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

Options

A. (3/2)r
B. (2/3)r
C. `(1/2)gr^(2)`
D. `v^(2)//2g`

Answer

B

Explanation

The body will lose constanct when centripetal acceleration becomes equal to the component of accelerationdue to gravity along the radius.

Velocity at `P,v = sqrt(2g(r-h)) ( :. V^(2)-u^(2)=2 gx)`
Centripetal acceleration will `v^(2)//r`. it should be equal to the component of g along PO. Hence.
`(v^(2))/(r)= g cos theta or (2g(r-h)/(r)= g xx(h)/(r)`
Solving we get, `h=((2r)/(3))`