Physics - Fluid Question with Solution | TestHub

The work done in stretching a liquid film is given by the product of surface tension and the change in surface area:

$W=T\times \Delta A$

Since a liquid film has two free surfaces (front and back), the total increase in surface area is:

$\Delta A=2\times L\times \Delta x$

Given:
Length of the wire, $L=10\text{ cm}=0.1\text{ m}$
Increase in distance, $\Delta x=1\text{ mm}=10^{-3}\text{ m}$
Surface tension, $T=72\text{ N/m}$

Substituting the values to find the change in area:

$\Delta A=2\times 0.1\times 10^{-3}=2\times 10^{-4}{\text{ m}}^2$

Now, calculating the work done:

$W=72\times 2\times 10^{-4}$

$W=144\times 10^{-4}\text{ J}$

$W=1.44\times 10^{-2}\text{ J}$

Answer: $1.44\times 10^{-2}\mathrm{J}$