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.

Step-by-step Explanation:

To find the value of the unknown mass $m$ that keeps the rod in horizontal equilibrium, we apply the principle of moments (rotational equilibrium) about the pivot (wedge).

For a system in rotational equilibrium, the sum of the counter-clockwise torques about any pivot point must equal the sum of the clockwise torques about that same point:

$\sum {\tau }_{\text{counter-clockwise}}=\sum {\tau }_{\text{clockwise}}$

Let's identify the forces acting on the rod and their respective distances from the pivot (wedge placed at the 40 cm mark):

1. Suspended mass on the left:
Mass $m_1=2\text{kg}$ is suspended at the 20 cm mark.
The distance from the pivot is:
$d_1=40\text{cm}-20\text{cm}=20\text{cm}$
This force produces a counter-clockwise torque about the pivot.

2. Mass of the uniform rod:
The rod is uniform, has a length of 200 cm, and a mass $M=500\text{g}=0.5\text{kg}$.
Its weight acts at its center of gravity, which is at the midpoint of the rod (100 cm mark).
The distance of the center of gravity from the pivot is:
$d_{\text{CM}}=100\text{cm}-40\text{cm}=60\text{cm}$ (to the right of the pivot).
This weight produces a clockwise torque about the pivot.

3. Unknown suspended mass on the right:
Mass $m$ is suspended at the 160 cm mark.
The distance from the pivot is:

This mass produces a clockwise torque about the pivot.

Equating the torques about the pivot:

$m_1·g·d_1=M·g·d_{\text{CM}}+m·g·d_2$

We can divide the entire equation by the acceleration due to gravity ($g$):

$m_1·d_1=M·d_{\text{CM}}+m·d_2$

Substituting the values into the equation:

$2·20=0.5·60+m·120$

$40=30+120m$

Subtracting 30 from both sides:

$10=120m$

Solving for $m$:

$m=\frac{10}{120}=\frac{1}{12}\text{kg}$

Thus, the value of the unknown mass $m$ required to balance the rod is $\frac{1}{12}\text{kg}$.