A soap bubble of radius $r_{1}$ is placed on another soap bubble of radius $r_{2} (r_{1} < r_{2})$ . The radius $R$ of the soap film separating the two bubbles is:

r_{1} + r_{2}

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\sqrt{r_{1}^{2} + r_{2}^{2}}

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(r_{1}^{3} + r_{2}^{3})

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

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Solution

The correct option is D $\frac{r_{2} r_{1}}{r_{2} - r_{1}}$
We know excess pressure in a soap film
$P_{e x c e s s} = \frac{4 T}{R}$

Equating the pressures, acting from both sides, at the interface
$∵ r_{2} > r_{1} \Rightarrow P_{2} < P_{1}$
$\Rightarrow \frac{4 T}{r_{1}} - \frac{4 T}{r_{2}} = \frac{4 T}{R}$

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

$∴ R = \frac{r_{2} r_{1}}{r_{2} - r_{1}}$

Hence, $(D)$ is the correct answer.

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