Question

A stone of mass m is pushed down a frictionless hemispherical bowl of radius r from a point O at a height H from the bottom. The stone just rises upto the half of other side of the bowl as shown in the figure, what is the initial speed of stone with which it was pushed?

• A. $\sqrt{g-(r-H)}$
  
• B. $\sqrt{2gH}$
  
• C. $\sqrt{g(r-2H)}$
  
• D. zero

Answer 1

The correct option is C $\sqrt{g(r-2H)}$


Let the velocity with which stone is pushed = u
We have,
$\frac{1}{2}m{u}^2+mgH=mg\frac{r}{2}u=\sqrt{2g(\frac{r}{2}-H)}$ or $u=\sqrt{g(r-2H)}$