Question

The ratio of time taken by a body to slide down along a rough inclined plane to an identical smooth inclined plane for angle of inclination and coefficient of friction are as follows:-

S. no.	Ratio of times	Angle of inclinations	Coefficient of frictions
1.	$t_1/t_2=2$	$\theta ={30}^0$	${\mu }_1$
2.	$t_1/t_2=4$	$\theta ={45}^0$	${\mu }_2$
3.	$t_1/t_2=3$	$\theta ={30}^0$	${\mu }_3$

Then the correct relation between ${\mu }_1,{\mu }_2,{\mu }_3$ is:

• A. ${\mu }_1<{\mu }_2<{\mu }_3$
• B. ${\mu }_1>{\mu }_2>{\mu }_3$
• C. ${\mu }_1<{\mu }_3<{\mu }_2$
• D. ${\mu }_1>{\mu }_3>{\mu }_2$

Answer 1

Answer: (C)

${\mu }_1<{\mu }_3<{\mu }_2$

Let $s$ be the length of inclined plane with inclined angle $\theta$. A mass takes $t_1$ time on rough surface with friction coefficient $\mu$ and takes $t_2$ time on smooth surface.
Acceleration of mass $m$ along inclined plane on smooth surface $=$ $gsin\theta$ and acceleration on rough surface $=$$gsin\theta -\mu gcos\theta$
so $s=\frac{gsin\theta {t_2}^2}{2}=\frac{(gsin\theta -\mu gcos\theta ){t_1}^2}{2}$
$\mu =\frac{sin\theta -\frac{sin\theta {t_2}^2}{{t_1}^2}}{cos\theta }$
$\mu =tan\theta (1-\frac{{t_2}^2}{{t_1}^2})$

After putting the value:
${\mu }_1=\frac{3}{4\sqrt{3}}=0.43,{\mu }_2=\frac{15}{16}=0.93,{\mu }_3=\frac{8}{9\sqrt{3}}=0.51$