Viscosity & Surface Tension
Why honey pours slowly, why raindrops don't hit like bullets, why drops are round and why water climbs up a thin tube. This is the second half of Mechanical Properties of Fluids, built up from zero to exam level.
Pressure, Pascal's law, buoyancy, continuity and Bernoulli are on Fluids I. Short on time? Read the formula cards, the red “Exam trap” boxes and the night-before checklist.
Viscosity: friction inside a fluid
Tip a jar of honey and a glass of water. Both are liquids, yet honey crawls and water rushes out. Stir both with a spoon: honey drags the spoon back far more. The property that makes the difference is viscosity.
Picture a flowing liquid as a stack of very thin sheets (layers) sliding over one another, like the pages of a book when you push the top cover sideways. The layer touching a solid surface sticks to it and does not move. Each layer above moves a little faster than the one below. Neighbouring layers rub against each other: the faster layer tries to drag the slower one forward, and the slower one tries to hold the faster one back. This internal friction between layers is viscosity.
This orderly layer-by-layer flow is called laminar (streamline) flow. Flow through a pipe is also laminar at low speed: the layer at the wall is at rest and the central layer is fastest.
Viscosity is the property of a fluid by which it opposes relative motion between its adjacent layers. The tangential force between layers is the viscous force. A fluid with zero viscosity (and incompressible) is called an ideal fluid.
Newton found experimentally that the viscous force \(F\) between two layers is proportional to the area of contact \(A\) and to how quickly the speed changes as you move from one layer to the next: the velocity gradient \(dv/dx\), where \(x\) is measured perpendicular to the flow.
Symbols and assumptions: F = signed viscous force (N), opposing relative layer motion; η = dynamic viscosity (Pa·s); A = sheared area (m²); v = flow velocity (m/s); x = distance across layers (m); dv/dx = velocity gradient (s⁻¹). Use |F| = ηA|dv/dx| for the magnitude.
Symbols and assumptions: v = moving-plate speed relative to the fixed plate (m/s); d = film thickness (m); x = perpendicular distance through the film (m); dv/dx = velocity gradient (s⁻¹). The v/d form assumes a linear velocity profile and no slip.
Symbols and assumptions: η = viscosity (Pa·s); F = viscous-force magnitude (N); A = plate area (m²); v = relative plate speed (m/s); d = film thickness (m). M, L and T in the dimensional expression mean mass, length and time.
In a solid, shear modulus = shear stress / shear strain. In a fluid, viscosity = shear stress / rate of shear strain. A fluid keeps deforming as long as the force acts, so the rate, not the strain, is what matters. (See shear modulus.)
Layers sliding over each other
Watch the purple dye line: it starts vertical and leans over, because higher layers move faster. Halve d and the gradient (and F) doubles.
Temperature and viscosity: the classic trap
Liquids: viscosity comes mainly from the attraction (cohesion) between neighbouring molecules. Heat a liquid and the molecules move apart and gain energy, so \(\eta\) decreases. Warm honey or engine oil flows easily.
Gases: molecules are far apart; viscosity comes from molecules darting between layers and carrying momentum with them. Hotter means faster molecules and more momentum transfer, so \(\eta\) increases with temperature.
| Fluid | Temperature | η (approx.) |
|---|---|---|
| Water | 20 °C | 1.0 × 10⁻³ Pa·s |
| Water | 100 °C | 0.3 × 10⁻³ Pa·s (falls) |
| Blood | 37 °C | ≈ 3 × 10⁻³ Pa·s |
| Castor oil / glycerine | 20 °C | ≈ 1 Pa·s |
| Air | 20 °C → 100 °C | 1.8 → 2.2 × 10⁻⁵ Pa·s (rises) |
- Gas viscosity increases with temperature. “Viscosity of gases decreases with temperature” is the incorrect statement (Q12). Liquids decrease, gases increase.
- Ideal fluid: η = 0 (non-viscous) and incompressible (Q42). Not “η = 1”, not “infinite”.
- For this test, choose a lubricant with sufficiently high viscosity (Q43; you chose low). A viscous oil film is not easily squeezed out from between moving parts, so the metal surfaces stay apart. An oil that is too thin may fail to maintain the film. In practice, viscosity must suit the load, speed and temperature; excessively viscous oil also increases resistance.
- Viscosity is a property of liquids and gases; solids do not have it.
Blood viscosity sets how hard the heart must pump; thicker blood means more work. Engine oil grades are viscosity grades. Shock absorbers and door closers use viscous oil to slow motion smoothly.
A metal plate of area 200 cm² rests on a 2 mm thick layer of glycerine (η = 8.5 poise) spread on a fixed table. What horizontal force keeps it moving at a steady 4 cm/s? (Set the simulation to these values to check.)
Show solution
- Convert to SI: A = 200 × 10⁻⁴ = 0.02 m², d = 2 × 10⁻³ m, v = 0.04 m/s, η = 8.5 poise = 0.85 Pa·s.
- Steady speed, so the applied force equals the viscous force: \(F = \eta A v/d\).
- \(F = 0.85 \times 0.02 \times 0.04 / 0.002 = 0.34\) N.
Answer: 0.34 N
A cube of side \(a\) and density \(\rho\) slides down an incline of angle \(\theta\) on a thin oil film of thickness \(t\), at constant speed \(v\). Find η. Then evaluate for a 10 cm wooden cube (ρ = 600 kg/m³), θ = 30°, t = 0.2 mm, v = 0.5 m/s.
Show solution
- Constant speed means zero net force along the incline: \(mg\sin\theta = \eta A \dfrac{v}{t}\).
- With \(m = \rho a^3\) and contact area \(A = a^2\): \(\rho a^3 g \sin\theta = \eta a^2 v/t\).
- \(\eta = \dfrac{\rho a g t \sin\theta}{v}\).
- Numbers: \(\eta = 600 \times 0.1 \times 9.8 \times 2\times10^{-4} \times 0.5 / 0.5 = 0.118\) Pa·s.
Answer: η = ρagt sinθ / v ≈ 0.12 Pa·s
A river is 8 m deep. Its surface water flows at 3 m/s and the water at the bed is at rest. Assuming the speed changes uniformly with depth, find the viscous stress between horizontal layers. (η of water = 10⁻³ Pa·s.)
Show solution
- Velocity gradient \(= 3/8 = 0.375\ \text{s}^{-1}\) (uniform, from 0 at the bed to 3 m/s at the top).
- Stress \(= F/A = \eta\, dv/dx = 10^{-3} \times 0.375\).
Answer: 3.75 × 10⁻⁴ N/m²
From your Test-06
Viscous flow through a pipe: Poiseuille's law
A tap needs a pressure difference to keep water moving through a long pipe. Viscosity transfers momentum between neighbouring layers, and the layer touching a stationary wall has zero speed in the no-slip model. The flow is fastest at the axis. For fully developed, steady laminar flow of an incompressible Newtonian fluid through a straight circular tube, the speed profile is parabolic.
Let r be the inner tube radius, ℓ its length and ΔP the inlet-to-outlet pressure drop for a horizontal tube. Do not confuse tube radius r with distance s from the axis.
Symbols and assumptions: Q = volume flow rate (m³/s); r = inner pipe radius (m); ΔP = pressure drop (Pa); η = viscosity (Pa·s); ℓ = pipe length (m). The pipe is horizontal, circular, straight and has fully developed laminar flow with no slip.
Symbols and assumptions: v(s) = axial speed (m/s) at radial distance s from the axis (m); r = pipe radius (m); ΔP = pressure drop (Pa); η = viscosity (Pa·s); ℓ = pipe length (m); v_max = axis speed; v̄ = section-average speed. At s = r, v = 0.
Radius matters to the fourth power
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In a constant-area horizontal pipe, speed may stay the same while pressure drops because viscosity dissipates mechanical energy. Ideal Bernoulli without a loss term cannot describe that pressure drop. Poiseuille's law does not apply to turbulent flow; a large predicted Reynolds number warns that the laminar prediction needs checking.
Water (η = 1.0 × 10⁻³ Pa·s) passes through a 1 m long tube of inner radius 1 mm under a 100 Pa pressure difference. Find Q in the laminar model.
Show solution
- \(Q=\pi(10^{-3})^4(100)/(8\times10^{-3}\times1)=3.93\times10^{-8}\) m³/s.
- 1 m³ = 10⁶ mL, so Q = 0.0393 mL/s.
- \(\bar v=Q/(\pi r^2)=0.0125\) m/s and \(R_e=\rho\bar v(2r)/\eta=25\), consistent with laminar flow.
0.0393 mL/s; mean speed 0.0125 m/s
This extension connects viscosity to real pipe flow. Reference: OpenStax: viscosity and turbulence.
Check your understanding
Stokes' law and terminal velocity
A raindrop falling from a cloud 1 km up would reach about 140 m/s if there were no air, fast enough to hurt badly. Real raindrops arrive at only a few metres per second. Why? As a body falls through a fluid, the fluid drags on it, and the drag grows with speed. Soon the drag balances the pull of gravity, the net force becomes zero, and the body falls at a constant speed: its terminal velocity.
For a small sphere moving slowly through a viscous fluid (laminar flow around it), Stokes showed that the drag is:
Symbols and assumptions: F_v = drag magnitude (N); η = liquid viscosity (Pa·s); r = sphere radius (m); v = sphere speed relative to the liquid (m/s). Drag opposes that relative motion. The sphere is rigid and sufficiently far from walls; the Reynolds number must be very small.
Symbols and assumptions: v_t = terminal settling speed (m/s); r = sphere radius (m); ρ = sphere density; σ = liquid density, both in kg/m³; g = gravitational acceleration (m/s²); η = viscosity (Pa·s). Here σ is a density, not stress or Poisson’s ratio. A negative result indicates upward motion; Stokes-flow assumptions must hold.
Where the formula comes from
Three forces act on a sphere of radius \(r\) falling through a fluid:
- Weight, down: \(W = \tfrac43\pi r^3 \rho g\)
- Buoyant force (upthrust), up: \(B = \tfrac43\pi r^3 \sigma g\)
- Viscous drag, opposite to motion (up): \(F_v = 6\pi\eta r v\), which starts at zero and grows as \(v\) grows
At first \(F_v = 0\), so the sphere accelerates. As \(v\) rises, \(F_v\) rises, the net force shrinks and the acceleration falls. When \(F_v = W - B\), the net force is zero and \(v\) stays fixed:
\[ 6\pi\eta r v_t = \tfrac43\pi r^3(\rho - \sigma)g \;\Rightarrow\; v_t = \frac{2r^2(\rho-\sigma)g}{9\eta} \]
Ball falling through a viscous liquid
Drop, then double r and drop again: the plateau rises 4×, not 2×. Make ρ smaller than σ to see the ball rise like an air bubble. The faint curves are your previous runs.
The v–t graph
Speed rises quickly at first, then levels off and approaches \(v_t\) without crossing it: \(v = v_t\,(1 - e^{-t/\tau})\) with \(\tau = 2\rho r^2/9\eta\). Acceleration starts at \(g(1-\sigma/\rho)\), not \(g\), because buoyancy acts from the start, and falls to zero. The distance–time graph curves at first and then becomes a straight line (constant speed). For small drops τ is only milliseconds, so they reach terminal velocity almost at once.
- \(v_t \propto r^2\), not \(r\) (Q24). The assertion “a bigger raindrop falls faster” is true; the reason “\(v_t \propto r\)” is false.
- \(v_t \propto r^2\) only when the density is fixed. If the mass is fixed (and buoyancy is ignored), \(mg = 6\pi\eta r v_t\) gives \(v_t \propto 1/r\). Same mass, double radius: speed halves (Q30).
- At a given speed, drag \(\propto r\). Since cross-section area \(A = \pi r^2\), drag \(\propto \sqrt{A}\), i.e. \(A^{1/2}\) (Q13).
Symbols and assumptions: F_v = upward viscous drag (N); W = weight (N); B = buoyant force (N), not bulk modulus; M = sphere mass (kg); r = sphere radius (m); ρ = sphere density; σ = liquid density (kg/m³); g = gravitational acceleration (m/s²). This form describes a sphere settling downward.
Symbols and assumptions: n = number of identical initial drops; r = each small radius; R = merged radius, in matching units; v = initial terminal speed; v′ = merged terminal speed (m/s). Densities and viscosity stay unchanged; both speeds must be in the Stokes regime.
Symbols and assumptions: v = steady upward speed (m/s); r = bubble radius (m); σ = solution density (kg/m³); g = gravitational acceleration (m/s²); η = solution viscosity (Pa·s). Air weight is neglected and the school model assumes Stokes-type drag; a clean mobile bubble interface can behave differently.
Drop a small steel ball into a tall jar of the liquid. Once it reaches its terminal velocity, time it between two marks to get \(v_t\), then use \(\eta = 2r^2(\rho-\sigma)g/9v_t\). Start timing well below the surface, after the ball has stopped speeding up.
A copper ball (8.9 g/cc) and an aluminium ball (2.7 g/cc) of the same radius fall through glycerine (1.3 g/cc). The copper ball's terminal speed is 0.38 m/s. Find the aluminium ball's.
Show solution
- Same r and η, so \(v_t \propto (\rho-\sigma)\). Do not use the density ratio itself.
- \(\dfrac{v_{Al}}{v_{Cu}} = \dfrac{2.7-1.3}{8.9-1.3} = \dfrac{1.4}{7.6}\).
- \(v_{Al} = 0.38 \times 1.4/7.6 = 0.07\) m/s.
Answer: 0.07 m/s
A steel ball of radius 2 mm (density 7800 kg/m³) reaches a terminal speed of 3 cm/s in an oil of density 900 kg/m³. Find the viscosity of the oil. (g = 9.8 m/s²)
Show solution
- \(\eta = \dfrac{2r^2(\rho-\sigma)g}{9v_t} = \dfrac{2 \times (2\times10^{-3})^2 \times 6900 \times 9.8}{9 \times 0.03}\).
- Numerator \(= 2 \times 4\times10^{-6} \times 6900 \times 9.8 = 0.541\); denominator \(= 0.27\).
Answer: η ≈ 2.0 Pa·s
(a) 27 identical rain droplets, each falling at terminal speed \(v\), merge into one drop. New terminal speed? (b) A ball of radius R and density 2σ falls in a liquid of density σ with terminal speed V. A ball of radius R/2 and density 5σ falls in the same liquid. Its terminal speed?
Show solution
- (a) \(R = 27^{1/3} r = 3r\). \(v_t \propto r^2\), so \(v' = 9v\).
- (b) \(v_t \propto r^2(\rho-\sigma)\). Old: \(R^2(2\sigma - \sigma) = R^2\sigma\). New: \((R/2)^2(5\sigma-\sigma) = R^2\sigma\). Same!
Answer: (a) 9v (b) V
Two spheres have the same mass, but one has three times the radius of the other. Ignoring buoyancy, compare their terminal speeds in the same liquid.
Show solution
- At terminal speed, \(mg = 6\pi\eta r v_t \Rightarrow v_t = mg/6\pi\eta r\).
- Same m, so \(v_t \propto 1/r\). The bigger sphere is less dense, which is why the \(r^2\) rule does not apply.
Answer: the larger sphere falls at one-third the speed
From your Test-06
Reynolds number: smooth or chaotic?
Open a tap a little: the stream is smooth and glassy. Open it fully: the water churns and turns white. Smoke from an incense stick rises in a smooth thread, then suddenly breaks into curls. Slow flow is laminar; fast flow becomes turbulent, full of eddies.
Which one you get depends on a contest between two effects. Inertia (ρ and v) keeps a moving lump of fluid going and lets disturbances grow. Viscosity (η) damps disturbances and keeps layers orderly. The Reynolds number measures the ratio of the two:
Symbols and assumptions: R_e = Reynolds number, dimensionless; ρ = fluid density (kg/m³); v = mean pipe speed (m/s); d = pipe diameter (m); η = viscosity (Pa·s). The force ratio represents a characteristic scaling, not a universal equality for every geometry.
Reynolds' dye experiment
A thin thread of dye is injected on the axis. At low Re it stays a straight line; above about 2000 it wobbles, breaks into eddies and mixes across the whole pipe.
“If the Reynolds number is greater than 2000, the flow is ___ and ___ forces dominate.” Answer: turbulent; inertial. A large \(R_e\) means inertia beats viscosity. Turbulence is more likely for fast flow, wide pipes, dense and less viscous fluids. Critical speed: \(v_c = R_e\,\eta/\rho d\).
Use the 1000/2000 convention for the questions in your coaching paper. These are approximate teaching ranges, not universal transition thresholds. The actual onset of turbulence depends on geometry and disturbances; follow the convention stated in a question.
(a) Water (ρ = 1000 kg/m³, η = 10⁻³ Pa·s) flows at 15 cm/s through a pipe of diameter 2 cm. What type of flow is it? (b) What is the largest speed for steady laminar flow (\(R_e\) = 1000) in a 1 cm pipe?
Show solution
- (a) \(R_e = \rho v d/\eta = 1000 \times 0.15 \times 0.02/10^{-3} = 3000 \gt 2000\), so the flow is turbulent.
- (b) \(v = R_e\eta/\rho d = 1000 \times 10^{-3}/(1000 \times 0.01) = 0.1\) m/s.
Answer: (a) turbulent (Re = 3000) (b) 10 cm/s
From your Test-06
Surface tension: a liquid's stretched skin
Lay a needle gently on water and it floats, even though steel is 8 times denser than water. Insects like water striders walk on ponds, and you can see the surface dimpling under their feet. Small drops are round. Wet paint-brush hairs spread out under water but clump together when you pull the brush out. All of these show that a liquid surface behaves like a stretched elastic skin.
9.6.1 · Surface energy: the molecular picture
Molecules in a liquid attract each other, but only over a short range (a few molecular diameters). A molecule deep inside is pulled equally in all directions, so the net force on it is zero. A molecule at the surface has neighbours below and beside it but almost none above, so it feels a net inward (downward) pull.
Bringing a molecule from inside up to the surface means working against this pull, and the molecule also loses some of its bonds. So surface molecules have extra energy compared with inside ones. A liquid therefore tends to keep its surface area as small as possible. For a given volume, a sphere has the least area, which is why drops are round.
Why surface molecules are pulled inward
Tap a highlighted molecule to select it, then tap anywhere in the liquid to move it (or drag with a mouse). The dashed circle is its range of attraction.
Surface tension \(T\) (or S) is the force per unit length acting on an imaginary line drawn on the liquid surface. The force is perpendicular to the line and tangent to the surface: \(T = F/L\). Equivalently, it is the surface energy per unit area (next section).
Symbols and assumptions: T = surface tension (N/m); F = tangential force (N) across a line of length L (m) on one liquid surface. The line lies in the surface; the force is perpendicular to that line. T here is not temperature.
Symbols and assumptions: F = force pulling the movable wire (N); T = surface tension (N/m); L = wire length in contact with one face of the film (m). Both faces contribute, giving the factor 2.
Soap film on a U-frame
The pull does not depend on how far out the slider is. Unlike a spring, the film just adds new surface; its tension stays T.
- Surface tension is a scalar (Q27). The force due to it has a direction (along the surface, perpendicular to the line), but T itself has no fixed direction. “Surface tension is a vector” is the wrong statement.
- Temperature: T decreases as temperature rises (cohesion weakens) and becomes zero at the critical temperature.
- Impurities: a highly soluble impurity such as NaCl raises T. Sparingly soluble ones such as soap or detergent lower T. Oil, grease or dust on the surface also lower it (Q26).
Minimum force to pull something off a liquid surface
When you lift an object off a liquid surface, the liquid clings to its edges and pulls down with \(T\) on every metre of edge the liquid touches. Count every contact line; a wire touched on both sides counts twice.
A solid disc has only its outer rim in contact: \(2\pi RT\). A disc with a hole has two rims: \(2\pi(R+r)T\). A wire frame is wetted on both sides of every wire: a square of side \(l\) gives \(8lT\), not \(4lT\). A needle of length \(l\) has a contact line on each side: \(2lT\).
A U-frame holds a water film. Its sliding wire is 5 cm long. The film just supports the slider plus a hanging weight, total 7.2 × 10⁻³ N. Find T.
Show solution
- The film has two surfaces: upward force \(= 2TL\).
- \(2T(0.05) = 7.2\times10^{-3} \Rightarrow T = 7.2\times10^{-3}/0.1\).
Answer: T = 0.072 N/m (if you forget the factor 2 you get 0.144, which is a common wrong option)
Show that the largest diameter of a steel needle that can float on water is \(d = \sqrt{8T/\pi\rho g}\), and evaluate it for ρ = 7800 kg/m³, T = 0.072 N/m. (Ignore buoyancy.)
Show solution
- Per unit length, weight \(= \dfrac{\pi d^2}{4}\rho g\).
- The maximum upward pull comes when the surface on both sides becomes vertical: \(2T\) per unit length.
- \(\dfrac{\pi d^2 \rho g}{4} = 2T \Rightarrow d = \sqrt{\dfrac{8T}{\pi\rho g}} = \sqrt{\dfrac{8 \times 0.072}{\pi \times 7800 \times 9.8}}\).
Answer: d ≈ 1.5 × 10⁻³ m = 1.5 mm
A thin disc of radius r floats on water. At its rim the water surface meets the disc making an angle θ with the vertical. The disc displaces water of weight W. What is the weight of the disc?
Show solution
- Upward forces: buoyancy W, plus surface tension along the rim. The rim length is \(2\pi r\), and the pull is along the surface, at θ to the vertical, so its vertical part is \(T\cos\theta\) per unit length.
- Equilibrium: weight \(= W + 2\pi r T\cos\theta\).
Answer: W + 2πrT cos θ
From your Test-06
Surface energy: making new surface costs work
Every surface molecule carries extra energy. To make more surface you must haul molecules up from inside, so you must do work. That work is stored as surface energy. Go back to the U-frame: pull the slider out by \(\Delta x\) against the force \(2TL\). The work is \(2TL\Delta x\), and the new area created is \(L\Delta x\) on each of the two faces, \(2L\Delta x\) in total. So
\[ W = T \times (\text{total new area}) \quad\Rightarrow\quad T = \frac{W}{\Delta A} \]
Surface tension is therefore also the surface energy per unit area. N/m and J/m² are the same unit.
Symbols and assumptions: W = work to increase surface area (J); T = constant surface tension (N/m); ΔA = increase in area of one face (m²). The total new area is 2ΔA.
Symbols and assumptions: W = increase in surface energy (J); T = surface tension (N/m); r₁, r₂ = initial/final bubble radii (m). This counts two surfaces of a thin soap film and assumes T stays constant.
Symbols and assumptions: E = surface energy (J); r = drop radius (m); T = surface tension (N/m). The drop has area 4πr² and one liquid–air interface; creating extra area increases E.
Symbols and assumptions: ΔE = positive energy released (J); n = number of equal small drops; r = each small radius (m); R = merged radius (m); T = surface tension (N/m). Volume conservation gives R³ = nr³; surface area decreases.
- When drops merge, total volume stays the same but total surface area decreases, so surface energy is liberated (the big drop warms slightly). Nothing about “mass is converted into energy”: mass is conserved here (Q7).
- Two equal drops of radius r merge: \(R = 2^{1/3}r\), energy released \(= T[2(4\pi r^2) - 4\pi R^2] = 4\pi r^2 T\,(2 - 2^{2/3})\) (Q15).
- Splitting one drop into many needs energy, since you are creating surface.
Energy released heats the big drop: \(\Delta\theta = \dfrac{3T}{\rho s}\left(\dfrac1r - \dfrac1R\right)\), where s is the specific heat (J kg⁻¹ K⁻¹).
A rectangular soap film is stretched from 8 cm × 5 cm to 8 cm × 10 cm. The work done is 5.76 × 10⁻⁴ J. Find T.
Show solution
- Change in area of one face: \(8\times10 - 8\times5 = 40\ \text{cm}^2 = 4\times10^{-3}\ \text{m}^2\).
- Two faces: total new area \(= 8\times10^{-3}\ \text{m}^2\).
- \(T = W/\Delta A_{total} = 5.76\times10^{-4}/8\times10^{-3}\).
Answer: T = 0.072 N/m
A soap bubble (T = 3 × 10⁻² N/m) is blown from radius 1 cm to radius 3 cm. Work done?
Show solution
- \(W = 2 \times 4\pi(r_2^2 - r_1^2) \times T = 8\pi T(r_2^2 - r_1^2)\).
- \(= 8\pi \times 0.03 \times (9 - 1)\times10^{-4} = 8\pi \times 0.03 \times 8\times10^{-4}\).
Answer: W ≈ 6.0 × 10⁻⁴ J (watch out: diameters in the question → halve them first)
A water drop of radius 1 mm (T = 0.072 N/m) is sprayed into 1000 identical droplets. How much work is needed?
Show solution
- \(r = R/1000^{1/3} = R/10\). New area \(= 1000 \times 4\pi (R/10)^2 = 10 \times 4\pi R^2\).
- Increase in area \(= 9 \times 4\pi R^2\), i.e. \(4\pi R^2(n^{1/3}-1)\).
- \(W = 9 \times 4\pi \times 10^{-6} \times 0.072 \approx 8.1\times10^{-6}\) J.
Answer: ≈ 8.1 × 10⁻⁶ J
From your Test-06
Angle of contact: does the liquid wet the solid?
Water dropped on clean glass spreads into a thin patch. Water on a lotus leaf or a waxy, greasy surface rolls into beads. Mercury on glass forms shiny balls. The difference is a tug-of-war between two attractions:
- Cohesion: liquid molecules attracting each other.
- Adhesion: liquid molecules attracting the solid.
If adhesion wins, the liquid spreads over the solid (it wets it). If cohesion wins, the liquid pulls itself into a ball.
The angle of contact θ is the angle between the tangent to the liquid surface at the point of contact and the solid surface, measured inside the liquid. It depends on the pair of materials (and on cleanliness), not on the amount of liquid.
Symbols and assumptions: S_la = liquid–air surface tension; S_sl = solid–liquid interfacial energy per area; S_sa = solid–air interfacial energy per area, all in N/m = J/m². θ = equilibrium contact angle measured through the liquid. This balance assumes a smooth equilibrium contact line.
Contact angle: drop shape and meniscus
| θ < 90° (acute) | θ = 90° | θ > 90° (obtuse) | |
|---|---|---|---|
| Attraction | Adhesion > cohesion | Balanced | Cohesion > adhesion |
| Wetting | Wets the solid | – | Does not wet |
| Meniscus in tube | Concave (curves down in the middle) | Flat | Convex (bulges up) |
| Capillary | Rises | No rise or fall | Falls (depression) |
| Example | Water–clean glass (≈ 0°) | Water–silver (often quoted) | Mercury–glass (≈ 140°) |
“The angle of contact for a non-wetting liquid is ___” → obtuse. “A convex meniscus means θ is ___” → greater than 90°. Concave ↔ acute ↔ rises. Convex ↔ obtuse ↔ falls.
Soaps and detergents lower surface tension and reduce θ, so water wets cloth and soaks into the tiny gaps where dirt sits; dirt and grease are then held in the water and washed away. Waterproofing agents do the opposite: they make θ large (obtuse), so water beads up and rolls off instead of soaking in.
Statement I: Adding detergent to water increases the angle of contact between water and cloth. Statement II: Detergents reduce the surface tension of water.
Show solution
- Detergent helps water wet the cloth, so θ decreases. Statement I is false.
- Detergent lowers T. Statement II is true.
Answer: Statement I false, Statement II true
From your Test-06
Drops and bubbles: excess pressure inside
A curved liquid surface acts like a stretched balloon skin and squeezes what is inside. So the pressure on the concave side (inside) of a curved surface is higher than on the convex side (outside). The more curved the surface, i.e. the smaller the radius, the bigger the squeeze. That is why a balloon is hardest to blow at the very start.
Derivation for a drop (energy method)
Let a drop of radius \(r\) have excess pressure \(P_{ex} = P_{in} - P_{out}\). Imagine it grows to \(r + \Delta r\):
- Work done by the excess pressure \(= P_{ex} \times \Delta V = P_{ex}\,4\pi r^2\Delta r\)
- Increase in surface energy \(= T\,\Delta A = T\,8\pi r\Delta r\)
Equating them gives \(P_{ex} = 2T/r\). A soap bubble has two surfaces, so its \(\Delta A\) doubles, giving \(4T/r\). An air bubble inside a liquid has only one surface (liquid–air), giving \(2T/r\). The simulation shows the same result by force balance on half the drop.
Symbols and assumptions: P_in, P_out = pressures inside/outside the drop (Pa); T = surface tension (N/m); r = spherical drop radius (m). Their difference is the excess pressure, not the full inside pressure.
Symbols and assumptions: P_in, P_out = inside/outside pressures (Pa); T = surface tension (N/m); r = bubble radius (m). A thin film has two surfaces, giving twice the drop’s excess pressure.
Symbols and assumptions: P_in = bubble inside pressure (Pa); P₀ = pressure on the liquid free surface (Pa); ρ = surrounding liquid density (kg/m³); g = gravitational acceleration (m/s²); h = bubble depth (m); T = liquid–air surface tension (N/m); r = bubble radius (m).
One surface or two? Excess pressure by force balance
Cut the object in half. The excess pressure pushes the right half to the right over area πr². Surface tension holds it back along the rim, 2πr long, once per surface.
| Liquid drop | Soap bubble (in air) | Air bubble (in liquid) | |
|---|---|---|---|
| Free surfaces | 1 | 2 | 1 |
| Excess pressure | 2T/r | 4T/r | 2T/r |
| Surface energy | 4πr²T | 8πr²T | 4πr²T |
| Work to grow r₁ → r₂ | 4πT(r₂² − r₁²) | 8πT(r₂² − r₁²) | 4πT(r₂² − r₁²) |
| Total pressure inside | P₀ + 2T/r | P₀ + 4T/r | P₀ + ρgh + 2T/r |
Two bubbles connected: the small one empties into the big one
Since \(P_{ex} = 4T/r\), the smaller bubble has the higher pressure. Connect two bubbles and air flows from small to large. The small bubble shrinks, its pressure rises further, and it keeps emptying. It stops only when it has become a shallow cap on its tube whose radius of curvature equals that of the big bubble.
Open the valve between two soap bubbles
T = 0.03 N/m. Notice the moment bubble 1 becomes a hemisphere: that is its smallest radius of curvature, so its pressure peaks there. After that it flattens into a cap.
- Q17 (A–R): “the bigger bubble grows and the smaller shrinks” is true, and “pressure inside is inversely proportional to radius” is the correct explanation.
- Q19: on equal tube mouths, a hemispherical bubble has the smallest possible radius of curvature (= the mouth radius), so it has the highest pressure. A less-than-hemisphere cap is flatter (bigger radius of curvature, lower pressure). Air flows from the hemisphere (1) to the other (2), and bubble 1 shrinks.
Symbols and assumptions: r = radius of curvature of the common film (m); r₁, r₂ = radii of the two bubbles (m), with r₁ < r₂; T = soap-film surface tension (N/m). For equal bubble radii, the separating film is flat, corresponding to infinite r.
Symbols and assumptions: n = amount of gas (mol); P = absolute inside pressure (Pa); V = gas volume (m³); R in PV/(RT_K) = universal gas constant 8.314 J/(mol·K); T_K = gas temperature (K); T = surface tension (N/m); r = bubble radius (m); P₀ = outside pressure (Pa). In the merger result, R instead denotes the final bubble radius: read each R in its local context.
An air bubble of radius 0.2 mm is 10 cm below the surface of water. The pressure inside it exceeds atmospheric pressure by 1700 Pa. Find T. (ρ = 1000 kg/m³, g = 10 m/s²)
Show solution
- \(P_{in} - P_0 = \rho g h + 2T/r\).
- \(\rho g h = 1000 \times 10 \times 0.1 = 1000\) Pa, so \(2T/r = 700\) Pa.
- \(T = 700 \times 2\times10^{-4}/2 = 0.07\) N/m.
Answer: T = 0.07 N/m
Soap bubbles A (radius 1 cm) and B (radius 2 cm) are inside a chamber kept at 4 N/m². T = 0.02 N/m, same temperature. Find \(n_B/n_A\).
Show solution
- \(P_A = 4 + \dfrac{4(0.02)}{0.01} = 12\) N/m². \(P_B = 4 + \dfrac{4(0.02)}{0.02} = 8\) N/m².
- \(n \propto PV \propto Pr^3\): \(\dfrac{n_B}{n_A} = \dfrac{8 \times 2^3}{12 \times 1^3} = \dfrac{64}{12}\).
Answer: 16/3 ≈ 5.3
Two soap bubbles of radii 4 cm and 6 cm stick together. Radius of curvature of the common film?
Show solution
- \(r = r_1r_2/(r_2 - r_1) = 24/2\).
Answer: 12 cm, bulging into the 6 cm bubble
A liquid drop of density ρ floats half-immersed in another liquid of density σ. The surface tension of the drop's liquid is T. Find the radius of the drop.
Show solution
- Weight (down): \(\tfrac43\pi r^3\rho g\).
- Buoyancy on the immersed hemisphere (up): \(\tfrac23\pi r^3\sigma g\).
- Surface tension acts along the waterline circle at the equator (length \(2\pi r\)) and pulls vertically up: \(2\pi r T\).
- \(\tfrac43\pi r^3\rho g = \tfrac23\pi r^3\sigma g + 2\pi rT \Rightarrow r^2 g\,\dfrac{2(2\rho-\sigma)}{3} = 2T\).
Answer: \(r = \sqrt{\dfrac{3T}{g(2\rho - \sigma)}}\)
From your Test-06
Capillary rise: water climbing a thin tube
Dip the corner of a paper towel in water and the water climbs up the towel. A lamp wick lifts oil to the flame. Blotting paper drinks ink. A sugar cube sucks up tea. Water rises through the tiny gaps in soil to plant roots. Each of these is a network of very thin tubes, and liquid rises in thin tubes. This is capillarity (capilla = hair, as in hair-thin).
Why it rises: excess pressure again
Water wets glass, so inside a thin glass tube its surface (meniscus) is concave, curving down in the middle. Just below a concave surface the pressure is lower than just above it (above is the inside of the curve, below is the outside). So:
- Just above the meniscus the pressure is atmospheric, \(P_0\).
- Just below the meniscus it is \(P_0 - 2T/R\), where R is the radius of the meniscus.
- Outside the tube, at the flat water surface, the pressure is \(P_0\). Liquid is pushed up the tube until the column's weight makes up the difference: \(h\rho g = 2T/R\).
- Geometry: the meniscus meets the wall at θ, so \(R = r/\cos\theta\) (r = inner radius of the tube). For water–glass, θ ≈ 0 and R = r (a hemisphere).
Symbols and assumptions: h = signed vertical rise above the outside liquid level (m); T = surface tension (N/m); θ = contact angle through the liquid; r = tube inner radius (m); ρ = liquid density (kg/m³); g = gravitational acceleration (m/s²). Negative h means capillary depression.
Symbols and assumptions: r = inner tube radius (m); h = vertical rise (m); ρ = liquid density (kg/m³); g = gravitational acceleration (m/s²); T = surface tension (N/m); θ = contact angle through the liquid. The left side is the raised column’s weight; the right is the upward component of force around its rim.
Symbols and assumptions: T = inferred surface tension (N/m); r = measured inner radius (m); h = vertical rise to the meniscus bottom (m); ρ = liquid density (kg/m³); g = gravitational acceleration (m/s²); θ = contact angle. The h + r/3 correction assumes a nearly hemispherical meniscus at θ ≈ 0.
Capillary tube lab
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- Depends on the inner radius of the tube, not its mass, wall thickness or outer radius (Q16; you chose mass of tube).
- Tilted tube: the vertical height stays h. The length of liquid along the tube is \(l = h/\cos\alpha\), with α measured from the vertical. At 60°, \(l = 2h\) (Q9: 2 cm → 4 cm). If the angle is given from the horizontal, use \(l = h/\sin\).
- Tube shorter than h: the liquid rises to the top but does not overflow. The meniscus becomes flatter (its radius grows) so that \(hR\) = constant: \(R' = Rh/l\) (Q14). No perpetual-motion fountain.
- Free fall / satellite: effective g = 0, so the liquid rises and fills the whole tube, whatever its length (Q10: fills all 30 cm). On the Moon (g/6) the rise is 6h if the tube is long enough. In a lift accelerating up, \(g_{eff} = g + a\) and h decreases.
- Mass raised ∝ r: double the radius, double the mass, even though h halves (Q34: 2M).
Water (θ = 0) rising in a capillary of radius 0.5 mm holds up a column weighing 2.26 × 10⁻⁴ N. Find T.
Show solution
- Weight held = upward surface-tension force \(= 2\pi r T\cos\theta = 2\pi r T\).
- \(T = 2.26\times10^{-4}/(2\pi \times 5\times10^{-4}) = 2.26\times10^{-4}/3.14\times10^{-3}\).
Answer: T ≈ 0.072 N/m
Water rises 3 cm in a vertical capillary. The tube is tilted to 53° from the vertical (cos 53° ≈ 0.6). (a) Vertical height of the water column? (b) Length of tube wetted?
Show solution
- (a) The pressure balance involves only the vertical height, so it is still 3 cm.
- (b) \(l = h/\cos\alpha = 3/0.6 = 5\) cm.
Answer: (a) 3 cm (b) 5 cm
In the capillary rise method, water rises 6.0 cm in a clean glass tube of bore radius 0.24 mm. Find T (θ = 0, g = 9.8 m/s²).
Show solution
- \(T = rh\rho g/2 = 0.24\times10^{-3} \times 0.06 \times 1000 \times 9.8/2\).
Answer: T ≈ 0.071 N/m
Water would rise 8 cm in a capillary, but only 5 cm of the tube sticks out above the water. What happens?
Show solution
- The water reaches the top of the tube and stops; it cannot overflow, because once it reaches the top the meniscus simply changes shape.
- \(h \propto 1/R\) (meniscus radius), so \(R' = R \times 8/5 = 1.6R\). The meniscus flattens.
Answer: rises to the top (5 cm), no overflow; meniscus radius becomes 1.6 times
From your Test-06
Determining surface tension by capillary rise
Try it in the virtual lab, with the effect of detergents (Experiment 6) →
The syllabus specifically asks for this method. Knowing that a narrow tube makes water rise is the starting point; the experiment measures the rise and the inner radius to infer surface tension.
For a long, vertical, clean tube with a nearly circular bore, balance upward force \(2\pi rT\cos\theta\) against the weight \(\pi r^2h\rho g\) of the raised column. For clean water–glass, take \(\theta\approx0\) unless the question states otherwise.
Symbols and assumptions: T = inferred surface tension (N/m); ρ = liquid density (kg/m³); g = gravitational acceleration (m/s²); r = inner tube radius (m); h = vertical rise above the outside liquid surface (m); θ = contact angle measured through the liquid.
Apparatus and procedure
- Use a clean capillary held vertically in a beaker of clean water, a travelling microscope or suitable height scale, and a means of measuring the inner bore. Let the water settle and keep temperature approximately constant.
- Read the position of the bottom of the meniscus inside the capillary, keeping the line of sight level to avoid parallax.
- Read the outside water level using a pointer or the microscope. Subtract the two readings to obtain h; immersion depth below the outside level is not the capillary rise.
- Measure the inner diameter in two perpendicular directions at a suitable tube end. If the bore is nearly circular, average the readings and divide by two for r.
- Convert all lengths to metres and calculate T. Repeat height readings and report the measured temperature and assumptions.
From two height readings to surface tension
This calculator infers T from measured r and h; its two sliders are independent measurements. It assumes θ = 0, ρ = 1000 kg/m³, g = 9.8 m/s² and neglects the meniscus correction. Tube width is exaggerated for visibility.
Errors, precautions and the small correction
For \(T=\rho grh/2\), if ρ and g are treated as exact, the first-order maximum fractional uncertainty is \(\Delta T/T\approx\Delta r/r+\Delta h/h\). Independent random uncertainties would normally combine differently; the sum is the school-level worst-case estimate.
Grease and impurities can change both surface tension and contact angle. A tilted tube gives a length along the tube rather than the vertical height. A tube shorter than its natural rise requires a changed meniscus, so it is unsuitable for the simple measurement.
For a nearly hemispherical meniscus at θ ≈ 0, the small extra liquid volume gives the refined expression \(T\approx\rho gr(h+r/3)/2\) when h is measured to the meniscus bottom. Use this correction only when requested; do not mix the corrected and uncorrected definitions of height.
Measured r = 0.24 mm and h = 6.0 cm for clean water–glass. Estimate T without the small correction. Radius and height uncertainties are 2% and 3%; treat ρ = 1000 kg/m³ and g = 9.8 m/s² as exact.
Show solution
- \(T=1000\times9.8\times0.24\times10^{-3}\times0.060/2=0.07056\) N/m.
- Maximum percentage uncertainty = 2 + 3 = 5%.
- Absolute uncertainty ≈ 0.05 × 0.07056 = 0.00353 N/m.
T ≈ (0.071 ± 0.004) N/m, with a 5% worst-case uncertainty
Check your understanding
Problem patterns & shortcuts
Almost every question in this half of the chapter is one of the patterns below. Spot the pattern, write the proportionality, and the numbers usually fall out in one line.
1 · Plate on a film / block on an incline
\(F = \eta A v/d\). On an incline at constant v: \(mg\sin\theta = \eta A v/t\). Convert poise → Pa·s (÷10) and cm² → m² (×10⁻⁴).
2 · Terminal velocity ratios
\(v_t \propto r^2(\rho-\sigma)/\eta\). Same ball in two liquids, or two balls in one liquid: use \((\rho-\sigma)\), never ρ alone.
3 · Merge / break drops (speed)
n drops → one: \(v \times n^{2/3}\) (8 → ×4, 27 → ×9, 64 → ×16). One → n: \(v/n^{2/3}\).
4 · Same mass, different size
No buoyancy: \(v_t = mg/6\pi\eta r \propto 1/r\). Double r → half v.
5 · Free fall, then enters liquid (Q22)
“Speed does not change on entering” means it enters at \(v_t\): \(\sqrt{2gh} = v_t \Rightarrow h = v_t^2/2g\). E.g. a 1 mm ball, ρ = 10.5 × 10³ in water: \(v_t \approx 20.7\) m/s, h ≈ 22 m.
6 · Viscous force at terminal speed (Q45)
\(F_v = W - B = Mg(1 - \sigma/\rho)\). Liquid half as dense as the ball: Mg/2.
7 · Film work: count two faces
\(W = 2T\Delta A\) for a film, \(8\pi T(r_2^2-r_1^2)\) for a bubble. Diameters given? Halve them first.
8 · Merging drops: energy
Released \(= 4\pi r^2 T\,n(1-n^{-1/3}) = 4\pi R^2T(n^{1/3}-1)\). Two drops: \(4\pi r^2T(2-2^{2/3})\). Temperature rise: \(\frac{3T}{\rho s}(\frac1r - \frac1R)\).
9 · Pull-off force
Needle 2lT · disc 2πRT · disc with hole 2π(R + r)T · thin ring 4πRT · square frame 8lT. Add the object's weight if asked for total force.
10 · Excess pressure
Drop and air bubble 2T/r, soap bubble 4T/r. At depth: \(P_0 + \rho gh + 2T/r\). Smaller bubble → higher pressure.
11 · Bubble combos
Interface \(r_1r_2/(r_2-r_1)\). Moles \(\propto (P_0 + 4T/r)r^3\). Merge in vacuum (isothermal): \(R^2 = r_1^2+r_2^2\).
12 · Capillary ratios
\(h \propto T\cos\theta/r\rho g\); \(hr\) = const; mass raised ∝ r; weight lifted \(= 2\pi rT\cos\theta\); tilted \(l = h/\cos\alpha\); g → 0 fills the tube; tube too short means no overflow.
| Quantity | Formula | SI unit | Dimensions |
|---|---|---|---|
| Coefficient of viscosity η | F d / (A v) | Pa·s = N s m⁻² | [M L⁻¹ T⁻¹] |
| Velocity gradient | dv/dx | s⁻¹ | [T⁻¹] |
| Reynolds number | ρvd/η | none | [M⁰ L⁰ T⁰] |
| Surface tension T | F/L = W/ΔA | N m⁻¹ = J m⁻² | [M T⁻²] |
| Surface energy | T × A | J | [M L² T⁻²] |
Assume the drag on a sphere depends on η, r and v as \(F = k\,\eta^a r^b v^c\). Find a, b, c.
Show solution
- \([\mathrm{MLT^{-2}}] = [\mathrm{ML^{-1}T^{-1}}]^a[\mathrm{L}]^b[\mathrm{LT^{-1}}]^c\).
- M: a = 1. T: −a − c = −2 ⇒ c = 1. L: −a + b + c = 1 ⇒ b = 1.
Answer: F = k η r v (Stokes found k = 6π)
Night-before checklist
- \(F = \eta A\,dv/dx\); η in Pa·s, 1 poise = 0.1 Pa·s; [ML⁻¹T⁻¹].
- Liquids: η ↓ with T. Gases: η ↑ with T.
- Ideal fluid η = 0. For Test-06, a lubricant needs enough viscosity to maintain its film; practical choice depends on operating conditions.
- Poiseuille: \(Q=\pi r^4\Delta P/(8\eta\ell)\), fully developed laminar flow; axis speed = 2 × mean speed.
- Stokes \(F = 6\pi\eta rv\); drag ∝ r ∝ √A, for creeping flow around a sphere.
- \(v_t = 2r^2(\rho-\sigma)g/9\eta\): ∝ r², ∝ (ρ − σ). Same mass: ∝ 1/r.
- n drops merge: v × n^{2/3}. At v_t: F_v = W − B.
- Re = ρvd/η. Your coaching convention: < 1000 laminar, > 2000 turbulent with inertia dominant; thresholds depend on the flow setup.
- T = F/L, N/m, [MT⁻²], scalar; T ↓ with temperature, 0 at critical temperature.
- NaCl ↑ T; soap/detergent ↓ T.
- Film: 2 surfaces → F = 2TL, W = 2TΔA. Frame side l: 8lT.
- Merging drops release energy (not mass → energy).
- Excess P: drop 2T/r, soap bubble 4T/r, air bubble 2T/r (+ρgh at depth). Small bubble empties into big one.
- θ < 90°: wets, concave, rises. θ > 90°: convex, falls. Hg–glass ≈ 140°.
- \(h = 2T\cos\theta/r\rho g\), inner r; hr const; mass ∝ r; tilted l = h/cos α; short tube: the meniscus changes shape. In ideal free fall, the gravity limit disappears; finite-tube behaviour depends on wetting and boundaries.
- Capillary experiment: \(T=\rho grh/2\) for θ ≈ 0; use the inner radius and vertical rise. Maximum relative errors in r and h add.
More practice: all Test-06 questions · quick formulas: Revise · first half of the chapter: Fluids I.