Mechanical Properties of Fluids

🧠 Concepts Map of the deep dives

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Pascal's Law and Hydraulic Machines 🔉✕⇢deep concept

The French scientist Blaise Pascal, after whom the SI unit of pressure is named, carried out pioneering studies of the pressure in fluids and observed that the pressure in a fluid at rest is the same at all points that are at the same height. There are two closely related statements that go by the name of Pascal's law, and it pays to keep both in view because problems draw on each. The first is about equal pressure at equal heights in a fluid at rest; the second, the more famous, is about how an externally applied pressure is transmitted through an enclosed fluid. Both follow from the same idea — that a fluid at rest cannot sustain a shearing stress, so it can only push normally on any surface — and both are used constantly in the design of machines and in the analysis of JEE problems. 🔉✕⇢

To establish the first statement, consider a small element of fluid at rest in the interior of the fluid, taken in the convenient shape of a right-angled prism whose faces are so small that the effect of gravity on the element can be neglected in comparison with the pressure forces. The rest of the fluid exerts only normal forces on the faces of this prism, since a fluid at rest exerts no tangential force. Writing the condition of equilibrium — that the net force on the element is zero — and using the geometry of the prism to relate the areas of its faces, one finds that the pressures on the three faces are equal, $P_a=P_b=P_c$. Because the orientation of the prism was arbitrary, the pressure at a point is the same in every direction. 🔉✕⇢

This is a result of the first importance: pressure is exerted equally in all directions in a fluid at rest. It shows once again that pressure is a scalar and not a vector — no single direction can be assigned to it, because at any point it acts equally whichever way we orient the surface we imagine there. The force the fluid exerts on a real surface is of course directed (it is normal to that surface), but the pressure itself is a scalar field that has a definite value at each point of the fluid. 🔉✕⇢

Now take, in the same fluid at rest, a horizontal bar-shaped element of uniform cross-section. Gravity acts vertically and so has no horizontal component, while the pressures on the two flat ends push horizontally and inward. For the bar to be in equilibrium the horizontal forces at its two ends must balance exactly, and since the two ends have the same area the pressures at the two ends must be equal. This proves that for a liquid in equilibrium the pressure is the same at all points lying in the same horizontal plane; were it not, there would be an unbalanced horizontal force and the fluid would begin to flow. In the absence of any flow, the pressure must be the same everywhere in a horizontal plane, however the vessel is shaped. 🔉✕⇢

For the second, transmission statement, consider a horizontal cylinder fitted at one end with a piston and carrying three vertical open tubes rising from different points along its length. The height to which the liquid stands in each tube is a direct measure of the pressure at the foot of that tube, and in equilibrium the liquid stands at the same height in all three tubes, consistent with the pressure being the same throughout a horizontal fluid. If we now push the piston inward with some extra force, the liquid level rises in every one of the tubes, and — this is the key observation — it rises by the same amount in each, so that the levels are once again equal to one another. The additional pressure applied at the piston has been distributed uniformly and undiminished throughout the whole of the enclosed fluid. 🔉✕⇢

This is the operational form of Pascal's law, and it is worth stating carefully: whenever an external pressure is applied to any part of a fluid enclosed in a vessel, that pressure is transmitted undiminished and equally in all directions to every point of the fluid and to the walls of the container. The word 'undiminished' is essential — the applied pressure is not weakened as it spreads, so a pressure raised at one small piston appears in full at a large piston elsewhere. A great many devices — the hydraulic lift, hydraulic brakes, the hydraulic press, the dentist's and barber's chair — are built directly on this principle, and it is the single idea that lets a modest applied force be turned into an enormous one. 🔉✕⇢

In a hydraulic lift, two pistons of very different sizes are connected by, and rest upon, an enclosed incompressible liquid such as oil. A relatively small force $F_1$ is applied to the piston of small cross-sectional area $A_1$, and this sets up in the liquid immediately beneath it an additional pressure $P=\dfrac{F_1}{A_1}$. By Pascal's law this same pressure $P$ is transmitted, undiminished, throughout the connected liquid until it acts on the underside of the larger piston, of area $A_2$. The upward force that the liquid then exerts on the large piston is the pressure times its area, $F_2=P A_2=F_1\dfrac{A_2}{A_1}$. 🔉✕⇢

Because the area $A_2$ of the output piston is larger than the area $A_1$ of the input piston, the output force $F_2$ is larger than the input force $F_1$ by the factor $\dfrac{A_2}{A_1}$, which is called the mechanical advantage of the machine. If the large piston has, say, one hundred times the area of the small one, then a force of a few hundred newtons applied to the small piston becomes tens of thousands of newtons at the large one — quite enough for a person's modest push to raise a car or a loaded truck. This multiplication of force is the whole purpose of a hydraulic machine, and it is achieved simply by choosing a large ratio of piston areas. 🔉✕⇢

It is essential to understand that the device does not create energy, and does not multiply force from nothing; it trades distance for force. Conservation of volume of the incompressible liquid makes this quantitative. Whatever volume of liquid is pushed out from under the small piston must reappear under the large piston, so if the small piston descends a distance $L_1$ and the large piston rises a distance $L_2$, then $A_1 L_1=A_2 L_2$, giving $L_2=L_1\dfrac{A_1}{A_2}$. The large piston therefore rises by exactly as much less as the force upon it is greater. 🔉✕⇢

Combining the two relations shows that the work done is conserved. The work put in at the small piston is $F_1 L_1=P A_1 L_1$, and the work delivered at the large piston is $F_2 L_2=P A_2 L_2$; since $A_1 L_1=A_2 L_2$, these are equal, $F_1 L_1=F_2 L_2$. The machine multiplies the force but divides the displacement in the same ratio, so the product — the work, and hence the energy — is unchanged, exactly as the conservation of energy demands. A hydraulic press is a force amplifier, never an energy source. 🔉✕⇢

Hydraulic brakes in a motor car work on precisely the same principle. When the driver presses the brake pedal, a small force moves a piston in the master cylinder, raising the pressure of the brake fluid. By Pascal's law this pressure is transmitted undiminished through the brake lines to cylinders of larger area at each of the wheels, where it pushes pistons that force the brake shoes or pads against the rotating drum or disc. A small force at the pedal thus becomes a large retarding force at the wheels; and because the same pressure reaches every wheel cylinder, the braking effort is distributed evenly, which is vital for stopping the vehicle in a straight line. 🔉✕⇢

The hydraulic press used in industry to compress materials, and the hydraulic jack used to lift vehicles, are further embodiments of the same law, differing only in the ratio of areas and in how the input force is generated. In every case the analysis is the same: find the pressure set up by the input force, transmit it undiminished by Pascal's law, and multiply it by the output area to get the output force. The shape and length of the connecting tubes never enter, because the pressure is transmitted equally and undiminished throughout the fluid whatever path it takes. 🔉✕⇢

A frequent JEE situation places known weights on both pistons and asks for the equilibrium condition or the extra load one side can bear. The method is always the same equal-pressure statement, applied at a common horizontal level in the connecting liquid: the pressure just under the left piston must equal the pressure just under the right piston at the same height. If the two pistons carry loads and sit at the same level, this reads $\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}$; if they sit at different heights, the hydrostatic term $\rho g\,\Delta h$ for the liquid column between them is added to the lower side. Setting up that single equality resolves almost every hydraulic-press problem. 🔉✕⇢

The same principle is at work in power steering and in the hydraulic suspension of heavy vehicles, where a small force generated by a pump is converted, through a large-area piston, into the large force needed to turn the wheels or support the chassis. A hydraulic accumulator stores energy by holding liquid under pressure against a heavy loaded piston, and releases it on demand; here the stored quantity is genuinely energy, since the loaded piston is free to move, but the force multiplication at the point of use is still governed entirely by Pascal's law and the ratio of areas. 🔉✕⇢

It is worth pausing on the microscopic meaning of 'transmitted undiminished'. At the molecular level, pushing the piston in slightly increases the frequency and vigour of molecular collisions in the layer of liquid just beneath it; because the liquid is nearly incompressible and the molecules are in constant contact, this increased collisional pressure is passed on from layer to layer throughout the fluid almost instantaneously and without loss, until every boundary feels the same increase. This is why the pressure increase is the same at the far piston as at the near one, and why the connecting geometry is irrelevant. 🔉✕⇢

It is instructive to ask why Pascal's law works so cleanly for liquids. The answer is incompressibility. A liquid changes its volume hardly at all under pressure, so when the piston is pushed in, the liquid cannot simply be squeezed into a smaller space; the applied pressure must instead be borne by the whole body of liquid and passed on to every boundary. A gas, being highly compressible, would first compress under the piston and only gradually build up pressure, which is why hydraulic machines use oil or another liquid rather than air. The near-incompressibility of the working liquid is what guarantees that the pressure is transmitted promptly and undiminished. 🔉✕⇢

The mechanical advantage of a hydraulic press is often compared with that of a lever, and the analogy is exact in its bookkeeping. A lever multiplies force by the ratio of its arm lengths while the load moves through a correspondingly smaller distance; a hydraulic machine multiplies force by the ratio of piston areas while the load moves through a correspondingly smaller distance. In both, the conserved quantity is the work, and in both the amplification of force is paid for by a reduction in displacement. Recognising this shared structure helps a student reason about a hydraulic problem without re-deriving it each time. 🔉✕⇢

A subtlety worth noting is that Pascal's transmission law refers to the extra, applied pressure. The total pressure at a point in the enclosed liquid is the sum of this transmitted applied pressure and the ordinary hydrostatic pressure due to the liquid's own weight, which still varies with depth as $\rho g h$. In most hydraulic-machine problems the two pistons are at nearly the same height and the column of working liquid is short, so the hydrostatic contribution is negligible and only the transmitted applied pressure matters; but when the pistons are at very different heights the $\rho g h$ term must be restored to the balance. 🔉✕⇢

Historically, Pascal demonstrated the surprising power of transmitted pressure with his celebrated 'barrel' experiment: by pouring a small quantity of water down a very tall, thin vertical tube inserted into a sealed, water-filled barrel, he was able to burst the barrel. The tall thin column added only a little weight of water, yet it raised the pressure throughout the barrel by $\rho g h$ with $h$ the great height of the tube, and this raised pressure, acting undiminished over the barrel's large inner area, produced a force large enough to split it. The experiment dramatises both that pressure depends on height of column, not on quantity of liquid, and that it is transmitted throughout the enclosed fluid. 🔉✕⇢

As a numerical feel for the mechanical advantage, suppose the input piston has a radius of one centimetre and the output piston a radius of ten centimetres. The areas are in the ratio of the squares of the radii, $100:1$, so a force of one hundred newtons applied to the small piston produces ten thousand newtons at the large one — enough to lift a mass of about a tonne. To raise that load by one centimetre, however, the small piston must be driven in by one metre, again showing that the great gain in force is balanced by an equal loss in distance and that no energy has been manufactured. 🔉✕⇢

One practical point recurs in problems. Atmospheric pressure acts on the free upper faces of both pistons in a lift, and therefore contributes equally to both sides and cancels out of the force balance; so in hydraulic-machine problems we work with the gauge pressure set up by the applied force, not the absolute pressure. The quantity that matters is the piston area, and the governing relation $\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}$ — equal pressure on both pistons — is the compact statement of Pascal's law that most JEE questions on this topic come down to. 🔉✕⇢

In summary, Pascal's law has the two faces we began with, and a complete answer keeps both: in a fluid at rest the pressure is the same at all points at the same height and is exerted equally in all directions, and an externally applied pressure is transmitted undiminished throughout an enclosed fluid. From the first follow the shape-independence of pressure and the equal-height rule used to relate two arms of a connected liquid; from the second follow the hydraulic lift, the hydraulic press and the hydraulic brake, each multiplying force by the ratio of areas while conserving work. Mastering when to invoke which face, and remembering to add the hydrostatic $\rho g h$ term only when the two points differ in height, is what turns this deceptively simple law into a reliable problem-solving tool. 🔉✕⇢

Bernoulli's Principle 🔉✕⇢deep concept

Bernoulli's principle is the energy statement for a flowing fluid, and it is the single most examined idea in this chapter. It relates the pressure, the speed and the height of a fluid in steady flow, and it expresses the conservation of energy for the fluid, just as the equation of continuity expresses the conservation of mass. The Swiss scientist Daniel Bernoulli obtained the result in the eighteenth century, long before the general law of conservation of energy was formulated, and it remains one of the most far-reaching statements in all of fluid mechanics — from it follow the working of the aeroplane wing, the flow meter, the atomiser and the escape of liquid from a tank. 🔉✕⇢

To derive it we consider an ideal fluid, one that is both incompressible and non-viscous, in steady flow through a tube of flow whose cross-section and height both vary along its length. We fix attention on a small slug of the fluid and follow it as it moves from a lower, wider section — of area $A_1$, speed $v_1$, height $h_1$ and pressure $P_1$ — to an upper, narrower section, of area $A_2$, speed $v_2$, height $h_2$ and pressure $P_2$. Because the flow is steady, whatever happens to this slug is representative of the flow as a whole, and we may apply the work–energy theorem to it: the net work done on the slug equals the change in its kinetic energy. 🔉✕⇢

Two kinds of work act on the slug. First, the surrounding fluid pushes on it through pressure. The fluid behind pushes it forward at the lower section, doing positive work, while the fluid ahead pushes back at the upper section, doing negative work. In a time $\Delta t$ the slug advances so that a volume $\Delta V$ enters the lower section and an equal volume $\Delta V$ (equal, by the continuity equation for an incompressible fluid) leaves the upper section. The work done by the pressure forces is then $P_1\Delta V$ at the inlet and $-P_2\Delta V$ at the outlet, giving a net pressure work $(P_1-P_2)\Delta V$. 🔉✕⇢

Second, gravity does work on the slug as it is, in effect, lifted from height $h_1$ to height $h_2$. The mass of the transported volume is $\rho\,\Delta V$, and raising it through $(h_2-h_1)$ costs an amount of work $-\rho\,\Delta V\,g(h_2-h_1)$ against gravity (negative because gravity opposes the rise). The change in the kinetic energy of the same mass, as its speed changes from $v_1$ to $v_2$, is $\tfrac12\rho\,\Delta V(v_2^2-v_1^2)$. 🔉✕⇢

The work–energy theorem now sets the total work equal to the change in kinetic energy: $(P_1-P_2)\Delta V-\rho g(h_2-h_1)\Delta V=\tfrac12\rho\,\Delta V(v_2^2-v_1^2)$. Every term contains the common factor $\Delta V$, which cancels, and rearranging so that all the quantities belonging to section 1 stand on one side and those belonging to section 2 on the other gives $P_1+\tfrac12\rho v_1^2+\rho g h_1=P_2+\tfrac12\rho v_2^2+\rho g h_2$. 🔉✕⇢

Since the two sections were arbitrary points along the same streamline, the combination on each side must have the same value everywhere along the flow. This is Bernoulli's equation, $P+\tfrac12\rho v^2+\rho g h=\text{constant}$. It states that the sum of the pressure, the kinetic energy per unit volume and the potential energy per unit volume of an ideal fluid in steady flow is a constant along a streamline — a compact and powerful statement of energy conservation for a moving fluid. 🔉✕⇢

Each term deserves a name and an interpretation, because problems often ask which term dominates. Every term has the dimensions of pressure, which are the same as the dimensions of energy per unit volume. The term $P$ is the pressure energy per unit volume associated with the work the fluid can do by virtue of its pressure; $\tfrac12\rho v^2$ is the kinetic energy per unit volume, sometimes called the dynamic pressure; and $\rho g h$ is the gravitational potential energy per unit volume. Bernoulli's equation says these three can be freely converted into one another along the flow, provided their sum is preserved. 🔉✕⇢

For the important special case of flow along a horizontal tube, the height $h$ does not change, the term $\rho g h$ is the same at every point and drops out of the balance, leaving $P+\tfrac12\rho v^2=\text{constant}$. This is the form that carries the principle's central and most surprising message. Where the fluid moves faster, the term $\tfrac12\rho v^2$ is larger, so the pressure $P$ must be smaller to keep the sum constant; where the fluid moves slower, the pressure is larger. Speed and pressure trade off against one another: fast flow is low-pressure flow. 🔉✕⇢

Combining this horizontal form with the equation of continuity closes the argument and produces the chapter's most useful chain of reasoning. Continuity, $Av=\text{constant}$, says that the fluid must speed up where the tube narrows; Bernoulli then says that the pressure must fall exactly there. So the pressure is lowest at the narrowest, fastest part of a horizontal flow — a conclusion that runs directly against the untrained intuition that a squeezed fluid is a high-pressure fluid, and one that lies at the counter-intuitive heart of the Venturi meter, the aerofoil and the atomiser. 🔉✕⇢

The conditions under which Bernoulli's equation holds are every bit as important as the equation itself, and this is precisely where most JEE errors are made. The equation was derived for a flow that is, first, steady; second, incompressible; third, non-viscous; and fourth, followed along a single streamline. If any one of these four conditions is violated the equation cannot be trusted. It may not be applied across different streamlines, nor to turbulent flow where the motion is unsteady and chaotic, nor to a strongly viscous fluid, nor to a compressible gas whose density changes appreciably along the flow. 🔉✕⇢

The non-viscous condition deserves special emphasis because it is the one most often forgotten. In any real fluid, viscosity — internal friction between the layers — does negative work on the flow, and some of the mechanical energy is steadily converted into heat. As a result the Bernoulli sum does not stay constant but decreases along the direction of flow. Bernoulli's equation is therefore an idealisation, an excellent one for low-viscosity fluids such as water and air over short distances, but a poor one for honey, for blood moving through fine capillaries, or for any flow in which viscous losses are large. In such cases it is viscosity, not Bernoulli, that governs the motion. 🔉✕⇢

A single worked wrong-answer case fixes the most common trap in the mind. Water flows steadily through a horizontal pipe that narrows to a throat, and a student is asked to compare the pressure in the wide part with the pressure in the throat. Reasoning loosely that 'the fluid is being squeezed into a smaller space, so the pressure there must be higher', the student picks the throat as the high-pressure region. This is exactly backwards. Continuity forces the speed to rise in the narrow throat, and Bernoulli then forces the pressure to fall there; the throat is the low-pressure region, not the high-pressure one. Confusing the everyday sense of 'squeezed' with an increase of pressure is the classic Bernoulli mistake, and it is worth a great many marks each year. 🔉✕⇢

The Venturi meter is the direct application of the horizontal form and the standard examination vehicle for the whole topic. A pipe is fitted with a gradual constriction, and the fluid, obeying continuity, speeds up as it enters the throat; by Bernoulli its pressure drops there. The pressure difference between the wide section and the throat is read off either as a difference in the heights of liquid in two vertical standpipes rising from the two sections, or as the reading of a manometer connected across them. From that measured pressure difference, together with the two known cross-sectional areas, Bernoulli and continuity together yield the flow speed and hence the volume flow rate. The same physics underlies the carburettor of an engine, the filter pump, the Bunsen burner and the spray of an atomiser or scent bottle. 🔉✕⇢

Dynamic lift on an aeroplane wing is the same principle in a different guise. The wing, or aerofoil, is shaped and tilted so that the air streaming over its more curved upper surface travels faster than the air passing along its flatter lower surface. By Bernoulli's principle the faster air above is at a lower pressure than the slower air below, so there is a net upward pressure difference across the wing; multiplied by the wing area this gives the upward force, the dynamic lift, that supports the weight of the aircraft. The curving flight of a spinning cricket ball or tennis ball — the Magnus effect — arises from the same pressure difference, there produced by the spin dragging air faster past one side than the other. 🔉✕⇢

It is useful to name the pressures that appear in the horizontal form, because JEE problems on the Pitot tube turn on the distinction. The ordinary pressure $P$ measured by an instrument that moves with the fluid, or reads through a hole in a wall parallel to the flow, is called the static pressure. The extra term $\tfrac12\rho v^2$ that appears when the fluid is brought to rest is the dynamic pressure. Their sum, $P+\tfrac12\rho v^2$, is the total or stagnation pressure — the pressure that would be read at a point where the flow is completely stopped. A Pitot tube exploits exactly this: it measures static and stagnation pressure at the same place, and the difference gives $\tfrac12\rho v^2$, from which the flow speed follows. This is how the airspeed of an aircraft is measured. 🔉✕⇢

Dividing Bernoulli's equation through by $\rho g$ recasts every term as a length, and this 'head' form is worth recognising: $\dfrac{P}{\rho g}+\dfrac{v^2}{2g}+h=\text{constant}$. Here $P/\rho g$ is the pressure head, $v^2/2g$ the velocity head and $h$ the elevation head, and each is measured in metres. The head form makes the energy bookkeeping visual — the constant is the total head of the flow — and it is the language in which hydraulics and civil-engineering problems are usually posed. For JEE it is enough to know that the three forms (per unit volume, per unit mass, and per unit weight) are the same statement scaled by a constant, so a problem may be worked in whichever is most convenient. 🔉✕⇢

The atomiser, the spray gun and the Bunsen burner are everyday devices that run on the horizontal form. In an atomiser a rubber bulb drives air at high speed across the open top of a narrow tube dipping into the liquid; the fast air has a low pressure, atmospheric pressure on the liquid surface below then pushes the liquid up the tube, and it is caught by the air stream and sprayed out as a fine mist. The filter pump and the carburettor of a petrol engine work the same way, a fast stream drawing in a second fluid because its pressure has dropped. Recognising this 'fast stream lowers pressure, atmosphere pushes the second fluid in' pattern converts a whole family of application questions into one idea. 🔉✕⇢

The same low-pressure-in-the-fast-stream effect explains a set of demonstrations that feel paradoxical. Two light balls hung side by side swing towards each other when you blow air between them, because the moving air between the balls is at a lower pressure than the still air on their outer sides. A sheet of paper held below the lips rises when you blow across its top surface, because the fast air above it is at reduced pressure while the still air beneath pushes up. Two ships steaming on parallel courses are drawn dangerously together for the same reason. In every case the rule is the same: where the fluid moves faster, the pressure is lower. 🔉✕⇢

In physiology the failure of Bernoulli's ideal assumptions is as instructive as its successes. In a healthy artery the flow is smooth and the equation gives a fair account of the pressure. But where an artery narrows because of plaque, continuity forces the blood to speed up through the constriction and Bernoulli predicts a drop in pressure there; if the vessel is soft it may then be squeezed further shut, and the flow can become turbulent, producing the sounds a doctor hears through a stethoscope. In an aneurysm, where the vessel bulges wider, the blood slows and the pressure rises, tending to expand the weak wall still further. These are qualitative applications, but they are exactly the kind of reasoning JEE rewards. 🔉✕⇢

The quantitative Venturi result is worth setting down because it recurs. Writing horizontal Bernoulli between the wide section and the throat and eliminating the throat speed with continuity $A_1v_1=A_2v_2$ gives the volume flow rate $Q=A_1v_1=A_1A_2\sqrt{\dfrac{2(P_1-P_2)}{\rho\,(A_1^2-A_2^2)}}$. Everything on the right is either a fixed dimension of the meter or the measured pressure difference, so a single manometer reading yields the flow rate. The key modelling steps — flow horizontal so the height term drops, incompressible so continuity applies, and the meter short enough that viscous losses are negligible — are precisely the assumptions Bernoulli requires, which is why the Venturi meter is the archetypal Bernoulli problem. 🔉✕⇢

Two cautions prevent most remaining errors. First, Bernoulli's constant is the same only along one streamline; comparing points on different streamlines is legitimate only in the special case of irrotational flow, which is beyond the JEE syllabus, so always follow a single streamline from a point where you know the conditions to the point you are asked about. Second, the pressure that enters the equation is the absolute (or at least consistently gauge) pressure of the fluid itself, not the pressure of the container walls; at a free surface open to the air, or at a hole discharging into the atmosphere, that pressure is simply atmospheric and the same on both sides, so it cancels — a simplification that is the key to the efflux problem. 🔉✕⇢

A reliable four-step method turns almost any Bernoulli question into routine work. Choose two points on the same streamline, one where all quantities are known; write $P+\tfrac12\rho v^2+\rho g h$ at each and set them equal; use continuity $Av=\text{constant}$ to relate the two speeds whenever an area is given; and identify which terms vanish — the height term for horizontal flow, the speed term at a broad reservoir surface, the pressure term where the fluid meets the atmosphere. Executed in that order the algebra is short, and the same template solves the Venturi meter, the Pitot tube, the siphon and Torricelli's law without any new physics. 🔉✕⇢

Finally, Torricelli's law of efflux — the speed with which a liquid escapes from a small hole in an open tank — is not a separate principle at all but simply Bernoulli's equation applied along the streamline that runs from the open top surface down to the hole. That so many distinct and important results — the flow meter, the aerofoil, the atomiser and the efflux law — all flow from this one equation is why it repays careful study. Mastering the exact statement of Bernoulli's equation, its four conditions of validity, and above all its combination with the equation of continuity is the single highest-value investment a student can make in the whole of this chapter. 🔉✕⇢

Viscosity and Stokes' Law 🔉✕⇢deep concept

Real fluids are not the ideal, frictionless fluids assumed in Bernoulli's principle. When layers of a real fluid slide past one another, they resist the relative motion, and this internal friction of a fluid is called viscosity. It is the property that makes honey pour slowly and water pour quickly, and it is the reason the Bernoulli sum decreases along a real flow. 🔉✕⇢

Consider a fluid held between a fixed lower plate and a movable upper plate, with the fluid in the form of thin layers. When the top plate is pulled sideways with a steady velocity, the layer of fluid in contact with it moves with it, while the layer at the bottom stays at rest. In between, the velocity increases steadily from the bottom layer to the top, so the fluid is sheared. 🔉✕⇢

The moving layers exert tangential forces on one another that oppose the relative sliding. To keep the top plate moving at constant velocity a steady force must be applied, and experiment shows this force is proportional to the area of the layer and to the velocity gradient across it: $F=\eta A\dfrac{dv}{dx}$, where $\dfrac{dv}{dx}$ is the rate of change of velocity with distance across the layers. 🔉✕⇢

The constant $\eta$ is the coefficient of viscosity of the fluid. Rearranged, it is defined as the ratio of the shearing stress $F/A$ to the strain rate (the velocity gradient) $dv/dx$. Its SI unit is the pascal-second ($\mathrm{Pa\,s}$), also written $\mathrm{N\,s\,m^{-2}}$, and its dimensions are $[\mathrm{ML^{-1}T^{-1}}]$. A more viscous fluid has a larger $\eta$. 🔉✕⇢

Viscosity varies enormously between fluids and depends strongly on temperature. The viscosity of liquids decreases as temperature rises — hot honey flows more freely than cold — whereas the viscosity of gases increases with temperature. This temperature dependence is important in lubrication, where the oil must keep a suitable viscosity across the operating range of an engine. 🔉✕⇢

A body moving through a viscous fluid drags the adjacent fluid with it and experiences a retarding force from the fluid's viscosity. For a small sphere moving slowly through a large body of fluid, the mathematician and physicist George Stokes found that this viscous drag force depends on the viscosity of the fluid, the radius of the sphere and its velocity. 🔉✕⇢

The result, $F=6\pi\eta a v$, is known as Stokes' law, where $\eta$ is the coefficient of viscosity, $a$ the radius of the sphere and $v$ its velocity relative to the fluid. The drag is proportional to the first power of the speed, a signature of streamline (low-Reynolds-number) motion; at high speeds, where the flow becomes turbulent, the drag grows faster, roughly as the square of the speed. 🔉✕⇢

Stokes' law can be understood on dimensional grounds. The drag must depend on $\eta$, $a$ and $v$, and the only combination of these with the dimensions of force is $\eta a v$; the numerical factor $6\pi$ comes from the full hydrodynamic calculation. The linear dependence on radius and speed is what makes the law so useful for small, slow spheres such as raindrops, mist droplets and oil drops. 🔉✕⇢

Stokes' law is the tool that resolves a puzzle of everyday experience: raindrops do not strike the ground at the enormous speeds free fall would predict, because the viscous drag of the air limits their speed. As a drop speeds up, the Stokes drag rises in proportion, until it balances the drop's weight and the drop falls at a constant, modest speed — the terminal velocity taken up in the next concept. 🔉✕⇢

The definition through the two-plate experiment rewards a closer look, because it fixes the sign and the direction of the viscous force. The fluid clings to each solid surface it touches and moves with it: this is the no-slip condition, an experimental fact that holds for ordinary fluids. So the layer touching the fixed plate is at rest and the layer touching the moving plate travels at the plate's speed, and a smooth gradient of velocity is set up across the gap. Each layer drags the slower layer just below it forward and is itself held back by that layer, an action–reaction pair; the net effect is a tangential, or shearing, force between adjacent layers that always opposes their relative sliding. 🔉✕⇢

The quantity $dv/dx$ in Newton's law of viscous flow is the velocity gradient, or equivalently the rate of shear strain, and it is the true measure of how hard the fluid is being sheared. The shearing stress $F/A$ needed to maintain the flow is proportional to it, and the constant of proportionality is the coefficient of viscosity $\eta$. Writing the relation as $\eta=(F/A)\big/(dv/dx)$ shows the definition in words: viscosity is the ratio of shearing stress to the strain rate. A fluid that obeys this simple linear law, with $\eta$ independent of the rate of shear, is called a Newtonian fluid; water, air and thin oils are Newtonian to good accuracy. 🔉✕⇢

Not every fluid is Newtonian, and the exceptions are worth knowing qualitatively. In some fluids the apparent viscosity changes with the rate of shear. Blood, paint, tomato ketchup and many polymer solutions are non-Newtonian: ketchup, for instance, is thick and reluctant to pour until it is shaken or squeezed, whereupon its apparent viscosity drops and it flows freely. A paste of cornflour and water does the opposite, thickening the harder it is stirred. These fluids do not obey a single constant $\eta$, and the simple relation $F=\eta A\,dv/dx$ applies only to the Newtonian idealisation used throughout this chapter. 🔉✕⇢

Two systems of units for viscosity are in common use, and JEE problems mix them. In the SI system the unit is the pascal-second, $\mathrm{Pa\,s}=\mathrm{N\,s\,m^{-2}}$. In the older CGS system the unit is the poise ($\mathrm{P}$), named after Poiseuille, with $1\,\mathrm{Pa\,s}=10\,\mathrm{poise}$. Water at room temperature has a viscosity of about $1\,\mathrm{mPa\,s}$, or one centipoise, which is a convenient reference value; air is about fifty times less viscous, and glycerine and honey hundreds to thousands of times more. Carrying units carefully, and converting poise to pascal-seconds before substituting into Stokes' law, avoids a common numerical error. 🔉✕⇢

The molecular origin of viscosity is different in liquids and in gases, and this explains their opposite responses to temperature. In a liquid the resistance comes chiefly from the cohesive attraction between neighbouring molecules; heating the liquid gives the molecules more energy to slip past one another, so the attraction is more easily overcome and the viscosity falls — hot honey pours easily. In a gas, by contrast, viscosity arises from molecules carrying momentum as they jump between faster and slower layers; heating the gas makes them jump more vigorously, transporting momentum more effectively, so the viscosity rises. This is why the viscosity of liquids decreases with temperature while that of gases increases. 🔉✕⇢

Viscosity also governs how a real fluid flows along a pipe, and here it produces the parabolic velocity profile. Because the fluid sticks to the walls (no-slip) but is free in the middle, the flow is fastest along the axis of the tube and falls smoothly to zero at the walls, tracing out a paraboloid. The steady flow of a viscous fluid through a narrow tube is described by Poiseuille's law, which states that the volume flow rate $Q$ is proportional to the pressure difference $\Delta P$ across the tube and to the fourth power of the tube radius, and inversely proportional to the length and the viscosity: $Q=\dfrac{\pi r^4\,\Delta P}{8\eta L}$. 🔉✕⇢

The fourth-power dependence on radius in Poiseuille's law is dramatic and has real consequences. Halving the radius of a tube reduces the flow rate for a given pressure to a sixteenth of its former value. This is why a small narrowing of an artery by deposits forces the heart to raise the pressure steeply to maintain the same blood flow, and why the fine tubing in an experiment dominates its flow resistance. Poiseuille's law is the viscous-flow counterpart of the ideal-fluid Bernoulli result: where Bernoulli neglects viscosity, Poiseuille is built entirely upon it. 🔉✕⇢

The distinction between streamline and turbulent flow is decided by viscosity through the Reynolds number, a dimensionless combination $\mathrm{Re}=\dfrac{\rho v d}{\eta}$, where $d$ is a characteristic size such as a pipe diameter. It compares the inertial tendency of the fluid to keep moving with the viscous tendency to damp disturbances out. When the Reynolds number is small, below roughly a thousand for pipe flow, viscosity wins and the flow is smooth and streamline; when it is large, above a few thousand, inertia wins and the flow becomes turbulent, breaking into eddies. The speed at which the change sets in is the critical velocity, and it is larger for more viscous fluids and narrower tubes. 🔉✕⇢

This Reynolds-number criterion is exactly why Stokes' law carries the qualification 'slow, small sphere'. Stokes' law $F=6\pi\eta a v$ was derived for streamline flow around the sphere, that is, for a small Reynolds number based on the sphere's radius. For a large or fast sphere the flow behind it becomes turbulent, a wake of eddies forms, and the drag no longer follows the simple linear law but grows roughly as the square of the speed. A falling raindrop sits near the edge of this regime, which is why the Stokes-law estimate of its terminal velocity is only approximate. 🔉✕⇢

It is instructive to contrast viscous drag with ordinary solid friction, because students often blur the two. Sliding friction between solid surfaces is, to a good approximation, independent of the sliding speed and of the contact area, depending only on the normal force. Viscous drag is the opposite in both respects: it depends directly on the area of the sheared layers and grows with the relative speed (linearly for streamline flow). There is also no static viscous force — a fluid offers no resistance to a vanishingly slow, steady shear beyond the force needed to keep it moving — whereas solids sustain a static friction up to a threshold. Viscosity is friction within a fluid, not friction between solid surfaces. 🔉✕⇢

The energy that the viscous force removes from the flow does not disappear; it is converted into heat. When a viscous fluid flows through a pipe, the pressure drops steadily along the length precisely because work must be done against the internal friction, and that work appears as a slight warming of the fluid. This continuous dissipation is the microscopic reason the Bernoulli sum decreases along a real flow rather than staying constant, and it is why maintaining a flow of a viscous fluid requires a continuous input of energy from a pump or a pressure head. 🔉✕⇢

Finally, viscosity is measured by turning these same relations around. In the falling-sphere or Stokes method a small sphere is dropped through the liquid and timed over a marked distance once it is moving at terminal velocity; inverting the terminal-velocity formula then yields $\eta$. In the capillary, or Ostwald, method the time for a fixed volume of liquid to flow through a fine tube under its own weight is measured and Poiseuille's law is applied. Both methods are staples of the laboratory and both are favourite settings for numerical JEE problems, because each ties viscosity directly to a measurable time. 🔉✕⇢

The no-slip condition has a further consequence worth naming: the boundary layer. When a fluid flows past a solid surface, the fluid right at the surface is stationary, and the velocity climbs from zero to the free-stream value across a thin region next to the wall called the boundary layer. It is within this layer that the velocity gradient is steep and viscous forces are concentrated, even when the bulk of the flow behaves almost like an ideal fluid. The boundary-layer picture reconciles the two halves of this chapter: away from surfaces the flow can be treated by Bernoulli, while close to them viscosity always matters. 🔉✕⇢

A useful mental model treats viscosity as the diffusion of momentum. Just as heat conduction carries thermal energy from a hot layer to a cold one, and just as diffusion carries molecules from a concentrated region to a dilute one, viscosity carries momentum from a fast-moving layer of fluid to a slow-moving one. The molecules that wander from the fast layer into the slow one speed it up, and those that wander the other way slow the fast layer down; the smoothing of the velocity profile is momentum being transported down its gradient. This analogy places viscosity beside conduction and diffusion as one of the three classical transport processes. 🔉✕⇢

One quantitative habit closes most viscosity problems cleanly: before applying any formula, decide which regime the flow is in. Estimate the Reynolds number from the given data; if it is small, streamline relations — Stokes' law for a sphere, Poiseuille's law for a pipe, the linear $F=\eta A\,dv/dx$ for shear — all apply, and if it is large they do not and the drag is quadratic instead. Then keep the three governing dependences in view: the drag or the force rises with the velocity gradient, the flow rate in a pipe scales as the fourth power of the radius, and the coefficient $\eta$ itself shifts with temperature. Checking the regime first prevents the frequent mistake of forcing a streamline formula onto a turbulent situation where it cannot hold. 🔉✕⇢

It is precisely because real fluids are viscous that Bernoulli's non-viscous idealisation must be applied with care. Over short distances in low-viscosity fluids like water and air the energy lost to viscosity is small and Bernoulli works well; in fine tubes, slow flows and viscous liquids the losses dominate, and viscosity, not Bernoulli, controls the motion. 🔉✕⇢

To measure the coefficient of viscosity directly, one lets a metal block rest on a thin film of liquid on a table, connected over a frictionless pulley to a small hanging mass. The tension in the string equals the weight of the suspended mass, and this shear force $F$ drives the block at a constant speed once the viscous resistance of the film just balances it. Dividing the force by the block area gives the shearing stress $F/A$, dividing the constant speed by the film thickness gives the strain rate, and their ratio is the coefficient of viscosity $\eta$. The same laminar picture governs a liquid flowing in a pipe, where the velocity is maximum along the axis of the tube and decreases gradually to zero at the walls, so the layers of liquid slide over one another like the pages of a book pushed flat on a table. 🔉✕⇢

Stokes law is an interesting example of a retarding force that is proportional to velocity, and its most important consequence is terminal velocity. A raindrop falling through air accelerates initially due to gravity, but as the velocity increases the viscous force $6\pi\eta a v$ also increases, until the viscous force plus the buoyant force becomes equal to the force due to gravity. The net force is then zero and so is the acceleration, and the sphere descends with a constant velocity. Setting $6\pi\eta a v_t=\tfrac{4}{3}\pi a^3(\rho-\sigma)g$ in this equilibrium, where $\rho$ and $\sigma$ are the mass densities of the sphere and the fluid, gives the terminal velocity $v_t=2a^2(\rho-\sigma)g/9\eta$. Because it grows as the square of the radius, a larger drop reaches its terminal velocity at a much higher speed than a fine one. 🔉✕⇢

Angle of Contact and Capillarity 🔉✕⇢deep concept

When the surface of a liquid meets a solid, the liquid surface near the wall is curved rather than flat, forming a meniscus. The shape of this meniscus is described by the angle of contact, and it is set by the competition between the forces that pull the liquid toward the solid and the forces that pull it back into the bulk of the liquid. 🔉✕⇢

The angle of contact is defined as the angle between the tangent to the liquid surface at the point of contact and the solid surface, measured inside the liquid. When a liquid meets a solid the angle so formed is termed the angle of contact, and its value depends on the particular pair of solid and liquid in contact. 🔉✕⇢

Two kinds of intermolecular force decide the angle. Cohesive forces act between the molecules of the liquid itself; adhesive forces act between the molecules of the liquid and those of the solid. When adhesion is strong compared with cohesion, the liquid clings to the solid and spreads, giving a small angle of contact and a concave meniscus — the liquid is said to wet the solid. 🔉✕⇢

When cohesion dominates, as for mercury on glass, the liquid pulls itself together and away from the solid, giving an angle of contact greater than a right angle and a convex meniscus; such a liquid does not wet the solid. Water on clean glass has a small angle of contact and wets it, whereas mercury on glass has an angle of contact of about $140^\circ$ and does not. 🔉✕⇢

The most important consequence of the angle of contact and surface tension together is capillarity — the rise or fall of a liquid in a fine tube. When a narrow tube, called a capillary, is dipped into a wetting liquid such as water, the liquid rises up in the narrow tube in spite of gravity; when it is dipped into a non-wetting liquid such as mercury, the liquid is depressed below the outside level. 🔉✕⇢

The rise happens because the concave meniscus of a wetting liquid has a lower pressure just beneath it than the flat surface outside (by the excess-pressure relation for a curved surface). This pressure deficit is made up by a column of liquid rising in the tube until the extra weight of the raised column restores the balance. 🔉✕⇢

Balancing the upward pull of surface tension around the circumference against the weight of the raised column gives the height of rise. For a tube of radius $a$ and a liquid of surface tension $S$, density $\rho$ and angle of contact $\theta$, the liquid rises to a height $h=\dfrac{2S\cos\theta}{\rho g a}$. 🔉✕⇢

The formula shows that the rise is inversely proportional to the radius of the tube: the narrower the capillary, the higher the liquid climbs. It is this that makes water climb a fine glass tube tens of centimetres while barely rising in a wide one, and it is the reason the effect is called capillarity, from the Latin for hair. 🔉✕⇢

The factor $\cos\theta$ carries the wetting behaviour. For a wetting liquid $\theta$ is less than a right angle, $\cos\theta$ is positive and $h$ is positive — the liquid rises. For a non-wetting liquid such as mercury $\theta$ is greater than a right angle, $\cos\theta$ is negative and $h$ is negative — the liquid is depressed, exactly as observed. 🔉✕⇢

It is worth understanding why the angle of contact takes the value it does, because this fixes the sign of the whole effect. At the line where liquid, solid and vapour meet, three surface tensions act along the three interfaces: the solid–liquid tension, the solid–vapour tension and the liquid–vapour tension $S$. The angle of contact settles at whatever value balances these three along the solid surface, a condition expressed by Young's relation. Qualitatively, when the solid attracts the liquid strongly — strong adhesion — the solid–liquid interface is energetically favourable, the liquid spreads, and $\theta$ is small; when the solid attracts it weakly the liquid beads up and $\theta$ is large. The angle is thus a property of the particular solid–liquid–vapour combination, not of the liquid alone. 🔉✕⇢

The angle of contact is also extremely sensitive to the cleanliness of the surfaces, a fact of great practical consequence. A trace of grease on glass, which water does not wet, sharply increases the angle of contact and can reduce or reverse the capillary rise; this is why glassware for surface-tension experiments must be scrupulously clean. Conversely, adding a detergent or soap to water lowers its surface tension and reduces its angle of contact against greasy fabric, which is precisely how detergents let water penetrate and wet dirty cloth that plain water runs off. The chemistry of wetting agents is the deliberate engineering of the angle of contact. 🔉✕⇢

There is a second, equivalent way to derive the capillary rise that makes its connection to curved-surface pressure explicit, and JEE problems use both. Just beneath the concave meniscus of a wetting liquid the pressure is lower than atmospheric by the excess-pressure amount $2S/R$, where $R$ is the radius of curvature of the meniscus. For a meniscus that meets the wall of a tube of radius $a$ at an angle of contact $\theta$, geometry gives $R=a/\cos\theta$, so the pressure deficit is $2S\cos\theta/a$. The liquid rises until the hydrostatic pressure $\rho g h$ of the raised column exactly cancels this deficit, giving $\rho g h=2S\cos\theta/a$ and hence the same $h=2S\cos\theta/(\rho g a)$ as the force balance. 🔉✕⇢

That the force-balance derivation and the pressure-balance derivation give identical answers is not a coincidence but a check that the physics is consistent: the upward pull of surface tension around the rim and the pressure deficit beneath the curved meniscus are two descriptions of the same underlying surface-tension force. A student who can move fluently between the two pictures — force per unit length around the contact circle, and pressure difference across the curved surface — can handle any capillary problem the examiners set, including those that give the radius of curvature of the meniscus rather than the angle of contact directly. 🔉✕⇢

The simple formula slightly overstates the rise because it neglects the weight of the liquid in the meniscus itself, above the flat portion of the column. A more careful treatment for a hemispherical meniscus replaces the column height $h$ by $h+a/3$, the extra $a/3$ accounting for the liquid held in the curved cap. For fine capillaries, where $a$ is much smaller than $h$, this correction is negligible and the standard formula is used; it is quoted here because Advanced problems occasionally ask for it, and because it shows that the elementary derivation is a leading approximation rather than an exact result. 🔉✕⇢

Jurin's law is the name given to the inverse-radius result $h\propto 1/a$, and stating it as a law emphasises what is being asserted: for a given liquid and tube material at a fixed temperature, the product of the rise and the radius is constant. This gives a quick way to compare tubes — if a liquid rises $8\,\text{cm}$ in a tube of a certain bore, it rises only $2\,\text{cm}$ in a tube of four times the radius — and it underlies the falling-head and rise-height methods of measuring surface tension, in which $S$ is extracted from a measured rise in a tube of known radius. 🔉✕⇢

Mercury in a glass tube illustrates the depression quantitatively and is a favourite exam contrast. With an angle of contact of about $140^\circ$, $\cos\theta$ is roughly $-0.77$, so the same formula gives a negative $h$: the mercury inside a fine tube stands below the level in the reservoir, and its meniscus bulges upward (convex) rather than dipping down. This depression is why the bore of a mercury barometer or thermometer must be corrected for capillarity, and why mercury, unlike water, has to be pushed rather than drawn into a fine tube. 🔉✕⇢

A subtle but frequently tested point concerns a capillary tube that is shorter than the height to which the liquid would otherwise rise. The liquid does not spurt out of the top like a fountain. Instead it rises to the top of the tube and stops, and the meniscus there adjusts its radius of curvature — flattening out to a larger radius — so that the reduced pressure deficit $2S/R$ just balances the smaller available column. The liquid never overflows, because overflow would demand a curvature the surface cannot sustain. Recognising that it is the meniscus curvature, not the rise height, that adjusts is the key to these problems. 🔉✕⇢

Surface tension also depends on temperature, and this feeds through to capillarity. The surface tension of a liquid decreases as the temperature rises, falling to zero at the critical temperature where the distinction between liquid and vapour disappears. Because the capillary rise is proportional to $S$, warm water rises less in a capillary than cold water, other things being equal. This temperature dependence is exploited in devices that sense temperature through changes in a liquid's wetting behaviour, and it is one reason surface-tension measurements must specify the temperature at which they were made. 🔉✕⇢

Related surface-tension effects, though not capillary rise in a tube, share the same physics and often appear alongside it. A small, dense object such as a steel needle or a razor blade can be made to float on water, unsupported by buoyancy alone, because the depressed surface acts like a stretched skin whose upward surface-tension force supports the object's weight. Water striders and other insects walk on ponds for the same reason, their water-repellent legs resting in dimples of the surface. These demonstrations make the reality of the 'stretched membrane' vivid and are common qualitative JEE questions. 🔉✕⇢

It is worth drawing together why the meniscus curves the way it does, because the direction of the effect follows from it. For a wetting liquid, strong adhesion pulls the edge of the surface up the wall, making the meniscus concave (curving upward at the sides); the surface then has its centre of curvature above the liquid, the pressure just beneath it is reduced, and the liquid is drawn up. For a non-wetting liquid, strong cohesion pulls the surface away from the wall, making the meniscus convex; the centre of curvature lies below the surface, the pressure just beneath it is raised, and the liquid is pushed down. The shape of the meniscus, the sign of the excess pressure and the direction of the capillary effect are three faces of the same fact. 🔉✕⇢

Two habits make capillary problems reliable. First, always read off $\theta$ from the physics of the particular pair — near zero for clean water on glass, about $140^\circ$ for mercury on glass — and carry its cosine with the correct sign, so that a depression emerges automatically for a non-wetting liquid. Second, keep the inverse-radius dependence in the front of the mind, since most numerical questions turn on comparing two tubes or on the effect of halving a bore. With the sign of $\cos\theta$ and the $1/a$ scaling secure, the single formula $h=2S\cos\theta/(\rho g a)$ answers the whole family of capillary questions the examination can pose. 🔉✕⇢

Capillarity is of enormous practical importance. It draws water up through the fine channels of soil and into the roots and stems of plants, carries oil up the wick of a lamp, and lets a towel or blotting paper soak up liquid through the tiny gaps between its fibres. In each case a narrow gap plus a wetting liquid plus surface tension lifts the liquid against gravity, precisely as the capillary-rise formula describes. 🔉✕⇢

The angle of contact is fixed by the balance of three interfacial tensions acting along the line where liquid, solid and air meet: the liquid-air tension $S_{la}$, the solid-air tension $S_{sa}$ and the solid-liquid tension $S_{sl}$. Resolving these along the solid surface in equilibrium gives the relation $S_{la}\cos\theta + S_{sl} = S_{sa}$. The angle of contact is an obtuse angle if $S_{sl}$ is larger than $S_{la}$: the molecules of the liquid are then attracted strongly to themselves and weakly to those of the solid, it costs a lot of energy to create a liquid-solid surface, and the liquid does not wet the solid, as with water on a waxy surface or mercury on glass. If instead the liquid is strongly attracted to the solid, $S_{sl}$ falls, $\cos\theta$ increases and the angle of contact becomes acute, as for water on clean glass. 🔉✕⇢

This is why soaps, detergents and dyeing substances act as wetting agents: added to water they make the angle of contact small so that the liquid can penetrate a fabric well and become effective. Water proofing agents do the opposite, creating a large angle of contact between the water and the fibres so that drops bead up and run off rather than soaking in. The same competition between adhesion to the solid and cohesion within the liquid decides whether the meniscus is concave or convex, and hence whether the liquid climbs a fine tube or is pushed down inside it. 🔉✕⇢

Because a liquid-air interface has energy proportional to its area, a free drop or bubble takes the shape of least area for a given volume, which is a sphere. If a spherical drop of radius $r$ increases its radius by $\Delta r$, the extra surface energy created is $8\pi r\,\Delta r\,S$, and in equilibrium this energy cost is balanced by the work $(P_i-P_o)4\pi r^2\,\Delta r$ done by the pressure difference between the inside and the outside. Equating the two gives the excess pressure inside a drop as $P_i-P_o = 2S/r$. A soap bubble differs from a drop in that it has two liquid-air interfaces, so applying the same argument doubles the result to $4S/r$, which is why you have to blow a little hard, but not too hard, to form a soap bubble. 🔉✕⇢

Capillary rise follows directly from this pressure difference across a curved surface. Where water wets glass the meniscus is concave, so the pressure of the water just at the meniscus is less than the atmospheric pressure by $2S\cos\theta/a$ for a tube of radius $a$. Considering a point A just below the meniscus inside the tube and a point B at the flat surface outside, both must be at the same pressure, so $P_0 + h\rho g = P_i$, and combining the two relations gives $h\rho g = 2S\cos\theta/a$. The capillary rise $h$ is therefore due to surface tension and is larger for a smaller radius $a$ — the origin of capilla, the Latin word for hair — and for water in a fine tube of radius 0.05 cm it works out to about 2.98 cm. 🔉✕⇢

If instead the meniscus is convex, as it is for mercury, which does not wet glass, then $\cos\theta$ is negative, the pressure just below the surface is higher than atmospheric, and the liquid is pushed down so that the level in the capillary falls below the level in the vessel. The single formula $h = 2S\cos\theta/\rho g a$ handles both cases through the sign of $\cos\theta$, giving a rise for a wetting liquid and a depression for a non-wetting one. Carrying the correct sign of the cosine is the single most reliable habit in solving capillary problems. 🔉✕⇢

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