Rotational Kinematics & Dynamics question from MHT CET 2026 (12 May, Shift 2)

Answer: (c)

1. $I = \frac{1}{2}MR^2 = \frac{1}{2}(100)(1)^2 = 50\text{ kg m}^2$
2. $\omega = 300\text{ rpm} = 10\pi\text{ rad/s}$; reversing gives $\Delta\omega = 20\pi\text{ rad/s}$.
3. $\alpha = \frac{20\pi}{50} = 0.4\pi\text{ rad/s}^2$
4. $\tau = I\alpha = 50 \times 0.4\pi = 20\pi\text{ N m}$
5. So the answer is (c) $20\pi$ N m.