A smooth inclined plane of length L, having an inclination `theta` with horizontal is inside a lift which is moving down with retardation a. The time taken by a block to slide down the inclined plane from rest will be

Options

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

Answer

D

Explanation

(d) Moving down with retardation a means, lift is accelerated upwards. With respect to lift, pseudo force on the block will be ma in downward direction, where m is the mass of block. So, downward force mg on the block will be replaced by m(g+a). Therefore, acceleration of block relative to plane will be,
`a_(r)=(g+a) sin theta`(down the plane)
From , `" " L=(1)/(2)a_(r)t^(2)" " (s=(1)/(2)at^(2))`
`t=sqrt((2L)/(a_(r)))=sqrt((2L)/((g+a)sin theta))`