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(D) For a soap bubble,the excess pressure is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension and $r$ is the radius.Let $P_1$ be th

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$2.4$

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(D) For a soap bubble,the excess pressure is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension and $r$ is the radius.Let $P_1$ be the pressure in the region between the two bubbles. For the inner bubble of radius $r_1 = 4 \ cm$:$P_2 - P_1 = \frac{4T}{4} \quad ...(i)$For the outer bubble of radius $r_2 = 6 \ cm$:$P_1 - P_0 = \frac{4T}{6} \quad ...(ii)$Adding equations $(i)$ and $(ii)$:$(P_2 - P_1) + (P_1 - P_0) = \frac{4T}{4} + \frac{4T}{6}$$P_2 - P_0 = 4T \left( \frac{1}{4} + \frac{1}{6} \right)$Let $r$ be the radius of a single bubble that has an excess pressure equal to $P_2 - P_0$:$P_2 - P_0 = \frac{4T}{r}$Equating the two expressions for $P_2 - P_0$:$\frac{4T}{r} = 4T \left( \frac{1}{4} + \frac{1}{6} \right)$$\frac{1}{r} = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}$$r = \frac{12}{5} = 2.4 \ cm$.

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