Question

A smooth inclined plane of length $L$ having an inclination 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

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

Here downward retardation means upward acceleration i.e $g^{\prime }=g+a$
The acceleration of the body along incline $a^{\prime }=g^{\prime }\sin \theta =(g+a)\sin \theta$
using $S=ut+\left(1/2\right)a{t}^2\Rightarrow L=0+\left(1/2\right)(g+a)\sin \theta t^2$
or $t=\sqrt{\frac{2L}{(g+a)\sin \theta }}$