The pressure inside two soap bubbles, A and B | vedclass

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(A) Outside pressure $P_0 = 1 \,atm$. Excess pressure inside a soap bubble is given by $\Delta P = \frac{4T}{r}$, where $T$ is surface tension and $r$

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$2: 1$

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(A) Outside pressure $P_0 = 1 \,atm$. Excess pressure inside a soap bubble is given by $\Delta P = \frac{4T}{r}$, where $T$ is surface tension and $r$ is the radius. Excess pressure for bubble $A$: $\Delta P_A = P_A - P_0 = 1.01 - 1 = 0.01 \,atm$. Excess pressure for bubble $B$: $\Delta P_B = P_B - P_0 = 1.02 - 1 = 0.02 \,atm$. Since $\Delta P \propto \frac{1}{r}$, we have $r \propto \frac{1}{\Delta P}$. Therefore, the ratio of radii is $\frac{r_A}{r_B} = \frac{\Delta P_B}{\Delta P_A} = \frac{0.02}{0.01} = \frac{2}{1}$.

The pressure inside two soap bubbles, $A$ and $B$, is $1.01 \,atm$ and $1.02 \,atm$ respectively. The ratio of their respective radii $(r_A : r_B)$ is (outside pressure $= 1 \,atm$).

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