A block can slide on a smooth inclined plane of inclination `theta` kept on the floor of a lift. When the lift is descending with a retardation a , the acceleration of the block relative to the incline is

Options

A. `sqrt((2L)/((g + a) sin theta))`
B. `sqrt((2L)/((g - a) sin theta))`
C. `sqrt((2L)/(a sin theta))`
D. `sqrt((2L)/(g sin theta))`

Answer

A

Explanation

Acceleration of block with respect to lift
`a_(r ) = (m (g + a) sin theta)/(m) (g + a) sin theta`
`s = (1)/(2) a_(r ) r^(2), t = sqrt((2s)/(a_(r))) = sqrt((2L)/((g + a) sin theta))`