Practice portal › Mechanical Properties of Fluids › Excess Pressure in Drops and Bubbles
Asked in JEE Main 3rd Sept 1st Shift 2020 · Excess pressure
Given: pressures 1.01 atm and 1.02 atm, against an outside pressure of 1 atm.
Idea: only the excess over the atmosphere is set by the curvature, so subtract first — the excesses are 0.01 atm and 0.02 atm.
For a soap bubble Δ p=(4T)/R, so (R₁)/(R₂)=(Δ p₂)/(Δ p₁)=(0.02)/(0.01)=2.
(V₁)/(V₂)=((R₁)/(R₂))³=2³.
(V₁)/(V₂)=8:1.
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