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A block sliding down a rough $45 °$ inclined plane has half the velocity it would have had, the inclined plane been smooth. The coefficient of sliding friction between block and the inclined plane is

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\frac{3}{4}

\frac{1}{\sqrt{2}}

\frac{1}{4}

\frac{1}{2 \sqrt{2}}

answer is B.

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Detailed Solution

v_{rough} = \frac{1}{2} v_{smooth}

$\sqrt{2 l g (\sin θ - μ \cos θ)} = \frac{1}{2} \sqrt{2 l g \sin θ}$

Substitute on both sides

$2 l g (\sin θ - μ \cos θ) = \frac{1}{4} 2 l g \sin θ$

$4 \sin θ - 4 μ \cos θ = \sin θ$

$3 \sin θ = 4 μ \cos θ$

$μ = \frac{3}{4} \tan θ$

$θ = 45 °$

$μ = \frac{3}{4}$

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