Centre of gravity can be defined

**A. ** as that point where the total gravitational torque on the body is greater than zero

**B. ** as that point where the total gravitational force on the body is zero

**C. ** as that point where the total gravitational torque on the body is zero

**D. ** as the center of mass

**Answer : ****Option C**

**Explaination / Solution: **

The point at which the entire weight of a body may be thought of as centered so that if supported at this point the body would balance perfectly, so it can also be defined as as that point where the total gravitational torque on the body is zero.

The point at which the entire weight of a body may be thought of as centered so that if supported at this point the body would balance perfectly, so it can also be defined as as that point where the total gravitational torque on the body is zero.

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Analogue of mass in rotational motion is

**A. ** angular acceleration

**B. ** moment of inertia

**C. ** rotary mass

**D. ** torque

**Answer : ****Option B**

**Explaination / Solution: **

The moment of inertia is a physical quantity which tells how easily a body can be rotated about a given axis. It is a rotational analogue of mass. It plays the same role in rotational motion as 'inertia' does in translational motion. Large moment of inertia also helps in keeping the rotation motion uniform.

The moment of inertia is a physical quantity which tells how easily a body can be rotated about a given axis. It is a rotational analogue of mass. It plays the same role in rotational motion as 'inertia' does in translational motion. Large moment of inertia also helps in keeping the rotation motion uniform.

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The radius of gyration of a body about an axis may be defined as the

**A. ** distance from the center of mass to the axis of rotation

**B. ** distance from the center of gravity to the center of mass

**C. ** distance from the center of gravity to the axis of rotation

**D. ** distance from the axis of a mass point whose mass is equal to the mass of the whole body and whose moment of inertia is equal to the moment of inertia of the body about the axis

**Answer : ****Option D**

**Explaination / Solution: **

It is the perpendicular distance from the axis of rotation to a point mass of mass m that gives an equivalent inertia to the original object (of mass, m)

if radius of gyration is k then

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The moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is.

**A. ** equal to the its moments of inertia about an axis perpendicular to the axis of rotation axis and lying in the plane of the body

**B. ** equal to the average of its moments of inertia about three perpendicular axes concurrent with perpendicular axis and lying in the plane of the body

**C. ** equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body

**D. ** equal to the difference of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body

**Answer : ****Option C**

**Explaination / Solution: **

According to perpendicular axis theorem "For a planar object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia of two perpendicular axes through the same point in the plane of the object".

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The moment of inertia of a body about any axis is

**A. ** equal to the difference of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

**B. ** equal to the sum of the moment of inertia of the body about any axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

**C. ** equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

**D. ** equal to the average of the moments of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

**Answer : ****Option C**

**Explaination / Solution: **

according to parallel axis theorem the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its Centre of mass and the product of its mass and the square of the distance between the two parallel axes.

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For rolling motion without slipping the kinetic energy of the body is

**A. ** equal to the kinetic energy of rotation
**B. ** equal to the kinetic energy of translation

**C. ** equal to the difference of kinetic energies of translation and rotation

**D. ** equal to the sum of kinetic energies of translation and rotation

**Answer : ****Option D**

**Explaination / Solution: **

Rolling motion without slipping involve both translation motion and rotation motion. Hence kinetic energy of body in rolling motion is the sum of both translation kinetic energy and rotational kinetic energy.

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A solid sphere of mass M and radius R spins about an axis passing through its centre making 600 rpm. Its kinetic energy of rotation is

**A. ** 2/5π2MR
**B. ** 80π2MR2
**C. ** 80πMR
**D. ** 2/5πM2R2
**Answer : ****Option B**

**Explaination / Solution: **

Kinetic energy of rotation

K =

revolution per second

angular velocity rad/sec

moment of inertia about an axis passing through centre of solid sphere

Kinetic energy of rotation

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A child swinging on a swing in a sitting position stands up. Then, the time period of the swing will

**A. ** Increase if the child is tall and decrease if the child is short.

**B. ** Increase

**C. ** Decrease

**D. ** Remains the same

**Answer : ****Option C**

**Explaination / Solution: **

As the child stand up then Centre of mass is shifted in upwards direction from the sitting position of child, thus decreasing the effective length of the swing, so as the time period is proportion to square root of length, thus time period decreases.

time period

Time period of swing decreases as effective length decreases.

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Angular momentum is

**A. ** A polar vector

**B. ** None of these

**C. ** A scalar

**D. ** An axial vector

**Answer : ****Option D**

**Explaination / Solution: **

Axial vector represent rotational effect and are always along the axis of rotation. Direction of angular momentum always along the axis of rotation in accordance with right hand screw rule. Hence angular momentum is an axial vector.

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Angular momentum of a body is defined as the product of

**A. ** Moment of inertia and angular velocity

**B. ** Centripetal force and radius

**C. ** Mass and angular velocity

**D. ** Linear velocity and angular velocity

**Answer : ****Option A**

**Explaination / Solution: **

if r is perpendicular to p (in case of circular motion)

L = mvr

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