Three particles of equal mass 'm' are situated at the vertices of an equilateral triangle of side `L`. The work done in increasing the side of the triangle to `2L` is

Options

A. `5.0xx10^(-12)J`
B. `2.25xx10^(-10)J`
C. `4.0xxx10^(-11)J`
D. `6.0xx10^(-15)J`

Answer

To find the work done in increasing the side of an equilateral triangle formed by three particles of equal mass 'm' from length `L` to `2L`, we need to calculate the change in gravitational potential energy of the system.

### Step-by-Step Solution:

1. **Understanding the System**:
We have three particles of mass `m` at the vertices of an equilateral triangle with side length `L`. The gravitational potential energy (U) between any two masses is given by the formula:
\[ U = -\frac{G m_1 m_2}{r} \]
where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between them.

2. **Calculating Initial Potential Energy (U_initial)**:
For three particles, the potential energy due to the pairwise interactions is:
\[ U_{\text{initial}} = U_{12} + U_{13} + U_{23} \]
Each pair has a distance of `L`, so:
\[ U_{12} = -\frac{G m m}{L}, \quad U_{13} = -\frac{G m m}{L}, \quad U_{23} = -\frac{G m m}{L} \]
Therefore, the total initial potential energy is:
\[ U_{\text{initial}} = 3 \left(-\frac{G m^2}{L}\right) = -\frac{3G m^2}{L} \]

3. **Calculating Final Potential Energy (U_final)**:
When the side length is increased to `2L`, the potential energy for each pair becomes:
\[ U_{12} = -\frac{G m m}{2L}, \quad U_{13} = -\frac{G m m}{2L}, \quad U_{23} = -\frac{G m m}{2L} \]
Thus, the total final potential energy is:
\[ U_{\text{final}} = 3 \left(-\frac{G m^2}{2L}\right) = -\frac{3G m^2}{2L} \]

4. **Calculating the Work Done (W)**:
The work done in changing the configuration of the system is equal to the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values we calculated:
\[ W = \left(-\frac{3G m^2}{2L}\right) - \left(-\frac{3G m^2}{L}\right) \]
Simplifying this gives:
\[ W = -\frac{3G m^2}{2L} + \frac{3G m^2}{L} = \frac{3G m^2}{L} - \frac{3G m^2}{2L} = \frac{3G m^2}{2L} \]

5. **Final Result**:
Therefore, the work done in increasing the side of the triangle from `L` to `2L` is:
\[ W = \frac{3G m^2}{2L} \]