Question

$A$ 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. $\left( g =10\, m / s ^{2}\right)$

• A. $\frac{1}{2} kg$
• B. $\frac{1}{3} kg$
• C. $\frac{1}{6} kg$
• D. $\frac{1}{12} kg$

Answer 1

The Correct Option is D

To solve this problem, we need to ensure that the rod is in rotational equilibrium. Therefore, the sum of the torques around the pivot point (the wedge at the 40 cm mark) must be zero.

Let's analyze the situation: 

• The length of the rod is \(200 \, \text{cm}\), and its center of mass is at the \(100 \, \text{cm}\) mark.
• The rod's mass is \(500 \, \text{g} = 0.5 \, \text{kg}\). Thus, its weight is \(0.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 5 \, \text{N}\).
• A mass of \(2 \, \text{kg}\) is suspended at the \(20 \, \text{cm}\) mark, creating a force of \(2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N}\).
• An unknown mass \(m\) is suspended at the \(160 \, \text{cm}\) mark.

We will calculate the torques around the pivot point (\(40 \, \text{cm}\) mark). The torque \((\tau)\) caused by a force \((F)\) at a distance \((d)\) from the pivot is given by:

\(\tau = F \times d\)

Let's calculate the torque for each force:

1. Torque due to the 2 kg mass at 20 cm mark:
   • Distance from pivot = \(40 \, \text{cm} - 20 \, \text{cm} = 20 \, \text{cm} = 0.2 \, \text{m}\)
   • Torque = \(20 \, \text{N} \times 0.2 \, \text{m} = 4 \, \text{Nm}\) (Clockwise)
2. Torque due to the rod's weight:
   • Distance from pivot = \(100 \, \text{cm} - 40 \, \text{cm} = 60 \, \text{cm} = 0.6 \, \text{m}\)
   • Torque = \(5 \, \text{N} \times 0.6 \, \text{m} = 3 \, \text{Nm}\) (Anticlockwise)
3. Torque due to the unknown mass \(m\) at 160 cm mark:
   • Distance from pivot = \(160 \, \text{cm} - 40 \, \text{cm} = 120 \, \text{cm} = 1.2 \, \text{m}\)
   • Torque = \(m \times 10 \, \text{m/s}^2 \times 1.2 \, \text{m} = 12m \, \text{Nm}\) (Anticlockwise)

For rotational equilibrium, the sum of the clockwise torques must equal the sum of anticlockwise torques:

\(4 \, \text{Nm} = 3 \, \text{Nm} + 12m \, \text{Nm}\)

Solving for \(m\):

\(4 - 3 = 12m\)

\(1 = 12m\)

\(m = \frac{1}{12} \, \text{kg}\)

Therefore, the mass \(m\) required for the rod to be in equilibrium is \(\frac{1}{12} \, \text{kg}\).