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Two soap bubbles having radii ' $r_1$ ' and ' $r_2$ ' has inside pressure ' $\mathrm{P}_1$ ' and '…
Two soap bubbles having radii ' $r_1$ ' and ' $r_2$ ' has inside pressure ' $\mathrm{P}_1$ ' and ' $\mathrm{P}_2$ ' respectively. If $\mathrm{P}_0$ is external pressure then ratio of their volume is
$\frac{\left(\mathrm{P}_1-\mathrm{P}_0\right)}{\left(\mathrm{P}_2-\mathrm{P}_0\right)}$ $\frac{\left(\mathrm{P}_2-\mathrm{P}_0\right)}{\left(\mathrm{P}_1-\mathrm{P}_0\right)}$ $\frac{\left(\mathrm{P}_2-\mathrm{P}_0\right)^3}{\left(\mathrm{P}_1-\mathrm{P}_0\right)^3}$ $\frac{\left(\mathrm{P}_1-\mathrm{P}_0\right)^3}{\left(\mathrm{P}_2-\mathrm{P}_0\right)^3}$ Solution
Excess pressure $=P_i-P_0=\frac{4 T}{R} \Rightarrow P_i-P_0 \propto \frac{1}{R}$
$\therefore \quad$ As per the equation,
$\begin{aligned}
& \frac{P_1-P_0}{P_2-P_0}=\frac{r_2}{r_1} \\
& \text { Volume }=\frac{4}{3} \pi R^3 \Rightarrow \quad \text { Volume } \propto R^3 \\
\therefore \quad & \frac{V_1}{V_2}=\left(\frac{r_1}{r_2}\right)^3=\left(\frac{P_2-P_0}{P_1-P_0}\right)^3
\end{aligned}$
Asked in: MHT CET 2024 (10 May Shift 2)
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