Question

A block moves down a smooth inclined plane of inclination $\theta $ . Its velocity on reaching the bottom is $v$ . If it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is $v / n$ , where $n$ is a number greater than $1$. The coefficient of friction is given by

• A. $\mu =\tan \theta \left(1 - \frac{1}{n^{2}}\right)$
• B. $\mu =\cot \theta \left(1 - \frac{1}{n^{2}}\right)$
• C. $\mu=\tan \theta\left(1-\frac{1}{n^{2}}\right)^{1 / 2}$
• D. $\mu = \text{cot} \theta \left(1 - \frac{1}{n^{2}}\right)^{\text{1/2}}$s

Answer 1

Answer: (A)

For smooth, $ \, v^{2}=2\left(g \, \sin \theta \right) \, s . . .\left(\right.1\left.\right)$
For rough, $ \, \left(\frac{v}{n}\right)^{2}= \, 2 \, g \, \left(\sin \theta - \mu \cos \theta \right)S. . .\left(\right.2\left.\right)$
on solving, $ \, \mu =\left(1 - \frac{1}{n^{2}}\right)tan\theta $