SM 15 16 XII Physics Unit-1 Section-B

44 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

Section - B : Multiple Choice Type Questions

1. Answer (1, 2, 3)

(i) Between inner and outer sphere E is absent hence potential is constant.

(ii) When we connect a body to earth, potential becomes zero.

(iii) Vinner sphere

= 0 = R

kQ

R

kq

2

++

q

R2R

Q

s

2

Qq

2. Answer (1, 2, 3)

(i) Charge will appear on outer surface of conductor.

+

+ +

(ii) At the surface of conductor surface charge density () )( curvature of radius

1

R

Surface charge density is not uniform for irregular shape

(iii) Electric field is zero inside conductor.

3. Answer (2, 4)

3 = 4y x

y

x

z

Electric field vector is perpendicular to the surface, there are infinite

planes which cut xy plane in line 3y = 4x. All vectors in form of

)ˆˆ3ˆ4( kcijCE are perpendicular.

4. Answer (1, 2, 3)

All motions straight line, circular and helical are possible. Examples are like these

Straight line If particle is at rest and electric field is uniform.

Circular Motion of negatively charged particle around positive infinite line charge.

Helical Motion of negatively charged particle around positive infinite line charge as shown in diagram

v1

v2

wireof axis along motion for eresponsibl

velocity tangential

2

1

⇒

⇒

v

v

45Success Magnet-Solutions (Part-II) Electromagnetism


5. Answer (1, 2, 3, 4)

Q1

Q2

E Point P

(0, 0, 0) ( , 0, 0)a

( /2, , 0)a a

y

If Ex at P is zero, then magnitude and sign of charges must be same.

Field at ⎟⎠

⎞⎜⎝

⎛0,0,

2

a

is zero while on point (2a, 0, 0) and (–2a, 0, 0) E field is along x axis while at point ⎟⎠

⎞⎜⎝

⎛a

a,0,

2 E

is along z axis.

6. Answer (2, 4)

z

vE

y

vE

x

vE zyx

;;

Ey and E

z are zero. So change in potential along y and z-axis are zero

∫∫ dxEvx

y

z

x

1 m

O2

2

0xEv + constant (non-linear)

Electric flux through the cubical volume 000 )0(1 EEE

Gauss theorem implies

0

0E

qE

00Eq

7. Answer (1, 2, 3)

At a point, two potentials are not possible.

8. Answer (1, 2, 4)

Induced charge per unit area of the outer surface, 21

4 b

Q

Electric field at the outer surface 2

00

1

4 b

Q

Electric field at inner surface = 2

04 a

Q

If inner or outer surface of the sphere is earthed, it comes to zero potential. If a charge is brought from infinity

to the sphere, no work should be done charge on outer surface of sphere is zero.



9. Answer (1, 3, 4)

As dielectric constant varies from left to right, strength of electric field will be difference at difference points on a

line to plates.

If we draw a gaussian surface, flux will not be zero so charge enclosed will also be non-zero.

10. Answer (2, 4)

Capacitance,

d

AC

0

If d increases, C decreases. Potential difference V is constant

Charge, Q = CV decreases

Electric field, d

VE decreases.

Energy stored in capacitor, 2

2

1CVU decreases. Energy flows back to the source of voltage and/or gets

dissipated as heat

Also, work done by the external agent goes to the source of voltage and/or gets dissipated as heat.

11. Answer (1, 2, 3)

+2 – –

O q1

– q1

q2

– q2 O

Option 1 in modified charge on outer surfaces of 1 and 3 plate will not be changed

O + q1 – q

2 + O = + Q [Conservation of charge on plates 1 and 3]

q1 – q

2 = Q ...(i)

V1 – V

3 = 0 =

1 2

0 0

q d q d

A A

...(iii)

By solving these two equations q1 = q

2 =

2

Q

12. Answer (3, 4)

Before removing cell,

Ceq

= 3

2C

Charge on each capacitor

3

2 C

After removing cell and reconnecting the two capacitors, charge on capacitor with capacitance C

02

CCC

QQ

Similarly charge on capacitor 2C = 0

So, all energy stored in capacitors is lost.

13. Answer (1, 4)

C1 = 4

0 (1.5) a = 6

0a

C2 = a

a

a

aa

aa

0

2

00 12)5.0(

)5.1()4(

)5.1(

5.14

C3 = a

a

aa

a

0

0

2

0 185.0

5.15.14

)5.1(

5.15.14

C3 > C

2 > C

1



14. Answer (1, 3)

15. Answer (1, 2, 3)

S is open S is closed

V 2V

+–

q CV =

V 2V

CV–CV +CV

+CV

–CV

Work done by V = CV2

Work done by 2V = 0

Heat dissipated

2

2 2

2

CVCV CV

2

2

CV

16. Answer (1, 2, 3)

Charge will flow only when the circuit is closed, i.e., all keys are closed.

Initial charges q1 = 100 C and q

2 = 80 C

If a charge q flows anticlockwise, new charges are 1

100 –q q and 2

40 –q q

KVL 100 – 40 –

01 2

q q q = 80 C

New charges are 20 C and – 40 C capacitor C2 does not lose energy finally

17. Answer (1, 2, 3)

By ohm’s law i

VR and if increase voltage across voltmeter, current divides and resistance of circuit gets

changed.

Voltmeter R1, Ammeter R

2

321

1RRR

RV

32

1

11

1

RRRr

rRRr

rRV

A

V

R3

r ))(( 3211

1

RRRrrR

rR

)(1 321

1

1

RRr

RR

R

⎟⎠

⎞⎜⎝

⎛

)( 321

321

1

RRr

RRRR

R

VV



18. Answer (2, 3)

For maximum current

5.22

m

n

nEI

2.5

nE

nE

m

mn

E

mn

mnE

5

90

5

2

For maximum current 0545

–2

m

m = 3 & n = 15

19. Answer (2, 3, 4)

If resistances of A and B is R, resistance of C is 2R.

Power consumption in C, R

VPC

2

2

Power consumption in B, R

V

R

VPB

4

)2/( 22

Power consumption in A, R

V

R

VPA

4

)2/( 22

20. Answer (1, 3)

∫ ∫ ⎟⎠

⎞⎜⎝

⎛

R L

A

dx

xL

LdR

0 0

0 )2ln(0

A

LR

i = current = )2ln(0L

VA

R

V

Current density, )2ln(0L

V

A

iJ

= constant

J = 6(sigma) E E = J

)(2ln)2ln(0

0

xL

V

L

V

xL

LE

⎟

⎠

⎞⎜⎝

⎛

at x = 0 ; L

VE

)2(ln

at x = L ; L

VE

)4(ln

21. Answer (1, 4)

Random motion of electron depends on temperature.

Average speed is non-zero for any moving object. In absence of current, net flow of charge is zero.

22. Answer (1, 4)

Points B and D are at same potential. Equivalent circuit diagram is as follows:

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

AB,DR/2

R/2

R/2O

R

R

C

Equivalent resistance between O and A,

15

7RR

eq



23. Answer (1, 3)

Given that after every one second, potential or charge falls by a factor of 0 × 8.

After 2 second, Q = 0.64Q0

U = (0.64)2U0 (∵ U Q2)

Loss by a factor 2 369

1– (0.64)625

Potential V = 0.64 × 100 V = 64 V

0 00.8

⎛ ⎞ ⎜ ⎟

⎝ ⎠

t

q Q Q e at t = 1 s

1

ln(5 / 4) second

24. Answer (1, 2, 3)

If K1 is closed, no current flows through this wire reading of ammeter does not change,

when both K1 and K

2 are closed, even then no current flows Ammeter reading remains same.

When all the keys are closed, the middle part of each wire is shorted Req.

decreases

reading of ammeter increases.

25. Answer (1, 3, 4)

A4

I = 0.3 A

1.5 V, r

1.50.3

4

r

r = 1 Cell is non-ideal (1) is correct

Now, with a 4 resistance, current through the cell

I 1.5 V, 1

A4

4

I

2

I

2

1.50.5 A

2 1

I

Current through ammeter is 0.25 A

Power dissipated in Ammeter = 0.252 × 4 = 0.25 W

in cell = 0.52 × 1 = 0.25 W



in 4 resistance = 0.252 × 4 = 0.25 W

One third power is dissipated in cell (3) is correct

Reading of voltmeter 1.5

4 1.2 V4

r

(4) is correct

26. Answer (2, 3, 4)

Accuracy is proportional to length and does not depend on material. Also, accuracy is independent of cross-

sectional area. Meter bridge works on Wheatstone bridge.

27. Answer (2, 3)

xxxxxx

0

V

B =

A

B

C

B = 0

q60º

B = 0

xxxxxxxxxxxxxxx

In regions (A) and (C) charged particle will go undeflected

In region (B)

Path will be helical

28. Answer (1, 4)

a = 1

2

2

12

m

T qB

mT

qB

b = 1

2

sin37º

3

sin53º 4

mv

R qB

mvR

qB

c = 1 1

2 2

cos37º 4

sin37º 3

P v T

P v T

3 4

1 14 3

abc

a bc

29. Answer (2, 3)

Magnetic field due to a moving charged particle is given by

3

0

4 r

rvqB

��

�

ir ˆ05.0�

; meter

v

�

should be such that rv

��

should point along positive y-axis.



30. Answer (2, 3)

If 21 andVV are parallel, then deviation produced by

21 and BB are opposite

2

22

1

11 sinsin

r

d

r

d ⇒

1V

2V

d1

d2

1 = –

2

sin1 = –sin

2

2

2

1

1

2

2

1

1

qB

mV

d

qB

mV

d

r

d

r

d ⇒

d1B

1 = d

2B

2

1

2

2

1

2

1

B

B

d

d

r

r

31. Answer (2, 3)

Force on the wire at the instant shown in figure,

iilBF ˆ;�

kBilriilBjr ˆˆˆ �

kBilr ˆ

external�

BilrkkBilr )ˆ()ˆ(Powerext

��

32. Answer (3, 4)

V

d

If charge particle is positive

For turning

d > x

d > 2

rr

qB

Vmd

2

d

x

30°

60°C

1

qB

Vm

qB

mV

2

VV 2



If it is negative

2

3

2

rrrx

For turning

d > x

2

3rd

d

x

rr/2

30°

qB

Vm

qB

mV 2

3

VV 3

2

32(a). Answer (1, 3, 4) IIT-JEE 2008

When radius r > l, particle will move to region III.

qB

mV > l V >

m

Blq

when V = m

Blq, the particle moves in the biggest semicircle possible in region II.

Time spent t = Bq

m in region II, provided particle returns to region I.

32(b). Answer (1, 3) JEE (Advanced)-2013

Mt

qB

Clearly θ = 30° = 6

C

F

x L=

ˆ2 j

�

v

ˆ2 3i

ˆ4ix = 03

100 50

6 36 10 10

M M MB

q Qq

B

�

must be in –z direction.

33. Answer (1, 4)

Take a cubical Gaussian surface and find flux through it

0∫

qsdE�

�

ixEE ˆ

0�

implies there is distribution of positive charge in the region which is possible

0∫ sdB�

�

(Gauss law for magnetism)

This is not valid for ixBB ˆ

0�

34. Answer (1, 2, 3)

qB

mKr

2

For He+2 and O+2 radius is same, neutron is neutral hence turning of neutron is not possible.



35. Answer (1, 4)

For equilibrium,

0 1 2 0 1 2 0 1 2,

2 2 2

i i i i i iF l Mg r

r Mg mg

⇒

�

Force on the moving charge is perpendicular either left or right.

36. Answer (1, 2, 3)

(Using Faraday's and Lenz's law)

37. Answer (1, 3, 4)

i = 10e–2t

–2

– 20

tdie

dt

At t = 0

i

A 5 V B2

At t

10

A 5 V B2 Ldi

dt

= 40 V

– 20 – 5 40 A B

V V

– –15 V

A BV V

At t 0, 0⎛ ⎞ ⎜ ⎟⎝ ⎠

dii

dt

VA – V

B = 5 V

U = 21

2Li = 0

38. Answer (1, 2, 3)

22

2

2

22

2

2

BbBaVV

BbVV

BaVV

AO

AD

DO

B C

A

b

a

OD

B

VOA

can never be zero.

2

)2( 2 dcaBVV CO and it is independent of ‘b’.



39. Answer (1, 3, 4)

Lxir

dt

diLir

rA

A

i r

C

QEVV

EC

QVV

AC

AB

2

1

2

A

B

BE

C

C

C1

i

E

B

Lx

Q1 = CE

Q2 = 2CE

1

2

1

2 Q

Q

40. Answer (1, 2, 3)

Bulb 2 dies as soon as key is switched into 1, because there

LR1

B1 B

2

21

is no induced e.m.f. Initially current in L is 2R

E

, total heat

developed = energy stored in solenoid =

2

22

1

⎟⎟⎠

⎞⎜⎜⎝

⎛

R

EL

41. Answer (1, 3, 4)

Loop tries to maximise its area and time required is very small and final shape is circular and Fnet

is zero on loop,

center of mass is at rest.

Ai = ab

2

⎟⎠

⎞⎜⎝

⎛

ba

Af

R

Bab

ba

R

ABQ

⎥⎥⎦

⎤

⎢⎢⎣

⎡

–

)( 2

42. Answer (2, 4)

R

Ei

Edt

diL

max

i

tFor Circuit 2 Circuit 1

dt

di

L

E 33

13

12

⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛

dt

di

dt

di

R

E2i

0i0

R

L

3

2

3 2

3

2

1



43. Answer (1, 2, 3)

i3

i1

i2

+q

–qF

Bi3

i1

i2

–q

+q

BR

Bvl

at = 0t

V

44. Answer (1, 3)

22a

dt

dBae

22

2a

a

a

R

ei (anticlockwise)

Electric field lines are circular and PR is along the diameter of circular field lines. Hence electric field along

PR is zero. So, no current flows in PR.

45. Answer (2, 3, 4)

If resistor is present in a circuit, power will be always dissipated, capacitor stores electrical energy, inductor

stores magnetic energy.

46. Answer (2, 3, 4)

For purely resistive circuit, instantaneous power is P = V0I0sin2t this expression has never zero average value

for any interval of time.

47. Answer (2, 3, 4)

222

222

)(500

)(

CLR

CLR

VVV

VVVV

R L C

VR < 500

VL ~ V

C < 500

Hence only first option is not possible

48. Answer (1, 2, 3)

In steady state

25 Ai

4

2 100 V

Shorted

–3

2

14 10 625

2L

U

= 1250 × 10–3

= 125 J

Potential difference across L2 at (t )

0di

Ldt

� � �

Documents

SM 15 16 XII Physics Unit-1 Section-B