Two soap bubbles having radii ' r 1 ' and ' r 2 ' has inside pressure ' P 1 ' and ' P 2 ' respectively. If P 0 is external pressure then ratio of thei

Question

Two soap bubbles having radii ' r 1 ​ ' and ' r 2 ​ ' has inside pressure ' P 1 ​ ' and ' P 2 ​ ' respectively. If P 0 ​ is external pressure then ratio of their volume is

TestHub · Accepted Answer

The excess pressure inside a soap bubble is given by $Δ P = \frac{4 T}{r}$ .

For the first bubble, $P_{1} - P_{0} = \frac{4 T}{r _{1}} \Rightarrow r_{1} = \frac{4 T}{P _{1} - P _{0}}$ .

For the second bubble, $P_{2} - P_{0} = \frac{4 T}{r _{2}} \Rightarrow r_{2} = \frac{4 T}{P _{2} - P _{0}}$ .

The volume of a spherical bubble is $V = \frac{4}{3} π r^{3}$ .

The ratio of their volumes is $\frac{V _{1}}{V _{2}} = (\frac{r _{1}}{r _{2}})^{3}$ .

Substituting the values of $r_{1}$ and $r_{2}$ , we get $\frac{V _{1}}{V _{2}} = \frac{\frac{4 T}{P _{1} - P _{0}}}{\frac{4 T}{P _{2} - P _{0}}}^{3} = \frac{( P _{2} - P _{0} ) ^{3}}{( P _{1} - P _{0} ) ^{3}}$ .

Answer: $\frac{( P _{2} - P _{0} ) ^{3}}{( P _{1} - P _{0} ) ^{3}}$

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