JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2021
- Hard
- Numerical
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)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
on balancing
\(m g \cos \theta=\frac{m v^{2}}{R}....(1)\)
\(\cos \theta=\frac{h}{R}\)
Energy conservation
\(m g\{R-h\}=\frac{1}{2} m v^{2}....(2)\)
From \((1)\) and \((2) \Rightarrow m g\left\{\frac{h}{R}\right\}=\frac{2 m g\{R-h\}}{R}\)
\(h=\frac{2 R}{3}=2\, m\)
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