RvK7TW3Y

1SCIENCE
2TEXTBOOK FOR CLASS IX
3First Edition
4February 2006 Phalguna 1927
5Reprinted
6November 2006 Kartika 1928
7November 2007 Kartika 1929
8January 2009 Magha 1930
9December 2009 Pausa 1931
10November 2010 Kartika 1932
11December 2011 Pausa 1933
12October 2012 Asvina 1934
13October 2013 Asvina 1935
14PD 750T MJ
15© National Council of Educational
16Research and Training, 2006
17` 110.00
18Printed on 80 GSM paper with NCERT
19watermark
20Published at the Publication Division
21by the Secretary, National Council of
22Educational Research and Training,
23Sri Aurobindo Marg, New Delhi 110 016 and
24printed at Shagun Offset Pvt. Ltd., B-3,
25Sector-65, Noida 201 301 (UP)
26ISBN 81-7450-492-3
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66Cover
67Nidhi Wadhwa
68Layout and Illustrations
69Digital Expressions
70FOREWORD
71The National Curriculum Framework (NCF), 2005, recommends that children’s
72life at school must be linked to their life outside the school. This principle
73marks a departure from the legacy of bookish learning which continues to shape
74our system and causes a gap between the school, home and community. The
75syllabi and textbooks developed on the basis of NCF signify an attempt to
76implement this basic idea. They also attempt to discourage rote learning and
77the maintenance of sharp boundaries between different subject areas. We hope
78these measures will take us significantly further in the direction of a childcentred
79system of education outlined in the National Policy on Education (1986).
80The success of this effort depends on the steps that school principals and
81teachers will take to encourage children to reflect on their own learning and to
82pursue imaginative activities and questions. We must recognise that, given
83space, time and freedom, children generate new knowledge by engaging with
84the information passed on to them by adults. Treating the prescribed textbook
85as the sole basis of examination is one of the key reasons why other resources
86and sites of learning are ignored. Inculcating creativity and initiative is possible
87if we perceive and treat children as participants in learning, not as receivers
88of a fixed body of knowledge.
89These aims imply considerable change in school routines and mode of
90functioning. Flexibility in the daily time-table is as necessary as rigour in
91implementing the annual calendar so that the required number of teaching
92days are actually devoted to teaching. The methods used for teaching and
93evaluation will also determine how effective this textbook proves for making
94children’s life at school a happy experience, rather than a source of stress or
95boredom. Syllabus designers have tried to address the problem of curricular
96burden by restructuring and reorienting knowledge at different stages with
97greater consideration for child psychology and the time available for teaching.
98The textbook attempts to enhance this endeavour by giving higher priority and
99space to opportunities for contemplation and wondering, discussion in small
100groups, and activities requiring hands-on experience.
101The National Council of Educational Research and Training (NCERT)
102appreciates the hard work done by the textbook development team responsible
103for this book. We wish to thank the Chairman of the advisory group in science
104and mathematics, Professor J.V. Narlikar and the Chief Advisor for this book,
105Professor Rupamanjari Ghosh, School of Physical Sciences, Jawaharlal Nehru
106University, New Delhi, for guiding the work of this committee. Several teachers
107contributed to the development of this textbook; we are grateful to them and
108their principals for making this possible. We are indebted to the institutions
109and organisations which have generously permitted us to draw upon their
110resources, material and personnel. We are especially grateful to the members
111of the National Monitoring Committee, appointed by the Department of Secondary
112and Higher Education, Ministry of Human Resource Development under the
113Chairmanship of Professor Mrinal Miri and Professor G.P. Deshpande, for their
114valuable time and contribution. As an organisation committed to systemic
115reform and continuous improvement in the quality of its products, NCERT
116welcomes comments and suggestions which will enable us to undertake further
117revision and refinement.
118Director
119New Delhi National Council of Educational
12020 December 2005 Research and Training
121(iv)
122TEXTBOOK DEVELOPMENT COMMITTEE
123CHAIRMAN, ADVISORY GROUP FOR TEXTBOOKS IN SCIENCE AND MATHEMATICS
124J.V. Narlikar, Emeritus Professor, Chairman, Advisory Committee Inter
125University Centre for Astronomy & Astrophysics (IUCCA), Ganeshbhind,
126Pune University, Pune
127CHIEF ADVISOR
128Rupamanjari Ghosh, Professor, School of Physical Sciences, Jawaharlal Nehru
129University, New Delhi
130MEMBERS
131Anjni Koul, Lecturer, Department of Education in Science and Mathematics
132(DESM), NCERT, New Delhi
133Anupam Pachauri, 1317, Sector 37, Faridabad, Haryana
134Anuradha Gulati, TGT, CRPF Public School, Rohini, Delhi
135Asfa M. Yasin, Reader, Pandit Sunderlal Sharma Central Institute of Vocational
136Education, NCERT, Bhopal
137Charu Maini, PGT, DAV School, Sector 14, Gurgaon, Haryana
138Dinesh Kumar, Reader, DESM, NCERT, New Delhi
139Gagan Gupta, Reader, DESM, NCERT, New Delhi
140H.L. Satheesh, TGT , DM School, Regional Institute of Education, Mysore
141Madhuri Mahapatra, Reader, Regional Institute of Education, Bhubaneswar,
142Orissa
143Puran Chand, Jt. Director, Central Institute of Educational Technology, NCERT,
144New Delhi
145S.C. Jain, Professor, DESM, NCERT, New Delhi
146Sujatha G.D., Assistant Mistress, V.V.S. Sardar Patel High School, Rajaji Nagar,
147Bangalore
148S.K. Dash, Reader, DESM, NCERT, New Delhi
149Seshu Lavania, Reader, Department of Botany, University of Lucknow, Lucknow
150Satyajit Rath, Scientist, National Institute of Immunology, JNU Campus, New
151Delhi
152Sukhvir Singh, Reader, DESM, Regional Institute of Education, Ajmer, Rajasthan
153Uma Sudhir, Eklavya, Indore
154MEMBER-COORDINATOR
155Brahm Parkash, Professor, DESM, NCERT, New Delhi
156ACKNOWLEDGEMENTS
157The National Council of Educational Research and Training is grateful to the
158members of the Textbook Development Team, whose names are given separately,
159for their contribution in the development of the Science textbook for Class IX.
160The Council also gratefully acknowledges the contribution of the participating
161members of the Review Workshop in the finalisation of the book: P.K.
162Bhattacharya, Professor, DESM, NCERT; Anita Julka, Reader, DEGSN, NCERT;
163Tausif Ahmad, PGT, New Era Sr. Sec. School, New Delhi; Samarketu, PGT in
164Physics, JNV, MESRA, Ranchi; Meenakshi Sharma, PGT in Biology, SVEM,
165Ankleshwar, Gujarat; Raji Kamlasanan, PGT in Biology, DTEA SNSU School,
166R.K. Puram, New Delhi; Meenambika Menon, TGT in Science, Cambridge School,
167Noida; Lalit Gupta, TGT in Science, Govt. Boys Sr. Sec. School No. 2, Uttam
168Nagar, New Delhi; Manoj Kumar Gupta, Lecturer in Chemistry, Mukherji
169Memorial Sr. Sec. School, Shahdara, Delhi; Vijay Kumar, Vice-Principal, Govt.
170Sarvodaya, Co. Edu. Sr. Sec. School, Anand Vihar, Delhi; Kanhaya Lal, Principal
171(Retd.), Deptt. of Education, GNCT of Delhi, Delhi; K.B. Gupta, Professor (Retd.),
172NCERT, New Delhi; Kuldeep Singh, TGT in Science, JNV, Meerut; R.A. Goel,
173Principal (Retd.), Delhi; Sumit Kumar Bhatnagar, Department of Education,
174GNCT of Delhi, Delhi.
175Acknowledgements are due to M. Chandra, Professor and Head,
176Department of Education in Science and Mathematics, NCERT, New Delhi for
177providing all academic and administrative support.
178The Council also gratefully acknowledges the support provided by the APC
179Office of DESM, administrative staff of DESM; Deepak Kapoor, Incharge
180Computer Centre, DESM; Saima, DTP Operator; Mohd. Qamar Tabrez,
181Copy Editor; Mathew John and Randhir Thakur, Proof Readers. The efforts of
182the Publication Department, NCERT are also highly appreciated.
183CONTENTS
184FOREWORD iii
185Chapter 1 MATTER IN OUR SURROUNDINGS 1
186Chapter 2 IS MATTER AROUND US PURE 14
187Chapter 3 ATOMS AND MOLECULES 31
188Chapter 4 STRUCTURE OF THE ATOM 46
189Chapter 5 THE FUNDAMENTAL UNIT OF LIFE 57
190Chapter 6 TISSUES 68
191Chapter 7 DIVERSITY IN LIVING ORGANISMS 80
192Chapter 8 MOTION 98
193Chapter 9 FORCE AND LAWS OF MOTION 114
194Chapter 10 GRAVITATION 131
195Chapter 11 WORK AND ENERGY 146
196Chapter 12 SOUND 160
197Chapter 13 WHY DO WE FALL ILL 176
198Chapter 14 NATURAL RESOURCES 189
199Chapter 15 IMPROVEMENT IN FOOD RESOURCES 203
200ANSWERS 216 – 218
201
202
203WE, THE PEOPLE OF INDIA,
204[SOVEREIGN SOCIALIST SECULAR
205DEMOCRATIC REPUBLIC]
206JUSTICE,
207LIBERTY
208EQUALITY
209FRATERNITY
210IN OUR CONSTITUENT ASSEMBLY
211HEREBY ADOPT, ENACT AND GIVE TO
212OURSELVES THIS CONSTITUTION.
213having
214solemnly resolved to constitute India into a
215and to secure
216to all its citizens :
217social, economic and
218political;
219of thought, expression, belief,
220faith and worship;
221of status and of opportunity;
222and to promote among them all
223assuring the dignity of
224the individual and the [unity and
225integrity of the Nation];
226this twenty-sixth day of November, 1949 do
2271
2282
2291. Subs. by the Constitution (Forty-second Amendment) Act, 1976, Sec.2,
230for "Sovereign Democratic Republic" (w.e.f. 3.1.1977)
2312. Subs. by the Constitution (Forty-second Amendment) Act, 1976, Sec.2,
232for "Unity of the Nation" (w.e.f. 3.1.1977)
233THE CONSTITUTION OF
234INDIA
235PREAMBLE
236As we look at our surroundings, we see a large
237variety of things with different shapes, sizes
238and textures. Everything in this universe is
239made up of material which scientists have
240named “matter”. The air we breathe, the food
241we eat, stones, clouds, stars, plants and
242animals, even a small drop of water or a
243particle of sand – every thing is matter. We
244can also see as we look around that all the
245things mentioned above occupy space and
246have mass. In other words, they have both
247mass* and volume**.
248Since early times, human beings have
249been trying to understand their surroundings.
250Early Indian philosophers classified matter in
251the form of five basic elements – the
252“Panch Tatva”– air, earth, fire, sky and water.
253According to them everything, living or nonliving,
254was made up of these five basic
255elements. Ancient Greek philosophers had
256arrived at a similar classification of matter.
257Modern day scientists have evolved two
258types of classification of matter based on their
259physical properties and chemical nature.
260In this chapter we shall learn about
261matter based on its physical properties.
262Chemical aspects of matter will be taken up
263in subsequent chapters.
2641.1 Physical Nature of Matter
2651.1.1 MATTER IS MADE UP OF PARTICLES
266For a long time, two schools of thought prevailed
267regarding the nature of matter. One school
268believed matter to be continuous like a block
269of wood, whereas, the other thought that matter
270was made up of particles like sand. Let us
271perform an activity to decide about the nature
272of matter – is it continuous or particulate?
273Activity ______________ 1.1
274• Take a 100 mL beaker.
275• Fill half the beaker with water and
276mark the level of water.
277• Dissolve some salt/ sugar with the help
278of a glass rod.
279• Observe any change in water level.
280• What do you think has happened to
281the salt?
282• Where does it disappear?
283• Does the level of water change?
284In order to answer these questions we
285need to use the idea that matter is made up
286of particles. What was there in the spoon, salt
287or sugar, has now spread throughout water.
288This is illustrated in Fig. 1.1.
2891.1.2 HOW SMALL ARE THESE PARTICLES
290OF MATTER?
291Activity ______________ 1.2
292• Take 2-3 crystals of potassium
293permanganate and dissolve them in
294100 mL of water.
295Fig. 1.1: When we dissolve salt in water, the particles
296of salt get into the spaces between particles
297of water.
298* The SI unit of mass is kilogram (kg).
299** The SI unit of volume is cubic metre (m3
300). The common unit of measuring volume is
301litre (L) such that 1L = 1 dm3
302, 1L = 1000 mL, 1 mL = 1 cm3
303.
3041
305MATTER IN OUR SURROUNDINGS
306Chapter
3072015-16 12.11.14
3082 SCIENCE
3092015-16 12.11.14
310• Take out approximately 10 mL of this
311solution and put it into 90 mL of clear
312water.
313• Take out 10 mL of this solution and
314put it into another 90 mL of clear water.
315• Keep diluting the solution like this 5 to
3168 times.
317• Is the water still coloured ?
3181.2.2 PARTICLES OF MATTER ARE
319CONTINUOUSLY MOVING
320Activity ______________ 1.3
321• Put an unlit incense stick in a corner
322of your class. How close do you have to
323go near it so as to get its smell?
324• Now light the incense stick. What
325happens? Do you get the smell sitting
326at a distance?
327• Record your observations.
328Activity ______________ 1.4
329• Take two glasses/beakers filled with
330water.
331• Put a drop of blue or red ink slowly
332and carefully along the sides of the first
333beaker and honey in the same way in
334the second beaker.
335• Leave them undisturbed in your house
336or in a corner of the class.
337• Record your observations.
338• What do you observe immediately after
339adding the ink drop?
340• What do you observe immediately after
341adding a drop of honey?
342• How many hours or days does it take
343for the colour of ink to spread evenly
344throughout the water?
345Activity ______________ 1.5
346• Drop a crystal of copper sulphate or
347potassium permanganate into a glass
348of hot water and another containing
349cold water. Do not stir the solution.
350Allow the crystals to settle at the
351bottom.
352• What do you observe just above the
353solid crystal in the glass?
354• What happens as time passes?
355• What does this suggest about the
356particles of solid and liquid?
357• Does the rate of mixing change with
358temperature? Why and how?
359From the above three activities (1.3, 1.4 and
3601.5), we can conclude the following:
361Fig. 1.2: Estimating how small are the particles of
362matter. With every dilution, though the colour
363becomes light, it is still visible.
364This experiment shows that just a few
365crystals of potassium permanganate can
366colour a large volume of water (about
3671000 L). So we conclude that there must be
368millions of tiny particles in just one crystal
369of potassium permanganate, which keep on
370dividing themselves into smaller and smaller
371particles.
372The same activity can be done using
3732 mL of Dettol instead of potassium
374permanganate. The smell can be detected
375even on repeated dilution.
376The particles of matter are very small –
377they are small beyond our imagination!!!!
3781.2 Characteristics of Particles of
379Matter
3801.2.1 PARTICLES OF MATTER HAVE SPACE
381BETWEEN THEM
382In activities 1.1 and 1.2 we saw that particles
383of sugar, salt, Dettol, or potassium
384permanganate got evenly distributed in water.
385Similarly, when we make tea, coffee or
386lemonade (nimbu paani ), particles of one type
387of matter get into the spaces between particles
388of the other. This shows that there is enough
389space between particles of matter.
390MATTER IN OUR S URROUNDING S 3
3912015-16 12.11.14
392• If we consider each student as a
393particle of matter, then in which group
394the particles held each other with the
395maximum force?
396Activity ______________ 1.7
397• Take an iron nail, a piece of chalk and
398a rubber band.
399• Try breaking them by hammering,
400cutting or stretching.
401• In which of the above three substances
402do you think the particles are held
403together with greater force?
404Activity ______________ 1.8
405• Open a water tap, try breaking the
406stream of water with your fingers.
407• Were you able to cut the stream of
408water?
409• What could be the reason behind the
410stream of water remaining together?
411The above three activities (1.6, 1.7 and
4121.8) suggest that particles of matter have force
413acting between them. This force keeps the
414particles together. The strength of this force
415of attraction varies from one kind of matter
416to another.
417uestions
4181. Which of the following are
419matter?
420Chair, air, love, smell, hate,
421almonds, thought, cold, colddrink,
422smell of perfume.
4232. Give reasons for the following
424observation:
425The smell of hot sizzling food
426reaches you several metres
427away, but to get the smell from
428cold food you have to go close.
4293. A diver is able to cut through
430water in a swimming pool. Which
431property of matter does this
432observation show?
4334. What are the characteristics of
434the particles of matter?
435Particles of matter are continuously
436moving, that is, they possess what we call
437the kinetic energy. As the temperature rises,
438particles move faster. So, we can say that with
439increase in temperature the kinetic energy of
440the particles also increases.
441In the above three activities we observe
442that particles of matter intermix on their own
443with each other. They do so by getting into
444the spaces between the particles. This
445intermixing of particles of two different types
446of matter on their own is called diffusion. We
447also observe that on heating, diffusion
448becomes faster. Why does this happen?
4491.2.3 PARTICLES OF MATTER ATTRACT
450EACH OTHER
451Activity ______________ 1.6
452• Play this game in the field— make four
453groups and form human chains as
454suggested:
455• The first group should hold each
456other from the back and lock arms
457like Idu-Mishmi dancers (Fig. 1.3).
458Fig. 1.3
459• The second group should hold hands
460to form a human chain.
461• The third group should form a chain
462by touching each other with only their
463finger tips.
464• Now, the fourth group of students
465should run around and try to break the
466three human chains one by one into
467as many small groups as possible.
468• Which group was the easiest to break?
469Why?
470Q
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4731.3 States of Matter
474Observe different types of matter around you.
475What are its different states? We can see that
476matter around us exists in three different
477states– solid, liquid and gas. These states of
478matter arise due to the variation in the
479characteristics of the particles of matter.
480Now, let us study about the properties of
481these three states of matter in detail.
4821.3.1 THE SOLID STATE
483Activity _____________ 1.9
484• Collect the following articles— a pen,
485a book, a needle and a piece of wooden
486stick.
487• Sketch the shape of the above articles
488in your notebook by moving a pencil
489around them.
490• Do all these have a definite shape,
491distinct boundaries and a fixed volume?
492• What happens if they are hammered,
493pulled or dropped?
494• Are these capable of diffusing into each
495other?
496• Try compressing them by applying
497force. Are you able to compress them?
498All the above are examples of solids. We
499can observe that all these have a definite
500shape, distinct boundaries and fixed volumes,
501that is, have negligible compressibility. Solids
502have a tendency to maintain their shape when
503subjected to outside force. Solids may break
504under force but it is difficult to change their
505shape, so they are rigid.
506Consider the following:
507(a) What about a rubber band, can it
508change its shape on stretching? Is it
509a solid?
510(b) What about sugar and salt? When
511kept in different jars these take the
512shape of the jar. Are they solid?
513(c) What about a sponge? It is a solid
514yet we are able to compress it. Why?
515All the above are solids as:
516• A rubber band changes shape under
517force and regains the same shape when
518the force is removed. If excessive force is
519applied, it breaks.
520• The shape of each individual sugar or
521salt crystal remains fixed, whether we
522take it in our hand, put it in a plate or in
523a jar.
524• A sponge has minute holes, in which
525air is trapped, when we press it, the air
526is expelled out and we are able to
527compress it.
5281.3.2 THE LIQUID STATE
529Activity _____________1.10
530• Collect the following:
531(a) water, cooking oil, milk, juice, a
532cold drink.
533(b) containers of different shapes. Put
534a 50 mL mark on these containers
535using a measuring cylinder from
536the laboratory.
537• What will happen if these liquids are
538spilt on the floor?
539• Measure 50 mL of any one liquid and
540transfer it into different containers one
541by one. Does the volume remain the
542same?
543• Does the shape of the liquid remain the
544same ?
545• When you pour the liquid from one
546container into another, does it flow
547easily?
548We observe that liquids have no fixed
549shape but have a fixed volume. They take up
550the shape of the container in which they are
551kept. Liquids flow and change shape, so they
552are not rigid but can be called fluid.
553Refer to activities 1.4 and 1.5 where we
554saw that solids and liquids can diffuse into
555liquids. The gases from the atmosphere
556diffuse and dissolve in water. These gases,
557especially oxygen and carbon dioxide, are
558essential for the survival of aquatic animals
559and plants.
560All living creatures need to breathe for
561survival. The aquatic animals can breathe
562under water due to the presence of dissolved
563oxygen in water. Thus, we may conclude that
564solids, liquids and gases can diffuse into
565liquids. The rate of diffusion of liquids is
566MATTER IN OUR S URROUNDING S 5
5672015-16 12.11.14
568higher than that of solids. This is due to the
569fact that in the liquid state, particles move
570freely and have greater space between each
571other as compared to particles in the solid
572state.
5731.3.3 THE GASEOUS STATE
574Have you ever observed a balloon seller filling
575a large number of balloons from a single
576cylinder of gas? Enquire from him how many
577balloons is he able to fill from one cylinder.
578Ask him which gas does he have in the cylinder.
579Activity _____________1.11
580• Take three 100 mL syringes and close
581their nozzles by rubber corks, as
582shown in Fig.1.4.
583• Remove the pistons from all the
584syringes.
585• Leaving one syringe untouched, fill
586water in the second and pieces of chalk
587in the third.
588• Insert the pistons back into the
589syringes. You may apply some vaseline
590on the pistons before inserting them
591into the syringes for their smooth
592movement.
593• Now, try to compress the content by
594pushing the piston in each syringe.
595We have observed that gases are highly
596compressible as compared to solids and
597liquids. The liquefied petroleum gas (LPG)
598cylinder that we get in our home for cooking
599or the oxygen supplied to hospitals in
600cylinders is compressed gas. Compressed
601natural gas (CNG) is used as fuel these days
602in vehicles. Due to its high compressibility,
603large volumes of a gas can be compressed
604into a small cylinder and transported easily.
605We come to know of what is being cooked
606in the kitchen without even entering there,
607by the smell that reaches our nostrils. How
608does this smell reach us? The particles of the
609aroma of food mix with the particles of air
610spread from the kitchen, reach us and even
611farther away. The smell of hot cooked food
612reaches us in seconds; compare this with the
613rate of diffusion of solids and liquids. Due to
614high speed of particles and large space
615between them, gases show the property of
616diffusing very fast into other gases.
617In the gaseous state, the particles move
618about randomly at high speed. Due to this
619random movement, the particles hit each
620other and also the walls of the container. The
621pressure exerted by the gas is because of this
622force exerted by gas particles per unit area
623on the walls of the container.
624Fig. 1.4
625• What do you observe? In which case
626was the piston easily pushed in?
627• What do you infer from your
628observations?
629Fig.1.5: a, b and c show the magnified schematic
630pictures of the three states of matter. The
631motion of the particles can be seen and
632compared in the three states of matter.
6336 SCIENCE
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6351.4.1 EFFECT OF CHANGE OF TEMPERATURE
636Activity _____________1.12
637• Take about 150 g of ice in a beaker and
638suspend a laboratory thermometer so
639that its bulb is in contact with the ice,
640as in Fig. 1.6.
641uestions
6421. The mass per unit volume of a
643substance is called density.
644(density = mass/volume).
645Arrange the following in order of
646increasing density – air, exhaust
647from chimneys, honey, water,
648chalk, cotton and iron.
6492. (a) Tabulate the differences in
650the characterisitcs of states
651of matter.
652(b) Comment upon the following:
653rigidity, compressibility,
654fluidity, filling a gas
655container, shape, kinetic
656energy and density.
6573. Give reasons
658(a) A gas fills completely the
659vessel in which it is kept.
660(b) A gas exerts pressure on the
661walls of the container.
662(c) A wooden table should be
663called a solid.
664(d) We can easily move our hand
665in air but to do the same
666through a solid block of wood
667we need a karate expert.
6684. Liquids generally have lower
669density as compared to solids.
670But you must have observed that
671ice floats on water. Find out why.
6721.4 Can Matter Change its State?
673We all know from our observation that water
674can exist in three states of matter–
675• solid, as ice,
676• liquid, as the familiar water, and
677• gas, as water vapour.
678What happens inside the matter during
679this change of state? What happens to the
680particles of matter during the change of
681states? How does this change of state take
682place? We need answers to these questions,
683isn’t it?
684Q
685(a)
686(b)
687Fig. 1.6: (a) Conversion of ice to water, (b) conversion
688of water to water vapour
689MATTER IN OUR S URROUNDING S 7
6902015-16 12.11.14
691• Start heating the beaker on a low flame.
692• Note the temperature when the ice
693starts melting.
694• Note the temperature when all the ice
695has converted into water.
696• Record your observations for this
697conversion of solid to liquid state.
698• Now, put a glass rod in the beaker and
699heat while stirring till the water starts
700boiling.
701• Keep a careful eye on the thermometer
702reading till most of the water has
703vaporised.
704• Record your observations for the
705conversion of water in the liquid state
706to the gaseous state.
707On increasing the temperature of solids,
708the kinetic energy of the particles increases.
709Due to the increase in kinetic energy, the
710particles start vibrating with greater speed.
711The energy supplied by heat overcomes the
712forces of attraction between the particles. The
713particles leave their fixed positions and start
714moving more freely. A stage is reached when
715the solid melts and is converted to a liquid.
716The temperature at which a solid melts to
717become a liquid at the atmospheric pressure
718is called its melting point.
719The melting point of a solid is an
720indication of the strength of the force of
721attraction between its particles.
722The melting point of ice is 273.16 K*. The
723process of melting, that is, change of solid
724state into liquid state is also known as fusion.
725When a solid melts, its temperature
726remains the same, so where does the heat
727energy go?
728You must have observed, during the
729experiment of melting, that the temperature
730of the system does not change after the
731melting point is reached, till all the ice melts.
732This happens even though we continue to
733heat the beaker, that is, we continue to supply
734heat. This heat gets used up in changing the
735state by overcoming the forces of attraction
736between the particles. As this heat energy is
737absorbed by ice without showing any rise in
738temperature, it is considered that it gets
739hidden into the contents of the beaker and is
740known as the latent heat. The word latent
741means hidden. The amount of heat energy
742that is required to change 1 kg of a solid into
743liquid at atmospheric pressure at its melting
744point is known as the latent heat of fusion.
745So, particles in water at 00 C (273 K) have
746more energy as compared to particles in ice
747at the same temperature.
748When we supply heat energy to water,
749particles start moving even faster. At a certain
750temperature, a point is reached when the
751particles have enough energy to break free
752from the forces of attraction of each other. At
753this temperature the liquid starts changing
754into gas. The temperature at which a liquid
755starts boiling at the atmospheric pressure is
756known as its boiling point. Boiling is a bulk
757phenomenon. Particles from the bulk of the
758liquid gain enough energy to change into the
759vapour state.
760For water this temperature is 373 K
761(100 0C = 273 + 100 = 373 K).
762Can you define the latent heat of
763vaporisation? Do it in the same way as we
764have defined the latent heat of fusion.
765Particles in steam, that is, water vapour at
766373 K (1000 C) have more energy than water
767at the same temperature. This is because
768particles in steam have absorbed extra energy
769in the form of latent heat of vaporisation.
770*Note: Kelvin is the SI unit of temperature, 00 C =273.16 K. For convenience, we take 00 C = 273 K
771after rounding off the decimal. To change a temperature on the Kelvin scale to the Celsius scale
772you have to subtract 273 from the given temperature, and to convert a temperature on the
773Celsius scale to the Kelvin scale you have to add 273 to the given temperature.
774So, we infer that the state of matter can
775be changed into another state by changing
776the temperature.
777We have learnt that substances around
778us change state from solid to liquid and from
779liquid to gas on application of heat. But there
7808 SCIENCE
7812015-16 12.11.14
782closer? Do you think that increasing or
783decreasing the pressure can change the state
784of matter?
785are some that change directly from solid state
786to gaseous state and vice versa without
787changing into the liquid state.
788Activity _____________1.13
789• Take some camphor or ammonium
790chloride. Crush it and put it in a china
791dish.
792• Put an inverted funnel over the china
793dish.
794• Put a cotton plug on the stem of the
795funnel, as shown in Fig. 1.7.
796* atmosphere (atm) is a unit of measuring pressure exerted by a gas. The unit of pressure is Pascal (Pa):
7971 atmosphere = 1.01 × 105 Pa. The pressure of air in atmosphere is called atmospheric pressure. The
798atmospheric pressure at sea level is 1 atmosphere, and is taken as the normal atmospheric pressure.
799Fig. 1.7: Sublimation of ammonium chloride
800Fig. 1.8: By applying pressure, particles of matter can
801be brought close together.
802Applying pressure and reducing
803temperature can liquefy gases.
804Have you heard of solid carbon dioxide
805(CO2
806)?It is stored under high pressure. Solid
807CO2
808 gets converted directly to gaseous state
809on decrease of pressure to 1 atmosphere*
810without coming into liquid state. This is the
811reason that solid carbon dioxide is also known
812as dry ice.
813Thus, we can say that pressure and
814temperature determine the state of a
815substance, whether it will be solid, liquid
816or gas.
817• Now, heat slowly and observe.
818• What do you infer from the above
819activity?
820A change of state directly from solid to
821gas without changing into liquid state (or vice
822versa) is called sublimation.
8231.4.2 EFFECT OF CHANGE OF PRESSURE
824We have already learnt that the difference in
825various states of matter is due to the
826difference in the distances between the
827constituent particles. What will happen when
828we start putting pressure and compress a gas
829enclosed in a cylinder? Will the particles come Fig. 1.9: Interconversion of the three states of matter
830MATTER IN OUR S URROUNDING S 9
8312015-16 12.11.14
832dish and keep it inside a cupboard or
833on a shelf in your class.
834• Record the room temperature.
835• Record the time or days taken for the
836evaporation process in the above cases.
837• Repeat the above three steps of activity
838on a rainy day and record your
839observations.
840• What do you infer about the effect of
841temperature, surface area and wind
842velocity (speed) on evaporation?
843You must have observed that the rate of
844evaporation increases with–
845• an increase of surface area:
846We know that evaporation is a surface
847phenomenon. If the surface area is
848increased, the rate of evaporation
849increases. For example, while putting
850clothes for drying up we spread them out.
851• an increase of temperature:
852With the increase of temperature, more
853number of particles get enough kinetic
854energy to go into the vapour state.
855• a decrease in humidity:
856Humidity is the amount of water vapour
857present in air. The air around us cannot
858hold more than a definite amount of
859water vapour at a given temperature. If
860the amount of water in air is already high,
861the rate of evaporation decreases.
862• an increase in wind speed:
863It is a common observation that clothes
864dry faster on a windy day. With the
865increase in wind speed, the particles of
866water vapour move away with the wind,
867decreasing the amount of water vapour
868in the surrounding.
8691.5.2 HOW DOES EVAPORATION CAUSE
870COOLING?
871In an open vessel, the liquid keeps on
872evaporating. The particles of liquid absorb
873energy from the surrounding to regain the
874energy lost during evaporation. This
875absorption of energy from the surroundings
876make the surroundings cold.
877uestions
8781. Convert the following
879temperature to celsius scale:
880a. 300 K b. 573 K.
8812. What is the physical state of
882water at:
883a. 250ºC b. 100ºC ?
8843. For any substance, why does the
885temperature remain constant
886during the change of state?
8874. Suggest a method to liquefy
888atmospheric gases.
8891.5 Evaporation
890Do we always need to heat or change pressure
891for changing the state of matter? Can you
892quote some examples from everyday life where
893change of state from liquid to vapour takes
894place without the liquid reaching the boiling
895point? Water, when left uncovered, slowly
896changes into vapour. Wet clothes dry up.
897What happens to water in the above two
898examples?
899We know that particles of matter are
900always moving and are never at rest. At a
901given temperature in any gas, liquid or solid,
902there are particles with different amounts of
903kinetic energy. In the case of liquids, a small
904fraction of particles at the surface, having
905higher kinetic energy, is able to break away
906from the forces of attraction of other particles
907and gets converted into vapour. This
908phenomenon of change of a liquid into
909vapours at any temperature below its boiling
910point is called evaporation.
9111.5.1 FACTORS AFFECTING EVAPORATION
912Let us understand this with an activity.
913Activity _____________1.14
914• Take 5 mL of water in a test tube and
915keep it near a window or under a fan.
916• Take 5 mL of water in an open china
917dish and keep it near a window or
918under a fan.
919• Take 5 mL of water in an open china
920Q
9211 0 SCIENCE
9222015-16 12.11.14
923What happens when you pour some
924acetone (nail polish remover) on your palm?
925The particles gain energy from your palm or
926surroundings and evaporate causing the
927palm to feel cool.
928After a hot sunny day, people sprinkle
929water on the roof or open ground because
930the large latent heat of vaporisation of water
931helps to cool the hot surface.
932Can you cite some more examples from
933daily life where we can feel the effect of cooling
934due to evaporation?
935Why should we wear cotton clothes in
936summer?
937During summer, we perspire more
938because of the mechanism of our body which
939keeps us cool. We know that during
940evaporation, the particles at the surface of
941the liquid gain energy from the surroundings
942or body surface and change into vapour. The
943heat energy equal to the latent heat of
944vaporisation is absorbed from the body
945leaving the body cool. Cotton, being a good
946absorber of water helps in absorbing the
947sweat and exposing it to the atmosphere for
948easy evaporation.
949Why do we see water droplets on the outer
950surface of a glass containing ice-cold
951water?
952Let us take some ice-cold water in a
953tumbler. Soon we will see water droplets on
954the outer surface of the tumbler. The water
955vapour present in air, on coming in contact
956with the cold glass of water, loses energy and
957gets converted to liquid state, which we see
958as water droplets.
959uestions
9601. Why does a desert cooler cool
961better on a hot dry day?
9622. How does the water kept in an
963earthen pot (matka) become cool
964during summer?
9653. Why does our palm feel cold
966when we put some acetone or
967petrol or perfume on it?
9684. Why are we able to sip hot tea or
969milk faster from a saucer rather
970than a cup?
9715. What type of clothes should we
972wear in summer?
973More to know
974Now scientists are talking of five states of matter: Solid, Liquid, Gas, Plasma and BoseEinstein
975Condensate.
976Plasma: The state consists of super energetic and super excited particles. These particles
977are in the form of ionised gases. The fluorescent tube and neon sign bulbs consist of
978plasma. Inside a neon sign bulb there is neon gas and inside a fluorescent tube there
979is helium gas or some other gas. The gas gets ionised, that is, gets charged when
980electrical energy flows through it. This charging up creates a plasma glowing inside
981the tube or bulb. The plasma glows with a special colour depending on the nature of
982gas. The Sun and the stars glow because of the presence of plasma in them. The plasma
983is created in stars because of very high temperature.
984Bose-Einstein Condensate: In 1920, Indian physicist Satyendra Nath Bose had done
985some calculations for a fifth state of matter. Building on his calculations, Albert Einstein
986predicted a new state of matter – the Bose-Einstein
987Condensate (BEC). In 2001, Eric A. Cornell, Wolfgang
988Ketterle and Carl E. Wieman of USA received the Nobel
989prize in physics for achieving “Bose-Einstein
990condensation”. The BEC is formed by cooling a gas of
991extremely low density, about one-hundred-thousandth
992the density of normal air, to super low temperatures.
993You can log on to www.chem4kids.com to get more
994information on these fourth and fifth states of matter.
995Q
996S.N. Bose
997(1894-1974)
998Albert Einstein
999(1879-1955)
1000MATTER IN OUR S URROUNDING S 1 1
10012015-16 12.11.14
1002What
1003you have
1004learnt
1005• Matter is made up of small particles.
1006• The matter around us exists in three states— solid, liquid
1007and gas.
1008• The forces of attraction between the particles are maximum in
1009solids, intermediate in liquids and minimum in gases.
1010• The spaces in between the constituent particles and kinetic
1011energy of the particles are minimum in the case of solids,
1012intermediate in liquids and maximum in gases.
1013• The arrangement of particles is most ordered in the case of
1014solids, in the case of liquids layers of particles can slip and
1015slide over each other while for gases, there is no order, particles
1016just move about randomly.
1017• The states of matter are inter-convertible. The state of matter
1018can be changed by changing temperature or pressure.
1019• Sublimation is the change of gaseous state directly to solid
1020state without going through liquid state, and vice versa.
1021• Boiling is a bulk phenomenon. Particles from the bulk (whole)
1022of the liquid change into vapour state.
1023• Evaporation is a surface phenomenon. Particles from the
1024surface gain enough energy to overcome the forces of attraction
1025present in the liquid and change into the vapour state.
1026• The rate of evaporation depends upon the surface area exposed
1027to the atmosphere, the temperature, the humidity and the
1028wind speed.
1029• Evaporation causes cooling.
1030• Latent heat of vaporisation is the heat energy required to change
10311 kg of a liquid to gas at atmospheric pressure at its boiling
1032point.
1033• Latent heat of fusion is the amount of heat energy required to
1034change 1 kg of solid into liquid at its melting point.
10351 2 SCIENCE
10362015-16 12.11.14
1037Exercises
10381. Convert the following temperatures to the celsius scale.
1039(a) 293 K (b) 470 K.
10402. Convert the following temperatures to the Kelvin scale.
1041(a) 25°C (b) 373°C.
10423. Give reason for the following observations.
1043(a) Naphthalene balls disappear with time without leaving any
1044solid.
1045(b) We can get the smell of perfume sitting several metres away.
10464. Arrange the following substances in increasing order of forces
1047of attraction between the particles— water, sugar, oxygen.
10485. What is the physical state of water at—
1049(a) 25°C (b) 0°C (c) 100°C ?
10506. Give two reasons to justify—
1051(a) water at room temperature is a liquid.
1052(b) an iron almirah is a solid at room temperature.
10537. Why is ice at 273 K more effective in cooling than water at the
1054same temperature?
10558. What produces more severe burns, boiling water or steam?
10569. Name A,B,C,D,E and F in the following diagram showing
1057change in its state
1058Quantity Unit Symbol
1059Temperature kelvin K
1060Length metre m
1061Mass kilogram kg
1062Weight newton N
1063Volume cubic metre m3
1064Density kilogram per cubic metre kg m–3
1065Pressure pascal Pa
1066• Some measurable quantities and their units to remember:
1067MATTER IN OUR S URROUNDING S 1 3
10682015-16 12.11.14
1069Group Activity
1070Prepare a model to demonstrate movement of particles in solids,
1071liquids and gases.
1072For making this model you will need
1073• A transparent jar
1074• A big rubber balloon or piece of stretchable rubber sheet
1075• A string
1076• Few chick-peas or black gram or dry green peas.
1077How to make?
1078• Put the seeds in the jar.
1079• Sew the string to the centre of the rubber sheet and put some
1080tape to keep it tied securely.
1081• Stretch and tie the rubber sheet on the mouth of the jar.
1082• Your model is ready. Now run your fingers up and down the
1083string by first tugging at it slowly and then rapidly.
1084Fig. 1.10: A model for happy converting of solid to liquid and liquid to gas.
1085Fig. 2.1: Some consumable items
1086Have you ever noticed the word ‘pure’
1087written on the packs of these consumables?
1088For a common person pure means having no
1089adulteration. But, for a scientist all these things
1090are actually mixtures of different substances
1091and hence not pure. For example, milk is
1092actually a mixture of water, fat, proteins etc.
1093When a scientist says that something is pure,
1094it means that all the constituent particles of
1095that substance are the same in their chemical
1096nature. A pure substance consists of a single
1097type of particles. In other words, a substance
1098is a pure single form of matter.
1099As we look around, we can see that most
1100of the matter around us exist as mixtures of
1101two or more pure components, for example,
1102sea water, minerals, soil etc. are all mixtures.
11032.1 What is a Mixture? 2.1 What is a Mixture?
1104Mixtures are constituted by more than one
1105kind of pure form of matter, known as a
1106substance. A substance cannot be separated
1107into other kinds of matter by any physical
1108process. We know that dissolved sodium
1109chloride can be separated from water by the
1110physical process of evaporation. However,
1111sodium chloride is itself a substance and
1112cannot be separated by physical process into
1113its chemical constituents. Similarly, sugar is
1114a substance because it contains only one kind
1115of pure matter and its composition is the same
1116throughout.
1117Soft drink and soil are not single
1118substances. Whatever the source of a
1119substance may be, it will always have the
1120same characteristic properties.
1121Therefore, we can say that a mixture
1122contains more than one substance.
11232.1.1 TYPES OF MIXTURES
1124Depending upon the nature of the
1125components that form a mixture, we can have
1126different types of mixtures.
1127Activity ______________ 2.1
1128• Let us divide the class into groups A,
1129B, C and D.
1130• Group A takes a beaker containing
113150 mL of water and one spatula full of
1132copper sulphate powder. Group B takes
113350 mL of water and two spatula full of
1134copper sulphate powder in a beaker.
1135• Groups C and D can take different
1136amounts of copper sulphate and
1137potassium permanganate or common
1138salt (sodium chloride) and mix the given
1139components to form a mixture.
1140• Report the observations on the
1141uniformity in colour and texture.
1142• Groups A and B have obtained a
1143mixture which has a uniform
1144composition throughout. Such
1145mixtures are called homogeneous
1146mixtures or solutions. Some other
1147examples of such mixtures are: (i) salt
1148How do we judge whether milk, ghee, butter,
1149salt, spices, mineral water or juice that we
1150buy from the market are pure?
11512
1152IS MATTER AROUND US PURE
1153Chapter
1154© NCERT
1155not to be republished
1156More to know
1157in water and (ii) sugar in water.
1158Compare the colour of the solutions
1159of the two groups. Though both the
1160groups have obtained copper sulphate
1161solution but the intensity of colour of
1162the solutions is different. This shows
1163that a homogeneous mixture can have
1164a variable composition.
1165• Groups C and D have obtained
1166mixtures, which contain physically
1167distinct parts and have non-uniform
1168compositions. Such mixtures are called
1169heterogeneous mixtures. Mixtures of
1170sodium chloride and iron filings, salt
1171and sulphur, and oil and water are
1172examples of heterogeneous mixtures.
1173Activity ______________ 2.2
1174• Let us again divide the class into four
1175groups – A, B, C and D.
1176• Distribute the following samples to
1177each group:
1178− Few crystals of copper sulphate to
1179group A.
1180− One spatula full of copper
1181sulphate to group B.
1182− Chalk powder or wheat flour to
1183group C.
1184− Few drops of milk or ink to
1185group D.
1186• Each group should add the given
1187sample in water and stir properly using
1188a glass rod. Are the particles in the
1189mixture visible?
1190• Direct a beam of light from a torch
1191through the beaker containing the
1192mixture and observe from the front.
1193Was the path of the beam of light
1194visible?
1195• Leave the mixtures undisturbed for a
1196few minutes (and set up the filtration
1197apparatus in the meantime). Is the
1198mixture stable or do the particles begin
1199to settle after some time?
1200• Filter the mixture. Is there any residue
1201on the filter paper?
1202Discuss the results and form an
1203opinion.
1204• Groups A and B have got a solution.
1205• Group C has got a suspension.
1206• Group D has got a colloidal solution.
1207Now, we shall learn about solutions,
1208suspensions and colloidal solutions in the
1209following sections.
1210uestions
12111. What is meant by a substance?
12122. List the points of differences
1213between homogeneous and
1214heterogeneous mixtures.
12152.2 What is a Solution? What is a Solution?
1216A solution is a homogeneous mixture of two
1217or more substances. You come across various
1218types of solutions in your daily life. Lemonade,
1219soda water etc. are all examples of solutions.
1220Usually we think of a solution as a liquid that
1221contains either a solid, liquid or a gas
1222dissolved in it. But, we can also have solid
1223solutions (alloys) and gaseous solutions (air).
1224In a solution there is homogeneity at the
1225particle level. For example, lemonade tastes the
1226same throughout. This shows that particles of
1227sugar or salt are evenly distributed in the
1228solution.
1229Q
1230Fig. 2.2: Filtration
1231Alloys: Alloys are mixtures of two or
1232more metals or a metal and a non-metal
1233and cannot be separated into their
1234components by physical methods. But
1235still, an alloy is considered as a mixture
1236because it shows the properties of its
1237constituents and can have variable
1238composition. For example, brass is a
1239mixture of approximately 30% zinc and
124070% copper.
1241IS MATTER AROUND US P URE 15
1242© NCERT
1243not to be republished
124416 SCIENCE
1245A solution has a solvent and a solute as
1246its components.The component of the solution
1247that dissolves the other component in it
1248(usually the component present in larger
1249amount) is called the solvent. The component
1250of the solution that is dissolved in the solvent
1251(usually present in lesser quantity) is called
1252the solute.
1253Examples:
1254(i) A solution of sugar in water is a solid
1255in liquid solution. In this solution,
1256sugar is the solute and water is the
1257solvent.
1258(ii) A solution of iodine in alcohol known
1259as ‘tincture of iodine’, has iodine (solid)
1260as the solute and alcohol (liquid) as
1261the solvent.
1262(iii) Aerated drinks like soda water etc., are
1263gas in liquid solutions. These contain
1264carbon dioxide (gas) as solute and
1265water (liquid) as solvent.
1266(iv) Air is a mixture of gas in gas. Air is a
1267homogeneous mixture of a number of
1268gases. Its two main constituents are:
1269oxygen (21%) and nitrogen (78%). The
1270other gases are present in very small
1271quantities.
1272Properties of a solution
1273• A solution is a homogeneous mixture.
1274• The particles of a solution are smaller
1275than 1 nm (10-9 metre) in diameter. So,
1276they cannot be seen by naked eyes.
1277• Because of very small particle size, they
1278do not scatter a beam of light passing
1279through the solution. So, the path of
1280light is not visible in a solution.
1281• The solute particles cannot be
1282separated from the mixture by the
1283process of filtration. The solute particles
1284do not settle down when left undisturbed,
1285that is, a solution is stable.
12862.2.1 CONCENTRATION OF A SOLUTION
1287In activity 2.2, we observed that groups A and
1288B obtained different shades of solutions. So,
1289we understand that in a solution the relative
1290proportion of the solute and solvent can be
1291varied. Depending upon the amount of solute
1292present in a solution, it can be called a dilute,
1293concentrated or a saturated solution. Dilute
1294and concentrated are comparative terms. In
1295activity 2.2, the solution obtained by group
1296A is dilute as compared to that obtained by
1297group B.
1298Activity ______________ 2.3
1299• Take approximately 50 mL of water
1300each in two separate beakers.
1301• Add salt in one beaker and sugar or
1302barium chloride in the second beaker
1303with continuous stirring.
1304• When no more solute can be dissolved,
1305heat the contents of the beaker to
1306raise the temperature by about 5°C.
1307• Start adding the solute again.
1308Is the amount of salt and sugar or barium
1309chloride, that can be dissolved in water at a
1310given temperature, the same?
1311At any particular temperature, a solution
1312that has dissolved as much solute as it is
1313capable of dissolving, is said to be a saturated
1314solution. In other words, when no more solute
1315can be dissolved in a solution at a given
1316temperature, it is called a saturated solution.
1317The amount of the solute present in the
1318saturated solution at this temperature is
1319called its solubility.
1320If the amount of solute contained in a
1321solution is less than the saturation level, it is
1322called an unsaturated solution.
1323What would happen if you were to take a
1324saturated solution at a certain temperature
1325and cool it slowly.
1326We can infer from the above activity that
1327different substances in a given solvent have
1328different solubilities at the same temperature.
1329The concentration of a solution is the amount
1330of solute present in a given amount (mass or
1331volume) of solution, or the amount of solute
1332dissolved in a given mass or volume of solvent.
1333Concentration of solution = Amount of solute/
1334 Amount of solution
1335Or
1336Amount of solute/Amount of solvent
1337© NCERT
1338not to be republished
1339IS MATTER AROUND US P URE 17
1340There are various ways of expressing the
1341concentration of a solution, but here we will
1342learn only two methods.
1343(i) Mass by mass percentage of a solution
1344Mass of solute = ×100
1345Mass of solution
1346(ii) Mass by volume percentage of a solution
1347Mass of solute = ×100
1348Volume of solution
1349Example 2.1 A solution contains 40 g of
1350common salt in 320 g of water.
1351Calculate the concentration in terms of
1352mass by mass percentage of the
1353solution.
1354Solution:
1355Mass of solute (salt) = 40 g
1356Mass of solvent (water) = 320 g
1357We know,
1358Mass of solution = Mass of solute +
1359Mass of solvent
1360= 40 g + 320 g
1361= 360 g
1362Mass percentage of solution
1363Mass of solute = ×100
1364Massof solution
1365 40 = ×100 =11.1%
1366360
13672.2.2 What is a suspension? 2.2.2 What is a suspension?
1368Non-homogeneous systems, like those
1369obtained by group C in activity 2.2, in which
1370solids are dispersed in liquids, are called
1371suspensions. A suspension is a heterogeneous
1372mixture in which the solute particles do not
1373dissolve but remain suspended throughout
1374the bulk of the medium. Particles of a
1375suspension are visible to the naked eye.
1376Properties of a Suspension
1377• Suspension is a heterogeneous
1378mixture.
1379• The particles of a suspension can be
1380seen by the naked eye.
1381• The particles of a suspension scatter a
1382beam of light passing through it and
1383make its path visible.
1384• The solute particles settle down when
1385a suspension is left undisturbed, that
1386is, a suspension is unstable. They can
1387be separated from the mixture by the
1388process of filtration. When the particles
1389settle down, the suspension breaks
1390and it does not scatter light any more.
13912.2.3 WHAT IS A COLLOIDAL SOLUTION?
1392The mixture obtained by group D in activity 2.2
1393is called a colloid or a colloidal solution. The
1394particles of a colloid are uniformly spread
1395throughout the solution. Due to the relatively
1396smaller size of particles, as compared to that of
1397a suspension, the mixture appears to be
1398homogeneous. But actually, a colloidal solution
1399is a heterogeneous mixture, for example, milk.
1400Because of the small size of colloidal
1401particles, we cannot see them with naked
1402eyes. But, these particles can easily scatter a
1403beam of visible light as observed in activity
14042.2. This scattering of a beam of light is called
1405the Tyndall effect after the name of the
1406scientist who discovered this effect.
1407Tyndall effect can also be observed when a
1408fine beam of light enters a room through a small
1409hole. This happens due to the scattering of light
1410by the particles of dust and smoke in the air.
1411Fig. 2.3: (a) Solution of copper sulphate does not show
1412Tyndall effect, (b) mixture of water and milk
1413shows Tyndall effect.
1414© NCERT
1415(a) (b)
1416not to be republished
141718 SCIENCE
1418Q
1419• They cannot be separated from the
1420mixture by the process of filtration. But,
1421a special technique of separation known
1422as centrifugation (perform activity 2.5),
1423can be used to separate the colloidal
1424particles.
1425The components of a colloidal solution are
1426the dispersed phase and the dispersion
1427medium. The solute-like component or the
1428dispersed particles in a colloid form the
1429dispersed phase, and the component in which
1430the dispersed phase is suspended is known
1431as the dispersing medium. Colloids are
1432classified according to the state (solid, liquid
1433or gas) of the dispersing medium and the
1434dispersed phase. A few common examples are
1435given in Table 2.1. From this table you can
1436see that they are very common everyday life.
1437uestions
14381. Differentiate between homogeneous
1439and heterogeneous mixtures
1440with examples.
14412. How are sol, solution and
1442suspension different from each
1443other?
14443. To make a saturated solution,
144536 g of sodium chloride is dissolved
1446in 100 g of water at 293 K.
1447Find its concentration at this
1448temperature.
1449Tyndall effect can be observed when
1450sunlight passes through the canopy of a dense
1451forest. In the forest, mist contains tiny droplets
1452of water, which act as particles of colloid
1453dispersed in air.
1454Fig. 2.4: The Tyndall effect
1455Properties of a colloid
1456• A colloid is a heterogeneous mixture.
1457• The size of particles of a colloid is too
1458small to be individually seen by naked
1459eyes.
1460• Colloids are big enough to scatter a
1461beam of light passing through it and
1462make its path visible.
1463• They do not settle down when left
1464undisturbed, that is, a colloid is quite
1465stable.
1466Table 2.1: Common examples of colloids
1467Dispersed Dispersing Type Example
1468phase Medium
1469Liquid Gas Aerosol Fog, clouds, mist
1470Solid Gas Aerosol Smoke, automobile exhaust
1471Gas Liquid Foam Shaving cream
1472Liquid Liquid Emulsion Milk, face cream
1473Solid Liquid Sol Milk of magnesia, mud
1474Gas Solid Foam Foam, rubber, sponge, pumice
1475Liquid Solid Gel Jelly, cheese, butter
1476Solid Solid Solid Sol Coloured gemstone, milky glass
1477© NCERT
1478not to be republished
1479IS MATTER AROUND US P URE 19
14802.3 Separating the Components Separating the Components
1481of a Mixture of a Mixture
1482We have learnt that most of the natural
1483substances are not chemically pure. Different
1484methods of separation are used to get
1485individual components from a mixture.
1486Separation makes it possible to study and
1487use the individual components of a mixture.
1488Heterogeneous mixtures can be separated
1489into their respective constituents by simple
1490physical methods like handpicking, sieving,
1491filtration that we use in our day-to-day life.
1492Sometimes special techniques have to be used
1493for the separation of the components of a
1494mixture.
14952.3.1 HOW CAN WE OBTAIN COLOURED
1496COMPONENT ( DYE) FROM BLUE/
1497BLACK INK?
1498Activity ______________ 2.4
1499• Fill half a beaker with water.
1500• Put a watch glass on the mouth of the
1501beaker (Fig. 2.5).
1502• Put few drops of ink on the watch glass.
1503• Now start heating the beaker. We do
1504not want to heat the ink directly. You
1505will see that evaporation is taking place
1506from the watch glass.
1507• Continue heating as the evaporation
1508goes on and stop heating when you do
1509not see any further change on the
1510watch glass.
1511• Observe carefully and record your
1512observations.
1513Now answer
1514• What do you think has got evaporated
1515from the watch glass?
1516• Is there a residue on the watch glass?
1517• What is your interpretation? Is ink a
1518single substance (pure) or is it a
1519mixture?
1520We find that ink is a mixture of a dye in
1521water. Thus, we can separate the volatile
1522component (solvent) from its non-volatile
1523solute by the method of evaporation.
15242.3.2 HOW CAN WE SEPARATE CREAM
1525FROM MILK?
1526Now-a-days, we get full-cream, toned and
1527double-toned varieties of milk packed in polypacks
1528or tetra packs in the market. These
1529varieties of milk contain different amounts
1530of fat.
1531Activity ______________ 2.5
1532• Take some full-cream milk in a test
1533tube.
1534• Centrifuge it by using a centrifuging
1535machine for two minutes. If a
1536centrifuging machine is not available
1537in the school, you can do this activity
1538at home by using a milk churner, used
1539in the kitchen.
1540• If you have a milk dairy nearby, visit it
1541and ask (i) how they separate cream
1542from milk and (ii) how they make
1543cheese (paneer) from milk.
1544Now answer
1545• What do you observe on churning the
1546milk?
1547• Explain how the separation of cream
1548from milk takes place.
1549Sometimes the solid particles in a liquid
1550are very small and pass through a filter paper.
1551For such particles the filtration technique
1552Fig. 2.5: Evaporation cannot be used for separation. Such mixtures
1553© NCERT
1554not to be republished
155520 SCIENCE
1556are separated by centrifugation. The principle
1557is that the denser particles are forced to the
1558bottom and the lighter particles stay at the
1559top when spun rapidly.
1560Applications
1561• Used in diagnostic laboratories for
1562blood and urine tests.
1563• Used in dairies and home to separate
1564butter from cream.
1565• Used in washing machines to squeeze
1566out water from wet clothes.
15672.3.3 HOW CAN WE SEPARATE A MIXTURE
1568OF TWO IMMISCIBLE LIQUIDS?
1569Activity ______________ 2.6
1570• Let us try to separate kerosene oil
1571from water using a separating funnel.
1572• Pour the mixture of kerosene oil and
1573water in a separating funnel (Fig. 2.6).
1574• Let it stand undisturbed for sometime
1575so that separate layers of oil and water
1576are formed.
1577• Open the stopcock of the separating
1578funnel and pour out the lower layer of
1579water carefully.
1580• Close the stopcock of the separating
1581funnel as the oil reaches the stop-cock.
1582Applications
1583• To separate mixture of oil and water.
1584• In the extraction of iron from its ore,
1585the lighter slag is removed from the
1586top by this method to leave the molten
1587iron at the bottom in the furnace.
1588The principle is that immiscible liquids
1589separate out in layers depending on their
1590densities.
15912.3.4 HOW CAN WE SEPARATE A MIXTURE
1592OF SALT AND AMMONIUM CHLORIDE?
1593We have learnt in chapter 1 that ammonium
1594chloride changes directly from solid to
1595gaseous state on heating. So, to separate such
1596mixtures that contain a sublimable volatile
1597component from a non-sublimable impurity
1598(salt in this case), the sublimation process is
1599used (Fig. 2.7). Some examples of solids which
1600sublime are ammonium chloride, camphor,
1601naphthalene and anthracene.
1602Fig. 2.7: Separation of ammonium chloride and salt
1603Fig. 2.6: Separation of immiscible liquids by sublimation
1604© NCERT
1605not to be republished
1606IS MATTER AROUND US P URE 21
16072.3.5 IS THE DYE IN BLACK INK A SINGLE
1608COLOUR?
1609Activity ______________ 2.7
1610• Take a thin strip of filter paper.
1611• Draw a line on it using a pencil,
1612approximately 3 cm above the lower
1613edge [Fig. 2.8 (a)].
1614• Put a small drop of ink (water soluble,
1615that is, from a sketch pen or fountain
1616pen) at the centre of the line. Let it dry.
1617• Lower the filter paper into a jar/glass/
1618beaker/test tube containing water so
1619that the drop of ink on the paper is just
1620above the water level, as shown in Fig.
16212.8(b) and leave it undisturbed.
1622• Watch carefully, as the water rises up
1623on the filter paper. Record your
1624observations.
1625This process of separation of components
1626of a mixture is known as chromatography.
1627Kroma in Greek means colour. This technique
1628was first used for separation of colours, so
1629this name was given. Chromatography is the
1630technique used for separation of those solutes
1631that dissolve in the same solvent.
1632With the advancement in technology,
1633newer techniques of chromatography have
1634been developed. You will study about
1635chromatography in higher classes.
1636Applications
1637To separate
1638• colours in a dye
1639• pigments from natural colours
1640• drugs from blood.
16412.3.6 HOW CAN WE SEPARATE A MIXTURE
1642OF TWO MISCIBLE LIQUIDS?
1643Activity ______________ 2.8
1644• Let us try to separate acetone and water
1645from their mixture.
1646• Take the mixture in a distillation flask.
1647Fit it with a thermometer.
1648• Arrange the apparatus as shown in
1649Fig. 2.9.
1650• Heat the mixture slowly keeping a close
1651watch at the thermometer.
1652• The acetone vaporises, condenses in
1653the condenser and can be collected
1654from the condenser outlet.
1655• Water is left behind in the distillation
1656flask.
1657Fig. 2.8: Separation of dyes in black ink using
1658chromatography
1659Now answer
1660• What do you observe on the filter paper
1661as the water rises on it?
1662• Do you obtain different colours on the
1663filter paper strip?
1664• What according to you, can be the
1665reason for the rise of the coloured spot
1666on the paper strip?
1667The ink that we use has water as the
1668solvent and the dye is soluble in it. As the
1669water rises on the filter paper it takes along
1670with it the dye particles. Usually, a dye is a
1671mixture of two or more colours. The coloured
1672component that is more soluble in water, rises
1673faster and in this way the colours get
1674separated. Fig.2.9: Separation of two miscible liquids by
1675distillation
1676© NCERT
1677not to be republished
167822 SCIENCE
1679Now answer
1680• What do you observe as you start
1681heating the mixture?
1682• At what temperature does the
1683thermometer reading become
1684constant for some time?
1685• What is the boiling point of acetone?
1686• Why do the two components separate?
1687This method is called distillation. It is used
1688for the separation of components of a mixture
1689containing two miscible liquids that boil
1690without decomposition and have sufficient
1691difference in their boiling points.
1692To separate a mixture of two or more
1693miscible liquids for which the difference in
1694boiling points is less than 25 K, fractional
1695distillation process is used, for example, for
1696the separation of different gases from air,
1697different factions from petroleum products
1698etc. The apparatus is similar to that for simple
1699distillation, except that a fractionating
1700column is fitted in between the distillation
1701flask and the condenser.
1702A simple fractionating column is a tube
1703packed with glass beads. The beads provide
1704surface for the vapours to cool and condense
1705repeatedly, as shown in Fig. 2.10.
17062.3.7 HOW CAN WE OBTAIN DIFFERENT
1707GASES FROM AIR ?
1708Air is a homogeneous mixture and can be
1709separated into its components by fractional
1710distillation. The flow diagram (Fig. 2.11)
1711shows the steps of the process.
1712Fig. 2.10: Fractional distillation
1713Fig. 2.11: Flow diagram shows the process of
1714obtaining gases from air
1715If we want oxygen gas from air (Fig. 2.12),
1716we have to separate out all the other gases
1717present in the air. The air is compressed by
1718increasing the pressure and is then cooled
1719by decreasing the temperature to get liquid
1720air. This liquid air is allowed to warm-up
1721slowly in a fractional distillation column,
1722where gases get separated at different heights
1723depending upon their boiling points.
1724Answer the following:
1725• Arrange the gases present in air in
1726increasing order of their boiling points.
1727• Which gas forms the liquid first as the
1728air is cooled?
1729© NCERT
1730not to be republished
1731IS MATTER AROUND US P URE 23
1732it. To remove these impurities, the process of
1733crystallisation is used. Crystallisation is a
1734process that separates a pure solid in the form
1735of its crystals from a solution. Crystallisation
1736technique is better than simple evaporation
1737technique as –
1738• some solids decompose or some, like
1739sugar, may get charred on heating to
1740dryness.
1741• some impurities may remain dissolved
1742in the solution even after filtration. On
1743evaporation these contaminate the
1744solid.
1745Applications
1746• Purification of salt that we get from sea
1747water.
1748• Separation of crystals of alum (phitkari)
1749from impure samples.
1750Thus, by choosing one of the above
1751methods according to the nature of the
1752components of a mixture, we get a pure
1753substance. With advancements in technology
1754many more methods of separation techniques
1755have been devised.
1756In cities, drinking water is supplied from
1757water works. A flow diagram of a typical water
1758works is shown in Fig. 2.13. From this figure
1759write down the processes involved to get the
1760supply of drinking water to your home from
1761the water works and discuss it in your class.
1762Fig. 2.12: Separation of components of air
17632.3.8 HOW CAN WE OBTAIN PURE COPPER
1764SULPHATE FROM AN IMPURE SAMPLE?
1765Activity ______________ 2.9
1766• Take some (approximately 5 g) impure
1767sample of copper sulphate in a china
1768dish.
1769• Dissolve it in minimum amount of
1770water.
1771• Filter the impurities out.
1772• Evaporate water from the copper
1773sulphate solution so as to get a
1774saturated solution.
1775• Cover the solution with a filter paper
1776and leave it undisturbed at room
1777temperature to cool slowly for a day.
1778• You will obtain the crystals of copper
1779sulphate in the china dish.
1780• This process is called crystallisation.
1781Now answer
1782• What do you observe in the china dish?
1783• Do the crystals look alike?
1784• How will you separate the crystals from
1785the liquid in the china dish?
1786The crystallisation method is used to
1787purify solids. For example, the salt we get
1788from sea water can have many impurities in
1789© NCERT
1790not to be republished
179124 SCIENCE
1792uestions
17931. How will you separate a mixture
1794containing kerosene and petrol
1795(difference in their boiling points
1796is more than 25ºC), which are
1797miscible with each other?
17982. Name the technique to separate
1799(i) butter from curd,
1800(ii) salt from sea-water,
1801(iii) camphor from salt.
18023. What type of mixtures are
1803separated by the technique of
1804crystallisation?
18052.4 Physical and Chemical Physical and Chemical
1806Changes Changes
1807To understand the difference between a pure
1808substance and a mixture, let us understand
1809the difference between a physical and a
1810chemical change.
1811In the previous chapter, we have learnt
1812about a few physical properties of matter. The
1813properties that can be observed and specified
1814like colour, hardness, rigidity, fluidity,
1815density, melting point, boiling point etc. are
1816the physical properites.
1817The interconversion of states is a physical
1818change because these changes occur without
1819a change in composition and no change in
1820the chemical nature of the substance.
1821Although ice, water and water vapour all look
1822different and display different physical
1823properties, they are chemically the same.
1824Q
1825Fig. 2.13: Water purification system in water works
1826Both water and cooking oil are liquid but
1827their chemical characteristics are different.
1828They differ in odour and inflammability. We
1829know that oil burns in air whereas water
1830extinguishes fire. It is this chemical property
1831of oil that makes it different from water.
1832Burning is a chemical change. During this
1833process one substance reacts with another
1834to undergo a change in chemical composition.
1835Chemical change brings change in the
1836chemical properties of matter and we get new
1837substances. A chemical change is also called
1838a chemical reaction.
1839During burning of a candle, both physical
1840and chemical changes take place. Can you
1841distinguish these?
1842uestions
18431. Classify the following as
1844chemical or physical changes:
1845• cutting of trees,
1846• melting of butter in a pan,
1847• rusting of almirah,
1848• boiling of water to form steam,
1849• passing of electric current,
1850through water and the water
1851breaking down into hydrogen
1852and oxygen gases,
1853• dissolving common salt in
1854water,
1855• making a fruit salad with raw
1856fruits, and
1857• burning of paper and wood.
18582. Try segregating the things
1859around you as pure substances
1860or mixtures.
1861Q
1862© NCERT
1863not to be republished
1864IS MATTER AROUND US P URE 25
18652.5 What are the Types of Pure What are the Types of Pure
1866Substances? Substances?
1867On the basis of their chemical composition,
1868substances can be classified either as
1869elements or compounds.
18702.5.1 ELEMENTS
1871Robert Boyle was the first scientist to use the
1872term element in 1661. Antoine Laurent
1873Lavoisier (1743-94), a French chemist, was
1874the first to establish an experimentally useful
1875definition of an element. He defined an
1876element as a basic form of matter that cannot
1877be broken down into simpler substances by
1878chemical reactions.
1879Elements can be normally divided into
1880metals, non-metals and metalloids.
1881Metals usually show some or all of the
1882following properties:
1883• They have a lustre (shine).
1884• They have silvery-grey or golden-yellow
1885colour.
1886• They conduct heat and electricity.
1887• They are ductile (can be drawn into
1888wires).
1889• They are malleable (can be hammered
1890into thin sheets).
1891• They are sonorous (make a ringing
1892sound when hit).
1893Examples of metals are gold, silver, copper,
1894iron, sodium, potassium etc. Mercury is the
1895only metal that is liquid at room temperature.
1896Non-metals usually show some or all of the
1897following properties:
1898• They display a variety of colours.
1899• They are poor conductors of heat and
1900electricity.
1901• They are not lustrous, sonorous or
1902malleable.
1903Examples of non-metals are hydrogen,
1904oxygen, iodine, carbon (coal, coke),
1905bromine, chlorine etc. Some elements have
1906intermediate properties between those of
1907metals and non-metals, they are
1908called metalloids; examples are boron,
1909silicon, germanium etc.
1910More to know
1911• The number of elements known at
1912present are more than 100.
1913Ninety-two elements are naturally
1914occurring and the rest are manmade.
1915• Majority of the elements are solid.
1916• Eleven elements are in gaseous
1917state at room temperature.
1918• Two elements are liquid at room
1919temperature–mercury and
1920bromine.
1921• Elements, gallium and cesium
1922become liquid at a temperature
1923slightly above room temperature
1924(303 K).
19252.5.2 COMPOUNDS
1926A compound is a substance composed of two
1927or more elements, chemically combined with
1928one another in a fixed proportion.
1929What do we get when two or more elements
1930are combined?
1931Activity _____________ 2.10
1932• Divide the class into two groups. Give
19335 g of iron filings and 3 g of sulphur
1934powder in a china dish to both the
1935groups.
1936Group I
1937• Mix and crush iron filings and sulphur
1938powder.
1939Group II
1940• Mix and crush iron filings and sulphur
1941powder. Heat this mixture strongly till
1942red hot. Remove from flame and let the
1943mixture cool.
1944Groups I and II
1945• Check for magnetism in the material
1946obtained. Bring a magnet near the
1947material and check if the material is
1948attracted towards the magnet.
1949• Compare the texture and colour of the
1950material obtained by the groups.
1951• Add carbon disulphide to one part of
1952the material obtained. Stir well and
1953filter.
1954• Add dilute sulphuric acid or dilute
1955hydrochloric acid to the other part of
1956© NCERT
1957not to be republished
195826 SCIENCE
1959the material obtained.(Note: teacher
1960supervision is necessary for this
1961activity).
1962• Perform all the above steps with both
1963the elements (iron and sulphur)
1964separately.
1965Now answer
1966• Did the material obtained by the two
1967groups look the same?
1968• Which group has obtained a material
1969with magnetic properties?
1970• Can we separate the components of the
1971material obtained?
1972• On adding dilute sulphuric acid or
1973dilute hydrochloric acid, did both the
1974groups obtain a gas? Did the gas in
1975both the cases smell the same or
1976different?
1977The gas obtained by Group I is hydrogen,
1978it is colourless, odourless and combustible–
1979it is not advised to do the combustion test for
1980hydrogen in the class. The gas obtained by
1981Group II is hydrogen sulphide. It is a colourless
1982gas with the smell of rotten eggs.
1983You must have observed that the products
1984obtained by both the groups show different
1985properties, though the starting materials were
1986the same. Group I has carried out the activity
1987involving a physical change whereas in case
1988of Group II, a chemical change (a chemical
1989reaction) has taken place.
1990• The material obtained by group I is a
1991mixture of the two substances. The
1992substances given are the elements– iron
1993and sulphur.
1994• The properties of the mixture are the
1995same as that of its constituents.
1996• The material obtained by group II is a
1997compound.
1998• On heating the two elements strongly we
1999get a compound, which has totally
2000different properties compared to the
2001combining elements.
2002• The composition of a compound is the
2003same throughout. We can also observe
2004that the texture and the colour of the
2005compound are the same throughout.
2006Thus, we can summarise the physical
2007and chemical nature of matter in the
2008following graphical organiser :
2009Table 2.2: Mixtures and Compounds
2010Mixtures Compounds
20111. Elements or compounds just mix 1. Elements react to form new compounds.
2012together to form a mixture and no
2013new compound is formed.
20142. A mixture has a variable composition. 2. The composition of each new substance
2015is always fixed.
20163, A mixture shows the properties of the 3. The new substance has totally different
2017constituent substances. properties.
20184. The constituents can be seperated 4. The constituents can be separated only
2019fairly easily by physical methods. by chemical or electrochemical
2020reactions.
2021© NCERT
2022not to be republished
2023IS MATTER AROUND US P URE 27
2024What
2025you have you have
2026learnt
2027• A mixture contains more than one substance (element and/or
2028compound) mixed in any proportion.
2029• Mixtures can be separated into pure substances using
2030appropriate separation techniques.
2031• A solution is a homogeneous mixture of two or more substances.
2032The major component of a solution is called the solvent, and
2033the minor, the solute.
2034• The concentration of a solution is the amount of solute present
2035per unit volume or per unit mass of the solution/solvent.
2036• Materials that are insoluble in a solvent and have particles
2037that are visible to naked eyes, form a suspension. A suspension
2038is a heterogeneous mixture.
2039• Colloids are heterogeneous mixtures in which the particle size
2040is too small to be seen with the naked eye, but is big enough to
2041scatter light. Colloids are useful in industry and daily life. The
2042particles are called the dispersed phase and the medium in
2043which they are distributed is called the dispersion medium.
2044• Pure substances can be elements or compounds. An element
2045is a form of matter that cannot be broken down by chemical
2046reactions into simpler substances. A compound is a substance
2047composed of two or more different types of elements, chemically
2048combined in a fixed proportion.
2049• Properties of a compound are different from its constituent
2050elements, whereas a mixture shows the properties of its
2051constituting elements or compounds.
2052© NCERT
2053not to be republished
205428 SCIENCE
2055Exercises Exercises Exercises
20561. Which separation techniques will you apply for the separation
2057of the following?
2058(a) Sodium chloride from its solution in water.
2059(b) Ammonium chloride from a mixture containing sodium
2060chloride and ammonium chloride.
2061(c) Small pieces of metal in the engine oil of a car.
2062(d) Different pigments from an extract of flower petals.
2063(e) Butter from curd.
2064(f) Oil from water.
2065(g) Tea leaves from tea.
2066(h) Iron pins from sand.
2067(i) Wheat grains from husk.
2068(j) Fine mud particles suspended in water.
20692. Write the steps you would use for making tea. Use the words
2070solution, solvent, solute, dissolve, soluble, insoluble, filtrate
2071and residue.
20723. Pragya tested the solubility of three different substances at
2073different temperatures and collected the data as given below
2074(results are given in the following table, as grams of substance
2075dissolved in 100 grams of water to form a saturated solution).
2076(a) What mass of potassium nitrate would be needed to
2077produce a saturated solution of potassium nitrate in
207850 grams of water at 313 K?
2079(b) Pragya makes a saturated solution of potassium chloride
2080in water at 353 K and leaves the solution to cool at room
2081temperature. What would she observe as the solution
2082cools? Explain.
2083(c) Find the solubility of each salt at 293 K. Which salt has
2084the highest solubility at this temperature?
2085(d) What is the effect of change of temperature on the
2086solubility of a salt?
2087Temperature in K
2088283 293 313 333 353
2089 Solubility
2090Potassium nitrate 21 32 62 106 167
2091Sodium chloride 36 36 36 37 37
2092Potassium chloride 35 35 40 46 54
2093Ammonium chloride 24 37 41 55 66
2094Substance Dissolved
2095© NCERT
2096not to be republished
2097IS MATTER AROUND US P URE 29
20984. Explain the following giving examples.
2099(a) saturated solution
2100(b) pure substance
2101(c) colloid
2102(d) suspension
21035. Classify each of the following as a homogeneous or
2104heterogeneous mixture.
2105soda water, wood, air, soil, vinegar, filtered tea.
21066. How would you confirm that a colourless liquid given to you is
2107pure water?
21087. Which of the following materials fall in the category of a “pure
2109substance”?
2110(a) Ice
2111(b) Milk
2112(c) Iron
2113(d) Hydrochloric acid
2114(e) Calcium oxide
2115(f) Mercury
2116(g) Brick
2117(h) Wood
2118(i) Air.
21198. Identify the solutions among the following mixtures.
2120(a) Soil
2121(b) Sea water
2122(c) Air
2123(d) Coal
2124(e) Soda water.
21259. Which of the following will show “Tyndall effect”?
2126(a) Salt solution
2127(b) Milk
2128(c) Copper sulphate solution
2129(d) Starch solution.
213010. Classify the following into elements, compounds and
2131mixtures.
2132(a) Sodium
2133(b) Soil
2134(c) Sugar solution
2135(d) Silver
2136(e) Calcium carbonate
2137(f) Tin
2138(g) Silicon
2139© NCERT
2140not to be republished
214130 SCIENCE
2142(h) Coal
2143(i) Air
2144(j) Soap
2145(k) Methane
2146(l) Carbon dioxide
2147(m) Blood
214811. Which of the following are chemical changes?
2149(a) Growth of a plant
2150(b) Rusting of iron
2151(c) Mixing of iron filings and sand
2152(d) Cooking of food
2153(e) Digestion of food
2154(f) Freezing of water
2155(g) Burning of a candle.
2156Group Activity Group Activity
2157Take an earthen pot (mutka), some pebbles and sand. Design a
2158small-scale filtration plant that you could use to clean muddy
2159water.
2160© NCERT
2161not to be republished
2162Ancient Indian and Greek philosophers have
2163always wondered about the unknown and
2164unseen form of matter. The idea of divisibility
2165of matter was considered long back in India,
2166around 500 BC. An Indian philosopher
2167Maharishi Kanad, postulated that if we go on
2168dividing matter (padarth), we shall get smaller
2169and smaller particles. Ultimately, a time will
2170come when we shall come across the smallest
2171particles beyond which further division will
2172not be possible. He named these particles
2173Parmanu. Another Indian philosopher,
2174Pakudha Katyayama, elaborated this doctrine
2175and said that these particles normally exist
2176in a combined form which gives us various
2177forms of matter.
2178Around the same era, ancient Greek
2179philosophers – Democritus and Leucippus
2180suggested that if we go on dividing matter, a
2181stage will come when particles obtained
2182cannot be divided further. Democritus called
2183these indivisible particles atoms (meaning
2184indivisible). All this was based on
2185philosophical considerations and not much
2186experimental work to validate these ideas
2187could be done till the eighteenth century.
2188By the end of the eighteenth century,
2189scientists recognised the difference between
2190elements and compounds and naturally
2191became interested in finding out how and why
2192elements combine and what happens when
2193they combine.
2194Antoine L. Lavoisier laid the foundation
2195of chemical sciences by establishing two
2196important laws of chemical combination.
21973.1 Laws of Chemical Combination
2198The following two laws of chemical
2199combination were established after much
2200experimentations by Lavoisier and Joseph
2201L. Proust.
22023.1.1 LAW OF CONSERVATION OF MASS
2203Is there a change in mass when a chemical
2204change (chemical reaction) takes place?
2205Activity ______________ 3.1
2206• Take one of the following sets, X and Y
2207of chemicals–
2208X Y
2209(i) copper sulphate sodium carbonate
2210(ii) barium chloride sodium sulphate
2211(iii) lead nitrate sodium chloride
2212• Prepare separately a 5% solution of any
2213one pair of substances listed under X
2214and Y in water.
2215• Take a little amount of solution of Y in
2216a conical flask and some solution of X
2217in an ignition tube.
2218• Hang the ignition tube in the flask
2219carefully; see that the solutions do not
2220get mixed. Put a cork on the flask
2221(see Fig. 3.1).
2222Fig. 3.1: Ignition tube containing solution of X, dipped
2223in a conical flask containing solution of Y.
22243
2225ATOMS AND MOLECULES
2226Chapter
222732 SCIENCE
2228• Weigh the flask with its contents
2229carefully.
2230• Now tilt and swirl the flask, so that the
2231solutions X and Y get mixed.
2232• Weigh again.
2233• What happens in the reaction flask?
2234• Do you think that a chemical reaction
2235has taken place?
2236• Why should we put a cork on the mouth
2237of the flask?
2238• Does the mass of the flask and its
2239contents change?
2240Law of conservation of mass states that
2241mass can neither be created nor destroyed in
2242a chemical reaction.
22433.1.2 LAW OF CONSTANT PROPORTIONS
2244Lavoisier, along with other scientists, noted
2245that many compounds were composed of two
2246or more elements and each such compound
2247had the same elements in the same
2248proportions, irrespective of where the
2249compound came from or who prepared it.
2250In a compound such as water, the ratio of
2251the mass of hydrogen to the mass of oxygen
2252is always 1:8, whatever the source of water.
2253Thus, if 9 g of water is decomposed, 1 g of
2254hydrogen and 8 g of oxygen are always
2255obtained. Similarly in ammonia, nitrogen and
2256hydrogen are always present in the ratio 14:3
2257by mass, whatever the method or the source
2258from which it is obtained.
2259This led to the law of constant proportions
2260which is also known as the law of definite
2261proportions. This law was stated by Proust
2262as “In a chemical substance the elements are
2263always present in definite proportions by
2264mass”.
2265The next problem faced by scientists was
2266to give appropriate explanations of these laws.
2267British chemist John Dalton provided the
2268basic theory about the nature of matter.
2269Dalton picked up the idea of divisibility of
2270matter, which was till then just a philosophy.
2271He took the name ‘atoms’ as given by the
2272Greeks and said that the smallest particles of
2273matter are atoms. His theory was based on
2274the laws of chemical combination. Dalton’s
2275atomic theory provided an explanation for the
2276law of conservation of mass and the law of
2277definite proportions.
2278John Dalton was born in
2279a poor weaver’s family in
22801766 in England. He
2281began his career as a
2282teacher at the age of
2283twelve. Seven years later
2284he became a school
2285principal. In 1793, Dalton
2286left for Manchester to
2287teach mathematics,
2288physics and chemistry in
2289a college. He spent most of his life there
2290teaching and researching. In 1808, he
2291presented his atomic theory which was a
2292turning point in the study of matter.
2293According to Dalton’s atomic theory, all
2294matter, whether an element, a compound or
2295a mixture is composed of small particles called
2296atoms. The postulates of this theory may be
2297stated as follows:
2298(i) All matter is made of very tiny particles
2299called atoms.
2300(ii) Atoms are indivisible particles, which
2301cannot be created or destroyed in a
2302chemical reaction.
2303(iii) Atoms of a given element are identical
2304in mass and chemical properties.
2305(iv) Atoms of different elements have
2306different masses and chemical
2307properties.
2308(v) Atoms combine in the ratio of small
2309whole numbers to form compounds.
2310(vi) The relative number and kinds of
2311atoms are constant in a given
2312compound.
2313You will study in the next chapter that all
2314atoms are made up of still smaller particles.
2315 uestions
23161. In a reaction, 5.3 g of sodium
2317carbonate reacted with 6 g of
2318ethanoic acid. The products were
23192.2 g of carbon dioxide, 0.9 g
2320water and 8.2 g of sodium Q ethanoate. Show that these
2321John Dalton
2322ATOMS AND MOLECULES 33
2323We might think that if atoms are so
2324insignificant in size, why should we care about
2325them? This is because our entire world is
2326made up of atoms. We may not be able to see
2327them, but they are there, and constantly
2328affecting whatever we do. Through modern
2329techniques, we can now produce magnified
2330images of surfaces of elements showing atoms.
2331observations are in agreement
2332with the law of conservation of
2333mass.
2334sodium carbonate + ethanoic acid
2335→ sodium ethanoate + carbon
2336dioxide + water
23372. Hydrogen and oxygen combine in
2338the ratio of 1:8 by mass to form
2339water. What mass of oxygen gas
2340would be required to react
2341completely with 3 g of hydrogen
2342gas?
23433. Which postulate of Dalton’s
2344atomic theory is the result of the
2345law of conservation of mass?
23464. Which postulate of Dalton’s
2347atomic theory can explain the law
2348of definite proportions?
23493.2 What is an Atom?
2350Have you ever observed a mason building
2351walls, from these walls a room and then a
2352collection of rooms to form a building? What
2353is the building block of the huge building?
2354What about the building block of an ant-hill?
2355It is a small grain of sand. Similarly, the
2356building blocks of all matter are atoms.
2357How big are atoms?
2358Atoms are very small, they are smaller than
2359anything that we can imagine or compare
2360with. More than millions of atoms when
2361stacked would make a layer barely as thick
2362as this sheet of paper.
2363Atomic radius is measured in nanometres.
23641/10 9 m = 1 nm
23651 m = 109
2366 nm
2367Relative Sizes
2368Radii (in m) Example
236910–10 Atom of hydrogen
237010–9 Molecule of water
237110–8 Molecule of haemoglobin
237210–4 Grain of sand
237310–2 Ant
237410–1 Watermelon
2375Fig. 3.2: An image of the surface of silicon
23763.2.1 WHAT ARE THE MODERN DAY
2377SYMBOLS OF ATOMS OF DIFFERENT
2378ELEMENTS?
2379Dalton was the first scientist to use the
2380symbols for elements in a very specific sense.
2381When he used a symbol for an element he
2382also meant a definite quantity of that element,
2383that is, one atom of that element. Berzilius
2384suggested that the symbols of elements be
2385made from one or two letters of the name of
2386the element.
2387Fig. 3.3: Symbols for some elements as proposed by
2388Dalton
238934 SCIENCE
2390In the beginning, the names of elements
2391were derived from the name of the place where
2392they were found for the first time. For example,
2393the name copper was taken from Cyprus.
2394Some names were taken from specific colours.
2395For example, gold was taken from the English
2396word meaning yellow. Now-a-days, IUPAC
2397(International Union of Pure and Applied
2398Chemistry) approves names of elements. Many
2399of the symbols are the first one or two letters
2400of the element’s name in English. The first
2401letter of a symbol is always written as a capital
2402letter (uppercase) and the second letter as a
2403small letter (lowercase).
2404For example
2405(i) hydrogen, H
2406(ii) aluminium, Al and not AL
2407(iii) cobalt, Co and not CO.
2408Symbols of some elements are formed
2409from the first letter of the name and a letter,
2410appearing later in the name. Examples are:
2411(i) chlorine, Cl, (ii) zinc, Zn etc.
2412Other symbols have been taken from the
2413names of elements in Latin, German or Greek.
2414For example, the symbol of iron is Fe from its
2415Latin name ferrum, sodium is Na from
2416natrium, potassium is K from kalium.
2417Therefore, each element has a name and a
2418unique chemical symbol.
2419Table 3.1: Symbols for some elements
2420Element Symbol Element Symbol Element Symbol
2421Aluminium Al Copper Cu Nitrogen N
2422Argon Ar Fluorine F Oxygen O
2423Barium Ba Gold Au Potassium K
2424Boron B Hydrogen H Silicon Si
2425Bromine Br Iodine I Silver Ag
2426Calcium Ca Iron Fe Sodium Na
2427Carbon C Lead Pb Sulphur S
2428Chlorine Cl Magnesium Mg Uranium U
2429Cobalt Co Neon Ne Zinc Zn
2430(The above table is given for you to refer
2431to whenever you study about elements. Do
2432not bother to memorise all in one go. With
2433the passage of time and repeated usage you
2434will automatically be able to reproduce the
2435symbols).
24363.2.2 ATOMIC MASS
2437The most remarkable concept that Dalton’s
2438atomic theory proposed was that of the atomic
2439mass. According to him, each element had a
2440characteristic atomic mass. The theory could
2441explain the law of constant proportions so well
2442that scientists were prompted to measure the
2443atomic mass of an atom. Since determining
2444the mass of an individual atom was a relatively
2445difficult task, relative atomic masses were
2446determined using the laws of chemical
2447combinations and the compounds formed.
2448Let us take the example of a compound,
2449carbon monoxide (CO) formed by carbon and
2450oxygen. It was observed experimentally that
24513 g of carbon combines with 4 g of oxygen to
2452form CO. In other words, carbon combines
2453with 4/3 times its mass of oxygen. Suppose
2454we define the atomic mass unit (earlier
2455abbreviated as ‘amu’, but according to the
2456latest IUPAC recommendations, it is now
2457written as ‘u’ – unified mass) as equal to the
2458mass of one carbon atom, then we would
2459assign carbon an atomic mass of 1.0 u and
2460oxygen an atomic mass of 1.33 u. However, it
2461is more convenient to have these numbers as
2462whole numbers or as near to a whole numbers
2463ATOMS AND MOLECULES 35
2464as possible. While searching for various
2465atomic mass units, scientists initially took 1/
246616 of the mass of an atom of naturally
2467occurring oxygen as the unit. This was
2468considered relevant due to two reasons:
2469• oxygen reacted with a large number of
2470elements and formed compounds.
2471• this atomic mass unit gave masses of
2472most of the elements as whole numbers.
2473However, in 1961 for a universally
2474accepted atomic mass unit, carbon-12 isotope
2475was chosen as the standard reference for
2476measuring atomic masses. One atomic mass
2477unit is a mass unit equal to exactly one-twelfth
2478(1/12th) the mass of one atom of carbon-12.
2479The relative atomic masses of all elements
2480have been found with respect to an atom of
2481carbon-12.
2482Imagine a fruit seller selling fruits without
2483any standard weight with him. He takes a
2484watermelon and says, “this has a mass equal
2485to 12 units” (12 watermelon units or 12 fruit
2486mass units). He makes twelve equal pieces of
2487the watermelon and finds the mass of each
2488fruit he is selling, relative to the mass of one
2489piece of the watermelon. Now he sells his fruits
2490by relative fruit mass unit (fmu), as in Fig.
24913.4.
2492Fig. 3.4 : (a) Watermelon, (b) 12 pieces, (c) 1/12 of
2493watermelon, (d) how the fruit seller can
2494weigh the fruits using pieces of
2495watermelon
2496Similarly, the relative atomic mass of the
2497atom of an element is defined as the average
2498mass of the atom, as compared to 1/12th the
2499mass of one carbon-12 atom.
2500Table 3.2: Atomic masses of
2501a few elements
2502Element Atomic Mass (u)
2503Hydrogen 1
2504Carbon 12
2505Nitrogen 14
2506Oxygen 16
2507Sodium 23
2508Magnesium 24
2509Sulphur 32
2510Chlorine 35.5
2511Calcium 40
25123.2.3 HOW DO ATOMS EXIST?
2513Atoms of most elements are not able to exist
2514independently. Atoms form molecules and
2515ions. These molecules or ions aggregate in
2516large numbers to form the matter that we can
2517see, feel or touch.
2518uestions
25191. Define the atomic mass unit.
25202. Why is it not possible to see an
2521atom with naked eyes?
25223.3 What is a Molecule?
2523A molecule is in general a group of two or
2524more atoms that are chemically bonded
2525together, that is, tightly held together by
2526attractive forces. A molecule can be defined
2527as the smallest particle of an element or a
2528compound that is capable of an independent
2529existence and shows all the properties of that
2530substance. Atoms of the same element or of
2531different elements can join together to form
2532molecules.
2533Q
253436 SCIENCE
25353.3.1 MOLECULES OF ELEMENTS
2536The molecules of an element are constituted
2537by the same type of atoms. Molecules of many
2538elements, such as argon (Ar), helium (He) etc.
2539are made up of only one atom of that element.
2540But this is not the case with most of the nonmetals.
2541For example, a molecule of oxygen
2542consists of two atoms of oxygen and hence it
2543is known as a diatomic molecule, O2
2544. If 3
2545atoms of oxygen unite into a molecule, instead
2546of the usual 2, we get ozone. The number of
2547atoms constituting a molecule is known as
2548its atomicity.
2549Metals and some other elements, such as
2550carbon, do not have a simple structure but
2551consist of a very large and indefinite number
2552of atoms bonded together.
2553Let us look at the atomicity of some
2554non-metals.
2555Table 3.3 : Atomicity of some
2556elements
2557Type of Name Atomicity
2558Element
2559Non-Metal Argon Monoatomic
2560Helium Monoatomic
2561Oxygen Diatomic
2562Hydrogen Diatomic
2563Nitrogen Diatomic
2564Chlorine Diatomic
2565Phosphorus Tetra-atomic
2566Sulphur Poly-atomic
2567Table 3.4 : Molecules of some
2568compounds
2569Compound Combining Ratio
2570Elements by
2571Mass
2572Water Hydrogen, Oxygen 1:8
2573Ammonia Nitrogen, Hydrogen 14:3
2574Carbon
2575dioxide Carbon, Oxygen 3:8
2576Activity ______________ 3.2
2577• Refer to Table 3.4 for ratio by mass of
2578atoms present in molecules and Table
25793.2 for atomic masses of elements. Find
2580the ratio by number of the atoms of
2581elements in the molecules of
2582compounds given in Table 3.4.
2583• The ratio by number of atoms for a
2584water molecule can be found as follows:
2585Element Ratio Atomic Mass Simplest
2586by mass ratio/ ratio
2587mass (u) atomic
2588mass
2589H 1 1
25901
25911
2592=1 2
2593O 8 16
25948
259516
2596=
25971
25982
25991
2600• Thus, the ratio by number of atoms for
2601water is H:O = 2:1.
26023.3.3 WHAT IS AN ION?
2603Compounds composed of metals and nonmetals
2604contain charged species. The charged
2605species are known as ions. An ion is a charged
2606particle and can be negatively or positively
2607charged. A negatively charged ion is called an
2608‘anion’ and the positively charged ion, a
2609‘cation’. Take, for example, sodium chloride
2610(NaCl). Its constituent particles are positively
2611charged sodium ions (Na+
2612) and negatively
2613charged chloride ions (Cl–
2614). Ions may consist
2615of a single charged atom or a group of atoms
26163.3.2 MOLECULES OF COMPOUNDS
2617Atoms of different elements join together in
2618definite proportions to form molecules of
2619compounds. Few examples are given in Table
26203.4.
2621ATOMS AND MOLECULES 37
2622that have a net charge on them. A group of
2623atoms carrying a charge is known as a
2624polyatomic ion (Table 3.6). We shall learn more
2625about the formation of ions in Chapter 4.
2626exercise, we need to learn the symbols and
2627combining capacity of the elements.
2628The combining power (or capacity) of an
2629element is known as its valency. Valency can
2630be used to find out how the atoms of an
2631element will combine with the atom(s) of
2632another element to for m a chemical
2633compound. The valency of the atom of an
2634element can be thought of as hands or arms
2635of that atom.
2636Human beings have two arms and an
2637octopus has eight. If one octopus has to catch
2638hold of a few people in such a manner that all
2639the eight arms of the octopus and both arms
2640of all the humans are locked, how many
2641humans do you think the octopus can hold?
2642Represent the octopus with O and humans
2643with H. Can you write a formula for this
2644combination? Do you get OH4 as the formula?
2645The subscript 4 indicates the number of
2646humans held by the octopus.
2647The valencies of some common ions are
2648given in Table 3.6. We will learn more about
2649valency in the next chapter.
2650* Some elements show more than one valency. A Roman numeral shows their valency in a bracket.
2651Table 3.5: Some ionic compounds
2652Ionic Constituting Ratio
2653Compound Elements by
2654Mass
2655Calcium oxide Calcium and
2656oxygen 5:2
2657Magnesium Magnesium
2658sulphide and sulphur 3:4
2659Sodium Sodium
2660chloride and chlorine 23:35.5
26613.4 Writing Chemical Formulae
2662The chemical formula of a compound is a
2663symbolic representation of its composition.
2664The chemical formulae of different
2665compounds can be written easily. For this
2666Table 3.6: Names and symbols of some ions
2667Vale- Name of Symbol Non- Symbol Polyatomic Symbol
2668ncy ion metallic ions
2669element
26701. Sodium Na+ Hydrogen H+ Ammonium NH4
2671+
2672Potassium K+ Hydride H- Hydroxide OH–
2673Silver Ag+ Chloride Cl- Nitrate NO3
2674–
2675Copper (I)* Cu+ Bromide Br- Hydrogen
2676Iodide I
2677– carbonate HCO3
2678–
26792. Magnesium Mg2+ Oxide O2- Carbonate CO3
26802–
2681Calcium Ca2+ Sulphide S
26822- Sulphite SO3
26832–
2684Zinc Zn2+ Sulphate SO4
26852–
2686Iron (II)* Fe2+
2687Copper (II)* Cu2+
26883. Aluminium Al3+ Nitride N
26893- Phosphate PO4
26903–
2691Iron (III)* Fe3+
269238 SCIENCE
26933. Formula of carbon tetrachloride
2694For magnesium chloride, we write the
2695symbol of cation (Mg2+) first followed by the
2696symbol of anion (Cl-
2697). Then their charges are
2698criss-crossed to get the formula.
26994. Formula of magnesium chloride
2700Formula : MgCl2
2701Thus, in magnesium chloride, there are
2702two chloride ions (Cl-
2703) for each magnesium
2704ion (Mg2+).The positive and negative charges
2705must balance each other and the overall
2706structure must be neutral. Note that in the
2707formula, the charges on the ions are not
2708indicated.
2709Some more examples
2710(a) Formula for aluminium oxide:
2711Formula : Al 2O3
2712(b) Formula for calcium oxide:
2713Here, the valencies of the two elements
2714are the same. You may arrive at the formula
2715Ca2O2
2716. But we simplify the formula as CaO.
2717The rules that you have to follow while writing
2718a chemical formula are as follows:
2719• the valencies or charges on the ion must
2720balance.
2721• when a compound consists of a metal and
2722a non-metal, the name or symbol of the
2723metal is written first. For example:
2724calcium oxide (CaO), sodium chloride
2725(NaCl), iron sulphide (FeS), copper oxide
2726(CuO) etc., where oxygen, chlorine,
2727sulphur are non-metals and are written
2728on the right, whereas calcium, sodium,
2729iron and copper are metals, and are
2730written on the left.
2731• in compounds formed with polyatomic ions,
2732the ion is enclosed in a bracket before writing
2733the number to indicate the ratio. In case
2734the number of polyatomic ion is one, the
2735bracket is not required. For example, NaOH.
27363.4.1 FORMULAE OF SIMPLE COMPOUNDS
2737The simplest compounds, which are made up
2738of two different elements are called binary
2739compounds. Valencies of some ions are given
2740in Table 3.6. You can use these to write
2741formulae for compounds.
2742While writing the chemical formulae for
2743compounds, we write the constituent elements
2744and their valencies as shown below. Then we
2745must crossover the valencies of the combining
2746atoms.
2747Examples
27481. Formula of hydrogen chloride
2749Formula of the compound would be HCl.
27502. Formula of hydrogen sulphide
2751ATOMS AND MOLECULES 39
2752following formulae:
2753(i) Al 2
2754(SO4
2755)
27563
2757(ii) CaCl2
2758(iii) K2
2759SO4
2760(iv) KNO3
2761(v) CaCO3
2762.
27633. What is meant by the term
2764chemical formula?
27654. How many atoms are present in a
2766(i) H2
2767S molecule and
2768(ii) PO4
27693– ion?
27703.5 Molecular Mass and Mole
2771Concept
27723.5.1 MOLECULAR MASS
2773In section 3.2.2 we discussed the concept of
2774atomic mass. This concept can be extended
2775to calculate molecular masses. The molecular
2776mass of a substance is the sum of the atomic
2777masses of all the atoms in a molecule of the
2778substance. It is therefore the relative mass of
2779a molecule expressed in atomic mass units (u).
2780Example 3.1 (a) Calculate the relative
2781molecular mass of water (H 2O).
2782(b) Calculate the molecular mass of
2783HNO3
2784.
2785Solution:
2786(a) Atomic mass of hydrogen = 1u,
2787oxygen = 16 u
2788So the molecular mass of water, which
2789contains two atoms of hydrogen and one
2790atom of oxygen is = 2 × 1+ 1×16
2791= 18 u
2792(b) The molecular mass of HNO3
2793 = the
2794atomic mass of H + the atomic mass of
2795N+ 3 × the atomic mass of O
2796= 1 + 14 + 48 = 63 u
27973.5.2 FORMULA UNIT MASS
2798The formula unit mass of a substance is a
2799sum of the atomic masses of all atoms in a
2800formula unit of a compound. Formula unit
2801mass is calculated in the same manner as we
2802calculate the molecular mass. The only
2803(c) Formula of sodium nitrate:
2804Formula : NaNO3
2805(d) Formula of calcium hydroxide:
2806Formula : Ca(OH)2
2807Note that the formula of calcium
2808hydroxide is Ca(OH)2
2809 and not CaOH2
2810. We use
2811brackets when we have two or more of the
2812same ions in the formula. Here, the bracket
2813around OH with a subscript 2 indicates that
2814there are two hydroxyl (OH) groups joined to
2815one calcium atom. In other words, there are
2816two atoms each of oxygen and hydrogen in
2817calcium hydroxide.
2818(e) Formula of sodium carbonate:
2819Formula : Na2
2820CO3
2821In the above example, brackets are not needed
2822if there is only one ion present.
2823(f) Formula of ammonium sulphate:
2824Formula : (NH4
2825)
28262SO4
2827uestions
28281. Write down the formulae of
2829(i) sodium oxide
2830(ii) aluminium chloride
2831(iii) sodium suphide
2832(iv) magnesium hydroxide
28332. Write down the names of Q compounds represented by the
283440 SCIENCE
2835difference is that we use the word formula
2836unit for those substances whose constituent
2837particles are ions. For example, sodium
2838chloride as discussed above, has a formula
2839unit NaCl. Its formula unit mass can be
2840calculated as–
28411 × 23 + 1 × 35.5 = 58.5 u
2842Example 3.2 Calculate the formula unit
2843mass of CaCl2
2844.
2845Solution:
2846Atomic mass of Ca
2847+ (2 × atomic mass of Cl)
2848= 40 + 2 × 35.5 = 40 + 71 = 111 u
2849uestions
28501. Calculate the molecular masses
2851of H2
2852, O2
2853, Cl2
2854, CO2
2855, CH4
2856, C2H6
2857,
2858C2H4
2859, NH3
2860, CH3
2861OH.
28622. Calculate the formula unit
2863masses of ZnO, Na2O, K2CO3
2864,
2865given atomic masses of Zn = 65 u,
2866Na = 23 u, K = 39 u, C = 12 u,
2867and O = 16 u.
28683.5.3 MOLE CONCEPT
2869Take an example of the reaction of hydrogen
2870and oxygen to form water:
28712H2
2872+ O2 → 2H2O.
2873The above reaction indicates that
2874(i) two molecules of hydrogen combine
2875with one molecule of oxygen to form
2876two molecules of water, or
2877(ii) 4 u of hydrogen molecules combine
2878with 32 u of oxygen molecules to form
287936 u of water molecules.
2880We can infer from the above equation that
2881the quantity of a substance can be
2882characterised by its mass or the number of
2883molecules. But, a chemical reaction equation
2884indicates directly the number of atoms or
2885molecules taking part in the reaction.
2886Therefore, it is more convenient to refer to the
2887quantity of a substance in terms of the
2888number of its molecules or atoms, rather than
2889their masses. So, a new unit “mole” was Q introduced. One mole of any species (atoms,
2890Fig. 3.5: Relationship between mole, Avogadro number and mass
2891ATOMS AND MOLECULES 41
2892molecules, ions or particles) is that quantity
2893in number having a mass equal to its atomic
2894or molecular mass in grams.
2895The number of particles (atoms, molecules
2896or ions) present in 1 mole of any substance is
2897fixed, with a value of 6.022 × 1023. This is an
2898experimentally obtained value. This number
2899is called the Avogadro Constant or Avogadro
2900Number (represented by N0
2901),named in honour
2902of the Italian scientist, Amedeo Avogadro.
29031 mole (of anything) = 6.022 × 1023 in number,
2904 as, 1 dozen = 12 nos.
29051 gross = 144 nos.
2906Besides being related to a number, a mole
2907has one more advantage over a dozen or a
2908gross. This advantage is that mass of 1 mole
2909of a particular substance is also fixed.
2910The mass of 1 mole of a substance is equal
2911to its relative atomic or molecular mass in
2912grams. The atomic mass of an element gives
2913us the mass of one atom of that element in
2914atomic mass units (u). To get the mass of
29151 mole of atom of that element, that is, molar
2916mass, we have to take the same numerical
2917value but change the units from ‘u’ to ‘g’. Molar
2918mass of atoms is also known as gram atomic
2919mass. For example, atomic mass of
2920hydrogen=1u. So, gram atomic mass of
2921hydrogen = 1 g.
29221 u hydrogen has only 1 atom of hydrogen
29231 g hydrogen has 1 mole atoms, that is,
29246.022 × 1023 atoms of hydrogen.
2925Similarly,
292616 u oxygen has only 1 atom of oxygen,
292716 g oxygen has 1 mole atoms, that is,
29286.022 × 1023 atoms of oxygen.
2929To find the gram molecular mass or molar
2930mass of a molecule, we keep the numerical
2931value the same as the molecular mass, but
2932simply change units as above from u to g. For
2933example, as we have already calculated,
2934molecular mass of water (H2O) is 18 u. From
2935here we understand that
293618 u water has only 1 molecule of water,
293718 g water has 1 mole molecules of water,
2938that is, 6.022 × 1023 molecules of water.
2939Chemists need the number of atoms and
2940molecules while carrying out reactions, and
2941for this they need to relate the mass in grams
2942to the number. It is done as follows:
29431 mole = 6.022 × 1023 number
2944= Relative mass in grams.
2945Thus, a mole is the chemist’s counting unit.
2946The word “mole” was introduced around
29471896 by Wilhelm Ostwald who derived the
2948term from the Latin word moles meaning a
2949‘heap’ or ‘pile’. A substance may be considered
2950as a heap of atoms or molecules. The unit
2951mole was accepted in 1967 to provide a simple
2952way of reporting a large number– the massive
2953heap of atoms and molecules in a sample.
2954Example 3.3
29551. Calculate the number of moles for the
2956following:
2957(i) 52 g of He (finding mole from
2958mass)
2959(ii) 12.044 × 1023 number of He atoms
2960(finding mole from number of
2961particles).
2962Solutions:
2963No. of moles = n
2964Given mass = m
2965Molar mass = M
2966Given number of particles = N
2967Avogadro number of particles = N0
2968(i) Atomic mass of He = 4 u
2969Molar mass of He = 4g
2970Thus, the number of moles
2971=
2972givenmass
2973molar mass
2974m 52
2975n 13
2976M 4
2977⇒ = = =
2978(ii) we know,
29791 mole = 6.022 × 1023
2980The number of moles
2981given number of particles
2982Avogadronumber
2983=
2984
298523
298623
2987o
2988N 12.044 10
2989n 2
2990N 6.022 10
2991× ⇒ = = =
2992×
299342 SCIENCE
2994Example 3.4 Calculate the mass of the
2995following:
2996(i) 0.5 mole of N2 gas (mass from mole
2997of molecule)
2998(ii) 0.5 mole of N atoms (mass from
2999mole of atom)
3000(iii) 3.011 × 1023 number of N atoms
3001(mass from number)
3002(iv) 6.022 × 1023 number of N2
3003molecules (mass from number)
3004Solutions:
3005(i) mass = molar mass × number of
3006moles
3007⇒ = × = × = m M n 28 0.5 14g
3008(ii) mass = molar mass × number of
3009moles
3010⇒ m = M × n = 14 × 0.5 = 7 g
3011(iii) The number of moles, n
30120
3013givennumber of particles N
3014Avogadro number N
3015= =
3016
301723
301823
30193.011 10
30206.022 10
3021×
3022=
3023×
3024
302523
302623
30273.011 10
3028m M n 14
30296.022 10
3030× ⇒ = × = ×
3031×
3032 = × = 14 0.5 7 g
3033(iv)
30340
3035N
3036n
3037N
3038=
303923
304023
30410
3042N 6.022 10 m M 28
3043N 6.022 10
3044× ⇒ = × = ×
3045×
3046 = × = 28 1 28 g
3047Example 3.5 Calculate the number of
3048particles in each of the
3049following:
3050(i) 46 g of Na atoms (number from
3051mass)
3052(ii) 8 g O2
3053 molecules (number of
3054molecules from mass)
3055(iii) 0.1 mole of carbon atoms (number
3056from given moles)
3057Solutions:
3058(i) The number of atoms
3059givenmass Avogadronumber
3060molar mass
3061= ×
30620
3063m
3064N N
3065M
3066⇒ = ×
306723
306823
306946
3070N 6.022 10
307123
3072N 12.044 10
3073⇒ = × ×
3074⇒ = ×
3075(ii) The number of molecules
30760
3077givenmass Avogadro number
3078molar mass
3079m
3080N N
3081M
3082atomic massof oxygen 16 u
3083= ×
3084⇒ = ×
3085=
3086∴ molar mass of O2
3087 molecules
3088= 16 × 2 = 32g
308923
309023
309123
30928
3093N 6.022 10
309432
3095N 1.5055 10
30961.51 10
3097⇒ = × ×
3098⇒ = ×
3099≃ ×
3100(iii) The number of particles (atom) =
3101number of moles of particles ×
3102Avogadro number
3103N = n × N0 = 0.1 x 6.022 × 1023
3104 = 6.022 × 1022
3105uestions
31061. If one mole of carbon atoms
3107weighs 12 grams, what is the
3108mass (in grams) of 1 atom of
3109carbon?
31102. Which has more number of
3111atoms, 100 grams of sodium or
3112100 grams of iron (given, atomic
3113mass of Na = 23 u, Fe = 56 u)?
3114Q
3115ATOMS AND MOLECULES 43
3116What
3117you have
3118learnt
3119• During a chemical reaction, the sum of the masses of the
3120reactants and products remains unchanged. This is known
3121as the Law of Conservation of Mass.
3122• In a pure chemical compound, elements are always present in
3123a definite proportion by mass. This is known as the Law of
3124Definite Proportions.
3125• An atom is the smallest particle of the element that cannot
3126usually exist independently and retain all its chemical
3127properties.
3128• A molecule is the smallest particle of an element or a compound
3129capable of independent existence under ordinary conditions.
3130It shows all the properties of the substance.
3131• A chemical formula of a compound shows its constituent
3132elements and the number of atoms of each combining element.
3133• Clusters of atoms that act as an ion are called polyatomic ions.
3134They carry a fixed charge on them.
3135• The chemical formula of a molecular compound is determined
3136by the valency of each element.
3137• In ionic compounds, the charge on each ion is used to
3138determine the chemical formula of the compound.
3139• Scientists use the relative atomic mass scale to compare the
3140masses of different atoms of elements. Atoms of carbon-12
3141isotopes are assigned a relative atomic mass of 12 and the
3142relative masses of all other atoms are obtained by comparison
3143with the mass of a carbon-12 atom.
3144• The Avogadro constant 6.022 × 1023 is defined as the number
3145of atoms in exactly 12 g of carbon-12.
3146• The mole is the amount of substance that contains the same
3147number of particles (atoms/ ions/ molecules/ formula units
3148etc.) as there are atoms in exactly 12 g of carbon-12.
3149• Mass of 1 mole of a substance is called its molar mass.
3150Exercises
31511. A 0.24 g sample of compound of oxygen and boron was found
3152by analysis to contain 0.096 g of boron and 0.144 g of oxygen.
3153Calculate the percentage composition of the compound by
3154weight.
31552. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of
3156carbon dioxide is produced. What mass of carbon dioxide will
315744 SCIENCE
3158be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen?
3159Which law of chemical combination will govern your answer?
31603. What are polyatomic ions? Give examples.
31614. Write the chemical formulae of the following.
3162(a) Magnesium chloride
3163(b) Calcium oxide
3164(c) Copper nitrate
3165(d) Aluminium chloride
3166(e) Calcium carbonate.
31675. Give the names of the elements present in the following
3168compounds.
3169(a) Quick lime
3170(b) Hydrogen bromide
3171(c) Baking powder
3172(d) Potassium sulphate.
31736. Calculate the molar mass of the following substances.
3174(a) Ethyne, C2H2
3175(b) Sulphur molecule,S8
3176(c) Phosphorus molecule, P4 (Atomic mass of phosphorus
3177= 31)
3178(d) Hydrochloric acid, HCl
3179(e) Nitric acid, HNO3
31807. What is the mass of—
3181(a) 1 mole of nitrogen atoms?
3182(b) 4 moles of aluminium atoms (Atomic mass of aluminium
3183= 27)?
3184(c) 10 moles of sodium sulphite (Na2
3185SO3
3186)?
31878. Convert into mole.
3188(a) 12 g of oxygen gas
3189(b) 20 g of water
3190(c) 22 g of carbon dioxide.
31919. What is the mass of:
3192(a) 0.2 mole of oxygen atoms?
3193(b) 0.5 mole of water molecules?
319410. Calculate the number of molecules of sulphur (S8
3195) present in
319616 g of solid sulphur.
319711. Calculate the number of aluminium ions present in 0.051 g of
3198aluminium oxide.
3199(Hint: The mass of an ion is the same as that of an atom of the
3200same element. Atomic mass of Al = 27 u)
3201ATOMS AND MOLECULES 45
3202Group Activity
3203Play a game for writing formulae.
3204Example1 : Make placards with symbols and valencies of the
3205elements separately. Each student should hold two
3206placards, one with the symbol in the right hand and
3207the other with the valency in the left hand. Keeping
3208the symbols in place, students should criss-cross their
3209valencies to form the formula of a compound.
3210Example 2 : A low cost model for writing formulae: Take empty
3211blister packs of medicines. Cut them in groups,
3212according to the valency of the element, as shown in
3213the figure. Now, you can make formulae by fixing one
3214type of ion into other.
3215For example:
3216Na+ SO4
32172- P04
32183-
3219Formula for sodium sulphate:
32202 sodium ions can be fixed on one sulphate ion.
3221Hence, the formula will be: Na2SO4
3222Do it yourself :
3223Now, write the formula of sodium phosphate.
3224In Chapter 3, we have learnt that atoms and
3225molecules are the fundamental building
3226blocks of matter. The existence of different
3227kinds of matter is due to different atoms
3228constituting them. Now the questions arise:
3229(i) What makes the atom of one element
3230different from the atom of another element?
3231and (ii) Are atoms really indivisible, as
3232proposed by Dalton, or are there smaller
3233constituents inside the atom? We shall find
3234out the answers to these questions in this
3235chapter. We will learn about sub-atomic
3236particles and the various models that have
3237been proposed to explain how these particles
3238are arranged within the atom.
3239A major challenge before the scientists at
3240the end of the 19th century was to reveal the
3241structure of the atom as well as to explain its
3242important properties. The elucidation of the
3243structure of atoms is based on a series of
3244experiments.
3245One of the first indications that atoms are
3246not indivisible, comes from studying static
3247electricity and the condition under which
3248electricity is conducted by different
3249substances.
32504.1 Charged Particles in Matter Charged Particles in Matter
3251For understanding the nature of charged
3252particles in matter, let us carry out the
3253following activities:
3254Activity ______________ 4.1
3255A. Comb dry hair. Does the comb then
3256attract small pieces of paper?
3257B. Rub a glass rod with a silk cloth and
3258bring the rod near an inflated balloon.
3259Observe what happens.
3260From these activities, can we conclude
3261that on rubbing two objects together, they
3262become electrically charged? Where does this
3263charge come from? This question can be
3264answered by knowing that an atom is divisible
3265and consists of charged particles.
3266Many scientists contributed in revealing
3267the presence of charged particles in an atom.
3268It was known by 1900 that the atom was
3269not a simple, indivisible particle but contained
3270at least one sub-atomic particle – the electron
3271identified by J.J. Thomson. Even before the
3272electron was identified, E. Goldstein in 1886
3273discovered the presence of new radiations in
3274a gas discharge and called them canal rays.
3275These rays were positively charged radiations
3276which ultimately led to the discovery of
3277another sub-atomic particle. This sub-atomic
3278particle had a charge, equal in magnitude but
3279opposite in sign to that of the electron. Its
3280mass was approximately 2000 times as that
3281of the electron. It was given the name of
3282proton. In general, an electron is represented
3283as ‘e–’ and a proton as ‘p+’. The mass of a proton
3284is taken as one unit and its charge as plus
3285one. The mass of an electron is considered to
3286be negligible and its charge is minus one.
3287It seemed highly likely that an atom was
3288composed of protons and electrons, mutually
3289balancing their charges. It also appeared that
3290the protons were in the interior of the atom,
3291for whereas electrons could easily be peeled
3292off but not protons. Now the big question was:
3293what sort of structure did these particles of
3294the atom form? We will find the answer to
3295this question below.
32964
3297STRUCTURE TRUCTURE OF THE ATOM
3298Chapter
3299© NCERT
3300not to be republished
3301uestions
33021. What are canal rays?
33032. If an atom contains one electron
3304and one proton, will it carry any
3305charge or not?
33064.2 The Structure of an Atom
3307We have learnt Dalton’s atomic theory in
3308Chapter 3, which suggested that the atom
3309was indivisible and indestructible. But the
3310discovery of two fundamental particles
3311(electrons and protons) inside the atom, led
3312to the failure of this aspect of Dalton’s atomic
3313theory. It was then considered necessary to
3314know how electrons and protons are arranged
3315within an atom. For explaining this, many
3316scientists proposed various atomic models.
3317J.J. Thomson was the first one to propose a
3318model for the structure of an atom.
33194.2.1 THOMSON’S MODEL OF AN ATOM
3320Thomson proposed the model of an atom to
3321be similar to that of a Christmas pudding.
3322The electrons, in a sphere of positive charge,
3323were like currants (dry fruits) in a spherical
3324Christmas pudding. We can also think of a
3325watermelon, the positive charge in the atom
3326is spread all over like the red edible part of
3327the watermelon, while the electrons are
3328studded in the positively charged sphere, like
3329the seeds in the watermelon (Fig. 4.1).
3330Thomson proposed that:
3331(i) An atom consists of a positively
3332charged sphere and the electrons are
3333embedded in it.
3334(ii) The negative and positive charges are
3335equal in magnitude. So, the atom as a
3336whole is electrically neutral.
3337Although Thomson’s model explained that
3338atoms are electrically neutral, the results of
3339experiments carried out by other scientists
3340could not be explained by this model, as we
3341will see below.
33424.2.2 RUTHERFORD’S MODEL OF AN ATOM
3343Ernest Rutherford was interested in knowing
3344how the electrons are arranged within an
3345atom. Rutherford designed an experiment for
3346this. In this experiment, fast moving alpha
3347(α)-particles were made to fall on a thin
3348gold foil.
3349• He selected a gold foil because he wanted
3350as thin a layer as possible. This gold foil
3351was about 1000 atoms thick.
3352• α-particles are doubly-charged helium
3353ions. Since they have a mass of 4 u, the
3354fast-moving α-particles have a
3355considerable amount of energy.
3356• It was expected that α-particles would be
3357deflected by the sub-atomic particles in
3358the gold atoms. Since the α-particles were
3359much heavier than the protons, he did
3360not expect to see large deflections.
3361Q
3362Fig.4.1: Thomson’s model of an atom
3363J.J. Thomson (1856-
33641940), a British
3365physicist, was born in
3366Cheetham Hill, a suburb
3367of Manchester, on
336818 December 1856. He
3369was awarded the Nobel
3370prize in Physics in 1906
3371for his work on the
3372discovery of electrons.
3373He directed the Cavendish Laboratory at
3374Cambridge for 35 years and seven of his
3375research assistants subsequently won
3376Nobel prizes.
3377STRUCTURE OF THE ATOM 47
3378© NCERT
3379not to be republished
338048 SCIENCE
3381Fig. 4.2: Scattering of α-particles by a gold foil
3382But, the α-particle scattering experiment
3383gave totally unexpected results (Fig. 4.2). The
3384following observations were made:
3385(i) Most of the fast moving α-particles
3386passed straight through the gold foil.
3387(ii) Some of the α-particles were deflected
3388by the foil by small angles.
3389(iii) Surprisingly one out of every 12000
3390particles appeared to rebound.
3391In the words of Rutherford, “This result
3392was almost as incredible as if you fire a
339315-inch shell at a piece of tissue paper and it
3394comes back and hits you”.
3395hear a sound when each stone strikes the
3396wall. If he repeats this ten times, he will hear
3397the sound ten times. But if a blind-folded
3398child were to throw stones at a barbed-wire
3399fence, most of the stones would not hit the
3400fencing and no sound would be heard. This
3401is because there are lots of gaps in the fence
3402which allow the stone to pass through them.
3403Following a similar reasoning, Rutherford
3404concluded from the α-particle scattering
3405experiment that–
3406(i) Most of the space inside the atom is
3407empty because most of the α-particles
3408passed through the gold foil without
3409getting deflected.
3410(ii) Very few particles were deflected from
3411their path, indicating that the positive
3412charge of the atom occupies very little
3413space.
3414(iii) A very small fraction of α-particles
3415were deflected by 1800,indicating that
3416all the positive charge and mass of the
3417gold atom were concentrated in a very
3418small volume within the atom.
3419From the data he also calculated that the
3420radius of the nucleus is about 105 times less
3421than the radius of the atom.
3422On the basis of his experiment,
3423Rutherford put forward the nuclear model of
3424an atom, which had the following features:
3425(i) There is a positively charged centre in
3426an atom called the nucleus. Nearly all
3427the mass of an atom resides in the
3428nucleus.
3429(ii) The electrons revolve around the
3430nucleus in circular paths.
3431(iii) The size of the nucleus is very small
3432as compared to the size of the atom.
3433Drawbacks of Rutherford’s model of
3434the atom
3435The revolution of the electron in a circular orbit
3436is not expected to be stable. Any particle in a
3437circular orbit would undergo acceleration.
3438During acceleration, charged particles would
3439radiate energy. Thus, the revolving electron
3440would lose energy and finally fall into the
3441nucleus. If this were so, the atom should be
3442highly unstable and hence matter would not
3443exist in the form that we know. We know that
3444atoms are quite stable.
3445E. Rutherford (1871-1937)
3446was born at Spring Grove
3447on 30 August 1871. He was
3448known as the ‘Father’ of
3449nuclear physics. He is
3450famous for his work on
3451radioactivity and the
3452discovery of the nucleus of an atom with
3453the gold foil experiment. He got the Nobel
3454prize in chemistry in 1908.
3455Let us think of an activity in an open field
3456to understand the implications of this
3457experiment. Let a child stand in front of a
3458wall with his eyes closed. Let him throw
3459stones at the wall from a distance. He will
3460© NCERT
3461not to be republished
3462STRUCTURE OF THE ATOM 49
34634.2.3 BOHR’S MODEL OF ATOM
3464In order to overcome the objections raised
3465against Rutherford’s model of the atom,
3466Neils Bohr put forward the following
3467postulates about the model of an atom:
3468(i) Only certain special orbits known as
3469discrete orbits of electrons, are allowed
3470inside the atom.
3471(ii) While revolving in discrete orbits the
3472electrons do not radiate energy.
3473uestions
34741. On the basis of Thomson’s model
3475of an atom, explain how the atom
3476is neutral as a whole.
34772. On the basis of Rutherford’s
3478model of an atom, which subatomic
3479particle is present in the
3480nucleus of an atom?
34813. Draw a sketch of Bohr’s model
3482of an atom with three shells.
34834. What do you think would be the
3484observation if the α-particle
3485scattering experiment is carried
3486out using a foil of a metal other
3487than gold?
34884.2.4 NEUTRONS
3489In 1932, J. Chadwick discovered another subatomic
3490particle which had no charge and a
3491mass nearly equal to that of a proton. It was
3492eventually named as neutron. Neutrons are
3493present in the nucleus of all atoms, except
3494hydrogen. In general, a neutron is
3495represented as ‘n’. The mass of an atom is
3496therefore given by the sum of the masses of
3497protons and neutrons present in the nucleus.
3498uestions
34991. Name the three sub-atomic
3500particles of an atom.
35012. Helium atom has an atomic mass
3502of 4 u and two protons in its
3503nucleus. How many neutrons
3504does it have?
35054.3 How are Electrons Distributed
3506in Different Orbits (Shells)? in Different Orbits (Shells)?
3507The distribution of electrons into different
3508orbits of an atom was suggested by Bohr and
3509Bury.
3510The following rules are followed for writing
3511the number of electrons in different energy
3512levels or shells:
3513(i) The maximum number of electrons
3514present in a shell is given by the
3515Neils Bohr (1885-1962)
3516was born in Copenhagen
3517on 7 October 1885. He was
3518appointed professor of
3519physics at Copenhagen
3520University in 1916. He got
3521the Nobel prize for his work
3522on the structure of atom in
35231922. Among Professor
3524Bohr’s numerous writings, three appearing
3525as books are:
3526(i) The Theory of Spectra and Atomic
3527Constitution, (ii) Atomic Theory and,
3528(iii) The Description of Nature.
3529These orbits or shells are called energy
3530levels. Energy levels in an atom are shown in
3531Fig. 4.3.
3532Q
3533Fig. 4.3: A few energy levels in an atom
3534These orbits or shells are represented by
3535the letters K,L,M,N,… or the numbers,
3536n=1,2,3,4,….
3537Q
3538© NCERT
3539not to be republished
354050 SCIENCE
3541formula 2n2, where ‘n’ is the orbit
3542number or energy level index, 1,2,3,….
3543Hence the maximum number of
3544electrons in different shells are as
3545follows:
3546first orbit or K-shell will be = 2 × 12
3547 = 2,
3548second orbit or L-shell will be = 2 × 22
3549= 8, third orbit or M-shell will be = 2 ×
355032 = 18, fourth orbit or N-shell will be
3551= 2 × 42
3552= 32, and so on.
3553(ii) The maximum number of electrons
3554that can be accommodated in the
3555outermost orbit is 8.
3556(iii) Electrons are not accommodated in a
3557given shell, unless the inner shells are
3558filled. That is, the shells are filled in a
3559step-wise manner.
3560Atomic structure of the first eighteen
3561elements is shown schematically in Fig. 4.4.
3562• The composition of atoms of the first
3563eighteen elements is given in Table 4.1.
3564uestions
35651. Write the distribution of electrons
3566in carbon and sodium atoms.
35672. If K and L shells of an atom are
3568full, then what would be the total
3569number of electrons in the atom?
35704.4 Valency Valency
3571We have learnt how the electrons in an atom
3572are arranged in different shells/orbits. The
3573electrons present in the outermost shell of
3574an atom are known as the valence electrons.
3575From the Bohr-Bury scheme, we also
3576know that the outermost shell of an atom can
3577Activity ______________ 4.2
3578• Make a static atomic model displaying
3579electronic configuration of the first
3580eighteen elements.
3581Fig.4.4: Schematic atomic structure of the first eighteen elements
3582Q
3583accommodate a maximum of 8 electrons. It
3584was observed that the atoms of elements,
3585having a completely filled outermost shell
3586show little chemical activity. In other words,
3587their combining capacity or valency is zero.
3588Of these inert elements, the helium atom has
3589© NCERT
3590not to be republished
3591STRUCTURE OF THE ATOM 51
3592Table 4.1: Composition of Atoms of the First Eighteen Elements
3593with Electron Distribution in Various Shells
3594two electrons in its outermost shell and all
3595other elements have atoms with eight
3596electrons in the outermost shell.
3597The combining capacity of the atoms of
3598other elements, that is, their tendency to react
3599and form molecules with atoms of the same
3600or different elements, was thus explained as
3601an attempt to attain a fully-filled outermost
3602shell. An outermost-shell, which had eight
3603electrons was said to possess an octet. Atoms
3604would thus react, so as to achieve an octet in
3605the outermost shell. This was done by
3606sharing, gaining or losing electrons. The
3607number of electrons gained, lost or shared
3608so as to make the octet of electrons in the
3609outermost shell, gives us directly the
3610combining capacity of the element, that is,
3611the valency discussed in the previous chapter.
3612For example, hydrogen/lithium/sodium
3613atoms contain one electron each in their
3614outermost shell, therefore each one of them
3615can lose one electron. So, they are said to
3616have valency of one. Can you tell, what is
3617valency of magnesium and aluminium? It is
3618two and three, respectively, because
3619magnesium has two electrons in its outermost
3620shell and aluminium has three electrons in
3621its outermost shell.
3622If the number of electrons in the
3623outermost shell of an atom is close to its full
3624capacity, then valency is determined in a
3625different way. For example, the fluorine atom
3626has 7 electrons in the outermost shell, and
3627its valency could be 7. But it is easier for
3628Name of Symbol Atomic Number Number Number ValeElement
3629Number of of of ncy
3630Protons Neutrons Electrons K L M N
3631Hydrogen H 1 1 - 1 1 - - - 1
3632Helium He 2 2 2 2 2 - - - 0
3633Lithium Li 3 3 4 3 2 1 - - 1
3634Beryllium Be 4 4 5 4 2 2 - - 2
3635Boron B 5 5 6 5 2 3 - - 3
3636Carbon C 6 6 6 6 2 4 - - 4
3637Nitrogen N 7 7 7 7 2 5 - - 3
3638Oxygen O 8 8 8 8 2 6 - - 2
3639Fluorine F 9 9 10 9 2 7 - - 1
3640Neon Ne 10 10 10 10 2 8 - - 0
3641Sodium Na 11 11 12 11 2 8 1 - 1
3642Magnesium Mg 12 12 12 12 2 8 2 - 2
3643Aluminium Al 13 13 14 13 2 8 3 - 3
3644Silicon Si 14 14 14 14 2 8 4 - 4
3645Phosphorus P 15 15 16 15 2 8 5 - 3,5
3646Sulphur S 16 16 16 16 2 8 6 - 2
3647Chlorine Cl 17 17 18 17 2 8 7 - 1
3648Argon Ar 18 18 22 18 2 8 8 0
3649Distribution of
3650Electrons
3651© NCERT
3652not to be republished
365352 SCIENCE
3654fluorine to gain one electron instead of losing
3655seven electrons. Hence, its valency is
3656determined by subtracting seven electrons
3657from the octet and this gives you a valency of
3658one for fluorine. Valency can be calculated in
3659a similar manner for oxygen. What is the
3660valency of oxygen that you get from this
3661calculation?
3662Therefore, an atom of each element has a
3663definite combining capacity, called its valency.
3664Valency of the first eighteen elements is given
3665in the last column of Table 4.1.
3666uestion
36671. How will you find the valency
3668of chlorine, sulphur and
3669magnesium?
36704.5 Atomic Number and Mass Atomic Number and Mass
3671Number
36724.5.1 ATOMIC NUMBER
3673We know that protons are present in the
3674nucleus of an atom. It is the number of
3675protons of an atom, which determines its
3676atomic number. It is denoted by ‘Z’. All atoms
3677of an element have the same atomic number,
3678Z. In fact, elements are defined by the number
3679of protons they possess. For hydrogen, Z = 1,
3680because in hydrogen atom, only one proton
3681is present in the nucleus. Similarly, for
3682carbon, Z = 6. Therefore, the atomic number
3683is defined as the total number of protons
3684present in the nucleus of an atom.
36854.5.2 MASS NUMBER
3686After studying the properties of the subatomic
3687particles of an atom, we can conclude
3688that mass of an atom is practically due to
3689protons and neutrons alone. These are
3690present in the nucleus of an atom. Hence
3691protons and neutrons are also called
3692nucleons. Therefore, the mass of an atom
3693resides in its nucleus. For example, mass of
3694carbon is 12 u because it has 6 protons and
36956 neutrons, 6 u + 6 u = 12 u. Similarly, the
3696mass of aluminium is 27 u (13 protons+14
3697neutrons). The mass number is defined as
3698the sum of the total number of protons and
3699neutrons present in the nucleus of an atom.
3700In the notation for an atom, the atomic
3701number, mass number and symbol of the
3702element are to be written as:
3703Mass Number
3704Q
3705Symbol of
3706element
3707Atomic Number
3708For example, nitrogen is written as 14
37097 N .
3710uestions
37111. If number of electrons in an atom
3712is 8 and number of protons is also
37138, then (i) what is the atomic
3714number of the atom? and (ii) what
3715is the charge on the atom?
37162. With the help of Table 4.1, find
3717out the mass number of oxygen
3718and sulphur atom.
37194.6 Isotopes Isotopes
3720In nature, a number of atoms of some
3721elements have been identified, which have the
3722same atomic number but different mass
3723numbers. For example, take the case of
3724hydrogen atom, it has three atomic species,
3725namely protium (1
37261H), deuterium ( 2
37271 H or D)
3728and tritium ( 3
37291H or T). The atomic number of
3730each one is 1, but the mass number is 1, 2
3731and 3, respectively. Other such examples are
3732(i) carbon, 12
37336 C and 14
37346 C, (ii) chlorine, 35
373517 Cl
3736and 37
373717 Cl, etc.
3738On the basis of these examples, isotopes
3739are defined as the atoms of the same element,
3740having the same atomic number but different
3741mass numbers. Therefore, we can say that
3742there are three isotopes of hydrogen atom,
3743namely protium, deuterium and tritium.
3744Q
3745© NCERT
3746not to be republished
3747STRUCTURE OF THE ATOM 53
3748Q
3749Many elements consist of a mixture of
3750isotopes. Each isotope of an element is a pure
3751substance. The chemical properties of
3752isotopes are similar but their physical
3753properties are different.
3754Chlorine occurs in nature in two isotopic
3755forms, with masses 35 u and 37 u in the ratio
3756of 3:1. Obviously, the question arises: what
3757should we take as the mass of chlorine atom?
3758Let us find out.
3759The mass of an atom of any natural
3760element is taken as the average mass of all
3761the naturally occuring atoms of that element.
3762If an element has no isotopes, then the mass
3763of its atom would be the same as the sum of
3764protons and neutrons in it. But if an element
3765occurs in isotopic forms, then we have to
3766know the percentage of each isotopic form
3767and then the average mass is calculated.
3768The average atomic mass of chlorine atom,
3769on the basis of above data, will be
377075 25 35 37
3771100 100
3772105 37 142 35 .5 u
377344 4
3774
3775This does not mean that any one atom of
3776chlorine has a fractional mass of 35.5 u. It
3777means that if you take a certain amount of
3778chlorine, it will contain both isotopes of
3779chlorine and the average mass is 35.5 u.
3780Applications
3781Since the chemical properties of all the
3782isotopes of an element are the same,
3783normally we are not concerned about
3784taking a mixture. But some isotopes have
3785special properties which find them useful
3786in various fields. Some of them are :
3787(i) An isotope of uranium is used as a fuel
3788in nuclear reactors.
3789(ii) An isotope of cobalt is used in the
3790treatment of cancer.
3791(iii) An isotope of iodine is used in the
3792treatment of goitre.
37934.6.1 ISOBARS
3794Let us consider two elements — calcium,
3795atomic number 20, and argon, atomic
3796number 18. The number of electrons in these
3797atoms is different, but the mass number of
3798both these elements is 40. That is, the total
3799number of nucleons is the same in the atoms
3800of this pair of elements. Atoms of different
3801elements with different atomic numbers,
3802which have the same mass number, are
3803known as isobars.
3804uestions
38051. For the symbol H,D and T
3806tabulate three sub-atomic
3807particles found in each of them.
38082. Write the electronic configuration
3809of any one pair of isotopes and
3810isobars.
3811What
3812you have you have
3813learnt
3814• Credit for the discovery of electron and proton goes to J.J.
3815Thomson and E.Goldstein, respectively.
3816• J.J. Thomson proposed that electrons are embedded in a
3817positive sphere.
3818© NCERT
3819not to be republished
382054 SCIENCE
3821• Rutherford’s alpha-particle scattering experiment led to the
3822discovery of the atomic nucleus.
3823• Rutherford’s model of the atom proposed that a very tiny
3824nucleus is present inside the atom and electrons revolve around
3825this nucleus. The stability of the atom could not be explained
3826by this model.
3827• Neils Bohr’s model of the atom was more successful. He
3828proposed that electrons are distributed in different shells with
3829discrete energy around the nucleus. If the atomic shells are
3830complete, then the atom will be stable and less reactive.
3831• J. Chadwick discovered presence of neutrons in the nucleus of
3832an atom. So, the three sub-atomic particles of an atom are:
3833(i) electrons, (ii) protons and (iii) neutrons. Electrons are
3834negatively charged, protons are positively charged and neutrons
3835have no charges. The mass of an electron is about
38361
38372000 times
3838the mass of an hydrogen atom. The mass of a proton and a
3839neutron is taken as one unit each.
3840• Shells of an atom are designated as K,L,M,N,….
3841• Valency is the combining capacity of an atom.
3842• The atomic number of an element is the same as the number
3843of protons in the nucleus of its atom.
3844• The mass number of an atom is equal to the number of nucleons
3845in its nucleus.
3846• Isotopes are atoms of the same element, which have different
3847mass numbers.
3848• Isobars are atoms having the same mass number but different
3849atomic numbers.
3850• Elements are defined by the number of protons they possess.
3851Exercises Exercises Exercises
38521. Compare the properties of electrons, protons and neutrons.
38532. What are the limitations of J.J. Thomson’s model of the atom?
38543. What are the limitations of Rutherford’s model of the atom?
38554. Describe Bohr’s model of the atom.
38565. Compare all the proposed models of an atom given in this
3857chapter.
38586. Summarise the rules for writing of distribution of electrons in
3859various shells for the first eighteen elements.
38607. Define valency by taking examples of silicon and oxygen.
3861© NCERT
3862not to be republished
3863STRUCTURE OF THE ATOM 55
38648. Explain with examples (i) Atomic number, (ii) Mass number,
3865(iii) Isotopes and iv) Isobars. Give any two uses of isotopes.
38669. Na+ has completely filled K and L shells. Explain.
386710. If bromine atom is available in the form of, say, two isotopes
386879
386935 Br (49.7%) and 81
387035 Br (50.3%), calculate the average atomic
3871mass of bromine atom.
387211. The average atomic mass of a sample of an element X is 16.2 u.
3873What are the percentages of isotopes 16
38748 X and 18
38758 X in the
3876sample?
387712. If Z = 3, what would be the valency of the element? Also, name
3878the element.
387913. Composition of the nuclei of two atomic species X and Y are
3880given as under
3881X Y
3882Protons = 6 6
3883Neutrons = 6 8
3884Give the mass numbers of X and Y. What is the relation between
3885the two species?
388614. For the following statements, write T for True and F for False.
3887(a) J.J. Thomson proposed that the nucleus of an atom
3888contains only nucleons.
3889(b) A neutron is formed by an electron and a proton
3890combining together. Therefore, it is neutral.
3891(c) The mass of an electron is about
38921
38932000 times that of proton.
3894(d) An isotope of iodine is used for making tincture iodine,
3895which is used as a medicine.
3896Put tick (9) against correct choice and cross (×) against
3897wrong choice in questions 15, 16 and 17
389815. Rutherford’s alpha-particle scattering experiment was
3899responsible for the discovery of
3900(a) Atomic Nucleus (b) Electron
3901(c) Proton (d) Neutron
390216. Isotopes of an element have
3903(a) the same physical properties
3904(b) different chemical properties
3905(c) different number of neutrons
3906(d) different atomic numbers.
390717. Number of valence electrons in Cl–
3908 ion are:
3909(a) 16 (b) 8 (c) 17 (d) 18
3910© NCERT
3911not to be republished
391256 SCIENCE
391318. Which one of the following is a correct electronic configuration
3914of sodium?
3915(a) 2,8 (b) 8,2,1 (c) 2,1,8 (d) 2,8,1.
391619. Complete the following table.
3917Atomic Mass Number Number Number Name of
3918Number Number of of of the Atomic
3919Neutrons Protons Electrons Species
39209 - 10 - - -
392116 32 - - - Sulphur
3922- 24 - 12 - -
3923-2 - 1 - -
3924-1 0 1 0 -
3925© NCERT
3926not to be republished
3927While examining a thin slice of cork, Robert
3928Hooke saw that the cork resembled the
3929structure of a honeycomb consisting of many
3930little compartments. Cork is a substance
3931which comes from the bark of a tree. This
3932was in the year 1665 when Hooke made this
3933chance observation through a self-designed
3934microscope. Robert Hooke called these boxes
3935cells. Cell is a Latin word for ‘a little room’.
3936This may seem to be a very small and
3937insignificant incident but it is very important
3938in the history of science. This was the very
3939first time that someone had observed that
3940living things appear to consist of separate
3941units. The use of the word ‘cell’ to describe
3942these units is used till this day in biology.
3943Let us find out about cells.
39445.1 What are Living Organisms What are Living Organisms
3945Made Up of? Made Up of?
3946Activity ______________ 5.1
3947• Let us take a small piece from an onion
3948bulb. With the help of a pair of forceps,
3949we can peel off the skin (called
3950epidermis) from the concave side (inner
3951layer) of the onion. This layer can be
3952put immediately in a watch-glass
3953containing water. This will prevent the
3954peel from getting folded or getting dry.
3955What do we do with this peel?
3956• Let us take a glass slide, put a drop of
3957water on it and transfer a small piece
3958of the peel from the watch glass to the
3959slide. Make sure that the peel is
3960perfectly flat on the slide. A thin camel
3961hair paintbrush might be necessary to
3962help transfer the peel. Now we put a
3963drop of safranin solution on this piece
3964followed by a cover slip. Take care to
3965avoid air bubbles while putting the
3966cover slip with the help of a mounting
3967needle. Ask your teacher for help. We
3968have prepared a temporary mount of
3969onion peel. We can observe this slide
3970under low power followed by high
3971powers of a compound microscope.
3972Fig. 5.1: Compound microscope
3973What do we observe as we look through
3974the lens? Can we draw the structures that
3975we are able to see through the microscope,
3976on an observation sheet? Does it look like
3977Fig. 5.2?
3978Eyepiece
3979Coarse adjustment
3980Fine adjustment
3981Arm
3982Objective lens
3983Stage
3984Swivel
3985Mirror
3986Base
3987Body tube
3988Clip
3989 Microscope slide
3990Condenser
3991Fig. 5.2: Cells of an onion peel
39925
3993THE FUNDAMENTAL UNDAMENTAL UNIT OF LIFE
3994Chapter
3995Nucleus
3996© NCERT
3997Cells
3998not to be republished
399958 SCIENCE
4000Chlamydomonas, Paramoecium and bacteria.
4001These organisms are called unicellular
4002organisms (uni = single). On the other hand,
4003many cells group together in a single body
4004and assume different functions in it to form
4005various body parts in multicellular organisms
4006(multi = many) such as some fungi, plants
4007and animals. Can we find out names of some
4008more unicellular organisms?
4009Every multi-cellular organism has come
4010from a single cell. How? Cells divide to
4011produce cells of their own kind. All cells thus
4012come from pre-existing cells.
4013Activity ______________ 5.2
4014• We can try preparing temporary
4015mounts of leaf peels, tip of roots of
4016onion or even peels of onions of different
4017sizes.
4018• After performing the above activity, let
4019us see what the answers to the following
4020questions would be:
4021(a) Do all cells look alike in terms of
4022shape and size?
4023(b) Do all cells look alike in structure?
4024(c) Could we find differences among
4025cells from different parts of a plant
4026body?
4027(d) What similarities could we find?
4028Some organisms can also have cells of
4029different kinds. Look at the following picture.
4030It depicts some cells from the human body.
4031Nerve Cell
4032Fat cell
4033Sperm
4034Bone
4035cell
4036Smooth
4037muscle
4038cell
4039Blood
4040cells
4041Ovum
4042Fig. 5.3: Various cells from the human body
4043More to know
4044We can try preparing temporary mounts
4045of peels of onions of different sizes. What do
4046we observe? Do we see similar structures or
4047different structures?
4048What are these structures?
4049These structures look similar to each other.
4050Together they form a big structure like an
4051onion bulb! We find from this activity that
4052onion bulbs of different sizes have similar
4053small structures visible under a microscope.
4054The cells of the onion peel will all look the
4055same, regardless of the size of the onion they
4056came from.
4057These small structures that we see are
4058the basic building units of the onion bulb.
4059These structures are called cells. Not only
4060onions, but all organisms that we observe
4061around are made up of cells. However, there
4062are also single cells that live on their own.
4063Cells were first discovered by
4064Robert Hooke in 1665. He observed
4065the cells in a cork slice with the help
4066of a primitive microscope.
4067Leeuwenhoek (1674), with the
4068improved microscope, discovered the
4069free living cells in pond water for the
4070first time. It was Robert Brown in
40711831 who discovered the nucleus in
4072the cell. Purkinje in 1839 coined the
4073term ‘protoplasm’ for the fluid
4074substance of the cell. The cell theory,
4075that all the plants and animals are
4076composed of cells and that the cell is
4077the basic unit of life, was presented
4078by two biologists, Schleiden (1838)
4079and Schwann (1839). The cell theory
4080was further expanded by Virchow
4081(1855) by suggesting that all cells
4082arise from pre-existing cells. With the
4083discovery of the electron microscope
4084in 1940, it was possible to observe and
4085understand the complex structure of
4086the cell and its various organelles.
4087The invention of magnifying lenses led to
4088the discovery of the microscopic world. It is
4089now known that a single cell may constitute
4090a whole organism as in Amoeba,
4091© NCERT
4092not to be republished
4093THE FUNDAMENTAL UNIT OF LIFE 59
4094The shape and size of cells are related to
4095the specific function they perform. Some cells
4096like Amoeba have changing shapes. In some
4097cases the cell shape could be more or less
4098fixed and peculiar for a particular type of cell;
4099for example, nerve cells have a typical shape.
4100Each living cell has the capacity to
4101perform certain basic functions that are
4102characteristic of all living forms. How does
4103a living cell perform these basic functions?
4104We know that there is a division of labour in
4105multicellular organisms such as human
4106beings. This means that different parts of
4107the human body perform different functions.
4108The human body has a heart to pump blood,
4109a stomach to digest food and so on. Similarly,
4110division of labour is also seen within a single
4111cell. In fact, each such cell has got certain
4112specific components within it known as cell
4113organelles. Each kind of cell organelle
4114performs a special function, such as making
4115new material in the cell, clearing up the
4116waste material from the cell and so on. A
4117cell is able to live and perform all its
4118functions because of these organelles. These
4119organelles together constitute the basic unit
4120called the cell. It is interesting that all cells
4121are found to have the same organelles, no
4122matter what their function is or what
4123organism they are found in.
4124uestions
41251. Who discovered cells, and how?
41262. Why is the cell called the
4127structural and functional unit of
4128life?
41295.2 What is a Cell Made Up of? What is a Cell Made Up of?
4130What is the Structural What is the Structural
4131Organisation of a Cell? Organisation of a Cell?
4132We saw above that the cell has special
4133components called organelles. How is a cell
4134organised?
4135If we study a cell under a microscope, we
4136would come across three features in almost
4137every cell; plasma membrane, nucleus and
4138cytoplasm. All activities inside the cell and
4139interactions of the cell with its environment
4140are possible due to these features. Let us see
4141how.
41425.2.1 PLASMA MEMBRANE OR CELL
4143MEMBRANE
4144This is the outermost covering of the cell that
4145separates the contents of the cell from its
4146external environment. The plasma membrane
4147allows or permits the entry and exit of some
4148materials in and out of the cell. It also
4149prevents movement of some other materials.
4150The cell membrane, therefore, is called a
4151selectively permeable membrane.
4152How does the movement of substances
4153take place into the cell? How do substances
4154move out of the cell?
4155Some substances like carbon dioxide or
4156oxygen can move across the cell membrane
4157by a process called diffusion. We have studied
4158the process of diffusion in earlier chapters.
4159We saw that there is spontaneous movement
4160of a substance from a region of high
4161concentration to a region where its
4162concentration is low.
4163Something similar to this happens in cells
4164when, for example, some substance like CO2
4165(which is cellular waste and requires to be
4166excreted out by the cell) accumulates in high
4167concentrations inside the cell. In the cell’s
4168external environment, the concentration of
4169CO2 is low as compared to that inside the
4170cell. As soon as there is a difference of
4171concentration of CO2 inside and outside a cell,
4172CO2 moves out of the cell, from a region of
4173high concentration, to a region of low
4174concentration outside the cell by the process
4175of diffusion. Similarly, O2 enters the cell by
4176the process of diffusion when the level or
4177concentration of O2
4178 inside the cell decreases.
4179Thus, diffusion plays an important role in
4180gaseous exchange between the cells as well
4181as the cell and its external environment.
4182Water also obeys the law of diffusion. The
4183movement of water molecules through such
4184a selectively permeable membrane is called
4185Q
4186© NCERT
4187not to be republished
418860 SCIENCE
4189osmosis. The movement of water across the
4190plasma membrane is also affected by the
4191amount of substance dissolved in water.
4192Thus, osmosis is the passage of water from a
4193region of high water concentration through a
4194semi-permeable membrane to a region of low
4195water concentration.
4196What will happen if we put an animal cell
4197or a plant cell into a solution of sugar or salt
4198in water?
4199One of the following three things could
4200happen:
42011. If the medium surrounding the cell has
4202a higher water concentration than the
4203cell, meaning that the outside solution
4204is very dilute, the cell will gain water
4205by osmosis. Such a solution is known
4206as a hypotonic solution.
4207Water molecules are free to pass
4208across the cell membrane in both
4209directions, but more water will come
4210into the cell than will leave. The net
4211(overall) result is that water enters the
4212cell. The cell is likely to swell up.
42132. If the medium has exactly the same
4214water concentration as the cell, there
4215will be no net movement of water
4216across the cell membrane. Such a
4217solution is known as an isotonic
4218solution.
4219Water crosses the cell membrane
4220in both directions, but the amount
4221going in is the same as the amount
4222going out, so there is no overall
4223movement of water. The cell will stay
4224the same size.
42253. If the medium has a lower
4226concentration of water than the cell,
4227meaning that it is a very concentrated
4228solution, the cell will lose water by
4229osmosis. Such a solution is known as
4230a hypertonic solution.
4231Again, water crosses the cell
4232membrane in both directions, but this
4233time more water leaves the cell than
4234enters it. Therefore the cell will shrink.
4235Thus, osmosis is a special case of diffusion
4236through a selectively permeable membrane.
4237Now let us try out the following activity:
4238Activity ______________ 5.3
4239Osmosis with an egg
4240(a) Remove the shell of an egg by dissolving
4241it in dilute hydrochloric acid. The shell
4242is mostly calcium carbonate. A thin
4243outer skin now encloses the egg. Put
4244the egg in pure water and observe after
42455 minutes. What do we observe?
4246The egg swells because water passes
4247into it by osmosis.
4248(b) Place a similar de-shelled egg in a
4249concentrated salt solution and observe
4250for 5 minutes. The egg shrinks. Why?
4251Water passes out of the egg solution
4252into the salt solution because the salt
4253solution is more concentrated.
4254We can also try a similar activity with dried
4255raisins or apricots.
4256Activity ______________ 5.4
4257• Put dried raisins or apricots in plain
4258water and leave them for some time.
4259Then place them into a concentrated
4260solution of sugar or salt. You will
4261observe the following:
4262(a) Each gains water and swells when
4263placed in water.
4264(b) However, when placed in the
4265concentrated solution it loses water,
4266and consequently shrinks.
4267Unicellular freshwater organisms and
4268most plant cells tend to gain water through
4269osmosis. Absorption of water by plant roots
4270is also an example of osmosis.
4271Thus, diffusion is important in exhange
4272of gases and water in the life of a cell. In
4273additions to this, the cell also obtains
4274nutrition from its environment. Different
4275molecules move in and out of the cell
4276through a type of transport requiring use
4277of energy.
4278The plasma membrane is flexible and is
4279made up of organic molecules called lipids
4280and proteins. However, we can observe the
4281structure of the plasma membrane only
4282through an electron microscope.
4283The flexibility of the cell membrane also
4284enables the cell to engulf in food and other
4285material from its external environment. Such
4286processes are known as endocytosis. Amoeba
4287acquires its food through such processes.
4288© NCERT
4289not to be republished
4290THE FUNDAMENTAL UNIT OF LIFE 61
4291Activity ______________ 5.5
4292• Find out about electron microscopes
4293from resources in the school library or
4294through the internet. Discuss it with
4295your teacher.
4296uestions
42971. How do substances like CO2 and
4298water move in and out of the cell?
4299Discuss.
43002. Why is the plasma membrane
4301called a selectively permeable
4302membrane?
43035.2.2 CELL WALL
4304Plant cells, in addition to the plasma
4305membrane, have another rigid outer covering
4306called the cell wall. The cell wall lies outside
4307the plasma membrane. The plant cell wall is
4308mainly composed of cellulose. Cellulose is a
4309complex substance and provides structural
4310strength to plants.
4311When a living plant cell loses water
4312through osmosis there is shrinkage or
4313contraction of the contents of the cell away
4314from the cell wall. This phenomenon is known
4315as plasmolysis. We can observe this
4316phenomenon by performing the following
4317activity:
4318Activity ______________ 5.6
4319• Mount the peel of a Rheo leaf in water
4320on a slide and examine cells under the
4321high power of a microscope. Note the
4322small green granules, called
4323chloroplasts. They contain a green
4324substance called chlorophyll. Put a
4325strong solution of sugar or salt on the
4326mounted leaf on the slide. Wait for a
4327minute and observe under a
4328microscope. What do we see?
4329• Now place some Rheo leaves in boiling
4330water for a few minutes. This kills the
4331cells. Then mount one leaf on a slide
4332and observe it under a microscope. Put
4333a strong solution of sugar or salt on
4334the mounted leaf on the slide. Wait for
4335a minute and observe it again. What
4336do we find? Did plasmolysis occur now?
4337What do we infer from this activity? It
4338appears that only living cells, and not dead
4339cells, are able to absorb water by osmosis.
4340Cell walls permit the cells of plants, fungi
4341and bacteria to withstand very dilute
4342(hypotonic) external media without bursting.
4343In such media the cells tend to take up water
4344by osmosis. The cell swells, building up
4345pressure against the cell wall. The wall exerts
4346an equal pressure against the swollen cell.
4347Because of their walls, such cells can
4348withstand much greater changes in the
4349surrounding medium than animal cells.
43505.2.3 NUCLEUS
4351Remember the temporary mount of onion peel
4352we prepared? We had put iodine solution on
4353the peel. Why? What would we see if we tried
4354observing the peel without putting the iodine
4355solution? Try it and see what the difference
4356is. Further, when we put iodine solution on
4357the peel, did each cell get evenly coloured?
4358According to their chemical composition
4359different regions of cells get coloured
4360differentially. Some regions appear darker
4361than other regions. Apart from iodine solution
4362we could also use safranin solution or
4363methylene blue solution to stain the cells.
4364We have observed cells from an onion; let
4365us now observe cells from our own body.
4366Activity ______________ 5.7
4367• Let us take a glass slide with a drop of
4368water on it. Using an ice-cream spoon
4369gently scrape the inside surface of the
4370cheek. Does any material get stuck on
4371the spoon? With the help of a needle
4372we can transfer this material and
4373spread it evenly on the glass slide kept
4374ready for this. To colour the material
4375we can put a drop of methylene blue
4376solution on it. Now the material is ready
4377for observation under microscope. Do
4378not forget to put a cover-slip on it!
4379• What do we observe? What is the shape
4380of the cells we see? Draw it on the
4381observation sheet.
4382Q
4383© NCERT
4384not to be republished
438562 SCIENCE
4386• Was there a darkly coloured, spherical
4387or oval, dot-like structure near the
4388centre of each cell? This structure is
4389called nucleus. Were there similar
4390structures in onion peel cells?
4391The nucleus has a double layered covering
4392called nuclear membrane. The nuclear
4393membrane has pores which allow the transfer
4394of material from inside the nucleus to its
4395outside, that is, to the cytoplasm (which we
4396will talk about in section 5.2.4).
4397The nucleus contains chromosomes,
4398which are visible as rod-shaped structures
4399only when the cell is about to divide.
4400Chromosomes contain information for
4401inheritance of features from parents to next
4402generation in the form of DNA (Deoxyribo
4403Nucleic Acid) molecules. Chromosomes are
4404composed of DNA and protein. DNA molecules
4405contain the information necessary for
4406constructing and organising cells. Functional
4407segments of DNA are called genes. In a cell
4408which is not dividing, this DNA is present as
4409part of chromatin material. Chromatin
4410material is visible as entangled mass of thread
4411like structures. Whenever the cell is about to
4412divide, the chromatin material gets organised
4413into chromosomes.
4414The nucleus plays a central role in cellular
4415reproduction, the process by which a single
4416cell divides and forms two new cells. It also
4417plays a crucial part, along with the
4418environment, in determining the way the cell
4419will develop and what form it will exhibit at
4420maturity, by directing the chemical activities
4421of the cell.
4422In some organisms like bacteria, the
4423nuclear region of the cell may be poorly
4424defined due to the absence of a nuclear
4425membrane. Such an undefined nuclear region
4426containing only nucleic acids is called a
4427nucleoid. Such organisms, whose cells lack
4428a nuclear membrane, are called prokaryotes
4429(Pro = primitive or primary; karyote ≈ karyon
4430= nucleus). Organisms with cells having a
4431nuclear membrane are called eukaryotes.
4432Prokaryotic cells (see Fig. 5.4) also lack
4433most of the other cytoplasmic organelles
4434present in eukaryotic cells. Many of the
4435functions of such organelles are also
4436performed by poorly organised parts of the
4437cytoplasm (see section 5.2.4). The chlorophyll
4438in photosynthetic prokaryotic bacteria is
4439associated with membranous vesicles (bag
4440like structures) but not with plastids as in
4441eukaryotic cells (see section 5.2.5).
4442Ribosomes Plasma
4443membrane
4444Cell wall
4445Nucleoid
4446Fig. 5.4: Prokaryotic cell
44475.2.4 CYTOPLASM
4448When we look at the temporary mounts of
4449onion peel as well as human cheek cells, we
4450can see a large region of each cell enclosed
4451by the cell membrane. This region takes up
4452very little stain. It is called the cytoplasm.
4453The cytoplasm is the fluid content inside the
4454plasma membrane. It also contains many
4455specialised cell organelles. Each of these
4456organelles performs a specific function for the
4457cell.
4458Cell organelles are enclosed by
4459membranes. In prokaryotes, beside the
4460absence of a defined nuclear region, the
4461membrane-bound cell organelles are also
4462absent. On the other hand, the eukaryotic
4463cells have nuclear membrane as well as
4464membrane-enclosed organelles.
4465The significance of membranes can be
4466illustrated with the example of viruses.
4467Viruses lack any membranes and hence do
4468not show characteristics of life until they enter
4469a living body and use its cell machinery to
4470multiply.
4471© NCERT
4472not to be republished
4473THE FUNDAMENTAL UNIT OF LIFE 63
4474uestion
44751. Fill in the gaps in the following
4476table illustrating differences
4477between prokaryotic and
4478eukaryotic cells.
4479Prokaryotic Cell Eukaryotic Cell
44801. Size : generally 1. Size: generally
4481small ( 1-10 μm) large ( 5-100 μm)
44821 μm = 10–6 m
44832. Nuclear region: 2. Nuclear region:
4484_______________ well defined and
4485_______________ surrounded by a
4486and known as__ nuclear membrane
44873. Chromosome: 3. More than one
4488single chromosome
44894. Membrane-bound 4. _______________
4490cell organelles _______________
4491absent _______________
44925.2.5 CELL ORGANELLES
4493Every cell has a membrane around it to keep
4494its own contents separate from the external
4495environment. Large and complex cells,
4496including cells from multicellular organisms,
4497need a lot of chemical activities to support
4498their complicated structure and function. To
4499keep these activities of different kinds
4500separate from each other, these cells use
4501membrane-bound little structures (or
4502‘organelles’) within themselves. This is one of
4503the features of the eukaryotic cells that
4504distinguish them from prokaryotic cells. Some
4505of these organelles are visible only with an
4506electron microscope.
4507We have talked about the nucleus in a
4508previous section. Some important examples
4509of cell organelles which we will discuss now
4510are: endoplasmic reticulum, Golgi apparatus,
4511lysosomes, mitochondria, plastids and
4512vacuoles. They are important because they
4513carry out some very crucial functions in cells.
45145.2.5 ( 5.2.5 (i) ENDOPLASMIC RETICULUM (ER)
4515The endoplasmic reticulum (ER) is a large
4516network of membrane-bound tubes and
4517sheets. It looks like long tubules or round or
4518oblong bags (vesicles). The ER membrane is
4519similar in structure to the plasma membrane.
4520There are two types of ER– rough endoplasmic
4521reticulum (RER) and smooth endoplasmic
4522reticulum (SER). RER looks rough under a
4523microscope because it has particles called
4524ribosomes attached to its surface. The
4525ribosomes, which are present in all active
4526cells, are the sites of protein manufacture.
4527The manufactured proteins are then sent to
4528various places in the cell depending on need,
4529using the ER. The SER helps in the
4530manufacture of fat molecules, or lipids,
4531important for cell function. Some of these
4532proteins and lipids help in building the cell
4533membrane. This process is known as
4534membrane biogenesis. Some other proteins
4535and lipids function as enzymes and
4536hormones. Although the ER varies greatly in
4537appearance in different cells, it always forms
4538a network system.
4539Q
4540Fig. 5.5: Animal cell
4541Thus, one function of the ER is to serve as
4542channels for the transport of materials
4543(especially proteins) between various regions
4544of the cytoplasm or between the cytoplasm
4545and the nucleus. The ER also functions as a
4546cytoplasmic framework providing a surface
4547© NCERT
4548not to be republished
454964 SCIENCE
4550Camillo Golgi was born at
4551Corteno near Brescia in
45521843. He studied
4553medicine at the
4554University of Pavia. After
4555graduating in 1865, he
4556continued to work in
4557Pavia at the Hospital of
4558St. Matteo. At that time
4559most of his investigations
4560were concerned with the nervous system,
4561In 1872 he accepted the post of Chief
4562Medical Officer at the Hospital for the
4563Chronically Sick at Abbiategrasso. He first
4564started his investigations into the nervous
4565system in a little kitchen of this hospital,
4566which he had converted into a laboratory.
4567However, the work of greatest importance,
4568which Golgi carried out was a revolutionary
4569method of staining individual nerve and cell
4570structures. This method is referred to as
4571the ‘black reaction’. This method uses a
4572weak solution of silver nitrate and is
4573particularly valuable in tracing the
4574processes and most delicate ramifications
4575of cells. All through his life, he continued
4576to work on these lines, modifying and
4577improving this technique. Golgi received
4578the highest honours and awards in
4579recognition of his work. He shared the
4580Nobel prize in 1906 with Santiago Ramony
4581Cajal for their work on the structure of the
4582nervous system.
45835.2.5 ( 5.2.5 (iii) LYSOSOMES
4584Lysosomes are a kind of waste disposal
4585system of the cell. Lysosomes help to keep
4586the cell clean by digesting any foreign material
4587as well as worn-out cell organelles. Foreign
4588materials entering the cell, such as bacteria
4589or food, as well as old organelles end up in
4590the lysosomes, which break them up into
4591small pieces. Lysosomes are able to do this
4592because they contain powerful digestive
4593enzymes capable of breaking down all organic
4594material. During the disturbance in cellular
4595metabolism, for example, when the cell gets
4596Fig. 5.6: Plant cell
4597for some of the biochemical activities of the
4598cell. In the liver cells of the group of animals
4599called vertebrates (see Chapter 7), SER plays
4600a crucial role in detoxifying many poisons and
4601drugs.
46025.2.5 ( 5.2.5 (ii) GOLGI APPARATUS
4603The Golgi apparatus, first described by
4604Camillo Golgi, consists of a system of
4605membrane-bound vesicles arranged
4606approximately parallel to each other in stacks
4607called cisterns. These membranes often have
4608connections with the membranes of ER and
4609therefore constitute another portion of a
4610complex cellular membrane system.
4611The material synthesised near the ER is
4612packaged and dispatched to various targets
4613inside and outside the cell through the Golgi
4614apparatus. Its functions include the storage,
4615modification and packaging of products in
4616vesicles. In some cases, complex sugars may
4617be made from simple sugars in the Golgi
4618apparatus. The Golgi apparatus is also
4619involved in the formation of lysosomes [see
46205.2.5 (iii)].
4621© NCERT
4622not to be republished
4623THE FUNDAMENTAL UNIT OF LIFE 65
4624damaged, lysosomes may burst and the
4625enzymes digest their own cell. Therefore,
4626lysosomes are also known as the ‘suicide
4627bags’ of a cell. Structurally, lysosomes are
4628membrane-bound sacs filled with digestive
4629enzymes. These enzymes are made by RER.
46305.2.5 ( 5.2.5 (iv) MITOCHONDRIA
4631Mitochondria are known as the powerhouses
4632of the cell. The energy required for various
4633chemical activities needed for life is released
4634by mitochondria in the form of ATP
4635(Adenosine triphopshate) molecules. ATP is
4636known as the energy currency of the cell. The
4637body uses energy stored in ATP for making
4638new chemical compounds and for mechanical
4639work. Mitochondria have two membrane
4640coverings instead of just one. The outer
4641membrane is very porous while the inner
4642membrane is deeply folded. These folds create
4643a large surface area for ATP-generating
4644chemical reactions.
4645Mitochondria are strange organelles in the
4646sense that they have their own DNA and
4647ribosomes. Therefore, mitochondria are able
4648to make some of their own proteins.
46495.2.5 ( 5.2.5 (V) PLASTIDS
4650Plastids are present only in plant cells. There
4651are two types of plastids – chromoplasts
4652(coloured plastids) and leucoplasts (white or
4653colourless plastids). Plastids containing the
4654pigment chlorophyll are known as
4655chloroplasts. Chloroplasts are important for
4656photosynthesis in plants. Chloroplasts also
4657contain various yellow or orange pigments in
4658addition to chlorophyll. Leucoplasts are
4659primarily organelles in which materials such
4660as starch, oils and protein granules are
4661stored.
4662The internal organisation of the plastids
4663consists of numerous membrane layers
4664embedded in a material called the stroma.
4665Plastids are similar to mitochondria in
4666external structure. Like the mitochondria,
4667plastids also have their own DNA and
4668ribosomes.
46695.2.5 ( 5.2.5 (vi) VACUOLES
4670Vacuoles are storage sacs for solid or liquid
4671contents. Vacuoles are small sized in animal
4672cells while plant cells have very large vacuoles.
4673The central vacuole of some plant cells may
4674occupy 50-90% of the cell volume.
4675In plant cells vacuoles are full of cell sap
4676and provide turgidity and rigidity to the cell.
4677Many substances of importance in the life of
4678the plant cell are stored in vacuoles. These
4679include amino acids, sugars, various organic
4680acids and some proteins. In single-celled
4681organisms like Amoeba, the food vacuole
4682contains the food items that the Amoeba has
4683consumed. In some unicellular organisms,
4684specialised vacuoles also play important roles
4685in expelling excess water and some wastes
4686from the cell.
4687uestions
46881. Can you name the two organelles
4689we have studied that contain
4690their own genetic material?
46912. If the organisation of a cell is
4692destroyed due to some physical
4693or chemical influence, what will
4694happen?
46953. Why are lysosomes known as
4696suicide bags?
46974. Where are proteins synthesised
4698inside the cell?
4699Each cell thus acquires its structure and
4700ability to function because of the organisation
4701of its membrane and organelles in specific
4702ways. The cell thus has a basic structural
4703organisation. This helps the cells to perform
4704functions like respiration, obtaining nutrition,
4705and clearing of waste material, or forming new
4706proteins.
4707Thus, the cell is the fundamental
4708structural unit of living organisms. It is also
4709the basic functional unit of life.
4710Q
4711© NCERT
4712not to be republished
471366 SCIENCE
4714What
4715you have you have
4716learnt
4717• The fundamental organisational unit of life is the cell.
4718• Cells are enclosed by a plasma membrane composed of lipids
4719and proteins.
4720• The cell membrane is an active part of the cell. It regulates the
4721movement of materials between the ordered interior of the cell
4722and the outer environment.
4723• In plant cells, a cell wall composed mainly of cellulose is located
4724outside the cell membrane.
4725• The presence of the cell wall enables the cells of plants, fungi
4726and bacteria to exist in hypotonic media without bursting.
4727• The nucleus in eukaryotes is separated from the cytoplasm by
4728double-layered membrane and it directs the life processes of
4729the cell.
4730• The ER functions both as a passageway for intracellular
4731transport and as a manufacturing surface.
4732• The Golgi apparatus consists of stacks of membrane-bound
4733vesicles that function in the storage, modification and packaging
4734of substances manufactured in the cell.
4735• Most plant cells have large membranous organelles called
4736plastids, which are of two types – chromoplasts and leucoplasts.
4737• Chromoplasts that contain chlorophyll are called chloroplasts
4738and they perform photosynthesis.
4739• The primary function of leucoplasts is storage.
4740• Most mature plant cells have a large central vacuole that helps
4741to maintain the turgidity of the cell and stores important
4742substances including wastes.
4743• Prokaryotic cells have no membrane-bound organelles, their
4744chromosomes are composed of only nucleic acid, and they have
4745only very small ribosomes as organelles.
4746Exercises Exercises Exercises
47471. Make a comparison and write down ways in which plant cells
4748are different from animal cells.
47492. How is a prokaryotic cell different from a eukaryotic cell?
47503. What would happen if the plasma membrane ruptures or breaks
4751down?
4752© NCERT
4753not to be republished
4754THE FUNDAMENTAL UNIT OF LIFE 67
47554. What would happen to the life of a cell if there was no Golgi
4756apparatus?
47575. Which organelle is known as the powerhouse of the cell? Why?
47586. Where do the lipids and proteins constituting the cell membrane
4759get synthesised?
47607. How does an Amoeba obtain its food?
47618. What is osmosis?
47629. Carry out the following osmosis experiment:
4763Take four peeled potato halves and scoos each one out to make
4764potato cups. One of these potato cups should be made from a
4765boiled potato. Put each potato cup in a trough containing water.
4766Now,
4767(a) Keep cup A empty
4768(b) Put one teaspoon sugar in cup B
4769(c) Put one teaspoon salt in cup C
4770(d) Put one teaspoon sugar in the boiled potato cup D.
4771Keep these for two hours. Then observe the four potato cups
4772and answer the following:
4773(i) Explain why water gathers in the hollowed portion of
4774B and C.
4775(ii) Why is potato A necessary for this experiment?
4776(iii) Explain why water does not gather in the hollowed out
4777portions of A and D.
4778© NCERT
4779not to be republished
4780From the last chapter, we recall that all living
4781organisms are made of cells. In unicellular
4782organisms, a single cell performs all basic
4783functions. For example, in Amoeba, a single
4784cell carries out movement, intake of food and
4785respiratory gases, respiration and excretion.
4786But in multi-cellular organisms there are
4787millions of cells. Most of these cells are
4788specialised to carry out a few functions. Each
4789specialised function is taken up by a different
4790group of cells. Since these cells carry out only
4791a particular function, they do it very
4792efficiently. In human beings, muscle cells
4793contract and relax to cause movement, nerve
4794cells carry messages, blood flows to transport
4795oxygen, food, hormones and waste material
4796and so on. In plants, vascular tissues conduct
4797food and water from one part of the plant to
4798other parts. So, multi-cellular organisms
4799show division of labour. Cells specialising in
4800one function are often grouped together in
4801the body. This means that a particular
4802function is carried out by a cluster of cells at
4803a definite place in the body. This cluster of
4804cells, called a tissue, is arranged and designed
4805so as to give the highest possible efficiency of
4806function. Blood, phloem and muscle are all
4807examples of tissues.
4808A group of cells that are similar in
4809structure and/or work together to achieve a
4810particular function forms a tissue.
48116.1 Are Plants and Animals Made Are Plants and Animals Made
4812of Same Types of Tissues? of Same Types of Tissues?
4813Let us compare their structure and functions.
4814Do plants and animals have the same
4815structure? Do they both perform similar
4816functions?
4817There are noticeable differences between
4818the two. Plants are stationary or fixed – they
4819don’t move. Most of the tissues they have are
4820supportive, which provides them with
4821structural strength. Most of these tissues are
4822dead, since dead cells can provide mechanical
4823strength as easily as live ones, and need less
4824maintenance.
4825Animals on the other hand move around
4826in search of food, mates and shelter. They
4827consume more energy as compared to plants.
4828Most of the tissues they contain are living.
4829Another difference between animals and
4830plants is in the pattern of growth. The growth
4831in plants is limited to certain regions, while
4832this is not so in animals. There are some
4833tissues in plants that divide throughout their
4834life. These tissues are localised in certain
4835regions. Based on the dividing capacity of the
4836tissues, various plant tissues can be classified
4837as growing or meristematic tissue and
4838permanent tissue. Cell growth in animals is
4839more uniform. So, there is no such
4840demarcation of dividing and non-dividing
4841regions in animals.
4842The structural organisation of organs and
4843organ systems is far more specialised and
4844localised in complex animals than even in very
4845complex plants. This fundamental difference
4846reflects the different modes of life pursued
4847by these two major groups of organisms,
4848particularly in their different feeding methods.
4849Also, they are differently adapted for a
4850sedentary existence on one hand (plants) and
4851active locomotion on the other (animals),
4852contributing to this difference in organ system
4853design.
4854It is with reference to these complex
4855animal and plant bodies that we will now talk
4856about the concept of tissues in some detail.
48576
4858TISSUES
4859Chapter
4860© NCERT
4861not to be republished
4862uestions
48631. What is a tissue?
48642. What is the utility of tissues in
4865multi-cellular organisms?
48666.2 Plant Tissues Plant Tissues
48676.2.1 MERISTEMATIC TISSUE
4868• From the above observations, answer
4869the following questions:
48701. Which of the two onions has longer
4871roots? Why?
48722. Do the roots continue growing even
4873after we have removed their tips?
48743. Why would the tips stop growing in
4875jar 2 after we cut them?
4876The growth of plants occurs only in certain
4877specific regions. This is because the dividing
4878tissue, also known as meristematic tissue, is
4879located only at these points. Depending on
4880the region where they are present,
4881meristematic tissues are classified as apical,
4882lateral and intercalary (Fig. 6.2). New cells
4883produced by meristem are initially like those
4884of meristem itself, but as they grow and
4885mature, their characteristics slowly change
4886and they become differentiated as
4887components of other tissues.
4888Fig. 6.1: Growth of roots in onion bulbs
4889Activity ______________ 6.1
4890• Take two glass jars and fill them with
4891water.
4892• Now, take two onion bulbs and place
4893one on each jar, as shown in
4894Fig. 6.1.
4895• Observe the growth of roots in both the
4896bulbs for a few days.
4897• Measure the length of roots on day 1,
48982 and 3.
4899• On day 4, cut the root tips of the onion
4900bulb in jar 2 by about 1 cm. After this,
4901observe the growth of roots in both the
4902jars and measure their lengths each
4903day for five more days and record the
4904observations in tables, like the table
4905below:
4906Length Day 1 Day 2 Day 3 Day 4 Day 5
4907Jar 1
4908Jar 2
4909Q
4910Apical meristem is present at the growing
4911tips of stems and roots and increases the
4912length of the stem and the root. The girth of
4913the stem or root increases due to lateral
4914meristem (cambium). Intercalary meristem is
4915the meristem at the base of the leaves or
4916internodes (on either side of the node)
4917on twigs.
4918Apical meristem
4919Intercalary meristem
4920Lateral meristem
4921Fig. 6.2: Location of meristematic tissue in plant body
4922Jar 1 Jar 2
4923TISSUES 69
4924© NCERT
4925not to be republished
492670 SCIENCE
4927As the cells of this tissue are very active,
4928they have dense cytoplasm, thin cellulose
4929walls and prominent nuclei. They lack
4930vacuoles. Can we think why they would lack
4931vacuoles? (You might want to refer to the
4932functions of vacuoles in the chapter on cells.)
49336.2.2 PERMANENT TISSUE
4934What happens to the cells formed by
4935meristematic tissue? They take up a specific
4936role and lose the ability to divide. As a result,
4937they form a permanent tissue. This process
4938of taking up a permanent shape, size, and a
4939function is called differentiation. Cells of
4940meristematic tissue differentiate to form
4941different types of permanent tissue.
4942• Now, answer the following on the basis
4943of your observation:
49441. Are all cells similar in structure?
49452. How many types of cells can
4946be seen?
49473. Can we think of reasons why there
4948would be so many types of cells?
4949• We can also try to cut sections of plant
4950roots. We can even try cutting sections
4951of root and stem of different plants.
49526.2.2 ( 6.2.2 (i) SIMPLE PERMANENT TISSUE
4953A few layers of cells form the basic packing
4954tissue. This tissue is parenchyma, a type of
4955permanent tissue. It consists of relatively
4956unspecialised cells with thin cell walls. They
4957are live cells. They are usually loosely packed,
4958Trichome
4959Mucilaginous canal
4960Cuticle
4961Epidermis
4962Hypodermis
4963Cortex
4964Endodermis
4965Pericycle
4966Phloem
4967Cambium
4968Vascular bundle
4969Pith
4970Medullary ray
4971Xylem
4972Fig. 6.3: Section of a stem
4973Activity ______________ 6.2
4974• Take a plant stem and with the help
4975of your teacher cut into very thin slices
4976or sections.
4977• Now, stain the slices with safranin.
4978Place one neatly cut section on a slide,
4979and put a drop of glycerine.
4980• Cover with a cover-slip and observe
4981under a microscope. Observe the
4982various types of cells and their
4983arrangement. Compare it with Fig. 6.3.
4984so that large spaces between cells
4985(intercellular spaces) are found in this tissue
4986[Fig. 6.4 a(i)]. This tissue provides support to
4987plants and also stores food. In some
4988situations, it contains chlorophyll and
4989performs photosynthesis, and then it is called
4990chlorenchyma. In aquatic plants, large air
4991cavities are present in parenchyma to give
4992buoyancy to the plants to help them float.
4993Such a parenchyma type is called
4994aerenchyma. The parenchyma of stems and
4995roots also stores nutrients and water.
4996© NCERT
4997not to be republished
4998TISSUES 71
4999Fig. 6.4: Various types of simple tissues: (a) Parenchyma (i) transverse section, (ii) longitudinal section;
5000(b) Collenchyma (i) transverse section, (ii) longitudinal section; (c) Sclerenchyma (i) transverse section,
5001(ii) longitudinal section.
5002b (i)
5003Wall thickenings
5004Nucleus
5005Vacuole
5006Cell wall
5007b (ii)
5008End wall
5009Primary cell wall
5010(thickened at corners)
5011Chloroplast
5012Nucleus
5013Vacuole
5014Cytoplasm
5015Intercellular space
5016Cytoplasm
5017Nucleus
5018Middle lamella
5019Chloroplast
5020Vacuole
5021Intercellular space
5022Primary cell wall
5023a (ii)
5024a (i)
5025Intercellular spaces
5026Narrow lumen
5027Lignified
5028thick wall
5029c (ii)
5030Simple
5031pit pair
5032c (i)
5033The flexibility in plants is due to another
5034permanent tissue, collenchyma. It allows
5035easy bending in various parts of a plant (leaf,
5036stem) without breaking. It also provides
5037mechanical support to plants. We can find
5038this tissue in leaf stalks below the epidermis.
5039The cells of this tissue are living, elongated
5040and irregularly thickened at the
5041corners. There is very little intercellular
5042space (Fig. 6.4 b).
5043© NCERT
5044not to be republished
504572 SCIENCE
5046Yet another type of permanent tissue is
5047sclerenchyma. It is the tissue which makes
5048the plant hard and stiff. We have seen the
5049husk of a coconut. It is made of
5050sclerenchymatous tissue. The cells of this
5051tissue are dead. They are long and narrow as
5052the walls are thickened due to lignin (a
5053chemical substance which acts as cement and
5054hardens them). Often these walls are so thick
5055that there is no internal space inside the cell
5056(Fig. 6.4 c). This tissue is present in stems,
5057around vascular bundles, in the veins of
5058leaves and in the hard covering of seeds and
5059nuts. It provides strength to the plant parts.
5060Activity ______________ 6.3
5061• Take a freshly plucked leaf of Rhoeo.
5062• Stretch and break it by applying
5063pressure.
5064• While breaking it, keep it stretched
5065gently so that some peel or skin
5066projects out from the cut.
5067• Remove this peel and put it in a petri
5068dish filled with water.
5069• Add a few drops of safranin.
5070• Wait for a couple of minutes and then
5071transfer it onto a slide. Gently place a
5072cover slip over it.
5073• Observe under microscope.
5074epidermis may be thicker since protection
5075against water loss is critical. The entire
5076surface of a plant has this outer covering of
5077epidermis. It protects all the parts of the plant.
5078Epidermal cells on the aerial parts of the plant
5079often secrete a waxy, water-resistant layer on
5080their outer surface. This aids in protection
5081against loss of water, mechanical injury and
5082invasion by parasitic fungi. Since it has a
5083protective role to play, cells of epidermal
5084tissue form a continuous layer without
5085intercellular spaces. Most epidermal cells are
5086relatively flat. Often their outer and side walls
5087are thicker than the inner wall.
5088We can observe small pores here and there
5089in the epidermis of the leaf. These pores are
5090called stomata (Fig. 6.5). Stomata are
5091enclosed by two kidney-shaped cells called
5092guard cells. They are necessary for
5093exchanging gases with the atmosphere.
5094Transpiration (loss of water in the form of
5095water vapour) also takes place through
5096stomata.
5097Think about which gas may be required
5098for photosynthesis.
5099Find out the role of transpiration in plants.
5100Epidermal cells of the roots, whose
5101function is water absorption, commonly bear
5102long hair-like parts that greatly increase the
5103total absorptive surface area.
5104In some plants like desert plants,
5105epidermis has a thick waxy coating of cutin
5106(chemical substance with waterproof quality)
5107on its outer surface. Can we think of a reason
5108for this?
5109Is the outer layer of a branch of a tree
5110different from the outer layer of a young stem?
5111As plants grow older, the outer protective
5112tissue undergoes certain changes. A strip of
5113secondary meristem replaces the epidermis
5114of the stem. Cells on the outside are cut off
5115from this layer. This forms the several-layer
5116thick cork or the bark of the tree. Cells of
5117cork are dead and compactly arranged
5118without intercellular spaces (Fig. 6.6). They
5119also have a chemical called suberin in their
5120walls that makes them impervious to gases
5121and water.
5122Fig. 6.5: Guard cells and epidermal cells: (a) lateral
5123view, (b) surface view
5124(a) (b)
5125Guard
5126cell
5127Stomata
5128Epidermal
5129cell
5130Guard
5131cells
5132What you observe is the outermost layer
5133of cells, called epidermis. The epidermis is
5134usually made of a single layer of cells. In some
5135plants living in very dry habitats, the
5136© NCERT
5137not to be republished
5138TISSUES 73
51396.2.2 ( 6.2.2 (ii) COMPLEX PERMANENT TISSUE
5140The different types of tissues we have
5141discussed until now are all made of one type
5142of cells, which look like each other. Such
5143tissues are called simple permanent tissue.
5144Yet another type of permanent tissue is
5145complex tissue. Complex tissues are made of
5146more than one type of cells. All these cells
5147coordinate to perform a common function.
5148Xylem and phloem are examples of such
5149complex tissues. They are both conducting
5150tissues and constitute a vascular bundle.
5151Vascular or conductive tissue is a distinctive
5152feature of the complex plants, one that has
5153made possible their survival in the terrestrial
5154environment. In Fig. 6.3 showing a section of
5155stem, can you see different types of cells in
5156the vascular bundle?
5157Xylem consists of tracheids, vessels,
5158xylem parenchyma (Fig. 6.7 a,b,c) and xylem
5159fibres. The cells have thick walls, and many
5160of them are dead cells. Tracheids and vessels
5161are tubular structures. This allows them to
5162transport water and minerals vertically. The
5163parenchyma stores food and helps in the
5164sideways conduction of water. Fibres are
5165mainly supportive in function.
5166Phloem is made up of four types of
5167elements: sieve tubes, companion cells,
5168phloem fibres and the phloem parenchyma
5169[Fig. 6.7 (d)]. Sieve tubes are tubular cells with
5170perforated walls. Phloem is unlike xylem in
5171that materials can move in both directions in
5172it. Phloem transports food from leaves to other
5173Cork cells Ruptured epidermis
5174Phloem
5175Fig. 6.6: Protective tissue
5176parts of the plant. Except for phloem fibres,
5177phloem cells are living cells.
5178Xylem
5179Sieve plate
5180Sieve tube
5181Phloem
5182parenchyma
5183Companion cell
5184Nucleus
5185Cytoplasm
5186Pits
5187Pit
5188(a) Tracheid (b) Vessel (c) Xylem parenchyma
5189Fig. 6.7: Types of complex tissue
5190(d) Section of phloem
5191© NCERT
5192not to be republished
519374 SCIENCE
5194uestions
51951. Name types of simple tissues.
51962. Where is apical meristem found?
51973. Which tissue makes up the husk
5198of coconut?
51994. What are the constituents of
5200phloem?
52016.3 Animal Tissues Animal Tissues
5202When we breathe we can actually feel the
5203movement of our chest. How do these body
5204parts move? For this we have specialised cells
5205called muscle cells (Fig. 6.8). The contraction
5206and relaxation of these cells result in
5207movement.
5208During breathing we inhale oxygen. Where
5209does this oxygen go? It is absorbed in the
5210lungs and then is transported to all the body
5211cells through blood. Why would cells need
5212oxygen? The functions of mitochondria we
5213studied earlier provide a clue to this question.
5214Blood flows and carries various substances
5215from one part of the body to the other. For
5216example, it carries oxygen and food to all cells.
5217It also collects wastes from all parts of the
5218body and carries them to the liver and kidney
5219for disposal.
5220Blood and muscles are both examples of
5221tissues found in our body. On the basis of
5222the functions they perform we can think of
5223different types of animal tissues, such as
5224epithelial tissue, connective tissue, muscular
5225tissue and nervous tissue. Blood is a type of
5226connective tissue, and muscle forms
5227muscular tissue.
52286.3.1 EPITHELIAL TISSUE
5229The covering or protective tissues in the
5230animal body are epithelial tissues. Epithelium
5231covers most organs and cavities within the
5232body. It also forms a barrier to keep different
5233body systems separate. The skin, the lining
5234of the mouth, the lining of blood vessels, lung
5235alveoli and kidney tubules are all made of
5236epithelial tissue. Epithelial tissue cells are
5237tightly packed and form a continuous sheet.
5238They have only a small amount of cementing
5239material between them and almost no
5240intercellular spaces. Obviously, anything
5241entering or leaving the body must cross at
5242least one layer of epithelium. As a result, the
5243permeability of the cells of various epithelia
5244play an important role in regulating the
5245exchange of materials between the body and
5246the external environment and also between
5247different parts of the body. Regardless of the
5248type, all epithelium is usually separated from
5249the underlying tissue by an extracellular
5250fibrous basement membrane.
5251Different epithelia (Fig. 6.9) show differing
5252structures that correlate with their unique
5253functions. For example, in cells lining blood
5254vessels or lung alveoli, where transportation
5255of substances occurs through a selectively
5256Q
5257Fig. 6.8: Location of muscle fibres
5258Smooth muscle fibres
5259Smooth muscle fibre
5260(Cell)
5261© NCERT
5262Nucleus
5263not to be republished
5264TISSUES 75
5265permeable surface, there is a simple flat kind
5266of epithelium. This is called the simple
5267squamous epithelium. Simple squamous
5268epithelial cells are extremely thin and flat and
5269form a delicate lining. The oesophagus and
5270the lining of the mouth are also covered with
5271squamous epithelium. The skin, which
5272protects the body, is also made of squamous
5273epithelium. Skin epithelial cells are arranged
5274in many layers to prevent wear and tear. Since
5275they are arranged in a pattern of layers, the
5276epithelium is called stratified squamous
5277epithelium.
5278Where absorption and secretion occur, as
5279in the inner lining of the intestine, tall
5280epithelial cells are present. This columnar
5281(meaning ‘pillar-like’) epithelium facilitates
5282movement across the epithelial barrier. In the
5283respiratory tract, the columnar epithelial
5284tissue also has cilia, which are hair-like
5285projections on the outer surfaces of epithelial
5286cells. These cilia can move, and their
5287movement pushes the mucus forward to clear
5288it. This type of epithelium is thus ciliated
5289columnar epithelium.
5290Cuboidal epithelium (with cube-shaped
5291cells) forms the lining of kidney tubules and
5292ducts of salivary glands, where it provides
5293mechanical support. Epithelial cells often
5294acquire additional specialisation as gland
5295cells, which can secrete substances at the
5296epithelial surface. Sometimes a portion of the
5297epithelial tissue folds inward, and a
5298multicellular gland is formed. This is
5299glandular epithelium.
53006.3.2 CONNECTIVE TISSUE
5301Blood is a type of connective tissue. Why
5302would it be called ‘connective’ tissue? A clue
5303is provided in the introduction of this chapter!
5304Now, let us look at this type of tissue in some
5305more detail. The cells of connective tissue are
5306loosely spaced and embedded in an
5307intercellular matrix (Fig. 6.10). The matrix
5308may be jelly like, fluid, dense or rigid. The
5309nature of matrix differs in concordance with
5310the function of the particular connective
5311tissue.
5312Take a drop of blood on a slide and observe
5313different cells present in it under a microscope.
5314(a) Squamous
5315(d) Stratified squamous
5316(c) Columnar (Ciliated)
5317(b) Cuboidal
5318Fig. 6.9: Different types of epithelial tissues
5319© NCERT
5320not to be republished
532176 SCIENCE
5322Blood has a fluid (liquid) matrix called
5323plasma, in which red blood cells (RBCs), white
5324blood cells (WBCs) and platelets are
5325suspended. The plasma contains proteins,
5326salts and hormones. Blood flows and
5327transports gases, digested food, hormones
5328and waste materials to different parts of the
5329body.
5330Bone is another example of a connective
5331tissue. It forms the framework that supports
5332the body. It also anchors the muscles and
5333supports the main organs of the body. It is a
5334strong and nonflexible tissue (what would be
5335the advantage of these properties for bone
5336functions?). Bone cells are embedded in a
5337hard matrix that is composed of calcium and
5338phosphorus compounds.
5339Two bones can be connected to each other
5340by another type of connective tissue called
5341the ligament. This tissue is very elastic. It has
5342considerable strength. Ligaments contain
5343very little matrix. Tendons connect muscles
5344to bones and are another type of connective
5345tissue. Tendons are fibrous tissue with great
5346strength but limited flexibility.
5347Another type of connective tissue,
5348cartilage, has widely spaced cells. The solid
5349matrix is composed of proteins and sugars.
5350Cartilage smoothens bone surfaces at joints
5351and is also present in the nose, ear, trachea
5352and larynx. We can fold the cartilage of the
5353ears, but we cannot bend the bones in our
5354arms. Think of how the two tissues are
5355different!
5356Areolar connective tissue is found between
5357the skin and muscles, around blood vessels
5358and nerves and in the bone marrow. It fills
5359the space inside the organs, supports internal
5360organs and helps in repair of tissues.
5361Where are fats stored in our body? Fatstoring
5362adipose tissue is found below the skin
5363and between internal organs. The cells of this
5364tissue are filled with fat globules. Storage of
5365fats also lets it act as an insulator.
53666.3.3 MUSCULAR TISSUE
5367Muscular tissue consists of elongated cells,
5368also called muscle fibres. This tissue is
5369responsible for movement in our body.
5370Reticular fibre Fibroblast
5371Macrophage
5372Collagen fibre
5373Mast cell Plasma cell
5374Fat droplet Nucleus
5375Adipocyte
5376(a)
5377(b) Haversian canal
5378(contains blood vessels
5379and nerve fibres)
5380Canaliculus (contains
5381slender process of bone
5382cell or osteocyte)
5383Hyaline matrix
5384Chondrocyte
5385Different white
5386blood corpuscles
5387Neutrophil
5388(polynuclear
5389leucocyte)
5390Eosinophil Basophil
5391Nucleus
5392Cytoplasm
5393Red blood
5394corpuscle
5395Lymphocyte Monocyte Platelets
5396(c)
5397(d)
5398(e)
5399Fig. 6.10: Types of connective tissues: (a) areolar
5400tissue, (b) adipose tissue, (c) compact
5401bone, (d) hyaline cartilage, (e) types of
5402blood cells
5403© NCERT
5404not to be republished
5405TISSUES 77
5406Muscles contain special proteins called
5407contractile proteins, which contract and relax
5408to cause movement.
5409to bones and help in body movement. Under
5410the microscope, these muscles show alternate
5411light and dark bands or striations when
5412stained appropriately. As a result, they are
5413also called striated muscles. The cells of this
5414tissue are long, cylindrical, unbranched and
5415multinucleate (having many nuclei).
5416The movement of food in the alimentary
5417canal or the contraction and relaxation of
5418blood vessels are involuntary movements. We
5419cannot really start them or stop them simply
5420by wanting to do so! Smooth muscles [Fig.
54216.11(b)] or involuntary muscles control such
5422movements. They are also found in the iris of
5423the eye, in ureters and in the bronchi of the
5424lungs. The cells are long with pointed ends
5425(spindle-shaped) and uninucleate (having a
5426single nucleus). They are also called
5427unstriated muscles – why would they be
5428called that?
5429The muscles of the heart show rhythmic
5430contraction and relaxation throughout life.
5431These involuntary muscles are called cardiac
5432muscles [Fig. 6.11(c)]. Heart muscle cells are
5433cylindrical, branched and uninucleate.
5434Compare the structures of different types
5435of muscular tissues. Note their shape,
5436number of nuclei and position of nuclei within
5437the cell.
54386.3.4 NERVOUS TISSUE
5439All cells possess the ability to respond to
5440stimuli. However, cells of the nervous tissue
5441are highly specialised for being stimulated
5442and then transmitting the stimulus very
5443rapidly from one place to another within the
5444body. The brain, spinal cord and nerves are
5445all composed of the nervous tissue. The cells
5446of this tissue are called nerve cells or neurons.
5447A neuron consists of a cell body with a
5448nucleus and cytoplasm, from which long thin
5449hair-like parts arise (Fig. 6.12). Usually each
5450neuron has a single long part, called the axon,
5451and many short, branched parts called
5452dendrites. An individual nerve cell may be up
5453to a metre long. Many nerve fibres bound
5454together by connective tissue make up
5455a nerve.
5456Nuclei
5457Striations
5458(a)
5459Spindle shaped
5460muscle cell
5461Nucleus
5462(b)
5463(c)
5464Striations
5465Nuclei
5466Fig. 6.11: Types of muscles fibres: (a) striated
5467muscle, (b) smooth muscle, (c) cardiac
5468muscle
5469We can move some muscles by conscious
5470will. Muscles present in our limbs move when
5471we want them to, and stop when we so decide.
5472Such muscles are called voluntary muscles
5473[Fig. 6.11(a)]. These muscles are also called
5474skeletal muscles as they are mostly attached
5475© NCERT
5476not to be republished
547778 SCIENCE
5478Nerve impulses allow us to move our
5479muscles when we want to. The functional
5480Nucleus
5481Dendrite
5482Axon
5483Nerve ending
5484Cell body
5485combination of nerve and muscle tissue is
5486fundamental to most animals. This
5487combination enables animals to move rapidly
5488in response to stimuli.
5489uestions
54901. Name the tissue responsible for
5491movement in our body.
54922. What does a neuron look like?
54933. Give three features of cardiac
5494muscles.
54954. What are the functions of areolar Q tissue?
5496What
5497you have you have
5498learnt
5499• Tissue is a group of cells similar in structure and function.
5500• Plant tissues are of two main types – meristematic and
5501permanent.
5502• Meristematic tissue is the dividing tissue present in the growing
5503regions of the plant.
5504• Permanent tissues are derived from meristematic tissue once
5505they lose the ability to divide. They are classified as simple and
5506complex tissues.
5507• Parenchyma, collenchyma and sclerenchyma are three types
5508of simple tissues. Xylem and phloem are types of complex
5509tissues.
5510• Animal tissues can be epithelial, connective, muscular and
5511nervous tissue.
5512• Depending on shape and function, epithelial tissue is classified
5513as squamous, cuboidal, columnar, ciliated and glandular.
5514• The different types of connective tissues in our body include
5515areolar tissue, adipose tissue, bone, tendon, ligament, cartilage
5516and blood.
5517• Striated, unstriated and cardiac are three types of muscle
5518tissues.
5519• Nervous tissue is made of neurons that receive and conduct
5520impulses.
5521Fig. 6.12: Neuron-unit of nervous tissue
5522© NCERT
5523not to be republished
5524TISSUES 79
5525Exercises Exercises Exercises
55261. Define the term “tissue”.
55272. How many types of elements together make up the xylem tissue?
5528Name them.
55293. How are simple tissues different from complex tissues in plants?
55304. Differentiate between parenchyma, collenchyma and
5531sclerenchyma on the basis of their cell wall.
55325. What are the functions of the stomata?
55336. Diagrammatically show the difference between the three types
5534of muscle fibres.
55357. What is the specific function of the cardiac muscle?
55368. Differentiate between striated, unstriated and cardiac muscles
5537on the basis of their structure and site/location in the body.
55389. Draw a labelled diagram of a neuron.
553910. Name the following.
5540(a) Tissue that forms the inner lining of our mouth.
5541(b) Tissue that connects muscle to bone in humans.
5542(c) Tissue that transports food in plants.
5543(d) Tissue that stores fat in our body.
5544(e) Connective tissue with a fluid matrix.
5545(f) Tissue present in the brain.
554611. Identify the type of tissue in the following: skin, bark of tree,
5547bone, lining of kidney tubule, vascular bundle.
554812. Name the regions in which parenchyma tissue is present.
554913. What is the role of epidermis in plants?
555014. How does the cork act as a protective tissue?
555115. Complete the table:
5552© NCERT
5553not to be republished
5554Have you ever thought of the multitude of
5555life-forms that surround us? Each organism
5556is different from all others to a lesser or
5557greater extent. For instance, consider yourself
5558and a friend.
5559• Are you both of the same height?
5560• Does your nose look exactly like your
5561friend’s nose?
5562• Is your hand-span the same as your
5563friend’s?
5564However, if we were to compare ourselves
5565and our friends with a monkey, what would
5566we say? Obviously, we and our friends have
5567a lot in common when we compare ourselves
5568with a monkey. But suppose we were to add
5569a cow to the comparison? We would then
5570think that the monkey has a lot more in
5571common with us than with the cow.
5572Activity ______________ 7.1
5573• We have heard of ‘desi’ cows and Jersey
5574cows.
5575• Does a desi cow look like a Jersey cow?
5576• Do all desi cows look alike?
5577• Will we be able to identify a Jersey cow
5578in a crowd of desi cows that don’t look
5579like each other?
5580• What is the basis of our identification?
5581In this activity, we had to decide which
5582characteristics were more important in
5583forming the desired category. Hence, we were
5584also deciding which characteristics could be
5585ignored.
5586Now, think of all the different forms in
5587which life occurs on earth. On one hand we
5588have microscopic bacteria of a few micrometre
5589in size. While on the other hand we have blue
5590whale and red wood trees of california
5591of approximate sizes of 30 metres and
5592100 metres respectively. Some pine trees live
5593for thousands of years while insects like
5594mosquitoes die within a few days. Life also
5595ranges from colourless or even transparent
5596worms to brightly coloured birds and flowers.
5597This bewildering variety of life around us
5598has evolved on the earth over millions of
5599years. However, we do not have more than a
5600tiny fraction of this time to try and
5601understand all these living organisms, so we
5602cannot look at them one by one. Instead, we
5603look for similarities among the organisms,
5604which will allow us to put them into different
5605classes and then study different classes or
5606groups as a whole.
5607In order to make relevant groups to study
5608the variety of life forms, we need to decide
5609which characteristics decide more
5610fundamental differences among organisms.
5611This would create the main broad groups of
5612organisms. Within these groups, smaller subgroups
5613will be decided by less important
5614characteristics.
5615uestions
56161. Why do we classify organisms?
56172. Give three examples of the range
5618of variations that you see in lifeforms
5619around you.
56207.1 What is the Basis of What is the Basis of
5621Classification? Classification?
5622Attempts at classifying living things into
5623groups have been made since time
5624immemorial. Greek thinker Aristotle classified
5625animals according to whether they lived on
5626Q
56277
5628DIVERSITY IVERSITY IN LIVING ORGANISMS RGANISMS
5629Chapter
5630© NCERT
5631not to be republished
5632land, in water or in the air. This is a very
5633simple way of looking at life, but misleading
5634too. For example, animals that live in the sea
5635include corals, whales, octopuses, starfish
5636and sharks. We can immediately see that
5637these are very different from each other in
5638numerous ways. In fact, their habitat is the
5639only point they share in common. This is not
5640an appropriate way of making groups of
5641organisms to study and think about.
5642We therefore need to decide which
5643characteristics to be used as the basis for
5644making the broadest divisions. Then we will
5645have to pick the next set of characteristics
5646for making sub-groups within these divisions.
5647This process of classification within each
5648group can then continue using new
5649characteristics each time.
5650Before we go on, we need to think about
5651what is meant by ‘characteristics’. When we
5652are trying to classify a diverse group of
5653organisms, we need to find ways in which
5654some of them are similar enough to be
5655thought of together. These ‘ways’, in fact, are
5656details of appearance or behaviour, in other
5657words, form and function.
5658What we mean by a characteristic is a
5659particular form or a particular function. That
5660most of us have five fingers on each hand is
5661thus a characteristic. That we can run, but
5662the banyan tree cannot, is also a
5663characteristic.
5664Now, to understand how some
5665characteristics are decided as being more
5666fundamental than others, let us consider how
5667a stone wall is built. The stones used will have
5668different shapes and sizes. The stones at the
5669top of the wall would not influence the choice
5670of stones that come below them. On the other
5671hand, the shapes and sizes of stones in the
5672lowermost layer will decide the shape and size
5673of the next layer and so on.
5674The stones in the lowermost layer are like
5675the characteristics that decide the broadest
5676divisions among living organisms. They are
5677independent of any other characteristics in
5678their effects on the form and function of the
5679organism. The characteristics in the next level
5680would be dependent on the previous one and
5681would decide the variety in the next level. In
5682this way, we can build up a whole hierarchy
5683of mutually related characteristics to be used
5684for classification.
5685Now-a-days, we look at many inter-related
5686characteristics starting from the nature of the
5687cell in order to classify all living organisms.
5688What are some concrete examples of such
5689characteristics used for a hierarchical
5690classification?
5691• A eukaryotic cell has membrane-bound
5692organelles, including a nucleus, which
5693allow cellular processes to be carried out
5694efficiently in isolation from each other.
5695Therefore, organisms which do not have
5696a clearly demarcated nucleus and other
5697organelles would need to have their
5698biochemical pathways organised in very
5699different ways. This would have an effect
5700on every aspect of cell design. Further,
5701nucleated cells would have the capacity
5702to participate in making a multicellular
5703organism because they can take up
5704specialised functions. Therefore, this is
5705a basic characteristic of classification.
5706• Do the cells occur singly or are they
5707grouped together and do they live as an
5708indivisible group? Cells that group
5709together to form a single organism use
5710the principle of division of labour. In such
5711a body design, all cells would not be
5712identical. Instead, groups of cells will
5713carry out specialised functions. This
5714makes a very basic distinction in the
5715body designs of organisms. As a result,
5716an Amoeba and a worm are very different
5717in their body design.
5718• Do organisms produce their own food
5719through the process of photosynthesis?
5720Being able to produce one’s own food
5721versus having to get food from outside
5722would make very different body designs
5723necessary.
5724• Of the organisms that perform
5725photosynthesis (plants), what is the level
5726of organisation of their body?
5727• Of the animals, how does the individual’s
5728body develop and organise its different
5729parts, and what are the specialised
5730organs found for different functions?
5731DIVERSITY IN LIVING ORGANISMS 81
5732© NCERT
5733not to be republished
573482 SCIENCE
5735We can see that, even in these few questions
5736that we have asked, a hierarchy is developing.
5737The characteristics of body design used for
5738classification of plants will be very different
5739from those important for classifying animals.
5740This is because the basic designs are different,
5741based on the need to make their own food
5742(plants), or acquire it (animals). Therefore,
5743these design features (having a skeleton, for
5744example) are to be used to make sub-groups,
5745rather than making broad groups.
5746uestions
57471. Which do you think is a more basic
5748characteristic for classifying
5749organisms?
5750(a) the place where they live.
5751(b) the kind of cells they are
5752made of. Why?
57532. What is the primary characteristic
5754on which the first division of
5755organisms is made?
57563. On what bases are plants and
5757animals put into different
5758categories?
57597.2 Classification and Evolution
5760All living things are identified and categorised
5761on the basis of their body design in form and
5762function. Some characteristics are likely to
5763make more wide-ranging changes in body
5764design than others. There is a role of time in
5765this as well. So, once a certain body design
5766comes into existence, it will shape the effects
5767of all other subsequent design changes,
5768simply because it already exists. In other
5769words, characteristics that came into
5770existence earlier are likely to be more basic
5771than characteristics that have come into
5772existence later.
5773This means that the classification of life
5774forms will be closely related to their evolution.
5775What is evolution? Most life forms that we
5776see today have arisen by an accumulation of
5777changes in body design that allow the
5778organism possessing them to survive better.
5779Charles Darwin first described this idea of
5780evolution in 1859 in his book, The Origin of
5781Species.
5782When we connect this idea of evolution to
5783classification, we will find some groups of
5784organisms which have ancient body designs
5785that have not changed very much. We will
5786also find other groups of organisms that have
5787acquired their particular body designs
5788relatively recently. Those in the first group
5789are frequently referred to as ‘primitive’ or ‘lower’
5790organisms, while those in the second group
5791are called ‘advanced’ or ‘higher’ organisms. In
5792reality, these terms are not quite correct since
5793they do not properly relate to the differences.
5794All that we can say is that some are ‘older’
5795organisms, while some are ‘younger’
5796organisms. Since there is a possibility that
5797complexity in design will increase over
5798evolutionary time, it may not be wrong to say
5799that older organisms are simpler, while Q younger organisms are more complex. More to know
5800Biodiversity means the diversity of life
5801forms. It is a word commonly used to
5802refer to the variety of life forms found
5803in a particular region. Diverse life forms
5804share the environment, and are
5805affected by each other too. As a result,
5806a stable community of different species
5807comes into existence. Humans have
5808played their own part in recent times
5809in changing the balance of such
5810communities. Of course, the diversity
5811in such communities is affected by
5812particular characteristics of land,
5813water, climate and so on. Rough
5814estimates state that there are about ten
5815million species on the planet, although
5816we actually know only one or two
5817millions of them. The warm and humid
5818tropical regions of the earth, between
5819the tropic of Cancer and the tropic of
5820Capricorn, are rich in diversity of plant
5821and animal life. This is called the region
5822of megadiversity. Of the biodiversity
5823of the planet, more than half is
5824concentrated in a few countries –
5825Brazil, Colombia, Ecuador, Peru,
5826Mexico, Zaire, Madagascar,
5827Australia, China, India, Indonesia and
5828Malaysia.
5829© NCERT
5830not to be republished
5831DIVERSITY IN LIVING ORGANISMS 83
5832uestions
58331. Which organisms are called
5834primitive and how are they
5835different from the so-called
5836advanced organisms?
58372. Will advanced organisms be the
5838same as complex organisms?
5839Why?
58407.3 The Hierarchy of Classification- The Hierarchy of Classification- The Hierarchy of ClassificationGroups
5841Biologists, such as Ernst Haeckel (1894),
5842Robert Whittaker (1959) and Carl Woese
5843(1977) have tried to classify all living
5844organisms into broad categories, called
5845kingdoms. The classification Whittaker
5846proposed has five kingdoms: Monera,
5847Protista, Fungi, Plantae and Animalia, and
5848is widely used. These groups are formed on
5849the basis of their cell structure, mode and
5850source of nutrition and body organisation.
5851The modification Woese introduced by
5852dividing the Monera into Archaebacteria (or
5853Archaea) and Eubacteria (or Bacteria) is also
5854in use.
5855Further classification is done by naming
5856the sub-groups at various levels as given in
5857the following scheme:
5858Kingdom
5859Phylum (for animals) / Division (for plants)
5860Class
5861Order
5862Family
5863Genus
5864Species
5865Thus, by separating organisms on the
5866basis of a hierarchy of characteristics into
5867smaller and smaller groups, we arrive at the
5868basic unit of classification, which is a
5869‘species’. So what organisms can be said to
5870belong to the same species? Broadly, a species
5871includes all organisms that are similar
5872enough to breed and perpetuate.
5873The important characteristics of the five
5874kingdoms of Whittaker are as follows:
58757.3.1 MONERA
5876These organisms do not have a defined
5877nucleus or organelles, nor do any of them
5878show multi-cellular body designs. On the
5879other hand, they show diversity based on
5880many other characteristics. Some of them
5881have cell walls while some do not. Of course,
5882having or not having a cell wall has very
5883different effects on body design here from
5884having or not having a cell wall in multicellular
5885organisms. The mode of nutrition of
5886organisms in this group can be either by
5887synthesising their own food (autotrophic) or
5888getting it from the environment
5889(heterotrophic). This group includes bacteria,
5890blue-green algae or cyanobacteria, and
5891mycoplasma. Some examples are shown
5892in Fig. 7.1.
5893Q
5894Bacteria
5895Resting
5896spore
5897Heterocyst
5898Anabaena
5899Fig. 7.1: Monera
59007.3.2 PROTISTA
5901This group includes many kinds of unicellular
5902eukaryotic organisms. Some of these
5903organisms use appendages, such as hair-like
5904cilia or whip-like flagella for moving around.
5905Their mode of nutrition can be autotrophic
5906or heterotrophic. Examples are unicellular
5907algae, diatoms and protozoans (see Fig. 7.2
5908for examples).
5909© NCERT
5910not to be republished
591184 SCIENCE
5912to become multicellular organisms at certain
5913stages in their lives. They have cell-walls made
5914of a tough complex sugar called chitin.
5915Examples are yeast and mushrooms (see Fig.
59167.3 for examples).
5917Water vacuole
5918Cilia
5919Macronucleus
5920Micronucleus
5921Oral groove
5922Cytosome
5923Food vacuole
5924Cytopyge
5925Waste
5926Flagellum (long)
5927Flagellum (short)
5928Eyespot
5929Photoreceptor
5930Contractile
5931vacuole Chloroplast
5932Nucleolus
5933Nucleus
5934Ectoplasm
5935Endoplasm
5936Mitochondria
5937Nucleus
5938Crystals
5939Food vacuole
5940Contractile vacuole
5941Advancing
5942pseudopod
5943Paramecium
5944Amoeba
5945Euglena
5946Fig. 7.2: Protozoa
59477.3.3 FUNGI
5948These are heterotrophic eukaryotic
5949organisms. They use decaying organic
5950material as food and are therefore called
5951saprophytes. Many of them have the capacity
5952Fig. 7.3: Fungi
5953Some fungal species live in permanent
5954mutually dependent relationships with bluegreen
5955algae (or cyanobacteria). Such
5956relationships are called symbiotic. These
5957symbiobic life forms are called lichens. We
5958have all seen lichens as the slow-growing
5959large coloured patches on the bark of trees.
59607.3.4 PLANTAE
5961These are multicellular eukaryotes with cell
5962walls. They are autotrophs and use
5963chlorophyll for photosynthesis. Thus, all
5964plants are included in this group. Since
5965plants and animals are most visible forms
5966of the diversity of life around us, we will look
5967at the subgroups in this category later
5968(section 7.4).
59697.3.5 ANIMALIA
5970These include all organisms which are
5971multicellular eukaryotes without cell walls.
5972They are heterotrophs. Again, we will look
5973at their subgroups a little later in
5974section 7.5.
5975Penicillium Agaricus
5976Aspergillus
5977© NCERT
5978not to be republished
5979DIVERSITY IN LIVING ORGANISMS 85
5980Fig. 7.4: The Five Kingdom classification
5981uestions
59821. What is the criterion for
5983classification of organisms as
5984belonging to kingdom Monera or
5985Protista?
59862. In which kingdom will you place
5987an organism which is singlecelled,
5988eukaryotic and
5989photosynthetic?
59903. In the hierarchy of classification,
5991which grouping will have the
5992smallest number of organisms
5993with a maximum of
5994characteristics in common and
5995which will have the largest
5996number of organisms?
5997Q 7.4 Plantae
5998The first level of classification among plants
5999depends on whether the plant body has welldifferentiated,
6000distinct components. The next
6001level of classification is based on whether the
6002differentiated plant body has special tissues
6003for the transport of water and other
6004substances within it. Further classification
6005looks at the ability to bear seeds and whether
6006the seeds are enclosed within fruits.
60077.4.1 THALLOPHYTA
6008Plants that do not have well-differentiated
6009body design fall in this group. The plants in
6010this group are commonly called algae. These
6011© NCERT
6012not to be republished
601386 SCIENCE
6014plants are predominantly aquatic. Examples
6015are Spirogyra, Ulothrix, Cladophora and Chara
6016(see Fig. 7.5).
60177.4.2 BRYOPHYTA
6018These are called the amphibians of the plant
6019kingdom. The plant body is commonly
6020differentiated to form stem and leaf-like
6021structures. However, there is no specialised
6022tissue for the conduction of water and other
6023substances from one part of the plant body
6024to another. Examples are moss (Funaria) and
6025Marchantia (see Fig. 7.6).
6026Fig. 7.5: Thallophyta – Algae
6027Cell-wall
6028Chloroplast
6029Pyrenoids
6030Nucleus
6031Cytoplasm
6032Ulothrix
6033Cladophora
6034Ulva
6035Spirogyra
6036Chara
6037Fig. 7.6: Some common bryophytes
60387.4.3 PTERIDOPHYTA
6039In this group, the plant body is differentiated
6040into roots, stem and leaves and has
6041specialised tissue for the conduction of water
6042and other substances from one part of the
6043plant body to another. Some examples are
6044Marsilea, ferns and horse-tails (see Fig. 7.7).
6045The thallophytes, the bryophytes and the
6046pteridophytes have naked embryos that are
6047called spores. The reproductive organs of
6048plants in all these three groups are very
6049inconspicuous, and they are therefore called
6050‘cryptogamae’, or ‘those with hidden
6051reproductive organs’.
6052Riccia
6053Marchantia Funaria
6054© NCERT
6055not to be republished
6056DIVERSITY IN LIVING ORGANISMS 87
6057On the other hand, plants with welldifferentiated
6058reproductive tissues that
6059ultimately make seeds are called
6060phanerogams. Seeds are the result of the
6061reproductive process. They consist of the
6062embryo along with stored food, which serves
6063for the initial growth of the embryo during
6064germination. This group is further classified,
6065based on whether the seeds are naked or
6066enclosed in fruits, giving us two groups:
6067gymnosperms and angiosperms.
60687.4.4 GYMNOSPERMS
6069This term is made from two Greek words:
6070gymno– means naked and sperma– means
6071seed. The plants of this group bear naked
6072seeds and are usually perennial, evergreen
6073and woody. Examples are pines and deodar
6074(see Fig. 7.8 for examples).
60757.4.5 ANGIOSPERMS
6076This word is made from two Greek words:
6077angio means covered and sperma– means
6078seed. The seeds develop inside an organ which
6079is modified to become a fruit. These are also
6080called flowering plants. Plant embryos in
6081seeds have structures called cotyledons.
6082Cotyledons are called ‘seed leaves’ because
6083in many instances they emerge and become
6084green when the seed germinates. Thus,
6085cotyledons represent a bit of pre-designed
6086plant in the seed. The angiosperms are
6087divided into two groups on the basis of the
6088number of cotyledons present in the seed.
6089Plants with seeds having a single cotyledon
6090are called monocotyledonous or monocots.
6091Plants with seeds having two cotyledons are
6092called dicots (see Figs. 7.9 and 7.10).
6093Marsilea Fern
6094Leaf
6095Sporocarp
6096Stem
6097Root
6098Fig. 7.7: Pteridophyta
6099Pinus
6100Fig. 7.8: Gymnosperms
6101Cycas
6102Fig. 7.9: Monocots – Paphiopedilum
6103Fig. 7.10: Dicots – Ipomoea
6104© NCERT
6105not to be republished
610688 SCIENCE
6107Activity ______________ 7.2
6108• Soak seeds of green gram, wheat,
6109maize, peas and tamarind. Once they
6110become tender, try to split the seed. Do
6111all the seeds break into two nearly
6112equal halves?
6113• The seeds that do are the dicot seeds
6114and the seeds that don’t are the
6115monocot seeds.
6116• Now take a look at the roots, leaves and
6117flowers of these plants.
6118• Are the roots tap-roots or fibrous?
6119• Do the leaves have parallel or reticulate
6120venation?
6121Fig. 7.11: Classification of plants
6122• How many petals are found in the
6123flower of these plants?
6124• Can you write down further
6125characteristics of monocots and dicots
6126on the basis of these observations?
6127uestions
61281. Which division among plants has
6129the simplest organisms?
61302. How are pteridophytes different
6131from the phanerogams?
61323. How do gymnosperms and
6133angiosperms differ from each Q other?
6134© NCERT
6135not to be republished
6136DIVERSITY IN LIVING ORGANISMS 89
61377.5 Animalia nimalia
6138These are organisms which are eukaryotic,
6139multicellular and heterotrophic. Their cells
6140do not have cell-walls. Most animals are
6141mobile.
6142They are further classified based on the
6143extent and type of the body design
6144differentiation found.
61457.5.1 PORIFERA
6146The word Porifera means organisms with
6147holes. These are non-motile animals attached
6148to some solid support. There are holes or
6149‘pores’, all over the body. These lead to a canal
6150system that helps in circulating water
6151throughout the body to bring in food and
6152oxygen. These animals are covered with a
6153hard outside layer or skeleton. The body
6154design involves very minimal differentiation
6155and division into tissues. They are commonly
6156called sponges, and are mainly found in
6157marine habitats. Some examples are shown
6158in Fig. 7.12.
6159layers of cells: one makes up cells on the
6160outside of the body, and the other makes the
6161inner lining of the body. Some of these species
6162live in colonies (corals), while others have a
6163solitary like–span (Hydra). Jellyfish and sea
6164anemones are common examples (see
6165Fig. 7.13).
6166Euplectelea Sycon
6167Spongilla
6168Fig. 7.12: Porifera
6169Tentacles
6170Sea anemone
6171Tentacles
6172Stinging cell
6173Mouth
6174Epidermis
6175Mesoglea
6176Gastrodermis
6177Gastrovascular
6178cavity
6179Foot
6180Fig. 7.13: Coelenterata
61817.5.3 PLATYHELMINTHES
6182The body of animals in this group is far more
6183complexly designed than in the two other
6184groups we have considered so far. The body
6185is bilaterally symmetrical, meaning that the
6186left and the right halves of the body have the
6187same design. There are three layers of cells
6188from which differentiated tissues can be
6189made, which is why such animals are called
6190triploblastic. This allows outside and inside
6191body linings as well as some organs to be
6192made. There is thus some degree of tissue
6193formation. However, there is no true internal
6194body cavity or coelom, in which welldeveloped
6195organs can be accommodated. The
6196body is flattened dorsiventrally, meaning from
6197top to bottom, which is why these animals
6198are called flatworms. They are either freeliving
6199or parasitic. Some examples are freeliving
6200animals like planarians, or parasitic
6201animals like liverflukes (see Fig. 7.14 for
6202examples).
6203Hydra
62047.5.2 COELENTERATA (CNIDARIA)
6205These are animals living in water. They show
6206more body design differentiation. There is a
6207cavity in the body. The body is made of two
6208© NCERT
6209not to be republished
621090 SCIENCE
62117.5.5 ANNELIDA
6212Annelid animals are also bilaterally
6213symmetrical and triploblastic, but in addition
6214they have a true body cavity. This allows true
6215organs to be packaged in the body structure.
6216There is, thus, extensive organ differentiation.
6217This differentiation occurs in a segmental
6218fashion, with the segments lined up one after
6219the other from head to tail. These animals
6220are found in a variety of habitats– fresh water,
6221marine water as well as land. Earthworms
6222and leeches are familiar examples (see
6223Fig. 7.16).
6224Acetabulum
6225Scolex Sucker
6226Neck
6227Liverfluke Tape worm
6228Eyes
6229Branched
6230gastrovascular
6231cavity
6232Pharynx
6233Mouth
6234and anus
6235Planaria
6236Fig. 7.14: Platyhelminthes
62377.5.4 NEMATODA
6238The nematode body is also bilaterally
6239symmetrical and triploblastic. However, the
6240body is cylindrical rather than flattened.
6241There are tissues, but no real organs,
6242although a sort of body cavity or a pseudocoelom,
6243is present. These are very familiar
6244as parasitic worms causing diseases, such
6245as the worms causing elephantiasis (filarial
6246worms) or the worms in the intestines
6247(roundworm or pinworms). Some examples
6248are shown in Fig. 7.15.
6249Female
6250Male
6251Ascaris Wuchereria
6252Fig. 7.15: Nematodes (Aschelminthes)
6253Tentacle
6254Palp
6255Parapodia
6256Parapodia
6257Nereis Earthworm Leech
6258Fig. 7.16: Annelida
62597.5.6 ARTHROPODA
6260This is probably the largest group of animals.
6261These animals are bilaterally symmetrical and
6262segmented. There is an open circulatory
6263system, and so the blood does not flow in welldefined
6264blood vessels. The coelomic cavity is
6265blood-filled. They have jointed legs (the word
6266‘arthropod’ means ‘jointed legs’). Some
6267familiar examples are prawns, butterflies,
6268houseflies, spiders, scorpions and crabs (see
6269Fig. 7.17).
6270Genital
6271papillae Anus
6272© NCERT
6273not to be republished
6274DIVERSITY IN LIVING ORGANISMS 91
6275Palamnaeus
6276(Scorpion)
62777.5.7 MOLLUSCA
6278In the animals of this group, there is bilateral
6279symmetry. The coelomic cavity is reduced.
6280There is little segmentation. They have an
6281open circulatory system and kidney-like
6282organs for excretion. There is a foot that is
6283used for moving around. Examples are snails
6284and mussels (see Fig. 7.18).
62857.5.8 ECHINODERMATA
6286In Greek, echinos means hedgehog, and
6287derma means skin. Thus, these are spiny
6288skinned organisms. These are exclusively
6289free-living marine animals. They are
6290triploblastic and have a coelomic cavity. They
6291also have a peculiar water-driven tube system
6292that they use for moving around. They have
6293hard calcium carbonate structures that they
6294use as a skeleton. Examples are starfish and
6295sea urchins (see Fig. 7.19).
6296Pariplaneta
6297(Cockroach)
6298Palaemon
6299(Prawn)
6300Aranea(Spider)
6301Fig. 7.17: Arthropoda
6302Chiton
6303Octopus
6304Pila
6305Unio
6306Fig. 7.18: Mollusca
6307Antedon
6308(feather star)
6309Holothuria
6310(sea cucumber)
6311Echinus (sea urchin) Asterias (star fish)
63127.5.9 PROTOCHORDATA
6313These animals are bilaterally symmetrical,
6314triploblastic and have a coelom. In addition,
6315they show a new feature of body design,
6316namely a notochord, at least at some stages
6317during their lives. The notochord is a long
6318rod-like support structure (chord=string) that
6319runs along the back of the animal separating
6320the nervous tissue from the gut. It provides a
6321place for muscles to attach for ease of
6322movement. Protochordates may not have a
6323proper notochord present at all stages in their
6324Fig. 7.19: Echinodermata
6325Musca
6326(House fly)
6327Butterfly
6328Scolopendra
6329(Centipede)
6330© NCERT
6331not to be republished
633292 SCIENCE
6333lives or for the entire length of the animal.
6334Protochordates are marine animals.
6335Examples are Balanoglossus, Herdmania and
6336Amphioxus (see Fig. 7.20).
63377.5.10 ( 7.5.10 (i) PISCES
6338These are fish. They are exclusively aquatic
6339animals. Their skin is covered with scales/
6340plates. They obtain oxygen dissolved in water
6341by using gills. The body is streamlined, and
6342a muscular tail is used for movement. They
6343are cold-blooded and their hearts have only
6344two chambers, unlike the four that humans
6345have. They lay eggs. We can think of many
6346kinds of fish, some with skeletons made
6347entirely of cartilage, such as sharks, and some
6348with a skeleton made of both bone and
6349cartilage, such as tuna or rohu [see examples
6350in Figs. 7.21 (a) and 7.21 (b)].
6351Proboscis
6352Collarette
6353Collar
6354Branchial region
6355Gill pores
6356Dorsally
6357curved
6358genital wings
6359Middosrsal
6360ridge
6361Hepatic caeca
6362Hepatic region
6363Posthepatic
6364region
6365Anus
6366Fig. 7.20: A Protochordata: Balanoglossus
63677.5.10 VERTEBRATA
6368These animals have a true vertebral column
6369and internal skeleton, allowing a completely
6370different distribution of muscle attachment
6371points to be used for movement.
6372Vertebrates are bilaterally symmetrical,
6373triploblastic, coelomic and segmented, with
6374complex differentiation of body tissues and
6375organs. All chordates possess the following
6376features:
6377(i) have a notochord
6378(ii) have a dorsal nerve cord
6379(iii) are triploblastic
6380(iv) have paired gill pouches
6381(v) are coelomate.
6382Vertebrates are grouped into five classes.
6383Caulophyryne jordani
6384(Angler fish)
6385Synchiropus splendidus
6386(Mandarin fish)
6387Pterois volitans
6388(Lion fish)
6389Spiracle Eye
6390Pelvic fin
6391Dorsal fin
6392Caudal fin
6393Tail
6394Electric ray (Torpedo)
6395Sting ray
6396Mouth
6397Dorsal fin
6398Eye Tail
6399Gills Pectoral
6400fin
6401Pelvic
6402fin
6403Fig. 7.21 (a): Pisces
6404Scoliodon (Dog fish)
6405© NCERT
6406not to be republished
6407DIVERSITY IN LIVING ORGANISMS 93
64087.5.10 ( 7.5.10 (ii) AMPHIBIA
6409These animals differ from the fish in the lack
6410of scales, in having mucus glands in the skin,
6411and a three-chambered heart. Respiration is
6412through either gills or lungs. They lay eggs.
6413These animals are found both in water and
6414on land. Frogs, toads and salamanders are
6415some examples (see Fig. 7.22).
64167.5.10 ( 7.5.10 (iii) REPTILIA
6417These animals are cold-blooded, have scales
6418and breathe through lungs. While most of
6419them have a three-chambered heart,
6420crocodiles have four heart chambers. They
6421lay eggs with tough coverings and do not need
6422to lay their eggs in water, unlike amphibians.
6423Snakes, turtles, lizards and crocodiles fall in
6424this category (see Fig. 7.23).
6425Eye Head
6426Nostril
6427Pectoral Mouth
6428fin Pelvic
6429fin
6430Caudal
6431fin
6432Labeo rohita (Rohu)
6433Pectoral
6434fin Mouth
6435Brood
6436pouch
6437Dorsal
6438fin
6439Tail
6440Male Hippocampus
6441(Sea horse)
6442Wing like pectoral
6443Tail Pelvic fin
6444Scales
6445Exocoetus (Flying fish)
6446Anabas (Climbing perch)
6447Fig. 7.21 (b): Pisces
6448Fig. 7.22: Amphibia
6449Rana tigrina
6450(Common frog)
6451Toad
6452Hyla (Tree frog)
6453Salamander
6454Turtle
6455Chameleon
6456King Cobra
6457House wall lizard
6458(Hemidactylus)
6459Flying lizard (Draco)
64607.5.10 ( 7.5.10 (iv) AVES
6461These are warm-blooded animals and have a
6462four-chambered heart. They lay eggs. There
6463is an outside covering of feathers, and two
6464forelimbs are modified for flight. They breathe
6465through lungs. All birds fall in this category
6466(see Fig. 7.24 for examples).
6467Fig. 7.23: Reptilia
6468© NCERT
6469not to be republished
647094 SCIENCE
64717.5.10 ( 7.5.10 (V) MAMMALIA
6472Mammals are warm-blooded animals with
6473four-chambered hearts. They have mammary
6474glands for the production of milk to nourish
6475their young. Their skin has hairs as well as
6476sweat and oil glands. Most mammals familiar
6477to us produce live young ones. However, a
6478few of them, like the platypus and the echidna
6479lay eggs, and some, like kangaroos give birth
6480to very poorly developed young ones. Some
6481examples are shown in Fig. 7.25.
6482The scheme of classification of animals is
6483shown in Fig. 7.26.
6484White Stork
6485(Ciconia ciconia)
6486Ostrich
6487(Struthio camelus)
6488Male Tufted Duck
6489(Aythya fuligula)
6490Pigeon
6491Sparrow
6492Crow
6493Fig. 7.24: Aves (birds)
6494Fig. 7.25: Mammalia
6495uestions
64961. How do poriferan animals differ
6497from coelenterate animals?
64982. How do annelid animals differ
6499from arthropods?
65003. What are the differences between
6501amphibians and reptiles?
65024. What are the differences between
6503animals belonging to the Aves
6504group and those in the mammalia
6505group?
6506Q
6507Carolus Linnaeus (Karl
6508von Linne) was born in
6509Sweden and was a doctor
6510by professsion. He was
6511interested in the study of
6512plants. At the age of 22,
6513he published his first
6514paper on plants. While
6515serving as a personal
6516physician of a wealthy
6517government official, he studied the
6518diversity of plants in his employer’s
6519garden. Later, he published 14 papers and
6520also brought out the famous book
6521Systema Naturae from which all
6522fundamental taxonomical researches have
6523taken off. His system of classification was
6524a simple scheme for arranging plants so
6525as to be able to identify them again.
6526Carolus Linnaeus
6527(1707-1778)
6528Whale
6529Human
6530Cat
6531Rat
6532Bat
6533© NCERT
6534not to be republished
6535DIVERSITY IN LIVING ORGANISMS 95
6536Fig. 7.26: Classification of animals
6537© NCERT
6538not to be republished
653996 SCIENCE
65407.6 Nomenclature Nomenclature
6541Why is there a need for systematic naming of
6542living organisms?
6543Activity ______________ 7.3
6544• Find out the names of the following
6545animals and plants in as many
6546languages as you can:
65471. Tiger 2. Peacock 3. Ant
65484. Neem 5. Lotus 6. Potato
6549As you might be able to appreciate, it
6550would be difficult for people speaking or
6551writing in different languages to know when
6552they are talking about the same organism.
6553This problem was resolved by agreeing upon
6554a ‘scientific’ name for organisms in the same
6555manner that chemical symbols and formulae
6556for various substances are used the world
6557over. The scientific name for an organism is
6558thus unique and can be used to identify it
6559anywhere in the world.
6560The system of scientific naming or
6561nomenclature we use today was introduced
6562by Carolus Linnaeus in the eighteenth
6563century. The scientific name of an organism
6564is the result of the process of classification
6565which puts it along with the organisms it is
6566most related to. But when we actually name
6567the species, we do not list out the whole
6568hierarchy of groups it belongs to. Instead, we
6569limit ourselves to writing the name of the
6570genus and species of that particular
6571organism. The world over, it has been agreed
6572that both these names will be used in Latin
6573forms.
6574Certain conventions are followed while
6575writing the scientific names:
65761. The name of the genus begins with a
6577capital letter.
65782. The name of the species begins with a
6579small letter.
65803. When printed, the scientific name is
6581given in italics.
65824. When written by hand, the genus
6583name and the species name have to
6584be underlined separately.
6585Activity ______________ 7.4
6586• Find out the scientific names of any
6587five common animals and plants. Do
6588these names have anything in common
6589with the names you normally use to
6590identify them?
6591What
6592you have you have
6593learnt
6594• Classification helps us in exploring the diversity of life forms.
6595• The major characteristics considered for classifying all
6596organisms into five major kingdoms are:
6597(a) whether they are made of prokaryotic or eukaryotic cells
6598(b) whether the cells are living singly or organised into multicellular
6599and thus complex organisms
6600(c) whether the cells have a cell-wall and whether they prepare
6601their own food.
6602• All living organisms are divided on the above bases into five
6603kingdoms, namely Monera, Protista, Fungi, Plantae and
6604Animalia.
6605• The classification of life forms is related to their evolution.
6606© NCERT
6607not to be republished
6608DIVERSITY IN LIVING ORGANISMS 97
6609• Plantae and Animalia are further divided into subdivisions
6610on the basis of increasing complexity of body organisation.
6611• Plants are divided into five groups: Thallophytes, Bryophytes,
6612Pteridophytes, Gymnosperms and Angiosperms.
6613• Animals are divided into ten groups: Porifera, Coelenterata,
6614Platyhelminthes, Nematoda, Annelida, Arthropoda, Mollusca,
6615Echinodermata, Protochordata and Vertebrata.
6616• The binomial nomenclature makes for a uniform way of
6617identification of the vast diversity of life around us.
6618• The binomial nomenclature is made up of two words – a generic
6619name and a specific name.
6620Exercises Exercises Exercises
66211. What are the advantages of classifying organisms?
66222. How would you choose between two characteristics to be used
6623for developing a hierarchy in classification?
66243. Explain the basis for grouping organisms into five kingdoms.
66254. What are the major divisions in the Plantae? What is the basis
6626for these divisions?
66275. How are the criteria for deciding divisions in plants different
6628from the criteria for deciding the subgroups among animals?
66296. Explain how animals in Vertebrata are classified into further
6630subgroups.
6631© NCERT
6632not to be republished
6633In everyday life, we see some objects at rest
6634and others in motion. Birds fly, fish swim,
6635blood flows through veins and arteries and
6636cars move. Atoms, molecules, planets, stars
6637and galaxies are all in motion. We often
6638perceive an object to be in motion when its
6639position changes with time. However, there
6640are situations where the motion is inferred
6641through indirect evidences. For example, we
6642infer the motion of air by observing the
6643movement of dust and the movement of leaves
6644and branches of trees. What causes the
6645phenomena of sunrise, sunset and changing
6646of seasons? Is it due to the motion of the
6647earth? If it is true, why don’t we directly
6648perceive the motion of the earth?
6649An object may appear to be moving for
6650one person and stationary for some other. For
6651the passengers in a moving bus, the roadside
6652trees appear to be moving backwards. A
6653person standing on the road–side perceives
6654the bus alongwith the passengers as moving.
6655However, a passenger inside the bus sees his
6656fellow passengers to be at rest. What do these
6657observations indicate?
6658Most motions are complex. Some objects
6659may move in a straight line, others may take
6660a circular path. Some may rotate and a few
6661others may vibrate. There may be situations
6662involving a combination of these. In this
6663chapter, we shall first learn to describe the
6664motion of objects along a straight line. We
6665shall also learn to express such motions
6666through simple equations and graphs. Later,
6667we shall discuss ways of describing circular
6668motion.
6669Activity ______________ 8.1
6670• Discuss whether the walls of your
6671classroom are at rest or in motion.
6672Activity ______________ 8.2
6673• Have you ever experienced that the
6674train in which you are sitting appears
6675to move while it is at rest?
6676• Discuss and share your experience.
6677Think and Act
6678We sometimes are endangered by the
6679motion of objects around us, especially
6680if that motion is erratic and
6681uncontrolled as observed in a flooded
6682river, a hurricane or a tsunami. On the
6683other hand, controlled motion can be a
6684service to human beings such as in the
6685generation of hydro-electric power. Do
6686you feel the necessity to study the
6687erratic motion of some objects and
6688learn to control them?
66898.1 Describing Motion
6690We describe the location of an object by
6691specifying a reference point. Let us
6692understand this by an example. Let us
6693assume that a school in a village is 2 km north
6694of the railway station. We have specified the
6695position of the school with respect to the
6696railway station. In this example, the railway
6697station is the reference point. We could have
6698also chosen other reference points according
6699to our convenience. Therefore, to describe the
6700position of an object we need to specify a
6701reference point called the origin.
67028
6703MOTION
6704Chapter
67058.1.1 MOTION ALONG A STRAIGHT LINE
6706The simplest type of motion is the motion
6707along a straight line. We shall first learn to
6708describe this by an example. Consider the
6709motion of an object moving along a straight
6710path. The object starts its journey from O
6711which is treated as its reference point
6712(Fig. 8.1). Let A, B and C represent the
6713position of the object at different instants. At
6714first, the object moves through C and B and
6715reaches A. Then it moves back along the same
6716path and reaches C through B.
6717displacement, are used to describe the overall
6718motion of an object and to locate its final
6719position with reference to its initial position
6720at a given time.
6721Activity ______________ 8.3
6722• Take a metre scale and a long rope.
6723• Walk from one corner of a basket-ball
6724court to its oppposite corner along its
6725sides.
6726• Measure the distance covered by you
6727and magnitude of the displacement.
6728• What difference would you notice
6729between the two in this case?
6730Activity ______________ 8.4
6731• Automobiles are fitted with a device
6732that shows the distance travelled. Such
6733a device is known as an odometer. A
6734car is driven from Bhubaneshwar to
6735New Delhi. The difference between the
6736final reading and the initial reading of
6737the odometer is 1850 km.
6738• Find the magnitude of the displacement
6739between Bhubaneshwar and New Delhi
6740by using the Road Map of India.
6741The total path length covered by the object
6742is OA + AC, that is 60 km + 35 km = 95 km.
6743This is the distance covered by the object. To
6744describe distance we need to specify only the
6745numerical value and not the direction of
6746motion. There are certain quantities which
6747are described by specifying only their
6748numerical values. The numerical value of a
6749physical quantity is its magnitude. From this
6750example, can you find out the distance of the
6751final position C of the object from the initial
6752position O? This difference will give you the
6753numerical value of the displacement of the
6754object from O to C through A. The shortest
6755distance measured from the initial to the final
6756position of an object is known as the
6757displacement.
6758Can the magnitude of the displacement
6759be equal to the distance travelled by an
6760object? Consider the example given in
6761(Fig. 8.1). For motion of the object from O to
6762A, the distance covered is 60 km and the
6763magnitude of displacement is also 60 km.
6764During its motion from O to A and back to B,
6765the distance covered = 60 km + 25 km = 85 km
6766Fig. 8.1: Positions of an object on a straight line path
6767while the magnitude of displacement = 35 km.
6768Thus, the magnitude of displacement (35 km)
6769is not equal to the path length (85 km).
6770Further, we will notice that the magnitude of
6771the displacement for a course of motion may
6772be zero but the corresponding distance
6773covered is not zero. If we consider the object
6774to travel back to O, the final position concides
6775with the initial position, and therefore, the
6776displacement is zero. However, the distance
6777covered in this journey is OA + AO = 60 km +
677860 km = 120 km. Thus, two different physical
6779quantities — the distance and the
6780MOTION 99
6781100 SCIENCE
6782uestions
67831. An object has moved through a
6784distance. Can it have zero
6785displacement? If yes, support
6786your answer with an example.
67872. A farmer moves along the
6788boundary of a square field of side
678910 m in 40 s. What will be the
6790magnitude of displacement of the
6791farmer at the end of 2 minutes 20
6792seconds from his initial position?
67933. Which of the following is true for
6794displacement?
6795(a) It cannot be zero.
6796(b) Its magnitude is greater than
6797the distance travelled by the
6798object.
67998.1.2 UNIFORM MOTION AND NONUNIFORM
6800MOTION
6801Consider an object moving along a straight
6802line. Let it travel 5 m in the first second,
68035 m more in the next second, 5 m in the
6804third second and 5 m in the fourth second.
6805In this case, the object covers 5 m in each
6806second. As the object covers equal distances
6807in equal intervals of time, it is said to be in
6808uniform motion. The time interval in this
6809motion should be small. In our day-to-day
6810life, we come across motions where objects
6811cover unequal distances in equal intervals
6812of time, for example, when a car is moving
6813on a crowded street or a person is jogging
6814in a park. These are some instances of
6815non-uniform motion.
6816Activity ______________ 8.5
6817• The data regarding the motion of two
6818different objects A and B are given in
6819Table 8.1.
6820• Examine them carefully and state
6821whether the motion of the objects is
6822uniform or non-uniform.
6823Q
6824(a)
6825(b)
6826Fig. 8.2
6827Table 8.1
6828Time Distance Distance
6829travelled by travelled by
6830object A in m object B in m
68319:30 am 10 12
68329:45 am 20 19
683310:00 am 30 23
683410:15 am 40 35
683510:30 am 50 37
683610:45 am 60 41
683711:00 am 70 44
68388.2 Measuring the Rate of Motion
6839MOTION 101
6840Look at the situations given in Fig. 8.2. If
6841the bowling speed is 143 km h–1 in Fig. 8.2(a)
6842what does it mean? What do you understand
6843from the signboard in Fig. 8.2(b)?
6844Different objects may take different
6845amounts of time to cover a given distance.
6846Some of them move fast and some move
6847slowly. The rate at which objects move can
6848be different. Also, different objects can move
6849at the same rate. One of the ways of
6850measuring the rate of motion of an object is
6851to find out the distance travelled by the object
6852in unit time. This quantity is referred to as
6853speed. The SI unit of speed is metre per
6854second. This is represented by the symbol
6855m s–1 or m/s.The other units of speed include
6856centimetre per second (cm s–1) and kilometre
6857per hour (km h–1). To specify the speed of an
6858object, we require only its magnitude. The
6859speed of an object need not be constant. In
6860most cases, objects will be in non-uniform
6861motion. Therefore, we describe the rate of
6862motion of such objects in terms of their
6863average speed. The average speed of an object
6864is obtained by dividing the total distance
6865travelled by the total time taken. That is,
6866average speed = Total distance travelled
6867Total time taken
6868If an object travels a distance s in time t then
6869its speed v is,
6870v = s
6871t (8.1)
6872Let us understand this by an example. A
6873car travels a distance of 100 km in 2 h. Its
6874average speed is 50 km h–1. The car might
6875not have travelled at 50 km h–1 all the time.
6876Sometimes it might have travelled faster and
6877sometimes slower than this.
6878Example 8.1 An object travels 16 m in 4 s
6879and then another 16 m in 2 s. What is
6880the average speed of the object?
6881Solution:
6882Total distance travelled by the object =
688316 m + 16 m = 32 m
6884Total time taken = 4 s + 2 s = 6 s
6885Average speed =
6886Total distance travelled
6887Total time taken
6888= 32 m
68896 s = 5.33 m s–1
6890Therefore, the average speed of the object
6891is 5.33 m s–1.
68928.2.1 SPEED WITH DIRECTION
6893The rate of motion of an object can be more
6894comprehensive if we specify its direction of
6895motion along with its speed. The quantity that
6896specifies both these aspects is called velocity.
6897Velocity is the speed of an object moving in a
6898definite direction. The velocity of an object
6899can be uniform or variable. It can be changed
6900by changing the object’s speed, direction of
6901motion or both. When an object is moving
6902along a straight line at a variable speed, we
6903can express the magnitude of its rate of
6904motion in terms of average velocity. It is
6905calculated in the same way as we calculate
6906average speed.
6907In case the velocity of the object is
6908changing at a uniform rate, then average
6909velocity is given by the arithmetic mean of
6910initial velocity and final velocity for a given
6911period of time. That is,
6912average velocity = initialvelocity + final velocity
69132
6914Mathematically, vav = u+v
69152
6916(8.2)
6917where vav is the average velocity, u is the initial
6918velocity and v is the final velocity of the object.
6919Speed and velocity have the same units,
6920that is, m s–1 or m/s.
6921Activity ______________ 8.6
6922• Measure the time it takes you to walk
6923from your house to your bus stop or
6924the school. If you consider that your
6925average walking speed is 4 km h–1,
6926estimate the distance of the bus stop
6927or school from your house.
6928102 SCIENCE
6929= 50
6930km 1000 m 1h
6931× ×
6932h 1km 3600s
6933= 13.9 m s–1
6934The average speed of the car is
693550 km h–1 or 13.9 m s–1.
6936Example 8.3 Usha swims in a 90 m long
6937pool. She covers 180 m in one minute
6938by swimming from one end to the other
6939and back along the same straight path.
6940Find the average speed and average
6941velocity of Usha.
6942Solution:
6943Total distance covered by Usha in 1 min
6944is 180 m.
6945Displacement of Usha in 1 min = 0 m
6946Average speed =
6947Total distance covered
6948Totaltimetaken
6949=
6950180m 180 m 1 min
6951= ×
69521min 1min 60 s
6953= 3 m s-1
6954Average velocity =
6955Displacement
6956Totaltimetaken
6957=
69580 m
695960 s
6960= 0 m s–1
6961The average speed of Usha is 3 m s–1
6962and her average velocity is 0 m s–1.
69638.3 Rate of Change of Velocity
6964During uniform motion of an object along a
6965straight line, the velocity remains constant
6966with time. In this case, the change in velocity
6967of the object for any time interval is zero.
6968However, in non-uniform motion, velocity
6969varies with time. It has different values at
6970different instants and at different points of
6971the path. Thus, the change in velocity of the
6972object during any time interval is not zero.
6973Can we now express the change in velocity of
6974an object?
6975Activity ______________ 8.7
6976• At a time when it is cloudy, there may
6977be frequent thunder and lightning. The
6978sound of thunder takes some time to
6979reach you after you see the lightning.
6980• Can you answer why this happens?
6981• Measure this time interval using a
6982digital wrist watch or a stop watch.
6983• Calculate the distance of the nearest
6984point of lightning. (Speed of sound in
6985air = 346 m s-1.)
6986uestions
69871. Distinguish between speed and
6988velocity.
69892. Under what condition(s) is the
6990magnitude of average velocity of
6991an object equal to its average
6992speed?
69933. What does the odometer of an
6994automobile measure?
69954. What does the path of an object
6996look like when it is in uniform
6997motion?
69985. During an experiment, a signal
6999from a spaceship reached the
7000ground station in five minutes.
7001What was the distance of the
7002spaceship from the ground
7003station? The signal travels at the
7004speed of light, that is, 3 × 108
7005m s–1.
7006Example 8.2 The odometer of a car reads
70072000 km at the start of a trip and
70082400 km at the end of the trip. If the
7009trip took 8 h, calculate the average
7010speed of the car in km h–1 and m s–1.
7011Solution:
7012Distance covered by the car,
7013s = 2400 km – 2000 km = 400 km
7014Time elapsed, t = 8 h
7015Average speed of the car is,
7016vav = 400 km
70178 h
7018s
7019t
7020= 50 km h–1
7021Q
7022MOTION 103
7023To answer such a question, we have to
7024introduce another physical quantity called
7025acceleration, which is a measure of the
7026change in the velocity of an object per unit
7027time. That is,
7028acceleration =
7029change in velocity
7030time taken
7031If the velocity of an object changes from
7032an initial value u to the final value v in time t,
7033the acceleration a is,
7034v–u
7035a =
7036t (8.3)
7037This kind of motion is known as
7038accelerated motion. The acceleration is taken
7039to be positive if it is in the direction of velocity
7040and negative when it is opposite to the
7041direction of velocity. The SI unit of
7042acceleration is m s–2 .
7043If an object travels in a straight line and
7044its velocity increases or decreases by equal
7045amounts in equal intervals of time, then the
7046acceleration of the object is said to be
7047uniform. The motion of a freely falling body
7048is an example of uniformly accelerated
7049motion. On the other hand, an object can
7050travel with non-uniform acceleration if its
7051velocity changes at a non-uniform rate. For
7052example, if a car travelling along a straight
7053road increases its speed by unequal amounts
7054in equal intervals of time, then the car is said
7055to be moving with non-uniform acceleration.
7056Activity ______________ 8.8
7057• In your everyday life you come across
7058a range of motions in which
7059(a) acceleration is in the direction of
7060motion,
7061(b) acceleration is against the
7062direction of motion,
7063(c) acceleration is uniform,
7064(d) acceleration is non-uniform.
7065• Can you identify one example each of
7066the above type of motion?
7067Example 8.4 Starting from a stationary
7068position, Rahul paddles his bicycle to
7069attain a velocity of 6 m s–1 in 30 s. Then
7070he applies brakes such that the velocity
7071of the bicycle comes down to 4 m s-1 in
7072the next 5 s. Calculate the acceleration
7073of the bicycle in both the cases.
7074Solution:
7075In the first case:
7076initial velocity, u = 0 ;
7077final velocity, v = 6 m s–1 ;
7078time, t = 30 s .
7079From Eq. (8.3), we have
7080v–u
7081a =
7082t
7083Substituting the given values of u,v and
7084t in the above equation, we get
7085 –1 –1 6 m s – 0 m s =
708630 s
7087a
7088= 0.2 m s–2
7089In the second case:
7090initial velocity, u = 6 m s–1;
7091final velocity, v = 4 m s–1;
7092time, t = 5 s.
7093Then, –1 –1 4 m s – 6 m s =
70945 s
7095a
7096= –0.4 m s–2 .
7097The acceleration of the bicycle in the
7098first case is 0.2 m s–2 and in the second
7099case, it is –0.4 m s–2.
7100uestions
71011. When will you say a body is in
7102(i) uniform acceleration? (ii) nonuniform
7103acceleration?
71042. A bus decreases its speed from
710580 km h–1 to 60 km h–1 in 5 s.
7106Find the acceleration of the bus.
71073. A train starting from a railway
7108station and moving with uniform
7109acceleration attains a speed
711040 km h–1 in 10 minutes. Find its
7111acceleration.
7112Q
7113104 SCIENCE
71148.4 Graphical Representation of
7115Motion
7116Graphs provide a convenient method to
7117present basic information about a variety of
7118events. For example, in the telecast of a
7119one-day cricket match, vertical bar graphs
7120show the run rate of a team in each over. As
7121you have studied in mathematics, a straight
7122line graph helps in solving a linear equation
7123having two variables.
7124To describe the motion of an object, we
7125can use line graphs. In this case, line graphs
7126show dependence of one physical quantity,
7127such as distance or velocity, on another
7128quantity, such as time.
71298.4.1 DISTANCE–TIME GRAPHS
7130The change in the position of an object with
7131time can be represented on the distance-time
7132graph adopting a convenient scale of choice.
7133In this graph, time is taken along the x–axis
7134and distance is taken along the y-axis.
7135Distance-time graphs can be employed under
7136various conditions where objects move with
7137uniform speed, non-uniform speed, remain
7138at rest etc.
7139Fig. 8.3: Distance-time graph of an object moving
7140with uniform speed
7141We know that when an object travels
7142equal distances in equal intervals of time, it
7143moves with uniform speed. This shows that
7144the distance travelled by the object is directly
7145proportional to time taken. Thus, for uniform
7146speed, a graph of distance travelled against
7147time is a straight line, as shown in Fig. 8.3.
7148The portion OB of the graph shows that the
7149distance is increasing at a uniform rate. Note
7150that, you can also use the term uniform
7151velocity in place of uniform speed if you take
7152the magnitude of displacement equal to the
7153distance travelled by the object along the
7154y-axis.
7155We can use the distance-time graph to
7156determine the speed of an object. To do so,
7157consider a small part AB of the distance-time
7158graph shown in Fig 8.3. Draw a line parallel
7159to the x-axis from point A and another line
7160parallel to the y-axis from point B. These two
7161lines meet each other at point C to form a
7162triangle ABC. Now, on the graph, AC denotes
7163the time interval (t2 – t1) while BC corresponds
7164to the distance (s2 – s1). We can see from the
7165graph that as the object moves from the point
7166A to B, it covers a distance (s2 – s1) in time
7167(t2 – t1
7168). The speed, v of the object, therefore
7169can be represented as
7170v = 2 1
71712 1
7172–
7173–
7174s s
7175t t (8.4)
7176We can also plot the distance-time graph
7177for accelerated motion. Table 8.2 shows the
7178distance travelled by a car in a time interval
7179of two seconds.
7180Table 8.2: Distance travelled by a
7181car at regular time intervals
7182Time in seconds Distance in metres
71830 0
71842 1
71854 4
71866 9
71878 16
718810 25
718912 36
7190MOTION 105
7191The distance-time graph for the motion
7192of the car is shown in Fig. 8.4. Note that the
7193shape of this graph is different from the earlier
7194distance-time graph (Fig. 8.3) for uniform
7195motion. The nature of this graph shows nonlinear
7196variation of the distance travelled by
7197the car with time. Thus, the graph shown in
7198Fig 8.4 represents motion with non-uniform
7199speed.
72008.4.2 VELOCITY-TIME GRAPHS
7201The variation in velocity with time for an
7202object moving in a straight line can be
7203represented by a velocity-time graph. In this
7204graph, time is represented along the x-axis
7205Fig. 8.4: Distance-time graph for a car moving with
7206non-uniform speed
7207Fig. 8.5: Velocity-time graph for uniform motion of
7208a car
7209and the velocity is represented along the
7210y-axis. If the object moves at uniform velocity,
7211the height of its velocity-time graph will not
7212change with time (Fig. 8.5). It will be a straight
7213line parallel to the x-axis. Fig. 8.5 shows the
7214velocity-time graph for a car moving with
7215uniform velocity of 40 km h–1.
7216We know that the product of velocity and
7217time give displacement of an object moving
7218with uniform velocity. The area enclosed by
7219velocity-time graph and the time axis will be
7220equal to the magnitude of the displacement.
7221To know the distance moved by the car
7222between time t1 and t
72232 using Fig. 8.5, draw
7224perpendiculars from the points corresponding
7225to the time t1
7226 and t2 on the graph. The velocity
7227of 40 km h–1 is represented by the height AC
7228or BD and the time (t2 – t 1
7229) is represented by
7230the length AB.
7231So, the distance s moved by the car in
7232time (t2 – t1
7233) can be expressed as
7234s = AC × CD
7235= [(40 km h–1) × (t
72362 – t1) h]
7237= 40 (t
72382
7239– t1) km
7240= area of the rectangle ABDC (shaded
7241in Fig. 8.5).
7242We can also study about uniformly
7243accelerated motion by plotting its velocity–
7244time graph. Consider a car being driven along
7245a straight road for testing its engine. Suppose
7246a person sitting next to the driver records its
7247velocity after every 5 seconds by noting the
7248reading of the speedometer of the car. The
7249velocity of the car, in km h–1 as well as in
7250m s–1, at different instants of time is shown
7251in table 8.3.
7252Table 8.3: Velocity of a car at
7253regular instants of time
7254Time Velocity of the car
7255 (s) (m s–1) (km h–1)
72560 00
72575 2.5 9
725810 5.0 18
725915 7.5 27
726020 10.0 36
726125 12.5 45
726230 15.0 54
7263106 SCIENCE
7264In this case, the velocity-time graph for the
7265motion of the car is shown in Fig. 8.6. The
7266nature of the graph shows that velocity
7267changes by equal amounts in equal intervals
7268of time. Thus, for all uniformly accelerated
7269motion, the velocity-time graph is a
7270straight line.
7271Fig. 8.6: Velocity-time graph for a car moving with
7272uniform accelerations.
7273You can also determine the distance
7274moved by the car from its velocity-time graph.
7275The area under the velocity-time graph gives
7276the distance (magnitude of displacement)
7277moved by the car in a given interval of time.
7278If the car would have been moving with
7279uniform velocity, the distance travelled by it
7280would be represented by the area ABCD
7281under the graph (Fig. 8.6). Since the
7282magnitude of the velocity of the car is
7283changing due to acceleration, the distance s
7284travelled by the car will be given by the area
7285ABCDE under the velocity-time graph
7286(Fig. 8.6).
7287That is,
7288s = area ABCDE
7289= area of the rectangle ABCD + area of
7290the triangle ADE
7291= AB × BC +
72921
72932 (AD × DE)
7294In the case of non-uniformly accelerated
7295motion, velocity-time graphs can have any
7296shape.
7297Fig. 8.7: Velocity-time graphs of an object in nonuniformly
7298accelerated motion.
7299Fig. 8.7(a) shows a velocity-time graph
7300that represents the motion of an object whose
7301velocity is decreasing with time while
7302Fig. 8.7 (b) shows the velocity-time graph
7303representing the non-uniform variation of
7304velocity of the object with time. Try to interpret
7305these graphs.
7306Activity ______________ 8.9
7307• The times of arrival and departure of a
7308train at three stations A, B and C and
7309the distance of stations B and C from
7310station A are given in table 8.4.
7311Table 8.4: Distances of stations B
7312and C from A and times of arrival
7313and departure of the train
7314Station Distance Time of Time of
7315from A arrival departure
7316(km) (hours) (hours)
7317A 0 08:00 08:15
7318B 120 11:15 11:30
7319C 180 13:00 13:15
7320• Plot and interpret the distance-time
7321graph for the train assuming that its
7322motion between any two stations is
7323uniform.
7324Velocity (km h–1)
7325MOTION 107
73268.5 Equations of Motion by
7327Graphical Method
7328When an object moves along a straight line
7329with uniform acceleration, it is possible to
7330relate its velocity, acceleration during motion
7331and the distance covered by it in a certain
7332time interval by a set of equations known as
7333the equations of motion. There are three such
7334equations. These are:
7335v = u + at (8.5)
7336s = ut + ½ at2 (8.6)
73372 a s = v2
7338 – u2 (8.7)
7339where u is the initial velocity of the object
7340which moves with uniform acceleration a for
7341time t, v is the final velocity, and s is the
7342distance travelled by the object in time t.
7343Eq. (8.5) describes the velocity-time relation
7344and Eq. (8.6) represents the position-time
7345relation. Eq. (8.7), which represents the
7346relation between the position and the velocity,
7347can be obtained from Eqs. (8.5) and (8.6) by
7348eliminating t. These three equations can be
7349derived by graphical method.
73508.5.1 EQ UATION FOR VELOCITY-TIME
7351RELATION
7352Consider the velocity-time graph of an object
7353that moves under uniform acceleration as
7354Activity _____________ 8.10
7355• Feroz and his sister Sania go to school
7356on their bicycles. Both of them start at
7357the same time from their home but take
7358different times to reach the school
7359although they follow the same route.
7360Table 8.5 shows the distance travelled
7361by them in different times
7362Table 8.5: Distance covered by Feroz
7363and Sania at different times on
7364their bicycles
7365Time Distance Distance
7366travelled travelled
7367by Feroz by Sania
7368(km) (km)
73698:00 am 0 0
73708:05 am 1.0 0.8
73718:10 am 1.9 1.6
73728:15 am 2.8 2.3
73738:20 am 3.6 3.0
73748:25 am – 3.6
7375Q
7376• Plot the distance-time graph for their
7377motions on the same scale and
7378interpret.
7379uestions
73801. What is the nature of the
7381distance-time graphs for uniform
7382and non-uniform motion of an
7383object?
73842. What can you say about the
7385motion of an object whose
7386distance-time graph is a straight
7387line parallel to the time axis?
73883. What can you say about the
7389motion of an object if its speedtime
7390graph is a straight line
7391parallel to the time axis?
73924. What is the quantity which is
7393measured by the area occupied
7394below the velocity-time graph? Fig. 8.8: Velocity-time graph to obtain the equations
7395of motion
7396108 SCIENCE
7397= OA × OC +
73981
73992 (AD × BD) (8.10)
7400Substituting OA = u, OC = AD = t and BD
7401= at, we get
7402s = u × t +
74031 ( ) 2
7404t ×at
7405or s = u t +
74061
74072 a t 2
74088.5.3 EQUATION FOR POSITION–VELOCITY
7409RELATION
7410From the velocity-time graph shown in
7411Fig. 8.8, the distance s travelled by the object
7412in time t, moving under uniform acceleration
7413a is given by the area enclosed within the
7414trapezium OABC under the graph. That is,
7415s = area of the trapezium OABC
7416= OA + BC ×OC
74172
7418Substituting OA = u, BC = v and OC = t,
7419we get
7420
74212
7422u+v t
7423s (8.11)
7424From the velocity-time relation (Eq. 8.6),
7425we get
7426 v–u
7427t =
7428a (8.12)
7429Using Eqs. (8.11) and (8.12) we have
7430 v+u v-u s =
74312a
7432or 2 a s = v2
7433 – u2
7434Example 8.5 A train starting from rest
7435attains a velocity of 72 km h–1 in
74365 minutes. Assuming that the
7437acceleration is uniform, find (i) the
7438acceleration and (ii) the distance
7439travelled by the train for attaining this
7440velocity.
7441shown in Fig. 8.8 (similar to Fig. 8.6, but now
7442with u ≠ 0). From this graph, you can see that
7443initial velocity of the object is u (at point A)
7444and then it increases to v (at point B) in time
7445t. The velocity changes at a uniform rate a.
7446In Fig. 8.8, the perpendicular lines BC and
7447BE are drawn from point B on the time and
7448the velocity axes respectively, so that the
7449initial velocity is represented by OA, the final
7450velocity is represented by BC and the time
7451interval t is represented by OC. BD = BC –
7452CD, represents the change in velocity in time
7453interval t.
7454Let us draw AD parallel to OC. From the
7455graph, we observe that
7456BC = BD + DC = BD + OA
7457Substituting BC = v and OA = u,
7458we get v = BD + u
7459or BD = v – u (8.8)
7460From the velocity-time graph (Fig. 8.8),
7461the acceleration of the object is given by
7462a =
7463Change in velocity
7464time taken
7465= BD BD = AD OC
7466Substituting OC = t, we get
7467a = BD
7468t
7469or BD = at (8.9)
7470Using Eqs. (8.8) and (8.9) we get
7471v = u + at
74728.5.2 EQUATION FOR POSITION-TIME
7473RELATION
7474Let us consider that the object has travelled
7475a distance s in time t under uniform
7476acceleration a. In Fig. 8.8, the distance
7477travelled by the object is obtained by the area
7478enclosed within OABC under the velocity-time
7479graph AB.
7480Thus, the distance s travelled by the object
7481is given by
7482s = area OABC (which is a trapezium)
7483= area of the rectangle OADC + area of
7484the triangle ABD
7485MOTION 109
7486Solution:
7487We have been given
7488u = 0 ; v = 72 km h–1 = 20 m s-1 and
7489t = 5 minutes = 300 s.
7490(i) From Eq. (8.5) we know that
7491 v–u
7492a =
7493t
7494–1 –1
7495–2
749620 m s – 0 ms = 300 s
74971
7498= ms
749915
7500(ii) From Eq. (8.7) we have
75012 a s = v2
7502 – u2 = v2 – 0
7503Thus,
75042
7505–1 2
7506–2
7507=
75082
7509(20 m s ) =
75102×(1/15) ms
7511v
7512s
7513a
7514= 3000 m
7515= 3 km
7516The acceleration of the train is
75171
751815 m s– 2
7519and the distance travelled is 3 km.
7520Example 8.6 A car accelerates uniformly
7521from 18 km h–1 to 36 km h–1 in 5 s.
7522Calculate (i) the acceleration and (ii) the
7523distance covered by the car in that time.
7524Solution:
7525We are given that
7526u = 18 km h–1 = 5 m s–1
7527v = 36 km h–1 = 10 m s–1 and
7528t = 5 s .
7529(i) From Eq. (8.5) we have
7530v–u
7531a =
7532t
7533=
7534-1 -1 10 m s – 5 m s
75355 s
7536= 1 m s–2
7537(ii) From Eq. (8.6) we have
7538s = u t +
75391
75402 a t 2
7541= 5 m s–1 × 5 s +
75421
75432 × 1 m s–2 × (5 s) 2
7544= 25 m + 12.5 m
7545= 37.5 m
7546The acceleration of the car is 1 m s–2
7547and the distance covered is 37.5 m.
7548Example 8.7 The brakes applied to a car
7549produce an acceleration of 6 m s-2 in
7550the opposite direction to the motion. If
7551the car takes 2 s to stop after the
7552application of brakes, calculate the
7553distance it travels during this time.
7554Solution:
7555We have been given
7556a = – 6 m s–2 ; t = 2 s and v = 0 m s–1.
7557From Eq. (8.5) we know that
7558v = u + at
75590 = u + (– 6 m s–2) × 2 s
7560 or u = 12 m s–1 .
7561From Eq. (8.6) we get
7562s = u t +
75631
75642 a t 2
7565= (12 m s–1 ) × (2 s) +
75661
75672 (–6 m s–2 ) (2 s)2
7568= 24 m – 12 m
7569= 12 m
7570Thus, the car will move 12 m before it
7571stops after the application of brakes.
7572Can you now appreciate why drivers
7573are cautioned to maintain some
7574distance between vehicles while
7575travelling on the road?
7576uestions
75771. A bus starting from rest moves
7578with a uniform acceleration of
75790.1 m s-2 for 2 minutes. Find (a)
7580the speed acquired, (b) the Q distance travelled.
7581110 SCIENCE
75822. A train is travelling at a speed of
758390 km h–1. Brakes are applied so
7584as to produce a uniform
7585acceleration of – 0.5 m s-2. Find
7586how far the train will go before it
7587is brought to rest.
75883. A trolley, while going down an
7589inclined plane, has an
7590acceleration of 2 cm s-2. What will
7591be its velocity 3 s after the start?
75924. A racing car has a uniform
7593acceleration of 4 m s-2. What
7594distance will it cover in 10 s after
7595start?
75965. A stone is thrown in a vertically
7597upward direction with a velocity
7598of 5 m s-1. If the acceleration of the
7599stone during its motion is 10 m s–2
7600in the downward direction, what
7601will be the height attained by the
7602stone and how much time will it
7603take to reach there?
76048.6 Uniform Circular Motion
7605When the velocity of an object changes, we
7606say that the object is accelerating. The change
7607in the velocity could be due to change in its
7608magnitude or the direction of the motion or
7609both. Can you think of an example when an
7610object does not change its magnitude of
7611velocity but only its direction of motion?
7612Let us consider an example of the motion
7613of a body along a closed path. Fig 8.9 (a) shows
7614the path of an athlete along a rectangular
7615track ABCD. Let us assume that the athlete
7616runs at a uniform speed on the straight parts
7617AB, BC, CD and DA of the track. In order to
7618keep himself on track, he quickly changes
7619his speed at the corners. How many times
7620will the athlete have to change his direction
7621of motion, while he completes one round? It
7622is clear that to move in a rectangular track
7623once, he has to change his direction of motion
7624four times.
7625Now, suppose instead of a rectangular
7626track, the athlete is running along a
7627hexagonal shaped path ABCDEF, as shown
7628in Fig. 8.9(b). In this situation, the athlete
7629will have to change his direction six times
7630while he completes one round. What if the
7631track was not a hexagon but a regular
7632octagon, with eight equal sides as shown by
7633ABCDEFGH in Fig. 8.9(c)? It is observed that
7634as the number of sides of the track increases
7635the athelete has to take turns more and more
7636often. What would happen to the shape of
7637the track as we go on increasing the number
7638of sides indefinitely? If you do this you will
7639notice that the shape of the track approaches
7640the shape of a circle and the length of each of
7641the sides will decrease to a point. If the athlete
7642moves with a velocity of constant magnitude
7643along the circular path, the only change in
7644his velocity is due to the change in the
7645direction of motion. The motion of the athlete
7646moving along a circular path is, therefore, an
7647example of an accelerated motion.
7648We know that the circumference of a circle
7649of radius r is given by 2 r . If the athlete takes
7650t seconds to go once around the circular path
7651of radius r, the velocity v is given by
76522 r
7653v =
7654t (8.13)
7655When an object moves in a circular path
7656with uniform speed, its motion is called
7657uniform circular motion.
7658(a) Rectangular track (b) Hexagonal track
7659(c) Octagonal shaped track (d) A circular track
7660Fig. 8.9: The motion of an athlete along closed tracks
7661of different shapes.
7662MOTION 111
7663If you carefully note, on being released
7664the stone moves along a straight line
7665tangential to the circular path. This is
7666because once the stone is released, it
7667continues to move along the direction it has
7668been moving at that instant. This shows that
7669the direction of motion changed at every point
7670when the stone was moving along the circular
7671path.
7672When an athlete throws a hammer or a
7673discus in a sports meet, he/she holds the
7674hammer or the discus in his/her hand and
7675gives it a circular motion by rotating his/her
7676own body. Once released in the desired
7677direction, the hammer or discus moves in the
7678direction in which it was moving at the time
7679it was released, just like the piece of stone in
7680the activity described above. There are many
7681more familiar examples of objects moving
7682under uniform circular motion, such as the
7683motion of the moon and the earth, a satellite
7684in a circular orbit around the earth, a cyclist
7685on a circular track at constant speed
7686and so on.
7687Activity _____________ 8.11
7688• Take a piece of thread and tie a small
7689piece of stone at one of its ends. Move
7690the stone to describe a circular path
7691with constant speed by holding the
7692thread at the other end, as shown in
7693Fig. 8.10.
7694Fig. 8.10: A stone describing a circular path with
7695a velocity of constant magnitude.
7696• Now, let the stone go by releasing the
7697thread.
7698• Can you tell the direction in which the
7699stone moves after it is released?
7700• By repeating the activity for a few times
7701and releasing the stone at different
7702positions of the circular path, check
7703whether the direction in which the
7704stone moves remains the same or not.
7705What
7706you have
7707learnt
7708• Motion is a change of position; it can be described in terms of
7709the distance moved or the displacement.
7710• The motion of an object could be uniform or non-uniform
7711depending on whether its velocity is constant or changing.
7712• The speed of an object is the distance covered per unit time,
7713and velocity is the displacement per unit time.
7714• The acceleration of an object is the change in velocity per unit
7715time.
7716• Uniform and non-uniform motions of objects can be shown
7717through graphs.
7718• The motion of an object moving at uniform acceleration can be
7719described with the help of three equations, namely
7720v = u + at
7721s = ut + ½ at2
77222as = v2 – u2
7723112 SCIENCE
7724where u is initial velocity of the object, which moves with
7725uniform acceleration a for time t, v is its final velocity and s is
7726the distance it travelled in time t.
7727• If an object moves in a circular path with uniform speed, its
7728motion is called uniform circular motion.
7729Exercises
77301. An athlete completes one round of a circular track of diameter
7731200 m in 40 s. What will be the distance covered and the
7732displacement at the end of 2 minutes 20 s?
77332. Joseph jogs from one end A to the other end B of a straight
7734300 m road in 2 minutes 30 seconds and then turns around
7735and jogs 100 m back to point C in another 1 minute. What are
7736Joseph’s average speeds and velocities in jogging (a) from A to
7737B and (b) from A to C?
77383. Abdul, while driving to school, computes the average speed for
7739his trip to be 20 km h–1. On his return trip along the same
7740route, there is less traffic and the average speed is
774130 km h–1. What is the average speed for Abdul’s trip?
77424. A motorboat starting from rest on a lake accelerates in a straight
7743line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the
7744boat travel during this time?
77455. A driver of a car travelling at 52 km h–1 applies the brakes and
7746accelerates uniformly in the opposite direction. The car stops
7747in 5 s. Another driver going at 3 km h–1 in another car applies
7748his brakes slowly and stops in 10 s. On the same graph paper,
7749plot the speed versus time graphs for the two cars. Which of
7750the two cars travelled farther after the brakes were applied?
77516. Fig 8.11 shows the distance-time graph of three objects A,B
7752and C. Study the graph and answer the following questions:
7753Fig. 8.11
7754MOTION 113
7755(a) Which of the three is travelling the fastest?
7756(b) Are all three ever at the same point on the road?
7757(c) How far has C travelled when B passes A?
7758(d) How far has B travelled by the time it passes C?
77597. A ball is gently dropped from a height of 20 m. If its velocity
7760increases uniformly at the rate of 10 m s-2, with what velocity
7761will it strike the ground? After what time will it strike the
7762ground?
77638. The speed-time graph for a car is shown is Fig. 8.12.
7764Fig. 8.12
7765(a) Find how far does the car travel in the first 4 seconds.
7766Shade the area on the graph that represents the distance
7767travelled by the car during the period.
7768(b) Which part of the graph represents uniform motion of the
7769car?
77709. State which of the following situations are possible and give
7771an example for each of these:
7772(a) an object with a constant acceleration but with zero velocity
7773(b) an object moving in a certain direction with an acceleration
7774in the perpendicular direction.
777510. An artificial satellite is moving in a circular orbit of radius
777642250 km. Calculate its speed if it takes 24 hours to revolve
7777around the earth.
7778In the previous chapter, we described the
7779motion of an object along a straight line in
7780terms of its position, velocity and acceleration.
7781We saw that such a motion can be uniform
7782or non-uniform. We have not yet discovered
7783what causes the motion. Why does the speed
7784of an object change with time? Do all motions
7785require a cause? If so, what is the nature of
7786this cause? In this chapter we shall make an
7787attempt to quench all such curiosities.
7788For many centuries, the problem of
7789motion and its causes had puzzled scientists
7790and philosophers. A ball on the ground, when
7791given a small hit, does not move forever. Such
7792observations suggest that rest is the “natural
7793state” of an object. This remained the belief
7794until Galileo Galilei and Isaac Newton
7795developed an entirely different approach to
7796understand motion.
7797In our everyday life we observe that some
7798effort is required to put a stationary object
7799into motion or to stop a moving object. We
7800ordinarily experience this as a muscular effort
7801and say that we must push or hit or pull on
7802an object to change its state of motion. The
7803concept of force is based on this push, hit or
7804pull. Let us now ponder about a ‘force’. What
7805is it? In fact, no one has seen, tasted or felt a
7806force. However, we always see or feel the effect
7807of a force. It can only be explained by
7808describing what happens when a force is
7809applied to an object. Pushing, hitting and
7810pulling of objects are all ways of bringing
7811objects in motion (Fig. 9.1). They move
7812because we make a force act on them.
7813From your studies in earlier classes, you
7814are also familiar with the fact that a force can
7815be used to change the magnitude of velocity
7816of an object (that is, to make the object move
7817faster or slower) or to change its direction of
7818motion. We also know that a force can change
7819the shape and size of objects (Fig. 9.2).
7820(a) The trolley moves along the
7821direction we push it.
7822(c) The hockey stick hits the ball forward
7823(b) The drawer is pulled.
7824Fig. 9.1: Pushing, pulling, or hitting objects change
7825their state of motion.
7826(a)
7827(b)
7828Fig. 9.2: (a) A spring expands on application of force;
7829(b) A spherical rubber ball becomes oblong
7830as we apply force on it.
78319
7832FORCE AND LAWS OF MOTION
7833Chapter
7834box with a small force, the box does not move
7835because of friction acting in a direction
7836opposite to the push [Fig. 9.4(a)]. This friction
7837force arises between two surfaces in contact;
7838in this case, between the bottom of the box
7839and floor’s rough surface. It balances the
7840pushing force and therefore the box does not
7841move. In Fig. 9.4(b), the children push the
7842box harder but the box still does not move.
7843This is because the friction force still balances
7844the pushing force. If the children push the
7845box harder still, the pushing force becomes
7846bigger than the friction force [Fig. 9.4(c)].
7847There is an unbalanced force. So the box
7848starts moving.
7849What happens when we ride a bicycle?
7850When we stop pedalling, the bicycle begins
7851to slow down. This is again because of the
7852friction forces acting opposite to the direction
7853of motion. In order to keep the bicycle moving,
7854we have to start pedalling again. It thus
7855appears that an object maintains its motion
7856under the continuous application of an
7857unbalanced force. However, it is quite
7858incorrect. An object moves with a uniform
7859velocity when the forces (pushing force and
7860frictional force) acting on the object are
7861balanced and there is no net external force
7862on it. If an unbalanced force is applied on
7863the object, there will be a change either in its
7864speed or in the direction of its motion. Thus,
7865to accelerate the motion of an object, an
7866unbalanced force is required. And the change
7867in its speed (or in the direction of motion)
7868would continue as long as this unbalanced
7869force is applied. However, if this force is
78709.1 Balanced and Unbalanced
7871Forces
7872Fig. 9.3 shows a wooden block on a horizontal
7873table. Two strings X and Y are tied to the two
7874opposite faces of the block as shown. If we
7875apply a force by pulling the string X, the block
7876begins to move to the right. Similarly, if we
7877pull the string Y, the block moves to the left.
7878But, if the block is pulled from both the sides
7879with equal forces, the block will not move.
7880Such forces are called balanced forces and
7881do not change the state of rest or of motion of
7882an object. Now, let us consider a situation in
7883which two opposite forces of different
7884magnitudes pull the block. In this case, the
7885block would begin to move in the direction of
7886the greater force. Thus, the two forces are
7887not balanced and the unbalanced force acts
7888in the direction the block moves. This
7889suggests that an unbalanced force acting on
7890an object brings it in motion.
7891Fig. 9.3: Two forces acting on a wooden block
7892What happens when some children try to
7893push a box on a rough floor? If they push the
7894(a) (b) (c)
7895Fig. 9.4
7896FORCE AND LAWS OF MOTION 115
7897116 SCIENCE
7898removed completely, the object would
7899continue to move with the velocity it has
7900acquired till then.
79019.2 First Law of Motion
7902By observing the motion of objects on an
7903inclined plane Galileo deduced that objects
7904move with a constant speed when no force
7905acts on them. He observed that when a marble
7906rolls down an inclined plane, its velocity
7907increases [Fig. 9.5(a)]. In the next chapter,
7908you will learn that the marble falls under the
7909unbalanced force of gravity as it rolls down
7910and attains a definite velocity by the time it
7911reaches the bottom. Its velocity decreases
7912when it climbs up as shown in Fig. 9.5(b).
7913Fig. 9.5(c) shows a marble resting on an ideal
7914frictionless plane inclined on both sides.
7915Galileo argued that when the marble is
7916released from left, it would roll down the slope
7917and go up on the opposite side to the same
7918height from which it was released. If the
7919inclinations of the planes on both sides are
7920equal then the marble will climb the same
7921distance that it covered while rolling down. If
7922the angle of inclination of the right-side plane
7923were gradually decreased, then the marble
7924would travel further distances till it reaches
7925the original height. If the right-side plane were
7926ultimately made horizontal (that is, the slope
7927is reduced to zero), the marble would continue
7928to travel forever trying to reach the same
7929height that it was released from. The
7930unbalanced forces on the marble in this case
7931are zero. It thus suggests that an unbalanced
7932(external) force is required to change the
7933motion of the marble but no net force is
7934needed to sustain the uniform motion of the
7935marble. In practical situations it is difficult
7936to achieve a zero unbalanced force. This is
7937because of the presence of the frictional force
7938acting opposite to the direction of motion.
7939Thus, in practice the marble stops after
7940travelling some distance. The effect of the
7941frictional force may be minimised by using a
7942smooth marble and a smooth plane and
7943providing a lubricant on top of the planes.
7944Fig. 9.5: (a) the downward motion; (b) the upward
7945motion of a marble on an inclined plane;
7946and (c) on a double inclined plane.
7947Newton further studied Galileo’s ideas on
7948force and motion and presented three
7949fundamental laws that govern the motion of
7950objects. These three laws are known as
7951Newton’s laws of motion. The first law of
7952motion is stated as:
7953An object remains in a state of rest or of
7954uniform motion in a straight line unless
7955compelled to change that state by an applied
7956force.
7957In other words, all objects resist a change
7958in their state of motion. In a qualitative way,
7959the tendency of undisturbed objects to stay
7960at rest or to keep moving with the same
7961velocity is called inertia. This is why, the first
7962law of motion is also known as the law of
7963inertia.
7964Certain experiences that we come across
7965while travelling in a motorcar can be
7966explained on the basis of the law of inertia.
7967We tend to remain at rest with respect to the
7968seat until the driver applies a braking force
7969to stop the motorcar. With the application of
7970brakes, the car slows down but our body
7971tends to continue in the same state of motion
7972because of its inertia. A sudden application
7973of brakes may thus cause injury to us by
7974FORCE AND LAWS OF MOTION 117
7975impact or collision with the panels in front.
7976Safety belts are worn to prevent such
7977accidents. Safety belts exert a force on our
7978body to make the forward motion slower. An
7979opposite experience is encountered when we
7980are standing in a bus and the bus begins to
7981move suddenly. Now we tend to fall
7982backwards. This is because the sudden start
7983of the bus brings motion to the bus as well
7984as to our feet in contact with the floor of the
7985bus. But the rest of our body opposes this
7986motion because of its inertia.
7987When a motorcar makes a sharp turn at
7988a high speed, we tend to get thrown to one
7989side. This can again be explained on the basis
7990of the law of inertia. We tend to continue in
7991our straight-line motion. When an
7992unbalanced force is applied by the engine to
7993change the direction of motion of the
7994motorcar, we slip to one side of the seat due
7995to the inertia of our body.
7996The fact that a body will remain at rest
7997unless acted upon by an unbalanced force
7998can be illustrated through the following
7999activities:
8000Activity ______________ 9.1
8001• Make a pile of similar carom coins on
8002a table, as shown in Fig. 9.6.
8003• Attempt a sharp horizontal hit at the
8004bottom of the pile using another carom
8005coin or the striker. If the hit is strong
8006enough, the bottom coin moves out
8007quickly. Once the lowest coin is
8008removed, the inertia of the other coins
8009makes them ‘fall’ vertically on the table.
8010Galileo Galilei was born
8011on 15 February 1564 in
8012Pisa, Italy. Galileo, right
8013from his childhood, had
8014interest in mathematics
8015and natural philosophy.
8016But his father
8017Vincenzo Galilei wanted
8018him to become a medical
8019doctor. Accordingly,
8020Galileo enrolled himself
8021for a medical degree at the
8022University of Pisa in 1581 which he never
8023completed because of his real interest in
8024mathematics. In 1586, he wrote his first
8025scientific book ‘The Little Balance [La
8026Balancitta]’, in which he described
8027Archimedes’ method of finding the relative
8028densities (or specific gravities) of substances
8029using a balance. In 1589, in his series of
8030essays – De Motu, he presented his theories
8031about falling objects using an inclined plane
8032to slow down the rate of descent.
8033In 1592, he was appointed professor of
8034mathematics at the University of Padua in
8035the Republic of Venice. Here he continued his
8036observations on the theory of motion and
8037through his study of inclined planes and the
8038pendulum, formulated the correct law for
8039uniformly accelerated objects that the
8040distance the object moves is proportional to
8041the square of the time taken.
8042Galileo was also a remarkable craftsman.
8043He developed a series of telescopes whose
8044optical performance was much better than
8045that of other telescopes available during those
8046days. Around 1640, he designed the first
8047pendulum clock. In his book ‘Starry
8048Messenger’ on his astronomical discoveries,
8049Galileo claimed to have seen mountains on
8050the moon, the milky way made up of tiny
8051stars, and four small bodies orbiting Jupiter.
8052In his books ‘Discourse on Floating Bodies’
8053and ‘Letters on the Sunspots’, he disclosed
8054his observations of sunspots.
8055Using his own telescopes and through his
8056observations on Saturn and Venus, Galileo
8057argued that all the planets must orbit the Sun
8058and not the earth, contrary to what was
8059believed at that time.
8060Galileo Galilei
8061(1564 – 1642)
8062Fig. 9.6: Only the carom coin at the bottom of a
8063pile is removed when a fast moving carom
8064coin (or striker) hits it.
8065118 SCIENCE
8066five-rupees coin if we use a one-rupee coin, we
8067find that a lesser force is required to perform
8068the activity. A force that is just enough to
8069cause a small cart to pick up a large velocity
8070will produce a negligible change in the motion
8071of a train. This is because, in comparison to
8072the cart the train has a much lesser tendency
8073to change its state of motion. Accordingly, we
8074say that the train has more inertia than the
8075cart. Clearly, heavier or more massive objects
8076offer larger inertia. Quantitatively, the inertia
8077of an object is measured by its mass. We may
8078thus relate inertia and mass as follows:
8079Inertia is the natural tendency of an object to
8080resist a change in its state of motion or of
8081rest. The mass of an object is a measure of
8082its inertia.
8083uestions
80841. Which of the following has more
8085inertia: (a) a rubber ball and a
8086stone of the same size? (b) a
8087bicycle and a train? (c) a fiverupees
8088coin and a one-rupee coin?
80892. In the following example, try to
8090identify the number of times the
8091velocity of the ball changes:
8092“A football player kicks a football
8093to another player of his team who
8094kicks the football towards the
8095goal. The goalkeeper of the
8096opposite team collects the football
8097and kicks it towards a player of
8098his own team”.
8099Also identify the agent supplying
8100the force in each case.
81013. Explain why some of the leaves
8102may get detached from a tree if
8103we vigorously shake its branch.
81044. Why do you fall in the forward
8105direction when a moving bus
8106brakes to a stop and fall
8107backwards when it accelerates
8108from rest?
81099.4 Second Law of Motion
8110The first law of motion indicates that when
8111an unbalanced external force acts on an
8112Activity ______________ 9.2
8113• Set a five-rupee coin on a stiff card
8114covering an empty glass tumbler
8115standing on a table as shown in
8116Fig. 9.7.
8117• Give the card a sharp horizontal flick
8118with a finger. If we do it fast then the
8119card shoots away, allowing the coin to
8120fall vertically into the glass tumbler due
8121to its inertia.
8122• The inertia of the coin tries to maintain
8123its state of rest even when the card
8124flows off.
8125Fig. 9.7: When the card is flicked with the
8126finger the coin placed over it falls in the
8127tumbler.
8128Activity ______________ 9.3
8129• Place a water-filled tumbler on a tray.
8130• Hold the tray and turn around as fast
8131as you can.
8132• We observe that the water spills. Why?
8133Observe that a groove is provided in a
8134saucer for placing the tea cup. It prevents
8135the cup from toppling over in case of sudden
8136jerks.
81379.3 Inertia and Mass
8138All the examples and activities given so far
8139illustrate that there is a resistance offered by
8140an object to change its state of motion. If it is
8141at rest it tends to remain at rest; if it is moving
8142it tends to keep moving. This property of an
8143object is called its inertia. Do all bodies have
8144the same inertia? We know that it is easier to
8145push an empty box than a box full of books.
8146Similarly, if we kick a football it flies away.
8147But if we kick a stone of the same size with
8148equal force, it hardly moves. We may, in fact,
8149get an injury in our foot while doing so!
8150Similarly, in activity 9.2, instead of a
8151Q
8152FORCE AND LAWS OF MOTION 119
8153object, its velocity changes, that is, the object
8154gets an acceleration. We would now like to
8155study how the acceleration of an object
8156depends on the force applied to it and how
8157we measure a force. Let us recount some
8158observations from our everyday life. During
8159the game of table tennis if the ball hits a player
8160it does not hurt him. On the other hand, when
8161a fast moving cricket ball hits a spectator, it
8162may hurt him. A truck at rest does not require
8163any attention when parked along a roadside.
8164But a moving truck, even at speeds as low as
81655 m s–1, may kill a person standing in its path.
8166A small mass, such as a bullet may kill a
8167person when fired from a gun. These
8168observations suggest that the impact
8169produced by the objects depends on their
8170mass and velocity. Similarly, if an object is to
8171be accelerated, we know that a greater force
8172is required to give a greater velocity. In other
8173words, there appears to exist some quantity
8174of importance that combines the object’s
8175mass and its velocity. One such property
8176called momentum was introduced by Newton.
8177The momentum, p of an object is defined as
8178the product of its mass, m and velocity, v.
8179That is,
8180p = mv (9.1)
8181Momentum has both direction and
8182magnitude. Its direction is the same as that
8183of velocity, v. The SI unit of momentum is
8184kilogram-metre per second (kg m s-1 ). Since
8185the application of an unbalanced force brings
8186a change in the velocity of the object, it is
8187therefore clear that a force also produces a
8188change of momentum.
8189Let us consider a situation in which a car
8190with a dead battery is to be pushed along a
8191straight road to give it a speed of 1 m s-1,
8192which is sufficient to start its engine. If one
8193or two persons give a sudden push
8194(unbalanced force) to it, it hardly starts. But
8195a continuous push over some time results in
8196a gradual acceleration of the car to this speed.
8197It means that the change of momentum of
8198the car is not only determined by the
8199magnitude of the force but also by the time
8200during which the force is exerted. It may then
8201also be concluded that the force necessary to
8202change the momentum of an object depends
8203on the time rate at which the momentum is
8204changed.
8205The second law of motion states that the
8206rate of change of momentum of an object is
8207proportional to the applied unbalanced force
8208in the direction of force.
82099.4.1 MATHEMATICAL FORMULATION OF
8210SECOND LAW OF MOTION
8211Suppose an object of mass, m is moving along
8212a straight line with an initial velocity, u. It is
8213uniformly accelerated to velocity, v in time, t
8214by the application of a constant force, F
8215throughout the time, t. The initial and final
8216momentum of the object will be, p1 = mu and
8217p2
8218 = mv respectively.
8219The change in momentum ∝ p2 – p1
8220∝ mv – mu
8221∝ m × (v – u).
8222The rate of change of momentum ∝ m vu
8223t
8224× − ( )
8225Or, the applied force,
8226F ∝ m vu
8227t
8228× − ( )
8229F km v u
8230t = × − ( ) (9.2)
8231= k m a
8232(9.3)
8233Here a [ = (v – u)/t ] is the acceleration,
8234which is the rate of change of velocity. The
8235quantity, k is a constant of proportionality. The
8236SI units of mass and acceleration are kg and
8237m s-2 respectively. The unit of force is so chosen
8238that the value of the constant, k becomes one.
8239For this, one unit of force is defined as the
8240amount that produces an acceleration of 1 m
8241s-2 in an object of 1 kg mass. That is,
82421 unit of force = k × (1 kg) × (1 m s-2 ).
8243Thus, the value of k becomes 1. From Eq. (9.3)
8244F = ma (9.4)
8245The unit of force is kg m s-2 or newton,
8246which has the symbol N. The second law of
8247120 SCIENCE
8248motion gives us a method to measure the force
8249acting on an object as a product of its mass
8250and acceleration.
8251The second law of motion is often seen in
8252action in our everyday life. Have you noticed
8253that while catching a fast moving cricket ball,
8254a fielder in the ground gradually pulls his
8255hands backwards with the moving ball? In
8256doing so, the fielder increases the time during
8257which the high velocity of the moving ball
8258decreases to zero. Thus, the acceleration of
8259the ball is decreased and therefore the impact
8260of catching the fast moving ball (Fig. 9.8) is
8261also reduced. If the ball is stopped suddenly
8262then its high velocity decreases to zero in a
8263very short interval of time. Thus, the rate of
8264change of momentum of the ball will be large.
8265Therefore, a large force would have to be
8266applied for holding the catch that may hurt
8267the palm of the fielder. In a high jump athletic
8268event, the athletes are made to fall either on
8269a cushioned bed or on a sand bed. This is to
8270increase the time of the athlete’s fall to stop
8271after making the jump. This decreases the
8272rate of change of momentum and hence the
8273force. Try to ponder how a karate player
8274breaks a slab of ice with a single blow.
8275The first law of motion can be
8276mathematically stated from the mathematical
8277expression for the second law of motion. Eq.
8278(9.4) is
8279F = ma
8280or F = mv u−
8281t
8282( )
8283(9.5)
8284or Ft = mv – mu
8285That is, when F = 0, v = u for whatever time, t
8286is taken. This means that the object will
8287continue moving with uniform velocity, u
8288throughout the time, t. If u is zero then v will
8289also be zero. That is, the object will remain
8290at rest.
8291Example 9.1 A constant force acts on an
8292object of mass 5 kg for a duration of
82932 s. It increases the object’s velocity
8294from 3 m s–1 to 7 m s-1. Find the
8295magnitude of the applied force. Now, if
8296the force was applied for a duration of
82975 s, what would be the final velocity of
8298the object?
8299Solution:
8300We have been given that u = 3 m s–1
8301and v = 7 m s-1, t = 2 s and m = 5 kg.
8302From Eq. (9.5) we have,
8303F = mv u−
8304t
8305( )
8306Substitution of values in this relation
8307gives
8308F = 5 kg (7 m s-1 – 3 m s-1 )/2 s = 10 N.
8309Now, if this force is applied for a
8310duration of 5 s (t = 5 s), then the final
8311velocity can be calculated by rewriting
8312Eq. (9.5) as
8313v u
8314Ft
8315m
8316= +
8317On substituting the values of u, F, m and
8318t, we get the final velocity,
8319v = 13 m s-1.
8320Fig. 9.8: A fielder pulls his hands gradually with the
8321moving ball while holding a catch.
8322FORCE AND LAWS OF MOTION 121
8323Example 9.2 Which would require a greater
8324force –– accelerating a 2 kg mass at 5 m
8325s–2 or a 4 kg mass at 2 m s-2 ?
8326Solution:
8327From Eq. (9.4), we have F = ma.
8328Here we have m1 = 2 kg; a1 = 5 m s-2
8329and m2 = 4 kg; a2 = 2 m s-2 .
8330Thus, F1 = m1a1 = 2 kg × 5 m s-2 = 10 N;
8331and F2 = m2a2
8332 = 4 kg × 2 m s-2 = 8 N.
8333⇒ F1
8334 > F2
8335.
8336Thus, accelerating a 2 kg mass at
83375 m s-2 would require a greater force.
8338Example 9.3 A motorcar is moving with a
8339velocity of 108 km/h and it takes 4 s to
8340stop after the brakes are applied.
8341Calculate the force exerted by the
8342brakes on the motorcar if its mass along
8343with the passengers is 1000 kg.
8344Solution:
8345The initial velocity of the motorcar
8346u = 108 km/h
8347= 108 × 1000 m/(60 × 60 s)
8348= 30 m s-1
8349and the final velocity of the motorcar
8350v = 0 m s-1.
8351The total mass of the motorcar along
8352with its passengers = 1000 kg and the
8353time taken to stop the motorcar, t = 4 s.
8354From Eq. (9.5) we have the magnitude
8355of the force (F) applied by the brakes as
8356m(v – u)/t.
8357On substituting the values, we get
8358F = 1000 kg × (0 – 30) m s-1/4 s
8359= – 7500 kg m s-2 or – 7500 N.
8360The negative sign tells us that the force
8361exerted by the brakes is opposite to the
8362direction of motion of the motorcar.
8363Example 9.4 A force of 5 N gives a mass
8364m1, an acceleration of 10 m s–2 and a
8365mass m2, an acceleration of 20 m s-2.
8366What acceleration would it give if both
8367the masses were tied together?
8368Solution:
8369From Eq. (9.4) we have m1 = F/a1
8370; and
8371m2 = F/a2. Here, a1 = 10 m s-2;
8372a2 = 20 m s-2 and F = 5 N.
8373Thus, m1 = 5 N/10 m s-2 = 0.50 kg; and
8374m2
8375 = 5 N/20 m s-2 = 0.25 kg.
8376If the two masses were tied together,
8377the total mass, m would be
8378m = 0.50 kg + 0.25 kg = 0.75 kg.
8379The acceleration, a produced in the
8380combined mass by the 5 N force would
8381be, a = F/m = 5 N/0.75 kg = 6.67 m s-2.
8382Example 9.5 The velocity-time graph of a
8383ball of mass 20 g moving along a
8384straight line on a long table is given in
8385Fig. 9.9.
8386Fig. 9.9
8387How much force does the table exert on
8388the ball to bring it to rest?
8389Solution:
8390The initial velocity of the ball is 20 cm s-1.
8391Due to the friction force exerted by the
8392table, the velocity of the ball decreases
8393down to zero in 10 s. Thus, u = 20 cm s–1;
8394v = 0 cm s-1 and t = 10 s. Since the
8395velocity-time graph is a straight line, it is
8396clear that the ball moves with a constant
8397acceleration. The acceleration a is
8398a
8399v u
8400t = −
8401= (0 cm s-1 – 20 cm s-1)/10 s
8402= –2 cm s-2 = –0.02 m s-2.
8403122 SCIENCE
8404The force exerted on the ball F is,
8405F = ma = (20/1000) kg × (– 0.02 m s-2)
8406 = – 0.0004 N.
8407The negative sign implies that the
8408frictional force exerted by the table is
8409opposite to the direction of motion of
8410the ball.
84119.5 Third Law of Motion
8412The first two laws of motion tell us how an
8413applied force changes the motion and provide
8414us with a method of determining the force.
8415The third law of motion states that when one
8416object exerts a force on another object, the
8417second object instantaneously exerts a force
8418back on the first. These two forces are always
8419equal in magnitude but opposite in direction.
8420These forces act on different objects and never
8421on the same object. In the game of football
8422sometimes we, while looking at the football
8423and trying to kick it with a greater force,
8424collide with a player of the opposite team.
8425Both feel hurt because each applies a force
8426to the other. In other words, there is a pair of
8427forces and not just one force. The two
8428opposing forces are also known as action and
8429reaction forces.
8430Let us consider two spring balances
8431connected together as shown in Fig. 9.10. The
8432fixed end of balance B is attached with a rigid
8433support, like a wall. When a force is applied
8434through the free end of spring balance A, it is
8435observed that both the spring balances show
8436the same readings on their scales. It means
8437that the force exerted by spring balance A on
8438balance B is equal but opposite in direction
8439to the force exerted by the balance B on
8440balance A. The force which balance A exerts
8441on balance B is called the action and the force
8442of balance B on balance A is called the
8443reaction. This gives us an alternative
8444statement of the third law of motion i.e., to
8445every action there is an equal and opposite
8446reaction. However, it must be remembered
8447that the action and reaction always act on two
8448different objects.
8449Fig. 9.10: Action and reaction forces are equal and
8450opposite.
8451Suppose you are standing at rest and
8452intend to start walking on a road. You must
8453accelerate, and this requires a force in
8454accordance with the second law of motion.
8455Which is this force? Is it the muscular effort
8456you exert on the road? Is it in the direction
8457we intend to move? No, you push the road
8458below backwards. The road exerts an equal
8459and opposite reaction force on your feet to
8460make you move forward.
8461It is important to note that even though
8462the action and reaction forces are always
8463equal in magnitude, these forces may not
8464produce accelerations of equal magnitudes.
8465This is because each force acts on a different
8466object that may have a different mass.
8467When a gun is fired, it exerts a forward
8468force on the bullet. The bullet exerts an equal
8469and opposite reaction force on the gun. This
8470results in the recoil of the gun (Fig. 9.11).
8471Since the gun has a much greater mass than
8472the bullet, the acceleration of the gun is much
8473less than the acceleration of the bullet. The
8474third law of motion can also be illustrated
8475when a sailor jumps out of a rowing boat. As
8476the sailor jumps forward, the force on the boat
8477moves it backwards (Fig. 9.12).
8478Fig. 9.11: A forward force on the bullet and recoil of
8479the gun.
8480FORCE AND LAWS OF MOTION 123
8481The cart shown in this activity can be
8482constructed by using a 12 mm or 18 mm thick
8483plywood board of about 50 cm × 100 cm with
8484two pairs of hard ball-bearing wheels (skate
8485wheels are good to use). Skateboards are not
8486as effective because it is difficult to maintain
8487straight-line motion.
84889.6 Conservation of Momentum
8489Suppose two objects (two balls A and B, say)
8490of masses mA and mB are travelling in the same
8491direction along a straight line at different
8492velocities uA and uB, respectively [Fig. 9.14(a)].
8493And there are no other external unbalanced
8494forces acting on them. Let uA > uB and the
8495two balls collide with each other as shown in
8496Fig. 9.14(b). During collision which lasts for
8497a time t, the ball A exerts a force FAB on ball B
8498and the ball B exerts a force FBA on ball A.
8499Suppose vA and vB are the velocities of the two
8500balls A and B after the collision, respectively
8501[Fig. 9.14(c)].
8502Activity ______________ 9.4
8503• Request two children to stand on two
8504separate carts as shown in Fig. 9.13.
8505• Give them a bag full of sand or some
8506other heavy object. Ask them to play a
8507game of catch with the bag.
8508• Does each of them receive an
8509instantaneous reaction as a result of
8510throwing the sand bag (action)?
8511• You can paint a white line on
8512cartwheels to observe the motion of
8513the two carts when the children throw
8514the bag towards each other.
8515Fig. 9.12: As the sailor jumps in forward direction,
8516the boat moves backwards.
8517Fig. 9.13
8518Now, place two children on one cart and
8519one on another cart. The second law of motion
8520can be seen, as this arrangement would show
8521different accelerations for the same force.
8522Fig. 9.14: Conservation of momentum in collision of
8523two balls.
8524From Eq. (9.1), the momenta (plural of
8525momentum) of ball A before and after the
8526collision are mAuA and mA
8527vA, respectively. The
8528rate of change of its momentum (or FAB, action)
8529during the collision will be m
8530v u
8531t A
8532A A ( ) − .
8533Similarly, the rate of change of momentum of
8534ball B (= FBA or reaction) during the collision
8535will be m
8536v u
8537t B
8538B B ( ) − .
8539According to the third law of motion, the
8540force FAB exerted by ball A on ball B (action)
8541124 SCIENCE
8542Activity ______________ 9.6
8543• Take a test tube of good quality glass
8544material and put a small amount of
8545water in it. Place a stop cork at the
8546mouth of it.
8547• Now suspend the test tube horizontally
8548by two strings or wires as shown in
8549Fig. 9.16.
8550• Heat the test tube with a burner until
8551water vaporises and the cork blows
8552out.
8553• Observe that the test tube recoils in
8554the direction opposite to the direction
8555and the force FBA exerted by the ball B on ball
8556A (reaction) must be equal and opposite to
8557each other. Therefore,
8558FAB = – FBA (9.6)
8559or m
8560v u
8561t A
8562A A ( ) − = – m
8563v u
8564t B
8565B B ( ) − .
8566This gives,
8567mAuA + mBuB = mAvA + mBvB (9.7)
8568Since (mAuA + mBuB) is the total momentum
8569of the two balls A and B before the collision
8570and (mAvA
8571 + mBvB
8572) is their total momentum
8573after the collision, from Eq. (9.7) we observe
8574that the total momentum of the two balls
8575remains unchanged or conserved provided no
8576other external force acts.
8577As a result of this ideal collision
8578experiment, we say that the sum of momenta
8579of the two objects before collision is equal to
8580the sum of momenta after the collision provided
8581there is no external unbalanced force acting
8582on them. This is known as the law of
8583conservation of momentum. This statement
8584can alternatively be given as the total
8585momentum of the two objects is unchanged
8586or conserved by the collision.
8587Activity ______________ 9.5
8588• Take a big rubber balloon and inflate
8589it fully. Tie its neck using a thread.
8590Also using adhesive tape, fix a straw
8591on the surface of this balloon.
8592• Pass a thread through the straw and
8593hold one end of the thread in your
8594hand or fix it on the wall.
8595• Ask your friend to hold the other end
8596of the thread or fix it on a wall at some
8597distance. This arrangement is shown
8598in Fig. 9.15.
8599• Now remove the thread tied on the
8600neck of balloon. Let the air escape
8601from the mouth of the balloon.
8602• Observe the direction in which the
8603straw moves.
8604Fig. 9.15
8605of the cork.
8606Fig. 9.16
8607• Also, observe the difference in the
8608velocity the cork appears to have and
8609that of the recoiling test tube.
8610Example 9.6 A bullet of mass 20 g is
8611horizontally fired with a velocity
8612150 m s-1 from a pistol of mass 2 kg.
8613What is the recoil velocity of the pistol?
8614Solution:
8615We have the mass of bullet,
8616m1 = 20 g (= 0.02 kg) and the mass of
8617the pistol, m2 = 2 kg; initial velocities of
8618the bullet (u1) and pistol (u2) = 0,
8619respectively. The final velocity of the
8620bullet, v1
8621 = + 150 m s-1 . The direction of
8622bullet is taken from left to right (positive,
8623by convention, Fig. 9.17). Let v be the
8624FORCE AND LAWS OF MOTION 125
8625recoil velocity of the pistol.
8626Total momenta of the pistol and bullet
8627before the fire, when the gun is at rest
8628= (2 + 0.02) kg × 0 m s–1
8629= 0 kg m s–1
8630Total momenta of the pistol and bullet
8631after it is fired
8632= 0.02 kg × (+ 150 m s–1)
8633 + 2 kg × v m s–1
8634= (3 + 2v) kg m s–1
8635According to the law of conservation of
8636momentum
8637Total momenta after the fire = Total
8638momenta before the fire
86393 + 2v = 0
8640⇒ v = − 1.5 m s–1.
8641Negative sign indicates that the direction
8642in which the pistol would recoil is
8643opposite to that of bullet, that is, right
8644to left.
8645Example 9.7 A girl of mass 40 kg jumps
8646with a horizontal velocity of 5 m s-1 onto
8647a stationary cart with frictionless
8648wheels. The mass of the cart is 3 kg.
8649What is her velocity as the cart starts
8650moving? Assume that there is no
8651external unbalanced force working in
8652the horizontal direction.
8653Solution:
8654Let v be the velocity of the girl on the
8655cart as the cart starts moving.
8656The total momenta of the girl and cart
8657before the interaction
8658= 40 kg × 5 m s–1 + 3 kg × 0 m s–1
8659= 200 kg m s–1.
8660Total momenta after the interaction
8661= (40 + 3) kg × v m s–1
8662= 43 v kg m s–1.
8663According to the law of conservation of
8664momentum, the total momentum is
8665conserved during the interaction. That
8666is,
866743 v = 200
8668⇒ v = 200/43 = + 4.65 m s–1.
8669The girl on cart would move with a
8670velocity of 4.65 m s–1 in the direction in
8671Fig. 9.17: Recoil of a pistol which the girl jumped (Fig. 9.18).
8672Fig. 9.18: The girl jumps onto the cart.
8673(a) (b)
8674126 SCIENCE
8675Example 9.8 Two hockey players of
8676opposite teams, while trying to hit a
8677hockey ball on the ground collide and
8678immediately become entangled. One
8679has a mass of 60 kg and was moving
8680with a velocity 5.0 m s–1 while the other
8681has a mass of 55 kg and was moving
8682faster with a velocity 6.0 m s–1 towards
8683the first player. In which direction and
8684with what velocity will they move after
8685they become entangled? Assume that
8686the frictional force acting between the feet
8687of the two players and ground is
8688negligible.
8689Solution:
8690If v is the velocity of the two entangled
8691players after the collision, the total
8692momentum then
8693= (m1
8694 + m2) × v
8695= (60 + 55) kg × v m s–1
8696= 115 × v kg m s–1.
8697Equating the momenta of the system
8698before and after collision, in accordance
8699with the law of conservation of
8700momentum, we get
8701v = – 30/115
8702= – 0.26 m s–1.
8703Thus, the two entangled players would
8704move with velocity 0.26 m s–1 from right
8705to left, that is, in the direction the
8706second player was moving before
8707the collision.
8708Fig. 9.19: A collision of two hockey players: (a) before collision and (b) after collision.
8709Let the first player be moving from left
8710to right. By convention left to right is
8711taken as the positive direction and thus
8712right to left is the negative direction (Fig.
87139.19). If symbols m and u represent the
8714mass and initial velocity of the two
8715players, respectively. Subscripts 1 and
87162 in these physical quantities refer to the
8717two hockey players. Thus,
8718m1
8719 = 60 kg; u1
8720 = + 5 m s-1 ; and
8721m2 = 55 kg; u2
8722 = – 6 m s-1 .
8723The total momentum of the two players
8724before the collision
8725= 60 kg × (+ 5 m s-1) +
872655 kg × (– 6 m s-1)
8727= – 30 kg m s-1
8728uestions
87291. If action is always equal to the
8730reaction, explain how a horse
8731can pull a cart.
87322. Explain, why is it difficult for a
8733fireman to hold a hose, which
8734ejects large amounts of water at
8735a high velocity.
87363. From a rifle of mass 4 kg, a bullet
8737of mass 50 g is fired with an
8738initial velocity of 35 m s–1 .
8739Calculate the initial recoil
8740velocity of the rifle.
8741Q
8742FORCE AND LAWS OF MOTION 127
87434. Two objects of masses 100 g and
8744200 g are moving along the same
8745line and direction with velocities
8746of 2 m s–1 and 1 m s–1 ,
8747respectively. They collide and
8748after the collision, the first object
8749moves at a velocity of 1.67 m s–1.
8750Determine the velocity of the
8751second object.
8752What
8753you have
8754learnt
8755• First law of motion: An object continues to be in a state of rest
8756or of uniform motion along a straight line unless acted upon
8757by an unbalanced force.
8758• The natural tendency of objects to resist a change in their state
8759of rest or of uniform motion is called inertia.
8760• The mass of an object is a measure of its inertia. Its SI unit is
8761kilogram (kg).
8762• Force of friction always opposes motion of objects.
8763• Second law of motion: The rate of change of momentum of an
8764object is proportional to the applied unbalanced force in the
8765direction of the force.
8766• The SI unit of force is kg m s–2. This is also known as newton
8767and represented by the symbol N. A force of one newton
8768produces an acceleration of 1 m s–2 on an object of mass 1 kg.
8769• The momentum of an object is the product of its mass and
8770velocity and has the same direction as that of the velocity.
8771Its SI unit is kg m s–1 .
8772• Third law of motion: To every action, there is an equal and
8773opposite reaction and they act on two different bodies.
8774• In an isolated system (where there is no external force), the
8775total momentum remains conserved.
8776CONSERVATION LAWS
8777All conservation laws such as conservation of momentum, energy, angular momentum,
8778charge etc. are considered to be fundamental laws in physics. These are based on
8779observations and experiments. It is important to remember that a conservation law cannot
8780be proved. It can be verified, or disproved, by experiments. An experiment whose result is
8781in conformity with the law verifies or substantiates the law; it does not prove the law. On
8782the other hand, a single experiment whose result goes against the law is enough to disprove
8783it.
8784The law of conservation of momentum has been deduced from large number of
8785observations and experiments. This law was formulated nearly three centuries ago. It is
8786interesting to note that not a single situation has been realised so far, which contradicts
8787this law. Several experiences of every-day life can be explained on the basis of the law of
8788conservation of momentum.
8789128 SCIENCE
8790Exercises
87911. An object experiences a net zero external unbalanced force. Is
8792it possible for the object to be travelling with a non-zero velocity?
8793If yes, state the conditions that must be placed on the
8794magnitude and direction of the velocity. If no, provide a reason.
87952. When a carpet is beaten with a stick, dust comes out of it.
8796Explain.
87973. Why is it advised to tie any luggage kept on the roof of a bus
8798with a rope?
87994. A batsman hits a cricket ball which then rolls on a level ground.
8800After covering a short distance, the ball comes to rest. The ball
8801slows to a stop because
8802(a) the batsman did not hit the ball hard enough.
8803(b) velocity is proportional to the force exerted on the ball.
8804(c) there is a force on the ball opposing the motion.
8805(d) there is no unbalanced force on the ball, so the ball would
8806want to come to rest.
88075. A truck starts from rest and rolls down a hill with a constant
8808acceleration. It travels a distance of 400 m in 20 s. Find its
8809acceleration. Find the force acting on it if its mass is
88107 tonnes (Hint: 1 tonne = 1000 kg.)
88116. A stone of 1 kg is thrown with a velocity of 20 m s–1 across
8812the frozen surface of a lake and comes to rest after travelling
8813a distance of 50 m. What is the force of friction between the
8814stone and the ice?
88157. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg,
8816along a horizontal track. If the engine exerts a force of 40000 N
8817and the track offers a friction force of 5000 N, then calculate:
8818(a) the net accelerating force;
8819(b) the acceleration of the train; and
8820(c) the force of wagon 1 on wagon 2.
88218. An automobile vehicle has a mass of 1500 kg. What must be
8822the force between the vehicle and road if the vehicle is to be
8823stopped with a negative acceleration of 1.7 m s–2?
88249. What is the momentum of an object of mass m, moving with a
8825velocity v?
8826(a) (mv)
88272 (b) mv2 (c) ½ mv2 (d) mv
882810. Using a horizontal force of 200 N, we intend to move a wooden
8829cabinet across a floor at a constant velocity. What is the
8830friction force that will be exerted on the cabinet?
883111. Two objects, each of mass 1.5 kg, are moving in the same
8832straight line but in opposite directions. The velocity of each
8833FORCE AND LAWS OF MOTION 129
8834object is 2.5 m s-1 before the collision during which they
8835stick together. What will be the velocity of the combined
8836object after collision?
883712. According to the third law of motion when we push on an object,
8838the object pushes back on us with an equal and opposite force.
8839If the object is a massive truck parked along the roadside, it
8840will probably not move. A student justifies this by answering
8841that the two opposite and equal forces cancel each other.
8842Comment on this logic and explain why the truck does not
8843move.
884413. A hockey ball of mass 200 g travelling at 10 m s–1 is struck by
8845a hockey stick so as to return it along its original path with a
8846velocity at 5 m s–1. Calculate the change of momentum occurred
8847in the motion of the hockey ball by the force applied by the
8848hockey stick.
884914. A bullet of mass 10 g travelling horizontally with a velocity of
8850150 m s–1 strikes a stationary wooden block and comes to rest
8851in 0.03 s. Calculate the distance of penetration of the bullet
8852into the block. Also calculate the magnitude of the force exerted
8853by the wooden block on the bullet.
885415. An object of mass 1 kg travelling in a straight line with a velocity
8855of 10 m s–1 collides with, and sticks to, a stationary wooden
8856block of mass 5 kg. Then they both move off together in the
8857same straight line. Calculate the total momentum just before
8858the impact and just after the impact. Also, calculate the velocity
8859of the combined object.
886016. An object of mass 100 kg is accelerated uniformly from a velocity
8861of 5 m s–1 to 8 m s–1 in 6 s. Calculate the initial and final
8862momentum of the object. Also, find the magnitude of the force
8863exerted on the object.
886417. Akhtar, Kiran and Rahul were riding in a motorcar that was
8865moving with a high velocity on an expressway when an insect
8866hit the windshield and got stuck on the windscreen. Akhtar
8867and Kiran started pondering over the situation. Kiran suggested
8868that the insect suffered a greater change in momentum as
8869compared to the change in momentum of the motorcar (because
8870the change in the velocity of the insect was much more than
8871that of the motorcar). Akhtar said that since the motorcar was
8872moving with a larger velocity, it exerted a larger force on
8873the insect. And as a result the insect died. Rahul while
8874putting an entirely new explanation said that both the
8875motorcar and the insect experienced the same force and a
8876change in their momentum. Comment on these suggestions.
887718. How much momentum will a dumb-bell of mass 10 kg transfer
8878to the floor if it falls from a height of 80 cm? Take its downward
8879acceleration to be 10 m s–2 .
8880130 SCIENCE
8881Additional
8882Exercises
8883A1. The following is the distance-time table of an object in motion:
8884Time in seconds Distance in metres
88850 0
88861 1
88872 8
88883 27
88894 64
88905 125
88916 216
88927 343
8893(a) What conclusion can you draw about the acceleration?
8894Is it constant, increasing, decreasing, or zero?
8895(b) What do you infer about the forces acting on the object?
8896A2. Two persons manage to push a motorcar of mass 1200 kg at a
8897uniform velocity along a level road. The same motorcar can be
8898pushed by three persons to produce an acceleration of
88990.2 m s-2. With what force does each person push the motorcar?
8900(Assume that all persons push the motorcar with the same
8901muscular effort.)
8902A3. A hammer of mass 500 g, moving at 50 m s-1, strikes a nail.
8903The nail stops the hammer in a very short time of 0.01 s. What
8904is the force of the nail on the hammer?
8905A4. A motorcar of mass 1200 kg is moving along a straight line
8906with a uniform velocity of 90 km/h. Its velocity is slowed down
8907to 18 km/h in 4 s by an unbalanced external force. Calculate
8908the acceleration and change in momentum. Also calculate the
8909magnitude of the force required.
8910In Chapters 8 and 9, we have learnt about
8911the motion of objects and force as the cause
8912of motion. We have learnt that a force is
8913needed to change the speed or the direction
8914of motion of an object. We always observe that
8915an object dropped from a height falls towards
8916the earth. We know that all the planets go
8917around the Sun. The moon goes around the
8918earth. In all these cases, there must be some
8919force acting on the objects, the planets and
8920on the moon. Isaac Newton could grasp that
8921the same force is responsible for all these.
8922This force is called the gravitational force.
8923In this chapter we shall learn about
8924gravitation and the universal law of
8925gravitation. We shall discuss the motion of
8926objects under the influence of gravitational
8927force on the earth. We shall study how the
8928weight of a body varies from place to place.
8929We shall also discuss the conditions for
8930objects to float in liquids.
893110.1 Gravitation
8932We know that the moon goes around the
8933earth. An object when thrown upwards,
8934reaches a certain height and then falls
8935downwards. It is said that when Newton was
8936sitting under a tree, an apple fell on him. The
8937fall of the apple made Newton start thinking.
8938He thought that: if the earth can attract an
8939apple, can it not attract the moon? Is the force
8940the same in both cases? He conjectured that
8941the same type of force is responsible in both
8942the cases. He argued that at each point of its
8943orbit, the moon falls towards the earth,
8944instead of going off in a straight line. So, it
8945must be attracted by the earth. But we do
8946not really see the moon falling towards the
8947earth.
8948Let us try to understand the motion of
8949the moon by recalling activity 8.11.
8950Activity _____________ 10.1
8951• Take a piece of thread.
8952• Tie a small stone at one end. Hold the
8953other end of the thread and whirl it
8954round, as shown in Fig. 10.1.
8955• Note the motion of the stone.
8956• Release the thread.
8957• Again, note the direction of motion of
8958the stone.
8959Fig. 10.1: A stone describing a circular path with a
8960velocity of constant magnitude.
8961Before the thread is released, the stone
8962moves in a circular path with a certain speed
8963and changes direction at every point. The
8964change in direction involves change in velocity
8965or acceleration. The force that causes this
8966acceleration and keeps the body moving along
8967the circular path is acting towards the centre.
8968This force is called the centripetal (meaning
8969‘centre-seeking’) force. In the absence of this
897010
8971GRAVITATION RAVITATION
8972Chapter
8973© NCERT
8974not to be republished
8975132 SCIENCE
897610.1.1 UNIVERSAL LAW OF GRAVITATION
8977Every object in the universe attracts every
8978other object with a force which is proportional
8979to the product of their masses and inversely
8980proportional to the square of the distance
8981between them. The force is along the line
8982joining the centres of two objects.
8983force, the stone flies off along a straight line.
8984This straight line will be a tangent to the
8985circular path. More to know
8986Tangent to a circle
8987A straight line that meets the circle
8988at one and only one point is called a
8989tangent to the circle. Straight line
8990ABC is a tangent to the circle at
8991point B.
8992The motion of the moon around the earth
8993is due to the centripetal force. The centripetal
8994force is provided by the force of attraction of
8995the earth. If there were no such force, the
8996moon would pursue a uniform straight line
8997motion.
8998It is seen that a falling apple is attracted
8999towards the earth. Does the apple attract the
9000earth? If so, we do not see the earth moving
9001towards an apple. Why?
9002According to the third law of motion, the
9003apple does attract the earth. But according
9004to the second law of motion, for a given force,
9005acceleration is inversely proportional to the
9006mass of an object [Eq. (9.4)]. The mass of an
9007apple is negligibly small compared to that of
9008the earth. So, we do not see the earth moving
9009towards the apple. Extend the same argument
9010for why the earth does not move towards the
9011moon.
9012In our solar system, all the planets go
9013around the Sun. By arguing the same way,
9014we can say that there exists a force between
9015the Sun and the planets. From the above facts
9016Newton concluded that not only does the
9017earth attract an apple and the moon, but all
9018objects in the universe attract each other. This
9019force of attraction between objects is called
9020the gravitational force.
9021G 2
9022Mm
9023F =
9024d
9025Fig. 10.2: The gravitational force between two
9026uniform objects is directed along the line
9027joining their centres.
9028Let two objects A and B of masses M and
9029m lie at a distance d from each other as shown
9030in Fig. 10.2. Let the force of attraction between
9031two objects be F. According to the universal
9032law of gravitation, the force between two
9033objects is directly proportional to the product
9034of their masses. That is,
9035F M × m (10.1)
9036And the force between two objects is inversely
9037proportional to the square of the distance
9038between them, that is,
90392
90401 F
9041d (10.2)
9042Combining Eqs. (10.1) and (10.2), we get
9043F 2
9044M m
9045d (10.3)
9046or, G 2
9047M×m
9048F =
9049d (10.4)
9050where G is the constant of proportionality and
9051is called the universal gravitation constant.
9052By multiplying crosswise, Eq. (10.4) gives
9053 F × d 2
9054 = G M × m
9055© NCERT
9056not to be republished
9057GRAVITATION 133
9058Isaac Newton was born
9059in Woolsthorpe near
9060Grantham, England.
9061He is generally
9062regarded as the most
9063original and
9064influential theorist in
9065the history of science.
9066He was born in a poor
9067farming family. But he
9068was not good at
9069farming. He was sent
9070to study at Cambridge
9071University in 1661. In
90721665 a plague broke
9073out in Cambridge and so Newton took a year
9074off. It was during this year that the incident of
9075the apple falling on him is said to have
9076occurred. This incident prompted Newton to
9077explore the possibility of connecting gravity
9078with the force that kept the moon in its orbit.
9079This led him to the universal law of
9080gravitation. It is remarkable that many great
9081scientists before him knew of gravity but failed
9082to realise it.
9083Newton formulated the well-known laws of
9084motion. He worked on theories of light and
9085colour. He designed an astronomical telescope
9086to carry out astronomical observations.
9087Newton was also a great mathematician. He
9088invented a new branch of mathematics, called
9089calculus. He used it to prove that for objects
9090outside a sphere of uniform density, the sphere
9091behaves as if the whole of its mass is
9092concentrated at its centre. Newton
9093transformed the structure of physical
9094science with his three laws of motion and the
9095universal law of gravitation. As the keystone
9096of the scientific revolution of the seventeenth
9097century, Newton’s work combined the
9098contributions of Copernicus, Kepler, Galileo,
9099and others into a new powerful synthesis.
9100It is remarkable that though the
9101gravitational theory could not be verified at
9102that time, there was hardly any doubt about
9103its correctness. This is because Newton based
9104his theory on sound scientific reasoning and
9105backed it with mathematics. This made the
9106theory simple and elegant. These qualities are
9107now recognised as essential requirements of a
9108good scientific theory.
9109Isaac Newton
9110(1642 – 1727)
9111How did Newton guess the
9112inverse-square rule?
9113There has always been a great interest
9114in the motion of planets. By the 16th
9115century, a lot of data on the motion of
9116planets had been collected by many
9117astronomers. Based on these data
9118Johannes Kepler derived three laws,
9119which govern the motion of planets.
9120These are called Kepler’s laws. These are:
91211. The orbit of a planet is an ellipse with
9122the Sun at one of the foci, as shown in
9123the figure given below. In this figure O
9124is the position of the Sun.
91252. The line joining the planet and the Sun
9126sweep equal areas in equal intervals
9127of time. Thus, if the time of travel from
9128A to B is the same as that from C to D,
9129then the areas OAB and OCD are
9130equal.
91313. The cube of the mean distance of a
9132planet from the Sun is proportional to
9133the square of its orbital period T. Or,
9134r3
9135/T2 = constant.
9136It is important to note that Kepler
9137could not give a theory to explain
9138the motion of planets. It was Newton
9139who showed that the cause of the
9140planetary motion is the gravitational
9141force that the Sun exerts on them. Newton
9142used the third law
9143of Kepler to
9144calculate the
9145gravitational force
9146of attraction. The
9147gravitational force
9148of the earth is
9149weakened by distance. A simple argument
9150goes like this. We can assume that the
9151planetary orbits are circular. Suppose the
9152orbital velocity is v and the radius of the
9153orbit is r. Then the force acting on an
9154orbiting planet is given by F v2/r.
9155If T denotes the period, then v = 2Ï€r/T,
9156so that v2 r 2
9157/T2.
9158We can rewrite this as v2 (1/r) ×
9159( r3
9160/T2
9161). Since r3/T2
9162 is constant by Kepler’s
9163third law, we have v2 1/r. Combining
9164this with F v2/ r, we get, F 1/ r2
9165.
9166A
9167B
9168C D
9169O
9170© NCERT
9171not to be republished
9172134 SCIENCE
9173From Eq. (10.4), the force exerted by
9174the earth on the moon is
9175G 2
9176M × m
9177F =
9178d
917911 2 -2 24 22
91808 2
91816.7 10 N m kg 6 10 kg 7.4 10 kg
9182(3.84 10 m)
9183
9184= 2.01 × 1020 N.
9185Thus, the force exerted by the earth on
9186the moon is 2.01 × 1020 N.
9187uestions
91881. State the universal law of
9189gravitation.
91902. Write the formula to find the
9191magnitude of the gravitational
9192force between the earth and an
9193object on the surface of the earth.
919410.1.2 IMPORTANCE OF THE UNIVERSAL
9195LAW OFGRAVITATION
9196The universal law of gravitation successfully
9197explained several phenomena which were
9198believed to be unconnected:
9199(i) the force that binds us to the earth;
9200(ii) the motion of the moon around the
9201earth;
9202(iii) the motion of planets around the Sun;
9203and
9204(iv) the tides due to the moon and the Sun.
920510.2 Free Fall Free Fall
9206Let us try to understand the meaning of free
9207fall by performing this activity.
9208Activity _____________ 10.2
9209• Take a stone.
9210• Throw it upwards.
9211• It reaches a certain height and then it
9212starts falling down.
9213We have learnt that the earth attracts
9214objects towards it. This is due to the
9215gravitational force. Whenever objects fall
9216towards the earth under this force alone, we
9217say that the objects are in free fall. Is there
9218or
92192
9220G
9221F d
9222M m (10.5)
9223The SI unit of G can be obtained by
9224substituting the units of force, distance and
9225mass in Eq. (10.5) as N m2 kg–2.
9226The value of G was found out by
9227Henry Cavendish (1731 – 1810) by using a
9228sensitive balance. The accepted value of G is
92296.673 × 10–11 N m2
9230 kg–2.
9231We know that there exists a force of
9232attraction between any two objects. Compute
9233the value of this force between you and your
9234friend sitting closeby. Conclude how you do
9235not experience this force!
9236The law is universal in the sense that
9237it is applicable to all bodies, whether
9238the bodies are big or small, whether
9239they are celestial or terrestrial.
9240Inverse-square
9241Saying that F is inversely
9242proportional to the square of d
9243means, for example, that if d gets
9244bigger by a factor of 6, F becomes
92451
924636 times smaller.
9247Example 10.1 The mass of the earth is
92486 × 1024 kg and that of the moon is
92497.4 1022 kg. If the distance between
9250the earth and the moon is 3.84105 km,
9251calculate the force exerted by the earth
9252on the moon. G = 6.7 10–11 N m2 kg-2.
9253Solution:
9254The mass of the earth, M = 6 1024 kg
9255The mass of the moon,
9256m = 7.4 1022 kg
9257The distance between the earth and the
9258moon,
9259d = 3.84 105 km
9260= 3.84 105 1000 m
9261= 3.84 108 m
9262G = 6.7 10–11 N m2 kg–2
9263More to know
9264Q
9265© NCERT
9266not to be republished
9267GRAVITATION 135
9268calculations, we can take g to be more or less
9269constant on or near the earth. But for objects
9270far from the earth, the acceleration due to
9271gravitational force of earth is given by
9272Eq. (10.7).
927310.2.1 TO CALCULATE THE VALUE OF g
9274To calculate the value of g, we should put
9275the values of G, M and R in Eq. (10.9),
9276namely, universal gravitational constant,
9277G = 6.7 × 10–11 N m2 kg-2, mass of the earth,
9278M = 6 × 1024 kg, and radius of the earth,
9279R = 6.4 × 106 m.
9280G 2
9281M
9282g =
9283R
9284-11 2 -2 24
92856 2
92866.7 10 N m kg 6 10 kg = (6.4 10 m)
9287
9288 = 9.8 m s–2.
9289Thus, the value of acceleration due to gravity
9290of the earth, g = 9.8 m s–2.
929110.2.2 MOTION OF OBJECTS UNDER THE
9292INFLUENCE OF GRAVITATIONAL
9293FORCE OF THE EARTH
9294Let us do an activity to understand whether
9295all objects hollow or solid, big or small, will
9296fall from a height at the same rate.
9297Activity _____________ 10.3
9298• Take a sheet of paper and a stone. Drop
9299them simultaneously from the first floor
9300of a building. Observe whether both of
9301them reach the ground simultaneously.
9302• We see that paper reaches the ground
9303little later than the stone. This happens
9304because of air resistance. The air offers
9305resistance due to friction to the motion
9306of the falling objects. The resistance
9307offered by air to the paper is more than
9308the resistance offered to the stone. If
9309we do the experiment in a glass jar from
9310which air has been sucked out, the
9311paper and the stone would fall at the
9312same rate.
9313any change in the velocity of falling objects?
9314While falling, there is no change in the
9315direction of motion of the objects. But due to
9316the earth’s attraction, there will be a change
9317in the magnitude of the velocity. Any change
9318in velocity involves acceleration. Whenever an
9319object falls towards the earth, an acceleration
9320is involved. This acceleration is due to the
9321earth’s gravitational force. Therefore, this
9322acceleration is called the acceleration due to
9323the gravitational force of the earth (or
9324acceleration due to gravity). It is denoted by
9325g. The unit of g is the same as that of
9326acceleration, that is, m s–2.
9327We know from the second law of motion
9328that force is the product of mass and
9329acceleration. Let the mass of the stone in
9330activity 10.2 be m. We already know that there
9331is acceleration involved in falling objects due
9332to the gravitational force and is denoted by g.
9333Therefore the magnitude of the gravitational
9334force F will be equal to the product of mass
9335and acceleration due to the gravitational
9336force, that is,
9337F = m g (10.6)
9338From Eqs. (10.4) and (10.6) we have
93392 = G
9340M m
9341m g
9342d
9343or G 2
9344M
9345g =
9346d (10.7)
9347where M is the mass of the earth, and d is
9348the distance between the object and the earth.
9349Let an object be on or near the surface of
9350the earth. The distance d in Eq. (10.7) will be
9351equal to R, the radius of the earth. Thus, for
9352objects on or near the surface of the earth,
9353G 2
9354M×m
9355mg =
9356R (10.8)
9357 G 2
9358M
9359g =
9360R (10.9)
9361The earth is not a perfect sphere. As the
9362radius of the earth increases from the poles
9363to the equator, the value of g becomes greater
9364at the poles than at the equator. For most
9365© NCERT
9366not to be republished
9367136 SCIENCE
9368We know that an object experiences
9369acceleration during free fall. From Eq. (10.9),
9370this acceleration experienced by an object is
9371independent of its mass. This means that all
9372objects hollow or solid, big or small, should
9373fall at the same rate. According to a story,
9374Galileo dropped different objects from the top
9375of the Leaning Tower of Pisa in Italy to prove
9376the same.
9377As g is constant near the earth, all the
9378equations for the uniformly accelerated
9379motion of objects become valid with
9380acceleration a replaced by g (see section 8.5).
9381The equations are:
9382v = u + at (10.10)
9383s = ut +
93841
93852 at2 (10.11)
9386v2 = u2
9387 + 2as (10.12)
9388where u and v are the initial and final
9389velocities and s is the distance covered in
9390time, t.
9391In applying these equations, we will take
9392acceleration, a to be positive when it is in the
9393direction of the velocity, that is, in the
9394direction of motion. The acceleration, a will
9395be taken as negative when it opposes the
9396motion.
9397Example 10.2 A car falls off a ledge and
9398drops to the ground in 0.5 s. Let
9399g = 10 m s–2 (for simplifying the
9400calculations).
9401(i) What is its speed on striking the
9402ground?
9403(ii) What is its average speed during the
94040.5 s?
9405(iii) How high is the ledge from the
9406ground?
9407Solution:
9408 Time, t = ½ second
9409Initial velocity, u = 0 m s–1
9410Acceleration due to gravity, g = 10 m s–2
9411Acceleration of the car, a = + 10 m s–2
9412 (downward)
9413(i) speed v = a t
9414 v = 10 m s–2 × 0.5 s
9415= 5 m s–1
9416(ii) average speed= 2
9417u +v
9418= (0 m s–1+ 5 m s–1)/2
9419= 2.5 m s–1
9420(iii) distance travelled, s = ½ a t2
9421= ½ × 10 m s–2 × (0.5 s)2
9422= ½ × 10 m s–2 × 0.25 s2
9423= 1.25 m
9424Thus,
9425(i) its speed on striking the ground
9426= 5 m s–1
9427(ii) its average speed during the 0.5 s
9428= 2.5 m s–1
9429(iii) height of the ledge from the ground
9430= 1.25 m.
9431Example 10.3 An object is thrown
9432vertically upwards and rises to a height
9433of 10 m. Calculate (i) the velocity with
9434which the object was thrown upwards
9435and (ii) the time taken by the object to
9436reach the highest point.
9437Solution:
9438Distance travelled, s = 10 m
9439Final velocity, v = 0 m s–1
9440Acceleration due to gravity, g = 9.8 m s–2
9441Acceleration of the object, a = –9.8 m s–2
9442(upward motion)
9443(i) v 2 = u2 + 2a s
94440 = u 2 + 2 × (–9.8 m s–2)× 10 m
9445–u 2 = –2 × 9.8 × 10 m2 s–2
9446u = 196 m s-1
9447 u = 14 m s-1
9448(ii) v = u + a t
94490 = 14 m s–1 – 9.8 m s–2 × t
9450t = 1.43 s.
9451Thus,
9452(i) Initial velocity, u = 14 m s–1, and
9453(ii) Time taken, t = 1.43 s.
9454uestions
94551. What do you mean by free fall?
94562. What do you mean by acceleration Q due to gravity?
9457© NCERT
9458not to be republished
9459GRAVITATION 137
946010.3 Mass
9461We have learnt in the previous chapter that
9462the mass of an object is the measure of its
9463inertia (section 9.3). We have also learnt that
9464greater the mass, the greater is the inertia. It
9465remains the same whether the object is on
9466the earth, the moon or even in outer space.
9467Thus, the mass of an object is constant and
9468does not change from place to place.
946910.4 Weight
9470We know that the earth attracts every object
9471with a certain force and this force depends
9472on the mass (m) of the object and the
9473acceleration due to the gravity (g). The weight
9474of an object is the force with which it is
9475attracted towards the earth.
9476We know that
9477F = m × a, (10.13)
9478that is,
9479F = m × g. (10.14)
9480The force of attraction of the earth on an
9481object is known as the weight of the object. It
9482is denoted by W. Substituting the same in
9483Eq. (10.14), we have
9484W = m × g (10.15)
9485As the weight of an object is the force with
9486which it is attracted towards the earth, the
9487SI unit of weight is the same as that of force,
9488that is, newton (N). The weight is a force acting
9489vertically downwards; it has both magnitude
9490and direction.
9491We have learnt that the value of g is
9492constant at a given place. Therefore at a given
9493place, the weight of an object is directly
9494proportional to the mass, say m, of the object,
9495that is, W m. It is due to this reason that
9496at a given place, we can use the weight of an
9497object as a measure of its mass. The mass of
9498an object remains the same everywhere, that
9499is, on the earth and on any planet whereas
9500its weight depends on its location.
950110.4.1 WEIGHT OF AN OBJECT ON
9502THE MOON
9503We have learnt that the weight of an object
9504on the earth is the force with which the earth
9505attracts the object. In the same way, the
9506weight of an object on the moon is the force
9507with which the moon attracts that object. The
9508mass of the moon is less than that of the
9509earth. Due to this the moon exerts lesser force
9510of attraction on objects.
9511Let the mass of an object be m. Let its
9512weight on the moon be Wm. Let the mass of
9513the moon be Mm and its radius be Rm.
9514By applying the universal law of
9515gravitation, the weight of the object on the
9516moon will be
95172 G m
9518m
9519m
9520M m W
9521R (10.16)
9522Let the weight of the same object on the
9523earth be We. The mass of the earth is M and
9524its radius is R.
9525Table 10.1
9526Celestial Mass (kg) Radius (m)
9527body
9528Earth 5.98 1024 6.37 106
9529Moon 7.36 1022 1.74 106
9530From Eqs. (10.9) and (10.15) we have,
9531e G 2
9532M m
9533W
9534R (10.17)
9535Substituting the values from Table 10.1 in
9536Eqs. (10.16) and (10.17), we get
9537
953822
95392 6
95407.36 10 kg G
95411.74 10 m
9542 m
9543m
9544W
954510 2.431 10 G × W m m (10.18a)
9546and 11 1.474 10 G × W m e (10.18b)
9547Dividing Eq. (10.18a) by Eq. (10.18b), we get
954810
954911
95502.431 10
95511.474 10
9552m
9553e
9554W
9555W
9556or
95571 0.165
95586
9559m
9560e
9561W
9562W = ≈ (10.19)
9563Weight of theobject onthe moon 1 = Weightof theobject onthe earth 6
9564Weight of the object on the moon
9565= (1/6) × its weight on the earth.
9566© NCERT
9567not to be republished
9568138 SCIENCE
9569Example 10.4 Mass of an object is 10 kg.
9570What is its weight on the earth?
9571Solution:
9572Mass, m = 10 kg
9573Acceleration due to gravity, g = 9.8 m s–2
9574W = m × g
9575W = 10 kg × 9.8 m s-2 = 98 N
9576Thus, the weight of the object is 98 N.
9577Example 10.5 An object weighs 10 N when
9578measured on the surface of the earth.
9579What would be its weight when
9580measured on the surface of the moon?
9581Solution:
9582We know,
9583Weight of object on the moon
9584= (1/6) × its weight on the earth.
9585That is,
958610
95876 6
9588e
9589m
9590W
9591W= = N.
9592= 1.67 N.
9593Thus, the weight of object on the
9594surface of the moon would be 1.67 N.
9595uestions
95961. What are the differences between
9597the mass of an object and its
9598weight?
95992. Why is the weight of an object on
9600the moon
96011
96026 th its weight on the
9603earth?
960410.5 Thrust and Pressure Thrust and Pressure
9605Have you ever wondered why a camel can run
9606in a desert easily? Why an army tank weighing
9607more than a thousand tonne rests upon a
9608continuous chain? Why a truck or a motorbus
9609has much wider tyres? Why cutting tools have
9610sharp edges? In order to address these
9611questions and understand the phenomena
9612involved, it helps to introduce the concepts
9613of the net force in a particular direction
9614(thrust) and the force per unit area (pressure)
9615acting on the object concerned.
9616Let us try to understand the meanings of
9617thrust and pressure by considering the
9618following situations:
9619Situation 1: You wish to fix a poster on a
9620bulletin board, as shown in Fig 10.3. To do
9621this task you will have to press drawing pins
9622with your thumb. You apply a force on the
9623surface area of the head of the pin. This force
9624is directed perpendicular to the surface area
9625of the board. This force acts on a smaller area
9626at the tip of the pin.
9627Q
9628Fig. 10.3: To fix a poster, drawing pins are pressed
9629with the thumb perpendicular to the board.
9630Situation 2: You stand on loose sand. Your
9631feet go deep into the sand. Now, lie down on
9632the sand. You will find that your body will
9633not go that deep in the sand. In both cases
9634the force exerted on the sand is the weight of
9635your body.
9636© NCERT
9637not to be republished
9638GRAVITATION 139
9639You have learnt that weight is the force
9640acting vertically downwards. Here the force
9641is acting perpendicular to the surface of the
9642sand. The force acting on an object
9643perpendicular to the surface is called thrust.
9644When you stand on loose sand, the force,
9645that is, the weight of your body is acting on
9646an area equal to area of your feet. When you
9647lie down, the same force acts on an area equal
9648to the contact area of your whole body, which
9649is larger than the area of your feet. Thus, the
9650effects of forces of the same magnitude on
9651different areas are different. In the above
9652cases, thrust is the same. But effects are
9653different. Therefore the effect of thrust
9654depends on the area on which it acts.
9655The effect of thrust on sand is larger while
9656standing than while lying. The thrust on unit
9657area is called pressure. Thus,
9658thrust
9659Pressure =
9660area
9661(10.20)
9662Substituting the SI unit of thrust and area in
9663Eq. (10.20), we get the SI unit of pressure as
9664N/m2
9665 or N m–2.
9666In honour of scientist Blaise Pascal, the
9667SI unit of pressure is called pascal, denoted
9668as Pa.
9669Let us consider a numerical example to
9670understand the effects of thrust acting on
9671different areas.
9672Example 10.6 A block of wood is kept on a
9673tabletop. The mass of wooden block is
96745 kg and its dimensions are 40 cm × 20
9675cm × 10 cm. Find the pressure exerted
9676by the wooden block on the table top if
9677it is made to lie on the table top with its
9678sides of dimensions (a) 20 cm × 10 cm
9679and (b) 40 cm × 20 cm.
9680Solution:
9681The mass of the wooden block = 5 kg
9682The dimensions
9683= 40 cm × 20 cm × 10 cm
9684Here, the weight of the wooden block
9685applies a thrust on the table top.
9686That is,
9687 Thrust = F = m × g
9688= 5 kg × 9.8 m s–2
9689= 49 N
9690Area of a side = length × breadth
9691= 20 cm × 10 cm
9692= 200 cm2 = 0.02 m2
9693From Eq. (10.20),
9694Pressure = 2
969549N
96960.02m
9697= 2450 N m-2.
9698When the block lies on its side of
9699dimensions 40 cm × 20 cm, it exerts
9700the same thrust.
9701Area= length × breadth
9702= 40 cm × 20 cm
9703= 800 cm2
9704 = 0.08 m2
9705From Eq. (10.20),
9706Pressure = 2
970749 N
97080.08 m
9709= 612.5 N m–2
9710The pressure exerted by the side 20 cm
9711× 10 cm is 2450 N m–2 and by the side
971240 cm × 20 cm is 612.5 N m–2.
9713Thus, the same force acting on a smaller
9714area exerts a larger pressure, and a smaller
9715pressure on a larger area. This is the reason
9716why a nail has a pointed tip, knives have sharp
9717edges and buildings have wide foundations.
971810.5.1 PRESSURE IN FLUIDS
9719All liquids and gases are fluids. A solid exerts
9720pressure on a surface due to its weight.
9721Fig. 10.4 Similarly, fluids have weight, and they also
9722© NCERT
9723not to be republished
9724140 SCIENCE
9725by the water on the bottle is greater than its
9726weight. Therefore it rises up when released.
9727To keep the bottle completely immersed,
9728the upward force on the bottle due to water
9729must be balanced. This can be achieved by
9730an externally applied force acting downwards.
9731This force must at least be equal to the
9732difference between the upward force and the
9733weight of the bottle.
9734The upward force exerted by the water on
9735the bottle is known as upthrust or buoyant
9736force. In fact, all objects experience a force of
9737buoyancy when they are immersed in a fluid.
9738The magnitude of this buoyant force depends
9739on the density of the fluid.
974010.5.3 WHY OBJECTS FLOAT OR SINK WHEN
9741PLACED ON THE SURFACE OF WATER?
9742Let us do the following activities to arrive at
9743an answer for the above question.
9744Activity _____________ 10.5
9745• Take a beaker filled with water.
9746• Take an iron nail and place it on the
9747surface of the water.
9748• Observe what happens.
9749The nail sinks. The force due to the
9750gravitational attraction of the earth on the
9751iron nail pulls it downwards. There is an
9752upthrust of water on the nail, which pushes
9753it upwards. But the downward force acting
9754on the nail is greater than the upthrust of
9755water on the nail. So it sinks (Fig. 10.5).
9756exert pressure on the base and walls of the
9757container in which they are enclosed.
9758Pressure exerted in any confined mass of fluid
9759is transmitted undiminished in all directions.
976010.5.2 BUOYANCY
9761Have you ever had a swim in a pool and felt
9762lighter? Have you ever drawn water from a
9763well and felt that the bucket of water is heavier
9764when it is out of the water? Have you ever
9765wondered why a ship made of iron and steel
9766does not sink in sea water, but while the same
9767amount of iron and steel in the form of a sheet
9768would sink? These questions can be answered
9769by taking buoyancy in consideration. Let us
9770understand the meaning of buoyancy by
9771doing an activity.
9772Activity _____________ 10.4
9773• Take an empty plastic bottle. Close the
9774mouth of the bottle with an airtight
9775stopper. Put it in a bucket filled with
9776water. You see that the bottle floats.
9777• Push the bottle into the water. You feel
9778an upward push. Try to push it further
9779down. You will find it difficult to push
9780deeper and deeper. This indicates that
9781water exerts a force on the bottle in the
9782upward direction. The upward force
9783exerted by the water goes on increasing
9784as the bottle is pushed deeper till it is
9785completely immersed.
9786• Now, release the bottle. It bounces
9787back to the surface.
9788• Does the force due to the gravitational
9789attraction of the earth act on this
9790bottle? If so, why doesn’t the bottle stay
9791immersed in water after it is released?
9792How can you immerse the bottle in
9793water?
9794The force due to the gravitational
9795attraction of the earth acts on the bottle in
9796the downward direction. So the bottle is
9797pulled downwards. But the water exerts an
9798upward force on the bottle. Thus, the bottle
9799is pushed upwards. We have learnt that
9800weight of an object is the force due to
9801gravitational attraction of the earth. When the
9802bottle is immersed, the upward force exerted
9803Fig. 10.5: An iron nail sinks and a cork floats when
9804placed on the surface of water.
9805© NCERT
9806not to be republished
9807GRAVITATION 141
9808• Observe what happens to elongation
9809of the string or the reading on the
9810balance.
9811You will find that the elongation of the
9812string or the reading of the balance decreases
9813as the stone is gradually lowered in the water.
9814However, no further change is observed once
9815the stone gets fully immersed in the water.
9816What do you infer from the decrease in the
9817extension of the string or the reading of the
9818spring balance?
9819We know that the elongation produced in
9820the string or the spring balance is due to the
9821weight of the stone. Since the extension
9822decreases once the stone is lowered in water,
9823it means that some force acts on the stone in
9824upward direction. As a result, the net force
9825on the string decreases and hence the
9826elongation also decreases. As discussed
9827earlier, this upward force exerted by water is
9828known as the force of buoyancy.
9829What is the magnitude of the buoyant
9830force experienced by a body? Is it the same
9831in all fluids for a given body? Do all bodies in
9832a given fluid experience the same buoyant
9833force? The answer to these questions is
9834Activity _____________ 10.6
9835• Take a beaker filled with water.
9836• Take a piece of cork and an iron nail of
9837equal mass.
9838• Place them on the surface of water.
9839• Observe what happens.
9840The cork floats while the nail sinks. This
9841happens because of the difference in their
9842densities. The density of a substance is
9843defined as the mass per unit volume. The
9844density of cork is less than the density of
9845water. This means that the upthrust of water
9846on the cork is greater than the weight of the
9847cork. So it floats (Fig. 10.5).
9848The density of an iron nail is more than
9849the density of water. This means that the
9850upthrust of water on the iron nail is less than
9851the weight of the nail. So it sinks.
9852Therefore objects of density less than that
9853of a liquid float on the liquid. The objects of
9854density greater than that of a liquid sink in
9855the liquid.
9856uestions
98571. Why is it difficult to hold a school
9858bag having a strap made of a thin
9859and strong string?
98602. What do you mean by buoyancy?
98613. Why does an object float or sink
9862when placed on the surface of
9863water?
986410.6 Archimedes’ Principle rchimedes’ Principle
9865Activity _____________ 10.7
9866• Take a piece of stone and tie it to one
9867end of a rubber string or a spring
9868balance.
9869• Suspend the stone by holding the
9870balance or the string as shown in
9871Fig. 10.6 (a).
9872• Note the elongation of the string or the
9873reading on the spring balance due to
9874the weight of the stone.
9875• Now, slowly dip the stone in the water
9876in a container as shown in
9877Fig. 10.6 (b).
9878Fig. 10.6: (a) Observe the elongation of the rubber
9879string due to the weight of a piece of stone
9880suspended from it in air. (b) The elongation
9881decreases as the stone is immersed
9882in water.
9883(a)
9884(b)
9885Q
9886© NCERT
9887not to be republished
9888142 SCIENCE
9889contained in Archimedes’ principle, stated as
9890follows:
9891When a body is immersed fully or partially
9892in a fluid, it experiences an upward force that
9893is equal to the weight of the fluid displaced
9894by it.
9895Now, can you explain why a further
9896decrease in the elongation of the string was
9897not observed in activity 10.7, as the stone
9898was fully immersed in water?
9899Archimedes was a Greek
9900scientist. He discovered the
9901principle, subsequently
9902named after him, after
9903noticing that the water in a
9904bathtub overflowed when he
9905stepped into it. He ran
9906through the streets shouting
9907“Eureka!”, which means “I
9908have got it”. This knowledge helped him to
9909determine the purity of the gold in the crown
9910made for the king.
9911His work in the field of Geometry and
9912Mechanics made him famous. His
9913understanding of levers, pulleys, wheelsand-axle
9914helped the Greek army in its war
9915with Roman army.
9916Archimedes’ principle has many
9917applications. It is used in designing ships and
9918submarines. Lactometers, which are used to
9919determine the purity of a sample of milk and
9920hydrometers used for determining density of
9921liquids, are based on this principle.
9922uestions
99231. You find your mass to be 42 kg
9924on a weighing machine. Is your
9925mass more or less than 42 kg?
99262. You have a bag of cotton and an
9927iron bar, each indicating a mass
9928of 100 kg when measured on a
9929weighing machine. In reality, one
9930is heavier than other. Can you
9931say which one is heavier
9932and why?
993310.7 Relative Density Relative Density
9934As you know, the density of a substance is
9935defined as mass of a unit volume. The unit of
9936density is kilogram per metre cube (kg m–3).
9937The density of a given substance, under
9938specified conditions, remains the same.
9939Therefore the density of a substance is one
9940of its characteristic properties. It is different
9941for different substances. For example, the
9942density of gold is 19300 kg m-3 while that of
9943water is 1000 kg m-3. The density of a given
9944sample of a substance can help us to
9945determine its purity.
9946It is often convenient to express density
9947of a substance in comparison with that of
9948water. The relative density of a substance is
9949the ratio of its density to that of water:
9950Densityof a substance Relativedensity= Density of water
9951Since the relative density is a ratio of
9952similar quantities, it has no unit.
9953Example 10.7 Relative density of silver is
995410.8. The density of water is 103
9955 kg m–3.
9956What is the density of silver in SI unit?
9957Solution:
9958Relative density of silver = 10.8
9959Relative density
9960=
9961Density of silver
9962Density of water
9963Density of silver
9964 = Relative density of silver
9965× density of water
9966= 10.8 × 103
9967 kg m–3.
9968Archimedes
9969© NCERT
9970Q
9971not to be republished
9972GRAVITATION 143
9973What
9974you have you have
9975learnt
9976• The law of gravitation states that the force of attraction between
9977any two objects is proportional to the product of their masses
9978and inversely proportional to the square of the distance between
9979them. The law applies to objects anywhere in the universe.
9980Such a law is said to be universal.
9981• Gravitation is a weak force unless large masses are involved.
9982• Force of gravitation due to the earth is called gravity.
9983• The force of gravity decreases with altitude. It also varies on
9984the surface of the earth, decreasing from poles to the equator.
9985• The weight of a body is the force with which the earth
9986attracts it.
9987• The weight is equal to the product of mass and acceleration
9988due to gravity.
9989• The weight may vary from place to place but the mass stays
9990constant.
9991• All objects experience a force of buoyancy when they are
9992immersed in a fluid.
9993• Objects having density less than that of the liquid in which
9994they are immersed, float on the surface of the liquid. If the
9995density of the object is more than the density of the liquid in
9996which it is immersed then it sinks in the liquid.
9997Exercises xercises xercises
99981. How does the force of gravitation between two objects change
9999when the distance between them is reduced to half ?
100002. Gravitational force acts on all objects in proportion to their
10001masses. Why then, a heavy object does not fall faster than a
10002light object?
100033. What is the magnitude of the gravitational force between the
10004earth and a 1 kg object on its surface? (Mass of the earth is
100056 × 1024 kg and radius of the earth is 6.4 × 106
10006 m.)
100074. The earth and the moon are attracted to each other by
10008gravitational force. Does the earth attract the moon with a force
10009that is greater or smaller or the same as the force with which
10010the moon attracts the earth? Why?
100115. If the moon attracts the earth, why does the earth not move
10012towards the moon?
10013© NCERT
10014not to be republished
10015144 SCIENCE
100166. What happens to the force between two objects, if
10017(i) the mass of one object is doubled?
10018(ii) the distance between the objects is doubled and tripled?
10019(iii) the masses of both objects are doubled?
100207. What is the importance of universal law of gravitation?
100218. What is the acceleration of free fall?
100229. What do we call the gravitational force between the earth and
10023an object?
1002410. Amit buys few grams of gold at the poles as per the instruction
10025of one of his friends. He hands over the same when he meets
10026him at the equator. Will the friend agree with the weight of gold
10027bought? If not, why? [Hint: The value of g is greater at the
10028poles than at the equator.]
1002911. Why will a sheet of paper fall slower than one that is crumpled
10030into a ball?
1003112. Gravitational force on the surface of the moon is only
100321
100336 as
10034strong as gravitational force on the earth. What is the weight
10035in newtons of a 10 kg object on the moon and on the earth?
1003613. A ball is thrown vertically upwards with a velocity of 49 m/s.
10037Calculate
10038(i) the maximum height to which it rises,
10039(ii) the total time it takes to return to the surface of the earth.
1004014. A stone is released from the top of a tower of height 19.6 m.
10041Calculate its final velocity just before touching the ground.
1004215. A stone is thrown vertically upward with an initial velocity of
1004340 m/s. Taking g = 10 m/s2, find the maximum height reached
10044by the stone. What is the net displacement and the total
10045distance covered by the stone?
1004616. Calculate the force of gravitation between the earth and the
10047Sun, given that the mass of the earth = 6 × 1024 kg and of the
10048Sun = 2 × 1030 kg. The average distance between the two is
100491.5 × 1011 m.
1005017. A stone is allowed to fall from the top of a tower 100 m high
10051and at the same time another stone is projected vertically
10052upwards from the ground with a velocity of 25 m/s. Calculate
10053when and where the two stones will meet.
1005418. A ball thrown up vertically returns to the thrower after 6 s.
10055Find
10056(a) the velocity with which it was thrown up,
10057(b) the maximum height it reaches, and
10058(c) its position after 4 s.
10059© NCERT
10060not to be republished
10061GRAVITATION 145
1006219. In what direction does the buoyant force on an object immersed
10063in a liquid act?
1006420. Why does a block of plastic released under water come up to
10065the surface of water?
1006621. The volume of 50 g of a substance is 20 cm3. If the density of
10067water is 1 g cm–3, will the substance float or sink?
1006822. The volume of a 500 g sealed packet is 350 cm3
10069. Will the packet
10070float or sink in water if the density of water is 1 g cm–3? What
10071will be the mass of the water displaced by this packet?
10072© NCERT
10073not to be republished
10074In the previous few chapters we have talked
10075about ways of describing the motion of
10076objects, the cause of motion and gravitation.
10077Another concept that helps us understand and
10078interpret many natural phenomena is ‘work’.
10079Closely related to work are energy and power.
10080In this chapter we shall study these concepts.
10081All living beings need food. Living beings
10082have to perform several basic activities to
10083survive. We call such activities ‘life processes’.
10084The energy for these processes comes from
10085food. We need energy for other activities like
10086playing, singing, reading, writing, thinking,
10087jumping, cycling and running. Activities that
10088are strenuous require more energy.
10089Animals too get engaged in activities. For
10090example, they may jump and run. They have
10091to fight, move away from enemies, find food
10092or find a safe place to live. Also, we engage
10093some animals to lift weights, carry loads, pull
10094carts or plough fields. All such activities
10095require energy.
10096Think of machines. List the machines that
10097you have come across. What do they need for
10098their working? Why do some engines require
10099fuel like petrol and diesel? Why do living
10100beings and machines need energy?
1010111.1 Work
10102What is work? There is a difference in the
10103way we use the term ‘work’ in day-to-day life
10104and the way we use it in science. To make
10105this point clear let us consider a few examples.
1010611.1.1 NOT MUCH ‘WORK’ IN SPITE OF
10107WORKING HARD!
10108Kamali is preparing for examinations. She
10109spends lot of time in studies. She reads books,
10110draws diagrams, organises her thoughts,
10111collects question papers, attends classes,
10112discusses problems with her friends, and
10113performs experiments. She expends a lot of
10114energy on these activities. In common
10115parlance, she is ‘working hard’. All this ‘hard
10116work’ may involve very little ‘work’ if we go by
10117the scientific definition of work.
10118You are working hard to push a huge rock.
10119Let us say the rock does not move despite all
10120the effort. You get completely exhausted.
10121However, you have not done any work on the
10122rock as there is no displacement of the rock.
10123You stand still for a few minutes with a
10124heavy load on your head. You get tired. You
10125have exerted yourself and have spent quite a
10126bit of your energy. Are you doing work on the
10127load? The way we understand the term ‘work’
10128in science, work is not done.
10129You climb up the steps of a staircase and
10130reach the second floor of a building just to
10131see the landscape from there. You may even
10132climb up a tall tree. If we apply the scientific
10133definition, these activities involve a lot of work.
10134In day-to-day life, we consider any useful
10135physical or mental labour as work. Activities
10136like playing in a field, talking with friends,
10137humming a tune, watching a movie, attending
10138a function are sometimes not considered to
10139be work. What constitutes ‘work’ depends
10140on the way we define it. We use and define
10141the term work differently in science. To
10142understand this let us do the following
10143activities:
10144Activity _____________ 11.1
10145• We have discussed in the above
10146paragraphs a number of activities
10147which we normally consider to be work
1014811
10149WORK AND ENERGY
10150Chapter
10151© NCERT
10152not to be republished
10153Activity _____________ 11.3
10154• Think of situations when the object is
10155not displaced in spite of a force acting
10156on it.
10157• Also think of situations when an object
10158gets displaced in the absence of a force
10159acting on it.
10160• List all the situations that you can
10161think of for each.
10162• Discuss with your friends whether
10163work is done in these situations.
1016411.1.3 WORK DONE BY A CONSTANT FORCE
10165How is work defined in science? To
10166understand this, we shall first consider the
10167case when the force is acting in the direction
10168of displacement.
10169Let a constant force, F act on an object.
10170Let the object be displaced through a
10171distance, s in the direction of the force (Fig.
1017211.1). Let W be the work done. We define work
10173to be equal to the product of the force and
10174displacement.
10175Work done = force × displacement
10176W = F s (11.1)
10177in day-to-day life. For each of these
10178activities, ask the following questions
10179and answer them:
10180(i) What is the work being done on?
10181(ii) What is happening to the object?
10182(iii) Who (what) is doing the work?
1018311.1.2 SCIENTIFIC CONCEPTION OF WORK
10184To understand the way we view work and
10185define work from the point of view of science,
10186let us consider some situations:
10187Push a pebble lying on a surface. The
10188pebble moves through a distance. You exerted
10189a force on the pebble and the pebble got
10190displaced. In this situation work is done.
10191A girl pulls a trolley and the trolley moves
10192through a distance. The girl has exerted a
10193force on the trolley and it is displaced.
10194Therefore, work is done.
10195Lift a book through a height. To do this
10196you must apply a force. The book rises up.
10197There is a force applied on the book and the
10198book has moved. Hence, work is done.
10199A closer look at the above situations
10200reveals that two conditions need to be
10201satisfied for work to be done: (i) a force should
10202act on an object, and (ii) the object must be
10203displaced.
10204If any one of the above conditions does
10205not exist, work is not done. This is the way
10206we view work in science.
10207A bullock is pulling a cart. The cart
10208moves. There is a force on the cart and the
10209cart has moved. Do you think that work is
10210done in this situation?
10211Activity _____________ 11.2
10212• Think of some situations from your
10213daily life involving work.
10214• List them.
10215• Discuss with your friends whether
10216work is being done in each situation.
10217• Try to reason out your response.
10218• If work is done, which is the force acting
10219on the object?
10220• What is the object on which the work
10221is done?
10222• What happens to the object on which
10223work is done?
10224Fig. 11.1
10225Thus, work done by a force acting on an
10226object is equal to the magnitude of the force
10227multiplied by the distance moved in the
10228direction of the force. Work has only
10229magnitude and no direction.
10230In Eq. (11.1), if F = 1 N and s = 1 m then
10231the work done by the force will be 1 N m.
10232Here the unit of work is newton metre (N m)
10233or joule (J). Thus 1 J is the amount of work
10234WORK AND ENERGY 147
10235© NCERT
10236not to be republished
10237148 SCIENCE
10238done on an object when a force of 1 N
10239displaces it by 1 m along the line of action of
10240the force.
10241Look at Eq. (11.1) carefully. What is the
10242work done when the force on the object is
10243zero? What would be the work done when
10244the displacement of the object is zero? Refer
10245to the conditions that are to be satisfied to
10246say that work is done.
10247Example 11.1 A force of 5 N is acting on an
10248object. The object is displaced through
102492 m in the direction of the force (Fig.
1025011.2). If the force acts on the object all
10251through the displacement, then work
10252done is 5 N × 2 m =10 N m or 10 J.
10253Fig. 11.4
10254Consider a situation in which an object is
10255moving with a uniform velocity along a
10256particular direction. Now a retarding force, F,
10257is applied in the opposite direction. That is,
10258the angle between the two directions is 180º.
10259Let the object stop after a displacement s. In
10260such a situation, the work done by the force,
10261F is taken as negative and denoted by the
10262minus sign. The work done by the force is
10263F × (–s) or (–F × s).
10264It is clear from the above discussion that
10265the work done by a force can be either positive
10266or negative. To understand this, let us do the
10267following activity:
10268Activity _____________ 11.4
10269• Lift an object up. Work is done by the
10270force exerted by you on the object. The
10271object moves upwards. The force you
10272exerted is in the direction of
10273displacement. However, there is the
10274force of gravity acting on the object.
10275• Which one of these forces is doing
10276positive work?
10277• Which one is doing negative work?
10278• Give reasons.
10279Work done is negative when the force acts
10280opposite to the direction of displacement.
10281Work done is positive when the force is in the
10282direction of displacement.
10283Example 11.2 A porter lifts a luggage of 15
10284kg from the ground and puts it on his
10285head 1.5 m above the ground. Calculate
10286the work done by him on the luggage.
10287Solution:
10288Mass of luggage, m = 15 kg and
10289displacement, s = 1.5 m.
10290Fig. 11.2
10291uestion
102921. A force of 7 N acts on an object.
10293The displacement is, say 8 m, in
10294the direction of the force
10295(Fig. 11.3). Let us take it that the
10296force acts on the object through
10297the displacement. What is the
10298work done in this case?
10299Fig. 11.3
10300Consider another situation in which the
10301force and the displacement are in the same
10302direction: a baby pulling a toy car parallel to
10303the ground, as shown in Fig. 11.4. The baby
10304has exerted a force in the direction of
10305displacement of the car. In this situation, the
10306work done will be equal to the product of the
10307force and displacement. In such situations,
10308the work done by the force is taken as positive.
10309Q
10310© NCERT
10311not to be republished
10312WORK AND ENERGY 149
10313raised hammer falls on a nail placed on a
10314piece of wood, it drives the nail into the wood.
10315We have also observed children winding a
10316toy (such as a toy car) and when the toy is
10317placed on the floor, it starts moving. When a
10318balloon is filled with air and we press it we
10319notice a change in its shape. As long as we
10320press it gently, it can come back to its original
10321shape when the force is withdrawn. However,
10322if we press the balloon hard, it can even
10323explode producing a blasting sound. In all
10324these examples, the objects acquire, through
10325different means, the capability of doing work.
10326An object having a capability to do work is
10327said to possess energy. The object which does
10328the work loses energy and the object on which
10329the work is done gains energy.
10330How does an object with energy do work?
10331An object that possesses energy can exert a
10332force on another object. When this happens,
10333energy is transferred from the former to the
10334latter. The second object may move as it
10335receives energy and therefore do some work.
10336Thus, the first object had a capacity to do
10337work. This implies that any object that
10338possesses energy can do work.
10339The energy possessed by an object is thus
10340measured in terms of its capacity of doing
10341work. The unit of energy is, therefore, the same
10342as that of work, that is, joule (J). 1 J is the
10343energy required to do 1 joule of work.
10344Sometimes a larger unit of energy called kilo
10345joule (kJ) is used. 1 kJ equals 1000 J.
1034611.2.1FORMS OF ENERGY
10347Luckily the world we live in provides energy
10348in many different forms. The various forms
10349include mechanical energy (potential energy
10350+ kinetic energy), heat energy, chemical
10351energy, electrical energy and light energy.
10352Think it over !
10353How do you know that some entity is a
10354form of energy? Discuss with your friends
10355and teachers.
10356Work done, W = F × s = mg × s
10357= 15 kg × 10 m s-2 × 1.5 m
10358= 225 kg m s-2 m
10359= 225 N m = 225 J
10360Work done is 225 J.
10361uestions
103621. When do we say that work is
10363done?
103642. Write an expression for the work
10365done when a force is acting on
10366an object in the direction of its
10367displacement.
103683. Define 1 J of work.
103694. A pair of bullocks exerts a force
10370of 140 N on a plough. The field
10371being ploughed is 15 m long.
10372How much work is done in
10373ploughing the length of the field?
1037411.2 Energy
10375Life is impossible without energy. The demand
10376for energy is ever increasing. Where do we
10377get energy from? The Sun is the biggest
10378natural source of energy to us. Many of our
10379energy sources are derived from the Sun. We
10380can also get energy from the nuclei of atoms,
10381the interior of the earth, and the tides. Can
10382you think of other sources of energy?
10383Activity _____________ 11.5
10384• A few sources of energy are listed above.
10385There are many other sources of
10386energy. List them.
10387• Discuss in small groups how certain
10388sources of energy are due to the Sun.
10389• Are there sources of energy which are
10390not due to the Sun?
10391The word energy is very often used in our
10392daily life, but in science we give it a definite
10393and precise meaning. Let us consider the
10394following examples: when a fast moving
10395cricket ball hits a stationary wicket, the wicket
10396is thrown away. Similarly, an object when
10397raised to a certain height gets the capability
10398to do work. You must have seen that when a
10399Q
10400© NCERT
10401not to be republished
10402150 SCIENCE
10403Fig. 11.5
10404• The trolley moves forward and hits the
10405wooden block.
10406• Fix a stop on the table in such a
10407manner that the trolley stops after
10408hitting the block. The block gets
10409displaced.
10410• Note down the displacement of the
10411block. This means work is done on the
10412block by the trolley as the block has
10413gained energy.
10414• From where does this energy come?
10415• Repeat this activity by increasing the
10416mass on the pan. In which case is the
10417displacement more?
10418• In which case is the work done more?
10419• In this activity, the moving trolley does
10420work and hence it possesses energy.
10421A moving object can do work. An object
10422moving faster can do more work than an
10423identical object moving relatively slow. A
10424moving bullet, blowing wind, a rotating wheel,
10425a speeding stone can do work. How does a
10426bullet pierce the target? How does the wind
10427move the blades of a windmill? Objects in
10428motion possess energy. We call this energy
10429kinetic energy.
10430A falling coconut, a speeding car, a rolling
10431stone, a flying aircraft, flowing water, blowing
10432wind, a running athlete etc. possess kinetic
10433energy. In short, kinetic energy is the energy
10434possessed by an object due to its motion. The
10435kinetic energy of an object increases with its
10436speed.
10437How much energy is possessed by a
10438moving body by virtue of its motion? By
10439definition, we say that the kinetic energy of a
10440body moving with a certain velocity is equal to
10441the work done on it to make it acquire that
10442velocity.
1044311.2.2 KINETIC ENERGY
10444Activity _____________11.6
10445• Take a heavy ball. Drop it on a thick
10446bed of sand. A wet bed of sand would
10447be better. Drop the ball on the sand
10448bed from height of about 25 cm. The
10449ball creates a depression.
10450• Repeat this activity from heights of
1045150 cm, 1m and 1.5 m.
10452• Ensure that all the depressions are
10453distinctly visible.
10454• Mark the depressions to indicate the
10455height from which the ball was
10456dropped.
10457• Compare their depths.
10458• Which one of them is deepest?
10459• Which one is shallowest? Why?
10460• What has caused the ball to make a
10461deeper dent?
10462• Discuss and analyse.
10463Activity _____________ 11.7
10464• Set up the apparatus as shown in
10465Fig. 11.5.
10466• Place a wooden block of known mass
10467in front of the trolley at a convenient
10468fixed distance.
10469• Place a known mass on the pan so
10470that the trolley starts moving.
10471James Prescott
10472Joule was an
10473outstanding
10474British physicist.
10475He is best known
10476for his research in
10477electricity and
10478thermodynamics.
10479Amongst other
10480things, he
10481formulated a law
10482for the heating
10483effect of electric
10484current. He also
10485verified experimentally the law of
10486conservation of energy and discovered
10487the value of the mechanical equivalent
10488of heat. The unit of energy and work
10489called joule, is named after him.
10490James Prescott Joule
10491(1818 – 1889)
10492© NCERT
10493not to be republished
10494WORK AND ENERGY 151
10495Let us now express the kinetic energy of
10496an object in the form of an equation. Consider
10497an object of mass, m moving with a uniform
10498velocity, u. Let it now be displaced through a
10499distance s when a constant force, F acts on it
10500in the direction of its displacement. From
10501Eq. (11.1), the work done, W is F s. The work
10502done on the object will cause a change in its
10503velocity. Let its velocity change from u to v.
10504Let a be the acceleration produced.
10505In section 8.5, we studied three equations
10506of motion. The relation connecting the initial
10507velocity (u) and final velocity (v) of an object
10508moving with a uniform acceleration a, and
10509the displacement, s is
10510v2 – u2 = 2a s (8.7)
10511This gives
105122 2 v –u
10513s =
105142a (11.2)
10515From section 9.4, we know F = m a. Thus,
10516using (Eq. 11.2) in Eq. (11.1), we can write
10517the work done by the force, F as
10518
10519
10520
105212 2 v -u
10522W =m a
105232a
10524or
10525 1
105262
105272 2 W = m v –u (11.3)
10528If the object is starting from its stationary
10529position, that is, u = 0, then
105301
105312
105322 W = m v (11.4)
10533It is clear that the work done is equal to the
10534change in the kinetic energy of an object.
10535If u = 0, the work done will be
105361
105372
105382 m v .
10539Thus, the kinetic energy possessed by an
10540object of mass, m and moving with a uniform
10541velocity, v is
105421
105432
105442 E = m v k (11.5)
10545Example 11.3 An object of mass 15 kg is
10546moving with a uniform velocity of 4
10547m s–1 . What is the kinetic energy
10548possessed by the object?
10549Solution:
10550Mass of the object, m = 15 kg, velocity
10551of the object, v = 4 m s–1.
10552From Eq. (11.5),
105531
105542
105552 E = m v k
10556= 1
105572 × 15 kg × 4 m s–1 × 4 m s–1
10558= 120 J
10559The kinetic energy of the object is 120 J.
10560Example 11.4 What is the work to be done
10561to increase the velocity of a car from
1056230 km h–1 to 60 km h–1 if the mass of
10563the car is 1500 kg?
10564Solution:
10565Mass of the car, m =1500 kg,
10566initial velocity of car, u = 30 km h–1
10567= 30 ×1000 m
1056860 × 60s
10569= 8.33 m s–1.
10570Similarly, the final velocity of the car,
10571v = 60 km h–1
10572= 16.67 m s–1.
10573Therefore, the initial kinetic energy of
10574the car,
10575Eki
105761
105772
105782 = m u
10579= 1
105802 × 1500 kg × (8.33 m s–1)2
10581= 52041.68 J.
10582The final kinetic energy of the car,
10583Ekf = 1
105842 × 1500 kg × (16.67 m s–1)
105852
10586= 208416.68 J.
10587Thus, the work done = Change in
10588kinetic energy
10589= Ekf – Eki
10590= 156375 J.
10591© NCERT
10592not to be republished
10593152 SCIENCE
10594uestions
105951. What is the kinetic energy of an
10596object?
105972. Write an expression for the kinetic
10598energy of an object.
105993. The kinetic energy of an object of
10600mass, m moving with a velocity
10601of 5 m s–1 is 25 J. What will be its
10602kinetic energy when its velocity
10603is doubled? What will be its
10604kinetic energy when its velocity
10605is increased three times?
1060611.2.3 POTENTIAL ENERGY
10607Activity _____________ 11.8
10608• Take a rubber band.
10609• Hold it at one end and pull from the
10610other. The band stretches.
10611• Release the band at one of the ends.
10612• What happens?
10613• The band will tend to regain its original
10614length. Obviously the band had
10615acquired energy in its stretched
10616position.
10617• How did it acquire energy when
10618stretched?
10619Activity _____________11.9
10620• Take a slinky as shown below.
10621• Ask a friend to hold one of its ends.
10622You hold the other end and move away
10623from your friend. Now you release the
10624slinky.
10625Activity ___________ 11.11
10626• Lift an object through a certain
10627height. The object can now do work.
10628It begins to fall when released.
10629• This implies that it has acquired some
10630energy. If raised to a greater height it
10631can do more work and hence possesses
10632more energy.
10633• From where did it get the energy? Think
10634and discuss.
10635In the above situations, the energy gets
10636stored due to the work done on the object.
10637The energy transferred to an object is stored
10638as potential energy if it is not used to cause a
10639change in the velocity or speed of the object.
10640You transfer energy when you stretch a
10641rubber band. The energy transferred to the
10642band is its potential energy. You do work while
10643winding the key of a toy car. The energy
10644transferred to the spring inside is stored as
10645potential energy. The potential energy
10646possessed by the object is the energy present
10647in it by virtue of its position or configuration.
10648Activity ____________11.12
10649• Take a bamboo stick and make a bow
10650as shown in Fig. 11.6.
10651• Place an arrow made of a light stick on
10652it with one end supported by the
10653stretched string.
10654• Now stretch the string and release the
10655arrow.
10656• Notice the arrow flying off the bow.
10657Notice the change in the shape of the
10658bow.
10659• The potential energy stored in the bow
10660due to the change of shape is thus used
10661in the form of kinetic energy in
10662throwing off the arrow.
10663Q
10664Fig.11.6: An arrow and the stretched string
10665on the bow.
10666• What happened?
10667• How did the slinky acquire energy when
10668stretched?
10669• Would the slinky acquire energy when
10670it is compressed?
10671Activity ____________11.10
10672• Take a toy car. Wind it using its key.
10673• Place the car on the ground.
10674• Did it move?
10675• From where did it acquire energy?
10676• Does the energy acquired depend on
10677the number of windings?
10678• How can you test this?
10679© NCERT
10680not to be republished
10681WORK AND ENERGY 153
1068211.2.4 POTENTIAL ENERGY OF AN OBJECT
10683AT A HEIGHT
10684An object increases its energy when raised
10685through a height. This is because work is
10686done on it against gravity while it is being
10687raised. The energy present in such an object
10688is the gravitational potential energy.
10689The gravitational potential energy of an
10690object at a point above the ground is defined
10691as the work done in raising it from the ground
10692to that point against gravity.
10693It is easy to arrive at an expression for
10694the gravitational potential energy of an object
10695at a height.
10696Fig. 11.7
10697Consider an object of mass, m. Let it be
10698raised through a height, h from the ground.
10699A force is required to do this. The minimum
10700force required to raise the object is equal to
10701the weight of the object, mg. The object gains
10702energy equal to the work done on it. Let the
10703work done on the object against gravity be
10704W. That is,
10705work done, W = force × displacement
10706= mg × h
10707= mgh
10708Since work done on the object is equal to
10709mgh, an energy equal to mgh units is gained
10710by the object. This is the potential energy (EP)
10711of the object.
10712Ep
10713 = mgh (11.7)
10714More to know
10715The potential energy of an object at
10716a height depends on the ground level
10717or the zero level you choose. An
10718object in a given position can have a
10719certain potential energy with respect
10720to one level and a different value of
10721potential energy with respect to
10722another level.
10723It is useful to note that the work done by
10724gravity depends on the difference in vertical
10725heights of the initial and final positions of
10726the object and not on the path along which
10727the object is moved. Fig. 11.8 shows a case
10728where a block is raised from position A to B
10729by taking two different paths. Let the height
10730AB = h. In both the situations the work done
10731on the object is mgh.
10732Fig. 11.8
10733Example 11.5 Find the energy possessed
10734by an object of mass 10 kg when it is at
10735a height of 6 m above the ground. Given,
10736g = 9.8 m s–2.
10737Solution:
10738Mass of the object, m = 10 kg,
10739displacement (height), h = 6 m, and
10740acceleration due to gravity, g = 9.8 m s–2.
10741From Eq. (11.6),
10742Potential energy = mgh
10743= 10 kg × 9.8 m s–2 × 6 m
10744= 588 J.
10745The potential energy is 588 J.
10746© NCERT
10747not to be republished
10748154 SCIENCE
10749Example 11.6 An object of mass 12 kg is
10750at a certain height above the ground.
10751If the potential energy of the object is
10752480 J, find the height at which the
10753object is with respect to the ground.
10754Given, g = 10 m s–2.
10755Solution:
10756Mass of the object, m = 12 kg,
10757potential energy, Ep
10758 = 480 J.
10759 Ep = mgh
10760480 J = 12 kg × 10 m s–2 × h
10761 h = –2
10762480 J
10763120 kg m s = 4 m.
10764The object is at the height of 4 m.
1076511.2.5 ARE VARIOUS ENERGY FORMS
10766INTERCONVERTIBLE?
10767Can we convert energy from one form to
10768another? We find in nature a number of
10769instances of conversion of energy from one
10770form to another.
10771Activity ____________11.13
10772• Sit in small groups.
10773• Discuss the various ways of energy
10774conversion in nature.
10775• Discuss following questions in your
10776group:
10777(a) How do green plants produce food?
10778(b) Where do they get their energy from?
10779(c) Why does the air move from place
10780to place?
10781(d) How are fuels, such as coal and
10782petroleum formed?
10783(e) What kinds of energy conversions
10784sustain the water cycle?
10785Activity ___________ 11.14
10786• Many of the human activities and the
10787gadgets we use involve conversion of
10788energy from one form to another.
10789• Make a list of such activities and
10790gadgets.
10791• Identify in each activity/gadget the
10792kind of energy conversion that takes
10793place.
1079411.2.6 LAW OF CONSERVATION OF ENERGY
10795In activities 11.13 and 11.14, we learnt that
10796the form of energy can be changed from one
10797form to another. What happens to the total
10798energy of a system during or after the
10799process? Whenever energy gets transformed,
10800the total energy remains unchanged. This is
10801the law of conservation of energy. According
10802to this law, energy can only be converted from
10803one form to another; it can neither be created
10804or destroyed. The total energy before and after
10805the transformation remains the same. The
10806law of conservation of energy is valid
10807in all situations and for all kinds of
10808transformations.
10809Consider a simple example. Let an object
10810of mass, m be made to fall freely from a
10811height, h. At the start, the potential energy is
10812mgh and kinetic energy is zero. Why is the
10813kinetic energy zero? It is zero because its
10814velocity is zero. The total energy of the object
10815is thus mgh. As it falls, its potential energy
10816will change into kinetic energy. If v is the
10817velocity of the object at a given instant, the
10818kinetic energy would be ½mv2. As the fall of
10819the object continues, the potential energy
10820would decrease while the kinetic energy would
10821increase. When the object is about to reach
10822the ground, h = 0 and v will be the highest.
10823Therefore, the kinetic energy would be the
10824largest and potential energy the least.
10825However, the sum of the potential energy and
10826kinetic energy of the object would be the same
10827at all points. That is,
10828potential energy + kinetic energy = constant
10829 or
108301
10831constant.
108322
108332 mgh + mv = (11.7)
10834The sum of kinetic energy and potential energy
10835of an object is its total mechanical energy.
10836We find that during the free fall of the object,
10837the decrease in potential energy, at any point
10838in its path, appears as an equal amount of
10839increase in kinetic energy. (Here the effect of
10840air resistance on the motion of the object has
10841been ignored.) There is thus a continual
10842transformation of gravitational potential
10843energy into kinetic energy.
10844© NCERT
10845not to be republished
10846WORK AND ENERGY 155
10847A stronger person may do certain work in
10848relatively less time. A more powerful vehicle
10849would complete a journey in a shorter time
10850than a less powerful one. We talk of the power
10851of machines like motorbikes and motorcars.
10852The speed with which these vehicles change
10853energy or do work is a basis for their
10854classification. Power measures the speed of
10855work done, that is, how fast or slow work is
10856done. Power is defined as the rate of doing
10857work or the rate of transfer of energy. If an
10858agent does a work W in time t, then power is
10859given by:
10860Power = work/time
10861or
10862W
10863P =
10864t (11.8)
10865The unit of power is watt [in honour of
10866James Watt (1736 – 1819)] having the symbol
10867W. 1 watt is the power of an agent, which
10868does work at the rate of 1 joule per second.
10869We can also say that power is 1 W when the
10870rate of consumption of energy is 1 J s–1.
108711 watt = 1 joule/second or 1 W = 1 J s–1.
10872We express larger rates of energy transfer in
10873kilowatts (kW).
108741 kilowatt = 1000 watts
108751 kW = 1000 W
108761 kW = 1000 J s–1.
10877The power of an agent may vary with time.
10878This means that the agent may be doing work
10879at different rates at different intervals of time.
10880Therefore the concept of average power is
10881useful. We obtain average power by dividing
10882the total energy consumed by the total time
10883taken.
10884Example 11.7 Two girls, each of weight 400
10885N climb up a rope through a height of 8
10886m. We name one of the girls A and the
10887other B. Girl A takes 20 s while B takes
1088850 s to accomplish this task. What is the
10889power expended by each girl?
10890Solution:
10891(i) Power expended by girl A:
10892Weight of the girl, mg = 400 N
10893Displacement (height), h = 8 m
10894Activity ___________ 11.15
10895• An object of mass 20 kg is dropped
10896from a height of 4 m. Fill in the blanks
10897in the following table by computing
10898the potential energy and kinetic
10899energy in each case.
10900Height at Potential Kinetic Ep + Ek
10901which object energy energy
10902is located (Ep
10903= mgh) (Ek = mv2/2)
10904m J JJ
109054
109063
109072
109081
10909Just above
10910the ground
10911• For simplifying the calculations, take
10912the value of g as 10 m s–2.
10913Think it over !
10914What would have happened if nature had
10915not allowed the transformation of energy?
10916There is a view that life could not have
10917been possible without transformation of
10918energy. Do you agree with this?
1091911.3 Rate of Doing Work
10920Do all of us work at the same rate? Do
10921machines consume or transfer energy at the
10922same rate? Agents that transfer energy do
10923work at different rates. Let us understand this
10924from the following activity:
10925Activity ___________ 11.16
10926• Consider two children, say A and B.
10927Let us say they weigh the same. Both
10928start climbing up a rope separately.
10929Both reach a height of 8 m. Let us say
10930A takes 15 s while B takes 20 s to
10931accomplish the task.
10932• What is the work done by each?
10933• The work done is the same. However,
10934A has taken less time than B to do
10935the work.
10936• Who has done more work in a given
10937time, say in 1 s?
10938© NCERT
10939not to be republished
10940156 SCIENCE
10941uestions
109421. What is power?
109432. Define 1 watt of power.
109443. A lamp consumes 1000 J of
10945electrical energy in 10 s. What is
10946its power?
109474. Define average power.
1094811.3.1 COMMERCIAL UNIT OF ENERGY
10949The unit joule is too small and hence is
10950inconvenient to express large quantities of
10951energy. We use a bigger unit of energy called
10952kilowatt hour (kW h).
10953What is 1 kW h? Let us say we have a
10954machine that uses 1000 J of energy every
10955second. If this machine is used continuously
10956for one hour, it will consume 1 kW h of energy.
10957Thus, 1 kW h is the energy used in one hour
10958at the rate of 1000 J s–1 (or 1 kW).
109591 kW h = 1 kW ×1 h
10960= 1000 W × 3600 s
10961= 3600000 J
109621 kW h = 3.6 × 106 J.
10963The energy used in households, industries
10964and commercial establishments are usually
10965expressed in kilowatt hour. For example,
10966electrical energy used during a month is
10967expressed in terms of ‘units’. Here, 1 ‘unit’
10968means 1 kilowatt hour.
10969Example 11.9 An electric bulb of 60 W is
10970used for 6 h per day. Calculate the ‘units’
10971of energy consumed in one day by the
10972bulb.
10973Solution:
10974Power of electric bulb = 60 W
10975= 0.06 kW.
10976Time used, t = 6 h
10977Energy = power × time taken
10978= 0.06 kW × 6 h
10979= 0.36 kW h
10980= 0.36 ‘units’.
10981The energy consumed by the bulb is
109820.36 ‘units’.
10983Time taken, t = 20 s
10984From Eq. (11.8),
10985Power, P = Work done/time taken
10986= mgh
10987t
10988=
10989400 N × 8 m
1099020s
10991= 160 W.
10992(ii) Power expended by girl B:
10993Weight of the girl, mg = 400 N
10994Displacement (height), h = 8 m
10995Time taken, t = 50 s
10996Power, P = mgh
10997t
10998=
10999400 N × 8 m
1100050 s
11001= 64 W.
11002Power expended by girl A is 160 W.
11003Power expended by girl B is 64 W.
11004Example 11.8 A boy of mass 50 kg runs
11005up a staircase of 45 steps in 9 s. If the
11006height of each step is 15 cm, find his
11007power. Take g = 10 m s–2.
11008Solution:
11009Weight of the boy,
11010mg = 50 kg × 10 m s–2 = 500 N
11011Height of the staircase,
11012h = 45 × 15/100 m = 6.75 m
11013Time taken to climb, t = 9 s
11014From Eq. (11.8),
11015power, P = Work done/time taken
11016= mgh
11017t
11018= 500 N × 6 .75 m
110199s
11020= 375 W.
11021Power is 375 W.
11022Q
11023© NCERT
11024not to be republished
11025WORK AND ENERGY 157
11026Activity ___________ 11.17
11027• Take a close look at the electric meter
11028installed in your house. Observe its
11029features closely.
11030• Take the readings of the meter each
11031day at 6.30 am and 6.30 pm.
11032• How many ‘units’ are consumed
11033during day time?
11034• How many ‘units’ are used during
11035night?
11036• Do this activity for about a week.
11037• Tabulate your observations.
11038• Draw inferences from the data.
11039• Compare your observations with
11040the details given in the monthly
11041electricity bill.
11042What
11043you have you have
11044learnt
11045• Work done on an object is defined as the magnitude of the
11046force multiplied by the distance moved by the object in the
11047direction of the applied force. The unit of work is joule:
110481 joule = 1 newton × 1 metre.
11049• Work done on an object by a force would be zero if the
11050displacement of the object is zero.
11051• An object having capability to do work is said to possess energy.
11052Energy has the same unit as that of work.
11053• An object in motion possesses what is known as the kinetic
11054energy of the object. An object of mass, m moving with velocity
11055v has a kinetic energy of 2 mv
110561
110572 .
11058• The energy possessed by a body due to its change in position
11059or shape is called the potential energy. The gravitational
11060potential energy of an object of mass, m raised through a height,
11061h from the earth’s surface is given by m g h.
11062• According to the law of conservation of energy, energy can
11063only be transformed from one form to another; it can neither
11064be created nor destroyed. The total energy before and after
11065the transformation always remains constant.
11066• Energy exists in nature in several forms such as kinetic
11067energy, potential energy, heat energy, chemical energy etc.
11068The sum of the kinetic and potential energies of an object is
11069called its mechanical energy.
11070• Power is defined as the rate of doing work. The SI unit of
11071power is watt. 1 W = 1 J/s.
11072• The energy used in one hour at the rate of 1kW is called 1 kW
11073h.
11074© NCERT
11075not to be republished
11076158 SCIENCE
11077Exercises Exercises Exercises
110781. Look at the activities listed below. Reason out whether or not
11079work is done in the light of your understanding of the term
11080‘work’.
11081• Suma is swimming in a pond.
11082• A donkey is carrying a load on its back.
11083• A wind-mill is lifting water from a well.
11084• A green plant is carrying out photosynthesis.
11085• An engine is pulling a train.
11086• Food grains are getting dried in the sun.
11087• A sailboat is moving due to wind energy.
110882. An object thrown at a certain angle to the ground moves in a
11089curved path and falls back to the ground. The initial and the
11090final points of the path of the object lie on the same horizontal
11091line. What is the work done by the force of gravity on the object?
110923. A battery lights a bulb. Describe the energy changes involved
11093in the process.
110944. Certain force acting on a 20 kg mass changes its velocity from
110955 m s–1 to 2 m s–1. Calculate the work done by the force.
110965. A mass of 10 kg is at a point A on a table. It is moved to a point
11097B. If the line joining A and B is horizontal, what is the work
11098done on the object by the gravitational force? Explain your
11099answer.
111006. The potential energy of a freely falling object decreases
11101progressively. Does this violate the law of conservation of
11102energy? Why?
111037. What are the various energy transformations that occur when
11104you are riding a bicycle?
111058. Does the transfer of energy take place when you push a
11106huge rock with all your might and fail to move it? Where is
11107the energy you spend going?
111089. A certain household has consumed 250 units of energy during
11109a month. How much energy is this in joules?
1111010. An object of mass 40 kg is raised to a height of 5 m above the
11111ground. What is its potential energy? If the object is allowed to
11112fall, find its kinetic energy when it is half-way down.
1111311. What is the work done by the force of gravity on a satellite
11114moving round the earth? Justify your answer.
1111512. Can there be displacement of an object in the absence of any
11116force acting on it? Think. Discuss this question with your
11117friends and teacher.
11118© NCERT
11119not to be republished
11120WORK AND ENERGY 159
1112113. A person holds a bundle of hay over his head for 30 minutes
11122and gets tired. Has he done some work or not? Justify your
11123answer.
1112414. An electric heater is rated 1500 W. How much energy does it
11125use in 10 hours?
1112615. Illustrate the law of conservation of energy by discussing the
11127energy changes which occur when we draw a pendulum
11128bob to one side and allow it to oscillate. Why does the bob
11129eventually come to rest? What happens to its energy
11130eventually? Is it a violation of the law of conservation of
11131energy?
1113216. An object of mass, m is moving with a constant velocity, v.
11133How much work should be done on the object in order to bring
11134the object to rest?
1113517. Calculate the work required to be done to stop a car of 1500 kg
11136moving at a velocity of 60 km/h?
1113718. In each of the following a force, F is acting on an object of
11138mass, m. The direction of displacement is from west to east
11139shown by the longer arrow. Observe the diagrams carefully
11140and state whether the work done by the force is negative,
11141positive or zero.
1114219. Soni says that the acceleration in an object could be zero
11143even when several forces are acting on it. Do you agree with
11144her? Why?
1114520. Find the energy in kW h consumed in 10 hours by four
11146devices of power 500 W each.
1114721. A freely falling object eventually stops on reaching the
11148ground. What happenes to its kinetic energy?
11149© NCERT
11150not to be republished
11151Everyday we hear sounds from various
11152sources like humans, birds, bells, machines,
11153vehicles, televisions, radios etc. Sound is a
11154form of energy which produces a sensation
11155of hearing in our ears. There are also other
11156forms of energy like mechanical energy, heat
11157energy, light energy etc. We have talked about
11158mechanical energy in the previous chapters.
11159You have been taught about conservation of
11160energy, which states that we can neither
11161create nor destroy energy. We can just
11162change it from one form to another. When
11163you clap, a sound is produced. Can you
11164produce sound without utilising your energy?
11165Which form of energy did you use to produce
11166sound? In this chapter we are going to learn
11167how sound is produced and how it is
11168transmitted through a medium and received
11169by our ear.
1117012.1 Production of Sound
11171Activity _____________ 12.1
11172• Take a tuning fork and set it vibrating
11173by striking its prong on a rubber pad.
11174Bring it near your ear.
11175• Do you hear any sound?
11176• Touch one of the prongs of the vibrating
11177tuning fork with your finger and share
11178your experience with your friends.
11179• Now, suspend a table tennis ball or a
11180small plastic ball by a thread from a
11181support [Take a big needle and a
11182thread, put a knot at one end of the
11183thread, and then with the help of the
11184needle pass the thread through the
11185ball]. Touch the ball gently with the
11186prong of a vibrating tuning
11187fork (Fig. 12.1).
11188• Observe what happens and discuss
11189with your friends.
11190Activity _____________ 12.2
11191• Fill water in a beaker or a glass up to
11192the brim. Gently touch the water surface
11193with one of the prongs of the vibrating
11194tuning fork, as shown in Fig. 12.2.
11195• Next dip the prongs of the vibrating
11196tuning fork in water, as shown in Fig.
1119712.3.
11198• Observe what happens in both the
11199cases.
11200• Discuss with your friends why this
11201happens.
11202Fig. 12.1: Vibrating tuning fork just touching the
11203suspended table tennis ball.
11204Fig. 12.2: One of the prongs of the vibrating tuning
11205fork touching the water surface.
1120612
11207SOUND
11208Chapter
11209© NCERT
11210not to be republished
11211plucked vibrates and produces sound. If you
11212have never done this, then do it and observe
11213the vibration of the stretched rubber band.
11214Activity _____________ 12.3
11215• Make a list of different types of musical
11216instruments and discuss with your
11217friends which part of the instrument
11218vibrates to produce sound.
1121912.2 Propagation of Sound
11220Sound is produced by vibrating objects. The
11221matter or substance through which sound
11222is transmitted is called a medium. It can be
11223solid, liquid or gas. Sound moves through a
11224medium from the point of generation to the
11225listener. When an object vibrates, it sets the
11226particles of the medium around it vibrating.
11227The particles do not travel all the way from
11228the vibrating object to the ear. A particle of
11229the medium in contact with the vibrating
11230object is first displaced from its equilibrium
11231position. It then exerts a force on the adjacent
11232particle. As a result of which the adjacent
11233particle gets displaced from its position of
11234rest. After displacing the adjacent particle the
11235first particle comes back to its original
11236position. This process continues in the
11237medium till the sound reaches your ear. The
11238disturbance created by a source of sound in
11239Fig. 12.3: Both the prongs of the vibrating tuning
11240fork dipped in water.
11241From the above activities what do you
11242conclude? Can you produce sound without
11243a vibrating object?
11244In the above activities we have produced
11245sound by striking the tuning fork. We can
11246also produce sound by plucking, scratching,
11247rubbing, blowing or shaking different objects.
11248As per the above activities what do we do to
11249the objects? We set the objects vibrating and
11250produce sound. Vibration means a kind of
11251rapid to and fro motion of an object. The
11252sound of the human voice is produced due
11253to vibrations in the vocal cords. When a bird
11254flaps its wings, do you hear any sound? Think
11255how the buzzing sound accompanying a bee
11256is produced. A stretched rubber band when
11257Fig. 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.
11258Can sound make a light spot dance?
11259Take a tin can. Remove both ends to make it a hollow cylinder. Take a balloon and stretch
11260it over the can, then wrap a rubber band around the balloon. Take a small piece of mirror.
11261Use a drop of glue to stick the piece of mirror to the balloon. Allow the light through a slit
11262to fall on the mirror. After reflection the light spot is seen on the wall, as shown in Fig.
1126312.4. Talk or shout directly into the open end of the can and observe the dancing light spot
11264on the wall. Discuss with your friends what makes the light spot dance.
11265SOUND 161
11266© NCERT
11267not to be republished
11268162 SCIENCE
11269the medium travels through the medium and
11270not the particles of the medium.
11271A wave is a disturbance that moves
11272through a medium when the particles of the
11273medium set neighbouring particles into
11274motion. They in turn produce similar motion
11275in others. The particles of the medium do not
11276move forward themselves, but the
11277disturbance is carried forward. This is what
11278happens during propagation of sound in a
11279medium, hence sound can be visualised as a
11280wave. Sound waves are characterised by the
11281motion of particles in the medium and are
11282called mechanical waves.
11283Air is the most common medium through
11284which sound travels. When a vibrating object
11285moves forward, it pushes and compresses the
11286air in front of it creating a region of high
11287pressure. This region is called a compression
11288(C), as shown in Fig. 12.5. This compression
11289starts to move away from the vibrating object.
11290When the vibrating object moves backwards,
11291it creates a region of low pressure called
11292rarefaction (R), as shown in Fig. 12.5. As the
11293object moves back and forth rapidly, a series
11294of compressions and rarefactions is created
11295in the air. These make the sound wave that
11296propagates through the medium.
11297Compression is the region of high pressure
11298and rarefaction is the region of low pressure.
11299Pressure is related to the number of particles
11300of a medium in a given volume. More density
11301of the particles in the medium gives more
11302pressure and vice versa. Thus, propagation
11303of sound can be visualised as propagation of
11304density variations or pressure variations in
11305the medium.
11306uestion
113071. How does the sound produced by
11308a vibrating object in a medium
11309reach your ear?
1131012.2.1 SOUND NEEDS A MEDIUM TO TRAVEL
11311Sound is a mechanical wave and needs a
11312material medium like air, water, steel etc. for
11313its propagation. It cannot travel through
11314vacuum, which can be demonstrated by the
11315following experiment.
11316Take an electric bell and an airtight glass
11317bell jar. The electric bell is suspended inside
11318the airtight bell jar. The bell jar is connected
11319to a vacuum pump, as shown in Fig. 12.6. If
11320you press the switch you will be able to hear
11321the bell. Now start the vacuum pump. When
11322the air in the jar is pumped out gradually,
11323the sound becomes fainter, although the
11324same current is passing through the bell.
11325After some time when less air is left inside
11326the bell jar you will hear a very feeble sound.
11327What will happen if the air is removed
11328completely? Will you still be able to hear the
11329sound of the bell?
11330Fig. 12.5: A vibrating object creating a series of
11331compressions (C) and rarefactions (R) in
11332the medium.
11333Q
11334Fig. 12.6: Bell jar experiment showing sound cannot
11335travel in vacuum.
11336© NCERT
11337not to be republished
11338SOUND 163
11339waves. In these waves the individual particles
11340of the medium move in a direction parallel to
11341the direction of propagation of the
11342disturbance. The particles do not move from
11343one place to another but they simply oscillate
11344back and forth about their position of rest.
11345This is exactly how a sound wave propagates,
11346hence sound waves are longitudinal waves.
11347There is also another type of wave, called
11348a transverse wave. In a transverse wave
11349particles do not oscillate along the line of
11350wave propagation but oscillate up and down
11351about their mean position as the wave travels.
11352Thus a transverse wave is the one in which
11353the individual particles of the medium move
11354about their mean positions in a direction
11355perpendicular to the direction of wave
11356propagation. Light is a transverse wave but
11357for light, the oscillations are not of the
11358medium particles or their pressure or density
11359– it is not a mechanical wave. You will come
11360to know more about transverse waves in
11361higher classes.
1136212.2.3 CHARACTERISTICS OF A SOUND WAVE
11363We can describe a sound wave by its
11364• frequency
11365• amplitude and
11366• speed.
11367A sound wave in graphic form is shown
11368in Fig. 12.8(c), which represents how density
11369and pressure change when the sound wave
11370moves in the medium. The density as well as
11371the pressure of the medium at a given time
11372varies with distance, above and below the
11373average value of density and pressure.
11374Fig. 12.8(a) and Fig. 12.8(b) represent the
11375density and pressure variations, respectively,
11376as a sound wave propagates in the medium.
11377Compressions are the regions where
11378particles are crowded together and
11379represented by the upper portion of the curve
11380in Fig. 12.8(c). The peak represents the region
11381of maximum compression. Thus,
11382compressions are regions where density as
11383well as pressure is high. Rarefactions are the
11384regions of low pressure where particles are
11385spread apart and are represented by the
11386uestions
113871. Explain how sound is produced
11388by your school bell.
113892. Why are sound waves called
11390mechanical waves?
113913. Suppose you and your friend are
11392on the moon. Will you be able to
11393hear any sound produced by
11394your friend?
1139512.2.2 SOUND WAVES ARE LONGITUDINAL
11396WAVES
11397Activity _____________ 12.4
11398• Take a slinky. Ask your friend to hold
11399one end. You hold the other end.
11400Now stretch the slinky as shown in
11401Fig. 12.7 (a). Then give it a sharp push
11402towards your friend.
11403• What do you notice? If you move your
11404hand pushing and pulling the slinky
11405alternatively, what will you observe?
11406• If you mark a dot on the slinky, you
11407will observe that the dot on the slinky
11408will move back and forth parallel to the
11409direction of the propagation of the
11410disturbance.
11411Fig. 12.7: Longitudinal wave in a slinky.
11412The regions where the coils become closer
11413are called compressions (C) and the regions
11414where the coils are further apart are called
11415rarefactions (R). As we already know, sound
11416propagates in the medium as a series of
11417compressions and rarefactions. Now, we can
11418compare the propagation of disturbance in a
11419slinky with the sound propagation in the
11420medium. These waves are called longitudinal
11421Q
11422(a)
11423(b)
11424© NCERT
11425not to be republished
11426164 SCIENCE
11427Frequency tells us how frequently an
11428event occurs. Suppose you are beating a
11429drum. How many times you are beating the
11430drum per unit time is called the frequency of
11431your beating the drum. We know that when
11432sound is propagated through a medium, the
11433density of the medium oscillates between a
11434maximum value and a minimum value. The
11435change in density from the maximum value
11436to the minimum value, again to the maximum
11437value, makes one complete oscillation. The
11438number of such oscillations per unit time is
11439the frequency of the sound wave. If we can
11440count the number of the compressions or
11441rarefactions that cross us per unit time, we
11442will get the frequency of the sound wave. It is
11443usually represented by ν (Greek letter, nu).
11444Its SI unit is hertz (symbol, Hz).
11445The time taken by two consecutive
11446compressions or rarefactions to cross a fixed
11447point is called the time period of the wave. In
11448other words, we can say that the time taken
11449for one complete oscillation in the density of
11450the medium is called the time period of the
11451valley, that is, the lower portion of the curve
11452in Fig. 12.8(c). A peak is called the crest and
11453a valley is called the trough of a wave.
11454The distance between two consecutive
11455compressions (C) or two consecutive
11456rarefactions (R) is called the wavelength, as
11457shown in Fig. 12.8(c), The wavelength is
11458usually represented by λ (Greek letter
11459lambda). Its SI unit is metre (m).
11460Heinrich Rudolph Hertz
11461was born on 22 February
114621857 in Hamburg,
11463Germany and educated at
11464the University of Berlin. He
11465confirmed J.C. Maxwell’s
11466electromagnetic theory by
11467his experiments. He laid the
11468foundation for future
11469development of radio, telephone, telegraph
11470and even television. He also discovered the
11471photoelectric effect which was later
11472explained by Albert Einstein. The SI unit
11473of frequency was named as hertz in his
11474honour.
11475Fig. 12.8: Sound propagates as density or pressure variations as shown in (a) and (b), (c) represents
11476graphically the density and pressure variations.
11477© NCERT
11478H. R. Hertz
11479not to be republished
11480SOUND 165
11481sound wave. It is represented by the symbol
11482T. Its SI unit is second (s). Frequency and
11483time period are related as follows:
11484. 1
11485T
11486A violin and a flute may both be played
11487at the same time in an orchestra. Both
11488sounds travel through the same medium,
11489that is, air and arrive at our ear at the same
11490time. Both sounds travel at the same speed
11491irrespective of the source. But the sounds
11492we receive are different. This is due to the
11493different characteristics associated with the
11494sound. Pitch is one of the characteristics.
11495How the brain interprets the frequency of
11496an emitted sound is called its pitch. The faster
11497the vibration of the source, the higher is
11498the frequency and the higher is the pitch,
11499as shown in Fig. 12.9. Thus, a high pitch
11500sound corresponds to more number of
11501compressions and rarefactions passing a
11502fixed point per unit time.
11503Objects of different sizes and conditions
11504vibrate at different frequencies to produce
11505sounds of different pitch.
11506The magnitude of the maximum
11507disturbance in the medium on either side of
11508the mean value is called the amplitude of the
11509wave. It is usually represented by the letter
11510A, as shown in Fig. 12.8(c). For sound its
11511unit will be that of density or pressure.
11512The loudness or softness of a sound is
11513determined basically by its amplitude. The
11514amplitude of the sound wave depends upon
11515the force with which an object is made to
11516vibrate. If we strike a table lightly, we hear a
11517soft sound because we produce a sound wave
11518of less energy (amplitude). If we hit the table
11519hard we hear a loud sound. Can you tell why?
11520Loud sound can travel a larger distance as it
11521is associated with higher energy. A sound
11522wave spreads out from its source. As it moves
11523away from the source its amplitude as well
11524as its loudness decreases. Fig. 12.10 shows
11525the wave shapes of a loud and a soft sound
11526of the same frequency.
11527Fig. 12.9: Low pitch sound has low frequency and
11528high pitch of sound has high frequency.
11529Fig. 12.10: Soft sound has small amplitude and
11530louder sound has large amplitude.
11531The quality or timber of sound is that
11532characteristic which enables us to
11533distinguish one sound from another having
11534the same pitch and loudness. The sound
11535which is more pleasant is said to be of a rich
11536© NCERT
11537not to be republished
11538166 SCIENCE
11539quality. A sound of single frequency is called
11540a tone. The sound which is produced due to
11541a mixture of several frequencies is called a
11542note and is pleasant to listen to. Noise is
11543unpleasant to the ear! Music is pleasant to
11544hear and is of rich quality.
11545uestions
115461. Which wave property determines
11547(a) loudness, (b) pitch?
115482. Guess which sound has a higher
11549pitch: guitar or car horn?
11550The speed of sound is defined as the
11551distance which a point on a wave, such as a
11552compression or a rarefaction, travels per unit
11553time.
11554We know,
11555speed, v = distance / time
11556= T
11557Here λ is the wavelength of the sound wave.
11558It is the distance travelled by the sound wave
11559in one time period (T) of the wave. Thus,
11560v = λ ν
115611
11562T
11563
11564 ∵
11565or v = λ ν
11566That is, speed = wavelength × frequency.
11567The speed of sound remains almost the
11568same for all frequencies in a given medium
11569under the same physical conditions.
11570Example 12.1 A sound wave has a
11571frequency of 2 kHz and wave length
1157235 cm. How long will it take to travel
115731.5 km?
11574Solution:
11575Given,
11576Frequency, ν = 2 kHz = 2000 Hz
11577Wavelength, λ = 35 cm = 0.35 m
11578We know that speed, v of the wave
11579= wavelength × frequency
11580v = λ ν
11581= 0.35 m × 2000 Hz = 700 m/s
11582The time taken by the wave to travel a
11583distance, d of 1.5 km is
11584Thus sound will take 2.1 s to travel a
11585distance of 1.5 km.
11586uestions
115871. What are wavelength, frequency,
11588time period and amplitude of a
11589sound wave?
115902. How are the wavelength and
11591frequency of a sound wave
11592related to its speed?
115933. Calculate the wavelength of a
11594sound wave whose frequency is
11595220 Hz and speed is 440 m/s in
11596a given medium.
115974. A person is listening to a tone of
11598500 Hz sitting at a distance of
11599450 m from the source of the
11600sound. What is the time interval
11601between successive compressions
11602from the source?
11603The amount of sound energy passing each
11604second through unit area is called the
11605intensity of sound. We sometimes use the
11606terms “loudness” and “intensity”
11607interchangeably, but they are not the same.
11608Loudness is a measure of the response of the
11609ear to the sound. Even when two sounds are
11610of equal intensity, we may hear one as louder
11611than the other simply because our ear detects
11612it better.
11613uestion
116141. Distinguish between loudness
11615and intensity of sound.
1161612.2.4 SPEED OF SOUND IN DIFFERENT
11617MEDIA
11618Sound propagates through a medium at a
11619finite speed. The sound of a thunder is heard
11620a little later than the flash of light is seen.
11621Q Q
11622© NCERT
11623Q
11624not to be republished
11625SOUND 167
11626So, we can make out that sound travels with
11627a speed which is much less than the speed
11628of light. The speed of sound depends on the
11629properties of the medium through which it
11630travels. You will learn about this dependence
11631in higher classes. The speed of sound in a
11632medium depends on temperature of the
11633medium. The speed of sound decreases when
11634we go from solid to gaseous state. In any
11635medium as we increase the temperature the
11636speed of sound increases. For example, the
11637speed of sound in air is 331 m s–1 at 0 ºC
11638and 344 m s–1 at 22 ºC. The speeds of sound
11639at a particular temperature in various media
11640are listed in Table 12.1. You need not
11641memorise the values.
11642Sonic boom: When the speed of any object
11643exceeds the speed of sound it is said to be
11644travelling at supersonic speed. Bullets, jet
11645aircrafts etc. often travel at supersonic
11646speeds. When a sound, producing source
11647moves with a speed higher than that of
11648sound, it produces shock waves in air.
11649These shock waves carry a large amount
11650of energy. The air pressure variation
11651associated with this type of shock waves
11652produces a very sharp and loud sound
11653called the “sonic boom”. The shock waves
11654produced by a supersonic aircraft have
11655enough energy to shatter glass and even
11656damage buildings.
1165712.3 Reflection of Sound
11658Sound bounces off a solid or a liquid like a
11659rubber ball bounces off a wall. Like light, sound
11660gets reflected at the surface of a solid or liquid
11661and follows the same laws of reflection as you
11662have studied in earlier classes. The directions
11663in which the sound is incident and is reflected
11664make equal angles with the normal to the
11665reflecting surface at the point of incidence, and
11666the three are in the same plane. An obstacle of
11667large size which may be polished or rough is
11668needed for the reflection of sound waves.
11669Activity _____________ 12.5
11670• Take two identical pipes, as shown in
11671Fig. 12.11. You can make the pipes
11672using chart paper. The length of the
11673pipes should be sufficiently long
11674as shown.
11675• Arrange them on a table near a wall.
11676• Keep a clock near the open end of one
11677of the pipes and try to hear the sound
11678of the clock through the other pipe.
11679• Adjust the position of the pipes so that
11680you can best hear the sound of
11681the clock.
11682• Now, measure the angles of incidence
11683and reflection and see the relationship
11684between the angles.
11685• Lift the pipe on the right vertically
11686to a small height and observe
11687what happens.
11688Table 12.1: Speed of sound in
11689different media at 25 ºC
11690State Substance Speed in m/s
11691Solids Aluminium 6420
11692Nickel 6040
11693Steel 5960
11694Iron 5950
11695Brass 4700
11696Glass (Flint) 3980
11697Liquids Water (Sea) 1531
11698Water (distilled) 1498
11699Ethanol 1207
11700Methanol 1103
11701Gases Hydrogen 1284
11702Helium 965
11703Air 346
11704Oxygen 316
11705Sulphur dioxide 213
11706uestion
117071. In which of the three media, air,
11708water or iron, does sound travel
11709the fastest at a particular Q temperature?
11710© NCERT
11711not to be republished
11712168 SCIENCE
1171312.3.1ECHO
11714If we shout or clap near a suitable reflecting
11715object such as a tall building or a mountain,
11716we will hear the same sound again a little
11717later. This sound which we hear is called an
11718echo. The sensation of sound persists in our
11719brain for about 0.1 s. To hear a distinct echo
11720the time interval between the original sound
11721and the reflected one must be at least 0.1s.
11722If we take the speed of sound to be 344 m/s
11723at a given temperature, say at 22 ºC in air,
11724the sound must go to the obstacle and reach
11725back the ear of the listener on reflection after
117260.1s. Hence, the total distance covered by the
11727sound from the point of generation to the
11728reflecting surface and back should be at least
11729(344 m/s) × 0.1 s = 34.4 m. Thus, for hearing
11730distinct echoes, the minimum distance of the
11731obstacle from the source of sound must be
11732half of this distance, that is, 17.2 m. This
11733distance will change with the temperature of
11734air. Echoes may be heard more than once
11735due to successive or multiple reflections. The
11736rolling of thunder is due to the successive
11737reflections of the sound from a number of
11738reflecting surfaces, such as the clouds and
11739the land.
1174012.3.2 REVERBERATION
11741A sound created in a big hall will persist by
11742repeated reflection from the walls until it is
11743reduced to a value where it is no longer
11744audible. The repeated reflection that results
11745in this persistence of sound is called
11746reverberation. In an auditorium or big hall
11747excessive reverberation is highly undesirable.
11748To reduce reverberation, the roof and walls of
11749the auditorium are generally covered with
11750sound-absorbent materials like compressed
11751fibreboard, rough plaster or draperies. The
11752seat materials are also selected on the basis
11753of their sound absorbing properties.
11754Example 12.2 A person clapped his hands
11755near a cliff and heard the echo after
117565 s. What is the distance of the cliff
11757from the person if the speed of the
11758sound, v is taken as 346 m s–1?
11759Solution:
11760Given,
11761Speed of sound, v = 346 m s–1
11762Time taken for hearing the echo,
11763t = 5 s
11764Distance travelled by the sound
11765= v × t = 346 m s–1 × 5 s = 1730 m
11766In 5 s sound has to travel twice the
11767distance between the cliff and the
11768person. Hence, the distance between
11769the cliff and the person
11770= 1730 m/2 = 865 m.
11771uestion
117721. An echo returned in 3 s. What is
11773the distance of the reflecting
11774surface from the source, given
11775that the speed of sound is
11776342 m s–1?
1177712.3.3 USES OF MULTIPLE REFLECTION
11778OF SOUND
117791. Megaphones or loudhailers, horns,
11780musical instruments such as trumpets
11781and shehanais, are all designed to
11782send sound in a particular direction
11783without spreading it in all directions,
11784as shown in Fig 12.12.
11785Fig. 12.11: Reflection of sound
11786Q
11787© NCERT
11788not to be republished
11789SOUND 169
11790soundboard may be placed behind the
11791stage so that the sound, after reflecting
11792from the sound board, spreads evenly
11793across the width of the hall (Fig 12.15).
11794Fig 12.12: A megaphone and a horn.
11795In these instruments, a tube followed
11796by a conical opening reflects sound
11797successively to guide most of the sound
11798waves from the source in the forward
11799direction towards the audience.
118002. Stethoscope is a medical instrument
11801used for listening to sounds produced
11802within the body, chiefly in the heart or
11803lungs. In stethoscopes the sound of the
11804patient’s heartbeat reaches the doctor’s
11805ears by multiple reflection of sound, as
11806shown in Fig.12.13.
11807Megaphone
11808Horn
11809Fig.12.13: Stethoscope
118103. Generally the ceilings of concert halls,
11811conference halls and cinema halls are
11812curved so that sound after reflection
11813reaches all corners of the hall, as shown
11814in Fig 12.14. Sometimes a curved
11815Fig. 12.14: Curved ceiling of a conference hall.
11816Fig. 12.15: Sound board used in a big hall.
11817uestion
118181. Why are the ceilings of concert
11819halls curved?
1182012.4 Range of Hearing ange of Hearing
11821The audible range of sound for human beings
11822extends from about 20 Hz to 20000 Hz (one
11823Hz = one cycle/s). Children under the age of
11824© NCERT
11825Q
11826not to be republished
11827170 SCIENCE
11828five and some animals, such as dogs can hear
11829up to 25 kHz (1 kHz = 1000 Hz). As people
11830grow older their ears become less sensitive to
11831higher frequencies. Sounds of frequencies
11832below 20 Hz are called infrasonic sound or
11833infrasound. If we could hear infrasound we
11834would hear the vibrations of a pendulum just
11835as we hear the vibrations of the wings of a
11836bee. Rhinoceroses communicate using
11837infrasound of frequency as low as 5 Hz.
11838Whales and elephants produce sound in the
11839infrasound range. It is observed that some
11840animals get disturbed before earthquakes.
11841Earthquakes produce low-frequency
11842infrasound before the main shock waves
11843begin which possibly alert the animals.
11844Frequencies higher than 20 kHz are called
11845ultrasonic sound or ultrasound. Ultrasound
11846is produced by dolphins, bats and porpoises.
11847Moths of certain families have very sensitive
11848hearing equipment. These moths can hear
11849the high frequency squeaks of the bat and
11850know when a bat is flying nearby, and are
11851able to escape capture. Rats also play games
11852by producing ultrasound.
11853Hearing Aid: People with hearing loss may
11854need a hearing aid. A hearing aid is an
11855electronic, battery operated device. The
11856hearing aid receives sound through a
11857microphone. The microphone converts the
11858sound waves to electrical signals. These
11859electrical signals are amplified by an
11860amplifier. The amplified electrical signals
11861are given to a speaker of the hearing aid.
11862The speaker converts the amplified
11863electrical signal to sound and sends to the
11864ear for clear hearing.
11865uestions
118661. What is the audible range of the
11867average human ear?
118682. What is the range of frequencies
11869associated with
11870(a) Infrasound?
11871(b) Ultrasound?
1187212.5 Applications of Ultrasound
11873Ultrasounds are high frequency waves.
11874Ultrasounds are able to travel along welldefined
11875paths even in the presence of
11876obstacles. Ultrasounds are used extensively
11877in industries and for medical purposes.
11878• Ultrasound is generally used to clean
11879parts located in hard-to-reach places,
11880for example, spiral tube, odd shaped
11881parts, electronic components etc.
11882Objects to be cleaned are placed in a
11883cleaning solution and ultrasonic
11884waves are sent into the solution. Due
11885to the high frequency, the particles of
11886dust, grease and dirt get detached and
11887drop out. The objects thus get
11888thoroughly cleaned.
11889• Ultrasounds can be used to detect
11890cracks and flaws in metal blocks.
11891Metallic components are generally
11892used in construction of big structures
11893like buildings, bridges, machines and
11894also scientific equipment. The cracks
11895or holes inside the metal blocks, which
11896are invisible from outside reduces the
11897strength of the structure. Ultrasonic
11898waves are allowed to pass through the
11899metal block and detectors are used to
11900detect the transmitted waves. If there
11901is even a small defect, the ultrasound
11902gets reflected back indicating the
11903presence of the flaw or defect, as shown
11904in Fig. 12.16.
11905Q Fig 12.16: Ultrasound is reflected back from the
11906defective locations inside a metal block.
11907© NCERT
11908not to be republished
11909SOUND 171
11910Ordinary sound of longer wavelengths
11911cannot be used for such purpose as it will
11912bend around the corners of the defective
11913location and enter the detector.
11914• Ultrasonic waves are made to reflect
11915from various parts of the heart and
11916form the image of the heart. This technique
11917is called ‘echocardiography’.
11918• Ultrasound scanner is an instrument
11919which uses ultrasonic waves for
11920getting images of internal organs of
11921the human body. A doctor may image
11922the patient’s organs such as the liver,
11923gall bladder, uterus, kidney, etc. It
11924helps the doctor to detect
11925abnormalities, such as stones in the
11926gall bladder and kidney or tumours
11927in different organs. In this technique
11928the ultrasonic waves travel through
11929the tissues of the body and get
11930reflected from a region where there is
11931a change of tissue density. These
11932waves are then converted into
11933electrical signals that are used to
11934generate images of the organ. These
11935images are then displayed on a
11936monitor or printed on a film. This
11937technique is called ‘ultrasonography’.
11938Ultrasonography is also used for
11939examination of the foetus during
11940pregnancy to detect congenial defects
11941and growth abnormalities.
11942• Ultrasound may be employed to break
11943small ‘stones’ formed in the kidneys
11944into fine grains. These grains later get
11945flushed out with urine.
1194612.5.1SONAR
11947The acronym SONAR stands for SOund
11948Navigation And Ranging. Sonar is a device
11949that uses ultrasonic waves to measure the
11950distance, direction and speed of underwater
11951objects. How does the sonar work? Sonar
11952consists of a transmitter and a detector and is
11953installed in a boat or a ship, as shown in Fig.
1195412.17.
11955Fig.12.17: Ultrasound sent by the transmitter and
11956received by the detector.
11957The transmitter produces and transmits
11958ultrasonic waves. These waves travel through
11959water and after striking the object on the
11960seabed, get reflected back and are sensed
11961by the detector. The detector converts the
11962ultrasonic waves into electrical signals which
11963are appropriately interpreted. The distance
11964of the object that reflected the sound wave
11965can be calculated by knowing the speed of
11966sound in water and the time interval between
11967transmission and reception of the
11968ultrasound. Let the time interval between
11969transmission and reception of ultrasound
11970signal be t and the speed of sound through
11971seawater be v. The total distance, 2d travelled
11972by the ultrasound is then, 2d = v × t.
11973The above method is called echo-ranging.
11974The sonar technique is used to determine the
11975depth of the sea and to locate underwater
11976hills, valleys, submarine, icebergs, sunken
11977ship etc.
11978Example 12.3 A ship sends out ultrasound
11979that returns from the seabed and is
11980detected after 3.42 s. If the speed of
11981ultrasound through seawater is
119821531 m/s, what is the distance of the
11983seabed from the ship?
11984Solution:
11985Given,
11986Time between transmission and
11987detection, t = 3.42 s.
11988© NCERT
11989not to be republished
11990172 SCIENCE
11991Speed of ultrasound in sea water,
11992v = 1531 m/s
11993Distance travelled by the ultrasound
11994= 2 × depth of the sea = 2d
11995where d is the depth of the sea.
119962d = speed of sound × time
11997= 1531 m/s × 3.42 s = 5236 m
11998d = 5236 m/2 = 2618 m.
11999Thus, the distance of the seabed from
12000the ship is 2618 m or 2.62 km.
12001uestion
120021. A submarine emits a sonar pulse,
12003which returns from an
12004underwater cliff in 1.02 s. If the
12005speed of sound in salt water is
120061531 m/s, how far away is the
12007cliff?
12008As mentioned earlier, bats search out prey
12009and fly in dark night by emitting and
12010detecting reflections of ultrasonic waves. The
12011high-pitched ultrasonic squeaks of the bat
12012are reflected from the obstacles or prey and
12013returned to bat’s ear, as shown in Fig. 12.18.
12014The nature of reflections tells the bat where
12015the obstacle or prey is and what it is like.
12016Porpoises also use ultrasound for navigation
12017and location of food in the dark.
1201812.6 Structure of Human Ear Structure of Human Ear
12019How do we hear? We are able to hear with
12020the help of an extremely sensitive device
12021called the ear. It allows us to convert pressure
12022variations in air with audible frequencies into
12023electric signals that travel to the brain via the
12024auditory nerve. The auditory aspect of human
12025ear is discussed below.
12026Q
12027Fig. 12.18: Ultrasound is emitted by a bat and it is
12028reflected back by the prey or an obstacle.
12029The outer ear is called ‘pinna’. It collects
12030the sound from the surroundings. The
12031collected sound passes through the auditory
12032canal. At the end of the auditory canal there
12033is a thin membrane called the ear drum or
12034tympanic membrane. When a compression
12035of the medium reaches the eardrum the
12036pressure on the outside of the membrane
12037increases and forces the eardrum inward.
12038Similarly, the eardrum moves outward when
12039a rarefaction reaches it. In this way the
12040eardrum vibrates. The vibrations are
12041amplified several times by three bones (the
12042hammer, anvil and stirrup) in the middle ear.
12043The middle ear transmits the amplified
12044pressure variations received from the sound
12045wave to the inner ear. In the inner ear, the
12046pressure variations are turned into electrical
12047signals by the cochlea. These electrical signals
12048are sent to the brain via the auditory nerve,
12049and the brain interprets them as sound.
12050Fig. 12.19: Auditory parts of human ear.
12051© NCERT
12052not to be republished
12053SOUND 173
12054What
12055you have you have
12056learnt
12057• Sound is produced due to vibration of different objects.
12058• Sound travels as a longitudinal wave through a material
12059medium.
12060• Sound travels as successive compressions and rarefactions in
12061the medium.
12062• In sound propagation, it is the energy of the sound that travels
12063and not the particles of the medium.
12064• Sound cannot travel in vacuum.
12065• The change in density from one maximum value to the
12066minimum value and again to the maximum value makes one
12067complete oscillation.
12068• The distance between two consecutive compressions or two
12069consecutive rarefactions is called the wavelength, λ.
12070• The time taken by the wave for one complete oscillation of the
12071density or pressure of the medium is called the time period, T.
12072• The number of complete oscillations per unit time is called the
12073frequency (ν),
120741
12075T .
12076• The speed v, frequency ν, and wavelength λ, of sound are related
12077by the equation, v = λν.
12078• The speed of sound depends primarily on the nature and the
12079temperature of the transmitting medium.
12080• The law of reflection of sound states that the directions in which
12081the sound is incident and reflected make equal angles with
12082the normal to the reflecting surface at the point of incidence
12083and the three lie in the same plane.
12084• For hearing a distinct sound, the time interval between the
12085original sound and the reflected one must be at least 0.1 s.
12086• The persistence of sound in an auditorium is the result of
12087repeated reflections of sound and is called reverberation.
12088• Sound properties such as pitch, loudness and quality are
12089determined by the corresponding wave properties.
12090• Loudness is a physiological response of the ear to the intensity
12091of sound.
12092• The amount of sound energy passing each second through
12093unit area is called the intensity of sound.
12094• The audible range of hearing for average human beings is
12095in the frequency range of 20 Hz – 20 kHz.
12096© NCERT
12097not to be republished
12098174 SCIENCE
12099• Sound waves with frequencies below the audible range are
12100termed “infrasonic” and those above the audible range are
12101termed “ultrasonic”.
12102• Ultrasound has many medical and industrial applications.
12103• The SONAR technique is used to determine the depth of the
12104sea and to locate under water hills, valleys, submarines,
12105icebergs, sunken ships etc.
12106Exercises Exercises Exercises
121071. What is sound and how is it produced?
121082. Describe with the help of a diagram, how compressions and
12109rarefactions are produced in air near a source of sound.
121103. Cite an experiment to show that sound needs a material
12111medium for its propagation.
121124. Why is sound wave called a longitudinal wave?
121135. Which characteristic of the sound helps you to identify your
12114friend by his voice while sitting with others in a dark room?
121156. Flash and thunder are produced simultaneously. But thunder
12116is heard a few seconds after the flash is seen, why?
121177. A person has a hearing range from 20 Hz to 20 kHz. What are
12118the typical wavelengths of sound waves in air corresponding
12119to these two frequencies? Take the speed of sound in air as
12120344 m s–1.
121218. Two children are at opposite ends of an aluminium rod. One
12122strikes the end of the rod with a stone. Find the ratio of times
12123taken by the sound wave in air and in aluminium to reach the
12124second child.
121259. The frequency of a source of sound is 100 Hz. How many times
12126does it vibrate in a minute?
1212710. Does sound follow the same laws of reflection as light does?
12128Explain.
1212911. When a sound is reflected from a distant object, an echo is
12130produced. Let the distance between the reflecting surface and
12131the source of sound production remains the same. Do you
12132hear echo sound on a hotter day?
1213312. Give two practical applications of reflection of sound waves.
1213413. A stone is dropped from the top of a tower 500 m high into a
12135pond of water at the base of the tower. When is the splash
12136heard at the top? Given, g = 10 m s–2 and speed of sound =
12137340 m s–1.
1213814. A sound wave travels at a speed of 339 m s–1. If its
12139wavelength is 1.5 cm, what is the frequency of the wave?
12140© NCERT
12141not to be republished
12142SOUND 175
12143Will it be audible?
1214415. What is reverberation? How can it be reduced?
1214516. What is loudness of sound? What factors does it depend on?
1214617. Explain how bats use ultrasound to catch a prey.
1214718. How is ultrasound used for cleaning?
1214819. Explain the working and application of a sonar.
1214920. A sonar device on a submarine sends out a signal and receives
12150an echo 5 s later. Calculate the speed of sound in water if the
12151distance of the object from the submarine is 3625 m.
1215221. Explain how defects in a metal block can be detected using
12153ultrasound.
1215422. Explain how the human ear works.
12155© NCERT
12156not to be republished
12157Activity _____________ 13.1
12158• We have all heard of the earthquakes
12159in Latur, Bhuj, Kashmir etc. or the
12160cyclones that attack the coastal
12161regions. Think of as many different
12162ways as possible in which people’s
12163health would be affected by such a
12164disaster if it took place in our
12165neighbourhood.
12166• How many of these ways we can think of
12167are events that would occur when the
12168disaster is actually happening?
12169• How many of these health-related events
12170would happen long after the actual
12171disaster, but would still be because of the
12172disaster?
12173• Why would one effect on health fall into
12174the first group, and why would another
12175fall into the second group?
12176When we do this exercise, we realise that
12177health and disease in human communities
12178are very complex issues, with many
12179interconnected causes. We also realise that
12180the ideas of what ‘health’ and ‘disease’ mean
12181are themselves very complicated. When we
12182ask what causes diseases and how we prevent
12183them, we have to begin by asking what these
12184notions mean.
12185We have seen that cells are the basic units
12186of living beings. Cells are made of a variety of
12187chemical substances – proteins, carbohydrates,
12188fats or lipids, and so on. Although
12189the pictures look quite static, in reality the
12190living cell is a dynamic place. Something or
12191the other is always happening. Cells move
12192from place to place. Even in cells that do not
12193move, there is repair going on. New cells are
12194being made. In our organs or tissues, there
12195are various specialised activities going on –
12196the heart is beating, the lungs are breathing,
12197the kidney is filtering urine, the brain is
12198thinking.
12199All these activities are interconnected. For
12200example, if the kidneys are not filtering urine,
12201poisonous substances will accumulate. Under
12202such conditions, the brain will not be able to
12203think properly. For all these interconnected
12204activities, energy and raw material are needed
12205from outside the body. In other words, food
12206is a necessity for cell and tissue functions.
12207Anything that prevents proper functioning of
12208cells and tissues will lead to a lack of proper
12209activity of the body.
12210It is in this context that we will now look
12211at the notions of health and disease.
1221213.1 Health and its Failure Health and its Failure
1221313.1.1 THE SIGNIFICANCE OF ‘HEALTH’
12214We have heard the word ‘health’ used quite
12215frequently all around us. We use it ourselves
12216as well, when we say things like ‘my
12217grandmother’s health is not good’. Our
12218teachers use it when they scold us saying ‘this
12219is not a healthy attitude’. What does the word
12220‘health’ mean?
12221If we think about it, we realise that it
12222always implies the idea of ‘being well’. We can
12223think of this well-being as effective
12224functioning. For our grandmothers, being able
12225to go out to the market or to visit neighbours
12226is ‘being well’, and not being able to do such
12227things is ‘poor health’. Being interested in
12228following the teaching in the classroom so that
12229we can understand the world is called a
12230‘healthy attitude’; while not being interested
12231is called the opposite. ‘Health’ is therefore a
12232state of being well enough to function well
12233physically, mentally and socially.
1223413
12235WHY DO WE FALL ILL
12236Chapter
12237© NCERT
12238not to be republished
12239We need food for health, and this food will
12240have to be earned by doing work. For this,
12241the opportunity to do work has to be available.
12242Good economic conditions and jobs are
12243therefore needed for individual health.
12244We need to be happy in order to be truly
12245healthy, and if we mistreat each other and
12246are afraid of each other, we cannot be happy
12247or healthy. Social equality and harmony are
12248therefore necessary for individual health. We
12249can think of many other such examples of
12250connections between community issues and
12251individual health.
1225213.1.3 DISTINCTIONS BETWEEN ‘HEALTHY’
12253AND ‘DISEASE-FREE’
12254If this is what we mean by ‘health’, what do
12255we mean by ‘disease’? The word is actually
12256self-explanatory – we can think of it as
12257‘disease’ – disturbed ease. Disease, in other
12258words, literally means being uncomfortable.
12259However, the word is used in a more limited
12260meaning. We talk of disease when we can find
12261a specific and particular cause for discomfort.
12262This does not mean that we have to know the
12263absolute final cause; we can say that
12264someone is suffering from diarrhoea without
12265knowing exactly what has caused the loose
12266motions.
12267We can now easily see that it is possible
12268to be in poor health without actually suffering
12269from a particular disease. Simply not being
12270diseased is not the same as being healthy.
12271‘Good health’ for a dancer may mean being
12272able to stretch his body into difficult but
12273graceful positions. On the other hand, good
12274health for a musician may mean having enough
12275breathing capacity in his/her lungs to control
12276the notes from his/her flute. To have the
12277opportunity to realise the unique potential
12278in all of us is also necessary for real health.
12279So, we can be in poor health without there
12280being a simple cause in the form of an
12281identifiable disease. This is the reason why,
12282when we think about health, we think about
12283societies and communities. On the other
12284hand, when we think about disease, we think
12285about individual sufferers.
1228613.1.2 PERSONAL AND COMMUNITY ISSUES
12287BOTH MATTER FOR HEALTH
12288If health means a state of physical, mental
12289and social well-being, it cannot be something
12290that each one of us can achieve entirely on
12291our own. The health of all organisms will
12292depend on their surroundings or their
12293environment. The environment includes the
12294physical environment. So, for example, health
12295is at risk in a cyclone in many ways.
12296But even more importantly, human beings
12297live in societies. Our social environment,
12298therefore, is an important factor in our
12299individual health. We live in villages, towns
12300or cities. In such places, even our physical
12301environment is decided by our social
12302environment.
12303Consider what would happen if no agency
12304is ensuring that garbage is collected and
12305disposed. What would happen if no one takes
12306responsibility for clearing the drains and
12307ensuring that water does not collect in the
12308streets or open spaces?
12309So, if there is a great deal of garbage
12310thrown in our streets, or if there is open drainwater
12311lying stagnant around where we live,
12312the possibility of poor health increases.
12313Therefore, public cleanliness is important for
12314individual health.
12315Activity _____________ 13.2
12316• Find out what provisions are made by
12317your local authority (panchayat/
12318municipal corporation) for the supply
12319of clean drinking water.
12320• Are all the people in your locality able
12321to access this?
12322Activity _____________ 13.3
12323• Find out how your local authority
12324manages the solid waste generated in
12325your neighbourhood.
12326• Are these measures adequate?
12327• If not, what improvements would you
12328suggest?
12329• What could your family do to reduce
12330the amount of solid waste generated
12331during a day/week?
12332WHY DO WE FALL ILL 177
12333© NCERT
12334not to be republished
12335178 SCIENCE
12336uestions
123371. State any two conditions
12338essential for good health.
123392. State any two conditions
12340essential for being free of disease.
123413. Are the answers to the above
12342questions necessarily the same
12343or different? Why?
1234413.2 Disease and Its Causes Disease and Its Causes
1234513.2.1 WHAT DOES DISEASE LOOK LIKE?
12346Let us now think a little more about diseases.
12347In the first place, how do we know that there
12348is a disease? In other words, how do we know
12349that there is something wrong with the body?
12350There are many tissues in the body, as we
12351have seen in Chapter 6. These tissues make
12352up physiological systems or organ systems
12353that carry out body functions. Each of the
12354organ systems has specific organs as its parts,
12355and it has particular functions. So, the
12356digestive system has the stomach and
12357intestines, and it helps to digest food taken
12358in from outside the body. The musculoskeletal
12359system, which is made up of bones and
12360muscles, holds the body parts together and
12361helps the body move.
12362When there is a disease, either the
12363functioning or the appearance of one or more
12364systems of the body will change for the worse.
12365These changes give rise to symptoms and
12366signs of disease. Symptoms of disease are the
12367things we feel as being ‘wrong’. So we have a
12368headache, we have cough, we have loose
12369motions, we have a wound with pus; these
12370are all symptoms. These indicate that there
12371may be a disease, but they don’t indicate what
12372the disease is. For example, a headache may
12373mean just examination stress or, very rarely,
12374it may mean meningitis, or any one of a dozen
12375different diseases.
12376Signs of disease are what physicians will
12377look for on the basis of the symptoms. Signs
12378will give a little more definite indication of
12379the presence of a particular disease.
12380Physicians will also get laboratory tests done
12381to pinpoint the disease further.
1238213.2.2 ACUTE AND CHRONIC DISEASES
12383The manifestations of disease will be different
12384depending on a number of factors. One of the
12385most obvious factors that determine how we
12386perceive the disease is its duration. Some
12387diseases last for only very short periods of
12388time, and these are called acute diseases. We
12389all know from experience that the common
12390cold lasts only a few days. Other ailments can
12391last for a long time, even as much as a lifetime,
12392and are called chronic diseases. An example
12393is the infection causing elephantiasis, which
12394is very common in some parts of India.
12395Activity _____________ 13.4
12396• Survey your neighbourhood to find out:
12397(1) how many people suffered from
12398acute diseases during the last three
12399months,
12400(2) how many people developed chronic
12401diseases during this same period,
12402(3) and finally, the total number of
12403people suffering from chronic
12404diseases in your neighbourhood.
12405• Are the answers to questions (1) and
12406(2) different?
12407• Are the answers to questions (2) and
12408(3) different?
12409• What do you think could be the reason
12410for these differences? What do you think
12411would be the effect of these differences
12412on the general health of the population?
1241313.2.3 CHRONIC DISEASES AND POOR
12414HEALTH
12415As we can imagine, acute and chronic
12416diseases have different effects on our health.
12417Any disease that causes poor functioning of
12418some part of the body will affect our general
12419health as well. This is because all functions
12420of the body are necessary for general health.
12421But an acute disease, which is over very soon,
12422will not have time to cause major effects on
12423general health, while a chronic disease will
12424do so.
12425As an example, think about a cough and
12426cold, which all of us have from time to time.
12427Most of us get better and become well within
12428a week or so. And there are no bad effects on
12429Q
12430© NCERT
12431not to be republished
12432WHY DO WE FALL ILL 179
12433would not lead to loose motions. But they do
12434become contributory causes of the disease.
12435Why was there no clean drinking water
12436for the baby? Perhaps because the public
12437services are poor where the baby’s family
12438lives. So, poverty or lack of public services
12439become third-level causes of the baby’s
12440disease.
12441It will now be obvious that all diseases
12442will have immediate causes and contributory
12443causes. Also, most diseases will have many
12444causes, rather than one single cause.
1244513.2.5 INFECTIOUS AND NON-INFECTIOUS
12446CAUSES
12447As we have seen, it is important to keep public
12448health and community health factors in mind
12449when we think about causes of diseases. We
12450can take that approach a little further. It is
12451useful to think of the immediate causes of
12452disease as belonging to two distinct types. One
12453group of causes is the infectious agents,
12454mostly microbes or micro-organisms.
12455Diseases where microbes are the immediate
12456causes are called infectious diseases. This is
12457because the microbes can spread in the
12458community, and the diseases they cause will
12459spread with them.
12460Things to ponder
124611. Do all diseases spread to people
12462coming in contact with a sick person?
124632. What are the diseases that are not
12464spreading?
124653. How would a person develop those
12466diseases that don’t spread by contact
12467with a sick person?
12468On the other hand, there are also diseases
12469that are not caused by infectious agents. Their
12470causes vary, but they are not external causes
12471like microbes that can spread in the
12472community. Instead, these are mostly
12473internal, non-infectious causes.
12474For example, some cancers are caused by
12475genetic abnormalities. High blood pressure
12476can be caused by excessive weight and lack
12477of exercise. You can think of many other
12478diseases where the immediate causes will not
12479be infectious.
12480our health. We do not lose weight, we do not
12481become short of breath, we do not feel tired
12482all the time because of a few days of cough
12483and cold. But if we get infected with a chronic
12484disease such as tuberculosis of the lungs,
12485then being ill over the years does make us
12486lose weight and feel tired all the time.
12487We may not go to school for a few days if
12488we have an acute disease. But a chronic
12489disease will make it difficult for us to follow
12490what is being taught in school and reduce
12491our ability to learn. In other words, we are
12492likely to have prolonged general poor health
12493if we have a chronic disease. Chronic diseases
12494therefore, have very drastic long-term effects
12495on people’s health as compared to acute
12496diseases.
1249713.2.4 CAUSES OF DISEASES
12498What causes disease? When we think about
12499causes of diseases, we must remember that
12500there are many levels of such causes. Let us
12501look at an example. If there is a baby suffering
12502from loose motions, we can say that the cause
12503of the loose motions is an infection with a
12504virus. So the immediate cause of the disease
12505is a virus.
12506But the next question is – where did the
12507virus come from? Suppose we find that the
12508virus came through unclean drinking water.
12509But many babies must have had this unclean
12510drinking water. So, why is it that one baby
12511developed loose motions when the other
12512babies did not?
12513One reason might be that this baby is not
12514healthy. As a result, it might be more likely
12515to have disease when exposed to risk, whereas
12516healthier babies would not. Why is the baby
12517not healthy? Perhaps because it is not well
12518nourished and does not get enough food. So,
12519lack of good nourishment becomes a secondlevel
12520cause of the disease the baby is suffering
12521from. Further, why is the baby not well
12522nourished? Perhaps because it is from a
12523household which is poor.
12524It is also possible that the baby has some
12525genetic difference that makes it more likely
12526to suffer from loose motions when exposed
12527to such a virus. Without the virus, the genetic
12528difference or the poor nourishment alone
12529© NCERT
12530not to be republished
12531180 SCIENCE
12532The ways in which diseases spread, and
12533the ways in which they can be treated and
12534prevented at the community level would be
12535different for different diseases. This would
12536depend a lot on whether the immediate
12537causes are infectious or non-infectious.
12538uestions
125391. List any three reasons why you
12540would think that you are sick and
12541ought to see a doctor. If only one
12542of these symptoms were present,
12543would you still go to the doctor?
12544Why or why not?
125452. In which of the following case do
12546you think the long-term effects on
12547your health are likely to be most
12548unpleasant?
12549• if you get jaundice,
12550• if you get lice,
12551• if you get acne.
12552Why?
1255313.3 Infectious Diseases Infectious Diseases
1255413.3.1 INFECTIOUS AGENTS
12555We have seen that the entire diversity seen in
12556the living world can be classified into a few
12557groups. This classification is based on
12558common characteristics between different
12559organisms. Organisms that can cause disease
12560are found in a wide range of such categories
12561of classification. Some of them are viruses,
12562some are bacteria, some are fungi, some are
12563single-celled animals or protozoans. Some
12564diseases are also caused by multicellular
12565organisms, such as worms of different kinds.
12566Peptic ulcers and the Nobel prize
12567For many years, everybody used to think
12568that peptic ulcers, which cause acidity–
12569related pain and bleeding in the stomach
12570and duodenum, were because of lifestyle
12571reasons. Everybody thought that a
12572stressful life led to a lot of acid secretion
12573in the stomach, and eventually caused
12574peptic ulcers.
12575Then two Australians made a discovery
12576that a bacterium, Helicobacter pylori, was
12577responsible for peptic ulcers. Robin Warren
12578(born 1937), a pathologist from Perth,
12579Australia, saw these small curved bacteria
12580in the lower part of the stomach in many
12581patients. He noticed that signs of
12582inflammation were always present around
12583these bacteria. Barry Marshall (born 1951),
12584a young clinical fellow, became interested
12585in Warren’s findings and succeeded in
12586cultivating the bacteria from these sources.
12587In treatment studies, Marshall and
12588Warren showed that patients could be
12589cured of peptic ulcer only when the
12590bacteria were killed off from the stomach.
12591Thanks to this pioneering discovery by
12592Marshall and Warren, peptic ulcer disease
12593is no longer a chronic, frequently disabling
12594condition, but a disease that can be cured
12595by a short period of treatment with
12596antibiotics.
12597Q
12598For this achievement, Marshall and
12599Warren (seen in the picture) received the
12600Nobel prize for physiology and medicine
12601in 2005.
12602© NCERT
12603not to be republished
12604WHY DO WE FALL ILL 181
12605Fig. 13.1(b): Picture of staphylococci, the bacteria
12606which can cause acne. The scale of the
12607image is indicated by the line at top left,
12608which is 5 micrometres long.
12609Fig. 13.1(d): Picture of Leishmania, the protozoan
12610organism that causes kala-azar. The
12611organisms are oval-shaped, and each
12612has one long whip-like structure. One
12613organism (arrow) is dividing, while a cell
12614of the immune system (lower right) has
12615gripped on the two whips of the dividing
12616organism and is sending cell processes
12617up to eat up the organism. The immune
12618cell is about ten micrometres in diameter.
12619Fig. 13.1(e): Picture of an adult roundworm (Ascaris
12620lumbricoides is the technical name) from
12621the small intestine. The ruler next to it
12622shows four centimetres to give us an
12623idea of the scale.
12624Fig. 13.1(a): Picture of SARS viruses coming out (see
12625arrows for examples) of the surface of
12626an infected cell. The white scale line
12627represents 500 nanometres, which is
12628half a micrometre, which is onethousandth
12629of a millimetre. The scale line
12630gives us an idea of how small the things
12631we are looking at are.
12632Courtesy: Emerging Infectious
12633Deseases, a journal of CDC, U.S.
12634Fig. 13.1(c): Picture of Trypanosoma, the protozoan
12635organism responsible for sleeping
12636sickness. The organism is lying next to
12637a saucer-shaped red blood cell to give
12638an idea of the scale.
12639Copyright: Oregon Health and Science
12640University, U.S.
12641© NCERT
12642not to be republished
12643182 SCIENCE
12644Common examples of diseases caused by
12645viruses are the common cold, influenza,
12646dengue fever and AIDS. Diseases like typhoid
12647fever, cholera, tuberculosis and anthrax are
12648caused by bacteria. Many common skin
12649infections are caused by different kinds of
12650fungi. Protozoan microbes cause many
12651familiar diseases, such as malaria and kalaazar.
12652All of us have also come across intestinal
12653worm infections, as well as diseases like
12654elephantiasis caused by diffferent species of
12655worms.
12656Why is it important that we think of these
12657categories of infectious agents? The answer
12658is that these categories are important factors
12659in deciding what kind of treatment to use.
12660Members of each one of these groups –
12661viruses, bacteria, and so on – have many
12662biological characteristics in common.
12663All viruses, for example, live inside host
12664cells, whereas bacteria very rarely do. Viruses,
12665bacteria and fungi multiply very quickly, while
12666worms multiply very slowly in comparison.
12667Taxonomically, all bacteria are closely related
12668to each other than to viruses and vice versa.
12669This means that many important life
12670processes are similar in the bacteria group
12671but are not shared with the virus group. As a
12672result, drugs that block one of these life
12673processes in one member of the group is likely
12674to be effective against many other members
12675of the group. But the same drug will not work
12676against a microbe belonging to a different
12677group.
12678As an example, let us take antibiotics.
12679They commonly block biochemical pathways
12680important for bacteria. Many bacteria, for
12681example, make a cell-wall to protect
12682themselves. The antibiotic penicillin blocks
12683the bacterial processes that build the cellwall.
12684As a result, the growing bacteria become
12685unable to make cell-walls, and die easily.
12686Human cells don’t make a cell-wall anyway,
12687so penicillin cannot have such an effect on
12688us. Penicillin will have this effect on any
12689bacteria that use such processes for making
12690cell-walls. Similarly, many antibiotics work
12691against many species of bacteria rather than
12692simply working against one.
12693But viruses do not use these pathways at
12694all, and that is the reason why antibiotics do
12695not work against viral infections. If we have a
12696common cold, taking antibiotics does not
12697reduce the severity or the duration of the
12698disease. However, if we also get a bacterial
12699infection along with the viral cold, taking
12700antibiotics will help. Even then, the antibiotic
12701will work only against the bacterial part of
12702the infection, not the viral infection.
12703Activity _____________ 13.5
12704• Find out how many of you in your class
12705had cold/cough/fever recently.
12706• How long did the illness last?
12707• How many of you took antibiotics (ask
12708your parents if you had antibiotics)?
12709• How long were those who took
12710antibiotics ill?
12711• How long were those who didn’t take
12712antibiotics ill?
12713• Is there a difference between these two
12714groups?
12715• If yes, why? If not, why not?
1271613.3.2 MEANS OF SPREAD
12717How do infectious diseases spread? Many
12718microbial agents can commonly move from
12719an affected person to someone else in a variety
12720of ways. In other words, they can be
12721‘communicated’, and so are also called
12722communicable diseases.
12723Such disease-causing microbes can
12724spread through the air. This occurs through
12725the little droplets thrown out by an infected
12726person who sneezes or coughs. Someone
12727standing close by can breathe in these
12728droplets, and the microbes get a chance to
12729start a new infection. Examples of such
12730diseases spread through the air are the
12731common cold, pneumonia and tuberculosis.
12732We all have had the experience of sitting
12733near someone suffering from a cold and
12734catching it ourselves. Obviously, the more
12735crowded our living conditions are, the more
12736likely it is that such airborne diseases will
12737spread.
12738© NCERT
12739not to be republished
12740WHY DO WE FALL ILL 183
12741Fig. 13.2: Air-transmitted diseases are easier to
12742catch the closer we are to the infected
12743person. However, in closed areas, the
12744droplet nuclei recirculate and pose a risk
12745to everybody. Overcrowded and poorly
12746ventilated housing is therefore a major
12747factor in the spread of airborne diseases.
12748Diseases can also be spread through
12749water. This occurs if the excreta from someone
12750suffering from an infectious gut disease, such
12751as cholera, get mixed with the drinking water
12752used by people living nearby. The choleracausing
12753microbes will enter new hosts
12754through the water they drink and cause
12755disease in them. Such diseases are much
12756more likely to spread in the absence of safe
12757supplies of drinking water.
12758The sexual act is one of the closest
12759physical contact two people can have with
12760each other. Not surprisingly, there are
12761microbial diseases such as syphilis or AIDS
12762that are transmitted by sexual contact from
12763one partner to the other. However, such
12764sexually transmitted diseases are not spread
12765by casual physical contact. Casual physical
12766contacts include handshakes or hugs or
12767sports, like wrestling, or by any of the other
12768ways in which we touch each other socially.
12769Other than the sexual contact, the AIDS virus
12770can also spread through blood-to-blood
12771contact with infected people or from an
12772infected mother to her baby during pregnancy
12773or through breast feeding.
12774We live in an environment that is full of
12775many other creatures apart from us. It is
12776inevitable that many diseases will be
12777transmitted by other animals. These animals
12778carry the infecting agents from a sick person
12779to another potential host. These animals are
12780thus the intermediaries and are called
12781vectors. The commonest vectors we all know
12782are mosquitoes. In many species of
12783mosquitoes, the females need highly
12784nutritious food in the form of blood in order
12785to be able to lay mature eggs. Mosquitoes feed
12786on many warm-blooded animals, including
12787us. In this way, they can transfer diseases
12788from person to person.
12789Fig. 13.3: Common methods of transmission of
12790diseases.
1279113.3.3 ORGAN- SPECIFIC AND TISSUESPECIFIC
12792MANIFESTATIONS
12793The disease-causing microbes enter the body
12794through these different means. Where do they
12795go then? The body is very large when
12796compared to the microbes. So there are many
12797possible places, organs or tissues, where they
12798could go. Do all microbes go to the same tissue
12799or organ, or do they go to different ones?
12800Different species of microbes seem to have
12801evolved to home in on different parts of the
12802body. In part, this selection is connected to
12803their point of entry. If they enter from the air
12804via the nose, they are likely to go to the lungs.
12805© NCERT
12806not to be republished
12807184 SCIENCE
12808This is seen in the bacteria causing
12809tuberculosis. If they enter through the mouth,
12810they can stay in the gut lining like typhoidcausing
12811bacteria. Or they can go to the liver,
12812like the viruses that cause jaundice.
12813But this needn’t always be the case. An
12814infection like HIV, that comes into the body
12815via the sexual organs, will spread to lymph
12816nodes all over the body. Malaria-causing
12817microbes, entering through a mosquito bite,
12818will go to the liver, and then to the red blood
12819cells. The virus causing Japanese
12820encephalitis, or brain fever, will similarly enter
12821through a mosquito bite. But it goes on to
12822infect the brain.
12823The signs and symptoms of a disease will
12824thus depend on the tissue or organ which
12825the microbe targets. If the lungs are the
12826targets, then symptoms will be cough and
12827breathlessness. If the liver is targeted, there
12828will be jaundice. If the brain is the target, we
12829will observe headaches, vomiting, fits or
12830unconsciousness. We can imagine what the
12831symptoms and signs of an infection will be if
12832we know what the target tissue or organ is,
12833and the functions that are carried out by this
12834tissue or organ.
12835In addition to these tissue-specific effects
12836of infectious disease, there will be other
12837common effects too. Most of these common
12838effects depend on the fact that the body’s
12839immune system is activated in response to
12840infection. An active immune system recruits
12841many cells to the affected tissue to kill off the
12842disease-causing microbes. This recruitment
12843process is called inflammation. As a part of
12844this process, there are local effects such as
12845swelling and pain, and general effects such
12846as fever.
12847In some cases, the tissue-specificity of the
12848infection leads to very general-seeming
12849effects. For example, in HIV infection, the
12850virus goes to the immune system and
12851damages its function. Thus, many of the
12852effects of HIV-AIDS are because the body can
12853no longer fight off the many minor infections
12854that we face everyday. Instead, every small
12855cold can become pneumonia. Similarly, a
12856minor gut infection can produce major
12857diarrhoea with blood loss. Ultimately, it is
12858these other infections that kill people
12859suffering from HIV-AIDS.
12860It is also important to remember that the
12861severity of disease manifestations depend on
12862the number of microbes in the body. If the
12863number of microbes is very small, the disease
12864manifestations may be minor or unnoticed.
12865But if the number is of the same microbe
12866large, the disease can be severe enough to be
12867life-threatening. The immune system is a
12868major factor that determines the number of
12869microbes surviving in the body. We shall look
12870into this aspect a little later in the chapter.
1287113.3.4 PRINCIPLES OF TREATMENT
12872What are the steps taken by your family when
12873you fall sick? Have you ever thought why you
12874sometimes feel better if you sleep for some
12875time? When does the treatment involve
12876medicines?
12877Based on what we have learnt so far, it
12878would appear that there are two ways to treat
12879an infectious disease. One would be to reduce
12880the effects of the disease and the other to kill
12881the cause of the disease. For the first, we can
12882provide treatment that will reduce the
12883symptoms. The symptoms are usually
12884because of inflammation. For example, we can
12885take medicines that bring down fever, reduce
12886pain or loose motions. We can take bed rest so
12887that we can conserve our energy. This will
12888enable us to have more of it available to focus
12889on healing.
12890But this kind of symptom-directed
12891treatment by itself will not make the infecting
12892microbe go away and the disease will not be
12893cured. For that, we need to be able to kill off
12894the microbes.
12895How do we kill microbes? One way is to
12896use medicines that kill microbes. We have seen
12897earlier that microbes can be classified into
12898different categories. They are viruses, bacteria,
12899fungi or protozoa. Each of these groups of
12900organisms will have some essential
12901biochemical life process which is peculiar to
12902© NCERT
12903not to be republished
12904WHY DO WE FALL ILL 185
12905we can get some easy answers. For airborne
12906microbes, we can prevent exposure by
12907providing living conditions that are not
12908overcrowded. For water-borne microbes, we
12909can prevent exposure by providing safe
12910drinking water. This can be done by treating
12911the water to kill any microbial contamination.
12912For vector-borne infections, we can provide
12913clean environments. This would not, for
12914example, allow mosquito breeding. In other
12915words, public hygiene is one basic key to the
12916prevention of infectious diseases.
12917In addition to these issues that relate to
12918the environment, there are some other general
12919principles to prevent infectious diseases. To
12920appreciate those principles, let us ask a
12921question we have not looked at so far.
12922Normally, we are faced with infections
12923everyday. If someone is suffering from a cold
12924and cough in the class, it is likely that the
12925children sitting around will be exposed to the
12926infection. But all of them do not actually suffer
12927from the disease. Why not?
12928This is because the immune system of our
12929body is normally fighting off microbes. We
12930have cells that specialise in killing infecting
12931microbes. These cells go into action each time
12932infecting microbes enter the body. If they are
12933successful, we do not actually come down
12934with any disease. The immune cells manage
12935to kill off the infection long before it assumes
12936major proportions. As we noted earlier, if the
12937number of the infecting microbes is
12938controlled, the manifestations of disease will
12939be minor. In other words, becoming exposed
12940to or infected with an infectious microbe does
12941not necessarily mean developing noticeable
12942disease.
12943So, one way of looking at severe infectious
12944diseases is that it represents a lack of success
12945of the immune system. The functioning of the
12946immune system, like any other system in our
12947body, will not be good if proper and sufficient
12948nourishment and food is not available.
12949Therefore, the second basic principle of
12950prevention of infectious disease is the
12951availability of proper and sufficient food for
12952everyone.
12953that group and not shared with the other
12954groups. These processes may be pathways for
12955the synthesis of new substances or respiration.
12956These pathways will not be used by us
12957either. For example, our cells may make new
12958substances by a mechanism different from
12959that used by bacteria. We have to find a drug
12960that blocks the bacterial synthesis pathway
12961without affecting our own. This is what is
12962achieved by the antibiotics that we are all
12963familiar with. Similarly, there are drugs that
12964kill protozoa such as the malarial parasite.
12965One reason why making anti-viral
12966medicines is harder than making antibacterial
12967medicines is that viruses have few
12968biochemical mechanisms of their own. They
12969enter our cells and use our machinery for
12970their life processes. This means that there
12971are relatively few virus-specific targets to aim
12972at. Despite this limitation, there are now
12973effective anti-viral drugs, for example, the
12974drugs that keep HIV infection under control.
1297513.3.5 PRINCIPLES OF PREVENTION
12976All of what we have talked about so far deals
12977with how to get rid of an infection in someone
12978who has the disease. But there are three
12979limitations of this approach to dealing with
12980infectious disease. The first is that once
12981someone has a disease, their body functions
12982are damaged and may never recover
12983completely. The second is that treatment will
12984take time, which means that someone
12985suffering from a disease is likely to be
12986bedridden for some time even if we can give
12987proper treatment. The third is that the person
12988suffering from an infectious disease can serve
12989as the source from where the infection may
12990spread to other people. This leads to the
12991multiplication of the above difficulties. It is
12992because of such reasons that prevention of
12993diseases is better than their cure.
12994How can we prevent diseases? There are
12995two ways, one general and one specific to each
12996disease. The general ways of preventing
12997infections mostly relate to preventing
12998exposure. How can we prevent exposure to
12999infectious microbes?
13000If we look at the means of their spreading,
13001© NCERT
13002not to be republished
13003186 SCIENCE
13004Activity _____________13.6
13005• Conduct a survey in your locality.
13006Talk to ten families who are well-off
13007and ten who are very poor (in your
13008estimation). Both sets of families
13009should have children who are below
13010five years of age. Measure the heights
13011of these children. Draw a graph of the
13012height of each child against its age
13013for both sets of families.
13014• Is there a difference between the
13015groups? If yes, why?
13016• If there is no difference, do you think
13017that your findings mean that being
13018well-off or poor does not matter for
13019health?
13020These are the general ways of preventing
13021infections. What are the specific ways? They
13022relate to a peculiar property of the immune
13023system that usually fights off microbial
13024infections. Let us cite an example to try and
13025understand this property.
13026These days, there is no smallpox
13027anywhere in the world. But as recently as a
13028hundred years ago, smallpox epidemics were
13029not at all uncommon. In such an epidemic,
13030people used to be very afraid of coming near
13031someone suffering from the disease since they
13032were afraid of catching the disease.
13033However, there was one group of people
13034who did not have this fear. These people would
13035provide nursing care for the victims of
13036smallpox. This was a group of people who had
13037had smallpox earlier and survived it, although
13038with a lot of scarring. In other words, if you
13039had smallpox once, there was no chance of
13040suffering from it again. So, having the disease
13041once was a means of preventing subsequent
13042attacks of the same disease.
13043This happens because when the immune
13044system first sees an infectious microbe, it
13045responds against it and then remembers it
13046specifically. So the next time that particular
13047microbe, or its close relatives enter the body,
13048the immune system responds with even
13049greater vigour. This eliminates the infection
13050even more quickly than the first time around.
13051This is the basis of the principle of
13052immunisation.
13053Immunisation
13054Traditional Indian and Chinese medicinal
13055systems sometimes deliberately rubbed the
13056skin crusts from smallpox victims into the
13057skin of healthy people. They thus hoped
13058to induce a mild form of smallpox that
13059would create resistance against the
13060disease.
13061Famously, two centuries ago, an
13062English physician
13063named Edward
13064Jenner, realised
13065that milkmaids
13066who had had
13067cowpox did not
13068catch smallpox
13069even during
13070epidemics.
13071Cowpox is a very
13072mild disease.
13073Jenner tried
13074deliberately giving
13075cowpox to people
13076(as he can be seen doing in the picture), and
13077found that they were now resistant to
13078smallpox. This was because the smallpox
13079virus is closely related to the cowpox virus.
13080‘Cow’ is ‘vacca’ in Latin, and cowpox is
13081‘vaccinia’. From these roots, the word
13082‘vaccination’ has come into our usage.
13083We can now see that, as a general principle,
13084we can ‘fool’ the immune system into
13085developing a memory for a particular infection
13086by putting something, that mimics the microbe
13087we want to vaccinate against, into the body.
13088This does not actually cause the disease but
13089this would prevent any subsequent exposure
13090to the infecting microbe from turning into
13091actual disease.
13092Many such vaccines are now available for
13093preventing a whole range of infectious
13094diseases, and provide a disease-specific
13095means of prevention. There are vaccines
13096against tetanus, diphtheria, whooping cough,
13097measles, polio and many others. These form
13098the public health programme of childhood
13099© NCERT
13100not to be republished
13101WHY DO WE FALL ILL 187
13102immunisation for preventing infectious
13103diseases.
13104Of course, such a programme can be
13105useful only if such health measures are
13106available to all children. Can you think of
13107reasons why this should be so?
13108Some hepatitis viruses, which cause
13109jaundice, are transmitted through water.
13110There is a vaccine for one of them, hepatitis
13111A, in the market. But the majority of children
13112in many parts of India are already immune
13113to hepatitis A by the time they are five years
13114old. This is because they are exposed to the
13115virus through water. Under these
13116circumstances, would you take the vaccine?
13117Activity _____________ 13.7
13118• Rabies virus is spread by the bite of
13119infected dogs and other animals. There
13120are anti-rabies vaccines for both
13121humans and animals. Find out the
13122plan of your local authority for
13123Q
13124the control of rabies in your
13125neighbourhood. Are these measures
13126adequate? If not, what improvements
13127would you suggest?
13128uestions
131291. Why are we normally advised to
13130take bland and nourishing food
13131when we are sick?
131322. What are the different means by
13133which infectious diseases are
13134spread?
131353. What precautions can you take
13136in your school to reduce the
13137incidence of infectious diseases?
131384. What is immunisation?
131395. What are the immunisation
13140programmes available at the
13141nearest health centre in your
13142locality? Which of these diseases
13143are the major health problems in
13144your area?
13145What
13146you have you have
13147learnt
13148• Health is a state of physical, mental and social well-being.
13149• The health of an individual is dependent on his/her physical
13150surroundings and his/her economic status.
13151• Diseases are classified as acute or chronic, depending on their
13152duration.
13153• Disease may be due to infectious or non-infectious causes.
13154• Infectious agents belong to different categories of organisms
13155and may be unicellular and microscopic or multicellular.
13156• The category to which a disease-causing organism belongs
13157decides the type of treatment.
13158• Infectious agents are spread through air, water, physical contact
13159or vectors.
13160• Prevention of disease is more desirable than its successful
13161treatment.
13162• Infectious diseases can be prevented by public health hygiene
13163measures that reduce exposure to infectious agents.
13164© NCERT
13165not to be republished
13166188 SCIENCE
13167• Infectious diseases can also be prevented by using
13168immunisation.
13169• Effective prevention of infectious diseases in the community
13170requires that everyone should have access to public hygiene
13171and immunisation.
13172Exercises Exercises Exercises
131731. How many times did you fall ill in the last one year? What were
13174the illnesses?
13175(a) Think of one change you could make in your habits in order
13176to avoid any of/most of the above illnesses.
13177(b) Think of one change you would wish for in your
13178surroundings in order to avoid any of/most of the above
13179illnesses.
131802. A doctor/nurse/health-worker is exposed to more sick people
13181than others in the community. Find out how she/he avoids
13182getting sick herself/himself.
131833. Conduct a survey in your neighbourhood to find out what the
13184three most common diseases are. Suggest three steps that could
13185be taken by your local authorities to bring down the incidence
13186of these diseases.
131874. A baby is not able to tell her/his caretakers that she/he is
13188sick. What would help us to find out
13189(a) that the baby is sick?
13190(b) what is the sickness?
131915. Under which of the following conditions is a person most likely
13192to fall sick?
13193(a) when she is recovering from malaria.
13194(b) when she has recovered from malaria and is taking care of
13195someone suffering from chicken-pox.
13196(c) when she is on a four-day fast after recovering from malaria
13197and is taking care of someone suffering from chicken-pox.
13198Why?
131996. Under which of the following conditions are you most likely to
13200fall sick?
13201(a) when you are taking examinations.
13202(b) when you have travelled by bus and train for two days.
13203(c) when your friend is suffering from measles.
13204© NCERT
13205Why?
13206not to be republished
13207Our planet, Earth is the only one on which
13208life, as we know it, exists. Life on Earth is
13209dependent on many factors. Most life-forms
13210we know need an ambient temperature,
13211water, and food. The resources available on
13212the Earth and the energy from the Sun are
13213necessary to meet the basic requirements of
13214all life-forms on the Earth.
13215What are these resources on the Earth?
13216These are the land, the water and the air.
13217The outer crust of the Earth is called the
13218lithosphere. Water covers 75% of the Earth’s
13219surface. It is also found underground. These
13220comprise the hydrosphere. The air that covers
13221the whole of the Earth like a blanket, is called
13222the atmosphere. Living things are found
13223where these three exist. This life-supporting
13224zone of the Earth where the atmosphere, the
13225hydrosphere and the lithosphere interact and
13226make life possible, is known as the biosphere.
13227Living things constitute the biotic
13228component of the biosphere. The air, the
13229water and the soil form the non-living or
13230abiotic component of the biosphere. Let us
13231study these abiotic components in detail in
13232order to understand their role in sustaining
13233life on Earth.
1323414.1 The Breath of Life: Air The Breath of Life: Air
13235We have already talked about the composition
13236of air in the first chapter. It is a mixture of
13237many gases like nitrogen, oxygen, carbon
13238dioxide and water vapour. It is interesting to
13239note that even the composition of air is the
13240result of life on Earth. In planets such as
13241Venus and Mars, where no life is known to
13242exist, the major component of the atmosphere
13243is found to be carbon dioxide. In fact, carbon
13244dioxide constitutes up to 95-97% of the
13245atmosphere on Venus and Mars.
13246Eukaryotic cells and many prokaryotic
13247cells, discussed in Chapter 5, need oxygen to
13248break down glucose molecules and get energy
13249for their activities. This results in the
13250production of carbon dioxide. Another process
13251which results in the consumption of oxygen
13252and the concomitant production of carbon
13253dioxide is combustion. This includes not just
13254human activities, which burn fuels to get
13255energy, but also forest fires.
13256Despite this, the percentage of carbon
13257dioxide in our atmosphere is a mere fraction
13258of a percent because carbon dioxide is ‘fixed’
13259in two ways: (i) Green plants convert carbon
13260dioxide into glucose in the presence of
13261Sunlight and (ii) many marine animals use
13262carbonates dissolved in sea-water to make
13263their shells.
1326414.1.1 THE ROLE OF THE ATMOSPHERE IN
13265CLIMATE CONTROL
13266We have talked of the atmosphere covering
13267the Earth, like a blanket. We know that air is
13268a bad conductor of heat. The atmosphere
13269keeps the average temperature of the Earth
13270fairly steady during the day and even during
13271the course of the whole year. The atmosphere
13272prevents the sudden increase in temperature
13273during the daylight hours. And during the
13274night, it slows down the escape of heat into
13275outer space. Think of the moon, which is
13276about the same distance from the Sun that
13277the Earth is. Despite that, on the surface of
13278the moon, with no atmosphere, the
13279temperature ranges from –190º C to 110º C.
1328014
13281NATURAL RESOURCES ESOURCES
13282Chapter
13283© NCERT
13284not to be republished
13285190 SCIENCE
13286the heating of water bodies and the activities
13287of living organisms. The atmosphere can be
13288heated from below by the radiation that is
13289reflected back or re-radiated by the land or
13290water bodies. On being heated, convection
13291currents are set up in the air. In order to gain
13292some understanding of the nature of
13293convection currents, let us perform the
13294following activity:
13295Activity _____________ 14.2
13296• Place a candle in a beaker or widemouthed
13297bottle and light it. Light an
13298incense stick and take it to the mouth
13299of the above bottle (Figure 14.1).
13300• Which way does the smoke flow when
13301the incense stick is kept near the edge
13302of the mouth?
13303• Which way does the smoke flow when
13304the incense stick is kept a little above
13305the candle?
13306• Which way does the smoke flow when
13307the incense stick is kept in other
13308regions?
13309Activity _____________ 14.1
13310• Measure the temperature of the
13311following :
13312Take (i) a beaker full of water, (ii) a
13313beaker full of soil/sand and (iii) a closed
13314bottle containing a thermometer. Keep
13315them in bright Sunlight for three hours.
13316Now measure the temperature of all 3
13317vessels. Also, take the temperature
13318reading in shade at the same time.
13319Now answer
133201. Is the temperature reading more in
13321activity (i) or (ii)?
133222. Based on the above finding, which
13323would become hot faster – the land or
13324the sea?
133253. Is the thermometer reading of the
13326temperature of air (in shade) the same
13327as the temperature of sand or water?
13328What do you think is the reason for
13329this? And why does the temperature
13330have to be measured in the shade?
133314. Is the temperature of air in the closed
13332glass vessel/bottle the same as the
13333temperature taken in open air? (i) What
13334do you think is the reason for this?
13335(ii) Do we ever come across this
13336phenomenon in daily life?
13337As we have seen above, sand and water
13338do not heat up at the same rate. What do you
13339think will be their rates of cooling? Can we
13340think of an experiment to test the prediction?
1334114.1.2 THE MOVEMENT OF AIR: WINDS
13342We have all felt the relief brought by cool
13343evening breezes after a hot day. And
13344sometimes, we are lucky enough to get rains
13345after some days of really hot weather. What
13346causes the movement of air, and what decides
13347whether this movement will be in the form of
13348a gentle breeze, a strong wind or a terrible
13349storm? What brings us the welcome rains?
13350All these phenomena are the result of
13351changes that take place in our atmosphere
13352due to the heating of air and the formation of
13353water vapour. Water vapour is formed due to
13354Fig. 14.1: Air currents being caused by the uneven
13355heating of air.
13356The patterns revealed by the smoke show
13357us the directions in which hot and cold air
13358move. In a similar manner, when air is heated
13359by radiation from the heated land or water, it
13360rises. But since land gets heated faster than
13361water, the air over land would also be heated
13362faster than the air over water bodies.
13363So, if we look at the situation in coastal
13364regions during the day, the air above the land
13365© NCERT
13366not to be republished
13367NATURAL RESOURCES 191
13368gets heated faster and starts rising. As this
13369air rises, a region of low pressure is created
13370and air over the sea moves into this area of
13371low pressure. The movement of air from one
13372region to the other creates winds. During the
13373day, the direction of the wind would be from
13374the sea to the land.
13375At night, both land and sea start to cool.
13376Since water cools down slower than the land,
13377the air above water would be warmer than
13378the air above land.
13379On the basis of the above discussion, what
13380can you say about:
133811. the appearance of areas of low and
13382high pressure in coastal areas at night?
133832. the direction in which air would flow
13384at night in coastal areas?
13385Similarly, all the movements of air
13386resulting in diverse atmospheric phenomena
13387are caused by the uneven heating of the
13388atmosphere in different regions of the Earth.
13389But various other factors also influence these
13390winds – the rotation of the Earth and the
13391presence of mountain ranges in the paths of
13392the wind are a couple of these factors. We
13393will not go into these factors in detail in this
13394chapter, but think about this: how do the
13395presence of the Himalayas change the flow of
13396a wind blowing from Allahabad towards the
13397north?
1339814.1.3 RAIN
13399Let us go back now to the question of how
13400clouds are formed and bring us rain. We could
13401start by doing a simple experiment which
13402demonstrates some of the factors influencing
13403these climatic changes.
13404Activity _____________ 14.3
13405• Take an empty bottle of the sort in
13406which bottled water is sold. Pour about
134075-10 mL of water into it and close the
13408bottle tightly. Shake it well or leave it
13409out in the Sun for ten minutes. This
13410causes the air in the bottle to be
13411saturated with water vapour.
13412• Now, take a lighted incense stick. Open
13413the cap of the bottle and allow some of
13414the smoke from the incense stick to
13415enter the bottle. Quickly close the bottle
13416once more. Make sure that the cap is
13417fitting tightly. Press the bottle hard
13418between your hands and crush it as
13419much as possible. Wait for a few
13420seconds and release the bottle. Again
13421press the bottle as hard as you can.
13422Now answer
134231. When did you observe that the air
13424inside seemed to become ‘foggy’?
134252. When does this fog disappear?
134263. When is the pressure inside the bottle
13427higher?
134284. Is the ‘fog’ observed when the pressure
13429in the bottle is high or when it is low?
134305. What is the need for smoke particles
13431inside the bottle for this experiment?
134326. What might happen if you do the
13433experiment without the smoke from the
13434incense stick? Now try it and check if
13435the prediction was correct. What might
13436be happening in the above experiment
13437in the absence of smoke particles?
13438The above experiment replicates, on a very
13439small scale, what happens when air with a
13440very high content of water vapour goes from
13441a region of high pressure to a region of low
13442pressure or vice versa.
13443When water bodies are heated during the
13444day, a large amount of water evaporates and
13445goes into the air. Some amount of water
13446vapour also get into the atmosphere because
13447of various biological activities. This air also
13448gets heated. The hot air rises up carrying the
13449water vapour with it. As the air rises, it
13450expands and cools. This cooling causes the
13451water vapour in the air to condense in the
13452form of tiny droplets. This condensation of
13453water is facilitated if some particles could act
13454as the ‘nucleus’ for these drops to form
13455around. Normally dust and other suspended
13456particles in the air perform this function.
13457Once the water droplets are formed, they
13458grow bigger by the ‘condensation’ of these
13459water droplets. When the drops have grown
13460big and heavy, they fall down in the form of
13461rain. Sometimes, when the temperature of air
13462© NCERT
13463not to be republished
13464192 SCIENCE
1346514.1.4 AIR POLLUTION
13466We keep hearing of the increasing levels of
13467oxides of nitrogen and sulphur in the news.
13468People often bemoan the fact that the quality
13469of air has gone down since their childhood.
13470How is the quality of air affected and how
13471does this change in quality affect us and other
13472life forms?
13473The fossil fuels like coal and petroleum
13474contain small amounts of nitrogen and
13475sulphur. When these fuels are burnt, nitrogen
13476and sulphur too are burnt and this produces
13477different oxides of nitrogen and sulphur. Not
13478only is the inhalation of these gases
13479dangerous, they also dissolve in rain to give
13480rise to acid rain. The combustion of fossil fuels
13481also increases the amount of suspended
13482particles in air. These suspended particles
13483could be unburnt carbon particles or
13484substances called hydrocarbons. Presence of
13485high levels of all these pollutants cause
13486visibility to be lowered, especially in cold
13487weather when water also condenses out of
13488air. This is known as smog and is a visible
13489indication of air pollution. Studies have
13490shown that regularly breathing air that
13491contains any of these substances increases
13492the incidence of allergies, cancer and heart
13493diseases. An increase in the content of these
13494harmful substances in air is called air
13495pollution.
13496is low enough, precipitation may occur in the
13497form of snow, sleet or hail.
13498Rainfall patterns are decided by the
13499prevailing wind patterns. In large parts of
13500India, rains are mostly brought by the southwest
13501or north-east monsoons. We have also
13502heard weather reports that say ‘depressions’
13503in the Bay of Bengal have caused rains in
13504some areas (Figure 14.2).
13505Activity _____________ 14.4
13506• Collect information from newspapers
13507or weather reports on television about
13508rainfall patterns across the country.
13509Also find out how to construct a raingauge
13510and make one. What precautions
13511are necessary in order to get reliable
13512data from this rain-gauge? Now answer
13513the following questions :
13514• In which month did your city/town/
13515village get the maximum rainfall?
13516• In which month did your state/union
13517territory get the maximum rainfall?
13518• Is rain always accompanied by thunder
13519and lightning? If not, in which season
13520do you get more of thunder and
13521lightning with the rain?
13522Activity _____________ 14.5
13523• Find out more about monsoons and
13524cyclones from the library. Try and find
13525out the rainfall pattern of any other
13526country. Is the monsoon responsible for
13527rains the world over?
13528Fig. 14.2: Satellite picture showing clouds over India.
13529Fig. 14.3: Lichen
13530© NCERT
13531not to be republished
13532NATURAL RESOURCES 193
13533people are forced to spend considerable
13534amounts of time in fetching water from faraway
13535sources.
13536Activity _____________ 14.7
13537• Many municipal corporations are trying
13538water-harvesting techniques to
13539improve the availability of water.
13540• Find out what these techniques are and
13541how they would increase the water that
13542is available for use.
13543But why is water so necessary? And do
13544all organisms require water? All cellular
13545processes take place in a water medium. All
13546the reactions that take place within our body
13547and within the cells occur between
13548substances that are dissolved in water.
13549Substances are also transported from one
13550part of the body to the other in a dissolved
13551form. Hence, organisms need to maintain the
13552level of water within their bodies in order to
13553stay alive. Terrestrial life-forms require fresh
13554water for this because their bodies cannot
13555tolerate or get rid of the high amounts of
13556dissolved salts in saline water. Thus, water
13557sources need to be easily accessible for
13558animals and plants to survive on land.
13559Activity _____________ 14.8
13560• Select a small area (say, 1 m2) near a
13561water-body, it may be a river, stream,
13562lake or pond. Count the number of
13563different animals and plants in this
13564area. Also, check the number of
13565individuals of each type or species.
13566• Compare this with the number of
13567individuals (both animals and plants)
13568found in an area of the same size in a
13569dry, rocky region.
13570• Is the variety of plant and animal life
13571the same in both these areas?
13572Activity _____________ 14.9
13573• Select and mark out a small area (about
135741 m2) in some unused land in or near
13575your school.
13576• As in the above activity, count the
13577number of different animals and plants
13578in this area and the number of
13579individuals of each species.
13580Activity _____________ 14.6
13581• Organisms called lichens are found to
13582be very sensitive to the levels of
13583contaminants like sulphur dioxide in
13584the air. As discussed earlier in section
135857.3.3, lichens can be commonly found
13586growing on the barks of trees as a thin
13587greenish-white crust. See if you can
13588find lichen growing on the trees in your
13589locality.
13590• Compare the lichen on trees near busy
13591roads and trees some distance away.
13592• On the trees near roads, compare the
13593incidence of lichen on the side facing
13594the road and on the side away from the
13595road.
13596What can you say about the levels of
13597polluting substances near roads and away
13598from roads on the basis of your findings
13599above?
13600uestions
136011. How is our atmosphere different
13602from the atmospheres on Venus
13603and Mars?
136042. How does the atmosphere act as
13605a blanket?
136063. What causes winds?
136074. How are clouds formed?
136085. List any three human activities
13609that you think would lead to air
13610pollution.
1361114.2 Water: A Wonder Liquid
13612Water occupies a very large area of the Earth’s
13613surface and is also found underground. Some
13614amount of water exists in the form of water
13615vapour in the atmosphere. Most of the water
13616on Earth’s surface is found in seas and oceans
13617and is saline. Fresh water is found frozen in
13618the ice-caps at the two poles and on snowcovered
13619mountains. The underground water
13620and the water in rivers, lakes and ponds is
13621also fresh. However, the availability of fresh
13622water varies from place to place. Practically
13623every summer, most places have to face a
13624shortage of water. And in rural areas, where
13625water supply systems have not been installed,
13626Q
13627© NCERT
13628not to be republished
13629194 SCIENCE
13630• Remember to do this in the same place
13631twice in a year, once during summer
13632or the dry season and once after it has
13633rained.
13634Now answer
136351. Were the numbers similar both times?
136362. In which season did you find more
13637variety of plants and animals?
136383. In which season did you find more
13639number of individuals of each variety?
13640After compiling the results of the above
13641two activities, think if there is any relationship
13642between the amount of available water and
13643the number and variety of plants and animals
13644that can live in a given area. If there is a
13645relationship, where do you think you would
13646find a greater variety and abundance of life –
13647in a region that receives 5 cm of rainfall in a
13648year or a region that receives 200 cm of
13649rainfall in a year? Find the map showing
13650rainfall patterns in the atlas and predict
13651which States in India would have the
13652maximum biodiversity and which would have
13653the least. Can we think of any way of checking
13654whether the prediction is correct?
13655The availability of water decides not only
13656the number of individuals of each species that
13657are able to survive in a particular area, but it
13658also decides the diversity of life there. Of
13659course, the availability of water is not the only
13660factor that decides the sustainability of life
13661in a region. Other factors like the temperature
13662and nature of soil also matter. But water is
13663one of the major resources which determine
13664life on land.
1366514.2.1 WATER POLLUTION
13666Water dissolves the fertilisers and pesticides
13667that we use on our farms. So some percentage
13668of these substances are washed into the water
13669bodies. Sewage from our towns and cities and
13670the waste from factories are also dumped into
13671rivers or lakes. Specific industries also use
13672water for cooling in various operations and
13673later return this hot water to water-bodies.
13674Another manner in which the temperature of
13675the water in rivers can be affected is when water
13676is released from dams. The water inside the
13677deep reservoir would be colder than the water
13678at the surface which gets heated by the Sun.
13679All this can affect the life-forms that are
13680found in these water bodies in various ways.
13681It can encourage the growth of some life-forms
13682and harm some other life-forms. This affects
13683the balance between various organisms which
13684had been established in that system. So we
13685use the term water-pollution to cover the
13686following effects:
136871. The addition of undesirable
13688substances to water-bodies. These
13689substances could be the fertilisers and
13690pesticides used in farming or they
13691could be poisonous substances, like
13692mercury salts which are used by
13693paper-industries. These could also be
13694disease-causing organisms, like the
13695bacteria which cause cholera.
136962. The removal of desirable substances
13697from water-bodies. Dissolved oxygen
13698is used by the animals and plants that
13699live in water. Any change that reduces
13700the amount of this dissolved oxygen
13701would adversely affect these aquatic
13702organisms. Other nutrients could also
13703be depleted from the water bodies.
137043. A change in temperature. Aquatic
13705organisms are used to a certain range
13706of temperature in the water-body
13707where they live, and a sudden marked
13708change in this temperature would be
13709dangerous for them or affect their
13710breeding. The eggs and larvae of
13711various animals are particularly
13712susceptible to temperature changes.
13713uestions
137141. Why do organisms need water?
137152. What is the major source of fresh
13716water in the city/town/village
13717where you live?
137183. Do you know of any activity
13719which may be polluting this water Q source?
13720© NCERT
13721not to be republished
13722NATURAL RESOURCES 195
1372314.3 Mineral Riches in the Soil Mineral Riches in the Soil
13724Soil is an important resource that decides the
13725diversity of life in an area. But what is the
13726soil and how is it formed? The outermost layer
13727of our Earth is called the crust and the
13728minerals found in this layer supply a variety
13729of nutrients to life-forms. But these minerals
13730will not be available to the organisms if the
13731minerals are bound up in huge rocks. Over
13732long periods of time, thousands and millions
13733of years, the rocks at or near the surface of
13734the Earth are broken down by various
13735physical, chemical and some biological
13736processes. The end product of this breaking
13737down is the fine particles of soil. But what
13738are the factors or processes that make soil?
13739• The Sun: The Sun heats up rocks
13740during the day so that they expand.
13741At night, these rocks cool down and
13742contract. Since all parts of the rock
13743do not expand and contract at the
13744same rate, this results in the
13745formation of cracks and ultimately the
13746huge rocks break up into smaller
13747pieces.
13748• Water: Water helps in the formation
13749of soil in two ways. One, water could
13750get into the cracks in the rocks formed
13751due to uneven heating by the Sun. If
13752this water later freezes, it would cause
13753the cracks to widen. Can you think
13754why this should be so? Two, flowing
13755water wears away even hard rock over
13756long periods of time. Fast flowing water
13757often carries big and small particles
13758of rock downstream. These rocks rub
13759against other rocks and the resultant
13760abrasion causes the rocks to wear
13761down into smaller and smaller
13762particles. The water then takes these
13763particles along with it and deposits it
13764further down its path. Soil is thus
13765found in places far away from its
13766parent-rock.
13767• Wind: In a process similar to the way
13768in which water rubs against rocks and
13769wears them down, strong winds also
13770erode rocks down. The wind also
13771carries sand from one place to the
13772other like water does.
13773• Living organisms also influence the
13774formation of soil. The lichen that we
13775read about earlier, also grows on the
13776surface of rocks. While growing, they
13777release certain substances that cause
13778the rock surface to powder down and
13779form a thin layer of soil. Other small
13780plants like moss, are able to grow on
13781this surface now and they cause the
13782rock to break up further. The roots of
13783big trees sometimes go into cracks in
13784the rocks and as the roots grow bigger,
13785the crack is forced bigger.
13786Activity ____________14.10
13787• Take some soil and put it into a beaker
13788containing water. The water should be
13789at least five times the amount of soil
13790taken. Stir the soil and water vigorously
13791and allow the soil to settle down.
13792Observe after some time.
13793• Is the soil at the bottom of the beaker
13794homogenous or have layers formed?
13795• If layers have formed, how is one layer
13796different from another?
13797• Is there anything floating on the
13798surface of the water?
13799• Do you think some substances would
13800have dissolved in the water? How would
13801you check?
13802As you have seen, soil is a mixture. It
13803contains small particles of rock (of different
13804sizes). It also contains bits of decayed living
13805organisms which is called humus. In addition,
13806soil also contains various forms of
13807microscopic life. The type of soil is decided
13808by the average size of particles found in it
13809and the quality of the soil is decided by the
13810amount of humus and the microscopic
13811organisms found in it. Humus is a major
13812factor in deciding the soil structure because
13813it causes the soil to become more porous and
13814allows water and air to penetrate deep
13815underground. The mineral nutrients that are
13816found in a particular soil depends on the
13817rocks it was formed from. The nutrient
13818content of a soil, the amount of humus
13819present in it and the depth of the soil are
13820© NCERT
13821not to be republished
13822196 SCIENCE
13823some of the factors that decide which plants
13824will thrive on that soil. Thus, the topmost
13825layer of the soil that contains humus and
13826living organisms in addition to the soil
13827particles is called the topsoil. The quality of
13828the topsoil is an important factor that decides
13829biodiversity in that area.
13830Modern farming practices involve the use
13831of large amounts of fertilizers and pesticides.
13832Use of these substances over long periods of
13833time can destroy the soil structure by killing
13834the soil micro-organisms that recycle
13835nutrients in the soil. It also kills the
13836Earthworms which are instrumental in
13837making the rich humus. Fertile soil can
13838quickly be turned barren if sustainable
13839practices are not followed. Removal of useful
13840components from the soil and addition of
13841other substances, which adversely affect the
13842fertility of the soil and kill the diversity of
13843organisms that live in it, is called soil pollution.
13844The soil that we see today in one place
13845has been created over a very long period of
13846time. However, some of the factors that
13847created the soil in the first place and brought
13848the soil to that place may be responsible for
13849the removal of the soil too. The fine particles
13850of soil may be carried away by flowing water
13851or wind. If all the soil gets washed away and
13852the rocks underneath are exposed, we have
13853lost a valuable resource because very little
13854will grow on the rock.
13855Activity ____________14.11
13856• Take two identical trays and fill them
13857with soil. Plant mustard or green gram
13858or paddy in one of the trays and water
13859both the trays regularly for a few days,
13860till the first tray is covered by plant
13861growth. Now, tilt both the trays and fix
13862them in that position. Make sure that
13863both the trays are tilted at the same
13864angle. Pour equal amount of water
13865gently on both trays such that the water
13866flows out of the trays (Fig. 14.4).
13867• Study the amount of soil that is carried
13868out of the trays. Is the amount the
13869same in both the trays?
13870• Now pour equal amounts of water on
13871both the trays from a height. Pour three
13872or four times the amount that you
13873poured earlier.
13874• Study the amount of soil that is
13875carried out of the trays now. Is the
13876amount the same in both the trays?
13877• Is the amount of soil that is carried out
13878more or less or equal to the amount
13879washed out earlier?
13880Fig. 14.4: Effect of flowing water on the top-soil
13881The roots of plants have an important role
13882in preventing soil erosion. The large-scale
13883deforestation that is happening all over the
13884world not only destroys biodiversity, it also
13885leads to soil erosion. Topsoil that is bare of
13886vegetation, is likely to be removed very
13887quickly. And this is accelerated in hilly or
13888mountainous regions. This process of soil
13889erosion is very difficult to reverse. Vegetative
13890cover on the ground has a role to play in the
13891percolation of water into the deeper
13892layers too.
13893uestions
138941. How is soil formed?
138952. What is soil erosion?
138963. What are the methods of
13897preventing or reducing soil
13898erosion?
1389914.4 Biogeochemical Cycles Biogeochemical Cycles
13900A constant interaction between the biotic and
13901abiotic components of the biosphere makes
13902it a dynamic, but stable system. These
13903interactions consist of a transfer of matter
13904and energy between the different components
13905of the biosphere. Let us look at some
13906processes involved in the maintenance of the
13907above balance.
13908Q
13909© NCERT
13910not to be republished
13911NATURAL RESOURCES 197
1391214.4.1 THE WATER-CYCLE
13913You have seen how the water evaporates from
13914the water bodies and subsequent
13915condensation of this water vapour leads to
13916rain. But we don’t see the seas and oceans
13917drying up. So, how is the water returning to
13918these water bodies? The whole process in
13919which water evaporates and falls on the land
13920as rain and later flows back into the sea via
13921rivers is known as the water-cycle. This cycle
13922is not as straight-forward and simple as this
13923statement seems to imply. All of the water
13924that falls on the land does not immediately
13925flow back into the sea. Some of it seeps into
13926the soil and becomes part of the underground
13927reservoir of fresh-water. Some of this
13928underground water finds its way to the
13929surface through springs. Or we bring it to
13930the surface for our use through wells or tubewells.
13931Water is also used by terrestrial animals
13932and plants for various life-processes
13933(Fig. 14.5).
13934water. Thus rivers carry many nutrients from
13935the land to the sea, and these are used by
13936the marine organisms.
1393714.4.2 THE NITROGEN-CYCLE
13938Nitrogen gas makes up 78% of our
13939atmosphere and nitrogen is also a part of
13940many molecules essential to life like proteins,
13941nucleic acids (DNA and RNA) and some
13942vitamins. Nitrogen is found in other
13943biologically important compounds such as
13944alkaloids and urea too. Nitrogen is thus an
13945essential nutrient for all life-forms and life
13946would be simple if all these life-forms could
13947use the atmospheric nitrogen directly.
13948However, other than a few forms of bacteria,
13949life-forms are not able to convert the
13950comparatively inert nitrogen molecule into
13951forms like nitrates and nitrites which can be
13952taken up and used to make the required
13953molecules. These ‘nitrogen-fixing’ bacteria
13954may be free-living or be associated with some
13955species of dicot plants. Most commonly, the
13956nitrogen-fixing bacteria are found in the roots
13957of legumes (generally the plants which give
13958us pulses) in special structures called rootnodules.
13959Other than these bacteria, the only
13960other manner in which the nitrogen molecule
13961is converted to nitrates and nitrites is by a
13962physical process. During lightning, the high
13963temperatures and pressures created in the
13964air convert nitrogen into oxides of nitrogen.
13965These oxides dissolve in water to give nitric
13966and nitrous acids and fall on land along with
13967rain. These are then utilised by various lifeforms.
13968What happens to the nitrogen once it is
13969converted into forms that can be taken up
13970and used to make nitrogen-containing
13971molecules? Plants generally take up nitrates
13972and nitrites and convert them into amino
13973acids which are used to make proteins. Some
13974other biochemical pathways are used to make
13975the other complex compounds containing
13976nitrogen. These proteins and other complex
13977compounds are subsequently consumed by
13978animals. Once the animal or the plant dies,
13979other bacteria in the soil convert the various
13980compounds of nitrogen back into nitrates and
13981Fig. 14.5: Water-cycle in nature
13982Let us look at another aspect of what
13983happens to water during the water-cycle. As
13984you know, water is capable of dissolving a
13985large number of substances. As water flows
13986through or over rocks containing soluble
13987minerals, some of them get dissolved in the
13988© NCERT
13989not to be republished
13990198 SCIENCE
13991nitrites. A different type of bacteria converts
13992the nitrates and nitrites into elemental
13993nitrogen. Thus, there is a nitrogen-cycle in
13994nature in which nitrogen passes from its
13995elemental form in the atmosphere into simple
13996molecules in the soil and water, which get
13997converted to more complex molecules in living
13998beings and back again to the simple nitrogen
13999molecule in the atmosphere.
1400014.4.3 THE CARBON-CYCLE
14001Carbon is found in various forms on the
14002Earth. It occurs in the elemental form as
14003diamonds and graphite. In the combined
14004state, it is found as carbon dioxide in the
14005atmosphere, as carbonate and hydrogencarbonate
14006salts in various minerals, while all
14007life-forms are based on carbon-containing
14008molecules like proteins, carbohydrates, fats,
14009Fig.14.6: Nitrogen-cycle in nature
14010nucleic acids and vitamins. The endoskeletons
14011and exoskeletons of various animals are also
14012formed from carbonate salts. Carbon is
14013incorporated into life-forms through the basic
14014process of photosynthesis which is performed
14015in the presence of Sunlight by all life-forms that
14016contain chlorophyll. This process converts
14017carbon dioxide from the atmosphere or
14018dissolved in water into glucose molecules.
14019These glucose molecules are either converted
14020into other substances or used to provide
14021energy for the synthesis of other biologically
14022important molecules (Fig. 14.7).
14023The utilisation of glucose to provide energy
14024to living things involves the process of
14025respiration in which oxygen may or may not
14026be used to convert glucose back into carbon
14027dioxide. This carbon dioxide then goes back
14028into the atmosphere. Another process that
14029© NCERT
14030not to be republished
14031NATURAL RESOURCES 199
14032adds to the carbon dioxide in the atmosphere
14033is the process of combustion where fuels are
14034burnt to provide energy for various needs like
14035heating, cooking, transportation and
14036industrial processes. In fact, the percentage
14037of carbon dioxide in the atmosphere is said
14038to have doubled since the industrial
14039revolution when human beings started
14040burning fossil fuels on a very large scale.
14041Carbon, like water, is thus cycled repeatedly
14042through different forms by the various
14043physical and biological activities.
1404414.4.3 ( 14.4.3 (i) THE GREENHOUSE EFFECT
14045Recall the reading taken by you under (iii) in
14046Activity 14.1. Heat is trapped by glass, and
14047hence the temperature inside a glass
14048enclosure will be much higher than the
14049surroundings. This phenomenon was used
14050to create an enclosure where tropical plants
14051Fig. 14.7: Carbon-cycle in nature
14052could be kept warm during the winters in
14053colder climates. Such enclosures are called
14054greenhouses. Greenhouses have also lent
14055their name to an atmospheric phenomenon.
14056Some gases prevent the escape of heat from
14057the Earth. An increase in the percentage of
14058such gases in the atmosphere would cause
14059the average temperatures to increase worldwide
14060and this is called the greenhouse effect.
14061Carbon dioxide is one of the greenhouse
14062gases. An increase in the carbon dioxide
14063content in the atmosphere would cause more
14064heat to be retained by the atmosphere and
14065lead to global warming.
14066Activity ____________14.12
14067• Find out what the consequences of
14068global warming would be.
14069• Also, find out the names of some other
14070greenhouse gases.
14071© NCERT
14072not to be republished
14073200 SCIENCE
1407414.4.4 THE OXYGEN-CYCLE
14075Oxygen is a very abundant element on our
14076Earth. It is found in the elemental form in
14077the atmosphere to the extent of 21%. It also
14078occurs extensively in the combined form in
14079the Earth’s crust as well as also in the air in
14080the form of carbon dioxide. In the crust, it is
14081found as the oxides of most metals and
14082silicon, and also as carbonate, sulphate,
14083nitrate and other minerals. It is also an
14084essential component of most biological
14085molecules like carbohydrates, proteins,
14086nucleic acids and fats (or lipids).
14087But when we talk of the oxygen-cycle, we
14088are mainly referring to the cycle that
14089maintains the levels of oxygen in the
14090atmosphere. Oxygen from the atmosphere is
14091used up in three processes, namely
14092combustion, respiration and in the formation
14093of oxides of nitrogen. Oxygen is returned to
14094the atmosphere in only one major process,
14095that is, photosynthesis. And this forms the
14096broad outline of the oxygen-cycle in nature
14097(Fig. 14.8).
14098bacteria, are poisoned by elemental oxygen.
14099In fact, even the process of nitrogen-fixing by
14100bacteria does not take place in the presence
14101of oxygen.
1410214.5 Ozone Layer Ozone Layer
14103Elemental oxygen is normally found in the
14104form of a diatomic molecule. However, in the
14105upper reaches of the atmosphere, a molecule
14106containing three atoms of oxygen is found.
14107This would mean a formula of O3
14108 and this is
14109called ozone. Unlike the normal diatomic
14110molecule of oxygen, ozone is poisonous and
14111we are lucky that it is not stable nearer to
14112the Earth’s surface. But it performs an
14113essential function where it is found. It absorbs
14114harmful radiations from the Sun. This
14115prevents those harmful radiations from
14116reaching the surface of the Earth where they
14117may damage many forms of life.
14118Recently it was discovered that this ozone
14119layer was getting depleted. Various man-made
14120compounds like CFCs (carbon compounds
14121having both fluorine and chlorine which are
14122very stable and not degraded by any biological
14123process) were found to persist in the
14124atmosphere. Once they reached the ozone
14125layer, they would react with the ozone
14126molecules. This resulted in a reduction of the
14127ozone layer and recently they have discovered
14128a hole in the ozone layer above the Antartica.
14129It is difficult to imagine the consequences for
14130life on Earth if the ozone layer dwindles
14131further, but many people think that it would
14132be better not to take chances. These people
14133advocate working towards stopping all further
14134damage to the ozone layer.
14135Fig. 14.8: Oxygen-cycle in nature
14136Though we usually think of oxygen as
14137being necessary to life in the process of
14138respiration, it might be of interest to you to
14139learn that some forms of life, especially
14140Fig. 14.9: Satellite picture showing the hole (magenta
14141colour) in the ozone layer over Antartica
14142October
141431980
14144October
141451985
14146October
141471990
14148© NCERT
14149not to be republished
14150NATURAL RESOURCES 201
14151Activity ____________14.13
14152• Find out which other molecules are
14153thought to damage the ozone layer.
14154• Newspaper reports often talk about the
14155hole in the ozone layer.
14156• Find out whether the size of this hole
14157is changing and in what manner
14158scientists think this would affect life
14159on Earth (Fig. 14.9).
14160Q
14161What
14162you have you have
14163learnt
14164• Life on Earth depends on resources like soil, water and air,
14165and energy from the Sun.
14166• Uneven heating of air over land and water-bodies causes winds.
14167• Evaporation of water from water-bodies and subsequent
14168condensation give us rain.
14169• Rainfall patterns depend on the prevailing wind patterns in an
14170area.
14171• Various nutrients are used again and again in a cyclic fashion.
14172This leads to a certain balance between the various components
14173of the biosphere.
14174• Pollution of air, water and soil affect the quality of life and
14175harm the biodiversity.
14176• We need to conserve our natural resources and use them in a
14177sustainable manner.
14178Exercises Exercises Exercises
141791. Why is the atmosphere essential for life?
141802. Why is water essential for life?
141813. How are living organisms dependent on the soil? Are organisms
14182that live in water totally independent of soil as a resource?
141834. You have seen weather reports on television and in newspapers.
14184How do you think we are able to predict the weather?
14185uestions
141861. What are the different states in
14187which water is found during the
14188water cycle?
141892. Name two biologically important
14190compounds that contain both
14191oxygen and nitrogen.
141923. List any three human activities
14193which would lead to an increase
14194in the carbon dioxide content of air.
141954. What is the greenhouse effect?
141965. What are the two forms of
14197oxygen found in the atmosphere?
14198© NCERT
14199not to be republished
14200202 SCIENCE
142015. We know that many human activities lead to increasing
14202levels of pollution of the air, water-bodies and soil. Do you
14203think that isolating these activities to specific and limited
14204areas would help in reducing pollution?
142056. Write a note on how forests influence the quality of our air,
14206soil and water resources.
14207© NCERT
14208not to be republished
14209216 SCIENCE
14210Chapter 3
142114. (a) MgCl
142122
14213(b) CaO
14214(c) Cu (NO3
14215)
142162
14217(d) AlCl3
14218(e) CaCO3
142195. (a) Calcium, oxygen
14220(b) Hydrogen, bromine
14221(c) Sodium, hydrogen, carbon and oxygen
14222(d) Potassium, sulphur and oxygen
142236. (a) 26 g
14224(b) 256 g
14225(c) 124 g
14226(d) 36.5 g
14227(e) 63 g
142287. (a) 14 g
14229(b) 108 g
14230(c) 1260 g
142318. (a) 0.375 mole
14232(b) 1.11 mole
14233(c) 0.5 mole
142349. (a) 3.2 g
14235(b) 9.0 g
1423610. 3.76 × 1022 molecules
1423711. 6.022 × 1020 ions
14238Chapter 4
1423910. 80.006
1424011. 16
142418 × =90% , 18
142428 × = 10%
1424312. Valency = 1, Name of the element is lithium,
1424413. Mass number of X =12, Y=14, Relationship is Isotope.
1424514. (a) F (b) F (c) T (d) F
1424615. (a) ü (b) × (c) × (d) ×
1424716. (a) × (b) × (c) ü (d) ×
14248Answers
14249ANSWERS 217
1425017. (a) × (b) ü (c) × (d) ×
1425118. (a) × (b) × (c) × (d) ü
1425219.
14253Atomic Mass Number Number Number Name of the
14254Number Number of of of Atomic
14255Neutrons Protons Electrons Species
142569 19 10 9 9 Fluorine
1425716 32 16 16 16 Sulphur
1425812 24 12 12 12 Magnesium
1425901 2 01 1 01 Deuterium
1426001 1 0 1 0 Protium
14261Chapter 8
142621. (a) distance = 2200 m; displacement = 200 m.
142632. (a) average speed = average velocity = 2.00 m s–1
14264(b) average speed = 1.90 m s–1 ; average velocity = 0.952 m s–1
142653. average speed = 24 km h
14266–1
142674. distance travelled = 96 m
142687. velocity = 20 m s–1; time = 2 s
1426910. speed = 3.07 km s–1
14270Chapter 9
142714. c
142725. 14000 N
142736. – 4 N
142747. (a) 35000 N
14275(b) 1.944 m s
14276–2
14277(c) 15556 N
142788. 2550 N in a direction opposite to the motion of the vehicle
142799. d
1428010. 200 N
1428111. 0 m s–1
1428213. 3 kg m s
14283–1
1428414. 2.25 m; 50 N
1428515. 10 kg m s–1; 10 kg m s–1; 5/3 m s–1
1428616. 500 kg m s–1; 800 kg m s–1; 50 N
1428718. 40 kg m s
14288–1
14289A2. 240 N
14290A3. 2500 N
14291A4. 5 m s–2; 2400 kg m s–1; 6000 N
14292ANSWERS 217
14293218 SCIENCE
14294Chapter 10
142953. 9.8 N
1429612. Weight on earth is 98 N and on moon is 16.3 N.
1429713. Maximum height is 122.5 m and total time is 5 s + 5 s = 10 s.
1429814. 19.6 m/s
1429915. Maximum height = 80 m, Net displacement = 0, Total distance covered = 160 m.
1430016. Gravitational force = 3.56 × 1022 N.
1430117. 4 s, 80 m from the top.
1430218. Initial velocity = 29.4 m s–1, height = 44.1 m. After 4 s the ball will be at a distance of 4.9 m
14303from the top or 39.2 m from the bottom.
1430421. The substance will sink.
1430522. The packet will sink. The mass of water displaced is 350 g.
14306Chapter 11
143072. Zero
143084. 210 J
143095. Zero
143109. 9 × 108
14311 J
1431210. 2000 J, 1000 J
1431311. Zero
1431414. 15 kWh (Unit)
1431517. 208333.3 J
1431618. (i) Zero
14317(ii) Positive
14318(iii) Negative
1431920. 20 kWh
14320Chapter 12
143217. 17.2 m, 0.0172 m
143228. 18.55
143239. 6000
1432413. 11.47 s
1432514. 22,600 Hz
1432620. 1450 ms-1