Q: A force of 50\text{ N} is applied perpendicular to a rod at a distance of 0.4\text{ m} from the pivot. The torque produced is:

20\text{ N}\cdot\text{m}. \tau = Fr\sin\theta with \theta = 90^\circ, so \tau = 50 \times 0.4 = 20\text{ N}\cdot\text{m}.

Q: A body of moment of inertia 8\text{ kg}\cdot\text{m}^2 is given an angular acceleration of 2.5\text{ rad/s}^2. The torque applied is:

20\text{ N}\cdot\text{m}. \tau = I\alpha = 8 \times 2.5 = 20\text{ N}\cdot\text{m}.

Q: A force of 30\text{ N} acts at a point 1.2\text{ m} from the axis, making an angle of 30^\circ with the position vector. The magnitude of the torque is:

18\text{ N}\cdot\text{m}. \tau = Fr\sin\theta = 30 \times 1.2 \times \sin 30^\circ = 30 \times 1.2 \times 0.5 = 18\text{ N}\cdot\text{m}.

Q: In rotational motion, torque is the analogue of which quantity in linear motion?

Force. Comparing \tau = I\alpha with F = ma shows torque plays the role of force, moment of inertia the role of mass, and angular acceleration the role of linear acceleration.

Q: A body of moment of inertia 5\text{ kg}\cdot\text{m}^2 rotates with angular velocity 4\text{ rad/s}. Its angular momentum is:

20\text{ kg}\cdot\text{m}^2/\text{s}. L = I\omega = 5 \times 4 = 20\text{ kg}\cdot\text{m}^2/\text{s}.

Q: A constant torque of 15\text{ N}\cdot\text{m} acts on a body for 2\text{ seconds}. The change in its angular momentum is:

30\text{ kg}\cdot\text{m}^2/\text{s}. Angular impulse equals change in angular momentum: \Delta L = \tau\,\Delta t = 15 \times 2 = 30\text{ kg}\cdot\text{m}^2/\text{s}.

Q: The angular momentum of a rotating body remains constant when:

The net external torque acting on it is zero. Since \tau = \dfrac{dL}{dt}, a zero net external torque gives \dfrac{dL}{dt} = 0, hence L is conserved — even if I and \omega individually change.

Question 1

A force of 50	ext{ N} is applied perpendicular to a rod at a distance of 0.4	ext{ m} from the pivot. The torque produced is:

Accepted Answer

20	ext{ N}\cdot	ext{m}. 	au = Fr\sin	heta with 	heta = 90^\circ, so 	au = 50 	imes 0.4 = 20	ext{ N}\cdot	ext{m}.

Question 2

A body of moment of inertia 8	ext{ kg}\cdot	ext{m}^2 is given an angular acceleration of 2.5	ext{ rad/s}^2. The torque applied is:

Accepted Answer

20	ext{ N}\cdot	ext{m}. 	au = I\alpha = 8 	imes 2.5 = 20	ext{ N}\cdot	ext{m}.

Question 3

A force of 30	ext{ N} acts at a point 1.2	ext{ m} from the axis, making an angle of 30^\circ with the position vector. The magnitude of the torque is:

Accepted Answer

18	ext{ N}\cdot	ext{m}. 	au = Fr\sin	heta = 30 	imes 1.2 	imes \sin 30^\circ = 30 	imes 1.2 	imes 0.5 = 18	ext{ N}\cdot	ext{m}.

Question 4

In rotational motion, torque is the analogue of which quantity in linear motion?

Accepted Answer

Force. Comparing 	au = I\alpha with F = ma shows torque plays the role of force, moment of inertia the role of mass, and angular acceleration the role of linear acceleration.

Question 5

The torque acting on a rotating body is equal to:

Accepted Answer

The rate of change of its angular momentum. Newton's second law for rotation is \vec{	au} = \dfrac{d\vec{L}}{dt}, the rotational counterpart of \vec{F} = \dfrac{d\vec{p}}{dt}.

Question 6

A body of moment of inertia 5	ext{ kg}\cdot	ext{m}^2 rotates with angular velocity 4	ext{ rad/s}. Its angular momentum is:

Accepted Answer

20	ext{ kg}\cdot	ext{m}^2/	ext{s}. L = I\omega = 5 	imes 4 = 20	ext{ kg}\cdot	ext{m}^2/	ext{s}.

Question 7

A constant torque of 15	ext{ N}\cdot	ext{m} acts on a body for 2	ext{ seconds}. The change in its angular momentum is:

Accepted Answer

30	ext{ kg}\cdot	ext{m}^2/	ext{s}. Angular impulse equals change in angular momentum: \Delta L = 	au\,\Delta t = 15 	imes 2 = 30	ext{ kg}\cdot	ext{m}^2/	ext{s}.

Question 8

The angular momentum of a rotating body remains constant when:

Accepted Answer

The net external torque acting on it is zero. Since 	au = \dfrac{dL}{dt}, a zero net external torque gives \dfrac{dL}{dt} = 0, hence L is conserved — even if I and \omega individually change.

Question 9

A ballet dancer spinning with angular velocity \omega pulls her arms inward, reducing her moment of inertia by 40\%. Her new angular velocity is:

Accepted Answer

1.67\omega. No external torque acts, so I_1\omega_1 = I_2\omega_2. With I_2 = 0.6I_1: \omega_2 = \dfrac{I_1\omega}{0.6I_1} = \dfrac{5}{3}\omega \approx 1.67\omega.

Question 10

The dimensional formula of angular momentum is:

Accepted Answer

[M^1L^2T^{-1}]. L = mvr gives [M][LT^{-1}][L] = [M^1L^2T^{-1}]. (Note [M^1L^2T^{-2}] is the dimension of energy and of torque.)

Question 11

The SI unit of angular momentum is:

Accepted Answer

ext{kg}\cdot	ext{m}^2/	ext{s}. L = I\omega has units 	ext{kg}\cdot	ext{m}^2 	imes 	ext{s}^{-1} = 	ext{kg}\cdot	ext{m}^2/	ext{s}, which is the same as 	ext{J}\cdot	ext{s}.

Question 12

A planet revolves around the Sun in an elliptical orbit. Its angular momentum about the Sun is conserved because:

Accepted Answer

The gravitational force is central, so its torque about the Sun is zero. Gravity always acts along the line joining planet and Sun, so \vec{r} and \vec{F} are antiparallel and \vec{	au} = \vec{r} 	imes \vec{F} = 0. This is exactly Kepler's second law.

Question 13

If the moment of inertia of a rotating body is halved while its angular momentum stays constant, its angular velocity:

Accepted Answer

Doubles. With L = I\omega constant, \omega \propto \dfrac{1}{I}. Halving I therefore doubles \omega.

Question 14

A disc of moment of inertia 2	ext{ kg}\cdot	ext{m}^2 rotating at 10	ext{ rad/s} is suddenly coupled to a coaxial stationary disc of moment of inertia 3	ext{ kg}\cdot	ext{m}^2. Their common angular velocity is:

Accepted Answer

4	ext{ rad/s}. Angular momentum is conserved: I_1\omega_1 = (I_1 + I_2)\omega. So \omega = \dfrac{2 	imes 10}{2 + 3} = \dfrac{20}{5} = 4	ext{ rad/s}.

Question 15

The angular momentum of a particle of mass m moving with velocity v along a circle of radius r is:

Accepted Answer

mvr. For circular motion the velocity is perpendicular to the radius vector, so L = |\vec{r} 	imes m\vec{v}| = mvr.

Question 16

The direction of the torque vector \vec{	au} = \vec{r} 	imes \vec{F} is:

Accepted Answer

Perpendicular to the plane containing \vec{r} and \vec{F}. A cross product is always perpendicular to both its factors, so \vec{	au} is normal to the plane of \vec{r} and \vec{F}, with the sense given by the right-hand rule.

Question 17

A couple acting on a rigid body produces:

Accepted Answer

Pure rotation with no translation, since the net force is zero. A couple consists of two equal, opposite, non-collinear forces. Their vector sum is zero (so the centre of mass does not accelerate) but their moments add, producing a pure turning effect.

Question 18

A torque acts on a body of moment of inertia 10	ext{ kg}\cdot	ext{m}^2 producing an angular acceleration of 3	ext{ rad/s}^2. The torque is:

Accepted Answer

30	ext{ N}\cdot	ext{m}. 	au = I\alpha = 10 	imes 3 = 30	ext{ N}\cdot	ext{m}.

Question 19

A disc of moment of inertia 4	ext{ kg}\cdot	ext{m}^2 speeds up from 5	ext{ rad/s} to 15	ext{ rad/s} in 2	ext{ s}. The torque acting on it is:

Accepted Answer

20	ext{ N}\cdot	ext{m}. \alpha = \dfrac{15 - 5}{2} = 5	ext{ rad/s}^2, so 	au = I\alpha = 4 	imes 5 = 20	ext{ N}\cdot	ext{m}.

Question 20

A body of moment of inertia 0.5	ext{ kg}\cdot	ext{m}^2 rotates at 20	ext{ rad/s}. Its angular momentum is:

Accepted Answer

10	ext{ kg}\cdot	ext{m}^2/	ext{s}. L = I\omega = 0.5 	imes 20 = 10	ext{ kg}\cdot	ext{m}^2/	ext{s}.

Torque & Angular Momentum — MH-CET Physics MCQs with Solutions

Questions with solutions