Regent Physics 50 MCQ

MULTIPLE CHOICE QUESTIONS 1. A hollow metal sphere of radius 15 cm is charged such that the potential on its surface is 20 V. The potential at the centre of the sphere is a. 20 V b. 30 V c. 5 V d. 0 V Sol: Correct option is (a) For a hollow metal sphere the potential at centre of sphere is equal to the potential on the surface of the sphere. Therefore potential at the centre of sphere is 20 V. 2. A uniform electric field having a magnitude E and direction along the positive y-axis exists. If the potential V is zero at y=0, then its value at y=+y will be a. yE b. 2yE c. –yE d. -2yE Sol: Correct option is (c) We know that Vo  0 Also Vy is potential at y=+y then Vo  V y  yE  Vy   yE 3. An electric dipole of momentum Po is placed in the position of stable equilibrium in uniform electric field of intensity Eo . The couple required to rotate it through angle  from the initial position is a. Po Eo b. Po Eo sin  c. 2 Po Eo d. 2 Po Eo sin  Sol: Correct option is (b) Couple = qEo  a sin  Where q  a  Po  Couple required = Po Eo sin  4. A given charge Qo is situated at a certain distance from an electric dipole in the end on position, experiences a force Fo . If the distance of charge is halved the force acting on the charge Qo will be a. F b. 2F c. 3F d. 8F Sol: Correct option is (d) 2Kp r3 2Kp Q  F r3 Electric field at a distance r  Hence force at a distance r  Also force at a distance r 2 Kp on charge Q =  Q  8F 2 2 r 2   Hence if distance of charge is halved force acting = 8

* 5. xyz is an equilateral triangle. Charges -2q are placed at each corner. The electric intensity at A will be a. 1 2q 2 4 o r 2 1 q 4 o 2r 2 b. c. 0 d. 1 q2 4o 2r Sol: Correct option is (c) Electric intensity at A is Where E = 1 ( 2 q ) 4o r 1  120  Enet  E  2  E  cos    E  2 E   0 2  2  Enet  0 6. Work done in carrying a charge q one around the circle of radius R with a charge Q at the centre is a. 1 qQ 4o r 2 b. 0 c. 1 q 2Q 4o r 1 qQ 2 d. 4o r 2 Sol: Correct option is (b) The force acting perpendicular to the displacement of the charge. Hence work done = 0 7. Two conducting spheres of radii R1 and R2 are at the same potential. The ratio of their charges is a. R1R2 b. R2 R1 c. R12 R2 d. R1 R2 Correct option is (d) Sol: Potential of sphere 1, V1  1 Q1 4o R1 Potential of sphere 2, V2  As V1  V2 1 Q2 4 o R2 1 Q1 1 Q2  4 o R1 4o R2  Q1 R1  Q2 R2 8. An electric cell does 25 joules of work in carrying 5 coulombs charges around the close circuit. The electromotive force at the cell is a. 3 volts b. 8 volts c. 5 volts d. 9 volts Sol: Correct option is (c) Work 25 Joules   5volts Ch arg e 5coulombs Electromotive force = 9. Two conducting spheres of radii R1 and R2 are at the same chartged. The ratio of their potential is a. R1R2 b. R12 R2 c. R2 R1 d. R12 R2 2 Sol: Correct option is (c) 1 Q1 4o R1 Potential of sphere 1, V1  Potential of sphere 2, V2  As Q1  Q2 1 Q2 4 o R2  V1 R2  V2 R1 10. Two conducting spheres of radii R1 and R2 have same electric field near their surface. The ratio of their electric potential is a. R2 R1 b. R2 2 R12 c. R2 R1 R1 R2 Correct option is (d) d. Sol: Electric field of sphere 1, E1  1 Q1 V1  4o R12 R1 1 Q2 V2  4o R2 2 R2 Electric field of sphere 2, E2  As E1  E2 V1 V2  R1 R2  V1 R1  V2 R2 11. If the force exerted by a small spherical charged object on another charged object at 8 m is 5 N, what will be the force exerted when the second object is moved to 16 m. a. 3 N b. 2 N c. 2.25 N d. 1.25 N Sol: F= Correct option is (d) 1 q1q2 4o (8)2 Also F=5 5 1 q1q2 4 o (8)2 1 q1q2  1.25 N 4 o (16) 2 Now 12. If the force of an electric file don 10 coulomb charge is 25 N, The intensity of field is a. 3 N/C b. 2.5 N/C c. 3.8 ev d. 2.5 ev Sol: Correct option is (b) Force ch arg e Electric filed intensity = 25 N = 2.5 N/C 10Coulomb = 13. Two plates are 2 cm apart and the potential difference between them is 200 volts. The electric field intensity between the plates is a. 300 N/C b. 100 N/C c. 10000 N/C d. 1000 N/C Sol: Correct option is (c) V d Electric field intensity = V= 200 Volts d= 2 10 2 m E 200  10000 N/C 2  102 14. Electrons are caused to fall through a potential difference of 250 volts. If they were initially at rest, their final speed is a. 9.37 106 m / s b. 8.24 106 m / s c. 5.9 107 m / s d. 9.91 108 m / s Sol: Correct option is (a) 1 me ve 2  eV 2  ve  2eV me 2  1.6  1019  250 9.1 1031 800 1012 9.1  ve   ve   ve  9.37 106 m / s 15. The capacitors of a parallel plate condenser does not depend upon a. medium between plates b. distance between plates. c. area of plates d. metal of plates. Sol: Correct option is (d) o A d Capacitance = Capacitance is independent of metal of plates of conductor. 16. In a charged capacitor, the energy is stored in a. metal plates of capacitor b. positive charges c. negative charges d. filed between the plates Sol: Correct option is (d) Energy is stored in the form of field present between the plates. 17. Which of the following relations is correct? a. V  q2 C b. q  CV c. qCV  1 d. q  Sol: C V Correct option is (b) If q is the charge on the body and V is potential then Capacitance C   q  CV q V 18. The capacitance of the parallel plate condenser is given by a. V  C Q o d A o A d b. C  c. C  d. C  Sol: d o A Correct option is (c) o d A Capaci tan ce  Where o = constant A= Area of plates d= distance between the plates. 19. The capacitance of a parallel plate condenser is Co . Its capacity when the separation between the plates is doubled will be a. 2Co b. 3Co c. Co 3 Co 2 d. Sol: Correct option is (d) o d A Capacitance = Co  When d '  2d C'  o A o A Co   d' 2d 2 20. Two condensers of capacitance C1 and C2 respectively are connected in parallel. The equivalent capacitance of system is a. C1C2 b. C12C1 c. C1  C2 d. C1 C2 Sol: Correct option is (c) Ceq  C1  C2 Equivalent capacitance is equal to the summation of capacitance of each capacitor connected in parallel. 21. Two condensers of capacitance C1 and C2 are connected in series. The equivalent capacitance of the system is a. C1 C2 b. C1C2 (C1  C2 ) C2 C1 c. d. C1  C2 Sol: Correct option is (b) When two condensers are connected in parallel then let Ceq be their equivalent capacitance 1 1 1 CC    Ceq  1 2 Ceq C1 Ceq C1  C2 22. A condenser of capacity 30  F is charged to 20 volts. The energy stored in the condenser is a. 6 103 joules b. 8  104 joules c. 9  103 joules d. 17  103 joules Sol: Correct option is (a) Energy stored = 1  C V 2 2 1   30  106  (20)2 2 1   30  400  106 2  6000 106  6  103 joules 23. A 30  F capacitor has charge on each plate 6 103 coulomb, and then the energy stored is a. 30  104 joules b. 25 102 joules c. 87.5 103 joules d. 60 102 joules Sol: Correct option is (d) Q= 6 103 coulomb C= 30  106 F We know that Energy = 1 CV 2 2 Also Q=CV  V   V  200 volts Q 6  103  C 30  106 Energy = 1 CV 2 2 1   30  106  4 10 4 2 1   30  106  (200)2 2 60 102 joules 24. Two capacitors 6  F and 3  F are connected in series, the effective capacitance of the system will be a. 3  F b. 2  F c. 6  F d. 1  F Sol: Correct option is (b) 1 1 1   Ceq 6  F 3 F 6  3 1012  2  106 F 6 9 10 Ceq  2  F  Ceq  25. If a 6  F condenser is charged to 100 volts and its plates are connected by a wire, the heat produced in the wire is a. 8  103 joules b. 3.8  102 joules c. 3  102 joules d. 7.6 102 joules Sol: Correct option is (c) Heat produced = Energy stored in condenser = 1 CV 2 2 1   6 106  (100) 2 2  3 102 joules 26. Two capacitors each of capacitance 2  F are connected in parallel. To this combination, a third capacitor of capacitance 4  F is connected in series. The equivalent capacitance of the system is a. 2  F b. 7  F c. 12  F d. 8.3  F Sol: Correct option is (a) 1 1 1   Ceq 2  F  2  F 4  F  Ceq  4 F  4 F 8 F Ceq  2  F 27. A condenser is charged through a potential difference of 100 volts possesses charge of 0.2 coulomb, when the condenser of discharged it would release an energy a. 1 joules b. 8 joules c. 7 joules d. 10 joules Sol: Correct option is (d) Q = 9.2 coulomb V = 100 volts We know that Q = CV C Q 0.2 C   2 10 3 F V 100 1 CV 2 2 Now energy released = 1   2  103  (100)2  10 joules 2 28. A parallel plate capacitor has a capacitance of 30  F in air and 45  F when immersed in another medium. The dielectric constant K of the medium is a. 2 b. 1.5 c. 2.8 d. 3.5 Sol: Cair  Correct option is (b) o A d o A d Cmedium  K K Cmedium 45 F   1.5 30  F Cair 29. How many number of electrons are to be put on a spherical conductor of radius 0.3 m to produce an electric field of 0.012 N/C just above its surface a. 6 105 b. 7.2  104 c. 7.5 105 d. 0 Sol: Correct option is (c) Electronic field just above the surface of spherical conductor E KQ 9 109  Q  r2 (0.3) 2 0.012  9  109  (ne) 9  102 ne  0.012  1011 12 10 14 16 10 20 n  7.5 105 n 30. What is the angle between the electric dipole moment and the electric field strength due to it on the axial line? a. 30o b. 0o c. 60o d. 90o Sol: Correct option is (b) Both of the dipole moment and electric field are in same direction, hence the angle between them is 0o 31. What is the angle between the electric dipole moment and the electric field strength due to it on the equatorial line? a. 180o b. 60o c. 30o d. 90o Sol: Correct option is (a) On equatorial line the electric field strength is in the opposite direction to that of electric dipole moment, hence angle between them is 180o . 32. Three charges 1C, -2C, 1C are placed at three vertices of an equilateral triangle. Which of the statements is true for the centre O of triangle? a. At O field and potential both are zero. b. At O field is non-zero but potential is zero. c. At O field and potential both are non-zero. d. At O field is zero but potential is non-zero. Sol: Correct option is (b) At O potential due to three charges is zero but there exists a net electric field. 33. 64 identical drops of mercury are charged simultaneously to the same potential of 5 units. Assuming the drops are made to combine to form one large drop, then its potential is a. 40 units b. 10 units c. 15 units d. 80 units Sol: Correct option is (d) Let radius of small drops be r and that of a big drop is R, then 4 4 64   r 3   R 3 3 3  R  4r Let potential of small drops = 1 q 4o r Also 1 q  5 units 4o r  1 64  q 4 o 4r Then potential of big drop  5  16  80 units 34. Two particles of masses M and 3M with charges q and q are placed in an uniform electric field E and allowed to move for same time. The ratio of their kinetic energies will be a. 1:2 b. 2:1 c. 3:1 d. 1:3 Sol: Correct option is (c) For particle 1 a1  For particle 2 a2  qE M qE 3M  a1  2a2 1 1 Also K1   M  (a1t ) 2  Ma12t 2 2 2 1 1 and K 2   3M  (a2t ) 2  3Ma2 2t 2 2 2 K1 Ma12 M 9a2 2 3     K 2 3Ma2 2 3M a2 2 1 35. When two capacitors one of 4  F and other 8  F are connected in series and the combination is charged to a potential difference of 90 volts. The potential difference across 4  F capacitor is a. 70 volts b. 60 volts c. 10 volts d. 90 volts Sol: Correct option is (b) Let C1  4  F and C2  8 F Also let V1 is potential difference across C1 and V2 is potential difference across C2 C1V1  C2V2  V1  4  F  8 F  V2  V1  2V2 V1  V2  90 Also  3V2  90  V2  30  V1  60 volts 36. Ten capacitors each of capacitance 1  F are to be connected to obtain a 9 capacitance of  F . Which of the following combination is correct. 10 a. 5 in series 5 in parallel. b. 3 in series and 7 in parallel. c. 9 in parallel and 1 in series. d. All 10 in series. Sol: Correct option is (c) 1 1 1   Ceq 1  1  1  1  1  1  1  1  1 1 1 1 10  1  9 Ceq 9 Ceq  9 F 10 37. In order to obtain a time constant of 5 seconds in a RC circuit containing a resistance of 200  the capacity of a condenser should be a. 0.75 F b. 0.1 F c. 0.625 F d. 0.025 F Sol: Correct option is (d) T=RC= 5 200 x C = 5  C   C  0.025 F 5 1  200 40 38. A 6  F capacitor is charged to 100 volts and then its plates are joined through a resistance of 2 K  . The heat produced in the resistance is a. 3  102 joules b. 6 103 joules c. 8  102 joules d. 7 10 3 joules Sol: Correct option is (a) Heat produced = Energy stored in the capacitor 1  CV 2 2 1   6 106  (100) 2 2  3 10 2 joules 39. The resistance of a wire of uniform radius R and length l is x. the resistance of another wire of same material but radius 2R and length 2l will be a. 2x b. 3x c. x 2 x 3 d. Sol: x x'  Correct option is (c)  R2  2l l  (2 R) 2  R l 2  1 x  2 2 40. Two resistances R1 and R2 are connected in series and resistance R3 is connected in parallel to this series connection. The equivalent resistance of the combination will be a. R1 R2 R3 ( R1  R2 ) R3 R1 c. ( R2  R3 ) b. d. ( R1  R2 ) R3 R1  R2  R3 Sol: Correct option is (d) Equivalent of series connection = R1  R2 1 1 1   Req R1  R2 R3 Req  ( R1  R2 ) R3 R1  R2  R3 41. The resistance of a wire is R  . The wire is stretched to double its length. What is the new resistance of wire in ohm? a. 3R b. R c. 8R d. 4R Sol: R Correct option is (d) l A l '  2l Now we know that volume of wire is constant, therefore Al  A'l '  Al  A' (2l )  A'  A 2 R '  l   (2l )  4 R  ' A' A 2 When a d.c. battery of emf V and internal resistance R delivers R maximum power to an external resistance R ' , then ratio ' equals R a. 1 b. 2 c. 4 d. 3 Sol: Correct option is (a) 42. For the battery to deliver maximum power the internal resistance should be R equal to the external resistance, hence R  R '  '  1 R 43. A current of 3.2 A is flowing in a conductor. The number of electrons flowing per second through the conductor will be a. 8  1017 electrons b. 2  1019 electrons c. 4 1019 electrons d. 3.5  1019 electrons Sol: Correct option is (b) Q  3.2 A t Current = Q= 3.2 coulombs n 1.6 1019  3.2C  n  2  1019 electrons 44. The resistance of 10 cm long wire is 2 ohm. The wire is stretched to uniform wire of 20 cm length. The resistance of stretched wire will be a. 6 ohm b. 4 ohm c. 8 ohm d. 10 ohm Sol: R Correct option is (c) L A L'  2l Here  A1  A 2  R1  4 R  R '  4  2  8 ohm 45. Three 5 ohm resistors are connected to form a triangle. The resistance between any two corners is a. 5 ohm 3 b. 8 ohm c. 10 ohm d. 10 ohm 3 Sol: Correct option is (d) 1 1 1 1 1     Req 5  5 5 10 5 1 15 50 10   Req   ohm Req 50 15 3 46. A battery of 10 volts is connected to the terminals of a 5 meter long wire of uniform thickness and resistance of the order of 1000 ohm. The difference of potential difference between two points separated by 1 m on the wire will be a. 2 volts b. 5 volts c. 3 volts d. 7 volts Sol: Correct option is (a) 10volts  2v / m 5m Potential gradient = Potential difference between two points 1 m apart = 2 v/m x 1 m = 2 volts 47. A wire has resistance of 3  . A second wire of same material is having length half and radius of cross section half that of the wire. The resistance of second wire is a. 4  b. 1.5  c. 3  d. 6  Sol: R Correct option is (d)  L  3  l A  r2 Now l '  l r and r '  2 2  R'    l 2  r  2 2   r2 l  2  3  2  6 48. A wire has resistance 16 ohms. It is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is a. 2 ohms b. 8 ohms c. 4 ohms d. 16 ohms Sol: Correct option is (c) When wire of 16 ohms is bent in the form of a circle, then the resistance of circle, then the resistance of semicircle is 80 ohms. Req  8  8 64   4 8  8 16 49. The smallest resistance which can be obtained by connecting 100 resistances of 2 ohm each is a. 4 ohms b. 1 ohms 50 c. 50 ohms d. 200 ohms Sol: Correct option is (b) Equivalent resistance is smallest when all the resistance are connected in parallel. For parallel connection Req  2 1   100 50 50. A cell of negligible resistance and e.m.f. of 30 volts is connected to series combination of 6,3 and 1 ohm resistance. The potential difference in volts between the terminals of 1 ohm resistance is a. 6 volts b. 8 volts c. 3 volts d. 18 volts Sol: Correct option is (c) 30  3A 6  3 1 Current = Potential difference across 10 ohm = 3 A x 1 Ohm = 3 volts