A film of soap solution is formed between two straight parallel wires of length \(10 \mathrm{~cm}\) each separated by \(0.5 \mathrm{~cm}\). If their separation is increased by \(1 \mathrm{~mm}\) while still maintaining their parallelism. How much work will have to be done?
(surface tension of solution \(=65 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) )

The increase in surface area of the film is, \(\Delta \mathrm{A}=\mathrm{A}_2-\mathrm{A}_1\)
\(\mathrm{A}_1=2 \times l \times \mathrm{b}=2 \times 10 \times 10^{-2} \times 0.5 \times 10^{-2}\)
\(\mathrm{A}_2=2 \times l \times(\mathrm{b}+1)=2 \times 10 \times 10^{-2} \times(0.5+0.1)\)
\(\times 10^{-2}\)
\(A_2-A_1=\left[2 \times 10 \times 10^{-2} \times(0.5+0.1) \times 10^{-2}\right]\)
\(-\left[2 \times 10 \times 10^{-2} \times 0.5 \times 10^{-2}\right]\)
\(=2 \times 10^{-4} \mathrm{~m}^2\)
\(\therefore \quad\) Work done \(=\) Increase in surface energy
\(=\mathrm{TdA}\)
\(=\left(65 \times 10^{-2}\right) \times\left(2 \times 10^{-4}\right)\)
\(=1.3 \times 10^{-4}\)
\(=13 \times 10^{-5} \mathrm{~J}\)