Class 9 Science_clone

Revised and Enlarged

Approved by Curriculum Development
Centre, Government of Nepal as an
additional material for schools.

Times’ Crucial

SCIENCE
and ENVIRONMENT

9Grade

Authors/Editors:
Kamal Prasad Sapkota
Rajan Kumar Shrestha

Surendra Ojha Editors: Dipak Kumar Dangi
Sagar Dahal Hemanta Dhungana
Santosh Pokharel Kamal Kanta Dhungel
Govinda Paudel Mani Ram Rai

Times’ Crucial

SCIENCE
and ENVIRONMENT
9Grade

Published by:
Times International Publication Pvt. Ltd.

Authors/Editors:
Kamal Prasad Sapkota
Rajan Kumar Shrestha

Edition:
First: 2070 B.S.
Revised Edition: 2074 B.S.
Revised Edition: 2075 B.S.
Revised Edition: 2077 B.S.
Revised Edition: 2078 B.S.

Layout and Design:
Ramesh Maharjan

© Copyright: Publisher

Printed in Nepal

Preface

Human life is always in progress. Newer technologies and equipmen are
developed or discovered in every second. Lengthy and time consuming work of past
are now done within a few moments. People can travel far off distances within short
time. Several fatal diseases have been eradicated. The achievements, that we are
enjoying today are the results of advancement in science and technology. Hence,
science has become an integral part of our education system.

Times' Crucial Science and Environment is a series of text books for the
school level students of grades LKG to class Ten. The series has been prepared for
the young learners emphasizing on student-centred teaching techniques, Learning
Circle Based Activities RTCEVKECDNG CEVKXKVKGU UEKGPVKſE CRRTQCEJGU CPF KPPQXCVKXG
learning techniques. The text of each lesson is preceded by a warm up activity that
encourages students to take part actively in the learning process. The series includes
the teaching techniques and methods for the teachers under the title Note to the
Teacher. It is based on the latest syllabus prescribed by CDC Government of Nepal.
Hence, this series acts as the foundation of science for the curious and inquisitive
young minds.

'CEJ DQQM QH VJKU UGTKGU EQXGTU VJG U[NNCDWU QH 5EKGPEG 6JG ſTUV RCTV EQPVCKPU
chapters of Science and the second part, chapters of Environment Science. All the
chapters of Science and Environment are provided with a wide variety of exercises
which encourage the students in learning and sharpening their mind. A Project Work
is asked at the end of each lesson so that student can apply their knowledge to
solve the problems of their day-to-day life. In the beginning of each lesson, thought
provoking questions are given under the heading Mind Openers. This activity will
encourage students to use their brain.

We feel delighted to extend our sincere gratitude to all the editors for their
painstaking contribution in editing the book and making it more simpler and crucial.
We are equally thankful to our publishers Mr. Kamal Pokharel and Mr. Purushottam
Dahal without whom this series would not have been possible in the present form.
Moreover, We are thankful to Mr. Santosh Pokharel for his effort to make this book
error-free. We are also thankful to Mr. Ramesh Maharjan for the kind cooperation and
innovative skills in providing the outstanding form to the books.

Finally, it will be our pleasure to receive constructive suggestions and
recommendations from the teachers, students and well wishers for the further
improvement of the books.

- Authors

Contents

Chapter Page No.

Physics 1-146

1 Measurement 1

2 Force 17
3 Simple Machine 46
4 Work, Energy and Power 65
5 Light 82
6 Sound 102
7 Current Electricity and Magnetism 126

Chemistry 147-256

8 Valency and Molecular Formula 147
9 Chemical Reaction 168
10 Solubility 178
11 Some Gases 191
12 Metals 214
13 Carbon and Its Compounds 226
14 Water 234
15 Chemical Fertilizers Used in Agriculture 251

Biology 257-376

16 &ODVVLÀFDWLRQ RI /LYLQJ 7KLQJV 259
17 Adaptation of Organisms 295
18 Systems 313
19 Sense Organs 344
20 Evolution 356
21 Nature and Environment 368
377-407
Geology and Astronomy 379
22 Natural Hazards 389
23 Green House 397
24 7KH (DUWK LQ WKH 8QLYHUVH 410
411
6SHFLÀFDWLRQ *ULG
Model Test Paper

Chapter

1 0HDVXUHPHQW Louis Essen

Objectives He is known for the precise
measurement of time and the
determination of the speed of light.

ƐƟŵĂƚĞĚ WĞƌŝŽĚƐ͗ ϲ

ƚ ƚŚĞ ĞŶĚ ŽĨ ƚŚĞ ůĞƐƐŽŶ͕ ƐƚƵĚĞŶƚƐ ǁŝůů ďĞ ĂďůĞ ƚŽ͗

• GHÀQH IXQGDPHQWDO DQG GHULYHG TXDQWLWLHV
• VKRZ WKH UHODWLRQVKLS EHWZHHQ IXQGDPHQWDO DQG GHULYHG XQLWV
• PHDVXUH VRPH IXQGDPHQWDO TXDQWLWLHV DQG GHULYHG TXDQWLWLHV

Mind Openers

• What is measurement?
• :KDW LV SK\VLFDO TXDQWLW\"
• :KDW DUH IXQGDPHQWDO DQG GHULYHG XQLWV"
• How do you measure mass and time? Discuss.

Introduction
In our practical life, we need to measure various quantities. We
measure volume of water and amount of rice at the time of cooking
food. Measurement of length of cloth and size of human body is done
before sewing clothes. When we go to a shop for buying things, the
VKRSNHHSHU PHDVXUHV WKH TXDQWLW\ RI WKLQJV DQG JLYHV WR XV 7KXV
measurement is common and general human activity.
We can measure some quantities whereas some other quantities
cannot be measured. 7KH TXDQWLWLHV ZKLFK FDQ EH PHDVXUHG DUH
called physical quantities. Length, mass, time, volume, speed,
acceleration, etc are physical quantities. But sorrow, happiness,
VDGQHVV EHDXW\ HWF FDQQRW EH PHDVXUHG 7KHUHIRUH WKH\ DUH FDOOHG
non-physical quantities.
For measuring any physical quantity, we need certain standard
RU NQRZQ TXDQWLW\ FDOOHG XQLW 7KH PHDVXUHPHQW RI DQ\ XQNQRZQ
quantity can be done by comparing it with a known or standard
TXDQWLW\ 7KXV PHDVXUHPHQW LV GHÀQHG DV WKH SURFHVV RI FRPSDULVRQ
of an unknown quantity with a known or standard quantity.

1 Times' Crucial Science Book - 9

Importance of measurement
1. It is needed for selling, buying and exchanging of goods.
,W LV QHHGHG IRU SHUIRUPLQJ VFLHQWLÀF H[SHULPHQWV
3. It is needed for preparing medicines and treating patients.
4. It is needed for preparing food.
,W LV QHHGHG IRU JOREDO XQGHUVWDQGLQJ RI WKH TXDQWLW\ RI D VXEVWDQFH

Unit
Unit is a known or standard quantity in terms of which other
physical quantities are measured. For examples, kilogram, metre,
second, etc. Metre is a unit because it is used to measure the
physical quantities like length, area, volume, velocity, etc and value
of 1 meter is known. If the value of 1 meter is unknown, we don’t
understand the value of 6 meter, 7 meter and so on. When we say
length of a bench is 6 metres, it means the length of the bench is 6
times of one metre. For complete measurement, we need numerical
value and unit. If we express the measurement in numerical value
only, it is meaningless. For example, if we say length of a bench is
6, it does not give full sense. It may be 6 feet, 6 inch, or 6 metre.

Physical quantity = Numerical value × Unit
Example: Length = 5 × m = 5m

7KHUH DUH GLIIHUHQW XQLWV IRU PHDVXULQJ GLIIHUHQW SK\VLFDO TXDQWLWLHV
7KH XQLWV KDYLQJ IROORZLQJ IHDWXUHV DUH VXLWDEOH IRU PHDVXUHPHQW

7KH XQLW VKRXOG EH ZHOO GHÀQHG DQG DFFHSWHG
2. The unit should be reproducible.
3. The unit should not change with the change in

temperature, pressure, time, etc.
4. The unit should be of suitable size.
Sometimes, some fraud devices are used for measuring physical
TXDQWLWLHV 7R PRQLWRU WKH PHDVXULQJ GHYLFHV DQG WR SXQLVK WKH
wrong doers, the government of Nepal has a separate department.
7KH Nepal Bureau of Standards and Metrology monitors the
measuring devices in every two years in the markets and seizes
fraud devices. Culprits are punished according to law.
7KH 1HSDO %XUHDX RI 6WDQGDUGV DQG 0HWURORJ\ KDV UHJLRQDO RIÀFHV
in different parts of the country. One of them lies in Kathmandu
valley whereas seven others are outside the valley.

Times' Crucial Science Book - 9 2

Activity 1.1 Mark two points in a white sheet of paper. Measure the distance between

these two points using the ruler (scale) of each student of your class. Did you get the same
length in each case? Why do these scales give different measurements? Discuss in class.

)XQGDPHQWDO TXDQWLWLHV

7KH TXDQWLWLHV ZKLFK DUH LQGHSHQGHQW RI RWKHU TXDQWLWLHV DUH
called fundamental quantities. For examples, length, mass, time,
temperature, electric current, luminous intensity and amount of
substance are fundamental quantities.

'HULYHG TXDQWLWLHV

7KH TXDQWLWLHV ZKLFK DUH GHULYHG IURP IXQGDPHQWDO TXDQWLWLHV
are called derived quantities. For examples, velocity, area, speed,
acceleration, density, etc.

6SHHG LV FDOFXODWHG DV GLVWDQFH WUDYHOOHG E\ D ERG\ SHU XQLW WLPH.

Speed = Distance
7LPH

Distance means the length DQG KHQFH VSHHG FDQ EH H[SUHVVHG DV

Speed = Length
7LPH

7KXV VSHHG LV GHULYHG IURP WZR IXQGDPHQWDO TXDQWLWLHV length and
WLPH 7KHUHIRUH LW LV D GHULYHG TXDQWLW\ ,Q WKH VDPH ZD\ DUHD LV
derived from two lengths, i.e. Area = (length)2. 7KHUHIRUH DUHD LV D
GHULYHG TXDQWLW\ 7KHUH DUH WKRXVDQGV RI GHULYHG TXDQWLWLHV
Systems of Measurement
Different traditional and local systems of units and measurements
are still in use at various places. Dharni, Ser, Chhatak, etc. are used
IRU WKH PHDVXUHPHQW RI PDVV )HHW KDQG VSDQ ÀQJHUV \DUG HWF
are used for the measurement of length. Mana, pathi, muri, etc are
XVHG IRU WKH PHDVXUHPHQW RI YROXPH 7KRVH XQLWV FDQ EH XQGHUVWRRG
by the local people only. Moreover their values differ from person to
SHUVRQ DQG SODFH WR SODFH 7R PDLQWDLQ XQLIRUPLW\ LQ WKH XQLWV DQG
PHDVXUHPHQW IROORZLQJ V\VWHPV RI PHDVXUHPHQW DUH XVHG
a) FPS System: 7KH V\VWHP RI PHDVXUHPHQW LQ ZKLFK OHQJWK LV
measured in foot, mass in pound and time in second is called FPS
system.

3 Times' Crucial Science Book - 9

b) CGS System: 7KH V\VWHP RI PHDVXUHPHQW LQ ZKLFK OHQJWK LV
measured in centimeter, mass in gram and time in second is called
CGS system.
c) MKS System: 7KH V\VWHP RI PHDVXUHPHQW LQ ZKLFK OHQJWK LV
measured in meter, mass in kilogram and time in second is called
MKS system.
d) SI System: 7KH IXOO IRUP RI 6, LV V\VWHP LQWHUQDWLRQDO GH XQLWV
In 1960, an international convention of scientists from the various
countries was held in France. All the scientists agreed to use the
VDPH XQLWV WKURXJKRXW WKH ZRUOG 7KHVH XQLWV DUH FDOOHG 6, XQLWV
7KXV 6, V\VWHP LV WKH V\VWHP RI XQLWV ZKLFK LV UHFRPPHQGHG E\
the international convention of scientists in 1960 and is accepted
throughout the world.
It is the extended form of MKS system. In this system, some other
quantities are also added as fundamental quantities. Following
table shows the fundamental quantities and their units.

SN Fundamental quantities Fundamental units Symbol

1 Length Meter m

2 Mass Kilogram kg

3 7LPH Second s

4 7HPSHUDWXUH Kelvin K

5 Electric Current Ampere A

6 Luminous intensity Candela cd

7 Amount of substance Mole mol

Importance of SI units
,W EULQJV XQLIRUPLW\ LQ PHDVXUHPHQW DOO RYHU WKH ZRUOG
,W HOLPLQDWHV WKH LQDFFXUDF\ DQG VROYHV WKH SUREOHPV GXH WR
local system of measurement.
,W KHOSV LQ FDUU\LQJ RXW GLIIHUHQW VFLHQWLÀF H[SHULPHQWV

Differences between MKS and SI system

MKS system SI system

1. It is the metric system of measurement. ,W LV WKH UHÀQHG DQG H[WHQGHG IRUP RI 0.6 V\VWHP

,W LV EDVHG RQ WKUHH IXQGDPHQWDO TXDQWLWLHV 2. It is based on seven fundamental quantities
mass, length and time. (including three quantities of MKS system).

3. It has a limited use. 3. It is used all over the world.

Times' Crucial Science Book - 9 4

Fundamental units
7KH XQLWV ZKLFK DUH XVHG WR PHDVXUH IXQGDPHQWDO TXDQWLWLHV DUH
called fundamental units. In other words, fundamental units are
the units which are independent of other units. Meter, second,
kilogram, Kelvin, Ampere, candela, etc are some fundamental units

Fact Reason

Kelvin is fundamental unit, why?
Kelvin is fundamental unit because it is an independent unit
and it is not derived from any other fundamental unit.

Derived units
7KH XQLWV RI GHULYHG TXDQWLWLHV DUH FDOOHG GHULYHG XQLWV In other
words, the units which are derived from fundamental units
are called derived units. For examples, units of velocity, area,
acceleration, force, volume, density, etc. are derived units because
they are derived from two or more fundamental units.

Fact Reason

7KH XQLW RI GHQVLW\ LV GHULYHG XQLW ZK\"
7KH XQLW RI GHQVLW\ LV NJ P3 which is derived from kg and m.
Hence, the unit of density is derived unit.

7KH GHULYHG XQLWV RI VRPH SK\VLFDO TXDQWLWLHV DUH JLYHQ EHORZ

S.N. Derived Related Units Symbol Funuintsdainmveolnvteadl
1. quantities formulae Metre per
ms-1 P V
2. Velocity Distance second
3. 7LPH Meter per ms-2 m
4. Acceleration square second m2 s×s
5. Area Velocity Square metre m3
6. 7LPH Cubic metre N m×m
7. Volume J
8. Force length × breadth Newton W m×m×m
9. Work length × breadth×
10. Power Joule NJ P3 kg×m
11. height s×s
Density Watt Pa
12 mass × acceleration Kilogram per Hz kg×m×m
Pressure s×s
Frequency Force × distance cubic metre V
Potential Pascal kg×m×m
difference Work s×s×s
7LPH Hertz kg
Moment Mass
Volume Volt m×m×m
Force Newton × kg
Area
Velocity Meter s×s×m
Wave length
work done &\FOH V
charge
kg×m×m
Force × distance s×s×s×A

Nm kgm2 V2

5 Times' Crucial Science Book - 9

:K\ LV XQLW RI YHORFLW\ D GHULYHG XQLW"
7KH XQLWV WKDW DUH H[SUHVVHG LQ WHUPV RI WZR RU PRUH IXQGDPHQWDO XQLWV DUH FDOOHG
derived units. In case of velocity,
:H KDYH

Velocity = Displacement = Length m
7LPH 7LPH =s

Since the unit of velocity contains two fundamental units, i.e. metre and second,
it is a derived unit.

Similarly, unit of density is a derived unit. It is derived from two fundamental units
kilogram (kg) and metre (m).

Density = Mass = kg
Volume m3

7KXV XQLW RI GHQVLW\ LV NJ P3.

Similarly, the unit of acceleration is a derived unit.

Acceleration = Change in velocity = P V = m
7LPH s s2

7KXV XQLW RI DFFHOHUDWLRQ LV P V2, which is derived from two
IXQGDPHQWDO XQLWV PHWUH P DQG VHFRQG V DQG LV D GHULYHG XQLW

Solved Numerical Problem 1.1

What are the fundamental units involved in Watt?

6ROXWLRQ :DWW LV WKH XQLW RI SRZHU

We know that,

Power = Work done
7LPH WDNHQ

= Force × Displacement
7LPH

Mass × Acceleration × Displacement
= 7LPH

= Mass × Velocity × Displacement
7LPH ð 7LPH

Mass × Displacement × Displacement
= 7LPH ð 7LPH ð 7LPH

Mass, displacement and time are fundamental quantities. Now putting their
SI units in their places,

= Kg × m × m = kgm2s-3
s×s×s

7KXV WKH IXQGDPHQWDO XQLWV LQYROYHG LQ ZDWW DUH NJP2s-3.

Times' Crucial Science Book - 9 6

Measurement of length

/HQJWK LV GHÀQHG DV WKH GLVWDQFH EHWZHHQ DQ\ WZR SRLQWV Its SI
unit is metre. It is also measured in the units such as centimeter,
kilometer, feet, mile, etc.

2QH VWDQGDUG PHWUH LV GHÀQHG DV WKH GLVWDQFH EHWZHHQ WZR ÀQH
gold lines marked on Platinum-Iridium alloy rod at temperature
of 0°C kept at the International Bureau of weights and measures
at Sevres near Paris. Copies of one standard meter are made and
distributed to standardizing agencies of different countries. In
Nepal standard metre has been kept in Nepal Bureau of Standards
and Metrology, Kathmandu. Commercial standard metres are made
based on the standard metre kept by the standardizing agency.
7KH FRPPHUFLDO VWDQGDUG metres are checked periodically by this
metrology department.
Multiples and sub multiples of metre Conversion of Units

10 millimetre (mm) = 1 centimetre (cm) 1 cm = 10ï m
10 centimetre = 1 decimetre (dm)
10 decimetre = 1 metre (m) 1mm = 10ï m
10 metre = 1 decametre (dam)
10 decametre = 1 hectometre (hm) 1Pm = 10ï m
10 hectometre = 1kilometre (km) = 10ï m
1nm

1Å = 10ï m

Pm = micrometer, nm = nanometre

Å = angstrom

Correct method of measuring length

Following precautions should be taken for correct measurement of
length.

D $YRLG JDS EHWZHHQ WKH VFDOH DQG WKH REMHFW WR EH PHDVXUHG
b) Put the scale in such a way that its zero reading lies at one end

RI WKH REMHFW
F )L[ \RXU H\HV SHUSHQGLFXODU WR WKH VFDOH DW WKH HQG RI WKH REMHFW
d) Use error free scale.

7 Times' Crucial Science Book - 9

Fact Reason

7KH OHQJWK RI D URSH LV P :KDW GRHV LW PHDQ"
It means that the length of the rope is 3 times the standard
meter kept at the International Bureau of Weight and Measures
in France.

Measurement of mass

7KH WRWDO DPRXQW RI PDWWHU 1 kilogram mass of platinum
contained in a body is called its iridium alloy
mass. 7KH PDVV RI DQ REMHFW GHSHQGV
upon the number of molecules and
size of the molecules contained in
it. Its SI unit is kilogram (kg). It
is also measured in gram, quintal,
ton, etc.

2QH VWDQGDUG NLORJUDP LV GHÀQHG DV WKH PDVV RI 3ODWLQXP ,ULGLXP
alloy cylinder of diameter equal to its height kept at International
Bureau of Weights and Measures at Sevres in France. A copy of
one standard kilogram is kept in the standardizing agency of every
country.

Multiples and sub-multiples of gram Conversion of Units

10 milligram (mg) = 1 centigram (cg) 1mg = 10ï g

10 centigram = 1 decigram (dg) 1g = 10ï kg

10 decigram = 1 gram (g)

10 gram = 1 decagram (dag) 1kg = 103 g

10 decagram = 1 hectogram (hg) 1mg = 10ï kg

10 hectogram = 1 kilogram (kg) 1quintal = 102kg

100 kilogram = 1 quintal

1000 kilogram = 1 metric tonne 1metric tonne = 103kg

Fact Reason

7KH PDVV RI D ERG\ LV NJ :KDW GRHV LW PHDQ"
It means that the mass of the body is 5 times the standard
kilogram kept at the International Bureau of Weight and
Measures in France.

Measurement of time
7KH LQWHUYDO EHWWHU WR ZULWH LQVWHDG RI GXUDWLRQ
between any two events is called time. Its SI unit
LV VHFRQG 7LPH LV DOVR PHDVXUHG LQ PLQXWH KRXU

Times' Crucial Science Book - 9 8

day, week, month, year, decade, etc.
2QH VWDQGDUG VHFRQG LV GHÀQHG DV WKH WLPH WDNHQ E\ &DHVLXP
atoms to make 9192631770 vibrations.
7KH WLPH WDNHQ E\ WKH HDUWK WR PDNH RQH FRPSOHWH URWDWLRQ DURXQG
its axis is one solar day. A solar day is divided into 24 equal parts.
Each part is called an hour. One hour is divided into 60 equal parts.
Each such part is called minute. One minute is again divided into
HTXDO SDUWV HDFK SDUW LV FDOOHG VHFRQG 7KXV

1 solar day = 24 hrs
1 hour = 60 minutes
1 minute = 60 seconds
1 solar day = 24 × 60 × 60 seconds
1 solar day = 86400 seconds
? VHFRQG
th of a solar day

7KXV VHFRQG PD\ DOVR EH GHÀQHG DV th part of a mean
solar day.
7LPH LV PHDVXUHG E\ GLIIHUHQW W\SHV RI ZDWFKHV RU FORFNV
a. Mechanical watch: 7KLV ZDWFK ZRUNV RQ WKH EDVLV RI oscillation
RI D VLPSOH SHQGXOXP 7KH OHQJWK RI D pendulum changes due to
the change in temperature thereby changing the time period of
oscillation. So, it cannot measure time accurately.

b. Quartz watch: It works due to vibrations of quartz crystals. It
measures the time more accurately than mechanical watch.

c. Atomic watch: 7KLV ZDWFK ZRUNV RQ WKH EDVLV RI vibrations of
Caesium-133 atom. It measures time the most accurately.

Activity 1.2 7R REVHUYH WKH RVFLOODWLRQ RI D SHQGXOXP

0DWHULDOV UHTXLUHG

A thread, a metallic bob, stop watch, etc.

Procedure:

7DNH D PHWDOOLF ERE DQG WLH LW ZLWK RQH HQG RI A B
a thread of about 50 cm length.

9 Times' Crucial Science Book - 9

7LH DQRWKHU HQG RI WKH WKUHDG WR D QDLO À[HG RQ WKH ZDOO
3. Pull the bob upto the point A and then release it.
7KH PRYHPHQW RI WKH ERE IURP SRLQW $ WR % WKHQ EDFN WR $ LV FDOOHG

oscillation.
5. Note the time required for 20 oscillations and then calculate the time

required for 1 oscillation.

'LIIHUHQFHV EHWZHHQ SHQGXOXP FORFN DQG TXDUW] FORFN

Pendulum clock Quartz clock

1. It works on the basis of oscillation 1. It works on the basis of
of a simple pendulum. oscillation of quartz crystals.

7KH WLPH VKRZQ E\ D SHQGXOXP 7KH WLPH VKRZQ E\ D TXDUW]
FORFN ÁXFWXDWHV E\ D IHZ VHFRQGV WR FORFN ÁXFWXDWHV E\ D IHZ VHFRQGV
minutes in a day. in a month.

Zenith
When one stands at a particular place, the point in the sky directly
above the head is called zenith. If the sun is at a particular zenith,
it takes 24 hours to come back to the same zenith.
Measurement of area
7KH VSDFH RFFXSLHG E\ D VXUIDFH RI D ERG\ LV FDOOHG LWV DUHD Its SI
unit is square metre (m2).
7KH RWKHU XQLWV RI DUHD DUH VTXDUH FHQWLPHWHU (cm2), square feet
(ft2), etc.
Areas of regular objects are calculated by different formulae.

breadth radius
height
length base

Area of a rectangle = length × breadth

A=l×b

Area of right angled triangle = 1 × base × height
×b×h 2
1
A = 2

$UHD RI FLUFOH › UDGLXV 2

Times' Crucial Science Book - 9 10

$ ›U2 [ [22
:KHUH › 7

Area of a square = length × length
A= l×l
A = l2

Area of an irregular body is calculated by using a graph paper.

Activity 1.3 7R ÀQG WKH DUHD RI D OHDI

0DWHULDOV UHTXLUHG

Graph paper, a leaf, pencil, etc.

Procedure:

7DNH D OHDI DQG SODFH LW RQ D JUDSK SDSHU

2. Mark the boundary of the leaf with a pencil.

3. Remove the leaf and count the number of complete squares and then
the number of incomplete squares within the boundary of the leaf.
&DOFXODWH WKH DUHD RI WKH OHDI E\ XVLQJ IROORZLQJ IRUPXOD
Area of the leaf

= (No. of complete squares + 1 × No. of incomplete squares) × Area of
a unit square in graph 2

Measurement of volume
7RWDO VSDFH RFFXSLHG E\ D ERG\ LV FDOOHG LWV YROXPH 7KH 6, XQLW RI
volume is cubic metre (m3 7KH RWKHU XQLWV RI YROXPH DUH FXELF IHHW
cubic centimeter, cubic kilometer, etc. Volume of regular objects is
calculated by different formulae.

50 0

height r 40 0

30 0

length breadth 20 0
10 0

Volume of a cuboid body = length × breadth × height

V=l×b×h

Volume of a cube = length × length × length

=l×l×l ? V = l3

9ROXPH RI D F\OLQGULFDO ERG\ › ð UDGLXV 2 × height

9 ›U2h

Volume of a sphere = 4 › UDGLXV 3 ? V = 4 ›U3
3 3

11 Times' Crucial Science Book - 9

Volume of liquid is determined by using a measuring cylinder.
Similarly, volume of irregular objects is calculated by using
measuring cylinder. In this method, a principle is used, in which

Volume of irregular body = Volume of the displaced liquid.

Activity 1.4 7R ÀQG WKH YROXPH RI DQ LUUHJXODU REMHFW

0DWHULDOV UHTXLUHG

Measuring cylinder, water, a piece of stone, etc.

Procedure:

7DNH D PHDVXULQJ F\OLQGHU DQG SRXU
water in it upto 50cm3.

7LH D SLHFH RI VWRQH WR D WKUHDG DQG LPPHUVH LW LQWR WKH ZDWHU RI WKH F\OLQGHU

3. Note the new volume of the water.

4. Calculate the volume of the stone by using formula.

9ROXPH RI VWRQH 9ROXPH RI WKH ZDWHU DIWHU LPPHUVLQJ VWRQH ï LQLWLDO YROXPH RI ZDWHU

= V2 ï91

ï IURP WKH ÀJXUHV

= 20cm3.

7KXV WKH YROXPH RI WKH LUUHJXODU REMHFW L H D SLHFH RI VWRQH LV FP3.

6FLHQWLÀF QRWDWLRQ
Very small and very large numbers can be expressed in power of
WHQ ZKLFK LV NQRZQ DV VFLHQWLÀF QRWDWLRQ )ROORZLQJ SRLQWV VKRXOG
EH NHSW LQ PLQG ZKLOH H[SUHVVLQJ QXPEHUV LQ SRZHU RI WHQ
i) While shifting decimals to the left, the power of 10 should be

increased by one in each shift.
Example

3750000
Here, decimal is at last, shifting decimal to left by one shift, then

3750000 = 375000.0 × 101
Similarly, 375000 = 3.75 × 101 × 105

3750000 = 3.75 × 106
ii) While shifting decimals to right, the power of ten should be

decreased by one in each shift.
Example

0.000754
= 7.54 × 10ï

Times' Crucial Science Book - 9 12

1RWH 7KH QXPEHU ZKLFK LV NHSW LQ SRZHU RI O VKRXOG QRW EH PRUH
than 10 and less than 1.
Solved Numerical Problem 1.2

a. Express 98410000 in power of 10.
98410000
= 9841000.0 × 101 [Shifting decimal to the left by one shift)
= 9.841000 × 101 × 106 [Shifting decimal to the left by 6 shift)
= 9.841 × 107

?98410000 = 9.841 × 107
b. Express 0.0000732 in power of 10.

0.0000732
= 00.000732 × 10ï [Shifting decimal to the right by one shift)
= 7.32 × 10ï × 10ï [Shifting decimal to the right by four shifts)
= 7.32 × 10ï
? 0.0000732 = 7.32 × 10ï

Solved Numerical Problem 1.3

&DOFXODWH WKH YROXPH RI D VSKHUH ZKRVH UDGLXV LV FP

*LYHQ,

Radius (r) = 6cm

Volume of sphere (V) =?

:H KDYH

V= 4 ›U3 = 4 × 22 × 63 = 905.14cm3
3 3 7

7KHUHIRUH YROXPH RI WKH VSKHUH FP3

Learn and Write

1. SI system is extended form of MKS system. Why?
In MKS System, only three quantities, i.e. length, mass and time
are used as fundamental quantities. But in SI system, four more
quantities, i.e. temperature, current, luminous intensity and
amount of substance are also included as fundamental quantities.
It shows that, the units of length, mass and time are same in both
WKH V\VWHPV 7KXV WKH 6, V\VWHP LV WKH H[WHQGHG IRUP RI 0.6
system.

13 Times' Crucial Science Book - 9

7KH XQLW RI ZRUN LV GHULYHG XQLW :K\"
7KH TXDQWLW\ RI ZRUN GRQH LV REWDLQHG E\ XVLQJ IRUPXOD
Work = Force × displacement
= Mass × acceleration × displacement

= Mass × Velocity × displacement
7LPH

= Mass × Displacement × displacement
7LPHð 7LPH

= kg×m×m [Writing the units of quantities]
s×s

= kgm2 V2

Since the unit of work involves three fundamental units, i.e. kg, m
and s. Hence, it is a derived unit.
7KH 6, XQLWV PDNH WKH PHDVXUHPHQW V\VWHPDWLF :K\"
Before the declaration of SI units, there were various local systems
RI PHDVXUHPHQW 7KH XQLWV XVHG LQ RQH SODFH FRXOG QRW EH XQGHUVWRRG
in another place or their amounts used to be different. When SI
units were declared, they started to be used throughout the world.
7KXV WKH PHDVXUHPHQW V\VWHP EHFDPH XQLIRUP DQG V\VWHPDWLF
4. Write any two advantages of SI system of measurement.
7KH 6, V\VWHP RI PHDVXUHPHQW LV WKH PRUH DFFXUDWH DQG VFLHQWLILF
V\VWHP RI PHDVXUHPHQW ,W KDV WKH IROORZLQJ DGYDQWDJHV
a. It includes the units of seven fundamental quantities. Using

these fundamental quantities, we can determine the units of
any derived quantity.
b. It has brought uniformity in measurement all over the world.
7KH ZHLJKW DQG OHQJWK PHDVXULQJ GHYLFHV LQ WKH PDUNHW DUH
checked by the Nepal Bureau of Standards and Metrology once in
every two years. Why?
7KH FRUUHFW PHDVXULQJ GHYLFHV PD\ JHW HUURUV GXH WR GLIIHUHQW
factors such as weather, wear and tear due to excessive use, etc.
Sometimes, the ill-minded and corrupt businesspersons may use
IDOVH GHYLFHV IRU WKH PHDVXUHPHQW SXUSRVH 7R FRUUHFW HUURUV DQG WR
remove the false devices from the market, the measuring devices
are checked by Nepal Bureau of Standards and Metrology once in
every two years.
6. What is meant by scientific notation or standard notation?
7KH VKRUW UHSUHVHQWDWLRQ RI ODUJH ILJXUHV RI QXPEHUV XVLQJ WKH
power of 10 is called scientific notation. It is also known as standard
notation.

Times' Crucial Science Book - 9 14

Glossary Wrongdoers
A local unit to measure the volume of grains
&XOSULWV Eight manas is equal to one Pathi
0DQD Related to commerce and trade, available in the market
3DWKL Unit of volume of dry object
&RPPHUFLDO
6HU

Main points to remember

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RU VWDQGDUG TXDQWLW\ LV FDOOHG PHDVXUHPHQW

8QLW LV WKH NQRZQ RU VWDQGDUG TXDQWLW\ LQ WHUPV RI ZKLFK RWKHU
SK\VLFDO TXDQWLWLHV DUH PHDVXUHG

3. 7KH TXDQWLWLHV WKDW FDQ EH PHDVXUHG DUH FDOOHG SK\VLFDO TXDQWLWLHV
7KH SK\VLFDO TXDQWLWLHV ZKLFK DUH LQGHSHQGHQW RI RWKHU SK\VLFDO

TXDQWLWLHV DUH FDOOHG IXQGDPHQWDO TXDQWLWLHV
7KH TXDQWLWLHV ZKLFK DUH GHULYHG IURP IXQGDPHQWDO TXDQWLWLHV

DUH GHULYHG TXDQWLWLHV
0.6 &*6 )36 DQG 6, V\VWHPV DUH YDULRXV V\VWHPV RI PHDVXUHPHQW
/HQJWK LV GHÀQHG DV WKH GLVWDQFH EHWZHHQ DQ\ WZR SRLQWV
7KH GXUDWLRQ EHWZHHQ DQ\ WZR HYHQWV LV FDOOHG WLPH
2QH VWDQGDUG PHWHU LV GHÀQHG DV WKH GLVWDQFH EHWZHHQ WZR JROG OLQHV

marked on Platinum-Iridium rod at temperature of 0°C kept at the
,QWHUQDWLRQDO %XUHDX RI :HLJKWV DQG 0HDVXUHV DW 6HYUHV LQ )UDQFH
2QH VWDQGDUG NLORJUDP LV GHÀQHG DV WKH PDVV RI 3ODWLQXP ,ULGLXP
DOOR\ F\OLQGHU RI GLDPHWHU HTXDO WR LWV KHLJKW NHSW DW ,QWHUQDWLRQDO
%XUHDX RI :HLJKWV DQG 0HDVXUHV DW 6HYUHV LQ )UDQFH
11. 2QH VWDQGDUG VHFRQG LV GHÀQHG DV th part of a mean solar day.
7KH VSDFH RFFXSLHG E\ WKH VXUIDFH RI DQ REMHFW LV FDOOHG DUHD
7RWDO VSDFH RFFXSLHG E\ D ERG\ LV FDOOHG LWV YROXPH

Exercise

1. Choose the best alternative in each case.

a. Which of the following is a physical quantity?

i. Tension ii. Grief iii. Heat iv. Happiness

b. Which of the following is a fundamental quantity?

i. Speed ii. Area iii. Volume iv. Length

c. Which of the following is not the derived unit?

i. Mole ii. Newton iii. Pascal iv. Watt

d. When was the SI system of measurement brought into use?

i. 1995 ii. 1960 iii. 1980 iv. 2000

e. Which of the following is a fundamental unit?

i. Candela ii. Ampere iii. Mole iv. All of these

15 Times' Crucial Science Book - 9

2. Answer these questions in brief.

a. Define:

i) Measurement ii) Physical quantity iii) One standard metre

iv) Zenith v) One standard second

b. Why is measurement important?

c. What is fundament unit? Give some examples.

d. What is unit? What are the features of good units?

e. What are various systems of measurement?

f. What is SI unit? Why is it better system of units?

g. Write differences between:

i. Pendulum clock and quartz clock

ii. Fundamental unit and derived unit

h. What are the fundamental units involved in the units of following
quantities

i. Volume ii. Work iii. Power

iv. Density v. Potential difference

i. What is area? How do you measure the area of an irregular body?

j. What is volume? How do you measure volume of an irregular body?
*LYH UHDVRQV

i. Watt is a derived unit.

ii. SI system is the extended form of MKS system.

iii. Pendulum clock is not accurate clock for the measurement of time.
4. Numerical problems

a. Convert the following.

i) 2.4 millimetres into metre ii) 4 hours into seconds

iii) 200 grams into kilogram iv) 124 milligram into gram

v) 300 cm2 into m2 vi) 4 km2 into m2
b. A book has dimensions of 24cm × 15cm × 4cm and weight 1.5 kg. It

consists of 500 sheets of papers. Find

i. the volume of the book ii. thickness of each sheet

iii. the area of each sheet

c. Find the volume of a cylinder whose length is 150 cm and diameter

is 60 cm.

d. Express the following numbers in power of 10. Simplify if necessary.

i. 753000000 ii. 0.000835 iii. 3.75 × 10−2× 4.2 × 106

Answers 4.a) i. 0.0024 m ii. 1.44 × 104 s iii. 0.2 kg iv. 0.124 g
v. 0.03 m2 vi. 4 × 106m2 b) i. 1440 cm3 ii. 8 × 10–3 cm iii. 360 cm2
c) 4.24 × 105 cm3 d) i. 7.53 × 108 ii. 8.35 × 10–4 iii. 1.575 × 105

Project Work

Observe the different devices that are being used to measure different physicalquantities
LQ \RXU KRPH DQG ORFDOLW\ 7KHQ ZULWH \RXU ILQGLQJV LQ WKH WDEOH

Times' Crucial Science Book - 9 16

Chapter )RUFH

2 Sir Isaac Newton

+H GLVFRYHUHG 1HZWRQLDQ PHFKDQLFV 8QLYHUVDO
JUDYLWDWLRQ &DOFXOXV 1HZWRQ
V ODZV RI PRWLRQ
2SWLFV %LQRPLDO VHULHV 1HZWRQ
V PHWKRG HWF

Objectives ƐƟŵĂƚĞĚ WĞƌŝŽĚƐ͗ ϭϮ

ƚ ƚŚĞ ĞŶĚ ŽĨ ƚŚĞ ůĞƐƐŽŶ͕ ƐƚƵĚĞŶƚƐ ǁŝůů ďĞ ĂďůĞ ƚŽ͗

• WHOO HIIHFWV RI IRUFH
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• GHÀQH DFFHOHUDWLRQ DQG FDOFXODWH WKH DFFHOHUDWLRQ RI D PRYLQJ REMHFW
• H[SODLQ HTXDWLRQV RI PRWLRQ DQG VROYH VLPSOH SUREOHPV UHODWHG WR WKH PRWLRQ
• H[SODLQ 1HZWRQ·V ODZV RI PRWLRQ DQG WHOO WKHLU XVHV

Mind Openers

• $ EDOO HDVLO\ PRYHV EXW QRW D EXV ZKHQ ZH SXVK WKHP :K\"
• :K\ GR SDVVHQJHUV RI D EXV IDOO EDFNZDUG ZKHQ WKH EXV PRYHV VXGGHQO\"
• :KDW LV DFFHOHUDWLRQ" +RZ FDQ \RX FDOFXODWH DFFHOHUDWLRQ RI DQ REMHFW" 'LVFXVV

Introduction

Suppose, a book is lying on a table. It remains in the same position
until someone displaces it by pushing or pulling. When anyone pulls
or pushes, force is applied.

1RZ FRQVLGHU D EDOO LV UROOLQJ RQ D JURXQG 7KH EDOO ÀQDOO\ VWRSV
after covering some distance. But, the ball covers more distance
if more force is applied to the ball in the same direction of motion.
7KH EDOO FRYHUV PRUH GLVWDQFH ZKHQ WKH URXJKQHVV RI WKH JURXQG
GHFUHDVHV 7KH UHDVRQ WKDW WKH EDOO VWRSV DIWHU FRYHULQJ VRPH
distance is the opposing force called friction created between the
ball and the rough surface.

Consider a big stone is at rest. Apply force to it by pushing and
SXOOLQJ +HUH \RXU IRUFH PD\ QRW EH VXIÀFLHQW WR PRYH LW 6LPLODUO\
D PRYLQJ ERG\ GRHV QRW VWRS ZKHQ WKH DSSOLHG IRUFH LV LQVXIÀFLHQW
It is clear from the above discussion that force is used to change the
position or state of a body. But, it is not always sure that force can
FKDQJH WKH VWDWH 7KXV IRUFH LV GHÀQHG DV WKH SXOO RU SXVK ZKLFK

17 Times' Crucial Science Book - 9

changes or tends to change the state of rest or motion of an object.
7KH SI unit of force is Newton. It is measured in Dyne in CGS
system.
Relationship between newton and dyne
QHZWRQ NJ ð P V2 (? F= m × a)
J ð FP V2
JFP V2
=105JFP V2
=105dynes
?1N=105dynes

Effects of force

a. Force can change the state of rest or motion of a body.
b. It can change the speed of a moving body.
c. It can change direction of a moving body.
d. It can change the shape of a body.

Vectors and scalars

3K\VLFDO TXDQWLWLHV DUH FODVVLÀHG LQWR WZR FDWHJRULHV 7KH\ DUH

i. Vectors ii. Scalars

9HFWRU TXDQWLWLHV

7KH TXDQWLWLHV ZKLFK KDYH ERWK PDJQLWXGH DQG GLUHFWLRQ DUH FDOOHG
vector quantities. Displacement, force, velocity, acceleration, etc
are vector quantities. A vector quantity is generally represented
E\ D OLQH ZLWK DUURZKHDG 7KH OHQJWK RI WKH OLQH UHSUHVHQWV WKH
magnitude and the arrowhead represents the direction.

For example,

A body covers displacement of 10m from point A to B towards east.
It is represented by a straight line AB with an arrowhead towards
A 10m B
east, i.e. it is represented as AB .

If the displacement of 20m is covered from point C to D due east,
it is represented by a straight line CD with an arrowhead towards

east, i.e. it is represented as CD .

C 20m D

9HFWRUV FDQQRW EH DGGHG E\ VLPSOH DOJHEUDLF PHWKRGV 7KHLU
additions are done by the laws of vectors such as triangle law,
parallelogram law, etc.

Times' Crucial Science Book - 9 18

Example:
1. Consider two forces 1 N and 3 N are applied on a body in
WKH VDPH GLUHFWLRQ 7KH UHVXOWDQW IRUFH RI WKRVH WZR IRUFHV LV
given by the addition of these two forces. Mathematically,
Resultant force = (1N + 3N = 4N)
1N 4N
3N .
2. Consider 1N force is applied to an object towards one
direction and 3 N force is applied to the same object in
RSSRVLWH GLUHFWLRQ DV VKRZQ LQ WKH ILJXUH 7KH UHVXOWDQW IRUFH
is calculated by subtracting one force from another force.
Resultant force = 3N —1N = 2N.

1N 3N 2N

3. Consider 2N force is applied to an object towards one
direction and another 2N force is applied to the same object
LQ RSSRVLWH GLUHFWLRQ 7KH UHVXOWDQW IRUFH ZLOO EH ]HUR EHFDXVH
two forces cancel each other.
Resultant force = 2N —2N = 0

2N 2N

6FDODU TXDQWLWLHV
7KH TXDQWLWLHV ZKLFK KDYH RQO\ PDJQLWXGH EXW QR GLUHFWLRQ DUH
called scalar quantities. Mass, time, volume, area, temperature, etc
DUH VFDODU TXDQWLWLHV 7KH\ DUH DGGHG E\ VLPSOH DOJHEUDLF UXOHV )RU
example, 5 kg of sugar when added to 6 kg sugar, the resultant
mass will be 11 kg. Similarly, 5 pens when added to 20 pens, the
total pens will be 25 pens and so on.

Fact Reason

Speed is scalar quantity but velocity is vector quantity, why?
Speed has only magnitude but not direction whereas velocity
has both magnitude and direction. Hence, speed is scalar
quantity but velocity is vector quantity.

19 Times' Crucial Science Book - 9

'LIIHUHQFHV EHWZHHQ YHFWRUV DQG VFDODUV

Vector TXDQWLWLHV Scalar TXDQWLWLHV

7KH TXDQWLW\ ZKLFK KDV ERWK 7KH TXDQWLW\ ZKLFK KDV RQO\
magnitude and direction is magnitude but no direction is
called vector quantity. called scalar quantity.

2. It is represented by a straight 2. It is represented by a straight
line with an arrowhead. line without an arrowhead.

7KH VXP RI WZR YHFWRUV PD\ 7KH VXP RI WZR VFDODUV LV
be positive, negative or zero. always positive.

4. Vector quantities cannot be 7KH VFDODU TXDQWLWLHV FDQ
added or subtracted by simple be added by simple algebraic
algebraic methods. methods.

Distance and displacement

When a person has to move 3
IURP SRLQW $ WR SRLQW % +H 2

She can go along the paths A 1 B
1, 2, 3 or 4 as shown in the 4
ÀJXUH

When the person travels through paths 1, 2, 3 and 4, suppose that
he has to travel 20m, 30m, 40m and 50 meters respectively.

Here, the actual length of path 1, 2, 3 and 4 is 20m, 30m, 40m and
P UHVSHFWLYHO\ 7KHUHIRUH WKH GLVWDQFH RI SDWK DQG LV
20m, 30m, 40m and 50m respectively. Here, between the same
points A and B, the actual lengths of different paths are different.

7KH DFWXDO OHQJWK RI D SDWK WUDYHOOHG C
by a body is called distance. 7KH 6,
unit of distance is meter (m). Distance
is a scalar quantity. 5m 3m

But, the shortest distance between
the points A and B is the length of
the path 1. It is displacement. 7KH A 4m B
shortest distance between initial and
ÀQDO SRVLWLRQV RI DQ REMHFW LV FDOOHG
displacement. ,W FDQ DOVR EH GHÀQHG DV WKH VKRUWHVW GLVWDQFH EHWZHHQ
DQ\ WZR SRLQWV LQ D SDUWLFXODU GLUHFWLRQ 7KH 6, XQLW RI GLVSODFHPHQW
is meter (m). Displacement is a vector quantity.

Times' Crucial Science Book - 9 20

Suppose a person moves from point A to B then to C. Let, the distance
EHWZHHQ $ DQG % LV P DQG GLVWDQFH EHWZHHQ % DQG & LV P 7KXV WKH
total distance he travels is 7m. But the shortest distance between A
DQG & LV WKH OHQJWK RI $& ,W LV P 7KXV WKH DFWXDO GLVWDQFH WUDYHOOHG
by the person is 7m but the displacement is only 5m.
Differences between distance and displacement

Distance Displacement

1. It is the actual length of a 1. It is the shortest distance
path travelled by an object. between any two points in a
particular direction.

2. It is a scalar quantity. 2. It is a vector quantity.

3. It can be added or subtracted 3. It is added or subtracted by
by simple mathematical rule. using vector laws.

4. Its value is always positive. 4. Its value may be positive,
negative or zero.

Speed
Suppose a body moves from one place to another either in a straight
or curved path and covers 20 meter distance in 2 seconds. In such
case, we can say that it covers 10m distance in 1 second and its
VSHHG LV P V

6SHHG LV GHÀQHG DV WKH GLVWDQFH WUDYHOOHG E\ DQ REMHFW LQ XQLW WLPH

Distance covered (d) ?v = dt
Speed = 7LPH WDNHQ W

Since the distance is measured in meter and time in second, speed is
PHDVXUHG LQ PHWHU SHU VHFRQG P V ,W LV DOVR PHDVXUHG LQ NLORPHWHU
SHU KRXU NP KU 6SHHG LV D VFDODU TXDQWLW\

Velocity

When a body moves from one point to another in a straight path in
a particular direction, its speed is called velocity.

7KHUHIRUH YHORFLW\ FDQ EH GHÀQHG DV WKH VSHHG RI D ERG\ LQ D
particular direction.

Since distance in a particular direction is called displacement,
YHORFLW\ FDQ EH GHÀQHG DV GLVSODFHPHQW RI DQ REMHFW LQ XQLW WLPH

Displacement ?v = st
Velocity = 7LPH WDNHQ

21 Times' Crucial Science Book - 9

Since the SI unit of displacement is meter and that of time is second, the
6, XQLW RI YHORFLW\ LV PHWHU SHU VHFRQG P V 9HORFLW\ LV D YHFWRU TXDQWLW\

Solved Numerical Problem 2.1

$ PRWRU F\FOLVW WUDYHOV D GLVWDQFH RI NLORPHWHUV LQ PLQXWHV &DOFXODWH KLV VSHHG

Solution

Distance (d) = 4.2 km = 4200 m

7LPH W PLQXWH ð V V

Speed (v) =?

We have, Distance travelled (d)
7LPH WDNHQ W
Speed= = 4224000 P V

7KHUHIRUH WKH VSHHG RI WKH PRWRU F\FOLVW LV P V

'LIIHUHQFHV EHWZHHQ VSHHG DQG YHORFLW\

Speed Velocity

1. It is the distance travelled by 1. It is the displacement of an
an object in unit time. object per unit time.

2. It is a scalar quantity. 2. It is a vector quantity.

3. Speed of a moving body is 3. Velocity of a moving body may
always positive. be positive, negative or zero.

4. It can be added or subtracted 4. It is added or subtracted by
by simple mathematical rule. vector rules

Uniform and variable motion

When a body covers equal distance in equal interval of time, its
motion is called uniform motion.
0 sec
1 sec 1 sec 1 sec 1 sec

a 5m b 5m c 5m d 5m e

,Q WKH DERYH ÀJXUH D FDU LV FRYHULQJ P GLVWDQFH LQ WKH ÀUVW RQH
second. It further covers 5m distance in the next one second and so
on. Its motion is uniform motion because it travels equal distance
in equal interval of time. Motion of hands of watch, motion of the
planets, motion of the moon, etc. are examples of uniform motion.
When a body covers unequal distance in equal interval of time, its
motion is called variable motion. 7KH YDULDEOH PRWLRQ LV DOVR NQRZQ
as non-uniform motion.

Times' Crucial Science Book - 9 22

0 sec 1 sec 1 sec 1 sec

a 5m b 10m c 4m d

,Q WKH DERYH ÀJXUH WKH FDU WUDYHOOHG P LQ WKH ÀUVW RQH VHFRQG
P LQ WKH QH[W RQH VHFRQG DQG P LQ WKH WKLUG RQH VHFRQG 7KXV
LW FRYHUV GLIIHUHQW GLVWDQFHV LQ HTXDO LQWHUYDO RI WLPH 7KHUHIRUH
its motion is called variable motion. Motion of animals, motion of
vehicles, motion of water in river, motion of human beings, etc are
examples of variable motion.

Fact Reason

7KH DFFHOHUDWLRQ RI D ERG\ PRYLQJ ZLWK XQLIRUP YHORFLW\ LV ]HUR ZK\"
Acceleration is the rate of change in velocity and there is no
change in velocity of the body moving with uniform velocity.
Hence, the acceleration of a body moving with uniform velocity
is zero.

Acceleration

When a body moves in variable motion, its velocity keeps on
FKDQJLQJ 7KH UDWH E\ ZKLFK WKH YHORFLW\ RI D ERG\ FKDQJHV LV FDOOHG
its acceleration. 7KXV the rate of change of velocity of a moving
body is called acceleration.

6XSSRVH D ERG\ LV PRYLQJ IURP D SRLQW ,WV YHORFLW\ LV P V LQ WKH
EHJLQQLQJ 6XSSRVH LWV YHORFLW\ VORZO\ LQFUHDVHV DQG EHFRPHV P V
after 5 seconds.

X P V Y P V
0 sec 5 sec

t=5sec

+HUH WKH YHORFLW\ RI WKH FDU LQFUHDVHG IURP P V WR P V LQ V
7KH LQFUHDVH LQ WKH YHORFLW\ RI WKH FDU LQ V LV P V 7KHUHIRUH
LWV YHORFLW\ LV LQFUHDVHG E\ P V LQ HYHU\ RQH VHFRQG 7KXV LWV
DFFHOHUDWLRQ LV P V2.
Suppose, a body is moving with initial velocity ‘u’ and its velocity
increases to ‘v’ in time ‘t’, then the acceleration ‘a’ is calculated by
the formula,

Acceleration = Final Velocity - Initial Velocity ?a = v - u
7LPH taken t

23 Times' Crucial Science Book - 9

Unit of acceleration

In SI unit, the change in velocity is measured in meter per second
DQG WLPH LQ VHFRQG 7KHUHIRUH WKH 6, XQLW RI DFFHOHUDWLRQ ZLOO EH
PHWHU SHU VHFRQG SHU VHFRQG RU P V2.

Acceleration = Change in Velocity
7LPH taken

P V m m
=s = s × s = s2

If the velocity of a moving body decreases, its acceleration will
EH QHJDWLYH ,W LV FDOOHG UHWDUGDWLRQ 7KXV the rate of decrease of
velocity of a body is called retardation. It is also a vector quantity
DQG LWV XQLW LV P V2.

Solved Numerical Problem 2.2

$ EXV LV PRYLQJ ZLWK D YHORFLW\ RI P V ,W LQFUHDVHV LWV YHORFLW\ WR P V DIWHU
second. Find its acceleration.

Solution

*LYHQ,

,QLWLDO 9HORFLW\ X P V

)LQDO 9HORFLW\ Y P V

7LPH W VHF

Acceleration (a) =?

We have, a = v - u or, a = 30 - 15 or,a = 15 P V2
t 5 5

7KHUHIRUH WKH DFFHOHUDWLRQ RI WKH EXV LV P V2.

$YHUDJH YHORFLW\
When a body moves in non-uniform motion, its velocity changes
continuously. In such a situation, we use average velocity. Average
velocity of a body is the total displacement travelled by the body
divided by total time.

Average velocity(v–) = 7RWDO GLVSODFHPHQW
7RWDO WLPH WDNHQ

,WV 6, XQLW LV P V DQG &*6 XQLW LV FP V

Times' Crucial Science Book - 9 24

If a body gains a total displacement of 50 m in 10 seconds, its average
velocity is calculated as,

Average velocity(v–) = 7RWDO GLVSODFHPHQW = 50 P V
7RWDO WLPH WDNHQ 10

If the velocity of a body changes with a constant rate, its average
velocity is calculated by using a formula,

Average velocity(v–) = Initial velocity + Final velocity (v–) = u + v
2 2

Initial velocity is the velocity of a body at the beginning of the time
and is denoted by u. Final velocity is the velocity of a body at the
end of time and is denoted by v.

,I D ERG\ LQFUHDVHV LWV YHORFLW\ IURP P V WR P V ZLWK D FRQVWDQW
UDWH LWV DYHUDJH YHORFLW\ LV FDOFXODWHG E\

Average velocity(v–) = Initial velocity + Final velocity = 20 + 30 P V
2 2

(TXDWLRQV RI PRWLRQ

If a body moves with a constant acceleration, its initial velocity
u ÀQDO YHORFLW\ v, distance travelled s, time taken t and uniform
acceleration a are related, the relation can be shown by equations.
7KHVH HTXDWLRQV DUH FDOOHG equations of motion. 7KH\ DUH

1. v = u + at

2. s= u v ×t
2

u v 3. v2 = u2 + 2as
s
t 4. s = ut + 1 at2
2

To prove: v = u + at
We know that, acceleration is the rate of change of velocity.

Acceleration = Final Velocity - Initial Velocity
7LPH taken

25 Times' Crucial Science Book - 9

v-u
a= t

or, at = v —u

?v = u + at
To prove: s = u + v × t

2
We have,

Initial velocity + Final velocity
Average velocity =

2

= u + v....................(i)
2
Again,
7RWDO GLVSODFHPHQW
Average velocity = = s ...................(ii)
t
7RWDO WLPH WDNHQ

From equations (i) and (ii), we have

s u+v
t= 2

u+v
?s = 2 × t

To prove: v2 = u2 + 2as

We have, Final Velocity - Initial Velocity
Acceleration =
7LPH taken
Again,
v-u v-u
a= t or, t = a ................(i)

7RWDO GLVSODFHPHQW $YHUDJH YHORFLW\ ð 7LPH

Initial velocity + Final velocity
s = ð 7LPH

2

Times' Crucial Science Book - 9 26

s= u + v × t............................(ii)
2
6XEVWLWXWLQJ WKH YDOXH RI ¶W· IURP HTXDWLRQ L LQ HTXDWLRQ LL ZH KDYH

s= u+v × v-u or, s = v2 - u2
2 a 2a

or, 2as = v2 ï X2

?v2 = u2 + 2as

To prove: s = ut + 1
We have, 2 at2

Acceleration = Final Velocity - Initial Velocity
7LPH

v-u
a= t
or, at = Y - u

? Y X DW L

Again,

7RWDO GLVWDQFH $YHUDJH YHORFLW\ ð 7LPH

s= Initial velocity + Final velocity ð 7LPH
2

s= u+v × t............................(ii)
2

6XEVWLWXWLQJ WKH YDOXH RI ¶Y· IURP HTXDWLRQ L LQ HTXDWLRQ LL ZH KDYH

s= u+(u+at) ×t = 2u+at × t = 2ut + at2
2 2 2
2

1
? s = ut + 2 at2
Special conditions
1. When a body starts from rest, u = 0.
:KHQ D PRYLQJ ERG\ FRPHV WR UHVW Y
:KHQ D ERG\ PRYHV XQGHU WKH LQÁXHQFH RI JUDYLW\ D J
:KHQ D ERG\ LV WKURZQ YHUWLFDOO\ XSZDUGV Y
:KHQ D ERG\ LV DOORZHG WR IDOO YHUWLFDOO\ GRZQZDUGV X

27 Times' Crucial Science Book - 9

Solved Numerical Problem 2.3

,I D ERG\ JDLQV P GLVSODFHPHQW LQ V )LQG LWV DYHUDJH YHORFLW\
Given,
7RWDO GLVSODFHPHQW G P
7RWDO WLPH W V
Average velocity =?
We have,
7RWDO GLVSODFHPHQW 500
Average velocity= 7LPH WDNHQ = 10 P V

7KHUHIRUH DYHUDJH YHORFLW\ RI WKH ERG\ LV P V

Solved Numerical Problem 2.4

$ ERG\ FKDQJHV LWV YHORFLW\ IURP NP KU WR NP KU LQ V )LQG LWV
DFFHOHUDWLRQ DQG GLVWDQFH WUDYHOOHG
Given,
54km 54 × 1000m
Initial velocity (u) NP KU = P V
1hr 3600s

72km 72 × 1000m P V
Final velocity (v) NP KU =
1hr 3600s

Acceleration (a) = ?
Distance (s) = ?
We have, v = u + at
or, 20 = 15 + a × 2
or, 2a = 5

?D P V2
Again,

s= u+v ×t
2

15+20

= × 2 = 35m

2

7KHUHIRUH WKH DFFHOHUDWLRQ LV P V2 and distance travelled is 35m.

Times' Crucial Science Book - 9 28

Solved Numerical Problem 2.5

$ EXV LQFUHDVHV LWV YHORFLW\ IURP P V WR P V DQG FRYHUV P
distance. Find.

L WLPH UHTXLUHG DQG LL DFFHOHUDWLRQ
Given,

,QLWLDO YHORFLW\ X P V

)LQDO YHORFLW\ Y P V

Distance travelled (s)= 300m

7LPH W "

Acceleration (a)= ?

We have,

S= u+v ×t or, 300= 30+45 ×t or, t= 600 ? t = 8s
Again, 2 2 75

v = u + at

45 = 30 + a × 8

a = ï
8
15
or, a = 8 ? D P V2

7KHUHIRUH WLPH WDNHQ W V DQG DFFHOHUDWLRQ D P V2.

Solved Numerical Problem 2.6

,I D EXV WKDW VWDUWHG IURP UHVW DWWDLQV DQ DFFHOHUDWLRQ RI P V2 while
WUDYHOOLQJ D GLVWDQFH RI NLORPHWHUV ZKDW ZLOO EH WKH ÀQDO YHORFLW\ RI WKH
EXV" $OVR FDOFXODWH WKH WLPH WDNHQ E\ EXV WR WDYHO WKLV GLVWDQFH.
*LYHQ

Initial velocity (u) = 0
$FFHOHUDWLRQ D P V2

Distance travelled (s) = 2 km = 2×1000m = 2000m
Final velocity (v) = ?

29 Times' Crucial Science Book - 9

using the formula:

v2 = u2+2as

= 02+2×0.4×2000

= 1600

?v = 1600 P V

+HQFH WKH ÀQDO YHORFLW\ RI WKH EXV LV P V

Again,

)LQDO YHORFLW\ Y P V

Initial Velocity (u) = 0

7LPH WDNHQ W "

1RZ ZH KDYH

v = u+at

v-u 40- 0
t= a = 0.4 =100s
7KHUHIRUH the time required by the bus is 100 seconds..

Inertia and mass
A book lying on a table does not change its position itself. Similarly,
a body moving in a particular direction does not stop itself. Here,
IRUFH LV WR EH DSSOLHG WR FKDQJH WKHLU VWDWH RI UHVW RU PRWLRQ 7KLV
property of a body is called inertia.
Inertia is the property of a body due to which it remains or tends to
remain in its previous state of rest or uniform motion in a straight
line.
Inertia of a body depends upon the mass of the body. An object
having more mass has more inertia and vice versa.

Fact Reason

,W LV GLIÀFXOW WR PRYH D ODUJH VWRQH WKDQ D VPDOO RQH ZK\"
Inertia is directly proportional to mass due to which the inertia
RI UHVW LV PRUH LQ ODUJH VWRQH 6R LW LV GLIÀFXOW WR PRYH D ODUJH
stone than a small one.

Times' Crucial Science Book - 9 30

Activity 2.1 7R VWXG\ WKH UHODWLRQVKLS EHWZHHQ PDVV DQG LQHUWLD

Materials required:
7ZR FDQV VDQG WKUHDG RU URSH HWF
Procedure:
7DNH WZR FDQV RI HTXDO VL]H DQG ÀOO RQH RI WKHP ZLWK VDQG
6XVSHQG ERWK WKH FDQV DV VKRZQ LQ WKH ÀJXUH.
3. Apply force to each of them to

bring into motion. Which one is
easier to bring into motion?
Observation:
0RUH IRUFH LV UHTXLUHG IRU WKH FDQ ÀOOHG
with sand to bring into motion.
Conclusion:
Inertia increases with increase in mass.

Types of inertia
,QHUWLD LV RI WKUHH W\SHV 7KH\ DUH
i. Inertia of rest ii. Inertia of motion iii. Inertia of direction
Inertia of rest
It is the property of a body by virtue of which it remains or tries to
remain in the state of rest until external force is applied.

Activity 2.2 7R VWXG\ WKH LQHUWLD RI UHVW

Materials required:
Card board, a glass, a coin, etc.
Procedure:
7DNH D JODVV DQG SODFH D FDUG ERDUG RYHU WKH JODVV DV VKRZQ LQ WKH ÀJXUH
2. Put a coin on the cardboard just over the mouth of the glass.

*LYH D VXGGHQ SXVK WR WKH FDUGERDUG ZLWK D ÀQJHU :KDW KDSSHQV"

31 Times' Crucial Science Book - 9

Observation:
7KH FDUGERDUG FRPHV LQWR PRWLRQ DQG IDOOV WR WKH JURXQG EXW WKH FRLQ IDOOV LQWR
the glass.
Explanation:
7KH FRLQ WULHV WR UHPDLQ LQ WKH UHVW VWDWH GXH WR LQHUWLD RI UHVW EXW WKH FDUG ERDUG
FRPHV LQWR PRWLRQ GXH WR WKH DSSOLFDWLRQ RI IRUFH 7KXV WKH FRLQ IDOOV LQWR WKH JODVV

Examples of inertia of rest
1. Passengers of a bus fall backward when a bus suddenly starts

WR PRYH In the beginning, both the bus and the passengers are
at rest. When the bus suddenly moves, the lower parts of body
of the passengers also come in motion along with the bus, but
the upper parts of the passengers’ body try to remain at rest
GXH WR LQHUWLD RI UHVW 7KHUHIRUH WKH SDVVHQJHUV IDOO EDFNZDUG
when a bus suddenly starts moving.
2. When a carpet is beaten by a stick, the dust flies away or falls
GRZQ In the beginning both carpet and dust particles are at
rest. When the carpet is beaten by a stick, the carpet comes
into motion but the dust tries to remain at rest due to inertia
RI UHVW 7KXV WKH GXVW JHWV VHSDUDWHG DQG IOLHV DZD\ RU IDOOV
down when the carpet is beaten by a stick.
0DQJR IUXLWV IDOO IURP D WUHH ZKHQ WKH WUHH LV VKDNHQ When
a mango tree is shaken, the branches of the tree come into
motion due to the effect of force applied. But, the mango try
WR UHPDLQ DW UHVW GXH WR LQHUWLD RI UHVW 7KXV WKH PDQJR JHW
separated from the mango tree and fall down.
Inertia of motion
It is the property of a body by virtue of which a moving body tries
to remain in the state of motion in the same direction until the
external force is applied.
Examples of inertia of motion
1. Passengers of a bus jerk forward when a moving bus suddenly
VWRSV When a bus is moving, both the bus and the passengers
are in motion state. When the bus suddenly stops, the lower
body parts of the body of passengers come into rest along

Times' Crucial Science Book - 9 32

with the bus. But, the upper parts of the body try to remain
LQ PRWLRQ GXH WR LQHUWLD RI PRWLRQ 7KXV WKH SDVVHQJHUV MHUN
forward when the moving bus suddenly stops.
$Q DWKOHWH UXQV VRPH GLVWDQFH EHIRUH WDNLQJ D ORQJ MXPS When
an athlete runs some distance before taking a long jump, the
velocity gained by him at the time of running is added to the
YHORFLW\ WDNHQ E\ KLP DW WKH WLPH RI MXPS 7KXV WKH DWKOHWH
becomes able to take longer jump.
3. When a passenger jumps out from a moving bus, he falls to the
GLUHFWLRQ RI WKH EXV When a bus is moving, the passengers and
the bus both are in motion. When a passenger jumps from the
bus, his legs come in contact with the ground and come to rest
but his upper parts try to remain in motion in the same direction
GXH WR LQHUWLD RI PRWLRQ 7KXV WKH SDVVHQJHU IDOOV RQ WKH JURXQG
LQ WKH GLUHFWLRQ RI WKH EXV 7R SUHYHQW IURP IDOOLQJ WKH SDVVHQJHU
has to run some distance in the direction of the bus.
4. When wheels of a vehicle rotate at high speed, the mud
VWLFNLQJ WR WKHP IOLHV DZD\ WDQJHQWLDOO\ 7KLV LV GXH WR LQHUWLD
of direction.
Momentum
If you attempt to stop a moving body, you need to apply some force.
Similarly, you have to apply force to move a stationary object.
More force is required to stop a moving body which has more mass.
In the same way, more force is required to stop a body moving with
KLJK YHORFLW\ 7KHVH REVHUYDWLRQV DUH GXH WR WKH TXDQWLW\ RI PRWLRQ
SUHVHQW LQ D ERG\ 7KH TXDQWLW\ RI PRWLRQ LQ D ERG\ GHSHQGV RQ WKH
mass (m) and velocity (v) of the moving body.
7KH SURGXFW RI PDVV DQG YHORFLW\ RI D PRYLQJ ERG\ LV FDOOHG LWV
momentum.
Mathematically,

Momentum= Mass × Velocity
?p = m × v

Momentum is a vector quantity. Since, the SI unit of mass is kg and
that of velocity is P V, the SI unit of momentum is NJP V.

33 Times' Crucial Science Book - 9

A stationary object has zero momentum. It is because the velocity
of a stationary object is zero.
As stated above, momentum of a body depends on its mass and
velocity. In other words, the momentum of a body is directly
proportional to the product of mass and velocity. Hence, the
momentum of a body with more mass and velocity is more than
that having less mass and velocity. It is the reason why a person
receives more injury if struck by a cricket ball than that with a
shuttle cock.
Newton’s laws of motion
Sir Isaac Newton (1642-1727 AD), a British scientist made detailed
study on WKH PRWLRQ RI ERGLHV DQG JDYH WKUHH ODZV 7KH\ DUH NQRZQ
as Newton’s laws of motion.
1HZWRQ·V ÀUVW ODZ RI PRWLRQ
It states that everybody continues in its state of rest or uniform
motion in a straight line unless it is acted upon by some external
forces.
Explanation
According to this law, a body at rest remains
in the same state and a body in motion
continues moving in the same direction until
H[WHUQDO IRUFH LV DSSOLHG 7KLV LQDELOLW\ RI
a body to change its state of rest or motion
E\ LWVHOI LV FDOOHG LQHUWLD 7KXV WKH ILUVW ODZ
of motion has the same concept as that of
LQHUWLD 7KHUHIRUH the first law of motion is called law of inertia.
From the first law of motion, it is clear that a body changes its state
of rest or motion in a particular direction due to application of force.
7KXV the first law of motion gives the definition of force. According
to it, force is an external agency that changes or tends to change
state of rest or motion of a body in a particular direction.

Fact Reason

:K\ LV 1HZWRQV ÀUVW ODZ RI PRWLRQ FDOOHG ODZ RI LQHUWLD" 1HZWRQ·V
ÀUVW ODZ RI PRWLRQ VWDWHV WKDW ´HYHU\ ERG\ FRQWLQXHV WR UHPDLQ
in the position of rest or uniform motion unless external force
LV DSSOLHG WR LW µ +HUH 1HZWRQ·V ÀUVW ODZ RI PRWLRQ KDV GHÀQHG
LQHUWLD 6R 1HZWRQV ÀUVW ODZ RI PRWLRQ LV FDOOHG ODZ RI LQHUWLD

Times' Crucial Science Book - 9 34

([DPSOHV RI ÀUVW ODZ RI PRWLRQ
a. When a mango tree is shaken, mango fruits fall down.
E 3DVVHQJHUV RI D EXV IDOO EDFNZDUG ZKHQ WKH EXV VXGGHQO\ PRYHV
F 7KH EODGHV RI D IDQ FRQWLQXH WR PRYH IRU VRPHWLPH HYHQ WKRXJK
it is switched off. When the fan is switched off, the blades of the
IDQ WU\ WR UHPDLQ LQ WKH SUHYLRXV VWDWH RI PRWLRQ GXH WR LQHUWLD
RI PRWLRQ 7KXV WKH EODGHV RI WKH IDQ URWDWH IRU VRPHWLPH HYHQ
after the fan is switched off.
G $ PDQ MXPSLQJ IURP D PRYLQJ EXV PD\ IDOO GRZQ :KHQ D
PDQ MXPSV IURP D PRYLQJ EXV KLV IHHW FRPH LQ FRQWDFW ZLWK
the ground and come to the rest state. But, his upper parts try
WR UHPDLQ LQ WKH SUHYLRXV VWDWH RI PRWLRQ LQ WKH GLUHFWLRQ RI WKH
PRYLQJ EXV 7KXV WKH PDQ IDOOV LQ WKH GLUHFWLRQ RI WKH EXV

Newton’s second law of motion

It states that acceleration produced in a body is directly proportional
to the force applied and inversely proportional to its mass.

Explanation

When a force is applied to a body continuously, acceleration is
produced. When more force is applied, the acceleration produced is
also more. It means acceleration is directly proportional to the force
applied, i.e. a D F.

Similarly, when two bodies, one lighter and another heavier are
acted by equal amount of force, the lighter body moves faster. It
means the acceleration produced on the lighter body is more than
ai.cec.ealeDram1tiofnor’ath’ eisgpivroednufcoerdceo. n
that on heavier body, a body having
Let us suppose that,
mass ’m’ due to the application of force ’F’. According to Newton’s
second law of motion,

a D F………….(i)

a D 1 ……….(ii)
m
&RPELQLQJ HTXDWLRQV L DQG LL ZH KDYH

a D F × 1 or, a D F or, F D m × a
m m

or, F = km × a……(iii) :KHUH N LV D SURSRUWLRQDOLW\ FRQVWDQW

1 Newton force is defined as that force which when applied to a body
RI PDVV NJ SURGXFHV DQ DFFHOHUDWLRQ RI P V2.

35 Times' Crucial Science Book - 9

,W PHDQV ,I ) 1 DQG P NJ WKHQ D P V2. Putting their
values in equation (iii),

1=k×1×1 ?k=1

Substituting the value of k LQ HTXDWLRQ LLL ZH KDYH

F=1×m×a ?F = ma

7KXV WKH DPRXQW RI IRUFH LV HTXDO WR WKH SURGXFW RI PDVV DQG
acceleration. If the value of acceleration and mass are known,
WKH DPRXQW RI IRUFH FDQ EH FDOFXODWHG XVLQJ WKLV HTXDWLRQ 7KXV
Newton’s second law of motion gives the measurement of force.

Solved Numerical Problem 2.7

:KDW DPRXQW RI IRUFH LV UHTXLUHG WR PRYH D ERG\ RI PDVV NJ ZLWK DQ
DFFHOHUDWLRQ RI P V2?
Solution:

Mass (m) = 5 kg
$FFHOHUDWLRQ D P V2

Force (F) = ?
We have,

F = m × a = 5 × 4 = 20 N
? 7KH DPRXQW RI IRUFH UHTXLUHG LV 1

Solved Numerical Problem 2.8

:KDW DPRXQW RI IRUFH LV UHTXLUHG WR GLVSODFH D ERG\ RI PDVV NJ WKURXJK
a distance of 20m in 4s?

Solution: Mass (m) = 5 kg
Distance (s) = 20 m
7LPH W V
,QLWLDO YHORFLW\ X P V
Acceleration (a) = ?
Force (F) = ?

We have,

s = ut + 1 at2 or, 20 = 0 × 4 + 1 × a × 42
2 2

Times' Crucial Science Book - 9 36

or, 20 = 1 × a × 16 or, 20 = 8a or, a = 20 P V2
2 8
Again,

F = ma = 5 × 2.5 = 12.5 N

?Amount of force required = 12.5 N.

Solved Numerical Problem 2.9

$ FDU RI PDVV NJ PRYLQJ ZLWK D VSHHG RI NP KU VWRSV DIWHU
seconds by applying brakes. Calculate:

D WKH GLVWDQFH WUDYHOOHG E\ WKH FDU
b. the amount of force applied

Solution:

Mass (m) = 5000kg

72km 72 × 1000m P V
Initial Velocity (u) NP KU =
1hr 3600s
)LQDO YHORFLW\ Y P V

7LPH W V

Distance travelled (s) = ?

Amount of force (F) = ?

:H KDYH u+v 20+0
$JDLQ s= 2 ×t
= 2 × = 60m

Y X DW RU Dð RU D ï

ï ï
or, a = 6 or, a = 3 P V2
$FFRUGLQJ WR 1HZWRQ
V VHFRQG ODZ RI PRWLRQ
( )= 500 ×
F = ma ï ï 1
3

Here, negative force means force is applied against the motion of the
body.

7KHUHIRUH the amount of force applied is 1666.67N

37 Times' Crucial Science Book - 9

$SSOLFDWLRQ RI 1HZWRQ·V VHFRQG ODZ RI PRWLRQ

A cricket player moves his hands backwards while catching the
cricket ball.

In the beginning, the velocity of the ball is u. When the cricket
SOD\HU FDWFKHV WKH EDOO WKH ÀQDO YHORFLW\ EHFRPHV ]HUR

In the equation,

F = ma

or, F=m× v–u
t
or , F = PYïPX
t
or, F= 1
t PYïPX

7KH TXDQWLW\ PYïPX LV FRQVWDQW 7KXV IRUFH LV LQYHUVHO\

proportional to the time, i.e. F D 1 .
t

It means when more time is taken to stop the ball, less force is
VXIÀFLHQW WR bring the ball to the rest. For increasing the time, the
player moves his hands backwards and saves his hands from injury.

Activity 2.3 7R VKRZ WKDW WKH DFWLRQ DQG UHDFWLRQ DUH HTXDO

Materials required: A B
7ZR VSULQJ EDODQFHV

Procedure:

7DNH WZR VSULQJ EDODQFHV
2. Put hook of one balance to the hook of another balance as
VKRZQ LQ ÀJXUH
+ROG RQH RI WKH EDODQFHV ÀUPO\ RQ D VXSSRUW DQG SXOO WKH

another balance. What readings do you see in both balances?

Observation:

Reading in both balances will be same.

Explanation:

When the spring balance A applies force to the spring balance B , it
LV DFWLRQ 7KH VSULQJ EDODQFH ·%· DOVR DSSOLHV HTXDO DQG RSSRVLWH IRUFH
to the balance B as reaction. Since action and reaction are equal, the
reading in both balances become equal.

Times' Crucial Science Book - 9 38

Activity 2.4 7R VKRZ WKDW DFWLRQ DQG UHDFWLRQ DFW LQ RSSRVLWH GLUHFWLRQ

Materials required:

A balloon, cello tape, paper, a Balloon pushes on air Air pushes on balloon
long thread, etc.

Procedure:
1. Make a paper spool by rolling a paper.
7DNH D ORQJ WKUHDG DQG SDVV RQH HQG RI WKH WKUHDG WKURXJK WKH VSRRO
7LH WZR HQGV RI WKH WKUHDG RQ RSSRVLWH ZDOOV RI WKH URRP WR PDNH

WKH WKUHDG VWUHWFKHG DV VKRZQ LQ ÀJXUH
,QÁDWH D EDOORRQ DQG WLH LWV PRXWK
5. Fix the balloon in the spool with the help of pieces of tape.
6. Release the air from the balloon by opening its mouth. What

happens?

Observation:

7KH EDOORRQ PRYHV EDFNZDUG ZKHQ WKH DLU FRPHV RXW

Explanation:

Coming out of air from the balloon is action. As a reaction, the
balloon moves backward.

1HZWRQ·V WKLUG ODZ RI PRWLRQ
It states that ”to every action, there is equal and opposite reaction”.
Explanation
When a force is applied to a body, the body also produces another
force. When a body A exerts force on a body B, the body B also
exerts equal and opposite force on the body A 7KXV D VLQJOH IRUFH
is impossible to exist. Force always exists in pair. 7KH SDLU RI WKH
forces involved are called action and reaction. 7KH IRUFH action is the
cause and the force reaction is the effect.

Examples of third law of motion
a. When a man swims, he pushes the water in the backward
direction as an action. The water pushes the man in the
forward direction as a reaction. Thus, swimming is possible.
b. When a person walks, he pushes the ground backward. Here,
the force applied by the person to the ground backward is
action. As a reaction, the ground applies force to the feet in the
forward direction. Thus, he can walk.

39 Times' Crucial Science Book - 9

F ,Q URFNHWV WKH EXUQW JDVHV FRPH RXW YHUWLFDOO\ GRZQZDUGV
ZLWK D JUHDW IRUFH DV DFWLRQ $V D UHDFWLRQ WKH JDVHV JLYH HTXDO
DQG RSSRVLWH IRUFH WR WKH URFNHWV 7KXV WKH URFNHW ÁLHV XS

G $ EXOOHW LV ÀUHG IURP D JXQ DV DQ DFWLRQ 7KH EXOOHW JLYHV HTXDO
and opposite force to the gun as reaction. Thus, the gun recoils
EDFNZDUGV ZKHQ EXOOHW LV ÀUHG IURP LW

e. While rowing a boat, a person pushes water backward as
DFWLRQ $V D UHDFWLRQ WKH ZDWHU JLYHV HTXDO DQG RSSRVLWH IRUFH
WR WKH ERDW 7KXV WKH ERDW PRYHV IRUZDUG

Balanced and unbalanced forces
)RUFHV RFFXU LQ SDLU 7KH IRUFHV FDQ EH HLWKHU EDODQFHG RU XQEDODQFHG
If the resultant magnitude of all the forces acting up on a body
is zero, the forces are called balanced forces. 7KH EDODQFHG IRUFHV
FDQQRW EULQJ D ERG\ LQWR PRWLRQ 7KH EDODQFHG IRUFHV DUH HTXDO
in magnitude but opposite in direction. Sometimes, the resultant
magnitude of more than two forces can be zero. For example, in tug
of war, players pull the rope in two different directions. If the total
IRUFHV RI ERWK WKH VLGHV DUH HTXDO WKH URSH GRHV QRW PRYH 7KLV LV DQ
example of balanced forces.

100N

100N

If the resultant magnitude of all the forces acting up on a bady is not
zero, the forces are called unbalanced forces. 7KH XQEDODQFHG IRUFHV
bring an object into motion. Kicking a football on the ground, pushing
a small box on the table, pushing a piece of stone forward, etc are
the examples of unbalanced forces. While pushing a small box on the
table, the force of friction tries to oppose the motion of box. However,
the force applied by you overcomes the frictional force and the box
moves on the table. Here, the force applied by you and the frictional
force are two different forces that act in opposite directions. As their
resultant magnitude is more than zero, the box moves forward.

Times' Crucial Science Book - 9 40

Learn and Write

1. It is easier to catch a tennis ball than a cricket ball. Why?

According to Newton’s second law of motion, F = ma Suppose,
v-u
t is same for both the tennis and cricket ball. It means the
tennis and cricket ball have equal velocity u while catching
and they are brought to rest (v = 0) in equal time (t). 7KHQ WKH
force applied is directly proportional to the mass. i.e. ) Ĵ P
7KXV WKH FULFNHW EDOO QHHGV PRUH IRUFH WKDQ WKH WHQQLV EDOO WR
bring to the rest due to more mass of the cricket ball.

2. An object can be accelerated without change in its speed. How?
When an object is moving in a circular path with a constant
speed, its direction keeps on changing and hence velocity too.
'XH WR FKDQJH LQ YHORFLW\ WKH ERG\ KDV DFFHOHUDWLRQ 7KXV D
body with constant speed possesses acceleration.

3. A boat is pushed backward when a man jumps out from the
boat. Why?
When a man jumps out from a boat as an action, the force
is applied by the man to the boat as reaction. As a result of
reaction, the boat moves backwards.

4. Action and reaction are equal and opposite but they do not
cancel each other. Why?
Action and reaction are equal and opposite but they do not act
on a same object. Action acts on one object whereas reaction acts
on another object. Due to this, they do not cancel each other.

5. What is the relationship between mass and inertia of a body?
7KH PDVV RI D ERG\ LV WKH PHDVXUH RI LWV LQHUWLD 7KH JUHDWHU

the mass of a body, the greater is its inertia. It means that
inertia of a body is directly proportional to its mass.

7KH DFFHOHUDWLRQ RI D FDU LV P V2. What does it mean?
Acceleration means the rate change of velocity of a moving
ERG\ ,I WKH DFFHOHUDWLRQ RI D FDU LV P V2, it means that the
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7. How does Newton’s first law of motion define force?
According to the Newton’s first law of motion, a body at rest
remains at rest and a body in motion continues to be in uniform

41 Times' Crucial Science Book - 9

motion in a straight line unless an external force acts upon it.
7KH ERG\ DW UHVW GRHV QRW PRYH LI WKH IRUFH LV QRW VXIILFLHQW
for moving it. It means that force is an external agency which
changes or tries to change the state of rest or uniform motion
of a body in a straight line.
8. A gun recoils when a bullet is fired from it. Why?

When a bullet is fired from a gun, the gun exerts some force
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move forward. According to Newton’s third law of motion, the
bullet also exerts an equal and opposite force on the gun as
reaction. Due to this reaction force, the gun recoils.

Glossary

Cover travel
Magnitude amount, quantity or relative size
Retardation decreasing acceleration
Proportional having a constant ratio
Jerk pull or move with a sudden movement
Displacement change in position

Main points to remember

1. Force is an external agency which changes or tends to change
state of rest or of motion of a body in a particular direction.

2. Inertia is the property of a body due to which it remains or tends
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or tries to remain in the state of rest until the external force is
applied.

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remains or tries to remain in the state of motion in the same
direction until the external force is applied.

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FDOOHG YHFWRU TXDQWLWLHV

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FDOOHG VFDODU TXDQWLWLHV

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Times' Crucial Science Book - 9 42

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YHORFLW\

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of rest or uniform motion in a straight line unless it is acted by
some external forces.

12. The second law of motion states that acceleration produced on a
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proportional to its mass.

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and opposite reaction”.

Exercise
1. Choose the best alternative in each case.

a. Which of the following is a vector quantity?

i. Force ii. Velocity iii. Acceleration iv. All of these

b. What is the SI unit of retardation?

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c. If a body starts moving from rest, its initial velocity is

L X LL Y LLL ]HUR LY P V

d. If the velocity time graph of a moving body is a straight
line parallel to the time axis, the velocity of the body is

i. Uniform ii. Increasing iii. Decreasing iv. Zero

e. Which of the following is defined by Newton's first law of motion?

i. Motion ii. Velocity iii. Force iv. Inertia

2. Answer these questions in brief.

a. What is force? What are its effects?
b. What is inertia? What are the factors upon which the

inertia depends?
c. What is inertia of rest? Give some examples of inertia of rest.
d. What is inertia of motion? Give some examples of inertia

of motion.
e. What are vector quantities? Give some examples.
f. What are scalar quantities? Give some examples.

43 Times' Crucial Science Book - 9

'LVWLQJXLVK EHWZHHQ

i. Vectors and scalars

ii. Distance and displacement

iii. Speed and velocity

iv. Uniform motion and non-uniform motion

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1 v2 = u2 + 2as
i. s = ut + 2 at2 ii.
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i. Newton’s first law of motion

ii. Newton’s third law of motion

6. State Newton’s second law of motion. Prove F = ma.

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i. A swimmer pushes water backwards.
ii. An athlete runs before taking a long jump.
iii. We beat blanket with a stick to remove dust particles.
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v. A cricket player lowers his hands while catching the ball.
vi. Passengers of a bus jerk forward when the moving bus

suddenly stops.
vii. When we jump on a concrete surface, the feet get injured.
viii. A gun recoils when a bullet is fired from it.
8. Numerical problems

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8 seconds. Find the acceleration produced in the body.

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seconds when brakes are applied. Find the retardation of
the car and distance travelled by the car.

F $ ERG\ PRYLQJ ZLWK DQ LQLWLDO YHORFLW\ RI P V KDV DQ
DFFHOHUDWLRQ RI P V2. Find the velocity of the body after 4s
and distance travelled in that time.

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travelled and amount of force applied.

Times' Crucial Science Book - 9 44

H $ EXV VWDUWV IURP UHVW DQG DWWDLQV D YHORFLW\ RI P V
after 10 seconds. Find the acceleration and the distance
travelled in 5 seconds and in 10 seconds.

I $ YHKLFOH ZDV PRYLQJ ZLWK D VSHHG RI NP KU 2Q VHHLQJ
a baby 40m ahead on the road, the driver jammed on the
brakes and it came to rest in 5 seconds. Whether accident
occurred or not?

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a baby on the road 65 m a head, the driver jammed the
brakes and the bus stopped with uniform retardation of
P V2. Calculate the distance travelled by the bus and the
time to stop.

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hr is brought to rest over a distance of 40 m. Find the
retardation and force applied on brakes.

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sees a child and brings the vehicle to rest in 5 seconds.
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of the vehicle is 500 kg and the mass of the driver is 70 kg)

Answers D P V2 E P V2 P F P V P

G P V2 P 1 H P V2, 56.25 m, 225 m

f. Distance travelled 50 m, accident occurred g. 4s, 40 m

K P V2, 1406.25 N i. 1710 N

Project Work

Use kinetic trolley and spring balance to test and verify newton’s 2nd law of
motion. Perform this experiment taking help from your teacher.

45 Times' Crucial Science Book - 9

Chapter

3 6LPSOH Archimedes
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Objectives

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• introduce simple machine and explain its types,
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Mind Openers

• Do you know what simple machine means?
• How do simple machines make work easier?
• How many types of simple machine do you know? What are they? Discuss.

Introduction
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useful to us in many ways. For instance, we use a pulley to draw
water from well, we use a nail-cutter to trim our nails, we use a
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they make our work easier and faster. Due to their simple structure,
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in structure and make our work easier and faster are called simple
machines. 1DLO FXWWHU EURRP VFLVVRUV ÀUHWRQJV EHDP EDODQFH
crowbar, wheel barrow, etc are some examples of simple machines.
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are formed by the combination of large number of simple machines.
Such machines are called complex machines. For example, sewing
machine, motor engine, water mill, windmill, etc.
Advantages of simple machines
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Times' Crucial Science Book - 9 46