Q118 / 180 · AIIMS 2010 · Physics

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

  • A
    $\frac{(a+b)}{2}$
  • B
    $\frac{a b}{a+b}$
  • Correct answer
    C
    $\sqrt{a^2+b^2}$
  • D
    $a+b$

Solution

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 $$ 4 \pi a^2+4 \pi b^2=4 \pi R^2 \text { or } R=\sqrt{a^2+b^2} $$

Hence, the correct answer is option: 'C'.

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