A film of soap solution is formed between two straight parallel wires of length $10 \mathrm{~cm}$ each…
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 $=65 \times 10^{-2} \mathrm{~N} / \mathrm{m}$ )
$7.22 \times 10^{-6} \mathrm{~J}$
$13.0 \times 10^{-5} \mathrm{~J}$
$2.88 \times 10^{-5} \mathrm{~J}$
$5.76 \times 10^{-5} \mathrm{~J}$
Solution
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
$=\mathrm{TdA}$
$=\left(65 \times 10^{-2}\right) \times\left(2 \times 10^{-4}\right)$
$=1.3 \times 10^{-4}$
$=13 \times 10^{-5} \mathrm{~J}$