Chapter 2 Mechanical Properties of Fluids
1) Multiple Choice Questions
i) A hydraulic lift is designed to lift heavy objects of
maximum mass 2000 kg. The area of cross-section of piston carrying
the load is 2.25 × 10-2 m2. What is the maximum
pressure the piston would have to bear?
(A) 0.8711 × 106 N/m2
(B) 0.5862 × 107 N/m2
(C) 0.4869 × 105 N/m2
(D) 0.3271 × 104 N/m2
Answer:
(A) 0.8711 × 106 N/m2
ii) Two capillary tubes of radii 0.3 cm and 0.6 cm are
dipped in the same liquid. The ratio of heights through which the liquid will
rise in the tubes is
(A) 1:2
(B) 2:1
(C) 1:4
(D) 4:1
Answer:
(B) 2:1
iii) The energy stored in a soap bubble of diameter 6 cm
and T = 0.04 N/m is nearly
(A) 0.9 × 10-3 J
(B) 0.4 × 10-3 J
(C) 0.7 × 10-3 J
(D) 0.5 × 10-3 J
Answer:
(A) 0.9 × 10-3 J
iv) Two hail stones with radii in the ratio of 1:4 fall
from a great height through the atmosphere. Then the ratio of their terminal
velocities is
(A) 1:2
(B) 1:12
(C) 1:16
(D) 1:8
Answer:
(C) 1:16
v) In Bernoulli’s theorem, which of the following is
conserved?
(A) linear momentum
(B) angular momentum
(C) mass
(D) energy
Answer:
(D) energy
2) Answer in brief.
i) Why is the surface tension of paints and lubricating
oils kept low?
Answer:
For better wettability (surface coverage), the surface tension and angle of
contact of paints and lubricating oils must below.
ii) How much amount of work is done in forming a soap
bubble of radius r?
Answer:
Let T be the surface tension of a soap solution. The initial surface area of
soap bubble = 0
The final surface area of soap bubble = 2 × 4πr2
∴
The increase in surface area = 2 × 4πr2-
The work done in blowing the soap bubble is W = surface tension × increase in
surface area = T × 2 × 4πr2 = 8πr2T
iii) What is the basis of the Bernoulli’s principle?
Answer:
Conservation of energy.
iv) Why is a low density liquid used as a manometric
liquid in a physics laboratory?
Answer:
An open tube manometer measures the gauge pressure, p — p0 =
hpg, where p0 is the pressure being measured, p0 is
the atmospheric pressure, h is the difference in height between the manometric
liquid of density p in the two arms. For a given pressure p, the product hp is
constant. That is, p should be small for h to be large. Therefore, for
noticeably large h, laboratory manometer uses a low density liquid.
v) What is an incompressible fluid?
Answer:
An incompressible fluid is one which does not undergo change in volume for a
large range of pressures. Thus, its density has a constant value throughout the
fluid. In most cases, all liquids are incompressible.
Question 3.
Why two or more mercury drops form a single drop when brought in contact with
each other?
Answer:
A spherical shape has the minimum surface area- to-volume ratio of all
geometric forms. When two drops of a liquid are brought in contact, the
cohesive forces between their molecules coalesces the drops into a single
larger drop. This is because, the volume of the liquid remaining the same, the
surface area of the resulting single drop is less than the combined surface
area of the smaller drops. The resulting decrease in surface energy is released
into the environment as heat.
Proof : Let n droplets each of radius r coalesce to form a
single drop of radius R. As the volume of the liquid remains constant, volume
of the drop = volume of n droplets
Surface area of n droplets = n × πR2
Surface area of the drop = 4πR2 = n2/3 × πR2
∴
The change in the surface area = surface area of drop –
surface area of n droplets
= πR2(n2/3 – n)
Since the bracketed term is negative, there is a decrease in surface area and a
decrease in surface energy.
Question 4.
Why does velocity increase when water flowing in broader pipe enters a narrow
pipe?
Answer:
When a tube narrows, the same volume occupies a greater length, as
schematically shown in below figure. A1 is the cross section of
the broader pipe and that of narrower pipe is A2. By the equation of
continuity, V2 = (A1/A2)V1
Since A1/A2 > v2 > v1.
For the same volume to pass points 1 and 2 in a given time, the speed must be
greater at point 2.
The process is exactly reversible. If the fluid flows in the opposite
direction, its speed decreases when the tube widens.
Question 5.
Why does the speed of a liquid increase and its pressure decrease when a liquid
passes through constriction in a horizontal pipe?
Answer:
Consider a horizontal constricted tube.
Let A1 and A2 be the cross-sectional areas at
points 1 and 2, respectively. Let v1 and v2 be
the corresponding flow speeds, ρ is the density of the fluid in the pipeline.
By the equation of continuity,
Therefore, the speed of the liquid increases as it passes through the
constriction. Since the meter is assumed to be horizontal, from Bernoulli’s
equation we get,
Again, since A1 > A2, the bracketed term is
positive so that p1 > p2. Thus, as the fluid
passes through the constriction or throat, the higher speed results in lower
pressure at the throat.
Question 6.
Derive an expression of excess pressure inside a liquid drop.
Answer:
Consider a small spherical liquid drop with a radius R. It has a convex
surface, so that the pressure p on the concave side (inside the liquid) is
greater than the pressure p0 on the convex side (outside the liquid). The
surface area of the drop is
A = 4πR2 … (1)
Imagine an increase in radius by an infinitesimal amount dR from the
equilibrium value R. Then, the differential increase in surface area would be
dA = 8πR ∙ dR …(2)
The increase in surface energy would be equal to the work required to increase
the surface area :
dW = T∙dA = 8πTRdR …..(3)
We assume that dR is so small that the pressure inside remains the same, equal
to p. All parts of the surface of the drop experience an outward force per unit
area equal to ρ — ρ0. Therefore, the work done by this outward
pressure-developed force against the surface tension force during the increase
in radius dR is
which is called Laplace’s law for a spherical membrane (or Young-Laplace
equation in spherical form).
[Notes : (1) The above method is called the principle of virtual work. (2)
Equation (5) also applies to a gas bubble within a liquid, and the excess
pressure in this case is also called the gauge pressure. An air or gas bubble
within a liquid is technically called a cavity because it has only one
gas-liquid interface. A bubble, on the other hand, such as a soap bubble, has
two gas-liquid interfaces.]
Question 7.
Obtain an expression for conservation of mass starting from the equation of
continuity.
Answer:
Question 8.
Explain the capillary action.
Answer:
(1) When a capillary tube is partially immersed in a wetting liquid, there is
capillary rise and the liquid meniscus inside the tube is concave, as shown in
below figure.
Consider four points A, B, C, D, of which point A is just
above the concave meniscus inside the capillary and point B is just below it.
Points C and D are just above and below the free liquid surface outside.
Let PA, PB, PC and PD be
the pressures at points A, B, C and D, respectively.
Now, PA = PC = atmospheric pressure
The pressure is the same on both sides of the free surface of a liquid, so that
The pressure on the concave side of a meniscus is always greater than that on
the convex side, so that
PA > PB
∴
PD > PB (∵ PA = PD)
The excess pressure outside presses the liquid up the
capillary until the pressures at B and D (at the same horizontal level)
equalize, i.e., PB becomes equal to PD. Thus, there
is a capillary rise.
(2) For a non-wetting liquid, there is capillary depression
and the liquid meniscus in the capillary tube is convex, as shown in above
figure.
Consider again four points A, B, C and D when the meniscus
in the capillary tube is at the same level as the free surface of the liquid.
Points A and B are just above and below the convex meniscus. Points C and D are
just above and below the free liquid surface outside.
The pressure at B (PB) is greater than that at A
(PA). The pressure at A is the atmospheric pressure H and at D, PD ≃ H = PA. Hence, the hydrostatic pressure at the
same levels at B and D are not equal, PB > PD.
Hence, the liquid flows from B to D and the level of the liquid in the
capillary falls. This continues till the pressure at B’ is the same as that D’,
that is till the pressures at the same level are equal.
Question 9.
Derive an expression for capillary rise for a liquid having a concave meniscus.
Answer:
Consider a capillary tube of radius r partially immersed into a wetting liquid
of density p. Let the capillary rise be h and θ be the angle of contact at the
edge of contact of the concave meniscus and glass. If R is the radius of
curvature of the meniscus then from the figure, r = R cos θ.
Surface tension T is the tangential force per unit length acting along the
contact line. It is directed into the liquid making an angle θ with the
capillary wall. We ignore the small volume of the liquid in the meniscus. The
gauge pressure within the liquid at a depth h, i.e., at the level of the free
liquid surface open to the atmosphere, is
Equations (3) and (4) are also valid for capillary depression h of a
non-wetting liquid. In this case, the meniscus is convex and θ is obtuse. Then,
cos θ is negative but so is h, indicating a fall or depression of the liquid in
the capillary. T is positive in both cases.
[Note : The capillary rise h is called Jurin height, after James Jurin who
studied the effect in 1718. For capillary rise, Eq. (3) is also called the
ascent formula.]
Question 10.
Find the pressure 200 m below the surface of the ocean if pressure on the free
surface of liquid is one atmosphere. (Density of sea water = 1060 kg/m3)
[Ans. 21.789 × 105 N/m2]
Answer:
Data : h = 200 m, p = 1060 kg/m3,
p0 = 1.013 × 105 Pa, g = 9.8 m/s2
Absolute pressure,
p = p0 + hρg
= (1.013 × 103) + (200)(1060)(9.8)
= (1.013 × 105) + (20.776 × 105)
= 21.789 × 105 = 2.1789 MPa
Question 11.
In a hydraulic lift, the input piston had surface area 30 cm2 and
the output piston has surface area of 1500 cm2. If a force of 25 N
is applied to the input piston, calculate weight on output piston. [Ans. 1250
N]
Answer:
Question 12.
Calculate the viscous force acting on a rain drop of diameter 1 mm, falling
with a uniform velocity 2 m/s through air. The coefficient of viscosity of air
is 1.8 × 10-5 Ns/m2.
[Ans. 3.393 × 10-7 N]
Answer:
Data : d = 1 mm, v0 = 2 m / s,
η = 1.8 × 10-5 N.s/m2
r = d/2 = 0.5 mm = 5 × 10-4 m
By Stokes’ law, the viscous force on the raindrop is f = 6πηrv0
= 6 × 3.142 (1.8 × 10-5 N.s/m2 × 5 × 10-4 m)(2
m/s)
= 3.394 × 10-7 N
Question 13.
A horizontal force of 1 N is required to move a metal plate of area 10-2 m2 with
a velocity of 2 × 10-2 m/s, when it rests on a layer of oil 1.5
× 10-3 m thick. Find the coefficient of viscosity of oil. [Ans.
7.5 Ns/m2]
Answer:
Question 14.
With what terminal velocity will an air bubble 0.4 mm in diameter rise in a
liquid of viscosity 0.1 Ns/m2 and specific gravity 0.9? Density
of air is 1.29 kg/m3. [Ans. – 0.782 × 10-3 m/s, The
negative sign indicates that the bubble rises up]
Answer:
Data : d = 0.4 mm, η = 0.1 Pa.s, ρL = 0.9 × 103 kg/m3 =
900 kg/m3, ρair = 1.29 kg/m3, g = 9.8 m/s2.
Since the density of air is less than that of oil, the air bubble will rise up
through the liquid. Hence, the viscous force is downward. At terminal velocity,
this downward viscous force is equal in magnitude to the net upward force.
Viscous force = buoyant force – gravitational force
Question 15.
The speed of water is 2m/s through a pipe of internal diameter 10 cm. What
should be the internal diameter of nozzle of the pipe if the speed of water at
nozzle is 4 m/s?
[Ans. 7.07 × 10-2m]
Answer:
Data : d1 = 10 cm = 0.1 m, v1 = 2 m/s, v2 =
4 m/s
By the equation of continuity, the ratio of the speed is
Question 16.
With what velocity does water flow out of an orifice in a tank with gauge
pressure 4 × 105 N/m2 before the flow starts?
Density of water = 1000 kg/m3. [Ans. 28.28 m/s]
Answer:
Data : ρ — ρ0 = 4 × 105 Pa, ρ = 103 kg/m3
If the orifice is at a depth h from the water surface in a tank, the gauge
pressure there is
ρ – ρ0 = hρg … (1)
By Toricelli’s law of efflux, the velocity of efflux,
Substituting for h from Eq. (1),
Question 17.
The pressure of water inside the closed pipe is 3 × 105 N/m2.
This pressure reduces to 2 × 105 N/m2 on
opening the value of the pipe. Calculate the speed of water flowing through the
pipe. (Density of water = 1000 kg/m3). [Ans. 14.14 m/s]
Answer:
Data : p1 = 3 × 105 Pa, v1 = 0,
p2 = 2 × 105 Pa, ρ = 103 kg/m3
Assuming the potential head to be zero, i.e., the pipe to be horizontal, the
Bernoulli equation is
Question 18.
Calculate the rise of water inside a clean glass capillary tube of radius 0.1
mm, when immersed in water of surface tension 7 × 10-2 N/m. The
angle of contact between water and glass is zero, density of water = 1000 kg/m3,
g = 9.8 m/s2.
[Ans. 0.1429 m]
Answer:
Data : r = 0.1 mm = 1 × 10-4m, θ = 0°,
T = 7 × 10-2 N/m, r = 103 kg/m3, g =
9.8 m/s2
cos θ = cos 0° = 1
= 0.143 m
Question 19.
An air bubble of radius 0.2 mm is situated just below the water surface.
Calculate the gauge pressure. Surface tension of water = 7.2 × 10-2 N/m.
[Ans. 720 N/m2]
Answer:
Question 20.
Twenty seven droplets of water, each of radius 0.1 mm coalesce into a single
drop. Find the change in surface energy. Surface tension of water is 0.072 N/m.
[Ans. 1.628 × 10-7 J = 1.628 erg]
Answer:
Data : r = 1 mm = 1 × 10-3 m, T = 0.472 J/m2
Let R be the radius of the single drop formed due to the coalescence of 8
droplets of mercury.
Volume of 8 droplets = volume of the single drop as the volume of the liquid
remains constant.
∴ 8r3 = R3
∴
2r = R
Surface area of 8 droplets = 8 × 4πr2
Surface area of single drop = 4πR2
∴
Decrease in surface area = 8 × 4πr2 – 4πR2
= 4π(8r2 – R2)
= 4π[8r2 – (2r)2]
= 4π × 4r2
∴
The energy released = surface tension × decrease
in surface area = T × 4π × 4r2
= 0.472 × 4 × 3.142 × 4 × (1 × 10-3)2
= 2.373 × 10-5 J
The decrease in surface energy = 0.072 × 4 × 3.142 × 18 × (1 × 10-4)2
= 1.628 × 10-7 J
Question 21.
A drop of mercury of radius 0.2 cm is broken into 8 identical droplets. Find
the work done if the surface tension of mercury is 435.5 dyne/cm. [Ans. 2.189 ×
10-5J]
Answer:
Let R be the radius of the drop and r be the radius of each droplet.
Data : R = 0.2 cm, n = 8, T = 435.5 dyn/cm
As the volume of the liquid remains constant, volume of n droplets = volume of
the drop
Surface area of the drop = 4πR2
Surface area of n droplets = n × 4πR2
∴
The increase in the surface area = surface area of n droplets-surface area of
drop
= 4π(2 — 1)R2 = 4πR2
∴
The work done
= surface tension × increase in surface area
= T × 4πR2 =
435.5 × 4 × 3.142 × (0.2)2
= 2.19 × 102 ergs = 2.19 × 10-5 J
Question 22.
How much work is required to form a bubble of 2 cm radius from the soap
solution having surface tension 0.07 N/m.
[Ans. 0.7038 × 10-3 J]
Answer:
Data : r = 4 cm = 4 × 10-2 m, T = 25 × 10-3 N/m
Initial surface area of soap bubble = 0
Final surface area of soap bubble = 2 × 4πr2
∴
Increase in surface area = 2 × 4πr2
The work done
= surface tension × increase in surface area
= T × 2 × 4πr2
= 25 × 10-3 × 2 × 4 × 3.142 × (4 × 10-2)2
= 1.005 × 10-3 J
The work done = 0.07 × 8 × 3.142 × (2 × 10-2)2
= 7.038 × 10-4 J
Question 23.
A rectangular wire frame of size 2 cm × 2 cm, is dipped in a soap solution and
taken out. A soap film is formed, if the size of the film is changed to 3 cm ×
3 cm, calculate the work done in the process. The surface tension of soap film
is 3 × 10-2 N/m. [Ans. 3 × 10-5 J]
Answer:
Data : A1 = 2 × 2 cm2 = 4 × 10-4 m2,
A2 = 3 × 3 cm2 =9 × 10-4 m2,
T = 3 × 10-2 N/m
As the film has two surfaces, the work done is W = 2T(A2 – A1)
= 2(3 × 10-2)(9 × 10-4 × 10-4)
= 3.0 × 10-5 J = 30 µJ
Questions and Answers
Can you tell? (Textbook Page No. 27)
Question 1.
Why does a knife have a sharp edge or a needle has a sharp tip ?
Answer:
For a given force, the pressure over which the force is exerted depends
inversely on the area of contact; smaller the area, greater the pressure. For
instance, a force applied to an area of 1 mm2 applies a
pressure that is 100 times as great as the same force applied to an area of 1
cm2. The edge of a knife or the tip of a needle has a small area of
contact. That is why a sharp needle is able to puncture the skin when a small
force is exerted, but applying the same force with a finger does not.
Use your brain power
Question 1.
A student of mass 50 kg is standing on both feet. Estimate the pressure exerted
by the student on the Earth. Assume reasonable value to any quantity you need;
justify your assumption. You may use g = 10 m/s2, By what factor
will it change if the student lies on back ?
Answer:
According to the most widely used Du Bois formula for body
surface area (BSA), the student’s BSA = 1.5 m2, so that the area of
his back is less than half his BSA, i.e., < 0.75 m2. When the
student lies on his back, his area of contact is much smaller than this. So,
estimating the area of contact to be 0.3 m2, i.e., 10 times more
than the area of his feet, the pressure will be less by a factor of 10 or more,
[Du Bois formula : BSA = 0.2025 × W0.425 × H0.725,
where W is weight in kilogram and H is height in metre.]
Can you tell? (Textbook Page No. 30)
Question 1.
The figures show three containers filled with the same oil. How will the
pressures at the reference compare ?
Answer:
Filled to the same level, the pressure is the same at the bottom of each
vessel.
Use Your Brain Power (Textbook Page 35)
Question 1.
Prove that equivalent SI unit of surface tension is J/m2.
Answer:
The SI unit of surface tension =
Try This (Textbook Page No. 36)
Question 1.
Take a ring of about 5 cm in diameter. Tie a thread slightly loose at two
diametrically opposite points on the ring. Dip the ring into a soap solution
and take it out. Break the film on any one side of the thread. Discuss what
happens.
Answer:
On taking the ring out, there is a soap film stretched over the ring, in which
the thread moves about quite freely. Now, if the film is punctured with a pin
on one side-side A in below figure-then immediately the thread is pulled taut
by the film on the other side as far as it can go. The thread is now part of a
perfect circle, because the surface tension on the side F of the film acts
everywhere perpendicular to the thread, and minimizes the surface area of the
film to as small as possible.
Can You Tell ? (Textbook Page No. 38)
Question 1.
How does a waterproofing agent work ?
Answer:
Wettability of a surface, and thus its propensity for penetration of water,
depends upon the affinity between the water and the surface. A liquid wets a
surface when its contact angle with the surface is acute. A waterproofing
coating has angle of contact obtuse and thus makes the surface hydrophobic.
Brain Teaser (Textbook Page No. 41)
Question 1.
Can you suggest any method to measure the surface tension of a soap solution?
Will this method have any commercial application?
Answer:
There are more than 40 methods for determining equilibrium surface tension at
the liquid-fluid and solid-fluid boundaries. Measuring the capillary rise (see
Unit 2.4.7) is the laboratory method to determine surface tension.
Among the various techniques, equilibrium surface tension is
most frequently measured with force tensiometers or optical (or the drop
profile analysis) tensiometers in customized measurement setups.
[See https: / / www.biolinscientific.com /measurements /surface-tension]
Question 2.
What happens to surface tension under different gravity (e.g., aboard the
International Space Station or on the lunar surface)?
Answer:
Surface tension does not depend on gravity.
[Note : The behaviour of liquids on board an orbiting spacecraft is mainly
driven by surface tension phenomena. These make predicting their behaviour more
difficult than under normal gravity conditions (i.e., on the Earth’s surface).
New challenges appear when handling liquids on board a spacecraft, which are
not usually present in terrestrial environments. The reason is that under the
weightlessness (or almost weightlessness) conditions in an orbiting spacecraft,
the different inertial forces acting on the bulk of the liquid are almost zero,
causing the surface tension forces to be the dominant ones.
In this ‘micro-gravity’ environment, the surface tension
forms liquid drops into spheres to minimize surface area, causes liquid columns
in a capillary rise up to its rim (without over flowing). Also, when a liquid
drop impacts on a dry smooth surface on the Earth, a splash can be observed as
the drop disintegrates into thousands of droplets. But no splash is observed as
the drop hits dry smooth surface on the Moon. The difference is the atmosphere.
As the Moon has no atmosphere, and therefore no gas surrounding a falling drop,
the drop on the Moon does not splash.
(See http://mafija.fmf.uni-Ij.si/]
Can you tell? (Textbook Page No. 45)
Question 1.
What would happen if two streamlines intersect?
Answer:
The velocity of a fluid molecule is always tangential to the streamline. If two
streamlines intersect, the velocity at that point will not be constant or
unique.
Activity
Question 1.
Identify some examples of streamline flow and turbulent flow in everyday life.
How would you explain them ? When would you prefer a streamline flow?
Answer:
Smoke rising from an incense stick inside a wind-less room, air flow around a
car or aeroplane in motion are some examples of streamline flow, Fish,
dolphins, and even massive whales are streamlined in shape to reduce drag.
Migratory birds species that fly long distances often have particular features
such as long necks, and flocks of birds fly in the shape of a spearhead as that
forms a streamlined pattern.
Turbulence results in wasted energy. Cars and aeroplanes are
painstakingly streamlined to reduce fluid friction, and thus the fuel
consumption. (See ‘Disadvantages of turbulence’ in the following box.)
Turbulence is commonly seen in washing machines and kitchen mixers. Turbulence
in these devices is desirable because it causes mixing. (Also see ‘Advantages
of turbulence’ in the following box.) Recent developments in high-speed
videography and computational tools for modelling is rapidly advancing our understanding
of the aerodynamics of bird and insect flights which fascinate both physicists
and biologists.
Use your Brain power (Textbook Page No. 46)
Question 1.
The CGS unit of viscosity is the poise. Find the relation between the poise and
the SI unit of viscosity.
Answer:
SI unit : the pascal second (abbreviated Pa.s), 1 Pa.s = 1 N.m-2.s
CGS unit: dyne.cm-2.s, called the poise [symbol P, named after Jean
Louis Marie Poiseuille (1799 -1869), French physician].
[Note : Thè most commonly used submultiples are the millipascalsecond (mPa.s)
and the centipoise (cP). 1 mPa.s = 1 cP.]
Use your Brain power (Textbook Page No. 49)
Question 1.
A water pipe with a diameter of 5.0 cm is connected to another pipe of diameter
2.5 cm. How would the speeds of the water flow compare ?
Answer:
Water is an incompressible fluid (almost). Then, by the equation of
continuity, the ratio of the speeds, is
Do you know? (Textbook Page No. 50)
Question 1.
How does an aeroplane take off?
Answer:
A Venturi meter is a horizontal constricted tube that is used to measure the
flow speed through a pipeline. The constricted part of the tube is called the
throat. Although a Venturi meter can be used for a gas, they are most commonly
used for liquids. As the fluid passes through the throat, the higher speed
results in lower pressure at point 2 than at point 1. This pressure difference
is measured from the difference in height h of the liquid levels in the U-tube
manometer containing a liquid of density ρm. The following treatment
is limited to an incompressible fluid.
Let A1 and A2 be the cross-sectional areas at
points 1 and 2, respectively. Let v1 and v2 be
the corresponding flow speeds. ρ is the density of the fluid in the pipeline.
By the equation of continuity,
v1A1 = v2A2 … (1)
Since the meter is assumed to be horizontal, from Bernoufli’s equation we get,
The pressure difference is equal to ρmgh, where h is the differences
in liquid levels in the manometer.
Then,
Equation (3) gives the flow speed of an incompressible fluid in the pipeline.
The flow rates of practical interest are the mass and volume flow rates through
the meter.
Volume flow rate =A1v1 and mass flow rate = density
× volume flow
rate = ρA1v1
[Note When a Venturi meter is used in a liquid pipeline, the pressure
difference is measured from the difference in height h of the levels of the
same liquid in the two vertical tubes, as shown in the figure. Then, the
pressure difference is equal to ρgh.
The flow meter is named after Giovanni Battista Venturi (1746—1822), Italian
physicist.]
Question 2.
Why do racer cars and birds have typical shape ?
Answer:
The streamline shape of cars and birds reduce drag.
Question 3.
Have you experienced a sideways jerk while driving a two wheeler when a heavy
vehicle overtakes you ?
Answer:
Suppose a truck passes a two-wheeler or car on a highway. Air passing between
the vehicles flows in a narrower channel and must increase its speed according
to Bernoulli’s principle causing the pressure between them to drop. Due to
greater pressure on the outside, the two-wheeler or car veers towards the
truck.
When two ships sail parallel side-by-side within a distance considerably less
than their lengths, since ships are widest toward their middle, water moves
faster in the narrow gap between them. As water velocity increases, the
pressure in between the ships decreases due to the Bernoulli effect and draws
the ships together. Several ships have collided and suffered damage in the
early twentieth century. Ships performing At-sea refueling or cargo transfers
performed by ships is very risky for the same reason.
Question 4.
Why does dust get deposited only on one side of the blades of a fan ?
Answer:
Blades of a ceiling/table fan have uniform thickness (unlike that of an
aerofoil) but are angled (cambered) at 8° to 12° (optimally, 10°) from their
plane. When they are set rotating, this camber causes the streamlines
above/behind a fan blade to detach away from the surface of the blade creating
a very low pressure on that side. The lower/front streamlines however follow
the blade surface. Dust particles stick to a blade when it is at rest as well
as when in motion both by intermolecular force of adhesion and due to static
charges. However, they are not dislodged from the top/behind surface because of
complete detachment of the streamlines.
The lower/front surface retains some of the dust because
during motion, a thin layer of air remains stationary relative to the blade.
Question 5.
Why helmets have specific shape?
Answer:
Air drag plays a large role in slowing bike riders (especially, bicycle) down.
Hence, a helmet is aerodynamically shaped so that it does not cause too much
drag.
Use your Brain power (Textbook Page No. 52)
Question 1.
Does the Bernoulli’s equation change when the fluid is at rest ? How ?
Answer:
Bernoulli’s principle is for fluids in motion. Hence, it is pointless to apply
it to a fluid at rest. Nevertheless, for a fluid is at rest, the Bernoulli
equation gives the pressure difference due to a liquid column.
For a static fluid, v1 = v2 =
0. Bernoulli’s equation in that case is p1 + ρgh1 =
ρ2 + ρgh2
Further, taking h2 as the reference height
of zero, i.e., by setting h2 = 0, we get p2 = p1 +
ρgh1
This equation tells us that in static fluids, pressure
increases with depth. As we go from point 1 to point 2 in the fluid, the depth
increases by h1 and consequently, p2 is greater
than p1 by an amount ρgh1.
In the case, p1 = p0, the
atmospheric pressure at the top of the fluid, we get the familiar gauge
pressure at a depth h1 = ρgh1. Thus, Bernoulli’s
equation confirms the fact that the pressure change due to the weight of a
fluid column of length h is ρgh.