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

A smooth inclined plane of length L, having an inclination theta with

1.	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
Answer» `sqrt((2L)/(sqrt(a sin theta)))` `sqrt((2L)/(g sin theta))` `sqrt((2L)/((g-a) sin theta))` `sqrt((2L)/((g+a) sin theta))` SOLUTION :(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))`