A block placed on a rough inclined plane of inclination left(θ=30^{circ}right) can just be pushed u

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4-1.Newton's Laws of Motion

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A block placed on a rough inclined plane of inclination $\left(\theta=30^{\circ}\right)$ can just be pushed upwards by applying a force " $F$ " as shown. If the angle of inclination of the inclined plane is increased to $\left(\theta=60^{\circ}\right)$, 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 is

A

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

B

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

C

$\frac{2 \sqrt{3}-1}{\sqrt{3}+1}$

D

none of these

Solution

$F=m g \sin 30^{\circ}+\mu m g \cos 30^{\circ}$

$=\frac{m g}{2}[1+\mu \sqrt{3}]$

$F+f=m g \sin 60^{\circ}$

$F=\frac{m g}{2}[\sqrt{3}-\mu]$

Now $(1)=(2)$

$1+\mu \sqrt{3}=\sqrt{3}-\mu$

$\mu=\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}$

Standard 11

Physics

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