The excess pressure inside the first soap bub | vedclass

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Question

Excess pressure in first soap bubble,

$\mathrm{p}_{1}=\frac{4 \mathrm{T}}{\mathrm{r}_{1}}$

excess pressure inside second bubble,

$\mathrm{p}_

Vedclass · Accepted Answer

$1 : 27$

Vedclass · Answer

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,

$\frac{{\frac{4}{3}\pi {\rm{r}}_1^3}}{{\frac{4}{3}\pi {\rm{r}}_2^3}} = \frac{{{{\rm{v}}_1}}}{{{{\rm{v}}_2}}} \Rightarrow \frac{{{{\rm{v}}_1}}}{{{{\rm{v}}_2}}} = \frac{1}{{27}}$

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

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