Question

If the excess pressure inside a soap bubble is balanced by oil column of height $ 2\,mm, $ then the surface tension of soap solution will be ( $ r=1\,cm $ and density $ d=0.8\, g/\,cc $ )

• A. $ 3.9\,N/m $
• B. $ 3.9\times 10^{-1}N/m $
• C. $ 3.9\times 10^{-2}N/m $
• D. $ 3.9\,dyne/m $

Answer 1

Answer: (C)

Excess pressure inside a soap bubble of radius $R$ is
$=\frac{4 T}{R}$ ... (i)
where $T$ is surface tension of liquid film. Pressure due to oil column
$=h \rho g$ ... (ii)
where $h$ is height of column, $p$ the density and $g$ the gravity.
From Eqs. (i) and (ii), we get
$\frac{4 T}{R}=h \rho g$
$\Rightarrow T=\frac{h \rho g R}{4}$
Given $/ t=2\, mm =0.2\, cm,\, g=980\, cm\, s ^{-2}$
$\rho=0.8\, g / cc, R=1\, cm$
$=T=\frac{0.2 \times 0.8 \times 980}{4}$
$=3.92 \times 10$ dyne $cm^{-1}$
In $Nm^{-1},\, T=3.9 \times 10 \times \frac{10^{-5}}{10^{-2}}$
$=3.9 \times 10^{-2} Nm ^{-1}$