Question

Consider a soap film on a rectangular frame of wire of area $4\times 4c{m}^2$. If the area of the soap film is increased to $4\times 5c{m}^2$, the work done in the process will be (The surface tension of the soap film is $3\times {10}^{-2}N/m$)

• A. $12\times {10}^{-6}J$
• B. $2.4\times {10}^{-5}J$
• C. $60\times {10}^{-6}J$
• D. $96\times {10}^{-6}J$

Answer 1

The correct option is B $2.4\times {10}^{-5}J$

Consider the problem

Given,

initial area

$\begin{array}{c}{A}_1=4cm\times 4cm=16c{m}^2\\ =16\times {10}^{-4}{m}^2\end{array}$

Final area

$\begin{array}{c}{A}_2=4cm\times 5cm=20c{m}^2\\ =20\times {10}^{-4}{m}^2\end{array}$

Surface tension $T=3\times {10}^{-2}N/m$

Work done $=?$

In case of rectangular frame the water wets it from two sides,

Hence, the change in area $=dA=2$

$\begin{array}{c}\left(A_2-A_1\right)=2\times \left(20\times {10}^{-4}-16\times {10}^{-4}\right)\\ \therefore dA=2\left(A_2-A_1\right)=2\times 4\times {10}^{-4}\\ =8\times {10}^{-4}{m}^2\end{array}$

Work done $=$change in surface energy

$\begin{array}{c}=TdA=3\times {10}^{-2}\times 8\times {10}^{-4}\\ =2.4\times {10}^{-5}J\end{array}$

Hence, Option $B$ is the correct answer.