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(D) Let $P_0$ be the atmospheric pressure,$P_2$ be the pressure in the region between the two bubbles,and $P_1$ be the pressure inside the smaller bub

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$2$

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(D) Let $P_0$ be the atmospheric pressure,$P_2$ be the pressure in the region between the two bubbles,and $P_1$ be the pressure inside the smaller bubble.For the outer bubble of radius $R_2 = 6\,cm$,the excess pressure is $P_2 - P_0 = \frac{4T}{R_2} = \frac{4T}{6}$.For the inner bubble of radius $R_1 = 3\,cm$,the excess pressure is $P_1 - P_2 = \frac{4T}{R_1} = \frac{4T}{3}$.Adding these two equations,we get the total excess pressure inside the smaller bubble relative to the atmosphere:$P_1 - P_0 = (P_1 - P_2) + (P_2 - P_0) = \frac{4T}{3} + \frac{4T}{6} = \frac{8T + 4T}{6} = \frac{12T}{6} = 2T$.For a single soap bubble of radius $r$,the excess pressure is $P_{excess} = \frac{4T}{r}$.Equating the two,we have $\frac{4T}{r} = 2T$,which gives $r = \frac{4}{2} = 2\,cm$.

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