A solid steel ball of diameter 3.6 mm acquires a terminal velocity 2.45×10⁻² m s⁻¹ while falling under gravity through an oil of density 925 kg m⁻³. Taking the density of steel as 7825 kg m⁻³ and g=9.8 m s⁻², the viscosity of the oil in SI units is

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(3) 1.99. Given: r=1.8 mm=1.8×10⁻³ m, v=2.45×10⁻² m s⁻¹, ρₛ=7825 and ρₒ=925 kg m⁻³, g=9.8 m s⁻². At the terminal velocity, 4/3π r³(ρₛ-ρₒ)g=6πη rv, so η=(2r²(ρₛ-ρₒ)g)/(9v). Density difference =7825-925=6900 kg m⁻³, and r²=3.24×10⁻⁶ m². η=(2×3.24×10⁻⁶×6900×9.8)/(9×2.45×10⁻²)=(0.438)/(0.2205). η=1.99 Pa s.

A solid steel ball of diameter 3.6 mm acquires a terminal velocity 2.45×10⁻² m s⁻¹ while falling under gravity through an oil of density 925 kg m⁻³. Taking the density of steel as 7825 kg m⁻³ and g=9.8 m s⁻², the viscosity of the oil in SI units is

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