Question

A rectangular film of liquid is extended from $(4\text{cm}\times 2\text{cm})$ to $(5\text{cm}\times 4\text{cm})$. If the work done is $3\times {10}^{-4}\text{J}$, then the value of the surface tension of the liquid is

• A. $0.250\text{N/m}$
• B. $0.125\text{N/m}$
• C. $0.500\text{N/m}$
• D. $0.725\text{N/m}$

Answer 1

The correct option is B $0.125\text{N/m}$

Given,
Initial area of rectangular film, $A_i=4\text{cm}\times 2\text{cm}=8{\text{cm}}^2$
$=8\times {10}^{-4}{\text{m}}^2$
Final area of rectangular film, $A_f=5\text{cm}\times 4\text{cm}=20{\text{cm}}^2$
$=20\times {10}^{-4}{\text{m}}^2$
Work done, $W=3\times {10}^{-4}\text{J}$
We know that, for liquid film, number of film surfaces, $n=2$
Work done $=$ Surface Tension $\times$ change in surface area $\times$ number of film surfaces
$\Rightarrow W=T\times \Delta A\times n$
$\Rightarrow 3\times {10}^{-4}=T\times (20-8)\times {10}^{-4}\times 2$
$\Rightarrow T=0.125\text{N/m}$