SM 15 16 XII Physics Unit-1 Section-E

7/25/2019 SM 15 16 XII Physics Unit-1 Section-E

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78 Electromagnetism Success Magnet-Solutions (Part-II)

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Section - E : Matrix-Match Type

1. Answer A(p, q), B(p, r, s, t), C(p, r, s, t), D(r, s, t)

(A) At (a, 0, 2a ) and (a, 2a , 0) ipE 0. Ep

(B) At (a, 0, 0) and (0, 0, a) jpE and hence Ep is maximum

(C) E at (a, 0, 0) is in +x direction

E at (0, 0, a) is in x direction

(D) At (a, 0, 0) and (0, 0, a) E is parallel to ip , hence Ep is zero.

(E) At (0, a, 0)

E

is in xdirection

2. Answer A(r), B(r), C(q), D(q)

Case I : Saturation charge = C 2E= 2 CE

Maximum energy stored in capacitor = 2 CE2

Heat generated in resistor = 2 CE2

Energy supplied by battery = 4 CE2

Case II : Saturation charge = 2 CE

Maximum energy in capacitor = 2 CE2

Energy supplied by cells = CE2+ CE. 2E = 3 CE2

Heat generated in resistor = 3 CE2 2 CE2= CE2

3. Answer A(p, t), B(q, s), C(p, s), D(r, s)

R

OA

Work done in rotating the dipole equals change in potential energy of the system. Potential of the centre should

be zero, as metal is earthed. Electrons will flow from or to earth to satisfy this.

4. Answer A(p), B(p), C(r, s), D(p)

C2= kC

2(increases)

V2< V

2, V

1> V

1

E1> E

1. p

Force between plates =0

2

2 A

Q

C> CQ> Q

Force increases

Potential difference between pointsXand Yis equal to e.m.f. of cell


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79Success Magnet-Solutions (Part-II) Electromagnetism


Ratio of energy stored =2

1

C

C

C2> C

2 Ratio increases

5. Answer A(p, r, s), B(t), C(p, r), D(p, q, r)(A) Plane is equatorial

30

1 14 2

b

a

kp qd r dr

a br

(B) E ds

(C) y= 0 is equatorial

(D) x= 0 is axial and some points are between charges.

5a. Answer (1) [JEE (Advanced)-2014]

P. Not along +y Q. Not along +x

Q1

Q2

Q3

Q4

Q1

Q2

Q3

Q4

R. Not along y S. Not along x

Q1

Q2

Q3

Q4 Q

1 Q

2 Q

3 Q

4

6. Answer A(p, q, s), B(r, t), C(s), D(p, r, t)

RC

t

RC

t

eR

EeII

0

RC

tlI 0lnln

RC

1cot

(A) > RC> RC

(B) < RC< RC r, t

I0> I

0 R< Rs

I0< I

0 R> Rp, r, t


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7. Answer A(q, r, t), B(q, s, t), C(q, s, t), D(q, s, t)

(A) Electric field inside parallel plate capacitor is uniform. So, energy density 20

2

1EUE is non-zero and

uniform.

(B) Electric field between the shells is due to charge on inner sphere which behaves

like a point charge. So, its electric field is non-uniform.

(C) Electric field between the shells is due to charge which has flowed from earth to

inner sphere. It also behaves like a point charge whose electric field is non-uniform.

q q 1+q

1

q1

(D) In the region outside the outer shell, there is electric field due to charge (q q1)

which behaves like a point charge.

8. Answer A(p), B(p, q, r, t), C(p, t), D(p, q, r, t)

(A) VA

= VB

BA

R1 i

1

V1

V3

V2

C1

R2 i

2

R3

R4

(B) Depending on values of parameters shown VA V

B

and VB VCmay be +ve, ve, zero.(C) As potential difference across the plates is same, V

A V

B= 0 is V

B V

C> 0

(D) Depending on values of k1, k

2, k

3V

A V

Bmay be +ve, ve, zero V

B V

C> 0

8a. (1, 4) [JEE (Advanced)-2014]

1 2C C C

0 0

1 2

/ 3 2 / 3,

K A AC C

d d

0( 2)

3

K AC

d

1

2C K

C K

Also,1 2

,

VE E

d where Vis potential difference between the plates.

9. Answer A(p, q, r, t), B(p, r, t), C(p, q, r, s), D(q, t)

5

R

20

2.5 V

Q

i

+QQ

P

ai 1.025

5.2

V251.05.2 QP VV

p, q, r, t

2 V

R

i

Q

P

i= 0

VP V

Q= 2V

p, r, t


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10

0 V0 V2 V 2 V

Q+ Q2 3 1R1

PQ

i

3 V 1 V

Ai 1.010

12

VP V

Q= 2V

Current through R= 0

Q= CV = 2 1 = 2 C

p, q, r, s

+ Q Q

Q

i

P

i= 0.1 A

VP V

Q= 4 4

0 0.1 = 0 V

Q = CV = 0

= 2 2 = 4C

10. Answer A(r, s), B(q, r, s), C(q, r, s), D(p, q, r, s, t)

(A) is property of material, for conductors it decreases with rise in temperature.

(B) Thermal energy generated in unit volume =

2E

(C) J= E

(D)

AVAEI

11. Answer A(q, r), B(q, r), C(q), D(q)

(A) In parallel = 2x

In series =x

(B) In parallel =x

In series = 2x

(C) Key closed = 0

Key open =x

(D) Key closed =x

Key open =x

12. Answer A(p, s, t), B(s), C(p, q, r, t), D(q)

(A) For current in E2to be zero 1

2

1

E RER r

(B) For maximum power in R, 1 2

1 2

r rR

r r

(C) V1= V

2always

(D) For current in E1to be zero 21

2

E RE

R r


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12a. Answer (1, 2, 4) [JEE (Advanced)-2014]

Using KVL, inABCDEFA, we get

1 2 3 1

0iR V iR V

1 2

1 3

V Vi

R R

Using KVL inABCDA,

1

1 1

1

0 0 V

V iR i R

V1

V2

R1

R2

R3

B

A D

F E

iC

1 1 2

1 1 3

V V V

R R R

1 1 1 3 1 1 2 1

V R V R V R V R

1 2

1 3

V V

R R

Now possible answers are 1, 2, 4.

13. Answer A(r, s, t), B(r, t), C(q), D(p, q, r, s)Thermal power in maximum when R = r.

Maximum power =r

E

4

2

r R= R

P

For any other value Power mg sin respectively.


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(C) jilBF 0

A component of this force is down along the incline. Case is similar to part (A).

(D) jilBF

0

A component of this force is up along the incline. Case is similar to part (B).

23. Answer A(p), B(s, t), C(q), D(r)

(A)

2

0 .4

tan.4

sin.4 L

L

(B)L

L2

2

0

(C) 3

2

2

2

10

2

RRL

(D) L= does not depend on

(p)2

2

0

2

LL

(q)3

2

2

2

10

4

RRL

(r) L= does not depends on

(s)

L

L

2

2

0

(t) Area of cross section of smaller solenoid l2

24. Answer A(p, t), B(p, t), C(q, s), D(q, r)

(p) Flux through the loop is not changing

(q) Flux through the loop is not changing

(r) Flux through the loop is increasing

(s) Flux through the loop is decreasing

(t) Area of cross section of smaller solenoid 2

25. Answer A(p, q, r, s, t), B(q, r, t), C(p, q, r, s, t), D(q)

26. Answer A(p), B(p, q, r, t), C(p), D(q, t)

(A) VA V

C= Zero

(B) Field is nonconservative

Field Lines

a b

cd

(C) Electrostatic field = 0

(D) Electric field at C, positive


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26a. Answer (4) (IIT-JEE 2009)

According to Lenz law I1 is from ato band I

2from cto d.

27. Answer A(p, s), B(r), C(q), D(r, s)Since no current flows, rod is in uniform motion

28. Answer A(p, r, s, t), B(p, r, s, t), C(p, r, s, t), D(p, r, s, t)

In all the physical situations given in columnI, all the possibilities arise.

(A) = 60, VA

VB= 0

> 60 VA

VB= +ve

< 60 VA

VB= ve

(B) Depending on field is uniform or non-uniform induced current imay be positive, negative, zero.

(C) The rod may be moving in any direction or may be moving with constant speed.dt

dqmay accordingly

be +ve, ve or zero.

(D) Induced emf is such a coil varies sinusoidally, may be zero at an instant and may have any polaritydepending on sense of rotation of the loop.

29. Answer A(p), B(q, t), C(r), D(s)

(A) 104

40

R

VI

R

AI 2100

(B) V0= I

0. Z= 250 V

(C) V502

0 V

V

VL= 40

VC

VR= 40

V2 = (VL V

C)2+ V

R

2

VC= 10 V

(D) 110

10

I

VX

c

c

Vc= 10 volts because V= 50 volts

12 2,

2L C

X X

2 2

100L

,

1

'C

XC

1 100

2C

2

100C

1 1

2 2 2 22

100 100

CL

125Hz

22

100


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30. Answer A(p, t), B(q), C(p, t), D(q)

e.m.f. leads current

XL> X

C

CL

1

On decreasing ,XLX

Cdecreases

Zdecreases

(A)Z

VI

0

0 increases

(B)

Z

R1cos decreases

(C) P =Z

V2

increases

(D) Zdecreases

Incase of resonance, current amplitude and therefore power developed is maximum

31. Answer A(p, r, s, t), B(q, r, s), C(p, r, s, t), D(p, r, s, t)

(A) XL=X

C resonance

A110

10

Z

VIRZ , I

1= 1A

VL= I X

L= 10 V

VC= IX

C= 10 V

(B) ZXX CL

VR= 0 But V

L=VC= 10 V

I= 0, I= 0

(C) All potential is dropped across Land CVL= V

C= 10 V

22

11

1

2

1

RXX CL

XL= X

C

Z = R

A110

10

Z

VI

VL= V

C= V

R= 10 V, 1

101

10I A


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(D) XL= X

C

Equivalent to case in part - A

I1= 1A

31(a) Answer A(r, s, t), B(q, r, s, t), C(p, q), D(q, r, s, t) IIT-JEE 2010

(p) V1= 0, V

2= V(XL= 0)

(q) XL= 0, V

1= 0, V

2= V

(r) XL= L= 2fL= 1.884

V1= IX

L

V2= IR

V1< V

2

(s) XL= 1.884

XC=

31 1010

3L

XC>> X

L

V2= IX

C, V

1= IX

L

(t) R= 1 k

31010

3CX

XC> RV

2> V

1

Also, V2= IXL, V1= IR.

SM 15 16 XII Physics Unit-1 Section-E

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Transcript of SM 15 16 XII Physics Unit-1 Section-E