Class 9 Science

CONTENTS

FOREWORD iii

Chapter 1 MATTER IN OUR SURROUNDINGS 1

Chapter 2 IS MATTER AROUND US PURE 14

Chapter 3 ATOMS AND MOLECULES 31

Chapter 4 STRUCTURE OF THE ATOM 46

Chapter 5 THE FUNDAMENTAL UNIT OF LIFE 57

Chapter 6 TISSUES 68

Chapter 7 DIVERSITY IN LIVING ORGANISMS 80

Chapter 8 MOTION 98

Chapter 9 FORCE AND LAWS OF MOTION 114

Chapter 10 GRAVITATION 131

Chapter 11 WORK AND ENERGY 146

Chapter 12 SOUND 160

Chapter 13 WHY DO WE FALL ILL 176

Chapter 14 NATURAL RESOURCES 189

Chapter 15 IMPROVEMENT IN FOOD RESOURCES 203

ANSWERS 216 – 218

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/book_publishing/Class 9/Science/ANSWER.pdf

As we look at our surroundings, we see a largevariety of things with different shapes, sizesand textures. Everything in this universe ismade up of material which scientists havenamed “matter”. The air we breathe, the foodwe eat, stones, clouds, stars, plants andanimals, even a small drop of water or aparticle of sand– each thing is matter. We canalso see as we look around that all the thingsmentioned above occupy space, that is,volume* and have mass.

Since early times, human beings havebeen trying to understand their surroundings.Early Indian philosophers classified matter inthe form of five basic elements – the“Panch Tatva”– air, earth, fire, sky and water.According to them everything, living or non-living, was made up of these five basicelements. Ancient Greek philosophers hadarrived at a similar classification of matter.

Modern day scientists have evolved twotypes of classification of matter based on theirphysical properties and chemical nature.

In this chapter we shall learn aboutmatter based on its physical properties.Chemical aspects of matter will be taken upin subsequent chapters.

1.1 Physical Nature of Matter1.1.1 MATTER IS MADE UP OF PARTICLES

For a long time, two schools of thoughtprevailed regarding the nature of matter. Oneschool believed matter to be continuous likea block of wood, whereas, the other thoughtthat matter was made up of particles likesand. Let us perform an activity to decideabout the nature of matter – is it continuousor particulate?

Activity ______________ 1.1• Take a 100 mL beaker.• Fill half the beaker with water and

mark the level of water.• Dissolve some salt/ sugar with the help

of a glass rod.• Observe any change in water level.• What do you think has happened to

the salt?• Where does it disappear?• Does the level of water change?

In order to answer these questions weneed to use the idea that matter is made upof particles. What was there in the spoon, saltor sugar, has now spread throughout water.This is illustrated in Fig. 1.1.

1.1.2 HOW SMALL ARE THESE PARTICLES

OF MATTER?Activity ______________ 1.2• Take 2-3 crystals of potassium

permanganate and dissolve them in100 mL of water.

Fig. 1.1: When we dissolve salt in water, the particlesof salt get into the spaces between particlesof water.

* The SI unit of volume is cubic metre (m3). The common unit of measuring volume is litre (L) such that 1L = 1 dm3, 1L = 1000 mL, 1 mL = 1 cm3.

11111MMMMMATTERATTERATTERATTERATTER INININININ O O O O OURURURURUR S S S S SURROUNDINGSURROUNDINGSURROUNDINGSURROUNDINGSURROUNDINGS

Chapter

SCIENCE2

• Take out approximately 10 mL of thissolution and put it into 90 mL of clearwater.

• Take out 10 mL of this solution andput it into another 90 mL of clear water.

• Keep diluting the solution like this 5 to8 times.

• Is the water still coloured ?

1.2.2 PARTICLES OF MATTER ARE

CONTINUOUSLY MOVING

Activity ______________ 1.3

• Put an unlit incense stick in a cornerof your class. How close do you have togo near it so as to get its smell?

• Now light the incense stick. Whathappens? Do you get the smell sittingat a distance?

• Record your observations.

Activity ______________ 1.4

• Take two glasses/beakers filled withwater.

• Put a drop of blue or red ink slowlyand carefully along the sides of the firstbeaker and honey in the same way inthe second beaker.

• Leave them undisturbed in your houseor in a corner in the class.

• Record your observations.• What do you observe immediately after

adding the ink drop?• What do you observe immediately after

adding a drop of honey?• How many hours or days does it take

for the colour of ink to spread evenlythroughout the water?

Activity ______________ 1.5• Drop a crystal of copper sulphate or

potassium permanganate into a glassof hot water and another containingcold water. Do not stir the solution.Allow the crystals to settle at thebottom.

• What do you observe just above thesolid crystal in the glass?

• What happens as time passes?• What does this suggest about the

particles of solid and liquid?• Does the rate of mixing change with

temperature? Why and how?

From the above three activities (1.3, 1.4 and1.5), we can conclude the following:

Fig. 1.2: Estimating how small are the particles ofmatter. With every dilution, though the colourbecomes light, it is still visible.

This experiment shows that just a fewcrystals of potassium permanganate cancolour a large volume of water (about1000 L). So we conclude that there must bemillions of tiny particles in just one crystalof potassium permanganate, which keep ondividing themselves into smaller and smallerparticles. Ultimately a stage is reached whenthe particles cannot divide further intosmaller particles.

The same activity can be done using2 mL of Dettol instead of potassiumpermanganate. The smell can be detectedeven on repeated dilution.

The particles of matter are very small –they are small beyond our imagination!!!!

1.2 Characteristics of Particles ofMatter

1.2.1 PARTICLES OF MATTER HAVE SPACEBETWEEN THEM

In activities 1.1 and 1.2 we saw that particlesof sugar, salt, Dettol, or potassiumpermanganate got evenly distributed in water.Similarly, when we make tea, coffee orlemonade (nimbu paani ), particles of one typeof matter get into the spaces between particlesof the other. This shows that there is enoughspace between particles of matter.

MATTER IN OUR SURROUNDINGS 3

• If we consider each student as a particleof matter, then in which group theparticles held each other with themaximum force?

Activity ______________ 1.7• Take an iron nail, a piece of chalk and

a rubber band.• Try breaking them by hammering,

cutting or stretching.• In which of the above three substances

do you think the particles are heldtogether with greater force?

Activity ______________ 1.8• Open a water tap, try breaking the

stream of water with your fingers.• Were you able to cut the stream of

water?• What could be the reason behind the

stream of water remaining together?

The above three activities (1.6, 1.7 and1.8) suggest that particles of matter have forceacting between them. This force keeps theparticles together. The strength of this forceof attraction varies from one kind of matterto another.

uestions1. Which of the following are

matter?Chair, air, love, smell, hate,almonds, thought, cold, cold-drink, smell of perfume.

2. Give reasons for the followingobservation:The smell of hot sizzling foodreaches you several metresaway, but to get the smell fromcold food you have to go close.

3. A diver is able to cut throughwater in a swimming pool. Whichproperty of matter does thisobservation show?

4. What are the characteristics ofthe particles of matter?

Particles of matter are continuouslymoving, that is, they possess what we callthe kinetic energy. As the temperature rises,particles move faster. So, we can say that withincrease in temperature the kinetic energy ofthe particles also increases.

In the above three activities we observethat particles of matter intermix on their ownwith each other. They do so by getting intothe spaces between the particles. Thisintermixing of particles of two different typesof matter on their own is called diffusion. Wealso observe that on heating, diffusionbecomes faster. Why does this happen?

1.2.3 PARTICLES OF MATTER ATTRACT

EACH OTHER

Activity ______________ 1.6• Play this game in the field— make four

groups and form human chains assuggested:

• The first group should hold each otherfrom the back and lock arms like Bihudancers (Fig. 1.3).

Fig. 1.3

• The second group should hold handsto form a human chain.

• The third group should form a chainby touching each other with only theirfinger tips.

• Now, the fourth group of studentsshould run around and try to break thethree human chains one by one intoas many small groups as possible.

• Which group was the easiest to break?Why?

Q

SCIENCE4

1.3 States of MatterObserve different types of matter around you.What are its different states? We can see thatmatter around us exists in three differentstates– solid, liquid and gas. These states ofmatter arise due to the variation in thecharacteristics of the particles of matter.

Now, let us study about the properties ofthese three states of matter in detail.

1.3.1 THE SOLID STATE

Activity _____________ 1.9• Collect the following articles— a pen, a

book, a needle and a piece of thread.• Sketch the shape of the above articles

in your notebook by moving a pencilaround them.

• Do all these have a definite shape,distinct boundaries and a fixedvolume?

• What happens if they are hammered,pulled or dropped?

• Are these capable of diffusing into eachother?

• Try compressing them by applyingforce. Are you able to compress them?

All the above are examples of solids. Wecan observe that all these have a definiteshape, distinct boundaries and fixed volumes,that is, have negligible compressibility. Solidshave a tendency to maintain their shape whensubjected to outside force. Solids may breakunder force but it is difficult to change theirshape, so they are rigid.

Consider the following:

(a) What about a rubber band, can itchange its shape on stretching? Is ita solid?

(b) What about sugar and salt? Whenkept in different jars these take theshape of the jar. Are they solid?

(c) What about a sponge? It is a solidyet we are able to compress it. Why?

All the above are solids as:• A rubber band changes shape under

force and regains the same shape when

the force is removed. If excessive force isapplied, it breaks.

• The shape of each individual sugar orsalt crystal remains fixed, whether wetake it in our hand, put it in a plate or ina jar.

• A sponge has minute holes, in whichair is trapped, when we press it, the airis expelled out and we are able tocompress it.

1.3.2 THE LIQUID STATE

Activity _____________1.10• Collect the following:

(a) water, cooking oil, milk, juice, acold drink.

(b) containers of different shapes. Puta 50 mL mark on these containersusing a measuring cylinder fromthe laboratory.

• What will happen if these liquids arespilt on the floor?

• Measure 50 mL of any one liquid andtransfer it into different containers oneby one. Does the volume remain thesame?

• Does the shape of the liquid remain thesame ?

• When you pour the liquid from onecontainer into another, does it floweasily?

We observe that liquids have no fixedshape but have a fixed volume. They take upthe shape of the container in which they arekept. Liquids flow and change shape, so theyare not rigid but can be called fluid.

Refer to activities 1.4 and 1.5 where wesaw that solids and liquids can diffuse intoliquids. The gases from the atmospherediffuse and dissolve in water. These gases,especially oxygen and carbon dioxide, areessential for the survival of aquatic animalsand plants.

All living creatures need to breathe forsurvival. The aquatic animals can breatheunder water due to the presence of dissolvedoxygen in water. Thus, we may conclude thatsolids, liquids and gases can diffuse intoliquids. The rate of diffusion of liquids is


higher than that of solids. This is due to thefact that in the liquid state, particles movefreely and have greater space between eachother as compared to particles in the solidstate.

1.3.3 THE GASEOUS STATE

Have you ever observed a balloon seller fillinga large number of balloons from a singlecylinder of gas? Enquire from him how manyballoons is he able to fill from one cylinder.Ask him which gas does he have in the cylinder.

Activity _____________1.11• Take three 100 mL syringes and close

their nozzles by rubber corks, asshown in Fig.1.4.

• Remove the pistons from all thesyringes.

• Leaving one syringe untouched, fillwater in the second and pieces of chalkin the third.

• Insert the pistons back into thesyringes. You may apply some vaselineon the pistons before inserting theminto the syringes for their smoothmovement.

• Now, try to compress the content bypushing the piston in each syringe.

We have observed that gases are highlycompressible as compared to solids andliquids. The liquefied petroleum gas (LPG)cylinder that we get in our home for cookingor the oxygen supplied to hospitals incylinders is compressed gas. Compressednatural gas (CNG) is used as fuel these daysin vehicles. Due to its high compressibility,large volumes of a gas can be compressedinto a small cylinder and transported easily.

We come to know of what is being cookedin the kitchen without even entering there,by the smell that reaches our nostrils. Howdoes this smell reach us? The particles of thearoma of food mix with the particles of airspread from the kitchen, reach us and evenfarther away. The smell of hot cooked foodreaches us in seconds; compare this with therate of diffusion of solids and liquids. Due tohigh speed of particles and large spacebetween them, gases show the property ofdiffusing very fast into other gases.

In the gaseous state, the particles moveabout randomly at high speed. Due to thisrandom movement, the particles hit eachother and also the walls of the container. Thepressure exerted by the gas is because of thisforce exerted by gas particles per unit areaon the walls of the container.

Fig. 1.4

• What do you observe? In which casewas the piston easily pushed in?

• What do you infer from yourobservations?

Fig.1.5: a, b and c show the magnified schematicpictures of the three states of matter. Themotion of the particles can be seen andcompared in the three states of matter.

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1.4.1 EFFECT OF CHANGE OF TEMPERATURE

Activity _____________1.12• Take about 150 g of ice in a beaker and

suspend a laboratory thermometer sothat its bulb is in contact with the ice,as in Fig. 1.6.

uestions1. The mass per unit volume of a

substance is called density.(density = mass/volume).Arrange the following in order ofincreasing density – air, exhaustfrom chimneys, honey, water,chalk, cotton and iron.

2. (a) Tabulate the differences inthe characterisitcs of statesof matter.

(b) Comment upon the following:rigidity, compressibility,fluidity, filling a gascontainer, shape, kineticenergy and density.

3. Give reasons(a) A gas fills completely the

vessel in which it is kept.(b) A gas exerts pressure on the

walls of the container.(c) A wooden table should be

called a solid.(d) We can easily move our hand

in air but to do the samethrough a solid block of woodwe need a karate expert.

4. Liquids generally have lowerdensity as compared to solids.But you must have observed thatice floats on water. Find out why.

1.4 Can Matter Change its State?We all know from our observation that watercan exist in three states of matter–

• solid, as ice,• liquid, as the familiar water, and• gas, as water vapour.What happens inside the matter during

this change of state? What happens to theparticles of matter during the change ofstates? How does this change of state takeplace? We need answers to these questions,isn’t it?

Q

(a)

(b)

Fig. 1.6: (a) Conversion of ice to water, (b) conversionof water to water vapour


• Start heating the beaker on a low flame.• Note the temperature when the ice

starts melting.• Note the temperature when all the ice

has converted into water.• Record your observations for this

conversion of solid to liquid state.• Now, put a glass rod in the beaker and

heat while stirring till the water startsboiling.

• Keep a careful eye on the thermometerreading till most of the water hasvaporised.

• Record your observations for theconversion of water in the liquid stateto the gaseous state.

On increasing the temperature of solids,the kinetic energy of the particles increases.Due to the increase in kinetic energy, theparticles start vibrating with greater speed.The energy supplied by heat overcomes theforces of attraction between the particles. Theparticles leave their fixed positions and startmoving more freely. A stage is reached whenthe solid melts and is converted to a liquid.The temperature at which a solid melts tobecome a liquid at the atmospheric pressureis called its melting point.

The melting point of a solid is anindication of the strength of the force ofattraction between its particles.

The melting point of ice is 273.16 K*. Theprocess of melting, that is, change of solidstate into liquid state is also known as fusion.When a solid melts, its temperatureremains the same, so where does the heatenergy go?

You must have observed, during theexperiment of melting, that the temperatureof the system does not change after themelting point is reached, till all the ice melts.This happens even though we continue toheat the beaker, that is, we continue to supplyheat. This heat gets used up in changing the

state by overcoming the forces of attractionbetween the particles. As this heat energy isabsorbed by ice without showing any rise intemperature, it is considered that it getshidden into the contents of the beaker and isknown as the latent heat. The word latentmeans hidden. The amount of heat energythat is required to change 1 kg of a solid intoliquid at atmospheric pressure at its meltingpoint is known as the latent heat of fusion.So, particles in water at 00 C (273 K) havemore energy as compared to particles in iceat the same temperature.

When we supply heat energy to water,particles start moving even faster. At a certaintemperature, a point is reached when theparticles have enough energy to break freefrom the forces of attraction of each other. Atthis temperature the liquid starts changinginto gas. The temperature at which a liquidstarts boiling at the atmospheric pressure isknown as its boiling point. Boiling is a bulkphenomenon. Particles from the bulk of theliquid gain enough energy to change into thevapour state.

For water this temperature is 373 K(100 0C = 273 + 100 = 373 K).

Can you define the latent heat ofvaporisation? Do it in the same way as wehave defined the latent heat of fusion.Particles in steam, that is, water vapour at373 K (1000 C) have more energy than waterat the same temperature. This is becauseparticles in steam have absorbed extra energyin the form of latent heat of vaporisation.

*Note: Kelvin is the SI unit of temperature, 00 C =273.16 K. For convenience, we take 00 C = 273 Kafter rounding off the decimal. To change a temperature on the Kelvin scale to the Celsius scaleyou have to subtract 273 from the given temperature, and to convert a temperature on theCelsius scale to the Kelvin scale you have to add 273 to the given temperature.

So, we infer that the state of matter canbe changed into another state by changingthe temperature.

We have learnt that substances aroundus change state from solid to liquid and fromliquid to gas on application of heat. But there

SCIENCE8

closer? Do you think that increasing ordecreasing the pressure can change the stateof matter?

are some that change directly from solid stateto gaseous state and vice versa withoutchanging into the liquid state.

Activity _____________1.13• Take some camphor or ammonium

chloride. Crush it and put it in a chinadish.

• Put an inverted funnel over the chinadish.

• Put a cotton plug on the stem of thefunnel, as shown in Fig. 1.7.

* atmosphere (atm) is a unit of measuring pressure exerted by a gas. The unit of pressure is Pascal (Pa):1 atmosphere = 1.01 × 105 Pa. The pressure of air in atmosphere is called atmospheric pressure. Theatmospheric pressure at sea level is 1 atmosphere, and is taken as the normal atmospheric pressure.

Fig. 1.7: Sublimation of ammonium chloride

Fig. 1.8: By applying pressure, particles of mattercan be brought close together.

Applying pressure and reducingtemperature can liquefy gases.

Have you heard of solid carbon dioxide(CO2)? It is stored under high pressure. SolidCO2 gets converted directly to gaseous stateon decrease of pressure to 1 atmosphere*without coming into liquid state. This is thereason that solid carbon dioxide is also knownas dry ice.

Thus, we can say that pressure andtemperature determine the state of asubstance, whether it will be solid, liquidor gas.

• Now, heat slowly and observe.• What do you infer from the above

activity?

A change of state directly from solid togas without changing into liquid state (or viceversa) is called sublimation.

1.4.2 EFFECT OF CHANGE OF PRESSURE

We have already learnt that the difference invarious states of matter is due to thedifference in the distances between theconstituent particles. What will happen whenwe start putting pressure and compress a gasenclosed in a cylinder? Will the particles come Fig. 1.9: Interconversion of the three states of matter


dish and keep it inside a cupboard oron a shelf in your class.

• Record the room temperature.• Record the time or days taken for the

evaporation process in the above cases.• Repeat the above three steps of activity

on a rainy day and record yourobservations.

• What do you infer about the effect oftemperature, surface area and windvelocity (speed) on evaporation?

You must have observed that the rate ofevaporation increases with–• an increase of surface area:

We know that evaporation is a surfacephenomenon. If the surface area isincreased, the rate of evaporationincreases. For example, while puttingclothes for drying up we spread them out.

• an increase of temperature:With the increase of temperature, morenumber of particles get enough kineticenergy to go into the vapour state.

• a decrease in humidity:Humidity is the amount of water vapourpresent in air. The air around us cannothold more than a definite amount ofwater vapour at a given temperature. Ifthe amount of water in air is already high,the rate of evaporation decreases.

• an increase in wind speed:It is a common observation that clothesdry faster on a windy day. With theincrease in wind speed, the particles ofwater vapour move away with the wind,decreasing the amount of water vapourin the surrounding.

1.5.2 HOW DOES EVAPORATION CAUSE

COOLING?In an open vessel, the liquid keeps onevaporating. The particles of liquid absorbenergy from the surrounding to regain theenergy lost during evaporation. Thisabsorption of energy from the surroundingsmake the surroundings cold.

uestions1. Convert the following temperature

to celsius scale:a. 300 K b. 573 K.

2. What is the physical state ofwater at:a. 250ºC b. 100ºC ?

3. For any substance, why does thetemperature remain constantduring the change of state?

4. Suggest a method to liquefyatmospheric gases.

1.5 EvaporationDo we always need to heat or change pressurefor changing the state of matter? Can youquote some examples from everyday life wherechange of state from liquid to vapour takesplace without the liquid reaching the boilingpoint? Water, when left uncovered, slowlychanges into vapour. Wet clothes dry up.What happens to water in the above twoexamples?

We know that particles of matter arealways moving and are never at rest. At agiven temperature in any gas, liquid or solid,there are particles with different amounts ofkinetic energy. In the case of liquids, a smallfraction of particles at the surface, havinghigher kinetic energy, is able to break awayfrom the forces of attraction of other particlesand gets converted into vapour. Thisphenomenon of change of a liquid intovapours at any temperature below its boilingpoint is called evaporation.

1.5.1 FACTORS AFFECTING EVAPORATION

Let us understand this with an activity.

Activity _____________1.14

• Take 5 mL of water in a test tube andkeep it near a window or under a fan.

• Take 5 mL of water in an open chinadish and keep it near a window orunder a fan.

• Take 5 mL of water in an open china

Q

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What happens when you pour someacetone (nail polish remover) on your palm?The particles gain energy from your palm orsurroundings and evaporate causing thepalm to feel cool.

After a hot sunny day, people sprinklewater on the roof or open ground becausethe large latent heat of vaporisation of waterhelps to cool the hot surface.

Can you cite some more examples fromdaily life where we can feel the effect of coolingdue to evaporation?Why should we wear cotton clothes insummer?

During summer, we perspire morebecause of the mechanism of our body whichkeeps us cool. We know that duringevaporation, the particles at the surface ofthe liquid gain energy from the surroundingsor body surface and change into vapour. Theheat energy equal to the latent heat ofvaporisation is absorbed from the bodyleaving the body cool. Cotton, being a goodabsorber of water helps in absorbing thesweat and exposing it to the atmosphere foreasy evaporation.

Why do we see water droplets on the outersurface of a glass containing ice-coldwater?

Let us take some ice-cold water in atumbler. Soon we will see water droplets onthe outer surface of the tumbler. The watervapour present in air, on coming in contactwith the cold glass of water, loses energy andgets converted to liquid state, which we seeas water droplets.

uestions1. Why does a desert cooler cool

better on a hot dry day?2. How does the water kept in an

earthen pot (matka) become coolduring summer?

3. Why does our palm feel coldwhen we put some acetone orpetrol or perfume on it?

4. Why are we able to sip hot tea ormilk faster from a saucer ratherthan a cup?

5. What type of clothes should wewear in summer?

Mor

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Now scientists are talking of five states of matter: Solid, Liquid, Gas, Plasma andBose-Einstein Condensate.

Plasma: The state consists of super energetic and super excited particles. Theseparticles are in the form of ionised gases. The fluorescent tube and neon sign bulbsconsist of plasma. Inside a neon sign bulb there is neon gas and inside a fluorescenttube there is helium gas or some other gas. The gas gets ionised, that is, gets chargedwhen electrical energy flows through it. This charging up creates a plasma glowinginside the tube or bulb. The plasma glows with a special colour depending on thenature of gas. The Sun and the stars glow because of the presence of plasma inthem. The plasma is created in stars because of very high temperature.

Bose-Einstein Condensate: In 1920, Indian physicist Satyendra Nath Bose haddone some calculations for a fifth state of matter. Building on his calculations, Albert

Einstein predicted a new state of matter – the Bose-Einstein Condensate (BEC). In 2001, Eric A. Cornell,Wolfgang Ketterle and Carl E. Wieman of USA receivedthe Nobel prize in physics for achieving “Bose-Einsteincondensation”. The BEC is formed by cooling a gas ofextremely low density, about one-hundred-thousandththe density of normal air, to super low temperatures.You can log on to www.chem4kids.com to get moreinformation on these fourth and fifth states of matter.

Q

S.N. Bose(1894-1974)

Albert Einstein(1879-1955)


Whatyou havelearnt

• Matter is made up of small particles.

• The matter around us exists in three states— solid, liquidand gas.

• The forces of attraction between the particles are maximum insolids, intermediate in liquids and minimum in gases.

• The spaces in between the constituent particles and kineticenergy of the particles are minimum in the case of solids,intermediate in liquids and maximum in gases.

• The arrangement of particles is most ordered in the case ofsolids, in the case of liquids layers of particles can slip andslide over each other while for gases, there is no order, particlesjust move about randomly.

• The states of matter are inter-convertible. The state of mattercan be changed by changing temperature or pressure.

• Sublimation is the change of gaseous state directly to solidstate without going through liquid state, and vice versa.

• Boiling is a bulk phenomenon. Particles from the bulk (whole)of the liquid change into vapour state.

• Evaporation is a surface phenomenon. Particles from thesurface gain enough energy to overcome the forces of attractionpresent in the liquid and change into the vapour state.

• The rate of evaporation depends upon the surface area exposedto the atmosphere, the temperature, the humidity and thewind speed.

• Evaporation causes cooling.

• Latent heat of vaporisation is the heat energy required to change1 kg of a liquid to gas at atmospheric pressure at its boilingpoint.

• Latent heat of fusion is the amount of heat energy required tochange 1 kg of solid into liquid at its melting point.

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Exercises1. Convert the following temperatures to the Celsius scale.

(a) 300 K (b) 573 K.

2. Convert the following temperatures to the Kelvin scale.

(a) 25°C (b) 373°C.

3. Give reason for the following observations.

(a) Naphthalene balls disappear with time without leaving anysolid.

(b) We can get the smell of perfume sitting several metres away.

4. Arrange the following substances in increasing order of forcesof attraction between the particles— water, sugar, oxygen.

5. What is the physical state of water at—

(a) 25°C (b) 0°C (c) 100°C ?

6. Give two reasons to justify—

(a) water at room temperature is a liquid.

(b) an iron almirah is a solid at room temperature.

7. Why is ice at 273 K more effective in cooling than water at thesame temperature?

8. What produces more severe burns, boiling water or steam?

9. Name A,B,C,D,E and F in the following diagram showing changein its state

Quantity Unit Symbol

Temperature kelvin KLength metre mMass kilogram kgWeight newton NVolume cubic metre m3

Density kilogram per cubic metre kg m–3

Pressure pascal Pa

• Some measurable quantities and their units to remember:


Group ActivityPrepare a model to demonstrate movement of particles in solids,liquids and gases.

For making this model you will need• A transparent jar

• A big rubber balloon or piece of stretchable rubber sheet

• A string

• Few chick-peas or black gram or dry green peas.

How to make?• Put the seeds in the jar.

• Sew the string to the centre of the rubber sheet and put sometape to keep it tied securely.

• Stretch and tie the rubber sheet on the mouth of the jar.

• Your model is ready. Now run your fingers up and down thestring by first tugging at it slowly and then rapidly.

Fig. 1.10: A model for happy converting of solid to liquid and liquid to gas.

Fig. 2.1: Some consumable items

Have you ever noticed the word ‘pure’written on the packs of these consumables?For a common person pure means having noadulteration. But, for a scientist all thesethings are actually mixtures of differentsubstances and hence not pure. For example,milk is actually a mixture of water, fat,proteins etc. When a scientist says thatsomething is pure, it means that all theconstituent particles of that substance arethe same in their chemical nature. A puresubstance consists of a single type ofparticles.

As we look around, we can see that mostof the matter around us exist as mixtures oftwo or more pure components, for example,sea water, minerals, soil etc. are all mixtures.

2.1 What is a Mixture?Mixtures are constituted by more than onekind of pure form of matter, known as asubstance. A substance cannot be separatedinto other kinds of matter by any physicalprocess. We know that dissolved sodiumchloride can be separated from water by the

physical process of evaporation. However,sodium chloride is itself a substance andcannot be separated by physical process intoits chemical constituents. Similarly, sugar isa substance because it contains only one kindof pure matter and its composition is the samethroughout.

Soft drink and soil are not singlesubstances. Whatever the source of asubstance may be, it will always have thesame characteristic properties.

Therefore, we can say that a mixturecontains more than one substance.

2.1.1 TYPES OF MIXTURES

Depending upon the nature of thecomponents that form a mixture, we can havedifferent types of mixtures.

Activity ______________ 2.1• Let us divide the class into groups A,

B, C and D.• Group A takes a beaker containing

50 mL of water and one spatula full ofcopper sulphate powder. Group B takes50 mL of water and two spatula full ofcopper sulphate powder in a beaker.

• Groups C and D can take differentamounts of copper sulphate andpotassium permanganate or commonsalt (sodium chloride) and mix the givencomponents to form a mixture.

• Report the observations on theuniformity in colour and texture.

• Groups A and B have obtained amixture which has a uniformcomposition throughout. Suchmixtures are called homogeneousmixtures or solutions. Some otherexamples of such mixtures are: (i) salt

How do we judge whether milk, ghee, butter,salt, spices, mineral water or juice that webuy from the market are pure?

22222IIIIISSSSS M M M M MATTERATTERATTERATTERATTER A A A A AROUNDROUNDROUNDROUNDROUND U U U U USSSSS P P P P PUREUREUREUREURE

Chapter

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in water and (ii) sugar in water.Compare the colour of the solutions ofthe two groups. Though both thegroups have obtained copper sulphatesolution but the intensity of colour ofthe solutions is different. This showsthat a homogeneous mixture can havea variable composition.

• Groups C and D have obtainedmixtures, which contain physicallydistinct parts and have non-uniformcompositions. Such mixtures are calledheterogeneous mixtures. Mixtures ofsodium chloride and iron filings, saltand sulphur, and oil and water areexamples of heterogeneous mixtures.

Activity ______________ 2.2• Let us again divide the class into four

groups – A, B, C and D.• Distribute the following samples to

each group:− Few crystals of copper sulphate to

group A.− One spatula full of copper

sulphate to group B.− Chalk powder or wheat flour to

group C.− Few drops of milk or ink to

group D.• Each group should add the given

sample in water and stir properly usinga glass rod. Are the particles in themixture visible?

• Direct a beam of light from a torchthrough the beaker containing themixture and observe from the front.Was the path of the beam of lightvisible?

• Leave the mixtures undisturbed for afew minutes (and set up the filtrationapparatus in the meantime). Is themixture stable or do the particles beginto settle after some time?

• Filter the mixture. Is there any residueon the filter paper?Discuss the results and form anopinion.

• Groups A and B have got a solution.• Group C has got a suspension.• Group D has got a colloidal solution.

Now, we shall learn about solutions,suspensions and colloidal solutions in thefollowing sections.

uestions1. What is meant by a pure

substance?2. List the points of differences

between homogeneous andheterogeneous mixtures.

2.2 What is a Solution?A solution is a homogeneous mixture of twoor more substances. You come across varioustypes of solutions in your daily life. Lemonade,soda water etc. are all examples of solutions.Usually we think of a solution as a liquid thatcontains either a solid, liquid or a gasdissolved in it. But, we can also have solidsolutions (alloys) and gaseous solutions (air).In a solution there is homogeneity at theparticle level. For example, lemonade tastesthe same throughout. This shows thatparticles of sugar or salt are evenly distributedin the solution.

QFig. 2.2: Filtration

Alloys: Alloys are homogeneousmixtures of metals and cannot beseparated into their components byphysical methods. But still, an alloyis considered as a mixture becauseit shows the properties of itsconstituents and can have variablecomposition. For example, brass isa mixture of approximately 30%zinc and 70% copper.

IS MATTER AROUND US PURE 15

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A solution has a solvent and a solute asits components. The component of thesolution that dissolves the other componentin it (usually the component present in largeramount) is called the solvent. The componentof the solution that is dissolved in the solvent(usually present in lesser quantity) is calledthe solute.

Examples:

(i) A solution of sugar in water is a solidin liquid solution. In this solution,sugar is the solute and water is thesolvent.

(ii) A solution of iodine in alcohol knownas ‘tincture of iodine’, has iodine (solid)as the solute and alcohol (liquid) asthe solvent.

(iii) Aerated drinks like soda water etc., aregas in liquid solutions. These containcarbon dioxide (gas) as solute andwater (liquid) as solvent.

(iv) Air is a mixture of gas in gas. Air is ahomogeneous mixture of a number ofgases. Its two main constituents are:oxygen (21%) and nitrogen (78%). Theother gases are present in very smallquantities.

Properties of a solution

• A solution is a homogeneous mixture.• The particles of a solution are smaller

than 1 nm (10-9 metre) in diameter. So,they cannot be seen by naked eyes.

• Because of very small particle size, theydo not scatter a beam of light passingthrough the solution. So, the path oflight is not visible in a solution.

• The solute particles cannot beseparated from the mixture by theprocess of filtration. The solute particlesdo not settle down when left undisturbed,that is, a solution is stable.

2.2.1 CONCENTRATION OF A SOLUTION

In activity 2.2, we observed that groups A andB obtained different shades of solutions. So,we understand that in a solution the relative

proportion of the solute and solvent can bevaried. Depending upon the amount of solutepresent in a solution, it can be called a dilute,concentrated or a saturated solution. Diluteand concentrated are comparative terms. Inactivity 2.2, the solution obtained by groupA is dilute as compared to that obtained bygroup B.

Activity ______________ 2.3• Take approximately 50 mL of water

each in two separate beakers.• Add salt in one beaker and sugar or

barium chloride in the second beakerwith continuous stirring.

• When no more solute can be dissolved,heat the contents of the beaker.

• Start adding the solute again.

Is the amount of salt and sugar or bariumchloride, that can be dissolved in water at agiven temperature, the same?

At any particular temperature, a solutionthat has dissolved as much solute as it iscapable of dissolving, is said to be a saturatedsolution. In other words, when no more solutecan be dissolved in a solution at a giventemperature, it is called a saturated solution.The amount of the solute present in thesaturated solution at this temperature iscalled its solubility.

If the amount of solute contained in asolution is less than the saturation level, it iscalled an unsaturated solution.

What would happen if you were to take asaturated solution at a certain temperatureand cool it slowly.

We can infer from the above activity thatdifferent substances in a given solvent havedifferent solubilities at the same temperature.

The concentration of a solution is theamount of solute present in a given amount(mass or volume) of solution, or the amountof solute dissolved in a given mass or volumeof solvent.Concentration of solution = Amount of solute/

Amount of solutionOr

Amount of solute/Amount of solvent


There are various ways of expressing theconcentration of a solution, but here we willlearn only two methods.

(i) Mass by mass percentage of a solution

Mass of solute= ×100

Mass of solution

(ii) Mass by volume percentage of a solution


Volume of solution

Example 2.1 A solution contains 40 g ofcommon salt in 320 g of water.Calculate the concentration in terms ofmass by mass percentage of thesolution.

Solution:

Mass of solute (salt) = 40 gMass of solvent (water) = 320 g

We know,Mass of solution = Mass of solute +

Mass of solvent= 40 g + 320 g= 360 g

Mass percentage of solution


Massof solution

40

= ×100 =11.1%360

2.2.2 What is a suspension?Non-homogeneous systems, like thoseobtained by group C in activity 2.2, in whichsolids are dispersed in liquids, are calledsuspensions. A suspension is a heterogeneousmixture in which the solute particles do notdissolve but remain suspended throughoutthe bulk of the medium. Particles of asuspension are visible to the naked eye.

Properties of a Suspension

• Suspension is a heterogeneousmixture.

• The particles of a suspension can beseen by the naked eye.

• The particles of a suspension scatter abeam of light passing through it andmake its path visible.

• The solute particles settle down whena suspension is left undisturbed, thatis, a suspension is unstable. They canbe separated from the mixture by theprocess of filtration.

2.2.3 WHAT IS A COLLOIDAL SOLUTION?The mixture obtained by group D in activity2.2 is called a colloid or a colloidal solution.The particles of a colloid are uniformly spreadthroughout the solution. Due to the relativelysmaller size of particles, as compared to thatof a suspension, the mixture appears to behomogeneous. But actually, a colloidalsolution is a heterogeneous mixture, forexample, milk.

Because of the small size of colloidalparticles, we cannot see them with nakedeyes. But, these particles can easily scatter abeam of visible light as observed in activity2.2. This scattering of a beam of light is calledthe Tyndall effect after the name of thescientist who discovered this effect.

Tyndall effect can also be observed whena fine beam of light enters a room through asmall hole. This happens due to the scatteringof light by the particles of dust and smoke inthe air.

Fig. 2.3: (a) Solution of copper sulphate does notshow Tyndall effect, (b) mixture of water

and milk shows Tyndall effect.

(a) (b)

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• They cannot be separated from themixture by the process of filtration. But,a special technique of separation knownas centrifugation (perform activity 2.5),can be used to separate the colloidalparticles.

The components of a colloidal solution arethe dispersed phase and the dispersionmedium. The solute-like component or thedispersed particles in a colloid form thedispersed phase, and the component in whichthe dispersed phase is suspended is knownas the dispersing medium. Colloids areclassified according to the state (solid, liquidor gas) of the dispersing medium and thedispersed phase. A few common examples aregiven in Table 2.1. From this table you cansee that they are very common everyday life.

uestions1. Differentiate between homogen-

eous and heterogeneous mixtureswith examples.

2. How are sol, solution andsuspension different from eachother?

3. To make a saturated solution,36 g of sodium chloride is dissolvedin 100 g of water at 293 K.Find its concentration at thistemperature.

Tyndall effect can be observed whensunlight passes through the canopy of adense forest. In the forest, mist contains tinydroplets of water, which act as particles ofcolloid dispersed in air.

Fig. 2.4: The Tyndall effect

Properties of a colloid

• A colloid is a heterogeneous mixture.• The size of particles of a colloid is too

small to be individually seen by nakedeyes.

• Colloids are big enough to scatter abeam of light passing through it andmake its path visible.

• They do not settle down when leftundisturbed, that is, a colloid is quitestable.

Table 2.1: Common examples of colloids

Dispersed Dispersing Type Examplephase Medium

Liquid Gas Aerosol Fog, clouds, mist

Solid Gas Aerosol Smoke, automobile exhaust

Gas Liquid Foam Shaving cream

Liquid Liquid Emulsion Milk, face cream

Solid Liquid Sol Milk of magnesia, mud

Gas Solid Foam Foam, rubber, sponge, pumice

Liquid Solid Gel Jelly, cheese, butter

Solid Solid Solid Sol Coloured gemstone, milky glass


2.3 Separating the Componentsof a Mixture

We have learnt that most of the naturalsubstances are not chemically pure. Differentmethods of separation are used to getindividual components from a mixture.Separation makes it possible to study anduse the individual components of a mixture.

Heterogeneous mixtures can be separatedinto their respective constituents by simplephysical methods like handpicking, sieving,filtration that we use in our day-to-day life.Sometimes special techniques have to be usedfor the separation of the components of amixture.

2.3.1 HOW CAN WE OBTAIN COLOUREDCOMPONENT (DYE) FROM BLUE/BLACK INK?

Activity ______________ 2.4• Fill half a beaker with water.• Put a watch glass on the mouth of the

beaker (Fig. 2.5).• Put few drops of ink on the watch glass.• Now start heating the beaker. We do

not want to heat the ink directly. Youwill see that evaporation is taking placefrom the watch glass.

• Continue heating as the evaporationgoes on and stop heating when you donot see any further change on thewatch glass.

• Observe carefully and record yourobservations.

Now answer• What do you think has got evaporated

from the watch glass?

• Is there a residue on the watch glass?

• What is your interpretation? Is ink asingle substance (pure) or is it amixture?

We find that ink is a mixture of a dye inwater. Thus, we can separate the volatilecomponent (solvent) from its non-volatilesolute by the method of evaporation.

2.3.2 HOW CAN WE SEPARATE CREAM

FROM MILK?Now-a-days, we get full-cream, toned anddouble-toned varieties of milk packed in poly-packs or tetra packs in the market. Thesevarieties of milk contain different amountsof fat.

Activity ______________ 2.5• Take some full-cream milk in a test

tube.• Centrifuge it by using a centrifuging

machine for two minutes. If acentrifuging machine is not availablein the school, you can do this activityat home by using a milk churner, usedin the kitchen.

• If you have a milk dairy nearby, visit itand ask (i) how they separate creamfrom milk and (ii) how they makecheese (paneer) from milk.

Now answer• What do you observe on churning the

milk?

• Explain how the separation of creamfrom milk takes place.

Sometimes the solid particles in a liquidare very small and pass through a filter paper.For such particles the filtration techniquecannot be used for separation. Such mixturesFig. 2.5: Evaporation

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are separated by centrifugation. The principleis that the denser particles are forced to thebottom and the lighter particles stay at thetop when spun rapidly.

Applications

• Used in diagnostic laboratories forblood and urine tests.

• Used in dairies and home to separatebutter from cream.

• Used in washing machines to squeezeout water from wet clothes.

2.3.3 HOW CAN WE SEPARATE A MIXTURE

OF TWO IMMISCIBLE LIQUIDS?

Activity ______________ 2.6• Let us try to separate kerosene oil from

water using a separating funnel.• Pour the mixture of kerosene oil and

water in a separating funnel (Fig. 2.6).• Let it stand undisturbed for sometime

so that separate layers of oil and waterare formed.

• Open the stopcock of the separatingfunnel and pour out the lower layer ofwater carefully.

• Close the stopcock of the separatingfunnel as the oil reaches the stop-cock.

Applications

• To separate mixture of oil and water.• In the extraction of iron from its ore,

the lighter slag is removed from the topby this method to leave the molten ironat the bottom in the furnace.

The principle is that immiscible liquidsseparate out in layers depending on theirdensities.


OF SALT AND AMMONIUM CHLORIDE?We have learnt in chapter 1 that ammoniumchloride changes directly from solid togaseous state on heating. So, to separate suchmixtures that contain a sublimable volatilecomponent from a non-sublimable impurity(salt in this case), the sublimation process isused (Fig. 2.7). Some examples of solids whichsublime are ammonium chloride, camphor,naphthalene and anthracene.

Fig. 2.7: Separation of ammonium chloride and saltby sublimationFig. 2.6: Separation of immiscible liquids


2.3.5 IS THE DYE IN BLACK INK A SINGLECOLOUR?

Activity ______________ 2.7• Take a thin strip of filter paper.• Draw a line on it using a pencil,

approximately 3 cm above the loweredge [Fig. 2.8 (a)].

• Put a small drop of ink (water soluble,that is, from a sketch pen or fountainpen) at the centre of the line. Let it dry.

• Lower the filter paper into a jar/glass/beaker/test tube containing water sothat the drop of ink on the paper is justabove the water level, as shown in Fig.2.8(b) and leave it undisturbed.

• Watch carefully, as the water rises upon the filter paper. Record yourobservations.

This process of separation of componentsof a mixture is known as chromatography.Kroma in Greek means colour. This techniquewas first used for separation of colours, sothis name was given. Chromatography is thetechnique used for separation of those solutesthat dissolve in the same solvent.

With the advancement in technology,newer techniques of chromatography havebeen developed. You will study aboutchromatography in higher classes.

Applications

To separate• colours in a dye• pigments from natural colours• drugs from blood.


OF TWO MISCIBLE LIQUIDS?

Activity ______________ 2.8• Let us try to separate acetone and water

from their mixture.• Take the mixture in a distillation flask.

Fit it with a thermometer.• Arrange the apparatus as shown in

Fig. 2.9.• Heat the mixture slowly keeping a close

watch at the thermometer.• The acetone vaporises, condenses in

the condenser and can be collectedfrom the condenser outlet.

• Water is left behind in the distillationflask.

Fig. 2.8: Separation of dyes in black ink using

chromatography

Now answer• What do you observe on the filter paper

as the water rises on it?

• Do you obtain different colours on thefilter paper strip?

• What according to you, can be thereason for the rise of the coloured spoton the paper strip?

The ink that we use has water as thesolvent and the dye is soluble in it. As thewater rises on the filter paper it takes alongwith it the dye particles. Usually, a dye is amixture of two or more colours. The colouredcomponent that is more soluble in water, risesfaster and in this way the colours getseparated.

Fig.2.9: Separation of two miscible liquids bydistillation

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Now answer• What do you observe as you start

heating the mixture?

• At what temperature does thethermometer reading become constantfor some time?

• What is the boiling point of acetone?

• Why do the two components separate?

This method is called distillation. It is usedfor the separation of components of a mixturecontaining two miscible liquids that boilwithout decomposition and have sufficientdifference in their boiling points.

To separate a mixture of two or moremiscible liquids for which the difference inboiling points is less than 25 K, fractionaldistillation process is used, for example, forthe separation of different gases from air,different factions from petroleum productsetc. The apparatus is similar to that for simpledistillation, except that a fractionatingcolumn is fitted in between the distillationflask and the condenser.

A simple fractionating column is a tubepacked with glass beads. The beads providesurface for the vapours to cool and condenserepeatedly, as shown in Fig. 2.10.

2.3.7 HOW CAN WE OBTAIN DIFFERENT

GASES FROM AIR ?Air is a homogeneous mixture and can beseparated into its components by fractionaldistillation. The flow diagram (Fig. 2.11)shows the steps of the process.

Fig. 2.10: Fractional distillation

Fig. 2.11: Flow diagram shows the process ofobtaining gases from air

If we want oxygen gas from air (Fig. 2.12),we have to separate out all the other gasespresent in the air. The air is compressed byincreasing the pressure and is then cooledby decreasing the temperature to get liquidair. This liquid air is allowed to warm-upslowly in a fractional distillation column,where gases get separated at different heightsdepending upon their boiling points.

Answer the following:

• Arrange the gases present in air inincreasing order of their boiling points.

• Which gas forms the liquid first as theair is cooled?


it. To remove these impurities, the process ofcrystallisation is used. Crystallisation is aprocess that separates a pure solid in theform of its crystals from a solution.Crystallisation technique is better thansimple evaporation technique as –

• some solids decompose or some, likesugar, may get charred on heating todryness.

• some impurities may remain dissolvedin the solution even after filtration. Onevaporation these contaminate thesolid.

Applications

• Purification of salt that we get from seawater.

• Separation of crystals of alum (phitkari)from impure samples.

Thus, by choosing one of the abovemethods according to the nature of thecomponents of a mixture, we get a puresubstance. With advancements in technologymany more methods of separation techniqueshave been devised.

In cities, drinking water is supplied fromwater works. A flow diagram of a typical waterworks is shown in Fig. 2.13. From this figurewrite down the processes involved to get thesupply of drinking water to your home fromthe water works and discuss it in your class.

Fig. 2.12: Separation of components of air

2.3.8 HOW CAN WE OBTAIN PURE COPPER

SULPHATE FROM AN IMPURE SAMPLE?

Activity ______________ 2.9• Take some (approximately 5 g) impure

sample of copper sulphate in a chinadish.

• Dissolve it in minimum amount ofwater.

• Filter the impurities out.• Evaporate water from the copper

sulphate solution so as to get asaturated solution.

• Cover the solution with a filter paperand leave it undisturbed at roomtemperature to cool slowly for a day.

• You will obtain the crystals of coppersulphate in the china dish.

• This process is called crystallisation.

Now answer• What do you observe in the china dish?

• Do the crystals look alike?

• How will you separate the crystals fromthe liquid in the china dish?

The crystallisation method is used topurify solids. For example, the salt we getfrom sea water can have many impurities in

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uestions1. How will you separate a mixture

containing kerosene and petrol(difference in their boiling pointsis more than 25ºC), which aremiscible with each other?

2. Name the technique to separate(i) butter from curd,(ii) salt from sea-water,(iii) camphor from salt.

3. What type of mixtures areseparated by the technique ofcrystallisation?

2.4 Physical and ChemicalChanges

To understand the difference between a puresubstance and a mixture, let us understandthe difference between a physical and achemical change.

In the previous chapter, we have learntabout a few physical properties of matter. Theproperties that can be observed and specifiedlike colour, hardness, rigidity, fluidity,density, melting point, boiling point etc. arethe physical properites.

The interconversion of states is a physicalchange because these changes occur withouta change in composition and no change inthe chemical nature of the substance.Although ice, water and water vapour all lookdifferent and display different physicalproperties, they are chemically the same.

QFig. 2.13: Water purification system in water works

Both water and cooking oil are liquid buttheir chemical characteristics are different.They differ in odour and inflammability. Weknow that oil burns in air whereas waterextinguishes fire. It is this chemical propertyof oil that makes it different from water.Burning is a chemical change. During thisprocess one substance reacts with anotherto undergo a change in chemical composition.Chemical change brings change in thechemical properties of matter and we get newsubstances. A chemical change is also calleda chemical reaction.

During burning of a candle, both physicaland chemical changes take place. Can youdistinguish these?

uestions1. Classify the following as

chemical or physical changes:• cutting of trees,• melting of butter in a pan,• rusting of almirah,• boiling of water to form steam,• passing of electric current,

through water and the waterbreaking down into hydrogenand oxygen gases,

• dissolving common salt inwater,

• making a fruit salad with rawfruits, and

• burning of paper and wood.2. T ry segregating the things

around you as pure substancesor mixtures.

Q


2.5 What are the Types of PureSubstances?

On the basis of their chemical composition,substances can be classified either aselements or compounds.

2.5.1 ELEMENTS

Robert Boyle was the first scientist to use theterm element in 1661. Antoine LaurentLavoisier (1743-94), a French chemist, wasthe first to establish an experimentally usefuldefinition of an element. He defined anelement as a basic form of matter that cannotbe broken down into simpler substances bychemical reactions.

Elements can be normally divided intometals, non-metals and metalloids.

Metals usually show some or all of thefollowing properties:

• They have a lustre (shine).• They have silvery-grey or golden-yellow

colour.• They conduct heat and electricity.• They are ductile (can be drawn into

wires).• They are malleable (can be hammered

into thin sheets).• They are sonorous (make a ringing

sound when hit).Examples of metals are gold, silver,

copper, iron, sodium, potassium etc. Mercuryis the only metal that is liquid at roomtemperature.

Non-metals usually show some or all ofthe following properties:

• They display a variety of colours.• They are poor conductors of heat and

electricity.• They are not lustrous, sonorous or

malleable.Examples of non-metals are hydrogen,

oxygen, iodine, carbon (coal, coke),bromine, chlorine etc. Some elements haveintermediate properties between those ofmetals and non-metals, they arecalled metalloids; examples are boron,silicon, germanium etc.

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• The number of elements knownat present are more than 100.Ninety-two elements are naturallyoccurring and the rest are man-made.

• Majority of the elements are solid.• Eleven elements are in gaseous

state at room temperature.• Two elements are liquid at room

temperature–mercury andbromine.

• Elements, gallium and cesiumbecome liquid at a temperatureslightly above room temperature(303 K).

2.5.2 COMPOUNDS

A compound is a substance composed of twoor more elements, chemically combined withone another in a fixed proportion.What do we get when two or more elementsare combined?

Activity _____________2.10• Divide the class into two groups. Give

50 g of iron filings and 3 g of sulphurpowder in a china dish to both thegroups.

Group I• Mix and crush iron filings and sulphur

powder.

Group II• Mix and crush iron filings and sulphur

powder. Heat this mixture strongly tillred hot. Remove from flame and let themixture cool.

Groups I and II• Check for magnetism in the material

obtained. Bring a magnet near thematerial and check if the material isattracted towards the magnet.

• Compare the texture and colour of thematerial obtained by the groups.

• Add carbon disulphide to one part ofthe material obtained. Stir well andfilter.

• Add dilute sulphuric acid or dilutehydrochloric acid to the other part of

SCIENCE26

the material obtained.(Note: teachersupervision is necessary for thisactivity).

• Perform all the above steps with boththe elements (iron and sulphur)separately.

Now answer• Did the material obtained by the two

groups look the same?

• Which group has obtained a materialwith magnetic properties?

• Can we separate the components of thematerial obtained?

• On adding dilute sulphuric acid ordilute hydrochloric acid, did both thegroups obtain a gas? Did the gas inboth the cases smell the same ordifferent?

The gas obtained by Group I is hydrogen,it is colourless, odourless and combustible–it is not advised to do the combustion test forhydrogen in the class. The gas obtained byGroup II is hydrogen sulphide. It is acolourless gas with the smell of rotten eggs.

You must have observed that the productsobtained by both the groups show differentproperties, though the starting materials werethe same. Group I has carried out the activityinvolving a physical change whereas in caseof Group II, a chemical change (a chemicalreaction) has taken place.

• The material obtained by group I is amixture of the two substances. Thesubstances given are the elements– ironand sulphur.

• The properties of the mixture are thesame as that of its constituents.

• The material obtained by group II is acompound.

• On heating the two elements stronglywe get a compound, which has totallydifferent properties compared to thecombining elements.

• The composition of a compound is thesame throughout. We can also observethat the texture and the colour of thecompound are the same throughout.Thus, we can summarise the physicaland chemical nature of matter in thefollowing graphical organiser :

Table 2.2: Mixtures and Compounds

Mixtures Compounds

1. Elements or compounds just mix 1. Elements react to form new compounds.together to form a mixture and nonew compound is formed.

2. A mixture has a variable composition. 2. The composition of each new substanceis always fixed.

3, A mixture shows the properties of the 3. The new substance has totally differentconstituent substances. properties.

4. The constituents can be seperated 4. The constituents can be separated onlyfairly easily by physical methods. by chemical or electrochemical

reactions.


Whatyou havelearnt• A mixture contains more than one substance (element and/or

compound) mixed in any proportion.• Mixtures can be separated into pure substances using

appropriate separation techniques.• A solution is a homogeneous mixture of two or more substances.

The major component of a solution is called the solvent, andthe minor, the solute.

• The concentration of a solution is the amount of solute presentper unit volume or per unit mass of the solution/solvent.

• Materials that are insoluble in a solvent and have particlesthat are visible to naked eyes, form a suspension. A suspensionis a heterogeneous mixture.

• Colloids are heterogeneous mixtures in which the particle sizeis too small to be seen with the naked eye, but is big enough toscatter light. Colloids are useful in industry and daily life. Theparticles are called the dispersed phase and the medium inwhich they are distributed is called the dispersion medium.

• Pure substances can be elements or compounds. An elementis a form of matter that cannot be broken down by chemicalreactions into simpler substances. A compound is a substancecomposed of two or more different types of elements, chemicallycombined in a fixed proportion.

• Properties of a compound are different from its constituentelements, whereas a mixture shows the properties of itsconstituting elements or compounds.

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Exercises1. Which separation techniques will you apply for the separation

of the following?

(a) Sodium chloride from its solution in water.

(b) Ammonium chloride from a mixture containing sodiumchloride and ammonium chloride.

(c) Small pieces of metal in the engine oil of a car.

(d) Different pigments from an extract of flower petals.

(e) Butter from curd.

(f) Oil from water.

(g) Tea leaves from tea.

(h) Iron pins from sand.

(i) Wheat grains from husk.

(j) Fine mud particles suspended in water.

2. Write the steps you would use for making tea. Use the wordssolution, solvent, solute, dissolve, soluble, insoluble, filtrateand residue.

3. Pragya tested the solubility of three different substances atdifferent temperatures and collected the data as given below(results are given in the following table, as grams of substancedissolved in 100 grams of water to form a saturated solution).

(a) What mass of potassium nitrate would be needed toproduce a saturated solution of potassium nitrate in50 grams of water at 313 K?

(b) Pragya makes a saturated solution of potassium chloridein water at 353 K and leaves the solution to cool at roomtemperature. What would she observe as the solutioncools? Explain.

(c) Find the solubility of each salt at 293 K. Which salt hasthe highest solubility at this temperature?

(d) What is the effect of change of temperature on thesolubility of a salt?

Temperature in K

283 293 313 333 353

Potassium nitrate 21 32 62 106 167

Sodium chloride 36 36 36 37 37

Potassium chloride 35 35 40 46 54

Ammonium chloride 24 37 41 55 66

Substance Dissolved


4. Explain the following giving examples.

(a) saturated solution

(b) pure substance

(c) colloid

(d) suspension

5. Classify each of the following as a homogeneous orheterogeneous mixture.

soda water, wood, air, soil, vinegar, filtered tea.

6. How would you confirm that a colourless liquid given to you ispure water?

7. Which of the following materials fall in the category of a “puresubstance”?

(a) Ice

(b) Milk

(c) Iron

(d) Hydrochloric acid

(e) Calcium oxide

(f) Mercury

(g) Brick(h) Wood(i) Air.

8. Identify the solutions among the following mixtures.(a) Soil(b) Sea water(c) Air

(d) Coal(e) Soda water.

9. Which of the following will show “Tyndall effect”?(a) Salt solution(b) Milk(c) Copper sulphate solution(d) Starch solution.

10. Classify the following into elements, compounds and mixtures.(a) Sodium(b) Soil(c) Sugar solution(d) Silver

(e) Calcium carbonate(f) Tin(g) Silicon

SCIENCE30

(h) Coal

(i) Air

(j) Soap

(k) Methane

(l) Carbon dioxide

(m) Blood

11. Which of the following are chemical changes?

(a) Growth of a plant

(b) Rusting of iron

(c) Mixing of iron filings and sand

(d) Cooking of food

(e) Digestion of food

(f) Freezing of water

(g) Burning of a candle.

Group ActivityTake an earthen pot (mutka), some pebbles and sand. Design asmall-scale filtration plant that you could use to clean muddywater.

Ancient Indian and Greek philosophers havealways wondered about the unknown andunseen form of matter. The idea of divisibilityof matter was considered long back in India,around 500 BC. An Indian philosopherMaharishi Kanad, postulated that if we go ondividing matter (padarth), we shall get smallerand smaller particles. Ultimately, a time willcome when we shall come across the smallestparticles beyond which further division willnot be possible. He named these particlesParmanu. Another Indian philosopher,Pakudha Katyayama, elaborated this doctrineand said that these particles normally existin a combined form which gives us variousforms of matter.

Around the same era, ancient Greekphilosophers – Democritus and Leucippussuggested that if we go on dividing matter, astage will come when particles obtainedcannot be divided further. Democritus calledthese indivisible particles atoms (meaningindivisible). All this was based onphilosophical considerations and not muchexperimental work to validate these ideascould be done till the eighteenth century.

By the end of the eighteenth century,scientists recognised the difference betweenelements and compounds and naturallybecame interested in finding out how and whyelements combine and what happens whenthey combine.

Antoine L. Lavoisier laid the foundationof chemical sciences by establishing twoimportant laws of chemical combination.

3.1 Laws of Chemical CombinationThe following two laws of chemicalcombination were established after much

experimentations by Lavoisier andJoseph L. Proust.

3.1.1 LAW OF CONSERVATION OF MASS

Is there a change in mass when a chemicalchange (chemical reaction) takes place?

Activity ______________ 3.1• Take one of the following sets, X and Y

of chemicals–X Y

(i) copper sulphate sodium carbonate(ii) barium chloride sodium sulphate(iii) lead nitrate sodium chloride

• Prepare separately a 5% solution of anyone pair of substances listed under Xand Y in water.

• Take a little amount of solution of Y ina conical flask and some solution of Xin an ignition tube.

• Hang the ignition tube in the flaskcarefully; see that the solutions do notget mixed. Put a cork on the flask(see Fig. 3.1).

Fig. 3.1: Ignition tube containing solution of X, dippedin a conical flask containing solution of Y.

33333AAAAATOMSTOMSTOMSTOMSTOMS ANDANDANDANDAND M M M M MOLECULESOLECULESOLECULESOLECULESOLECULES

Chapter

SCIENCE32

• Weigh the flask with its contentscarefully.

• Now tilt and swirl the flask, so that thesolutions X and Y get mixed.

• Weigh again.• What happens in the reaction flask?• Do you think that a chemical reaction

has taken place?• Why should we put a cork on the

mouth of the flask?• Does the mass of the flask and its

contents change?

Law of conservation of mass states thatmass can neither be created nor destroyedin a chemical reaction.

3.1.2 LAW OF CONSTANT PROPORTIONS

Lavoisier, along with other scientists, notedthat many compounds were composed of twoor more elements and each such compoundhad the same elements in the sameproportions, irrespective of where thecompound came from or who prepared it.

In a compound such as water, the ratio ofthe mass of hydrogen to the mass of oxygenis always 1:8, whatever the source of water.Thus, if 9 g of water is decomposed, 1 g ofhydrogen and 8 g of oxygen are alwaysobtained. Similarly in ammonia, nitrogen andhydrogen are always present in the ratio 14:3by mass, whatever the method or the sourcefrom which it is obtained.

This led to the law of constant proportionswhich is also known as the law of definiteproportions. This law was stated by Proustas “In a chemical substance the elements arealways present in definite proportions bymass”.

The next problem faced by scientists wasto give appropriate explanations of these laws.British chemist John Dalton provided thebasic theory about the nature of matter.Dalton picked up the idea of divisibility ofmatter, which was till then just a philosophy.He took the name ‘atoms’ as given by theGreeks and said that the smallest particlesof matter are atoms. His theory was basedon the laws of chemical combination. Dalton’s

atomic theory provided an explanation for thelaw of conservation of mass and the law ofdefinite proportions.

John Dalton was born ina poor weaver’s family in1766 in England. Hebegan his career as ateacher at the age oftwelve. Seven years laterhe became a schoolprincipal. In 1793, Daltonleft for Manchester toteach mathematics,physics and chemistry ina college. He spent most of his life thereteaching and researching. In 1808, hepresented his atomic theory which was aturning point in the study of matter.

According to Dalton’s atomic theory, allmatter, whether an element, a compound ora mixture is composed of small particlescalled atoms. The postulates of this theorymay be stated as follows:

(i) All matter is made of very tiny particlescalled atoms.

(ii) Atoms are indivisible particles, whichcannot be created or destroyed in achemical reaction.

(iii) Atoms of a given element are identicalin mass and chemical properties.

(iv) Atoms of different elements havedif ferent masses and chemicalproperties.

(v) Atoms combine in the ratio of smallwhole numbers to form compounds.

(vi) The relative number and kinds ofatoms are constant in a givencompound.

You will study in the next chapter that allatoms are made up of still smaller particles.

uestions1. In a reaction, 5.3 g of sodium

carbonate reacted with 6 g ofethanoic acid. The products were2.2 g of carbon dioxide, 0.9 gwater and 8.2 g of sodiumethanoate. Show that these

Q

John Dalton

ATOMS AND MOLECULES 33

We might think that if atoms are soinsignificant in size, why should we careabout them? This is because our entire worldis made up of atoms. We may not be able tosee them, but they are there, and constantlyaffecting whatever we do. Through moderntechniques, we can now produce magnifiedimages of surfaces of elements showingatoms.

observations are in agreementwith the law of conservation ofmass.sodium carbonate + ethanoic acid→ sodium ethanoate + carbondioxide + water

2. Hydrogen and oxygen combine inthe ratio of 1:8 by mass to formwater. What mass of oxygen gaswould be required to reactcompletely with 3 g of hydrogengas?

3. Which postulate of Dalton’satomic theory is the result of thelaw of conservation of mass?

4. Which postulate of Dalton’satomic theory can explain the lawof definite proportions?

3.2 What is an Atom?Have you ever observed a mason buildingwalls, from these walls a room and then acollection of rooms to form a building? Whatis the building block of the huge building?What about the building block of an ant-hill?It is a small grain of sand. Similarly, thebuilding blocks of all matter are atoms.

How big are atoms?

Atoms are very small, they are smaller thananything that we can imagine or comparewith. More than millions of atoms whenstacked would make a layer barely as thickas this sheet of paper.

Atomic radius is measured in nanometres.1/10 9 m = 1 nm

1 m = 109 nm

Relative Sizes

Radii (in m) Example10–10 Atom of hydrogen

10–9 Molecule of water

10–8 Molecule of haemoglobin

10–4 Grain of sand

10–2 Ant

10–1 Watermelon

Fig. 3.2: An image of the surface of silicon

3.2.1 WHAT ARE THE MODERN DAY

SYMBOLS OF ATOMS OF DIFFERENT

ELEMENTS?Dalton was the first scientist to use thesymbols for elements in a very specific sense.When he used a symbol for an element healso meant a definite quantity of that element,that is, one atom of that element. Berziliussuggested that the symbols of elements bemade from one or two letters of the name ofthe element.

Fig. 3.3: Symbols for some elements as proposed byDalton

SCIENCE34

In the beginning, the names of elementswere derived from the name of the place wherethey were found for the first time. Forexample, the name copper was taken fromCyprus. Some names were taken fromspecific colours. For example, gold was takenfrom the English word meaning yellow.Now-a-days, IUPAC (International Union ofPure and Applied Chemistry) approves namesof elements. Many of the symbols are the firstone or two letters of the element’s name inEnglish. The first letter of a symbol is alwayswritten as a capital letter (uppercase) and thesecond letter as a small letter (lowercase).

For example

(i) hydrogen, H(ii) aluminium, Al and not AL(iii) cobalt, Co and not CO.

Symbols of some elements are formedfrom the first letter of the name and a letter,appearing later in the name. Examples are:(i) chlorine, Cl, (ii) zinc, Zn etc.

Other symbols have been taken from thenames of elements in Latin, German or Greek.For example, the symbol of iron is Fe from itsLatin name ferrum, sodium is Na fromnatrium, potassium is K from kalium.Therefore, each element has a name and aunique chemical symbol.

Table 3.1: Symbols for some elements

Element Symbol Element Symbol Element Symbol

Aluminium Al Copper Cu Nitrogen NArgon Ar Fluorine F Oxygen OBarium Ba Gold Au Potassium KBoron B Hydrogen H Silicon SiBromine Br Iodine I Silver AgCalcium Ca Iron Fe Sodium NaCarbon C Lead Pb Sulphur SChlorine Cl Magnesium Mg Uranium U

Cobalt Co Neon Ne Zinc Zn

(The above table is given for you to referto whenever you study about elements. Donot bother to memorise all in one go. With

the passage of time and repeated usage youwill automatically be able to reproduce thesymbols).

3.2.2 ATOMIC MASS

The most remarkable concept that Dalton’satomic theory proposed was that of the atomicmass. According to him, each element had acharacteristic atomic mass. The theory couldexplain the law of constant proportions sowell that scientists were prompted to measurethe atomic mass of an atom. Sincedetermining the mass of an individual atomwas a relatively difficult task, relative atomicmasses were determined using the laws ofchemical combinations and the compoundsformed.

Let us take the example of a compound,carbon monoxide (CO) formed by carbon andoxygen. It was observed experimentally that3 g of carbon combines with 4 g of oxygen toform CO. In other words, carbon combineswith 4/3 times its mass of oxygen. Supposewe define the atomic mass unit (earlierabbreviated as ‘amu’, but according to thelatest IUPAC recommendations, it is nowwritten as ‘u’ – unified mass) as equal to themass of one carbon atom, then we would

assign carbon an atomic mass of 1.0 u andoxygen an atomic mass of 1.33 u. However, itis more convenient to have these numbers


as whole numbers or as near to a wholenumbers as possible. While searching forvarious atomic mass units, scientists initiallytook 1/16 of the mass of an atom of naturallyoccurring oxygen as the unit. This wasconsidered relevant due to two reasons:• oxygen reacted with a large number of

elements and formed compounds.• this atomic mass unit gave masses of

most of the elements as whole numbers.However, in 1961 for a universally

accepted atomic mass unit, carbon-12 isotopewas chosen as the standard reference formeasuring atomic masses. One atomic massunit is a mass unit equal to exactly one-twelfth (1/12th) the mass of one atom ofcarbon-12. The relative atomic masses of allelements have been found with respect to anatom of carbon-12.

Imagine a fruit seller selling fruits withoutany standard weight with him. He takes awatermelon and says, “this has a mass equalto 12 units” (12 watermelon units or 12 fruitmass units). He makes twelve equal pieces ofthe watermelon and finds the mass of eachfruit he is selling, relative to the mass of onepiece of the watermelon. Now he sells hisfruits by relative fruit mass unit (fmu), as inFig. 3.4.

Fig. 3.4 : (a) Watermelon, (b) 12 pieces, (c) 1/12 ofwatermelon, (d) how the fruit seller canweigh the fruits using pieces of watermelon

Similarly, the relative atomic mass of theatom of an element is defined as the average

mass of the atom, as compared to 1/12th themass of one carbon-12 atom.

Table 3.2: Atomic masses ofa few elements

Element Atomic Mass (u)

Hydrogen 1

Carbon 12

Nitrogen 14

Oxygen 16

Sodium 23

Magnesium 24

Sulphur 32

Chlorine 35.5

Calcium 40

3.2.3 HOW DO ATOMS EXIST?Atoms of most elements are not able to existindependently. Atoms form molecules andions. These molecules or ions aggregate inlarge numbers to form the matter that we cansee, feel or touch.

uestions1. Define the atomic mass unit.2. Why is it not possible to see an

atom with naked eyes?

3.3 What is a Molecule?A molecule is in general a group of two ormore atoms that are chemically bondedtogether, that is, tightly held together byattractive forces. A molecule can be definedas the smallest particle of an element or acompound that is capable of an independentexistence and shows all the properties of thatsubstance. Atoms of the same element or ofdifferent elements can join together to formmolecules.

Q

SCIENCE36

3.3.1 MOLECULES OF ELEMENTS

The molecules of an element are constitutedby the same type of atoms. Molecules of manyelements, such as argon (Ar), helium (He) etc.are made up of only one atom of that element.But this is not the case with most of the non-metals. For example, a molecule of oxygenconsists of two atoms of oxygen and hence itis known as a diatomic molecule, O2. If 3atoms of oxygen unite into a molecule, insteadof the usual 2, we get ozone. The number ofatoms constituting a molecule is known asits atomicity.

Molecules of metals and some otherelements, such as carbon, do not have asimple structure but consist of a very largeand indefinite number of atoms bondedtogether.

Let us look at the atomicity of someelements.

Table 3.3 : Atomicity of someelements

Types of Name AtomicityElement

Non-Metal Argon Monoatomic

Helium Monoatomic

Oxygen Diatomic

Hydrogen Diatomic

Nitrogen Diatomic

Chlorine Diatomic

Phosphorus Tetra-atomic

Sulphur Poly-atomic

Metal Sodium Monoatomic

Iron Monoatomic

Aluminium Monoatomic

Copper Monoatomic

Table 3.4 : Molecules of somecompounds

Compound Combining RatioElements by

Mass

Water Hydrogen, Oxygen 1:8

Ammonia Nitrogen, Hydrogen 14:3

CarbonDioxide Carbon, Oxygen 3:8

Activity ______________ 3.2• Refer to Table 3.4 for ratio by mass of

atoms present in molecules and Table3.2 for atomic masses of elements. Findthe ratio by number of the atoms ofelements in the molecules ofcompounds given in Table 3.4.

• The ratio by number of atoms for awater molecule can be found as follows:

Element Ratio Atomic Mass Simplestby mass ratio/ ratio

mass (u) atomicmass

H 1 11

1=1 2

O 8 168

16=

1

21

• Thus, the ratio by number of atoms forwater is H:O = 2:1.

3.3.3 WHAT IS AN ION?Compounds composed of metals and non-metals contain charged species. The chargedspecies are known as ions. An ion is a chargedparticle and can be negatively or positivelycharged. A negatively charged ion is calledan ‘anion’ and the positively charged ion, a‘cation’. Take, for example, sodium chloride(NaCl). Its constituent particles are positivelycharged sodium ions (Na+) and negativelycharged chloride ions (Cl–). Ions may consistof a single charged atom or a group of atoms

3.3.2 MOLECULES OF COMPOUNDS

Atoms of different elements join together indefinite proportions to form molecules ofcompounds. Few examples are given inTable 3.4.


that have a net charge on them. A group ofatoms carrying a charge is known as apolyatomic ion. We shall learn more aboutthe formation of ions in Chapter 4.

exercise, we need to learn the symbols andcombining capacity of the elements.

The combining power (or capacity) of anelement is known as its valency. Valency canbe used to find out how the atoms of anelement will combine with the atom(s) ofanother element to form a chemicalcompound. The valency of the atom of anelement can be thought of as hands or armsof that atom.

Human beings have two arms and anoctopus has eight. If one octopus has to catchhold of a few people in such a manner thatall the eight arms of the octopus and botharms of all the humans are locked, how manyhumans do you think the octopus can hold?Represent the octopus with O and humanswith H. Can you write a formula for thiscombination? Do you get OH4 as the formula?The subscript 4 indicates the number ofhumans held by the octopus.

The valencies of some simple andpolyatomic ions are given in Table 3.6. Wewill learn more about valency in the nextchapter.

* Some elements show more than one valency. A Roman numeral shows their valency in a bracket.

Table 3.5: Some ionic compounds

Ionic Constituting RatioCompound Elements by

Mass

Calcium oxide Calcium andoxygen 5:2

Magnesium Magnesiumsulphide and sulphur 3:4

Sodium Sodiumchloride and chlorine 23:35.5

3.4 Writing Chemical FormulaeThe chemical formula of a compound is asymbolic representation of its composition.The chemical formulae of dif ferentcompounds can be written easily. For this

Table 3.6: Some common, simple and polyatomic ions

Vale- Name of Symbol Non- Symbol Polyatomic Symbolncy ion metallic ions

element

1. Sodium Na+ Hydrogen H+ Ammonium NH4+

Potassium K+ Hydride H- Hydroxide OH–

Silver Ag+ Chloride Cl- Nitrate NO3–

Copper (I)* Cu+ Bromide Br- HydrogenIodide I– carbonate HCO3

–

2. Magnesium Mg2+ Oxide O2- Carbonate CO32–

Calcium Ca2+ Sulphide S2- Sulphite SO32–

Zinc Zn2+ Sulphate SO42–

Iron (II)* Fe2+

Copper (II)* Cu2+

3. Aluminium Al3+ Nitride N3- Phosphate PO43–

Iron (III)* Fe3+

SCIENCE38

3. Formula of carbon tetrachloride

The formulae of ionic compounds are simplythe whole number ratio of the positive tonegative ions in the structure.

For magnesium chloride, we write thesymbol of cation (Mg2+) first followed by thesymbol of anion (Cl-). Then their charges arecriss-crossed to get the formula.

4. Formula of magnesium chloride

Formula : MgCl2

Thus, in magnesium chloride, there aretwo chloride ions (Cl-) for each magnesiumion (Mg2+). The positive and negative chargesmust balance each other and the overallstructure must be neutral. Note that in theformula, the charges on the ions are notindicated.

Some more examples(a) Formula for aluminium oxide:

Formula : Al2O

3

(b) Formula for calcium oxide:

Here, the valencies of the two elementsare the same. You may arrive at the formulaCa

2O

2. But we simplify the formula as CaO.

The rules that you have to follow while writinga chemical formula are as follows:• the valencies or charges on the ion must

balance.• when a compound consists of a metal

and a non-metal, the name or symbol ofthe metal is written first. For example:calcium oxide (CaO), sodium chloride(NaCl), iron sulphide (FeS), copper oxide(CuO) etc., where oxygen, chlorine,sulphur are non-metals and are writtenon the right, whereas calcium, sodium,iron and copper are metals, and arewritten on the left.

• in compounds formed with polyatomicions, the ion is enclosed in a bracketbefore writing the number to indicate theratio.

3.4.1 FORMULAE OF SIMPLE COMPOUNDS

The simplest compounds, which are made upof two different elements are called binarycompounds. Valencies of some ions are givenin Table 3.6. You can use these to writeformulae for compounds.

While writing the chemical formulae forcompounds, we write the constituentelements and their valencies as shown below.Then we must crossover the valencies of thecombining atoms.

Examples

1. Formula of hydrogen chloride

Formula of the compound would be HCl.

2. Formula of hydrogen sulphide


following formulae:(i) Al

2(SO

4)3

(ii) CaCl2

(iii) K2SO

4

(iv) KNO3

(v) CaCO3.

3. What is meant by the termchemical formula?

4. How many atoms are present in a(i) H

2S molecule and

(ii) PO4

3– ion?

3.5 Molecular Mass and MoleConcept

3.5.1 MOLECULAR MASS

In section 3.2.2 we discussed the concept ofatomic mass. This concept can be extendedto calculate molecular masses. The molecularmass of a substance is the sum of the atomicmasses of all the atoms in a molecule of thesubstance. It is therefore the relative mass ofa molecule expressed in atomic mass units (u).

Example 3.1 (a) Calculate the relativemolecular mass of water (H2O).(b) Calculate the molecular mass ofHNO3.

Solution:

(a) Atomic mass of hydrogen = 1u,oxygen = 16 u

So the molecular mass of water, whichcontains two atoms of hydrogen andone atom of oxygen is = 2 × 1+ 1×16

= 18 u(b) The molecular mass of HNO3 = the

atomic mass of H + the atomic mass ofN+ 3 × the atomic mass of O

= 1 + 14 + 48 = 63 u

3.5.2 FORMULA UNIT MASS

The formula unit mass of a substance is asum of the atomic masses of all atoms in aformula unit of a compound. Formula unitmass is calculated in the same manner aswe calculate the molecular mass. The only

(c) Formula of sodium nitrate:

Formula : NaNO3

(d) Formula of calcium hydroxide:

Formula : Ca(OH)2

Note that the formula of calciumhydroxide is Ca(OH)

2 and not CaOH

2. We use

brackets when we have two or more of thesame ions in the formula. Here, the bracketaround OH with a subscript 2 indicates thatthere are two hydroxyl (OH) groups joined toone calcium atom. In other words, there aretwo atoms each of oxygen and hydrogen incalcium hydroxide.

(e) Formula of sodium carbonate:

Formula : Na2CO

3

In the above example, brackets are notneeded if there is only one ion present.

(f) Formula of ammonium sulphate:

Formula : (NH4)2SO

4

uestions1. Write down the formulae of

(i) sodium oxide(ii) aluminium chloride(iii) sodium suphide(iv) magnesium hydroxide

2. Write down the names ofcompounds represented by the

Q

SCIENCE40

difference is that we use the word formulaunit for those substances whose constituentparticles are ions. For example, sodiumchloride as discussed above, has a formulaunit NaCl. Its formula unit mass can becalculated as–

1 × 23 + 1 × 35.5 = 58.5 u

Example 3.2 Calculate the formula unitmass of CaCl2.

Solution:

Atomic mass of Ca+ (2 × atomic mass of Cl)= 40 + 2 × 35.5 = 40 + 71 = 111 u

uestions1. Calculate the molecular masses

of H2, O

2, Cl

2, CO

2, CH

4, C

2H

6,

C2H

4, NH

3, CH

3OH.

2. Calculate the formula unitmasses of ZnO, Na

2O, K

2CO

3,

given atomic masses of Zn = 65 u,Na = 23 u, K = 39 u, C = 12 u,and O = 16 u.

3.5.3 MOLE CONCEPT

Take an example of the reaction of hydrogenand oxygen to form water:

2H2+ O2 → 2H2O.

The above reaction indicates that(i) two molecules of hydrogen combine

with one molecule of oxygen to formtwo molecules of water, or

(ii) 4 u of hydrogen molecules combinewith 32 u of oxygen molecules to form36 u of water molecules.

We can infer from the above equation thatthe quantity of a substance can becharacterised by its mass or the number ofmolecules. But, a chemical reaction equationindicates directly the number of atoms ormolecules taking part in the reaction.Therefore, it is more convenient to refer tothe quantity of a substance in terms of thenumber of its molecules or atoms, rather thantheir masses. So, a new unit “mole” wasintroduced. One mole of any species (atoms,

Q

Fig. 3.5: Relationship between mole, Avogadro number and mass


molecules, ions or particles) is that quantityin number having a mass equal to its atomicor molecular mass in grams.

The number of particles (atoms, moleculesor ions) present in 1 mole of any substanceis fixed, with a value of 6.022 × 1023. This isan experimentally obtained value. Thisnumber is called the Avogadro Constant orAvogadro Number (represented by N0), namedin honour of the Italian scientist,Amedeo Avogadro.1 mole (of anything) = 6.022 × 1023 in number,

as, 1 dozen = 12 nos.1 gross = 144 nos.

Besides being related to a number, a molehas one more advantage over a dozen or agross. This advantage is that mass of 1 moleof a particular substance is also fixed.

The mass of 1 mole of a substance is equalto its relative atomic or molecular mass ingrams. The atomic mass of an element givesus the mass of one atom of that element inatomic mass units (u). To get the mass of1 mole of atom of that element, that is, molarmass, we have to take the same numericalvalue but change the units from ‘u’ to ‘g’.Molar mass of atoms is also known as gramatomic mass. For example, atomic mass ofhydrogen=1u. So, gram atomic mass ofhydrogen = 1 g.1 u hydrogen has only 1 atom of hydrogen1 g hydrogen has 1 mole atoms, that is,6.022 × 1023 atoms of hydrogen.Similarly,16 u oxygen has only 1 atom of oxygen,16 g oxygen has 1 mole atoms, that is,6.022 × 1023 atoms of oxygen.

To find the gram molecular mass or molarmass of a molecule, we keep the numericalvalue the same as the molecular mass, butsimply change units as above from u to g.For example, as we have already calculated,molecular mass of water (H2O) is 18 u. Fromhere we understand that18 u water has only 1 molecule of water,18 g water has 1 mole molecules of water,that is, 6.022 × 1023 molecules of water.

Chemists need the number of atoms andmolecules while carrying out reactions, and

for this they need to relate the mass in gramsto the number. It is done as follows:1 mole = 6.022 × 1023 number

= Relative mass in grams.Thus, a mole is the chemist’s counting unit.

The word “mole” was introduced around1896 by Wilhelm Ostwald who derived theterm from the Latin word moles meaning a‘heap’ or ‘pile’. A substance may be consideredas a heap of atoms or molecules. The unitmole was accepted in 1967 to provide a simpleway of reporting a large number– the massiveheap of atoms and molecules in a sample.

Example 3.31. Calculate the number of moles for the

following:(i) 52 g of He (finding mole from

mass)(ii) 12.044 × 1023 number of He atoms

(finding mole from number ofparticles).

Solutions:

No. of moles = nGiven mass = mMolar mass = MGiven number of particles = NAvogadro number of particles = N0

(i) Atomic mass of He = 4 uMolar mass of He = 4g

Thus, the number of moles

= given mass

molar mass

m 52n 13

M 4⇒ = = =

(ii) we know,1 mole = 6.022 × 1023

The number of moles

given number of particles

Avogadro number=

23

23o

12.044 10Nn 2

N 6.022 10

×⇒ = = =

×

SCIENCE42

Example 3.4 Calculate the mass of thefollowing:(i) 0.5 mole of N2 gas (mass from mole

of molecule)(ii) 0.5 mole of N atoms (mass from

mole of atom)(iii) 3.011 × 1023 number of N atoms

(mass from number)(iv) 6.022 × 1023 number of N2

molecules (mass from number)

Solutions:

(i) mass = molar mass × number ofmoles

m M n 28 0.5 14 g⇒ = × = × =

(ii) mass = molar mass × number ofmoles

⇒ m = M × n = 14 × 0.5 = 7 g

(iii) The number of moles, n

0

given number of particles N

Avogadro number N= =

23

23

3.011 10

6.022 10

×=

×

23

23

3.011 10m M n 14

6.022 10

×⇒ = × = ×

×

14 0.5 7 g= × =

(iv)0

Nn

N=

23

230

6.022 10Nm M 28

N 6.022 10

×⇒ = × = ×

×

28 1 28 g= × =

Example 3.5 Calculate the number ofparticles in each of thefollowing:

(i) 46 g of Na atoms (number frommass)

(ii) 8 g O2 molecules (number ofmolecules from mass)

(iii) 0.1 mole of carbon atoms (numberfrom given moles)

Solutions:

(i) The number of atoms

given massAvogadro number

molar mass= ×

0

mN N

M⇒ = ×

23

23

46N 6.022 10

23N 12.044 10

⇒ = × ×

⇒ = ×

(ii) The number of molecules

0

given massAvogadro number

molar mass

mN N

Matomic massof oxygen 16 u

= ×

⇒ = ×

=

∴ molar mass of O2 molecules= 16 × 2 = 32g

23

23

23

8N 6.022 10

32N 1.5055 10

1.51 10

⇒ = × ×

⇒ = ×

×;

(iii) The number of particles (atom) =number of moles of particles ×Avogadro numberN = n × N0 = 0.1 x 6.022 × 1023

= 6.022 × 1022

uestions1. If one mole of carbon atoms

weighs 12 gram, what is themass (in gram) of 1 atom ofcarbon?

2. Which has more number ofatoms, 100 grams of sodium or100 grams of iron (given, atomicmass of Na = 23 u, Fe = 56 u)?

Q


Whatyou havelearnt• During a chemical reaction, the sum of the masses of the

reactants and products remains unchanged. This is knownas the Law of Conservation of Mass.

• In a pure chemical compound, elements are always present ina definite proportion by mass. This is known as the Law ofDefinite Proportions.

• An atom is the smallest particle of the element that can existindependently and retain all its chemical properties.

• A molecule is the smallest particle of an element or a compoundcapable of independent existence under ordinary conditions.It shows all the properties of the substance.

• A chemical formula of a compound shows its constituentelements and the number of atoms of each combining element.

• Clusters of atoms that act as an ion are called polyatomic ions.They carry a fixed charge on them.

• The chemical formula of a molecular compound is determinedby the valency of each element.

• In ionic compounds, the charge on each ion is used to determinethe chemical formula of the compound.

• Scientists use the relative atomic mass scale to compare themasses of different atoms of elements. Atoms of carbon-12isotopes are assigned a relative atomic mass of 12 and therelative masses of all other atoms are obtained by comparisonwith the mass of a carbon-12 atom.

• The Avogadro constant 6.022 × 1023 is defined as the numberof atoms in exactly 12 g of carbon-12.

• The mole is the amount of substance that contains the samenumber of particles (atoms/ ions/ molecules/ formula unitsetc.) as there are atoms in exactly 12 g of carbon-12.

• Mass of 1 mole of a substance is called its molar mass.

Exercises1. A 0.24 g sample of compound of oxygen and boron was found

by analysis to contain 0.096 g of boron and 0.144 g of oxygen.Calculate the percentage composition of the compound byweight.

2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g ofcarbon dioxide is produced. What mass of carbon dioxide will

SCIENCE44

be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen?Which law of chemical combination will govern your answer?

3. What are polyatomic ions? Give examples.4. Write the chemical formulae of the following.

(a) Magnesium chloride(b) Calcium oxide(c) Copper nitrate(d) Aluminium chloride(e) Calcium carbonate.

5. Give the names of the elements present in the followingcompounds.

(a) Quick lime(b) Hydrogen bromide(c) Baking powder(d) Potassium sulphate.

6. Calculate the molar mass of the following substances.(a) Ethyne, C2H2

(b) Sulphur molecule, S8

(c) Phosphorus molecule, P4

(Atomic mass of phosphorus= 31)

(d) Hydrochloric acid, HCl(e) Nitric acid, HNO3

7. What is the mass of—(a) 1 mole of nitrogen atoms?(b) 4 moles of aluminium atoms (Atomic mass of aluminium

= 27)?(c) 10 moles of sodium sulphite (Na

2SO

3)?

8. Convert into mole.(a) 12 g of oxygen gas(b) 20 g of water

(c) 22 g of carbon dioxide.9. What is the mass of:

(a) 0.2 mole of oxygen atoms?(b) 0.5 mole of water molecules?

10. Calculate the number of molecules of sulphur (S8) present in16 g of solid sulphur.

11. Calculate the number of aluminium ions present in 0.051 g ofaluminium oxide.

(Hint: The mass of an ion is the same as that of an atom of thesame element. Atomic mass of Al = 27 u)


Group ActivityPlay a game for writing formulae.

Example1 : Make placards with symbols and valencies of theelements separately. Each student should hold twoplacards, one with the symbol in the right hand andthe other with the valency in the left hand. Keepingthe symbols in place, students should criss-cross theirvalencies to form the formula of a compound.

Example 2 : A low cost model for writing formulae: Take emptyblister packs of medicines. Cut them in groups,according to the valency of the element, as shown inthe figure. Now, you can make formulae by fixing onetype of ion into other.

For example:

Na+ SO42- P04

3-

Formula for sodium sulphate:

2 sodium ions can be fixed on one sulphate ion.

Hence, the formula will be: Na2SO

4

Do it yourself :

Now, write the formula of sodium phosphate.

In Chapter 3, we have learnt that atoms andmolecules are the fundamental buildingblocks of matter. The existence of differentkinds of matter is due to different atomsconstituting them. Now the questions arise:(i) What makes the atom of one elementdifferent from the atom of another element?and (ii) Are atoms really indivisible, asproposed by Dalton, or are there smallerconstituents inside the atom? We shall findout the answers to these questions in thischapter. We will learn about sub-atomicparticles and the various models that havebeen proposed to explain how these particlesare arranged within the atom.

A major challenge before the scientists atthe end of the 19th century was to reveal thestructure of the atom as well as to explain itsimportant properties. The elucidation of thestructure of atoms is based on a series ofexperiments.

One of the first indications that atoms arenot indivisible, comes from studying staticelectricity and the condition under whichelectricity is conducted by differentsubstances.

4.1 Charged Particles in MatterFor understanding the nature of chargedparticles in matter, let us carry out thefollowing activities:

Activity ______________ 4.1A. Comb dry hair. Does the comb then

attract small pieces of paper?B. Rub a glass rod with a silk cloth and

bring the rod near an inflated balloon.Observe what happens.

From these activities, can we concludethat on rubbing two objects together, theybecome electrically charged? Where does thischarge come from? This question can beanswered by knowing that an atom is divisibleand consists of charged particles.

Many scientists contributed in revealingthe presence of charged particles in an atom.

It was known by 1900 that the atom wasnot a simple, indivisible particle but containedat least one sub-atomic particle – the electronidentified by J.J. Thomson. Even before theelectron was identified, E. Goldstein in 1886discovered the presence of new radiations ina gas discharge and called them canal rays.These rays were positively charged radiationswhich ultimately led to the discovery ofanother sub-atomic particle. This sub-atomicparticle had a charge, equal in magnitude butopposite in sign to that of the electron. Itsmass was approximately 2000 times as thatof the electron. It was given the name ofproton. In general, an electron is representedas ‘e–’ and a proton as ‘p+’. The mass of a protonis taken as one unit and its charge as plusone. The mass of an electron is considered tobe negligible and its charge is minus one.

It seemed highly likely that an atom wascomposed of protons and electrons, mutuallybalancing their charges. It also appeared thatthe protons were in the interior of the atom,for whereas electrons could easily be peeledoff but not protons. Now the big question was:what sort of structure did these particles ofthe atom form? We will find the answer tothis question below.

44444SSSSSTRUCTURETRUCTURETRUCTURETRUCTURETRUCTURE OFOFOFOFOF THETHETHETHETHE A A A A ATOMTOMTOMTOMTOM

Chapter

uestions1. What are canal rays?2. If an atom contains one electron

and one proton, will it carry anycharge or not?

4.2 The Structure of an AtomWe have learnt Dalton’s atomic theory inChapter 3, which suggested that the atomwas indivisible and indestructible. But thediscovery of two fundamental particles(electrons and protons) inside the atom, ledto the failure of this aspect of Dalton’s atomictheory. It was then considered necessary toknow how electrons and protons are arrangedwithin an atom. For explaining this, manyscientists proposed various atomic models.J.J. Thomson was the first one to propose amodel for the structure of an atom.

4.2.1 THOMSON’S MODEL OF AN ATOM

Thomson proposed the model of an atom tobe similar to that of a Christmas pudding.The electrons, in a sphere of positive charge,were like currants (dry fruits) in a sphericalChristmas pudding. We can also think of awatermelon, the positive charge in the atomis spread all over like the red edible part ofthe watermelon, while the electrons arestudded in the positively charged sphere, likethe seeds in the watermelon (Fig. 4.1).

Thomson proposed that:(i) An atom consists of a positively

charged sphere and the electrons areembedded in it.

(ii) The negative and positive charges areequal in magnitude. So, the atom as awhole is electrically neutral.

Although Thomson’s model explained thatatoms are electrically neutral, the results ofexperiments carried out by other scientistscould not be explained by this model, as wewill see below.

4.2.2 RUTHERFORD’S MODEL OF AN ATOM

Ernest Rutherford was interested in knowinghow the electrons are arranged within anatom. Rutherford designed an experiment forthis. In this experiment, fast moving alpha(α)-particles were made to fall on a thingold foil.• He selected a gold foil because he wanted

as thin a layer as possible. This gold foilwas about 1000 atoms thick.

• α-particles are doubly-charged heliumions. Since they have a mass of 4 u, thefast-moving α-particles have aconsiderable amount of energy.

• It was expected that α-particles would bedeflected by the sub-atomic particles inthe gold atoms. Since the α-particles weremuch heavier than the protons, he didnot expect to see large deflections.

Q

Fig.4.1: Thomson’s model of an atom

J.J. Thomson (1856-1940), a Britishphysicist, was born inCheetham Hill, a suburbof Manchester, on18 December 1856. Hewas awarded the Nobelprize in Physics in 1906for his work on thediscovery of electrons.He directed the Cavendish Laboratory atCambridge for 35 years and seven of hisresearch assistants subsequently wonNobel prizes.

STRUCTURE OF THE ATOM 47

SCIENCE48

Fig. 4.2: Scattering of α-particles by a gold foil

But, the α-particle scattering experimentgave totally unexpected results (Fig. 4.2). Thefollowing observations were made:

(i) Most of the fast moving α-particlespassed straight through the gold foil.

(ii) Some of the α-particles were deflectedby the foil by small angles.

(iii) Surprisingly one out of every 12000particles appeared to rebound.

In the words of Rutherford, “This resultwas almost as incredible as if you fire a15-inch shell at a piece of tissue paper and itcomes back and hits you”.

hear a sound when each stone strikes thewall. If he repeats this ten times, he will hearthe sound ten times. But if a blind-foldedchild were to throw stones at a barbed-wirefence, most of the stones would not hit thefencing and no sound would be heard. Thisis because there are lots of gaps in the fencewhich allow the stone to pass through them.

Following a similar reasoning, Rutherfordconcluded from the α-particle scatteringexperiment that–

(i) Most of the space inside the atom isempty because most of the α-particlespassed through the gold foil withoutgetting deflected.

(ii) Very few particles were deflected fromtheir path, indicating that the positivecharge of the atom occupies very littlespace.

(iii) A very small fraction of α-particleswere deflected by 1800, indicating thatall the positive charge and mass of thegold atom were concentrated in a verysmall volume within the atom.

From the data he also calculated that theradius of the nucleus is about 105 times lessthan the radius of the atom.

On the basis of his experiment,Rutherford put forward the nuclear model ofan atom, which had the following features:

(i) There is a positively charged centre inan atom called the nucleus. Nearly allthe mass of an atom resides in thenucleus.

(ii) The electrons revolve around thenucleus in well-defined orbits.

(iii) The size of the nucleus is very smallas compared to the size of the atom.

Drawbacks of Rutherford’s model ofthe atom

The orbital revolution of the electron is notexpected to be stable. Any particle in acircular orbit would undergo acceleration.During acceleration, charged particles wouldradiate energy. Thus, the revolving electronwould lose energy and finally fall into thenucleus. If this were so, the atom should behighly unstable and hence matter would notexist in the form that we know. We know thatatoms are quite stable.

E. Rutherford (1871-1937)was born at Spring Groveon 30 August 1871. He wasknown as the ‘Father’ ofnuclear physics. He isfamous for his work onradioactivity and the

discovery of the nucleus of an atom withthe gold foil experiment. He got the Nobelprize in chemistry in 1908.

Let us think of an activity in an open fieldto understand the implications of thisexperiment. Let a child stand in front of awall with his eyes closed. Let him throwstones at the wall from a distance. He will


4.2.3 BOHR’S MODEL OF ATOM

In order to overcome the objections raisedagainst Rutherford’s model of the atom,Neils Bohr put forward the followingpostulates about the model of an atom:

(i) Only certain special orbits known asdiscrete orbits of electrons, are allowedinside the atom.

(ii) While revolving in discrete orbits theelectrons do not radiate energy.

uestions1. On the basis of Thomson’s model

of an atom, explain how the atomis neutral as a whole.

2. On the basis of Rutherford’smodel of an atom, which sub-atomic particle is present in thenucleus of an atom?

3. Draw a sketch of Bohr’s modelof an atom with three shells.

4. What do you think would be theobservation if the α-particlescattering experiment is carriedout using a foil of a metal otherthan gold?

4.2.4 NEUTRONS

In 1932, J. Chadwick discovered another sub-atomic particle which had no charge and amass nearly equal to that of a proton. It waseventually named as neutron. Neutrons arepresent in the nucleus of all atoms, excepthydrogen. In general, a neutron isrepresented as ‘n’. The mass of an atom istherefore given by the sum of the masses ofprotons and neutrons present in the nucleus.

uestions1. Name the three sub-atomic

particles of an atom.2. Helium atom has an atomic mass

of 4 u and two protons in itsnucleus. How many neutronsdoes it have?

4.3 How are Electrons Distributedin Different Orbits (Shells)?

The distribution of electrons into differentorbits of an atom was suggested by Bohr andBury.

The following rules are followed for writingthe number of electrons in different energylevels or shells:

(i) The maximum number of electronspresent in a shell is given by the

Neils Bohr (1885-1962)was born in Copenhagenon 7 October 1885. He wasappointed professor ofphysics at CopenhagenUniversity in 1916. He gotthe Nobel prize for his workon the structure of atom in1922. Among Professor

Bohr’s numerous writings, three appearingas books are:(i) The Theory of Spectra and AtomicConstitution, (ii) Atomic Theory and,(iii) The Description of Nature.

These orbits or shells are called energylevels. Energy levels in an atom are shown inFig. 4.3.

Q

Fig. 4.3: A few energy levels in an atom

These orbits or shells are represented bythe letters K,L,M,N,… or the numbers,n=1,2,3,4,….

Q

SCIENCE50

formula 2n2, where ‘n’ is the orbitnumber or energy level index, 1,2,3,….Hence the maximum number ofelectrons in different shells are asfollows:first orbit or K-shell will be = 2 × 12 = 2,second orbit or L-shell will be = 2 × 22

= 8, third orbit or M-shell will be = 2 ×32 = 18, fourth orbit or N-shell will be= 2 × 42= 32, and so on.

(ii) The maximum number of electronsthat can be accommodated in theoutermost orbit is 8.

(iii) Electrons are not accommodated in agiven shell, unless the inner shells arefilled. That is, the shells are filled in astep-wise manner.

Atomic structure of the first eighteenelements is shown schematically in Fig. 4.4.

• The composition of atoms of the firsteighteen elements is given in Table 4.1.

uestions1. Write the distribution of electrons

in carbon and sodium atoms.2. If K and L shells of an atom are

full, then what would be the totalnumber of electrons in the atom?

4.4 ValencyWe have learnt how the electrons in an atomare arranged in different shells/orbits. Theelectrons present in the outermost shell ofan atom are known as the valence electrons.

From the Bohr-Bury scheme, we alsoknow that the outermost shell of an atom can

Activity ______________ 4.2

• Make a static atomic model displayingelectronic configuration of the firsteighteen elements.

Fig.4.4: Schematic atomic structure of the first eighteen elements

Q

accommodate a maximum of 8 electrons. Itwas observed that the atoms of elements,having a completely filled outermost shellshow little chemical activity. In other words,their combining capacity or valency is zero.Of these inert elements, the helium atom has


Table 4.1: Composition of Atoms of the First Eighteen Elementswith Electron Distribution in Various Shells

two electrons in its outermost shell and allother elements have atoms with eightelectrons in the outermost shell.

The combining capacity of the atoms ofother elements, that is, their tendency to reactand form molecules with atoms of the sameor different elements, was thus explained asan attempt to attain a fully-filled outermostshell. An outermost-shell, which had eightelectrons was said to possess an octet. Atomswould thus react, so as to achieve an octet inthe outermost shell. This was done bysharing, gaining or losing electrons. Thenumber of electrons gained, lost or sharedso as to make the octet of electrons in theoutermost shell, gives us directly thecombining capacity of the element, that is,

the valency discussed in the previous chapter.For example, hydrogen/lithium/sodiumatoms contain one electron each in theiroutermost shell, therefore each one of themcan lose one electron. So, they are said tohave valency of one. Can you tell, what isvalency of magnesium and aluminium? It istwo and three, respectively, becausemagnesium has two electrons in its outermostshell and aluminium has three electrons inits outermost shell.

If the number of electrons in theoutermost shell of an atom is close to its fullcapacity, then valency is determined in adifferent way. For example, the fluorine atomhas 7 electrons in the outermost shell, andits valency could be 7. But it is easier for

Name of Symbol Atomic Number Number Number Vale-Element Number of of of ncy

Protons Neutrons Electrons K L M N

Hydrogen H 1 1 - 1 1 - - - 1Helium He 2 2 2 2 2 - - - 0

Lithium Li 3 3 4 3 2 1 - - 1

Beryllium Be 4 4 5 4 2 2 - - 2

Boron B 5 5 6 5 2 3 - - 3

Carbon C 6 6 6 6 2 4 - - 4

Nitrogen N 7 7 7 7 2 5 - - 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 - - 0

Sodium 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 - 4

Phosphorus P 15 15 16 15 2 8 5 - 3,5

Sulphur S 16 16 16 16 2 8 6 - 2

Chlorine Cl 17 17 18 17 2 8 7 - 1

Argon Ar 18 18 22 18 2 8 8 0

Distribution ofElectrons

SCIENCE52

fluorine to gain one electron instead of losingseven electrons. Hence, its valency isdetermined by subtracting seven electronsfrom the octet and this gives you a valency ofone for fluorine. Valency can be calculated ina similar manner for oxygen. What is thevalency of oxygen that you get from thiscalculation?

Therefore, an atom of each element has adefinite combining capacity, called its valency.Valency of the first eighteen elements is givenin the last column of Table 4.1.

uestion1. How will you find the valency

of chlorine, sulphur andmagnesium?

4.5 Atomic Number and MassNumber

4.5.1 ATOMIC NUMBER

We know that protons are present in thenucleus of an atom. It is the number ofprotons of an atom, which determines itsatomic number. It is denoted by ‘Z’. All atomsof an element have the same atomic number,Z. In fact, elements are defined by the numberof protons they possess. For hydrogen, Z = 1,because in hydrogen atom, only one protonis present in the nucleus. Similarly, forcarbon, Z = 6. Therefore, the atomic numberis defined as the total number of protonspresent in the nucleus of an atom.

4.5.2 MASS NUMBER

After studying the properties of the sub-atomic particles of an atom, we can concludethat mass of an atom is practically due toprotons and neutrons alone. These arepresent in the nucleus of an atom. Henceprotons and neutrons are also callednucleons. Therefore, the mass of an atomresides in its nucleus. For example, mass of

carbon is 12 u because it has 6 protons and6 neutrons, 6 u + 6 u = 12 u. Similarly, themass of aluminium is 27 u (13 protons+14neutrons). The mass number is defined asthe sum of the total number of protons andneutrons present in the nucleus of an atom.In the notation for an atom, the atomicnumber, mass number and symbol of theelement are to be written as:

Mass Number

QSymbol ofelement

Atomic Number

For example, nitrogen is written as 147 N .

uestions1. If number of electrons in an atom

is 8 and number of protons is also8, then (i) what is the atomicnumber of the atom? and (ii) whatis the charge on the atom?

2. With the help of Table 4.1, findout the mass number of oxygenand sulphur atom.

4.6 IsotopesIn nature, a number of atoms of someelements have been identified, which have thesame atomic number but different massnumbers. For example, take the case ofhydrogen atom, it has three atomic species,

namely protium (11 H), deuterium ( 2

1 H or D)

and tritium ( 31H or T). The atomic number of

each one is 1, but the mass number is 1, 2and 3, respectively. Other such examples are

(i) carbon, 126 C and 14

6 C, (ii) chlorine, 3517 Cl

and 3717 Cl, etc.

On the basis of these examples, isotopesare defined as the atoms of the same element,having the same atomic number but differentmass numbers. Therefore, we can say thatthere are three isotopes of hydrogen atom,namely protium, deuterium and tritium.

Q


Q

Many elements consist of a mixture ofisotopes. Each isotope of an element is a puresubstance. The chemical properties ofisotopes are similar but their physicalproperties are different.

Chlorine occurs in nature in two isotopicforms, with masses 35 u and 37 u in the ratioof 3:1. Obviously, the question arises: whatshould we take as the mass of chlorine atom?Let us find out.

The mass of an atom of any naturalelement is taken as the average mass of allthe naturally occuring atoms of that element.If an element has no isotopes, then the massof its atom would be the same as the sum ofprotons and neutrons in it. But if an elementoccurs in isotopic forms, then we have toknow the percentage of each isotopic formand then the average mass is calculated.

The average atomic mass of chlorine atom,on the basis of above data, will be

75 2535 37

100 100× + ×

⎡⎛ ⎞⎜ ⎟⎢⎝ ⎠⎣

105 37 14235.5 u

4 4 4+ = =

⎤⎛ ⎞= ⎜ ⎟ ⎥⎝ ⎠ ⎦

This does not mean that any one atom ofchlorine has a fractional mass of 35.5 u. Itmeans that if you take a certain amount ofchlorine, it will contain both isotopes ofchlorine and the average mass is 35.5 u.

Applications

Since the chemical properties of all theisotopes of an element are the same,normally we are not concerned abouttaking a mixture. But some isotopes havespecial properties which find them usefulin various fields. Some of them are :

(i) An isotope of uranium is used as a fuelin nuclear reactors.

(ii) An isotope of cobalt is used in thetreatment of cancer.

(iii) An isotope of iodine is used in thetreatment of goitre.

4.6.1 ISOBARS

Let us consider two elements — calcium,atomic number 20, and argon, atomicnumber 18. The number of electrons in theseatoms is different, but the mass number ofboth these elements is 40. That is, the totalnumber of nucleons is the same in the atomsof this pair of elements. Atoms of differentelements with different atomic numbers,which have the same mass number, areknown as isobars.

uestions1. For the symbol H,D and T

tabulate three sub-atomicparticles found in each of them.

2. Write the electronic configurationof any one pair of isotopes andisobars.

Whatyou havelearnt• Credit for the discovery of electron and proton goes to J.J.

Thomson and E.Goldstein, respectively.

• J.J. Thomson proposed that electrons are embedded in apositive sphere.

SCIENCE54

• Rutherford’s alpha-particle scattering experiment led to thediscovery of the atomic nucleus.

• Rutherford’s model of the atom proposed that a very tinynucleus is present inside the atom and electrons revolve aroundthis nucleus. The stability of the atom could not be explainedby this model.

• Neils Bohr’s model of the atom was more successful. Heproposed that electrons are distributed in different shells withdiscrete energy around the nucleus. If the atomic shells arecomplete, then the atom will be stable and less reactive.

• J. Chadwick discovered presence of neutrons in the nucleus ofan atom. So, the three sub-atomic particles of an atom are:(i) electrons, (ii) protons and (iii) neutrons. Electrons arenegatively charged, protons are positively charged and neutrons

have no charges. The mass of an electron is about 1

2000 times

the mass of an hydrogen atom. The mass of a proton and aneutron is taken as one unit each.

• Shells of an atom are designated as K,L,M,N,….

• Valency is the combining capacity of an atom.

• The atomic number of an element is the same as the numberof protons in the nucleus of its atom.

• The mass number of an atom is equal to the number of nucleonsin its nucleus.

• Isotopes are atoms of the same element, which have differentmass numbers.

• Isobars are atoms having the same mass number but differentatomic numbers.

• Elements are defined by the number of protons they possess.

Exercises1. Compare the properties of electrons, protons and neutrons.

2. What are the limitations of J.J. Thomson’s model of the atom?

3. What are the limitations of Rutherford’s model of the atom?

4. Describe Bohr’s model of the atom.

5. Compare all the proposed models of an atom given in thischapter.

6. Summarise the rules for writing of distribution of electrons invarious shells for the first eighteen elements.

7. Define valency by taking examples of silicon and oxygen.


8. Explain with examples (i) Atomic number, (ii) Mass number,(iii) Isotopes and iv) Isobars. Give any two uses of isotopes.

9. Na+ has completely filled K and L shells. Explain.

10. If bromine atom is available in the form of, say, two isotopes7935 Br (49.7%) and 81

35 Br (50.3%), calculate the average atomic

mass of bromine atom.

11. The average atomic mass of a sample of an element X is 16.2 u.

What are the percentages of isotopes 168 X and 18

8 X in the

sample?

12. If Z = 3, what would be the valency of the element? Also, namethe element.

13. Composition of the nuclei of two atomic species X and Y aregiven as under

X Y

Protons = 6 6

Neutrons = 6 8

Give the mass numbers of X and Y. What is the relation betweenthe two species?

14. For the following statements, write T for True and F for False.

(a) J.J. Thomson proposed that the nucleus of an atomcontains only nucleons.

(b) A neutron is formed by an electron and a protoncombining together. Therefore, it is neutral.

(c) The mass of an electron is about 1

2000 times that of proton.

(d) An isotope of iodine is used for making tincture iodine,which is used as a medicine.

Put tick ( ) against correct choice and cross (×) againstwrong choice in questions 15, 16 and 17

15. Rutherford’s alpha-particle scattering experiment wasresponsible for the discovery of

(a) Atomic Nucleus (b) Electron

(c) Proton (d) Neutron

16. Isotopes of an element have

(a) the same physical properties

(b) different chemical properties

(c) different number of neutrons

(d) different atomic numbers.

17. Number of valence electrons in Cl– ion are:

(a) 16 (b) 8 (c) 17 (d) 18

SCIENCE56

18. Which one of the following is a correct electronic configurationof sodium?

(a) 2,8 (b) 8,2,1 (c) 2,1,8 (d) 2,8,1.

19. Complete the following table.

Atomic Mass Number Number Number Name ofNumber Number of of of the Atomic

Neutrons Protons Electrons Species

9 - 10 - - -

16 32 - - - Sulphur

- 24 - 12 - -

- 2 - 1 - -

- 1 0 1 0 -

While examining a thin slice of cork, RobertHooke saw that the cork resembled thestructure of a honeycomb consisting of manylittle compartments. Cork is a substancewhich comes from the bark of a tree. Thiswas in the year 1665 when Hooke made thischance observation through a self-designedmicroscope. Robert Hooke called these boxescells. Cell is a Latin word for ‘a little room’.

This may seem to be a very small andinsignificant incident but it is very importantin the history of science. This was the veryfirst time that someone had observed thatliving things appear to consist of separateunits. The use of the word ‘cell’ to describethese units is used till this day in biology.

Let us find out about cells.

5.1 What are Living OrganismsMade Up of ?

Activity ______________ 5.1• Let us take a small piece from an onion

bulb. With the help of a pair of forceps,we can peel of f the skin (calledepidermis) from the concave side (innerlayer) of the onion. This layer can beput immediately in a watch-glasscontaining water. This will prevent thepeel from getting folded or getting dry.What do we do with this peel?

• Let us take a glass slide, put a drop ofwater on it and transfer a small pieceof the peel from the watch glass to theslide. Make sure that the peel isperfectly flat on the slide. A thin camelhair paintbrush might be necessary tohelp transfer the peel. Now we put adrop of iodine solution on this piecefollowed by a cover slip. Take care to

avoid air bubbles while putting thecover slip with the help of a mountingneedle. Ask your teacher for help. Wehave prepared a temporary mount ofonion peel. We can observe this slideunder low power followed by highpowers of a compound microscope.

Fig. 5.1: Compound microscope

What do we observe as we look throughthe lens? Can we draw the structures thatwe are able to see through the microscope,on an observation sheet? Does it look likeFig. 5.2?

Eyepiece

Coarse adjustment

Fine adjustment

ArmObjective lens

StageSwivel

Mirror

Base

Body tube

Clip Microscope slide

Condenser

Fig. 5.2: Cells of an onion peel

55555TTTTTHEHEHEHEHE F F F F FUNDAMENTALUNDAMENTALUNDAMENTALUNDAMENTALUNDAMENTAL U U U U UNITNITNITNITNIT OFOFOFOFOF L L L L LIFEIFEIFEIFEIFE

Chapter

SCIENCE58

Chlamydomonas, Paramoecium and bacteria.These organisms are called unicellularorganisms (uni = single). On the other hand,many cells group together in a single bodyand assume different functions in it to formvarious body parts in multicellular organisms(multi = many) such as some fungi, plantsand animals. Can we find out names of somemore unicellular organisms?

Every multi-cellular organism has comefrom a single cell. How? Cells divide toproduce cells of their own kind. All cells thuscome from pre-existing cells.

Activity ______________ 5.2• We can try preparing temporary

mounts of leaf peels, tip of roots ofonion or even peels of onions of differentsizes.

• After performing the above activity, letus see what the answers to the followingquestions would be:(a) Do all cells look alike in terms of

shape and size?(b) Do all cells look alike in structure?(c) Could we find differences among

cells from different parts of a plantbody?

(d) What similarities could we find?

Some organisms can also have cells ofdifferent kinds. Look at the following picture.It depicts some cells from the human body.

Nerve Cell

Fat cell

Sperm

Bonecell

Smoothmuscle

cell

Bloodcells

Ovum

Fig. 5.3: Various cells from the human body

Mor

e to

kno

w

We can try preparing temporary mountsof peels of onions of different sizes. What dowe observe? Do we see similar structures ordifferent structures?

What are these structures?

These structures look similar to each other.Together they form a big structure like anonion bulb! We find from this activity thatonion bulbs of different sizes have similarsmall structures visible under a microscope.The cells of the onion peel will all look thesame, regardless of the size of the onion theycame from.

These small structures that we see arethe basic building units of the onion bulb.These structures are called cells. Not onlyonions, but all organisms that we observearound are made up of cells. However, thereare also single cells that live on their own.

Cells were first discovered byRobert Hooke in 1665. He observedthe cells in a cork slice with the helpof a primitive microscope.Leeuwenhoek (1674), with theimproved microscope, discovered thefree living cells in pond water for thefirst time. It was Robert Brown in1831 who discovered the nucleus inthe cell. Purkinje in 1839 coined theterm ‘protoplasm’ for the fluidsubstance of the cell. The cell theory,that all the plants and animals arecomposed of cells and that the cell isthe basic unit of life, was presentedby two biologists, Schleiden (1838)and Schwann (1839). The cell theorywas further expanded by Virchow(1855) by suggesting that all cellsarise from pre-existing cells. With thediscovery of the electron microscopein 1940, it was possible to observe andunderstand the complex structure ofthe cell and its various organelles.

The invention of magnifying lenses led tothe discovery of the microscopic world. It isnow known that a single cell may constitutea whole organism as in Amoeba,

THE FUNDAMENTAL UNIT OF LIFE 59

The shape and size of cells are related tothe specific function they perform. Some cellslike Amoeba have changing shapes. In somecases the cell shape could be more or lessfixed and peculiar for a particular type of cell;for example, nerve cells have a typical shape.

Each living cell has the capacity toperform certain basic functions that arecharacteristic of all living forms. How does aliving cell perform these basic functions? Weknow that there is a division of labour inmulticellular organisms such as humanbeings. This means that different parts of thehuman body perform different functions. Thehuman body has a heart to pump blood, astomach to digest food and so on. Similarly,division of labour is also seen within a singlecell in many cases. In fact, each such cellhas got certain specific components within itknown as cell organelles. Each kind of cellorganelle performs a special function, suchas making new material in the cell, clearingup the waste material from the cell and soon. A cell is able to live and perform all itsfunctions because of these organelles. Theseorganelles together constitute the basic unitcalled the cell. It is interesting that all cellsare found to have the same organelles, nomatter what their function is or whatorganism they are found in.

uestions1. Who discovered cells, and how?2. Why is the cell called the

structural and functional unit oflife?

5.2 What is a Cell Made Up of ?What is the StructuralOrganisation of a Cell?

We saw above that the cell has specialcomponents called organelles. How is a cellorganised?

If we study a cell under a microscope, wewould come across three features in almost

every cell; plasma membrane, nucleus andcytoplasm. All activities inside the cell andinteractions of the cell with its environmentare possible due to these features. Let us seehow.

5.2.1 PLASMA MEMBRANE OR CELL

MEMBRANE

This is the outermost covering of the cell thatseparates the contents of the cell from itsexternal environment. The plasma membraneallows or permits the entry and exit of somematerials in and out of the cell. It alsoprevents movement of some other materials.The cell membrane, therefore, is called aselectively permeable membrane.

How does the movement of substancestake place into the cell? How do substancesmove out of the cell?

Some substances like carbon dioxide oroxygen can move across the cell membraneby a process called diffusion. We have studiedthe process of diffusion in earlier chapters.We saw that there is spontaneous movementof a substance from a region of highconcentration to a region where itsconcentration is low.

Something similar to this happens in cellswhen, for example, some substance like CO

2

(which is cellular waste and requires to beexcreted out by the cell) accumulates in highconcentrations inside the cell. In the cell’sexternal environment, the concentration ofCO2 is low as compared to that inside thecell. As soon as there is a difference ofconcentration of CO2 inside and outside a cell,CO2 moves out of the cell, from a region ofhigh concentration, to a region of lowconcentration outside the cell by the processof diffusion. Similarly, O2 enters the cell bythe process of diffusion when the level orconcentration of O2 inside the cell decreases.Thus, diffusion plays an important role ingaseous exchange between the cells as wellas the cell and its external environment.

Water also obeys the law of diffusion. Themovement of water molecules through sucha selectively permeable membrane is called

Q

SCIENCE60

osmosis. The movement of water across theplasma membrane is also affected by theamount of substance dissolved in water.Thus, osmosis is the passage of water from aregion of high water concentration through asemi-permeable membrane to a region of lowwater concentration.

What will happen if we put an animal cellor a plant cell into a solution of sugar or saltin water?

One of the following three things couldhappen:

1. If the medium surrounding the cell hasa higher water concentration than thecell, meaning that the outside solutionis very dilute, the cell will gain waterby osmosis. Such a solution is knownas a hypotonic solution.

Water molecules are free to passacross the cell membrane in bothdirections, but more water will comeinto the cell than will leave. The net(overall) result is that water enters thecell. The cell is likely to swell up.

2. If the medium has exactly the samewater concentration as the cell, therewill be no net movement of wateracross the cell membrane. Such asolution is known as an isotonicsolution.

Water crosses the cell membranein both directions, but the amountgoing in is the same as the amountgoing out, so there is no overallmovement of water. The cell will staythe same size.

3. If the medium has a lowerconcentration of water than the cell,meaning that it is a very concentratedsolution, the cell will lose water byosmosis. Such a solution is known asa hypertonic solution.

Again, water crosses the cellmembrane in both directions, but thistime more water leaves the cell thanenters it. Therefore the cell will shrink.

Thus, osmosis is a special case of diffusionthrough a selectively permeable membrane.Now let us try out the following activity:

Activity ______________ 5.3Osmosis with an egg

(a) Remove the shell of an egg by dissolvingit in dilute hydrochloric acid. The shellis mostly calcium carbonate. A thinouter skin now encloses the egg. Putthe egg in pure water and observe after5 minutes. What do we observe?The egg swells because water passesinto it by osmosis.

(b) Place a similar de-shelled egg in aconcentrated salt solution and observefor 5 minutes. The egg shrinks. Why?Water passes out of the egg solutioninto the salt solution because the saltsolution is more concentrated.

We can also try a similar activity with driedraisins or apricots.

Activity ______________ 5.4• Put dried raisins or apricots in plain

water and leave them for some time.Then place them into a concentratedsolution of sugar or salt. You willobserve the following:

(a) Each gains water and swells whenplaced in pure water.

(b) However, when placed in theconcentrated solution it loses water,and consequently shrinks.

Unicellular freshwater organisms andmost plant cells tend to gain water throughosmosis. Absorption of water by plant rootsis also an example of osmosis.

Thus, diffusion is important in exhangeof gases and water in the life of a cell. Inadditions to this, the cell also obtainsnutrition from its environment. Differentmolecules move in and out of the cell througha type of transport requiring use of energy inthe form of ATP.

The plasma membrane is flexible and ismade up of organic molecules called lipidsand proteins. However, we can observe thestructure of the plasma membrane onlythrough an electron microscope.

The flexibility of the cell membrane alsoenables the cell to engulf in food and othermaterial from its external environment. Suchprocesses are known as endocytosis. Amoebaacquires its food through such processes.


Activity ______________ 5.5• Find out about electron microscopes

from resources in the school library orthrough the internet. Discuss it withyour teacher.

uestions1. How do substances like CO

2 and

water move in and out of the cell?Discuss.

2. Why is the plasma membranecalled a selectively permeablemembrane?

5.2.2 CELL WALL

Plant cells, in addition to the plasmamembrane, have another rigid outer coveringcalled the cell wall. The cell wall lies outsidethe plasma membrane. The plant cell wall ismainly composed of cellulose. Cellulose is acomplex substance and provides structuralstrength to plants.

When a living plant cell loses waterthrough osmosis there is shrinkage orcontraction of the contents of the cell awayfrom the cell wall. This phenomenon is knownas plasmolysis. We can observe thisphenomenon by performing the followingactivity:

Activity ______________ 5.6• Mount the peel of a Rheo leaf in water

on a slide and examine cells under thehigh power of a microscope. Note thesmall green granules, calledchloroplasts. They contain a greensubstance called chlorophyll. Put astrong solution of sugar or salt on themounted leaf on the slide. Wait for aminute and observe under amicroscope. What do we see?

• Now place some Rheo leaves in boilingwater for a few minutes. This kills thecells. Then mount one leaf on a slideand observe it under a microscope. Puta strong solution of sugar or salt onthe mounted leaf on the slide. Wait fora minute and observe it again. Whatdo we find? Did plasmolysis occur now?

What do we infer from this activity? Itappears that only living cells, and not deadcells, are able to absorb water by osmosis.

Cell walls permit the cells of plants, fungiand bacteria to withstand very dilute(hypotonic) external media without bursting.In such media the cells tend to take up waterby osmosis. The cell swells, building uppressure against the cell wall. The wall exertsan equal pressure against the swollen cell.Because of their walls, such cells canwithstand much greater changes in thesurrounding medium than animal cells.

5.2.3 NUCLEUS

Remember the temporary mount of onion peelwe prepared? We had put iodine solution onthe peel. Why? What would we see if we triedobserving the peel without putting the iodinesolution? Try it and see what the differenceis. Further, when we put iodine solution onthe peel, did each cell get evenly coloured?

According to their chemical compositiondifferent regions of cells get coloureddifferentially. Some regions appear darkerthan other regions. Apart from iodine solutionwe could also use safranin solution ormethylene blue solution to stain the cells.

We have observed cells from an onion; letus now observe cells from our own body.

Activity ______________ 5.7• Let us take a glass slide with a drop of

water on it. Using an ice-cream spoongently scrape the inside surface of thecheek. Does any material get stuck onthe spoon? With the help of a needlewe can transfer this material andspread it evenly on the glass slide keptready for this. To colour the materialwe can put a drop of methylene bluesolution on it. Now the material is readyfor observation under microscope. Donot forget to put a cover-slip on it!

• What do we observe? What is the shapeof the cells we see? Draw it on theobservation sheet.

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SCIENCE62

• Was there a darkly coloured, sphericalor oval, dot-like structure near thecentre of each cell? This structure iscalled nucleus. Were there similarstructures in onion peel cells?

The nucleus has a double layered coveringcalled nuclear membrane. The nuclearmembrane has pores which allow the transferof material from inside the nucleus to itsoutside, that is, to the cytoplasm (which wewill talk about in section 5.2.4).

The nucleus contains chromosomes,which are visible as rod-shaped structuresonly when the cell is about to divide.Chromosomes contain information forinheritance of features from parents to nextgeneration in the form of DNA (DeoxyriboNucleic Acid) molecules. Chromosomes arecomposed of DNA and protein. DNA moleculescontain the information necessary forconstructing and organising cells. Functionalsegments of DNA are called genes. In a cellwhich is not dividing, this DNA is present aspart of chromatin material. Chromatinmaterial is visible as entangled mass of threadlike structures. Whenever the cell is about todivide, the chromatin material gets organisedinto chromosomes.

The nucleus plays a central role in cellularreproduction, the process by which a singlecell divides and forms two new cells. It alsoplays a crucial part, along with theenvironment, in determining the way the cellwill develop and what form it will exhibit atmaturity, by directing the chemical activitiesof the cell.

In some organisms like bacteria, thenuclear region of the cell may be poorlydefined due to the absence of a nuclearmembrane. Such an undefined nuclear regioncontaining only nucleic acids is called anucleoid. Such organisms, whose cells lacka nuclear membrane, are called prokaryotes(Pro = primitive or primary; karyote ≈ karyon= nucleus). Organisms with cells having anuclear membrane are called eukaryotes.

Prokaryotic cells (see Fig. 5.4) also lackmost of the other cytoplasmic organelles

present in eukaryotic cells. Many of thefunctions of such organelles are alsoperformed by poorly organised parts of thecytoplasm (see section 5.2.4). The chlorophyllin photosynthetic prokaryotic bacteria isassociated with membranous vesicles (baglike structures) but not with plastids as ineukaryotic cells (see section 5.2.5).

RibosomesPlasmamembrane

Cell wall

Nucleoid

Fig. 5.4: Prokaryotic cell

5.2.4 CYTOPLASM

When we look at the temporary mounts ofonion peel as well as human cheek cells, wecan see a large region of each cell enclosedby the cell membrane. This region takes upvery little stain. It is called the cytoplasm.The cytoplasm is the fluid content inside theplasma membrane. It also contains manyspecialised cell organelles. Each of theseorganelles performs a specific function for thecell.

Cell organelles are enclosed bymembranes. In prokaryotes, beside theabsence of a defined nuclear region, themembrane-bound cell organelles are alsoabsent. On the other hand, the eukaryoticcells have nuclear membrane as well asmembrane-enclosed organelles.

The significance of membranes can beillustrated with the example of viruses.Viruses lack any membranes and hence donot show characteristics of life until they entera living body and use its cell machinery tomultiply.


uestion1. Fill in the gaps in the following

table illustrating differencesbetween prokaryotic andeukaryotic cells.

Prokaryotic Cell Eukaryotic Cell

1. Size : generally 1. Size: generallysmall ( 1-10 μm) large ( 5-100 μm)1 μm = 10–6 m

2. Nuclear region: 2. Nuclear region:_______________ well defined and_______________ surrounded by aand known as__ nuclear membrane

3. Chromosome: 3. More than onesingle chromosome

4. Membrane-bound 4. _______________cell organelles _______________absent _______________

5.2.5 CELL ORGANELLES

Every cell has a membrane around it to keepits own contents separate from the externalenvironment. Large and complex cells,including cells from multicellular organisms,need a lot of chemical activities to supporttheir complicated structure and function. Tokeep these activities of different kindsseparate from each other, these cells usemembrane-bound little structures (or‘organelles’) within themselves. This is one ofthe features of the eukaryotic cells thatdistinguish them from prokaryotic cells. Someof these organelles are visible only with anelectron microscope.

We have talked about the nucleus in aprevious section. Some important examplesof cell organelles which we will discuss noware: endoplasmic reticulum, Golgi apparatus,lysosomes, mitochondria, plastids andvacuoles. They are important because theycarry out some very crucial functions in cells.

5.2.5 (i) ENDOPLASMIC RETICULUM (ER)The endoplasmic reticulum (ER) is a largenetwork of membrane-bound tubes andsheets. It looks like long tubules or round oroblong bags (vesicles). The ER membrane issimilar in structure to the plasma membrane.There are two types of ER– rough endoplasmicreticulum (RER) and smooth endoplasmicreticulum (SER). RER looks rough under amicroscope because it has particles calledribosomes attached to its surface. Theribosomes, which are present in all activecells, are the sites of protein manufacture.The manufactured proteins are then sent tovarious places in the cell depending on need,using the ER. The SER helps in themanufacture of fat molecules, or lipids,important for cell function. Some of theseproteins and lipids help in building the cellmembrane. This process is known asmembrane biogenesis. Some other proteinsand lipids function as enzymes andhormones. Although the ER varies greatly inappearance in different cells, it always formsa network system.

Q

Fig. 5.5: Animal cell

Thus, one function of the ER is to serveas channels for the transport of materials(especially proteins) between various regionsof the cytoplasm or between the cytoplasmand the nucleus. The ER also functions as acytoplasmic framework providing a surface

SCIENCE64

Camillo Golgi was bornat Corteno near Bresciain 1843. He studiedmedicine at theUniversity of Pavia. Aftergraduating in 1865, hecontinued to work inPavia at the Hospital ofSt. Matteo. At that timemost of his investigationswere concerned with the nervous system,In 1872 he accepted the post of ChiefMedical Officer at the Hospital for theChronically Sick at Abbiategrasso. He firststarted his investigations into the nervoussystem in a little kitchen of this hospital,which he had converted into a laboratory.However, the work of greatest importance,which Golgi carried out was a revolutionarymethod of staining individual nerve and cellstructures. This method is referred to asthe ‘black reaction’. This method uses aweak solution of silver nitrate and isparticularly valuable in tracing theprocesses and most delicate ramificationsof cells. All through his life, he continuedto work on these lines, modifying andimproving this technique. Golgi receivedthe highest honours and awards inrecognition of his work. He shared theNobel prize in 1906 with Santiago RamonyCajal for their work on the structure of thenervous system.

5.2.5 (iii) LYSOSOMES

Lysosomes are a kind of waste disposalsystem of the cell. Lysosomes help to keepthe cell clean by digesting any foreign materialas well as worn-out cell organelles. Foreignmaterials entering the cell, such as bacteriaor food, as well as old organelles end up inthe lysosomes, which break them up intosmall pieces. Lysosomes are able to do thisbecause they contain powerful digestiveenzymes capable of breaking down all organicmaterial. During the disturbance in cellularmetabolism, for example, when the cell gets

Fig. 5.6: Plant cell

for some of the biochemical activities of thecell. In the liver cells of the group of animalscalled vertebrates (see Chapter 7), SER playsa crucial role in detoxifying many poisons anddrugs.

5.2.5 (ii) GOLGI APPARATUS

The Golgi apparatus, first described byCamillo Golgi, consists of a system ofmembrane-bound vesicles arrangedapproximately parallel to each other in stackscalled cisterns. These membranes often haveconnections with the membranes of ER andtherefore constitute another portion of acomplex cellular membrane system.

The material synthesised near the ER ispackaged and dispatched to various targetsinside and outside the cell through the Golgiapparatus. Its functions include the storage,modification and packaging of products invesicles. In some cases, complex sugars maybe made from simple sugars in the Golgiapparatus. The Golgi apparatus is alsoinvolved in the formation of lysosomes [see5.2.5 (iii)].


damaged, lysosomes may burst and theenzymes digest their own cell. Therefore,lysosomes are also known as the ‘suicidebags’ of a cell. Structurally, lysosomes aremembrane-bound sacs filled with digestiveenzymes. These enzymes are made by RER.

5.2.5 (iv) MITOCHONDRIA

Mitochondria are known as the powerhousesof the cell. The energy required for variouschemical activities needed for life is releasedby mitochondria in the form of ATP(Adenosine triphopshate) molecules. ATP isknown as the energy currency of the cell. Thebody uses energy stored in ATP for makingnew chemical compounds and for mechanicalwork. Mitochondria have two membranecoverings instead of just one. The outermembrane is very porous while the innermembrane is deeply folded. These folds createa large surface area for ATP-generatingchemical reactions.

Mitochondria are strange organelles in thesense that they have their own DNA andribosomes. Therefore, mitochondria are ableto make some of their own proteins.

5.2.5 (V) PLASTIDS

Plastids are present only in plant cells. Thereare two types of plastids – chromoplasts(coloured plastids) and leucoplasts (white orcolourless plastids). Plastids containing thepigment chlorophyll are known aschloroplasts. Chloroplasts are important forphotosynthesis in plants. Chloroplasts alsocontain various yellow or orange pigments inaddition to chlorophyll. Leucoplasts areprimarily organelles in which materials suchas starch, oils and protein granules arestored.

The internal organisation of the plastidsconsists of numerous membrane layersembedded in a material called the stroma.Plastids are similar to mitochondria inexternal structure. Like the mitochondria,plastids also have their own DNA andribosomes.

5.2.5 (vi) VACUOLES

Vacuoles are storage sacs for solid or liquidcontents. Vacuoles are small sized in animalcells while plant cells have very large vacuoles.The central vacuole of some plant cells mayoccupy 50-90% of the cell volume.

In plant cells vacuoles are full of cell sapand provide turgidity and rigidity to the cell.Many substances of importance in the life ofthe plant cell are stored in vacuoles. Theseinclude amino acids, sugars, various organicacids and some proteins. In single-celledorganisms like Amoeba, the food vacuolecontains the food items that the Amoeba hasconsumed. In some unicellular organisms,specialised vacuoles also play important rolesin expelling excess water and some wastesfrom the cell.

uestions1. Can you name the two organelles

we have studied that containtheir own genetic material?

2. If the organisation of a cell isdestroyed due to some physicalor chemical influence, what willhappen?

3. Why are lysosomes known assuicide bags?

4. Where are proteins synthesisedinside the cell?

Each cell thus acquires its structure andability to function because of the organisationof its membrane and organelles in specificways. The cell thus has a basic structuralorganisation. This helps the cells to performfunctions like respiration, obtaining nutrition,and clearing of waste material, or forming newproteins.

Thus, the cell is the fundamentalstructural unit of living organisms. It is alsothe basic functional unit of life.

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SCIENCE66

Whatyou havelearnt• The fundamental organisational unit of life is the cell.

• Cells are enclosed by a plasma membrane composed of lipidsand proteins.

• The cell membrane is an active part of the cell. It regulates themovement of materials between the ordered interior of the celland the outer environment.

• In plant cells, a cell wall composed mainly of cellulose is locatedoutside the cell membrane.

• The presence of the cell wall enables the cells of plants, fungiand bacteria to exist in hypotonic media without bursting.

• The nucleus in eukaryotes is separated from the cytoplasm bydouble-layered membrane and it directs the life processes ofthe cell.

• The ER functions both as a passageway for intracellulartransport and as a manufacturing surface.

• The Golgi apparatus consists of stacks of membrane-boundvesicles that function in the storage, modification and packagingof substances manufactured in the cell.

• Most plant cells have large membranous organelles calledplastids, which are of two types – chromoplasts and leucoplasts.

• Chromoplasts that contain chlorophyll are called chloroplastsand they perform photosynthesis.

• The primary function of leucoplasts is storage.

• Most mature plant cells have a large central vacuole that helpsto maintain the turgidity of the cell and stores importantsubstances including wastes.

• Prokaryotic cells have no membrane-bound organelles, theirchromosomes are composed of only nucleic acid, and they haveonly very small ribosomes as organelles.

Exercises1. Make a comparison and write down ways in which plant cells

are different from animal cells.

2. How is a prokaryotic cell different from a eukaryotic cell?

3. What would happen if the plasma membrane ruptures or breaksdown?


4. What would happen to the life of a cell if there was no Golgiapparatus?

5. Which organelle is known as the powerhouse of the cell? Why?

6. Where do the lipids and proteins constituting the cell membraneget synthesised?

7. How does an Amoeba obtain its food?

8. What is osmosis?

9. Carry out the following osmosis experiment:

Take four peeled potato halves and scoos each one out to makepotato cups. One of these potato cups should be made from aboiled potato. Put each potato cup in a trough containing water.Now,

(a) Keep cup A empty

(b) Put one teaspoon sugar in cup B

(c) Put one teaspoon salt in cup C

(d) Put one teaspoon sugar in the boiled potato cup D.

Keep these for two hours. Then observe the four potato cupsand answer the following:

(i) Explain why water gathers in the hollowed portion ofB and C.

(ii) Why is potato A necessary for this experiment?

(iii) Explain why water does not gather in the hollowed outportions of A and D.

From the last chapter, we recall that all livingorganisms are made of cells. In unicellularorganisms, a single cell performs all basicfunctions. For example, in Amoeba, a singlecell carries out movement, intake of food andrespiratory gases, respiration and excretion.But in multi-cellular organisms there aremillions of cells. Most of these cells arespecialised to carry out a few functions. Eachspecialised function is taken up by a differentgroup of cells. Since these cells carry out onlya particular function, they do it veryefficiently. In human beings, muscle cellscontract and relax to cause movement, nervecells carry messages, blood flows to transportoxygen, food, hormones and waste materialand so on. In plants, vascular tissues conductfood and water from one part of the plant toother parts. So, multi-cellular organismsshow division of labour. Cells specialising inone function are often grouped together inthe body. This means that a particularfunction is carried out by a cluster of cells ata definite place in the body. This cluster ofcells, called a tissue, is arranged and designedso as to give the highest possible efficiency offunction. Blood, phloem and muscle are allexamples of tissues.

A group of cells that are similar instructure and/or work together to achieve aparticular function forms a tissue.

6.1 Are Plants and Animals Madeof Same Types of Tissues?

Let us compare their structure and functions.Do plants and animals have the samestructure? Do they both perform similarfunctions?

There are noticeable differences betweenthe two. Plants are stationary or fixed – theydon’t move. Most of the tissues they have aresupportive, which provides them withstructural strength. Most of these tissues aredead, since dead cells can provide mechanicalstrength as easily as live ones, and need lessmaintenance.

Animals on the other hand move aroundin search of food, mates and shelter. Theyconsume more energy as compared to plants.Most of the tissues they contain are living.

Another difference between animals andplants is in the pattern of growth. The growthin plants is limited to certain regions, whilethis is not so in animals. There are sometissues in plants that divide throughout theirlife. These tissues are localised in certainregions. Based on the dividing capacity of thetissues, various plant tissues can be classifiedas growing or meristematic tissue andpermanent tissue. Cell growth in animals ismore uniform. So, there is no suchdemarcation of dividing and non-dividingregions in animals.

The structural organisation of organs andorgan systems is far more specialised andlocalised in complex animals than even in verycomplex plants. This fundamental differencereflects the different modes of life pursuedby these two major groups of organisms,particularly in their different feeding methods.Also, they are differently adapted for asedentary existence on one hand (plants) andactive locomotion on the other (animals),contributing to this difference in organ systemdesign.

It is with reference to these complexanimal and plant bodies that we will now talkabout the concept of tissues in some detail.

66666TTTTTISSUESISSUESISSUESISSUESISSUES

Chapter

uestions1. What is a tissue?2. What is the utility of tissues in

multi-cellular organisms?

6.2 Plant Tissues6.2.1 MERISTEMATIC TISSUE

• From the above observations, answerthe following questions:1. Which of the two onions has longer

roots? Why?2. Do the roots continue growing even

after we have removed their tips?3. Why would the tips stop growing in

jar 2 after we cut them?

The growth of plants occurs only in certainspecific regions. This is because the dividingtissue, also known as meristematic tissue, islocated only at these points. Depending onthe region where they are present,meristematic tissues are classified as apical,lateral and intercalary (Fig. 6.2). New cellsproduced by meristem are initially like thoseof meristem itself, but as they grow andmature, their characteristics slowly changeand they become dif ferentiated ascomponents of other tissues.

Fig. 6.1: Growth of roots in onion bulbs

Activity ______________ 6.1• Take two glass jars and fill them with

water.• Now, take two onion bulbs and place

one on each jar, as shown inFig. 6.1.

• Observe the growth of roots in both thebulbs for a few days.

• Measure the length of roots on day 1,2 and 3.

• On day 4, cut the root tips of the onionbulb in jar 2 by about 1 cm. After this,observe the growth of roots in both thejars and measure their lengths eachday for five more days and record theobservations in tables, like the tablebelow:

Length Day 1 Day 2 Day 3 Day 4 Day 5

Jar 1

Jar 2

Q

Apical meristem is present at the growingtips of stems and roots and increases thelength of the stem and the root. The girth ofthe stem or root increases due to lateralmeristem (cambium). Intercalary meristem isthe meristem at the base of the leaves orinternodes (on either side of the node)on twigs.

Apical meristem

Intercalary meristem

Lateral meristem

Fig. 6.2: Location of meristematic tissue in plant body

Jar 1 Jar 2

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SCIENCE70

As the cells of this tissue are very active,they have dense cytoplasm, thin cellulosewalls and prominent nuclei. They lackvacuoles. Can we think why they would lackvacuoles? (You might want to refer to thefunctions of vacuoles in the chapter on cells.)

6.2.2 PERMANENT TISSUE

What happens to the cells formed bymeristematic tissue? They take up a specificrole and lose the ability to divide. As a result,they form a permanent tissue. This processof taking up a permanent shape, size, and afunction is called differentiation. Cells ofmeristematic tissue differentiate to formdifferent types of permanent tissue.

• Now, answer the following on the basisof your observation:1. Are all cells similar in structure?2. How many types of cells can

be seen?3. Can we think of reasons why there

would be so many types of cells?• We can also try to cut sections of plant

roots. We can even try cutting sectionsof root and stem of different plants.

6.2.2 (i) SIMPLE PERMANENT TISSUE

A few layers of cells form the basic packingtissue. This tissue is parenchyma, a type ofpermanent tissue. It consists of relativelyunspecialised cells with thin cell walls. Theyare live cells. They are usually loosely packed,

Trichome

Mucilaginous canalCuticle

Epidermis

Hypodermis

Cortex

Endodermis

Pericycle

PhloemCambium

Metaxylem

Protoxylem

Vascular bundlePith

Medullary ray

Xylem

Fig. 6.3: Section of a stem

Activity ______________ 6.2• Take a plant stem and with the help

of your teacher cut into very thin slicesor sections.

• Now, stain the slices with safranin.Place one neatly cut section on a slide,and put a drop of glycerine.

• Cover with a cover-slip and observeunder a microscope. Observe thevarious types of cells and theirarrangement. Compare it with Fig. 6.3.

so that large spaces between cells(intercellular spaces) are found in this tissue[Fig. 6.4 a(i)]. This tissue provides support toplants and also stores food. In somesituations, it contains chlorophyll andperforms photosynthesis, and then it is calledchlorenchyma. In aquatic plants, large aircavities are present in parenchyma to givebuoyancy to the plants to help them float.Such a parenchyma type is calledaerenchyma. The parenchyma of stems androots also stores nutrients and water.

TISSUES 71

Fig. 6.4: Various types of simple tissues: (a) Parenchyma (i) transverse section, (ii) longitudinal section;(b) Collenchyma (i) transverse section, (ii) longitudinal section; (c) Sclerenchyma (i) transverse section,(ii) longitudinal section.

b (i)

Wall thickenings

Nucleus

VacuoleCell wall

b (ii)

End wall

Primary cell wall(thickened at corners)

Chloroplast

Nucleus

Vacuole

Cytoplasm

Intercellular space

Cytoplasm

Nucleus

Middle lamella

Chloroplast

Vacuole

Intercellular space

Primary cell wall

a (ii)

a (i)

Intercellular spaces

Narrow lumen

Lignifiedthick wall

c (ii)

Simplepit pair

c (i)

The flexibility in plants is due to anotherpermanent tissue, collenchyma. It allowseasy bending in various parts of a plant (leaf,stem) without breaking. It also providesmechanical support to plants. We can find

this tissue in leaf stalks below the epidermis.The cells of this tissue are living, elongatedand irregularly thickened at thecorners. There is very little intercellularspace (Fig. 6.4 b).

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Yet another type of permanent tissue issclerenchyma. It is the tissue which makesthe plant hard and stiff. We have seen thehusk of a coconut. It is made ofsclerenchymatous tissue. The cells of thistissue are dead. They are long and narrow asthe walls are thickened due to lignin (achemical substance which acts as cement andhardens them). Often these walls are so thickthat there is no internal space inside the cell(Fig. 6.4 c). This tissue is present in stems,around vascular bundles, in the veins ofleaves and in the hard covering of seeds andnuts. It provides strength to the plant parts.

Activity ______________ 6.3• Take a freshly plucked leaf of Rheo.• Stretch and break it by applying

pressure.• While breaking it, keep it stretched

gently so that some peel or skinprojects out from the cut.

• Remove this peel and put it in a petridish filled with water.

• Add a few drops of safranin.• Wait for a couple of minutes and then

transfer it onto a slide. Gently place acover slip over it.

• Observe under microscope.

epidermis may be thicker since protectionagainst water loss is critical. The entiresurface of a plant has this outer covering ofepidermis. It protects all the parts of the plant.Epidermal cells on the aerial parts of the plantoften secrete a waxy, water-resistant layer ontheir outer surface. This aids in protectionagainst loss of water, mechanical injury andinvasion by parasitic fungi. Since it has aprotective role to play, cells of epidermaltissue form a continuous layer withoutintercellular spaces. Most epidermal cells arerelatively flat. Often their outer and side wallsare thicker than the inner wall.

We can observe small pores here and therein the epidermis of the leaf. These pores arecalled stomata (Fig. 6.5). Stomata areenclosed by two kidney-shaped cells calledguard cells. They are necessary forexchanging gases with the atmosphere.Transpiration (loss of water in the form ofwater vapour) also takes place throughstomata.

Think about which gas may be requiredfor photosynthesis.Find out the role of transpiration in plants.

Epidermal cells of the roots, whosefunction is water absorption, commonly bearlong hair-like parts that greatly increase thetotal absorptive surface area.

In some plants like desert plants,epidermis has a thick waxy coating of cutin(chemical substance with waterproof quality)on its outer surface. Can we think of a reasonfor this?

Is the outer layer of a branch of a treedifferent from the outer layer of a young stem?

As plants grow older, the outer protectivetissue undergoes certain changes. A strip ofsecondary meristem replaces the epidermisof the stem. Cells on the outside are cut offfrom this layer. This forms the several-layerthick cork or the bark of the tree. Cells ofcork are dead and compactly arrangedwithout intercellular spaces (Fig. 6.6). Theyalso have a chemical called suberin in theirwalls that makes them impervious to gasesand water.

Fig. 6.5: Guard cells and epidermal cells: (a) lateralview, (b) surface view

(a) (b)

Guardcell

Stomata

Epidermalcell

Guardcells

What you observe is the outermost layerof cells, called epidermis. The epidermis isusually made of a single layer of cells. In someplants living in very dry habitats, the

TISSUES 73

6.2.2 (ii) COMPLEX PERMANENT TISSUE

The different types of tissues we havediscussed until now are all made of one typeof cells, which look like each other. Suchtissues are called simple permanent tissue.Yet another type of permanent tissue iscomplex tissue. Complex tissues are made ofmore than one type of cells. All these cellscoordinate to perform a common function.Xylem and phloem are examples of suchcomplex tissues. They are both conductingtissues and constitute a vascular bundle.Vascular or conductive tissue is a distinctivefeature of the complex plants, one that hasmade possible their survival in the terrestrialenvironment. In Fig. 6.3 showing a section ofstem, can you see different types of cells inthe vascular bundle?

Xylem consists of tracheids, vessels,xylem parenchyma (Fig. 6.7 a,b,c) and xylemfibres. The cells have thick walls, and manyof them are dead cells. Tracheids and vesselsare tubular structures. This allows them totransport water and minerals vertically. Theparenchyma stores food and helps in thesideways conduction of water. Fibres aremainly supportive in function.

Phloem is made up of four types ofelements: sieve tubes, companion cells,phloem fibres and the phloem parenchyma[Fig. 6.7 (d)]. Sieve tubes are tubular cells withperforated walls. Phloem is unlike xylem inthat materials can move in both directions init. Phloem transports food from leaves to other

Cork cells Ruptured epidermis

Phloem

Fig. 6.6: Protective tissue

parts of the plant. Except for phloem fibres,phloem cells are living cells.

Xylem

Sieve plate

Sieve tube

Phloemparenchyma

Companion cell

Nucleus

CytoplasmPits

Pit

(a) Tracheid (b) Vessel (c) Xylem parenchyma

Fig. 6.7: Types of complex tissue

(d) Section of phloem

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uestions1. Name types of simple tissues.2. Where is apical meristem found?3. Which tissue makes up the husk

of coconut?4. What are the constituents of

phloem?

6.3 Animal TissuesWhen we breathe we can actually feel themovement of our chest. How do these bodyparts move? For this we have specialised cellscalled muscle cells (Fig. 6.8). The contractionand relaxation of these cells result inmovement.

During breathing we inhale oxygen. Wheredoes this oxygen go? It is absorbed in thelungs and then is transported to all the bodycells through blood. Why would cells needoxygen? The functions of mitochondria westudied earlier provide a clue to this question.Blood flows and carries various substancesfrom one part of the body to the other. Forexample, it carries oxygen and food to all cells.It also collects wastes from all parts of thebody and carries them to the liver and kidneyfor disposal.

Blood and muscles are both examples oftissues found in our body. On the basis ofthe functions they perform we can think ofdifferent types of animal tissues, such asepithelial tissue, connective tissue, musculartissue and nervous tissue. Blood is a type ofconnective tissue, and muscle formsmuscular tissue.

6.3.1 EPITHELIAL TISSUE

The covering or protective tissues in theanimal body are epithelial tissues. Epitheliumcovers most organs and cavities within thebody. It also forms a barrier to keep differentbody systems separate. The skin, the liningof the mouth, the lining of blood vessels, lungalveoli and kidney tubules are all made ofepithelial tissue. Epithelial tissue cells aretightly packed and form a continuous sheet.They have only a small amount of cementingmaterial between them and almost nointercellular spaces. Obviously, anythingentering or leaving the body must cross atleast one layer of epithelium. As a result, thepermeability of the cells of various epitheliaplay an important role in regulating theexchange of materials between the body andthe external environment and also betweendifferent parts of the body. Regardless of thetype, all epithelium is usually separated fromthe underlying tissue by an extracellularfibrous basement membrane.

Different epithelia (Fig. 6.9) show differingstructures that correlate with their uniquefunctions. For example, in cells lining bloodvessels or lung alveoli, where transportationof substances occurs through a selectively

Q

Fig. 6.8: Location of muscle fibres

Smooth muscle fibres

Smooth muscle fibre

(Cell)

Nucleus

TISSUES 75

permeable surface, there is a simple flat kindof epithelium. This is called the simple

squamous epithelium. Simple squamousepithelial cells are extremely thin and flat andform a delicate lining. The oesophagus andthe lining of the mouth are also covered withsquamous epithelium. The skin, whichprotects the body, is also made of squamousepithelium. Skin epithelial cells are arrangedin many layers to prevent wear and tear. Sincethey are arranged in a pattern of layers, theepithelium is called stratified squamousepithelium.

Where absorption and secretion occur, asin the inner lining of the intestine, tallepithelial cells are present. This columnar(meaning ‘pillar-like’) epithelium facilitatesmovement across the epithelial barrier. In therespiratory tract, the columnar epithelialtissue also has cilia, which are hair-likeprojections on the outer surfaces of epithelialcells. These cilia can move, and theirmovement pushes the mucus forward to clearit. This type of epithelium is thus ciliatedcolumnar epithelium.

Cuboidal epithelium (with cube-shapedcells) forms the lining of kidney tubules andducts of salivary glands, where it providesmechanical support. Epithelial cells oftenacquire additional specialisation as glandcells, which can secrete substances at theepithelial surface. Sometimes a portion of theepithelial tissue folds inward, and amulticellular gland is formed. This isglandular epithelium.

6.3.2 CONNECTIVE TISSUE

Blood is a type of connective tissue. Whywould it be called ‘connective’ tissue? A clueis provided in the introduction of this chapter!Now, let us look at this type of tissue in somemore detail. The cells of connective tissue areloosely spaced and embedded in anintercellular matrix (Fig. 6.10). The matrixmay be jelly like, fluid, dense or rigid. Thenature of matrix differs in concordance withthe function of the particular connectivetissue.

Take a drop of blood on a slide and observedifferent cells present in it under a microscope.

(a) Squamous

(d) Stratified squamous

(c) Columnar (Ciliated)

(b) Cuboidal

Fig. 6.9: Different types of epithelial tissues

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Blood has a fluid (liquid) matrix calledplasma, in which red blood cells (RBCs), whiteblood cells (WBCs) and platelets aresuspended. The plasma contains proteins,salts and hormones. Blood flows andtransports gases, digested food, hormonesand waste materials to different parts of thebody.

Bone is another example of a connectivetissue. It forms the framework that supportsthe body. It also anchors the muscles andsupports the main organs of the body. It is astrong and nonflexible tissue (what would bethe advantage of these properties for bonefunctions?). Bone cells are embedded in ahard matrix that is composed of calcium andphosphorus compounds.

Two bones can be connected to each otherby another type of connective tissue calledthe ligament. This tissue is very elastic. It hasconsiderable strength. Ligaments containvery little matrix. Tendons connect bones tomuscles and are another type of connectivetissue. Tendons are fibrous tissue with greatstrength but limited flexibility.

Another type of connective tissue,cartilage, has widely spaced cells. The solidmatrix is composed of proteins and sugars.Cartilage smoothens bone surfaces at jointsand is also present in the nose, ear, tracheaand larynx. We can fold the cartilage of theears, but we cannot bend the bones in ourarms. Think of how the two tissues aredifferent!

Areolar connective tissue is found betweenthe skin and muscles, around blood vesselsand nerves and in the bone marrow. It fillsthe space inside the organs, supports internalorgans and helps in repair of tissues.

Where are fats stored in our body? Fat-storing adipose tissue is found below the skinand between internal organs. The cells of thistissue are filled with fat globules. Storage offats also lets it act as an insulator.

6.3.3 MUSCULAR TISSUE

Muscular tissue consists of elongated cells,also called muscle fibres. This tissue isresponsible for movement in our body.

Reticular fibre Fibroblast

Macrophage

Collagen fibreMast cell Plasma cell

Fat droplet Nucleus

Adipocyte

(a)

(b)Haversian canal(contains blood vessels

and nerve fibres)

Canaliculus (containsslender process of bone

cell or osteocyte)

Hyaline matrix

Chondrocyte

Different whiteblood corpuscles

Neutrophil(polynuclearleucocyte)

BasophilEosinophil

Nucleus

Cytoplasm

Red bloodcorpuscle

Lymphocyte Monocyte Platelets

(c)

(d)

(e)

Fig. 6.10: Types of connective tissues: (a) areolartissue, (b) adipose tissue, (c) compactbone, (d) hyaline cartilage, (e) types ofblood cells

TISSUES 77

Muscles contain special proteins calledcontractile proteins, which contract and relaxto cause movement.

to bones and help in body movement. Underthe microscope, these muscles show alternatelight and dark bands or striations whenstained appropriately. As a result, they arealso called striated muscles. The cells of thistissue are long, cylindrical, unbranched andmultinucleate (having many nuclei).

The movement of food in the alimentarycanal or the contraction and relaxation ofblood vessels are involuntary movements. Wecannot really start them or stop them simplyby wanting to do so! Smooth muscles [Fig.6.11(b)] or involuntary muscles control suchmovements. They are also found in the iris ofthe eye, in ureters and in the bronchi of thelungs. The cells are long with pointed ends(spindle-shaped) and uninucleate (having asingle nucleus). They are also calledunstriated muscles – why would they becalled that?

The muscles of the heart show rhythmiccontraction and relaxation throughout life.These involuntary muscles are called cardiacmuscles [Fig. 6.11(c)]. Heart muscle cells arecylindrical, branched and uninucleate.

Compare the structures of different typesof muscular tissues. Note their shape,number of nuclei and position of nuclei withinthe cell.

6.3.4 NERVOUS TISSUE

All cells possess the ability to respond tostimuli. However, cells of the nervous tissueare highly specialised for being stimulatedand then transmitting the stimulus veryrapidly from one place to another within thebody. The brain, spinal cord and nerves areall composed of the nervous tissue. The cellsof this tissue are called nerve cells or neurons.A neuron consists of a cell body with anucleus and cytoplasm, from which long thinhair-like parts arise (Fig. 6.12). Usually eachneuron has a single long part, called the axon,and many short, branched parts calleddendrites. An individual nerve cell may be upto a metre long. Many nerve fibres boundtogether by connective tissue make upa nerve.

Nuclei

Striations

(a)Spindle shaped

muscle cell

Nucleus

(b)

(c)

Striations

Nuclei

Fig. 6.11: Types of muscles fibres: (a) striatedmuscle, (b) smooth muscle, (c) cardiacmuscle

We can move some muscles by consciouswill. Muscles present in our limbs move whenwe want them to, and stop when we so decide.Such muscles are called voluntary muscles[Fig. 6.11(a)]. These muscles are also calledskeletal muscles as they are mostly attached

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Nerve impulses allow us to move ourmuscles when we want to. The functional

Nucleus

Dendrite

AxonNerve ending

Cell body

combination of nerve and muscle tissue isfundamental to most animals. Thiscombination enables animals to move rapidlyin response to stimuli.

uestions1. Name the tissue responsible for

movement in our body.2. What does a neuron look like?3. Give three features of cardiac

muscles.4. What are the functions of areolar

tissue?

QWhatyou havelearnt• Tissue is a group of cells similar in structure and function.

• Plant tissues are of two main types – meristematic andpermanent.

• Meristematic tissue is the dividing tissue present in the growingregions of the plant.

• Permanent tissues are derived from meristematic tissue oncethey lose the ability to divide. They are classified as simple andcomplex tissues.

• Parenchyma, collenchyma and sclerenchyma are three typesof simple tissues. Xylem and phloem are types of complextissues.

• Animal tissues can be epithelial, connective, muscular andnervous tissue.

• Depending on shape and function, epithelial tissue is classifiedas squamous, cuboidal, columnar, ciliated and glandular.

• The different types of connective tissues in our body includeareolar tissue, adipose tissue, bone, tendon, ligament, cartilageand blood.

• Striated, unstriated and cardiac are three types of muscletissues.

• Nervous tissue is made of neurons that receive and conductimpulses.

Fig. 6.12: Neuron-unit of nervous tissue

TISSUES 79

Exercises1. Define the term “tissue”.

2. How many types of elements together make up the xylem tissue?Name them.

3. How are simple tissues different from complex tissues in plants?

4. Differentiate between parenchyma, collenchyma andsclerenchyma on the basis of their cell wall.

5. What are the functions of the stomata?

6. Diagrammatically show the difference between the three typesof muscle fibres.

7. What is the specific function of the cardiac muscle?

8. Differentiate between striated, unstriated and cardiac muscleson the basis of their structure and site/location in the body.

9. Draw a labelled diagram of a neuron.

10. Name the following.

(a) Tissue that forms the inner lining of our mouth.

(b) Tissue that connects muscle to bone in humans.

(c) Tissue that transports food in plants.

(d) Tissue that stores fat in our body.

(e) Connective tissue with a fluid matrix.

(f) Tissue present in the brain.

11. Identify the type of tissue in the following: skin, bark of tree,bone, lining of kidney tubule, vascular bundle.

12. Name the regions in which parenchyma tissue is present.

13. What is the role of epidermis in plants?

14. How does the cork act as a protective tissue?

15. Complete the table:

Have you ever thought of the multitude oflife-forms that surround us? Each organismis different from all others to a lesser orgreater extent. For instance, consider yourselfand a friend.• Are you both of the same height?• Does your nose look exactly like your

friend’s nose?• Is your hand-span the same as your

friend’s?However, if we were to compare ourselves

and our friends with a monkey, what wouldwe say? Obviously, we and our friends havea lot in common when we compare ourselveswith a monkey. But suppose we were to adda cow to the comparison? We would thenthink that the monkey has a lot more incommon with us than with the cow.

Activity ______________ 7.1• We have heard of ‘desi’ cows and Jersey

cows.• Does a desi cow look like a Jersey cow?• Do all desi cows look alike?• Will we be able to identify a Jersey cow

in a crowd of desi cows that don’t looklike each other?

• What is the basis of our identification?

In this activity, we had to decide whichcharacteristics were more important informing the desired category. Hence, we werealso deciding which characteristics could beignored.

Now, think of all the different forms inwhich life occurs on earth. On one hand wehave microscopic bacteria of a few micrometrein size. While on the other hand we have bluewhale and red wood trees of californiaof approximate sizes of 30 metres and100 metres repectively. Some pine trees live

for thousands of years while insects likemosquitoes die within a few days. Life alsoranges from colourless or even transparentworms to brightly coloured birds and flowers.

This bewildering variety of life around ushas evolved on the earth over millions ofyears. However, we do not have more than atiny fraction of this time to try andunderstand all these living organisms, so wecannot look at them one by one. Instead, welook for similarities among the organisms,which will allow us to put them into differentclasses and then study different classes orgroups as a whole.

In order to make relevant groups to studythe variety of life forms, we need to decidewhich characteristics decide morefundamental differences among organisms.This would create the main broad groups oforganisms. Within these groups, smaller sub-groups will be decided by less importantcharacteristics.

uestions1. Why do we classify organisms?2. Give three examples of the range

of variations that you see in life-forms around you.

7.1 What is the Basis ofClassification?

Attempts at classifying living things intogroups have been made since timeimmemorial. Greek thinker Aristotle classifiedanimals according to whether they lived on

Q

77777DDDDDIVERSITYIVERSITYIVERSITYIVERSITYIVERSITY INININININ L L L L LIVINGIVINGIVINGIVINGIVING O O O O ORGANISMSRGANISMSRGANISMSRGANISMSRGANISMS

Chapter

land, in water or in the air. This is a verysimple way of looking at life, but misleadingtoo. For example, animals that live in the seainclude corals, whales, octopuses, starfishand sharks. We can immediately see thatthese are very different from each other innumerous ways. In fact, their habitat is theonly point they share in common. This is nogood as a way of making groups of organismsto study and think about.

We therefore need to decide whichcharacteristics to be used as the basis formaking the broadest divisions. Then we willhave to pick the next set of characteristicsfor making sub-groups within these divisions.This process of classification within eachgroup can then continue using newcharacteristics each time.

Before we go on, we need to think aboutwhat is meant by ‘characteristics’. When weare trying to classify a diverse group oforganisms, we need to find ways in whichsome of them are similar enough to bethought of together. These ‘ways’, in fact, aredetails of appearance or behaviour, in otherwords, form and function.

What we mean by a characteristic is aparticular form or a particular function. Thatmost of us have five fingers on each hand isthus a characteristic. That we can run, butthe banyan tree cannot, is also acharacteristic.

Now, to understand how somecharacteristics are decided as being morefundamental than others, let us consider howa stone wall is built. The stones used will havedifferent shapes and sizes. The stones at thetop of the wall would not influence the choiceof stones that come below them. On the otherhand, the shapes and sizes of stones in thelowermost layer will decide the shape and sizeof the next layer and so on.

The stones in the lowermost layer are likethe characteristics that decide the broadestdivisions among living organisms. They areindependent of any other characteristics intheir effects on the form and function of theorganism. The characteristics in the next levelwould be dependent on the previous one andwould decide the variety in the next level. In

this way, we can build up a whole hierarchyof mutually related characteristics to be usedfor classification.

Now-a-days, we look at many inter-relatedcharacteristics starting from the nature of thecell in order to classify all living organisms.What are some concrete examples of suchcharacteristics used for a hierarchicalclassification?• A eukaryotic cell has membrane-bound

organelles, including a nucleus, whichallow cellular processes to be carried outefficiently in isolation from each other.Therefore, organisms which do not havea clearly demarcated nucleus and otherorganelles would need to have theirbiochemical pathways organised in verydifferent ways. This would have an effecton every aspect of cell design. Further,nucleated cells would have the capacityto participate in making a multicellularorganism because they can take upspecialised functions. Therefore, this isa basic characteristic of classification.

• Do the cells occur singly or are theygrouped together and do they live as anindivisible group? Cells that grouptogether to form a single organism usethe principle of division of labour. In sucha body design, all cells would not beidentical. Instead, groups of cells willcarry out specialised functions. Thismakes a very basic distinction in thebody designs of organisms. As a result,an Amoeba and a worm are very differentin their body design.

• Do organisms produce their own foodthrough the process of photosynthesis?Being able to produce one’s own foodversus having to get food from outsidewould make very different body designsnecessary.

• Of the organisms that performphotosynthesis (plants), what is the levelof organisation of their body?

• Of the animals, how does the individual’sbody develop and organise its differentparts, and what are the specialisedorgans found for different functions?

DIVERSITY IN LIVING ORGANISMS 81

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We can see that, even in these fewquestions that we have asked, a hierarchy isdeveloping. The characteristics of body designused for classification of plants will be verydifferent from those important for classifyinganimals. This is because the basic designsare different, based on the need to make theirown food (plants), or acquire it (animals).Therefore, these design features (having askeleton, for example) are to be used to makesub-groups, rather than making broad groups.

uestions1. Which do you think is a more basic

characteristic for classifyingorganisms?(a) the place where they live.(b) the kind of cells they are

made of. Why?2. What is the primary characteristic

on which the first division oforganisms is made?

3. On what bases are plants andanimals put into differentcategories?

7.2 Classification and EvolutionAll living things are identified and categorisedon the basis of their body design in form andfunction. Some characteristics are likely tomake more wide-ranging changes in bodydesign than others. There is a role of time inthis as well. So, once a certain body designcomes into existence, it will shape the effectsof all other subsequent design changes,simply because it already exists. In otherwords, characteristics that came intoexistence earlier are likely to be more basicthan characteristics that have come intoexistence later.

This means that the classification of lifeforms will be closely related to their evolution.What is evolution? Most life forms that wesee today have arisen by an accumulation ofchanges in body design that allow theorganism possessing them to survive better.Charles Darwin first described this idea ofevolution in 1859 in his book, The Origin ofSpecies.

When we connect this idea of evolution toclassification, we will find some groups oforganisms which have ancient body designsthat have not changed very much. We willalso find other groups of organisms that haveacquired their particular body designsrelatively recently. Those in the first groupare frequently referred to as ‘primitive’ or ‘lower’organisms, while those in the second groupare called ‘advanced’ or ‘higher’ organisms. Inreality, these terms are not quite correct sincethey do not properly relate to the differences.All that we can say is that some are ‘older’organisms, while some are ‘younger’organisms. Since there is a possibility thatcomplexity in design will increase overevolutionary time, it may not be wrong to saythat older organisms are simpler, whileyounger organisms are more complex.Q

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Biodiversity means the diversity of lifeforms. It is a word commonly used torefer to the variety of life forms foundin a particular region. Diverse life formsshare the environment, and areaffected by each other too. As a result,a stable community of different speciescomes into existence. Humans haveplayed their own part in recent timesin changing the balance of suchcommunities. Of course, the diversityin such communities is affected byparticular characteristics of land,water, climate and so on. Roughestimates state that there are about tenmillion species on the planet, althoughwe actually know only one or twomillions of them. The warm and humidtropical regions of the earth, betweenthe tropic of Cancer and the tropic ofCapricorn, are rich in diversity of plantand animal life. This is called the regionof megadiversity. Of the biodiversityof the planet, more than half isconcentrated in a few countries –Brazil, Colombia, Ecuador, Peru,Mexico, Zaire, Madagascar,Australia, China, India, Indonesia andMalaysia.


uestions1. Which organisms are called

primitive and how are theydifferent from the so-calledadvanced organisms?

2. Will advanced organisms be thesame as complex organisms?Why?

7.3 The Hierarchy of Classification-Groups

Biologists, such as Ernst Haeckel (1894),Robert Whittaker (1959) and Carl Woese(1977) have tried to classify all livingorganisms into broad categories, calledkingdoms. The classification Whittakerproposed has five kingdoms: Monera,Protista, Fungi, Plantae and Animalia, andis widely used. These groups are formed onthe basis of their cell structure, mode andsource of nutrition and body organisation.The modification Woese introduced bydividing the Monera into Archaebacteria (orArchaea) and Eubacteria (or Bacteria) is alsoin use.

Further classification is done by namingthe sub-groups at various levels as given inthe following scheme:

KingdomPhylum (for animals) / Division (for plants)

ClassOrder

FamilyGenus

SpeciesThus, by separating organisms on the

basis of a hierarchy of characteristics intosmaller and smaller groups, we arrive at thebasic unit of classification, which is a‘species’. So what organisms can be said tobelong to the same species? Broadly, a speciesincludes all organisms that are similarenough to breed and perpetuate.

The important characteristics of the fivekingdoms of Whittaker are as follows:

7.3.1 MONERA

These organisms do not have a definednucleus or organelles, nor do any of themshow multi-cellular body designs. On theother hand, they show diversity based onmany other characteristics. Some of themhave cell walls while some do not. Of course,having or not having a cell wall has verydifferent effects on body design here fromhaving or not having a cell wall in multi-cellular organisms. The mode of nutrition oforganisms in this group can be either bysynthesising their own food (autotrophic) orgetting it from the environment(heterotrophic). This group includes bacteria,blue-green algae or cyanobacteria, andmycoplasma. Some examples are shownin Fig. 7.1.

Q

Bacteria

Restingspore

Heterocyst

Anabaena

Fig. 7.1: Monera

7.3.2 PROTISTA

This group includes many kinds of unicellulareukaryotic organisms. Some of theseorganisms use appendages, such as hair-likecilia or whip-like flagella for moving around.Their mode of nutrition can be autotrophicor heterotrophic. Examples are unicellularalgae, diatoms and protozoans (see Fig. 7.2for examples).

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to become multicellular organisms at certainstages in their lives. They have cell-walls madeof a tough complex sugar called chitin.Examples are yeast and mushrooms (see Fig.7.3 for examples).

Water vacuole

Cilia

MacronucleusMicronucleus

Oral groove

Cytosome

Food vacuole

Cytopyge

Waste

Flagellum (long)

Flagellum (short)

Eyespot

Photoreceptor

Contractilevacuole Chloroplast

Photoreceptor

Nucleolus

Nucleus

EctoplasmEndoplasm

MitochondriaNucleus

Crystals

Food vacuole

Contractile vacuole

Advancingpseudopod

Paramecium

Amoeba

Euglena

Fig. 7.2: Protozoa

7.3.3 FUNGI

These are heterotrophic eukaryoticorganisms. They use decaying organicmaterial as food and are therefore calledsaprophytes. Many of them have the capacity

Fig. 7.3: Fungi

Some fungal species live in permanentmutually dependent relationships with blue-green algae (or cyanobacteria). Suchrelationships are called symbiotic. Thesesymbiobic life forms are called lichens. Wehave all seen lichens as the slow-growinglarge coloured patches on the bark of trees.

7.3.4 PLANTAE

These are multicellular eukaryotes with cellwalls. They are autotrophs and usechlorophyll for photosynthesis. Thus, allplants are included in this group. Sinceplants and animals are most visible formsof the diversity of life around us, we will lookat the subgroups in this category later(section 7.4).

7.3.5 ANIMALIA

These include all organisms which aremulticellular eukaryotes without cell walls.They are heterotrophs. Again, we will lookat their subgroups a little later insection 7.5.

Penicillium Agaricus

Aspergillus


Fig. 7.4: The Five Kingdom classification

uestions1. What is the criterion for

classification of organisms asbelonging to kingdom Monera orProtista?

2. In which kingdom will you placean organism which is single-celled, eukaryotic andphotosynthetic?

3. In the hierarchy of classification,which grouping will have thesmallest number of organismswith a maximum ofcharacteristics in common andwhich will have the largestnumber of organisms?

Q7.4 PlantaeThe first level of classification among plantsdepends on whether the plant body has well-differentiated, distinct components. The nextlevel of classification is based on whether thedifferentiated plant body has special tissuesfor the transport of water and othersubstances within it. Further classificationlooks at the ability to bear seeds and whetherthe seeds are enclosed within fruits.

7.4.1 THALLOPHYTA

Plants that do not have well-differentiatedbody design fall in this group. The plants inthis group are commonly called algae. These

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plants are predominantly aquatic. Examplesare Spirogyra, Ulothrix, Cladophora and Chara(see Fig. 7.5).

7.4.2 BRYOPHYTA

These are called the amphibians of the plantkingdom. The plant body is commonlydifferentiated to form stem and leaf-likestructures. However, there is no specialisedtissue for the conduction of water and othersubstances from one part of the plant bodyto another. Examples are moss (Funaria) andMarchantia (see Fig. 7.6).

Fig. 7.5: Thallophyta – Algae

Cell-wall

Chloroplast

Pyrenoids

Nucleus

Cytoplasm

Ulothrix

Cladophora

Ulva

Spirogyra

Chara

Fig. 7.6: Some common bryophytes

7.4.3 PTERIDOPHYTA

In this group, the plant body is differentiatedinto roots, stem and leaves and hasspecialised tissue for the conduction of waterand other substances from one part of theplant body to another. Some examples areMarsilea, ferns and horse-tails (see Fig. 7.7).

The thallophytes, the bryophytes and thepteridophytes have naked embryos that arecalled spores. The reproductive organs ofplants in all these three groups are veryinconspicuous, and they are therefore called‘cryptogamae’, or ‘those with hiddenreproductive organs’.

Riccia

Marchantia Funaria


On the other hand, plants with well-differentiated reproductive tissues thatultimately make seeds are calledphanerogams. Seeds are the result of thereproductive process. They consist of theembryo along with stored food, which servesfor the initial growth of the embryo duringgermination. This group is further classified,based on whether the seeds are naked orenclosed in fruits, giving us two groups:gymnosperms and angiosperms.

7.4.4 GYMNOSPERMS

This term is made from two Greek words:gymno– means naked and sperma– meansseed. The plants of this group bear nakedseeds and are usually perennial, evergreenand woody. Examples are pines, such asdeodar (see Fig. 7.8 for examples).

7.4.5 ANGIOSPERMS

This word is made from two Greek words:angio means covered and sperma– meansseed. The seeds develop inside an organ whichis modified to become a fruit. These are alsocalled flowering plants. Plant embryos inseeds have structures called cotyledons.Cotyledons are called ‘seed leaves’ becausein many instances they emerge and becomegreen when the seed germinates. Thus,cotyledons represent a bit of pre-designedplant in the seed. The angiosperms aredivided into two groups on the basis of thenumber of cotyledons present in the seed.Plants with seeds having a single cotyledonare called monocotyledonous or monocots.Plants with seeds having two cotyledons arecalled dicots (see Figs. 7.9 and 7.10).

Marsilea Fern

Leaf

Sporocarp

Stem

Root

Fig. 7.7: Pteridophyta

Pinus

Fig. 7.8: Gymnosperms

Cycas

Fig. 7.9: Monocots – Paphiopedilum

Fig. 7.10: Dicots – Ipomoea

SCIENCE88

Activity ______________ 7.2• Soak seeds of green gram, wheat,

maize, peas and tamarind. Once theybecome tender, try to split the seed. Doall the seeds break into two nearlyequal halves?

• The seeds that do are the dicot seedsand the seeds that don’t are themonocot seeds.

• Now take a look at the roots, leaves andflowers of these plants.

• Are the roots tap-roots or fibrous?• Do the leaves have parallel or reticulate

venation?

Fig. 7.11: Classification of plants

• How many petals are found in theflower of these plants?

• Can you write down furthercharacteristics of monocots and dicotson the basis of these observations?

uestions1. Which division among plants has

the simplest organisms?2. How are pteridophytes different

from the phanerogams?3. How do gymnosperms and

angiosperms differ from eachother?

Q


7.5 AnimaliaThese are organisms which are eukaryotic,multicellular and heterotrophic. Their cellsdo not have cell-walls. Most animals aremobile.

They are further classified based on theextent and type of the body designdifferentiation found.

7.5.1 PORIFERA

The word means organisms with holes. Theseare non-motile animals attached to some solidsupport. There are holes or ‘pores’, all overthe body. These lead to a canal system thathelps in circulating water throughout thebody to bring in food and oxygen. Theseanimals are covered with a hard outside layeror skeleton. The body design involves veryminimal differentiation and division intotissues. They are commonly called sponges,and are mainly found in marine habitats.Some examples are shown in Fig. 7.12.

layers of cells: one makes up cells on theoutside of the body, and the other makes theinner lining of the body. Some of these specieslive in colonies (corals), while others have asolitary like–span (Hydra). Jellyfish and seaanemones are common examples (seeFig. 7.13).

SyconEuplectelia

Spongilla

Fig. 7.12: Porifera

Tentacles

Sea anemone

Tentacles

Stinging cell

Mouth

EpidermisMesoglea

Gastrodermis

Gastrovascularcavity

Foot

Fig. 7.13: Coelenterata

7.5.3 PLATYHELMINTHES

The body of animals in this group is far morecomplexly designed than in the two othergroups we have considered so far. The bodyis bilaterally symmetrical, meaning that theleft and the right halves of the body have thesame design. There are three layers of cellsfrom which differentiated tissues can bemade, which is why such animals are calledtriploblastic. This allows outside and insidebody linings as well as some organs to bemade. There is thus some degree of tissueformation. However, there is no true internalbody cavity or coelom, in which well-developed organs can be accommodated. Thebody is flattened dorsiventrally, meaning fromtop to bottom, which is why these animalsare called flatworms. They are either free-living or parasitic. Some examples are free-living animals like planarians, or parasiticanimals like liverflukes (see Fig. 7.14 forexamples).

Hydra

7.5.2 COELENTERATA

These are animals living in water. They showmore body design differentiation. There is acavity in the body. The body is made of two

SCIENCE90

7.5.5 ANNELIDA

Annelid animals are also bilaterallysymmetrical and triploblastic, but in additionthey have a true body cavity. This allows trueorgans to be packaged in the body structure.There is, thus, extensive organ differentiation.This differentiation occurs in a segmentalfashion, with the segments lined up one afterthe other from head to tail. These animalsare found in a variety of habitats– fresh water,marine water as well as land. Earthwormsand leeches are familiar examples (seeFig. 7.16).

Acetabulum

Scolex SuckerNeck

Tape wormLiverfluke

EyesBranchedgastrovascularcavity

PharynxMouth

and anus

Planareia

Fig. 7.14: Platyhelminthes

7.5.4 NEMATODA

The nematode body is also bilaterallysymmetrical and triploblastic. However, thebody is cylindrical rather than flattened.There are tissues, but no real organs,although a sort of body cavity or a pseudo-coelom, is present. These are very familiaras parasitic worms causing diseases, suchas the worms causing elephantiasis (filarialworms) or the worms in the intestines(roundworm or pinworms). Some examplesare shown in Fig. 7.15.

Female

Male

Ascaris Wuchereria

Fig. 7.15: Nematodes (Aschelminthes)

Tentacle

Palp

Parapodia

Parapodia

Nereis Earthworm Leech

Fig. 7.16: Annelida

7.5.6 ARTHROPODA

This is probably the largest group of animals.These animals are bilaterally symmetrical andsegmented. There is an open circulatorysystem, and so the blood does not flow in well-defined blood vessels. The coelomic cavity isblood-filled. They have jointed legs (the word‘arthropod’ means ‘jointed legs’). Somefamiliar examples are prawns, butterflies,houseflies, spiders, scorpions and crabs (seeFig. 7.17).

Genitalpapillae

Anus


Palamnaeus(Scorpion)

7.5.7 MOLLUSCA

In the animals of this group, there is bilateralsymmetry. The coelomic cavity is reduced.There is little segmentation. They have anopen circulatory system and kidney-likeorgans for excretion. There is a foot that isused for moving around. Examples are snailsand mussels (see Fig. 7.18).

7.5.8 ECHINODERMATA

In Greek, echinos means hedgehog, andderma means skin. Thus, these are spinyskinned organisms. These are exclusivelyfree-living marine animals. They aretriploblastic and have a coelomic cavity. Theyalso have a peculiar water-driven tube systemthat they use for moving around. They havehard calcium carbonate structures that theyuse as a skeleton. Examples are starfish andsea urchins (see Fig. 7.19).

Pariplaneta(Cockroach)

Palaemon(Prawn)

Aranea(Spider)

Fig. 7.17: Arthropoda

ChitonOctopus

Pila

Unio

Fig. 7.18: Mollusca

Antedon(feather star)

Holothuria(sea cucumber)

Echinus (sea urchin) Asterias (star fish)

7.5.9 PROTOCHORDATA

These animals are bilaterally symmetrical,triploblastic and have a coelom. In addition,they show a new feature of body design,namely a notochord, at least at some stagesduring their lives. The notochord is a longrod-like support structure (chord=string) thatruns along the back of the animal separatingthe nervous tissue from the gut. It provides aplace for muscles to attach for ease ofmovement. Protochordates may not have aproper notochord present at all stages in their

Fig. 7.19: Echinodermata

Musca(House fly)

Butterfly

Scolopendra(Centipede)

SCIENCE92

lives or for the entire length of the animal.Protochordates are marine animals.Examples are Balanoglossus, Herdemaniaand Amphioxus (see Fig. 7.20).

7.5.10 (i) PISCES

These are fish. They are exclusively water-living animals. Their skin is covered withscales/plates. They obtain oxygen dissolvedin water by using gills. The body isstreamlined, and a muscular tail is used formovement. They are cold-blooded and theirhearts have only two chambers, unlike thefour that humans have. They lay eggs. Wecan think of many kinds of fish, some withskeletons made entirely of cartilage, such assharks, and some with a skeleton made ofboth bone and cartilage, such as tuna or rohu[see examples in Figs. 7.21 (a) and 7.21 (b)].

Proboscis

Collarette

Collar

Branchial region

Gill pores

Dorsallycurvedgenital wings

Middosrsalridge

Hepatic caeca

Hepatic region

Posthepaticregion

Anus

Fig. 7.20: A Protochordata: Balanoglossus

7.5.10 VERTEBRATA

These animals have a true vertebral columnand internal skeleton, allowing a completelydifferent distribution of muscle attachmentpoints to be used for movement.

Vertebrates are bilaterally symmetrical,triploblastic, coelomic and segmented, withcomplex differentiation of body tissues andorgans. All chordates possess the followingfeatures:

(i) have a notochord(ii) have a dorsal nerve cord(iii) are triploblastic(iv) have paired gill pouches(v) are coelomate.

Vertebrates are grouped into five classes.

Caulophyryne jordani(Angler fish)

Synchiropus splendidus(Mandarin fish)

Pterois volitans(Lion fish)

SpiracleEye

Pelvic finDorsal fin

Caudal fin

Tail

Electric ray (Torpedo)

Sting ray

Mouth

Dorsal finTailEye

Gills Pectoralfin

Pelvicfin

Fig. 7.21 (a): Pisces

Scoliodon (Dog fish)


7.5.10 (ii) AMPHIBIA

These animals differ from the fish in the lackof scales, in having mucus glands in the skin,and a three-chambered heart. Respiration isthrough either gills or lungs. They lay eggs.These animals are found both in water andon land. Frogs, toads and salamanders aresome examples (see Fig. 7.22).

7.5.10 (iii) REPTILIA

These animals are cold-blooded, have scalesand breathe through lungs. While most ofthem have a three-chambered heart,crocodiles have four heart chambers. Theylay eggs with tough coverings and do not needto lay their eggs in water, unlike amphibians.Snakes, turtles, lizards and crocodiles fall inthis category (see Fig. 7.23).

Eye Head

Nostril

MouthPectoralfinPelvic

fin

Caudalfin

Labeo rohita (Rohu)

PectoralfinMouth

Broodpouch

Dorsalfin

TailMale Hippocampus

(Sea horse)

Wing like pectoral

TailPelvic fin

Scales

Exocoetus (Flying fish)

Anabas (Climbing perch)

Fig. 7.21 (b): Pisces

Fig. 7.22: Amphibia

Rana tigrina(Common frog)

Toad

Hyla (Tree frog)

Salamander

Turtle

Chameleon

King Cobra

House wall lizard(Hemidactylus)

Flying lizard (Draco)

7.5.10 (iv) AVES

These are warm-blooded animals and have afour-chambered heart. They lay eggs. Thereis an outside covering of feathers, and twoforelimbs are modified for flight. They breathethrough lungs. All birds fall in this category(see Fig. 7.24 for examples).

Fig. 7.23: Reptilia

SCIENCE94

7.5.10 (V) MAMMALIA

Mammals are warm-blooded animals withfour-chambered hearts. They have mammaryglands for the production of milk to nourishtheir young. Their skin has hairs as well assweat and oil glands. Most mammals familiarto us produce live young ones. However, afew of them, like the platypus and the echidnalay eggs, and some, like kangaroos give birthto very poorly developed young ones. Someexamples are shown in Fig. 7.25.

The scheme of classification of animals isshown in Fig. 7.26.

White Stork(Ciconia ciconia)

Ostrich(Struthio camelus)

Male Tufted Duck(Aythya fuligula)

Pigeon

Sparrow

Crow

Fig. 7.24: Aves (birds)

Fig. 7.25: Mammalia

uestions1. How do poriferan animals differ

from coelenterate animals?2. How do annelid animals differ

from arthropods?3. What are the differences between

amphibians and reptiles?4. What are the differences between

animals belonging to the Avesgroup and those in the mammaliagroup?

QCarolus Linnaeus (Karlvon Linne) was born inSweden and was a doctorby professsion. He wasinterested in the study ofplants. At the age of 22,he published his firstpaper on plants. Whileserving as a personalphysician of a wealthygovernment official, he studied thediversity of plants in his employer’sgarden. Later, he published 14 papers andalso brought out the famous bookSystema Naturae from which allfundamental taxonomical researches havetaken off. His system of classification wasa simple scheme for arranging plants soas to be able to identify them again.

Carolus Linnaeus(1707-1778)

Whale

Human

Cat

Rat

Bat


Fig. 7.26: Classification of animals

SCIENCE96

7.6 NomenclatureWhy is there a need for systematic naming ofliving organisms?

Activity ______________ 7.3• Find out the names of the following

animals and plants in as manylanguages as you can:1. Tiger 2. Peacock 3. Ant4. Neem 5. Lotus 6. Potato

As you might be able to appreciate, itwould be difficult for people speaking orwriting in different languages to know whenthey are talking about the same organism.This problem was resolved by agreeing upona ‘scientific’ name for organisms in the samemanner that chemical symbols and formulaefor various substances are used the worldover. The scientific name for an organism isthus unique and can be used to identify itanywhere in the world.

The system of scientific naming ornomenclature we use today was introducedby Carolus Linnaeus in the eighteenthcentury. The scientific name of an organismis the result of the process of classification

which puts it along with the organisms it ismost related to. But when we actually namethe species, we do not list out the wholehierarchy of groups it belongs to. Instead, welimit ourselves to writing the name of thegenus and species of that particularorganism. The world over, it has been agreedthat both these names will be used in Latinforms.

Certain conventions are followed whilewriting the scientific names:

1. The name of the genus begins with acapital letter.

2. The name of the species begins with asmall letter.

3. When printed, the scientific name isgiven in italics.

4. When written by hand, the genusname and the species name have tobe underlined separately.

Activity ______________ 7.4• Find out the scientific names of any

five common animals and plants. Dothese names have anything in commonwith the names you normally use toidentify them?

Whatyou havelearnt• Classification helps us in exploring the diversity of life forms.

• The major characteristics considered for classifying allorganisms into five major kingdoms are:

(a) whether they are made of prokaryotic or eukaryotic cells

(b) whether the cells are living singly or organised into multi-cellular and thus complex organisms

(c) whether the cells have a cell-wall and whether they preparetheir own food.

• All living organisms are divided on the above bases into fivekingdoms, namely Monera, Protista, Fungi, Plantae andAnimalia.

• The classification of life forms is related to their evolution.


• Plantae and Animalia are further divided into subdivisions onthe basis of increasing complexity of body organisation.

• Plants are divided into five groups: Thallophytes, Bryophytes,Pteridophytes, Gymnosperms and Angiosperms.

• Animals are divided into ten groups: Porifera, Coelenterata,Platyhelminthes, Nematoda, Annelida, Arthropoda, Mollusca,Echinodermata, Protochordata and Vertebrata.

• The binomial nomenclature makes for a uniform way ofidentification of the vast diversity of life around us.

• The binomial nomenclature is made up of two words – a genericname and a specific name.

Exercises1. What are the advantages of classifying organisms?

2. How would you choose between two characteristics to be usedfor developing a hierarchy in classification?

3. Explain the basis for grouping organisms into five kingdoms.

4. What are the major divisions in the Plantae? What is the basisfor these divisions?

5. How are the criteria for deciding divisions in plants differentfrom the criteria for deciding the subgroups among animals?

6. Explain how animals in Vertebrata are classified into furthersubgroups.

In everyday life, we see some objects at restand others in motion. Birds fly, fish swim,blood flows through veins and arteries andcars move. Atoms, molecules, planets, starsand galaxies are all in motion. We oftenperceive an object to be in motion when itsposition changes with time. However, thereare situations where the motion is inferredthrough indirect evidences. For example, weinfer the motion of air by observing themovement of dust and the movement of leavesand branches of trees. What causes thephenomena of sunrise, sunset and changingof seasons? Is it due to the motion of theearth? If it is true, why don’t we directlyperceive the motion of the earth?

An object may appear to be moving forone person and stationary for some other. Forthe passengers in a moving bus, the roadsidetrees appear to be moving backwards. Aperson standing on the road–side perceivesthe bus alongwith the passengers as moving.However, a passenger inside the bus sees hisfellow passengers to be at rest. What do theseobservations indicate?

Most motions are complex. Some objectsmay move in a straight line, others may takea circular path. Some may rotate and a fewothers may vibrate. There may be situationsinvolving a combination of these. In thischapter, we shall first learn to describe themotion of objects along a straight line. Weshall also learn to express such motionsthrough simple equations and graphs. Later,we shall discuss ways of describing circularmotion.

Activity ______________ 8.1• Discuss whether the walls of your

classroom are at rest or in motion.

Activity ______________ 8.2• Have you ever experienced that the

train in which you are sitting appearsto move while it is at rest?

• Discuss and share your experience.

Think and ActWe sometimes are endangered by themotion of objects around us, especiallyif that motion is erratic anduncontrolled as observed in a floodedriver, a hurricane or a tsunami. On theother hand, controlled motion can be aservice to human beings such as in thegeneration of hydro-electric power. Doyou feel the necessity to study theerratic motion of some objects andlearn to control them?

8.1 Describing MotionWe describe the location of an object byspecifying a reference point. Let usunderstand this by an example. Let usassume that a school in a village is 2 km northof the railway station. We have specified theposition of the school with respect to therailway station. In this example, the railwaystation is the reference point. We could havealso chosen other reference points accordingto our convenience. Therefore, to describe theposition of an object we need to specify areference point called the origin.

88888MMMMMOTIONOTIONOTIONOTIONOTION

Chapter

8.1.1 MOTION ALONG A STRAIGHT LINE

The simplest type of motion is the motionalong a straight line. We shall first learn todescribe this by an example. Consider themotion of an object moving along a straightpath. The object starts its journey from Owhich is treated as its reference point(Fig. 8.1). Let A, B and C represent theposition of the object at different instants. Atfirst, the object moves through C and B andreaches A. Then it moves back along the samepath and reaches C through B.

displacement, are used to describe the overallmotion of an object and to locate its finalposition with reference to its initial positionat a given time.

Activity ______________ 8.3• Take a metre scale and a long rope.• Walk from one corner of a basket-ball

court to its oppposite corner along itssides.

• Measure the distance covered by youand magnitude of the displacement.

• What difference would you noticebetween the two in this case?

Activity ______________ 8.4• Automobiles are fitted with a device

that shows the distance travelled. Sucha device is known as an odometer. Acar is driven from Bhubaneshwar toNew Delhi. The difference between thefinal reading and the initial reading ofthe odometer is 1850 km.

• Find the magnitude of the displacementbetween Bhubaneshwar and New Delhiby using the Road Map of India.

The total path length covered by the objectis OA + AC, that is 60 km + 35 km = 95 km.This is the distance covered by the object. Todescribe distance we need to specify only thenumerical value and not the direction ofmotion. There are certain quantities whichare described by specifying only theirnumerical values. The numerical value of aphysical quantity is its magnitude. From thisexample, can you find out the distance of thefinal position C of the object from the initialposition O? This difference will give you thenumerical value of the displacement of theobject from O to C through A. The shortestdistance measured from the initial to the finalposition of an object is known as thedisplacement.

Can the magnitude of the displacementbe equal to the distance travelled by anobject? Consider the example given in(Fig. 8.1). For motion of the object from O toA, the distance covered is 60 km and themagnitude of displacement is also 60 km.During its motion from O to A and back to B,the distance covered = 60 km + 25 km = 85 km

Fig. 8.1: Positions of an object on a straight line path

while the magnitude of displacement = 35 km.Thus, the magnitude of displacement (35 km)is not equal to the path length (85 km).Further, we will notice that the magnitude ofthe displacement for a course of motion maybe zero but the corresponding distancecovered is not zero. If we consider the objectto travel back to O, the final position concideswith the initial position, and therefore, thedisplacement is zero. However, the distancecovered in this journey is OA + AO = 60 km +60 km = 120 km. Thus, two different physicalquantities — the distance and the

MOTION 99

SCIENCE100

uestions1. An object has moved through a

distance. Can it have zerodisplacement? If yes, supportyour answer with an example.

2. A farmer moves along theboundary of a square field of side10 m in 40 s. What will be themagnitude of displacement of thefarmer at the end of 2 minutes20 seconds?

3. Which of the following is true fordisplacement?

(a) It cannot be zero.(b) Its magnitude is greater than

the distance travelled by theobject.

8.1.2 UNIFORM MOTION AND NON-UNIFORM MOTION

Consider an object moving along a straightline. Let it travel 50 km in the first hour,50 km more in the second hour, 50 km in thethird hour and 50 km in the fourth hour. Inthis case, the object covers 50 km in eachhour. As the object covers equal distances inequal intervals of time, it is said to be inuniform motion. The time interval in thismotion may be small or big. In ourday-to-day life, we come across motions whereobjects cover unequal distances in equalintervals of time, for example, when a car ismoving on a crowded street or a person isjogging in a park. These are some instancesof non-uniform motion.

Activity ______________ 8.5

• The data regarding the motion of twodifferent objects A and B are given inTable 8.1.

• Examine them carefully and statewhether the motion of the objects isuniform or non-uniform.

Q

(a)

(b)

Fig. 8.2

Table 8.1

Time Distance Distancetravelled by travelled by

object A in m object B in m

9:30 am 10 12

9:45 am 20 19

10:00 am 30 23

10:15 am 40 35

10:30 am 50 37

10:45 am 60 41

11:00 am 70 44

8.2 Measuring the Rate of Motion

MOTION 101

Look at the situations given in Fig. 8.2. Ifthe bowling speed is 143 km h–1 in Fig. 8.2(a)what does it mean? What do you understandfrom the signboard in Fig. 8.2(b)?

Different objects may take differentamounts of time to cover a given distance.Some of them move fast and some moveslowly. The rate at which objects move canbe different. Also, different objects can moveat the same rate. One of the ways ofmeasuring the rate of motion of an object isto find out the distance travelled by the objectin unit time. This quantity is referred to asspeed. The SI unit of speed is metre persecond. This is represented by the symbolm s–1 or m/s. The other units of speed includecentimetre per second (cm s–1) and kilometreper hour (km h–1). To specify the speed of anobject, we require only its magnitude. Thespeed of an object need not be constant. Inmost cases, objects will be in non-uniformmotion. Therefore, we describe the rate ofmotion of such objects in terms of theiraverage speed. The average speed of an objectis obtained by dividing the total distancetravelled by the total time taken. That is,

average speed = Total distance travelled

Total time taken

If an object travels a distance s in time t thenits speed v is,

v = s

t(8.1)

Let us understand this by an example. Acar travels a distance of 100 km in 2 h. Itsaverage speed is 50 km h–1. The car mightnot have travelled at 50 km h–1 all the time.Sometimes it might have travelled faster andsometimes slower than this.

Example 8.1 An object travels 16 m in 4 sand then another 16 m in 2 s. What isthe average speed of the object?

Solution:

Total distance travelled by the object =16 m + 16 m = 32 mTotal time taken = 4 s + 2 s = 6 s

Average speed = Total distance travelled

Total time taken

= 32 m6 s = 5.33 m s–1

Therefore, the average speed of the objectis 5.33 m s–1.

8.2.1 SPEED WITH DIRECTION

The rate of motion of an object can be morecomprehensive if we specify its direction ofmotion along with its speed. The quantity thatspecifies both these aspects is called velocity.Velocity is the speed of an object moving in adefinite direction. The velocity of an objectcan be uniform or variable. It can be changedby changing the object’s speed, direction ofmotion or both. When an object is movingalong a straight line at a variable speed, wecan express the magnitude of its rate ofmotion in terms of average velocity. It iscalculated in the same way as we calculateaverage speed.

In case the velocity of the object ischanging at a uniform rate, then averagevelocity is given by the arithmetic mean ofinitial velocity and final velocity for a givenperiod of time. That is,

average velocity = initial velocity + final velocity

2

Mathematically, vav

= u + v

2(8.2)

where vav

is the average velocity, u is the initialvelocity and v is the final velocity of the object.

Speed and velocity have the same units,that is, m s–1 or m/s.

Activity ______________ 8.6• Measure the time it takes you to walk

from your house to your bus stop orthe school. If you consider that youraverage walking speed is 4 km h–1,estimate the distance of the bus stopor school from your house.

SCIENCE102

= 50km 1000m 1h

× ×h 1km 3600s

= 13.9 m s–1

The average speed of the car is50 km h–1 or 13.9 m s–1.

Example 8.3 Usha swims in a 90 m longpool. She covers 180 m in one minuteby swimming from one end to the otherand back along the same straight path.Find the average speed and averagevelocity of Usha.

Solution:Total distance covered by Usha in 1 minis 180 m.Displacement of Usha in 1 min = 0 m

Average speed = Total distance covered

Total timetaken

=180m 180 m 1 min

= ×1min 1min 60s

= 3 m s-1

Average velocity = Displacement

Total timetaken

= 0m

60 s

= 0 m s–1

The average speed of Usha is 3 m s–1

and her average velocity is 0 m s–1.

8.3 Rate of Change of VelocityDuring uniform motion of an object along astraight line, the velocity remains constantwith time. In this case, the change in velocityof the object for any time interval is zero.However, in non-uniform motion, velocityvaries with time. It has different values atdifferent instants and at different points ofthe path. Thus, the change in velocity of theobject during any time interval is not zero.Can we now express the change in velocity ofan object?

Activity ______________ 8.7• At a time when it is cloudy, there may

be frequent thunder and lightning. Thesound of thunder takes some time toreach you after you see the lightning.

• Can you answer why this happens?• Measure this time interval using a

digital wrist watch or a stop watch.• Calculate the distance of the nearest

point of lightning. (Speed of sound inair = 346 m s-1.)

uestions1. Distinguish between speed and

velocity.2. Under what condition(s) is the

magnitude of average velocity ofan object equal to its averagespeed?

3. What does the odometer of anautomobile measure?

4. What does the path of an objectlook like when it is in uniformmotion?

5. During an experiment, a signalfrom a spaceship reached theground station in five minutes.What was the distance of thespaceship from the groundstation? The signal travels at thespeed of light, that is, 3 × 108

m s–1.

Example 8.2 The odometer of a car reads2000 km at the start of a trip and2400 km at the end of the trip. If thetrip took 8 h, calculate the averagespeed of the car in km h–1 and m s–1.

Solution:

Distance covered by the car,s = 2400 km – 2000 km = 400 kmTime elapsed, t = 8 hAverage speed of the car is,

vav

= 400 km

8 h=

s

t

= 50 km h–1

Q

MOTION 103

To answer such a question, we have tointroduce another physical quantity calledacceleration, which is a measure of thechange in the velocity of an object per unittime. That is,

acceleration =change in velocity

time taken

If the velocity of an object changes froman initial value u to the final value v in time t,the acceleration a is,

v – ua =

t(8.3)

This kind of motion is known asaccelerated motion. The acceleration is takento be positive if it is in the direction of velocityand negative when it is opposite to thedirection of velocity. The SI unit ofacceleration is m s–2 .

If an object travels in a straight line andits velocity increases or decreases by equalamounts in equal intervals of time, then theacceleration of the object is said to beuniform. The motion of a freely falling bodyis an example of uniformly acceleratedmotion. On the other hand, an object cantravel with non-uniform acceleration if itsvelocity changes at a non-uniform rate. Forexample, if a car travelling along a straightroad increases its speed by unequal amountsin equal intervals of time, then the car is saidto be moving with non-uniform acceleration.

Activity ______________ 8.8• In your everyday life you come across

a range of motions in which(a) acceleration is in the direction of

motion,(b) acceleration is against the

direction of motion,(c) acceleration is uniform,(d) acceleration is non-uniform.

• Can you identify one example each ofthe above type of motion?

Example 8.4 Starting from a stationaryposition, Rahul paddles his bicycle to

attain a velocity of 6 m s–1 in 30 s. Thenhe applies brakes such that the velocityof the bicycle comes down to 4 m s-1 inthe next 5 s. Calculate the accelerationof the bicycle in both the cases.

Solution:

In the first case:initial velocity, u = 0 ;final velocity, v = 6 m s–1 ;time, t = 30 s .From Eq. (8.3), we have

v – ua =

t

Substituting the given values of u,v andt in the above equation, we get

( )–1 –16m s – 0m s=

30 sa

= 0.2 m s–2

In the second case:initial velocity, u = 6 m s–1;final velocity, v = 4 m s–1;time, t = 5 s.

Then, ( )–1 –14m s – 6m s

=5 s

a

= –0.4 m s–2 .The acceleration of the bicycle in thefirst case is 0.2 m s–2 and in the secondcase, it is –0.4 m s–2.

uestions1. When will you say a body is in

(i) uniform acceleration? (ii) non-uniform acceleration?

2. A bus decreases its speed from80 km h–1 to 60 km h–1 in 5 s.Find the acceleration of the bus.

3. A train starting from a railwaystation and moving with uniformacceleration attains a speed40 km h–1 in 10 minutes. Find itsacceleration.

Q

SCIENCE104

8.4 Graphical Representation ofMotion

Graphs provide a convenient method topresent basic information about a variety ofevents. For example, in the telecast of aone-day cricket match, vertical bar graphsshow the run rate of a team in each over. Asyou have studied in mathematics, a straightline graph helps in solving a linear equationhaving two variables.

To describe the motion of an object, wecan use line graphs. In this case, line graphsshow dependence of one physical quantity,such as distance or velocity, on anotherquantity, such as time.

8.4.1 DISTANCE–TIME GRAPHS

The change in the position of an object withtime can be represented on the distance-timegraph adopting a convenient scale of choice.In this graph, time is taken along the x–axisand distance is taken along the y-axis.Distance-time graphs can be employed undervarious conditions where objects move withuniform speed, non-uniform speed, remainat rest etc.

Fig. 8.3: Distance-time graph of an object movingwith uniform speed

We know that when an object travelsequal distances in equal intervals of time, itmoves with uniform speed. This shows that

the distance travelled by the object is directlyproportional to time taken. Thus, for uniformspeed, a graph of distance travelled againsttime is a straight line, as shown in Fig. 8.3.The portion OB of the graph shows that thedistance is increasing at a uniform rate. Notethat, you can also use the term uniformvelocity in place of uniform speed if you takethe magnitude of displacement equal to thedistance travelled by the object along they-axis.

We can use the distance-time graph todetermine the speed of an object. To do so,consider a small part AB of the distance-timegraph shown in Fig 8.3. Draw a line parallelto the x-axis from point A and another lineparallel to the y-axis from point B. These twolines meet each other at point C to form atriangle ABC. Now, on the graph, AC denotesthe time interval (t

2 – t

1) while BC corresponds

to the distance (s2 – s

1). We can see from the

graph that as the object moves from the pointA to B, it covers a distance (s

2 – s

1) in time

(t2 – t

1). The speed, v of the object, therefore

can be represented as

v = 2 1

2 1

–

–

s s

t t(8.4)

We can also plot the distance-time graphfor accelerated motion. Table 8.2 shows thedistance travelled by a car in a time intervalof two seconds.

Table 8.2: Distance travelled by acar at regular time intervals

Time in seconds Distance in metres

0 0

2 1

4 4

6 9

8 16

10 25

12 36

MOTION 105

The distance-time graph for the motionof the car is shown in Fig. 8.4. Note that theshape of this graph is different from the earlierdistance-time graph (Fig. 8.3) for uniformmotion. The nature of this graph shows non-linear variation of the distance travelled bythe car with time. Thus, the graph shown inFig 8.4 represents motion with non-uniformspeed.

8.4.2 VELOCITY-TIME GRAPHS

The variation in velocity with time for anobject moving in a straight line can berepresented by a velocity-time graph. In thisgraph, time is represented along the x-axis

Fig. 8.4: Distance-time graph for a car moving withnon-uniform speed

Fig. 8.5: Velocity-time graph for uniform motion ofa car

and the velocity is represented along they-axis. If the object moves at uniform velocity,the height of its velocity-time graph will notchange with time (Fig. 8.5). It will be a straightline parallel to the x-axis. Fig. 8.5 shows thevelocity-time graph for a car moving withuniform velocity of 40 km h–1.

We know that the product of velocity andtime give displacement of an object movingwith uniform velocity. The area enclosed byvelocity-time graph and the time axis will beequal to the magnitude of the displacement.

To know the distance moved by the carbetween time t

1 and t2 using Fig. 8.5, draw

perpendiculars from the points correspondingto the time t

1 and t2 on the graph. The velocity

of 40 km h–1 is represented by the height ACor BD and the time (t

2 – t

1) is represented by

the length AB.So, the distance s moved by the car in

time (t2 – t

1) can be expressed as

s = AC × CD= [(40 km h–1) × (t

2 – t

1) h]

= 40 (t2– t

1) km

= area of the rectangle ABDC (shadedin Fig. 8.5).

We can also study about uniformlyaccelerated motion by plotting its velocity–time graph. Consider a car being driven alonga straight road for testing its engine. Supposea person sitting next to the driver records itsvelocity after every 5 seconds by noting thereading of the speedometer of the car. Thevelocity of the car, in km h–1 as well as inm s–1, at different instants of time is shownin table 8.3.

Table 8.3: Velocity of a car atregular instants of time

Time Velocity of the car (s) (m s–1) (km h

–1)

0 0 05 9 2.510 18 5.015 27 7.520 36 10.025 45 12.530 54 15.0

SCIENCE106

In this case, the velocity-time graph forthe motion of the car is shown in Fig. 8.6.The nature of the graph shows that velocitychanges by equal amounts in equal intervalsof time. Thus, for all uniformly acceleratedmotion, the velocity-time graph is astraight line.

Fig. 8.6: Velocity-time graph for a car moving withuniform accelerations.

You can also determine the distancemoved by the car from its velocity-time graph.The area under the velocity-time graph givesthe distance (magnitude of displacement)moved by the car in a given interval of time.If the car would have been moving withuniform velocity, the distance travelled by itwould be represented by the area ABCDunder the graph (Fig. 8.6). Since themagnitude of the velocity of the car ischanging due to acceleration, the distance stravelled by the car will be given by the areaABCDE under the velocity-time graph(Fig. 8.6).That is,

s = area ABCDE= area of the rectangle ABCD + area of

the triangle ADE

= AB × BC + 1

2(AD × DE)

In the case of non-uniformly acceleratedmotion, velocity-time graphs can have anyshape.

Fig. 8.7: Velocity-time graphs of an object in non-uniformly accelerated motion.

Fig. 8.7(a) shows a velocity-time graphthat represents the motion of an object whosevelocity is decreasing with time whileFig. 8.7 (b) shows the velocity-time graphrepresenting the non-uniform variation ofvelocity of the object with time. Try to interpretthese graphs.

Activity ______________ 8.9• The times of arrival and departure of a

train at three stations A, B and C andthe distance of stations B and C fromstation A are given in table 8.4.

Table 8.4: Distances of stations Band C from A and times of arrivaland departure of the train

Station Distance Time of Time offrom A arrival departure(km) (hours) (hours)

A 0 08:00 08:15B 120 11:15 11:30C 180 13:00 13:15

• Plot and interpret the distance-timegraph for the train assuming that itsmotion between any two stations isuniform.

MOTION 107

8.5 Equations of Motion byGraphical Method

When an object moves along a straight linewith uniform acceleration, it is possible torelate its velocity, acceleration during motionand the distance covered by it in a certaintime interval by a set of equations known asthe equations of motion. There are three suchequations. These are:

v = u + at (8.5)s = ut + ½ at2 (8.6)

2 a s = v2 – u2 (8.7)where u is the initial velocity of the objectwhich moves with uniform acceleration a fortime t, v is the final velocity, and s is thedistance travelled by the object in time t.Eq. (8.5) describes the velocity-time relationand Eq. (8.6) represents the position-timerelation. Eq. (8.7), which represents therelation between the position and the velocity,can be obtained from Eqs. (8.5) and (8.6) byeliminating t. These three equations can bederived by graphical method.

8.5.1 EQUATION FOR VELOCITY-TIME

RELATION

Consider the velocity-time graph of an objectthat moves under uniform acceleration as

Activity _____________8.10• Feroz and his sister Sania go to school

on their bicycles. Both of them start atthe same time from their home but takedifferent times to reach the schoolalthough they follow the same route.Table 8.5 shows the distance travelledby them in different times

Table 8.5: Distance covered by Ferozand Sania at different times ontheir bicycles

Time Distance Distancetravelled travelledby Feroz by Sania

(km) (km)

8:00 am 0 0

8:05 am 1.0 0.8

8:10 am 1.9 1.6

8:15 am 2.8 2.3

8:20 am 3.6 3.0

8:25 am – 3.6

Q• Plot the distance-time graph for their

motions on the same scale andinterpret.

uestions1. What is the nature of the

distance-time graphs for uniformand non-uniform motion of anobject?

2. What can you say about themotion of an object whosedistance-time graph is a straightline parallel to the time axis?

3. What can you say about themotion of an object if its speed-time graph is a straight lineparallel to the time axis?

4. What is the quantity which ismeasured by the area occupiedbelow the velocity-time graph?

Fig. 8.8: Velocity-time graph to obtain the equationsof motion

SCIENCE108

= OA × OC + 1

2 (AD × BD) (8.10)

Substituting OA = u, OC = AD = t andBD = at, we get

s = u × t + 1

( )2

t ×at

or s = u t + 12

a t 2

8.5.3 EQUATION FOR POSITION–VELOCITY

RELATION

From the velocity-time graph shown inFig. 8.8, the distance s travelled by the objectin time t, moving under uniform accelerationa is given by the area enclosed within thetrapezium OABC under the graph. That is,

s = area of the trapezium OABC

= ( )OA + BC ×OC

2

Substituting OA = u, BC = v and OC = t,we get

( )2

=u + v t

s (8.11)

From the velocity-time relation (Eq. 8.6),we get

t =v u

a( )

(8.12)

Using Eqs. (8.11) and (8.12) we have

( ) ( )×v + u v - us=

2a

or 2 a s = v2 – u2

Example 8.5 A train starting from restattains a velocity of 72 km h–1 in5 minutes. Assuming that theacceleration is uniform, find (i) theacceleration and (ii) the distancetravelled by the train for attaining thisvelocity.

shown in Fig. 8.8 (similar to Fig. 8.6, but nowwith u ≠ 0). From this graph, you can see thatinitial velocity of the object is u (at point A)and then it increases to v (at point B) in timet. The velocity changes at a uniform rate a.In Fig. 8.8, the perpendicular lines BC andBE are drawn from point B on the time andthe velocity axes respectively, so that theinitial velocity is represented by OA, the finalvelocity is represented by BC and the timeinterval t is represented by OC. BD = BC –CD, represents the change in velocity in timeinterval t.

Let us draw AD parallel to OC. From thegraph, we observe that

BC = BD + DC = BD + OASubstituting BC = v and OA = u,we get v = BD + uor BD = v – u (8.8)From the velocity-time graph (Fig. 8.8),

the acceleration of the object is given by

a = Change in velocity

time taken

=BD BD

=AD OC

Substituting OC = t, we get

a = BD

t

or BD = at (8.9)

Using Eqs. (8.8) and (8.9) we getv = u + at

8.5.2 EQUATION FOR POSITION-TIME

RELATION

Let us consider that the object has travelleda distance s in time t under uniformacceleration a. In Fig. 8.8, the distancetravelled by the object is obtained by the areaenclosed within OABC under the velocity-timegraph AB.

Thus, the distance s travelled by the objectis given by

s = area OABC (which is a trapezium)= area of the rectangle OADC + area of

the triangle ABD

MOTION 109

Solution:

We have been givenu = 0 ; v = 72 km h–1 = 20 m s-1 andt = 5 minutes = 300 s.(i) From Eq. (8.5) we know that

( )v – ua =

t

–1 –1

–2

20 m s – 0 m s=

300s

1= m s

15

(ii) From Eq. (8.7) we have2 a s = v2 – u2 = v2 – 0Thus,

2

–1 2

–2

=2

(20 ms )=

2×(1/15) ms

vs

a

= 3000 m= 3 km

The acceleration of the train is 1

15m s– 2

and the distance travelled is 3 km.

Example 8.6 A car accelerates uniformlyfrom 18 km h

–1 to 36 km h

–1 in 5 s.

Calculate (i) the acceleration and (ii) thedistance covered by the car in that time.

Solution:

We are given thatu = 18 km h–1 = 5 m s–1

v = 36 km h–1 = 10 m s–1 andt = 5 s .

(i) From Eq. (8.5) we have

v – ua =

t

= -1 -110 cm s – 5 m s

5s

= 1 m s–2

(ii) From Eq. (8.6) we have

s = u t + 1

2 a t 2

= 5 m s–1 × 5 s + 1

2× 1 m s–2 × (5 s)2

= 25 m + 12.5 m

= 37.5 mThe acceleration of the car is 1 m s

–2

and the distance covered is 37.5 m.

Example 8.7 The brakes applied to a carproduce an acceleration of 6 m s-2 inthe opposite direction to the motion. Ifthe car takes 2 s to stop after theapplication of brakes, calculate thedistance it travels during this time.

Solution:

We have been givena = – 6 m s–2 ; t = 2 s and v = 0 m s–1.From Eq. (8.5) we know that

v = u + at0 = u + (– 6 m s–2) × 2 s

or u = 12 m s–1 .From Eq. (8.6) we get

s = u t + 1

2 a t 2

= (12 m s–1 ) × (2 s) + 1

2 (–6 m s–2 ) (2 s)2

= 24 m – 12 m= 12 m

Thus, the car will move 12 m before itstops after the application of brakes.Can you now appreciate why driversare cautioned to maintain somedistance between vehicles whiletravelling on the road?

uestions1. A bus starting from rest moves

with a uniform acceleration of0.1 m s-2 for 2 minutes. Find (a)the speed acquired, (b) thedistance travelled.Q

SCIENCE110

2. A train is travelling at a speed of90 km h–1. Brakes are applied soas to produce a uniformacceleration of – 0.5 m s-2. Findhow far the train will go before itis brought to rest.

3. A trolley, while going down aninclined plane, has anacceleration of 2 cm s-2. What willbe its velocity 3 s after the start?

4. A racing car has a uniformacceleration of 4 m s-2. Whatdistance will it cover in 10 s afterstart?

5. A stone is thrown in a verticallyupward direction with a velocityof 5 m s-1. If the acceleration of thestone during its motion is 10 m s–2

in the downward direction, whatwill be the height attained by thestone and how much time will ittake to reach there?

8.6 Uniform Circular MotionWhen the velocity of an object changes, wesay that the object is accelerating. The changein the velocity could be due to change in itsmagnitude or the direction of the motion orboth. Can you think of an example when anobject does not change its magnitude ofvelocity but only its direction of motion?

Let us consider an example of the motionof a body along a closed path. Fig 8.9 (a) showsthe path of an athlete along a rectangulartrack ABCD. Let us assume that the athleteruns at a uniform speed on the straight partsAB, BC, CD and DA of the track. In order tokeep himself on track, he quickly changeshis speed at the corners. How many timeswill the athlete have to change his directionof motion, while he completes one round? Itis clear that to move in a rectangular trackonce, he has to change his direction of motionfour times.

Now, suppose instead of a rectangulartrack, the athlete is running along ahexagonal shaped path ABCDEF, as shownin Fig. 8.9(b). In this situation, the athletewill have to change his direction six timeswhile he completes one round. What if thetrack was not a hexagon but a regularoctagon, with eight equal sides as shown byABCDEFGH in Fig. 8.9(c)? It is observed thatas the number of sides of the track increasesthe athelete has to take turns more and moreoften. What would happen to the shape ofthe track as we go on increasing the numberof sides indefinitely? If you do this you willnotice that the shape of the track approachesthe shape of a circle and the length of each ofthe sides will decrease to a point. If the athletemoves with a velocity of constant magnitudealong the circular path, the only change inhis velocity is due to the change in thedirection of motion. The motion of the athletemoving along a circular path is, therefore, anexample of an accelerated motion.

We know that the circumference of a circleof radius r is given by π2 r . If the athlete takest seconds to go once around the circular pathof radius r, the velocity v is given by

π2 rv =

t(8.13)

When an object moves in a circular pathwith uniform speed, its motion is calleduniform circular motion.

(a) Rectangular track (b) Hexagonal track

(d) A circular track(c) Octagonal shaped track

Fig. 8.9: The motion of an athlete along closed tracksof different shapes.

MOTION 111

If you carefully note, on being releasedthe stone moves along a straight linetangential to the circular path. This isbecause once the stone is released, itcontinues to move along the direction it hasbeen moving at that instant. This shows thatthe direction of motion changed at every pointwhen the stone was moving along the circularpath.

When an athlete throws a hammer or adiscus in a sports meet, he/she holds thehammer or the discus in his/her hand andgives it a circular motion by rotating his/herown body. Once released in the desireddirection, the hammer or discus moves in thedirection in which it was moving at the timeit was released, just like the piece of stone inthe activity described above. There are manymore familiar examples of objects movingunder uniform circular motion, such as themotion of the moon and the earth, a satellitein a circular orbit around the earth, a cycliston a circular track at constant speedand so on.

Activity _____________8.11• Take a piece of thread and tie a small

piece of stone at one of its ends. Movethe stone to describe a circular pathwith constant speed by holding thethread at the other end, as shown inFig. 8.10.

Fig. 8.10: A stone describing a circular path witha velocity of constant magnitude.

• Now, let the stone go by releasing thethread.

• Can you tell the direction in which thestone moves after it is released?

• By repeating the activity for a few timesand releasing the stone at differentpositions of the circular path, checkwhether the direction in which thestone moves remains the same or not.

Whatyou havelearnt• Motion is a change of position; it can be described in terms of

the distance moved or the displacement.• The motion of an object could be uniform or non-uniform

depending on whether its velocity is constant or changing.• The speed of an object is the distance covered per unit time,

and velocity is the displacement per unit time.• The acceleration of an object is the change in velocity per unit

time.• Uniform and non-uniform motions of objects can be shown

through graphs.• The motion of an object moving at uniform acceleration can be

described with the help of three equations, namelyv = u + at

s = ut + ½ at2

2as = v2 – u2

SCIENCE112

where u is initial velocity of the object, which moves withuniform acceleration a for time t, v is its final velocity and s isthe distance it travelled in time t.

• If an object moves in a circular path with uniform speed, itsmotion is called uniform circular motion.

Exercises1. An athlete completes one round of a circular track of diameter

200 m in 40 s. What will be the distance covered and thedisplacement at the end of 2 minutes 20 s?

2. Joseph jogs from one end A to the other end B of a straight300 m road in 2 minutes 50 seconds and then turns aroundand jogs 100 m back to point C in another 1 minute. What areJoseph’s average speeds and velocities in jogging (a) from A toB and (b) from A to C?

3. Abdul, while driving to school, computes the average speed forhis trip to be 20 km h–1. On his return trip along the sameroute, there is less traffic and the average speed is40 km h–1. What is the average speed for Abdul’s trip?

4. A motorboat starting from rest on a lake accelerates in a straightline at a constant rate of 3.0 m s–2 for 8.0 s. How far does theboat travel during this time?

5. A driver of a car travelling at 52 km h–1 applies the brakes andaccelerates uniformly in the opposite direction. The car stopsin 5 s. Another driver going at 3 km h–1 in another car applieshis brakes slowly and stops in 10 s. On the same graph paper,plot the speed versus time graphs for the two cars. Which ofthe two cars travelled farther after the brakes were applied?

6. Fig 8.11 shows the distance-time graph of three objects A,Band C. Study the graph and answer the following questions:

Fig. 8.11

MOTION 113

(a) Which of the three is travelling the fastest?

(b) Are all three ever at the same point on the road?

(c) How far has C travelled when B passes A?

(d) How far has B travelled by the time it passes C?

7. A ball is gently dropped from a height of 20 m. If its velocityincreases uniformly at the rate of 10 m s-2, with what velocitywill it strike the ground? After what time will it strike theground?

8. The speed-time graph for a car is shown is Fig. 8.12.

Fig. 8.12

(a) Find how far does the car travel in the first 4 seconds.Shade the area on the graph that represents the distancetravelled by the car during the period.

(b) Which part of the graph represents uniform motion of thecar?

9. State which of the following situations are possible and givean example for each of these:

(a) an object with a constant acceleration but with zero velocity

(b) an object moving in a certain direction with an accelerationin the perpendicular direction.

10. An artificial satellite is moving in a circular orbit of radius42250 km. Calculate its speed if it takes 24 hours to revolvearound the earth.

In the previous chapter, we described themotion of an object along a straight line interms of its position, velocity and acceleration.We saw that such a motion can be uniformor non-uniform. We have not yet discoveredwhat causes the motion. Why does the speedof an object change with time? Do all motionsrequire a cause? If so, what is the nature ofthis cause? In this chapter we shall make anattempt to quench all such curiosities.

For many centuries, the problem ofmotion and its causes had puzzled scientistsand philosophers. A ball on the ground, whengiven a small hit, does not move forever. Suchobservations suggest that rest is the “naturalstate” of an object. This remained the beliefuntil Galileo Galilei and Isaac Newtondeveloped an entirely different approach tounderstand motion.

In our everyday life we observe that someeffort is required to put a stationary objectinto motion or to stop a moving object. Weordinarily experience this as a muscular effortand say that we must push or hit or pull onan object to change its state of motion. Theconcept of force is based on this push, hit orpull. Let us now ponder about a ‘force’. Whatis it? In fact, no one has seen, tasted or felt aforce. However, we always see or feel the effectof a force. It can only be explained bydescribing what happens when a force isapplied to an object. Pushing, hitting andpulling of objects are all ways of bringingobjects in motion (Fig. 9.1). They movebecause we make a force act on them.

From your studies in earlier classes, youare also familiar with the fact that a force canbe used to change the magnitude of velocityof an object (that is, to make the object movefaster or slower) or to change its direction ofmotion. We also know that a force can changethe shape and size of objects (Fig. 9.2).

(a) The trolley moves along thedirection we push it.

(c) The hockey stick hits the ball forward

(b) The drawer is pulled.

Fig. 9.1: Pushing, pulling, or hitting objects changetheir state of motion.

(a)

(b)

Fig. 9.2: (a) A spring expands on application of force;(b) A spherical rubber ball becomes oblong

as we apply force on it.

99999FFFFFORCEORCEORCEORCEORCE ANDANDANDANDAND L L L L LAWSAWSAWSAWSAWS OFOFOFOFOF M M M M MOTIONOTIONOTIONOTIONOTION

Chapter

box with a small force, the box does not movebecause of friction acting in a directionopposite to the push [Fig. 9.4(a)]. This frictionforce arises between two surfaces in contact;in this case, between the bottom of the boxand floor’s rough surface. It balances thepushing force and therefore the box does notmove. In Fig. 9.4(b), the children push thebox harder but the box still does not move.This is because the friction force still balancesthe pushing force. If the children push thebox harder still, the pushing force becomesbigger than the friction force [Fig. 9.4(c)].There is an unbalanced force. So the boxstarts moving.

What happens when we ride a bicycle?When we stop pedalling, the bicycle beginsto slow down. This is again because of thefriction forces acting opposite to the directionof motion. In order to keep the bicycle moving,we have to start pedalling again. It thusappears that an object maintains its motionunder the continuous application of anunbalanced force. However, it is quiteincorrect. An object moves with a uniformvelocity when the forces (pushing force andfrictional force) acting on the object arebalanced and there is no net external forceon it. If an unbalanced force is applied onthe object, there will be a change either in itsspeed or in the direction of its motion. Thus,to accelerate the motion of an object, anunbalanced force is required. And the changein its speed (or in the direction of motion)would continue as long as this unbalancedforce is applied. However, if this force is

9.1 Balanced and UnbalancedForces

Fig. 9.3 shows a wooden block on a horizontaltable. Two strings X and Y are tied to the twoopposite faces of the block as shown. If weapply a force by pulling the string X, the blockbegins to move to the right. Similarly, if wepull the string Y, the block moves to the left.But, if the block is pulled from both the sideswith equal forces, the block will not move.Such forces are called balanced forces anddo not change the state of rest or of motion ofan object. Now, let us consider a situation inwhich two opposite forces of differentmagnitudes pull the block. In this case, theblock would begin to move in the direction ofthe greater force. Thus, the two forces arenot balanced and the unbalanced force actsin the direction the block moves. Thissuggests that an unbalanced force acting onan object brings it in motion.

Fig. 9.3: Two forces acting on a wooden block

What happens when some children try topush a box on a rough floor? If they push the

(a) (b) (c)

Fig. 9.4

FORCE AND LAWS OF MOTION 115

SCIENCE116

removed completely, the object wouldcontinue to move with the velocity it hasacquired till then.

9.2 First Law of MotionBy observing the motion of objects on aninclined plane Galileo deduced that objectsmove with a constant speed when no forceacts on them. He observed that when a marblerolls down an inclined plane, its velocityincreases [Fig. 9.5(a)]. In the next chapter,you will learn that the marble falls under theunbalanced force of gravity as it rolls downand attains a definite velocity by the time itreaches the bottom. Its velocity decreaseswhen it climbs up as shown in Fig. 9.5(b).Fig. 9.5(c) shows a marble resting on an idealfrictionless plane inclined on both sides.Galileo argued that when the marble isreleased from left, it would roll down the slopeand go up on the opposite side to the sameheight from which it was released. If theinclinations of the planes on both sides areequal then the marble will climb the samedistance that it covered while rolling down. Ifthe angle of inclination of the right-side planewere gradually decreased, then the marblewould travel further distances till it reachesthe original height. If the right-side plane wereultimately made horizontal (that is, the slopeis reduced to zero), the marble would continueto travel forever trying to reach the sameheight that it was released from. Theunbalanced forces on the marble in this caseare zero. It thus suggests that an unbalanced(external) force is required to change themotion of the marble but no net force isneeded to sustain the uniform motion of themarble. In practical situations it is difficultto achieve a zero unbalanced force. This isbecause of the presence of the frictional forceacting opposite to the direction of motion.Thus, in practice the marble stops aftertravelling some distance. The effect of thefrictional force may be minimised by using asmooth marble and a smooth plane andproviding a lubricant on top of the planes.

Fig. 9.5: (a) the downward motion; (b) the upwardmotion of a marble on an inclined plane;and (c) on a double inclined plane.

Newton further studied Galileo’s ideas onforce and motion and presented threefundamental laws that govern the motion ofobjects. These three laws are known asNewton’s laws of motion. The first law ofmotion is stated as:

An object remains in a state of rest or ofuniform motion in a straight line unlesscompelled to change that state by an appliedforce.

In other words, all objects resist a changein their state of motion. In a qualitative way,the tendency of undisturbed objects to stayat rest or to keep moving with the samevelocity is called inertia. This is why, the firstlaw of motion is also known as the law ofinertia.

Certain experiences that we come acrosswhile travelling in a motorcar can beexplained on the basis of the law of inertia.We tend to remain at rest with respect to theseat until the drives applies a braking forceto stop the motorcar. With the application ofbrakes, the car slows down but our bodytends to continue in the same state of motionbecause of its inertia. A sudden applicationof brakes may thus cause injury to us by


impact or collision with the panels in front.Safety belts are worn to prevent suchaccidents. Safety belts exert a force on ourbody to make the forward motion slower. Anopposite experience is encountered when weare standing in a bus and the bus begins tomove suddenly. Now we tend to fallbackwards. This is because the sudden startof the bus brings motion to the bus as wellas to our feet in contact with the floor of thebus. But the rest of our body opposes thismotion because of its inertia.

When a motorcar makes a sharp turn ata high speed, we tend to get thrown to oneside. This can again be explained on the basisof the law of inertia. We tend to continue inour straight-line motion. When anunbalanced force is applied by the engine tochange the direction of motion of themotorcar, we slip to one side of the seat dueto the inertia of our body.

The fact that a body will remain at restunless acted upon by an unbalanced forcecan be illustrated through the followingactivities:

Activity ______________ 9.1• Make a pile of similar carom coins on

a table, as shown in Fig. 9.6.• Attempt a sharp horizontal hit at the

bottom of the pile using another caromcoin or the striker. If the hit is strongenough, the bottom coin moves outquickly. Once the lowest coin isremoved, the inertia of the other coinsmakes them ‘fall’ vertically on the table.

Galileo Galilei was bornon 15 February 1564 inPisa, Italy. Galileo, rightfrom his childhood, hadinterest in mathematicsand natural philosophy.But his fatherVincenzo Galilei wantedhim to become a medicaldoctor. Accordingly,Galileo enrolled himselffor a medical degree at theUniversity of Pisa in 1581 which he nevercompleted because of his real interest inmathematics. In 1586, he wrote his firstscientific book ‘The Little Balance [LaBalancitta] ’, in which he describedArchimedes’ method of finding the relativedensities (or specific gravities) of substancesusing a balance. In 1589, in his series ofessays – De Motu, he presented his theoriesabout falling objects using an inclined planeto slow down the rate of descent.

In 1592, he was appointed professor ofmathematics at the University of Padua inthe Republic of Venice. Here he continued hisobservations on the theory of motion andthrough his study of inclined planes and thependulum, formulated the correct law foruniformly accelerated objects that thedistance the object moves is proportional tothe square of the time taken.

Galileo was also a remarkable craftsman.He developed a series of telescopes whoseoptical performance was much better thanthat of other telescopes available during thosedays. Around 1640, he designed the firstpendulum clock. In his book ‘StarryMessenger’ on his astronomical discoveries,Galileo claimed to have seen mountains onthe moon, the milky way made up of tinystars, and four small bodies orbiting Jupiter.In his books ‘Discourse on Floating Bodies’and ‘Letters on the Sunspots’, he disclosedhis observations of sunspots.

Using his own telescopes and through hisobservations on Saturn and Venus, Galileoargued that all the planets must orbit the Sunand not the earth, contrary to what wasbelieved at that time.

Galileo Galilei(1564 – 1642)

Fig. 9.6: Only the carom coin at the bottom of apile is removed when a fast moving caromcoin (or striker) hits it.

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five-rupees coin if we use a one-rupee coin, wefind that a lesser force is required to performthe activity. A force that is just enough tocause a small cart to pick up a large velocitywill produce a negligible change in the motionof a train. This is because, in comparison tothe cart the train has a much lesser tendencyto change its state of motion. Accordingly, wesay that the train has more inertia than thecart. Clearly, heavier or more massive objectsoffer larger inertia. Quantitatively, the inertiaof an object is measured by its mass. We maythus relate inertia and mass as follows:Inertia is the natural tendency of an object toresist a change in its state of motion or ofrest. The mass of an object is a measure ofits inertia.

uestions1. Which of the following has more

inertia: (a) a rubber ball and astone of the same size? (b) abicycle and a train? (c) a five-rupees coin and a one-rupee coin?

2. In the following example, try toidentify the number of times thevelocity of the ball changes:“A football player kicks a footballto another player of his team whokicks the football towards thegoal. The goalkeeper of theopposite team collects the footballand kicks it towards a player ofhis own team”.Also identify the agent supplyingthe force in each case.

3. Explain why some of the leavesmay get detached from a tree ifwe vigorously shake its branch.

4. Why do you fall in the forwarddirection when a moving busbrakes to a stop and fallbackwards when it acceleratesfrom rest?

9.4 Second Law of MotionThe first law of motion indicates that whenan unbalanced external force acts on an

Activity ______________ 9.2• Set a five-rupee coin on a stiff playing

card covering an empty glass tumblerstanding on a table as shown inFig. 9.7.

• Give the card a sharp horizontal flickwith a finger. If we do it fast then thecard shoots away, allowing the coin tofall vertically into the glass tumbler dueto its inertia.

• The inertia of the coin tries to maintainits state of rest even when the cardflows off.

Fig. 9.7: When the playing card is flicked with thefinger the coin placed over it falls in thetumbler.

Activity ______________ 9.3• Place a water-filled tumbler on a tray.• Hold the tray and turn around as fast

as you can.• We observe that the water spills. Why?

Observe that a groove is provided in asaucer for placing the tea cup. It preventsthe cup from toppling over in case of suddenjerks.

9.3 Inertia and MassAll the examples and activities given so farillustrate that there is a resistance offered byan object to change its state of motion. If it isat rest it tends to remain at rest; if it is movingit tends to keep moving. This property of anobject is called its inertia. Do all bodies havethe same inertia? We know that it is easier topush an empty box than a box full of books.Similarly, if we kick a football it flies away.But if we kick a stone of the same size withequal force, it hardly moves. We may, in fact,get an injury in our foot while doing so!Similarly, in activity 9.2, instead of a

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object, its velocity changes, that is, the objectgets an acceleration. We would now like tostudy how the acceleration of an objectdepends on the force applied to it and howwe measure a force. Let us recount someobservations from our everyday life. Duringthe game of table tennis if the ball hits a playerit does not hurt him. On the other hand, whena fast moving cricket ball hits a spectator, itmay hurt him. A truck at rest does not requireany attention when parked along a roadside.But a moving truck, even at speeds as low as5 m s–1, may kill a person standing in its path.A small mass, such as a bullet may kill aperson when fired from a gun. Theseobservations suggest that the impactproduced by the objects depends on theirmass and velocity. Similarly, if an object is tobe accelerated, we know that a greater forceis required to give a greater velocity. In otherwords, there appears to exist some quantityof importance that combines the object’smass and its velocity. One such propertycalled momentum was introduced by Newton.The momentum, p of an object is defined asthe product of its mass, m and velocity, v.That is,

p = mv (9.1)

Momentum has both direction andmagnitude. Its direction is the same as thatof velocity, v. The SI unit of momentum iskilogram-metre per second (kg m s-1). Sincethe application of an unbalanced force bringsa change in the velocity of the object, it istherefore clear that a force also produces achange of momentum.

Let us consider a situation in which a carwith a dead battery is to be pushed along astraight road to give it a speed of 1 m s-1,which is sufficient to start its engine. If oneor two persons give a sudden push(unbalanced force) to it, it hardly starts. Buta continuous push over some time results ina gradual acceleration of the car to this speed.It means that the change of momentum ofthe car is not only determined by themagnitude of the force but also by the timeduring which the force is exerted. It may thenalso be concluded that the force necessary to

change the momentum of an object dependson the time rate at which the momentum ischanged.

The second law of motion states that therate of change of momentum of an object isproportional to the applied unbalanced forcein the direction of force.

9.4.1 MATHEMATICAL FORMULATION OF

SECOND LAW OF MOTION

Suppose an object of mass, m is moving alonga straight line with an initial velocity, u. It isuniformly accelerated to velocity, v in time, tby the application of a constant force, Fthroughout the time, t. The initial and finalmomentum of the object will be, p

1 = mu and

p2 = mv respectively.

The change in momentum ∝ p2 – p

1

∝ mv – mu∝ m × (v – u).

The rate of change of momentum ∝ ( )× −m v u

tOr, the applied force,

F ∝ ( )× −m v u

t

( )× −=

km v u

tF (9.2)

= k m a (9.3)

Here a [ = (v – u)/t ] is the acceleration,which is the rate of change of velocity. Thequantity, k is a constant of proportionality.The SI units of mass and acceleration are kgand m s-2 respectively. The unit of force is sochosen that the value of the constant, kbecomes one. For this, one unit of force isdefined as the amount that produces anacceleration of 1 m s-2 in an object of 1 kgmass. That is,

1 unit of force = k × (1 kg) × (1 m s-2).

Thus, the value of k becomes 1. From Eq. (9.3)

F = ma (9.4)

The unit of force is kg m s-2 or newton,which has the symbol N. The second law of

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motion gives us a method to measure theforce acting on an object as a product of itsmass and acceleration.

The second law of motion is often seen inaction in our everyday life. Have you noticedthat while catching a fast moving cricket ball,a fielder in the ground gradually pulls hishands backwards with the moving ball? Indoing so, the fielder increases the time duringwhich the high velocity of the moving balldecreases to zero. Thus, the acceleration ofthe ball is decreased and therefore the impactof catching the fast moving ball (Fig. 9.8) isalso reduced. If the ball is stopped suddenlythen its high velocity decreases to zero in avery short interval of time. Thus, the rate ofchange of momentum of the ball will be large.Therefore, a large force would have to beapplied for holding the catch that may hurtthe palm of the fielder. In a high jump athleticevent, the athletes are made to fall either ona cushioned bed or on a sand bed. This is toincrease the time of the athlete’s fall to stopafter making the jump. This decreases therate of change of momentum and hence theforce. Try to ponder how a karate playerbreaks a slab of ice with a single blow.

The first law of motion can bemathematically stated from the mathematicalexpression for the second law of motion. Eq.(9.4) is

F = ma

or F( )−

=m v u

t(9.5)

or Ft = mv – mu

That is, when F = 0, v = u for whatever time,t is taken. This means that the object willcontinue moving with uniform velocity, uthroughout the time, t. If u is zero then v willalso be zero. That is, the object will remainat rest.

Example 9.1 A constant force acts on anobject of mass 5 kg for a duration of2 s. It increases the object’s velocityfrom 3 m s–1 to 7 m s-1. Find themagnitude of the applied force. Now, ifthe force was applied for a duration of5 s, what would be the final velocity ofthe object?

Solution:

We have been given that u = 3 m s–1

and v = 7 m s-1, t = 2 s and m = 5 kg.From Eq. (9.5) we have,

F ( )−

=m v u

t

Substitution of values in this relationgives

F = 5 kg (7 m s-1 – 3 m s-1)/2 s = 10 N.

Now, if this force is applied for aduration of 5 s (t = 5 s), then the finalvelocity can be calculated by rewritingEq. (9.5) as

= +Ft

v um

On substituting the values of u, F, mand t, we get the final velocity,

v = 13 m s-1.Fig. 9.8: A fielder pulls his hands gradually with themoving ball while holding a catch.


Example 9.2 Which would require agreater force –– accelerating a 2 kg massat 5 m s–2 or a 4 kg mass at 2 m s-2 ?

Solution:

From Eq. (9.4), we have F = ma.Here we have m

1 = 2 kg; a

1 = 5 m s-2

and m2 = 4 kg; a

2 = 2 m s-2.

Thus, F1 = m

1a

1 = 2 kg × 5 m s-2 = 10 N;

and F2 = m

2a

2 = 4 kg × 2 m s-2 = 8 N.

⇒ F1 > F

2.

Thus, accelerating a 2 kg mass at5 m s-2 would require a greater force.

Example 9.3 A motorcar is moving with avelocity of 108 km/h and it takes 4 s tostop after the brakes are applied.Calculate the force exerted by thebrakes on the motorcar if its mass alongwith the passengers is 1000 kg.

Solution:

The initial velocity of the motorcaru = 108 km/h

= 108 × 1000 m/(60 × 60 s)= 30 m s-1

and the final velocity of the motorcarv = 0 m s-1.The total mass of the motorcar alongwith its passengers = 1000 kg and thetime taken to stop the motorcar, t = 4s. From Eq. (9.5) we have the magnitudeof the force applied by the brakes F asm(v – u)/t.On substituting the values, we getF = 1000 kg × (0 – 30) m s-1/4 s

= – 7500 kg m s-2 or – 7500 N.The negative sign tells us that the forceexerted by the brakes is opposite to thedirection of motion of the motorcar.

Example 9.4 A force of 5 N gives a massm

1, an acceleration of 10 m s–2 and a

mass m2, an acceleration of 20 m s-2.

What acceleration would it give if boththe masses were tied together?

Solution:

From Eq. (9.4) we have m1 = F/a

1; and

m2 = F/a

2. Here, a

1 = 10 m s-2;

a2 = 20 m s-2 and F = 5 N.Thus, m

1 = 5 N/10 m s-2 = 0.50 kg; and

m2 = 5 N/20 m s-2 = 0.25 kg.

If the two masses were tied together,the total mass, m would bem = 0.50 kg + 0.25 kg = 0.75 kg.The acceleration, a produced in thecombined mass by the 5 N force wouldbe, a = F/m = 5 N/0.75 kg = 6.67 m s-2.

Example 9.5 The velocity-time graph of aball of mass 20 g moving along astraight line on a long table is given inFig. 9.9.

Fig. 9.9

How much force does the table exerton the ball to bring it to rest?

Solution:

The initial velocity of the ball is 20 cm s-1.Due to the friction force exerted by thetable, the velocity of the ball decreasesdown to zero in 10 s. Thus, u = 20 cm s–1;v = 0 cm s-1 and t = 10 s. Since thevelocity-time graph is a straight line, it isclear that the ball moves with a constantacceleration. The acceleration a is,

−=

v ua

t

= (0 cm s-1 – 20 cm s-1)/10 s= –2 cm s-2 = –0.02 m s-2.

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The force exerted on the ball F is,F = ma

= (20/1000) kg × (– 0.02 m s-2)

= – 0.0004 N.The negative sign implies that thefrictional force exerted by the table isopposite to the direction of motion ofthe ball.

9.5 Third Law of MotionThe first two laws of motion tell us how anapplied force changes the motion and provideus with a method of determining the force.The third law of motion states that when oneobject exerts a force on another object, thesecond object instantaneously exerts a forceback on the first. These two forces are alwaysequal in magnitude but opposite in direction.These forces act on different objects and neveron the same object. In the game of footballsometimes we, while looking at the footballand trying to kick it with a greater force,collide with a player of the opposite team.Both feel hurt because each applies a forceto the other. In other words, there is a pair offorces and not just one force. The twoopposing forces are also known as action andreaction forces.

Let us consider two spring balancesconnected together as shown in Fig. 9.10. Thefixed end of balance B is attached with a rigidsupport, like a wall. When a force is appliedthrough the free end of spring balance A, it isobserved that both the spring balances showthe same readings on their scales. It meansthat the force exerted by spring balance A onbalance B is equal but opposite in directionto the force exerted by the balance B onbalance A. The force which balance A exertson balance B is called the action and the forceof balance B on balance A is called thereaction. This gives us an alternativestatement of the third law of motion i.e., toevery action there is an equal and oppositereaction. However, it must be rememberedthat the action and reaction always act ontwo different objects.

Fig. 9.10: Action and reaction forces are equal andopposite.

Suppose you are standing at rest andintend to start walking on a road. You mustaccelerate, and this requires a force inaccordance with the second law of motion.Which is this force? Is it the muscular effortyou exert on the road? Is it in the directionwe intend to move? No, you push the roadbelow backwards. The road exerts an equaland opposite reaction force on your feet tomake you move forward.

It is important to note that even thoughthe action and reaction forces are alwaysequal in magnitude, these forces may notproduce accelerations of equal magnitudes.This is because each force acts on a differentobject that may have a different mass.

When a gun is fired, it exerts a forwardforce on the bullet. The bullet exerts an equaland opposite reaction force on the gun. Thisresults in the recoil of the gun (Fig. 9.11).Since the gun has a much greater mass thanthe bullet, the acceleration of the gun is muchless than the acceleration of the bullet. Thethird law of motion can also be illustratedwhen a sailor jumps out of a rowing boat. Asthe sailor jumps forward, the force on the boatmoves it backwards (Fig. 9.12).

Fig. 9.11: A forward force on the bullet and recoil ofthe gun.


The cart shown in this activity can beconstructed by using a 12 mm or 18 mm thickplywood board of about 50 cm × 100 cm withtwo pairs of hard ball-bearing wheels (skatewheels are good to use). Skateboards are notas effective because it is difficult to maintainstraight-line motion.

9.6 Conservation of MomentumSuppose two objects (two balls A and B, say)of masses m

A and m

B are travelling in the same

direction along a straight line at differentvelocities u

A and u

B, respectively [Fig. 9.14(a)].

And there are no other external unbalancedforces acting on them. Let u

A > u

B and the

two balls collide with each other as shown inFig. 9.14(b). During collision which lasts fora time t, the ball A exerts a force F

AB on ball B

and the ball B exerts a force FBA

on ball A.Suppose v

A and v

B are the velocities of the

two balls A and B after the collision,respectively [Fig. 9.14(c)].

Activity ______________ 9.4• Request two children to stand on two

separate carts as shown in Fig. 9.13.• Give them a bag full of sand or some

other heavy object. Ask them to play agame of catch with the bag.

• Does each of them receive aninstantaneous reaction as a result ofthrowing the sand bag (action)?

• You can paint a white line oncartwheels to observe the motion of thetwo carts when the children throw thebag towards each other.

Fig. 9.12: As the sailor jumps in forward direction,the boat moves backwards.

Fig. 9.13

Now, place two children on one cart andone on another cart. The second law of motioncan be seen, as this arrangement would showdifferent accelerations for the same force.

Fig. 9.14: Conservation of momentum in collision oftwo balls.

From Eq. (9.1), the momenta (plural ofmomentum) of ball A before and after thecollision are m

Au

A and m

Av

A, respectively. The

rate of change of its momentum (or FAB

, action)

during the collision will be ( )−A A

A

v um

t.

Similarly, the rate of change of momentumof ball B (= F

BA or reaction) during the collision

will be ( )−B B

B

v um

t.

According to the third law of motion, theforce F

AB exerted by ball A on ball B (action)

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Activity ______________ 9.6• Take a test tube of good quality glass

material and put a small amount ofwater in it. Place a stop cork at themouth of it.

• Now suspend the test tube horizontallyby two strings or wires as shown inFig. 9.16.

• Heat the test tube with a burner untilwater vaporises and the cork blows out.

• Observe that the test tube recoils inthe direction opposite to the directionof the cork.

and the force FBA

exerted by the ball B on ballA (reaction) must be equal and opposite toeach other. Therefore,

FAB

= – FBA

(9.6)

or( )−A A

A

v um

t= –

( )−B BB

v um

t.

This gives,

mAu

A + m

Bu

B = m

Av

A + m

Bv

B(9.7)

Since (mAu

A + m

Bu

B) is the total momentum

of the two balls A and B before the collisionand (m

Av

A + m

Bv

B) is their total momentum

after the collision, from Eq. (9.7) we observethat the total momentum of the two ballsremains unchanged or conserved provided noother external force acts.

As a result of this ideal collisionexperiment, we say that the sum of momentaof the two objects before collision is equal tothe sum of momenta after the collisionprovided there is no external unbalancedforce acting on them. This is known as thelaw of conservation of momentum. Thisstatement can alternatively be given as thetotal momentum of the two objects isunchanged or conserved by the collision.

Activity ______________ 9.5• Take a big rubber balloon and inflate

it fully. Tie its neck using a thread. Alsousing adhesive tape, fix a straw on thesurface of this balloon.

• Pass a thread through the straw andhold one end of the thread in your handor fix it on the wall.

• Ask your friend to hold the other endof the thread or fix it on a wall at somedistance. This arrangement is shownin Fig. 9.15.

• Now remove the thread tied on the neckof balloon. Let the air escape from themouth of the balloon.

• Observe the direction in which thestraw moves.

Fig. 9.15

Fig. 9.16

• Also, observe the difference in thevelocity the cork appears to have andthat of the recoiling test tube.

Example 9.6 A bullet of mass 20 g ishorizontally fired with a velocity150 m s-1 from a pistol of mass 2 kg.What is the recoil velocity of the pistol?

Solution:

We have the mass of bullet,m

1 = 20 g (= 0.02 kg) and the mass of

the pistol, m2 = 2 kg; initial velocities of

the bullet (u1) and pistol (u

2) = 0,

respectively. The final velocity of thebullet, v

1 = + 150 m s-1. The direction

of bullet is taken from left to right(positive, by convention, Fig. 9.17). Letv be the recoil velocity of the pistol.


Total momenta of the pistol and bulletbefore the fire, when the gun is at rest

= (2 + 0.02) kg × 0 m s–1

= 0 kg m s–1

Total momenta of the pistol and bulletafter it is fired

= 0.02 kg × (+ 150 m s–1) + 2 kg × v m s–1

= (3 + 2v) kg m s–1

According to the law of conservation ofmomentumTotal momenta after the fire = Totalmomenta before the fire

3 + 2v = 0⇒ v = − 1.5 m s–1.

Negative sign indicates that thedirection in which the pistol wouldrecoil is opposite to that of bullet, thatis, right to left.

Example 9.7 A girl of mass 40 kg jumpswith a horizontal velocity of 5 m s-1 ontoa stationary cart with frictionlesswheels. The mass of the cart is 3 kg.What is her velocity as the cart startsmoving? Assume that there is noexternal unbalanced force working inthe horizontal direction.

Solution:

Let v be the velocity of the girl on thecart as the cart starts moving.The total momenta of the girl and cartbefore the interaction

= 40 kg × 5 m s–1 + 3 kg × 0 m s–1

= 200 kg m s–1.

Total momenta after the interaction

= (40 + 3) kg × v m s–1

= 43 v kg m s–1.

According to the law of conservation ofmomentum, the total momentum isconserved during the interaction.That is,

43 v = 200⇒ v = 200/43 = + 4.65 m s–1.

The girl on cart would move with avelocity of 4.65 m s–1 in the direction inwhich the girl jumped (Fig. 9.18).Fig. 9.17: Recoil of a pistol

Fig. 9.18: The girl jumps onto the cart.

(a) (b)

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Example 9.8 Two hockey players ofopposite teams, while trying to hit ahockey ball on the ground collide andimmediately become entangled. Onehas a mass of 60 kg and was movingwith a velocity 5.0 m s–1 while the otherhas a mass of 55 kg and was movingfaster with a velocity 6.0 m s–1 towardsthe first player. In which direction andwith what velocity will they move afterthey become entangled? Assume thatthe frictional force acting between thefeet of the two players and ground isnegligible.

Solution:

If v is the velocity of the two entangledplayers after the collision, the totalmomentum then

= (m1 + m

2) × v

= (60 + 55) kg × v m s–1

= 115 × v kg m s–1.Equating the momenta of the systembefore and after collision, in accordancewith the law of conservation ofmomentum, we get

v = – 30/115= – 0.26 m s–1.

Thus, the two entangled players wouldmove with velocity 0.26 m s–1 from rightto left, that is, in the direction thesecond player was moving beforethe collision.

Fig. 9.19: A collision of two hockey players: (a) before collision and (b) after collision.

Let the first player be moving from leftto right. By convention left to right istaken as the positive direction and thusright to left is the negative direction (Fig.9.19). If symbols m and u represent themass and initial velocity of the twoplayers, respectively. Subscripts 1 and2 in these physical quantities refer tothe two hockey players. Thus,

m1 = 60 kg; u

1 = + 5 m s-1; and

m2 = 55 kg; u

2 = – 6 m s-1.

The total momentum of the two playersbefore the collision

= 60 kg × (+ 5 m s-1) +55 kg × (– 6 m s-1)

= – 30 kg m s-1

uestions1. If action is always equal to the

reaction, explain how a horse canpull a cart.

2. Explain, why is it difficult for afireman to hold a hose, whichejects large amounts of water ata high velocity.

3. From a rifle of mass 4 kg, a bulletof mass 50 g is fired with aninitial velocity of 35 m s–1.Calculate the initial recoil velocityof the rifle.

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4. Two objects of masses 100 g and200 g are moving along the sameline and direction with velocitiesof 2 m s–1 and 1 m s–1, respectively.

They collide and after the collision,the first object moves at a velocityof 1.67 m s–1. Determine thevelocity of the second object.

Whatyou havelearnt• First law of motion: An object continues to be in a state of rest

or of uniform motion along a straight line unless acted uponby an unbalanced force.

• The natural tendency of objects to resist a change in their stateof rest or of uniform motion is called inertia.

• The mass of an object is a measure of its inertia. Its SI unit iskilogram (kg).

• Force of friction always opposes motion of objects.• Second law of motion: The rate of change of momentum of an

object is proportional to the applied unbalanced force in thedirection of the force.

• The SI unit of force is kg m s–2. This is also known as newtonand represented by the symbol N. A force of one newtonproduces an acceleration of 1 m s–2 on an object of mass 1 kg.

• The momentum of an object is the product of its mass andvelocity and has the same direction as that of the velocity. ItsSI unit is kg m s–1.

• Third law of motion: To every action, there is an equal andopposite reaction and they act on two different bodies.

• In an isolated system, the total momentum remains conserved.

CONSERVATION LAWS

All conservation laws such as conservation of momentum, energy, angular momentum,charge etc. are considered to be fundamental laws in physics. These are based onobservations and experiments. It is important to remember that a conservation law cannotbe proved. It can be verified, or disproved, by experiments. An experiment whose result isin conformity with the law verifies or substantiates the law; it does not prove the law. Onthe other hand, a single experiment whose result goes against the law is enough to disproveit.

The law of conservation of momentum has been deduced from large number ofobservations and experiments. This law was formulated nearly three centuries ago. It isinteresting to note that not a single situation has been realised so far, which contradictsthis law. Several experiences of every-day life can be explained on the basis of the law ofconservation of momentum.

SCIENCE128

Exercises1. An object experiences a net zero external unbalanced force. Is

it possible for the object to be travelling with a non-zero velocity?If yes, state the conditions that must be placed on themagnitude and direction of the velocity. If no, provide a reason.

2. When a carpet is beaten with a stick, dust comes out of it.Explain.

3. Why is it advised to tie any luggage kept on the roof of a buswith a rope?

4. A batsman hits a cricket ball which then rolls on a level ground.After covering a short distance, the ball comes to rest. The ballslows to a stop because

(a) the batsman did not hit the ball hard enough.

(b) velocity is proportional to the force exerted on the ball.

(c) there is a force on the ball opposing the motion.

(d) there is no unbalanced force on the ball, so the ball wouldwant to come to rest.

5. A truck starts from rest and rolls down a hill with a constantacceleration. It travels a distance of 400 m in 20 s. Find itsacceleration. Find the force acting on it if its mass is 7 metrictonnes (Hint: 1 metric tonne = 1000 kg.)

6. A stone of 1 kg is thrown with a velocity of 20 m s–1 across thefrozen surface of a lake and comes to rest after travelling adistance of 50 m. What is the force of friction between thestone and the ice?

7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg,along a horizontal track. If the engine exerts a force of 40000 Nand the track offers a friction force of 5000 N, then calculate:

(a) the net accelerating force;

(b) the acceleration of the train; and

(c) the force of wagon 1 on wagon 2.

8. An automobile vehicle has a mass of 1500 kg. What must bethe force between the vehicle and road if the vehicle is to bestopped with a negative acceleration of 1.7 m s–2?

9. What is the momentum of an object of mass m, moving with avelocity v?

(a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv

10. Using a horizontal force of 200 N, we intend to move a woodencabinet across a floor at a constant velocity. What is the frictionforce that will be exerted on the cabinet?

11. Two objects, each of mass 1.5 kg, are moving in the samestraight line but in opposite directions. The velocity of each


object is 2.5 m s-1 before the collision during which they sticktogether. What will be the velocity of the combined object aftercollision?

12. According to the third law of motion when we push on an object,the object pushes back on us with an equal and opposite force.If the object is a massive truck parked along the roadside, itwill probably not move. A student justifies this by answeringthat the two opposite and equal forces cancel each other.Comment on this logic and explain why the truck does notmove.

13. A hockey ball of mass 200 g travelling at 10 m s–1 is struck bya hockey stick so as to return it along its original path with avelocity at 5 m s–1. Calculate the change of momentum occurredin the motion of the hockey ball by the force applied by thehockey stick.

14. A bullet of mass 10 g travelling horizontally with a velocity of150 m s–1 strikes a stationary wooden block and comes to restin 0.03 s. Calculate the distance of penetration of the bulletinto the block. Also calculate the magnitude of the force exertedby the wooden block on the bullet.

15. An object of mass 1 kg travelling in a straight line with a velocityof 10 m s–1 collides with, and sticks to, a stationary woodenblock of mass 5 kg. Then they both move off together in thesame straight line. Calculate the total momentum just beforethe impact and just after the impact. Also, calculate the velocityof the combined object.

16. An object of mass 100 kg is accelerated uniformly from a velocityof 5 m s–1 to 8 m s–1 in 6 s. Calculate the initial and finalmomentum of the object. Also, find the magnitude of the forceexerted on the object.

17. Akhtar, Kiran and Rahul were riding in a motorcar that wasmoving with a high velocity on an expressway when an insecthit the windshield and got stuck on the windscreen. Akhtarand Kiran started pondering over the situation. Kiran suggestedthat the insect suffered a greater change in momentum ascompared to the change in momentum of the motorcar (becausethe change in the velocity of the insect was much more thanthat of the motorcar). Akhtar said that since the motorcar wasmoving with a larger velocity, it exerted a larger force on theinsect. And as a result the insect died. Rahul while putting anentirely new explanation said that both the motorcar and theinsect experienced the same force and a change in theirmomentum. Comment on these suggestions.

18. How much momentum will a dumb-bell of mass 10 kg transferto the floor if it falls from a height of 80 cm? Take its downwardacceleration to be 10 m s–2.

SCIENCE130

AdditionalExercisesA1. The following is the distance-time table of an object in motion:

Time in seconds Distance in metres

0 0

1 1

2 8

3 27

4 64

5 125

6 216

7 343

(a) What conclusion can you draw about the acceleration? Isit constant, increasing, decreasing, or zero?

(b) What do you infer about the forces acting on the object?

A2. Two persons manage to push a motorcar of mass 1200 kg at auniform velocity along a level road. The same motorcar can bepushed by three persons to produce an acceleration of0.2 m s-2. With what force does each person push the motorcar?(Assume that all persons push the motorcar with the samemuscular effort.)

A3. A hammer of mass 500 g, moving at 50 m s-1, strikes a nail.The nail stops the hammer in a very short time of 0.01 s. Whatis the force of the nail on the hammer?

A4. A motorcar of mass 1200 kg is moving along a straight linewith a uniform velocity of 90 km/h. Its velocity is slowed downto 18 km/h in 4 s by an unbalanced external force. Calculatethe acceleration and change in momentum. Also calculate themagnitude of the force required.

A5. A large truck and a car, both moving with a velocity of magnitudev, have a head-on collision and both of them come to a haltafter that. If the collision lasts for 1 s:

(a) Which vehicle experiences the greater force of impact?

(b) Which vehicle experiences the greater change inmomentum?

(c) Which vehicle experiences the greater acceleration?

(d) Why is the car likely to suffer more damage than the truck?

In Chapters 8 and 9, we have learnt aboutthe motion of objects and force as the causeof motion. We have learnt that a force isneeded to change the speed or the directionof motion of an object. We always observe thatan object dropped from a height falls towardsthe earth. We know that all the planets goaround the Sun. The moon goes around theearth. In all these cases, there must be someforce acting on the objects, the planets andon the moon. Isaac Newton could grasp thatthe same force is responsible for all these.This force is called the gravitational force.

In this chapter we shall learn aboutgravitation and the universal law ofgravitation. We shall discuss the motion ofobjects under the influence of gravitationalforce on the earth. We shall study how theweight of a body varies from place to place.We shall also discuss the conditions forobjects to float in liquids.

10.1 GravitationWe know that the moon goes around theearth. An object when thrown upwards,reaches a certain height and then fallsdownwards. It is said that when Newton wassitting under a tree, an apple fell on him. Thefall of the apple made Newton start thinking.He thought that: if the earth can attract anapple, can it not attract the moon? Is the forcethe same in both cases? He conjectured thatthe same type of force is responsible in boththe cases. He argued that at each point of itsorbit, the moon falls towards the earth,instead of going off in a straight line. So, itmust be attracted by the earth. But we donot really see the moon falling towards theearth.

Let us try to understand the motion ofthe moon by recalling activity 8.11.

Activity _____________10.1• Take a piece of thread.• Tie a small stone at one end. Hold the

other end of the thread and whirl itround, as shown in Fig. 10.1.

• Note the motion of the stone.• Release the thread.• Again, note the direction of motion of

the stone.

Fig. 10.1: A stone describing a circular path with avelocity of constant magnitude.

Before the thread is released, the stonemoves in a circular path with a certain speedand changes direction at every point. Thechange in direction involves change in velocityor acceleration. The force that causes thisacceleration and keeps the body moving alongthe circular path is acting towards the centre.This force is called the centripetal (meaning‘centre-seeking’) force. In the absence of this

1010101010GGGGGRAVITATIONRAVITATIONRAVITATIONRAVITATIONRAVITATION

Chapter

SCIENCE132

10.1.1 UNIVERSAL LAW OF GRAVITATION

Every object in the universe attracts everyother object with a force which is proportionalto the product of their masses and inverselyproportional to the square of the distancebetween them. The force is along the linejoining the centres of two objects.

force, the stone flies off along a straight line.This straight line will be a tangent to thecircular path.

Mor

e to

kno

w

Tangent to a circle

A straight line that meets the circleat one and only one point is called atangent to the circle. Straight lineABC is a tangent to the circle atpoint B.

The motion of the moon around the earthis due to the centripetal force. The centripetalforce is provided by the force of attraction ofthe earth. If there were no such force, themoon would pursue a uniform straight linemotion.

It is seen that a falling apple is attractedtowards the earth. Does the apple attract theearth? If so, we do not see the earth movingtowards an apple. Why?

According to the third law of motion, theapple does attract the earth. But accordingto the second law of motion, for a given force,acceleration is inversely proportional to themass of an object [Eq. (9.4)]. The mass of anapple is negligibly small compared to that ofthe earth. So, we do not see the earth movingtowards the apple. Extend the same argumentfor why the earth does not move towards themoon.

In our solar system, all the planets goaround the Sun. By arguing the same way,we can say that there exists a force betweenthe Sun and the planets. From the above factsNewton concluded that not only does theearth attract an apple and the moon, but allobjects in the universe attract each other. Thisforce of attraction between objects is calledthe gravitational force.

G2

MmF =

d

Fig. 10.2: The gravitational force between twouniform objects is directed along the linejoining their centres.

Let two objects A and B of masses M andm lie at a distance d from each other as shownin Fig. 10.2. Let the force of attraction betweentwo objects be F. According to the universallaw of gravitation, the force between twoobjects is directly proportional to the productof their masses. That is,

F∝M × m (10.1)And the force between two objects is inverselyproportional to the square of the distancebetween them, that is,

∝2

1F

d(10.2)

Combining Eqs. (10.1) and (10.2), we get

F ∝ 2

×M m

d(10.3)

or, G2

M × mF =

d(10.4)

where G is the constant of proportionality andis called the universal gravitation constant.By multiplying crosswise, Eq. (10.4) gives

F × d 2 = G M × m

GRAVITATION 133

Isaac Newton was bornin Woolsthorpe nearGrantham, England.He is generallyregarded as the mostoriginal andinfluential theorist inthe history of science.He was born in a poorfarming family. But hewas not good atfarming. He was sentto study at CambridgeUniversity in 1661. In1665 a plague broke

out in Cambridge and so Newton took a yearoff. It was during this year that the incident ofthe apple falling on him is said to haveoccurred. This incident prompted Newton toexplore the possibility of connecting gravitywith the force that kept the moon in its orbit.This led him to the universal law ofgravitation. It is remarkable that many greatscientists before him knew of gravity but failedto realise it.

Newton formulated the well-known laws ofmotion. He worked on theories of light andcolour. He designed an astronomical telescopeto carry out astronomical observations.Newton was also a great mathematician. Heinvented a new branch of mathematics, calledcalculus. He used it to prove that for objectsoutside a sphere of uniform density, the spherebehaves as if the whole of its mass isconcentrated at its centre. Newtontransformed the structure of physicalscience with his three laws of motion and theuniversal law of gravitation. As the keystoneof the scientific revolution of the seventeenthcentury, Newton’s work combined thecontributions of Copernicus, Kepler, Galileo,and others into a new powerful synthesis.

It is remarkable that though thegravitational theory could not be verified atthat time, there was hardly any doubt aboutits correctness. This is because Newton basedhis theory on sound scientific reasoning andbacked it with mathematics. This made thetheory simple and elegant. These qualities arenow recognised as essential requirements of agood scientific theory.

Isaac Newton(1642 – 1727)

How did Newton guess theinverse-square rule?

There has always been a great interestin the motion of planets. By the 16thcentury, a lot of data on the motion ofplanets had been collected by manyastronomers. Based on these dataJohannes Kepler derived three laws,which govern the motion of planets.These are called Kepler’s laws. These are:1. The orbit of a planet is an ellipse with

the Sun at one of the foci, as shown inthe figure given below. In this figure Ois the position of the Sun.

2. The line joining the planet and the Sunsweep equal areas in equal intervalsof time. Thus, if the time of travel fromA to B is the same as that from C to D,then the areas OAB and OCD areequal.

3. The cube of the mean distance of aplanet from the Sun is proportional tothe square of its orbital period T. Or,r3/T2 = constant.It is important to note that Kepler

could not give a theory to explainthe motion of planets. It was Newtonwho showed that the cause of theplanetary motion is the gravitationalforce that the Sun exerts on them. Newtonused the third lawof Kepler tocalculate thegravitational forceof attraction. Thegravitational forceof the earth isweakened by distance. A simple argumentgoes like this. We can assume that theplanetary orbits are circular. Suppose theorbital velocity is v and the radius of theorbit is r. Then the force acting on anorbiting planet is given by F∝v2/r.

If T denotes the period, then v = 2πr/T,so that v2 ∝ r2/T2.

We can rewrite this as v2 ∝ (1/r) ×( r3/T2). Since r3/T2 is constant by Kepler’sthird law, we have v2 ∝ 1/r. Combiningthis with F ∝ v2/ r, we get, F ∝ 1/ r2.

A

B

CD

O

SCIENCE134

From Eq. (10.4), the force exerted bythe earth on the moon is

G 2

M × mF =

d

11 2 -2 24 22

8 2

6.7 10 N m kg 6 10 kg 7.4 10 kg

(3.84 10 m)

−× × × × ×=

×

= 2.01 × 1020 N.

Thus, the force exerted by the earth onthe moon is 2.01 × 1020 N.

uestions1. State the universal law of

gravitation.2. Write the formula to find the

magnitude of the gravitationalforce between the earth and anobject on the surface of the earth.

10.1.2 IMPORTANCE OF THE UNIVERSALLAW OF GRAVITATION

The universal law of gravitation successfullyexplained several phenomena which werebelieved to be unconnected:

(i) the force that binds us to the earth;(ii) the motion of the moon around the

earth;(iii) the motion of planets around the Sun;

and(iv) the tides due to the moon and the Sun.

10.2 Free FallLet us try to understand the meaning of freefall by performing this activity.

Activity _____________10.2• Take a stone.• Throw it upwards.• It reaches a certain height and then it

starts falling down.

We have learnt that the earth attractsobjects towards it. This is due to thegravitational force. Whenever objects falltowards the earth under this force alone, wesay that the objects are in free fall. Is there

or2

G =×

F d

M m(10.5)

The SI unit of G can be obtained bysubstituting the units of force, distance andmass in Eq. (10.5) as N m2 kg–2.

The value of G was found out byHenry Cavendish (1731 – 1810) by using asensitive balance. The accepted value of G is6.673 × 10–11 N m2 kg–2.

We know that there exists a force ofattraction between any two objects. Computethe value of this force between you and yourfriend sitting closeby. Conclude how you donot experience this force!

The law is universal in the sense thatit is applicable to all bodies, whetherthe bodies are big or small, whetherthey are celestial or terrestrial.

Inverse-square

Saying that F is inverselyproportional to the square of dmeans, for example, that if d getsbigger by a factor of 6, F becomes

1

36times smaller.

Example 10.1 The mass of the earth is6 × 1024 kg and that of the moon is7.4 × 1022 kg. If the distance betweenthe earth and the moon is 3.84×105 km,calculate the force exerted by the earthon the moon. G = 6.7 × 10–11 N m2 kg-2.

Solution:

The mass of the earth, M = 6 × 1024 kgThe mass of the moon,

m = 7.4 × 1022 kgThe distance between the earth and themoon,

d = 3.84 × 105 km= 3.84 × 105 × 1000 m= 3.84 × 108 m

G = 6.7 × 10–11 N m2 kg–2

Mor

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kno

w

Q

GRAVITATION 135

calculations, we can take g to be more or lessconstant on or near the earth. But for objectsfar from the earth, the acceleration due togravitational force of earth is given byEq. (10.7).

10.2.1 TO CALCULATE THE VALUE OF gTo calculate the value of g, we should putthe values of G, M and R in Eq. (10.9),namely, universal gravitational constant,G = 6.7 × 10–11 N m2 kg-2, mass of the earth,M = 6 × 1024 kg, and radius of the earth,R = 6.4 × 106 m.

G2

Mg =

R

-11 2 -2 24

6 2

6.7 10 N m kg 6 10 kg=

(6.4 10 m)

× × ××

= 9.8 m s–2.

Thus, the value of acceleration due to gravityof the earth, g = 9.8 m s–2.

10.2.2 MOTION OF OBJECTS UNDER THE

INFLUENCE OF GRAVITATIONAL

FORCE OF THE EARTH

Let us do an activity to understand whetherall objects hollow or solid, big or small, willfall from a height at the same rate.

Activity _____________10.3• Take a sheet of paper and a stone. Drop

them simultaneously from the first floorof a building. Observe whether both ofthem reach the ground simultaneously.

• We see that paper reaches the groundlittle later than the stone. This happensbecause of air resistance. The air offersresistance due to friction to the motionof the falling objects. The resistanceoffered by air to the paper is more thanthe resistance offered to the stone. Ifwe do the experiment in a glass jar fromwhich air has been sucked out, thepaper and the stone would fall at thesame rate.

any change in the velocity of falling objects?While falling, there is no change in thedirection of motion of the objects. But due tothe earth’s attraction, there will be a changein the magnitude of the velocity. Any changein velocity involves acceleration. Whenever anobject falls towards the earth, an accelerationis involved. This acceleration is due to theearth’s gravitational force. Therefore, thisacceleration is called the acceleration due tothe gravitational force of the earth (oracceleration due to gravity). It is denoted byg. The unit of g is the same as that ofacceleration, that is, m s–2.

We know from the second law of motionthat force is the product of mass andacceleration. Let the mass of the stone inactivity 10.2 be m. We already know that thereis acceleration involved in falling objects dueto the gravitational force and is denoted by g.Therefore the magnitude of the gravitationalforce F will be equal to the product of massand acceleration due to the gravitationalforce, that is,

F = m g (10.6)

From Eqs. (10.4) and (10.6) we have

2= G

×M mm g

d

or G2

Mg =

d(10.7)

where M is the mass of the earth, and d isthe distance between the object and the earth.

Let an object be on or near the surface ofthe earth. The distance d in Eq. (10.7) will beequal to R, the radius of the earth. Thus, forobjects on or near the surface of the earth,

G2

M × mmg =

R(10.8)

G2

Mg =

R(10.9)

The earth is not a perfect sphere. As theradius of the earth increases from the polesto the equator, the value of g becomes greaterat the poles than at the equator. For most

SCIENCE136

We know that an object experiencesacceleration during free fall. From Eq. (10.9),this acceleration experienced by an object isindependent of its mass. This means that allobjects hollow or solid, big or small, shouldfall at the same rate. According to a story,Galileo dropped different objects from the topof the Leaning Tower of Pisa in Italy to provethe same.

As g is constant near the earth, all theequations for the uniformly acceleratedmotion of objects become valid withacceleration a replaced by g (see section 8.5).The equations are:

v = u + at (10.10)

s = ut + 12

at2 (10.11)

v2 = u2 + 2as (10.12)

where u and v are the initial and finalvelocities and s is the distance covered intime, t.

In applying these equations, we will takeacceleration, a to be positive when it is in thedirection of the velocity, that is, in thedirection of motion. The acceleration, a willbe taken as negative when it opposes themotion.

Example 10.2 A car falls off a ledge anddrops to the ground in 0.5 s. Letg = 10 m s–2 (for simplifying thecalculations).(i) What is its speed on striking the

ground?(ii) What is its average speed during the

0.5 s?(iii) How high is the ledge from the

ground?

Solution:

Time, t = ½ secondInitial velocity, u = 0 m s–1

Acceleration due to gravity, g = 10 m s–2

Acceleration of the car, a = + 10 m s–2

(downward)(i) speed v = a t

v = 10 m s–2 × 0.5 s= 5 m s–1

(ii) average speed=2

u +v

= (0 m s–1+ 5 m s–1)/2= 2.5 m s–1

(iii) distance travelled, s = ½ a t2

= ½ × 10 m s–2 × (0.5 s)2

= ½ × 10 m s–2 × 0.25 s2

= 1.25 mThus,(i) its speed on striking the ground

= 5 m s–1

(ii) its average speed during the 0.5 s= 2.5 m s–1

(iii) height of the ledge from the ground= 1.25 m.

Example 10.3 An object is thrownvertically upwards and rises to a heightof 10 m. Calculate (i) the velocity withwhich the object was thrown upwardsand (ii) the time taken by the object toreach the highest point.

Solution:

Distance travelled, s = 10 mFinal velocity, v = 0 m s–1

Acceleration due to gravity, g = 9.8 m s–2

Acceleration of the object, a = –9.8 m s–2

(upward motion)(i) v 2 = u2 + 2a s

0 = u 2 + 2 × (–9.8 m s–2) × 10 m–u 2 = –2 × 9.8 × 10 m2 s–2

u = 196 m s-1

u = 14 m s-1

(ii) v = u + a t0 = 14 m s–1 – 9.8 m s–2 × tt = 1.43 s.

Thus,(i) Initial velocity, u = 14 m s–1, and(ii) Time taken, t = 1.43 s.

uestions1. What do you mean by free fall?2. What do you mean by acceleration

due to gravity?Q

GRAVITATION 137

10.3 MassWe have learnt in the previous chapter thatthe mass of an object is the measure of itsinertia (section 9.3). We have also learnt thatgreater the mass, the greater is the inertia. Itremains the same whether the object is onthe earth, the moon or even in outer space.Thus, the mass of an object is constant anddoes not change from place to place.

10.4 WeightWe know that the earth attracts every objectwith a certain force and this force dependson the mass (m) of the object and theacceleration due to the gravity (g). The weightof an object is the force with which it isattracted towards the earth.We know that

F = m × a, (10.13)that is,

F = m × g. (10.14)The force of attraction of the earth on an

object is known as the weight of the object. Itis denoted by W. Substituting the same inEq. (10.14), we have

W = m × g (10.15)As the weight of an object is the force with

which it is attracted towards the earth, theSI unit of weight is the same as that of force,that is, newton (N). The weight is a force actingvertically downwards; it has both magnitudeand direction.

We have learnt that the value of g isconstant at a given place. Therefore at a givenplace, the weight of an object is directlyproportional to the mass, say m, of the object,that is, W ∝ m. It is due to this reason thatat a given place, we can use the weight of anobject as a measure of its mass. The mass ofan object remains the same everywhere, thatis, on the earth and on any planet whereasits weight depends on its location.

10.4.1 WEIGHT OF AN OBJECT ONTHE MOON

We have learnt that the weight of an objecton the earth is the force with which the earth

attracts the object. In the same way, theweight of an object on the moon is the forcewith which the moon attracts that object. Themass of the moon is less than that of theearth. Due to this the moon exerts lesser forceof attraction on objects.

Let the mass of an object be m. Let itsweight on the moon be W

m. Let the mass of

the moon be Mm and its radius be R

m.

By applying the universal law ofgravitation, the weight of the object on themoon will be

2G

×= m

m

m

M mW

R (10.16)

Let the weight of the same object on theearth be W

e. The mass of the earth is M and

its radius is R.

Table 10.1

Celestial Mass (kg) Radius (m)body

Earth 5.98 × 1024 6.37 × 106

Moon 7.36 × 1022 1.74 × 106

From Eqs. (10.9) and (10.15) we have,

2G×

=e

M mW

R (10.17)

Substituting the values from Table 10.1 inEqs. (10.16) and (10.17), we get

( )22

26

7.36 10 kgG

1.74 10 m

× ×=

×m

mW

102.431 10 G × = ×mW m (10.18a)

and 111.474 10 G × = ×eW m (10.18b)Dividing Eq. (10.18a) by Eq. (10.18b), we get

10

11

2.431 10

1.474 10

×=

×m

e

W

W

or 1

0.1656

m

e

W

W= ≈ (10.19)

Weight of theobject on the moon 1=

Weight of theobject on theearth 6

Weight of the object on the moon= (1/6) × its weight on the earth.

SCIENCE138

Example 10.4 Mass of an object is 10 kg.What is its weight on the earth?

Solution:

Mass, m = 10 kgAcceleration due to gravity, g = 9.8 m s–2

W = m × gW = 10 kg × 9.8 m s-2 = 98 N

Thus, the weight of the object is 98 N.

Example 10.5 An object weighs 10 N whenmeasured on the surface of the earth.What would be its weight whenmeasured on the surface of the moon?

Solution:

We know,Weight of object on the moon

= (1/6) × its weight on the earth.That is,

10

6 6e

m

WW = = N.

= 1.67 N.Thus, the weight of object on thesurface of the moon would be 1.67 N.

uestions1. What are the differences between

the mass of an object and itsweight?

2. Why is the weight of an object on

the moon 16

th its weight on the

earth?

10.5 Thrust and PressureHave you ever wondered why a camel can runin a desert easily? Why an army tank weighingmore than a thousand tonne rests upon acontinuous chain? Why a truck or a motorbushas much wider tyres? Why cutting tools havesharp edges? In order to address thesequestions and understand the phenomenainvolved, it helps to introduce the concepts

of the net force in a particular direction(thrust) and the force per unit area (pressure)acting on the object concerned.

Let us try to understand the meanings ofthrust and pressure by considering thefollowing situations:Situation 1: You wish to fix a poster on abulletin board, as shown in Fig 10.3. To dothis task you will have to press drawing pinswith your thumb. You apply a force on thesurface area of the head of the pin. This forceis directed perpendicular to the surface areaof the board. This force acts on a smaller areaat the tip of the pin.

QFig. 10.3: To fix a poster, drawing pins are pressed

with the thumb perpendicular to the board.

Situation 2: You stand on loose sand. Yourfeet go deep into the sand. Now, lie down onthe sand. You will find that your body willnot go that deep in the sand. In both casesthe force exerted on the sand is the weight ofyour body.

GRAVITATION 139

You have learnt that weight is the forceacting vertically downwards. Here the forceis acting perpendicular to the surface of thesand. The force acting on an objectperpendicular to the surface is called thrust.

When you stand on loose sand, the force,that is, the weight of your body is acting onan area equal to area of your feet. When youlie down, the same force acts on an area equalto the contact area of your whole body, whichis larger than the area of your feet. Thus, theeffects of forces of the same magnitude ondifferent areas are different. In the abovecases, thrust is the same. But effects aredifferent. Therefore the effect of thrustdepends on the area on which it acts.

The effect of thrust on sand is larger whilestanding than while lying. The thrust on unitarea is called pressure. Thus,

thrustPressure=

area(10.20)

Substituting the SI unit of thrust and area inEq. (10.20), we get the SI unit of pressure asN/m2 or N m–2.

In honour of scientist Blaise Pascal, theSI unit of pressure is called pascal, denotedas Pa.

Let us consider a numerical example tounderstand the effects of thrust acting ondifferent areas.

Example 10.6 A block of wood is kept on atabletop. The mass of wooden block is5 kg and its dimensions are 40 cm × 20cm × 10 cm. Find the pressure exerted

by the wooden block on the table top ifit is made to lie on the table top with itssides of dimensions (a) 20 cm × 10 cmand (b) 40 cm × 20 cm.

Solution:

The mass of the wooden block = 5 kgThe dimensions

= 40 cm × 20 cm × 10 cmHere, the weight of the wooden blockapplies a thrust on the table top.That is, Thrust = F = m × g

= 5 kg × 9.8 m s–2

= 49 NArea of a side = length × breadth

= 20 cm × 10 cm= 200 cm2 = 0.02 m2

From Eq. (10.20),

Pressure = 2

49N

0.02m

= 2450 N m-2.When the block lies on its side ofdimensions 40 cm × 20 cm, it exertsthe same thrust.Area= length × breadth

= 40 cm × 20 cm= 800 cm2 = 0.08 m2

From Eq. (10.20),

Pressure = 2

49 N

0.08 m

= 612.5 N m–2

The pressure exerted by the side 20 cm× 10 cm is 2450 N m–2 and by the side40 cm × 20 cm is 612.5 N m–2.

Thus, the same force acting on a smallerarea exerts a larger pressure, and a smallerpressure on a larger area. This is the reasonwhy a nail has a pointed tip, knives have sharpedges and buildings have wide foundations.

10.5.1 PRESSURE IN FLUIDS

All liquids and gases are fluids. A solid exertspressure on a surface due to its weight.Similarly, fluids have weight, and they alsoFig. 10.4

SCIENCE140

by the water on the bottle is greater than itsweight. Therefore it rises up when released.

To keep the bottle completely immersed,the upward force on the bottle due to watermust be balanced. This can be achieved byan externally applied force acting downwards.This force must at least be equal to thedifference between the upward force and theweight of the bottle.

The upward force exerted by the water onthe bottle is known as upthrust or buoyantforce. In fact, all objects experience a force ofbuoyancy when they are immersed in a fluid.The magnitude of this buoyant force dependson the density of the fluid.

10.5.3 WHY OBJECTS FLOAT OR SINK WHEN

PLACED ON THE SURFACE OF WATER?Let us do the following activities to arrive atan answer for the above question.

Activity _____________10.5• Take a beaker filled with water.• Take an iron nail and place it on the

surface of the water.• Observe what happens.

The nail sinks. The force due to thegravitational attraction of the earth on theiron nail pulls it downwards. There is anupthrust of water on the nail, which pushesit upwards. But the downward force actingon the nail is greater than the upthrust ofwater on the nail. So it sinks (Fig. 10.5).

exert pressure on the base and walls of thecontainer in which they are enclosed.Pressure exerted in any confined mass of fluidis transmitted undiminished in all directions.

10.5.2 BUOYANCY

Have you ever had a swim in a pool and feltlighter? Have you ever drawn water from awell and felt that the bucket of water is heavierwhen it is out of the water? Have you everwondered why a ship made of iron and steeldoes not sink in sea water, but whether thesame amount of iron and steel in the form ofa sheet would sink? These questions can beanswered by taking buoyancy inconsideration. Let us understand themeaning of buoyancy by doing an activity.

Activity _____________10.4• Take an empty plastic bottle. Close the

mouth of the bottle with an airtightstopper. Put it in a bucket filled withwater. You see that the bottle floats.

• Push the bottle into the water. You feelan upward push. Try to push it furtherdown. You will find it difficult to pushdeeper and deeper. This indicates thatwater exerts a force on the bottle in theupward direction. The upward forceexerted by the water goes on increasingas the bottle is pushed deeper till it iscompletely immersed.

• Now, release the bottle. It bouncesback to the surface.

• Does the force due to the gravitationalattraction of the earth act on thisbottle? If so, why doesn’t the bottle stayimmersed in water after it is released?How can you immerse the bottle inwater?

The force due to the gravitationalattraction of the earth acts on the bottle inthe downward direction. So the bottle ispulled downwards. But the water exerts anupward force on the bottle. Thus, the bottleis pushed upwards. We have learnt thatweight of an object is the force due togravitational attraction of the earth. When thebottle is immersed, the upward force exerted

Fig. 10.5: An iron nail sinks and a cork floats whenplaced on the surface of water.

GRAVITATION 141

• Observe what happens to elongationof the string or the reading on thebalance.

You will find that the elongation of thestring or the reading of the balance decreasesas the stone is gradually lowered in the water.However, no further change is observed oncethe stone gets fully immersed in the water.What do you infer from the decrease in theextension of the string or the reading of thespring balance?

We know that the elongation produced inthe string or the spring balance is due to theweight of the stone. Since the extensiondecreases once the stone is lowered in water,it means that some force acts on the stone inupward direction. As a result, the net forceon the string decreases and hence theelongation also decreases. As discussedearlier, this upward force exerted by water isknown as the force of buoyancy.

What is the magnitude of the buoyantforce experienced by a body? Is it the samein all fluids for a given body? Do all bodies ina given fluid experience the same buoyantforce? The answer to these questions is

Activity _____________10.6• Take a beaker filled with water.• Take a piece of cork and an iron nail of

equal mass.• Place them on the surface of water.• Observe what happens.

The cork floats while the nail sinks. Thishappens because of the difference in theirdensities. The density of a substance isdefined as the mass per unit volume. Thedensity of cork is less than the density ofwater. This means that the upthrust of wateron the cork is greater than the weight of thecork. So it floats (Fig. 10.5).

The density of an iron nail is more thanthe density of water. This means that theupthrust of water on the iron nail is less thanthe weight of the nail. So it sinks.

Therefore objects of density less than thatof a liquid float on the liquid. The objects ofdensity greater than that of a liquid sink inthe liquid.

uestions1. Why is it difficult to hold a school

bag having a strap made of a thinand strong string?

2. What do you mean by buoyancy?3. Why does an object float or sink

when placed on the surface ofwater?

10.6 Archimedes’ Principle

Activity _____________10.7• Take a piece of stone and tie it to one

end of a rubber string or a springbalance.

• Suspend the stone by holding thebalance or the string as shown inFig. 10.6 (a).

• Note the elongation of the string or thereading on the spring balance due tothe weight of the stone.

• Now, slowly dip the stone in the waterin a container as shown inFig. 10.6 (b).

Fig. 10.6: (a) Observe the elongation of the rubberstring due to the weight of a piece of stonesuspended from it in air. (b) The elongationdecreases as the stone is immersedin water.

(a)

(b)

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SCIENCE142

contained in Archimedes’ principle, stated asfollows:

When a body is immersed fully or partiallyin a fluid, it experiences an upward force thatis equal to the weight of the fluid displacedby it.

Now, can you explain why a furtherdecrease in the elongation of the string wasnot observed in activity 10.7, as the stonewas fully immersed in water?

Archimedes was a Greekscientist. He discovered theprinciple, subsequentlynamed after him, afternoticing that the water in abathtub overflowed when hestepped into it. He ranthrough the streets shouting“Eureka!”, which means “I

have got it”. This knowledge helped him todetermine the purity of the gold in the crownmade for the king.

His work in the field of Geometry andMechanics made him famous. Hisunderstanding of levers, pulleys, wheels-and-axle helped the Greek army in its warwith Roman army.

Archimedes’ principle has manyapplications. It is used in designing ships andsubmarines. Lactometers, which are used todetermine the purity of a sample of milk andhydrometers used for determining density ofliquids, are based on this principle.

uestions1. You find your mass to be 42 kg

on a weighing machine. Is yourmass more or less than 42 kg?

2. You have a bag of cotton and aniron bar, each indicating a massof 100 kg when measured on aweighing machine. In reality, oneis heavier than other. Can yousay which one is heavierand why?

10.7 Relative DensityAs you know, the density of a substance isdefined as mass of a unit volume. The unit ofdensity is kilogram per metre cube (kg m–3).The density of a given substance, underspecified conditions, remains the same.Therefore the density of a substance is oneof its characteristic properties. It is differentfor different substances. For example, thedensity of gold is 19300 kg m-3 while that ofwater is 1000 kg m-3. The density of a givensample of a substance can help us todetermine its purity.

It is often convenient to express densityof a substance in comparison with that ofwater. The relative density of a substance isthe ratio of its density to that of water:

Density of a substanceRelativedensity =

Density of water

Since the relative density is a ratio ofsimilar quantities, it has no unit.

Example 10.7 Relative density of silver is10.8. The density of water is 103 kg m–3.What is the density of silver in SI unit?

Solution:

Relative density of silver = 10.8

Relative density

= Density of silver

Density of water

Density of silver = Relative density of silver

× density of water

= 10.8 × 103 kg m–3.

Archimedes

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GRAVITATION 143

Whatyou havelearnt• The law of gravitation states that the force of attraction between

any two objects is proportional to the product of their massesand inversely proportional to the square of the distance betweenthem. The law applies to objects anywhere in the universe.Such a law is said to be universal.

• Gravitation is a weak force unless large masses are involved.

• Force of gravitation due to the earth is called gravity.

• The force of gravity decreases with altitude. It also varies onthe surface of the earth, decreasing from poles to the equator.

• The weight of a body is the force with which the earthattracts it.

• The weight is equal to the product of mass and accelerationdue to gravity.

• The weight may vary from place to place but the mass staysconstant.

• All objects experience a force of buoyancy when they areimmersed in a fluid.

• Objects having density less than that of the liquid in whichthey are immersed, float on the surface of the liquid. If thedensity of the object is more than the density of the liquid inwhich it is immersed then it sinks in the liquid.

Exercises1. How does the force of gravitation between two objects change

when the distance between them is reduced to half ?

2. Gravitational force acts on all objects in proportion to theirmasses. Why then, a heavy object does not fall faster than alight object?

3. What is the magnitude of the gravitational force between theearth and a 1 kg object on its surface? (Mass of the earth is6 × 1024 kg and radius of the earth is 6.4 × 106 m.)

4. The earth and the moon are attracted to each other bygravitational force. Does the earth attract the moon with a forcethat is greater or smaller or the same as the force with whichthe moon attracts the earth? Why?

5. If the moon attracts the earth, why does the earth not movetowards the moon?

SCIENCE144

6. What happens to the force between two objects, if

(i) the mass of one object is doubled?

(ii) the distance between the objects is doubled and tripled?

(iii) the masses of both objects are doubled?

7. What is the importance of universal law of gravitation?

8. What is the acceleration of free fall?

9. What do we call the gravitational force between the earth andan object?

10. Amit buys few grams of gold at the poles as per the instructionof one of his friends. He hands over the same when he meetshim at the equator. Will the friend agree with the weight of goldbought? If not, why? [Hint: The value of g is greater at thepoles than at the equator.]

11. Why will a sheet of paper fall slower than one that is crumpledinto a ball?

12. Gravitational force on the surface of the moon is only 1

6 as

strong as gravitational force on the earth. What is the weightin newtons of a 10 kg object on the moon and on the earth?

13. A ball is thrown vertically upwards with a velocity of 49 m/s.Calculate

(i) the maximum height to which it rises,

(ii) the total time it takes to return to the surface of the earth.

14. A stone is released from the top of a tower of height 19.6 m.Calculate its final velocity.

15. A stone is thrown vertically upward with an initial velocity of40 m/s. Taking g = 10 m/s2, find the maximum height reachedby the stone. What is the net displacement and the totaldistance covered by the stone?

16. Calculate the force of gravitation between the earth and theSun, given that the mass of the earth = 6 × 1024 kg and of theSun = 2 × 1030 kg. The average distance between the two is1.5 × 1011 m.

17. A stone is allowed to fall from the top of a tower 100 m highand at the same time another stone is projected verticallyupwards from the ground with a velocity of 25 m/s. Calculatewhen and where the two stones will meet.

18. A ball thrown up vertically returns to the thrower after 6 s.Find

(a) the velocity with which it was thrown up,

(b) the maximum height it reaches, and

(c) its position after 4 s.

GRAVITATION 145

19. In what direction does the buoyant force on an object immersedin a liquid act?

20. Why does a block of plastic released under water come up tothe surface of water?

21. The volume of 50 g of a substance is 20 cm3. If the density ofwater is 1 g cm–3, will the substance float or sink?

22. The volume of a 500 g sealed packet is 350 cm3. Will the packetfloat or sink in water if the density of water is 1 g cm–3? Whatwill be the mass of the water displaced by this packet?

In the previous few chapters we have talkedabout ways of describing the motion ofobjects, the cause of motion and gravitation.Another concept that helps us understand andinterpret many natural phenomena is ‘work’.Closely related to work are energy and power.In this chapter we shall study these concepts.

All living beings need food. Living beingshave to perform several basic activities tosurvive. We call such activities ‘life processes’.The energy for these processes comes fromfood. We need energy for other activities likeplaying, singing, reading, writing, thinking,jumping, cycling and running. Activities thatare strenuous require more energy.

Animals too get engaged in activities. Forexample, they may jump and run. They haveto fight, move away from enemies, find foodor find a safe place to live. Also, we engagesome animals to lift weights, carry loads, pullcarts or plough fields. All such activitiesrequire energy.

Think of machines. List the machines thatyou have come across. What do they need fortheir working? Why do some engines requirefuel like petrol and diesel? Why do livingbeings and machines need energy?

11.1 WorkWhat is work? There is a difference in theway we use the term ‘work’ in day-to-day lifeand the way we use it in science. To makethis point clear let us consider a few examples.

11.1.1 NOT MUCH ‘WORK’ IN SPITE OFWORKING HARD!

Kamali is preparing for examinations. Shespends lot of time in studies. She reads books,

draws diagrams, organises her thoughts,collects question papers, attends classes,discusses problems with her friends, andperforms experiments. She expends a lot ofenergy on these activities. In commonparlance, she is ‘working hard’. All this ‘hardwork’ may involve very little ‘work’ if we go bythe scientific definition of work.

You are working hard to push a huge rock.Let us say the rock does not move despite allthe effort. You get completely exhausted.However, you have not done any work on therock as there is no displacement of the rock.

You stand still for a few minutes with aheavy load on your head. You get tired. Youhave exerted yourself and have spent quite abit of your energy. Are you doing work on theload? The way we understand the term ‘work’in science, work is not done.

You climb up the steps of a staircase andreach the second floor of a building just tosee the landscape from there. You may evenclimb up a tall tree. If we apply the scientificdefinition, these activities involve a lot of work.

In day-to-day life, we consider any usefulphysical or mental labour as work. Activitieslike playing in a field, talking with friends,humming a tune, watching a movie, attendinga function are sometimes not considered tobe work. What constitutes ‘work’ dependson the way we define it. We use and definethe term work differently in science. Tounderstand this let us do the followingactivities:

Activity _____________11.1• We have discussed in the above

paragraphs a number of activitieswhich we normally consider to be work

1111111111WWWWWORKORKORKORKORK ANDANDANDANDAND E E E E ENERGYNERGYNERGYNERGYNERGY

Chapter

Activity _____________11.3• Think of situations when the object is

not displaced in spite of a force actingon it.

• Also think of situations when an objectgets displaced in the absence of a forceacting on it.

• List all the situations that you canthink of for each.

• Discuss with your friends whetherwork is done in these situations.

11.1.3 WORK DONE BY A CONSTANT FORCE

How is work defined in science? Tounderstand this, we shall first consider thecase when the force is acting in the directionof displacement.

Let a constant force, F act on an object.Let the object be displaced through adistance, s in the direction of the force (Fig.11.1). Let W be the work done. We define workto be equal to the product of the force anddisplacement.

Work done = force × displacementW = F s (11.1)

in day-to-day life. For each of theseactivities, ask the following questionsand answer them:(i) What is the work being done on?(ii) What is happening to the object?(iii) Who (what) is doing the work?

11.1.2 SCIENTIFIC CONCEPTION OF WORK

To understand the way we view work anddefine work from the point of view of science,let us consider some situations:

Push a pebble lying on a surface. Thepebble moves through a distance. You exerteda force on the pebble and the pebble gotdisplaced. In this situation work is done.

A girl pulls a trolley and the trolley movesthrough a distance. The girl has exerted aforce on the trolley and it is displaced.Therefore, work is done.

Lift a book through a height. To do thisyou must apply a force. The book rises up.There is a force applied on the book and thebook has moved. Hence, work is done.

A closer look at the above situationsreveals that two conditions need to besatisfied for work to be done: (i) a force shouldact on an object, and (ii) the object must bedisplaced.

If any one of the above conditions doesnot exist, work is not done. This is the waywe view work in science.

A bullock is pulling a cart. The cartmoves. There is a force on the cart and thecart has moved. Do you think that work isdone in this situation?

Activity _____________11.2• Think of some situations from your

daily life involving work.• List them.• Discuss with your friends whether

work is being done in each situation.• Try to reason out your response.• If work is done, which is the force acting

on the object?• What is the object on which the work

is done?• What happens to the object on which

work is done?

Fig. 11.1

Thus, work done by a force acting on anobject is equal to the magnitude of the forcemultiplied by the distance moved in thedirection of the force. Work has onlymagnitude and no direction.

In Eq. (11.1), if F = 1 N and s = 1 m thenthe work done by the force will be 1 N m.Here the unit of work is newton metre (N m)or joule (J). Thus 1 J is the amount of work

WORK AND ENERGY 147

SCIENCE148

done on an object when a force of 1 Ndisplaces it by 1 m along the line of action ofthe force.

Look at Eq. (11.1) carefully. What is thework done when the force on the object iszero? What would be the work done whenthe displacement of the object is zero? Referto the conditions that are to be satisfied tosay that work is done.

Example 11.1 A force of 5 N is acting onan object. The object is displacedthrough 2 m in the direction of the force(Fig. 11.2). If the force acts on the objectall through the displacement, thenwork done is 5 N × 2 m =10 N m or10 J.

Fig. 11.4

Consider a situation in which an objectis being displaced by the action of forces andwe identify one of the forces, F acting oppositeto the direction of the displacement s, thatis, the angle between the two directions is180º. In such a situation, the work done bythe force, F is taken as negative and denotedby the minus sign. The work done by the forceis F × (–s) or (–F × s).

It is clear from the above discussion thatthe work done by a force can be either positiveor negative. To understand this, let us do thefollowing activity:

Activity _____________11.4• Lift an object up. Work is done by the

force exerted by you on the object. Theobject moves upwards. The force youexerted is in the direction ofdisplacement. However, there is theforce of gravity acting on the object.

• Which one of these forces is doingpositive work?

• Which one is doing negative work?• Give reasons.

Work done is negative when the force actsopposite to the direction of displacement.Work done is positive when the force is in thedirection of displacement.

Example 11.2 A porter lifts a luggage of15 kg from the ground and puts it onhis head 1.5 m above the ground.Calculate the work done by him on theluggage.

Solution:

Mass of luggage, m = 15 kg anddisplacement, s = 1.5 m.

Fig. 11.2

uestion1. A force of 7 N acts on an object.

The displacement is, say 8 m, inthe direction of the force(Fig. 11.3). Let us take it that theforce acts on the object throughthe displacement. What is thework done in this case?

Fig. 11.3

Consider another situation in which theforce and the displacement are in the samedirection: a baby pulling a toy car parallel tothe ground, as shown in Fig. 11.4. The babyhas exerted a force in the direction ofdisplacement of the car. In this situation, thework done will be equal to the product of theforce and displacement. In such situations,the work done by the force is taken as positive.

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WORK AND ENERGY 149

raised hammer falls on a nail placed on apiece of wood, it drives the nail into the wood.We have also observed children winding a toy(such as a toy car) and when the toy is placedon the floor, it starts moving. When a balloonis filled with air and we press it we notice achange in its shape. As long as we press itgently, it can come back to its original shapewhen the force is withdrawn. However, if wepress the balloon hard, it can even explodeproducing a blasting sound. In all theseexamples, the objects acquire, throughdifferent means, the capability of doing work.An object having a capability to do work issaid to possess energy. The object which doesthe work loses energy and the object on whichthe work is done gains energy.

How does an object with energy do work?An object that possesses energy can exert aforce on another object. When this happens,energy is transferred from the former to thelatter. The second object may move as itreceives energy and therefore do some work.Thus, the first object had a capacity to dowork. This implies that any object thatpossesses energy can do work.

The energy possessed by an object is thusmeasured in terms of its capacity of doingwork. The unit of energy is, therefore, the sameas that of work, that is, joule (J). 1 J is theenergy required to do 1 joule of work.Sometimes a larger unit of energy called kilojoule (kJ) is used. 1 kJ equals 1000 J.

11.2.1 FORMS OF ENERGY

Luckily the world we live in provides energyin many different forms. The various formsinclude potential energy, kinetic energy, heatenergy, chemical energy, electrical energy andlight energy.

Think it over !

How do you know that some entity is aform of energy? Discuss with your friendsand teachers.

Work done, W = F × s = mg × s= 15 kg × 10 m s-2 × 1.5 m= 225 kg m s-2 m= 225 N m = 225 J

Work done is 225 J.

uestions1. When do we say that work is

done?2. Write an expression for the work

done when a force is acting onan object in the direction of itsdisplacement.

3. Define 1 J of work.4. A pair of bullocks exerts a force

of 140 N on a plough. The fieldbeing ploughed is 15 m long.How much work is done inploughing the length of the field?

11.2 EnergyLife is impossible without energy. The demandfor energy is ever increasing. Where do weget energy from? The Sun is the biggestnatural source of energy to us. Many of ourenergy sources are derived from the Sun. Wecan also get energy from the nuclei of atoms,the interior of the earth, and the tides. Canyou think of other sources of energy?

Activity _____________11.5• A few sources of energy are listed above.

There are many other sources ofenergy. List them.

• Discuss in small groups how certainsources of energy are due to the Sun.

• Are there sources of energy which arenot due to the Sun?

The word energy is very often used in ourdaily life, but in science we give it a definiteand precise meaning. Let us consider thefollowing examples: when a fast movingcricket ball hits a stationary wicket, the wicketis thrown away. Similarly, an object whenraised to a certain height gets the capabilityto do work. You must have seen that when a

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SCIENCE150

Fig. 11.5

• The trolley moves forward and hits thewooden block.

• Fix a stop on the table in such amanner that the trolley stops afterhitting the block. The block getsdisplaced.

• Note down the displacement of theblock. This means work is done on theblock by the trolley as the block hasgained energy.

• From where does this energy come?• Repeat this activity by increasing the

mass on the pan. In which case is thedisplacement more?

• In which case is the work done more?• In this activity, the moving trolley does

work and hence it possesses energy.

A moving object can do work. An objectmoving faster can do more work than anidentical object moving relatively slow. Amoving bullet, blowing wind, a rotating wheel,a speeding stone can do work. How does abullet pierce the target? How does the windmove the blades of a windmill? Objects inmotion possess energy. We call this energykinetic energy.

A falling coconut, a speeding car, a rollingstone, a flying aircraft, flowing water, blowingwind, a running athlete etc. possess kineticenergy. In short, kinetic energy is the energypossessed by an object due to its motion. Thekinetic energy of an object increases with itsspeed.

How much energy is possessed by amoving body by virtue of its motion? Bydefinition, we say that the kinetic energy of abody moving with a certain velocity is equalto the work done on it to make it acquire thatvelocity.

11.2.2 KINETIC ENERGY

Activity _____________11.6• Take a heavy ball. Drop it on a thick

bed of sand. A wet bed of sand wouldbe better. Drop the ball on the sandbed from height of about 25 cm. Theball creates a depression.

• Repeat this activity from heights of50 cm, 1m and 1.5 m.

• Ensure that all the depressions aredistinctly visible.

• Mark the depressions to indicate theheight from which the ball wasdropped.

• Compare their depths.• Which one of them is deepest?• Which one is shallowest? Why?• What has caused the ball to make a

deeper dent?• Discuss and analyse.

Activity _____________11.7• Set up the apparatus as shown in

Fig. 11.5.• Place a wooden block of known mass

in front of the trolley at a convenientfixed distance.

• Place a known mass on the pan so thatthe trolley starts moving.

James PrescottJoule was ano u t s t a n d i n gBritish physicist.He is best knownfor his research inelectricity andthermodynamics.Amongst otherthings, heformulated a lawfor the heatingeffect of electriccurrent. He also

verified experimentally the law ofconservation of energy and discoveredthe value of the mechanical equivalentof heat. The unit of energy and workcalled joule, is named after him.

James Prescott Joule(1818 – 1889)

WORK AND ENERGY 151

Let us now express the kinetic energy ofan object in the form of an equation. Consideran object of mass, m moving with a uniformvelocity, u. Let it now be displaced through adistance s when a constant force, F acts on itin the direction of its displacement. FromEq. (11.1), the work done, W is F s. The workdone on the object will cause a change in itsvelocity. Let its velocity change from u to v.Let a be the acceleration produced.

In section 8.5, we studied three equationsof motion. The relation connecting the initialvelocity (u) and final velocity (v) of an objectmoving with a uniform acceleration a, andthe displacement, s is

v2 – u2 = 2a s (8.7)

This gives2 2v – u

s =2a

(11.2)

From section 9.4, we know F = m a. Thus,using (Eq. 11.2) in Eq. (11.1), we can writethe work done by the force, F as

×⎛ ⎞⎜ ⎟⎝ ⎠

2 2v - uW = m a

2a

or

( )1

22 2W = m v – u (11.3)

If the object is starting from its stationaryposition, that is, u = 0, then

12

2W = m v (11.4)

It is clear that the work done is equal to thechange in the kinetic energy of an object.

If u = 0, the work done will be 1

22m v .

Thus, the kinetic energy possessed by anobject of mass, m and moving with a uniformvelocity, v is

1

22

kE = m v (11.5)

Example 11.3 An object of mass 15 kg ismoving with a uniform velocity of4 m s–1. What is the kinetic energypossessed by the object?

Solution:

Mass of the object, m = 15 kg, velocityof the object, v = 4 m s–1.From Eq. (11.5),

1

22

kE = m v

= 12

× 15 kg × 4 m s–1 × 4 m s–1

= 120 JThe kinetic energy of the object is 120 J.

Example 11.4 What is the work to be doneto increase the velocity of a car from30 km h–1 to 60 km h–1 if the mass ofthe car is 1500 kg?

Solution:

Mass of the car, m =1500 kg,initial velocity of car, u = 30 km h–1

= 30 ×1000m

60 × 60s

= 8.33 m s–1.

Similarly, the final velocity of the car,

v = 60 km h–1

= 16.67 m s–1.Therefore, the initial kinetic energy ofthe car,

Eki

1

22= m u

= 1

2× 1500 kg × (8.33 m s–1)2

= 52041.68 J.

The final kinetic energy of the car,

Ekf

= 12

× 1500 kg × (16.67 m s–1)2

= 208416.68 J.

Thus, the work done = Change inkinetic energy

= Ekf – E

ki

= 156375 J.

SCIENCE152

uestions1. What is the kinetic energy of an

object?2. Write an expression for the kinetic

energy of an object.3. The kinetic energy of an object of

mass, m moving with a velocityof 5 m s–1 is 25 J. What will be itskinetic energy when its velocityis doubled? What will be itskinetic energy when its velocityis increased three times?

11.2.3 POTENTIAL ENERGY

Activity _____________11.8• Take a rubber band.• Hold it at one end and pull from the

other. The band stretches.• Release the band at one of the ends.• What happens?• The band will tend to regain its original

length. Obviously the band hadacquired energy in its stretchedposition.

• How did it acquire energy whenstretched?

Activity _____________11.9• Take a slinky as shown below.• Ask a friend to hold one of its ends.

You hold the other end and move awayfrom your friend. Now you release theslinky.

Activity ____________11.11• Lift an object through a certain height.

The object can now do work. It beginsto fall when released.

• This implies that it has acquired someenergy. If raised to a greater height itcan do more work and hence possessesmore energy.

• From where did it get the energy? Thinkand discuss.

In the above situations, the energy getsstored due to the work done on the object.The energy transferred to an object is storedas potential energy if it is not used to cause achange in the velocity or speed of the object.

You transfer energy when you stretch arubber band. The energy transferred to theband is its potential energy. You do work whilewinding the key of a toy car. The energytransferred to the spring inside is stored aspotential energy. The potential energypossessed by the object is the energy presentin it by virtue of its position or configuration.

Activity ____________11.12• Take a bamboo stick and make a bow

as shown in Fig. 11.6.• Place an arrow made of a light stick on

it with one end supported by thestretched string.

• Now stretch the string and release thearrow.

• Notice the arrow flying off the bow.Notice the change in the shape of thebow.

• The potential energy stored in the bowdue to the change of shape is thus usedin the form of kinetic energy inthrowing off the arrow.

Q

Fig.11.6: An arrow and the stretched string

on the bow.

• What happened?• How did the slinky acquire energy when

stretched?• Would the slinky acquire energy when

it is compressed?

Activity ____________11.10• Take a toy car. Wind it using its key.• Place the car on the ground.• Did it move?• From where did it acquire energy?• Does the energy acquired depend on

the number of windings?• How can you test this?

WORK AND ENERGY 153

11.2.4 POTENTIAL ENERGY OF AN OBJECT

AT A HEIGHT

An object increases its energy when raisedthrough a height. This is because work isdone on it against gravity while it is beingraised. The energy present in such an objectis the gravitational potential energy.

The gravitational potential energy of anobject at a point above the ground is definedas the work done in raising it from the groundto that point against gravity.

It is easy to arrive at an expression forthe gravitational potential energy of an objectat a height.

Fig. 11.7

Consider an object of mass, m. Let it beraised through a height, h from the ground.A force is required to do this. The minimumforce required to raise the object is equal tothe weight of the object, mg. The object gainsenergy equal to the work done on it. Let thework done on the object against gravity beW. That is,

work done, W = force × displacement= mg × h= mgh

Since work done on the object is equal tomgh, an energy equal to mgh units is gainedby the object. This is the potential energy (E

P)

of the object.E

p = mgh (11.7)

Mor

e to

kno

w

The potential energy of an object ata height depends on the groundlevel or the zero level you choose.An object in a given position canhave a certain potential energy withrespect to one level and a differentvalue of potential energy withrespect to another level.

It is useful to note that the work done bygravity depends on the difference in verticalheights of the initial and final positions ofthe object and not on the path along whichthe object is moved. Fig. 11.8 shows a casewhere a block is raised from position A to Bby taking two different paths. Let the heightAB = h. In both the situations the work doneon the object is mgh.

Fig. 11.8

Example 11.5 Find the energy possessedby an object of mass 10 kg when it is ata height of 6 m above the ground. Given,g = 9.8 m s–2.

Solution:

Mass of the object, m = 10 kg,displacement (height), h = 6 m, andacceleration due to gravity, g = 9.8 m s–2.From Eq. (11.6),Potential energy = mgh

= 10 kg × 9.8 m s–2 × 6 m= 588 J.

The potential energy is 588 J.

SCIENCE154

Example 11.6 An object of mass 12 kg isat a certain height above the ground.If the potential energy of the object is480 J, find the height at which theobject is with respect to the ground.Given, g = 10 m s–2.

Solution:

Mass of the object, m = 12 kg,potential energy, E

p = 480 J.

Ep

= mgh480 J = 12 kg × 10 m s–2 × h

h = –2

480 J

120 kg ms = 4 m.

The object is at the height of 4 m.

11.2.5 ARE VARIOUS ENERGY FORMS

INTERCONVERTIBLE?Can we convert energy from one form toanother? We find in nature a number ofinstances of conversion of energy from oneform to another.

Activity ____________11.13• Sit in small groups.• Discuss the various ways of energy

conversion in nature.• Discuss following questions in your

group:(a) How do green plants produce food?(b) Where do they get their energy from?(c) Why does the air move from place

to place?(d) How are fuels, such as coal and

petroleum formed?(e) What kinds of energy conversions

sustain the water cycle?

Activity ____________11.14• Many of the human activities and the

gadgets we use involve conversion ofenergy from one form to another.

• Make a list of such activities andgadgets.

• Identify in each activity/gadget thekind of energy conversion that takesplace.

11.2.6 LAW OF CONSERVATION OF ENERGY

In activities 11.13 and 11.14, we learnt thatthe form of energy can be changed from oneform to another. What happens to the totalenergy of a system during or after theprocess? Whenever energy gets transformed,the total energy remains unchanged. This isthe law of conservation of energy. Accordingto this law, energy can only be converted fromone form to another; it can neither be createdor destroyed. The total energy before and afterthe transformation remains the same. Thelaw of conservation of energy is validin all situations and for all kinds oftransformations.

Consider a simple example. Let an objectof mass, m be made to fall freely from aheight, h. At the start, the potential energy ismgh and kinetic energy is zero. Why is thekinetic energy zero? It is zero because itsvelocity is zero. The total energy of the objectis thus mgh. As it falls, its potential energywill change into kinetic energy. If v is thevelocity of the object at a given instant, thekinetic energy would be ½mv2. As the fall ofthe object continues, the potential energywould decrease while the kinetic energy wouldincrease. When the object is about to reachthe ground, h = 0 and v will be the highest.Therefore, the kinetic energy would be thelargest and potential energy the least.However, the sum of the potential energy andkinetic energy of the object would be the sameat all points. That is,potential energy + kinetic energy = constant

or

1constant.

22mgh + mv = (11.7)

The sum of kinetic energy and potentialenergy of an object is its total mechanicalenergy.

We find that during the free fall of theobject, the decrease in potential energy, atany point in its path, appears as an equalamount of increase in kinetic energy. (Herethe effect of air resistance on the motion ofthe object has been ignored.) There is thus acontinual transformation of gravitationalpotential energy into kinetic energy.

WORK AND ENERGY 155

A stronger person may do certain work inrelatively less time. A more powerful vehiclewould complete a journey in a shorter timethan a less powerful one. We talk of the powerof machines like motorbikes and motorcars.The speed with which these vehicles changeenergy or do work is a basis for theirclassification. Power measures the speed ofwork done, that is, how fast or slow work isdone. Power is defined as the rate of doingwork or the rate of transfer of energy. If anagent does a work W in time t, then power isgiven by:

Power = work/time

orW

P =t

(11.8)

The unit of power is watt [in honour ofJames Watt (1736 – 1819)] having the symbolW. 1 watt is the power of an agent, whichdoes work at the rate of 1 joule per second.We can also say that power is 1 W when therate of consumption of energy is 1 J s–1.

1 watt = 1 joule/second or 1 W = 1 J s–1.We express larger rates of energy transfer inkilowatts (kW).

1 kilowatt = 1000 watts1 kW = 1000 W1 kW = 1000 J s–1.

The power of an agent may vary with time.This means that the agent may be doing workat different rates at different intervals of time.Therefore the concept of average power isuseful. We obtain average power by dividingthe total energy consumed by the total timetaken.

Example 11.7 Two girls, each of weight400 N climb up a rope through a heightof 8 m. We name one of the girls A andthe other B. Girl A takes 20 s while Btakes 50 s to accomplish this task.What is the power expended byeach girl?

Solution:

(i) Power expended by girl A:Weight of the girl, mg = 400 N

Activity ____________11.15• An object of mass 20 kg is dropped from

a height of 4 m. Fill in the blanks inthe following table by computing thepotential energy and kinetic energy ineach case.

Height at Potential Kinetic Ep + Ekwhich object energy energy

is located (Ep= mgh) (Ek = mv2/2)

m J J J

4

3

2

1

Just abovethe ground

• For simplifying the calculations, takethe value of g as 10 m s–2.

Think it over !What would have happened if nature hadnot allowed the transformation of energy?There is a view that life could not havebeen possible without transformation ofenergy. Do you agree with this?

11.3 Rate of Doing WorkDo all of us work at the same rate? Domachines consume or transfer energy at thesame rate? Agents that transfer energy dowork at different rates. Let us understand thisfrom the following activity:

Activity ____________11.16• Consider two children, say A and B.

Let us say they weigh the same. Bothstart climbing up a rope separately.Both reach a height of 8 m. Let us sayA takes 15 s while B takes 20 s toaccomplish the task.

• What is the work done by each?• The work done is the same. However,

A has taken less time than B to dothe work.

• Who has done more work in a giventime, say in 1 s?

SCIENCE156

uestions1. What is power?2. Define 1 watt of power.3. A lamp consumes 1000 J of

electrical energy in 10 s. What isits power?

4. Define average power.

11.3.1 COMMERCIAL UNIT OF ENERGY

The unit joule is too small and hence isinconvenient to express large quantities ofenergy. We use a bigger unit of energy calledkilowatt hour (kW h).

What is 1 kW h? Let us say we have amachine that uses 1000 J of energy everysecond. If this machine is used continuouslyfor one hour, it will consume 1 kW h of energy.Thus, 1 kW h is the energy used in one hourat the rate of 1000 J s–1 (or 1 kW).

1 kW h = 1 kW ×1 h= 1000 W × 3600 s= 3600000 J

1 kW h = 3.6 × 106 J.The energy used in households, industries

and commercial establishments are usuallyexpressed in kilowatt hour. For example,electrical energy used during a month isexpressed in terms of ‘units’. Here, 1 ‘unit’means 1 kilowatt hour.

Example 11.9 An electric bulb of 60 W isused for 6 h per day. Calculate the‘units’ of energy consumed in one dayby the bulb.

Solution:

Power of electric bulb = 60 W= 0.06 kW.

Time used, t = 6 hEnergy = power × time taken

= 0.06 kW × 6 h= 0.36 kW h= 0.36 ‘units’.

The energy consumed by the bulb is0.36 ‘units’.

Displacement (height), h = 8 mTime taken, t = 20 sFrom Eq. (11.8),

Power, P = Work done/time taken

= mgh

t

=400 N × 8 m

20s

= 160 W.

(ii) Power expended by girl B:Weight of the girl, mg = 400 NDisplacement (height), h = 8 mTime taken, t = 50 s

Power, P =mgh

t

=400 N × 8 m

50s

= 64 W.

Power expended by girl A is 160 W.Power expended by girl B is 64 W.

Example 11.8 A boy of mass 50 kg runsup a staircase of 45 steps in 9 s. If theheight of each step is 15 cm, find hispower. Take g = 10 m s–2.

Solution:

Weight of the boy,mg = 50 kg × 10 m s–2 = 500 N

Height of the staircase,h = 45 × 15/100 m = 6.75 m

Time taken to climb, t = 9 sFrom Eq. (11.8),power, P = Work done/time taken

= mgh

t

= 500 N × 6.75 m

9s

= 375 W.

Power is 375 W.

Q

WORK AND ENERGY 157

Activity ____________11.17• Take a close look at the electric meter

installed in your house. Observe itsfeatures closely.

• Take the readings of the meter eachday at 6.30 am and 6.30 pm.

• How many ‘units’ are consumed duringday time?

• How many ‘units’ are used duringnight?

• Do this activity for about a week.• Tabulate your observations.• Draw inferences from the data.• Compare your observations with

the details given in the monthlyelectricity bill.

Whatyou havelearnt• Work done on an object is defined as the magnitude of the

force multiplied by the distance moved by the object in thedirection of the applied force. The unit of work is joule:1 joule = 1 newton × 1 metre.

• Work done on an object by a force would be zero if thedisplacement of the object is zero.

• An object having capability to do work is said to possess energy.Energy has the same unit as that of work.

• An object in motion possesses what is known as the kineticenergy of the object. An object of mass, m moving with velocity

v has a kinetic energy of 2mv

12 .

• The energy possessed by a body due to its change in positionor shape is called the potential energy. The gravitationalpotential energy of an object of mass, m raised through a height,h from the earth’s surface is given by m g h.

• According to the law of conservation of energy, energy can onlybe transformed from one form to another; it can neither becreated nor destroyed. The total energy before and after thetransformation always remains constant.

• Energy exists in nature in several forms such as kinetic energy,potential energy, heat energy, chemical energy etc. The sum ofthe kinetic and potential energies of an object is called itsmechanical energy.

• Power is defined as the rate of doing work. The SI unit of poweris watt. 1 W = 1 J/s.

• The energy used in one hour at the rate of 1kW is called 1 kW h.

SCIENCE158

Exercises1. Look at the activities listed below. Reason out whether or not

work is done in the light of your understanding of the term‘work’.

• Suma is swimming in a pond.

• A donkey is carrying a load on its back.

• A wind-mill is lifting water from a well.

• A green plant is carrying out photosynthesis.

• An engine is pulling a train.

• Food grains are getting dried in the sun.

• A sailboat is moving due to wind energy.

2. An object thrown at a certain angle to the ground moves in acurved path and falls back to the ground. The initial and thefinal points of the path of the object lie on the same horizontalline. What is the work done by the force of gravity on the object?

3. A battery lights a bulb. Describe the energy changes involvedin the process.

4. Certain force acting on a 20 kg mass changes its velocity from5 m s–1 to 2 m s–1. Calculate the work done by the force.

5. A mass of 10 kg is at a point A on a table. It is moved to a pointB. If the line joining A and B is horizontal, what is the workdone on the object by the gravitational force? Explain youranswer.

6. The potential energy of a freely falling object decreasesprogressively. Does this violate the law of conservation ofenergy? Why?

7. What are the various energy transformations that occur whenyou are riding a bicycle?

8. Does the transfer of energy take place when you push a hugerock with all your might and fail to move it? Where is theenergy you spend going?

9. A certain household has consumed 250 units of energy duringa month. How much energy is this in joules?

10. An object of mass 40 kg is raised to a height of 5 m above theground. What is its potential energy? If the object is allowed tofall, find its kinetic energy when it is half-way down.

11. What is the work done by the force of gravity on a satellitemoving round the earth? Justify your answer.

12. Can there be displacement of an object in the absence of anyforce acting on it? Think. Discuss this question with your friendsand teacher.

WORK AND ENERGY 159

13. A person holds a bundle of hay over his head for 30 minutesand gets tired. Has he done some work or not? Justify youranswer.

14. An electric heater is rated 1500 W. How much energy does ituse in 10 hours?

15. Illustrate the law of conservation of energy by discussing theenergy changes which occur when we draw a pendulum bob toone side and allow it to oscillate. Why does the bob eventuallycome to rest? What happens to its energy eventually? Is it aviolation of the law of conservation of energy?

16. An object of mass, m is moving with a constant velocity, v.How much work should be done on the object in order to bringthe object to rest?

17. Calculate the work required to be done to stop a car of 1500 kgmoving at a velocity of 60 km/h?

18. In each of the following a force, F is acting on an object ofmass, m. The direction of displacement is from west to eastshown by the longer arrow. Observe the diagrams carefullyand state whether the work done by the force is negative,positive or zero.

19. Soni says that the acceleration in an object could be zero evenwhen several forces are acting on it. Do you agree with her?Why?

20. Find the energy in kW h consumed in 10 hours by four devicesof power 500 W each.

21. A freely falling object eventually stops on reaching the ground.What happenes to its kinetic energy?

Everyday we hear sounds from varioussources like humans, birds, bells, machines,vehicles, televisions, radios etc. Sound is aform of energy which produces a sensationof hearing in our ears. There are also otherforms of energy like mechanical energy, heatenergy, light energy etc. We have talked aboutmechanical energy in the previous chapters.You have been taught about conservation ofenergy, which states that we can neithercreate nor destroy energy. We can justchange it from one form to another. Whenyou clap, a sound is produced. Can youproduce sound without utilising your energy?Which form of energy did you use to producesound? In this chapter we are going to learnhow sound is produced and how it istransmitted through a medium and receivedby our ear.

12.1 Production of Sound

Activity _____________12.1• Take a tuning fork and set it vibrating

by striking its prong on a rubber pad.Bring it near your ear.

• Do you hear any sound?• Touch one of the prongs of the vibrating

tuning fork with your finger and shareyour experience with your friends.

• Now, suspend a table tennis ball or asmall plastic ball by a thread from asupport [Take a big needle and athread, put a knot at one end of thethread, and then with the help of theneedle pass the thread through theball]. Touch the ball gently with theprong of a vibrating tuningfork (Fig. 12.1).

• Observe what happens and discusswith your friends.

Activity _____________12.2• Fill water in a beaker or a glass up to

the brim. Gently touch the water surfacewith one of the prongs of the vibratingtuning fork, as shown in Fig. 12.2.

• Next dip the prongs of the vibratingtuning fork in water, as shown in Fig.12.3.

• Observe what happens in both thecases.

• Discuss with your friends why thishappens.

Fig. 12.1: Vibrating tuning fork just touching thesuspended table tennis ball.

Fig. 12.2: One of the prongs of the vibrating tuningfork touching the water surface.

1212121212SSSSSOUNDOUNDOUNDOUNDOUND

Chapter

plucked vibrates and produces sound. If youhave never done this, then do it and observethe vibration of the stretched rubber band.

Activity _____________12.3• Make a list of different types of musical

instruments and discuss with yourfriends which part of the instrumentvibrates to produce sound.

12.2 Propagation of SoundSound is produced by vibrating objects. Thematter or substance through which soundis transmitted is called a medium. It can besolid, liquid or gas. Sound moves through amedium from the point of generation to thelistener. When an object vibrates, it sets theparticles of the medium around it vibrating.The particles do not travel all the way fromthe vibrating object to the ear. A particle ofthe medium in contact with the vibratingobject is first displaced from its equilibriumposition. It then exerts a force on the adjacentparticle. As a result of which the adjacentparticle gets displaced from its position ofrest. After displacing the adjacent particle thefirst particle comes back to its originalposition. This process continues in themedium till the sound reaches your ear. Thedisturbance created by a source of sound in

Fig. 12.3: Both the prongs of the vibrating tuning

fork dipped in water.

From the above activities what do youconclude? Can you produce sound withouta vibrating object?

In the above activities we have producedsound by striking the tuning fork. We canalso produce sound by plucking, scratching,rubbing, blowing or shaking different objects.As per the above activities what do we do tothe objects? We set the objects vibrating andproduce sound. Vibration means a kind ofrapid to and fro motion of an object. Thesound of the human voice is produced dueto vibrations in the vocal cords. When a birdflaps its wings, do you hear any sound? Thinkhow the buzzing sound accompanying a beeis produced. A stretched rubber band when

Fig. 12.4: A beam of light from a light source is made to fall on a mirror. The reflected light is falling on the wall.

Can sound make a light spot dance?

Take a tin can. Remove both ends to make it a hollow cylinder. Take a balloon and stretchit over the can, then wrap a rubber band around the balloon. Take a small piece of mirror.Use a drop of glue to stick the piece of mirror to the balloon. Allow the light through a slitto fall on the mirror. After reflection the light spot is seen on the wall, as shown in Fig.12.4. Talk or shout directly into the open end of the can and observe the dancing light spoton the wall. Discuss with your friends what makes the light spot dance.

SOUND 161

SCIENCE162

the medium travels through the medium andnot the particles of the medium.

A wave is a disturbance that movesthrough a medium when the particles of themedium set neighbouring particles intomotion. They in turn produce similar motionin others. The particles of the medium do notmove forward themselves, but thedisturbance is carried forward. This is whathappens during propagation of sound in amedium, hence sound can be visualised as awave. Sound waves are characterised by themotion of particles in the medium and arecalled mechanical waves.

Air is the most common medium throughwhich sound travels. When a vibrating objectmoves forward, it pushes and compresses theair in front of it creating a region of highpressure. This region is called a compression(C), as shown in Fig. 12.5. This compressionstarts to move away from the vibrating object.When the vibrating object moves backwards,it creates a region of low pressure calledrarefaction (R), as shown in Fig. 12.5. As theobject moves back and forth rapidly, a seriesof compressions and rarefactions is createdin the air. These make the sound wave thatpropagates through the medium.Compression is the region of high pressureand rarefaction is the region of low pressure.Pressure is related to the number of particlesof a medium in a given volume. More densityof the particles in the medium gives morepressure and vice versa. Thus, propagationof sound can be visualised as propagation ofdensity variations or pressure variations inthe medium.

uestion1. How does the sound produced by

a vibrating object in a mediumreach your ear?

12.2.1 SOUND NEEDS A MEDIUM TO TRAVEL

Sound is a mechanical wave and needs amaterial medium like air, water, steel etc. forits propagation. It cannot travel throughvacuum, which can be demonstrated by thefollowing experiment.

Take an electric bell and an airtight glassbell jar. The electric bell is suspended insidethe airtight bell jar. The bell jar is connectedto a vacuum pump, as shown in Fig. 12.6. Ifyou press the switch you will be able to hearthe bell. Now start the vacuum pump. Whenthe air in the jar is pumped out gradually,the sound becomes fainter, although thesame current is passing through the bell.After some time when less air is left insidethe bell jar you will hear a very feeble sound.What will happen if the air is removedcompletely? Will you still be able to hear thesound of the bell?

Fig. 12.5: A vibrating object creating a series ofcompressions (C) and rarefactions (R) inthe medium.

Q

Fig. 12.6: Bell jar experiment showing sound cannottravel in vacuum.

SOUND 163

waves. In these waves the individual particlesof the medium move in a direction parallel tothe direction of propagation of thedisturbance. The particles do not move fromone place to another but they simply oscillateback and forth about their position of rest.This is exactly how a sound wave propagates,hence sound waves are longitudinal waves.

There is also another type of wave, calleda transverse wave. In a transverse waveparticles do not oscillate along the line ofwave propagation but oscillate up and downabout their mean position as the wave travels.Thus a transverse wave is the one in whichthe individual particles of the medium moveabout their mean positions in a directionperpendicular to the direction of wavepropagation. Light is a transverse wave butfor light, the oscillations are not of themedium particles or their pressure or density– it is not a mechanical wave. You will cometo know more about transverse waves inhigher classes.

12.2.3 CHARACTERISTICS OF A SOUND WAVE

We can describe a sound wave by its• frequency• amplitude and• speed.

A sound wave in graphic form is shownin Fig. 12.8(c), which represents how densityand pressure change when the sound wavemoves in the medium. The density as well asthe pressure of the medium at a given timevaries with distance, above and below theaverage value of density and pressure.Fig. 12.8(a) and Fig. 12.8(b) represent thedensity and pressure variations, respectively,as a sound wave propagates in the medium.

Compressions are the regions whereparticles are crowded together andrepresented by the upper portion of the curvein Fig. 12.8(c). The peak represents the regionof maximum compression. Thus,compressions are regions where density aswell as pressure is high. Rarefactions are theregions of low pressure where particles arespread apart and are represented by the

uestions1. Explain how sound is produced

by your school bell.2. Why are sound waves called

mechanical waves?3. Suppose you and your friend are

on the moon. Will you be able tohear any sound produced byyour friend?

12.2.2 SOUND WAVES ARE LONGITUDINAL

WAVES

Activity _____________12.4• Take a slinky. Ask your friend to hold

one end. You hold the other end.Now stretch the slinky as shown inFig. 12.7 (a). Then give it a sharp pushtowards your friend.

• What do you notice? If you move yourhand pushing and pulling the slinkyalternatively, what will you observe?

• If you mark a dot on the slinky, youwill observe that the dot on the slinkywill move back and forth parallel to thedirection of the propagation of thedisturbance.

Fig. 12.7: Longitudinal wave in a slinky.

The regions where the coils become closerare called compressions (C) and the regionswhere the coils are further apart are calledrarefactions (R). As we already know, soundpropagates in the medium as a series ofcompressions and rarefactions. Now, we cancompare the propagation of disturbance in aslinky with the sound propagation in themedium. These waves are called longitudinal

Q

(a)

(b)

SCIENCE164

Frequency tells us how frequently anevent occurs. Suppose you are beating adrum. How many times you are beating thedrum per unit time is called the frequency ofyour beating the drum. We know that whensound is propagated through a medium, thedensity of the medium oscillates between amaximum value and a minimum value. Thechange in density from the maximum valueto the minimum value, again to the maximumvalue, makes one complete oscillation. Thenumber of such oscillations per unit time isthe frequency of the sound wave. If we cancount the number of the compressions orrarefactions that cross us per unit time, wewill get the frequency of the sound wave. It isusually represented by ν (Greek letter, nu).Its SI unit is hertz (symbol, Hz).

The time taken by two consecutivecompressions or rarefactions to cross a fixedpoint is called the time period of the wave. Inother words, we can say that the time takenfor one complete oscillation in the density ofthe medium is called the time period of the

valley, that is, the lower portion of the curvein Fig. 12.8(c). A peak is called the crest anda valley is called the trough of a wave.

The distance between two consecutivecompressions (C) or two consecutiverarefactions (R) is called the wavelength, asshown in Fig. 12.8(c), The wavelength isusually represented by λ (Greek letterlambda). Its SI unit is metre (m).

Heinrich Rudolph Hertzwas born on 22 February1857 in Hamburg,Germany and educated atthe University of Berlin. Heconfirmed J.C. Maxwell’selectromagnetic theory byhis experiments. He laid thefoundation for future

development of radio, telephone, telegraphand even television. He also discovered thephotoelectric effect which was laterexplained by Albert Einstein. The SI unitof frequency was named as hertz in hishonour.

Fig. 12.8: Sound propagates as density or pressure variations as shown in (a) and (b), (c) representsgraphically the density and pressure variations.

H. R. Hertz

SOUND 165

sound wave. It is represented by the symbolT. Its SI unit is second (s). Frequency andtime period are related as follows:

.1

=νT

A violin and a flute may both be playedat the same time in an orchestra. Bothsounds travel through the same medium,that is, air and arrive at our ear at the sametime. Both sounds travel at the same speedirrespective of the source. But the soundswe receive are different. This is due to thedifferent characteristics associated with thesound. Pitch is one of the characteristics.

How the brain interprets the frequencyof an emitted sound is called the pitch. Thefaster the vibration of the source, the higheris the frequency and the higher is the pitch,as shown in Fig. 12.9. Thus, a high pitchsound corresponds to more number ofcompressions and rarefactions passing afixed point per unit time.

Objects of different sizes and conditionsvibrate at different frequencies to producesounds of different pitch.

The magnitude of the maximumdisturbance in the medium on either side ofthe mean value is called the amplitude of thewave. It is usually represented by the letterA, as shown in Fig. 12.8(c). For sound itsunit will be that of density or pressure.

The loudness or softness of a sound isdetermined basically by its amplitude. Theamplitude of the sound wave depends uponthe force with which an object is made tovibrate. If we strike a table lightly, we hear asoft sound because we produce a sound waveof less energy (amplitude). If we hit the tablehard we hear a loud sound. Can you tell why?Loud sound can travel a larger distance as itis associated with higher energy. A soundwave spreads out from its source. As it movesaway from the source its amplitude as wellas its loudness decreases. Fig. 12.10 showsthe wave shapes of a loud and a soft soundof the same frequency.

Fig. 12.9: Low pitch sound has low frequency andhigh pitch of sound has high frequency.

Fig. 12.10: Soft sound has small amplitude andlouder sound has large amplitude.

The quality or timber of sound is thatcharacteristic which enables us todistinguish one sound from another havingthe same pitch and loudness. The soundwhich is more pleasant is said to be of a rich

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quality. A sound of single frequency is calleda tone. The sound which is produced due toa mixture of several frequencies is called anote and is pleasant to listen to. Noise isunpleasant to the ear! Music is pleasant tohear and is of rich quality.

uestions1. Which wave property determines

(a) loudness, (b) pitch?2. Guess which sound has a higher

pitch: guitar or car horn?

The speed of sound is defined as thedistance which a point on a wave, such as acompression or a rarefaction, travels per unittime.We know,

speed, v = distance / time

= λT

Here λ is the wavelength of the sound wave.It is the distance travelled by the sound wavein one time period (T) of the wave. Thus,

v = λ ν 1T

⎛ ⎞=⎜ ⎟⎝ ⎠Q ν

or v = λ νThat is, speed = wavelength × frequency.

The speed of sound remains almost thesame for all frequencies in a given mediumunder the same physical conditions.

Example 12.1 A sound wave has afrequency of 2 kHz and wave length35 cm. How long will it take to travel1.5 km?

Solution:

Given,Frequency, ν = 2 kHz = 2000 HzWavelength, λ = 35 cm = 0.35 mWe know that speed, v of the wave

= wavelength frequencyv = λ ν

= 0.35 m × 2000 Hz = 700 m/s

The time taken by the wave to travel adistance, d of 1.5 km is

Thus sound will take 2.1 s to travel adistance of 1.5 km.

uestions1. What are wavelength, frequency,

time period and amplitude of asound wave?

2. How are the wavelength andfrequency of a sound waverelated to its speed?

3. Calculate the wavelength of asound wave whose frequency is220 Hz and speed is 440 m/s ina given medium.

4. A person is listening to a tone of500 Hz sitting at a distance of450 m from the source of thesound. What is the time intervalbetween successive compressionsfrom the source?

The amount of sound energy passing eachsecond through unit area is called theintensity of sound. We sometimes use theterms “loudness” and “intensity”interchangeably, but they are not the same.Loudness is a measure of the response of theear to the sound. Even when two sounds areof equal intensity, we may hear one as louderthan the other simply because our ear detectsit better.

uestion1. Distinguish between loudness

and intensity of sound.

12.2.4 SPEED OF SOUND IN DIFFERENT

MEDIA

Sound propagates through a medium at afinite speed. The sound of a thunder is hearda little later than the flash of light is seen.

Q Q

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SOUND 167

So, we can make out that sound travels witha speed which is much less than the speedof light. The speed of sound depends on theproperties of the medium through which ittravels. You will learn about this dependencein higher classes. The speed of sound in amedium depends also on temperature andpressure of the medium. The speed of sounddecreases when we go from solid to gaseousstate. In any medium as we increase thetemperature the speed of sound increases.For example, the speed of sound in air is331 m s–1 at 0 ºC and 344 m s–1 at 22 ºC. Thespeeds of sound at a particular temperaturein various media are listed in Table 12.1. Youneed not memorise the values.

Sonic boom: When the speed of any objectexceeds the speed of sound it is said to betravelling at supersonic speed. Bullets, jetaircrafts etc. often travel at supersonicspeeds. When a sound, producing sourcemoves with a speed higher than that ofsound, it produces shock waves in air.These shock waves carry a large amountof energy. The air pressure variationassociated with this type of shock wavesproduces a very sharp and loud soundcalled the “sonic boom”. The shock wavesproduced by a supersonic aircraft haveenough energy to shatter glass and evendamage buildings.

12.3 Reflection of SoundSound bounces off a solid or a liquid like arubber ball bounces off a wall. Like light,sound gets reflected at the surface of a solidor liquid and follows the same laws ofreflection as you have studied in earlierclasses. The directions in which the soundis incident and is reflected make equal angleswith the normal to the reflecting surface, andthe three are in the same plane. An obstacleof large size which may be polished or roughis needed for the reflection of sound waves.

Activity _____________12.5• Take two identical pipes, as shown in

Fig. 12.11. You can make the pipesusing chart paper. The length of thepipes should be sufficiently longas shown.

• Arrange them on a table near a wall.• Keep a clock near the open end of one

of the pipes and try to hear the soundof the clock through the other pipe.

• Adjust the position of the pipes so thatyou can best hear the sound ofthe clock.

• Now, measure the angles of incidenceand reflection and see the relationshipbetween the angles.

• Lift the pipe on the right verticallyto a small height and observewhat happens.

Table 12.1: Speed of sound indifferent media at 25 ºC

State Substance Speed in m/s

Solids Aluminium 6420

Nickel 6040

Steel 5960

Iron 5950

Brass 4700

Glass (Flint) 3980

Liquids Water (Sea) 1531

Water (distilled) 1498

Ethanol 1207

Methanol 1103

Gases Hydrogen 1284

Helium 965

Air 346

Oxygen 316

Sulphur dioxide 213

uestion1. In which of the three media, air,

water or iron, does sound travelthe fastest at a particulartemperature?Q

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12.3.1 ECHO

If we shout or clap near a suitable reflectingobject such as a tall building or a mountain,we will hear the same sound again a littlelater. This sound which we hear is called anecho. The sensation of sound persists in ourbrain for about 0.1 s. To hear a distinct echothe time interval between the original soundand the reflected one must be at least 0.1s.If we take the speed of sound to be 344 m/sat a given temperature, say at 22 ºC in air,the sound must go to the obstacle and reachback the ear of the listener on reflection after0.1s. Hence, the total distance covered by thesound from the point of generation to thereflecting surface and back should be at least(344 m/s) × 0.1 s = 34.4 m. Thus, for hearingdistinct echoes, the minimum distance of theobstacle from the source of sound must behalf of this distance, that is, 17.2 m. Thisdistance will change with the temperature ofair. Echoes may be heard more than oncedue to successive or multiple reflections. Therolling of thunder is due to the successivereflections of the sound from a number ofreflecting surfaces, such as the clouds andthe land.

12.3.2 REVERBERATION

A sound created in a big hall will persist byrepeated reflection from the walls until it isreduced to a value where it is no longeraudible. The repeated reflection that resultsin this persistence of sound is calledreverberation. In an auditorium or big hall

excessive reverberation is highly undesirable.To reduce reverberation, the roof and wallsof the auditorium are generally covered withsound-absorbent materials like compressedfibreboard, rough plaster or draperies. Theseat materials are also selected on the basisof their sound absorbing properties.

Example 12.2 A person clapped his handsnear a cliff and heard the echo after5 s. What is the distance of the clifffrom the person if the speed of thesound, v is taken as 346 m s–1?

Solution:

Given,Speed of sound, v = 346 m s–1

Time taken for hearing the echo,t = 5 sDistance travelled by the sound

= v × t = 346 m s–1 × 5 s = 1730 mIn 5 s sound has to travel twice thedistance between the cliff and theperson. Hence, the distance betweenthe cliff and the person

= 1730 m/2 = 865 m.

uestion1. An echo returned in 3 s. What is

the distance of the reflectingsurface from the source, giventhat the speed of sound is342 m s–1?

12.3.3 USES OF MULTIPLE REFLECTION

OF SOUND

1. Megaphones or loudhailers, horns,musical instruments such as trumpetsand shehanais, are all designed tosend sound in a particular directionwithout spreading it in all directions,as shown in Fig 12.12.

Fig. 12.11: Reflection of sound

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SOUND 169

curved soundboard may be placedbehind the stage so that the sound,after reflecting from the sound board,spreads evenly across the width of thehall (Fig 12.15).

Fig 12.12: A megaphone and a horn.

In these instruments, a tube followedby a conical opening reflects soundsuccessively to guide most of the soundwaves from the source in the forwarddirection towards the audience.

2. Stethoscope is a medical instrumentused for listening to sounds producedwithin the body, chiefly in the heart orlungs. In stethoscopes the sound of thepatient’s heartbeat reaches thedoctor’s ears by multiple reflection ofsound, as shown in Fig.12.13.

Megaphone

Horn

Fig.12.13: Stethoscope

3. Generally the ceilings of concert halls,conference halls and cinema halls arecurved so that sound after reflectionreaches all corners of the hall, asshown in Fig 12.14. Sometimes a

Fig. 12.14: Curved ceiling of a conference hall.

Fig. 12.15: Sound board used in a big hall.

uestion1. Why are the ceilings of concert

halls curved?

12.4 Range of HearingThe audible range of sound for human beingsextends from about 20 Hz to 20000 Hz (oneHz = one cycle/s). Children under the age of

Q

SCIENCE170

five and some animals, such as dogs can hearup to 25 kHz (1 kHz = 1000 Hz). As peoplegrow older their ears become less sensitiveto higher frequencies. Sounds of frequenciesbelow 20 Hz are called infrasonic sound orinfrasound. If we could hear infrasound wewould hear the vibrations of a pendulum justas we hear the vibrations of the wings of abee. Rhinoceroses communicate usinginfrasound of frequency as low as 5 Hz.Whales and elephants produce sound in theinfrasound range. It is observed that someanimals get disturbed before earthquakes.Earthquakes produce low-frequencyinfrasound before the main shock wavesbegin which possibly alert the animals.Frequencies higher than 20 kHz are calledultrasonic sound or ultrasound. Ultrasoundis produced by dolphins, bats and porpoises.Moths of certain families have very sensitivehearing equipment. These moths can hearthe high frequency squeaks of the bat andknow when a bat is flying nearby, and areable to escape capture. Rats also play gamesby producing ultrasound.

Hearing Aid: People with hearing loss mayneed a hearing aid. A hearing aid is anelectronic, battery operated device. Thehearing aid receives sound through amicrophone. The microphone converts thesound waves to electrical signals. Theseelectrical signals are amplified by anamplifier. The amplified electrical signalsare given to a speaker of the hearing aid.The speaker converts the amplifiedelectrical signal to sound and sends to theear for clear hearing.

uestions1. What is the audible range of the

average human ear?2. What is the range of frequencies

associated with(a) Infrasound?(b) Ultrasound?

12.5 Applications of UltrasoundUltrasounds are high frequency waves.Ultrasounds are able to travel along well-defined paths even in the presence ofobstacles. Ultrasounds are used extensivelyin industries and for medical purposes.

• Ultrasound is generally used to cleanparts located in hard-to-reach places,for example, spiral tube, odd shapedparts, electronic components etc.Objects to be cleaned are placed in acleaning solution and ultrasonicwaves are sent into the solution. Dueto the high frequency, the particles ofdust, grease and dirt get detached anddrop out. The objects thus getthoroughly cleaned.

• Ultrasounds can be used to detectcracks and flaws in metal blocks.Metallic components are generallyused in construction of big structureslike buildings, bridges, machines andalso scientific equipment. The cracksor holes inside the metal blocks, whichare invisible from outside reduces thestrength of the structure. Ultrasonicwaves are allowed to pass through themetal block and detectors are used todetect the transmitted waves. If thereis even a small defect, the ultrasoundgets reflected back indicating thepresence of the flaw or defect, asshown in Fig. 12.16.

Q Fig 12.16: Ultrasound is reflected back from thedefective locations inside a metal block.

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Ordinary sound of longer wavelengthscannot be used for such purpose as it willbend around the corners of the defectivelocation and enter the detector.

• Ultrasonic waves are made to reflectfrom various parts of the heart andform the image of the heart. This tech-nique is called ‘echocardiography’.

• Ultrasound scanner is an instrumentwhich uses ultrasonic waves forgetting images of internal organs ofthe human body. A doctor may imagethe patient’s organs such as the liver,gall bladder, uterus, kidney, etc. Ithelps the doctor to detectabnormalities, such as stones in thegall bladder and kidney or tumoursin different organs. In this techniquethe ultrasonic waves travel throughthe tissues of the body and getreflected from a region where there isa change of tissue density. Thesewaves are then converted intoelectrical signals that are used togenerate images of the organ. Theseimages are then displayed on amonitor or printed on a film. Thistechnique is called ‘ultrasonography’.Ultrasonography is also used forexamination of the foetus duringpregnancy to detect congenial defectsand growth abnormalities.

• Ultrasound may be employed to breaksmall ‘stones’ formed in the kidneysinto fine grains. These grains later getflushed out with urine.

12.5.1SONAR

The acronym SONAR stands for SOundNavigation And Ranging. Sonar is a devicethat uses ultrasonic waves to measure thedistance, direction and speed of underwaterobjects. How does the sonar work? Sonarconsists of a transmitter and a detector andis installed in a boat or a ship, as shown inFig. 12.17.

Fig.12.17: Ultrasound sent by the transmitter andreceived by the detector.

The transmitter produces and transmitsultrasonic waves. These waves travel throughwater and after striking the object on theseabed, get reflected back and are sensedby the detector. The detector converts theultrasonic waves into electrical signals whichare appropriately interpreted. The distanceof the object that reflected the sound wavecan be calculated by knowing the speed ofsound in water and the time interval betweentransmission and reception of theultrasound. Let the time interval betweentransmission and reception of ultrasoundsignal be t and the speed of sound throughseawater be v. The total distance, 2d travelledby the ultrasound is then, 2d = v × t.

The above method is called echo-ranging.The sonar technique is used to determine thedepth of the sea and to locate underwaterhills, valleys, submarine, icebergs, sunkenship etc.

Example 12.3 A ship sends out ultrasoundthat returns from the seabed and isdetected after 3.42 s. If the speed ofultrasound through seawater is1531 m/s, what is the distance of theseabed from the ship?

Solution:

Given,Time between transmission anddetection, t = 3.42 s.

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Speed of ultrasound in sea water,v = 1531 m/s

Distance travelled by the ultrasound= 2 × depth of the sea = 2d

where d is the depth of the sea.2d = speed of sound × time

= 1531 m/s × 3.42 s = 5236 md = 5236 m/2 = 2618 m.

Thus, the distance of the seabed fromthe ship is 2618 m or 2.62 km.

uestion1. A submarine emits a sonar pulse,

which returns from anunderwater cliff in 1.02 s. If thespeed of sound in salt water is1531 m/s, how far away is thecliff?

As mentioned earlier, bats search out preyand fly in dark night by emitting anddetecting reflections of ultrasonic waves. Thehigh-pitched ultrasonic squeaks of the batare reflected from the obstacles or prey andreturned to bat’s ear, as shown in Fig. 12.18.The nature of reflections tells the bat wherethe obstacle or prey is and what it is like.Porpoises also use ultrasound for navigationand location of food in the dark.

12.6 Structure of Human EarHow do we hear? We are able to hear withthe help of an extremely sensitive devicecalled the ear. It allows us to convert pressurevariations in air with audible frequencies intoelectric signals that travel to the brain viathe auditory nerve. The auditory aspect ofhuman ear is discussed below.

Q

Fig. 12.18: Ultrasound is emitted by a bat and it isreflected back by the prey or an obstacle.

The outer ear is called ‘pinna’. It collectsthe sound from the surroundings. Thecollected sound passes through the auditorycanal. At the end of the auditory canal thereis a thin membrane called the ear drum ortympanic membrane. When a compressionof the medium reaches the eardrum thepressure on the outside of the membraneincreases and forces the eardrum inward.Similarly, the eardrum moves outward whena rarefaction reaches it. In this way theeardrum vibrates. The vibrations areamplified several times by three bones (thehammer, anvil and stirrup) in the middle ear.The middle ear transmits the amplifiedpressure variations received from the soundwave to the inner ear. In the inner ear, thepressure variations are turned into electricalsignals by the cochlea. These electricalsignals are sent to the brain via the auditorynerve, and the brain interprets them assound.

Fig. 12.19: Auditory parts of human ear.

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Whatyou havelearnt• Sound is produced due to vibration of different objects.• Sound travels as a longitudinal wave through a material

medium.• Sound travels as successive compressions and rarefactions in

the medium.• In sound propagation, it is the energy of the sound that travels

and not the particles of the medium.• Sound cannot travel in vacuum.• The change in density from one maximum value to the

minimum value and again to the maximum value makes onecomplete oscillation.

• The distance between two consecutive compressions or twoconsecutive rarefactions is called the wavelength, λ.

• The time taken by the wave for one complete oscillation of thedensity or pressure of the medium is called the time period, T.

• The number of complete oscillations per unit time is called the

frequency (ν), 1

=νT

.

• The speed v, frequency ν, and wavelength λ, of sound are relatedby the equation, v = λν.

• The speed of sound depends primarily on the nature and thetemperature of the transmitting medium.

• The law of reflection of sound states that the directions in whichthe sound is incident and reflected make equal angles withthe normal to the reflecting surface and the three lie in thesame plane.

• For hearing a distinct sound, the time interval between theoriginal sound and the reflected one must be at least 0.1 s.

• The persistence of sound in an auditorium is the result ofrepeated reflections of sound and is called reverberation.

• Sound properties such as pitch, loudness and quality aredetermined by the corresponding wave properties.

• Loudness is a physiological response of the ear to the intensityof sound.

• The amount of sound energy passing each second throughunit area is called the intensity of sound.

• The audible range of hearing for average human beings is inthe frequency range of 20 Hz – 20 kHz.

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• Sound waves with frequencies below the audible range aretermed “infrasonic” and those above the audible range aretermed “ultrasonic”.

• Ultrasound has many medical and industrial applications.

• The SONAR technique is used to determine the depth of thesea and to locate under water hills, valleys, submarines,icebergs, sunken ships etc.

Exercises1. What is sound and how is it produced?2. Describe with the help of a diagram, how compressions and

rarefactions are produced in air near a source of sound.3. Cite an experiment to show that sound needs a material

medium for its propagation.4. Why is sound wave called a longitudinal wave?5. Which characteristic of the sound helps you to identify your

friend by his voice while sitting with others in a dark room?6. Flash and thunder are produced simultaneously. But thunder

is heard a few seconds after the flash is seen, why?7. A person has a hearing range from 20 Hz to 20 kHz. What are

the typical wavelengths of sound waves in air correspondingto these two frequencies? Take the speed of sound in air as344 m s–1.

8. Two children are at opposite ends of an aluminium rod. Onestrikes the end of the rod with a stone. Find the ratio of timestaken by the sound wave in air and in aluminium to reach thesecond child.

9. The frequency of a source of sound is 100 Hz. How many timesdoes it vibrate in a minute?

10. Does sound follow the same laws of reflection as light does?Explain.

11. When a sound is reflected from a distant object, an echo isproduced. Let the distance between the reflecting surface andthe source of sound production remains the same. Do youhear echo sound on a hotter day?

12. Give two practical applications of reflection of sound waves.13. A stone is dropped from the top of a tower 500 m high into a

pond of water at the base of the tower. When is the splashheard at the top? Given, g = 10 m s–2 and speed of sound =340 m s–1.

14. A sound wave travels at a speed of 339 m s–1. If its wavelengthis 1.5 cm, what is the frequency of the wave? Will it be audible?

SOUND 175

15. What is reverberation? How can it be reduced?

16. What is loudness of sound? What factors does it depend on?

17. Explain how bats use ultrasound to catch a prey.

18. How is ultrasound used for cleaning?

19. Explain the working and application of a sonar.

20. A sonar device on a submarine sends out a signal and receivesan echo 5 s later. Calculate the speed of sound in water if thedistance of the object from the submarine is 3625 m.

21. Explain how defects in a metal block can be detected usingultrasound.

22. Explain how the human ear works.

Activity _____________13.1• We have all heard of the earthquakes

in Latur, Bhuj, Kashmir etc. or thecyclones that attack the coastalregions. Think of as many differentways as possible in which people’shealth would be affected by such adisaster if it took place in ourneighbourhood.

• How many of these ways we can think ofare events that would occur when thedisaster is actually happening?

• How many of these health-related eventswould happen long after the actualdisaster, but would still be because of thedisaster?

• Why would one effect on health fall intothe first group, and why would anotherfall into the second group?

When we do this exercise, we realise thathealth and disease in human communitiesare very complex issues, with manyinterconnected causes. We also realise thatthe ideas of what ‘health’ and ‘disease’ meanare themselves very complicated. When weask what causes diseases and how we preventthem, we have to begin by asking what thesenotions mean.

We have seen that cells are the basic unitsof living beings. Cells are made of a variety ofchemical substances – proteins, carbo-hydrates, fats or lipids, and so on. Althoughthe pictures look quite static, in reality theliving cell is a dynamic place. Something orthe other is always happening. Cells movefrom place to place. Even in cells that do notmove, there is repair going on. New cells arebeing made. In our organs or tissues, thereare various specialised activities going on –the heart is beating, the lungs are breathing,

the kidney is filtering urine, the brain isthinking.

All these activities are interconnected. Forexample, if the kidneys are not filtering urine,poisonous substances will accumulate. Undersuch conditions, the brain will not be able tothink properly. For all these interconnectedactivities, energy and raw material are neededfrom outside the body. In other words, foodis a necessity for cell and tissue functions.Anything that prevents proper functioning ofcells and tissues will lead to a lack of properactivity of the body.

It is in this context that we will now lookat the notions of health and disease.

13.1 Health and its Failure13.1.1 THE SIGNIFICANCE OF ‘HEALTH’We have heard the word ‘health’ used quitefrequently all around us. We use it ourselvesas well, when we say things like ‘mygrandmother’s health is not good’. Ourteachers use it when they scold us saying ‘thisis not a healthy attitude’. What does the word‘health’ mean?

If we think about it, we realise that italways implies the idea of ‘being well’. We canthink of this well-being as ef fectivefunctioning. For our grandmothers, being ableto go out to the market or to visit neighboursis ‘being well’, and not being able to do suchthings is ‘poor health’. Being interested infollowing the teaching in the classroom so thatwe can understand the world is called a‘healthy attitude’; while not being interestedis called the opposite. ‘Health’ is therefore astate of being well enough to function wellphysically, mentally and socially.

1313131313WWWWWHYHYHYHYHY D D D D DOOOOO W W W W WEEEEE F F F F FALLALLALLALLALL I I I I ILLLLLLLLLL

Chapter

We need food for health, and this food willhave to be earned by doing work. For this,the opportunity to do work has to be available.Good economic conditions and jobs aretherefore needed for individual health.

We need to be happy in order to be trulyhealthy, and if we mistreat each other andare afraid of each other, we cannot be happyor healthy. Social equality and harmony aretherefore necessary for individual health. Wecan think of many other such examples ofconnections between community issues andindividual health.

13.1.3 DISTINCTIONS BETWEEN ‘HEALTHY’AND ‘DISEASE-FREE’

If this is what we mean by ‘health’, what dowe mean by ‘disease’? The word is actuallyself-explanatory – we can think of it as‘disease’ – disturbed ease. Disease, in otherwords, literally means being uncomfortable.However, the word is used in a more limitedmeaning. We talk of disease when we can finda specific and particular cause for discomfort.This does not mean that we have to know theabsolute final cause; we can say thatsomeone is suffering from diarrhoea withoutknowing exactly what has caused the loosemotions.

We can now easily see that it is possibleto be in poor health without actually sufferingfrom a particular disease. Simply not beingdiseased is not the same as being healthy.‘Good health’ for a dancer may mean beingable to stretch his body into difficult butgraceful positions. On the other hand, goodhealth for a musician may mean having enoughbreathing capacity in his/her lungs to controlthe notes from his/her flute. To have theopportunity to realise the unique potentialin all of us is also necessary for real health.

So, we can be in poor health without therebeing a simple cause in the form of anidentifiable disease. This is the reason why,when we think about health, we think aboutsocieties and communities. On the otherhand, when we think about disease, we thinkabout individual sufferers.

13.1.2 PERSONAL AND COMMUNITY ISSUES

BOTH MATTER FOR HEALTH

If health means a state of physical, mentaland social well-being, it cannot be somethingthat each one of us can achieve entirely onour own. The health of all organisms willdepend on their surroundings or theirenvironment. The environment includes thephysical environment. So, for example, healthis at risk in a cyclone in many ways.

But even more importantly, human beingslive in societies. Our social environment,therefore, is an important factor in ourindividual health. We live in villages, townsor cities. In such places, even our physicalenvironment is decided by our socialenvironment.

Consider what would happen if no agencyis ensuring that garbage is collected anddisposed. What would happen if no one takesresponsibility for clearing the drains andensuring that water does not collect in thestreets or open spaces?

So, if there is a great deal of garbagethrown in our streets, or if there is open drain-water lying stagnant around where we live,the possibility of poor health increases.Therefore, public cleanliness is important forindividual health.

Activity _____________13.2• Find out what provisions are made by

your local authority (panchayat/municipal corporation) for the supplyof clean drinking water.

• Are all the people in your locality ableto access this?

Activity _____________13.3• Find out how your local authority

manages the solid waste generated inyour neighbourhood.

• Are these measures adequate?• If not, what improvements would you

suggest?• What could your family do to reduce

the amount of solid waste generatedduring a day/week?

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uestions1. State any two conditions

essential for good health.2. State any two conditions

essential for being free of disease.3. Are the answers to the above

questions necessarily the sameor different? Why?

13.2 Disease and Its Causes13.2.1 WHAT DOES DISEASE LOOK LIKE?Let us now think a little more about diseases.In the first place, how do we know that thereis a disease? In other words, how do we knowthat there is something wrong with the body?There are many tissues in the body, as wehave seen in Chapter 6. These tissues makeup physiological systems or organ systemsthat carry out body functions. Each of theorgan systems has specific organs as its parts,and it has particular functions. So, thedigestive system has the stomach andintestines, and it helps to digest food takenin from outside the body. The musculoskeletalsystem, which is made up of bones andmuscles, holds the body parts together andhelps the body move.

When there is a disease, either thefunctioning or the appearance of one or moresystems of the body will change for the worse.These changes give rise to symptoms andsigns of disease. Symptoms of disease are thethings we feel as being ‘wrong’. So we have aheadache, we have cough, we have loosemotions, we have a wound with pus; theseare all symptoms. These indicate that theremay be a disease, but they don’t indicate whatthe disease is. For example, a headache maymean just examination stress or, very rarely,it may mean meningitis, or any one of a dozendifferent diseases.

Signs of disease are what physicians willlook for on the basis of the symptoms. Signswill give a little more definite indication ofthe presence of a particular disease.Physicians will also get laboratory tests doneto pinpoint the disease further.

13.2.2 ACUTE AND CHRONIC DISEASES

The manifestations of disease will be differentdepending on a number of factors. One of themost obvious factors that determine how weperceive the disease is its duration. Somediseases last for only very short periods oftime, and these are called acute diseases. Weall know from experience that the commoncold lasts only a few days. Other ailments canlast for a long time, even as much as a lifetime,and are called chronic diseases. An exampleis the infection causing elephantiasis, whichis very common in some parts of India.

Activity _____________13.4• Survey your neighbourhood to find out:

(1) how many people suffered fromacute diseases during the last threemonths,

(2) how many people developed chronicdiseases during this same period,

(3) and finally, the total number ofpeople suffering from chronicdiseases in your neighbourhood.

• Are the answers to questions (1) and(2) different?

• Are the answers to questions (2) and(3) different?

• What do you think could be the reasonfor these differences? What do you thinkwould be the effect of these differenceson the general health of the population?

13.2.3 CHRONIC DISEASES AND POOR

HEALTH

As we can imagine, acute and chronicdiseases have different effects on our health.Any disease that causes poor functioning ofsome part of the body will affect our generalhealth as well. This is because all functionsof the body are necessary for general health.But an acute disease, which is over very soon,will not have time to cause major effects ongeneral health, while a chronic disease willdo so.

As an example, think about a cough andcold, which all of us have from time to time.Most of us get better and become well withina week or so. And there are no bad effects on

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would not lead to loose motions. But they dobecome contributory causes of the disease.

Why was there no clean drinking waterfor the baby? Perhaps because the publicservices are poor where the baby’s familylives. So, poverty or lack of public servicesbecome third-level causes of the baby’sdisease.

It will now be obvious that all diseaseswill have immediate causes and contributorycauses. Also, most diseases will have manycauses, rather than one single cause.

13.2.5 INFECTIOUS AND NON-INFECTIOUSCAUSES

As we have seen, it is important to keep publichealth and community health factors in mindwhen we think about causes of diseases. Wecan take that approach a little further. It isuseful to think of the immediate causes ofdisease as belonging to two distinct types. Onegroup of causes is the infectious agents,mostly microbes or micro-organisms.Diseases where microbes are the immediatecauses are called infectious diseases. This isbecause the microbes can spread in thecommunity, and the diseases they cause willspread with them.

Things to ponder

1. Do all diseases spread to peoplecoming in contact with a sick person?

2. What are the diseases that are notspreading?

3. How would a person develop thosediseases that don’t spread by contactwith a sick person?

On the other hand, there are also diseasesthat are not caused by infectious agents. Theircauses vary, but they are not external causeslike microbes that can spread in thecommunity. Instead, these are mostlyinternal, non-infectious causes.

For example, some cancers are caused bygenetic abnormalities. High blood pressurecan be caused by excessive weight and lackof exercise. You can think of many otherdiseases where the immediate causes will notbe infectious.

our health. We do not lose weight, we do notbecome short of breath, we do not feel tiredall the time because of a few days of coughand cold. But if we get infected with a chronicdisease such as tuberculosis of the lungs,then being ill over the years does make uslose weight and feel tired all the time.

We may not go to school for a few days ifwe have an acute disease. But a chronicdisease will make it difficult for us to followwhat is being taught in school and reduceour ability to learn. In other words, we arelikely to have prolonged general poor healthif we have a chronic disease. Chronic diseasestherefore, have very drastic long-term effectson people’s health as compared to acutediseases.

13.2.4 CAUSES OF DISEASES

What causes disease? When we think aboutcauses of diseases, we must remember thatthere are many levels of such causes. Let uslook at an example. If there is a baby sufferingfrom loose motions, we can say that the causeof the loose motions is an infection with avirus. So the immediate cause of the diseaseis a virus.

But the next question is – where did thevirus come from? Suppose we find that thevirus came through unclean drinking water.But many babies must have had this uncleandrinking water. So, why is it that one babydeveloped loose motions when the otherbabies did not?

One reason might be that this baby is nothealthy. As a result, it might be more likelyto have disease when exposed to risk, whereashealthier babies would not. Why is the babynot healthy? Perhaps because it is not wellnourished and does not get enough food. So,lack of good nourishment becomes a second-level cause of the disease the baby is sufferingfrom. Further, why is the baby not wellnourished? Perhaps because it is from ahousehold which is poor.

It is also possible that the baby has somegenetic difference that makes it more likelyto suffer from loose motions when exposedto such a virus. Without the virus, the geneticdifference or the poor nourishment alone

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The ways in which diseases spread, andthe ways in which they can be treated andprevented at the community level would bedifferent for different diseases. This woulddepend a lot on whether the immediatecauses are infectious or non-infectious.

uestions1. List any three reasons why you

would think that you are sick andought to see a doctor. If only oneof these symptoms were present,would you still go to the doctor?Why or why not?

2. In which of the following case doyou think the long-term effects onyour health are likely to be mostunpleasant?• if you get jaundice,• if you get lice,• if you get acne.

Why?

13.3 Infectious Diseases

13.3.1 INFECTIOUS AGENTS

We have seen that the entire diversity seen inthe living world can be classified into a fewgroups. This classification is based oncommon characteristics between differentorganisms. Organisms that can cause diseaseare found in a wide range of such categoriesof classification. Some of them are viruses,some are bacteria, some are fungi, some aresingle-celled animals or protozoans. Somediseases are also caused by multicellularorganisms, such as worms of different kinds.

Peptic ulcers and the Nobel prize

For many years, everybody used to thinkthat peptic ulcers, which cause acidity–related pain and bleeding in the stomachand duodenum, were because of lifestylereasons. Everybody thought that astressful life led to a lot of acid secretionin the stomach, and eventually causedpeptic ulcers.

Then two Australians made a discoverythat a bacterium, Helicobacter pylori, wasresponsible for peptic ulcers. Robin Warren(born 1937), a pathologist from Perth,Australia, saw these small curved bacteriain the lower part of the stomach in manypatients. He noticed that signs ofinflammation were always present aroundthese bacteria. Barry Marshall (born 1951),a young clinical fellow, became interestedin Warren’s findings and succeeded incultivating the bacteria from these sources.

In treatment studies, Marshall andWarren showed that patients could becured of peptic ulcer only when thebacteria were killed off from the stomach.Thanks to this pioneering discovery byMarshall and Warren, peptic ulcer diseaseis no longer a chronic, frequently disablingcondition, but a disease that can be curedby a short period of treatment withantibiotics.

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For this achievement, Marshall andWarren (seen in the picture) received theNobel prize for physiology and medicinein 2005.


Fig. 13.1(b): Picture of staphylococci, the bacteriawhich can cause acne. The scale of theimage is indicated by the line at top left,which is 5 micrometres long.

Fig. 13.1(d): Picture of Leishmania, the protozoanorganism that causes kala-azar. Theorganisms are oval-shaped, and eachhas one long whip-like structure. Oneorganism (arrow) is dividing, while a cellof the immune system (lower right) hasgripped on the two whips of the dividingorganism and is sending cell processesup to eat up the organism. The immunecell is about ten micrometres in diameter.

Fig. 13.1(e): Picture of an adult roundworm (Ascarislumbricoides is the technical name) fromthe small intestine. The ruler next to itshows four centimetres to give us anidea of the scale.

Fig. 13.1(a): Picture of SARS viruses coming out (seearrows for examples) of the surface ofan infected cell. The white scale linerepresents 500 nanometres, which ishalf a micrometre, which is one-thousandth of a millimetre. The scale linegives us an idea of how small the thingswe are looking at are.Courtesy: Emerging InfectiousDeseases, a journal of CDC, U.S.

Fig. 13.1(c): Picture of Trypanosoma, the protozoanorganism responsible for sleepingsickness. The organism is lying next toa saucer-shaped red blood cell to givean idea of the scale.Copyright: Oregon Health and ScienceUniversity, U.S.

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Common examples of diseases caused byviruses are the common cold, influenza,dengue fever and AIDS. Diseases like typhoidfever, cholera, tuberculosis and anthrax arecaused by bacteria. Many common skininfections are caused by different kinds offungi. Protozoan microbes cause manyfamiliar diseases, such as malaria and kala-azar. All of us have also come across intestinalworm infections, as well as diseases likeelephantiasis caused by diffferent species ofworms.

Why is it important that we think of thesecategories of infectious agents? The answeris that these categories are important factorsin deciding what kind of treatment to use.Members of each one of these groups –viruses, bacteria, and so on – have manybiological characteristics in common.

All viruses, for example, live inside hostcells, whereas bacteria very rarely do. Viruses,bacteria and fungi multiply very quickly, whileworms multiply very slowly in comparison.Taxonomically, all bacteria are closely relatedto each other than to viruses and vice versa.This means that many important lifeprocesses are similar in the bacteria groupbut are not shared with the virus group. As aresult, drugs that block one of these lifeprocesses in one member of the group is likelyto be effective against many other membersof the group. But the same drug will not workagainst a microbe belonging to a differentgroup.

As an example, let us take antibiotics.They commonly block biochemical pathwaysimportant for bacteria. Many bacteria, forexample, make a cell-wall to protectthemselves. The antibiotic penicillin blocksthe bacterial processes that build the cell-wall. As a result, the growing bacteria becomeunable to make cell-walls, and die easily.Human cells don’t make a cell-wall anyway,so penicillin cannot have such an effect onus. Penicillin will have this effect on anybacteria that use such processes for makingcell-walls. Similarly, many antibiotics workagainst many species of bacteria rather thansimply working against one.

But viruses do not use these pathways atall, and that is the reason why antibiotics donot work against viral infections. If we have acommon cold, taking antibiotics does notreduce the severity or the duration of thedisease. However, if we also get a bacterialinfection along with the viral cold, takingantibiotics will help. Even then, the antibioticwill work only against the bacterial part ofthe infection, not the viral infection.

Activity _____________13.5• Find out how many of you in your class

had cold/cough/fever recently.• How long did the illness last?• How many of you took antibiotics (ask

your parents if you had antibiotics)?• How long were those who took

antibiotics ill?• How long were those who didn’t take

antibiotics ill?• Is there a difference between these two

groups?• If yes, why? If not, why not?

13.3.2 MEANS OF SPREAD

How do infectious diseases spread? Manymicrobial agents can commonly move froman affected person to someone else in a varietyof ways. In other words, they can be‘communicated’, and so are also calledcommunicable diseases.

Such disease-causing microbes canspread through the air. This occurs throughthe little droplets thrown out by an infectedperson who sneezes or coughs. Someonestanding close by can breathe in thesedroplets, and the microbes get a chance tostart a new infection. Examples of suchdiseases spread through the air are thecommon cold, pneumonia and tuberculosis.

We all have had the experience of sittingnear someone suffering from a cold andcatching it ourselves. Obviously, the morecrowded our living conditions are, the morelikely it is that such airborne diseases willspread.


Fig. 13.2: Air-transmitted diseases are easier tocatch the closer we are to the infectedperson. However, in closed areas, thedroplet nuclei recirculate and pose a riskto everybody. Overcrowded and poorlyventilated housing is therefore a majorfactor in the spread of airborne diseases.

Diseases can also be spread throughwater. This occurs if the excreta from someonesuffering from an infectious gut disease, suchas cholera, get mixed with the drinking waterused by people living nearby. The cholera-causing microbes will enter new hoststhrough the water they drink and causedisease in them. Such diseases are muchmore likely to spread in the absence of safesupplies of drinking water.

The sexual act is one of the closestphysical contact two people can have witheach other. Not surprisingly, there aremicrobial diseases such as syphilis or AIDSthat are transmitted by sexual contact fromone partner to the other. However, suchsexually transmitted diseases are not spreadby casual physical contact. Casual physicalcontacts include handshakes or hugs orsports, like wrestling, or by any of the otherways in which we touch each other socially.Other than the sexual contact, the AIDS viruscan also spread through blood-to-bloodcontact with infected people or from aninfected mother to her baby during pregnancyor through breast feeding.

We live in an environment that is full ofmany other creatures apart from us. It is

inevitable that many diseases will betransmitted by other animals. These animalscarry the infecting agents from a sick personto another potential host. These animals arethus the intermediaries and are calledvectors. The commonest vectors we all knoware mosquitoes. In many species ofmosquitoes, the females need highlynutritious food in the form of blood in orderto be able to lay mature eggs. Mosquitoes feedon many warm-blooded animals, includingus. In this way, they can transfer diseasesfrom person to person.

Fig. 13.3: Common methods of transmission ofdiseases.

13.3.3 ORGAN-SPECIFIC AND TISSUE-SPECIFIC MANIFESTATIONS

The disease-causing microbes enter the bodythrough these different means. Where do theygo then? The body is very large whencompared to the microbes. So there are manypossible places, organs or tissues, where theycould go. Do all microbes go to the same tissueor organ, or do they go to different ones?

Different species of microbes seem to haveevolved to home in on different parts of thebody. In part, this selection is connected totheir point of entry. If they enter from the airvia the nose, they are likely to go to the lungs.

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This is seen in the bacteria causingtuberculosis. If they enter through the mouth,they can stay in the gut lining like typhoid-causing bacteria. Or they can go to the liver,like the viruses that cause jaundice.

But this needn’t always be the case. Aninfection like HIV, that comes into the bodyvia the sexual organs, will spread to lymphnodes all over the body. Malaria-causingmicrobes, entering through a mosquito bite,will go to the liver, and then to the red bloodcells. The virus causing Japaneseencephalitis, or brain fever, will similarly enterthrough a mosquito bite. But it goes on toinfect the brain.

The signs and symptoms of a disease willthus depend on the tissue or organ whichthe microbe targets. If the lungs are thetargets, then symptoms will be cough andbreathlessness. If the liver is targeted, therewill be jaundice. If the brain is the target, wewill observe headaches, vomiting, fits orunconsciousness. We can imagine what thesymptoms and signs of an infection will be ifwe know what the target tissue or organ is,and the functions that are carried out by thistissue or organ.

In addition to these tissue-specific effectsof infectious disease, there will be othercommon effects too. Most of these commoneffects depend on the fact that the body’simmune system is activated in response toinfection. An active immune system recruitsmany cells to the affected tissue to kill off thedisease-causing microbes. This recruitmentprocess is called inflammation. As a part ofthis process, there are local effects such asswelling and pain, and general effects suchas fever.

In some cases, the tissue-specificity of theinfection leads to very general-seemingeffects. For example, in HIV infection, thevirus goes to the immune system anddamages its function. Thus, many of theeffects of HIV-AIDS are because the body canno longer fight off the many minor infectionsthat we face everyday. Instead, every smallcold can become pneumonia. Similarly, a

minor gut infection can produce majordiarrhoea with blood loss. Ultimately, it isthese other infections that kill peoplesuffering from HIV-AIDS.

It is also important to remember that theseverity of disease manifestations depend onthe number of microbes in the body. If thenumber of microbes is very small, the diseasemanifestations may be minor or unnoticed.But if the number is of the same microbelarge, the disease can be severe enough to belife-threatening. The immune system is amajor factor that determines the number ofmicrobes surviving in the body. We shall lookinto this aspect a little later in the chapter.

13.3.4 PRINCIPLES OF TREATMENT

What are the steps taken by your family whenyou fall sick? Have you ever thought why yousometimes feel better if you sleep for sometime? When does the treatment involvemedicines?

Based on what we have learnt so far, itwould appear that there are two ways to treatan infectious disease. One would be to reducethe effects of the disease and the other to killthe cause of the disease. For the first, we canprovide treatment that will reduce thesymptoms. The symptoms are usuallybecause of inflammation. For example, we cantake medicines that bring down fever, reducepain or loose motions. We can take bed rest sothat we can conserve our energy. This willenable us to have more of it available to focuson healing.

But this kind of symptom-directedtreatment by itself will not make the infectingmicrobe go away and the disease will not becured. For that, we need to be able to kill offthe microbes.

How do we kill microbes? One way is touse medicines that kill microbes. We have seenearlier that microbes can be classified intodifferent categories. They are viruses, bacteria,fungi or protozoa. Each of these groups oforganisms will have some essentialbiochemical life process which is peculiar to


we can get some easy answers. For airbornemicrobes, we can prevent exposure byproviding living conditions that are notovercrowded. For water-borne microbes, wecan prevent exposure by providing safedrinking water. This can be done by treatingthe water to kill any microbial contamination.For vector-borne infections, we can provideclean environments. This would not, forexample, allow mosquito breeding. In otherwords, public hygiene is one basic key to theprevention of infectious diseases.

In addition to these issues that relate tothe environment, there are some other generalprinciples to prevent infectious diseases. Toappreciate those principles, let us ask aquestion we have not looked at so far.Normally, we are faced with infectionseveryday. If someone is suffering from a coldand cough in the class, it is likely that thechildren sitting around will be exposed to theinfection. But all of them do not actually sufferfrom the disease. Why not?

This is because the immune system of ourbody is normally fighting off microbes. Wehave cells that specialise in killing infectingmicrobes. These cells go into action each timeinfecting microbes enter the body. If they aresuccessful, we do not actually come downwith any disease. The immune cells manageto kill off the infection long before it assumesmajor proportions. As we noted earlier, if thenumber of the infecting microbes iscontrolled, the manifestations of disease willbe minor. In other words, becoming exposedto or infected with an infectious microbe doesnot necessarily mean developing noticeabledisease.

So, one way of looking at severe infectiousdiseases is that it represents a lack of successof the immune system. The functioning of theimmune system, like any other system in ourbody, will not be good if proper and sufficientnourishment and food is not available.Therefore, the second basic principle ofprevention of infectious disease is theavailability of proper and sufficient food foreveryone.

that group and not shared with the othergroups. These processes may be pathways forthe synthesis of new substances or respiration.

These pathways will not be used by useither. For example, our cells may make newsubstances by a mechanism different fromthat used by bacteria. We have to find a drugthat blocks the bacterial synthesis pathwaywithout affecting our own. This is what isachieved by the antibiotics that we are allfamiliar with. Similarly, there are drugs thatkill protozoa such as the malarial parasite.

One reason why making anti-viralmedicines is harder than making anti-bacterial medicines is that viruses have fewbiochemical mechanisms of their own. Theyenter our cells and use our machinery fortheir life processes. This means that thereare relatively few virus-specific targets to aimat. Despite this limitation, there are noweffective anti-viral drugs, for example, thedrugs that keep HIV infection under control.

13.3.5 PRINCIPLES OF PREVENTION

All of what we have talked about so far dealswith how to get rid of an infection in someonewho has the disease. But there are threelimitations of this approach to dealing withinfectious disease. The first is that oncesomeone has a disease, their body functionsare damaged and may never recovercompletely. The second is that treatment willtake time, which means that someonesuffering from a disease is likely to bebedridden for some time even if we can giveproper treatment. The third is that the personsuffering from an infectious disease can serveas the source from where the infection mayspread to other people. This leads to themultiplication of the above difficulties. It isbecause of such reasons that prevention ofdiseases is better than their cure.

How can we prevent diseases? There aretwo ways, one general and one specific to eachdisease. The general ways of preventinginfections mostly relate to preventingexposure. How can we prevent exposure toinfectious microbes?

If we look at the means of their spreading,

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Activity _____________13.6• Conduct a survey in your locality.

Talk to ten families who are well-offand ten who are very poor (in yourestimation). Both sets of familiesshould have children who are belowfive years of age. Measure the heightsof these children. Draw a graph of theheight of each child against its agefor both sets of families.

• Is there a difference between thegroups? If yes, why?

• If there is no difference, do you thinkthat your findings mean that beingwell-off or poor does not matter forhealth?

These are the general ways of preventinginfections. What are the specific ways? Theyrelate to a peculiar property of the immunesystem that usually fights off microbialinfections. Let us cite an example to try andunderstand this property.

These days, there is no smallpoxanywhere in the world. But as recently as ahundred years ago, smallpox epidemics werenot at all uncommon. In such an epidemic,people used to be very afraid of coming nearsomeone suffering from the disease since theywere afraid of catching the disease.

However, there was one group of peoplewho did not have this fear. These people wouldprovide nursing care for the victims ofsmallpox. This was a group of people who hadhad smallpox earlier and survived it, althoughwith a lot of scarring. In other words, if youhad smallpox once, there was no chance ofsuffering from it again. So, having the diseaseonce was a means of preventing subsequentattacks of the same disease.

This happens because when the immunesystem first sees an infectious microbe, itresponds against it and then remembers itspecifically. So the next time that particularmicrobe, or its close relatives enter the body,the immune system responds with evengreater vigour. This eliminates the infectioneven more quickly than the first time around.This is the basis of the principle of

immunisation.Immunisation

Traditional Indian and Chinese medicinalsystems sometimes deliberately rubbed theskin crusts from smallpox victims into theskin of healthy people. They thus hopedto induce a mild form of smallpox thatwould create resistance against thedisease.

Famously, two centuries ago, anEnglish physiciannamed EdwardJenner, realisedthat milkmaidswho had hadcowpox did notcatch smallpoxeven duringe p i d e m i c s .Cowpox is a verymild disease.Jenner trieddeliberately givingcowpox to people

(as he can be seen doing in the picture), andfound that they were now resistant tosmallpox. This was because the smallpoxvirus is closely related to the cowpox virus.‘Cow’ is ‘vacca’ in Latin, and cowpox is‘vaccinia’. From these roots, the word

‘vaccination’ has come into our usage.

We can now see that, as a general principle,we can ‘fool’ the immune system intodeveloping a memory for a particular infectionby putting something, that mimics the microbewe want to vaccinate against, into the body.This does not actually cause the disease butthis would prevent any subsequent exposureto the infecting microbe from turning intoactual disease.

Many such vaccines are now available forpreventing a whole range of infectiousdiseases, and provide a disease-specificmeans of prevention. There are vaccinesagainst tetanus, diphtheria, whooping cough,measles, polio and many others. These formthe public health programme of childhood


immunisation for preventing infectiousdiseases.

Of course, such a programme can beuseful only if such health measures areavailable to all children. Can you think ofreasons why this should be so?

Some hepatitis viruses, which causejaundice, are transmitted through water.There is a vaccine for one of them, hepatitisA, in the market. But the majority of childrenin many parts of India are already immuneto hepatitis A by the time they are five yearsold. This is because they are exposed to thevirus through water. Under thesecircumstances, would you take the vaccine?

Activity _____________13.7• Rabies virus is spread by the bite of

infected dogs and other animals. Thereare anti-rabies vaccines for both

Qhumans and animals. Find out theplan of your local authority forthe control of rabies in yourneighbourhood. Are these measuresadequate? If not, what improvementswould you suggest?

uestions1. Why are we normally advised to

take bland and nourishing foodwhen we are sick?

2. What are the different means bywhich infectious diseases arespread?

3. What precautions can you takein your school to reduce theincidence of infectious diseases?

4. What is immunisation?5. What are the immunisation

programmes available at thenearest health centre in your

locality? Which of these diseases are the major health problems in your area?

Whatyou havelearnt• Health is a state of physical, mental and social well-being.

• The health of an individual is dependent on his/her physicalsurroundings and his/her economic status.

• Diseases are classified as acute or chronic, depending on theirduration.

• Disease may be due to infectious or non-infectious causes.

• Infectious agents belong to different categories of organismsand may be unicellular and microscopic or multicellular.

• The category to which a disease-causing organism belongsdecides the type of treatment.

• Infectious agents are spread through air, water, physical contactor vectors.

• Prevention of disease is more desirable than its successfultreatment.

• Infectious diseases can be prevented by public health hygiene

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measures that reduce exposure to infectious agents.• Infectious diseases can also be prevented by using

immunisation.• Effective prevention of infectious diseases in the community

requires that everyone should have access to public hygieneand immunisation.

Exercises1. How many times did you fall ill in the last one year? What were

the illnesses?(a) Think of one change you could make in your habits in order

to avoid any of/most of the above illnesses.(b) Think of one change you would wish for in your

surroundings in order to avoid any of/most of the aboveillnesses.

2. A doctor/nurse/health-worker is exposed to more sick peoplethan others in the community. Find out how she/he avoidsgetting sick herself/himself.

3. Conduct a survey in your neighbourhood to find out what thethree most common diseases are. Suggest three steps that couldbe taken by your local authorities to bring down the incidenceof these diseases.

4. A baby is not able to tell her/his caretakers that she/he issick. What would help us to find out(a) that the baby is sick?(b) what is the sickness?

5. Under which of the following conditions is a person most likelyto fall sick?(a) when she is recovering from malaria.(b) when she has recovered from malaria and is taking care of

someone suffering from chicken-pox.(c) when she is on a four-day fast after recovering from malaria

and is taking care of someone suffering from chicken-pox.Why?

6. Under which of the following conditions are you most likely tofall sick?(a) when you are taking examinations.(b) when you have travelled by bus and train for two days.(c) when your friend is suffering from measles.

Why?

Our planet, Earth is the only one on whichlife, as we know it, exists. Life on Earth isdependent on many factors. Most life-formswe know need an ambient temperature,water, and food. The resources available onthe Earth and the energy from the Sun arenecessary to meet the basic requirements ofall life-forms on the Earth.

What are these resources on the Earth?

These are the land, the water and the air.The outer crust of the Earth is called thelithosphere. Water covers 75% of the Earth’ssurface. It is also found underground. Thesecomprise the hydrosphere. The air that coversthe whole of the Earth like a blanket, is calledthe atmosphere. Living things are foundwhere these three exist. This life-supportingzone of the Earth where the atmosphere, thehydrosphere and the lithosphere interact andmake life possible, is known as the biosphere.

Living things constitute the bioticcomponent of the biosphere. The air, thewater and the soil form the non-living orabiotic component of the biosphere. Let usstudy these abiotic components in detail inorder to understand their role in sustaininglife on Earth.

14.1 The Breath of Life: AirWe have already talked about the compositionof air in the first chapter. It is a mixture ofmany gases like nitrogen, oxygen, carbondioxide and water vapour. It is interesting tonote that even the composition of air is theresult of life on Earth. In planets such asVenus and Mars, where no life is known toexist, the major component of the atmosphereis found to be carbon dioxide. In fact, carbon

dioxide constitutes up to 95-97% of theatmosphere on Venus and Mars.

Eukaryotic cells and many prokaryoticcells, discussed in Chapter 5, need oxygen tobreak down glucose molecules and get energyfor their activities. This results in theproduction of carbon dioxide. Another processwhich results in the consumption of oxygenand the concomitant production of carbondioxide is combustion. This includes not justhuman activities, which burn fuels to getenergy, but also forest fires.

Despite this, the percentage of carbondioxide in our atmosphere is a mere fractionof a percent because carbon dioxide is ‘fixed’in two ways: (i) Green plants convert carbondioxide into glucose in the presence ofSunlight and (ii) many marine animals usecarbonates dissolved in sea-water to maketheir shells.

14.1.1 THE ROLE OF THE ATMOSPHERE IN

CLIMATE CONTROL

We have talked of the atmosphere coveringthe Earth, like a blanket. We know that air isa bad conductor of heat. The atmospherekeeps the average temperature of the Earthfairly steady during the day and even duringthe course of the whole year. The atmosphereprevents the sudden increase in temperatureduring the daylight hours. And during thenight, it slows down the escape of heat intoouter space. Think of the moon, which isabout the same distance from the Sun thatthe Earth is. Despite that, on the surface ofthe moon, with no atmosphere, thetemperature ranges from –190º C to 110º C.

1414141414NNNNNATURALATURALATURALATURALATURAL R R R R RESOURCESESOURCESESOURCESESOURCESESOURCES

Chapter

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the heating of water bodies and the activitiesof living organisms. The atmosphere can beheated from below by the radiation that isreflected back or re-radiated by the land orwater bodies. On being heated, convectioncurrents are set up in the air. In order to gainsome understanding of the nature ofconvection currents, let us perform thefollowing activity:

Activity _____________14.2• Place a candle in a beaker or wide-

mouthed bottle and light it. Light anincense stick and take it to the mouthof the above bottle (Figure 14.1).

• Which way does the smoke flow whenthe incense stick is kept near the edgeof the mouth?

• Which way does the smoke flow whenthe incense stick is kept a little abovethe candle?

• Which way does the smoke flow whenthe incense stick is kept in otherregions?

Activity _____________14.1• Measure the temperature of the

following :Take (i) a beaker full of water, (ii) abeaker full of soil/sand and (iii) a closedbottle containing a thermometer. Keepthem in bright Sunlight for three hours.Now measure the temperature of all 3vessels. Also, take the temperaturereading in shade at the same time.

Now answer1. Is the temperature reading more in

activity (i) or (ii)?

2. Based on the above finding, whichwould become hot faster – the land orthe sea?

3. Is the thermometer reading of thetemperature of air (in shade) the sameas the temperature of sand or water?What do you think is the reason forthis? And why does the temperaturehave to be measured in the shade?

4. Is the temperature of air in the closedglass vessel/bottle the same as thetemperature taken in open air? (i) Whatdo you think is the reason for this?(ii) Do we ever come across thisphenomenon in daily life?

As we have seen above, sand and waterdo not heat up at the same rate. What do youthink will be their rates of cooling? Can wethink of an experiment to test the prediction?

14.1.2 THE MOVEMENT OF AIR: WINDS

We have all felt the relief brought by coolevening breezes after a hot day. Andsometimes, we are lucky enough to get rainsafter some days of really hot weather. Whatcauses the movement of air, and what decideswhether this movement will be in the form ofa gentle breeze, a strong wind or a terriblestorm? What brings us the welcome rains?

All these phenomena are the result ofchanges that take place in our atmospheredue to the heating of air and the formation ofwater vapour. Water vapour is formed due to

Fig. 14.1: Air currents being caused by the unevenheating of air.

The patterns revealed by the smoke showus the directions in which hot and cold airmove. In a similar manner, when air is heatedby radiation from the heated land or water, itrises. But since land gets heated faster thanwater, the air over land would also be heatedfaster than the air over water bodies.

So, if we look at the situation in coastalregions during the day, the air above the land

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gets heated faster and starts rising. As thisair rises, a region of low pressure is createdand air over the sea moves into this area oflow pressure. The movement of air from oneregion to the other creates winds. During theday, the direction of the wind would be fromthe sea to the land.

At night, both land and sea start to cool.Since water cools down slower than the land,the air above water would be warmer thanthe air above land.

On the basis of the above discussion, whatcan you say about:

1. the appearance of areas of low andhigh pressure in coastal areas at night?

2. the direction in which air would flowat night in coastal areas?

Similarly, all the movements of airresulting in diverse atmospheric phenomenaare caused by the uneven heating of theatmosphere in different regions of the Earth.But various other factors also influence thesewinds – the rotation of the Earth and thepresence of mountain ranges in the paths ofthe wind are a couple of these factors. Wewill not go into these factors in detail in thischapter, but think about this: how do thepresence of the Himalayas change the flow ofa wind blowing from Allahabad towards thenorth?

14.1.3 RAIN

Let us go back now to the question of howclouds are formed and bring us rain. We couldstart by doing a simple experiment whichdemonstrates some of the factors influencingthese climatic changes.

Activity _____________14.3• Take an empty bottle of the sort in

which bottled water is sold. Pour about5-10 mL of water into it and close thebottle tightly. Shake it well or leave itout in the Sun for ten minutes. Thiscauses the air in the bottle to besaturated with water vapour.

• Now, take a lighted incense stick. Openthe cap of the bottle and allow some ofthe smoke from the incense stick to

enter the bottle. Quickly close the bottleonce more. Make sure that the cap isfitting tightly. Press the bottle hardbetween your hands and crush it asmuch as possible. Wait for a fewseconds and release the bottle. Againpress the bottle as hard as you can.

Now answer1. When did you observe that the air

inside seemed to become ‘foggy’?

2. When does this fog disappear?

3. When is the pressure inside the bottlehigher?

4. Is the ‘fog’ observed when the pressurein the bottle is high or when it is low?

5. What is the need for smoke particlesinside the bottle for this experiment?

6. What might happen if you do theexperiment without the smoke from theincense stick? Now try it and check ifthe prediction was correct. What mightbe happening in the above experimentin the absence of smoke particles?

The above experiment replicates, on a verysmall scale, what happens when air with avery high content of water vapour goes froma region of high pressure to a region of lowpressure or vice versa.

When water bodies are heated during theday, a large amount of water evaporates andgoes into the air. Some amount of watervapour also get into the atmosphere becauseof various biological activities. This air alsogets heated. The hot air rises up carrying thewater vapour with it. As the air rises, itexpands and cools. This cooling causes thewater vapour in the air to condense in theform of tiny droplets. This condensation ofwater is facilitated if some particles could actas the ‘nucleus’ for these drops to formaround. Normally dust and other suspendedparticles in the air perform this function.

Once the water droplets are formed, theygrow bigger by the ‘condensation’ of thesewater droplets. When the drops have grownbig and heavy, they fall down in the form ofrain. Sometimes, when the temperature of air

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14.1.4 AIR POLLUTION

We keep hearing of the increasing levels ofoxides of nitrogen and sulphur in the news.People often bemoan the fact that the qualityof air has gone down since their childhood.How is the quality of air affected and howdoes this change in quality affect us and otherlife forms?

The fossil fuels like coal and petroleumcontain small amounts of nitrogen andsulphur. When these fuels are burnt, nitrogenand sulphur too are burnt and this producesdifferent oxides of nitrogen and sulphur. Notonly is the inhalation of these gasesdangerous, they also dissolve in rain to giverise to acid rain. The combustion of fossil fuelsalso increases the amount of suspendedparticles in air. These suspended particlescould be unburnt carbon particles orsubstances called hydrocarbons. Presence ofhigh levels of all these pollutants causevisibility to be lowered, especially in coldweather when water also condenses out ofair. This is known as smog and is a visibleindication of air pollution. Studies haveshown that regularly breathing air thatcontains any of these substances increasesthe incidence of allergies, cancer and heartdiseases. An increase in the content of theseharmful substances in air is called airpollution.

is low enough, precipitation may occur in theform of snow, sleet or hail.

Rainfall patterns are decided by theprevailing wind patterns. In large parts ofIndia, rains are mostly brought by the south-west or north-east monsoons. We have alsoheard weather reports that say ‘depressions’in the Bay of Bengal have caused rains insome areas (Figure 14.2).

Activity _____________14.4• Collect information from newspapers

or weather reports on television aboutrainfall patterns across the country.Also find out how to construct a rain-gauge and make one. What precautionsare necessary in order to get reliabledata from this rain-gauge? Now answerthe following questions :

• In which month did your city/town/village get the maximum rainfall?

• In which month did your state/unionterritory get the maximum rainfall?

• Is rain always accompanied by thunderand lightning? If not, in which seasondo you get more of thunder andlightning with the rain?

Activity _____________14.5• Find out more about monsoons and

cyclones from the library. Try and findout the rainfall pattern of any othercountry. Is the monsoon responsible forrains the world over?

Fig. 14.2: Satellite picture showing clouds over India.

Fig. 14.3: Lichen


people are forced to spend considerableamounts of time in fetching water from far-away sources.

Activity _____________14.7• Many municipal corporations are trying

water -harvesting techniques toimprove the availability of water.

• Find out what these techniques are andhow they would increase the water thatis available for use.

But why is water so necessary? And doall organisms require water? All cellularprocesses take place in a water medium. Allthe reactions that take place within our bodyand within the cells occur betweensubstances that are dissolved in water.Substances are also transported from onepart of the body to the other in a dissolvedform. Hence, organisms need to maintain thelevel of water within their bodies in order tostay alive. Terrestrial life-forms require freshwater for this because their bodies cannottolerate or get rid of the high amounts ofdissolved salts in saline water. Thus, watersources need to be easily accessible foranimals and plants to survive on land.

Activity _____________14.8• Select a small area (say, 1 m2) near a

water-body, it may be a river, stream,lake or pond. Count the number ofdifferent animals and plants in thisarea. Also, check the number ofindividuals of each type or species.

• Compare this with the number ofindividuals (both animals and plants)found in an area of the same size in adry, rocky region.

• Is the variety of plant and animal lifethe same in both these areas?

Activity _____________14.9• Select and mark out a small area (about

1 m2) in some unused land in or nearyour school.

• As in the above activity, count thenumber of different animals and plantsin this area and the number ofindividuals of each species.

Activity _____________14.6• Organisms called lichens are found to

be very sensitive to the levels ofcontaminants like sulphur dioxide inthe air. As discussed earlier in section7.3.3, lichens can be commonly foundgrowing on the barks of trees as a thingreenish-white crust. See if you canfind lichen growing on the trees in yourlocality.

• Compare the lichen on trees near busyroads and trees some distance away.

• On the trees near roads, compare theincidence of lichen on the side facingthe road and on the side away from theroad.

What can you say about the levels ofpolluting substances near roads and awayfrom roads on the basis of your findingsabove?

uestions1. How is our atmosphere different

from the atmospheres on Venusand Mars?

2. How does the atmosphere act asa blanket?

3. What causes winds?4. How are clouds formed?5. List any three human activities

that you think would lead to airpollution.

14.2 Water: A Wonder LiquidWater occupies a very large area of the Earth’ssurface and is also found underground. Someamount of water exists in the form of watervapour in the atmosphere. Most of the wateron Earth’s surface is found in seas and oceansand is saline. Fresh water is found frozen inthe ice-caps at the two poles and on snow-covered mountains. The underground waterand the water in rivers, lakes and ponds isalso fresh. However, the availability of freshwater varies from place to place. Practicallyevery summer, most places have to face ashortage of water. And in rural areas, wherewater supply systems have not been installed,

Q

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• Remember to do this in the same placetwice in a year, once during summeror the dry season and once after it hasrained.

Now answer1. Were the numbers similar both times?

2. In which season did you find morevariety of plants and animals?

3. In which season did you find morenumber of individuals of each variety?

After compiling the results of the abovetwo activities, think if there is any relationshipbetween the amount of available water andthe number and variety of plants and animalsthat can live in a given area. If there is arelationship, where do you think you wouldfind a greater variety and abundance of life –in a region that receives 5 cm of rainfall in ayear or a region that receives 200 cm ofrainfall in a year? Find the map showingrainfall patterns in the atlas and predictwhich States in India would have themaximum biodiversity and which would havethe least. Can we think of any way of checkingwhether the prediction is correct?

The availability of water decides not onlythe number of individuals of each species thatare able to survive in a particular area, but italso decides the diversity of life there. Ofcourse, the availability of water is not the onlyfactor that decides the sustainability of lifein a region. Other factors like the temperatureand nature of soil also matter. But water isone of the major resources which determinelife on land.

14.2.1 WATER POLLUTION

Water dissolves the fertilisers and pesticidesthat we use on our farms. So some percentageof these substances are washed into the waterbodies. Sewage from our towns and cities andthe waste from factories are also dumped intorivers or lakes. Specific industries also usewater for cooling in various operations andlater return this hot water to water-bodies.Another manner in which the temperature of

the water in rivers can be affected is when wateris released from dams. The water inside thedeep reservoir would be colder than the waterat the surface which gets heated by the Sun.

All this can affect the life-forms that arefound in these water bodies in various ways.It can encourage the growth of some life-formsand harm some other life-forms. This affectsthe balance between various organisms whichhad been established in that system. So weuse the term water-pollution to cover thefollowing effects:

1. The addition of undesirablesubstances to water-bodies. Thesesubstances could be the fertilisers andpesticides used in farming or theycould be poisonous substances, likemercury salts which are used bypaper-industries. These could also bedisease-causing organisms, like thebacteria which cause cholera.

2. The removal of desirable substancesfrom water-bodies. Dissolved oxygenis used by the animals and plants thatlive in water. Any change that reducesthe amount of this dissolved oxygenwould adversely affect these aquaticorganisms. Other nutrients could alsobe depleted from the water bodies.

3. A change in temperature. Aquaticorganisms are used to a certain rangeof temperature in the water-bodywhere they live, and a sudden markedchange in this temperature would bedangerous for them or affect theirbreeding. The eggs and larvae ofvarious animals are particularlysusceptible to temperature changes.

uestions1. Why do organisms need water?2. What is the major source of fresh

water in the city/town/villagewhere you live?

3. Do you know of any activitywhich may be polluting this watersource?Q


14.3 Mineral Riches in the SoilSoil is an important resource that decides thediversity of life in an area. But what is thesoil and how is it formed? The outermost layerof our Earth is called the crust and theminerals found in this layer supply a varietyof nutrients to life-forms. But these mineralswill not be available to the organisms if theminerals are bound up in huge rocks. Overlong periods of time, thousands and millionsof years, the rocks at or near the surface ofthe Earth are broken down by variousphysical, chemical and some biologicalprocesses. The end product of this breakingdown is the fine particles of soil. But whatare the factors or processes that make soil?

• The Sun: The Sun heats up rocksduring the day so that they expand.At night, these rocks cool down andcontract. Since all parts of the rockdo not expand and contract at thesame rate, this results in theformation of cracks and ultimately thehuge rocks break up into smallerpieces.

• Water: Water helps in the formationof soil in two ways. One, water couldget into the cracks in the rocks formeddue to uneven heating by the Sun. Ifthis water later freezes, it would causethe cracks to widen. Can you thinkwhy this should be so? Two, flowingwater wears away even hard rock overlong periods of time. Fast flowing wateroften carries big and small particlesof rock downstream. These rocks rubagainst other rocks and the resultantabrasion causes the rocks to weardown into smaller and smallerparticles. The water then takes theseparticles along with it and deposits itfurther down its path. Soil is thusfound in places far away from itsparent-rock.

• Wind: In a process similar to the wayin which water rubs against rocks andwears them down, strong winds alsoerode rocks down. The wind also

carries sand from one place to theother like water does.

• Living organisms also influence theformation of soil. The lichen that weread about earlier, also grows on thesurface of rocks. While growing, theyrelease certain substances that causethe rock surface to powder down andform a thin layer of soil. Other smallplants like moss, are able to grow onthis surface now and they cause therock to break up further. The roots ofbig trees sometimes go into cracks inthe rocks and as the roots grow bigger,the crack is forced bigger.

Activity ____________14.10• Take some soil and put it into a beaker

containing water. The water should beat least five times the amount of soiltaken. Stir the soil and water vigorouslyand allow the soil to settle down.Observe after some time.

• Is the soil at the bottom of the beakerhomogenous or have layers formed?

• If layers have formed, how is one layerdifferent from another?

• Is there anything floating on thesurface of the water?

• Do you think some substances wouldhave dissolved in the water? How wouldyou check?

As you have seen, soil is a mixture. Itcontains small particles of rock (of differentsizes). It also contains bits of decayed livingorganisms which is called humus. In addition,soil also contains various forms ofmicroscopic life. The type of soil is decidedby the average size of particles found in itand the quality of the soil is decided by theamount of humus and the microscopicorganisms found in it. Humus is a majorfactor in deciding the soil structure becauseit causes the soil to become more porous andallows water and air to penetrate deepunderground. The mineral nutrients that arefound in a particular soil depends on therocks it was formed from. The nutrientcontent of a soil, the amount of humuspresent in it and the depth of the soil are

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some of the factors that decide which plantswill thrive on that soil. Thus, the topmostlayer of the soil that contains humus andliving organisms in addition to the soilparticles is called the topsoil. The quality ofthe topsoil is an important factor that decidesbiodiversity in that area.

Modern farming practices involve the useof large amounts of fertilizers and pesticides.Use of these substances over long periods oftime can destroy the soil structure by killingthe soil micro-organisms that recyclenutrients in the soil. It also kills theEarthworms which are instrumental inmaking the rich humus. Fertile soil canquickly be turned barren if sustainablepractices are not followed. Removal of usefulcomponents from the soil and addition ofother substances, which adversely affect thefertility of the soil and kill the diversity oforganisms that live in it, is called soil pollution.

The soil that we see today in one placehas been created over a very long period oftime. However, some of the factors thatcreated the soil in the first place and broughtthe soil to that place may be responsible forthe removal of the soil too. The fine particlesof soil may be carried away by flowing wateror wind. If all the soil gets washed away andthe rocks underneath are exposed, we havelost a valuable resource because very littlewill grow on the rock.

Activity ____________14.11• Take two identical trays and fill them

with soil. Plant mustard or green gramor paddy in one of the trays and waterboth the trays regularly for a few days,till the first tray is covered by plantgrowth. Now, tilt both the trays and fixthem in that position. Make sure thatboth the trays are tilted at the sameangle. Pour equal amount of watergently on both trays such that the waterflows out of the trays (Fig. 14.4).

• Study the amount of soil that is carriedout of the trays. Is the amount thesame in both the trays?

• Now pour equal amounts of water onboth the trays from a height. Pour threeor four times the amount that youpoured earlier.

• Study the amount of soil that iscarried out of the trays now. Is theamount the same in both the trays?

• Is the amount of soil that is carried outmore or less or equal to the amountwashed out earlier?

Fig. 14.4: Effect of flowing water on the top-soil

The roots of plants have an important rolein preventing soil erosion. The large-scaledeforestation that is happening all over theworld not only destroys biodiversity, it alsoleads to soil erosion. Topsoil that is bare ofvegetation, is likely to be removed veryquickly. And this is accelerated in hilly ormountainous regions. This process of soilerosion is very difficult to reverse. Vegetativecover on the ground has a role to play in thepercolation of water into the deeperlayers too.

uestions1. How is soil formed?2. What is soil erosion?3. What are the methods of

preventing or reducing soilerosion?

14.4 Biogeochemical CyclesA constant interaction between the biotic andabiotic components of the biosphere makesit a dynamic, but stable system. Theseinteractions consist of a transfer of matterand energy between the different componentsof the biosphere. Let us look at someprocesses involved in the maintenance of theabove balance.

Q


14.4.1 THE WATER-CYCLE

You have seen how the water evaporates fromthe water bodies and subsequentcondensation of this water vapour leads torain. But we don’t see the seas and oceansdrying up. So, how is the water returning tothese water bodies? The whole process inwhich water evaporates and falls on the landas rain and later flows back into the sea viarivers is known as the water-cycle. This cycleis not as straight-forward and simple as thisstatement seems to imply. All of the waterthat falls on the land does not immediatelyflow back into the sea. Some of it seeps intothe soil and becomes part of the undergroundreservoir of fresh-water. Some of thisunderground water finds its way to thesurface through springs. Or we bring it tothe surface for our use through wells or tube-wells. Water is also used by terrestrial animalsand plants for various life-processes(Fig. 14.5).

water. Thus rivers carry many nutrients fromthe land to the sea, and these are used bythe marine organisms.

14.4.2 THE NITROGEN-CYCLE

Nitrogen gas makes up 78% of ouratmosphere and nitrogen is also a part ofmany molecules essential to life like proteins,nucleic acids (DNA and RNA) and somevitamins. Nitrogen is found in otherbiologically important compounds such asalkaloids and urea too. Nitrogen is thus anessential nutrient for all life-forms and lifewould be simple if all these life-forms coulduse the atmospheric nitrogen directly.However, other than a few forms of bacteria,life-forms are not able to convert thecomparatively inert nitrogen molecule intoforms like nitrates and nitrites which can betaken up and used to make the requiredmolecules. These ‘nitrogen-fixing’ bacteriamay be free-living or be associated with somespecies of dicot plants. Most commonly, thenitrogen-fixing bacteria are found in the rootsof legumes (generally the plants which giveus pulses) in special structures called root-nodules. Other than these bacteria, the onlyother manner in which the nitrogen moleculeis converted to nitrates and nitrites is by aphysical process. During lightning, the hightemperatures and pressures created in theair convert nitrogen into oxides of nitrogen.These oxides dissolve in water to give nitricand nitrous acids and fall on land along withrain. These are then utilised by various life-forms.

What happens to the nitrogen once it isconverted into forms that can be taken upand used to make nitrogen-containingmolecules? Plants generally take up nitratesand nitrites and convert them into aminoacids which are used to make proteins. Someother biochemical pathways are used to makethe other complex compounds containingnitrogen. These proteins and other complexcompounds are subsequently consumed byanimals. Once the animal or the plant dies,other bacteria in the soil convert the variouscompounds of nitrogen back into nitrates and

Fig. 14.5: Water-cycle in nature

Let us look at another aspect of whathappens to water during the water-cycle. Asyou know, water is capable of dissolving alarge number of substances. As water flowsthrough or over rocks containing solubleminerals, some of them get dissolved in the

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nitrites. A different type of bacteria convertsthe nitrates and nitrites into elementalnitrogen. Thus, there is a nitrogen-cycle innature in which nitrogen passes from itselemental form in the atmosphere into simplemolecules in the soil and water, which getconverted to more complex molecules in livingbeings and back again to the simple nitrogenmolecule in the atmosphere.

14.4.3 THE CARBON-CYCLE

Carbon is found in various forms on theEarth. It occurs in the elemental form asdiamonds and graphite. In the combinedstate, it is found as carbon dioxide in theatmosphere, as carbonate and hydrogen-carbonate salts in various minerals, while alllife-forms are based on carbon-containingmolecules like proteins, carbohydrates, fats,

Fig.14.6: Nitrogen-cycle in nature

nucleic acids and vitamins. The endoskeletonsand exoskeletons of various animals are alsoformed from carbonate salts. Carbon isincorporated into life-forms through the basicprocess of photosynthesis which is performedin the presence of Sunlight by all life-forms thatcontain chlorophyll. This process convertscarbon dioxide from the atmosphere ordissolved in water into glucose molecules.These glucose molecules are either convertedinto other substances or used to provideenergy for the synthesis of other biologicallyimportant molecules (Fig. 14.7).

The utilisation of glucose to provide energyto living things involves the process ofrespiration in which oxygen may or may notbe used to convert glucose back into carbondioxide. This carbon dioxide then goes backinto the atmosphere. Another process that


adds to the carbon dioxide in the atmosphereis the process of combustion where fuels areburnt to provide energy for various needs likeheating, cooking, transportation andindustrial processes. In fact, the percentageof carbon dioxide in the atmosphere is saidto have doubled since the industrialrevolution when human beings startedburning fossil fuels on a very large scale.Carbon, like water, is thus cycled repeatedlythrough different forms by the variousphysical and biological activities.

14.4.3 (i) THE GREENHOUSE EFFECT

Recall the reading taken by you under (iii) inActivity 14.1. Heat is trapped by glass, andhence the temperature inside a glassenclosure will be much higher than thesurroundings. This phenomenon was usedto create an enclosure where tropical plants

Fig. 14.7: Carbon-cycle in nature

could be kept warm during the winters incolder climates. Such enclosures are calledgreenhouses. Greenhouses have also lenttheir name to an atmospheric phenomenon.Some gases prevent the escape of heat fromthe Earth. An increase in the percentage ofsuch gases in the atmosphere would causethe average temperatures to increase world-wide and this is called the greenhouse effect.Carbon dioxide is one of the greenhousegases. An increase in the carbon dioxidecontent in the atmosphere would cause moreheat to be retained by the atmosphere andlead to global warming.

Activity ____________14.12• Find out what the consequences of

global warming would be.• Also, find out the names of some other

greenhouse gases.

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14.4.4 THE OXYGEN-CYCLE

Oxygen is a very abundant element on ourEarth. It is found in the elemental form inthe atmosphere to the extent of 21%. It alsooccurs extensively in the combined form inthe Earth’s crust as well as also in the air inthe form of carbon dioxide. In the crust, it isfound as the oxides of most metals andsilicon, and also as carbonate, sulphate,nitrate and other minerals. It is also anessential component of most biologicalmolecules like carbohydrates, proteins,nucleic acids and fats (or lipids).

But when we talk of the oxygen-cycle, weare mainly referring to the cycle thatmaintains the levels of oxygen in theatmosphere. Oxygen from the atmosphere isused up in three processes, namelycombustion, respiration and in the formationof oxides of nitrogen. Oxygen is returned tothe atmosphere in only one major process,that is, photosynthesis. And this forms thebroad outline of the oxygen-cycle in nature(Fig. 14.8).

bacteria, are poisoned by elemental oxygen.In fact, even the process of nitrogen-fixing bybacteria does not take place in the presenceof oxygen.

14.5 Ozone LayerElemental oxygen is normally found in theform of a diatomic molecule. However, in theupper reaches of the atmosphere, a moleculecontaining three atoms of oxygen is found.This would mean a formula of O3 and this iscalled ozone. Unlike the normal diatomicmolecule of oxygen, ozone is poisonous andwe are lucky that it is not stable nearer tothe Earth’s surface. But it performs anessential function where it is found. It absorbsharmful radiations from the Sun. Thisprevents those harmful radiations fromreaching the surface of the Earth where theymay damage many forms of life.

Recently it was discovered that this ozonelayer was getting depleted. Various man-madecompounds like CFCs (carbon compoundshaving both fluorine and chlorine which arevery stable and not degraded by any biologicalprocess) were found to persist in theatmosphere. Once they reached the ozonelayer, they would react with the ozonemolecules. This resulted in a reduction of theozone layer and recently they have discovereda hole in the ozone layer above the Antartica.It is difficult to imagine the consequences forlife on Earth if the ozone layer dwindlesfurther, but many people think that it wouldbe better not to take chances. These peopleadvocate working towards stopping all furtherdamage to the ozone layer.

Fig. 14.8: Oxygen-cycle in nature

Though we usually think of oxygen asbeing necessary to life in the process ofrespiration, it might be of interest to you tolearn that some forms of life, especially

Fig. 14.9: Satellite picture showing the hole (magentacolour) in the ozone layer over Antartica

October1980

October1985

October1990


Activity ____________14.13• Find out which other molecules are

thought to damage the ozone layer.• Newspaper reports often talk about the

hole in the ozone layer.• Find out whether the size of this hole

is changing and in what mannerscientists think this would affect lifeon Earth (Fig. 14.9).

QWhatyou havelearnt• Life on Earth depends on resources like soil, water and air,

and energy from the Sun.

• Uneven heating of air over land and water-bodies causes winds.

• Evaporation of water from water-bodies and subsequentcondensation give us rain.

• Rainfall patterns depend on the prevailing wind patterns in anarea.

• Various nutrients are used again and again in a cyclic fashion.This leads to a certain balance between the various componentsof the biosphere.

• Pollution of air, water and soil affect the quality of life andharm the biodiversity.

• We need to conserve our natural resources and use them in asustainable manner.

Exercises1. Why is the atmosphere essential for life?

2. Why is water essential for life?

3. How are living organisms dependent on the soil? Are organismsthat live in water totally independent of soil as a resource?

4. You have seen weather reports on television and in newspapers.How do you think we are able to predict the weather?

uestions1. What are the different states in

which water is found during thewater cycle?

2. Name two biologically importantcompounds that contain bothoxygen and nitrogen.

3. List any three human activitieswhich would lead to an increasein the carbon dioxide content of air.

4. What is the greenhouse effect?5. What are the two forms of

oxygen found in the atmosphere?

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5. We know that many human activities lead to increasinglevels of pollution of the air, water-bodies and soil. Do youthink that isolating these activities to specific and limitedareas would help in reducing pollution?

6. Write a note on how forests influence the quality of our air,soil and water resources.

We know that all living organisms need food.Food supplies proteins, carbohydrates, fats,vitamins and minerals, all of which we requirefor body development, growth and health.Both plants and animals are major sourcesof food for us. We obtain most of this foodfrom agriculture and animal husbandry.

We read in newspapers that efforts arealways being made to improve productionfrom agriculture and animal husbandry. Whyis this necessary? Why we cannot make dowith the current levels of production?

India is a very populous country. Ourpopulation is more than one billion people,and it is still growing. As food for this growingpopulation, we will soon need more than aquarter of a billion tonnes of grain every year.This can be done by farming on more land.But India is already intensively cultivated. Asa result, we do not have any major scope forincreasing the area of land under cultivation.Therefore, it is necessary to increase ourproduction efficiency for both crops andlivestock.

Efforts to meet the food demand byincreasing food production have led to somesuccesses so far. We have had the greenrevolution, which contributed to increasedfood-grain production. We have also had thewhite revolution, which has led to better andmore efficient use as well as availability of milk.

However, these revolutions mean that ournatural resources are getting used moreintensively. As a result, there are morechances of causing damage to our naturalresources to the point of destroying theirbalance completely. Therefore, it is importantthat we should increase food productionwithout degrading our environment anddisturbing the balances maintaining it.

Hence, there is a need for sustainablepractices in agriculture and animalhusbandry.

Also, simply increasing grain productionfor storage in warehouses cannot solve theproblem of malnutrition and hunger. Peopleshould have money to purchase food. Foodsecurity depends on both availability of foodand access to it. The majority of ourpopulation depends on agriculture for theirlivelihood. Increasing the incomes of peopleworking in agriculture is therefore necessaryto combat the problem of hunger. Scientificmanagement practices should be undertakento obtain high yields from farms. Forsustained livelihood, one should undertakemixed farming, intercropping, and integratedfarming practices, for example, combineagriculture with livestock/poultry/fisheries/bee-keeping.

The question thus becomes – how do weincrease the yields of crops and livestock?

15.1 Improvement in Crop YieldsCereals such as wheat, rice, maize, milletsand sorghum provide us carbohydrate forenergy requirement. Pulses like gram (chana),pea (matar), black gram (urad), green gram(moong), pigeon pea (arhar), lentil (masoor),provide us with protein. And oil seedsincluding soyabean, ground nut, sesame,castor, mustard, linseed and sunflowerprovide us with necessary fats (Fig. 15.1).Vegetables, spices and fruits provide a rangeof vitamins and minerals in addition to smallamounts of proteins, carbohydrates and fats.In addition to these food crops, fodder cropslike berseem, oats or sudan grass are raisedas food for the livestock.

1515151515IIIIIMPROVEMENTMPROVEMENTMPROVEMENTMPROVEMENTMPROVEMENT INININININ F F F F FOODOODOODOODOOD R R R R RESOURCESESOURCESESOURCESESOURCESESOURCES

Chapter

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the kharif season from the month of June toOctober, and some of the crops are grown inthe winter season, called the rabi season fromNovember to April. Paddy, soyabean, pigeonpea, maize, cotton, green gram and blackgram are kharif crops, whereas wheat, gram,peas, mustard, linseed are rabi crops.

In India there has been a four timesincrease in the production of food grains from1960 to 2004 with only 25% increase in thecultivable land area. How has this increasein production been achieved? If we think ofthe practices involved in farming, we can seethat we can divide it into three stages. Thefirst is the choice of seeds for planting. Thesecond is the nurturing of the crop plants.The third is the protection of the growing andharvested crops from loss. Thus, the majorgroups of activities for improving crop yieldscan be classified as:

• Crop variety improvement• Crop production improvement• Crop protection management.

15.1.1 CROP VARIETY IMPROVEMENT

This approach depends on finding a cropvariety that can give a good yield. Varieties orstrains of crops can be selected by breedingfor various useful characteristics such asdisease resistance, response to fertilisers,product quality and high yields. One way ofincorporating desirable characters into cropvarieties is by hybridisation. Hybridisationrefers to crossing between geneticallydissimilar plants. This crossing may beintervarietal (between different varieties),interspecific (between two different species ofthe same genus) or intergeneric (betweendifferent genera). Another way of improvingthe crop is by introducing a gene that wouldprovide the desired characteristic. Thisresults in genetically modified crops.

For new varieties of crops to be accepted,it is necessary that the variety produces highyields under different conditions that arefound in different areas. Farmers would needto be provided with good quality seeds of aparticular variety, that is, the seeds should

Fig. 15.1: Different types of crops

uestion1. What do we get from cereals,

pulses, fruits and vegetables?

Different crops require different climaticconditions, temperature and photoperiods fortheir growth and completion of their life cycle.Photoperiods are related to the duration ofsunlight. Growth of plants and flowering aredependent on sunlight. As we all know, plantsmanufacture their food in sunlight by theprocess of photosynthesis. There are somecrops, which are grown in rainy season, called

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IMPROVEMENT IN FOOD RESOURCES 205

all be of the same variety and germinate underthe same conditions.

Cultivation practices and crop yield arerelated to weather, soil quality and availabilityof water. Since weather conditions such asdrought and flood situations areunpredictable, varieties that can be grown indiverse climatic conditions are useful.Similarly, varieties tolerant to high soil salinityhave been developed. Some of the factors forwhich variety improvement is done are:

• Higher yield: To increase theproductivity of the crop per acre.

• Improved quality: Qualityconsiderations of crop products varyfrom crop to crop. Baking quality isimportant in wheat, protein quality inpulses, oil quality in oilseeds andpreserving quality in fruits andvegetables.

• Biotic and abiotic resistance: Cropsproduction can go down due to biotic(diseases, insects and nematodes) andabiotic (drought, salinity, waterlogging, heat, cold and frost) stressesunder different situations. Varietiesresistant to these stresses can improvecrop production.

• Change in maturity duration: Theshorter the duration of the crop fromsowing to harvesting, the moreeconomical is the variety. Such shortdurations allow farmers to growmultiple rounds of crops in a year.Short duration also reduces the costof crop production. Uniform maturitymakes the harvesting process easyand reduces losses during harvesting.

• Wider adaptability: Developingvarieties for wider adaptability willhelp in stabilising the crop productionunder dif ferent environmentalconditions. One variety can then begrown under dif ferent climaticconditions in different areas.

• Desirable agronomic characteristics:Tallness and profuse branching aredesirable characters for fodder crops.Dwarfness is desired in cereals, so

that less nutrients are consumed bythese crops. Thus developing varietiesof desired agronomic characters helpgive higher productivity.

uestions1. How do biotic and abiotic factors

affect crop production?2. What are the desirable agronomic

characteristics for cropimprovements?

15.1.2 CROP PRODUCTION MANAGEMENT

In India, as in many other agriculture-basedcountries, farming ranges from small to verylarge farms. Different farmers thus have moreor less land, money and access to informationand technologies. In short, it is the money orfinancial conditions that allow farmers to takeup dif ferent farming practices andagricultural technologies. There is acorrelation between higher inputs and yields.Thus, the farmer’s purchasing capacity forinputs decides cropping system andproduction practices. Therefore, productionpractices can be at different levels. Theyinclude ‘no cost’ production, ‘low cost’production and ‘high cost’ productionpractices.

15.1.2 (i) NUTRIENT MANAGEMENT

Just as we need food for development, growthand well-being, plants also require nutrientsfor growth. Nutrients are supplied to plantsby air, water and soil. There are sixteennutrients which are essential for plants. Airsupplies carbon and oxygen, hydrogen comesfrom water, and soil supplies the otherthirteen nutrients to plants. Amongst thesethirteen nutrients, six are required in largequantities and are therefore called macro-nutrients. The other seven nutrients are usedby plants in small quantities and are thereforecalled micro-nutrients (Table 15.1).

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SCIENCE206

our environment from excessive use offertilizers. Using biological waste material isalso a way of recycling farm waste. Based onthe kind of biological material used, manurecan be classified as:

(i) Compost and vermi-compost: Theprocess in which farm waste materiallike livestock excreta (cow dung etc.),vegetable waste, animal refuse,domestic waste, sewage waste, straw,eradicated weeds etc. is decomposedin pits is known as composting. Thecompost is rich in organic matter andnutrients. Compost is also preparedby using earthworms to hasten theprocess of decomposition of plant andanimal refuse. This is called vermi-compost.

(ii) Green manure: Prior to the sowing ofthe crop seeds, some plants like sunhemp or guar are grown and thenmulched by ploughing them into thesoil. These green plants thus turn intogreen manure which helps inenriching the soil in nitrogen andphosphorus.

FERTILIZERS

Fertilizers are commercially produced plantnutrients. Fertilizers supply nitrogen,phosphorus and potassium. They are usedto ensure good vegetative growth (leaves,branches and flowers), giving rise to healthyplants. Fertilizers are a factor in the higheryields of high-cost farming.

Fertilizers should be applied carefully interms of proper dose, time, and observing pre-and post-application precautions for theircomplete utilisation. For example, sometimesfertilizers get washed away due to excessiveirrigation and are not fully absorbed by theplants. This excess fertilizer then leads towater pollution.

Also, as we have seen in the previouschapter, continuous use of fertilizers in anarea can destroy soil fertility because theorganic matter in the soil is not replenishedand micro-organisms in the soil are harmedby the fertilizers used. Short-term benefits ofusing fertilizers and long-term benefits of

Deficiency of these nutrients affectsphysiological processes in plants includingreproduction, growth and susceptibility todiseases. To increase the yield, the soil canbe enriched by supplying these nutrients inthe form of manure and fertilizers.

uestions1. What are macro-nutrients and

why are they called macro-nutrients?

2. How do plants get nutrients?

MANURE

Manure contains large quantities of organicmatter and also supplies small quantities ofnutrients to the soil. Manure is prepared bythe decomposition of animal excreta and plantwaste. Manure helps in enriching soil withnutrients and organic matter and increasingsoil fertility. The bulk of organic matter inmanure helps in improving the soil structure.This involves increasing the water holdingcapacity in sandy soils. In clayey soils, thelarge quantities of organic matter help indrainage and in avoiding water logging.

In using manure we use biological wastematerial, which is advantageous in protecting

Table 15.1: Nutrients suppliedby air, water and soil

Source Nutrients

Air carbon, oxygen

Water hydrogen, oxygen

Soil (i) Macronutrients:nitrogen, phosphorus,potassium, calcium,magnesium, sulphur

(ii) Micronutrients:iron, manganese, boron,zinc, copper,molybdenum, chlorine

Q


India has a wide variety of water resourcesand a highly varied climate. Under suchconditions, several different kinds of irrigationsystems are adopted to supply water toagricultural lands depending on the kinds ofwater resources available. These includewells, canals, rivers and tanks.

• Wells: There are two types of wells,namely dug wells and tube wells. In adug well, water is collected from waterbearing strata. Tube wells can tapwater from the deeper strata. Fromthese wells, water is lifted by pumpsfor irrigation.

• Canals: This is usually an elaborateand extensive irrigation system. In thissystem canals receive water from oneor more reservoirs or from rivers. Themain canal is divided into branchcanals having further distributaries toirrigate fields.

• River Lift Systems: In areas wherecanal flow is insufficient or irregulardue to inadequate reservoir release,the lift system is more rational. Wateris directly drawn from the rivers forsupplementing irrigation in areasclose to rivers.

• Tanks: These are small storagereservoirs, which intercept and storethe run-off of smaller catchmentareas.

Fresh initiatives for increasing the wateravailable for agriculture include rainwaterharvesting and watershed management. Thisinvolves building small check-dams whichlead to an increase in ground water levels.The check-dams stop the rainwater fromflowing away and also reduce soil erosion.

15.1.2 (iii) CROPPING PATTERNS

Different ways of growing crops can be usedto give maximum benefit.

Mixed cropping is growing two or morecrops simultaneously on the same piece ofland, for example, wheat + gram, or wheat +mustard, or groundnut + sunflower. Thisreduces risk and gives some insuranceagainst failure of one of the crops.

using manure for maintaining soil fertility haveto be considered while aiming for optimumyields in crop production.

uestion1. Compare the use of manure and

fertilizers in maintaining soilfertility.

Organic farming is a farming system withminimal or no use of chemicals as fertilizers,herbicides, pesticides etc. and with amaximum input of organic manures, recycledfarm-wastes (straw and livestock excreta), useof bio-agents such as culture of blue greenalgae in preparation of biofertilizers, neemleaves or turmeric specifically in grain storageas bio-pesticides, with healthy croppingsystems [mixed cropping, inter-cropping andcrop rotation as discussed below in15.1.2.(iii)]. These cropping systems arebeneficial in insect, pest and wheat controlbesides providing nutrients.

15.1.2 (ii) IRRIGATION

Most agriculture in India is rain-fed, that is,the success of crops in most areas isdependent on timely monsoons and sufficientrainfall spread through most of the growingseason. Hence, poor monsoons cause cropfailure. Ensuring that the crops get water atthe right stages during their growing seasoncan increase the expected yields of any crop.Therefore, many measures are used to bringmore and more agricultural land underirrigation.

QM

ore

to k

now

Droughts occur because of scarcity orirregular distribution of rains. Droughtposes a threat to rain-fed farmingareas, where farmers do not useirrigation for crop production anddepend only on rain. Light soils haveless water retention capacity. In areaswith light soils, crops get adverselyaffected by drought conditions.Scientists have developed some cropvarieties which can tolerate droughtconditions.

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Inter-cropping is growing two or morecrops simultaneously on the same field in adefinite pattern (Fig.15.2). A few rows of onecrop alternate with a few rows of a secondcrop, for example, soyabean + maize, or fingermillet (bajra) + cowpea (lobia). The crops areselected such that their nutrientrequirements are different. This ensuresmaximum utilisation of the nutrientssupplied, and also prevents pests anddiseases from spreading to all the plantsbelonging to one crop in a field. This way,both crops can give better returns.

(gokhroo), Parthenium (gajar ghas),Cyperinus rotundus (motha). They competefor food, space and light. Weeds take upnutrients and reduce the growth of the crop.Therefore, removal of weeds from cultivatedfields during the early stages of crop growth isessential for a good harvest.

Generally insect pests attack the plantsin three ways: (i) they cut the root, stem andleaf, (ii) they suck the cell sap from variousparts of the plant, and (iii) they bore into stemand fruits. They thus affect the health of thecrop and reduce yields.

Diseases in plants are caused bypathogens such as bacteria, fungi and viruses.These pathogens can be present in andtransmitted through the soil, water and air.

Weeds, insects and diseases can becontrolled by various methods. One of themost commonly used methods is the use ofpesticides, which include herbicides,insecticides and fungicides. These chemicalsare sprayed on crop plants or used for treatingseeds and soil. However, excessive use ofthese chemicals creates problems, since theycan be poisonous to many plant and animalspecies and cause environmental pollution.

Weed control methods also includemechanical removal. Preventive methodssuch as proper seed bed preparation, timelysowing of crops, intercropping and croprotation also help in weed control. Some otherpreventive measures against pests are the useof resistant varieties, and summer ploughing,in which fields are ploughed deep in summersto destroy weeds and pests.

uestion1. Which of the following conditions

will give the most benefits? Why?(a) Farmers use high-quality

seeds, do not adoptirrigation or use fertilizers.

(b) Farmers use ordinaryseeds, adopt irrigation anduse fertilizer.

(c) Farmers use quality seeds,adopt irrigation, usefertilizer and use cropprotection measures.

QFig. 15.2 : Intercropping

The growing of different crops on a pieceof land in a pre-planned succession is knownas crop rotation. Depending upon theduration, crop rotation is done for differentcrop combinations. The availability ofmoisture and irrigation facilities decide thechoice of the crop to be cultivated after oneharvest. If crop rotation is done properly thentwo or three crops can be grown in a yearwith good harvests.

15.1.3 CROP PROTECTION MANAGEMENT

Field crops are infested by a large number ofweeds, insect pests and diseases. If weeds andpests are not controlled at the appropriatetime then they can damage the crops so muchthat most of the crop is lost.

Weeds are unwanted plants in thecultivated field, for example, Xanthium


Activity _____________15.1• Visit a weed-infested field in the month

of July or August and make a list ofthe weeds and insect pests in the field.

STORAGE OF GRAINS

Storage losses in agricultural produce can bevery high. Factors responsible for such lossesare biotic— insects, rodents, fungi, mites andbacteria, and abiotic— inappropriatemoisture and temperatures in the place ofstorage. These factors cause degradation inquality, loss in weight, poor germinability,discolouration of produce, all leading to poormarketability. These factors can be controlledby proper treatment and by systematicmanagement of warehouses.

Preventive and control measures are usedbefore grains are stored for future use. Theyinclude strict cleaning of the produce beforestorage, proper drying of the produce first insunlight and then in shade, and fumigationusing chemicals that can kill pests.

uestions1. Why should preventive measures

and biological control methods bepreferred for protecting crops?

2. What factors may be responsiblefor losses of grains duringstorage?

Activity _____________15.2• Make a herbarium of cereals, pulses

and oil seeds and identify the seasonsof their sowing and harvesting.

15.2 Animal HusbandryAnimal husbandry is the scientificmanagement of animal livestock. It includesvarious aspects such as feeding, breeding anddisease control. Animal-based farmingincludes cattle, goat, sheep, poultry and fishfarming. As the population increases and asliving standards increase, the demand formilk, eggs and meat is also going up. Also,the growing awareness of the need forhumane treatment of livestock has broughtin new limitations in livestock farming. Thus,livestock production also needs to beimproved.

15.2.1 CATTLE FARMING

Cattle husbandry is done for two purposes—milk and draught labour for agricultural worksuch as tilling, irrigation and carting. Indiancattle belong to two different species, Bosindicus, cows, and Bos bubalis, buffaloes.Milk-producing females are called milchanimals (dairy animals), while the ones usedfor farm labour are called draught animals.

Milk production depends, to some extent,on the duration of the lactation period,meaning the period of milk production after

Q

Table 15.2: Nutritional values of animal products

AnimalProducts

Fat Protein Sugar Minerals Water Vitamins

Milk (Cow) 3.60 4.00 4.50 0.70 87.20 B1, B2, B12, D, E

Egg 12.00 13.00 * 1.00 74.00 B2, D

Meat 3.60 21.10 * 1.10 74.20 B2, B12

Fish 2.50 19.00 * 1.30 77.20 Niacin, D, A

*Present in very small amounts

Per cent (%) Nutrients

SCIENCE210

the birth of a calf. So, milk production can beincreased by increasing the lactation period.Exotic or foreign breeds (for example, Jersey,Brown Swiss) are selected for long lactationperiods, while local breeds (for example, RedSindhi, Sahiwal) show excellent resistance todiseases. The two can be cross-bred to getanimals with both the desired qualities.

uestion1. Which method is commonly used

for improving cattle breeds andwhy?

Activity _____________15.3• Visit a livestock farm. Note the

following:(1) Number of cattle and number of

different breeds.(2) The amount of daily milk production

from the different breeds.

Proper cleaning and shelter facilities forcows and buffaloes are required for humanefarming, for the health of the animals andfor production of clean milk as well. Animalsrequire regular brushing to remove dirt and

loose hair. They should be sheltered underwell-ventilated roofed sheds that protect themfrom rain, heat and cold. The floor of the cattleshed needs to be sloping so as to stay dryand to facilitate cleaning.

The food requirements of dairy animalsare of two types: (a) maintenancerequirement, which is the food required tosupport the animal to live a healthy life, and(b) milk producing requirement, which is thetype of food required during the lactationperiod. Animal feed includes: (a) roughage,which is largely fibre, and (b) concentrates,which are low in fibre and contain relativelyhigh levels of proteins and other nutrients.Cattle need balanced rations containing allnutrients in proportionate amounts. Besidessuch nutritious food material, certain feedadditives containing micronutrients promotethe health and milk output of dairy animals.

Cattle suffer from a number of diseases.The diseases, besides causing death, reducemilk production. A healthy animal feedsregularly and has a normal posture. Theparasites of cattle may be both externalparasites and internal parasites. The externalparasites live on the skin and mainly causeskin diseases. The internal parasites likeworms, affect stomach and intestine whileflukes damage the liver. Infectious diseasesare also caused by bacteria and viruses.Vaccinations are given to farm animalsagainst many major viral and bacterialdiseases.

15.2.2 POULTRY FARMING

Poultry farming is undertaken to raisedomestic fowl for egg production and chickenmeat. Therefore, improved poultry breeds aredeveloped and farmed to produce layers foreggs and broilers for meat.

The cross-breeding programmes betweenIndian (indigenous, for example, Aseel) andforeign (exotic, for example, Leghorn) breedsfor variety improvement are focused on todevelop new varieties for the followingdesirable traits—

(i) number and quality of chicks;

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Fig. 15.3: Indigenous milch breed of cattle


(ii) dwarf broiler parent for commercialchick production;

(iii) summer adaptation capacity/tolerance to high temperature;

(iv) low maintenance requirements;(v) reduction in the size of the egg-laying

bird with ability to utilise morefibrous cheaper diets formulatedusing agricultural by-products.

The ration (daily food requirement) for broilersis protein rich with adequate fat. The level ofvitamins A and K is kept high in the poultryfeeds.

Poultry fowl suffer from a number ofdiseases caused by virus, bacteria, fungi,parasites, as well as from nutritionaldeficiencies. These necessitate propercleaning, sanitation, and spraying ofdisinfectants at regular intervals. Appropriatevaccination can prevent the occurrence ofinfectious diseases and reduce loss of poultryduring an outbreak of disease.

uestions1. What management practices are

common in dairy and poultryfarming?

2. What are the differences betweenbroilers and layers and in theirmanagement?

Activity _____________15.4• Visit a local poultry farm. Observe types

of breeds and note the type of ration,housing and lighting facilities given tothem. Identify the growers, layers andbroilers.

15.2.3 FISH PRODUCTION

Fish is a cheap source of animal protein forour food. Fish production includes the finnedtrue fish as well as shellfish such as prawnsand molluscs. There are two ways of obtainingfish. One is from natural resources, which iscalled capture fishing. The other way is byfish farming, which is called culture fishery.

The water source of the fish can be eitherseawater or fresh water, such as in rivers andponds. Fishing can thus be done both bycapture and culture of fish in marine andfreshwater ecosystems.

15.2.3 (i) MARINE FISHERIES

India’s marine fishery resources include7500 km of coastline and the deep seas

Fig. 15.4

Aseel Leghorn

uestion1. Discuss the implications of the

following statement:“It is interesting to note thatpoultry is India’s most efficientconverter of low fibre food stuff(which is unfit for humanconsumption) into highlynutritious animal protein food.”

EGG AND BROILER PRODUCTION

Broiler chickens are fed with vitamin-richsupplementary feed for good growth rate andbetter feed efficiency. Care is taken to avoidmortality and to maintain feathering andcarcass quality. They are produced as broilersand sent to market for meat purposes.

For good production of poultry birds, goodmanagement practices are important. Theseinclude maintenance of temperature andhygienic conditions in housing and poultryfeed, as well as prevention and control ofdiseases and pests.

The housing, nutritional andenvironmental requirements of broilers aresomewhat different from those of egg layers.

QQ

SCIENCE212

beyond it. Popular marine fish varietiesinclude pomphret, mackerel, tuna, sardines,and Bombay duck. Marine fish are caughtusing many kinds of fishing nets from fishingboats. Yields are increased by locating largeschools of fish in the open sea using satellitesand echo-sounders.

Some marine fish of high economic valueare also farmed in seawater. This includesfinned fishes like mullets, bhetki, and pearlspots, shellfish such as prawns (Fig. 15.5),mussels and oysters as well as seaweed.Oysters are also cultivated for the pearlsthey make.

Macrobrachium rosenbergii(fresh water)

Peneaus monodon(marine)

Fig. 15.5 : Fresh water and marine prawns

As marine fish stocks get further depleted,the demand for more fish can only be met bysuch culture fisheries, a practice calledmariculture.

15.2.3 (ii) INLAND FISHERIES

Fresh water resources include canals, ponds,reservoirs and rivers. Brackish waterresources, where seawater and fresh watermix together, such as estuaries and lagoonsare also important fish reservoirs. Whilecapture fishing is also done in such inlandwater bodies, the yield is not high. Most fishproduction from these resources is throughaquaculture.

Fish culture is sometimes done incombination with a rice crop, so that fish aregrown in the water in the paddy field. Moreintensive fish farming can be done incomposite fish culture systems. Both localand imported fish species are used in suchsystems.

In such a system, a combination of five orsix fish species is used in a single fishpond.These species are selected so that they donot compete for food among them havingdifferent types of food habits. As a result, thefood available in all the parts of the pond isused. As Catlas are surface feeders, Rohusfeed in the middle-zone of the pond, Mrigalsand Common Carps are bottom feeders, andGrass Carps feed on the weeds, together thesespecies (Fig. 15.6) can use all the food in thepond without competing with each other. Thisincreases the fish yield from the pond.

(a)(b)

(c)(d)

Fig. 15.6: (a) Catla (b) Silver carp (c) Rohu (d) GrassCarp (e) Mrigal (f) Common Carp

(e)

(f)

One problem with such composite fishculture is that many of these fish breed onlyduring monsoon. Even if fish seed is collectedfrom the wild, it can be mixed with that ofother species as well. So, a major problem infish farming is the lack of availability of good-quality seed. To overcome this problem, wayshave now been worked out to breed these fishin ponds using hormonal stimulation. Thishas ensured the supply of pure fish seed indesired quantities.


uestions1. How are fish obtained?2. What are the advantages of

composite fish culture?

Activity _____________15.5• Visit a fish farm in fish breeding

season and note the following:(1) Varieties of fish in the ponds(2) Types of ponds(3) Feed ingredients being used in

the farm(4) Find out what the production

capacity of the farm is

15.2.4 BEE-KEEPING

Honey is widely used and therefore bee-keeping for making honey has become anagricultural enterprise. Since bee-keepingneeds low investments, farmers use it as anadditional income generating activity. Inaddition to honey, the beehives are a sourceof wax which is used in various medicinalpreparations.

The local varieties of bees used forcommercial honey production are Apis ceranaindica, commonly known as the Indian bee,A. dorsata, the rock bee and A. florae, thelittle bee. An Italian bee variety, A. mellifera,has also been brought in to increase yield ofhoney. This is the variety commonly used forcommercial honey production.

The Italian bees have high honey collectioncapacity. They sting somewhat less. They stayin a given beehive for long periods, and breedvery well. For commercial honey production,bee farms or apiaries are established.

The value or quality of honey dependsupon the pasturage, or the flowers availableto the bees for nectar and pollen collection.In addition to adequate quantity of pasturage,the kind of flowers available will determinethe taste of the honey.

uestions1. What are the desirable

characters of bee varietiessuitable for honey production?

2. What is pasturage and how is itrelated to honey production?

Q(a) (b)

Fig. 15.7: (a) Arrangement of beehive in an apiary(b) honey extractor

QWhatyou havelearnt• There are thirteen nutrients essential for crops. Of these, six

are required in large quantities and are known as macro-nutrients whereas seven nutrients are required in smallquantities and are known as micro-nutrients.

• Manure and fertilizers are the main sources of nutrient supplyto crops.

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• Organic farming is a farming system with minimal or nouse of chemicals as fertilizers, herbicides, pesticides etc.and with a maximum input of organic manures, recyledfarm wastes, and bio-agents, with healthy cropping systems.

• Mixed farming is a system of farming on a particular farmwhich includes crop production, raising of livestock etc.

• Mixed cropping is growing of two or more crops simultaneouslyon the same piece of land.

• Growing two or more crops in definite row patterns is knownas inter-cropping.

• The growing of different crops on a piece of land in pre-plannedsuccession is called crop rotation.

• Varietal improvement is required for higher yield, good quality,biotic and abiotic resistance, shortening the maturity duration,wider adaptability and desirable agronomic characteristics.

• Farm animals require proper care and management such asshelter, feeding, breeding and disease control. This is calledanimal husbandry.

• Poultry farming is done to raise domestic fowls. Poultryproduction includes egg production and broiler production forpoultry meat.

• To enhance poultry production, cross breeding is done betweenIndian and exotic breeds for variety improvement.

• Fish may be obtained from marine resources as well as inlandresources.

• To increase production of fish, they can be cultured in marineand inland ecosystems.

• Marine fish capture is done by fishing nets guided by echo-sounders and satellites.

• Composite fish culture system is commonly used for fishfarming.

• Bee-keeping is done to get honey and wax.

Exercises1. Explain any one method of crop production which ensures

high yield.

2. Why are manure and fertilizers used in fields?

3. What are the advantages of inter-cropping and crop rotation?

4. What is genetic manipulation? How is it useful in agriculturalpractices?


5. How do storage grain losses occur?

6. How do good animal husbandry practices benefit farmers?

7. What are the benefits of cattle farming?

8. For increasing production, what is common in poultry, fisheriesand bee-keeping?

9. How do you differentiate between capture fishing, maricultureand aquaculture?

SCIENCE216

Chapter 34. (a) MgCl2

(b) CaO

(c) Cu (NO3)2(d) AlCl

3

(e) CaCO3

5. (a) Calcium, oxygen

(b) Hydrogen, bromine

(c) Sodium, hydrogen, carbon and oxygen

(d) Potassium, sulphur and oxygen

6. (a) 26 g

(b) 256 g

(c) 124 g

(d) 36.5 g

(e) 63 g

7. (a) 14 g

(b) 108 g

(c) 1260 g

8. (a) 0.375 mole

(b) 1.11 mole

(c) 0.5 mole

9. (a) 3.2 g

(b) 9.0 g

10. 3.76 × 1022 molecules

11. 6.022 × 1020 ions

Chapter 4

10. 80.006

11. 168 × =90% , 18

8 × = 10%

12. Valency = 1, Name of the element is lithium,

13. Mass number of X =12, Y=14, Relationship is Isotope.

14. (a) F (b) F (c) T (d) F

15. (a) (b) × (c) × (d) ×16. (a) × (b) × (c) (d) ×

Answers

ANSWERS 217

17. (a) × (b) (c) × (d) ×18. (a) × (b) × (c) × (d)

19.

Atomic Mass Number Number Number Name of theNumber Number of of of Atomic

Neutrons Protons Electrons Species

9 19 10 9 9 Fluorine

16 32 16 16 16 Sulphur

12 24 12 12 12 Magnesium

01 2 01 1 01 Deuterium

01 1 0 1 0 Protium

Chapter 81. (a) distance = 2200 m; displacement = 200 m.

2. (a) average speed = average velocity = 2.00 m s–1

(b) average speed = 1.90 m s–1

; average velocity = 0.952 m s–1

3. average speed = 24 km h–1

4. distance travelled = 96 m

7. velocity = 20 m s–1

; time = 2 s

10. speed = 3.07 km s–1

Chapter 94. c

5. 14000 N

6. – 4 N

7. (a) 35000 N

(b) 3.5 m s-2

(c) 28000 N

8. 2550 N in a direction opposite to the motion of the vehicle

9. d

10. 200 N

11. 0 m s-1

13. 3 kg m s-1

14. 2.25 m; 50 N

15. 10 kg m s-1; 10 kg m s-1; 5/3 m s-1

16. 500 kg m s-1; 800 kg m s-1; 50 N

18. 40 kg m s-1

A2. 240 N

A3. 2500 N

A4. 5 m s-2; 2400 kg m s-1; 6000 N

ANSWERS 217

SCIENCE218

Chapter 10

3. 9.8 N

12. Weight on earth is 98 N and on moon is 16.3 N.

13. Maximum height is 122.5 m and total time is 5 s + 5 s = 10 s.

14. 19.6 m/s

15. Maximum height = 80 m, Net displacement = 0, Total distance covered = 160 m.

16. Gravitational force = 53.36 × 1032 N.

17. 4 s, 80 m from the top.

18. Initial velocity = 29.4 m s–1, height = 44.1 m. After 4 s the ball will be at a distance of 4.9 mfrom the top or 39.2 m from the bottom.

21. The substance will sink.

22. The packet will sink. The mass of water displaced is 350 g.

Chapter 11

2. Zero

4. 210 J

5. Zero

9. 9 × 108 J

10. 2000 J, 1000 J

11. Zero

14. 15 kWh (Unit)

17. 208333.3 J

18. (i) Zero

(ii) Positive

(iii) Negative

20. 20 kWh

Chapter 12

7. 17.2 m, 0.0172 m

8. 18.55

9. 6000

13. 11.47 s

14. 22,600 Hz

20. 1450 ms-1

Class 9 Science

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