Q 11-09-152JEE MainJEE Main 2018 (16 Apr, Shift 1)Medium
A small soap bubble of radius $4\ \text{cm}$ is trapped inside another bubble of radius $6\ \text{cm}$ without any contact. Let $P_2$ be the pressure inside the inner bubble and $P_0$, the pressure outside the outer bubble. Radius of another bubble with pressure difference $P_2 - P_0$ between its inside and outside would be:
Answer: (A) $2.4\ \text{cm}$
Excess pressure inside a soap bubble (two surfaces) is $\dfrac{4T}{r}$. Across both bubbles:
$$P_2 - P_0 = \frac{4T}{4} + \frac{4T}{6} = 4T\left(\frac14 + \frac16\right) = \frac{4T}{12/5}$$
A single bubble with this excess pressure has radius
$$r = \frac{12}{5} = 2.4\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics