When two soap bubbles of radii a and b(b>a) coalesce,

Question:

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

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

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

  3. (3) $\frac{\mathrm{b}-\mathrm{a}}{\mathrm{ab}}$

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


Correct Option: 1

Solution:

(1)

Excess pressure at common surface is given by

$P_{e x}=4 T\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{4 T}{r}$

$\therefore \frac{1}{\mathrm{r}}=\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}$

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

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