A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be :

Options

A. `8piR^2T`
B. `3piR^2T`
C. `(1/8)piR^2T`
D. `4piR^2T`

Answer

To solve the problem of calculating the work done when a small liquid drop of radius \( R \) is divided into 27 identical smaller drops, we will follow these steps:

### Step 1: Understand the Problem
We have a single liquid drop of radius \( R \) which is divided into 27 smaller drops. We need to calculate the work done in this process, which is related to the change in surface energy due to the change in surface area.

### Step 2: Calculate the Initial Surface Area
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4 \pi R^2 \]
For the initial drop of radius \( R \), the surface area \( A_{\text{initial}} \) is:
\[ A_{\text{initial}} = 4 \pi R^2 \]

### Step 3: Determine the Radius of the Smaller Drops
Since the original drop is divided into 27 smaller drops, we can use the principle of conservation of volume to find the radius \( r \) of the smaller drops. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
The volume of the original drop is:
\[ V_{\text{initial}} = \frac{4}{3} \pi R^3 \]
The total volume of the 27 smaller drops is:
\[ V_{\text{final}} = 27 \left(\frac{4}{3} \pi r^3\right) = \frac{4}{3} \pi (27 r^3) \]
Setting the initial volume equal to the final volume gives:
\[ \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (27 r^3) \]
Cancelling \( \frac{4}{3} \pi \) from both sides, we have:
\[ R^3 = 27 r^3 \]
Taking the cube root of both sides, we find:
\[ R = 3r \quad \Rightarrow \quad r = \frac{R}{3} \]

### Step 4: Calculate the Surface Area of the Smaller Drops
Now we can calculate the surface area of one of the smaller drops:
\[ A_{\text{small}} = 4 \pi r^2 = 4 \pi \left(\frac{R}{3}\right)^2 = 4 \pi \frac{R^2}{9} = \frac{4 \pi R^2}{9} \]
Since there are 27 smaller drops, the total surface area of the smaller drops is:
\[ A_{\text{final}} = 27 \times A_{\text{small}} = 27 \times \frac{4 \pi R^2}{9} = 12 \pi R^2 \]

### Step 5: Calculate the Change in Surface Area
The change in surface area \( \Delta A \) is given by:
\[ \Delta A = A_{\text{final}} - A_{\text{initial}} = 12 \pi R^2 - 4 \pi R^2 = 8 \pi R^2 \]

### Step 6: Calculate the Work Done
The work done \( W \) in the process, which is equal to the change in surface energy, is given by:
\[ W = T \cdot \Delta A \]
Substituting the change in surface area:
\[ W = T \cdot (8 \pi R^2) = 8 \pi R^2 T \]

### Final Answer
Thus, the work done in the process of dividing the drop is:
\[ \boxed{8 \pi R^2 T} \]