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# 2012 Sample Physics Solution 2

Apr 03, 2018

## Documents

Yash Paunikar
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• 7/28/2019 2012 Sample Physics Solution 2

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I ITJ EE PHYSICS SAMPLE PAPER - ISOLUTIONS

SECTIONI

Straight Objective Type

1. uM

vM

Tdt22

C

Tdt

v

..... (i)

gll

gu2

2

MvTdt v

Tdt

A

T1

M ..... (ii)

MaMgT1

Mg

T1

B

a..... (iii)

al

v2

From (i) and (ii) uM

vM

22

3

3

uv

999

2 g

l

gl

l

ua .

(b)

2. All the equipotential surfaces of the field between the

sphere and the plate are convex down ward. Hence on

any straight line parallel to plate the points farther from

the sphere will have potential lower than those closer to

sphere. 1 2

V3V2 V1

(b)

3. Torque about O is zero as well angular momentum hence = 0.

(d)

4. )(2

1)(

2

1

2vvMyHMg 21

2

1Mv

wIvM cc2

1)(

2

1 2 .....(i)

cMvMv12 ..... (ii)

vvc2 ..... (iii)

yvv1

Solving (i), (ii) and (iii) )(4

3 2

yHgv g

H

tyH

g

dt

dy

v

3

)(3

4

(d)

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5. Both upper half and lower half will have same effective area of

2

2

R so charge in flux will be same and induced emf will have

some value. But since the resistance is different due to which

current must be different but ring is as a whole is closed circuit

so electric field will be generated to make the current flow in

both parts to be same.

010riRE ..... (i)

0riRE ..... (ii)

092 irRE

R

ir

E 2

9

r

bR

r

E

i 1111

2

r

bR

R

rE

112

9 2= Rb

22

9

10 r

i10r

E

+

E

+

(b)

6. twwtatwaE 00 coscoscos )]cos()[cos(2

cos 000 wwwwa

twa

Highest possible energy for photon corresponds to frequency 0ww hence.

Qww

hKE

2

)( 0max

(c)

7. ghhrWST2 ,

2

2 ghhrWg

Heat = Q = gST WW =2

22 ghrand

gr

Sh

2

g

SQ

22

(a)

8.svhat

22

1 sp vh

vat2

21 tt

1

2

2

s

p

s

p

v

v

hv

v

a

(a)

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SECTION II

Reasoning Type

9. PV= nRT

V

nR

T

P

)(

1

slopeV

P

T

12

v2 = constant

v1 = constant

(b)

10. Potential atEandKare different due to which current flows betweenEandKmake current

flow betweenAB and CD also possible.

(d)

11. Assertion and Reason correct and correct explanation.

(a)

12. Acceleration relative to cart parallel to incline is always zero only the acceleration

perpendicular to incline will change in different situation due to which change in tension but

angle will remain same and string always remain perpendicular to the incline.

(d)

SECTION III

Passage-IWhen the temperature of rods in increased there will be increase in their lengths and thereby

the springs are compressed, let21 , xx and 3x be the compression in the three springs

respectively. ThenL

Kx1 2Kx2 2Kx2 3Kx3

2212

xxxTL

TL

21 2KxKx and 32 32 KxKx

321 32 xxx

TLxx

x2

3

32

111

TLx

11

91

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2

2

1111

9

2

1

2

1TLKKxE 222 )(

242

81TKL

Similarly

222

2484

81TKLE

222

3242

27TKLE

13. (b)

14. (a)

15. (a)

Passage-II16. The charge q on the small disc can he calculated by applying Gauss law

qdsE 0

Since one side of the small disc is in contact with plate

xvVd

rrEq

2

0

2

0 d

rx 0

2

.

(b)

17. Let sv be the steady velocity of the small disc just after its collision with the bottom plate

then the steady state kinetic energy sk of the disc just above the bottom plate is given by

2

2

1ss mvk

for each round trip disc gains electrostatic energy by

U= 2qV

afterbeforeloss kkk beforeke )1(2

afterk112

.

Since sk is the energy after collision at the bottom plate and )( mgdqVks is the

energy before the collision at the top plates total energy loss can be written as

)()1(1

1 22 mgdqVkkk sstot

In its steady state Ushould be compensated by totalk

)()1(11

2 22

mgdqVkkqV ss

])1()1[(1

22

4

2

mgdqVks

mgdqVmvs 2

2

2

22

112

1

)2(1

2

1 22

2

2gd

m

xVvs

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2Vvs .

(c)18. The disc will lose kinetic energy eventually cease to move when the disc cannot reach the

top plate. In other words, the threshold voltage cV can be determined from the condition

that velocity of the disc at the top plate is zero. In order to have velocity zero at top plate

kinetic energy at top plate must satisfy the relation

0mgdqVkK css

sk is steady state kinetic energy at the bottom plate

011 2

2

2

2

mgdqVmgde

eqV

e

ecc

xmgd

eeVc 2

2

11

(d)

SECTION IV

Matrix Match Type

1.

Concave Convex

(A) m < 0 |m| < 1

Object as well image is real in case of concave mirror object and image is virtual

incase of convex.

(B) m < 0 |m| > 1

Both real in case of concave both virtual incase of convex.

(C) |m| < 1 m > 0

Concave object virtual image real.

Convexobject real image virtual.

(D) |m| > 1 m > 0

Concave object real image virtual.

Convex object virtual, image real.

(A)2, 4; (B)1, 3; (C)2, 3; (D) 2, 3

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2.

q, v, m

y0O

F

B

2B

x

R/2

MA

Ry0B

R

C

D

qB

mvR ; 1cos 0

R

yR

= 2 ;)2( Bq

mtAB ;

)(Bq

mtBC

))1((cos3)3( 1

qB

m

qB

mttT BCAB

sinRXA ; sin2RABXX AB

sin4RBCXX BC

sin5RCDXX CD

When velocity be come parallel

qB

m

qB

m

Bq

m

qB

mtt AMFA

2

33

22

(A)2; (B)1; (C)3, 4; (D)1, 2

SECTION V

Subjective or Numerical Problems

1. 2Tcos = W+ 2w

2Ncos (90

) = W2Nsin = W

T AB sin 2 = cossin2

1OBNABw

T 4 sin 2 = w 2 sin +N rcot

(AB = 4 m ; OB = rcosec )

Solving from above r= 3m

D

2

90

CA

90

O

NN

w w

W

B

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2.a

xay seclog ; w

td

d(Constant)

a

x

dx

dytan ;

a

x

yd 22

2

sec1

2

2

2 23

1

dx

yd

dx

dy

=a

xa sec

a

x

dx

dytantan

a

x; x = a

awdt

da

dt

dx; 0

2

2

dt

xd; aw

a

x

dt

dx

dx

dy

dt

dytan

dxaa

xaw

dt

yd 1sec

2

2

2

a

xaw 22 sec

Now resultant acceleration

2

2

22

2

2

dt

yd

dt

xda

a

xwa 442 sec0

22

22 sec Ra

w

a

xaw

a

a

4sec

2

1

2

1

22 22

= 2m/sec424

18

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