Physics Test 157

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Physics Test 157

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Question 1
4 / -1

A 2 m rod of mass 1 kg rotates at an angular speed of 15 rad/s about its ends. Then K.E. is

A
150 J

B
1500 J

C
100 J

D
1.5 J

Solution

Given,

l=2m

m=1kg

ω=15rd//s

K.E.=1/2I ω

=1/2x(ml²/3) ω²

=1/2x(1x4/3)(15)²

=150J
Question 2
4 / -1

Three rings, each of mass P and radius Q are arranged as shown in the figure. The moment of inertia of the arrangement about YY' axis will be

A

B

C

D

Solution

For ring 1

(MOI)1 = MOI about diameter + PQ²

MOI about diameter = ½ PQ²

(MOI)1 = 3/2 PQ²

Similarly, (MOI)2 = 3/2 PQ²

(MOI)3 = ½ PQ²

Total MOI = 7/2 PQ²
Question 3
4 / -1

Consider the following statements

Assertion (A) : The moment of inertia of a rigid body reduces to its minimum value as compared to any other parallel axis when the axis of rotation passes through its centre of mass.

Reason (R) : The weight of a rigid body always acts through its centre of mass in uniform gravitational field. Of these statements :

A
both A and R are true and R is the correct explanation of A

B
both A and R are true but R is not a correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

Moment of inertia about its centroidal axis has a minimum value as the centroidal axis has mass evenly distributed around it thereby providing minimum resistance to rotation as compared to any other axis. Also the statement-2 is correct but is not the correct explanation for statement-1.
Question 4
4 / -1

A disc of radius b and mass m rolls down an inclined plane of vertical height h. the translational speed when it reaches the bottom of the plane will be

A

B

C

D

Solution

The difference in the potential of the body when it rolls down through a vertical height h, is mgh.

As the KE at the top point is zero and let say KE at bottom is ½ mv² + ½ Iw²

Where m is its mass, I is its moment of inertia, I = ½ mr²

Where r is its radius, v is its gained translational speed and w is its gained angular speed.

w = v/r

Hence equating PE and KE gives

mgh = ½ mv² + ½ Iw²

That is mgh = ½ mv² + ½ mr².(v/r)²

We get mgh = ½ mv² + ¼ mv²

Thus we get v = √4gh/3
Question 5
4 / -1

The kinetic energy T of a particle of mass m moving in a circle of radius r varies with the distance traced, S as T = KS². The tangential acceleration of the particle is

A

B

C

D

Solution
Question 6
4 / -1

Water falls from a height of 60m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? (g = 10 m/s²)

A
7.0 kW

B
10.2 kW

C
8.1 kW

D
12.3 kW

Solution
Question 7
4 / -1

A force F = 20 + 10y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is :

A
5 J

B
25 J

C
20 J

D
30 J

Solution

Work done under the given variable force is :
Question 8
4 / -1

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

A
D

B
3/2D

C
7/5D

D
5/4D

Solution

Since from the given figure, in order to complete the vertical circle, velocity at the bottom of vertical circle should be .

As body is at rest initially, i.e., speed = 0.

At point A, speed = v.

As track is frictionless, so total mechanical energy will remain constant.

∴ 0 + mgh = 1/2mv² + 0

⇒ h = v² / 2g

For completing the vertical circle,
Question 9
4 / -1

A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively -

A

B

C

D

Solution

PE + KE = mgs

at given point

KE = 3PE

So, 4PE = mgs

4mgh = mgs

H = s/4
Question 10
4 / -1

A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic enegy of the particle becomes equal to 8 × 10⁻⁴ J by the end of the second revoluation after the beginning of the motion ?

A
0.18 m/s²

B
0.2 m/s²

C
0.1 m/s²

D
0.15 m/s²

Solution

Given: Mass of particle, M = 10g = 10/1000 kg

radius of circle R = 6.4 cm

Kinetic energy E of particle = 8 × 10–4J

acceleration at = ?