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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.Let the radii of the

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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.Let the radii of the two soap bubbles be $r_1$ and $r_2$ respectively.The excess pressure in the first bubble is $\Delta P_1 = \frac{4T}{r_1}$ and in the second bubble is $\Delta P_2 = \frac{4T}{r_2}$.According to the problem,$\Delta P_1 = 3 \Delta P_2$.Substituting the expressions,we get $\frac{4T}{r_1} = 3 \left( \frac{4T}{r_2} \right)$.This simplifies to $\frac{1}{r_1} = \frac{3}{r_2}$,which means $r_2 = 3r_1$.The volume of a spherical bubble is $V = \frac{4}{3} \pi r^3$.The ratio of the volumes is $\frac{V_1}{V_2} = \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \left( \frac{r_1}{r_2} \right)^3$.Substituting $r_2 = 3r_1$,we get $\frac{V_1}{V_2} = \left( \frac{r_1}{3r_1} \right)^3 = \left( \frac{1}{3} \right)^3 = \frac{1}{27}$.Thus,the ratio is $1: 27$.

The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is:

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