Question

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

• A. \(\frac{1}{12}\) kg
• B. \(\frac{1}{2}\)kg
• C. \(\frac{1}{2}\)kg
• D. \(\frac{1}{2}\)kg

Answer 1

The Correct Option is A

To find the value of the unknown mass \( m \) such that the rod is in equilibrium, we need to apply the principles of moments (torques). A system is in equilibrium when the sum of the clockwise moments about any pivot is equal to the sum of the counterclockwise moments.

Let's analyze the problem step by step: 

1. The rod is 200 cm long and has a mass of 500 g (0.5 kg). The center of mass of the rod is at its midpoint, i.e., at 100 cm from either end. However, since it's pivoted at the 40 cm mark, the distance of the center of mass from the pivot is \( 60 \, \text{cm} \) (100 cm - 40 cm).
2. The downward force due to the weight of the rod is \( 0.5 \times 10 = 5 \, \text{N} \).
3. Given:
   • A 2 kg mass is hung at the 20 cm mark, creating a downward force of \( 2 \times 10 = 20 \, \text{N} \). The distance from the pivot is \( 40 \, \text{cm} - 20 \, \text{cm} = 20 \, \text{cm} \).
   • An unknown mass \( m \) is hung at the 160 cm mark. The downward force is \( m \times 10 \). The distance from the pivot is \( 160 \, \text{cm} - 40 \, \text{cm} = 120 \, \text{cm} \).
4. The moment due to the 2 kg mass is \( 20 \, \text{N} \times 20 \, \text{cm} = 400 \, \text{Nm} \) (counterclockwise).
5. The moment due to the weight of the rod is \( 5 \, \text{N} \times 60 \, \text{cm} = 300 \, \text{Nm} \) (clockwise).
6. The moment due to mass \( m \) is \( (m \times 10) \times 120 \, \text{cm} \) (clockwise).

For equilibrium, the sum of clockwise moments should equal the sum of counterclockwise moments:

\(20 \times 20 = 5 \times 60 + m \times 10 \times 120\) 
 

Simplifying:

\(400 = 300 + 1200m\) 
\(100 = 1200m\) 
\(m = \frac{100}{1200} = \frac{1}{12} \, \text{kg}\) 
 

Thus, the value of the unknown mass \( m \) is \(\frac{1}{12}\) kg.