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Excess Pressure in Bubbles

Two soap bubb...

Question

Two soap bubbles of different radii $R_{A}$ and $R_{B} (< R_{A})$ coalesces to form an interface of radii $R$ . The correct relation of $R$ with $R_{A}$ and $R_{B}$ is

A

R = R_{A} - R_{B}

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B

R = \frac{R_{A} + R_{B}}{2}

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C

\frac{1}{R} = \frac{1}{R_{B}} - \frac{1}{R_{A}}

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D

\frac{1}{R} = \frac{1}{R_{A}} + \frac{1}{R_{B}}

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Solution

The correct option is C $\frac{1}{R} = \frac{1}{R_{B}} - \frac{1}{R_{A}}$
We know excess pressure in soap bubble is
$Δ p = \frac{4 T}{R}$
$p_{2} - p_{1} = \frac{4 T}{R} . . . (1)$

According to question

For Bubble $A$ , $p_{A} - p_{o} = \frac{4 T}{R_{A}}$ $p_{A} = \frac{4 T}{R_{A}} + p_{o} . . . (2)$

For Bubble $B$ , $p_{B} - p_{o} = \frac{4 T}{R_{B}}$ $p_{B} = \frac{4 T}{R_{B}} + p_{o} . . . (3)$
For Interface, $p_{B} - p_{A} = \frac{4 T}{R}$
$\Rightarrow \frac{4 T}{R_{B}} - \frac{4 T}{R_{A}} = \frac{4 T}{R}$
$\Rightarrow \frac{1}{R} = \frac{1}{R_{B}} - \frac{1}{R_{A}}$

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Excess Pressure in Bubbles

Standard XII Physics

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