What is the terminal velocity of a spherical steel ball of radius r falling through a viscous oil of viscosity eta according to Stokes' Law?
Answer and explanation
Correct answer: A. Directly proportional to the square of the radius (r^2)
- Step 1: Net force at terminal velocity: gravitational force minus buoyancy equals Stokes drag: (4/3) pi r^3 (rho - sigma) g = 6 pi eta r v_t. Step 2. Solving for v_t: v_t = [2 r^2 (rho - sigma) g] / (9 eta). Step 3. Thus, terminal velocity is proportional to r^2. Answer: Directly proportional to the square of the radius (r^2). Why not the others: Stokes drag force is linear in r, but mass scales as r^3, yielding r^2 for velocity.
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