Question

A rectangular soap film is $5cm$ long and $4cm$ wide. If the work done in increasing its area $7cm\times 5cm$ is $0.06J$, the surface tension of the soap solution is:

• A. $0.2J/m^2$
• B. $2J/m^2$
• C. $20J/m^2$
• D. $72J/m^2$

Answer 1

The correct option is C $20J/m^2$

Work done by tension,

$W=T\left(2\Delta A\right)$

$\Rightarrow T=d\frac{W}{2\Delta A}=\frac{0.06}{2(7\times 5-5\times 4)\times {10}^{-4}}=20J{m}^{-2}$

Hence, tension of soap solution is $20J{m}^{-2}$.