Question

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 a body to slide down the inclined plane, from rest, will be

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

Answer 1

The correct option is A $t=\sqrt{\frac{2L}{(g+a)sin\theta }}$

Downward retardation means upward acceleration.
g'= g + a
Now $t=\sqrt{\frac{2L}{g^{\prime }sin\theta }}$ or $t=\sqrt{\frac{2L}{(g+a)sin\theta }}$