The surface tension of a soap solution is `30 xx 10^(-3) Nm^(-1)`. The work done in stretching a bubble of this solution of surface area , 5cm x 5cm , to an area of 10cm x 10cm , is

Options

A. S
B. `6.0 xx 10^(-4) J`
C. `4.5 xx 10^(-5) J`
D. `7.5 xx 10^(-4) J`

Explanation

To solve the problem of calculating the work done in stretching a soap bubble from a surface area of \(5 \, \text{cm} \times 5 \, \text{cm}\) to an area of \(10 \, \text{cm} \times 10 \, \text{cm}\), we will follow these steps:

### Step 1: Calculate the Initial and Final Surface Areas
1. **Initial Surface Area (A1)**:
\[ A_1 = 5 \, \text{cm} \times 5 \, \text{cm} = 25 \, \text{cm}^2 \]

2. **Final Surface Area (A2)**:
\[ A_2 = 10 \, \text{cm} \times 10 \, \text{cm} = 100 \, \text{cm}^2 \]

### Step 2: Convert Surface Areas to Square Meters
Since the surface tension is given in \( \text{N/m} \), we need to convert the areas from \( \text{cm}^2 \) to \( \text{m}^2 \):
1. **Convert \( \text{cm}^2 \) to \( \text{m}^2 \)**:
\[ A_1 = 25 \, \text{cm}^2 = 25 \times 10^{-4} \, \text{m}^2 = 0.0025 \, \text{m}^2 \]
\[ A_2 = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 0.01 \, \text{m}^2 \]

### Step 3: Calculate the Change in Surface Area
1. **Change in Surface Area (ΔA)**:
\[ \Delta A = A_2 - A_1 = 0.01 \, \text{m}^2 - 0.0025 \, \text{m}^2 = 0.0075 \, \text{m}^2 \]

### Step 4: Calculate the Work Done
The work done (W) in stretching the bubble is given by the formula:
\[ W = \text{Surface Tension} \times \Delta A \]
Substituting the values:
1. **Given Surface Tension**:
\[ \text{Surface Tension} = 30 \times 10^{-3} \, \text{N/m} \]
2. **Calculate Work Done**:
\[ W = 30 \times 10^{-3} \, \text{N/m} \times 0.0075 \, \text{m}^2 \]
\[ W = 30 \times 0.0075 \times 10^{-3} \, \text{J} \]
\[ W = 0.000225 \, \text{J} = 2.25 \times 10^{-4} \, \text{J} \]

### Final Answer
The work done in stretching the bubble is:
\[ \boxed{2.25 \times 10^{-4} \, \text{J}} \]