﻿ The moment of inertia of a uniform cylinder of length from Physics Class 12 JEE Year 2017 Free Solved Previous Year Papers

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# JEE Physics 2017 Exam Questions

#### Multiple Choice Questions

1.

A slender uniform rod of mass M and length

C.

Taking torque about pivot τ = Iα
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2.

The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius)

B.

Where M is mass of earth

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3.

The moment of inertia of a uniform cylinder of length

• 1

C.

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4.

A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

A.

Velocity at any time t is given by

v = u + at

v = v0 + (–g)t
v = v0 – gt

Straight line negative slope.

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5.

A time-dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be

• 9J

• 18 J

• 4.5 J

• 22 J

C.

4.5 J

F = 6t = ma
⇒ a = 6t

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6.

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is

D.

By conservation of linear momentum
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7.

The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the numbers of molecules in the room before and after heating, then nf – ni will be

• 2.5 x 1025

• -2.5 x 1025

• -1.61 x 1023

• 1.38 x 1023

B.

-2.5 x 1025

Using ideal gas equation
PV = nRT
(N is number of moles)
P0V0 = niR × 290 ...... (1)
[Ti = 273 + 17 = 290 K]
After heating
P0V0 = NfR × 300 ...... (2)
[Tf = 273 + 27 = 300 K]
from equation (1) and (2)

putting P0 = 105 PA and V0 = 30 m3
Number of molecules nf – ni = – 2.5 × 1025

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8.

The following observations were taken for determining surface tension T of water by the capillary method :
Diameter of capillary, D = 1.25 × 10–2 m rise of water, h = 1.45 × 10–2 m
Using g = 9.80 m/s2 and the simplified relation T = (rhg/2) x103 n/m, the possible error in surface tension is closest to:

• 2.4%

• 10%

• 0.15%

• 1.5%

D.

1.5%

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9.

A particle is executing simple harmonic motion with a time period T. AT time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like

B.

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10.

A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is 1/8 mv02,the value of k will be

• 10-4 kg m-1

• 10–1 kg m–1 s–1

• 10-3 kg m-1

• 10-3 kg s-1

A.

10-4 kg m-1

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