Stokes' Law in a Narrow Cylinder: The Wall Correction
During the viscosity practical, a student drops small steel balls of radius 2.0 mm and density 7.8 × 10³ kg m⁻³ into a measuring cylinder of castor oil of density 0.96 × 10³ kg m⁻³. The handbook value of the oil's viscosity at room temperature is 0.99 Pa s. The cylinder has an internal diameter of only 3.0 cm. The student times a ball between two marks 0.20 m apart, both well below the oil surface, and obtains 4.4 s. Using Stokes' law he calculates a viscosity close to 1.3 Pa s, about 30% above the handbook value, and concludes that the oil has been contaminated. A second student uses a 10 cm diameter vessel, the same oil and the same method, and obtains a value much closer to 0.99 Pa s. The teacher says the discrepancy is a correction term, not contamination.
1. Calculate the viscosity from the first student's data and identify the physical reason for the discrepancy. Apply the wall correction $\eta_{\text{true}}=\eta_{\text{measured}}/(1+2.4r/R)$ and check whether it resolves the problem.
Read the worked answer
The terminal speed is $v=0.20/4.4=4.55\times10^{-2}\ \text{m s}^{-1}$. Stokes' law with buoyancy gives
$\eta=\dfrac{2r^2(\rho-\sigma)g}{9v}$
$=\dfrac{2(2.0\times10^{-3})^2(7800-960)(9.8)}{9(4.55\times10^{-2})}$
$=\dfrac{0.536}{0.409}=1.31\ \text{Pa s}$.
The formula assumes an infinite extent of fluid. In a cylinder of radius $R=1.5$ cm the oil must flow back around the ball, and the cylinder wall retards it, so the measured terminal speed is lower than the free-fluid value and the computed viscosity is too high. The correction with $r/R=0.2/1.5=0.133$ gives a factor $1+2.4(0.133)=1.32$, so
$\eta_{\text{true}}=1.31/1.32=0.99\ \text{Pa s}$,
in exact agreement with the handbook. The oil is not contaminated; the apparatus was too narrow.
2. Why does the second student's 10 cm vessel give a more reliable answer, and why does it matter that both sets of marks lie well below the oil surface?
Read the worked answer
With $R=5$ cm, the correction factor becomes $1+2.4(0.2/5)=1.10$, so the uncorrected error is only about 10% instead of 32%; as $R\to\infty$ the factor tends to 1 and Stokes' law applies as written. A wider vessel therefore needs a smaller (and less uncertain) correction.
The marks matter because a ball released at the surface accelerates until the viscous drag plus buoyancy balances its weight. Only after that transient does it move at constant terminal velocity, which is the $v$ that appears in Stokes' law. Timing a segment close to the surface averages in slower, accelerating motion, giving a $v$ that is too small and hence a viscosity that is too large. Placing both marks well below the surface ensures the ball is already at terminal speed, so the timing measures the correct quantity. The near-surface region also has meniscus and surface-tension effects that are best avoided.
3. The student proposes repeating the experiment with balls of radius 4.0 mm to get a longer, more accurate time. Predict the effect on the terminal velocity and on the reliability of the result, using the $r^2$ dependence and the wall correction.
Read the worked answer
Since $v=\dfrac{2r^2(\rho-\sigma)g}{9\eta}$, doubling the radius should increase the terminal speed about fourfold, from $4.55\times10^{-2}$ to roughly $0.18\ \text{m s}^{-1}$. The fall time between the marks would drop to about 1.1 s, which is actually less accurate to time, not more.
The wall correction becomes worse: $1+2.4(0.4/1.5)=1.64$, a 64% correction in the narrow tube. More seriously, the Reynolds number becomes
$Re=\dfrac{\sigma v(2r)}{\eta}=\dfrac{960\times0.18\times0.008}{0.99}\approx1.4$,
which is at or beyond the limit where Stokes' law is valid (the flow behind the ball begins to separate). The larger ball would therefore give a systematically wrong viscosity even after the wall correction. Better practice is to keep the ball small, use a wide vessel, and time a long fall rather than enlarge the ball.