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Surface Tension

A bubble of r...

Question

A bubble of radius $r_{1}$ is inside the other of radius $r_{2} (r_{2} > r_{1})$ . The radius of a bubble which has same pressure difference as the pressure difference of inside the smaller and outside the larger bubble is:

A

\frac{r_{1} r_{2}}{r_{2} + r_{1}}

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B

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

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C

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

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D

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

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Solution

The correct option is A $\frac{r_{1} r_{2}}{r_{2} + r_{1}}$
$P_{1} = P_{O} + Δ P_{1}$
$P_{2} = P_{0} + Δ P_{1} + Δ P_{2}$
$= P_{o} + \frac{2 T}{r_{1}} + \frac{2 T}{r_{2}}$
$P_{2} - P_{0} = 2 T (\frac{1}{r_{1}} + \frac{1}{r_{2}}) = \frac{2 T}{R}$
So the radius is $R = \frac{r_{1} r_{2}}{r_{1} + r_{2}}$

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