Conservation of Energy question from JEE Main 2025 (3 Apr, Shift 1)

Answer: (c)

1. Total energy $mgS = K + U$ and $K = 3U$, so $4U = mgS$.
2. $mgh = \frac{mgS}{4} \Rightarrow h = \frac{S}{4}$
3. Fall distance $= \frac{3S}{4}$: $v = \sqrt{2g \cdot \frac{3S}{4}} = \sqrt{\frac{3gS}{2}}$
4. So the answer is (c) $\frac{S}{4}, \sqrt{\frac{3gS}{2}}$.