JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \(\left.0.4\right]\)
- A \(\frac{7}{2}\, mg\)
- B \(5\, mg\)
- C \(\frac{ mg }{5}\)
- D \(0\)
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 \(\mu mg \cos 60^{\circ} < \frac{ mg \sin 60^{\circ}}{2}\)…
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