Blooming science-8 part-1- press

Approved by Government of Nepal, Ministry of Education, Curriculum
Development Center (CDC).

Blooming

SCIENCE
&

ENVIRONMENT

Book

8

Authors
Raj Kumar Dhakal
Purushottam Devkota

Shubharambha Publication Pvt. Ltd.

Kathmandu, Nepal

Published by:

Shubharambha Publication Pvt. Ltd.

Kathmandu, Nepal
URL: www.shubharambhapublication.com
E-mail: [email protected]

www.facebook.com/shubharambhapublication

Blooming Science and Environment Book-8

Authors : Raj Kumar Dhakal

Purushottam Devkota

Video Content : Laxmi Nand Dhakal

Layout Design : Ram Malakar

Language Editor : Krishna Prasad Regmi

Copyright © : Publisher

Edition : 2077

© : Publisher

No part of this book may be reproduced or transmitted by means (electronic, photocopying,

recording or otherwise) without prior written permission from the publisher. Any breach

of this condition will entail legal action and prosecution.

What if books could talk to you?

That was the question we asked ourselves.
We believe education should be interactive and enjoyable. We’re living in the IT
era and education is the only sector lagging behind when it comes to
integrating new technologies inside and outside the classroom.
That’s why we teamed with Freelancer Company Pvt. Ltd. and Mr. Laxmi
Nand Dhakal to create an immersive experience to redefine how we learn. With
Laxmi Sir’s brains and the technological excellence of Freelancer Company,
we’ve created in-depth video materials that aid the topics covered in this textbook.
We hope to make learning fun again! With a few clicks, you will be able to learn

more about any topic in this book via practical experiment videos.

Freela nc er Company Pvt. Ltd. Laxmi Nand Dhakal
IT Company in Kathmandu Scientist, Teacher, HOD (R&D Dept)
https://creativescienceproject.com
https://freelancercompany.com.np Laxmi Nand Dhakal is the creative force
Freelancer Company took this project who made this project possible.
from concept and made it into reality His 20+ years of teaching experience
with their technical expertise. along with his dedication to the craft of
They handled everything from scripting, Science has added immense value to this
camera work, audio recording, video project.
editing, website design & development, We’re proud to have such a talented
QR code generation to the overall individual in our team.
management of the project.

Sc an & Enjoy! Shubharambha Publication Pvt. Ltd.
Thank you. Kathmandu, Nepal
Dinesh Adhikari
Academic Director

Find the step-by-step guide to using QR-code on the next page.

How to use QR codes
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be fully visible on the screen.
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video.
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it to watch the video.

Preface

The series Blooming Science and Environment has been brought out as an
indispensable resource for school level students and has intended to provide concise
and comprehensible explanation of key concepts, facts and principles across science
disciplines. Organized around the National Science curriculum prescribed by
Curriculum Development Centre, Sanothimi, Bhaktapur, the series presents solid
overviews of the most commonly encountered school science topics with sound
academic and fun activities.

The clear and accessible definitions, concise language, helpful diagrams and
illustrations and other science activities offered in this series will nonetheless help
teachers understand science concepts to the degree to which they can develop
rich and exciting inquiry approaches to exploring these concepts with students in
the classroom. As the series has been brought out considering the age and other
psychological factors of children, the learning materials in this series appeal to the
sense of the children and they are related to the world of young learners. Activities
with varieties of questions in this series are meant to assess and evaluate the level of
students’ inquisitiveness.

As each unit begins with its objectives and estimated teaching periods which help
teachers to complete the course in time. Moreover, each lesson in the series ends
with Let's Learn, Points to Remember and Boost Up Exercises; these two sections
are meant to provide good review to students and enhance their ability to solve the
exercise questions. Each lesson has Multiple Choices Questions, Project Work that are
meant to arouse more creativity and interest in the students for better understanding
and adjustment with their scientific world.

We are grateful to students, parents and principals who shared their valuable
suggestions in materializing this series. Any constructive suggestions and
recommendations for the betterment of this series will be highly acknowledged.

AUTHORS

Physics Contents

1. Measurement.................................................................................................7

2. Force and Motion........................................................................................19

3. Simple Machine..........................................................................................35

4. Pressure.......................................................................................................44

5. Work, Energy and Power............................................................................58

6. Heat.............................................................................................................73

7. Light............................................................................................................85

8. Sound.........................................................................................................101

9. Magnetism.................................................................................................109

10. Electricity..................................................................................................114

Chemistry

11. Matters......................................................................................................123

12. Separation of Mixture...............................................................................148

13. Metals and Non-metals.............................................................................156

14. Acid, Base and Salt...................................................................................169

15. Some useful Chemiclas.............................................................................179

Biology

16. Living Organism.......................................................................................187

17. Cell and Tissue.........................................................................................214

18. Life Process...............................................................................................226

Astronomy / Geology

19. Structure of the Earth................................................................................247

20. Climate and Weather................................................................................259

21. The Earth and the Universe.......................................................................268

Environment Science

22. Environmental and Its Balance.................................................................277

23. Environmental Degradation and Conservation.........................................301

24. Sustainable development and Biodiversity 317

Practical Work Sheet 331

Model Questions 334

List of Video Experiments 340

Chapter Physics

1 Measurement

Learning Outcomes Estimated Periods: 3+1

On the completion of this unit, the students will be able to:

 define fundamental and derived units.
 identify the units of length, mass and time, area and volume in different

measurement systems.
 introduce density and relative density with their formulae.
 solve some simple numerical problems related to density.

The term measurement is common and very important in our daily life, because we
perform different types of measurement in our every day activities. For example,
we measure the mass of vegetables by using beam balance, length of any object by
using scale, temperature of the body by using thermometer, etc. So, mass, length,
temperature, etc. are physical quantities. A physical quantity is that which can be
measured directly or indirectly. Density, area, volume, time, etc. are some more
examples of it. Feeling, kindness, innocence, etc. are not physical quantities because
they cannot be measured. Physical quantities are of two types. They are fundamental
quantity and derived quantity.

In our daily life we take different measurements. We use different kinds of
measurements while buying goods in the market, making play ground for table
tennis, foot-ball, volley ball, high jump, long jump, race etc. and studying different
subjects in the class. We also measure things at home. We use measuring devices
like measuring tape or scale for the measurement of length, weights and weighting
balance for measuring masses and clock or watch for measurement of time. We come
to know the exact quantity of a substance only when we measure the substance with a
standard measurement. In physics, quantities which are available in our surroundings
and can be measured are called physical quantities.

While studying physics, we have to do different kinds of activities and in many
activities we have to measure things correctly. Correct observation, correct
measurement and correct conclusion or results are necessary in science. Many
activities of our life are also based on correct measurement. We measure length,
time, mass, area, volume, density etc. Measurement is necessary in our daily life.

The process of comparing a physical quantity with known standard quantity of the
same kind is called measurement. Measurement is essential to get the exact quantity
of the substance in our daily life.

Blooming Science & Environment Book 8 7

Importance of Measurement
1. The measurement helps in selling and buying goods.
2. It is important in performing scientific experiments to establish truth about the

physical phenomenon.
3. It helps to obtain the accurate results about physical quantities.
4. It helps in performing experiment for making our daily food easier.

Fundamental Physical Quantities
Those physical quantities, which do not depend upon other quantities are called
fundamental quantities. For example, mass of a body is fundamental quantity
because it cannot be expressed in terms of other quantities. How many meters equal
1 kg? doesn’t make any sense. Mass is not depended with length. Length, time,
temperature, electric current, amount of substance, luminous intensity, etc. are the
examples of fundamental quantities.

Derived Physical Quantities
Those physical quantities which are to be expressed in terms of other fundamental
quantities are called derived quantities. And they cannot be measured directly. Area
is a derived quantity as it depends on fundamental quantity, the length. Some other
examples of derived quantity are velocity, acceleration, area, volume, density, etc.

Fundamental and Derived Units

Both types of physical quanties can be measured by using various units to know the
exact quantity. Unknown physical quantities are compared with the known standard
quantity and expressed interms of unit. Unit is a fixed definite quantity which is
used as standard of reference to measure the quantity of same kind. For eg. metre,
centimetre etc are the units of length and gram, kilogram etc are the units of the mass.
There are two types of units:
1. Fundamental units
2. Derived units

8 Blooming Science & Environment Book 8

1. Fundamental units

The units which are used to measure the fundamental physical quanties are
called fundamental units. These units do not depend on other units. For example;
metre (m), kilogram (kg) and second (s) are fundamental units. The following are the
fundamental units used in physics.

Fundamental Quantity Unit Symbol

Length metre m
Mass kilogram kg
Time second s
Temperature Kelvin K
Electric current ampere A
Luminous intensity candela cd
Amount of substance mol
Mole

2. Derived units

The units which are used to measure derived physical quantities are called derived
units. These units are formed by the combination of two or more fundamental units.
For example; the unit of area (m2), volume (m3), density (kg/m3), work done(Joule),
power (watt), etc are some derived units.

The unit of density is kg/m3 which is derived unit. This is because it is expressed with
the help of fundamental units kg (unit of mass) and metre (unit of length). Similarly,
the unit of volume is m3, which is derived unit as it depends on the fundamental unit,
metre (unit of length). Unit of area, speed, velocity, acceleration, etc. are some other
examples of derived units.

Standard System of Measurement

People in different places of our country still use the local units like haat, bitta, mana,
pathi, muri, dharni, etc. for the measurement of length and mass. These units are
understood by the people in a certain place and may vary from place to place and
from person to person. In order to maintain the uniformity in measurements, some
other standard system of measurement have been used.

System of Units

Different units of fundamental units combine together to form a system of units.
There are four system of units used in measurement.

(i) CGS system (ii) MKS system

(iii) FPS system and (iv) SI system

Blooming Science & Environment Book 8 9

(i) CGS System: The full form of this system is Centimetre-Gram-Second system.
In this system, the fundamental units are centimeter, gram and second.

(ii) MKS System: The full form of this system is Metre-Kilogram-Second system.
The system in which length is measured in metre (m), mass in kilogram (kg)
and time in second (s) is called MKS system. In this system, the fundamental
units are metre, kilogram and second.

(iii) FPS System: The full form of this system is Foot-Pound-Second system. In
this system, the fundamental unit are foot, pound and second.

(iv) SI System: The full form of this system is System Internal De’ Unit. This
system is a modified form of MKS system and is called international system of
units and in short it is written as SI. In this system, the fundamental units are
metre, kilogram and second. This system includes four units such as the unit of
electricity-Ampere (A), unit of temperature-kelvin (K), unit of light intensity-
candela (cd) and unit of mass of matter-Mole (mol).

Measurement of Mass

Mass is the quantity of matter contained in a body. The mass of an object is measured
in kilogram (kg). The mass of an object is measured by beam balance. A beam balance
is shown in the figure below. The objects whose mass is to be found is placed on one
pan and the standard weights are placed on the other pan. Mass of a body depends
on the number and the size of atoms or molecules in it. It is the measure of inertia of
a body.

One Kilogram mass 250gm 1/2 Kg 1 Kg 5 Kg

Kilogram is the SI unit of mass which is same
all over the world. The mass of standard
weight (Dhak) made up of platinum and
iridium alloy kept at International Bureau
of Weight and Measurement in France is
considered as 1 kg and standard weight of
1 kg used in other countries are made by
comparing with that standard weight.

The metric table of mass is given below:

10 milligram (mg) = 1 centigram (cg)

10 centigram (cg) = 1 decigram (dg)

10 decigram (dm) = 1 gram (gm)

10 gram (gm) = 1 decagram (dag)

10 Blooming Science & Environment Book 8

10 decagram ( dag) = 1 hectogram (hg) Scan for practical experiment
10 hectogram (hg) = 1 kilogram(kg)
100 kilogram (kg) = 1 quintal (q)
10 quintal (q) = 1 ton (t)

Measurement of Time visit: csp.codes/c08e01

Time is the interval between two events. In ancient time, people didn't have watch.
By studying regular events such as sunset, sunrise, change in seasons, etc. they used
to calculate time. They called the time interval between two consecutive full moons
as one month and the time interval between two consecutive sunrises or sunsets

as a day.

Wall Clock Pendulum Clock Stop Watch

Nowadays, people use a watch to measure time. There are different types of watches
such as digital watch, quartz watch, pendulum watch, stop watch, etc. to measure
time. The digital watch and the quartz watch give the accurate time. A stop watch can
be started or stopped according to our desire and is used to measure the time taken
by the events.

The SI unit of time is second. We also use bigger units such as minute, hour, day,
year, etc. to measure the longer period of time. These bigger units of time are called
multiples of second.

We also use smaller units of time such as microsecond and millisecond to measure
the smaller period of time. These are called sub-multiples of second.

We know that the earth rotates about its own axis as well as it revolves around the
sun. The time taken by the earth to rotate once about its own axis is called a day. If
a day is equally divided into 24 equal parts, each parts is called an hour. If one hour
is subdivided into 60 equal parts, each part is called a second. One second may be
defined as the 1 th part of a solar day.

86,400

The time taken by the earth to revolve once around the sun is called one year. The
earth takes about 365.25 days to revolve once around the sun. Hence, one year equals
365.25 days.

Blooming Science & Environment Book 8 11

Measurement of Weight Weight

The weight of a body is the force with which the earth pulls it. The
weight of a body does not remain the same at all places but it is different
at different places. The weight of a body is measured by a spring balance
but not by physical balance because a physical balance does not measure
the amount of pull of gravity i.e. weight of body. A spring balance is
shown in figure below. Thus a spring balance tells us only the amount of
force of gravity acting on the body at a place but it does not tell us the
mass of the body. So, a spring balance only measures the force of gravity
i.e. weight of the body. The SI unit of weight is newton (N). Weight of a
body can be measured as the product of mass of body and acceleration
due to gravity i.e.

Weight (w) = m × g

Differences between Mass and Weight

S.N. Mass S.N. Weight

1. It is the quantity of matter 1. It is the force with which a body

contained in a body. is attracted towards its centre of a

planet/satellite.

2. Its value is the same everywhere 2. Its value differs from place to

in the universe. place.

3. It is a scalar quantity. 3. It is a vector quantity.

4. It is measured in kilogram. 4. It is measured in newton.

5. It is measured by using physical 5. It is measured by using a spring

balance. balance.

Density

In physics, the word density refers to the lightness or heaviness of different materials.
The density of a substance is defined as its mass per unit volume. Generally, the
symbol used for density is ρ(Rho).

∴ Density Mass
= Volume

M
or, ρ = V

The SI unit of mass and volume are kilogram (kg) and cubic metre (m3) respectively.
Hence, the SI unit of density is kg/m3. Its other unit is g/cm3, when the mass of a
body is measured in gram (g) and volume in cubic centimetre (cm3). Density does

12 Blooming Science & Environment Book 8

not depend on shape and size of a body. It is same for a given material whatever be
its shape and size. But, density of a body decreases with the increase in temperature.

We say: Iron is heavier than wood. But we have seen that a wooden chair is heavier
than an iron nail.

If we take 1 kg iron and 1 kg of wood, their weight will be exactly the same but their
volume will be different. Again, if we take equal volumes of iron and wood, the iron
will be many times heavier than the wood.

The density of water is 1g/cm3 or 1000 kg/m3 and ice is 920 kg/m3 or 0.92 g/cm3.

Relative Density (or Specific Gravity)

Relative density means comparative density i.e. when the density of a body is
compared with that of another body.

For instance, the relative density of a body A with respect to that of body B is given

as, Relative density = Density of body A .
Density of body B

Water is chosen, as a standard substance because it is the most common liquid and
its density is 1g/cm3 at 4oC.

Thus, when we divide the density of a body by that of water, we get the relative
density of the substance. The division is simple and the answer comes as the same
as that of the density.

Thus, the relative density of a body is defined as the ratio between the densities of the
body to the density of water at 4oC.

Density of Substance
So, Relative density = Density of Water at 4oC

Let’s calculate the relative density of iron whose density is 7.9 g/cm3 and density of
water is 1 g/cm3.

Density of Iron = 7.9g/cm3 = 7.9
Relative density of iron = Density of water at 4oC 1g/cm3

R.D. of a substance is measured by using hydrometer.

Unit of Relative Density: Since the relative density is a ratio between the density of
a body and the density of water, it is a pure number and hence has no unit.

Relative density signifies how many times a given substance is heavier than water
when the volume of a substance and water are same.

If the relative density of a substance is more than one, the substance is heavy and
sinks in water. Conversely, if the relative density is less than one, then the substance
is lighter than water and hence will float in water.

Blooming Science & Environment Book 8 13

Differences between the Density and R.D of a Substance

Density Relative density (R.D or S.G)

1. It measures the mass of a body in 1. It measures the density of a body
unit volume. w.r.t. the density of water at 4oc.

2. Density (r) = mass/ volume 2. R.D. = density of substances/ density
of water at 4oc

3. Its SI unit is kg/m3 3. It is unit less.

4. After measuring mass and volume of 4. It can be measured by using hydrom-
substance only it can be measured by eter.
using formula r = m/v

Solved Numerical Problem

Density of milk is 1.03 g/cm3 and density of water is 1g/cm3. Find the relative
density of milk?

Solution: Here, Density of milk = 1.03 g/cm3

Density of water = 1 g/cm3

Relative density of milk = ?

We know,

Relative density of milk = Density of milk = 1.03g/cm3
∴ Density of water 1g/cm3

The relative density of milk = 1.03

Floating and Sinking

The solid bodies sink in liquids if their density is greater than that of the liquids.
Unlike it, solids floating, if they have lower density than that of the liquids. For
example, a steel ball sinks in water, but floats in mercury. Cork and wood float on
water but metals like iron and steel sink in it.

So, the floating and sinking of a body depends upon the density of the liquid and the
density of the body. The body which sinks in liquid can be made to float by giving
a proper shape such that it displaces liquid equal to its own weight. Iron ball sinks
in water but the ship made of iron floats in water. A ship is given a proper shape to
displace water equal to its own weight.

14 Blooming Science & Environment Book 8

Activity

Materials required: Beaker, water, cork, iron piece, kerosene

Method:

1. Take a beaker partially filled with water.
2. Place a piece of iron on the water surface. Does it sink or float?
3. Now, place a piece of wooden cork in the water. What happens now?
4. Now pour a drop of kerosene over the water and observe.

We know that the density of water is less than Cork floats Iron sinks
that of iron, but the density of water is higher
than that of cork. This activity confirms that
those substances whose density is less than
that of liquid float on it and those substances
whose density is greater than that of liquid
sink in the liquid.

Density and R.D of some substances.

Substance Density in kg/m3 gm/m3 R.D
1 1
Water 1000 0.8 0.8

Kerosene 800 0.92 0.92
0.6-0.8 0.6-0.8
Ice 920
7.8 7.8
Wood 600-800 13.6 13.6
2.7 2.7
Iron 7800

Mercury 13600

Aluminium 2700

Principle of Floatation

When a body is immersed in liquid, it experiences an up-thrust. If the up-thrust given
by the liquid is more than the weight of the body, then it floats on liquid. According
to the law of floatation, a body floats on liquid only when the weights of the displaced
liquid is equal to weight of the object.

Weight of a floating body = Weight of water displaced by it.

Solved Numerical Problem

Calculate density of a body of mass 50 kg if it has volume 5m3.
Solution:

Here, Volume of a body (v) = 5 m3
Mass (m) = 50 kg
Density (D) = ?

Blooming Science & Environment Book 8 15

We have, D = M = 50 kg/5m3 = 10 kg/m3
V

∴ Density of a body is 10 kg/m3.

Main Points to Remember

1. Those quantities, which can be measured are called physical quantities.
2. The physical quantities, which are independent of each other are called

fundamental quantities.
3. The physical quantity, which depends on fundamental quantities is derived

quantity.
4. Measurement means comparision of an unknown quantity with a known

quantity.
5. Those units, which are independent of each other are called fundamental units.
6. Those units which depend on fundamental units are called derived units.
7. CGS, MKS, FPS and SI Systems are the standard systems of units.
8. Mass per unit volume is called density. Its unit is kg/m3 or g/cm3.
9. The ratio of density of a substance to density of water at 4oC is called Relative

density.

PRO J ECTWORK

Take a wooden block. Find its mass and dimensions to calculate its density.

Exercise
1. Fill in the blanks.

a. Square metre is the SI unit of ……………………..
b. Acceleration is a ……………. quantity.
c. 1 day is equal to …………………. seconds.
d. The time is determined by using a …………………
e. Pendulum clock is used to determine the ……………..

2. Tick (√) for the correct and cross (×) for the incorrect statements.
a. Kg/m3 is a derived unit.
b. There are 86,400 seconds in a day.
c. A measuring cylinder is used to determine the time.
d. Volume is not a derived quantity.
e. The RD has no unit.

16 Blooming Science & Environment Book 8

3. Answer the following questions.
a. What is the standard unit used to measure the mass?
b. What is the SI unit of temperature?
c. Write some formula used to calculate the area of a regular object.
d. Name the fundamental units involved in the of density unit.
e. Encircle the SI units from the give list of units given:
Minute, Kelvin, mole, bitta, dharni, metre.
f. What do you mean by measurement?
g. What is the importance of measurement?
h. Define fundamental and derived quantities with two examples of each.
i. The unit of the area is derived unit. Why?
j. Distinguish between fundamental and derived unit.
k. What is SI system? Write the differences between SI system and CGS
system.
l. What is mass? Write its SI unit.
m. What do you mean by physical quantity?
n. What is density? Define relative density.
o. Distinguish between mass and weight.
p. Distinguish between density and relative density.

4. Define.

a. FPS system b. Standard unit c. Unit

d. MKS system e. Derived unit f. CGS system

5. Numerical Problems.

a. Convert the following:

a. 340 cm into m b. 67 kg into g

c. 2 days into seconds d. 86,400 seconds into day.

b. The length of a book is 20 cm, its breadth is 15 cm and its height is 25 mm.
Find its volume and area of the greatest face.

c. The volume, length and height of a room are 81m3, 500 cm and 3.5 m
respectively. Find its breadth.

d. The area of a surface is 42000 mm2. If its breadth is 20 cm, find its length
in cm, mm and m.

Blooming Science & Environment Book 8 17

e. Study the give diagram and find the greatest surface area. Also calculate the
volume of the brick.
10 cm
6 cm

5 cm

6. Study the give figure and answer these questions:
a. Write the level of A and B.
b. Why does the body immerse in water?
c. What is the volume of the body that is immersed in water?

7. Calculate density of a body if its mass is 20 kg and its volume is 4 m3.

8. Calculate relative density of a metal if density of metal is A
13.6 g/cm3 and density water at 4oc is 1 g/cm3.
60
Answers 50
40
(2) 750 cm3, 300cm2 (3) 4.83m
30 B
(4) 21 cm, 210 mm, 0.21m
20
10

(5) 300 cm3, 60cm2 (6) 60, 30 cm3

(7) 5 kg/m3 (8) 13.6

Glossary

Density : the ratio of mass and volume of a substance.

Metric system : the system of measurement that uses the metre, the kilogram
and the second as basic units

Current : the flow of electricity through a wire

Luminous intensity : measure of the light-emiting ability of a source of light

Metrology : the scientific study of measurement

Boundary layer : the thin layer of fluid formed around a solid body

Immerse : to put something into a liquid so that it is completely covered

18 Blooming Science & Environment Book 8

Chapter Force and Motion

2

Learning Outcomes Estimated Periods: 4+1

On the completion of this unit, the students will be able to:
 describe average and relative velocity.
 define acceleration and retardation
 write and use equations of motion
 solve some simple numerical related to velocity and acceleration

Rest and Motion

In our daily life we see many objects. Some of them are at rest and others are in
motion. We compare one object with other objects in the surroundings to determine
whether it is at rest or in motion. The trees on the roadside, buildings, etc. are at rest.
They do not change their position with respect to other objects. The flying birds,
planes in the sky, flying kite, vehicles on the road are said to be in motion. These
bodies change their position with respect to their surroundings.

A body is said to be at rest if it does not change its position with respect to other
objects in its surrounding. A body is said to be in motion if it changes its position
with regard to objects in its surroundings. For determining rest and motion, we should
consider a fixed point or reference frame.

Reference Frame:

What is reference Point?

The reference point or frame is the point
about which we study the motion of
the object. For example, when we are
travelling by bus, we are in motion with
respect to a nearby tree. Here, the tree
acts as a reference point or frame.

Blooming Science & Environment Book 8 19

Uniform and Variable Motion
Let us consider a car moving on the road as shown in fig.

4 sec 3 sec 2 sec 1 sec
20 m 20 m 20 m 20 m

In the above figure a car moves from position A to position D. The distance covered
by the vehicle in every second can be understood from the figure. In every second, it
covers 20m. This means the car is moving with uniform motion.

If a body can cover equal distances in equal intervals of time then the body is said to
move with uniform motion.

Let us study the variable motion of a body in the figure.

11 sec 9 sec 5 sec 2 sec

20 m 20 m 20 m 20 m
D C B A

In figure, we can see that the car is not covering equal distance in equal time. That
means it is slow or fast at the same given time. It covers different distance in the
same time. Thus, a body is said to move with variable motion if it covers different
distance in the same interval of time.

The motion of the body cannot be same all the time. The motion of a body up a steep
road is generally slow. The bicycles can run fast down slopes.

Speed

When you travel in a bus, you want to know how far your destination is and how long

it will take to reach that destination. These two quantities, distance and time, are

related to each other and give you the idea of speed. Speed is the distance covered in

unit time. Speed is defined as the rate of change of distance with time.

Speed (m/s) = distance travelled (m) Scan for practical experiment
time taken (s)

Speed is measured in m/s or Km/hr.

When a body travels equal distance in equal interval of time
the body is said to be moving with uniform speed.

visit: csp.codes/c08e02

20 Blooming Science & Environment Book 8

Solved Numerical Problems

1. If a person takes 10 minutes to cover a distance of 1200m, how far does s/
he move every second?

Solution:

Distance = 1200m

Time = 10 min = 10 × 60 = 600 seconds

Distance travelled in one second = distance
time

= 1200
600

= 2m/s

∴ The speed is 2m/s.

2. If a vehicle covers 3km in 5 minutes, how far does it move in 1 second?

Solution:

Distance = 3km = 3 × 1000 = 3000m

Time = 5 min = 5 × 60 = 300 seconds

Distance traveled in 1 second = distance
time

= 3000
300

= 10m

∴Distance travelled in 1 second = 10m, so its speed is 10m/s.

Average Speed

A car travelling on a straight road which is clear of traffic, might possibly keep a

steady speed, and in 1 hour it could travel 40 km, in which case the speedometer

would indicate 40 km/hr throughout that hour. On most roads however, the speed

would vary, and in such a case we could talk of an average speed of 40km/hr, although

at various times during that hour the speedometer may indicate 80 km/hr, 20 km/hr,

10 km/hr, and zero. If the car had travelled at a constant speed of 40 km/hr, it would

have covered the same distance in 1 hour.

total distance travelled
Average speed = total time taken
D
or, Speed = t

Blooming Science & Environment Book 8 21

Vector and Scalar Quantities
Physical quantities are classified into two types. They are vector and scalar quantities.

Vector

Suppose a car is moving with a velocity of 50m/s from east to west. It means that
it covers 50m in one second towards the west. Velocity has both magnitude and
direction. So, it is called a vector quantity. Similarly displacement, force, momentum
etc are vector quantities.

Vector is a physical quantity that has both magnitude and direction.

Scalar

Physical quantities like mass, temperature and speed have only magnitude but no
direction. Speed tells us distance covered by a body per unit time in any direction. It
means it is a scalar quantity.

Scalar is a physical quantity which has only magnitude but no direction.

The speed has only magnitude and no direction. Magnitude can also be said as
quantity. Thus a quantity which has magnitude and no direction is called a scalar
quantity. Speed is a scalar quantity.

Velocity

In the given fig the car covers 10m in each 10 m/s B
second. That is, it covers equal distances in A
equal intervals of time in definite direction.
Thus, velocity of the car is 10 m/s.

The term speed and velocity are often used

synonymously and are both represented by D C
the symbol ‘v’. The velocity is the speed
10 m/s
in a particular direction, i.e. it is the rate of
Fig: Measurement of speed
change of distance in a particular direction.

As the distance in a particular direction is called displacement, velocity is defined as

the rate of change of displacement with time.

Velocity = change of displacement
time taken

It is measured in m/s or cm/s or km/hr.

Velocity (v) = distance (m)
22 time (s)

Blooming Science & Environment Book 8

Differences between Speed and Velocity

S.N. Speed S.N. Velocity

1. Rate of change of distance is 1. Rate of change of distance in

called speed. a particular direction is called

velocity.

2. Speed is a scalar quantity since 2. Velocity is a vector quantity since it

it has only magnitude. has both magnitude and direction.

3. Speed = distance travelled ( d ) 3. Velocity = displacement ( s )
time t time t

Variable Velocity

If a body does not cover equal distance in equal interval of time, the body is said
to move with variable velocity. In this case, the body covers different distances in
different interval of time.

Solved Numerical Problems

1. If a man reaches a 25 kilometre distance in 30 minutes in his car, find the
average velocity.

Solution:
Time = 30 min = 30 × 60 = 1800 seconds

Distance (s) = 25 km = 25 × 1000 = 25000 m

Average Velocity =?

We have,

Average velocity distance (m)
= time (s)
25000
= 1800

Hence, Average velocity = 13.89 m/s

2. A vehicle is moving with a velocity of 20m/s. How far does it move in one
hour?

Solution: = 20m/s
Velocity (u) = 3600 seconds
Time (t) = 1 hour =?
Distance (s)
We have,


Blooming Science & Environment Book 8 23

s = Average velocity × time (u × t)

= 20 × 3600

= 72000 m

= 72 km

Hence, Distance covered = 72 km

Relative Velocity

The velocity of a body described with respect to a reference point is called relative
velocity. Point R represents a reference point in the following examples.

Case 1

Let the car A and car B start from point M. The velocity of car A becomes 10 m/s and
the velocity of car B becomes 15 m/s.

After 1 second, car A has travelled 10 m to the left and car B has travelled 15 m to the
right of point M. From car A, vehicle B appears to have travelled 25 m in 1 second.
Here relative to car A, the velocity of car B is 25m/s.

Thus, if one object is moving in the direction opposite to the other, then the velocity
of A relative to B is calculated by using the formula.

VAB = VA + VB

= 10 m/s + 15 m/s

= 25 m/s

The velocity of B relative to A is calculated by using the formula:

VBA = VB + VA

= 15 + 10

= 25 m/s

AB

10 m/s 15 m/s

A 10 m 15 m B
10 m/s 15 m/s
After 1 second

M

24 Blooming Science & Environment Book 8

Case 2

Let car A and car B are at the same initial position. If the direction to the right is
considered ‘+’, then velocity of car A becomes + 10 m/s and the velocity of car B
becomes + 15 m/s

After 1 second, car A covers 10 m and car B 15 m.

Now, if two objects A and B are moving in the same direction on the same path with
velocities VA and VB respectively, then the velocity of A relative to B is calculated by
using the formula.

VAB = VA - VB
= 10 - 15

= -5 m/s

The velocity of B relative to A is calculated by using formula

VBA = VB - VA

= 15 - 10

= 5m/s

A +ve
+ve 15 m

10 m/s 10 m M

After 1 second
B 15 m/s

Case 3

Car A and car B are at the same initial position. If the direction to the right is
considered ‘+’, then the velocity of car A becomes + 10m’s and also the velocity of
car B becomes + 10m/s. After 1 second, car A covers 10m and car B also covers 10
m from point M.

Now, if two objects A and B are moving in the same direction on the same path with
same velocity, then the relative to object A, object B has zero velocity.

Blooming Science & Environment Book 8 25

A +ve A +ve
10 m/s 10 m/s
10 m

B After 1 second B
10 m/s 10 m/s

10 m

M

Solved Numerical Problems

1 Two vehicles are running in the same direction. The speed of the first
vehicle is 16 m/s and that of the second vehicle is 4 m/s. What is the velocity
of the first vehicle relative to the second vehicle? What would be their
relative velocity if they were moving in opposite direction?

Solution:

Here, Speed of first vehicle (VA) = 16m/s

Speed of second vehicle (VB) = 4 m/s

Relative velocity of the first vehicle to second vehicle when both are

moving in the same direction (VAB) = ?

VAB = VA - VB

= 16 - 4

= 12 m/s

∴ The relative velocity = 12 m/s.

When they move in opposite direction, the relative velocity (VAB) = ?

VAB = VA + VB

= 16 + 4

= 20 m/s

2. The velocity of car A is 20 m/s towards the east and the velocity of the car B
is 15 m/s towards the west. If both the cars start from the same point at
the same time, what will be the distance between them after 2 minutes?
What distance will each of them have travelled during time?

Solution:

Here, Velocity of car A towards east (VA) = 20 m/s
Velocity of car B towards west (VB) = 15 m/s

26 Blooming Science & Environment Book 8

Time taken = 2 minutes

= (2 × 60) s

= 120s

Relative velocity of the car A and B (VAB) =?
VAB = VA + VB
= 20 + 15

= 35 m/s

Distance between them after 2 minutes (S) ?

SAB = Relative velocity x time [∴ v = s ]
t

= 35 x 120

= 4200 m

Again to find the distance covered by car A and car B,

Distance covered by car A moving towards east (SA) = VA × t
= 20 × 120

= 2400 m east

Distance covered by car B moving towards west (SB) = VB × t

= 15 × 120

= 1800 m west

Acceleration Scan for practical experiment

It is practically not possible for a body to move with uniform visit: csp.codes/c08e03
velocity. The velocity of an object changes either by increase
or decrease of speed, or by a change of direction. The
velocity also changes when both direction and speed change.
Whenever velocity of an object changes with respect to time,
it is said to have acceleration. Acceleration is defined as the
rate of change of velocity with time.

Let a body moves from initial velocity ‘u’. It travels for ‘t’ seconds and gains next
velocity of final velocity ‘v’. Then,

Change in velocity = Final velocity - Initial velocity

Rate of change in velocity = Final Velocity - Initial velocity
Time

Blooming Science & Environment Book 8 27

Final velocity - Initial velocity

Acceleration (a) = Time taken

or, a = v - u …………..(i)
Unit t

If the change in velocity is measured in m/s and the time in second, the acceleration
is measured in metre per second per second. For example, the acceleration of a bus
4m/s2 means in each second, the velocity of the bus increases by 4m/s.

If the velocity increases after every second, the acceleration is said to be positive.
If the velocity decreases after every second, the acceleration is said to be negative.

Retardation is the rate of decrease in velocity. A negative acceleration is also known
as retardation or deceleration.

Solved Numerical Problems

1. A man leaves home on a bicycle and reaches his office in 50 minutes. His
office is 10 km away from his home. Calculate the average speed of his
bicycle.

Solution: Here, Time (t) = 50 minutes

= (50 × 60)s

= 3000s

Distance (s) = 10 km

= 10 × 1000 m

= 10000m

Average velocity (v) =?

According to the formula,

v = s
t

= 10000
300

= 3.33 m/s

Therefore, the average velocity of bicycle is 3.33 m/s.

28 Blooming Science & Environment Book 8

2. If a school bus runs with a speed of 50 m/s, how far does it reach in 8
minute?

Solution: Here, Speed (u) = 50 m/s

Time (t) = 8 min

= 8 × 60

= 480 s

Distance (s) =?

According to the formula,

u = s
s t

= ut

= 50 × 480

= 24,000 m

= 24 km

Therefore, the school bus travels a distance of 24 km.

Equation of Motion in a Straight Line

A car is moving along a straight line with its initial velocity ‘u’, distance ‘s’,
acceleration ‘a’, time ‘t’ and final velocity ‘v’.

According to the definition of acceleration.

Acceleration = final velocity - initial velocity
time
v-u
a = t

or, t = v - u …………………… (i)
t

or, v - u = at

v = u + at ………………... (ii)

Average velocity = v + u……………….. (iii)
2

Average distance = average velocity × t

s = u + v × t ……………. (iv)
2

Substituting the value of v from (ii) to (iii),
u + v
s = 2 × t

Blooming Science & Environment Book 8 29

s = u + u + at × t
2

= 2u + at × t
2

= 2ut + at2
2 2

s = ut + 1 at2
2

v2 = u2 + 2as ………………... (vii)

Again,

Substituting the value of t from (i) to (iv)

s = v + u × v-u
or, s 2 a

= v2 - u2
2a

or, v2 - u2 = 2as………………….. (vi)

Solved Numerical Problems

1. A bus starts to move from rest and acquires a velocity of 30 m/s in
10 seconds. Calculate its acceleration.

Solution:

Here, Initial velocity (u) = 0m/s

Final velocity (v) = 30 m/s

Time (t) = 10 sec

Acceleration (a) = ?

We have,

a = v-u
t

= 30 - 0
10

= 30
10

= 3m/s2

Therefore, the acceleration of bus is 3m/s2.

30 Blooming Science & Environment Book 8

2. A car starts moving from the rest. If the acceleration of the car remains
4m/s2 for 10 seconds, what will be its final velocity?

Solution:

Here, Initial velocity (u) = 0 m/s

Time (t) = 10 s

Acceleration (a) = 4m/s2

Final velocity (v) = ?

We have,

a= v-u
t

or, v = u + at

= 0 + 4 × 10

= 40 m/s

Therefore, final velocity of a car will be 40m/s.

3. A car is moving with a velocity of 10 m/s. The driver applies brakes and
brings the car to rest in 5 seconds. Calculate its retardation.

Solution: Initial velocity (u) = 10 m/s
Here, Final velocity (v) = 0 m/s
Time taken (t) = 5 seconds
Retardation (-a) =?


According the formula,

a = v-u
t

= 0 - 10
5

= 10
5

or, = -2m/s2

or, -a = 2m/s2

Thus, the acceleration of the car is -2m/s2. It is negative in sign, but the negative
acceleration is known as retardation, so the car has a retardation (-a) of 2m/s2.

Blooming Science & Environment Book 8 31

Main Points to Remember

1. A body is said to be at rest if it does not change its position with respect to the
surroundings.

2. A body is said to be at motion if it changes its position with respect to the
surrounding.

3. Rest and motion are relative terms.
4. Displacement is the shortest distance between initial and final position of a

body.
5. Those physical quantities which have both magnitude and direction are called

vector quantities whereas those with only magnitude are called scalar quantities.
6. Speed is distance travelled by a body per unit time in any direction.
7. Velocity is a distance travelled by a body per unit time in particular direction.
8. Acceleration is the rate of change in velocity.
9. The negative acceleration is known as deceleration or retardation.
10. Relative motion is the motion of a body described with respect to reference

point.
11. The average speed is calculated by dividing the total distance by time.
12. If a body covers equal distance in an equal interval of time, it is called uniform

motion.
13. If a body covers different distances in an equal interval of time, it is called

variable motion.

PRO J ECTWORK

Go to the playground of your school with measuring tape and stopwatch. Make
some group of your friends. Mark the required points and calculate the velocity
and relative velocity of some of your friends.

Exercise

1. State whether “True” or “False” for the following statements.
a. Rest and motion are relative terms.
b. Everything on the earth is at rest.
c. The total distance moved by a body is called displacement.
d. Speed and velocity give the same meaning.
e. Arithmetic mean of initial and final velocity is known as average velocity.
f. Rate of increase in velocity is called acceleration.

32 Blooming Science & Environment Book 8

2. Answer the following questions.
a. Define rest and motion.
b. What is displacement.
c. What is the difference between speed and velocity?
d. What do you understand by average speed?
e. What is acceleration?
f. What is uniform motion? How do you differentiate it from variable motion?
g. What is the point of reference?
h. What is uniform velocity?
i. What is relative velocity?
j. The velocity of a car is 15m/s but its velocity is zero. How do you express
such a situation of a moving body?
k. The velocity of a car is 10m/s after 10 seconds. Is it moving with uniform
velocity? Explain.
l. What is a reference frame?
m. Distinguish between displacement and distance travelled.

3. Prove mathematically: (a) s = ut + 1 at2 (b) v2 - u2 = 2as
2

4. Define the following.

a. Velocity b. Uniform-velocity c. Variable-velocity

d. Object at rest e. Object in motion f. Acceleration

g. Retardation

5. Solve the following numerical problems.

a. A car starts from rest. If the acceleration of the car is 0.2 m/s2, what will

be its velocity at the end of 3 minutes? And what distance will it cover at

that time? (Ans: 36m/s, 3240m)

b. A car is moving with at a velocity of 45 km/hr. The driver applies the
brakes and the car comes to rest in 3 seconds. What is its retardation?
How far does it move before coming to rest? (Ans. 4.16m/s2, 93.75m)

c. A train is moving with a velocity of 72 km/hr. When the brakes are
applied, the acceleration is reduced to -0.5 m/s2. Calculate the distance
covered by the train before coming to rest. (Ans. 40 seconds; 400m)

d. Two vehicles are moving in the same direction. The first one has the

velocity 12 m/s and the second has a velocity 8m/s. Find their relative

velocity. If they move in opposite direction, what would be the relative

velocity? (Ans. 4m/s, 20 m/s)

Blooming Science & Environment Book 8 33

e. A car moves with a velocity 20 m/s to east and the other with a velocity

15 m/s to the west. If they move from the same point, what will be the

distance between them after two minutes? How far will each of them

travel in 2 minutes? (Ans: 4200m, 2400m East, 1800m West)

f. The velocity of a car is 30m/s and that of next car is 20m/s. If both the
car begin to move from the same place towards the same direction, what
would be the distance between them after two minutes? What distance
do they cover in this time?

g. A car starts from rest. If the acceleration of the car remains 2 m/s2 for 2
seconds, calculate the final velocity of the car.

h. A bus starts to move from the rest with an acceleration of 0.25m/s2. Find
its final velocity after 3 minutes.

i. Find the acceleration of a car, if its velocity increases from 20 m/s to 50
m/s in 5 seconds.

j. A ball is thrown vertically upward with a velocity of 40m/s. Calculate
maximum height and time to return to the thrower. (Ans. 80m, 8 sec)

Glossary

Rest : condition of not changing position

Motion : condition of changing position.

Distance : the total length travelled by a body in any direction.

Displacement : the shortest distance travelled by a body in a particular
direction.

Speed : the distance travelled by a body in unit time.

Velocity : the displacement made by a moving body in a unit time.

Uniform motion : a body covers equal distances in equal interval of time.

Variable motion : a body covers unequal distance in equal interval of time.

Scalar : quantity having only magnitude.

Vector : quantity having both magnitude and direction.

Relative velocity : the velocity of an object with respect to other object.

34 Blooming Science & Environment Book 8

Chapter Simple Machine

3

Learning Outcomes Estimated Periods: 3+1

On the completion of this unit, the students will be able to:
 give example of lever and introduce lever with its working principle.
 introduce MA, VR and efficiency.
 Solve simple numerical problems related to MA, VR and efficiency of lever.

We use different types of devices in our daily life to make our work easier. A spade
is used to dig field, a broom is used to sweep, a hammer is used to pass a nail and a
plank is used to drag a load at height. All these simple things or devices are used to
make our work easier and convenient to do. Many such devices, which are used in
our daily life are called simple machines. The combination of simple machines forms
a compound machine.

Simple machine is defined as a device, which is used in our daily life to make our
work easier, faster and more comfortable.

Simple machines make our work easier in the following ways:

1. They transfer force from one point to another point. In the given figure a knife
is used to open the lid of a can. The applied effort at ‘A’ is transferred by ‘B’.

Effort
B

A

2. They accelerate rate of doing work. For example, when
one part of a dhiki is pushed for a less distance, the other
part of it is lifted for more distance.

3. They multiply force, i.e. more load is lifted by applying
less effort. For example, a heavy stone can be turned
with a crowbar by applying less effort.

4. They change the direction of force. For example, in
pulleys, when effort is applied downwards at one end of the rope, the load is
lifted upwards.

Blooming Science & Environment Book 8 35

Machines help us to do works showing one or more
properties as mentioned above. Scan for practical experiment

Effort is applied Load is lifted
downwards upwards

Lever visit: csp.codes/c08e04

“Give me a point for fulcrum, I will turn the earth”. This

statement is of Archimedes who discovered Effort
lever in 240 B.C. The statement shows that

how efficient a lever is. Load Ed

Lever is a rigid bar, may be straight or bend

which is capable to rotate about a fixed Ld
point called fulcrum or pivot. The distance Fulcrum
between fulcrum and load is called as load

distance and the distance between fulcrum and effort is called effort distance. Out of

all the simple machines, only lever can show all the properties of a simple machine.

On the basis of the location of effort, load and fulcrum on a lever, it is classified in
three types. They are, first class lever, second class lever and third class lever.

Look at the figure below. The load of the

stone is 500N and is being lifted by an 20 N
Ed
effort 20N. It is possible because the effort 500 N
distance is longer than the load distance.

If effort distance is longer than the load Ld
distance, a small effort can lift or balance a
Fulcrum
heavier load.

Working Principle of Lever

The distance between load and fulcrum is called load distance and that between
effort and fulcrum is called effort distance.

The effort and the load always act in opposite direction to each other. In the given
figure, the effort acts in clockwise direction and the load acts in anticlockwise
direction.

When the product of effort and effort distance is equal to the product of load and load
distance, a lever stays in the state of equilibrium.

i.e. at the state of equilibrium in a lever,

36 Blooming Science & Environment Book 8

effort × effort distance = load × load distance
∴ It is also called as the principle of lever.

Mechanical Advantage, Velocity Ratio and Efficiency of Lever

Mechanical Advantage

The ratio of the load moved by a simple machine to the effort applied on it is called
mechanical advantage.

So, mechanical advantage (MA) = Load (L)
Effort (E)

Mechanical advantage of a simple machine indicates the number of times a load
is moved by the effort applied on it. For example, if mechanical advantage of a
machine is 5, that means load moved by effort is 5 times the effort.

The mechanical advantage of a simple machine is affected by:

(a) Friction and (b) Weight of the simple machine.

(a) Friction: Friction reduces the mechanical advantage. Each and every machine
has friction. Some machines have less friction than the others. Some of the
effort applied on the machine is used to overcome the friction produced in it
that makes the waste of effort.

(b) Weight of the Simple Machine: Weight of a simple machine affects the
mechanical advantage of the simple machine in some cases. In the case of
wheel and axle, inclined plane and screw, the weight of simple machine does
not affect the mechanical advantage. Some of the effort is used up to overcome
the weight of movable pulley in the combined pulley system. In the case of
lever, if effort arm is longer, the extra weight of effort arm helps to lift the load.

Velocity Ratio L2

When we use effort on a

simple machine load is L1 20 cm 50 cm E1
moved to some distance.
F
At the same time effort
will also be moved .To L1 D
move a load heavier than 200 N

the effort, it has to move

a longer distance than

that moved by the load.

In the figure a lever is E2
used to lift the load of
37
Blooming Science & Environment Book 8

200N by applying the effort 80N. A small load could lift a heavy load because effort
distance (50cm) is longer than the load distance (20 cm). But when the effort moves
from E1 to E2 the load rises from L1 to L2. The distance E1E2 is longer than the
distance L1L2. So, the effort has to move a longer distance to move the load by a
small distance.

The ratio of the distance moved by effort to the distance moved by load is called
velocity ratio of the machine.

i.e. Velocity ratio (V.R.) = Distance moved by effort (Effort distance)
Distance moved by load( Load distance)

In a lever, when the effort arm is longer then its effort distance is also more.

So, velocity ratio in a lever is the ratio of the effort arm to the load arm.
Celocity Ratio (V.R) = Effort / Load

Velocity ratio in a machine is neither affected by the friction nor by the weight of the
simple machine. Velocity ratio indicates the maximum mechanical advantage that
can be gained in a simple machine when the machine is frictionless. But in practice,
it is not possible to have the mechanical advantage equal to the velocity ratio. So, the
velocity ratio is always greater than the mechanical advantage.

Efficiency [η]

When an effort is applied in a machine, some work is done, which is called work
input. The machine, in turn, does some work which is called work output.

The work done by effort or work done in a machine is called work input (Wi) and the
work done by load or work done by the machine is called work output (Wo).
The ratio of work output to work input in a machine is called efficiency of that
machine. It is expressed in percentage.

Work output (joule)
Mathematically, Efficiency (η) = Work input (joule) × 100%

In a simple machine, the value of output is always less than input, because no machine
is frictionless and the friction wastes some of the input energy in the form of heat.
Due to this reason, a real machine does not have the efficiency of 100% or more.
Relation among MA, VR and η

We know,

Work output
η = Work input × 100%

38 Blooming Science & Environment Book 8

or, η = L × Ld × 100% [work output =L× Ld and work input = E × Ed]
E × Ed

or, η L/E
= Ed/Ld × 100%

MA L Ed
∴ η = VR × 100% [ MA = E and VR = Ld ]

Due to the friction, the value of MA is lesser than the value of VR. Hence, the
efficiency of a machine is never 100% or more. A machine with 100% efficiency is
called an ideal or perfect machine . But perfect machine is not in our practical life.

On the basis of position of effort, load and fulcrum, lever are classified into three

classes.

i) First Class Lever ii) Second Class Lever iii) Third Class Lever

First Class Lever

In this lever fulcrum lies in between load and effort. A see-saw, a beam balance, a
crow bar, a pair of scissors, a pair of pliers, etc. are the examples of first class lever.

In the given figure ‘a’ is the load distance and ‘b’ is the effort distance. According to
the principle of lever. a × L = b × E

In this lever load distance and effort distance can be changed as desired. So, in the
first class lever, effort distance can be made longer than or equal to or shorter than
the load distance.

In this type of lever all the three advantages of a simple machine can be gained.

LF E

ab
Fulcrum

Load (L)

Effort (E)

Blooming Science & Environment Book 8 39

For example,
• If effort distance is longer than the load distance, a small effort can lift a heavy

load. This type of lever magnifies the effort applied. The load is always greater than
the effort applied, so that the mechanical advantage is also greater than one.
• When effort distance is equal to the load distance the load balanced will be
equal to the effort applied. So, the mechanical advantage is equal to one. This
kind of lever does not magnify the effort applied.
• If load distance is longer than the effort distance, greater effort is necessary to
lift a small load. It cannot magnify the effort. So, at this condition mechanical
advantage is less than one. But this type of lever increases the speed of work.
For example, scissors cut cloth faster than a tin cutter does.
• The direction of effort is also changed with the first class lever. In this effort
acts downward to raise the load.

Second Class Lever
In a lever of the second class, the load is put in between fulcrum and effort.

ab

Fulcrum Load(L) Effort (E)

The lever of class second works on the same principle as the first class lever does.

For equilibrium, a × L = b × E

Where ‘a’ and ‘b’ are load distance and effort distance respectively.

In the second class lever, effort distance is always longer than load distance. Therefore,
the effort used is always less than the load raised and the mechanical advantage is
always greater than one.

To make the work more easier, the load can be shifted towards the fulcrum in this lever.

Third Class Lever

Effort (E) lies in between load and fulcrum in this lever. Effort distance (b) is always
shorter than load distance in this type of lever. Again some resistance against the
effort is produced at the fulcrum which makes waste of some effort. Due to these two
reasons the load raised is always less than the effort used in this lever

Fulcrum Load (L)

Effort

ba

40 Blooming Science & Environment Book 8

The mechanical advantage is always less than one in third class lever.
This lever works on the same principle as class I and II.
For equilibrium, a × L = b × E.

Solved Numerical Problems

1. An effort of 60 N is applied to lift a load of 240N by using a crow bar as
first class lever. If the effort is applied at 205m away from fulcrum and
load is kept at 50cm from fulcrum, calculate its M.A, V.R. and efficiency.

Solution,

Given,

Load = 241 N

Effort = 60 N

Effort arm = 2.5 m

Load are = 50 cm = 0.5 m

M.a. = ?

V.R. = ?

Efficeincy (η) =?

we know that, Load
Effort
M.A. =

= 26400NN

= 4 m
= 02..55 m


∴ V.R = 5

Again M.A
V.R
η = × 100%

= 54 × 100%

= 80 %

∴ η = 80%

Therefore, M.A = 4

V.R = 5

Efficiency = 80%

Blooming Science & Environment Book 8 41

Main Points to Remember

1. A simple machine is used to:

(a) make work easier,

(b) change the direction of force and

(c) increase the speed of work.
2. If effort distance is longer than the load distance, a small effort can lift or

balance a heavier load.
3. At the state of equilibrium in a lever, effort × effort distance = load × load distance.

4. The ratio of the load moved by a simple machine to the effort applied on it is
load
called mechanical advantage. M.A. = effort

5. The mechanical advantage is reduced by the friction and the weight of the machine.

6. The ratio of the distance moved by the effort to the distance moved by the load
is called velocity ratio.

velocity ratio = Distance moved by effort
Distance moved by load

8. (a) In a first class lever, either the effort or the load is bigger than the other.

(b) In a second class lever, load is always greater than effort.

i.e. MA is always greater than one.

(c) In a third class lever, effort is always greater than load.

i.e. MA is always less than one.

Exercise
1. Answer the following questions in brief.

(a) What is a simple machine? What are the three advantages of a simple
machine?

(b) What factors affect the mechanical advantage of a machine? How can it
be increased?

(c) Define velocity ratio. What does a V.R. of a machine indicate?

(d) Give the advantages of a first class lever.

(e) What is a perfect machine?

(f) Draw a labelled diagram to show Principle of second class lever.

(g) What is efficiency of a machine? It is not 100%, why?

(h) What is principle of lever?

42 Blooming Science & Environment Book 8

2. Give reasons:

(a) Velocity ratio is always greater than mechanical advantage in a simple
machine.

(b) Mechanical advantage is always greater than one in a second class lever.
(c) Mechanical advantage is always less than one in a third class lever.

3. Classify lever with 2/2 examples.

4. Solve the following numericals.

a. How much effort is needed to lift a load of 100N placed at a distance of 20 cm
from fulcrum, if effort is applied at 60 cm from the fulcrum on opposite side of
the load? Calculate mechanical advantage and velocity ratio of the lever.

(33.33N, 3, 3)

b. To lift a load of 300N in a second class lever an effort of 50 N is applied

at a distance 90 cm from the fulcrum, at what distance should the load be

put? Draw a diagram of the lever system. (15 cm)

c. Two girls Rita weighig 500N and sita 400N are playing see-saw. At

d. which point Rita has to sit to balance Sita at 4m? (At 3:2m)

1.5m long crow bar is used as first class lever to lift a load of 500N a

shown in the figure. Calculate the effort applied, M.A, V.R. and

efficiency of given lever Effort

D500N ( 250N, 2, 2,
50 cm 100%)

Glossary

Anti clockwise : In a curve opposite in direction to the movement of the hands of
a clock

Clockwise : in a curve corresponding in direction to the movement of the
hands of a clock

Crane : a large tall machine used for moving heavy objects by suspending
them from a projecting arm

Blooming Science & Environment Book 8 43

Chapter Pressure

4

Learning Outcomes Estimated Periods: 4+1

On the completion of this unit, the students will be able to:
• introduce atmospheric pressure and its importance
• introduce liquid pressure.
• derive equation for liquid pressure (P = dgh)
• solve simple simple numerical problems related to pressure.

Introduction

You know that the cutting edge of a knife is sharp. Why is it made so? This is due to
the fact that this sharp edge exerts a force on a small area. When force is exerted on
a small area, it cuts easily.

The wheel of the tractors are made wider because wider wheels exerts less force on
the large area. Here force is exerted on the wide area, so wide area saves the tyre from
dipping down. So pressure is a physical quantity that tells how effective a force is.

Pressure is defined as the force per unit area.

If the force of different magnitude acts on the same area, the pressure varies. Pressure
is more when more force acts on the body. Similarly, if same force acts on two
different areas, the pressure varies. Pressure is more when the area is small. For a
force ‘F’ acting on a surface of area ‘A’, pressure is calculated by the formula given
below.

Pressure = Force or, P = F
Area A

The above expression shows that pressure depends upon amount of force and the

area on which it is applied. For a given force, if area is decreased, pressure increases

and vice-versa. Pressure is directly proportional with applied force and inversely

proportional to the area.

Mathematically,

P∝F.....................(i)

P∝ 1 ....................(ii)
A

44 Blooming Science & Environment Book 8

Combining equation (i) and (ii)

P∝ F or P = kAF where k is a constant.
A

If P = 1 Pa, F = 1 N, A = 1m2, then k = 1

\ P = F
A

The unit of force is Newton (N). One Newton is the amount of force required to

produce an acceleration of 1m/s2 on a body of mass 1 kg. Force is expressed as,

Force = mass x acceleration

Newton (N) = Kilogram (kg) × metre per square second (m/s2). The unit of force is
Newton and the unit of area is square metre (m2). Now, the unit of pressure is Newton
per square metre (N/m2). It is also called Pascal (Pa). The pressure is also expressed
in mm of Hg or cm of Hg. The standard atmospheric pressure at sea level is 760 mm
of Hg.

Solved Numerical Problems

1. A sack of rice weighting 500N is kept first in a big box having an area of
20m2. It is then transferred to a small box with an area of 10m2. Calculate
the pressure in each case.

Solution:

For big box

Here, Force (F1) = 500N

Area (A1) = 20m2

Pressure (P1) = ?

According to the formula,

P1 = F1
A1

= 500
20

= 25 N/m2 or pascal

For small box.

Force (F2) = 500N

Area (A2) = 10m2

Pressure (P2) = ?

Blooming Science & Environment Book 8 45

Accordion to the formula,

P2 = F2 = 500 = 50N/m2 or Pa
A2 10

2. Study the figure given below and find the maximum and minimum
pressure exerted by a brick of dimensions 20 cm × 10 cm × 5 cm of mass
1.2 kg. (Acceleration due to gravity (g) = 9.8 m/s2)

5 cm

10 cm 20 cm 20 cm
10 cm
5 cm

Soft surface Soft surface

Case ‘A’ (Minimum Pressure)

The brick is placed on the foam with broad surface in contact with it.

Solution:

Here, Area (A1) = 20 cm × 10 cm = 0.2 × 0.1 = 0.02 m2

Weight (F1) = 1.2 kg × 9.8 m/s2 = 11.76N

Pressure (P1) = ?

According to the formula,

P1 = F1 = 101..0726 = 588 N/m2 or Pa
A1

Case B (Maximum Pressure)

The brick is placed on the foam with narrow surface in contact with it.

Solution:

Here, Area (A2) = 10 cm × 5 cm
= 50 cm2

= 5 × 10-3m2

Weight (F2) = 1.2 kg × 9.8 m/s2

= 11.76 N

46 Blooming Science & Environment Book 8

Pressure (P2) = ?

According to the formula,

P2 = F2 = 11.76 = 2352 N/m2
A2 5 × 10-3

Both the bricks have same weight. The pressure exerted by the brick with its
narrow surface in contact is more.

Pressure in Liquids

Matter occupies space. Matter has weight and it can exert pressure as well. The
pressure of a body depends on the weight as well as on the surface area occupied by
it. This has been proved in previous sections. Let us study about liquid pressure in
this section.

The general concept of liquid is water. But there are many examples of liquid. It is
a state of matter. Since matter occupies space and has weight, liquid also occupies
space and has weight. A body which has weight can exert pressure. Therefore, liquids
can also exert pressure.

We know that,

P = F
A
mass(m) × acceleration due to gravity (g)
or, P = A ( F = mg)

or, P= density (d) × volume (V) × g ( d = mV, m = d × V) A
A
h

or, P = d × V × g = d×A×h×g ( V = A × h)
A A

Hence, P = dhg

Hence, Pressure exerted by the liquid is equal to the product of density, depth of the

liquid and acceleration due to gravity. If the density of the liquid increases or depth

increases, the pressure also increases. Scan for practical experiment

\ P = dhg ………………… (1)

Here, P = Liquid pressure

d = Density of liquid

h = Depth of liquid

g = Acceleration due to gravity visit: csp.codes/c08e05

Thus, pressure exerted by a liquid depends on (i) density (d) of the liquid, (ii) height
(h) of the liquid from its free surface and (iii) acceleration due to gravity (g). Liquid
pressure is directly proportional to these factors.

Blooming Science & Environment Book 8 47

Solved Numerical Problems

1. What is the pressure exerted by water in the given figure? Density of water
is 1000 kg/m3, g = 10 m/s2 .

Solution:

Given,

Density of water (d) = 1000 kg
m2
h = 2m
Depth of the liquid (h) = 2m

Acceleration due to gravity (g) = 10 m/s2

Pressure (P) = ?

We have,

Pressure (P) = h × d × g

= 2 × 1000 × 10

= 20,000N/m2

Hence, Pressure (P) = 20,000N/m2

2. What is the pressure at 6m depth for a liquid that has density equal to 800
kg/m3?

Solution:

Given,

Density (d) = 800 Kg/m3

Depth (h) = 6 m

Acceleration due to gravity (g) = 10 m/s2

Pressure (P) = ?

We have,

Pressure (P) = h × d × g

= 6 × 800 × 10

= 48,000 N/m2 or Pascal

48 Blooming Science & Environment Book 8

3. The length, breadth and height of a water tank are 3m, 2m and 2m
respectively. If half of it is filled with water, what pressure is exerted at its
bottom?

Density of water = 1000 kg/m3, g = 10 m/s2.

Solution:

Given,

Density (d) = 1000 kg/m3

Depth (h) = 1m

Acc. due to gravity (g) = 10 m/s2

Pressure (P) = ?

We have,

Pressure (P) = h × d × g = 1 x 1000 x 10

= 10,000 N/m2

Hence, Pressure (P) = 10,000 Pascal

4. A drum is filled with liquid. The height of the liquid in the drum is 2m and
its pressure is 500 N/m2. What is the density of this liquid?

Solution:

Given,

Pressure (P) = 500 N/m2

Height (h) = 2m

Acc. due to gravity (g) = 10 m/s2

Density (d) = ?

We have,

Pressure (P) = h × d × g

or, d = = 25 kg/m3

Density of the liquid = 25 kg/m3

Blooming Science & Environment Book 8 49

Liquids Exert Pressure
Activity

To show liquid exerts pressure

Take a small tube open on both the sides. Fix a rubber membrane
at one end of the tube and put some water in it. You notice that
the rubber membrane begins to extend. The extension in the
rubber membrane is due to the liquid above it.

Rubber membrane

Liquid has its own characteristics. The characteristics of liquid pressure can be
summarized in points below:

(a) The pressure of liquid at a point within it is directly
proportional to the depth of the point from the free
surface of the liquid or the height of the liquid column
above the point.

Take water in a cylinder or a long tin-can having three

holes fitted with stop-corks. When the stop-corks are

opened, water emerges out of the holes as shown in Fig: Pressure increase as

figure. height increase

Water from hole 1 emerges with the least force whereas water from hole 3
emerges with the greatest force. It means pressure at hole 3 is more than at
holes 1 and 2. Moreover, this pressure acts on the sides of the vessel which is
called lateral pressure. This shows that pressure at a point increases with height
of the liquid above the point.

(b) The pressure at a point is directly proportional to the density of the liquid.
P ∝ d

As the density of the liquid increases, the pressure of the liquid also increases.

(c) The pressure and thrust of a liquid at rest are always perpendicular to the
surface in contact with it.

(d) The pressure at any point within the liquid is the same in all directions. That is,
the liquid exerts pressure equally in all directions.

(e) The pressure at a point within the liquid does not depend on the volume of the
liquid in the vessel but it depends only on the height of the liquid above the point.

(f) The pressure at a point is independent of the shape of the vessel.

50 Blooming Science & Environment Book 8