A rectangular soap film is formed on a wire frame. The work done in...

Initial area of the soap film, $A_1 = 10\text{ cm} \times 5\text{ cm} = 50\text{ cm}^2 = 50 \times 10^{-4}\text{ m}^2$

Final area of the soap film, $A_2 = 10\text{ cm} \times 8\text{ cm} = 80\text{ cm}^2 = 80 \times 10^{-4}\text{ m}^2$

Change in area, $\Delta A = A_2 - A_1 = 80 \times 10^{-4} - 50 \times 10^{-4} = 30 \times 10^{-4}\text{ m}^2$

Since a soap film has two free surfaces, the total increase in surface area is $2\Delta A$.

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

$W = 0.04 \times 2 \times 30 \times 10^{-4}$

$W = 2.4 \times 10^{-4}\text{ J}$

Answer: $2.4 \times 10^{-4}\text{ J}$