Q 11-05-137JEE MainJEE Main 2021 (27 Jul, Shift 2)Easy
A small block slides down from the top of hemisphere of radius $R = 3$ m as shown in the figure. The height $h$ at which the block will lose contact with the surface of the sphere is ______ m. (Assume there is no friction between the block and the hemisphere)
Numerical value type. Enter your answer.
Answer: 2
At angle $\theta$ from the vertical, speed: $v^2 = 2gR(1 - \cos\theta)$.
Losing contact ($N = 0$): $mg\cos\theta = \dfrac{mv^2}{R} = 2mg(1 - \cos\theta) \Rightarrow \cos\theta = \dfrac23$.
$h = R\cos\theta = \dfrac23\times3 = 2$ m.
Solution by Sreeraj P, M.Sc Physics