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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)

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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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