Chapter 10 · Fluid Mechanics · Full book

Fluid Mechanics: The Physics of Flow and Float

Core intuition: Chapter Roadmap

A fluid is matter that cannot resist a shear at rest—so it flows. Everything in this chapter follows from that one property, combined with the three great conservation laws applied to a continuous medium.

The defining property. A fluid yields to any shear stress, however small; this is what separates liquids and gases from solids and forces us to speak of pressure (a force per unit area) rather than tension or compression along a direction.
Hydrostatics. A fluid at rest obeys \(dp/dz = -\rho g\), integrating to \(p = p_0 + \rho g h\)—pressure set by depth alone. From it follow the barometer, the manometer, and Pascal's principle.
Buoyancy. Archimedes' principle—the upthrust equals the weight of displaced fluid—decides what floats, how deep, and how much it can carry.
Continuity. Mass conservation in a tube of flow gives \(A v = \text{const}\): squeeze the channel and the fluid must speed up.
Bernoulli. Energy conservation along a streamline ties pressure, speed, and height together; its applications—Venturi, Torricelli, the Pitot tube, dynamic lift—are where the surprises live.
Viscosity and Poiseuille flow. Real fluids resist shear rate; this internal friction sets the flow rate through a pipe (\(Q \propto r^4\)) and the terminal velocity of a falling sphere.
Surface tension and capillarity. The energy cost of an interface curves menisci, raises liquids in fine tubes, and sets the excess pressure inside drops and bubbles.
The Reynolds number. A single dimensionless ratio decides when the orderly laminar picture gives way to turbulence—and when all the clean formulas above stop applying.

Master the move from “a fluid cannot hold a shear” to “mass, energy, and momentum are conserved in what flows,” and the whole subject organises itself.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Steel Ship Paradox: A steel cargo ship weighing fifty thousand tonnes floats without effort in the sea. A solid steel coin, a million times lighter, is dropped into the same water and sinks immediately. Both objects are made of the same material. Steel is denser than water. Nothing about the material has changed between the two objects. What decides whether something floats?
Speed Up, Pressure Down: Water enters a wide pipe at low speed and high pressure. It flows into a narrow constriction, speeds up, and—according to every measurement—its pressure drops. Intuitively, squeezing a fluid ought to raise its pressure. Yet the narrow section, where the fluid is most “squeezed together,” has the lowest pressure. Where has the intuition failed?
The Curve Ball: A spinning cricket ball curves sideways through still air. No sideways force is applied by the bowler after release. The ball leaves the hand moving in a straight line; the only forces on it are gravity (downward) and air resistance. Yet the ball curves horizontally. How does spinning generate a sideways aerodynamic force?
The Shower Curtain: When a shower runs, the freely hanging curtain swings inward, toward the running water, and presses against your legs. No draft is blowing inward. The water is falling downward, not blowing sideways. What force is pulling the curtain toward the running water?

Each question targets a trap that catches students every year.
By the end of this chapter, every one of them will be transparent.

What Is a Fluid?

◈ Pascal’s hydraulic press

live bench

Push the small piston and the pressure rises everywhere at once, so \(F_2=F_1\,A_2/A_1\). Set the area ratio and watch a light effort raise a heavy load — with the work bookkeeping that forbids a free lunch.

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↗ open in the Simulations page

Every previous chapter of this book dealt with rigid bodies or point particles: objects whose shape does not change under applied forces, or whose internal structure we could ignore entirely. A block on a ramp, a rotating disc, a satellite in orbit—none of these required us to think about what the object is made of at the molecular level.

Fluids are different. A fluid is a substance that deforms continuously and without limit under any shear stress, no matter how small. This single property—the inability to sustain a static shear—is what distinguishes every fluid from every solid, and it has sweeping consequences for every problem in this chapter.

Shear: The Distinguishing Test

Push the top face of a rubber block sideways while holding the bottom face fixed. The block deforms, but resists: an elastic shear stress pushes back against your hand, and when you release, the block springs back. A solid sustains a static shear.

Do the same with water in a container. Push the top layer sideways. The water offers no sustained resistance to a static displacement—it flows without limit. There is a viscous resistance to the rate of shear (more on this later in the chapter), but none to a static shear displacement.

This is why water takes the shape of its container. A solid maintains its shape because molecular bonds resist both compression and shear. A liquid resists compression (it has a large bulk modulus) but offers zero resistance to shear. A gas resists neither compression strongly nor shear at all.

Core intuition: The One Property That Defines a Fluid

A fluid is a substance that cannot sustain any static shear stress. Under any sustained lateral force, however small, a fluid deforms indefinitely—it does not reach a static equilibrium. This is a qualitative distinction, not a quantitative one: it is what separates the entire class of fluid problems from all of solid mechanics.

Liquids and Gases: Two Kinds of Fluid

Both liquids and gases are fluids—they share zero static shear modulus—but they differ in compressibility.

A liquid is nearly incompressible. Its molecules are in near-contact; enormous pressure is needed to change the volume appreciably. For all problems in this chapter, liquids are treated as perfectly incompressible: density \(\rho\) is constant regardless of pressure.

A gas is highly compressible: pressure and volume are strongly coupled. However, when flow speeds are well below the speed of sound, the pressure differences in a gas are small enough that density changes are negligible. This is the regime of incompressible flow, valid for virtually all JEE, NEET, and AP Fluid Mechanics problems.

Throughout this chapter, fluid means an incompressible fluid unless explicitly stated otherwise.

Pressure: The Language of Fluid Forces

Fluids transmit force through pressure, not through shear or tension. Pressure is defined as the normal (perpendicular) force per unit area that the fluid exerts on any surface in contact with it: \[ p = \frac{F_\perp}{A} \]

Pressure is a scalar. At any given point in a fluid at rest, the pressure exerted on a surface element is the same regardless of the orientation of that surface. This isotropy of static pressure—which we will derive below—is the foundation of Pascal's principle.

The SI unit is the pascal: \(1\,\mathrm{Pa} = 1\,\mathrm{N\,m^{-2}}\). Standard atmospheric pressure is \(p_0 = 1.013\times 10^5\,\mathrm{Pa} \approx 10^5\,\mathrm{Pa}\).

Connections to Earlier Chapters

Newton's laws (Chapter 3). Fluid statics is \(\sum\vec{F} = 0\) applied to a fluid element. Fluid dynamics is \(\sum\vec{F} = m\vec{a}\) applied to a fluid parcel. No new law of physics is introduced.

Work–energy theorem (Chapter 4). Bernoulli's equation, the central result of fluid dynamics, is the work–energy theorem applied to a fluid parcel moving along a streamline. It has no physical content beyond energy conservation.

Gravitation (Chapter 8). The hydrostatic equation \(dp/dz = -\rho g\) is gravity pulling on each fluid layer. The pressure increase with depth is the weight of the fluid column above.

Waves. The speed of sound \(v = \sqrt{B/\rho}\) was quoted without proof in the Waves volume. In this chapter we introduce the bulk modulus \(B\) fully and justify that formula.

What This Chapter Covers

We first derive the hydrostatic pressure equation and Pascal's law. Archimedes' principle is treated next, answering the steel ship question. We then develop fluid dynamics: the continuity equation, Bernoulli's equation, and applications. Viscosity, surface tension, pitfalls, and the exercise bank follow.

Misconception alert

“Pressure acts in the direction of the applied force.” Wrong. Pressure is isotropic in a static fluid: at any point, the pressure is the same regardless of the orientation of the test surface. The force exerted by pressure on a surface element depends on the surface's orientation, not on how the pressure was created.
“A denser fluid always settles to the bottom.” Only for static, immiscible fluids. In static equilibrium, denser immiscible layers do settle below lighter ones (oil floats on water). In flowing systems, and with miscible fluids, density differences drive more complex convective behaviour. Do not apply the settling rule outside static, immiscible contexts.

✓ Checkpoint · Before We Begin

A rubber eraser, a blob of modelling clay, and a body of water (confined to a box) are each subjected to a small lateral (shear) force. The eraser deforms and springs back; the clay deforms and holds the new shape; the water flows. Using only the definition in this section, classify each as a solid or a fluid.
Eraser: solid—it resists static shear and is elastic. Clay: solid at the JEE level—it sustains static shear below its yield stress (plastic solid). Water: fluid—it cannot sustain any static shear and deforms without limit.
Two identical cylinders are filled to the same height \(H\), one with fresh water (\(\rho = 1000\,\mathrm{kg\,m^{-3}}\)) and one with mercury (\(\rho = 13{,}600\,\mathrm{kg\,m^{-3}}\)). Which exerts a greater pressure on the base, and by what factor?
Mercury: pressure at base \(= \rho g H\); since height is equal, the ratio is \(13{,}600/1000 = 13.6\). Mercury exerts \(13.6\) times the pressure.
Explain in one sentence—using only the concept of shear resistance—why a liquid takes the shape of its container but a solid does not.
A liquid cannot sustain any static shear stress, so it deforms continuously under any unbalanced lateral force until it fills the container; a solid's shear modulus resists such deformation and maintains its shape.

Pressure and the Hydrostatic Equation

◈ Pressure with depth

live bench

Pick a fluid and lower the probe: pressure climbs linearly as \(p=p_0+\rho g h\). The gauge and absolute readouts and the p–depth graph move together — depth alone sets the pressure, not the shape of the vessel.

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A fluid at rest is in static equilibrium: every infinitesimal fluid element has zero net force on it. Applying Newton's first law to such an element yields, with almost no algebra, the single equation that governs all of hydrostatics.

Pressure Is Isotropic: The Pascal Proof

Before asking how pressure varies in space, we must establish that pressure at a single point is a well-defined scalar, independent of how we orient the test surface.

Consider a small right-triangular wedge of fluid at rest. Let the three faces have outward unit normals: horizontal (area \(dy \cdot 1\)), vertical (area \(dx \cdot 1\)), and hypotenuse (area \(ds \cdot 1\)) at angle \(\theta\) from horizontal. Let \(p_x\), \(p_y\), \(p_n\) be the pressures on the vertical, horizontal, and hypotenuse faces respectively.

Horizontal equilibrium: \[ p_x\,dy = p_n\,ds\sin\theta = p_n\,dy \quad\Longrightarrow\quad p_x = p_n \]

Vertical equilibrium (weight of the wedge \(\to 0\) as \(dx,dy\to 0\)): \[ p_y\,dx = p_n\,ds\cos\theta = p_n\,dx \quad\Longrightarrow\quad p_y = p_n \]

Therefore \(p_x = p_y = p_n\) for any angle \(\theta\). Pressure at a point is the same on every orientation of test surface. This is the microscopic content of Pascal's principle.

Deriving the Hydrostatic Equation

Fig 11.1 A thin fluid slab of area \(A\), thickness \(dz\), at depth \(h\). The upward push on its base beats the downward push on its top by exactly the slab’s weight — giving \(dp/dz=-\rho g\), which integrates to \(p=p_0+\rho g h\).

Take a thin horizontal slab of fluid at height \(z\) (measured upward from any reference), with cross-sectional area \(A\) and thickness \(dz\). Mass of slab: \(dm = \rho A\,dz\).

Forces acting on it:

Upward pressure from below: \(p(z)\cdot A\)
Downward pressure from above: \(p(z+dz)\cdot A\)
Weight downward: \(\rho g A\,dz\)

Vertical equilibrium, \(\sum F_z = 0\): \[ p(z)\cdot A\;-\;p(z+dz)\cdot A\;-\;\rho g A\,dz = 0 \] \[ \frac{dp}{dz} = -\rho g \]

Birth of the formula: The Hydrostatic Pressure Equation

For an incompressible fluid (\(\rho = \) const) in a uniform gravitational field, integrating \(dp/dz = -\rho g\) from the free surface (where \(p = p_0\), absolute atmospheric pressure) to a depth \(h\) below the surface: \[ \boxed{p = p_0 + \rho g h} \] \(h\) is measured downward from the free surface. Equivalently, for any two points in the same connected static fluid: \[ p_1 + \rho g z_1 = p_2 + \rho g z_2 \] where \(z\) is measured upward. The quantity \(p + \rho g z\) is constant throughout a connected, static, incompressible fluid.

Physical Interpretation

This equation is not a formula—it is a statement that the pressure at any depth must support the weight of the fluid column above per unit area. This yields three immediate consequences.

1. Pressure depends only on depth, not container shape. The pressure at depth \(h\) in a narrow tube, a wide tank, and an irregularly shaped vessel are identical, provided the vessels are filled with the same fluid and their surfaces are open to the same atmosphere. This is the hydrostatic paradox: a narrow column of water can exert the same base pressure as a vast ocean, if their depths are equal.

2. The free surface of a connected fluid is horizontal at rest. If it were not, the hydrostatic equation would give different pressures on the two sides just below the surface, driving a flow from high to low pressure until the surface levels.

3. Gauge pressure vs. absolute pressure. The term \(\rho g h\) is the gauge pressure: pressure in excess of atmospheric. Tyre pressures, blood pressures, and most engineering readings are gauge pressures. Absolute pressure = gauge pressure + \(p_0\). This distinction trips up JEE candidates regularly.

Layered Fluids and Pressure Matching

When immiscible fluids of different densities occupy the same container, the pressure is continuous across their interface (but its gradient changes). At every horizontal level, the pressure is the same everywhere on that level within the connected fluid.

The strategy for all layered-fluid problems is:

Identify the interface between fluids.
Match pressures at the interface from both sides.
Use \(p = p_0 + \rho g h\) within each layer.

Pascal's Law and the Hydraulic Press

Fig 11.2 Pascal’s law: a force \(F_1\) on the small piston raises the pressure \(F_1/A_1\) everywhere at once, so the large piston pushes back with \(F_2=F_1\,A_2/A_1\). A hydraulic press multiplies force by the area ratio — the small piston simply moves farther.

Pressure is transmitted undiminished throughout a connected fluid. A force \(F_1\) on a small piston of area \(A_1\) creates pressure \(\Delta p = F_1/A_1\) everywhere in the fluid. On a large piston of area \(A_2\), this produces force: \[ F_2 = \Delta p \cdot A_2 = F_1\,\frac{A_2}{A_1} \] A small force produces a large force. Energy is conserved because the small piston must move distance \(d_1 = d_2\,(A_2/A_1)\), so \(F_1 d_1 = F_2 d_2\).

Example 1

A U-tube manometer contains two immiscible fluids: oil (\(\rho_1 = 800\,\mathrm{kg\,m^{-3}}\)) on the left arm and water (\(\rho_2 = 1000\,\mathrm{kg\,m^{-3}}\)) on the right arm. The oil column extends \(20\,\mathrm{cm}\) above the oil–water interface. Find the height difference between the oil surface (left) and the water surface (right). \((g = 9.8\,\mathrm{m\,s^{-2}})\)

Solution. Let the oil–water interface be the reference level (height zero) on both sides. Let the oil surface be at height \(H_1 = 0.20\,\mathrm{m}\) and the water surface at height \(H_2\) above the interface. At the interface, pressures from both arms must be equal: \[ p_0 + \rho_1 g H_1 = p_0 + \rho_2 g H_2 \] \[ H_2 = \frac{\rho_1}{\rho_2}\,H_1 = \frac{800}{1000}\times 0.20 = 0.16\,\mathrm{m} \] The oil surface is \(0.20\,\mathrm{m}\) above the interface; the water surface is \(0.16\,\mathrm{m}\) above it. Height difference = \(0.20 - 0.16 = \mathbf{0.04\,m = 4\,cm}\), with the oil surface higher. Physical check: oil is less dense, so it must stand taller to produce the same pressure at the interface. ✓

✓ Checkpoint · Pressure and Hydrostatics

A large cylindrical tank of base area \(A\) is filled with water to height \(H\). A narrow vertical tube of cross-section \(a \ll A\) is connected to the base and water is pumped up until the level in the tube reaches \(2H\). State the pressure at the base of the tank and explain the “hydrostatic paradox”: why the shape of the container is irrelevant.
Pressure at the base \(= p_0 + \rho g(2H)\), determined by the connected free-surface height \(2H\) in the tube, not by the water level in the tank. The pressure at any depth depends only on the height of the connected free surface above—not on the column geometry, volume, or shape. This is the hydrostatic paradox: a pencil-thin tube can raise the base pressure of a vast tank to any value.
A hydraulic press has pistons of diameters \(2\,\mathrm{cm}\) and \(20\,\mathrm{cm}\). A force of \(200\,\mathrm{N}\) is applied to the small piston. Find the load the large piston can support, and the distance the small piston must be pushed down to raise the large piston by \(1\,\mathrm{cm}\).
Area ratio \((20/2)^2 = 100\). Force \(= 200\times 100 = 20{,}000\,\mathrm{N}\). Volume conservation: \(A_1 d_1 = A_2 d_2\), so \(d_1 = 100\times 1\,\mathrm{cm} = 100\,\mathrm{cm} = 1\,\mathrm{m}\).
(Trap) A student calculates the pressure on a diver at \(10\,\mathrm{m}\) depth in seawater (\(\rho = 1025\,\mathrm{kg\,m^{-3}}\)) as \(\rho g h = 1025\times 9.8\times 10 \approx 100.5\,\mathrm{kPa}\) and reports this as the pressure on the diver. Identify the error and correct the answer.
The student computed gauge pressure only, omitting atmospheric pressure \(p_0 \approx 101.3\,\mathrm{kPa}\). Absolute pressure \(= 101.3 + 100.5 \approx 201.8\,\mathrm{kPa}\), roughly twice atmospheric. This is why a diver feels ear pressure at \(10\,\mathrm{m}\) and requires a compressed-air supply. JEE frequently asks for absolute pressure when students compute gauge, or vice versa.

Archimedes' Principle and Buoyancy

◈ Archimedes buoyancy

live bench

Slide the body’s density against the fluid’s: the submerged fraction settles at \(\rho_b/\rho_f\). Buoyancy and weight arrows compete and the verdict — float, sink or hover — follows.

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↗ open in the Simulations page

Around 250 BCE, King Hiero II of Syracuse handed his goldsmith a lump of pure gold and commissioned a royal crown. When the crown was returned, it had the correct weight. But the king was suspicious: had the goldsmith substituted cheaper silver for some of the gold, making up the mass deficit with extra metal? The crown could not be melted down to check—it was, after all, a crown.

Hiero handed the problem to Archimedes of Syracuse, the greatest mathematician and engineer of the ancient world. The story—told by the Roman architect Vitruvius three centuries later, and possibly embellished—is that Archimedes was lowering himself into a full bath when he noticed water sloshing over the rim. He realised, in that instant, that the volume of water displaced equalled the volume of his body submerged, and that this gave him a way to measure the crown's volume without cutting it. Gold is denser than silver; an adulterated crown of the same mass would have a larger volume and displace more water.

The legend continues that Archimedes leapt from the bath and ran naked through the streets of Syracuse shouting Héureka!—“I have found it!”

Whether or not he ran naked down the street, the physics he discovered that day is precise, general, and profound enough to appear on every competitive examination two and a half millennia later.

Observation challenge

How would you actually do the experiment? You have the crown, a bucket of water, a scale, and nothing else. Before reading on: design the measurement that catches the goldsmith. The key is that you need the crown's density, not just its mass or volume separately.

Buoyancy requires no new physics. It is a direct consequence of the hydrostatic pressure equation: the pressure at the bottom of a submerged object is greater than the pressure at the top, and this pressure difference produces a net upward force. Everything else in this section is a consequence of that fact.

The Origin of Buoyancy: A Pressure Argument

Fig 11.3 Pressure grows with depth, so the upward force on the bottom face exceeds the downward force on the top; the horizontal pushes cancel. The leftover is the buoyant force — Archimedes’ upthrust, equal to the weight of displaced fluid.

Submerge an object of arbitrary shape in a fluid. Every point on the object's surface is in contact with fluid at some pressure. The net fluid force is the integral of \(p\,d\vec{A}\) over the entire surface (inward normals).

Now consider the following thought experiment: remove the object and replace it with a “ghost” volume of fluid of exactly the same shape, filled with the same fluid as the surroundings. The surface pressure distribution is unchanged (it depends only on depth, not on the material inside). This ghost volume is in equilibrium—it is just fluid in fluid. Therefore the net pressure force on the ghost volume exactly equals its weight, directed upward.

But the pressure force on the real object has the same magnitude as on the ghost volume (same surface, same pressures). Therefore:

Birth of the formula

The buoyant force on any object, fully or partially submerged, equals the weight of the fluid it displaces: \[ \boxed{F_\mathrm{buoy} = \rho_f\,V_\mathrm{sub}\,g} \] where \(\rho_f\) is the fluid density and \(V_\mathrm{sub}\) is the submerged volume of the object. Direction: vertically upward, always. Point of application: the centre of buoyancy—the centroid of the displaced fluid volume (not the object's centre of mass).

The buoyant force depends only on the fluid density and the displaced volume. It does not depend on the object's density, mass, material, or shape. A wooden block and a steel block of equal volume experience identical buoyant forces when fully submerged in water.

Floating, Sinking, and Neutral Buoyancy

Fig 11.4 A floating body settles until buoyancy balances weight. Equating \(\rho_f g V_{sub}=\rho_b g V\) fixes the submerged fraction \(V_{sub}/V=\rho_b/\rho_f\) — the ratio of densities, and nothing else, sets how deep it rides.

Compare \(F_\mathrm{buoy}\) with the object's weight \(mg = \rho_\mathrm{obj} V_\mathrm{obj} g\):

\(\rho_\mathrm{obj} \gt \rho_f\): Even at full submersion, \(F_\mathrm{buoy} \lt mg\). The object sinks. A solid steel cube sinks in water because \(\rho_\mathrm{steel} = 7800\,\mathrm{kg\,m^{-3}} \gt 1000\,\mathrm{kg\,m^{-3}}\).

\(\rho_\mathrm{obj} \lt \rho_f\): The object floats with only part of its volume submerged. Equilibrium requires \(F_\mathrm{buoy} = mg\): \[ \rho_f\,V_\mathrm{sub}\,g = \rho_\mathrm{obj}\,V_\mathrm{obj}\,g \] \[ \frac{V_\mathrm{sub}}{V_\mathrm{obj}} = \frac{\rho_\mathrm{obj}}{\rho_f} \]

An iceberg (\(\rho \approx 920\,\mathrm{kg\,m^{-3}}\), \(\rho_\mathrm{sea} \approx 1025\,\mathrm{kg\,m^{-3}}\)) submerges \(920/1025 \approx 89.8\%\) of its volume: almost \(90\%\) is hidden.

\(\rho_\mathrm{obj} = \rho_f\): The object is in neutral buoyancy and hovers at any depth. Submarines adjust ballast tank flooding to achieve this.

Resolving the Steel Ship Question

A solid steel cube of volume \(V\) has average density \(\rho_\mathrm{steel} = 7800\,\mathrm{kg\,m^{-3}}\). Its weight exceeds the buoyant force at full submersion. It sinks.

Now take the same mass of steel and hammer it into a hollow hull of total enclosed volume \(V'\) (steel + air inside). The total mass is still \(m = \rho_\mathrm{steel} V_\mathrm{steel}\) (the air mass is negligible), but the total enclosed volume \(V' \gg V_\mathrm{steel}\).

The average density of the closed system is: \[ \bar{\rho} = \frac{m}{V'} = \frac{\rho_\mathrm{steel}\,V_\mathrm{steel}}{V'} \] By making the hull large enough (enclosing enough air), we can make \(\bar\rho\) as small as we wish—well below \(\rho_\mathrm{water} = 1000\,\mathrm{kg\,m^{-3}}\).

Core intuition: What Actually Determines Whether Something Floats

An object floats if and only if the average density of its total closed volume—hull material plus all enclosed contents (air, cargo, water in ballast tanks)—is less than the fluid density. The material of the hull is not the deciding factor; the fraction of the enclosed volume that is air (or other low-density material) is. A steel ship, a sealed empty bottle, and a hollow sphere all float by this one principle.

Apparent Weight

When an object is weighed while submerged, the scale reads the apparent weight: \[ W_\mathrm{app} = mg - F_\mathrm{buoy} = (\rho_\mathrm{obj} - \rho_f)\,V_\mathrm{obj}\,g \]

This answers the puzzlebox challenge. Weigh the crown in air: read \(W_\mathrm{air} = mg\). Weigh it again while fully submerged in water: read \(W_\mathrm{app}\). The difference \(W_\mathrm{air} - W_\mathrm{app} = \rho_w V g\) gives the crown's volume \(V\). Now compute the crown's density: \(\rho_\mathrm{crown} = m/V = W_\mathrm{air}/(g \cdot V)\). If \(\rho_\mathrm{crown} \lt \rho_\mathrm{gold} = 19{,}300\,\mathrm{kg\,m^{-3}}\), the goldsmith cheated. No melting, no cutting, no destruction of the crown. Just a bucket of water and a scale—and the insight that the buoyant force measures displaced volume directly. History does not record what happened to the goldsmith.

Misconception alert

“The buoyant force equals the weight of the object.” Wrong—unless the object is floating in equilibrium. The buoyant force always equals the weight of the displaced fluid. For a floating object, equilibrium makes these numerically equal. For a submerged sinking object, \(F_\mathrm{buoy} \lt mg\). For an object held under by a string (which would otherwise float), \(F_\mathrm{buoy} \gt mg\) and the string pulls downward. Always compute \(\rho_f V_\mathrm{sub} g\) first; then compare with \(mg\) to determine the net force or the constraint force.

Example 2

A wooden block of mass \(m = 2.0\,\mathrm{kg}\) and volume \(V = 3.0\times10^{-3}\,\mathrm{m^3}\) is held fully submerged in water (\(\rho_w = 1000\,\mathrm{kg\,m^{-3}}\)) by a vertical string attached to the container floor. (a) Find the tension in the string. (b) The string is cut. Find the fraction of the block's volume that protrudes above the water surface at equilibrium. \((g = 9.8\,\mathrm{m\,s^{-2}})\)

Solution. (a) Buoyant force (upward): \[ F_\mathrm{buoy} = \rho_w V g = 1000\times3.0\times10^{-3}\times9.8 = 29.4\,\mathrm{N} \] Weight of block (downward): \[ mg = 2.0\times9.8 = 19.6\,\mathrm{N} \] Since \(F_\mathrm{buoy} \gt mg\), the block would rise; the string pulls it downward. Equilibrium: \(F_\mathrm{buoy} = mg + T\): \[ T = 29.4 - 19.6 = \mathbf{9.8\,\mathrm{N}} \] (b) Density of block: \[ \rho_\mathrm{block} = \frac{m}{V} = \frac{2.0}{3.0\times10^{-3}} \approx 667\,\mathrm{kg\,m^{-3}} \] Since \(\rho_\mathrm{block} \lt \rho_w\), the block floats. Using the equation above: \[ \frac{V_\mathrm{sub}}{V} = \frac{667}{1000} \approx 0.667 \] Fraction above water \(= 1 - 0.667 = \mathbf{1/3}\).

✓ Checkpoint · Archimedes' Principle and Buoyancy

A metal block is weighed in air (reading \(W_\mathrm{air}\)) and then while fully submerged in oil of density \(\rho_\mathrm{oil}\) (reading \(W_\mathrm{oil}\)). Derive an expression for the volume of the block in terms of \(W_\mathrm{air}\), \(W_\mathrm{oil}\), \(\rho_\mathrm{oil}\), and \(g\).
\(F_\mathrm{buoy} = W_\mathrm{air} - W_\mathrm{oil}\). Also \(F_\mathrm{buoy} = \rho_\mathrm{oil} V g\). Therefore \(V = (W_\mathrm{air} - W_\mathrm{oil})/(\rho_\mathrm{oil}\,g)\). This is the operating principle of the Archimedes hydrostatic balance—density measured without melting or cutting.
An ice cube of mass \(m\) floats in a glass of fresh water. When the ice melts completely, does the water level rise, fall, or remain unchanged? Justify rigorously using the equation above.
Unchanged. Floating ice displaces water of volume \(V_\mathrm{sub} = m/\rho_w\) (from \(\rho_w V_\mathrm{sub} g = mg\)). When it melts, it produces mass \(m\) of water occupying volume \(m/\rho_w\)—exactly the displaced volume. The water level is unchanged. Caveat: if the ice contains trapped air bubbles, the argument fails slightly; JEE has tested this variant.
(Olympiad level) A uniform solid cylinder (\(\rho_c = 800\,\mathrm{kg\,m^{-3}}\), height \(H\), base area \(A\)) is placed in a cylindrical container of base area \(4A\), initially half-filled with water. Find the height by which the cylinder protrudes above the final water surface.
Let \(x\) = submerged depth of cylinder at equilibrium. Buoyancy condition: \(\rho_c H A g = \rho_w x A g\), so \(x = (\rho_c/\rho_w) H = 0.8H\). The cylinder displaces water of volume \(0.8HA\), which spreads over the remaining container area \(3A\), raising the water level by \(0.8H/3\). The protrusion above the final water surface \(= H - x = H - 0.8H = 0.2H\), independent of the water-level rise. (The rise shifts both the cylinder and the surface upward by the same amount.)

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Gauge versus absolute pressure on a diver

A scuba diver descends to a depth of \(30\,\mathrm{m}\) in seawater (\(\rho = 1025\,\mathrm{kg\,m^{-3}}\)). Find (a) the gauge pressure and (b) the absolute pressure at that depth. \((g = 9.8\,\mathrm{m\,s^{-2}},\ p_0 = 1.013\times10^5\,\mathrm{Pa})\)

Solution. Given: \(h = 30\,\mathrm{m}\), \(\rho = 1025\,\mathrm{kg\,m^{-3}}\), \(p_0 = 1.013\times10^5\,\mathrm{Pa}\).
Find: gauge pressure \(p_g\) and absolute pressure \(p\).
Setup: Gauge pressure is the fluid column's own contribution, \(p_g = \rho g h\); absolute pressure adds the atmosphere pressing on the surface, \(p = p_0 + \rho g h\).
Solve: \[ p_g = \rho g h = 1025 \times 9.8 \times 30 = 3.01\times10^5\,\mathrm{Pa}, \] \[ p = p_0 + p_g = 1.013\times10^5 + 3.01\times10^5 = 4.03\times10^5\,\mathrm{Pa}. \] Answer: \(\boxed{p_g \approx 3.01\times10^5\,\mathrm{Pa},\quad p \approx 4.03\times10^5\,\mathrm{Pa}}\)
Check: Rule of thumb: each \(10\,\mathrm{m}\) of seawater adds about one atmosphere, since \(\rho g(10) \approx 1.0\times10^5\,\mathrm{Pa}\). Thus \(30\,\mathrm{m}\) gives about \(3\) atm of gauge pressure and about \(4\) atm absolute, matching the figures found. ✓

Example 2 · Hydraulic press

A hydraulic press has input and output pistons of areas \(A_1 = 2.0\,\mathrm{cm^2}\) and \(A_2 = 40\,\mathrm{cm^2}\). A force \(F_1 = 30\,\mathrm{N}\) is applied to the small piston. Find the force on the large piston.

Solution. Given: \(A_1 = 2.0\,\mathrm{cm^2}\), \(A_2 = 40\,\mathrm{cm^2}\), \(F_1 = 30\,\mathrm{N}\).
Find: the output force \(F_2\).
Setup: By Pascal's principle the applied pressure is transmitted undiminished, so \(F_1/A_1 = F_2/A_2\).
Solve: \[ F_2 = F_1\,\frac{A_2}{A_1} = 30 \times \frac{40}{2.0} = 30 \times 20 = 600\,\mathrm{N}. \] Answer: \(\boxed{F_2 = 600\,\mathrm{N}}\)
Check: Pressure route: the applied pressure is \(p = F_1/A_1 = 30/(2.0\times10^{-4}) = 1.5\times10^5\,\mathrm{Pa}\); then \(F_2 = pA_2 = 1.5\times10^5 \times 40\times10^{-4} = 600\,\mathrm{N}\), the same value. The press multiplies force by the area ratio \(20\), at the cost of moving the large piston \(20\) times less far. ✓

Example 3 · Floating fraction of an oak block

A cube of oak (\(\rho_b = 750\,\mathrm{kg\,m^{-3}}\)) floats in fresh water (\(\rho_w = 1000\,\mathrm{kg\,m^{-3}}\)). What fraction of the cube is submerged, and what fraction stands above the surface?

Solution. Given: \(\rho_b = 750\,\mathrm{kg\,m^{-3}}\), \(\rho_w = 1000\,\mathrm{kg\,m^{-3}}\).
Find: submerged fraction \(f\) and the fraction above water.
Setup: For a floating body the buoyant force equals the weight: \(\rho_w g (fV) = \rho_b g V\), so \(f = \rho_b/\rho_w\) by the equation above.
Solve: \[ f = \frac{\rho_b}{\rho_w} = \frac{750}{1000} = 0.75, \qquad 1 - f = 0.25. \] Answer: \(\boxed{f_\mathrm{sub} = 0.75,\quad f_\mathrm{above} = 0.25}\)
Check: Limiting cases: as \(\rho_b \to \rho_w\), \(f \to 1\) (the block floats flush, on the verge of sinking); as \(\rho_b \to 0\), \(f \to 0\) (a weightless block rides on top). The intermediate value \(0.75\) for a wood three-quarters as dense as water is sensible. ✓

Example 4 · Torricelli efflux speed

A large open tank is filled with water. A small hole is opened \(5.0\,\mathrm{m}\) below the free surface. Find the speed at which water leaves the hole. \((g = 9.8\,\mathrm{m\,s^{-2}})\)

Solution. Given: depth of hole below surface \(h = 5.0\,\mathrm{m}\); tank open and wide.
Find: the efflux speed \(v\).
Setup: Apply Bernoulli between the free surface (speed \(\approx 0\), pressure \(p_0\)) and the hole (pressure \(p_0\)). The pressure terms cancel, leaving Torricelli's law: \(v = \sqrt{2gh}\).
Solve: \[ v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5.0} = \sqrt{98} = 9.9\,\mathrm{m\,s^{-1}}. \] Answer: \(\boxed{v \approx 9.9\,\mathrm{m\,s^{-1}}}\)
Check: Free-fall analogue: \(\sqrt{2gh}\) is exactly the speed a body acquires falling freely through \(5.0\,\mathrm{m}\). Efflux speed depends only on the depth of the hole, not on the tank's width or the liquid's density—Torricelli's theorem. ✓

Example 5 · Speeding up at a nozzle

Water flows through a garden hose of internal diameter \(2.0\,\mathrm{cm}\) at \(1.5\,\mathrm{m\,s^{-1}}\). It exits through a nozzle of diameter \(0.50\,\mathrm{cm}\). Find the exit speed.

Solution. Given: \(D_1 = 2.0\,\mathrm{cm}\), \(v_1 = 1.5\,\mathrm{m\,s^{-1}}\), \(D_2 = 0.50\,\mathrm{cm}\).
Find: the exit speed \(v_2\).
Setup: Water is effectively incompressible, so the volume flow rate is conserved: \(A_1 v_1 = A_2 v_2\) (continuity). With circular cross-sections, \(A \propto D^2\).
Solve: \[ v_2 = v_1\,\frac{A_1}{A_2} = v_1\left(\frac{D_1}{D_2}\right)^{\!2} = 1.5 \times \left(\frac{2.0}{0.50}\right)^{\!2} = 1.5 \times 16 = 24\,\mathrm{m\,s^{-1}}. \] Answer: \(\boxed{v_2 = 24\,\mathrm{m\,s^{-1}}}\)
Check: Flow-rate consistency: narrowing the diameter by a factor of \(4\) cuts the area by \(16\), so the speed must rise by \(16\) to push the same volume per second through. Both ends carry \(Q = A_1 v_1 = A_2 v_2\). ✓

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

Pressure at Depth
A submarine descends to a depth of \(250\,\mathrm{m}\) in seawater (\(\rho = 1025\,\mathrm{kg\,m^{-3}}\)). Find (a) the gauge pressure and (b) the absolute pressure at that depth. \((g = 9.8\,\mathrm{m\,s^{-2}},\; p_0 = 1.013\times10^5\,\mathrm{Pa})\)
(a) \(\rho g h = 1025\times9.8\times250 = 2.51\times10^6\,\mathrm{Pa} \approx 2.51\,\mathrm{MPa}\). (b) \(p_0 + 2.51\times10^6 \approx 2.61\,\mathrm{MPa}\), about \(25.8\) atmospheres.
Hydraulic Press
A hydraulic press has pistons of areas \(A_1 = 5\,\mathrm{cm^2}\) and \(A_2 = 500\,\mathrm{cm^2}\). A force of \(50\,\mathrm{N}\) is applied to the small piston. Find (a) the force on the large piston and (b) the distance the small piston must move to raise the large piston by \(2\,\mathrm{mm}\).
(a) \(F_2 = F_1(A_2/A_1) = 50\times100 = 5000\,\mathrm{N}\). (b) \(d_1 = d_2(A_2/A_1) = 2\times10^{-3}\times100 = 0.20\,\mathrm{m}\).
Floating Iceberg
An iceberg of density \(\rho_i = 917\,\mathrm{kg\,m^{-3}}\) floats in seawater (\(\rho_w = 1025\,\mathrm{kg\,m^{-3}}\)). What fraction of its volume lies above the surface?
Submerged fraction \(= \rho_i/\rho_w = 917/1025 = 0.895\); fraction above \(= 1-0.895 = \mathbf{0.105}\) (about \(10.5\%\)).
U-Tube with Three Fluids
A U-tube holds mercury (\(\rho_m = 13600\,\mathrm{kg\,m^{-3}}\)) in its lower bend. Oil (\(\rho_o = 800\,\mathrm{kg\,m^{-3}}\)) is poured into the left arm to a height of \(12\,\mathrm{cm}\) above the oil–mercury interface, and water (\(\rho_w = 1000\,\mathrm{kg\,m^{-3}}\)) into the right arm. Find the height of water needed so that the mercury stands at the same level in both arms.
Equal mercury levels \(\Rightarrow\) equal pressure at that level: \(\rho_o g(0.12) = \rho_w g\,h_w \Rightarrow h_w = (800/1000)(0.12) = \mathbf{9.6\,\mathrm{cm}}\).
Submerged Cylinder under an Oil Layer
A wooden cylinder (\(\rho_c = 700\,\mathrm{kg\,m^{-3}}\), height \(H = 20\,\mathrm{cm}\), base area \(A\)) is held fully submerged in water by a string to the container floor. A \(5\,\mathrm{cm}\) layer of oil (\(\rho_o = 800\,\mathrm{kg\,m^{-3}}\)) floats on the water, and the cylinder's top face sits exactly at the oil–water interface. (a) Find the buoyant force. (b) Find the string tension. (c) Describe the final equilibrium after the string is cut.
(a) The cylinder is entirely in water, so \(F_b = \rho_w A H g = 1000(0.20)(9.8)A = \mathbf{1960A\,\mathrm{N}}\). (b) Weight \(= \rho_c A H g = 1372A\,\mathrm{N}\); since \(F_b\gt W\) the string pulls down: \(T = F_b - W = \mathbf{588A\,\mathrm{N}}\). (c) Cut: the cylinder rises. As \(\rho_c\lt \rho_o\lt \rho_w\) it would float wholly in oil only if the oil were \(\geq 0.875H = 17.5\,\mathrm{cm}\) deep; it is only \(5\,\mathrm{cm}\), so the cylinder straddles. With \(5\,\mathrm{cm}\) in oil and \(x_w\) in water, \(1000\,x_w + 800(0.05) = 700(0.20) \Rightarrow x_w = \mathbf{0.10\,\mathrm{m}}\); the remaining \(\mathbf{5\,\mathrm{cm}}\) stands above the oil.

Fluid Mechanics: The Physics of Flow and Float

Perplexing Questions

What Is a Fluid?

Shear: The Distinguishing Test

Liquids and Gases: Two Kinds of Fluid

Pressure: The Language of Fluid Forces

Connections to Earlier Chapters

What This Chapter Covers

Pressure and the Hydrostatic Equation

Pressure Is Isotropic: The Pascal Proof

Deriving the Hydrostatic Equation

Physical Interpretation

Layered Fluids and Pressure Matching

Pascal's Law and the Hydraulic Press

Archimedes' Principle and Buoyancy

The Origin of Buoyancy: A Pressure Argument

Floating, Sinking, and Neutral Buoyancy

Resolving the Steel Ship Question

Apparent Weight

Solved examples

Problem bank

Chapter test

The chapter continues.