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A body of mass Mkg is on the top point o...

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 ms^(−1)` At this instant the angle made by the radius vector of the body with the vertical is ( Take `g=10 m//s^(2)`)

A

`30^@`

B

`60^(@)`

C

`45^(@)`

D

`20^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B
`mg cos theta = ( mv ^(2))/(r )`
` therefore cos theta =(v^(2))/( rg) = (1)/(2) `
` therefore theta = 60^(@)`
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