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PhysicsCh 9 Mechanical Properties of Fluids

Mechanical Properties of Fluids: Class 11 Physics · Chapter 9

Physics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 110 min
  • 1 simulation
  • 29 practice questions
  • 27 MCQs

What you will learn in Mechanical Properties of Fluids

  • Explain what makes liquids and gases fluids, and contrast their compressibility and response to shear with those of solids
  • Define pressure and density, prove Pascal's law, derive P = Pa + ρgh and apply it to barometers, manometers and hydraulic machines
  • Describe steady, streamline and turbulent flow, and use the equation of continuity
  • Derive Bernoulli's equation from the work–energy theorem and apply it to efflux (Torricelli's law), Venturi flow and dynamic lift
  • Define the coefficient of viscosity, use Stokes' law and derive the terminal velocity of a falling sphere
  • Explain surface energy, surface tension and the angle of contact, and derive the excess pressure in drops and bubbles and the height of capillary rise

Original companion text aligned to NCERT Physics Class 11, Chapter 9, pp. 180–201. Not affiliated with or endorsed by NCERT.

9.1 Introduction

Tip a steel tumbler of water into a glass and the water takes the glass's shape at once. When a pressure cooker whistles, steam rushes out of its vent. Water and steam look nothing alike, yet they share one ability: both flow. Any substance that can flow is called a fluid (तरल), so the word covers liquids and gases alike.

Air, the blood your heart pumps (about five litres a minute at rest), the sap in a neem tree and the monsoon clouds over the Western Ghats are all fluids on the move.

What sets a fluid apart from a solid? A solid holds its own shape; a fluid has none and settles into whatever holds it. Liquids and gases then part ways over volume. A litre of milk is still a litre in a jug or in a bottle, but a gas spreads to fill every corner of its container. In Chapter 8 you saw that pressure changes the volume of any material. For solids and liquids the change is tiny, while a gas gives way easily: gases are far more compressible.

The decisive difference is how a material answers a shearing force, one that acts along a surface. Glue an eraser to your desk and push its top sideways: it leans a little and then holds, because the solid develops a shear stress that balances your push. Push the top of a layer of liquid and it keeps sliding, however gentle the push. A fluid at rest can support essentially no shear stress, and that is exactly why it flows (Fig. 9.1).

3/3The solid holds its lean, but the liquid keeps deforming for as long as the push lasts.

Fig. 9.1A solid resists a shearing push; a fluid gives way to it.NCERT p. 180

Left, a solid block between a fixed base and a top plate; right, a liquid layer under a floating plate with a vertical dye streak. Step 1: nothing is pushed. Step 2: both plates are pushed sideways; the solid tilts a little and the dye streak leans as the liquid starts to slide. Step 3: the solid keeps its small tilt, while the liquid's plate has slid much further and the streak leans far over, because the liquid keeps flowing.

SimulationSolid, liquid and gas in a closed box

A closed box of water particles with a thermometer and a temperature slider from −20 °C to 120 °C. Below 0 °C the particles only jiggle about fixed places (ice). Between 0 °C and 100 °C they slide past one another but stay together at the bottom of the box (liquid water). Above 100 °C they fly apart and spread through the whole box (steam).

Fig. 9.2Drag the temperature slider and watch whether the particles keep together or fill the whole box.
Try this
  • At 25 °C the particles slide past one another. Do they fill the box, or keep a fixed volume at the bottom?
  • Heat the box past 100 °C. How much of the box do the particles occupy now?
  • Cool it below 0 °C. Can the particles still move past each other, or only jiggle in place?

We first study fluids at rest: pressure, how it grows with depth and the machines that use it. Then we set fluids moving and use energy conservation to reach Bernoulli's principle. Two properties of real fluids close the chapter: viscosity, a kind of internal friction, and surface tension, which makes a liquid surface behave like a stretched skin.

Quick check
1 of 2
Which property puts liquids and gases together in one class, fluids?
Show answer

Answer (D) At rest they cannot support a shear stress, so they flow

A fluid gives way to any shear stress, so it flows. Only gases fill their container and only gases compress easily, and many liquids (mercury, for one) are denser than many solids.

9.2 Pressure

Press a drawing pin into a soft board with your thumb. The point goes in, yet the flat head does not hurt you, although the same force acts at both ends. A heavy truck spreads its load over ten or more tyres, and a camel's broad padded feet keep it from sinking into desert sand. In each case the effect is decided by the force divided by the area that carries it.

Now lower a stone into a bucket of still water. The water pushes on every part of the stone and of the bucket, and each push is perpendicular to the surface it acts on. Why? A fluid at rest cannot hold a shear stress (Section 9.1). If the water pressed on a wall at a slant, it would feel a slanted reaction whose part along the wall is a shear, and it would slide. Still water can only push straight on (Fig. 9.3, first step).

To measure the push at a point, picture a tiny sensor: a sealed, airless can closed by a light piston of area A, held out by a calibrated spring. Placed in the fluid, the piston moves in until the spring's force balances the fluid's normal force F on it. The ratio gives the average push per unit area over the piston:

Average pressure
Pav=FA
(9.1)

Shrinking the piston towards a point defines the pressure (दाब) there:

Pressure at a point
P=lim⁡ΔA→0ΔFΔA
(9.2)
Definition
Pressure

The pressure at a point in a fluid is the normal force per unit area on a small surface placed there. It is a scalar, measured in pascal (1 Pa=1 N m−2).

ExampleA 1 kg bottle of water standing on a 50 cm2 base presses on the table with about 2.0×103 Pa.

The SI unit of pressure, 1 N m−2, is called the pascal (Pa) after the French scientist Blaise Pascal (1623–1662), whose work on fluids we meet next; larger pressures are often quoted in standard atmospheres, 1 atm=1.013×105 Pa, close to the air pressure at sea level. The dimensions of pressure are [ML−1T−2]. Notice that pressure carries no direction: (9.2) uses only the force normal to the little surface, and that surface may face any way. Pressure is a scalar.

Because a fluid has no fixed shape, we describe how much matter it packs into a space by its density, the mass per unit volume:

Density
ρ=mV
(9.3)

Density has dimensions [ML−3] and unit  kg m−3. Dividing by the density of water at 4 °C, 1.000×103 kg m−3, gives the relative density, a pure number. Mercury's is about 13.6, so a litre of mercury has a mass of 13.6 kg (Table 9.1). Squeezing hardly changes a liquid's density, while a gas becomes denser in proportion to the pressure on it.

Table 9.1Approximate densities of some fluids (liquids near 20 °C unless stated; gases at 20 °C and 1 atm)NCERT Table 9.1 · p. 181
FluidDensity (kg m−3)Relative density
Water (4 °C)10001.000
Water (20 °C)9980.998
Sea water≈ 1025≈ 1.03
Milk≈ 1030≈ 1.03
Kerosene≈ 800≈ 0.80
Petrol≈ 740≈ 0.74
Glycerine12601.26
Mercury≈ 13 60013.6
Air1.200.0012
Helium0.170.00017
Example 9.1NCERT Example 9.1 · p. 182

Standing on four tyres

A car of mass 1500 kg rests on four identical tyres. Each tyre flattens against the road over a patch about 12 cm×15 cm. Estimate the average pressure the tyres exert on the road. Take g=9.8 m s−2.

Try it yourself first. Then reveal the solution one step at a time.

9.2.1 Pascal's Law

Float a metre under the surface of a pool and turn your head slowly. Your eardrums feel the same squeeze whichever way they face. Experiments of this kind led Pascal to two conclusions about a fluid at rest, now called Pascal's law: the pressure at a point acts equally in every direction, and it is the same at all points on one horizontal level. Both follow from Newton's laws.

LawPascal's law (fluid at rest)

In a fluid at rest, the pressure at a point is the same in every direction, and points at the same horizontal level within one connected body of the fluid are at the same pressure.

DerivationWhy pressure has no direction

Isolate a tiny wedge of the fluid shaped like a right-angled prism (Fig. 9.3). Its cross-section is a right triangle with a vertical face of area Ax, a horizontal face of area Ay and a sloping face of area As that makes an angle θ with the horizontal. The surrounding fluid pushes normally on the faces with forces Fx, Fy and Fs. The wedge's weight scales as the cube of its size and the face forces as the square, so for a small enough wedge the weight is negligible.

The wedge is in equilibrium. The push on the slope has a horizontal part Fssin⁡θ and a vertical part Fscos⁡θ, which the other two pushes must cancel. The triangle's geometry links the areas by the same factors:

Fx=Fssin⁡θ,Ax=Assin⁡θFy=Fscos⁡θ,Ay=Ascos⁡θ

Dividing each force by its area,

FxAx=FyAy=FsAs
(9.4)

that is, Px=Py=Ps. The angle θ was arbitrary, so the sloping face could point any way at all: at a point, the pressure is the same in every direction. Whatever the orientation of a surface inside a fluid at rest, the fluid pushes on it normally with this one pressure.

4/4Fssin⁡θ balances Fx and Fscos⁡θ balances Fy; dividing by the areas gives one pressure.

Fig. 9.3Still water pushes at right angles on every surface, and the wedge shows that the push is the same in all directions.NCERT Fig. 9.2 · p. 182

Left, a tank of still water. Step 1: a tilted block sits in the water, with arrows meeting each of its faces and each wall at right angles. Step 2: a tiny right-angled wedge of the water is enlarged on the right, faces labelled x (vertical), y (base) and s (slope), with the slope angle θ. Step 3: a force arrow presses normally on each face. Step 4: the slope's force is split into horizontal and vertical parts that balance the other two forces.

For the second part of the law, take a thin horizontal rod of the fluid with flat ends of area A. Gravity acts across the rod, so along it only the end pushes PLA and PRA act. The rod stays at rest, so PLA=PRA and PL=PR. Any two points on one level, joined through the same fluid, share one pressure; if they did not, fluid would flow from the high-pressure end towards the other.

9.2.2 Variation of Pressure with Depth

Dive towards the bottom of a deep pool and your ears start to hurt. The deeper you go, the more water lies above you, and that water must be held up by the pressure beneath it. The argument below turns this into a formula.

DerivationPressure at a depth

Picture a vertical cylinder of the fluid itself, with flat ends of area A: its top face at level 1 and its bottom face a height h lower, at level 2 (Fig. 9.4). The sideways pushes on its curved surface cancel in pairs. Vertically, three forces act: P1A down on the top, P2A up on the bottom and the weight mg of the fluid inside. With the cylinder at rest,

(P2−P1)A=mg
(9.5)

The mass inside is m=ρV=ρAh. Substituting and cancelling A,

P2−P1=ρgh
(9.6)

Now raise level 1 to the free surface, where the air presses with atmospheric pressure Pa, and call the pressure at depth h simply P:

Pressure at depth h
P=Pa+ρgh
(9.7)
Definition
Gauge pressure

The gauge pressure at a point is the amount by which the pressure there exceeds atmospheric pressure, P−Pa. At depth h in a liquid open to the air it equals ρgh.

ExampleTwo metres under water the gauge pressure is about 2.0×104 Pa.

Look at what (9.7) leaves out. The area A cancelled, and nothing in the result describes the vessel. The pressure at a point depends only on how far below the free surface it lies, not on the width or shape of the container or on how much liquid it holds. A tall thin pipe of water and a broad tank filled to the same height press equally hard on each square centimetre of their bases. This surprising result is called the hydrostatic paradox (Fig. 9.4, last step).

Join vessels of different shapes at the bottom and pour in water: it settles to one level in all of them, since only a common level gives each height in the joining pipe a single pressure. Masons use exactly this when they mark a level line on two walls with a clear plastic pipe full of water; the water surfaces at the two ends always lie at the same height.

4/4Same depth, same pressure at the base, whatever the vessel's shape or volume.

Fig. 9.4Pressure grows with depth and depends on nothing else about the container.NCERT Fig. 9.3 · p. 183

A tank of water. Step 1: an imaginary vertical cylinder of water is outlined between level 1 and level 2, a height h lower. Step 2: arrows show P1A down on its top, P2A up on its bottom, its weight mg and equal sideways pushes. Step 3: the cylinder's top moves up to the open surface, where the air presses with Pa, giving P = Pa + ρgh. Step 4: three differently shaped vessels filled to one level appear, each with an identical pressure dial at its base.

Example 9.2NCERT Example 9.2 · p. 183

A dive off Swaraj Dweep

A scuba diver explores a reef off Swaraj Dweep (Havelock Island) in the Andamans, 15 m below the surface. Take the density of sea water as 1025 kg m−3, g=9.8 m s−2 and Pa=1.013×105 Pa. Find (a) the gauge pressure and (b) the absolute pressure on the diver. (c) What depth of sea water adds one atmosphere?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
At what depth in fresh water (ρ=1000 kg m−3) is the gauge pressure equal to 1.013×105 Pa? Take g=9.8 m s−2. Give the depth in metres.
m
Show answer

Answer 10.34 m

h=P/(ρg)=1.013×105/(1000×9.8)=10.3 m. A barometer filled with water would need a tube more than 10 m tall.

9.2.3 Atmospheric Pressure and Gauge Pressure

We live at the bottom of an ocean of air. Every square metre of ground carries the weight of all the air piled above it, right up to where the atmosphere fades out, and that weight per unit area is the atmospheric pressure. Near sea level its value is about 1.013×105 Pa: the air above each square metre weighs as much as a loaded 10-tonne truck. We are not crushed because our body fluids push outward almost as hard.

Evangelista Torricelli (1608–1647), who had worked as Galileo's assistant, found how to measure it in 1643. He filled a glass tube about a metre long, sealed at one end, with mercury, covered the open end, upended it in a dish of mercury and then uncovered it. The mercury sank until its column stood about 76 cm above the level in the dish, leaving an almost empty space at the top. This instrument is the mercury barometer (Fig. 9.5).

Why does the column stop there? The space above it holds only a trace of mercury vapour, whose pressure is negligible. A point in the tube at the level of the dish's surface and a point on that open surface lie on one level in the same mercury, so they share one pressure, which is Pa at the open surface. The column of height h above the first point must supply it:

Mercury barometer
Pa=ρgh
(9.8)

Here ρ is the density of mercury. With ρ=1.36×104 kg m−3 and h=0.760 m, (9.8) gives 1.013×105 Pa; a water barometer would need a tube over 10 m tall. Doctors still quote blood pressure in millimetres of mercury; the pressure of a 1 mm column of mercury, about 133 Pa, is called one torr in Torricelli's honour. Weather scientists use the bar (105 Pa) and the millibar; weather charts mark pressure in hectopascals, and 1 hPa is one millibar.

Sea-level pressure rises and falls by a few percent with the weather; a steady fall of the barometer warns of an approaching low-pressure system, such as a Bay of Bengal cyclone.

To measure a pressure difference we use an open-tube manometer: a U-tube partly filled with a liquid, one arm connected to the gas being tested and the other open to the air (Fig. 9.5, last step). Points at one level in the liquid share one pressure, so the gas pressure P exceeds Pa by the extra liquid column h in the open arm: P−Pa=ρgh. A manometer therefore reads gauge pressure directly. For small differences a light liquid such as water or oil gives a readable height; for large ones, dense mercury keeps the tube short.

2/3The column falls to about 760 mm; the air's push on the dish holds it there.

Fig. 9.5The atmosphere holds up a 76 cm column of mercury; a manometer shows gauge pressure as a difference in height.NCERT Fig. 9.5 · p. 183

Left, a glass tube closed at the top stands upside down in a dish of mercury. Step 1: the tube is full of mercury. Step 2: the column has dropped to 760 mm above the dish level, with a near-vacuum above it and arrows showing the air pressing on the dish. Step 3: the barometer fades and a U-tube manometer appears on the right, one arm joined to a gas bulb, the other open, with the liquid higher by h in the open arm.

Example 9.3NCERT Example 9.3 · p. 184

An atmosphere of constant density

Air at sea level on a mild day has a density of about 1.20 kg m−3. Suppose, unrealistically, that the density kept this value all the way up. How high would the atmosphere reach, for a sea-level pressure of 1.013×105 Pa? Compare your answer with the height of Mount Everest, 8849 m. Take g=9.8 m s−2.

Try it yourself first. Then reveal the solution one step at a time.

Example 9.4NCERT Example 9.4 · p. 184

Six kilometres down

India's Deep Ocean Mission is building Matsya 6000, a submersible designed to carry three people to a depth of 6000 m. At that depth, taking sea water's density as a constant 1030 kg m−3, g=9.8 m s−2 and Pa=1.013×105 Pa, find (a) the gauge pressure, (b) the absolute pressure and (c) the net force on a circular window 10 cm in diameter if the cabin is kept at 1 atm.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
A blood-pressure reading of 120 mm of mercury is a gauge pressure of how many kilopascals? (ρHg=13 600 kg m−3, g=9.8 m s−2)
kPa
Show answer

Answer 16 kPa

P=ρgh=(13 600)(9.8)(0.120)=1.60×104 Pa=16.0 kPa above atmospheric pressure.

9.2.4 Hydraulic Machines

Squeeze the bottom of a closed toothpaste tube and paste appears at the nozzle; poke a water balloon and it bulges everywhere at once. A push on one part of an enclosed fluid is felt throughout it. Try it with a plastic bottle filled to the brim with water, capped, and pierced with pinholes at several places: squeeze it gently and every jet grows stronger at the same moment, whichever way the hole faces. The extra pressure from your hand has reached all points of the water equally.

LawPascal's law (transmission of pressure)

Any change of pressure applied to an enclosed fluid at rest reaches every part of the fluid, and the walls holding it, without any loss.

A hydraulic lift puts this to work (Fig. 9.6). Two pistons, a small one of area A1 and a large one of area A2, close the two ends of a sealed, liquid-filled system. A force F1 on the small piston adds a pressure F1/A1, which reaches the large piston unchanged and pushes it up with a force

Hydraulic lift
F2=F1A1 A2=A2A1 F1

The force is multiplied by A2/A1, the mechanical advantage. Energy is not created, though. The liquid is practically incompressible, so the volume pushed out of the small cylinder, A1d1, equals the volume added under the large piston, A2d2. The large piston moves a distance smaller by the same factor, and F1d1=F2d2: for frictionless pistons, the work put in equals the work got out.

2/3The extra pressure F1/A1 reaches every point and lifts the large piston with F2=(A2/A1)F1.

Fig. 9.6A small force on a small piston lifts a large load, which rises less than the small piston falls.NCERT Fig. 9.6 · p. 185

A narrow and a wide cylinder are joined at the bottom by a liquid-filled channel; the wide piston carries a car. Step 1: nothing is pushed. Step 2: a force presses on the narrow piston, arrows show the pressure spreading through the liquid, and a large upward force acts on the wide piston. Step 3: the narrow piston has moved down a long way and the car has risen a little; dashed bands show the volume swept below the narrow piston equals the thin layer added under the wide one.

Car brakes use the same idea. At each wheel, a piston in a wheel cylinder presses the brake pads or shoes against a spinning disc or drum. All four wheel cylinders are joined by narrow pipes full of brake fluid to a single master cylinder under the pedal, whose piston is much smaller. A modest push on the pedal therefore becomes a large force at every wheel, and because the fluid carries one pressure throughout, the four wheels are braked equally, which keeps the car from slewing sideways. Excavator arms, a barber's adjustable chair and a mechanic's bottle jack are other hydraulic machines.

Example 9.5NCERT Example 9.5 · p. 185

A syringe-powered lifting arm

For a science fair, Meera builds a lifting arm from two syringes joined by a water-filled tube. The plunger of the small syringe has a diameter of 1.2 cm and that of the large one 3.0 cm. (a) What upward force does the large plunger exert when she pushes the small one with 8.0 N? (b) If she pushes the small plunger in by 5.0 cm, how far does the large plunger move out? Ignore friction.

Try it yourself first. Then reveal the solution one step at a time.

Example 9.6NCERT Example 9.6 · p. 186

Lifting a car at a service station

At a car service station, compressed air pushes on a small piston of radius 4.0 cm, and oil passes the pressure on to a large piston of radius 20 cm carrying a car of mass 1100 kg. (a) What force must the air exert on the small piston to hold the car up? (b) What gauge pressure of air does this need? Take g=9.8 m s−2 and ignore the weights of the pistons.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 2
A hydraulic press has pistons of diameters 2.0 cm and 20 cm. What force, in newtons, does the large piston exert when 50 N is applied to the small one?
N
Show answer

Answer 5000 N

The area ratio is (20/2.0)2=100, so F2=100×50 N=5000 N. Using the diameter ratio (10) instead of the area ratio is the usual slip.

9.3 Streamline Flow

Light an agarbatti in a still room and watch its smoke. For the first few centimetres it rises as a smooth, unbroken thread; higher up the thread wavers and breaks into curls and puffs. The same air behaves in two quite different ways. The study of fluids in motion, fluid dynamics, starts by describing the smooth kind precisely.

To describe a flow we can watch fixed places rather than individual particles. Imagine a camera fixed on one spot in a canal, recording the velocity of whatever water passes that spot. If the reading never changes, at that spot and at every other, the flow is steady. Steady does not mean uniform: water may run faster through a narrow stretch than a wide one, so a given particle can speed up as it travels. What stays fixed is the velocity at each place.

Definition
Streamline

A streamline is the path followed by fluid particles in steady flow; its tangent at any point gives the direction of the fluid's velocity there.

ExampleThe smooth thread of smoke just above an incense stick traces streamlines.

In steady flow the pattern of streamlines is a fixed map of the motion. Two streamlines can never cross: at the crossing a particle would have two possible velocities, and the flow could not be steady. A bundle of neighbouring streamlines forms a tube of flow; fluid enters it at one end and leaves at the other, but never crosses its sides.

DerivationThe equation of continuity

Take a tube of flow, or a pipe, whose cross-section changes from A1 to A2, with speeds v1 and v2 there (Fig. 9.7). In a short time Δt the fluid crossing section 1 fills a slab of length v1Δt and mass ρ1A1v1Δt. In steady flow no fluid piles up between the sections, so the same mass crosses section 2 in the same time. Cancelling Δt,

ρ1A1v1=ρ2A2v2
(9.9)

A liquid is practically incompressible, so ρ1=ρ2 and

Equation of continuity
A1v1=A2v2
(9.10)

The two sections were any two along the tube, so

Av=constant
(9.11)

The product Av is the volume flow rate, the volume passing each second, in  m3 s−1. The equation of continuity says it is the same all along a pipe; it is the law of conservation of mass for an incompressible fluid. Where a pipe narrows the fluid must speed up, and where it widens the fluid slows down. On a streamline picture, crowded lines mean fast flow. Partly closing the end of a garden hose with your thumb uses this: if the open area falls to a fifth while the flow rate holds steady, the water leaves five times faster and reaches much farther.

Steady flow survives only at low enough speeds. Above a critical speed, which depends on the fluid, the pipe and any obstacles, the flow becomes turbulent (विक्षुब्ध): the velocity at a point changes erratically and the fluid churns into eddies of many sizes. The breaking agarbatti smoke, the foaming rapids of the Ganga near Rishikesh and the swirling wake behind a speeding bus are all turbulent. Smooth flow in which neighbouring layers slide past one another without mixing is called laminar flow.

Compare

Streamline (laminar) flow and turbulent flow

Streamline (laminar)

  • Velocity at each point steady in time
  • Streamlines never cross
  • Layers slide past without mixing
  • Slow flows, narrow pipes, thick liquids

Turbulent

  • Velocity at each point fluctuates
  • Eddies; no fixed streamline pattern
  • Layers mix strongly
  • Fast flows and flow past obstacles

2/3In time Δt the same volume crosses both sections: A1v1Δt=A2v2Δt.

Fig. 9.7Where streamlines crowd together the fluid moves faster; the same volume passes every section each second.NCERT Fig. 9.7 · p. 187

A horizontal pipe narrows from a wide section on the left to a narrow section on the right. In the first step five streamlines run along the pipe and squeeze together in the narrow part, with a short velocity arrow in the wide part and a long one in the narrow part. In the second step two shaded slabs mark the fluid that crosses each section in the same short time: a short, wide slab of length v1Δt on the left and a long, narrow slab of length v2Δt on the right, holding the same volume. In the third step the streamlines in the narrow part are replaced by swirling eddies, showing turbulent flow.

Quick check
1 of 3
Water flows at 0.60 m s−1 through a pipe of cross-section 12 cm2, which then narrows to 3.0 cm2. What is the speed in the narrow part, in m/s?
m/s
Show answer

Answer 2.4 m/s

v2=v1A1/A2=0.60×12/3.0=2.4 m s−1. A quarter of the area means four times the speed.

9.4 Bernoulli's Principle

Water entering the narrow part of a pipe speeds up, so something must push it forward. Only the surrounding fluid can, through a pressure that is higher behind the water than ahead of it. Faster flow should therefore go with lower pressure. In his book Hydrodynamica (1738) the Swiss physicist Daniel Bernoulli turned this idea into an exact relation by applying energy conservation to a moving fluid.

DerivationBernoulli's equation from the work–energy theorem

Take an ideal fluid: incompressible, free of viscosity and in steady flow through a pipe whose cross-section and height both change (Fig. 9.8). At the lower section 1 the area is A1, the speed v1, the pressure P1 and the height h1; at section 2 they are A2, v2, P2 and h2. Follow the fluid lying between cross-sections K and M. In a short time Δt its rear face moves from K to L, a distance v1Δt, and its front face from M to N, a distance v2Δt. By continuity the two slivers have equal volumes, ΔV=A1v1Δt=A2v2Δt.

The fluid behind pushes the rear face forward through v1Δt, doing work P1A1v1Δt=P1ΔV. The fluid ahead pushes back on the front face while it moves forward through v2Δt, doing work −P2ΔV. The net work done on our fluid is

W=(P1−P2) ΔV

In steady flow the stretch from L to M looks the same before and after, so the net change is as if the sliver of mass Δm=ρ ΔV had been carried from section 1 to section 2. Its kinetic and potential energies change by

ΔK=12Δm (v22−v12)ΔU=Δm g (h2−h1)

By the work–energy theorem (Chapter 5), W=ΔK+ΔU. Dividing through by ΔV, with Δm/ΔV=ρ,

P1−P2=12ρ(v22−v12)+ρg(h2−h1)

and collecting each section's terms on its own side,

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2
(9.12)

Sections 1 and 2 were any two places along the flow, so

Bernoulli's equation
P+12ρv2+ρgh=const.
(9.13)

3/3Pressure forces do net work (P1−P2)ΔV, which raises the sliver's kinetic and potential energy.

Fig. 9.8The net effect of the flow is to move a sliver of volume ΔV from the low, wide section to the high, narrow one.NCERT Fig. 9.9 · p. 188

A pipe rises from a wide section low on the left to a narrow section high on the right. Step 1: the fluid between cross-sections K and M is shaded, with a thin sliver next to K in a second colour. Step 2: the fluid now lies between L and N; the left sliver has gone and an equal-volume, longer, thinner sliver has appeared beyond M. Step 3: arrows show P1A1 pushing forward at the left end and P2A2 pushing back at the right end, with heights h1 and h2 measured from a reference line.

PrincipleBernoulli's principle

In the steady flow of an incompressible, non-viscous fluid, the quantity P+12ρv2+ρgh, that is, pressure plus kinetic energy per unit volume plus potential energy per unit volume, has a single value all along any one streamline.

Each term of Bernoulli's equation is measured in  J m−3, which is the same as a pascal, so pressure itself can be read as a store of energy per unit volume. Put v=0 everywhere and (9.13) gives back P1−P2=ρg(h2−h1), the hydrostatic result (9.6): fluid statics is a special case.

The equation is only as good as its assumptions. It needs steady flow: in turbulent flow the speed and pressure at a point fluctuate and it fails. It treats the fluid as incompressible, ignoring energy stored by squeezing it, so a gas qualifies only when its density barely changes. And it assumes no viscosity (Section 9.5); in a real fluid some mechanical energy turns into heat, and the pressure downstream comes out lower than (9.12) predicts. For water and for air at modest speeds it is nonetheless remarkably useful.

ExampleNCERT p. 188

A Venturi meter

A horizontal water main of diameter 6.0 cm narrows smoothly to a throat of diameter 3.0 cm and widens again. Pressure taps show that the pressure in the throat is 4.5 kPa lower than in the main pipe. Find the speed in the main pipe and the flow rate. Treat the water as an ideal fluid of density 1000 kg m−3.

Try it yourself first. Then reveal the solution one step at a time.

A scent atomiser or a paint spray gun blows air fast across the top of a thin tube dipping into the liquid; the pressure there drops, so the liquid is pushed up and torn into droplets. A Bunsen burner draws in air at its base for the same reason.

InteractiveVenturi tube lab

A horizontal pipe with a narrow throat in the middle and two vertical manometer tubes, one over the wide part and one over the throat. Sliders set the speed in the wide part and the throat's area as a fraction of the pipe's. Readouts give the throat speed from continuity, the pressure drop from Bernoulli's equation and the two column heights; the column over the throat falls as the throat narrows, and a warning appears if the throat pressure drops below atmospheric. Play sends tracer dots along the pipe, moving faster through the throat.

Fig. 9.9Set the flow speed and the size of the throat, then press Play. Compare the speeds, the pressures and the two manometer columns.

9.4.1 Speed of Efflux: Torricelli's Law

Efflux means outflow. Torricelli noticed that liquid spurting from a hole in a tank leaves with the speed it would gain by falling freely from the level of the liquid's surface down to the hole. Bernoulli's equation shows why.

DerivationSpeed of efflux

A tank of liquid, density ρ, has a small hole of area A1 in its side at height y1 above the base and the free surface, of area A2, at height y2 (Fig. 9.10). The gas above the surface is at pressure P, and the jet leaves into the atmosphere at Pa. Continuity gives v2=(A1/A2) v1. When the tank is much wider than the hole (A2≫A1) the surface sinks so slowly that we may set v2≈0. Applying (9.12) between the hole (1) and the surface (2),

Pa+12ρv12+ρgy1=P+ρgy2

With h=y2−y1, the depth of the hole below the surface,

v1=2gh+2(P−Pa)ρ
(9.14)

If the tank is open to the air, P=Pa and

Torricelli's law
v1=2gh
(9.15)

This is exactly the speed of a stone dropped from rest through a height h, and for the same reason: potential energy ρgh per unit volume becomes kinetic energy 12ρv2. A drum of water with a tap 0.80 m below its surface delivers water at 2(9.8)(0.80)≈4.0 m s−1. In the opposite extreme, when the gas above the liquid is at a pressure far above atmospheric, the 2gh term hardly matters and the container pressure alone sets the jet speed. Fire extinguishers, spray cans and, in a more complicated way, rocket engines work in this regime.

2/3The jet leaves at v1=2gh, the speed of a stone dropped through h.

Fig. 9.10Water leaves a hole at depth h with the speed of a body that has fallen freely through h.NCERT Fig. 9.10 · p. 189

A tank of water stands on legs, with a small hole in its side a depth h below the water surface. In the first step the hole is marked 1 and the surface 2, with h shown between them. In the second step a jet curves out of the hole and, beside the tank, a stone falls from the level of the water surface to the level of the hole, reaching the same speed as the jet. In the third step a lid traps gas at a pressure P above the water, and the jet now lands much farther from the tank.

Quick check
1 of 3
A large open tank has a small hole 1.25 m below the water surface. With what speed, in m/s, does water leave the hole? (g=9.8 m s−2)
m/s
Show answer

Answer 4.95 m/s

v=2gh=2×9.8×1.25=4.95 m s−1, the speed of a stone dropped through 1.25 m.

9.4.2 Dynamic Lift

A fluid streaming past a moving body can push it sideways, at right angles to its motion. This dynamic lift keeps a soaring eagle aloft with barely a wingbeat, holds up aircraft and bends the path of a spinning ball. Bernoulli's principle explains at least part of it.

Start with a ball that only moves forward. Seen from the ball, air streams past it, and the pattern of streamlines above the ball is a mirror image of the pattern below (Fig. 9.11). Equal speeds mean equal pressures, so the air pushes neither up nor down; it only drags the ball back.

Spin breaks the symmetry. A spinning ball carries a thin skin of air round with it. With backspin, that skin moves along with the oncoming air over the top of the ball and against it underneath, so the air speeds up above and slows below; the streamlines crowd together on top and spread out underneath. Bernoulli's principle then gives a lower pressure above than below, and the ball is pushed upward. Lift caused by spin is called the Magnus effect. A spin bowler's drift and a footballer's curling free kick show it. (A fast bowler's swing has a different cause: the seam makes air cling differently to the two sides.)

An aircraft wing has a special cross-section, an aerofoil: rounded at the front, curved on top and tapering to a sharp rear edge. Tilted slightly up into the oncoming air, it makes the air over its upper surface flow faster than the air beneath. The lower pressure on top gives an upward lift that, in level flight, balances the aircraft's weight. Seen another way, the wing deflects air downwards and, by Newton's third law, the air pushes the wing up; both pictures describe the same flow.

3/3A wing makes the air above it flow faster than the air below, so the pressure difference lifts it.

Fig. 9.11Faster air on one side means lower pressure there and a lift force towards that side.NCERT Fig. 9.11 · p. 190

Air streams from right to left. Step 1: a ball without spin sits among streamlines that bend symmetrically above and below it; there is no lift. Step 2: the ball has backspin, shown by a curved arrow; streamlines crowd together above it and spread out below, and an arrow shows upward lift. Step 3: an aircraft wing appears on the right, streamlines crowded over its curved top, with a lift arrow and labels for fast, low-pressure air above and slower, higher-pressure air below.

Example 9.7NCERT Example 9.7 · p. 190

Holding up a cargo aircraft

A cargo aircraft of mass 1.5×105 kg has a total wing area of 600 m2. It flies level at 720 km h−1 at a height where the density of air is 0.90 kg m−3. (a) Estimate how much higher the average pressure is under the wings than above them. (b) Estimate by what fraction the air over the wings must move faster than the air beneath them. Take g=9.8 m s−2.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 2
A table-tennis ball is hit to the right with backspin, so that its top surface moves to the left. The air pushes it
Show answer

Answer (A) upward

Relative to the ball the air streams to the left. Backspin carries air along over the top, so the air above moves faster and its pressure is lower: the Magnus force points up, and the ball stays in the air longer.

9.5 Viscosity

Stir a jar of honey and a glass of water with the same spoon at the same speed: the honey fights back far more. Real fluids resist the sliding of one layer over another, a kind of internal friction called viscosity (श्यानता). It acts only while layers move relative to each other; a fluid at rest feels none.

The cleanest way to see viscosity is to shear a thin film of liquid steadily. Spread oil between two flat plates a small distance l apart, keep one plate fixed and slide the other parallel to it at a constant speed v (Fig. 9.12). The liquid clings to both plates (the oil touching each plate moves with it), so across the gap the velocity climbs in a straight line from zero to v. Keeping the plate going needs a steady force F, because each layer drags on its neighbours. Swap the oil for honey and F shoots up for the same v: honey is more viscous. Smooth, layered flow of this kind is laminar.

The same clinging shapes flow in a pipe. The liquid at the wall stays put, so the velocity profile is a rounded curve, zero at the wall and largest on the axis: the liquid moves in nested cylindrical sleeves, each sliding inside the next (last step of Fig. 9.12).

A solid under a steady shear stress takes up a fixed strain. A fluid cannot: a small block of the liquid that starts out rectangular is sheared into a parallelogram that leans further and further for as long as the plate moves. In a time Δt its top shifts by Δx=v Δt relative to its bottom, so the shear strain Δx/l keeps growing. For water, oils and many other liquids, experiment shows that the stress F/A on a plate of area A is proportional not to the strain but to its rate, Δx/(l Δt)=v/l. The constant of proportionality defines the viscosity:

1/3The speed rises evenly from zero at the fixed plate to v at the moving plate.

Fig. 9.12In laminar flow the layers slide over one another; the speed is zero at a fixed wall.NCERT Fig. 9.12 · p. 191

Left, oil between a fixed lower plate and an upper plate moving right. Step 1: arrows in the oil grow steadily longer from bottom to top, showing speed rising evenly from zero to v. Step 2: a small rectangular block of oil leans over into a parallelogram, its top shifted by Δx across the thickness l. Step 3: the plates fade and a pipe appears, with speed arrows longest on the axis and zero at the walls, their tips on a curved profile.

Coefficient of viscosity
η=F/Av/l=FlvA
(9.16)
Definition
Coefficient of viscosity

The coefficient of viscosity η (eta) of a fluid is the shear stress in it divided by the rate of shear strain. Its SI unit is the pascal second,  Pa s= N s m−2, also called the poiseuille (Pl).

ExampleWater at 20 °C has η≈1.0×10−3 Pa s; glycerine is more than a thousand times as viscous.

Viscosity has dimensions [ML−1T−1]. Runny liquids such as water, petrol and alcohol have low viscosity; glycerine, honey, engine oil and tar have high viscosity. Blood, at a few millipascal seconds, is thicker than water. Table 9.2 gives some values.

Table 9.2Viscosities of some fluids (approximate; 1 mPa s = 0.001 Pa s)NCERT Table 9.2 · p. 192
FluidTemperature (°C)Viscosity (mPa s)
Water01.79
Water201.00
Water400.65
Water1000.28
Castor oil20≈ 990
Glycerine20≈ 1400
Air200.018
Air1000.022

Temperature matters a great deal. Heat a liquid and its molecules break free of their neighbours more easily, so its viscosity falls: warm ghee pours where cold ghee barely moves, and water at 100 °C is between three and four times less viscous than at 20 °C. In a gas the trend reverses. Gas molecules hardly attract one another; a gas's viscosity comes from molecules darting between layers and carrying momentum with them. Heating makes them faster, so a gas becomes more viscous when hot.

Example 9.8NCERT Example 9.8 · p. 191

A plate sliding on an oil film

A flat steel plate of area 0.040 m2 rests on a film of oil 0.40 mm thick spread on a level table. A light string tied to the plate runs over an ideal pulley at the table's edge to a hanging mass of 50 g. Once released, the plate slides at a steady 2.0 cm s−1. Find the coefficient of viscosity of the oil. (g=9.8 m s−2)

Try it yourself first. Then reveal the solution one step at a time.

9.5.1 Stokes' Law

A marble sinking through a bottle of shampoo, a dust grain settling in still air and a food pellet drifting down through aquarium water each drag the nearby fluid along, so the fluid holds them back. For a small sphere moving slowly, George Gabriel Stokes (1819–1903) showed in 1851 that this drag is proportional to the sphere's speed, its radius and the fluid's viscosity:

Stokes' law
F=6πηav
(9.17)
where
a
radius of the sphere
v
its speed relative to the fluid

We shall not derive it, but its dimensions check: [η][a][v]=[ML−1T−1][L][LT−1]=[MLT−2], a force. The law holds while the flow round the sphere stays smooth, which means small, slow spheres or very viscous fluids. In Chapter 5 you calculated the work done by resistive forces; Stokes' law tells us what such a force is for a small sphere.

DerivationTerminal velocity

Drop a sphere of radius a and density ρ into a fluid of density σ. Three forces act: its weight 43πa3ρg downward, the buoyant force 43πa3σg upward and the viscous drag 6πηav upward. At first v=0 and the sphere accelerates. As it gains speed the drag grows, until drag plus buoyancy equals the weight. From then on the net force is zero and the sphere falls at a constant terminal velocity vt:

6πηavt=43πa3(ρ−σ)g
Terminal velocity
vt=2a2(ρ−σ)g9η
(9.18)

The terminal velocity grows as the square of the radius and falls as the viscosity rises. A cloud droplet 10 μm across settles at only about 3 mm s−1, so the gentlest updraft keeps clouds afloat. Large raindrops fall at several metres per second; that is too fast for Stokes' law itself, but they too reach a steady terminal speed. If ρ<σ, (9.18) gives a negative vt: the sphere rises at a steady speed, like an air bubble in a bottle of honey.

Example 9.9NCERT Example 9.9 · p. 192

Timing a ball in glycerine

In a falling-ball experiment, a steel ball bearing of radius 1.5 mm is dropped into a tall jar of glycerine at 20 °C. Once it moves steadily, it takes 8.8 s to fall between two marks 20.0 cm apart. Find the viscosity of glycerine. Densities: steel 7.8×103 kg m−3, glycerine 1.26×103 kg m−3; g=9.8 m s−2.

Try it yourself first. Then reveal the solution one step at a time.

InteractiveFalling-ball lab

A tall jar of liquid with a ball you choose (steel, glass or nylon) and a radius slider; the liquid can be glycerine or castor oil. Play drops the ball and plots its speed against time on a graph covering the first few milliseconds, where the speed rises and levels off at the terminal value. Readouts give the terminal velocity from Stokes' law, the time to fall 20 cm, and the weight, buoyancy and drag at terminal speed. A nylon ball in glycerine rises instead of sinking, and a note warns when the ball moves too fast for Stokes' law to be accurate.

Fig. 9.13Pick a ball and a liquid, set the radius and press Play. Watch the speed level off at the terminal velocity.
Quick check
1 of 3
Which is the SI unit of the coefficient of viscosity?
Show answer

Answer (D) Pa s

η is stress divided by strain rate: pascal divided by s−1, which is Pa s. N m−1 is the unit of surface tension and N m−2 is the pascal itself.

9.6 Surface Tension

A pond skater strides across a pond without sinking. Dewdrops on a spider's web are tiny spheres. A towel with one corner dipped in a bucket slowly turns damp far above the water line. A steel paper clip laid gently on water floats, although steel is nearly eight times as dense as water. All of these happen at the free surface of a liquid, which behaves rather like a stretched elastic skin with extra energy of its own. The effect is called surface tension (पृष्ठ तनाव). Gases have no free surface, so it is a property of liquids.

9.6.1 Surface Energy

Why should a surface cost energy? Compare two molecules of the same liquid (Fig. 9.14): one buried deep inside, the other at the surface with air above it. Each is attracted by its neighbours, and every attraction lowers its potential energy, rather as a ball resting in a hollow has less energy than one perched on the rim. The buried molecule has neighbours all round; their pulls cancel and its energy is as low as it gets. The surface molecule has neighbours only beside and below it, roughly half as many, so it is less tightly bound and feels a net pull into the liquid.

How much energy is involved? Pulling all the molecules apart, that is, boiling the liquid away, takes about 41 kJ per mole of water at its boiling point. A surface molecule has lost about half its bonds, so it carries, very roughly, half of that per-molecule energy in excess. Making new surface therefore costs energy, and a liquid left alone adopts the smallest area that its surroundings allow. Most surface effects follow from this single fact.

Seen at the molecular scale, the edge of a liquid is fuzzy: over a thickness of just a few molecules, the density falls from its full value to nearly zero.

4/4Pulling the wire out by d adds area 2ld, because the film has two faces.

Fig. 9.14Surface molecules carry extra energy, so stretching a film needs a steady force.NCERT Fig. 9.15 · p. 193

Left, a grid of liquid molecules below the surface. Step 1: a molecule deep inside is highlighted, pulled equally by neighbours all round. Step 2: a surface molecule is highlighted; it has neighbours only beside and below it, and a thick arrow shows its net inward pull. Step 3: a soap film appears on the right, stretched across a wire frame closed by a sliding wire. Step 4: the slider has been pulled out a distance d by a force F, adding a strip of film.

9.6.2 Surface Energy and Surface Tension

To measure the extra energy, stretch a soap film on a wire frame with one side free to slide: a straight wire of length l (Fig. 9.14, last steps). The film pulls the slider inward, so you must hold it with a force F. Pull it out slowly through a distance d; the work Fd becomes surface energy of the new film. A film has two surfaces, front and back, so the new area is 2ld. If S is the surface energy per unit area,

S (2ld)=Fd
(9.19)
Surface tension
S=F2l
(9.20)

Read one way, S is the surface energy per unit area, in  J m−2; read the other, it is the force per unit length with which the surface pulls on its edge, in  N m−1. The two units are the same.

Definition
Surface tension

The surface tension S of a liquid is the force per unit length acting in the plane of its surface, at right angles to any line drawn in it. It equals the extra energy per unit area of the surface.

ExampleWater at 20 °C: S≈0.073 N m−1.

Two remarks sharpen the picture. First, slit the surface in your mind along any line, as you might cut a stretched drum skin. Each side would pull on the cut edge with a force S per unit length, at right angles to the line and in the plane of the surface. Inside the surface these pulls balance; only at a free edge, like the slider, is there a net inward pull. Second, the energy is a property of the boundary between two substances and changes if either of them is changed, so we write Sla for a liquid–air surface, Ssl for solid–liquid and Ssa for solid–air.

Table 9.3Surface tensions of some liquids (approximate)NCERT Table 9.3 · p. 194
LiquidTemperature (°C)Surface tension (N m−1)
Water200.073
Water1000.059
Ethanol200.022
Glycerine200.063
Soap solution20≈ 0.03
Mercury20≈ 0.48

Surface tension, like a liquid's viscosity, usually falls as the liquid warms. Whether a liquid clings to a solid depends on the interface energies. A drop sticks when replacing a patch of solid–air surface and a patch of liquid–air surface by a single solid–liquid interface lowers the energy, that is, when Ssl<Ssa+Sla.

A liquid's pull can also be weighed. If the straight lower edge of a thin glass plate, of length l, just touches a liquid, the surface tugs down on both faces of the plate with a total force 2Slal. So hang the plate from one arm of a balance, counterpoise it, and bring the liquid up to touch its edge; then add a mass m to the other pan until the plate just tears free. Then mg=2Slal, or

Sla=mg2l
(9.21)

9.6.3 Angle of Contact

Where a liquid surface meets a solid it is usually curved. The angle of contact (स्पर्श कोण) θ is the angle between the solid surface and the tangent to the liquid surface at the line of contact, measured through the liquid. Rain beads up on a taro (arbi) leaf with θ well above 90∘, while a drop of water on clean glass spreads into a thin lens with a small θ (Fig. 9.15).

At the line where the drop's edge meets the solid, three interfaces come together: liquid–air, solid–liquid and solid–air, with tensions Sla, Ssl and Ssa. Each pulls the line along its own surface. Along the solid, Ssa pulls outward, away from the drop, while Ssl and the part of Sla parallel to the solid pull inward. (The solid balances the part of Sla at right angles to it.) For the line to stay put,

Slacos⁡θ+Ssl=Ssa
(9.22)

Read it as cos⁡θ=(Ssa−Ssl)/Sla. A liquid attracted more strongly to the solid than to itself makes Ssl small, so cos⁡θ is large and θ is acute: the liquid wets the solid, as water wets clean glass and as cooking oil creeps across a steel plate. When the liquid's own molecules hold together more strongly, the solid–liquid interface is costly (Ssl>Ssa), cos⁡θ<0 and θ is obtuse: the liquid refuses to wet the solid and pulls into beads, like rain on a freshly waxed car or mercury on glass.

Washing powders and soaps lower the contact angle, so that water soaks into the fibres of cloth instead of sitting on them as drops, and carries the dirt away. The water-repellent coating on a raincoat or an umbrella does the opposite: it raises the contact angle so that drops roll straight off.

3/3Along the solid, Slacos⁡θ+Ssl=Ssa (arrows not to scale).

Fig. 9.15The angle of contact is measured through the liquid; it is obtuse where the liquid does not wet the solid.NCERT Fig. 9.17 · p. 195

Two water drops. Step 1: the left drop, on a waxy leaf, is almost a ball; a dashed tangent at its edge makes an obtuse angle θ with the leaf, measured through the water. Step 2: the right drop, on clean glass, is a flat cap whose tangent makes an acute angle θ. Step 3: three arrows act at the right drop's edge: liquid–air tension along the tangent, solid–liquid tension along the glass towards the drop, and solid–air tension away from it.

Quick check
1 of 3
The units N m−1 and J m−2 both describe
Show answer

Answer (C) surface tension

Surface tension measures a pull per unit length and, equally, an energy per unit area; 1 J m−2=1 N m m−2=1 N m−1.

9.6.4 Drops and Bubbles

Tiny droplets, such as dew on a cobweb or the mist from a plant sprayer, are almost perfect spheres, and so is a drifting soap bubble. The reason is the energy cost of surface: of all shapes that hold a given volume, the sphere has the least area. (A sphere of volume 1 cm3 has a surface of about 4.84 cm2; a cube of the same volume has 6 cm2.) Only gravity and air resistance stop larger drops from being round.

Surface tension also makes the pressure inside a drop higher than outside. Take a spherical drop of radius r in equilibrium, with pressure Pi inside and Po outside, and imagine it swelling slightly to radius r+Δr.

DerivationExcess pressure inside a drop

The surface area grows by

ΔA=4π(r+Δr)2−4πr2≈8πr Δr

neglecting (Δr)2, so the extra surface energy is

ΔE=8πr Δr Sla
(9.23)

At equilibrium this energy is supplied by the pressure difference pushing the surface out, as the volume grows by 4πr2Δr:

W=(Pi−Po) 4πr2 Δr
(9.24)

Setting W=ΔE,

Drop or cavity
Pi−Po=2Slar
(9.25)

The general rule is that across a curved liquid surface the pressure is higher on the concave side. An air bubble inside a liquid (a cavity) has one surface, so its excess pressure is also 2S/r. A soap bubble is different: it is a thin film with two surfaces, outer and inner, each contributing 2S/r:

Soap bubble
Pi−Po=4Slar
(9.26)

Smaller bubbles have the larger excess pressure. That is why blowing a soap bubble takes a firm puff at the start and gets easier as it grows, and why, if a small and a large soap bubble are joined by a tube, the small one empties into the large one.

3/3Two surfaces: Pi−Po=4S/r.

Fig. 9.16Pressure is higher on the concave side of a liquid surface; a soap bubble has two surfaces.NCERT Fig. 9.18 · p. 196

Three spheres of the same radius r side by side, each with its excess pressure written below. In the first step the liquid drop in air is highlighted: it has one curved surface, and its inside pressure exceeds the outside by 2S/r. In the second step the air bubble inside a liquid is highlighted: again one surface, and an excess pressure of 2S/r. In the third step the soap bubble is highlighted: a thin film with an outer and an inner surface, so its excess pressure is 4S/r.

Example 9.10NCERT Example 9.10 · p. 197

An aquarium bubbler

An aquarium air pump pushes air down a narrow tube whose open end, of internal radius 0.50 mm, sits 12.0 cm below the water surface. What air pressure must the pump supply to push out a bubble that is exactly a hemisphere at the tube's mouth, and what is the excess pressure inside that bubble? Take S=7.28×10−2 N m−1, ρ=1000 kg m−3, g=9.8 m s−2 and Pa=1.013×105 Pa.

Try it yourself first. Then reveal the solution one step at a time.

9.6.5 Capillary Rise

Dip one end of a narrow glass tube in water and the water climbs inside it, above the outside level; the narrower the tube, the higher it goes. This capillary rise takes its name from the Latin for hair, since in a hair-thin tube the rise is large. Its cause is the pressure difference across the curved water surface, the meniscus, in the tube.

DerivationHeight of capillary rise

Let the tube have radius a and let water meet the glass at a contact angle θ (acute). The meniscus is then concave upward, part of a sphere of radius R, and the geometry of Fig. 9.17 gives R=a/cos⁡θ. The concave side, the air above, is at the higher pressure, so just below the meniscus the water's pressure Pin is less than atmospheric by

Pa−Pin=2SR=2Scos⁡θa
(9.27)

Now take a point in the tube level with the outside water surface. It lies on one level with points on that open surface, in the same connected water, so its pressure is Pa. It is a height h below the meniscus, so

Pin+ρgh=Pa
(9.28)

Combining (9.27) and (9.28),

Capillary rise
h=2Scos⁡θρga
(9.29)

The rise is inversely proportional to the tube's radius and comes entirely from surface tension. For water in clean glass, θ≈0. When cos⁡θ is negative, as for mercury in glass (θ≈140∘), (9.29) gives a negative h: the meniscus is convex and the liquid stands below the outside level.

2/3The meniscus is part of a sphere of radius R=a/cos⁡θ; just below it the pressure is 2S/R below Pa.

Fig. 9.17Water climbs a narrow tube; mercury, which does not wet glass, is pushed down.NCERT Fig. 9.19 · p. 197

Left, a narrow glass tube stands in a beaker of water; the water inside has risen a height h, ending in a concave meniscus. Step 2: an enlarged view shows the meniscus as part of a circle of radius R, centre above it, meeting the wall of the tube (radius a) at the contact angle θ, so R = a/cos θ. Step 3: the enlarged view is replaced by a tube in mercury, where the level inside stands below the outside surface with a convex top.

ExampleNCERT p. 197

How high, and how low?

A clean glass capillary of internal radius 0.25 mm stands upright in a dish. Find the change in level inside it (a) for water, with S=0.073 N m−1, θ=0, ρ=1000 kg m−3; (b) for mercury, with S=0.48 N m−1, θ=140∘, ρ=1.36×104 kg m−3. Take g=9.8 m s−2.

Try it yourself first. Then reveal the solution one step at a time.

Capillarity draws oil up the cotton wick of a diya and lets blotting paper and towels soak up water. It helps water climb the narrow vessels of plants, though in a tall tree most of the lift comes from evaporation at the leaves.

Quick check
1 of 3
Water rises 3.0 cm in a clean glass capillary. How high, in cm, will it rise in a tube of half that radius?
cm
Show answer

Answer 6 cm

h∝1/a, so halving the radius doubles the rise, to 6.0 cm.
Recap

Mechanical Properties of Fluids: key points and summary

  1. Liquids and gases flow because at rest they cannot support a shear stress; liquids keep their volume and are nearly incompressible, gases fill their container (Section 9.1).
  2. Pressure is normal force per unit area, P=lim⁡ΔF/ΔA: a scalar measured in pascal, with 1 atm=1.013×105 Pa; density is ρ=m/V (Section 9.2).
  3. Pascal's law: in a fluid at rest the pressure at a point acts equally in all directions and is the same at all points on one level (Section 9.2.1).
  4. P=Pa+ρgh: pressure depends only on depth, not on the vessel's shape; P−Pa is the gauge pressure (Section 9.2.2).
  5. A barometer balances Pa against about 76 cm of mercury; a manometer reads gauge pressure; 1 torr=1 mm of mercury ≈133 Pa and 1 bar=105 Pa (Section 9.2.3).
  6. An enclosed fluid transmits pressure undiminished; a hydraulic lift multiplies force by A2/A1 while the work in equals the work out (Section 9.2.4).
  7. In steady flow streamlines never cross, and Av is constant (continuity, mass conservation); beyond a critical speed the flow turns turbulent (Section 9.3).
  8. Bernoulli: P+12ρv2+ρgh is constant along a streamline of an ideal fluid in steady flow; it is energy conservation per unit volume (Section 9.4).
  9. Torricelli's law v=2gh gives the efflux speed from an open tank (Section 9.4.1); faster flow on one side of a body produces dynamic lift, as in the Magnus effect and aerofoils (Section 9.4.2).
  10. Viscosity: η is shear stress divided by the rate of shear strain, in Pa s; it falls with temperature for liquids and rises for gases (Section 9.5).
  11. Stokes' law F=6πηav leads to the terminal velocity vt=2a2(ρ−σ)g/9η (Section 9.5.1).
  12. Surface tension S can be read as a pull per unit length of surface or as extra energy per unit area; a film gives S=F/2l, and at a contact line Slacos⁡θ+Ssl=Ssa (Section 9.6.2, Section 9.6.3).
  13. Excess pressure is 2S/r in a drop or cavity and 4S/r in a soap bubble; capillary rise is h=2Scos⁡θ/ρga (Section 9.6.4, Section 9.6.5).
Practice · 29 questions

Practice questions with answers (NCERT “Exercises”)

  1. Short answerEasyNCERT Exercises Q9.1

    Explain the following.

    1. A nurse measures blood pressure with the cuff on the upper arm, level with the heart. Why would a reading taken with the arm raised high above the head come out lower?

      Show solution

      Blood is a fluid, so within it pressure changes with height as ρgh. Raising the arm puts the cuff above the heart, where the pressure is lower by ρgh; for 40 cm that is about (1060)(9.8)(0.40)≈4 kPa, roughly 30 mm of mercury. Readings are taken at heart level so that they can be compared.

    2. On the summit of Mount Everest the air pressure is only about a third of its sea-level value, even though traces of air are found hundreds of kilometres higher. How can two-thirds of the air's weight lie in the lowest 9 km?

      Show solution

      Air is compressible. The lower layers are squeezed by the weight of all the air above them, so they are much denser than the upper layers. Most of the atmosphere's mass is therefore packed into the lowest few kilometres; the thin air above Everest, though it extends far up, weighs only about a third of the total.

    3. A classmate says, 'Pressure is force divided by area, and force is a vector, so pressure must be a vector.' What is wrong with this?

      Show solution

      Only the component of force normal to the surface enters the definition, and at a point in a fluid at rest the pressure is the same whichever way the small surface faces (Pascal's law). A quantity with no preferred direction is a scalar; the direction of the push comes from the surface it acts on.

  2. Short answerMediumNCERT Exercises Q9.2

    Explain the following in terms of surface energy and the angle of contact.

    1. Water spreads out on clean glass but rolls into beads on a waxed car bonnet, while mercury forms beads even on clean glass.

      Show solution

      Water is strongly attracted to glass, so the solid–liquid interface has low energy (Ssl<Ssa), cos⁡θ>0 and θ is acute: water wets glass and spreads. Wax attracts water weakly (Ssl>Ssa), so θ is obtuse and water beads up. Mercury atoms attract one another far more strongly than they attract glass, so mercury's contact angle with glass is obtuse (about 140∘) and it forms drops.

    2. Why does the surface tension of a liquid not depend on how large its surface is?

      Show solution

      Surface tension is energy per unit area (or force per unit length). It depends on how each surface molecule is bound, which is the same however many molecules are at the surface. Doubling the area doubles the surface energy but leaves S unchanged.

    3. Why is detergent added to washing water, and what does it do to the angle of contact?

      Show solution

      Detergent molecules gather at the interfaces and lower the liquid–air and solid–liquid tensions. The contact angle becomes small, so the water wets the fibres and soaks into the cloth instead of standing on it as drops, and it carries dirt away.

    4. A blob of water released inside an orbiting space station becomes a nearly perfect sphere. Why?

      Show solution

      In free fall nothing flattens the blob, so only its surface energy matters. For a given volume a sphere has the least surface area, hence the least surface energy, and the liquid pulls itself into that shape.

  3. Quick answerEasyNCERT Exercises Q9.3 (a)–(d)

    Decide whether each statement is true or false.

    1. Hot water has a lower surface tension than cold water.

      Show answer

      Answer True

      Surface tension generally falls as a liquid warms.
    2. When heated, a gas becomes more viscous while a liquid becomes less viscous.

      Show answer

      Answer True

      Faster gas molecules carry more momentum between layers; warm liquid molecules slip past one another more easily.
    3. A steady sideways push on a layer of fluid produces a fixed shear strain, just as it does in a solid.

      Show answer

      Answer False

      In a solid the shear stress is proportional to the shear strain; in a fluid it is proportional to the rate of shear strain, so the strain keeps growing.
    4. The speeding-up of water where a pipe narrows follows from conservation of mass alone, without Bernoulli's principle.

      Show answer

      Answer True

      Av= constant is the continuity equation, a statement of mass conservation. Bernoulli's principle is needed only to find the pressure there.
  4. Quick answerMediumNCERT Exercises Q9.3 (e)

    A one-tenth-scale model of an aircraft is tested in a wind tunnel. Compared with the real aircraft, at what air speed does the flow round the model become turbulent?

    Show answer

    Answer (C) at a higher speed

    Turbulence sets in when a combination of speed and size, ρvL/η (the Reynolds number), passes a critical value. A model one-tenth the size needs about ten times the speed to reach it.
  5. Short answerMediumNCERT Exercises Q9.4 (a)–(d)

    Use continuity or Bernoulli's principle to explain the following.

    1. When an express train rushes past a platform, people standing close to the edge feel pulled towards the train.

      Show solution

      The train drags the air beside it along at high speed, so the pressure of that air drops below that of the still air behind the people. The pressure difference pushes them towards the train, which is why we must stand behind the yellow line.

    2. A gardener fits a tapering nozzle to a hose, and the water now reaches the far end of the garden.

      Show solution

      The flow rate Av is set by the supply. The nozzle reduces the area A at the outlet, so the speed v rises in proportion, and a faster jet travels farther.

    3. In a hospital drip, the nurse sets the flow rate by squeezing the thin tube with a roller clamp rather than by raising or lowering the bottle. Why is the clamp so much more effective?

      Show solution

      In the thin tube viscosity controls the flow. For such smooth flow the rate rises as the fourth power of the tube's radius (Poiseuille's relation, used in Question 14) but only in proportion to the pressure difference, which the bottle's height sets. Halving the bore cuts the flow sixteen-fold; halving the height only halves it.

    4. A garden sprinkler spins as water jets out of its bent arms, and a can of water on a smooth trolley rolls backwards when water squirts from a hole near its base.

      Show solution

      Each jet carries momentum away from the vessel every second. By Newton's third law the vessel gets an equal backward push, a thrust, which turns the sprinkler and drives the trolley backwards, just as exhaust drives a rocket.

  6. NumericalEasyNCERT Exercises Q9.5

    A 60 kg sprinter stands still on one foot in spiked shoes. Only the six spikes touch the track, each over a circle of diameter 2.0 mm. What pressure do the spikes exert on the track? (g=9.8 m s−2)

    Pa
    Show solution

    Force: F=mg=(60)(9.8)=588 N. Area: 6π(1.0×10−3 m)2=1.885×10−5 m2.

    P=FA=3.1×107 Pa

    That is about 300 atmospheres, which is why spikes grip a track so well.

  7. NumericalEasyNCERT Exercises Q9.6

    How tall would a barometer filled with glycerine (density 1260 kg m−3) have to be at normal atmospheric pressure, 1.013×105 Pa? Why is mercury the practical choice? (g=9.8 m s−2)

    m
    Show solution
    h=Paρg=1.013×105(1260)(9.8) m=8.2 m

    Mercury is about 10.8 times denser than glycerine, so its column is 10.8 times shorter (76 cm), and its vapour pressure is tiny, so the space above the column is almost empty.

  8. NumericalMediumNCERT Exercises Q9.7

    A valve on an undersea oil pipeline rests on the sea floor 2.5 km down. Its casing is rated to withstand a pressure of 5.0×107 Pa. Find the pressure on it and say whether it is safe. (Sea water 1030 kg m−3, g=9.8 m s−2, Pa=1.013×105 Pa; ignore currents.)

    Pa
    Show solution
    P=Pa+ρgh=2.53×107 Pa

    This is about half the rated 5.0×107 Pa, so the casing is safe, with a safety factor of about 2.

  9. NumericalMediumNCERT Exercises Q9.8

    A two-post hydraulic lift is rated for vehicles of up to 2500 kg. The piston under the load has a cross-section of 350 cm2. What is the greatest pressure the small piston, and the oil, must withstand? (g=9.8 m s−2)

    Pa
    Show solution
    P=mgA2=(2500)(9.8)0.0350 Pa=7.0×105 Pa

    The same pressure acts on the small piston whatever its size; only the force on it is smaller.

  10. NumericalMediumNCERT Exercises Q9.9

    A U-tube holds mercury at the bottom. Water is poured into one arm and kerosene into the other until the mercury surfaces in the two arms are level; the water column is then 12.0 cm tall and the kerosene column 15.0 cm. Find the relative density of kerosene.

    Show solution

    The mercury surfaces are level, so the pressures on them are equal and the two liquid columns weigh the same per unit area:

    ρwg(0.120)=ρkg(0.150)

    so ρk/ρw=0.80.

  11. NumericalHardNCERT Exercises Q9.10

    Now 10.0 cm more water is added on the water side of the U-tube in Question 10, and 10.0 cm more kerosene on the kerosene side. Find the difference between the two mercury levels and say which is higher. (Relative density of mercury: 13.6.)

    cm
    Show solution

    The columns are now 22.0 cm of water and 25.0 cm of kerosene. The extra water pushes harder, so the mercury sinks on the water side. At the level of that lower mercury surface the pressures are equal. Above it stand 22.0 cm of water on one side, and mercury of height h topped by 25.0 cm of kerosene on the other. In centimetres of water:

    22.0=13.6 h+0.80(25.0)h=2.013.6=0.147 cm

    The mercury stands about 1.5 mm higher in the kerosene arm.

  12. Short answerEasyNCERT Exercises Q9.11

    A friend wants to estimate the water pressure inside the foaming rapids of the Ganga at Shivpuri, where rafters ride, using Bernoulli's equation. Will that work? Give reasons.

    Show solution

    No. In the rapids the flow is turbulent: the speed and pressure at each point change erratically, and mechanical energy is constantly lost to eddies and heat. Bernoulli's equation needs steady flow along streamlines with no such losses.

  13. Short answerMediumNCERT Exercises Q9.12

    Riya subtracts atmospheric pressure from every pressure in her calculation with Bernoulli's equation, working only with gauge pressures. Will the speed she finds come out wrong? When could this shortcut fail?

    Show solution

    Gauge pressure is absolute pressure minus Pa. Subtracting the same constant from both sides of (9.12) changes nothing, so the answer is the same. It would matter if the two points were under noticeably different atmospheric pressures (a large height difference in air), or where a formula needs the absolute pressure itself, such as the density of a gas.

  14. NumericalHardNCERT Exercises Q9.13

    Castor oil (density 960 kg m−3, viscosity 0.99 Pa s) flows steadily along a level tube 1.2 m long with an inner radius of 0.80 cm. Oil is collected at the outlet at 3.0×10−3 kg s−1. For laminar flow the volume flow rate is given by Poiseuille's relation, Q=πr4ΔP/(8ηl). Find the pressure difference between the ends. Is laminar flow a fair assumption?

    Pa
    Show solution

    Volume flow rate: Q=3.0×10−3/960=3.13×10−6 m3 s−1.

    ΔP=8ηlQπr4=8(0.99)(1.2)(3.13×10−6)π(0.0080)4 Pa
    ΔP≈2.3×103 Pa

    The average speed is Q/πr2≈1.6 cm s−1. So slow a flow of so viscous an oil is laminar: the Reynolds number ρv(2r)/η≈0.2, far below the value of about 2000 at which flow in a pipe turns turbulent.

  15. NumericalMediumNCERT Exercises Q9.14

    In a wind-tunnel test, air moves at 52 m s−1 over the top of a model wing and at 45 m s−1 under it. The wing's area is 1.8 m2 and the air's density 1.2 kg m−3. Find the lift on the wing.

    N
    Show solution
    ΔP=12ρ(vu2−vl2)=12(1.2)(2704−2025) Pa=407 Pa
    F=ΔP A=407×1.8 N≈733 N
  16. Quick answerMediumNCERT Exercises Q9.15

    Two sketches show an ideal liquid flowing steadily through a level pipe with a narrow middle section. Each has one open vertical tube over the wide part and one over the narrow part. In sketch P the liquid stands lower in the tube over the narrow part; in sketch Q it stands higher there. Which sketch is wrong?

    Show answer

    Answer (B) sketch Q

    In the narrow part the liquid moves faster (continuity), so by Bernoulli's equation its pressure is lower and the liquid stands lower in the tube above it. Sketch Q shows the opposite, so it is wrong.
  17. NumericalMediumNCERT Exercises Q9.16

    The tube of a garden sprayer has a cross-section of 6.0 cm2, and its nozzle has 25 holes, each 0.80 mm in diameter. The liquid moves along the tube at 2.4 m per minute. With what speed does it leave the holes?

    m/s
    Show solution

    Speed in the tube: 2.4 m per minute =0.040 m s−1. Total hole area: 25π(4.0×10−4 m)2=1.257×10−5 m2.

    v=A1v1Aholes=(6.0×10−4)(0.040)1.257×10−5=1.9 m s−1
  18. NumericalEasyNCERT Exercises Q9.17

    A soap film is formed in a wire frame closed at the bottom by a light sliding wire 8.0 cm long. The film holds up the slider together with a small weight hung from it, a total of 4.0×10−3 N. Find the surface tension of the soap solution.

    N/m
    Show solution

    The film pulls the slider up along its length on both faces, with total force 2Sl, which balances the weight:

    S=W2l=4.0×10−32(0.080) N m−1=0.025 N m−1
  19. Short answerMediumNCERT Exercises Q9.18

    A soap film in a rectangular frame, closed at the bottom by a light sliding wire, holds up a weight of 3.0×10−2 N hung from the wire. Two more frames have sliding wires of the same length, but one frame is twice as tall and the other has a pointed top instead of a flat one. Using the same soap solution at the same temperature, what weight can each of these films hold? Explain.

    Show solution

    The film pulls up on the slider with a force 2Sl, which depends only on the surface tension and the slider's length, not on the area or shape of the film above. Both films therefore hold the same 3.0×10−2 N. Stretching a film creates more surface but does not change S.

  20. NumericalMediumNCERT Exercises Q9.19

    A drop of mercury has a radius of 2.00 mm. Taking the surface tension of mercury as 0.48 N m−1 and atmospheric pressure as 1.013×105 Pa, find the excess pressure inside the drop and the total pressure inside it.

    Pa
    Show solution
    2Sr=2(0.48)2.00×10−3 Pa=480 Pa
    Pi=Pa+480 Pa=1.018×105 Pa
  21. NumericalMediumNCERT Exercises Q9.20

    A soap bubble has a radius of 4.00 mm, and the soap solution's surface tension is 0.0300 N m−1. Atmospheric pressure is 1.013×105 Pa and g=9.8 m s−2.

    1. What is the excess pressure inside the soap bubble, in Pa?

      Pa
      Show solution
      4Sr=4(0.0300)4.00×10−3 Pa=30 Pa
    2. An air bubble of the same radius forms 25.0 cm deep in a container of the same solution, whose relative density is 1.10. What is the pressure inside it, in Pa?

      Pa
      Show solution

      One surface, so the excess is 2S/r=15 Pa. The liquid outside is at Pa+ρgh with ρgh=(1100)(9.8)(0.250)=2695 Pa:

      Pi=(101 300+2695+15) Pa=1.040×105 Pa
  22. Quick answerMedium

    Three vessels, a straight cylinder, a funnel that widens upwards and a flask that narrows upwards, have equal base areas and are filled with water to the same height. Which statement is true?

    Show answer

    Answer (C) The water exerts the same force on each base, although the weights of water differ

    Same depth gives the same pressure, and on equal areas the same force ρghA. The funnel's sloping walls hold up its extra water; in the flask the walls push down on the water and make up for the missing weight.
  23. Quick answerMedium

    Assertion (A): Two streamlines in a steady flow can never cross. Reason (R): At a crossing point a fluid particle would have two different velocities.

    Show answer

    Answer (A) Both A and R are true, and R correctly explains A

    Both are true, and R is exactly why A holds: the velocity at a point of a steady flow is unique, so the tangent directions of two crossing streamlines would contradict it.
  24. Quick answerMedium

    Eight identical tiny raindrops each fall through still air at a terminal speed of 1.0 cm s−1. They merge into a single drop. Assuming Stokes' law still applies, find the new terminal speed in cm/s.

    cm/s
    Show answer

    Answer 4 cm/s

    The volume is 8 times larger, so the radius doubles; vt∝a2, so the speed becomes 4 times larger, 4.0 cm s−1.
  25. Quick answerMedium

    A capillary tube stands in water inside a lift. When the lift accelerates upward, the height of the water column in the tube

    Show answer

    Answer (B) decreases

    In the lift's frame the effective gravity is g+a0, where a0 is the lift's acceleration. Since h∝1/g in (9.29), the column gets shorter. In free fall it would fill the whole tube.
  26. Quick answerMedium

    How much work is needed to blow a soap bubble of radius 3.0 cm from a very small one, if the surface tension of the solution is 0.030 N m−1? Give the answer in joules.

    J
    Show answer

    Answer 0.00068 J

    W=S×2(4πr2)=0.030×8π(0.030)2=6.8×10−4 J.
  27. Quick answerMedium

    Assertion (A): The viscosity of a gas increases as the gas is heated. Reason (R): Faster gas molecules carry more momentum from one layer of the gas to the next.

    Show answer

    Answer (A) Both A and R are true, and R correctly explains A

    Both are true, and R explains A: a gas's viscosity arises from molecules moving between layers and transferring momentum, which they do more vigorously when hot.
  28. Quick answerHard

    A thousand identical water droplets, each of radius 0.10 mm, merge into one drop. How much surface energy is released, in joules? (S=0.073 N m−1)

    J
    Show answer

    Answer 0.00000826 J

    R=10r=1.0 mm. Area lost =4π(1000r2−R2)=4π(1.0×10−5−1.0×10−6)=1.13×10−4 m2; energy released =S ΔA=8.3×10−6 J, which slightly warms the drop.
  29. Quick answerMediumNCERT Exercises Q9.4 (e)

    A spin bowler's delivery drifts sideways in the air before it pitches. Which explanation is best?

    Show answer

    Answer (D) The spinning ball drags air round with it, making the air faster on one side; the lower pressure there pushes the ball sideways

    This is the Magnus effect: the spin speeds up the air on one side and slows it on the other, and the resulting pressure difference pushes the ball off its parabolic path.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbookPractice question here
NCERT Exercises Q9.1Practice Q1
NCERT Exercises Q9.2Practice Q2
NCERT Exercises Q9.3 (a)–(d)Practice Q3
NCERT Exercises Q9.3 (e)Practice Q4
NCERT Exercises Q9.4 (a)–(d)Practice Q5
NCERT Exercises Q9.5Practice Q6
NCERT Exercises Q9.6Practice Q7
NCERT Exercises Q9.7Practice Q8
NCERT Exercises Q9.8Practice Q9
NCERT Exercises Q9.9Practice Q10
NCERT Exercises Q9.10Practice Q11
NCERT Exercises Q9.11Practice Q12
NCERT Exercises Q9.12Practice Q13
NCERT Exercises Q9.13Practice Q14
NCERT Exercises Q9.14Practice Q15
NCERT Exercises Q9.15Practice Q16
NCERT Exercises Q9.16Practice Q17
NCERT Exercises Q9.17Practice Q18
NCERT Exercises Q9.18Practice Q19
NCERT Exercises Q9.19Practice Q20
NCERT Exercises Q9.20Practice Q21
NCERT Exercises Q9.4 (e)Practice Q29

Quick check: MCQs on Mechanical Properties of Fluids

  1. 9.1 Introduction2 questions
  2. 9.2.2 Variation of Pressure with Depth3 questions
  3. 9.2.3 Atmospheric Pressure and Gauge Pressure3 questions
  4. 9.2.4 Hydraulic Machines2 questions
  5. 9.3 Streamline Flow3 questions
  6. 9.4.1 Speed of Efflux: Torricelli's Law3 questions
  7. 9.4.2 Dynamic Lift2 questions
  8. 9.5.1 Stokes' Law3 questions
  9. 9.6.3 Angle of Contact3 questions
  10. 9.6.5 Capillary Rise3 questions
Glossary · 14

Key terms (English–Hindi)

angle of contactस्पर्श कोणsparsh kon
The angle between a solid surface and the tangent to a liquid surface where they meet, measured through the liquid.First seen in the chapter →
Bernoulli's principleबरनौली का सिद्धांतbarnauli ka siddhaant
Along a streamline of an ideal fluid in steady flow, P+12ρv2+ρgh is constant.First seen in the chapter →
capillary riseकेशिकत्वkeshikatva
Rise or fall of a liquid in a narrow tube caused by surface tension: h=2Scos⁡θ/ρga.First seen in the chapter →
equation of continuityसांतत्य समीकरणsaantatya sameekaran
For steady flow of an incompressible fluid, Av is the same at every cross-section; it expresses mass conservation.First seen in the chapter →
fluidतरलtaral
A substance that can flow: a liquid or a gas. At rest it cannot support a shear stress.First seen in the chapter →
gauge pressureप्रमापी दाबpramaapi daab
Pressure above atmospheric pressure, P−Pa; what tyre gauges and manometers read.First seen in the chapter →
Pascal's lawपास्कल का नियमpaaskal ka niyam
In a fluid at rest pressure acts equally in all directions, is the same on a level, and a change in it is passed on undiminished.First seen in the chapter →
pressureदाबdaab
Normal force per unit area at a point in a fluid; a scalar, measured in pascal (Pa).First seen in the chapter →
relative densityआपेक्षिक घनत्वaapekshik ghanatva
Density of a substance divided by the density of water at 4 °C; it has no unit.First seen in the chapter →
streamlineधारारेखाdhaararekha
The path of fluid particles in steady flow; its tangent gives the fluid's velocity at each point.
surface tensionपृष्ठ तनावprishth tanaav
Force per unit length in a liquid surface, equal to its extra energy per unit area; unit N m−1.First seen in the chapter →
terminal velocityसीमांत वेगseemaant veg
The steady speed a body reaches falling through a fluid, when drag and buoyancy balance its weight.First seen in the chapter →
turbulent flowविक्षुब्ध प्रवाहvikshubdh pravaah
Irregular flow beyond a critical speed, in which the velocity at a point fluctuates and eddies form.First seen in the chapter →
viscosityश्यानताshyaanata
Internal friction between fluid layers in relative motion; η is shear stress divided by strain rate, in Pa s.First seen in the chapter →