Q 11-05-068JEE MainJEE Main 2024 (4 Apr, Shift 2)Medium
A body of mass $m$ kg slides from rest along the curve of a vertical circle from point $A$ to $B$ on a frictionless path, as shown. The velocity of the body at $B$ is (given $R = 14\ \text{m}$, $g = 10\ \text{m/s}^2$ and $\sqrt2 = 1.4$)
Answer: (D) $21.9\ \text{m/s}$
Height of $A$ above $B$: $h = R + R\sin45^\circ = 14 + \dfrac{14}{1.4} = 14 + 10 = 24\ \text{m}$ (with $\sin45^\circ = 1/\sqrt2$ and $\sqrt2 = 1.4$).
$$v = \sqrt{2gh} = \sqrt{2\times10\times24} = \sqrt{480} \approx 21.9\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics