Question

A thin film of water is formed between two straight parallel wires each of length 8 cm separated by distance of 0.6 cm. The work done to increase the distance between the wires to 0.8 cm is (Surface tension of water = 0.07 Nm$^{-1}$)

• A. 33.6 $\mu$J
• B. 22.4 $\mu$J
• C. 11.2 $\mu$J
• D. 44.8 $\mu$J

Hint:

When calculating work done against surface tension for a thin film (like a soap film), remember that the film has two surfaces. The total change in surface area is always twice the change in the geometric area.

Answer 1

The Correct Option is B

The work done in increasing the surface area of a liquid film is given by:
$W = T \times \Delta A_{total}$, where T is the surface tension and $\Delta A_{total}$ is the total change in surface area.
A thin film has two surfaces (top and bottom), so the total surface area is twice the geometric area.
Let the length of the wires be $L = 8$ cm = $0.08$ m.
Initial separation $d_1 = 0.6$ cm = $0.006$ m.
Final separation $d_2 = 0.8$ cm = $0.008$ m.
Initial geometric area of the film: $A_1 = L \times d_1$.
Final geometric area of the film: $A_2 = L \times d_2$.
The change in geometric area is $\Delta A_{geom} = A_2 - A_1 = L(d_2 - d_1)$.
$\Delta A_{geom} = (0.08 \text{ m}) \times (0.008 \text{ m} - 0.006 \text{ m}) = 0.08 \times 0.002 = 0.00016 \text{ m}^2$.
The total change in surface area, considering both surfaces, is:
$\Delta A_{total} = 2 \times \Delta A_{geom} = 2 \times 0.00016 \text{ m}^2 = 0.00032 \text{ m}^2$.
Now, calculate the work done.
The surface tension is given as $T = 0.07$ N/m.
$W = T \times \Delta A_{total} = 0.07 \times 0.00032$.
$W = 7 \times 10^{-2} \times 32 \times 10^{-5} = 224 \times 10^{-7}$ J.
$W = 22.4 \times 10^{-6}$ J.
Since $1 \mu\text{J} = 10^{-6}$ J, the work done is $22.4 \mu$J.