If two soap bubbles A and B of radii r_1 and | vedclass

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(B) For a soap bubble,the excess pressure inside is given by $P_{ex} = \frac{4T}{r}$,where $T$ is the surface tension and $r$ is the radius.The total

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$r_1^3: r_2^3$

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(B) For a soap bubble,the excess pressure inside is given by $P_{ex} = \frac{4T}{r}$,where $T$ is the surface tension and $r$ is the radius.The total pressure inside the bubble is $P = P_{atm} + P_{ex}$. Since the bubbles are in a vacuum,$P_{atm} = 0$.Thus,$P = \frac{4T}{r}$.Using the ideal gas law $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is the mass of air,$M$ is the molar mass of air),we have $P = \frac{mRT}{MV}$.Substituting $P = \frac{4T}{r}$ and $V = \frac{4}{3}\pi r^3$,we get $\frac{4T}{r} = \frac{mRT}{M(\frac{4}{3}\pi r^3)}$.Rearranging for mass $m$,we get $m = \frac{4T}{r} \cdot \frac{M \cdot 4\pi r^3}{3RT} = \frac{16\pi TM}{3RT} \cdot r^2$.Since $T$,$M$,$R$,and the temperature are constant,$m \propto r^2$.Therefore,the ratio of masses is $\frac{m_1}{m_2} = \frac{r_1^2}{r_2^2}$.

If two soap bubbles $A$ and $B$ of radii $r_1$ and $r_2$ respectively are kept in vacuum at constant temperature,then the ratio of masses of air inside the bubbles $A$ and $B$ is

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