Question

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

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

Answer 1

Answer: (B)

$0.125{Nm}^{-1}$

Work done=Surface tension of film $\times$ Change in area of the film

or, $W=T\times \Delta A$

Here,

$\begin{array}{c}{A}_1=4cm\times 2cm=8c{m}^2\\ A_2=5cm\times 4cm=20c{m}^2\\ \Delta A=2\left(A_2-A_1\right)=24c{m}^2\\ =24\times {10}^4{m}^2\\ W=3\times {10}^{-4}J\\ \therefore T=\frac{W}{\Delta A}=\frac{3\times {10}^{-4}}{24\times {10}^{-4}}\\ =\frac{1}{8}=0.125N{m}^{-1}\end{array}$