JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Fluid Mechanics

Two spherical soap bubbles of radii

r_{1}

and

r_{2}

in vacuum combine under isothermal conditions. The resulting bubble has radius equal to :

(A) $r_{1} r_{2}$
(B) $r_{1}^{2} + r_{2}^{2}$ ✓ Correct
(C) $\frac{r _{1} r _{2}}{r _{1} + r _{2}}$
(D) $\frac{r _{1} + r _{2}}{2}$

Answer: (B)

Solution

For a soap bubble, the excess pressure is

P = \frac{4 T}{r}

. Under isothermal conditions,

P V =

constant. For the two bubbles combining:

P_{1} V_{1} + P_{2} V_{2} = P V

, which gives

\frac{4 T}{r _{1}} \cdot \frac{4}{3} π r_{1}^{3} + \frac{4 T}{r _{2}} \cdot \frac{4}{3} π r_{2}^{3} = \frac{4 T}{r} \cdot \frac{4}{3} π r^{3}

. Simplifying:

r_{1}^{2} + r_{2}^{2} = r^{2}

, so

r = r_{1}^{2} + r_{2}^{2}

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option

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