Physics Solution FULL TEST Date: 01-4-2015

1.

Sol. (d)LC

f2

1

LC does not represent the dimension of frequency

2.

Sol. (d) Let the car accelerate at rate for time 1t then maximum velocity attained,

110 ttv

Now, the car decelerates at a rate for time )( 1tt and finally comes to rest. Then,

)(0 1ttv 110 ttt

tt

1

tv

3.

Sol. (a) Normal reaction at the highest point

mgrmv

R 2

Reaction is inversely proportional to the radius of the curvature of path and radius isminimum for path depicted in (a).

4.

Sol. (b)

5.

Sol. (b) The diode in lower branch is forward biased and diode in upper branch is reversebiased

Ai505

30205

.

6. (c)Initially the tube was open at both ends and then it is closed. Further

fo =o

v2l

and fc =c

v4l

Since, tube is half dipped in water, lc = ol2

fci =

o o

v vl 2l42

= f0 = f

correct option is (C).7.

A

Cv

v

B v

v

vv

A

B

C

A

B

2v

2vCPure translation Pure Rotation Rolling without

Slipping

+ =

+ =

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Sol. (c) AZ

AZ XX 1

01

8.

Sol. (c) By law of conservation of momentum

221122110 vmvmvmvm

– ve sign indicates that both he particles are moving in opposite direction. Now de-Broglie wavelengths

111 vm

h and

222 vm

h ; 1

11

22

2

1 vmvm

9.

Sol. (d)

10.

Sol. (c) Because induced e.m.f. is given by11.

Sol. (d)

12.

Sol. (d)R

i

Ri

B

2

2

4)2(

400

Ri

83 0

13.

Sol. (d) QQQ 21 ..... (i) and 221

rQQ

kF .....(ii)

From (i) and (ii) 211 )(

rQQkQ

F

For F to be maximum 01

dQdF

221Q

QQ

14.

Sol. (a)

15.

Sol. (c) Velocity of sound in gasMRT

v

MT

v

53

2835

457

2

22

R

R

MM

v

v

H

He

He

N

He

N

16.

Sol. (b)l

AKtQ )( 2111

1

andl

AKtQ )( 2122

2

given21

tQ

tQ 2211 AKAK

17.

Sol. (b) Pressure at bottom of the lake = ghP 0

Pressure at half the depth of a lake gh

P 20

According to given condition

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)(32

21

00 ghPghP ghP 61

31

0

mgP

h 20101010223

50

.

18.Solution: (B) By conservation of energy

21mgR mv v 2gR2

19.

Solution: y = ax- bx2 ;

For maximum y, 2

2

dxyd&0

dxdy < 0

dxdy

= a – 2bx = 0 x =b2a

0b2dx

yd2

2

for x =b2a

, y is maximum.

ymax = ab4

ab2ab

b2a 22

.

20.21.Solution:

60

F 1rF

mgFsin60

Fcos60

N

N = F sin60 + Mg (1)

F cos60 =1r

F = N = (F sin60 + Mg)

F =Mg

cos60 sin60

=

1 3 15 42 3 20N

11 1 32 22 3

=5 4 20N

1

Ans. (A)22.Sol. (B) Angular momentum = (momentum) × (perpendicular distance of the line of action

of momentum from the axis of rotation)Angular momentum about O

mvL h2

… (i)

Now,2 2 2V sin Vh2g 4g

[ = 45°] ... (ii)

From (i) and (ii)

3mL (2 gh)h m 2gh2

Also from (i) and (ii)

600

F

m 3kg

12 3

V2

45V

hh

O

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2 3mV V mVL4g2 4 2g

23.

Solution: (A)

2

escapee e

1 2GM GMmK.E. m2 R R

( )bodyinitially e

1 GMmK.E.2 R=

By Law of Conservation of Energy

Total Initial Total FinalMechanical Mechanical

Energy Energy

( ) ( )surface at height hK.E. P.E. K.E. P.E.+ = +

e e e

1 GMm GMm GMm02 R R R h

{ velocity at maximum height is zero}

eh R24.

Solution:(C) 2Fl 1 1y or l ; l

l D

22 1

2 121 2

l D 4 or l 4 l 4cml D

25.

Solution:(B)In this case, T- mg cos 20 = centripetal force

T> mg cos 20

26.

Ans. (C)

Solution: TnR27dTnCQ P ,

R

27CP

TnR25TnCU V ,

R

25CV

and TnRUQW

2:5:7W:U:Q

Hence, C is correct27.

Solution: P.d. across the points = r.E

V2 – V1 = )k3j2i).(k4j3i2( = 2 6 + 12 = 4 volts.

(C)28.

Solution : Since volume of the wire does not change.

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1A1 = 2 A2 where 1 and A1 are the initial length and cross-section of the wire, and

2 and A2 are the final length and cross-section.

1

2

2

1

AA

l

l. . . (1)

R1 = 1

1

A

land R2 =

2

2

A

l

2

11

l

l

l

l

l

l

21

2

2

1

2

1

AA

RR

. . . (2)

22

21

l

l

2

1

RR

as 2 = 31

R2 = 9R1 = (9 4) = 36Ans. (d)

29.

Sol. (B) When a charged particle is moving at right angle to the magnetic field than a forceacts on it which behaves as a centripetal force and moves the particle in circularmotion.

2

A AA

m v q.v B2r

A Am v qB2r

Similarly for second particle moving with half radius ascompared to first we have

B Bm v qBr

B B B Bm v m v2r r

mA vA = 2mBvB

mAvA > mBvB

30.Solution: (A)

BA5mH15 V1

B AV e 15 I 1 V+ - + ´ = VB – VA = 15 – e – I

Here, I = 5A, 3 3d dIe L 5 10 10 5V

dt dt

VB – VA = 15V

A B