alleryThe terminal velocity v_t of a small sphere of radius r falling in a fluid of viscosity η is given by Stokes' law as:
Av_t ∝ r
Bv_t ∝ r²
Cv_t ∝ 1 / r
Dv_t ∝ 1 / r²
Answer & Solution
Correct answer: B. v_t ∝ r²
1. At terminal velocity, weight = buoyancy + viscous drag.
2. Stokes' viscous drag: F = 6 π η r v_t.
3. Net weight (downward) = (4/3)π r³ (ρ_s − ρ_l) g.
4. Setting them equal gives v_t = 2 r² (ρ_s − ρ_l) g / (9 η).
5. So v_t ∝ r².
_Source: syllabus Physics Ch 9 §9.7 Viscosity + Terminal Velocity_
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