Terminal settling velocity by Stokes’ law (fine particle in water) A soil particle of diameter 0.05 cm has specific gravity Gs = 2.67. Estimate its terminal settling velocity in distilled water of dynamic viscosity 0.01 poise, assuming Stokes’ law applies (laminar regime).

Introduction / Context:
For very fine, approximately spherical particles settling slowly in a viscous fluid, Stokes’ law provides the terminal velocity. This is used in sedimentation analysis of fine soils (hydrometer method) where laminar conditions prevail.

Given Data / Assumptions:

• Specific gravity of particle, Gs = 2.67.
• Dynamic viscosity of water, μ = 0.01 poise (i.e., 0.01 g/(cm·s)).
• Gravitational acceleration g = 981 cm/s^2.
• Particle diameter d is interpreted in Stokes’ law as the sphere diameter.
• Laminar (creeping) flow assumption holds.

Concept / Approach:

Stokes’ law (cgs units) for a spherical particle is v = [(ρ_s − ρ_w) * g * d^2] / (18 * μ). With ρ_s = Gs * ρ_w and ρ_w = 1 g/cm^3, the density difference becomes (Gs − 1).

Step-by-Step Solution:

Verification / Alternative check:

Dimensional analysis confirms cm/s. The result matches typical hydrometer-scale velocities for silt-size particles.

Why Other Options Are Wrong:

Values 0.2200–0.2300 cm/s bracket the computed figure; 0.2275 cm/s is the most accurate with given constants.

Common Pitfalls:

Using diameter in mm without converting to cm in cgs; applying Stokes’ law to sand-size particles where Reynolds number is too high; forgetting that proportionality is to d^2.

Final Answer:

0.2275 cm/s