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

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$1 : 27$

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

Column-$I$	Column-$II$
$(a)$ Liquid drop in air	$(i)$ $\frac{4T}{R}$
$(b)$ Bubble of liquid in air	$(ii)$ $\frac{2T}{R}$
	$(iii)$ $\frac{2R}{T}$

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

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