Physics - Fluid Question with Solution | TestHub

Let the radius of the big soap bubble be $R$ and the radius of each small soap bubble be $r$.

Assuming the volume of the air inside is conserved:
$\frac{4}{3}\pi R^3=27\times \frac{4}{3}\pi r^3$
$R^3=27r^3\Rightarrow R=3r$

By conservation of charge, the total charge $Q$ on the big bubble is distributed equally among the 27 small bubbles. Let $q$ be the charge on each small bubble:
$Q=27q\Rightarrow q=\frac{Q}{27}$

The surface charge density of the big bubble is:
${\sigma }_1=\frac{Q}{4\pi R^2}$

The surface charge density of each small bubble is:
${\sigma }_2=\frac{q}{4\pi r^2}=\frac{\frac{Q}{27}}{4\pi {\left(\frac{R}{3}\right)}^2}=\frac{Q}{27}\times \frac{9}{4\pi R^2}=\frac{1}{3}\left(\frac{Q}{4\pi R^2}\right)=\frac{{\sigma }_1}{3}$

The mechanical force acting per unit area (electrostatic pressure) on a charged surface is given by $f=\frac{{\sigma }^2}{2{ϵ}_0}$.

Therefore, the ratio of the mechanical force per unit area of the big bubble to that of the small bubble is:
$\frac{f_1}{f_2}={\left(\frac{{\sigma }_1}{{\sigma }_2}\right)}^2={\left(\frac{{\sigma }_1}{\frac{{\sigma }_1}{3}}\right)}^2=(3{)}^2=\frac{9}{1}$

Answer: $\frac{9}{1}$