A rectangular wire frame of size 2 cm `xx` 2 cm, is dipped in a soap solution and taken out. A soap film is formed, it the size of the film is changed to 3 cm x 3 cm, Calculate the work done in the process. The surface tension of soap film is `3 xx 10^(-2)` N/m.

Answer

To calculate the work done in changing the size of the soap film from 2 cm x 2 cm to 3 cm x 3 cm, we will follow these steps:

### Step 1: Calculate the initial and final areas of the soap film.

1. **Initial Area (A1)**:
\[ A1 = 2 \, \text{cm} \times 2 \, \text{cm} = 4 \, \text{cm}^2 \]
Convert to square meters:
\[ A1 = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \]

2. **Final Area (A2)**:
\[ A2 = 3 \, \text{cm} \times 3 \, \text{cm} = 9 \, \text{cm}^2 \]
Convert to square meters:
\[ A2 = 9 \, \text{cm}^2 = 9 \times 10^{-4} \, \text{m}^2 \]

### Step 2: Calculate the change in area (dA).

3. **Change in Area (dA)**:
\[ dA = A2 - A1 = (9 \times 10^{-4} \, \text{m}^2) - (4 \times 10^{-4} \, \text{m}^2) = 5 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Consider the soap film structure.

Since the soap film has two surfaces (front and back), the effective change in area for work done is:
\[ dA_{\text{effective}} = 2 \times dA = 2 \times (5 \times 10^{-4} \, \text{m}^2) = 10 \times 10^{-4} \, \text{m}^2 = 1 \times 10^{-3} \, \text{m}^2 \]

### Step 4: Calculate the work done (W).

4. **Work Done (W)**:
The work done is given by the formula:
\[ W = T \times dA_{\text{effective}} \]
Where \( T \) is the surface tension of the soap film, which is given as \( 3 \times 10^{-2} \, \text{N/m} \).

Substituting the values:
\[ W = (3 \times 10^{-2} \, \text{N/m}) \times (1 \times 10^{-3} \, \text{m}^2) \]
\[ W = 3 \times 10^{-5} \, \text{J} \]

### Final Answer:
The work done in the process is \( 3 \times 10^{-5} \, \text{J} \).

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