Interpret the figure/setup

A uniform rod $CD$ of length $1\,\text{m}$ is hinged at $C$ to a vertical wall. A mass of $8\,\text{kg}$ hangs from the free end $D$. The rod itself has mass $2\,\text{kg}$.

The rod is supported by a cable $AB$. From the standard equilibrium configuration for this question, the rod is horizontal and the cable makes an angle such that its vertical component provides the balancing torque. The figure corresponds to a $3\!:\!4\!:\!5$ triangle, so $\sin\theta = \frac{3}{5}$ where $\theta$ is the angle between the cable and the rod.

List the forces on the rod

• Weight of hanging object at $D$: $W_1 = 8g = 8\times 10 = 80\,\text{N}$
• Weight of uniform rod acting at its center: $W_2 = 2g = 2\times 10 = 20\,\text{N}$
• Tension $T$ in the cable at the end of the rod.
• Hinge reaction at $C$ (its torque about $C$ is zero, so we need not resolve it).

Take moments about the hinge $C$

Since the system is in equilibrium, net torque about $C$ must be zero.

The clockwise torques due to weights are:

• Due to the $8\,\text{kg}$ mass at distance $1\,\text{m}$: $\tau_1 = 80\times 1 = 80\,\text{N m}$
• Due to the rod's weight at its center, distance $0.5\,\text{m}$: $\tau_2 = 20\times 0.5 = 10\,\text{N m}$

Total clockwise torque: $\tau_{\text{cw}} = 80 + 10 = 90\,\text{N m}$

Torque due to tension

Only the component of tension perpendicular to the rod contributes.

If the cable makes angle $\theta$ with the rod, then perpendicular component is $T\sin\theta$

Hence anticlockwise torque due to tension is $\tau_T = T\sin\theta \cdot 1 = T\sin\theta$

Using $\sin\theta = \frac{3}{5}$, $\tau_T = T\cdot \frac{3}{5}$

Equilibrium condition

$T\cdot \frac{3}{5} = 90$

Therefore, $T = 90\cdot \frac{5}{3} = 150\,\text{N}$

Compare with options

The calculated tension is $T = 150\,\text{N}$

This value is not present among the given options $90, 240, 30, 300$.

Check stored answer $D = 300\,\text{N}$

If $T=300\,\text{N}$, then its perpendicular component would be $T\sin\theta = 300\times \frac{3}{5} = 180\,\text{N}$ giving torque $180\,\text{N m}$, which is double the required balancing torque of $90\,\text{N m}$. So $300\,\text{N}$ is inconsistent with equilibrium.

Conclusion

Based on standard torque balance, the tension should be $\boxed{150\,\text{N}}$

Since this is not among the options, there is likely an error either in the figure/options or in the stored answer.