An inclined plane of height h and length l have the angle of inclination `theta`. The time taken by a body to come from the top of the bottom of this inclined plane will be

Options

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

Answer

B

Explanation

( b) Force down the the plane `=mg sin theta`
`therefore` Acceleration down the plane `=g sin theta`

Using the relation, `h=ut+(1)/(2)gt^(2)`
`l=0+(1)/(2)(g sin theta)t^(2)`
`t^(2)=(2l)/(g sin theta)=(2h)/(g sin^(2) theta)" "(therefore sin theta=(h)/(l))`
or `t=(1)/( sin theta) sqrt((2h)/(g))`