Question

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

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

Answer 1

Answer: (B)

Net acceleration in downward motion decreases.
The free body diagram of the set up is shown.
Net acceleration is $(g-a) \sin \theta$

Using equation of motion
$s=u t+\frac{1}{2} a^{\prime} t^{2}$
where, $s$ is displacement, $u$ is initial velocity, $a$ is acceleration and $t$ is time.
We have, $u=0, a^{\prime}=(g-a) \sin \theta, s=L$
$\therefore L=0+\frac{1}{2}(g-a) \sin \theta \cdot t^{2}$
$\Rightarrow t=\sqrt{\frac{2 L}{(g-a) \sin \theta}}$