Question

A particle is kept at rest at the top of a sphere of diameter 42 m. When disturbed slightly, it slides down. At what height 'h' from the bottom, the particle will leave the sphere.

a)14 m
b)28 m
c)35 m
d)7 m

Answer 1

answer is option (C) ..see attachment.....hope it will help

Answer 2

Given :

The diameter of the sphere = 42m

To Find :

The height from bottom where the particle leaves the sphere = ?

Solution :

• From equations of motion, $v^{2}-u^{2}=2as$

• By substituting the values  

         $v^{2}-u^{2}=2as$

         $v^{2}-u^{2}=2g(R-RCos\theta)$

         $v^{2}-u^{2}=2gR(1-Cos\theta)\rightarrow Equation(1)$

• From the free body diagram of the particle

        $mgCos\theta-N=\frac{mv^{2} }{R}$

        $mgCos\theta-N=2mg(1-Cos\theta) \rightarrow Equation(2)$

        $mgCos\theta = 2mg(1-cos\theta)\:\:\:\:[N=0]$

        $Cos\theta = \frac{2}{3}$

• Height of particle leaving the sphere = $2R-h$

       $=2R-R(1-Cos\theta)$

       $= 2R-R(1-\frac{2}{3})$

       $=\frac{5R}{3} =\frac{5}{3}\times 21$

       =$35m$

The height where the particle leaves the sphere is 35m from the bottom