The time taken by a body to slide down a rough $45 °$ inclined plane is twice that required to slide down a smooth $45 °$ inclined plane. The coefficient of kinetic friction between the object and rough plane is given by

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$\frac{3}{4}$

$\sqrt{\frac{2}{3}}$

$\frac{1}{3}$

$\sqrt{\frac{3}{4}}$

answer is B.

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Detailed Solution

$t = \sqrt{\frac{2 S}{a}} \propto \frac{1}{\sqrt{a}} ∴ \frac{t_{1}}{t_{2}} = \sqrt{\frac{a_{2}}{a_{1}}} \frac{2}{1} = \sqrt{\frac{g \sin θ}{g \sin θ - μ_{K} g \cos θ}} = \sqrt{\frac{1}{1 - μ_{K}}}$

As, $\sin θ = \cos θ a t 45 °$

On solving the above equation, we get

$μ_{K} = \frac{3}{4}$

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