The moment of inertia of a metre scale of mass 0.6 kg about an axis perpendicular to the scale and passing through 30 cm position on the scale is given by (Breadth of'the scale is negligible)

Options

A. `"0.104 kg - m"^(2)`
B. `"0.208 kg - m"^(2)`
C. `"0.070 kg - m"^(2)`
D. `"0.148 kg - m"^(2)`

Explanation

To find the moment of inertia of a meter scale of mass 0.6 kg about an axis perpendicular to the scale and passing through the 30 cm position, we can use the parallel axis theorem. Here’s a step-by-step solution:

### Step 1: Understand the Configuration
The meter scale is 1 meter long, and we need to find the moment of inertia about an axis that is perpendicular to the scale and located at the 30 cm mark.

### Step 2: Identify the Parameters
- Mass of the scale (m) = 0.6 kg
- Length of the scale (L) = 1 m = 100 cm
- Distance from the center of the scale to the axis at 30 cm = 50 cm - 30 cm = 20 cm = 0.2 m

### Step 3: Calculate the Moment of Inertia about the Center
The moment of inertia of a uniform rod about its center is given by the formula:
\[ I_{center} = \frac{1}{12} m L^2 \]
Substituting the values:
\[ I_{center} = \frac{1}{12} \times 0.6 \, \text{kg} \times (1 \, \text{m})^2 = \frac{0.6}{12} = 0.05 \, \text{kg m}^2 \]

### Step 4: Apply the Parallel Axis Theorem
The parallel axis theorem states:
\[ I = I_{center} + m d^2 \]
where \(d\) is the distance from the center of mass to the new axis. Here, \(d = 0.2 \, \text{m}\).

Substituting the values:
\[ I = 0.05 \, \text{kg m}^2 + 0.6 \, \text{kg} \times (0.2 \, \text{m})^2 \]
Calculating \(0.6 \times (0.2)^2\):
\[ 0.6 \times 0.04 = 0.024 \, \text{kg m}^2 \]

### Step 5: Final Calculation
Now, substituting back into the equation:
\[ I = 0.05 \, \text{kg m}^2 + 0.024 \, \text{kg m}^2 = 0.074 \, \text{kg m}^2 \]

### Conclusion
The moment of inertia of the meter scale about the axis perpendicular to the scale and passing through the 30 cm position is:
\[ \boxed{0.074 \, \text{kg m}^2} \]