Two soap bubbles of radii R and r come in con | vedclass

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(B) The excess pressure inside a soap bubble of radius $r$ is given by $P = \frac{4S}{r}$,where $S$ is the surface tension.For the bubble with radius

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$\frac{Rr}{R-r}$

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(B) The excess pressure inside a soap bubble of radius $r$ is given by $P = \frac{4S}{r}$,where $S$ is the surface tension.For the bubble with radius $R$,the pressure inside is $P_1 = P_{atm} + \frac{4S}{R}$.For the bubble with radius $r$,the pressure inside is $P_2 = P_{atm} + \frac{4S}{r}$.Since $R > r$,the pressure $P_2$ is greater than $P_1$.The common surface will bulge towards the bubble with the larger radius $(R)$.The pressure difference across the common surface is $\Delta P = P_2 - P_1 = \frac{4S}{r} - \frac{4S}{R}$.Let $r_c$ be the radius of curvature of the common surface. Then $\Delta P = \frac{4S}{r_c}$.Equating the two expressions for $\Delta P$:$\frac{4S}{r_c} = 4S \left( \frac{1}{r} - \frac{1}{R} \right)$.$\frac{1}{r_c} = \frac{R - r}{Rr}$.Therefore,$r_c = \frac{Rr}{R - r}$.

Two soap bubbles of radii $R$ and $r$ come in contact. $R$ is more than $r$. The radius of curvature of the common surface is

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