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JEE Mains · Physics · STD 11 - Friction
  • JEE Mains 2021
  • Hard
  • Numerical

As shown in the figure, a block of mass \(\sqrt{3}\, kg\) is kept on a horizontal rough surface of coefficient of friction \(\frac{1}{3 \sqrt{3}}\). The critical force to be applied on the vertical surface as shown at an angle \(60^{\circ}\) with horizontal such that it does not move, will be \(3 x\). The value of \(3x\) will be \(\left[ g =10 m / s ^{2} ; \sin 60^{\circ}=\frac{\sqrt{3}}{2} ; \cos 60^{\circ}=\frac{1}{2}\right]\)

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Answer & Solution

Correct Answer

(B) \(10\)

Step-by-step Solution

Detailed explanation

\(F \cos 60^{\circ}=\mu N\) or \(\frac{ F }{2}=\frac{1}{3 \sqrt{3}} N\) \(\ldots\) (1)

\(\& N=\sin 60^{\circ}+\sqrt{3} g\) \(\ldots(2)\)

From equation

\((1)\;and\;(2)\)

\(\frac{F}{2}=\frac{1}{3 \sqrt{3}}\left(\frac{F \sqrt{3}}{2}+\sqrt{3} g\right)\)

\(\Rightarrow F = g =10\) Newton \(=3 x\)

So \(x=\frac{10}{3}=3.33\)

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