Q. A smooth inclined plane of length $L$ having inclination $\theta$ with the horizontal is inside a lift which is moving down with retardation $a$. The time taken by $\alpha$ 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) $\sqrt{\frac{2L}{{g\ sin}\theta}}$
D) $\sqrt{\frac{2L}{\alpha\sin\theta}}$
Explanation
(a) Downward retardation means upward acceleration
$g^{'} = g + a$
Now, $t = \sqrt{\frac{2L}{g^{'}\sin\theta}}$ or $t = \sqrt{\frac{2L}{(g + a)\sin\theta}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.