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Question: Answered & Verified by Expert
Two spherical soap bubbles of radii $a$ and $b$ in vacuum coaleasce under isothermal conditions. The resulting bubbles has a radius given by
PhysicsMechanical Properties of FluidsAIIMSAIIMS 2010
Options:
  • A $\frac{(a+b)}{2}$
  • B $\frac{a b}{a+b}$
  • C $\sqrt{a^2+b^2}$
  • D $a+b$
Solution:
2593 Upvotes Verified Answer
The correct answer is: $\sqrt{a^2+b^2}$
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
$$
4 \pi a^2+4 \pi b^2=4 \pi R^2 \text { or } R=\sqrt{a^2+b^2}
$$

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