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A box of mass 4 Kg is placed on a rough inclined plane of inclination $6 0^{0}$ . Its downward motion can be prevented by applying an upward pull F. And it can be made to slide upwards by applying a force 3F. The coefficient of friction between the box and inclined plane is

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

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

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

$\frac{1}{2}$

answer is B.

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Detailed Solution

$\frac{F}{3 F} = \frac{m g (\sin θ - μ \cos θ)}{m g (\sin θ + μ \cos θ)}$

$\sin θ + μ \cos θ = 3 (\sin θ - μ \cos θ)$

$μ = \frac{1}{2} \tan θ$

$a s θ = 60^{0} μ = \frac{\sqrt{3}}{2}$

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