Question

A rectangular film of liquid is extended from $(2\text{cm}\times 8\text{cm})$ to $(4\text{cm}\times 16\text{cm})$. If work done is $5\times {10}^{-4}\text{J}$, value of the surface tension of the liquid is

• A. $0.06\text{N/m}$
• B. $0.08\text{N/m}$
• C. $0.05\text{N/m}$
• D. $0.09\text{N/m}$

Answer 1

The correct option is C $0.05\text{N/m}$

Given , work done $\left(W\right)=5\times {10}^{-4}\text{J}$
Change in area $\left(\Delta A\right)=(4\text{cm}\times 16\text{cm})-(2\text{cm}\times 8\text{cm})$
$=48{\text{cm}}^2=48\times {10}^{-4}{\text{m}}^2$
We know from the definition of surface energy that
$\text{Surface energy}\left(S\right)=\frac{\text{Work done}\left(W\right)}{\text{change in area}\left(\Delta A\right)}$
Since the film has two surfaces, the change in area
$\Delta A=2\times 48\times {10}^{-4}{\text{m}}^2$
$\Rightarrow S=\frac{5\times {10}^{-4}}{2\times 48\times {10}^{-4}}$
$\Rightarrow S=0.05{\text{N/m}}^2$
The value of surface energy $\left(S\right)$ of a liquid is numerically equal to the value of surface tension.
$\therefore$ we can say that, surface tension of the liquid is $0.05\text{N/m}$
Hence, option (c) is the correct answer.