Two air bubbles of radii $2 m m$ and $4 m m$ , formed in the same liquid come together to form a big bubble. If the surface tension of the liquid is $0.70 N m^{- 1}$ , the radius of curvature of common interface to both bubbles will be:

6 m m

with concave surface towards smaller bubble.

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2 m m

with concave surface towards bigger bubble.

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4 m m

with concave surface towards smaller bubble.

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4 m m

with concave surface towards bigger bubble.

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Solution

The correct option is C $4 m m$ with concave surface towards smaller bubble.
If $p_{o}$ is the atmospheric pressure, then $p_{1}$ inside the smaller bubble and $p_{2}$ inside the bigger bubble are given by
$p_{1} - p_{o} = \frac{4 σ}{r_{1}}$ and $p_{2} - p_{o} = \frac{4 σ}{r_{2}}$
The pressure is more inside the smaller bubble and thus when and double bubble is formed, the interface will be concave towards the smaller bubble side. The excess pressure p corresponding the this surface is given as
$p = p_{1} - p_{2} = \frac{4 σ}{R}$
here R is radius of curvature of the internal film. On substituting
$(p_{o} + \frac{4 σ}{r_{1}}) - (p_{o} + \frac{4 σ}{r_{2}}) = \frac{4 σ}{R}$
or
$\frac{1}{r_{1}} - \frac{1}{r_{2}} = \frac{1}{R}$
or
$R = \frac{r_{1} r_{2}}{r_{2} - r_{1}} = \frac{2 \times 4}{4 - 2} = 4 m m$

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Two air bubbles of radii 2mm and 4mm, formed in the same liquid come together to form a big bubble. If the surface tension of the liquid is 0.70Nm−1, the radius of curvature of common interface to both bubbles will be:

Two air bubbles of radii $2 m m$ and $4 m m$ , formed in the same liquid come together to form a big bubble. If the surface tension of the liquid is $0.70 N m^{- 1}$ , the radius of curvature of common interface to both bubbles will be: