For a sphere falling where the drag coefficient is essentially constant (Newton’s regime, high Reynolds number), the terminal settling velocity varies with diameter d according to which proportionality?

Introduction / Context:Settling behavior transitions across regimes. In the Newton (constant C_d) regime at high Reynolds numbers, drag force is quadratic in velocity and proportional to cross-sectional area, while weight–buoyancy scales with volume.

Given Data / Assumptions:

• Constant drag coefficient C_d.
• Sphere diameter = d; area ∝ d^2; volume ∝ d^3.

Concept / Approach:At terminal velocity v_t, drag balances net weight: (1/2) C_d ρ_f A v_t^2 = (ρ_p − ρ_f) g V. Since A ∝ d^2 and V ∝ d^3, rearranging gives v_t ∝ sqrt(d), i.e., v_t ∝ d^0.5.

Step-by-Step Solution:Write force balance: drag = net weight.Substitute A and V scalings with d.Solve for v_t ∝ (d^3 / d^2)^0.5 = d^0.5.

Verification / Alternative check:In contrast, in Stokes (laminar) regime, v_t ∝ d^2; the constant-C_d regime clearly yields the square-root dependence, supporting the selected option.

Why Other Options Are Wrong:d or d^2 correspond to other regimes/assumptions; 1/d and d^3 are inconsistent with Newton-law drag.

Common Pitfalls:Using Stokes-law dependence outside its low-Re validity range.

Final Answer:v_t ∝ d^0.5