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Mechanical Properties Of Fluids

Question
CBSEENPH11018391

Two soap bubbles of radii 'a' and 'b' of same liquid come together to form a double bubble. Find the radius of curvature of common interface of two bubbles.

Take a < b and surface tension of soap liquid 'a' is T.

Solution
Given, two bubbles of radii 'a' and 'b'. 
Let,
p1be the excess of pressure inside the bubble of radius 'a', and
p2 be the excess of pressure inside the bubble of radius 'b'.

Therefore, we have
P1=4Ta  and P2=4Tb
The two bubbles come together to form a big bubble.
Then let r be the radius of common interface.
Now the difference of pressure on the two sides at interface is given by, 
increment p space equals space p subscript 1 space minus space p subscript 2 space
space space space space space space space
space space space space space space space equals space 4 T space open parentheses 1 over a minus 1 over b close parentheses space

A l s o comma space increment p space equals space fraction numerator 4 T over denominator r end fraction

rightwards double arrow space fraction numerator 4 T over denominator r end fraction space equals space 4 T space left parenthesis fraction numerator b minus a over denominator a b end fraction right parenthesis

rightwards double arrow space r space equals space open parentheses fraction numerator b a over denominator b minus a end fraction close parentheses   
'r' is the radius of curvature of common interface of two bubbles.