IX Physics EM Title Free.pmd - SCERT Telangana

SCERT TELANGANA

INSPIRE AWARDS – MANAK

‘Innovation in Science Pursuit for Inspired Research’ (INSPIRE)scheme is one of the flagship programmes of Department of Science &Technology (DST), Government of India. The INSPIRE Awards - MANAK(Million Minds Augmenting National Aspirations and Knowledge) aims to

• Motivate students in the age group of 10-15 years and studying inclasses 6 – 10.

• Attract intelligent students towards Sciences.

• Identifying intelligent students and encourage them to study Science from early age.

• Develop complex human resources to promote scientific, technologicaldevelopment and research.

The objective of the scheme is to target one million original ideas/innovations rootedin science and societal applications to foster a culture of creativity and innovative thinkingamong school children.

Schools can nominate best innovative ideas online through the website:www.inspireawards- dst-gov.in.

Each selected student is awarded with Rs.10,000/- (Rupees Ten Thousand). One shouldutilize the amount for making project / model and for display at District Level Exhibitionand Project Competition.

Participate in INSPIRE Awards – MANAK Scheme – Develop our Country

Step Description

A. Use Android mobile phone or tablet to view content linked to QR Code:

1. Click on Play Store on your mobile/ tablet.

2. In the search bar type DIKSHA.

3.

will appear on your screen.

4. Click Install

5. After successful download and installation, Click Open

6. Choose your prefered Language - Click English

7. Click Continue

8. Select Student/ Teacher (as the case may be) and Click on Continue

9. On the top right, click on the QR code scanner icon and scan a QR code

printed in your book

OR

Click on the search icon and type the code printed below the QR code,

in the search bar.

10. A list of linked topics is displayed

11. Click on any link to view the desired content

B. Use Computer to view content linked to QR code:

1. Go to https://diksha.gov.in/telangana

2. Click on Explore Content

3. Enter the code printed below the QR code in the browser search bar

4. A list of linked topics is displayed

5. Click on any link to view the desired content

Let us know how to use QR codes

In this textbook, you will see many printed QR (Quick Response) codes, such as

Use your mobile phone or tablet or computer to see interesting lessons, videos, documents,etc. linked to the QR code.

SCERT TELANGANA

iFree distribution by T.S. Government 2020-21

Sri M. Ramabrahmam, Lecturer,Govt. IASE, Masabtank, Hyderabad.

Dr. P. Shankar, Asst. Professor,IASE, O.U., Hyderabad.

Prof. Kamal Mahendroo,Vidya Bhawan Education Resource Centre,

Udaipur, Rajastan.

Dr. TVS Ramesh,Co-ordinator, C&T Dept.,

SCERT, Hyderabad.

Miss. Preeti Misra,Vidya Bhawan Education Resource Centre,

Udaipur, Rajastan.

Mr Kishore Darak,Vidya Bhawan Education Resource Centre,

Udaipur, Rajastan.

Co-ordinators

Academic Support

Editors

Published by Government of Telangana, Hyderabad.

Respect the LawGet the Rights

Grow by EducationBehave Humbly

PHYSICAL SCIENCESPHYSICAL SCIENCESPHYSICAL SCIENCESPHYSICAL SCIENCESPHYSICAL SCIENCES

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. NannuruUpendar Reddy,Professor & Head C&T Dept.,

SCERT., Hyderabad.

Prof. V. SudhakarDept of Education, EFLU, Hyderabad.

SCERT TELANGANA

ii

© Government of Telangana, Hyderabad.

First Published 2013New Impressions 2014, 2015, 2016, 2017, 2018, 2019, 2020

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, Telangana.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. Maplitho,Title Page 200 G.S.M. White Art Card

Free Distribution by Government of Telangana 2020-21

Printed in Indiaat the Telangana Govt. Text Book Press,

Mint Compound, Hyderabad,Telangana.

SCERT TELANGANA

iiiFree distribution by T.S. Government 2020-21

Sri K. Sudhakara Chary, SGT,UPS Neelikurthy, Warangal.

Sri Kishan Thatoju, Computer Operator,

C&T Dept.,SCERT, 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, Asst. Professor,IASE, O.U., Hyderabad.

Dr.K. Suresh, SA,ZPHS Pasaragonda, Warangal.

Sri V. Gurunadha Rao, SA,ZPPSS Parvathagiri, Warangal.

Sri Dandala 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., Hyderabad.

Sri B. Sudhakar, Director,Govt. Textbook printing press, Hyderabad.

Dr. Nannuru Upendar Reddy,Professor & Head C&T Dept., S.C.E.R.T., Hyderabad.

Sri M. Ramabrahmam, Lecturer,Govt. IASE, Masabtank, Hyderabad.

Sri A. Nagaraja Sekhar, SA,ZPHS, Chathakonda, Khammam.

Sri Y. Venkat Reddy, SA,ZPHS Chivemla, Nalgonda.

Sri Kurra Suresh Babu, B.Tech., MA., MPhill

Mana Media Graphics, Hyderabad.Sri Md. Ayyub Ahmed, S.A.,

Z.P. H.S U/M, Atmakur, Mahbubnagar.

SCERT TELANGANA

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 bythemselves. Everything has its own prominence. Human being is only a part of thenature. The aspect which distinguishes the humans from all other organisms andexclusive for them is their extraordinary thinking power. Thinking transforms aperson as a unique entity from rest of the nature. Though it usually appearssimple and normal, the intricacies of the very nature often challenges us to untiethe tough knots of its hidden secrets, day in and day out.

The human being intuitionally contemplates and searches solutions for all thecritical challenges, all around,relentlessly. Curiously, the questions and answers areconcealed in the nature itself. The role of science, in fact, is to find them out. Forthis sake, some questions, some more thoughts, and some other investigationsare 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 iswhy, Galileo Galilei, the Italian astronomer,emphasized that scientific learning isnothing 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 tocomprehend and appreciate the laws governing the nature in designing tremendousdiversity found around here and everywhere. Scientific learning is not just disclosingnew things. It is also essential to go ahead with deep understanding of the nature’sintrinsic principles;without interrupting the harmony of interrelation andinterdependence 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.

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 importanceto field experience/ experiments in science teaching.

SCERT TELANGANA

vFree distribution by T.S. Government 2020-21

There is a great need for compulsory implementation of instructions of theNational Curriculum Framework- 2005 which emphasizes linking of the science teachingwith local environment. The Right to Education Act- 2009 also suggested thatpriority 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 anew generation with scientific thinking.The key aspect of science teaching is tomake the children understand the thinking process of scientists and their effortsbehind each and every discovery. The State Curriculum Framework- 2011 statedthat children should be able to express their own ideas and opinions on variousaspects.All the genuine concepts should culminate into efficacious science teaching,make the teaching-learning interactions in the classroom, laboratory and fieldveryeffective and really become useful for the children to face the life challengesefficiently.

We thank the Vidya Bhawan Society, Udaipur (Rajasthan), Dr. Desh PandayRtd Prof. College of Engineering Osmania University and Sri D.R. Varaprasad formerLecturer ELTC Hyderabad for their cooperation in developing these new textbooks,the writers for preparing the lessons, the editors for checking the textualmatters 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 textbook so as to inculcate scientific thinking process and inspire scientific approach inthe children.

Energized Text Books facilitate the students in understanding the conceptsclearly, accurately and effectively. This book has been “Energized” with QR (QuickResponse) Codes. Content in the QR Codes can be read with the help of anysmart phone or can as well be presented on the Screen with LCD projector/K-Yanprojector. The content in the QR Codes is mostly in the form of videos, animationsand slides, and is an additional information to what is already there in the textbooks.

This additional content will help the students understand the concepts clearlyand will also help the teachers in making their interaction with the students moremeaningful.

At the end of each chapter, questions are provided in a separate QR Codewhich can assess the achievement of learning outcomes by the students.

We expect the students and the teachers to use the content available in theQR Codes optimally and make their class room interaction more enjoyable andeducative.

Director,SCERT, Hyderabad

SCERT TELANGANA

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 information of relevant website addresses and pass on to students so that they canutilize internet 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.

SCERT TELANGANA

viiFree distribution by T.S. Government 2020-21

We believe, you must have realized that 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.

SCERT TELANGANA

viii

ACADEMIC STANDARDSACADEMIC STANDARDS

S.No. Academic Standard Explanation

1.

2.

3.

4.

5.

6.

7.

Conceptual understanding

Asking questions andmaking hypothesis

Experimentation and fieldinvestigation.

Information skills andProjects

Communication throughdrawing, model making

Appreciation andaesthetic sense, 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 understandconcepts, to clarify doubts about the concepts and toparticipate in discussions. They are able to guess theresults of on issue with proper reasoning, able to predictthe results of experiments.

Children are able to do the experiments given in thetext book and developed on their own. Able to arrangethe apparatus, record the observational findings, suggestalternative apparatus, takes necessary precautions whiledoing the experiments, able to do to alternateexperiments by changing variables. They are able toparticipate in field investigation and prepare reports.

Children are able to collect information related to theconcepts given in the text book by using various methods(interviews, checklist questionnaire) analyse theinformation and interpret it. Able to conduct project works.

Children are able to communicate their conceptualunderstanding by the way of drawing pictures labellingthe parts of the diagram by drawing graphs, flow chartsand making models.

Children are able to appreciate the nature and efforts ofscientists and human beings in the development ofscience and have aesthetic sense towards nature. Theyare also able to follow constitutional values.

Children are able to apply the knowledge of scientificconcept they learned, to solve the problem faced in dailylife situations. Recognise the importance of biodiversityand takes measures to protect the biodiversity.

SCERT TELANGANA

ixFree distribution by T.S. Government 2020-21

22222

33333

55555

11111 Matter around us

Motion

Laws of motion

Gravitation

INDEX

06 June 1

11 June/July 11

10 July 31

10 August 47

10 Aug/Sept 66

10 September 81

12 Oct/Nov 99

10 November 127

08 November 147

11 December 163

10 January 188

10 February 207

Page No.Periods Month

44444 Refraction of light at plane surfaces

66666 Is matter pure

77777Atoms and molecules andchemical reactions

88888 Floating bodies

99999 What is inside atom

1010101010 Work and energy

1111111111 Heat

1212121212 Sound

1234567

v

N

Revision March

SCERT TELANGANA

x

OUR NATIONAL ANTHEM

- Rabindranath Tagore

Jana-gana-mana-adhinayaka, jaya he

Bharata-bhagya-vidhata.

Punjab-Sindh-Gujarat-Maratha

Dravida-Utkala-Banga

Vindhya-Himachala-Yamuna-Ganga

Uchchhala-jaladhi-taranga.

Tava shubha name jage,

Tava shubha asisa mage,

Gahe tava jaya gatha,

Jana-gana-mangala-dayaka jaya he

Bharata-bhagya-vidhata.

Jaya he! jaya he! jaya he!

Jaya jaya jaya, jaya he!!

PLEDGE

- Paydimarri Venkata Subba Rao

“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.”

SCERT TELANGANA

1Telangana Government Free Distribution 2020-21

The word "matter" has very specificmeaning in science. Let us try to understandthe concept of matter.

You had read about metals, non-metals;synthetic and natural fibres, acids and basesetc., in previous classes. These are allexamples of matter. All the things aroundus which exist in a variety of shapes, sizesand texture are also examples of ‘matter’.The water we drink, our food, clothes andvarious things that we use in our day to daylife, the air we breathe, even our body etc.,are examples of matter.

Anything in this world that occupiesspace and has mass is considered as matter.

States of matterIn previous classes, you had learnt that

water can exist as a solid (ice), a liquid oras a gas (water vapour).We say that solidliquid and gas are three different states ofmatter. Water can be found in all thesestates.

Is there any substance which can befound in three states like water?

Chapter

MATTER AROUND US

Now look carefully at different objectsaround you. You can classify them, intoone of the three states of matter.

For example, you can say that wood andcoal are solids and petrol is a liquid.

Milk is also 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 and gases

Shape and volume

Do solids have definite shape and fixedvolume?Take two solid objects, say a pen and a

book, and put them in different containers.Do you find any change in their shape orvolume?

You might have seen a wide range ofsolids in your surroundings.

SCERT TELANGANA

2 Matter Around Us

Imagine dropping a book or a pen onthe floor. It does not flow but remains rigidwith a definite shape, distinct boundariesand a fixed volume. This shows that solidshave a definite shape and a fixed volume.

Activity-1Identifying the shape and volumeof liquids

For doing this activity, we need ameasuring jar (cylinder) and containers ofdifferent shapes as shown in fig.--1.

Note: It is not compulsory to collectsame containers as shown in fig.- 1. Youcan collect the containers of differentshapes available to you.

We also need some liquids like wateroil and milk.

Take some water in one of thecontainers using the measuring jar.Examine the shape of water in thecontainer. Pour the same water in anothercontainer and have a look at the shape, again.Repeat the process till you completepouring of water in all containers.

What is the shape of the water indifferent containers?

Is the shape of water same or differentin all the above cases?

What shape does water take if it spillson the floor?

Take 50ml of water with the measuringjar and pour it in a tumbler. Mark the levelof water on the tumbler and remove waterfrom it.

Now measure 50 ml of the milk withthe measuring jar and pour it in the sametumbler. Mark the level of the milk on it. Are the levels of water and milk same?

Remove the milk from the tumbler. Nowpour oil into it up to the level marked for water. Can you guess the volume of oil?

This activity may seem very simple butwe observe two important properties ofliquids from this activity. The shape of the liquid depends on the

shape of the container. Though liquid takes different shapes

depending on the shape of the containerits volume remains same.

Liquids can flow easily. Hence, they arealso called "fluids".

Look into in a dictionary of science toget its meaning.

You may find that gases have no fixedshape like liquids. Gases also flow likeliquids. Hence both gases and liquids arecalled fluids. Then what are the differencesbetween liquids and gases?

Fig -1: Different shaped containers havingliquid of same volume

SCERT TELANGANA


Activity-2

Do the gases have a definite shapeand a fixed volume ?

You might have heard about CNG(Compressed Natural Gas). Go to a CNGpump and ask them where they store CNG.Also see where CNG is stored in a CNGrun vehicle. Lastly see how CNG from thepump is transferred to vehicles. Does CNG have a fixed volume? Does CNG have 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 fixed volume.

Compressibility

Activity-3Observing the compressibility ofdifferent materials

Take a 50ml syringe. Draw the pistonto suck in air. Place your finger on thenozzle and press. Observe depth of pistonmoved into syringe. Is it easy or hard topress?

Do you find any change in the volumeof air in the syringe?Now fill water in the syringe and press

the piston. When 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 fuel filling station

Fig - 2: CNG cylinder in a car

Fig - 5

SCERT TELANGANA

4 Matter Around Us

What do you observe when you pressthe wood?

Is there 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. Now a daysCNG is used in many auotmobiles. For allthese purposes, large volume of gas iscompressed into cylinders of smallvolume to make it portable.

Think and discuss

Let us stretch a rubber band. Is there achange in its shape?

Is rubber band solid or liquid? Why?(What will happen if the stretching isstopped ? What will 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 that salt is a liquid on the basis

of change in its shape? Justify your answer.Take a sponge. Observe its shape.

Can you compress it? Is it a soild? Why?Think. Is anything coming out from thesponge when it is compressed.

Why can't you able to compress awooden 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 your friend lights the incensestick, the scent in the vapour form andsmoke mixes with air and moves across theroom and reach our nose.

In this case, smoke, vapour of scent andair are gases and are highly mobile.

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 deodorant spray reachsomeone on the other end at the sametime?

Activity-5Observing the diffusion of liquids

Take 250 ml round bottomed flask with23 water in it. Use a dropper and put a few

drops of blue or red ink or Potassiumpermanganate solution slowly along the sideof flask.

What do you observe after adding the

Fig - 6: Diffusion of potassiumpermanganate in water

SCERT TELANGANA


Diffusion of two gases

Aim: To observe the speed of diffusion oftwo gases.Material required: Long glass tube withscale, liquid Ammonia, Hydrochloric acid,pieces of cotton, two rubber corks and pairof tongs.

Note: Teacher should take care ofhandling hydrochloric acid and preventthe children from touching the acid.

Procedure: Take a one meter long narrowglass tube.

Take two pieces of cotton. Soak one inhydrochloric acid solution and another inammonia solution. Insert them separatelyat the two ends of the tube with the help oftongs. At the same time close the ends ofthe glass tube with rubber cork and observe.

The hydrochloric acid gives offhydrogen chloride gas and ammoniasolution gives off ammonia gas.

Both gases react together to form awhite fumes of ammonium chloride.

Observe the white ring in the tube dueto formation of ammonium chloride.Explain. How did the two gases travel along the

tube? Which gas travelled faster?

drop of ink or Potassiumpermanganate?

Can you observe that liquids alsodiffuse into each other like gases?

How much time does it take the colourto spread evenly throughout water?

What do you conclude from thisactivity?

Activity-6Observing the diffusion ofparticles of solids into liquids

Take a beaker full of water and add afew crystals of potassium permanganate toit and observe the changes.

Repeat the experiment with crystalsof copper sulphate. Do you observe diffusion? Is it faster or slower than that observed

in the activities 4 and 5?From activities 4, 5, and 6 it is clear

that solids and liquids diffuse into liquidsand gases diffuse in to gases. Diffusion isthe movement of a liquid or gas from anarea of higher concentration to an area oflower consentration. The particles will riseuntil they are evenly distributed.

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.

During respiration oxygen diffusesfrom lungs into blood. Carbon dioxidediffuses from blood into lungs.

Solids, liquids and gases diffuse intoliquids and rate of diffusion of gases ishigher than that of liquids or solids.

Lab Activity

Fig - 7HCl NH3

SCERT TELANGANA

6 Matter Around Us

Do thisSo far you have studied some

properties that can be used to distinguishbetween solids, liquids and gases. Fill thefollowing table 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 three states.You must have seen many other

materials that can exist in different states.For example, coconut oil is usually

liquid. But on cooling 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 balls(naphthalene) being placed in clothes. Thesmell remains for some time even after theballs disappear. This is because the mothballs have changed from solid state togaseous state.

But there are some substances thatchange directly from solid state to gaseousstate and vice versa without passingthrough the liquid state. We have readabout sublimation which is one suchchange.

Solids, liquids and gases are statesof matter but you need to think about, whyare the properties of same matter differentin different states?

When does water change into ice andthen into vapour?

Why do gases diffuse faster than solidsor liquids?Scientists have tried to explain these facts

by examining the physical nature of matter.

What is matter made up of ?All matter is made of very tiny

particles. This looks as a simple statementbut it is very difficult to explain andunderstand about matter.

For this we need more details about theparticles and their arrangement insidevarious forms of matter.

Activity - 7

How small are the particles ofmatter?

Take a beaker with water and add 1 or2 crystals of potassium permanganate anddissolve them in water.

What colour do you observe ?

Now take out approximately 10ml ofthis solution and add it to 90ml of clearwater in another beaker.

What does happen to the colour of waterin second beaker ?

Fig - 8

SCERT TELANGANA


Again take out 10ml of this solutionand add to another 90ml of clear water.Carryout this process 4, 5 times as shownin fig.- 8 and observe changes in intensityof colour of the solution.

Is the water in the last beaker still coloured? How is it possible for two small

crystals of potassium permanganate tocolour a large volume of water?

What do you 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 - 8There exists space betweenparticles

Take a graduated beaker and fill it withsome water 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 the level ofwater.

Does the level of water change? Where did the salt go? Can you see it in the water?

From the activities 7 & 8 we canconclude that liquid particles in a liquidhave 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 - 9

Observing the force of attractionbetween the particles of matter

Open a water tap and allow the water toreach the ground. Now try to stop thestream of water with your finger.

Fig - 9

SCERT TELANGANA

8 Matter Around Us

Are you able to move your fingerthrough the stream of water any wherefrom the tap to ground?

What is the reason behind the streamof water remaining together?

Now try to move your finger throughan iron nail, as in the stream of water.

Are you able to do it? If yes, does it rejoin?

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 and different in differentforms 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 gases candiffuse 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 reasons forhigher rate of diffusion of gases.

1. Speed of gas particles is very high.

2. The space between gas particles is veryhigh.Similarly the greater diffusion rate in

liquids compared to solids is becauseparticles in liquids move freely and havegreater space between them whencompared to particles of solids.

Observe the 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. Fewerparticles of gas 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 weallow Bromine to diffuse in vaccum, itdiffuses faster into vaccum, because thereare no particles to obstruct in its way.

Fig - 10

Solid

Liquid

Gas

SCERT TELANGANA


Key words

Matter, states of matter, solid, liquid, gas, particles, diffusion, compressibility,forces of attraction, compressed natural gas.

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.

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 inliquids and minimum in gases.

The particles are arranged orderly in the case of solids while particles move randomlyin gases.

Diffusion is possible only when particles of matter move continously.

Rate of diffusion of gases is higher than that of liquids (or) solids.

Reflections on Concepts1. Explian diffusion phenomena based on the

states of matter. (AS1)

2. Mention the properties of solids (AS1)

3. Mention the properties of liquids (AS1)

4. Explain "fluid" with one example. (AS1)

5. Mention the properties of gases. (AS1)

6. Give two daily life situation where you observe the diffusion. (AS1)

What we have learnt

Let us Improve our learning

SCERT TELANGANA

10 Matter Around Us

Application of Concepts

1. Mention the applications of compressibility in our daily life? (AS1) 2. Mention the situtions where we use diffusion in our day-to-day life (AS1) 3. How can we smell perfume sitting several meters away from the source?(AS1) 4. How do you prove that the speed of diffusion of ammonia is more than that of the

speed of diffusion of hydrochloric acid? (AS3) 5. Give examples that the matter which will be available in different states. (AS1)

Higher Order Thinking questions1. We can't rejoin the broken chalk easily. Give reason. (AS1)2. Is the space between the particles in the matter influence the speed of diffusion?

Explain. (AS2)

Multiple choice questions1. Which of the following is available in three states in our daily life (at normal conditions)

(a) Petrol (b) Water (c) Milk (d) Kerosine [ ]2. Which of the following can be easily compressed to less volume. [ ]

(a) Iron (b) Water (c) Air (d) Wooden piece

Suggested experiments1. Conduct an experiment to observe the speed of diffusion of two substances.2. Conduct an experiment to show the space between the particles of matter and

write the report.

Suggested projects1. Make a model to explain the structure of particles in solids, liquids and gases.2. What are the factors influencing diffusion, whether the arrangement of atoms in

the substance that diffuse or the arrangement of atoms of the medium in whichthe substance is kept.

3. Some solids diffuse in liquids but not in gasses, some solids diffuse in gasses butnot in liquids. Why?

SCERT TELANGANA


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 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 questionswe need to understand the terms ‘relative’and 'relative motion'.

Great progress in understandingmotion occurred when Galileo undertookhis study of rolling balls on inclinedplanes. To understand motion, we needto understand the meaning of the word'relative', which plays an important rolein explaining motion.

What is relative ?We use many statements in our daily

life to express our views. The meaning ofa statement depends on the context inwhich it is made.

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 “Thiswater is triangular’’ can hardly be given anymeaning.

A statement has a meaning only whenthere is a relation between words.

Similarly there exists other situationsin our daily life where we use statementshaving meaning depending upon thesituation. Let us observe the followingexample.

Right and Left

As shown in the fig.-1, two persons sayA and B are moving opposite to each otheron a road.

Chapter

MOTION

v

N

SCERT TELANGANA

12 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 onthe left side of 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 his left and right sides.

Is day or night just now ?The answer depends on where the

question is being asked. When it isdaytime in Hyderabad, it is night in New-York. The simple fact is that day and nightare relative notions and our questioncannot be answered without indicating theplace on the globe where the question isbeing asked.Up and down

Can orientations like 'up' and 'down'be the same for all persons at all places?Observe the following fig.- 2.

For the person standing at A on theglobe, his position appears up and theorientation of person standing at B appears

down but for the person standing at B itappears exactly opposite. Similarly for thepersons standing at the points C and D,the directions of up and down are notsame. They change with the point ofobservation on the globe.Observe the fig-2 by inverting the book Why do we observe these changes?

Fig-2We know that earth is a sphere, the

upward direction of the vertical position onits surface depends upon the place on theearth’s surface, where the vertical 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.

AB

A

B

DC

SCERT TELANGANA


To understand the idea of motion, let us take the following hypothetical activity.Observe the fig.- 3 and follow the conversation between Srinu and Somesh who stand

beside a road as shown in the fig.-3.

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 likethe car.

Srinu : How do you decide thatthe 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: Motion in view of Passenger

West

Passenger

Driver

Fig-3: Motion in view of Srinu and Somesh

Somesh

Srinu

East

SCERT TELANGANA

14 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 answer may 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' or 'rest' of an objectdepends on the observer. So motion is acombined property of the observer and thebody which is being observed.

Now we are able to define motion ofan object.

A body is said to be in motion whenits position is changing continuously withtime relative to an observer.

Note: Any object can be taken as apoint of 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.

Fig-5:Distance - displacement

Fig.- 5 shows the path taken by the ballwhen it was thrown into air. In the aboveexample (Fig-5), A to S to B shows the pathtravelled by the ball and is the actualdistance travelled by it. AB is the straightline drawn from the initial position to thefinal position. It shows the distance as wellas the direction of the displacement fromA to B and is called displacement vector.

To describe a physical situation, somequantities are specified with magnitude aswell as direction. Such a physical quantityis called vector. So, displacement is avector. The physical quantity which doesnot require any direction for itsspecification is called scalar. So distanceis a scalar.

SCERT TELANGANA


Sirpur Kaghaz Nagar

Kazipet

Hyderabad

Think and discuss

A vector can be represented as adirected line segment. It’s length indicatesmagnitude and arrow indicates it’sdirection. Point ‘A’ is called tail and initialpoint ‘B’ is called head or end point.

We represent the displacement vectorby AB

where A to B straight arrow is thedirection and the shortest (straight line)distance AB is the magnitude of thedisplacement vector AB

. The SI unit of distance or displacement

is metre denoted by ‘m’. Other units like kilometre, centimetre

etc. are also used to express thisquantity.1 km = 1000 m1 m = 100 cm

Activity-2

Drawing the displacementvectors

A car moves along different paths asshown in figures 6(a) and 6(b). The pointsA and B are the initial and final positionsof the car.

Draw the displacement vectors for twosituations given below.

Fig-6(a) Fig-6(b)

Generally, the distance covered anddisplacement are time dependentquantities.

What is the displacement of the bodyif it returns to the same point from

where it started? Give one examplefrom daily life.

When do the distance and magnitudeof displacement become equal?

Average speed and averagevelocity

A train named Telangana express startsat 2.00 pm from Sirpur Kaghaz Nagar andreaches Hyderabad at 8.00 pm the sameday as shown in fig.- 7.

Fig-7Draw displacement vectors from

Sirpur Kaghaz Nagar to Kazipet, Kazipet toHyderabad and from Sirpur Kaghaz Nagarto Hyderabad.

Let the distance of the entire trip fromSirpur Kaghaz Nagar to Hyderabad be 300km.

The journey time is 6 h. What is thedistance covered by the train in each hour?

It is equal to 300km/6h = 50km/h

SCERT TELANGANA

16 Motion

Think and discuss

Can you say that the train has coveredexactly 50 km in each hour?

Obviously the answer is “No”, becausethere may be some variations in distancecovered by the train each hour. So we takethe average 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 in theabove example be 120 km due South–West.What is the displacement in each hour?

The displacement per hour = 120 km/ 6h South - West = 20km / h South - 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 donot give any information about the motionof the 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 25h. What is his average speedduring the journey?

Can you measure the average speed and average velocity? How can you differentiate speed and average velocity? Let us do some activities to undersandabout speed and velocity.

Activity-3Measuring the average speed

Choose two points (say A & B) 50meters apart in the school play ground.Ask some students to stand at point A. Askanother group of students with stopwatches to stand at B.

When you clap your hand, thestudents at A start running towards thepoint B in any path. At the same time thestudents at B start their stop watches.

See that for each runner there is astudent at B to measure the time taken forcompleting the race. Note the time takenby each student to cover the distancebetween the points A and B in table.

Table -1Student Time taken average

to reach B (sec) speed m/s

A1 t1= s1=A2 t2= s2=A3 t3= s3=

The student who took the least time to reachB (from A) is said to be the fastest runner, i.e.,he/she has the greatest average speed.

Measuring the average velocityRepeat the whole activity after drawing

a set of parallel straight lines from A to Band ask each student to run along a line (This

Total distance Time taken

Average speed =

Total Displacement Time takenAverage velocity =

SCERT TELANGANA


ensures that each student is covering thesame distance along a straight line specifiedfor him/her from A to B)

Measure the time taken by each studentand note it in a table as shown above andcalculate the average velocity of eachstudent. The student who took the least timeto reach B from A along the line is said tohave run with the greatest average velocity.What difference did you notice betweenthe two activities? Why are we calling the ratio of distanceand time as average speed in first activityand as average velocity in second activity?Discuss with your teacher.

Speed and VelocityObjects in motion often have

variations in their speeds. For example acar which travels along a street at 50 km/h, get slowed down to 0km/h at a red lightand then attain a speed of 30 km/h due totraffic on the road. Can you find the speed of the car at a

particular instant of time?You can tell the speed of the car at any

instant by looking at its speedometer. The speedat any instant is called instantaneous speed.

We can describe the motion of a carmoving along a straight road with varyingspeed using a distance – vs – time graph.

Along the horizontal axis we plot thetime elapsed in seconds, and along thevertical axis the distance covered inmetres.

A general case of motion with varyingspeed is shown in fig.- 8.

Fig-8: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. It is

Then we calculate average speed for avery short time interval encompassing thetime at an instant t3. This interval is so shortinterval, that the value of average speedwould not change materially if it was madeeven shorter.The instantaneous speed isrepresented by the slope of the curve at agiven instant of time. The slope of the curvegives speed of the car at that instant. If theslope is large, speed is high and if the slopeis 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 =

SCERT TELANGANA

18 Motion

Think and discuss

km/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.

Activity-4Observing the direction ofmotion of a body

Carefully whirl a small object tied atthe end of the string in the horizontalplane. Release the object while it iswhirling on the string. In what direction does it move?

Try to release the object at differentpoints 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 somecases the velocity of an object which is inmotion is constant but in other cases itcontinuously changes. Which motion is called uniform?

Why?Let us find out.

Velocity is the speed of an object in aspecified direction.

For Example :A car moves with 15 m/s due east. Here

15 m/s is speed and 15 m/s due east is velocity

Fig-9Velocity gives the idea of how fast the

body moves in specified direction. Velocityis a vector. It can be represented by a directedline segment. Its length indicates speed andarrow gives the direction of motion.

If a body moves in a curved path, thetangent drawn at a point on the curve givesdirection of velocity at that instant.Observe the following diagram and try todraw tangents to the curve at differentpoints. Does the direction of velocity ofbody remain constant or not?

Very often you must have seen trafficpolice stopping motorists or scooterdrivers who drive fast and fine them.Does fine for speeding depend onaverage speed or instantaneousspeed? Explain.

One airplane travels due north at 300

Fig-10:Direction of velocity at a point of path

v

N

SCERT TELANGANA


Uniform motionActivity-5

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 table2.

Table -2Time Distance

(t in seconds) (s in metres)0 0

1 4

2 8

3 12

4 16

-- --

What is the shape of the graph?

You will get a graph which resemblesthe graph shown in fig-11.

Fig-11

The straight line graph shows that thecyclist 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 beuniform when its velocity is constant."

Non uniform motion

In our daily life in many situationswhen 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 thevalues given in table 3.

Table-3

Time Distance(t in seconds) (s in metres)

0 01 12 43 94 16-- --

SCERT TELANGANA

20 Motion

What is the shape of the graph? Is it a straight line or not? Why?

Activity-6

Observing the motion of a ballon an inclined plane

Fig-12:Ball moving downthe inclined plane

Set up an inclined plane as shown in fig.-12. Take a ball and release it from the topof the inclined plane. The positions of theball at various times are shown in fig.-13.

What is the path of the ball on theinclined plane?

How does the velocity of the ballchange?

Draw velocity vectors in fig.- 12 at timest=0s, 1s and 2s

On close observation we find that whenthe ball moves down the inclined plane itsspeed increases gradually, and the directionof motion remains constant.

Set up an inclined plane as shown in fig.-13. Take a ball and push it with a speedfrom the bottom of the inclined plane sothat it moves up.

What is the path of the ball?

What happens to its speed?

Draw velocity vectors at times t=0s, 1s,2s in fig.--13.

In above two situations of activity-6,we observe that the speed changes but thedirection of motion remains constant.

Activity-7

Observing uniform circularmotion

Whirl a stone which is tied to the endof a string continuously. Draw its path ofmotion and velocity vectors at differentpositions as shown in the fig.- 14. Assumethat the speed of the stone is constant.

What is the path of the stone?

t=0st=1s

t=2s

t=2st=1s

t=0s

Fig-13:Ball moving upthe inclined plane

v

v

v

v

v

v

v

v

Fig-14

SCERT TELANGANA


Think and discuss

It is clear that the path is a circle andthe 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 examples for motion

of an object where its speed remainsconstant but velocity changes?

Activity-8

Observing the motion of anobject thrown into air

Throw a stone into the air while 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 withspeed remaining constant.

3. Both direction and speed changesimultaneously."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 butno change in direction of motion.

AccelerationWe can change the velocity of an

object by changing its speed or its directionof motion or both. In either case the bodyis said to be accelerated. Accelerationgives an idea how quickly the velocity ofa body is changing.

What is acceleration? How can weknow that a body is accelerating?

We experience acceleration manytimes 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 theseat due to the acceleration.

SCERT TELANGANA

22 Motion

Suppose we are driving a car. Let us

steadily increase the velocity from 30 km/

h to 35 km/h in 1sec and then 35km/h to

40km/h in the next second and so on.

In the above case the velocity of the

car is increasing 5 km/hr per second.

This rate of change of velocity of an

object is called acceleration.

Acceleration is uniform when equal

changes of velocity occur in equal

intervals of time.

Uniform acceleration is the ratio of

change in velocity to time taken.

The term acceleration not only applies

to increasing velocity but also to decreasing

velocity. For example when we apply

brakes to a car in motion, its velocity

decreases continuously. We call this as

deceleration. We can observe the

deceleration of a stone thrown up vertically

into the air and similarly we can experience

deceleration when a moving train comes

to rest.

Let us suppose that we are moving in

a curved path in a bus. We experience

acceleration that pushes us towards the

outer part of the curve.

Observe fig-15. The motion of an

object in a curved path at different instants

is shown as a motion diagram. The length

of the vector at a particular point

corresponds to the magnitude of velocity

(speed) at that point and arrow indicates

direction of motion at every instant.

At which point is the speed maximum?

Does the object in motion possess

acceleration or not?

We distinguish speed and velocity for

this reason and define 'acceleration' as the

rate at which velocity changes, there by

encompassing changes both in speed and

direction.

Acceleration is also a vector and is

directed along the direction of change in

velocity.

The SI unit of acceleration is m/s2

A

B C

Fig-15: Motion diagram

SCERT TELANGANA


Think and discuss

What is the acceleration of a 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“Acceleration gives an idea of how fastthe position changes.”

Equations of uniformaccelerated motion

Consider the motion of object alongstraight line with constant acceleration.(Uniform acceleration).

Then,

'' - denotes changesLet u be the velocity at the time t = 0

and v be the velocity at the time t and let sbe the displacement covered by the bodyduring time “ t” as shown in fig.- 16.

From the definition of uniformacceleration,

Acceleration,

Since the acceleration of the body isconstant.

But we know

From here onwards we manipulatethe 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 =

vt

constanta

Fig-16

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+atv – u a

t =SCERT TELANGANA

24 Motion

The equations of uniform accelerated motion are,

v = u + ats = ut + ½ at2

v2 – u2 = 2as

NOTE:

1. If the speed of an object increases, thedirection of velocity and accelerationare one and the same.

2. If the speed of an object decreases, thedirection of velocity and accelerationare in opposite directions. In such acase, at a certain instant the speedbecomes zero.

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 the air.)

v – u a

sv + u 2 =( (( (

v2 – u2 2as................. (4)=

Fig-17(a)Sign of displacement depends on position

of particle

Sign Convention RulesThese rules helps to signify the direction of displacement (s), velocity (v) and

acceleration (a)

Fig-17(b)Sign of velocity is independent of position,

depends on the directioin of motion

– +s = +ve

OOrigin

– +O

– +O

s = –ve

s = 0

– +O

– +O

– +O

– +O

Originv = +ve

v = –ve

v = +ve

v = –ve

SCERT TELANGANA


Aim To find the acceleration and velocity of

an object moving on an inclined track.Materials required

Glass marbles, book, digital clock, longplastic 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 a bookand the other end on the floor, as shown infig.--18.

Keep a steel plate on the floor at thebottom of the track. Consider the readingat the bottom of the track to be zero.

Take a marble having adequate size totravel in the track freely. Now release themarble freely from a certain distance say40cm. Start the digital clock when themarble is released. It moves down on thetrack and strikes the steel plate. Stop thedigital clock when a sound is produced.Repeat the same experiment for the samedistance 2 to 3 times and note the valuesof times in table-4.

Table-4

Repeat the same experiment for variousdistances.

Find average time and 2S/t2 for everytrail. Will it be constant and equal toacceleration? Why?

Draw distance vs time (S-t) graph forthe 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 the acceleration and drawthe S-t graph.

Give your explanation for variousaccelerations related to slopes.

The values found in this experimentare approximate.

Lab Activity

Fig-18Books

Track

Steel plate

Distance,S (cm)

Timet (s)

Averagetimet

2S/t2

t1 t2 t3

SCERT TELANGANA

26 Motion

Example 1A car is moving with the initial velocity

15 m/s. Car stoped after 5s by applicationof breaks. Find the retardation(Decelaration).

Solution

t = 5 s

v = 0 m/s

u = 15 m/s

a = ?

Substituting the values in the followingequation.

v u at0 15 a 5

15a5

a –3 m/s2

Example 2A bus is moving with the initial velocity

of 'u' m/s. After applying the breaks, itsretardation is 0.5 m/s2 and it stoped after12s. Find the initial velocity (u) anddistance travel by the bus after applying thebreaks.

Solution

a = –0.5 m/s2

v = 0 m/s

t = 12 s

u = ?

v u at0 u 0.5 120 u 6

u 6 m/s

Initial velocity of the bus 6 m/s.

2

2

1s ut at2

112 6 0.5 122

172 722

36 m

Bus has stoped 36 m distance afterapplying the break

Example 3At a distance L= 400m away from the

signal light,brakes are applied to alocomotive moving with a velocity, u = 54km/h.Determine the position of rest 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 ,

v = u + at

Here u = 54 km/h = 54 x 5/18 =15 m/s

Let v = 0 at time 't' and given

a = -0.3 m/s2

From v = u+at we get v uta

We get, 15t0.3

= 50 s

SCERT TELANGANA


During which it will cover a distance

2us2a

215

2 0.3

2250.6

= 375 m

Thus in 1 minute after the applicationof brakes the locomotive will be at adistance l = L – s = 400 – 375 = 25 m fromthe signal light.

Example 4What is the speed of the body moving

with uniform acceleration at the midpointof two points on a straight line, where thespeeds are u and v respectively?

SolutionLet 'a' be the constant acceleration and

s be the distance between the two points,From equation of motion

v2 – u2 = 2as................. (1)

Let v0 be the speed of the body atmidpoint ‘M’ of the given points.

Applying the same equation used above,we get

2 20

sv u 2a2

From (1), we get

2 22 20

v uv u2

2 22 20

v uv u2

2 2 220

v u 2uv2

2 2

0v uv

2

Example 5A car travels from rest with a constant

acceleration 'a' for 't' seconds. What is theaverage speed of the car for its journey ifthe car moves along a straight road?

SolutionThe car starts from rest, so u = 0The distance covered in time t

21s at2

Average speed = TotaldistanceTime taken

2at2

vt

at2

s

u M v0v

Fig-19SCERT TELANGANA

28 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 aspecified direction.

Average speed is distance covered per unit time and average velocity is displacement ina specified direction per unit time.

Speed at an instant is instantaneous speed which gives the idea of how fast the positionof 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

2

2 2

v u at

1s ut a t

2

v u 2as

= +

= +

− =

Reflections on concepts

1. Distinguish between speed and velocity. (AS1)

2. Briefly explain about constant acceleration?

3. How can you say that a body is in motion? Is it a common property? (AS1)

4. Are average speed and average velocity same? If not explain why? (AS1)

5. How do you measure instantaneous speed? (AS1)

6. Explain acceleration with an example ? (AS1)

What we have learnt


Key wordsRelative, distance, displacement, average speed, average velocity,

instantaneous speed (speed), velocity, uniform motion, acceleration,uniformacceleration, rectilinear motion, vectors, scalars.

SCERT TELANGANA


Application of concepts

1. In the given figure distance vs time graphs showing motion oftwo cars A and B are given. Which car moves fast ? (AS1)

2. A train of length 50m is moving with a constant speed of 10m/s.Calculate the time taken by the train to cross an electric poleand a bridge of length 250 m. (5s , 30s) (AS1)

3. Draw the distance vs time graph when the speed of a body increases uniformly. (AS5)

4. Draw the distance – time graph when its speed decreases uniformly.(AS5)

5. What is the average speed of a Cheetah that sprints 100m in 4sec. ? What if it sprints50m in 2sec? (25 m/s)( AS7)

6. A car travels at a speed of 80 km/h during the first half of its running time and at40 km/h during the other half. Find the average speed of the car. (60 km/h) ( AS7)

7. 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. (AS7)

(7/6 m/s, 1/3 m /s2, 8.33m)

Higher Order Thinking questions

1. 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 )

2. 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 wokeup and ran, but rabbitrealized that the tortoise had won the race. Draw distance vs time graph for this story.(AS5)

Multiple choice questions1. The distance travelled by an object in a specified direction is [ ]

(a) Speed (b) Displacement

(c) Velocity (d) Acceleration

S

A

B

t

SCERT TELANGANA

30 Motion

2. If an object is moving with constant velocity then the motion is [ ]

(a) Motion with Non uniform acceleration

(b) Motion with Uniform Acceleration

(c) Uniform Motion (d) Non uniform Motion

3. If there is change in the velocity of the object then the state of object with respectto motion is [ ]

(a) State of Constant Speed (b) State of Constant velocity

(c) State of Uniform Motion (d) State of Non uniform Motion

4. If the acceleration of a moving object is constant then the motion is saidto be [ ]

(a) Motion with Constant Speed (b) Motion with Uniform Acceleration

(c) Motion with Uniform Velocity (d) Motion with Non Uniform acceleration

Suggested experiments1. Conduct an expermiment to find acceleration and velocity of an object moving on an

inclined plane and write a report.

Suggested projects1. Calculate the avarage speeds of students of your class we who have participated in

100 meters and 200 meters running race. Write a report.

2. Suppose that the three balls shown in figure below start simultaneously from the topsof the hills. Which one reaches the bottom first ? Explain. (AS2)

SCERT TELANGANA


We observe the changes in motion ofmany objects around us. We discussed theconcepts of velocity and acceleration inthe chapter ‘Motion’.

Philosophers of the ancient worldwere very much interested in the study ofmotion. One question always popping upin their mind was, what is the natural stateof an object if left to itself? Ourcommonsense tells that every movingobject on earth if it is left free for sometime, gradually comes to rest by itself.What happens if you stop peddling yourbicycle? It slows down gradually andfinally stops.

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 stated that an objectin motion will remain in same motion aslong as no external force is applied on it.

Galileo came up with two thoughtfulexperiments. He did his experiments on

inclined planes with smooth surfaces andobserved that the smoother the surface, thefarther the ball travelled. He extended thisargument and concluded that if the surfacewas perfectly smooth, the ball will travelindefinitely, until encountered by anotherobject. (In real world such a surface ofcourse does not exist).

Fig-1: (a) downward motion (b) upwardmotion (c) motion on a plane surface

As shown in fig.- 1 (a) he observed thatwhen a marble rolls down a slope, it picksup speed due to the force of gravity of theearth.

In fig.- 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 fig.- 1(c) it hasno reason to speed up or slow down. So itwill continue to move with constantvelocity.

Chapter

LAWS OF MOTION

1(a)1(b)

1(c)

SCERT TELANGANA

32 Laws of Motion

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 surfaceGalileo observed that, as shown in fig.-

2 (a), the marble released from its initialheight rolled down due to the force of gravityand then moves up the slope until it reachedits initial height. Then he reduced the angleof the upward slope and did the sameexperiment as in fig.- 2 (b). The marblerolled up the same height, but it had to gofarther in this instance. That means thedistance travelled by it is greater. He madehis observation by further reduction in theangle of the upward slope, he got the sameresults. To reach the same height the ballhad to go farther each time.

Then a question arose in his mind,“How far must it has to move to reach thesame height if it has no slope to go up”?Since it has no slope to go up as shown infig.- 2(c), obviously it should keep onmoving forever along the level surface at

constant velocity. He concluded that thenatural state of a moving object, if it is freeof external 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 force acts on it.

Galileo imagined a world where thereis no friction. But as we learnt in classVIII this is not possible in reality becausefriction, which affects the motion of anobject plays an important role in our lives.For example, if there were no friction wewould not have been able to walk onground, we would not have been able tostop a fast moving car etc. It is verydifficult to perform many physicalactivities without friction. Built uponideas primarily developed by Aristotle andGalileo, Sir Isaac Newton proposed histhree fundamental laws which explain theconnection between force and a changein motion. These three laws are popularlyknown as Newton's laws of motion.

First law of motionThe first law of motion can be stated

as follows: "Every object will remain atrest or 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 forceacts on 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)

SCERT TELANGANA


Do you know?

Activity-1

Observing the motion of a coinkept on thick paper

Take a thick paper. keep it on a glasstumbler. Now take a coin and keep it onthe centre of the paper. As shown infig.- 3. Push the paper with your finger asfast as you can. What do you observe? What happens to the coin?

Fig-3: Fast pushing of thick paperkept on a glass tumbler

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 therate of descent.

Galileo was also a remarkablecraftsman. He developed a series oftelescopes whose optical performance wasmuch better than that of other telescopesavailable during those 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, contraryto what was believed at that time.

SCERT TELANGANA

34 Laws of Motion

Similarly when you are travelling ina bus, the sudden stop of the bus makes youfall forward, due to dynamic inertia of thebody.

The body which is in motion alwaystries to move in same direction until somenet force act on it. This property is knownas dynamic 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, untila force slows it down or stops it. If the ballgoes high, the force of gravity slows it down.If the ball rolls on the ground, the force offriction makes the ball slow down and 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 of an objectis either zero or constant.

Activity-2

Observing the motion of thecoins hit by a striker

Fig-4: Hitting the stack of coinswith a striker

Make a stack of carrom coins on thecarrom board.Give a sharp hit at thebottom of the stack with striker. You canfind that the bottom coin will be removedfrom the stack, and the others in the stackwill slide down as shown in fig.- 4.

What are your observations from the above activities? Why does the coin fall inside the

glass tumbler? Why does the stack of carrom coins

fall down vertically?To understand this, we have to discuss

some more examples which we face in ourdaily lives.

For example when the bus whichis at rest begins to move suddenly, theperson standing in the bus falls backwardbecause of static inertia of the body. Theobject at rest will try to remain at rest. untillwe apply an external force there will beno change in the position of the object.This is known as static inertia.

SCERT TELANGANA


Thus when the net force acting on abody is zero, we say that the body is inequilibrium. Newton's first Law of motionis also known as the Law of Inertia.Inertia and mass

We have learnt that inertia is the propertyof an object that resists changes in its stateof motion. All objects have this tendency. Do all the bodies have the same inertia? What factors can decide the inertia of

a body?Which is easy for you, to push a

bicycle or a car? You can observe that it isdifficult to push the car. We say car hasgreater inertia than the bicycle. Why doesthe car possess 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 blockswith same 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 the 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 atable cloth is jerked from a table,leaving the dishes that were on thecloth nearly in their originalpositions. What do you need to perform this

successfully? Which cloth should we use? Is it

cloth 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?

SCERT TELANGANA

36 Laws of Motion

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 what

happens to an object when non-zero netforce acts on it.

Place a ball on the verandah and pushit gently. 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 the forceapplied on it and how we measure a force.

Linear momentumLet us recall our observations from 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 toits high speed. We all know that a heavytruck causes more damage than a bicycleif both hit a wall. These can be explainedby a concept called 'momentum' which isusually denoted by 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 phrase “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 becausevelocity is a vector. Hence, the directionof momentum is in the direction ofvelocity.

The SI unit of momentum is kg – m/sor N-s

SCERT TELANGANA


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 onthis ice block and observe the acceleration.

Fig-8: Same forces applied on abjects ofdifferent masses

In both the cases, the objectaccelerates. But we can observe in the

second case, it does not speed up as quicklyas before.From the above examples what have you

noticed?The larger the net force the greater the

acceleration, if the mass of the body isconstant, and also larger the mass lesserthe acceleration, if a constant net force isapplied.

According to Newton's 'Princpia'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 Change in momentum

Time

netpFt

p is the change in momentum of a

particle or a system of particles broughtabout by the net force in a time interval t.

When the symbol of proportionality isremoved, a constant is inserted in theequation.

netpF kt

(k is constrant)

The SI units of momentum and timeare ‘kg- m/s’ and‘s’ respectively. The unitof force is so chosen that the value ofconstant ‘k’ becomes 1. So that,

netpFt

We know p=mvso that,p mv)If the mass of the body is constant

during its motion then,

F F

SCERT TELANGANA

38 Laws of Motion

p = mvNow we have,

Fnet = mvt

We know that vt

= 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= pt

is a universal formula that can

be applied for any system whereasFnet= ma can be applied only for asystem with constant 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’)

Example 2Atwood machine

Fig-9

Atwood used the system to proveNewtons laws of motion. Atwood machineconsists of two loads of masses m1 and m2attached to the ends of a limp ofinextensible string as shown in the fig.- 9.The string runs over a pulley. Find theacceleration of each load and tension in thestring (m1>m2)

SolutionFrom fig.- 10 we know that tension of

string always tries to pull the bodies up.

Fig-10

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)

1 Tm g

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.

SCERT TELANGANA


From the FBD of mass m2

Fnet = T- m2g = m2a ------------ (2)

2T m g

Solving (1) and (2) equations, we get

1 2

1 2

( )m m gam m

1 2

1 2

2m m gTm m

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 fig.- 12. Pull the springbalances in opposite directions as shownin fig.- 12.

Fig-12: 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 show different readingsby pulling them simultaneously inopposite directions? Why not?

According to third law of motion, whenan object exerts a force on an other object,the second object also exerts a force onthe 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 youthinking that the ground also exerts someforce in the opposite direction on you?

Think and discuss

Observe the following diagram.

Fig- 11What is the upper limit of weight thata strong man of mass 80kg can lift asshown in figure?

What is the momentum of a ceiling fanwhen it is rotating?

Is it possible to move in a curved pathin the absence of a net force?

SCERT TELANGANA

40 Laws of Motion

A

B

magnitude and opposite in direction.When a fish swims in water, the fish

pushes 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 fig.--14

Fig-14 : 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 fig.- 15.Hold one end of the thread and ask yourfriend to hold the other end of thethread.

Is it not surprising to hear that whenyou push a wall then the wall pushes youback!

Fig-13: Action and reaction forcesIf two objects interact, the force FBA

exerted by the object ‘A’ on the object ‘B’is equal in magnitude and opposite indirection to the force FAB 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, whereasNewton’s third law of motion applies to aninteraction between the two bodies. Notethat the two forces in Newton’s third lawnever act 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 in

SCERT TELANGANA


Now remove your fingers from theballoon’s neck so as to release the airfrom the balloon.

What happens now ?

Fig-15: 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 fig.-.15

This is balanced by the elastic forcesof the balloon.

Fig-16: The forces on the inside wall of aballoon

When you release the neck of the balloonto allow air to escape from the balloon, whathappens? The balloon moves in a directionopposite to that of escaping air. Themomentum of the balloon and air is zerobegin with, when the air escapes with samevelocity, the balloon moves in the oppositedirection to balance the momentum of theescaping air.

Aim: To show the action and reactionforces acting on two different objects.Material required: Test tube, rubber corkcap, Bunsen burner, laboratory stand, waterand thread.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 fig.- 17.

Heat the test tube with a Bunsen burneruntil water vaporizes and cork capblows out.

Fig-17

Observe the movement of the test tubewhen rubber cork cap blows out.Comparethe directions of movement of test tube aswell as rubber cork cap. Observe thedifference in the velocity of rubber corkcap and that of recoiling test tube. What do you infer from above

experiment?

Lab Activity

Test tube

Water

String

cork cap

SCERT TELANGANA

42 Laws of Motion

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 equaland opposite? 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 momentumLet two objects with masses m1 and m2

are traveling with different velocities u1 andu2 respectively in the same direction alonga straight line. If u1>u2 they collide witheach other and the collision lasts for time‘t’, which is very small.During the collisionthe first marble exerts a force on the secondmarble F21 and the second marble exerts aforce on the first marble F12. Let v1 and v2

be the velocities of the marblesrespectively after collision.

Fig-18: 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 tpt

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

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 of"Conservation of Momentum" statesthat in the absence of a net external forceon the system, the momentum of thesystem remains unchanged".

SCERT TELANGANA


ImpulseIt will be surprising if anybody says that

the fall doesn’t hurt, but it is the sudden stopat the end that hurts you. Is it true?

Why does a pole vault jumper land onthick mats of foam?

Is it safe to jump on sand rather than acement floor? Why?

A softer and more cushioned landingsurface 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= pt

In order to minimize Fnet , you have tomaximize the stopping time.

We get Fnet t = p

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 observe thefollowing 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-19: (a) Fig-19: (b)Fig-19: (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 = Fnet1 t1

When we drop the egg on a cushionedpillow, it doesn’t break because a smallerforce acts on the egg for longer time.

p = Fnet t2

Even if the p is the same in both thecases, the magnitude of the net force (Fnet)acting on the egg determines whether theegg will break 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 longer

SCERT TELANGANA

44 Laws of Motion

Key words

What we have learnt

From the conservation of linearmomentum, We get,

m1v1+m2v2 = 0m1v1 = - 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.

Laws of motion, Inertia, Mass,Linear momentum, Conservation ofmomentum, Impulse, Impulsive force

First Law of Motion: A body continuesits state of rest or of uniform motionunless a net force acts on it.

The natural tendency of objects to resista change in their state of rest or ofuniform motion is called inertia.

The mass of an object is a measure ofinertia. SI unit of mass is Kilogram(kg).

Second Law of Motion: The rate ofchange of momentum of a body isdirectly proportional to the net forceacting on it and it takes place in thedirection of net force.

time. The ball stops only when your handsstop. This shows that 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 cars for safety.Why?

Example 3A 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.Solution

Since the pressure of the powder gasesin the bore of the cannon is an internalforce the net external force acting oncannon during 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

SCERT TELANGANA


Linear momentum of a body is theproduct of its mass and velocity.p = mv

One ‘Newton’ is the force which whenacting on a body of mass 1 kg, producesan acceleration of 1 m/s2

1 newton (N) = 1kg x 1 ms-2

Third Law of Motion: If one objectexerts a force on other object, thesecond object exerts a force on the firstone with equal magnitude but inopposite direction.

Reflections on Concepts1) Explain the reasons for the following. (AS1)

a) Why dust comes out of a carpet when it is beaten with a stick?b) Luggage kept on the roof of a bus is tied with a rope.c) Why a pace bowler in cricket runs from a long distance before he bowls?

2) Illustrate an example of each of the three laws of motion.(AS1)3) Explain the following (AS1)

a) Static Inertia b) Inertia of motion c) momentumd) impulse e)impulsive force

Application of Concepts

1) Two objects have masses 8 kg and 25 kg. Which one has more inertia? Why?(AS1)

2) What is the momentum of a 6.0 kg ball bowling with a velocity of 2.2 m/s?(Ans: 13.2 kg m/s2)(AS1)

3) 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)4) A man of mass 30 kg uses a rope to climb which bears only 450 N. What is the

maximum acceleration with which he can climb safely? ( Ans: 15 m/s2 ) (AS7)

Higher Order Thinking questions

1. A 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 motion of the vehicle) (AS7)


SCERT TELANGANA

46 Laws of Motion

2. 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: 3 m/s in opposite direction) (AS7)

3. If a fly collides with the windshield of a fast-moving bus, (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?

Multipule Choice Questions

1. The scientist who said that "an object in motion will remain in same motion as long asno external force is applied" is [ ]a) Aristotle b) Galileo c) Newton d) Dalton

2. If the net force acting on an object is zero, then the body is said to be in the state of[ ]

a) Equilibrium b) Motion c) Inertia of motion d) Uniform motion

3. The inertia of a body depends on [ ]a) Shape b) Volume c) Mass d) Area

4. Newton used the word ‘mass in motion’ to represent [ ]a) Linear momentum b) Inertia of motionc) Velocity d) Inertia at rest

5. The S.I unit of momentum is [ ]a) m/sec b) Kg-m c) k.g.m/sec d) Kg. m/sec2

Suggested Experiments

1. Conduct an experiment to prove Newton’s first law of motion and write a report.

2. Conduct an experiment to show the action and reaction forces acting on two

different objects.

Suggested Projects

1. Observe some daily life examples for Newton’s first law of motion and explainthe situaions. Write a report on your observations.

2. Write a report on the action and reaction in the systems that you have observed inyour daily life which are the evident of Newton’s third law of motion.

SCERT TELANGANA

47Free distribution by T.S. Government 2020-21

Refraction of Light atPlane Surfaces

4Chapter

We have learnt about the reflection oflight on plane surfaces in class 7 and 8.Beauty of the nature is made apparent withlight. Light exhibits many interestingphenomena.

Let us try to explore a few of them.You might have observed that a coin kept

at the bottom of a vessel filled with waterappears to be raised. Similarly, a lemonkept in a glass of water appears to be biggerthan its size. When a thick glass slab isplaced over some printed letters, the lettersappear raised when viewed through the glassslab.• What could be the reasons for the above

observations?

Activity 1Take some water in a glass tumbler.

Keep a pencil in it. Look at the pencil fromone side of the glass and also from the topof the glass.• How does it look?• Do you find any difference between

the two views?

Activity 2Go to a long wall (of length about 30

feet) facing the Sun. Go to one end of awall and ask your friend to bring a brightmetal object near the other end of the wall.When the object is a few inches from thewall it appears distorted and you will seea reflected image in the wall as though thewall were a mirror.• Why is there an image of the object

on the wall?To answer the above questions and to

give reasons for the situations mentionedwe need to understand the phenomenonof refraction of light.

Refraction

Activity 3Take a shallow vessel with opaque

walls such as a mug. (A tin or a pan issuitable). Place a coin at the bottom of thevessel. Move away from the vessel untilyou cannot see the coin.

SCERT TELANGANA

Refraction at Plane Surfaces48 IX Class

See fig.- 1(b).

Stand there. Ask your friend to fill thevessel with water. When the vessel is filledwith water the coin comes back into view.See fig.- 1(c).• Why are you able to see the coin when

the vessel is filled with water?You know that the ray of light

originating from the coin, doesn’t reachyour eye when the vessel is empty (see fig.-1b). Hence you couldn’t see the coin. Butthe coin becomes visible to you after thevessel is filled with water.• How is it possible?• Do you think that the ray reaches your

eye when the vessel is filled with water?

If yes, draw a ray diagram from the cointo the eye. Keep in mind that the light raytravelling in a medium takes a straight linepath.• What happens to the light ray at the

interface between water and air?• What could be the reason for this

bending of the light ray in the secondinstance?The above questions can be answered

by Fermat’s principle, the principle statesthat the light ray always travels between twopoints in a path which needs the shortestpossible time to cover. Let us apply thisprinciple to our activity.

By observing the path of the ray, it isclear that the light ray changes its directionat the interface separating the two mediai.e, water and air. This path is chosen bythe light ray so as to minimize time oftravel between coin and eye. This ispossible only if the speed of the lightchanges at interface of two media. Thuswe can conclude that the speed of the lightchanges when light propagates from onemedium to another medium.

“The process of changing speed at aninterface when light travels from onemedium to another resulting in a change indirection is refraction of light. The processof refraction involves bending of light rayexcept when it is incident normally”.

Consider that light travels frommedium 1with speed v1 to medium 2 withspeed v2 as shown in figures-2(a) and (b).

fig-1(a)

Eye Light ray

Opaquewalls

Coin

Coin

Light ray

Opaquewalls

Water

Eye

fig-1(c)

Coin

Light ray

Opaquewalls

Eye

fig-1(b)

SCERT TELANGANA


• What difference do you notice in fig2(a) and Fig 2(b) with the respect torefracted rays?

• Is there any relation betweenbehaviour of refracted rays and speedsof the light?Experiments have showed that change

in the direction of light is due to changein the speed of the light in the medium.

If v2 is less than v1 then medium 2 is saidto be denser with respect to medium 1.

If v2 is greater than v1 then medium 2 issaid to be rarer with respect to medium 1.

If light ray enters from rarer mediumto denser medium then refracted raymoves towards the normal drawn at the

interface of separation of two media. Whenit travels from denser medium to rarermedium it bends away from normal. Wehave seen that the ray of light deviates fromits path at the interface. Draw a normal atthe point of incidence as shown in fig.- (3).

Let ‘i’ be the angle made by incidentray with normal and ‘r’ be the angle madeby refracted ray with the normal. Theseangles are called angle of incidence andangle of refraction respectively.

To explain the process of refraction weneed to know about a constant calledrefractive index which is the property of atransparent medium. Let us learn about it.

The extent of the change in directionthat takes place when a light ray propagatesthrough one medium to another medium isexpressed in terms of refractive index.Refractive index

Light travels in vacuum with a speednearly equal to 3x108 m/s (denoted byletter ‘c’). The speed of light is smallerthan ‘c’ in other transparent media.

Let ‘v’ be the speed of light in a certainmedium. Then the ratio of speed of lightin vacuum to the speed of light in that

V1

V2

N

N

Denser

Rarer

fig-2(b)

fig-2(a)

V1

V2

N

NN-normal

Rarer

Denser

Interface

N

i

r

N

N

fig-3

SCERT TELANGANA


Material medium Refractive index Material medium Refractive indexAir 1.0003 Canada balsam 1.53Ice 1.31 Rock salt 1.54Water 1.33 Carbon Diasulphide 1.63Kerosene 1.44 Dense flint glass 1.65Fused quartz 1.46 Ruby 1.71Turpentine oil 1.47 Sapphire 1.77Crown glass 1.52 Diamond 2.42Benzene 1.50

medium is defined as refractive index ‘n’.It is called absolute refractive index.Absolute refractive index (n)

speed of light in vacuum = speed of light in medium

n = cv ...............(1)

It is a dimensionless quantity becauseit is a ratio of the same physical quantities.Refractive index gives us an idea of howfast or how slow light travels in a medium.

The speed of light in a medium is low whenrefractive index of the medium is high andvice versa. The refractive index ‘n’ meansthat the speed of light in that medium is nth

part of speed of light in vacuum.For example the refractive index of

glass is 32 .Then the speed of light in glass

is 8

83 10 2 1032

c mv sn×

= = = × .

Table:1 Refractive indices of some material media.

Note : From table-1, you know thatan optically denser medium may notpossess greater mass density. For example,kerosene with high refractive index isoptically denser than water although itsmass density is less than water.• Why do different material media

possess different values of refractiveIndices?

• On what factors does the refractiveindex of a medium depend?Refractive index depends on the

following factors.(1) Nature of material (2) Wavelength

of light used. (You will learn about this inyour higher classes).

Relative refractive indexThe refractive index of a medium with

respect to another medium is defined as theratio of speed of light in the first mediumto the speed of light in the second medium.Let v1 and v2 be the speeds of light in thefirst and second media respectively. Then,

Refractive index of second mediumwith respect to first medium is given by

n21 = speed of light in medium - 1speed of light in medium - 2

n21 = 1

2

vv

Dividing both numerator anddenominator by c we get

SCERT TELANGANA


Lab Activity 1

= 1 21 2

1 22 1

v nv v 1 1c c n nv n

= = =

n21 = 2

1

nn ....................(2)

This is called relative refractive Index.We define relative Refractive index asfollowRelative refractive index, n21 =

2

1

Refractive index of second medium (n )Refractive index of first medium (n )

Aim: Obtaining a relation betweenangle of incidence and angle of refraction.

Materials required: A plank, whitechart, protractor, scale, small black paintedplank, a semi circular glass disc of thicknessnearly 2cm, pencil and laser light.

Procedure :Take a wooden plank which is covered

with white chart. Draw two perpendicularlines, passing through the middle of thepaper as shown in the fig.- 4(a). Let thepoint of intersection be O. Mark one lineas NN which is normal to the another linemarked as MM. Here MM represents theline drawn along the interface of twomedia and NN represents the normaldrawn to this line at ‘O’.

Take a protractor and place it along NNin such way that its centre coincides withO as shown in fig.- 4(a). Then mark theangles from 00 to 900 on both sides of theline NN as shown in fig.- 4(a). Repeat thesame on the other side of the line NN. Theangles should be indicated on the curvedline.

Now place a semi-circular glass discso that its diameter coincides with theinterface line (MM) and its centercoincides with the point O. Point a laserlight along NN in such a way that the lightpropagates from air to glass through theinterface at point O and observe the pathof laser light coming from other side ofdisc as shown in fig.- 4 (b). (If you cannotobserve the path of laser light put a black-coloured plank against the curved line andobserve the light point and imagine the lightpath).• Is there any deviation?

Send Laser light along a line whichmakes150 (angle of incidence) with NNand see that it passes through point O.Measure its corresponding angle ofrefraction, by observing laser light comingfrom the other side (Circular side) of the

fig-4(a)

fig-4(b)

SCERT TELANGANA


fig-5(a)

glass slab. Note these values in table-2. Dothe same for the angles of incidence such

as 200,300,400,500 and 600 and note thecorresponding angles of refraction.

i r Sin i Sin r Sin i / Sin r

Table 2

Find sin i, sin r for every ‘i’ and ‘r’ andevaluate sin i/ sin r for every incident angle‘i’.

Note : Take the help of your teacherto find the values of sin i and sin r foreach case.

Finally, we will get the ratio sinsin

ir as a

constant.• Is this ratio equal to refractive index

of glass? Why?This ratio gives the value of refractive

index of glass. In the above experimentyou might have noticed that ‘r’ is less than‘i’ in all cases and the refracted ray bendstowards normal in each case.• What do you conclude from these

observations?From the above experiment we can

conclude that when light ray travels froma rarer medium (air) to a denser medium(glass) the value of r is less than the valueof ‘i’ and the refracted ray bends towardsthe normal.• Can you guess what happen when light

ray travels from a denser medium to ararer medium?

Let us take up another activity to findthis.

Actitivy 4Take a metal disk. Use a protractor and

mark angles along its edge as shown inthe fig.- 5(a). Arrange two straws at thecentre of the disk in such a way that theycan be rotated freely about the centre ofthe disc.

Adjust one of the straws to make anangle 100. Immerse half of the discvertically into the water, filled in atransparent vessel. While dipping, verifythat the straw at 100 is inside the water.From the top of the vessel try to view thestraw which is inside the water as shownin fig.- 5(b). Then adjust the other strawwhich is outside the water until both strawsappear to be in a single straight line.

SCERT TELANGANA


Then take the disc out of the water andobserve the two straws on it. You will findthat they are not in a single straight line.

• Why do the straws appear to be in astraight line when we view them fromthe top?Measure the angle between the normal

and second straw. Draw table-2 again inyour notebook and note the value. Do thesame for various angles. Find thecorresponding angles of refraction andnote them in the table drawn. Use the datain the table and find refractive index ofwater. Do not take up this activity forangles of incidence greater than 48degrees. You will learn the reasons for thisin the following sections.

You will observe that in the aboveactivity, ‘r’ is greater than ‘i’ in all cases

that means when light travels from water(denser) to air (rarer). It behaves in anopposite way to that we observed in labactivity 1.

From this activity we can generalizethat when the light ray travels from denserto rarer, it bends away from the normaland r > i.• Is there any relation between the angle

of incidence and the angles ofrefraction?The relation between angle of

incidence and angle of refraction can begiven by n1sin i = n2sin r.

This is called Snell’s law2

1

nsin isin r n

⇒ =

we know 2 1

1 2

n vn v

=

1

2

sin i vsin r v

⇒ =

When light travels from onemedium to another the ratio of speeds is

1

2

vv and the ratio of refractive indices is 2

1

nn

.Hence angle of incident and angle ofrefraction will follow 1

2

sin i vsin r v

=

fig5(b)

Straw

Straw

Derivation of Snell’s Law :

Consider the following analogy to derive it.Let us imagine that a person has fallen out of

boat and is screaming for help in the water at pointB as shown in fig.- 6(a).

The line marked through point ‘X’ is the shoreline. Let us assume that we are at a point ‘A’ onthe the shore and we saw the accident. In order tosave the person we need to travel a certain distanceon land and a certain distance in water.

fig-6(a)

A

B

XSCERT TELANGANA


i

rF

D

E

r

i

C X

Bfig-6(c)

A

r

At D and C angle of refraction istaken as r because the points Cand D are assumed to be close

N

N

Y

We know that, we can run faster on land thanwe can swim in water.• What do we do to save the person?• Which path enables us to save the person in the shortest possible time?• Do we go in a straight line?

By careful thought we would realize that itwould be advantageous to travel a greater distance on the land in order to decrease thedistance in water because we go much slower in water. For whatever speeds on land andin water, the final path that one has to follow to reach the person is ACB, and that thispath takes the shortest time of all the possible paths (see fig.- 6c). If we take any otherroute, it will be longer. If we plot a graph for the time taken to reach the girl against theposition of any point when we cross the shore line, we would get a curve somethinglike that shown in fig.- 6(b). In this graph, the distance from point Y to the points like D,C are taken as values of Z.

Where ‘C’, the point on shore line, corresponds to the shortest of all possibletimes. Let us consider a point ‘D’ on shore line which is very close to point ‘C’ suchthat there is essentially no change in time between path ACB and ADB.

Let us try to calculate how long it would take to go from A to B by the two pathsone through point D and another through point C (see fig.- 6c). First look at the pathson the land as shown in fig.- 6(c). If we draw a perpendicular DE; between two paths atD, we see that the path (AD) on land is shortened by the amount EC. On the other hand,in the water, by drawing a corresponding perpendicular CF we find that we have to gothe extra distance DF in water. In other words, we gain a time that is equal to go throughdistance EC on land but we lose the time that is equal to go extra distance DF in water.These times must be equal since we assumed there no change in time between the twopaths.

Let the time taken to travel from E to C and D to F be Δt and v1 and v2 be the speedsof running and swimming. From fig.- 6(c) we get,

C

time

t

Zfig-6(b)

SCERT TELANGANA


Above experiments and activities showthat refraction of light occurs according tocertain laws.

Following are the laws of refraction.1. The incident ray, the refracted ray

and the normal to interface of twotransparent media at the point ofincidence all lie in the same plane.

2. During refraction, light followsSnell’s law

n1 sin i = n2 sin r (or) sin i constantsin r

= .

• Is there any chance that angle ofrefraction is equal to 900? When doesthis happen?Let us find out

Total Internal Reflection

Activity 5Use the same materials as used in lab

activity 1. Place the semi circular glassdisc in such a way that its diametercoincides with interface line MM and itscentre coincides with point ‘O’ as we have

EC = v1 Δt and DF = v2 Δt1

1

EC vDF v

= ————(3)

Let i and r be the angles measured between the path ACB and normal NN,perpendicular to shore line X.• Can you find sin i and sin r from fig.- 6(c)?

Note : Sin of any acute angle in a right angled triangle can be defined as a ratio ofopposite side of that angle to the Hypotenuse.

sin θ = BCAC

From fig.- 6(c), we get;

sin i = ECDC and sin r =

DFDC

Therefore,

sin isin r =

ECDF ——————(4)

from equations (3) and (4), we have sin i / sin r = v1 / v2 ––—————(5)Thus to save the person, one should take such a path to satisfy the above equation.

We used the principle of least time to derive the above result. Hence we can apply thesame for the light ray also. From (5) we get,

1 1

2 2

sin i v nsin r v n

= = , (since 1 1

2 2

v nv n

= )

n1 sin i = n2 sin r.This is called Snell’s law.

A B

C

θ

SCERT TELANGANA


done in lab activity. Now send light fromthe curved side of the semicircular glassdisc. This means that we are making thelight travel from denser medium to rarermedium. Start with angle of incidence (i)equal to 00 i.e., along the normal and lookfor the refracted ray on the other side ofthe disc.• Where do you find the refracted ray?• Does it deviate from its path when it

enters the rarer medium?You might have noticed that it doesn’t

deviate.Send laser light along angles of

incidence 50, 100, 150 etc.., and measurethe angle of refraction. Tabulate the resultsin table-3 as shown below and note thevalues ‘i’ and ‘r’ in the table.

• At what angle of incidence do younotice that the refracted ray grazes theinterface separating the two media (airand glass)?You will observe that at a certain angle

of incidence the refracted ray does notcome out but grazes the interface separatingair and glass. Measure the angle ofincidence for to this situation. This angleof incidence is known as critical angle.

Let us consider the light ray that travelsfrom medium 1 with refractive index n1 to

medium 2 with refractive index n2. Seefig.-7.

It is already found that the angle ofrefraction is more than angle of incidencewhen a light ray travels from denser(n1)to rarer medium (n2).

For the angle of incidence i, let r bethe angle of refraction.

From Snell’s law

n1 sin i = n2 sin r 1

2

n sin rn sin i

=

we know that, 1

2

nn is greater than 1, so

that sin r/ sin i is greater than 1.So weconclude that the angle of refraction isgreater than the angle of incidence, i.e, ris greater than i.

The angle of incidence at which thelight ray, travelling from denser to rarermedium, grazes the interface is calledcritical angle for denser medium. It isshown in fig.-7.

Let C be the critical angle. Then rbecomes 900

we get, o

1

2

n sin 90n sin c

=

1

2

n 1n sin c

⇒ =

NRarer

Denser N

C

fig-7

ri

Table 3

SCERT TELANGANA


We get sin c = 2

1

nn . We know that 1

2

nn

i.e., n12 is called refractive index of densermedium with respect to rarer medium

sin c = 12

1n

• Can you find the critical angle of waterusing the above equation? Discuss itin your class.

• What happens to light when the angleof incidence is greater than criticalangle?When the angle of the incidence is

greater than critical angle, the light ray getsreflected into the denser medium at theinterface i.e., light never enters the rarermedium. This phenomenon is called totalinternal reflection. It is shown in fig.- 7.

Discuss these ideas in your class andfind out the critical angle of water.Let us see an example on total internal

reflection.

Activity 6Take a transparent glass tumbler and

coin. Place the coin on a table and placeglass tumbler on the coin. Observe the coinfrom the side of the glass.• Can you see the coin?

Now fill the glass tumbler with waterand observe the coin from the side of theglass tumbler.• Can you see the coin?• Explain why the coin disappears from

view.

Activity 7Take a cylindrical transparent vessel

(you may use 1 L beaker). Place a coin at

the bottom of the vessel. Now pour wateruntil you get the image of the coin on thewater surface (look at the surface of waterfrom a side). See fig.- 8.

• Can you explain why the image of the coinis formed?There are many interesting situations

around us which involve the phenomenonof total internal reflection. One of that isa mirage which we witness while drivingor while walking on a road during a hotsummer day.

MiragesMirage is an optical illusion where it

appears that water has collected on theroad at a distant place but when we getthere, we don’t find any water.

• Do you know the reason why it appearsso?

fig-9(a)

fig-8

Coin atbottom

Image ofthe coinon thewater

surface

SCERT TELANGANA


Think and discuss

• Why should you see a mirage asa flowing water?

• Can you take a photo of a mirage?

The formation of a mirage is the bestexample where refractive index of amedium varies throughout the medium.

During a hot summer day, air just abovethe road surface is very hot and the air aboveis less warmer. As a result density of airincreases with height above road. We knowthat refractive index of air increases withdensity. Thus the refractive index of airincreases with height. So, the cooler air atthe top has greater refractive index thanhotter air just above the road. Light travelsfaster through the thinner hot air thanthrough the denser cool air above it.

When the light from a tall object suchas tree or from the sky passes through amedium just above the road, whoserefractive index decreases towards ground,it suffers, refraction and takes a curvedpath because of total internal reflection.See fig.- 9(c).

This refracted light reaches theobserver in a direction shown in Fig.--9c.This appears to the observer as if the ray isreflected from the ground. Hence we feelthe illusion of water being present on road(shown in fig.-9a) which is the virtual imageof the sky (mirage) and an inverted imageof tree on the road (shown in fig.-9c).

Applications of total internalreflectioni) Brilliance of diamonds: Total internal

reflection is the main reason forbrilliance of diamonds. The criticalangle of a diamond is very low (24.4o)due to higher refractive index. So if alight ray enters a diamond it is verylikely to undergo multiple total internalreflection which makes the diamondshine from its multiple edges.

ii) Optical fibres: Total internalreflection is the basic principle behindworking of optical fibre. An opticalfibre is very thin fibre made of glass(or) plastic having radius about amicrometer (10-6 m). A bunch of suchthin fibres form a light pipe.

fig-9(b): The paths of light rays when thereis no change in density of air

fig-9(c)

Refractive indexdecreases with depth

fig-10(a)

SCERT TELANGANA


Fig.- 10(a) shows the principle of lighttransmission by an optical fibre. Fig.- 10(b)sketches a optical fibre cable. Because ofthe small radius of the fibre, light going intoit makes a nearly glancing incidence on thewall. The angle of incidence is greater thanthe critical angle and hence total internalreflection takes place. The light is thustransmitted along the fibre.

All organs of the human body are notaccessible to the naked eye of the doctor,for example intestines. The doctor insertslight pipe into the stomach through themouth. Light is sent down through one setof fibres in the pipe. This illuminates theinside of the stomach. The light from theinside travels back through another set offibres in the pipe and the viewer gets theimage at the outer end (generally fed to thecomputer screen).

The other important application of fibreoptics is to transmit communicationsignals through light pipes. For example,about 2000 telephone signals,appropriately modulated with light waves,may be simultaneously transmitted througha typical optical fibre. The clarity of thesignals transmitted in this way is muchbetter than other conventional methods.

• How does light behave when a glass slabis introduced in its path?Let us see

Refraction Through a Glass SlabA thin glass slab is formed when a

medium is isolated from its surroundingsby two plane surfaces parallel to each other.Let us determine position and nature of theimage formed when the slab is placed infront of an object. Let us do an activity.

Aim: Determination of refracted raytracking by a glass slab and lateral shift.

Material required: Drawing board,chart paper, clamps, scale, pencil, thinglass slab and pins.

Procedure:Place a piece of chart (paper) on a

drawing board. Clamp it. Place a glass slabin the middle of the paper. Draw borderline along the edges of the slab by using apencil. Remove it. You will get a figure ofa rectangle. Name the vertices of therectangle as A, B, C and D.

Draw a perpendicular at a point on thelonger side (AB) of the rectangle. Againkeep the slab on paper such that itcoincides with the sides of the rectangleABCD. Take two pins. Stick them on theperpendicular line to AB. Take two morepins and stick them on the other side ofthe slab in such a way that all pins cometo view along a straight line. Remove theslab from its place. Take out the pins. Drawa straight line by using the dots formed bythe pins such that it reaches first edge (AB)

fig-10(b)

Lab Activity 2

SCERT TELANGANA


of the rectangle. You will get a long straightline.• What does it mean?

The light ray that falls perpendicular toone side of the slab surface comes out without any deviation.

Now take another piece of white charton the plank. Clamp it. Place a glass slab inthe middle of the paper. Again draw a borderline along the edges of the slab by using apencil. Remove the slab and name thevertices of the rectangle formed as A,B,Cand D. Draw a perpendicular at a point onthe longer side AB. Now draw a line, fromthe point of intersection where side AB ofrectangle and perpendicular meet, in sucha way that it makes 300 angle with normal.This line represents the incident ray fallingon the slab and the angle it makes withnormal represents the angle of incidence.

Now place the slab on the paper in suchway that it fits in the rectangle drawn. Fix twoidentical pins on the line making 300 anglewith normal such that they stand verticallywith equal height. By looking at the two pinsfrom the other side of the slab, fix two pins

in such a way that all pins appear to be along astraight line. Remove slab and take out pins.Draw a straight line by joining the dots formedby the pins up to the edge CD of the rectangle.This line represents emergent ray of the light.

Draw a perpendicular ON to the lineCD where our last line drawn meets the lineCD. Measure the angle between emergentray and normal. This is called angle ofemergence. (Check your drawing with thefig.- 11).• Is the line formed a straight line?• Are the angles of incidence and

emergence equal?• Are the incident and emergent rays

parallel?You will notice an important result that

the incident and emergent rays are parallel.• Can you find the distance between the

parallel rays?The distance between the parallel rays

is called lateral shift. Measure this shift.Repeat the experiment for different anglesof incidence and tabulate the values of angleof incidence and shift corresponding to eachangle of incidence in table (4).

• Can you find any relation between theangle of incidence and shift?

• Can you find the refractive index of theslab?Let us find the refractive index of the

glass slab.

ShiftAgle of InciddeneTable 4

fig-11

i

A B

CD

i

rr

• pins

SCERT TELANGANA


on it. Place slab again in the rectangleABCD. Take a pin. Place at a point P in sucha way that its length is parallel to the AB onthe perpendicular line at a distance of 15cmfrom the slab. Now take another pin and bylooking at the first pin from the other sideof the slab try to place the pin so that itforms a straight line with the first pin.Note : While doing this, keep your eyefixed along the edge of the glass slab andfirst pin is seen through the glass slab,where as second pin is seen through air i.e.outside of the glass slab.

Remove the slab and observe thepositions of the pins.• Are they in the same line?

Draw a perpendicular line from thesecond pin to the line on which first pin isplaced. Call the intersection point Q. Findthe distance between P and Q. We may callit vertical shift.• Is this shift independent of distance of

first pin from slab?To find this, do the same activity for

another distance of the pin from the slab.You will get the same vertical shift. Wecould use a formula to find out refractiveindex of the glass.

R.I = Thickness of the slab

(thickness of slab vertical shift)−

Activity 8Let us take a glass slabe and measure

the thickness of it. Note it in your note-book. Take a white chart and fix it on thetable. Take the slab and place it in themiddle of the chart. Draw its boundary.Remove the slab from its place. The linesform a rectangle. Name the vertices asA,B,C and D. Draw a perpendicular to thelonger line AB of the rectangle at any point

Key words

Refraction, Incident ray, Refracted ray, Angle of incidence, Angle ofRefraction, Absolute refractive index, Relative refractive index, Snell’s law,

Critical angle, Total internal Reflection, Mirage, Shift, Optical fibre.

fig-12

B

Glassslab

SCERT TELANGANA


• When light travels from one medium to another, its direction changes at the interface. Thephenomenon of changing direction at the interface of the two media is known as refraction.

• Absolute refractive index = Speed of light in vacuum/ Speed of light in medium n=c/v.

• Relative refractive index , n21 = 1 2

2 1

v nv n

= .

• Snell’s law is given by, n1 sin i = n2 sin r .• The angle of incidence, at which the light ray travelling from denser to rarer medium grazes the

interface, is called the critical angle for those media. sin C = 2

1

nn , where n1 is the refractive index of

denser medium and n2 is the refractive index of rarer medium. (n1>n2)• When the angle of incidence is greater than the critical angle, the light ray is reflected into denser

medium at interface. This phenomenon is called total internal reflection.

I. Reflections on concepts

1. The speed of the light in a diamond is 1, 24, 000km/s. Find the refractive index of diamond if thespeed of light in air is 3,00,000 km/s. (AS1)(Ans: 2.42)

2. Refractive index of glass relative to water is 9/8. What is the refractive index of water relativeto glass? (AS1) (Ans: 8/9)

3. The absolute refractive index of water is 4/3. What is the critical angle? (AS1)

(Ans: sin C = ¾)

4. Determine the refractive index of benzene if the critical angle of benzene with respect to air is420.(AS1) (Ans: 1.51)

5. Explain the formation of mirage? (AS1)

6. Explain the refraction of light through a glass slab with a neat ray diagram. (AS5)

7. Why do stars appear twinkling? (AS7)

What we have learnt


SCERT TELANGANA


II. Application of concepts1. A light ray is incident on air-liquid interface at 450and is refracted at 300. What is the refractive

index of the liquid? (AS7) (Ans: 1.414)2. In what cases does a light ray not deviate at the interface of two media?(AS7)3. Place an object on the table. Look at the object through the transparent glass slab. You will

observe that it will appear closer to you. Draw a ray diagram to show the passage of light ray inthis situation. (AS5)

4. Why does a diamond shine more than a glass piece cut to the same shape? (AS7)

III. Higher Order Thinking Questions1. Why is it difficult to shoot a fish swimming in water? (AS1)2. Explain why a test tube immersed at a certain angle in a tumbler of water appears to have a

mirror surface for a certain viewing position? (AS7)3. When we sit at a camp fire, objects beyond the fire are seen swaying. Give the reason for it.

(AS7)

1. Which of the following is Snell’s law. [ ]a) n1 sin i = sin r/ n2 b) n1/n2 = sin r / sin ic) n2 / n1 = sin r / sin i d) n2 sin i = constant

2. The refractive index of glass with respect to air is 2. Then the critical angle of glass-air interfaceis .................. [ ]a) 0o b) 45o c) 30o d) 60o

3. Total internal reflection takes place when the light ray travels from............... [ ]a) rarer to denser medium b) rarer to rarer mediumc) denser to rarer medium d) denser to denser medium

4. If the angle of incidence is equal to critical angle, then the angle of refraction is [ ]a) 0o b) 20o c) 90o d) 180o

5. Mirage is a best example for the phenomenon of [ ]a) Reflection b) Refraction c) Total internal reflection d) Shift

6. Refractive indices of Ice, Benzene, Ruby and Kerosene are 1.31, 1.50, 1.71 and 1.44respectively. In which of the above media, light travels showly ? [ ]a) Ice b) Benzene c) Ruby d) Kerosene

Multiple choice questions

SCERT TELANGANA


7. The relative refractive index of water with respect to air is 43

. Then relative refractive

index of air with respect to water is [ ]

a) 4 b) 3 c) 43 d)

34

8. In an experiment to trace the path of ray through a glass slab, Shiva traced as shown in [ ]the figure. The teacher asked him to idenfity the emerging ray. Whichof the following would Shiva indentify.

a) AB b) BCc) CD d) N2


1. Verify experimentally that sin isin r

is a constant.

2. Organise some activities to understand the phenomenon of total internal reflection

3. Conduct an experiment to find the relation between angle of incidence and angle of refraction,when light rays travel from denser to rarer medium.

4. Take a bright metal ball and make it black with soot in a candle flame. Immerse it in water. Howdoes it appear and why? (Make hypothesis and do the above experiment).

5. Take a glass vessel and pour some glycerine into it and then pour water up to the brim. Take aquartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel.

• What changes do you notice? • What could be the reasons for these changes?

6. Conduct the activity-7 again. How can you find critical angle of water? Explain your stepsbriefly.

7. Find the critical angle of glass and water with respect to air using activity - 5.

Suggested Project works

1. Collect the refractive indices of the following media. Compare them with those are given intable 1. Findout the pairs of media in which light travels almost with same speed.

Coconut oil, Cooking oil, Hydrogen gas, petrol, diesel, Glycerine, Vinegar, Hydrochloric acid,Transparent plastic.

2. Collect information on working of optical fibres.

3. Prepare a report about various uses of optical fibres in our daily life.

A

B

C

DP Q

RS

N1

N2

SCERT TELANGANA


4. Take a thin thermocol sheet. Cut it in circular discs of differentradii like 2cm, 3cm, 4cm, 4.5cm, 5cm etc and mark centerswith sketch pen. Now take needles of length nearly 6cm. Pina needle to each disc at its centre vertically. Take water in alarge opaque tray and place the disc with 2cm radius in such away that the needle is inside the water as shown in fig-P4.Now try to view the free end (head) of the needle from surface of the water.• Are you able to see the head of the needle?Now do the same with other discs of different radii. Try to see the head of the needle, eachtime.Note: the position of your eye and the position of the disc on water surface should not be changedwhile repeating the activity with other discs.• At what maximum radius of disc, were you not able to see the free end of the needle?• Why were you not able to view the head of the nail for certain radii of the discs?• Does this activity help you to find the critical angle of the medium (water)?• Draw a diagram to show the passage of light ray from the head of the nail in different situations.

fig-P4

SCERT TELANGANA

66 Gravitation

We have learnt about uniformlyaccelerated motion in the chapter 'motion'.In this chapter let us study about uniformcircular motion which is an example ofaccelerated 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 themoon moves around the earth. In all thesecases there must be some force acting onthese objects to make them move aroundanother object, instead of moving in astraight 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 motion

Activity-1

Observing the motion of anobject moving in a circular path

Take an electric motor (which is usedin toys) and fix a disc to the shaft of theelectric motor. Stick a small wooden blockon the disc at the edge as shown in fig.- 1(a). Switch on the motor. Find the timerequired to complete ten revolutions by theblock and repeat the same two to threetimes. Begin counting revolutions a fewseconds after starting the motor.

Chapter

5 GRAVITATION

Fig-1: (a) motion of wooden block on acircular plate (b) top view of wooden block

(a) (b)

WoodenBlock

Electric MotorCircular

Plate

Battery

SCERT TELANGANA


• Is the time of revolution constant?• Is the speed of the block constant?• What is the shape of its path?The wooden block moves in a circular

path with a constant speed. So this motionof the wooden block, is uniform circularmotion.

"Uniform circular motion is motionof the body with a constant speed in acircular 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-2

Take a piece of thread and tie asmall stone at one end. Hold the other endof the thread and rotate it round as shownin the Fig-2.

• What is the direction of motion ofthe stone?

Now release the thread and observethe diredction of motion of the stone.

• What is the direction of motion ofthe stone?

Before the thread is released thestone moves in a circular path, with acertain speed.

When the thread is released, thestone moves in linear motion.

This shows that the originaldirection of velocity of stone is linear andwhile in circular motion, the direction ofvelocity and acceleration always changesthus the stone remains in circular motion.It is clear that the change in direction ofvelocity makes an object to move incircular motion. This change is caused byan external force.

• Where from the stone gets thisforce?

• What will be the direction of thatforce?

The force that causes this change invelocity keeps the stone moving along thecircular path is acting towards the centre.This force is called centripetal force.

In the absence of this force (whenthe thread is released) the stone flies offalong a straight line. This line is a tangentto the circular path.Fig-2

SCERT TELANGANA

68 Gravitation

Think and discuss

The centripetal acceleration is given

by the formula, ac = 2v

rHence the centripetal force-

Fc = 2mv

r (∵Fc=mac )

Where m = mass of the objectv = Velocity of the objectr = Radius of the circular path.

Note : Centripetal force is a net force.It always directed to the centre.

• Can an object move along a curvedpath if no force acts on it?

• As a car speeds up when rounding acurve, does its centripetalacceleration increase? Use aboveequation to justify your answer.

• Calculate the tension in a string thatwhirls a 2 kg mass toy in a horizon-tal circle of radius 2.5 m, when it ismoving at a speed of 3m/s.

Universal law of gravitationOnce while Sir Isaac Newton was sitting

under a tree, an apple fell to the ground.

• Do you know what questions arosein his mind from this observation?

• Why did the apple fall to the ground?

• Why does the moon not fall to theground?

• What makes the moon to move in acircular orbit around the earth?

With the above observations Newtonassumed that there exists a force betweenany two objects in the Universe. This forceis named as Gravitational force. Thegeneralised law of gravitation as stated byNewton is "Eevery body in the universeattracts other body with a force which isdirectly proportional to the product of theirmasses and inversely proportional to the squareof the distance between them". The directionof the force of attraction is along the linejoining the centers of the two bodies.

Let two bodies of masses M1 and M2 beseparated by a distance of 'd', and the forceof gravitation between them is Fgrav.

Force of attraction ∝ (mass)1 (mass)2

Fgrav. ∝ m1 m2 and

Fgrav. ∝ 2

1d

Tangent to a CircleThe straight line which meets the circleat one and only one point is called thetanget to the circle at that point. Thepoint is called tangential point. Thetangent is prependicular to the radius ofthe circle at that point.

Fig-3

M1 M2Fgrav Fgrav

dFig-4

AC Tangent↔↔↔↔↔

OB Radius

SCERT TELANGANA


1 2grav 2

M MFd

∝

1 2grav 2

GM MFd

=

G is a proportionality constant, calleduniversal gravitational constant and foundexperimentally by 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 to be a point object.

Example 1What is the time periodof satellite near theearth's surface?(neglect height of theorbit of satellite fromthe surface of theearth)?

SolutionThe force on the satellite due to earth is

given by F = 2

GmMR

M-Mass of earth = 6 x 1024kgm-mass of satellite,R-radius of earth = 6.4 x 106 mLet v be the speed of the satellite

v = 2πR

T ⇒ T = 2πR

v

Required centripetal force is providedto satellite by the gravitational force hence

FC= 2mv

R.

But FC= 2

GMmR according to Newton's

law of gravitation.

i.e., 2

GMmR =

2

2

m(2πR)T R

⇒ T2 = 2 34π R

GMAs mass of the earth (M) and G are

constants the value of T depends only onthe radius of the earth.

⇒ T2 α 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.

Deduction of the UniversalLaw of Gravitation

Newton knew that the motion ofmoon around the earth is approximately

V m

M

Fig-5

Fig-6: Comparing the motions of themoon and apple

MoonV

Appleae = g

aC = am

Rm

ReSCERT TELANGANA

70 Gravitation

uniform circular motion. So certain netforce, which we call centripetal force,is required to maintain the uniformcircular motion.

So he introduced the idea of forceof attraction between the moon andearth. He proposed that earth attractsmoon and termed it as gravitationalforce. This gravitational force acts as acentripetal force and makes the moonrevolve around the earth in uniformcircular motion. Newton knew thefollowing data. The distance of themoon from center of the earth is384400 km = 3.844x1010 cm. Themoon takes 27.3 days or 2.35x106 s fora complete revolution around the earth.

You can calculate the speed of themoon using the equation,

v = 2πR

TThus the acceleration of the moon

towards the centre of the earth

am = 2 2

2

v 4π RR T

=

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 981 cm/s2. Thus accelerationof an apple, approximately is equal to981cm/s2.

He compared the both the accelerationof an apple, ae to the acceleration of themoon, am.

We get, e

m

a 981a 0.27

= ≅ 3640.-- (1)

Newton knew that the radius of Earth,Re and the distance of the moon fromthe centre of the Earth, Rm are 6371 kmand 3,84,400 km respectively. We get

m

e

R 384400R 6371

= ≅ 60.32

m

e

RR

⎛ ⎞⎜ ⎟⎝ ⎠

=(60.3)2 ≅ 3640----- (2)

From 1,2 equations it is clear that e

m

aa

= 2

m

e

RR

⎛ ⎞⎜ ⎟⎝ ⎠

So we get,

2

1aR

∝ ----- (3) and thus,

Force of attraction

2

1FR

∝ ----- (4)

Thus it became clear that the forceof gravity decreases with increase indistance of the object from the centerof the earth.

According to Newton's third law theforce on the apple by the earth is equalto the force on the earth by the apple.We get the force on an object by theearth by using the second law of motionand equation-1.

From Newton's second law of motion

F = ma, and from equation-1, 2

1aR

∝

⇒ a = 2

kR (where k is

proportionality constant)

Thus we get, F = 2

kmR

SCERT TELANGANA


Think and discuss

• According to the equation forgravitational 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?

Free fall

Activity-3Acceleration is independent ofmasses

Place a small paper on a book. Releasethe book with the paper from certain heightfrom the ground.

• What is your observation? Nowdrop the book and paper separately,what happens?

A body is said to be in free fall when onlythe gravitational force acts on that body.

Fig-8

Therefore the force on the apple by

the earth = 2

KmR ---- (5)

Where 'm' is the mass of apple and'R' is the radius of the earth.

Force on the earth by the apple = 2

KMR′

---- (6)

Where M is the mass of the earth.

The above forces are equal inmagnitude only when the followingcondition is satisfied.

K=GM and K' = Gm ---- (7)

Where G is a constant.

From equations (5) & (7) we have

force on apple by the earth, F = 2

GMmR

We conclude that gravitational forcebetween the masses is directlyproportional to the product of theirmasses.

• In fig.- 7, we see that the moon 'falls'around earth rather than straight intoit. If the magnitude of velocity werezero, how would it move?

Fig-7

m

gR

M

SCERT TELANGANA

72 Gravitation

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 = 2

GMmR ⇒

Fm = 2

GMR

From Newton's second law, F/m is equalto acceleration. Here this acceleration isdenoted by 'g'

Hence, g = 2

GMR

From the above equation you canconclude that 'g' is the independent of thebody's mass.

If there were no air friction orresistance, all the bodies would fall withthe same acceleration. This acceleration,produced due to gravitational force of theearth near the surface, is called free- fallacceleration.

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 indistances 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 above

equations.When we use these equations, you must

follow 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 a stop clock.

• What happens to its speed when itmoves up and down?

• What is the direction ofacceleration?

When a stone moves up, the speeddecreases. When a stone moves down, thespeed increases. So the free-fall accelerationis vertically downwards. No matter how youthrow objects, " g " is directed vertically downas shown in fig.-9. (Actually the body tendsto move towards the centre of the earth).

Observe the following table-1 to knowthe direction of acceleration.

Table-1

Fig-10

g Surface of the earth

Fig-9 direction of gravity

SCERT TELANGANA


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 fig.-12.

Then, u= 20 m/s

a = g = -10 m/s2

s = h = -25 mFig-11

SCERT TELANGANA

74 Gravitation

From equation of motion s = ut + ½ at2

- 25 = 20t - ½ x 10 x t2

-25 = 20t - 5 t2

-5 = 4t - t2

⇒ t2 - 4t - 5 = 0

Solving this equation,

We get, ( t - 5 ) ( t + 1 ) = 0

t= 5 or -1

t = 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 = u0 = 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-5Can 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

SCERT TELANGANA


Think and discuss

• What changes do you notice in thereadings of spring balance in abovetwo instances?

• Are the readings same? If not why?Some of you might have the experience

of diving into a swimming pool from certainheight.

• How do you feel during free-fall ofyour body from a 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 the tray 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 as theforce due to gravity upon it. When in

g

equilibrium on a firm surface, weight isbalanced by a support force or when insuspension, by a supporting tension. Ineither case with no acceleration, weightequals to mg. A support force can occurwithout regard to gravity. The definition ofthe weight of something is the net force itexerts against a support.

• When is your weight equal to mg?• Give example of when your weight

is zero?

Centre of gravity

Activity-7

Balancing of spoon and fork

Fasten a fork, spoon, and wooden matchstick together as shown. The combinationwill balance nicely - on the edge of theglass. Why?

Fig-16: Balancing of fork

Fig-15(a) Fig-15 (b)

SCERT TELANGANA

76 Gravitation

Acivity-8Can 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 a ladder on your shoulder?

When does it happen?

We need to introduce the idea of"Centre of gravity" to understand thisobservation.

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 gravity

Take a meter scale. Suspend it fromvarious 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 a 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 at this point.

• How to find the center of gravity ofan object?

Balanced Upwardforce due to finger

Weight

Fig-17

Fig-18SCERT TELANGANA


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-11Identifying 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 fig.-20 (a). Try this again when standing againsta wall as shown in fig.- 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)

SCERT TELANGANA

78 Gravitation

Think and discuss

Key words

• Where does the centre of gravity of a sphere and triangular lamina lie?• Can an object have more than one centres 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 incircular motion is called centripetal acceleration and it is always directed towardsthe center of the circle.

• The net force required to keep a body in uniform circular motion is called "Centripetal

force". FC = 2Mv

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

SCERT TELANGANA



1. How do you explain that an object is inuniform circular motion (AS1)

2. Calculate the acceleration of the moon towards earth’s center. (AS1)

3. Explain universal law of gravitation. (AS1)

4. Explain some situations where the center of gravity of man lies outside the body.(AS1)

5. Explain why a long pole is more beneficial to the tight rope walker if the pole hasslight bending. (AS7)


1. A car moves with constant speed of 10 m/s in a circular path of radius 10m. Themass of the car is 1000 kg. How much is the required centripetal force for the car?

(Ans:104N) (AS1)

2. 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)

3. 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.Newton)

4. A ball is dropped from a height. If it takes 0.2s to cross the last 6m before hitting theground, find the height from which it is dropped. Take g = 10m/ s2 (AS1)

(Ans: 48.05m)

5. What path will the moon take when the gravitational interaction between the moonand earth disappears? (AS2)

6. Why is it easier to carry the same amount of water in two buckets, one in each handrather than in a single bucket? (AS7)

Higer Order Thinking questions

1. 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)


SCERT TELANGANA

80 Gravitation


2. An apple falls from a tree. An insect in the apple finds that the earth is falling towardsit with an acceleration 'g'. Who exerts the force needed to accelerate the earth withthis acceleration? (AS7)

1. The acceleration which can change only the direction of velocity of a body is called[ ]

a) Acceleration due to gravity b) Uniform accelerationc) Centripetal acceleration d) Deceleration

2. The distance between the Earth and the Moon is [ ]a) 3,84,400 Km b) 3,84,400 cm c) 84,000 Km d) 86,000 Km

3. The value of Universal Gravitational Constant is [ ]a)6.67 x 10-11N.m2Kg-2 b) 9.8 m/ Sec 2

c) 6.67 x 10-12N.m2Kg-2 d) 981 m/ Sec2

4. The weight of an object whose mass is 1 Kg is [ ]a) 1 Kg/m2 b) 9.8 m/Sec2 c) 9.8 N d) 9.8 N/m2


1. Conduct an experiment to find the Centre of Gravity of an object and write a report.

2. Conduct an experiment to find 2

2st value for a freely falling body and also find the

value of 'g'.

Suggested Projects

1. Collect the information about the base area and stability of some objects with differentshapes and write a report.

2. Collect information about the path of revolution of moon around the earth and writea report.SCERT TELA

NGANA


liquids are spun rapidly. Commercially forseparating the cream from milk, a machinecalled centrifuge is used. It follows thesame principle. Centrifugation is also usedin a diagnostic laboratory, to test blood andurine samples. The sample is taken in a

centrifugation tube and thecentrifugation tube isplaced in a centrifugationmachine. On rotation ofthe tubes with machine,the heavier particles settle

to the bottom and the lighter particles cometo the top of the test tube.

Think and discuss

How does a laundry dryer squeeze outwater from wet clothes?

What is a mixture?Many things that we call pure are

actually mixtures of different substances.Juice is a mixture of water, sugar and fruitpulp. Even water we drink contains somesalts and minerals. All the matter aroundus can be classified into two groups – puresubstances and mixtures.

When a scientist says that somethingis pure, he means that the composition of

Chapter

6 IS MATTER PURE?

You might have gone to the market manytimes with your parents to purchase rice,salt, milk, ghee and other provisions. Youmust have tried to ensure that you got thepurest possible milk and pure ghee etc. Inour day to day language, ‘pure’ meanssomething with no adulteration. But inchemistry pure means something different.

Let us find out what is pure in chemistry.

Activity-1Is full cream milk pure?

Take some milk in a vessel. Spin itwith a milk churner for some time. Seefig-1

Fig-1: Churning of milkAfter some time, you observe

separation of a paste like solid out of themilk. This paste like solid called cream. Sofull cream milk contains more than onecomponent like fat and water. It is thereforea mixture. We have already studied aboutmixtures in previous classes. Let us learnmore about them.

Churning makes the lighter componentsto come to the surface when a mixture of

Fig-1(A): Centrifuge

SCERT TELANGANA

82 Is Mater Pure ?

the substance doesn’t change, no matterwhich part of the substance you take forexamination.

For example, whichever part of a puregold biscuit is taken as a sample, thecomposition is found to be samethroughout. (See fig.-.2)

Fig-2: Pure gold biscuit

But, the composition in mixtures is notalways same. The composition in somemixtures change, depending on the partyou have 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 fig.- 4aand 4b?

Fig-4: (a) Pure substance 4(b) Mixture

Types of mixturesYou have learnt about a mixture. Do

you know the types of mixtures? What arethey? Let’s find out.

Mixtures can be in solid, liquidgaseous states or the combination of thesethree states.

Activity-2

Finding out homogeneous andheterogeneous mixtures

Take two test tubes, Now add one teaspoon of salt to both the test tubes. Fillone test tube with water and other withkerosene 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 of mixture are uniformlydistributed throughout the bulk. Thecomponents of a homogeneous mixture aretoo intimately mixed up that it will bedifficult to distinguish them from oneanother by visual observation. For exampleair is a homogeneous 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 thecomponents separately. We call suchmixtures as homogeneous mixtures.

SCERT TELANGANA


• Can you give few more examples of thiskind?You have observed in the above activity

the salt added to kerosene has not dissolvedin it. Mixtures of this type are calledheterogeneous mixtures. Heterogeneousmixture is a mixture made up of differentsubstances. Which are not uniformlydistributed in it. For example the mixtures“oil and vinegar’’, “naphthalene and water’’are heterogeneous mixtures.

Thus we can conclude that mixtures are oftwo types homogeneous and heterogeneous. Doyou know that these can again be classified intodifferent kinds? Let us 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 of filterationis called a solution. Solutions can be inthe form of solids, liquids, or gases. Asolution has minimum two components, asolvent and a solute.The component of thesolution that dissolves the othercomponent in it (usually the componentpresent in larger quantity) is calledsolvent. The component of solution usuallythe component present in lesser quantitythat is dissolved in the solvent is calledsolute.

A solution of sugar is prepared bydissolving the sugar in water. In this solution,sugar (solid) is the solute and water (liquid)is the solvent. In the solution of iodine inalcohol (tincture of iodine) iodine (solid)is the solute and alcohol (liquid) is the

solvent. All the aerated drinks are liquidsolutions containing carbon dioxide (gas) assolute and water as solvent.

Can you give some more examples forsolutions and tell the solute and solventpresent 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,liquid or a gas dissolved in it. But,we can have solid solutions. Canyou give some examples?

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 andhence the path of light is not visible in asolution.

• Can you prove this with an experiment?

• If the solution is diluted, can the pathof light be visible?One more interesting property of

solution is that, the solute particles do notsettle down when left undisturbed. Canyou give a reason? If the solute particlesare settling down in a solution can we callit as a 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?

SCERT TELANGANA

84 Is Mater Pure ?

Concentration of a solutionCan we dissolve as much solute as we

want in a given solvent? How do you decidethe amount of solute that dissolves in asolvent.

Solubility is the number of grams of asolute that desolves in 100g. of solvent toform a saturated solution at a giventemperature.

For example, take one gram of sugarand add 50ml of water to it. Also take30gm of sugar and add the same amountof water to it in another beaker. Which oneof the above solutions is called as diluteand which one is called 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 inthe cup and stir until it dissolves. 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 dissolved inthe solution at a certain temperature, it issaid to be a saturated solution. In a

saturated solution, equilibrium with theundissolved solute at a certain temperature.If the amount of solute present in a solutionis less than that in the saturated solution, iscalled an unsaturated solution.Now take the solution prepared by you intoa beaker and heat it slowly by 5° to 6oCabove the Room Temperature (do not boil).The undissolved solute dissolves. Addsome more sugar to this solution. Younotice that more sugar dissolves in it easilywhen it is heated.

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 water inthe second beaker and warm the thirdbeaker.

What do you observe from the abovethree situations? 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 do you observe?

SCERT TELANGANA


SolutionMass of solute (salt) = 50gMass of solvent (water) = 200g

Mass of solution = Mass of solute + Massof solvent

= 50g + 200g = 250gMass percentage of a solution =

Example - 280ml of solution contains 20 g of solute.Calculate the concentration in terms of massby volume percentage of the solution.

Suspensions and Colloidal Solutions

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 the above solutions heterogeneousor homogeneous?)

Now do the following steps andwrite your 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? Doesthe solute settle down after some time?

Mass of solute

Mass of solutionX 100

50250

= X 100 = 20%

Mass of soluteMass of solution

Volume of soluteVolume ofsolution

What are the factors that affect the rateof the solubility of a solute?

From this activity we can conclude thatthe temperature size of the solute particles,and stirring are some of the factors thataffect the rate of solubility of solute in asolvent.

You know that solubility is themeasurement of amount of solute thatdissolves in a solvent at a certaintemperature. If the amount of solute presentis little, the solution is said to be dilute,and if the amount of solute present is more,the solution is said to be concentrated.

There are many ways of expressing theconcentration of a solution, we learn onlyabout two of those.

The concentration of a solution canbe defined as the amount in grams(mass)of solute present in 100 grams of (mass)solution or the amount (mass) in gramsof solute present in 100 ml of the solution.

(i) Mass percentage of a

solute = X 100

(ii) Volume percentage of a

solute = X 100

(iii) Mass by volume percentage of a

solution = X 100

Example - 1A solution contains 50g of common salt

in 200g of water. Calculate theconcentration in terms of mass by masspercentage of the solution?

Mass of soluteVolume ofsolution

SCERT TELANGANA

86 Is Mater Pure ?

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 paper visible? (Yes/No) (Yes/No)

(Yes/No)

Chalk mixture

Milk mixture

We find that the particles of chalk didnot dissolve but remained suspendedthroughout the volume of the water. Sothis is a heterogeneous mixture becausethe solute particles didn’t dissolve and theparticles are visible to naked eye. Suchheterogeneous mixtures are calledsuspensions. “Suspensions are theheterogeneous mixtures of a solid and aliquid, in which the solids do not dissolve,like mixture of soil and water”.

In activity-5, particles of milk in thesecond test tube are uniformly spreadthroughout the mixture. Due to smallersize of milk particles it appears to behomogeneous but it is a heterogeneousmixture. These particles easily scatter abeam of visible light. Such mixtures arecalled colloids or colloidal solutions.These mixtures possess the characteristicsin between a solution and a suspension.They are also called as colloidaldispersions. Colloidal dispersions may

appear homogeneous but are actuallyheterogeneous.

A large number of substances such asmilk, butter, cheese, cream, gel, boot polish,and clouds in the sky etc. are some moreexamples of colloids.

Think and discuss

• Have you ever observed carefully thesyrup that you take for cough? Why doyou shake it before consuming?

• Is it a suspension or colloidal solution?

Colloidal solutions are heterogeneousin nature and always consist of at least twophases; the disperse phase and thedispersion medium. Disperse phase is thesubstance that present in small proportionand consists of particles of colloidal size(1nm to 100nm). Dispersion medium is themedium in which the colloidal particles aredispersed. These two phases can be in theform of solid, liquid or a gas. Thus, differenttypes of colloidal solutions are possibledepending upon the physical state of thetwo phases.

Here are some common examples ofcolloids from our daily life. (See table-2)Don’t try to memorize this table-2, it isgiven only for your information.

We studied that the particles in a col-loidal solution can easily scatter a beam

SCERT TELANGANA


of visible light. This scattering of a beamof light is called the Tyndall effect, namedafter the scientist who discovered it. Youmay observe this effect in your day to daylife 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 leftopen between 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 when the smoke from the stoveis exposed to sun light.

• Did you ever observe this phenomenonin the cinema halls?

• Have you ever got 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: Type of Colloids and their Examples based on dispersion medium and dispersed phase

Dispersion 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

SCERT TELANGANA

88 Is Mater Pure ?

Fig- 8: Ice creamIs ice-cream a colloid ?

Ice cream is made by churning a mixtureof milk, sugar and flavours. This mixtureis slowly chilled to form ice cream. Thechurning process disperses air bubbles intothe mixture by foaming and breakup thelarge ice crystals into tiny particles. Theresult is a complex substance which

Table-3: Properties of suspension and colloidsSuspensions Colloids

Suspensions are heterogeneous mixtures. Colloids are heterogeneous mixtures.The particles of suspensions can be seen with The size of particles of colloid are too small to benaked eyes. seen by naked eyes..The particles of suspensions scatter a beam of The particles of colloids are big enough to scatterlight passing through it and make its a beam of light passing through it which makes itspath visible. path visible.The solute particles settle down when a suspension The particles don’t settle down when the colloidis kept undisturbed. When the particles settle left undisturbed. i.e., colloid is quite stable. down 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 the process of filtration or decantation.of 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 into theirrespective constituents?

Usually heterogeneous mixtures can beseparated into their respective constituents

by simple physical methods likehandpicking, sieving, filtration etc., as weuse in our day to day life.

Sometimes special techniques have tobe used for the separation of the componentsof mixture. We have learnt in class VI, howto separate mixtures in various ways like,flotation, filtration, crystallization,chromatography etc. Let us see more.

SCERT TELANGANA


Why do we use different separationtechniques for mixtures like grain andhusk as well as ammonium chloride andsalt though both of them areheterogeneous mixtures?

What is the basis for choosing aseparation technique to separatemixtures?

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 fig.--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 components ofmixture by sublimation

Fig- 9: Separating ammonium chloride andsaltTake one table spoon of common salt, onetablespoon of ammonium chloride andmix them.

Is the mixture heterogeneous? Givereasons.

How do we separate the salt andammonium chloride?

Take the mixture in a china dish. Takea glass funnel which is big enough to coverthe dish. Plug the mouth of the funnel withcotton and invert it over the dish as shownin fig.- 9. Keep the dish on the stand of stoveand heat for some time and observe the wallsof the 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

SCERT TELANGANA

90 Is Mater Pure ?

Procedure : Draw a thick line just abovethe bottom of the filter paper using themarker. Pour some water in the beaker andhang the paper strip with the help of a penciland cello tape in such a way that it shouldjust touch the surface of water as shown infig.- 11.

Make sure that the ink line or markdoes not touch the water.

Allow the water to move up thepaper for 5 minutes and then remove thestrip from 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 orderand in the same location on all thesamples?

Instead of non permanent marker usea permanent marker. What will you observe?

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? Will your results change in eachcase?

Is chromatography used only toseperate components of colouredcompounds?

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? Is water miscible in alcohol?

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 you nameany such liquids from your dailyobservation?

Lab Activity

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 dyes arethere in ink? How can we find out those?Is there any technique to separate thedifferent components in the ink? That iswhere chromatography would helpChromatography is a laboratory techniquefor the separation of mixtures into itsindividual components. We can usechromatography to separate componentsin ink. The process can also be used toseparate the coloured pigments in plantsand flowers. It is used to determine thecomposition of mixtures.

Paper Chromatography

Aim: Separating the components of inkusing paper chromatography.Material required: Beaker, rectangularshaped filter paper, black marker (non-permanent), water, pencil and cello tape.

Pencil

Cello tape

Strip offilter paper Beaker

Marker Line

Fig-11: Separating the components of ink

SCERT TELANGANA


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 do youobserve? How do you separate the twocomponents?

Take a separating funnel and pourthe mixture of kerosene and water in it.Let it stand undisturbed for some time. Sothat separate layers of oil and water areformed. Open the stopcock of theseparating funnel and pour out the lowerlayer (water) carefully. Close the stopcockof the separating funnel as the oil reachesthe stop-clock.

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 homogeneous

solution. Water and ethanol, for example,are miscible because they mix in allproportions. How can we separate suchmixtures?Distillation

Activity-9

Separation of two miscible liquidsby distillation

Acetone and water are also miscible.Take a mixture of acetone and water in adistillation flask. Fit it with a thermometerand clamp it to stand. Attach the condenserto the flask and on the other side of thecondenser keep a beaker to collectdistillate. Heat the mixture slowly keepinga close watch on the thermometer. Theacetone 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.

Thermometer

Water Outlet

Watercondenser

Cold water in

Acetone

Mixture ofAcetoneandWater

DistillationFlask

Clamp

Fig-13: Separating the mixture ofAcetone and water by distillation

Separatingfunnel

Kerosene Oil

Stop Cock

Water

SCERT TELANGANA

92 Is Mater Pure ?

What if the boiling points of thetwo liquids are close to each other?

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 fractionating column is a tubepacked with glass beads. The beadsprovide maximum possible surface areafor the vapours to cool and condenserepeatedly as shown in Fig.14

Fig-14: Fractional distillation

• Can you give any examples where weuse this 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 temperatures

Points Oxygen Argon NitrogenBoiling points (0C) -183 -186 -196

% air by volume 20.9 0.9 78.1

Flow chart shows the process of obtain-ing gases from air

If we want oxygen gas from air(fig.--15), we have to separate out all the othergases present in the air. The air is compressedby increasing the pressure and is then cooledby decreasing the temperature to get liquid air.This liquid air is allowed to warm up slowly ina fractional distillation column where gasesseparated at different temperatures dependingupon their boiling points.

Thermometer

Water Outlet

Watercondenser

Cold water in

Pure liquidcomponent

Mixture

DistillationFlask

Clamp

Cork

Fractionatingcolumn

Cork

Air in

Filter Freezingcold water in

Carbon dioxideout as dry ice Expansion

jet

LiquidAir

Fractionaldistillation

column

Nitrogenout

Argonout

Liquidoxygen

Separator

ColdCompressed air

Water out

Hot Air

Airunder

pressure

Fig-15: Separation of componants of air

SCERT TELANGANA


Think and discuss

• Arrange the gases present in air inincreasing order of their boillingpoints. What do you observe?

• Which gas forms the liquid first asthe air is cooled?

Types of pure substancesSo far we have studied about mixtures

example – substances whose componentscan be separated by physical methods. Whatabout substance that cannot be separatedfurther by any of the methods of separation?We call them as pure substances. Let usexplore further about them.

Activity-10

Can we separate Sulphur andOxygen from Copper sulphate

Take concentrated solution of coppersulphate in beaker and drop a piece ofaluminum foil in it. After some time youwill observe a layer of copper deposited onthe aluminum foil. The solution becomescolourless. Why did this happen? (recallactivities of the chapter on Metals andNon-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.

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 havealready studied properties of metals andnon-metals. Can you write down the namesof some elements that you know?

Elements have been used since theearly days of civilisation. Metals such asiron, lead, copper, helped in thedevelopment of civilizations. For thousandsof years, alchemists up to and includingIsaac Newton attempted to unearth newelements, and study their properties.

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.

SCERT TELANGANA

94 Is Mater Pure ?

Hennig Brand, a German alchemist, boiledurine to discover phosphorus in 1669. Butit was not until the late 18th century thatour knowledge of the elements really tookoff, as chemists developed new ways topurify and 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 intosimpler forms by chemical reactions.

If any substance can be separatedinto two or more constituent parts by achemical reaction, that substance isdefinitely a compound.

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 near

the material and check if the material isattracted towards the magnet.

Activity for group-2:

Mix and crush iron fillings andsulphur powder. Heat this mixture stronglyin china dish on a spirit lamp till it becomesblack. Remove it from flame and let themixture cool. Check for magnetism in thematerial obtained. Compare the texture andcolour of the material obtained by the twogroups.

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?

Group-1 has carried out the activityinvolving a physical change. Where asgroup-2 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 isthe same throughout its volume.

SCERT TELANGANA


Key words

What we have learnt

Matter (Solid,liquid or gas)

Pure substance(Fixed composition)

Can’t be separated by normalphysical methods

Mixtures(No fixed composition)

Can be separated by normalphysical methods

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

The chemical and physical nature of the matter is better understood by the followingflow chart.

Pure substances, Mixture, Heterogeneous mixture, Homogeneous mixture, Solution,Suspension, colloids, solvent, solute, concentration of solution, Tyndall effect, Evaporation,Centrifuge, Immiscible liquids, Miscible liquids, Chromatography, distillation, Fractionaldistillation, Elements, Compounds. Disperse phase, dispersion medium.

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 in grams present per 100 mlor per 100 g of the solution.

Materials that are insoluble in a solvent and have particles that are visible to nakedeye, is called a suspension. A suspension is a heterogeneous mixture.

SCERT TELANGANA

96 Is Mater Pure ?


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.

Reflections on concepts1. 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) Mixture of oil and water.(d) Fine mud particles suspended in water.

2. Explain the following giving examples.(AS1)(a) Saturated solution (b) Pure substance (c) colloid (d) Suspension

3. 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) Methane (k) Carbon dioxide (l) Sea water

Application of concepts1. Determine the mass by mass percentage concentration of a 100g salt solution which

contains 20g salt? (AS1) Ans: (20% )

2. 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%)

3. Classify the following substances in the below given table. (AS1)

SCERT TELANGANA


Ink, soda water, brass, fog, blood, aerosol sprays, fruit salad, black coffee, oil and water,boot polish, air, nail polish, starch solution, milk.

Solution Suspension Colloidal dispersion

Higher Order Thinking Questions1. Use the words given below and write the steps for making tea (AS7)

Solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

1. The machine used to separate the masive particles and light particles from a mixture is[ ]

a) Atwood machine b) Centrifuge c) Filter paper d) Separating funnel

2. Which is not formed by the physical mixing of two substances [ ]a) Mixture b) Compound c) Colloid d) Suspension

3. The substance which is relatively less in quantity in a solution is [ ]a) Solute b) Solvent c) Dispersion phase d) Dispersion medium

4. The amount of solute present in a saturated in 100g. at constant temperature is knownas its [ ]a) Solubility b) Concentration c) Volume percentage d)Weight percentage

5. If the quantity of solute is more in a solution then the solution is said to be[ ]a) Saturated solution b) Dilute Solutionc) Concentrated solution d) Unsaturated Solution

6. The phenomenon of scattering of a visible light by the particles of colloid is known as[ ]

a) Tyndall effect b) Chromotography c) Sublimation d) Reflection


SCERT TELANGANA

98 Is Mater Pure ?

7. Immiscible liquids can be separated by [ ]a) Distillation process b) Fractional distillationc) Chromotography d) Separating funnel

8. Miscible liquids can be separated by [ ]a) Distillation process b) Fractional distillationc) Chromotography d) Separating funnel


1. Which of the following will show Tyndall effect? How can you observe the Tyndalleffect in them?

a) salt solution b) milk c) solution of copper sulphate d) starch solution

2. Take a solution, a colloid, and a suspension in three different beakers. Pass a lightbeam from the side of the beaker and test whether they show Tyndall effect or not?

Suggested Projects

1. Make a list of solids, liquids and gases from your surroundings. (These substance maybe organic or chemical). Separate mixtures from them and classify them into solutions,colloids and suspensions.

SCERT TELANGANA


In the chapter "Is matter around uspure?" we used the terms elements andcompounds. You learned 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 theobservations made in previous classes likethe rusting of iron rod kept outside, 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 of charcoalgo?

• Wet clothes dry after some time -Where does the water go?

These questions and several othersimilar questions fascinated scientists formany years. Burning and combustionreactions, puzzled them to a greater extent.

Recall the chapter "Metals and non-metals."

• What happens to magnesium onburning it in air ?

• What happens to Sulphur on burningit in air?

• Are the weights of the reactants andproducts the same or different?

ATOMS, MOLECULES ANDCHEMICAL REACTIONS

Chapter

7

Do you know?

Antoine Lavoisier(1743-1794) was aFrench nobleman. Hemade many importantcontributions tochemistry and some

call him as the Father of ModernChemistry.

Lavoisier studied combustion reactions

in detail. He was able to find the masses

of reactants and products accurately

irrespective of their physical states.

Based on his observations he proposed

the law of conservation of mass.

SCERT TELANGANA

100 Atoms and Molecules

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 carry on a lab activity to observewhat will happen to weights of reactantsand products during a chemical reaction.

Aim: To find out the change in the massbefore and after a chemical reaction.Materials required: Lead nitrate,potassium iodide, distilled water, twoconical flasks, spring balance, small testtube, retort stand, rubber cork, thread, etc.Procedure

1. Prepare a solution of lead nitrate bydissolving approximately 2 grams oflead nitrate in 100 ml of distilledwater in a 250 ml conical flask.

2. Prepare a potassium iodide solutionby dissolving approximately 2 gmof Potassium iodide in 100 mlwater in another conical flask

3. Take 4ml solution of potassiumiodide in a small test tube from theabove prepared solution.

4. Hang the test tube containing 4ml ofpotassium iodide solution in theconical flask containing lead nitratesolution carefully, without mixingthe solutions. Close the flask witha cork.(see fig.- 1)

5. Weigh the flask with its contentscarefully using spring balance.

6. Now tilt and swirl the flask, so thatthe two solutions mix. (see fig.- 2).

7. Weigh the flask again using the samespring balance as shown in fig.- 3.

8. Record your observations:Weight of flask and contents beforemixing = ........................Weight of flask and contents aftermixing = .........................

Now, try to answer these questions:• Did you observe any precipitate in

the reaction?• Do you think that a chemical reaction

has taken place in the flask? Givereasons.

Lab Activity

Fig - 1

Fig - 2

Fig - 3

SCERT TELANGANA


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 experimentperformed by Landolt. Ask your teacherabout Landolt experiment.

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?

• Do the weights, of the flask and itscontents change during the activity?

• What are your conclusions?

Result:

A chemical reaction takes place and themass remains same before and after thechemical reaction. Therefore, mass isneither created nor destroyed in a chemicalreaction.

• Do you get the same result if theconical flask is not closed?

Law of conservation of mass

Earlier, it was thought that mass ofcharcoal decreases on burning. ButLavoisier, carried out the burning ofcharcoal in a closed apparatus and foundno change in mass.

Antoine Lavoisier on the basis of hisexperiment proposed the law ofconservation of mass. It states that,"Matter is neither created nordestroyed during a chemical reaction".More simply, "the mass of products isequal to the mass of the reactants in achemical reaction".

• Recall the burning of the Magnesiumribbon in air. Do you think mass isconserved during this reaction?

Naturalsample

Syntheticsample

Weight percentage

Copper 51.35 51.35 Carbon 38.91 38.91 Oxygen 9.74 9.74

ElementSCERT TELANGANA


Think and discuss

• What difference do you observe inthe percentage of copper, carbonand oxygen in two samples?

Similarly, Proust took water fromdifferent sources, and found that thepercentage of oxygen and hydrogen wasthe same in all samples. There was norelation between the place from where thesample came and its composition.

Based on his experiments, Proust putforward the law of constant (or definite)proportions. It states that, "a givenchemical substance always contains thesame elements combined in a fixedproportions by mass." 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.Do these 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? Whycannot the elements combine in anyarbitrary proportion ?

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, thenall elements must be made up ofextremely small particles, calledatoms.

2. If law of constant proportion is tobe followed, the particles of samesubstance can not be different.

Based on the above laws, Dalton proposed"a new system of Chemical Philosophy".

Dalton's atomic theory

Postulates of Dalton's Atomic theory :1. Matter consists of indivisible

particles called atoms.2. Atoms are neither created nor

destroyed in a chemical reaction.Reorganisation of atoms occur inchemical reactions.

John DaltonSCERT TELANGANA


Do you know?

Think and discuss

Atoms and moleculesVery often you may have heard that atoms

are the building blocks of all matter. But whatdoes 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 lakhs, along with itsthickness.

• Are elements also made of atoms?

We know that substances are made upof atoms or molecules. Atoms are the mostfundamental of all particles that can havean independent existence. Sometimes twoor 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 may be atomor molecule.

There are many elements whose smallestparticle is an atom. Iron, copper, zinc,aluminium, silver, gold, etc are examplesof substances in which the smallest particleis an atom.

3. All the atoms of a same element haveidentical mass and identical chemicalproperties. Atoms of differentelements have different masses anddifferent chemical properties.

4. Compounds are formed when atomsof different elements combine in simplewhole number ratios. That is, chemicalchange is the union or separation ofatoms as in whole numbers.

5. When atoms of different elementscombine in different whole numberof ratios they form differentcompounds. eg : carbon monoxide(CO); carbondioxide (CO2). Hence'C' and 'O' combine in 1:1 and 1: 2ratios respectively to give twodifferent compounds

• Which postulate of Dalton's theory isthe result of the law of conservationof 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 a Greekword 'a-tomio' (means- indivisible)

Do you know?

SCERT TELANGANA


Do you know?

Oxygen and nitrogen are examples ofsubstances in which the particles are acombination of two identical atoms. Thesmallest particles of elements that arestable are known as molecules. Forexample one sodium molecule has onesodium atom but one oxygen molecule hastwo oxygen 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 of asubstance that has independent existenceand retains all the properties of thatsubstance.

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 that 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.

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 reacts with 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 'helio'), 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.

Do you know?John Berzelius

suggested that initialletter of an element fromits name in english writtenin capitals should be the

symbol of that element, Eg. 'O' foroxygen, 'H' for Hydrogen and so on.

SCERT TELANGANA


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 usesome shortcuts. Using short forms orsymbols for naming the elements is onesolution.

Over a 118 elements have beendiscovered so far. How do we decide theirsymbols?

Table-2: Symbols for some elements

Usually, the first letter of the name ofthe element in English becomes the symbolof that element and is always written as acapital letter (upper case).

How do we write the symbols forCalcium, 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-1Some elements and their possible

symbols 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

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

SCERT TELANGANA


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, which can be seen in Table-4. Certain elements havesymbols based on their Latin names (or older names in other languages).

• Would you be able to recognise 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 in their molecules

We have learnt that several elementshave more than one atom in their smallestconstituent particles. It means theseelements contains two or more atomscombined together to form a molecule.Oxygen, hydrogen and nitrogen areexamples of such elements.

For example, a molecule of oxygenhas two atoms. We need a formula torepresent such a molecule in a simple way.The formula for oxygen molecule is O2.

Why don't we write it as 2 O ? Writinga formula in this way indicates twoseparate atoms of oxygen. Hence firstwrite the symbol for oxygen, and then write2 as a subscript after the letter O.

Subscript number indicates number ofatoms of Oxygen combined to form itsmolecule.

You might 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 fromharmful ultra violet 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 same element.

Element Sodium Silver Tungsten Potassium Copper Gold Iron Lead Mercury

Latin name Natrium Argentum Wolfram Kalium Cuprum Aurum Ferrum Plumbum Hydrargyrum

Symbol

Table-4

SCERT TELANGANA


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),

ValencyTill now, there are over 118 elements

known. These elements react with eachother to form compounds. Every elementhas a definite combining capacity, thatdetermines the atomicity of its molecules.Every element reacts with atoms of otherelement according to its combiningcapacity number. This combining capacitynumber is called valency.

Argon(Ar) exist as single atom. Hencethey are known 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 bemonoatomic?

• Why do some elements formdiatomic or triatomic molecules?

• Why do elements have differentatomicities ?

To understand the atomicities ofmolecules 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

Aluminium Monoatomic

Copper Monoatomic

Hydrogen H2 Diatomic

Oxygen Diatomic

Nitrogen Diatomic

Chlorine Diatomic

Ozone O3 Triatomic

Phosphorus Tetratomic

Sulphur S8 Octatomic

Table-5

SCERT TELANGANA


What is an ion?Compounds formed by metals and non

metals contain charged particles . Thecharged particles are known as ions. Anegatively charged ion is called anion andthe positive charge ion is cation.

For example sodium chloride does notcontain discrete molecules as itsconstituent units. Its constituent particlesare positively charged sodium ions (Na+)and negatively charged chloride ions (Cl–).

Ions may be a charged independentatoms or a group of atoms (polyatomic)that have a net charge on them. Hence ionsare charged particles.So atoms of the elements have power to

combine with atoms of other elements.This is known as its valency.

* elements which show variable valency.

Element Valency

Helium 0

Hydrogen 1

Fluorine 1

Chlorine 1

Nitrogen 3

Carbon 4

Table-6Valencies of some elements

Oxygen 2

Table-7: Some common, simple and polyatomic ionsNet Charge Cation Symbol Anion SymbolHydrogen H+ Hydride H -

Sodium Na+ Chloride Cl -

Potassium K+ Bromide Br -

Silver Ag + Iodide I -

Copper* Cu+ Hydroxide OH -

Ammonium NH4+ Nitrate NO3

-

Magnesium Mg+2 Oxide O-2

Calcium Ca+2 Sulphide S -2

Zinc Zn+2 Sulphate SO4-2

Copper* Cu+2 Carbonate CO3-2

Iron* Fe+2 Dichromate Cr2O7-2

Aluminium Al+3 Nitride N -3

Iron* Fe+3 Phosphate PO4-3

1 unit

2 units

3 units

Table-7: Some common, simple and poly atomic ions.

SCERT TELANGANA


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"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 some other element.

In 1961, it was universally accepted thatmass of carbon-12 atom would be used asa standard reference for measuring atomicmasses of other elements.

Observe the diagram (fig-4). Let usassume the circle in the diagram representsatomic mass of carbon-12. It is dividedinto 12 equal parts as shown in thefig.--4, and each part represents 1/12 ofatomic mass of carbon-12.

One atomic mass unit (amu) is definedas the mass exactly one twelth the atomicmass of Carbon-12 isotope.

The number of times one atom of givenelement is heavier than 1/12th of atomicmass of carbon-12 is called its atomicmass.

The atomic mass of an element isdefined as the average mass of all theisotopes of the element as compared to1/12th of the mass of one carbon -12 atom.

Atomic mass of an element is a ratio.Hence it has no units, but it is expressed inamu. According to latest IUPAC(International Union of Pure and AppliedChemistry) recommendations the amu hasbeen replaced by 'u', which is known asunified mass.

12

34567

8 9 10 1112

Fig - 4

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

SCERT TELANGANA


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 may contain one or more

atoms. A molecule may be formed by thecombination of atoms of same element ordifferent elements. For example, amolecule of water is formed by thecombination of two atoms of hydrogen andone atom of oxygen. All the molecules ofwater 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 inthe molecule must be in fixed numbers.If this number is not fixed, how could allthe particles of water be identical?

Each molecule of water contains 2atoms of hydrogen and 1 atom of oxygen.Chemical formulae ofcompounds

While writing the formula of acompound 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 carbonmonoxide contains one atom of carbonand one atom of oxygen.

SCERT TELANGANA


• 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 formulaeof compound in criss - cross method byusing valency.

For eg. Sodium carbonate.The followingsteps should be taken while writing thechemical formula of sodium carbonate.

1. Write the symbols of each atom orgroup of atoms side by side, usuallythe cation first and anion next -

Na CO3

2. Write the valency of each atom orgroup of atoms the top of its symbolNa1 (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 as the subscript to rightside of the constituents as theirsubscripts. Na2 (CO3)1

5. If any constituent receives thenumber 1, ignore it while writing theformula.

6. If a constituent has more than oneatom enclose it with in brackets, ifit carries 2 or more mention assubscripts. [Look at the formula ofaluminium sulphate].

Hence the formula for the sodiumcarbonate is Na2CO3.

ExamplesFormula of hydrogen chloride

formula: HClFormula of magnesium chloride

formula: MgCl2

Formula of calcium oxide Ca2 O2

Ca1 O1 formula: CaO

Formula of aluminium sulphate is Al2 (SO4) 3

Table-9: Formulae of some compounds

Compound FormulaSodium Carbonate Na2CO3

Sodium bicarbonate NaHCO3

Sodium hydroxide NaOHCopper Sulphate CuSO4

Silver Nitrate AgNO3

Hydrochloric Acid HClSulphuric Acid H2SO4

Nitric Acid HNO3

Ammonium Chloride NH4ClPotasium Dichromate K2Cr2O7

Potasium Permanganate KMnO4

Molecular massWe have already discussed the concepts

of atomic mass. This concept can beextended to calculate molecular masses.

The molecular mass of a substance is thesum of the atomic masses of all the atomspresent in a molecule of the substance. Asatomic mass is a relative, it is therefore themoleculer mass is also relative. Mass of amolecule expressed in unified Mass (u).

H1 Cl1

H1 Cl1

Mg2 Cl1

Mg1 Cl2

Ca1 O1

1.

2.

3.

4.

SCERT TELANGANA


Do you know?

For Example: calculate the molecularmass of H2SO4

Solution

2 (atomic mass of hydrogen) + (atomicmass of sulphur) + 4(atomic mass ofoxygen) = (2X1)+32+(4X16)=98 u

Formula unit mass

One formula unit of NaCl, means oneNa+ (Sodium ion) and one Cl- (Cholorineion), similarly one formula unit of MgBr2

means one Mg2+ ion and two Br- ions, andone formula unit of H2O means one H2Omolecule. The formula unit mass of asubstance is a sum of the atomic massesof all atoms in a formula unit of acompound. Formula 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 ions. Sodium Chloride has aformula unit NaCl. The formula unit masscan be calculated as:

1 x 23 + 1 x 35.5 = 58.5u

Mole concept

We have learnt that atoms and moleculesare extremely small in size and theirnumber is really very large. Even in a smallamount of any substance we find very largenumber of atoms or molecules.

How many molecules are there in 18grams of water?

How many atoms are there in 12 gramsof carbon?

You will be surprised to know that thenumber of molecules in 18 grams of waterand no.of atoms in 12 grams of carbon asthe same. This number is very large. Tohandle such large numbers, a unit calledmole is introduced. This is a numericalquantity.

One mole of a substance is the amountof the substance which contains as manyparticles (atoms, molcules, ions ....etc) orentities that are equal to the atoms presentin exactly 12 grams of 12C isotope.

The number of particles ( atoms ormolecules) present in one mole of anysubstance has a fixed value of 6.022X1023.This number is called Avogadro constant(NA) named in honour of the Italianscientist, 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.

SCERT TELANGANA


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 and molecular mass arenumerically equal but molar mass has unitsgrams and molecular mass has unified massunits.

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.

Chemical ReactionsWe have learnt about elements,

molecules and compounds. We also learntsymbols and formulae of some elementsand molecules. We know how to write aformula to a compound. • Have you observed occuring different

colours when fire crackers are burnt?• Have you observed the change when a

piece of plastic is burnt?• Have you observed the change in colour

of litmus paper when dipped in acid orbase?

These changes are called chemicalchanges. The chemical substances whichare participating in chemical reaction arecalled reactants and the new substancesformed are called products.

Why these changes are occuring?In chemical reactions atoms are

neither created nor destroyed. A chemicalreaction is a process that is usuallycharacterized by a chemical change in which

Fig-5: Diagram on concept of mole

MOLE

1g of hydrogenatoms

6.022 x 1023

atoms of hydrogen

16 g of oxygenatoms

32 g of oxygenmolecules

6.022 x 1023

atoms of oxygen6.022 x 1023

molecules of oxygen

2 g of hydrogenmolecules

6.022 x 1023

molecules of hydrogen

SCERT TELANGANA


the starting materials (reactants) are differentfrom the products. Chemical reactions occurwith the formation and breaking of chemicalbonds. (you will learn about chemicalbonding in class 10).

These chemical reactions areclassified into four types based on the waythat chemical reaction takes place.

Types of Chemical Reactions

Chemical Combination

Activity 3

(This activity needs Teacher’s assistance)- Take a small piece (about 3 cm long)

of magnesium ribbon.- Rub the magnesium ribbon with sand

paper.- Hold it with a pair of tongs.- Burn it with a spirit lamp or burner.

• What do you observe?

You will notice that, magnesium burns in oxygenby producing dazzling white flame and changesinto white powder. The white powder ismagnesium oxide.2Mg(s) + O2 (g) 2MgO (s)Magnesium Oxygen Magnesium oxide

In this reaction magnesium andoxygen combine to form a new substancemagnesium oxide. A reaction in whichsingle product is formed from two or morereactants is known as chemicalcombination reaction.

You will also notice release ofenormous amount of heat energy whenmagnesium is burnt in air.

Let us discuss some more examplesof combination reactions.i. Burning of Coal: When coal is burnt

in oxygen, carbon dioxide is produced.

C (s) + O2 (g) CO2 (g) + Q (heat energy)

ii. Slaked lime is prepared by adding waterto quick lime.

Ca O(s) + H2O(l) Ca(OH)2(aq)+ Q(heatenergy)

Large amount of heat energy isreleased on reaction of water with CaO(s).If you touch the walls of the container youwill feel the hotness. Such reactions arecalled exothermic reactions.

A solution of slaked lime producedin the reaction equation (ii) is used to whitewash the walls. Calcium hydroxide reacts

fig-6: Burning of magnesiumribbon

fig-7: Formation of slaked lime by the reaction of CaO with water

Beaker

Water

CaOSCERT TELANGANA


In the above activity, on heatingcalcium carbonate decomposes to calciumoxide and carbon dioxide.

HeatCaCO3 (s) ––––– CaO(s) + CO2 (g)Lime stone quick lime

It is a thermal decompositionreaction. When a decomposition reactionis carried out by heating, it is calledthermal decomposition reaction.

Activity 5Arrange the apparatus as shown infig.- 9.

- Take about 0.5g of lead nitrate powderin a boiling test tube.

- Hold the boiling tube with a test tubeholder.

- Heat the boiling tube over a flame. (seefig.- 9)

- Note down the change.- What do you observe?

fig-9:Heating of lead nitrate andemission of nitrogen dioxide

Stand

Clamp

Delivery tube

NO2

Gascollectingjar

Lead nitrateBoiling

tube

Bunsenburner

slowly with the carbon dioxide in air to form a thin layer of calcium carbonate on the walls.It gives a white shiny finish to the walls.Decompostion Reaction

Activity 4

- Take a pinch of calcium carbonate (lime stone) in a boiling tube.

- Arrange the apparatus as shown in figure 8.

- Heat the boiling tube over the flame of spirit lamp or burner.

- Now bring a burning match stick near the evolved gas as shown in the figure.

- What do you observe?

You will notice that match stick would be put off.

fig-8: Heating of calcium carbonate and testing the gas evolved with burning match stick

StandBunsenburner

Clamp

Delivery tube

Boiling tube

Calcium carbonate

Burning match stick

Put off match stick

SCERT TELANGANA


fig-11(a): Silver bromide(light yellow colour)

On heating lead nitrate decomposesto lead oxide, oxygen and nitrogen dioxide.You observe the brown fumes liberating inthe boiling tube. These brown fumes are ofnitrogen dioxide (NO2).

Heat2Pb(NO3)2

––––– 2PbO(s)+ 4NO2 (g)+ O2(g)Lead Nitrate lead oxide Nitrogen Oxygen

dioxide

This is also a thermaldecomposition reaction. Let us performsome more decomposition reactions

Activity 6- Take a plastic mug. Drill two holes at

its base.- Fit two ‘one holed rubber stoppers’ in

these holes.- Insert two graphite electrodes in these

rubber stoppers.- Connect the electrodes to 9V battery

as shown in fig.- Fill the mug with water, so that the

electrodes are immersed.- Add few drops of dilute sulphuric acid

to water.- Take two test tubes filled with water and

invert them over the two graphiteelectrodes.

- Switch on the current and leave theapparatus undisturbed for some time.

• What do you observe in the test tubes?You will notice the liberation of gas

bubbles at both the electrodes. Thesebubbles displace the water in the test tubes.Is the volume of gas collected in both thetest tubes same?

Once the test tubes are filled withgases take them out carefully. Test boththe gases separately by bringing a burningcandle near the mouth of each test tube.

• What do you observe in each case?• Can you predict the gas present in each

test tube?In the above activity on passing the

electricity, water dissociates to hydrogenand oxygen. This is called electrolyticcomposition reaction.

Electrolysis2H2O (l) ––––– 2 H2 (g) + O2 (g)

Activity 7- Take some quantity of silver bromide

on a watch glass.- Observe the colour of silver bromide.- Place the watch glass in sunlight for

some time.- Now observe the colour of silver

bromide.• What changes do you notice?• Did the colour of the silver bromide

change?

fig-10: Electrolysis of water

O2 H2

+ –Anode Cathode

9 Vbattery

Plastic mug

Test tubes

Switch

Graphite rodsAcidifiedwater

SCERT TELANGANA


Silver bromide decomposes to silver andbromine in sunlight. Light yellow colouredsilver bromide turns to gray due tosunlight. sunlight2AgBr(s) ––––– 2 Ag (s) + Br2 (g)

This decomposition reaction occursin presence of sunlight and such reactionsare called photochemical reactions.

All the above decompositionreactions require energy in the form ofheat, light or electricity for converting thereactants to products. All these reactionsare endothermic.Carry out the following Activities:i) Take a pinch of AgCl in a watch glass.

Keep it in sunlight for some time andobserve the change.

ii) Take some ferrous sulphate crystals ina boiling tube. Heat it over spirit lamp.

iii) Take about 2 gm of barium hydroxide ina test tube. Add about 1 gm ofammonium chloride and mix with glassrod. Touch the test tube with your palm.

What do you observe?

Displacement reaction.In displacement reaction one element

displaces another element from itscompound and takes its place there in.

Displacement of hydrogen fromacids by metals:

Generally metals which are moreactive than hydrogen displace it from anacid.

Let us observe the reaction infollowing activity.

Activity 8- Take a small quantity of zinc dust in

a conical flask.- Add dilute hydrochloric acid slowly.

- Now take a balloon and tie it to themouth of the conical flask.

- Closely observe the changes in theconical flask and balloon.

- What do you notice?You can see the gas bubbles coming outfrom the solution and the balloon bulgesout (fig.-12b). Zinc pieces react with dilutehydrochloric acid and liberate hydrogengas as shown below.Zn (s) + 2HCl (aq) ZnCl2(aq ) + H2 (g)

In the above reaction the element zinchas displaced hydrogen from hydrochloricacid. This is displacement reaction.

fig-11(b): when exposed tosunlight (gray colour) silver metal

fig-12(a)

fig-12(b)

Dil. HClConical flask

Zinc dust

Conical flask

Zinc dust

Balloonwith H2

gas

Dil. HCl

SCERT TELANGANA


Activity 9

- Take two iron nails and clean them byrubbing with sand paper.

- Take two test tubes and mark them as Aand B.

- Take about 10ml of copper sulphatesolution in each test tube. Dip one iron nailin copper sulphate solution of test tube Aand keep it undisturbed for 20 minutes.

- Keep the other iron nail and test tube aside.

- Now take out the iron nail from coppersulphate solution and compare with theother iron nail that has been kept aside. (seefig13-a)

- Compare the colours of the solutions inthe test tubes. (see fig13-b)

- What changes do you observe?

You will find the iron nail dipped incopper sulphate solution becoming brown.The blue colour of copper sulphate solutionin test tube ‘A’ fades.

The chemical reaction in thisactivity is:

Fe (s) + CuSO4 (aq) FeSO4 (aq) + Cu (s)

Iron is more reactive than copper,so it displaces copper from coppersulphate. This is another example ofdisplacement reaction.

Other examples of displacementreaction are:

i) Zn(s) + 2AgNO3(aq) Zn (NO3)2(aq)+ 2Ag(s)

ii) Pb (s) + CuCl2 (aq) PbCl2 (aq) + Cu (s)

fig-13(a):Iron nail dipped incopper sulphate solution

fig-13(b): Iron nail and coppersulphate solutions compared before

and after the experiment

A

B

SCERT TELANGANA


Double displacement reaction

Activity 10- Take a pinch of lead nitrate and dissolve

in 5.0ml of distilled water in a test tube.- Take a pinch of potassium iodide in a

test tube and dissolve in distilled water.

- Mix lead nitrate solution withpotassium iodide solution.

- What do you observe?A yellow coloured substance which

is insoluble in water, is formed asprecipitate. The precipitate is lead iodide.

Pb(NO3)2(aq)+2KI(aq) PbI2(s)+ 2KNO3(aq)lead nitrate potassium Iodide lead iodide potassium nitrate

This reaction is doubledisplacement reaction. If two reactantsexchange their constituents chemically andform two products, then the reaction iscalled as double displacement reaction.

Other examples of doubledisplacement reactions are:1) Sodium sulphate solution on mixing with

barium chloride solution forms a whiteprepitate of barium sulphate andsoluble sodium chloride.

Na2SO4 (aq) + BaCl2 (aq) BaSO4(s)+ 2NaCl(aq)

2) Sodium hydroxide reacts withhydrochloric acid to form sodiumchloride and water.

NaOH (aq) + HCl(aq) NaCl(aq) + H2O(l)

3) Sodium chloride spontaneouslycombines with silver nitrate in solutiongiving silver chloride precipitate andSodium nitrate.

NaCl (aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)

Oxidation and Reduction‘Oxidation’ is a chemical reaction

that involves the addition of oxygen orremoval of hydrogen.

‘Reduction’ is a chemical reactionthat involves the addition of hydrogen orremoval of oxygen.

Let us try to understand moreclearly doing this experiment.

Activity 11- Take about 1.0g of copper powder in a

china dish.

- Keep the china dish on a tripod standcontaining wire gauge.

- Heat it with a bunsen burner or with aspirit lamp.

- Do you find any change in colour ofcopper?

You will notice that the surface layerof copper becomes black.

• Why does the colour of copper change?• What is the black colour product

formed on the surface of copper?

fig-14: formation of lead iodide andpotassium nitrate

Pb(NO3)2

KI

Lead Iodide

KNO3

SCERT TELANGANA


In the activity on heating copperit reacts with oxygen present in theatmosphere to form copper oxide.

The reaction is shown below. Heat

Cu(s) + O2(g) ––––– 2 CuO(S)

• What do you notice?

• Is there any change in black colour ofcopper oxide?

You will notice that the black coatingon copper turns brown because copperoxide loses oxygen to form copper. In thisprocess oxygen is lost and the process iscalled Reduction.

HeatCuO(s) + H2(g) ––––– Cu(s) + H2O (g)

In the above reaction hydrogen isgained; such reaction is called reductionreaction.

Generally oxidation and reductionoccur in the same reaction. If one reactantgets oxidized, the other gets reduced. Such

fig-15(a):Copper Oxide change

into black colourfig-15(b): Oxidation of copper to copper

oxide

Chind dishcontaining

Copperpowder

Wire gauze

Tripodstand

Bunsenburner

fig-16: Reduction of copper oxide to copper

H2 gas H2O(g)

Bunsenburner

Stand

Glass tube

Black copper oxide Cork

Here copper combines with oxygen to form copper oxide. Here oxygen is gained and theprocess is called oxidation.

Now pass hydrogen gas over hot copper oxide obtained in above activity and observethe change (see fig. 16).

SCERT TELANGANA


reactions are called oxidation-reductionreactions or redox reactions.

In the CuO, H2 reaction CuO isreduced and H2 is oxidized.

Some other examples of redox reactionsare:

i) 2 Fe2O3(s) + 3C(s) 4 Fe(s) + 3CO2 (g)

ii) 2 PbO (3) + C(s) 2Pb(s) + CO2 (g)

Have you observe the effects of oxidationreactions in daily lifeCorrosion:

You must have observed that afreshly cut apple turns brown after sometime. The shining iron articles graduallybecome reddish brown when left for sometime. Burning of crackers produce dazzlinglight with white fumes.

• How do these changes occur?They are all the examples of the

process called oxidation.

Let us know how?Oxidation is the reaction of oxygen

molecules with different substancesstarting from metal to living tissue whichmay come in contact with it.

Apples, pears, bananas, potatoesetc., contain enzyme called polyphenoloxidase or tyrosinase, which reacts withoxygen and changes the colour on the cutsurface of the fruit.

The browning of iron, when left forsometime in moist air, is a processcommonly known as rusting of iron. Thisprocess is basically oxidation reactionwhich requires both oxygen and water.Rusting does not occur in oxygen freewater or dry air.

Burning of crackers is alsooxidation process of variety of chemicals,like Magnesium and Sulphur.• Did you notice the colour coating on

copper articles?When some metals are exposed to

moisture, acids etc., they tarnish due to theformation of respective metal oxide ontheir surface. This process is calledcorrosion.Look at the following example: Green coating on copper (see fig-18)

2Cu + O2 2CuO

Corrosion causes damage to carbodies, bridges, iron railings, ships etc.,and to all other objects that are made of

fig-17: Rusting of ironfig-18: Corrosion of copper

SCERT TELANGANA


metals. Especially corrosion of iron is aserious problem.

Corrosion can be prevented or atleast minimized by shielding the metalsurface from oxygen and moisture. It canbe prevented by painting, oiling, greasing,galvanizing, chrome plating or makingalloys. Galvanizing is a method ofprotecting iron from rusting by coatingthem a thin layer of Zinc.

Alloying is also a very good methodof improving properties of metal.Generally pure form of iron is very softand stretches easily when hot. Iron is mixedwith carbon, nickel and chromium to getan alloy called stainless steel. The stainlesssteel is hard and does not rust.

A metallic substance made bymixing and fusing two or more metals, or ametal and a nonmetal, to obtain desirablequalities such as hardness, lightness, andstrength is known as alloy.

For example: Brass, bronze, and steel.

Some more effects of oxidation oneveryday life

1. Combustion is the most commonexample for oxidation reactions.

For example: burning of woodinvolves release of carbon dioxide, watervapour along with huge amount of energy.

2. Rising of dough with yeast depends onoxidation of sugars to carbondioxide and water.

3. Bleaching of coloured objects usingmoist chlorine

Cl2 + H2O HOCl + HCl

HOCl HCl + [O]

Coloured object + [O] Colourless object.

Some times during rainy season thepower supply to our home from the electricpole will be interrupted due to formationof the metal oxide layer on the electricwire. This metal oxide is an electricalinsulator. On removing the metal oxidelayer formed on the wire with a sand paper,supply of electricity can be restored.

4. Rancidity• Have you ever tasted or smelt the

fat/oil containing food materialsleft for a long time?When fats and oils are oxidized they

become rancid. Their smell and tastechange.

Thus we can say that oxidationreactions in food material that were leftfor a long period are responsible forspoiling of food.

Rancidity is an oxidation reaction.• How can we prevent the spoiling

of food?The spoilage of food can be

prevented by adding preservatives likeVitamin C and Vitamin E.

Usually substances which preventoxidation (Antioxidants) are added to foodcontaining fats and oil. Keeping food in airtight containers helps to slow downoxidation process.

Manufacturers of potato chips flushbags of chips with nitrogen gas to preventthe chips from getting oxidized.

SCERT TELANGANA


• 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 a chemicalreaction and retain all its properties.

• A molecule is the smallest particle of an element or a compound that is capable ofindependent existence and retains 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 value of 6.022 x 1023.

• Mass of one mole of a substance is called its molar mass.

• In a combination reaction two or more substances combine to form a new singlesubstance.

• In a decomposition reaction a single substance decomposes to give two or moresubstances.

• Reactions in which heat energy is absorbed by the reactants are endothermicreactions.

Key words

Law of conservation of mass, Law of constant proportion. Atom, Symbol, Atomicmass, Atomic mass unit (amu), Unified mass (u), Molecule, Molecules ofelements, Molecules of compounds, Formula, Ion (cation, anion), Atomicity,Valency, Molecular mass, Formula unit mass, Mole, Avogadro constant, Molarmass, Chemical combination, Chemical decomposition, Displacement reaction,Double Displacement reaction, Oxidation, Reduction, Corrosion, Rancidity,Antioxidants.

What we have learnt

SCERT TELANGANA


• In exothermic reaction heat energy is released by the reactants.

• A displacement reaction occurs, when an element displaces another element fromits compound.

• Two different atoms or ions are exchanged in double displacement reactions.

• Oxidation is the gain of Oxygen or loss of Hydrogen.

• Loss of oxygen or gain of Hydrogen is Reduction.

• Corrosion causes damage to iron appliances.

• When fats and oils are oxidized, they become rancid.

• Precipitate is an insoluble substance.

Reflections on concepts1. Explain the process and precautions in verifying

law of conservation of mass. (AS-3)2. In a class, a teacher asked students to write the

molecular formula of oxygen Shamita wrote the formula as O2 and Priyanka as O.which one is correct? State the reason. (AS-1)

3. Find out the chemical names and formulae for the following common householdsubstances. (AS-1)a) common salt b) baking soda c) washing soda d) vinegar

4. Calculate the mass of the following (AS-1)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.

5. Convert into mole (AS-1)a) 12g of O2 gas. b) 20g of water. c) 22g of carbon dioxide.

6. Write the valencies of Fe in FeCl2 and FeCl3 (AS-1)7. Calculate the molar mass of Sulphuric acid (H2SO4) and glucose (C6H12O6)(AS-1)

8. 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) (AS-1)


SCERT TELANGANA


9. Fill the following table

10. Write an equation for decomposition reaction where energy is supplied in the formof Heat / light / electricity. (AS-1)

11. How chemical displacement reactions differ from chemical decompositionreaction? Explain with an example for each. (AS-1)

12. Name the reactions taking place in the presence of sunlight? (AS-1)

13. Give two examples for oxidation-reduction reaction. (AS-6)

14. Draw the digarm to show the experimental setup to verify the law of conservationof mass. (AS-5)

Application of concepts1. Why do we apply paint on iron articles? (AS-1)2. What is the use of keeping food in air tight containers? (AS-6)

Higher Order Thinking Questions

1. 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? (AS-1)

2. Carbon dioxide is added to 112g of calcium oxide. The product formed is 200g ofcalcium carbonate. Calculate the mass of carbon dioxide used. Which law of chemicalcombination will govern your answer. (AS-1)

3. Imagine what would happen if we do not have standard symbols for elements?(AS-2)

Sl.No. Name Symbol/formula Molar mass Number of particlespresent in molar mass.

1 Oxygen Atom 16g 6.022 x 1023 atoms ofoxygen

2 Oxygen molecule

3 Sodium

4 Sodium ion 23g

5 Sodium chloride 6.022 x 1023 units ofsodium chloride

6 water

SCERT TELANGANA



1. Fe2O3 + 2Al Al2O3 + 2 Fe. [ ] The above reaction is an example of:

a) Combination reaction b) Decomposition reactionc) Displacement reaction d) Double decomposition reaction

2. What happens when dil. hydrochloric acid is added to iron filings? Choose the correctanswer.a) Hydrogen gas and iron chloride are produced. [ ]b) Chlorine gas and iron hydroxide are produced.c) No reaction takes place.d) Iron salt and water are produced.

3. 2 PbO(s) + C (s) 2Pb(s) + CO2 (g) [ ]Which of the following statements are correct for the above chemical reaction?a) Lead oxide is reduced b) Carbon dioxide is oxidizedc) Carbon is oxidized d) (a) and (c) are correct

4. The chemical equation BaCl2 + Na2SO4 BaSO4 + 2NaCl represents

following type of chemical reaction. [ ]

i) displacement ii) combination

iii) decomposition iv) double-displacement

5. The reaction of formation hydrogen chloride from hydrogen and chloride

represents following type of chemical reaction [ ]

i) decomposition ii) displacement

iii) combination iv) double-displacement


1. Do an experiment to understand the changes in weight of reactants and products in

a chemical reaction, write a report.

Suggested Project

1. Collect the information about the symbols, atomic weights of first thirty elements

in the periodic table and write a report.

SCERT TELANGANA


You must have noticed thatsome things float on the surface of the waterand some things sink in it. Did youparticipate in the activity “floating &sinking” in the lesson on “materials” inclass 6? If so, you might have wonderedas to why some objects, which youexpected would sink, float on water. Didyou take one of those objects that floats onwater, and tried to see if it floats onkerosene or coconut 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 butfloat on water?Which objects sink in water?Draw a diagram of the tube, showingthe results of your activity.Why did different objects behavedifferently?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 whilethe piece 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 onwater even though it is heavier than amarble?What do we mean by ‘heavy’, what dowe mean by ‘light’?

Chapter

8 FLOATING BODIES

SCERT TELANGANA

128 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 boththese sentences? In science, we try toensure that each word we use, has the samemeaning for everyone. So let’s see in whatway these two 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 themeaning of the second sentence?

In second sentence, when we say ironis heavier than wood, it means if we takea piece 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.

Fill one to the brim with water and theother with oil.

Which will weigh more? Which liquid is denser?

Take two equal sized blocks made ofwood and rubber.

Which of these two blocks isheavier?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 blockis 30 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 more dense each solidobject is compared to water. This is knownas relative density of that object.

massvolume

gmcm3

kgm3

SCERT TELANGANA


Fig-2

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 twoweights are then compared.

Let us perform an activity to understandhow 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 electronic weighing machine, rubber erasers, wooden blocks, glass slides,iron nails, plastic cubes, piece of aluminium sheet, glass marbles, stones, corks etc (note:whatever object you take, ensure that its volume is more than 20 cc and it should not behollow). Record the 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. Pour waterin the overflow vessel until it starts drippingfrom its beak. When water stops drippingfrom the beak, place the 50ml measuringcylinder under it. Slip the object gently intothe overflow vessel as shown in fig.- 2,ensuring that water does not splash out. Once

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

the object is in the overflow vessel, waterflows out of the beak and collects in the 50ml. cylinder. Wait till the flow stops. (Theobject should be fully immersed in water. Ifthe object is not fully immersed, push it into the water with a pin. see fig.- 2)

Weigh the cylinderwith the water thatoverflowed and recordthe weight in column 4.

SCERT TELANGANA

130 Floating Bodies

If we subtract the weight of themeasuring cylinder from this weight, we getthe weight of water (column 5 of thetable-1). This is the weight of water equalto the volume 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 equalvolume of water (column 5). This tells ushow many times denser the object is,compared to water.

Relative density of an object =

Find the relative densities of all objectsthat you collected.

Based on table-1, answer the followingquestions.

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?Do objects that have a relative densityless than 1 sink in water or float on it?Do the objects that sink in water havea relative density less than1 or morethan 1?Classify the above materials as denserthan stone and less dense than stone.What relationship do you find betweenthe relative density of objects andfloating-sinking of the objects?One interesting aspect of relative

density 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 the relativedensity of liquids. For this, we need to findthe weight of a fixed volume of the liquid andthe weight of an equal volume of water. Theformula for finding the relative density of aliquid is:

Relative density of a liquid =

( Note: Here mass and weight areconsidered as equivalent)

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 electronic weighing machine and milk,groundnut oil, kerosene about 50 ml. eachin different containers.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

SCERT TELANGANA


Repeat this for other liquids and recordthe weights in column 3. Calculate theweight of each liquid by subtracting theweight of the empty bottle and record itin column 4. Calculate the relative density

of each liquid by comparing the weightof the liquid with the weight of samevolume of water and record these valuesin column 5.

Answer the following questions bycomparing Table-1 and Table-2

Which liquid will float on top ifgroundnut 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?How can you say ?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 therelative 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 tubeand fill it with water.

Put the refill in with the metallic pointinside the water as shown in fig.- 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 ofthe boiling tube. Think what to do to makethe refill to stand as shown in fig.- 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)

SCERT TELANGANA

132 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 ρ1 and ρ2

i) When they are taken with same mass ‘m’ and their volumes are V1 and V2, the

mass of water m = ρ1V1 ; V1= 1

mρ and the mass of milk m = ρ2V2 ; V2=

2

mρ

Total mass of water and milk is m + m = 2m

Total volume of water and milk is V1 + V2= 1

mρ +

2

mρ

= m 1 2

1 1ρ ρ

⎡ ⎤+⎢ ⎥

⎣ ⎦

= ( )1 2

1 2 m ρ ρ

ρ ρ+

Fig -3: Improvisedlactometer

SCERT TELANGANA


The effective density of the mixture ( ρρρρρeff) = Total mass

Total volume

=

=

=

ii) when they are taken with same volume ‘V’ and their masses are m1 and m2

the volume of water V = 1

1

mρ That is m1 = V ρ1

and the volume of milk V = 2

2

mρ That is m2 = V ρ2

Total mass of water and milk is m1 + m2 = V ρ1 + V ρ2

= V (ρ1 + ρ2)Total volume of water and milk is V + V = 2V

The effective density of the mixture ( ( ( ( ( ρρρρρeff ) = Total mass

Total volumeρ

eff = V (ρ1 + ρ2) 2V

= (ρ1 + ρ2)12

2 mm (ρ1 + ρ2) / ρ1 ρ2

2 (ρ1 + ρ2) / ρ1 ρ2

2 ρ1 ρ2

ρ1 + ρ2

When do objects float on water?

Activity-3

Do objects denser than waterfloat in it ?

Collect small objects as you did forLab Activity 1. Place them one by one in aglass of water and observe whether they sinkor float in water? Record your observationsin Table-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

SCERT TELANGANA

134 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 inwater. But in activity-3, we observed thatsubstances with a relative density greaterthan 1 sometimes float on water.

So it seems we cannot judge whethera substance will sink or float only on thebasis of its relative density. There isdefinitely some other factor which weneed to take into 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 weightof the water displaced by it to find itsrelative density. In that activity, weimmersed the substance fully in water andcollected the displaced 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 outletof the overflow jar. Then take the beakerfrom the weighing balance and place itbelow the outlet of the overflow jar. Takea wooden block, moisten it with water andthen drop it gently into the overflow jar.Don’t forcefully submerge the woodenblock in the water. Also, ensure that it doesnot block the outlet of the overflow jar.Water will flow out of the overflow jarand collect in the beaker kept under theoutlet of overflow jar.

Do you think the weight of the waterdisplaced by the wooden block will be lessthan or equal to or more than the weightof the wooden block? Make a guess. Placethe beaker containing the displaced wateron one pan of the weighing balance. Takethe wooden block, wipe it to clean offwater and place it on the other pan alongwith the weights that equal to the weightof empty beaker as shown in Fig.- 4.

Fig. 4

SCERT TELANGANA


Do the two pans balance?Is the weight of the water displaced bythe wooden block less than, equal to ormore than the weight of the woodenblock?Repeat this experiment with several

other substances that float or sink. ThingsTable -4

S. 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 byArchimedes. You will know about thisfurther in this chapter.)

Can you think of a way to make ironfloat on water? Perhaps, the followingactivity will give you some ideas abouthow you 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 iscorrect or not.

Now unfold the aluminium foil and makeit as a small bowl. Place this bowl in thewater 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.

SCERT TELANGANA

136 Floating Bodies

How much water did the bowl ofaluminium foil displace?Is the water displaced by foldedaluminium foil and bowl made usingsame aluminium foil the same?Explain why aluminium bowl floats on

the basis of your theory of floating substances.Can you now explain why large shipsmade of iron and steel float on waterwhile a small block of iron sinks in water?Why does the metal bowl displace largeramount of water than a metal piece?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 forceto balance the force of gravity. Thisupward force must come from water. Ifthe gravitational force on the object ismore than the upward force of water, theobject will sink in water. Let us do a simpleactivity to 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 fig.- 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 it bouncesback to the surface of water! So the upwardforce of water is a real, observable force(Buoyancy). This force acting on unit areaof the surface of an object (bottle) is calledstatic pressure of the water.

Pressure of Air

Activity -7Observing air pressure

Take a glass tumbler. Stick somecotton at the bottom of it. Immerse itinversely in water up to the bottom of thecontainer as shown in fig.- 6.

Take out the tumbler from water. Is thecotton attached to its bottom wet? Why?This is due to the force of air which isapplied on water by the air present in thetumbler and stops water from entering thetumbler. This force on unit area of water ispressure of air.

Fig. 6SCERT TELANGANA


24 cm

76 cm

Glass Tube

Stand

Mercury

GlassVessel

Vacuum

Atmospheric pressureAll objects on the surface of the earth are subject to constant atmospheric pressure.

Atmospheric pressure = Force of the atmospheresurface area of the earth =

FA

Atmospheric pressure = Weight of the atmospheresurface area of the earth = A

w

Atmospheric pressure = (Mass of the atmosphere) g(surface area of the earth)

× = Am g×

(average density of the atmosphere) (Volume of the atmosphere) gAtmospheric pressure =(surface area of the earth)

× ×

= Av gρ × ×

surface area of the earth Height of the atmosphere gThus Atmospheric pressure =surface area of the earth

ρ × × ×

= A

Ah gρ × × ×

Atmospheric pressure = ρ (Height of the atmosphere) g = ρ × h × gAtmospheric pressure = ρ h g

Po = ρ h g

Measuring atmospheric pressureWe do not directly experience this

atmospheric pressure, but we can measureit with barometers. The first barometer was

invented by ‘Torricelli’ using mercury. (Seefig.- 7)

At normal atmospheric pressure themercury barometer shows a mercurycolumn of 76 cm height in the glass tubeabove the surface of the mercury in thebowl. This is known as 1 atmosphericpressure.

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

SCERT TELANGANA

138 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) ρ g= Ahρg

Let ‘Po’ be the atmospheric pressure.Force on the column due to the atmospheric pressure = Po Athen,A hρg = Po APo = ρgh (of mercury)Since ρ,g are constants, the height of the mercury column depends on atmospheric

pressure.We can now calculate the value of atmospheric pressure ‘Po’, by substituting the values

of height of the mercury column ‘h’ , density of the mercury ‘ρ’ and acceleration due togravity ‘g’.Height of the mercury column h = 76cm = 76 10-2 mDensity of the mercury ρ= 13.6 gm/cc = 13.6 X 103 kg/m3

Acceleration due to gravity g = 9.8 m/s2

Po = hρgPo = (76 10-2 m) (13.6 103 kg/m3) (9.8 m/s2)Po = 1.01 105 kg.m/m2.s2

hence, Po = 1.01 105 N/m2 (∵1 kg.m/s2 = 1 newton)This value is also known as one atmospheric pressure. (1 atm)

1 atm = 1.01 105 N/m2 = 1.01 × 105 pascal (∵ 1 N/m2 is called pascal)

The mass of air that would occupy a cylindrical tube with cross sectional area of1cm2 and that extends 30km upto the top of the atmosphere is about 1kg. The weightapplied on the surface area of 1cm2 on the earth is the 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.

SCERT TELANGANA


Think and discuss

What would happen if the Toricelli experiment is done on moon? A stopper is inserted in a small hole of the glass tube of the mercury barometer belowthe top level of the mercury in it. What happens when you pull out the stopper from theglass tube?Why don’t we use water instead of mercury in Torricelli experiment? If we are readyto do this experiment, what length of tube is needed?Find the weight of the atmosphere around the earth (take the radius of earth as 6400km.)

Pressure at a depth “h” in a liquid Let us consider a container which contains a liquid of density “ρ”. Consider a cylindrical column of height ‘h’ from the surface of the liquid of cross

sectional area “A”. See the fig.- 8. The volume of the liquid column V = Ah Mass = Volume density m = Ah ρWeight W = mg = Ah ρ gYou know that from Newton’s law, the net force on it is zero,

because it is at rest. What are the forces acting on that watercolumn?

There are three forces acting, which are,i) Weight (W), vertically downii) Force on top surface due to atmospheric pressure(Po A), acting vertically downiii) Force on the bottom surface of the column due to static pressure of liquid (PA),

acting vertically up.From Newton's laws we getPA = Po

A + WPA = Po A + h ρ g AWhere P is the pressure at the depth “ h” from the surface of the liquid and Po is the

atmospheric pressure. PA = Po A + h ρ g A

P = Po + h ρ g ....... (1)This means that the pressure inside the liquid at a constant depth is constant.

Fig. 8

SCERT TELANGANA

140 Floating Bodies

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 ρ be the density of the liquid. See the fig.- 9.What is the pressure P1 in the liquid at depth h1 ?From equation (1) we get,

P 1 = Po + h1 ρ g ....... (2)Similarly, pressure P2 at depth h2 is given byP2 = Po + h2 ρ g ....... (3)From equations (3) and (2) we get

P2 - P1 = h2 ρ g - h1 ρ gP2 - P1= ρ g (h2 - h1)

from the figure h = h1 - h2 sowe have, P2 - P1 = hρ g

The pressure difference between two levels in that liquid = h ρ g.Here density of the liquid ‘ ρ’ and ‘g’ are constants, so the pressure difference increaseswith a increase in depth. What happens if we replace this cylindrical liquid column with anotherobjectwhich 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 ρg (values of the liquid)

P2- P1 = h × mV × g

P2 - P1 = h × mAh × g

P2 - P1 = mA × 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 thedisplaced liquid.

BuoyancyThe force applied on the object in upward direction is called “Buoyancy”. As per the above

equation this buoyant force is equal to the weight of the liquid displaced by the object.

Fig 9

SCERT TELANGANA


Measuring the force of buoyancyWe know 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 is immersed in water? Youmay notice that the stone, when immersed,appears to lose some weight.

Why does the stone lose weight whenit 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 lossof weight 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 noticethat in every case an immersed objectappears to lose weight.

When the object floats on the surfaceof water, it appears as if it has lost all itsweight, that is, the spring balance showszero reading for floating bodies! Forobjects that float on a liquid surface, theforce of buoyancy balances the force ofgravity at the 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 theimmersed stone

Suspend a stone from a spring balance.(it is better to take a stone that is morethan 300 gm) . Note the reading on thespring balance. The reading gives the weightof the stone. Take an overflow vessel withwater and place a graduated beaker belowthe beak. (Fig.- 10).

Now immerse the stone in the water.Note the reading on the spring balance and

Fig. 10

SCERT TELANGANA

142 Floating Bodies

Do you know?

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.

let's look at the story that is associatedwith this observation.

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 the

volume of water displaced ,the density of the crown could be obtained. This densitywould be lower than that of gold if cheaper and less dense metals had been added.Archimedes then took to the streets naked, so excited by his discovery that he hadforgotten to dress, crying "Eureka!" (meaning "I have found it!").

measure the volume of water thatoverflows into 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).

SCERT TELANGANA


Do you know?

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.

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. Uponwhich is the buoyant force greater?A piece of iron when placed on a blockof wood, this makes the wood to floatlower in the water. If the iron piece issuspended beneath the wood block,would it float at the same depth? Orlower or higher?You know that pressure difference at

different levels inside the liquid causesbuoyancy.

Can we increase the pressure inside theliquid?

It is only possible when the liquid isenclosed. A scientist named Pascal made aprinciple about what happens when anexternal 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 Fig.- 12. Here, we have anenclosed volume of fluid in a U-shapedtube. The fluid is enclosed in the tube by

How did Archimedes solve the King's problem? A simple arrangementcould be used to determine whether a golden crown is less dense than gold.The crown and a bar of gold, of the same mass as the crown, are suspendedfrom the two arms of a simple balance as shown in the figure. The balanceis lowered into a vessel of water. If the crown (left) is less dense than thegold bar (right), it definitely posses a larger volume than that of pure goldand will displace more water and thus experience a larger upward buoyantforce, causing the balance to tilt towards the gold bar. This would indicate that the crown was notof pure gold!

Note: This experiment holds good only when the crown doesn’t have any covered hollowportion in it. Think why?

Fig-12: Application of Pascal's principle(bramah press)

F1F2

P1 = 1

1

FA

P2 = 2

2

FA

V

V

Fluid

SCERT TELANGANA

144 Floating Bodies

F1 F2

Fluid

Key words

two leak-proof pistons in each arm. Thearea of cross-section of the right and lefttubes is A2 , A1 (A2 >A1).

When a force F1 is applied to the leftpiston the excess pressure acting on the

fluid volume is 1

1

FA .

According to Pascal's principle, thisexcess pressure is transmitted equallythroughout the fluid volume. That is, everyunit area of the fluid 'experiences' this

excess pressure of 1

1

FA .

The excess pressure in the right-sidetube (of cross section area A2) is also

1

1

FA and since its area is A2, the upward

force acting on the right piston is

F2 = 2 1

1

A FA×

; which is much larger in

magnitude than F1.

Thus the application of Pascal'sprinciple results in a large upward force(thrust) on the right piston when a smalldownward force is applied on the leftpiston.

Fig. 13: Hydraulic jack

This principle is used in the design andworking of hydraulic jacks/lifts (Fig-13)which you can see in automobileworkshops. A small downward forceapplied by the hand of the operator helps olift 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 ofbuoyancy).

What we have learnt

↑↑↑↑↑↓↓↓↓↓

SCERT TELANGANA


The apparent loss of weight of an object, which is immersed in a fluid, is equal tothe weight of fluid displaced by the object. (Archimedes' principle)

When an object floats on the surface of a fluid, it displaces a weight of fluid

equal 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)

Reflections on concepts1. Why do some objects float on the water? And

some sink? (AS1)2. Explain density and relative density and write their formulae. (AS1)3. Explain buoyancy in your own words. (AS1)4. How can you find the relative density of a liquid? (AS3)5. Draw the diagram of a mercury barometer. (AS5)

Application of concepts1. A solid sphere has a radius of 2 cm and a mass of 0.05 kg. What is the relative

density of 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

filled with 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 water filled in glass tumbler (density of

ice = 0.9 g/cm3). Will the water level in the glass rise? When the ice melts completely(AS1)

4. Find the pressure at a depth of 10m in water if the atmospheric pressure is 100kPa.[1Pa=1N/m2] [100kPa = 105 Pa = 105 N/m2 = 1 atm.] (AS1) [Ans: 198 kPa]

Higher Order Thinking Questions1. Can you make iron to float in water? How? (AS3)2. Where do you observe Archimedes principle in daily life? Give two examples.(AS7)3. Do all objects that sink in water, sink in oil? Give reason. (AS1)


SCERT TELANGANA

146 Floating Bodies

1. Unit of relative density is [ ]

a) g/cm3 b) cm/g3 c) N/m2 d) No units

2. The instrument used to measure the purity of milk is [ ]a) Barometer b) Hygrometer c) Lactometer d) Speedometer

3. If P0 = Pressure, ñ = Density, h= height, and g = accelaration due to gravity thenthe atmospheric pressure = [ ]

a) P0 = ñhg b) P0 = mgh c) P0 = vgh d) P0 = ½ mgh

4. The first barometer with mercury was invented by [ ]

a) Pascal b) Archimedis c) Newton d) Torecelli

5. The hydraulic jack which is used in automobile work shops, works on the principleof [ ]

a) Archimedes b) Pascal c) Torecelli d) Newton

6. The density of water at 250C is [ ]

a) 1g/cm3 b) 2g/cm3 c) 3g/cm3 d) 0.99g/cm3

Suggested Experiments1. Conduct an experiment to find the relative densities of different substances and

write a report.2. Conduct an experiment to understand the phenomenon that a stone immersed in

water loses its weight. Write a report on it.

Suggested Projects.1. The oil brakes in vehicles works on the Pascal’s principle. Collect the information

on working of air brakes in vehicles and write a report.2. Find the relative densities of different fruits and vegetables and write a report.


SCERT TELANGANA


In previous chapters we have learnt thatall matter is made up of atoms. Accordingto Dalton, atoms were indivisible. Thatmeans that they could not be divided intostill smaller parts. Atoms of an element areall identical to each other and different fromthe atoms of other elements.

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 Faradaydiscovered the presence of chargedparticles.

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 there

Chapter

9WHAT IS INSIDE

THE ATOM?

must exist some tiny particles in atom whichare responsible for atom to behavesometimes as a charged particle. As atomis considered 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 ParticlesIn science, theories change when

scientists discover new facts or clues.Sometimes, an idea or model must bechanged as new information is gathered.Dalton proposed that atoms could not bedivided. Experimental evidence began toshow that atoms were divisible and are madeup 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-atomic

SCERT TELANGANA

148 What is inside the atom?

particles, one is positively charged andanother is negatively charged. In fact, threedifferent subatomic particles werediscovered. Let us see how ideas aboutatoms have changed over the time with thediscovery of sub-atomic particles.

Electron, Proton and NeutronYou 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 werealways identical. 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 negligible

and 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 are atomsconsidered to be 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 approximately1836 times greater than that of theelectron. It is represented as 'p+' and itscharge is taken as one unit positive.

In 1932, James Chadwick discoveredanother sub-atomic particle which had nocharge and had a mass nearly equal to thatof a proton. It was eventually named asneutron. In general, a neutron isrepresented as 'n0'.

From this discussion we can concludethat atoms are made of small particles calledprotons, neutrons, and electrons. Each ofthese particles is described in terms ofmeasurable properties, like mass and charge.The proton and electron have equal, butopposite, electrical charges. A neutron has

SCERT TELANGANA


no electrical charge. The mass of theelectron is about 1

1836th that of a proton.

Fig-1: neutron, proton and electron

• If an atom consist of sub-atomicparticles like protons, neutrons andelectrons, how are they arranged inthe atom?

Structure of an Atom

Activity-1

Sketch the structure of atom asyou imagine.

You learnt about electron, proton andneutron. Suppose you want to arrange themin an atom, how do you do it?

Many arrangements are possible. Thinkthat atom looks like a room. You can arrangethe particles in alternating rows. Can youdraw and show how it will look? Imagine anddraw another arrangement of sub-atomicparticles inside a spherical shape keeping inview the nature of sub-atomic particles.

• In how many ways can you arrangethese sub-atomic particles in aspherical shape?

Discuss with your friends and try toprepare a model to show various ways ofarranging the sub-atomic particles in theatom.

To understand the structure of an atom,scientists 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:

1. An atom is considered to be asphere of uniform positive chargeand electrons are embedded in it, asshown in figure 2(a).

Fig-2(a) Fig-2(b)

2. The mass of the atom is consideredto be uniformly distributed throughout the atom.

3. The negative and the positivecharges are supposed to balance outand the atom as a whole iselectrically neutral.

SCERT TELANGANA


Do You Know?

amount of energy. The stream of alpha (α)

particles is directed towards a very thin

gold foil.

The gold foil was placed inside a

detector in such a way that the detector

would show a flash of light when an alpha

(α) particle strikes it (see fig-3). The entire

arrangement was kept in a vaccum chamber.

Think of Thomson's model of the atom.

When the alpha particle hit the foil,

Rutherford expected that they all would be

deflected by the positive charge spread

evenly throughout the gold atoms.

Rutherford's Observations

Fig-4

Scattering of alpha particles

A more familiar example that

represents Thomson's atomic model is

watermelon (figure-2(b)). The positive

charge is spread throughout the atom like

the red part of watermelon. The black seeds

distributed through out the red part

represents electrons. Thomson's model

was modified by one of his students. What

is the reason for its modification? The

reason was that some of the experiments

carried out by Rutherford, a student of

Thomson, gave results, which were not

infavour of Thomson's model.

Besides winning the Nobel Prize in

Physics himself, seven of Thomson's

research students and even his own son,

George, won Nobel Prizes in Physics.

One of Thomson's students was Ernest

Rutherford.

Rutherford's ααααα particlesscattering experiment

Ernest Rutherford was a scientist born

in New Zealand. In 1909, he did some

experiments using gold foil and alpha (α)

particles. Alpha (α) particles consist of two

protons and two neutrons bound together.

Since they do not have any electrons, they

are positively charged with two units of

charge. Let us see the experimental set up

and understand Rutherford's experiment.

There is a source of the fast moving

alpha particles which have a considerable

stream of alpha (α) particlesalpha (α)

particle source

gold foil

detectorFig-3

SCERT TELANGANA


Do You Know?

(i) Most of the space inside the atomis empty because most of the alphaparticles passed through the gold foilas shown 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) As shown in Fig-4 all the positivelycharged particles in an atom formeda small dense centre, called thenucleus of the atom. The electronswere not a part of nucleus.

ii) He also proposed that the negativelycharged electrons revolve around thenucleus 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 the Sun.

iii) The size of the nucleus is very smallas compared to the size of the atom.

Try to sketch Rutherford's model of theatom.

But, it was found that most of the alphaparticles passed straight through the atomswithout any deflection, some particles deflectedin small angles.Only very few particles weredeflected through large angles and a very verysmall number of particles were reflected rightback as shown in figure 4.

On average, for every 20000 alphaparticles that were fired at the gold foilduring Rutherford's famous experiment,only one was reflected back.

Let us try to understand the results ofRutherford's experiments.

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.

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 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 :

SCERT TELANGANA


Think and discuss

Compare Rutherford and Thomson'smodels of the atom on the followingbasis:

• Where is the positive chargeplaced?

• How are the electrons placed?

• Are they stationary inside the atomor moving?

Limitations of Rutherford'satomic model

• Did you observe any defect withRutherford's model of the atom?

Think of an atom like hydrogen with oneelectron and one proton. The electron isattracted by the proton in the nucleus. Even

in circular motion around the nucleus, theelectron should has an acceleration.

An accelerated charged particlemoving in a circular path would alwaysradiate energy continuously. Thus, therevolving electron would lose energycontinuously and get directed towards thepositively charged nucleus as shown in fig-5 and eventually crash into the nucleus.

If this is true, the atoms would becomehighly 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 other arrange-ment of sub-atomic particles in theatom which prevents the revolvingelectron to fall into the nucleus?

In 1913, a Danish scientist Neils Bohrsuggested an atomic model to explainstructure of atom and to overcomeRutherford's limitations.

Bohr's Model of the AtomIn order to overcome the limitations

of Rutherford's model, in 1913, NielsBohr put forward a thought that electronscan be found only in certain energy levels,or regions, around the nucleus. Electronsmust gain energy to move to a higherenergy level or they must lose energy tomove to a lower level.

Fig-5

SCERT TELANGANA


Think of books arranged in a bookshelf.They can be placed on a higher shelf or lowershelf but never between shelves.

Restricting the path of electron insidethe atom, Neils Bohr made the followingpostulates about the model of an atom:

1. The electrons revolve round thenucleus in discrete fixed circularorbits of the atom. These orbits orshells are called stationary energylevels.

2. While revolving in these discreteorbits the electrons do not radiateenergy and this is the reason whyelectrons do not fall into thenucleus.

3. The electron orbits or shells arerepresented by the letters K, L, M,N… or the numbers, n=1, 2, 3, 4,….. as shown in the figure 6.

• Do you think that Bohr's model isthe final model of the atom?

Niels Bohr could successfully explainthe properties of a hydrogen atom like theatomic spectra emitted by hydrogen atomemploying this model but this model couldnot predict the spectra of atoms or ionswith more than one electron.

You must have noticed that none of theatomic models that have been studied sofar, have mentioned about neutrons. This isbecause neutron was discovered later in1932. Until Rutherford's and Bohr's timethe neutron had not been discovered.Neutron was discovered nearly twodecades later. Except hydrogen atom, theatoms of all other elements containneutrons in their nuclei.

We studied that the mass of neutronsand proton is almost equal and is about 1836times heavier than the mass of the electron.So the mass of the entire atom is due toprotons and neutrons. Later, it wasdiscovered that most of the mass wasconcentrated inside the nucleus andtherefore the neutrons are also presentinside the nucleus.

The model of the atom, as we know ittoday, is the contribution of manyscientists. Let us observe the history of theAtom in the given chart.

Fig-6: Energy levels of an atom

SCERT TELANGANA


History of Atom

442 B.C.

1803 A.D.

1886 A.D.

1898 A.D.

1909 A.D.

1913 A.D.

1913 A.D.

1931 A.D.

Democritus

Metter is composedwith Indivisible

elements

First Atomic Theory

Gold stain

Discovery of CannelRays

J.J.Thomson

Discovery of electron incathode rays experiment

EarnestRutherford

Discovery of Nucleus

Neils Bohr

Electrons in Stationaryorbitals – Introductionof energy levels

HenryMosley

Atomic Number

James Chadwick

Discovery of Neutron

John Dalton

SCERT TELANGANA


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.

Table - 1

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.

Fig-7: Arrangement of electrons for the first eighteen elements

Distribution of electrons indifferent orbits (Shells)

According to atomic models, electronsmove around the nucleus of atom in variousshells. Electrons in different shells havedifferent energies. Each shell is representedby a number 'n' which is known as a shellnumber or energy level index. The shellclosest to the nucleus (and has the lowestenergy) is called the K-shell (n = 1), theshell farther away (and has higher energythan K-shell) is called the L-shell (n = 2),etc. Similarly other shells are denoted as(M, N, .....)

• 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 ofelectrons in different energy levels or shellsof an atom.

Bohr and Bury proposed the followingrules for electron distribution.

Shell number (n) The maximum numberof electrons in a shell(2n2)

K-shell (1) 2 (1)2 = 2

L-shell (2) 2 (2)2 = 8

M-shell (3) 2 (3)2 =18

N-shell (4) 2 (4)2 =32

SCERT TELANGANA


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.

Arrangement of electrons for thefirst eighteen 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, the

number of electrons present in outermostorbit of an atom is called its valency.

Valency of an atom explains thecombining capacity of the element withother elements in the above example thevalency of carbon atom is 4.

Let us consider some more examples,if you consider the atoms like hydrogen/lithium/sodium, it contains one electron inits outermost shell. Therefore valency ofhydrogen, lithium or sodium is one.Canyou tell, valencies of magnesium andaluminium? They are two and three,respectively, because magnesium has twoelectrons in its outermost shell andaluminium has three electrons in itsoutermost shell.

If the number of electrons in the outershell of an atom is nearly close to its fullcapacity, then valency is determined in adifferent way.

For example, the fluorine atomcontains 7 electrons in outermost shell, andits valency could be 7. But it is easier forfluorine to gain one electron than to loseseven electrons for obtaining 8 electronsin its outer most shell. Hence, its valencyis determined by subtracting sevenelectrons from eight and which gives avalency 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?

Observe the following Table . Valency ofthe first eighteen elements is given in thelast column of Table 2.

Electron

Proton

Neutron

SCERT TELANGANA


K L M NHe 2Ne 2 8Ar 2 8 8

Think and discuss

• Phosphorus and sulphur showmultiple valency. See table 2. Whysome elements show multipleValency? Discuss with your friendsand teachers.

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. Thesethree gases are very stable and have lowreactivity. Scientists, studying the

distribution of electrons in different shellsconcluded that the special arrangementsof electrons in He, Ne and Ar make themstable or un reactive with other elements.They do not react with other elements toform compounds. In other words we cansay that these gases are chemically inactiveand also known as inert or noble gases.

Atoms of all the noble gases exceptthose of helium have 8 electrons' in theirouter most shell. Thus an atom with eightelectrons or an octet in their outer most

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 - - - 0Lithium Li 3 3 4 3 2 1 - - 1

Berilium 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 - - 4Nitrogen N 7 7 7 7 2 5 - - 3

Oxygen 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 - - 0Sodium Na 11 11 12 11 2 8 1 - 1

Magnesium 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 - 4Phosphorus* P 15 15 16 15 2 8 5 - 5,3

Sulphur* 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

* Multiple valency elements

SCERT TELANGANA


Atomic mass number• Should we consider the number of

neutrons as a characteristic of an atom?The mass of an atom which is a

characteristic of an atom depends on thenumber of neutrons and protons its nucleuscontains. Number of protons in a nucleusis denoted by Z (atomic number) andnumber of neutrons of a nucleus is denotedby N.

The number of nucleons, is the totalnumber of protons and neutrons in anatom. It is called the atomic mass numberand is denoted by the 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.

shell, 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 outermostshell. From the above discussion, we canconclude that when an element reacts to formcompounds their atoms must combine insuch way that they can attain the stableelectron distribution of noble gases.

An atom can achieve an octet by twoways. 1) by transfer of electrons 2) by sharingof electrons. Both the processes result in theformation of bonds between atoms.

Let us go back to the question of whydo atoms of different elements aredifferent? How can you distinguishbetween the atoms of one element fromthe atoms of other element? An elementcan be recognised by certain characteristicsof its atoms.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 ofprotons in their atomic 'nuclei'. This valueis called the element's atomic number andis denoted by the letter Z.Atomic number is the number ofprotons in the nucleus of an atom

Writing symbols of atomsIn standard notation to represent an

atom, the atomic number, mass number andsymbol of the element are written as:

atomic mass number

atomic number

number of protons +no.of neutrons

number of protons

199 F↓

↓

Read as F - 9-19

SCERT TELANGANA


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 the fluorine has 19nucleons (protons + neutrons).

Therefore the number of neutronspresent in fluorine is equal to 19 - 9 = 10neutrons. (N = A - Z).Isotopes

It must be clear to you that everyelement 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 nucleonin its nucleus, an atom of deuterium hastwo neucleons in its nucleus, and tritiumhas three. Since atoms of hydrogen,deuterium and tritium have only oneproton in their nuclei, they only have oneelectron. But number of neutrons presentin hydrogen atom is not same in all cases.

The atoms of the same element whichhave the same number of protons but have

different 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 themass 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 istaken as an average mass of all thenaturally occurring atoms of the sampleelement.

126C

136C

146C

Did You Know?Two elements share the record for

the highest number of knownisotopes. Both xenon and cesiumhave 36 isotopes.

Protium (11H) Deuterium (1

2H) Tritium (13H)

Hydrogen

SCERT TELANGANA


The average atomic mass of chlorineatom, on the basis of above data, will be

Applications of isotopesSome isotopes are used for solving

chemical and medical mysteries. Isotopes

are also commonly used in the laboratoryto investigate the steps of a chemicalreaction.

• The isotope of uranium is used as afuel in nuclear reactors.

• The isotope of iodine is used in thetreatment of goitre (thyroid).

• The isotope of cobalt is used in thetreatment of cancer.

Key words

Atom, sub-atomic particle, electron, proton, neutron, nucleus, atomic number(Z), Atomic mass number (A), valency, isotopes

• An atom is the smallest particle of an element that retains the identity of the element.

• John Dalton's atomic theory described elements in terms of atoms, which he believedto be small, indivisible particles that make up all matter. He stated that all the atomsof the same element are identical in mass and size, but atoms of different elementsare different.

• The three sub-atomic particles of an atom are: (i) electron, (ii) proton (iii) neutron.• Electron is a negatively charged particle of the atom.• Proton, a positively charged particle is the part of atomic nucleus.• Neutron, an uncharged particle is the 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, which are

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 the atomicnucleus.

• 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

(35 x + 37 x )75100

25100

= ( + ) = = 35.5u1054

374

1424

SCERT TELANGANA


• 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

Reflections on concepts1. What are the three subatomic particles? (AS1)2. What were the three major observations

Rutherford made in the gold foil experiment?(AS1)

3. Give the main postulates of Bohr’s model of an atom. (AS1)4. State the valencies of magnesium and sodium (AS1)

Application of concepts1. Compare the sub-atomic particles electron, proton and neutron.(AS1)2. What are the limitations of J.J. Thomson’s model of the atom?(AS1)3. Define valency by taking examples of nitrogen and boron.(AS1)4. What is the main difference among the isotopes of the same element?(AS1)5. Fill in the missing information in the following table.(AS1)

Name Symbol Atomic Mass Number of Number ofNumber (Z) Number (A) Neutrons Electrons

Oxygen 168O 8 16 8 8

7 734

16SBeryllium 9

12 24

12 25


SCERT TELANGANA


Higher Order Thinking Questions1. Cl - has completely filled K, L & M shells. Explain it based on Bhor-Bury theory. (AS1)2. Explain the efforts made by scientists to explain the structure of atom by developing

various atomic models?(AS6)

1. Electron was invented by [ ]a) Thomson b) Chadwick c) Goldstein d) Stoney

2. Proton was invented by [ ]b) Thomson b) Chadwick c) Goldstein d) Stoney

3. Neutron was invented by [ ]a) Thomson b) Chadwick c) Goldstein d) Stoney

4. α - particles are made up of the following primary particles [ ]a) 2 protons and 2 neutrons b) 2 Protons and 2 Electronsc) 2 Neutrons and 2 Positrons d) 2 Protons and 2 Neutrinos

5. Which model of atom is known as Planetary model [ ]a) Thomson’s model b) Rutherford’s modelc) Bohr’s model d) Modern atomic model

6. Valency of Aluminium is [ ]a) 1 b) 2 c) 3 d) 4

7. The gas which is stable without octet configuration is [ ]a) Neon b) Argon c) Radon d) Helium

8. The sum of the number of protons and neutrons in an atom is knownas its [ ]a) Mass number b) Atomic number c) Valency d) Ion number

9. Deuterium and Tritium are the Isotopes of — [ ]a) Nitrogen b) Oxygen c) Hydrogen d) Helium

10. The electronic configuration of Sodium is [ ]a) 2,8 b) 8,2,1 c) 2,1,8 d) 2,8,1

Suggested Projects1. Write a report on the history of unveiling the structure of atom from John Dalton to

Neils Bohr.


SCERT TELANGANA


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. 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 theday-to-day activities.

Similarly you might have alsoobserved that people employ machinesin their homes to carry out different taskslike blowing of air by fan, pumping of waterby electric motor, heating of water byelectric heater, etc.

Washing machines and vacuumcleaners are used for cleaning of clothesand cleaning the house respectively.• How are these works being done?

• What do you need to do these works?Both human beings and machines need

energy to do work. Generally this energyis derived by human beings from food theyeat and for machines through the fuel orelectricity 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 identify the forceemployed for doing the work and the objecton which the work was done. Discuss withyour friends about possibility of energytransfer during work.

Chapter

10 WORK AND ENERGY

SCERT TELANGANA

164 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 waywe use it in 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 ofa 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 labour 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.Situation - 1

Fig - 1 A man lifts cement bags from the

ground and keeps them one by one in alorry.

SCERT TELANGANA


Situation - 2

Fig - 2

A girl is pulling a toy car on the groundand the trolley moves a distance.

Situation - 3

A boy is trying to push a huge rock lyingin a play ground.

Fig - 3

Situation - 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-1 given in the page 169in your note book.

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 table-1.

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 insituation -1 the position of the bag ischanged from ground to the height of thelorry, and similarly in situation - 2 thetoycar moved through a distance bychanging 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

SCERT TELANGANA

166 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 Let us consider the followingexample:

Table - 1

SituationWhether work

has beendone or not ?

(Yes/No)

Who is doingthe work?(Name the

force)

Object on whichwork has

been done?

Reason to saythat work

has been done

The changes younotice in objectson which workhas been done.

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

lorry2

3

4

situation -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 insituation - 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 situations 1 to 4 aredoing work but according to science onlythe forces applied by persons mentionedin situations 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.

Fig - 5

SCERT TELANGANA


Suppose that a constant force (F) actson an object and object is moved through adistance (s) along the direction of the force(F) as shown in fig.- 5.

In science, we define work to be equalto the product of the force (F) and thedisplacement (s) moved along the directionof the force.

Work done = Force x Displacement.

W = FS

This formula for work is used inonly translatory motion of the object.

Work has only magnitude but nodirection. So work is a scalar.

We measure force (F) in newtons(N) and distance (S) in meters (m). Inequation W=FS, if F=1 and S=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 1 newton(1N) displaces it by 1m along the line ofaction of the force.

Look at the equation W = FS

• What would be the work done whenthe force on the object is zero?

• What would be the work done whenthe displacement 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 = FS

= 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=9.8 m/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.

SCERT TELANGANA

168 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 = FS = 4.9 x 1.5

=7.35 JIn the situation mentioned in fig.- 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 thespeed of the ballwhile it moves up?

• What is the speed atits maximum height?

Positions of ball at variousinstants while moving up

• What happens to the speed of the ballduring 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 = – FSIf 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

Direction of motion

Fig - 6:

SCERT TELANGANA


The force and displacement are inopposite directions. Hence work done onthe box is negative.

That is, W = – FS

= – 100 x 4

= – 400 J

Example 4

A ball of mass 0.5 kg thrown upwardsreaches 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 = 5N

Displacement 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 daily life in various occasions like ‘Heis more energetic’ 'I am so tired and lostmy energy’, ‘Today I am feeling moreenergetic than yesterday’ 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 it restson the surface of the plate but is able to dosome work when it is raised to a height.

Similarly in case -2 you may noticethat the toy is at rest before winding thekey but the same toy moved, when thekey attached to it is wound up. Here thework done is due to energy.

Children may not be able to lift 25 kgrice bag but elders can do that.

• What could be the reason?

You may observe many suchsituations where the capacity to do workchanges from person to person.

SCERT TELANGANA

170 Work and Energy

Fig - 10

Now compress the spring with yourpalm 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.

Think and discuss

• What would happen if nature doesnot allow the transfer of energy?Disscuss with few examples.

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 the

object doing the work and the objecton which work has been done?

• Can any force do work without energytransfer?In science, we consider work has been

done 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.

Activity-2

Understanding the increase anddecrease in energy of an object

Take a hard spring and keep it on thetable as shown in the fig.- 10.

SCERT TELANGANA


Fig – 11 (a) Fig - 11 (b)

Consider a metal ball and a hollowplastic block which are kept on a table sideby side as shown in the fig.- 11 (a). Nowsuppose that the ball is separated from theblock and brought to one end of the tableas shown in the fig.- 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 fig.- 11 (b).Thus a moving ball is more energetic thanthe ball at rest because the moving ball isable to do the work on plastic block to pushit forward, whereas the same ball cannotdo any work when it is at rest. In otherwords, a body possess more energy whenit is moving than when it is at rest.

Kinetic energyActivity-3

Understanding the energy of moving objects

Repeat this activity by pushing themetal ball with more force so as toincrease its speed and observe the changein position of the plastic block on the table.You may notice that the increase in thespeed of the ball increases its capacity ofdoing work on the block.

Thus we can conclude that a movingobject can do work. An object movingfaster can do more work than an identicalobject moving 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 example:• When a fast moving cricket ball hits

the wickets, it makes the wickets totumble but if the swinging bat in thehands of a batsman hits the ball itreaches the boundary.

SCERT TELANGANA

172 Work and Energy

• A fast moving bullet pierces the target.• Blowing wind moves the blades of the

wind mill.Objects like a falling coconut, a

speeding car, a rolling stone, a flyingaircraft, flowing water and running athleteetc, also possess kinetic energy.• How can we find out as to how much

energy is possessed by a moving body?

Mathematical 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 fig.- 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’,final velocity ‘v’, acceleration ‘a’ anddisplacement ‘s’ is given by

v 2_ u2 = 2 a s or s =

2 2(v - u ) 2a

——(2)

We know by Newton’s second law ofmotion

Fnet= ma ————— (3) From equations (1), (2) and (3)

W = ma x2 2(v - u ) 2a

W = ½ m (v 2_ u2)

This is called work - energy theorem.As we have assumed that object is at

rest, 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

SCERT TELANGANA


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 from10 m/s to 20 m/s or when it goesfrom 20 m/s to 30 m/s.?

Example 5

Find the kinetic energy of a ball of 250 gmass, moving at a velocity of 40 cm /s.

Solution

Mass of the ball, m = 250g = 0.25 kg

Speed of the ball, v = 40 cm / s =0.4 m /s

Kinetic Energy,

K.E. = ½ (0.25) x (0.4)2 = 0.02 J

Example 6

The mass of a cyclist together with thebicycle is 90 kg. Calculate the work doneby cyclist if the speed increases from6km /h to 12 km / h.

Solution

Mass 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-4Understanding potential energy

Take a bamboo stick and make a bow asshown in the fig.- 13 (a). Place an arrowmade of a light stick on it with one end

Fig-13a Fig-13bSCERT TELANGANA

174 Work and Energy

supported by the string of the bow as shownin fig.-13 (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 fig.-13 (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,

fig.-13(a), when you release the arrow itgets separated from the bow and falls downon ground. But in second instance, fig.-13(b), you will notice the arrow flies withgreat 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 thearrow in the first case?

• Can we increase the potentialenergy of 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-5

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?

SCERT TELANGANA


Activity-6

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 sometimesdue to change 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

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 the object against gravityacting on it. The energy of such an object isknown as gravitational potential energy.

Fig-14 The gravitational potential energy of

an 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.

SCERT TELANGANA

176 Work and Energy

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 of these energiesgives the total mechanical energy of theaeroplane in flight.Conservation of energy

We find in nature a number of instancesof 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-7Listing 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.

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

SCERT TELANGANA


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 animal sources which in turntake plant 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.

SCERT TELANGANA

178 Work and Energy

Activity - 8Conservation 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 fig.- 15.

Fig-15Now pull the object or bob of the

pendulum to one side to the position A1

(Fig-15) and release it.What do you notice?

• You may notice that, the bob swingstowards opposite side and reaches thepoint A2. It repeats this motion over andover again.

• The potential energy of the bob isminimum 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 A2 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.

Thus energy can neither be created nordestroyed. It can only be changed from oneform 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 bodypossess both potential energy and kineticenergy during its fall to the ground.

Does the conservation of energy applyfor the system of free-fall body? How?

Let us find

Activity-9

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

SCERT TELANGANA


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 of

system of freely falling body?• Is the mechanical energy conserved in

the system?

Think and discuss

• Someone wants 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 in theneighbours house.

• Do all of us do work at the same rate?• Is the energy spent by the force doing

work the same every time?• Do machines consume or transfer

energy at the same rate every timewhile doing a particular work?

Let us consider the following example.• Raheem wanted to do some repairs in

the first floor of his building. Hebrought 100 bricks as advised by themason. A labourer was asked to shiftthese bricks from ground floor to firstfloor. The labourer completed thework of shifting bricks to first floor inone hour and asked Rs 150/- for thework.Next day the mason asked Raheem to

bring 100 bricks more to complete therepair work. Raheem brought another 100bricks and assigned the work of shiftingbricks to some another labourer. He

SCERT TELANGANA

180 Work and Energy

completed the work and demanded Rs 300/- for completing the assigned work.Raheem said that he paid Rs 150/- to thelabourer the day before. But the labourerargued that he worked more hours hencehe was eligible 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 as in the first caseof above mentioned example, may docertain work in relatively less timecompared to another person, similarly apowerful machine can do a work assignedto it in relatively short time compared withanother machine.

We talk about the power of machineslike motorbikes, motorcar, water pumpingmotors etc., the speed at which thesemachines do work is the basis for theirclassification. Power is a measure of therate of doing work, that is how fast or howslow 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 = WorkTime

P = wt

The unit of power is ‘watt’ and denotedby symbol ‘W’

1 watt is the power of an agent, whichdoes work at the rate of 1 joule per second.

We express larger rate of energytransfer 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 = 300s

Power delivered, P = wt

= 420300 = 1.4 W

Example 10A woman does 250 J of work in 10

seconds and a boy does 100 J of work in 4seconds. Who delivers more power?Solution

Power, P = wt

Power delivered by woman=25010 =25W

Power delivered by boy = 100

4 = 25W

SCERT TELANGANA


Both woman and boy deliver samepower. That is, rate of doing the work bywoman and boy are equal.

SOURCES OF ENERGYIn activity 8, we have learnt that

energy can be transformed from one formto another. Energy is of different forms andone form can be converted to another form.Example, observe fall of a coconut from acoconut tree. Potential energy of coconutgets converted into Kinetic energy. In thiscase the source for energy conversion isgravitational force. From this example wealso understand that the energytransformations need a source responsiblefor their transformations.

A source of energy is one which canprovide adequate amount of energy in aconvenient form over a long period of time.

• What is a good source of energy?

Good source of energy may beunderstood based on the following features:

That does a large amount of workper unit volume or unit mass ofthe source.

That is cheap and easily available.

That which is easier to use, storeand transport.

That is most economical and thatcauses no or minimum pollution.

Fuels:• What is the source of energy to

cook food?

• What is the source of energy to rideyour vehicle?

• What is the source of energy torun a thermal power plant?

• What do you call these sources ofenergy?

We use LPG, Kerosene, Wood,Petroleum, coal, etc..as sources of energyfor doing the above mentioned works. Thesesources of energy are known as fuels.

• Where do we get most of thesefuels?

We know that most of these fuelscan be obtained from earth crust. Thesefuels are known as fossil fuels.

• How are these fossil fuelsformed?

Plants, animals and other livingorganism after their death got buried underthe soil due to floods and other naturaldisaster for long period of the time. Duringcourse of time more soil, mud and rocksediments deposited over them. Due toabsence of oxygen, high pressure, heat andaction of bacteria, this organic matterconvert into fossil fuels.

• What could be the main source ofenergy which helps to form fossilfuels?

We know that plants and animalsgrow using solar energy. This solar energywas trapped in this organic matter throughnatural processes millions of years ago.

SCERT TELANGANA

182 Work and Energy

solar energy i.e. nearly 47% is onlyreaching the earth and the remaining isreflected back into atmosphere. India isreceiving energy more than 5000 trillionKWH during a year. Under clear cloudlesssky conditions, the daily average of solarenergy varies from 4 to 7 KWH/m2 in ourcountry.

Scientists have developed devices touse solar energy as a source of energy forcooking, electrical needs, etc. The devicesare mainly solar cooker, solar water heater,solar cells, etc.

Solar cells

A solar cell converts energy fromsunlight into electrical energy. A solar cellis made up of sandwiching a Silicon –boron layer and Silicon-arsenic layer. Thiscell can trap a very little amount ofelectricity. Hence a large number of cellsare joined together in series to form a solarpanel.

• Discuss various uses of solarpanel.

• What are the advantages anddisadvantages of solar energy?

• Do you think that these fuels getexhausted, if we use themcontinuously for a long time.

• What are other alternatives if thesefuels are exhausted?

These fossil fuels cannot bereproduced by any artificial methods. Theycannot be quickly replaced if onceexhausted. So these are called non-renewable sources of energy.

Think and Discuss:

Is the firewood obtained by cuttingof trees renewable or non renewablesource of energy? Why?

Renewable sources of energy

• Is there any alternate source ofenergy which cannot be exhausted?

• What could be the methods to getenergy from these sources?

We know that Sun is the main sourceof energy. We already discussed that thefossil fuels are also formed due to the solarenergy trapped in fossils.

1. Solar Energy:

Sun is the main source of energy.The energy received from Sun is known assolar energy. The sun has been radiating anenormous amount of energy at the presentrate nearly 5 billion years and will continueradiating energy at that rate for about 5billion years more. Only a little amount of

Fig. 16

SCERT TELANGANA


2. Bio mass energyBio mass is organic material that

comes from plants and animals and it is arenewable source of energy.

Bio mass contains stored energyfrom the Sun. Plants absorbe the Sun'senergy in a process called photosynthesis.When bio mass is burned, the chemicalenergy in bio mass is released as heat. Biomass can be burned directly. It can also beconverted into Coal, Petroleum, Cow dungCakes, Bio gass etc. Fuels like Coal,Petroleum are called as fossil fuels.

2(a). Bio gas

‘Biogas’ is a renewable source ofenergy. It is produced mainly from cowdung, sewage, crop residues, vegetablewastes etc. It contains about 65% ofmethane and widely used as fuel forcooking. The slurry obtained as a by-product, left in the biogas plant after thebiogas is used up can be used as a manure,which is rich in nitrogen and phosphorous.

3. Energy from SeaThe energy from the sea can be

obtained mainly in two forms. (a) Tidalenergy, and (b) Ocean Thermal Energy.

3(a). Tidal EnergyDuring the high tide, the water from

sea can be made to sent into a reservoir ofthe barrage and turns turbines. The turbinesthen turn the generators to produceelectricity.

• Discuss the advantages anddisadvantages of tidal energy.

3(b). Ocean Thermal Energy (OTE)

Heat from the sun is absorbed by thewater on the surface of ocean, but at deeperlevels of ocean, the temperature is veryless. So, there is temperature differencebetween the water ‘at the surface of ocean’and ‘at deeper levels’. This difference intemperature is called Ocean ThermalEnergy (OTE). The OTE can be convertedinto electrical energy by using OceanThermal Energy Conversion (OTEC)plants.

4. Geo Thermal Energy

The interior of the earth is very hot.Water seeping down deep into the earth isturned into steam and this can be suppliedto homes for heating purposes and forgenerating electricity. Electricity produced

Fig. 17

Fig. - 18

SCERT TELANGANA

184 Work and Energy

in this way is very less cost than to othersources, and also pollution free.

5. Wind energy

Moving air is called wind. Wind haskinetic energy. The energy of wind isharnessed by using wind mills.

Fig. 20

A wind mill consists a big fan likeblades fixed to tall poles at very higherplaces, such that they can rotate freelywhen wind blows. A dynamo is attached tothe shaft of blades, so the mechanicalenergy produced by rotating wind mill willbe converted into electrical energy by thedynamo.

The electrical energy produced bythese wind mills is pollution free.

6. Atomic EnergyAtomic energy is also known as

nuclear energy. The physical reaction whichinvolves changes in nucleus of an atom iscalled a nuclear reaction. The energyreleased during a nuclear reaction is callednuclear energy. This nuclear energy can beobtained by two types of nuclear reactions.1. Nuclear fission and 2. Nuclear fusion.

I. Nuclear fissionThe process in which heavy nucleus

of a radioactive atom (Ex: Uranium) splitsup into smaller nuclei when bombardedwith low energy neutrons is called nuclearfission. A small change in the nucleus ofheavy atoms releases a tremendous amountof energy.

235U92+ 1n0 139Ba56+ 94Kr36 + 31n0 + Energy

Fig. 21This energy is in the form of heat.

This heat energy is used to produceelectricity in nuclear power plants. In India,the nuclear power plants are situated atTarapur (Maharashtra), Rana pratap sagar(Rajasthan), Kalpakam (Tamilnadu), Narora(Uttar Pradesh), Kaprapur (Gujarat), andKaiga (Karnataka).

Fig. 19

SCERT TELANGANA


What we have learnt

Key words

II. Nuclear fusionThe process in which two nuclei of light elements combine to form a heavy nucleus

is called nuclear fusion. 2H1 +

2H1 3He2 +

1n0 + Energy

The energy released in this form can not be controlled. So this is not used as sourcefor producing electricity. These nuclear fusion reactions which are taking place in Sun's core arethe main source of Sun's energy.

Work, Energy, Transfer of energy, Sources of energy, Conservation of energy,Kinetic energy, Potential energy, Mechanical energy, Gravitational potential,Renewable sourcess of energy.

• 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.

SCERT TELANGANA

186 Work and Energy

• 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 the law of conservation of energy.

• Power is defined as the rate of doing work or rate of transfer of energy.

Reflections on concepts1. What is work according to science and

write its units. (AS1)2. Give few examples where displacement of an object is in the direction opposite to

the force acting on the object. (AS1)3. Write few daily life examples in which you observe conservation of energy. (AS6)4. Give some examples for renewable sourcess of energy (AS1)

Application of concepts1. 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)2. A 10 kg ball is dropped from a height of 10m. Find (a) the initial potential energy of

the ball , (b) the kinetic energy just before it reaches the ground, and (c) the speedjust before it reaches the ground. (AS1) (Ans: 980J, 980J, 14m/s)

3. 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) (Ans: 196N-m)

4. Find the mass of a body which has 5J of kinetic energy while moving at a speed of2m/s. (AS1) (Ans: 2.5 kg)

5. A cycle together with its rider weighs 100kg. How much work is needed to set itmoving at 3 m/s. (AS1) (Ans: 450J)

6. Which of the renewable sourcess of energy would you think soutable to producedin you native place. Why? (AS7)

Higher Order Thinking Questions1. When you lift a box from the floor and put it on an almirah the potential energy of

the box increases but there is no change in its kinetic energy. Is it violation ofconservation of energy? Explain. (AS7)

2. When an apple falls from a tree what happens to its gravitational potential energyjust as it reaches the ground? After it strikes the ground? (AS7)


SCERT TELANGANA


1. S.I. unit of work [ ]a) N-m b) Kg-m c) N/m d) N-m2

2. The energy possessed by a body by virtue of its motion is called as [ ]a) Potential energy b) Kinetic energyc) Attractive energy d) Gravitational energy

3. A person is climbing a ladder with a suitcase on his head. Then the work done by thatperson on that suitcase is [ ]a) Positive b) Negative c) Zero d) Can not be defined

4. If you have lifted a suitcase and kept it on a table., then the work done by you willdepend on [ ]a) The path of the motion of the suitcase b) The time taken by you to do the workc) Weight of the suitcase d) Your weight.


1. Conduct an experiment to prove the conservation of mechanical energy and write areport on it.

2. Conduct an experiment to calculate the total energy of a freely falling body at differentheights.

Suggested Projects

1. How will the increasing energy needs and conservation of energy influenceinternational peace, cooperation and security? Collect information on this and writea report.

2. Collect information about different sources of energy and write a report on theadvantages and disadvantages in harnessing energy from these sources.

3. Make different models showing the harnessing energy from different energy sources.


SCERT TELANGANA

188 Is Mater Pure ?

of heat energy from our body to theobject?

When you touch the metal or woodenpiece, you feel that they are cold. Thismeans that heat energy is being transferredfrom your finger to the pieces. When youremove your finger, you don’t get a feelingof ‘coldness’. This means that when heatenergy flows out of your body you get thefeeling of ‘coldness’ and when heat energyenters your body you get a feeling of‘hotness’. You can test this by bringing yourfinger near the flame of a matchstick!

So, if you feel that the metal pieceis ‘colder’ than the wooden piece, it mustmean that more heat energy flows out ofyour body when you touch the metalpiece as compared to the wooden piece.In other words, the ‘degree of coldness’of the metal piece is greater than that ofthe wooden piece.

The conventional definition oftemperature is “the degree of hotness orcoldness”.

We say that the metal piece is at alower ‘temperature’ as compared to the

Chapter

11 HEAT

Recall the experiments you did inClass 7 with the glass tumblers containingof cold water, lukewarm water and hotwater. We understood that ‘hot’ and ‘cold’are relative terms and that heat was a formof energy. We use the terms “Temperatureand Heat” to describe these observations.These words, technically, have specialmeanings. In order to understand theirmeanings let us do some activities.

Activity-1Take a piece of wood and a piece of

metal and keep them in a fridge or icebox. After 15 minutes, take them out andask your friend to touch them.

• Which is colder? Why?When we keep materials in a fridge,

they become cold i.e., they lose heatenergy. The iron and wooden pieces werekept in the fridge for the same period oftime but we feel that the metal piece iscolder than the wooden piece.

• What could be the reason for thisdifference in coldness?

• Does it have any relation to the transfer

SCERT TELANGANA


wooden piece, when they are taken out ofthe fridge.

• Why does transfer of heat energy takeplace between objects ?

• Does transfer of heat take place in allsituations?

• What are the conditions for transferof heat energy?

Let us find out

Thermal equilibrium-heat andtemperature

When two bodies are placed in thermalcontact, heat energy will be transferred fromthe ‘hotter’ body to the ‘colder’ body. Thistransfer of heat energy continues till bothbodies attain the same degree of hotness(or) coldness. At this stage, we say that thebodies have achieved ‘thermal equilibrium’.Thus, the state of thermal equilibriumdenotes a state of a body where it neitherreceives nor gives out heat energy.

If you are not feeling either hot orcold in your surroundings, then your bodyis said to be in thermal equilibrium withthe surrounding atmosphere. Similarly, thefurniture in the room is in thermalequilibrium with air in the room. So we cansay that the furniture and the air in the roomare at the same temperature.

Heat

• What is temperature?

• How can you differentiate it from heat?

Let us find out

Activity-2Take two glass tumblers and fill one

of them with hot water and another withcold water. Now take a laboratorythermometer, observe the mercury levelin it and note it in your book. Keep it inhot water. Observe changes in the mercurylevel. Note the reading.

• What change did you notice in thereading of the thermometer? (mercurylevel)?

• Did the mercury level increase ordecrease?

Now place the thermometer in coldwater and observe changes in the mercurylevel. Did the level decrease or increase?

We know that bodies in contactachieve thermal equilibrium due totransfer of heat energy. When you keepthe thermometer in hot water you observethat there is a rise in mercury level. Thishappens because heat got transferred fromthe hotter body (hot water) to the colderbody (mercury in thermometer). Similarlyin the second case you will observe thatthe mercury level comes down from itsinitial level because of the transfer of heatfrom mercury (hotter body) to water(colder body). Thus we define heat asfollows:

“Heat is a form of energy in transit,that flows from a body at highertemperature to a body at lowertemperature.”

SCERT TELANGANA

190 Is Mater Pure ?

The steadiness of the mercury columnof the thermometer indicates that flow of heatbetween the thermometer liquid (mercury)and water, has stopped. Thermal equilibriumhas been attained between the water andthermometer liquid (mercury). Thethermometer reading at thermal equilibriumgives the “temperature”. Thus ‘temperature’is a measure of thermal equilibrium.

If two different systems, A and B inthermal contact, are in thermal equilibriumindividually with another system C(thermal contact with A and B), will thesystems A and B be in thermal equilibriumwith each other?

We know that if A is in thermalequilibrium with C, then they both havethe same temperature. Similarly, B and Chave the same temperature. Thus A and Bwill have the same temperature and wouldtherefore be in thermal equilibrium with eachother. (A,B and C are in thermal contact).

The SI unit of heat is joule (J) andCGS unit is calorie (cal).The amount ofheat required to raise the temperature of1gram of water by 10C is called calorie.

1cal = 4.186 joules

The SI unit of temperature is kelvin(K). It can also be expressed as degreecelsius (0C).

00C = 273K

• How would you convert degreecelsius to kelvin?

Temperature in kelvin = 273 +temperature in degree celsius

K = 273 + T

Add 273 to the value of temperaturein degree celsius to get the temperatureon the kelvin scale.

Note: Temperature measured onKelvin scale is called absolute temperature.

Temperature and Kinetic energy

Activity-3Take two bowls one with hot water

and second with cold water. Gentlysprinkle grains of food colour on thesurface of the water in both bowls. .Observethe motion of the small grains of foodcolour.

• How do they move?• Why do they move randomly?• Why do the grains in hot water move

more rapidly than the grains in coldwater?

You will notice that the grains offood colour jiggle (move randomly). Thishappens because the molecules of waterin both bowls are in random motion. Weobserve that the jiggling of the grains offood colour in hot water is more whencompared to the jiggling in cold water.

We know that bodies possess kineticenergy when they are in motion.

As the speed of motion of particles(grain of food colour) in the bowls of wateris different, we can say that they have

SCERT TELANGANA


different kinetic energies.Thus weconclude that the average kinetic energyof molecules / particles of a hotter bodyis greater than that of a colder body. Sowe can say that the temperature of a bodyis an indicator of the average kinetic energyof molecules of that body.

“The average kinetic energy of themolecules is directly proportional to theabsolute temperature” (KE T)∝

Activity-4Take water in a container and heat it

to 60°C. Take a cylindrical transparentglass jar and fill half of it with this hotwater. Very gently pour coconut oil overthe surface of the water. (Take care thatthe water and oil do not mix). Put a lidwith two holes on the top of the glass jar.Take two thermometers and insert themthrough the holes of the lid in such a waythat the bulb of one thermometer lies onlyinside the water and other lies only insidethe coconut oil as shown in figure 1.

Now observethe readings of thetwo thermometers.The reading of thethermometer keptin water decreases,while, at the sametime, the readingof the thermo-meter kept in oilincreases.

• Why does this happen?Because the average kinetic energy

of the molecules of oil increases, whilethe average kinetic energy of the moleculesof water decreases. In other words, thetemperature of oil increases while thetemperature of water decreases.

• Can you say that the water losesenergy?

From the above discussion it is clearthat, water loses energy while oil gains energy;because of the temperature differencebetween the water and oil. Thus some heatenergy flows from water to oil. This means,the kinetic energy of the molecules of thewater decreases while the kinetic energy ofthe molecules of oil increases.

• Can you now differentiate betweenheat and temperature based on thediscussion we made of the aboveactivities?

With activities 2, 3 and 4 we candifferentiate heat and temperature as follows:

Heat is the energy that flows from ahotter body to a colder body. Temperatureis a quantity that denotes which body ishotter and which is colder. So temperaturedetermines direction of heat (energy) flow,whereas heat is the energy that flows.

Specific Heat

Activity 5Take a large jar with water and heat

it up to 80oC.Take two identical boilingfig-1 Hot water

Coconutoil

Thermometer

LidSCERT TELANGANA

192 Is Mater Pure ?

test tubes with single-holed corks. One ofthem is filled with 50g of water and theother with 50g of oil, both at roomtemperature. Insert two thermometersthrough holes of the corks, one each intotwo test tubes. Now clamp them to a retortstand and place them in a jar of hot wateras shown in figure 2.

Observe the readings ofthermometers every three minutes .Notethe readings in your notebook.

• In which test tube does the temperaturerise quickly?

• Are the amounts of heat given to thewater and oil same? How can youassume this?

We believe that the same amount ofheat is supplied to water and oil becausethey are kept in the jar of hot water for thesame interval of time.

We observe that the rate of rise intemperature of the oil is higher than thatof the rise in temperature of the water.• Why does this happen?

We conclude that the rate of rise intemperature depends on the nature of thesubstance.

Activity 6Take two beakers of equal volume

and take 250 grams of water in one beakerand 1 kg of water in another beaker. Notedown their initial temperatures using athermometer (initial temperatures shouldbe the same). Now heat both beakers tillthe temperature of water in the two beakersrises to 60 0C. Note down the time requiredto raise the temperature of water to 60 0Cin each beaker.

• Which beaker needed more time?You will notice that you need more

time to raise the temperature of 1 kg ofwater when compared to 250 grams ofwater. That means you need to supplymore heat energy to greater quantity ofwater than lesser quantity of water for samechange in temperature.

For same change in temperature theamount of heat (Q) absorbed by a substanceis directly proportional to its mass (m)

Q ∝ m ( when ΔT is constant ) ….(1)

Now take 1 litre of water in a beakerand heat it over a constant flame. Notethe temperature changes (ΔT ) for everytwo minutes.

• What do you notice?You will notice that the change in

temperature rise with time is proportional,

fig-2

Hot waterat 80oC

WaterOil

SCERT TELANGANA


that means, for the same mass (m) of waterthe change in temperature is proportionalto amount of heat (Q) absorbed by it.

Q ∝ΔT ( when ‘m’ is constant ) ….(2)

From equation (1) and (2), we get

Q ∝ mΔT Q = msΔT

Where ‘s’ is a constant for a givensubstance. This constant is called“specific heat” of the substance.

Q s = ––––

mΔTThe specific heat of a substance is

the amount of heat required to raise thetemperature of unit mass of the substanceby one unit.

• How much heat energy is required toraise the temperature of unit mass ofsubstance by 1°C ?

CGS unit of specific heat is cal/g-oCand SI unit of it is J / kg - K

1 cal/g- oC = 1 kcal /kg-K

= 4.2 x 103J/kg-K

We have seen that the rise intemperature depends on the nature of thesubstance; hence the specific heat of asubstance depends on its nature. If thespecific heat is high, the rate of rise (orfall) in temperature is low for samequantity of heat supplied. It gives us anidea of the degree of ‘reluctance’ of asubstance to change its temperature.

• Why is the specific heat different fordifferent substances?

Let us find out.

We know that the temperature of abody is directly proportional to the averagekinetic energy of particles of the body.Themolecules of the system (body or substance)have different forms of energies such aslinear kinetic energy, rotational kineticenergy, vibrational energy and potentialenergy between molecules. The total energyof the system is called internal energy ofthe system. When we supply heat energyto the system the heat energy given to itwill be shared by the molecules among thevarious forms of energy.

This sharing will vary fromsubstance to substance. The rise intemperature is high for a substance, if themaximum share of heat energy is utilised

Substance Specific heatIn cal/g-oC In J/kg-K

Lead 0.031 130Mercury 0.033 139Brass 0.092 380Zinc 0.093 391Copper 0.095 399Iron 0.115 483Glass(flint) 0.12 504Aluminum 0.21 882Kerosene oil 0.50 2100Ice 0.50 2100Water 1 4180Sea water 0.95 3900

SCERT TELANGANA

194 Is Mater Pure ?

for increasing its linear kinetic energy. Thissharing of heat energy of the system alsovaries with temperature .That is why thespecific heat is different for differentsubstances.

If we know the specific heat of asubstance, we can determine how much heat(Q) is needed to raise the temperature of acertain mass of the substance through certaindegrees by using the equation Q = msΔT

Applications of Specific heat capacity1. The sun delivers a large amount of

energy to the Earth daily. The watersources on Earth, particularly theoceans, absorb this energy formaintaining a relatively constanttemperature. The oceans behave like“heat store houses” for the earth.They can absorb large amounts ofheat at the equator withoutappreciable rise in temperature dueto high specific heat of water..Therefore, oceans moderate thesurrounding temperature near theequator. Ocean water transports theheat away from the equator to areascloser to the north and south poles.This transported heat helps moderatethe climates in parts of the Earth thatare far from the equator.

2. Water melon brought out from therefrigerator retains its coolness for alonger time than any other fruitbecause it contains a large percentageof water. (water has greater specificheat).

3. A samosa appears to be cool outsidebut it is hot when we eat it becausethe curry inside the samosa containsingredients with higher specific heats.

Method of mixtures

Activity - 7Situation – 1: Take two beakers of thesame size and pour 200 ml of water in eachof them. Now heat the water in bothbeakers till they attain the sametemperature. If you pour this water fromthese two beakers into a larger beaker, whattemperature could you expect the mixtureto be? Measure the temperature of themixture.• What do you observe?• What could be the reason for the fact

you observed?Situation – 2: Now heat the water in onebeaker to 90oC and the other to 60oC . Mixthe water from these beakers in a largerbeaker.

• What will the temperature of themixture be?

• Measure temperature of the mixture.What did you notice?

• Can you give reasons for the change intemperature?

Situation – 3: Now take 100 ml of waterat 90oC and 200 ml of water at 60oC andmix the two.

• What is the temperature of the mixture?• What difference do you notice in the

change of temperature?Let us find out.

SCERT TELANGANA


Let the initial temperatures of thesamples of masses m1 and m2 be T1 and T2

(the higher of the two temperatures iscalled T1, the lower is called T2). Let T bethe final temperature of the mixture.

The temperature of the mixture islower than the temperature of the hottersample but higher than the temperature ofthe colder sample. This means that the hotsample has lost heat, and the cold samplehas gained heat.

The amount of heat lost by the hottersample Q1 is m1S(T1- T).

The amount of heat gained by thecooler sample Q2 is m2S(T - T2).

Since heat lost by the hotter sampleis equal to the heat gained by the coolersample (assuming no loss of heat) i.e

Q1 = Q2

which can be written asm1S(T1 - T) = m2S(T - T2)

which can be simplified to

T = 1 1 2 2

1 2

T Tm mm m

++

You will notice the temperatures ofmixtures in situation – 2 and situation – 3are not equal.

• Can you guess the reason for this?• Can we find temperature of the mixture

using a thermometer?

Principle of method of mixturesWhen two or more bodies at different

temperatures are brought into thermal

contact, then net heat lost by the hot bodiesis equal to net heat gained by the cold bodiesuntil they attain thermal equilibrium. (If heatis not lost by any other process)

Net heat loss = Net heat gain

This is known as principle of methodof mixtures.

Determination of Specific heat of asolid

Aim: To find the specific heat of givensolid.

Material required: calorimeter,thermometer, stirrer, water, steam heater,wooden box and lead shots.

Procedure: Measure the mass of thecalorimeter along with stirrer.

Mass of the calorimeter, m1 = ................

Now fill one third of the volume ofcalorimeter with water. Measure its massand its temperature.

Mass of the calorimeter plus water,m2 = ................Mass of the water, m2–m1 = ...............

Temperature of water in calorimeter,T1 = ...........

Note: Calorimeter and water are atsame temperature.

Lab Activity

SCERT TELANGANA

196 Is Mater Pure ?

Take a few lead shots and place themin hot water or steam heater. Heat themupto a temperature 100oC. Let thistemperature be T2.

Transfer the hot lead shots quicklyinto the calorimeter (with minimum lossof heat). You will notice that the mixturesettles to a certain temperature after sometime.

Measure this temperature T3 and massof the calorimeter along with contents(water and lead shots).

Mass of the calorimeter along withcontents, m3 = .........

Mass of the lead shots,m3 – m2 = ............

Since there is no loss of heat tosurroundings, we can assume that the entireheat lost by the solid (lead shots) istransferred to the calorimeter and waterto reach the final temperature.

Let the specific heats of thecalorimeter, lead shots and water be Sc, Sl

and Sw respectively. According to themethod of mixtures, we know;

Knowing the specific heats ofcalorimeter and water, we can calculate thespecific heat of the solid (lead shots).

EvaporationWhen wet clothes dry, you will notice

that water in the clothes disappears.

• Where does the water go?Similarly, when the floor of a room

is washed with water, the water on the floordisappears within minutes and the floorbecomes dry.

• Why does water on the floor disappearafter some time?

Let us see.

Activity 8Take a few drops of spirit on your

palm using a droper.

• Why does your skin become colder?Take a few drops of spirit (say 1 ml)

in two petri dishes (a shallow glass orplastic cylindrical lidded dish used in thelaboratory) separately. Keep one of thedishes containing spirit under a ceiling fanand switch on the fan . Keep another dishwith its lid closed. Observe the quantityof spirit in both dishes after 5 minutes.

Heat lost by the solid = Heat gain by the calorimeter + Heat gain by the water

(m3- m2 ) Sl(T2-T3) = m1 Sc(T3-T1)+(m2-m1) Sw(T3-T1)

[m1Sc+(m2-m1)Sw] (T3-T1) Sl = ––––––––––––––––––––

(m3-m2)(T2-T3)

SCERT TELANGANA


• What do you notice?You will notice that spirit in the dish

that is kept under the ceiling fandisappears, where as you will find somespirit left in the dish that is kept in the liddeddish.

• What could be the reason for thischange?

To answer the above questions, youneed to understand the process ofevaporation. The molecules of spirit thatis kept in petri dish, continuously movewith random speeds in various directions.As a result these molecules collide withother molecules.

During the collision they transferenergy to other molecules. When themolecules inside the liquid collide withmolecules at the surface, the moleculesat the surface acquire energy and may flyoff from the surface.

Some of these escaping molecules maybe directed back into liquid when theycollide with the particles of air. If the numberof escaping molecules is greater than thenumber returned, then the number ofmolecules in the liquid decreases. Thus whena liquid is exposed to air, the molecules atthe surface keep on escaping from thesurface till the entire liquid disappears intoair. This process is called evaporation.

During the process of evaporation,the energy of the molecules inside the

liquid decreases and they slow down. Theytransfer this energy to escaping moleculesduring the collisions.

“The process of escaping ofmolecules from the surface of a liquid atany temperature is called evaporation”

Let us determine the reason for fasterevaporation of spirit kept under the fan. Ifair is blown over the liquid surface in anopen pan or petri dish, the number ofmolecules returned is reduced to a largeextent. This is because any moleculeescaping from the surface is blown awayfrom the vicinity of the liquid. Thisincreases the rate of evaporation. This isthe reason why the spirit in petri dish, thatis kept under ceiling fan evaporatesquickly when compared to that keptclosed. You will notice that clothes dryfaster when a wind is blowing.

It means that the temperature of asystem falls during evaporation.Evaporation is a surface phenomenon.

Activity 9

Effect of surface area,Humidity and wind speed onevaporation

Take 5ml of water each in a test tubeand in a china dish separately and keepthem under the fan. Take 5ml of water inanother china dish and keep it in thecupboard.

SCERT TELANGANA

198 Is Mater Pure ?

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 evaporation is fast?

• What do you infer about the effect ofsurface area and wind speed onevaporation?

You must have noticed that the rate ofevaporation in china dish is faster.

Since evaporation is a surfacephenomena, during evaporation processthe particles escape from the surface ofliquid. The increase in the surface areaprovides more scope for particles toescape from the surface. Hence, increasesthe rate of evaporation.

Humidity is another factor that effectsevaporation. The amount of water vapourpresent in air is called humidity.

The air around us cannot hold morethan a definite amount of water vapour ata given temperature.

If the amount of water vapour is highin air 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 withthe wind, decreasing the amount of watervapour in the surroundings.

We can also define evaporation as“the change of phase from liquid to gasthat occurs at the surface of the liquid”. Itis a cooling process, because the particlesof liquid continuously give up their energyto the particles that are escaping from thesurface.

Let us look at the following example.

• Why do we sweat while doing work?When we do work, we spend our

energy mostly in the form of heat energyfrom the body. As a result the temperatureof the skin becomes higher and the waterin the sweat glands starts evaporating. Thisevaporation cools the body.

Rate of evaporation of a liquiddepends on its surface area, temperatureand amount of vapour already present inthe surrounding air.

• Does the reverse process ofevaporation take place?

• When and how does it take place?Let us find out.

Condensation

Activity 10Place a glass tumbler on the table.

Pour cold water up to half its height.

• What do you observe on the outersurface of the tumbler?

• Why do water droplets form on theouter side of the glass?

SCERT TELANGANA


We know that the temperature ofsurrounding air is higher than thetemperature of the cold water.

Air contains water molecules in theform of vapour.

When the molecules of water in air,during their motion, strike the surface of theglass tumbler which is cool; they lose theirkinetic energy which lowers their temperatureand they get converted into droplets.

The energy lost by the watermolecules in air is gained by the moleculesof the glass tumbler. Hence the averagekinetic energy of the glass moleculesincreases. In turn the energy is transferredfrom glass molecules to the watermolecules in the glass.

In this way, the average kineticenergy of water molecules in the tumblerrises. Hence we can conclude that thetemperature of the water in glassincreases. This process is called‘condensation’. It is a warming process.

Condensation can also be defined as“the phase change from gas to liquid”.

Let us examine a situation:

You feel warm after you finish yourbath under the shower on a hot day. In thebathroom, the number of vapour moleculesper unit volume is greater than the numberof vapour molecules per unit volumeoutside the bathroom. When you try todry yourself with a towel, the vapour

molecules surrounding you condense onyour skin and this condensation makesyou feel warm.

HumiditySome vapour is always present in

air. This vapour may come fromevaporation of water from the surfaces ofrivers, lakes, ponds and from the dryingof wet clothes, sweat and so on. Thepresence of vapour molecules in air issaid to make the atmosphere humid. Theamount of water vapour present in air iscalled humidity.

Dew and FogIn early morning, during winter, you

might have noticed that water dropletsform on window panes, flowers, grass etc.

• How are these water droplets formed?Let us find out.

During winter nights, the atmospherictemperature goes down. The surfaces ofwindow-panes, flower, grass etc, becomestill colder. The air near them becomessaturated with vapour and condensationbegins. The water droplets condensed onsuch surfaces are known as dew.

If the temperature falls further, thewhole atmosphere in that region containsa large amount of vapour. So the watermolecules present in vapour condense onthe dust particles in air and form smalldroplets of water. These droplets keepfloating in the air and form a thick mist

SCERT TELANGANA

200 Is Mater Pure ?

which restricts visibility. This thick mist iscalled fog.

• Does the temperature of the water risecontinuously if heat is supplied to itcontinuously?

Boiling

Activity 11Take a beaker of water, keep it on

the burner .Note the readings ofthermometer for every 2 minutes.

• Did you see any rise or fall in the levelof the surface of the water, in thebeaker? Why?

• Does the temperature risecontinuously?

• When does the rise in temperature ofwater stop?

You will notice that, the temperatureof the water rises continuously, till itreaches 100°C. Beyond 100°C no furtherrise of temperature of water is seen. At100 0C, though supply of heat continues,the temperature does not increase further.We also observe a lot of bubbling at thesurface of water at 1000C. This is what wecall boiling of water

• Why does this happen?Water is a solution, there are many

impurities dissolved in it including somegases. When water or any liquid is heated,the solubility of gases it contains reduces.As a result, bubbles of gas are formed inthe liquid (at the bottom and on walls of

the vessel). Evaporation of water moleculesfrom the surrounding causes these bubbles,to become filled with saturated vapour,whose pressure increases as we increasethe temperature of the liquid by heating. Ata certain temperature, the pressure of thesaturated vapour inside the bubblesbecomes equal to the pressure exerted onthe bubbles from the outside (this pressureis equal to the atmospheric pressure plusthe pressure of the layer of water above thebubble). As a result, these bubbles riserapidly to the surface and collapse at thesurface releasing vapour present in bubblesinto air at the surface. This process ofconverting the liquid into vapour (gas)continues as long as you supply heat. Thisappears as boiling of water for us.

“Boiling is a process in which theliquid phase changes to gaseous phase ata constant temperature at a givenpressure.” This temperature is calledboiling point of the liquid.

• Are the processes of evaporation andboiling the same?

As you have seen in activity – 8 and10, the boiling of a liquid differsessentially from evaporation. Note thatevaporation takes place at anytemperature, while boiling occurs at adefinite temperature called the boilingpoint. Let us recall your observation inactivity – 10 that, when boiling processstarts, the temperature of the liquid cannot

SCERT TELANGANA


be raised further, no matter how long wecontinue to heat it. The temperatureremains constant at the boiling point untilall of the liquid has boiled away.

In activity – 10, you have noticed that,while heating the water in the beaker, thetemperature of water rises continuously tillit reaches 100 0C. But once boiling gotstarted, no further rise of temperature isseen though supply of heat continues.• Where does the heat energy supplied

go?This heat energy is used to change

the state of water from liquid to vapour(gas). This is called latent heat ofvapourization.

The heat energy required to change1gm of liquid to gas at constanttemperature is called latent heat ofvapourization.

Consider a liquid of mass ‘m’ thatrequires heat energy of ‘Q’ calories tochange from its state from liquid phase togaseous phase. Then Latent heat of

vaporization is Qm . Latent heat of

vaporization is denoted by ‘L’.CGS unit of latent heat of vaporization

is cal/gm and SI unit is J/kg.The boiling point of water at standard

atmospheric pressure (1atm) is 100°C or373K and Latent heat of vaporization ofwater is 540 cal/gm.

Let us now consider thetransformation of ice into water.

• Why does an ice cube get convertedinto water?

Melting

Activity 12Take small ice cubes in a beaker.

Insert the thermometer into ice cubes inthe beaker. Observe the reading of thethermometer. Now start heating the beakerkeeping it on a burner. Observe changesin the thermometer reading every 1 minutetill the ice completely melts and getsconverted into water.

• What changes do you notice in thereading of thermometer as time passesby?

• Does the temperature of the icechange during the process of melting?

You will observe that the temperatureof ice at the beginning is equal to or below0oC. If the temperature of ice is below 0oC,it goes on changing till it reaches 00C. Whenice starts melting, you will notice no changein temperature though you are supplyingheat continuously.• Why does this happen?

Thermometerglass rod

icecubes

spirit lamp

beaker

water

Fig - 3

SCERT TELANGANA

202 Is Mater Pure ?

Do you know?The heat energy supplied to the iceincreases the internal energy of themolecules of the ice. This increase ininternal energy of molecules weakens thebonds as well as breaks the bonds betweenthe molecules (H2O) in the ice. That is whythe ice (in solid phase) becomes water (inliquid phase).This process takes place at aconstant temperature 0°C or 273K. Thistemperature is called melting point. Thisprocess of converting solid into a liquid iscalled “Melting”.

The temperature of the ice does notchange during melting because the heat energygiven to the ice is totally utilized in breakingthe bonds between the water molecules.

The process in which solid phasechanges to liquid phase at a constanttemperature is called melting. This constanttemperature is called melting point.

• How much heat energy is required toconvert 1gm of ice to liquid?

The Heat energy required to convert1gm of solid completely into liquid at aconstant temperature is called Latent heatof fusion.

Consider a solid of mass m. Let heatenergy Q be required to change it from thesolid phase to liquid phase. The heatrequired to change 1gm of solid into liquid

is Qm .

Latent heat of fusion L = Qm . The

value of Latent heat of fusion of ice is80cal/gm

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 water. Hence ice floats onwater rather than sinking. This is veryimportant for survival of marine lifewhich lives in ponds in the colder areas.In extremely cold weather the water atthe top become colder and colder, untilit freezes. While the ice floats on thetop, the animals continue to live in thewater below, which does not freeze andremains at 40C. The ice on the top ofthe pond insulates water below it and itstops the water from losing the heat toair.

Freezing

You might have observed coconutoil and ghee getting converted from liquidstate to solid state during winter season.

• What could be the reason for thischange?

• What happens to water kept in arefrigerator?

• How does it get converted from liquidphase to solid phase?

We know that the water that is keptin a refrigerator converts to solid ice. You

SCERT TELANGANA


know that initial temperature of water ismore compared to the temperature of ice.It means that during the process ofconversion from liquid to solid, theinternal energy of the water decreases sothat it becomes a solid ice. This process iscalled freezing.

“The process in which a substance inliquid phase changes to solid phase bylosing some of its energy is calledfreezing.”

Freezing of water takes place at 0°Ctemperature and at one atmosphericpressure.

• Are the volumes of water and iceformed with same amount of waterequal? Why?

Let us find out.

Activity 13Take small glass bottle with a tight

lid .Fill it with water completely withoutany gaps and fix the lid tightly in such away that water does not come out of it.Put the bottle into the deep freezer of arefrigerator for a few hours. Take it outfrom the fridge and you will observe thatthe glass bottle breaks.

• Why did the glass bottle break?We know that the volume of the water

poured into the glass bottle is equal to thevolume of the bottle. When the waterfreezes to ice, the bottle is broken .Thismeans that the volume of the ice shouldbe greater than the volume of the waterfilled in the bottle.

In short, we say that water ‘expands’(increases in volume) on freezing!

Thus the density of ice is less thanthat of water and this explains why icefloats on water.

Key words

Temperature, Heat, Thermal equilibrium, Specific heat, Evaporation,Condensation, Humidity, Dew, Fog, Boiling, Latent heat of vaporization,

Melting, Freezing.

Think and discuss

• Why do we wear cotton clothes insummer?

• Why do we see water droplets onouter surface of a glass containingice - cold water?

• Why do pigs toil in the mud duringhot summer?

• Why do we store water in matkas(earthern pots)?

SCERT TELANGANA

204 Is Mater Pure ?

• Heat is a form of energy, in transit, that flows from a body at higher temperature to a body at lowertemperature.

• SI unit of heat is joul. Its CGS unit is calore.

1 calorie = 4.186 J.

• When two or more bodies are at different temperature are brought into thermal contact, then netheat lost by the hot bodies is equal to net heat gained by the cold bodies until they attain thermalequilibrium.

• The average kinetic energy of the molecules is directly proportional to the absolute temperature.

• The specific heat of a substance is the amount of heat required to raise the temperature of unit massof the substance by one unit.

S=Q

m tΔ

• The process of escaping of molecules from the surface of a liquid at any temperature is calledevaporation and it is a cooling process.

• Condensation is the reverse process of evaporation.

• Boiling is the process in which the liquid phase changes to gaseous phase at a constant temperatureand constant pressure.

• The heat energy is used to change the state of water from liquid to vapour is called Latent heat ofvapourisation.

• The heat energy required to convert 1 gm of solid completely into liquid at a constant temperatureis called latent heat of fusion.

Reflections on concepts1. Why do we get dew on the surface of a cold soft

drink bottle kept in open air? (AS1)

2. Water can evaporate at any temperature Explain with an example? (AS3)

3. What role does specific heat play in keeping a watermelon cool for a long time after removing itfrom a fridge on a hot day? (AS7)

4. Equal amounts of water are kept in a cap and in a dish. Which will evaporate faster? Why? (AS3)

5. Why specific heat is different for different substances? Explain (AS1)


What we have learnt

SCERT TELANGANA



1. Using the concept of evaporation explain why dogs pant during hot summer days? (AS1)

2. If 50g of water at 20oC temperature and 50 g of water 40oC temperature are mixed, what is thefinal temperature of the mixture of? (AS1)

3. What do you observe in the surroundings in terms of cooling or heating when water vapour isgetting condensed (AS1)

Higher Order Thinking Questions1. Answer these (AS1)

(a) How much energy is transferred when 1 g. of boiling water at 100oC condenses to waterat 100oC?

(b) How much energy is transferred when 1 g. of boiling water 100oC cools to water at 0oC?(c) How much energy is released or absorbed when 1 g. of water at 0oC freezes to ice at

0oC?(d) How much energy released or absorbed when 1 g. of steam at 100oC cools to ice at 0oC?

2. Suppose that 1 L of water is heated for a certain time to rise its temperature for 2oC. If 2L ofwater is heated for the same time, how much of its temperature would rise?(AS7)

1. Which of the following is a warming process [ ]

a) Evaporation b) condensation c) boiling d) all the above

2. Melting is a process in which solid phase changes to [ ]

a) liquid phase b) liquid phase at constant temperature

c) gaseous phase d) Gaseous phase at constant temperature

3. Three bodies A, B and C are in thermal equilibrium. The temperature of B is 45oC. then thetemperature of C is [ ]

a) 45oC b) 50oC c) 40oC d) any temperature

4. The temperature of a steel rod is 330K. Its temperature in oC is [ ]

a) 55oC b) 57oC c) 59oC d) 53oC

5. When ice melts, its temperature [ ]

a) remains constant b) increases

c) decreases d) first decrease and then increase


SCERT TELANGANA

206 Is Mater Pure ?


1. Find the specific heat of solids experimentally write a report.

2. Perform an experiment to prove that the rate of evaporation of a liquid depends on its surface areaand vapor already present in surrounding air write a report.

3. Take different metal pieces of same size and heat them to same temperature, dip them immediatelyin the beakers containing water of same level. Observe the temperature differences in the water.Write your observations

Suggested Projects1. Take 2kg of ice is at -5oC. Supply heat is continuously to ice. Till it starts boiling. Note the

temperature every minute. Draw a graph between temperature and time using the values you get.What do you understand from the graph. Write the conclusions. ( You know that ice melts at 0ocand boils at l00°C.

2. Observe the evaporation process and write the report in the form of a table for the followingsubstances’ are given in the table for given conditions.

Substance Petrol, kerosene, alcohol, water, glycerin, camphor

Conditions Inside the room, outside the room, exposing to sunlight, outside shadow

3. Observe the evaporation process of water which is kept inside ,and outside the house andrepeatedly do this activity with different shaped dishes. Write the report.

SCERT TELANGANA


In class 8, we have learnt about soundwill be produced by vibrating bodies andalso learnt how sound is transmittedthrough a medium and received by our ears.In this chapter we will study about natureof sound, 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

12

Activity-1

Sound is a form of energyTake a tin can and remove both ends to

make a hollow cylinder as shown infig.- 1. Take a balloon and stretch it overthe can. Wrap 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 fig.-1. 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

SOUND

SCERT TELANGANA

208 Sound

To see vibrations, attach a small pieceof steel wire to one of its prongs as shownin fig.- 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 fig.- 2. Repeat theexperiment when the fork is not vibratingand observe the difference in the line.formed.

Fig - 2

• What do you conclude from the aboveactivity?

• Can you produce sound withoutvibration in the body?

In above activity we have producedvibrations in a tuning fork by striking itwith a rubber hammer. We observe that avibrating tuning fork produces sound. Thussound is produced by vibrating bodies.

• Give some examples of vibratingbodies which produce sound.

• Which part of our body vibrates whenwe speak?

• 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 traveller, explained thatsound travels 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?

SCERT TELANGANA


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 through 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 (waves) in air and they strikeour ears. The other explanation is that someparticles are shot off from the source ofsound and they 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.

SCERT TELANGANA

210 Sound

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. Fig.- 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 in fig.- 3? It dragsback the layer of air near it, decreasingthe density of air there. The particles ofair in the next layer on the right move andfill this less dense area. As a result, itsown density reduces. In the same way, thedensity of air in successive layers on theright decrease one after the other. We saythat a rarefaction pulse 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(C) andrarefactions (R) in a slinky

1. Take a slinky, it is a spring – shapedtoy which can be extended orcompressed very easily. It is veryflexible and can be put into many shapeseasily. You can send continuous waveson a slinky. Lay it down on a table orthe floor as shown in the fig.- 4 and aska friend to hold one end. Pull the otherend to stretch the slinky, and then moveit to and fro along 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.

SCERT TELANGANA


Fig-5: Transverse waves in a slinky

2. Hang a slinky from a fixed support. Holdit gently at the lower end and quicklymove your hand sideways and back.What do you observe? This will causea hump on the slinky near the lower end.This hump travels upwards on the slinkyas shown in fig.- 5. What is travellingupwards? The part of the slinky thatwas at the bottom in the beginning isstill at the bottom. Similarly no otherpart of the slinky has moved up. Onlythe disturbance has moved up. Hencewe may say that a wave has travelledup through 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. Fig.- 6shows the variation in the density of airnear the source at a particular time and thevariation in the density of air with positionis also shown by a graph in Fig.- 6. Sincethe pressure of air is proportional todensity at a given temperature, the plot of

SCERT TELANGANA

212 Sound

density versus position 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. position 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 fig.- 6) or minimumdensity (rarefactions) like B and D asshown in fig.- 6 remain the same. So thesevalues get repeated after a fixed distance.This distance is called the wave length ofthe wave. It is denoted by the Greek letter

λ (read as ‘lambda’). We can definewavelength as follows.

The distance between two consecutivecompressions or two consecutiverarefactions is called the wave length of asound wave. (Fig 7)

Being a length, wavelength is measuredin metres. SI unit of wave length is metres(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.

1

Den

sity Normal

density

Position

λλλλλ

λλλλλ

Den

sity

A B C D E F GCrest

TroughP Q R

Normaldensity

Position

λλλλλ10

-1

10

-1

SCERT TELANGANA


“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 υυυυυ (readas ‘nu’) to denote frequency.

Relation between frequency and timeperiod

Let us find the relationship betweenfrequency and time period.Let the time taken for υ oscillations = 1sThe time taken for one oscillation=(1/υ) s

But the time taken for one oscillationis called the time period (T) and the numberof oscillations per second is called thefrequency (υ).

Hence Frequency and time period arerelated as T = 1/υ or υ = 1/T

The SI unit of frequency is hertz (Hz).It is named after Heinrich Rudolph Hertz.

Heinrich Rudolf 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 wasnamed hertz 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 a minimum value.

Dis

plac

emen

t Ampiltude

Position

1

0

-1

SCERT TELANGANA

214 Sound

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/ υ =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 compression or rarefaction,travels in unit time is called speed of soundwave.

Let the distance travelled by a wave inT seconds = λ metres

The distance travelled by a wave in

1second = Tλ

metres

Thus by definition, speed of sound

wave v = Tλ

——————— (1)

We know that frequency υ = 1T —(2)

From equation (1) & (2) we get v = υ λSpeed 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, speed of sound refers tothe speed of sound waves in air. Howeverspeed of sound varies from substance tosubstance. Sound travels faster in liquidsand nonporous solids than it does in air. Ittravels about 4.3 times faster in water(1484 m/s) and nearly 15 times as fast iniron (5120 m/s) than in air at 200 C. In dryair at 200 C the speed of sound is343.2 m/s. this is 1236 km/hr or about 1kmin 3s.

Think and discuss

• During a thunderstrom if you noticea 3 second delay between the flashof lightning and sound of thunder.What is 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

SCERT TELANGANA


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?Actually pitch of sound is the

sensation conveyed to our brain by thesound waves falling on our ears whichdepends directly on the frequency of theincident sound waves. The greater thefrequency of a musical note, higher is thepitch.

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(υ) = 40,000 Hz

Wave length (λ) = distance betweentwo consecutive compression orrarefaction pulses.

λ = 1cmFrom v= υ λ = 40,000 Hz 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?

Dis

plac

emen

t

Fig-9(a)Sound of lower pitch

Position

Fig-9(b)Sound of higher pitch

Dis

plac

emen

t

Position

10

-1

10

-1

SCERT TELANGANA

216 Sound

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 withtime is shown as a graph for two soundswith different amplitudes.

The amplitude of the sound wave in fig.-10(a) is greater than the amplitude of soundwave in fig.-10(b). So the graph in fig.-10(a) represents a louder sound and thegraph in fig.- 10(b) represents a soft 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

Frequency256 288 320 341.3 384 426.7 480 512

(Hz)

Dis

plac

emen

t

Position

Dis

plac

emen

t

Position

1

0

-1

10

-1

SCERT TELANGANA


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 at one instanceand amplitude at 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 andplace them on a table near a wall. Askyour friend to speak softly into one tubewhile you use the other tube to listen.Adjust the tube until you hear the bestsound. You will find that you hear yourfriend’s voice best when the tubes makeequal angles with a normal to the wall asshown in fig.--12. Why?

Reflection of sound follows the same lawsas the reflection of light. When sound isreflected. i.e., the directions in which thesound is incident and reflected make equalangles with the normal to the reflecting surface.• 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 the characteristicwhich enables us to distinguish betweenmusical notes emitted by different musicalinstruments or voices even though they havethe same pitch and loudness. It is becausedifferent wave forms are produced bydifferent musical instruments. Hence thequality of a note depends on its wave form.

Fig-11

Fig.--11 shows graphical representationof sound waves produced by a tuning fork, aviolin and a piano playing the same note(fundamental frequency = 440Hz) with equalloudness (amplitude).

SCERT TELANGANA

218 Sound

You will not be able to hear your friend’svoice clearly. Think about the planes of thetube carrying incident sound and reflectedsound. What will happen to these planes ifwe lift one of the pipes? If you lift one of thepipes then the pipe carrying incident soundand the pipe carrying reflected sound willnot be in the same plane. Hence we cannothear 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, to 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

SCERT TELANGANA


Do you know?

prolonged sound wave.In an auditorium or big hall excessive

reverberation is highly undesirable. Toreduce reverberation, the roof and walls ofthe auditorium are generally covered withsound – absorbent materials likecompressed fibre 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 = 2dt

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 =2dt

V = 2640.8

ms = 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,

SCERT TELANGANA

220 Sound

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 infig.- 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 range of audibility is 20Hz- 20 KHz. We cannot hear sounds offrequencies less than 20 Hz or more than20 KHz. These limits vary from person toperson and with age. Children can hearsounds of somewhat higher frequencies,say up to 30 KHz. With age, our ability tohear high-frequency sound diminishes. Forthe elders, the upper limit often falls to 10-

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 the sounds producedwithin the body, mainly 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 fig.- 15.

Fig-15

SCERT TELANGANA


Ultrasound is used extensively in industriesand 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 vibrationsknock 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?

12 KHz. However, we take 20-20,000 Hzas the audible range for an average person.

Even in the audible range, the humanears are 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 Hz . 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 ultrasoundUltrasounds are high frequency sound

waves. They are able to travel along a welldefined path in gaseous and liquid media.

SCERT TELANGANA

222 Sound

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 (ECG) is atechnique in which ultrasonic waves,reflected from various parts of the heart,form an image of the heart.

Ultrasonography (USG) is routinelyused to show doctors, the images of apatient’s organs such as the liver, gallbladder, uterus, etc. It helps doctors todetect abnormalities such as stones in thegall 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 Sound Navigation

And Ranging. This is a method fordetecting 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 receiver which areinstalled in the “Observation Centre” onboard of a ship. From the observationcentre on board a ship, ultrasonic waves ofhigh frequencies, say 100 KHz, are sent inall directions under the water through

SCERT TELANGANA


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

SCERT TELANGANA

224 Sound

What we have learnt

Key words

Mechanical energy, tuning fork, longitudinal wave, transverse wave, compression,rarefaction, crest, trough, density of medium, pressure, wavelength, amplitude,frequency, pitch, loudness, quality of sound, echo, reverberation, infrasonic, audiblerange, 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.

• Loudness of sound is defined as the degree of sensation produced in the ear.

• The quality of sound is the characteristic which enables us to distinguish betweenmusical notes emitted by different musical instruments.

SCERT TELANGANA


• 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.


1. Explain the following terms

a) amplitude b) wavelength c) frequency (AS1)

2. Write the relation between wavelength, frequency and speed of sound (AS1)

3. Which has larger frequency – infrasonic sound or ultrasonic sound? (AS2)

4. Why is soft furnishing avoided in concert halls? (AS7)


1. Does the sound follow same laws of reflection as light does? (AS1)

2. Two sources A and B vibrate with the same amplitude. They produce sounds of

frequencies 1kHz and 30kHz respectively. Which of the two waves will have larger

power? (AS1)

3. With the help of a diagram describe how compression and rarefaction pulses are

produced in air near a source of sound. (AS5)

4. How are multiple reflections of sound helpful to doctors and engineers?

(AS7)


SCERT TELANGANA

226 Sound

Higher Order Thinking Questions

1. Explain the working and applications of SONAR. (AS1)

2. How do echoes in a normal room affect the quality of the sounds that we hear?

(AS7)

1. When can you say that the sound is propagating thourgh a medium [ ]

a) If the medium is travelling b) The particles of a medium are travelling

c) When the source of sound is travelling d) When the disturbance is travelling.

2. The unit for the number of waves produced in a second are [ ]

a) hertz b) joule c) meter d) pascal

3. The sounds of frequency less than 20 Hz are known as [ ]

a) Audible range b) Ultra sounds c) Infra sounds d) Sonic boom

4. The sound limit between the frequency of 20Hz – 20000Hz is called as [ ]

a) Audible range b) Ultra sound range c) Infra sound range d) Sonic boom


1. Conduct an experiment to listen the reflected sound and write a report.

Suggested Projects

1. Collect the information about the animals which communicate through Infrasonics

or ultrasonics. Collect their pictures and write a report on their communication

technique.

2. “We know that the sound is a form of energy. So, the large amount of energy produced

due to the sound pollution in cosmopolitan cities can be used to our day to day

needs of energy. It also helps us to protect biodiversity in urban areas”. Do you

agree with this statement? Collect information onthis and write a report.


SCERT TELANGANA

IX Physics EM Title Free.pmd - SCERT Telangana

Documents

Transcript of IX Physics EM Title Free.pmd - SCERT Telangana