A solid cylinder roll up without slipping an inclined plane of inclination `theta` with an initial velocity v. How for does the cylinder go up the plane :-

Options

A. `(3v^2)/(2g sin theta)`
B. `v^2/(4g sintheta)`
C. `(3v^2)/(g sin theta)`
D. `(3v^2)/(4v sin theta )`

Answer

D

Explanation

`1/2mv^2 (1+1/2)=mgh`
`h=(3v^2)/(4g)`
s sin `theta=h rArr s=h/(sin theta)=(3v^2)/(4 g sin theta)`