An air bubble of diameter 6 mm rises steadily through a solution of density 1750 kg m⁻³ at 0.35 cm s⁻¹. The coefficient of viscosity of the solution is ______ Pa s. [Neglect density of air, g = 10 m s⁻²]
Physics · Circular Motion
JEE Mains 8 April 2023 Shift 1 Morning Shift

Question

An air bubble of diameter 6 mm rises steadily through a solution of density 1750 kg m⁻³ at 0.35 cm s⁻¹. The coefficient of viscosity of the solution is ______ Pa s. [Neglect density of air, g = 10 m s⁻²]

Physics · Circular Motion

JEE Mains 8 April 2023 Shift 1 Morning Shift

G1 Physics · Accepted Answer

At terminal speed, buoyant force equals viscous drag because the density of air inside the bubble is neglected. Buoyant force: F(B) = ρ(4/3)πr³g Stokes drag: F(v) = 6πηrv Equating: ρ(4/3)πr³g = 6πηrv Therefore, η = 2ρr²g/(9v) Given: ρ = 1750 kg m⁻³ r = 3 mm = 3×10⁻³ m g = 10 m s⁻² v = 0.35 cm s⁻¹ = 3.5×10⁻³ m s⁻¹ Thus, η = [2×1750×(3×10⁻³)²×10] / [9×3.5×10⁻³] = 10 Pa s Final value: 10