When two soap bubbles of radius r_1 and r_2 ( | vedclass

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(C) Let $P_1$ and $P_2$ be the excess pressures inside the two soap bubbles of radii $r_1$ and $r_2$ respectively.For a soap bubble,the excess pressur

Vedclass · Accepted Answer

$\frac{r_1 r_2}{r_2 - r_1}$

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(C) Let $P_1$ and $P_2$ be the excess pressures inside the two soap bubbles of radii $r_1$ and $r_2$ respectively.For a soap bubble,the excess pressure is given by $P = \frac{4T}{r}$.Thus,$P_1 = \frac{4T}{r_1}$ and $P_2 = \frac{4T}{r_2}$.When they coalesce,the common interface acts as a surface with radius of curvature $r$. The pressure difference across this common interface is $\Delta P = P_1 - P_2$ (since $r_1  P_2$).Therefore,$\frac{4T}{r} = \frac{4T}{r_1} - \frac{4T}{r_2}$.Dividing by $4T$,we get $\frac{1}{r} = \frac{1}{r_1} - \frac{1}{r_2}$.$\frac{1}{r} = \frac{r_2 - r_1}{r_1 r_2}$.Hence,$r = \frac{r_1 r_2}{r_2 - r_1}$.

When two soap bubbles of radius $r_1$ and $r_2$ $(r_2 > r_1)$ coalesce,the radius of curvature of the common surface is:

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