Two mercury drops (each of radius `'r'`) merge to from bigger drop. The surface energy of the bigger drop, if `'T'` is the surface tension, is :

Options

A. `4etar^(2)T`
B. `2etar^(2)T`
C. `2^(8//3) pir^(2)T`
D. `2^(5//3)etar^(2)T`

Explanation

To find the surface energy of a bigger mercury drop formed by merging two smaller drops, we can follow these steps:

### Step 1: Understand the Volume Conservation
When two mercury drops of radius \( r \) merge to form a bigger drop, the volume of the bigger drop must equal the combined volume of the two smaller drops.

The volume \( V \) of a single drop is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, the volume of two drops is:
\[ V_{\text{total}} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

### Step 2: Set Up the Volume Equation for the Bigger Drop
Let the radius of the bigger drop be \( R \). The volume of the bigger drop is:
\[ V_{\text{bigger}} = \frac{4}{3} \pi R^3 \]
Setting the two volumes equal gives:
\[ \frac{4}{3} \pi R^3 = \frac{8}{3} \pi r^3 \]

### Step 3: Solve for the Radius of the Bigger Drop
Canceling \( \frac{4}{3} \pi \) from both sides:
\[ R^3 = 2 r^3 \]
Taking the cube root of both sides:
\[ R = r \cdot 2^{1/3} \]

### Step 4: Calculate the Surface Area of the Bigger Drop
The surface area \( A \) of the bigger drop is given by:
\[ A = 4 \pi R^2 \]
Substituting \( R = r \cdot 2^{1/3} \):
\[ A = 4 \pi (r \cdot 2^{1/3})^2 = 4 \pi r^2 \cdot 2^{2/3} \]

### Step 5: Calculate the Surface Energy
The surface energy \( U \) is given by the product of surface tension \( T \) and surface area \( A \):
\[ U = T \cdot A = T \cdot 4 \pi r^2 \cdot 2^{2/3} \]
Thus:
\[ U = 4 \pi T r^2 \cdot 2^{2/3} \]

### Step 6: Final Expression for Surface Energy
We can simplify this expression:
\[ U = 2 \cdot 2^{2/3} \cdot 2 \pi T r^2 = 2^{1 + 2/3} \cdot 2 \pi T r^2 = 2^{5/3} \cdot 2 \pi T r^2 \]
This leads us to the final expression:
\[ U = 2^{8/3} \pi T r^2 \]

### Conclusion
The surface energy of the bigger drop is:
\[ U = 2^{8/3} \pi T r^2 \]