Conservation of Energy question from JEE Main 2024 (30 Jan, Shift 1)

Answer: (c)

1. Track is frictionless, so loss of PE = gain of KE: $v = \sqrt{2g\,\Delta h}$.
2. Height drop from A (1 m) to B (0.5 m): $\Delta h = 0.5\text{ m}$.
3. $v = \sqrt{2 \times 10 \times 0.5} = \sqrt{10}\text{ m/s}$
4. So the answer is (c) $\sqrt{10}\text{ m/s}$.