Three equal masses each of mass 'm' are palced at the three corners of an equilateral of side a
If a fourth particle of equal mass is placed at the centre of triangle then net force acting on it is equal to .

Options

A. `(Gm^(2))/(a^(2))`
B. `(4Gm^(2))/(3a^(2))`
C. `(3Gm^(2))/(a^(2))`
D. zero

Answer

To solve the problem, we need to determine the net gravitational force acting on a mass placed at the center of an equilateral triangle formed by three other equal masses at its corners.

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have three equal masses (each of mass 'm') placed at the vertices of an equilateral triangle with side length 'a'. A fourth mass (also of mass 'm') is placed at the center of the triangle.

2. **Finding the Distance from the Center to the Vertices**:
The distance from the center of the triangle to any vertex can be calculated using geometry. For an equilateral triangle, this distance (let's call it 'r') can be derived as:
\[ r = \frac{a}{\sqrt{3}} \]
This is derived from the properties of the triangle and using the cosine of 30 degrees.

3. **Calculating the Gravitational Field Due to One Mass**:
The gravitational field (E) at the center due to one of the masses at the vertex is given by:
\[ E = \frac{G \cdot m}{r^2} \]
Substituting 'r' from the previous step:
\[ E = \frac{G \cdot m}{\left(\frac{a}{\sqrt{3}}\right)^2} = \frac{G \cdot m \cdot 3}{a^2} = \frac{3Gm}{a^2} \]

4. **Direction of the Gravitational Fields**:
The gravitational field due to each of the three masses acts towards the respective mass. Since the triangle is symmetric, the components of the gravitational fields in the horizontal direction will cancel out.

5. **Resolving the Gravitational Fields**:
Each gravitational field can be resolved into two components (vertical and horizontal). The horizontal components will cancel each other out due to symmetry, while the vertical components will add up. The vertical component of the gravitational field from each mass is:
\[ E_{vertical} = E \cdot \cos(60^\circ) = E \cdot \frac{1}{2} \]
Therefore, the total vertical component from all three masses is:
\[ E_{net} = 3 \cdot E \cdot \frac{1}{2} = \frac{3E}{2} \]

6. **Calculating the Net Gravitational Force**:
The net gravitational force acting on the mass at the center is given by:
\[ F_{net} = m \cdot E_{net} \]
Substituting for \(E\):
\[ F_{net} = m \cdot \frac{3 \cdot \frac{3Gm}{a^2}}{2} = \frac{9Gm^2}{2a^2} \]

7. **Conclusion**:
The net gravitational force acting on the mass placed at the center of the triangle is:
\[ F_{net} = \frac{9Gm^2}{2a^2} \]

### Final Answer:
The net force acting on the mass at the center of the triangle is \( \frac{9Gm^2}{2a^2} \).

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