Q 11-05-085JEE MainJEE Main 2024 (30 Jan, Shift 1)Easy
A particle is placed at the point $A$ of a frictionless track $ABC$ as shown in the figure. It is gently pushed towards the right. The speed of the particle when it reaches the point $B$ is: (Take $g = 10\ \text{m s}^{-2}$)
Answer: (B) $\sqrt{10}\ \text{m s}^{-1}$
The particle starts from rest at $A$ (height 1 m) and falls to $B$ (height 0.5 m). With no friction,
$$v_B = \sqrt{2g\,\Delta h} = \sqrt{2\times10\times0.5} = \sqrt{10}\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics