Q 11-09-023NEETJEE MainHard
Two soap bubbles of radii $3$ cm and $4$ cm join isothermally in vacuum to form a single bubble. The radius of the new bubble is
Answer: (A) $5$ cm
In vacuum the pressure inside a bubble is $\dfrac{4T}{r}$. At constant temperature $pV$ (the amount of gas) is conserved:
$$\frac{4T}{R}\cdot\frac{4}{3}\pi R^3 = \frac{4T}{r_1}\cdot\frac{4}{3}\pi r_1^3 + \frac{4T}{r_2}\cdot\frac{4}{3}\pi r_2^3$$
So $R^2 = r_1^2 + r_2^2 = 9 + 16 = 25$, and $R = 5$ cm.
Solution by Sreeraj P, M.Sc Physics