In the question number 51 , the potential at the centre is

Options

A. `-2(Gm)/(l)`
B. `3sqrt(2)(Gm)/(l)`
C. `-2sqrt(2)(Gm)/(l)`
D. `-4sqrt(2)(Gm)/(l)`

Explanation

To find the gravitational potential at the center of a square with four equal masses placed at its vertices, we can follow these steps:

### Step 1: Understand the Setup
We have a square with side length \( L \) and four masses, each of mass \( m \), positioned at the vertices of the square. We need to calculate the gravitational potential at the center of this square.

### Step 2: Determine the Distance from the Center to a Vertex
The distance from the center of the square to any vertex can be calculated using the Pythagorean theorem. The center divides the square into four equal parts, and the distance from the center to a vertex (let's denote it as \( r \)) is given by:
\[ r = \frac{L}{\sqrt{2}} \]
This is because each half of the square forms a right triangle with sides of length \( \frac{L}{2} \).

### Step 3: Write the Formula for Gravitational Potential
The gravitational potential \( V \) due to a single mass \( m \) at a distance \( r \) is given by:
\[ V = -\frac{Gm}{r} \]
where \( G \) is the gravitational constant.

### Step 4: Calculate the Total Potential at the Center
Since there are four masses contributing to the potential at the center, the total potential \( V_{\text{total}} \) is the sum of the potentials due to each mass:
\[ V_{\text{total}} = V_A + V_B + V_C + V_D \]
Substituting the expression for potential:
\[ V_{\text{total}} = -\frac{Gm}{r} - \frac{Gm}{r} - \frac{Gm}{r} - \frac{Gm}{r} = -4 \cdot \frac{Gm}{r} \]
Now substituting \( r = \frac{L}{\sqrt{2}} \):
\[ V_{\text{total}} = -4 \cdot \frac{Gm}{\frac{L}{\sqrt{2}}} = -4 \cdot \frac{Gm \sqrt{2}}{L} \]

### Step 5: Final Expression for the Potential
Thus, the potential at the center of the square is:
\[ V_{\text{total}} = -\frac{4\sqrt{2}Gm}{L} \]

### Conclusion
The potential at the center of the square formed by four masses at its vertices is:
\[ V_{\text{center}} = -\frac{4\sqrt{2}Gm}{L} \]