Question

A body of mass Mkg is on the top point of a smooth hemisphere of radius 5m . It is released to slide down the surface of hemisphere it leaves the surface when its velocity is 5 $m{s}^{-1}$ At this instant the angle made by the radius vector of the body with the vertical is (g= 10 $m{s}^{-2}$)

• A. ${\cos }^{-1}\left(\frac{3}{4}\right)$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{3}{4}\right)$

$\begin{array}{c}mg\left(R-R\cos \theta \right)=\frac{1}{2}m{v}^2\\ \Rightarrow v^2=2gR\left(1-\cos \theta \right)\\ \Rightarrow 25=2\times 10\times 5\left(1-\cos \theta \right)\\ \Rightarrow 25=100\left(1-\cos \theta \right)\\ \frac{1}{4}=1-\cos \theta \\ \cos \theta =1-\frac{1}{4}\\ =\frac{3}{4}\\ \theta ={\cos }^{-1}\left(\frac{3}{4}\right)\\ Hence,\\ optionAiscorrectanswer.\end{array}$