq kq kq, , , ,  ......, 1, 2, 4, 8, , x=0, , q, , –q, , 1, , 2, , q, 3, , 4, , ⎡ 1 1 1, ⎤, kq ⎢1     .....⎥, 2, 4, 8, ⎣, ⎦, , kq.1, V, ⎛ 1⎞, 1 ⎜ ⎟, ⎝ 2⎠, V, , 2kq, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 61, , 29. A pellet carrying a charge of 0.5 coulomb is accelerated through a potential of 2000 volts. It attains some kinetic, energy equal to, (1) 1000 erg, , (2) 1000 joule, , (3) 1000 kWh, , (4) 500 erg, , Sol. Answer (2), KE = qV, KE = (0.5) (2000) = 1000 J, 30. A cylindrical capacitor has two co-axial cylinders of length 20 cm and radii 2r and r. Inner cylinder is given a, charge 10 C and outer cylinder a charge of –10 C. The potential difference between the two cylinders will, be, (1), , 0.1 In 2, mV, 4 0, , In 2, (2) 4 m V, 0, , (3), , 10 In 2, mV, 4 0, , (4), , 0.01 In 2, mV, 4 0, , Sol. Answer (1), C , , 2  0 l, ⎛ b⎞, ln ⎜ ⎟, ⎝ a⎠, , ⎛ 20 ⎞, 2  0 ⎜, ⎝ 100 ⎟⎠, C, ln 2, , 10  106 ln3, ⎛ 20 ⎞, 20 ⎜, ⎝ 100 ⎟⎠, , V , , V , , 10 4 ln2, 4 0, , V , , 0.1ln2, mV, 40, , 31. A charge q is distributed uniformly on the surface of a sphere of radius R. It is covered by a concentric hollow, conducting sphere of radius 2R. Charge on the outer surface of the hollow sphere will be, if it is earthed, (1), , q, 2, , (2) 2q, , (3) 4q, , (4) Zero, , Sol. Answer (4), If outer surface is earthed its charge becomes 0., 32. There exists an electric field of magnitude E in x-direction. If the work done in moving a charge of 0.2 C through, a distance of 2 m along a line making an angle 60° with x-axis is 4 J, then the value of E is, (1), , 3 N/C, , (2) 4 N/C, , (3) 5 N/C, , (4) 20 N/C, , Sol. Answer (4), E, , W = qEd cos, 4   0.2  E  2 , , 1, 2, , E = 20 N/C, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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62, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 33. In a region of space, suppose there exists a uniform electric field, , with a velocity v  –2 jˆ, its potential energy, ,  , ⎛ V ⎞, E  1 0 iˆ ⎜, ⎝ m ⎟⎠, , . If a positive charge moves, , (1) Increases, , (2) Decreases, , (3) Does not change, , (4) Initially increases then decreases, , >, , Sol. Answer (3), , 2 j, E = 10, , , As charge moves perpendicular to E , no change in energy occurs., , 34. ABC is a right angled triangle situated in a uniform electric field E which is in the plane of the triangle. The, points A and B are at the same potential of 15 V while the point C is at a potential of 20 V. AB = 3cm and, BC = 4 cm. The magnitude of electric field is (in S.I. Units), A, 3 cm, B, (1) 100, , C, , 4 cm, , (2) 125, , (3) 167, , (4) 208, , 15 V, , Sol. Answer (2), , ⎛ 4 ⎞, 5  E⎜, ⎝ 100 ⎟⎠, , 3, , E = 125 N/C, , 15 V, , 4, , 20 V, , 35. A hollow spherical conductor of radius r potential of 100 V at its outer surface. The potential inside the hollow, r, from its centre is, at a distance of, 2, (1) 100 V, (2) 50 V, (3) 200 V, (4) Zero, Sol. Answer (1), Inside the conductor, the field is 0 and potential is constant V = 100 V., 36. A spherical conductor having charge q and radius r is placed at the centre of a spherical shell of radius R, and having charge Q (R > r). The potential difference between the two is, (1) Proportional to Q, , (2) Proportional to q, , (3) Dependent on both Q and q, , (4) Independent of both Q and q, , Sol. Answer (2), ⎡1 1 ⎤, VA  VB  kq ⎢  ⎥, ⎣r R ⎦, , V  q, , r, , R, , q, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 63, , 37. The work which is required to be done to make an arrangement of four particles each having a charge +q such, that the particles lie at the four corners of a square of side a is, (1), , , , 4 2, , , , kq 2, a, , (2) 4, , Sol. Answer (1), U, , kq 2, a, , (3), , q, , q, , q, , q, , , , 2 2, , , , kq 2, a, , (4) 2, , kq 2, a, , 4kq 2 2kq 2, , a, a 2, , W=U, , 38. The net capacitance of a system of capacitance as shown in the figure between points A and B is, 1 F, A, , 1 F, , 1 F, , 1 F, , (1) 1 F, , (2) 2 F, , Sol. Answer (2), Cnet = 2 F, , (3) 3 F, , 1, , It is balanced W.S.B, , B, , 1 F, , (4) 4 F, , 1, 1, , A, , B, 1, , 1, , SECTION - C, Previous Years Questions, 1., , A parallel plate air capacitor has capacity C, distance of separation between plates is d and potential difference, V is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is, [Re-AIPMT-2015], (1), , C 2V 2, 2d 2, , (2), , C 2V 2, 2d, , (3), , CV 2, 2d, , (4), , CV 2, d, , Sol. Answer (3), Force between the plates, , 2., , F, , q2, 1,  qE, 2 A0 2, , F, , 1, V 1 V2, CV  C, 2, d 2 d, , A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of, the sphere respectively are, [AIPMT-2014], (1) Zero and, , Q, 4 0 R, , 2, , Q, (2) 4 R and zero, 0, , Q, Q, (3) 4 R and, (4) Both are zero, 40 R 2, 0, , Sol. Answer (2), Electric field inside a conductor is always zero and conductor is a equipotential body. The value of electric, potential at the surface will be at the centre., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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64, 3., , Electrostatic Potential and Capacitance, , Solution of Assignment, , In a region, the potential is represented by V (x, y, z) = 6x – 8xy – 8y + 6yz, where V is in volts and x, y, z are, in meters. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is, [AIPMT-2014], (1) 6 5 N, , (2) 30 N, , (4) 4 35 N, , (3) 24 N, , Sol. Answer (4), Electric force F = qE, E, , ⎡V ˆ V ˆ V, V,  ⎢, i , j, r, y, z, ⎣ x, , ⎤, kˆ⎥, ⎦, , V, , , 6 x  8 xy  8y  6yz  6  8y, x x, , V,  8 x  8  6z, y, V,  6y, z, E(1, 1, 1)   ⎡⎣(6  8)iˆ  ( 8  8  6) ˆj  (6)kˆ⎤⎦,   ⎡⎣2iˆ  10 ˆj  6kˆ⎤⎦, , | E |  2 35 N/C, F  qE  (2)(2 35)  4 35 N, Two thin dielectric slabs of dielectric constants K1 and K2 (K1 < K2) are inserted between plates of a parallel, plate capacitor, as shown in the figure. The variation of electric field E between the plates with distance d as, measured from plate P is correctly shown by, [AIPMT-2014], , P+, , –, , +, , K1, , K2, , Q, , –, , 4., , E, , E, (1), , (2), , 0, , d, , E, , E, (3), , 0, , d, , (4), , 0, , d, , 0, , d, , Sol. Answer (3), In vacuum electric field between parallel plate capacitor is given by, , E, , Q, A 0, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 65, , In medium,, E , , Q, kA0, , as K2 > K1,  Electric field will be less in K2., 5., , A, B and C are three points in a uniform electric field. The electric potential is, A, , B, , [NEET-2013], , E, , C, , (1) Maximum at B, , (2) Maximum at C, , (3) Same at all the three points A, B and C, , (4) Maximum at A, , Sol. Answer (1), , A, , B, C, VB > VC > VA, 6., , An electric dipole of moment p is placed in an electric field of intensity E . The dipole acquires a position such, that the axis of the dipole makes an angle  with the direction of the field. Assuming that the potential energy, of the dipole to be zero when  = 90°, the torque and the potential energy of the dipole will respectively be, [AIPMT (Prelims)-2012], (1) pE sin, 2pE cos, , (2) pE cos, –pE sin, , I  pE sin , , , , U  – pE cos , , E, , Four point charges –Q, –q, 2q and 2Q are placed, one at each corner of the square. The relation between Q, and q for which the potential at the centre of the square is zero is, [AIPMT (Prelims)-2012], (1) Q = q, , (2) Q =, , 1, q, , Sol. Answer (3), , Vp , , k  –Q , r, , , , k  –q , r, , , , k  2q , r, , , , k  2Q , r, , (3) Q = – q, , (4) Q = –, , –Q, r, r, 2Q, , q = –Q, , 1, q, , –q, , 0, , kQ kq, , 0, r, r, , 8., , (4) pE sin, –2pE cos, , , P, , Sol. Answer (3), , 7., , (3) pE sin, –pE cos, , r, P, , r, 2q, , A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between, the plates is d and area of each plate is A, the energy stored in the capacitor is, [AIPMT (Mains-2012 & Prelims-2011)], (1), , 1, 0 E 2, 2, , 2 Ad, (2) E , 0, , (3), , 1, 0 E 2 Ad, 2, , (4) 0EAd, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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66, , Electrostatic Potential and Capacitance, , Sol. Answer (3), , 9., , C, , A0, d, , U , , 1,  E 2 ·Ad, 2 0, , Solution of Assignment, , d, , A, , Two metallic spheres of radii 1 cm and 3 cm are given charges of –1 × 10–2 C and 5 × 10–2 C, respectively. If, these are connected by a conducting wire, the final charge on the bigger sphere is, [AIPMT (Mains)-2012], (1) 2 × 10–2 C, , (2) 3 × 10–2 C, , (3) 4 × 10–2 C, , (4) 1 × 10–2 C, , Sol. Answer (2), q 4  10 –2 – q, , 3, 1, , q = 12 × 10–2 – 3q, 4q = 12 × 10–2, q = 3 × 10–2, 10. Four electric charges +q, +q, –q and –q are placed at the corners of a square of side 2L (see figure). The, electric potential at point A, midway between the two charges +q and +q, is, [AIPMT (Prelims)-2011], , +q, , –q, , A, –q, , +q, (1) Zero, , (2), , 1 2q, (1 5 ), 40 L, , (3), , 1 2q ⎛, 1 ⎞, 1, ⎟, 40 L ⎜⎝, 5⎠, , (4), , 1 2q ⎛, 1 ⎞, 1–, ⎟, 40 L ⎜⎝, 5⎠, , Sol. Answer (4), , q, , kq kq, kq, kq, V , , –, –, L, L L 5 L 5, , , 2L, , –q, , 2L, , –q, , L, L, , kq ⎡⎣2 5 – 2⎤⎦, , q, , L 5, , 11. Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC, 2a. D and, E are the mid points of BC and CA.The work done in taking a charge Q from D to E is [AIPMT (Mains)-2011], , A, E, B, (1) Zero, , 3qQ, (2) 4 a, 0, , D, , C, 3qQ, (3) 8 a, 0, , qQ, (4) 4 a, 0, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 67, , Sol. Answer (1), , q, a, E, , q, , a, , D, 2a, , 2a, a, , a, , q, , VD = VE, W = q(VD – VE) = 0, 12. The electric potential V at any, point (x, y, z), in metres in space is given by V = 4x2 volt. The electric field at, the point (1, 0, 2) in volt/metre, is, [AIPMT (Mains)-2011], (1) 16 along positive x-axis, , (2) 8 along negative x-axis, , (3) 8 along positive x-axis, , (4) 16 along negative x-axis, , Sol. Answer (2), V = 4x2, , Ex  –, , dV,  –8x, dx, , Ex = –8 N/C, 4 V., 13. A series combination of n1 capacitors, each of value C1, is charged by a source of potential difference, When another parallel combination of n2 capacitors, each of value 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, [AIPMT (Prelims)-2010], , 2C1, (1) n n, 1 2, , n2, (2) 16 n C1, 1, , n2, (3) 2 n C1, 1, , (4), , 16C1, n1n2, , Sol. Answer (4), Cnet , , C1, n1, , U, , 1 C1 2 2, ·4 V, 2 n1, , U, , 8C1V 2, n1, , Cnet = n2C2, , U, , 1, n2C2V 2, 2, , 8C1V 2 n2C2V 2, , 2, n1, C2 , , 16C1, n1n2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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68, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 14. A condenser of capacity C is charged to a potential difference of V1. The plates of the condenser are then, connected to an ideal inductor of inductance L. The current through the inductor when the potential difference, across the condenser reduces to V2 is, [AIPMT (Mains)-2010], ⎛ C V  V 2, 1, 2, (1) ⎜⎜, L, ⎝, , 1/2, , ⎞, ⎟, ⎟, ⎠, , (2), , C, , , , V12, ,  V22, , , , L, , (3), , C, , , , V12, ,  V22, , , , ⎛ C V12  V22, ⎜, (4) ⎜, L, ⎝, , , , L, ,  ⎞⎟, , 1/2, , ⎟, ⎠, , Sol. Answer (4), 15. Two parallel metal plates having charges +Q and –Q face each other at a certain distance between them. If the, plates are now dipped in kerosene oil tank, the electric field between the plates will [AIPMT (Mains)-2010], (1) Become zero, , (2) Increase, , (3) Decrease, , (4) Remain same, , Sol. Answer (3), , q =C V, constant increases decreases, E decreases, 16. Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities , – and,  respectively. If VA, VB and VC denote the potentials of the three shells, then for c = a + b, we have, [AIPMT (Prelims)-2009], (1) VC = VB  VA, , (2) VC  VB  VA, , (4) VC = VA  VB, , (3) VC = VB = VA, , Sol. Answer (4), , VA , , k  4a2 k  4b2 k  4c 2, –, , a, b, c, , , , VA = 4k(a – b + c), , VB , , 2, , –, , 2, , k  4a, k  4b, k  4c, –, , b, b, c, , , a, , b, c, , 2, , ⎡ a2, ⎤, VB  k 4 ⎢ – b  c ⎥, ⎢⎣ b, ⎥⎦, VC , , k 4a 2 k 4b 2 k 4c 2, –, , c, c, c, , ⎡ a2 – b2, ⎤, VC  k 4 ⎢,  c⎥, ⎥⎦, ⎣⎢ c, If VA = VC, , a–bc , , a2 – b2, c, c, , c=a+b, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 69, , 17. Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and, breakdown voltage of the combination will be, [AIPMT (Prelims)-2009], (1) 3C,, , V, 3, , (2), , C, , 3V, 3, , (3) 3C, 3V, , (4), , C V, ,, 3 3, , Sol. Answer (2), Net capacitance , , C, 3, , Breakdown voltage = 3 V, , V, , V, , V, , C, , C, , C, , , 18. The electric potential at a point (x, y, z) is given by V = –x2y – xz3 + 4. The electric field E at that point is, [AIPMT (Prelims)-2009], , (1) E  iˆ(2 xy  z3 )  ˆjxy 2  kˆ 3z 2 x, , , ˆ 3  jxyz, ˆ, ˆ 2, (2) E  iz,  kz, , , (3) E  iˆ(2 xy  z3 )  ˆjx 2  kˆ 3 xz 2, , , (4) E  iˆ2 xy  ˆj ( x 2  y 2 )  kˆ(3 xz  y 2 ), , Sol. Answer (3), V = –x2y – xz3 + 4, , Ex  –, , dV,   2xy  z3, dx, , Ey  –, , dV,  x2, dy, , Ez = + 3xz2, , E  2 xy  z 3 i  x 2 j  3 xz 2 , k, , , , , , 19. The mean free path of electrons in a metal is 4 × 10–8 m. The electric field which can given on an average 2 eV, energy to an electron in the metal will be in units of V/m, [AIPMT (Prelims)-2009], (1) 5 × 10–11, , (2) 8 × 10–11, , (3) 5 × 107, , (4) 8 × 107, , Sol. Answer (3), 2 = E (4 × 10–8), E, , 2,  108, 4, , E = 0.5 × 108 N/C, E = 5 × 107, 20. The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section, A such that the uniform electric field between the plates is E, is, [AIPMT (Prelims)-2008], (1), , 1,  0 E 2 Ad, 2, , (2), , 1,  0 E 2 / A.d, 2, , (3)  0 E 2 / Ad, , (4)  0 E 2 Ad, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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70, , Electrostatic Potential and Capacitance, , Solution of Assignment, , Sol. Answer (1), U, , 1, CV 2, 2, , U, , 1 A 0,  Ed 2, 2 d, , U, , A0E 2d, 2, , 21. The electric potential at a point in free space due to a charge Q coulomb is Q × 1011 volts. The electric field at, that point is, [AIPMT (Prelims)-2008], (1) 12 0Q × 1022 volt/m, , (2) 4 0Q × 1022 volt/m, , (3) 12 0Q × 1020 volt/m, , (4) 4 0Q × 1020 volt/m, , Sol. Answer (2), , 9  109 Q,  Q  1011, r, r = 9 × 10–2, E, , E, , 9  109  Q, 81 10 –4, 1,  1013 Q, 9, , KQ,  Q  1011, r, , r, , E, , k, 1011, kQ, k2, , 1022, , E = 4Q × 1022, 22. Charges +q and –q are placed at points A and B respectively which are a distance 2L apart, C is the midpoint, between A and B. The work done in moving a charge +Q along the semicircle CRD is [AIPMT (Prelims)-2007], , R, , A, qQ, (1)  6 L, 0, , qQ, (2)  4 L, 0, , C, , B, , D, , qQ, (3) 2 L, 0, , qQ, (4)  2 L, 0, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 71, , Sol. Answer (1), , R +Q, 2L, A, (+q), , C, , B, (–q), , VC , , kq kq, –, 0, L, L, , VD , , kq kq –2kq, –, , 3L L, 3L, , D, , W = +Q(VC – VD), W Q·, , 2kq, 3L, , 23. Two condensers, one of capacity C and other of capacity C/2 are connected to a V-volt battery, as shown. The, work done in charging fully both the condensers is, [AIPMT (Prelims)-2007], , C, , V, , (1), , 1, CV2, 2, , (2) 2CV2, , C/2, , (3), , 1, CV2, 4, , (4), , 3, CV2, 4, , Sol. Answer (4), , W , , 1, 1C 2, CV 2 , V, 2, 2 2, , W , , 3, CV 2, 4, , V, , C, , C/2, , 24. A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery, the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential, difference between the plates, [AIPMT (Prelims)-2006], (1) Decreases, , (2) Does not change, , (3) Becomes zero, , (4) Increases, , Sol. Answer (4), U, , 1, CV 2, 2, , q =C V, constant decreases, , d increases, C decreases, increases, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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72, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 25. A network of four capacitors of capacity equal to C1 = C, C2 = 2C, C3 = 3C and C4 = 4C are connected to a battery, [AIPMT (Prelims)-2005], as shown in the figure. The ratio of the charges on C2 and C4 is, , C2, C3, , C1, , C4, , V, (1), , 22, 3, , (2), , 3, 22, , Sol. Answer (2), q1 24C·11, , q2, 36C, , 7, 4, , (3), , (4), , 4, 7, , 2C q2, 3C, , C, 4C, q1, , q1 22, , q2, 3, , 26. As per this diagram a point charge +q is placed at the origin O. Work done in taking another point charge –Q, from the point A [Coordinates (0, a)] to another point B [Coordinates (a, 0)] along the straight path AB is, [AIPMT (Prelims)-2005], y, A, , O, , (1) Zero, , ⎛ qQ 1 ⎞, (2) ⎜ 4 2 ⎟ 2a, 0 a ⎠, ⎝, , B, , x, , ⎛ qQ 1 ⎞ a, (3) ⎜ 4 2 ⎟ , 2, 0 a ⎠, ⎝, , ⎛ qQ 1 ⎞, (4) ⎜ 4 2 ⎟ 2a, 0 a ⎠, ⎝, , y, , Sol. Answer (1), A, (0, a), , VA = VB, W = –Q(VB – VA), , | VA = VB, (+q)O, , W=0, , B, (a, 0), , x, , 27. As a result of change in the magnetic flux linked to the closed loop shown in the figure, an emf V volt is induced, in the loop. The work done (joules) in taking a charge Q coulomb once along the loop is, [AIPMT (Prelims)-2005], , (1) QV, , (2) Zero, , (3) 2QV, , (4), , QV, 2, , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 73, , 28. Two charges q1 and q2 are placed 30 cm apart, as shown in the figure. A third charge q3 is moved along the arc, q3, of a circle of radius 40 cm from C to D. The change in the potential energy of the system is 4 k , where k, 0, is, [AIPMT (Prelims)-2005], q3, C, , 40 cm, q1, A, (1) 8q2, , 30 cm B D, , (2) 8q1, , Sol. Answer (1), VC , , q2, , k  q1   100, 40, , , , k  q2  100, 50, , (3) 6q2, , C, , q3, , 40 cm, , VC  k  2.5q1  2q2 , VD , , kq1 100 , 40, , , , (4) 6q1, , q2, , q1, , kq2 100 , , 10, cm, , 30, cm, , D, , 10, , VD = k (2.5q1 + 10q2), U = q3 (VD – VC), U = q3 (8kq2), U = kq3 (8q2), 29. An electric dipole of dipole moment p is aligned parallel to a uniform electric field E. The energy required to, rotate the dipole by 90° is, (1) pE2, , (2) p2E, , (3) pE, , (4) Infinity, , Sol. Answer (3), , P, , E, , , P, , Ui  – pE, , U  0, , W = U = 0 – (–pE), W = pE, 30. Which of the following is not true?, (1) For a point charge, the electrostatic potential varies as 1/r, (2) For a dipole, the potential depends on the position vector and dipole moment vector, (3) The electric dipole potential varies as 1/r at large distance, (4) For a point charge, the electrostatic field varies as 1/r2, Sol. Answer (3), Option (3) is wrong for dipole, potential varies as, , 1, r2, , ., , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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74, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 31. How many 1 F capacitors must be connected in parallel to store a charge of 1 C with a potential of 110 V, across the capacitors?, (1) 990, , (2) 900, , (3) 9090, , (4) 909, , Sol. Answer (3), For n capacitors, Cnet = nC, 1 = n(10–6)(110), n, , 1000000,  9090, 110, , 32. A hollow metallic sphere of radius 10 cm is charged such that potential of its surface is 80 V. The potential, at the centre of the sphere would be, (1) 80 V, , (2) 800 V, , (3) Zero, , (4) 8 V, , Sol. Answer (1), , B, A, VA = VB = 80 V, 33. Charge q2 is at the centre of a circular path with radius r. Work done in carrying charge q1 once around this, equipotential path, would be, q1q 2, 1, (2) 4  r, 0, , q1q 2, 1, (1) 4  2, r, 0, , Sol. Answer (3), , (3) Zero, , (4) Infinite, , r, , W = q(Vf – Vi), , q2, , kq, Vf  Vi  2, r, W=0, , 34. An electric dipole of moment an electric is placed in the position of stable equilibrium in uniform electric field, of intensity E. This is rotated through an angle  from the initial position. The potential energy of the electric, dipole in the final position is, (1) – pE cos , , (2) pE (1 – cos ), , (3) pE cos , , (4) pE sin , , Sol. Answer (1), , P, , P, , , , E, , E, , U = –pEcos, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 75, , 35. There is an electric field E in x-direction. If the work done on moving a charge of 0.2 C through a distance of, 2 m along a line making an angle 60° with x-axis is 4 J, then what is the value of E?, (1) 5 N/C, , (2) 20 N/C, , (3), , 3 N/C, , (4) 4 N/C, , Sol. Answer (2), q = 0.2 C, , 2m, , r=2m, , E, ,  = 60°, , ⎛ 1⎞, 4   0.2 E  2  ⎜ ⎟, ⎝2⎠, E = 20 N/C, 36. Two metallic spheres of radii 1 cm and 2 cm are given charges 10–2 C and 5 × 10–2 C respectively. If they, are connected by a conducting wire, the final charge on the smaller sphere is, (1) 3 × 10–2 C, , (2) 2 × 10–2 C, , (3) 1 × 10–2 C, , (4) 6 × 10–2 C, , Sol. Answer (2), , 10–2, , –2, , q 6  10 – q, , 1, 2, , 5 × 10–2, , 1, , 2, , q, , 6 × 10–2 –q, , 2q = 6 × 10–2 – q, q = 2 × 10–2 C, , 37. The energy stored in a capacitor of capacity C and potential V is given by, (1), , CV, 2, , (2), , C 2V 2, 2, , (3), , C 2V, 2, , (4), , CV 2, 2, , Sol. Answer (4), U, , 1, CV 2, 2, , 38. An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. Its, final velocity will be, (1), , 2eV, m, , (2), , eV, m, , (3), , eV, 2m, , (4), , eV, m, , Sol. Answer (1), eV , , 1, mv 2, 2, , 2eV,  v2, m, v, , 2eV, m, , 39. When a proton is accelerated through 1 V, then its kinetic energy will be, (1) 1 eV, , (2) 13.6 eV, , (3) 1840 eV, , (4) 0.54 eV, , Sol. Answer (1), KE = 1 × 1 = 1 eV, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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76, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 40. A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2) has a capacitance C., If the oil is removed, then capacitance of the capacitor becomes, (1), , C, 2, , (2) 2C, , (3), , 2C, , (4), , C, 2, , Sol. Answer (4), , C , , KC = C,, , C, 2, , 41. In bringing an electron towards another electron, the electrostatic potential energy of the system, (1) Becomes zero, , (2) Increases, , (3) Decreases, , (4) Remains same, , Sol. Answer (2), As 'r' decreases so, PE increases., 42. What is the effective capacitance between points X and Y?, , C1= 6  F, C3= 6  F C5=20  F C2= 6  F, X, , C, , A, , D, , B, , Y, , C4= 6  F, (1) 12 F, , (2) 18 F, , (3) 24 F, , (4) 6 F, , Sol. Answer (4), , B, 6 F, , C 20, X, , 6, B, , 6, , Y, , ⇒, , 6 F, , X, , Y, 6 F, , 6, , , , 6 F, , VB = VC, , 6 F, C, , Above is the diagram of a balanced W0B0, Cnet = 6 F, 43. A capacitor is charged with a battery and energy stored is U. After disconnecting battery another capacitor, of same capacity is connected in parallel to the first capacitor. Then energy stored in each capacitor is, (1) U/2, , (2) U/4, , (3) 4U, , (4) 2U, , Sol. Answer (2), U, , q2, 2C, , q/2, 2, , ⎛ q⎞, ⎜⎝ ⎟⎠, 2, U , 2C, , q2, U, U , , 8C 4, , q/2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Electrostatic Potential and Capacitance, , 77, , 44. Energy per unit volume for a capacitor having area A and separation d kept at potential difference V is given by, (1), , 1 V2, 2 0 d2, , 1 V2, (2) 2, 2, 0 d, , (3), , 1 CV 2, 2, , (4), , Q2, 2C, , Sol. Answer (1), , 1 ⎛V ⎞, 1, 2, Energy density   0 E   0 ⎜ ⎟, 2 ⎝d⎠, 2, , 2, , , , 1 V2, , 2 0 d2, , 45. Some charge is being given to a conductor. Then its potential is, (1) Maximum at surface, (2) Maximum at centre, (3) Remain same throughout the conductor, (4) Maximum somewhere between surface and centre, Sol. Answer (3), The potential of a conductor is same throughout its interior and at its surface., 46. A capacitor of capacity C1 is charged upto V volt and then connected to an uncharged capacitor of capacity, C2. The final potential difference across each will be, C 2V, (1) C  C, 1, 2, , C1V, (2) C  C, 1, 2, , C, ⎛, (3) ⎜⎜1  2, C1, ⎝, , ⎞, ⎟⎟ V, ⎠, , C, ⎛, (4) ⎜⎜1  2, C1, ⎝, , ⎞, ⎟⎟ ·V, ⎠, , Sol. Answer (2), C1, V, , C2, 0, , Vnet , , C1V, C1  C2, , 47. Identical charges (–q) are placed at each corners of cube of side b then electrostatic potential energy of charge, (+q) which is placed at centre of cube will be, (1), ,  8 3q 2, 40 b, , (2), , Sol. Answer (3), , r , , 3b, 2, , ⎡, ⎤, ⎢ k  q  – q  ⎥, ⎥, U  8⎢, ⎢ b 3 ⎥, ⎢⎣, ⎥⎦, 2, , U , U , , –q, , –q, ,  8 2 q2,  0 b, –q, –q, , (3), ,  4q 2, 3  0 b, , (4), , 8 2 q2, 4 0 b, , r, –q, , q –q, , –q, , –q, , –16 q 2, 4  0 3 b, –4 q 2,  0 3 b, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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78, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 48. Three capacitors each of capacity 4 F are to be connected in such a way that the effective capacitance is, 6 F. This can be done by, (1) Connecting all of them in series, , (2) Connecting them in parallel, , (3) Connecting two in series and one in parallel, , (4) Connecting two in parallel and one in series, , Sol. Answer (3), 4 F, , A, , 4 F, , B, , 4 F, , Cnet = 6 F, 49. An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform, electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy, are respectively, (1) 2qE and minimum, , (2) qE and pE, , (3) Zero and minimum, , (4) qE and maximum, , Sol. Answer (3), F=0, , P, , E, , U = –pE, , Minimum, 50. A bullet of mass 2 kg is having a charge of 2 mC. Through what potential difference must it be accelerated,, starting from rest, to acquire a speed of 10 m/s?, (1) 5 kV, , (2) 50 kV, , (3) 5 V, , (4) 50 V, , Sol. Answer (2), ,  2  10 V  21 · 2 ·10,  2  10 V  100, –3, , 2, , –3, , V = 50 × 103 = 50 kV, SECTION - D, Assertion-Reason Type Questions, 1., , A : For a given potential function, electric intensity function can be uniquely derived., R : For a given electric intensity function, electric potential function can be uniquely derived., , Sol. Answer (3), Reason is wrong. Electric potential difference can be derived but not electric potential function., 2., , A : An equipotential surface is normal to electric field lines., R : Potential increases in the direction of electric field., , Sol. Answer (3), Reason is wrong. Electric potential decreases in the direction of electric field., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 3., , Electrostatic Potential and Capacitance, , 79, , A : One may have zero potential but non-zero electric field at a point in space., R : Potential is a scalar quantity., , Sol. Answer (2), Both are correct but reason is not the correct explanation of assertion., 4., , A : Absolute value of potential is not defined., R : Two equipotential lines cannot intersect each other., , Sol. Answer (2), Both are correct but reason is not the correct explanation of assertion., 5., , A : Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. The magnitude, of electrostatic force between them is exactly given by Q1 Q2/40r2, where r is the distance between their, centres., R : Here charges Q1 and Q2 can be assumed to be concentrated at the centres of their respective spheres., , Sol. Answer (4), Both are wrong., 6., , A : Work done by the field of a nucleus in a complete orbit of the electron is zero even if the orbit is elliptical., R : Electrostatic force is conservative in nature., , Sol. Answer (1), Both are correct and reason is correct explanation of assertion., 7., , A : Electric field is discontinuous across the surface of a charged conductor., R : Electric potential is discontinuous across the surface of a charged conductor., , Sol. Answer (3), Reason is wrong. Potential is continuous across the surface of charged conductor., 8., , A : Water has a much greater dielectric constant than any other ordinary substance., R : Water has permanent dipole moment., , Sol. Answer (1), Both are correct and reason is the correct explanation of assertion., 9., , A : Electric potential of a positively charged body may be negative., R : The potential of a conductor does not depend on the charge of the conductor., , Sol. Answer (3), Reason is wrong. Potential of a conductor depends on the charge of the conductor., 10. A : The potential difference between two concentric spherical shells depends only on the charge of inner shell., R : The electric field in the region in between two shells depends on the charge of inner shell and electric field, is the negative of potential gradient., Sol. Answer (1), Both are correct and reason is the correct explanation of assertion., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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80, , Electrostatic Potential and Capacitance, , Solution of Assignment, , 11. A : If E be electric field at a point, in free space then energy density at that point will be, , 1,  0E 2 ., 2, , R : Electrostatic field is a conservative field., Sol. Answer (2), Both are correct but reason is not the correct explanation of assertion., 12. A : A capacitor is a device which stores electric energy in the form of electric field., R : Net charge on the capacitor is always zero., Sol. Answer (2), Both are correct but reason is not the correct explanation of assertion., 13. A : When two conductors charged to different potentials are connected to each other, the negative charge, always flows from lower potential to higher potential., R : In the charging process, there is always a flow of electrons only., Sol. Answer (2), Both are correct but reason is not the correct explanation of assertion., , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Chapter, , 18, , Current Electricity, Solutions, SECTION - A, Objective Type Questions, 1., , Electric current has both magnitude and direction. It is a, (1) Vector quantity, , (2) Scalar quantity, , (3) Tensor quantity, , (4) None of these, , Sol. Answer (2), Current is a scalar quantity., 2., , Identical piece of Ge and Cu are taken and cooled, then, (1) Resistivity of both increases, , (2) Resistivity of both decreases, , (3) Resistivity of Cu increases and Ge decreases, , (4) Resistivity of Cu decreases and Ge increases, , Sol. Answer (4), Ge is a semiconductor so its resistivity decreases with increase in temperature., 3., , A current of 10 A is maintained in a conductor of cross-section 1 cm2. If the number density of free electrons, be 9 × 1028 m–3, the drift velocity of free electrons is, (1) 6.94 × 10–6 m/s, , (2) 5.94 × 10–2 m/s, , (3) 1.94 × 10–3 m/s, , (4) 2.94 × 10–4 m/s, , Sol. Answer (1), i = neAVd, 10 = (9 × 1028) (1.6 × 10–19) (10–4) VD, Solving, we get, VD = 6.94 × 10–6 m/s, 4., , A potential difference of 5 V is applied across a conductor of length 10 cm. If drift velocity of electrons is, 2.5 × 10–4 m/s, then electron mobility will be, (1) 5 × 10–4 m2 V–1 s–1, , (2) 5 × 10–6 m2 V–1 s–1, , (3) 5 × 10–2 m2 V–1 s–1 (4) Zero, , Sol. Answer (2), , , VD, E, , 2.5  104  0.1, 5, –6,  = 5 × 10 m2V–1s–1, , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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82, 5., , Current Electricity, , Solution of Assignment, , A potential difference of 10 V is applied across a conductor of 1000 . The number of electrons flowing through, the conductor in 300 s is, (1) 1.875 × 1016, , (2) 1.875 × 1017, , (3) 1.875 × 1022, , (4) 1.875 × 1019, , Sol. Answer (4), , 6., , i, , 10,  0.01 A, 1000, , n, , 0.01, 1.6  1019, , If n, e,  and m are representing electron density, charge, relaxation time and mass of an electron respectively,, then the resistance of a wire of length l and cross-sectional area A is given by, , (1), , ml, 2, , ne  A, , (2), , mA, 2, , ne l, , (3), , ne 2  A, ml, , (4), , ne 2 A, m l, , Sol. Answer (1), , 7., , R, , Pl, A, , R, , ml, ne 2IA, , Ohm's law fails in, (1) Diode, , (2) Thyristor, , (3) PN junction system (4) All of these, , Sol. Answer (4), Ohm's law is not applicable for diode, thyristor as well as PN junction, 8., , The resistance of a rectangular block of copper of dimensions 2 mm × 2 mm × 5 m between two square faces, is 0.02 . What is the resistivity of copper?, (1) 1.6 × 10–8 , , (2) 1.6 × 10–6 -m, , (3) 1.6 × 10–8 -m, , (4) Zero, , Sol. Answer (3), R, , l, A, , 0.02 , , , , (5), 4  10 6, , 8  108, 5, , = 1.6 × 10–8 -m, 9., , If a copper wire is stretched to make its radius decrease by 0.1%, then the percentage increase in resistance, is nearly, (1) 0.1%, , (2) 0.8%, , (3) 0.4%, , (4) 0.2%, , Sol. Answer (3), For x% decrease in radius, increase in resistance is 4x%,  0.4 %, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 83, , 10. A certain piece of copper is to be shaped into a conductor of minimum resistance. Its length and diameter, should respectively be, (2) 2L,, , (1) L, D, , D, 2, , (3), , L, , 2D, 2, , (4) L,, , D, 2, , Sol. Answer (3), For (L / 2, D), resistance is minimum, 11. A wire of resistance x ohm is drawn out, so that its length is increased to twice its original length, and its, new resistance becomes 20 , then x will be, (1) 5 , , (2) 10 , , (3) 15 , , (4) 20 , , Sol. Answer (1), x, , l, A, , 20 , , (2l ), ( A / 2), , x, 1, , 20 4, x=5, 12. A piece of wire is cut into four equal parts and the pieces are bundled together side by side to form a thicker, wire. Compared with that of the original wire, the resistance of the bundle is, (1) The same, , (2), , 1, as much, 16, , (3), , 1, as much, 8, , (4), , 1, as much, 4, , Sol. Answer (2), , R/4, R/4, R/4, R/4, Rnet , , R, 16, , 13. Two wires A and B of the same material, having radii in the ratio 1 : 2 carry currents in the ratio, 4 : 1. The ratio of drift speed of electrons in A and B is, (1) 16 : 1, , (2) 1 : 16, , (3) 1 : 4, , (4) 4 : 1, , Sol. Answer (1), 4i = neA VD, i = ne(4A) V'D, 4, VD, VD', , VD, , 4VD',  16 : 1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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84, , Current Electricity, , Solution of Assignment, , 14. The temperature co-efficient of resistance of a wire at 0°C is 0.00125 °C–1. At 25°C its resistance is one ohm., The resistance of the wire will be 1.2 ohm at, (1) 225 K, , (2) 190°C, , (3) 260°C, , (4) 185 K, , Sol. Answer (2), 1 = R0(1 + (0.00125) (25)), 1.2 = R0(1 + (0.00125) ), 12 1  0.00125, , 10, 1.03125, , Solving, we get,  = 190°C, 15. A conductor behaves as a superconductor, (1) Above critical temperature, , (2) At critical temperature, , (3) At 100°C, , (4) At boiling point of that metal, , Sol. Answer (2), A conductor behaves as a superconductor at critical temperature., 16. A carbon resistor has coloured strips as shown in figure. What is its resistance?, , Violet, , Yellow, (1) 410  ± 2%, , Brown, , (2) 470  ± 5%, , Gold, (3) 420  ± 3%, , (4) 405  ± 2%, , Sol. Answer (2), Y, , V, , B, , G, , 4, 7, 1, 47 × 10 ± 5%, , 5, , 470 ± 5%, , 6,  . One of the resistance wire is broken and the, 5, effective resistance becomes 2 ohms. Then the resistance (in ohm) of the wire that got broken is, , 17. Two resistors are joined in parallel whose resultant is, , (1), , 6, 5, , (2) 2, , (3), , 3, 5, , (4) 3, , Sol. Answer (4), R1R2, 6, , R1  R2 5, , R1 = 2, , 2R2, 6, , 2  R2 5, 5R2 = 6 + 3R2, 2R2 = 6, R2 = 3 , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 85, , 18. A wire has resistance 12 ohm. It is bent in the form of a circle. The effective resistance between the two points, on any diameter of the circle is, (1) 12 , , (2) 24 , , (3) 3 , , (4) 6 , , Sol. Answer (3), , 6, , A, , B, , 6, R=3, 19. A technician has only two resistance coils. By using them singly, in series or in parallel, he is able to obtain, the resistance 3, 4, 12 and 16 ohms. The resistance of the two coils are, (1) 6 and 10 ohms, , (2) 4 and 12 ohms, , (3) 7 and 9 ohms, , (4) 4 and 16 ohms, , Sol. Answer (2), R1R2, 3, R1  R2, , R1 + R2 = 16, R1R2 = 48, R1(16 – R1)= 48, R12 – 16R1 + 48 = 0, R1 = 4, 12, 20. Two resistances r1 and r2 (r1 < r2) are joined in parallel. The equivalent resistance R is such that, (1) R > r1 + r2, , (2) R  r1 r2, , (3) r1 < R < r2, , (4) R < r1, , Sol. Answer (4), In parallel the equivalent resistance is less than both the resistances., 21. The resultant resistance value of n resistances each of r ohms and connected in series is x. When those n, resistances are connected in parallel, the resultant value is, (1), , x, n, , (2), , x, n2, , (3) n2 x, , (4) n x, , Sol. Answer (2), For series, Req = x = nr, x, n, for parallel, , r=, , 1, n, , Req, r, , Req =, , x, r, x, Req = n n = 2, , , n, n, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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86, , Current Electricity, , Solution of Assignment, , 22. Twelve wires of equal resistance R are connected to form a cube. The effective resistance between two diagonal, ends A and E will be, D, C, , A, , B, E, , H, G, (1), , 5R, 6, , (2), , F, , 6R, 5, , (3) 12 R, , (4) 3 R, , Sol. Answer (1), The resistance between the body diagonal ends are, , 5R, 6, , 23. According to this diagram, the potential difference across the terminals is (internal resistance of, cell = r), E,r, i, + –, (1) V = E – i r, , (2) V = E + i r, , (3) V = E, , (4) Zero, , Sol. Answer (2), VA  E  ir  VB, , E, , r, , Ai, , B, , VA – VB = E + ir, 24. A current of 2 A flows in a system of conductors shown in figure. The potential difference VA – VB will be, , A, 2, , 2A, , 3, , D, , C, 2, , 3, B, (1) +2 volt, , (2) –1 volt, , Sol. Answer (3), VA + 2 – 3 = VB, VA – VB = 1, , 2, , (3) +1 volt, , A, 3, , 2A, 1A, 3, , (4) –2 volt, , 2, B, , 25. Reading of the ideal voltmeter in the circuit below is, , V, , 12 V, , 6V, , 2, , (1) 10 V, , (2) 8 V, , 4, , (3) 6 V, , (4) Zero, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 87, , Sol. Answer (1), , 12 V, , V, , 2, i, , 6V, 4, , E1r2  E2 r1 48  12 60, , ,  10 V, r1  r2, 6, 6, , Aliter V =, , 6,  1A, 6, , V  6  1(4), , V = 10 V, 26. Potential of the point B in the circuit below is, 2, , B, , 1, , 3, , 12V, 2, 1, , (1) 5 V, , (2) 6 V, , (3) 7 V, 2, , Sol. Answer (3), , i, , 9, 1A, 9, , 3V, , | Vp = 0 Volt, , VA – 3 – 3 – 1 = VB, , (4) 8 V, B, 3, , 1, 12 V, , VA – VB = 7, , 2, P, , 1, , 3V, , 27. Reading of the ammeter in the circuit below is, 6, 6, 24 V, , A, , 6, , 6, , (1) 16 A, , (2) 3 A, , (3) 4 A, , (4) 12 A, , 6, , Sol. Answer (2), Rnet = 8 , , 6, , 24, 8, i=3A, i, , A, , 6, , 24 V, 6, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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88, , Current Electricity, , Solution of Assignment, , 28. The ammeter reading in the circuit below is, 2, 1, , 11 A, (1) 2 A, , A, 3, , (2) 3 A, , 11 A, , (3) 6 A, , (4) 5 A, , Sol. Answer (3), , i, 1.2, , 11  i, 1, i = 13.2 – 1.2i, i, , 2, 11 A, , 1 1, , 13.2 132, , 2.2, 22, , A, , 11 A, , 3, , i=6A, 29. Thousand cells of same emf E and same internal resistance r are connected in series in same order without an, external resistance. The potential drop across 399 cells is found to be, (1) Zero, , (2) 399 E, , (3) 601 E, , (4) 1000 E, , Sol. Answer (1), Current through the circuit i =, , 1000E, E, =, 1000r, r, , ⎛E⎞, Potential drop across one cell = E – ir = E  ⎜⎝ ⎟⎠ r = 0, r,  For 399 cells, total potential drop is zero, 30. Five cells each of e.m.f. E and internal resistance r are connected in series. If due to over sight, one cell is, connected wrongly, then the equivalent e.m.f. of the combination is, (1) 5 E, , (2) 2 E, , (3) 3 E, , (4) 4 E, , Sol. Answer (3), Net Emf = E + E + E + E – E, = 3E, 31. Two batteries of different e.m.f.'s and internal resistance connected in series with each other and with an external, load resistor. The current is 3.0 A. When the polarity of one battery is reversed, the current becomes 1.0 A., The ratio of the e.m.f.'s of the two batteries is, (1) 2.5, , (2) 2, , (3) 1.5, , (4) 1, , Sol. Answer (2), , 3, , E1  E2, R  r1  r2, , 1, , E1  E2, R  r1  r2, , 3, , E1  E2, E1  E2, , 2, , E1, E2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 89, , 32. In figure, the e.m.f. of the cell is 2 V and internal resistance is negligible. The resistance of the voltmeter is 80, ohm. The reading of the voltmeter will be, 2V, + –, V, 80 , 20 , , (1) 2 volt, , (2) 1.33 volt, , Sol. Answer (2), 2, 1, , A, 60 30, , i, , 1, A, 60, , (3) 1.60 volt, , (4) 0.80 volt, , 2V, , Rnet = 60 , i, , 80 , , 80 , V, , 20 , , 80 4, V ,  V, 60 3, , 80 , , 33. Calculate the current shown by the ammeter A in the circuit diagram, , 5, 10, , , 1, , 0, , 10, , , , 10, , A, , 5, , +, –, , 0.4 V, (1) 0.1 A, , (2) 0.2 A, , Sol. Answer (2), , (3) 0.3 A, , (4) 0.4 A, , 5, , Rnet = 2 , , , 10, , 10, , , 0.4 = i(2), i = 0.2 A, , 10, , , , 10, , 5, , A, , 0.4 V, 34. In the circuit shown, R 1A B, A, , 12 V, 4, , 2, , (1) R = 8 ohms, , (2) R = 6 ohms, , (3) R = 10 ohms, , (4) Potential difference between A and B is 2 V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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90, , Current Electricity, , Solution of Assignment, , Sol. Answer (2), , 1A, , R, , 0 – 2 + 12 – R – 4 = 0, , 12 V, , 6–R=0, , 2, , 4, , R=6, , 35. Five identical lamps each of resistance R = 1100  are connected to 220 V as shown in figure. The reading, of ideal ammeter A is, , R, , 220 V, , R, , R, , R, , R, , A, , (1), , 1, A, 5, , (2), , 2, A, 5, , (3) 3 A, 5, , (4) 1 A, , Sol. Answer (3), , Rnet =, i, , 1100,  220 , 5, , 220 V, , 220,  1A, 220, , 1, 5, , 1, 5, , 1, 5, , A, , 3, iA  A, 5, , 1, 5, , 1, 5, , 3, 5, , 36. In the circuit shown, R1 is increased. What happens to the reading of the voltmeter (ideal)?, V, R1, , R2, R3, V, , (1) Increases, , (2) Decreases, , (3) First increases then decreases, , (4) Does not change, , Sol. Answer (4), , V, R1, , R2, R3, , V, Voltmeter reading is always V, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 91, , 37. In the meter bridge shown, the resistance X has a negative temperature coefficient of resistance. Neglecting, the variation in other resistors, when current is passed for some time, in the cirucit, balance point should shift, towards., R, , X, , G, A, , (1) A, , B, , C, , (2) B, , (3) First A then B, , (4) It will remain at C, , Sol. Answer (2), When current is passed, temperature increases, so resistance decreases thus, balance point shift towards, 38. A voltmeter is connected in parallel with a variable resistance R which is in series with an ammeter and a cell, as shown in the figure. For one value of R, the meters read 0.3 A and 0.9 V. For another value of R the, readings are 0.25 A and 1.0 V. What is the internal resistance of the cell?, , + –, R, V, (1) 0.5 , , (2) 2 , , (3) 1.2 , , (4) 1 , , Sol. Answer (2), –, , –, , +, , 0.9 = E – 0.3r, 1 = E – 0.25r, 0.1 = 0.05r, , r , , A, , E, r, R, , 10,  2r, 5, , V, , 39. A galvanometer of resistance 100  gives full scale deflection at 10 mA current. What should be the value of, shunt so that it can measure a current of 100 mA?, (1) 11.11 , , (2) 1.1 , , (3) 9.9 , , (4) 4.4 , , Sol. Answer (1), , S, , 10  103  100, (100  10)  10, , 3, , , , 1000,  11.11 , 90, , 40. Two cells of e.m.f. E1 and E2 are joined in series and the balancing length of the potentiometer wire is 625, cm. If the terminals of E 1 are reversed, the balancing length obtained is 125 cm. Given, E2 > E1, the ratio E1 : E2 will be, (1) 2 : 3, , (2) 5 : 1, , (3) 3 : 2, , (4) 1 : 5, , Sol. Answer (1), E1  E2 625, , 5, E1  E2 125, E1 3, , E2 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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92, , Current Electricity, , Solution of Assignment, , 41. A 10 m long potentiometer wire is connected to a battery having a steady voltage. A Leclanche cell is balanced, at 4 m length of the wire. If the length is kept the same, but its cross-section is doubled, the null point will, be obtained at, (1) 8 m, , (2) 4 m, , (3) 2 m, , (4) None of these, , Sol. Answer (2), If length is kept same, potential gradient remains same, so null point does not change., 42. Of the two bulbs in a house hold circuit, one glows brighter than the other, Which of the two bulbs has a large, resistance?, (1) The bright bulb, (2) The dim bulb, (3) Both have the same resistance, (4) The brightness does not depend upon the resistance, Sol. Answer (2), , 1, P, Thus, bulb which glows dimmer has larger resistance., R, , 43. Two electric bulbs whose resistance are in the ratio of 1 : 2, are connected in parallel to a constant voltage, source. The power dissipated in them has the ratio, (1) 2 : 1, , (2) 1 : 1, , (3) 1 : 4, , (4) 1 : 2, , Sol. Answer (1), P, , V2, R, , P, , 1, R, , P1 R2 2, , , P2 R1 1, , 44. The same mass of aluminium is drawn into two wires 1 mm and 2 mm thick. Two wires are connected in series, and current is passed through them. Heat produced in the wire is in the ratio, (1) 16 : 1, , (2) 8 : 32, , (3) 8 : 2, , (4) 4 : 2, , Sol. Answer (1), , 4l, , l, , A, , 4A, , H1 : H2 = R1 : R2 =, , 4l l, :, A 4A, , = 16 : 1, , 45. How many 60 W bulbs may be safely run on 220 V using a 5 A fuse?, (1) 18, , (2) 16, , (3) 14, , (4) 12, , Sol. Answer (1), P = (220) (5) = 1100, nP1 = 1100, n60= 1100, n = 18.3, n = 18 bulbs, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 93, , 46. Three identical bulbs B1, B2 and B3 are connected to the mains as shown in figure. If B3 is disconnected from, the circuit by opening switch S, then incandescence of bulb B1 will, , B1, , S, B2, , B3, , V, , (1) Increase, , (2) Decrease, , (3) Become zero, , (4) No change, , Sol. Answer (2), As net resistance increases, so current decreases, so bulb B1 gets dimer., 47. A 50 W bulb connected in series with a heater coil is put to an AC mains. Now the bulb is replaced by a, 100 W bulb. The heater output will, (1) Double, , (2) Halve, , Sol. Answer (3), , (3) Increase, , R, , R, , (4) Decrease, , R/2, , R, , 2, , ⎛ E ⎞, P1  ⎜, ⎟ R', ⎝R R'⎠, , P2 , , ~, , ER ', ⎛R, ⎞, ⎜  R '⎟, ⎝2, ⎠, , ~, , 2, , P2 > P1, 48. In a circuit shown in figure, the heat produced in 3 ohm resistor due to a current flowing in it is 12 J. The heat, produced in 4 ohm resistor is, 2, , 4, , 3, , (1) 2 J, , (2) 4 J, , Sol. Answer (2), 12 = i12(3), i1 = 2, 2 6, , i2 3, , (3) 64 J, , , , (4) 32 J, , 4, , i2, , 3, , i1, , i2 = 1, H = 124 = 4 J, 49. Refer to the circuit shown. What will be the total power dissipation in the circuit if P is the power dissipated, in R1? It is given that R2 = 4 R1 and R3 = 12 R1, , I, R2, , R1, , R3, (1) 4P, , (2) 7P, , (3) 13P, , (4) 17P, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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94, , Current Electricity, , Solution of Assignment, , Sol. Answer (1), E, R3  12R1, 16R, Rnet = 4R1, , i, , I, , E, 4R1, , P, , E 2R1, , 16R12, , 3E, 16R, , E2, , 16R1, , E, 16R, , 2, , P2 , , 9E, 9P, .4R1 , 2, 4, 256R1, , P3 , , E2, 3E 2 3P, .12, R, , , 1, 64R1, 4, 256R12, , P, , E, R1, R2 = 4R1, R3 = 12R1, , 9P 3P 16P, , , 4, 4, 4, = 4P, , 50. Three identical resistors R1 = R2 = R3 are connected as shown to a battery of constant e.m.f. The power dissipated, is, R2, , R1, , (1) The least in R1, , R3, , (2) Greatest in R1, (3) In the ratio 1:2 in resistance R1 and R2 respectively, (4) The same in R1 and in the parallel combination of R2 and R3, , V, , Sol. Answer (2), Power dissipated is maximum in R1., , 51. Four equal resistance dissipated 5 W of power together when connected in series to a battery of negligible, internal resistance. The total power dissipated in these resistance when connected in parallel across the same, battery would be, (1) 125 W, , (2) 80 W, , (3) 20 W, , (4) 5 W, , Sol. Answer (2), P, 5, 4, P = 20, , 4(20) = 80 W, Net power in parallel, 52. Two heater coils separately take 10 minute and 5 minute to boil certain amount of water. If both the coils are, connected in series, the time taken will be, (1) 15 min, , (2) 7.5 min, , (3) 3.33 min, , (4) 2.5 min, , Sol. Answer (1), t = t1 + t2, t = 10 + 5, t = 15 minute, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 95, , 53. A cell sends a current through a resistance R for time t. Now the same cell sends current through another, resistance r for the same time. If same amount of heat is developed in both the resistance, then the internal, resistance of cell is, (1), , (R  r ), 2, , (2), , (R  r ), 2, , (3), , (Rr ), 2, , (4), , (Rr ), , Sol. Answer (4), H1 , , E 2R, (R  r1 )2, , H2 , , E 2R, (r  r1)2, , R, ,  R  r1 , , 2, , , , r, ,  r  r1 2, , r  r1, r, , R  r1, R, , Rr1  r R  r r1  R r, r1  R  r   Rr, , , , R r, , r1  Rr, , SECTION - B, Objective Type Questions, 1., , The charge in the 2 F capacitor at steady state is, , 1V, , 2, , 1.5 V, , 2 F, , 2V, (1) Zero, , 2, , (2) 2 C, , (3) 4 C, , (4) 6 C, , Sol. Answer (1), i, , 1V, , 1,  0.25 A, 4, , VB  0.5  1  1.5  Vc, , VB = VC, V = 0, q=0, , A, , 2, , C, , 1.5 V, 2V, , 2 F, , B, , 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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96, 2., , Current Electricity, , Solution of Assignment, , In the following diagram, the lengths of wires AB and BC are equal, but the radius of wire AB is double that, of BC. The ratio of potential gradient on wires AB and on BC will be (wires are made of same material), E, A, , (1) 4 : 1, , C, , B, , (2) 1 : 4, , (3) 2 : 1, , (4) 1 : 1, , Sol. Answer (2), =, , R, 4, , i, , 4E, 5R, , V1 , , E, 5, , 4E, V2 , 5, , l, A, , l, B, , 2r, , r, R, , C, , V1 = V2, =1:4, 3., , In the circuit shown, the thermal power dissipated in R1 is P. The thermal power dissipated in R2 is, , R2= R, i, , i, R1= R, R3= 2R, , (1) P, , (2), , 4P, 9, , (3), , i2R, , 4i 2, 4P, P2 , R, 9, 9, , (4), , P, 9, , R2 = R, , Sol. Answer (2), P=, , 2P, 3, , 2i, j, R1 = R, 1 R = 2R, 3, 3, , 4., , Consider the combination of resistors as shown in figure and pick out the correct statement, R2, R1, R3, , A, , R6, R4, , B, , (1) R1 & R4 are connected in parallel, , R5, (2) R1 & R2 are connected in series, , (3) R2 & R3 are connected in parallel, , (4) R6 & R4 are connected in parallel, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 97, , Sol. Answer (3), , R2, R1, R3, , A, , R6, R4, , B, R5, , R2 and R3  Parallel, 5., , Select the correct statement, (1) Electric current is a vector quantity, (2) Resistivity of a conductor decreases with increase in temperature, (3) Resistance is the opposition to the flow of current, (4) Current density is a scalar quantity, , Sol. Answer (3), Resistance is the opposition to the flow of current, 6., , In the circuit shown in figure, all cells are ideal. The current through 2  resistor is, 10, 6V, , 4V, 8V, 2, , (1) 5 A, , (2) 1 A, , (3) 0.2 A, , (4) Zero, , 10 , , Sol. Answer (1), VA – 4 – 6 = VB, VA – VB = w, , 6V, , 10 = i(2), , 4V, , 8V, , i=5A, , 2, A, , 7., , B, , The effective resistance of the network between points A & B is, 2r, A, , r, , r, , 3, , B, r, , (1) r, , r, , (2) 2r, , r, , r, , (3), , 4r, 3, , (4), , 7r, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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98, , Current Electricity, , Solution of Assignment, , Sol. Answer (2), r, , r, , r, , 2r, 3, r, , r, , 2r, 3, , r, , 2r, 3, , Rnet = 2r, 8., , The following circuit consist of a 5 F capacitor, having charge 50 C as shown. The switch is closed at, t = 0. The value of current in 2 M resistor at t = 0 is, C = 5 µF, , 2M , , q = 50  C, , S, , (1) 1 A, Sol. Answer (3), , (2) 2 A, 5 F, , (3) 5 A, , (4) 5 A, , 2 M, , 50, V ,  10V, 5, 10 = i(2 × 106), , 50 C, , i = 5 A, 9., , There are a large number of cells available, each marked (6 V, 0.5 ) to be used to supply current to a device of, resistance 0.75 , requiring 24 A current. How should the cells be arranged, so that power is transmitted to the, load using minimum number of cells?, (1) Six rows, each containing four cells, , (2) Four rows, each containing six cells, , (3) Four rows, each containing four cells, , (4) Six rows, each containing six cells, , Sol. Answer (2), E = 6V, r = 0.5 , R = 0.75 , i = 24 , S(0.5) = P(0.75), 2s = 3p, , i, , PSE, Sr  PR, , 3, P ⎛⎜ P ⎞⎟ 6, 24  ⎝ 2 ⎠, 15P, P = 4 rows, S = 6 cells, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 99, , 10. A circuit containing five resistors is connected to a battery with a 12 V emf as shown in figure. The potential, difference across 4 resistor is, 5, 4, 20, , 2, , (1) 3 V, , (2) 6 V, , 5, , 12,  3A, 4, , 6, , (3) 9 V, , Sol. Answer (2), i, , 12 V, , 1.5, , (4) 12 V, , 4, 1.5, , 20 , , V4  6V, , 1.5, 2, 3A, , 6, , 12 V, , 11. The temperature coefficient of resistance of tungsten is 4.5 × 10 –3 °C –1 and that of germanium is, –5 × 10–2 °C–1. A tungsten wire of resistance 100  is connected in series with a germanium wire of resistance R., The value of R for which the resistance of combination does not change with temperature is, (1) 9 , , (2) 1111 , , (3) 0.9 , , (4) 111.1 , , Sol. Answer (1), R11 + R22 = 0, (100) (4.5 × 10–3) = R(5 × 10–2), 0.9 × 10 = R, R = 9r, 12. Consider the ladder network shown in figure. What should be the value of resistance R, so that effective resistance, between A & B becomes independent of number of elements in the combination?, 2, , 2, , 2, , 2, , 2, , A, 8, , 8, , 8, , 8, , R, , B, 2, , (1) 2 , , 2, , (2) 4 , , 2, , 2, , (3) 8 , , 2, , (4) 16 , , Sol. Answer (2), For R = 4r, the sequence repeats itself., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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100, , Current Electricity, , Solution of Assignment, , 13. Three identical bulbs are connected as shown in figure. When switch S is closed, the power consumed in bulb B, is P. What will be the power consumed by the same bulb when switch S is opened?, , A, , S, , E, , 9P, 4, , (1), , (2), , B, , 16P, 9, , C, , 9P, 16, , (3), , (4), , 4P, 9, , Sol. Answer (1), 2, , E ⎞, P  ⎛⎜, ⎟ R, ⎝ 2R ⎠, , E2, 4R, When K is closed, P, , i, , k, , A, , E, , B, , C, , 2E, 3R, , P' , , E2, .R, 9R 2, , P' , , E2, 9R, , P 9, , P' 4, P' , , 4P, 9, , 14. In the circuit shown in figure, if ammeter and voltmeter are ideal, then the power consumed in 9  resistor will be, 6, , 9, 20 , , V, , 10 , 30 , , A, , 30 V, , (1) 3.33 W, , (2) 4 W, , (3) 1.44 W, , Sol. Answer (2), , 2/3 9 , , No current goes through ideal voltmeter, 30, i, 1A, 30, 4, P . 94W, 9, , 20 , , 10 , , 1A, A, , (4) 500 W, 6, V, , 1/3 30 , 1A, , 30 V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 101, , 15. Two identical bulbs are connected in parallel across an ideal source of emf E. The ammeter A and voltmeter V are, ideal. If bulb B2 gets fused, then, (1) Reading of A will increase but that of V will remain same, A, , (2) Reading of A will decrease but that of V will increase, , V, , (3) Reading of A will decrease but that of V will remain same, (4) Reading of A will increase and reading of V will also increase, , B1, , B2, , E, , Sol. Answer (3), , A, , V, B1, , B2, , If B2 gets fused, Rnet increases, i decreases, but reading of V remains same., 16. In the network shown in figure, power dissipated in 3  is 12 W. Power dissipated in 4  will be, 2, , 4, 3, , (1) 4 W, , (2) 2 W, , Sol. Answer (1), , (3) 64 W, 1, , 12 = i2 (3), , 2, , (4) 32 W, , 4, , 2, , i=2, P' = 12(4) = 4 W, , 3, , 17. Effective resistance across AB in the network shown in, 2, 4, , A, 2, , 4, B, , (1) 6 , , (2) 3 , , (3) 5 , , (4) 8 , , Sol. Answer (2), 2, , Rnet = 3 , , 4, , A, 2, , 4, B, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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102, , Current Electricity, , Solution of Assignment, , 18. Potential difference across AB in the network shown is, E, , A, , r, E, , r, , r, , E, r, , (1) Zero, , B, , E, , (3) E , , (2) E, , Sol. Answer (1), , A, , (4) E – 2Ir, , r, , E, , E, R, VA – E+ ir – E + ir = VB, i, , Ir, 2, , E, , r, , r, , E, , VA – E + E – E + E = VB, , E, , r, , VA = VB, , B, , 19. Current through 10  resistor shown in figure is, , 5V, , 1, , 10 V, , 2, , 10 , (1) Zero, , (2) 1 A, , (3) 1.5 A, , Sol. Answer (1), , Enet, , 5 10, , = 1 2, 1, 1, 2, , 5V, , 1, , 10 V, , 2, , (4) 2 A, , 10 , , Enet = 0, i=0, , 20. Three identical cells are connected in parallel across AB. Net emf across AB is, , A, (1) 10 V, , (2) 30 V, , 10 V, , 3, , 10 V, , 3, , 10 V, , 3, B, (3) 15 V, , (4) 12 V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 103, , Sol. Answer (1), , Enet, , 10 V, , 3, , 10 V, , 3, , 10 V, , 3, , 10 10 10, , , 3, 3  10 volt,  3, 1 1 1,  , 3 3 3, , 21. When current supplied by a cell to a circuit is 0.3 A, its terminal potential difference is 0.9 V. When the current, supplied becomes 0.25 A, its terminal potential difference becomes 1.0 V. The internal resistance of the cell, is, (1) 0.5 , , (2) 2 , , (3) 1.2 , , (4) 1 , , Sol. Answer (2), 0.9 = E – 0.3 r, 1 = E – 0.25 r, 0.1 = 0.05 r, r=2r, 22. Coefficient of linear expansion of material of resistor is . Its temperature coefficient of resistivity and resistance, are  and R respectively, then correct relation is, (1) R =  – , , (2) R =  + , , (3) R =  + 3, , (4) R =  – 3, , Sol. Answer (1), , R, , l, A, , R P l DA, , , , R, P, l, A, , R = P + R – 2, R = P – , 23. A current of 10 A is maintained in a conductor of cross-section 1 cm2. If the free electron density in the, conductor is 9 × 1028 m–3, then drift velocity of free electrons is, (1) 6.94 × 10–6 m/s, , (2) 5.94 × 10–2 m/s, , (3) 1.94 × 10–3 m/s, , (4) 2.94 × 10–4 m/s, , Sol. Answer (1), 10 = 9 × 1028 × 1.6 × 10–19 × 10–4 × VD, Solving, we get, V0 = 6.94 × 10–6 m/s, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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104, , Current Electricity, , Solution of Assignment, , 24. Current I versus time t graph through a conductor is shown in the figure. Average current through the conductor, in the interval 0 to 15 s is, I, 10 A, , 5s, , (1) 1 A, , (2) 10 A, , t, , 15 s, , (3) 7.5 A, , (4) 5 A, , Sol. Answer (4), q = Area (I / t), q , , 1, .10  15  75 C, 2, , iavg , , 75, 5A, 15, , 25. Ten 60 W, 220 V bulbs are connected in series to 220 V supply. Power consumed in the circuit is, (1) 6 W, , (2) 12 W, , (3) 180 W, , (4) 600 W, , Sol. Answer (1), P, , V2, R, , R, , V 2  220 , , P, 60, , PTotal , , 2, , V2,  2202  60, =6W, Req = 10 220 2,  , , 26. Potential difference VA – VB in the network shown is, , A, 2, , 3, , 2A, , 2A, 3, , 2, B, , (1) 1 V, , (2) –1 V, , (3) 2 V, , A, , Sol. Answer (1), VA + 2 – 3 = VB, , 3, , 2, , VA – 1 = VB, VA – VB = 1, , (4) –2 V, , 2A, , 2A, , 1, 1, 2, , 3, B, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 27. Potential difference across AB i.e., VA – VB is, , 105, , A, , 12 V, , 6V, , 2, , 4, B, , (1) 10 V, , (2) 8 V, , (3) 6 V, , (4) Zero, A, , Sol. Answer (1), 6 = 6i, i=1A, , 12 V, , V = 6 + 4, , 2, , V = 10, , 6V, 4, , B, , 28. Potential difference VB – VA in the network shown is, , B, , 2, , 1, , 3, , 12 V, 2, , 1, , 3V, , A, , (1) 7 V, , (2) 6 V, , (3) 5 V, , (4) 8 V, , Sol. Answer (1), Rnet = 9, V=9V, i=1A, 29. Five cells each of emf E and internal resistance r are connected in series. Due to oversight one cell is, connected wrongly. The equivalent internal resistance of the combination is, (1) 3r, , (2) 2r, , (3) 5r, , (4) 4r, , Sol. Answer (3), All the internal resistances will be still in series, there will be no impact of polarity on the equivalent resistance., 30. Current I in the network shown in figure is, , 6, 6, I, , 6, , 24 V, 6, , (1) 16 A, , (2) 3 A, , (3) 4 A, , (4) 12 A, , Sol. Answer (2), Rnet = 8 , V = 24 V, 24 = i(8), i=3A, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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106, , Current Electricity, , Solution of Assignment, , 31. Value of the resistance R in the figure is, R, , 1A, 12 V, , 4, , (1) 6 , , 2, , (2) 8 , , (3) 10 , , (4) 12 , , Sol. Answer (1), 0 – 2(1) + 12 – 1(R) – 4(1) = 0, 6–R=0, R=6, 32. Current through the 25  resistor as shown in figure is, 5, 25 , , 10 , A, 20 , , B, 10 , , 10 V, , (1) 1 A, , (2) 2 A, , (3) 2.5 A, , (4) Zero, , Sol. Answer (4), (15)(30),  10 , 45, Current through 25  resistance is zero in balanced wheatstone Bridge., Rnet , , 33. Resistance across AB as shown in figure is, , 2, , A, , (1) 2 , , (2) 4 , , Sol. Answer (1), C, , 4, , 4, , C, 4, , 4, , 3, , 4, , 4, , B, , 4, , 2, , (3) 6 , , (4) 12 , , 2, B, , 4, , 3, , 2, A, , 2, , D, , 2, , 2, , 2, , 2, , 2, , B, , 2  net resistance, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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108, 3., , Current Electricity, , Solution of Assignment, , A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the, ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be, [Re-AIPMT-2015], (1) 1 A, , (2) 0.5 A, , (3) 0.25 A, , Sol. Answer (2), , (4) 2 A, , 30 V, , 30 V, Reading of Ammeter I =  40.8    480 || 20 , , I, , 30 V, =  40.8  19.2  , , 480 , 20 , , 40.8 , , A, , = 0.5 A, 4., , A, B and C are voltmeters of resistance R, 1.5R and 3R respectively as shown in the figure. When some, potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively, then, [AIPMT-2015], , X, (1) V A  VB  VC, , B, , A, , Y, , C, , (2) V A  VB  VC, , (3) V A  VB  VC, , (4) V A  VB  VC, , Sol. Answer (2), R ZY , , 1.5R  3R, R, 1.5  3  R, , 1.5R, X, , R, , R XZ  R ZY  R, , Z, , Y, 3R, ,  V XZ  VZY,  V A  VB  VC, 5., , A potentiometer wire has length 4 m and resistance 8 . The resistance that must be connected in series, with the wire and an accumulator of e.m.f. 2 V, so as to get a potential gradient 1 mV per cm on the wire is, [AIPMT-2015], (1) 48 , , (2) 32 , , (3) 40 , , (4) 44 , , Sol. Answer (2), P.D. across the wire = 1mV/cm × 400 cm, V0 = 0.4 V, Current in the wire I , , R, , 0.4, 1, A, A, 8, 20, , V  V0 2  0.4, ,  32 , I, ⎛ 1 ⎞, ⎜ 20 ⎟, ⎝, ⎠, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 6., , Current Electricity, , 109, , A resistance R draws power P when connected to an AC source. If an inductance is now placed in series, with the resistance, such that the impedance of the circuit becomes Z, the power drawn will be, [AIPMT-2015], , ⎛R⎞, (2) P ⎜ ⎟, ⎝Z⎠, , (1) P, , 2, , R, Z, , (3) P, , ⎛R ⎞, (4) P ⎜ ⎟, ⎝Z⎠, , Sol. Answer (2), In case of only R, Power (P ) , , 2, Vrms, R, , ...(1), , When inductance is connected in series with resistance., P '  Vrms irms cos , ⎛V ⎞⎛ R ⎞,  Vrms ⎜ rms ⎟ ⎜ ⎟, ⎝ Z ⎠⎝ Z ⎠, , , 7., , 2, Vrms, R, Z2, , P , , (PR ), 2, R(Vrms,  PR ), Z2, , P' , , PR 2, Z2, , Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall, of potential per km is 8 volt and the average resistance per km is 0.5 . The power loss in the wire is, [AIPMT-2014], (1) 19.2 W, , (2) 19.2 kW, , (3) 19.2 J, , (4) 12.2 kW, , Sol. Answer (2), ⎛  82 ⎞, 640,  19.2 kW, Total power loss = 150 ⎜, ⎟ = 150 , 5, ⎝ 0.5 ⎠, , 8., , The resistances in the two arms of the meter bridge are 5  and R , respectively. When the resistance R is, shunted with an equal resistance, the new balance point is at 1.6 l1. The resistance R, is, [AIPMT-2014], , 5, , R, , G, A, (1) 10 , , (2) 15 , , l1, , 100 – l1, , B, , (3) 20 , , (4) 25 , , Sol. Answer (2), When R is not shunted, l1, 5, , R 100  l1, , …(i), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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110, , Current Electricity, , Solution of Assignment, , When R is shunted with R (i.e., a resistance R is connected in parallal), 1.6l1, 5, , R, 100  1.6l1, 2, ,  , , …(ii), , Solve (i) and (ii) for R, 9., , A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used, across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire, itself is 4 m long. When the resistance, R, connected across the given cell, has values of (i) Infinity, (ii) 9.5 ,, the 'balancing lengths', on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of, internal resistance of the cell is, [AIPMT-2014], (1) 0.25 , , (2) 0.95 , , (3) 0.5 , , (4) 0.75 , , Sol. Answer (3), Potential gradient  k  , , V 2, V,   0.5, l, 4, m, , When R  , No current will flow through R., , v, , So at the balance point, , 4m, , (0.5) (3) = E,  E = 1.5 volt, , …(i), , A, , When R = 9.5 , , B, , Using KVL in loop ABCD, E, Rr, and E – ir = (0.5) (2.85), i, , E, , r, R, , G, D, C, , …(ii), …(iii), , From (i) E = 1.5 volt and R = 9.5 , Solving (ii) and (iii) r = 0.5 , 10. In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the, resistance of ammeter will be, [AIPMT-2014], (1), , 1, G, 499, , (2), , 499, G, 500, , (3), , 1, G, 500, , (4), , 500, G, 499, , Sol. Answer (3), 11. A wire of resistance 4  is stretched to twice its original length. The resistance of stretched wire would be, [NEET-2013], (1) 4 , , (2) 8 , , (3) 16 , , (4) 2 , , Sol. Answer (3), 4, , l, A, , R, , (2l ),  A / 2, , 4, 4, R, R=1, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 111, , 12. The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10  is, [NEET-2013], (1) 0.5 , , (2) 0.8 , , (3) 1.0 , , (4) 0.2 , , Sol. Answer (1), 2.1 = 0.2(10 + r), 10.5 = 10 + r, r = 0.5, 13. The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and 90, ohm, respectively. The e.m.f and internal resistance of the cell are 7 volt and 5 ohm respectively. If the, galvanometer resistance is 50 ohm, the current drawn from the cell will be:, [NEET-2013], (1) 0.2 A, , (2) 0.1 A, , (3) 2.0 A, , (4) 1.0 A, , Sol. Answer (1), , 30, , 10, G, 30, , 90, , 5, , 7V, i, , 7, A, 37, , 14. In the circuit shown the cells A and B have negligible resistances. For VA = 12 V, R1 = 500  and R = 100 , [AIPMT (Prelims)-2012], the galvanometer (G) shows no deflection. The value of VB is, R1, VA, , (1) 12 V, , (2) 6 V, , G, R, , VB, , (3) 4 V, , (4) 2 V, , Sol. Answer (4), , 500 , , 100 , , 12 V, , i, , G, , VB, , 12, 1, , A, 600 50, , VB , , 1, .100  2 V, 50, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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112, , Current Electricity, , Solution of Assignment, , 15. A ring is made of a wire having a resistance R0 = 12  . Find the points A and B, as shown in the figure, at, which a current carrying conductor should be connected so that the resistance R of the sub circuit between, 8, [AIPMT (Prelims)-2012], these points is equal to ., 3, l1, B, , A, , l2, , l1 3, (1) l  8, 2, , l1 1, (2) l  2, 2, , l1 5, (3) l  8, 2, , Sol. Answer (2), l1  l 2 l1  l 2 3, , , 12l1, 12l 2, 8, ,  l1  l2 2, 12l1l 2, , , , l1, , 12l, l1  l 2, , l2, , 12l 2, l1  l 2, , l1 1, (4) l  3, 2, , 3, 8, , 2(l12  l22  2l1 l2 )  9, , 2l12  l 22  5l1l 2  0, 2l12  4l1l 2  l1l 2  2l 2  0, 2l1  l1  2l 2   l 2 (l1  2l 2 )  0, l1 1, , l2 2, , 16. A millivoltmeter of 25 millivolt range is to be converted into an ammeter of 25 ampere range. The value (in ohm), of necessary shunt will be, [AIPMT (Prelims)-2012], (1) 1, , (2) 0.05, , (3) 0.001, , (4) 0.01, , Sol. Answer (3), 25,  25R, 1000, , R = 0.001 , 17. If voltage across a bulb rated 220 V – 100 W drops by 2.5% of its rated value, the percentage of the rated value, by which the power would decrease is, [AIPMT (Prelims)-2012], (1) 5%, , (2) 10%, , (3) 20%, , (4) 2.5%, , Sol. Answer (1), , P, , V2, R, , P, V, 2, P, IV, 5% = 2.5%, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 18. The power dissipated in the circuit shown in the figure is 30 W. The value of R is, , 113, , [AIPMT (Mains)-2012], , R, 5, , 10 V, (1) 20 , , (2) 15 , , (3) 10 , , (4) 30 , , Sol. Answer (3), , R, , 100 100, ,  30, 5, R, , 5, , 100,  10, R, R = 10 , , 10 V, , 19. A cell having an emf  and internal resistance r is connected across a variable external resistance R. As the, resistance R is increased, the plot of potential difference V across R is given by, [AIPMT (Mains)-2012], , , (1), , V, , (2), , O, , , , , , V, , (3) V, , O, , R, , O, , R, , V, (4), , O, , R, , R, , Sol. Answer (3), V = E – ir, i, , E, Rr, , V , V , , ER, Rr, , V, , E, 1, , r, R, , 20. If power dissipated in the 9  resistor in the circuit shown is 36 W, the potential difference across the 2 , resistor is, [AIPMT (Prelims)-2011], 9, , 6, , V, (1), , 2V, , (2), , 4V, , 2, (3), , 8V, , (4), , 10 V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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114, , Current Electricity, , Solution of Assignment, , Sol. Answer (4), , i, , 9, , i, , 6, , V, , 2, , 36 = i29, i2 = 4, i=2, , 2 6, , i1 9, i1 = 3, V2 = (5) (2) = 10 V, , 21. A current of 2 A flows through a 2  resistor when connected across a battery. The same battery supplies a, current of 0.5 A when connected across a 9  resistor. The internal resistance of the battery is, [AIPMT (Prelims)-2011], (1) 1 , , (2) 0.5 , , (3) 1/3 , , (4) 1/4 , , Sol. Answer (3), Z, , E, 2r, , 0.5 , 4, , E, 9r, , 9r, 2r, , 8 + 4r = 9 + r, 22. The rate of increase of thermo e.m.f. with temperature at the neutral temperature of a thermocouple, [AIPMT (Prelims)-2011], (1) Is negative, (2) Is positive, (3), , Is zero, , (4) Depends upon the choice of the two materials of the thermocouple, Sol. Answer (3), 23. A galvanometer of resistance, G is shunted by a resistance S ohm. To keep the main current in the circuit, unchanged, the resistance to be put in series with the galvanometer is, [AIPMT (Mains)-2011], (1), , G2, (S  G ), , (2), , G, (S  G ), , (3), , S2, (S  G ), , (4), , SG, (S  G ), , Sol. Answer (1), E, E, , GS, G, R, G S, , G, , GS, R, GS, , R, , G2, GS, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 115, , 24. A thermocouple of negligible resistance produces an e.m.f. of 40 V/°C in the linear range of temperature. A, galvanometer of resistance 10 ohm whose sensitivity is 1 A/div, is employed with the thermocouple. The smallest, value of temperature difference that can be detected by the system will be, (1) 0.1°C, , (2) 0.25°C, , (3) 0.5°C, , [AIPMT (Mains)-2011], (4) 1°C, , Sol. Answer (2), 25. In the circuit shown in the figure, if the potential at point A is taken to be zero, the potential at point B is, [AIPMT (Mains)-2011], , R1, 1A, R2, , 1V, , (3) –1 V, , 2V, , R1, , R = 12R, R=2, , R2, , VA + 1 + 2 – 2 = VB, , 2A, , 1 AC 2 A, , (2) +1 V, , Sol. Answer (2), , VA + 1 = VB, , B, 2, , A, (1) –2 V, , 2V, , D, , R, , 2W, , (4) +2 V, , B, 2A, , 1A, A, , VA – VB = –1, , 1A 2A, , 26. Consider the following two statements :, (A) Kirchhoff's junction law follows from the conservation of charge., (B) Kirchhoff's loop law follows from the conservation of energy., Which of the following is correct?, , [AIPMT (Prelims)-2010], , (1) Both (A) and (B) are correct, , (2) Both (A) and (B) are wrong, , (3) (A) is correct and (B) is wrong, , (4) (A) is wrong and (B) is correct, , Sol. Answer (1), Both A and B are correct based on theory, 27. A galvanometer has a coil of resistance 100 and gives a full scale deflection for 30 mA current. If it is to work, as a voltmeter of 30 V range, the resistance required to be added will be, (1) 1000 , , (2) 900 , , (3) 1800 , , [AIPMT (Prelims)-2010], (4) 500 , , Sol. Answer (2), , M, , 30,  1000  100 = 900 , 30, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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116, , Current Electricity, , Solution of Assignment, , 28. A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k volt/cm, and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points,, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths l1 cm and, l2 cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal, respectively, to, [AIPMT (Prelims)-2010], , +, , –, , (), , A, , 2, –, A, +, (1) k(l2 – l1) and kl2, , B, , 1, , R, , G, 3, X, , (), , (2) kl1 and k(l2 – l1), , (3) k(l2 – l1) and kl1, , (4) kl1 and kl2, , Sol. Answer (2), 1R = kl1, 1(R + x) = kl2, x = kl2 – kl1, R = kl1, 29. In producing chlorine by electrolysis 100 kW power at 125 V is being consumed. How much chlorine per minute, is liberated? (ECE of chlorine is 0.367 × 10–6 kg/C), [AIPMT (Prelims)-2010], (1) 1.76 × 10–3 kg, , (2) 9.67 × 10–3 kg, , (3) 17.61 × 10–3 kg, , (4) 3.67 × 10–3 kg, , Sol. Answer (3), 30. The thermo e.m.f. E in volts of a certain thermo-couple is found to vary with temperature difference  in °C, between the two junctions according to the relation, E = 30 –, couple will be, (1) 450°C, , (2) 400°C, , 2, . The neutral temperature for the thermo15, [AIPMT (Mains)-2010], , (3) 225°C, , (4) 30°C, , Sol. Answer (3), 31. See the electric circuit shown in this figure. Which of the following equations is a correct equation for it?, [AIPMT (Prelims)-2009], , R, 1, , i1, , r1, , i2, r2, , 2, , (1) 2 – i2 r2 – 1 – i1r1 = 0, , (2) –2 – (i1 + i2)R + i2r2 = 0, , (3) 1 – (i1 + i2)R + i1r1 = 0, , (4) 1 – (i1 + i2)R – i1r1 = 0, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 117, , Sol. Answer (4), , R, i1 E1, , r1, E2, , r2, , i2, , –(i1 + i2)R – i1r1 + E1 = 0, 32. A wire of resistance 12 ohms per metre is bent to form a complete circle of radius 10 cm. The resistance, between its two diametrically opposite points. A and B as shown in the figure, is, [AIPMT (Prelims)-2009], , A, , (1) 3, , (2) 6, , B, , (3) 6, , (4) 0.6, , Sol. Answer (4), , 12, , 12, Rnet , , , 10, , , 10, , 6,  0.6 , 10, , 33. A galvanometer having a coil resistance of 60  shows full scale deflection when a current of 1 A passes through, it. It can be converted into an ammeter to read currents upto 5 A by, [AIPMT (Prelims)-2009], (1) Putting in series a resistance of 15 , , (2) Putting in series a resistance of 240 , , (3) Putting in parallel a resistance of 15 , , (4) Putting in parallel a resistance of 240 , , Sol. Answer (3), 34. A student measures the terminal potential difference (V) of a cell (of emf  and internal resistance r) as a function, of the current (I) flowing through it. The slope and intercept, of the graph between V and I, then, respectively,, equal, [AIPMT (Prelims)-2009], (1) – r and , , (2) r and –, , (3) – and r, , (4)  and – r, , Sol. Answer (1), , , VE, i, , tan  = r, slope = –r, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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118, , Current Electricity, , Solution of Assignment, , 35. A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being, short circuited through a resistance of 10 . Its internal resistance is, [AIPMT (Prelims)-2008], (2) 1 , , (1) Zero, , (3) 0.5 , , (4) 2 , , Sol. Answer (2), Let the potential gradient of the potentiometer is k, E = k(110), , …(i), , iR = k(100), i, , E, E, , R  r 10  r, , ⎛ E ⎞, ⎜⎝, ⎟ 10  k 100, 10  r ⎠, , …(ii), , divide (i) and (ii), r=1, 36. A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance, become respectively, [AIPMT (Prelims)-2008], (1) 1.1 times, 1.1 times, , (2) 1.2 times, 1.1 times, , (3) 1.21 times, same, , (4) Both remain the same, , Sol. Answer (3), Al  A ', , 1 1l, 10, , A' , , 10 A, 11, , R, , l, A, , R' , , 11l, 100 A, , R 100, , R ' 121, , R ' = 1.21R, 37. In the circuit shown, the current through the 4  resistor is 1 A when the points P and M are connected to a, d.c. voltage source. The potential difference between the points M and N is, [AIPMT (Prelims)-2008], 4, , P, , 3, , M, , 0.5 , N, , 1, , 0.5 , , (1) 3.2 V, , (2) 1.5 V, , (3) 1.0 V, , (4) 0.5 V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 119, , Sol. Answer (1), , 1A, , 4A, 3, , P, , 4, , 3, , M, , 0.5 , i, , 0.5 , , N 1, , 7, 2, , 3i 12, , i=2, V = 2 V, 38. An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature, 20°C? The temperature of boiling water is 100°C., [AIPMT (Prelims)-2008], (1) 4.2 min, , (2) 6.3 min, , (3) 8.4 min, , (4) 12.6 min, , Sol. Answer (2), t, , 1 4200  80, 880  60, , t = 6.3 minute, 39. A current of 3 A. flows through the 2  resistor shown in the circuit. The power dissipated in the 5  resistor, is, [AIPMT (Prelims)-2008], 2, 4, , 1, , (1) 5 watt, , (2) 4 watt, , 5, , (3) 2 watt, , (4) 1 watt, , Sol. Answer (1), , 3A, , 2, , i1=1.5 4 , i2=1, , 1, , 5, , 3 4, , i1 2, , i1 = 1.5, , 3 6, , i2 2, , i2 = 1, , P = (12) (5) = 5 W, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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120, , Current Electricity, , Solution of Assignment, , 40. A galvanometer of resistance 50  is connected to a battery of 3 V along with a resistance of 2950  in series., A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20, divisions, the resistance in series should be, [AIPMT (Prelims)-2008], (1) 4450 , , (2) 5050 , , (3) 5550 , , (4) 6050 , , Sol. Answer (1), i, , 3, A, 3000, , 1 division =, Required , , 1, A, 30000, 20, 2, A =, A, 30000, 3000, , 2, 3, , 3000 3000  R, 6000 + 2R = 9000, R = 1500 , 41. The total power dissipated in watt in the circuit shown here is, , [AIPMT (Prelims)-2007], , 6, 3, 4, , (1) 4, , (2) 16, , 18 V, , (3) 40, , (4) 54, , Sol. Answer (4), 6, 3, 4, , 18 V, , 18 ,  54W, P , 6, 42. A steady current of 1.5 A flows through a copper voltameter for 10 minutes. If the electrochemical equivalent of, copper is 30 × 10–5 g coulomb–1, the mass of copper deposited on the electrode will be, [AIPMT (Prelims)-2007], (1) 0.27 g, , (2) 0.40 g, , (3) 0.50 g, , (4) 0.67 g, , Sol. Answer (1), 43. If the cold junction of a thermo-couple is kept at 0°C and the hot junction is kept at T°C then the relation between, [AIPMT (Prelims)-2007], neutral temperature (Tn) and temperature of inversion (Ti) is, (1) Tn = Ti + T, , (2) Tn = Ti + T/2, , (3) Tn = 2Ti, , (4) Tn = Ti – T, , Sol. Answer (2), Ti = 2Tn, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 121, , 44. Three resistances P, Q, R each of 2  and an unknown resistance S form the four arms of a Wheatstone bridge, circuit. When a resistance of 6  is connected in parallel to S the bridge gets balanced. What is the value of, S?, [AIPMT (Prelims)-2007], (1) 1 , Sol. Answer (3), 2 2(6  S), , 2, 6S, , (2) 2 , , (3) 3 , , 2, , 2, , 2, , S, , 3S = 6 + S, 2S = 6, , (4) 6 , , 6, , S=3, 45. The resistance of an ammeter is 13  and its scale is graduated for a current upto 100 A. After an additional, shunt has been connected to this ammeter it becomes possible to measure currents upto 750 A by this meter., The value of shunt-resistance is, [AIPMT (Prelims)-2007], (1) 2 k, , (2) 20 , , (3) 2 , , (4) 0.2 , , Sol. Answer (3), 46. In producing chlorine through electrolysis 100 W power at 125 V is being consumed. How much chlorine per, min is liberated? ECE of chlorine is 0.367 × 10–6 kg/C, [AIPMT (Prelims)-2006], (1) 17.6 mg, , (2) 21.3 mg, , (3) 24.3 mg, , (4) 13.6 mg, , Sol. Answer (1), 47. In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will, [AIPMT (Prelims)-2006], , A, 4, , 4, , 1, , 3, B, V, , (1) Flow from A to B, , (2) Flow in the direction which will be decided by the value of V, (3) Be zero, (4) Flow from B to A, Sol. Answer (4), 12V 3V, i, , 32, 8, i, 4, , 3V, i 8, 8, , 2i , i, , 3V, i, 8, , A, V, 8, 2V, 8, , 4, , 4, , P, 3, , 1, B, V, , V, 8, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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122, , Current Electricity, , VP – VA =, , V, 2, , VP – VB =, , V, 4, , Solution of Assignment, , VB > VA, Current flows from B to A, 48. Two cells, having the same emf, are connected in series through an external resistance R. Cells have internal, resistances r1 and r2 (r1 > r2) respectively. When the circuit is closed, the potential difference across the first, cell is zero, The value of R is, [AIPMT (Prelims)-2006], (1) r1 – r2, , (2), , r1  r2, 2, , (3), , r1  r2, 2, , (4) r1 + r2, , Sol. Answer (1), 2E, E, , r1  r2  R r1, , 2r1 = r1 + r2 + R, R = r1 – r2, 49. Power dissipated across the 8  resistor in the circuit shown here is 2 W. The power Ù dissipated in watt units, across the 3  resistor is, [AIPMT (Prelims)-2006], , 1, , 3, i, 8, , (1) 2, , (2) 1, , (3) 0.5, , (4) 3, , Sol. Answer (4), 2 = i28, , i, , 1, 2, , 0.5 4, , i, 8, , 1, , 3  i1, i, 8  0.5, , i=2, P=3W, 50. Kirchhoff’s first and second laws for electrical circuits are consequences of, , [AIPMT (Prelims)-2006], , (1) Conservation of energy, (2) Conservation of electric charge and energy respectively, (3) Conservation of electric charge, (4) Conservation of energy and electric charge respectively, Sol. Answer (2), Theory, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 123, , 51. Two batteries, one of emf 18 V and internal resistance 2  and the other of emf 12 V and internal resistance 1, , are connected as shown. The voltmeter V will record a reading of, [AIPMT(Prelims)-2005], , V, 2, 18 V, 1, 12 V, (1) 15 V, , (2) 30 V, , (3) 14 V, , (4) 18 V, , Sol. Answer (3), , V, 18 V, , 2, , 12 V 1 , , i, , 6,  2A, 3, , V = 18 – 2(2), V = 14 V, 52. For the network shown in the figure, the value of the current i is, , [AIPMT (Prelims)-2005], , 4, , i, , 4, , 2, 3, , 6, V, , (1), , 9V, 35, , (2), , 5V, 18, , (3), , 18, 5, , 18, V i, 5, i, , (4), , 18V, 5, , 2, , Sol. Answer (2), Rnet , , 5V, 9, , 4, 3, 6, , 5V, 18, , V, , 53. The temperature of inversion of a thermocouple is 620°C and the neutral temperature is 300°C. What is the, temperature of cold junction ?, [AIPMT (Prelims)-2005], (1) 20°C, , (2) 320°C, , (3) –20°C, , (4) 40°C, , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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124, , Current Electricity, , Solution of Assignment, , 54. When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance, between two of diametrically opposite points will be, [AIPMT (Prelims)-2005], (1), , R, 4, , R, 8, , (2), , (3) 4R, , (4), , R, 2, , Sol. Answer (1), , R, 2, , Rnet , , R, 4, , 55. A 5 A fuse wire can withstand a maximum power of 1 W in circuit. The resistance of the fuse wire is, [AIPMT (Prelims)-2005], (1) 0.2 , , (2) 5 , , (3) 0.4 , , (4) 0.04 , , Sol. Answer (4), 1 = 52R, , R, , 1,  0.04, 25, , 56. A 12 cm wire is given a shape of a right angled triangle ABC having sides 3 cm, 4 cm and 5 cm as shown, in the figure. The resistance between two ends (AB, BC, CA) of the respective sides are measured one by, one by a multi-meter. The resistances will be in the ratio, A, , 3 cm, , 5 cm, , B, , (1) 3 : 4 : 5, , C, , 4 cm, , (2) 9 : 16 : 25, , (3) 27 : 32 : 35, , (4) 21 : 24 : 25, , Sol. Answer (3), , A, , A, , 5, , 3, , B, , RAB , , 27, 12, , 5, , 3, , C, , 4, , A, 5, , 3, B, , C, , 4, , RBC , , B, , 32, 12, , 4, , RAC , , C, , 35, 12, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 125, , 57. Two rods are joined end to end, as shown. Both have a cross-sectional area of 0.01 cm2. Each is, 1 meter long. One rod is a copper with a resistivity of 1.7 × 10–6 ohm-centimeter, the other is of iron with a, resistivity of 10–5 ohm-centimeter., How much voltage is required to produce a current of 1 ampere in the rods?, , V, , Cu, (1) 0.117 V, , (2) 0.00145 V, , Fe, (3) 0.0145 V, , (4) 1.7 × 10–6 V, , Sol. Answer (1), 5  100, , R  1.17  10, 0.01, , R = 0.117 , V = (1) (0.117), V = 0.117 V, 58. Ten identical cells connected in series are needed to heat a wire of length one meter and radius 'r' by 10ºC, in time 't'. How many cells will be required to heat the wire of length two meter of the same radius by the, same temperature in time 't'?, (1) 10, , (2) 20, , (3) 30, , (4) 40, , Sol. Answer (2), 2, , ⎛ 10E ⎞, ⎜, ⎟ rt  MC (10), ⎝ r ⎠, ⎛ nE ⎞ 2rt  2MC(10), ⎜, ⎟, ⎝ 2r ⎠, 100(4) 1, , 2, 2n 2, , n = 20, 59. The mobility of charge carriers increases with, (1) Increase in the average collision time, , (2) Increase in the electric field, , (3) Increase in the mass of the charge carriers, , (4) Decrease in the charge of the mobile carriers, , Sol. Answer (1), Theory, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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126, , Current Electricity, , Solution of Assignment, , 60. A 1250 W heater operates at 115 V. What is the resistance of the heating coil?, (1) 1.6 , , (2) 13.5 , , (3) 1250 , , (4) 10.6 , , Sol. Answer (4), , R, , (115)2, 1250, , R = 10.58 , 61. A wire 50 cm long and 1 mm2 in cross-section carries a current of 4 A when connected to a 2 V battery. The, resistivity of the wire is, (1) 4 × 10–6 -m, , (2) 1 × 10–6 -m, , (3) 2 × 10–7 -m, , (4) 5 × 10–7 -m, , Sol. Answer (2), 2 = 4R, R, , 1 P (0.5), , 2, 10 6, , , , = 10–6, , 62. Six resistors of 3  each are connected along the sides of a hexagon and three resistors of 6  each are, connected along AC, AD and AE as shown in the figure. The equivalent resistance between A and B is equal, to, E 3, D, 3, 3, , , , F, , , , 3, A, (1) 2 , , C, , , , (2) 6 , , 3, , 3, B, , (3) 3 , , (4) 9 , , Sol. Answer (1), , E, , 3, , 3, F, , 6, , D, 3, , 6, , 3, , C, , 6, , 3, , A, 3, , B, , Rnet = 2 , 63. Identify the set in which all the three materials are good conductors of electricity?, (1) Cu, Hg and NaCl, , (2) Cu, Ge and Hg, , (3) Cu, Ag and Au, , (4) Cu, Si and diamond, , Sol. Answer (3), Cu, Ag, Au, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 127, , 64. A flow of 107 electrons per second in a conducting wire constitutes a current of, (1) 1.6 × 10–12 A, , (2) 1.6 × 1026 A, , (3) 1.6 × 10–26 A, , (4) 1.6 × 1012 A, , Sol. Answer (1), i = 107 × 1.6 × 10–19, i = 1.6 × 10–12, 65. In the network shown in the figure, each of the resistance is equal to 2 . The resistance between the points, A and B is, , 2, , 2, , 2, , A, , 2, , B, , 2, (1) 3 , , (2) 4 , , (3) 1 , , (4) 2 , , Sol. Answer (4), , 2, , 2, , A, 2, , 2, , 2, , B, , Rnet = 2 , 66. Two wires of the same metal have same length, but their cross-sectional areas are in the ratio, 3 : 1. They are joined in series. The resistance of thicker wire is 10 . The total resistance of the combination, will be, , (1) 40 , , (2) 100 , , (3), , 5, , 2, , (4), , 40, , 3, , Sol. Answer (1), , l, , l, , 3A A, l,  10, 3A, l,  30, A, , Rnet = 40 , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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128, , Current Electricity, , Solution of Assignment, , 67. When the key K is pressed at time t = 0, then which of the following statement about the current I in the, resistor AB of resistance 1000  of the given circuit is true?, , 2V, K, , (2) At t = 0, / = 2 mA and with time it goes to 1 mA, (3) I = 1 mA at all t, , B, , A, , 1000 , , C, , 1 F, , 1000 , , (1) I oscillates between 1 mA and 2 mA, , (4) I = 2 mA at all t, Sol. Answer (2), , K, , t=0, , i, , 2,  2 mA, 1000, , 2V, , 1 F, , t=, i, , 100 , 1000 , , 2,  1mA, 1000, , 68. Three copper wires have lengths and cross-sectional areas as (/, A), (2l, A/2) and (l/2, 2A). Resistance is, minimum in, (1) Wire of cross-sectional area 2A, , (2) Wire of cross-sectional area 1/2 A, , (3) Wire of cross-sectional area A, , (4) Same in all three cases, , Sol. Answer (1), , R1 , , l, A, , R2 , , 2l, .2, A, , R3 , , l, 2.2a, , R3 is least, 69. Kirchhoff’s first law, i.e.  i = 0 at a junction, deals with the conservation of, (1) Momentum, , (2) Angular momentum, , (3) Charge, , (4) Energy, , Sol. Answer (3), Theory, 70. A galvanometer having a resistance of 8 ohms is shunted by a wire of resistance 2 ohms. If the total current, is 1 A, the part of it passing through the shunt will be, (1) 0.2 A, , (2) 0.8 A, , (3) 0.25 A, , (4) 0.5 A, , Sol. Answer (2), , 2, , iG (8), 1  iG, , 2 – 2iG = 8iG, iG = 0.2, i – iG = 0.8 A, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 129, , 71. In a meter bridge, the balancing length from the left end (standard resistance of one ohm is in the right gap), is found to be 20 cm. The value of the unknown resistance is, (1) 0.8 , , (2) 0.5 , , (3) 0.4 , , (4) 0.25 , , Sol. Answer (4), x, 1, , 20 80, , x, , 1, , 4, , 72. A potentiometer consists of a wire of length 4 m and resistance 10 . It is connected to a cell of e.m.f. 2 V., The potential difference per unit length of the wire will be, (1) 5 V/m, , (2) 2 V/m, , (3) 0.5 V/m, , (4) 10 V/m, , Sol. Answer (3), V 2,   0.5 V/m, l, 4, , 73. Calculate the net resistance of the circuit between A and B, , 3, A, , 8, 3, , (2), , 14 , 3, , B, , 7, 6, , (1), , 4, , 8, (3), , 16 , 3, , (4), , 22 , 3, , Sol. Answer (2), , 3, , 4, , A, , B, 6, , Rnet =, , 8, , 7  14 14, , , 21, 3, , 74. A car battery of emf 12 V and internal resistance 5 × 10–2 , receives a current of 60 A, from external source,, then terminal potential difference of battery is, (1) 12 V, , (2) 9 V, , (3) 15 V, , (4) 20 V, , Sol. Answer (3), V  12 , , 5, .60, 100, , V = 15 V, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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130, , Current Electricity, , Solution of Assignment, , 75. The potentiometer is best for measuring voltage as, (1) It has a sensitive galvanometer, , (2) It has wire of high resistance, , (3) It measures p.d. in closed circuit, , (4) It measures p.d in open circuit, , Sol. Answer (4), Theory, 76. If specific resistance of a potentiometer wire is 10–7 m and current flow through it is 0.1 A, cross-sectional, area of wire is 10–6 m2 then potential gradient will be, (1) 10–2 volt/m, , (2) 10–4 volt/m, , (3) 10–6 volt/m, , (4) 10–8 volt/m, , Sol. Answer (1),  = 10–7, i = 0.1, , R, , l, 107 l, , 6, 10, 10, , V  0.1, , l, 10, , V,  0.01, l, , 77. Specific resistance of a conductor increases with, (1) Increase in temperature, , (2) Increase in cross-sectional area, , (3) Increase in cross-sectional and decrease in length (4) Decrease in cross-sectional area, Sol. Answer (1), Theory, 78. For a cell, terminal potential difference is 2.2 V when circuit is open and reduces to1.8 V when cell is connected, to a resistance of R = 5 . Determine internal resistance of cell (r), (1), , 10 , 9, , (2), , 9 , 10, , (3), , 11 , 9, , (4), , 5, 9, , Sol. Answer (1), 1.8 = 2.2 – ir, ir = 0.4, , i, , 2.2, 5r, , 2.2,  0.4, 5r, , 22r = 20 + 4r, 18r = 20, r , , 10, 9, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Current Electricity, , 131, , 79. The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and, 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the, galvanometer resistance is 50 ohm, the current drawn from the cell will be, (1) 0.2 A, , (2) 0.1 A, , (3) 2.0 A, , (4) 1.0 A, , Sol. Answer (1), 30, , 10, G, 30, , 7V, , i, , 90, , 5, , 7, A, 37, , 80. A heating coil is labelled 100 W, 220 V. The coil is cut in half and the two pieces are joined in parallel to the, same source. The energy now liberated per second is, (1) 200 J, , (2) 400 J, , (3) 25 J, , (4) 50 J, , Sol. Answer (2), , R, , (220)2, = 484 , 60, , Rnet = 121 , P=, , 220  220,  220 2, =, = 400 W, 121, 121, , 81. A (100 W, 200 V) bulb is connected to a 160 volt supply. The power consumption would be, (1) 100 W, , (2) 125 W, , (3) 64 W, , (4) 80 W, , Sol. Answer (3), 100 ⎛ 200 ⎞, ⎜, ⎟, P, ⎝ 160 ⎠, , 2, , 100 25, , P, 16, , P = 64 W, 82. If two bulbs, whose resistances are in the ratio of 1 : 2 are connected in series, the power dissipated in them, has the ratio of, (1) 2 : 1, , (2) 1 : 4, , (3) 1 : 1, , (4) 1 : 2, , Sol. Answer (4), P1 : P2 = R1 : R2 = 1 : 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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132, , Current Electricity, , Solution of Assignment, , 83. Two bulbs of (40 W, 200 V), and (100 W, 200 V). Then correct relation for their resistances, (1) R40 < R100, , (2) R40 > R100, , (3) R40 = R100, , (4) No relation can be predicted, , Sol. Answer (2), , R40 , , (200)2, 40, , R60 , , (200)2, 100, , R40 > R60, 84. Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is, connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will, be, (1) 50 watt, 100 watt, , (2) 100 watt, 50 watt, , (3) 200 watt, 150 watt, , (4) 50 watt, 200 watt, , Sol. Answer (4), Series, , P, , 100  100,  50 W, 200, , Parallel, P = 100 + 100 = 200 W, , SECTION - D, Assertion-Reason Type Questions, 1., , A : For a given conductor, electric current does not vary even if its cross sectional area varies., R : A conductor remains uncharged when current flows through it., , Sol. Answer (2), 2., , A : When a steady current flows through a conductor of non-uniform cross-section, the current density, electric, field and drift velocity do not remain constant., R : For a constant current the current density, electric field and drift velocity are inversely proportional to crosssectional area., , Sol. Answer (1), 3., , A : To a metal wire of diameter d and length L when the applied voltage is doubled, drift velocity gets doubled., R : For a constant voltage when the length is doubled, drift velocity will be halved but drift velocity is, independent of diameter., , Sol. Answer (2), 4., , A : Kirchhoff's Current law is applicable at any junction or node in the circuit., R : Kirchhoff's laws are general in nature., , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 5., , Current Electricity, , 133, , A : Voltage across a resistor decreases in the direction of current and increases opposite to the direction of, current., R : Voltage drop or gain across a capacitor depends on the direction of current., , Sol. Answer (3), 6., , A : The voltage across a battery may be less, equal or more than the emf of the battery., R : Voltage across a battery also depends on the magnitude and direction of current., , Sol. Answer (1), 7., , A : Practically a voltmeter will measure the voltage across the battery but not its EMF., R : EMF of a cell is measured with the help of a potentiometer., , Sol. Answer (2), 8., , A : A potentiometer can act as an ideal voltmeter., R : An ideal voltmeter has infinite resistance., , Sol. Answer (2), 9., , A : Ohm's law is universally applicable for all conducting elements., R : All conducting elements show straight line graphic variation on (I – V) plot., , Sol. Answer (4), 10. A : A low voltage supply, from which high currents are to be withdrawn, must have very low internal resistance., R : Maximum current drawn from a source is inversely proportional to internal resistance., Sol. Answer (1), 11. A : High voltage (high tension) supply must have very large internal resistance, R : If the circuit is accidentally shorted, then the current drawn will not exceed safety limits if internal, resistance is high., Sol. Answer (1), 12. A : Alloys of metals usually have greater resistivity than that of their constituent metals., R : Alloys usually have much lower thermal coefficient of resistance than pure metals., Sol. Answer (2), 13. A : Current density is a vector quantity., R : Electric current, passing through a given area is the flux of current density through that area., Sol. Answer (2), 14. A : When two cells of equal EMF and equal internal resistances are connected in parallel with positive plate of, one to the positive plate of the other then, the net EMF of the combination will be equal to the EMF of each, cell., R : Effective internal resistance of the parallel combination of two identical cells will be half of the internal resistance, of each cell., Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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134, , Current Electricity, , Solution of Assignment, , 15. A : The drift velocity of electrons in a conductor is very small still current in a conductor is established almost, instantaneously on closing the switch., R : Electric field in the conductor sets up with speed of light., Sol. Answer (1), 16. A : When temperature of a metallic wire is increased, its resistance increases., R : As the temperature is increased, average relaxation time increases., Sol. Answer (3), 17. A : The potentiometer wire should have uniform cross sectional area., R : On the potentiometer wire the jockey is gently touched, not pressed hard., Sol. Answer (2), ,   , , https://t.me/NEET_StudyMaterial, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456