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(C) Let the solid cylinder be in equilibrium.The forces acting along the inclined plane are the tension $T$ and the friction $f$. The component of gra

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$\frac{mg}{5}$

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(C) Let the solid cylinder be in equilibrium.The forces acting along the inclined plane are the tension $T$ and the friction $f$. The component of gravity along the plane is $mg \sin 60^{\circ}$.For translational equilibrium: $T + f = mg \sin 60^{\circ} \quad ......(i)$For rotational equilibrium about the center of the cylinder: $TR - fR = 0 \implies T = f \quad ......(ii)$Substituting $(ii)$ into $(i)$: $2f = mg \sin 60^{\circ} \implies f = \frac{mg \sin 60^{\circ}}{2} = \frac{mg \sqrt{3}}{4} \approx 0.433 mg$.The limiting friction is $f_{L} = \mu_{s} N = \mu_{s} mg \cos 60^{\circ} = 0.4 	imes mg 	imes 0.5 = 0.2 mg$.Since the required friction $(0.433 mg)$ is greater than the limiting friction $(0.2 mg)$,the cylinder will not remain in static equilibrium and will roll down.The friction acting will be kinetic friction: $f_{k} = \mu_{k} N$. Assuming $\mu_{k} = \mu_{s} = 0.4$,we get $f_{k} = 0.4 	imes mg 	imes 0.5 = 0.2 mg = \frac{mg}{5}$.

$A$ 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 $0.4$]

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