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 \(\left[ g =10 m / s ^{2} ; \sin 60^{\circ}=\frac{\sqrt{3}}{2} ; \cos 60^{\circ}=\frac{1}{2}\right]\)

Pick an option to check your answer.
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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