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The excess pressure inside the first soap bubble is three times that inside the second bubble then, the ratio of volume of the first to the second bubble will be
$1 : 27$
$3 : 1$
$1 : 3$
$1 : 9$
Solution
Excess pressure in first soap bubble,
$\mathrm{p}_{1}=\frac{4 \mathrm{T}}{\mathrm{r}_{1}}$
excess pressure inside second bubble,
$\mathrm{p}_{2}=\frac{4 \mathrm{T}}{\mathrm{r}_{2}}$
On dividing these, we get
$\frac{p_{1}}{p_{2}}=\frac{r_{2}}{r_{1}}$
but $\mathrm{p}_{1}=3 \mathrm{p}_{2} \Rightarrow \frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{1}{3}$
$\Rightarrow\left(\frac{r_{1}}{r_{2}}\right)^{3}=\frac{1}{27}$
So, ratio of their volumes is,
$\frac{{\frac{4}{3}\pi {\rm{r}}_1^3}}{{\frac{4}{3}\pi {\rm{r}}_2^3}} = \frac{{{{\rm{v}}_1}}}{{{{\rm{v}}_2}}} \Rightarrow \frac{{{{\rm{v}}_1}}}{{{{\rm{v}}_2}}} = \frac{1}{{27}}$
