The excess pressure inside the first soap bub | vedclass

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(A) The excess pressure inside a soap bubble of radius $r$ is given by $P = \frac{4T}{r}$, where $T$ is the surface tension.Given that the excess pres

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$1$:$27$

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(A) The excess pressure inside a soap bubble of radius $r$ is given by $P = \frac{4T}{r}$, where $T$ is the surface tension.Given that the excess pressure in the first bubble $(P_1)$ is three times that in the second bubble $(P_2)$:$P_1 = 3P_2$$\frac{4T}{r_1} = 3 	imes \frac{4T}{r_2}$$\frac{1}{r_1} = \frac{3}{r_2} \implies r_2 = 3r_1$The volume of a spherical soap bubble is $V = \frac{4}{3}\pi r^3$.The ratio of the volumes is:$\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \left(\frac{r_1}{r_2}ight)^3$Substituting $r_2 = 3r_1$:$\frac{V_1}{V_2} = \left(\frac{r_1}{3r_1}ight)^3 = \left(\frac{1}{3}ight)^3 = \frac{1}{27}$Thus, the ratio of the volume of the first soap bubble to the volume of the second soap bubble is $1:27$.

The excess pressure inside the first soap bubble is three times that inside the second soap bubble. The ratio of the volume of the first soap bubble to the volume of the second soap bubble is

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