Physics - Fluid Question with Solution | TestHub

Let the radius of curvature of the common internal film surface of the double bubble formed by two bubbles$\mathrm{A}$and$\mathrm{B}$be$r$.



Excess of pressure as compared to the atmosphere inside$\mathrm{A}$is

$p_1=\frac{4T}{r_1}=\frac{4T}{0.03}$

Excess of pressure inside$\mathrm{B}$is

$p_2=\frac{4T}{r_2}=\frac{4T}{0.04}$

In the double bubble, the pressure difference between$\mathrm{A}$and$\mathrm{B}$on either side of the common surface is

$\frac{4T}{0.03}-\frac{4T}{0.04}=\frac{4T}{r}$

$\Rightarrow \frac{1}{0.03}-\frac{1}{0.04}=\frac{1}{r}$

$\Rightarrow r=\frac{0.03\times 0.04}{0.01}=\mathrm{0.12 }\mathrm{m}$