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. The time taken by the body to reach the bottom is

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

Answer 1

The correct option is C $\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}$

Applying motion in straight line

$s=ut+\frac{1}{2}a{t}^{2}l=t+\frac{1}{2}g\sin \theta t^2\frac{2h}{g\left({\sin }^2\theta \right)}=t^{2}t=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}$