Q. Two spherical soap bubbles of radii $a$ and $b$ in vacuum coalesce under isothermal conditions. The resulting bubble has a radius given by
A) $\frac{(a + b)}{2}$
B) $\frac{ab}{a + b}$
C) $\sqrt{a^{2} + b^{2}}$
D) $a + b$
Explanation
(c) Since, the bubbles coalesce in vacuum and there is no change in temperature, hence its surface energy does not change. This means that the surface area remains unchanged. Hence,
$4\ \pi{\ a}^{2} + 4\ \pi\ b^{2} = 4\pi\ R^{2}$ or $R = \sqrt{a^{2} + b^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.