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.For the first bubble,$p_1 =

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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.For the first bubble,$p_1 = \frac{4T}{r_1}$.For the second bubble,$p_2 = \frac{4T}{r_2}$.Given that $p_1 = 3p_2$,we have $\frac{4T}{r_1} = 3 	imes \frac{4T}{r_2}$.This simplifies to $\frac{1}{r_1} = \frac{3}{r_2}$,which implies $\frac{r_1}{r_2} = \frac{1}{3}$.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,we get $\frac{V_1}{V_2} = \left(\frac{1}{3}ight)^3 = \frac{1}{27}$.

The excess pressure inside the first soap bubble is three times that inside the second bubble. Then,the ratio of the volume of the first to the second bubble will be:

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