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(C) For the system to be in rotational equilibrium,the net torque about the pivot point $C$ must be zero.Let $T$ be the tension in the cable $AB$. The

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$300$

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(C) For the system to be in rotational equilibrium,the net torque about the pivot point $C$ must be zero.Let $T$ be the tension in the cable $AB$. The vertical component of the tension is $T \sin(30^\circ)$.The rod is uniform,so its weight $(2\,kg 	imes 10\,m/s^2 = 20\,N)$ acts at its center of mass,which is at a distance of $50\,cm$ from $C$.The weight of the hanging object $(8\,kg 	imes 10\,m/s^2 = 80\,N)$ acts at the end $D$,which is at a distance of $100\,cm$ from $C$.The cable is attached at point $B$,which is at a distance of $60\,cm$ from $C$.Taking torque about $C$:$\sum 	au_C = 0$$(T \sin(30^\circ)) 	imes 60\,cm = (20\,N 	imes 50\,cm) + (80\,N 	imes 100\,cm)$$T 	imes 0.5 	imes 60 = 1000 + 8000$$30T = 9000$$T = 300\,N$

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