When two soap bubbles of radii a and b (b a | vedclass

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(A) Let $P_1$ and $P_2$ be the excess pressures inside the soap bubbles of radii $a$ and $b$ respectively.The excess pressure inside a soap bubble is

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$\frac{ab}{b-a}$

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(A) Let $P_1$ and $P_2$ be the excess pressures inside the soap bubbles of radii $a$ and $b$ respectively.The excess pressure inside a soap bubble is given by $P = \frac{4T}{r}$,where $T$ is the surface tension.For the two bubbles,$P_1 = \frac{4T}{a}$ and $P_2 = \frac{4T}{b}$.When they coalesce,they form a common interface with radius of curvature $r$.The pressure difference across this common interface is $\Delta P = P_1 - P_2$ (since $a  P_2$).Thus,$\frac{4T}{r} = \frac{4T}{a} - \frac{4T}{b}$.Dividing by $4T$,we get $\frac{1}{r} = \frac{1}{a} - \frac{1}{b}$.$\frac{1}{r} = \frac{b-a}{ab}$.Therefore,$r = \frac{ab}{b-a}$.

When two soap bubbles of radii $a$ and $b$ $(b > a)$ coalesce,the radius of curvature of the common surface is:

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