Side of an equilateral triangle is I . Three point masses, each of magnitude m, are palced at the three vertices of the triangle . Momment of inertia of this system about one side of the triangle as axis is given by

Options

A. `(3ml^2)/4`
B. `4/3 mt^2`
C. `3/2 ml^2`
D. `2/3 ml^2`

Explanation

To find the moment of inertia of a system of three point masses placed at the vertices of an equilateral triangle about one of its sides, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have an equilateral triangle with side length \( L \).
- Three point masses, each of mass \( m \), are placed at the vertices of the triangle.

2. **Identify the Axis of Rotation**:
- The axis of rotation is along one of the sides of the triangle. Let's denote the vertices of the triangle as \( A \), \( B \), and \( C \) with the side \( AB \) being the axis.

3. **Calculate the Moment of Inertia for Each Mass**:
- The moment of inertia \( I \) about an axis 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.

4. **Calculate for Mass at Vertex A and B**:
- The masses at vertices \( A \) and \( B \) are on the axis of rotation. Therefore, their distances from the axis are zero:
\[ I_A = m \cdot 0^2 = 0 \]
\[ I_B = m \cdot 0^2 = 0 \]

5. **Calculate for Mass at Vertex C**:
- The distance of mass \( m \) at vertex \( C \) from the axis \( AB \) can be calculated using the height of the triangle. The height \( h \) of an equilateral triangle is given by:
\[ h = \frac{\sqrt{3}}{2} L \]
- The distance from the side \( AB \) to vertex \( C \) is \( h \):
\[ r_C = h = \frac{\sqrt{3}}{2} L \]
- Therefore, the moment of inertia for mass \( C \) is:
\[ I_C = m \cdot \left(\frac{\sqrt{3}}{2} L\right)^2 = m \cdot \frac{3}{4} L^2 \]

6. **Total Moment of Inertia**:
- Now, we can sum up the moments of inertia:
\[ I_{total} = I_A + I_B + I_C = 0 + 0 + m \cdot \frac{3}{4} L^2 = \frac{3}{4} m L^2 \]

### Final Result:
The moment of inertia of the system about one side of the triangle is:
\[ I = \frac{3}{4} m L^2 \]