When a body slides down from rest along a smooth inclined plane making an angle of $45^{\circ}$ with the horizontal,it takes time $T$. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance,it is seen to take time $pT$,where $p$ is some number greater than $1$. Calculate the coefficient of friction between the body and the rough plane.

(D) For a smooth inclined plane:
The acceleration is $a = g \sin \theta$.
Given $\theta = 45^{\circ}$,so $a = g \sin 45^{\circ} = \frac{g}{\sqrt{2}}$.
The distance $d$ covered in time $T$ starting from rest is $d = \frac{1}{2} a T^2 = \frac{1}{2} \left( \frac{g}{\sqrt{2}} \right) T^2 = \frac{g T^2}{2\sqrt{2}}$.
For a rough inclined plane:
The net force is $F_{net} = mg \sin \theta - f_k = mg \sin \theta - \mu mg \cos \theta = mg(\sin \theta - \mu \cos \theta)$.
The acceleration is $a' = g(\sin \theta - \mu \cos \theta)$.
Since $\theta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,so $a' = g \left( \frac{1 - \mu}{\sqrt{2}} \right)$.
The distance $d$ covered in time $pT$ is $d = \frac{1}{2} a' (pT)^2 = \frac{1}{2} g \left( \frac{1 - \mu}{\sqrt{2}} \right) p^2 T^2$.
Equating the two expressions for $d$:
$\frac{g T^2}{2\sqrt{2}} = \frac{g p^2 T^2 (1 - \mu)}{2\sqrt{2}}$
$1 = p^2 (1 - \mu)$
$1 - \mu = \frac{1}{p^2}$
$\mu = 1 - \frac{1}{p^2} = \frac{p^2 - 1}{p^2}$.