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SM 15 16 XII Physics Unit-1 Section-D

Transcript

71Success Magnet-Solutions (Part-II) Electromagnetism

Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Section - D : Assertion - Reason Type

1. Answer (2)

Potential energy of a system of charges and an individual point charge may be zero, but the statements are

independent.

2. Answer (1)

At any boundary change in normal component of E is 0

. Consider examples of a long sheet of charge or

a charged long metal plate.

3. Answer (2)

Loss of electrostatic potential energy takes place in form of heat in connecting wires.

E decreases and energy density decreases at all points, but it is not the correct justification for loss in en-

ergy.

4. Answer (4)

In the arrangement, no charge can stay on A.

A

B

Transfer of negative charge takes place from low potential to high potential.

5. Answer (3)

Field induced in dipole is opposite to applied field, because of dipoles induced in the same direction. Thus

net field inside dielectric is less than applied.

6. Answer (2)

EqE electric

, does not depend on V . Charge is invariant. Both statements are independent.

7. Answer (1)

A negative charge is induced on face of ball B closer to A and is attracted. As the attraction is spontaneous,

potential energy of the system must be decreasing.

8. Answer (2)

As an example, potential is not zero in this case

q

+

+

+

Net charge zero

Second statement is correct but independent as metal is equipotential.

9. Answer (2)

Both statements are independently correct.

10. Answer (3)

On introducing any dielectric or metal piece between plates of capacitor, its capacitance increases. E poten-

tial energy may increase or decrease depending on whether capacitor is connected to battery or not.

11. Answer (1)

If a charge is brought from infinity to a point in electric field on equitorial line, work done is zero, as on the

line potential is zero.

72 Electromagnetism Success Magnet-Solutions (Part-II)

Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

12. Answer (4)

VE . V function should be known to determine E . For point charge r

V

r

kqE

||2 .

13. Answer (1)

Due to electrostatic shielding, 0E inside car.14. Answer (1)

Work done by external force = R

QU

0

2

8 15. Answer (1)

In some of liquid dielectrics with polar molecules, k decreases with increase in temperature. C = kC alsodecreases.

16. Answer (3)

Earth is taken as reference of potential, but earth has net negative charge.

17. Answer (1)

Energy density = 2

02

1E .

18. Answer (1)

Electric potential energy may provide the energy to separate two charged metal drops.

19. Answer (2)

Both statements are independently correct.

InfiniteLine charge

20. Answer (1)

Acceleration of a particle cannot be infinite, as it has finite mass.

21. Answer (3)

Electric field near electrodes of cell is not zero, but resistance of air is very large.

22. Answer (3)

Due to electrostatic repulsion, radius of a charged bubble increases and excess pressure equal to r

T4

decreases. T does not change.

23. Answer (1)

Whether voltmeter is ideal or not, both capacitors are in parallel after key is closed.

24. Answer (2)

25. Answer (1)

Work done by battery of lower emf is negative, as it is charged.

26. Answer (2)

If potential difference already across capacitor is more than emf of battery, it looses its extra charge. Heat is

generated during charging and discharging.

73Success Magnet-Solutions (Part-II) Electromagnetism

Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

27. Answer (4)

28. Answer (1)

Statement-2 supports and justifies statement-1.

29. Answer (1)

Statement-2 justifies statement-1.

30. Answer (4)

31. Answer (1)

The correct explanation is that the magnetic field due to a magnet is non-uniform and a dipole experiences a net

force in non-uniform magnetic field.

32. Answer (3)

Internal conductance of cell can be measured by the arrangement as,

V = ir V = ri +

x intercept : r

y intercept : r

x

y int

int

33. Answer (2)

Both statements are independently correct.

34. Answer (3)

35. Answer (1)

E

J . J is same at every section E .

36. Answer (1)

Current in both charging and discharging circuits decreases with time.

37. Answer (4)

38. Answer (1)

Statement-2 is correct and supportive.

39. Answer (2)

R

V V

R

VP

2

P PR R

1 2

1 2

>

<

R1

R2

P1

P2

Bulb 2 glows more brightly.Voltage drop of lesser power rating (larger R) is more.

74 Electromagnetism Success Magnet-Solutions (Part-II)

40. Answer (1)

RAB

< RCD

i1 > i

2

B4

Bi1

A

B3

Di2

RiRi2

2

2

1 B

4 glows brighter

41. Answer (3)

Time constants of both the circuits = CRR

RR.

21

21

but that does not ensure same charge at any time on ca-pacitors or same current in branchs of the capacitor.

42. Answer (4)

i = neAv. i = constant. As v increases, n decreases

43. Answer (3)

However the currents are arranged, force on any wire due to any other wire will be in the plane containing the

wires. Hence force due to any two wires on any of the wire will be in different planes and will not cancel.

44. Answer (4)

Adjacent coils in spring carry current in same direction and attract spring, therefore, tends to contract. Spring

may already be extended and its potential energy may thus be decreased.

45. Answer (1)

If magnetic fields are drawn in the region around a current carrying loop their density is greatest near the cen-

tre of the loop indicating maximum strength at the centre.

46. Answer (2)

A charged particle does not accelerate in a region with electric and magnetic field, if net force on it is zero,

not just the force due to magnetic field. Also, force due to magnetic field may or may not be zero depending

on the direction of motion of charge.

47. Answer (4)

Area of a given flexible loop depends on its shape and hence its magnetic moment depends on its shape.

48. Answer (4)

Restoring torque is proportional to sin (not ) and hence oscillatory motion of the loop will not be simple har-monic.

49. Answer (4)

dlB . may not be zero, if electric field in the space is varying, even when current crossing the loop is zero.50. Answer (4)

Magnetic force )( BVq is zero, as V for electron is zero, but net electromagnetic force will not be zero.

75Success Magnet-Solutions (Part-II) Electromagnetism

51. Answer (3)

An electron being in motion at any temperature in a wire experiences force, but net force on the wire is zero.

52. Answer (4)

Magnetic field does not loose its energy to a charged particle, but requires energy to set up.

53. Answer (1)

Electromagnetic induction leads to repulsion in this case.

53a. Answer (2, 4) (IIT-JEE 2009)

(1) A >

B

iA < i

B

and mA = m

B

hA < h

B

(2) A <

B

iA > i

B

and mA = m

B

hA > h

B

(3) A >

B

iA < i

B

and mA > m

B

hA < h

B

(4) rA < r

B and m

A < m

B

hB > h

A

54. Answer (1)

A time varying magnetic field results in electric fields lines in form of closed loops, which can tangentially ac-

celerate a particle.

55. Answer (4)

Direction of the currents are mutually opposite since in first case magnet and loop are coming closer and in

second case moving apart.

56. Answer (1)

Both statements are correct and statement-2 is the correct reason.

57. Answer (3)

In ballistic galvanometer, the coil is wound on a non-metallic frame, but the justification given is incorrect.

58. Answer (4)

Whatever may be the direction of current, field due to wire will be perpendicular to magnetic meridian and net

field in plane of oscillation of magnet will be same in magnitude.

59. Answer (1)

It is due to electromagnetic induction, current in the current does not want to instantaneously reduce to zero,

if the circuit is unplugged and hence a spark is seen. While plugging there is no initial current in the circuit,

rather the circuit opposes the initial direct increase in current due to electromagnetic induction.

76 Electromagnetism Success Magnet-Solutions (Part-II)

60. Answer (1)

By Lenzs law, eddy currents generated in aluminium pipe are such that they oppose the cause of change in

flux, which in this case is moving magnet. Hence motion of magnet is dampened.

61. Answer (1)

Inductance can be said to be analogous to mass, as it poses inertia to current change in the electrical circuit.

62. Answer (3)

Potential of a charge changes when it is moved along the field line.

63. Answer (1)

1 2

2 M

E M

E

F v vF F

F c

64. Answer (4)

Faster pulling out of sheet out of magnetic field means larger eddy currents in the sheet and larger opposing

force on the sheet. Force will be required to oppose the magnetic force to pull out the sheet (even if it is moved

at constant speed)

65. Answer (4)

Although magnetic force on an electron is zero in the moving frame, but electromagnetic force on it is non-

zero. Therefore, there will be induced e.m.f. in the rod.

66. Answer (1)

Ferrites, being oxides of ferromagnetic materials of high resistivity, are used in transformers as core to minimise

heat losses due to eddy currents.

67. Answer (4)

As the key is switched off, both bulbs come in series and both will die gradually as the current through inductor

does not change instantly.

68. Answer (2)

Induced e.m.f. dt

diL in inductor does not depend on i, but

dt

di

dt

di, may be non-zero even if i is zero. For

alternating current i = i0 sin (t + ), induced e.m.f. is L i

0 cos (t + ), which are completely out of

phase.

Both statements are independently correct.

69. Answer (3)

R

LT . R may be different for the coils.

L of wires does not depends on radius of wire. L = 0n2V. Statement2 is incorrect

70. Answer (1)

XL = .L

Statement 1 is correctStatement-2 is correct explanation of statement-1

77Success Magnet-Solutions (Part-II) Electromagnetism

71. Answer (1)

The correct explanation is that the magnetic field due to a magnet is non-uniform and a dipole experiences a net

force in non-uniform magnetic field.

72. Answer (4)

I leads XL < X

C

LC

1 or < 0

At resonant frequency XL = X

C current and voltages are in phase.

73. Answer (1)

XL V

L > V

C X

L > X

C

CL

1

74. Answer (1)

Load resistance R > internal resistance r power is maximum at R = r, and is larger for smaller R r, step up

transformer decreases R r