FLUID DYNAMICS

LEARNING OBJECTIVES
At the end of this chapter the students will be able to:
1. Understand that viscous forces in a fluid cause a retarding force on an object moving through it.
2. Use Stokes’ law to derive an expression for terminal velocity of a spherical body falling through
a viscous fluid under laminar conditions.
3. Understand the terms steady (laminar, streamline) flow, incompressible flow, non-viscous flow
as applied to the motion of an ideal fluid.
4. Appreciate that at a sufficiently high velocity, the flow of viscous fluid undergoes a transition
from laminar to turbulence conditions.
5. Appreciate the equation of continuity Av = Constant for the flow of an ideal and incompressible
fluid.
6. Appreciate that the equation of continuity is a form of the principle of conservation of mass.
7. Understand that the pressure difference can arise from different rates of flow of a fluid
(Bernoulli effect).
8. Derive Bernoulli’s equation in form P + ½ρv2 + ρgh = constant.
9. Explain how Bernoulli effect is applied in the filter pump, atomizers, in the flow of air over an
aerofoil, Venturimeter and in blood physics.
10. Give qualitative explanations for the swing of a spinning ball.

INTRODUCTION:
The study of fluids in motion is relatively complicated, but analysis can be simplified by making
a few assumptions. The analysis is further simplified by the use of two important conservation
principles; the conservation of mass and the conservation of energy. The law of conservation of mass
gives us the equation of continuity while the law of conservation of energy is the basis of Bernoulli’s
equation. The equation of continuity and the Bernoulli’s equation along with their applications in
aeroplane and blood circulation are discussed in this chapter.

FLUID:
“Any thing that can flow is called fluid”.
Example:
Water and air are the examples of fluid. S.I unit of η is Poiseuille or Pl.

FLUID DYNAMICS:
“The branch of physics that deals with fluid in motion and their effects is called fluid
dynamics”.
243
244 Physics Intermediate Part-I

Conservation Principle of Fluid Dynamics:

Following are the two principles that deal with motion of fluid.
Law of Conservation of Mass:
Law of conservation of mass of fluid is given by the equation of continuity.
Law of Conservation of Energy:
Law of conservation of energy of fluid is given by Bernoulli’s equation.
VISCOUS DRAG AND STOKE’S LAW:
Viscosity and Viscous:
“The frictional force existing between the different layers of fluid that opposes its motion is
called viscosity”.
The substance possessing this property is said to be viscous.

Co-efficient of Viscosity (η
η):
“The tangential force that produces a velocity difference of 1 ms−1 between the adjacent layers
that are one metre apart is known as co-efficient of viscosity”. It is represented by η.
Substances having Large Co-efficient of Viscosity:
Substances that cannot flow easily are said to have large co- Viscosities of Liquids and
efficient of viscosity. Gases at 30°C
Viscosity
Examples: Material
10− 3 (Nsm− 2)
Honey and thick tar are the substances having large co-efficient Air 0.019
of viscosity. Acetone 0.295
Methanol 0.510
Substances having Small Co-efficient of Viscosity: Benzene 0.564
Water 0.801
Substances that flow easily are said to have small co-efficient of Ethanol 1.000
viscosity. Plasma 1.6
Glycerin 6.29
Examples:
Air, water, milk are the examples of substances having small co-efficient of viscosity.
Drag Force (Fluid Friction):
“When a substance moves through a fluid it experiences a retarding force that opposes its
motion. This retarding force is called drag force”.
Effect of Speed:
The drag force increases as the speed of an object increases through a fluid but at very high
speed drag force may no longer be proportional to the speed.
Examples:
(i) The force offered by the air to the motion of cyclist.
(ii) The retarding force offered by air to the car moving at high speed.
[Chapter-6] Fluid Dynamics 245
Stoke’s Law:
The law that expresses the drag force in mathematical form, according to which drag force ‘Fd’
on a sphere of radius ‘r’ moving with velocity ‘v’ through a fluid of viscosity ‘η’ is given by
Fd = 6πηrv
Limitation:
The expression of drag force is valid for slow moving objects. At high speed drag force is
proportional to squared value of speed and is given by
1
Fd = ρC Av2
2 D
Where, ρ = Density
CD = Drag co-efficient
A = Frontal area
TERMINAL VELOCITY:
“The maximum constant velocity of an object falling vertically downward through a fluid
under the action of gravity is called terminal velocity”.
Symbol:
It is represented by ‘vt’.
Mathematical Form:
Mathematically terminal velocity is given by
2ρgr2
vt =
9η
Derivation:
Suppose a fog droplet of radius ‘r’ is falling down due to its
weight ‘W’, through a medium of co-efficient of viscosity ‘η’ and it
experiences a drag force ‘Fd’ in opposite direction to that of weight. A table tennis ball can be made
The net force acting on the droplet is suspended in the stream of air
coming from the nozzle of a hair
F = ω – Fd …… (1) dryer.

Now F = ma , ω = mg and Fd = 6 πηrv

So eq. (1) becomes Drag force
ma = mg − 6πηrv …… (2)
As the speed of droplet continues to increase, the drag force ‘Fd’ Fog droplet
also increases till it attains a magnitude which will be equal to its
weight. In this case the velocity of fog droplet will be maximum and Weight
constant so acceleration ‘a’ will be zero. So eq. (2) becomes
m(0) = mg – 6 πηrvt
0 = mg – 6 πηrvt
6 πηrvt = mg
mg
vt = …… (3)
6πηr
246 Physics Intermediate Part-I

As we know that:
m = ρV …… (a) Viscosity of gases rises with
temperature.
Where m = Mass of droplet
ρ = Density of droplet
4 3
Volume of sphere (droplet) = πr
3
So eq. (a) becomes
4 3
m = ρ πr
3
Put value of ‘m’ in eq. (3)
4 3
ρ πr g
3
vt =
6πηr
2
vt =   (ρgr2)
9η
2ρgr2
vt =
9η

vt ∝ r2
Result:
The above expression shows that terminal velocity is directly proportional to square of radius of
sphere.
FLUID FLOW:
The fluid passing through the cross-sectional area of pipe per unit time is called fluid flow.
Types of Fluid Flow:
Following are the two types of fluid flow:
Laminar Flow Turbulent Flow
The regular and steady flow of fluid is called The irregular and un-steady flow of fluid is called
laminar or streamline flow. turbulent flow.
OR OR
The flow in which every particle follows the same The flow in which each particle does not
path that has already been followed by other necessarily follow the same path that has already
particles is called laminar flow. Velocity of been followed by other particles. Velocity of fluid
streamlines remains same. changes abruptly.
[Chapter-6] Fluid Dynamics 247

Properties of Ideal Fluid:

It has following properties:
Non-viscous:
There is no internal frictional force between adjacent layers of
fluid.
Formula one racing cars have a
Incompressible: streamlined design.

It has constant density.

Steady Flow:
The fluid motion is steady.
Dolphins have streamlined bodies
EQUATION OF CONTINUITY: to assist their movement in water.

The product of the cross-sectional area of a pipe and the speed of an ideal fluid at any point
along the pipe is constant.
(OR)
Speed of fluid through any pipe is inversely proportional to cross-sectional area provided that
the fluid is incompressible.
Mathematical Representation:
Av = Constant
or A1v1 = A2v2
1
or v ∝
A

Derivation:
Consider an ideal fluid is passing through a non-uniform pipe of variable height as shown in
figure.
At end X:
Suppose ∆m1 is the mass of the fluid at end “X” and ∆m2 is the mass of fluid at end “Y”.
Therefore, v2

A2
∆m1 = Volume × Density
Y
∆m1 = A1∆x1 × ρ1 …… (1) ∆x2
v1
A1 = Cross-sectional area
A1
∆x1 = Length of pipe at end X X
∆x1
ρ1 = Density of fluid
248 Physics Intermediate Part-I

Similarly at end Y:
∆m2 = A2∆x2ρ2 …… (2)
A2 = Cross-sectional area
∆x2 = Length of pipe at end Y
ρ2 = Density of fluid
Since, the fluid is incompressible. By applying law of
conservation of mass.
Mass of fluid entering at point ‘X’=Mass of fluid leaving at point ‘Y’
∆m1 = ∆m2 …… (3)
Using eq. (1) and (2) in eq. (3)
A1∆x1ρ1 = A2∆x2ρ2 …… (4)
Since the fluid passing through the pipe is incompressible. So, As the water falls, its speed increases
and so its cross sectional area
ρ1 = ρ2 = ρ decreases as mandated by the
continuity equation.

So eq. (4) becomes:

A1∆x1ρ = A2∆x2ρ
A1∆x1 = A2∆x2
Dividing both sides by ∆t:
∆x1 ∆x2 S.I unit of flow rate is m3/s.
A1   = A2   …… (5)
 ∆t   ∆t 
∆ x1 ∆ x2
As = v1 and = v2
∆t ∆t
So eq. (5) becomes
A1v1 = A2v2
or Av = Constant …… (a)
1
or v ∝ …… (b)
A

Eq. (a) and (b) are mathematical form of equation of continuity and constant is called flow rate.
Volume
Flow rate =
Time
BERNOULLI’S EQUATION (1738):
Bernoulli described an equation in (1738).
Statement:
“Sum of static and dynamic pressure of an ideal fluid through a non-uniform pipe is always
constant. It is one of the fundamental equations of fluid dynamics that relates pressure of fluid to its
speed and height”.
[Chapter-6] Fluid Dynamics 249
(OR)
“For an incompressible, non-viscous, and steady flow of the fluid through a non-uniform
pipe, the sum of pressure, kinetic energy per unit volume and the potential energy per unit volume
always remains constant”.
Mathematical Form:
Bernoulli’s equation can be written mathematically as:
1
P+ ρv2 + ρgh = Constant
2
1 1
P1 + ρv12 + ρgh1 = P2 + ρv22 + ρgh2
2 2
Derivation:
Consider an ideal fluid passing through a non-uniform pipe of different heights from surface S
and areas of cross-section. The fluid is flowing from end A to end B with velocity v1 at end A and v2 at
end B. The force experienced by the fluid at end A is
∆x1

P1
A1 v1
∆x2
∆m
v2 P2
A2
h1 ∆m
h2

F1
As P1 =
A1
F1 = A1P1
F1 = P1A1 …… (1)
Here P1 = Pressure of fluid at end A
W1 = Work Done by the fluid at end A
So, W1 = F1 ∆x1 …… (2)
Using the value of F1 from eq. (1) in eq. (2):
W1 = P1A1∆x1 …… (3)
Similarly, work done by the fluid at end ‘B’ is:
W2 = −P2A2∆x2 …… (4)
Negative sign shows that the work is done against the fluid force.
Total work done will be equal to the sum of work done at end ‘A’ and end ‘B’.
250 Physics Intermediate Part-I

W = W1 + W2 …… (5)
or W = P1A1∆x1 − P2A2∆x2 nozzle
P1A1∆x1 P2A2 ∆x2 air
W = ×t − ×t …… (6)
t t
∆ x1 ∆x2
As = v1 and = v2
t t
paint
So eq. (6) becomes
W = P1A1v1t − P2A2v2t
From equation of continuity
A stream of air passing over a tube
A1v1 = A2v2 = Av dipped in a liquid will cause the
liquid to rise in the tube as shown.
W = P1Avt – P2Avt This effect is used in perfume
bottles and paint sprayers.
W2 = (P1 – P2) Avt
As Avt = V = Volume occupied by fluid
W = (P1 – P2)V
m
Also V =
ρ
m
So, W = (P1 – P2) …… (a)
ρ
The work is used to change the kinetic energy and potential
energy of the fluid. So by law of conservation of energy.
W = ∆K.E + ∆P.E …… (7)
Now, ∆K.E = K.Ef − K.Ei
1 1
= mv22 − mv12 …… (b)
2 2
Also ∆P.E = P.Ef – P.Ei
= mgh2 – mgh1 …… (c)
Using value of (a)(b)(c) in eq. (7): A chimney works best when it is
tall and exposed to air currents,
m 1 1 which reduces the pressure at the
(P1 – P2) = mv22 − mv12 + mgh2 – mgh1 top and force the upward flow of
ρ 2 2 smoke.
m 1 1
(P1 – P2) = m ( v22 − v21 + gh2 – gh1)
ρ 2 2
1 1
(P1 – P2) = ρ ( v22 − v21 + gh2 – gh1)
2 2
1 1
(P1 – P2) = ρv22 − ρv12 + ρgh2 – ρgh1
2 2
1 2 1 2
P1 + ρv1 + ρgh1 = ρv2 + ρgh2 + P2
2 2
1 1
or P1 + 2 ρv21 + ρgh1 = P2+2 ρv22 + ρgh2

This expression is known as Bernoulli’s equation.

[Chapter-6] Fluid Dynamics 251
APPLICATIONS OF BERNOULLI’S EQUATION:
Torricelli’s Theorem:
Torricelli’s theorem gives us the speed of efflux of fluid.
Speed of Efflux:
“The velocity gained by a fluid while falling vertically downward under the action of gravity is
called speed of efflux”.
Mathematical Form:
Speed of efflux can be given by
v2 = 2g (h1 − h2)
Derivation:
Consider a liquid flowing out of a tank at two points ‘A’ and ‘B’. Let ‘C’ is the point from where
the flow of fluid starts. Let v1 is the velocity of fluid at point ‘C’ which is nearly equal to zero because
 1
the speed of fluid is very small due to large cross sectional area ∴ v ∝ .
 A
The velocity at point A is v2. Let h1 and h2 are heights of h1 C v1
point C and A Bernoulli’s equation at C is
1 2
P1 + ρgh1 + ρv = Constant …… (1)
2 1 h1 − h2
Similarly Bernoulli’s equation at point ‘A’ can also be
written as
1 2 v3 A v2
P2 + ρgh2 + ρv = Constant …… (2)
2 2 B h2
Comparing eq. (1) and (2)
1 2 1
P1 + ρgh1 + ρv = P2 + ρgh2 + ρv22 …… (3)
2 1 2
Taking pressure equal to atmospheric pressure i.e.,
P1 = P2 = Patm
So eq. (3) becomes:
1 1
Patm + ρgh1 + ρv12 = Patm + ρgh2 + ρv22
2 2
1 1
ρv21 + ρgh1 = ρgh2 + ρv22 …… (4)
2 2
By equation of continuity:
1 The speed of efflux will maximum
v ∝
A when the whole is at the bottom
of the tank.
As, A1 >> A2
v1 << v2
So, v1 = 0
252 Physics Intermediate Part-I

So eq. (4) become:

1 2 Water
ρgh1 = ρgh2 + ρv
2 2
1 2
gh1 = gh2 + v
2 2 Air Flow

1 2
v = gh1 – gh2
2 2
v22 = 2g(h1 – h2)
Taking square root on both sides

v2 = 2g(h1 − h2) A filter pump has a constriction in

the centre, so that a jet of water
If the orifice is pointed upward as at point ‘B’. Its kinetic from the tap flows faster here.
energy would allow the liquid to rise to the level of water tank. But due This cause a drop in pressure
near it and air, therefore, flows in
to viscous energy losses the water level at point ‘B’ is bit lower. from the side tube. The air and
water together are expelled through
The speed of efflux of liquid is the same as the speed of ball the lower part of the pump.
that falls through height (h1 – h2).
Relation between Speed and Pressure of Fluid (Bernoulli’s Principle):
The speed of a fluid will be low when its pressure is high and vice versa.
Explanation:
Consider a fluid is flowing through a horizontal pipe of non-uniform cross sectional area as
shown in figure.
Suppose vA, PA and vB, PB are the velocities and pressure of fluid at point A and B. Now we
apply Bernoulli’s theorem at point ‘A’ and ‘B’ Bernoulli’s equation at point B is
1
PB + ρghB + ρv2B = Constant …… (1)
2
Bernoulli’s equation at point A is
1 2
PA + ρghA + ρv = Constant …… (2)
2 A
Comparing eq. (1) and (2)
1 1
PB + ρghB + ρv2B = PA + ρghA + ρvA2
2 2
Since the tube is horizontal i.e.,
hA = hB = h
Therefore,
1 1 2
PA + ρgh + ρA2A = PB + ρgh + ρv P P
2 2 B A C
P B
1 2 1 2
PA + ρvA = PB + ρvB …… (3) A1
2 2
A2
1
or P + ρv2 = Constant …… (A)
2
[Chapter-6] Fluid Dynamics 253
From eq. (3)
1 1
PA – PB = ρv2B − ρv2A
2 2
1
PA – PB = ρ[vB2 − vA2 ] …… (4)
2
From eq. (4), for water
ρ = 1000 kgm−3
Suppose vA = 0.2 ms−1
vB = 2 ms−1
So eq. (4) becomes:
1
PA – PB = (1000) [(2)2 – (0.2)2]
2
= 500 [4 – 0.04]
PA – PB = 1980 Nm−2
Result:
It shows the pressure in the narrow pipe where streamlines are closer together is much smaller
than in the wider pipe.
Applications of Bernoulli’s Principle:
(a) Lift on an Aeroplane: Faster air which has
low pressure
The flow of air around aeroplane wing is shown in figure.
The wing is designed to deflect the air such that the
streamlines are closer together above the wing than below the
Direction of plane
wing. Thus air is traveling faster on the upper side of the wing
than on the lower side of the wing. The pressure will be lower at Slower air which
upward force
has high pressure
the top of the wing and the aeroplane will be forced upward.

(b) Swing in the Tennis Ball: Faster air, Deflection force

lower pressure
When tennis ball is hit by racket in such a way that it spins
as well as moves forward. The velocity of air on one side of the
ball increases due to the spinning and air speed (streamlines) are
in same direction and hence pressure decreases. This gives an
extra curvature to the ball known as swing which deceives an
opponent player. Slower air,
higher pressure
Venturi Relation:
Spinning ball
An expression which shows the effect of decrease in
pressure with increase in the speed of fluid in a horizontal pipe is
known as venturi relation.
Mathematical Form:
Mathematically it is given by
1
P1 − P2 = ρv22
2
254 Physics Intermediate Part-I

Derivation:
Suppose an ideal fluid passes through a horizontal pipe of non-uniform cross-sectional area as
shown in figure.
Suppose cross sectional area of the pipe at point one is A1, A1
where the speed of fluid is v1 and cross- sectional area at point 2 A2
is A2, where speed of fluid is v2. Let the heights of pipe from a
v1 v2
levelled surface at point 1 and 2 are h1 and h2. Now Bernoulli’s
equation at point 1. P2

1 P1 h
P1 + ρg h1 + ρv22 = Constant …… (1)
2
Similarly Bernoulli’s equation at point 2 is given by
1
P2 + ρgh2 + 2 ρv22 = Constant …… (2)

Comparing eq. (1) and (2)

1 1
P1 + ρgh1 + ρv21 = P2 + ρgh2 + ρv22
2 2
Since tube is placed horizontally i.e.,
h1 = h2 = h
1 2 1
P1 + ρgh + ρv = P2 + ρgh + ρv22
2 1 2
1 1
P1 + ρV12 = P2 + ρv21 …… (3)
2 2
From equation of continually
A1v1 = A2v2
A2v2
v1 =
A1
Since A1 >>> A2 so v1 is very small, neglecting v1 i.e., v1 ≈ 0.
So, eq. (3) becomes:
1
P1 + 0 = P2 + ρv22
2
The range (distance covered) of
1 fluid will be maximum when the
or P1 – P 2 = ρv22 whole is at middle of the tank.
2
Application:
Venturimeter:
It is device which is used to measure the speed of liquid flow.

2(P1 − P2)
v =
ρ
[Chapter-6] Fluid Dynamics 255

Blood Flow:
Nature of Blood: Air

Following are common properties of blood:

Atmospheric
(i) Blood is incompressible fluid i.e., its density is constant. pressure
(ii) The density of blood is nearly equal to that of water. Low
pressure
(iii) Viscosity of blood is greater, 3 to 5 times than viscosity of Gas
water, because of nearly 50% concentration of red blood
cells.
Nature of Blood Vessels: The carburetor of a car engine
uses a Venturi duct to feed the
Blood vessels have the following properties: correct mix of air and petrol to the
cylinders. Air is drawn through
(i) Blood vessels are not rigid. the duct and along a pipe to the
cylinders. A tiny inlet at the side of
(ii) They stretch like rubber hose. duct is fed with petrol. The air
through the duct moves very fast,
creating low pressure in the duct,
(iii) They remain inflated all the times. which draws petrol vapour into
the air stream.
(iv) Even during the relaxation of heart, due to tension in the
walls of the blood vessels the pressure of blood inside is
greater than the external atmospheric pressure.

Blood Pressure:
The force exerted by the blood per unit area of the blood vessels in known as blood pressure.
Types of Blood Pressure:
Following are the types of blood pressure
(i) Systolic pressure (High)
(ii) Diastolic Pressure (Low)
Systolic Pressure Diastolic Pressure
• The pressure exerted by the blood per unit • The pressure exerted by the blood per unit
area of the blood vessels during the area of the blood vessels during ventricular
ventricular contraction is called systolic relaxation is called diastolic pressure.
pressure.
• Its value is 120 torr (1 torr = 133.3 Nm−2). • It ranges from 75 torr to 80 torr.
• Blood flow is turbulent during the systolic • Blood flow is streamline inside the blood
pressure. vessels during diastolic pressure.

Sphygmomanometer:
It is an instrument which measures the blood pressure dynamically.
Stethoscope:
A device that measures rate of pulsation.
256 Physics Intermediate Part-I

Measurement of Blood Pressure:

An inflatable bag is wound around the arm of patient and external pressure on the arm is
increased by inflating the bag. The effect is to squeeze the arm and compress the blood vessels inside.
When the external pressure becomes larger than systolic pressure, the vessels collapse cutting off the
flow of blood. By opening the release valve on the bag, the external pressure is gradually decreased. A
stethoscope detects the instant at which the external pressure becomes equal to the systolic pressure.
When the systolic pressure becomes equal to the external applied pressure gurgle appears on the
stethoscope. The reading at this instant on the metre of stethoscope gives us the systolic pressure.
When external applied pressure is further decreased it becomes equal to the diastolic pressure
and gurgle disappears on the stethoscope. In this case the measuring meter of the stethoscope indicates
the diastolic pressure.

300 Air bulb

280 Release valve
260
240 220
200
180
160
140
120 bag Stethoscope
100
80
60
40
20

mm Hg (Torr)

Time (s)

For Your Information

Flow Rate:
From equation of continuity
ρAv = Constant
Now
mAv  m
ρAv = ∴ ρ = v 
A∆x  
 m  ∆x  ∆x 
ρAv =     V = 
∆x  t   t 
m
ρAv = = Mass flow rate
t
m
ρAv =
t
m 1
Av = ×
t ρ
m V  m
Av = × ∴ ρ = v 
t m  
V
Av = = Flow rate
t
Note: 1 atmosphere = 1.013 × 105 Pa
[Chapter-6] Fluid Dynamics 257

SOLVED EXAMPLES
EXAMPLE 6.1

A tiny water droplet of radius 0.010 cm descends through air from a high building.
–6 –1 –1
Calculate its terminal velocity. Given that η for air = 19 x 10 kg m s and density of water
–3
ρ = 1000 kgm .

Data:
Radius of water droplet = r = 0.010 cm

= 0.010 x 10–2 m

= 1 x 10–4 m

For air = η = 19 x 10–6 kg m–1 s–1

Density of water = ρ = 1000 kg m–3

To Find:
Terminal velocity = vt = ?

SOLUTION

2 g r2 ρ
Using: vt =
9η

2 (9.8) (1 x 10–4)2 1000

=
9 x 19 x 10–6

2000 x 9.8 x 10–8

=
9 x 19 x 10–6

9.8 x 2 x 10–5 x 106

= 9 x 19

2 x 9.8 x 10
=
9 x 19

196
=
171

= 1.1 m/s

Result:
Terminal velocity = vt = 1.1 m/s
258 Physics Intermediate Part-I

EXAMPLE 6.2
A water hose with an internal diameter of 20mm at the outlet discharges 30 kg of water in
–3
60 s. Calculate the water speed at the outlet. Assume the density of water is 1000 kgm and its
flow is steady.
Data:
Diameter of hose = d = 20mm
= 20 x 10–3 m
= 2 x 10–2 m
Mass of water = m = 30 kg
Time = t = 60 sec
Density of water = ρ = 1000 kg m–3
To Find:
Speed of water = v = ?

SOLUTION
From equation of continuity
m = ρAv∆t
m
Also v =
ρA∆t
Since area of cross section A = π r2
m
∴ v =
ρ π r2 ∆ t
30  d
= 2  ∴ r = 2
d  
1000 x 3.14   60
2
30
= 2
d 
1000 x 3.14   x 60
4
30 x 4
=
1000 x 3.14 x 60 (2 x 10–2)2
120
=
1000 x 3.14 x 60 x 4 x 10–4
2
=
3.14 x 4 x 10–1
10
=
6.28
= 1.6 m/s
Result:
Speed of water = v = 1.6 m/s
[Chapter-6] Fluid Dynamics 259

EXAMPLE 6.3
Water flows down hill through a closed vertical funnel. The flow speed at the top is 12.0
–1
cms . The flow speed at the bottom is twice the speed at the top. If the funnel is 40 cm long and
5 –2
the pressure at the top is 1.103 x 10 Nm , what is the pressure at the bottom?
Data:
Flow speed at top = v1 = 12 cm/s P1 v1

= 0.12 m/s
Flow speed at bottom = v2 = 24 cm/s
h1 − h2 h1
= 0.24 m/s
Length of funnel = h = h1 – h2
v2
= 40 cm P2 h2 Ground
= 0.4 m
Pressure at top = P1 = 1.013 x 105 Nm–2
To Find:
Pressure at bottom = P2 = ?

SOLUTION
Using Bernoulli’s equation:
1 2 1 2
P1 + ρ v1 + ρ g h1 = P2 + ρ v2 + ρ g h2
2 2
1 2 1 2
P2 = P1 + ρ v1 – ρ v2 + ρ g h1 – ρ g h2
2 2
1 2 2
= P1 + ρ (v1 – v2) + ρ g (h1 – h2)
2
1
= 1.013 x 105 + 2 (1000) [(0.12)2 – (0.24)2] + 1000 x 9.8 x 0.4

= 1.013 x 105 + 500 (0.0144 – 0.0576) + 980 x 0.4

= 1.013 x 105 + 500 (– 0.0432) + 3920
= 1.013 x 105 – 21.6 + 3920
= 1.013 x 105 + 3898.4
= 1.013 x 105 + 0.038984 x 105
= 105 (1.013 + 0.03898)
= 1.05 x 105 Nm–2
Result:
Pressure at bottom = P2 = 1.05 × 105 Nm−2
260 Physics Intermediate Part-I

SHORT QUESTION ANSWERS

6.1 Explain what do you understand by term viscosity?
Ans. Viscosity: The property of fluids due to which they oppose relative motion between their
different layers.
(OR)
Viscosity is the kind of internal frictional force in fluids which opposes the motion of fluids
when they flow.
Fd
Fd = 6πηvr ⇒ η =
6πvr
6.2 What is meant by drag force? What are the factors upon which drag force acting upon a
small sphere of radius ‘r’, moving down through liquid, depends?
Ans. Drag Force: All objects moving through fluid experience a retarding force called a drag force.
The drag force ‘Fd’ on a sphere of radius ‘r’, moving slowly with the speed ‘v’ through a fluid of
viscosity η is given by Stoke’s law
Fd = 6πηvr
So, the drag force depends upon:
• Viscosity of medium
• Radius of sphere
• Speed of sphere
6.3 Why fog droplets appear to be suspended in air?
Ans. When the drag force is equal to the weight of droplet i.e., Fd = w, then net force Fd
acting on the droplet is zero so, the fog droplet appears to be suspended in air
because of its very minute value of terminal velocity given by
2ρgr2
vt = ⇒ vt ∝ r2
9η
vt is smaller due to smaller value of r, whose square becomes further smaller. w

6.4 Explain the difference between laminar flow and turbulent flow.

Ans. Laminar Flow Turbulent Flow

(1) A regular, steady and smooth flow is (1) The irregular and unsteady flow is
known as laminar flow. turbulent flow.

(2) Direction of flow of particles is always (2) Direction of flow of particles is not
along direction of overall flow of fluid. along direction of overall flow of
fluid.
[Chapter-6] Fluid Dynamics 261

(3) All particles passing through specific (3) All particles passing through specific
point have same velocities. point do not have same velocities.

(4) Examples: (4) Examples:

(i) Flow of gentle breeze. (i) Very strongly flowing winds.
(ii) Flow of water around dolphin. (ii) Flow of water in the form of
(iii) Flow of water in wide and smooth water falls.
river. (iii) Flow of water from top of
mountain.

6.5 State Bernoulli’s relation for a liquid in motion and describe some of its applications.
Ans. Bernoulli’s equation states that in a steady, irrotational and in compressible fluid, the pressure
‘P’, the fluid speed ‘v’ and elevation ‘h’ at any point are related by
1 2
P+ ρv + ρgh = Constant
2
Applications:
• Torricelli’s theorem:

v = 2g(h1 − h2)
• Relation between speed and pressure:
1 2
P+ ρv = Constant
2

• Venturi relation:

2(P1 − P2)
v =
ρ
6.6 A person is standing near train. Is there any danger that he will fall towards it?
Ans. Yes he is in danger. Train

Reason.As speed high where pressure is low. Due to fast Fast speed (Low P)
speed of train, air between person and train will have less
pressure than the air on other side of person. So, a person
may experience a push from a high pressure to low
pressure which may cause him to fall towards train.
Low speed (high P)

6.7 Identify the correct answer. What do you infer from Bernoulli’s theorem?
(i) Where the speed of fluid is high pressure will be low.
(ii) Where the speed of fluid is high pressure will be high.
(iii) This theorem is valid only for turbulent flow of liquid.
Ans. (i) Where speed of fluid is high, pressure will be low.
262 Physics Intermediate Part-I

6.8 Two row boats moving parallel in same direction are pulled towards each other. Explain?
Ans. As speed high where pressure is low. When two boats moving parallel in same direction, the
velocity of water between the two boats becomes higher due to which pressure between the boats
decreases than on the other side. So, boats experience a push from high pressure to low pressure
which may cause them to pull towards each other.
Boat A Boat B
Low speed Low speed
(High P) (High P)

High speed
(Low P)

6.9 Explain, how the swing is produced in fast moving cricket ball?
Ans. During the motion of spinning ball, the spin High speed (Low P)
motion strengthens the motion of air on one side
and cancels the motion of wind on other side. This
causes the difference in pressure which results into
a force from high pressure to low pressure i.e.,
from slow air to fast air. Hence, straight line
motion of ball changes into curved motion of ball,
as shown in figure.

Low speed (High P)

6.10 Explain the working of a carburetor of a motorcar using Bernoulli’s principle?

Ans. The carburetor of a car uses a venturi duct which Air
feeds the correct mixture of air and petrol to the
cylinders. Air is drawn through duct and along a
Atmospheric
pipe to the cylinders. A tiny inlet at the side of duct pressure High speed
is fed with petrol. The air through duct moves at a
Low
greater speed producing low pressure in the duct pressure
so, the petrol is pushed along with air stream in the Gas
form of vapours.

6.11 For which position will the maximum blood pressure in body have the smallest value?
(a) Standing up (b) Sitting (c) Lying horizontally (d) Standing on one’s head
Ans. (c) Lying horizontally
Reason: As all the body parts are at same level with the heart. Hence heart does not have to do
work in pumping the blood against the gravity.
6.12 In an orbiting space station, would the blood pressure in major arteries in the leg ever be
greater than the B.P in the major arteries in the neck?
Ans. No, there will be no change in blood pressure in major arteries in the leg and in the major arteries
in the neck, due to weightlessness in orbiting space station.
[Chapter-6] Fluid Dynamics 263

EXTRA SHORT QUESTIONS

Q. What is hydrodynamics?
Ans. It is the branch of science that deals with the motion of the fluids (liquids and gases).
Q. In case of liquids what does the term ‘streamline’ refer to, how does it differ from a tube of flow?
Ans. A streamline may be defined as the path straight or curved. Such that the tangent to it at any
point indicates the direction of flow of the liquid at that point. A group of stream- lines is said to
constitute a tube of flow.
Q. What is fluidity?
Ans. The reciprocal of viscosity is called fluidity.
Q. Why do machine parts get jammed in winter?
Ans. Due to the low temperature in winter. The coefficient of viscosity of the engine oil and the
lubricants increases. The machine parts get jammed due to this reason.
264 Physics Intermediate Part-I

PROBLEMS WITH SOLUTIONS

PROBLEM 6.1
Formatted
Certain globular protein particle has a density of 1246 kg m−3. It falls through pure water
η = 8.0 × 10−4 Nm− 2s) with a terminal speed of 3.0 cm h− 1. Find the radius of the particle.
(η Formatted
Formatted
Data:
ρ = 1246 kgm−3

η = 8 × 10−4 Nm−2s

vt = 3.0 cmh−1
3
=
100 × 3600

= 8.33 × 10−6 m/s

Required:
Radius of particle = r = ?

Calculations:
2gr2ρ
As, vt =
9η

9η vt
r2 =
2ρg

(9)(8 × 10−4)(8.33 × 10−6)

r2 =
(2)(1246)(9.8)

599.76 × 10−10
r2 =
24421.6

r2 = 0.02455 × 10−10

r = 0.02455 × 10−10

r = 0.156 × 10−5

r = 1.6 × 10−6 m

Result:
Radius of particle = r = 1.6 × 10−6 m
[Chapter-6] Fluid Dynamics 265

PROBLEM 6.2
Formatted
Water flows through a hose, whose internal diameter is 1 cm, at a speed of 1 ms−1. What
Formatted
should be the diameter of the nozzle if the water is to emerge at 21 ms−1?
Formatted
Data: Formatted
Formatted
Internal diameter of hose = d1 = 1 cm = 10−2 m
Formatted
Speed of water in hose = v1 = 1 ms−1 Formatted
−1
Speed of water emerging = v2 = 21 ms
Required:
Diameter of nozzle = d2 = ?
Calculations:
According to equation of continuity:
A1v1 = A2v2
2 2
πr1v1 = πr2v2
2 2
d1 d2  d
v = v r = 2
4 1 4 2  
2 2
d1v1 = d2v2
2
2 d1v1
d2 =
v2
2 (10−2)2(1)
d2 =
21
2
d2 = 0.047 × 10−4
2
d2 = 0.047 × 10−4
d2 = 0.2 × 10−2 m

d2 = 0.2 cm

Result:
Diameter of nozzle = d2 = 0.2 cm

PROBLEM 6.3
Formatted
The pipe near the lower end of a large water storage tank develops a small leak and a
stream of water shoots from it. The top of water in the tank is 15 m above the point of leak. Formatted

(i) With what speed does the water rush from the hole?
Formatted
(ii) If the hole has an area of 0.060 cm2, how much water flows out in one second?
266 Physics Intermediate Part-I

Data:
Height of water = h = 15 m
Area of hole = A = 0.06 cm2
= 0.06 × 10−4 m2
t = 1s
Required:
(i) Speed of water = v = ?
V
(ii) Volume of water = = ?
t
Calculations:
For Speed:
According to Torricelli’s theorem:
v = 2g(h1 − h2)
v = 2gh
v = 2(9.8)(15)
v = 294
v = 17.14 m/s
v = 17 m/s
For Volume of Flow Rate:
Rate of flow = Volume of water flowing out per sec. = Av
Rate of flow = Av
= 0.06 × 10−4 × 17
= 102 × 10−6 m3s−1
Hence, volume of water flowing out per second i.e.,

Rate of flow = 102 cm3/sec.

V
= 102 cm3 s−1
t
As t = 1s
V = 102 cm3 s−1 × 1 s
V = 102 cm3
Result:
(a) v = 17 m/s
(b) V = 102 cm3
[Chapter-6] Fluid Dynamics 267

PROBLEM 6.4
Formatted
Water is flowing smoothly through a closed pipe system. At one point the speed of water is 3
Formatted
ms , while at another point 3 m higher, the speed is 4.0 ms−1. If the pressure is 80 kPa at the lower
−1
Formatted
point, what is pressure at the upper point?
Formatted
Data:
Speed of water (Point 1) = v1 = 3 ms−1
Speed of water (Point 2) = v2 = 4 ms−1
Height of upper point = h2 = 3m
Height of lower point = h1 = 0m
Pressure at lower point = P1 = 80 kPa = 80,000 Pa
Required:
Pressure at upper point = P2 = ?
Calculations:
According to Bernoulli’s theorem:
1 2 1 2
P1 + ρv + ρgh1 = P2 + ρv2 + ρgh2
2 1 2
1 2 1 2
P2 = P1 + ρv − ρv + ρgh1 − ρgh2
2 1 2 2
1  1 
P2 = (80,000) +  × 1000 × 32 −  × 1000 × 42
2  2 
+ (1000 × 9.8 × 0) − (1000 × 9.8 × 3)
= 80,000 + (500 × 9) − (500 × 16) − 29400
= 80,000 + 4,500 − 8,000 − 29,400
= 47,100
= 47.1 × 103 Pa

P2 = 47 KPa

Result:
Pressure at upper point = P2 = 47 kPa

PROBLEM 6.5
Formatted
An airplane wing is designed so that when the speed of the air across the top of the wing is
450 ms , the speed of air below the wing is 410 ms−1. What is the pressure difference between the top
−1
Formatted
and bottom of the wings?
Formatted
(Density of air = 1.29 kgm−3). Formatted
268 Physics Intermediate Part-I

Data:
v at upper surface = v1 = 450 ms−1
v at lower surface = v2 = 410 ms−1
Density of air = ρair = 1.29 kg m−3
Required:
Pressure difference = P2 − P1
∆P = ?
Calculations:
According to venturi relation:
1 2 2
P2 − P1 = ρ(v1 − v2)
2
1
=   (1.29)[(450)2 − (410)2]
2
= (0.645)(202500 − 168100)
= (0.645)(34400)
= 22188 Pa

P2 − P1 = 22.188 kPa

Result:
Pressure difference = P2 − P1 = 22 kPa

PROBLEM 6.6
Formatted
The radius of the aorta is about 1.0 cm and the blood flowing through it has a speed of
about 30 cms−1. Calculate the average speed of the blood in the capillaries using the fact that
although each capillary has a diameter of about 8 × 10−4 cm, there are literally millions of them so
that their total cross section is about 2000 cm2. Formatted

Data:
v1 = 30 cms−1
= 30 × 10−2 m/s
r1 = 1 cm = 1 × 10−2 m
A2 = 2000 cm2 = 0.2 m2
d2 = 8 × 10−4 cm
= 8 × 10−6 m
Required:
Average speed of blood = v2 = ?
[Chapter-6] Fluid Dynamics 269
Calculations:
By equation of continuity:
A1v1 = A2v2
A1v1
v2 = A2
2
πr1 × v1
v2 =
A2
3.14 × 10−2 × 10−2 × 30 × 10−2
v2 = 0.2

v2 = 4.71 × 10−4 ms−1

Result:
Average speed of blood
v2 = 4.71 × 10−4 ms−1
v2 = 5 × 10−4 ms−1 (approx)

PROBLEM 6.7
Formatted
How large must a heating duct be if air moving 3.0 ms−1 along it can replenish the air in a
Formatted
room of 300m3 volume every 15 min? Assume the air’s density remains constant.
Data:
v = 3 ms−1
Volume = V = 300 m3
Time = t = 15 min.
= 15 × 60 = 900 sec.
Required:
Size of duct = r = ?
Calculations:
As,
Volume
Rate of flow = Time
V
= = Av
t
V
t = πr2v

V
r2 =
t×π×v
300
r2 =
900 × 3.14 × 3
270 Physics Intermediate Part-I

300
r2 =
8478
r2 = 0.0354
r = 0.188 m (By square root)
r = 18.8 cm

r ≈ 19 cm

Result:
Size / Radius of duct = r = 19 cm

PROBLEM 6.8
Formatted
An airplane design calls for a “lift” due to the net force of the moving air on the wing of
about 1000 Nm− 2 of wing area. Assume that air flows past the wing of an aircraft with streamline
flow. If the speed of flow past the lower wing surface is 160 ms−1? What is the required speed over Formatted

the upper surface to give a “lift” of 1000 Nm−2. The density of air is 1.29 kgm− 3 and assume Formatted
maximum thickness of wing be one meter. Formatted
Formatted
Data:
Pressure difference = P1 − P2 = 1000 N/m2
Speed = v1 = 160 m/s
Air density = ρair = 1.29 kg/m3
Thickness of wing = h1 − h2 = 1 m
Required:
Speed over upper surface = v2 = ?
Calculations:
Using venturi relation:
1 2 1 2
P1 + ρv = P2 + ρv2
2 1 2
1 2 2
P1 − P2 = 2 ρ(v2 − v1)

Putting the values:

1 2
1000 =  (1.29)(v2 − (160)2)
2
1 2
1000 =  (1.29)(v2 − 25600)
2
2 × 1000 2
= v2 − 25600
1.29
2 2000
v2 = + 25600
1.29
[Chapter-6] Fluid Dynamics 271
2
v2 = 1550 + 25600
2
v2 = 27150
v2 = 27150
v2 = 164.77 ms−1

v2 = 165 ms−1

Result:
Speed of air over the upper surface wing = v2 = 165 ms−1

PROBLEM 6.9
Formatted
What gauge pressure is required in the city mains for a stream from a fire hose connected
Formatted
to the mains to reach a vertical height of 15 m?
Formatted
Data:
Vertical height = h = h1 − h2
h1 − h2 = 15 m
Density of water = ρ = 1000 kg m−3
Required:
Pressure difference = P2 − P1 = ∆P = ?
Calculations:
Using Bernoulli’s equation:
1 2 1 2
P1 + ρv + ρgh1 = P2 + ρv2 + ρgh2
2 1 2
As the speed of streams remains same throughout its motion. So,
v1 = v2 = v
P1 + ρgh1 = P2 + ρgh2
P2 − P1 = ρgh1 − ρgh2
P2 − P1 = ρg(h1 − h2)
P2 − P1 = (1000)(9.8)(15)

∆P = 1.47 × 105 Pa

Result:
Pressure difference = ∆P = 1.47 × 105 Pa
272 Physics Intermediate Part-I

MULTIPLE CHOICE QUESTIONS

Note: You have four choices for each objective type question as a, b, c and d. Tick the
correct choice.
1. Which of the following is not unit of Coefficient of viscosity?
(a) Ns/m2 (b) kg m−1s−1 (c) Poise (d) None of these
2. With increase in temperature the viscosity of:
(a) Gas decreases and liquid increases (b) A gas increases and liquid decreases
(c) Both gas and liquid decreases (d) Both gas and liquid increases
3. Viscosity of an ideal fluid is:
(a) Zero (b) 0.5 (c) 1.0 (d) Infinite
4. Viscosity is the property of:
(a) Liquids only (b) Gases only
(c) Solids and liquids only (d) Liquids and gases only
5. The cause of viscosity of liquid is:
(a) Gravitational force (b) Cohesive force (c) Adhesive force (d) Diffusion
6. Viscosity is the property by virtue of which a liquid:
(a) Becomes spherical in shape (b) Occupies minimum surface area
(c) Oppose relative motion between layers (d) Regains its deformed position
7. SI unit of flow rate is:
(a) m2/s (b) m3/s (c) m/s (d) m/s2
8. Value of η for air at 30°°C is  × 10−5 Nsm− 2.
(a) 1.9 (b) 0.19 (c) 0.019 (d) 19
9. A two meter high tank if a hole appears at its middle than the speed of water will be
 m/s.
(a) 4.42 (b) 5.42 (c) 6.42 (d) None of these
10. Rate of flow = Constant, is in according to:
(a) Bernoulli theorem (b) Continuity equation
(c) Venture relation (d) None of these
11. The speed of water through a pipe of non-uniform thickness is 2 ms−1 where diameter of
pipe is 6 cm. The speed of water at a where diameter of pipe is 3 cm will be:
(a) 1 ms−1 (b) 4 ms−1 (c) 8 ms−1 (d) 16 ms−1
12. The clouds float in atmosphere because of their:
(a) Low temperature (b) Low density (c) Low viscosity (d) All of these
13. The maximum drag force on a falling sphere is 9.8 N its real weight is:
(a) 0N (b) 9.8 N (c) 4.9 N (d) 98 N
[Chapter-6] Fluid Dynamics 273
14. Laminar flow occurs usually at:
(a) High speed (b) Low speed (c) Very high speed (d) None of these
15. The force of viscosity is:
(a) Electromagnetic (b) Gravitational (c) Nuclear (d) None of these
16. The law of conservation of mass is given by:
(a) Continuity equation (b) Bernoulli equation
(c) Stokes Theorem (d) None of these
17. When the magnitude of drag force becomes equal to weight the net force acting on the
droplet is:
(a) Maximum (b) Minimum (c) Zero (d) None of these
18. When droplet of water has terminal velocity the acceleration is:
(a) Positive (b) Negative (c) Becomes zero (d) None of these
19. For an ideal fluid in steady flow, the stream lines are:
(a) Parallel (b) Anti-parallel (c) Perpendicular (d) Intersect each other
20. A liquid flows through a pipe of non-uniform cross-section. If A1 and A2 are the cross-
section of the portions of the pipe at two points the ratio of velocities of liquid at these
points will be:
A1 1 A2
(a) A1A2 (b) (c) (d)
A2 A1A2 A1
21. If the radius of the sphere falling vertically is doubled, its terminal velocity:
(a) Becomes double (b) Becomes four times
(c) Becomes zero (d) Reduces to half
22. Fluid is said to be ideal if it:
(a) is non-viscous (b) is in-compressible (c) has steady flow (d) all of these
23. The fluid is said to be incompressible if its density is:
(a) Zero (b) Very high (c) Very small (d) Constant
24. Flow rate can be defined as:
(a) Volume flow per unit time (b) Mass flow per unit time
(c) Mass flow per unit area (d) Volume flow per unit area
25. A pipe has a fluid flowing with speed v which is exerting a pressure P on the walls of the
pipe. If the speed is increased, the pressure on wall:
(a) Remains same (b) Increases (c) Decreases (d) None of these
26. Mass rate flow of a fluid is equal to .
Av
(a) Av (b) ρ Av (c) (d) None of these
ρ
27. When the streamlines are far apart from each other, the pressure will be:
(a) Low (b) High (c) Remains same (d) becomes zero
274 Physics Intermediate Part-I

28. The law of conservation of energy is given by:

(a) Continuity equation (b) Bernoulli equation
(c) Stokes theorem (d) None of these
29. Terminal velocity is directly proportional to:
(a) Radius of object (b) Square of radius of object
(c) Coefficient of viscosity (d) Both (b) and (c)
30. If velocity of particle at different points does not change with time, flow is:
(a) Streamline (b) Laminar (c) Steady (d) All
31. As a water bubble comes from bottom of the lake to top, its radius:
(a) Increases (b) Decreases (c) Remains same (d) None of these
32. If radius of droplet is halved then its terminal velocity in viscous medium:
(a) Doubles (b) Becomes halved
(c) Becomes one fourth (d) Becomes four times
33. Two fog droplets have radii 2 : 3. Their terminal velocities, if all other factors are kept
constant will be:
(a) 4:6 (b) 4:9 (c) 2:9 (d) 4:3
1
34. As Bernoulli’s equation is P + ρv2 + ρgh = Constant. Here ρgh has same dimensions as
2
that of:
(a) Work (b) Pressure (c) Momentum (d) Power
35. Stoke’s law is applicable to:
(a) Very fast moving objects only (b) Spherical shape object
(c) Slow moving objects (d) Both (b) and (c)
36. A tank containing water has an orifice in one vertical side. If the centre of the orifice is 10
m below the surface level in tank, then velocity of efflux is:
(a) 44 ms−1 (b) 14 ms−1 (c) 10 ms−1 (d) 49 ms−1
37. Water is flowing through a pipe under constant pressure. At some place, the pipe becomes
narrow, the pressure of water at this place:
(a) Increases (b) Decreases (c) Remains same (d) None of these
38. If two ping pong balls are suspended near each other and a fast stream of air is produced
in the space between the balls, then the balls:
(a) Come closer (b) Move farther
(c) Remain in original position (d) Fall down
39. Bernoulli’s equation is the fundamental equation in fluid dynamics that relates pressure to
fluid speed and:
(a) Acceleration due to gravity (b) Height
(c) Coefficient of viscosity of medium (d) Drag force
[Chapter-6] Fluid Dynamics 275
40. When the terminal velocity is reached the acceleration of a body moving through a viscous
medium is:
(a) Increases (b) Decreases (c) Becomes zero (d) None of these
41. When the magnitude of the drag force becomes equal to the weight, the net force acting on
the droplet:
(a) Minimum (b) Zero (c) Maximum (d) None of these
42. For a incompressible fluid its:
(a) Density is constant (b) No internal friction is present
(c) Velocity is constant (d) All of these
43. As compared to external atmospheric pressure, the pressure of blood inside vessels is:
(a) Greater (b) Lesser (c) Zero (d) Infinity
44. The viscosity of blood is greater  times to that of water.
(a) 1 to 3 (b) 3 to 5 (c) 5 to 7 (d) 2 to 4
45. Density of blood is equal to:
(a) Milk (b) Water (c) Honey (d) 1 thick tar
46. Sphygmomanometer is used to measure:
(a) Blood pressure (b) Blood flow (c) Blood density (d) All of these
47. Blood is  fluid.
(a) Compressible (b) Incompressible (c) Ideal (d) Non-viscous
48. 1 torr =
(a) 1.33 Nm2 (b) 13.3 Nm2 (c) 133.3 Nm2 (d) 13.3 Nm
49. Which of the following fluid has highest viscosity at 30°°C?
(a) Glycerin (b) Plasma (c) Acetone (d) Water
50. Diastolic pressure of normal healthy person is:
(a) 70 to 75 torr (b) 75 to 80 torr (c) 70 to 80 torr (d) None of these
51. A piece of ice is floating in a jar containing water. When ice melts the level of water:
(a) Falls (b) Rises (c) Remains same (d) None of these
52. A device used to measure the speed of fluid flow is:
(a) Venturi-meter (b) Odometer
(c) Speedo-meter (d) Sphygmomano-meter
53. During the landing of an aeroplane the speed of air across the top surface of the wings is:
(a) Constant (b) Smaller than at lower surface
(c) Same as at lower surface (d) Greater than at lower surface
54. The effect used in perfume bottles and paint sprays is based upon:
(a) Bernoulli’s theorem (b) Equation of continuity
(c) Pascal’s law (d) Archimedes principle
276 Physics Intermediate Part-I

55. Substances which cannot flow easily have:

(a) Large viscosity (b) Zero viscosity (c) Small viscosity (d) None of these
56. For which position, will the max blood pressure in the body have the smallest value:
(a) Standing upright (b) Sitting (c) Lying horizontally (d) Standing on head
57. Volume rate flow of a fluid is equal to .
Av
(a) Av (b) ρAv (c) (d) None of these
ρ
58. Pressure energy per unit volume of liquid is equal to:
P
(a) P (b) (c) ρ×P (d) ρAv
V
59. If the velocity of the liquid is greater than its critical velocity than flow of the liquid will be:
(a) Streamline (b) Turbulent (c) Steady (d) None of these
60. A fog droplet falling vertically down through air with an acceleration:
(a) Equal to g (b) Less than g (c) Greater than g (d) Equal to zero

ANSWERS
Sr. Ans. Sr. Ans. Sr. Ans. Sr. Ans. Sr. Ans.
1. (d) 2. (b) 3. (a) 4. (d) 5. (b)
6. (c) 7. (b) 8. (a) 9. (a) 10. (b)
11. (c) 12. (b) 13. (b) 14. (b) 15. (a)
16. (a) 17. (c) 18. (c) 19. (a) 20. (d)
21. (b) 22. (d) 23. (d) 24. (a) 25. (c)
26. (b) 27. (b) 28. (b) 29. (b) 30. (d)
31. (a) 32. (c) 33. (b) 34. (b) 35. (d)
36. (b) 37. (b) 38. (a) 39. (b) 40. (c)
41. (b) 42. (a) 43. (a) 44. (b) 45. (b)
46. (a) 47. (b) 48. (c) 49. (a) 50. (b)
51. (c) 52. (a) 53. (b) 54. (a) 55. (a)
56. (c) 57. (a) 58. (a) 59. (b) 60. (b)