NotesShala Class 11 · Physics

Chapter09

Mechanical Properties of Fluids

Hydrostatics · Surface Tension · Hydrodynamics and Viscosity

Reading key Definition Note Remember Derivation Strategy Formulas sit on graph paper · coloured words are the key terms

Part 1

Hydrostatics

Fluids give very little resistance to a shear stress. Even a very small shear stress changes their shape. This is the key property of fluids. In this part we study fluids at rest.

1.1Thrust

Take a fluid kept in a container. Its molecules are always in random motion. They move because of their thermal velocities.

So the molecules keep hitting the walls and bouncing back. In each hit, the momentum of a molecule changes in the direction normal to the wall. This normal momentum passes on to the wall.

The liquid pushes normally on the walls and the base

Definition

The thrust of a fluid is the normal momentum that its molecules give to the container walls in unit time.

1.2Liquid in Equilibrium

Concept

For a liquid in equilibrium, the net force on it always acts perpendicular to its surface. The proof is given below.

Setup — Given Information

A liquid is kept in a vessel. It is at rest.
Suppose the net force $F$ makes an angle $\theta$ with the direction parallel to the surface.
Split $F$ into two components, as in the figure.

F

has a part

F\cos\theta

along the surface and a part

F\sin\theta

normal to it

iTangential Component

$F\cos\theta$ acts along (parallel to) the liquid surface.

iiNormal Component

$F\sin\theta$ acts perpendicular to the liquid surface.

Derivation

The liquid is at rest, so it does not flow. Hence the part of the force along the surface must be zero:

$F\cos\theta = 0 \qquad \text{…(i)}$
$F \neq 0, \text{ so } \cos\theta = 0 \;\Rightarrow\; \theta = 90^\circ$

Result

When a liquid is in equilibrium at rest, the net force always acts perpendicular to the liquid surface.

1.3Pressure

Definition

The pressure at a point in a liquid at rest is the thrust (normal force) that the liquid exerts per unit area around that point.

Let a total force $F$ act normally on a flat area $A$ . Then the pressure is:

Pressure

p = \dfrac{\text{thrust}}{\text{area}} = \dfrac{F}{A}

where, $F$ = total normal force on the surface, $A$ = area on which it acts

CGS Unit

$\text{dyne cm}^{-2}$

SI Unit

$\text{N m}^{-2}$ (pascal)

Dimensions

$[\mathrm{M L^{-1} T^{-2}}]$

Good to know

One Pascal (1 Pa)

A pressure of $1\ \text{N m}^{-2}$ is called 1 pascal. The unit is named after the French scientist Blaise Pascal (1623–1662). He did the first major studies of pressure in fluids.

1\ \text{Pa} = 1\ \text{N m}^{-2} = 10\ \text{dyne cm}^{-2}

Atmosphere (atm)

The atmosphere (atm) is another common unit of pressure. 1 atm is the pressure of the air at sea level.

1\ \text{atm} = 1.013 \times 10^{5}\ \text{Pa}

Force Not Spread Evenly

If the force is not uniform, the pressure is different at different points. Let a small normal force $\Delta F$ act on a small area $\Delta A$ around a point. Then:

p = \displaystyle\lim_{\Delta A \to 0}\dfrac{\Delta F}{\Delta A} = \dfrac{dF}{dA}

Note

Pressure is a Scalar

At a given point in a liquid, the pressure has the same value in every direction. So no definite direction goes with pressure. Hence pressure is a scalar quantity.

Practical Applications of Pressure

For the same force, a smaller area gives a larger pressure.

iKnife

A sharp knife cuts more easily than a blunt knife. Its edge has a very small area, so the pressure is large.

iiCamel

A camel has broad feet. So it presses less on sand and walks easily. A man finds walking on sand difficult.

iiiRailway Tracks

Tracks rest on large wooden or iron sleepers. They spread the weight and reduce the pressure on the ground.

ivNeedle

A sharp needle goes into the skin easily, but a dull one does not. This happens even when we apply the same force.

Good to know

Ice-skating

The sharp metal edges of skates press hard on the ice. This pressure melts a little ice under them. The tiny water drops act like rigid ball bearings. So the skater glides smoothly.

1.4Density and Relative Density

Definition

The density of a substance is its mass per unit volume. It is written as $\rho$ .

Let a substance have mass $M$ and volume $V$ . Then its density is:

Density

\rho = \dfrac{M}{V}

where, $M$ = mass of the substance, $V$ = its volume

SI unit of density

\text{kg m}^{-3}

Dimensions of density

[\mathrm{M L^{-3}}]

Relative Density (Specific Gravity)

Definition

The relative density (also called specific gravity) of a substance is its density divided by the density of water at 4°C.

Relative Density

\text{Relative density} = \dfrac{\text{Density of substance}}{\text{Density of water at } 4^\circ\text{C}}

Unit and Nature

It is a ratio of two densities. So it has neither a unit nor dimensions. It is a positive scalar.

Water at 4°C

Water is densest at 4°C. Its density then is $1000\ \text{kg m}^{-3}$ .

1.5Pascal's Law

Law

Pascal's law: In a fluid at rest, the pressure is the same at all points that are at the same height.

This was first noticed by the French scientist Blaise Pascal.

Proof of Pascal's Law

The proof uses two ideas:

iNormal Force

In a fluid at rest, the force on any layer is normal to it.

iiNewton's First Law

The small element of fluid is in equilibrium. So the forces on it balance.

Setup — Given Information

Take a tiny element ABC-DEF inside a fluid at rest. It has the shape of a right-angled prism.
The prism is so small that all its parts are at the same depth. So gravity has the same effect at all its points.
Pressures $p_a$ , $p_b$ , $p_c$ act on the faces BEFC, ADFC and ADEB.
The normal forces on these faces are $F_a$ , $F_b$ , $F_c$ . Their areas are $A_a$ , $A_b$ , $A_c$ .

Tiny prism

ABC\text{-}DEF

in a fluid at rest (left) and the two components of

F_b

(right)

Derivation

In the right triangle ABC, $\angle ACB = \theta$ . Resolve the normal forces along the horizontal and the vertical:

$F_b\sin\theta = F_c \quad\text{and}\quad F_b\cos\theta = F_a \qquad \text{…(i)}$
From the geometry of the prism, the face areas are related as:
$A_b\sin\theta = A_c \quad\text{and}\quad A_b\cos\theta = A_a \qquad \text{…(ii)}$
Divide each force equation in (i) by the matching area equation in (ii):
$\dfrac{F_b\sin\theta}{A_b\sin\theta} = \dfrac{F_c}{A_c} \quad\text{and}\quad \dfrac{F_b\cos\theta}{A_b\cos\theta} = \dfrac{F_a}{A_a}$
$\therefore\ \dfrac{F_a}{A_a} = \dfrac{F_b}{A_b} = \dfrac{F_c}{A_c}$

Result of Pascal's Law

p_a = p_b = p_c

since $p = \dfrac{F}{A}$ , all three faces feel the same pressure

Note

So a fluid at rest presses on a body inside it equally in all directions. No direction can be given to this pressure. Hence pressure is not a vector quantity.

1.6Variation of Pressure with Depth

Setup — Given Information

A fluid of density $\rho$ is at rest in a cylindrical vessel.
Two points A and B are a vertical distance $h$ apart.
Imagine a cylinder of fluid with cross-section area $A$ . Point A is on its top face and point B on its bottom face.

Imaginary fluid cylinder between A (top) and B (bottom)

Derivation

The weight of the fluid cylinder acts downward:
$w = mg = \text{volume} \times \text{density} \times g = Ah\rho g \qquad [\because V = A \times h]$
The fluid is at rest. The horizontal forces cancel. So the total downward force equals the total upward force:
$F_1 + w = F_2 \quad\Rightarrow\quad F_2 - F_1 = w$
Here $F_1$ acts downward on the top face and $F_2$ acts upward on the bottom face.
Write each force as pressure × area:
$p_2 A - p_1 A = Ah\rho g \quad\Rightarrow\quad A(p_2 - p_1) = A(h\rho g)$

Pressure Difference with Depth

p_2 - p_1 = h\rho g

Note

This pressure difference depends on three things: the vertical height between A and B, the density $\rho$ of the fluid, and the value of $g$ .

Now move point A up to the free surface, which is open to the air. Put $p_1 = p_a$ (atmospheric pressure) and $p_2 = p$ :

Pressure at a Depth h

p = p_a + h\rho g

At a depth $h$ below an open liquid surface, the pressure is more than the atmospheric pressure by $h\rho g$ .

Definition

The gauge pressure at a point is the total pressure there minus the atmospheric pressure.

p - p_a = h\rho g

1.7Hydrostatic Paradox

Vessels A, B and C have different shapes. They are joined at the bottom. Each one holds a different amount of liquid. Yet the liquid level is the same in all three. This is called the hydrostatic paradox.

Vessels A, B and C hold different amounts of liquid, but the level is the same

Concept

Hydrostatic Paradox

Liquid pressure depends only on how tall the liquid column is. The shape of the vessel does not matter.

Liquid pressure and the height of the liquid columnDepends on it

Liquid pressure and the shape of the vesselDoes not depend

1.8Atmospheric Pressure and Gauge Pressure

Definition

The layer of air around the Earth is called the Earth's atmosphere. The pressure exerted by this air is called the atmospheric pressure ( $p_a$ ). It equals the force of the air column on each unit area of the Earth's surface.

Atmospheric pressure at sea level = 1 atmosphere (1 atm)

\approx 1.013 \times 10^{5}\ \text{N m}^{-2}

The atmosphere exerts a very large pressure. The methods below show this.

Mercury Barometer

The Italian scientist Evangelista Torricelli (1608–1647) made the first instrument to measure atmospheric pressure. It is called a barometer.
Fill a long glass tube (one end closed) with mercury. Then turn it upside down into a trough of mercury.
The mercury in the tube settles about 76 cm higher than the mercury in the trough.

Mercury barometer: A and B are at the same level

Note

Above the mercury column, the tube holds only mercury vapour. Its pressure is very small, so we neglect it.

Points A and B are at the same level, so their pressures are equal. So the atmospheric pressure equals the pressure of the mercury column of height $h$ :

Atmospheric Pressure from a Barometer

p_a = h\rho g

where, $h$ = height of the mercury column, $\rho$ = density of mercury

Example

Put $h = 0.76$ m, $\rho = 13.6 \times 10^{3}\ \text{kg m}^{-3}$ and $g = 9.8\ \text{m s}^{-2}$ :

p_a = 0.76 \times 13.6 \times 10^{3} \times 9.8 = 1.013 \times 10^{5}\ \text{N m}^{-2} = 1.013 \times 10^{5}\ \text{Pa}

This value is the sea-level atmospheric pressure.

Pressure of a 1 mm column of mercury1 torr

Open Tube Manometer

We use an open tube manometer to measure pressure differences. The figure shows the set-up.

It is a U-shaped tube with some liquid in it.
One arm is open to the air. The other arm is joined to a vessel (flask) whose pressure we want to find.

Open tube manometer: A and B are at the same level

Low-density liquid (oil or water)Small pressure differences

High-density liquid (mercury)Large pressure differences

Note

Pressure Relation

Points A and B are at the same level in the same liquid. So by Pascal's law, the pressure $p$ at A equals the pressure at B:

p = p_A = p_B = p_a + h\rho g

Absolute Pressure and Gauge Pressure

Absolute Pressure

Absolute pressure is the full, actual pressure $p$ at a point.

Gauge Pressure

The gauge pressure is the absolute pressure at a point minus the atmospheric pressure.

Gauge Pressure

p_g = p - p_a = h\rho g

So the gauge pressure is proportional to $h$ .

Good to know

Many pressure gauges read the gauge pressure directly. Examples: a gauge that checks tyre pressure, and a blood pressure gauge (sphygmomanometer).

Height of the Atmosphere

Standard atmospheric pressure is $1.013 \times 10^{5}$ Pa. Suppose the air had a uniform density $\rho = 1.29\ \text{kg m}^{-3}$ all the way up. Let $h$ be the height of the air column that gives this pressure. Then:

h\rho g = 1.013 \times 10^{5} \;\Rightarrow\; h = \dfrac{1.013 \times 10^{5}}{\rho g} = \dfrac{1.013 \times 10^{5}}{1.29 \times 9.8}\ \text{m}

Height of the Atmosphere (uniform density)

h \approx 8.01 \times 10^{3}\ \text{m} \approx 8\ \text{km}

Note

In reality, the density of air falls with height. The value of $g$ also falls. In fact, the atmosphere reaches a height of nearly 100 km.

Different Units of Pressure

iAtmosphere (atm)

It equals the pressure of a mercury column 76 cm high, at 0°C, at 45° latitude and at mean sea level.

iiBar and Millibar

Used in meteorology (weather science).

1\ \text{bar} = 10^{5}\ \text{Pa} = 10^{6}\ \text{dyne cm}^{-2}

1\ \text{mbar} = 10^{-3}\ \text{bar} = 100\ \text{Pa}

iiiTorr

Named after Torricelli.

1\ \text{torr} = 1\ \text{mm of Hg} = 133\ \text{Pa}

Unit Relation

1\ \text{atm} = 1.013\ \text{bar} = 760\ \text{torr}

Note

If the atmospheric pressure falls by 10 mm of Hg or more, a storm is likely on its way.
Gauge pressure can be positive ( $p > p_a$ ) or negative ( $p < p_a$ ).
In inflated tyres and in the human circulatory system, the absolute pressure is more than the atmospheric pressure. So the gauge pressure is positive. It is called over pressure.
When we suck a drink through a straw, the pressure in our lungs is less than the atmospheric pressure. So the gauge pressure there is below zero.

Tyres and blood vessels (

p > p_a

)Positive gauge pressure

Drinking with a straw (

p < p_a

)Negative gauge pressure

1.9Pascal's Law for Transmission of Fluid Pressure

Law

Pascal's law (transmission of pressure): Apply an external pressure to any part of a fluid that is enclosed in a vessel and at rest. The fluid passes this pressure on to every part of it. The pressure reaches every point without any loss and acts equally in all directions.

Pressure applied to one part of a confined fluidPassed equally in all directions

Note

This law has many uses in daily life. Some of them are given in the next topic.

1.10Applications of Pascal's Law

(i) Hydraulic Lift

Concept

A hydraulic lift is used to raise heavy loads. It is based on Pascal's law and works as a force multiplier.

Setup — Given Information

Two cylinders $C_1$ and $C_2$ are joined by a pipe E.
Each has a tight, frictionless piston. The piston areas are different.
A liquid fills the cylinders and the pipe.

Hydraulic lift:

f

on area

a

F

on area

A

Derivation

Push the smaller piston (area $a$ ) down with a force $f$ . The pressure on the liquid is:
$p = \dfrac{f}{a}$
By Pascal's law, this pressure reaches the larger piston (area $A$ ). So the upward force on it is:
$F = p \times A = \dfrac{f}{a} \times A = \dfrac{A}{a} \times f$

Upward Force on the Larger Piston

F = \dfrac{A}{a}\,f

Since

A > a

, we get

F > f

. A small force gives a large forceHeavy load lifted

(ii) Hydraulic Brakes

Hydraulic brakes also work on Pascal's law.

Construction

A tube T is filled with brake oil.
One end of T goes to the wheel cylinder. It has two pistons, $P_1$ and $P_2$ . These pistons are joined to the brake shoes $S_1$ and $S_2$ .
The other end of T goes to the master cylinder. A lever system links its piston P to the brake pedal.

Hydraulic brakes: master cylinder (

P

), tube

T

, wheel cylinder (

P_1

P_2

) and brake shoes (

S_1

S_2

)

Cross-section area: wheel cylinder compared with master cylinderWheel cylinder is larger

Working of Hydraulic Brakes

Pedal pressed: When the driver presses the brake pedal, the lever system pushes piston P. By Pascal's law, the pressure reaches pistons $P_1$ and $P_2$ .
Shoes pushed out: Pistons $P_1$ and $P_2$ push the brake shoes apart. The shoes press on the inner rim of the wheel and slow down the wheel.
Force multiplied: The wheel-cylinder pistons have a larger area than piston P. So a small force on the pedal gives a large thrust on the wheels. This makes the brakes work.
Pedal released: When the driver releases the pedal, a spring pulls the shoes back from the rim to their normal place. This pushes the oil back into the master cylinder.
All wheels: The master cylinder is joined by tubes to all the wheels. So equal pressure reaches every wheel.
Result: So a small force on the pedal puts a far larger force on the drums of the wheels. This helps the driver keep the vehicle under control.

1.11Buoyancy

Definition

When a body is partly or fully immersed in a fluid at rest, it pushes aside (displaces) some fluid. The fluid then pushes the body upward.

This upward force on the body is called the upthrust or buoyant force. The effect is called buoyancy.

Where the Weight Acts

The weight acts at the centre of gravity of the body.

Where the Upthrust Acts

The buoyant force acts through the centre of buoyancy. This point is the centre of gravity of the displaced liquid.

1.12Archimedes' Principle

Principle

Archimedes' principle gives the size of the buoyant force. A body fully or partly immersed in a liquid at rest feels an upthrust. The upthrust equals the weight of the liquid that the immersed part of the body pushes aside (displaces). This force passes through the centre of gravity of the fluid displaced.

Proof of Archimedes' Principle

Setup — Given Information

A cylinder of height $h$ and area $A$ is immersed in a fluid.
Upward force $F_2$ on its bottom is larger than downward force $F_1$ on its top.
$p_1$ , $p_2$ = pressures on the top and bottom faces.

Forces on an immersed body

Derivation

The net upward force of the fluid on the body is:
$F_2 - F_1 = (p_2 - p_1)A$
From the pressure difference with depth:
$p_2 - p_1 = h\rho g$
So the net upward force is:
$(p_2 - p_1)A = h\rho g A = mg \qquad [\because m = A\rho h]$
Here $\rho$ = density of the fluid and $A$ = cross-section area of the body. So $m$ is the mass of the fluid displaced.

Result

The fluid pushes the body up with a force equal to the weight of the fluid it displaces, that is, $mg$ .

Laws of Floatation

A body placed in a fluid can behave in three ways. Let $W_b$ be its weight and $F_B$ the upthrust when it is fully immersed.

iWeight > Upthrust

$W_b > F_B$ : the body sinks.

iiWeight = Upthrust

$W_b = F_B$ : the weight is just balanced. The body floats fully immersed.

iiiWeight < Upthrust

$W_b < F_B$ : the body floats partly immersed.

Law

Law of Floatation

Let a floating body have total volume $V_b$ . Let a part $V_p$ of it be under the fluid. At equilibrium, the weight of the body equals the weight of the fluid it displaces:

V_b\,\rho_b\,g = V_p\,\rho_f\,g

Law of Floatation

\dfrac{\rho_b}{\rho_f} = \dfrac{V_p}{V_b}

where, $\rho_b$ = density of the body, $\rho_f$ = density of the fluid

Part 2

Surface Tension

2.1Surface Tension

Concept

A liquid at rest has some extra potential energy at its free surface. Because of this, the surface tries to keep its area as small as possible. So the surface acts like a stretched membrane.

e.g. Steel is denser than water. Still, a steel needle can be made to float on water. The water surface acts like a stretched elastic sheet and holds the needle up. This property is called surface tension.

Imagine a line AB drawn on a liquid surface. The surface on each side pulls this line. The two pulls are equal and opposite. They act perpendicular to the line. So the line stays in equilibrium.

The pull is proportional to the length of the line. Let $l$ be its length and $F$ the force on either side. Then:

F \propto l \;\Rightarrow\; F = Sl \;\Rightarrow\; S = \dfrac{F}{l}

Equal and opposite pulls

F

on the line

AB

Definition

Surface tension ( $S$ ) is the force per unit length on an imaginary line drawn on the free surface of a liquid. The force acts at right angles to the line and lies along the surface (it is tangential). Surface tension is a scalar quantity.

Surface Tension

S = \dfrac{F}{l} = \dfrac{\text{Force}}{\text{Length}}

SI Unit

$\text{N m}^{-1}$

CGS Unit

$\text{dyne cm}^{-1}$

Dimensions

$\dfrac{[\mathrm{MLT^{-2}}]}{[\mathrm{L}]} = [\mathrm{ML^{0}T^{-2}}]$

Factors Affecting Surface Tension

1 · Temperature

Surface tension falls as the temperature rises. It becomes zero at a certain temperature. This is called the critical temperature of the liquid.

For small changes in temperature, it falls almost linearly: $\;S_t = S_0(1 - \alpha t)$

Here $S_t$ = surface tension at $t^\circ$ C, $S_0$ = surface tension at $0^\circ$ C and $\alpha$ = temperature coefficient of surface tension.

Soup tastes nicer when hot than when cold. Hot soup has a lower surface tension, so it spreads over more of the tongue.

2 · Impurities

Adding impurities changes surface tension a lot.

Sodium chloride dissolves easily (it is highly soluble). Adding it raises the surface tension of water.
Phenol or soap dissolves only a little (sparingly soluble). Adding it lowers the surface tension of water.

Detergents and Surface Tension

Detergents used to wash dirty clothes show a common use of surface tension.
Oily dirt on clothes does not get wet by plain water. So water by itself cannot remove the dirt.
A detergent molecule sticks to both water and dirt molecules. So washing carries the dirt away.

Applications of Surface Tension

iRound Drops

Rain drops, and drops of mercury on a glass plate, are spherical.

iiBrush Hair

The hair of a shaving or painting brush spread out in water. They cling together as soon as the brush is taken out.

iiiFloating Needle

A greased needle, laid gently on water, stays afloat.

ivOil Drops

An oil drop spreads out on cold water. On hot water, it keeps its shape.

Measuring the Surface Tension of a Liquid

A thin, flat glass plate is held vertical. It hangs from one arm of a balance. Weights on the other arm balance it. A vessel of the liquid is kept below the plate.

The vessel is raised until the liquid just touches the plate. Surface tension then pulls the plate down. Now weights are added until the plate just leaves the liquid. Let this extra weight be $mg$ . The liquid touches both faces of the plate edge, so the total length is $2l$ .

A glass plate pulled down by surface tension, balanced by weights

Surface Tension (Plate Method)

S = \dfrac{mg}{2l}

where, $m$ = extra mass needed, $l$ = length of the plate edge

2.2Surface Energy

Definition

A liquid surface always tries to shrink to the smallest possible area. To increase this area at constant temperature, work must be done. This work is stored in the surface film as potential energy.

The surface energy is this potential energy stored per unit area of the surface film. In other words, it is the work needed to increase the surface area by one unit.

Surface Energy

\text{Surface energy} = \dfrac{\text{Work done in increasing the surface area}}{\text{Increase in surface area}}

SI unit of surface energy: $\text{J m}^{-2}$

How Surface Energy Relates to Surface Tension

Setup — Given Information

Take a rectangular wire frame PQRS. Its side QR can slide.
Dip the frame in soap solution to form a soap film on it. Surface tension pulls the wire QR inward.
QR is pulled outward through a small distance against this force. This stretches the film.

Derivation

Surface tension is force per unit length. A soap film has two surfaces, so the total length on QR is $2l$ :
$S = \dfrac{F'}{2l} \quad\Rightarrow\quad F' = S \times 2l$
An outside force $F$ (equal to $F'$ ) moves QR very slowly through a distance $x$ . The work done is:
$W = \text{Force} \times \text{distance} = S \times 2l \times x$
The film has two sides, so its area grows by $2l \times x$ . Hence:
$\text{Surface energy} = \dfrac{\text{Work done}}{\text{Surface area}} = \dfrac{S \cdot 2lx}{2lx} = S$

Soap film on frame

PQRS

: (a) in equilibrium, (b) after

QR

is pulled out by

x

Surface Energy and Surface Tension

\text{Surface energy} = S

Surface energy and surface tension of a liquid have the same numerical value.

2.3Angle of Contact

Definition

Near the place where a liquid touches another medium, the liquid surface is curved. The angle of contact ( $\theta$ ) is the angle that the tangent to the liquid surface, drawn at the point of contact, makes with the solid surface inside the liquid.

The angle of contact depends on these factors:

iMaterials

The nature of the solid and the liquid in contact.

iiCleanliness

How clean the surfaces in contact are.

iiiMedium

The medium (like air) above the liquid surface.

ivTemperature

How hot the liquid is (its temperature).

e.g. Water forms drops on a lotus leaf. But it spreads out on a clean plastic plate.

Three surfaces meet at the line of contact. Each pair has its own surface tension:

$S_{sa}$ : surface tension between the solid and air
$S_{la}$ : surface tension between the liquid and air
$S_{sl}$ : surface tension between the solid and liquid

Drops with (a) an obtuse and (b) an acute angle of contact

\theta

At the line of contact, the three surface forces are in equilibrium. Split $S_{la}$ into two parts: $S_{la}\cos\theta$ along the solid surface and $S_{la}\sin\theta$ perpendicular to it. For balance along the solid, we get:

Angle of Contact

S_{sa} = S_{sl} + S_{la}\cos\theta \quad\Rightarrow\quad \cos\theta = \dfrac{S_{sa} - S_{sl}}{S_{la}}

So there are two cases:

S_{sl} > S_{sa}

$\cos\theta$ is negative, so $\theta > 90^\circ$ (obtuse). Such liquids do not wet the solid.

e.g. Mercury on glass: $\theta > 90^\circ$ , non-wetting.

S_{sl} < S_{sa}

$\cos\theta$ is positive, so $\theta < 90^\circ$ (acute). Such liquids wet the solid.

e.g. Water on glass: $\theta < 90^\circ$ , wetting.

2.4Excess Pressure Inside Drops and Bubbles

Excess Pressure: Liquid Drop

Setup — Given Information

Take a liquid drop of radius $R$ . It is a sphere. Its surface tension is $S$ .
Because it is curved, the pressure inside is more than outside. This excess pressure $p$ acts normally outward.
$p = p_i - p_o$ , where $p_i$ = pressure inside and $p_o$ = atmospheric pressure outside.
Let this pressure increase the radius from $R$ to $R + dR$ .

Liquid drop: excess pressure

p = p_i - p_o

Derivation

First surface area $= 4\pi R^2$ . When the radius becomes $R + dR$ , the new area is:
$4\pi(R+dR)^2 = 4\pi R^2 + 8\pi R\,dR \qquad [dR^2 \text{ neglected}]$
So the area grows by $8\pi R\,dR$ . Work done = gain in surface energy = (gain in area) × surface tension:
$W = (8\pi R\,dR)\,S \qquad \text{…(i)}$
This work is also done by the excess pressure. Work = excess pressure × area × change in radius:
$W = p \times 4\pi R^2 \times dR \qquad \text{…(ii)}$
Equate (i) and (ii):
$p \times 4\pi R^2 \times dR = 8\pi R\,dR\,S$

Excess Pressure Inside a Liquid Drop

p_i - p_o = \dfrac{2S}{R}

Excess Pressure: Soap Bubble

A soap bubble is a thin film. It has two free surfaces, one inside and one outside. So the real gain in surface area is doubled:

\text{Total gain in area} = 2 \times 8\pi R\,dR = 16\pi R\,dR

Equate the work done to the gain in surface energy:

p \times 4\pi R^2 \times dR = 16\pi R\,dR\,S

Soap bubble: a thin film with two surfaces

Excess Pressure Inside a Soap Bubble

p_i - p_o = \dfrac{4S}{R}

Air Bubble Inside a Liquid

Inside a liquid, an air bubble behaves like a liquid drop in air. It has just one free surface. So the extra pressure inside it is:

Excess Pressure Inside an Air Bubble

p = \dfrac{2S}{R}

Note

Let an air bubble of radius $R$ be at a depth $h$ below the free surface of a liquid. The liquid has density $\rho$ and surface tension $S$ . Then the pressure inside the bubble is more than the atmospheric pressure by:

p = \dfrac{2S}{R} + h\rho g

2.5Capillarity

Definition

Capilla is a Latin word. It means hair. A tube with a very fine bore, as thin as a hair, is called a capillary tube.

Dip a glass capillary in water: the water rises in it. Dip it in mercury: the mercury level falls in it. This rise or fall of a liquid in a capillary tube is called capillarity.

Water rises in glass tubes (concave surface). Mercury falls (convex surface)

Some Examples

We use towels to dry our skin.
In trees, sap rises through fine vessels. This is like capillary action.

Capillary Rise

(a) Water rises by

h

in a capillary. (b) Concave surface: radius of curvature

r

, tube radius

R

, angle of contact

\theta

Setup — Given Information

A pressure difference exists across a curved liquid surface. One result of this is that water rises in a narrow tube (capillary) against gravity.
A capillary of radius $R$ is dipped in a vessel of water.
The water surface in the capillary becomes concave. So there is a pressure difference across the two sides of this surface (meniscus).

Derivation

Let $r$ be the radius of curvature of the concave surface. From figure (b), $r = R\sec\theta$ . The pressure $p_a$ on the air side is more than the pressure $p_0$ just below the surface:
$(p_a - p_0) = \dfrac{2S}{r} = \dfrac{2S}{R\sec\theta} = \dfrac{2S}{R}\cos\theta \qquad \text{…(i)}$
Points A and B are at the same level. By Pascal's law, they have the same pressure. So $p_0 + \rho g h = p_a$ , which gives:
$p_a - p_0 = \rho g h \qquad \text{…(ii)} \qquad [p_a = \text{atmospheric pressure}]$
From equations (i) and (ii):
$\rho g h = \dfrac{2S}{R}\cos\theta$

Height of Liquid Rise in a Capillary

h = \dfrac{2S\cos\theta}{\rho R g}

Liquids that wet glass (like water) rise in the capillary. Liquids that do not wet glass fall in it.

Blotting paper soaks up inkCapillarity

A towel soaks up waterCapillary action

Part 3

Hydrodynamics and Viscosity

The study of moving fluids is called hydrodynamics or fluid dynamics.

3.1Streamline, Laminar and Turbulent Flow

Streamline Flow

Definition

In streamline flow, every particle that reaches a given point moves exactly like the particle that passed there before it. It follows the same path and has the same velocity.

Definition

A streamline is a curve drawn in the flow. Its tangent at any point shows the direction of the fluid's velocity there.

A streamline is the path of a fluid particle in steady flow.

Velocity is along the tangent at every point

Properties of Streamlines

iThey Never Cross

Two streamlines can never cross. At a crossing point, a particle would have two directions of motion at once. Then the flow would not be steady.

iiCrowding Means Speed

Where the streamlines come closer together, the liquid moves faster. Where they are spread out, it moves slower.

A region of streamline flow

Laminar Flow

Sometimes a liquid flows over a horizontal surface as layers moving with different velocities. Such a flow is called laminar flow. No particle leaves its own layer to move into another. Laminar flow is usually a streamline flow.

Laminar flow: layers slide with different speeds

Turbulent Flow

In rivers and canals, the water may move very fast. The boundary surfaces may also cause sudden changes in velocity. Then the flow becomes irregular and disordered. Such a flow is called turbulent flow.

Turbulent flow: eddies form behind a pillar

3.2Equation of Continuity

Law

Equation of continuity: Let a non-viscous, incompressible fluid flow in a streamline way through a pipe whose cross-section changes. Then the product of the cross-section area and the fluid velocity ( $av$ ) is the same at every point of the pipe.

Setup — Given Information

A liquid of density $\rho$ flows through a tube AB. The liquid is non-viscous and incompressible. The cross-section of AB changes along its length.
At A: area $a_1$ , velocity $v_1$ .
At B: area $a_2$ , velocity $v_2$ .

Fluid flow through pipe

AB

Derivation

In time $\Delta t$ , the mass of fluid entering at A and the mass leaving at B are:
$m_1 = a_1 v_1 \Delta t\,\rho \quad\text{and}\quad m_2 = a_2 v_2 \Delta t\,\rho$
By conservation of mass, $m_1 = m_2$ . The fluid is incompressible, so $\rho$ stays the same:
$a_1 v_1 \Delta t\,\rho = a_2 v_2 \Delta t\,\rho \;\Rightarrow\; a_1 v_1 = a_2 v_2$

Equation of Continuity

a_1 v_1 = a_2 v_2 = av = \text{constant}

In a narrow part of the tube the fluid speeds up. In a wide part it slows down.

3.3Bernoulli's Theorem

Principle

Bernoulli's theorem: Take an incompressible, non-viscous fluid in streamline, irrotational flow. The total of its pressure energy, kinetic energy and potential energy, each per unit volume, is the same at every cross-section along the flow.

The theorem comes from the law of conservation of energy.

Bernoulli's Equation

p + \dfrac{1}{2}\rho v^2 + \rho g h = \text{constant}

$p$ = pressure energy, $\tfrac{1}{2}\rho v^2$ = kinetic energy, $\rho g h$ = potential energy, all per unit volume ( $\rho$ = density). Daniel Bernoulli gave it in 1738.

Setup — Given Information

The fluid is ideal, that is, non-viscous and incompressible. It has streamline flow through a pipe whose cross-section changes.
At A: pressure $p_1$ , area $a_1$ , velocity $v_1$ , height $h_1$ .
At B: pressure $p_2$ , area $a_2$ , velocity $v_2$ , height $h_2$ .

An ideal fluid flowing through a pipe whose cross-section changes

Derivation

The pressure forces do net work on the fluid. Using $a_1 v_1 = a_2 v_2$ , this work is:
$W = \left(\dfrac{p_1}{\rho} - \dfrac{p_2}{\rho}\right)m$
This work increases the kinetic and potential energy of the fluid. The work-energy theorem then gives:
$\dfrac{p_1}{\rho} + \dfrac{1}{2}v_1^2 + g h_1 = \dfrac{p_2}{\rho} + \dfrac{1}{2}v_2^2 + g h_2$

Per Unit Volume

p + \dfrac{1}{2}\rho v^2 + \rho g h = \text{constant}

In Terms of Heads (÷ ρg)

\dfrac{p}{\rho g} + \dfrac{v^2}{2g} + h = \text{constant}

Here $\dfrac{p}{\rho g}$ is the pressure head, $\dfrac{v^2}{2g}$ is the velocity head and $h$ is the gravitational head.

Note

Bernoulli's Equation for a Fluid at Rest

For a fluid at rest, $v_1 = v_2 = 0$ . So the velocity terms drop out:

p_1 - p_2 = \rho g (h_2 - h_1)

Limitations of Bernoulli's Theorem

Zero viscosity: it strictly holds only for non-viscous fluids.
Incompressible: the fluid must be incompressible. Its elastic energy is not counted.
Streamline only: it is not valid when the flow is turbulent or unsteady.

3.4Applications of Bernoulli's Theorem

Torricelli's Law: Speed of Efflux

The speed of efflux is the speed of a liquid coming out of a small hole (orifice) in a tank. The orifice is at depth $h$ below the free surface. Bernoulli's equation for the surface and the orifice gives:

Torricelli's Law

v = \sqrt{2gh}

Same speed as a body falling freely through height $h$ .

Liquid coming out of an orifice at depth

h

Venturimeter

A venturimeter (venturi tube) measures the flow speed of an incompressible fluid in a pipe.

This horizontal tube has a wide part (area $a_1$ ) and a narrow neck (area $a_2$ ). A manometer joins them. Continuity and Bernoulli's equation link the speed to the pressure difference:

p_1 - p_2 = \dfrac{1}{2}\rho v_1^2\left(\dfrac{a_1^2}{a_2^2} - 1\right)

Venturimeter with a manometer

Atomizer or Sprayer

The atomizer is based on Bernoulli's principle. Pressing the bulb (the forward stroke of the piston) sends a fast stream of air past the top of the tube.

So the pressure there ( $p_2$ ) is low. Atmospheric pressure ( $p_1$ ) on the liquid pushes it up the tube. The liquid meets the fast air and breaks into a fine spray.

Atomizer (sprayer)

Aerofoil (Aircraft Wing) Lift

The wing of an aircraft has a special shape, called an aerofoil. Air flows faster over its top than along its bottom.

By Bernoulli's principle, the pressure above the wing is lower than below it. This pressure difference pushes the wing up. The upward force is called dynamic lift.

Air flow around a wing gives an upward lift

Spinning and Non-spinning Ball

Ball Without Spin

The air moves past the ball with the same speed above and below it. So there is no pressure difference. The ball feels no force up or down.

Same flow above and below: no lift

Ball With Spin

A spinning ball drags air with it. So the air speed is higher on one side of the ball. The resulting pressure difference bends its path into a curve. This is called the Magnus effect.

Crowded streamlines above: net force upward

3.5Viscosity

Definition

Let one layer of a liquid move over the next layer. A tangential friction force then acts between them. This backward dragging force opposes their relative motion.

Viscosity is the property of a fluid that opposes the relative motion between its layers. It is also called fluid friction.

e.g. Machine parts get jammed in winter, because the lubricating oil becomes more viscous (thicker) in the cold.

Coefficient of Viscosity

Take a layer of fluid that is being deformed by a shearing stress. The coefficient of viscosity $\eta$ is the shearing stress divided by the rate of change of strain.

\eta = \dfrac{\text{shearing stress}}{\text{rate of change of strain}}

A fluid layer deformed by a shearing force

Coefficient of Viscosity

\eta = \dfrac{F/A}{v/l} = \dfrac{Fl}{vA} = \dfrac{F}{A\,\dfrac{dv}{dx}}

where, $F$ = tangential force, $A$ = area of the layer, $\dfrac{dv}{dx}$ = velocity gradient. Dimensions: $[\mathrm{ML^{-1}T^{-1}}]$

CGS Unit: Poise

1\ \text{poise} = 1\ \text{dyne s cm}^{-2}

$\eta = 1$ poise when a tangential force of $1$ dyne per cm² of area keeps a relative velocity of $1\ \text{cm s}^{-1}$ between two layers $1$ cm apart.

SI Unit: Decapoise

1\ \text{decapoise} = 10\ \text{poise} = 1\ \text{N s m}^{-2}

The SI unit $\text{N s m}^{-2}$ (or Pa s) is also called the decapoise.

Fluid Friction Force (Viscous Force)

Between two fluid layers, the viscous force varies directly as the area $A$ of the layers. It also varies directly as the velocity gradient $\dfrac{dv}{dx}$ between them.

Viscous (Fluid Friction) Force

F = \eta A \dfrac{dv}{dx}

where, $\eta$ = coefficient of viscosity, $A$ = area of the layers, $\dfrac{dv}{dx}$ = velocity gradient. In vector form a minus sign is added. It shows that the force opposes the motion.

Velocity gradient

\dfrac{dv}{dx}

between two fluid layers

3.6Critical Velocity

Definition

Critical velocity is the limiting speed of flow of a liquid. Up to this speed, the flow stays streamline. Above it, the flow becomes turbulent.

The critical velocity $v_c$ of a liquid flowing in a tube depends on:

iViscosity

Coefficient of viscosity of the liquid $(\eta)$

iiDensity

Density of the liquid $(\rho)$

iiiRadius

Radius of the tube $(r)$

Write $v_c = k\,\eta^a \rho^b r^c$ , where $k$ is a dimensionless constant. Comparing dimensions on both sides gives $a = 1$ , $b = -1$ and $c = -1$ . So:

Critical Velocity

v_c = \dfrac{k\eta}{\rho r}

For streamline flow, $v_c$ should be large. So $\eta$ should be large, and $\rho$ and $r$ should be small.

Tends to be Streamline

Liquids with high viscosity and low density, in narrow pipes.

Tends to be Turbulent

Liquids with low viscosity and high density, in broad pipes.

3.7Reynolds Number

Definition

The Reynolds number is a dimensionless number whose value tells whether the flow of a liquid in a pipe is steady (streamline) or turbulent. It was given by Osborne Reynolds (1842–1912).

Reynolds Number

R_e = \dfrac{\rho v D}{\eta}

where, $\rho$ = density of the liquid, $v$ = its velocity, $\eta$ = its coefficient of viscosity, $D$ = diameter of the pipe

R_e < 1000

The flow is streamline (laminar).

1000

2000

The flow is unsteady. It may switch between laminar and turbulent.

R_e > 2000

The flow is turbulent.

Note

The Reynolds number at which turbulence just starts is called the critical Reynolds number. The Reynolds number compares two forces, each taken per unit area. It is the inertial force divided by the viscous force.

R_e = \dfrac{\text{Inertial force}}{\text{Force of viscosity}}

3.8Stokes' Law

Concept

Let a small sphere fall through a viscous fluid at rest. The fluid layers touching the sphere move along with it. The layers far away stay at rest. This relative motion produces a backward dragging (viscous) force that opposes the motion of the sphere.

Law

Stokes' law: A small sphere of radius $r$ moves with velocity $v$ through a fluid of viscosity $\eta$ . The backward dragging force on it is $F = 6\pi\eta r v$ . This law was given by Sir George G. Stokes (1819–1903).

Stokes' Law (Viscous Drag Force)

F = 6\pi\eta r v

Stokes worked out by calculation that the constant is $k = 6\pi$ .

So the viscous drag depends on three things: (i) the coefficient of viscosity $(\eta)$ of the fluid, (ii) the velocity $(v)$ of the body and (iii) the radius $(r)$ of the spherical body.

Good to know

Examples: a falling raindrop, a swinging pendulum bob.
Applications: finding the charge of an electron (Millikan's oil-drop experiment), how clouds form and separate, and how a parachute works.

3.9Terminal Velocity

Definition

The terminal velocity is the maximum constant velocity that a body gains while falling through a viscous fluid.

A raindrop first speeds up as it falls. Its viscous force grows with speed. Soon its weight = upthrust + viscous force. From then on it falls at a constant (terminal) velocity.

Setup — Given Information

Let a sphere (radius $r$ , density $\rho$ ) fall through a fluid of density $\sigma$ and viscosity $\eta$ . Its terminal velocity is $v$ .
Weight of the sphere (downward) $= \tfrac{4}{3}\pi r^3 \rho g$
Upthrust = weight of fluid displaced (upward) $= \tfrac{4}{3}\pi r^3 \sigma g$
Viscous force (upward, by Stokes' law) $= 6\pi\eta r v$

Derivation

At terminal velocity, weight = upthrust + viscous force:
$\tfrac{4}{3}\pi r^3 \rho g = \tfrac{4}{3}\pi r^3 \sigma g + 6\pi\eta r v$
Keep the viscous term on one side:
$6\pi\eta r v = \tfrac{4}{3}\pi r^3 (\rho - \sigma) g$

Terminal Velocity

v = \dfrac{2}{9}\cdot\dfrac{r^2 (\rho - \sigma) g}{\eta}

where, $r$ = radius of the body, $\rho$ = its density, $\sigma$ = density of the fluid, $\eta$ = coefficient of viscosity

3.10Poiseuille's Formula

Concept

Poiseuille's formula gives the volume of liquid that flows each second through a narrow horizontal tube (capillary tube). The tube has length $L$ and radius $r$ . A pressure difference $p$ is applied across its ends.

Poiseuille's Formula

V = \dfrac{Q}{t} = \dfrac{\pi p\, r^4}{8\eta L}

where, $p$ = pressure difference across the ends, $r$ = radius of the tube, $L$ = length of the tube, $\eta$ = coefficient of viscosity

Rate of flow

V

and radius of the tube

\left(V \propto r^4\right)

Depends very strongly

Rate of flow

V

and length

\left(V \propto \tfrac{1}{L}\right)

and viscosity

\left(V \propto \tfrac{1}{\eta}\right)

Inversely

End of Chapter 9 · notesshala.in