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 correct option is 1/12 kg.

To find the value of the unknown mass $m$ for the rod to be in rotational equilibrium, we apply the principle of moments about the wedge (pivot point).

1. Identify the given parameters from the statement and the figure:
- Length of the uniform rod: $L=200\mathrm{cm}$
- Mass of the rod: $M=500g=0.5\mathrm{kg}$
- Position of the wedge (pivot point): $x_p=40\mathrm{cm}$ mark
- First suspended mass: $m_1=2\mathrm{kg}$ at position $x_1=20\mathrm{cm}$ mark
- Second suspended mass: $m_2=m$ at position $x_2=160\mathrm{cm}$ mark

2. Position of the center of gravity of the rod:
Since the rod is uniform, its weight acts at its geometric center (midpoint), which is at:
$x_g=\frac{200}{2}=100\mathrm{cm}$ mark

3. Calculate the distance of each force from the pivot (40 cm mark):
- Distance of the 2 kg mass:
$d_1=40-20=20\mathrm{cm}$ (to the left of the pivot)
- Distance of the rod's center of gravity (0.5 kg):
$d_g=100-40=60\mathrm{cm}$ (to the right of the pivot)
- Distance of the unknown mass $m$:
$d_2=160-40=120\mathrm{cm}$ (to the right of the pivot)

4. Apply the Principle of Moments:
For the rod to remain in rotational equilibrium, the sum of the anticlockwise moments about the pivot must equal the sum of the clockwise moments:
$\text{Anticlockwise Moments}=\text{Clockwise Moments}$
The 2 kg mass on the left creates an anticlockwise torque. The weight of the rod and the mass $m$ on the right create clockwise torques:
$m_1\cdot g\cdot d_1=M\cdot g\cdot d_g+m\cdot g\cdot d_2$
Dividing both sides by the acceleration due to gravity ($g$):
$m_1\cdot d_1=M\cdot d_g+m\cdot d_2$

5. Substitute the values into the equation:
$2\cdot 20=0.5\cdot 60+m\cdot 120$
$40=30+120m$
Subtract 30 from both sides:
$10=120m$
Solve for $m$:
$m=\frac{10}{120}=\frac{1}{12}\mathrm{kg}$

Thus, the value of the unknown mass $m$ required to keep the rod in equilibrium is 1/12 kg.