Rotational Kinetic Energy, Work & Power question from JEE Main 2026 (24 Jan, Shift 2)

Answer: (b)

1. Pivot is $\frac{L}{3}$ from the lower end, so the centre of mass is $\frac{L}{2} - \frac{L}{3} = \frac{L}{6}$ above the pivot.
2. When the rod lies horizontally on the table, the centre of mass is level with the pivot, so it falls $\frac{L}{6}$.
3. $I_{pivot} = \frac{ML^2}{12} + M\left(\frac{L}{6}\right)^2 = \frac{ML^2}{9}$
4. $Mg\frac{L}{6} = \frac{1}{2}\cdot\frac{ML^2}{9}\omega^2 \Rightarrow \omega^2 = \frac{3g}{L}$
5. So the answer is (b) $\sqrt{\frac{3g}{L}}$.