If two soap bubbles of different radii $r_{1}$ and $r_{2}$ ( $r_{1} > r_{2}$ )coalesce to form a single double bubble having a common surface, then the radius of curvature of the interface is given by

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Solution

Let say $P_{1}$ and $P_{2}$ are the excess pressure on the two sides of the interface then the resultant excess pressure is $P$
$\Rightarrow P = P_{2} - P_{1} \Rightarrow \frac{4 T}{r} = \frac{4 T}{r_{2}} - \frac{4 T}{r_{1}} \Rightarrow \frac{1}{r} = \frac{1}{r_{2}} - \frac{1}{r_{1}} \Rightarrow r = \frac{r_{1} r_{1}}{r_{1} - r_{2}}$

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When two soap bubbles of radius $r_{1}$ and $r_{2}$ ( $r_{2}$ > $r_{1}$ ) coalesce, the radius of curvature of common surface is

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