When a body slides down from rest along a smooth inclined plane making an angle of $30°$ with the horizontal, it takes time 20 s. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it takes time 20p s, where p is some number greater than 1. The coefficient of friction between the body and the rough plane is 

 

The acceleration of the body sliding down smooth inclined plane is $a=g \sin \theta \dots \dots \left(1\right)$

The acceleration of the same body sliding down a rough inclined plane is 

$$\begin{gathered} a\text{'}=g\sin \theta -\mu g\cos \theta \\ =g(\sin \theta -\mu \cos \theta ) \dots \dots \left(2\right) \\ d=\frac{1}{2}a{t}^2=\frac{1}{2}a\text{'}t{\text{'}}^2 (\because u=0) \\ \frac{a}{a\text{'}}=\frac{t{\text{'}}^2}{t^2}=p^2 (\because t\text{'}=20p s, t=20 s) \\ \frac{g\sin \theta }{g(\sin \theta -\mu \cos \theta )}=p^2 [U\sin g (1) and (2\left)\right] \\ \sin \theta =p^2\sin \theta -\mu p^2\cos \theta ; \\ \mu p^2\cos \theta =p^2\sin \theta -\sin \theta \\ \mu \cos \theta =\frac{\left(p^2-1\right)}{p^2}\sin \theta ; \\ \mu =\tan \theta \left(1-\frac{1}{p^2}\right) \\ Here, \theta =30° \\ \therefore \mu =\left(1-\frac{1}{p^2}\right)\frac{1}{\sqrt{3}} \end{gathered}$$