When two soap bubbles of radius $\Gamma_1$ and $\Gamma_{2}$ $(\Gamma_{2} > \Gamma_{1})$ coalesce, the radius of curvature of common surface is
(a) $ r_2 - r_1 $ (b) $\frac{r_2 - r_1}{r_1 r_2}$ (c) $\frac{r_1 r_2}{r_2 - r_1}$ (d) $ r_2 + r_1 $
Text Solution
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For soap bubble of radius $\Gamma_{1}$ . $P_1 - P_0 = \frac{4\sigma}{r_1}$ , where $P_1$ is the pressure inside it and $P_0$ is the
pressure of the atmosphere.
Similarly, $P_2 - P_0 = \frac{4\sigma}{r_2}$
When bubbles coalesce and form a common surface,
$\mathrm{P}_1 - \mathrm{P}_2 = \frac{4\sigma}{r_1} - \frac{4\sigma}{r_2} = \frac{4\sigma}{R} \Rightarrow \frac{1}{R} = \frac{1}{r_1} - \frac{1}{r_2} \text{ or } R = \frac{r_1 \cdot r_2}{r_2 - r_1}$
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