JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Rotational Mechanics

Taking torques about the pivot C removes the hinge reaction. The cable pulls at B, a distance $60\text{cm}$ from C, at $30^{\circ }$ to the rod, so its torque is $T\sin 30^{\circ }\times 60$. Against it act the rod's weight $20\text{N}$ at its mid-point $50\text{cm}$ from C and the hanging load $80\text{N}$ at the far end $100\text{cm}$ from C. Balancing, $T\sin 30^{\circ }\times 60=20\times 50+80\times 100$, that is $30T=9000$. The tension in the cable is $T=300\text{N}$.