Physics - Fluid Question with Solution | TestHub

Initial area of the film, $A_1=6\text{ cm}\times 5\text{ cm}=30{\text{ cm}}^2$

Final area of the film, $A_2=8\text{ cm}\times 9\text{ cm}=72{\text{ cm}}^2$

Change in area, $\Delta A=A_2-A_1=72-30=42{\text{ cm}}^2$

Since a liquid film has two free surfaces, the total increase in surface area is:

$\Delta S=2\times \Delta A=2\times 42{\text{ cm}}^2=84{\text{ cm}}^2=84\times 10^{-4}{\text{ m}}^2$

The work done in expanding the film is given by:

$W=T\times \Delta S$

Substituting the given values:

$4\times 10^{-4}=T\times 84\times 10^{-4}$

$T=\frac{4}{84}=\frac{1}{21}\approx 0.0476\text{ N/m}$

Rounding to two decimal places, the surface tension is approximately $0.05\text{ N/m}$.

Answer: $0.05\frac{N}{m}$