A body is released from the top of a smooth inclined plane of inclination `theta`. It reaches the bottom with velocity `v`. If the angle of inclina-tion is doubled for the same length of the plane, what will be the velocity of the body on reach ing the ground .

Options

A. v
B. 2v
C. `[2 cos theta]^(1//2)v`
D. `[2 sin theta]^(1//2)v`

Answer

C

Explanation

by III EOM,
`v^(2)=O^(2)+2(g sin theta)s`
`v=sqrt(2gs sin theta)`,
`v=sqrt(2gs sin 2theta)=sqrt(2gs 2sin theta cos theta)`
`v=sqrt(2gs sin theta)sqrt(2cos theta)=vxx(2cos theta)^(1//2)`