Q 11-09-119JEE MainJEE Main 2021 (25 Jul, Shift 2)Medium
Two spherical soap bubbles of radii $r_1$ and $r_2$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:
Answer: (C) $\sqrt{r_1^2 + r_2^2}$
In vacuum the pressure inside a soap bubble is $P = \dfrac{4T}{r}$. Isothermal, so the total $PV$ (amount of air) is conserved:
$$\frac{4T}{r_1}\cdot\frac{4}{3}\pi r_1^3 + \frac{4T}{r_2}\cdot\frac{4}{3}\pi r_2^3 = \frac{4T}{R}\cdot\frac{4}{3}\pi R^3$$
$$r_1^2 + r_2^2 = R^2 \Rightarrow R = \sqrt{r_1^2 + r_2^2}$$
Solution by Sreeraj P, M.Sc Physics