Stokes-Law Viscosity by Terminal Speed
For a small sphere moving slowly through an effectively unbounded Newtonian liquid, Stokes drag is 6 pi eta R v; balancing drag with weight minus buoyancy gives terminal speed proportional to R squared and inversely proportional to viscosity.
Why this shows up in the exam
Falling-ball viscometers · Comparing liquid viscosities · Estimating particle settling rates
Learn the idea
A falling sphere reaches terminal speed when viscous drag balances effective weight. A larger sphere initially has more downward effective weight, but drag grows with speed; after transients die out, the forces balance and the steady speed reveals the liquid's viscosity.
🧠 Memory hook: At terminal speed, effective weight equals viscous drag.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- 6 pi eta R v_t = (4/3)pi R³(rho_s-rho_l)g — terminal force balance under Stokes-flow assumptions
- v_t = 2R²(rho_s-rho_l)g/(9 eta) — viscosity relation after terminal speed is established
How to approach it
- 1Check that terminal motion is specified
- 2Include buoyancy through the density difference
- 3Use the R-squared scaling to test statements or graphs
Common slip-ups that cost marks
- •Using measurements before terminal speed
- •Ignoring liquid temperature
- •Claiming launch speed changes the material viscosity
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
In a meter bridge, resistance 2 ohm is in the left gap and balance occurs at 40 cm from the left end. Find the right-gap resistance.
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