Question

A body slides on an inclined plane. If height of inclined plane is '$h$' and length is '$l$' and angle of inclination is $\theta$ then time is taken for travelling from upper points to lower point 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

Answer: (C)

$\frac{1}{sin\theta }\sqrt{\frac{2h}{g}}$

The length of incline plane is $l$.

The distance travelled by the particle $=l$

And the acceleration is sin component of $g$ is $a=gsin\theta$

By using newtons equation of motion:

$S=ut+\frac{1}{2}a{t}^2$

Here $u=0$ because particle start from rest.

$S=l=\frac{1}{2}gsin\theta t^2$ (equation 1)

For right angle triangle:

$sin\theta =\frac{perpendicular}{hypoteneous}=\frac{height}{inclinelenght}=\frac{h}{l}$

So, $l=\frac{h}{sin\theta }$ (equation 2)

By solving equation 1 and 2:

$\frac{h}{sin\theta }=\frac{1}{2}gsin\theta t^{2}t=\frac{1}{sin\theta }\sqrt{\frac{2h}{g}}$