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 A to that of \(B\) is

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