Q 11-09-166JEE MainMedium
A soap bubble of radius $r$ is formed inside another soap bubble of radius $6$ cm, the two being concentric. The pressure inside the inner bubble exceeds the atmospheric pressure by the same amount as the pressure inside an isolated soap bubble of radius $2$ cm. The value of $r$ is
Answer: (A) $3$ cm
Concept. A soap bubble has two surfaces, so the excess pressure across it is $\dfrac{4T}{R}$. For nested bubbles the excess pressures add.
Step 1. Pressure in the inner bubble.
$$P_{in} - P_0 = \frac{4T}{6} + \frac{4T}{r}$$
Step 2. Equate to a single bubble of radius 2 cm.
$$\frac{4T}{6} + \frac{4T}{r} = \frac{4T}{2} \;\Rightarrow\; \frac1r = \frac12 - \frac16 = \frac13$$
Answer. $3$ cm (Option A).
Solution by Sreeraj P, M.Sc Physics