10. Stokes' Method — Viscosity by Terminal Velocity of a Falling Sphere¶

Module

Fluid Properties: Viscosity & Surface Tension · Module 3

Try it live

Explore this on the interactive simulator with real-time sliders and graphs: 🔬 Launch in Simulator

Aim¶

To determine the coefficient of viscosity of a highly viscous liquid by measuring the terminal velocity of small spheres falling freely through it.

Theoretical Background¶

When a small sphere falls through a viscous fluid, it experiences three forces: its weight acting downward, the upward buoyant (upthrust) force due to the displaced fluid, and a viscous drag force that opposes its motion and increases with speed. Sir George Stokes showed that for a sphere falling slowly enough that the flow around it remains laminar, this drag force is directly proportional to the sphere's radius, its speed, and the fluid's viscosity — a result now called Stokes' law.

Immediately after release the sphere accelerates, since its weight initially exceeds the (still small) sum of buoyancy and drag; but as its speed increases, the drag force grows until the three forces exactly balance, at which point the sphere continues to fall at a constant speed called its terminal velocity. Because this terminal velocity depends on the sphere's radius and density and on the fluid's density and viscosity, measuring it gives a direct route to the fluid's viscosity if the other quantities are known.

In this experiment, small spheres of a known material (hence known density) and various radii are dropped, one at a time, into a tall glass jar of the liquid under study, and the time taken to fall a marked distance well below the point where terminal velocity is reached is measured. Repeating for several sphere sizes and plotting the terminal velocity against the square of the radius gives a straight line whose slope yields the viscosity.

Highly viscous liquids such as glycerine or castor oil are typically studied by this method, since terminal velocity is reached quickly and can be sustained over an easily measurable distance, unlike in a low-viscosity liquid such as water where a sphere would reach the bottom of the jar before ever attaining a genuinely steady terminal speed.

Governing Formulas¶

Stokes' law — terminal velocity:

\[ v = \dfrac{2}{9}\, \dfrac{r^{2} g (\rho-\sigma)}{\eta} \]

Coefficient of viscosity:

\[ \eta = \dfrac{2}{9}\, \dfrac{r^{2} g (\rho-\sigma)}{v} \]

where \(r\) is the sphere's radius, \(\rho\) its density, \(\sigma\) the density of the liquid, and \(v\) the terminal (constant) velocity.

Labelled apparatus layout — original schematic prepared for this guide (not a photograph). A tall, wide glass jar filled with the liquid under study, with two marked rings on the outside a known distance apart, well below the surface, and a mechanism for gently releasing small spheres one at a time along the central axis.

Simulator Controls¶

Control	Symbol	Range	Unit
Sphere radius	r	0.5 – 3.0	mm
Sphere material (density)	ρ	2700 – 8900	kg/m³
Liquid (density)	σ	1000 – 1300	kg/m³

Readouts:

Quantity	Symbol	Unit
Terminal velocity	v	cm/s

Procedure¶

Launch the Stokes' Method panel; select a sphere material and a liquid.
Vary the sphere radius r over the available range and record the terminal velocity v shown on the readout for each size.
Plot v against r² and verify that the points lie on a straight line through the origin.
Repeat for a different liquid (different density/viscosity) and compare the slope of the resulting graph.
Calculate the coefficient of viscosity η from the slope of the v vs r² graph, using the known ρ, σ.

Observation Table¶

Table: Terminal velocity for spheres of increasing radius

S. No.	r (mm)	r² (mm²)	ρ (kg/m³)	Terminal velocity v (cm/s)
1
2
3
4
5
6

Graph¶

Plot Terminal velocity, v (y-axis) against r² (x-axis).

Expected trend: a straight line through the origin (v ∝ r²).

v vs r², expected to be a straight line through the origin; slope = 2g(ρ−σ)/9η

15×15 grid — choose your own scale per square; plot from your observation table

Calculation¶

Compute the slope of the v vs r² graph: slope = 2g(ρ−σ)/9η.
Rearrange to find η = 2g(ρ−σ)/(9 × slope), using the known ρ and σ.
Repeat for a different sphere material or liquid as a cross-check, and compare the two independently obtained values of η.

Result¶

Table: Coefficient of viscosity of the liquid

Liquid	σ (kg/m³)	Sphere material	ρ (kg/m³)	Slope (cm/s per mm²)	η (poise)
1
2

Maximum Permissible Error¶

The sphere radius enters squared, and since ρ and σ are usually close in value for typical sphere/liquid combinations, the small difference (ρ−σ) can itself carry a relatively large fractional uncertainty.

Precautions¶

Drop each sphere gently along the central axis of the jar, well clear of the walls, since wall proximity introduces an extra drag correction not accounted for in the simple Stokes' law formula.
Mark the timing points well below the point of release, to be certain terminal velocity has genuinely been reached before timing begins.
Use small spheres and a sufficiently viscous liquid so the flow around the sphere remains laminar (low Reynolds number), as Stokes' law assumes.

Self-Check Quiz¶

1. A sphere falling through a viscous liquid reaches terminal velocity when:

A) Its weight becomes zero
B) The net force on it becomes zero, i.e. weight equals the sum of buoyancy and drag
C) The drag force becomes zero
D) It touches the bottom of the jar

Answer: B — Terminal velocity is precisely the constant speed at which the three forces (weight, buoyancy, drag) balance, giving zero net force and zero acceleration.

2. If the sphere's radius is doubled (all else fixed), Stokes' law predicts the terminal velocity:

A) Doubles
B) Increases fourfold
C) Increases eightfold
D) Halves

Answer: B — v ∝ r², so doubling r increases the terminal velocity by a factor of 4.

3. Why is this method suited to highly viscous liquids rather than water?

A) Stokes' law is only mathematically valid for viscous liquids
B) In a low-viscosity liquid the sphere reaches the bottom of the jar before ever settling into a steady terminal velocity that can be measured
C) Water is too dense to use
D) Spheres float in water

Answer: B — High viscosity keeps the terminal velocity low and the settling distance to reach it short, giving a long, easily measurable stretch of genuinely constant-velocity fall — this is impractical in a low-viscosity liquid like water.

Viva-Voce Questions¶

State Stokes' law and identify the three forces that act on a falling sphere in a viscous liquid.
Why must the sphere be released well above the marked timing points?
What correction becomes necessary if the sphere's diameter is not small compared to the diameter of the containing jar?
How would the terminal velocity change if a sphere of the same radius but greater density were used?
Why does this method work well for glycerine but poorly for water?
What is the Reynolds number, and why does Stokes' law require it to be small?