The diameter of an air bubble which was initially 2 mm, rises steadily through a of density 1750 kg m⁻³ at the rate of 0.35 cms⁻¹. The coefficient of viscosity of the is _________ poise (in nearest integer). (the density of air is negligible). Numerical answer type. Work it out, then open the answer.

Answer: 11. A bubble rising at terminal speed: v = 2r²(ρ - ρair)g9η ≈ (2r²ρ g)/(9η) (air density negligible). r = 1 mm = 10⁻³ m, v = 0.35 cm/s = 3.5×10⁻³ m/s, ρ = 1750 kg m⁻³. η = (2r²ρ g)/(9v) = 2×10⁻⁶×1750×109×3.5×10⁻³ = (0.035)/(0.0315) ≈ 1.11 Pa s 1 Pa s = 10 poise ⇒ η ≈ 11.1 poise So the answer is 11 poise.

The diameter of an air bubble which was initially 2 mm, rises… | JEE Main 2022 (28 Jul, Shift 1)

Q41JEE Main 2022 · 28 JUL · SHIFT 1Medium

Terminal Velocity

The diameter of an air bubble which was initially $2\text{ mm}$ , rises steadily through a of density $1750\text{ kg m}^{-3}$ at the rate of $0.35\text{ cms}^{-1}$ . The coefficient of viscosity of the is _________ poise (in nearest integer). (the density of air is negligible).

Numerical answer type. Work it out, then open the answer.

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Answer: 11

A bubble rising at terminal speed: $v = \dfrac{2r^2(\rho - \rho_{air})g}{9\eta} \approx \dfrac{2r^2\rho g}{9\eta}$ (air density negligible).
$r = 1\text{ mm} = 10^{-3}\text{ m}$ , $v = 0.35\text{ cm/s} = 3.5\times10^{-3}\text{ m/s}$ , $\rho = 1750\text{ kg m}^{-3}$ .
$\eta = \dfrac{2r^2\rho g}{9v} = \dfrac{2\times10^{-6}\times1750\times10}{9\times3.5\times10^{-3}} = \dfrac{0.035}{0.0315} \approx 1.11\text{ Pa s}$
$1\text{ Pa s} = 10\text{ poise} \Rightarrow \eta \approx 11.1\text{ poise}$
So the answer is $11$ poise.

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