Question

The work done in increasing the size of a soap film from $10\text{cm}\times 6\text{cm}$ to $10\text{cm}\times 11\text{cm}$ is $3\times {10}^{–4}\text{J}$. The surface tension of the film is

• A. $3.0\times {10}^{-2}\text{N/m}$
• B. $6.0\times {10}^{-2}\text{N/m}$
• C. $1.5\times {10}^{-2}\text{N/m}$
• D. $11.0\times {10}^{-2}\text{N/m}$

Answer 1

The correct option is A $3.0\times {10}^{-2}\text{N/m}$

Since, the film is formed on both the surfaces of soap film,
$\Delta U=2T\Delta A$
Substituting the values here -
$(10\times 11-10\times 6)\times {10}^{-4}\times 2T=\Delta U$
$50\times {10}^{-4}\times 2T=3\times {10}^{-4}$
$\Rightarrow T=\frac{3}{100}=3\times {10}^{-2}\text{N/m}$