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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2021
  • Hard
  • MCQ

A solid cylinder of mass \(m\) is wrapped with an inextensible light string and, is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is: [The coefficient of static friction, \(\mu_{ s },\) is \(\left.0.4\right]\)

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Answer & Solution

Correct Answer

(C) \(\frac{ mg }{5}\)

Step-by-step Solution

Detailed explanation

Let's take solid cylinder is in equilibrium

\(T + f = mg \sin 60......(i)\)

\(TR - fR =0......(ii)\)

Solving we get

\(T = f _{ seq }=\frac{ mg \sin \theta}{2}\)

But limiting friction \( < \) required friction

\(\mu mg \cos 60^{\circ} < \frac{ mg \sin 60^{\circ}}{2}\)

\(\therefore\) Hence cylinder will not remain in equilibrium

Hence \(f =\) kinetic

\(=\mu_{ k } N\)

\(=\mu_{ k } mg \cos 60^{\circ}\)

\(=\frac{ mg }{5}\)

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