Two partcles of masses 1 kg and 2 kg are placed at a distance of 3 m. Moment of inertia of the particles about an axis passing through their centre of mass and perpendicular to the line joining them is (in `kg-m^(2))` a) 6 b) 9 c) 8 d) 12

Options

A. 6
B. 9
C. 8
D. 12

Explanation

To find the moment of inertia of two particles about an axis passing through their center of mass and perpendicular to the line joining them, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses and Distance**:
- Let the masses of the two particles be \( m_1 = 1 \, \text{kg} \) and \( m_2 = 2 \, \text{kg} \).
- The distance between the two particles is \( d = 3 \, \text{m} \).

2. **Calculate the Center of Mass (CM)**:
- The position of the center of mass \( x_{CM} \) can be calculated using the formula:
\[ x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
- Assuming \( x_1 = 0 \) (position of \( m_1 \)) and \( x_2 = 3 \, \text{m} \) (position of \( m_2 \)):
\[ x_{CM} = \frac{(1 \, \text{kg} \cdot 0) + (2 \, \text{kg} \cdot 3 \, \text{m})}{1 \, \text{kg} + 2 \, \text{kg}} = \frac{6 \, \text{kg-m}}{3 \, \text{kg}} = 2 \, \text{m} \]

3. **Calculate the Distances from the Center of Mass**:
- The distance of \( m_1 \) from the center of mass:
\[ r_1 = x_{CM} - x_1 = 2 \, \text{m} - 0 \, \text{m} = 2 \, \text{m} \]
- The distance of \( m_2 \) from the center of mass:
\[ r_2 = x_2 - x_{CM} = 3 \, \text{m} - 2 \, \text{m} = 1 \, \text{m} \]

4. **Calculate the Moment of Inertia**:
- The moment of inertia \( I \) about the center of mass is given by:
\[ I = m_1 r_1^2 + m_2 r_2^2 \]
- Substituting the values:
\[ I = (1 \, \text{kg} \cdot (2 \, \text{m})^2) + (2 \, \text{kg} \cdot (1 \, \text{m})^2) \]
\[ I = (1 \cdot 4) + (2 \cdot 1) = 4 + 2 = 6 \, \text{kg-m}^2 \]

5. **Final Answer**:
- The moment of inertia of the particles about the axis passing through their center of mass is \( 6 \, \text{kg-m}^2 \).

### Conclusion:
The correct option is **(a) 6**.