Rolling Motion question from JEE Main 2021 (18 Mar, Shift 2)

Answer: (d)

1. The string touches the cylinder on the top side and is fixed, so that point is at rest; for equilibrium the torques give $T = f$, and $T + f = mg\sin 60^\circ \Rightarrow f = \dfrac{\sqrt{3}}{4}mg \approx 0.43mg$.
2. Maximum static friction $= \mu N = 0.4 \times mg\cos 60^\circ = 0.2mg$, which is less, so the cylinder cannot stay at rest.
3. It turns about the string point and the contact point slips, so kinetic friction acts: $f = \mu N = 0.2mg = \dfrac{mg}{5}$.
4. So the answer is (d) $\frac{mg}{5}$.