PHYSICS

mk MOCK TEST (MAINS) PAGE # 1

http://manishkumarphysics.in/

MOCK TEST-2PART-A

Q.1[Sol. r F

; r and F

So . r 0 and .F = 0. ]

Q.2

[Sol.

130

10m/s

10m

Range R =2u sin 2

g 2(10) sin 60 3

10 5 310 2

. ]

Q.3[Sol. T = m (g a) ; T = 4.9 (10 5) ; T = 15 4.9 ; T = 24N. ]

Q.4[Sol. Momentum = mv

[p] = [M1L1T1]

Plank constant P = h

; h = P[h] = [M1L2T1]. ]

Q.5

[Sol.0 0

1 c

; 20 0

1c

. ]

Q.6[Sol. v2 = u2 + 2as

0 = u2 + 2(a)s

s =2u

2a; s u2

So, distance travelled before coming to rest = 24m. ]

Q.7[Sol. Reduce the energy loss due to eddy current. ]

Q.8[Sol. Escape velocity does not depend on angle. ]

Q.9[Sol. Both will have same reading. ]

PHYSICS

mk MOCK TEST (MAINS) PAGE # 2

http://manishkumarphysics.in/Q.10[Sol. Normal reaction N = 10 Newton

Maximum friction = N = 0.2 10 = 2Nfor equilibium f = weightweight = 2N. ]

Q.11

[Sol. 21K I2

; 2K 1 II '(2 ) I '2 2 8

So, L' = I''I LL ' (2 )8 4

. ]

Q.12

[Sol. v = u + at ; 0 = 6 + ( 10) 10 ; 6 0.06100

. ]

Q.13

[Sol. W = 12

k (x22 x1

2)

W = 12

5 103 (102 52) 104 = 18.75 N m. ]

Q.14

[Sol. T 2 g

Finally 1.21T ' 2g

; T = 1.1 TT

% increment = T T 100%T

= 10%. ]

Q.15[Sol. Rate of cooling Rc = C ( 0) = C()

So n = 1. ]

Q.16[Sol. Second law of thermodynamics. ]

Q.17[Sol. x = 4 cos t + 4 sin t

Amplitude 2 2 2 21 2a a a 4 4 = 4 2 . ]

Q.18

[Sol. 6 540T

25 10 10

T = 16C. ]

PHYSICS

mk MOCK TEST (MAINS) PAGE # 3

http://manishkumarphysics.in/Q.19[Sol. P T3

In adiabatic process P T 1

31

;

32

. ]

Q.20

[Sol.2

2VP VR

Power consumed = 250 watt. ]

Q.21[Sol. Outside magnet N to S

Inside magnet S to N. ]

Q.22[Sol. Fm v Workdone = 0. ]

Q.23

[Sol. R = 2 V 3I 1.5AR 2

. ]

Q.24

[Sol. V = Vshell + VQkq KQR (R / 2)

.]

Q.25

[Sol. g gg

I RS

I I

. ]

Q.26[Sol. m = 9.8 g/m = 9 103 kg/m

T = 10 kg-wt = 10 gNL = 1m

3

1 T 1 10gn2L m 2 1 9 10

n = 50 Hz. ]

Q.27

[Sol. By Gauss Theorem 2 10

q

. ]

PHYSICS

mk MOCK TEST (MAINS) PAGE # 4

http://manishkumarphysics.in/Q.28[Sol. n = 256Hz

Piano wire frequency may be256 +5

256 5or

On increasing tension frequency decreases 2 beats per sec.

T n If256 +5 no. of beats

no. of beats256 5So answer is 256 5. ]

Q.29[Sol. Remains unchanged. ]

Q.30

[Sol. Velocity of wave = 600 2n2 2

= 300 m/sec. ]

Q.31[Sol. Trust = Mass Acceleration = 3.5 104 10 = 3.5 105N. ]

Q.32[Sol. The earth radiates in the infra-red region of the spectrum. The spectrum is correctly given by

Wiens law.]

Q.33[Sol. Protons are not emitted in radioactive decay. ]

Q.34

[Sol. S NS NS NS N

IT 2 2secMB

I 'T 2M 'B

I / 9 T 22 secMB 3 3

. ]

Q.35[Sol. There are 8 particles

4 particles2+ particles

So atomic no. is decreased by = 8 2 = 16and atomic no. is increased by = 4 1 = 4and atomic no. is decreased by = 2 1 = 2So finally atomic no. is decreased by

= 16 4 + 2 = 14Final Z = 92 14 = 78. ]

PHYSICS

mk MOCK TEST (MAINS) PAGE # 5

http://manishkumarphysics.in/Q.36[Sol: Conserving angular momentum, we have

mvR = mvr ...........(1) [v = speed when spaceship is just touching the planet]Conserving energy, we have

21

mv2 = 21

mv2 rGMm

.....................(2)

Solving (1) & (2), we get, R =2/1

2 2

rGMv

vr

]

Q.37[Sol: mu = 2mV2 mV1

u = 2V2 V1

uV2V2u

VV21

1212

Solving 0V1 , 2uV2

time ur4

2/ur2t (B) ]

Q.38[Sol: x = displacement of rod w.r.t. sphere

2R

222R

y = displacement of sphere w.r.t. ground

My = M(x y) 4R

2xy Ans: B ]

Q.39[Sol: mg37cosT0

00 maT37sinT

0maT37tanmg

min0 Ta43gmT

0max a4

39mT

5.21.025Aa 20 Ratio = 2 Ans: A ]

PHYSICS

mk MOCK TEST (MAINS) PAGE # 6

http://manishkumarphysics.in/Q.40

[Sol: Initial charging rate = initial current in the line of capacitor = R5E2

Steady state p.d. across capacitor

V0 = 32

E q0 = CV0 = 32

EC

t = iq0 =

R5E2

32 EC

= 35

RC ]

Q.41[Sol: Velocity of swimmer as seen from ground will be equal to velocity of river and for the car and

swimmer to reach same point in same time car's speed should be same as river speed.

Q.42

[Sol:

tAB = ax2

tAC = a

x22tAD =

a

x32

tAB : tBC : tCD

1 : 2 1 : 3 2 ]

Q.43[Sol: 1.5 sin = 1.6 sin x

sinx = 85

sin (B) ]

Q.45

[Sol: PV = nRT2

2

1

1

nP

nP

But P1 < P2, hence n1 < n2 or m1 < m2(C) ]