Rotational Kinetic Energy, Work & Power question from MHT CET 2026 (19 May, Shift 2)

Answer: (c)

1. Rod pivoted at end B: $I = \frac{ml^2}{3}$; centre of mass falls by $\frac{l}{2}(1 - \cos\theta)$.
2. $\frac{1}{2}\cdot\frac{ml^2}{3}\omega^2 = mg\frac{l}{2}(1 - \cos\theta)$
3. $\omega^2 = \frac{3g}{l}(1 - \cos\theta) = \frac{6g}{l}\sin^2\frac{\theta}{2}$
4. So the answer is (c) $\sqrt{\frac{6g}{l}}\sin\frac{\theta}{2}$.