In remeasurement of mass of a given object by the principle of moments, the meter scale is hung from its midd -point. A known weight of mass 2kg is hung at one end of meter scale at distance 50 cm and unknown weight of mass m kg is bung at 20cm from the centre on other side. The Value of m is

Options

A. 2kg
B. 5kg
C. 2.5 kg
D. 0.8 kg

Answer

To solve the problem of finding the unknown mass \( m \) using the principle of moments, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- A meter scale is hung from its midpoint.
- A known mass of \( 2 \, \text{kg} \) is hung at one end (50 cm from the center).
- An unknown mass \( m \) is hung at a distance of \( 20 \, \text{cm} \) from the center on the other side.

2. **Identify the Forces**:
- The weight of the known mass \( 2 \, \text{kg} \) is \( W_1 = 2g \) (where \( g \) is the acceleration due to gravity).
- The weight of the unknown mass \( m \) is \( W_2 = mg \).

3. **Apply the Principle of Moments**:
- For the system to be in equilibrium (balanced), the clockwise moments about the pivot point must equal the counterclockwise moments.
- The moment due to the known mass is given by:
\[ \text{Moment}_1 = W_1 \times \text{Distance from center} = (2g) \times 0.5 \, \text{m} = 2g \times 0.5 = g \]
- The moment due to the unknown mass is given by:
\[ \text{Moment}_2 = W_2 \times \text{Distance from center} = (mg) \times 0.2 \, \text{m} = mg \times 0.2 \]

4. **Set the Moments Equal**:
- Since the scale is balanced, we can set the two moments equal to each other:
\[ g = mg \times 0.2 \]

5. **Cancel \( g \) from Both Sides**:
- Assuming \( g \neq 0 \), we can divide both sides by \( g \):
\[ 1 = m \times 0.2 \]

6. **Solve for \( m \)**:
- Rearranging the equation gives:
\[ m = \frac{1}{0.2} = 5 \, \text{kg} \]

### Final Answer:
The value of the unknown mass \( m \) is \( 5 \, \text{kg} \).

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