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(C) Let $P_0$ be the atmospheric pressure. The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = \frac{4T}{R}$.For the two bu

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$4$

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(C) Let $P_0$ be the atmospheric pressure. The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = \frac{4T}{R}$.For the two bubbles with radii $R_1 = 2 \, cm$ and $R_2 = 4 \, cm$,the pressures inside them are:$P_1 = P_0 + \frac{4T}{R_1}$$P_2 = P_0 + \frac{4T}{R_2}$The pressure difference across the common interface is $\Delta P_{int} = P_1 - P_2 = \frac{4T}{R_1} - \frac{4T}{R_2}$.If $R$ is the radius of curvature of the interface,then $\Delta P_{int} = \frac{4T}{R}$.Equating the two expressions:$\frac{4T}{R} = 4T \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$\frac{1}{R} = \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{2} - \frac{1}{4} = \frac{2-1}{4} = \frac{1}{4}$Therefore,$R = 4 \, cm$.

Two soap bubbles of radii $2 \, cm$ and $4 \, cm$ join to form a double bubble in air. The radius of curvature of the interface is .......... $cm$.

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