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

Explanation

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.
`implies :. p_(A)=(4T)/R_(A)` and `p_(B)=(4T)/(R_(B))`

If `R` is radius of common interface, then we must have
`p_(A)=p_(B)=(4T)/R`
or `(4T)/(R_(A))-(4T)/(R_(B))=(4T)/R`
This gives
`R=(R_(A)R_(B))/(R_(B)-R_(A)) =(0.002xx0.004)/(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.