A spherical liquid drop of radius `R` is divided into eight equal droplets. If the surface tension is `T`, then the work done in this process will be

Options

A. `2piR^(2)T`
B. `3piR^(2)T`
C. `4piR^(2)T`
D. `2piRT^(2)`

Explanation

To solve the problem of calculating the work done when a spherical liquid drop of radius \( R \) is divided into eight equal droplets, we can follow these steps:

### Step 1: Understand the Volume Conservation
The volume of the original drop must equal the total volume of the smaller droplets after division. The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
Let the radius of each smaller droplet be \( r \). Since there are 8 droplets, the total volume of the smaller droplets can be expressed as:
\[ 8 \times \frac{4}{3} \pi r^3 \]
Setting the volumes equal gives:
\[ \frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3 \]

### Step 2: Simplify the Volume Equation
Cancelling \( \frac{4}{3} \pi \) from both sides, we have:
\[ R^3 = 8 r^3 \]
Taking the cube root of both sides, we find:
\[ R = 2r \]

### Step 3: Calculate the Surface Areas
Next, we calculate the surface area of the original drop and the total surface area of the smaller droplets. The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
For the original drop with radius \( R \):
\[ A_{\text{original}} = 4 \pi R^2 \]
Substituting \( R = 2r \):
\[ A_{\text{original}} = 4 \pi (2r)^2 = 16 \pi r^2 \]

For the 8 smaller droplets:
\[ A_{\text{small}} = 8 \times 4 \pi r^2 = 32 \pi r^2 \]

### Step 4: Determine the Change in Surface Area
The change in surface area \( \Delta A \) when the drop is divided is:
\[ \Delta A = A_{\text{small}} - A_{\text{original}} = 32 \pi r^2 - 16 \pi r^2 = 16 \pi r^2 \]

### Step 5: Calculate the Work Done
The work done \( W \) in increasing the surface area is given by the product of the surface tension \( T \) and the change in surface area:
\[ W = T \cdot \Delta A = T \cdot (16 \pi r^2) \]

### Step 6: Substitute Back to Capital R
Since \( r = \frac{R}{2} \), we can express \( r^2 \) in terms of \( R \):
\[ r^2 = \left(\frac{R}{2}\right)^2 = \frac{R^2}{4} \]
Substituting this into the work done expression:
\[ W = T \cdot (16 \pi \cdot \frac{R^2}{4}) = 4 \pi R^2 T \]

### Final Answer
Thus, the work done in this process is:
\[ \boxed{4 \pi R^2 T} \]