Chapter 9 of 14

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IGCSE Class 11 Physics★ Unit VII · part of the 20-mark block covering Units VII to IX in CBSE Class 11 Physics (70 marks total); Bernoulli and viscosity numericals are frequent in JEE Main

Mechanical Properties of Fluids

A solid resists being deformed by a fixed amount; a fluid resists only the speed of the deformation. That one distinction opens up everything here — why a stiletto heel damages a floor a heavier person in flat shoes does not, why a spinning cricket ball refuses to travel in a parabola, and why sap climbs to the top of a tall tree against gravity. CBSE Class 11 Physics Chapter 9.

⏱ 22 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Explain why pressure is a scalar and derive p = p0 + rho g h
2Apply Pascal's law to hydraulic lifts and hydraulic brakes
3Use the equation of continuity, and separate it from Bernoulli's principle
4Apply Bernoulli's equation to Torricelli's law and dynamic lift
5Say why heating raises the viscosity of a gas but lowers that of a liquid
6Apply Poiseuille's equation and explain its fourth-power dependence on radius
7Use Stokes' law to find the terminal velocity of a sphere
8Compute excess pressure correctly for a drop, a submerged bubble, and a soap bubble
9Explain angle of contact and derive the capillary rise formula

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Why this chapter matters

A solid resists being deformed by a fixed amount; a fluid resists only how fast you deform it. That single difference explains why a stiletto heel dents a floor a far heavier person in flat shoes does not, why a spinning cricket ball swerves off a parabola, and why sap climbs to the top of a tree against gravity.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 11 Physics: Mechanical Properties of Solids

Stress, strain and shear — the comparison this chapter draws between a solid and a fluid

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Class 11 Physics: Work, Energy and Power

Energy conservation, which Bernoulli's equation is a fluid version of

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Class 11 Physics: Laws of Motion

Newton's second and third laws — behind Stokes' law and the backward thrust of a leaking vessel

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Class 9 Science: Gravitation

Weight and density, used throughout every pressure calculation here

Mechanical Properties of Fluids

1. What this chapter covers

A fluid is anything that flows — liquids and gases together. The previous chapter asked how solids resist being deformed. This one asks what happens when a material has no fixed shape to defend.

Textbook section	Topic
9.1	Introduction
9.2	Pressure — Pascal's law, depth, atmospheric and gauge pressure, hydraulic machines
9.3	Streamline flow and the equation of continuity
9.4	Bernoulli's principle, Torricelli's law, dynamic lift
9.5	Viscosity, Stokes' law, terminal velocity
9.6	Surface tension, surface energy, angle of contact, drops, bubbles, capillary rise

Not in the 2026-27 chapter

Topic	Status
Archimedes' principle	Zero occurrences in the chapter, and CBSE does not list it here
Venturi-meter	Zero occurrences, and not listed — CBSE names only Torricelli's law and dynamic lift as Bernoulli applications
Reynolds number	Zero occurrences — but see the warning below

The Reynolds number one needs care. CBSE's Unit VII entry for this chapter lists critical velocity. The chapter mentions the idea once, calling it critical speed — "beyond a limiting value, called critical speed, this flow loses steadiness and becomes turbulent" — but gives no formula and no Reynolds number, which is what older editions used to define it with.

So you can be asked about critical velocity, and the book will give you the concept without the arithmetic. Worse, Exercise 9.13 asks you to check whether a flow is laminar, which cannot be done without the Reynolds number. Section 8 below supplies it.

2. What makes a fluid different from a solid

The sharpest way to put it is a comparison Exercise 9.3 draws out:

	Solid	Fluid
Shearing stress is proportional to	shear strain	rate of shear strain
Resists	being deformed	the speed of the deformation
Sustains a shear indefinitely?	Yes	No — it just keeps flowing

Push sideways on a block of rubber and it settles at a new shape and stays there. Push sideways on water and it does not settle anywhere — it keeps moving for as long as you push. That is the whole definition of a fluid, and it is why a fluid has a viscosity rather than a shear modulus.

3. Pressure in a fluid at rest

$p = \frac{F}{A}$

Pressure is a scalar, and this gets asked directly. Force is a vector and area has an orientation, so the ratio looks as though it should have a direction. But in a fluid at rest the pressure at a point is the same in every direction — turn a tiny test surface any way you like and the force per unit area is unchanged. A quantity with no preferred direction cannot be a vector.

Pascal's law

Blaise Pascal observed that the pressure in a fluid at rest is the same at all points at the same height. Everything in this section follows from it.

How pressure varies with depth

$p = p_{0} + ρ g h$

The extra term is just the weight of the fluid column above you, spread over its area.

Two consequences worth carrying:

Blood pressure is higher at your feet than at your brain, by roughly $ρ g h$ over the 1.5 m between them.
Pressure depends on depth, not on the shape of the vessel. A narrow tube and a wide tank filled to the same level have the same pressure at the bottom.

Absolute against gauge pressure

Term	Meaning
Absolute pressure	The total, $p_{0} + ρ g h$
Gauge pressure	The excess over atmospheric, $ρ g h$ — what a tyre gauge reads

Where students go wrong: in Bernoulli's equation it makes no difference which you use, because the constant atmospheric term appears on both sides and cancels. But you must use the same one at every point. Mixing them leaves an uncancelled $1 0^{5}$ Pa in the working. Exercise 9.12 is exactly this question.

Why the atmosphere thins so fast

Atmospheric pressure halves in about the first 6 km, even though the atmosphere is over 100 km deep. This is not gravity weakening — over 6 km that change is negligible.

It is because air is compressible while a liquid is not. The lower layers are squeezed by the weight above them, so most of the atmosphere's mass sits close to the ground. Pressure counts the weight above you, not the height above you.

Torricelli's barometer

Evangelista Torricelli (1608-1647) devised the first method of measuring atmospheric pressure. A long tube closed at one end is filled with mercury and inverted into a trough. The space above the column contains only mercury vapour, at negligible pressure, so:

$p_{a t m} = ρ g h$

which gives the familiar 0.76 m of mercury.

Why mercury, and not something cheaper? Since $h = p / (ρ g)$ , the height needed is inversely proportional to density. Pascal repeated the experiment with wine at 984 kg m⁻³ and needed a column over 10 metres tall — that is Exercise 9.6. The densest convenient liquid gives the shortest, most practical instrument.

4. Pascal's law at work: hydraulic machines

Apply pressure to an enclosed fluid and it is transmitted undiminished to every part of the fluid and to the walls of the container.

Put a small piston of area $a$ and a large one of area $A$ into the same closed fluid. The pressure is the same at both, so:

$\frac{F _{o u t}}{A} = \frac{F _{in}}{a} ⟹ F_{o u t} = F_{in} \times \frac{A}{a}$

Say this precisely, because the exam wants the distinction: the pressure is transmitted unchanged; the force is multiplied, by the ratio of the areas. A hydraulic lift is not a way of getting free energy — the large piston moves through a correspondingly smaller distance.

Application	How it uses the law
Hydraulic lift	A small force on a small piston raises a car on a large one
Hydraulic brakes	One pedal force is transmitted equally to all four wheel cylinders, so the braking is even

5. Streamline flow and continuity

Streamline (steady) flow means the velocity at any fixed point does not change with time. Every particle passing through a given point follows the same path — the streamline.

Turbulent flow is the opposite: irregular, eddying, with the velocity at a point changing constantly. The chapter's example is a fast stream hitting rocks, forming white-water rapids.

Between them lies the critical speed — "beyond a limiting value, called critical speed, this flow loses steadiness and becomes turbulent."

The equation of continuity

$A_{1} v_{1} = A_{2} v_{2} or A v = constant$

This is conservation of mass, nothing more. Whatever volume enters one end per second must leave the other end per second, so a narrower pipe forces a higher speed.

Keep this separate from Bernoulli. Continuity tells you the fluid speeds up at a constriction; Bernoulli then tells you the pressure there drops. Exercise 9.3(d) turns on exactly this division of labour — the speeding-up follows from conservation of mass, not from Bernoulli.

Partly block a tap with your fingers and the water jets out fast for the same reason: you cut $A$ , so $v$ must rise.

6. Bernoulli's principle

For steady, streamline, non-viscous, incompressible flow:

$p + \frac{1}{2} ρ v^{2} + ρ g h = constant$

It is conservation of energy per unit volume — a pressure term, a kinetic term and a potential term.

Those four assumptions are examinable. Exercise 9.11 asks whether Bernoulli applies to a river rapid, and the answer is no: a rapid is turbulent and unsteady, and it dissipates energy as heat and sound, so the total is not conserved along the path.

Torricelli's law — speed of efflux

For a liquid escaping through a small hole a depth $h$ below the surface:

$v = 2 g h$

That is exactly the speed of a body dropped from height $h$ . The liquid behaves as though it had simply fallen that far, which is a memorable way to hold the result.

Dynamic lift

The air moves faster over the curved upper surface of an aerofoil than under it. By Bernoulli, faster means lower pressure, so the higher pressure underneath pushes the wing up.

The spinning ball is the same idea. A spinning cricket ball drags air around with it — on one side that dragged air adds to the airflow and on the other it opposes it. Unequal speeds mean unequal pressures, so there is a sideways force, and the ball swerves off its parabola. This is the Magnus effect.

And blowing over a sheet of paper lifts it for the same reason: fast air above, ordinary still air below, so the pressure underneath wins.

7. Viscosity

Viscosity is internal friction in a fluid — the resistance of one layer sliding over the next. For a fluid,

$shearing stress = η \times rate of shear strain$

where $η$ is the coefficient of viscosity, measured in Pa s.

Temperature does opposite things to liquids and gases

	Mechanism	Effect of heating
Liquids	Intermolecular attraction between layers	Viscosity decreases
Gases	Molecules crossing between layers, carrying momentum	Viscosity increases

This reversal is a favourite one-mark question. Heating a liquid weakens the bonds that resist sliding. Heating a gas makes its molecules dart between layers faster, transferring more momentum, so the resistance goes up.

Poiseuille's equation

For steady laminar flow through a narrow tube:

$Q = \frac{π Δ p r ^{4}}{8 η L}$

Look at that fourth power. Halving the radius cuts the flow to one sixteenth. This is why the needle size of a syringe controls the flow rate far more effectively than the doctor's thumb pressure — a point Exercise 9.4(c) makes directly.

8. Stokes' law, terminal velocity, and the missing Reynolds number

Stokes' law

A sphere of radius $a$ moving at speed $v$ through a fluid of viscosity $η$ feels a retarding force:

$F = 6 π η a v$

Terminal velocity

As the sphere accelerates, the drag grows until it balances the weight minus the upthrust. After that the sphere falls at constant speed:

$6 π η a v_{t} = \frac{4}{3} π a^{3} (ρ - σ) g ⟹ v_{t} = \frac{2 a ^{2} ( ρ - σ ) g}{9 η}$

where $ρ$ is the sphere's density and $σ$ the fluid's.

Read what it depends on: terminal velocity goes as the square of the radius and inversely as the viscosity. Double the radius and the sphere falls four times faster; double the viscosity and it falls half as fast.

The Reynolds number, which the chapter no longer has

This is not in the 2026-27 chapter, but CBSE lists critical velocity and Exercise 9.13 asks you to check that a flow is laminar. Here is what you need:

$R_{e} = \frac{ρ v D}{η}$

Value	Flow
Below about 1000	Laminar
Above about 2000	Turbulent

It is a pure number with no units — a ratio of inertial to viscous forces.

It also answers Exercise 9.3(e): turbulence begins at a fixed Reynolds number, which depends on the product of speed and size. A wind-tunnel model is smaller than the real aircraft, so it must be tested at a greater speed to reach the same Reynolds number.

9. Surface tension

The molecules at a liquid's surface have neighbours below and beside them but not above, so they are pulled inward. Creating new surface therefore costs energy, and the liquid behaves as though its surface were an elastic skin under tension.

Two equivalent definitions, and both get used:

As	Definition	Units
Surface tension	Force per unit length of a line in the surface	N m⁻¹
Surface energy	Energy per unit area of surface	J m⁻²

They are numerically identical, which is worth checking as a dimensional exercise.

Surface tension does not depend on the area of the surface. Both definitions are per-unit ratios, so making the surface bigger changes nothing. Exercise 9.18 turns entirely on this: three films of different heights and shapes but the same edge length all support the same weight.

And it falls as temperature rises, because faster molecules feel a weaker net inward pull.

Why a free drop is spherical

With no external force, surface tension acts to minimise surface area. For a given volume the sphere has the least possible surface area, so that is what a free drop becomes.

Excess pressure — get the factor right

This is the most error-prone corner of the chapter, because the number of surfaces changes.

Case	Surfaces	Excess pressure
Liquid drop (e.g. mercury in air)	1	$2 S / r$
Air bubble inside a liquid	1	$2 S / r$
Soap bubble in air	2	$4 S / r$

A soap bubble is the only one with two. It is a thin film with air on both sides, so it has an inner and an outer surface. A drop of liquid, or a bubble of air inside a liquid, has a single interface. Exercise 9.20 deliberately puts both cases in one question.

Notice $r$ is in the denominator: smaller bubbles have larger excess pressure. Connect a small bubble to a large one and the small one blows into the large one.

10. Angle of contact and capillary rise

Every liquid-solid pair has two competing forces:

Cohesion — between liquid molecules
Adhesion — between liquid and solid molecules

Which one wins decides everything.

	Water on glass	Mercury on glass
Stronger force	Adhesion	Cohesion
Angle of contact	Acute	Obtuse
Behaviour	Spreads, wets the glass	Beads up, does not wet
In a capillary tube	Rises	Falls

Detergents work by lowering the angle of contact. Water only penetrates the narrow gaps between fibres if it wets them, and wetting means a small angle of contact — which is what Exercise 9.2(d) is asking.

Capillary rise

The Latin capilla means hair — a hair-thin tube gives a very large rise.

$h = \frac{2 S c o s θ}{ρ g a}$

The rise is inversely proportional to the radius, so halving the tube's bore doubles the height the liquid climbs. This is why sap reaches the top of tall trees, and why the hairs of a paintbrush draw together into a fine tip when wet.

If $θ$ is obtuse, as for mercury, $cos θ$ is negative and $h$ comes out negative — the liquid is depressed below the outside level rather than raised.

Summary

A solid resists a shear strain; a fluid resists only the rate of shear strain, which is why fluids flow indefinitely under a steady push.
Pressure is a scalar: at a point in a fluid at rest it is the same in every direction.
Pascal's law — pressure in a fluid at rest is the same at all points at the same height.
$p = p_{0} + ρ g h$ ; pressure depends on depth, never on the shape of the vessel.
Gauge pressure is the excess over atmospheric. Either convention works in Bernoulli, provided it is used at every point.
The atmosphere halves in 6 km because air is compressible, not because gravity weakens.
Torricelli's barometer gives $p_{a t m} = ρ g h$ ; mercury is used because a lighter liquid needs an impractically tall column.
In a hydraulic machine the pressure is transmitted undiminished and the force is multiplied by the area ratio.
Continuity, $A v =$ constant, is conservation of mass — it, not Bernoulli, is what makes fluid speed up at a constriction.
Bernoulli: $p + \frac{1}{2} ρ v^{2} + ρ g h =$ constant, valid only for steady, streamline, non-viscous, incompressible flow.
Torricelli's law: $v = 2 g h$ — the same speed as a body dropped through $h$ .
Dynamic lift, the Magnus effect on a spinning ball, and paper rising when you blow over it are all the same Bernoulli argument.
Heating decreases the viscosity of a liquid but increases that of a gas — different mechanisms entirely.
Poiseuille's flow rate goes as $r^{4}$ , which is why needle size beats thumb pressure.
Stokes' law $F = 6 π η a v$ gives $v_{t} = 2 a^{2} (ρ - σ) g /9 η$ — terminal velocity goes as the square of the radius.
Reynolds number is absent from the 2026-27 chapter but CBSE lists critical velocity and Exercise 9.13 needs it: $R_{e} = ρ v D / η$ , laminar below ~1000.
Surface tension is force per unit length and equals surface energy per unit area. It is independent of the area and decreases with temperature.
Excess pressure is $2 S / r$ for a drop or a submerged air bubble, but $4 S / r$ for a soap bubble, which has two surfaces.
Adhesion beating cohesion gives an acute contact angle and wetting; cohesion winning gives an obtuse angle and beading.
Capillary rise $h = 2 S cos θ / (ρ g a)$ — inversely proportional to the tube radius, and negative for mercury.
Archimedes' principle and the Venturi-meter are in neither the 2026-27 chapter nor the syllabus for it.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Pressure

p = F/A,  in Pa

A scalar — the pressure at a point in a fluid at rest is the same in every direction, so it has no direction of its own to be a vector.

Pressure with depth

p = p0 + rho g h

Depends on depth only, never on the shape of the container. Gauge pressure is just the rho g h part.

Pascal's law in a hydraulic machine

F_out/A = F_in/a,  so F_out = F_in x (A/a)

Pressure is transmitted undiminished; force is multiplied by the ratio of the piston areas.

Equation of continuity

A1 v1 = A2 v2

Conservation of mass. This, not Bernoulli, is why fluid speeds up at a constriction.

Bernoulli's equation

p + (1/2) rho v^2 + rho g h = constant

Valid only for steady, streamline, non-viscous, incompressible flow. A river rapid fails every one of these conditions.

Torricelli's law

v = sqrt(2 g h)

Speed of efflux from a hole at depth h — the same speed as a body simply dropped through h.

Poiseuille's equation

Q = pi (dp) r^4 / (8 eta L)

Flow rate goes as the FOURTH power of the radius. Halving the radius cuts the flow to one sixteenth.

Stokes' law

F = 6 pi eta a v

Viscous drag on a sphere of radius a moving at speed v through a fluid of viscosity eta.

Terminal velocity

v_t = 2 a^2 (rho - sigma) g / (9 eta)

Goes as the square of the radius, inversely with viscosity. Doubling the radius quadruples the terminal speed.

Reynolds number (not in the chapter, but needed)

Re = rho v D / eta

Laminar below about 1000, turbulent above about 2000. CBSE lists critical velocity and Exercise 9.13 needs this even though the 2026-27 chapter omits it.

Excess pressure in a drop or a submerged bubble

excess p = 2S/r

One surface: a liquid drop in air, or an air bubble inside a liquid.

Excess pressure in a soap bubble in air

excess p = 4S/r

Two surfaces, inner and outer, because it is a thin film with air on both sides. The most common mix-up in the chapter.

Capillary rise

h = 2 S cos(theta) / (rho g a)

Inversely proportional to the tube's radius a. Negative for an obtuse contact angle, giving depression rather than rise, as with mercury.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Treating pressure as a vector

✓ At a point in a fluid at rest, pressure is the same in every direction, so it has no direction to point in. It is a scalar, even though it is defined as force over area.

WATCH OUT

✗ Thinking gravity gets weaker by 6 km altitude, explaining why pressure halves

✓ Gravity barely changes over 6 km. Air is compressible, so most of the atmosphere's mass is squeezed into the lowest few kilometres, which is why pressure falls so fast near the ground.

WATCH OUT

✗ Mixing gauge and absolute pressure within one Bernoulli calculation

✓ Either convention works, because the constant atmospheric term cancels — but only if the same one is used at every point in the equation. Mixing them leaves an uncancelled 10^5 Pa in the working.

WATCH OUT

✗ Crediting Bernoulli's principle for the speed increase at a constriction

✓ The equation of continuity, conservation of mass, is what forces the speed up. Bernoulli's principle then says what happens to the pressure as a consequence.

WATCH OUT

✗ Applying Bernoulli's equation to turbulent or unsteady flow

✓ Bernoulli assumes steady, streamline, non-viscous, incompressible flow. A river rapid is turbulent and dissipates energy as heat and sound, so the equation does not apply to it.

WATCH OUT

✗ Assuming heating always lowers viscosity

✓ It lowers the viscosity of a liquid, where viscosity comes from intermolecular attraction that heat weakens. It RAISES the viscosity of a gas, where viscosity comes from molecules crossing between layers, and heat speeds that up.

WATCH OUT

✗ Believing thumb pressure controls the flow rate of a syringe more than the needle does

✓ Poiseuille's equation makes flow rate proportional to the fourth power of the radius. Halving the needle's radius cuts the flow to one sixteenth, an effect no reasonable change in thumb pressure can match.

WATCH OUT

✗ Using 2S/r for a soap bubble's excess pressure

✓ A soap bubble in air has two surfaces, inner and outer, so its excess pressure is 4S/r. A liquid drop or an air bubble submerged in a liquid has only one surface and takes 2S/r.

WATCH OUT

✗ Thinking a larger film supports more weight

✓ Surface tension is force per unit length, not per unit area, so the weight a film supports depends only on the length of its supporting edge, not on how large the film is.

WATCH OUT

✗ Assuming mercury rises in a capillary tube like water

✓ Mercury has an obtuse angle of contact with glass, so cos(theta) is negative in the capillary rise formula, and the mercury level is DEPRESSED below the outside level rather than raised.

WATCH OUT

✗ Quoting Archimedes' principle or a Venturi-meter as chapter material

✓ Neither term appears in the 2026-27 NCERT chapter and neither is listed by CBSE for it. The listed Bernoulli applications are Torricelli's law and dynamic lift only.

WATCH OUT

✗ Trying to check laminar flow without the Reynolds number

✓ Reynolds number Re = rho v D / eta has been removed from the chapter text, but Exercise 9.13 still requires it to confirm an assumption of laminar flow. Learn the formula and the laminar threshold of about 1000, even though the book no longer states them.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s

Exercises

Pressure and Pascal's law, hydraulic lifts, U-tube manometers, Bernoulli's equation and dynamic lift, Poiseuille's law and viscosity, and surface tension of drops, bubbles and films (Q9.1-9.20)

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Mechanical Properties of Fluids?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min worth ~20 marks in IGCSE exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•A fluid resists the RATE of shear strain, not the strain itself — that is what makes it a fluid rather than a solid.
•Pressure is a scalar: at a point in a fluid at rest it is the same in every direction.
•Pascal's law: pressure in a fluid at rest is the same at all points at the same height.
•p = p0 + rho g h. Pressure depends on depth only, never on the container's shape.
•The atmosphere thins fast because air is compressible, not because gravity weakens with height.
•Torricelli's barometer gives p_atm = rho g h; a denser liquid needs a shorter, more practical column.
•In a hydraulic machine, pressure is transmitted undiminished; force is multiplied by the ratio of piston areas.
•Gauge and absolute pressure differ by a constant that cancels in Bernoulli's equation, provided you are consistent.
•The equation of continuity, Av = constant, is conservation of mass, not Bernoulli's principle.
•Bernoulli's equation needs steady, streamline, non-viscous, incompressible flow — a rapid satisfies none of these.
•Torricelli's law: v = sqrt(2gh), the same speed as a body dropped through height h.
•Dynamic lift and the Magnus effect on a spinning ball both follow from faster flow meaning lower pressure.
•Heating decreases a liquid's viscosity but increases a gas's viscosity — opposite mechanisms.
•Poiseuille's flow rate goes as the fourth power of the radius, which is why needle size dominates thumb pressure.
•Stokes' law: F = 6 pi eta a v. Terminal velocity v_t = 2a^2(rho-sigma)g/9eta, going as the square of the radius.
•Reynolds number Re = rho v D/eta is not in the 2026-27 chapter but is needed for Exercise 9.13; laminar below about 1000.
•Surface tension is force per unit length and equals surface energy per unit area; it does not depend on the area of the surface.
•Surface tension decreases as temperature rises.
•A free drop is spherical because a sphere has the least surface area for a given volume.
•Excess pressure is 2S/r for a drop or a submerged bubble, but 4S/r for a soap bubble in air, which has two surfaces.
•Smaller bubbles have larger excess pressure — connect a small one to a large one and it empties into the large one.
•An acute angle of contact means wetting and capillary rise; an obtuse one means beading and capillary depression.
•Detergents work by lowering the angle of contact so water can wet and penetrate narrow fibres.
•Capillary rise h = 2S cos(theta)/(rho g a) is inversely proportional to the tube's radius.
•Archimedes' principle and the Venturi-meter are absent from the 2026-27 chapter and not listed by CBSE.

IGCSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit VII sits inside the 20-mark block covering Units VII to IX (CBSE Class 11 Physics, 70 marks)

Question type	Marks each	Typical count	What it tests
Pressure	2-3	1	Pascal's law, depth dependence, hydraulic machines
Bernoulli's principle	3-5	1	Continuity, Torricelli's law, dynamic lift, and the four assumptions
Viscosity	2-3	1	Poiseuille's equation, Stokes' law, terminal velocity
Surface tension	3-5	1	Excess pressure in drops, bubbles and films; capillary rise

Prep strategy

Keep continuity and Bernoulli separate — one gives the speed, the other the pressure
Memorise the four assumptions behind Bernoulli's equation; a stated exception like a rapid tests exactly these
Drill the surface-count rule for excess pressure: one surface for a drop or submerged bubble, two for a soap bubble in air
Learn the Reynolds number even though the chapter omits it — Exercise 9.13 needs it
Practise the fourth-power reasoning in Poiseuille's equation; it appears in several disguised forms

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Hydraulic brakes and lifts

Pascal's law lets one modest push on a small piston or a brake pedal become a large, evenly distributed force at every wheel or on a car's full weight.

Why aircraft wings are curved on top

The shape speeds up the air above the wing, and Bernoulli's principle turns that speed difference into the pressure difference that lifts the plane.

Why a doctor changes the needle, not just the pressure

Poiseuille's fourth-power law means the needle's bore controls the injection rate far more than how hard the plunger is pushed.

Detergents and wetting

A detergent works by lowering water's angle of contact so it can wet and penetrate the narrow gaps between fibres in cloth, rather than beading on the surface.

How trees move water to their highest leaves

Capillary rise through narrow channels in the wood, combined with the fact that rise is inversely proportional to tube radius, lets water reach heights gravity alone would never allow.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

State which of continuity or Bernoulli you are using and why — many marks are lost by quoting the right formula for the wrong reason.

Before applying Bernoulli, check the four conditions: steady, streamline, non-viscous, incompressible. If a question describes turbulence or friction, say so and explain why it does not apply.

For any excess-pressure question, count the surfaces first — one for a drop or a submerged bubble, two for a soap bubble in air.

Learn Reynolds number and the laminar threshold even though the current chapter omits them; Exercise 9.13 and similar JEE questions still expect it.

In viscosity problems, check whether you have been given a mass flow rate or a volume flow rate before using Poiseuille's equation, which needs volume.

Remember gauge and absolute pressure give the same Bernoulli result only if used consistently at every point in the same calculation.

For capillary problems, check the sign of cos(theta) before reporting rise or depression — an obtuse angle gives a negative height.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Derive the drag force and terminal velocity for a sphere falling through a fluid using dimensional analysis, and compare the result with Stokes' law.

STRETCH

Analyse the flow rate and pressure drop for capillaries in parallel, and compare with the series case worked in the JEE Advanced problem above.

STRETCH

Derive the Reynolds number from dimensional analysis as the ratio of inertial to viscous forces, and use it to estimate the onset of turbulence in a garden hose.

STRETCH

Work out the shape of a rotating liquid surface using Bernoulli's equation in a rotating frame, and relate it to a parabolic mirror.

STRETCH

Investigate the physics of a Pitot tube, and derive how it measures flow speed from a pressure difference.

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainTerminal velocity and coalescenceSingle correct

Two identical small water drops, each with terminal velocity $v$ , coalesce to form a single larger drop. Assuming the terminal velocity depends on the drop radius exactly as in Stokes' law, the new terminal velocity is:

(a) $v$ (b) $2 v$ (c) $2^{2/3} v$ (d) $4 v$

Stuck? Show the approach

Terminal velocity goes as the square of the radius. Work out the new radius first by keeping the volume conserved when the two drops merge, then apply that scaling.

Show the full solution

Step 1 — Find the radius of the merged drop. Volume is conserved when the drops coalesce. Two drops of radius $a$ have combined volume $2 \times \frac{4}{3} π a^{3}$ , equal to the volume of one drop of radius $R$ :

$\frac{4}{3} π R^{3} = 2 \times \frac{4}{3} π a^{3} ⟹ R^{3} = 2 a^{3} ⟹ R = 2^{1/3} a$

Step 2 — Apply the terminal velocity scaling. From Stokes' law, $v_{t} \propto a^{2}$ . So

$\frac{v _{t}^{'}}{v _{t}} = (\frac{R}{a})^{2} = (2^{1/3})^{2} = 2^{2/3}$

Step 3 — Write the new terminal velocity.

$v_{t}^{'} = 2^{2/3} v$

Answer: (c) 2^(2/3) v

The trap

Reaching for $2 v$ by assuming terminal velocity doubles when the drop doubles. It goes as $a^{2}$ , not linearly, and the radius itself only grows as the cube root of the volume — two separate factors that combine to $2^{2/3}$ , not $2$ .

JEE MainBernoulli's principleNumerical

A tank is filled with water to a height of $H = 4$ m. A small hole is made at height $h$ above the base of the tank. Find the value of $h$ that gives the maximum horizontal range of the water jet after it leaves the hole.

Stuck? Show the approach

Write the range as a function of the depth below the surface, then treat it as an optimisation problem — take the derivative and set it to zero, or use the symmetry of the resulting expression.

Show the full solution

Step 1 — Set up the two motions. The hole sits at a depth $(H - h)$ below the free surface. By Torricelli's law, the exit speed is

$v = 2 g (H - h)$

After leaving the hole, the water is a horizontal projectile falling the remaining height $h$ to the ground:

$h = \frac{1}{2} g t^{2} ⟹ t = \frac{2 h}{g}$

Step 2 — Write the range.

$x = v t = 2 g (H - h) \times \frac{2 h}{g} = 2 h (H - h)$

Step 3 — Maximise it. $x$ is largest when the product $h (H - h)$ is largest. This is a downward parabola in $h$ with roots at $h = 0$ and $h = H$ , so its maximum sits exactly halfway between them:

$h = \frac{H}{2}$

Step 4 — Substitute.

$h = \frac{4}{2} = 2 m$

And the maximum range itself is $x_{ma x} = 2 (H /2) (H /2) = H = 4$ m — the maximum range always equals the full height of the water.

Answer: h = H/2 = 2 m (giving a maximum range equal to H, 4 m)

The trap

Trying calculus on the wrong variable, or forgetting that the depth below the surface is $(H - h)$ , not $h$ itself. Writing $v = 2 g h$ instead of $2 g (H - h)$ silently swaps which end of the tank the hole is measured from and gives the wrong optimum entirely.

JEE AdvancedExcess pressure and coalescenceNumerical

1000 identical water droplets, each of radius $1.0$ mm, coalesce to form a single large drop. Surface tension of water is $S = 0.072$ N m⁻¹. Find the energy released in this process, and state where that energy goes.

Stuck? Show the approach

Total surface area falls when small drops merge into one big one, because a single sphere has less surface area per unit volume than many small ones. The energy released equals surface tension times the decrease in total area.

Show the full solution

Step 1 — Find the radius of the combined drop. Volume is conserved:

$\frac{4}{3} π R^{3} = n \times \frac{4}{3} π r^{3} ⟹ R = n^{1/3} r$

With $n = 1000$ and $r = 1.0 \times 1 0^{- 3}$ m:

$R = 100 0^{1/3} \times (1.0 \times 1 0^{- 3}) = 10 \times 1.0 \times 1 0^{- 3} = 1.0 \times 1 0^{- 2} m$

Step 2 — Find the total surface area before and after.

$A_{i} = n \times 4 π r^{2} = 1000 \times 4 π (1.0 \times 1 0^{- 3})^{2} = 1.257 \times 1 0^{- 2} m^{2}$

$A_{f} = 4 π R^{2} = 4 π (1.0 \times 1 0^{- 2})^{2} = 1.257 \times 1 0^{- 3} m^{2}$

Step 3 — Find the decrease in area, and the energy released.

$Δ A = A_{i} - A_{f} = 1.257 \times 1 0^{- 2} - 1.257 \times 1 0^{- 3} = 1.131 \times 1 0^{- 2} m^{2}$

Surface energy is energy per unit area, so the energy released is

$E = S \times Δ A = 0.072 \times 1.131 \times 1 0^{- 2} \approx 8.1 \times 1 0^{- 4} J$

Step 4 — Where the energy goes. Coalescing 1000 small drops into one large one reduces the total surface area by a factor of ten (since $A \propto n^{2/3}$ for fixed total volume, and $n^{2/3} = 100$ here — a tenfold reduction). That surface energy does not vanish; it converts into heat, very slightly warming the combined drop.

Answer: E ≈ 8.1 × 10⁻⁴ J, released as heat in the merged drop

The trap

Computing $A_{i}$ using the combined volume's surface instead of summing $n$ separate small-sphere areas, or forgetting the factor of $n$ altogether when finding $A_{i}$ . The area of 1000 small spheres is far larger than the area of one sphere holding the same total volume — that gap is precisely the energy source.

JEE AdvancedSeries flow through capillariesNumerical

Two capillary tubes of equal length but radii $r$ and $2 r$ are connected end to end (in series), and a liquid flows through them under a total pressure difference $Δ p$ . Find what fraction of $Δ p$ appears across the narrower tube.

Stuck? Show the approach

In series, the volume flow rate $Q$ through both tubes must be equal — whatever enters the first tube each second must leave the second tube each second. Use Poiseuille's equation to write the pressure drop across each tube in terms of the common $Q$ , then take the ratio.

Show the full solution

Step 1 — State what is common to both tubes. Since the tubes are in series, the same liquid volume flows through both every second: $Q_{1} = Q_{2} = Q$ .

Step 2 — Write Poiseuille's equation for each tube.

$Q = \frac{π Δ p _{1} r ^{4}}{8 η L} = \frac{π Δ p _{2} ( 2 r ) ^{4}}{8 η L}$

Since $Q$ , $η$ and $L$ are the same for both,

$Δ p_{1} r^{4} = Δ p_{2} (2 r)^{4} = 16 Δ p_{2} r^{4} ⟹ Δ p_{1} = 16 Δ p_{2}$

Step 3 — Use the total pressure difference. The two pressure drops must add up to the total:

$Δ p_{1} + Δ p_{2} = Δ p$

Substituting $Δ p_{1} = 16Δ p_{2}$ :

$16Δ p_{2} + Δ p_{2} = Δ p ⟹ Δ p_{2} = \frac{Δ p}{17}, Δ p_{1} = \frac{16Δ p}{17}$

Step 4 — State the fraction asked for. The narrower tube (radius $r$ ) carries fraction

$\frac{Δ p _{1}}{Δ p} = \frac{16}{17} \approx 0.94$

Answer: 16/17 ≈ 94% of the total pressure drop occurs across the narrower tube

The trap

Assuming the pressure drop splits according to the tubes' areas, or splits equally since the lengths are equal. It is neither — it follows the inverse fourth power of the radius, which is why almost all the resistance concentrates in the thinner tube even though it looks like a minor difference in size.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh

JEE Main and Advanced (Fluid Mechanics)High

NEET PhysicsMedium

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Force is a vector and points somewhere, but the pressure at a point inside a fluid at rest does not. If you place a tiny test surface at that point and rotate it to face any direction at all, the force per unit area on it stays exactly the same. A quantity that behaves identically in every direction has no direction of its own to be described by, and that is what makes it a scalar rather than a vector, despite being built out of a force.

They answer different questions and one leads to the other. The equation of continuity, Av = constant, is simply conservation of mass: whatever volume of fluid enters a pipe each second must also leave it each second, so the fluid speeds up wherever the pipe narrows. Bernoulli's principle is conservation of energy along a streamline, and it is what then tells you what happens to the pressure once you know the speed has changed. Continuity supplies the speed; Bernoulli supplies the pressure.

The formula 2S/r comes from a single curved surface pulling inward. A raindrop, or an air bubble sitting inside a liquid, has exactly one such surface — the boundary between the two substances. A soap bubble floating in air, however, is a thin film of liquid with air both inside and outside it, so it has two surfaces, an inner one and an outer one, each contributing its own 2S/r. Adding the two gives 4S/r. Whenever you see a soap bubble specifically, check whether the question means the bubble itself or an air pocket submerged in the soap solution, because the two cases use different formulas.

The Reynolds number was present in older editions of this chapter and has since been removed from the main text, along with any formula for critical velocity. CBSE's syllabus was not updated to match, so it still lists critical velocity as examinable, and Exercise 9.13 explicitly asks you to check whether an assumption of laminar flow holds — something that cannot be verified without it. In practice this means learning Re = rho v D / eta and the rough rule that flow is laminar below a Reynolds number of about 1000, even though your chapter itself never states either.

Viscosity comes from a different physical mechanism in the two cases. In a liquid, molecules are held close together and viscosity arises from the attractive forces between neighbouring layers; heating gives the molecules more energy to overcome that attraction, so the liquid flows more freely and its viscosity falls. In a gas, molecules are far apart and viscosity arises instead from molecules randomly crossing between layers moving at different speeds and carrying momentum with them; heating makes the molecules move faster and cross more often, transferring more momentum and increasing the viscosity. The same word, opposite causes.

It comes down to the angle of contact, which in turn comes down to whether adhesion to the glass or cohesion within the liquid wins. Water adheres to glass more strongly than it holds to itself, giving an acute angle of contact, and the capillary rise formula then gives a positive height — the water climbs the tube. Mercury holds to itself far more strongly than it adheres to glass, giving an obtuse angle of contact, which makes the cosine in the formula negative, so the calculated height comes out negative too. A negative height means depression rather than rise, which is exactly what you see: mercury sits lower inside a capillary tube than the level outside it.

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