Question

The time taken by a body to slide down a rough ${30}^{\circ }$ inclined plane is twice of that required to slide down a smooth ${30}^{\circ }$ inclined plane. The coefficient of kinetic friction between the object and the rough plane is given by

• A. $\frac{8}{9}$
• B. $\frac{\sqrt{3}}{4}$
• C. $\sqrt{\frac{3}{4}}$
• D. $\sqrt{\frac{2}{4}}$

Answer 1

The correct option is B $\frac{\sqrt{3}}{4}$

Acceleration of block moving down rough plane $=a_1=g\sin \theta -\mu g\cos \theta$
Acceleration of block moving down smooth surface $=a_2=g\sin \theta$

Time taken to slide down the plane a distance of $s$ is found by
$s=\frac{1}{2}a{t}^2$
$\Rightarrow t=\sqrt{\frac{2s}{a}}$

Thus $t_1=\sqrt{\frac{2s}{g\sin {30}^{\circ }-\mu g\cos {30}^{\circ }}}...\left(i\right)t_2=\sqrt{\frac{2s}{g\sin {30}^{\circ }}}...\left(ii\right)$

As it is given that the time taken by a body to slide down a rough ${30}^{\circ }$ inclined plane is twice that required to slide down a smooth ${30}^{\circ }$ inclined plane.
$t_1=2t_2$

By squaring and dividing the equation $\left(i\right)/\left(ii\right)$ we get,

$\Rightarrow 2^2=\frac{g\sin {30}^{\circ }}{g\sin {30}^{\circ }-\mu g\cos {30}^{\circ }}$
$\Rightarrow 4=\frac{g}{g-\sqrt{3}\mu g}$
$\Rightarrow 4-4\sqrt{3}\mu =1$
$\Rightarrow \mu =\frac{\sqrt{3}}{4}$

Hence option B is the correct answer