Stokes law

Drop a small steel ball into a tall jar of thick oil or glycerine and watch closely. For a fraction of a second it speeds up, exactly as any falling body should under gravity — and then, strangely, it stops accelerating. From that point on it sinks at a fixed, unhurried speed all the way to the bottom, as if some invisible hand has capped how fast it is allowed to fall. Nothing outside is holding it back. The fluid itself is doing the braking, and it does so more and more firmly the faster the ball tries to go, until braking and pulling exactly cancel. That invisible hand has a name and a formula. Stokes' law, worked out by the Irish physicist George Gabriel Stokes in 1851, gives the exact viscous drag force a slow-moving, small sphere feels from a fluid, and from it we can predict the steady "terminal velocity" every falling sphere eventually settles into — whether it is a raindrop drifting down through air, a red blood cell settling in a test tube, a dust grain sinking through the ocean, or the steel ball in your jar of glycerine. It is one of the cleanest force-balance results in the whole "Mechanical Properties of Fluids" chapter: three forces, one equation, one elegant $r^2$ result that predicts how radius controls fall speed. This section builds Stokes' law from the physical picture of viscous drag, derives the terminal-velocity formula by balancing weight, buoyancy and drag, and works through the same steel-ball-in-glycerine numbers you will meet in your textbook and exams — twice, at board level and at competitive-exam speed.

Key takeaways

Stokes' law gives the viscous drag force on a small sphere moving slowly through a fluid: $F = 6\pi\eta r v$ , valid only for smooth, streamline flow around the sphere.
The drag force grows with the sphere's radius $r$ , its speed $v$ , and the fluid's viscosity $\eta$ — three separate levers, all multiplying together.
A falling sphere feels three forces: weight (down), buoyant upthrust (up), and viscous drag (up, growing with speed) — it accelerates only until these balance.
Terminal velocity $v_t = \dfrac{2r^2(\rho-\sigma)g}{9\eta}$ is the constant speed reached once net force becomes zero; after that instant the sphere moves at constant velocity, not constant acceleration.
Terminal velocity scales as $r^2$ , not $r$ — doubling the radius quadruples $v_t$ , a favourite examiner trap.
The sign of $(\rho - \sigma)$ tells you the direction: a sphere denser than the fluid sinks with positive $v_t$ ; a sphere less dense than the fluid (like a rising air bubble in water) has the drag act downward as it rises.
Key idea: Stokes' law states the viscous drag on a slow-moving small sphere as F = 6πηrv, and balancing weight against buoyancy and this drag gives a terminal velocity that grows as the square of the sphere's radius, v_t = 2r²(ρ − σ)g / 9η.

The core insight

A small sphere falling through a viscous fluid speeds up only until the upward viscous drag $6\pi\eta r v$ plus buoyancy exactly balances its weight, after which it falls at a constant terminal velocity $v_t = \dfrac{2r^2(\rho-\sigma)g}{9\eta}$ that depends on the square of its radius.

Where it fits in the syllabus

Stokes' law is the direct continuation of viscosity in the CBSE Class 11 "Mechanical Properties of Fluids" unit: having defined viscosity as the internal friction of a fluid, the syllabus asks you to apply that idea to a concrete, examinable situation — a small sphere moving through a viscous fluid — and derive the resulting drag force and terminal velocity by a force balance. Both the drag-force formula $F = 6\pi\eta r v$ and the terminal-velocity formula $v_t = \frac{2r^2(\rho-\sigma)g}{9\eta}$ are prescribed results you are expected to state, derive, and use in numericals; the underlying Navier–Stokes-level fluid dynamics and Poiseuille's flow-rate law are explicitly outside the board syllabus, so keep the reasoning at the force-balance level shown here.

The physical picture

When a fluid flows smoothly past a solid surface — in this case around a small moving sphere — layers of fluid closer to the sphere are dragged along with it, while layers farther away lag behind, exactly as in ordinary viscous flow between two plates. That internal friction between fluid layers is what produces a net backward force on the sphere. Stokes worked out this force exactly for a sphere in slow, streamline flow (no turbulence, no eddies) and found it depends on only three quantities: the fluid's viscosity, the sphere's radius, and its speed relative to the fluid.

F = 6\pi \eta r v

where: $F$ is the viscous drag force (newtons, N), $\eta$ is the coefficient of viscosity of the fluid (pascal-seconds, Pa·s), $r$ is the radius of the sphere (metres), and $v$ is the speed of the sphere relative to the fluid (m/s). The physical reading is that each factor pulls in the same direction: a thicker (more viscous) fluid drags harder, a bigger sphere presents more surface for the fluid to grip, and a faster-moving sphere meets more resistance — drag is not constant, it grows with speed. This growing-with-speed behaviour is exactly what eventually stops the acceleration of a falling sphere. Stokes' law holds strictly only when the flow around the sphere stays streamline (laminar) rather than turbulent — that is, for small spheres moving slowly enough that the fluid closes smoothly behind them without eddies. The boundary between the two regimes is set qualitatively by the Reynolds number, a dimensionless measure of how disturbed the flow becomes; you are only expected to know this qualitatively, not to compute a Reynolds number. Now picture a small sphere of density $\rho$ and radius $r$ released from rest inside a still, viscous fluid of density $\sigma$ . Three forces act on it as it begins to fall:

W = \frac{4}{3}\pi r^3 \rho g, \qquad F_B = \frac{4}{3}\pi r^3 \sigma g, \qquad F = 6\pi \eta r v

where: $W$ is the sphere's weight acting downward, $F_B$ is the buoyant upthrust acting upward (Archimedes' principle — the weight of fluid displaced by the sphere's volume $\tfrac{4}{3}\pi r^3$ ), and $F$ is the Stokes viscous drag acting upward once the sphere is moving downward, since drag always opposes relative motion. At the very first instant of release, $v = 0$ , so the drag term vanishes and only weight and buoyancy act; if $\rho > \sigma$ the sphere begins to accelerate downward. As it speeds up, the drag term grows in direct proportion to $v$ , gradually eating into the net downward force, until the three forces balance exactly and the acceleration drops to zero — the sphere has reached terminal velocity.

Figure

A falling sphere in a viscous fluid with three labelled force arrows, weight pointing down, buoyancy and viscous drag pointing up, alongside a speed versus time graph that rises and then flattens out at the terminal velocity

Derivation ABalancing weight, buoyancy and drag to find terminal velocity

The goal is to find the constant speed $v_t$ that a small sphere of density $\rho$ and radius $r$ eventually reaches while falling through a still viscous fluid of density $\sigma$ and viscosity $\eta$ , using nothing beyond Archimedes' principle, Stokes' law, and Newton's first law applied at the instant net force vanishes. The physical picture already tells us the fluid can never fully stop the sphere or push it back up (as long as $\rho > \sigma$ ); it can only asymptotically cap its speed. Turning that qualitative picture into a number is a matter of writing all three forces algebraically and setting their sum to zero.

Step 1. Write each force in terms of the sphere's volume. A sphere of radius $r$ has volume $V = \frac{4}{3}\pi r^3$ . Its weight is $W = V\rho g = \frac{4}{3}\pi r^3 \rho g$ , acting vertically downward. By Archimedes' principle, the upward buoyant force equals the weight of fluid displaced, which is the same volume $V$ but the fluid's density $\sigma$ : $F_B = V\sigma g = \frac{4}{3}\pi r^3 \sigma g$ , acting upward. The third force is the Stokes viscous drag $F = 6\pi\eta r v$ , acting upward once the sphere moves downward with speed $v$ , since drag always opposes the relative motion between sphere and fluid.

Step 2. State the terminal-velocity condition. At terminal velocity the sphere is no longer accelerating — its net force is zero, so by Newton's first law the three forces must sum to zero along the vertical: the single downward force (weight) must exactly equal the sum of the two upward forces (buoyancy plus drag).

\frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi \eta r v_t

Step 3. Isolate the drag term and simplify. Move the buoyancy term to the left-hand side and factor out the common volume term:

6\pi \eta r v_t = \frac{4}{3}\pi r^3 g(\rho - \sigma)

Both sides carry a factor of $\pi r$ , which cancels, leaving $6\eta v_t = \frac{4}{3}r^2 g(\rho-\sigma)$ .

Step 4. Solve for $v_t$ by dividing through by $6\eta$ .

v_t = \frac{4 r^2 g(\rho - \sigma)}{18\eta} = \frac{2r^2(\rho - \sigma)g}{9\eta}

Step 5. Read the result physically. Terminal velocity is proportional to $r^2$ , not $r$ — this is the single most important, and most often mis-stated, feature of the formula, since it means a sphere twice as large (same material, same fluid) falls four times, not twice, as fast at terminal speed. It is also proportional to the density difference $(\rho - \sigma)$ : if the sphere is denser than the fluid ( $\rho > \sigma$ ), $v_t$ is positive and the sphere sinks steadily; if the sphere is less dense than the fluid ( $\rho < \sigma$ , as for an air bubble rising in water), the same algebra with the drag direction reversed gives a rise velocity of the same magnitude and form, now directed upward. Finally, $v_t$ is inversely proportional to viscosity $\eta$ — a thicker fluid caps the fall at a lower speed, which is exactly why the steel ball creeps through glycerine but would streak through water. Physical check: setting $\rho = \sigma$ (sphere and fluid of equal density) correctly gives $v_t = 0$ — with no density difference there is no net driving force, so the sphere neither sinks nor rises, matching everyday experience of a neutrally buoyant object simply hanging in a fluid.

Worked examples

Board level — viscous drag force on a falling steel ball

A small steel ball of radius $1.0\times10^{-3}\ \text{m}$ is dropped into a tall jar of glycerine of viscosity $\eta = 1.5\ \text{Pa}\,\cdot\,\text{s}$ . Find the viscous drag force on the ball at the instant it is moving with speed $0.02\ \text{m/s}$ .

Step 1. Identify the formula. Since the ball is small and moving slowly through a viscous fluid in streamline flow, Stokes' law applies directly: $F = 6\pi\eta r v$ .

Step 2. Substitute the given values, $\eta = 1.5\ \text{Pa}\,\cdot\,\text{s}$ , $r = 1.0\times10^{-3}\ \text{m}$ , $v = 0.02\ \text{m/s}$ .

F = 6 \times \pi \times 1.5 \times (1.0\times10^{-3}) \times 0.02

Step 3. Compute step by step: $6 \times 3.1416 = 18.85$ ; then $18.85 \times 1.5 = 28.27$ ; then $28.27 \times 1.0\times10^{-3} = 0.02827$ ; finally $0.02827 \times 0.02 = 5.65\times10^{-4}$ .

The viscous drag force on the ball at this speed is $F \approx 5.65\times10^{-4}\ \text{N}$ .

Competitive level — terminal velocity, and checking it against the drag law

The same steel ball (radius $1.0\times10^{-3}\ \text{m}$ , density $\rho = 7800\ \text{kg/m}^3$ ) is released from rest in the glycerine ( $\sigma = 1260\ \text{kg/m}^3$ , $\eta = 1.5\ \text{Pa}\,\cdot\,\text{s}$ , $g = 9.8\ \text{m/s}^2$ ). Find its terminal velocity, and verify that the drag force at that speed exactly balances the net downward force.

Step 1. Write the terminal-velocity formula derived above, $v_t = \dfrac{2r^2(\rho-\sigma)g}{9\eta}$ , and compute the pieces separately: $r^2 = (1.0\times10^{-3})^2 = 1.0\times10^{-6}\ \text{m}^2$ , and $\rho - \sigma = 7800 - 1260 = 6540\ \text{kg/m}^3$ .

Step 2. Substitute into the numerator: $2 \times 1.0\times10^{-6} \times 6540 \times 9.8$ . Multiply in order: $2 \times 1.0\times10^{-6} = 2.0\times10^{-6}$ ; then $2.0\times10^{-6} \times 6540 = 0.01308$ ; then $0.01308 \times 9.8 = 0.1282$ .

Step 3. Divide by the denominator $9\eta = 9 \times 1.5 = 13.5$ : $v_t = 0.1282 / 13.5 \approx 9.5\times10^{-3}\ \text{m/s}$ , i.e. about $9.5\ \text{mm/s}$ .

Step 4. Verify by plugging $v_t$ back into Stokes' law: $F = 6\pi\eta r v_t = 6\times 3.1416\times 1.5 \times (1.0\times10^{-3}) \times (9.5\times10^{-3}) \approx 2.69\times10^{-4}\ \text{N}$ . Compare this to the net downward force, weight minus buoyancy, $\frac{4}{3}\pi r^3 (\rho-\sigma)g = \frac{4}{3}\times 3.1416 \times (1.0\times10^{-3})^3 \times 6540 \times 9.8 \approx 2.68\times10^{-4}\ \text{N}$ — the two values agree (small rounding only), confirming the force balance.

The ball reaches a terminal velocity of about $9.5\ \text{mm/s}$ , at which point the upward viscous drag of roughly $2.7\times10^{-4}\ \text{N}$ exactly balances the net downward pull of weight minus buoyancy. Timing note: aim to finish both computations in under three minutes in an exam. The chain is always the same — plug into $v_t = 2r^2(\rho-\sigma)g/9\eta$ directly rather than re-deriving the balance from scratch every time; keep the verification step (Step 4 here) only if the question explicitly asks you to check consistency, since it doubles the arithmetic for no extra marks otherwise.

Common traps

Forgetting to subtract buoyancy (upthrust) from weight before balancing against viscous drag — omitting the buoyant force gives a wrong, too-high terminal velocity, since $\rho$ alone (instead of $\rho - \sigma$ ) is used.
Assuming terminal velocity is proportional to $r$ instead of $r^2$ — a very common algebra slip; doubling the sphere's radius quadruples its terminal velocity, not merely doubles it.
Applying Stokes' law to spheres that are too large or moving too fast, where the flow around the sphere is no longer smooth and streamline — Stokes' law strictly holds only for small spheres at low, streamline-flow speeds, below the critical velocity where turbulence sets in.
Mixing up $\rho$ (density of the sphere) and $\sigma$ (density of the fluid) in the terminal-velocity formula, which flips the sign of $(\rho - \sigma)$ and reverses the predicted direction of motion.
Treating the initial instant of release (where $v = 0$ and drag is zero) as though the sphere were already at terminal velocity — the sphere genuinely accelerates for a short interval before settling into constant speed; only after that transient does $v = v_t$ hold.

Common trapTerminal velocity is not reached instantly, and it scales as r-squared, not r

The two biggest scoring errors in this topic both come from oversimplifying the falling sphere's motion. First, a sphere released from rest does not move at terminal velocity from the very first moment — it genuinely accelerates for a brief interval while drag is still small, and only reaches constant speed once drag has grown enough to balance weight minus buoyancy; sketching the speed-versus-time graph as a flat line from $t=0$ is wrong, it must rise and then level off. Second, resist the urge to treat $v_t \propto r$ by analogy with the drag force $F \propto r$ — the terminal-velocity formula has an extra factor of $r$ hidden inside the sphere's volume (weight and buoyancy both scale as $r^3$ , drag only as $r$ , and solving for $v_t$ leaves $r^2$ behind). Always re-derive the power of $r$ from the balance equation rather than guessing it from the drag law alone.

You should be able to…

State Stokes' law, $F = 6\pi\eta r v$ , and identify what each symbol represents and why the flow must be streamline for it to hold.
Derive the terminal-velocity formula from scratch by writing weight, buoyancy and drag and setting their sum to zero.
Explain why terminal velocity depends on $r^2$ , not $r$ , and why it is inversely proportional to viscosity.
Predict whether a small body sinks or rises in a given fluid from the sign of $(\rho - \sigma)$ .
Solve numericals for viscous drag force and for terminal velocity given radius, densities, viscosity and $g$ .
Sketch and interpret the speed-versus-time graph of a sphere released from rest in a viscous fluid, including the initial accelerating phase and the flat terminal-velocity phase.

Ask the tutor: Want to walk through the force-balance derivation once more with a different sphere and fluid, or try a problem where a bubble rises through oil instead of a ball sinking through glycerine? Ask the AI tutor for extra practice.

Why does the ball not just keep speeding up forever as it falls?

Because the force resisting its motion, viscous drag, is not fixed — it grows in direct proportion to speed, $F = 6\pi\eta r v$ . At low speed the drag is small and weight (minus buoyancy) wins, so the ball accelerates. As speed increases, drag increases right along with it, eating into the net downward force. Eventually drag plus buoyancy exactly equals weight, the net force drops to zero, and by Newton's first law an object with zero net force moves at constant velocity. That constant, self-limiting speed is the terminal velocity — the fluid does not need to actively slow the ball down; it simply stops helping it speed up any further.

Does Stokes law work for any size of sphere, at any speed?

No — it is a formula for small spheres moving slowly enough that the fluid flows smoothly (streamline flow) around them, without turbulence or eddies forming behind the sphere. Once the sphere is too large, too fast, or the fluid too thin, the flow becomes turbulent and the simple $F = 6\pi\eta r v$ relation breaks down; a more complex, speed-squared drag law takes over instead, which is beyond the CBSE Class 11 syllabus. The qualitative boundary between the two regimes is described by the Reynolds number and a critical velocity, both of which you only need to know exist, not calculate.

Why does the ball fall so much more slowly in glycerine than it would in water or air?

Look at where viscosity $\eta$ sits in the terminal-velocity formula, $v_t = \frac{2r^2(\rho-\sigma)g}{9\eta}$ — it is in the denominator, so a larger $\eta$ directly shrinks $v_t$ . Glycerine has a viscosity roughly a thousand times that of water, so for the same steel ball the terminal velocity in glycerine works out to only millimetres per second, while in water or air (both far less viscous) the same ball would reach a much higher terminal speed before the much weaker drag could catch up with its weight. This is exactly why viscosity-measuring instruments drop a small sphere through the test fluid and time its steady fall — a slower fall directly signals a more viscous liquid.

Is there a similar formula if the sphere is rising instead of falling?

Yes, and it comes from exactly the same force balance, just with the sign of $(\rho - \sigma)$ flipped. If the sphere is less dense than the surrounding fluid, such as an air bubble in water or an oil droplet in a denser liquid, buoyancy exceeds weight and the object accelerates upward instead of downward; viscous drag now acts downward, opposing the upward motion, and grows with speed exactly as before. The same algebra gives a terminal rise velocity of the same magnitude and form, $v_t = \frac{2r^2(\sigma-\rho)g}{9\eta}$ , directed upward. The physics — drag growing with speed until it balances the net buoyant force — is identical either way.