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 $3\times {10}^{-4}\text{J},$ the value of surface tension of the liquid is

• A. $0.250{\text{Nm}}^{-1}$
• B. $0.2{\text{Nm}}^{-1}$
• C. $8.0{\text{Nm}}^{-1}$
• D. $0.125{\text{Nm}}^{-1}$

Answer 1

The correct option is D $0.125{\text{Nm}}^{-1}$

Given.
Initial area $A_i=4\times 2=8{\text{cm}}^2$

Final area $A_f=5\times 4=20{\text{cm}}^2$

Work done $W=3\times {10}^{-4}\text{J}$

Change in area $\left(\Delta A\right)=A_f-A_i=12c{m}^2=12\times {10}^{-4}{\text{m}}^2$

The work done for soap film is given by $W=S\left(2\Delta A\right)$

$\therefore 3\times {10}^{-4}=S\times 24\times {10}^{-4}$
$S=\frac{3}{24}=\frac{1}{8}{\text{Nm}}^{-1}=0.125{\text{Nm}}^{-1}$

Hence, option (c) is the correct answer.