The excess pressure inside the first soap bubble of radius R₁ is three times that inside the second soap bubble of radius R₂. The ratio of volumes of the first bubble to second bubble is (a) 1: 3 (b) 1: 6 (c) 1: 27 (d) 1: 9

Answer: (c). Δ P ∝ 1/R; Δ P₁ = 3Δ P₂ ⇒ R₁ = R₂/3. (V₁)/(V₂) = (1/3)³ = 1/27 So the answer is (c) 1: 27.

The excess pressure inside the first soap bubble of radius R₁ is… | MHT CET 2026 (15 Apr, Shift 1)

Q12MHT CET 2026 · 15 APR · SHIFT 1Medium

Excess Pressure: Drops & Bubbles

The excess pressure inside the first soap bubble of radius $R_1$ is three times that inside the second soap bubble of radius $R_2$ . The ratio of volumes of the first bubble to second bubble is

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Answer: (c)

$\Delta P \propto \frac{1}{R}$ ; $\Delta P_1 = 3\Delta P_2 \Rightarrow R_1 = \frac{R_2}{3}$ .
$\frac{V_1}{V_2} = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$
So the answer is (c) $1 : 27$ .

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