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 perpedicular to the line joining them is (in `kg-m^(2))`

Options

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

Explanation

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

### Step 1: Identify the masses and their positions
Let:
- Mass \( m_1 = 1 \, \text{kg} \) (located at \( x_1 = 0 \, \text{m} \))
- Mass \( m_2 = 2 \, \text{kg} \) (located at \( x_2 = 3 \, \text{m} \))

### Step 2: Calculate the position of the center of mass (CM)
The center of mass \( x_{cm} \) can be calculated using the formula:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
Substituting the values:
\[ x_{cm} = \frac{(1 \, \text{kg} \cdot 0 \, \text{m}) + (2 \, \text{kg} \cdot 3 \, \text{m})}{1 \, \text{kg} + 2 \, \text{kg}} = \frac{0 + 6}{3} = 2 \, \text{m} \]

### Step 3: Calculate the distances of each mass from the center of mass
Now, we need to find the distances of each mass from the center of mass:
- Distance of \( m_1 \) from \( x_{cm} \):
\[ d_1 = x_{cm} - x_1 = 2 \, \text{m} - 0 \, \text{m} = 2 \, \text{m} \]
- Distance of \( m_2 \) from \( x_{cm} \):
\[ d_2 = x_{cm} - x_2 = 2 \, \text{m} - 3 \, \text{m} = -1 \, \text{m} \quad (\text{taking absolute value: } 1 \, \text{m}) \]

### Step 4: Calculate the moment of inertia
The moment of inertia \( I \) about the center of mass is given by:
\[ I = m_1 d_1^2 + m_2 d_2^2 \]
Substituting the values:
\[ I = (1 \, \text{kg} \cdot (2 \, \text{m})^2) + (2 \, \text{kg} \cdot (1 \, \text{m})^2) \]
Calculating each term:
\[ I = (1 \cdot 4) + (2 \cdot 1) = 4 + 2 = 6 \, \text{kg-m}^2 \]

### Final Answer
The moment of inertia of the particles about the axis passing through their center of mass and perpendicular to the line joining them is \( 6 \, \text{kg-m}^2 \).

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