Terminal settling velocity of a small sphere (Stokes’ law form) A sphere of radius r and specific weight γ_s falls slowly through a viscous liquid of viscosity μ and specific weight γ. For creeping flow (very low Reynolds number), the terminal velocity v is:

Introduction / Context:
Stokes’ law gives terminal settling velocity for tiny particles where viscous forces dominate inertial forces (Re << 1). This is fundamental in sedimentation, centrifugation scaling, and particle sizing in process engineering and environmental engineering.

Given Data / Assumptions:

• Rigid sphere of radius r.
• Very low Reynolds number (creeping flow, no wake).
• Fluid is Newtonian with dynamic viscosity μ.
• Specific weights: particle γ_s, fluid γ; gravity acts downward.

Concept / Approach:

At terminal velocity, net force is zero: buoyant-reduced weight equals viscous drag. Balance: (γ_s − γ) * (4/3) * pi * r^3 = 6 * pi * μ * r * v, leading to the Stokes formula. Using γ = ρ g automatically incorporates g.

Step-by-Step Solution:

Verification / Alternative check:

Dimensional check: [γ] = N/m^3, so r^2 * γ / μ gives m/s as required. The dependence v ∝ r^2 is characteristic of Stokes settling.

Why Other Options Are Wrong:

Options with r, r^3, or sum (γ_s + γ) do not follow from the drag–weight balance; (9/2) factor is inverted.

Common Pitfalls:

Using density difference without g when specific weight is already used; applying Stokes’ law at higher Reynolds numbers where it no longer holds.

Final Answer:

v = (2/9) * r^2 * (γ_s − γ) / μ