Question

A body starts from rest and slides down the inclined surface of a smooth inclined plane having inclination $\theta$ with the horizontal. 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. $\sin \theta \sqrt{\frac{2h}{g}}$

Answer 1

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

Body will move along along the inclined plane. Net force ($F)$ along the inclined plane is $mg\sin \theta$


Hence, acceleration of body along the inclined surface is,
$a=\frac{F}{m}=\frac{mg\sin \theta }{m}$
$a=g\sin \theta ...\left(i\right)$
When body reaches the bottom, its displacement $\left(s\right)$ along the inclined surface is $l$.
$\sin \theta =\frac{h}{l}$

Applying equation of motion along the inclined surface,
$s=ut+\frac{1}{2}a{t}^2$
Here, $s=l,u=0$ and $a=g\sin \theta$
$\Rightarrow l=0+\frac{1}{2}\times g\sin \theta t^2$
$\Rightarrow t^2=\frac{2l}{g\sin \theta }$
$[\because l=\frac{h}{\sin \theta }]$
$\Rightarrow t^2=\frac{2h}{g{\sin }^2\theta }$
$\therefore t=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}$