JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 8)
A block of mass $m$ slides down the plane inclined at angle $$30^{\circ}$$ with an acceleration $$\frac{g}{4}$$. The value of coefficient of kinetic friction will be:
$$\frac{2 \sqrt{3}-1}{2}$$
$$\frac{\sqrt{3}}{2}$$
$$\frac{1}{2 \sqrt{3}}$$
$$\frac{2 \sqrt{3}+1}{2}$$
Explanation
$\mathrm{Mg} \sin 30^{\circ}-\mu \mathrm{mg} \cos 30^{\circ}=\mathrm{ma}$
$$ \frac{g}{2}-\frac{\sqrt{3}}{2} \cdot \mu g=\frac{g}{4} $$
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$$ \begin{aligned} & \frac{\sqrt{3}}{2} \mu=\frac{1}{4} \\\\ & \mu=\frac{1}{2 \sqrt{3}} \end{aligned} $$
$$ \frac{g}{2}-\frac{\sqrt{3}}{2} \cdot \mu g=\frac{g}{4} $$
$$ \begin{aligned} & \frac{\sqrt{3}}{2} \mu=\frac{1}{4} \\\\ & \mu=\frac{1}{2 \sqrt{3}} \end{aligned} $$
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