Under isothermal conditions, two soap bubbles of radii r₁ and r₂ coalesce to form a big drop. The radius of the big drop is (a) (r₁ + r₂)1/2 (b) (r₁ + r₂)² (c) (r₁² + r₂²)1/2 (d) (r₁ + r₂)³

Answer: (c). Isothermal, pressure terms from surface tension only (vacuum / neglect outside pressure): PV is conserved. For a soap bubble P = 4T/r, so PV = 4T/r · 4/3π r³ ∝ r². 4T/RR³ = 4T/r₁r₁³ + 4T/r₂r₂³ ⇒ R² = r₁² + r₂² So the answer is (c) (r₁² + r₂²)1/2.

Under isothermal conditions, two soap bubbles of radii r₁ and r₂… | MHT CET 2025 (26 Apr, Shift 2)

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Excess Pressure: Drops & Bubbles

Under isothermal conditions, two soap bubbles of radii $r_1$ and $r_2$ coalesce to form a big drop. The radius of the big drop is

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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) $\left(r_1^2 + r_2^2\right)^{1/2}$ .

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