Work of \(3.0 \times 10^{-4}\) joule is required to be done in increasing the size of a soap film from \(10…
Work of \(3.0 \times 10^{-4}\) joule is required to be done in increasing the size of a soap film from \(10 \mathrm{~cm} \times 6 \mathrm{~cm}\) to \(10 \mathrm{~cm} \times 11 \mathrm{~cm}\). The surface tension of the film is
\(5 \times 10^{-2} \mathrm{~N} / \mathrm{m}\)
\(3 \times 10^{-2} \mathrm{~N} / \mathrm{m}\)
\(1.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}\)
\(1.2 \times 10^{-2} \mathrm{~N} / \mathrm{m}\)
Solution
Surface tension, \(S=\frac{\text { work done }}{\text { increase in area }}\) As the soap film has two surfaces.
$\begin{aligned}
S & = \frac{3.0 \times 10^{-4} \, \mathrm{J}}{2 \times (10 \times 11 - 10 \times 6) \times 10^{-4} \, \mathrm{m}^2} \\
& = 3 \times 10^{-2} \, \mathrm{N} / \mathrm{m} .
\end{aligned}$