A body of mass m is on the top point of a smooth hemisphere of radius 5 m It is released to slide down the surface when its velocity is `5 m//s` . At this instant the angle made by the radius vector of the body with verticle is `( g=10 m//s^(2))`

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

C

Explanation

`mv^(2)/(r)=mg cos theta `
`v^(2)=r g cos theta `
`therefore cos theta =(v^(2))/(rg)=(25)/(5xx10)=(1)/(2)`
` theta =60^(@)`