Question

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. $\sqrt{\frac{2 l}{g}}$
• B. $\sqrt{\frac{2 l}{g-a}}$
• C. $\sqrt{\frac{2 l}{g+a}}$
• D. $\sqrt{\frac{2 l}{(g-a) \sin \theta}}$

Answer 1

Answer: (D)

Along $x$ -axis: $F=m g \sin \theta-m a \sin \theta$
$\frac{F}{m}=(g-a) \sin \theta$
Acceleration along the inclined plane $=(g-a) \sin \theta$
Using $S=u t+\frac{1}{2} a t^{2}$
$l=0+\frac{1}{2}(g-a) \sin \theta t^{2}$ or $t=\sqrt{\frac{2 l}{(g-a) \sin \theta}}$