Question

Two spherical soap bubbles of radii $a$ and $b$ in vacuum coaleasce under isothermal conditions. The resulting bubbles has a radius given by

• A. $\frac{\left(a+b\right)}{2}$
• B. $\frac{ab}{a+b}$
• C. $\sqrt{a^{2}+b^{2}}$
• D. $a+b$

Answer 1

Answer: (C)

$\sqrt{a^{2}+b^{2}}$

Answer 2

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 $R=\sqrt{a^{2}+b^{2}}$