Assertion (A): Inertia and moment of inertia are same quantities.
Reason (R): Moment of inertia represents the capacity of a rigid body to oppose its state of oscillatory motion.
Inertia (mass) measures resistance to translational motion, while moment of inertia measures resistance to rotational motion. They are distinct quantities. Moment of inertia opposes changes in a body's state of \(\text{rotational}\) motion, not oscillatory motion. Therefore, both Assertion (A) and Reason (R) are false.
Assertion (A): For the purpose of calculation of moment of inertia, body’s mass can be assumed to be concentrated at its centre of mass.
Reason (R): Moment of inertia of a rigid about an axis passing through its centre of mass is zero.
Moment of inertia depends critically on the distribution of mass relative to the axis of rotation, so mass cannot generally be assumed concentrated at the center of mass (A is false). Also, the moment of inertia of a rigid body about an axis passing through its center of mass is generally not zero (e.g., a disc has \(I = \frac{1}{2}MR^2\)). Thus, (R) is false. Both statements are incorrect.
Assertion (A): If total external torque on a rigid system is zero, its angular momentum remains constant.
Reason (R): The change in angular momentum is equal to the angular impulse of the resultant torque.
Assertion (A) is true, stating the conservation of angular momentum. Reason (R) is true, defining the angular impulse-momentum theorem \(\Delta \vec{\text{L}} = int \vec{\tau} \text{dt}\). If \(\vec{\tau}_{ext} = 0\), then \(\Delta \vec{\text{L}} = 0\), so \(\vec{\text{L}}\) is constant. (R) correctly explains (A).
Assertion (A): For a system of particles under central force field, the total angular momentum is conserved.
Reason (R): The torque acting on such a system is zero.
Assertion (A) is true, angular momentum is conserved when net torque is zero. Reason (R) is true. For a central force \(\vec{F}\) acting along \(\vec{r}\) (position vector), the torque \(\vec{\tau} = \vec{r} \times \vec{F} = 0\). Since \(\vec{\tau}=0\), \(\frac{\text{d}\vec{\text{L}}}{\text{dt}} = 0\), hence \(\vec{\text{L}}\) is conserved. (R) correctly explains (A).
Assertion (A): If two different axes are at same distance from the centre of mass of a rigid body then moment of inertia of the given rigid body about both the axes will always be equal.
Reason (R): According to perpendicular axis theorem \(\text{I} = \text{I}_{\text{cm}} + \text{Md}^2\) where symbols have their usual meaning.
Assertion (A) is false. Moment of inertia depends on both the distance and the orientation of the axis. Reason (R) is false. The given formula is for the parallel axis theorem, not the perpendicular axis theorem.
Assertion (A): A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling).
Reason (R): For pure rolling, work done against frictional force is zero.
Assertion (A) is true; friction provides the torque for rolling. Without friction, the wheel slips. Reason (R) is true; in pure rolling, the contact point is stationary, so static friction does no work. However, (R) does not explain (A).
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)
Under zero external torque, angular momentum is conserved: \( I_1 \omega_1 = I_2 \omega_2 \). If \( I_2 = I_1/4 \), then \( \omega_2 = 4\omega_1 \). The ratio of kinetic energy is \( \frac{K_2}{K_1} = \frac{\frac{1}{2}I_2\omega_2^2}{\frac{1}{2}I_1\omega_1^2} = \frac{1}{4} \times 16 = 4 \implies 4 : 1 \).
A constant torque of 100 N m turns a wheel of moment of inertia \(300 \text{kg} \text{m}^2\) about an axis passing through its centre. Starting from rest, its angular velocity after 3 s is
Angular acceleration \(\alpha = \frac{\tau}{I} = \frac{100}{300} = \frac{1}{3} \text{rad/s}^2\). The angular velocity is \(\omega = \omega_0 + \alpha t = 0 + \left(\frac{1}{3}\right)(3) = 1 \text{rad/s}\).
The ratio of the radius of gyration of a thin uniform disc about an axis passing through it centre and normal to its plane to the radius of gyration of the disc about its diameter is:
(2022)
Radius of gyration about center normal axis is $k_1 = R/\sqrt{2}$ and about diameter is $k_2 = R/2$. The ratio $k_1/k_2$ simplifies to $\sqrt{2}:1$.
From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a $90^\circ$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times $MR^2$. Then the value of ‘K’ is:
(2021)
Since mass is uniformly distributed, removing a $90^\circ$ sector removes $1/4$ of the mass, leaving $3/4$ of the mass at distance $R$. Thus $I = \frac{3}{4}MR^2$, making $K = \frac{3}{4}$.