Question

A block released from rest from the top of a smooth inclined plane of inclination ${45}^{\circ }$, takes time $t$ to reach the bottom. The same block released from rest, from top of a rough inclined plane of same inclination takes time $2t$ to reach the bottom. The coefficient of friction is

• A. $0.75$
• B. $0.5$
• C. $0.25$
• D. $0.4$

Answer 1

The correct option is A $0.75$

On smooth plane, acceleration is $g\sin \theta$. Then,

$l=\frac{1}{2}\left(g\sin \theta \right)t^2$$=\frac{1}{2}\left(g\sin {45}^{\circ }\right)t^2$$=\frac{5}{\sqrt{2}}{t}^2$

In case of rough surface,

$a_2=\frac{mg\sin \theta -f}{m}$$=\frac{mg\sin \theta -\mu mgc\cos \theta }{m}$$=g\sin \theta -\mu g\cos \theta$

In this case,

$l=\frac{1}{2}\left(a_2\right)(2t{)}^2$

Since, $l=\frac{5}{\sqrt{2}}{t}^2$

$\frac{5}{\sqrt{2}}{t}^2=\frac{1}{2}(10\times \frac{1}{\sqrt{2}}-\mu \times 10\times \frac{1}{\sqrt{2}})\left(4t^2\right)$

$1=4-4\mu$

$\mu =\frac{3}{4}=0.75$