Chapter 9 · Mechanical Properties of Fluids

9.1 Pressure, Pascal's Law and Hydraulic Machines

A fluid at rest pushes on every surface at right angles with a pressure that grows steadily with depth, $p=p_a+\rho gh$. Squeeze an enclosed fluid and the extra pressure reaches every part of it — the principle behind hydraulic lifts and brakes.

Walk across soft snow in ordinary boots and you sink to your knees; strap on wide snowshoes and you stay on top, although you weigh exactly the same. Press a drawing pin into a board and its sharp point goes in, while its broad head does not hurt your thumb. In both cases what decides the effect is not the force itself but how that force is shared out over an area. For fluids — liquids and gases, which flow and take the shape of their container — this idea, pressure, is everything: a fluid cannot be pushed with a point or pulled with a hook, it can only press on surfaces.

This section builds the physics of a fluid at rest. You will define pressure and density, see why pressure at a point acts equally in all directions, and derive how it increases with depth — the reason dams are thick at the bottom and divers' ears hurt. You will meet atmospheric pressure, measure it with a barometer, and learn to distinguish gauge pressure from absolute pressure. Finally Pascal's law shows that pressure applied to a trapped liquid is passed on undiminished, so a small force on a small piston can hold up a car on a large one.

Fluids, pressure and density

A solid keeps its shape because it resists shear: push the top of a book sideways and it pushes back. A fluid at rest cannot do this; any shearing force, however small, makes it flow. So the only force a fluid at rest can exert on a surface is a push at right angles to that surface. A stationary fluid can never exert a tangential force on a wall — if it did, the wall would push back along itself and the fluid next to it would start to slide.

📘 Definition 9.1 — Fluid and pressure

A fluid is a substance that flows — it cannot sustain a shearing stress while at rest. Liquids and gases are fluids.

If a force of magnitude $F$ acts normally (perpendicularly) on a surface of area $A$, the average pressure on the surface is

$$p_\text{av}=\frac{F}{A}.$$

The pressure at a point is the limit of $\dfrac{\Delta F}{\Delta A}$ for a small area $\Delta A$ shrinking round that point. Pressure is a scalar: the force on a surface element is $p\,\Delta A$ along its normal, whichever way the surface faces. SI unit: the pascal, $1\ \text{Pa}=1\ \text{N/m}^2$; dimensions $[\text{M L}^{-1}\text{T}^{-2}]$. Other units: $1\ \text{bar}=10^5\ \text{Pa}$, $1\ \text{atm}=1.013\times10^5\ \text{Pa}$.

📘 Definition 9.2 — Density and relative density

The density of a substance of mass $m$ and volume $V$ is $\rho=\dfrac{m}{V}$ (SI unit $\text{kg/m}^3$). Its relative density is the ratio of its density to the density of water at $4\ ^\circ\text{C}$ ($1.00\times10^3\ \text{kg/m}^3$); it is a pure number.

Liquids are nearly incompressible — their density hardly changes with pressure (the bulk modulus of water is about $2\times10^9\ \text{Pa}$) — so in this chapter $\rho$ of a liquid is taken as constant. Gases are easily compressed and their density changes with pressure.

A few densities to keep in mind (at room temperature): air $1.2\ \text{kg/m}^3$; petrol about $7\times10^2$; olive oil $9.2\times10^2$; water $1.00\times10^3$; sea water $1.03\times10^3$; blood $1.06\times10^3$; glycerine $1.26\times10^3$; aluminium $2.7\times10^3$; iron $7.8\times10^3$; mercury $13.6\times10^3\ \text{kg/m}^3$.

Does the pressure at a point in a fluid depend on the direction of the surface you place there? Imagine a tiny triangular wedge of fluid, at rest, with a right angle between its two smaller faces $B$ (horizontal) and $C$ (vertical), and a sloping face $A$ at angle $\theta$ to the vertical. The surrounding fluid pushes on each face normally. Because the wedge is in equilibrium, the horizontal and vertical forces balance: $F_C=F_A\cos\theta$ and $F_B=F_A\sin\theta$ (the wedge's weight is proportional to its volume and becomes negligible compared with these surface forces as the wedge shrinks). The face areas are related in the same way, $A_C=A_A\cos\theta$ and $A_B=A_A\sin\theta$. Dividing, $\dfrac{F_A}{A_A}=\dfrac{F_B}{A_B}=\dfrac{F_C}{A_C}$.

📐 Law 9.1 — Pascal's law for a fluid at rest

In a fluid at rest,

the pressure at a point is the same in every direction — it does not depend on which way a small surface at that point faces;
the pressure is the same at all points at the same horizontal level, provided they are joined through the same fluid at rest.

Part 2 follows by taking a thin horizontal cylinder of fluid between two points at the same level: only the end forces act along its axis, so they must be equal.

✏️ Example 1 — A heel and an elephant

(a) A $60\ \text{kg}$ dancer balances for a moment on one stiletto heel of area $1.0\ \text{cm}^2$. Find the pressure under the heel. (b) A $5000\ \text{kg}$ elephant stands on four feet, each of area $0.12\ \text{m}^2$. Find the pressure under its feet. (c) Which is more likely to damage a wooden floor?

💡 Show solution▶

(a) $F=mg=60\times9.8=588\ \text{N}$, $A=1.0\times10^{-4}\ \text{m}^2$: $p=\dfrac{588}{1.0\times10^{-4}}=5.9\times10^6\ \text{Pa}$.

(b) $F=5000\times9.8=4.9\times10^4\ \text{N}$ over $4\times0.12=0.48\ \text{m}^2$: $p=1.0\times10^5\ \text{Pa}$ — about one atmosphere.

Check Units: $\text{N/m}^2=\text{Pa}$ ✓. Areas must be in $\text{m}^2$: $1\ \text{cm}^2=10^{-4}\ \text{m}^2$, not $10^{-2}$.

Variation of pressure with depth

Now include gravity. Take a vertical cylinder of liquid of cross-section $A$ inside a liquid of density $\rho$ at rest, with its top face at depth $h_1$ and its bottom face at depth $h_2=h_1+h$. The forces on its curved side are horizontal and cancel. Vertically, the liquid below pushes up with $p_2A$, the liquid above pushes down with $p_1A$, and the cylinder's weight is $mg=\rho Ahg$. Equilibrium gives

$$p_2A-p_1A=\rho Ahg\quad\Longrightarrow\quad p_2-p_1=\rho gh .$$

📐 Result 9.2 — Pressure at a depth

In a liquid of constant density $\rho$ at rest, with uniform $g$, the pressures at two points a vertical distance $h$ apart differ by

$$p_2-p_1=\rho gh ,$$

the deeper point having the higher pressure. If the liquid's surface is open to the atmosphere at pressure $p_a$, the pressure at depth $h$ below the surface is

$$p=p_a+\rho gh .$$

The pressure depends only on the depth — not on the shape of the vessel, the area of the surface or the total amount of liquid.

That last sentence is the hydrostatic paradox of the chapter teaser: vessels of very different shapes, filled to the same depth, have the same pressure on equal bases, even though they hold very different weights of liquid. The sloping walls take up the difference. It is also why water finds its own level in connected vessels, why a town's water tank sits on a hill or a tower, and why a dam must be much thicker at the bottom, where the pressure is greatest.

The atmosphere is itself a fluid, a "sea" of air many kilometres deep, so it presses on everything at the Earth's surface. Air is so light that its pressure hardly changes over a few metres, but over kilometres it falls a lot (and, air being compressible, not linearly).

📘 Definition 9.3 — Atmospheric, gauge and absolute pressure

Atmospheric pressure $p_a$ is the pressure of the air at a place; at sea level it is about $1.013\times10^5\ \text{Pa}=1\ \text{atm}$, the pressure under a column of mercury $76\ \text{cm}$ high.

The absolute pressure $p$ at a point is its total pressure. The gauge pressure is the excess over atmospheric pressure:

$$p_\text{gauge}=p-p_a .$$

At depth $h$ in a liquid open to the air, $p_\text{gauge}=\rho gh$. Tyre gauges, blood-pressure meters and most pressure gauges read gauge pressure; a gauge pressure can be negative (partial vacuum), an absolute pressure cannot.

The mercury barometer. Fill a long glass tube with mercury and turn it upside down in a dish of mercury. The column falls until it stands about $76\ \text{cm}$ above the level in the dish, with an almost perfect vacuum above it (the pressure there is only that of the mercury vapour, negligible). The point in the tube level with the dish surface has the same pressure as the surface (Law 9.1), which is $p_a$; and it is a depth $h$ below the vacuum. So $p_a=\rho_\text{Hg}gh$. A pressure is therefore often quoted as a length of mercury: $1\ \text{atm}=76\ \text{cm of Hg}=760\ \text{mm Hg}$ ("torr").

The open-tube manometer. A U-tube containing a liquid of density $\rho_m$ has one arm joined to a vessel of gas and the other open to the air. If the liquid stands a height $h$ higher in the open arm, the gas has gauge pressure $p-p_a=\rho_mgh$; if lower, the gauge pressure is negative.

✏️ Example 2 — A diver's pressure

A diver is $10.0\ \text{m}$ below the surface of the sea ($\rho=1.03\times10^3\ \text{kg/m}^3$, $p_a=1.013\times10^5\ \text{Pa}$). (a) Find the gauge pressure and the absolute pressure on her. (b) At what depth would the absolute pressure be $3.00\ \text{atm}$?

💡 Show solution▶

(a) $p_\text{gauge}=\rho gh=1030\times9.8\times10.0=1.01\times10^5\ \text{Pa}$ (to 3 s.f.; $1.0094\times10^5$ unrounded).

Absolute $p=p_a+\rho gh=1.013\times10^5+1.009\times10^5=2.02\times10^5\ \text{Pa}$, almost exactly $2.00\ \text{atm}$: every $10\ \text{m}$ of sea water adds about one atmosphere.

(b) Gauge pressure needed $=2.00\ \text{atm}=2.026\times10^5\ \text{Pa}$, so $h=\dfrac{2.026\times10^5}{1030\times9.8}=20.1\ \text{m}$.

Check Twice the depth, twice the gauge pressure ✓ — gauge pressure is proportional to depth, absolute pressure is not. Part (a) is the default of Interactive 1.

Drag the probe up and down the tank (or use the slider). The graph beside the tank, drawn to the same depth scale, shows gauge pressure $\rho gh$ (red) and absolute pressure $p_a+\rho gh$ (blue). Change the liquid and both lines change slope; change the air pressure above the surface and only the blue line moves. Press Play to send the probe on a dive.

liquid:above the surface:depth:10.0 m

depth $h$—

gauge pressure $\rho gh$—

absolute pressure—

absolute, in atm—

✏️ Example 3 — Barometers on the plain and in the hills

Take $\rho_\text{Hg}=13.6\times10^3\ \text{kg/m}^3$. (a) Find the height of the mercury column in a barometer when $p_a=1.013\times10^5\ \text{Pa}$. (b) How tall would a water barometer have to be? (c) At a hill station the mercury column is $70.0\ \text{cm}$. Find the atmospheric pressure there, and estimate the height of the hill station above sea level, treating the air as having a constant density of $1.2\ \text{kg/m}^3$.

💡 Show solution▶

(a) $h=\dfrac{p_a}{\rho_\text{Hg}g}=\dfrac{1.013\times10^5}{13\,600\times9.8}=0.760\ \text{m}=76.0\ \text{cm}$.

(b) $h=\dfrac{1.013\times10^5}{1000\times9.8}=10.3\ \text{m}$ — $13.6$ times taller, which is why barometers use mercury.

(c) $p_a=13\,600\times9.8\times0.700=9.33\times10^4\ \text{Pa}$. The drop from sea level is $6.0\ \text{cm}$ of mercury, $\Delta p=13\,600\times9.8\times0.060=8.0\times10^3\ \text{Pa}$. With $\Delta p=\rho_\text{air}gH$: $H=\dfrac{8.0\times10^3}{1.2\times9.8}\approx6.8\times10^2\ \text{m}$.

Check The air above thins out with height, so its true average density over $680\ \text{m}$ is a little less than at sea level and the true height a little more — the estimate is reasonable for a few hundred metres but fails for mountains. Ratio check: $\dfrac{H}{6.0\ \text{cm}}=\dfrac{\rho_\text{Hg}}{\rho_\text{air}}=1.1\times10^4$ ✓.

✏️ Example 4 — Oil and water in a U-tube

A U-tube open at both ends contains water. Oil is poured into the left arm until the oil column is $15.0\ \text{cm}$ long; the water in the right arm then stands $12.0\ \text{cm}$ above the oil–water interface. Find the density of the oil and the difference in height between the two free surfaces.

💡 Show solution▶

Idea Choose the level of the interface. Below it there is only water, joined through the bend, so by Law 9.1 the pressures at that level in the two arms are equal. (Points at the same level are not at equal pressure higher up, where the arms hold different liquids.)

Equation $p_a+\rho_\text{oil}g(0.150)=p_a+\rho_wg(0.120)$, so $\rho_\text{oil}=1000\times\dfrac{12.0}{15.0}=800\ \text{kg/m}^3$.

Levels The oil surface is $15.0-12.0=3.0\ \text{cm}$ higher than the water surface.

Check The lighter liquid needs the taller column: $\rho h$ is the same on both sides, $800\times0.150=1000\times0.120=120\ \text{kg/m}^2$ ✓. This is a neat way to measure the density of any liquid that does not mix with water.

Buoyancy revisited. Result 9.2 also explains Archimedes' principle from Class 9. A body immersed in a liquid has a larger pressure on its lower surfaces than on its upper ones; for a block of height $h$ and horizontal area $A$, the net upward force is $(\rho gh)A=\rho Vg$, the weight of the liquid displaced. This upthrust appears again in Section 9.3 for a sphere falling through a liquid.

Pascal's law and hydraulic machines

Result 9.2 has a striking consequence. If the pressure at the surface of a liquid at rest is raised by $\Delta p$ — by a piston, or by the air above it — then the pressure at every depth, $p_\text{surface}+\rho gh$, rises by the same $\Delta p$. The liquid passes the change on to every part of itself and to the walls.

📐 Principle 9.3 — Pascal's principle of transmission of pressure

A change in pressure applied to an enclosed incompressible fluid at rest is transmitted undiminished to every point of the fluid and to the walls of its container.

(For a gas the same holds once it has settled into equilibrium; liquids are used in machines because they are nearly incompressible, so the pistons move without the fluid soaking up the motion.)

A hydraulic machine uses this. Two cylinders, one of small cross-section $A_1$ and one of large cross-section $A_2$, are joined by a pipe and filled with oil, each with a close-fitting piston. A force $F_1$ on the small piston raises the pressure by $\dfrac{F_1}{A_1}$; the same increase acts on the large piston, which is pushed with $F_2=\dfrac{F_1}{A_1}A_2$.

📐 Result 9.4 — The hydraulic lift (press)

For pistons at the same level, with negligible friction and an incompressible liquid,

$$\frac{F_1}{A_1}=\frac{F_2}{A_2},\qquad F_2=\frac{A_2}{A_1}F_1 .$$

The force is multiplied by $\dfrac{A_2}{A_1}$, the mechanical advantage. Because the volume of liquid pushed out of one cylinder enters the other, $A_1d_1=A_2d_2$: the large piston moves a distance $d_2=\dfrac{A_1}{A_2}d_1$, so

$$F_1d_1=F_2d_2 .$$

The work done on the small piston equals the work done by the large one — a hydraulic machine multiplies force, never energy.

Hydraulic jacks in garages, the presses that forge car bodies, excavator arms, barber's chairs and the brakes of every car work this way. In hydraulic brakes, the brake pedal pushes a small piston in a master cylinder; the pressure travels along thin pipes full of brake fluid to wider pistons at each wheel, which press the brake pads. Since the pressure is the same everywhere in the fluid, wheels with equal cylinders get equal braking forces. An air bubble in the line spoils everything: air is compressible, so the pedal goes "spongy" and the pressure does not build up.

✏️ Example 5 — A garage car hoist

In a hydraulic hoist the small piston has a radius of $1.5\ \text{cm}$ and the large piston a radius of $15\ \text{cm}$. A $1200\ \text{kg}$ car sits on the large piston. (a) What force on the small piston holds the car up? (b) What is the gauge pressure of the oil? (c) How far must the small piston move in total to lift the car by $20\ \text{cm}$, and how much work is done? (Neglect the masses of the pistons and the difference in their levels.)

💡 Show solution▶

(a) $\dfrac{A_2}{A_1}=\left(\dfrac{15}{1.5}\right)^2=100$. $F_2=1200\times9.8=1.176\times10^4\ \text{N}$, so $F_1=\dfrac{F_2}{100}=118\ \text{N}$ — about the weight of a $12\ \text{kg}$ bag.

(b) $p=\dfrac{F_2}{A_2}=\dfrac{1.176\times10^4}{\pi(0.15)^2}=1.66\times10^5\ \text{Pa}$ (the same as $\dfrac{118}{\pi(0.015)^2}$ ✓).

(c) $d_1=\dfrac{A_2}{A_1}d_2=100\times0.20=20\ \text{m}$ — for example $100$ strokes of $20\ \text{cm}$, refilling the small cylinder through a valve each time. Work $=F_2d_2=1.176\times10^4\times0.20=2.4\times10^3\ \text{J}$, equal to $F_1d_1=117.6\times20$ ✓.

Check Force $\times100$, distance $\div100$: energy is conserved. Interactive 2 shows these default values, stroke by stroke.

Choose the piston radii and the load. The force $F_1$ shown on the small piston is just enough to hold the load: the extra pressure $F_1/A_1$ is the same everywhere in the oil. Press Play to pump: each $20\ \text{cm}$ stroke of the small piston pushes oil through a one-way valve and raises the load by only $20\ \text{cm}\times A_1/A_2$. Watch the work-in and work-out readouts stay equal.

small $r_1$:1.5 cmlarge $r_2$:15 cmload:1200 kg

force needed $F_1$—

oil gauge pressure—

strokes / total lift—

work in = work out—

✏️ Example 6 — Hydraulic brakes

A driver presses the brake pedal with a force of $200\ \text{N}$. The pedal is a lever that multiplies this force by $4.0$ and pushes on a master-cylinder piston of area $1.8\ \text{cm}^2$. Each wheel cylinder has a piston of area $6.0\ \text{cm}^2$. Find (a) the pressure in the brake fluid, (b) the force on each wheel piston.

💡 Show solution▶

(a) Force on the master piston $4.0\times200=800\ \text{N}$; $p=\dfrac{800}{1.8\times10^{-4}}=4.4\times10^6\ \text{Pa}$ (gauge), about $44$ atmospheres.

(b) $F=pA=4.44\times10^6\times6.0\times10^{-4}=2.7\times10^3\ \text{N}$ on each wheel piston.

Why it works By Principle 9.3 the same pressure reaches all four wheels through the pipes, whatever their length or route, so the braking is even.

Check Force ratio $\dfrac{2667}{800}=3.33=\dfrac{6.0}{1.8}$ ✓, the ratio of areas.

⚠️ Common mistakes

1. Gauge or absolute? $\rho gh$ is the gauge pressure; the absolute pressure adds $p_a$. Read the question: "pressure on a diver" usually means absolute, "reading of a tyre gauge" means gauge.

2. Depth, not length. $h$ in $\rho gh$ is the vertical depth below the free surface. In a sloping pipe or an inclined barometer tube, use the vertical height.

3. Same level, same pressure — only through the same liquid. In a U-tube with two liquids, equate pressures at the level of the interface (below it there is one liquid), not at the top.

4. Radius versus area. The force ratio in a hydraulic machine is the ratio of areas, $\left(\dfrac{r_2}{r_1}\right)^2$. Doubling the radius multiplies the force by four.

5. "Free energy" from hydraulics. The bigger force moves a smaller distance; $F_1d_1=F_2d_2$.

6. Units. $1\ \text{cm}^2=10^{-4}\ \text{m}^2$; $76\ \text{cm of Hg}$ is a pressure only when multiplied out, $\rho_\text{Hg}gh$.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the force on a dam wall

In JEE Main, NEET and JEE Advanced (as part of pressure in a fluid); not required by the CBSE syllabus.

JEE problems often ask for the total force of a liquid on a vertical wall. The gauge pressure on a vertical wall of width $w$ holding liquid of depth $H$ rises linearly from $0$ at the surface to $\rho gH$ at the bottom. On a thin horizontal strip of height $\Delta y$ at depth $y$ the force is $\rho gy\,w\,\Delta y$; adding the strips is finding the area under the straight-line graph of $p$ against $y$ (Chapter 2), so

$$F=\tfrac12\rho gwH^2 ,$$

the area $wH$ times the average gauge pressure $\tfrac12\rho gH$. Because the pressure is greatest at the bottom, the resultant acts at a height $\tfrac{H}{3}$ above the base, where it tends to tip the dam. Example. A dam $50\ \text{m}$ wide holds water $20\ \text{m}$ deep: $F=\tfrac12(1000)(9.8)(50)(20)^2=9.8\times10^7\ \text{N}$, acting $6.7\ \text{m}$ above the base. Doubling the depth quadruples the force. (Atmospheric pressure acts on both faces of the dam and cancels.)

Exercises

Take $g=9.8\ \text{m/s}^2$, $\rho_\text{water}=1.00\times10^3\ \text{kg/m}^3$, $\rho_\text{Hg}=13.6\times10^3\ \text{kg/m}^3$ and $p_a=1.013\times10^5\ \text{Pa}$ unless stated otherwise.

Pressure and density.

1 A thumb pushes a drawing pin with a force of $20\ \text{N}$. The point has an area of $0.10\ \text{mm}^2$ and the head an area of $1.0\ \text{cm}^2$. Find the pressure on the board and on the thumb.
2 A metal block measures $0.20\ \text{m}\times0.10\ \text{m}\times0.050\ \text{m}$ and has mass $8.0\ \text{kg}$. Find its density, and the least and greatest pressures it can exert when resting on a table on one face.
3 Express $1\ \text{atm}$ in bar and in mm of mercury, and a tyre gauge pressure of $2.5\ \text{bar}$ in pascals.
4 A $250\ \text{g}$ piece of metal has a volume of $25\ \text{cm}^3$. Find its density in $\text{kg/m}^3$ and its relative density.
5 Find the dimensional formula of pressure and show that it is the same as that of energy per unit volume.
6 Explain why the force exerted by a fluid at rest on any surface is always perpendicular to the surface.
7 A $50\ \text{kg}$ hiker stands on both feet in snowshoes, each of area $0.18\ \text{m}^2$, and then in boots, each of area $0.030\ \text{m}^2$. Find the pressure on the snow in each case and their ratio.

💡 Show solutions▶

1 Board: $\dfrac{20}{1.0\times10^{-7}}=2.0\times10^8\ \text{Pa}$ ($1\ \text{mm}^2=10^{-6}\ \text{m}^2$). Thumb: $\dfrac{20}{1.0\times10^{-4}}=2.0\times10^5\ \text{Pa}$ — a thousand times less.

2 $V=1.0\times10^{-3}\ \text{m}^3$, $\rho=8.0\times10^3\ \text{kg/m}^3$. Weight $78.4\ \text{N}$. Least on the largest face ($0.020\ \text{m}^2$): $3.9\times10^3\ \text{Pa}$; greatest on the smallest face ($0.0050\ \text{m}^2$): $1.6\times10^4\ \text{Pa}$.

3 $1\ \text{atm}=1.013\ \text{bar}=760\ \text{mm Hg}$; $2.5\ \text{bar}=2.5\times10^5\ \text{Pa}$.

4 $\rho=\dfrac{0.250}{25\times10^{-6}}=1.0\times10^4\ \text{kg/m}^3$; relative density $10$.

5 $[p]=\dfrac{[\text{M L T}^{-2}]}{[\text{L}^2]}=[\text{M L}^{-1}\text{T}^{-2}]$; $\dfrac{[\text{energy}]}{[\text{volume}]}=\dfrac{[\text{M L}^2\text{T}^{-2}]}{[\text{L}^3]}=[\text{M L}^{-1}\text{T}^{-2}]$ ✓. (This is why $\rho gh$ and, in Section 9.2, $\tfrac12\rho v^2$ can be added to a pressure.)

6 A fluid at rest cannot sustain a shearing (tangential) stress. If it pushed along a wall, by Newton's third law the wall would push back along itself on the fluid, and the fluid layer would flow — contradicting "at rest". So only the normal component survives.

7 Weight $490\ \text{N}$. Snowshoes: $\dfrac{490}{0.36}=1.4\times10^3\ \text{Pa}$; boots: $\dfrac{490}{0.060}=8.2\times10^3\ \text{Pa}$; ratio $6.0$.

Pressure and depth, barometers and manometers.

1 Find the gauge pressure and the absolute pressure at the bottom of a swimming pool $2.5\ \text{m}$ deep.
2 At what depth in fresh water is the absolute pressure twice atmospheric pressure?
3 The deepest point of the ocean is about $11\ \text{km}$ down. Treating sea water ($1.03\times10^3\ \text{kg/m}^3$) as incompressible, estimate the gauge pressure there in pascals and in atmospheres.
4 A mercury barometer reads $74.0\ \text{cm}$. What is the atmospheric pressure in pascals?
5 An open-tube mercury manometer is joined to a gas cylinder. The mercury in the open arm stands $25.0\ \text{cm}$ higher than in the arm joined to the cylinder, and the barometer reads $76.0\ \text{cm}$ of mercury. Find the gauge pressure and the absolute pressure of the gas in cm of Hg and in pascals.
6 A U-tube contains mercury. Water is poured into one arm until the water column is $13.6\ \text{cm}$ long. How far above the water–mercury interface does the mercury stand in the other arm, and how far apart are the two free surfaces?
7 A blood-pressure reading of $120\ \text{mm Hg}$ is a gauge pressure. Convert it to pascals.
8 A tank contains $1.2\ \text{m}$ of water with a $0.50\ \text{m}$ layer of oil (density $800\ \text{kg/m}^3$) floating on top. Find the gauge pressure at the oil–water boundary and at the bottom.

💡 Show solutions▶

1 $\rho gh=1000\times9.8\times2.5=2.45\times10^4\ \text{Pa}$; absolute $1.013\times10^5+0.245\times10^5=1.26\times10^5\ \text{Pa}$.

2 Need $\rho gh=p_a$: $h=\dfrac{1.013\times10^5}{9800}=10.3\ \text{m}$.

3 $1030\times9.8\times1.1\times10^4=1.1\times10^8\ \text{Pa}$, about $1.1\times10^3\ \text{atm}$. (Sea water is compressed by about $5\%$ at this depth, so the true value is a few per cent higher.)

4 $13\,600\times9.8\times0.740=9.86\times10^4\ \text{Pa}$.

5 Gauge $25.0\ \text{cm Hg}=13\,600\times9.8\times0.250=3.33\times10^4\ \text{Pa}$. Absolute $76.0+25.0=101.0\ \text{cm Hg}=1.35\times10^5\ \text{Pa}$.

6 At the interface level: $\rho_wg(0.136)=\rho_\text{Hg}gh$, so $h=0.136\times\dfrac{1000}{13\,600}=0.0100\ \text{m}=1.00\ \text{cm}$. The water surface is $13.6-1.0=12.6\ \text{cm}$ above the mercury surface.

7 $13\,600\times9.8\times0.120=1.60\times10^4\ \text{Pa}$.

8 Boundary: $800\times9.8\times0.50=3.9\times10^3\ \text{Pa}$ ($3920$). Bottom: $3920+1000\times9.8\times1.2=3920+11\,760=1.57\times10^4\ \text{Pa}$.

III

Pascal's law and hydraulic machines.

1 The pistons of a hydraulic press have diameters $2.0\ \text{cm}$ and $20\ \text{cm}$. What force does the large piston exert when $50\ \text{N}$ is applied to the small one?
2 In part 1 the small piston moves down $10\ \text{cm}$. How far does the large piston move? Compare the work done on the two pistons.
3 A hydraulic lift must raise a $3000\ \text{kg}$ truck with a force of at most $500\ \text{N}$ on a small piston of radius $2.0\ \text{cm}$. Find the least radius of the large piston.
4 State Pascal's principle of transmission of pressure, and use it to explain why hydraulic brakes give equal braking forces at wheels with equal cylinders.
5 In a service station, compressed air pushes on oil that lifts a $1500\ \text{kg}$ car on a piston of radius $12\ \text{cm}$. What gauge pressure must the air have?
6 A hydraulic jack multiplies force by $100$. Does it violate the conservation of energy? Explain.
7 A force of $600\ \text{N}$ acts on a master-cylinder piston of area $2.0\ \text{cm}^2$. The four wheel cylinders each have piston area $8.0\ \text{cm}^2$. Find the fluid pressure and the force on each wheel piston.

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1 Area ratio $\left(\dfrac{20}{2.0}\right)^2=100$: $F_2=5.0\times10^3\ \text{N}$.

2 $d_2=\dfrac{10}{100}=0.10\ \text{cm}=1.0\ \text{mm}$. Work in $50\times0.10=5.0\ \text{J}$; work out $5000\times0.0010=5.0\ \text{J}$ — equal.

3 $\dfrac{A_2}{A_1}=\dfrac{3000\times9.8}{500}=58.8$, so $r_2=2.0\sqrt{58.8}=15\ \text{cm}$ ($15.3\ \text{cm}$).

4 A change of pressure applied to an enclosed incompressible fluid is transmitted undiminished to every part of the fluid and the walls. The pedal raises the pressure in the brake fluid; this same pressure acts on every wheel piston, so equal piston areas give equal forces $pA$.

5 $p=\dfrac{1500\times9.8}{\pi(0.12)^2}=\dfrac{14\,700}{0.0452}=3.2\times10^5\ \text{Pa}$ (gauge), about $3$ atmospheres above atmospheric.

6 No: the output piston moves $\tfrac1{100}$ of the input distance, so $F_1d_1=F_2d_2$; the work out never exceeds the work in (it is a little less because of friction).

7 $p=\dfrac{600}{2.0\times10^{-4}}=3.0\times10^6\ \text{Pa}$; each wheel piston: $3.0\times10^6\times8.0\times10^{-4}=2.4\times10^3\ \text{N}$.

Applications and harder problems.

1 (Modelling) A snorkeller breathes air at atmospheric pressure through a tube while her chest is $1.5\ \text{m}$ below the surface. Find the extra pressure of the water on her chest, and the extra force on a chest area of $0.080\ \text{m}^2$. Why are snorkels never made a few metres long?
2 The water level in a town's water tower is $25\ \text{m}$ above a ground-floor tap. Find the gauge pressure at the closed tap, and the force needed to hold a thumb over its opening of area $1.0\ \text{cm}^2$.
3 If the atmosphere had a uniform density of $1.29\ \text{kg/m}^3$ all the way up, how high would it reach?
4 A tall tube closed at the top and full of water stands with its open lower end in a bowl of water. What is the greatest height of water column the atmosphere can hold up? Why does a suction pump fail to lift water from a well deeper than this?
5 A cylinder holds water $0.40\ \text{m}$ deep under a light, freely sliding piston of area $0.050\ \text{m}^2$ that carries a $100\ \text{N}$ load. Find the gauge pressure at the bottom. If the load is doubled, by how much does the pressure at the bottom, and at mid-depth, increase?
6 (Modelling) A pressure sensor lowered into a lake reads an absolute pressure of $3.5\times10^5\ \text{Pa}$ when it touches the bottom. The air pressure at the lake is $1.00\times10^5\ \text{Pa}$. How deep is the lake?

💡 Show solutions▶

1 $\Delta p=1000\times9.8\times1.5=1.47\times10^4\ \text{Pa}$; force $1.47\times10^4\times0.080=1.2\times10^3\ \text{N}$ — like a $120\ \text{kg}$ mass on the chest, while the lungs are only at atmospheric pressure. The breathing muscles cannot expand the chest against much more, so a long snorkel is useless (and dangerous).

2 $1000\times9.8\times25=2.45\times10^5\ \text{Pa}$ (gauge); force $2.45\times10^5\times1.0\times10^{-4}=24.5\ \text{N}\approx25\ \text{N}$.

3 $H=\dfrac{p_a}{\rho g}=\dfrac{1.013\times10^5}{1.29\times9.8}=8.0\times10^3\ \text{m}$, about $8\ \text{km}$. (The real atmosphere thins gradually and has no top.)

4 $h=\dfrac{p_a}{\rho g}=10.3\ \text{m}$. A suction pump only lowers the pressure above the water; the atmosphere pushes the water up, and even a perfect vacuum lets it push water only $10.3\ \text{m}$ high. Deeper wells need a pump at the bottom that pushes.

5 $\dfrac{100}{0.050}+1000\times9.8\times0.40=2000+3920=5.9\times10^3\ \text{Pa}$. Doubling the load adds $\dfrac{100}{0.050}=2.0\times10^3\ \text{Pa}$ — at the bottom, at mid-depth and everywhere else (Principle 9.3).

6 $h=\dfrac{3.5\times10^5-1.00\times10^5}{1000\times9.8}=26\ \text{m}$ ($25.5$).

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Distinguish between absolute pressure and gauge pressure. Why is the wall of a dam made much thicker at the bottom than at the top?

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Definitions Absolute pressure is the total pressure at a point; gauge pressure is the excess over atmospheric pressure, $p_\text{gauge}=p-p_a$, which at depth $h$ in a liquid is $\rho gh$. [1]

Dam The pressure of the water increases linearly with depth, $p=p_a+\rho gh$, so the force on each metre of the wall is largest at the bottom; the wall must be strongest there. [1]

Exam tip Mention the formula: an answer that says "more water at the bottom" without $\rho gh$ loses the mark — it is depth, not quantity, that matters.

E2. NEET The absolute pressure at the bottom of a tank of water is $3p$, where $p$ is atmospheric pressure. If water is drawn off until the depth is reduced by one-fifth, the absolute pressure at the bottom becomes (a) $2p$ (b) $\dfrac{13p}{5}$ (c) $\dfrac{8p}{5}$ (d) $\dfrac{12p}{5}$

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Idea $3p=p+\rho gH$, so $\rho gH=2p$. New depth $\tfrac45H$: gauge pressure $\tfrac45(2p)=\tfrac{8p}{5}$.

Answer $p+\tfrac{8p}{5}=\tfrac{13p}{5}$: (b).

Exam tip (c) forgets the atmosphere; (d) scales the absolute pressure $3p$ by $\tfrac45$. Only the gauge part is proportional to depth.

E3. JEE Main (Numerical value) A U-tube open at both ends contains water. Oil of density $900\ \text{kg/m}^3$ is poured into one arm until the oil column is $10.0\ \text{cm}$ long. Find the difference between the levels of the two free surfaces, in mm.

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Set-up Equate pressures at the level of the oil–water interface: $\rho_\text{oil}g(10.0\ \text{cm})=\rho_wgh$, so $h=10.0\times0.900=9.0\ \text{cm}$ of water above the interface in the other arm.

Answer Difference $10.0-9.0=1.0\ \text{cm}=10\ \text{mm}$ (the oil surface is higher).

Exam tip Equating pressures at the bottom of the U, or at the level of the oil surface, gives nonsense — use the interface, below which there is only one liquid.

E4. JEE Advanced (One or more correct) In a hydraulic lift the small piston has area $1.0\ \text{cm}^2$ and the large piston area $50\ \text{cm}^2$; the large piston is $0.50\ \text{m}$ higher than the small one. The oil has density $800\ \text{kg/m}^3$, the pistons are light and frictionless, and a force of $20\ \text{N}$ on the small piston holds a load on the large piston at rest. Which statements are correct? (A) The load weighs $1000\ \text{N}$. (B) The load weighs about $980\ \text{N}$. (C) If the small piston is pushed down $10\ \text{cm}$, the large piston rises $2.0\ \text{mm}$. (D) The gauge pressure under the large piston is lower than that under the small piston by about $3.9\ \text{kPa}$.

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Pressures Under the small piston $p_1=\dfrac{20}{1.0\times10^{-4}}=2.0\times10^5\ \text{Pa}$. Going up $0.50\ \text{m}$ in the oil lowers the pressure by $\rho gh=800\times9.8\times0.50=3920\ \text{Pa}$: $p_2=1.96\times10^5\ \text{Pa}$. So (D) ✓.

Load $W=p_2A_2=1.9608\times10^5\times50\times10^{-4}=980\ \text{N}$: (B) ✓, (A) ✗ — (A) ignores the height difference.

Volumes $A_1d_1=A_2d_2$: $d_2=\dfrac{1.0}{50}\times10\ \text{cm}=0.20\ \text{cm}=2.0\ \text{mm}$: (C) ✓.

Answer (B), (C), (D).

Exam tip "Pressure is transmitted undiminished" applies to changes of pressure; the hydrostatic difference $\rho gh$ between two levels is still there.

E5. JEE Main NEET On a day when a vertical mercury barometer reads $76\ \text{cm}$, the barometer tube is tilted to make an angle of $60^\circ$ with the vertical (its top still well above the mercury). The length of the mercury column in the tube is now (a) $38\ \text{cm}$ (b) $76\ \text{cm}$ (c) $88\ \text{cm}$ (d) $152\ \text{cm}$

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Idea Pressure balance fixes the vertical height, $76\ \text{cm}$. Along a tube at $60^\circ$ to the vertical, $L\cos60^\circ=76$.

Answer $L=\dfrac{76}{0.5}=152\ \text{cm}$: (d).

Exam tip (a) multiplies by $\cos60^\circ$ instead of dividing; (c) uses $\sin60^\circ$ (correct only for $60^\circ$ to the horizontal).

E6. Board · Case study A garage uses a hydraulic jack whose small piston has a diameter of $2.0\ \text{cm}$ and whose large piston has a diameter of $10\ \text{cm}$. It is used to lift a $750\ \text{kg}$ car. (i) Find the least force on the small piston that will lift the car. [2] (ii) How far must the small piston move in total to raise the car by $5.0\ \text{cm}$? [1] (iii) Why must the oil in the jack be free of air bubbles? [1]

💡 Show solution▶

(i) By Pascal's principle $\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}$; $\dfrac{A_2}{A_1}=\left(\dfrac{10}{2.0}\right)^2=25$. $F_1=\dfrac{750\times9.8}{25}=294\ \text{N}\approx2.9\times10^2\ \text{N}$. [2]

(ii) $d_1=25\times5.0=125\ \text{cm}=1.25\ \text{m}$ (in several strokes). [1]

(iii) Air is compressible: pushing the piston would squeeze the bubbles instead of transmitting the pressure, so the load would not rise as expected. [1]

Exam tip Diameters are given — the area ratio is the square of the diameter ratio, $25$, not $5$.

🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. Why is blood pressure always measured with the cuff at the level of the heart — and how different is the pressure of the blood in your feet, your heart and your head?

Why it matters. High blood pressure (hypertension) affects more than a billion adults worldwide and is a leading cause of stroke and heart disease. Doctors in every country still quote blood pressure in millimetres of mercury, a unit from the barometer. A cuff held too low or too high on the arm gives a wrong reading, purely because of the $\rho gh$ of the blood column — an error clinical guidelines warn about.

What to do. Take the density of blood as $1.06\times10^3\ \text{kg/m}^3$ and a typical gauge pressure in the large arteries at heart level of $100\ \text{mm Hg}$ (an average over the heartbeat).

Convert $100\ \text{mm Hg}$ to pascals.
For a standing adult, take the feet $1.3\ \text{m}$ below the heart and the brain $0.40\ \text{m}$ above it. Treating the blood as a fluid at rest (the flow changes the result only a little in large arteries), find the arterial gauge pressure in the feet and in the brain, in mm Hg.
By how many mm Hg is a reading wrong if the cuff is $10\ \text{cm}$ below heart level? Compare with the $5\ \text{mm Hg}$ accuracy a clinic aims for.
Why do the ankles of people who stand all day swell, and why does a giraffe (brain about $2\ \text{m}$ above its heart) need a heart pressure of about $250\ \text{mm Hg}$?
Find out how an automatic wrist monitor tells you to hold your wrist, and explain the instruction with Result 9.2.

Tools: Calculator or spreadsheet; a tape measure; if available, a home blood-pressure monitor to compare readings with the arm raised, level and lowered (only as a demonstration, not a medical test).

Hand in: A labelled sketch of a standing person with the pressures marked, your calculations, and a short patient leaflet explaining in plain language why the arm must be at heart height.

Going further. In flowing blood the pressure also falls along the vessels because of viscosity (Section 9.3). Look up how the mean arterial pressure falls from the aorta to the capillaries and which kind of vessel causes most of the drop.

Instructor note: $100\ \text{mm Hg}$ is $13\,600\times9.8\times0.100$ $=1.33\times10^4\ \text{Pa}$. Feet: $1060\times9.8\times1.3$ $=1.35\times10^4\ \text{Pa}$, i.e. $+101\ \text{mm Hg}$, so about $200\ \text{mm Hg}$. Brain: $1060\times9.8\times0.40$ $=4.2\times10^3\ \text{Pa}$, i.e. $-31\ \text{mm Hg}$, so about $69\ \text{mm Hg}$. Cuff $10\ \text{cm}$ low: $+7.8\ \text{mm Hg}$ (each cm about $0.78\ \text{mm Hg}$), larger than the target accuracy. Giraffe: $2\ \text{m}$ of blood is about $156\ \text{mm Hg}$, so the heart must supply roughly that plus a working pressure at the brain.

🧪 Self-check quiz

0/5 correct

Q1. One pascal is equal to

Q2. The gauge pressure $5.0\ \text{m}$ below the surface of a fresh-water lake is about

Q3. A narrow vase and a wide bowl, with bases of equal area, are filled with water to the same depth. The pressure on the base is

Q4. The pistons of a hydraulic lift have radii in the ratio $1:10$. The force on the large piston is greater than the force on the small piston by a factor of

Q5. A mercury barometer is carried from sea level to the top of a mountain. The mercury column

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