When a body slides down from rest along a smooth inclined plane… | JEE Main 2021 (1 Sep, Shift 2)

Q19JEE Main 2021 · 1 SEP · SHIFT 2Medium

Friction on an Incline

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 = __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

From rest over the same distance $s = \frac{1}{2}at^2$ , so $t \propto \frac{1}{\sqrt{a}}$ .
Smooth: $a_1 = g\sin\theta$ . Rough: $a_2 = g(\sin\theta - \mu\cos\theta)$ .
$t_2 = \alpha t_1 \Rightarrow a_1 = \alpha^2 a_2 \Rightarrow \sin\theta - \mu\cos\theta = \frac{\sin\theta}{\alpha^2}$
$\mu = \tan\theta\left(1 - \frac{1}{\alpha^2}\right) = \frac{1}{\sqrt{3}}\left(\frac{\alpha^2-1}{\alpha^2}\right)$ for $\theta = 30^\circ$
So $x = 3$ .

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