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 value of the surface tension of the liquid is

• A. $0.250{\text{N m}}^{-1}$
• B. $0.125{\text{N m}}^{-1}$
• C. $0.275{\text{N m}}^{-1}$
• D. $0.325{\text{N m}}^{-1}$

Answer 1

The correct option is B $0.125{\text{N m}}^{-1}$

Work done $=$ surface tension of film $\times$ change in surface area
$\Rightarrow W=T\Delta A$
$\Rightarrow T=\frac{W}{\Delta A}$ ..............(1)
Here, $A_i=4\times 2=8{\text{cm}}^2$ and
$A_f=5\times 4=20{\text{cm}}^2$
$\therefore \Delta A=2(A_f-A_i)=2(20-8)=24{\text{cm}}^2$
$=24\times {10}^{-4}{\text{m}}^2$
Change in surface area has been taken double because two surfaces of thin film are formed on any rectangular loop.
Now putting the data in (1),
$T=\frac{3\times {10}^{-4}}{24\times {10}^{-4}}=\frac{1}{8}=0.125{\text{Nm}}^{-1}$