Two soap bubbles having radii r_1 and r_2 hav | vedclass

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(C) The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = P_i - P_0 = \frac{4T}{R}$,where $T$ is the surface tension of the s

Vedclass · Accepted Answer

$\frac{(P_2-P_0)^3}{(P_1-P_0)^3}$

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(C) The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = P_i - P_0 = \frac{4T}{R}$,where $T$ is the surface tension of the soap solution.From this,we can see that $(P_i - P_0) \propto \frac{1}{R}$,which implies $R \propto \frac{1}{(P_i - P_0)}$.For the two bubbles,we have the ratio of radii as $\frac{r_1}{r_2} = \frac{(P_2 - P_0)}{(P_1 - P_0)}$.The volume $V$ of a spherical bubble is given by $V = \frac{4}{3} \pi R^3$,so $V \propto R^3$.Therefore,the ratio of their volumes is $\frac{V_1}{V_2} = \left( \frac{r_1}{r_2} \right)^3$.Substituting the ratio of radii,we get $\frac{V_1}{V_2} = \left( \frac{P_2 - P_0}{P_1 - P_0} \right)^3$.

Two soap bubbles having radii $r_1$ and $r_2$ have inside pressures $P_1$ and $P_2$ respectively. If $P_0$ is the external pressure,then the ratio of their volumes is:

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