3 particles of mass 10 gm each are placed at the corners of an equilateral triangle of side 5 cm. The M.I. of the system about a perpendicular axis to the plane passing through a corner of the triangle will be : (in kg-`m^(2)`)

Options

A. `5xx10^(-2)`
B. `5xx10^(-3)`
C. `5xx10^(-4)`
D. `5xx10^(-5)`

Answer

To find the moment of inertia (M.I.) of a system of three particles located at the corners of an equilateral triangle about an axis perpendicular to the plane of the triangle and passing through one of its corners, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses and Geometry**:
- Each particle has a mass \( m = 10 \, \text{g} = 0.01 \, \text{kg} \) (since we need the mass in kg).
- The side of the equilateral triangle is \( a = 5 \, \text{cm} = 0.05 \, \text{m} \).

2. **Position of the Particles**:
- Let’s denote the corners of the triangle as \( A \), \( B \), and \( C \).
- The axis of rotation is perpendicular to the plane of the triangle and passes through point \( A \).

3. **Calculate the Distances**:
- The distance from particle \( A \) to the axis is \( r_A = 0 \) (since it is on the axis).
- The distance from particle \( B \) to the axis is \( r_B = a = 0.05 \, \text{m} \).
- The distance from particle \( C \) to the axis can be calculated using the height of the triangle. The height \( h \) of an equilateral triangle is given by:
\[ h = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} \times 0.05 \, \text{m} \approx 0.0433 \, \text{m} \]
- Therefore, the distance from \( C \) to the axis is \( r_C = h = 0.0433 \, \text{m} \).

4. **Moment of Inertia Calculation**:
- The moment of inertia \( I \) about the axis through \( A \) is given by:
\[ I = m_A r_A^2 + m_B r_B^2 + m_C r_C^2 \]
- Substituting the values:
\[ I = 0.01 \times 0^2 + 0.01 \times (0.05)^2 + 0.01 \times (0.0433)^2 \]
- Simplifying this:
\[ I = 0 + 0.01 \times 0.0025 + 0.01 \times 0.00187689 \]
\[ I = 0.01 \times (0.0025 + 0.00187689) = 0.01 \times 0.00437689 \]
\[ I \approx 0.0000437689 \, \text{kg m}^2 \]

5. **Convert to Standard Form**:
- Converting this to a more readable form:
\[ I \approx 4.37689 \times 10^{-5} \, \text{kg m}^2 \]

### Final Answer:
Thus, the moment of inertia of the system about the specified axis is approximately:
\[ I \approx 4.38 \times 10^{-5} \, \text{kg m}^2 \]