Mechanical Properties of Fluids: Class 11 Physics · Chapter 9
Physics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗
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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.
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.
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).
- 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.
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Answer (D) At rest they cannot support a shear stress, so they flow
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Answer False
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 , 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 on it. The ratio gives the average push per unit area over the piston:
Shrinking the piston towards a point defines the pressure (दाब) there:
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 ().
ExampleA bottle of water standing on a base presses on the table with about .
The SI unit of pressure, , 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, , close to the air pressure at sea level. The dimensions of pressure are . 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 has dimensions and unit . Dividing by the density of water at 4 °C, , 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.
| Fluid | Density () | Relative density |
|---|---|---|
| Water (4 °C) | 1000 | 1.000 |
| Water (20 °C) | 998 | 0.998 |
| Sea water | ≈ 1025 | ≈ 1.03 |
| Milk | ≈ 1030 | ≈ 1.03 |
| Kerosene | ≈ 800 | ≈ 0.80 |
| Petrol | ≈ 740 | ≈ 0.74 |
| Glycerine | 1260 | 1.26 |
| Mercury | ≈ 13 600 | 13.6 |
| Air | 1.20 | 0.0012 |
| Helium | 0.17 | 0.00017 |
Standing on four tyres
A car of mass rests on four identical tyres. Each tyre flattens against the road over a patch about . Estimate the average pressure the tyres exert on the road. Take .
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.
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.
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 , a horizontal face of area and a sloping face of area that makes an angle with the horizontal. The surrounding fluid pushes normally on the faces with forces , and . 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 and a vertical part , which the other two pushes must cancel. The triangle's geometry links the areas by the same factors:
Dividing each force by its area,
that is, . The angle 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.
4/4 balances and balances ; dividing by the areas gives one pressure.
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 . Gravity acts across the rod, so along it only the end pushes and act. The rod stays at rest, so and . 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.
Picture a vertical cylinder of the fluid itself, with flat ends of area : its top face at level 1 and its bottom face a height lower, at level 2 (Fig. 9.4). The sideways pushes on its curved surface cancel in pairs. Vertically, three forces act: down on the top, up on the bottom and the weight of the fluid inside. With the cylinder at rest,
The mass inside is . Substituting and cancelling ,
Now raise level 1 to the free surface, where the air presses with atmospheric pressure , and call the pressure at depth simply :
The gauge pressure at a point is the amount by which the pressure there exceeds atmospheric pressure, . At depth in a liquid open to the air it equals .
ExampleTwo metres under water the gauge pressure is about .
Look at what (9.7) leaves out. The area 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.
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.
A dive off Swaraj Dweep
A scuba diver explores a reef off Swaraj Dweep (Havelock Island) in the Andamans, below the surface. Take the density of sea water as , and . 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.
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Answer 10.34 m
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Answer (B)
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Answer False
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 : 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 at the open surface. The column of height above the first point must supply it:
Here is the density of mercury. With and , (9.8) gives ; 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 , is called one torr in Torricelli's honour. Weather scientists use the bar () and the millibar; weather charts mark pressure in hectopascals, and 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 exceeds by the extra liquid column in the open arm: . 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.
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.
An atmosphere of constant density
Air at sea level on a mild day has a density of about . Suppose, unrealistically, that the density kept this value all the way up. How high would the atmosphere reach, for a sea-level pressure of ? Compare your answer with the height of Mount Everest, . Take .
Try it yourself first. Then reveal the solution one step at a time.
Six kilometres down
India's Deep Ocean Mission is building Matsya 6000, a submersible designed to carry three people to a depth of . At that depth, taking sea water's density as a constant , and , find (a) the gauge pressure, (b) the absolute pressure and (c) the net force on a circular window in diameter if the cabin is kept at 1 atm.
Try it yourself first. Then reveal the solution one step at a time.
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Answer 16 kPa
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Answer (A) Its vertical height stays about 76 cm, so it fills a longer length of tube
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Answer (C) kerosene
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.
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 and a large one of area , close the two ends of a sealed, liquid-filled system. A force on the small piston adds a pressure , which reaches the large piston unchanged and pushes it up with a force
The force is multiplied by , the mechanical advantage. Energy is not created, though. The liquid is practically incompressible, so the volume pushed out of the small cylinder, , equals the volume added under the large piston, . The large piston moves a distance smaller by the same factor, and : for frictionless pistons, the work put in equals the work got out.
2/3The extra pressure reaches every point and lifts the large piston with .
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.
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 and that of the large one . (a) What upward force does the large plunger exert when she pushes the small one with ? (b) If she pushes the small plunger in by , how far does the large plunger move out? Ignore friction.
Try it yourself first. Then reveal the solution one step at a time.
Lifting a car at a service station
At a car service station, compressed air pushes on a small piston of radius , and oil passes the pressure on to a large piston of radius carrying a car of mass . (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 and ignore the weights of the pistons.
Try it yourself first. Then reveal the solution one step at a time.
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Answer 5000 N
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Answer (C) the same
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.
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.
Take a tube of flow, or a pipe, whose cross-section changes from to , with speeds and there (Fig. 9.7). In a short time the fluid crossing section 1 fills a slab of length and mass . In steady flow no fluid piles up between the sections, so the same mass crosses section 2 in the same time. Cancelling ,
A liquid is practically incompressible, so and
The two sections were any two along the tube, so
The product is the volume flow rate, the volume passing each second, in . 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.
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 the same volume crosses both sections: .
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.
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Answer 2.4 m/s
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Answer False
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Answer (D) where the fluid moves fastest
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.
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 , the speed , the pressure and the height ; at section 2 they are , , and . Follow the fluid lying between cross-sections K and M. In a short time its rear face moves from K to L, a distance , and its front face from M to N, a distance . By continuity the two slivers have equal volumes, .
The fluid behind pushes the rear face forward through , doing work . The fluid ahead pushes back on the front face while it moves forward through , doing work . The net work done on our fluid is
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 had been carried from section 1 to section 2. Its kinetic and potential energies change by
By the work–energy theorem (Chapter 5), . Dividing through by , with ,
and collecting each section's terms on its own side,
Sections 1 and 2 were any two places along the flow, so
3/3Pressure forces do net work , which raises the sliver's kinetic and potential energy.
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.
In the steady flow of an incompressible, non-viscous fluid, the quantity , 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 , which is the same as a pascal, so pressure itself can be read as a store of energy per unit volume. Put everywhere and (9.13) gives back , 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.
A Venturi meter
A horizontal water main of diameter narrows smoothly to a throat of diameter and widens again. Pressure taps show that the pressure in the throat is 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 .
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.
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.
A tank of liquid, density , has a small hole of area in its side at height above the base and the free surface, of area , at height (Fig. 9.10). The gas above the surface is at pressure , and the jet leaves into the atmosphere at . Continuity gives . When the tank is much wider than the hole () the surface sinks so slowly that we may set . Applying (9.12) between the hole (1) and the surface (2),
With , the depth of the hole below the surface,
If the tank is open to the air, and
This is exactly the speed of a stone dropped from rest through a height , and for the same reason: potential energy per unit volume becomes kinetic energy . A drum of water with a tap below its surface delivers water at . In the opposite extreme, when the gas above the liquid is at a pressure far above atmospheric, the 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 , the speed of a stone dropped through .
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.
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Answer 4.95 m/s
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Answer (D) lower, because the water moves faster
- Find and of the sliver
- Net work is
- Divide by
- Equal volumes pass both ends
- Apply
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- Equal volumes pass both ends
- Net work is
- Find and of the sliver
- Apply
- Divide by
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.
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.
Holding up a cargo aircraft
A cargo aircraft of mass has a total wing area of . It flies level at at a height where the density of air is . (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 .
Try it yourself first. Then reveal the solution one step at a time.
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Answer (A) upward
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Answer False
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 apart, keep one plate fixed and slide the other parallel to it at a constant speed (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 . Keeping the plate going needs a steady force , because each layer drags on its neighbours. Swap the oil for honey and shoots up for the same : 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 its top shifts by relative to its bottom, so the shear strain keeps growing. For water, oils and many other liquids, experiment shows that the stress on a plate of area is proportional not to the strain but to its rate, . The constant of proportionality defines the viscosity:
1/3The speed rises evenly from zero at the fixed plate to at the moving plate.
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.
The coefficient of viscosity (eta) of a fluid is the shear stress in it divided by the rate of shear strain. Its SI unit is the pascal second, , also called the poiseuille (Pl).
ExampleWater at 20 °C has ; glycerine is more than a thousand times as viscous.
Viscosity has dimensions . 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.
| Fluid | Temperature (°C) | Viscosity (mPa s) |
|---|---|---|
| Water | 0 | 1.79 |
| Water | 20 | 1.00 |
| Water | 40 | 0.65 |
| Water | 100 | 0.28 |
| Castor oil | 20 | ≈ 990 |
| Glycerine | 20 | ≈ 1400 |
| Air | 20 | 0.018 |
| Air | 100 | 0.022 |
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.
A plate sliding on an oil film
A flat steel plate of area rests on a film of oil 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 . Once released, the plate slides at a steady . Find the coefficient of viscosity of the oil. ()
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:
- radius of the sphere
- its speed relative to the fluid
We shall not derive it, but its dimensions check: , 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.
Drop a sphere of radius and density into a fluid of density . Three forces act: its weight downward, the buoyant force upward and the viscous drag upward. At first 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 :
The terminal velocity grows as the square of the radius and falls as the viscosity rises. A cloud droplet across settles at only about , 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 , (9.18) gives a negative : the sphere rises at a steady speed, like an air bubble in a bottle of honey.
Timing a ball in glycerine
In a falling-ball experiment, a steel ball bearing of radius is dropped into a tall jar of glycerine at 20 °C. Once it moves steadily, it takes to fall between two marks apart. Find the viscosity of glycerine. Densities: steel , glycerine ; .
Try it yourself first. Then reveal the solution one step at a time.
Show answer
Answer (D)
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Answer 8 cm/s
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Answer (D) faster, because the viscosity has fallen
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 adds area , because the film has two faces.
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 (Fig. 9.14, last steps). The film pulls the slider inward, so you must hold it with a force . Pull it out slowly through a distance ; the work becomes surface energy of the new film. A film has two surfaces, front and back, so the new area is . If is the surface energy per unit area,
Read one way, is the surface energy per unit area, in ; read the other, it is the force per unit length with which the surface pulls on its edge, in . The two units are the same.
The surface tension 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: .
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 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 for a liquid–air surface, for solid–liquid and for solid–air.
| Liquid | Temperature (°C) | Surface tension () |
|---|---|---|
| Water | 20 | 0.073 |
| Water | 100 | 0.059 |
| Ethanol | 20 | 0.022 |
| Glycerine | 20 | 0.063 |
| Soap solution | 20 | ≈ 0.03 |
| Mercury | 20 | ≈ 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 .
A liquid's pull can also be weighed. If the straight lower edge of a thin glass plate, of length , just touches a liquid, the surface tugs down on both faces of the plate with a total force . 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 to the other pan until the plate just tears free. Then , or
9.6.3 Angle of Contact
Where a liquid surface meets a solid it is usually curved. The angle of contact (स्पर्श कोण) 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 well above , while a drop of water on clean glass spreads into a thin lens with a small (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 , and . Each pulls the line along its own surface. Along the solid, pulls outward, away from the drop, while and the part of parallel to the solid pull inward. (The solid balances the part of at right angles to it.) For the line to stay put,
Read it as . A liquid attracted more strongly to the solid than to itself makes small, so is large and 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 (), and 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, (arrows not to scale).
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.
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Answer (C) surface tension
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Answer 0.03 N/m
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Answer (B) obtuse (more than 90°)
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 has a surface of about ; a cube of the same volume has .) 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 in equilibrium, with pressure inside and outside, and imagine it swelling slightly to radius .
The surface area grows by
neglecting , so the extra surface energy is
At equilibrium this energy is supplied by the pressure difference pushing the surface out, as the volume grows by :
Setting ,
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 . A soap bubble is different: it is a thin film with two surfaces, outer and inner, each contributing :
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: .
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.
An aquarium bubbler
An aquarium air pump pushes air down a narrow tube whose open end, of internal radius , sits 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 , , and .
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.
Let the tube have radius and let water meet the glass at a contact angle (acute). The meniscus is then concave upward, part of a sphere of radius , and the geometry of Fig. 9.17 gives . The concave side, the air above, is at the higher pressure, so just below the meniscus the water's pressure is less than atmospheric by
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 . It is a height below the meniscus, so
Combining (9.27) and (9.28),
The rise is inversely proportional to the tube's radius and comes entirely from surface tension. For water in clean glass, . When is negative, as for mercury in glass (), (9.29) gives a negative : the meniscus is convex and the liquid stands below the outside level.
2/3The meniscus is part of a sphere of radius ; just below it the pressure is below .
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.
How high, and how low?
A clean glass capillary of internal radius stands upright in a dish. Find the change in level inside it (a) for water, with , , ; (b) for mercury, with , , . Take .
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.
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Answer 6 cm
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Answer (B) falls below the outside level, with a convex surface
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Answer (A) The soap bubble has and the air bubble
Mechanical Properties of Fluids: key points and summary
- 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).
- Pressure is normal force per unit area, : a scalar measured in pascal, with ; density is (Section 9.2).
- 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).
- : pressure depends only on depth, not on the vessel's shape; is the gauge pressure (Section 9.2.2).
- A barometer balances against about 76 cm of mercury; a manometer reads gauge pressure; mm of mercury and (Section 9.2.3).
- An enclosed fluid transmits pressure undiminished; a hydraulic lift multiplies force by while the work in equals the work out (Section 9.2.4).
- In steady flow streamlines never cross, and is constant (continuity, mass conservation); beyond a critical speed the flow turns turbulent (Section 9.3).
- Bernoulli: is constant along a streamline of an ideal fluid in steady flow; it is energy conservation per unit volume (Section 9.4).
- Torricelli's law 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).
- Viscosity: is shear stress divided by the rate of shear strain, in ; it falls with temperature for liquids and rises for gases (Section 9.5).
- Stokes' law leads to the terminal velocity (Section 9.5.1).
- Surface tension can be read as a pull per unit length of surface or as extra energy per unit area; a film gives , and at a contact line (Section 9.6.2, Section 9.6.3).
- Excess pressure is in a drop or cavity and in a soap bubble; capillary rise is (Section 9.6.4, Section 9.6.5).
Practice questions with answers (NCERT “Exercises”)
Explain the following.
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?
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Blood is a fluid, so within it pressure changes with height as . Raising the arm puts the cuff above the heart, where the pressure is lower by ; for that is about , roughly 30 mm of mercury. Readings are taken at heart level so that they can be compared.
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?
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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.
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?
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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.
Explain the following in terms of surface energy and the angle of contact.
Water spreads out on clean glass but rolls into beads on a waxed car bonnet, while mercury forms beads even on clean glass.
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Water is strongly attracted to glass, so the solid–liquid interface has low energy (), and is acute: water wets glass and spreads. Wax attracts water weakly (), so 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 ) and it forms drops.
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 unchanged.
Why is detergent added to washing water, and what does it do to the angle of contact?
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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.
A blob of water released inside an orbiting space station becomes a nearly perfect sphere. Why?
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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.
Decide whether each statement is true or false.
Hot water has a lower surface tension than cold water.
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Answer True
Surface tension generally falls as a liquid warms.When heated, a gas becomes more viscous while a liquid becomes less viscous.
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Answer True
Faster gas molecules carry more momentum between layers; warm liquid molecules slip past one another more easily.A steady sideways push on a layer of fluid produces a fixed shear strain, just as it does in a solid.
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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.The speeding-up of water where a pipe narrows follows from conservation of mass alone, without Bernoulli's principle.
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Answer True
constant is the continuity equation, a statement of mass conservation. Bernoulli's principle is needed only to find the pressure there.
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, (the Reynolds number), passes a critical value. A model one-tenth the size needs about ten times the speed to reach it.Use continuity or Bernoulli's principle to explain the following.
When an express train rushes past a platform, people standing close to the edge feel pulled towards the train.
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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.
A gardener fits a tapering nozzle to a hose, and the water now reaches the far end of the garden.
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The flow rate is set by the supply. The nozzle reduces the area at the outlet, so the speed rises in proportion, and a faster jet travels farther.
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.
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.
A sprinter stands still on one foot in spiked shoes. Only the six spikes touch the track, each over a circle of diameter . What pressure do the spikes exert on the track? ()
PaShow solution
Force: . Area: .
That is about 300 atmospheres, which is why spikes grip a track so well.
How tall would a barometer filled with glycerine (density ) have to be at normal atmospheric pressure, ? Why is mercury the practical choice? ()
mShow solution
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.
A valve on an undersea oil pipeline rests on the sea floor down. Its casing is rated to withstand a pressure of . Find the pressure on it and say whether it is safe. (Sea water , , ; ignore currents.)
PaShow solution
This is about half the rated , so the casing is safe, with a safety factor of about 2.
A two-post hydraulic lift is rated for vehicles of up to . The piston under the load has a cross-section of . What is the greatest pressure the small piston, and the oil, must withstand? ()
PaShow solution
The same pressure acts on the small piston whatever its size; only the force on it is smaller.
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 tall and the kerosene column . 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:
so .
Now more water is added on the water side of the U-tube in Question 10, and 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.)
cmShow 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 topped by 25.0 cm of kerosene on the other. In centimetres of water:
The mercury stands about 1.5 mm higher in the kerosene arm.
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.
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?
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Gauge pressure is absolute pressure minus . 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.
Castor oil (density , viscosity ) flows steadily along a level tube long with an inner radius of . Oil is collected at the outlet at . For laminar flow the volume flow rate is given by Poiseuille's relation, . Find the pressure difference between the ends. Is laminar flow a fair assumption?
PaShow solution
Volume flow rate: .
The average speed is . So slow a flow of so viscous an oil is laminar: the Reynolds number , far below the value of about 2000 at which flow in a pipe turns turbulent.
In a wind-tunnel test, air moves at over the top of a model wing and at under it. The wing's area is and the air's density . Find the lift on the wing.
NShow solution
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.The tube of a garden sprayer has a cross-section of , and its nozzle has 25 holes, each in diameter. The liquid moves along the tube at per minute. With what speed does it leave the holes?
m/sShow solution
Speed in the tube: per minute . Total hole area: .
A soap film is formed in a wire frame closed at the bottom by a light sliding wire long. The film holds up the slider together with a small weight hung from it, a total of . Find the surface tension of the soap solution.
N/mShow solution
The film pulls the slider up along its length on both faces, with total force , which balances the weight:
A soap film in a rectangular frame, closed at the bottom by a light sliding wire, holds up a weight of 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 , 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 . Stretching a film creates more surface but does not change .
A drop of mercury has a radius of . Taking the surface tension of mercury as and atmospheric pressure as , find the excess pressure inside the drop and the total pressure inside it.
PaShow solution
A soap bubble has a radius of , and the soap solution's surface tension is . Atmospheric pressure is and .
What is the excess pressure inside the soap bubble, in Pa?
PaShow solution
An air bubble of the same radius forms deep in a container of the same solution, whose relative density is 1.10. What is the pressure inside it, in Pa?
PaShow solution
One surface, so the excess is . The liquid outside is at with :
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 . 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.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.
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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.Eight identical tiny raindrops each fall through still air at a terminal speed of . They merge into a single drop. Assuming Stokes' law still applies, find the new terminal speed in cm/s.
cm/sShow answer
Answer 4 cm/s
The volume is 8 times larger, so the radius doubles; , so the speed becomes 4 times larger, .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 , where is the lift's acceleration. Since in (9.29), the column gets shorter. In free fall it would fill the whole tube.How much work is needed to blow a soap bubble of radius from a very small one, if the surface tension of the solution is ? Give the answer in joules.
JShow answer
Answer 0.00068 J
.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.A thousand identical water droplets, each of radius , merge into one drop. How much surface energy is released, in joules? ()
JShow answer
Answer 0.00000826 J
. Area lost ; energy released , which slightly warms the drop.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 textbook | Practice question here |
|---|---|
| NCERT Exercises Q9.1 | Practice Q1 |
| NCERT Exercises Q9.2 | Practice 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 Q9.5 | Practice Q6 |
| NCERT Exercises Q9.6 | Practice Q7 |
| NCERT Exercises Q9.7 | Practice Q8 |
| NCERT Exercises Q9.8 | Practice Q9 |
| NCERT Exercises Q9.9 | Practice Q10 |
| NCERT Exercises Q9.10 | Practice Q11 |
| NCERT Exercises Q9.11 | Practice Q12 |
| NCERT Exercises Q9.12 | Practice Q13 |
| NCERT Exercises Q9.13 | Practice Q14 |
| NCERT Exercises Q9.14 | Practice Q15 |
| NCERT Exercises Q9.15 | Practice Q16 |
| NCERT Exercises Q9.16 | Practice Q17 |
| NCERT Exercises Q9.17 | Practice Q18 |
| NCERT Exercises Q9.18 | Practice Q19 |
| NCERT Exercises Q9.19 | Practice Q20 |
| NCERT Exercises Q9.20 | Practice Q21 |
| NCERT Exercises Q9.4 (e) | Practice Q29 |
Quick check: MCQs on Mechanical Properties of Fluids
- 9.1 Introduction2 questions
- 9.2.2 Variation of Pressure with Depth3 questions
- 9.2.3 Atmospheric Pressure and Gauge Pressure3 questions
- 9.2.4 Hydraulic Machines2 questions
- 9.3 Streamline Flow3 questions
- 9.4.1 Speed of Efflux: Torricelli's Law3 questions
- 9.4.2 Dynamic Lift2 questions
- 9.5.1 Stokes' Law3 questions
- 9.6.3 Angle of Contact3 questions
- 9.6.5 Capillary Rise3 questions
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, is constant.First seen in the chapter →
- capillary riseकेशिकत्वkeshikatva
- Rise or fall of a liquid in a narrow tube caused by surface tension: .First seen in the chapter →
- equation of continuityसांतत्य समीकरणsaantatya sameekaran
- For steady flow of an incompressible fluid, 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, ; 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 .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; is shear stress divided by strain rate, in Pa s.First seen in the chapter →