Mechanical Properties of Fluids — NEET Notes

Pressure in fluids

What separates a fluid from a solid? A solid resists shear — apply a tangential force to a brick and it deforms only slightly before snapping back when released. A fluid, by contrast, offers almost no resistance to shear: pour honey onto a tilted plate and it flows, however slowly. NCERT puts a number on this distinction — the shearing stress required to deform a fluid is about a million times smaller than the equivalent stress for a solid. Liquids and gases differ from each other only in compressibility: liquids have a definite volume that hardly changes with pressure, while gases fill whatever container they are given and respond strongly to pressure changes. Both flow, and both are fluids.

When a body is submerged in a fluid at rest, the fluid pushes against every part of the body's surface — and the push is always perpendicular to the surface. NCERT supplies a neat argument for why: if any component of the force were parallel to the surface, Newton's third law would demand an equal parallel push from the body on the fluid, which would set the fluid in motion. Since the fluid is at rest, no such tangential force can exist. The normal force per unit area is what we call pressure. A simple measuring device makes this concrete: an evacuated chamber fitted with a calibrated spring-loaded piston of area A, dropped into the fluid. The inward force F on the piston, balanced by the outward spring force, divided by A, gives the average pressure at that point. In the limit ΔA → 0, the limit ΔF/ΔA defines pressure rigorously at a single point.

P = F / A  ·  SI unit: 1 Pa = 1 N m⁻²

Pressure — normal force per unit area, named after Blaise Pascal

Pressure is a scalar, not a vector. The numerator F refers to the component of force normal to the surface; once we have taken that component, no direction information remains. The dimensions of pressure are [ML⁻¹T⁻²]. Useful conversions: 1 atm = 1.013 × 10⁵ Pa, 1 bar = 10⁵ Pa, 1 torr = 1 mm Hg = 133 Pa. The density of a fluid, often paired with pressure in the same equations, is ρ = m/V, with SI unit kg m⁻³ and dimensions [ML⁻³]. Water at 4 °C has ρ = 1.0 × 10³ kg m⁻³; this is the reference density for all relative-density measurements. The relative density of aluminium, for instance, is 2.7 — meaning aluminium is 2.7 times as dense as water. Densities of liquids depend only weakly on pressure (they are nearly incompressible); densities of gases depend strongly on it.

A worked NCERT example bolts down the intuition. The two femurs of a 40 kg human, each with a 10 cm² cross-section, together carry a downward force of about 400 N. The pressure they sustain is therefore P = 400 N / (20 × 10⁻⁴ m²) = 2 × 10⁵ Pa — roughly twice atmospheric. This is why a sharp needle, concentrating the same force on a tiny area, pierces skin that a blunt object cannot. Both force and contact area matter; the ratio is what the body actually feels.

Atmospheric pressure — Torricelli and the manometer

The atmosphere itself exerts pressure on everything beneath it — the weight of a column of air of unit cross-section, extending from sea level to the top of the atmosphere. At sea level this column weighs in at 1.013 × 10⁵ Pa. Evangelista Torricelli, in 1643, devised the first instrument to measure this directly. A long glass tube, closed at one end, is filled completely with mercury and inverted into an open trough of mercury. Some mercury falls out, leaving an almost perfect vacuum above the column. The atmospheric pressure pressing on the open mercury surface supports the column — and at sea level it supports a height of 76 cm. Applying P = ρgh with ρ = 13.6 × 10³ kg m⁻³ recovers 1.013 × 10⁵ Pa exactly. The space above the mercury — the "Torricellian vacuum" — contains only a trace of mercury vapour at negligible pressure; that vacuum is essential, because any gas above the column would press the mercury down and ruin the measurement.

An open-tube manometer measures pressure differences. A U-tube containing a liquid — oil for small differences, mercury for large — has one arm open to the atmosphere and the other connected to the vessel under test. The difference in liquid heights between the two arms, multiplied by ρg, gives the gauge pressure: the excess of the system's pressure over atmospheric. Most everyday pressure instruments — tyre gauges, sphygmomanometers, fuel-tank gauges — measure gauge pressure, not absolute. A tyre at "32 psi" is in fact at roughly 46 psi absolute, the extra 14.7 psi being contributed by the atmosphere itself.

How high does the atmosphere extend? If air had a uniform density of 1.29 kg m⁻³ (its sea-level value), Eq. ρgh = 1.01 × 10⁵ Pa would give h ≈ 7989 m, or about 8 km. The real atmosphere is denser near the surface and thinner higher up, so the cover actually extends well over 100 km — but most of the mass sits in the lowest 10 km. Atmospheric pressure varies with altitude, weather, and season: a barometer reading is one of the oldest weather instruments because a falling reading correlates so reliably with storms moving in.

Pascal's law & hydraulic machines

Pascal observed that pressure in a fluid at rest is the same at all points at the same height. The proof in NCERT considers a small prismatic element of fluid in equilibrium: balancing the normal forces on its three faces — using the geometric relations among the face areas — gives P_a = P_b = P_c, regardless of orientation. The pressure cannot favour any direction; if it did, the fluid would not be in equilibrium.

There is a second, operational form of the same law: any change in pressure applied to an enclosed fluid is transmitted undiminished and equally in all directions to every point of the fluid and the walls of the container. This is the form that drives every hydraulic machine on the planet.

A hydraulic lift exploits this transmission. Two pistons, of areas A₁ (small) and A₂ (large), are connected by an incompressible fluid. A force F₁ applied to the smaller piston creates a pressure P = F₁/A₁. By Pascal's law this pressure shows up unchanged at the larger piston, producing an upward force F₂ = PA₂ = F₁ · A₂/A₁. The mechanical advantage is the area ratio A₂/A₁: a force of 1.5 kN at a 5 cm piston can lift a 1350 kg car at a 15 cm piston (Example 9.6 of NCERT). Hydraulic brakes work identically — a light pedal-push at the master cylinder transmits undiminished pressure to the four wheel cylinders, distributing the braking force equally to all four wheels.

There is no free energy here — Pascal's law magnifies force, not work. Because the fluid is incompressible, the volume displaced at the small piston equals the volume displaced at the large piston: A₁d₁ = A₂d₂. So if the small piston travels d₁ = 6 cm, a piston nine times larger (A₂/A₁ = 9) advances only d₂ = 6 cm / 9 ≈ 0.67 cm. You trade distance for force — exactly as you do with a lever or a pulley. The product F · d (work) is conserved; only the force is amplified. A jack lifts a tonne with the force of a hand because each pump of the handle moves the load a fraction of a millimetre.

Variation of pressure with depth

Consider a cylindrical fluid element of base area A, extending from depth h₂ to h₁ below the surface, with h = h₂ − h₁ > 0. Vertical equilibrium of this cylinder demands that the upward force at the bottom face exceed the downward force at the top by exactly the weight of the column. Writing the weight as ρ · Ah · g and dividing through by A gives P₂ − P₁ = ρgh. Setting point 1 at the open surface (where P = P₀, atmospheric), the pressure at depth h below the surface is

P = P₀ + ρgh

Hydrostatic pressure — the deeper you go, the more pressure

The cross-sectional area A does not appear in this equation — the pressure at a given depth depends only on the vertical height of fluid above it, never on the container's shape or its base area. This is the hydrostatic paradox: three vessels of utterly different shapes, connected at the bottom, will fill to the same level when poured with water, even though they hold vastly different volumes. The water at the bottom doesn't care what shape the container is — it only feels the column of water above each section.

Two NCERT examples crystallise the size of the effect. At a depth of 10 m in a lake, P ≈ 2 atm — a 100% rise from the surface. At 1 km below the ocean surface, the absolute pressure climbs to about 104 atm, and on a 20 cm × 20 cm submarine window the net force is roughly 4.12 × 10⁵ N. The same equation explains why blood pressure in your feet is higher than in your brain — the standing column of blood adds a ρgh that the heart must overcome. NEET 2017 used the depth equation on a U-tube with oil and water: balancing the pressure at the common interface, ρ_oil g (140 mm) = ρ_water g (130 mm), gave ρ_oil = 1000 × 13/14 ≈ 928 kg m⁻³.

Archimedes' principle & buoyancy

The pressure at the bottom face of a submerged body exceeds the pressure at its top face by ρ_fluid · g · h, where h is the body's vertical extent. The net upward force from this pressure imbalance is exactly ρ_fluid · V_displaced · g — the weight of the fluid displaced. This is Archimedes' principle, and it follows directly from the hydrostatic depth equation. A body floats when this buoyant force balances its own weight; it sinks if its weight exceeds the maximum possible buoyant force (i.e. when the body's average density exceeds the fluid's).

For a partially submerged body, only the fraction of the volume below the waterline contributes to the displaced volume — and that fraction adjusts itself until buoyancy exactly balances weight. The volume-fraction that floats above the water is therefore (1 − ρ_body/ρ_fluid), which is why nearly 90% of an iceberg lies below the waterline (ρ_ice/ρ_seawater ≈ 0.9). The buoyant force acts at the centre of buoyancy — the centre of mass of the displaced fluid, not of the body itself — and the gap between centre of buoyancy and centre of gravity determines the stability of a floating object such as a ship.

NEET 2016 used exactly this idea: a cylinder of length L and density d floats between two non-mixing liquids of densities ρ and nρ, with length pL submerged in the denser liquid. Vertical equilibrium gives LAdg = (1 − p)LAρg + pLA(nρ)g, which solves to d = ρ {1 + (n − 1)p}. The principle is also the basis of the hydrometer (which measures liquid density by how deep it sinks) and the lactometer (which measures milk's relative density to detect adulteration).

Streamline vs turbulent flow

Set fluids in motion and a richer set of behaviours appears. A water tap turned on gently produces a smooth, glassy stream; open it further and the stream breaks into eddies and froth. The smooth regime is streamline flow (also called laminar): at any fixed point the fluid velocity is constant in time, and every fluid particle that passes that point follows the same trajectory afterwards. The paths traced out are streamlines, defined as curves whose tangent at every point matches the fluid's velocity at that point. No two streamlines can ever cross — if they did, a fluid particle arriving at the intersection would have two possible velocities, which violates steady flow. Steady flow does not mean that all particles have the same velocity, only that the velocity at any fixed point in space does not change with time; a particle accelerating from a wide section into a narrow nozzle is still part of a steady flow.

Above a critical speed the steady picture breaks down. Velocities at a point fluctuate randomly in time, energy is dissipated to eddies, and the flow is turbulent. Between the two regimes lies a transitional band in which both behaviours coexist. The three regimes are catalogued below by the Reynolds number Re = ρvD/η — the dimensionless ratio of inertial forces to viscous forces, with D the pipe diameter.

Equation of continuity

For an incompressible fluid in steady flow, the mass crossing any cross-section per unit time is the same everywhere along the pipe — mass cannot accumulate or disappear. Consider three sections P, R, Q of a pipe with cross-sectional areas A_P, A_R, A_Q and fluid speeds v_P, v_R, v_Q. In a small interval Δt, the mass crossing each section is ρAvΔt; mass conservation demands ρ_PA_Pv_PΔt = ρ_RA_Rv_RΔt = ρ_QA_Qv_QΔt. For an incompressible fluid the density is the same everywhere, and the equation collapses to A_Pv_P = A_Rv_R = A_Qv_Q. If A is the cross-sectional area and v the speed at that section, then ρAv = constant along the streamlines; for an incompressible fluid ρ is itself constant, so

A v = constant

Equation of continuity — mass conservation for incompressible flow

The quantity Av is the volume flux (m³ s⁻¹). The narrower a pipe, the faster the fluid: you see this every time you cover part of a hose nozzle with your thumb and watch the water jet farther. Streamlines drawn close together signify high speed; streamlines spread out signify low speed.

Bernoulli's principle

Daniel Bernoulli's 1738 result is the chapter's central equation. It relates pressure, speed, and height along a streamline for an ideal fluid — incompressible, non-viscous, and in steady flow. The derivation is the work-energy theorem applied to a moving slug of fluid: pressure does net work on the slug, that work goes into changing the slug's kinetic energy and potential energy, and the result is a conservation law that holds along each streamline.

The picture: consider a tube of varying cross-section with a fluid flowing through it. At one end the cross-section is A₁, fluid speed v₁, height h₁, pressure P₁; at the other, A₂, v₂, h₂, P₂. In a small time Δt, a slug of length v₁Δt sweeps through the inlet and an equal volume ΔV = A₁v₁Δt = A₂v₂Δt sweeps past the outlet (by continuity, since the fluid is incompressible). The slug at the inlet is pushed forward by pressure P₁ doing positive work P₁A₁(v₁Δt) = P₁ΔV; the slug at the outlet is pushed back by pressure P₂ doing negative work −P₂ΔV. Net work on the fluid: (P₁ − P₂)ΔV. This must equal the change in the slug's kinetic plus potential energy, which is ½ρΔV(v₂² − v₁²) + ρgΔV(h₂ − h₁). Dividing through by ΔV and rearranging gives Bernoulli's equation in symmetric form.

P + ½ ρv² + ρgh = constant

Bernoulli's equation — along a streamline, ideal fluid

Read it slowly. The three terms are pressure energy per unit volume, kinetic energy per unit volume, and gravitational potential energy per unit volume. Together they form the total mechanical energy of a unit volume of fluid, and Bernoulli's claim is that this total is conserved along the flow. When the fluid is at rest (v = 0), the equation collapses to P + ρgh = constant — exactly the hydrostatic depth equation. Bernoulli's equation, then, is hydrostatics with a kinetic-energy term tacked on. Equivalently, the units of each term — pascal — are the natural units of energy density (J m⁻³ = N m⁻²), which is why pressure can be combined directly with kinetic-energy density and gravitational-energy density in a single equation.

Bernoulli's equation has limits. It assumes the fluid has zero viscosity (no internal friction), is incompressible (so elastic energy is not stored), and flows steadily (no time-fluctuating pressures). In real fluids viscosity dissipates energy to heat, so the downstream P₂ falls below the Bernoulli prediction. For turbulent flow the equation fails altogether — the velocity at a point is no longer single-valued, so the relation P + ½ρv² + ρgh = constant has no clean meaning.

A worked numerical illustrates the size of the lift. A fully loaded Boeing of mass 3.3 × 10⁵ kg with 500 m² of wing must generate an upward pressure difference of ΔP = mg/A ≈ 6.5 × 10³ N m⁻². At a cruising speed of 960 km h⁻¹ ≈ 267 m s⁻¹ through air of density 1.2 kg m⁻³, the required speed difference between upper and lower wing surfaces is only about 8% of v_avg — a small fractional change in airflow yielding 330 tonnes of lift. The shape of the wing does the rest of the work.

Bernoulli applications — aerofoils, atomisers, Venturi-meters, spinning balls

The applications all follow from the same rule — wherever you can engineer a difference in flow speed, you create a difference in pressure.

One more application worth knowing: Torricelli's law of efflux. A small hole punched in the side of a tank at depth h below the free surface gives a jet of speed v = √(2gh) — exactly the speed a body in free fall would acquire after falling through h. This too is just Bernoulli applied at the free surface and the hole. The derivation: take point 1 at the hole (speed v₁, pressure P_atm) and point 2 at the open surface (speed v₂ ≈ 0 if the tank cross-section ≫ hole area, pressure P_atm). Bernoulli gives ½ρv₁² = ρgh, hence v₁ = √(2gh). If the tank is sealed and pressurised above the liquid (as in a rocket fuel tank), the speed of efflux is determined by the container pressure: v = √(2(P − P_atm)/ρ), and the gravitational term becomes negligible. This is how rocket exhaust works.

Viscosity & Stokes' law

Real fluids resist relative motion between their layers — an internal friction that NCERT calls viscosity. Imagine oil sandwiched between two parallel plates, the bottom plate fixed and the top plate dragged at velocity v. The fluid layer touching the moving plate moves at v, the layer touching the fixed plate is at rest, and intermediate layers move at intermediate velocities. Each layer experiences a forward pull from the layer above and a backward pull from the layer below — those tangential drags are viscous forces. The fluid in contact with a solid surface always moves at the same velocity as the surface; this "no-slip" condition is empirical, but universal for ordinary fluids on ordinary solids.

Inside a pipe, the same idea produces a parabolic velocity profile: the fluid touching the walls is stationary, the fluid at the centre moves fastest, and every concentric cylindrical shell in between moves at an intermediate speed. The streamlines are still parallel, but the speeds are different at different radii. This is what makes blood flow in arteries so layered, and why the central column of a slow stream moves fastest while the edges crawl.

The coefficient of viscosity η (eta) is defined as the ratio of shearing stress to strain rate:

η = (F/A) ÷ (v/ℓ)  ·  SI unit: N s m⁻² (poiseuille)

Coefficient of viscosity — internal friction per unit velocity gradient

Dimensions of η are [ML⁻¹T⁻¹]. Thin liquids (water, alcohol) have low η; thick liquids (coal tar, glycerine, honey, blood) have high η. For reference, water at 20 °C has η ≈ 1.0 mPa s, machine oil ~100 mPa s, glycerine ~1500 mPa s, honey ~10 000 mPa s, and air ~0.018 mPa s. Blood is about three times as viscous as water, but interestingly its relative viscosity (η/η_water) stays roughly constant between 0 °C and 37 °C — handy for a fluid that must work over a range of body temperatures.

The viscosity of liquids falls with temperature (thermal motion overcomes intermolecular cohesion), while the viscosity of gases rises with temperature (faster molecules transfer more momentum across layers). This temperature asymmetry is a classic NEET fact, and it has a clean microscopic explanation: in liquids, viscosity comes from cohesive bonds between molecules that thermal agitation disrupts; in gases, viscosity comes from momentum transfer between layers via molecular collisions, and faster molecules transfer momentum more efficiently. The same molecule, viewed as a liquid or as a gas, gives opposite trends.

Stokes' law gives the viscous drag on a sphere of radius r moving at velocity v through a fluid of viscosity η:

F = 6 π η r v

Stokes' law — viscous drag on a small sphere

Note that drag is linear in v (unlike the v² drag of high-Reynolds regimes). For a sphere falling under gravity through a fluid, three forces act: gravity ρ V g downward (with ρ the sphere density), buoyancy σ V g upward (with σ the fluid density), and viscous drag 6πηrv upward. The sphere accelerates until drag plus buoyancy balances weight; from then on it falls at constant terminal velocity:

v_t = 2r²(ρ − σ)g / 9η

Terminal velocity — sphere falling in a viscous fluid

Read the dependence: v_t grows as r², so larger raindrops fall faster than smaller ones; v_t falls as 1/η, so a ball drops slowly through glycerine and almost instantly through air; if ρ < σ the terminal velocity is negative — the sphere rises, like an air bubble in water. NEET 2022 tested the v(t) curve directly: a ball dropped into a viscous column accelerates from rest, then plateaus at v_t. NEET 2018 asked the rate of heat production at terminal velocity — since power = F · v = 6πηr · v² and v ∝ r², the rate of heat production scales as r⁵. NEET 2021 ran the same physics with buoyancy switched on and asked for the viscous force directly: F_v = mg − F_B.

An NCERT numerical: a 2 mm copper ball (ρ = 8.9 × 10³ kg m⁻³) falls through oil of density 1.5 × 10³ kg m⁻³ at a terminal velocity of 6.5 cm s⁻¹. Inverting v_t = 2r²(ρ − σ)g / 9η gives η ≈ 0.99 Pa s — about 1000 times the viscosity of water, which is why heavy machinery oil "feels thick" relative to water. Viscosity is also why a paint brush dipped in water and pulled out keeps its hairs gathered to a point (the water resists internal shear), and why honey takes longer than tea to drain from a tilted cup.

Reynolds number

The Reynolds number Re = ρvD/η decides which regime a flow is in. It is dimensionless: a check of units gives [kg m⁻³] · [m s⁻¹] · [m] ÷ [kg m⁻¹ s⁻¹] = 1. Below about 1000 the flow is firmly laminar; above about 3000 it is firmly turbulent; in between, the picture is unstable and depends on disturbances. The same Re predicts the same flow pattern in geometrically similar systems — which is why engineers can test small-scale models of aircraft, ships, and pipelines in wind tunnels and water channels and trust the results at full scale. So long as the model's Re matches the full-scale Re, the flow patterns are dynamically similar.

The physical reading of Re is that it compares two competing forces: the inertial force per unit volume scales as ρv²/D (mass density × kinetic-energy density divided by a length), while the viscous force per unit volume scales as ηv/D². Their ratio is exactly ρvD/η. When viscosity dominates (low Re), small disturbances die out and the flow stays smooth; when inertia dominates (high Re), small disturbances amplify and the flow tumbles into turbulence.

NIOS gives the classic example: blood flow in the aorta during the resting part of the heart cycle is laminar (Re ~ a few hundred), but during the contracting part it can briefly become turbulent. The "lub-dub" sounds heard through a stethoscope owe their character partly to these transient turbulent bursts. Smooth aircraft wings, river engineering, gas pipelines, and even paint sprays all live or die by the Reynolds number — keep it low and your design is predictable; let it soar and your design must absorb chaos.

Surface tension & surface energy

Look at a small drop of water at the tip of a tap: it pulls itself into a near-perfect sphere. Watch a needle laid carefully across a still water surface: it floats, despite being denser than water. Observe water rising through the fibres of a cotton wick against gravity, or a paint brush forming a fine tip when pulled out of water. All these everyday observations are surface-tension phenomena, and they share a single molecular origin.

A molecule in the bulk of a liquid is surrounded on all sides by other molecules and feels a net attractive force of zero. A molecule at the surface, however, has neighbours only on the liquid side; the air above offers almost nothing. So a surface molecule has higher potential energy than a bulk molecule — roughly half the heat of vaporisation per molecule, since to remove the surface molecule entirely you would have to break the remaining bonds. Liquids therefore minimise their surface area; a free drop, with no other forces, takes the shape with the smallest surface for its volume — a sphere. The same reasoning explains why mercury beads up into nearly perfect droplets on glass — its strong cohesion gives it a high surface tension, and the shape that minimises surface area for a given volume is the sphere.

To increase the area of a liquid surface, work must be done against the inward pull of cohesion. That work is stored in the liquid as additional surface energy. From this viewpoint, surface tension is not a force in the ordinary sense — it is an emergent property of the energetics of having molecules at an interface, and it has units of joules per square metre as much as it has units of newtons per metre. The two views are equivalent and the SI value is the same: N m⁻¹ = J m⁻².

Quantitatively, if we stretch a liquid film mounted on a wire frame by a small distance d, the area increases (the film has two surfaces) and work F · d must be done against an internal force F. The work goes into the film as extra surface energy S · (2 ℓ d), where ℓ is the length of the moving wire. Equating, S = F/(2ℓ). This quantity S — the surface tension — is simultaneously the force per unit length across a line in the surface and the surface energy per unit area of the interface.

T = F / L  ·  SI unit: N m⁻¹ = J m⁻²

Surface tension — force per unit length, or energy per unit area

Angle of contact θ is the angle between the tangent to the liquid surface at the line of contact and the solid surface, measured inside the liquid. It is determined by three interfacial tensions — solid-air S_sa, solid-liquid S_sl, and liquid-air S_la — through S_la cos θ + S_sl = S_sa. If θ < 90° the liquid wets the solid (water on clean glass, kerosene on most surfaces); if θ > 90° the liquid does not wet (water on wax, mercury on glass). Soaps and detergents reduce θ so the liquid penetrates fibres better; water-proofing agents do the opposite. Beads of water on a lotus leaf sit at θ ≈ 110°; a drop on a clean plastic plate spreads to θ near 0°. Both pictures are familiar — both are determined entirely by the three interfacial tensions.

Why are free drops and bubbles spherical? For a fixed volume, the sphere is the shape with the least surface area, so it minimises the surface energy. Gravity distorts large drops, but small ones — like the spray from an atomiser or the droplets in a fog — remain essentially perfect spheres. Surface tension also keeps a soap bubble stable: the inward pressure exerted by the curved film, exactly enough to balance the gas inside, holds the structure together.

One more useful result: the excess pressure inside a curved interface is 2T/r for a drop or cavity (one interface) and 4T/r for a bubble (two interfaces). The factor of two comes from the bubble having both an inner and outer surface, each contributing its own 2T/r. NEET 2023 used the soap-bubble surface-energy formula (energy = T · ΔA, with ΔA = 2 · 4πr² for the two surfaces) to ask the energy required to form a 2 cm bubble — about 3 × 10⁻⁴ J. The same factor-of-two trap appeared in NEET 2022: as a bubble expands, both 4T/r terms shrink, and the excess pressure inside the bubble falls — even though the bubble is "tighter" visually, the gas inside is at lower pressure.

Capillary action

"Capilla" is Latin for hair: a capillary tube is a tube as thin as a hair. Dip a clean glass capillary into a beaker of water and the water climbs up inside it, rising several centimetres above the free surface in the beaker — against gravity. The phenomenon is a direct consequence of surface tension and the angle of contact, and the height reached is given by

h = 2 T cos θ / (ρ g r)

Capillary rise — Jurin's law

Here T is surface tension, θ is the angle of contact, ρ is liquid density, g is gravity, and r is the capillary radius. Read off the dependencies: smaller radius ⇒ greater rise (the narrower the tube, the higher water climbs); higher surface tension ⇒ greater rise; greater contact angle (closer to 90°) ⇒ smaller rise; if θ > 90° (e.g. mercury in glass), cos θ is negative and the liquid falls below the outside level — capillary depression. NCERT plugs in numbers for a 0.05 cm-radius glass capillary and finds h ≈ 2.98 cm for water.

The derivation is short. Inside the capillary, the concave meniscus has effective radius r/cos θ, so the pressure just below the meniscus is less than atmospheric by 2T cos θ/r. Outside the capillary, the free surface is flat and the pressure just below it is atmospheric. The two pressures must match at the same height in the connected liquid, so the column inside must rise high enough for its weight (ρgh per unit area) to make up the pressure deficit. Equating ρgh = 2T cos θ/r yields Jurin's law.

NEET 2020 leveraged this directly: doubling the capillary radius — so the new tube has radius 2r — gives a new height h/2 (from h ∝ 1/r) but a new water mass that scales as r² · h ∝ r² · (1/r) = r. So the mass in the 2r tube is twice the mass in the original tube. A 5 g column in the original becomes 10 g in the wider tube — the right answer. Capillary action is also behind sap rising in trees, water creeping up cotton wicks against gravity, and the fact that towels work. The very thin xylem vessels of trees give heights well above what unaided capillarity could achieve, but the principle is identical — narrow tubes draw water up against gravity, no pump required.

A worked NCERT example: with T = 0.073 N m⁻¹, ρ = 10³ kg m⁻³, and a glass capillary of radius a = 0.05 cm = 5 × 10⁻⁴ m (taking cos θ ≈ 1), h = 2 × 0.073 / (10³ × 9.8 × 5 × 10⁻⁴) ≈ 2.98 × 10⁻² m, or roughly 3 cm. Halve the radius and the rise doubles. Push the radius down to a few micrometres and the rise climbs to many metres — which is, in part, how the tallest trees move water hundreds of feet up to their leaves.