In an arrangement four particles, each of mass 2 gram are situated at the coordinate points (3, 2, 0), (1, –1, 0), (0, 0, 0) and (–1, 1, 0). The moment of inertia of this arrangement about the Z-axis will be

Options

A. 8 units
B. 16 units
C. 43 units
D. 34 units

Answer

To find the moment of inertia of the given arrangement of four particles about the Z-axis, we will follow these steps:

### Step 1: Identify the mass and coordinates of the particles
Each particle has a mass \( m = 2 \) grams. The coordinates of the particles are:
- Particle 1: \( (3, 2, 0) \)
- Particle 2: \( (1, -1, 0) \)
- Particle 3: \( (0, 0, 0) \)
- Particle 4: \( (-1, 1, 0) \)

### Step 2: Calculate the distance \( r \) from the Z-axis for each particle
The moment of inertia about the Z-axis is given by the formula:
\[ I_z = \sum m_i r_i^2 \]
where \( r_i \) is the distance from the Z-axis, calculated as:
\[ r = \sqrt{x^2 + y^2} \]

Now, we calculate \( r \) for each particle:

- For Particle 1 at \( (3, 2, 0) \):
\[ r_1 = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \]

- For Particle 2 at \( (1, -1, 0) \):
\[ r_2 = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

- For Particle 3 at \( (0, 0, 0) \):
\[ r_3 = \sqrt{0^2 + 0^2} = 0 \]

- For Particle 4 at \( (-1, 1, 0) \):
\[ r_4 = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 3: Calculate \( r_i^2 \) for each particle
Now we will square the distances:
- \( r_1^2 = 13 \)
- \( r_2^2 = 2 \)
- \( r_3^2 = 0 \)
- \( r_4^2 = 2 \)

### Step 4: Substitute into the moment of inertia formula
Now we can substitute these values into the moment of inertia formula:
\[ I_z = m \left( r_1^2 + r_2^2 + r_3^2 + r_4^2 \right) \]
Substituting \( m = 2 \) grams:
\[ I_z = 2 \left( 13 + 2 + 0 + 2 \right) = 2 \left( 17 \right) = 34 \text{ gram cm}^2 \]

### Conclusion
The moment of inertia of this arrangement about the Z-axis is \( 34 \text{ gram cm}^2 \).

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