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 \(\left.0.4\right]\)

Pick an option to check your answer.
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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