The work done in increasing the size of a rectangular soap film with dimensions 8 cm x 3.75 cm to 10 cm x 6 cm is `2 xx 10^(-4) J`. The surface tension of the film in `(N m^(-1))` is

Options

A. `1.65xx10^(-2)`
B. `3.3xx10^(-2)`
C. `6.6xx10^(-2)`
D. `8.25xx10^(-2)`

Answer

To find the surface tension of the soap film, we can follow these steps:

### Step 1: Calculate the initial and final areas of the soap film.
- **Initial dimensions**: 8 cm x 3.75 cm
- **Final dimensions**: 10 cm x 6 cm

Convert the dimensions from centimeters to meters:
- Initial dimensions in meters:
- Length = 8 cm = 0.08 m
- Width = 3.75 cm = 0.0375 m
- Final dimensions in meters:
- Length = 10 cm = 0.1 m
- Width = 6 cm = 0.06 m

Now, calculate the areas:
- Initial area (A_initial) = Length_initial × Width_initial = 0.08 m × 0.0375 m = 0.003 m²
- Final area (A_final) = Length_final × Width_final = 0.1 m × 0.06 m = 0.006 m²

### Step 2: Calculate the change in area (ΔA).
\[ \Delta A = A_{final} - A_{initial} = 0.006 m² - 0.003 m² = 0.003 m² \]

### Step 3: Calculate the work done in terms of surface tension.
The work done (W) in increasing the area of the soap film is given as:
\[ W = 2 \times T \times \Delta A \]
Where \( T \) is the surface tension.

Rearranging the formula to find surface tension:
\[ T = \frac{W}{2 \times \Delta A} \]

### Step 4: Substitute the known values into the equation.
Given:
- Work done \( W = 2 \times 10^{-4} \) J
- Change in area \( \Delta A = 0.003 \) m²

Now substitute these values into the equation:
\[ T = \frac{2 \times 10^{-4}}{2 \times 0.003} \]

### Step 5: Simplify the equation.
\[ T = \frac{2 \times 10^{-4}}{0.006} = \frac{2 \times 10^{-4}}{6 \times 10^{-3}} = \frac{2}{6} \times 10^{-1} = \frac{1}{3} \times 10^{-1} \]

### Step 6: Calculate the final value of surface tension.
\[ T = 0.333 \times 10^{-1} \text{ N/m} = 3.33 \times 10^{-2} \text{ N/m} \]

### Final Answer:
The surface tension of the soap film is \( 3.33 \times 10^{-2} \) N/m.

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