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

Vedclass · Accepted Answer

$1:64$

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(A) The excess pressure $\Delta P$ inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension.Given that $\Delta P_1 = 4 \Delta P_2$,we have $\frac{4T}{r_1} = 4 	imes \frac{4T}{r_2}$.This simplifies to $\frac{1}{r_1} = \frac{4}{r_2}$,which implies $\frac{r_2}{r_1} = 4$ or $\frac{r_1}{r_2} = \frac{1}{4}$.The volume $V$ of a spherical bubble is given by $V = \frac{4}{3} \pi r^3$.Therefore,the ratio of the volumes is $\frac{V_1}{V_2} = \left( \frac{r_1}{r_2} ight)^3 = \left( \frac{1}{4} ight)^3 = \frac{1}{64}$.Thus,the ratio is $1:64$.

The excess pressure inside one soap bubble is four times that of another. What is the ratio of the volume of the first bubble to the second bubble?

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