Three equal masses each of mass 'm' are placed at the three-corner of an equilateral triangle 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. **Understand the Configuration**:
We have three equal masses (each of mass 'm') located at the vertices of an equilateral triangle with side length 'a'. A fourth mass (also of mass 'm') is placed at the center of this triangle.

2. **Determine the Distance from the Center to a Vertex**:
The distance from the center of the triangle to any vertex can be calculated using geometry. For an equilateral triangle, this distance (R) can be derived as:
\[ R = \frac{a}{\sqrt{3}} \]
This is because the center divides the median in a 2:1 ratio, and using trigonometric relationships, we find that the distance from the center to a vertex is \( \frac{a}{\sqrt{3}} \).

3. **Calculate the Gravitational Field at the Center**:
The gravitational field (E) due to one mass at the center is given by the formula:
\[ E = \frac{Gm}{R^2} \]
Substituting \( R = \frac{a}{\sqrt{3}} \):
\[ E = \frac{Gm}{\left(\frac{a}{\sqrt{3}}\right)^2} = \frac{Gm \cdot 3}{a^2} = \frac{3Gm}{a^2} \]
Since there are three masses, the gravitational fields due to each mass will be directed towards the respective masses.

4. **Components of the Gravitational Fields**:
The gravitational fields due to the three masses will have both horizontal and vertical components. Due to symmetry, the horizontal components will cancel each other out. The vertical components will add up.

If we denote the gravitational field due to one mass as \( E \):
- The vertical component of each gravitational field (pointing towards the center) can be calculated as:
\[ E_{vertical} = E \cdot \sin(30^\circ) = \frac{3Gm}{a^2} \cdot \frac{1}{2} = \frac{3Gm}{2a^2} \]

5. **Net Gravitational Field at the Center**:
Since there are three masses, the total vertical component of the gravitational field at the center will be:
\[ E_{net} = 3 \cdot E_{vertical} = 3 \cdot \frac{3Gm}{2a^2} = \frac{9Gm}{2a^2} \]
However, since the horizontal components cancel out, the net gravitational field at the center is:
\[ E_{net} = 0 \]

6. **Calculate the Gravitational Force on the Fourth Mass**:
The gravitational force (F) acting on the mass at the center due to the net gravitational field is given by:
\[ F = m \cdot E_{net} = m \cdot 0 = 0 \]

### Final Answer:
The net force acting on the fourth mass placed at the center of the triangle is **0**.