question_answer

A small body of mass m slides down from the top of a hemisphere of radius r (figure). The surface of block and hemisphere are frictionless. The height at which the body loose contact with the surface of sphere is

A)  \[\frac{3}{2}\,r\]                                  

B)  \[\frac{2}{3}\,r\]

C)  \[\frac{1}{2}\,g{{t}^{2}}\]                 

D)  \[\frac{{{v}^{2}}}{2g}\]

Correct Answer: B

Solution :

The body will loose contact when centripetal acceleration becomes equal to the component of acceleration due to gravity along the radius. Velocity at \[v=\,\,\sqrt{2g(r-h)}\,(\because \,\,{{v}^{2}}-{{u}^{2}}=2gx)\] Centripetal acceleration will be \[\frac{{{v}^{2}}}{r}\]. It should be equal to the component of g along PO. Hence, \[\frac{{{v}^{2}}}{r}=\,\,g\,\cos \theta \] or         \[\frac{2g\,\,(r-h)}{r}=g\,\,\times \,\,\frac{h}{r}\] Solving, we get \[h=\frac{2r}{3}\]