Two spherical soap bubbles of radii r1 and r2 in vacuum combine under… | JEE Main 2021 (25 Jul, Shift 2)

Q48JEE Main 2021 · 25 JUL · SHIFT 2Medium

Excess Pressure: Drops & Bubbles

Two spherical soap bubbles of radii r1 and r2 in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:

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Answer: (c)

Isothermal, pressure terms from surface tension only (vacuum / neglect outside pressure): $PV$ is conserved.
For a soap bubble $P = \frac{4T}{r}$ , so $PV = \frac{4T}{r} \cdot \frac{4}{3}\pi r^3 \propto r^2$ .
$\frac{4T}{R}R^3 = \frac{4T}{r_1}r_1^3 + \frac{4T}{r_2}r_2^3 \Rightarrow R^2 = r_1^2 + r_2^2$
So the answer is (c) $\sqrt{r_1^2 + r_2^2}$ .

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