A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass ‘m’ is suspended from the rod at 160 cm mark as shown in the figure. Find the value of ‘m’ such that the rod is in equilibrium. (g=10 m/s²)

The problem provides a uniform rod of length $200\backslash text\{\; cm\}$ and mass $500\backslash text\{\; g\}=0.5\backslash text\{\; kg\}$. The rod is supported at the $40\backslash text\{\; cm\}$ mark (the wedge). A known mass of $2\backslash text\{\; kg\}$ hangs at the $20\backslash text\{\; cm\}$ mark, and an unknown mass $m$ hangs at the $160\backslash text\{\; cm\}$ mark. We must find $m$ so that the rod is in equilibrium (net torque about the support is zero). The acceleration due to gravity is $g=10\backslash \; \backslash text\{m/s\}^2$.

First, identify the distances of each force from the pivot (the wedge at $40\backslash text\{\; cm\}$).

• Mass $2\backslash text\{\; kg\}$ is at $20\backslash text\{\; cm\}$. Distance from pivot: $40-20=20\backslash text\{\; cm\}$ (to the left).
• The rod’s own weight acts at its centre of mass, which is at the midpoint of the rod: $200/2=100\backslash text\{\; cm\}$. Distance from pivot: $100-40=60\backslash text\{\; cm\}$ (to the right).
• Unknown mass $m$ is at $160\backslash text\{\; cm\}$. Distance from pivot: $160-40=120\backslash text\{\; cm\}$ (to the right).

All forces act downward, producing torques about the pivot. torques that tend to rotate the rod clockwise are taken as positive; those that tend to rotate it anticlockwise are taken as negative.

Calculate each torque (force × distance). Use $F=mg$ for each mass.

Clockwise torques (right‑hand side):

Anticlockwise torque (left‑hand side):

For equilibrium the sum of clockwise torques equals the sum of anticlockwise torques:

Solve for $m$:

Thus the unknown mass that balances the rod is 1/12 kg, which matches the given correct option.