Two masses of 500 gram and 600 grams are attached to the 10 cm and 80 cm marks respectively of a light metre scale. The moment of inertia of this system about an axis passing through the centre of the scale will be

Options

A. `"0.134 kg-m"^(2)`
B. `"2 kg-m"^(2)`
C. `"0.56 kg-m"^(2)`
D. `"4.5 kg-m"^(2)`

Explanation

To find the moment of inertia of the system about an axis passing through the center of the scale, we can follow these steps:

### Step 1: Identify the masses and their positions
We have two masses:
- Mass \( m_1 = 500 \, \text{g} = 0.5 \, \text{kg} \) at the 10 cm mark (0.1 m)
- Mass \( m_2 = 600 \, \text{g} = 0.6 \, \text{kg} \) at the 80 cm mark (0.8 m)

### Step 2: Determine the position of the axis
The axis of rotation is at the center of the meter scale, which is at the 50 cm mark (0.5 m).

### Step 3: Calculate the distances from the axis
- For mass \( m_1 \):
\[ r_1 = 0.5 \, \text{m} - 0.1 \, \text{m} = 0.4 \, \text{m} \]

- For mass \( m_2 \):
\[ r_2 = 0.8 \, \text{m} - 0.5 \, \text{m} = 0.3 \, \text{m} \]

### Step 4: Use the formula for moment of inertia
The moment of inertia \( I \) about the axis is given by the formula:
\[ I = m_1 \cdot r_1^2 + m_2 \cdot r_2^2 \]

### Step 5: Substitute the values into the formula
Substituting the values we have:
\[ I = (0.5 \, \text{kg}) \cdot (0.4 \, \text{m})^2 + (0.6 \, \text{kg}) \cdot (0.3 \, \text{m})^2 \]

Calculating each term:
- First term:
\[ 0.5 \cdot (0.4)^2 = 0.5 \cdot 0.16 = 0.08 \, \text{kg m}^2 \]

- Second term:
\[ 0.6 \cdot (0.3)^2 = 0.6 \cdot 0.09 = 0.054 \, \text{kg m}^2 \]

### Step 6: Add the contributions to find the total moment of inertia
\[ I = 0.08 + 0.054 = 0.134 \, \text{kg m}^2 \]

### Final Answer
The moment of inertia of the system about the axis passing through the center of the scale is:
\[ \boxed{0.134 \, \text{kg m}^2} \]

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