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JEE MAIN 2021
01-09-21 S2
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 $\alpha \mathrm{T}$, 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 $x=$ $\_\_\_\_$
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Solution
On smooth incline $$ a=g \sin 30^{\circ} $$ by $\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}{ }^2$ $$ \mathrm{S}=\frac{1}{2} \frac{\mathrm{~g}}{2} \mathrm{~T}^2=\frac{\mathrm{g}}{4} \mathrm{~T}^2 $$ On rough incline $$ a=g \sin 30^{\circ}-\mu g \cos 30^{\circ} $$ by $\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2$ $$ S-{ }_4^1 g(1-\sqrt{3} \mu)(\alpha T)^2 $$ By (i) and (ii) $$ \begin{aligned} & \frac{1}{4} \mathrm{gT}^2=\frac{1}{4} \mathrm{~g}(1-\sqrt{3} \mu) \alpha^2 \mathrm{~T}^2 \\ & \Rightarrow \quad 1-\sqrt{3} \mathrm{~g}=\frac{1}{\alpha^2} \Rightarrow \mathrm{~g}=\left(\frac{\alpha^2-1}{\alpha^2}\right) \cdot \frac{1}{\sqrt{3}} \\ & \Rightarrow \mathrm{x}=3.00 \end{aligned} $$
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Question Tags
JEE Main
Physics
Easy
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