Pressure inside two soap bubbles is $1.01 \mathrm{~atm}$ and $1.03 \mathrm{~atm}$. The ratio between their…
Pressure inside two soap bubbles is $1.01 \mathrm{~atm}$ and $1.03 \mathrm{~atm}$. The ratio between their volumes is (Pressure outside the soap bubble is $1 \mathrm{~atm}$ )
9:1
27:1
81:1
3:1
Solution
Excess pressure inside a soap bubble is given by
$\Delta \mathrm{P}=\mathrm{P}_{\mathrm{i}}-\mathrm{P}_0=\frac{4 \mathrm{~T}}{\mathrm{r}}$
$\begin{aligned} & \Delta \mathrm{P}_1=1.01 \mathrm{~atm}-1 \mathrm{~atm}=0.01 \mathrm{~atm} \\ & \Delta \mathrm{P}_2=1.03 \mathrm{~atm}-1 \mathrm{~atm}=0.03 \mathrm{~atm}\end{aligned}$
$\begin{aligned} & \therefore \frac{\Delta \mathrm{P}_1}{\Delta \mathrm{P}_2}=\frac{\mathrm{r}_2}{\mathrm{r}_1} \\ & \frac{0.03}{0.01}=3=\frac{\mathrm{r}_2}{\mathrm{r}_1} \\ & \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^3=(3)^3=27\end{aligned}$