The moment of inertia of a meter scale of mass 0.6 kg about an axis perpendicular to the scale and located at the 20 cm position on the scale in kg m is (Breadth of the scale is negligible)

Options

A. 0.074
B. 0.104
C. 0.148
D. 0.208

Answer

To find the moment of inertia of a meter scale of mass 0.6 kg about an axis perpendicular to the scale and located at the 20 cm position, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Mass of the meter scale, \( m = 0.6 \, \text{kg} \)
- Length of the scale, \( L = 1 \, \text{m} \)
- Position of the axis from one end, \( d = 0.2 \, \text{m} \) (20 cm)

2. **Moment of Inertia about the Center**:
- The moment of inertia of a uniform rod about its center is given by the formula:
\[ I_C = \frac{1}{12} m L^2 \]
- Substituting the values:
\[ I_C = \frac{1}{12} \times 0.6 \, \text{kg} \times (1 \, \text{m})^2 = \frac{0.6}{12} = 0.05 \, \text{kg m}^2 \]

3. **Using the Parallel Axis Theorem**:
- To find the moment of inertia about the axis located at 20 cm (0.2 m), we use the parallel axis theorem:
\[ I = I_C + m d^2 \]
- Here, \( d \) is the distance from the center of the scale to the new axis. The center of the scale is at 0.5 m, so:
\[ d = 0.5 \, \text{m} - 0.2 \, \text{m} = 0.3 \, \text{m} \]
- Now substituting into the equation:
\[ I = 0.05 \, \text{kg m}^2 + 0.6 \, \text{kg} \times (0.3 \, \text{m})^2 \]
- Calculate \( m d^2 \):
\[ m d^2 = 0.6 \times 0.09 = 0.054 \, \text{kg m}^2 \]

4. **Final Calculation**:
- Now, adding the two components:
\[ I = 0.05 + 0.054 = 0.104 \, \text{kg m}^2 \]

### Conclusion:
The moment of inertia of the meter scale about the specified axis is:
\[ \boxed{0.104 \, \text{kg m}^2} \]