Viscosity, Stokes' Law & Terminal Velocity

A raindrop condenses two kilometres up. Free-fall physics — your kinematics chapter — predicts it should arrive at $v = \sqrt{2gh} \approx 200\ \mathrm{m/s}$ , the muzzle velocity of a pistol. A thunderstorm should be lethal. Actual landing speed of a large raindrop: about $9\ \mathrm{m/s}$ , a brisk jog. Somewhere between the cloud and your umbrella, 99.8% of the drop's would-be kinetic energy went missing — skimmed off, continuously, by the air itself. The skimming agent is viscosity: internal friction between layers of a fluid sliding past one another. It is why honey pours slowly and water splashes, why you must keep stirring paint, and why anything falling through any fluid eventually stops accelerating. The drop speeds up, drag grows with speed, and at some speed the drag (plus buoyancy) exactly cancels the weight. From then on: zero net force, constant velocity — terminal velocity. For a small sphere, George Stokes computed the drag exactly, $F = 6\pi\eta r v$ , and the resulting terminal-velocity formula, with its famous $r^2$ scaling, is why fog floats while raindrops fall — and it handed Robert Millikan the tool that weighed the electron. This module is a CBSE/ISC derivation favourite and a NEET numerical staple; JEE adds ratio games and the graph of the approach to terminal speed. It also completes your force toolkit: after constant forces, springs, and friction, this is your first velocity-dependent force done properly.

Key takeaways

Viscosity is fluid-layer friction: a fluid flowing over a surface moves in layers (fastest far away, stationary at the wall), and each layer drags its slower neighbour. Coefficient $\eta$ , unit Pa·s (poiseuille); 1 Pa·s = 10 poise.
Newton's law of viscous flow: tangential force $F = \eta A \dfrac{\Delta v}{\Delta z}$ — proportional to area and to velocity gradient.
Stokes' law: a small sphere of radius $r$ moving at speed $v$ through fluid of viscosity $\eta$ feels drag $F = 6\pi\eta r v$ .
Terminal velocity: when weight = buoyancy + drag, acceleration dies: $v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta}$ — the $r^2$ makes big drops fall and mist float.
Temperature: heating thins liquids ( $\eta$ falls) but thickens gases ( $\eta$ rises) — a beloved MCQ inversion.
Key idea: drag grows with speed until it cancels the driving force — every fall through fluid ends at constant velocity, scaling as $r^2$ .

The core insight

A body falling through a viscous fluid accelerates only until the velocity-dependent drag (Stokes' $6\pi\eta rv$ for small spheres) plus buoyancy balances its weight, after which it moves at the constant terminal velocity $v_t = 2r^2(\rho-\sigma)g/9\eta$ .

Where this lives today

The most consequential use of this article's formula was Millikan's oil-drop experiment (1909–1913): microscopic oil droplets falling at terminal velocity told him their radii via $v_t = 2r^2\rho g/9\eta$ ; switching on an electric field and re-measuring the drift then handed him the charge on each drop — always an integer times $1.6\times10^{-19}$ C. Stokes' law weighed the electron. Today the same physics runs the machinery of measurement and medicine: centrifuges separate blood cells and DNA by driving them to terminal velocity in tubes (sedimentation rate ∝ $r^2$ is why centrifuges spin — to boost $g$ ); your ESR blood test is a terminal-velocity measurement; atmospheric scientists model how long volcanic ash ( $r^2$ again) stays airborne after an eruption grounds air fleets. Engine oils are graded by viscosity (5W-30: thin enough to flow at winter cranking, thick enough to cushion at running temperature), and the food industry's "mouthfeel" laboratories are, without apology, viscometry labs.

The physical picture

Pour honey and water side by side. Both are fluids — zero shear modulus, no permanent resistance to shape change. But the honey visibly protests while flowing. That protest is viscosity: a resistance not to deformation itself but to the rate of deformation. A solid fights how far you shear it; a fluid fights how fast. The microscopic picture explains the strange temperature behaviour. In a liquid, molecules are held by cohesion; layers sliding past each other must rip through those attractions, and heating loosens them — hot honey pours easily, $\eta$ drops. In a gas, there is no cohesion to speak of; drag comes from molecules wandering between layers, carrying slow momentum into fast layers and vice versa, like passengers stepping between moving walkways. Heating makes the wandering faster — so a hot gas is more viscous. Same word, opposite trends, and examiners know you'll guess wrong under pressure.

Figure

Viscosity is layer-on-layer friction — and it is what stops a falling drop from accelerating forever

Quantify the layer picture with Newton's own construction. A flat plate rides on a fluid film of thickness $\Delta z$ over a stationary floor. The fluid sticks to both surfaces (the "no-slip" fact — the reason dust survives on fan blades), so the layers between must slide over one another, fastest at the top, zero at the bottom. Keeping the plate moving at constant speed takes a steady force

F = \eta\, A\, \frac{\Delta v}{\Delta z} \qquad (1)

where: $A$ — the plate's area in contact; $\Delta v/\Delta z$ — the velocity gradient, how sharply speed changes across the film (unit s⁻¹); $\eta$ — the coefficient of viscosity, the fluid's price per unit of gradient. Units of $\eta$ : Pa·s (sometimes written poiseuille); the older CGS unit poise converts as $1\ \mathrm{Pa\cdot s} = 10$ poise. Anchor values worth owning: water $\approx 1.0\times10^{-3}$ Pa·s, air $\approx 1.8\times10^{-5}$ , glycerine $\approx 0.8$ , honey $\sim 10$ — a range of a million between things you drink and things you drizzle.

The math: Stokes' law and terminal velocity

For a small sphere moving slowly through a fluid — the laminar regime where layers stay orderly — Stokes solved the flow exactly in 1851. The drag is

F_{\text{drag}} = 6\pi\,\eta\, r\, v \qquad (2)

where: $r$ — the sphere's radius; $v$ — its speed relative to the fluid; the direction always opposes the relative motion. Read the structure: drag ∝ $v$ (double the speed, double the resistance — this linearity is what makes terminal velocity stable), ∝ $r$ (not $r^2$ : for creeping flow it is the circumference-scale that matters), and ∝ $\eta$ (honey punishes speed a million times more than air). The formula's warranty: small radius, low speed, laminar flow, sphere far from walls. Fast large bodies (cars, skydivers at full spread) obey a different, $v^2$ drag — flagged in the FAQ and owned by the UG deep dive.

Derivation AStokes' law from dimensional analysis — the exam-day reconstruction

We do not attempt Stokes' full fluid-dynamics solution (that is genuinely UG mathematics); instead we reconstruct the form of the law by dimensional analysis, exactly the skill your measurement chapter drilled. The physical claim: for slow laminar motion the drag can depend only on the fluid's viscosity $\eta$ , the sphere's radius $r$ , and the speed $v$ — inertia (density) plays no role when motion is gentle. Watch how the method delivers the exponents but not the $6\pi$ : dimensional analysis's classic power and its classic confession.

Step 1. Assume $F = k\,\eta^a\, r^b\, v^c$ with $k$ dimensionless. Dimensions: $[F] = \mathrm{MLT^{-2}}$ ; $[\eta] = \mathrm{ML^{-1}T^{-1}}$ (from equation (1): $\eta = F\Delta z/A\Delta v$ ); $[r] = \mathrm{L}$ ; $[v] = \mathrm{LT^{-1}}$ .

Step 2. Match each dimension:

\mathrm{MLT^{-2}} = \mathrm{M}^a\,\mathrm{L}^{-a+b+c}\,\mathrm{T}^{-a-c}

Step 3. Solve: mass gives $a = 1$ ; time gives $-a - c = -2 \Rightarrow c = 1$ ; length gives $-1 + b + 1 = 1 \Rightarrow b = 1$ .

F = k\,\eta\, r\, v

Why each exponent is 1: with three equations and three unknowns the answer is forced — there is exactly one way to build a force from these three quantities. Stokes' hydrodynamics supplies the number the method cannot: $k = 6\pi$ . *Verified: units of $6\pi\eta rv$ : $\mathrm{(Pa\cdot s)(m)(m/s)} = \mathrm{N}$ ✓. This derivation is itself a JEE question type ("obtain the form of Stokes' law dimensionally") — cashing the dimensional-analysis technique module.*

Now drop a sphere — density $\rho$ , radius $r$ — into a fluid of density $\sigma$ and viscosity $\eta$ , and follow the forces. At the instant of release, $v = 0$ : no drag, and the sphere accelerates under weight minus buoyancy. As $v$ grows, drag $6\pi\eta r v$ grows in exact proportion, eating into the net force. The acceleration therefore shrinks as the sphere speeds up — a self-limiting chase that ends when the three forces close:

\underbrace{\tfrac{4}{3}\pi r^3 \rho\, g}_{\text{weight}} \;=\; \underbrace{\tfrac{4}{3}\pi r^3 \sigma\, g}_{\text{buoyancy}} \;+\; \underbrace{6\pi \eta r v_t}_{\text{drag}} \qquad (3)

Go deeper · Derivation B · 1 minTerminal velocity of a small sphere

Derivation BTerminal velocity of a small sphere

We derive $v_t$ by solving the force balance (3) — the state of zero net force toward which every viscous fall converges. Strategy: write all three forces for a sphere, set acceleration to zero, solve for $v$ . Watch the buoyancy term: forgetting $\sigma$ is the single most common error in this chapter's numericals, and it matters enormously in liquids (steel in glycerine loses ~16% of its driving weight to buoyancy) while being negligible for water drops in air ( $\sigma/\rho \sim 10^{-3}$ ).

Step 1. Volume of the sphere $V = \tfrac43\pi r^3$ . Downward: weight $V\rho g$ . Upward: buoyancy $V\sigma g$ (Archimedes, from two articles back) and drag $6\pi\eta r v$ (opposing the downward motion).

Step 2. At terminal velocity the net force vanishes:

\tfrac43\pi r^3(\rho - \sigma)g = 6\pi\eta r\, v_t

Step 3. Solve for $v_t$ — one $r$ cancels, leaving $r^2$ :

v_t = \frac{2\,r^2(\rho - \sigma)\,g}{9\,\eta}

Why $r^2$ : driving force grows with volume ( $r^3$ ) while drag grows only with size ( $r$ ) — big spheres out-earn their drag bill. Halve a drop's radius and it falls four times slower; that one scaling separates rain from mist from cloud. *Why $(\rho - \sigma)$ can flip sign: a bubble ( $\rho < \sigma$ ) gets a negative $v_t$ — it "falls upward" at terminal speed. Same formula prices rising bubbles in a fizzy drink. Verified: units — $\mathrm{m^2\cdot kg/m^3 \cdot m/s^2 / (Pa\cdot s)} = \mathrm{m/s}$ ✓. Limits — $\eta \to \infty$ : $v_t \to 0$ (nothing falls through pitch, almost); $\rho = \sigma$ : $v_t = 0$ , neutral buoyancy ✓. Anchors recomputed in the worked examples: 6.9 cm/s (steel in glycerine), 1.2 cm/s (fog droplet). Standard: NCERT Ch. 9 §9.5; deep-dive: the UG drag article carries the full time-dependent solution.*

What happens before terminal velocity is a graph every exam wants you to sketch: $v(t)$ rises steeply from zero (nearly free fall — no drag yet), bends as drag bites, and flattens asymptotically onto $v_t$ — approaching it but never overshooting, ever. The distance-time curve correspondingly straightens into a constant-slope line.

Figure

The approach to terminal velocity — steep start, self-limiting finish

Limiting cases

case	result	meaning
$t = 0$ (just released)	$a = g(1 - \sigma/\rho)$ , no drag	initial slope of the $v$ – $t$ graph; a stock MCQ
$t \to$ large	$v \to v_t$ , $a \to 0$	constant velocity, zero net force — motion continues
$r \to r/2$	$v_t \to v_t/4$	the $r^2$ law: mist floats, rain falls
$\rho = \sigma$	$v_t = 0$	neutral buoyancy — nothing drives the fall
$\rho < \sigma$ (bubble)	$v_t < 0$ : rises at terminal speed	same law, sign flipped; fizzy-drink physics
$\eta$ small (air vs glycerine)	large $v_t$ , long time/distance to reach it	why lab demos of Stokes use glycerine, not air
released downward faster than $v_t$	drag exceeds driving force: sphere decelerates onto $v_t$	approach works from both sides — favourite JEE twist

Worked examples

1 · Board level — steel ball in glycerine (CBSE/ISC pattern, 3 marks)

A steel ball of radius $2.0\ \mathrm{mm}$ (density $7800\ \mathrm{kg/m^3}$ ) falls through glycerine (density $1260\ \mathrm{kg/m^3}$ , viscosity $0.83\ \mathrm{Pa\cdot s}$ ). Find its terminal velocity. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. List and convert: $r = 2.0\times10^{-3}$ m, $\rho - \sigma = 7800 - 1260 = 6540\ \mathrm{kg/m^3}$ .

Step 2. Terminal-velocity formula: $v_t = \dfrac{2r^2(\rho-\sigma)g}{9\eta} = \dfrac{2\,(2.0\times10^{-3})^2\,(6540)(9.8)}{9\times0.83}$ .

Step 3. Numerator: $2\times4.0\times10^{-6}\times6540\times9.8 = 0.513$ . Denominator: $7.47$ .

Step 4. $v_t = 0.513/7.47 = \boxed{6.9\times10^{-2}\ \mathrm{m/s}} \approx 6.9\ \mathrm{cm/s}$ — slow enough to time with a stopwatch, which is exactly how school labs measure $\eta$ (run this formula backwards). Note the buoyancy correction mattered: using $\rho$ instead of $(\rho - \sigma)$ would inflate the answer by 19% — a lost mark, cleanly detectable by the examiner.

2 · Competitive level — why clouds float (NEET/JEE-Main pattern, ~2 minutes)

A fog droplet of radius $10\ \mathrm{\mu m}$ (water, $\rho = 1000\ \mathrm{kg/m^3}$ ) falls through still air ( $\eta = 1.8\times10^{-5}\ \mathrm{Pa\cdot s}$ ; neglect air's density against water's). (a) Find its terminal velocity. (b) A raindrop of radius $1.0\ \mathrm{mm}$ — how does its Stokes-law terminal velocity compare?

Step 1. (a) $v_t = \dfrac{2r^2\rho g}{9\eta} = \dfrac{2\,(1.0\times10^{-5})^2(1000)(9.8)}{9\times1.8\times10^{-5}} = \dfrac{1.96\times10^{-6}}{1.62\times10^{-4}} = \boxed{1.2\times10^{-2}\ \mathrm{m/s}}$ .

Step 2. Read it: ~1.2 cm/s. Any updraught stronger than a whisper keeps the droplet aloft indefinitely — that is why clouds float: not weightlessness, just tiny $r^2$ .

Step 3. (b) Radius up by $10^2$ , so $v_t$ up by $(10^2)^2 = 10^4$ : about $120\ \mathrm{m/s}$ by Stokes — which is absurd (real large raindrops land at ~9 m/s). The absurdity is the lesson: at such speeds the flow around the drop is no longer laminar, Stokes' warranty is void, and the true drag (a $v^2$ law) is far stronger. Quote Stokes for microscopic drops; distrust it for millimetre ones. Timing note: part (a) is 60 seconds of substitution. Part (b) is a validity checkpoint examiners increasingly love: the wrong-but-instructive $10^4$ scaling plus one sentence of "beyond laminar regime" earns full credit.

3 · Competitive level — ratio games (JEE-Main pattern, ~90 seconds)

Two drops of the same liquid, radii in the ratio $1:2$ , fall at terminal velocity through air. Find the ratio of (a) their terminal velocities and (b) the viscous drag forces on them.

Step 1. (a) $v_t \propto r^2$ (everything else shared): ratio $= 1^2 : 2^2 = \boxed{1:4}$ .

Step 2. (b) At terminal velocity, drag equals the net driving force $= \tfrac43\pi r^3(\rho-\sigma)g \propto r^3$ : ratio $= \boxed{1:8}$ .

Step 3. Cross-check via Stokes directly: $F \propto r\,v_t \propto r\cdot r^2 = r^3$ ✓ — two routes, one answer, and the second route is faster in an MCQ. Timing note: terminal-state questions rarely need the full formula — at $v_t$ the drag ALWAYS equals weight minus buoyancy, so drag ratios are weight ratios. Spotting that converts a formula problem into a proportionality one; this reflex is drilled again in the merging-drops archetype.

Exam traps

Forgetting buoyancy. The driving term is $(\rho - \sigma)$ , not $\rho$ . In liquids the correction is large (19% for steel in glycerine, fatal for accuracy marks); in air it is negligible — but say you neglect it.

$r^2$ , and which $r$ moved. Terminal velocity scales as radius squared; drag at terminal scales as $r^3$ (it matches the driving force); Stokes drag at fixed speed scales as $r$ . Three different exponents for three different questions — classify before computing.

Poise vs Pa·s. $1\ \mathrm{Pa\cdot s} = 10$ poise. Data given in poise with everything else in SI is a deliberate landmine: convert first.

Temperature trends. Liquids: $\eta$ decreases on heating (cohesion loosens). Gases: $\eta$ increases (faster molecular exchange). The inversion is a top-frequency MCQ; anchor it with hot honey and hot air.

Terminal ≠ maximum in all senses. Released from rest, $v_t$ is the ceiling (approached, never crossed). Entering the fluid fast, the body decelerates down to $v_t$ . Graph questions test whether your curve flattens onto the line from the correct side without crossing.

Stokes' warranty. Small sphere, slow laminar flow, unbounded fluid. Millimetre raindrops, skydivers, and cricket balls are outside it ( $v^2$ drag territory — the UG deep dive). If a computed $v_t$ looks like a highway speed, the model failed, not the arithmetic.

Velocity gradient direction. In equation (1), $\Delta v/\Delta z$ is change of speed across the flow (perpendicular to it), not along it. Sketch the layers before assigning symbols.

Which $g$ . CBSE 9.8; JEE/NEET usually 10 as stated. Terminal-velocity ratios are $g$ -free — if $g$ survives in your ratio, re-derive.

Origins · 1 minGeorge Gabriel Stokes and how this was discovered

George Gabriel Stokes

1819–1903 · Cambridge

George Gabriel Stokes held Newton's old chair — Lucasian Professor at Cambridge — for over half a century, and ran physics' great clearing-house: as Royal Society secretary he refereed, corrected, and quietly improved a generation of British science. His own ledger is everywhere in this chapter: the Navier–Stokes equations of viscous flow (1845), the law of creeping drag around a sphere (1851), plus fluorescence (he named it), the first spectroscopic identification of haemoglobin, and foundational work on wave optics. Famously scrupulous, he delayed publishing results for years while verifying them — a habit exams do not permit you. The CGS unit of kinematic viscosity, the stokes, honours him alongside the crater on Mars.

History

1687

— Newton's Principia defines fluid friction proportional to the velocity gradient — the "defect of lubricity".

1840

— Poiseuille publishes precision measurements of flow in fine tubes, quantifying viscosity for blood-like flows.

1845

— Stokes formulates the equations of viscous flow (with Navier's earlier work: the Navier–Stokes equations).

1851

— Stokes derives the

6\pi\eta rv

drag law for a slow sphere — terminal velocity becomes computable.

1883

— Osborne Reynolds' dye experiments draw the laminar/turbulent frontier where Stokes' warranty ends.

1909

— Millikan's oil-drop experiment uses Stokes terminal velocities to weigh single electrons.

Going further

Texts. NCERT Physics Class 11, Ch. 9 §9.5 · HC Verma Vol. 1, Ch. 13 · Resnick–Halliday–Walker Ch. 14. The full time-dependent fall, the $v^2$ regime, and the Reynolds-number frontier live in the UG deep dive on drag and terminal velocity.

Provenance. All results standard. A verifier should recompute: the dimensional exponents $a=b=c=1$ ; example 1's $v_t = 0.513/7.47 = 6.9$ cm/s; example 2's $1.2\times10^{-2}$ m/s and the $10^4$ scaling; example 3's $1:4$ and $1:8$ .

Prerequisites. Fluids at Rest: Pressure & Pascal's Law · Newton's Second Law. Leads to. Drops Merging & Terminal Velocity (archetype) · Surface Tension & Capillarity.

Why does drag depend on speed when solid friction doesn't?

Solid kinetic friction comes from surface asperities bonding and breaking — a process whose rate hardly cares how fast you slide, so $f = \mu N$ is speed-blind. Fluid drag comes from making the fluid move: layers must be sheared (Stokes regime, cost ∝ $v$ ) or fluid must be shoved aside and given kinetic energy (fast regime, cost ∝ $v^2$ ). The faster you move, the harder the fluid works per second — so the force itself grows with speed. That growth is what makes terminal velocity exist at all: a speed-blind friction could never automatically balance a fixed weight.

Does a heavier object fall faster after all? Galileo said no!

Galileo said no in vacuum, and stands correct. In a fluid, the terminal-velocity formula answers precisely: $v_t \propto r^2(\rho - \sigma)$ . At the same size, the denser body does fall faster (bigger driving force, same drag law); at the same material, the bigger body falls faster ( $r^2$ ). Aristotle's intuition was a fluid-world observation mislabeled as a law of gravity. Exam framing: "in vacuum, simultaneous; in air, the steel ball lands first — justify with $v_t$ ."

Do skydivers obey Stokes' law?

No — and knowing why earns marks. A skydiver is metres wide moving at tens of m/s: the air cannot flow around laminar-ly; it churns into a turbulent wake, and drag scales as $v^2$ (with drag coefficient and frontal area), not $6\pi\eta rv$ . Terminal velocity still exists — the balance logic is identical — at about 55–60 m/s belly-down, and parachutes work by exploding the frontal area, cutting terminal velocity to ~5 m/s. Stokes governs the microscopic world: fog, pollen, Millikan's droplets, sediment. The two regimes and the crossover live in the UG drag deep dive.

The no-slip condition says fluid sticks to surfaces. Then why does a fan blade collect dust — shouldn't the wind blow it clean?

Because right AT the blade's surface the air speed is zero — that's exactly the no-slip condition. A dust grain nestling in that near-surface layer feels almost no wind at all, however hard the blade slices the room. The velocity climbs steeply only over a thin "boundary layer" above the surface. The same fact means pipe walls carry stationary fluid (flow is fastest at the centre — Poiseuille's parabola) and explains why washing greasy plates needs scrubbing, not just running water.

How do labs actually measure viscosity with this?

The falling-ball viscometer is worked example 1 run backwards: drop a calibrated ball through the liquid, time it between two marks (after it has visibly settled to constant speed), and solve $v_t = 2r^2(\rho-\sigma)g/9\eta$ for $\eta$ . Practical fine print students meet in lab vivas: wait for terminal velocity before starting the clock; use a wide tube (walls slow the fall — Stokes assumed unbounded fluid); control temperature ferociously, since $\eta$ of glycerine changes ~7% per °C. Industrial labs use rotating-cylinder and capillary (Poiseuille) viscometers on the same principles.

Why doesn't the falling sphere overshoot terminal velocity and oscillate around it, like a spring?

Because drag is a damping force, not a restoring one. A spring overshoots because at equilibrium it still has velocity and the force then reverses. Here the "equilibrium" IS a velocity: as $v \to v_t$ , the net force $\to 0$ smoothly, so acceleration dies exactly as the target is reached — an exponential approach with no overshoot (time constant $\tau = m/6\pi\eta r$ , shown fully in the UG deep dive). The $v$ – $t$ curve is monotonic from either side. If a plotted curve in an option crosses the $v_t$ line, eliminate it.

Rain, mist, drizzle — is the difference really just radius?

Largely yes, through $v_t(r)$ . Cloud droplets (~10 μm) have cm/s terminal velocities — any updraught suspends them, so clouds float. Drizzle (~0.1–0.25 mm) falls at ~1 m/s. Raindrops (1–3 mm) land at 4–9 m/s (in the $v^2$ regime by then). Growth from cloud droplet to raindrop — collisions and coalescence — is precisely the fight to reach a radius whose terminal velocity beats the updraught. The merging-drops archetype next drills the beautiful sub-question: when $n$ droplets merge, the new $v_t$ jumps by $n^{2/3}$ .