Two soap bubbles having radii ' r₁ ' and ' r₂ ' has inside pressure ' P₁ ' and ' P₂ ' respectively. If P₀ is external pressure then ratio of their volume is (a) ((P₁ - P₀))/((P₂ - P₀)) (b) ((P₂ - P₀))/((P₁ - P₀)) (c) ((P₂ - P₀)³)/((P₁ - P₀)³) (d) ((P₁ - P₀)³)/((P₂ - P₀)³)

Answer: (c). Excess pressure P - P₀ = 4T/r, so r ∝ (1)/(P - P₀). (r₁)/(r₂) = (P₂ - P₀)/(P₁ - P₀) (V₁)/(V₂) = ((r₁)/(r₂))³ = ((P₂ - P₀)³)/((P₁ - P₀)³) So the answer is (c).

Two soap bubbles having radii ' r₁ ' and ' r₂ ' has inside pressure '… | MHT CET 2024 (10 May, Shift 2)

Q27MHT CET 2024 · 10 MAY · SHIFT 2Medium

Excess Pressure: Drops & Bubbles

Two soap bubbles having radii ' $r_1$ ' and ' $r_2$ ' has inside pressure ' $P_1$ ' and ' $P_2$ ' respectively. If $P_0$ is external pressure then ratio of their volume is

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Answer: (c)

Excess pressure $P - P_0 = \frac{4T}{r}$ , so $r \propto \frac{1}{P - P_0}$ .
$\frac{r_1}{r_2} = \frac{P_2 - P_0}{P_1 - P_0}$
$\frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^3 = \frac{(P_2 - P_0)^3}{(P_1 - P_0)^3}$
So the answer is (c).

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Excess Pressure: Drops & Bubbles question from MHT CET 2024 (10 May, Shift 2)

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