Question

A particle, placed at the highest point of a frictionless hemispherical surface, is allowed to slide down. The angular displacement $\theta$ at which it will leave the surface is

• A. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{2}\right)$
• C. ${\cos }^{-1}\left(\frac{2}{3}\right)$
• D. ${\cos }^{-1}\left(\frac{3}{4}\right)$

Answer 1

The correct option is C ${\cos }^{-1}\left(\frac{2}{3}\right)$

$\begin{array}{c}\text{Let mars of particle be 'm'}\\ \text{For equilibriun}mg\cos \theta =\frac{{mv}^2}{R}\end{array}$

$\begin{array}{c}\text{From conservation of mechanical energy}\\ \frac{1}{2}m{v}^2=mgR\left(1\cos \theta \right)-1\\ \text{when particle is leaving}N=0\end{array}$

$mg\cos \theta =\frac{m{v}^2}{R}-\theta$

Solving 1 &2 $\cos \theta =\frac{2}{3}$

$\begin{array}{c}\Rightarrow \theta ={\cos }^{-1}\left(\frac{2}{3}\right)\\ \text{Hence option}C\text{is correct}\end{array}$