The excess pressure inside the first soap bub | vedclass

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(C) 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 press

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$1: 27$

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(C) 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 of the first bubble $(P_1)$ is three times that of the second bubble $(P_2)$,we have $P_1 = 3P_2$.Substituting the formula for excess pressure: $\frac{4T}{r_1} = 3 	imes \frac{4T}{r_2}$.This simplifies to $\frac{1}{r_1} = \frac{3}{r_2}$,which means $r_2 = 3r_1$ or $\frac{r_1}{r_2} = \frac{1}{3}$.The volume of a spherical bubble is $V = \frac{4}{3}\pi r^3$.The ratio of the volumes is $\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \left(\frac{r_1}{r_2}ight)^3$.Substituting the ratio of radii: $\frac{V_1}{V_2} = \left(\frac{1}{3}ight)^3 = \frac{1}{27}$.Thus,the ratio of the volumes is $1: 27$.

The excess pressure inside the first soap bubble is three times that of a second soap bubble. What is the ratio of the volumes of the first bubble to the second bubble?

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