Rotational Kinetic Energy, Work & Power question from JEE Main 2018 (15 Apr, Shift 2)

Answer: (a)

1. When the rod has turned by $\alpha$, the centre of mass has fallen $\frac{L}{2}\sin\alpha$.
2. $\frac{1}{2}\cdot\frac{ML^2}{3}\omega^2 = Mg\frac{L}{2}\sin\alpha \Rightarrow \omega = \sqrt{\frac{3g\sin\alpha}{L}}$
3. Speed of end M: $v = \omega L = \sqrt{3gL\sin\alpha} \propto \sqrt{\sin\alpha}$
4. So the answer is (a) $\sqrt{\sin\alpha}$.