The work done in increasing the size of a soap film from `10cmxx6cm` to `10cmxx11cm` is `3xx10^-4` Joule. The surface tension of the film is

Options

A. `1.5xx10^-2(N)/(m)
B. `3.0xx10^-2(N)/(m)
C. `6.0xx10^-2(N)/(m)
D. `11.0xx10^-2(N)/(m)

Explanation

To find the surface tension of the soap film given the work done in increasing its size, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Dimensions**:
- The initial dimensions of the soap film are \(10 \, \text{cm} \times 6 \, \text{cm}\).
- The final dimensions of the soap film are \(10 \, \text{cm} \times 11 \, \text{cm}\).

2. **Calculate the Initial and Final Areas**:
- Initial Area, \(A_i = 10 \, \text{cm} \times 6 \, \text{cm} = 60 \, \text{cm}^2\).
- Final Area, \(A_f = 10 \, \text{cm} \times 11 \, \text{cm} = 110 \, \text{cm}^2\).

3. **Determine the Change in Area**:
- Change in Area, \(\Delta A = A_f - A_i = 110 \, \text{cm}^2 - 60 \, \text{cm}^2 = 50 \, \text{cm}^2\).
- Since the soap film has two surfaces (top and bottom), the total change in area is:
\[ \Delta A_{\text{total}} = 2 \times \Delta A = 2 \times 50 \, \text{cm}^2 = 100 \, \text{cm}^2. \]

4. **Convert Area to Square Meters**:
- Convert \(100 \, \text{cm}^2\) to square meters:
\[ 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 1 \times 10^{-2} \, \text{m}^2. \]

5. **Relate Work Done to Surface Tension**:
- The work done in increasing the area of the soap film is given by the formula:
\[ W = \text{Surface Tension} \times \Delta A_{\text{total}}. \]
- Given that \(W = 3 \times 10^{-4} \, \text{J}\), we can set up the equation:
\[ 3 \times 10^{-4} = \text{Surface Tension} \times 1 \times 10^{-2}. \]

6. **Solve for Surface Tension**:
- Rearranging the equation gives:
\[ \text{Surface Tension} = \frac{3 \times 10^{-4}}{1 \times 10^{-2}} = 3 \times 10^{-2} \, \text{N/m}. \]

### Final Answer:
The surface tension of the soap film is \(3 \times 10^{-2} \, \text{N/m}\).