Fluid mechanics

Class 11 Physics, unit 6, book pages 88–111

Practise 80 MCQs Take the unit quiz Review 37 cards

This unit studies fluids, that is liquids and gases, both at rest and in motion. It starts with the up-thrust on an immersed body and Archimedes' principle, then the viscous drag on a moving body, Stokes' law and terminal velocity. For flowing fluids it builds two ideas from conservation laws: the equation of continuity (conservation of mass) and Bernoulli's equation (conservation of energy). These explain the atomizer, filter pump, Torricelli's theorem, venturi meter, aerofoil lift and the Magnus effect, and the unit ends with superfluidity.

Learning outcomes (12)

Justify and use Archimedes's principle of flotation.
Justify how ships are engineered to float in the sea.
Define and apply the terms: steady (streamline or laminar) flow, incompressible flow and non-viscous flow as applied to the motion of an ideal fluid.
Use equation of continuity to solve problems.
Explain that squeezing the end of a rubber pipe results in increase in flow velocity.
Justify that the continuity is a form of the principle of conservation of mass.
Justify that the pressure difference can arise from different rates of flow of a fluid [Bernoulli effect].
Explain and apply Bernoulli's equation for horizontal and vertical fluid flow.
Explain why real fluids are viscous fluids.
Describe how viscous forces in a fluid cause a retarding force on an object moving through it.
Describe super fluidity [As the state in which a liquid will experience zero viscosity. Students should know the implications of this state e.g. this allows for super fluids to creep over the walls of containers to 'empty' themselves. It also implies that if you stir a superfluid, the vortices will keep spinning indefinitely.]
Analyze the real world applications of the Bernoulli effect [For example, atomisers in perfume bottles, the swinging trajectory of a spinning cricket ball and the lift of a spinning golf ball (the magnus effect), the use of Venturi ducts in filter pumps and car engineers to adjust the flow of fluid, etc.]

Introduction

Liquids and gases are fluids. The air we breathe, the water we drink and the blood flowing in our veins are all fluids, and life would not exist without the behaviour that fluids show. Fluid mechanics lets us answer questions such as: why does rising smoke curl and twist, how does a nozzle increase the speed of water leaving a hose, and how does an aeroplane gain lift while accelerating on the runway?

Fluid mechanics: The branch of physics in which we study fluids, either static or dynamic.
Fluid dynamics: The study of fluids in motion.

In Grade IX we dealt with many situations in which fluids are at rest. In this unit we deal mainly with the flow of fluids. The study of fluids in motion is relatively complex, but the analysis is simplified by making a few assumptions: the fluid is non-viscous, it is incompressible and its motion is steady.

The analysis is simplified further by two conservation principles:

Conservation of mass gives the equation of continuity.
Conservation of energy is the basis of Bernoulli's equation.

The equation of continuity and Bernoulli's equation, with their many applications in everyday life, sport, transport and technology, are the subject of this unit.

6.1Upthrust and Archimedes' principleOften examined

Why do heavy ships float on water? How does a balloon fly in the sky? Why does a mug filled with water feel lighter under water but heavy as soon as we take it out? Why do some objects float while others do not? The answers to all these questions rest on one fact: pressure increases with depth in a fluid.

Because of this, the upward force on the bottom of an object in a fluid is greater than the downward force on its top. There is therefore a net upward force on any object in any fluid. This force is called the buoyant force or up-thrust.

Up-thrust (buoyant force): The net upward force exerted by a fluid on an object placed in it, because the pressure at the bottom of the object is greater than at the top.

What happens to the object depends on how the up-thrust compares with the object's weight:

If the up-thrust is greater than the weight, the object rises to the surface and floats.
If the up-thrust is less than the weight, the object sinks.
If the up-thrust equals the weight, the object stays suspended at that depth.

The up-thrust is always present, whether the object floats, sinks or is suspended in the fluid. This is why an air-filled balloon released under water shoots up to the surface at once.

Archimedes' principle

Archimedes' principle: When an object is totally or partially immersed in a liquid, an up-thrust acts on it equal to the weight of the liquid it displaces.

Derivation Up-thrust on a cylinder in a liquid

Aim. Find the net upward force of a liquid on a solid cylinder and show that it equals the weight of the liquid displaced.

Assumptions.

The cylinder is completely immersed in a liquid of density $ρ$ , with its axis vertical.
The cylinder has cross-sectional area $A$ and height $h$ .
The surface of the liquid is at rest and the pressure at depth $d$ is $ρ g d$ .

Figure 6.2: Up-thrust acts on a cylinder immersed in a liquid. The liquid pushes down on the top face with force

F_{1}

and up on the bottom face with the larger force

F_{2}

Let the top face be at depth $h_{1}$ and the bottom face at depth $h_{2}$ below the surface. The height of the cylinder is the difference of the two depths.
$h_{2} - h_{1} = h$
The pressure of a liquid at a depth is $ρ g \times$ depth, so the pressures on the top and bottom faces are:
$P_{1} = ρ g h_{1} and P_{2} = ρ g h_{2}$
Force is pressure times area. The liquid pushes down on the top face and up on the bottom face. (The sideways forces on the curved wall are equal and opposite at the same depth, so they cancel.)
$F_{1} = P_{1} A = ρ g h_{1} A and F_{2} = P_{2} A = ρ g h_{2} A$
The net up-thrust $F$ is the upward force minus the downward force.
$F = F_{2} - F_{1}$
Put in the two forces.
$Up-thrust = ρ g h_{2} A - ρ g h_{1} A$
Take $ρ g A$ common.
$Up-thrust = ρ g A (h_{2} - h_{1}) = ρ g A h$
The product $A h$ is the volume $V$ of the cylinder, and this equals the volume of liquid the cylinder displaces.
$Up-thrust = ρ g V (6.1)$
The mass of the displaced liquid is $ρ V$ , so $ρ g V$ is the weight of the liquid displaced.
$Up-thrust = weight of the liquid displaced$

Result

Up-thrust = ρ g V = weight of the liquid displaced

Equation (6.1) is the mathematical form of Archimedes' principle. The up-thrust depends only on the density of the liquid and the volume immersed, not on what the object is made of. Archimedes' principle is also helpful for calculating the density of an object.

Ships

Ships are engineered to float in the sea by using several fundamental principles and design elements: buoyancy, stability and structural reliability.

Buoyancy. A ship is designed to displace a volume of water whose weight equals the weight of the ship, so that it floats. This is achieved by a hull shape that maximises displacement while minimising weight. The hull is designed to withstand the water pressure, which increases with depth. The hull displaces water and creates an upward buoyant force $F_{b}$ equal to the weight of the ship $W$ .
Stability. Ships are engineered to resist overturning by distributing the weight low, using ballast tanks, and designing a stable hull shape.
Structural reliability. The ship's structure is engineered to withstand stresses, loads and vibrations, using strong, lightweight materials such as steel and aluminium.

F_{b} = ρ V g = W

a ship floats when the up-thrust equals its weight

By combining these principles and design elements, ships are engineered to float and operate safely in the sea.

Submarines

A submarine can travel over as well as under water. It floats on the surface when the weight of water equal to its volume is greater than its own weight. Under this condition it is like a ship and travels over the water. It has a system of tanks that can be filled with and emptied of seawater.

When the tanks are filled with seawater, the weight of the submarine increases. As soon as the weight becomes greater than the up-thrust, it sinks and remains under water.
To return to the surface, the tanks are emptied of seawater.

Example 6.1: up-thrust on a wooden cube

A cubic block of wood is completely dipped in water. Calculate the up-thrust of water acting on it if each side of the block is

10 cm

long.

Given

$L = 10 cm = \frac{10}{100} m = 0.1 m$ length of each side

To find

Up-thrust $= ?$

By Archimedes' principle the up-thrust is $ρ g V$ .
$Up-thrust = ρ g V$
For a cube of side $L$ the volume is $V = L^{3}$ .
$Up-thrust = ρ g L^{3}$
For water use $ρ = 1000 kg m^{- 3}$ and $g = 9.8 m s^{- 2}$ .
$Up-thrust = 1000 \times 9.8 \times (0.1)^{3}$
Since $(0.1)^{3} = 1 0^{- 3}$ , the factor $1000$ cancels it.
$Up-thrust = 9.8 N$

Up-thrust = 9.8 N

Assignment 6.1: iron object in water

An iron object with density

7.8 g c m^{- 3}

appears

200 N

lighter in water than in air. Calculate (a) the volume of the object and (b) the weight of the object in air.

Given

$ρ_{iron} = 7.8 g c m^{- 3} = 7800 kg m^{- 3}$
$loss of weight = 200 N$ equal to the up-thrust of water
$ρ_{water} = 1000 kg m^{- 3}$

To find

(a) Volume $V$
(b) Weight in air $W$

The apparent loss of weight in water is the up-thrust. The object is fully immersed, so the volume of water displaced is the volume $V$ of the object.
$ρ_{water} g V = 200 N$
(a) Solve for $V$ .
$V = \frac{200}{1000 \times 9.8} = 2.04 \times 1 0^{- 2} m^{3}$
(b) The weight in air is mass times $g$ , with mass $= ρ_{iron} V$ .
$W = ρ_{iron} V g = 7800 \times 2.04 \times 1 0^{- 2} \times 9.8$
Evaluate. (Check: $W /200 = ρ_{iron} / ρ_{water} = 7.8$ , so $W = 7.8 \times 200$ .)
$W \approx 1.56 \times 1 0^{3} N$

V = 2.04 \times 1 0^{- 2} m^{3} (20.4 litres), W = 1560 N

Worked example: how much of a floating block is under water

A uniform wooden block of volume

0.020 m^{3}

and density

600 kg m^{- 3}

floats in water. Find its weight, the up-thrust on it, the volume under water and the fraction submerged.

Given

$V = 0.020 m^{3}$
$ρ_{b} = 600 kg m^{- 3}$
$ρ_{w} = 1000 kg m^{- 3}$

To find

$W$ , up-thrust, $V_{s u b}$ , fraction submerged

Weight of the block.
$W = ρ_{b} V g = 600 \times 0.020 \times 9.8 = 117.6 N$
A floating body is in equilibrium, so the up-thrust equals the weight.
$F_{b} = W = 117.6 N$
The up-thrust is the weight of water displaced, $F_{b} = ρ_{w} V_{s u b} g$ , so find the displaced (submerged) volume.
$V_{s u b} = \frac{F _{b}}{ρ _{w} g} = \frac{117.6}{1000 \times 9.8} = 0.012 m^{3}$
Fraction under water.
$\frac{V _{s u b}}{V} = \frac{0.012}{0.020} = 0.60 = \frac{ρ _{b}}{ρ _{w}}$

Figure: a floating block. The up-thrust (accent) equals the weight, and 60% of the block is under water.

W = F_{b} = 117.6 N, V_{s u b} = 0.012 m^{3}, 60% submerged

6.2Viscous drag and terminal velocityOften examined

Viscosity

Viscosity is a measure of a fluid's resistance to flow. A more viscous fluid pours slowly, while a less viscous fluid pours easily. Imagine a Styrofoam cup with a hole in the bottom. If you pour honey into the cup, it drains very slowly. If you fill the same cup with water, it drains much more quickly. This is because the viscosity of honey is much higher than that of water. Hence viscosity determines the rate of flow of a liquid.

Viscosity: The resistance offered by different layers of a fluid to its flow.

Gases also have viscosity, although it is a little harder to notice in ordinary circumstances. In most liquids viscosity decreases as temperature increases, whereas in most gases viscosity increases as temperature increases. It is therefore important to always measure the temperature of a fluid when determining its viscosity.

The coefficient of viscosity is represented by $η$ . Its unit is $N s m^{- 2}$ or $kg m^{- 1} s^{- 1}$ or $Pa s$ . Substances that flow easily have a small coefficient of viscosity, while those that do not flow easily have a large coefficient of viscosity. For example, water and air have a low coefficient of viscosity, whereas honey and engine oil have a high one.

Table 6.1: Viscosity of some fluids at $2 0^{\circ} C$ .
Substance	Viscosity (mPa s)
Water	1.0016
Whole milk	2.12
Honey	2000–10000
Glycerin	1410
Mercury	1.55
Methanol	0.5940
Ethanol	1.144

Drag force

A drag force acts on a solid object when it moves in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. If you stretch your hand out of the window of a fast-moving car, you can easily recognise that a considerable drag force has to be exerted on your hand as it moves through the air. You feel a smaller drag force when you tilt your hand. The drag force becomes larger as the speed of the car increases.

Drag force (viscous drag): The retarding force (resistance) experienced by an object moving through a fluid.
Aerodynamic drag (air resistance): The drag force when the fluid is a gas, such as air.
Hydrodynamic drag: The drag force when the fluid is a liquid, such as water.

The drag force depends on:

the size, shape and orientation of the object,
the viscosity of the fluid,
the relative speed of the object with respect to the fluid.

Stokes' law

For small objects the drag force is given by Stokes' law. According to Stokes' law, the drag force $F_{d}$ on a spherical object is directly proportional to the radius $r$ of the object, its velocity $v$ and the coefficient of viscosity $η$ of the fluid. The mathematical form of Stokes' law is:

F_{d} = 6 π η r v

(6.2)

6.2.1Terminal velocityOften examined

Terminal velocity: The maximum constant velocity acquired by a freely falling object in a viscous medium.

Terminal velocity is attained when the weight of the object is balanced by the upward drag force. Take a raindrop. At first it accelerates because of gravity, and there is no drag force. As its velocity increases, the retarding force also increases. As the drop falls faster and faster, the drag force keeps increasing, so

Net force = weight - drag force

Finally, when the viscous drag becomes equal to the force due to gravity, the net force becomes zero and so does the acceleration. The raindrop then falls with constant velocity, the terminal velocity $v_{t}$ .

Derivation Terminal velocity of a falling sphere

Aim. Find an expression for the terminal velocity $v_{t}$ of a sphere of radius $r$ and density $ρ$ falling through a fluid of viscosity $η$ .

Assumptions.

The object is a sphere, so Stokes' law applies.
The fluid is a gas or light fluid, so its up-thrust on the falling sphere is neglected, as in the book.
$ρ$ is the (uniform) density of the sphere.

Figure 6.4: A rain droplet falling freely under gravity. At first the drag is small and the weight is larger. At terminal velocity the drag

F_{d}

equals the weight

W

At terminal velocity the acceleration is zero, so the net force is zero: the weight $w$ is balanced by the drag $F_{d}$ .
$0 = w - F_{d} ⟹ F_{d} = w$
Use Stokes' law for the drag, $F_{d} = 6 π η r v_{t}$ , and $w = m g$ for the weight.
$6 π η r v_{t} = m g$
Solve for $v_{t}$ .
$v_{t} = \frac{m g}{6 π η r} (6.3)$
For a sphere of uniform density the mass is density times volume, and the volume of a sphere is $\frac{4}{3} π r^{3}$ .
$m = ρ V = ρ (\frac{4}{3} π r^{3}) (6.4)$
Put equation (6.4) into equation (6.3).
$v_{t} = \frac{ρ ( \frac{4}{3} π r ^{3} ) g}{6 π η r}$
Cancel $π$ and one factor of $r$ , and simplify the numbers: $\frac{4}{3} \div 6 = \frac{4}{18} = \frac{2}{9}$ .
$v_{t} = \frac{2 ρ g r ^{2}}{9 η} (6.5)$

Result

v_{t} = \frac{2 ρ g r ^{2}}{9 η}

Since $ρ$ , $g$ and $η$ are constants for a given drop and fluid, $v_{t} \propto r^{2}$ : the terminal velocity depends on the square of the radius of the sphere and is inversely proportional to the viscosity of the medium. The smaller drop has the smaller terminal speed and reaches it sooner than the larger one.

Worked example: terminal velocity of droplets of two sizes

Water droplets of density

1000 kg m^{- 3}

fall through air of viscosity

1.8 \times 1 0^{- 5} kg m^{- 1} s^{- 1}

. Find the terminal velocity for radius

0.05 mm

and for radius

0.15 mm

(neglect the up-thrust of air), and explain the shape of the graph.

Terminal velocity against radius

0.05

0.1

0.15

0.2

0.5

1.5

2.5

r / mm

v_{t} / m s^{- 1}

(0.05, 0.30)

(0.15, 2.72)

A parabola: tripling the radius makes the terminal velocity nine times larger.

Use $v_{t} = \frac{2 ρ g r ^{2}}{9 η}$ with $r = 0.05 mm = 5 \times 1 0^{- 5} m$ .
$v_{t} = \frac{2 \times 1000 \times 9.8 \times ( 5 \times 1 0 ^{- 5} ) ^{2}}{9 \times 1.8 \times 1 0 ^{- 5}} = \frac{4.9 \times 1 0 ^{- 5}}{1.62 \times 1 0 ^{- 4}} = 0.30 m s^{- 1}$
The second radius is $3$ times the first. Since $v_{t} \propto r^{2}$ , the speed is $3^{2} = 9$ times larger.
$v_{t} = 9 \times 0.3025 = 2.7 m s^{- 1}$
Why a parabola: the weight grows as $r^{3}$ but the drag grows as $r$ times $v$ , so the balance gives $v_{t} \propto r^{2}$ .
Common mistake: doubling the radius doubles the speed. It does not; it makes it four times as large.

v_{t} = 0.30 m s^{- 1} (r = 0.05 mm), 2.7 m s^{- 1} (r = 0.15 mm)

6.2.2Paratrooper's jump

A paratrooper who jumps out of a plane without opening the parachute first falls with a large acceleration and attains a high terminal velocity (Fig. 6.5 a). To land safely he opens his parachute (Fig. 6.5 b). The parachute provides a large surface area, which produces a large drag force (greater than his weight) opposite to the direction of motion. This slows the skydiver down: he decelerates. After some time the drag force F_d and the weight of the paratrooper become equal again, and he falls at a small steady speed to the ground.

Consider the typical graph of Fig. 6.5 (c):

At t=0\,\mathrm{s} the paratrooper jumps out of the plane. He accelerates and reaches a steady terminal speed of about 50\,\mathrm{m\,s^{-1}} when the parachute is not opened.
After 42\,\mathrm{s} he opens the parachute and a large deceleration develops.
Within the next 5\,\mathrm{s} he falls at a very moderate steady speed, the new terminal speed of 8\,\mathrm{m\,s^{-1}}.
He lands safely on the ground at time 70\,\mathrm{s}.

Velocity against time for a paratrooper

Figure 6.5 (c): A typical graph of velocity against time for a paratrooper (drawn as a sketch of the book's graph). The two flat parts are the two terminal velocities.

Worked example on Stokes' law

Example 6.2: drag on a small sphere in air

A spherical body of radius 2\,\mathrm{mm} is passing through air with velocity 2\,\mathrm{m\,s^{-1}}. Find the drag force on the body due to the air. The viscosity of air is 1.9\times10^{-5}\,\mathrm{kg\,m^{-1}\,s^{-1}}.

Given

r = 2\,\mathrm{mm}=2\times10^{-3}\,\mathrm{m}
v = 2\,\mathrm{m\,s^{-1}}
\eta = 1.9\times10^{-5}\,\mathrm{kg\,m^{-1}\,s^{-1}}

To find

Drag force F_d=\,?

The body is a sphere, so use Stokes' law.
F_d=6\pi\eta r v
Put in the values, with the radius in metres.
F_d=6\times3.14\times1.9\times10^{-5}\times2\times10^{-3}\times2
Multiply the numbers: 6\times3.14\times1.9\times2\times2=143.2, and the powers of ten give 10^{-5}\times10^{-3}=10^{-8}.
F_d=143.2\times10^{-8}\,\mathrm{N}

F_d=1.43\times10^{-6}\,\mathrm{N}

Assignment 6.2: terminal velocity of a raindrop

Find the terminal velocity obtained by a raindrop of radius 0.3\,\mathrm{mm} falling through air of viscosity 1.8\times10^{-5}\,\mathrm{kg\,m^{-1}\,s^{-1}}.

Given

r = 0.3\,\mathrm{mm}=3\times10^{-4}\,\mathrm{m}
\eta = 1.8\times10^{-5}\,\mathrm{kg\,m^{-1}\,s^{-1}}
\rho = 1000\,\mathrm{kg\,m^{-3}} density of water (the drop)
g = 9.8\,\mathrm{m\,s^{-2}}

To find

Terminal velocity v_t=\,?

Use the terminal velocity of a sphere, equation (6.5).
v_t=\dfrac{2\rho g r^{2}}{9\eta}
Put in the values.
v_t=\dfrac{2\times1000\times9.8\times(3\times10^{-4})^{2}}{9\times1.8\times10^{-5}}
Numerator: 2\times1000\times9.8\times9\times10^{-8}=1.764\times10^{-3}. Denominator: 9\times1.8\times10^{-5}=1.62\times10^{-4}.
v_t=\dfrac{1.764\times10^{-3}}{1.62\times10^{-4}}\approx10.9\,\mathrm{m\,s^{-1}}

v_t\approx10.9\,\mathrm{m\,s^{-1}}

6.3Fluid flow

A fluid is a substance that can flow, such as a liquid or a gas. The flow rate of a fluid is the volume of fluid passing a given point in a pipe per unit time. A fluid can flow in two ways: streamline flow or turbulent flow.

Streamline flow: If every particle passing a particular point moves along exactly the same smooth path followed by previous particles that have passed that point earlier, the flow is called streamline flow.

Streamline flow is also known as steady or laminar flow. It is a smooth flow of a fluid through a tube (Fig. 6.6 a). By smooth flow we mean that all particles of the fluid follow the same uniform path, called a streamline.

The streamlines (paths of different particles) do not cross each other.
Every fluid particle arriving at a given point has the same velocity.
It usually occurs at lower velocities and with streamlined objects.
In laminar flow the middle layer of the fluid tends to flow faster.

Figure 6.6: (a) Streamline flow: the layers slide smoothly past each other along parallel paths. (b) Turbulent flow: the flow is irregular, with small whirlpool-like regions.

Above a certain critical speed, streamline flow becomes turbulent. Turbulent flow is irregular flow characterised by small whirlpool-like regions (Fig. 6.6 b). It is also known as non-laminar flow. It is the unpredictable flow of a fluid resulting from excessive speed of the flow, or from sudden changes in the direction or size of the tube or pipe. Turbulent flow occurs at higher velocities or with non-streamlined objects, where the flow lines become disordered and mixed.

Turbulent flow: Irregular flow of a fluid.

Streamline (laminar, steady) flow	Turbulent flow
Particles follow smooth paths; streamlines do not cross	Irregular flow with small whirlpool-like regions
Every particle at a given point has the same velocity	Flow is unpredictable and mixed
Occurs at lower velocities and with streamlined objects	Occurs above the critical speed, at higher velocities, with non-streamlined objects or sudden changes of the tube

Ideal fluid

The motion of real fluids is very complex and not fully understood, so we make some simplifying assumptions. In the model of ideal fluid flow we assume that:

The fluid is non-viscous. In a non-viscous fluid, internal friction is neglected.
The flow is laminar. In laminar flow all particles passing through a point have the same velocity.
The fluid is incompressible. The density of an incompressible fluid is constant. Liquids are generally incompressible, while gases are compressible.
The flow is irrotational.

In real fluids there is always some viscosity, which means that real fluids are viscous. Viscosity is the thickness of the fluid, or the friction that the fluid faces while it flows. Ideal fluids have zero viscosity, so they do not experience an opposing force and have a non-viscous flow, while real fluids have a viscous flow.

6.4Equation of continuityOften examined

The equation of continuity is an important concept of fluid dynamics. It is used in pipes, tubes and ducts with flowing fluids or gases, rivers, power plants and so on.

Derivation Equation of continuity

Aim. Show that for the steady flow of an ideal fluid through a tube, the product of the cross-sectional area and the fluid speed is the same at every point.

Assumptions.

The flow is steady and the fluid is ideal.
The tube has a single entry and a single exit, and no fluid is added or removed in between (no source or sink).
A_1, v_1, \rho_1 are at the lower end and A_2, v_2, \rho_2 at the upper end.

Figure 6.7: Fluid flowing through a tube of varying cross-sectional area. In the same short time \Delta t the fluid moves \Delta x_1=v_1\Delta t at the narrow end and \Delta x_2=v_2\Delta t at the wide end.

In a short time \Delta t the fluid at the lower end covers a distance \Delta x_1 with velocity v_1.
\Delta x_1=v_1\,\Delta t
The volume of fluid that flows into the pipe at the lower end is the area times this distance.
V=A_1\,\Delta x_1=A_1v_1\,\Delta t
Mass is density times volume, so the mass of fluid in the region \Delta x_1 is:
\Delta m_1=\rho V=\rho_1A_1v_1\,\Delta t\qquad(6.6)
In the same way, in the same time \Delta t the mass of fluid passing the upper end, through \Delta x_2 with speed v_2 and area A_2, is:
\Delta m_2=\rho_2A_2v_2\,\Delta t\qquad(6.7)
For an ideal fluid the total mass is conserved: the mass that enters equals the mass that leaves. Equate (6.6) and (6.7).
\Delta m_1=\Delta m_2\quad\Longrightarrow\quad\rho_1A_1v_1=\rho_2A_2v_2\qquad(6.8)
If the fluid is incompressible, its density stays constant in steady flow, so \rho_1=\rho_2 cancels from both sides.
A_1v_1=A_2v_2\qquad(6.9)
Since the two points are any two points on the pipe, the product is the same everywhere.
Av=\text{constant}\qquad(6.10)

ResultA_1v_1=A_2v_2\quad\text{or}\quad Av=\text{constant}

For an ideal fluid, the product of the cross-sectional area of the pipe and the fluid speed at any point along the pipe is always constant (if there is no source or sink). This product is the volume flow per second, or simply the flow rate.

From the equation of continuity we can show that

v\propto\dfrac{1}{A}

The smaller the cross-sectional area, the greater the velocity of the fluid, and vice versa.

Applications of the equation of continuity

Water falling freely. When water falls freely its speed increases, so its cross-sectional area decreases (Fig. 6.8 a), in accordance with v\propto1/A.
Squeezing the end of a rubber pipe. By squeezing the end of a rubber pipe (Fig. 6.8 b) its cross-sectional area A decreases. As a result the velocity of the water increases, as shown by v\propto1/A. Reducing the area of the opening causes the velocity to increase.

Example 6.3: blood flow from the aorta

In a normal adult, the average speed of the blood through the aorta (radius r=0.8\,\mathrm{cm}) is 0.33\,\mathrm{m\,s^{-1}}. From the aorta the blood goes into major arteries, which are 30 in number, each of radius 0.4\,\mathrm{cm}. Calculate the speed of the blood through the arteries.

Given

r_1 = 0.8\,\mathrm{cm}=0.8\times10^{-2}\,\mathrm{m} radius of aorta
v_1 = 0.33\,\mathrm{m\,s^{-1}} velocity of blood in the aorta
r_2 = 0.4\,\mathrm{cm}=0.4\times10^{-2}\,\mathrm{m} radius of each artery

To find

Velocity of blood in the arteries v_2=\,?

The blood from the aorta is shared among 30 equal arteries, each with cross-sectional area A_2. So the equation of continuity has a factor 30 on the right.
A_1v_1=30\,A_2v_2
The vessels are circular, so A=\pi r^{2}.
\pi r_1^{2}v_1=30\,\pi r_2^{2}v_2
Cancel \pi.
r_1^{2}v_1=30\,r_2^{2}v_2
Solve for v_2.
v_2=\dfrac{1}{30}\left(\dfrac{r_1}{r_2}\right)^{2}v_1
Put in the values. The ratio r_1/r_2 has no units, so centimetres may be used.
v_2=\dfrac{1}{30}\left(\dfrac{0.8\,\mathrm{cm}}{0.4\,\mathrm{cm}}\right)^{2}\times0.33=\dfrac{4\times0.33}{30}
Evaluate.
v_2=0.044\,\mathrm{m\,s^{-1}}

The blood flows through the arteries at a speed much lower than in the aorta, because the total combined cross-sectional area of the 30 arteries is greater than that of the aorta. The slower speed of the blood in the arteries is favourable for the exchange of gases.

v_2=0.044\,\mathrm{m\,s^{-1}}

Assignment 6.3: deep and shallow water in rivers

Why does deep water flow slowly compared with shallow water in rivers?

The volume of water passing every cross-section of the river per second is the same (equation of continuity, Av=\text{constant}).
A_1v_1=A_2v_2
Where the river is deep, the cross-sectional area A of the flowing water is large. Since v\propto1/A, the speed is small there.
Where the river is shallow, the cross-sectional area is small, so the water has to flow faster to carry the same volume per second.

Worked example: why a tap stream narrows as it falls

Water leaves a tap with speed 1.0\,\mathrm{m\,s^{-1}} in a stream of radius 0.50\,\mathrm{cm}. Find the speed and the radius of the stream after it has fallen 0.20\,\mathrm{m}.

Water in free fall gains speed: v^2=v_0^2+2gh.
v^2=1.0^2+2(9.8)(0.20)=4.92\ \Rightarrow\ v=2.2\,\mathrm{m\,s^{-1}}
The same volume per second passes every level, so by the equation of continuity A_1v_1=A_2v_2, i.e. \pi r_1^2v_1=\pi r_2^2v_2.
r_2=r_1\sqrt{\dfrac{v_1}{v_2}}=0.50\sqrt{\dfrac{1.0}{2.218}}=0.34\,\mathrm{cm}
The faster water must pass through a smaller cross-section, so the stream gets thinner.

Figure: the stream from a tap narrows because the water speeds up as it falls (the flow rate stays the same).

v=2.2\,\mathrm{m\,s^{-1}},\quad r=0.34\,\mathrm{cm}

6.5Bernoulli's equationOften examined

Bernoulli's equation is an approximate relation between pressure, velocity and elevation for the flow of an ideal fluid. We derive it by applying the principle of conservation of energy to flowing fluids.

Derivation Bernoulli's equation

Aim. Derive the relation between pressure, speed and height at two points of the steady flow of an ideal fluid, using conservation of energy.

Assumptions.

The fluid is ideal: non-viscous, incompressible, with steady (laminar) flow.
The speed at the lower point is v_1 and at the upper point v_2; the heights are h_1 and h_2 and the pressures P_1 and P_2.
The work done on the fluid by the pressure forces changes its kinetic and potential energy.

Figure 6.9: Fluid flowing through a tube of varying cross-sectional area. The fluid is pushed in at the lower end by the force F_1 and resisted at the upper end by F_2; h_1 and h_2 are the heights of the two ends.

In a short time \Delta t, the fluid at the lower end covers the distance \Delta x_1 and at the upper end \Delta x_2.
\Delta x_1=v_1\,\Delta t\qquad(6.11)\qquad\Delta x_2=v_2\,\Delta t\qquad(6.12)
At the lower end the force F_1 moves the fluid through \Delta x_1. The work done on the fluid is force times distance.
W_1=F_1\,\Delta x_1\qquad(6.13)
Force is pressure times area, F=PA.
W_1=P_1A_1\,\Delta x_1\qquad(6.14)
At the upper end the force F_2 is opposite to the motion, so the work done on the fluid there is negative.
W_2=-P_2A_2\,\Delta x_2\qquad(6.15)
The net work done W is the sum of (6.14) and (6.15).
W=W_1+W_2=P_1A_1\,\Delta x_1-P_2A_2\,\Delta x_2\qquad(6.16)
Put in \Delta x_1=v_1\Delta t and \Delta x_2=v_2\Delta t.
W=P_1A_1v_1\,\Delta t-P_2A_2v_2\,\Delta t\qquad(6.17)
By the equation of continuity, A_1v_1\Delta t=A_2v_2\Delta t=V, the volume of fluid that flows in time \Delta t. Replace both products by V.
W=P_1V-P_2V=(P_1-P_2)V\qquad(6.18)
The volume is mass over density, V=m/\rho.
W=(P_1-P_2)\dfrac{m}{\rho}\qquad(6.19)
By conservation of energy, this work changes the kinetic energy and the potential energy of the fluid.
W=\Delta\text{K.E.}+\Delta\text{P.E.}
Write the two changes out, with h_1 and h_2 the heights of the lower and upper ends.
(P_1-P_2)\dfrac{m}{\rho}=\left(\tfrac12mv_2^{2}-\tfrac12mv_1^{2}\right)+(mgh_2-mgh_1)
Cancel the mass m from both sides.
(P_1-P_2)\dfrac{1}{\rho}=\left(\tfrac12v_2^{2}-\tfrac12v_1^{2}\right)+(gh_2-gh_1)
Multiply by \rho, then collect all the point-1 terms on the left and all the point-2 terms on the right.
P_1+\tfrac12\rho v_1^{2}+\rho gh_1=P_2+\tfrac12\rho v_2^{2}+\rho gh_2\qquad(6.20)
The subscripts only mark two different points along the pipe, so the general equation is:
P+\tfrac12\rho v^{2}+\rho gh=\text{constant}\qquad(6.21)

ResultP+\tfrac12\rho v^{2}+\rho gh=\text{constant}

Bernoulli's equation states that the sum of the pressure, the kinetic energy per unit volume and the potential energy per unit volume remains constant along a streamline of an ideal fluid. It is an important tool in investigating technical systems such as power generation stations, plumbing systems and the flight of an aeroplane.

The three terms of Bernoulli's equation; each has the unit \mathrm{J\,m^{-3}}=\mathrm{Pa}.
Term	Meaning	Name
P	pressure	pressure energy per unit volume
\tfrac12\rho v^{2}	\dfrac{\tfrac12mv^{2}}{V}	kinetic energy per unit volume
\rho gh	\dfrac{mgh}{V}	potential energy per unit volume

Worked example: pressure in a narrowing pipe, level and raised

Water (1000\,\mathrm{kg\,m^{-3}}) flows at 2.0\,\mathrm{m\,s^{-1}} with pressure 150\,\mathrm{kPa} in a pipe of cross-section 40\,\mathrm{cm^2}. It then enters a section of 10\,\mathrm{cm^2}. Find the pressure there (a) if the pipe stays level and (b) if this section is 1.5\,\mathrm{m} higher.

Speed in the narrow section from continuity.
v_2=\dfrac{A_1v_1}{A_2}=\dfrac{40\times2.0}{10}=8.0\,\mathrm{m\,s^{-1}}
(a) Level pipe, so the \rho gh terms cancel in P_1+\tfrac12\rho v_1^2=P_2+\tfrac12\rho v_2^2.
P_2=150000+\tfrac12(1000)(2.0^2-8.0^2)=150000-30000=120\,\mathrm{kPa}
(b) Now h_2-h_1=1.5\,\mathrm{m}, so subtract also \rho g\Delta h.
P_2=150000-30000-1000\times9.8\times1.5=105\,300\,\mathrm{Pa}\approx105\,\mathrm{kPa}
The pressure falls in the narrow part because the water speeds up, and falls further when it is also lifted.

Figure: along a level pipe the sum P+\tfrac12\rho v^2 stays the same, so pressure falls by exactly what the kinetic energy term gains.

P_2=120\,\mathrm{kPa}\ \text{(level)},\quad P_2\approx105\,\mathrm{kPa}\ \text{(raised)}

6.6Applications of Bernoulli's equation

According to Bernoulli's equation, slow air flow creates a high pressure while fast air flow creates a low pressure. This effect is used in many fields of life: in the filter pump, the venturi meter, atomizers, the flow of air over an aerofoil and in blood flow. The following sections discuss these applications.

6.6.1Atomizer

An atomizer is a device used to emit liquid droplets as a fine spray. It works on Bernoulli's principle.

Figure 6.10: A simple atomizer. Squeezing the bulb sends fast air across the top of the vertical tube; the lower pressure there pushes liquid up the tube and the air stream breaks it into drops.

How an atomizer works

When you squeeze the rubber bulb, high-speed air passes horizontally over the top of a vertical tube.
By Bernoulli's principle, fast air means low pressure, so the pressure there is lower than the pressure inside the container.
This pressure difference pushes the liquid from the reservoir up through the vertical narrow tube and into the moving stream of air.
The nozzle at the end of the horizontal tube makes the liquid break up into small drops, mixing it with the air, and it is carried away with the stream of air.

6.6.2Filter pump

A filter pump is designed on the basis of Bernoulli's effect. It has a jet in the middle (Fig. 6.11).

Figure 6.11: Filter pump. The fast jet of water lowers the pressure near it, so air is drawn in from the side tube and carried away with the water.

When water from the tap reaches the jet, its speed increases, so there is a pressure drop near it.
According to Bernoulli's principle, the pressure of the moving fluid decreases as its speed increases.
Air therefore flows in from the side tube to which the vessel is connected.
The air and the water together are forced to the bottom of the filter pump.

6.6.3Torricelli's theoremOften examined

Torricelli's theorem: The speed of flow of fluid from an orifice at depth h below the top surface of a liquid is equal to the speed gained by the fluid in free falling through the height h.

Derivation Speed of efflux from an orifice

Aim. Find the speed of a liquid leaving a small orifice in a tank at depth h below the top surface.

Assumptions.

The tank is filled with an ideal liquid and has an orifice (slit) near the bottom.
The top surface is at height h_1 and the orifice at height h_2 above the base.
The tank is wide, so the speed v_1 of the top surface is negligible compared with the speed of efflux v_2.

Figure 6.12: A tank filled with water with an orifice near the bottom. The top surface of the water comes down with almost zero speed v_1, while the water leaves the orifice with speed v_2. The depth of the orifice below the surface is h=h_1-h_2.

Apply Bernoulli's equation between the top surface (point 1) and the orifice (point 2).
P_1+\tfrac12\rho v_1^{2}+\rho gh_1=P_2+\tfrac12\rho v_2^{2}+\rho gh_2\qquad(6.22)
The speed v_1 of the top surface is negligible compared with v_2, so drop the second term on the left-hand side.
P_1+\rho gh_1+0=P_2+\tfrac12\rho v_2^{2}+\rho gh_2\qquad(6.23)
Write v_2=v and bring v to one side.
\tfrac12\rho v^{2}=P_1-P_2+\rho gh_1-\rho gh_2
Both the top surface and the orifice are open to the atmosphere, so P_1 and P_2 are both atmospheric pressure and P_1-P_2=0.
\tfrac12\rho v^{2}=\rho g(h_1-h_2)
Cancel \rho and solve for v.
\tfrac12v^{2}=g(h_1-h_2)\quad\Longrightarrow\quad v=\sqrt{2g(h_1-h_2)}
The depth of the orifice below the top surface is h=h_1-h_2.
v=\sqrt{2gh}\qquad(6.24)

Resultv=\sqrt{2gh}

The speed of the water leaving the orifice depends only on its depth h below the surface, not on the density of the liquid.

Equation (6.24) is also true for the velocity gained by an object falling from a height h. Let an object fall freely from height h, with initial velocity v_i=0 and final velocity v_f=v after falling the height h. Using the third equation of motion:

Speed gained in free fall

Write the third equation of motion.
2gh=v_f^{2}-v_i^{2}
Put in v_i=0 and v_f=v.
2gh=v^{2}-0^{2}
Take the square root.
v=\sqrt{2gh}

This is the same result as equation (6.24): the speed of flow of fluid from an orifice at depth h below the top surface of a liquid is equal to the speed gained by the fluid in free fall from the height h.

Worked example: efflux speed at different depths

Water leaks from small holes in the side of an open tank, one 0.50\,\mathrm{m} and one 2.0\,\mathrm{m} below the water surface. Find the speed from each hole, the flow rate through the lower hole if its area is 2.0\,\mathrm{cm^2}, and say what happens as the tank drains.

Efflux speed against depth

v = √(2gh) rises quickly at first and then more slowly.

Torricelli's theorem: v=\sqrt{2gh}.
v_1=\sqrt{2\times9.8\times0.50}=3.1\,\mathrm{m\,s^{-1}}
The deeper hole.
v_2=\sqrt{2\times9.8\times2.0}=6.3\,\mathrm{m\,s^{-1}}
The depth is 4 times as great, so the speed is only \sqrt4=2 times as great (see the graph).
Flow rate =Av with A=2.0\times10^{-4}\,\mathrm{m^2}.
Q=2.0\times10^{-4}\times6.26=1.3\times10^{-3}\,\mathrm{m^3\,s^{-1}}=1.3\,\mathrm{L\,s^{-1}}
The speed does not depend on the density of the liquid or on the size of the hole. As the tank drains, h falls and the speed decreases.

v_1=3.1\,\mathrm{m\,s^{-1}},\ v_2=6.3\,\mathrm{m\,s^{-1}},\ Q=1.3\,\mathrm{L\,s^{-1}}

6.6.4Venturi meterOften examined

A venturi meter (or flow meter) is used to measure the flow speed of a liquid through a tube. The liquid, which has high pressure and low velocity, is converted to low pressure and high velocity at a particular point, and again reaches high pressure and low velocity. The point where the fluid has low pressure and high velocity is the place where the venturi flow meter is used.

The venturi meter is constructed as in Fig. 6.13. It is designed to ensure streamline flow of the liquid through a pipe section of area A_1 with flow velocity v_1 and pressure P_1. In the narrow section (the throat) of area A_2 the fluid flows with speed v_2 and has pressure P_2. As a result, the pressure difference between points 1 and 2 appears as a height difference h.

Derivation Flow speed from a venturi meter

Aim. Find an expression for the speed v_1 in the wide section in terms of A_1, A_2 and the height difference h.

Assumptions.

The flow is streamline and the liquid is ideal.
The pipe is horizontal, so h_1=h_2 at points 1 and 2.
P_1-P_2=\rho gh, where h is the height difference of the liquid in the two vertical tubes.

Figure 6.13: Venturi meter. The pressure difference between the wide part (point 1) and the throat (point 2) appears as the height difference h of the liquid in the two vertical tubes.

Apply Bernoulli's equation to points 1 and 2.
P_1+\tfrac12\rho v_1^{2}+\rho gh_1=P_2+\tfrac12\rho v_2^{2}+\rho gh_2
The heights are equal, h_1=h_2, so the \rho gh terms cancel from both sides.
P_1+\tfrac12\rho v_1^{2}=P_2+\tfrac12\rho v_2^{2}
Rearrange for the pressure difference.
P_1-P_2=\tfrac12\rho\left(v_2^{2}-v_1^{2}\right)\qquad(6.25)
From the equation of continuity A_1v_1=A_2v_2, so v_2 in terms of v_1 is:
v_2=\dfrac{A_1}{A_2}\,v_1\qquad(6.26)
Put (6.26) in (6.25).
P_1-P_2=\tfrac12\rho\left[\left(\dfrac{A_1}{A_2}v_1\right)^{2}-v_1^{2}\right]
Take v_1^{2} common.
P_1-P_2=\tfrac12\rho v_1^{2}\left(\dfrac{A_1^{2}}{A_2^{2}}-1\right)\qquad(6.27)
The pressure difference is also measured by the height difference of the liquid in the two tubes.
P_1-P_2=\rho gh\qquad(6.28)
Equate (6.27) and (6.28).
\rho gh=\tfrac12\rho v_1^{2}\left(\dfrac{A_1^{2}}{A_2^{2}}-1\right)
Cancel \rho and multiply by 2.
2gh=v_1^{2}\left(\dfrac{A_1^{2}}{A_2^{2}}-1\right)
Solve for v_1.
v_1=\sqrt{\dfrac{2gh}{\dfrac{A_1^{2}}{A_2^{2}}-1}}\qquad(6.29)

Resultv_1=\sqrt{\dfrac{2gh}{\dfrac{A_1^{2}}{A_2^{2}}-1}}

If A_1, A_2 and h are known, equation (6.29) gives v_1, and the rate of flow can then be calculated using the equation of continuity, A_1v_1.

6.6.5Aerofoil

An aerofoil is the cross-sectional shape of an object that creates a lift when moved through a fluid such as air. Aerofoils are employed on aircraft as wings to produce lift perpendicular to the air flow (Fig. 6.14).

Figure 6.14: Cross-section of an aeroplane's wing. The air over the top travels faster (low pressure) than the air below (high pressure), giving a net upward force, the lift.

How an aerofoil produces lift

When a wing shaped like an aerofoil moves in air, the air flowing over the top travels faster, creating a region of low pressure.
The flow of air below the wing is slower, resulting in a region of higher pressure.
The pressure difference between the upper side and the lower side of the wing causes a net upward force, called lift, which helps the plane to take off.

6.6.6The Magnus effect

The Magnus effect is a phenomenon related to a spinning object moving through a fluid. In the Magnus effect, a spinning ball curves away from its path of flight (Fig. 6.15). The path of the spinning object is deflected in a manner that is not present when the object is not spinning. The Magnus effect depends on the rotational speed of the object. The deflection is caused by the pressure difference of the fluid on opposite sides of the spinning object, due to the varying fluid velocities.

Figure 6.15: Magnus effect. The spinning ball drags air with it, so the air moves faster over one side (lower pressure) and slower on the other (higher pressure); the ball is pushed towards the low-pressure side.

In the case of a ball spinning through the air, the spinning ball drags some of the surrounding air with it. A spinning object moving through a fluid deviates from its straight path because of pressure differences that develop in the fluid as a result of velocity changes induced by the spinning body.

Worked examples on Bernoulli's equation

Example 6.4: pressure in a pipe

Water is flowing in streamline flow through a closed pipe system. The speed of water at one point is 4\,\mathrm{m\,s^{-1}} and the pressure is 47.1\,\mathrm{kPa}, while at another point 3\,\mathrm{m} lower, the speed is 3\,\mathrm{m\,s^{-1}}. Find the pressure at the lower point.

Given

v_1 = 4\,\mathrm{m\,s^{-1}} speed at the higher point
P_1 = 47.1\,\mathrm{kPa}=47.1\times10^{3}\,\mathrm{Pa} pressure at the higher point
h_1 = 3.0\,\mathrm{m} height of the higher point
v_2 = 3\,\mathrm{m\,s^{-1}} speed at the lower point
h_2 = 0\,\mathrm{m} height of the lower point
\rho = 1000\,\mathrm{kg\,m^{-3}} density of water

To find

Pressure at the lower point P_2=\,?

Apply Bernoulli's equation between the two points.
P_1+\tfrac12\rho v_1^{2}+\rho gh_1=P_2+\tfrac12\rho v_2^{2}+\rho gh_2
Rearrange for P_2.
P_2=P_1+\tfrac12\rho\left(v_1^{2}-v_2^{2}\right)+\rho g\,(h_1-h_2)
Put in the values.
P_2=(47.1\times10^{3})+\tfrac12(1000)\left[(4)^{2}-(3)^{2}\right]+(1000)(9.8)(3-0)
Evaluate each term: the kinetic term is \tfrac12(1000)(7)=3500\,\mathrm{Pa} and the height term is 1000\times9.8\times3=29400\,\mathrm{Pa}.
P_2=47.1\times10^{3}+3.5\times10^{3}+29.4\times10^{3}
Add.
P_2=80\,000\,\mathrm{Pa}

P_2=80\,000\,\mathrm{Pa}=80\,\mathrm{kPa}

Assignment 6.4: lift on an aeroplane wing

The speed of air across the top and bottom of an aeroplane's wing is 450\,\mathrm{m\,s^{-1}} and 410\,\mathrm{m\,s^{-1}} respectively. Calculate the lift on the wing if the wing is 15\,\mathrm{m} long and 3\,\mathrm{m} wide (\rho=1.29\,\mathrm{kg\,m^{-3}} for air).

Given

v_{\text{top}} = 450\,\mathrm{m\,s^{-1}}
v_{\text{bottom}} = 410\,\mathrm{m\,s^{-1}}
A = 15\times3=45\,\mathrm{m^{2}} area of the wing
\rho = 1.29\,\mathrm{kg\,m^{-3}}

To find

Lift F=\,?

The wing is thin, so the heights of its top and bottom are almost the same and the \rho gh terms cancel in Bernoulli's equation. The pressure difference is then given by the speeds only.
P_{\text{bottom}}-P_{\text{top}}=\tfrac12\rho\left(v_{\text{top}}^{2}-v_{\text{bottom}}^{2}\right)
Put in the values.
\Delta P=\tfrac12(1.29)\left(450^{2}-410^{2}\right)=0.645\times(202\,500-168\,100)
Subtract and multiply.
\Delta P=0.645\times34\,400=2.22\times10^{4}\,\mathrm{Pa}
Lift is this pressure difference acting over the area of the wing, F=\Delta P\times A.
F=2.22\times10^{4}\times45\approx1.0\times10^{6}\,\mathrm{N}

F\approx1.0\times10^{6}\,\mathrm{N}

6.7Super fluidity

Super fluidity: A state in which a liquid experiences zero viscosity, hence flows without friction through any surface.

A superfluid therefore flows without any loss of kinetic energy. When it is stirred, a superfluid forms vortices that continue to spin indefinitely. It can creep over the walls of a container to 'empty' itself (Fig. 6.16).

Superfluidity is observed in liquid helium at temperatures near absolute zero, as well as in electrons within a superconducting solid. The neutron fluid in a neutron star may also be a superfluid. The unusual behaviour of a superfluid arises from quantum mechanical effects.

How a superfluid state is made

To create superfluid states, helium gas is cooled to a few degrees above absolute zero, as shown by the graph in Fig. 6.17. This is achieved by compressing the gas and then expelling it through a small nozzle. As the gas expands, it rapidly cools.

Figure 6.17: Pressure-temperature graph for helium. Below about 2.17\,\mathrm{K} (and below roughly 2.5\,\mathrm{MPa} to 3\,\mathrm{MPa}) liquid helium is in the superfluid state.

Superfluid in a rotating container

If a superfluid is placed in a rotating container, then instead of rotating uniformly with the container, the rotating state consists of quantised vortices.

When the container is rotated below a certain velocity (the critical angular velocity), the liquid remains perfectly stationary.
Once the critical angular velocity is reached, the superfluid forms a vortex.
The vortex strength of a superfluid is quantised: a superfluid can spin only at certain 'allowed' angular velocities. Rotation in a normal fluid (like water) is not quantised.
If the rotation speed is increased, more and more quantised vortices are formed, which arrange themselves in wonderful patterns.

Fluid mechanics and medical technology

Medical technology uses the heartbeat and the sound of blood flow to find the blood pressure. A fluid exerts pressure on the walls of its container. Blood, as a fluid, also exerts pressure on the walls of the heart, arteries, vessels and capillaries that make up the circulatory system. The sphygmomanometer is a well-known medical instrument used for the measurement of the blood pressure of a person.

Measuring blood pressure

The sphygmomanometer cuff (air bag) is placed around the upper arm of a person, inflated, and then deflated, while the meter measures the pressure of the blood passing through the arteries of the arm. The external pressure is increased by pressing the air bulb repeatedly. The sound of the blood flow is heard through a stethoscope.
The gradual increase in pressure squeezes the arm and compresses the blood vessels under the air bag. When the pressure in the air bag becomes greater than the systolic pressure (120\,\mathrm{torr}), the vessels collapse and cut off the flow of blood. At this instant the stethoscope detects no sound.
The pressure in the bag is then decreased by slowly opening the release valve of the air bulb. When the external pressure becomes equal to the systolic pressure, the blood flows (turbulent) with high speed through the narrow contracted vessel. The stethoscope detects sound at this instant, and the dial of the meter gives the systolic pressure.
On further opening the release valve, the pressure in the air bag drops enough that the vessels no longer remain compressed and the blood is now in laminar flow. At this instant no gurgling sounds are heard by the stethoscope, and this is the signal to record the diastolic pressure.

For a normal person the blood pressure varies between the systolic (120\,\mathrm{torr}) and the diastolic (75 to 80\,\mathrm{torr}) values. These values change with age.

Formula sheet

Up-thrust (Archimedes' principle)

\text{Up-thrust}=\rho g V

\rho: density of the fluid; g: acceleration due to gravity; V: volume of fluid displaced (equal to the immersed volume of the body)

Unit: \mathrm{N}

Floating ship

F_b=\rho V g=W

F_b: buoyant force; W: weight of the ship

Unit: \mathrm{N}

Stokes' law

F_d=6\pi\eta r v

\eta: coefficient of viscosity; r: radius of the sphere; v: velocity of the sphere

Unit: \mathrm{N}

Terminal velocity (from weight and drag)

v_t=\dfrac{mg}{6\pi\eta r}

m: mass of the sphere; g: acceleration due to gravity

Unit: \mathrm{m\,s^{-1}}

Terminal velocity of a sphere

v_t=\dfrac{2\rho g r^{2}}{9\eta}

\rho: density of the sphere; r: its radius; \eta: viscosity of the medium

Unit: \mathrm{m\,s^{-1}}

Equation of continuity

A_1v_1=A_2v_2\quad\text{or}\quad Av=\text{constant}

A: cross-sectional area of the pipe; v: speed of the fluid (the product is the flow rate)

Unit: \mathrm{m^{3}\,s^{-1}}

Speed and area

v\propto\dfrac{1}{A}

smaller area gives greater speed

Bernoulli's equation

P_1+\tfrac12\rho v_1^{2}+\rho gh_1=P_2+\tfrac12\rho v_2^{2}+\rho gh_2

P: pressure; \rho: density; v: speed; h: height at points 1 and 2

Unit: \mathrm{Pa}

Bernoulli's equation (general form)

P+\tfrac12\rho v^{2}+\rho gh=\text{constant}

pressure plus kinetic energy per unit volume plus potential energy per unit volume

Unit: \mathrm{Pa}

Torricelli's theorem

v=\sqrt{2gh}

h: depth of the orifice below the top surface of the liquid

Unit: \mathrm{m\,s^{-1}}

Venturi meter: pressure difference

P_1-P_2=\tfrac12\rho\left(v_2^{2}-v_1^{2}\right)=\tfrac12\rho v_1^{2}\left(\dfrac{A_1^{2}}{A_2^{2}}-1\right)

points 1 and 2 are in the wide part and the throat of the pipe

Unit: \mathrm{Pa}

Venturi meter: manometer reading

P_1-P_2=\rho gh

h: height difference of the liquid in the two tubes

Unit: \mathrm{Pa}

Venturi meter: speed in the wide part

v_1=\sqrt{\dfrac{2gh}{\dfrac{A_1^{2}}{A_2^{2}}-1}}

A_1, A_2: areas of the wide part and the throat

Unit: \mathrm{m\,s^{-1}}

Key terms

Fluid mechanics: The branch of physics in which we study fluids, either static or dynamic.
Fluid dynamics: The study of fluids in motion.
Up-thrust (buoyant force): The net upward force on an object in a fluid, because the pressure at the bottom of the object is greater than at its top.
Archimedes' principle: When an object is totally or partially immersed in a liquid, an up-thrust acts on it equal to the weight of the liquid it displaces.
Viscosity: The resistance offered by different layers of a fluid to its flow.
Coefficient of viscosity: The quantity \eta that measures viscosity; its unit is \mathrm{N\,s\,m^{-2}}, \mathrm{kg\,m^{-1}\,s^{-1}} or \mathrm{Pa\,s}.
Drag force (viscous drag): The retarding force (resistance) experienced by an object moving through a fluid.
Aerodynamic drag: The drag force when the fluid is a gas such as air (air resistance).
Hydrodynamic drag: The drag force when the fluid is a liquid such as water.
Stokes' law: The drag force on a spherical object is directly proportional to its radius, its velocity and the coefficient of viscosity of the fluid: F_d=6\pi\eta r v.
Terminal velocity: The maximum constant velocity acquired by a freely falling object in a viscous medium.
Streamline (laminar, steady) flow: Flow in which every particle passing a point follows exactly the same smooth path as the particles before it.
Turbulent flow: Irregular flow of a fluid, with small whirlpool-like regions.
Ideal fluid: A fluid that is non-viscous, incompressible and irrotational, in streamline flow.
Flow rate: The volume of fluid passing a given point in a pipe per unit time.
Equation of continuity: For an ideal fluid, the product of the cross-sectional area of the pipe and the fluid speed is the same at every point: Av=\text{constant}.
Bernoulli's equation: The sum of the pressure, the potential energy per unit volume and the kinetic energy per unit volume remains constant: P+\tfrac12\rho v^{2}+\rho gh=\text{constant}.
Atomizer: A device that emits liquid droplets as a fine spray, working on Bernoulli's principle.
Torricelli's theorem: The speed of flow of fluid from an orifice at depth h below the top surface of a liquid is equal to the speed gained by the fluid in free falling through the height h.
Venturi meter: A flow meter used to measure the flow speed (flow rate) of a liquid through a tube.
Aerofoil: The cross-sectional shape of an object that creates a lift when moved through a fluid such as air.
Lift: The net upward force on a wing caused by the pressure difference between its upper and lower sides.
Magnus effect: The curving of a spinning ball away from its path of flight, caused by the pressure difference of the fluid on opposite sides of the spinning object.
Super fluidity: A state in which a liquid experiences zero viscosity, hence flows without friction past any surface.

Multiple choice questionsp. 108–109

The book prints no answer key for these questions. Each key below was worked out from the methods of the unit.

Principle of floatation helps us to find:

ADensityAnswer
BVelocity
CArea
DPressure

Why

A floating body is in equilibrium, so its weight equals the up-thrust, and the up-thrust equals the weight of the liquid it displaces: \rho_{\text{object}}\,V g=\rho_{\text{liquid}}\,V_{\text{sub}}\,g. This ties the density of the object to the fraction of its volume that is submerged, so floatation gives the density (Section 6.1).

Velocity, area and pressure do not appear in this balance, so B, C and D are wrong.

According to Archimedes' principle, the up-thrust on an object equal to:

Aweight of the displaced liquidAnswer
Bvolume of the displaced liquid
Cmass of the displaced liquid
Ddensity of the displaced liquid

Why

The up-thrust is F=\rho gV (equation 6.1), and \rho gV is the weight of the liquid displaced, so the up-thrust equals the weight of the displaced liquid.

B is a volume, C is a mass and D is a density. None of these is a force, and the mass \rho V differs from the weight only by the factor g.

In the equation for Stoke's law, the sphere must be:

Amoving with a low, non-zero acceleration
Bmoving with a constant velocityAnswer
Cat rest in the fluid
Dof the same density as the fluid

Why

Stokes's law F_d=6\pi\eta rv gives the drag at a steady speed v. For the drag to balance the other forces, the sphere must move with constant velocity, so its acceleration is zero.

A sphere at rest has no velocity and so no drag (C). The density of the sphere does not appear in the equation (D), and a non-zero acceleration (A) means the speed is still changing, so the law does not apply.

Two identical spherical drops of water are falling through air with a steady velocity of 20\,\mathrm{cm\,s^{-1}}. If the drops combine to form a single drop, what would be the terminal velocity of the single drop?

A10\,\mathrm{cm\,s^{-1}}
B20\,\mathrm{cm\,s^{-1}}
C32\,\mathrm{cm\,s^{-1}}Answer
D40\,\mathrm{cm\,s^{-1}}

Why

Terminal velocity v_t=\dfrac{2\rho gr^{2}}{9\eta} is proportional to r^{2}. Two equal drops have twice the volume of one drop, so the combined drop has radius R=2^{1/3}r\approx1.26\,r.

The new speed is v'=v\left(\dfrac{R}{r}\right)^{2}=20\times2^{2/3}\approx20\times1.59\approx32\,\mathrm{cm\,s^{-1}}. B assumes the speed does not change, and D doubles the radius and takes v\propto r, which overstates the speed.

Spherical balls of radius 'r' are falling in a viscous fluid of viscosity '\eta' with a velocity 'v'. The retarding viscous force acting on the spherical ball is:

Ainversely proportional to 'r' but directly proportional to velocity 'v'
Bdirectly proportional to both radius 'r' and velocity 'v'Answer
Cinversely proportional to both radius 'r' and velocity 'v'
Ddirectly proportional to 'r' but inversely proportional to 'v'

Why

Stokes's law gives F_d=6\pi\eta rv: the drag is directly proportional to the radius and to the velocity (and to the viscosity).

A, C and D each contain an inverse relation, which is the opposite of the law. Viscous drag grows as the ball becomes larger and as it moves faster.

Two rain drops reach the Earth with their terminal velocities in the ratio 4:9. The ratio of their radii is:

A4:9
B2:3Answer
C3:2
D9:4

Why

Terminal velocity is proportional to r^{2} (equation 6.5), so \dfrac{v_1}{v_2}=\dfrac{r_1^{2}}{r_2^{2}}=\dfrac{4}{9}. Taking square roots gives \dfrac{r_1}{r_2}=\dfrac{2}{3}, that is 2:3.

A is the ratio of the velocities, not of the radii, and it has not been square-rooted. C and D invert the ratio.

A liquid flows through a pipe with a diameter of 10\,\mathrm{cm} at a velocity of 9\,\mathrm{cm\,s^{-1}}. If the diameter of the pipe decreases to 6\,\mathrm{cm}, the new velocity of the liquid will be:

A12\,\mathrm{cm\,s^{-1}}
B15\,\mathrm{cm\,s^{-1}}
C21\,\mathrm{cm\,s^{-1}}
D25\,\mathrm{cm\,s^{-1}}Answer

Why

By the equation of continuity A_1v_1=A_2v_2, with A\propto d^{2}: v_2=v_1\left(\dfrac{d_1}{d_2}\right)^{2}=9\times\left(\dfrac{10}{6}\right)^{2}=9\times\dfrac{25}{9}=25\,\mathrm{cm\,s^{-1}}.

B (15) uses the diameter ratio instead of its square. A and C do not satisfy A_1v_1=A_2v_2.

If a pipe with flowing water has a cross-sectional area nine times greater at point 2 than at point 1, what would be the relation of flow speed at the two points?

AThe flow speed at point 2 is nine times that at point 1
BThe flow speed at point 2 is three times that at point 1
CThe flow speed at point 1 is nine times that at point 2Answer
DThe flow speed at point 1 is three times that at point 2

Why

Continuity gives A_1v_1=A_2v_2. With A_2=9A_1, v_1=9v_2: the speed at point 1 is nine times the speed at point 2.

The wide end has the slower flow, so the speed at point 2 cannot be larger (A and B). D gives a factor of three, which would correspond to a square root of nine; the continuity equation gives the area ratio directly.

With an increase in temperature, the viscosity of liquids and gases, respectively will

Aincrease and increase
Bincrease and decrease
Cdecrease and increaseAnswer
Ddecrease and decrease

Why

In most liquids the viscosity decreases as the temperature rises (honey warms and pours more easily), whereas in most gases it increases with temperature (Section 6.2). So the liquid's viscosity decreases and the gas's increases.

A treats both as increasing and D treats both as decreasing. B reverses the behaviour of both.

According to equation of continuity, when water falls its speed increases, while its cross-sectional area ____.

Aremains same
BdecreasesAnswer
Cincreases
Dcan't be predicted

Why

The equation of continuity Av=\text{constant} gives A\propto 1/v. As the water speeds up while falling, the stream must become narrower (Fig. 6.8a).

The continuity equation predicts the area exactly, so A and D are wrong. C has the wrong direction.

The figure shows two balloons are hung by a string. When the air is blown harder through the straw, the balloons will:

Aremain stationary
Bmove closely to each otherAnswer
Cmove far apart to each other
Dburst

Why

A fast stream of air between the balloons has a low pressure (Bernoulli's principle). The air on the outer sides is at higher pressure, so the balloons are pushed towards the low-pressure gap and move closer together.

This is the effect of the two A4 sheets in the Activity of Section 6.6: the faster the air flows between them, the closer they come together.

Note. The printed photograph shows two balloons hung on strings with a straw blowing air between them. It is not reproduced here.

Which will produce the greatest increase in flow velocity through a tube?

Ahalving the tube radiusAnswer
Bdoubling the viscosity of the liquid
Cdoubling the tube area
Ddoubling the tube radius

Why

For a fixed flow rate, Q=Av gives v\propto 1/A\propto 1/r^{2}. Halving the radius quarters the area and multiplies the speed by four. Doubling the area halves the speed, and doubling the radius divides it by four.

Doubling the viscosity makes the liquid flow more slowly, not faster, so B is wrong.

Where does the Venturi effect specifically occur?

AWhere a stream is expanded
BWhere a stream flows downwards with gravity
CWhere a stream constrictsAnswer
DIn the exact center of a flowing stream

Why

The Venturi effect is the fall in pressure where a stream narrows. At a constriction the continuity equation makes the speed higher, and Bernoulli's principle then makes the pressure lower (Section 6.6.4).

Where a stream expands the speed falls and the pressure rises (A). Falling under gravity (B) is not the Venturi effect, and the centre of a stream (D) has no special pressure.

The end of a hose has a diameter of 4\,\mathrm{cm}. If one wants the velocity of the water coming out to be 4 times higher, what should be the diameter of the nozzle on the end?

A1/4\,\mathrm{cm}
B4\,\mathrm{cm}
C2\,\mathrm{cm}Answer
D1/2\,\mathrm{cm}

Why

Continuity gives A_1v_1=A_2v_2, so \dfrac{v_2}{v_1}=\dfrac{A_1}{A_2}=\left(\dfrac{d_1}{d_2}\right)^{2}. For a fourfold speed, \left(\dfrac{4}{d_2}\right)^{2}=4, so d_2=\dfrac{4}{2}=2\,\mathrm{cm}.

B gives no change in speed. A and D make the nozzle far too narrow: d_2 of 0.25\,\mathrm{cm} and 0.5\,\mathrm{cm} multiply the speed by 256 and 64, so they are wrong.

Short questionsp. 109–110

6.1

Why do athletes, such as swimmers and bicyclists, wear body suits in competition.

Solution

Athletes wear close-fitting suits to reduce the drag force, the retarding force of the fluid (air for cyclists, water for swimmers). Drag depends on the size, shape and orientation of the body, so a smooth, streamlined outline reduces it.

A tight suit reduces the effective size of the body and keeps the flow around it smooth. Less energy is lost pushing the fluid aside, so the athlete keeps a higher speed for the same effort. The same principle is why fast cars and aircraft are streamlined (Section 6.2).

6.22026 Annual I, Set B, Q.2 (iv) OR

Distinguish between turbulent and streamline flow.

Solution

Streamline (steady or laminar) flow is smooth flow in which every particle passing a point follows the same path as the particles before it. Turbulent (non-laminar) flow is irregular, with small whirlpool-like regions, and the flow lines become disordered and mixed.

Feature	Streamline flow	Turbulent flow
Path of particles	Smooth, the same path as earlier particles	Irregular, with whirlpool-like regions
Streamlines	Do not cross each other	Disordered and mixed
Velocity at a point	The same for every particle arriving there	Not predictable
Where it occurs	Lower velocities and streamlined objects	Higher velocities, sudden changes in the size or direction of the pipe, and non-streamlined objects

6.3

When there is a change in the width of the river. The speed of the water decreases in wider regions whereas the speed of water increases in the narrower regions. Why?

Solution

Water is an incompressible fluid in steady flow, so the volume flowing past any cross-section each second is the same. The equation of continuity therefore applies to the river:

A_1v_1=A_2v_2=\text{constant}

In a wide region the cross-sectional area A is large, so the speed v=\text{constant}/A is small. In a narrow region the area is small, so the same volume per second must move faster. Hence v\propto 1/A.

If the river is twice as wide at the same depth, the speed is halved.

6.4

It is dangerous to stand close to rail tracks when a rapidly moving train passes. Explain why?

Solution

When a train passes quickly, the air between the train and a person on the platform moves very fast. By Bernoulli's principle, P+\tfrac12\rho v^{2}=\text{constant} at the same height, so faster air has a lower pressure.

The pressure on the side of the person facing the train is therefore lower than the pressure on the outer side. The net force pushes the person towards the train, which is why one must stand well back from the rails.

6.5

Verify that pressure has units of energy per unit volume.

Solution

Pressure is force per unit area, so its SI unit is \mathrm{N\,m^{-2}}.

Energy per unit volume has the unit of energy (joule, \mathrm{N\,m}) divided by the unit of volume (\mathrm{m^{3}}):

\frac{\text{energy}}{\text{volume}}=\frac{\mathrm{N\,m}}{\mathrm{m^{3}}}=\mathrm{N\,m^{-2}}\quad(=\mathrm{J\,m^{-3}})

Both quantities have the unit \mathrm{N\,m^{-2}}, so pressure has units of energy per unit volume. This is also seen from P=\dfrac{F}{A}=\dfrac{Fd}{Ad}=\dfrac{W}{V}, where W is work and V volume.

6.6

A perfume bottle or atomizer sprays a fluid from inside the bottle when it is pressed. How does the fluid rise up in the vertical tube in the bottle?

Solution

Squeezing the bulb blows a fast stream of air horizontally across the top of the vertical tube. Fast-moving air has a low pressure (Bernoulli's principle), so the pressure at the top of the tube falls below the pressure of the air inside the bottle.

The higher pressure in the bottle pushes the liquid up the tube. At the nozzle the liquid is broken into tiny droplets, which the fast air carries away as a fine spray (Section 6.6.1).

6.7

If you lower the window on a car while moving, an empty plastic bag can sometimes fly out the window. Why does this happen?

Solution

When the window is lowered, air rushes past the opening at high speed. The air moving across the outside of the opening is fast, so its pressure is low. The air inside the car is almost still, so its pressure is higher (Bernoulli's principle).

The higher pressure inside pushes the bag out through the window, so it flies out.

6.8

Explain how an up-thrust is produced when an aeroplane is running on runway.

Solution

The wing has the shape of an aerofoil (Section 6.6.5). As the aeroplane runs along the runway, air flows faster over the curved upper surface than under the flatter lower surface, so the pressure is lower above the wing than below it.

P_{\text{bottom}}-P_{\text{top}}=\tfrac12\rho\left(v_{\text{top}}^{2}-v_{\text{bottom}}^{2}\right)

This pressure difference gives a net upward force on the wing, the lift, which helps the aeroplane leave the ground once it exceeds the weight of the aeroplane.

6.9

Why does a ball placed in a vertical air jet become suspended?

Solution

The air in the centre of the jet moves fastest, so its pressure is lowest (Bernoulli's principle). The slower air around the jet is at a higher pressure. The pressure difference gives a net upward force on the ball, which balances its weight.

If the ball drifts sideways out of the jet, it enters slower air at higher pressure, and this pressure pushes it back into the fast central stream. The ball is therefore held in equilibrium, suspended in the jet.

Note. The printed figure shows a ball in a vertical air jet; it is not reproduced here.

6.10

Can you increase the flow velocity of water in a rubber pipe by squeezing? Explain briefly.

Solution

Yes. Squeezing the end of the pipe reduces the cross-sectional area of the opening. The flow rate Av must be the same at every section, so the speed must increase:

A_1v_1=A_2v_2\quad\Rightarrow\quad v_2=\dfrac{A_1}{A_2}\,v_1

Since A_2<A_1, v_2>v_1. Halving the area of the opening doubles the speed of the water.

6.112026 Annual I, Set B, Q.2 (xii) OR

What is meant by superfluid? How is superfluid made?

Solution

A superfluid is a liquid with zero viscosity. It flows without friction through any surface and so loses no kinetic energy. When stirred it forms vortices that keep spinning indefinitely, and it can creep over the walls of a container to 'empty' itself.

Superfluidity is observed in liquid helium-4 below 2.17 K (helium-3 becomes superfluid below 0.0025 K). To make it, helium gas cooled to a few degrees above absolute zero is compressed and then expelled through a small nozzle. As the gas expands rapidly it cools into the superfluid state (Fig. 6.17).

Comprehensive questionsp. 110

6.1

What is up thrust? Explain with the help of examples.

Solution

Up-thrust

The up-thrust (buoyant force) is the upward force that a fluid exerts on any object placed in it, totally or partially. It arises because the pressure in a fluid increases with depth, so the fluid pushes up harder on the bottom of the object than it pushes down on the top.

Derivation Up-thrust on a cylinder

Aim. Show that the up-thrust on a body equals the weight of the liquid it displaces: F=\rho gV.

Assumptions.

The cylinder has uniform cross-sectional area A and height h, and is completely immersed in a liquid of density \rho at rest.
The sideways forces on the cylinder cancel each other.

Figure 6.2: Up-thrust acts on the cylinder immersed in liquid. The net up-thrust is F=F_2-F_1.

Pressure at the top face (depth h_1) and the bottom face (depth h_2).
P_1=\rho gh_1,\qquad P_2=\rho gh_2
The force on each face is the pressure times the area A. The top force acts downwards and the bottom force upwards.
F_1=P_1A=\rho gh_1A,\qquad F_2=P_2A=\rho gh_2A
The net upward force is the bottom force minus the top force.
F=F_2-F_1=\rho g\left(h_2-h_1\right)A
From the figure, h_2-h_1=h, the height of the cylinder, and Ah=V is its volume.
F=\rho g\,Ah=\rho gV
The liquid displaced has volume V and mass \rho V, so \rho gV is its weight.
F=\text{weight of the liquid displaced}

ResultF=\rho gV=\text{weight of the liquid displaced}

This is Archimedes\' principle (equation 6.1). The up-thrust depends on the volume displaced and the density of the fluid. It does not depend on the depth at which the object is immersed, or on its own weight.

Examples

A ship floats because its hull displaces a weight of water equal to the weight of the ship, so the up-thrust balances its weight.
A helium balloon rises because the up-thrust of the air is greater than the weight of the balloon and its gas.
A mug full of water feels lighter under water, but heavy as soon as it is lifted out, because the water supports part of its weight.
An air-filled balloon released under water shoots up to the surface because the up-thrust on it is greater than its weight.

The up-thrust is present whether the object floats, sinks or is suspended. It is greater than the weight when the object rises, equal to it when the object stays suspended, and less than it when the object sinks.

6.2

Explain Archimedes' principle of flotation.

Solution

Archimedes' principle

When an object is totally or partially immersed in a liquid, an up-thrust acts on it equal to the weight of the liquid it displaces: F_b=\rho_{\text{liquid}}\,V_{\text{sub}}\,g, where V_{\text{sub}} is the volume of the object below the surface.

Principle of floatation

A body floats when its weight is balanced by the up-thrust. A floating body is only partly below the surface: the volume it displaces is the submerged volume V_{\text{sub}}, which is less than its total volume V.

Derivation Fraction of a floating body that is submerged

Aim. Show that a body of density \rho_b floating in a liquid of density \rho_f has the fraction \rho_b/\rho_f of its volume below the surface.

Assumptions.

The body floats at rest, so its weight equals the up-thrust.
The body is uniform, with total volume V.

Weight of the body.
W=\rho_b\,V\,g
The up-thrust equals the weight of the displaced liquid, whose volume is the submerged volume.
F_b=\rho_f\,V_{\text{sub}}\,g
Equilibrium means the up-thrust equals the weight. Cancel g.
\rho_f\,V_{\text{sub}}\,g=\rho_b\,V\,g
Solve for the submerged fraction.
\dfrac{V_{\text{sub}}}{V}=\dfrac{\rho_b}{\rho_f}

Result\dfrac{V_{\text{sub}}}{V}=\dfrac{\rho_b}{\rho_f}

The fraction below the surface is \rho_b/\rho_f, so a body floats only if it is less dense than the liquid. Measuring the submerged fraction gives the density of the body (Section 6.1).

Application: ships and submarines

Buoyancy: a ship's hull is shaped to displace a volume of water whose weight equals the weight of the ship, so the up-thrust balances its weight.
Stability: weight is kept low with ballast tanks and a stable hull shape, so the ship does not overturn.
Structural reliability: strong, light materials such as steel and aluminium withstand stresses, loads and vibrations.

A submarine has tanks that can be filled with seawater. Filling them increases its weight, and when the weight becomes greater than the up-thrust it sinks. Emptying the tanks reduces the weight and it returns to the surface.

6.3

What is viscosity? Explain with the help of examples.

Solution

Viscosity

The resistance offered by the different layers of a fluid to their relative motion, that is to its flow, is called viscosity. A fluid with a large coefficient of viscosity flows slowly, and a fluid with a small coefficient flows easily.

\eta=\text{coefficient of viscosity},\qquad \text{SI unit: }\mathrm{N\,s\,m^{-2}}=\mathrm{kg\,m^{-1}\,s^{-1}}=\mathrm{Pa\,s}

Table 6.1: Viscosity of some fluids at 20^\circ\mathrm{C}.
Substance	Viscosity at 20^\circ\mathrm{C} (\mathrm{mPa\,s})
Water	1.0016
Whole milk	2.12
Honey	2000 to 10000
Glycerin	1410
Mercury	1.55
Methanol	0.5940
Ethanol	1.144

Examples

Honey poured from a jar flows slowly, while water flows easily. A Styrofoam cup with a hole in the bottom drains water quickly, but drains honey very slowly, because honey has a much higher viscosity. Viscosity therefore decides the rate of flow of a liquid.

Effect of temperature

In most liquids the viscosity decreases as the temperature rises, but in most gases it increases. The temperature must always be measured when a viscosity is quoted. Water and air have low coefficients of viscosity, while honey and engine oil have high ones.

6.4

What is viscous drag? Explain with the help of examples.

Solution

Viscous drag

The retarding force experienced by an object moving through a fluid is called drag force or viscous drag. In a gas such as air it is called aerodynamic drag or air resistance, and in a liquid such as water it is called hydrodynamic drag.

The drag depends on the size, shape and orientation of the object, on the viscosity of the fluid, and on the speed of the object relative to the fluid.

Examples

Hand out of a car window. A large drag is felt when the hand is stretched out of a fast-moving car, and a smaller drag when it is tilted. The drag increases as the speed of the car increases.
Swimmers and skydivers bend, twist or stretch their arms and legs. This changes their effective size and orientation, and so controls their speed and direction through the drag.
Geese fly in a V-formation on long migrations. The shape reduces the drag and the energy used by each bird.
Fast vehicles are designed as streamlined bodies to reduce drag. In a wind tunnel a high-speed airstream flows across a model to show the interaction.

Drag on a small sphere

For a small spherical object moving slowly, the drag is given by Stokes's law: F_d=6\pi\eta rv (equation 6.2). The drag is directly proportional to the radius r, the speed v and the coefficient of viscosity \eta.

6.5

Explain Stoke's law. Derive its mathematical formula by using dimensional analysis.

Solution

Stokes's law

For a small sphere moving slowly through a viscous fluid, the drag force is directly proportional to the radius r, the velocity v and the coefficient of viscosity \eta. Stokes's law is written F_d=6\pi\eta rv, where 6\pi is a constant found by experiment and theory.

Derivation Stokes's law by dimensional analysis

Aim. Find how the drag force F_d on a sphere depends on r, v and \eta.

Assumptions.

The drag depends only on F_d, r, v and \eta.
The relation has the form F_d=k\,\eta^{a}r^{b}v^{c}, where k is a pure number and a, b, c are unknown powers. M, L and T stand for mass, length and time.

Dimensions of the drag force (force is mass times acceleration).
[F_d]=\mathrm{MLT^{-2}}
Dimensions of the other quantities, from their SI units: \eta is in \mathrm{kg\,m^{-1}\,s^{-1}}, r in metres and v in \mathrm{m\,s^{-1}}.
[\eta]=\mathrm{ML^{-1}T^{-1}},\quad [r]=\mathrm{L},\quad [v]=\mathrm{LT^{-1}}
Substitute these into the assumed form.
\mathrm{MLT^{-2}}=\left(\mathrm{ML^{-1}T^{-1}}\right)^{a}\left(\mathrm{L}\right)^{b}\left(\mathrm{LT^{-1}}\right)^{c}
Collect the powers of M, L and T on the right.
\mathrm{MLT^{-2}}=\mathrm{M^{a}L^{\,b+c-a}T^{-a-c}}
Equate the powers. M gives a=1; T gives -a-c=-2, so c=1; L gives b+c-a=1, so b=1.
a=1,\qquad b=1,\qquad c=1
Substitute the powers back into the assumed form.
F_d=k\,\eta r v

ResultF_d\propto\eta rv\quad\Rightarrow\quad F_d=6\pi\eta rv

Dimensional analysis gives only the proportionality. The constant k=6\pi cannot be found this way; it comes from experiment and theory. Doubling the radius or the speed doubles the drag, and a more viscous fluid gives a larger drag.

6.62025 Annual I, Set D, Q.4

Explain terminal velocity and derive its equation by using Stoke's law.

Solution

Terminal velocity

The terminal velocity is the maximum constant velocity acquired by a freely falling object in a viscous medium. A raindrop accelerates at first because there is little drag. As its speed grows the drag grows, the net force falls, and at terminal velocity the drag balances the weight, so the drop falls at constant speed (Fig. 6.4).

Derivation Terminal velocity of a sphere

Aim. Derive v_t=\dfrac{2\rho g r^{2}}{9\eta} for a small sphere of density \rho and radius r, using Stokes's law.

Assumptions.

The sphere is small and moves through a fluid of viscosity \eta, so Stokes's law applies.
The up-thrust of the air is negligible compared with the weight, as in the book's derivation.

Forces on a sphere at terminal velocity: the drag F_d balances the weight, so the net force and the acceleration are zero.

At terminal velocity the acceleration is zero, so the weight and the drag balance.
W-F_d=0
Use W=mg and Stokes's law F_d=6\pi\eta rv_t.
mg=6\pi\eta rv_t
The mass comes from the density and the volume V=\dfrac43\pi r^{3}.
m=\rho V=\rho\left(\dfrac43\pi r^{3}\right)
Substitute for the mass.
\rho\left(\dfrac43\pi r^{3}\right)g=6\pi\eta r v_t
Solve for v_t and cancel \pi r.
v_t=\dfrac{\dfrac43\pi r^{3}\rho g}{6\pi\eta r}=\dfrac{4\rho g r^{2}}{18\eta}=\dfrac{2\rho g r^{2}}{9\eta}

Resultv_t=\dfrac{2\rho g r^{2}}{9\eta}

Equation 6.5. v_t\propto r^{2}: a drop with twice the radius falls four times faster, and v_t is inversely proportional to the viscosity. The paratrooper's graph (Fig. 6.5c) shows the same idea: the terminal speed falls from 50 m s^{-1} to 8 m s^{-1} when the parachute opens.

6.72025 Annual I, Set B, Q.5

State the equation of continuity and derive its mathematical form. Also write one example from daily life, where equation of continuity applies.

Solution

Equation of continuity

The equation of continuity is the statement of conservation of mass for the steady flow of an ideal (incompressible, non-viscous) fluid. It states that the product of the cross-sectional area and the speed of the fluid is constant at every point of a pipe: Av=\text{constant}.

Derivation Equation of continuity

Aim. Show that A_1v_1=A_2v_2 for a fluid flowing steadily through a tube of varying cross-section.

Assumptions.

The flow is steady: the speed at each point does not change with time.
The fluid is incompressible, so its density is the same at both ends: \rho_1=\rho_2=\rho.
No fluid enters or leaves through the sides of the tube.

Figure 6.7: Fluid flowing through a tube of varying cross-sectional areas. The narrow section has the higher speed.

In a short time \Delta t the fluid at the wide end moves a distance \Delta x_1=v_1\Delta t, so the volume entering the tube is A_1\Delta x_1.
\Delta V_1=A_1v_1\,\Delta t
The mass entering in this time is the density times the volume.
\Delta m_1=\rho_1A_1v_1\,\Delta t
Similarly, the mass leaving at the narrow end in the same time is
\Delta m_2=\rho_2A_2v_2\,\Delta t
Mass is conserved, so the mass entering equals the mass leaving.
\Delta m_1=\Delta m_2
Equate the two and cancel \Delta t.
\rho_1A_1v_1=\rho_2A_2v_2
The density is constant, so cancel \rho.
A_1v_1=A_2v_2

ResultA_1v_1=A_2v_2\qquad\text{or}\qquad Av=\text{constant}

Hence v\propto 1/A. The product Av is the volume flow rate (volume per second), which is the same at every section of a pipe.

Example from daily life

When water falls from a tap, its speed increases as it falls under gravity, so the stream becomes narrower (Fig. 6.8a). Squeezing the end of a rubber pipe reduces the area of the opening and makes the water leave faster (Fig. 6.8b).

6.8

Derive the Bernoulli's equation. Also, discuss its application in daily life.

Solution

Bernoulli's equation

Bernoulli's equation is the statement of conservation of energy for the steady flow of an ideal fluid. The work done on the fluid by the pressure forces at its two ends equals the change in its kinetic and potential energy.

Derivation Bernoulli's equation

Aim. Derive P+\tfrac12\rho v^{2}+\rho gh=\text{constant} for a fluid flowing through a tube of varying cross-section and height.

Assumptions.

The fluid is ideal (non-viscous, incompressible) and flows steadily.
Fluid enters at the lower end (area A_1, speed v_1, height h_1, pressure P_1) and leaves at the upper end (area A_2, speed v_2, height h_2, pressure P_2).
The same volume V (mass m=\rho V) moves from one end to the other in time t.

Figure 6.9: Fluid flowing through a tube of varying cross-sectional area, from height h_1 to height h_2.

Work done on the fluid at the lower end, where the force is F_1=P_1A_1 and the displacement is \Delta x_1.
W_1=P_1A_1\,\Delta x_1
At the upper end the force opposes the motion, so the work is negative.
W_2=-P_2A_2\,\Delta x_2
Net work done on the fluid.
W=W_1+W_2=P_1A_1\,\Delta x_1-P_2A_2\,\Delta x_2
The continuity equation gives A_1\Delta x_1=A_2\Delta x_2=V, the volume that flows through the tube.
W=P_1V-P_2V=(P_1-P_2)V
Work-energy theorem: net work equals the change in kinetic energy plus the change in potential energy.
(P_1-P_2)V=\left(\tfrac12 mv_2^{2}-\tfrac12 mv_1^{2}\right)+\left(mgh_2-mgh_1\right)
Write m=\rho V and divide every term by V.
P_1-P_2=\tfrac12\rho v_2^{2}-\tfrac12\rho v_1^{2}+\rho gh_2-\rho gh_1
Collect the terms at point 1 on the left and those at point 2 on the right.
P_1+\tfrac12\rho v_1^{2}+\rho gh_1=P_2+\tfrac12\rho v_2^{2}+\rho gh_2

ResultP+\tfrac12\rho v^{2}+\rho gh=\text{constant}

This is equation 6.21, along a streamline. The sum of the pressure, the kinetic energy per unit volume and the potential energy per unit volume stays constant. Where the speed is high in horizontal flow, the pressure is low.

Applications in daily life

Atomizer: a fast stream of air over the top of a vertical tube lowers the pressure there, so the liquid rises and is sprayed (Section 6.6.1).
Filter pump: water from the tap speeds up at the jet, its pressure drops, and air is drawn in from the side tube (Section 6.6.2).
Aerofoil: air moves faster over the curved top of a wing, giving lower pressure above it, so there is a net upward lift (Section 6.6.5).
Venturi meter and carburettor: the fast air in the narrow throat has a low pressure, which is used to measure flow or to draw petrol into the air stream (Sections 6.6.4 and Auto Engineering box).

6.92026 Annual I, Set B, Q.4

What is Torricelli's theorem. Explain.

Solution

Torricelli's theorem

Torricelli's theorem states that the speed of flow of fluid from an orifice at depth h below the top surface of a liquid is equal to the speed gained by the fluid in free fall through the height h.

Derivation Speed of efflux from an orifice

Aim. Derive v=\sqrt{2gh} for the speed of the liquid leaving an orifice at depth h.

Assumptions.

The liquid is ideal, so Bernoulli's equation applies.
The top surface (point 1) moves very slowly, so v_1 is negligible compared with the speed v_2 at the orifice.
Both the top surface and the orifice are open to the atmosphere, so P_1=P_2.

Figure 6.12: A tank filled with water with an orifice near the bottom. The top surface comes down with almost zero speed, and h is the depth of the orifice below the top surface.

Bernoulli's equation between the top surface (1) and the orifice (2).
P_1+\tfrac12\rho v_1^{2}+\rho gh_1=P_2+\tfrac12\rho v_2^{2}+\rho gh_2
Neglect v_1 and put P_1=P_2.
\rho gh_1=\tfrac12\rho v_2^{2}+\rho gh_2
Rearrange with h=h_1-h_2, the depth of the orifice below the top surface.
\tfrac12\rho v_2^{2}=\rho g\left(h_1-h_2\right)=\rho gh
Solve for the speed.
v_2=\sqrt{2gh}

Resultv=\sqrt{2gh}

This is equation 6.24. It is the speed an object gains in falling freely through the height h, from v^{2}=u^{2}+2gh with u=0. It does not depend on the density of the liquid or the size of the hole, and it increases with the depth of the orifice.

Torricelli's theorem gives the speed of water leaving a hole at the bottom of a bucket of depth h as \sqrt{2gh} (numerical problem 6.8).

6.10

What is venturi meter? Explain.

Solution

Venturi meter

A venturi meter (or flow meter) measures the flow speed of a liquid in a tube. The tube narrows to a throat of area A_2 between a wider section of area A_1. The fluid speeds up in the throat and its pressure falls, so the pressure difference between the two sections appears as a height difference h between two vertical tubes (Fig. 6.13).

Derivation Flow speed in a venturi meter

Aim. Express the flow speed v_1 in the wide section in terms of h, A_1 and A_2.

Assumptions.

The tube is horizontal, so h_1=h_2.
The flow is steady and the liquid is ideal and incompressible.
The height difference h of the liquid columns gives the pressure difference: P_1-P_2=\rho gh (equation 6.28).

Figure 6.13: Venturi meter. The pressure difference between points 1 and 2 appears as the height difference h between the two columns.

Bernoulli's equation between points 1 and 2 with h_1=h_2.
P_1+\tfrac12\rho v_1^{2}=P_2+\tfrac12\rho v_2^{2}
Rearrange for the pressure difference.
P_1-P_2=\tfrac12\rho\left(v_2^{2}-v_1^{2}\right)
Continuity between the wide section and the throat.
v_2=\dfrac{A_1}{A_2}\,v_1
Substitute for v_2.
P_1-P_2=\tfrac12\rho v_1^{2}\left(\dfrac{A_1^{2}}{A_2^{2}}-1\right)
Put in the pressure difference from the manometer, P_1-P_2=\rho gh.
\rho gh=\tfrac12\rho v_1^{2}\left(\dfrac{A_1^{2}}{A_2^{2}}-1\right)
Cancel \rho and solve for v_1.
v_1=\sqrt{\dfrac{2gh}{\dfrac{A_1^{2}}{A_2^{2}}-1}}

Resultv_1=\sqrt{\dfrac{2gh}{\left(A_1^{2}/A_2^{2}\right)-1}}

This is equation 6.29. If A_1, A_2 and h are known, the speed and so the flow rate Q=A_1v_1 can be found from one height reading. The larger the ratio A_1/A_2, the smaller the height difference for a given speed.

The carburettor of an engine is a venturi tube. Air is sucked through the narrow throat, the low pressure there draws petrol into the air stream, and the petrol mixes with the air before it enters the cylinder.

6.112025 Annual I, Set D, Q.2 (xiii) OR

Describe super fluidity.

Solution

Super fluidity

Super fluidity is the state in which a liquid experiences zero viscosity, so it flows without friction through any surface and without any loss of kinetic energy. Because of this, a superfluid creeps over the walls of a container to 'empty' itself (Fig. 6.16).

Properties

When stirred, a superfluid forms vortices that continue to spin indefinitely.
In a rotating container the superfluid does not rotate uniformly. Below a critical angular velocity the liquid stays stationary, and above it the superfluid forms quantised vortices, which can spin only at certain allowed angular velocities.
The behaviour arises from quantum mechanical effects.

Where superfluidity occurs

Liquid helium-4 becomes a superfluid below 2.17 K, and helium-3 below 0.0025 K. Electrons within a superconducting solid, and the neutron fluid in a neutron star, may also be superfluid.

How superfluid is made

Helium gas is cooled to a few degrees above absolute zero by compressing it and then expelling it through a small nozzle. As the gas expands rapidly it cools into the superfluid state, as the pressure-temperature graph of helium shows (Fig. 6.17).

Application

Superfluids are used in high-precision devices such as gyroscopes, which allow the measurement of some theoretically predicted gravitational effects. Superfluids have also been used to trap and slow the speed of light.

Numerical problemsp. 110–111

6.1

The settling speed of a small fog droplet in air is found to be 2.98\,\mathrm{mm\,s^{-1}}. Find out the radius of the droplet. The coefficient of viscosity of air is 1.9\times10^{-5}\,\mathrm{kg\,m^{-1}\,s^{-1}}.

Solution

Given

v_t = 2.98\times10^{-3}\,\mathrm{m\,s^{-1}}
\eta = 1.9\times10^{-5}\,\mathrm{kg\,m^{-1}\,s^{-1}}
\rho = 1000\,\mathrm{kg\,m^{-3}} density of water, taken for the droplet (not given in the question)
g = 9.8\,\mathrm{m\,s^{-2}}

To find

Solve the terminal velocity equation (6.5) for r.
v_t=\dfrac{2\rho gr^{2}}{9\eta}\quad\Rightarrow\quad r=\sqrt{\dfrac{9\eta v_t}{2\rho g}}
Substitute.
r=\sqrt{\dfrac{9\times1.9\times10^{-5}\times2.98\times10^{-3}}{2\times1000\times9.8}}=\sqrt{\dfrac{5.10\times10^{-7}}{1.96\times10^{4}}}
Evaluate.
r=\sqrt{2.60\times10^{-11}}=5.1\times10^{-6}\,\mathrm{m}

r\approx5.1\times10^{-6}\,\mathrm{m}

Book answer. 5.1\times10^{-6}\,\mathrm{m}

Note. The density of the droplet is not given. The book's answer uses water, \rho=1000\,\mathrm{kg\,m^{-3}}, which is used here, and the result agrees with the book.

6.2

Water is flowing at a velocity of 2.00\,\mathrm{m\,s^{-1}} through a hose with an internal diameter of 1.60\,\mathrm{cm}. (a) What is the flow rate in litres per second? (b) The fluid velocity in this hose's nozzle is 15.0\,\mathrm{m\,s^{-1}}. What is the nozzle's inside diameter?

Solution

Given

v_1 = 2.00\,\mathrm{m\,s^{-1}} speed in the hose
d_1 = 1.60\,\mathrm{cm}=1.60\times10^{-2}\,\mathrm{m} internal diameter of the hose
v_2 = 15.0\,\mathrm{m\,s^{-1}} speed in the nozzle

To find

(a) Q
(b) d_2

(a) Flow rate in litres per second.

Cross-sectional area of the hose.
A_1=\pi\left(\dfrac{d_1}{2}\right)^{2}=\pi\left(0.80\times10^{-2}\right)^{2}=2.01\times10^{-4}\,\mathrm{m^{2}}
Flow rate Q=A_1v_1.
Q=2.01\times10^{-4}\times2.00=4.02\times10^{-4}\,\mathrm{m^{3}\,s^{-1}}
Convert, since 1\,\mathrm{m^{3}}=1000\,\mathrm{L}.
Q=4.02\times10^{-4}\times10^{3}\,\mathrm{L\,s^{-1}}=0.402\,\mathrm{L\,s^{-1}}

Q\approx0.402\,\mathrm{L\,s^{-1}}

(b) Inside diameter of the nozzle.

The same flow rate passes the nozzle, so A_2=Q/v_2.
A_2=\dfrac{Q}{v_2}=\dfrac{4.02\times10^{-4}}{15.0}=2.68\times10^{-5}\,\mathrm{m^{2}}
Solve A_2=\pi d_2^{2}/4 for the diameter.
d_2=\sqrt{\dfrac{4A_2}{\pi}}=\sqrt{\dfrac{4\times2.68\times10^{-5}}{3.14}}
Evaluate.
d_2=5.84\times10^{-3}\,\mathrm{m}=0.584\,\mathrm{cm}

d_2\approx0.584\,\mathrm{cm}

Book answer. (a) 0.402\,\mathrm{L\,s^{-1}}, (b) 0.584\,\mathrm{cm}

6.3

Blood is pumped from the heart at a rate of 5.0\,\mathrm{L\,min^{-1}} into the aorta (of radius 1.0\,\mathrm{cm}). Determine the speed of blood through the aorta.

Solution

Given

Q = 5.0\,\mathrm{L\,min^{-1}}=\dfrac{5.0\times10^{-3}\,\mathrm{m^{3}}}{60\,\mathrm{s}}
r = 1.0\,\mathrm{cm}=1.0\times10^{-2}\,\mathrm{m} radius of the aorta

To find

Flow rate in SI units.
Q=\dfrac{5.0\times10^{-3}}{60}=8.33\times10^{-5}\,\mathrm{m^{3}\,s^{-1}}
Cross-sectional area of the aorta.
A=\pi r^{2}=\pi\left(1.0\times10^{-2}\right)^{2}=3.14\times10^{-4}\,\mathrm{m^{2}}
Speed from Q=Av.
v=\dfrac{Q}{A}=\dfrac{8.33\times10^{-5}}{3.14\times10^{-4}}=0.265\,\mathrm{m\,s^{-1}}
Convert to centimetres per second.
v=26.5\,\mathrm{cm\,s^{-1}}\approx27\,\mathrm{cm\,s^{-1}}

v\approx27\,\mathrm{cm\,s^{-1}}

Book answer. 27\,\mathrm{cm\,s^{-1}}

Note. The value 26.5\,\mathrm{cm\,s^{-1}} is rounded to two significant figures, which gives the book's answer.

6.4

Water flows through a pipe with an internal diameter 4\,\mathrm{cm} at a speed of 2\,\mathrm{m\,s^{-1}}. What should be the diameter of the nozzle, if the water is to emerge out with a speed of 4\,\mathrm{m\,s^{-1}}?

Solution

Given

d_1 = 4\,\mathrm{cm}
v_1 = 2\,\mathrm{m\,s^{-1}}
v_2 = 4\,\mathrm{m\,s^{-1}} speed at the nozzle

To find

Continuity, with A\propto d^{2}: \dfrac{\pi d_1^{2}}{4}v_1=\dfrac{\pi d_2^{2}}{4}v_2.
d_2^{2}=d_1^{2}\,\dfrac{v_1}{v_2}
Make d_2 the subject and substitute.
d_2=d_1\sqrt{\dfrac{v_1}{v_2}}=4\,\mathrm{cm}\times\sqrt{\dfrac{2}{4}}
Evaluate.
d_2=4\times0.707=2.83\,\mathrm{cm}

d_2\approx2.8\,\mathrm{cm}

Book answer. 2.8\,\mathrm{cm}

6.5

A hose lying on the ground has water coming out of it at a speed of 5.4\,\mathrm{m\,s^{-1}}. You lift the nozzle of the hose to a height of 1.3\,\mathrm{m} above the ground. At what speed does the water now come out of the hose?

Solution

Given

v_1 = 5.4\,\mathrm{m\,s^{-1}} speed at the nozzle on the ground
h_1 = 0 nozzle on the ground
h_2 = 1.3\,\mathrm{m} height of the raised nozzle
P_1, P_2 = \text{atmospheric} both nozzles open to the air

To find

The same hose before and after its nozzle is lifted to height h. The speed of the water leaving the nozzle is lower after lifting.

Bernoulli's equation between the two nozzles. The pressures are equal and h_1=0, and the density cancels.
\tfrac12v_1^{2}=\tfrac12v_2^{2}+gh_2
Solve for v_2.
v_2=\sqrt{v_1^{2}-2gh_2}
Substitute.
v_2=\sqrt{(5.4)^{2}-2\times9.8\times1.3}=\sqrt{29.16-25.48}=\sqrt{3.68}
Evaluate.
v_2=1.92\,\mathrm{m\,s^{-1}}

v_2\approx1.9\,\mathrm{m\,s^{-1}}

Book answer. 1.9\,\mathrm{m\,s^{-1}}

Note. The speed of 5.4 m s^{-1} is taken as the speed of the water at the nozzle on the ground, as the question implies.

6.6

A pipe has two different cross-sectional areas, A_1=25\,\mathrm{cm^{2}} and A_2=4\,\mathrm{cm^{2}}. The volumetric flow rate of water through the pipe is 5\times10^{-3}\,\mathrm{m^{3}\,s^{-1}}. Determine: (a) the speed of water in both cross-sectional areas of the pipe; (b) the pressure difference between them (\rho=10^{3}\,\mathrm{kg\,m^{-3}}).

Solution

Given

A_1 = 25\,\mathrm{cm^{2}}=2.5\times10^{-3}\,\mathrm{m^{2}}
A_2 = 4\,\mathrm{cm^{2}}=4.0\times10^{-4}\,\mathrm{m^{2}}
Q = 5\times10^{-3}\,\mathrm{m^{3}\,s^{-1}}
\rho = 10^{3}\,\mathrm{kg\,m^{-3}}

To find

(a) v_1, v_2
(b) P_1-P_2

(a) Speeds in both cross-sections.

From Q=Av in the wide section.
v_1=\dfrac{Q}{A_1}=\dfrac{5\times10^{-3}}{2.5\times10^{-3}}=2.0\,\mathrm{m\,s^{-1}}
From Q=Av in the narrow section.
v_2=\dfrac{Q}{A_2}=\dfrac{5\times10^{-3}}{4.0\times10^{-4}}=12.5\,\mathrm{m\,s^{-1}}

v_1=2\,\mathrm{m\,s^{-1}},\quad v_2=12.5\,\mathrm{m\,s^{-1}}

(b) Pressure difference between the two sections.

The pipe is horizontal, so h_1=h_2. Bernoulli's equation gives the pressure difference.
P_1-P_2=\tfrac12\rho\left(v_2^{2}-v_1^{2}\right)
Substitute.
P_1-P_2=\tfrac12\times10^{3}\left(12.5^{2}-2.0^{2}\right)=500\times152.25
Evaluate.
P_1-P_2=7.6\times10^{4}\,\mathrm{Pa}

P_1-P_2=7.6\times10^{4}\,\mathrm{Pa}

Book answer. (a) 2\,\mathrm{m\,s^{-1}} and 12.5\,\mathrm{m\,s^{-1}}, (b) 7.6\times10^{4}\,\mathrm{Pa}

Note. The pipe is taken as horizontal, as the question implies, so no height correction is needed.

6.7

The speed of water in a hose increases from 1.96\,\mathrm{m\,s^{-1}} to 25.5\,\mathrm{m\,s^{-1}} as it moves from the hose to the nozzle. Calculate the pressure in the hose, given that the absolute pressure in the nozzle is 1.01\times10^{5}\,\mathrm{N\,m^{-2}} (atmospheric, as it must be) and assuming level and frictionless flow.

Solution

Given

v_1 = 1.96\,\mathrm{m\,s^{-1}} speed in the hose
v_2 = 25.5\,\mathrm{m\,s^{-1}} speed at the nozzle
P_2 = 1.01\times10^{5}\,\mathrm{N\,m^{-2}} pressure at the nozzle
\rho = 1000\,\mathrm{kg\,m^{-3}} density of water, not given in the question

To find

Bernoulli's equation for level flow (h_1=h_2).
P_1+\tfrac12\rho v_1^{2}=P_2+\tfrac12\rho v_2^{2}
Solve for P_1.
P_1=P_2+\tfrac12\rho\left(v_2^{2}-v_1^{2}\right)
Substitute, with v_2^{2}=650.25 and v_1^{2}=3.84.
P_1=1.01\times10^{5}+\tfrac12\times1000\times\left(650.25-3.84\right)
Evaluate.
P_1=1.01\times10^{5}+3.23\times10^{5}=4.24\times10^{5}\,\mathrm{N\,m^{-2}}

P_1\approx4.24\times10^{5}\,\mathrm{N\,m^{-2}}

Book answer. 4.24\times10^{5}\,\mathrm{N\,m^{-2}}

Note. The density of water is not given. \rho=1000\,\mathrm{kg\,m^{-3}} is assumed, and the book's answer is consistent with it.

6.8

A hole is drilled at the bottom of a full bucket of water of height h. What will be the velocity of water exiting from the hole?

Solution

Given

h = height of the water above the hole
g = 9.8\,\mathrm{m\,s^{-2}}

To find

Torricelli's theorem: the speed of efflux equals the speed gained in free fall through h. The top surface moves slowly and both ends are at atmospheric pressure.
v=\sqrt{2gh}
Check with h=1.0\,\mathrm{m} as an example.
v=\sqrt{2\times9.8\times1.0}=4.4\,\mathrm{m\,s^{-1}}

v=\sqrt{2gh}

Book answer. \sqrt{2gh}

6.9

The efflux speed of oil from a narrow hole of a tank is 33.8\,\mathrm{m\,s^{-1}}, at pressure of 4.12\times10^{5}\,\mathrm{Pa}. Calculate the density of the oil?

Solution

Given

v = 33.8\,\mathrm{m\,s^{-1}} efflux speed
P = 4.12\times10^{5}\,\mathrm{Pa} pressure driving the flow above atmospheric

To find

\rho

Bernoulli between the oil in the tank (speed negligible) and the jet at the hole. The excess pressure P equals the kinetic energy per unit volume.
P=\tfrac12\rho v^{2}
Solve for the density.
\rho=\dfrac{2P}{v^{2}}
Substitute.
\rho=\dfrac{2\times4.12\times10^{5}}{(33.8)^{2}}=\dfrac{8.24\times10^{5}}{1142.4}
Evaluate.
\rho=7.21\times10^{2}\,\mathrm{kg\,m^{-3}}

\rho\approx7.2\times10^{2}\,\mathrm{kg\,m^{-3}}

Book answer. 0.72\,\mathrm{kg\,m^{-3}}

Note. Misprint in the book: 0.72\,\mathrm{kg\,m^{-3}} should read 0.72\times10^{3}\,\mathrm{kg\,m^{-3}} (about 721\,\mathrm{kg\,m^{-3}}), which is the value the calculation gives. The pressure is taken as the excess over atmospheric, which is the only reading that gives this density.

6.10

If the speed of air across the top and bottom of a small aeroplane's wing is 30\,\mathrm{m\,s^{-1}} and 20\,\mathrm{m\,s^{-1}} respectively. Calculate the pressure difference between the top and bottom of wings if \rho=1.29\,\mathrm{kg\,m^{-3}}, for air.

Solution

Given

v_{\text{top}} = 30\,\mathrm{m\,s^{-1}}
v_{\text{bottom}} = 20\,\mathrm{m\,s^{-1}}
\rho = 1.29\,\mathrm{kg\,m^{-3}} density of air

To find

P_{\text{bottom}}-P_{\text{top}}

Bernoulli at the same height: the pressure difference is half the density times the difference of the squares of the speeds.
P_{\text{bottom}}-P_{\text{top}}=\tfrac12\rho\left(v_{\text{top}}^{2}-v_{\text{bottom}}^{2}\right)
Substitute.
=\tfrac12\times1.29\times\left(30^{2}-20^{2}\right)=\tfrac12\times1.29\times500
Evaluate.
=322.5\,\mathrm{Pa}

P_{\text{bottom}}-P_{\text{top}}=322.5\,\mathrm{Pa}

Book answer. 322.5\,\mathrm{Pa}

6.11

An empty metrological balloon has a weight of 80\,\mathrm{N}. How much maximum contents the balloon can lift besides its own weight if it is filled with 10\,\mathrm{m^{3}} of hydrogen? (Density of hydrogen =0.09\,\mathrm{kg\,m^{-3}}, density of air =1.3\,\mathrm{kg\,m^{-3}}.)

Solution

Given

W_b = 80\,\mathrm{N} weight of the empty balloon
V = 10\,\mathrm{m^{3}} volume of the balloon
\rho_H = 0.09\,\mathrm{kg\,m^{-3}} density of hydrogen
\rho_{\text{air}} = 1.3\,\mathrm{kg\,m^{-3}} density of air

To find

W_L, the extra load it can lift

Up-thrust from the displaced air (Archimedes' principle).
F_b=\rho_{\text{air}}Vg=1.3\times10\times9.8=127.4\,\mathrm{N}
Weight of the hydrogen inside the balloon.
W_H=\rho_HVg=0.09\times10\times9.8=8.82\,\mathrm{N}
Balance: the up-thrust supports the balloon, the hydrogen and the extra load.
F_b=W_b+W_H+W_L\quad\Rightarrow\quad W_L=F_b-W_b-W_H
Substitute.
W_L=127.4-80-8.82
Evaluate.
W_L=38.58\,\mathrm{N}

W_L=38.58\,\mathrm{N}

Book answer. 38.58\,\mathrm{N}

Note. Assumes the balloon's full volume of 10 m^{3} displaces 10 m^{3} of air.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 61 mark

In a steady, incompressible flow, if the cross-sectional area of a pipe decreases to one-third of its original size, what happens to the fluid velocity?

AIt decreases to one-third
BIt remains the same
CIt triplesAnswer
DIt becomes nine times larger

Why

By the equation of continuity, for steady flow of an incompressible fluid, A_1v_1=A_2v_2. Here A_2=\dfrac{A_1}{3}, so

v_2=\dfrac{A_1}{A_2}\,v_1=3v_1

The fluid speeds up and its velocity triples. Option A is the tempting mistake of making the velocity change in the same way as the area; the speed is inversely proportional to the area, not directly. Option D would need the area to change by a factor of nine.

2025 Annual I, Set B, MCQ 71 mark

Maximum drag force on a 1 kg falling sphere is:

A9.8\,\mathrm{N}Answer
B1\,\mathrm{N}
C98\,\mathrm{N}
D4.9\,\mathrm{N}

Why

A falling sphere speeds up until the drag equals its weight; this is the terminal velocity, and the drag cannot grow beyond this in a steady fall. So the maximum drag is F_d=mg=1\times9.8=9.8\,\mathrm{N}. Option B (1\,\mathrm{N}) is only the mass in kilograms, not a force, and the drag cannot be larger than the weight in a steady fall.

2025 Annual I, Set B, Q.2 (iii) OR3 marks

What is meant by Magnus effect? Explain with the help of an example.

Solution

Magnus effect (1.5 marks). When a spinning object moves through a fluid, its path is deflected sideways, in a way that does not happen when it is not spinning. The spinning ball drags some of the surrounding air with it, so the air moves faster on one side and slower on the other. By Bernoulli's principle the faster side has lower pressure, and the ball is pushed towards the low-pressure side. The deflection depends on the rotational speed of the object.

Example (1.5 marks). A spinning cricket ball or golf ball. As the ball spins, the air moves faster over the side where the surface moves with the air stream, giving a lower pressure there. The pressure difference gives a sideways force, so the ball curves (swings) away from its straight path in flight, or a golf ball gets extra lift.

2026 Annual I, Set B, Q.2 (iv) OR3 marks

Distinguish between turbulent and streamline flow.

Solution

Feature	Streamline flow	Turbulent flow
Path of particles	Smooth, the same path as earlier particles	Irregular, with whirlpool-like regions
Streamlines	Do not cross each other	Disordered and mixed
Velocity at a point	The same for every particle arriving there	Not predictable
Where it occurs	Lower velocities and streamlined objects	Higher velocities, sudden changes in the size or direction of the pipe, and non-streamlined objects

2026 Annual I, Set B, Q.2 (vii) OR3 marks

Describe the working of Atomizers in perfume bottles.

Solution

An atomizer is a device that emits a liquid as a fine spray. It works on Bernoulli's principle: where a fluid moves fast its pressure is low, and where it moves slowly its pressure is high.

When the rubber bulb is squeezed, high-speed air passes horizontally over the top of a vertical tube whose lower end dips in the perfume.
By Bernoulli's principle fast air has low pressure, so the pressure at the top of the vertical tube becomes lower than the (atmospheric) pressure on the liquid in the bottle.
This pressure difference pushes the perfume up the narrow vertical tube and into the moving stream of air.
The nozzle at the end of the horizontal tube breaks the liquid into tiny drops, which the air stream carries away as a fine spray.

Figure 6.10: A simple atomizer. Squeezing the bulb sends fast air across the top of the vertical tube; the lower pressure there pushes liquid up the tube and the air stream breaks it into drops.

The same arrangement is used to spray paint and in the carburettor of an engine.

2026 Annual I, Set B, Q.2 (xii) OR1.5+1.5 = 3 marks

What is meant by superfluid? How is superfluid made?

Solution

2026 Annual I, Set B, Q.2 (xiii)1+1+1 = 3 marks

Water flows through a 1\,\mathrm{cm} diameter pipe with a speed of 1\,\mathrm{ms^{-1}}. What should be the diameter of the nozzle if the water is to emerge at 2.1\,\mathrm{ms^{-1}}?

Solution

Given

d_1 = 1\,\mathrm{cm} diameter of the pipe
v_1 = 1\,\mathrm{m\,s^{-1}} speed in the pipe
v_2 = 2.1\,\mathrm{m\,s^{-1}} speed at the nozzle

To find

Formula (1 mark). Water is incompressible and flows steadily, so the equation of continuity holds. The area is A=\pi d^{2}/4, so A\propto d^{2}.
A_1v_1=A_2v_2\ \Rightarrow\ \dfrac{\pi d_1^{2}}{4}v_1=\dfrac{\pi d_2^{2}}{4}v_2
**Make d_2 the subject (1 mark).**
d_2=d_1\sqrt{\dfrac{v_1}{v_2}}
Substitution and answer (1 mark).
d_2=1\,\mathrm{cm}\times\sqrt{\dfrac{1}{2.1}}=1\times0.690=0.69\,\mathrm{cm}

d_2=0.69\,\mathrm{cm}

2026 Annual I, Set B, Q.41+3+2 = 6 marks

a. State Bernoulli's equation. b. Derive velocity of efflux using Torricelli's theorem. Draw diagram as well.

Solution

Bernoulli's equation (1 mark)

Statement. For the steady flow of an ideal fluid (non-viscous, incompressible, laminar), the sum of the pressure, the kinetic energy per unit volume and the potential energy per unit volume has the same value at every point along a streamline:

\begin{aligned}P+\tfrac12\rho v^{2}+\rho gh&=\text{constant}\\ \text{or}\quad P_1+\tfrac12\rho v_1^{2}+\rho gh_1&=P_2+\tfrac12\rho v_2^{2}+\rho gh_2\end{aligned}

Bernoulli's equation is the law of conservation of energy for a flowing fluid. It shows that where the fluid moves fast the pressure is low.

Velocity of efflux, Torricelli's theorem (3 marks) and diagram (2 marks)

Torricelli's theorem states that the speed of flow of fluid from an orifice at depth h below the top surface of a liquid is equal to the speed gained by the fluid in free fall through the height h.

Derivation Speed of efflux from an orifice

Aim. Derive v=\sqrt{2gh} for the speed of the liquid leaving an orifice at depth h.

Assumptions.

The liquid is ideal, so Bernoulli's equation applies.
The top surface (point 1) moves very slowly, so v_1 is negligible compared with the speed v_2 at the orifice.
Both the top surface and the orifice are open to the atmosphere, so P_1=P_2.

Figure 6.12: A tank filled with water with an orifice near the bottom. The top surface comes down with almost zero speed, and h is the depth of the orifice below the top surface.

Bernoulli's equation between the top surface (1) and the orifice (2).
P_1+\tfrac12\rho v_1^{2}+\rho gh_1=P_2+\tfrac12\rho v_2^{2}+\rho gh_2
Neglect v_1 and put P_1=P_2.
\rho gh_1=\tfrac12\rho v_2^{2}+\rho gh_2
Rearrange with h=h_1-h_2, the depth of the orifice below the top surface.
\tfrac12\rho v_2^{2}=\rho g\left(h_1-h_2\right)=\rho gh
Solve for the speed.
v_2=\sqrt{2gh}

Resultv=\sqrt{2gh}

Torricelli's theorem gives the speed of water leaving a hole at the bottom of a bucket of depth h as \sqrt{2gh} (numerical problem 6.8).

2025 Annual I, Set B, Q.51+5+1 = 7 marks

State the equation of continuity and derive its mathematical form. Or, what basis is the equation of continuity derived?

Solution

Equation of continuity

Derivation Equation of continuity

Aim. Show that A_1v_1=A_2v_2 for a fluid flowing steadily through a tube of varying cross-section.

Assumptions.

The flow is steady: the speed at each point does not change with time.
The fluid is incompressible, so its density is the same at both ends: \rho_1=\rho_2=\rho.
No fluid enters or leaves through the sides of the tube.

Figure 6.7: Fluid flowing through a tube of varying cross-sectional areas. The narrow section has the higher speed.

In a short time \Delta t the fluid at the wide end moves a distance \Delta x_1=v_1\Delta t, so the volume entering the tube is A_1\Delta x_1.
\Delta V_1=A_1v_1\,\Delta t
The mass entering in this time is the density times the volume.
\Delta m_1=\rho_1A_1v_1\,\Delta t
Similarly, the mass leaving at the narrow end in the same time is
\Delta m_2=\rho_2A_2v_2\,\Delta t
Mass is conserved, so the mass entering equals the mass leaving.
\Delta m_1=\Delta m_2
Equate the two and cancel \Delta t.
\rho_1A_1v_1=\rho_2A_2v_2
The density is constant, so cancel \rho.
A_1v_1=A_2v_2

ResultA_1v_1=A_2v_2\qquad\text{or}\qquad Av=\text{constant}

Hence v\propto 1/A. The product Av is the volume flow rate (volume per second), which is the same at every section of a pipe.

Example from daily life

Basis (the 'or' part). The equation of continuity is derived on the basis of the law of conservation of mass: for an ideal fluid in steady flow the mass entering a pipe in a given time equals the mass leaving it (\rho_1A_1v_1=\rho_2A_2v_2), and for an incompressible fluid this becomes A_1v_1=A_2v_2.

2025 Annual I, Set D, MCQ 61 mark

The fluid is said to be incompressible, if its density is:

AConstantAnswer
BVery high
CVery small
DZero

Why

An incompressible fluid is one whose density stays constant, whatever the pressure (liquids are generally of this kind). Option D (zero density) would mean no fluid at all, and a very high or very small density says nothing about whether the density changes.

2026 Annual I, Set D, MCQ 71 mark

Continuity equation is based on:

AForce
BEnergy conservation
CMomentum
DMass conservationAnswer

Why

The equation of continuity, \rho_1A_1v_1=\rho_2A_2v_2 (which becomes A_1v_1=A_2v_2 for an incompressible fluid), says that the mass of fluid entering a pipe in a given time equals the mass leaving it, so it expresses the law of conservation of mass. The tempting option B, energy conservation, is the basis of Bernoulli's equation, not of the equation of continuity.

2025 Annual I, Set D, MCQ 71 mark

Equation of continuity is based upon law of conservation of:

AMomentum
BEnergy
CMassAnswer
DCharge

Why

The equation of continuity \rho_1A_1v_1=\rho_2A_2v_2 says that the mass entering a pipe equals the mass leaving it, so it expresses the law of conservation of mass. Bernoulli's equation is the one based on conservation of energy (option B).

2025 Annual I, Set D, Q.2 (ii) OR3 marks

When water falls from a tap, its cross-sectional area decreases as it comes down. Explain why?

Solution

Explanation. The falling water is a stream of an (almost) incompressible fluid in steady flow, so the equation of continuity applies: the volume of water passing every cross-section per second is the same.

A_1v_1=A_2v_2\quad\Rightarrow\quad v\propto\dfrac{1}{A}

The falling stream: the speed increases from v_1 to v_2, so the area decreases from A_1 to A_2.

As the water falls under gravity its speed increases (v_2>v_1). For the same flow rate, a larger speed needs a smaller area, so A_2<A_1 and the stream becomes narrower as it comes down. This is the conservation of mass: the same mass of water must pass each cross-section in each second.

2025 Annual I, Set D, Q.2 (vii)3 marks

What is meant by Magnus effect? Give its causes and effects.

Solution

Magnus effect (1 mark). The deviation of a spinning object from its straight path when it moves through a fluid.

Cause (1 mark). The spinning ball drags some of the surrounding air with it, so the air moves faster on one side of the ball and slower on the other. By Bernoulli's principle the faster air has lower pressure. The pressure difference across the ball produces a net force on it, towards the low-pressure side, at right angles to its motion.

Effects (1 mark). The path of the ball is deflected in a way that is not present when it is not spinning: a cricket ball swings in flight and a spinning golf ball gets lift. The deflection depends on the rotational speed of the ball.

2026 Annual I, Set D, Q.2 (vii) OR2+1 = 3 marks

Briefly explain Magnus effect (an application of Bernoulli Effect). Draw diagram as well.

Solution

Magnus effect (2 marks). The Magnus effect is the deviation of a spinning object from its straight path when it moves through a fluid. The spinning ball drags some of the surrounding air with it, so on one side of the ball the air moves faster and on the other side slower. By Bernoulli's principle the faster air has the lower pressure. The pressure difference across the ball gives a net force on it towards the low-pressure side, at right angles to its motion, and the ball curves away from its straight path. The deflection depends on the rotational speed of the ball.

Example (1 mark for the diagram). A spinning cricket or tennis ball swings or dips in flight, and a spinning golf ball gets lift, because of this pressure difference.

2026 Annual I, Set D, Q.2 (viii) OR1x3 = 3 marks

Briefly discuss any three properties of an ideal fluid.

Solution

In the model of ideal fluid flow the motion of a real fluid is simplified by the following assumptions (any three, 1 mark each):

The fluid is non-viscous. There is no internal friction (zero viscosity), so the fluid does not oppose its own flow and no energy is lost to friction.
The fluid is incompressible. Its density is constant everywhere. Liquids are generally incompressible, while gases are compressible.
The flow is laminar (steady, streamline). Every particle passing through a point has the same velocity, and the streamlines do not cross.
The flow is irrotational. There are no whirlpools or eddies in the flow.

Real fluids are viscous, so Bernoulli's equation and the equation of continuity apply to them only approximately.

2025 Annual I, Set D, Q.2 (xiii) OR2+1 = 3 marks

What is super fluidity? Give some of its applications.

Solution

Super fluidity (2 marks). Super fluidity is a state in which a liquid has zero viscosity, so it flows without friction through any surface and without any loss of kinetic energy. It is observed in liquid helium at temperatures near absolute zero (^{4}\mathrm{He} below 2.17\,\mathrm{K}). A superfluid can creep over the walls of its container to 'empty' itself, and when it is stirred, its vortices keep spinning indefinitely.

Applications (1 mark). Superfluids are used in high-precision devices such as gyroscopes, which allow the measurement of some theoretically predicted gravitational effects. They have also been used to trap and slow the speed of light.

2026 Annual I, Set D, Q.3 OR1+5 = 6 marks

a. State Archimedes principle. b. A water storage tank in a residential area develops a small leak near its lower end. The top of the water in the tank is 10\,\mathrm{m} above the point of the leak. (i) With what speed does the water rush from the hole? (ii) If the hole has an area of 0.080\,\mathrm{cm^2}, how much water flows out in one second?

Solution

a. Archimedes' principle (1 mark)

Archimedes' principle: when an object is totally or partially immersed in a liquid, an up-thrust acts on it equal to the weight of the liquid it displaces.

F_b=\rho Vg

Here \rho is the density of the liquid and V is the volume of the liquid displaced (the volume of the immersed part of the object). The up-thrust arises because the pressure at the bottom of the object is greater than at the top.

b. Leaking tank (5 marks)

Given

h = 10\,\mathrm{m} depth of the leak below the top of the water
A = 0.080\,\mathrm{cm^{2}}=0.080\times10^{-4}\,\mathrm{m^{2}}=8.0\times10^{-6}\,\mathrm{m^{2}} area of the hole
g = 9.8\,\mathrm{m\,s^{-2}}

To find

(i) speed of efflux v
(ii) volume flowing out per second, Q

(i) Speed with which the water rushes from the hole.

Torricelli's theorem: the speed of efflux from a hole at depth h below the top surface equals the speed gained in free fall through h (the top surface falls very slowly and both the surface and the hole are at atmospheric pressure).
v=\sqrt{2gh}
Substitute.
v=\sqrt{2\times9.8\times10}=\sqrt{196}=14\,\mathrm{m\,s^{-1}}

v=14\,\mathrm{m\,s^{-1}}

(ii) Volume of water that flows out in one second.

The volume flowing per second (flow rate) is the area of the hole times the speed of the water.
Q=Av
Substitute.
Q=8.0\times10^{-6}\times14=1.12\times10^{-4}\,\mathrm{m^{3}\,s^{-1}}
Convert, since 1\,\mathrm{m^{3}}=1000\,\mathrm{L}.
Q=0.112\,\mathrm{L\,s^{-1}}\approx1.1\times10^{-4}\,\mathrm{m^{3}\,s^{-1}}

Q\approx1.1\times10^{-4}\,\mathrm{m^{3}\,s^{-1}}

v=14\,\mathrm{m\,s^{-1}},\quad Q\approx1.1\times10^{-4}\,\mathrm{m^{3}\,s^{-1}}

2025 Annual I, Set D, Q.41+2+3 = 6 marks

State and explain terminal velocity. Derive its equation by using stoke's law.

Solution

Terminal velocity

Stokes' law. The viscous drag on a small sphere of radius r moving with speed v through a fluid of viscosity \eta is F_d=6\pi\eta rv.

Derivation Terminal velocity of a sphere

Aim. Derive v_t=\dfrac{2\rho g r^{2}}{9\eta} for a small sphere of density \rho and radius r, using Stokes's law.

Assumptions.

The sphere is small and moves through a fluid of viscosity \eta, so Stokes's law applies.
The up-thrust of the air is negligible compared with the weight, as in the book's derivation.

Forces on a sphere at terminal velocity: the drag F_d balances the weight, so the net force and the acceleration are zero.

At terminal velocity the acceleration is zero, so the weight and the drag balance.
W-F_d=0
Use W=mg and Stokes's law F_d=6\pi\eta rv_t.
mg=6\pi\eta rv_t
The mass comes from the density and the volume V=\dfrac43\pi r^{3}.
m=\rho V=\rho\left(\dfrac43\pi r^{3}\right)
Substitute for the mass.
\rho\left(\dfrac43\pi r^{3}\right)g=6\pi\eta r v_t
Solve for v_t and cancel \pi r.
v_t=\dfrac{\dfrac43\pi r^{3}\rho g}{6\pi\eta r}=\dfrac{4\rho g r^{2}}{18\eta}=\dfrac{2\rho g r^{2}}{9\eta}

Resultv_t=\dfrac{2\rho g r^{2}}{9\eta}

Exam focus

Bernoulli's equation: statement and derivation; Torricelli's theorem, v=\sqrt{2gh}, with the diagram. (long)
Q.4 (6 marks) in 2026 Set B (state Bernoulli, derive efflux, diagram) and Q.3 OR (b) in 2026 Set D (leaking tank, v=\sqrt{2gh}). Long questions 8 and 9 of the book and numericals 6.8 and 6.9.
Go to answer
Terminal velocity: state and derive v_t=\dfrac{2\rho gr^{2}}{9\eta} using Stokes' law. (long)
Q.4 (6 marks) in 2025 Set D and MCQ 7 in 2025 Set B; not in 2026. Long question 6 of the book.
Go to answer
Equation of continuity, A_1v_1=A_2v_2: statement, derivation, basis (conservation of mass) and numericals. (long)
Q.5 (7 marks) in 2025 Set B; MCQ 7 in 2026 Set D and 2025 Set D (conservation of mass), MCQ 6 in 2026 Set B (area to one-third) and Q.2 (xiii) in 2026 Set B (nozzle diameter). Long question 7 of the book.
Go to answer
Magnus effect: meaning, causes, effects, with the diagram. (short)
Q.2 (vii) OR in 2026 Set D, Q.2 (iii) OR in 2025 Set B and Q.2 (vii) in 2025 Set D: asked in three of the last four papers.
Go to answer
Super fluidity: meaning, how it is made, properties and applications. (short)
Q.2 (xii) OR in 2026 Set B and Q.2 (xiii) OR in 2025 Set D; long question 11 and short question 11 of the book.
Go to answer
Atomizer (perfume bottle): working on Bernoulli's principle, with the figure. (short)
Q.2 (vii) OR in 2026 Set B; short question 6 of the book.
Go to answer
Streamline and turbulent flow; the ideal fluid and its properties. (short)
Q.2 (iv) OR in 2026 Set B (distinguish the two flows), Q.2 (viii) OR in 2026 Set D (three properties of an ideal fluid) and MCQ 6 in 2025 Set D; short question 2 of the book.
Go to answer
Archimedes' principle and up-thrust, F_b=\rho Vg. (short)
Q.3 OR (a), 1 mark, in 2026 Set D; long questions 1 and 2 of the book.
Go to answer
Why does the stream from a tap narrow as it falls? (short)
Q.2 (ii) OR in 2025 Set D (not in 2026); MCQ 10 of the book.
Go to answer