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A body takes $1 \frac{1}{3}$ times as much time to slide down a rough inclined plane as it takes to slide down an identical bust smooth inclined plane. If the angle of inclination is $45 °$ , find the coefficient of friction.

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\frac{3}{16}

\frac{1}{16}

\frac{5}{16}

\frac{7}{16}

answer is D.

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Detailed Solution

t_{rough} = \frac{4}{3} t_{smooth}

$\sqrt{\frac{2 l}{g (\sin θ - μ \cos θ)}} = \frac{4}{3} \sqrt{\frac{2 l}{g \sin θ}}$

Substitute on both sides

$\frac{2 l}{g (\sin θ - μ \cos θ))} = \frac{16}{9} \times \frac{2 l}{g \sin θ}$

$9 \sin θ = 16 \sin θ - 16 μ \cos θ$

$16 μ \cos θ = 7 \sin θ$

$μ = \frac{7}{16} \tan θ$

$θ = 45 °$

$μ = \frac{7}{16}$

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