A solid sphere of radius r and density ρₛ falls through a liquid of density ρₗ and viscosity η, where ρₛ > ρₗ. Assuming Stokes' law is valid, what is the terminal speed?
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Question

A solid sphere of radius r and density ρₛ falls through a liquid of density ρₗ and viscosity η, where ρₛ > ρₗ. Assuming Stokes' law is valid, what is the terminal speed?

G1 Physics · Accepted Answer

At terminal speed, acceleration is zero. Therefore, weight = buoyant force + viscous drag. Weight: W = (4πr³/3)ρₛg. Buoyant force: B = (4πr³/3)ρₗg. Stokes drag: F(viscous) = 6πηrvₜ. Thus, (4πr³/3)(ρₛ−ρₗ)g = 6πηrvₜ. Cancel common factors: vₜ = 2r²(ρₛ−ρₗ)g/(9η). Correct option: B.