The excess pressure inside the first soap bubble of radius R 1 is three times that inside the second soap bubble of radius R 2 . The ratio of volumes of the first to second bubble is

Question

The excess pressure inside the first soap bubble of radius R 1 ​ is three times that inside the second soap bubble of radius R 2 ​ . The ratio of volumes of the first to second bubble is

TestHub · Accepted Answer

The excess pressure inside a soap bubble of radius $R$ is given by $Δ P = \frac{4 T}{R}$ , where $T$ is the surface tension.

Given that the excess pressure inside the first bubble is three times that inside the second bubble:

$Δ P_{1} = 3Δ P_{2}$

$\frac{4 T}{R _{1}} = 3 (\frac{4 T}{R _{2}})$

$\frac{R _{1}}{R _{2}} = \frac{1}{3}$

The volume of a spherical bubble is $V = \frac{4}{3} π R^{3}$ . The ratio of their volumes is:

$\frac{V _{1}}{V _{2}} = (\frac{R _{1}}{R _{2}})^{3}$

$\frac{V _{1}}{V _{2}} = (\frac{1}{3})^{3} = \frac{1}{27}$

The ratio of volumes of the first to the second bubble is $1 : 27$ .

Answer: $1 : 27$

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