Correct option is (3) \(\frac{mg}5\)
Let's take solid cylinder is in equilibrium
T + f = mg sin60° ....(i)
TR – fR = 0 ....(ii)
Solving we get
T = freq = \(\frac{mgsin\,\theta}2\)
But limiting friction < required friction
μmgcos60° < \(\frac{mg\,sin\,60^o}2\)
\(\therefore\) Hence cylinder will not remain in equilibrium
Hence f = kinetic
= μkN
= μkmgcos 60°
= \(\frac{mg}5\)