The work done is increasing the size of a soap film from 10 cm `xx` 6 cm to 10 cm `xx` 11 cm is ` 3 xx 10^(-4)` joule. The surface tension of the film is

Options

A. `1.5 xx 10^(-2)` N/m
B. `3.0 xx 10^(-2)` N/m
C. `6.0 xx 10^(-2)` N/m
D. `11.0 xx 10^(-2)` N/m

Answer

To solve the problem, we need to find the surface tension of the soap film given the work done in increasing its size. Here's a step-by-step solution:

### Step 1: Understand the given data
- Initial dimensions of the soap film: 10 cm x 6 cm
- Final dimensions of the soap film: 10 cm x 11 cm
- Work done (W) to increase the size: \(3 \times 10^{-4}\) Joules

### Step 2: Convert dimensions to meters
- Initial dimensions in meters:
- Width = 10 cm = 0.1 m
- Height = 6 cm = 0.06 m
- Final dimensions in meters:
- Width = 10 cm = 0.1 m
- Height = 11 cm = 0.11 m

### Step 3: Calculate the initial and final areas
- Initial area (A_initial):
\[ A_{\text{initial}} = \text{Width} \times \text{Height} = 0.1 \, \text{m} \times 0.06 \, \text{m} = 0.006 \, \text{m}^2 \]
- Final area (A_final):
\[ A_{\text{final}} = \text{Width} \times \text{Height} = 0.1 \, \text{m} \times 0.11 \, \text{m} = 0.011 \, \text{m}^2 \]

### Step 4: Calculate the change in area (ΔA)
- Change in area (ΔA):
\[ \Delta A = A_{\text{final}} - A_{\text{initial}} = 0.011 \, \text{m}^2 - 0.006 \, \text{m}^2 = 0.005 \, \text{m}^2 \]

### Step 5: Account for the surface tension acting on both sides of the film
Since the surface tension acts on both sides of the film, the effective change in area for the calculation of work done will be:
\[ \Delta A_{\text{effective}} = 2 \times \Delta A = 2 \times 0.005 \, \text{m}^2 = 0.01 \, \text{m}^2 \]

### Step 6: Use the work done formula to find surface tension (T)
The work done (W) is related to surface tension (T) and the change in area (ΔA) by the formula:
\[ W = T \times \Delta A_{\text{effective}} \]
Rearranging to find surface tension:
\[ T = \frac{W}{\Delta A_{\text{effective}}} \]
Substituting the values:
\[ T = \frac{3 \times 10^{-4} \, \text{J}}{0.01 \, \text{m}^2} = 3 \times 10^{-2} \, \text{N/m} \]

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

---