A smooth inclined plane is inclined at an angle `theta` with the horizontal. A body starts from rest and slides down the inclined surface. The time taken by the body to reach the bottom is

Options

A. `sqrt((2h)/g)`
B. `sqrt((2l)/g)`
C. `sqrt((2l)/(g sin theta))`
D. `1/(sin theta)sqrt(h/(2g))`

Answer

3

Explanation

`l=1/2 g sin thetat^(2)`
`t=sqrt((2l).(g sin theta)) =sqrt((2h)/g)xx1/(sin theta)`