Question

An object is placed on the surface of a smooth inclined plane of inclination $\theta .$ It takes time $t$ to reach the bottom. If the same object is allowed to slide down a rough inclined plane of inclination $\theta,$ it takes time $n t$ to reach the bottom where $n$ is the number greater than one. The coefficient of friction $\mu$ 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. $\mu=\cot \theta\left[1-\frac{1}{n^{2}}\right]^{1 / 2}$

Answer 1

Answer: (A)

On smooth inclined plane, the acceleration of body down the plane $=g \sin \theta .$ For rough inclined plane, the downward acceleration of body $=g \sin \theta-\mu g \cos$ $\theta$. Let $l$ be the length of inclined plane, then
$l=\frac{1}{2}(g \sin \theta) t^{2} \text { and } l=\frac{1}{2}(g \sin \theta-\mu g \cos \theta)(n t)^{2} $
$\therefore \frac{1}{2}(g \sin \theta) t^{2}=\frac{1}{2}(g \sin \theta-\mu g \cos \theta)(n t)^{2}$
or $\sin \theta=(\sin \theta-\mu \cos \theta) n^{2}=n^{2} \sin \theta-\mu n^{2} \cos \theta$
or $\quad \mu n^{2} \cos \theta=\left(n^{2}-1\right) \sin \theta$
or $\mu=\frac{\left(n^{2}-1\right) \sin \theta}{n^{2} \cos \theta}=\tan \theta\left(1-\frac{1}{n^{2}}\right)$