A water film is formed between two parallel wires of 10 cm length. The distance of 0.5 cm between the wires is increased by 1 mm . The work done in the process is (surface tension of water \(=72 \mathrm{~N} / \mathrm{m}\) ).

Surface area of film,
\(\mathrm{A}_1=l \times \mathrm{d}=10 \times 0.5=5 \mathrm{~cm}^2\)
The distance between wires ' \(d\) ' is increased by
\(1 \mathrm{~mm}=0.1 \mathrm{~cm}\)
Surface area of film becomes,
\(\mathrm{A}_2=10 \times(0.5+0.1)=6 \mathrm{~cm}^2\)
Increase in the surface area of the film is,
\(\begin{aligned}
& \Delta \mathrm{A}=\mathrm{A}_2-\mathrm{A}_1 \\
& \Delta \mathrm{~A}=6-5=1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2
\end{aligned}\)
Work done,
\(\begin{aligned}
& \mathrm{W}=2 \mathrm{~T} . \Delta \mathrm{A} \quad \ldots(\text { where } \mathrm{T} \text { is surface tension) } \\
& \mathrm{W}=2 \times 72 \times 10^{-4} \mathrm{~J} \\
& \mathrm{~W}=1.44 \times 10^{-2} \mathrm{~J}
\end{aligned}\)