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(D) The excess pressure $\Delta P$ inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension.This impl

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$1:27$

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(D) The excess pressure $\Delta P$ inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension.This implies that $\Delta P \propto \frac{1}{r}$.Given that the excess pressure in the first bubble is thrice that in the second bubble,we have $\Delta P_1 = 3 \Delta P_2$.Therefore,$\frac{r_1}{r_2} = \frac{\Delta P_2}{\Delta P_1} = \frac{1}{3}$.The volume $V$ of a spherical bubble is given by $V = \frac{4}{3} \pi r^3$,so $V \propto r^3$.The ratio of their volumes is $\frac{V_1}{V_2} = \left( \frac{r_1}{r_2} \right)^3 = \left( \frac{1}{3} \right)^3 = \frac{1}{27}$.

The excess pressure in a soap bubble is thrice that in another one. Then the ratio of their volumes is:

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