Four masses 1, 2, 3 and 4 kg each are placed on four corners A, B, C and D of a square of side `sqrt(2)` m. The moment of inertia of this system about an axis passing through the point of inter-section of diagonals and perpendicular to the plane of the square will be:

Options

A. `5kg m^(2)`
B. `7.5kgm^(2)`
C. `10kg m^(2)`
D. `12.5kg m^(2)`

Answer

To find the moment of inertia of the system about an axis passing through the intersection of the diagonals of the square and perpendicular to the plane of the square, we can follow these steps:

### Step 1: Understand the Configuration
We have a square ABCD with masses placed at its corners:
- Mass at A (1 kg)
- Mass at B (2 kg)
- Mass at C (3 kg)
- Mass at D (4 kg)

The side length of the square is given as \( \sqrt{2} \) m.

### Step 2: Determine the Coordinates of the Masses
We can place the square in a coordinate system:
- A (0, 0)
- B (\( \sqrt{2} \), 0)
- C (\( \sqrt{2} \), \( \sqrt{2} \))
- D (0, \( \sqrt{2} \))

The point of intersection of the diagonals (the center of the square) is at:
- O (\( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \))

### Step 3: Calculate the Distance from the Center to Each Mass
The distance \( r \) from the center O to each mass is calculated as follows:

1. For mass at A (1 kg):
\[ r_A = \sqrt{\left(0 - \frac{\sqrt{2}}{2}\right)^2 + \left(0 - \frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1} = 1 \text{ m} \]

2. For mass at B (2 kg):
\[ r_B = \sqrt{\left(\sqrt{2} - \frac{\sqrt{2}}{2}\right)^2 + \left(0 - \frac{\sqrt{2}}{2}\right)^2} = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2} = \sqrt{1} = 1 \text{ m} \]

3. For mass at C (3 kg):
\[ r_C = \sqrt{\left(\sqrt{2} - \frac{\sqrt{2}}{2}\right)^2 + \left(\sqrt{2} - \frac{\sqrt{2}}{2}\right)^2} = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{1} = 1 \text{ m} \]

4. For mass at D (4 kg):
\[ r_D = \sqrt{\left(0 - \frac{\sqrt{2}}{2}\right)^2 + \left(\sqrt{2} - \frac{\sqrt{2}}{2}\right)^2} = \sqrt{\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{1} = 1 \text{ m} \]

### Step 4: Calculate the Moment of Inertia
The moment of inertia \( I \) is given by the formula:
\[ I = \sum m_i r_i^2 \]
Where \( m_i \) is the mass and \( r_i \) is the distance from the axis of rotation.

Calculating for each mass:
- For mass at A:
\[ I_A = 1 \cdot (1^2) = 1 \text{ kg m}^2 \]
- For mass at B:
\[ I_B = 2 \cdot (1^2) = 2 \text{ kg m}^2 \]
- For mass at C:
\[ I_C = 3 \cdot (1^2) = 3 \text{ kg m}^2 \]
- For mass at D:
\[ I_D = 4 \cdot (1^2) = 4 \text{ kg m}^2 \]

Now, summing these contributions:
\[ I = I_A + I_B + I_C + I_D = 1 + 2 + 3 + 4 = 10 \text{ kg m}^2 \]

### Final Answer
The moment of inertia of the system about the specified axis is \( 10 \text{ kg m}^2 \).

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