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 action between the cylinder and the inclined plane is: 

[The coefficient of static friction, ${\mathrm{\mu }}_{\mathrm{S}}$ is 0.4]

Let's assume that the cylinder is in equilibrium.

Therefore

$\mathrm{T}+\mathrm{f}=\mathrm{mgsin}60°\mathrm{\_\_\_\_\_}\left(1\right)$

$\mathrm{TR}– \mathrm{fR}=0\mathrm{\_\_\_\_\_\_}\left(2\right)$

From equations 1 and 2,

$\mathrm{T}={\mathrm{f}}_{\mathrm{req}}=\frac{\mathrm{mgsin}{60}^{°}}{2}$ is the required friction for the cylinder to be in equilibrium.

But the limiting friction is 

${\mathrm{f}}_{\mathrm{L}}=\mu \mathrm{mgcos}60°$ is less than the required friction.

This means that the cylinder won't be in equilibrium and the friction will be 

$$\begin{gathered} \mathrm{f} = \mathrm{\mu K} \mathrm{N} \\ =\mathrm{\mu }\times \mathrm{mgcos}60° \\ =0.4\times \mathrm{mg}\times (½) \\ =\frac{\mathrm{mg}}{5} \end{gathered}$$

Hence the correct answer is $\frac{mg}{5}.$