When two soap bubbles of radii $a$ and $b$ $(b > a)$ coalesce, the radius of curvature of the common surface is

$\frac{b - a}{a b}$

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$\frac{a b}{(b - a)}$

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$\frac{a b}{(a + b)}$

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$\frac{(a + b)}{a b}$

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Solution

The correct option is B
$\frac{a b}{(b - a)}$

Explanation for the correct option:
Step $1$ : Finding the excess pressure on each bubble
The excess pressure at the bubble $1$ is,
$P_{1} = P_{0} + \frac{4 S}{b}$ , where $S$ is the surface tension

The excess pressure at the bubble $2$ is,
$P_{2} = P_{0} + \frac{4 S}{a}$ , where $S$ is the surface tension
Step $2$ : Finding the radius of curvature of the common surface
the excess pressure at the common surface is,
$P_{2} - P_{1} = P_{0} + \frac{4 S}{a} - (P_{0} + \frac{4 S}{b})$
$\Rightarrow$ $\frac{4 S}{R} = \frac{4 S}{a} - \frac{4 S}{b}$ where $R$ is the radius of curvature
$\Rightarrow$ $\frac{1}{R} = \frac{1}{a} - \frac{1}{b}$
$\Rightarrow$ $\frac{1}{R} = \frac{b - a}{a b}$
$\Rightarrow$ $R = \frac{a b}{b - a}$
Hence, the correct option is $B) \frac{a b}{(b - a)}$

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