This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Centripetal Force”.

1. In stable equilibrium, a body has maximum potential.

a) True

b) False

View Answer

Explanation: A body is said to be in stable equilibrium if it tends to regain its equilibrium position after being slightly displaced and released. In stable equilibrium, a body has minimum potential energy.

2. On being displaced, the centre of mass of unstable equilibrium goes ___________

a) Higher

b) Neither higher nor lower

c) First higher and then lower

d) Lower

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Explanation: A body is said to be in unstable equilibrium if it gets further displaced from its equilibrium position after being slightly displaced and released. In unstable equilibrium, a body possesses maximum potential energy and its centre of mass goes lower on being slightly displaced.

3. Which of the following stays in equilibrium even after being slightly displaced?

a) Stable equilibrium

b) Unstable equilibrium

c) Neutral equilibrium

d) Rigid body

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Explanation: If a body stays in equilibrium position even after being slightly displaced and released, it is said to be in neutral equilibrium. When a body is slightly displaced, its centre of mass is neither raised nor lowered and its potential energy remains constant.

4. Moment of inertia can be regarded as the measure of the rotational inertia of the body.

a) True

b) False

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Explanation: The mass of a body resists change in its state of linear motion, it is a measure of its inertia in linear motion, Similarly, the moment of inertia of a body about an axis of rotation resists a change in its rotational motion. The greater the moment of inertia of body, the greater is the torque required to change its state of rotation. Thus moment of inertia of a body can be regarded as the measure of the rotational inertia of the body.

5. The flying wheel attached to the shaft of steam engine works on the principle of ___________

a) Centripetal action

b) Moment of inertia

c) Newton’s third law of motion

d) Conservation of momentum

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Explanation: A flywheel is attached to the shaft of an engine. Because of its large moment of inertia, the flywheel opposes the sudden increase or decrease in the speed of the vehicle. It allows a gradual change in the speed and prevents jerky motions and hence ensures a smooth ride for the passengers.

6. A wheel of mass 8kg and radius of gyration 25cm is rotating at 300rpm. What is its moment of inertia?

a) 0.5 kgm^{2}

b) 10 kgm^{2}

c) 5 kgm^{2}

d) 0.25 kgm^{2}

View Answer

Explanation: M = 8kg, K = 25cm = 0.25m

Therefore, l = MK

^{2}= 8×0.25

^{2}= 0.5 kgm

^{2}.

7. The moment of inertia of a uniform circular disc about its diameter is 100gcm^{2}. What is its moment of inertia about its tangent?

a) 200 gcm^{2}

b) 100 gcm^{2}

c) 900 gcm^{2}

d) 500 gcm^{2}

View Answer

Explanation: By the theorem of parallel axes, moment of inertia about a tangent parallel to the diameter,

I = I

_{d}+MR

^{2}= 1/4 MR

^{2}+MR

^{2}=5/4 MR

^{2}

I = 5×100 = 500 gcm

^{2}.

8. The moment of inertia of a uniform circular disc about its diameter is 100 gcm^{2}. What is its moment of inertia about an axis perpendicular to its plane.

a) 500 gcm^{2}

b) 100 gcm^{2}

c) 200 gcm^{2}

d) 700 gcm^{2}

View Answer

Explanation: By theorem of perpendicular axes, moment of inertia of the disc about an axis perpendicular to its plane,

I = Sum of the moments of inertia about two perpendicular diameters

I = I

_{d}+I

_{d}=2×1/4×MR

^{2}=2×1000= 200 gcm

^{2}.

9. Calculate the moment of inertia of the earth about its diameter, taking it to be a sphere of 10^{25}kg and diameter 12800km.

a) 1.64 kgm^{2}

b) 16.4×10^{38} kgm^{2}

c) 1.64×10^{38} kgm^{2}

d) 0

View Answer

Explanation: M = 10

^{25}kg, R = 6400km = 6.4×10

^{6}m

Moment of inertia of the earth about its diameter I = 2/5 MR

^{2}= 2/5×10

^{25}×(6.4×10

^{6})

^{2}

I = 1.64×10

^{38}kgm

^{2}.

10. A torque of 2×10^{-4} Nm is applied to produce an angular acceleration of 4rad/s^{2} in a rotating body. What is the moment of inertia of the body?

a) 0.5 kgm^{2}

b) 5×10^{4} kgm^{2}

c) 0.5×10^{-4} kgm^{2}

d) 0.5×10^{4} kgm^{2}

View Answer

Explanation: Torque = Iα

I = Torque/α = (2×10

^{-4})/4=0.5×10

^{-4}kgm

^{2}.

**Sanfoundry Global Education & Learning Series – Engineering Physics.**

To practice all areas of Engineering Physics, __here is complete set of 1000+ Multiple Choice Questions and Answers__.