A block placed on a rough inclined plane of inclination $(θ = 30 °)$ can just be pushed upwards by applying a force "F" as shown. If the angle of inclination of the inclined plane is increased to $(θ = 60 °)$ , the same block can just be prevented from sliding down by application of a force of same magnitude. The coefficient of friction between the block and the inclined plane

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\frac{\sqrt{7} - 1}{\sqrt{3} + 1}

\frac{\sqrt{3} - 1}{\sqrt{3} + 1}

\frac{\sqrt{5} - 1}{\sqrt{3} + 1}

answer is A.

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Detailed Solution

$N = mgcos θ$

$F = f + mg \sin θ$

$F = mg$

$(μ \cos θ + \sin θ) \Rightarrow F = mg [μ \times \frac{\sqrt{3}}{2} + \frac{1}{2}]$

$F = \frac{mg}{2} [\sqrt{3} μ + 1] \to (I) when θ = 60 °$

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$F = mg (\sin θ - μ \cos θ)$

$F = \frac{mg}{2} [\sqrt{3} - μ] \to (II)$

Comparing (I) & (II)

$\sqrt{3} μ + 1 = \sqrt{3} - μ$

$(\sqrt{3} + 1) μ = (\sqrt{3} - 1)$

$μ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

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