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A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is ____________ cm.

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Detailed Solution

From the figure for the larger bubble,

$P_{1} - P_{0} = \frac{4 T}{r_{1}}______(1)$

similarly for the smaller bubble,

$P_{2} - P_{1} = \frac{4 T}{r_{2}}______(2)$

Now adding equation 1 and equation 2

$P_{2} - P_{0} = \frac{4 T}{r_{1}} + \frac{4 T}{r_{2}}$

$4 Tr = 4 {Tr}_{1} + 4 {Tr}_{2}$

$\frac{1}{r} = \frac{1}{r_{1}} + \frac{1}{r_{2}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2} \Rightarrow r = 2$

Hence the correct answer is 2cm.

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