JEE Mains · Physics · STD 11 - 4.2 friction
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 \(\left[ g =10 m / s ^{2} ; \sin 60^{\circ}=\frac{\sqrt{3}}{2} ; \cos 60^{\circ}=\frac{1}{2}\right]\)
- A \(20\)
- B \(10\)
- C \(40\)
- D \(25\)
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 \((1)\;and\;(2)\) \(\frac{F}{2}=\frac{1}{3 \sqrt{3}}\left(\frac{F \sqrt{3}}{2}+\sqrt{3} g\right)\) \(\Rightarrow F = g =10\)…
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