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iFree distribution by A.P. Government
Sri M. Ramabrahmam, Lecturer,Govt. IASE, Masabtank, Hyderabad.
Dr. P. Shnakar, Lecturer,DIET Hanamakonda, Warangal.
Prof. Kamal Mahendroo,Vidya Bhavan Educational Resource Centre,
Udaipur, Rajastan.
Dr. TVS Ramesh,Co-ordinator, C&T Dept.,SCERT, AP, Hyderabad.
Miss. Preeti Misra,Vidya Bhavan Educational Resource Centre,
Udaipur, Rajastan.
Mr Kishore Darak,Vidya Bhavan Educational Resource Centre,
Udaipur, Rajastan.
Co-ordinators
Academic Support
Editors
Published by Government of Andhra Pradesh, Hyderbad.
Respect the LawGet the Rights
Grow by EducationBehave Humbly
PHYSICAL SCIENCEPHYSICAL SCIENCEPHYSICAL SCIENCEPHYSICAL SCIENCEPHYSICAL SCIENCE
CLASS IXCLASS IXCLASS IXCLASS IXCLASS IX
Dr.B. Krishnarajulu Naidu,Retd., Professor of Physics
Osmania University, Hyderabad.
Dr.M. Adinarayana,Retd., Professor of Chemistry
Osmania University, Hyderabad.
Dr. N. Upendar Reddy,Professor & Head C&T Dept.,
SCERT., A.P., Hyderabad.
Prof. V. SudhakarDept of Education, EFLU, Hyderabad.
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ii
© Government of Andhra Pradesh, Hyderabad.
New Edition
First Published 2013
All rights reserved.
No part of this publication may be reproduced, storedin a retrieval system, or transmitted, in any form or byany means without the prior permission in writing of thepublisher, nor be otherwise circulated in any form ofbinding or cover other than that in which it is publishedand without a similar condition including this conditionbeing imposed on the subsequent purchaser.
The copy right holder of this book is the Directorof School Education, Hyderabad, Andhra Pradesh.We have used some photographs which are undercreative common licence. They are acknowledge atthe end of the book.
This Book has been printed on 70 G.S.M. S.S. Map litho,Title Page 200 G.S.M. White Art Card
Free Distribution by Government of Andhra Pradesh
Printed in Indiaat the Andhra Pradesh Govt. Text Book Press,
Mint Compound, Hyderabad,Andhra Pradesh.
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iiiFree distribution by A.P. Government
Sri K. Sudhakara Chary, SGT,UPS Neelikurthy, Warangal.
Sri Kishan Thatoju, Computer Operator,C&T Dept.,SCERT, AP, Hyderabad.
Sri R. Ananda Kumar, SA,ZPHS Laxmipuram, Visakhapatnam.
Sri K.V.K. Srikanth, SA,GTWAHS S.L.Puram, Srikakulam.
Sri M. Eswara Rao, SA,GHS Sompeta, Srikakulam
Sri S. Naushad Ali, SA,ZPHS G.D. Nellore, Chittoor.
Dr. P. Shankar, Lecturer,DIET Hanamakonda, Warangal.
Dr.K. Suresh, SA,ZPHS Pasaragonda, Warangal.
Sri V. Gurunadha Rao, SA,ZPPSS Parvathagiri, Warangal.
Sri D. Madhusudhana Reddy, SA,ZPHS Munagala, Nalgonda.
Text Book Development Committee
Writers
Cover page, Graphics & Designing
Sri A. Satyanarayana Reddy, Director,S.C.E.R.T. , A.P., Hyderabad.
Sri B. Sudhakar, Director,Govt. Textbook printing press,
A.P., Hyderabad.Dr. N. Upendar Reddy,
Professor & Head C&T Dept.,S.C.E.R.T., A.P., Hyderabad.
Sri Kurra Suresh Babu, B.Tech, MA., MPhill.Mana Media Graphics, Hyderabad.
Sri M. Ramabrahmam, Lecturer,Govt. IASE, Masabtank, Hyderabad.
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iv
Intro ...
The nature is life source for all living organisms. Rocks, water, hills and
valleys, trees, animals etc. embedded in it… each of them are unique by
themselves. Everything has its own prominence. Human being is only a part of the
nature. The aspect which distinguishes the humans from all other organisms and
exclusive for them is their extraordinary thinking power. Thinking transforms a
person as a unique entity from rest of the nature. Though it usually appears
simple and normal, the intricacies of the very nature often challenges us to untie
the tough knots of its hidden secrets, day in and day out.
The human being intuitionally contemplates and searches solutions for all the
critical challenges, all around,relentlessly. Curiously, the questions and answers are
concealed in the nature itself. The role of science, in fact, is to find them out. For
this sake, some questions, some more thoughts, and some other investigations
are quite necessary. Scientific study is to move on systematically in different ways,
until discovering concrete solutions. Essence of the investigations lies in inquiring i.e.
identifying questions, asking them and deriving adequate and apt answers. That is
why, Galileo Galilei, the Italian astronomer,emphasized that scientific learning is
nothing but improving the ability of questioning.
The teaching of science has to encourage children to think and work scientifically.
Also, it must enhance their love towards the nature. Even it should enable them to
comprehend and appreciate the laws governing the nature in designing tremendous
diversity found around here and everywhere. Scientific learning is not just disclosing
new things. It is also essential to go ahead with deep understanding of the nature’s
intrinsic principles;without interrupting the harmony of interrelation and
interdependence in the nature.
It is also necessary to step forward without interrupting the interrelationshipand interdependency along with understanding of the nature’s intrinsic principles.HighSchool children possess cognitive capacity of comprehending the nature andcharacteristics of the transforming world surrounding them. And they are able toanalyze abstract concepts.
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vFree distribution by A.P. Government
At this level, we cannot quench their sharp thinking capability with the dryteaching of mere equations and theoretic principles. For that, we should create alearning environment in the classroom which provides an opportunity for them toapply the scientific knowledge, explore multiple alternatives in solving problems andestablish new relations.
Scientific learning is not just confined to the four walls of classroom. It has adefinite connection to lab and field as well. Therefore, there is a lot of importance
to field experience/ experiments in science teaching.
There is a great need for compulsory implementation of instructions of the
National Curriculum Framework- 2005 which emphasizes linking of the science teaching
with local environment. The Right to Education Act- 2009 also suggested that
priority should be given to the achievement of learning competencies among children.
Likewise, science teaching should be in such a way that it would help cultivate a
new generation with scientific thinking.The key aspect of science teaching is to
make the children understand the thinking process of scientists and their efforts
behind each and every discovery. The State Curriculum Framework- 2011 stated
that children should be able to express their own ideas and opinions on various
aspects.All the genuine concepts should culminate into efficacious science teaching,
make the teaching-learning interactions in the classroom, laboratory and field
veryeffective and really become useful for the children to face the life challenges
efficiently.
We thank the VidyaBhavan Society, Rajasthan, Dr. Desh Panday Rtd Prof.
College of Engineering Osmania University and Sri D.R. Varaprasad former Lecturer
ELTC Hyderabad for their cooperation in developing these new text books,the
writers for preparing the lessons, the editors for checking the textual matters and
the DTP group for cutely composing the text book.
Teachers play a pivotal role in children’s comprehensive use of the text book.
We hope, teachers will exert their consistent efforts in proper utilization of the text
book so as to inculcate scientific thinking process and inspire scientific approach in
the children.
Director,
SCERT, AP, Hyderabad
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vi
Dear teachers...New Science Text Books are prepared in such a way that they develop children’s observation
power and research enthusiasm. It is a primary duty of teachers to devise teaching- learningprocesses which arouse children’s natural interest of learning things. The official documents ofNational& State Curriculum Frameworks and Right to Education Act are aspiring to bring grassroot changes in science teaching. These textbooks are adopted in accordance with such anaspiration. Hence, science teachers need to adapt to the new approach in their teaching. In viewof this, let us observe certain Dos and Don’ts:
• Read the whole text book and analyze each and every concept in it in depth.
• In the text book, at the beginning and ending of an activity, a few questions are given.Teacher need to initiate discussion while dealing with them in the classroom, attempt toderive answers; irrespective of right or wrong responses, and so try to explain concept.
• Develop/Plan activities for children which help them to understand concepts presented intext.
• Textual concepts are presented in two ways: one as the classroom teaching and the otheras the laboratory performance.
• Lab activities are part and parcel of a lesson. So, teachers must make the children conductall such activities during the lesson itself, but not separately.
• Children have to be instructed to follow scientific steps while performing lab activities andrelevant reports can be prepared and displayed.
• In the text some special activities as boxed items- ‘think and discuss, let us do, conductinterview, prepare report, display in wall magazine, participate in Theatre Day, do fieldobservation, organize special days’ are presented. To perform all of them is compulsory.
• ‘Ask your teacher, collect information from library or internet’- such items must also beconsidered as compulsory.
• If any concept from any other subject got into this text, the concerned subject teacher hasto be invited into the classroom to elucidate it.
• Collect info of relevant website addresses and pass on to students so that they can utilizeinternet services for learning science.
• Let there be science magazines and science books in the school library.
• Motivate every student to go through each lesson before it is being actually taught andencourage everyone to understand and learn independently, with the help of activities suchas Mind Mapping and exciting discussions.
• Plan and execute activities like science club, elocution, drawing, writing poetry on science,making models etc.to develop positive attitude among children environment, biodiversity,ecological balance etc.
• As a part of continuous comprehensive evaluation, observe and record children’s learningabilities during various activities conducted in classroom, laboratory and field.
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viiFree distribution by A.P. Government
We believe, you must have realizedthat the learning of science and scientific thinking arenot mere drilling of the lessons but, in fact, a valuable exercise in motivating the children toexplore solutions to problems all around by themselves systematically and preparing them tomeet life challenges properly.
Dear Students...
Learning science does not mean scoring good marks in the subject. Competencies likethinking logically and working systematically, learned through it,have to be practiced in dailylife. To achieve this, instead of memorizing the scientific theories by rote, one must be able tostudy them analytically. That means, in order to understand the concepts of science, you needto proceed by discussing, describing, conducting experiments to verify, making observations,confirming with your own ideas and drawing conclusions. This text helps you to learn in thatway.
What you need to do to achieve such things:• Thoroughly go through each lesson before the teacher actually deals with it.• Note down the points you came across so that you can grasp the lesson better.• Think of the principles in the lesson. Identify the concepts you need to know further,
to understand the lesson in depth.• Do not hesitate to discuss analytically about the questions given under the sub-heading
‘Think and Discuss’ with your friends or teachers.• You may get some doubts while conducting an experiment or discussing about a lesson.
Express them freely and clearly.• Plan to implement experiment/lab periods together with teachers, to understand the
concepts clearly. While learning through the experiments you may come to knowmany more things.
• Find out alternatives based on your own thoughts.• Relate each lesson to daily life situations.• Observe how each lesson is helpful to conserve nature. Try to do so.• Work as a group during interviews and field trips. Preparing reports and displaying
them is a must.• List out the observations regarding each lesson to be carried through internet, school
library and laboratory.• Whether in note book or exams, write analytically,expressing your own opinions.• Read books related to your text book, as many as you can.• You organize yourself the Science Club programs in your school.• Observe problems faced by the people in your locality and find out what solutions you
can suggest through your science classroom.• Discuss the things you learned in your science class with farmers, artisans etc.
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viii
ACADEMIC STANDARDS
S.No. Academic Standard Explanation
1.
2.
3.
4.
5.
6.
7.
Conceptual understanding
Asking questions and makinghypothesis
Expermentation and fieldinvestigation.
Inforamtion skills and Projects
Communication throughdrawing, model making
Appriciation and aestheticsence, values
Application to daily life,concern to bio diversity.
Children are able to explain, cite examples, give reasons,and give comparison and differences, explain the processof given concepts in the textbook. Children are able todevelop their own brain mappings.
Children are able to ask questions to understand, toclarify the concepts and to participate in discussions.They are able to make hypothesis on given issues.
To understand given concepts in the textbook childrenare able to do experiments on their own. They are ableto participate in field investigation and making reportson them.
Children are able to collect information (by usinginterviews, internet etc.) and analyses systematically.They are able to conduct their own project works.
Children are able to explain their conceptualunderstanding by drawing figures and making models.Able to platting graphs by using given information orcollected data.
Children are able to appreciate man power and nature,and have aesthetic sense towards nature. They are alsoable to follow constitutional values.
Children are able to utilize scientific concept to facetheir daily life situations. They are able to show concerntowards bio diversity.
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ixFree distribution by A.P. Government
1010101010
66666
22222
33333
44444
55555
77777
88888
99999
11111 Matter aournd us
Motion
Laws of motion
Is matter pure
Atoms and molecules
What is inside atom
Gravitation
Floating bodies
Work and energy
Sound
INDEX
12 June 1
14 July 15
13 July 34
12 August 52
12 September 70
10 October 89
12 November 104
12 December 119
10 January 139
12 February 162
Page No.Periods Month
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x
OUR NATIONAL ANTHEM
- Rabindranath Tagore
Jana gana mana adhinayaka Jaya heBharatha bhagya-vidhata
Punjab Sindhu Gujaratha MarathaDravida Utkala Banga.
Vindhya Himachala Yamuna GangaUchchala Jaladhi taranga,
Tava shubha name jageTava shubha asisha mage
Gahe tava jaya gatha
Jana gana mangala-dayaka jaya he,Bharatha bhagya –vidhatha,
Jaya he, jaha he, jaya he,Jaya jaya jaya jaya he
PLEDGE
“India is my country; all Indians are my brothers and sisters.I love my country, and I am proud of its rich and varied heritage.
I shall always strive to be worthy of it.
I shall give my parents, teachers and all elders respect,and treat everyone with courtesy. I shall be kind to animals.
To my country and my people, I pledge my devotion.
In their well-being and prosperity alone lies my happiness.”
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You may have heard the phrases like‘what is the matter?' "The Matter wasclosed". Have you ever wondered what this‘matter’ is? Meaning of this term is verydifferent for scientists that of from acommon man.
We have read about metals, non-metals; synthetic and natural materials,acids and bases etc., in previous classes.These are all examples of ‘matter’.
All the things around us which exist ina variety of shapes, sizes and texture arealso examples of ‘matter’.
The water we drink is a matter.Similarly our food, clothes and variousthings that we use in our day to day life,the air we breathe, even our body etc., areexamples of matter.
In a simple way anything in this worldthat occupies space and has mass isconsidered as matter.
States of matterIn previous classes, you learnt that
water can exist either as a solid (ice), as aliquid or as a gas (water vapour).
Chapter
� MATTER AROUND US
We say that solids, liquids and gases arethree different states of matter. Water canbe found in all these states.
� Is there any substance which can befound in three states like water?
Now let us look carefully at differentobjects which we find around us. We canclassify most of them, quite easily into oneof the three states of matter.
For example, you can say that wood andcoal are solids and petrol is a liquid.
Milk also is a liquid like petrol. But theproperties of petrol and milk are quitedifferent from each other.
� What are the properties that lead us toconsider petrol or milk as liquids?
Let us do some activities to understandthe properties of solids, liquids and gases.
Properties of solids, liquids andgases
� Do solids have definite shape and fixedvolume?If we take two solid objects, say a pen
and a book, and put them in differentcontainers, do you find any change in their
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2 Matter Around Us
shape or volume? Imagine dropping a bookon the floor. It does not flow but remainsrigid with a definite shape, distinctboundaries and a fixed volume. This showsthat solids have a definite shape, a fixedvolume.
Activity-1Identifying the shape and volumeof liquids
For doing this activity, we need ameasuring jar (cylinder) and containers ofdifferent shapes as shown in figure 1.
Note: It is not compulsory to collectsame containers as shown in figure 1. Youcan collect the containers of differentshapes available to you.
You also need some liquids like wateroil and milk.
Take some water in one of the containerusing measuring jar. Examine shape ofwater in the container. Pour the same waterin another container and have a look at theshape, again. Repeat the process till youcomplete pouring of water in all containers.
� What is the shape of the water indifferent containers?
� Is it same in all cases?
� What shape does water take if it spillon the floor?
Take 50ml of water with the measuringjar and pour it in glass. Mark the level ofwater on the glass and throw it out.
Now measure 50 mL of the milk withmeasuring jar and pour it in the same glass.Mark the level of the milk on it.
� Are the levels of water and milk same?
Remove the milk from the glass. Nowpour oil into the glass up to the level markedfor water.
� Can you guess volume of oil?
This activity may seem very simple butwe observe two important properties ofliquids from this activity.
One, the shape of the liquid dependson the shape of the container and secondthough liquid takes different shapesdepending on the shape of the container itsvolume remains same. Liquids can floweasily from one container to another. Theyare also called fluids.
� Can you tell, what does a fluid mean?
Look up in a dictionary of science toget its meaning.
So liquids have no fixed shape but havea fixed volume.
Fig -1 Diffrent shapes of liquid of samevolume
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Activity-2
Do the gases have a definite shapeand a fixed volume ?
You know about CNG (CompressedNatural Gas). Go to a petrol pump and askthem where they store CNG. Also seewhere CNG is stored in a CNG run vehicle.Lastly see how CNG from the pump istransferred to vehicles.� Does CNG has a fixed volume?� Does CNG has a definite shape?
From the observations in the aboveactivity and with our daily life experiences,we can find that CNG and all other gasesneither have a fixed shape nor volume.
Compressibility
Activity-3
Observing the compressibility ofdifferent materials
Take a 100ml syringe.Draw the piston to suck in air. Place
your finger on the nozzle and press.Observe depth of piston moved intosyringe. Is it easy or hard to press?
� Do you find any change in volume ofair in the syringe?Now fill water in the syringe and do the
press the piston as said above.� Is it easier to press the syringe with
water or air.Now take a piece of wood and press it
with your thumb.Fig - 3 CNG gas filling station
Fig - 4 CNG tank at petrol pump
Fig - 2 CNG cylinder in a car
Fig - 5
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4 Matter Around Us
� What do you observe when you pressthe wood?
� Is their any change in its volume?From the above observations, you find
that gases are highly compressible ascompared to liquids and solids.
In our houses liquefied petroleum gas(LPG) is used for cooking. Nowadays CNGis used in many auotmobiles. In all thesepurposes, large volume of gas iscompressed into cylinders of smallervolume to make it portable.
Think and discuss
� Let us stretch a rubber band. Is there achange in its shape?
� Is it solid or liquid? Why?(What does happen if the stretching isstopped ? What does happen if thestretching is too much?)
Take some finely powdered salt (notcrystals) and keep it in two different jars.� Which shape does the powdered salt
take?� Can you say salt as a liquid based on
change in its shape?Justify your observations.Take a sponge. Observe its shape.
� Can you compress it? Is it a soild? Why?� Why do'nt you able compress a wooden
block?
DiffusionActivity-4
Observing the diffusion of gasesAsk your friend to hold an unlit incense
stick and stand in one corner of the room.
Then you go and stand in the other corner.
� Can you smell anything?Now ask your friend to light the
incense stick.� Can you smell anything now?
When you light the incense stick, thescent in the vapour form and smoke mixwith air and move across the room to reachour nose.
The movement of air, vapours of scentand smoke is known as diffusion. In thiscase, smoke, vapour of scent and air arehighly mobile and are in the form of gases.
If you spray a perfume or deodorant inone corner of the room, it spreads soon toall directions.� Does the smell from burning incense
stick and deodarant spray reachsomeone on the other end at the sametime?
Activity-5Observing the diffusion of liquids
Take two 250 ml beakers and fill themwith water. Use a dropper and put a drop ofblue or red ink or KMnO4 solution slowlyalong the sides of first beaker.
Fig - 6 Diffusion of potassiumpermanganate in water
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Diffusion of two gases
Aim: To observe the speed of diffusion oftwo gasses.Material required: Glass tube with scale,Ammonium Solution, Hydrochloric acid,pieces of cotton and rubber cork.
Note: Teacher should take care ofhandling hydrochloric acid and preventthe children from touching the acid.
Procedure: Take a 1 m long narrow glasstube.
Take two pieces of cotton wool; soakone in hydrochloric acid solution andanother in ammonia solution.
Put them separately at the two ends ofthe tube. Block the ends of the tube andobserve.
The hydrochloric acid gives offhydrogen chloride gas and ammoniasolution gives off ammonia gas.
Both gases react together to form awhite substance called ammoniumchloride.� Observe where the ammonium chloride
is formed in the tube.Explain.� How the two gases traveled along the
tube?� Which gas traveled faster?
Do this
� What do you observe after adding theink drop or KMnO4 drop?You can observe that liquids also diffuse
into each other like gases.� How much time does it take the colour
of ink to spread evenly throughoutwater?
� What do you conclude from thisactivity?
Activity-6Observing the diffusion ofparticles of solids into liquids
Take a beaker full of water and add asmall potassium permanganate crystal to itand observe the changes.
Repeat the experiment with a crystalof copper sulphate.� Do you observe diffusion?� Is it faster or slower than that observed
in other two activities?From activities 4, 5, and 6 it is clear
that solids and liquids diffuse into liquidsand gases diffuse in to gases.
Certain gases from atmosphereparticularly oxygen and carbon dioxide,diffuse and dissolve in water. and supportthe survival of aquatic animals and plants,
Diffusion therefore is a very importantprocess for living things.
Oxygen diffuses from lungs into blood.Carbon dioxide diffuses from blood intolungs.
Solids, liquids and gases diffuse intoliquids and rate of diffusion of gases ishigher than that of liquids or solids.
Fig - 7
Lab Activity
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6 Matter Around Us
So far you have studied someproperties that can be used to distinguishbetween solids, liquids and gases. Fill thetable given below based on your knowledge.
Property Solid Liquid GasShape fixedVolume fixed
Compressibilty
Diffusion
Can matter change its state?We started our discussion by recalling
that water exists in all the three states.You must have seen many other
materials that can exist in different states.For example, coconut oil is usually
liquid. But if it is too cold it becomes solid.Camphor is a solid but if we leave it in
the open air for some time it directlychanges to gas.
You may have seen moth (naphthalene)balls being placed in clothes. The smellremains even when they disappear. This isbecause the solid balls have changed fromsolid state to gaseous state.
Solids, liquids and gases are states ofmatter but you need to think about, why arethe properties of matter different indifferent states?� When does water change into ice and
then into vapour?� Why do gases diffuse faster than solids
or liquids?Scientists have tried to explain these
facts by examining the physical nature ofmatter.
What is matter made up of ?All matter is made of very tiny particles.
This looks as simple statement but it is verydifficult to explain and understand aboutmatter.
For this we need more details aboutthese particles and their arrangement insidevarious forms of matter.
Activity - 8
How small are the particles of amatter?
Take a beaker with water. Mark the levelof water. Add 1 or 2 crystals of potassiumpermanganate and dissolve them in water.
� What does happen to the colour ofwater?
Now take out approximately 10ml ofthis solution and add it to 90ml of clearwater in another beaker.
� What does happen to the colour ofwater?
Again take out 10ml of this solutionand add to another 90ml of clear water.Carryout this process 4, 5 times as shownin figure 8 and observe changes in colourof the solution.
Fig - 8
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Andhra Pradesh Government Free Distribution 7
� Is the water still coloured?� How is it possible for two small
crystals of potassium permanganate tocolour a large volume of water?
� What do understand from this activity?Repeat the activity by taking a few
crystals of copper sulphate instead ofpotassium permanganate.
Several interesting conclusions can bedrawn from the above activity.
We can conclude that there must beseveral tiny particles in just one crystal ofpotassium permanganate, which areuniformly distributed in water to changeits colour.
Similarly a few crystals of coppersulphate too has several tiny particles whichare distributed in large quantity of water togive colour.
So both solid and liquids (includingwater) are made up of tiny particles.� How do the particles of the solid
distribute in the liquid?Let us find
Activity - 9There exists space betweenparticles
Take a beaker and fill it with somewater and mark the water level.
Add some salt and stir it thoroughlywith a glass rod. Observe if there is anychange in water level. Add some more saltand stir it again.
Observe the change in level of water.
� Does the level of water change?� What did happen to the salt?� Can you see it in the water?
From the activities 8 & 9 we canconclude that both solid and liquid particleshave some space between them and thesolid particles enter in to the spacebetween the liquid particles on dissolvingsolid in liquid.
Recall the incense stick activity. Doyou agree that gas is also made up ofparticles and they have large space betweenthem.
Particles of matter attract eachother
Activity - 10
Observing the force of attractionbetween the particles of thematter
Open a water tap and allow the water toreach the ground. Now try to break thestream of water with your finger.� Can you break the stream permanently
or momentarily?� Are you able to break the stream of
water any where from the tap to ground?
Fig - 9
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8 Matter Around Us
Fig - 10
� What is the reason behind the streamof water remaining together?
� Now try to break a piece of iron nailwith your hands.
� Are you able to do it? or Does it rejoin?� What about a piece of chalk?
From the above observations we can saythat particles of the matter have forcesacting between them that keeps theparticles together.
It is also clear that this force is notequally strong in all the forms of matter.
How diffusion takes place?We have already carried out several
activities to explain diffusion of particlesof solids, liquids and gases. Diffusion canbe possible only when the particles ofmatter move continuously.
In the incense stick activity, theparticles responsible for scent move andenter the space between the air particles.The scent particles quickly spread acrossthe room.
Particles of solids, liquids and gasescan diffuse into liquids and gases. Rate ofdiffusion of gases is higher than the liquids,while the rate of diffusion of liquids ishigher than solids. There are two reasonsfor higher rate of diffusion of gases. Oneis due to the higher speed of gas particlesand another due to greater space betweenthem.
Similarly the greater diffusion rate inliquids compared to solids is becauseparticles in liquids move freely and have
greater space between them whencompared to particles of solids.
Observe following diagram whichshows the difference in arrangement ofparticles in solids, liquids and gases.
In a gas the particles are not as closetogether as in a liquid. If a coloured gas ismixed with a colourless gas, the colourspreads evenly in it. This happens faster ingas than in a liquid, because of large gapsbetween the partcicles of gas make andfewer particles obstruct in the way ofspreading.
You can see the diffusion of brominewhen it diffuses through air. Bromine is abrownish coloured gas. Hence its diffusionin colourless air can be seen clearly. If wealow Bromine to diffuse in vaccume, itdiffuses faster into vaccume, because thereare no particles to obstruct in its way.
Now you understand about matter andstates of matter, you also understand thatmatter is made up of particles. You alsounderstand that it is the characteristics ofthe particles that decides the propertiesof matter and states of matter. You may becurious to know:
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Do you know?
� What chages do occur inside the matterduring a change of state?
� How does this change of state takeplace?
� What does happen to the particles ofmatter during a change of state?Let’s try to understand the factors
effecting the change of state.Effect of temperature
Do you guess what happen if water isheated during the diffusion of liquids as inactivity 5. You can try it yourself. You findthat diffusion becomes faster.� How does this experience help you to
discuss about the effect of temperatureon change of states in matter?Let us find,
Activity-11Effect of temperature on changeof state
Take about 100g of ice in a beaker.Keep the bulb of a laboratory thermometerin contact with the ice. Set up the beaker asshown in the figure 11.
Note the temperature. Heat the beakerslowly. Record the change in temperatureafter every 30 seconds.
Fig - 11
Let the ice melt completely.Now place a glass rod in beaker and
continue heating till water starts boiling.After sometime all the water will bevaporized.�� Does the temperature change
continuously?� Does the temperature remain constant
for a while at any time during the periodof heating?
� At what temperature does this happen?� Did you notice the changes in
temperature, at various stages like juststart melting, during melting and atcompletion of melting?
Strange behaviour of waterA liquid usually expands when it is
heated but water behaves differently.Between 0oC to 4 oC, its volume shrinks.Same amount of water in solid iceoccupies more volume than liquid water.Thus density of ice is less than thedensity of same amount of water. Henceice floats on water rather than sinking.This is very important for survival ofmarine life which lives in ponds in thecolder areas. In extremely cold weatherthe water at the top become colder andcolder, until it freezes. While the icefloats on the top, the animals Continueto live in the water below, which does notfreeze and remains at 40C. The ice on thetop of the pond insulates water below itand it stops the water from losing the heatto air.
The melting point of ice is 0OCelcius.
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10 Matter Around Us
Do you know?
1 kg of a solid, completely into liquid atatmospheric pressure at its melting point.� What does happen if the heating of
water continued?On supplying heat energy to water
particles they start moving faster. At acertain temperature, the particles haveenough energy to become free from theforces of attraction among the particles andthe liquid starts changing into gas. Thetemperature at which a liquid starts boilingat the atmospheric pressure is known as itsboiling point. Boiling point of water is100OC.
Particles in water vapour at 100OC havemore energy than the particles in liquidwater at the same temperature. This isbecause particles in water vapour haveabsorbed additional energy in the form oflatent heat of vaporization.
So we can infer that one state of mattercan be changed into another state byobsorbing or radiating heat energy.
Now, we have learnt that substancesaround us change state from solid to liquidand from liquid to gas on supply of heat.
But there are some substances thatchange directly from solid state to gaseousstate and vice versa without passing throughthe liquid state. We have read aboutsublimation which is one such change.
We can use another unit called Kelvinto measure temperature. The meltingpoint of ice is 0OCelcius. 00C i.e. 273K.and Boiling point of water is 1000C (or)273+100=373K
The temperature of the mixture of solidice and liquid water does not change forsometime when it reaches melting point.The temperature starts changing only whenice completely melts.
On heating the water in the beaker, theparticles gain energy. Due to additional heatenergy, the particles move more freely byovercoming the force of attraction betweenthe particles.
At a certain temperature the solidmelts and changes into liquid. Thetemperature at which a solid melts tobecome a liquid is called ‘melting point’.The process of changing liquid to solid isknown as ‘fusion’.
The melting point of a substancedepends on the strength of the force ofattraction among the particles.
The higher the force of attractionamong the particles the higher will be themelting point.
In above activity, the heat energyabsorbed by ice without showing any risein temperature is used up in changing thestate by overcoming the forces of attractionamong the particles.
Particles in liquid water at 00C havemore energy as compared to particles inice at the same temperature. Because theparticles in water absorb heat energy duringthe process of coversion from ice to liquidwater.
The amount of heat energy that isrequired to overcome the force ofattraction among the particles is given bythe latent heat of the substance.
Latent heat of fusion is defined as theamount of heat energy required to change
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You have exerienced drying of wetclothes in air. In this process water directlychanges into vapour form without reachingits boiling point.� Can you give few more examples for
this type of change?� What could be the reasons for this type
of changes in states?The particles of matter irrespective of
the state at the given temperatures possessdifferent energies.
For example, in liquids the particles atthe surface posses higher energy thanparticles in the bulk of water. The particleson the surface are able to break away fromthe force of attraction of other particlesand change into vapour state.
This phenomena of change of a liquidinto vapours at any temperature below itsboiling point is called evaporation.
Activity-12
Effect of surface area, Humidityand wind speed on evaporation
Take 5ml of water in a test tube and in achina dish separately and keep them underthe fan. Take 5ml of water in another chinadish and keep it in the cupboard.
Record room temperatures and timetaken for evaporation of water in all threecases. If possible, repeat the activity on arainy day and record your observations.� In which case evoparation is fast?� What do you infer about the effect of
surface area and wind speed onevaporation?
Effects of change of pressureWe already know that various kinds of
states are due to the differences in thedistances between the constituent particles.� What will happen when we apply
pressure and compress a gas enclosedin a cylinder?
� Will the particles come closer?� Do you think that increasing or
decreasing the pressure can change thestate of matter?
��� Can we liquify gases by applyingpressure or reducing temperature?Recall that solid carbon dioxide (from
the chapter "Combustion, fuel and flame"of last year) is stored under high pressures.
Solid carbon dioxide converts directlyin to gaseous state when the pressure isdecreased to 1 atmosphere. Due to thisreason solid carbon dioxide is also knownas dry ice.
Thus we can say that pressure andtemperature determine the state of asubstance - whether it will be solid, liquidor gas.
Let us use our understanding of particlenature of matter to explain some day to dayobservations.
Evaporation� Do we always need to supply heat or
change the pressure for changing thestate of matter?
� Can the change of state from liquid tovapour take place without the liquidreaching its boiling point?
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12 Matter Around Us
Think and discuss
Experience with evaporation� Why do you feel cooler after sweating?
When you get sweat, on working orphysical excercise the sweat evaporatesfrom surface of your body by absorbing theheat from your body. The particles of liquidabsorb energy required for evaporationfrom your body and escape to thesurroundings. This make you to feel cool.� Can you give some more examples
from daily life where we can feel theeffect of evaporation?
� Why do we store water in matkas(earthern pots)?
Discuss these topics with your friends.
� Why do we wear cotton clothes insummer?
� Why do we see water droplets on outersurface of a glass containing ice - coldwater?
� Why do pigs toil in the mud in hot days?
You must have noticed that the rate ofevaporation in china dish is faster.
Since evaporation is a surfacephenomena, during evoparation process theparticles escape from the surface of liquid.The increase in the surface area providesmore scope for particles to escape fromthe surface. Hence, increases the rate ofevaporation.
Humidity is another factor that effectsevaporation. The amount of water presentin air is called humidity.
The air around us cannot hold more thana definite amount of water vapour at a giventemperature.
If the amount of water vapour is high inair the rate of evaporation will decrease.So clothes dry slowly during rainy seasonbut fast on a sunny and windy day.
Because of increase in wind speed,water vapour particles move away with thewind, decreasing the amount of watervapour in the surroundings.
Key words
Matter, states of matter, solid, liquid, gas, particles, diffusion, forces ofattraction, evaporation, compressed natural gas, melting point, fusion, boilingpoint, sublimation, evaporation, latent heat of fusion.
What we have learnt
� Matter is made up of particles� The particles of matter are very small-they are small beyond our imagination� Particles of matter have space between them
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Improve your learning
� Particles of matter move continuously in liquids and gases.� Matter exists in three states i.e., solid, liquid and gas� The force of attraction between the particles are maximum in solids, intermediate in
liquids and minimum in gases.� The particles are arranged orderly in the case of solids while particles move randomly.
in gasses.� Diffusion is possible only when particles of matter move continously.� Rate of diffusion of gases is higher than that of liquids (or) solids.� The states of matter are not permanent. The state of matter can be changed by changing
temperature or pressure.� Evoparation is a surface phenomenon because particles from the surface gain enough
energy to overcome the forces of attraction present in the liquid and change into thevapour state.
� Boiling is a bulk phenomenon because particles of the entire liquid change into vapourstate.
� Changes in temperature, surface area and wind speed effect rates of evaporation.� Humidity is the amount of water vapour present in air.
1. Describe an activity which provides the evidence for (A,S1)a) the motion of particlesb) attraction between particlesc) inter-particle space
2. Name the characteristics of matter that are demonstrated by diffusion.(A,S1)3. "When sugar is dissolved in water there is no increase in volume". Is it true or false?
Comment on the statement keeping in mind the amount of sugar, amount of wateretc.(A,S1)
4. Is there any change in mass when a substance changes its state? Explain withexample.(A,S1)
5. Do all substances change from solid to liquid and liquid to gas on heating? Explain.(A,S1)6. Define the following terms: (A,S1)
a) melting point b) boiling point c) evaporation7. Correct the following statements. (A,S1)
a) Water boils at 100oC under atmospheric pressure.
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14 Matter Around Us
b) a liquid evaporates above its boiling pointc) solids have the largest inter-particle space.d) gases have the strongest inter-particle forces.
8. Why do we prefer to sip hot tea with a saucer rather than a cup?(A,S1)9. When water solidifies to ice then heat is (A,S1)
a) Liberatedb) Absorbedc) No changed) Depending on the condition of heat absorbed or liberated.
10. Convert the following temperatures to the Celsius scale.(A,S1)(a) 283k (b) 570k
11. Convert the following temperatures to the Kelvin scale.(A,S1)(a) 270c (b) 3670c
12. How can we smell perfume sitting several metres away from the source?(A,S2)13. Steam produces more severe burns than boiling water. Think why?(A,S2)14. Fill in the blanks.(A,S1)
a) Temperature does not change while a solid substance is ……...................... or a liquidsubstance is…….
b) A vapour on cooling changes into …..............….and on further cooling changesinto…....................................................
c) Matter changes from one state to another either raising the …..........................…….or lowering the…...............................………
d) A change in which a solid on heating directly changes into vapour state is called……................….
e) The inter particle spaces are....................…..in gaseous and….....................…..in solids.15. Match the following.(A,S1)
a) conversion of liquid into gas ( ) (i) gasb) Non- compressible ( ) (ii) solidc) maximum expansion ( ) (iii) particled) constituents of matter ( ) (iv) evaporation
16. How do you appreciate sweating mechanism of human body to control the temperatureof the body?(A,S6)
17. Make a model to explain the structure of particles in solids, liquids and gases.(A,S5)
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Andhra Pradesh Government Free Distribution 15
We are familiar with the ideaof motion. We see several examples ofmotion around us like motion of people,vehicles, trains, aeroplanes, birds, raindrops, objects thrown into air, etc. Weknow that it is due to the motion of theearth that the phenomena like sunrise,sunset, changes in the seasons etc occur.
� If earth is in motion why don’t wedirectly perceive the motion of theearth?
� Are the walls of your class room atrest or in motion? Why?
� Have you ever experienced that thetrain in which you sit appears to movewhen it is at rest? Why?
To give answers to these questions weneed to understand the terms ‘relative’ and'motion'.
Great progress in understandingmotion occurred when Galileo under tookhis study about rolling balls on inclinedplanes. To understand motion, we need tounderstand the meaning of the word'relative', which plays an important role inexplaining various ways of motion.
What is relative ?We use many statements in our daily
life to express our views. The meaning of astatement depends on the relation betweenthe words used in it.
Does every statement have a meaning?
Evidently the answer is ‘no’. Even ifyou choose perfectly sensible words andput them together according to all the rulesof grammar, you may still get completenon- sense. For instance the statement“This water is triangular “can hardly begiven any meaning.
A statement has a meaning only whenthere is a relation between words.
Similarly there may exist othersituations in our daily life where we usestatements having meaning depending uponthe situation. Let us observe the followingexample.
Right and Left
As shown in the figure 1, two personssay A and B are moving opposite to eachother on a road.
Chapter
�A C E O F D B
MOTION
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16 Motion
Fig-1Examine the meaning of the following
sentence.Question: On which side of the road is
the house? Is it on the right side or on theleft side to the road?
There are two answers for the abovequestion. For person A, the house is on theright and for the person B, the house is onthe left. Thus the position of the house isrelative to the observer i.e., clearly whenspeaking of left and right by a person, hehas to assume a direction based on whichhe can decide the left and right sides ofhimself.Is day or night just now ?
The answer depends on where thequestion is being asked. When it is daytimein Hyderabad, it is night in New- York. Thesimple fact is that day and night are relativenotions and our question cannot beanswered without indicating the point onthe globe where the question is being asked.Up and down
Can the orientations like up and downbe the same for all persons at all places?Observe the following figure 2.
For the person standing at A on theglobe, his position appears up and theorientation of person standing at B appearsdown but for the person standing at B it
appears exactly opposite. Similarly for thepersons standing at the points C and D thedirections of up and down are not same.They change with the point of observationon the globe.� Why do we observe these changes?
Fig-2We know that earth is a sphere, the
upward direction of the vertical positionon its surface decisively depends up on theplace on the earth’s surface, where thevertical is drawn.
Hence the notions “up and down” haveno meaning unless the point on the Earth’ssurface, to which they refer, is defined.
Discuss the meaning of the terms“longer and shorter” with few examples.� Are these terms relative or not?
Motion is relativeLike the terms right and left, up and
down, larger and shorter etc., ‘motion’ isalso relative to the observer. Let us examinethis.
AB
A
B
DC
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To understand the idea of motion, let us take the following hypothetical activity.Observe the figure 3 and follow the conversation between Srinu and Somesh who stand
beside a road as shown in the figure3.
Srinu : What is the state ofmotion of the tree?
Somesh : It is at rest.Srinu : What is the state of
motion of the car?Somesh : It is moving due east.Srinu : What is the state of
motion of the driver andthe passenger in the car?
Somesh : They are also moving like
the car.Srinu : How do you decide that
the car, the passenger andthe driver are moving?
Somesh : With respect to us, theposition of the car,thepassenger and the driverare changing with time.So, they are in motion.
Fig-4: Passenger’s point of view
West
Passenger
Driver
Fig-3: Somesh’s point of view
Somesh
Srinu
East
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18 Motion
Now follow the conversation of thedriver and the passenger in moving car.
Driver : What is the state ofmotion of the tree?
Passenger : It is moving due west
Driver : What is the state ofmotions of both thepersons beside the road?
Passenger : They are also moving duewest.
Driver : What is my state ofmotion?
Passenger : You are at rest.
Driver : What is the state ofmotion of the car?
� What answers should the passenger giveto the driver? Discuss with yourfriends.
From the above discussion, it is clearthat the tree is at rest with respect toSomesh and it is moving due west withrespect to passenger.
The motion of an object depends on theobserver. So motion is a combined propertyof the observer and the body which is beingobserved.
Now we are able to define motion ofan object.
A body is said to be in motion when itsposition is changing continuously with timerelative to an observer.
Note: Any object can be taken as a pointof observation.
���How do we understand motion?
Distance and displacement
Activity-1
Drawing path and distinguishingbetween distance anddisplacement
Take a ball and throw it into the air withsome angle to the horizontal. Observe itspath and draw it on paper.
Figure 5 shows the path taken by theball when it was thrown into air. “Distance”is the length of the path traversed by anobject in a given time interval anddisplacement is the shortest distancecovered by the object in a specifieddirection.
Fig-5:Distance - displacement
Observe the difference betweendistance and displacement from figure 5.
So, Displacement is a vector. Todescribe a physical situation, somequantities are specified with magnitude aswell as direction. Such a physical quantityis called a vector. The physical quantitywhich do not require any direction for itsexplanation is called scalar. So distance isa scalar.
S
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Think and discuss
A vector can be represented as adirected line segment. It’s lengthindicates magnitude and arrow indicatesit’s direction. Point ‘A’ is called tail andpoint ‘B’ is called head
In the above example (Fig-5)ASB shows the actual distance coveredby an object and AB is a displacementwhich is a straight line drawn frominitial position to final position of themotion.
The SI unit of distance or displacementis metre denoted by ‘m’. Other units likekilometre, centimtre etc. are also used toexpress this quantity.
1 km = 1000 m1 m = 100 cm
Activity-2
Drawing displacement vectorsA car moves along different paths as
shown in figures 6(a) and 6(b). The pointsA and B are the initial and final positionsof the car.
Draw displacement vectors for twosituations.
Fig-6(a) Fig-6(b)
Generally the distance covered anddisplacement are time dependentquantities.
� What is the displacement of the bodyif returns to same point where it isstarted? Give one example in dailylife.
� When do the distance and magnitudeof displacement become equal?
Average speed and averagevelocity
A train named Godavari express startsat 5.00 pm from Visakhapatnam andreaches Hyderabad at 5.00 am the next dayas shown in figure 7.
Fig-7Draw displacement vectors from
Visakhapatnam to Vijayawada, Vijayawadato Hyderabad and from Visakhapatnam toHyderabad.
Let the distance of the entire trip fromVisakhapatnam to Hyderabad be 720 km.
The journey time is 12 h. What is thedistance covered by the train in each hour?
It is equal to 720km/12h = 60km/h
A
B
Visakhapatnam
Vijayawada
Hyderabad
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20 Motion
Think and discuss
Can you say that the train has coveredexactly 60 km in each hour?
Obviously the answer is “No”, becausethere may be some variations in distancecovered by train for each hour. So we takeaverage of distances covered by the trainfor each hour to decide its average speed.The distance covered by an object in unittime is called average speed.
Let the displacement of the trip inabove example be 360 km due North–West.What is the displacement in each hour?
The displacement per hour = 360 km/ 12h north - west = 30km / h north - westThe displacement of an object per unit
time is called average velocity. AverageVelocity is a vector and is along thedirection of displacement.
The quantities average speed andaverage velocity explain the motion of abody in a given time interval. They do notgive any information about the motion ofthe body at a particular instant of time.
� What is the average speed of the carif it covers 200 km in 5 h?
� When does the average velocitybecome zero?
� A man used his car. The initial andfinal odometer readings are 4849 and5549 respectively. The journey timeis 70h. What his average speed duringthe journey?
Speed and velocityThings in motion often have variations
in their speeds. For example a car maytravel along a street at 50 km/h get slowdown to 0km/h at red light and then attain aspeed of 30 km/h due to traffic on the road.� Are you able to find the pead of a car at
a particular instant of time?You can tell the speed of the car at any
instant by looking at its speedometer. Thespeed at any instant is called instantaneousspeed.
Let us suppose that a body is movingrectilinearly (along a straight line) but withvarying speed.
How can we calculate the instantaneousspeed of the body at a certain point “O” ofits path?
Fig-8
We take a small segment AB on thispath, which contains the point O as shownfigure 8. We denote this small distancecovered by the body as S1, and the smalltime interval over which it covers thatdistance is taken as t1. We obtain theaverage speed over this segment by dividingS1 to t1. The speed varies continuously andhence it is different at different points ofsegment AB.
Total distance Time takenAverage speed =
Displacement Time takenAverage velocity = A C E O F D B
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Let us now reduce the length ofsegment AB. We choose a segment CDwhich also includes point O, the ratio S2/t2
gives us average speed for this smallersegment CD. If we take a segment EF whichalso includes point O and is smaller thansegments AB and CD, the change in speedover this segment will be still smaller.Dividing the distance S3 by the time intervalt3, we again obtain the average speed on thissmall part of the path.
If we reduce the time interval graduallyover which the distance travelled by thebody is being considered, then the distancecovered decreases continuously andultimately the segment of path having thispoint O contracts closer and closer to reachthe reference point O. Thus average speedapproaches a certain value. This value iscalled the instantaneous speed at point O.
“The instantaneous speed at a givenpoint is equal to the ratio betweensufficiently small distances over a part ofthe path which contains the reference pointand the corresponding small time interval.”
Let us understand this in a graphicalway.
Consider a car moving along a straightroad with varying speed.
A more useful way of describingmotion of any body along a straight line isdistance – vs – time graph.
Along the horizontal axis we plot thetime elapsed in seconds, and along thevertical axis the distance covered in metres.
A general case of motion with varyingspeed is shown in figure 9.
Fig-9:Distance vs time graph
� What is the speed of the car at theinstant of time ‘ t3’ for given motion? We know how to find average speed
during the time interval from t1 to t2, whichincludes the instant t3 is
Then we calculate average speed for avery short time interval encompassing thetime at an instant t3– which is so shortinterval, that value of average speed wouldnot change materially if it was made evenshorter.The instantaneous speed isrepresented by the slope of the curve at agiven instant of time. We can find slope ofthe curve at any point on it by drawing atangent to the curve at that point. The slopeof the curve gives speed of the car at thatinstant. If the slope is large, speed is highand if the slope is small, speed is low.
Speed gives the idea of how fast thebody moves. In general, bodies move in aparticular direction at an instant of interestand this direction may not be constantthroughout the journey. So we need todefine another quantity called “Velocity”.
time
Dist
ance
t1 t3 t2
s2
s1
s3
S2–S1
t2–t1
Average speed =
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22 Motion
� In what direction does it move?Try to release the object at different
points on the circle and observe thedirection of motion of object after it hasbeen released from the string.
You will notice that the object moveson a straight-line along the tangent to thecircle at the point where you released it.The direction of velocity is tangent to thepath at a point of interest.
The SI unit of velocity is metre/sec.In our daily life we must have observed
many motions where, in some cases thevelocity of object which is in motion isconstant but in other cases it continuouslychanges.� Which motion is called uniform? Why?
Let us find out.
Uniform motionActivity-4
Understanding uniform motionConsider a cyclist moving on a straight
road. The distance covered by him withrespect to time is given in the followingtable. Draw distance vs time graph for thegiven values in the table1.
Table -1Time Distance
(t in seconds) (s in metres)0 01 42 83 124 16-- --
Velocity is the speed of an object in aspecified direction.
Fig-10Velocity gives the idea of how fast the
body moves in specified direction. Velocityis a vector. It can be represented by adirected line segment. Its length indicatesspeed and arrow gives the direction ofmotion.
If a body moves in a curved path, thetangent drawn at a point on the curve givesdirection of velocity at that instant.Observe that following diagram and try todraw tangents to the curve at differentpoints. Does the direction of velocity ofbody remain constant or not?
Activity-3Observing the direction ofmotion of a body
Carefully whirl a small object on theend of the string in the horizontal plane.Release the object while it is whirling onthe string.
Fig-11:Direction of velocity at apoint of path
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Think and discuss
� What is the shape of the graph?You will get a graph which resembles
the graph shown in fig-12.
Fig-12The straight line graph shows that the
cyclist covers equal distances in equalintervals of time. From the graph you canunderstand that the instantaneous speed isequal to average speed. If the direction ofmotion of the cyclist is assumed asconstant then we conclude that velocity isconstant.
The motion of the body is said to bein uniform when its velocity is constant.
� Very often you must have seen trafficpolice stopping motorists andscooter drivers who drive fast andfine them. Does fine for speedingdepend on average speed orinstantaneous speed? Explain.
� One airplane travels due north at 300km/h and another airplane travels duesouth at 300 km/h Are their speedsthe same? Are their velocities thesame? Explain.
� The speedometer of the car indicatesa constant reading. Is the car inuniform motion? Explain.
Example-1A man standing under a street lamp of
height ‘H’ above the ground starts runningwith a constant speed ‘v’ in a constantdirection. The light from the lamp fallingon the man form a shadow of him. Find thevelocity with which the edge of the shadowof the man’s head moves over the ground ifhis height is “h”.Solution
Here we need to compare the motionsof the man and the edge of the shadow ofthe man’s head. To solve this problem, acommon starting point is necessary. Letthis point be ‘O’ which is taken exactlyunder the lamp as shown in the fig-13.
Let ‘s’ and ‘S’ be the distances coveredin a time interval ‘t’ from the point ‘O’ bythe man and the edge of the shadow of theman’s head respectively.
Fig-13
As shown in figure 13 we get twosimilar triangles namely �ABD and � ACO
From these two similar triangles
But we know,
s
t
Ss
h
H-h
O
A
D B
C
S H =
s H – h
s vt=
OC AO =
DBAD
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24 Motion
We know that S/t is the speed of the
shadow of the man’s head,
Then,
Non uniform motionIn our daily life in many situations
when a body is in motion, its velocitychanges with time. Let us observe thefollowing example.
Consider a cyclist moving on a straightroad. The distance covered by him withrespect to time is given in the followingtable. Draw distance vs time graph for thegiven values in the table2.
Table-2
Time Distance(t in seconds) (s in metres)
0 01 12 43 94 16-- --
� What is the shape of the graph?� Is it a straight line or not? Why?
Activity-5
Observing the motion of a ballon a inclined plane
Fig-14:Ball moving downthe inclined plane
Set up an inclined plane as shown infigure 14. Take a ball and release it fromthe top of the inclined plane. The positionsof the ball at various times are shown infigure14.� What is the path of the ball on the
inclined plane?� How does the velocity of the ball
change?On close observation we find that when
the ball moves down on the inclined planeits speed increases gradually. And thedirection of motion remains constant oninclined plane.
Set up an inclined plane as shown infigure 15. Take a ball and push it with aspeed from the bottom of the inclined planeso that it moves up.
t=0 t=1st=2s
t=3s
t=3s t=2st=1s
t=0s
S H =
vt H – h
Hv H–hV =
Fig-15:Ball moving upthe inclined plane
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Andhra Pradesh Government Free Distribution 25
Think and discuss
� What is the path of the ball?� What happens to its speed?
In above two situations of activity-5,we observe that the speed changes but thedirection of motion remains constant.
Activity-6
Observing uniform circularmotion
Whirl a stone which is tied to the endof the string continuously. Draw its path ofmotion and velocity vectors at differentpositions as shown in the figure 16.Assume that the speed of the stone isconstant.
� What is the path of the stone?It is clear that the path is a circle and
the direction of velocity changes at everyinstant of time but the speed is constant.
Hence in this activity we observe thatthough speed remains constant, its velocitychanges.� Can you give few example for motion
of an object where its speed remainsconstant but velocity changes?
Activity-7
Observing the motion of anobject thrown into air
Throw a stone into the air by makingsome angle with the horizontal. Observethe path taken by it. Draw a diagram toshow its path and velocity vectors.� Is the speed of the stone uniform?
Why?� Is the direction of motion constant?
How?In the above activity you might have
noticed that the speed and direction ofmotion both change continuously.� Can you give some more examples
where speed and directionsimultaneously change?From the above three activities you can
conclude that the change in velocity takesplace in three ways.1. Speed changes with direction remaining
constant.2. Direction of motion changes with
speed remaining constant.3. Both direction and speed change
simultaneously.Motion of an object is said to be non-
uniform when its velocity is changing.
� An ant is moving on the surface of aball. Does it’s velocity change or not?Explain.
� Give an example of motion wherethere is a change only in speed but nochange in direction of motion.
v
v
v
v
v
v
v
v
Fig-16
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26 Motion
AccelerationWe can change the velocity of an object
by changing its speed or its direction ofmotion or both. In either case the body issaid to be accelerated. Acceleration givesan idea how quickly velocity of a body ischanging.
� What is acceleration? How can weknow that a body is in acceleration?We experience acceleration many
times in our day to day activities. Forexample, if we are travelling in a bus or acar, when the driver presses the accelerator,the passengers sitting in the bus experienceacceleration. Our bodies press against theseats due to the acceleration.
Suppose we are driving a car. Let ussteadily increase the velocity from30 km/h to 35 km/h in 1sec and then35km/h to 40km/h in the next second andso on.
In the above case the velocity of car isincresing 5 km/hr per second.
This type of rate of change of velocityof an object is called acceleration.
Acceleration is uniform when in equalintervals of time, equal changes of velocityoccur.
Uniform acceleration is the ratio ofchange in velocity to time taken.
The term acceleration not only appliesto increasing velocity but also todecreasing velocity. For example when weapply brakes to a car in motion, its velocitydecreases continuously. We call this asdeceleration. We can observe thedeceleration of a stone thrown up vertically
into air and similarly we can experiencedeceleration when a train comes to rest.
Let us suppose that we are moving in acurved path in a bus. We experienceacceleration that pushes us towards theouter part of the curve.
Observe the following fig-17. Themotion of an object in a curved path atdifferent instants is shown as a motiondiagram. The length of the vector at aparticular point corresponds to themagnitude of velocity (speed) at that pointand arrow indicates direction of motion atevery instant.
� At which point is the speed maximum?
� Does the object in motion possessacceleration or not?
We distinguish speed and velocity forthis reason and define “acceleration” as therate at which velocity changes, there byencompassing changes both in speed anddirection.
Acceleration is also a vector and isdirected along the direction of change invelocity.
The SI unit of acceleration is m/s2
A
B C
Fig-17: Motion diagram
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Andhra Pradesh Government Free Distribution 27
Think and discuss
� What is the acceleration of the race carthat moves at constant velocity of 300km/h ?
� Which has the greater acceleration, anairplane, that goes from 1000 km/h to1005 km/h in 10s or a skateboard thatgoes from zero to 5km/h in 1 second?
� What is the deceleration of a vehiclemoving in a straight line that changesits velocity from 100 km/h to a deadstop in 10s?
� Correct your friend who says “Theacceleration gives the idea of how fastthe position changes.”
Equations of uniformaccelerated motion
Consider the motion of object alongstraight line with constant acceleration.
Then,
Let u be the velocity at the time t = 0and v be the velocity at the time t and let sbe the displacement covered by the bodyduring time “ t” shown in figure 18.
From the definition of uniformacceleration,
Acceleration,
Since the acceleration of the body isconstant.
But we know
From here onwards we aremanipulating the equations (1) and (2).
Put in equation (2), we have
ut + ½ a t2 s ................. (3)
From equation , we get
Substitute the value of t in equation(2) , we haveu
s
vat 't'at t= 0s
v – u t
a =
Change in velocity Time taken
Acceleration =
�v�t
constant�a �
Fig-18
v – uat �v ................. (1)u+at �
v + u 2
Average velocity =
Displacement Time takenAverage velocity =
v + u 2
=st
................. (2)
v u+at�
u + at + u2
=st
2u + at2
=st
=
v u+at�v – u a
t =
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28 Motion
The equations of motion are,
NOTE:
1. If the speed of an object increases, thedirection of velocity and accelerationare one and the same.
2. If the speed of the object decreases, thedirection of velocity and accelerationare in opposite directions. In such acase, at a certain instant speed becomeszero.
3. If there exists an acceleration of a bodyat a point where its speed becomes zerofor an instant; then the body 'returns' inthe direction of acceleration and movescontinuously. (like in the case of stonethrown vertically up into air.)
NOTE:Care must be taken toremember the following points while weare using equations of motion.
� Choose origin on a straight line. Thequantities directed due right are takenas positive and due left are taken asnegative.
� Expressing the displacement withproper sign, is important. Displacementis positive while measured along thepositive direction and is negative whilemeasured along negative direction.
Aim� To find the acceleration and velocity of
an object moving on an inclined track.� To draw the graph between distance and
time.Materials required
Glass marbles, identical books, digitalclock, long plastic tubes and steel plate.Procedure
Take a long plastic tube of length nearly200cm and cut it in half along the length ofthe tube. Use these tube parts as tracks.Mark the readings in cm along the track.Place the one end of the tube on the bookor books and the other end on the floor.
Keep a steel plate on the floor at thebottom of the track. Consider the readingat the bottom of the track as zero.
Take a marble having enough size totravel in the track freely. Now releasemarble freely from a certain distance say40cm. Start the digital clock when the
– O +
Lab Activity
v – u a
sv + u 2 =( (( (
v2 – u2 2as................. (4)=
v u+at�s ut + ½ a t2�
v2 – u2 2as=
Fig-19
Fig-20Books
Track
Steel plate
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Andhra Pradesh Government Free Distribution 29
marble is released. It moves down on thetrack and strikes the steel plate. Stop thedigital clock when sound is produced.Repeat the same experiment for the samedistance 2 to 3 times and note the valuesof times in the below table-3.
Table-3
Repeat the same experiment for variousdistances as said above.
Find average time and 2S/t2 for everytrail. Will it be constant and equal toacceleration? Why?
Draw distance vs time (S-t) graph forabove values in the table.
Do the same experiment for variousslopes of the track and find accelerationsin each case.� Is there any relation between the slope
and acceleration?� What do you notice from the distance
time graphs for various slopes?Do the same experiment with a small
iron block. Find acceleration and draw theS-t graph.
Give your explanation for variousaccelerations related to slopes.
The values found in this experimentare approximate one.Example 2
At a distance L= 400m away from thesignal light,brakes are applied to a
locomotive moving at a velocity u = 54km/h.Determine the position of thelocomotive relative to the signal light after1 min of the application of the brakes if itsacceleration a = - 0.3 m/s2
SolutionSince the locomotive moves with a
constant deceleration after the applicationof brakes, it will come to rest in “t” sec .
We know ,
Here u = 54 km/h = 54 x 5/18 =15 m/sv = 0 for given situation and givena = -0.3 m/s2
From v = u+at we get t =
We get,
During which it will cover a distance
= 375 mThus in 1 min after the application of
brakes the locomotive will be at a distancel = L – s = 400 – 375 = 25 m from thesignal light.
Example 3What is the speed of the body moving
with uniform accelerated motion atmidpoint of two points on the straight line,where the speeds are u and v respectively?
SolutionLet “a” be the constant acceleration.
And s be the distance between the twopoints,
From equation of motionv2 – u2 = 2as................. (1)
Distance,S (cm)
Timet (s)
Averagetimet
2S/t2
t1 t2 t3 v u+at�
–15–0.3
=t = 50s
–u2
2as =
v – ua
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30 Motion
Let v0 be the speed of the body atmidpoint ‘M’ of the given points.
Applying the same equation used above,we get
From (1),
After some algebra we get,
Example 4A car travels from rest with a constant
acceleration “a” for “t” seconds. What isthe average speed of the car for its journeyif the car moves along a straight road?
SolutionThe car starts from rest, so u = 0The distance covered in time t
Example 5A particle moving with constant
acceleration of 2m/s2 due west has an initial
velocity of 9 m/s due east. Find the distancecovered in the fifth second of its motion.
SolutionInitial velocity u = +9 m/sAcceleration a = -2 m/s2
In this problem, acceleration’sdirection is opposite to the velocity’sdirection.
Let “ t” be the time taken by the particleto, reach a point where it makes a turn alongthe straight line.
We have,
We get, t = 4.5sNow let us find the distance covered in
1/2 second i.e. from 4.5 to 5 secondLet u = 0 at t = 4.5 sec.Then distance covered in 1/2s.
= 1/4mTotal distance covered in fifth second
of its motion is given by S0 = 2s = 2 ( ¼ ) = ½ m.
s
u=9m/s
t=5s t=4.5s
t=4.5sa=2m/s2
rest
s
u M v0v
v02 – u2 2as/2=
Fig-21
v02 – u2 =
v2 – u2
2
�v0 = v2 + u2
2
=12
s a t2
Total distance Time takenAverage speed =
a t2/2t
V=
(( =a t2
Fig.22 :Motion of the particle
v u+at�
0 9 – 2t�
=12
s a t2=12
s a t2
[ ]=12
s x 2 x 12
2
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Andhra Pradesh Government Free Distribution 31
Key words Relative, distance, displacement, average speed, average velocity,
instantaneous speed (speed), velocity, acceleration, rectilinear motion
� Motion is relative. Motion of an object depends on the observer.� Distance is the path length traversed and displacement is the shortest distance in a
specified direction.� Average speed is distance covered per unit time and average velocity is displacement in
a specified direction per unit time.� Speed at an instant is instantaneous speed which gives the idea of how fast the position
of the body changes.� Velocity is speed in specified direction.� The motion is uniform when the velocity is constant.� A body has acceleration when the velocity of the body changes.� Acceleration is the rate of change of velocity.� The motion is said to be uniform accelerated motion if acceleration is constant.� The equations of uniform acclerated motion are given by
1. As shown in figure23, a point traverses the curved path. Draw the displacement vectorfrom given points A to B (AS5)
What we have learnt
Improve your learning
v u+at�s ut + ½ a t2�
v2 – u2 2as=
B
A
Fig-23
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32 Motion
2. “She moves at a constant speed in a constant direction.” . Rephrase the same sentencein fewer words using concepts related to motion.(AS1)
3. What is the average speed of a Cheetah that sprints 100m in 4sec. ? What if it sprints50m in 2sec? (25 m/s)(AS1, AS7)
4. Correct your friend who says, “ The car rounded the curve at a constant velocity of70 km/h”.(AS1)
5. Suppose that the three ball’s shown in figure start simultaneously from the tops of thehills. Which one reaches the bottom first ? Explain. See figure 24.(AS2 ,AS1)
6. Distance vs time graphs showing motion of two cars A and B are given. Which carmoves fast ? See figure 25.(AS1)
7. Draw the distance vs time graph when the speed of a body increases uniformly.(AS5)
8. Draw the distance – time graph when its speed decreases uniformly.(AS5)
9. A car travels at a velocity of 80 km/h during the first half of its running time and at 40km/h during the other half. Find the average speed of the car. (60 km/h) (AS1 , AS7)
10. A car covers half the distance at a speed of 50 km/h and the other half at 40km/h.Findthe average speed of the car. (44.44 km/h) (AS1 , AS7)
11. Derive the equation for uniform accelerated motion for the displacement covered inits nth second of its motion. (Sn = u + a (n - ½ )) (AS1)
12. A particle covers 10m in first 5s and 10m in next 3s. Assuming constant acceleration.Find initial speed, acceleration and distance covered in next 2s. (AS1 , AS7)
(7/6 m/s, 1/3 m /s2, 8.33m)
13. A car starts from rest and travels with uniform acceleration “���” for some time andthen with uniform retardation “ �” and comes to rest. The time of motion is “ t” . Findthe maximum velocity attained by it. ( �� t / (��+)) (AS1 , AS7)
14. A man is 48m behind a bus which is at rest. The bus starts accelerating at the rate of 1m/s2, at the same time the man starts running with uniform velocity of 10 m/s. What isthe minimum time in which the man catches the bus? (8s) (AS1 , AS7)
S A
B
t
Fig-24 Fig-25
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Andhra Pradesh Government Free Distribution 33
15. A body leaving a certain point “ O” moves with an a constant acceleration. At the end ofthe 5 th second its velocity is 1.5 m/s. At the end of the sixth second the body stops andthen begins to move backwards. Find the distance traversed by the body before it stops.Determine the velocity with which the body returns to point “ O “ ? (27m, -9 m/s)(AS1)
16. Distinguish between speed and velocity. (AS1)
17. What do you mean by constant acceleration?(AS1)
18. When the velocity is constant, can the average velocity over any time interval differfrom instantaneous velocity at any instant? If so, give an example; if not explain why.(AS2 , AS1)
19. Can the direction of velocity of an object reverse when it’s acceleration is constant? Ifso give an example; if not, explain why.(AS2 , AS1)
20. A point mass starts moving in a straight line with constant acceleration “a”. At a time tafter the beginning of motion, the acceleration changes sign, without change inmagnitude. Determine the time t0 from the beginning of the motion in which the pointmass returns to the initial position. ((2 + �2) t) (AS1)
21. Consider a train which can accelerate with an acceleration of 20cm/s2 and slow downwith deceleration of 100cm/s2.Find the minimum time for the train to travel betweenthe stations 2.7 km apart. (180 s) (AS1)
22. You may have heard the story of the race between the rabbit and tortoise. They startedfrom same point simultaneously with constant speeds. During the journey, rabbit tookrest somewhere along the way for a while. But the tortoise moved steadily with lesserspeed and reached the finishing point before rabbit. Rabbit awoke and ran, but rabbitrealized that the tortoise had won the race. Draw distance vs time graph for this story.(AS5)
23. A train of length 50m is moving with a constant speed of 10m/s. Calculate the timetaken by the train to cross an electric pole and a bridge of length 250 m. (5s , 30s) (AS1)
24. Two trains, each having a speed of 30km/h, are headed at each other on the same track.A bird flies off one train to another with a constant speed of 60km/h when they are60km apart till before they crash. Find the distance covered by the bird and how manytrips the bird can make from one train to other before they crash? (60km infinity,)(AS1)
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34 Laws of Motion
We observe the changes in motion ofmany objects around us. We discussed theconcepts of velocity and acceleration in thechapter ‘Motion’.
Philosophers of the ancient worldwere very much interested in the study ofmotion. One question agitating their mindwas, what is the natural state of an objectif it is left to itself? Our commonsense tellsthat every moving object on earth if it isleft free for some time gradually comesto rest by itself. What happens if you stoppeddling your bicycle? It slows downgradually and stops finally.
We wonder to know that Aristotle, thegreat philosopher of that time alsoconcluded that the natural state of anearthly object is to be at rest. He thoughtthat the object at rest requires noexplanation as any moving body naturallycomes to rest.
Galileo Galilee gave birth to modernscience by stating that an object in motionwill remain in same motion as long as noexternal force is applied on it.
Galileo came up with two ingeneousthought experiments. He did his
experiments on inclined planes withsmooth surfaces and observed that thesmoother the surface, the farther the balltravelled. He extended this argument andconcluded that if the surface was perfectlysmooth, the ball will travel indefinitely,until encountered by another object. (In realworld such a surface of course does notexist).
1 (a) 1 (b) 1 (c)
Fig-1 (a) downward motion (b) upwardmotion (c) motion on a plane surfaceAs shown in figure 1 (a) he observed
that when a marble rolls down a slope, itpicks up speed due to the force of gravityof the earth.
In figure 1 (b) when the object rolls upan inclined plane, its speed decreases. Nowlet us assume that a marble is moving on alevel surface as shown in figure 1(c) it hasno reason to speed up or slow down. So itwill continue to move with constantvelocity.
Chapter
� LAWS OF MOTION
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Andhra Pradesh Government Free Distribution 35
By this experiment, Galileo came to aconclusion which was in contrast toAristotle’s belief that the natural state ofan object is ‘rest’.
Fig-2 (a) (b) motion along inclined planeswith different slopes. (c) Motion from
inclined surface to plane surface
Galileo observed that, as shown infigugre 2 (a), the marble released from itsinitial height rolled down due to the forceof gravity and then moves up the slope untilit reached its initial height. Then he reducedthe angle of the upward slope and did thesame experiment as in figure 2 (b). Themarble rolled up the same height, but it hadto go farther in this instance. That meansthe distance travelled by it is greater. Hemade his observation by further reductionin the angle of the upward slope, he got thesame results. To reach the same height theball had to go farther each time.
Then a question arose in his mind, “howfar must it have to move to reach the sameheight if it has no slope to go up”? Since ithas no slope to go up as shown in figure2(c), obviously it should keep on moving
forever along the level surface at constantvelocity. He concluded that the naturalstate of a moving object, if it is free ofexternal influences, is uniform motion .What do you think of these experiments?Is any external force required to stop amoving object? From this experiment wecan say that an object will remain inuniform motion unless a net force acts onit.
Galileo imagined a world where thereis no friction. But as we learnt in class VIIIthis is not possible in reality becausefriction which affects the motion of theobject plays an important role in our life.For example if there were no friction wecould not be able to walk on ground, wewould not be able to stop a fast moving caretc. It is very difficult to perform manyphysical activities without friction. Builtupon ideas primarily developed by Aristotleand Galileo, Sir Isaac Newton proposed histhree fundamental laws which explain theconnection between force and a change inmotion. These three laws are popularlycalled as Newton's laws of motion.
First law of motionThe first law of motion can be stated
as follows: "Every object will remain at restor in a state of uniform motion unlesscompelled to change its state by the actionof a net force".
"Newton’s first law explains whathappens to an object when no net force actson it."
It either remains at rest or moves in astraight line with constant speed (that isuniform motion). Let’s discuss.
2 (a)
2 (b)
2 (c)
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36 Laws of Motion
Pen Cap
Paperhoop
Bottle
�
�
�
Do you know?
Activity-1
Observing the motion of a pencap kept on thick paper ring.
Make a circular strip from a thick paper.Balance the hoop on the center of the mouthof the bottle. Now balance a pen cap on thepaper hoop aligning it on the center of thebottle’s mouth as shown in figure 3. Pullthe paper hoop with your finger as fast asyou can.� What do you observe? Fig-3 Fast pulling of paper hoop kept on a bottle
� What happens to the pen cap?
Galileo Galilei was born on 15 February1564 in Pisa, Italy. Galileo has been calledthe “father of modern science”.
In 1589, in his series of essays, hepresented his theories about falling objectsusing an inclined plane to slow down the rateof descent.
Galileo was also a remarkable craftsman.He developed a series of telescopes whoseoptical performance was much better thanthat of other telescopes available duringthose days.
Around 1640, he designed the firstpendulum clock. In his book ‘StarryMessenger’ on his astronomical discoveries, Galileo claimed to have seen mountainson the moon, the Milky Way made up of tiny stars, and four small bodies orbitingJupiter. In his books ‘Discourse on Floating Bodies’ and ‘Letters on the Sunspots’,he disclosed his observations of sunspots.
Using his own telescopes and through his observations on Saturn and Venus,Galileo argued that all the planets must orbit the Sun and not the earth, contrary towhat was believed at that time.
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Andhra Pradesh Government Free Distribution 37
accelerated suddenly and moved forward,But the body of person tends to remain inthe same state because of inertia whichresulted in backward motion
In the second case when you aretravelling in a bus your body is alsotravelling with the velocity equal to that ofthe bus. If the bus stops suddenly your bodytends to continue in the same state ofmotion due to inertia. Newton’s First Lawof motion is also known as law of inertia.
With our day to day experiences, weall know that we must exert some force onan object to keep it moving. As far as theobject is concerned the force applied byus is just one of the several forces actingon it. The other forces might be friction,air resistance or gravity. Thus it is clear thatit is the net force which determines thechange in motion of an object.
Let us consider a football placed at reston the ground. The law of inertia tells usthat the football will remain in the samestate unless something moves it.
If you kick the ball, it will fly in thedirection you kicked, with certain speed,until a force slows it down or stops it. Ifthe ball went high, the force of gravity slowsit down. If the ball rolls on the ground theforce of friction make the ball slow downand stop.
If the net force acting on an object iszero, the object which is at rest remains atrest or if the object is already moving witha certain velocity it continues to move withthe same velocity. Thus we can representthe first law of motion as:
If Fnet = 0 then the velocity an object iseither zero or constant.
Activity-2
Observing the motion of thecoins hit by a striker
Fig-4 Hitting the stock of coinswith a striker
Make a stack of carom coins on thecarom board.Give a sharp hit at the bottomof the stack with striker. You can find thatthe bottom coin will be removed from thestack, and the others in the stack will slidedown as shown in figure 4.
� What are your observations from theabove activities?
� Why does the pen cap fall inside thebottle?
� Why does the stack of carom coinsfall down vertically?
To understand this, we have to discusssome more examples which we face in ourdaily life.
When the bus which is at rest beginsto move suddenly, the person standing inthe bus falls backward. Similarly when youare travelling in bus, the sudden stop of thebus makes you fall forward. Why does ithappen so? These changes can be describedonly with the word: Inertia.
In simple language we can say thatinertia means not accepting the change.Things tend to keep doing what they arealready doing. In the first case the bus
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38 Laws of Motion
Thus when the net force acting on abody is zero, we say that the body is inequilibrium.Inertia and mass
We have learnt that inertia is theproperty of an object that resists changesin its state of motion. All objects have thistendency.� Do all the bodies have the same inertia?� What factor can decide the inertia of a
body?Which is easy for you, to push a bicycle
or a car? You can observe that it is difficultto push the car. We say car has greaterinertia than the bicycle. Why does the carpossess more inertia than a bicycle?
Inertia is a property of matter thatresists changes in its state of motion orrest. It depends on the mass of the object.The car has more inertia than a bicyclebecause of its mass.
Mass of an object is considered as themeasure of inertia. We know that SI unitof mass is ‘kg’.
Activity-3
Pushing two wooden boxes withsame force
Take two rectangular wooden blockswith different masses and place them on astraight line drawn on a floor as shown infig-5. Give the same push at the same timeto both the blocks with the help of awooden scale.
� What do you find?� Which one goes farther? Why?� Which block accelerates more?
Fig-5 pushing wooden boxeswith same force
Through your observations you can tellthat the greater the mass of an object, themore it resists changes in its state ofmotion.
From the above examples we canconclude that some objects have moreinertia than others. Mass is a property ofan object that specifies how much inertiathe object has.
Think and discuss
� You may have seen the trick where atablecloth is jerked from a table,leaving the dishes that were on thecloth nearly in their originalpositions.
����� What do you need to perform thissuccessfully? ,
����� Which cloth should we use? Is itcloth made of thick cotton or thinsilk?
����� Should the dishes possess largemass or small mass?
����� Is it better to pull the cloth with alarge force or pull it with a gentleand steady force?
� What is the velocity of a small objectthat has separated from a rocketmoving in free space with velocity10km/s?
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Example 1
A body of mass ‘m’ is kept on thehorizontal floor and it is pushed in thehorizontal direction with a force of 10Ncontinuously, so that it moves steadily.a) Draw FBD(a diagram showing all the
forces acting on the body at a point oftime)
b) What is the value of friction?Solution
Fig-6 Free body diagram
Given that the body is moving steadily,Hence the net force on the body is zeroboth in horizontal and vertical directions.
Forces acting on it along horizontaldirection are force of friction (f), force ofpush (F)
We know that Fnet,x = 0 F+ (-f) = 0 F = fHence the value of force of friction is10N.
Second law of motionNewton’s second law explains us what
happens to an object when non-zero netforce acts on it.
Place a ball on the veranda and push itgently. Then the ball accelerates from rest.
Thus, we can say that force is an actionwhich produces acceleration.
A non zero net force acting on a bodydisturbs the state of equilibrium.
Now we are going to discuss how theacceleration of an object depends on theforce applied on it and how we measure aforce.
Linear momentumLet us recall our observations form our
everyday life. If a badminton ball and acricket ball hit you with same speed, whichone hurts you more? A small bullet firedfrom gun damages the wall only due to itshigh speed. We all know that a heavy truckdamages more than a bicycle if both hit awall. These can be explained by a conceptcalled momentum which is usually denotedby the symbol ‘p’.
From the above examples we can saythat the momentum depends on two factors:one is mass of an object and the other is itsvelocity. Newton used the word “mass inmotion” to represent the meaning ofmomentum. The momentum (p) of a bodyis simply defined as the product of its mass(m) and its velocity (v) : i.e.
Momentum = (mass) x (velocity) p = mvIt can be stated as mass in motion. As
all objects have mass, if an object is inmotion, then it acquires momentum.
Momentum is a vector because velocityis a vector. Hence, the direction ofmomentum is in the direction of velocity.
The SI unit of momentum is kg – m/sor N-s
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40 Laws of Motion
F
F
F F
Activity-4
Larger the net force greater theacceleration
Gently push a block of ice on a smoothsurface and observe how the object speedsup, in other words how it accelerates. Nowincrease the net force and observe changein its speed.
• Is the acceleration increased?
Fig-7 Different forces applied on sameobject
Activity-5
Larger the mass smaller theacceleration
Apply a force on an ice block. Itundergoes some acceleration.
Now take a block of ice with greatermass, but apply almost the same force onthe ice block which has greater mass andobserve the acceleration.
Fig-8 Same forces applied on abjects ofdifferent masses
In both the cases the object accelerates,but we can observe in the second case, itwill not speed up as quickly as before.
From the above examples what have younoticed? Larger the net force greater theacceleration if the mass of the body isconstant and also larger the mass lesser theacceleration if a constant net force isapplied.
According to Newton’s ‘Principia’:Second law states that the rate of changeof momentum of an object is proportionalto the net force applied on the object inthe direction of net force.
Thus net force Fnet �� change inmomentum / time
Fnet ����p
�������������t�p is the change in momentum of a
particle or a system of particles broughtby the net force in a time interval of �t.
When a symbol of proportionality isremoved, a constant is inserted in theequation.
Fnet � k���p
���������tThe SI units of momentum and time
are ‘kg- m/s’ and‘s’ respectively. The unitof force is so chosen that the value ofconstant ‘k’ becomes 1. So that,
Fnet ����p
������tWe know p=mvso that,�p ��� �mv
If the mass of the body is constantduring its motion then,
�p = m�v
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Andhra Pradesh Government Free Distribution 41
Now we have,
Fnet = m �v
������tWe know that �v/�t = a, is called
uniform acceleration.Then Fnet = maThe above formula says that the net
force produces acceleration in a body inthe direction of force.
SI units of force are kg.m/s2. This unithas been named as Newton (N) and1N=1kg.m/s2.
Note:����� Fnet= �p/�t is a universal formula that
can be applied for any system where asFnet= ma can be applied only forconstant mass.
����� To solve problems by using Newton’ssecond law, the weight of the body istaken as ‘mg’ vertically down. (You learnmore about this in the chapter‘Gravitation’)
Think and discuss
���Observe the following diagram.
Fig- 9What is the upper limit of weight thata strong man of mass 80kg can lift asshown in figure?
� What is the momentum of ceiling fanwhen it is rotating?
� Is it possible to move in a curved pathin the absence of a net force?
� Prove that the tension throughout thestring is uniform when the mass ofstring is considered to be zero.Example 2
A mat of mass 1kg and length 1m is placed on the floor. One end of the mat is pulledwith a constant speed of 1m/s towards the other end till the other end comes to motion(tillthe mat is reversed). How much force is required to do this?Solution
Fig-10
As shown in figure, a mat is beingpulled with a constant speed of v = 1m/s,so that the mass of the part of the mat iscontinuously increasing. Hence here themass is a variable.
The time required for bringing theentire mat in motion is given by
�t = distance covered by the end
speed= 2/1 =2s
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42 Laws of Motion
(Distance covered by theend=1m+1m=2m)
From Newton’s second law of motion,
Fnet = �p
= �(mv)
�t �����������tHere v is constant, so we get
Fnet = v �m�t
Where, �m is the change of mass in �ttime.
The change of mass in 2s is equal toentire mass of mat.
Fnet = (1m/s) x (1kg)2
=1
N2In the horizontal direction only one
force is acting. Hence the required forceis 1/2 NExample 3
Atwood machine
Fig-11Atwood machine consists of two loads
of masses m1 and m2 attached to the endsof a limp of inextensible string as shownin the figure 11. The string runs over apulley. Find the acceleration of each loadand tension in the string (m1>m2)
SolutionFrom figure 12 we know that tension
of string always tries to pull the bodies up.
Fig-12
From the FBD of the mass m1, thereexist two forces on the load of mass m1,one is tension of the string acting in upwarddirection and weight of the load (m1g)acting in downward direction.
The net force on m1 Fnet = m1a
m1g – T = m1a ------------ (1)
Thus the net force (Fnet) acting on massm1 produces an acceleration ‘a’ in it.
When m1 moves down, m2 moves up.So the magnitudes of acceleration aresame.
Fig-13
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Andhra Pradesh Government Free Distribution 43
the first one which is equal in magnitudebut opposite in direction.
The two opposing forces are knownas action and reaction pair.
Newton’s third law explains whathappens when one object exerts a force onanother object.
If you are walking on the ground, ateach step, you know that your feet exertsome force on the ground. Are you thinkingthat the ground also exerts some force inthe opposite direction on you?
Is it not surprising to hear that whenyou push a wall then the wall pushes youback!
Fig-15 Action and reaction forces
If two objects interact, the force FAB
exerted by the object ‘A’ on the object ‘B’is equal in magnitude and opposite indirection to the force FBA exerted by object‘B’ on the object ‘A’.
FAB = -FBA
The negative sign indicates that thereaction force is acting in a directionopposite to that of action force. This statesthat no single isolated force exists.
Newton’s first and second laws ofmotion apply to a single body, where asNewton’s third law of motion applies to an
From the FBD of mass m2
Fnet = T- m2g = m2a ------------ (2)
Solving (1) and (2) equations, we get
a = (m1-m2)g (m1+m2)
and
T = 2 m1m2g (m1+m2)
Third law of motion
Activity- 6
Pulling two spring balancesLet’s take two spring balances of equal
calibrations. Connect the two springbalances as shown in figure 14. Pull thespring balances in opposite directions asshown in figure 14.
Fig-14 Forces applied in oppositedirection.
� What do you notice from thereadings in the spring balances?
� Are the readings of two springbalances the same?
� Are we able to make the springbalances to show different readingsby pulling them simultaneously inopposite directions? Why?
According to third law of motion, whenan object exerts a force on the other object,the second object also exerts a force on
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44 Laws of Motion
interaction between the two bodies. Notethat two forces in Newton’s third law neveract on the same body.
The action-reaction pair in Newton’sthird law always represents forces actingon two different bodies simultaneously.
Let us consider the followingexamples.
When birds fly, they push the airdownwards with their wings, and the airpushes back the bird in opposite upwarddirection. Thus the force applied by thewings of bird on air and an opposite forceapplied by the air on wings are equal inmagnitude and opposite in direction.
When a fish swims in water, the fishpushes the water back and the water pushesthe fish with equal force but in oppositedirection. The force applied by the watermakes the fish to move forward.
A rocket accelerates by expelling gasat high velocity. The reaction force of thegas on the rocket accelerates the rocket ina direction opposite to the expelled gases.It is shown in figure-16
Fig-16 Motion of rocket
� Does the rocket exert a force on thegas expelled from it?
Activity-7
Balloon rocketInflate a balloon and press its neck withfingers to prevent air escaping from it.Pass a thread through a straw and tapethe balloon on the straw as shown inthe figure 17.Hold one end of the thread and ask yourfriend to hold the other end of thethread.Now remove your fingers from theballoon’s neck so as to release the airfrom the balloon.
� What happens now ?� How would you explain this situation
with Newton’s third law of motion?
Fig-17 Balloon rocket
Inflate a balloon and tie its neck. Theair within the balloon exerts force on thewalls of the balloon equally in all thedirections as shown in figure.18
Fig-18 The forces on the inside wall of aballoon
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Andhra Pradesh Government Free Distribution 45
When you release the neck of theballoon to allow air to escape from theballoon, what happens? When the airescapes there will be no force acting onthe wall at the neck. But still there is a forceexerted by air inside balloon which is actingin a direction opposite to the neck. Hencethe balloon accelerates in that direction, onwhich direction the net force is acting.
Aim: To show the action and reactionforces acting on two different objects.Material required: Test tube, cork cap,Bunsen burner and laboratory stand.Procedure����� Take a test tube of good quality glass
and put small amount of water in it.Place a cork cap at its mouth to closeit.
����� Now suspend the test tube horizontallywith the help of two strings as shownin the figure 19.
����� Heat the test tube with a Bunsen burneruntil water vaporize and cork cap blowsout.
Fig-19
Observe the movement of test tubewhen cork cap blows out.Compare thedirections of movement of test tube as wellas cork cap. Observe the difference in thevelocity of cork cap and that of recoilingtest tube.� What do you infer from above
experiment?
Think and discuss
�� The force exerted by the earth on theball is 8N. What is the force on theearth by the ball?
�� A block is placed on the horizontalsurface. There are two forces actingon the block. One, the downward pullof gravity and other a normal forceacting on it. Are these forces equal andopposite? Do they form action –reaction pair? Discuss with yourfriends.
�� Why is it difficult for a fire fighter tohold a hose that ejects large amountof water at high speed?
Conservation of momentum andimpulse
Let two objects with masses m1 and m2
travel with different velocities u1 and u2
respectively in the same direction along astraight line. As the velocities are differentthey collide with each other and thecollision lasts for time ‘t’, which is verysmall. During the collision the first marbleexerts a force on the second marble F21 andthe second marble exerts a force on thefirst marble F12. Let v1 and v2 be thevelocities of the marbles respectively aftercollision.
Lab Activity
Test tube
Water
String
cork cap
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46 Laws of Motion
m1u1+m2u2 is the total momentum ofthe two marbles before collision andm1v1+m2v2 is the total momentum of thetwo marbles after collision.
From the above equation we observethat the total momentum is unchangedbefore and after collisions. We can say thatthe momentum is conserved. Law ofconservation of momentum states in theabsence of a net external force on thesystem, the momentum of the systemremains unchanged.
A system is said to be isolated whennet external force acting on it is zero.
It will be surprising if anybody saysthat the fall doesn’t hurt, but the sudden stopat the end that hurts you. Is it true?� Why does a pole vault jumper land on
thick mats of foam?� Is it safe to jump on sand rather than a
cement floor? Why?A softer and more cushioned landing
surface provides a greater stopping distancebecause of the longer time taken to stop.That’s why the fielder pulls back his handswhile catching a fast moving cricket ball.In this situation, the fielder is trying toincrease the time to decrease its velocity.
Thus, the rate of change of momentumwill be less so that the force of impact ofthe ball on hands will be reduced.
As we expressed the second law as
Fnet
=
�p�t
It is important to minimize Fnet , youhave to maximize the stopping time.
We get Fnet �t = �p
Fig-20 Conservation of momentum
What are the momenta of the marblesbefore and after collision? Let’s knowfrom the table.
Marble 1 Marble 2
Momentum m1u1 m2u2
before collisionMomentum m1v1 m2v2
after collisionChange in m1v1-m1u1 m2v2-m2u2
momentum,�pRate of change (m1v1-m1u1) (m2v2-m2u2)in momentum t t��������p��������t
According to Newton’s third law ofmotion, the force exerted by first marbleon the second is equal to the force exertedby the second marble on the first one.
Hence F12 = -F21
Hence we get,��p)1 = - (�p)2
����t ���t m1v1-m1u1 =
- (m2v2-m2u2) t tAfter solving this, we getm1u1+m2u2 = m1v1+m2v2
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Andhra Pradesh Government Free Distribution 47
�p = Fnet �����tEven if the �p is the same in both cases,
the magnitude of the net force (Fnet) actingon the egg determines whether the egg willbreak or not.
Why does a fielder catch a fast movingcricket ball by pulling back his arms whilecatching it? If he doesn’t pull his hands backwhat would happen? The ball definitelyhurts him. When he pulls back his hands heexperiences a smaller force for a longertime. The ball stops only when your handsstop. This shows the change in themomentum not only depends on themagnitude of the force but also on the timeduring which force is exerted on thatobject.
Think and discuss
�� A meteorite burns in the atmospherebefore it reaches the earth’s surface.What happens to its momentum?
�� As you throw a heavy ball upward, isthere any change in the normal forceon your feet?
�� When a coconut falls from a tree andstrikes the ground without bouncing.What happens to its momentum?
�� Air bags are used in the cars for safety.Why?
Example 5A cannon of mass m1 = 12000 kg
located on a smooth horizontal platformfires a shell of mass m2 = 300 kg inhorizontal direction with a velocity v2 =400m/s. Find the velocity of the cannonafter it is shot.
From the above equation we know thatthe product of net force and interaction timeis called impulse of net force. Impulse isequivalent to the change in momentum thatan object experiences during an interaction.Forces exerted over a limited time arecalled impulsive forces. Often themagnitude of an impulsive force is so largethat its effect is appreciable, even thoughits duration is short. Let us observefollowing activity.
Activity-8
Dropping eggsTake two eggs and drop them from a
certain height such that one egg falls on aconcrete floor and another egg falls on acushioned pillow.
� What changes do you notice in botheggs after they are dropped? Why?
Fig-21 (a) fall of an egg on a concrete floor(b) fall of an egg on a cushioned pillow.
When we drop the egg on the concretefloor, it will break, because a large forceacts on the egg for the short interval oftime.
�p = Fnet �tWhen we drop the egg on a cushioned
pillow it doesn’t break because a smallerforce acts on the egg for longer time.
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48 Laws of Motion
SolutionSince the pressure of the powder gases
in the bore of the cannon is an internal forceso, the net external force acting on cannonduring the firing is zero.
Let v1 be the velocity of the cannon aftershot. The initial momentum of system iszero.
The final momentum of the system= m1v1+m2v2
From the conservation of linearmomentum, We get,
m1v1+m2v2 = 0
m1v1 = - m2v2
v1 = - m2v2 /m1
Substituting the given values in theabove equation, we get
v1 = - (300kg) x (400m/s)
12000kg= -10 m/s.
Thus the velocity of cannon is 10m/safter the shot.
Here ‘-’ sign indicates that the canonmoves in a direction opposite to the motionof the bullet.
Key words
What we have learnt
Laws of motion, Inertia, Mass, Linear Momentum Conservation of momentum,Impulse, Impulsive force
� First Law of Motion: A body continues its state of rest or of uniform motion unless anet force acts on it.
� The natural tendency of objects to resist a change in their state of rest or of uniformmotion is called inertia.
� The mass of an object is a measure of inertia. SI unit of mass is Kilogram (kg).� Second Law of Motion: The rate of change of momentum of a body is directly
proportional to the net force acting on it and it takes place in the direction of net force.� Linear momentum of a body is the product of its mass and velocity. p = mv� One ‘Newton’ is the force which when acting on a body of mass 1 kg, produces an
acceleration of 1 m/s2
1 newton (N) = 1kg x 1 ms-2
� Third Law of Motion: If one object exerts a force on the other object, the second objectexerts a force on the first one with equal magnitude but in opposite direction.
� In an isolated system, that means where there is no net force, the total momentum isconserved.
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1) Explain the reasons for the following. (AS1)
a) When a carpet is beaten with a stick, dust comes out of it.
b) Luggage kept on the roof of a bus is tied with a rope.
c) A pace bowler in cricket runs in from a long distance before he bowls.
2) Two objects have masses 8 kg and 25 kg. Which one has more inertia? Why?(AS1)
3) Keep a small rectangular shaped piece of paper on the edge of a table and place anold five rupee coin on its surface vertically as shown in the figure below. Now givea quick push to the paper with your finger. How do you explain inertia with thisexperiment? (AS3)
4) If a car is traveling westwards with a constant speed of 20 m/s, what is the resultantforce acting on it? (Ans:Zero)(AS1 ,AS7)
5) What is the momentum of a 6.0 kg bowling ball with a velocity of 2.2 m/s?(Ans: 13.2 kg m/s2)(AS1)
6) Two people push a car for 3 s with a combined net force of 200 N. (AS1)
(a) Calculate the impulse provided to the car.
(b) If the car has a mass of 1200 kg, what will be its change in velocity?
(Ans: (a) 600 N.s (b) 0.5 m/s)
7) What force is required to produce an acceleration of 3 m/s2 in an object of mass 0.7 kg?(Ans: 2.1 N)(AS1)
8) A force acts for 0.2 sec on an object having mass 1.4 kg initially at rest. The force stopsto act but the object moves through 4m in the next 2 sec. Find the magnitude of theforce? ( Ans: 14 N )(AS1)
Improve your learning
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50 Laws of Motion
9) An object of mass 5 kg is moving with a velocity of 10 ms-1. A force is applied so that in20 s, it attains a velocity of 25 ms-1. What is the force applied on the object?
( Ans: 5 N )(AS1)
10) Find the acceleration of body of mass 2kg from the figuresshown.
(Ans: 5m/s2, 2m/s2)(AS1)
11) Take some identical marbles. Make a path or a track keepingyour notebooks on either side so as to make a path in which marbles can move. Nowuse one marble to hit the other marbles. Take two, three marbles and make them to hitthe other marbles. What can you explain from your observations? (AS5)
12) A man of mass 30 kg uses a rope to climb which bears only 450 N. What is the maximumacceleration with which he can climb safely? ( Ans: 15 m/s2 ) (AS1 ,AS7)
13) An vehicle has a mass of 1500 kg. What must be the force between the vehicle and theroad if the vehicle is to be stopped with a negative acceleration of 1.7 ms-2? ( Ans:2550 N in a direction opposite to that of the vehicle) (AS1 ,AS7)
14) If a fly collides with the windshield of a fast-moving bus, (AS1 ,AS2)
(a) Is the impact force experienced, same for the fly and the bus? Why?
(b) Is the same acceleration experienced by the fly and the bus? Why?
15) A truck is moving under a hopper with a constant speed of 20m/s. Sand falls on thetruck at a rate 20 kg/s. What is the force acting on the truck due to falling of sand?(Ans: 400N opposite to the motion) (AS1 ,AS7)
16) Two rubber bands stretched to the standard length cause an object to accelerate at2 m/s2. Suppose another object with twice the mass is pulled by four rubber bandsstretched to the standard length. What is the acceleration of the second object? (Ans:2 m/s2) (AS1)
17) Illustrate an example of each of the three laws of motion.(AS1)
18) Two ice skaters initially at rest, push of each other. If one skater whose mass is 60 kghas a velocity of 2 m/s. What is the velocity of other skater whose mass is 40 kg?
(Ans: 30 m/s in opposite direction) (AS1 ,AS7)19) A passenger in moving train tosses a coin which falls behind him. It means that the
motion of the train is (AS7)a) Accelerated b) Uniformc) Retarded d) circular motion
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Andhra Pradesh Government Free Distribution 51
20) A horse continues to apply a force in order to move a cart with a constant speed.Explain. (AS1)
21) A force of 5N produces an acceleration of 8 ms-2 on a mass m1 and an accelerationof 24 ms–2 on a mass m2. What acceleration would the same force provide if both themasses are tied together? (AS1) (Ans: 6 m/s2)
22) A hammer of mass 400 g, moving at 30 m s-1, strikes a nail. The nail stops the hammerin a very short time of 0.01 s. What is the force of the nail on the hammer? (AS1)
(Ans: 1200 N)
23) System is shown in figure. (AS1)
Find the acceleration of the blocks and tension in thestring. Take g=10m/s2. (Ans: 5m/s2, 15N)
24) Three identical blocks, each of mass 10kg, are pulledas shown on the horizontal frictionless surface. If thetension (F) in the rope is 30N. What is the acceleration of each block? And what arethe tensions in the other ropes? (Neglect the masses of the ropes) (AS1)
(Ans: a = 1m/s2 , T1 =10N, T2 = 20N
25) A ball of mass ’m’ moves perpendicularly to a wall with a speed v, strikes it and reboundswith the same speed in the opposite direction. What is the direction and magnitude ofthe average force acting on the ball due to the wall? (AS7)
(Ans: 2mv/t away from the wall.)
26) Divya observed a horse pulling a cart. She thought that cart also pulls the horse withsame force in opposite direction. As per third law of motion the cart should not moveforward. But her observation of moving cart raised some questions in her mind. Canyou guess what questions are raised in her mind? (AS2)
27) How do you appreciate Galileo’s thought of “any moving body continues in the stateonly until some external force acts on it” which is a contradiction to the Aristatile’sbelief of “any moving body naturally comes to rest”.(AS6)
T1 T2
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52 Is Mater Pure ?
As per the principle of churning, thelighter components stay at the top when amixture of liquids are spun rapidly.Commercially for separating the creamfrom milk, a machine called centrifuge isused. It follows the same principle.Centrifugation is also used in a diagnosticlaboratory, to test blood and urine samples.The sample is taken in a test tube and thetest tube is placed in a centrifugationmachine. The heavier particles settle to thebottom and the lighter particles stay at thetop of the test tube.
Think and discuss
How does a washing machine squeezesout water from wet clothes?
What is a mixture?Many things that we call pure are
actually mixtures of different substances.Juice is mixture of water, sugar and fruitpulp. Even water contains some salts andminerals. All the matter around us can beclassified into two groups – puresubstances and mixtures.
When a scientist says that somethingis pure, he means that the substance is
Chapter
� IS MATTER PURE?
You might have gone to the marketmany times with your parents to purchaserice, salt, milk, ghee and other provisions.You must have tried to ensure that you gotthe purest possible milk and pure ghee etc.In our day to day language, ‘pure’ meanssomething with no adulteration. But inchemistry pure means something different.
Let us find out what is pure inchemistry?
Activity-1Is full cream pure?
Take some milk in a vessel. Spin itwith a milk churner for some time. Seefigure 1
Fig-1 Churning of milk
After some time, you observeseparation of a paste like solid out of themilk. This paste like solid called creamcontains more than one component in it. Itis therefore a mixture. We have alreadystudied about mixtures in previous classes,let us learn more about them.
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homogeneous i.e.,the composition doesn’tchange, no matter which part of thesubstance you take for examination.
For example, whichever part of a puregold biscuit is taken as a sample, thecomposition is found to be samethroughout. (See figure.2)
Fig-2 Pure gold biscuit
But, mixtures are not alwayshomogeneous. The composition in somemixtures change, depending on the part youhave taken as a sample.
Fig-3 Mixture
A mixture is generally made of two ormore components that are not chemicallycombined. The substances in a mixtureretain their own properties, and they canbe physically separated. What would you notice from the figure 4?
Fig-4 (a) Pure substance 4(b) Mixture
Types of mixturesYou have learnt what a mixture is. Do
you know the types of mixtures? What arethey? Let’s find out.
Mixtures can be in solid, in liquid or ingaseous states or the combination of thesethree states.
Activity-2
Finding out homogeneous andheterogeneous mixtures
Take two test tubes. Fill one test tubewith water and other with kerosene. Nowadd one tea spoon of salt in both the testtubes and stir them.
What do you notice ?In the first test tube you can observe
that the salt dissolves completely. Suchtypes of mixtures are called homogeneousmixtures. In other test tube salt is notdissolved. What do you conclude fromthis? Think!
In a homogeneous mixture thecomponents mixture are uniformlydistributed throughout it. The componentsof a homogeneous mixture are toointimately combined that it will be difficultto distinguish them from one another byvisual observation. For example air is ahomogeneous mixture of many gases.
We all prepare a drink ‘lemonade’ andenjoy its taste. It is a mixture of water,sugar, lemon juice and salt. Is ithomogeneous or not? If you taste aspoonful of lemonade, it tastes the samethroughout. The particles of sugar lemonjuice and salt are evenly distributed in thissolution and we cannot see the componentsseparately. We call such mixtures ashomogeneous mixtures.
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54 Is Mater Pure ?
• Can you give few more examples of thiskind?You have observed in the above activity
that the salt added to kerosene does notdissolve in it. It is a heterogeneousmixture. A heterogeneous mixture is amixture made up of different substances,or the same substance in different stateswhich are not uniformly distributed in it.
For example the mixtures “oil andwater’’, “naphthalene and water’’ areheterogeneous mixtures.
Thus we can conclude that mixtures areof two types homogeneous andheterogeneous. Do you know that these canagain be classified into different kinds? Letus find out.
SolutionsAll of us enjoy drinking soda water and
lemonade. We know that they areexamples of homogeneous mixtures. Thehomogeneous mixture of two or moresubstances from which we can not seperateits components by the process offilteration is called a solution. Solutionscan be in the form of solids, liquids, orgases. A solution has two components, asolvent and a solute.The component of thesolution that dissolve the other componentin it (usually the component present inlarger quantity) is called solvent. Thecomponent of solution that is dissolved inthe solvent (usually the component presentin lesser quantity) is called solute.
A solution of sugar is prepared bydissolving the sugar in water. In thissolution sugar (solid) is the solute andwater (liquid) is the solvent. In the solutionof iodine in alcohol (tincture of iodine)iodine (solid) is the solute and alcohol
(liquid) is the solvent. All the aerated drinksare liquid solutions containing carbondioxide (gas) as solute and water as solvent.
Can you give some more examples forsolutions and tell what the solute andsolvent present in those solutions?
Think and discuss
•“All the solutions are mixtures, but notall mixtures are solutions”. Discussabout the validity of the statement andgive reasons to support your argument.
•Usually we think of a solution as aliquid that contains either a solid, liquidor a gas dissolved in it. But, we can havesolid solutions. Can you give someexamples?
Properties of a solutionIn a solution the particles are so small
in size that we cannot see them with ournaked eyes. They do not scatter a beam oflight passing through the solution and hencethe path of light is not visible in a solution.
• Can you prove this with an expariment?
• If the solution is diluted, can the path oflight be visible?One more interesting property of
solution is that, the solute particles do notsettle down when left undisturbed. Can yougive a reason? If the solute particles aresettling down in a solution can we call it asa homogeneous mixture?
• What would happen if you add a littlemore solute to a solvent?
• How do you determine the percentageof the solute present in a solution?
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Concentration of a solutionCan we dissolve as much solute as we
want in a given solution? How can youdeside how much solute is allowed todissolve in a solvent.
The amount of solute present in asaturated solution at a certain temperatureis called its solubility at that temperature.
For example, take one gram of sugarand add 50ml of water to it. Also take 30gmof sugar and add the same amount of waterto it in another beaker. Which of thesolutions can be called as dilute and whichone as concentrated?
Activity-3
Preparation of saturated andunsaturated solutions
Take 50 ml of water in an empty cup.Add one spoon of sugar to the water in thecup and stir it till it get dissolved. Keep onadding sugar to the cup and stir till no moresugar can be dissolved in it. How manyspoons of sugar is added ?
Fig.5 Adding Sugar to water
When no more solute can be dissolvedin the solution at a certain temperature, itis said to be a saturated solution. Asaturated solution cannot hold any more
solute at a certain temperature. If theamount of solute present in a solution isless than the saturation level, it is called anunsaturated solution.
Can you tell what saturation level is?Now take the solution prepared by you
into a beaker and heat it slowly. (do notboil) Add some more sugar to this solution.You notice that the solution allows todissolve more sugar in it easily when it isheated.
Fig.6 Adding more sugar to water
Find out whether this is true for the saltsolution also?
Activity-4Factors affecting the rate ofdissolving
Take three glass beakers and fill eachof them with 100 ml of water. Add twospoons of salt to each beaker. Place thefirst beaker undisturbed, stir the solutionin the second beaker and warm the thirdbeaker.
What will you observe from the abovethree activities? Which method allows thesolute to dissolve in the solvent easily? Ifyou increase the temperature of the thirdbeaker, what will happen? Repeat theactivity by using salt crystals instead of saltpowder. What change can you observe?
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56 Is Mater Pure ?
Mass of soluteMass of solution
Mass of soluteVolume of solution
SolutionMass of solute (salt) = 50gMass of solvent (water) = 200gMass of solution = Mass of solute +
Mass of solvent = 50g + 200g = 250g
Mass percentage of a solution =
Suspensions and ColloidalSolutions
Activity-5
Finding of heterogeneousmixtures- suspensions andcolloids
Take some chalk powder in a test tube.Take a few drops of milk in another testtube. Add water to these samples and stirwith a glass rod. Observe whether theparticles in the mixtures are visible. Canyou call these mixtures as solutions? (Hint:Are your samples heterogeneous orhomogeneous?)
Now do the following steps and writeyour observations in the table 1.
• Direct a beam of light from a torch or alaser beam on the test tubes. Is the pathof the light beam visible in the liquid?
• Leave the mixture undisturbed for sometime. What changes do you observe?Does the solute settle down after sometime?
Mass of solute
Mass of solutionX 100
50250
= X 100 = 20%
What are the factors that affect thesolubility of a solute?
From this activity we can conclude thatthe temperature of the water, size of thesalt particles, and stiring of the solutionsare some of the factors that affect the rateof solubility of solute in a solvent.
You know that solubility is themeasurement of amount of solute thatdissolves in a solvent. If the amount ofsolute present is little, the solution is saidto be dilute, and if the amount of solutepresent is more, the solution is said to beconcentrated.
The concentration of a solution can bedefined as the amount (mass) of solutepresent in a given amount (mass) ofsolution or the amount (mass) of solutedissolved in a given volume of the solution.
There are many ways of expressingthe concentration of a solution, but herewe learn only about two of those.
(i) Mass by mass percentage of a
solution = X 100
(ii) Mass by volume percentage of a
solution = X 100Example
A solution contains 50g of common saltin 200g of water. Calculate theconcentration in terms of mass by masspercentage of the solution?
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Mixture
• Filter the mixtures. Did you find anyresidue on the filter papers?
Record your observations in the table-1
Table-1Is the path Residue isof the light Did solute seen on the
beam settle down?filter papervisible? (Yes/No)
Chalk mixture
Milk mixture
We find that the particles of chalk don’tdissolve but remained suspendedthroughout the volume of the water. So themixture we got is a heterogeneous mixturebecause the solute particles didn’t dissolveand the particles are visible to naked eye.Such heterogeneous mixtures are calledsuspensions. Suspensions are theheterogeneous mixtures of a solid and aliquid, in which the solids do not dissolve,like mixtures of soil and water.
Let us consider the mixtures of oil inwater and kerosene in water. These arespecial kinds of suspensions, calledemulstions. These mixtures consist of twoliquids that do not mix and the settle intolayers when they left undisturbed.
Give some more examples ofemulations which you see in your daily life.
Think and discuss
Have you ever observed carefully thesyrup that you take for cough? Why doyou shake them before consuming?
• It is a suspension (or) emulstion?
In activity-5, particles of milk in thesecond test tube are uniformly spreadthroughout the mixture. Due to smaller sizeof milk particles it appears to behomogeneous but it is a heterogeneousmixture. These particles easily scatter abeam of visible light. Such mixtures arecalled a colloids or colloidal solutions.These mixtures possess the characteristicsin between a solution and a suspension.They are also called as colloidaldispersions. Colloidal dispersions mayappear homogeneous but are actuallyheterogeneous.
A large number of substances such asmilk, butter, cheese, creams, gels, bootpolish, and clouds in the sky etc. are somemore examples of colloids.
Colloidal solutions areheterogeneous in nature and always consistof at least two phases; the disperse phaseand the dispersion medium. Dispersephase is the substance that present in smallproportion and consists of particles ofcolloidal size (1 to 100mm). Dispersionmedium is the medium in which thecolloidal particles are dispersed. These twophases can be in the form of solid, liquidor a gas. Thus, different types of colloidalsolutions are possible depending upon thephysical state of the two phases.
Here are some common examples ofcolloids from our daily life. (See table 2)Don’t try to memorize this table-2, it hasgiven only for your information.
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58 Is Mater Pure ?
We studied that the particles in a col-loidal solution can easily scatter a beam ofvisible light. This scattering of a beam oflight is called the Tyndall effect namedafter the scientist who discovered it. Youmay observe this effect in your day-todaylife when a fine beam of light enters a roomthrough a small hole or slit. You can try tosee Tyndall effect at your home.
Select a room where the sunlight fallsdirectly through a window. Close thewindows in such a way that a slit is left openbetween the windows. (Don’t closecompletely). What do you see?
You can also observe thisphenomenon while walking on a roadhaving a lot of trees on both sides. Whenthe sunlight passes through branches andleaves you see the path of dust particles.
Try to observe the Tyndall effect inthe kitchen, the smoke from the stove isexposed to sun light.
• Did you ever observe this phenomenonin the cinema halls?
• Have you ever get an opportunity ofgoing through deep forests? If you gothrough deep forest you canexperience this effect.
Fig.7 Tyndall effect in the forest
When sunlight passes through thecanopy of a dense forest; mist contains tinydroplets of water, which act as particles ofcolloid dispersed in air.
Table-2: Examples of dispersing medium and disersed phase
Dispersing Medium Dispersed Phase Colloid type Examples
Gas Liquid aerosol Fog, clouds, mist
Gas solid Aerosol Smoke, automobile exhaust
Liquid Gas Foam Shaving cream
Liquid Liquid Emulsion Milk, face cream
Liquid solid Sol Mud, milk of magnesia
Solid Gas Foam Foam, rubber,sponge, pumice stone
Solid liquid Gel Jelly, cheese, butter
Solid solid Solid sol Coloured gem stone,milky glass
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Fig. 8Is ice-cream a colloid ?
Ice cream is made by churning a mixtureof milk, sugar and flavours. This mixture isslowly chilled to form ice cream. Thechurning process disperses air bubbles intothe mixture by foaming and breakup thelarge ice crystals in to tiny particles. Theresult is a complex substance which
Table 3 Properties of suspension and colloidsSuspensions Colloids
Suspensions are heterogeneous mixtures. Colloid is a heterogeneous mixture.The particles of suspensions can be seen with The size of particles of a colloid is too small to benaked eyes. individually seen by naked eyes..The particles of a suspension scatter a beam of Colloids are big enough to scatter a beam of light passing through it and make its path visible of light passing through it which makes its path
visible.The solute particles settle down when a They don’t settle down when the suspension iskept undisturbed. When the particles left undisturbed. i.e., colloid is quite stable.settle down the suspension breaksand it does not scatter light any more.Suspension is unstable. The components can be The components cannot be separated from theseparated from the mixture by the process mixture by process of filtration.of the filtration or decantation. Centrifugation technique is used in separation.
contains solids (milk fats and milkproteins), liquids (water) and gases (airbubbles). Now can you guess whether icecream a colloid or not?
Think and discuss
Is there any difference between a truesolution and colloidal solution? If youfind the differences, what are thosedifferences?
Can you explain now in a comparativeway about suspensions and colloids? Letus see.
Separating the components of amixture
Till now we have discussed the typesof mixtures. Do you know techniques toseparate these mixtures in to theirrespective constituents?
Usually heterogeneous mixtures can beseparated into their respective constituentsby simple physical methods like
handpicking, sieving, filtration etc., as weuse in our day to day life.
Sometimes special techniques have tobe used for the separation of thecomponents of mixture. We have learnt inclass VI how to separate mixtures in variousways like, flotation, filtration,crystallization, chromatography etc. Let ussee more.
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60 Is Mater Pure ?
Why do we use different separationtechniques for mixtures like grain and huskas well as ammonium chloride and saltthough both of them are heterogeneousmixtures?� What is the basis for choosing a
separation technique to seperatemixtures?
EvaporationActivity-7
Process of evaporation of water
Fig-10 Evaporation of water
Take a beaker and fill it to half itsvolume with water. Keep a watch glass onthe mouth of the beaker as shown in figure-10. Put few drops of ink on the watch glass.Heat the beaker and observe the watchglass. Continue heating till you do notobserve any further change on the watchglass.
What is evaporated from the watchglass? Is there any residue on the watchglass?
We know that ink is a mixture of adye in water. We can separate thecomponents in the ink using evaporation.
Vapours
InkWatch Glass
Beaker
Water
Inverted funnel
Solidified ammoniumchloride
China dishMixture ofammonium chlorideand salt
Vapours ofammoniumchloride
Cotton Plug
SublimationActivity-6
Separation of mixtures bysublimation
Fig- 9 Separating ammonium chloride and saltTake one tablespoon of common salt onetablespoon of ammonium chloride and mixthem.
� Is the mixture heterogeneous? Givereasons.
� How do we separate the salt andammonium chloride?
Take the mixture in a china dish. Take aglass funnel which is big enough to cover thedish. Plug the mouth of the funnel with cottonand invert it over the dish as shown in figure9. Keep the dish on the stand of stove andheat for some time and observe the walls ofthe funnel. Initially you find vapours ofammonium chloride and then solidifiedammonium chloride on the walls of thefunnel.
Try it for mixtures that have camphoror naphthalene.
Think and discuss
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Procedure : Draw a thick line just abovethe bottom of the filter paper using themarker. Pour a small amount of water inthe beaker and hang the paper strip with helpof a pencil and tape in such a way that itshould just touch the surface of water asshown in figure 11.
Make sure that the ink line or markdoes not touch the water.
Allow the water to move up the paperfor 5 minutes and then remove the stripfrom water. Let it dry.
What colours did you observe in theblack ink sample?
Take two more paper strips andmarkers as samples and do the experiment.Do the colours occur in the same order andin the same location on all the samples?
Instead of non permanent marker usea permanent marker. What will youobserve?
Now touch the marker line to water.What will you notice?
Instead of thick line, draw a thin lineon the paper strip with non- permanentmarker? Do your results change in eachcase?
� Is chromatography used only toseperate components of colouredliquids?
Separation of immiscible andmiscible liquids
A liquid is said to be miscible if itdissolve completely in another liquid. Forexample alcohol is miscible in water. Canyou give some more examples for miscibleliquids.
An immiscible liquid is one whichdoesn’t dissolve but forms a layer overanother liquid and can be separated easilylike oil is immiscible in water. Can youname any such liquids from your dailyobservation?
Pencil
Cello tape
Strip ofBeaker
Marker
Lab Activity
Fig.11 Separating the components of ink
Think and discuss
Is it possible to find outadulteration of Kerosene in petrol withthis technique?
In activity-7 we saw that ink is amixture of solute and solvent. Is the dye inink a single colour? How many solutes arethere in ink? How can we find out those?Is there any technique to separate thedifferent components of dye in the ink?That is where chromatography would help?
Chromatography is a laboratorytechnique for the separation of mixturesinto its individual components. We can usechromatography to separate componentsof dyes in ink. The process can also be usedto separate the coloured pigments in plantsand flowers or used to determine thechemical composition of many substances.Paper Chromatography
Aim: Separating the components of inkusing paper chromatography.Material required: Beaker, rectangularshaped filter papers, black marker (non-permanent), water, pencil and cello tape.
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62 Is Mater Pure ?
Separatingfunnel
Kerosene Oil
Stop Cock
Water
Do you know how to separateimmiscible liquids?
Activity-8
Separation of immiscible liquids
Fig.12 Separating funnel
You must have seen a mixture of oiland water. How many layers can beobserved? How can you separate the twocomponents?
Take a separating funnel and pour themixture of kerosene oil / caster oil andwater in it. Let it stand undisturbed forsome time. So that separate layers of oiland water are formed. Open the stopcockof the separating funnel and pour out thelower layer of water carefully. Close thestopcock of the separating funnel as the oilreaches the stop-cock.
The underlying principle is that theimmiscible liquids separate out in to layersdepending on their densities. Did youobserve in your village how caster oil isprepare by boiling castor seeds? If not askyour parents and teachers.
Diesel is immiscible in water. Thebright rainbow patterns seen often duringrainy season when the diesel drops falls onthe road, are the result of thin film of dieselon water.
Fig.13 Drops of diesel on wet road
Separation of a mixture of twomiscible liquids
Sometimes a homogeneous solutionis formed by the mixing of liquids. Someliquids have the property of mixing in allproportions, forming a homogeneoussolution. This is known as miscibility.Water and ethanol, for example, aremiscible because they mix in allproportions. How can we separate suchmixtures?Distillation
Activity-9
Separation of two miscible liquidsby distilation
Thermometer
Water OutletClamp
Watercondenser
Cold water in
Acetone
Mixture ofAcetoneandWater
DistillationFlask
Clamp
Fig-14 Separating the mixture ofAcetone and water by distillation
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Acetone and water are also miscible.Take a mixture of acetone and water in adistillation flask. Fit it with a thermometerand arrange to stand. Attach the condenserof the flask on one side and other side keepa beaker to collect. Heat the mixture slowlykeeping a close watch on the thermometer.The acetone vaporizes and condenses in thecondenser. Acetone can be collected fromthe condenser outlet. Water remains in thedistillation flask.
The separation technique used aboveis called distillation. Distillation is usedin the separation of components of amixture containing two miscible liquids.But there should be a large difference inthe boiling points of the two liquids.
What if the boiling points of thetwo liquids are close to eachother?
To separate two or more miscibleliquids when the difference in their boilingpoints is less than 25oC, fractionaldistillation process is used. If thedifference in boiling points is greater than250C, a simple distillation is used.
Do you know what process offractional distillation is?
The apparatus is similar to that forsimple distillation except that afractionating column is fitted in betweenthe distillation flask and the condenser. Asimple fractionation column is a tubepacked with glass beads. The beads providemaximum possible surface area for thevapours to cool and condense repeatedlyas shown in Fig.15
Fig-15 Fractional distillation
� Can you give any examples where we usethis technique?
� How can we obtain different gases fromair ?
We have learnt that air is a homogeneousmixture. Can it be separated into itscomponents?
Let’s see the flow chart which gives thesteps of the process.
Air
Compress and cool by increasing pres-sure and decreasing temperature
Liquid air
Allow to warm up slowly in fractionaldistillation column
Gases get separated at different heights
Points Oxygen Argon Nitrogen
Boiling points (0C) -183 -186 -196
% air by volume 20.9 0.9 78.1
Fig. 16 Flow diagram shows the processof obtaining gases from air
Thermometer
Water Outlet
Watercondenser
Coldwater in
Pure liquidcomponent
Mixture
DistillationFlask
Clamp
Cork
Fractionatingcolumn
Cork
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64 Is Mater Pure ?
If we want oxygen gas from air(figure-17), we have to separate out all the othergases present in the air. The air is compressedby increasing the pressure and is then cooled bydecreasing the temperature to get liquid air. Thisliquid air is allowed to warm up slowly in afractional distillation column where gasesseparated at different temperatures dependingupon their boiling points.
Think and discuss
� Arrange the gases present in air inincreasing order of their boillingpoints. What do you observe?
� Which gas forms the liquid first as theair is cooled?
Types of pure substancesSo far we have studied about mixtures
viz – substances whose components can beseparated by physical methods. What aboutsubstances that cannot be separated furtherby any of the methods of separation? Wecall these as pure substances. Let us explorefurther about them.
Activity-10
Can we separate mixture ofCopper sulphate and Aluminum
Take a concentrated solution ofcopper sulphate in to beaker and drop apiece of aluminum foil in it. After sometime you will observe a layer of copperdeposited on the aluminum foil. Thesolution becomes colorless. Why did thishappen? (recall activities of the chapteron Metals and Non-metals)
We know that a chemical reactiontakes place among the copper ions presentin the solution with aluminum and coppermetal is separated. Does it mean thatcopper sulphate is a mixture? No it is not.
Here copper cannot be separatedfrom sulpher and oxygen by any physicalprocess. It can be separated only by achemical reaction. Substances such ascopper sulphate are called compounds.
Table-4 : Mixtures and compoundsMixtures Compounds
1. Elements or compounds just mix together to 1. Elements react to form new compounds.form a mixture and no new compound is formed.
2. A mixture has a variable composition. 2. The composition of each new substance isalways fixed.
3. A mixture shows the properties of the 3. The new substance has totally differentconstituent substances. properties.
4. The constituents can be separated fairly easily by 4. The constituents can be separated only byphysical methods. chemical or electrochemical reactions.
Air in
Filter Freezingcold water in
Carbon dioxideout as dry ice Expansion
jet
LiquidAir
Fractionaldistillationcolumn
Nitrogenout
Argonout
Liquidoxygen
Separator
ColdCompressed
air
Water out
Hot Air
Airunder
pressure
Fig.17 Separation of componants of air
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We can define compounds as puresubstances that can be separated into twoor more components only by means of achemical reaction.
We now have two types of puresubstances - compounds and elements.
Elements can be divided into metals,non-metals and metalloids. We have alreadystudied properties of metals and non-metals. Write down the names of someelements that you know.
Elements have been used since theearly days of civilisation viz – metals suchas iron, lead, copper, helped in thedevelopment of civilizations. For thousandsof years, alchemists – up to and includingIsaac Newton – attempted to unearth newelements, and study their properties.Hennig Brand, a German alchemist, boileddown urine to discover phosphorus in 1669.But it was not until the late 1700s that ourknowledge of the elements really took off,as chemists developed new ways to purifyand isolate elements.
Sir Humphry Davy, was extremelysuccessful in discovering many elements -sodium, magnesium, boron, chlorine andmany more. Robert Boyle used the termelement and Lavoisier was the first toestablish a useful definition of element. Hedefined an element as a basic form ofmatter that cannot be broken down in tosimpler substances by chemical reactions.
If any substance can be separated intotwo or more constituent parts by a chemicalreaction, that substance is definitely acompound.
What do we get when two or moreelements are combined? We canunderstand through an activity.
Activity-11
Understanding the nature ofelements, compounds andmixtures
Divide the class into two groups.Give 5g of iron fillings and 3g of sulphurpowder in a china dish to both the groups.
Activity for group-1:
Mix and crush iron fillings andsulphur powder. Check for magnetism inthe material obtained. Bring a magnet nearthe material and check if the material isattracted towards the magnet.
Activity for group-2:
Mix and crush iron fillings andsulphur powder. Heat this mixture stronglytill it becomes red hot. Remove it fromflame and let the mixture cool. Check formagnetism in the material obtained.Compare the texture and colour of thematerial obtained by the two groups.
The next part can be done if you havea lab set up in the school.
Each group should divide thematerial obtained into two parts. Addcarbon disulphide to one part. Stir well andfilter.
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66 Is Mater Pure ?
Add dilute sulphuric acid or dilutehydrochloric acid to the other part.
Do the same reactions separatelywith sulphur and iron. Observe the changes.
Now answer the following� Did the material obtained by the two
groups look the same?� Which group has obtained a material
with magnetic property?� Can we separate the components of the
material obtained?� On adding dilute sulphuric acid or dilute
hydrochloric acid did both the groupsobtain a gas?
� Did the gas in both the cases smell thesame or different?
The gas obtained by group 1 due toreaction of material with dilute HCl (or)H2SO4 solution is hydrogen. It iscolourless, odorless and combustible.
The gas obtained by group 2 due toreaction of material with dilute HCl (or)H2SO4 solution is hydrogen sulphide. It isa colourless gas with the smell of rotten
eggs. You must have observed that theproduct of the both groups shows differentproperties though the starting material wassame.
Group 1 has carried out the activityinvolving a physical change. Where asgroup examined a chemical change. Thematerial obtained by group 1 is a mixtureof two substances. i.e. iron and sulphur,which are elements.
The properties of mixture are thesame as that of its constituents’. Thematerial obtained by the group 2 is acompound. On heating the two elementsstrongly we get compound, which hastotally different properties, compared tothe properties of the combining elements.The composition of a compound is thesame throughout. We can also observe thatthe texture and colour of the compound arethe same throughout its volume.
The chemical and Physical nature ofthe matter is better understood by thefollowing flow chart.
Matter (Solid,liquid or gas)
Pure substance Mixtures(No fixed composition)
Compounds...........................................
Have fixed composition canbe brokendown into elements by chemical or
electrochemical reactions...............................................
for example, water, methane, sugar,salt, etc
Heterogeneous.............................
Non uniform composition.............................
for example, sand and salt, sugarand salt, water in oil etc
Homogeneous.................................Uniform composition
.....................................for example, sugar in water,
salt inwater, sulphur in carbondisulphide,water in alcohol etc
Elements...........................................
cannot be broken down to simplersubstances
...............................................for example, copper, oxygen, iron,
hydrogen, mercury, etc
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What we have learnt
Key words
Pure substances, Mixture, Heterogeneous mixture, Homogeneous mixture,Solution, Suspension, emulsions, colloidal dispersions, solvent, solute, concentrationof solution, Tyndall effect, Evaporation, Centrifuge, Immiscible liquids, Miscibleliquids, Cromatography, distillation, Fractional distillation, Elements Compounds.
� A mixture contains more than one substance (element and/or compound) mixed inany proportion.
� Mixtures can be separated into pure substances using appropriate separationtechniques.
� A solution is a homogeneous mixture of two or more substances. The majorcomponent of a solution is called the solvent, and the minor, the solute.
� The concentration of a solution is the amount of solute present per unit volume orper unit mass of the solution.
� Materials that are insoluble in a solvent and have particles that are visible to nakedeyes, form a suspension. A suspension is a heterogeneous mixture.
� Colloids are heterogeneous mixtures in which the particle size is too small to beseen with the naked eye, but is big enough to scatter light. Colloids are useful inindustry and daily life. The colloid has the dispersed phase and the medium in whichthey are distributed is called the dispersion medium.
� Pure substances can be elements or compounds. An element is a form of matter thatcannot be broken down by chemical reactions into simpler substances. A compoundis a substance composed of two or more different types of elements, chemicallycombined in a fixed proportion.
� Properties of a compound are different from its constituent elements, whereas amixture shows the properties of its constituting elements or compounds.
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68 Is Mater Pure ?
1. Which separation techniques will you apply for the separation of the following?(AS1)
(a) Sodium chloride from its solution in water.
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
(c) Small pieces of metal in the engine oil of a car.
(d) Different pigments from an extract of flower petals.
(e) Butter from curd.
(f) Oil from water.
(g) Tea leaves from tea.
(h) Iron pins from sand.
(i) Wheat grains from husk.
(j) Fine mud particles suspended in water.
2. Write the steps you would use for making tea. Use the words given below and writethe steps for making tea? (AS7)
Solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
3. Explain the following giving examples.(AS1)
(a) Saturated solution (b)Pure substance (c) colloid (d) Suspension
4. Classify each of the following as a homogeneous or heterogeneous mixture. Givereasons.(AS1)
Soda water, wood, air, soil, vinegar, filtered tea.
5. How would you confirm that a colourless liquid given to you is pure water? (AS1)
6. Which of the following materials fall in the category of a “pure substance”? Givereasons (AS1)
(a) Ice (b) Milk (c) Iron (d) Hydrochloric acid (e) Calcium oxide(f) Mercury (g) Brick (h) Wood (i) Air.
Improve your learning
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7. Identify the solutions among the following mixtures. (AS1)
(a) Soil (b) Sea water (c) Air (d) Coal (e) Soda water.
8. Which of the following will show “Tyndall effect”? How can you demonstrate Tyndalleffect in them? (AS1 AS3)
(a) Salt solution (b) Milk (c) Copper sulphate solution (d) Starch solution.
9. Classify the following into elements, compounds and mixtures. (AS1)(a) Sodium (b) Soil (c) Sugar solution (d) Silver(e) Calcium carbonate (f) Tin (g) Silicon (h) Coal(i) Air (j) Soap (k) Methane (l) Carbon dioxide(m) Blood
10. Classify the following substances in the below given table. (AS1)
Ink, soda water, brass, fog, blood, aerosol sprays, fruit salad, black coffee, oil andwater, boot polish, air, nail polish, starch solution, milk.
Solution Suspension Emulsion Colloidal dispersion
11. Take a solution, a suspension, a colloidal dispersion in different beakers. Test whethereach of these mixtures shows the Tyndall effect by focusing a light at the side of thecontainer.(AS3)
12. Draw the figures of arrangement of apparatus for distillation and fractional distillation.What do you find the major difference in these apparatus? (AS5)
13. Determine the mass by mass percentage concentration of a 100g salt solution whichcontains 20g salt? (AS1)
Ans: (20% )
14. Calculate the concentration interms of mass by volume percentage of the solutioncontaining. 2.5g potassium chloride in 50ml of potassium chloride (KCl)solution? (AS1) Ans: (5%)
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70 Atoms and Molecules
In the chapter "Is matter around us pure?"we used the terms elements andcompounds. You learnt about the role ofseparation techniques in identifyingelements. The pure components obtainedafter separation (or purification) are eitherelements or compounds.
In this chapter, we can use thisknowledge to explain some of observationsmade in previous classes like theobservation of rusting of iron rod keptoutside, etc.
• Does the weight of iron rod increaseor decrease, on rusting?
We notice that on burning charcoal, itleaves ash at the end ?
• Where does the matter charcoal go?• Wet clothes dry after some time -
where does the water go?These and other similar questions had
also fascinated scientists for many years;in particular, burning or combustionreactions.
Recall the chapter "Metals and non-metals."
• What does happen to Magnesium onburning it in air ?
• What does happen to Sulphur onburning it in air?
Think about the weight of the reactantsand products.
ATOMS AND MOLECULESChapter
�
Do you know?
Antoine Lavoisier(1743-1794) was aFrench nobleman. Hemade many importantcontributions tochemistry and some
call him the Father of ModernChemistry.
Lavoisier studied combustion reactionsin detail. For example, duringcombustion reactions, he not onlycarefully weighed the solid reactants,but also took into account the gasesinvolved. He perfected chemicalapparatus which ensured that gases didnot escape during the reactions. This ledto the law of conservation of mass.
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In this chapter, we use the followingterms frequently - elements, compounds,reactants and products. Discuss with yourfriends about meaning of these terms.Think of different examples for each term.
Let us start our investigations by doinga lab activity to see what will happen toweight of reactants and products during areaction.
Aim: To understand change in massbefore and after a chemical reaction.Material required: Lead nitrate,potassium iodide, distilled water, conicalflask, spring balance, test tube, stand,rubber cork, thread etc.Procedure
1. Prepare a solution by dissolvingapproximately 2 grams of leadnitrate in 100 ml of distilled water.
2. Prepare another solution bydissolving approximately 2 gm ofPotassium iodide in 100 ml water.
3. Take 100ml solution of lead nitratein 250ml conical flask.
4. Also take 4ml solution of potassiumiodide in test tube.
5. Hang the test tube in the flaskcarefully, without mixing thesolutions. Put a cork on theflask.(see figure 1)
6. Weigh the flask with its contentscarefully by spring balance.
7. Now tilt and swirl the flask, so thatthe two solutions mix.
(see figure 2).
8. Weigh the flask again by the samespring balance as shown in figure 3.
9. Record your observations:Weight of flask and contents beforemixing =Weight of flask and contents aftermixing =
Now, try to answer these questions:• What happens in the reaction flask?• Do you think that a chemical
reaction has taken place? Givereason.
Lab Activity
Fig - 1
Fig - 2
Fig - 3
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72 Atoms and Molecules
Do you know?
Think and discuss
Think and discuss
Though the law of conservation ofmass was proposed by Lavoisier, It wasexperimentally verfied by Landolt. Theexperiment carried out by us is amodified form of the experiment doneby Landolt.
Law of constant proportionsFrom the experiments on law of
conservation of mass, we saw that massdoes not change during a chemical reaction.
Now let us look at the results of someexperiments carried out by the Joseph L.Proust between 1798 and 1808.
Proust took two samples of coppercarbonate - a compound of copper, carbonand oxygen. He took a sample from natureand another sample prepared in the lab anddecomposed it chemically to findpercentage of copper, carbon and oxygenin the two samples.
The results obtained are given intable- 1
Table-1
• What do you observe from the table?
• Does the weight of the flask and itscontents change?
• What is your conclusion?
Result:
• A chemical reaction took place andthe mass remained same before andafter chemical reaction. Therefore,mass was neither created nordestroyed in the chemical reaction.
• Do you get the same result if theconical flask is not closed?
Law of conservation of mass
Antoine Lavoisier carried out muchexperimentation and he established theimportant law of conservation of mass. Itstates, "Matter is neither created nordestroyed during a chemical reaction.More simply, the mass of reactants isequal to the mass of the products ofchemical reaction. Earlier, it was thoughtthat mass is lost during burning of charcoal.But when Lavoisier, carried out the burningin a closed set up, he found no change inweight.
• Recall the burning of the Magnesiumribbon in air. Do you think mass isconserved during this reaction?
Naturalsample
Syntheticsample
Weightpercentage of
Copper 51.35 51.35 Carbon 38.91 38.91 Oxygen 9.74 9.74
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Think and discuss
• What difference do you observe inpercentage of copper, carbon andoxygen in two samples?
Similarly, Proust took water fromdifferent sources, and found that thepercentage of oxygen and hydrogen was thesame in all samples. There was no relationbetween the place from where the samplecame and its composition.
Based on his experiments, Proust putforwared the law of constant (or definite)proportions. It states that, "a givenchemical substance always contains thesame elements combined in a fixedproportions by weight." This means thatthe relative proportion of elements in acompound is independent of the source ormethod of preparation.
• 100 g of mercuric oxide decomposeto give 92.6 g of mercury and 7.4 g ofoxygen. Let us assume that 10 g ofoxygen reacts completely with 125 gof mercury to give mercuric oxide. Dothese values agree with the law ofconstant proportions?
• Discuss with your friends if the carbondioxide that you breathe out and thecarbon dioxide they breathe out areidentical. Is the composition of thecarbon dioxide of different sourcessame?
Why are the laws valid?By early 19th century, the scientists
knew some laws governing chemical
reactions. Why are these laws valid? Whycouldn't the elements combine in anyproportion ?
Many scientists tried to give appropriateexplanations. The English school teacherJohn Dalton proposed the basic theoryabout the nature of matter. Dalton reasonedhis proposals as mentioned below.
1. If mass was to be conserved, then allelements must be made up ofextremely small particles, calledatoms.
2. If law of constant proportion is tobe followed, the particles of samesubstance couldn't be dissimilar.
Based on the above laws, Daltonproposed A new system of ChemicalPhilosophy of atomic theory.
Dalton's atomic theoryThe following are the main postulates
of the theory:
1. Matter consists of indivisibleparticles called atoms.
2. Atoms are neither created nordestroyed in a chemical reaction.
Fig - 4John Dalton
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74 Atoms and Molecules
Think and discuss
Atoms and moleculesVery often you may have heard that atoms
are the building blocks of all matter. Butwhat does it mean? It means that matter iscomposed of tiny particles known as atoms.
These atoms are so small that we cannotsee them even with a high-poweredmicroscope. The number of atoms presenteven in a small amount of matter is verylarge.
The aluminium foil that we use topack food might seem thin to you. Butit has atoms in thousands.
• Are elements also made of atoms?
We know that substances are made upof - atoms or molecules. Atoms are themost fundamental of all particles that canhave an independent existence. Sometimestwo or more atoms combine to form a bigparticle. When atoms combine, they formmolecules. When the particles of asubstance contain only one type of atoms,that substance is called an element. Inelements the smallest particle that existmay be atoms or molecules.
There are many elements whose smallestparticle is an atom. Iron, copper, zinc,aluminium, silver, gold etc are examples ofsubstances in which the smallest particleis an atom.
Do you know?
Chemical reactions involvereorganization of atoms.
3. All the atoms of a given elementhave identical mass and chemicalproperties. Atoms of differentelements have different masses andchemical properties.
4. Compounds are formed when atomsof different elements combine insimple whole number ratios. That is,chemical change is the union orseparation of atoms as a whole.
• Which postulate of Dalton's theoryis the result of the law ofconservation of mass?
• Which postulate of Dalton's theorycan explain the law of constantproportions?
About 2600 years ago, an Indian sage(Rishi) called Kanada also postulatedatoms in his VAISHESIKA SUTRA. Theactual name of Kanada was Kasyapa -he was renamed after his KANASIDHANTHA. He proposed that allforms of matter are composed of verysmall particles known as anu and eachanu may be made up of still smallerparticles called parmanu.
The word 'atom' is derived from aGreek word 'a-tomio' (means -indivisible)
Do you know?
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Oxygen and nitrogen are examples ofsubstances in which the particles are acombination of two identical atoms.
Atoms of same elements or of differentelements can join together to formmolecules. If atoms of different elementsjoin together they form a new substanceknown as compound.
So we can have molecules of elementsand molecules of compounds. A moleculecan be defined as the smallest particle ofmatter that is capable of an independentexistence and retained all the properties ofthat substance.
Why do we name elements?
Do you know what gold is called in yourlanguage? But in other languages it wouldhave a different name. There are so manylanguages in the world so it is not possibleto know the different names of eachelement in different languages. To helpscientists communicate without confusion,we must have one name for each elementthat is accepted by everyone.
Do you know?
How elements like hydrogen andoxygen got their names?
Sometimes elements are named basedon their property. For example, the Latinword for water is 'hydro'. So the elementthat combined with oxygen to give waterwas named hydrogen.
At one time people believed that anysubstance that contained oxygen wouldbe acidic in nature. The Latin word foracid is 'oxy'. Hence the gas was calledoxygen , meaning 'gas that forms acid'. Itwas later discovered that the acidicproperty was not related to oxygen.However, by then the name had come intocommon use so it was not changed.
Place of discovery of element can alsoplay a role in its naming. For example,the gas which was first discovered in thesun (Greek name for Sun is 'helios'), wasnamed helium.
Can you guess the orgin of names ofpolonium and californium?
Sometimes elements were named tohonour the scientists. For example:Einsteinium, Fermium, Rutherfordiumand Mendelevium.
Symbols of elementsYou must have realized that chemistry
involves a lot of reactions. It will be a wasteof time to write the full name of theelements and compounds every time todescribe a reaction. To avoid this we use
Do you know?
John Berzeliussuggested thatinitial letter of anelement written incapitals should
represent that particular element, suchas 'O' for oxygen, 'H' for Hydrogen andso on.
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76 Atoms and Molecules
some shortcuts. Using short forms orsymbols for naming the elements is onesolution.
Over a 100 elements have beendiscovered so far. How do we decide theirsymbols?
Table-2: Symbols for some elements
Element Name Symbol
Hydrogen H
Oxygen O
Nitrogen N
Sulphur S
Carbon C
Calcium Ca
Chlorine Cl
Chromium Cr
Boron B
Barium Ba
Bromine Br
Beryllium Be
Aluminium Al
Iron Fe
Gold Au
Sodium Na
Potassium K
Do you see any problem with thismethod? We have 26 alphabets in Englishbut there are over 100 known elements.How do we write the symbols for calcium,Chlorine, Chromium?
We have already used the letter 'C' forCarbon. Look at the elements after Carbonand before Aluminium in the table.
Discuss with your teacher and friendshow the symbols have been decided forthese elements. Notice the following:
• A symbol can have either one or twoletters of English.
• The first letter of the symbol isalways upper case and the secondletter is always lower case.
Activity-1
Some elements and their possiblesymbols are given in table-3. Correctthem and give reasons for your corrections.
Table-3
Element Possible symbol
Aluminium al
Carbon c
Chromium Chr
Chlorine CL
Beryllium Be
Usually, the first letter of the name ofthe element in English became the symbolof that element and is always written as acapital letter (upper case).
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Some unusual symbolsThis is not the end of the problem. We observe that symbols for some elements come
from their names but some don't. If you are told that certain elements have symbols basedon their Latin names (or older names in other languages).
• Would you be able to guess the elements of the table-2, have symbols of thiscategory?
Activity-2
Write the symbols for given elementsLook up a periodic table and try to find the symbols for the given elements in table 4 and
write them against their names.
Elements with more than oneatom
Several elements have more than oneatom in their smallest constituent particle.Each particle in these elements containstwo atoms combined together to form amolecule. Oxygen, hydrogen and nitrogenare examples of such elements.
For example, a molecule of oxygen hastwo atoms. We need a formula to representsuch a molecule in a simple way. Theformula for oxygen molecule is O2.
Why did not we write it as 2 O ? Writinga formula in this way indicates two separateatoms of oxygen. Hence first write thesymbol for oxygen. and then write 2 as asubscript after the letter O.
Subscript number indicates number ofatoms of Oxygen combined to form itsmolecule.
You may have heard about Ozone gas.This gas is found in large quantities in theupper layers of the earth's atmosphere. Itprotects us by shielding the earth fromsome harmful rays of the sun. Everymolecule of ozone has three atoms ofoxygen. Can you write the formula ofozone?
AtomicityMolecules of many elements, such as
Argon (Ar), Helium(He) etc are made upof only one atom of that element. But thisis not the case with the most of non metals.In non metals the molecules contain morethan two atoms of different elements.
Element Sodium Silver Tungsten Potassium Copper Gold Iron Lead
Other name Natrium Argentum Wolfram Kalium Cuprum Aurum Ferrum Plumbum
Symbol
Table-4
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78 Atoms and Molecules
The number of atoms constituting amolecule is known as its atomicity.
For example, a molecule of hydrogenconsists of two atoms of hydrogen. Herethe atomicity is two; hence it is known asdiatomic molecule. Helium (He), Argon
ValencyThere are over 100 elements known now.
These elements react with each other toform compounds. Every element has adefinite combining capacity, thatdetermines the atomicity of its molecules.Every element reacts with other elementaccording to its combining capacity, whichwe call as its valency.
(Ar) exist as single atom. Hence they areknown as monoatomic.
Observe the following table to knowatomicity of molecules of few elementsand try to write the symbol of moleculebased on its atomicity.
• Why do some elements monatomic?• Why do some elements form
diatomic or triatomic molecules?• Why do elements have different
atomicitiesTo understand the atomicity of
molecules of elements and compounds weneed to understand the concept of valency.
• What is valency?
Name of the element Formula Atomicity
Argon Ar Monoatomic
Helium Monoatomic
Sodium Na Monoatomic
Iron Monoatomic
Aluminum Monoatomic
Copper Monoatomic
Hydrogen H2 Diatomic
Oxygen Diatomic
Nitrogen Diatomic
Chlorine Diatomic
Ozone O3 Triatomic
Phosphorus Tetratomic
Sulphur S8 Octatomic
Table-5
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What is an ion?Compounds formed by metals and non
metals contain charged species. Thecharged species are known as ions. Anegatively charged ion is called anion andthe positive charge ion is cation.
For example sodium chloride do notcontain discrete molecules as theirconstituent units. Its constituent particlesare positively charged sodium ions (Na+)and negatively charged chloride ions (Cl–).
Ions may be a single charged atoms or agroup of atoms (polyatomic) that have a netcharge on them.
So atoms of the elements have power tocombine with atoms of other elements.This is known as its valency.
Element Valency
Hydrogen 1
Carbon 4
Oxygen 2
Chlorine 1
Helium 0
Argon 0
Fluorine 1
Table-6
Table-7: Some common, simple and polyatomic ions
Net Charge Cation Symbol Anion SymbolSodium Na+ Hydride H -
Potassium K+ Chloride Cl -
Silver Ag + Bromide Br -
Copper* Cu+ Iodide I -
Ammonium NH4+ Hydroxide OH -
Nitrate NO3-
Magnesium Mg+2 Oxide O 2-
Calacium Ca+2 Sulphide S 2-
Zinc Zn+2 Sulphate SO42-
Copper* Cu+2 Carbonate CO32-
Iron* Fe+2
Aluminium Al+3 Nitride N 3-
Iron* Fe+3 Phosphate PO43-
1 unit
2 units
3 units
* elements which show variable valency.
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80 Atoms and Molecules
Valency of an ion is equal to themagnitude of its charge. For Examplevalency of chloride ion (Cl–) is 1. Valencyof sulphate ion (SO4
–2) is 2.Now refer the table-7 and try to write
the valencies of some other ions.
Atomic massThe most remarkable concept that
Dalton's atomic theory proposed was thatof atomic mass. According to him eachelement had a characteristic atomic mass.
Since, atoms are extremely light andsmall, scientists find it difficult to measuretheir individual masses. Hence, the massof the atom is compared with a standardatomic mass of other element. In 1961, itwas universally accepted that to choose theatomic mass of carbon-12 as a standardreference for measuring atomic masses ofother elements.
Observe the following diagram (fig-5).Let us assume the circle in the diagramrepresents atomic mass of carbon-12. Itis divided into 12 equal parts as shown inthe figure, and each part represents 1/12of atomic mass of carbon-12.
One atomic mass unit is defined a massexactly one twelfth the atomic mass ofCarbon-12 isotope.
The number of times one atom of givenelement is heavier than 1/12th part ofatomic mass of carbon-12 is called itsatomic mass.
The relative atomic mass of the atom ofan element is defined as the average massof the atom, as compared to 1/12th of themass of one carbon -12 atom.
Atomic mass of an element is a ratio andhas no units but expressed in amu accordingto latest IUPAC recommendations the amuhas been replaced by 'u', which is known asunified mass.
12
34567
8 9 10 1112
Fig - 5
Element Atomic Mass (in u) Element Atomic Mass (in u)Hydrogen 1 Aluminium 27Carbon 12 Phosphorus 31Nitrogen 14 Sulphur 32Oxygen 16 Chlorine 35.5Sodium 23 Potassium 39Magnesium 24 Calcium 40
Table-8: Atomic masses of a few elements
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Do you know?.1. Atomic weights of elements were
determined in the beginning withreference to hydrogen by JohnDalton.While searching various atomic massunits scientists initially took 1/16thof the mass of an atom of naturallyoccurring oxygen as a unit. This wasconsidered relevant due to tworeasons.
• Oxygen reacted with a largenumber of elements and formedcompounds.
• This atomic mass unit gave massesof most of the elements as wholenumbers.
2. During nineteenth century there wereno facilities to determine the massof an atom. Hence, chemistsdetermined the mass of one atomrelative to another by experiments.Today, atomic mass of an atom canbe determined very accurately withthe help of an instrument called massspectrometer.
Molecules of compoundsA molecule is formed by the
combination of different kinds of atomsthat are chemicelly bonded together byattractive forces. For example, a moleculeof water is formed by the combination of
atoms of hydrogen and oxygen. All themolecules of water are identical.
Is it possible for any number of atomsof hydrogen to combine with any numberof atoms of oxygen to form a molecule ofwater?
For all the molecules of water to beidentical, it is essential that the atoms ofhydrogen and oxygen that are present in themolecule must be in fixed numbers. If thisnumber is not fixed, how could all theparticles of water be identical?
Each molecule of water contains 2atoms of hydrogen and 1 atom of oxygen.
Chemical formulae of compoundsWhile writing the formula of a
compound we must keep two things inmind. First, we must see the elementspresent in a molecule of the compound.Second, we must see the number of atomsof each element present in that molecule.2 atoms of hydrogen and one atom ofoxygen are present in a molecule of water,its formula is H2O.
Another rule is that if the molecule of asubstance contains only one atom subscriptneed not be written in the formula.
Now look at another example, amolecule of carbon dioxide contains oneatom of carbon and two atoms of oxygen.carbon and oxygen also react to formanother compound called carbonmonoxide. A molecule of carbon monoxidecontains one atom of carbon and one atomof oxygen.
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82 Atoms and Molecules
• Can you write the formula of carbondioxide and carbon monoxide? Tryto write formula for them as we havedone in case of water molecule.
Let us try to write chemical formulaeby using valency of elements in criss -cross method.
The following steps should be takenwhile attempting to write a chemicalformula. Take sodium carbonate as anexample.
1. Write the symbols of atoms orgroup of atoms side by side, usuallythe cation first -Na CO3
2. Write the valency of each atom orgroup of atoms on the top of itssymbol
Na1 (CO3)2
3. Divide the valency numbers by theirhighest common factor if any to getthe simple ratio. Na1 (CO3)
2
4. Inter change the valency and writethe numbers to the lower right of theconstituents as subscript'. Na2
(CO3)1
5. If any constituent receives thenumber 1, ignore it while writing theformula. Na2CO3
6. If a group of atoms receives thenumber more than 1 encloses it within brackets
Hence the formula for the sodiumcarbonate is Na2CO3.
ExamplesFormula of hydrogen chloride
H1 Cl1 formula: HClFormula of magnesium chloride
Mg2 Cl1 formula: MgCl2
Formula of calcium oxideCa2 O2
Ca1 O1 formula: CaOFormula of aluminium sulphateAl3 (SO4)
2 formula: Al2 (SO4) 3
Table-9: Formulae of some compoundsCompound Formula
Sodium Carbonate Na2CO3
Sodium bicarbonate NaHCO3
Sodium hydroxide NaOHCopper Sulphate CuSO4
Silver Nitrate AgNO3
Hydrochloric Acid HClSulphuric Acid H2SO4
Nitric Acid HNO3
Ammonium Chloride NH4Cl
Molecular massWe have already discussed the concepts
of atomic mass. This concept can beextended to calculate molecular masses.
The molecular mass of substances is thesum of the atomic masses of all the atomsin a molecule of a substance. It is thereforethe relative mass of a molecule expressedin unified mass units.(u)
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Do you know?
For Example: calculate the molecularmass of H2SO4
Solution
2 (atomic mass of hydrogen) + (atomicmass of sulphur) + (4 X atomic mass ofoxygen) = (2X1)+32+(4X16)=98 u
Formula unit mass
A formula unit, as the name implies ,isone unit of atom, ion or moleculecorresponding to a given formula. Oneformula unit of Nacl, means one Na+ ironand one Cl- iron, similarly one formula unitof MgBr2 means one Mg2+ ion and two Br-
ions, and one formula unit of H2o meansone H2O molecule. The formula unit massof a substance is a sum of the atomicmasses of all atoms in a formula unit of acompound. Fermula unit mass is calculatedin the same manner as the molecular mass.The only difference is that formula unit isused for the substances whose constituentsparticles are irons. SodiumChloride has aformula unit Nacl. The formula unit masscan be calculated. as -
1 x 23 + 1 x 35.5 = 58.5(u)
Mole concept
We have learnt that atoms and moleculesare extremely small in size and theirnumber is really very large. Even in smallamount of any substance we find very largenumber of atoms or molecules.
How many molecules are there in 18grams of water?
How many atoms there are in 12 gramsof carbon?
You will be surprised to know that thenumber of particles in 18 grams of waterand 12 grams of carbon is same. Thisnumber is very large. To handle such largenumbers, a unit called mole is defined. Thisis a number quantity.
One mole of any species (atoms,molecules, ions or particles) are thequantity which is expressed in a numberhaving a mass equal to its atomic ormolecular mass in grams.
The number of particles present in onemole of any substance has a fixed value of6.022X1023. This is experimently obtainedvalue. This is called Avogadro constant (NA)named in honour of the Italian scientist,Amedeo Avogadro.
The word "mole" was introduced byWilhelm Ostwald, who derived the termfrom the latin word "moles" meaning a'heap' or 'pile'. A mole substance maybe considered as a heap of atoms ormolecules. The unit mole was acceptedin 1967 to provide a simple way ofreporting a large number-the massiveheap of atoms and molecules in asample.
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84 Atoms and Molecules
Molar massHaving defined mole, it is easier to know
the mass of 1 mole of substance. The massof 1 mole of a substance which is expressedin grams is called its molar mass.
The molar mass is numerically equal toatomic/molecular /formula mass in unifiedmass (u) expressed in grams.
For example molecular mass of water(H2O) =18u.
Molar mass of water= 18 g18 u water has only one molecule of
water. But 18 g water has one molemolecules of water that is 6.022X1023
molecules.
Key words
Law of conservation of mass, Law of constant proportion. Atom, Symbol, Atomicmass, Atomic mass unit (amu), Unified mass (u), Molecule, Molecules of elements,Molecules of compounds, Formula, Ion (cation, anion), Valency, Molecular mass,Formula unit mass, Mole, Avogadro constant, Molar mass.
Fig-6: Diagram on concept of mole
MOLE
1gm of hydrogenatoms
6.022 x 1023
atoms of hydrogen
16gm of oxygenatoms
32gm of oxygenmolecules
6.022 x 1023
atoms of oxygen6.022 x 1023
molecules of oxygen
2gm of hydrogenmolecules
6.022 x 1023
molecules of hydrogen
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What we have learnt
• The total mass of the products formed in a chemical reaction is exactly equal to themass of the reactants. This is known as the law of conservation of mass.
• In a chemical substance the elements are always present in fixed proportions bymass. This is known as the law of constant proportion.
• An atom is the smallest particle of an element that can participate in chemical reactionand retain all its properties.
• A molecule is the smallest particle of an element or a compound that is capable ofindependent existence and retain all the properties of that substance.
• Symbols represents atoms and formula represents molecules and compounds.
• Scientists use the relative atomic mass scale to compare the masses of differentatoms of elements.
• The number of times one atom of a given element is heavier than 1/12th part of massof carbon -12 atom is called its atomic mass.
• By using criss - crosss method we can write the chemical formula of a compound.
• The number of particles present in one mole of any substance is called Avogadroconstant (NA). It is a fixed a value of 6.022 X 1023.
• Mass of one mole of a substance is called its molar mass.
1. Draw the digarm to show the experimental setup for the law of conservation ofmass. (AS5)
2. Explain the process and precautions in verifying law of conservation of mass. (AS1)
3. 15.9g. of copper sulphate and 10.6g of sodium carbonate react together to give14.2g of sodium sulphate and 12.3g of copper carbonate. Which law of chemicalcombination is obeyed? How? (AS1 , AS2)
4. Carbon dioxide is added to 112g of calcium oxide. The product formed is 200g ofcalcium carbonate. Calculate the mass carbon dioxide used. Which law of chemicalcombination will govern your answer.(AS1 , AS2)
Improve your learning
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86 Atoms and Molecules
5. 0.24g sample of compound of oxygen and boron was found by analysis to contain0.144g of oxygen and 0.096g of boron. Calculate the percentage composition ofthe compound by weight. (AS1)
6. In a class, a teacher asked to write the molecular formula of oxygen Shamita wrotethe formula as O2 and Priyanka as O. which one is correct? State the reason.(AS1,AS2)
7. Imagine what would happen if we do not have standard symbols for elements?(AS2)
8. Mohith said "H2 differs from 2H". Justify. (AS1)
9. Lakshmi gives a statement "CO and Co both represents element". Is it correct? Statereason.(AS1,AS2)
10. The formula of water molecule is H2O. What information you get from thisformula.(AS1)
11. How would you write 2 molecules of oxygen and 5 molecules of Nitrogen.(AS1)
12. The formula of a metal oxide is MO. Then write the formula of its chloride.(AS1)
13. Formula of calcium hydroxide is Ca (OH)2 and zinc phosphate is Zn3(PO4)2. Thenwrite the formula to calcium phosphate.(AS1)
14. Find out the chemical names and formulae for the following common householdsubstances. (AS1)
a) common salt b) baking soda c) washing soda d) vinegar
15. Calculate the mass of the following (AS1)
a) 0.5 mole of N2 gas. b) 0.5 mole of N atoms.
c) 3.011 X 1023 number of N atoms. d) 6.022 X 1023 number of N2 molecules.
16. Calculate the number of particles in each of the following (AS1)
a) 46g of Na b) 8g of O2 c) 0.1 mole of hydrogen
17. Convert into mole (AS1)
a) 12g of O2 gas. b) 20g of water. c) 22g of carbon dioxide.
18. Write the valencies of Fe in FeCl2 and FeCl3(AS1)
19. Calculate the molar mass of Sulphuric acid (H2SO4) and glucose (C6H12O6)(AS1)
20. Which has more number of atoms - 100g of sodium or 100g of iron? Justify youranswer. (atomic mass of sodium = 23u, atomic mass of iron = 56u) (AS1)
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21. Complete the following table.(AS1)
22. Fill the following table(AS1)
Sl.No. Name Symbol/formula Molar mass Number of particlespresent in molar mass.
1 Atomic oxygen 16g 6.022 X 1023 atoms ofoxygen
2 Molecular oxygen
3 Sodium
4 Sodium ion 23g
5 Sodium chloride 6.022 X 1023 units ofsodium chloride
6 water
Anions Chloride Hydroxide Nitrate Sulphate Carbonate PhosphateCationsSodium NaClMagnisium MgSO4
CalciumAluminumAmmonium (NH4)3PO4
Group activity (AS4)
1. Make placards with symbols and valencies of the atoms of the elements separately.
Each student should hold two placards, one with the symbol in the right hand and the
other with the valency in the left hand. Keeping the symbols in place, students should
criss-cross their valencies to form the formula of a compound.
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88 Atoms and Molecules
2. Take empty blister packs of medicines. Cut them into pieces having (AS4)
single hallow strip
double hallow strip
triple hallow strip
Divide them into groups according to the valancies. Assume that the number ofhallow rounds of strips represents valency of an ion.
For example strip represents single valency ions like Na+, Cl-, H+ etc.,
Simlarly the remaining strips and represents
double and triple valency ions. (see table 7) Now you can make the formulae by
fixing one type of strip into other.
For example two sodium ion strips (Single hallow strips) can be fixed in one carbonate
ion strip (double hallow strip) like
hence the formula of sodium carbonate will be Na2 CO3
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In previous chapters we have learnt thatall matter is made of atoms. The firstmodern atomic theory was proposed byJohn Dalton. According to Dalton, atomswere indivisible. That means that they couldnot be divided into further smaller parts.Atoms of an element are all identical toeach other and different from the atoms ofother elements. This naturally led scientiststo ask the following questions :
• Why are the atoms of differentelements different?
• Is there anything inside atoms thatmake them to be same or different?
• Are atoms indivisible?
Atoms are too small to be seen withnaked eye. Scientists relied on indirectevidence to prove the existence of atoms.Since they could not see the atoms, theycould find its properties on the basis ofexperiments. Very soon they realized thatatoms could gain or lose charges. Duringelectrolysis experiments, Michael Faraday
Chapter
�WHAT IS INSIDE
THE ATOM?
discovered that atoms were accquiringnegative charge during process ofelectrolysis.
Michael Faraday's discovery raisedfew questions about the indivisibility ofatoms.
How could a neutral atom becomeelectrically charged? It is a contradictionto Dalton's theory that the atom wasindivisible. This led to an idea that theremust exist some tiny particles in atom whichare responsible for atom to behavesometimes as a charged particle. As atomis considered as electrically neutral, itprobably had some positive constituentsand equal number of negative constituentsto maintain its electrical neutrality. Thisgave scope to think about sub-atomicparticles.
Sub-atomic Particles
In science, theories change whenscientists discover new facts or clues.Sometimes, an idea or model must be
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90 What is inside the atom?
changed as new information is gathered.Dalton proposed that atoms could not bedivided. Experimental evidence began toshow that atoms were divisible and aremade up of small particle(s). Since theseparticles are smaller than the atom and arepresent inside an atom, they are calledsub - atomic particles.
Since atoms are neutral they shouldhave at least two types of sub-atomicparticles, one is positively charged andanother is negatively charged. In fact, threedifferent subatomic particles have beendiscovered, the third one is a praticle without any charge. Let us see how ideas aboutatoms have changed over the time with thediscovery of sub-atomic particles.
Electrons, Protons and NeutronsYou read about Faraday's electrolysis
experiments earlier. Other experiments ongases were carried out in the latter part of19th century. Scientists studied the effectsof applying electric current to gases at lowpressure by using discharge tubes. Otherscientists did similar experiments invacuum tubes. In 1897, a British physicistJoseph John Thomson demonstrated on thebasis of these experiments that negativelycharged particles are present in the atoms.
Intially, Thomson thought that thenegative particles would be different foreach element. But after carrying outexperiments with many different materials,he found that the negatively chargedparticles from all the materials were
always indentical. He concluded that sametype of negatively charged particles arepresent in the atoms of all the elements.These particles had a very small mass andare now called electrons.
Electrons were the first sub-atomicparticles discovered and studied. Anelectron is represented as 'e-'. The mass ofan electron is considered to be negligibleand its charge is considered as to be oneunit negative.
Think and Discuss
An atom is electrically neutral. But theelectrons present in it are negativelycharged particles. If only negativecharges were present, the atom would notbe neutral.Then, why do we find atoms as neutral?
The atom must also contain somepositively charged particles so that theoverall charge on it becomes neutral. Thissub-atomic particle would have a chargethat balances the charge of the electrons.This sub-atomic particle was named asproton in 1920. Its mass was approximately2000 times to that of the electron. It isrepresented as 'p+' and its charge is takenas one unit positive.
In 1932, James Chadwick discoveredyet another sub-atomic particle it had nocharge and had a mass nearly equal to thatof a proton. It was eventually named asneutron. In general, a neutron isrepresented as 'n'.
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From the above discussion we canconclude that atoms are made of smallparticles called protons, neutrons, andelectrons. Each of these particles isdescribed in terms of measurableproperties, like mass and charge. Theproton and electron have equal, butopposite, electrical charges. A neutron doesnot have an electrical charge. The mass ofthe electron is about 2000 times less thanthat of a proton or neutron
Fig-1 neutron proton, and electron
• If an atom consist of sub-atomicparticles like protons, neutrons andelectrons, how are they arranged inthe atom?
Let us find,The Structure of an Atom
Activity-1
Sketch the structure of atom asyou imagine.
You learnt about electron, proton andneutron. Suppose you had to arrange themin an atom, how do you do it?
Many arrangements are possible. Thinkthat atom looks like a room. You canarrange the particles in alternating rows.Can you draw and show how it will look?
Imagine and draw another arrangementof subatomic particles inside a sphericalshape keeping in view the nature ofsubatomic particles.
• In how many ways can you arrangethese subatomic particles in aspherical shape?
Discuss with your friends and try toprepare a model to show various ways ofarranging the subatomic particles in theatom.
To understand the structure of an atom,scientist developed different atomicmodels.
Thomson's Model of the AtomThis atomic model was proposed by J.J.
Thomson in 1898. This model wascommonly called the plum puddingmodel, referring to the way the fruit piecesare distributed throughout a plum pudding.According to this model:
• An atom is considered to be asphere of uniform positive chargeand electrons are embedded into it,as shown in figure 2(a).
Fig-2(a) Fig-2(b)
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92 What is inside the atom?
protons and two neutrons bound together.Since they do not have any electrons, theyare positively charged with two units ofcharge. Let us see the experimental set upand understand Rutherford's experiment.
Fig-3
There is a source of the fast movingalpha particles which have a considerableamount of energy. The stream of alphaparticles is directed towards a very thingold foil.
The alpha particle emitter or source andthe gold foil which was placed inside adetector are arranged in such a way thatthe dector would show a flash of light whenan alpha particle struck it. (see fig-3) Theentire arrangement was kept in a vaccumechamber.
Think of Thomson's model of the atom.When the alpha particle hit the foil,Rutherford expected that they all would bedeflected only a little bit by the positivecharge spread evenly throughout the goldatoms. He did not expect to see largedeflections.
• The total mass of the atom isconsidered to be uniformlydistributed through out the atom.
• The negative and the positivecharges are supposed to balance outand the atom as a whole iselectrically neutral.
A more familiar example thatrepresents Thomson's atomic model iswatermelon (figure-2(b)). The positivecharge is spread throughout the atom likethe red part of watermelon. The black seedsdistributed through out the red partrepresents electrons. Thomson's modelwas modified by one of his students. Whatis the reason for its modification? Thereason was that some of the experimentscarried out by Rutherford, a student ofThomson, gave results. Which were notinfavour of Thomson's model.
Besides winning the Nobel Prize inPhysics himself, seven of Thomson'sresearch students and even his own son,George, won Nobel Prizes in Physics.One of Thomson's students was ErnestRutherford.
Rutherford's alpha particlesscattering experiment
Ernest Rutherford was a scientist bornin New Zealand. In 1909, he did someexperiments using gold foil and alphaparticles. Alpha particles consist of two
Do You Know?
stream of alpha particlesalpha particle
source
gold foil
detector
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Thomson's model was assumed that thepositive charge was uniformly distributedthroughout the atom and it was expectedthat all the alpha particles would bedeflected. Since the alpha particles are verybig, the deflection was expected throughsmall angles. But the Rutherford found thatmost of the particles passed through thegold foil like stones thrown to a fence ofbig gaps as mentioned in the aboveexample. This led Rutherford to think aboutnew atomic model.
Rutherford concluded from the alphaparticles scattering experiment that :
(i) Most of the space inside the atomis empty because most of the alphaparticles passed through the gold foilwere deflected to a large extent asshown in figure 4.
(ii) A very small fraction of alpha-particles that were deflected rightback indicated that they had met avery large positive charge and masswhich repelled the charge on thealpha particle. So, all the positivecharge must be concentrated in avery small space within the atom.
On the basis of his experiment,Rutherford put forward the nuclear modelof an atom, which had the followingfeatures:
i) All the positively charged materialin an atom formed a small densecentre, called the nucleus of theatom. The electrons were not a partof nucleus.
Do You Know?
Rutherford's Observations
Fig-4
Scattering of alpha particles
But it was found that most of the alphaparticles passed straight through the atomswithout any deflection Only very fewpraticles were deflected through largeangles and a very, very small number ofpraticals were reflected right back as shownin figure 4.
On average, for every 12000 alphaparticles that were fired at the gold foilduring Rutherford's famous experiment,only one was reflected backward.
Let us try to understand the results ofRutherford's experments.
Suppose you throw a small stone at asolid wall in a horizontal direction. It willnot go through. But if you throw stonesthrough a fence of considerably big gaps,lots of them would pass through the gaps.
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94 What is inside the atom?
ii) He also proposed that the negativelycharged electrons revolve aroundthe nucleus in well-defined orbits.Rutherford's model is sometimesreferred to as the planetary modelbecause the motion of the electronsaround the nucleus resembles themotion of the planets around theSun.
iii) The size of the nucleus is very smallas compared to the size of the atom.
Try to sketch Rutherford's model of theatom.
Think and discuss
Compare Rutherford and Thomson'smodels of the atom on the followingbasis:
• Where the positive charge isplaced?
• How the electrons are placed?• Are they stable inside the atom? Or
moving?
Limitations of Rutherford'satomic model
• Do you see any problems withRutherford's model of the atom?
Fig-5
Think of an atom like hydrogen with oneelectron and one proton. The electron isattracted by the proton in the nucleus. Evenin circular motion around the nucleus, theelectron constantly should lose energy,because any particle in circular motionacquires acceleration.
An accelerated charged particlemoving in a circular path would alwaysradiate energy continuously. Thus therevolving electron would lose energycontinuously and directed towards thepositively charged nucleus as shown in fig-5 and eventually crash into the nucleus.
If this is true, the atoms should behighly unstable and the matter would notexist in the form that we see it now. But weknow that atoms are stable.
So we need to ask: Why is atom stable?
• Can you suggest any otherarrangement of subatomic particilesin the atom which prevents therevolving electron to fall into thenucleus?
In 1913, a Danish scientist Neils Bohrgave a model to explain structure of atomthat overcome Rutherford's limitations.
Bohr's Model of the AtomIn order to overcome the limitations
of Rutherford model, in 1913, Niels Bohrput forward a thought that electrons can befound only in certain energy levels, orregions, around the nucleus. Electrons mustgain energy to move to a higher energy level
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or they must lose energy to move to alower level.
Fig-6 Energy levels of an atom.
Think of books arranged in a bookshelf.They can be placed on a higher shelf orlower shelf but never between shelves.
Restricting the path of electron insidethe atom Neils Bohr made the followingpostulates about the model of an atom:
1. Only certain special, discrete orbitsof electrons are allowed inside theatom. These orbits or shells arecalled energy levels.
2. While revolving in these discreteorbits the electrons do not radiateenergy and this helps that theelectrons do not crash into thenucleus.
3. These orbits or shells arerepresented by the letters K, L, M,N… or the numbers, n=1, 2, 3, …..as shown in the figure 6.
• Do you think that Bohr's model isthe final model of the atom?
Niels Bohr could succefully explainthe properties of a hydrogen atom like theatomic spectra emitted by hydrogen atomemploying this model but this model couldnot predict the spectra of heavier atoms.
You must have noticed that none of theatomic models that we have studied so far,have mentioned about neutrons. This isbecause neutrons were discovered muchlater in 1932. Until Rutherford's time theneutron had not been discovered. Neutronswere discovered nearly two decades later.Except hydrogen atom, the atoms of allother elements contain neutrons in theirnuclei.
The model of the atom, as we know ittoday, is the combined result of the workof many scientists. Let us observe below.
We studied that the mass of neutronsand protons is almost equal and 2000 timesheavier than the mass of the electron. Sothe mass of the entire atom is practicallydue to protons and neutrons alone. Later itwas discovered that most of the mass wasconcentrated inside the nucleus andtherefore the neutrons are also presentinside the nucleus.
Distribution of electrons indifferent orbits (Shells)
According to atomic models, electronsmove around the nucleus of atom inelectron shells. Electrons in differentshells have different energies. Each shellis represented by a number, 'n', which isknown as a shell number or energy levelindex.
1806First Atomic modelJohn Dolton
1886Proton DiscoveryE. Goldstien
1911Nuclius
18501800 1900 1950Electron DiscoveryJ.J. Thomson (1897-1904)
1932Discoveryof Nutron
1913Bohr'sModel
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96 What is inside the atom?
Rule 2: Each energy level or electron shellis further divided into sub shells. Themaximum number of electrons that can beaccommodated in each sub shell is 8.
Rule 3: Electrons cannot be filled in agiven shell unless the inner shells arecompletely filled i.e., shells are filled instepwise manner.
Let us take the example of oxygen whereZ=8. Since number of electrons is equalto number of protons, it has eight electrons.
Step1. The K shell can accomdatesmaximum 2 electrons so the first 2electrons fill the shell of n = 1.
Step 2. The other 6 electrons will fill thehigher shell n = 2 or the L shell.
Step 3. Then, the electronic structure foroxygen atom is 2, 6.
The shell closest to the nucleus (andhas the lowest energy) is called the K-shell (n = 1), the shell farther away (andhas higher energy than K-shell) is calledthe L-shell (n = 2), etc.
• How many electrons can beaccommodated in each shell of anatom?
• Can a particular shell has just oneelectron?
• What is the criteria to decidenumber of electrons in a shell?
After explaining the structure of atomwith different atomic models scientistsstarted describing the distribution ofelectron in different energy levels or shellsof an atom. Bohr and Bury proposed thefollowing rules for electron distribution.
Rule 1: The maximum number of electronspresent in a shell is given by the formula2n2, where 'n' is the shell number or energylevel index, which takes values 1, 2, 3….The maximum number of electrons that canbe accommodated in each shell is shownin the table 1.
Shell number (n) The maximum numberof electrons in a shell
1 K-shell 2 (1)2 = 2
2 L-shell 2 (2)2 = 8
3 M-shell 2 (3)2 =18
4 N-shell 2 (4)2 =32
Table - 1
Fig-7 Arrangement of electrons for the first eighteen elements
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Arrangement of electrons for thefirsteighteen elements is shownschematically in figure 7.
ValencyWe have learnt arrangement of the
electrons in an atom in different shells/orbits.
Fig-8
Let us consider a Carbon atom. Theatomic number of Carbon atom is 6. Henceit possesses 6 electrons, which surroundits nucleus, as shown in the figure 8.
According to Bohr -Bury rule, thereshould be two electrons in the innermostshell (n=1). Out of 6 electrons of carbontwo electrons occupy first shell (n=1). Theremaining four electrons occupy the outermost shell n=2. The electrons present inthe outermost shell of an atom are knownas the valence electrons. Therefore, thenumber of electrons present in outermostorbit of an atom is called its valency.Valency of an atom explains the combiningcapacity of the element with other element.In above example the valency of carbonatom is 4.
Let us consider some more examples,if you consider the atoms like hydrogen/lithium/sodium, they contain one electronin each of their outermost shell, thereforevalency of hydrogen, lithium and sodiumis one.Can you tell, what is the valency ofmagnesium and aluminium? It is two andthree, respectively, because magnesium hastwo electrons in its outermost shell andaluminium has three electrons in itsoutermost shell.
If the number of electrons in the outershell of an atom is close to its full capacity,then valency is determined in a differentway. For example, the fluorine atomcontains 7 electrons in the outermost shell,and its valency could be 7. But it is easierfor fluorine to gain one electron than toloose seven electrons for obtaining 8electrons in its outer most shell. Hence,its valency is determined by subtractingseven electrons from eight and which givesyou a valency of '1', for fluorine. In a similarmanner we can calculate the valency ofoxygen.
• What is the valency of oxygen thatyou can calculate by the methoddiscussed above?
Think and discuss
• Phosphourus and sulphur showmultiple valency. See table 2 Whysome elements show multipleValency? Discuss with your friendsand Teachers.
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98 What is inside the atom?
K L M NHe 2Ne 2 8Ar 2 8 8
Observe the following Table . Valency of the first eighteen elements is given in the lastcolumn of Table 2.
What is the importance ofvalency?
See electron distribution in helium infigure 7 and table '2'. You will notice thatthe shell of helium has two electrons inouter most shell and the shell is filled toits full capacity. Neon and Argon have 8electrons in their outer most shells. Allthree gases are very stable and have lowreactivity. Scientists, studying thedistribution of electrons in different shellsconcluded that the special arrangementsof electrons in He, Ne and Ar makes themstable or unwilling to mix with otherelements. They do not react with otherelements to form compounds. In otherwords we can say that these gases arechemically inactive and also known as inertor noble gases.
Atoms of all the noble gasses except
those of helium have 8 electrons' in theirouter most shell. Thus an atom with eightelectrons or an octet in their outer mostshell, is chemically stable, or does notcombine with other atoms. An atom withtwo electrons in its outer most shell, alsois more stable when there is only one shellpresent in it.
An outermost-shell which has eightelectrons is said to possess an octet. Atomsof an element thus react with other atoms,so as to achieve an octet in their outermost
Name of element Symbol Atomic Number of Number of Number of Distribution of electrons Valencynumber Protons Neutrons Electrons K L M N
Hydrogen H 1 1 - 1 1 - - - 1
Helium He 2 2 2 2 2 - - - 0
Lithium Li 3 3 4 3 2 1 - - 1Berilium Be 4 4 5 4 2 2 - - 2
Boran B 5 5 6 5 2 3 - - 3
Carbon C 6 6 6 6 2 4 - - 4
Nitrogen N 7 7 7 7 2 5 - - 3Oxygen O 8 8 8 8 2 6 - - 2
Fluorine F 9 9 10 9 2 7 - - 1
Neon Ne 10 10 10 10 2 8 - - 0
Sodium Na 11 11 12 11 2 8 1 - 1Magnesium Mg 12 12 12 12 2 8 2 - 2
Aluminium Al 13 13 14 13 2 8 3 - 3
Silicon Si 14 14 14 14 2 8 4 - 4
Phosphorus P 15 15 16 15 2 8 5 - 5,3Sulphur S 16 16 16 16 2 8 6 - 2,6
Chlorine Cl 17 17 18 17 2 8 7 - 1
Argon Ar 18 18 22 18 2 8 8 - 0
Table 2.
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Andhra Pradesh Government Free Distribution 99
nucleus is denoted as Z (atomic number)and number of neutrons of a nucleus isdenoted by N.
The number of nucleons, i.e. the totalnumber of protons and neutrons, is calledthe atomic mass number and is denoted bythe letter A.Atomic mass number = atomic number +neutron number
A = Z + N• Mass number is a nearest
numerical to the mass of anindividual atom.
• Mass number is the number ofprotons plus the number ofneutrons.
Writing symbols of atomsIn standard notation to represent an
atom, the atomic number, mass number andsymbol of the element are written as:
shell. From the above discussion, we canconclude that when an element reacts toform compounds their atoms must becombing in such way that they can attainthe stable electron distribution of noblegases or inert gases.
An atom can achieve an octet by twoways, one by transfer of electrons andother by sharing of electrons. Both theprocesses result in the formation of bondsbetween atoms.
Let us go back to the question of whydo atoms of different elements aredifferent. How can you distinguish betweenthe atoms of one element from the atomsof other element? An element can berecognised by certain characteristics of itsatoms.Atomic number
We know that, the nucleus is at thecentre of the atom and contains the protonsand neutrons. Elements differ from oneanother according to the number of protonsin their atoms 'nuclei'. This value is calledthe element's atomic number and is denotedby the letter Z.Atomic number is the number ofprotons in the nucleus of an atom
Atomic mass number• Should we consider the number of
neutrons a characteristic of an atom?The mass of an atom which is a
characteristic of an atom depends on thenumber of neutrons and protons that it'snucleus contains. Number of protons in a
F is the symbol of the elementfluorine, its atomic number is written atbottom left. It tells us that this atom has 9protons. The mass number is written at topleft. It tells us that this fluorine has 19nucleons (protons + neutrons).
Therefore the number of neutronspresent in fluorine is equal to 19 - 9 = 10neutrons. (N = A - Z).
atomic mass number
atomic number
number of nucieons
number of nucieons
199 F�
�
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100 What is inside the atom?
IsotopesIt must be clear to you that every
element has a unique atomic number, ornumber of protons.
What about mass number? Does everyelement has a unique mass number, whichis different from the mass number of otherelements?
No, the mass number of an element isnot unique because there are more than onetype of atoms of the same element presentin nature in certain cases. Observe thefollowing figure of different Hydrogenatoms? What do you understand?
Fig-9
An atom of hydrogen has one neucleonin its nucleus, an atom of deuterium has twoneucleons in its nucleus, and tritium hasthree. Since atoms of hydrogen, deuteriumand tritium have only one proton in theirnuclei, they only have one electron. Butnumber of neutrons present in hydrogenatom is not same in all cases.
The atoms of the same element whichhave the same number of protons but havedifferent number of neutrons are calledisotopes. Deuterium and tritium areisotopes of hydrogen. The chemicalproperties of isotopes are similar. But theirphysical properties are different.
For Example: Carbon has three stableisotopes. Isotopes can also be representedby their element name followed by the
mass number. See following notations.
Carbon-12, carbon-13, & carbon-14
How do we determine the atomic mass ofan element with isotopes?
In nature, most elements occur as amixture of two or more isotopes, eachisotope has a certain percentage of naturaloccurrence.
For example, consider the isotopes ofChlorine. It occurs in nature in twoisotopic forms, with masses 35 units and37 units. The isotope with mass 35 ispresent in 75% and isotope with mass 37is present in 25% in nature
The atomic mass of an element is takenas an average mass of all the naturallyoccurring atoms of the sample element.
The average atomic mass of chlorineatom, on the basis of above data, will be
126C
136C
146C
Did You Know?Two elements share the record for
the highest number of known isotopes.Both xenon and cesium have 36isotopes.
(35 x + 37 x )75100
55100
= ( + ) = = 35.5u1054
374
1424
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Andhra Pradesh Government Free Distribution 101
Applications of isotopesSome isotopes are used for solving
chemical and medical mysteries. Isotopesare also commonly used in the laboratoryto investigate the steps of a chemicalreaction.
i) The isotope of uranium is used as afuel in nuclear reactors.
ii) The isotope of iodine is used in thetreatment of goitre.
iii) The is otope of cobalt is used in thetreatment of cancer.
Key wordsAtom, sub atomic particle, electron, proton, neutron, nucleus, atomic mass,
molecular mass, formula unit mass, atomic number (Z), mass number (A)valency, isotopes
• An atom is the smallest particle of an element that retains the identity of theelement.
• John Dalton's atomic theory described elements in terms of atoms, which hebelieved to be small, indivisible particles that make up all matter. He stated thatall the atoms of the same element are identical in mass and size, but atoms ofdifferent elements are different.
• The three sub-atomic particles of an atom are: (i) electrons, (ii) protons and(iii) neutrons.
• Electron is a negatively charged particle of the atom
• Proton a positively charged particle that is part of atomic nucleus
• Neutron is an uncharged particle that is part of atomic nucleus
• Credit for the discovery of electron and neutron goes to J.J. Thomson and J.Chadwick respectively.
• Thomson determined that atoms contain negatively charged particles, whichare now called electrons. He developed a model of the atom that shows electronsembedded throughout the mass of positively charged material.
• Rutherford's alpha-particle scattering experiment led to the discovery of theatomic nucleus.
• Ernest Rutherford's model of the atom has large empty space, with a small,dense, positively charged nucleus in the centre.
What we have learnt
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102 What is inside the atom?
• Neils Bohr modified Rutherford's model of the atom by stating that electrons movein specific energy levels around the nucleus.
• The atomic number of an element is the same as the number of protons in the nucleusof its atom.
• The mass number of an atom is equal to the number of nucleons in its nucleus.• Valency is the combining capacity of an atom.• An atom with eight electrons or an octet in their outer most shell is chemically
stable, or does not combine with other atoms.• Isotopes are atoms which have the same number of protons, but a different number
of neutrons
1. What are the three subatomic particles? (AS1)2. Compare the characteristics of electrons, protons and neutrons.(AS1)3. What are the limitations of J.J. Thomson's model of the atom?(AS1)4. What were the three major observations Rutherford made in the gold foil
experiment?(AS1)5. Sketch Rutherford's atomic model. Why Rutherford's model of the atom is called
the planetary model?(AS5)6. Put tick (�) against correct choice and cross (×) against wrong choice: (AS1)
i) In Rutherford's gold foil experiment, majority of alpha particles passed directlythrough the gold foil. This observation leads to which conclusion?a) The positive region of the atom is very small.b) The majority of the atom must consist of empty space.c) The alpha particle makes a direct hit on the positive region.d) The positive region of the atom is very dense.
ii) In Rutherford's gold foil experiment, occasionally the alpha particle veered froma straight-line path. This observation leads to which conclusion?a) The positive region of the atom is very small.b) The majority of the atom must consist of empty space.c) The alpha particle makes a direct hit on the positive region.d) The positive region of the atom is very dense.
Improve your learning
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Andhra Pradesh Government Free Distribution 103
7. Which one of the following is a correct electronic configuration of sodium?(AS1)(a) 2,8 (b) 8,2,1 (c) 2,1,8 (d) 2,8,1.
8. Give the main postulates of Bohr's model of an atom. (AS1)9. Compare all the proposed models of an atom given in this chapter.(AS1)10. Define valency by taking examples of nitrogen and boron.(AS1)11. State the valencies of the following elements : Magnesium and Sodium (AS1)12. If Z = 5, what would be the valency of the element? (AS2)13. Write the atomic number and the symbol of an element which has mass number 32
and the number of neutrons 16 in the nucleus.(AS1)14. Cl- has completely filled K&L shells. Explain. (AS1)15. What is the main difference between isotopes of the same element?(AS1)16. For the following statements, write T for True and F for False.(AS1)
a. J.J. Thomson proposed that the nucleus of an atom contains only nucleons.b. A neutron is formed by an electron and a proton combining together. Therefore,
it is neutral.c. The mass of an electron is 1/2000 times that of proton.
17. Fill in the missing information in the following table.(AS4)
18. How do you appreciate the efforts made by scientists to explain the structure ofatom by developing various atomic models?(AS6)
19. Geeta got a doubt, "Why atomic nucleus contains proton and neutrons inside it?Why can't electrons and neutrons?" Can you help to clarify her doubt? Explain.(AS1)
20. Collect information about various experiments conducted and theories proposed byscientists starting from John Dalton to Niels Bohr and prepare a story with a title"History of atom".(AS4)
Name Symbol Atomic Mass Number of Number ofNumber Z Number A Neutrons Electrons
Oxygen 16O8 8 16 8 87 7
34SBeryllium 9
12 24
12 25
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104 Gravitation
We have learnt about uniformaccelerated motion in the chapter 'motion'.In this chapter let us study about uniformcircular motion which is an example of non-unifom accelerated motion.
We always observe that an objectdropped from certain height falls towardsthe earth. We know that all planets movearound the sun. We also know that the moonmoves around the earth. In all these casesthere must be some force acting on theseobjects to make them move around anotherobject, instead of moving in a straight line.
• What is that force?• Is the motion of the earth around the
sun uniform motion?• Is the motion of the moon around
the earth uniform motion?Newton explained the motion of moon
by using the concept of uniform circularmotion and then he developed the idea ofgravitation between any two masses.
In this chapter you will learn aboutgravitation and centre of gravity.
Uniform circular motionActivity-1
Observing the motion of anobject moving in a circular path
Take an electric motor and fix a disc tothe shaft of the electric motor. Place asmall wooden block on the disc at the edgeas shown in figure 1 (a). Switch on themotor. Find the time required to completeten revolutions by the block and repeat thesame two to three times. Begin countingof revolutions after few seconds of startof motor.
• Is the time of revolution constant?• Is the speed of the block constant?
Chapter
� GRAVITATION
CircularPlate
WoodenBlock
Electric Motor
Battery (a) (b)Fig-1 (a) motion of wooden block on a
circular plate (b) top view of wooden block
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Andhra Pradesh Government Free Distribution 105
• What is the shape of path?The wooden block moves in a circular
path with a constant speed. So this motionof wooden block is called uniform circularmotion.
"Uniform circular motion is a motionof the body with a constant speed incircular path"
• Does the velocity of the bodychange in uniform circular motion?Why?
• Does the body in uniform circularmotion have an acceleration? Whatis the direction of acceleration?
Activity-2Drawing velocity vectors for abody in uniform circular motion
Recall the motion and path taken by thewooden block in the above activity. Drawthe path of the wooden block and drawvelocity vectors at successive timeintervals as shown in the figure 2.
Use figure 2 and transfer tails of eachvelocity vector to coincide at a single pointas shown in figure 3(a) without changingtheir directions. Figure 3(a) represents thevelocity vectors of figure 2 and the directed
line joining two vectors represents changein velocity (�V).
Let a body move with a constant speed vin a circular path of radius 'R'. The velocityvector changes direction and appears torotate. If a velocity vector is rotatedthrough a small angle, the change invelocity (�V) will be represented by thebase of the isosceles triangle as shown infigure 3(a).
Let us consider the change in velocityduring the course of a complete revolutionof a body; the sum of the magnitudes ofthe changes in velocity during a completerevolution will be equal to the sum of thesides of the depicted polygon. But thedirection of velocity is changingcontinuously.
It is perfectly clear that the smaller wetake vertex angles of the small triangles,the less will be our error. The smaller thesides of our polygon , the closer they clingto the circle of radius v as shown in figure3(b) Consequently, the exact value of the
VV
V
VV
V
V
V
R
Fig-2: Velocity vectors atdifferent points
Fig-3(a) & (b): Transformed velocity vectors
3(a)V
�V
V
VVV
V
VV �V
�V
�V�V
�V
�V
�V
3(b)V
V�V
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106 Gravitation
Think and discuss
sum of the magnitudes of the changes invelocity of the body during the course ofrevolution, will be equal to thecircumference "2 v" of the circle.
We know that the magnitude of theacceleration is equal to the ratio ofmagnitude of change in velocity for onerevolution and time period.
Let aC be magnitude of acceleration ofthe body in uniform circular motion.
That is, aC=2 v/TWhere 'T' is time required to complete
revolution.We knew that T= 2 R/vSubstituting this expression in
preceding formula, we get, aC =v2/RAs the vertex angle of isosceles
triangles decreases, the angle between thechange in velocity and velocity vectorapproaches 900.
Therefore, the acceleration of a body inuniform circular motion is directedperpendicular to its velocity. But how arethe velocity and acceleration directedrelative to the path? Since the velocity istangent to the path; the acceleration of thebody is directed along the radius towardsthe centre of the circle.
The acceleration which can change onlythe direction of velocity of a body is called"centripetal acceleration".
Newton's second law says that net forceon a moving body produces an accelerationin it which is directed along the net force.
So in uniform circular motion, a net
force acts towards the centre. This net forceis called "Centripetal force".
The net force which can change only thedirection of the velocity of a body is called"centripetal force".
Let us find the value of centripetal force.
According to the Newton's second lawof motion,
Fnet = (mass) (acceleration)
FC = maC
FC = mv2/R (Since aC = v2/R)Where R is the radius of circle.In uniform circular motion, 'FC' always
directs towards the centre.Note: Centripetal force is a net force
directed towards the centre of thecircle.
• Can an object move along a curvedpath if no force acts on it?
• As a car speeds up when roundinga curve, does its centripetalacceleration increase? Use anequation to defend your answer.
• Calculate the tension in a string thatwhirls a 2 kg - toy in a horizontalcircle of radius 2.5 m when itmoves at 3m/s.
V V
V
aCaC
aC
Fig-4
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Andhra Pradesh Government Free Distribution 107
Universal law of gravitationOnce while Isaac Newton was sitting
under a tree, an apple fell to the ground.• Do you know what questions arose
in his mind from this observation?• Why did the apple fall to the ground?• Why does the moon not fall to the
ground?• What makes the moon to move in a
circular orbit around the earth?
Newton knew that the motion of moonaround the earth is approximately uniformcircular motion. So certain net force, whichwe call centripetal force, is required tomaintain the uniform circular motion.
So he introduced the idea of force ofattraction between the moon and earth. Heproposed that earth attracts moon andtermed it as gravitational force. Thisgravitational force acts as a centripetalforce and makes the moon to revolvearound the earth in uniform circularmotion. Newton knew the following data.
The distance of the moon from center ofthe earth is 384 400 km = 3.844x1010 cm.The moon takes 27.3 days or 2.35x106 sfor a complete revolution around the earth.
• What is the speed of the moon?You can calculate the speed of the moon
using the equation,v = 2 R/T
Thus the acceleration of the moontowards the centre of the earth
am = v2/R=4 2R/T2
Substituting the values of R and T inabove equation we can get
am = 0.27 cm/s2.Galileo found that the acceleration of
bodies acquired near the surface of earthis equal to 981cm/s2. Thus acceleration ofapple approximately is equal to 981cm/s2.
He compared the both the accelerationof apple, ae to the acceleration of the moon,am. We get, ae / am=981/0.27 � 3640.
Newton knew that the radius of Earth,Re and the distance of the moon from thecentre of the Earth, Rm are 6371 km and3,84,400 km respectively. We get
Rm/ Re =384400/6371 � 60.3(Rm/ Re )
2 =(60.3)2 ��� 3640From above discussion it is clear that
ae/ am = (Rm / Re )2
So we get, Acceleration �1/ (Distance)2
a � 1/R2 ----- (1)Force of attraction �1/(Distance)2
F � 1/R2 ----- (2)
V
Appleam = g
aC = am
Moon
Rm
Re
Fig-5: Comparing the motions of themoon and apple
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108 Gravitation
Thus it became clear that the force ofgravity decreases with increase in distanceof the object from the center of the earth.
According to the Newton's third lawforce on the apple by the earth is equal toforce on the earth by the apple. We get theforce on the object by the earth by usingsecond law of motion and equation-1.
From Newton's second law of motionF = ma, and from equation-1, a � 1/R2
a = k/R2 (where k is proportionalityconstant)
Thus we get, F = km/R2
Therefore the force on the apple by theearth = Km/R2 ---- (3)
Where 'm' is the mass of apple and 'R' isthe radius of the earth.
Force on the earth by the apple = K'M/R2
---- (4)
Where M is the mass of the earth.
The above forces are equal in magnitudeonly when the following condition issatisfied.
K=GM and K' = Gm ---- (5)
From equations (3) & (5) we have forceon apple by the earth, F = GMm/R2
We conclude that gravitational forcebetween the masses is directly proportionalto product of their masses.
Force of attraction ��� (mass)1 (mass)2
Newton generalized the force ofgravitation and said it acts on all bodies in
the universe. The universal law ofgravitation states that every body in theuniverse attracts other body with a forcewhich is directly proportional to theproduct of their masses and inverselyproportional to the square of the distancebetween them. The direction of the forceof attraction is along the line joining thecenters of the two bodies.
Let two bodies of masses M1 and M2 beseparated by a distance of 'd'. Then theforce of gravitation between them
Fgrav � M1 M2 /d2
Fgrav =G M1 M2 /d2
G is a proportionality constant, calleduniversal gravitational constant and foundby Henry Cavendish to be
G=6.67x10-11 Nm2Kg-2
The value of G is equal to the magnitudeof force between a pair of 1- kg masses thatare 1 metre apart.Note: This formula is applicable tospherical bodies. We use the above formulafor all bodies on the earth though they arenot spherical because, when compared tothe earth, any object on earth is very smalland it is assumed as a point object.
M1 M2Fgrav Fgrav
d
Fig-6
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Andhra Pradesh Government Free Distribution 109
Think and discuss• In figure 7, we see that the moon
'falls' around earth rather thanstraight into it. If the magnitude ofvelocity were zero, how would itmove?
Fig-7• According to the equation for
gravitational force, what happens tothe force between two bodies if themass of one of the bodies doubled?
• If there is an attractive forcebetween all objects, why do we notfeel ourselves gravitating towardmassive buildings in our vicinity?
• Is the force of gravity stronger on apiece of iron than on a piece of woodif both have the same mass?
• An apple falls because of thegravitational attraction of earth.What is the gravitational attractionof apple on the earth? Why?
Example 1What is the time
period of satellitenear the earth surfaceneglect height of theorbit of satellite fromthe surface ofground?
SolutionThe force on the satellite due to earth is
given by F =G m M /R2
M-Mass of earth,m-mass of satellite,R-radius of earth.Let v be the speed of the satellite
v=2 R/T T=2 R/vRequired centripetal force is provided
to satellite by the gravitational force henceFC=m v2/R.
But FC=GMm/R2 according to Newton'slaw of gravitation.
i.e., GMm/R2 = m(2 �R)2/T2R T2 = 4 2 R3/GM, as mass of the earth
(M) and G are constants the value of Tdepends only on the radius of the earth. T2 ��R3
Substituting the values of M, R and G inabove equation we get, T = 84.75 minutes.
Thus the satellite revolving around theearth in a circular path near to the earth'ssurface takes 1Hour and 24.7 minutesapproximately to complete one revolutionaround earth.
Free fall
Activity-3
Acceleration is independent ofmasses
Place a small paper on a book. Releasethe book with the paper from certain heightfrom the ground.
• What is your observation? Now
V m
M
Fig-8
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110 Gravitation
drop the book and paper separately,what happens?
A body is said to be free fall body whenonly one gravitational force acts on thatbody.
Fig-9
Let us drop a body of mass m near theearth's surface.
Let M be the mass of the earth and R bethe radius of the earth.
Now the force of attraction on the massis given by,
F = GMm/R2 F/m = GM/ R2
From Newton's second law, F/m is equalto acceleration. Here this acceleration isdenoted by 'g'
Hence, g = GM/ R2
from above equation you can concludethat 'g' is the independent of the bodiesmass.
If there were no air friction orresistance, all the bodies fall with sameacceleration. This acceleration, produceddue to gravitational force of the earth nearthe surface, is called free- fall acceleration.
Mass of the earth (M) = 6 x 1024 kgRadius of earth (R) = 6.4 x 106 kmPutting these values in the above
equation.We get g = 9.8 m/s2 (approximately)In general, this value of acceleration due
to gravity changes due to change in
m
gR
M
distances of objects from the center of theearth.
Since free - fall acceleration is constantnear the ground, the equations of uniformaccelerated motion can be used for the caseof free-fall body.
v = u + at,s = ut + ½ at2 ,v2 - u2 = 2as.While solving problems related to Free
fall objects we use 'g' instead of 'a' in aboveequations.
When we use these equations, you mustfollow the sign convention ( It is discussedin the chapter "motion")
Activity-4
What is the direction of 'g'Throw a stone vertically up. Measure the
time required for it to come back to earth'ssurface with stop clock.
• What happens to speed when itmoves up and down?
• What is the direction ofacceleration?
When a stone moves up, the speeddereases. When a stone moves down, thespeed increases. So the free-fallacceleration is vertically downwards. Nomatter how you throw objects, the " g " isdirected vertically down as shown infigure10. g
surface of the earth
Fig-10
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Andhra Pradesh Government Free Distribution 111
Think and discuss
+
_
u
hg
• Give an example for the motion ofan object of zero speed and withnon zero acceleration?
• Two stones are thrown into air withspeeds 20 m/s, 40m/srespectively? What areaccelerations possessed by theobjects?
Example 2A body is projected vertically up. What
is the distance covered in its last secondof upward motion? (g = 10 m/s2)
SolutionThe distance covered by the object in its
last second of its upward motion is equalto the distance covered in the first secondof its downward motion.
Hence s = ½ gt2= ½ x 10 x 1 = 5 mExample 3Two bodies fall freely from different
heights and reach the groundsimultaneously. The time of descent forthe first body is t1 = 2s and for the secondt2 = 1s. At what height was the first bodysituated when the other began to fall?( g = 10 m/s2 )
at t=1sh1=20m
h2=5m
SolutionThe second body takes 1 second to reach
ground. So, we need to find the distancetraveled by the first body in its first secondand in two seconds.
The distance covered by first body in 2s,h1 = ½ gt2 = ½ × 10 × 22 = 20m.The distance covered in 1s, h2 = 5 m.The height of the first body when the
other begin to fall h = 20 - 5 = 15m.Example 4A stone is thrown vertically up from the
tower of height 25m with a speed of 20 m/s.What time does it take to reach the ground?( g = 10 m/s2 )
Fig-12SolutionSign convention must be used to solve
this problem. It is shown in figure.We consider the upward direction as
positive and downward direction as negativewith respect to a point of reference. In theabove example the point of projection isconsidered as the point of reference.Fig-11
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112 Gravitation
Then, u= 20 m/sa = g = -10 m/s2
s = h = -25 mFrom equation of motion s = ut + ½ at2
- 25 = 20t - ½ x 10 x t2
-25 = 20t - 5 t2
-5 = 4t - t2
t2 - 4t - 5 = 0Solving this equation,We get, ( t - 5 ) ( t + 1 ) = 0
t= 5 or -1t = 5s
Example 5Find the time taken, by the body
projected vertically up with a speed of u,to return back to the ground.
SolutionLet us take the equation S = ut + ½ a t2
For entire motion S = 0
Fig-13a = - gu = uO = ut - ½gt2
½gt2 = utt = 2u/g
gu
up down
maximumheight point
WeightWeight of a body is the force of
attraction on the body due to earth.So, from Newton's second law of
motion.Fnet = maWe get,W = mgIt is measured in newtons1 kg body weighs 9.8 N2 kg body weighs 19.6 N10kg body weighs 98 N
Activity-5
Can we measure the weight offree-fall body
Let us find,
Fig-14 (a) Fig-14 (b)
Take a spring balance and suspend it tothe ceiling and put some weight to it. Notethe reading of the spring balance. Now dropthe spring balance with load from certainheight to fall freely. Carefully observe thechange in the position of indicator on thespring balance scale while it is in free-fall.
g
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Think and discuss
• What changes do you notice in thereadings of spring balance in abovetwo instances?
• Are they same? If not why?Some of you might have the experience
of diving into a swimming pool fromcertain height.
• How do you feel during the free-fallof your body from the height?
Activity-6
Observing the changes duringthe free-fall of a body
Take a transparent tray and make holeson opposite sides. Take two or three rubberbands and tie them tightly, close to eachother between the holes. Now place a stoneon the bands as shown in the figures 15(a)and 15(b).
• Do the bands bend? Now drop thetray with stone. Now what happens?
We get the following results in free fall.In spring - mass activity. The reading
becomes zero. In jumping, the man feelsweightlessness. In the activity-6, the bandsare straight. No stretch occurs in the rubberbands.
We treated the weight of an object asthe force due to gravity upon it. When inequilibrium on a firm surface, weight is
g
evidenced by a support force or when insuspension, by a supporting tension. Ineither case with no acceleration, weightequals mg. A support force can occurwithout regard to gravity. So broaderdefinition of the weight of something is theforce it exerts against a supporting floor.
When the body falls freely then itexperiences weightlessness. Even in thisweightless condition, there is still agravitational force acting on the body,causing downward acceleration. Butgravity now is not felt as weight becausethere is no support force.
• When is your weight equal to mg?• Give example of when your weight
is zero?
Centre of gravity
Activity-7Balancing of spoon and fork
Fig-16
Fig-15 (a) Fig-15 (b)
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114 Gravitation
Fasten a fork, spoon, and wooden matchstick together as shown. The combinationwill balance nicely - on the edge of theglass. Why?
Acivity-8
Can you get up without bending
Sit in a chair comfortably as shown infig-17. Try to get up from the chair withoutbending your body or legs.
• Are we able to do so? If not why?
Activity-9
Balancing a ladderTry to balance ladder on your shoulder?When does it happen?We need to introduce the idea of
"Centre of gravity" to understand the aboveactivities.
The center of gravity is simply theaverage position of weight distribution.
The point where total weight appears to actis called centre of gravity.
Activity-10
Locating centre of gravityTake a meter scale. Suspend it from
various points. What do you notice?Suspend it from its mid point. Whathappens?
The center of gravity of a uniform object,such as meter stick, is at its midpoint. Thestick behaves as if its entire weight wasconcentrated at that point. The supportgiven to that single point gives support tothe entire stick. Balancing an objectprovides a simple method of locating itscentre of gravity. The many small arrowsrepresent the pull of gravity all along themeter stick. All of these can be combinedinto resultant force acting through thecentre of gravity.
The entire weight of the stick may bethought of as acting at this single point.Hence we can balance the stick by applyinga single upward force in a direction thatpasses through this point.
Banalced Upwardforce due to finger
Weight
Fig-17
Fig-18
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• How to find the center of gravity ofan object?
The center of gravity of any freelysuspended object lies directly beneath thepoint of suspension.
If a vertical line is drawn through thepoint of suspension, the center of gravitylies somewhere along that line. Todetermine exactly where it lies along theline, we have only to suspend the objectfrom the some other point and draw asecond vertical line through that point ofsuspension. The center of gravity lieswhere the two lines intersect.
Activity-11
Identifying the center of gravityof a ring
Find the centre of gravity of ring usingthe method explained in the above example.
• Where does the center of gravity ofa ring lie?
• Does the center of gravity of a bodyexist outside the body?
• Does center of gravity of an objectexist at a point where there is nomass of the object?
StabilityThe location of the centre of gravity is
important for stability. If we draw a linestraight down from the centre of gravity ofan object of any shape and it falls insidethe base of the object, then the object willstable.
If the line through the center of gravityfalls outside the base then the object willbe unstable.
Activity-12
Shift of the center of gravity andits effects
When you stand erect, where is yourcentre of gravity?
Try to touch your toes as shown in figure20 (a). Try this gain when standing againsta wall as shown in figure 20 (b).
• Are you able to touch your toes insecond case as shown in fig-20(b)?If not why?
• What difference do you notice in thecenter of gravity of your body in theabove two positions?
Fig-19
Fig-20 (a) Fig-20 (b)
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116 Gravitation
Think and discuss
Key words
• Where does the center gravity of a sphere and triangular lamina lie?• Can an object have more than one centre of gravity?• Why doesn't the leaning tower of Pisa topple over?• Why must you bend forward when carrying a heavy load on your back?
Uniform circular motion, centripetal acceleration centripetal force, centre ofgravity, law of gravitation, weight, weightlessness, stability, free fall.
• Motion of body with constant speed in a circular path is called uniform circularmotion .
• The acceleration which causes changes only in direction of the velocity of a body iscalled centripetal acceleration and it is always directed towards the center of thecircle.
• The net force required to keep a body in uniform circular motion is called "Centripetalforce". FC = Mv2 / R.
• Every object in the universe attracts other bodies. The force of attraction betweenthe bodies is directly proportional to the product of masses and inversely proportionalto the square of the distance between them.
• All the bodies have the same acceleration (9.8m/s2) near the surface of the earth.But this acceleration slightly decreases as we move away from the surface of theearth.
• A body is said to be in free fall when only gravity acts on it. (its acceleration is 'g')
• Weight of an object is the force of gravity acting on it.
W = mg
During free fall condition, the body is in state of 'weightlessness'.
• The center of gravity is the point where total weight of the body acts.
• The body is in equilibrium when the weight vector goes through the base of the body.
What we have learnt
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1. What path will the moon take when the gravitational interaction between the moonand earth disappears? (AS2)
2. A car moves with constant speed of 10 m/s in a circular path of radius 10m. Themass of the car is 1000 kg. Who or what is providing the required centripetal forcefor the car? How much is it? (Ans:104N) (AS1)
3. A small metal washer is placed on the top of a hemisphere of radius R. Whatminimum horizontal velocity should be imparted to the washer to detach it from thehemisphere at the initial point of motion? (See figure) (AS1 , AS7)
(Ans: v =� )
4. Explain why a long pole is more beneficial to the tight rope walker if the pole hasslight bending. (AS1 , AS7)
5. Why is it easier to carry the same amount of water in two buckets, one in each handrather than in a single bucket? (AS7)
6. What is the speed of an apple dropped from a tree after 1.5 second? What distancewill it cover during this time? Take g=10m/s2 (AS1) (Ans: 15m/s; 11.25m)
7. A body is projected with a speed of 40 m/s vertically up from the ground. What isthe maximum height reached by the body? What is the entire time of motion? Whatis the velocity at 5 seconds after the projection? Take g=10m/s2 (AS1)
(Ans: 80m; 8s; 10m/s down)
8. A boy is throwing balls into the air one by one in such a way that when the first ballthrown reaches maximum height he starts to throw the second ball. He repeats thisactivity. To what height do the balls rise if he throws twice in a second? (AS1 .AS7)
(Ans: 1/4m)
9. A man is standing against a wall such that his right shoulder and right leg are incontact with the surface of the wall along his height. Can he raise his left leg at thisposition without moving his body away from the wall? Why? Explain.(AS7)
v
R
Improve your learning
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118 Gravitation
10. A ball is dropped from a height. If it takes 0.2s to cross the last 6m before hittingthe ground, find the height from which it is dropped. Take g = 10m/ s2 (AS1)(Ans: 54.45m)
11. A ball is dropped from a balloon going up at a speed of 5 m/s. If the balloon was ata height 60 m. at the time of dropping the ball, how long will the ball take to reachthe ground? (AS1, AS7) (Ans: 4s)
12. A ball is projected vertically up with a speed of 50 m/s. Find the maximum height ,the time to reach the maximum height, and the speed at the maximum height(g=10 m/ s2 ) (AS1) (Ans: 125m; 5s; zero)
13. Two cars having masses m1 and m2 move in circles of radii r1 and r2 respectively. Ifthey complete the circle in equal time. What is the ratio of their speeds andcentripetal accelerations? (AS1) (Ans: r1/r2, r1/r2)
14. Two spherical balls of mass 10 kg each are placed with their centers 10 cm apart.Find the gravitational force of attraction between them. (AS1) (Ans: 104G)
15. Find the free-fall acceleration of an object on the surface of the moon, if the radiusof the moon and its mass are 1740 km and 7.4 x 1022 kg respectively. Compare thisvalue with free fall acceleration of a body on the surface of the earth.
(AS1) (Ans: approximately 1.63 m/s2)16. Can you think of two particles which do not exert gravitational force on each other?
(AS2)17. An apple falls from a tree. An insect in the apple finds that the earth is falling
towards it with an acceleration g. Who exerts the force needed to accelerate theearth with this acceleration? (AS7)
18. A scooter weighing 150kg together with its rider moving at 36 km/hr is to take aturn of radius 30 m. What force on the scooter towards the center is needed tomake the turn possible? Who or what provides this? (AS1) (Ans: 500N)
19. The bob of a simple pendulum of length 1 m has mass 100g and a speed of 1.4 m/sat the lowest point in its path. Find the tension in the string at this moment.(AS1) (Ans: 1.076N)
20. How can you find the centre of gravity of a India map made of steel? Explain.(AS3)21. Explain some situations where the center of gravity of man lies out side the body.
(AS1)22. Where does the center of gravity of the atmosphere of the earth lie?(AS2)23. Where does the center of gravity lie, when a boy is doing sit-ups? Does weight
vector pass through the base or move away from the base? Explain.(AS7)
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You must have noticed thatsome things float on watersurface and some things sink in it. Did youparticipate in the activity “floating &sinking” in the lesson on “materials” inclass 6? If so, you might have wondered asto why some objects, which you expectedto sink, float on water. Did you take oneof those objects that floats on water, andtried to see if it floats on kerosene orcoconut oil?
Have a little funTake a boiling tube and fill about half
of it with water. Add 15 to 20 ml ofkerosene to the water. Drop in, one by one,plastic buttons, pins, matchsticks, smallpebbles, tiny paper balls, some sand, bitsof wax etc. Close the mouth of the boilingtube and shake it well. Wait for some timeand observe what happens.
Fig :1
� Did kerosene float above the water ordid water float above the kerosene?
� Which objects float in kerosene?� Which objects sink in kerosene but float
on water?� Which objects sink in water?� Draw a diagram of the tube, showing the
results of your activity.� Why did different objects behave
differently?We shall try to find answers to these
questions in this chapter.You know that, if a glass marble and a
small wooden piece are dropped in waterthe glass marble sinks in water but a smallpiece of wood floats on it. Do you knowwhy this happens? We think that a marblesinks in water because it is heavy while thepiece of wood floats, because it is light.
Now take a wooden block which isheavier than the marble and put it in water.What happens?� Why does the wooden block float on
water even though it is heavier than amarble?
� What do we mean by ‘heavy’, what dowe mean by ‘light’?
Chapter
FLOATING BODIES
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120 Floating Bodies
To understand the results of the aboveactivity you must understand the meaningof the term heavy. We use this word in oureveryday life in two ways. We say “twokilograms of wood is heavier than onekilogram of iron”. At the same time, wealso say “iron is heavier than wood”.
Can you explain the difference inmeaning of the word ‘heavier’ in both thesesentences? In science we try to ensure thateach word we use gives the same meaningfor everyone. So let’s see in what way thesetwo sentences differ.
The first sentence says that, if we keeptwo kilograms of wood in one pan of abalance and one kilogram of iron in theother, the beam of balance will tilt towardsthe pan with wood in it. What is the meaningof the second sentence?
In second sentence when we say ironis heavier than wood, it means if we take apiece of iron and a piece of wood of thesame size (that is, they have the samevolume) and weigh them, the iron willweigh more than the wood.
In the language of science, it may bestated as “the density of iron is more thanthat of the wood”. Density is defined asmass per unit volume.
Density =
Units for density is or
We therefore say, the denser object is‘heavy’ and the less dense object is ‘light’.
Comparing density – relativedensity
Activity-1Take two test-tubes of the same size and
fill one to the brim with water and the otherwith oil.
� Which will weigh more?� Which liquid is denser?Take two equal sized blocks made of
wood and rubber.� Which of these two blocks is
heavier?� Which one is denser?
Think and discuss
Let us suppose you have two blocksand you do not know what material theyare made of. The volume of one block is30 cm3 while the other is 60 cm3. Thesecond block is heavier than the first.Based on this information, can you tellwhich of the two blocks is denser?
When the volume of two objects isunknown, it is difficult to tell which objectis denser solely on the basis of theirweights. One way to compare the densityof objects is to take equal volumes of thetwo objects and compare their weights, butthis may not be possible for some solids.
For this we can use a simple method ofcomparing the density of each object withwater. In the following activity we shall findout how many times each solid object isdense compared to water. This is known asrelative density of that object.
massvolume
gmcm3
kgm3
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Relative density of an object =
To find the relative density of an object,we must first weigh the object and thenweigh an equal volume of water. The two
weights are then compared.Let us perform an activity to understand
how this is done. But first, check yourweighing instrument. We shall have toweigh objects several times, so yourinstrument should function properly.
Aim: Finding the relative density of different objectsMaterial required: overflow vessel, 50ml measuring cylinder, weighing balance andweights or spring balance, rubber erasers, wooden blocks, glass slides, iron nails, plasticcubes, piece of aluminium sheet, glass marbles, stones, cork etc (note: whatever objectyou take, ensure that its volume is more than 20 cc and it should not be hollow). Recordthe results of your activity in table 1. (Copy this table in your note book)
Table -1
Weigh the 50ml measuring cylinder andnote its weight here.Weight = …………….Procedure:Weigh the object, record this in column 3of the Table 1.
We need to find the weight of waterequal to the volume of the object. Pourwater in the overflow vessel until it startsdripping from its beak. When water stopsdripping from the beak, place the 50mlmeasuring cylinder under it. Slip the objectgently into the overflow vessel as shownin figure 2, ensuring that water does not
Fig-2:
Lab Activity 1
S.No. Name ofobject
Weight ofdisplaced
waterand cylinder
Weight ofwater
displaced bythe object
Relativedensityof theobject
Weight ofobject
(1) (2) (3) (4) (5) (6)
density of the object density of water
splash out. Once the object is in theoverflow vessel, water flows out of the beakand collects in the 50 ml. cylinder. Waittill the flow stops. (The object should befully immersed in water. If the object is notfully immersed, push it in to the water witha pin. see figure 2)
Weigh thecylinder with thewater thatoverflowed andrecord the weight incolumn 4.
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122 Floating Bodies
If we subtract the weight of themeasuring cylinder from this weight, we getthe weight of water (column 5 of the table1). This is the weight of water equal to thevolume of the object.
Now we can find the relative densityof the object (column 6) by taking theweight of the object (column 3) anddividing it by the weight of an equal volumeof water (column 5). This tells us howmany times denser the object is, comparedto water.
Relative density of an object =
Find the relative densities of all objectsyou collected.
Based on the table 1, answer thefollowing questions.� What is the relative density of wood?� What is the relative density of glass?� Which is denser, rubber or plastic?� Which is denser, wood or cork?� Classify the above materials as denser
than stone and less dense than stone.� Do objects that have a relative density
less than 1 sink in water or float on it?� Do the objects that sink in water have a
relative density less than1 or morethan1?What relationship do you find between
the relative density of objects and floating-sinking of the objects.
One interesting aspect of relativedensity is that it has no units. Because
relative density is a ratio of the densitiesof a material and water. It is a comparisonof quantities having the same units, so ithas no units.
Relative density of liquidsWe have discussed the relative density
of solid objects. We can also find therelative density of liquids. For this, we needto find the weight of a fixed volume of theliquid and the weight of an equal volume ofwater. The formula for finding the relativedensity of a liquid is:
Relative density of a liquid =
Aim: To find the relative density of milk,groundnut oil and kerosene.
Material required: Small bottle of 50 ml.capacity (the bottle should weigh not lessthan 10gm), weighing balance and weightsor spring balance and milk, groundnut oil,kerosene about 50 ml. each in differentcontainers.
Procedure: Find the values given below.Weight of empty bottle = ……..........……..
Weight of the bottle with 50ml of water = ............................
Weight of 50ml of water = …..........………
Weigh the bottle with milk in it. Recordthe weight in column 3 of the Table 2.
Lab Activity 2
weight of the objectweight of water equal to the volume of the object.
weight of the liquid weight of the same volume of water
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Andhra Pradesh Government Free Distribution 123
Repeat this for other liquids and record theweights in column 3. Calculate the weightof each liquid by subtracting the weight ofthe empty bottle and record it in column 4.
Calculate the relative density of each liquidby comparing the weight of the liquid withthe weight of same volume of water andrecord these values in column 5.
Answer the following questions bycomparing Table 1 and Table 2� Which liquid will float on top if
groundnut oil is poured over water ?� If we put a wooden block in kerosene,
will it float or sink? Give reasons foryour answer.
� A piece of wax floats in water but thesame piece sinks in a liquid say liquid‘X’. Will the relative density of liquid‘X’ be less than 1 or greater than 1?Why?Can we use relative density to find out
whether water has been added to milk?Let’s try and find out.� If we mix some water in milk, will the
relative density of the mixture be lessthan or more than the relative densityof milk? Refer to Table 2 to find theanswer.
� If we take two bottles of equal volumeand pour pure milk in one and milkmixed with water in the other, whichone will be heavier?We can use a simple instrument to find
this out. It is called lactometer.
Activity-2
Making of lactometerTake an empty ball pen refill. It should
have a metal point. Take a boiling tube andfill it with water.
Put the refill in with the metallic pointinside the water as shown in figure 3 (Therefill may not stand vertically in the wateras shown in figure, it may slant and causethe top of the refill to touch the wall of theboiling tube. Think what to do to make therefill to stand as shown in figure 3.)
Table -2
(1) (2) (3) (4) (5)
1 Milk
2 Groundnut oil
3 Kerosene
S.No.
Name of liquid Weight of theliquid (gm)
Relativedensity of the
liquid
Weight of thebottle filled with
liquid (gm)
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124 Floating Bodies
Did the refill sink completely or is some part above the water surface?Use a pen to mark the point on the refill to show the part which is
above the water surface.Pour out the water from the boiling tube and fill it with milk. Float
the refill in the milk.Did the refill sink up to the same mark as it sankin water? If not, did it sink more or less in milk than in water? Why didthis happen?
Put a second mark, on the refill, at the point showing the part whichis above the surface of the milk.
Now pour a mixture of milk and water in the boiling tube.If we put the refill in this mixture, to which point will it sink? Make
a guess.Test if your guess is correct by actually dipping the refill in the
milk-water mixture.Now, are you able to test whether water is added to milk or not
by using the above instrument?We can use a similar instrument, called a hydrometer/densitometer to find out the
density of any liquid.Example 1
What is the effective density of the mixture of water and milk when
i) they are taken with same massesii) they are taken with same volumes
Solution:Let us say the densities of water and milk are �1 and �2
i) When they are taken with same mass ‘m’ and their volumes are V1 and V2, the mass
of water m = �1V1 ; V1 = and the mass of milk m = �2V2 ; V2 =
Total mass of water and milk is m + m = 2m
Total volume of water and milk is V1 + V2= +
= m ( + )
=
Fig -3: Improvisedlactometer
m�
1
m�
2
m�
1
m�
2
1��
1
1��
2
m (�1 + �2)��
1 �2
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Andhra Pradesh Government Free Distribution 125
The effective density of the mixture( �����eff)= Total mass / Total volume
=
=
=
ii) when they are taken with same volume ‘V’ and their masses are m1 and m2the volume of water V = m1
/��1
That is m1 = V�1
and the volume of milk V = m2 /��2
That is m2 = V �2
Total mass of water and milk is m1 + m2 = V �1 + V �2
= V (�1 + �2)Total volume of water and milk is V + V = 2VThe effective density of the mixture ���������������eff ) = Total mass / Total volume
�eff = V (�1 + �2)
2V = (�1 + �2)
1�
2 mm (�1 + �2) / �1 �2
2 (�1 + �2) / �1 �2
2 �1 �2
�1 + �2
When do objects float on water?
Activity-3
Do the objects denser thanwater float in it ?
Collect small objects as you did for LabActivity 1. Place them one by one in a glassof water and observe whether they sink orfloat in water? Record your observations inTable 3.
Take the values of relative densitiesfrom table 1.
Table -3Object Relative density Floats / Sinks
Rubber eraser
Rubber ball
Plastic cube
Plastic pen
Iron nail
Geometry box
Glass marble
Wood
Stone
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126 Floating Bodies
� What do you observe in the aboveactivity?
� Why do some objects float in waterthough they are denser than water?
� List out the objects that float on watereven though they are made up ofmaterial which is denser than water.
We know that the substances with arelative density greater than 1 sink in water.But in activity 3, we observed thatsubstances with a relative density greaterthan 1 sometimes float on water.
So it seems we cannot judge whether asubstance will sink or float only on the basisof its relative density. There is definitelysome other factor which we need to takeinto account.
Let’s investigate that special property,which a substance that floats has but asubstance that sinks doesn’t have.
In lab activity 1, we had compared theweight of the substance with the weight ofthe water displaced by it to find its relativedensity. In that activity, we immersed thesubstance fully in water and collected thedisplaced water.
We shall now do the same activity, butwith a slight difference.
The substance will again be put in water.But this time, if it sinks we’ll let it sink andif it floats, we’ll let it float. We’ll thencompare the weight of water displaced byit with the weight of the substance.
Activity-4
Is the weight of an object andweight of water displaced by itequal ?
Take a beaker and weigh it. Note downits weight in your note book.
Fill water in an overflow jar. Wait untilthe water stops dripping from the outlet ofthe overflow jar. Then take the beaker fromthe weighing balance and place it below theoutlet of the overflow jar. Take a woodenblock, moisten it with water and then dropit gently into the overflow jar. Don’tforcefully submerge the wooden block inthe water. Also, ensure that it does notblock the outlet of the overflow jar. Waterwill flow out of the overflow jar and collectin the beaker kept under the outlet ofoverflow jar.
Do you think the weight of the waterdisplaced by the wooden block will be lessthan or equal to or more than the weight ofthe wooden block? Make a guess. Place thebeaker containing the displaced water onone pan of the weighing balance. Take thewooden block, wipe it to clean off waterand place it on the other pan along with theweights that equal to the weight of emptybeaker as shown in Figure 4.
Fig. 4
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Andhra Pradesh Government Free Distribution 127
� Do the two pans balance?� Is the weight of the water displaced by
the wooden block less than, equal to ormore than the weight of the wooden block?
Repeat this experiment with severalother substances that float or sink. Things
Table -4S. Name of the substance Weight of the substance Weight of displaced waterNo.1 Plastic bowl
2 Ball
3 Steel container
4 A fruit that floats
5 A fruit that sinks
6
7
8
On the basis of the table 4, explain therelationship between the weight of thesubstances that float and the weight of thewater displaced by them.
Can you express this special propertyof substances that makes the substance tofloat in the form of a principle?
(The special property of floatingsubstances that you identified in thisactivity was first discovered by Archimedes.You will know about this further in thischapter.)
Can you think of a way to make ironfloat on water? Perhaps, the followingactivity will give you some ideas about howyou can make iron float on water.
Activity-5Making aluminium to float
Take a small sheet of aluminium foil.Fold it four or five times, pressing the foiltight after each fold. You already know therelative density of aluminium from anearlier Lab activity 1. With this given valueof the relative density of aluminium, canyou guess whether the aluminium foil willfloat or sink in water?
Drop the folded aluminium foil in thewater and test whether your guess is corrector not.
Now unfold the aluminium foil andmake it as a small bowl. Place this bowl inthe water and see whether it floats or sinks.
that float could include a plastic bowl, a ball,a steel container, fruits etc.
In each case, check whether the weightof the water displaced is less than, equal toor more than the weight of the substance.Note your observations in the table 4.
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128 Floating Bodies
� How much water did the bowl ofaluminium foil displace?
� Is the water displaced by foldedaluminium foil and bowl of aluminiumfoil the same?Explain why aluminium bowl floats on
the basis of your theory of floatingsubstances.� Can you now explain why large ships
made of iron and steel float on waterwhile a small block of iron sinks inwater?
� Why does the metal bowl displacelanger amount of water than a metalpiece?To know this you must understand the
pressure in fluids.
Upward force in liquidsWhen we put an object on the surface
of water in a container, the force of gravity,exerted by the Earth, pulls the objectdownwards i.e. towards the bottom of thecontainer. However, for objects that floaton water, there must be an upward force tobalance the force of gravity. This upwardforce must come from water. If thegravitational force on the object is morethan the upward force of water, the objectwill sink in water. Let us do a simple activityto observe this upward force.
Activity-6
Observing the upward force ofliquids
Take an empty plastic bottle. Put thecap on it tightly. Place the bottle in a bucketof water. The bottle will float.
Push the bottle into the water by yourhand as shown in figure 5.
Fig. 5
Do you feel an upward force? Try topush it further down. Do you feel anyincrease in the upward thrust? In fact, theupward force of water keeps on increasingas you try to push the bottle down. Now,release the bottle and observe how itbounces back to the surface of water! Sothe upward force of water is a real,observable force. This force acting on unitarea of the surface of an object (bottle) iscalled pressure.
Pressure of Air
Activity -7Observing air pressure
Take a glass tumbler. Stick some cottonat the bottom of it. Immerse it inversely inwater up to the bottom of the container asshown in figure 6.
Take out the tumbler from water. Isthe cotton attached to its bottom wet? Why?
Fig. 6
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This is due to the force of air which is applied on water by the air present in the tumbler andstops water from entering the tumbler. This force on unit area of water is pressure ofair.
Atmospheric pressureAll objects on the surface of the earth are subject to constant atmospheric pressure.Atmospheric pressure = Force of the atmosphere / surface area of the earthAtmospheric pressure = Weight of the atmosphere / surface area of the earthAtmospheric pressure = (Mass of the atmosphere) � g / (surface area of the earth)
Atmospheric pressure = (average density of the atmosphere)�(Volume of the atmosphere)� g
(surface area of the earth )Thus
Atmospheric pressure = �� � Surface area of the earth � Height of the atmosphere�g
surface area of the earthAtmospheric pressure = � � (Height of the atmosphere) � gAtmospheric pressure = � h g
Po = � h g
Measuring atmospheric pressureWe cannot experience this
atmospheric pressure, but we can recognizeand measure it with barometers. The firstbarometer is invented by ‘Torricelli’ usingmercury. (See figure 7)
At normal atmospheric pressure themercury barometer shows mercurycolumn with 76 cm height in the glasstube above the surface of the mercuryin the bowl. This is known as 1atmospheric pressure.
� Why is the height of mercury columnnearly 76cm in the tube?
What is the state of the mercurycolumn in the tube? It is at rest, so net forceon it is zero. The weight of the column inthe tube is equal to the force applied on itby the mercury in the bowl due toatmospheric pressure. These two must beequal in magnitude and opposite indirection.
Fig-7: Barometer
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130 Floating Bodies
Do you know?
Weight of the mercury column (W) = mass of mercury (m) � g = (Volume) (density) g
= (cross sectional area of the tube) (height of the column) � g = Ah�g
Let ‘Po’ be the atmospheric pressure.Force on the column due to the atmospheric pressure = Po Athen,A h�g = Po APo = �gh (of mercury)�,g are constants. So the height of the mercury column depends on atmospheric pressure.we can calculate the value of atmospheric pressure ‘Po’, by substituting the values ofheight of the mercury column ‘h’ , density of the mercury ‘�’ and acceleration due togravity ‘g’.Height of the mercury column h = 76cm = 76 � 10-2 mDensity of the mercury �= 13.6 gr/cc = 13.6 X 103 kg/m3
Acceleration due to gravity g = 9.8 m/s2
Po = h�gPo = (76 � 10-2 m) � (13.6 � 103 kg/m3) � (9.8 m/s2)Po = 1.01 � 105 kg.m/m2.s2
1 kg.m/s2 = 1 Newtonhence, Po = 1.01� 105 N/m2
This value is called one atmospheric pressure.1 Atmosphere = 1.01� 105 N/m2
(1 N/m2 is called Pascal)
The mass of air that would occupy in a cylindrical tube with cross sectional area of1cm2 and that extends 30km up to the top of atmosphere is about 1kg. The weight appliedon the surface area of 1cm2 on the earth is nothing but atmospheric pressure.
Atmospheric pressurePo = mg/A = 1 kg X 10 m/s2 /1 cm2 =10 N/cm2 or 105 N/m2 (105 Pascal)This value is nearly equal to 1atm.
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Think and discuss
� What would happen if Toricelli experiment had been done on moon?� A stopper is inserted in the small hole of the glass tube of the mercury barometer
below the top level of the mercury in it. What happens when you pull out the stopperfrom the glass tube?
� Why don’t we use water instead of mercury in Toricelli experiment? If we are ready todo this experiment, what length of tube is needed?
� Find the weight of the atmosphere around the earth (take the radius of earth is 6400km.)
Pressure at a depth “h” in a liquid Let us consider a container which contains a liquid in it of density “�”. Consider a cylindrical column of height ‘h’ from the surface of liquid of cross
sectional area “A”. See the figure 8.
The volume of the liquid column V = Ah Mass = Volume � density m = Ah �Weight W = mg = Ah � gWhat is the state of the motion of the liquid column?
You know that from Newton’s law, the net force on it is zero, because it is at rest. Whatare the forces acting on that water column?
There are three forces acting, which are,
i) Weight (W), vertically down
ii) Force on top surface due to atmospheric pressure(Po A), acting vertically down
iii) Force on the bottom surface of the column due to static pressure of liquid (PA),
acting vertically up.
From Newton’s law we get
PA = Po A + W
PA = Po A + h � g A
Fig. 8
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132 Floating Bodies
Where P is the pressure at the depth “ h” from the surface of the liquid and Po is theatmospheric pressure.
PA = Po A + h � g AP = Po + h � g
This means that the pressure inside the liquid at a constant depth is constant.
Pressure difference at different levels of depth in fluidsLet us consider a cylindrical column of liquid of height ‘h’ with cross sectional area‘A’ and let � be the density of the liquid. See the figure 9.What is the pressure P1 in the liquid at depth h1 ?From equation (1) we get,
P 1 = Po + h1 � gSimilarly, pressure P2 at depth h2 is given by P2 = Po + h2 � gFrom (3)-(2) we get
P2 - P1 = h2 � g - h1 � gP2 - P1=�� g (h2 - h1)
from the figure h = h1 - h2 sowe have, P2 - P1 = h� g
The pressure difference between two levels in that liquid = h � g.Here density of the liquid ‘ �’ and ‘g’ are constants, so the pressure difference increaseswith a increase in depth. ��What happens if we replace this cylindrical liquid column with another objectwhich is made up of a material whose density is not equal to the density of liquid?The pressure difference in the liquid P2 - P1 = h �g (values of the liquid)
P2- P1 = h × m/V × g
P2 - P1 = h × m/Ah × g
P2 - P1 = m/A × g
(P2 - P1)A = m × g (Weight of the displaced liquid)
Since F= P×A and W= mg
we get F = W (values of the displaced liquid)
Here ‘F’ is the force applied on the object and ‘W’ is the weight of the displaced liquid.So the force applied on the object by the liquid is equal to the weight of the displacedliquid.
The force applied on the object in upward direction is called “Buoyancy”. As per theabove equation this buoyant force is equal to the weight of the liquid displaced by theobject.
Fig 9
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Measuring the force of buoyancyWe have seen that when an object is
immersed in water it experiences anupward force, the force of buoyancy. Canwe measure this upward force? Let's try.
Activity-8
Let us measure the force ofbuoyancy
Suspend a stone from a spring balance.Note the reading of the spring balance. Thereading gives the weight of the stone. Takea beaker half filled with water. Nowimmerse the stone in the water. Note thereading of the spring balance. The readingof the spring balance gives the 'weight' ofthe immersed stone. Do you notice anychange in the weight of the stone beforeand after it gets immersed in water? Youmay notice that the stone, when immersed,appears to lose some weight.� Why does the stone lose weight when
it is immersed?The immersed stone appears to lose
weight because the force of buoyancy,exerted on the stone by the water, in theupward direction, serves to reduce theforce of gravity. Thus the apparent loss ofweight must be equal to the force ofbuoyancy acting on the immersed stone.We can measure the force of buoyancyexerted by a liquid, by measuring theapparent loss of weight of an objectimmersed in that liquid. You will notice thatin every case an immersed object appearsto lose weight.
When the object floats on the surfaceof water, it appears as if it lost all its weight,that is, the spring balance shows zeroreading for floating bodies! For objects thatfloat on a liquid surface, the force ofbuoyancy balances the force of gravity atthe surface of the liquid.
Now let us repeat this activity andmeasure the weight of the water displacedby the immersed stone.
Activity-9
Measuring the weight of thewater displaced by the immersedstone
Suspend a stone from a spring balance.(it is better to take the stone more than 300gm) . Note the reading on the spring balance.The reading gives the weight of the stone.Take an overflow vessel with water andplace a graduated beaker below the beak.(Figure 10).
Now immerse the stone in the water.Note the reading on the spring balance and
Fig. 10
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134 Floating Bodies
Do you know?
measure the volume of water that overflowsinto the graduated beaker.
The reading of the spring balance givesthe weight of the immersed stone and thebeaker reading gives the volume of waterdisplaced by the stone (Figure 11).
� By how much does the weight of thestone appear to be decreased?(Apparent loss of weight of the stone)
� What is the weight of the displacedvolume of water?
� Do you observe any connectionbetween the two?The apparent loss of weight of the
immersed stone is equal to the weight ofwater displaced by the stone i.e., equal tothe force of buoyancy exerted by thewater.
This wonderful observation was madeby Archimedes, an ancient Greek scientist.
Archimedes' principleArchimedes' principle states that when
a body is immersed in a fluid it experiencesan upward force of buoyancy equal to theweight of fluid displaced by the immersedportion of the body.
Archimedes (287-212 BC)
Fig-11
Archimedes was a Greek scientist. At that time theKing had a crown made of gold. The King however, suspectedthat the crown made was not of pure gold and askedArchimedes to verify this. Archimedes had to solve theproblem without damaging the crown, so he could not meltit down into a regularly shaped body in order to calculate itsdensity. While taking a bath, he noticed that the level of thewater in the tub rose as he got in, and realized that this effectcould be used to determine the volume of the crown. Thesubmerged crown would displace an amount of water equalto its own volume. By dividing the mass of the crown by thevolume of water displaced ,the density of the crown could
be obtained. This density would be lower than that of gold if cheaper and less densemetals had been added. Archimedes then took to the streets naked, so excited by hisdiscovery that he had forgotten to dress, crying "Eureka!" (meaning "I have found it!").
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Let's look at the story that is associatedwith this observation.
Think and discuss
�� Why is it easier for you to float in saltwater than in fresh water?
� Why is there no horizontal buoyantforce on a submerged body?
� Two solid blocks of identical size aresubmerged in water. One block is ironand the other is aluminium. Upon whichis the buoyant force greater?
� A piece of iron is placed on a block ofwood, makes it to float lower in thewater. If the iron piece is suspendedbeneath the wood block, would it floatat the same depth? Or lower or higher?
You know that pressure difference atdifferent levels of height inside the liquidcauses buoyancy.� Can we increase the pressure inside the
liquid?
It is only possible when the liquid isenclosed. A scientist named Pascal made aprinciple about what happens when aexternal pressure is applied to a enclosedliquid. Let us know about it.
Pascal's principlePascal's principle states that external
pressure applied to a enclosed body of fluidis transmitted equally in all directionsthroughout the fluid volume and the wallsof the containing vessel.
Look at Figure 12. Here, we have anenclosed volume of fluid in a U-shapedtube. The fluid is enclosed in the tube by
THINK! How did Archimedes solve the King's problem? Asimple arrangement could be used to determine whether a goldencrown is less dense than gold. The crown and a bar of gold, of thesame mass as the crown, are suspended from the two arms of asimple balance as shown in the figure. The balance is lowered intoa vessel of water. If the crown (left) is less dense than the gold bar(right), it definitely posses a larger volume than that of pure goldand will displace more water and thus experience a larger upwardbuoyant force, causing the balance to tilt towards the gold bar. This would indicate thatthe crown was not of pure gold!
Note: This experiment holds good only when the crown doesn’t have any coveredhollow portion in it. Think why?
F1 F2P1 = F1 / A1 P2 = F2 / A2
V
V
Fluid
Fig-12: Application of Pascal's principle(bramha press)
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136 Floating Bodies
Key words
two leak-proof pistons in each arm. Theratio of cross-section areas of the rightand left tubes is A2 : A1 and A2 >A1.
When a force F1 is applied to the leftpiston the excess pressure acting on thefluid volume is F1/A1.
According to Pascal's principle, thisexcess pressure is transmitted equallythroughout the fluid volume. That is, everyunit area of the fluid 'experiences' thisexcess pressure of F1/A1.
The excess pressure in the right-sidetube (of cross section area A2) is alsoF1/A1 and since its area is A2, the upwardforce acting on the right piston isF2 = A2 x F1 / A1 ; which is much larger inmagnitude than F1.
Thus the application of Pascal'sprinciple results in a large upward force(thrust) on the right piston when a small
downward force is applied on the leftpiston.
Fig. 13: Hydraulic jackThis principle is used in the design and
working of hydraulic jacks/lifts (Fig-13)which you can see in automobileworkshops. A small downward forceapplied by the hand of the operator helpsto lift a heavy vehicle with ease!
Density, Relative density, Lactometer, Hydrometer / densitometer, atmosphericpressure, barometer, buoyancy.
� Objects having a density less than the liquid in which they are immersed, float onthe surface of the liquid.
� All objects experience a force of buoyancy when immersed in a fluid.� When an object is immersed in a fluid it appears to lose weight (because of
buoyancy).� The apparent loss of weight of an object, which is immersed in a liquid, is equal to
the weight of liquid displaced by the object. (Archimedes' principle)
What we have learnt
F1 F2
Fluid
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� When an object floats on the surface of a liquid, it displaces a weight of liquidequal to its own weight.
� The pressure exerted by a liquid increases with depth below the surface of liquid.� External pressure, applied to an enclosed volume of fluid, is transmitted equally in
all directions throughout the fluid volume.(Pascal's principle)
1. A solid sphere has a radius of 2 cm and a mass of 0.05 kg. What is the relative densityof the sphere? (AS1) [Ans: 1.49 ]
2. A small bottle weighs 20 g when empty and 22 g when filled with water. When it is filledwith oil it weighs 21.76 g. What is the density of oil ? (AS1) [ Ans: 0.88 g/cm3]
3. An ice cube floats on the surface of a glass of water (density of ice =0.9 g/cm3).Whenthe ice melts will the water level in the glass rise? (AS1)
4. The volume of 50g of a substance is 20 cm3. If the density of water is 1g/cm3,will thesubstance sink or float when placed on the surface of water ? What will be the mass ofwater displaced by the substance? (AS1) [Ans: 20 g]
5. Find the pressure at a depth of 10m in water if the atmospheric pressure is 100kPa.[1Pa=1N/m2] [100kPa = 105Pa = 105N/m2 = 1 atm.] (AS1) [Ans: 198 kPa]
6. Why some objects float on the water? And some sink? (AS1)
7. Explain density and relative density and write formulae. (AS1)
8. What is the value of density of water? (AS1)
9. Find the relative density of wood. Explain the process. (AS3)
10. Which is denser, water or milk? (AS2)
11. What is buoyancy? (AS1)
12. Classify the following things into substances having Relative Density> 1 and RelativeDensity< 1 Wood, iron, rubber, plastic, glass, stone, cork, air, coal, ice, wax, paper,milk, kerosene, groundnut oil, soap (AS1)
13. How can you appreciate the technology of making ships float, using the material whichsink in water ? (AS6)
14. Can you make iron to float? How? (AS3)
15. How can you find the relative density of a liquid? (AS3)
Improve your learning
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138 Floating Bodies
16. Find the relative density of different fruits and vegetables and write a list. (AS3)
17. Make a lactometer with refill. What you do to make the refill stand vertically straight?(AS5)
18. Draw the diagram of mercury barometer. (AS5)
19. How do you appreciate Pascal’s discovery in helping to make hydraulic jacks. (AS6)
20. How do you appreciate Archimedes discovery of force of buoyancy. (AS6)
21. You found the relative densities of some solids and some liquids by some activities.List the solids and liquids in increasing order of relative density.(AS4)
22. Iron sinks in water, wood floats in water. If we tied an iron piece to the same volume ofwood piece, and dropped them in water, will it sink or float? Make a guess and find outwhether your guess is correct or wrong with an experiment. Give reasons.(AS2 . (AS3)
23. Liquid brakes in automobiles follow principle of bramha press (Pascal’s principle).What about air brakes? Collect the information about the working process of airbrakes.(AS4)
24. Where do you observe Archimedes principle in our daily life? Give two examples.(AS7)
25. Where do you observe Pascal’s principle in our daily life? Give few examples.(AS7)
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In previous few chapters you havelearnt about various ways of describingthe motion of objects and causes of motion.In performing our day - to - day activitieswe use various words like work, energy, andpower which are closely related to eachother and sometimes we use these wordswithout paying much attention.. In thischapter you will examine these conceptsmore carefully.
People carry out various tasks in dailylife. For example, lifting of weights,carrying of weights, sweeping / cleaning ofhouse, cooking of food, and watering ofplants in the garden etc are some of the dayto day activities.
Similarly you might have also observedthat people employ machines in theirhomes to carry out different tasks likeblowing of air by fan, pumping of water byelectric motor, heating of water by electricheater etc.
Washing machines, vacuum cleanersare used for cleaning of clothes, andcleaning the house respectively.� How are these works being done?
� What do you need to do these works?Both human beings or machines need
energy to do work. Generally this energyis supplied to human beings through foodthey take in and for machines through theelectricity supplied to them.
In all the examples mentioned above,we notice that a person or a machine doingwork spends some energy to do work. Forexample to lift your school bag you spendsome energy. Similarly an electric fan usessome electric energy for blowing air.
� Where does the energy spent goultimately?
� Is there any transfer of energy whilework is being done?
� Can we do any work without transferof energy?
Think about various works youobserved and try to indentify the forceemployed for doing the work and the objecton which the work was done. Discuss withyour friends about possibility of energytransfer during work.
Chapter
WORK AND ENERGY
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140 Work and Energy
WorkIn our daily life, we use the term ‘work’
in various situations. The word 'work' takeson different meanings depending on thesituation. For example the statements like‘ I am working in a factory’ , ‘The Ramayanais a great work of Valmiki’ ‘The machineis in working condition’ ‘There are largenumber of worked out problems in thisbook’ ‘ Let us work out a plan for next year’etc have different meanings. There is adifference between the way we use the term‘work’ in our day to day life and the way weuse it Science.
Let us examine these situations.i) Priyanka is preparing for examinations.
She spends lot of time in studies. Shereads books, draws diagrams, organizesher thoughts, collects question papers,attends special classes, discussesproblems with her friends and performsexperiments etc.In our common view ‘she is workinghard’ but if we go by scientificdefinition of work the above mentionedactivities are not considered as work.
ii) Rangaiah is working hard to push thehuge rock. Let us say the rock does notmove despite all efforts by him. Hegets completely exhausted and tired.According to our common view heworked hard but as per science he hasnot done any work on the rock.
iii) Let us assume you climb up thestaircase and reach the second floor of
a building. You spend some energy todo this. In our common understandingyou are not doing any work but as perscience you have done a lot of work toreach the second floor of the building.In our daily life we consider any useful
physical or mental labor as ‘work’For example we consider cooking of
food, washing of clothes, sweeping, doinghome work, reading, writing etc. as works.But according to scientific definition ofwork all of these activities are notconsidered as work, only a few of them areconsidered as work.� What is work?� Why is there difference between
general view of work and scientificview of work?
Scientific meaning of the workTo understand the way we view work
and scientific meaning of work let usobserve the following examples.Example-1
Fig - 1
A man is lifting cement bags from theground and keeping them one by one in alorry.
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Example-2
Fig - 2
A girl is pulling a toy car on the groundand the trolley moves a distance. Example-3
A boy is trying to push a huge rock lyingin a play ground.
Fig - 3
Example-4A porter is waiting on the platform of a
railway station with luggage on his head.
Fig - 4
� Are all the people mentioned in theabove examples doing work?
� How do you define work?
To know the meaning of workaccording to science analyse the aboveexamples as per the table in the followingactivity 1.
Activity-1
Let us understand the meaning ofwork as per science
Copy the table given below in your notebook.
Discuss with your friends whetherwork is being done in each of the examplesmentioned above. What could be yourreason to say that work has been done?Write your reasons in the following table.
After careful comparison of all theabove mentioned examples, you canunderstand that in each case the personinvolved in action/ doing work, spends someenergy to perform the actions mentionedin the examples. In some cases there is achange in the position of the object onwhich work has been done, For instance inexample -1 the position of the bag ischanged from ground to the height of thelorry, and similarly the toycar movedthrough a distance by changing its position.
In some other cases though the personis doing the work and spends energy, thereis no change in the position of the objecton which work has been done. Like in
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142 Work and Energy
1. A force should act on the object.2. The object must be displaced or
there must be change in the position of theobject.
Let us define the term ‘work’
Definition of work in science
Fig - 5
Let us consider the following
Table - 1
SituationWhether work
has beendone or not ?
(Yes/No)
Who is doingthe work
(Name theforce)
Object on whichwork hasbeen done.
Reason to saythat work
has been done
The changes younotice in objectson which workhas been done
Example – 1 Yes
Man,Muscular
force
Cement bag
The personis lifting the bagfrom ground to
lorry usingmuscular force
The cementbag moved
from theground to theheight of the
lorryExample – 2
Example – 3
Example – 4
example -3, the boy who is trying to pushthe huge rock by applying force on it spendslot of energy though there is no change inthe position of the rock. Similarly inexample - 4 the porter standing on therailway platform with luggage on his headspends a lot of energy against thegravitational force acting on the luggagethough there is no change in the positionof luggage.
In our common perception of work allthe forces acting on bodies applied bypersons mentioned in examples 1 to 4 aredoing work but according to science onlythe forces applied by persons mentionedin examples 1 and 2 are doing the work.
According to the scientific concept ofthe work, two conditions need to besatisfied in order to say that work has beendone.
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example:Suppose that a constant force (F) acts
on an object and object is moved through adistance (s) along the direction of the force(F) as shown in figure 5
In science we define work to be equalto the product of the force (F) and thedistance (s) moved along the direction ofthe force.
Work done = Force x Displacement.
W = F s
This formula for work is used inonly translatory motion of the object.
Work has only magnitude, no direction.So work is a scalar.
We measure force (F) in Newtons(N) and distance (s) in meters (m). Inequation W=F s, if F=1 and N=1 then thework done by the force will be ‘1 N-m’.Hence the unit of work is ‘Newton-meter’(N-m) or ‘Joule’ (J)
Thus 1 Joule (J) is the amount of workdone on an object when a force of 1Newton displaces it by 1m along the lineof action of the force.
Look at the equation W = F s
� What would be the work done when theforce on the object is zero?
� What would be the work done when thedisplacement of the object is zero?
� Can you give some examples, wherethe displacement of the object is zero?
Think and discuss
� A wooden chair is dragged on the levelfloor and brought to the same place. Letthe distance covered be 's' and frictionalforce acted on the chair by the floor be'f' . What is the work done by thefrictional force ?
Example 1A boy pushes a book kept on a table by
applying a force of 4.5 N. Find the workdone by the force if the book is displacedthrough 30 cm along the direction of push.Solution Force applied on the book (F) = 4.5 NDisplacement (s) = 30 cm = (30/100) m
= 0.3m Work done, W = F s
= 4.5 x 0.3 = 1.35 J
Example 2Calculate the work done by a student inlifting a 0.5 kg book from the ground andkeeping it on a shelf of 1.5 m height. (g=9.8m/s2)SolutionThe mass of the book = 0.5 kgThe force of gravity acting on the book isequal to ‘mg’That is mg = 0.5 x 9.8 = 4.9 N
The student has to apply a force equalto that of force of gravity acting on thebook in order to lift it.
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144 Work and Energy
Dir
ectio
n of
mot
ion
Thus force applied by the student on thebook, F = 4.9 NDisplacement in the direction of force, s = 1.5mWork done, W = F s = 4.9 x 1.5
=7.35 JIn the situation mentioned in figure 5
the displacement of the object is in thedirection of the force. But there are certaincases where the displacement of the objectmay be in a direction opposite to the forceacting on it.
For example if a ballis thrown up (Fig – 6), themotion is in upwarddirection, where as theforce due to earth’sgravity is in downwarddirection.� What happens to the
speed of the ballwhile it moves up?
� What is the speed atits maximum height?Fig - 6 Positions of ball at various
instants while moving up� What happens to the speed of the ball
during its downward motion?Similarly the ball moving on plain
ground (Fig-7) will get stopped after sometime due to frictional force acting on it inopposite direction.
Fig - 7
If the force acting on an object anddisplacement are in opposite directionsthen the work done by the force is taken asnegative.
W = – F sIf work has positive value, the body on
which the work has been done would gainenergy.
If work has negative value, the body onwhich the work has been done loses energy
Think and discuss
Lift an object up from the ground.Work done by the force exerted by youon the object moves it in upwarddirection. Thus the force applied is in thedirection of displacement. Howeverthere exists a force of gravitation on theobject at the same time
� Which one of these forces is doingpositive work?
� Which one is doing negative work?
� Give reasons.
Example - 3 A box is pushed through a distance of 4macross a floor offering 100N resistance.How much work is done by the resistingforce?Solution The force of friction acting on the box, F = 100N The displacement of the box, s = 4m
The force and displacement are in
Direction of motion
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opposite directions. Hence work done onthe box is negative. That is, W = – F s
= – 100 x 4 = – 400 J
Example 4 A ball of mass 0.5 kg thrown upwards
reaches a maximum height of 5m.Calculate the work done by the force ofgravity during this vertical displacementconsidering the value of g = 10m/s2.Solution Force of gravity acting on the ball, F = mg = 0.5 x 10 = 5NDisplacement of the ball, s = 5m
The force and displacement are inopposite directions. Hence work done bythe force of gravity on the ball is negative.
W = – F s = – 5 x 5 = – 25 J
Idea of EnergyThe word Energy is very often used in
our daily in various occasions like ‘ He ismore energetic’ I am so tired and lost myenergy’, Today I am feeling more energeticthan yester day’ etc.� What is energy?� How can we decide that an object
possess energy or not?Let us consider the following cases
Case -1A metal ball kept in a ceramic plate is
raised to a certain height from the plate andallowed to fall on it.
� What will happen to the plate? Why?
Fig - 8
Case -2A toy car is placed on floor without
winding the key attached to it and the sametoy car placed on the floor after windingthe key attached to it.
Fig - 9 : A Toy Car
� What changes do you notice? Why? In case -1 you might have noticed
that the metal ball does no work when itrests on the surface of the plate but is ableto do some work when it is raised to aheight .
Similarly in case -2 you may notice thatthe toy is at rest before winding the keybut the same toy gets energy to move whenthe key attached to it is wound up.
Children may not be able to lift 25 kgrice bag but elders can do that.
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146 Work and Energy
Activity-2
Understanding the increase anddecrease in energy of an object
Take a hard spring and keep it on thetable as shown in the figure 10.
Fig - 10Now compress the spring with your
palm and release it after few seconds.Observe the changes in its position, stateand size after compressing the spring andreleasing it. You will notice that whenspring is being compressed there is achange in its size. When it is released itgains some energy and may even jump fromthe table. The work done by your palm onthe spring increases its energy and makesit to jump away from the table.
Thus we can conclude that the objectwhich does work loses energy and theobject on which work has been done gainsenergy. If negative work is done on anobject, its energy decreases. For example,when a ball moves on the ground the forceof friction does negative work on the ball(because it acts in a direction opposite tothe motion of the ball). This negative workdone on the ball decreases its kineticenergy and makes it comes to rest aftersome time.
� What could be the reason?
You may observe many suchsituations where the capacity to do workchanges from person to person.
Similarly the capacity of doing workby an object on another object depends onposition and state of the object which isdoing work. Thus an object acquires energythrough different means and is able to dowork.
Energy transfer and the work We learned in previous paragraphs,
that we need energy to do any work and aperson doing work spends some energywhile doing the work i.e. the person doingthe work loses some energy.
� Where does this energy go?
� Is there any energy transfer between theobject doing the work and the objecton which work has been done?
� Can any force do work without energytransfer?
In science, we consider work has beendone only when there is a change in theposition of the object. The object is ableto change its position due to the energytransferred to it by the force doing thework. Thus whenever work is done on anobject its energy either increases ordecreases.
For example if we push a wooden blockwhich is kept on a table, it starts movingdue to the work done on it and as a result itgains kinetic energy.
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Forms of energyEnergy can exists in several forms like
mechanical energy, light energy, thermalenergy, sound energy, electrical energy,magnetic energy etc..
Apart from these, electric batteries,cooking gas, coal, and petroleum etc. havechemical energy stored in them. Evenmatter itself is a concentrated form ofenergy and sometimes can be convertedinto other forms like nuclear energy, heatenergy etc.
Energy inside a human bodyThe human body is a complicated
system in which a variety of energyexchange takes place. The main source ofenergy for human body is food. Energy isused up when the muscles are stretched andheart pumps blood and so on.
Work done by the force inside the bodyin performing various activities reduces theenergy of the body. We feel tired in suchsituations.
� Do you know why a person gets tiredstanding at place for long time?
Though the person standing is not doingany work externally, a lot of work is beingdone inside the body.
The muscles of the body becomestretched when he stands for long time andheart has to pump more blood to muscles.This leads to loss of energy inside the bodyand hence he gets tired.
Think and discuss
� What would happen if nature doesnot allow the transfer of energy?Disscuss with few examples.
Sources of energy Life would be impossible without
energy. We need energy in different formslike muscular energy, thermal energy, lightenergy, electrical energy etc., to do our dayto day activities. The demand for energy isever increasing.
� Where do we get energy from?
The Sun is the biggest natural andprimary source of energy for us. Manyother secondary sources are derived fromthe Sun. We can also get energy from theinterior of the earth and from tides of thesea.� Can you think of other sources of
energy?
Activity-3
List the energy sources A few sources of energy are mentioned
above. There are many other sources ofenergy.
List them and identify which of themare sources of energy due to Sun. Discusswith your friends how these sources areconsidered as sources of energy due to theSun.� Are there sources of energy which are
not dependent on the Sun?
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148 Work and Energy
Consider a metal ball and a hollowplastic block which are kept on a table sideby side as shown in the figure 11 (a). Nowsuppose that the ball is separated from theblock and brought to one end of the tableas shown in the figure 11 (b) and pushed toroll on the table with speed ‘v’.
� What will happen to block?
� What changes do you notice in theposition and state of the ball and blockwhen the ball is allowed to roll on thetable?
You may notice that when the ball ispushed , it starts moving with speed of ‘v’and hits the plastic block and displaces itfrom point A to B as shown in figure 11(b). Thus a moving ball is more energeticthan the ball at rest because the moving ballis able to do the work on plastic block topush it forward, whereas the same ballcannot do any work when it is at rest. Inother words, a body possess more energywhen it is moving than when it is at rest.
Fig – 11 (a) Fig - 11 (b)
Repeat this activity by pushing the metalball with more force so as to increase itsspeed and observe the change in positionof the plastic block on the table. You maynotice that the increase in the speed of theball increases its capacity of doing work onthe block.
Thus we can conclude that a movingobject can do work. An object moving fastercan do more work than an identical objectmoving relatively slow.
The energy possessed by an object dueto its motion is called kinetic energy.
The Kinetic energy of an objectincreases with its speed.
We come across many situations in ourdaily life where objects with kinetic energydo work on other objects. For examplewhen a fast moving cricket ball hits thewickets, it makes the wickets to tumble butif the swinging bat in the hands of a batsmanhits the ball it reaches the boundary.
Kinetic energy
Activity-4
Understanding the energy of moving objects
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Similarly a fast moving bullet pieces thetarget, blowing wind moves the blades ofthe wind mill. Objects like a fallingcoconut, a speeding car, a rolling stone, aflying aircraft, flowing water and runningathlete etc, also possess kinetic energy.� How can we find out as to how much
energy is possessed by a moving body?
Numerical expression for kineticenergy
We know that the kinetic energy of abody at rest is zero, but the kinetic energyof a body moving with certain velocity isequal to the work done on it to make itacquire that velocity from rest.
Fig – 12
Let us assume that an object of mass(m) is at rest on a smooth horizontal planeas shown in figure 12. Let it be displacedthrough a distance ‘s’ from the point A toB by a force (F) acting upon it in thedirection of the displacement. In thehorizontal direction the net force Fnet isequal to the applied force F.
The work done on the object by the net forceW = Fnet s = F s — (1)
Let the work done on the object causea change in its velocity from ‘u’ to ‘v’ andthe acceleration produced be ‘a’.
In the chapter motion we studied aboutequations of uniform accelerated motion.The relation between initial velocity u, finalvelocity v, acceleration ‘a’ and displacement‘s’ is given by
v 2_ u2 = 2 a s or s = (v
2_ u2) / 2 a ——(2)
We know by Newton’s second law ofmotion
Fnet= ma ————— (3)
From equations (1), (2) and (3)
W = ma x (v 2_ u2) / 2a
W = ½ m (v 2_ u2)
This is called work - energy theorem.
As we have assumed that object is atrest, its initial velocity u = o, then
W = ½ m v2
Thus the work done on the object isequal to ½ m v2
We know that kinetic Energy of a bodymoving with certain velocity is equal towork done on the object to acquire thatvelocity from rest.
Thus the kinetic Energy (K.E.)possessed by an object of mass ‘m’ andmoving with velocity ‘v’ is equal to ½ m v2
K.E. = ½ m v2
2m
2m
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150 Work and Energy
Think and discuss
� Why is it easier to stop a lightlyloaded truck than heavier one thathas equal speed.
� Does the Kinetic energy of a carchange more when it goes from 10m/s to 20 m/s or when it goes from20 m/s to 30 m/s.?
� A person starts from rest and beginsto run. The runner puts a certainmomentum into himself. What is themomentum of ground? And therunner puts a certain amount ofkinetic energy into himself. What isthe kinetic energy of the ground.?
Example 5Find the kinetic energy of a ball of 250 gmass, moving at a velocity of 40 cm /s.
SolutionMass of the ball, m = 250g = 0.25 kgSpeed of the ball, v = 40 cm / s =0.4 m /sKinetic Energy,K.E. = ½ (0.25) x (0.4)2 = 0.02 J
Example 6The mass of a cyclist together with the
bicycle is 90 kg. Calculate the work doneby cyclist if the speed increases from6km /h to 12 km / h.
SolutionMass of cyclist together with bike,
m = 90 kg.Initial velocity, u = 6km/h = 6x(5/18) = 5/3 m/s
Final velocity, v = 12 km /h = 12x(5/18) = 10/3 m/s
Initial kinetic energy
K.E (i) = ½ mu2
= ½ (90) (5/3)2
= ½ (90) (5/3) (5/3)
= 125 J
Final kinetic energy,
K.E (f) = ½ m v2
= ½ (90) (10/3) 2
= ½ (90) (10/3)(10/3)
= 500 J
The work done by the cyclist = Change inkinetic energy = K.E (f) - K.E(i)
= 500 J – 125 J = 375 J.
Potential energy
Activity-5Understanding potential energy
Take a bamboo stick and make a bow asshown in the figure 13 (a). Place an arrowmade of a light stick on it with one end
Fig-13a Fig-13b
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Andhra Pradesh Government Free Distribution 151
supported by the string of the bow as shownin figure13 (a) and stretch the string gentlyand release the arrow.� What do you notice?
Now place the arrow on the bowwith one end supported by the string andstretch the string applying more force andrelease the arrow as shown in figure13 (b)� What differences do you notice in these
two instances with respect to motionof arrow?
� Is there any change in the shape of thebow when the string is stretched byapplying more force?You may notice that in the first instance,
figure13(a), when you release the arrow itgets separated from the bow and falls downon ground. But in second instance,figure13(b), you will notice the arrow flieswith great speed into the air.
From this activity we can conclude thatthe bow in normal shape is not able to pushthe arrow but when we stretch the string, itacquires energy to throw the arrow into airwith a great speed. The energy acquired bythe bow due to change in its shape is knownas its potential energy.
Where does the bow get this energyfrom?
Why is it not able to throw off the arrowin the first case?
Can we increase the potential energyof the bow?
Discuss with friends about changes youneed to make to the bow for increasing itspotential energy .
In the first instance of above activity,you have gently stretched the string of thebow. Thus the work done by the force onthe bow is negligible and energy transferredto the bow due to this work is alsonegligible. Hence the bow is not able topush the arrow.
In the second instance, you have appliedmore force on the string to stretch it. Thusthe work done by you on bow changed itsshape and it acquires large amount ofenergy. This energy is stored as potentialenergy in the bow which is responsible forthrowing the arrow into air with a greatspeed.
In our daily life we come across manysuch situations where the work done on anobject is stored as potential energy in it andused to do various other works.
For example the work done in windingthe key of toy car is stored as potentialenergy in it and forces the car to move onthe ground when released.
Do the following activities for thebetter understanding of potential energy.
Activity-6
Observing the energy instretched rubber band
Take a rubber band hold it at one endand pull it from the other end and thenrelease the rubber band at one of the ends.
What happens?
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152 Work and Energy
Potential energy of an object at height(or) Gravitational potential energy
An object increases its energy when itis raised through a height. This is becauseof the work done on object against gravityacting on it. The energy of such object isknown as gravitational potential energy.
Fig-14
The gravitational potential energy ofan object at a point above the ground isdefined as the work done in raising it fromthe ground to that point against gravity.
Consider an object of mass m raisedto height h from the ground. A force isrequired to do this. The minimum forcerequired to raise the object is equal to theweight of the object (mg). The object gainsenergy equal to the work done on it. Letthe work done on the object against gravitybe ‘W’.
That is Work done,
W = force x displacement = mg x h
= mgh.
Activity-7
Observing the energy in anobject at some height
Take a heavy ball. Drop it on a thickbed of wet sand from different heights from25cm to 1.5 m. Observe the depressioncreated by the ball on the bed of sand.Compare depths of these depressions.
� What do you notice?
� Is there any relation to the depth ofdepression and the height from whichball was dropped?
In above activity we can observe thatobjects also acquire energy sometimes due tochange in position.
Let us consider the following example
We use a hammer to drive nails into aplank. If a hammer is placed just on the tipof a nail, it hardly moves into the plank.
But if the hammer is raised to a certainheight and then allowed to fall on the nail,we observe that the nail is pushed somedistance into the plank.
Thus the energy of hammer is increasedwhen it is raised to a certain height. Thisenergy is due to the special position(height) of the hammer.
The energy possessed by an objectbecause of its position or shape is calledits potential energy
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Since the work done on the object isequal to ‘ mgh’ an energy equal to ‘mgh’units is gained by the object, This is thepotential energy of the object at a height‘h’.
P.E. = mgh.
Think and discuss
� Does the international space stationhave gravitational potential energy ?
Example 7A block of 2 kg is lifted up through 2m
from the ground. Calculate the potentialenergy of the block at that point.[Take g=9.8m/s2]
SolutionMass of the block, m = 2 kgHeight raised, h = 2 mAcceleration due to gravity, g = 9.8 m/s2
Potential Energy of the block,P.E. = m g h
= (2) (9.8) (2) = 39.2 J
Example 8A book of mss 1 kg is raised through a
height ‘h’. If the potential energy increasedby 49 J, find the height raised.
SolutionThe increase in potential energy = mghThat is, mgh = 49 J (1)(9.8)h = 49 JThe height raised, h = (49) / (1x 9.8) = 5m
Mechanical energyThe sum of the kinetic energy and the
potential energy of an object is called itsmechanical energy.
Consider the following exampleThe kinetic energy of an aeroplane at
rest is zero. Its potential energy when it ison ground is also considered as zero. Thusits mechanical energy is zero while it restson the ground. When the same aeroplaneflies it has kinetic energy as well aspotential energy, the sum these energiesgives the total mechanical energy of theaeroplane in flight.
Conversion of energyWe find in nature a number of instances
of conversion of energy from one form toanother form. Sun is the big source ofenergy in nature. The solar energy from theSun is converted into various forms likelight energy, heat energy, wind energy etc.
Apart from this, we observe variousinstances of energy conversions in day today activities, like conversion of electricalenergy into heat energy by Iron box,chemical energy into light energy by a torchetc.
Activity-8Listing the energy conversions innature and in day to day life
Discuss various ways of energyconversions in nature as well as in our dayto day activities and make a separate listof situations for natural conversions ofenergy and energy conversions in day-to-day life and write them in table-2, table-3.
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154 Work and Energy
Table -2: Energy conversions taking place in nature
S.No. Situation of energy conversion in Nature
1 Heat energy from the Sun used for preparing food by plants getsconverted into chemical energy.
2
3
4
Table -3 Energy conversions taking place in day to day activities.
S.No. Situation of energy conversion Gadgets/appliances used for energy conversion
1 Conversion of electrical energy Electric faninto mechanical energy
2
3
4
Discuss about the following questionswith your friends� How do green plants produce food?� How are fuels like coal and petroleum
formed?� What kind of energy conversions
sustain the water cycle in nature?We see other energy conversions in
nature, for example snow deposited ataltitudes melts and the water so formedflows down to seas. In the process, itspotential energy is converted into kineticenergy. We convert the kinetic energy ofwater to electrical energy in hydroelectricpower plants.
Dead plants buried deep below theearth’s surface for millions of years getconverted to fuels like petroleum and coal,which have chemical energy stored in them.
The food we eat comes from plantsources or an animal source which takesplant parts as their food.
When we eat food, several chemicalprocesses take place and the chemicalenergy stored in food gets converted in tovarious forms needed by the body. Forexample when we walk, run, exercise etc.,the energy from food is used for kineticenergy.
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Activity - 9Conservation of mechanicalenergy
Take a long thread say 50-60cm longand attach a small heavy object like a metalball at one end, tie other to a nail fixed tothe wall as shown in the figure 15.
Fig-15
Now pull the object or bob of thependulum to one side to the position A1
(Fig-15) and release it.What do you notice?You may notice that, the bob swings
towards opposite side and reaches the pointA2. It repeats this motion over and overagain.� The potential energy of the bob is
minimum at A and reaches maximum atA1 because the height of the bob ismaximum at that position.
� When the bob is released from thispoint (A1), its potential energydecreases and kinetic energy startsincreasing slowly.
� When the bob reaches the position Aits kinetic energy reaches maximum andpotential energy becomes minimum.
� As the bob proceeds from A to A1 itspotential energy increases slowly andbecomes maximum at A2
If we neglect small loss of energy dueto air resistance , the sum of potentialenergy and kinetic energy remains constantat any point on the path of motion duringthe oscillation of the pendulum.
Thus the total mechanical energy in thesystem of pendulum remains constant. Thisis called conservation of mechanicalenergy.
Similarly energy can neither be creatednor destroyed. It can only be changed fromone form to another.
This is called law of conservation ofenergy.
When a ball is dropped from a heightits gravitational potential energy decreases,but as the ball comes into motion, its kineticenergy increases. Thus a free-fall bodypossesses both potential energy as well askinetic energy during its fall to the ground.
Does the conservation of energy applyfor the system of free-fall body? How?
Let us find
Activity-10
Calculating the total energy offreefall at different heights
An object of mass 20 kg is droppedfrom a height of 4m. Compute the potentialand kinetic energy in each case given in thefollowing table and write the values inrespective column of the table. (Take g =10 ms-2.)
A
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156 Work and Energy
Height atwhich object
locatedin meters (m)
Velocity ofobject atdifferent
height [in m/s]
Potential energyEp= mgh
[in Joules (J)]
KineticenergyEk = mv2 /2
[in Joules (J)]
Total energy( Ep+ Ek)
[in Joules (J)]
Table-4
4.0 0
3.55 3
3.0 ��20
2.35 ��33
0.8 8
� What do you say about total energy ofsystem of freely falling body?
� Is the mechanical energy conserved inthe system?
Think and discuss
� Someone wanting to sell you a superball claims that it will bounce to aheight greater than the height fromwhich it is dropped. Would you buythis ball ? If yes explain, if notexplain.
� A ball, intially at the top of theinclined hill, is allowed to roll down.At the bottom its speed is 4 m/s. Nextthe ball is again rolled down the hill,but this time it does not start fromrest. It has an initial speed of 3 m/s.How fast is it going when it gets tothe bottom ?
PowerIn our day to day life we observe many
situtation where same activity is beingcompleted in different time intervals. For
example a hefty rickshaw puller is able toreach the destination in less time whencompared other thinner rickshaw puller.Sometimes we notice that a grinder in ourhome takes more time to grind 1kg of ‘ dal’when compared to the grinder inneighbours house.� Do all of us do the work at the same
rate?� Is the energy spent by the force doing
work same every time?� Do the machines consume or transfer
energy at same rate every time whiledoing a particular work?
Let us consider the following example.Raheem wanted to do some repairs in
the first floor of his building. He brought100 bricks as advised by the mason. Alabourer was asked to shift these bricksfrom ground floor to first floor. Thelabourer completed the work of shiftingbricks to first floor in one hour and askedRs 150/- for the work.
Next day the mason asked Raheem tobring 100 bricks more to complete the
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Andhra Pradesh Government Free Distribution 157
repair work. Raheem brought another 100bricks and assigned the work of shiftingbricks to some another labourer. Hecompleted the work and demanded Rs 300/- for completing the assigned work.Raheem said that he paid Rs 150/- to thelabourer yester day. But the labourer arguedthat he worked more hours hence he iseligible for more money.� Whose argument is correct?� Is work done in two cases same?� Why is there a change in rate of doing
work?In the above example the amount of
work done is same in both cases. But thetime taken to complete the work isdifferent. This means that rate of doingwork is different.
A stronger person like in the first caseof above mentioned example may do certainwork in relatively less time compared toother person, similarly a powerful machinecan do a work assigned to it in relativelyless time when compared with anothermachine.
We talk about the power of machineslike motorbikes, motorcar, water pumpingmotors etc., the speed at which thesemachines do the work is the basis for theirclassification. Power is a measure of thespeed of the work done, that is how fast orhow slow work is done.
Power is defined as the rate of doingwork or rate of transfer of energy.
If an agent does a work W in time t,then power is given by
Power = Work / time
P = W/tThe unit of power is ‘watt’ and denoted
by symbol ‘W’1 watt is the power of an agent, which
does work at the rate of 1 joule per second.We express larger rate of energy
transfer in kilowatts (kW)
1 kilowatt (kW) 1000 watts (W) 1kW 1000 J. s-1
Think and discuss
� The work done by a force F1 is largerthan the work done by another forceF2. Is it neccesary that powerdelivered by F1 is also larger than thatof F2 ? Why ?
Example 9A person performs 420 J of work in 5
minutes. Calculate the power delivered byhim.SolutionWork done by the person,W = 420 JTime taken to complete the work,
t = 5 min = 5 x 60s = 300sPower delivered, P = W / t = 420/300 = 1.4 WExample 10
A woman does 250 J of work in 10seconds and a boy does 100 J of work in 4seconds. Who delivers more power?SolutionPower, P = W / tPower delivered by woman=250/10=25WPower delivered by boy = 100/4 = 25W
Both woman and boy deliver samepower. That is, rate of doing the work bywoman and boy are equal.
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158 Work and Energy
What we have learnt
Key words
Work, Energy, Transfer of energy, Sources of energy, Conservation of energy,Kinetic energy, Potential energy, Mechanical energy, Gravitational potential.
� Two conditions need to be satisfied in order to say that work has taken place. One is, aforce should act on the object and another is, the object must be displaced or theremust be change in the position of the object.
� The work done by a force acting on an object is equal to the magnitude of force (F)multiplied by the distance moved (s). This formula for work is used in only translatorymotion of the object.
� Work has only magnitude, no direction. So work is a scalar.
� If the force acting on an object and displacement are in opposite directions then thework done by the force is taken as negative.
� If work has positive value, the body on which work has been done would gain energy. Ifwork has negative value, the body on which work has been done loses energy.
� Capability of doing work by an object or energy possessed by an object depends onposition and state of the object which is doing work.
� Whenever work has been done on an object its energy either increases or decreases.
� Sun is the biggest natural source of energy to us. Many other sources are derived fromit.
� The energy possessed by an object due to its motion is called kinetic energy.
� The energy possessed by an object because of its position or shape is called its potentialenergy.
� The sum of the kinetic energy and the potential energy of an object is called itsmechanical energy.
� Energy can neither be created nor destroyed. It can only be changed from one form toanother. This is called law of conservation of energy.
� Power is defined as the rate of doing work or rate of transfer of energy.
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1. Define work and write its units. (AS1)
2. Give few examples where displacement of an object is in the direction opposite to theforce acting on the object. (AS1)
3. Identify the wrong statement among the following. Rewrite them by making necessarycorrections. (AS1)
a) Work and energy have different units.
b) When an aeroplane takes off, the work done by its weight is positive.
c) The potential energy of spring increases when it is extended and decreases when it is compressed.
d) If the work done by external forces on a system is negative then the energy of the system decreases.
e) When a body is falling freely from a height, its kinetic energy remains constant.
f) The unit of power is watt.
4 What is mechanical energy? (AS1)
5 State the principle of conservation of energy. (AS1)
6 When you lift a box from the floor and put it on an almirah the potential energy of thebox increases but there is no change in its kinetic energy. Is it violation of conservationof energy? Explain. (AS7)
7 One person says that potential energy of a particular book kept in an almrah is 20 J andother says it is 30J . Is one of them necessarily wrong? Give reasons. (AS2 ,AS1)
8 In which of the following cases is the work done positive or zero or negative? (AS1)
a) Work done by the porter on a suitcase in lifting it from the platform on to his head.
b) Work done by the force of gravity on suitcase as the suitcase falls from porter’shead.
c) Work done by the porter standing on platform with suitcase on his head.
d) Work done by force of gravity on a ball thrown up vertically up into the sky.
e) Work done by force applied by hands of a man swimming in a pond.
Improve your learning
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160 Work and Energy
9. What is potential energy? Derive an equation for gravitational potential energy of abody of mass ‘m’ at a height ‘h’. (AS1)
10. When an apple falls from a tree what happens to its gravitational potential energy just asit reaches the ground? After it strikes the ground? (AS7)
11. Let us assume that you have lifted a suitcase and kept it on a table. On which of thefollowing does the work done by you depend or not depend? Why? (AS2 , AS1)
a) The path taken by the suitcase. b) The time taken by you in doing so.
c) The weight of the suitcase. d) Your weight.
12. When you push your bicycle up an incline, the potential energy of the bicycle and yourselfincrease. Where does this energy come from? (AS7)
13. Why does a person standing for a long time get tired when he does not appear to bedoing any work. (AS7)
14. What is kinetic energy? Derive an equation for the kinetic energy of a body of mass ‘m’moving at a speed ‘v’. (AS1)
15. A free-fall object eventually stops on reaching the ground. What happens to its kineticenergy? (AS1)
16. A man carrying a bag of total mass 25kg climbs up to a height of 10m in 50 seconds.Calculate the power delivered by him on the bag. (AS1) ( Ans : 49J)
17. A 10 kg ball is dropped from a height of 10m. Find (a) the initial potential energy of theball , (b) the kinetic energy just before it reaches the ground, and (c) the speed justbefore it reaches the ground. (AS1) (Ans: 980J, 980J, 14m/s)
18. Calculate the work done by a person in lifting a load of 20 kg from the ground andplacing it 1m high on a table. (AS1)
19. Find the mass of a body which has 5J of kinetic energy while moving at a speedof 2m/s. (AS1)
20. A cycle together with its rider weighs 100kg. How much work is needed to set it movingat 3 m/s. (AS1)
21. When the speed of a ball is doubled its kinetic energy (AS1)
(a) Remains same (b) gets doubled
(c) Becomes half (d) becomes 4 times
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22. Two bodies of unequal masses are dropped from the top of building. Which of thefollowing is equal for both bodies at any instance? (AS1 , AS2)
(a) Speed (b) force of gravity
(c) potential energy (d) kinetic energy
23. A man with a box on his head is climbing up a ladder. The work done by the man on thebox is (AS1)
(a) Positive (b) Negative
(c) Zero (d) Undefined.
24. A porter with a suitcase on his head is climbing up steps with uniform speed. The workdone by the “weight of the suitcase” on the suitcase is (AS1)
(a) Positive (b) Negative
(c) Zero (d) Undefined
25. How will the increasing energy needs and conservation of energy influence international peace, co-operation and security? Discuss. (AS7)
26. How do you appreciats role of energy conversion occuring naturally in maintaining ecologicalbalance of nature. (AS6)
27. Collect picturs showing various situation where potential energy possessed by object depends anits shape and position. Prepare a scrap book. (AS4)
28. Draw a digrame to show conservation of mechanical energy in cafe of free falling body. (AS5)
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162 Sound
In class 8 we have learnt about soundsproduced by vibrating bodies and alsolearnt how sound is transmitted through amedium and received by our ears. In thischapter we will study about nature ofsound, its production, propagation andcharacteristics.
Every day we hear sounds from varioussources like birds, bells, machines,vehicles, television and Radio etc. Our earshelp us to hear the sounds produced at adistance.
� How does sound reach our ears fromthe source of its production?
� Does it travel by itself or is there anyforce bringing it to our ears?
� What is sound? Is it a force or anenergy?
� Why don’t we hear sounds when ourears are closed?
Let us find out.
Chapter
�� SOUND
Activity-1
Sound is a form of energyTake a tin can and remove both ends to
make a hollow cylinder as shown infigure 1. Take a balloon and stretch it overthe can. Rap a rubber band around theballoon. Take a small piece of mirror andstick it on the balloon. Take a laser lightand let it fall on the mirror. After reflectionthe light spot is seen on the wall as shownin figure. Now shout directly into the openend of the can and observe the dancing light.
Fig-1 Observing vibrations of light� Why is the light ray dancing, after sound
is made in the tin?� What do you infer from this?� Can we say that sound is a form of
mechanical energy?Like the stretched rubber sheet in the
above activity, sound produced at a distance
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To see vibrations, attach a small pieceof steel wire to one of its prongs as shownin figure 2. While it is vibrating, try to drawa straight line on a piece of smoked glassas quick as possible with it. Keep the endof the wire in such a way that it just touchesthe glass. A line is formed in the form ofwave as shown in figure 2. Repeat theexperiment when the fork is not vibratingand observe the difference in the lineformed.
Fig - 2
� What do you conclude from the aboveactivity?
� Can you produce sound withoutvibration in the body?
In above activity we have producedvibrations in tuning fork by striking it witha hammer. We observe that vibrating tuningfork produces sound. Thus sound isproduced by vibrating bodies.
� Give some examples of vibrating bodieswhich produce sound.
� What part of our body vibrates when wespeak?
� Do all vibrating bodies necessarilyproduce sound?
travels through air and reaches our ears toproduce a sensation of hearing in our ears.
Do you know?
Glimpses of history of soundFrom the very early times the
question “How sound travels through air”,attracted the attention of philosophers.Pythagoras (around 570 B.C), a Greekscholar and traveler, explained that soundtravels in air due to the to and fromotion of the air particles, which actupon the ear and produce the sensationof sound. Galileo (1564-1642) andBacon (1561-1625) agreed with theabove theory but it was Newton who firstexplained the phenomenon ofpropagation of sound through air.
Production of sound
Activity-2
Observing the vibration of tuningfork
Take a tuning fork and strike one of itsprongs gently with a rubber hammer andbring it near your ear.
� Do you hear any sound?
Touch one of the prongs of the tuningfork with your finger. What do you feel?Share your feeling with your friends.
� Do you see any vibrations in the tuningfork?
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164 Sound
Do you know?
A tuning fork is an acoustic resonator; it is a steelbar, bent into a U- shape (Prongs), with a handle at thebend. It resonates at a specific constant pitch when setinto vibration by striking it with a rubber hammer. Thepitch of the tuning fork depends on the length of theprongs. It is mainly used as a standard of pitch to tuneother musical instruments.
Propagation of SoundWe know that sound is produced by
vibrating objects. The matter or thesubstance through which sound istransmitted is called the medium.
When a source of sound vibrates itcreates a disturbance in the medium nearit. This means that the condition of themedium near the source becomes differentfrom its normal condition. The disturbancecould be in the form of compression of themedium close to the source. This
How does sound travel?We know that sound is a form of
energy. It travels through the air and reachesour ears to give the sensation of sound.
If energy transfer takes place duringsound propagation, then in which form, doesit travel in air?
There may be two possible ways bywhich transfer of energy from the sourceof sound to our ears takes places. One isthat the source of sound producesdisturbances in air and they strike our ears.The other explanation is that some particlesare shot off from the source of sound andthey reach our ears.
If the second explanation is correct,vibrating body would gradually lose itsweight as particles are continuously shotoff from it, This certainly never happens,because it would lead to vanishing of theobject. Thus we can conclude that ‘soundtravels through disturbances in the form ofwaves’, can be taken as a correctexplanation.
� If sound travels in the form of awave then what is the pattern?
Membrane
Compression pulse
Rarefaction pulse
Fig-3
The device first invented in 1711 by a British musician John Shore.
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disturbance then travels in the medium. Letus see how it travels.
Consider a vibrating membrane ofmusical instrument like a drum or tabla. Asit moves back and forth, it produces asound. Figure 3 shows the membrane atdifferent instants and the condition of theair near it at those instants.
As the membrane moves forward(towards the right in Figure), it pushes theparticles of air in the layer in front of it.So, the particles of air in the layer getcloser to each other. Hence the density ofair increases locally and this layer of airpushes and compresses the layer next to it,which then compresses the next layer, andso on. In this way the disturbance movesforward. We call this type of disturbanceas compression pulse. The particles of themedium do not travel with the compressionpulse they only oscillate about a meanposition. It is the disturbance which travelsin the forward direction.
What happens when the membranemoves back (to the left)? It drags back thelayer of air near it, decreasing the densityof air there. The particles of air in the nextlayer on the right move in to fill this lessdense area. As a result, its own densityreduces. In the same way, the density of airin successive layers on the right decreaseone after the other. We say that a rarefactionpulse moves to the right.
As the membrane moves back andforth repeatedly, compression andrarefaction pulses are produced, one after
R C R C R C R C R C R C R C R C
the other. These two types of pulses travelone behind the other, carrying thedisturbance with it. This is how soundtravels in air.
Think and Discuss
Do compressions and rarefactions insound wave travel in same directions orin opposite directions ? Explain.
Types of waves
Activity-3
Demonstrating types of wavepropagation
Fig-4 Compressions and rarefactions in aslinky
1. Take a slinky, it is a spring – shaped toywhich can be extended or compressedvery easily. It is very flexible and canbe put into many shapes easily. You cansend continuous waves on a slinky. Layit down on a table or the floor as shownin the figure 4 and ask a friend to holdone end. Pull the other end to stretchthe slinky, and then move it to and froalong its length.You will see alternate compressionsand rarefactions of the coil. This issimilar to the pattern of varying densityproduced in a medium when soundpasses through it.
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166 Sound
Fig-5 Transvers waves in a slinky
2. Hang a slinky from a fixed support.Hold it gently at the lower end andquickly move your hand sideways andback. What do you observe? This willcause a hump on the slinky near thelower end.This hump travels upwards on the slinkyas shown in figure 5. What is travellingupwards? The part of the slinky that wasat the bottom in the beginning is still atthe bottom. Similarly no other part ofthe slinky has moved up. Only thedisturbance has moved up. Hence wemay say that a wave has travelled upthrough the slinky.We have discussed two examples of
wave propagation in a slinky. In the firstcase the vibrations are along the directionof wave motion and in the second case thevibrations are perpendicular to the directionof wave motion.
If the particles of the medium vibratealong the direction of wave, the wave iscalled a longitudinal wave.
If the particles of the medium vibrateperpendicular to the direction of wave, thenthe wave is called a transverse wave.
Longitudinal wave involves change inthe density of the medium, whereastransverse wave involves change in the'shape' of the medium.� What do you say about sound waves in
air by the above activity?� Are they longitudinal or transverse?
Sound waves are longitudinalAs we have seen, when a sound wave
passes through air, the layers in the mediumare alternately pushed and pulled. Thus theparticles of the medium move to and froalong the direction of propagation.Therefore sound waves in air arelongitudinal.
Characteristics of the soundwave
Four quantities play an important rolein describing the nature of a wave. Thesequantities are its wave length, amplitude,frequency and wave speed. They are calledcharacteristics of the wave. Let us learnabout these characteristics in the contextof the sound wave.
Let us consider a sound wave producedby a source such as a tuning fork. Figure 6shows the variation in the density of airnear the source at a particular time and thevariation in the density of air with distanceis also shown by a graph in Figure 6. Sincethe pressure of air is proportional todensity at a given temperature, the plot of
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density verses distance will also have thesame shape.
Fig-6It can be seen from the graph that in
portions like PQ, the density is more thanthe normal density, represents acompression. In portions like QR, thedensity is less than the normal, representsa rarefaction.
Thus the compressions are the regionswhere density as well as pressure is high.Rarefactions are the regions where thedensity as well as pressure is low. In abovedensity vs. distance graph, the peak of thegraph is called crest and valley of the graphis called trough.
1. Wave lengthAt any given instant, the density of air
is different at different places along thedirection in which sound is moving. For asource like a tuning fork, the distancebetween the consecutive position ofmaximum density (compressions) (like Cand E in the above figure 6) or minimumdensity (rarefactions) like B and E as shownin figure 6 remain the same. So these valuesget repeated after a fixed distance. Thisdistance is called the wave length of thewave. It is denoted by the Greek letter
�� (read as ‘lambda’). We can definewavelength as follows.
The distance between two consecutivecompressions or two consecutiverarefaction is called the wave length of asound wave.
Being a length, wavelength is measuredin meters. SI unit of wave length is meter(m).
Fig-7
2. AmplitudeThe amplitude of a sound wave in air
can be described in terms of the density ofair or pressure of air or the displacementof the layers of air. You know that whensound travels in air, the layers of air moveto and fro, causing compressions andrarefactions. As a result, the density andpressure of air at a place varies. Its valueincreases from the normal to reach amaximum and then reduces to a minimum.
The amplitude of density of medium isthe maximum variation in the density whensound wave passes through it. Similarly wecan define amplitude of pressure anddisplacement of the particle of a mediumwhen sound travels through it.
Den
sity
A B C D E F GCrest
TroughP Q R
Normaldensity
Distance
�����
1
Den
sity Normal
density
Distance
�����
�����
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168 Sound
“The time taken to complete oneoscillation of the density of the medium iscalled the time period of the sound wave”.It is represented by the symbol (T). Its SIunit is second (s).
Frequency is a quantity that is closelyrelated to time period. We can define thatthe frequency of sound wave as follows.
“The number of oscillations of thedensity of the medium at a place per unittime is called the frequency of the soundwave”.
We usually use the Greek letter � (readas ‘nu’) to denote frequency.
Relation between frequency and timeperiod
Let us find the relationship betweenfrequency and time period.Let the time taken for � oscillations = 1sThe time taken for one oscillation=(1/�) s
But the time taken for one oscillationis called the time period (T) and the numberof oscillations per second is called thefrequency (�).
Hence Frequency and time period arerelated as T = 1/� or ��� = 1/T
The SI unit of frequency is hertz (Hz).It is named after Heinrich Rudolph Hertz.
Distance
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Heinrich Rudolph Hertz was born on 22 February 1857 inHamburg, Germany and educated at the University of Berlin.Hewas the first to conclusively prove the existence ofelectromagnetic waves. He laid the foundation for futuredevelopment of radio, telephone, telegraph and even television.He also discovered the photoelectric effect which was laterexplained by Albert Einstein. The SI unit of frequency was namedHertz in his honour.
Fig-8 Ampiltude of a waveThus the maximum disturbance of
particles in the medium on either side ofmean position is called amplitude of wave.It is usually represented by a letter A. Theunits of amplitude depend on which termsthe amplitude is being described. Becauseif sound wave is moving through air wedescribe amplitude in terms of density andpressure. If sound wave is moving in solids,we describe amplitude in terms ofdisplacement of particles from their meanpositions.
Terms of Units ofdescribing amplitudeamplitudeDensity Kg/m3
Pressure PascalDisplacement Metre.
3. Time period and frequencyWe know that when sound is
propagating through a medium the densityof medium oscillates between a maximumvalue and minimum value.
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Larger units of frequency
Kilo Hertz (KHz) 103 HzMega Hertz (MHz) 106 HzGiga Hertz ( GHz) 109 HzTera Hertz (THz) 1012Hz
Example-1Find the time period of the wave whose
frequency is 500Hz?Solution from T = 1/ � =1/500 s
=0.002 s
Think and discuss
• Does the frequency of sound wavesdepend on the medium in which ittravels? How?
• The frequency of source of sound is10Hz. How many times does itvibrate in one minute?
• Gently strike a hanging bell (templebell) and try to listen to the soundproduced by it with a stethoscopekeeping it both at bottom portion andtop portion of the bell. Is the pitchand loudness of the sound same atthe two portions ? Why ?
4. Speed of Sound waveThe distance by which a point on the
wave, such as a compressions orrarefaction, travels in unit time is calledspeed of sound wave.
Let the distance travelled by a wave inT seconds = � metres
The distance travelled by a wave in
1second = �/T metresThus by definition, speed of sound
wave v = �/T ——————— (1)We know that frequency �� = 1/T —(2) From equation (1) & (2) we get v = � �Speed of asound wave = frequency x wave length.
The speed of a sound wave depends onthe properties such as the temperature andnature of the medium in which it istravelling. But the speed of sound remainsalmost the same for all frequencies in agiven medium under the same physicalconditions.
In common speech, speed of soundrefers to the speed of sound waves in air.However speed of sound varies fromsubstance to substance. Sound travelsfaster in liquids and nonporous solids thanit does in air. It travels about 4.3 timesfaster in water (1484 m/s) and nearly 15times as fast in iron (5120 m/s) than in airat 200 C. In dry air at 200 C the speed ofsound is 343.2 m/s. this is 1236 km/hr orabout 1km in 3s.
Think and discuss
• During a thunderstrom if you note a3 second delay between the flash oflightning and sound of thunder. Whatis the approximate distance ofthunderstrom from you.
Example-2
1. In a certain gas, a source produces40.000 compression and 40.000rarefaction pulses in 1 sec. When thesecond compression pulse is produced; the
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170 Sound
sounds and noises. The sounds whichproduce pleasing effect on the ear arecalled musical sounds while the soundswhich produce unpleasant effect are callednoises.
There are three characteristics bywhich we can distinguish a musical notefrom other.
They are 1. Pitch 2. Loudness3. Quality
1. Pitch• Sound of mosquito is shrill while sound
of lion is growl.• Female voice is shriller than male
voice.From the above examples, what
property of sound differentiates them?Pitch is the characteristic of sound
which distinguishes between a shrill soundand a growling sound. Actually pitch ofsound is the sensation conveyed to ourbrain by the sound waves falling on our earswhich depends directly on the frequencyof the incident sound waves. The greaterthe frequency of a musical note, higher isthe pitch.
first is 1cm away from the source.Calculate the wave speed.Solution
We know frequency is equal to numberof compression or rarefaction pulsestravelled per second, hence frequency(�) = 40,000 Hz
Wave length (�) = distance betweentwo consecutive compression orrarefaction pulses.
� = 1cmFrom v= � � = 40.000Hz x 1cm =
40.000cm/s = 400m/s
Sonic boomWhen a body moves with a speed
which is greater than the speed of soundin air, it is said to be travelling atsupersonic speed. Jet fighters, bulletsetc., often travel at supersonic speeds.
When a sound producing sourcemoves with a speed higher than that ofsound, it produces shock waves in air.These waves carry a large amount ofenergy. They produce a very sharp andloud sound called the sonic boom.
The sonic boom produced bysupersonic aircraft is accompanied bywaves that have enough energy to shatterglass and even damage buildings.
Characteristics of a musicalsound
In the previous class we learnt that allsounds can be roughly classified as musical
Do you know?
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Sound of lower pitch
Distance
Fig-9(a)
Sound of lower pitch
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Fig-9(b)
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The tuning fork set is prepared basedon the above frequencies.
2. LoudnessIf we strike a school bell lightly, we
hear a soft sound. If we hit the same bellhard we hear a loud sound. Can you guessthe reasons for this change? The reason forthis change in intensity of sound is due tothe another characteristic of sound calledloudness.
Loudness of sound is defined as thedegree of the sensation produced on the ear.
The loudness or softness of a sound isdetermined basically by its amplitude. Theamplitude of the sound wave depends uponthe 'force' with which the objects are madeto vibrate.
Fig-10(a) Louder sound
Fig-10(b) Soft Sound
In the above figures 10(a) and 10(b)the variation of wave disturbance with timeis shown as a graph for two sounds withdifferent amplitudes.
The amplitude of the sound wave infigure10(a) is greater than the amplitudeof sound wave in figure10(b). So the graphin figure 10(a) represents a louder soundand the graph in figure 10(b) represents asoft sound.
The loudness of the sound is measuredin decibels (dB). It signifies the soundpressure level. Human ears pickup soundsfrom 10 dB to 180 dB. The loudness ofsound is considered normal, if it is between50 dB to 60dB.
A normal human being can tolerateloudness of 80 dB. The sound above 80 dBis painful and causes various healthproblems. The decibel level of a jet enginetaking off is 120 dB.
Therefore people working near theairbase need to protect their ears by usingear plugs. Otherwise it may lead to hearingloss. Listening to very loud music throughearphones of MP3 player or mobile phonesalso leads to hearing loss because loudnessof sound means high energy is delivered to
In musical terms, the pitch of the note determines the position of the note on the musicalscale which is denoted as
Note: C D E F G A B C1
(sa) (re) (ga) (ma) (pa) (dha) (ni) (sa)1
Frequency 256 288 320 341.3 384 426.7 480 512(Hz)
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172 Sound
Think and discuss
� Two girls are playing on identicalstringed instruments. The strings of theboth instruments are adjusted to givenotes of same pitch. Will the quality oftwo notes be same? Justify your answer.
� What change, would you expect in thecharacteristic of a musical sound whenwe increase its frequency one instanceand amplitude in another instance?
Reflection of soundDoes sound get reflected at the surface
of a solid? Let us find out?
Activity-4Listening to reflected sound
Take two long, identical tubes and placethem on a table near a wall. Ask your friendto speak softly into one tube while you usethe other tube to listen. Adjust the tube untilyou hear the best sound. You will find thatyou hear your friend’s voice best when thetubes make equal angles with a normal tothe wall as shown in figure 12. Why?
Reflection of sound follows the samelaws as the reflection of light when soundis reflected. i.e., the directions in which thesound is incident and reflected make equalangles with the normal to the reflectingsurface.� What happens if you lift your tube
slightly above the table?� Are able to listen to the sound? If not
why?
our ears. Hence we should be very carefulwhen listening to music through ear phones.
3. QualityYou might have noticed different
sounds produced by different instrumentssuch as violin, piano, flute, etc. Todistinguish between two sounds, we needto learn about the quality of sound.
The quality of sound is thecharacteristic which enables us todistinguish between musical notes emittedby different musical instruments or voiceseven though they have the same pitch andloudness. It is because different waveforms are produced by different musicalinstruments. Hence the quality of a notedepends on its wave form.
Fig-11
figure 11 shows graphicalrepresentation of sound wave produced bya tuning fork, a violin and a piano playingthe same note (fundamental frequency =440Hz) with equal loudness.
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You will not be able to hear your friend’svoice clearly. Think about the planes of thetube carrying incident sound and reflectedsound. What will happens to these planesif we lift one of the pipes? If you lift oneof the pipes then the pipe carrying incidentsound, the pipe carrying reflected soundwill not be in the same plane. Hence wecannot hear a clear sound.
Fig-12
Repeat the experiment by placing flatobjects of different materials (steel andplastic trays, a card board, a tray draped withcloth, etc) against the wall and observechanges in sounds.
� Do hard surfaces reflect sound betterthan soft ones?
As you have seen in the second part ofthe activity, the reflection of sound isdependent on the reflecting surface.Generally, hard surfaces reflect soundbetter than soft surfaces. But unlike light,which is reflected well only from highlypolished surfaces, sound reflects quite wellfrom rough surfaces as well. For example,an un-plastered brick wall will reflectsound quite well.
Think and discuss
� What could be the reason for betterreflection of sound by rough surfacesthan polished surfaces?
EchoIf we shout or clap standing at a suitable
distance from a reflecting object such as atall building or a mountain, we will hear thesame sound again a little later. This soundwhich we hear is called an echo. Thesensation of sound persists in our brain forabout 0.1s. This is called persistence ofsound. To hear a distinct echo, the timeinterval between the original sound and thereflected sound must be at least 0.1s. Thismeans that if a sound produced by a sourceis reflected in less than 0.1 sec., the echowould not be heard. For sound to reflectafter 0.1 sec., what should be the minimumdistance between the source and theobstacle?
Let us now derive a formula for findingout speed of sound if we hear an echo.
Fig-13 Let the distance travelled by sound
from source to obstacle = dThen distance travelled by sound from
obstacle to source = d
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Do you know?
In an auditorium or big hall excessivereverberation is highly undesirable. Toreduce reverberation, the roof and walls ofthe auditorium are generally covered withsound – absorbent materials likecompressed fiber board, rough plaster ordraperies. The seat materials are alsoselected on the basis of their soundabsorbing properties.
Think and discuss
� In a closed box if you say hello, thesound heard will be Hellooooo…..What does it mean?
Relation between Echo andReverberation
Reverberation is quite different froman echo. A reflected sound, arriving at theposition of listener more than 0.1s afterthe direct sound is called an echo. Areflection of sound, arriving at the listenerin less than 0.1s after the direct sound iscalled reverberation.
Uses of multiple reflection ofsound1. A megaphone and a horn
Megaphones, horns, musicalinstruments such as trumpets, shehanai andloud speakers are all designed to send soundin a particular direction without spreadingit in all directions, as shown in fig 14.
In these instruments, a tube followedby a conical opening reflects soundsuccessively to guide most of the sound
Thus, total distance travelled by soundwave = 2d
Let echo time be ‘t’ Sec
Speed = total distance travelled/echo time = 2d/t
The roaring of thunder is due to thesuccessive reflections of the sound froma number of reflection surfaces, such asthe clouds and the land.
Think and discuss
� Why is an echo weaker than theoriginal sound ?
Example-3An echo is heard after 0.8s, when a boy
fires a cracker, 132m away from a tallbuilding. Calculate the speed of sound?Solution
Echo time (t) = 0.8sTotal distance travelled by sound wave 2d = 2x132 m = 264m
From, speed of sound V =2d/t V = 264m/0.8s = 330 m/s
ReverberationA reverberation is perceived when the
reflected sound wave reaches your ear inless than 0.1s after the original sound wave.Since the original sound and reflected soundwaves tend to combine, we get to hear one,prolonged sound wave.
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3. Designing of concert halls andcinema halls
Generally the ceilings of concert halls,conference halls and cinema halls aredesigned such that sound after reflectionreaches all corners of the hall, as shown infigure 16. In some halls a curved sealing isarranged in such a way that the sound afterreflecting from the ceiling spreads evenlyacross the hall.
Fig-16
Think and discuss
� Why do we put cushions on the chairs,carpet on the floor, straw materials onthe walls in cinema halls?
Range of hearingThe human ears are able to hear sound
in a frequency range of about 20 Hz to20,000 Hz. The above range of frequenciesis thus written as 20 Hz- 20 KHz. We cannothear sounds of frequencies less than 20 Hzor more than 20 KHz. These limits varyfrom person to person and with age.Children can hear sounds of somewhathigher frequencies, say up to 30 KHz. Withage, our ability to hear high-frequencysound diminishes. For the elders, the upperlimit often falls to 10-12 KHz. However,
waves from the source in the forwarddirection towards the audience
Fig-14
Think and discuss
� What is the advantage of havingconical openings in Horns,megaphones etc ?
2. StethoscopeStethoscope is a medical instrument
used for listening to sounds producedwithin the body, chiefly in the heart orlungs. In stethoscopes the sound of thepatient’s heartbeat reaches the doctor’sears by multiple reflection and amplifyingthe sound as shown in figure 15.
Fig-15
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176 Sound
Ultra sounds are used extensively inindustries and for medical purposes.
Industrial applications ofultrasonic waves 1. Drilling holes or making cuts of desired shape
Holes can also be drilled usingultrasonic vibrations produced in a metallicrod, called a horn, This acts like a hammer,hammering the plate about a hundredthousand times per second.
The shape of the hole is the same asthat of the tip of the horn. . Ultrasoniccutting and drilling are very effective forfragile materials like glass, for whichordinary methods might not succeed.
2. Ultrasonic cleaningWe normally clean dirty clothes, plates
or other large objects by dipping them in adetergent solution and then rubbing andwashing. Parts located in hard-to-reachplaces, cannot be easily cleaned by thismethod.
Ultrasonics help in cleaning suchobjects which are placed in a cleaningsolution and ultrasonic waves are sent intothe solution. This causes high-frequencyvibrations in the solution. These vibrationsknocks off all dirt and grease particles fromthe objects, which are then removed usingordinary water.3. Ultrasonic detection of defects in metals
Metallic components are used inbuildings, bridges, machines, scientificequipment, and so on.
Do you know?
we take 20-20,000 Hz as the audible rangefor an average person.
Even in the audible range, the humanear is not equally sensitive to allfrequencies,. It is the most sensitive tofrequencies around 2,000-3,000 Hz, whereit can hear even a very low-intensity sound.
Sound of frequency less than 20 Hz isknown as infrasonic sound or infrasound.Sound of frequency greater than 20 KHz isknown as ultrasonic sound or ultrasound.
Different animals have differentranges of audible frequencies. A dog canhear sounds of frequencies up to about50 KHz and a bat, up to about 100 KHz.Dolphins can hear sounds of even higherfrequencies. These animals can alsoproduce ultrasonic waves andcommunicate using them. Bats useultrasonic waves to navigate, as we shallsee a little later. Animals such aselephants and whales produce sounds offrequencies less than 20 KHz . Scientistshave found that elephants grieve overtheir dead by producing infrasonicsounds. Some fishes can hear sounds, offrequencies as low as 1-25 Hz.Rhinoceroses communicate usinginfrasonic sounds of frequency around5 Hz.
Applications of ultrasoundUltra sounds are high frequency sound
waves. They are able to travel along a welldefined path in gaseous and liquid media.
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If there are cracks or holes inside themetal used, the strength of the structure orcomponent is reduced and it can fail. Suchdefects are not visible from the outside.Ultrasonic waves can be used to detect suchdefects.
Medical applications ofultrasound 1. Imaging of organs
Ultrasonic waves have given doctorspowerful and safe tools for imaging humanorgans. Echocardiography is a technique inwhich ultrasonic waves, reflected fromvarious parts of the heart, form an imageof the heart.
Ultrasonography is routinely used toshow doctors, the images of a patient’sorgans such as the liver, gall bladder, uterus,etc. It helps doctors to detect abnormalitiessuch as stones in the gall bladder, tumors,etc.
It is also used to monitor the growthof a foetus inside the mother’s womb.
Ultrasonography is safer than older X-ray imaging technique. Repeated X-rays canharm tissues, especially those of a foetus.
Fig-17 Image of Ultrasound scanning
2. Surgical use of ultrasoundThe ability of ultrasonic waves to cause
molecules of materials to vibratevigorously and thus cause certain materialsto break into tiny pieces (emulsify) isemployed in ultrasound surgery. Cataractremoval is a very common example.
Ultrasound is also employed to breaksmall stones that form in the kidneys intofine grains. These grains get flushed outwith urine. This method has eliminated theneed to perform surgery.
Think and discuss
� What is the benefit of using ultrasoundover light waves in the aboveapplications?
SONARDo you know how we can measure the
depth of the sea? Let us find out.SONAR stands for Sonographic
Navigation And Ranging. This is a methodfor detecting and finding the distance ofobjects under water by means of reflectedultrasonic waves. The device used in thismethod is also called SONAR.� How does a SONAR system work?
SONAR system consists of atransmitter and a detector which areinstalled in the “Observation Centre” onboard of a ship. From the observationcentre on board a ship, ultrasonic waves ofhigh frequencies, say 1,000 KHz, are sentin all directions under the water through
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178 Sound
transmitter. These wavestravel in straight lines tillthey hit an object such as asubmarine, a sunken ship,a school of fish, etc. Thewaves are then reflected,and are received back bythe receiver at theobservation centre. Thedirection from which areflected wave comes tothe observation centretells the direction in whichthe object is located. Fromthe time between sendingthe ultrasonic wave andreceiving its echo, and the
Fig-18
speed of sound in sea water, the distanceof the object from the observation centreis calculated. Reflections from variousangles can be utilized to determine theshape and size of the object.
Let d be the distance between the sonarand an underwater object, t be time betweensending an ultrasonic wave and receiving itsecho from the object and u is the speed ofsound in water.
The total distance covered by the wavefrom the sonar to the object and back is 2d.
Using s = ut, Or 2d = ut
d= ut
2
This method of finding distances isalso called echo ranging. Marine geologistsuse this method to determine the depth of
the sea and to locate underwater hills andvalleys.
Example 4A research team sends a sonar signal
to confirm the depth of a sea. They heardan echo after 6s.Find the depth of the sea.If the speed of sound in sea water is 1500m/s?Solution
Let the depth of the sea = d mThen Total distancetravelled by sonar signal (s) = 2dSpeed of sound in sea water (u)
= 1500 m/sTotal time taken (t) = 6sFrom, s = ut, 2d = 1500 m/s x 6s d = 9000/2 m =4.5 km
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What we have learnt
Key words
Mechanical energy, tuning fork, longitudinal wave, transverse wave, compression,rarefaction, crest, trough, density of mediam, pressure, wavelength, amplitude,frequency, pitch, loudness, quality of sound, echo, reverberation, infrasonic, sonic,ultrasonic and SONAR.
� Sound is a form of mechanical energy which produces sensation of hearing.
� A tuning fork is an acoustic resonator, which resonates at constant pitch when set intovibration.
� If the particles of the medium move to and fro along the direction of propagation ofthe wave, the waves are called longitudinal waves.
� Sound waves are longitudinal.
� The region of high density of particles in the medium during propagation of sound iscalled compression and low density regions are called rarefaction.
� The distance between two consecutive compressions or rarefactions is calledwavelength.
� The maximum variation in density or pressure from the mean value is called amplitudeor the maximum disturbance of particles of a medium from their mean position iscalled amplitude.
� The time taken to complete one oscillation of density in the medium is called timeperiod of sound wave.
� The number of oscillations of the density of the medium at a place per unit time iscalled frequency.
� The distance by which a point of on the wave, such as a compression or rarefactiontravels per unit time is called speed of sound.
� Pitch is a characteristic of sound which distinguishes between a shrill sound and adeep, low sound.
� Loudness of sound is defined as the degree of sensation produced in the ear.
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180 Sound
� The quality of sound is the characteristic which enables us to distinguish between musicalnotes emitted by different musical instruments.
� A reflection of sound, arriving at the listener in more than 0.1s after direct sound iscalled an echo.
� A reflection of sound, arriving at the listener in less than 0.1s after direct sound iscalled reverberation.
� Sound of frequency between 20Hz – 20KHz is called sonic or audible limit.
� Sound of frequency less than 20Hz is known as infrasonic sounds.
� Sound of frequency higher than 20KHz is known as ultrasonic sounds.
� SONAR stands for Sound Navigation and Ranging.
I. Pick the correct answer
1. When we say sound travels in a medium (AS1) ( )
A) The medium travels
B) The particles of the medium travel
C) The source travels
D) The disturbance travels
2. A sound wave consists of (AS1) ( )
A) number of compression pulses only
B) number of rarefaction pulses only
C) number of compression and rarefaction pulses one after the other
D) vacuum only
3. Hertz stands for oscillations per (AS1) ( )
A) second B) minute C)hour D) milli second
Improve your learning
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4. When we increase the loudness of sound of a TV, the property of sound that changes isA) amplitude B) frequency C) wavelength D) speed (AS1)
5. The characteristic of the sound that describes how the brain interprets the frequency ofsound is called (AS1) ( )
A) Pitch B) loudness C) quality D) sound
6. In a stethoscope, sound of heart beats travel through stethoscopes tube (AS1) ( )
A) By bending along the tube B) In a straight line
C) Undergoing multiple reflections D) all of the above
7. Explain the following terms a) amplitude b) wavelength c) frequency (AS1)
8. Deduce the relation between wavelength, frequency and speed of sound (AS1)
9. How are multiple reflections of sound helpful to doctors and engineers? (AS7)
10. Name two quantities that vary periodically at a place in air as a sound wave travels throughit. (AS1)
11. Which has larger frequency – infrasonic sound or ultrasonic sound? (AS7 (AS2)
12. The grandparents and parents of two-year old girl are playing with her in a room. Asound source produces a 28-kHz sound. Who in the room is most likely to hear thesound? (AS2 , AS7)
13. Does the sound follow same laws of reflection as light does? (AS1)
14. Why is soft furnishing avoided in concert halls? (AS7)
15. Two sources A and B vibrate with the same amplitude. They produce sounds offrequencies 1kHz and 30kHz respectively. Which of the two waves will have largerpower? (AS1)
16. What do you understand by a sound wave? (AS1)
17. Define the wavelength of a sound wave. How is it related to the frequency and the wavespeed? (AS1)
18. Explain how echoes are used by bats to judge the distance of an obstacle in front ofthem. (AS1)
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182 Sound
19. With the help of a diagram describe how compression and rarefaction pulses are producedin air near a source of sound. (AS5)
20. How do echoes in a normal room affect the quality of the sounds that we hear? (AS7)
21. Explain the working and applications of SONAR. (AS1)
22. Find the period of a source of a sound wave whose frequency is 400Hz. (AS1)
23. A sound wave travels at a speed of 340 m/s. If its wavelength is 2cm, what is the frequencyof the wave? Will it be in the audible range? (AS1)
24. Given that sound travels in air at 340 m/s, find the wavelength of the waves in air producedby a 20kHz sound source. If the same source is put in a water tank, what would be thewavelength of the sound waves in water? Speed of sound in water = 1,480 m/s. (AS7)
25. A man is lying on the floor of a large, empty hemispherical hall, in such a way that hishead is at the centre of the hall. He shouts “Hello!” and hears the echo of his voice after0.2 s. What is the radius of the hall? (AS7) (Speed of sound in air = 340 m/s)
26. “We know that sound is a form of energy. So, the large amount of energy produced duethe sound pollution in cosmopolitan cities can be used to our day to day needs of energy.It also helps us to protect bio diversity in urban areas”. Do you agree with this statement?Explain. (AS7)
27. How do you appreciate efforts of a musician to produce melodious sound using a musicalinstrument by simultaneously controlling frequency and amplitude of the soundsproduced by it. (AS6)
28. You might have observed that sometimes your pet dog starts barking though no one isseen near in its surroundings or no disturbance heard nearby. Does this observationraise any doubts in your mind about the peculiar behavior of dog after your understandingabout ‘range of hearing the sound’. If yes, write them. (AS2)
Project work (AS4)
Find out the information about names of animals and their photographs from internet,which communicate using infra-sonic or ultra-sonic sound and prepare a scrap book.
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