A spherical ball of radius 1 cm and density 5500 kg/m³ is released from rest at a height 1.25 m above the surface of a liquid of density 1000 kg/m³. Immediately after entering the liquid, it continues with constant speed. Take g = 10 m/s² and assume Stokes' law. If the liquid viscosity is N/10 Pa·s, find N.
Physics · Properties of solid and fluid · viscosity and terminal velocity
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Question

A spherical ball of radius 1 cm and density 5500 kg/m³ is released from rest at a height 1.25 m above the surface of a liquid of density 1000 kg/m³. Immediately after entering the liquid, it continues with constant speed. Take g = 10 m/s² and assume Stokes' law. If the liquid viscosity is N/10 Pa·s, find N.

Physics · Properties of solid and fluid · viscosity and terminal velocity

G1 Physics · Accepted Answer

Speed just before entering the liquid: u = √(2gh) = √(2×10×1.25) = √25 = 5 m/s. The ball moves with constant speed immediately after entry, so vₜ = 5 m/s. For Stokes motion, vₜ = 2r²(ρₛ−ρₗ)g/ (9η). Substitute: r = 1 cm = 0.01 m, ρₛ−ρₗ = 5500−1000 = 4500 kg/m³, g = 10, vₜ = 5. Therefore, 5 = 2(0.01)²(4500)(10)/ (9η). The numerator is 9. Hence, 5 = 9/(9η) = 1/η. Therefore, η = 0.20 Pa·s = 2/10 Pa·s. Hence, N = 2.