Rotational Motion - NEET Physics Questions
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Rotational Motion

Question 41: easy

A uniform circular disc of mass \(2\text{ kg}\) and radius \(100\text{ cm}\) (hinged at centre) is subjected to a constant torque of \(4\text{ Nm}\). If the disc was initially at rest, then its angular speed after \(4\text{ s}\) will be

1. 16 rad/s
2. 4 rad/s
3. 2 rad/s
4. 8 rad/s
View Answer

Moment of inertia \(I = \frac{1}{2} M R^2 = \frac{1}{2} \times 2 \times 1^2 = 1\text{ kg m}^2\). Angular acceleration \(\alpha = \frac{\tau}{I} = \frac{4}{1} = 4\text{ rad/s}^2\). Angular speed \(\omega = \alpha t = 4 \times 4 = 16\text{ rad/s}\).

Question 42: easy

The force acting on a particle is \( \vec{F} = \hat{i} + 2\hat{j} + 3\hat{k} \text{ N} \). Find the torque (in N m) of this force about origin if position vector of the particle is \( \vec{r} = 7\hat{i} + 3\hat{j} + 5\hat{k} \text{ m} \).

1. \( \hat{i} + 16\hat{j} - 11\hat{k} \)
2. \( -\hat{i} - 16\hat{j} + 11\hat{k} \)
3. \( \hat{i} + 16\hat{j} + 11\hat{k} \)
4. \( -\hat{i} + 9\hat{j} + 11\hat{k} \)
View Answer

Torque \( \vec{\tau} \) is given by \( \vec{r} \times \vec{F} \). Evaluating the cross product: \( \vec{\tau} = \det \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \ 7 & 3 & 5 \ 1 & 2 & 3 \end{pmatrix} = -\hat{i} - 16\hat{j} + 11\hat{k} \text{ N m} \).

Question 43: easy

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is

1. \(\sqrt{2} : 1\)
2. \(1 : \sqrt{2}\)
3. \(2 : 1\)
4. \(1 : 2\)
View Answer

About central normal axis, \(I_1 = \frac{1}{2}MR^2 = MK_1^2 \Rightarrow K_1 = \frac{R}{\sqrt{2}}\). About diameter, \(I_2 = \frac{1}{4}MR^2 = MK_2^2 \Rightarrow K_2 = \frac{R}{2}\). The ratio is \(\frac{K_1}{K_2} = \sqrt{2} : 1\).

Question 44: easy

If moment of inertia of a spinning object drops to \(\left(\frac{1}{4}\right)^{\text{th}}\) of its initial value, the ratio of new rotational kinetic energy to initial rotational kinetic energy will be (Assume net external torque about the axis of rotation is zero)

1. 1 : 4
2. 4 : 1
3. 2 : 1
4. 1 : 2
View Answer

Since external torque is zero, angular momentum \(L = I\omega\) is conserved. Rotational kinetic energy is \(K = \frac{L^2}{2I}\). If \(I' = I/4\), then \(K' = 4K\), so \(K' : K = 4 : 1\).

Question 45: easy

Assertion (A): Value of radius of gyration of a body depends on axis of rotation.


Reason (R): Radius of gyration is rms distance of particles of the body from the axis of rotation.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Radius of gyration is defined as \( k = \sqrt{I/M} \), where \( I \) is the moment of inertia. Since \( I \) depends on the axis of rotation, \( k \) also depends on the axis of rotation. R correctly defines \( k \) as an RMS distance from the axis, which means its value depends on that axis. Both A and R are true, and R explains A.

Question 46: easy

Assertion (A): Kinetic energy of a rigid body can be greater than \( \frac{1}{2}mv^2 \), where \( m \) is mass of rigid body & \( v \) is speed of centre of mass of body.


Reason (R): Kinetic energy of a particle (point mass) cannot be greater than \( \frac{1}{2}mv^2 \), where \( m \) is mass of particle & \( v \) is speed of particle.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

The total kinetic energy of a rigid body is \( K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \), where the second term is rotational KE. For a particle, \( K = \frac{1}{2}mv^2 \) only. Therefore, a rigid body's KE can be greater than \( \frac{1}{2}mv^2 \).


Both A and R are true, and R explains A by highlighting the difference in KE components.

Question 47: easy

Assertion (A): A disc rolls without slipping on a fixed rough horizontal surface. Then there is no point on the disc whose velocity is in vertical direction.


Reason (R): Rolling motion can be taken as combination of translation and rotation. Due to the translational part of motion a velocity (translational component) exist in horizontal direction for any point on the disc rolling on a fixed rough horizontal surface.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

The velocity of any point on a rolling disc is \( \vec{v} = v_{CM}\hat{i} + (\vec{\omega} \times \vec{r}) \). For pure rolling, \( v_{CM} = \omega R \). If \( v_x = v_{CM} + \omega y = 0 \), this occurs only at the contact point (\( y = -R \)), where \( v_y = -\omega x = 0 \). Thus, no point has a purely vertical velocity. Both A and R are true, and R explains A.

Question 48: easy

Assertion (A): By definition, pure rolling of a body occurs when velocity of its point of contact is zero relative to the surface on which it rolls.


Reason (R): A body is purely rolling (rolling without slipping). The velocity of point of contact (of body) must be zero with respect to ground.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Pure rolling is fundamentally defined by the condition that there is no relative motion (slipping) between the contact point of the rolling body and the surface it rolls on. This means their relative velocity must be zero.


Both A and R state this definition/condition, with R reinforcing A. Thus, both A and R are true, and R correctly explains A.

Question 49: easy

Assertion (A): Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The solid cylinder will reach the bottom of the inclined plane first.


Reason (R): By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

The acceleration of a rolling body down an incline is \( a = \frac{g\sin\theta}{1 + I/(MR^2)} \). For a solid cylinder, \( I = \frac{1}{2}MR^2 \); for a hollow cylinder, \( I = MR^2 \). Since the solid cylinder has a smaller \( I/(MR^2) \) ratio, its acceleration is greater, and it reaches the bottom first.


By conservation of energy, \( Mgh \) converts to kinetic energy, so total KEs are identical if \( M \) and \( h \) are same.


Thus A and R are true, but R does not explain why one reaches first (which depends on the distribution of KE).

Question 50: easy

Assertion (A): When the body is rolling purely, the velocity of the point of contact should be zero relative to the surface in contact.


Reason (R): Friction is necessary for a body to roll purely on a level horizontal ground.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion A is the definition of pure rolling (no slipping at the contact point).


Reason R states that friction is required for a body to initiate or maintain pure rolling on a horizontal surface by providing the necessary torque/force.


Both A and R are true statements, but R describes a condition for pure rolling, not an explanation of its definition.