**Given:**
- Radius of the original liquid drop = 7 cm
- Number of equal droplets after division = 27
- Surface tension = 2 * 10^-3 N/m

**To find:**
The work done in the process.

**Explanation:**

1. **Surface Tension:**
- Surface tension is defined as the force acting per unit length along the surface of a liquid in contact with a gas or another liquid.
- It is denoted by the symbol 'γ' and its SI unit is N/m.

2. **Work Done in the Process:**
- When a liquid drop is divided into smaller droplets, work is done against the surface tension to increase the surface area of the droplets.
- The work done against surface tension is given by the formula:
Work = Change in surface area * Surface tension

3. **Change in Surface Area:**
- The original liquid drop has a radius of 7 cm, so its surface area is given by:
Surface area of original drop = 4πr^2 = 4π(7)^2 = 616π cm^2

- After division into 27 equal droplets, each droplet will have a radius of:
Radius of each droplet = (7 cm) / 3 = 2.33 cm

- The total surface area of all 27 droplets is given by:
Total surface area of 27 droplets = 27 * 4π(2.33)^2 = 1047.06π cm^2

- Therefore, the change in surface area is:
Change in surface area = Total surface area of 27 droplets - Surface area of original drop
= 1047.06π - 616π = 431.06π cm^2

4. **Work Done:**
- The work done against surface tension is given by:
Work = Change in surface area * Surface tension

- Substituting the given values:
Work = (431.06π cm^2) * (2 * 10^-3 N/m)
= 0.86212π N/m

- Approximating the value of π to 3.14:
Work = 0.86212 * 3.14 N/m
= 2.71 N/m

5. **Answer:**
- The work done in the process is approximately 2.71 N/m.
- None of the given options (1), (2), (3), or (4) match the calculated value.
- Therefore, the correct answer is not provided among the given options.

First find radius of smaller drop and then use ∆U=S∆A. and then W=-∆U
I hope this will help you