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 0.4]

Translational Equilibrium 
$mg\sin \theta =T+f$  ----------(1)
Rotational Equilibrium (about CM)
$$\begin{gathered} T\times R=f\times R \\ \Rightarrow T=f----\left(2\right) \end{gathered}$$ 
 $\Rightarrow f=\frac{mg\sin 60}{2}=\frac{\sqrt{3}}{4}mg=0.433mg$
But maximum frictional fore is 
$f_{\max }=\mu mg\cos 60$  =  $0.2mg$
Applied force in greater than maximum static frictional force.

So, friction = 0.2 mg$=\frac{1}{5}mg$