The pressure inside a soap bubble A is 1.01 atmosphere and that in a soap bubble B is 1.02 atmosphere. The ratio of volume of bubble A to that of \(B\) is [Surrounding pressure \(=1\) atmosphere \(]\)

Outside pressure \(=1 \mathrm{~atm}\)
Pressure inside first bubble \(=1.01 \mathrm{~atm}\)
Pressure inside second bubble \(=1.02 \mathrm{~atm}\)
\(\therefore \quad\) Excess pressures will be \(\Delta \mathrm{P}_1=1.01-1=0.01 \mathrm{~atm}\) and \(\Delta \mathrm{P}_2=1.02-1=0.02 \mathrm{~atm}\)
\(\begin{array}{ll} & \text { Now, } \Delta \mathrm{P} \propto \frac{1}{\mathrm{r}} \Rightarrow \mathrm{r} \propto \frac{1}{\Delta \mathrm{P}} \\ \therefore \quad & \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{\Delta \mathrm{P}_2}{\Delta \mathrm{P}_1}=\frac{0.02}{0.01}=\frac{2}{1} \\ & \text { Now, } \mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3 \\ & \Rightarrow \mathrm{~V} \propto \mathrm{r}^3 \\ \therefore \quad & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^3=\left(\frac{2}{1}\right)^3=\frac{8}{1}\end{array}\)