Laws of Motion

JEE Main level, single correct, medium

175 of 210

Question 175 of 210

A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2N2N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10N.10N. The coefficient of static friction between the block and the plane is: [Take g=10m/s2g=10m/{s}^{2} ]

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Answer: (D) 32\dfrac{\sqrt{3}}{2}

2+mgsin⁡⁡θ=μmgcos⁡⁡θ2+mg\sin ⁡\theta =\mu mg\cos ⁡\theta
mgsin⁡⁡θ−μmgcos⁡⁡θ=2........(1)mg\sin ⁡\theta -\mu mg\cos ⁡\theta =2........\left(1\right)
mgsin⁡⁡θ+μmgcos⁡⁡θ=10.......(2)mg\sin ⁡\theta +\mu mg\cos ⁡\theta =10.......\left(2\right)
equation (1) devided by (2)

mg(sin⁡⁡θ−μcos⁡⁡θ)mg(sin⁡⁡θ+μcos⁡⁡θ)=−210=−15\dfrac{mg\left(\sin ⁡\theta -\mu \cos ⁡\theta \right)}{mg\left(\sin ⁡\theta +\mu \cos ⁡\theta \right)}=\dfrac{-2}{10}=\dfrac{-1}{5}

(1−μ3)5=−(1+μ3)\left(1-\mu \sqrt{3}\right)5=-\left(1+\mu \sqrt{3}\right)

6=43μ6=4\sqrt{3}\mu

μ=32\mu =\dfrac{\sqrt{3}}{2}

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