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Question
A spherical soap bubble A of radius 2 cm is formed inside another bubble B of radius 4 cm. Show that the radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is lesser than the radius of both soap bubbles A and B.
Solution
From the excess pressure inside a soap bubble
ΔP = `(4"T")/"R"`
Here the two bubbles having the same pressure and temperature. So the radius of the combined bubbles,
`1/"R" = 1/"R"_1 + 1/"R"_2`
`1/"R" = 1/2 + 1/4 = (2 + 1)/4 = 3/4`
R = `4/3` = 1.33
∴ R = 1.33 cm
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