Question

A body takes time $t$ to reach the bottom of an inclined plane of angle $\theta$ with the horizontal. If the plane is made rough, time taken now is $2t$. The coefficient of friction of the rough surface is:

• A. $\frac{3}{4}\tan \theta$
• B. $\frac{2}{3}\tan \theta$
• C. $\frac{1}{4}\tan \theta$
• D. $\frac{1}{2}\tan \theta$

Answer 1

Answer: (A)

The free body diagram shows the various forces acting on the block. When plane is frictionless, then acceleration, $a=g\sin \theta$.
Let time taken is $t_1$ Since body starts from rest, initial velocity $u=0$
$\therefore s=ut+\frac{1}{2}a{t}_1^2$
$\Rightarrow s=\frac{1}{2}g\sin \theta t_1^2$
$\Rightarrow t_1=\sqrt{\frac{2s}{g\sin \theta }}$
When plane is made rough, then acceleration $a=g(\sin \theta -\mu \cos \theta )$, where $\mu$ is coefficient of friction.
$\therefore t_2=\sqrt{\frac{2s}{g(\sin \theta -\mu \cos \theta )}}$
Given, $t_2=2t_1$
$\therefore \sqrt{\frac{2s}{g(\sin \theta -\mu \cos \theta )}}$
$=2\sqrt{\frac{2s}{g\sin \theta }}$
$\therefore \frac{1}{\sin \theta -\mu \cos \theta }=\frac{4}{\sin \theta }$
$\Rightarrow \sin \theta =4\sin \theta -4\mu \cos \theta$
$\Rightarrow \mu =\frac{3}{4}\tan \theta$