Question

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

• A. $5\times {10}^{-2}\text{N/m}$
• B. $3\times {10}^{-2}\text{N/m}$
• C. $2\times {10}^{-2}\text{N/m}$
• D. $4\times {10}^{-2}\text{N/m}$

Answer 1

The correct option is B $3\times {10}^{-2}\text{N/m}$

Given,
Work done in increasing the size of soap film, $W=3\times {10}^{-4}\text{J}$
Initial area of film, $A_i=10\text{cm}\times 6\text{cm}=60{\text{cm}}^2$
$=60\times {10}^{-4}{\text{m}}^2$
Final area of film, $A_f=10\text{cm}\times 11\text{cm}=110{\text{cm}}^2$
$=110\times {10}^{-4}{\text{m}}^2$
Now, area increased $=A_f-A_i=50\times {10}^{-4}{\text{m}}^2$
As film has 2 sides,
$\therefore$ Total increased area $\Delta A=2\times 50\times {10}^{-4}{\text{m}}^2={10}^{-2}{\text{m}}^2$
We know that,
Work done $=$ surface tension $\times$ increase in surface area
$\Rightarrow W=T\Delta A$
$\Rightarrow 3\times {10}^{-4}=T\times {10}^{-2}$
$\Rightarrow T=3\times {10}^{-2}\text{N/m}$