Question

A soap film is formed on a frame of area $4\times {10}^{-3}{\text{m}}^2$. If the area of the film is reduced to half, what is the change in its potential energy?
(Surface tension $=40\times {10}^{-3}\text{N/m}$)

• A. $32\times {10}^{-5}\text{J}$
• B. $16\times {10}^{-5}\text{J}$
• C. $32\times {10}^{-6}\text{J}$
• D. $16\times {10}^{-6}\text{J}$

Answer 1

The correct option is B $16\times {10}^{-5}\text{J}$

Area of the frame $=4\times {10}^{-3}{\text{m}}^2$
Since, soap film has two interfaces,
So, area of the soap film $=2\times 4\times {10}^{-3}{\text{m}}^2$
Now, potential energy of the film
$U=$ surface tension $\times$ area of the film
$=40\times {10}^{-3}\times 2\times 4\times {10}^{-3}=32\times {10}^{-5}\text{J}$

When the area of the film is reduced to half, potential energy of film
$=40\times {10}^{-3}\times 2\times 2\times {10}^{-3}=16\times {10}^{-5}\text{J}$
$\therefore$ Change in potential energy $=(32\times {10}^{-5})-(16\times {10}^{-5})=16\times {10}^{-5}\text{J}$