Rolling Motion question from MHT CET 2025 (26 Apr, Shift 2)

Answer: (b)

1. Energy conservation for rolling: $mgh = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)$.
2. Solid sphere: $\frac{k^2}{R^2} = \frac{2}{5}$, so $mgh = \frac{7}{10}mv^2$
3. $v = \sqrt{\frac{10gh}{7}}$
4. So the answer is (b) $\left(\frac{10gh}{7}\right)^{1/2}$.