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 ${x}=..... .$
JEE MAIN 2021, Difficult
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On smooth incline
${a}={g} \sin 30^{\circ}$
by ${S}={ut}+\frac{1}{2} {at}^{2}$
${S}=\frac{1}{2} \frac{{g}}{2} {T}^{2}=\frac{{g}}{4} {T}^{2} \ldots \ldots (i)$
On rough incline
${a}={g} \sin 30^{\circ}-\mu {g} \cos 30^{\circ}$
${by} {S}={ut}+\frac{1}{2} {at}^{2}$
${S}=\frac{1}{4} {g}(1-\sqrt{3} \mu)(\alpha {T})^{2} \ldots \text { (ii) }$
$\text { By (i) and (ii) }$
$\frac{1}{4} {g} {T}^{2}=\frac{1}{4} {g}(1-\sqrt{3} \mu) \alpha^{2} {T}^{2}$
$\Rightarrow 1-\sqrt{3} {g}=\frac{1}{\alpha^{2}} \Rightarrow {g}=\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right) \cdot \frac{1}{\sqrt{3}}$
$\Rightarrow {x}=3.00$
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