There is a soap film on a rectangular frame of wire of area `4 xx 4 cm^(2)` . If the area of the frame is increased to `4 xx 5 cm^(2)`,find the work done in the process. Given surface tension of soap films `= 3 xx 10^(-2) Nm^(-1)`.

Explanation

To find the work done in increasing the area of the soap film from \(4 \times 4 \, \text{cm}^2\) to \(4 \times 5 \, \text{cm}^2\), we can follow these steps:

### Step 1: Calculate the Initial and Final Areas
1. **Initial Area \(A_1\)**:
\[ A_1 = 4 \, \text{cm} \times 4 \, \text{cm} = 16 \, \text{cm}^2 \]

2. **Final Area \(A_2\)**:
\[ A_2 = 4 \, \text{cm} \times 5 \, \text{cm} = 20 \, \text{cm}^2 \]

### Step 2: Calculate the Change in Area
3. **Change in Area \(\Delta A\)**:
\[ \Delta A = A_2 - A_1 = 20 \, \text{cm}^2 - 16 \, \text{cm}^2 = 4 \, \text{cm}^2 \]

### Step 3: Convert Change in Area to Square Meters
4. **Convert \(\Delta A\) to \(\text{m}^2\)**:
\[ \Delta A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \]

### Step 4: Use the Surface Tension to Calculate Work Done
5. **Given Surface Tension \(T\)**:
\[ T = 3 \times 10^{-2} \, \text{N/m} \]

6. **Work Done \(W\)**:
\[ W = T \times \Delta A = (3 \times 10^{-2} \, \text{N/m}) \times (4 \times 10^{-4} \, \text{m}^2) \]

### Step 5: Calculate the Work Done
7. **Final Calculation**:
\[ W = 3 \times 10^{-2} \times 4 \times 10^{-4} = 12 \times 10^{-6} \, \text{J} = 1.2 \times 10^{-5} \, \text{J} \]

Thus, the work done in the process is:
\[ \boxed{1.2 \times 10^{-5} \, \text{J}} \]

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