SECOND bubble = 1.02 atm Excess pressure \[\DELTA {{P}_{1}}=1.01-1=0.01\]atm Excess pressure \[\Delta {{P}_{2}}=1.02-1=0.02\] atm \[\Delta P\propto \frac{1}{r}\Rightarrow r\propto \frac{1}{\Delta P}\Rightarrow \frac{{{r}_{1}}}{{{r}_{2}}}=\frac{\Delta {{P}_{2}}}{\Delta {{P}_{1}}}=\frac{0.02}{0.01}=\frac{2}{1}\] SINCE \[V=\frac{4}{3}\PI {{r}^{3}}\ \ \therefore \ \ \frac{{{V}_{1}}}{{{V}_{2}}}={{\LEFT( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}={{\left( \frac{2}{1} \right)}^{3}}=\frac{8}{1}\]