Question Details

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).

Options

1/2 kg

1/3 kg

1/6 kg

1/12 kg

Show Answer

Correct Answer :

Option D

1/12 kg

Solution :

The correct option is 1/12 kg.

Step-by-Step Explanation:

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

1. Identify the given values from the question and the provided figure:
- Total length of the uniform rod = $200 cm$
- Mass of the uniform rod, $M = 500 g = 0.5 kg$
- Position of the wedge (pivot point) = $40 cm$ mark
- A suspended mass of $2 kg$ is at the $20 cm$ mark
- An unknown suspended mass $m$ is at the $160 cm$ mark
- Acceleration due to gravity, $g = 10 m / s^{2}$

2. Locate the Center of Gravity of the rod:
Since the rod is uniform, its weight acts at its geometric center (midpoint):
Center of gravity (C.G.) of the rod = $\frac{200}{2} = 100 cm$ mark

3. Calculate the distance of each force from the pivot (at the 40 cm mark):
- Distance to the $2 kg$ mass (on the left side):
$d_{1} = 40 cm - 20 cm = 20 cm$
- Distance to the weight of the rod (on the right side, at $100 cm$ ):
$d_{G} = 100 cm - 40 cm = 60 cm$
- Distance to the unknown mass $m$ (on the right side):
$d_{2} = 160 cm - 40 cm = 120 cm$

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:
$Anticlockwise Moments = Clockwise Moments$
$(2 kg \times g \times d_{1}) = (M \times g \times d_{G}) + (m \times g \times d_{2})$