In short

Viscosity is a fluid's resistance to shear — the internal friction between neighbouring fluid layers moving at different speeds. Quantitatively, for a fluid sheared between two parallel layers, the shear stress \tau needed to maintain a velocity gradient dv/dy is

\tau = \eta\,\frac{dv}{dy},

where \eta is the coefficient of (dynamic) viscosity, measured in Pa·s. This is Newton's law of viscosity, and fluids obeying it are called Newtonian. For a small sphere of radius r moving at speed v through a viscous fluid of viscosity \eta (in the slow, laminar regime), the drag is Stokes' law:

\boxed{\; F_\text{drag} = 6\pi\eta r v \;}

A body falling freely through a viscous fluid reaches terminal velocity when drag plus buoyancy balances gravity. For a sphere of radius r, density \rho falling through fluid of density \sigma and viscosity \eta,

v_\text{term} = \frac{2}{9}\frac{(\rho - \sigma)g r^2}{\eta}.

This is why a steel ball sinks fast in water but slowly in honey, why mist droplets hang in air, and how Millikan measured the charge of the electron.

Pour water from one glass to another and it splashes and gurgles in half a second. Pour honey the same way and it takes a slow, deliberate string of amber that seems almost reluctant to leave the jar. Both are liquids. Both obey Archimedes and Pascal and Bernoulli. What makes one pour like a waterfall and the other like treacle?

The answer is viscosity — the internal friction of a flowing fluid. In water, the molecules slide past each other easily; in honey, they cling. Any time one layer of a fluid tries to move relative to its neighbour, the two layers exert a frictional force on each other, and that force is what viscosity measures. Low viscosity means slippery, fast-pouring liquids like water, petrol, and air. High viscosity means sluggish, resisting liquids like honey, mustard oil, and glycerine.

In the last two chapters you worked with ideal fluids — non-viscous, frictionless, perfectly cooperative. That idealisation is clean but incomplete. Every real fluid has viscosity; without it, rivers would not carry silt, aircraft would not generate lift, and the raindrops falling outside your window would reach the ground at the speed of a freely-falling body (about 700 km/h from 2 km up, which would be catastrophic). Viscosity is the small but crucial correction that converts the perfect fluids of the textbook into the honest fluids of the world.

This article builds viscosity from the ground up — the velocity-gradient picture, Newton's law of viscosity, the coefficient \eta with real values you can calibrate against, Stokes' law for the drag on a small sphere, and the practical consequence you will meet in every JEE problem set: terminal velocity.

The velocity gradient — why layers slide unequally

Take two parallel glass plates, one fixed and one sliding. Between them is a thin layer of fluid — water, honey, whatever. Push the top plate at a steady speed v while the bottom plate stays still. Now zoom in to the fluid between them.

Fluid sheared between two parallel plates Two horizontal plates with fluid between them. The top plate moves right at speed v, the bottom plate is fixed. Five arrows inside the fluid show a linearly varying velocity profile: zero at the bottom plate, one quarter v at quarter height, half v at half height, three quarters v, and full v at the top plate. Labels mark the plate separation as y = d. top plate moves at $v$ bottom plate fixed $y = d$ velocity profile $v(y) = (y/d)\, v$
Fluid in Couette flow between two parallel plates. The top plate moves at speed $v$, the bottom plate is fixed. The fluid's velocity varies linearly from zero at the bottom to $v$ at the top — a uniform velocity gradient $dv/dy = v/d$.

The fluid layer right next to the bottom plate does not slip — it has the same speed as the plate, namely zero. The layer next to the top plate likewise does not slip — it moves at v. In between, each thin layer of fluid moves at some intermediate speed. Experiment (and theory) says the velocity varies linearly with height: v(y) = (y/d)v, where d is the gap between plates and y is the height above the bottom plate.

The "no slip" condition — that fluid in contact with a solid surface has the same velocity as the surface — is a cornerstone of viscous flow. It is not a derivation but a well-tested empirical fact, traceable to the attractive forces between fluid molecules and solid surfaces.

The quantity

\frac{dv}{dy} = \frac{v}{d}

is the velocity gradient — how fast the speed changes as you move across the fluid. In this linear profile, the gradient is constant. In more complex flows (like water flowing through a pipe) the gradient varies from point to point; the definition dv/dy handles both cases.

Newton's law of viscosity

What force does the top plate need to drag it at a steady speed? Newton proposed — and experiment confirms — that this force is proportional to the plate's area A and to the velocity gradient dv/dy:

F = \eta A \frac{dv}{dy},

or equivalently, dividing both sides by area:

\tau \equiv \frac{F}{A} = \eta\,\frac{dv}{dy}.

The constant of proportionality \eta is the coefficient of viscosity (or simply the viscosity). This is Newton's law of viscosity.

Why: experimentally, doubling the plate speed doubles the required force; doubling the gap halves the force (because the gradient halves). The ratio F/A \div dv/dy is a fluid-specific constant — the viscosity. The linearity of this relationship defines what a Newtonian fluid is. Honey, water, oil, air, blood (to a good approximation), glycerine — all Newtonian. Toothpaste, ketchup, cornflour-in-water — non-Newtonian, and outside the scope of this article.

The left-hand side \tau = F/A has units of force per area, which is pressure (Pa). It is called the shear stress — it is not an isotropic pressure like in a static fluid, but specifically the tangential force per unit area on the sliding surface.

The right-hand side: \eta has units of Pa·s (pascal-second); the velocity gradient has units (m/s)/m = s^{-1}. Check: Pa·s × s⁻¹ = Pa. Units balance.

The SI unit and the poise

The SI unit of viscosity is the pascal-second (Pa·s), also written as N·s/m². You will also see the poise in older textbooks and Indian classroom problems:

1 \text{ poise} = 0.1 \text{ Pa·s}.

The poise is a CGS unit, but it persists because the viscosity of water at 20°C is almost exactly 1 centipoise (0.001 Pa·s). For easy estimation, the poise is a useful unit — for quantitative physics, stick with Pa·s.

Reference values to calibrate your intuition

Fluid \eta at 20 °C (Pa·s) \eta at 20 °C (poise)
Air 1.8 \times 10^{-5} 1.8 \times 10^{-4}
Petrol 6 \times 10^{-4} 6 \times 10^{-3}
Water 1.0 \times 10^{-3} 1.0 \times 10^{-2}
Blood (plasma) 3 \times 10^{-3} 3 \times 10^{-2}
Mustard oil 5 \times 10^{-2} 5 \times 10^{-1}
Olive oil 8 \times 10^{-2} 8 \times 10^{-1}
Glycerine (100%) 1.4 14
Honey 2 – 10 20 – 100
Molten glass 10^{7} 10^{8}

A few things to notice. Honey is about 10,000 times more viscous than water. Air is about 50 times less viscous than water. Glycerine sits in between water and honey, which is why your lab experiments on viscosity usually use glycerine — it is slow enough to measure by stopwatch but not so slow that the experiment takes an hour.

Viscosity is strongly temperature-dependent. Water at 0 °C has \eta = 1.8 \times 10^{-3} Pa·s — almost twice its 20 °C value. Honey refrigerated for a winter morning in Shimla is almost unpourable. Warm it to summer midday temperature in Chennai and it pours like syrup. The physical reason: higher temperature means more thermal motion, which helps molecules slide past each other. (Gases behave differently — air's viscosity increases with temperature, because faster-moving gas molecules transfer more momentum across layers.)

Why viscosity exists — a one-sentence picture

At the molecular level, viscosity is momentum transfer between layers. A faster layer's molecules drift into a slower layer (by random thermal motion and by direct intermolecular forces), delivering momentum that speeds the slow layer up. Reciprocally, slower molecules drifting into the fast layer slow it down. The net effect is a tangential force between the two layers — exactly what Newton's law of viscosity quantifies. For gases this momentum-transfer picture is transparent; for liquids it is muddled by the fact that liquid molecules are almost always in contact with their neighbours, so the transfer is more like an ongoing tug-of-war. But the idea is the same: viscosity is friction at the molecular scale.

Stokes' law — the drag on a small sphere

Now push the problem forward. Instead of fluid between two flat plates, imagine a small sphere (a marble, a raindrop, a steel ball bearing) moving at speed v through a still fluid. As the sphere moves, it drags some fluid along with it. That fluid exerts a backward drag force on the sphere. What is the drag?

The full answer requires solving the Navier–Stokes equations around the sphere — complicated. But for a small sphere moving slowly (more precisely: low Reynolds number, which you will meet in the next chapter), the answer is compact and exact:

\boxed{\; F_\text{drag} = 6\pi\eta r v \;} \tag{Stokes' law}

where r is the radius of the sphere, v its speed through the fluid, and \eta the fluid's viscosity. This is Stokes' law.

The plausibility argument via dimensions

You cannot derive Stokes' law from dimensional analysis alone (the factor of 6\pi is a hard-won result of the full Navier–Stokes solution), but you can guess the form.

What could the drag force depend on? The sphere's radius r (bigger sphere, more fluid to shove aside), the sphere's speed v (faster sphere, more drag), the fluid's viscosity \eta (stickier fluid, more drag), and possibly the fluid's density \rho (heavier fluid, maybe more drag from the inertia of the fluid being shoved aside).

For slow flows, inertial effects from \rho are negligible compared to viscous effects from \eta — this is what "low Reynolds number" means. So drop \rho. The remaining quantities are r, v, \eta. Check dimensions:

The only combination that gives units of force (kg·m/s² = N) is

r \cdot v \cdot \eta = \text{m} \cdot \frac{\text{m}}{\text{s}} \cdot \frac{\text{kg}}{\text{m} \cdot \text{s}} = \frac{\text{kg} \cdot \text{m}}{\text{s}^2} = \text{N}.

So the drag must be proportional to \eta r v, times some dimensionless number. Stokes' full calculation gives that number as 6\pi \approx 18.85.

Why: dimensional analysis cannot fix the pure number 6\pi, but it can force the dependence on r, v, and \eta to be linear in each. The full answer requires solving the creeping-flow (Navier–Stokes) equations around a sphere with appropriate boundary conditions, which is done in JEE Advanced fluid mechanics texts and in any graduate fluid-mechanics course. For our purposes, the linear dependence is what matters: double the radius, double the drag; double the speed, double the drag.

When is Stokes' law valid?

Three conditions must hold.

  1. The sphere is spherical and small compared to the size of the container. Non-spherical shapes (coins, pebbles, raindrops at large size) need correction factors.

  2. The flow around the sphere is laminar (not turbulent) — which requires the Reynolds number \text{Re} = \rho v r / \eta \lesssim 1. This is a strong restriction. A marble falling through water has \text{Re} \sim 10^3 — well outside Stokes. Stokes works for mist droplets in air, cells sedimenting in water, dust motes settling, and small steel balls in glycerine.

  3. The sphere's density is not vastly different from the fluid's (otherwise the sphere accelerates too fast to be in the slow-flow regime).

When Stokes fails, you typically use empirical drag correlations that add a "form drag" term proportional to \rho v^2. For JEE and for all typical school problems, Stokes' law is the working formula and the Reynolds criterion is the flag for when to trust it.

Terminal velocity — drag, gravity, and buoyancy in balance

Drop a steel ball into a tall glass cylinder of glycerine and watch it fall. At first it accelerates — gravity exceeds drag. As it speeds up, the drag grows (because F_\text{drag} \propto v). Eventually drag plus buoyancy balances gravity, the net force vanishes, and the ball falls at a constant speed forever after (or until it hits the bottom). That constant speed is the terminal velocity, v_\text{term}.

Let the sphere have radius r, density \rho, and fall through a fluid of density \sigma and viscosity \eta. Three forces act on it:

Free body diagram of a sphere at terminal velocity A spherical ball at centre. Three force arrows emerge from the centre: a long downward arrow labelled weight W equals four thirds pi r cubed rho g. Two upward arrows, one labelled buoyancy F B equals four thirds pi r cubed sigma g, and one labelled drag F D equals six pi eta r v term. The two upward arrows together equal the single downward arrow. sphere $W = \tfrac{4}{3}\pi r^3 \rho g$ $F_\text{D} = 6\pi\eta r v$ $F_\text{B} = \tfrac{4}{3}\pi r^3 \sigma g$ $v$ direction of fall
Free body diagram of a sphere moving downward at speed $v$ through a viscous fluid. At terminal velocity, the downward weight $W$ equals the sum of the upward buoyancy $F_\text{B}$ and upward viscous drag $F_\text{D}$.

At terminal velocity, the net force is zero:

W = F_\text{B} + F_\text{D}.
\tfrac{4}{3}\pi r^3 \rho g = \tfrac{4}{3}\pi r^3 \sigma g + 6\pi \eta r v_\text{term}.

Bring the buoyancy to the left:

\tfrac{4}{3}\pi r^3 (\rho - \sigma) g = 6\pi \eta r v_\text{term}.

Cancel \pi and divide both sides by r:

\tfrac{4}{3} r^2 (\rho - \sigma) g = 6 \eta v_\text{term}.

Solve for v_\text{term}:

\boxed{\; v_\text{term} = \frac{2}{9}\cdot \frac{(\rho - \sigma) g r^2}{\eta} \;} \tag{TV}

Why: the 4/3 divided by 6 gives 4/18 = 2/9. The r^3 in the volume and the r^1 in Stokes' law leave r^2. The key physics lives in the three ratios: terminal velocity grows with the density excess (\rho - \sigma), with the square of the radius, and inversely with viscosity.

Read the formula — three consequences to remember

Terminal-velocity fall — animated

The simulation below shows a 3 mm steel ball (density \rho = 7800 kg/m³) released from rest at the top of a column of glycerine (viscosity \eta = 1.4 Pa·s, density \sigma = 1260 kg/m³). It accelerates for a brief moment, then settles at its terminal velocity. Watch the speed stabilise after a fraction of a second.

The terminal velocity here is

v_\text{term} = \frac{2}{9}\cdot\frac{(7800 - 1260) \times 9.8 \times (0.003)^2}{1.4} = \frac{2}{9}\cdot \frac{576.8}{1.4} \approx 0.092 \text{ m/s}.

The characteristic time to approach terminal velocity is \tau = m/(6\pi\eta r) \approx 0.005 s — much shorter than the animation duration, so the ball is essentially at terminal velocity for the whole fall.

Animated: a steel ball reaching terminal velocity in glycerine A red ball starts at the top of a glycerine column. It accelerates briefly then settles at the terminal velocity of approximately 0.092 m per second. Over 5 seconds it falls approximately 0.46 metres. Because the approach to terminal velocity is much faster than the total fall time, the motion is essentially uniform. glycerine (η = 1.4 Pa·s)
A 3 mm steel ball released at the top of a glycerine column. After a brief transient (under 0.01 s), the ball falls at its terminal velocity of 0.092 m/s. Ghost markers show its position at 1 s, 2 s, 3 s, 4 s — equally spaced in distance, the signature of uniform motion. Click replay to watch again.

Explore the dependence of terminal velocity

Drag the radius slider below to see how terminal velocity depends on ball size (for a steel ball in water, \eta = 1 \times 10^{-3} Pa·s, \rho = 7800 kg/m³, \sigma = 1000 kg/m³). Notice the quadratic rise — a 2 mm ball has four times the terminal velocity of a 1 mm ball.

Interactive: terminal velocity versus sphere radius A curve showing terminal velocity as a function of sphere radius for a steel ball in water. Values: at radius 1 mm, terminal velocity is 14.8 m per second. At radius 2 mm, 59 m per second. At radius 3 mm, 133 m per second. These values assume Stokes law, which does not actually hold at such radii because the Reynolds number is large. radius $r$ (mm) terminal velocity (m/s, Stokes) 0 40 80 120 160 1 2 3 drag the red point along the axis
Terminal velocity of a steel ball in water rises as the square of the radius. The Reynolds number readout is a warning: past $r \approx 0.05$ mm, Re exceeds 1 and Stokes' law begins to overestimate the real terminal velocity. Past $r \approx 1$ mm, turbulent-wake drag dominates and the real terminal velocity is much smaller than the Stokes prediction. The formula is valid in the slow-flow regime.

Worked examples

Example 1: A steel ball bearing in a glycerine-filled tube

A steel ball of radius r = 2.0 mm and density \rho = 7800 \text{ kg/m}^3 is dropped into a tall glass tube filled with glycerine (density \sigma = 1260 \text{ kg/m}^3, viscosity \eta = 1.4 \text{ Pa·s}). Find the terminal velocity. Check whether Stokes' law is valid by computing the Reynolds number \text{Re} = \rho_\text{fluid} v r / \eta.

Steel ball falling through a glycerine column A vertical glass tube filled with glycerine. A steel ball is shown at the top. Arrows on the ball show gravity pulling down, drag and buoyancy pushing up. A scale on the side marks the distance the ball has fallen. glycerine column steel ball, $r = 2$ mm $W$ $F_D$ $F_B$
A 2 mm steel ball falling through glycerine. Forces on the ball at terminal velocity: gravity $W$ down, drag $F_D$ and buoyancy $F_B$ both up.

Step 1. Apply the terminal-velocity formula.

v_\text{term} = \frac{2}{9}\cdot \frac{(\rho - \sigma) g r^2}{\eta}.
v_\text{term} = \frac{2}{9}\cdot \frac{(7800 - 1260)(9.8)(0.002)^2}{1.4}.

Why: direct substitution. Make sure the radius is in metres before squaring. The mass density difference (\rho - \sigma) is the "effective" density that gravity and buoyancy act on.

Step 2. Compute step by step.

\rho - \sigma = 7800 - 1260 = 6540 \text{ kg/m}^3.

(\rho - \sigma) g = 6540 \times 9.8 = 64{,}092 \text{ N/m}^3.

(\rho - \sigma) g r^2 = 64{,}092 \times 4.0 \times 10^{-6} = 0.2564 \text{ N/m}.

(\rho - \sigma) g r^2/\eta = 0.2564/1.4 = 0.1832 \text{ m/s}.

v_\text{term} = (2/9) \times 0.1832 \approx 0.0407 \text{ m/s} \approx 4.1 \text{ cm/s}.

Why: it is easier to track units if you compute intermediates one at a time. 4.0 \times 10^{-6} is (2 \times 10^{-3})^2. Keep two or three significant figures throughout.

Step 3. Check the Reynolds number.

\text{Re} = \frac{\sigma v_\text{term} r}{\eta} = \frac{1260 \times 0.0407 \times 0.002}{1.4} = \frac{0.1026}{1.4} \approx 0.073.

Why: Re is dimensionless — the numerator (\sigma v r) has units (\text{kg/m}^3)(\text{m/s})(\text{m}) = \text{kg/(m·s)}, same as \eta. The ratio is pure number. For Stokes' law to hold, Re should be well below 1.

Result: v_\text{term} \approx 4.1 cm/s. Reynolds number \approx 0.07, comfortably in the Stokes regime — the formula is trustworthy.

What this shows: A classic school demonstration: steel ball, glycerine, stopwatch, scale. Measure how long the ball takes to fall between two marks and you have an experimental measurement of glycerine's viscosity. This is the basis of the falling-sphere viscometer, the simplest instrument for measuring liquid viscosity.

Example 2: A raindrop falling through air — why mist hangs in the sky

A raindrop of radius r = 0.1 mm (a "mist" droplet) falls through air. Density of water \rho = 1000 \text{ kg/m}^3; density of air \sigma = 1.2 \text{ kg/m}^3; viscosity of air \eta = 1.8 \times 10^{-5} Pa·s. Find the terminal velocity and check Stokes' validity. Then redo the calculation for a larger raindrop of radius 1 mm.

Raindrops of two sizes falling through air Two droplets shown side by side. Left: a very small droplet labelled mist, 0.1 mm radius, with a short downward arrow labelled 1 cm per second. Right: a larger droplet labelled rain, 1 mm radius, with a longer downward arrow labelled 4 m per second in Stokes limit. The drastic difference in terminal velocity is highlighted. mist: $r = 0.1$ mm $v \approx 1$ cm/s rain: $r = 1$ mm $v \approx 4$ m/s (Stokes) (real: ≈ 6.5 m/s, not Stokes)
Two raindrops of very different sizes. The mist droplet falls at 1 cm/s — slow enough to hang in the air for minutes. The raindrop falls at several m/s — fast enough to reach the ground in seconds.

Step 1 — cloud droplet, r = 10 μm = 10^{-5} m. Start with a genuine Stokes-regime droplet. Terminal velocity:

v_\text{term} = \frac{2}{9} \cdot \frac{(\rho - \sigma) g r^2}{\eta} = \frac{2}{9} \cdot \frac{999 \times 9.8 \times (10^{-5})^2}{1.8 \times 10^{-5}} = \frac{2}{9} \cdot \frac{9.79 \times 10^{-7}}{1.8 \times 10^{-5}} \approx 0.012 \text{ m/s} = 1.2 \text{ cm/s}.

Reynolds number: \text{Re} = (1.2 \times 0.012 \times 10^{-5})/(1.8 \times 10^{-5}) \approx 0.008. Well within Stokes' regime — the answer is trustworthy.

Why: take the density excess \rho - \sigma \approx 999 \text{ kg/m}^3 (water is a thousand times denser than air, so the \sigma correction barely matters). A 10 μm droplet falls at about a centimetre per second — slow enough to drift for minutes in still air, and easily lofted by the faintest updraft. This is why clouds stay aloft.

Step 2 — mist droplet, r = 0.1 mm = 10^{-4} m. Naïve Stokes:

v_\text{term, Stokes} = \frac{2}{9} \cdot \frac{999 \times 9.8 \times (10^{-4})^2}{1.8 \times 10^{-5}} \approx 1.21 \text{ m/s}.

Reynolds check: \text{Re} = (1.2 \times 1.21 \times 10^{-4})/(1.8 \times 10^{-5}) \approx 8 — well outside the Stokes regime. At this Re, inertial effects around the droplet contribute significant extra drag, and the real terminal velocity is about 0.27 m/s, less than a quarter of the Stokes prediction. The Stokes formula overestimates by a factor of four.

Step 3 — rain droplet, r = 1 mm = 10^{-3} m, using Stokes (knowing it will fail).

v_\text{term, Stokes} = \frac{2}{9}\cdot \frac{999 \times 9.8 \times 10^{-6}}{1.8 \times 10^{-5}} \approx 121 \text{ m/s}.

That's supersonic, which is obviously absurd. The real terminal velocity of a 1 mm raindrop is about 6.5 m/s — 20 times less. Reynolds check: Re \approx (1.2 \times 121 \times 10^{-3})/(1.8 \times 10^{-5}) \approx 8000. Far, far outside the Stokes regime.

Why: at high Re, drag becomes quadratic in velocity, not linear: F_D \approx \tfrac{1}{2} C_D \rho A v^2. The terminal velocity then scales like \sqrt{r} instead of r^2. A 1 mm raindrop fall speed is set by this turbulent-drag regime, not Stokes.

Result: Tiny cloud droplets (10 μm) fall at about 1 cm/s — slow enough to be held aloft by even mild updrafts. Raindrop-sized droplets (1 mm) fall far outside Stokes and reach about 6.5 m/s in reality, an order of magnitude slower than Stokes predicts (because real drag at high Re is stronger).

What this shows: This example illustrates why the Reynolds-number check matters. Knowing Stokes' law is not enough — you must also know when to trust it. Cloud physics, aerosol science, pollution droplet dynamics, and inhaler medicine all live in the Stokes regime (small droplets). Rain, hail, and any droplet larger than a millimetre does not.

Example 3: Mustard oil dripping down a glass rod

A thin layer of mustard oil (viscosity \eta = 0.05 Pa·s, density \sigma = 920 \text{ kg/m}^3) coats a vertical glass rod of radius R = 5 mm. Far from the rod, the air is still. A tiny steel bead of radius r = 0.5 mm and density \rho = 7800 \text{ kg/m}^3 is embedded in the oil film and begins to sink. Assuming the film is thick enough that Stokes' law holds, find the terminal velocity of the bead through the oil and how long it takes to fall 10 cm.

Step 1. Terminal velocity. Applying (TV):

v_\text{term} = \frac{2}{9}\cdot \frac{(7800 - 920)(9.8)(0.0005)^2}{0.05}.

(\rho - \sigma)g = 6880 \times 9.8 = 67{,}424 \text{ N/m}^3.

(\rho - \sigma)g r^2 = 67{,}424 \times 2.5 \times 10^{-7} = 0.01686.

v_\text{term} = (2/9) \times (0.01686/0.05) = (2/9) \times 0.337 \approx 0.0749 m/s \approx 7.5 cm/s.

Why: substitute into the same master formula. Compared with the steel ball in glycerine from Example 1 (4 cm/s), this smaller bead in mustard oil falls faster — the r^2 factor is 16× smaller here but \eta is 28× smaller, so the net effect is a higher terminal velocity. Viscosity matters more than size in this comparison.

Step 2. Check the Reynolds number.

\text{Re} = \frac{\sigma v r}{\eta} = \frac{920 \times 0.0749 \times 5 \times 10^{-4}}{0.05} = \frac{0.0344}{0.05} \approx 0.69.

Re \approx 0.7 — borderline. Stokes' law gives a mild over-prediction; a 10–20% correction would refine it, but for an order-of-magnitude answer it is fine.

Step 3. Time to fall 10 cm at constant speed.

t = \frac{0.10}{0.0749} \approx 1.34 \text{ s}.

Why: once at terminal velocity, distance equals speed times time. The transient (approach to terminal velocity) is much shorter than 1 s, so this is an excellent approximation.

Result: The bead falls at about 7.5 cm/s and covers 10 cm in roughly 1.3 seconds.

What this shows: Mustard oil's viscosity (about 50 times water's, much less than glycerine's) produces a comfortable, watchable falling motion. This is why mustard oil is a better choice than glycerine for a school experiment where you want to see the fall happen in real time without slowing down to a crawl.

Common confusions

If you have Stokes' law, the terminal-velocity formula, and can check Reynolds validity, you have the working content of this article. What follows is the derivation of the approach to terminal velocity (the transient), the connection to the Millikan oil drop experiment, the extension to non-Newtonian fluids, and the Reynolds-number bridge to the next chapter.

How quickly does a sphere reach terminal velocity?

Before terminal velocity is reached, the sphere accelerates under net force F_\text{net} = W - F_B - 6\pi\eta r v. Newton's second law:

m\frac{dv}{dt} = (W - F_B) - 6\pi\eta r v = (W - F_B)\left(1 - \frac{v}{v_\text{term}}\right),

where v_\text{term} = (W - F_B)/(6\pi\eta r). This is a first-order linear ODE with solution

v(t) = v_\text{term}\left(1 - e^{-t/\tau}\right),

where the time constant is

\tau = \frac{m}{6\pi\eta r} = \frac{2\rho r^2}{9\eta}.

For the steel ball in glycerine (Example 1), \tau = 2 \times 7800 \times (0.002)^2/(9 \times 1.4) \approx 5 \times 10^{-3} s — five milliseconds. After 5 time constants (\sim 25 ms) the ball is within 1% of terminal velocity. This is why the earlier animation looks like pure uniform motion — the transient is over almost instantly.

For a cloud droplet in air (r = 10 μm), \tau \sim 10^{-6} s — a microsecond. For a raindrop (r = 1 mm, but in the turbulent-drag regime where Stokes fails), the effective time constant is longer — of order tenths of a second — because quadratic drag produces a different functional form for v(t).

Millikan's oil drop experiment — measuring the electron's charge with Stokes' law

In 1909, Robert Millikan used Stokes' law to perform one of the most beautiful measurements in physics: the charge of a single electron. Tiny oil droplets were sprayed into a chamber between two parallel plates. The droplets picked up a few excess electrons. By turning on an electric field E between the plates, an extra electrostatic force qE acted on each droplet (where q is the droplet's net charge).

By adjusting the field so the droplet hovered stationary:

qE = (W - F_B).

Meanwhile, with the field off, the droplet fell at terminal velocity, giving Millikan the droplet's radius (via r^2 = 9\eta v_\text{term}/[2(\rho - \sigma)g]). Once r was known, W - F_B could be computed from densities and dimensions, and from qE the charge q was known — and it was always an integer multiple of a fundamental unit e = 1.6 \times 10^{-19} C. Millikan measured the electron charge to within 1% using nothing but Stokes' law, a voltmeter, and a microscope.

Non-Newtonian fluids — when \tau = \eta\, dv/dy fails

Some fluids do not obey Newton's law of viscosity. Instead, their shear stress depends on shear rate in a non-linear way.

  • Shear-thinning (pseudoplastic): viscosity decreases with shear rate. Examples: paint, blood, ketchup, toothpaste when squeezed, mayonnaise.
  • Shear-thickening (dilatant): viscosity increases with shear rate. Example: cornflour and water (the famous "oobleck" — walk on it fast and it's solid; stand still and you sink).
  • Bingham plastic: behaves as solid below a yield stress, then flows like a Newtonian fluid above it. Example: toothpaste at rest vs being squeezed out, drilling mud, custard.
  • Viscoelastic: exhibits both viscous and elastic behaviour (solid-like for fast deformations, fluid-like for slow). Example: silly putty, polymer melts.

For these fluids, Stokes' law does not apply directly. The general framework is tensorial rheology, beyond the scope of school physics but important in chemical engineering, food science, and biomedicine.

Viscosity and the Reynolds number — bridge to the next chapter

The Reynolds number is the dimensionless ratio

\text{Re} = \frac{\rho v L}{\eta},

where L is a characteristic length scale (sphere radius, pipe diameter, wing chord). It measures the ratio of inertial forces (\rho v^2) to viscous forces (\eta v/L) in the flow.

  • Re \ll 1: viscous forces dominate; flow is laminar; Stokes' law holds.
  • Re \sim 1: transitional.
  • Re \gg 1: inertial forces dominate; flow is turbulent; drag becomes quadratic in velocity; Bernoulli-like energy arguments (with turbulent corrections) describe the flow.

The Reynolds criterion is universal — it applies to all Newtonian fluid flows and determines when to trust which approximation. Its history, the transition to turbulence, and its implications for everything from pipe design to aircraft lift are the subject of the next chapter.

Kinematic vs dynamic viscosity

There are two viscosities you will meet. The dynamic viscosity \eta (also called \mu) is what we have been using: \tau = \eta\, dv/dy, units Pa·s. The kinematic viscosity is

\nu = \frac{\eta}{\rho},

with units m²/s. This appears naturally in the Navier–Stokes equations in the form \nu\nabla^2\vec{v}, where it plays the role of a diffusivity for momentum. For water, \nu \approx 10^{-6} m²/s; for air, \nu \approx 1.5 \times 10^{-5} m²/s (interestingly, air has higher kinematic viscosity than water because its much lower density more than compensates for its lower dynamic viscosity).

An Indian footnote — the cave physics of Ajanta

The Ajanta cave paintings (2nd century BCE to 6th century CE) survived in part because of the slow, viscous flow of the mineral-rich binders used in the pigments. The colloidal suspension of pigment in an oil-like organic binder had a viscosity high enough that it did not drip off vertical rock faces during painting, and then set over years into a dried, highly stable film. The rheology of these paint mixtures — viscosity controlled by particle loading, organic solvent chemistry, and temperature — was implicit in the craft knowledge of the painters. Modern forensic analysis has identified the binder chemistry, but the viscosity that made the application possible was mastered empirically two thousand years ago.

Where this leads next