Physics - Fluid Question with Solution | TestHub

Initial area of the film, $A_1=5\text{ cm}\times 4\text{ cm}=20{\text{ cm}}^2=20\times 10^{-4}{\text{ m}}^2$

Final area of the film, $A_2=7\text{ cm}\times 8\text{ cm}=56{\text{ cm}}^2=56\times 10^{-4}{\text{ m}}^2$

Increase in area of the film, $\Delta A=A_2-A_1=36\times 10^{-4}{\text{ m}}^2$

Since a liquid film has two free surfaces, the total increase in surface area is $2\times \Delta A=2\times 36\times 10^{-4}{\text{ m}}^2=72\times 10^{-4}{\text{ m}}^2$

Work done, $W=T\times 2\Delta A$

$3\times 10^{-4}=T\times 72\times 10^{-4}$

$T=\frac{3}{72}=\frac{1}{24}\approx 0.0416\text{ N/m}$

The closest value is $0.04\text{ N/m}$.

Answer: $0.04\mathrm{N}/\mathrm{m}$