Question

A smooth inclined plane is inclined at an angle $\theta$ with the horizontal. A body starts from rest and slides down the inclined surface, then the time taken by the body to reach the bottom is

• A. $\sqrt{\frac{2 h}{g}}$
• B. $\sqrt{\frac{21}{g}}$
• C. $\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}$
• D. $\sin \theta \sqrt{\frac{2 h}{g}}$

Answer 1

Answer: (C)

Acceleration due to gravity along inclined plane
$g^{1}=g \cos \left(90^{\circ}-\theta\right)=g \sin \theta$
$\therefore $ Time taken, $t=\sqrt{\frac{2 s}{g^{\prime}}}=\sqrt{\frac{2 \ell}{g \sin \theta}}$
But $\sin \theta=\frac{ h }{\ell} \Rightarrow \ell=\frac{ h }{\sin \theta}$
Hence, $t=\sqrt{\frac{2}{g \sin \theta}-\frac{h}{\sin \theta}}$
$t=\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}$