Question

When a body slides down from rest along a smooth inclined plane making an angle of $30^{\circ }$ with the horizontal, it takes time $T$. When the same body slides down from the rest along a rough inclined plane making the same angle and through the same distance, it takes time $\alpha T$, where $\alpha$ is a constant greater than $1$ . The co-efficient of friction between the body and the rough plane is $\frac{1}{\sqrt{x}}\left(\frac{{\alpha }^2-1}{{\alpha }^2}\right)$ where $x=$, ...

Answer 1

Answer: 3

On smooth incline
$a=g\sin 30^{\circ }$
by $S=ut+\frac{1}{2}a{t}^2$
$S=\frac{1}{2}\frac{g}{2}{T}^2=\frac{g}{4}{T}^2\dots \dots .$ (i)

On rough incline
$a=g\sin 30^{\circ }-\mu g\cos 30^{\circ }$
by $S=ut+\frac{1}{2}a{t}^2$
$S=\frac{1}{4}g(1-\sqrt{3}\mu )(\alpha T{)}^2...(ii)$
By (i) and (ii)
$\frac{1}{4}g{T}^2=\frac{1}{4}g(1-\sqrt{3}\mu ){\alpha }^2{T}^2$
$\Rightarrow 1-\sqrt{3}g=\frac{1}{{\alpha }^2}$
$\Rightarrow g=\left(\frac{{\alpha }^2-1}{{\alpha }^2}\right).\frac{1}{\sqrt{3}}$
$\Rightarrow x=3.00$