SM 15 16 XII Physics Unit-1 Section-G

99Success 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 - G : Integer Answer Type

1. Answer (7)

Cavity can be treated as a combination of + and . However field, due to '' at the centre of cavity is zero.

R

3R Gaussian surface

Cavity

2R

Net field at the centre of cavity is simply due to +. Use Gauss' law3 3

2

0

4(2 )

34 (2 )

R R

E R

0

7

12

RE

n = 72. Answer (2)

Q Q

Q

2Q

1

(2 )0

2

KQ K QV

R R

2

0

( )

2 2 8

KQ K Q KQ QV

R R R R

x = 23. Answer (4)

q = CV

8. 2 4 A

4

dq dvi C

dt dt

4. Answer (2)

Initial K.E. = 21

2mv

Final K.E. =

2 2

1 1

2 2 2 2

v vm m

= 21

4mv

5. Answer (4)

1

3 10 6 3 16

3 6 3V

2

3 10 6 3 4

3 6 3V

V1 V

2 = 4 volt

100 Electromagnetism Success Magnet-Solutions (Part-II)


6. Answer (5)

The equivalent arrangement is shown 10 V 5 V

15 V

Q P S R

RQ

7. Answer (2)

0 04

3

2 2

A A

dd dd

K

k = 2

8. Answer (0)

I = 5 4 3 2 1

1.5 A1 2 3 4

V1 = 1 1.5 1 = 0.5 V

V2 = 2 1.5 2 = 1 V

V3 = 3 1.5 3 = 1.5 V

V4 = 4 4 1.5 = 2 V

V5 = 5 1.5 0 = 5 V

None of the cell has zero terminal voltage9. Answer (2)

Rupper

= 10 Rlower

= 5

So current in lower is i and in upper is 2

i

2

44

2

iH

H5 = i25

4

5

1

5

H

H

5

4

102

5 5

HH cal/sec

10. Answer (0)

11. Answer (4)

Consider as element at a distance x (of thickness dx). Charge on the element



As dq = dxand dI = fdq, f = frequency

We get,

0

2

dIdB

x

= 0 .2

fdq

x

=

0.

2

f dx

x

L

x

+

dx

O

L

+ + +

Net magnetic field is

2

0

2

L

L

f dxB

x

= 0

elog 2

2

f

= 0

log 24

e

0 (4) log 24

eB

0log 2

4e

x , x = 4

12. Answer (3)

Tsin T

T

(dm) R2Bd I

Consider a differential element of length dl on the ring, as shown.

dl = R(2) dm = (R2)

By Newton's second law,

2Tsin = B.R(2)I + R(2) R.2T(2) = B.RI (2) + R2.2 (2)[sin as is very small]

T = BIR + R22

As R = 1 metre, 1 kg/meter, rad/s, I = 0.4A, B = 5.0 tesla

2 215 0.4 1 1 ( )T = 2 + 1 = 3 N



12(a). Answer (3) IIT-JEE 2010

From FBD of a small element on wire

2 sin2

dT idlB

22

dT iRd B

idlB

T T

T = BiR

2

BilT

idlB

T T

13. Answer (3)

Fm = il

B = 1 2 6 = 12 N

fP = ma = 12 N

B

37

xFP = ma

mgcos37

N

ilB F=m

mgsin37

So acceleration of wire

= gsin37 (w.r.t. inclined plane)

= 6 m/s2

14. Answer (5)

M.I. of ring about diameter

2

2 2 2400 10 200 10 kgm 2 kgm2

I

at t = 0 M B maximum angular velocity when | |M B

21

2i f

I v v

= (MBcos90) (MBcos0)

25 rad/s

MB

I

15. Answer (2)

In critical condition is equal to gravitational torqueBx contribute torque

2A

x

Mgi

r B



16. Answer (4)

At any time t

1 2/( 2 )(1 )Rt L L M

VI e

R

at t =

202 A

10

VI

R

L1 = L

2 = M = 5 mH

k

at = 0, key

is closed

t kI

2 2 2

1 1 1

1 1

2 2U LI LI MI

= 3 2 3 2 3 21 15 10 (2) 5 10 (2) 5 10 (2)

2 2

= 10 103 + 10 103 + 20 103

= 40 103 J

U = 4 102 J

= x 102 J

x = 4

17. Answer (1)

X

2 m/s.R = 5

B = 2 tesla

l = 50 cm

x = 5 2 = 10 metre

L = 5 H

I = 1A

charge = B x LI

R

change in fluxCharge =

resistance

= 2 0.5 10 5 1

5

= 10 5

5

= 1 C

18. Answer (9)

Applying Kirchhoff's law

QBLv

C where C = equivalent capacitance

Q = BLCv

v = (2t + 1) m/s.



Q = BLC. (2t + 1)

B = 1.5 tesla

L = 20 cm

C = 10 Ft = 1 s

X30 F

30 F

30 F

V

X10 F VQ = 1.5 0.2 10 (3) C

= 9 C19. Answer (3)

Let q be the charge in capacitor of capacitance C, when current is maximum.

q q0 q+

2C C

L

+

imax

00 0

2

q q q diL

C C dt

0

3 qq3n

20. Answer (7)

3 C/s dqidt

In L-C oscillation total energy is constant.

2 2

2max1 1

2 2 2

q qLI

C C

qmax

= (q2 + LCI2)1/2

= 7 C

21. Answer (8)

24 2cosV t

024 volt

2rms

VV

Irms

= 12 A

In resonance, impedence of circuit is equal to resistance

242

12

rms

rms

VR

I



When connected with battery, the final current

124

2 1

VI

R r amp

Magnetic energy = 21

2LI

= 21 1 (4)

2

= 8 joule

22. Answer (5)

( )M NiA jM

C

8 M j 2 3 TB i j

16 M B kIn pure rolling; C will be instantaneous centre of rotation

So 27

165

MR

716 2 1

5

2

40 rad

7 s

23. Answer (2)

2

0

0

2A

T

T

I dt

dt

Documents

SM 15 16 XII Physics Unit-1 Section-G