CHAPTER TEN
MECHANICAL PROPERTIES OF FLUIDS
10.1 INTRODUCTION
In this chapter, we shall study some common physical
properties of liquids and gases. Liquids and gases can flow
and are therefore, called fluids. It is this property that
distinguishes liquids and gases from solids in a basic way.
Fluids are everywhere around us. Earth has an envelop of
air and two-thirds of its surface is covered with water. Water
is not only necessary for our existence; every mammalian
body constitute mostly of water. All the processes occurring
in living beings including plants are mediated by fluids. Thus
understanding the behaviour and properties of fluids is
important.
How are fluids different from solids? What is common in
liquids and gases? Unlike a solid, a fluid has no definite
shape of its own. Solids and liquids have a fixed volume,
whereas a gas fills the entire volume of its container. We
have learnt in the previous chapter that the volume of solids
can be changed by stress. The volume of solid, liquid or gas
depends on the stress or pressure acting on it. When we
talk about fixed volume of solid or liquid, we mean its volume
under atmospheric pressure. The difference between gases
and solids or liquids is that for solids or liquids the change
in volume due to change of external pressure is rather small.
In other words solids and liquids have much lower
compressibility as compared to gases.
Shear stress can change the shape of a solid keeping its
volume fixed. The key property of fluids is that they offer
very little resistance to shear stress; their shape changes by
application of very small shear stress. The shearing stress
of fluids is about million times smaller than that of solids.
10.2 PRESSURE
A sharp needle when pressed against our skin pierces it. Our
skin, however, remains intact when a blunt object with a
wider contact area (say the back of a spoon) is pressed against
it with the same force. If an elephant were to step on a man’s
chest, his ribs would crack. A circus performer across whose
10.1 Introduction
10.2 Pressure
10.3 Streamline flow
10.4 Bernoulli’s principle
10.5 Viscosity
10.6 Surface tension
Summary
Points to ponder
Exercises
Additional exercises
Appendix
2020-21
chest a large, light but strong wooden plank is
placed first, is saved from this accident. Such
everyday experiences convince us that both the
force and its coverage area are important. Smaller
the area on which the force acts, greater is the
impact. This impact is known as pressure.
When an object is submerged in a fluid at
rest, the fluid exerts a force on its surface. This
force is always normal to the object’s surface.
This is so because if there were a component of
for
ce parallel to the surface, the object will also
exert a force on the fluid parallel to it; as a
consequence of Newton’s third law. This force
will cause the fluid to flow parallel to the surface.
Since the fluid is at rest, this cannot happen.
Hence, the force exerted by the fluid at rest has
to be perpendicular to the surface in contact
with it. This is shown in Fig.10.1(a).
The normal force exerted by the fluid at a point
may be measured. An idealised form of one such
pressure-measuring device is shown in Fig.
10.1(b). It consists of an evacuated chamber with
a spring that is calibrated to measure the force
acting on the piston. This device is placed at a
point inside the fluid. The inward force exerted
by the fluid on the piston is balanced by the
outward spring force and is thereby measured.
If F is the magnitude of this normal force on the
piston of area A then the average pressure P
av
is defined as the normal force acting per unit
area.
P
F
A
av
=
(10.1)
In principle, the piston area can be made
arbitrarily small. The pressure is then defined
in a limiting sense as
P =
lim
∆
A 0
→
∆
∆
F
A
(10.2)
Pressure is a scalar quantity. We remind the
reader that it is the component of the force
normal to the area under consideration and not
the (vector) force that appears in the numerator
in Eqs. (10.1) and (10.2). Its dimensions are
[ML
–1
T
–2
]. The SI unit of pressure is N m
–2
. It has
been named as pascal (Pa) in honour of the
French scientist Blaise Pascal (1623-1662) who
carried out pioneering studies on fluid pressure.
A common unit of pressure is the atmosphere
(atm), i.e. the pressure exerted by the
atmosphere at sea level (1 atm = 1.013 × 10
5
Pa).
Another quantity, that is indispensable in
describing fluids, is the density
ρ
. For a fluid of
mass m occupying volume V,
ρ
=
m
V
(10.3)
The dimensions of density are [ML
–3
]. Its SI
unit is kg m
–3
. It is a positive scalar quantity. A
liquid is largely incompressible and its density
is therefore, nearly constant at all pressures.
Gases, on the other hand exhibit a large
variation in densities with pressure.
The density of water at 4
o
C (277 K) is
1.0 × 10
3
kg m
–3
. The relative density of a
substance is the ratio of its density to the
density of water at 4
o
C. It is a dimensionless
positive scalar quantity. For example the relative
density of aluminium is 2.7. Its density is
2.7 × 10
3
kg m
–3
.
The densities of some common
fluids are displayed in Table 10.1.
Table 10.1 Densities of some common fluids
at STP*
(a) (b)
Fig. 10.1 (a) The force exerted by the liquid in the
beaker on the submerged object or on the
walls is normal (perpendicular) to the
surface at all points.
(b) An idealised device for measuring
pressure.
* STP means standard temperature (0
0
C) and 1 atm pressure.
MECHANICAL PROPERTIES OF FLUIDS 251
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252 PHYSICS
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Example 10.1 The two thigh bones
(femurs), each of cross-sectional area10 cm
2
support the upper part of a human body of
mass 40 kg. Estimate the average pressure
sustained by the femurs.
Answer Total cross-sectional area of the
femurs is A = 2 × 10 cm
2
= 20 × 10
–4
m
2
. The
force acting on them is F = 40 kg wt = 400 N
(taking g = 10 m s
–2
). This force is acting
vertically down and hence, normally on the
femurs. Thus, the average pressure is
25
m N 10 2
−
×==
A
F
P
av
t
10.2.1 Pascal’s Law
The French scientist Blaise Pascal observed that
the pressure in a fluid at rest is the same at all
points if they are at the same height. This fact
may be demonstrated in a simple way.
Fig. 10.2 shows an element in the interior of
a fluid at rest. This element ABC-DEF is in the
form of a right-angled prism. In principle, this
prismatic element is very small so that every
part of it can be considered at the same depth
from the liquid surface and therefore, the effect
of the gravity is the same at all these points.
But for clarity we have enlarged this element.
The forces on this element are those exerted by
the rest of the fluid and they must be normal to
the surfaces of the element as discussed above.
Thus, the fluid exerts pressures P
a
, P
b
and P
c
on
this element of area corresponding to the normal
forces F
a
, F
b
and F
c
as shown in Fig. 10.2 on the
faces BEFC, ADFC and ADEB denoted by A
a
, A
b
and A
c
respectively. Then
F
b
sinθ = F
c
, F
b
cosθ = F
a
(by equilibrium)
A
b
sinθ = A
c
, A
b
cosθ = A
a
(by geometry)
Thus,
;
b c a
b c a
b c a
F F F
P P P
A A A
= = = =
(10.4)
Hence, pressure exerted is same in all
directions in a fluid at rest. It again reminds us
that like other types of stress, pressure is not a
vector quantity. No direction can be assigned
to it. The force against any area within (or
bounding) a fluid at rest and under pressure is
normal to the area, regardless of the orientation
of the area.
Now consider a fluid element in the form of a
horizontal bar of uniform cross-section. The bar
is in equilibrium. The horizontal forces exerted
at its two ends must be balanced or the
pressure at the two ends should be equal. This
proves that for a liquid in equilibrium the
pressure is same at all points in a horizontal
plane. Suppose the pressure were not equal in
different parts of the fluid, then there would be
a flow as the fluid will have some net force
acting on it. Hence in the absence of flow the
pressure in the fluid must be same everywhere
in a horizontal plane.
10.2.2 Variation of Pressure with Depth
Consider a fluid at rest in a container. In
Fig. 10.3 point 1 is at height h above a point 2.
The pressures at points 1 and 2 are P
1
and P
2
respectively. Consider a cylindrical element of
fluid having area of base A and height h. As the
fluid is at rest the resultant horizontal forces
should be zero and the resultant vertical forces
should balance the weight of the element. The
forces acting in the vertical direction are due to
the fluid pressure at the top (P
1
A) acting
downward, at the bottom (P
2
A) acting upward.
If mg is weight of the fluid in the cylinder we
have
(P
2
− P
1
) A = mg (10.5)
Now, if ρ is the mass density of the fluid, we
have the mass of fluid to be m = ρV= ρhA so
that
P
2
−
P
1
= ρgh (10.6)
Fig. 10.2 Proof of Pascal’s law. ABC-DEF is an
element of the interior of a fluid at rest.
This element is in the form of a right-
angled prism. The element is small so that
the effect of gravity can be ignored, but it
has been enlarged for the sake of clarity.
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MECHANICAL PROPERTIES OF FLUIDS 253
t
Fig.10.3 Fluid under gravity. The effect of gravity is
illustrated through pressure on a vertical
cylindrical column.
Pressure difference depends on the vertical
distance h between the points (1 and 2), mass
density of the fluid ρ and acceleration due to
gravity g. If the point 1 under discussion is
shifted to the top of the fluid (say, water), which
is open to the atmosphere, P
1
may be replaced
by atmospheric pressure (P
a
) and we replace P
2
by P. Then Eq. (10.6) gives
P =
P
a
+ ρgh (10.7)
Thus, the pressure P, at depth below the
surface of a liquid open to the atmosphere is
greater than atmospheric pressure by an
amount ρgh. The excess of pressure, P −
P
a
, at
depth h is called a gauge pressure at that point.
The area of the cylinder is not appearing in
the expression of absolute pressure in Eq. (10.7).
Thus, the height of the fluid column is important
and not cross-sectional or base area or the shape
of the container. The liquid pressure is the same
at all points at the same horizontal level (same
depth). The result is appreciated through the
example of hydrostatic paradox. Consider three
vessels A, B and C [Fig.10.4] of different shapes.
They are connected at the bottom by a horizontal
pipe. On filling with water, the level in the three
vessels is the same, though they hold different
amounts of water. This is so because water at
the bottom has the same pressure below each
section of the vessel.
Fig 10.4 Illustration of hydrostatic paradox. The
three vessels A, B and C contain different
amounts of liquids, all upto the same
height.
Example 10.2 What is the pressure on a
swimmer 10 m below the surface of a lake?
Answer Here
h = 10 m and
ρ
= 1000 kg m
-3
. Take g = 10 m s
–2
From Eq. (10.7)
P =
P
a
+ ρgh
= 1.01 × 10
5
Pa + 1000 kg m
–3
× 10 m s
–2
× 10 m
= 2.01 × 10
5
Pa
≈ 2 atm
This is a 100% increase in pressure from
surface level. At a depth of 1 km, the increase
in pressure is 100 atm! Submarines are designed
to withstand such enormous pressures. t
10.2.3 Atmospheric Pressure and
Gauge Pressure
The pressure of the atmosphere at any point is
equal to the weight of a column of air of unit
cross-sectional area extending from that point
to the top of the atmosphere. At sea level, it is
1.013 × 10
5
Pa (1 atm). Italian scientist
Evangelista Torricelli (1608–1647) devised for
the first time a method for measuring
atmospheric pressure. A long glass tube closed
at one end and filled with mercury is inverted
into a trough of mercury as shown in Fig.10.5 (a).
This device is known as ‘mercury barometer’.
The space above the mercury column in the tube
contains only mercury vapour whose pressure
P is so small that it may be neglected. Thus,
the pressure at Point A=0. The pressure inside
the coloumn at Point B must be the same as the
pressure at Point C, which is atmospheric
pressure, P
a
.
P
a
= ρgh (10.8)
where ρ is the density of mercury and h is the
height of the mercury column in the tube.
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254 PHYSICS
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t
In the experiment it is found that the mercury
column in the barometer has a height of about
76 cm at sea level equivalent to one atmosphere
(1 atm). This can also be obtained using the
value of ρ in Eq. (10.8). A common way of stating
pressure is in terms of cm or mm of mercury
(Hg). A pressure equivalent of 1 mm is called a
torr (after Torricelli).
1 torr = 133 Pa.
The mm of Hg and torr are used in medicine
and physiology. In meteorology, a common unit
is the bar and millibar.
1 bar = 10
5
Pa
An open tube manometer is a useful
instrument for measuring pressure differences.
It consists of a U-tube containing a suitable
liquid i.e., a low density liquid (such as oil) for
measuring small pressure differences and a
high density liquid (such as mercury) for large
pressure differences. One end of the tube is open
to the atmosphere and the other end is
connected to the system whose pressure we want
to measure [see Fig. 10.5 (b)]. The pressure P at
A is equal to pressure at point B. What we
normally measure is the gauge pressure, which
is P − P
a
, given by Eq. (10.8) and is proportional
to manometer height h.
Pressure is same at the same level on both
sides of the U-tube containing a fluid. For
liquids, the density varies very little over wide
ranges in pressure and temperature and we can
treat it safely as a constant for our present
purposes. Gases on the other hand, exhibits
large variations of densities with changes in
pressure and temperature. Unlike gases, liquids
are, therefore, largely treated as incompressible.
Example 10.3 The density of the
atmosphere at sea level is 1.29 kg/m
3
.
Assume that it does not change with
altitude. Then how high would the
atmosphere extend?
Answer We use Eq. (10.7)
ρgh = 1.29 kg m
–3
× 9.8 m s
2
× h m = 1.01 × 10
5
Pa
∴ h = 7989 m ≈ 8 km
In reality the density of air decreases with
height. So does the value of g. The atmospheric
cover extends with decreasing pressure over
100 km. We should also note that the sea level
atmospheric pressure is not always 760 mm of
Hg. A drop in the Hg level by 10 mm or more is a
sign of an approaching storm. t
Example 10.4 At a depth of 1000 m in an
ocean (a) what is the absolute pressure?
(b) What is the gauge pressure? (c) Find
the force acting on the window of area
20 cm × 20 cm of a submarine at this depth,
the interior of which is maintained at sea-
level atmospheric pressure. (The density of
sea water is 1.03 × 10
3
kg m
-3
,
g = 10 m s
–2
.)
(b) The open tube manometer
Fig 10.5 Two pressure measuring devices.
Fig 10.5 (a) The mercury barometer.
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MECHANICAL PROPERTIES OF FLUIDS 255
Answer Here h = 1000 m and ρ = 1.03 × 10
3
kg m
-3
.
(a) From Eq. (10.6), absolute pressure
P =
P
a
+ ρgh
= 1.01 × 10
5
Pa
+ 1.03 × 10
3
kg m
–3
× 10 m s
–2
× 1000 m
= 104.01 × 10
5
Pa
≈ 104 atm
(b) Gauge pressure is P −
P
a
= ρgh = P
g
P
g
= 1.03 × 10
3
kg m
–3
× 10 ms
2
× 1000 m
= 103 × 10
5
Pa
≈ 103 atm
(c) The pressure outside the submarine is
P =
P
a
+ ρgh and the pressure inside it is P
a
.
Hence, the net pressure acting on the
window is gauge pressure, P
g
= ρgh. Since
the area of the window is A = 0.04 m
2
, the
force acting on it is
F = P
g
A = 103 × 10
5
Pa × 0.04 m
2
= 4.12 × 10
5
N
t
10.2.4 Hydraulic Machines
Let us now consider what happens when we
change the pressure on a fluid contained in a
vessel. Consider a horizontal cylinder with a
piston and three vertical tubes at different
points [Fig. 10.6 (a)]. The pressure in the
horizontal cylinder is indicated by the height of
liquid column in the vertical tubes. It is necessarily
the same in all. If we push the piston, the fluid level
rises in all the tubes, again reaching the same level
in each one of them.
This indicates that when the pressure on the
cylinder was increased, it was distributed
uniformly throughout. We can say whenever
external pressure is applied on any part of a
fluid contained in a vessel, it is transmitted
undiminished and equally in all directions.
This is another form of the Pascal’s law and it
has many applications in daily life.
A number of devices, such as hydraulic lift
and hydraulic brakes, are based on the Pascal’s
law. In these devices, fluids are used for
transmitting pressure. In a hydraulic lift, as
shown in Fig. 10.6 (b), two pistons are separated
by the space filled with a liquid. A piston of small
cross-section A
1
is used to exert a force F
1
directly
on the liquid. The pressure P =
1
1
F
A
is
transmitted throughout the liquid to the larger
cylinder attached with a larger piston of area A
2
,
which results in an upward force of P × A
2
.
Therefore, the piston is capable of supporting a
large force (large weight of, say a car, or a truck,
Archemedes’ Principle
Fluid appears to provide partial support to the objects placed in it. When a body is wholly or partially
immersed in a fluid at rest, the fluid exerts pressure on the surface of the body in contact with the
fluid. The pressure is greater on lower surfaces of the body than on the upper surfaces as pressure in
a fluid increases with depth. The resultant of all the forces is an upward force called buoyant force.
Suppose that a cylindrical body is immersed in the fluid. The upward force on the bottom of the body
is more than the downward force on its top. The fluid exerts a resultant upward force or buoyant force
on the body equal to (P
2
– P
1
)
××
××
×
A (Fig. 10.3). We have seen in equation 10.4 that (P
2
-P
1
)A = ρghA. Now,
hA is the volume of the solid and ρhA is the weight of an equivaliant volume of the fluid. (P
2
-P
1
)A = mg.
Thus, the upward force exerted is equal to the weight of the displaced fluid.
The result holds true irrespective of the shape of the object and here cylindrical object is considered
only for convenience. This is Archimedes’ principle. For totally immersed objects the volume of the
fluid displaced by the object is equal to its own volume. If the density of the immersed object is more
than that of the fluid, the object will sink as the weight of the body is more than the upward thrust. If
the density of the object is less than that of the fluid, it floats in the fluid partially submerged. To
calculate the volume submerged, suppose the total volume of the object is V
s
and a part V
p
of it is
submerged in the fluid. Then, the upward force which is the weight of the displaced fluid is ρ
f
gV
p
,
which must equal the weight of the body; ρ
s
gV
s
= ρ
f
gV
p
or ρ
s
/ρ
f
= V
p
/V
s
The apparent weight of the
floating body is zero.
This principle can be summarised as; ‘the loss of weight of a body submerged (partially or fully) in
a fluid is equal to the weight of the fluid displaced’.
Fig 10.6 (a) Whenever external pressure is applied
on any part of a fluid in a vessel, it is
equally transmitted in all directions.
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256 PHYSICS
t
t
placed on the platform) F
2
= PA
2
=
1 2
1
F A
A
. By
changing the force at A
1
, the platform can be
moved up or down. Thus, the applied force has
been increased by a factor of
2
1
A
A
and this factor
is the mechanical advantage of the device. The
example below clarifies it.
Fig 10.6 (b) Schematic diagram illustrating the principle
behind the hydraulic lift, a device used to
lift heavy loads.
Example 10.5 Two syringes of different
cross-sections (without needles) filled with
water are connected with a tightly fitted
rubber tube filled with water. Diameters of
the smaller piston and larger piston are 1.0
cm and 3.0 cm respectively. (a) Find the
force exerted on the larger piston when a
force of 10 N is applied to the smaller piston.
(b) If the smaller piston is pushed in through
6.0 cm, how much does the larger piston
move out?
Answer (a) Since pressure is transmitted
undiminished throughout the fluid,
(
)
( )
2
–2
2
2 1
2
–2
1
3 /2 10 m
10 N
1/2 10 m
A
F F
A
π
π
×
= = ×
×
= 90 N
(b) Water is considered to be perfectly
incompressible. Volume covered by the
movement of smaller piston inwards is equal to
volume moved outwards due to the larger piston.
2211
ALAL
=
j 0.67 × 10
-2
m = 0.67 cm
Note, atmospheric pressure is common to both
pistons and has been ignored. t
Example 10.6 In a car lift compressed air
exerts a force F
1
on a small piston having
a radius of 5.0 cm. This pressure is
transmitted to a second piston of radius
15 cm (Fig 10.7). If the mass of the car to
be lifted is 1350 kg, calculate F
1
. What is
the pressure necessary to accomplish this
task? (g = 9.8 ms
-2
).
Answer Since pressure is transmitted
undiminished throughout the fluid,
= 1470 N
≈ 1.5 × 10
3
N
The air pressure that will produce this
force is
This is almost double the atmospheric
pressure. t
Hydraulic brakes in automobiles also work on
the same principle. When we apply a little force
on the pedal with our foot the master piston
Archimedes was a Greek philosopher, mathematician, scientist and engineer. He
invented the catapult and devised a system of pulleys and levers to handle heavy
loads. The king of his native city Syracuse, Hiero II, asked him to determine if his gold
crown was alloyed with some cheaper metal, such as silver without damaging the crown.
The partial loss of weight he experienced while lying in his bathtub suggested a solution
to him. According to legend, he ran naked through the streets of Syracuse, exclaiming “Eureka,
eureka!”, which means “I have found it, I have found it!”
Archimedes (287–212 B.C.)
2020-21
MECHANICAL PROPERTIES OF FLUIDS 257
moves inside the master cylinder, and the
pressure caused is transmitted through the
brake oil to act on a piston of larger area. A large
force acts on the piston and is pushed down
expanding the brake shoes against brake lining.
In this way, a small force on the pedal produces
a large retarding force on the wheel. An
important advantage of the system is that the
pressure set up by pressing pedal is transmitted
equally to all cylinders attached to the four
wheels so that the braking effort is equal on
all wheels.
10.3 STREAMLINE FLOW
So far we have studied fluids at rest. The study
of the fluids in motion is known as fluid
dynamics. When a water tap is turned on slowly,
the water flow is smooth initially, but loses its
smoothness when the speed of the outflow is
increased. In studying the motion of fluids, we
focus our attention on what is happening to
various fluid particles at a particular point in
space at a particular time. The flow of the fluid
is said to be steady if at any given point, the
velocity of each passing fluid particle remains
constant in time. This does not mean that the
velocity at different points in space is same. The
velocity of a particular particle may change as it
moves from one point to another. That is, at some
other point the particle may have a different
velocity, but every other particle which passes
the second point behaves exactly as the previous
particle that has just passed that point. Each
particle follows a smooth path, and the paths of
the particles do not cross each other.
Fig. 10.7 The meaning of streamlines. (a) A typical
trajectory of a fluid particle.
(b) A region of streamline flow.
The path taken by a fluid particle under a
steady flow is a streamline. It is defined as a
curve whose tangent at any point is in the
direction of the fluid velocity at that point.
Consider the path of a particle as shown in
Fig.10.7 (a), the curve describes how a fluid
particle moves with time. The curve PQ is like a
permanent map of fluid flow, indicating how the
fluid streams. No two streamlines can cross, for
if they do, an oncoming fluid particle can go
either one way or the other and the flow would
not be steady. Hence, in steady flow, the map of
flow is stationary in time. How do we draw closely
spaced streamlines ? If we intend to show
streamline of every flowing particle, we would
end up with a continuum of lines. Consider planes
perpendicular to the direction of fluid flow e.g.,
at three points P, R and Q in Fig.10.7 (b). The
plane pieces are so chosen that their boundaries
be determined by the same set of streamlines.
This means that number of fluid particles
crossing the surfaces as indicated at P, R and Q
is the same. If area of cross-sections at these
points are A
P
,A
R
and A
Q
and speeds of fluid
particles are v
P
, v
R
and v
Q
, then mass of fluid
∆m
P
crossing at A
P
in a small interval of time ∆t
is ρ
P
A
P
v
P
∆t. Similarly mass of fluid ∆m
R
flowing
or crossing at A
R
in a small interval of time ∆t is
ρ
R
A
R
v
R
∆t and mass of fluid ∆m
Q
is ρ
Q
A
Q
v
Q
∆t
crossing at A
Q
. The mass of liquid flowing out
equals the mass flowing in, holds in all cases.
Therefore,
ρ
P
A
P
v
P
∆t = ρ
R
A
R
v
R
∆t = ρ
Q
A
Q
v
Q
∆t (10.9)
For flow of incompressible fluids
ρ
P
= ρ
R
= ρ
Q
Equation (10.9) reduces to
A
P
v
P
= A
R
v
R
= A
Q
v
Q
(10.10)
which is called the equation of continuity and
it is a statement of conservation of mass in flow
of incompressible fluids. In general
Av = constant (10.11)
Av gives the volume flux or flow rate and
remains constant throughout the pipe of flow.
Thus, at narrower portions where the
streamlines are closely spaced, velocity
increases and its vice versa. From (Fig 10.7b) it
is clear that A
R
> A
Q
or v
R
< v
Q
, the fluid is
accelerated while passing from R to Q. This is
associated with a change in pressure in fluid
flow in horizontal pipes.
Steady flow is achieved at low flow speeds.
Beyond a limiting value, called critical speed,
this flow loses steadiness and becomes
turbulent. One sees this when a fast flowing
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258 PHYSICS
stream encounters rocks, small foamy
whirlpool-like regions called ‘white water
rapids are formed.
Figure 10.8 displays streamlines for some
typical flows. For example, Fig. 10.8(a) describes
a laminar flow where the velocities at different
points in the fluid may have different
magnitudes but their directions are parallel.
Figure 10.8 (b) gives a sketch of turbulent flow.
Fig. 10.8 (a) Some streamlines for fluid flow.
(b) A jet of air striking a flat plate placed
perpendicular to it. This is an example
of turbulent flow.
10.4 BERNOULLI’S PRINCIPLE
Fluid flow is a complex phenomenon. But we
can obtain some useful properties for steady
or streamline flows using the conservation
of energy.
Consider a fluid moving in a pipe of varying
cross-sectional area. Let the pipe be at varying
heights as shown in Fig. 10.9. We now suppose
that an incompressible fluid is flowing through
the pipe in a steady flow. Its velocity must
change as a consequence of equation of
continuity. A force is required to produce this
acceleration, which is caused by the fluid
surrounding it, the pressure must be different
in different regions. Bernoulli’s equation is a
general expression that relates the pressure
difference between two points in a pipe to both
velocity changes (kinetic energy change) and
elevation (height) changes (potential energy
change). The Swiss Physicist Daniel Bernoulli
developed this relationship in 1738.
Consider the flow at two regions 1 (i.e., BC)
and 2 (i.e., DE). Consider the fluid initially lying
between B and D. In an infinitesimal time
interval ∆t, this fluid would have moved. Suppose
v
1
is the speed at B and v
2
at D, then fluid initially
at B has moved a distance v
1
∆t to C (v
1
∆t is small
enough to assume constant cross-section along
BC). In the same interval ∆t the fluid initially at
D moves to E, a distance equal to v
2
∆t. Pressures
P
1
and P
2
act as shown on the plane faces of
areas A
1
and A
2
binding the two regions. The
work done on the fluid at left end (BC) is W
1
=
P
1
A
1
(v
1
∆t) = P
1
∆V. Since the same volume ∆V
passes through both the regions (from the
equation of continuity) the work done by the fluid
at the other end (DE) is W
2
= P
2
A
2
(v
2
∆t) = P
2
∆V or,
the work done on the fluid is –P
2
∆V. So the total
work done on the fluid is
W
1
– W
2
= (P
1
− P
2
) ∆V
Part of this work goes into changing the kinetic
energy of the fluid, and part goes into changing
the gravitational potential energy. If the density
of the fluid is ρ and ∆m = ρA
1
v
1
∆t = ρ∆V is the
mass passing through the pipe in time ∆t, then
change in gravitational potential energy is
∆U = ρg∆V (h
2
− h
1
)
The change in its kinetic energy is
∆K =
1
2
ρ ∆V (v
2
2
− v
1
2
)
We can employ the work – energy theorem
(Chapter 6) to this volume of the fluid and
this yields
(P
1
− P
2
) ∆V =
1
2
ρ ∆V (v
2
2
− v
1
2
) + ρg∆V (h
2
− h
1
)
We now divide each term by ∆V to obtain
(P
1
− P
2
) =
1
2
ρ (v
2
2
− v
1
2
) + ρg (h
2
− h
1
)
Daniel Bernoulli was a Swiss scientist and mathematician, who along with Leonard
Euler had the distinction of winning the French Academy prize for mathematics
10 times. He also studied medicine and served as a professor of anatomy and
botany for a while at Basle, Switzerland. His most well-known work was in
hydrodynamics, a subject he developed from a single principle: the conservation of
energy. His work included calculus, probability, the theory of vibrating strings,
and applied mathematics. He has been called the founder of mathematical physics.
Daniel Bernoulli (1700–1782)
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MECHANICAL PROPERTIES OF FLUIDS 259
We can rearrange the above terms to obtain
P
1
+
1
2
ρv
1
2
+ ρgh
1
= P
2
+
1
2
ρv
2
2
+ ρgh
2
(10.12)
This is Bernoulli’s equation. Since 1 and 2
refer to any two locations along the pipeline,
we may write the expression in general as
P +
1
2
ρv
2
+ ρgh = constant (10.13)
Fig. 10.9 The flow of an ideal fluid in a pipe of
varying cross section. The fluid in a
section of length v
1
∆
t moves to the section
of length v
2
∆
t in time
∆
t.
In words, the Bernoulli’s relation may be
stated as follows: As we move along a streamline
the sum of the pressure (P), the kinetic energy
per unit volume
ρv
2
2
and the potential energy
per unit volume (ρgh) remains a constant.
Note that in applying the energy conservation
principle, there is an assumption that no energy
is lost due to friction. But in fact, when fluids
flow, some energy does get lost due to internal
friction. This arises due to the fact that in a
fluid flow, the different layers of the fluid flow
with different velocities. These layers exert
frictional forces on each other resulting in a loss
of energy. This property of the fluid is called
viscosity and is discussed in more detail in a
later section. The lost kinetic energy of the fluid
gets converted into heat energy. Thus,
Bernoulli’s equation ideally applies to fluids with
zero viscosity or non-viscous fluids. Another
restriction on application of Bernoulli theorem
is that the fluids must be incompressible, as
the elastic energy of the fluid is also not taken
into consideration. In practice, it has a large
number of useful applications and can help
explain a wide variety of phenomena for low
viscosity incompressible fluids. Bernoulli’s
equation also does not hold for non-steady or
turbulent flows, because in that situation
velocity and pressure are constantly fluctuating
in time.
When a fluid is at rest i.e., its velocity is zero
everywhere, Bernoulli’s equation becomes
P
1
+ ρgh
1
= P
2
+ ρgh
2
(P
1
− P
2
) = ρg (h
2
− h
1
)
which is same as Eq. (10.6).
10.4.1 Speed of Efflux: Torricelli’s Law
The word efflux means fluid outflow. Torricelli
discovered that the speed of efflux from an open
tank is given by a formula identical to that of a
freely falling body. Consider a tank containing
a liquid of density ρ with a small hole in its side
at a height y
1
from the bottom (see Fig. 10.10).
The air above the liquid, whose surface is at
height y
2
, is at pressure P. From the equation
of continuity [Eq. (10.10)] we have
v
1
A
1
= v
2
A
2
v
A
A
v
2
1
2
=
1
Fig. 10.10 Torricelli’s law. The speed of efflux, v
1
,
from the side of the container is given by
the application of Bernoulli’s equation.
If the container is open at the top to the
atmosphere then
1
2 h
v g
=
.
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260 PHYSICS
If the cross-sectional area of the tank A
2
is
much larger than that of the hole (A
2
>>A
1
), then
we may take the fluid to be approximately at rest
at the top, i.e., v
2
= 0. Now, applying the Bernoulli
equation at points 1 and 2 and noting that at
the hole P
1
= P
a
, the atmospheric pressure, we
have from Eq. (10.12)
2
1 1 2
1
2
a
P v g y P g y
+ ρ + ρ = + ρ
Taking y
2
– y
1
= h we have
( )
2
2
a
1
P P
v g h
−
= +
ρ
(10.14)
When P >>P
a
and 2 g h may be ignored, the
speed of efflux is determined by the container
pressure. Such a situation occurs in rocket
propulsion. On the other hand, if the tank is
open to the atmosphere, then P = P
a
and
hgv
2
1
=
(10.15)
This is also the speed of a freely falling body.
Equation (10.15) represents Torricelli’s law.
10.4.2 Venturi-meter
The Venturi-meter is a device to measure the
flow speed of incompressible fluid. It consists of
a tube with a broad diameter and a small
constriction at the middle as shown in
Fig. (10.11). A manometer in the form of a
U-tube is also attached to it, with one arm at
the broad neck point of the tube and the other
at constriction as shown in Fig. (10.11). The
manometer contains a liquid of density ρ
m
. The
speed v
1
of the liquid flowing through the tube
at the broad neck area A is to be measured
from equation of continuity Eq. (10.10) the speed
at the constriction becomes
2 1
v v
=
A
a
. Then
using Bernoulli’s equation (Eq.10.12) for (h
1
=h
2
),
we get
P
1
+
1
2
ρv
1
2
= P
2
+
1
2
ρv
1
2
(A/a)
2
So that
P
1
- P
2
=
1
2
ρv
1
2
A
a
2
1–
(10.16)
This pressure difference causes the fluid in
the U-tube connected at the narrow neck to rise
in comparison to the other arm. The difference
in height h measure the pressure difference.
P
1
– P
2
= ρ
m
gh =
1
2
ρv
1
2
2
– 1
A
a
So that the speed of fluid at wide neck is
v
1
=
–½
2
2
– 1
m
gh
A
a
ρ
ρ
(10.17)
The principle behind this meter has many
applications. The carburetor of automobile has
a Venturi channel (nozzle) through which air
flows with a high speed. The pressure is then
lowered at the narrow neck and the petrol
(gasoline) is sucked up in the chamber to provide
the correct mixture of air to fuel necessary for
combustion. Filter pumps or aspirators, Bunsen
burner, atomisers and sprayers [See Fig. 10.12]
used for perfumes or to spray insecticides work
on the same principle.
Fig. 10.12 The spray gun. Piston forces air at high
speeds causing a lowering of pressure
at the neck of the container.
h
A
a
2
1
Fig. 10.11 A schematic diagram of Venturi-meter.
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MECHANICAL PROPERTIES OF FLUIDS 261
t
Example 10.7 Blood velocity: The flow of
blood in a large artery of an anesthetised
dog is diverted through a Venturi meter.
The wider part of the meter has a cross-
sectional area equal to that of the artery.
A = 8 mm
2
. The narrower part has an area
a = 4 mm
2
. The pressure drop in the artery
is 24 Pa. What is the speed of the blood in
the artery?
Answer We take the density of blood from Table
10.1 to be 1.06 × 10
3
kg m
-3
. The ratio of the
areas is
A
a
= 2. Using Eq. (10.17) we obtain
t
10.4.3 Blood Flow and Heart Attack
Bernoulli’s principle helps in explaining blood
flow in artery. The artery may get constricted
due to the accumulation of plaque on its inner
walls. In order to drive the blood through this
constriction a greater demand is placed on the
activity of the heart. The speed of the flow of
the blood in this region is raised which lowers
the pressure inside and the artery may
collapse due to the external pressure. The
heart exerts further pressure to open this
artery and forces the blood through. As the
blood rushes through the opening, the
internal pressure once again drops due to
same reasons leading to a repeat collapse.
This may result in heart attack.
10.4.4 Dynamic Lift
Dynamic lift is the force that acts on a body,
such as airplane wing, a hydrofoil or a spinning
ball, by virtue of its motion through a fluid. In
many games such as cricket, tennis, baseball,
or golf, we notice that a spinning ball deviates
from its parabolic trajectory as it moves through
air. This deviation can be partly explained on
the basis of Bernoulli’s principle.
(i) Ball moving without spin: Fig. 10.13(a)
shows the streamlines around a
non-spinning ball moving relative to a fluid.
From the symmetry of streamlines it is clear
that the velocity of fluid (air) above and below
the ball at corresponding points is the same
resulting in zero pressure difference. The air
therefore, exerts no upward or downward
force on the ball.
(ii) Ball moving with spin: A ball which is
spinning drags air along with it. If the
surface is rough more air will be dragged.
Fig 10.13(b) shows the streamlines of air
for a ball which is moving and spinning at
the same time. The ball is moving forward
and relative to it the air is moving
backwards. Therefore, the velocity of air
above the ball relative to the ball is larger
and below it is smaller (see Section 10.3).
The stream lines, thus, get crowded above
and rarified below.
This difference in the velocities of air results
in the pressure difference between the lower and
upper faces and there is a net upward force on
the ball. This dynamic lift due to spining is called
Magnus effect.
(a) (b) (c)
Fig 10.13 (a) Fluid streaming past a static sphere. (b) Streamlines for a fluid around a sphere spinning clockwise.
(c) Air flowing past an aerofoil.
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262 PHYSICS
t
Aerofoil or lift on aircraft wing: Figure 10.13
(c) shows an aerofoil, which is a solid piece
shaped to provide an upward dynamic lift when
it moves horizontally through air. The cross-
section of the wings of an aeroplane looks
somewhat like the aerofoil shown in Fig. 10.13 (c)
with streamlines around it. When the aerofoil
moves against the wind, the orientation of the
wing relative to flow direction causes the
streamlines to crowd together above the wing
more than those below it. The flow speed on top
is higher than that below it. There is an upward
force resulting in a dynamic lift of the wings and
this balances the weight of the plane. The
following example illustrates this.
Example 10.8 A fully loaded Boeing
aircraft has a mass of 3.3 × 10
5
kg. Its total
wing area is 500 m
2
. It is in level flight with
a speed of 960 km/h. (a) Estimate the
pressure difference between the lower and
upper surfaces of the wings (b) Estimate
the fractional increase in the speed of the
air on the upper surface of the wing relative
to the lower surface. [The density of air is ρ
= 1.2 kg m
-3
]
Answer (a) The weight of the Boeing aircraft is
balanced by the upward force due to the
pressure difference
∆P × A = 3.3 × 10
5
kg × 9.8
P
∆
= (3.3 × 10
5
kg × 9.8 m s
–2
) / 500 m
2
= 6.5 × 10
3
Nm
-2
(b) We ignore the small height difference
between the top and bottom sides in Eq. (10.12).
The pressure difference between them is
then
∆P v v=
( )
ρ
2
2
2
1
2
–
where v
2
is the speed of air over the upper
surface and v
1
is the speed under the bottom
surface.
v v
P
v v
2 1
2 1
2
–
(
)
=
+
( )
∆
ρ
Taking the average speed
v
av
= (v
2
+ v
1
)/2 = 960 km/h = 267 m s
-1
,
we have
v v v
P
v
2 1
2
– /
(
)
=
av
av
∆
ρ
≈ 0.08
The speed above the wing needs to be only 8
% higher than that below. t
10.5 VISCOSITY
Most of the fluids are not ideal ones and offer some
resistance to motion. This resistance to fluid motion
is like an internal friction analogous to friction when
a solid moves on a surface. It is called viscosity.
This force exists when there is relative motion
between layers of the liquid. Suppose we consider
a fluid like oil enclosed between two glass plates
as shown in Fig. 10.14 (a). The bottom plate is fixed
while the top plate is moved with a constant
velocity v relative to the fixed plate. If oil is
replaced by honey, a greater force is required to
move the plate with the same velocity. Hence
we say that honey is more viscous than oil. The
fluid in contact with a surface has the same
velocity as that of the surfaces. Hence, the layer
of the liquid in contact with top surface moves
with a velocity v and the layer of the liquid in
contact with the fixed surface is stationary. The
velocities of layers increase uniformly from
bottom (zero velocity) to the top layer (velocity
v). For any layer of liquid, its upper layer pulls
it forward while lower layer pulls it backward.
This results in force between the layers. This
type of flow is known as laminar. The layers of
liquid slide over one another as the pages of a
book do when it is placed flat on a table and a
horizontal force is applied to the top cover. When
a fluid is flowing in a pipe or a tube, then velocity
of the liquid layer along the axis of the tube is
maximum and decreases gradually as we move
towards the walls where it becomes zero, Fig.
10.14 (b). The velocity on a cylindrical surface
in a tube is constant.
On account of this motion, a portion of liquid,
which at some instant has the shape ABCD, take
the shape of AEFD after short interval of time
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MECHANICAL PROPERTIES OF FLUIDS 263
t
(∆t). During this time interval the liquid has
undergone a shear strain of
∆x/l. Since, the strain in a flowing fluid
increases with time continuously. Unlike a solid,
here the stress is found experimentally to depend
on ‘rate of change of strain’ or ‘strain rate’ i.e.
∆x/(l ∆t) or v/l instead of strain itself. The
coefficient of viscosity (pronounced ‘eta’) for a
fluid is defined as the ratio of shearing stress to
the strain rate.
η
= =
F A
v l
F l
v A
/
/
(10.18)
The SI unit of viscosity is poiseiulle (Pl). Its
other units are N s m
-2
or Pa s. The dimensions
of viscosity are [ML
-1
T
-1
]. Generally, thin liquids,
like water, alcohol, etc., are less viscous than
thick liquids, like coal tar, blood, glycerine, etc.
The coefficients of viscosity for some common
fluids are listed in Table 10.2. We point out two
facts about blood and water that you may find
interesting. As Table 10.2 indicates, blood is
‘thicker’ (more viscous) than water. Further, the
relative viscosity (η/η
water
) of blood remains
constant between 0
o
C and 37
o
C.
The viscosity of liquids decreases with
temperature, while it increases in the case of gases.
Example 10.9 A metal block of area 0.10 m
2
is connected to a 0.010 kg mass via a string
that passes over an ideal pulley (considered
massless and frictionless), as in Fig. 10.15.
A liquid with a film thickness of 0.30 mm
is placed between the block and the table.
When released the block moves to the right
with a constant speed of 0.085 m s
-1
. Find
the coefficient of viscosity of the liquid.
Answer The metal block moves to the right
because of the tension in the string. The tension
T is equal in magnitude to the weight of the
suspended mass m. Thus, the shear force F is
F = T = mg = 0.010 kg × 9.8 m s
–2
= 9.8 × 10
-2
N
Shear stress on the fluid = F/A = N/m
2
Strain rate =
η
=
stress
strain rate
s
-1
=
= 3.46 × 10
-3
Pa s
t
(a)
Fig. 10.15 Measurement of the coefficient of viscosity
of a liquid.
(b)
Fig 10.14 (a) A layer of liquid sandwiched between
two parallel glass plates, in which the
lower plate is fixed and the upper one is
moving to the right with velocity v
(b) velocity distribution for viscous flow in
a pipe.
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264 PHYSICS
t
where ρ and σ are mass densities of sphere and
the fluid, respectively. We obtain
v
t
= 2a
2
(ρ-σ)g / (9
η
) (10.20)
So the terminal velocity v
t
depends on the
square of the radius of the sphere and inversely
on the viscosity of the medium.
You may like to refer back to Example 6.2 in
this context.
Example 10.10 The terminal velocity of a
copper ball of radius 2.0 mm falling through
a tank of oil at 20
o
C is 6.5 cm s
-1
. Compute
the viscosity of the oil at 20
o
C. Density of
oil is 1.5 ×10
3
kg m
-3
, density of copper is
8.9 × 10
3
kg m
-3
.
Answer We have v
t
= 6.5 × 10
-2
ms
-1
, a = 2 × 10
-3
m,
g = 9.8 ms
-2
, ρ = 8.9 × 10
3
kg m
-3
,
σ =1.5 ×10
3
kg m
-3
. From Eq. (10.20)
= 9.9 × 10
-1
kg m
–1
s
–1
t
10.6 SURFACE TENSION
You must have noticed that, oil and water do
not mix; water wets you and me but not ducks;
mercury does not wet glass but water sticks to
it, oil rises up a cotton wick, inspite of gravity,
Sap and water rise up to the top of the leaves of
the tree, hair of a paint brush do not cling
together when dry and even when dipped in
water but form a fine tip when taken out of it.
All these and many more such experiences are
related with the free surfaces of liquids. As
liquids have no definite shape but have a
definite volume, they acquire a free surface when
poured in a container. These surfaces possess
some additional energy. This phenomenon is
known as surface tension and it is concerned
with only liquid as gases do not have free
surfaces. Let us now understand this
phenomena.
Table10.2 The viscosities of some fluids
Fluid T(
o
C) Viscosity (mPl)
Water 20 1.0
100 0.3
Blood 37 2.7
Machine Oil 16 113
38 34
Glycerine 20 830
Honey – 200
Air 0 0.017
40 0.019
10.5.1 Stokes’ Law
When a body falls through a fluid it drags the
layer of the fluid in contact with it. A relative
motion between the different layers of the fluid
is set and, as a result, the body experiences a
retarding force. Falling of a raindrop and
swinging of a pendulum bob are some common
examples of such motion. It is seen that the
viscous force is proportional to the velocity of
the object and is opposite to the direction of
motion. The other quantities on which the force
F depends are viscosity η of the fluid and radius
a of the sphere. Sir George G. Stokes (1819–
1903), an English scientist enunciated clearly
the viscous drag force F as
6
F av
η
= π
(10.19)
This is known as Stokes’ law. We shall not
derive Stokes’ law.
This law is an interesting example of retarding
force, which is proportional to velocity. We can
study its consequences on an object falling
through a viscous medium. We consider a
raindrop in air. It accelerates initially due to
gravity. As the velocity increases, the retarding
force also increases. Finally, when viscous force
plus buoyant force becomes equal to the force
due to gravity, the net force becomes zero and so
does the acceleration. The sphere (raindrop) then
descends with a constant velocity. Thus, in
equilibrium, this terminal velocity v
t
is given by
6π
η
av
t
= (4π/3) a
3
(ρ-σ)g
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MECHANICAL PROPERTIES OF FLUIDS 265
Fig. 10.16 Schematic picture of molecules in a liquid, at the surface and balance of forces. (a) Molecule inside
a liquid. Forces on a molecule due to others are shown. Direction of arrows indicates attraction of
repulsion. (b) Same, for a molecule at a surface. (c) Balance of attractive (AI and repulsive (R) forces.
terms of this fact. What is the energy required
for having a molecule at the surface? As
mentioned above, roughly it is half the energy
required to remove it entirely from the liquid
i.e., half the heat of evaporation.
Finally, what is a surface? Since a liquid
consists of molecules moving about, there cannot
be a perfectly sharp surface. The density of the
liquid molecules drops rapidly to zero around
z = 0 as we move along the direction indicated
Fig 10.16 (c) in a distance of the order of a few
molecular sizes.
and to disperse them far away from each other
in order to evaporate or vaporise, the heat of
evaporation required is quite large. For water it
is of the order of 40 kJ/mol.
Let us consider a molecule near the surface
Fig. 10.16(b). Only lower half side of it is
surrounded by liquid molecules. There is some
negative potential energy due to these, but
obviously it is less than that of a molecule in
bulk, i.e., the one fully inside. Approximately
it is half of the latter. Thus, molecules on a
liquid surface have some extra energy in
comparison to molecules in the interior. A
liquid, thus, tends to have the least surface
area which external conditions permit.
Increasing surface area requires energy. Most
surface phenomenon can be understood in
10.6.2 Surface Energy and Surface Tension
As we have discussed that an extra energy is
associated with surface of liquids, the creation
of more surface (spreading of surface) keeping
other things like volume fixed requires
additional energy. To appreciate this, consider
a horizontal liquid film ending in bar free to slide
over parallel guides Fig (10.17).
10.6.1 Surface Energy
A liquid stays together because of attraction
between molecules. Consider a molecule well
inside a liquid. The intermolecular distances are
such that it is attracted to all the surrounding
molecules [Fig. 10.16(a)]. This attraction results
in a negative potential energy for the molecule,
which depends on the number and distribution
of molecules around the chosen one. But the
average potential energy of all the molecules is
the same. This is supported by the fact that to
take a collection of such molecules (the liquid)
Fig. 10.17 Stretching a film. (a) A film in equilibrium;
(b) The film stretched an extra distance.
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266 PHYSICS
Suppose that we move the bar by a small
distance d as shown. Since the area of the
surface increases, the system now has more
energy, this means that some work has been
done against an internal force. Let this internal
force be F, the work done by the applied force is
F
.
d = Fd. From conservation of energy, this is
stored as additional energy in the film. If the
surface energy of the film is S per unit area, the
extra area is 2dl. A film has two sides and the
liquid in between, so there are two surfaces and
the extra energy is
S (2dl) = Fd (10.21)
Or, S=Fd/2dl = F/2l (10.22)
This quantity S is the magnitude of surface
tension. It is equal to the surface energy per unit
area of the liquid interface and is also equal to
the force per unit length exerted by the fluid on
the movable bar.
So far we have talked about the surface of one
liquid. More generally, we need to consider fluid
surface in contact with other fluids or solid
surfaces. The surface energy in that case depends
on the materials on both sides of the surface. For
example, if the molecules of the materials attract
each other, surface energy is reduced while if they
repel each other the surface energy is increased.
Thus, more appropriately, the surface energy is
the energy of the interface between two materials
and depends on both of them.
We make the following observations from
above:
(i) Surface tension is a force per unit length
(or surface energy per unit area) acting in
the plane of the interface between the plane
of the liquid and any other substance; it also
is the extra energy that the molecules at the
interface have as compared to molecules in
the interior.
(ii) At any point on the interface besides the
boundary, we can draw a line and imagine
equal and opposite surface tension forces S
per unit length of the line acting
perpendicular to the line, in the plane of the
interface. The line is in equilibrium. To be
more specific, imagine a line of atoms or
molecules at the surface. The atoms to the
left pull the line towards them; those to the
right pull it towards them! This line of atoms
is in equilibrium under tension. If the line
really marks the end of the interface, as in
Figure 10.16 (a) and (b) there is only the force
S per unit length acting inwards.
Table 10.3 gives the surface tension of various
liquids. The value of surface tension depends
on temperature. Like viscosity, the surface
tension of a liquid usually falls with
temperature.
Table 10.3 Surface tension of some liquids at the
temperatures indicated with the
heats of the vaporisation
Liquid Temp (
o
C) Surface Heat of
Tension vaporisation
(N/m) (kJ/mol)
Helium –270 0.000239 0.115
Oxygen –183 0.0132 7.1
Ethanol 20 0.0227 40.6
Water 20 0.0727 44.16
Mercury 20 0.4355 63.2
A fluid will stick to a solid surface if the
surface energy between fluid and the solid is
smaller than the sum of surface energies
between solid-air, and fluid-air. Now there is
attraction between the solid surface and the
liquid. It can be directly measured
experimentaly as schematically shown in Fig.
10.18. A flat vertical glass plate, below which a
vessel of some liquid is kept, forms one arm of
the balance. The plate is balanced by weights
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MECHANICAL PROPERTIES OF FLUIDS 267
on the other side, with its horizontal edge just
over water. The vessel is raised slightly till the
liquid just touches the glass plate and pulls it
down a little because of surface tension. Weights
are added till the plate just clears water.
Fig. 10.18 Measuring Surface Tension.
Suppose the additional weight required is W.
Then from Eq. 10.22 and the discussion given
there, the surface tension of the liquid-air
interface is
S
la
= (W/2l) = (mg/2l ) (10.23)
where m is the extra mass and l is the length of
the plate edge. The subscript (la) emphasises
the fact that the liquid-air interface tension is
involved.
10.6.3 Angle of Contact
The surface of liquid near the plane of contact,
with another medium is in general curved. The
angle between tangent to the liquid surface at
the point of contact and solid surface inside the
liquid is termed as angle of contact. It is denoted
by
θ
. It is different at interfaces of different pairs
of liquids and solids. The value of
θ
determines
whether a liquid will spread on the surface of a
solid or it will form droplets on it. For example,
water forms droplets on lotus leaf as shown in
Fig. 10.19 (a) while spreads over a clean plastic
plate as shown in Fig. 10.19(b).
(a)
(b)
Fig. 10.19 Different shapes of water drops with
interfacial tensions (a) on a lotus leaf (b)
on a clean plastic plate.
We consider the three interfacial tensions at
all the three interfaces, liquid-air, solid-air and
solid-liquid denoted by S
la
, S
sa
and S
sl
, respectively
as given in Fig. 10.19 (a) and (b). At the line of
contact, the surface forces between the three media
must be in equilibrium. From the Fig. 10.19(b) the
following relation is easily derived.
S
la
cos
θ
+ S
sl
= S
sa
(10.24)
The angle of contact is an obtuse angle if
S
sl
> S
la
as in the case of water-leaf interface
while it is an acute angle if S
sl
< S
la
as in the
case of water-plastic interface. When
θ
is an
obtuse angle then molecules of liquids are
attracted strongly to themselves and weakly to
those of solid, it costs a lot of energy to create a
liquid-solid surface, and liquid then does not
wet the solid. This is what happens with water
on a waxy or oily surface, and with mercury on
any surface. On the other hand, if the molecules
of the liquid are strongly attracted to those of
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268 PHYSICS
the solid, this will reduce S
sl
and therefore,
cos
θ
may increase or
θ
may decrease. In this
case
θ
is an acute angle. This is what happens
for water on glass or on plastic and for kerosene
oil on virtually anything (it just spreads). Soaps,
detergents and dying substances are wetting
agents. When they are added the angle of
contact becomes small so that these may
penetrate well and become effective. Water
proofing agents on the other hand are added to
create a large angle of contact between the water
and fibres.
10.6.4 Drops and Bubbles
One consequence of surface tension is that free
liquid drops and bubbles are spherical if effects
of gravity can be neglected. You must have seen
this especially clearly in small drops just formed
in a high-speed spray or jet, and in soap bubbles
blown by most of us in childhood. Why are drops
and bubbles spherical? What keeps soap
bubbles stable?
As we have been saying repeatedly, a liquid-
air interface has energy, so for a given volume
the surface with minimum energy is the one with
the least area. The sphere has this property.
Though it is out of the scope of this book, but
you can check that a sphere is better than at
least a cube in this respect! So, if gravity and
other forces (e.g. air resistance) were ineffective,
liquid drops would be spherical.
Another interesting consequence of surface
tension is that the pressure inside a spherical
drop Fig. 10.20(a) is more than the pressure
outside. Suppose a spherical drop of radius r is
in equilibrium. If its radius increase by ∆r. The
extra surface energy is
[4π(r + ∆r)
2
- 4πr
2
] S
la
= 8πr ∆r S
la
(10.25)
If the drop is in equilibrium this energy cost is
balanced by the energy gain due to
expansion under the pressure difference (P
i
– P
o
)
between the inside of the bubble and the outside.
The work done is
W = (P
i
– P
o
) 4πr
2
∆r (10.26)
so that
(P
i
– P
o
) = (2 S
la
/ r) (10.27)
In general, for a liquid-gas interface, the
convex side has a higher pressure than the
concave side. For example, an air bubble in a
liquid, would have higher pressure inside it.
See Fig 10.20 (b).
Fig. 10.20 Drop, cavity and bubble of radius r.
A bubble Fig 10.20 (c) differs from a drop
and a cavity; in this it has two interfaces.
Applying the above argument we have for a
bubble
(P
i
– P
o
) = (4 S
la
/ r) (10.28)
This is probably why you have to blow hard,
but not too hard, to form a soap bubble. A little
extra air pressure is needed inside!
10.6.5 Capillary Rise
One consequence of the pressure difference
across a curved liquid-air interface is the well-
known effect that water rises up in a narrow
tube in spite of gravity. The word capilla means
hair in Latin; if the tube were hair thin, the rise
would be very large. To see this, consider a
vertical capillary tube of circular cross section
(radius a) inserted into an open vessel of water
(Fig. 10.21). The contact angle between water
Fig. 10.21 Capillary rise, (a) Schematic picture of a
narrow tube immersed water.
(b) Enlarged picture near interface.
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MECHANICAL PROPERTIES OF FLUIDS 269
and glass is acute. Thus the surface of water in
the capillary is concave. This means that
there is a pressure difference between the
two sides of the top surface. This is given by
(P
i
– P
o
) =(2S/r) = 2S/(a sec
θ
)
= (2S/a) cos
θ
(10.29)
Thus the pressure of the water inside the
tube, just at the meniscus (air-water interface)
is less than the atmospheric pressure. Consider
the two points A and B in Fig. 10.21(a). They
must be at the same pressure, namely
P
0
+ h
ρ
g = P
i
= P
A
(10.30)
where
ρρ
ρρ
ρ
is the density of water and h is called
the capillary rise [Fig. 10.21(a)]. Using
Eq. (10.29) and (10.30) we have
h
ρ
g = (P
i
– P
0
) = (2S cos
θ
)/a (10.31)
The discussion here, and the Eqs. (10.26) and
(10.27) make it clear that the capillary rise is
due to surface tension. It is larger, for a smaller
a. Typically it is of the order of a few cm for fine
capillaries. For example, if a = 0.05 cm, using
the value of surface tension for water (Table
10.3), we find that
h = 2S/(
ρ
g a)
-1
3 -3 -2 -4
2 ×(0.073 N m )
=
(10 kg m ) (9.8 m s )(5 × 10 m)
= 2.98 × 10
–2
m = 2.98 cm
Notice that if the liquid meniscus is convex,
as for mercury, i.e., if cos
θ
is negative then from
Eq. (10.30) for example, it is clear that the liquid
will be lower in the capillary !
10.6.6 Detergents and Surface Tension
We clean dirty clothes containing grease and oil
stains sticking to cotton or other fabrics by
adding detergents or soap to water, soaking
clothes in it and shaking. Let us understand
this process better.
Washing with water does not remove grease
stains. This is because water does not wet greasy
dirt; i.e., there is very little area of contact
between them. If water could wet grease, the flow
of water could carry some grease away.
Something of this sort is achieved through
detergents. The molecules of detergents are
hairpin shaped, with one end attracted to water
and the other to molecules of grease, oil or wax,
thus tending to form water-oil interfaces. The result
is shown in Fig. 10.22 as a sequence of figures.
In our language, we would say that addition
of detergents, whose molecules attract at one
end and say, oil on the other, reduces drastically
the surface tension S (water-oil). It may even
become energetically favourable to form such
interfaces, i.e., globs of dirt surrounded by
detergents and then by water. This kind of
process using surface active detergents or
surfactants is important not only for cleaning,
but also in recovering oil, mineral ores etc.
Fig. 10.22 Detergent action in terms of what
detergent molecules do.
.
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270 PHYSICS
t
Example 10.11 The lower end of a capillary
tube of diameter 2.00 mm is dipped 8.00
cm below the surface of water in a beaker.
What is the pressure required in the tube
in order to blow a hemispherical bubble at
its end in water? The surface tension of
water at temperature of the experiments is
7.30×10
-2
Nm
-1
. 1 atmospheric pressure =
1.01 × 10
5
Pa, density of water = 1000 kg/m
3
,
g = 9.80 m s
-2
. Also calculate the excess
pressure.
Answer The excess pressure in a bubble of gas
in a liquid is given by 2S/r, where S is the
surface tension of the liquid-gas interface. You
should note there is only one liquid surface in
this case. (For a bubble of liquid in a gas, there
are two liquid surfaces, so the formula for
excess pressure in that case is 4S/r.) The
radius of the bubble is r. Now the pressure
outside the bubble P
o
equals atmospheric
pressure plus the pressure due to 8.00 cm of
water column. That is
P
o
= (1.01 × 10
5
Pa + 0.08 m × 1000 kg m
–3
× 9.80 m s
–2
)
= 1.01784 × 10
5
Pa
Therefore, the pressure inside the bubble is
P
i
= P
o
+ 2S/r
= 1.01784 × 10
5
Pa
+ (2 × 7.3 × 10
-2
Pa m/10
-3
m)
= (1.01784 + 0.00146) × 10
5
Pa
= 1.02 × 10
5
Pa
where the radius of the bubble is taken
to be equal to the radius of the capillary tube,
since the bubble is hemispherical ! (The answer
has been rounded off to three significant
figures.) The excess pressure in the
bubble is 146 Pa. t
SUMMARY
1. The basic property of a fluid is that it can flow. The fluid does not have any
resistance to change of its shape. Thus, the shape of a fluid is governed by the
shape of its container.
2. A liquid is incompressible and has a free surface of its own. A gas is compressible
and it expands to occupy all the space available to it.
3. If F is the normal force exerted by a fluid on an area A then the average pressure P
av
is defined as the ratio of the force to area
A
F
P
av
=
4. The unit of the pressure is the pascal (Pa). It is the same as N m
-2
. Other common
units of pressure are
1 atm = 1.01×10
5
Pa
1 bar = 10
5
Pa
1 torr = 133 Pa = 0.133 kPa
1 mm of Hg = 1 torr = 133 Pa
5. Pascal’s law states that: Pressure in a fluid at rest is same at all points which are at
the same height. A change in pressure applied to an enclosed fluid is transmitted
undiminished to every point of the fluid and the walls of the containing vessel.
6. The pressure in a fluid varies with depth h according to the expression
P = P
a
+ ρgh
where ρ is the density of the fluid, assumed uniform.
7. The volume of an incompressible fluid passing any point every second in a pipe of
non uniform crossection is the same in the steady flow.
v A = constant ( v is the velocity and A is the area of crossection)
The equation is due to mass conservation in incompressible fluid flow.
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MECHANICAL PROPERTIES OF FLUIDS 271
8. Bernoulli’s principle states that as we move along a streamline, the sum of the
pressure (P), the kinetic energy per unit volume (ρv
2
/2) and the potential energy per
unit volume (ρgy) remains a constant.
P + ρv
2
/2 + ρgy = constant
The equation is basically the conservation of energy applied to non viscuss fluid
motion in steady state. There is no fluid which have zero viscosity, so the above
statement is true only approximately. The viscosity is like friction and converts the
kinetic energy to heat energy.
9. Though shear strain in a fluid does not require shear stress, when a shear stress is
applied to a fluid, the motion is generated which causes a shear strain growing
with time. The ratio of the shear stress to the time rate of shearing strain is known
as coefficient of viscosity, η.
where symbols have their usual meaning and are defined in the text.
10. Stokes’ law states that the viscous drag force F on a sphere of radius a moving with
velocity v through a fluid of viscosity is, F = 6π
η
av.
11. Surface tension is a force per unit length (or surface energy per unit area) acting in
the plane of interface between the liquid and the bounding surface. It is the extra
energy that the molecules at the interface have as compared to the interior.
POINTS TO PONDER
1. Pressure is a scalar quantity. The definition of the pressure as “force per unit area”
may give one false impression that pressure is a vector. The “force” in the numerator of
the definition is the component of the force normal to the area upon which it is
impressed. While describing fluids as a concept, shift from particle and rigid body
mechanics is required. We are concerned with properties that vary from point to point
in the fluid.
2. One should not think of pressure of a fluid as being exerted only on a solid like the
walls of a container or a piece of solid matter immersed in the fluid. Pressure exists at
all points in a fluid. An element of a fluid (such as the one shown in Fig. 10.2) is in
equilibrium because the pressures exerted on the various faces are equal.
3. The expression for pressure
P = P
a
+ ρgh
holds true if fluid is incompressible. Practically speaking it holds for liquids, which
are largely incompressible and hence is a constant with height.
4. The gauge pressure is the difference of the actual pressure and the atmospheric pressure.
P – P
a
= P
g
Many pressure-measuring devices measure the gauge pressure. These include the tyre
pressure gauge and the blood pressure gauge (sphygmomanometer).
5. A streamline is a map of fluid flow. In a steady flow two streamlines do not intersect as
it means that the fluid particle will have two possible velocities at the point.
6. Bernoulli’s principle does not hold in presence of viscous drag on the fluid. The work
done by this dissipative viscous force must be taken into account in this case, and P
2
[Fig. 10.9] will be lower than the value given by Eq. (10.12).
7. As the temperature rises the atoms of the liquid become more mobile and the coefficient
of viscosity, η falls. In a gas the temperature rise increases the random motion of
atoms and η increases.
8. Surface tension arises due to excess potential energy of the molecules on the surface
in comparison to their potential energy in the interior. Such a surface energy is present
at the interface separating two substances at least one of which is a fluid. It is not the
property of a single fluid alone.
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272 PHYSICS
EXERCISES
10.1 Explain why
(a) The blood pressure in humans is greater at the feet than at the brain
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of
its value at the sea level, though the height of the atmosphere is more than
100 km
(c) Hydrostatic pressure is a scalar quantity even though pressure is force
divided by area.
10.2 Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water
with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the
same surface tends to form drops. (Put differently, water wets glass while
mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent disolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape
10.3 Fill in the blanks using the word(s) from the list appended with each statement:
(a) Surface tension of liquids generally ... with temperatures (increases / decreases)
(b) Viscosity of gases ... with temperature, whereas viscosity of liquids ... with
temperature (increases / decreases)
(c) For solids with elastic modulus of rigidity, the shearing force is proportional
to ... , while for fluids it is proportional to ... (shear strain / rate of shear
strain)
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows
(conservation of mass / Bernoulli’s principle)
(e) For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for
turbulence for an actual plane (greater / smaller)
10.4 Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it
(b) When we try to close a water tap with our fingers, fast jets of water gush
through the openings between our fingers
(c) The size of the needle of a syringe controls flow rate better than the thumb
pressure exerted by a doctor while administering an injection
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on
the vessel
(e) A spinning cricket ball in air does not follow a parabolic trajectory
10.5 A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with
a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor ?
2020-21
MECHANICAL PROPERTIES OF FLUIDS 273
10.6 Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density
984 kg m
–3
. Determine the height of the wine column for normal atmospheric
pressure.
10.7 A vertical off-shore structure is built to withstand a maximum stress of 10
9
Pa. Is
the structure suitable for putting up on top of an oil well in the ocean ? Take the
depth of the ocean to be roughly 3 km, and ignore ocean currents.
10.8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000
kg. The area of cross-section of the piston carrying the load is 425 cm
2
. What
maximum pressure would the smaller piston have to bear ?
10.9 A U-tube contains water and methylated spirit separated by mercury. The mercury
columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm
of spirit in the other. What is the specific gravity of spirit ?
10.10 In the previous problem, if 15.0 cm of water and spirit each are further poured into
the respective arms of the tube, what is the difference in the levels of mercury in
the two arms ? (Specific gravity of mercury = 13.6)
10.11 Can Bernoulli’s equation be used to describe the flow of water through a rapid in a
river ? Explain.
10.12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s
equation ? Explain.
10.13 Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0
cm. If the amount of glycerine collected per second at one end is 4.0 × 10
–3
kg s
–1
,
what is the pressure difference between the two ends of the tube ? (Density of glycerine
= 1.3 × 10
3
kg m
–3
and viscosity of glycerine = 0.83 Pa s). [You may also like to check
if the assumption of laminar flow in the tube is correct].
10.14 In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the
upper and lower surfaces of the wing are 70 m s
–1
and 63 m s
-1
respectively. What is
the lift on the wing if its area is 2.5 m
2
? Take the density of air to be 1.3 kg m
–3
.
10.15 Figures 10.23(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of
the two figures is incorrect ? Why ?
Fig. 10.23
10.16 The cylindrical tube of a spray pump has a cross-section of 8.0 cm
2
one end of
which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube
is 1.5 m min
–1
, what is the speed of ejection of the liquid through the holes ?
10.17 A U-shaped wire is dipped in a soap solution, and removed. The thin soap film
formed between the wire and the light slider supports a weight of 1.5 × 10
–2
N
(which includes the small weight of the slider). The length of the slider is 30 cm.
What is the surface tension of the film ?
10.18 Figure 10.24 (a) shows a thin liquid film supporting a small weight = 4.5 × 10
–2
N.
What is the weight supported by a film of the same liquid at the same temperature
in Fig. (b) and (c) ? Explain your answer physically.
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274 PHYSICS
Fig. 10.24
10.19 What is the pressure inside the drop of mercury of radius 3.00 mm at room
temperature ? Surface tension of mercury at that temperature (20 °C) is 4.65 ×
10
–1
N m
–1
. The atmospheric pressure is 1.01 × 10
5
Pa. Also give the excess pressure
inside the drop.
10.20 What is the excess pressure inside a bubble of soap solution of radius 5.00 mm,
given that the surface tension of soap solution at the temperature (20 °C) is 2.50 ×
10
–2
N m
–1
? If an air bubble of the same dimension were formed at depth of 40.0
cm inside a container containing the soap solution (of relative density 1.20), what
would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 10
5
Pa).
Additional Exercises
10.21 A tank with a square base of area 1.0 m
2
is divided by a vertical partition in the
middle. The bottom of the partition has a small-hinged door of area 20 cm
2
. The
tank is filled with water in one compartment, and an acid (of relative density 1.7) in
the other, both to a height of 4.0 m. compute the force necessary to keep the door
close.
10.22 A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a)
When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (b)
The liquid used in the manometers is mercury and the atmospheric pressure is 76
cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a)
and (b), in units of cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with
mercury) are poured into the right limb of the manometer ? (Ignore the small
change in the volume of the gas).
Fig. 10.25
10.23 Two vessels have the same base area but different shapes. The first vessel takes
twice the volume of water that the second vessel requires to fill upto a particular
common height. Is the force exerted by the water on the base of the vessel the same
in the two cases ? If so, why do the vessels filled with water to that same height give
different readings on a weighing scale ?
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MECHANICAL PROPERTIES OF FLUIDS 275
10.24 During blood transfusion the needle is inserted in a vein where the gauge pressure
is 2000 Pa. At what height must the blood container be placed so that blood may
just enter the vein ? [Use the density of whole blood from Table 10.1].
10.25 In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube
to its change in the potential and kinetic energy. (a) What is the largest average
velocity of blood flow in an artery of diameter 2 × 10
–3
m if the flow must remain
laminar ? (b) Do the dissipative forces become more important as the fluid velocity
increases ? Discuss qualitatively.
10.26 (a) What is the largest average velocity of blood flow in an artery of radius 2×10
–3
m
if the flow must remain lanimar? (b) What is the corresponding flow rate ? (Take
viscosity of blood to be 2.084 × 10
–3
Pa s).
10.27 A plane is in level flight at constant speed and each of its two wings has an area of
25 m
2
. If the speed of the air is 180 km/h over the lower wing and 234 km/h over
the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg
m
–3
).
10.28 In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop
of radius 2.0 × 10
–5
m and density 1.2 × 10
3
kg m
–3
. Take the viscosity of air at the
temperature of the experiment to be 1.8 × 10
–5
Pa s. How much is the viscous force
on the drop at that speed ? Neglect buoyancy of the drop due to air.
10.29 Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube
of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By
what amount does the mercury dip down in the tube relative to the liquid surface
outside ? Surface tension of mercury at the temperature of the experiment is 0.465
N m
–1
. Density of mercury = 13.6 × 10
3
kg m
–3
.
10.30 Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form
a U-tube open at both ends. If the U-tube contains water, what is the difference in
its levels in the two limbs of the tube ? Surface tension of water at the temperature
of the experiment is 7.3 × 10
–2
N m
–1
. Take the angle of contact to be zero and
density of water to be 1.0 × 10
3
kg m
–3
(g = 9.8 m s
–2
) .
Calculator/Computer – Based Problem
10.31 (a) It is known that density ρ of air decreases with height y as
0
o
y/y
e
−
ρ = ρ
where ρ
0
= 1.25 kg m
–3
is the density at sea level, and y
0
is a constant. This density
variation is called the law of atmospheres. Obtain this law assuming that the
temperature of atmosphere remains a constant (isothermal conditions). Also assume
that the value of g remains constant.
(b) A large He balloon of volume 1425 m
3
is used to lift a payload of 400 kg. Assume
that the balloon maintains constant radius as it rises. How high does it rise ?
[Take y
0
= 8000 m and ρ
He
= 0.18 kg m
–3
].
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276 PHYSICS
APPENDIX 10.1 : WHAT IS BLOOD PRESSURE ?
In evolutionary history there occurred a time when animals started spending a significant amount
of time in the upright position. This placed a number of demands on the circulatory system. The
venous system that returns blood from the lower extremities to the heart underwent changes. You
will recall that veins are blood vessels through which blood returns to the heart. Humans and
animals such as the giraffe have adapted to the problem of moving blood upward against gravity.
But animals such as snakes, rats and rabbits will die if held upwards, since the blood remains in
the lower extremities and the venous system is unable to move it towards the heart.
Fig. 10.26 Schematic view of the gauge pressures in the arteries in various parts of the human body while
standing or lying down. The pressures shown are averaged over a heart cycle.
Figure 10.26 shows the average pressures observed in the arteries at various points in the human body.
Since viscous effects are small, we can use Bernoulli’s equation, Eq. (10.13),
2
1
Constant
2
P v gy+ ρ + ρ =
to understand these pressure values. The kinetic energy term (
ρ
v
2
/2) can be ignored since the velocities in
the three arteries are small (≈ 0.1 m s
–1
) and almost constant. Hence the gauge pressures at the brain P
B
,
the heart P
H
, and the foot P
F
are related by
P
F
= P
H
+ ρ
g h
H
= P
B
+ ρ
g h
B
(10.34)
where ρ is the density of blood.
Typical values of the heights to the heart and the brain are h
H
= 1.3 m and h
B
= 1.7 m. Taking
ρ
= 1.06 × 10
3
kg m
–3
we obtain that P
F
= 26.8 kPa (kilopascals) and
P
B
= 9.3 kPa given that P
H
= 13.3 kPa.
Thus the pressures in the lower and upper parts of the body are so different when a person is standing,
but are almost equal when he is lying down. As mentioned in the text the units for pressure more
commonly employed in medicine and physiology are torr and mm of Hg. 1 mm of Hg = 1 torr = 0.133 kPa.
Thus the average pressure at the heart is P
H
= 13.3 kPa = 100 mm of Hg.
The human body is a marvel of nature. The veins in the lower extremities are equipped with valves,
which open when blood flows towards the heart and close if it tends to drain down. Also, blood is returned
at least partially by the pumping action associated with breathing and by the flexing of the skeletal muscles
during walking. This explains why a soldier who is required to stand at attention may faint because of
insufficient return of the blood to the heart. Once he is made to lie down, the pressures become equalized
and he regains consciousness.
An instrument called the sphygmomanometer usually measures the blood pressure of humans. It is a
fast, painless and non-invasive technique and gives the doctor a reliable idea about the patient’s health.
The measurement process is shown in Fig. 10.27. There are two reasons why the upper arm is used. First,
it is at the same level as the heart and measurements here give values close to that at the heart. Secondly,
the upper arm contains a single bone and makes the artery there (called the brachial artery) easy to
compress. We have all measured pulse rates by placing our fingers over the wrist. Each pulse takes a little
less than a second. During each pulse the pressure in the heart and the circulatory system goes through a
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MECHANICAL PROPERTIES OF FLUIDS 277
maximum as the blood is pumped by the heart (systolic pressure) and a minimum as the heart relaxes
(diastolic pressure). The sphygmomanometer is a device, which measures these extreme pressures. It
works on the principle that blood flow in the brachial (upper arm) artery can be made to go from
laminar to turbulent by suitable compression. Turbulent flow is dissipative, and its sound can be
picked up on the stethoscope.
The gauge pressure in an air sack wrapped around the upper arm is measured using a manometer or a
dial pressure gauge (Fig. 10.27). The pressure in the sack is first increased till the brachial artery is closed.
The pressure in the sack is then slowly reduced while a stethoscope placed just below the sack is used to
listen to noises arising in the brachial artery. When
the pressure is just below the systolic (peak)
pressure, the artery opens briefly. During this brief
period, the blood velocity in the highly constricted
artery is high and turbulent and hence noisy. The
resulting noise is heard as a tapping sound on the
stethoscope. When the pressure in the sack is
lowered further, the artery remains open for a longer
portion of the heart cycle. Nevertheless, it remains
closed during the diastolic (minimum pressure)
phase of the heartbeat. Thus the duration of the
tapping sound is longer. When the pressure in the
sack reaches the diastolic pressure the artery is
open during the entire heart cycle. The flow is
however, still turbulent and noisy. But instead of a
tapping sound we hear a steady, continuous roar
on the stethoscope.
The blood pressure of a patient is presented as the ratio of systolic/diastolic pressures. For a resting
healthy adult it is typically 120/80 mm of Hg (120/80 torr). Pressures above 140/90 require medical
attention and advice. High blood pressures may seriously damage the heart, kidney and other organs and
must be controlled.
Fig. 10.27 Blood pressure measurement using the
sphygmomanometer and stethoscope.
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