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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Solution

The correct option is C

Let 'r be the radius of common surface of two bubbles.

$I f Δ P_{1} a n d Δ P_{2} a r e t h e e x c e s s p r e s s u r e s i n t w o s o a p b u b b l e s o f r a d i u s r_{1} a n d r_{2} a n d Δ P b e t h e p r e s s u r e d i f f e r e n c e b e t w e e n t h e b u b b l e s a f t e r c o a l e s c e w i t h e a c h o t h e r . ∴ Δ P = Δ P_{1} - Δ P_{2} \frac{4 T}{r} = \frac{4 T}{r_{1}} - \frac{4 T}{r_{2}} \frac{1}{r} = \frac{1}{r_{1}} - \frac{1}{r_{2}} r = \frac{r_{1} r_{2}}{r_{2} - r_{1}}$

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