Two soap bubbles of radii $r_{1}$ and $r_{2}$ equal to $4 c m$ and $5 c m$ are touching each other over a common surface $S_{1} S_{2}$ (shown in figure). The radius of surface $S_{1} S_{2}$ will be:

4 c m

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20 c m

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5 c m

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4.5 c m

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Solution

The correct option is B $20 c m$
First soap bubble has the radius has the radius $r_{1} = 4 c m$ and another bubble has the radius $r_{2} = 5 c m$
The access pressure due to surface tension in first bubble is given by $ρ_{1} = \frac{4 T}{r_{1}}$
The access pressure due to surface tension in Second bubble is given by $ρ_{2} = \frac{4 T}{r_{2}}$
Now if two bubble are touching each other, then they have a common surface $S_{1} S_{2}$ , The access pressure due to Surface tension at surface $S_{1} S_{2}$ is $ρ = \frac{4 T}{R}$
As both surfaces are touching each - other, hence the pressure difference at the common surface would be the excess pressure due to tension for the surface $S_{1} S_{2}$
$\Rightarrow ρ = ρ_{1} - ρ_{2}$
$\Rightarrow \frac{4 T}{R} = \frac{4 T}{r_{1}} - \frac{4 T}{r_{2}}$
$R = \frac{r_{1} r_{2}}{r_{2} - r_{1}} = \frac{4 \times 5}{5 - 4} = 20 c m$
Hence Correct answer is $B$

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