Two soap bubbles, of radii 3 mm and 4 mm, are in contact with each other. Find the radius of curvature of the common interface (film) separating them.

1 mm
7 mm
12 mm
12/7 mm

Correct answer: 12 mm

Solution

Excess pressure inside a soap bubble is 4T/r, larger for the smaller bubble. The pressure difference across the common film equals 4T/r_common, where r_common is the radius of curvature of the interface. Setting 4T/r1 - 4T/r2 = 4T/r_common gives 1/r_common = 1/r1 - 1/r2, so r_common = r1*r2/(r2 - r1).

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