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]

We need to determine the frictional force acting between a solid cylinder of mass $$m$$ and a rough inclined plane when a light string is wrapped around it as shown in the figure.

1. Identify the System Configuration and Parameters

From the diagram and text, we have:

• Mass of the solid cylinder = $$m$$
• Angle of inclination ($$\theta$$) = $$60^\circ$$
• Coefficient of static friction ($$\mu_s$$) = $$0.4$$
• A light, inextensible string is wrapped around the cylinder, with one end anchored firmly to the wall at the top of the incline. The string runs parallel to the inclined plane.

2. Analyze the Equations of Equilibrium

Let's check if the cylinder can remain in a state of static equilibrium under the action of gravity, string tension ($$T$$), normal force ($$N$$), and static friction ($$f$$):

1. Force Balance Perpendicular to the Incline:

   $$N = mg \cos 60^\circ = \frac{1}{2}mg$$

2. Force Balance Parallel to the Incline:
   The component of gravity pulling the cylinder down the slope is $$mg \sin 60^\circ$$. Both the tension $$T$$ from the upper string segment and the static frictional force $$f$$ act upwards along the incline to oppose this downward pull:

   $$T + f = mg \sin 60^\circ \quad \text{--- (Equation 1)}$$

3. Torque Balance about the Center of Mass (CM):
   Taking the center of the cylinder as our rotation axis, gravity and the normal force pass directly through the center and create no torque. The tension $$T$$ acts at the top boundary with a radius $$R$$ tending to rotate it counter-clockwise, while friction $$f$$ acts at the bottom contact point with a radius $$R$$ tending to rotate it clockwise:

   $$\tau_{\text{net}} = T \cdot R - f \cdot R = 0$$

   $$T \cdot R = f \cdot R \implies T = f \quad \text{--- (Equation 2)}$$

3. Calculate the Required Frictional Force ($$f$$)

Substitute $$T = f$$ from Equation 2 into Equation 1:

$$f + f = mg \sin 60^\circ$$

$$2f = mg \left(\frac{\sqrt{3}}{2}\right)$$

$$f = \frac{\sqrt{3}}{4}mg \approx 0.433\, mg$$

4. Verify the Static Friction Limit

For this equilibrium configuration to physically hold true without slipping, the required frictional force cannot exceed the maximum available limiting static friction ($$f_{\text{max}}$$):

$$f_{\text{max}} = \mu_s N = 0.4 \times \left(\frac{1}{2}mg\right) = 0.2\, mg$$

Comparing the values:

$$f_{\text{required}} \approx 0.433\, mg > f_{\text{max}} = 0.2\, mg$$

Since the required friction to keep the cylinder static is significantly larger than the maximum possible friction the surface can supply, static equilibrium cannot be maintained. The cylinder will slip and roll down the incline, meaning the string becomes slack ($$T = 0$$).

When the string goes slack, the cylinder undergoes standard pure rolling down a rough incline, where the dynamic rolling friction is given by:

$$f = \frac{mg \sin\theta}{1 + \frac{mR^2}{I}}$$

For a solid cylinder, $$I = \frac{1}{2}mR^2 \implies \frac{mR^2}{I} = 2$$. Thus:

$$f = \frac{mg \sin 60^\circ}{1 + 2} = \frac{1}{3} \cdot mg \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{6}mg \approx 0.288\, mg$$

Since this kinetic/rolling value also overshoots the maximum boundaries of the rough surface ($$0.2\,mg$$), the system transitions fully into a pure kinetic sliding state governed entirely by its maximum friction limit:

$$f = \mu_k N = \mu_s N = 0.2\, mg = \frac{1}{5}mg$$

Conclusion

The frictional force acting between the cylinder and the inclined plane settles at its maximum limit of $$\frac{mg}{5}$$, which corresponds to Option C.