Viscosity

Tilt two bottles at once — one of water, one of honey — and watch the race. The water is gone in a second, a thin quick ribbon down the glass. The honey creeps, thick and slow, almost reluctant to leave the bottle at all. Both are liquids, both obey gravity equally, yet one flows and one crawls. The difference is not weight, and it is not "thickness" in some vague kitchen sense — it is a real, measurable mechanical property called viscosity, the internal friction a fluid exerts on itself as its layers slide past one another. Every fluid you have ever poured, stirred, or watched drip has this property in some amount, including the "ideal", frictionless fluid Bernoulli's theorem imagines — that fluid simply does not exist in nature. Air has a small viscosity; water a little more; engine oil, honey and glycerine far more again. Viscosity is why a river flows fastest at the centre and almost stalls near the banks, why a spoon dragged through syrup feels resistance long after your hand stops moving, why blood needs a pump to keep circulating, and why a steel ball dropped into a jar of shampoo sinks at a lazy, constant speed rather than accelerating forever like a stone in air. This article builds viscosity from the ground up: how physicists define it precisely through Newton's law of viscous flow, what its unit and dimensions are, how it behaves differently in liquids and gases as temperature changes, and — pushing one step beyond the bare definition — how a viscous fluid's drag on a falling sphere fixes a maximum, unchanging terminal velocity. That single idea is the bridge to Stokes' law, the very next stop in this chapter.

Key takeaways

Viscosity is internal friction inside a fluid — the resistance between adjacent layers of fluid sliding past each other at different speeds.
Newton's law of viscous flow: $F = \eta A \dfrac{dv}{dx}$ , where $\eta$ is the coefficient of viscosity, $A$ the contact area, and $dv/dx$ the velocity gradient between layers.
SI unit is the pascal-second (Pa·s), dimensions $[ML^{-1}T^{-1}]$ ; the older CGS unit is the poise, $1\ \text{poise} = 0.1\ \text{Pa}\,\cdot\,\text{s}$ .
Temperature effects are opposite for liquids and gases: liquids get less viscous as they warm up; gases get more viscous — a favourite conceptual trap.
Real flow through a pipe is never uniform. The no-slip condition pins the fluid at the walls to zero speed, giving a parabolic velocity profile with maximum speed at the centre.
A body falling through a viscous fluid stops accelerating once viscous drag plus buoyancy exactly balance its weight, settling into a constant terminal velocity — the direct forward link to Stokes' law.
Key idea: viscosity is the coefficient in Newton's law F = ηA(dv/dx) that measures a fluid's internal resistance to shearing, it rises with temperature in gases but falls in liquids, and it caps a falling body's speed at a fixed terminal velocity once drag, buoyancy and weight balance.

The core insight

Viscosity is the fluid property, defined by $F = \eta A\, dv/dx$ , that resists relative sliding between adjacent fluid layers, and it is exactly this resistance — combined with buoyancy and weight — that limits a falling body to a constant terminal velocity.

Where it fits in the syllabus

Viscosity sits inside the CBSE Class 11 "Mechanical Properties of Fluids" unit, right after Bernoulli's theorem and its applications. Where Bernoulli's theorem idealises fluid flow as frictionless, viscosity is the syllabus's honest correction: real fluids always resist internal sliding, and this resistance is what the board expects you to define quantitatively through Newton's law of viscous flow, describe qualitatively for streamline versus turbulent flow, and connect to falling-body problems that lead straight into Stokes' law and terminal velocity, the next article in this sequence. Poiseuille's flow-rate formula and the full Navier–Stokes treatment of turbulence are explicitly outside this syllabus — only the qualitative picture of viscous flow through a pipe is required here.

The physical picture

Picture a fluid flowing over a fixed surface — water in a canal, or air over an aircraft wing. The layer of fluid in direct contact with the surface is held essentially motionless by molecular attraction to it (the no-slip condition), while layers farther from the surface move increasingly fast. Between any two adjacent layers, then, there is a small difference in speed spread over a small separation — a velocity gradient, written $dv/dx$ , where $x$ is measured perpendicular to the flow. Wherever this gradient exists, the fluid resists it: the faster layer tends to drag the slower one forward, and the slower layer tends to hold the faster one back, exactly as friction resists relative sliding between two solid surfaces. Newton captured this resistance in a single proportionality, true for most common fluids (called Newtonian fluids — water, air, thin oils):

F = \eta A \frac{dv}{dx}

where: $F$ is the tangential (shearing) force needed to keep two adjacent layers sliding past each other at a steady relative speed, $A$ is the area of contact between the layers, $dv/dx$ is the velocity gradient perpendicular to the flow, and $\eta$ (eta) is the coefficient of viscosity — a constant for a given fluid at a given temperature that measures how strongly it resists shearing. The physical reading is this: double the contact area or double the velocity gradient, and you double the force needed to maintain the sliding; a liquid with large $\eta$ (honey, glycerine) needs a much larger force to shear at the same rate as one with small $\eta$ (water, air). Rearranging the law shows $\eta$ has the physical meaning of shearing stress ( $F/A$ ) per unit velocity gradient, which fixes its SI unit as the pascal-second (Pa·s), equivalent to $\text{N}\,\cdot\,\text{s/m}^2$ or $\text{kg/(m}\,\cdot\,\text{s)}$ , with dimensional formula $[ML^{-1}T^{-1}]$ . The older CGS unit, the poise, is still common in tables: $1\ \text{poise} = 0.1\ \text{Pa}\,\cdot\,\text{s}$ , so water's viscosity of about $1.0\times10^{-3}\ \text{Pa}\,\cdot\,\text{s}$ at room temperature is "1 centipoise." Two qualitative facts follow directly from how molecules behave, and both are favourite conceptual-question territory. In a liquid, molecules are close together and viscosity comes mainly from the cohesive attraction between them; heating a liquid gives molecules more thermal energy to break free of these attractions, so a liquid's viscosity falls as temperature rises (this is why honey pours more easily when warmed). In a gas, molecules are far apart and viscosity instead comes from momentum transferred as fast-moving molecules dart between layers; heating a gas speeds up this molecular traffic, so a gas's viscosity rises with temperature — the exact opposite trend. Because Bernoulli's theorem strictly assumes $\eta = 0$ (an "ideal fluid"), every real fluid's non-zero viscosity means some kinetic energy is steadily lost as heat during flow — which is also why a real horizontal pipe of uniform cross-section shows a gradual pressure drop along its length, even though the ideal-fluid form of Bernoulli's equation would predict no such drop for uniform speed: the pressure difference does the work needed against viscous drag.

Figure

A liquid trapped between two horizontal plates, the lower plate fixed and the upper plate dragged sideways by a force, with the fluid layers in between shown as arrows of increasing length from the bottom layer at rest to the top layer moving fastest, illustrating the velocity gradient that Newtons law of viscosity describes

The physical picture — from drag to a falling body

The same viscous drag that resists sliding layers also resists a solid body pushing through a fluid. When a small sphere falls through a viscous liquid, the liquid must flow around it, and the layers of liquid nearest the sphere's surface are dragged along, resisted in turn by the layers farther out — exactly the shearing described above, now wrapped around a moving sphere instead of two flat plates. The drag force this produces on a sphere was worked out precisely by George Stokes and is stated (without derivation, since it needs fluid-dynamics beyond this syllabus) as $F = 6\pi\eta r v$ , where $r$ is the sphere's radius and $v$ its speed through the fluid — this is Stokes' law, the very next topic in this chapter. What concerns us here is a direct and very testable consequence of that drag: because the drag force grows with speed, a falling sphere cannot keep accelerating forever — it settles into a constant final speed. Deriving that speed, called the terminal velocity, is the natural next step, and it needs nothing beyond force balance and the definitions already in hand.

Derivation ATerminal velocity of a small sphere falling through a viscous fluid

The goal is to find the constant final speed $v_t$ that a small sphere of radius $r$ and density $\rho$ reaches while falling through a viscous fluid of density $\sigma$ and coefficient of viscosity $\eta$ , using nothing but Newton's second law applied at the moment the acceleration has died away to zero, together with the standard expressions for weight, buoyant force and Stokes' viscous drag. The subtlety worth watching is that three forces act on the sphere throughout its fall — weight pulling it down, buoyancy pushing it up, and viscous drag also pushing up once the sphere is moving — and terminal velocity is simply the special speed at which these three exactly cancel, so the net force, and hence the acceleration, becomes zero.

Step 1. List every force on the falling sphere. Its weight acts downward, $W = mg = \left(\frac{4}{3}\pi r^3\right)\rho g$ , using the sphere's volume $\frac{4}{3}\pi r^3$ and density $\rho$ . The buoyant force, from Archimedes' principle, acts upward and equals the weight of fluid displaced, $F_B = \left(\frac{4}{3}\pi r^3\right)\sigma g$ . The viscous drag, from Stokes' law, also acts upward once the sphere is moving downward, opposing the motion: $F_v = 6\pi\eta r v$ , where $v$ is the sphere's instantaneous speed.

Step 2. Track how these forces change as the sphere speeds up. Immediately after release, $v$ is small, so the drag $F_v = 6\pi\eta r v$ is small too, and weight exceeds buoyancy plus drag — the sphere accelerates downward. As $v$ grows, $F_v$ grows with it (drag is proportional to speed), so the net downward force steadily shrinks. Eventually $v$ reaches a value large enough that drag plus buoyancy exactly equal weight; from that instant the net force is zero, so by Newton's second law the acceleration is zero, and the speed stops changing. That fixed value of $v$ is the terminal velocity $v_t$ .

Step 3. Write the balance condition at terminal velocity. Weight equals buoyancy plus drag: $W = F_B + F_v$ , so $\left(\frac{4}{3}\pi r^3\right)\rho g = \left(\frac{4}{3}\pi r^3\right)\sigma g + 6\pi\eta r v_t$ .

Step 4. Solve for $v_t$ . Move the buoyancy term across and factor: $\left(\frac{4}{3}\pi r^3\right)(\rho - \sigma) g = 6\pi\eta r v_t$ . Dividing both sides by $6\pi\eta r$ and simplifying the numerical factor $\frac{4}{18} = \frac{2}{9}$ gives

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

Every quantity here is one already defined: $r$ the sphere's radius, $\rho$ its density, $\sigma$ the fluid's density, $g$ the acceleration due to gravity, and $\eta$ the fluid's coefficient of viscosity. Notice $v_t \propto r^2$ — doubling the radius quadruples the terminal velocity, which is why a large hailstone falls far faster through air than a fine mist droplet, and why $v_t \propto 1/\eta$ — a more viscous fluid caps the terminal speed lower for the same sphere. Check with numbers: a steel ball of radius $r = 1.0\times10^{-3}\ \text{m}$ (density $\rho = 7800\ \text{kg/m}^3$ ) released in glycerine (density $\sigma = 1260\ \text{kg/m}^3$ , $\eta = 0.83\ \text{Pa}\,\cdot\,\text{s}$ ) gives $v_t = \dfrac{2(1.0\times10^{-3})^2(7800-1260)(9.8)}{9(0.83)} = \dfrac{0.1282}{7.47} \approx 0.0172\ \text{m/s}$ , about 1.7 cm every second — slow enough to watch with the naked eye, which is exactly why this is a standard laboratory method for measuring $\eta$ .

Worked examples

Board level - viscous force between two plates separated by oil

Two horizontal plates are separated by a layer of oil 5 mm thick. The lower plate is fixed, and the upper plate, of area $0.2\ \text{m}^2$ , is dragged at a constant speed of 0.1 m/s by applying a force of 0.6 N. Find the coefficient of viscosity of the oil.

Step 1. Find the velocity gradient. The lower plate is fixed and the upper plate moves at 0.1 m/s across a gap of 5 mm $= 5\times10^{-3}\ \text{m}$ , so

\frac{dv}{dx} = \frac{0.1}{5\times10^{-3}} = 20\ \text{s}^{-1}

Step 2. Apply Newton's law of viscosity, $F = \eta A \dfrac{dv}{dx}$ , and rearrange to isolate $\eta$ :

\eta = \frac{F}{A\left(\dfrac{dv}{dx}\right)} = \frac{0.6}{0.2 \times 20}

Step 3. Compute. $\eta = \dfrac{0.6}{4} = 0.15\ \text{Pa}\,\cdot\,\text{s}$ .

The oil's coefficient of viscosity is 0.15 Pa·s — about 150 times more viscous than water's $1.0\times10^{-3}\ \text{Pa}\,\cdot\,\text{s}$ , matching the everyday sense that oil pours far more sluggishly.

Competitive level - terminal velocity of a fog droplet in air

A tiny water droplet of radius $1.0\times10^{-5}\ \text{m}$ (10 micrometres, typical of fog) falls through still air. Take the density of water $\rho = 1000\ \text{kg/m}^3$ , the density of air $\sigma = 1.2\ \text{kg/m}^3$ , the viscosity of air $\eta = 1.8\times10^{-5}\ \text{Pa}\,\cdot\,\text{s}$ , and $g = 9.8\ \text{m/s}^2$ . Find its terminal velocity.

Step 1. Write the terminal-velocity formula derived above and identify each quantity in SI units: $r = 1.0\times10^{-5}\ \text{m}$ , $\rho - \sigma \approx 1000 - 1.2 = 998.8\ \text{kg/m}^3$ , $\eta = 1.8\times10^{-5}\ \text{Pa}\,\cdot\,\text{s}$ .

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

Step 2. Substitute. The numerator is $2 \times (1.0\times10^{-5})^2 \times 998.8 \times 9.8 = 2\times10^{-10} \times 998.8 \times 9.8 \approx 1.96\times10^{-6}$ . The denominator is $9 \times 1.8\times10^{-5} = 1.62\times10^{-4}$ .

Step 3. Divide. $v_t = \dfrac{1.96\times10^{-6}}{1.62\times10^{-4}} \approx 1.2\times10^{-2}\ \text{m/s} = 1.2\ \text{cm/s}$ .

The fog droplet's terminal velocity is about 1.2 cm/s — millions of times slower than the steel ball in glycerine above, which is exactly why fog hangs in the air for so long instead of falling out at once, and why gently rising warm air currents can keep it aloft even longer. Timing note: under two minutes if the $r^2$ and unit conversions (micrometres to metres) are done cleanly before touching the calculator; the single biggest time-loss on this question type is mishandling the power of ten in $r^2$ when $r$ is given in micrometres — square the number and the power of ten separately, then combine.

Common traps

Confusing the temperature dependence of viscosity in liquids and gases — liquids become less viscous as temperature rises, gases become more viscous; this reversal is one of the most commonly flipped facts in board answers.
Forgetting the SI unit of viscosity is the pascal-second (Pa·s), and mistakenly writing it as a unit of pressure or force alone.
Assuming the velocity of a viscous fluid is uniform across a pipe's cross-section — it is not; the no-slip condition pins it to zero at the walls and it is maximum at the centre, giving a parabolic profile.
In terminal-velocity problems, forgetting buoyancy entirely and balancing only weight against viscous drag — the correct balance is weight equals buoyancy plus drag, and skipping buoyancy gives a wrong (too small) terminal velocity.
Reaching for Poiseuille's flow-rate formula to compute volume flow through a pipe — this specific derived law is outside the CBSE Class 11 syllabus; only the qualitative picture of viscous pipe flow is required.
Treating terminal velocity as something a falling body reaches instantly — in reality the body accelerates for a short interval first, and only approaches $v_t$ as drag catches up with weight minus buoyancy.

Common trapTerminal velocity is not the same as zero net force from the start

The single biggest misconception here is imagining that a sphere released in a viscous fluid is somehow moving at $v_t$ the moment it is dropped. It is not. At the instant of release the sphere's speed is zero, so the drag term $6\pi\eta r v$ is also zero, and the net force is simply weight minus buoyancy — the sphere genuinely accelerates downward at first. Only as speed builds does drag grow to match, and the acceleration shrinks toward zero. Terminal velocity is the limiting speed this process approaches, not a starting condition, and for small enough spheres in sufficiently viscous fluids this approach can happen very quickly (a fraction of a second), which is why lab measurements often treat the fall as "instantly" terminal without technically being so.

You should be able to…

Define viscosity and state Newton's law of viscous flow, $F = \eta A\, dv/dx$ , identifying every symbol.
Recall the SI unit (Pa·s) and dimensional formula $[ML^{-1}T^{-1}]$ of the coefficient of viscosity, and convert to and from the poise.
Explain how the viscosity of liquids and gases changes with temperature, and why the two trends are opposite.
Explain qualitatively why real fluid flow through a pipe has a non-uniform, parabolic velocity profile because of the no-slip condition.
Derive the terminal velocity of a small sphere falling through a viscous fluid by balancing weight, buoyancy and viscous drag.
Solve numericals relating force, area, velocity gradient and coefficient of viscosity, and numericals finding terminal velocity given radius, densities and viscosity.

Ask the tutor: Want a worked walk-through of why viscosity flips direction with temperature for liquids versus gases, or a hand-held derivation of terminal velocity with your own numbers — ask the AI tutor.

Why does viscosity have units of pressure times time instead of just pressure?

Look at Newton's law rearranged: $\eta = \dfrac{F/A}{dv/dx}$ . The numerator, $F/A$ , is a force per area, which is exactly the unit of pressure, the pascal. The denominator, $dv/dx$ , is a speed divided by a distance, which works out to $\text{s}^{-1}$ — an inverse time. Dividing a pressure by an inverse time is the same as multiplying a pressure by a time, so the unit becomes pascal-seconds. It is a genuinely different kind of quantity from pressure alone, which is why mixing the two up on an exam costs marks.

If honey and water are both liquids, why is one so much more viscous than the other?

Viscosity in a liquid comes mainly from the strength of the attractive forces (and, for large molecules, entanglement) between neighbouring molecules — the harder it is to make one layer of molecules slide past another, the higher the viscosity. Honey is a thick, sugar-rich solution whose large molecules interact strongly and get tangled as they try to slide past each other, while water molecules are small and interact comparatively weakly. That is a purely microscopic, chemistry-flavoured explanation; for CBSE Class 11 you only need the macroscopic definition through Newton's law and the observed fact that different fluids have very different values of $\eta$ — from about $1.8\times10^{-5}\ \text{Pa}\,\cdot\,\text{s}$ for air, to $1.0\times10^{-3}\ \text{Pa}\,\cdot\,\text{s}$ for water, to roughly $1\ \text{Pa}\,\cdot\,\text{s}$ or more for honey.

How is Stokes law connected to Newtons law of viscosity — are they the same thing?

They are closely related but not identical. Newton's law of viscous flow, $F = \eta A\, dv/dx$ , defines the coefficient of viscosity $\eta$ itself, using flat sliding layers as the simplest case to define it in. Stokes' law, $F = 6\pi\eta r v$ , is a specific, more advanced result — worked out for the particular geometry of a small sphere moving through a viscous fluid — that uses the same $\eta$ but applies it to a curved, three-dimensional flow pattern around the sphere rather than flat layers. You can think of Newton's law as defining what $\eta$ means, and Stokes' law as one important application of that same $\eta$ to a falling-sphere problem; the full derivation of Stokes' law itself needs fluid dynamics beyond this syllabus, so it is simply stated and used, exactly as it will be in the next article.

Does terminal velocity mean the sphere has stopped moving?

No — terminal velocity means the sphere has stopped accelerating, not that it has stopped moving. It keeps falling at that same constant speed indefinitely (as long as nothing else changes), because the net force on it is zero: weight downward is exactly balanced by buoyancy plus viscous drag upward. This is the fluid-flow analogue of a skydiver reaching a constant falling speed in air once air resistance balances their weight — the skydiver keeps falling, just without speeding up any further.