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

4 cm

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20 cm

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5 cm

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4.5 cm

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Solution

The correct option is A 20 cm
The excess pressure inside any soap bubble is given by the expression,
$P_{e x c e s s} = \frac{4 T}{r}$

$P_{1} = \frac{4 T}{4}$
$P_{2} = \frac{4 T}{5}$

The difference between the 2 pressures will account for the radius of the new surface

$P_{1} - P_{2} = \frac{4 T}{r}$

$\frac{4 T}{4} - \frac{4 T}{5} = \frac{4 T}{r}$

Solving, gives r = 20 cm

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