Q. A frictionless inclined plane of length $l$ having inclination $\theta$ is placed inside a lift which is accelerating downward with an acceleration $a( < g)$. If a block is allowed to move, down the inclined plane, from rest, then the time taken by the block to slide from top of the inclined plane to the bottom of the inclined plane is
A) $\sqrt{\frac{2l}{g}}$
B) $\sqrt{\frac{2l}{g - a}}$
C) $\sqrt{\frac{2l}{g + a}}$
D) $\sqrt{\frac{2l}{(g - a)\sin\theta}}$
Explanation
(d) Effective value of acceleration due to gravity in the lift $= g - a$
Acceleration down the inclined plane
Using, $s = ut + \frac{1}{2}at^{2}$, we get
$l = \frac{1}{2}(g - a)\sin{{\theta t}^{2},}\ $we get
$t = \sqrt{\frac{2l}{(g - a)\sin\theta}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.