Two soap bubbles of radii 3r and 4r are in co | vedclass

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(C) The excess pressure inside a soap bubble of radius $R$ is given by $P = \frac{4T}{R}$,where $T$ is the surface tension.When two bubbles of radii $

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$12r$

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(C) The excess pressure inside a soap bubble of radius $R$ is given by $P = \frac{4T}{R}$,where $T$ is the surface tension.When two bubbles of radii $r_1$ and $r_2$ (where $r_2 > r_1$) are in contact,the pressure inside the smaller bubble $(P_1)$ is higher than the pressure inside the larger bubble $(P_2)$.The pressure difference across the common interface is $\Delta P = P_1 - P_2 = \frac{4T}{r_1} - \frac{4T}{r_2} = 4T \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.If $R$ is the radius of curvature of the common interface,then $\Delta P = \frac{4T}{R}$.Equating the two expressions for $\Delta P$: $\frac{4T}{R} = 4T \left( \frac{r_2 - r_1}{r_1 r_2} \right)$.Therefore,$R = \frac{r_1 r_2}{r_2 - r_1}$.Given $r_1 = 3r$ and $r_2 = 4r$,we have $R = \frac{(3r)(4r)}{4r - 3r} = \frac{12r^2}{r} = 12r$.

Two soap bubbles of radii $3r$ and $4r$ are in contact with each other. The radius of curvature of the interface between the bubbles is:

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