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Consider 2 inclined plane of same height 1^st has a smooth surface and angle of inclination is $45^{\circ}$ , other has a rough surface and angle of inclination is $60^{\circ}$ . If ratio of time taken to slide on then its ‘n’. Find the coefficient of friction of rough inclined plane

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$μ = 3 n^{2}$

$μ = \frac{3 - 2 n^{2}}{\sqrt{3}}$

$μ = \frac{3 - \sqrt{3} n^{2}}{2}$

$μ = \frac{2 n^{2}}{\sqrt{3}}$

answer is B.

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Detailed Solution

$(g \frac{\sqrt{3}}{2} - \frac{μ g}{2}) = g (\frac{\sqrt{3}}{2} - \frac{μ}{2})$

$t_{1} = \sqrt{\frac{2 l_{1}}{a}} t_{2} = \sqrt{\frac{2 l_{2}}{a}}$

$= \sqrt{\frac{2 \sqrt{2} h}{(\frac{g}{\sqrt{2}})}} = \sqrt{\frac{2 \times 2 h}{\sqrt{3} g (\frac{\sqrt{3}}{2} - \frac{μ}{2})}}$

$= \sqrt{\frac{4 h}{g}} = \sqrt{\frac{8 h}{g (3 - \sqrt{3} μ)}}$

$\frac{t_{1}}{t_{2}} = \sqrt{\frac{3 - \sqrt{3} μ}{2}} = n \Rightarrow 3 - \sqrt{3} μ = 2 n^{2}$

$\Rightarrow \sqrt{3} μ = 3 - 2 n^{2} \Rightarrow μ = (\frac{3 - 2 n^{2}}{\sqrt{3}})$

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