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

Answer

To find the net force acting on a mass placed at the center of an equilateral triangle formed by three equal masses at its corners, we can follow these steps:

### Step 1: Understand the Setup
We have three equal masses, each of mass \( m \), placed at the corners of an equilateral triangle with side length \( a \). A fourth mass of equal mass \( m \) is placed at the center of the triangle.

### Step 2: Calculate the Distance from the Center to a Corner
In an equilateral triangle, the distance from the center to any vertex (corner) can be calculated using the formula:
\[ d = \frac{a}{\sqrt{3}} \]
This distance will be used to calculate the gravitational force acting on the mass at the center due to each corner mass.

### Step 3: Calculate the Gravitational Force from One Corner Mass
The gravitational force \( F \) acting on the mass at the center due to one corner mass is given by Newton's law of gravitation:
\[ F = \frac{G m m}{d^2} = \frac{G m^2}{\left(\frac{a}{\sqrt{3}}\right)^2} = \frac{G m^2 \cdot 3}{a^2} \]
Where \( G \) is the gravitational constant.

### Step 4: Analyze the Forces Acting on the Center Mass
Since there are three corner masses, the forces \( F_1 \), \( F_2 \), and \( F_3 \) acting on the center mass from each corner mass will have the same magnitude \( F \) and will be directed radially outward from the center towards each corner.

### Step 5: Resolve the Forces
The forces \( F_1 \), \( F_2 \), and \( F_3 \) can be resolved into components. Due to symmetry, the horizontal components of the forces will cancel out. The vertical components will also cancel out in pairs.

### Step 6: Calculate the Net Force
Since the forces are symmetrically arranged, the net force acting on the mass at the center due to the three corner masses will be zero:
\[ F_{\text{net}} = F_1 + F_2 + F_3 = 0 \]

### Conclusion
The net force acting on the mass placed at the center of the triangle is:
\[ \text{Net Force} = 0 \]