Question

Two separate air bubbles (radii 0.002 cm. and 0.004 m) formed of the same liquid (surface tension 0.07 N/m) come together to form a double bubble. Find the radius and the sense of curvature of the internal film surface common to both the bubbles.

Answer 1

Let R be the radius of curvature of common surface when bubbles A and B of radii $R_{A}and{R}_B$ coalesce. The excess pressure.
in A and B are 4T.R_A and 4T.R_B, respectively.
$\therefore p_A=\frac{4T}{R_A}and{p}_B=\frac{4T}{R_B}$
If R is radius of common interface, then we must have
$P_A-p_B=\frac{4T}{R}$
Or$\frac{4T}{R_A}-\frac{4T}{R_B}=\frac{4T}{R}$
This gives $R=\frac{R_A{R}_B}{R_B-R_A}=\frac{0.002\times 0.004}{0.004-0.002}=0.004m$
As the excess pressure is always towards concave surface and pressure in smaller
bubble is greater than that in larger bubble, the common surface is concave towards the centre of the smaller bubble.