Question

The ratio of time taken by a body to slide down a smooth surface to rough identical inclined surface of angle $\theta$ is $‘n^{\prime }$. Find the coefficient of friction.

• A. $\frac{\tan \theta }{(1-n{)}^2}$
• B. $\cot \theta (1-n^2)$
• C. $\sin \theta (1-n^2)$
• D. $\tan \theta (1-n^2)$

Answer 1

The correct option is D $\tan \theta (1-n^2)$

Acceleration of the block on a smooth inclined plane:
$a=g\sin \theta$
Hence $s=\frac{1}{2}g\left(\sin \theta \right)t_1^2$;
Acceleration of the block on rough inclined plane: $a=g(\sin \theta -\mu \cos \theta )$
Hence, $s=\frac{1}{2}g(\sin \theta -\mu \cos \theta )t_2^2$;
Given:
$\frac{t_1}{t_2}=n;\frac{1}{2}g\left(\sin \theta \right)t_1^2=\frac{1}{2}g(\sin \theta -\mu \cos \theta )t_2^2\left(\sin \theta \right)n^2=(\sin \theta -\mu \cos \theta )\Rightarrow (1-n^2)\sin \theta =\mu \cos \theta$
$\Rightarrow \mu =\tan \theta (1-n^2)$