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

Question

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

Dr. Abubacker · Accepted Answer

(d) The excess of pressure inside the first bubble of radius $r_{1}$ is, $p_{1} = 4\ S/r_{1}$; and in the second bubble of radius $r_{2}$ is, $p_{2} = 4\ S/r_{2}$
$p = \frac{4S}{r} = \frac{4S}{r_{1}} - \frac{4S}{r_{2}}$
$\Rightarrow \frac{1}{r} = \frac{r_{2} - r_{1}}{r_{1}r_{2}}$
$\Rightarrow r = \frac{r_{1}r_{2}}{r_{2} - r_{1}}$
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Q. When two soap bubbles of radius $r_{1}$ and $r_{2}\ (r_{2} > r_{1})$ coalesce, the radius of curvature of common surface is

Explanation

Reviewed by Dr. Abubacker

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