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(B) The excess pressure $\Delta P$ inside a soap bubble of radius $r$ is given by the formula $\Delta P = \frac{4T}{r}$,where $T$ is the surface tensi

Vedclass · Accepted Answer

$4:1$

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(B) The excess pressure $\Delta P$ inside a soap bubble of radius $r$ is given by the formula $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension of the soap solution.From this relation,we can see that $\Delta P \propto \frac{1}{r}$.Let the radius of the first bubble be $r_1 = r$ and the radius of the second bubble be $r_2 = 4r$.The ratio of their excess pressures is $\frac{\Delta P_1}{\Delta P_2} = \frac{r_2}{r_1} = \frac{4r}{r} = \frac{4}{1}$.Wait,re-evaluating: If the first bubble has radius $r$ and the second has $4r$,then $\Delta P_1 = \frac{4T}{r}$ and $\Delta P_2 = \frac{4T}{4r} = \frac{T}{r}$.Thus,the ratio $\Delta P_1 : \Delta P_2 = 4:1$.

If the radius of a soap bubble is four times that of another,then the ratio of their excess pressures will be

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