Simple Pendulum question from JEE Main 2024 (29 Jan, Shift 2)

Answer: (d)

1. Released from horizontal, so the drop to the lowest point is $L = 10\text{ m}$.
2. Only $90\%$ of $mgL$ becomes KE: $\frac{1}{2}mv^2 = 0.9\,mgL$.
3. $v^2 = 2 \times 0.9 \times 10 \times 10 = 180 \Rightarrow v = \sqrt{180}$.
4. So the answer is (d) $6\sqrt{5}\text{ m s}^{-1}$.