Work done in forming a liquid drop of radius R is `W_(1)` and that of radius 3R is `W_(2)`. The ratio of work done is

Explanation

To find the ratio of work done in forming a liquid drop of radius R (denoted as \( W_1 \)) and that of radius \( 3R \) (denoted as \( W_2 \)), we can follow these steps:

### Step 1: Understand the Concept of Surface Energy
The work done in forming a liquid drop is related to the change in surface energy. The surface energy of a liquid drop is given by the formula:
\[ \text{Surface Energy} = \text{Surface Tension} \times \text{Surface Area} \]
Thus, the work done \( W \) in forming a drop is equal to the surface energy of that drop.

### Step 2: Calculate the Surface Area for Each Drop
The surface area \( A \) of a sphere (which is the shape of a drop) is given by:
\[ A = 4\pi r^2 \]
For a drop of radius \( R \):
\[ A_1 = 4\pi R^2 \]
For a drop of radius \( 3R \):
\[ A_2 = 4\pi (3R)^2 = 4\pi \times 9R^2 = 36\pi R^2 \]

### Step 3: Calculate the Work Done for Each Drop
Using the surface tension \( T \), we can express the work done in forming each drop:
- For the drop of radius \( R \):
\[ W_1 = T \times A_1 = T \times 4\pi R^2 \]
- For the drop of radius \( 3R \):
\[ W_2 = T \times A_2 = T \times 36\pi R^2 \]

### Step 4: Find the Ratio of Work Done
Now we can find the ratio of the work done:
\[ \frac{W_1}{W_2} = \frac{T \times 4\pi R^2}{T \times 36\pi R^2} \]
The \( T \), \( 4\pi \), and \( R^2 \) terms cancel out:
\[ \frac{W_1}{W_2} = \frac{4}{36} = \frac{1}{9} \]

### Conclusion
Thus, the ratio of work done in forming a liquid drop of radius \( R \) to that of radius \( 3R \) is:
\[ W_1 : W_2 = 1 : 9 \]