When two soap bubbles of radius r1 and r2 (r2>r1) coalesce, the radius of curvature of common surface is

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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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$\frac{r_{2} - r_{1}}{r_{1} r_{2}}$

$\frac{r_{1} r_{2}}{r_{2} - r_{1}}$

$r_{2} + r_{1}$

$r_{2} - r_{1}$

answer is C.

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Detailed Solution

Excess pressure in case of both bubbles= $\frac{4 T}{r_{1}}, \frac{4 T}{r_{2}}$ respectively

When they coalesce, net excess pressure= $\frac{4 T}{r_{1}} - \frac{4 T}{r_{2}} = 4 T (\frac{r_{2} - r_{1}}{r_{1} r_{2}})$

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