Page 1
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 twothirds 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
202223
Page 2
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 twothirds 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
202223
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
force 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
pressuremeasuring 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 (16231662) 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 ?. For a fluid of
mass m occupying volume V,
? =
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
202223
Page 3
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 twothirds 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
202223
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
force 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
pressuremeasuring 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 (16231662) 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 ?. For a fluid of
mass m occupying volume V,
? =
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
202223
252 PHYSICS
t
Example 10.1 The two thigh bones
(femurs), each of crosssectional 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 crosssectional 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
2 5
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 ABCDEF is in the
form of a rightangled 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
sin? = F
c
, F
b
cos? = F
a
(by equilibrium)
A
b
sin? = A
c
, A
b
cos? = 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 crosssection. 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 ? is the mass density of the fluid, we
have the mass of fluid to be m = ?V= ?hA so
that
P
2

P
1
= ?gh (10.6)
Fig. 10.2 Proof of Pascal’s law. ABCDEF 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.
202223
Page 4
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 twothirds 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
202223
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
force 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
pressuremeasuring 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 (16231662) 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 ?. For a fluid of
mass m occupying volume V,
? =
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
202223
252 PHYSICS
t
Example 10.1 The two thigh bones
(femurs), each of crosssectional 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 crosssectional 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
2 5
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 ABCDEF is in the
form of a rightangled 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
sin? = F
c
, F
b
cos? = F
a
(by equilibrium)
A
b
sin? = A
c
, A
b
cos? = 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 crosssection. 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 ? is the mass density of the fluid, we
have the mass of fluid to be m = ?V= ?hA so
that
P
2

P
1
= ?gh (10.6)
Fig. 10.2 Proof of Pascal’s law. ABCDEF 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.
202223
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 ? 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
+ ?gh (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 ?gh. 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 crosssectional 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 ? = 1000 kg m
3
. Take g = 10 m s
–2
From Eq. (10.7)
P =
P
a
+ ?gh
= 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
crosssectional 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
= ?gh (10.8)
where ? is the density of mercury and h is the
height of the mercury column in the tube.
202223
Page 5
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 twothirds 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
202223
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
force 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
pressuremeasuring 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 (16231662) 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 ?. For a fluid of
mass m occupying volume V,
? =
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
202223
252 PHYSICS
t
Example 10.1 The two thigh bones
(femurs), each of crosssectional 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 crosssectional 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
2 5
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 ABCDEF is in the
form of a rightangled 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
sin? = F
c
, F
b
cos? = F
a
(by equilibrium)
A
b
sin? = A
c
, A
b
cos? = 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 crosssection. 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 ? is the mass density of the fluid, we
have the mass of fluid to be m = ?V= ?hA so
that
P
2

P
1
= ?gh (10.6)
Fig. 10.2 Proof of Pascal’s law. ABCDEF 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.
202223
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 ? 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
+ ?gh (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 ?gh. 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 crosssectional 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 ? = 1000 kg m
3
. Take g = 10 m s
–2
From Eq. (10.7)
P =
P
a
+ ?gh
= 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
crosssectional 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
= ?gh (10.8)
where ? is the density of mercury and h is the
height of the mercury column in the tube.
202223
254 PHYSICS
t
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 ? 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 Utube 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 Utube 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)
?gh = 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.
202223
Read More