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 bottom, the particle will leave the sphere :

• A. 14 m
• B. 16 m
• C. 35 m
• D. 70 m

Answer 1

The correct option is C 35 m

Given,

$D=2R=42m$

$R=21m$

$h=R(1-cos\theta )$

From the 3rd equation of motion,

$v^2=u^2+2gh$

$v^2=2gR(1-cos\theta )$. . . . .(1)

From the free body diagram.

$mgcos\theta -N=\frac{m{v}^2}{R}=2mg(1-cos\theta )$. . . . . . .(2)

When contact lost,

$N=0$

equation (2) becomes

$mgcos\theta =2mg(1-cos\theta )$

$cos\theta =2-2cos\theta$

$3cos\theta =2$

$cos\theta =\frac{2}{3}$

$2R-h=2R-R(1-cos\theta )$

$2R-h=2R-R(1-\frac{2}{3})=\frac{5R}{3}=\frac{5}{3}\times 21=35m$

The correct option is C.