To Our Readers

FORTHCOMING COMPETITIVE EXAMS.

Dear Readers,

It gives us great pleasure and satisfaction to present to you the August issue ofyour favourite and frontline magazine ‘Competition Science Vision’. It is generallyclaimed by the toppers and high ranking successful candidates of all pre-medicalexaminations that each issue of this magazine is highly useful, unique and unbeatenin matter of its contents and way of presentation. Traditionally, we always try toimprove the extent and quality of the subject matter and make it more and moreexamination-oriented keeping in view the changes introduced in the examinationpattern.

CSV meets fully your requirements in all the four subjects. It has been markedas second to none in its field by its readers. It covers all pre-medical tests heldthroughout the country at present.

Hardwork under proper guidance, constant practice and revision have beenwidely claimed by successful candidates as the core elements of their success. Inmatters of guidance CSV stands matchless in the worthy estimation of our wisereaders.

Read CSV regularly and intelligently. It gives you the power to masteryour career and shape your destiny.

With best wishes for your all-round success.Sincerely yours,

Mahendra Jain (Editor)

C.S.V. / August / 2009 / 666

2009Chhattisgarh Sashastra Bal/Bharat Rakshit Vahini Constable

Recruitment Test (General Duty) (July 10)Haryana VLD Diploma Entrace Exam. (July 11)Gurgaon Gramin Bank Officers (Scale-I) Exam. (July 12)Jawahar Navodaya Vidyalaya Entrance Exam., 2009 (Class IX) (July 12)Ordnance and Ordnance Equipment Factories TradeApprentices Training Selection Test (July 12)Madhya Pradesh Jail Department Guard Recruitment Exam. (July 12)Rajasthan Sanskrit Education Department, TeachersGrade-III Exam. (July 22)Rajasthan Sanskrit Education Department, Senior TeachersExam. (July 23-24)Haryana School Teachers, Eligibility Test, 2009 for ElementaryTeachers (July 24)Haryana School Teachers, Eligibility Test, 2009 for Lecturers (July 25)R.A.S. (Mains) Examination (July 25–Aug. 12)UPSC Special Class Railway Apprentices Exam., 2009 (July 26)Haryana School Teachers, Eligibility Test, 2009 forMaster/Mistress (July 26)M.P. Guruji Eligibility Exam. (July 26)Chandigarh Police Constable (India Reserve Battalion)Recruitment Test (July 26)S.S.C. Data Entry Operator Exam., 2009 (Aug. 2)Uttarakhand B.T.C. Entrance Exam. (Aug. 2)Madhya Pradesh Assistant Seed Certified Officer SelectionTest, 2009 (Aug. 2)

Jharkhand High School Teachers Exam. (Aug. 9)Bank of India Clerk Exam. (Aug. 9)Central Bank of India Probationary Officer Examination (Aug. 16)IGNOU B.Ed. Entrance Exam., 2009 (Aug. 16)(Closing Date : 17 July, 2009)Delhi Fire Service, Fire Operator Exam. (Aug. 23 & 30)PNB Management Trainee Exam. (Aug. 30)UPSC National Defence Academy and Naval AcademyExamination (II), 2009 (Aug. 30)Indian Air Force Airman Selection Test Group ‘Y’ Trade (Aug.)SSC Central Police Organisation Sub-Inspectors Exam., 2009 (Sept. 6)EPFO Social Security Assistant Exam., 2009 (Sept. 6)(Closing Date : 8 July, 2009)Combined Defence Services Examination (II), 2009 (Sept. 13)PNB Clerical Cadre Exam. (Sept. 20)Delhi SSSB Trained Graduate Teachers (English, Mathe-matics, Natural Science) Exam. (Sept. 27)Bihar Telecom Technical Assistant Exam., 2008 (Oct. 4)(Closing Date : 20 July, 2009)Delhi SSSB Trained Graduate Teacher (Social Science) Exam. (Oct. 10)UPSC CPF Assistant Commandants Exam., 2009 (Oct. 11)Delhi SSSB Trained Graduate Teachers (Sanskrit, Hindi, Urdu,Punjabi) Exam. (Oct. 11)Indian Economic Service/Indian Statistical ServiceExamination, 2009 (Nov. 21)Rajasthan State Eligibility Test (SET) (Nov. 22)(Closing Date : 31 July, 2009)

C.S.V. / August / 2009 / 667 / 1A

There is always a section ofstudents which wants that theexamination dates be postponed forsometime so that they may make athorough revision of the whole courseor complete the chapters which havebeen left unfinished. Behind thisdemand there is only one argumentthat they did not have enough time forproper preparations. On the otherside, there is another section ofexaminees which is always ready totake their examination and wantsthat the examination may finish atthe earliest. It does not look nice thatthe examinations may be postponedonce a date has been fixed for them.

In our daily life also we find twosections of people—one section ofthe people seems to be having notime for anything, while the othersection of the people has time forevery useful thing. The people of thissection never complain of having notime for anything. These people havetime to complete all assignments suchas, complete their studies, attendgames, go to important meetings andalso join important functions of theirfriends and relations etc. There aresome people in this section who haveto earn their living as well.

As opposed to this the people ofthe section one have only one com-plaint to make that they had no time.They even make their business acause of great excuse and make itconvenient to forget many thingsunder their cover. If proper account iskept of work, they have done duringthe twenty four hours, it will be seenthat they have hardly devoted two tothree hours to serious work. It is nosurprise if such persons are notsuccessful in anything in their lives.

A student may lag behind in hispreparations on account of his limi-tations or circumstances but in hisheart of hearts he also wants to beone of those who are always readyto welcome the examinations. The

examinee who is always ready totake the examination is apparently fullof self confidence and as such isalways full of hope and optimism. Webelieve that you also would like tomake the best of your time and com-plete your course within the definiteperiod of time and keep alwaysready to take the examination.

To make the best of time meansthat we do not waste our time inuseless things and gossipping. Everyexaminee should first of all draw outhis time table as to for how manyhours he has to devote to his studiesand also the timings of his studies.He should be clear in his mind thatparticular hours he must devote to hisbooks and do nothing else during thatperiod. Such students may have toface criticism of many types such ashe does not talk to anybody, he triesto show himself, a studious man, heis very proud etc. but he has to ignoresuch things and follow his time tablewith full determination. Then and thenalone will he be able to make himselfready to take the examination.Develop the habit of leaving your bedin the morning at a particular time and

then sit for study for the period asfixed in your time table. This would doyou two benefits—there would remaina continuity in your study andsecondly you will be free from thetension which is the result of irregularstudy. One who makes regularstudies keeps also some hours forextra study which makes him full ofenthusiasm. Regular studies will doyou one more good that you will getpleasure in doing constructive workand will never feel depressed.

The continuous study createsboredom which needs to be removed.For this the candidate must make achange. For example, after readingfor sometime if one sits to write some-thing he will see that boredom isreleased to a great extent. Forrecreation it seems to be more properthat we develop the habit of reducingto writing what we read. This makeswhat you read a part of your per-sonality and you become exact inyour expression. As has been rightlysaid, “Reading makes a man perfectand writing makes a man exact.”

We want to emphasize that if wekeep a constant eye on our time tablewe shall see that all our things aredone in their own way and withoutresistance and then we will have nocomplaint of having no time. If wework regularly and with full devotionwe will always be ready to take theexamination and will never think of itspostponment. This is a known factthat without proper tests no one getsperfect knowledge. So you shouldmake up your mind to welcomeexaminations and as such be alwaysready for it. Also, do things syste-matically. Hotch-potch working is theresult of hotch-potch of ideas. Syste-matise your ideas so as to acquire asystem in your working and then yoursuccess is assured. That would giveyou confidence to go ahead and openthe gates for progress.

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TTHOUGHTS FOR THE MONTHTHOUGHTS FOR THE MONTH

❥ An independent judiciary is vital to equal opportunity.

❥ Truth is above everything, but higher still, is the living of truth.

❥ Everything appears coloured to the jaundiced eye.

❥ A vocal minority can create the impression of being the majority.

❥ Our greatest evils flow from ourselves.

❥ Genius is often perseverance in disguise.

❥ He that knows least commonly assumes most.

❥ Handsome is that handsome does.

❥ A highbrow is a person educated beyond his intelligence.

❥ They never die who live for others.

❥ Men without the minimum restraint, dignity and integrity cannot bearchitects of a new India.

❥ On the highest throne in the world, we still sit only on our ownbottom.

❥ Fear of death is worse than death itself.

❥ One crime is everything, two nothing.

❥ If you cannot be free, be as free as you can.

❥ The devil hath powers to assume pleasing shapes.

❥ First deserve, then desire.

C.S.V. / August / 2009 / 669

Climate Change—Imme-diate and Biggest Threat

to Health

Top experts of medical sciencesof U.K. have published a reportwarning that climate change is thebiggest threat to global health of the21st century. Rising global tempera-tures would have a catastrophic effecton human health and patterns ofinfection would change with insect-borne diseases such as malaria anddengue fever spreading more easily.

Insect Invasion : Patterns of infectionwould change, with insect-borne disea-ses such as malaria and dengue feverspreading more easily.

The report says that the poorestpeople in the world will be worstaffected. Although the carbon footprintof the poorest billion people is about 3per cent of the world’s total footprint,loss of life is expected to be 500times greater in Africa than in thewealthy countries. The impact of heatwaves, flooding and food shortageswill be felt globally.

Climate change is an immediatedanger. It is going to affect you and itwill inevitably affect your children.The report says the evidence ofgreenhouse gas emissions, tempera-ture and sea-level rises, the meltingof icesheets, ocean acidification andextreme climatic events suggest thatthe forecasts by the Intergovernmentalpanel on Climate Change might betoo conservative.

There is an awful lot we can do.Reducing carbon emissions wouldencourage people to cut use ofvehicles, and if that led to more

walking and cycling, it would tend tolower stress levels, reduce obesity,lessen heart and lung diseases andstroke risks.

Indian Space Shuttle willbe a Milestone

An Indian space shuttle will betest-fired from the spaceport atSriharikota within a year. ReusableLaunch Vehicle-Technology Demons-trator (RLV-TD), as it is called, will bea rocket-aircraft combination : theaircraft with a winged body, which isthe RLV, will sit vertically on therocket.

The engineering model of theaircraft is ready at the VikramSarabhai Space Centre (VSSC) inThiruvananthapuram. The first stageof the Satellite Launch Vehicle-3 willform the booster rocket. Weighingnine tonnes, it is called S-9. After ittakes off like a rocket, the booster willrelease the unmanned aircraft, whichwill go into space. At the end of themission, the aircraft will land on thesea.

Next year, the prototype of theRLV-TD will be ready for flight-testing.This will be a milestone for ISRO. TheRLV will open a new dimension in thelaunch vehicle technology and trans-portation system of ISRO. Groundtesting of the booster rocket is donerecently at Sriharikota.

The aircraft will stand over therocket, nosetip up, and its tail will beinterfaced with the rocket. In otherwords, the entire RLV will standvertically on top of the booster.

The booster rocket will take theRLV to a specific altitude, release theRLV and fall into the sea. On re-entryinto the Earth’s atmosphere, the RLVwill land in the sea, to be recovered.

First European FaceUnveiled

A scientist of U.K. succeeded toreconstruct the face of the firstanatomically modern human to live in

Europe, who inhabited the ancientforests of the Carpathian Mountainsin what is the present day Romania,about 35,000 years ago.

The reconstruction of the facethat could be male or female is basedon the skull and jaw-bone found in acave where bears were known tohibernate.

Face of an ancient European

The facial features indicate theclose affinity of these early Europeansto their immediate African ancestors,although it was not still possible todetermine the person’s sex.

Professor Richard Neave, whoreconstructed this face, based hisassessment on a careful and minutemeasurement of the bone fragmentsand his deep understanding of howthe soft tissues of the face are builtaround the bones of the skull.

The reconstruction is intended torecord, how human origins and evolu-tion from our birth place in Africa tothe long migratory routes, led us topopulate even the most distant partsof the globe.

It is impossible from the bones todetermine the skin colour of the indi-vidual, although scientists speculate itwas probably darker than modern dayEuropeans, reflecting a more recentAfrican origin.

Taken together, the material isthe first that securely documents whatmodern humans looked like whenthey spread into Europe.

C.S.V. / August / 2009 / 670

Extended Sleep ImprovesPerformance

Wondering how star athletes winworld-level championships as recentlyRoger Federer won the French OpenTennis Championship. The secretmay lie in increased sleep. In accor-dance with a study, by the researchersof Stanford University in California,the athletes, who extended their night-sleep, reported improvements invarious drills conducted after everyregular practice. The research statedthat many of the athletes, who parti-cipated in this study, for the first timerealized the importance of sleep andhow it affects their performance duringcompetitions.

While most athletes and coachingstaff may believe that sleep is animportant contributing factor in sportsand peak performance can only occurwhen an athlete’s sleep and sleephabits are optimal.

Five healthy students betweenthe age-group of 18 to 21 participatedin the study. A record of their sleep/wake pattern for a three-week-periodwas recorded. Athletic’s performance,including sprinting and hitting drills,was recorded after every practice.Athletes extended their sleep to 10hours a night for six weeks. Mood anddaytime sleepiness were recorded.Furthermore, daily sleep/wake acti-vities were monitored using sleepactigraphy.

Results suggested that sleepextension in athletes was associatedwith a faster sprinting drill, increasedhitting depth in drill and increasedhitting accuracy. According to theresearch team the findings of thisstudy could be relevant in all walks oflife. The results of this study werepresented at the 23rd Annual Meetingof Associated Professional SleepSocieties.

As Alaska Glaciers Melt,Sea Level Falls

Global warming conjures imagesof rising seas that threaten coastalareas, but in Juneau, Alaska, asalmost nowhere else in the world,climate change is having the oppositeeffect. As the glaciers melt there, theland is rising, causing the sea toretreat.

The geology is complex, but itboils down to this : Relieved ofbillions of tons of glacial weight, theland has risen much as a cushion.The land is ascending so fast that therising seas—a ubiquitous byproductof global warming—cannot keeppace.

As a result, the relative sea-levelis falling, at a rate ‘among the highestever recorded’, according to a reportby a panel of experts.

The Ape Laughs LikeHumans

What happens if you tickle agorilla ? The ape laughs like humans,according to a research, whichsuggests that the origins of laughtercan be traced back more than 16million years. Findings of a researchteam of the University of Portsmouth(Britain) indicate that humans haveinherited ability to laugh from the lastcommon ancestor from which theyand great apes evolved. This investi-gation is the first phylogenetic test ofthe evolutionary continuity of a humanemotional expression. It supports theidea that there is laughter in apes.

Humour in Evolution : Origin oflaughter can be traced back to more

than 16m years.

Professor Marina Davila Ross ledthe research team. The experts whocarried out the research on gorillas,chimpanzees and orangutans, believethat it started out as a ‘grunt like’ noisewith our distant ancestors andgradually turned into the more sophi-sticated chuckles and guffaws, weknow today. The research proves thatlaughter evolved gradually over thelast 10 to 16 million years of primateevolutionary history.

Super Cereal FightsObesity, Cancer

A new study has shown that soycereal, developed by a professor of

food and nutrition, not only helpstackle obesity but cuts cancer riskalso. Professor Soo-Yeun Lee fromthe University of Illinois has come upwith a cinnamon-flavoured soy cerealthat can cut the risk of prostate andbreast cancer.

Giant Space TornadoesPower the Northern

Lights

Different cultures have attributedtheir spectacular light show to fire-breathing dragons, dancing gods andghostly clans at war. Now researchhas found that Northern Lights oraurora borealis, are powered by giant,electrical tornadoes spinning at morethan a million miles an hour andstretching thousands of miles intospace.

Scientists used a set of fivesatellites designed to measure theEarth’s magnetic field to generate thefirst images of the whirling vertices.They show how vast quantities ofcharged particles emitted by the Sunfirst pile up in huge clouds about40,000 miles above the night side ofEarth. Then as the energy they holdbecomes too great, the particlesexplode downwards towards Earth,spinning as they go.

‘‘When these space tornadoesreach the upper atmosphere, theirenormous energy heats the air sostrongly that it starts glowing. That iswhat generates the aurorae’’, saidProfessor Karl Heinz Glassmeier(Germany), the leader of researchteam.

The Northern lights and theirSouthern equivalent, the auroraaustralis, create spectacular movingdisplays of different shades andcolours of light.

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UPKAR PRAKASHAN, AGRA-2E-mail : publisher@upkar.in Website : www.upkar.in

By : Dr. Lal, Mishra & Kumar Code No. 1624 Rs. 250/-

Useful for Various Competitive Exams.

C.S.V. / August / 2009 / 671 / 2

AWARDS/HONOURS

10th IIFA Awards, 2009

The 10th IIFA (InternationalIndian Film Academy) Awards cere-mony for 2009 was concluded inMacau, China. Major names fromBollywood transcended on this beauti-ful island of China. The function tookplace in the Venetian Macao ResortHotel, Macau. The event continuedfor three days.

Bollywood director AshutoshGovarikar’s historical love epic JodhaaAkbar swept the 10th IIFA awards,taking home trophies for the bestpicture, best director and best actor.

Awardees are—

Best Film—Jodhaa AkbarBest Actor in a Leading Role

(Male)—Hrithik Roshan—JodhaaAkbar

Best Actor in a Leading Role(Female)—Priyanka Chopra—Fashion

Best Actor in Supporting Role(Male)—Arjun Rampal—Rock On

Best Actor in a SupportingRole (Female)—Kangana Ranaut—Fashion

Best Actor in a Comic Role—Abhishek Bachchan—Dostana

Best Actor in a NegativeRole—Akshaye Khanna—Race

Best Debutant Star (Male)—Farhan Akhtar

Best Debutant Star (Female)—Asin

Best Director—Ashutosh Gowa-riker—Jodhaa Akbar

Best Story—Neeraj Pandey—AWednesday

Best Music—A. R. Rahman—Jodhaa Akbar

Best Lyrics—Javed Akhtar—Jashn-E-Bahara (Jodhaa Akbar)

Best Playback Singer (Male)—Javed Ali—Jashn-E-Bahara (JodhaaAkbar)

Best Playback Singer (Female)—Shreya Ghoshal—Teri Ore (Singhis King)

Outstanding Achievement byan Indian in International Cinema—Aishwarya Rai Bachchan

Lifetime Achievement Award—Rajesh Khanna

Best Dialogue—Manu Rishi—Oye Lucky ! Lucky Oye !

Best Screenplay—NeerajPandey—A Wednesday !

Sound Editing—Ghajini

Sound Recording—Ghajini

Best Action—Ghajini

Special Effects—Ghajini

Best Sound Recording—RockOn !!

Best Editing—Jodhaa Akbar

Best Make-up—Jodhaa AkbarArt Direction—Jodhaa AkbarBest Costume Award—Jodhaa

Akbar

Best Background Score—A. R.Rahman

Star of the Decade (Male)—Shahrukh Khan

Star of the Decade (Female)—Aishwarya Rai Bachchan

Music Director of the Decade—A. R. Rahman

Man Booker InternationalAward, 2009

Canadian short story writer, AliceMunro, beat Mahasweta Devi andhost of other literary heavyweights,including Nobel Laureate V. S.Naipaul, to win the £ 60,000 ManBooker International Prize.

Ms. Munro (78) is regarded asone of Canada’s most celebratedwriters. She was amazed anddelighted to win the prize.

The prize, different from theannual Booker Prize for Fiction, isawarded once every two years to aliving author for his or her lifetimeachievements.

BOOKS

Twilight of the Tigers—G. H.Peiris (The book deals with whatwould be the next for Tamils in SriLanka. The dismissal of devolution ofpower as a solution has dark fore-boding. After three decades holding agun to Sri Lanka’s head, theLiberation Tigers of Tamil Eelam hasfinally been defeated militarily and itsleader Velupillai Prabhakaran is dead.The decisive battles between theLTTE and the Sri Lankan military werefought since January 2009 till themiddle of May 2009. The simple pointis that in Sri Lanka, Sinhalese are 73per cent and Tamils are 12 per cent.Sri Lankan Tamils are liberated fromthe LTTE, but are afraid of not havinga strong voice to speak up for them.The book deals with these facts).

World of Work—Pub. by Inter-national Institute for Labour Studies,Geneva (The book gives the detailsthat globalisation has resulted in thewidening of income–inequality acrossand within the countries.)

India’s Energy Security—Ed. byLigia Noronha and Anant Sudarshan(The book draws the attention to theneed for formulating holistic policieson energy.)

Democracy and Human Deve-lopment in India—Naresh Gupta(The book explains how the demo-cracy and human development areinterrelated.)

DAYS

July 1—Doctors’ DayJuly 6—Zoonoses DayJuly 11—World Population Day

Ustad Akbar Ali Khan—Sarodmaestro Ustad Akbar Ali Khan (88)passed away in Francisco (U.S.A.) onJune 19, 2009 after a prolongedkidney ailment. He is survived by his

C.S.V. / August / 2009 / 672

wife Mary, three sons and a daughter.He was born in 1922 in Comilladistrict, now in Bangladesh.

A recipient of Padma Bhushanand Padma Vibhushan, the Ustadwas a colossus in the world of Indianclassical music for the last fivedecades. He was admired by botheastern as well as western musiciansfor his brilliant compositions and hismastery of the 25 string instrument.He also composed music for Indianfilms of Chetan Anand, Satyajit, TapanSinha and others.

Habib Tanvir—Noted playwrightHabib Tanvir passed away on June 8,2009. He was one of the greateststalwarts of the Indian stage, knownfor blending theatre, folk art andpoetry in his works. He left an indeliblemark on the minds of the viewers. Hewas born on September 1, 1923 inRaipur. He took M.A. degree fromAligarh Muslim University and joinedAll India Radio, Bombay as Directorin 1945. His full name was HabibAhmed Khan and he had adopted thepen name ‘Tanvir’. He also wrotesongs for Hindi films. He won SangeetNatak Akademi award in 1969. Hewas honoured with Bhushan. He hadbeen a Member of Rajya Sabha(1972–1978).

Rajeev Motwani—Rajiv Motwani(47) was talented mathematicianwhose contribution to the world ofcomputer science influenced thedevelopment of algorithm-basedsearch technology. He passed awayin a freak drowning accident in hisswimming pool on June 7, 2009. InAtherton, California. He wasprofessor at Stanford University. Heis survived by his wife Asha Jadejaand the daughters Naitri and Anya.

Google’s founders Sergey Brinand Laurence page, were mentoredby Dr. Motwani as he conductedresearch to launch what wouldbecome an iconic service. His contri-bution to frame the ‘world wide web(www)’ is incomparable.

G. E. Vahanvati (New AttorneyGen.)—Goolam E. Vahanvati (60) tillrecently the Solicitor-General of India,is appointed as the next Attorney-General of India for a period of threeyears. He will succeed Milon K.

Banerjee. Prior to being appointed theSolicitor-General, he was Advocate-General of Maharashtra till June 2004.

Mr. Vahanvati is the first Muslimto occupy the top law officer’s post inthe country in the last six decades.

Pradeep Vasant Naik (NewChief of the Air Staff)—Air ChiefMarshal Pradeep Vasant Naik (60)took over as the 19th Chief of theIndian Air Force succeeding Air ChiefMarshal Fali Homi Major. With morethan four decades of distinguishedservices, Air Chief Marshal Naik tookserious part in Indo-Pak war in 1971.He was honoured with the VishishtSeva Medal. He was commissionedinto the IAF in 1969 as a fighter pilot.So far he has clocked more than3000 hours of flying.

P. K. Barbora (New Vice-Chiefof Air Staff)—Air Marshal P. K.Barbora took over as the new Vice-Chief of the Air Staff.

Nirmal Kumar Verma (NewNavy Chief)—Vice-Admiral, NirmalKumar Verma, will be the next Chiefof the Naval Staff. He will take chargefrom Admiral Sureesh Mehta who willretire on August 31, 2009. Vice-Admiral Verma, currently Flag OfficerCommanding-in-Chief, Eastern NavalCommand, commanded aircraft carrierINS Viraat. He was decorated with theParam Vishisht Seva Medal.

R. Khullar (New CommerceSec.)—Union Government, on June 5,2009, appointed DisinvestmentSecretary, Rahul Khullar asCommerce Secretary. He will replaceG. K. Pillai a 1975 batch IAS Officer,Mr. Khullar was shifted to theDepartment of Disinvestment fromthe Commerce Ministry.

Father’s Name Late Shri Jagjivan Ram

Mother’s Name Smt. Indrani DeviDate of Birth 31.03.1945Place of Birth Patna (Bihar)Marital Status MarriedDate of Marriage 29 Nov., 1968Spouse’s Name Shri Manjul KumarNo. of Son 1No. of Daughters 2Educational Qualifications M.A., LL.B., Advanced Diploma in Spanish Educated

at Indraprastha College and Miranda House, Delhi(Delhi)

Profession Social Worker, Advocate, Civil ServantPermanent Address D-1029, New Friends Colony, New Delhi–110 065

(011) 26910618, 26910639, 9810630165 (M)

Fax—91-11-26910618

A. K. Kembhavi (New Director,IUCAA)—Ajit Keshav Kembhavi hasbeen appointed as the Director ofInter-University Centre for Astronomyand Astrophysics (IUCAA), Pune.IUCAA is one of the five inter-university research centres of theUniversity Grants Commission. Prof.Khembhavi will take charge on Sep-tember 1, 2009 from Professor NareshDadhich.

Professor Kembhavi took hisPh.D. degree in 1979 under theguidance of Professor Jayant Narlikar.His research interests includedquasars and other active galaxies,intermediate red-shift galaxies, galaxymorphology, tidal capture binary starsin globular clusters, astronomical databases and the virtual observatory(VO).

P. Varghese (New AustralianEnvoy)—Australia has named PeterVarghese as the next High Commis-sioner to India with concurrent acredi-tation to Bhutan. He succeeds JohnMcCarthy. Mr. Varghese will takecharge in August 2009. Mr. Vargheseserved as the High Commissioner inMalaysia and had been also a part ofAustralian missions Vienna, Washing-ton and Tokyo.

Meira Kumar—History was madeon June 3, 2009 when diplomat-turned-politician Meira Kumar becamethe first womanSpeaker of the LokSabha, with theotherwise fractiousHouse setting asideits differences toelect the Dalit leaderunanimously. The Meira Kumar

C.S.V. / August / 2009 / 673

name of the 64-year-old Congressleader from Bihar was proposed byparty President Sonia Gandhi andseconded by the Prime MinisterManmohan Singh and BJP leaderL. K. Advani together led her to thepodium and later paid tribunes to awoman.

This is Ms. Meira Kumar’s fifthterm in Lok Sabha. At present sherepresents Sasaram (Bihar).

Karia Munda—Seven-timeMember of Parliament, Karia Munda,was unanimously elected DeputySpeaker of the Lok Sabha. A dozen ofnominations werefiled in Mr. Munda’sfavour. The motionwas first moved bythe oppositionleader L. K. Advaniand seconded byBJP President Karia Munda

Rajnath Singh. Subsequently, leadersrepresenting other parties, includingCongress, moved similar motions. A32-year-old unbroken tradition ofhaving the Deputy Speaker from theopposition—which began in 1977, thevery year Mr. Munda entered the LokSabha—has been preserved with theunanimous election of Mr. Munda.This sentiment was echoed by theentire house. Mr. Munda is elected tothis 15th Lok Sabha from Khunti inJharkhand on BJP ticket.

Mr. Karia Munda was born onApril 20, 1936 in Ranchi district ofJharkhand. He took his M.A. degreefrom Ranchi University. He is a seniorleader of Bhartiya Janta Party andhad been a Cabinet Minister inVajpayee’s Government.

‘IRAN’ on the Boil—Thirty yearsafter momentous events broughtAyatollah Ruhollah Khomeini to thefore as Iran’s man of destiny, hisIslamic revolution has skidded intouncharted territory. The officialdeclaration of incumbent PresidentMohamoud Ahmadinejad as a runaway winner in a hotly disputed andpossibly rigged presidential electionhas generated a white heat of popularanger that has no precedent in Iran’spost-revolution history.

Iran’s Political Turmoil

The purported victor’s triumpha-lism at a huge victory rally wasimmediately challenged on the streetsby close to a million supporters of MirHosain Mausavi, the principalopponent. They braved a ban orderand choked a nine-kilometre stretchleading to Azadi (freedom) Square, aprominent Tehran landmark. Theunrest has now spread to other cities,including Tabriz, Shiraj and others.

Only a fair, free and crediblymonitored fresh election can heal anation that is treading a thin line bet-ween fear of theocratic authoritarian-ism and hope of genuine reform thatis not in conflict with revolution’sfundamentals.

Lalgarh (West Bengal)—Socialunrest is known to manifest itself asproblems of law and order. But thereverse can also be true sometimes,politically inspired violence seeks thecover of socio-economic grievance.Maoist, outlaws who went on therampage in the Lalgarh of WestMidnapore district of West Bengal,carrying out murderous attacks on theruling Communist Party of India(Marxist) workers and destroying theirhouses and party offices, were lookingto settle political scores in the guiseof protesting ‘police atrocities’. Usingthe neighbouring State of Jharkhandas the base, they established reignof terror and drove out security

personnel and CPI(M) workers andsympathisers. With tribal folk ashuman shield, they have now shoughtto create ‘liberated zones’ in thedistrict. The offensive, time to takeadvantage of the electoral debacle ofCPI(M) in the recently concluded LokSabha election, would not have beenpossible without the support of themain opposition party, the TrinamoolCongress. Either directly or indirectly,the Police Santrosh BirodhiJanashadharaner Committee or thePeople’s Committee against pol iceatrocities, which spread heads theagitation, has drawn substance fromthe opportunism of the TrinamoolCongress and its partner theCongress. Union Government isreluctant to extend full assistance toWest Bengal Government.

The situation in West Midnaporeis too serious to allow such crasspoliticking. Manmohan Singh Govern-ment must not lose any more time incoming to the aid of West Bengal’sLeft Front Government to tackling theMaoists and their surrogates. Elsethere will be a heavier cost to pay.

SPORTS

Badminton

Indonesian Open—Ace Indianshuttler, Saina Nehwal, on June 21,2009 in Jakarta, scripted history by

C.S.V. / August / 2009 / 674

becoming the first Indian to win aSuper Series tournament after sheclinched the Indonesian Open titlewith a stunning victory over the higherrankeded Chinese Lin Wang. Sainabeat Wang 12-21, 21-18, 21-9 in athriller that lasted for 49 minutes.

Big Moment : Saina Nehwal scripted aremarkable victory over her higher-ranked opponent China’s Wang Lin inthe Indonesian Open.

In the final, World No. 8 Sainacame from behind to outsmart WorldNo. 3 Wang and avenge her lastweek’s Singapore Open Super Seriesloss to her.

Hailing Saina’s victory as animportant milestone, BadmintonAssociation of India announced acash prize of Rs. 2 lakh as a recog-nition of here feat.

The results (finals) : Women :Saina Nehwal (Ind) bt Lin Wang (Chn)12-21, 21-18, 21-9.

Doubles : Eei Hui Chin & Pei TtyWong (Mas) bt Shu Cheng & YunlieZhao (Chn) 21-16, 21-16.

Tennis

French Open Tennis Tourna-ment 2009—Roger Federer gloriouslyequalled Pete Sampras’ record of 14Grand Slam title, when he won hismaiden French Open title in Paris onJune 7, 2009.

Top Five Grand Slam Winners

14-Pete Sampras (U.S.) and RogerFederer (Switzerland), 12-Roy Emerson(Australia), 11-Bjorn Borg (Sweden)and Rod Laver (Australia), 10-BillTilden (U.S.), 8-Andre Agassi (U.S.),Jimmy Connors (U.S.), Ivan Lendl(Czechoslovakia), Fred Perry (Britain)and Ken Rosewall (Australia).

Federer achieved this victorywithout his having to combat theformidable rivals such as RafaelNadal, Andy Murrray and NovakDjokovic, all of whom suffered shockdefeats.

French Open 2009

Men’s Singles—Switzerland’sRoger Federer beat Robin Soderling ofSweden 6-1, 7-6, 6-4 in the men’ssingles final of the French Open 2009 inParis (France) on June 7, 2009. Thisvictory took him level with great friendPete Sampras of USA as the holder of14 Grand Slam titles. With this titleFederer has joined the elite club of fiveothers—Fred Perry (Britain), Don Budge(USA), Rod Laver (Australia), RoyEmerson (Australia) and Andre Agassi(USA)—who have won all four GrandSlams.

Women’s Singles—Russia’sSvetlana Kuznetsova claimed thewomen’s singles title by defeating hercompatriot Dinara Safina 6-4, 6-2 onJune 6, 2009. By defeating her sheclaimed her second Grand Slam title.

Men’s Doubles—India’s LeanderPaes and the Czech Republic’s LukasDlouhy won the men’s doubles titleoutclassing South Africa’s WesleyMoodie and Belgium’s Dick Norman.

Women’s Doubles—AnabelMedina Garrigues and Virginia RuanoPascual of Spain clinched the women’sdoubles crown by defeating VictoriaAzarenka of Belarus and Elena Vesninaof Russia.

Mixed Doubles—In the mixeddoubles, Liezel Huber and Bob Bryan ofUSA beat Vania King of USA andMarcelo Melo of Brazil.

Cricket

ICC World Twenty–20 CricketChampionship—Final of this tourna-ment was played between Pakistanand Sri Lanka on June 21, 2009 atLord’s in London. Pakistan won thefinal by eight wickets.

Man of the match—Shahid Afridi

Man of the series—TillkaratneDilshan

New Union Government’sAgenda—While addressing the joint

session of Parliament on June 4,2009, the President Pratibha Patilunveiled the new Union Government’sagenda—

Manmohan’s Agenda

● Internal security and preservation ofcommunal harmony.

● Economic growth in agriculturemanufacturing and services.

● Consolidation of the existing flag-ship programmes.

● Concerted action for the welfare ofwomen, youth, children, other back-ward classes, SCs, STs, minorities,differently-abled and elderly.

● Governance reform.

● Creation and modernisation ofinfrastructure and capacity addition.

● Prudent fiscal management.

● Energy security and environmentprotection.

● Constructive and creative engage-ment with the world.

● Promotion of a culture of enterpriseand innovation.

‘‘The yearning of our young peoplefor inclusiveness—economic, socialand cultural—and the rejection ofthe forces of divisiveness…con-tinues as both its (govt’s) inspiringvision and unfinished business.

—President, Pratibha Patil

New Apex Body for HigherEducation—The creation of aNational Commission for HigherEducation and Research that willsubsume as many as 13 existingprofession councils and regulatoryagencies, including the UniversityGrants Commission, Medical Councilof India and the AICTE, is a keyrecommendation of a committeeheaded by well-known educationist,Professor Yashpal. A draft legislationand constitutional amendment arerecommended.

The proposal autonomous statu-tory body will comprise six membersand a chairman appointed by thePresident. State Higher EducationCouncils, along the lines of thoseexisting in West Bengal, Kerala andAndhra Pradesh, will form the secondtier of the system.

●●●

C.S.V. / August / 2009 / 675

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Inspiring Young Talent ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

‘‘My hard working, teachers’ guidance, support of family members andbelief in God are the elements of my success.’’

—Rashmi SinghTopper—U.P.-CPMT 2009 (1st Position)

[‘Competition Science Vision’ arranged an exclusive interview with Miss Rashmi Singh who has thecredit of standing first on the list of successful candidates in U.P.-CPMT, 2009. In addition she has alsocleared AIPMT (CBSE) and BHU Pre-medical Test with high ranks. For her brilliant success, she deservesall praise and our heartiest congratulations. This important interview is presented here in its original form.]

CSV—Congratulations on yourbrilliant success.

Rashmi—Thank you.CSV—Before knowing your result

what did you think about those whoachieve top positions ?

Rashmi—I used to think thatGod has given some special featurein them. But now I am convinced thatnothing is special in them.

CSV—Achieving top position hascome as surprise to you or were youconfident of achieving it ?

Rashmi—It was really surprisingto me. I had never thought of becom-ing topper. But after giving examina-tion of U.P.-CPMT and calculating mymarks, I was expecting my position intop ten and not of a topper.

CSV—What do you think is thesecret of your success ?

Rashmi—My hard working,teachers’ guidance, my brother familymembers support as well belief inGod.

CSV—In how many attempts didyou get this success ?

Rashmi—This was my 2ndattempt.

CSV—What were the shortcom-ings in your preparation for earlierattempts ? How did you make up forthem this time ?

Rashmi—In earlier attempts, I didnot prepare well any subject, I justgave the exam for knowing patternetc. This time I had joined coachingand from the starting of session Istarted preparing for competitiveexam.

CSV—From where did you getthe inspiration of choosing a medicalcareer ?

Rashmi—I belong to rural area. Iused to see that many poor peoplewere not getting proper treatment.So, for them I thought that I have tobe a doctor.

—CSV is really very helpfulin PMT examinations. It helps tounderstand basic concepts. Itsscience tips, multiple choicequestions and assertion-reasontype questions are really wonder-ful for these tests.

—Rashmi Singh

CSV—What planning did youmake for preparation ? Please tellsomething in detail.

Rashmi—First, I used to makelist of work. I have to do in a day. Iused to use my full effort to completethose pieces of work. I planned toread and solve problems of NCERTtext book at least for 3 times. Apartfrom this I used to solve objectiveafter reading text book.

CSV—How much time did youdevote daily and regularly for Physics,Chemistry, Zoology and Botany ?

Rashmi—For me it was notpossible to read all the four subjectsdaily. But I used to read three subjectsatleast. I devoted 3 hours for Botany,3 hours for Zoology, 3 hours forBotany. If time remained then 1 or 2hours for Physics.

CSV—Out of the above foursubjects, to which subject did you givemore weightage and why ?

Rashmi—To both Botany andZoology I have given more weightagebecause they together make 50% ofour question paper. These arescorable subjects but Physics andChemistry gave rank in exam.

CSV—Did you make completestudy of all topics or of some selectivetopics ?

Rashmi—I studied all the topics.

CSV—How did you give finaltouches to your preparation ?

Rashmi—Last time I preparedthe topics but I used to forget what Iread.

Bio-DataName—Rashmi Singh

Father’s Name—Sri BhishmaPitamah Singh

Mother’s Name—Smt. ChandaSingh

Educational Qualifications—

H.S./Std. X—85·4% (Raj EnglishSchool, Varanasi), 2006.

Inter/Std. XII—87% (Raj EnglishSchool, Varanasi), 2008.

Special achievements—

● 1st rank in U.P.-CPMT, 2009.

● AIPMT (CBSE) Main—AIR-110(OBC-18)

● BHU Screening—AIR 26th (OBC-5)

CSV—Did you prepare notes ?

Rashmi—Yes, In coaching myteachers helped me to prepare notes.

CSV—What was your attitude forsolving numerical questions ? Whatweightage did you give them ?

Rashmi—2 or 3 hours beforecommencement of any exam I usedto solve numerical questions randomlyof any chapter. Due to this step andtricks of numerical striked immediatelyin mind. I gave much importance tothem because they helped me inbecoming topper.

C.S.V. / August / 2009 / 676

CSV—How much time is suffi-cient for preparing for this exami-nation ?

Rashmi—Time is not an impor-tant factor. I did planning for a day.Whatever the topics I had to completethem, I used to complete in specifiedtime whether it took 2 hours or 10hours.

Personal QualitiesHobby—Watching cartoon

Ideal Person—Dr. A.P.J. AbdulKalam

Strong Point—Hard work

Weak Point—Nervousness

CSV—From what level of educa-tion should an aspirant begin pre-paring for it ?

Rashmi—You must know basicconcepts of all subject. You should tryto feel the subjects.

CSV—What was your order ofpreference for various branches forwhich this test is held ?

Rashmi—MBBS, BDS, BAMS,BHMS.

CSV—Please mention variousbooks in each subject and magazineson which you based your preparation.

Rashmi—NCERT Text Book andCompetition Science Vision magazine.

CSV—Did you take coaching inyour preparation ?

Rashmi—Yes, JRS Tutorial,Varanasi. I was very much impressedwith the director of JRS Tutorial Mr.A. K. Jha who taught me Physics invery scientific manner. I was alsoimpressed with Professor of BotanyMr. Diwedi.

CSV—What help do the sciencemagazines render in the preparationsfor this examination ?

Rashmi—They helped me under-standing basic concept in manytopics.

CSV—What will be your criterionfor selecting a magazine for theseexamination ?

Rashmi—Select those magazinewhich has contents of your syllabusand latest G.K.

CSV—What is your opinion aboutour Competition Science Vision ? Howmuch helpful and useful do you findit ?

Rashmi— It was really veryhelpful to me. It helped me under-standing basic concepts. ScienceTips, multiple choice questions aswell as assertion-reason were verywonderful for the medical students.

CSV—Please suggest in whatway CSV can be made more usefulfor medical aspirants.

Rashmi—Latest discovery invarious medical areas should bepublished. For AIIMS somethingshould be specially published.

CSV—Please mention your posi-tion in the merit list as well as themarks obtained in different subjects.What was your aggregate percentageof marks ?

Rashmi—Ist Position.Physics—48/50Chemistry—48/50Zoology—46/50Botany—50/50Total—192/200, i.e., 96%.CSV—What books/magazines/

newspapers did you read for G.K.preparations ?

Rashmi—Competition ScienceVision.

CSV—Whom would you like togive credit for your success ?

Rashmi—I would like to givecredit of my success to my family,relatives as well as my teachers.

CSV—Please tell us somethingabout your family.

Rashmi—I have mummy, papa,elder brother and sister. My fathername is Mr. Bhishma Pitamah Singhwho is engineer in DRDA (Son-bhadra) and my brother is doingMBBS, Allahabad and sister alsodoing MBBS from Jhansi, my motheris housewife.

CSV—What in your frank opinionhas been the biggest mistake in yourpreparation for this test ?

Rashmi—After giving any com-petitive exam. I used to rest foratleast 3 days and I did not solvePhysics problem much.

CSV—What message would youlike to give for our readers of CSV ?

Rashmi—Please read CSV veryseriously for easy selection in PMT asit is very useful for medical studentsbecause all the four subjects aregiven together in this magazine.

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C.S.V. / August / 2009 / 677

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Inspiring Young Talent ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

‘‘Continuous hard work, Long practice and faith & belief in God are theelements of my success.’’

—Anjali SinghTopper—Uttarakhand, PMT–2009 (1st Position)

[‘Competition Science Vision’ arranged an exclusive interview with Miss Anjali Singh who has thecredit of standing first on the list of successful candidates in Uttarakhand PMT, 2009 in the very firstattempt. In addition, she has also cleared AFMC in the written examination. For this brilliant success shedeserves all praise and our heartiest congratulations. This important interview is presented here in itsoriginal form.]

CSV—Congratulations on yourbrilliant success.

Anjali—Thank you, sir.CSV—Before knowing your result

what did you think about those whoachieve top positions ?

Anjali—Same as now othersthink of me. As brilliant and very hard-working personalities.

CSV—Achieving top position hascome as surprise to you or were youconfident of achieving it ?

Anjali—As a surprise. But I wasconfident of achieving a good rank inthis PMT.

CSV—What do you think is thesecret of your success ?

Anjali—Continuous hardwork,long practice and faith & belief in God.

CSV—In how many attempts didyou get this success ?

Anjali—In Uttarakhand PMT, Ihad appeared 1st time but after 10 + 2(2005), I had prepared for medicalentrance exam for four years.

CSV—What were the shortcom-ings in your preparation for earlierattempts ? How did you make up forthem this time ?

Anjali—Earlier, I had not pre-pared some topics (Like Animal taxo-nomy, families of angiosperms etc.).Also, I was lazy at the last time after alot of hardwork in the beginning. Butthis year from the very starting. I pre-pared these topics well and continuedmy studies until the last exam.

CSV—From where did you getthe inspiration of choosing a medicalcareer ?

Anjali—It was a dream for mesince my childhood. But when mycousins and friends were selected, itbecame an aim for me, which I had toachieve anyway. In childhood I likedthe clinic of paediatricians very much.

—It is a very good magazineand has influenced me much. Itcontains good MCQs as well assound and brief material forrevision. Interviews of toppersand high ranking candidates arevery inspiring.

—Anjali Singh

CSV—From when did you startthe preparation for it ?

Anjali—I started my preparationof medical entrance exam afterintermediate. In XI and XII Std., I wasnot so sincere and passed inter-mediate with both Maths and Bio. asan average student.

CSV—What planning did youmake for preparation ? Please tellsomething in detail.

Anjali—First I prepared my weaktopics in all subjects and had acommand on them. I did a lot ofpractice by solving MCQs of allsubjects. I gave a lot of attention toBotany and Physics. At the time ofexams, instead of solving MCQs, Iconcentrated on the revision of topics,thoroughly.

CSV—How much time did youdevote daily and regularly for Physics,Chemistry, Zoology and Botany ?

Anjali—My study hours were notfixed. I studied all the time but when Ifelt exhausted. I liked to take rest or a

sleep. Atleast 2 hours are required foreach subject and extra time isneeded for solving MCQs.

CSV—Out of the above foursubjects, to which subject did you givemore weightage and why ?

Anjali—I payed more attention toPhysics and Botany. I gave maximumtime to Biology as a whole as I think Iam a little weak in Biology (Botanyespecially). I had a good command onPhysics, but maintaining the same, itrequires a lot of practice and a lot oftime.

Bio-Data

Name—Anjali SinghFather’s Name—Mr. Desh Raj

SinghMother’s Name—Mrs. Vinay SinghEducational Qualifications—H.S./Std. X—75% (St. Francis Sec.

School, Agra), 2003Inter/Std. XII—77% (Holy Public

School, Agra), 2005.Special achievements—

● 1st rank in state PMT● I had been selected in AFMC

(written) also called for an inter-view.

CSV—Did you make completestudy of all topics or of some selectivetopics ?

Anjali—I studied all the topicscompletely in the starting and solvedall MCQs but at the time of exams, Ileft few topics of low weightage (Like,Aging, Virus etc.)

CSV—How did you give finaltouches to your preparation ?

Anjali—By solving more andmore multiple choice questions andrevising my topics frequently.

CSV—Did you prepare notes ?Anjali—I had taken coaching. So

I already had notes. I wrote all theformulae and important points in

C.S.V. / August / 2009 / 678

Physics and Physical Chemistry. Butdid not prepared any notes in othersubjects, because biology, organicand inorganic chemistry don’t haveany certainty from where questionsmay be asked.

Personal Qualities

Hobbies—Listening music (filmysongs), cooking

Ideal Person—My mother and allthe selected students

Strong Point—My continuous hard-work when I feel it is a requirement.

Weak Point—My sleep andsilly mistakes while solving MCQs.

CSV—What was your attitude forsolving numerical questions ? Whatweightage did you give them ?

Anjali—I had Maths in XII, so Ican do well in numerical questions. Igave them as much time as requiredby using formulae and practisingMCQs.

CSV—How much time is suffi-cient for preparing for this examina-tion ?

Anjali—Two years study duringXI and XII is sufficient if we studyproperly and seriously.

CSV—From what level of educa-tion should an aspirant begin pre-paring for it ?

Anjali—According to me, astudent preparing for medicalentrance exam should begin the pre-paration from XI Std.

CSV—What was your order ofpreference for various branches forwhich this test is held ?

Anjali—Only MBBS.

CSV—Please mention variousbooks in each subject and magazineson which you based your preparation.

Anjali—Elementary Biology,Pradeep Physics and ComprehensiveChemistry for subjective study. Forpreparing MCQs, I used objectiveDinesh for Biology, CSV for allsubjects and Pradeep for Chemistry.

CSV—Did you take coaching inyour preparation ?

Anjali—Yes, I had taken coach-ing for first two years (after XII) andthen for one year at Agra.

CSV—What help do the sciencemagazines render in the preparationsfor this examination ?

Anjali—I used science maga-zines CSV for solving MCQs and lastyears exam questions.

CSV—What will be your criterionfor selecting a magazine for theseexamination ?

Anjali—I had not given muchimportance to this aspect. But I admitthat for quick revision, magazines arehelpful.

CSV—What is your opinion aboutour Competition Science Vision ? Howmuch helpful and useful do you findit ?

Anjali—It contains good MCQsas well as a sound and brief materialfor revision. It contain interviews of toprankers, which is very much inspiringfor the students preparing for thesame. It influenced me very much.

CSV—Please suggest in whatway CSV can be made more usefulfor medical aspirants.

Anjali—By adding NCERT basedtopics. Many new points had beenadded in NCERT. So, I think, CSVshould contain a separate NCERTcorner containing extra points forrevision.

CSV—Please mention your posi-tion in the merit list as well as themarks obtained in different subjects.What was your aggregate percentageof marks ?

Anjali—I got 1st rank inUttarakhand PMT (general) with 44marks in Physics, 43 in Chemistry, 41each in Zoology and Botany. All areout of 50 on aggregate 169 marks outof 200, i.e., 84·5%.

CSV—What books/magazines/newspapers did you read for G. K.preparations ?

Anjali—I had read only dailynewspaper (Amar Ujala) as only 20questions in G. K. are asked in AIIMS.So, I payed only a negligible impor-tance to G. K.

CSV—Whom would you like togive credit for your success ?

Anjali—I would like to give myregards to God and my parents, whohad been always with me during mypreparation. I am also very muchthankful to all my respected teachers.

CSV—Please tell us somethingabout your family.

Anjali—My father is a Chemistrylecturer. My mother is housewife. Ihave one sister and two brothers, allare younger to me. My sister (ShilpiSingh) had also qualified in CBSEthis year.

CSV—What in your frank opinionhas been the biggest mistake in yourpreparation for this test ?

Anjali—My unlimited sleep andlaziness. When I lost my interest instudies in between and found it diffi-cult to regain it. When I got goodranks in tests (in watching), I becamequite loose.

CSV—What message would youlike to give for our readers of CSV ?

Anjali—First make your conceptsclear. Read the theory deeply, solveMCQs, complete a book wholly,instead of solving few topics fromdifferent books. Nothing is impossibleif we have patience and strong willpower. Impossible word itself says‘‘I am possible’’. Best of luck to all.

●●●

Compiled by : Dr. N. K. Singh

Cod

e N

o. 1

642

Rs.

65/

-

(Useful for Various Competitive Examinations)

New Release

● E-mail : publisher@upkar.in ● Website : www.upkar.in Upkar Prakashan, AGRA-2

MathematicalFormulae

Code Code 248248Rs. 76/-Rs. 76/-

Code 248Rs. 76/-

HINDIEDITION

C.S.V. / August / 2009 / 679 / 3

Physics

1. Smaller is the potential gradient along a potentio-meter wire, the more is the

➠Sensitivity of potentiometer

2. The standard equation of trajectory of a projectile interms of range (R) is

➠ y = x ( )1 – xR tan αααα

3. Can Kirchhoff's laws be applied to both the direct andalternating currents ?

➠Yes

4. What are the limitations of the Newton’s laws ?

➠ (a) The relation →→→→F m

→→→→a would not hold good

in case m does not remain constant.

(b) The first two laws do not hold good ineach and every frame of reference. It isonly in a very special frame of referencethat the two laws of motion hold good.Such a frame of reference is called aninertial frame of reference.

5. What limits the energy that can be provided by acell ?

➠Amount of reactants

(chemicals used in the cell)

6. Excess pressure inside a liquid drop; inside a liquidbubble, inside an air bubble in a liquid are

➠ 2TR ;

4TR ;

2TR

7. Why is it necessary to keep the concentration ofCuSO4 solution constant in Daniell cell ?

➠This helps to get a steady e.m.f.

8. What is the equation of a plane progressive simpleharmonic wave ?

➠ y = a sin 2ππππ

λλλλ (vt – x ) = a sin 2ππππ ( )t

T – xλλλλ

With usual meaning of notations9. How are the coils wound in a resistance box ?

➠The resistance coil is doubly wound on abobbin to avoid electromagnetic induction

10. The equation of a stationary wave when the wave isreflected from a rigid boundary is

➠ y = 2a cos 2ππππx

λλλλ cos

2ππππtT

11. Board of trade unit is defined as the amount of workdone when a power of one kilowatt is consumed for

➠One hour

12. The magnifying power of a telescope for image atinfinity is

➠ M = – fofe

13. The three characteristics of a heating wire are➠High resistivity, high melting point,

heat resistant14. Intensity of electric field due to a point charge ‘q ’ at a

distance r

➠ →→→→E =

14ππππεεεε0

. qr

2 r̂

15. Which of the two has more resistance : 100Wtungsten bulb or 1000W heater, both marked for220 V.

➠100W tungsten bulb16. What is the one basic difference between Biot-

Savart’s law and Coulomb’s law ?➠ The charge element appearing in Coulomb’s

law is a SCALAR but the current element (idl )appearing in Biot-Savart’s law is a VECTOR

(i →→→→dl)

17. Kepler's third law is➠ (Period of planet)2 ∝∝∝∝ (Distance from sun)3

18. The force between two magnetic poles of polestrengths m and m′ separated by distance r is givenby

➠ F = μμμμ0

4ππππ ( )mm ′′′′

r 2

19. Planets are natural satellites of the➠ Sun

20. What is the unit of ‘Magnetomotive force’ ?➠ Gilbert (Gb.)

Chemistry

21. The true use of chemistry is not to make gold, but toprepare medicines. Who said this ?

➠Paracelsus (1493–1541)

22. The first super-conductor of family 1-2-3, discoveredin 1987 is

➠ Y Ba2Cu3O7

23. First modern text book of chemistry was written by :➠Berzelius

24. The reactions in which a single reagent undergoesboth oxidation and reduction are known as

➠ Disproportionation reactions

25. Meson theory of nuclear stability was given by :➠Yukawa

C.S.V. / August / 2009 / 680

26. Isothermal gel-sol transformation brought about byshaking is termed as

➠ Thixothropy

27. The quantitative aspect of dealing with the mass andvolume relation among the reactants and products isknown as :

➠Stoichiometry

28. Fuel value of hydrogen is➠ 142 kJ/gm

29. A mole is Chemists unit for counting atoms, mole-cules, ions and other microscopic species. A molemeans :

➠A collection of 6·022 ×××× 1023 particles

30. An instrument used to detect and measure radiationby the fluorescence is known as

➠ Scintillation counter

31. The ratio of charge (e) of the electron to its mass wasfound to be 1·76 × 108 coulomb/gm (5·28× 1017

esu/gm). It was measured for first time by :➠Sir J. J. Thomson (1887)

32. Two liquids which mix in all proportions are called➠ Miscible

33. The amount of radiant energy in a photon isproportional to the frequency of the radiation. Thiswas the observation of :

➠Max Planck (1900)

34. Bonding in which the bonding electrons are relativelyfree to move throughout the three-dimensionalstructure, is called

➠ Metallic-bonding

35. The theory of quantum mechanics was independentlyand simultaneously formulated by

➠Werner Heisenberg and Erwin

Schrodinger (1925-26)36. A substance capable of behaving as either an acid or

a base is called➠ Amphoteric

37. Cyclic aliphatic hydrocarbons are known as :➠Alicyclic hydrocarbons

38. A substance which acts as a proton donor is called➠ Bronsted-Lowry acid

39. Organic compounds containing atoms such as O, S,N etc, as part of the ring system are known as :

➠Heterocyclic compounds

40. A cell used to obtain sodium metal by electrolysis ofmolten NaCl, is known as

➠ Downs cell

Zoology

41. DNA that contains genes from more than one source➠Recombinant DNA

42. Who first discovered the lethal genes ?➠ Cuenot

43. An union or joining of 2 or more arteries, veins orother vessels

➠Anastamosis

44. What is called a unit of recombination ?➠ Recon

45. Phenomenon in which synthesis of fatty acids andketo acids takes place from amino acids

➠Ketogenesis

46. What type of RNA carries a sequence of codons toribosomes ?

➠ m RNA

47. Structures in which a bone fits into a socket of theother.

➠Gomphosis

48. Which serous membrane lines the body cavity andcovers the organs therein in many animals ?

➠ Peritoneum

49. A hormone that stimulates growth of seminiferoustubules and spermatogenesis in men

➠Follicle stimulating Hormone

50. What is called the duct which derives from mesodermand conveys gametes or excretory products from thecodon to the exterior ?

➠ Coelomoduct

51. Honey bees repair, filling and cementing cracks andcrevices called

➠Propolis

52. What is called an animal that warms his body mainlyby the heat of cellular respiration ?

➠ An Ednotherm

53. Abiotically polymerized amino acids that are joined ina preferred manner

➠Proteinoid

54. What is called the product made from swim bladderof fishes containing 90% gelatin ?

➠ Isinglass

55. A group of structural and regulating genes thatfunctions as a single unit

➠Operon

56. In which naturally occurring phospholipid complex,choline is bound to phosphoric acid group ?

➠ Lecithins

57. A blood substance that produces vaso contraction➠Angiotensin

58. Which form of coelom is formed by the splitting ofembryonic mesoderm ?

➠ Schizocoel

59. A pelagia free swimming, colonial luminiscent thalia-cean is called

➠Pyrosoma

60. Which portion of internal ear is coiled and containsorgan of hearing ?

➠ Cochlea

C.S.V. / August / 2009 / 681

Botany

61. What are bacteriophages attacking E. coli called ?➠Coliphages

62. What often glucose metabolism provides ?➠ Energy in the form of ATP

63. How many capsomeres are present in a virion ofφ ×174 ?

➠Twelve64. What causes separation of the complementary chains

of a DNA molecule ?➠ Denaturation

65. What is the time between infection of host andassembly of new phages called ?

➠Eclipse period66. What is the chief means of achieving the genetic

variation ?➠ Mutation

67. Who gave the concept of pure line ?➠W. L. Johanssen

68. Which factor causes most mutation in nature ?➠ Radiation

69. Which acid was first produced by fermentation ?➠Lactic acid

70. Who gave the concept of pure line ?➠ W. L. Johanssen

71. What is Triticale ?➠A man-made cereal

72. What term is applied for specific sequence of deve-lopment of a community relating to particular set ofphysical and chemical conditions ?

➠ Sere73. What happens when a cell is placed in strong salt

solution ?➠Water comes out by exosmosis,

as a result the cell is shrinked74. CO2, CFC, NO2 and CH4 causes global warming of

atmosphere which is called➠ Green house effect

75. What type of inflorescence is found in Ocimum san-ctum ?

➠Verticillaster76. What are climbers growing on large trees and

becoming woody due to secondary growth called ?➠ Lianas

77. What refers an allelomorph ?➠Pairs of determiners

78. Where the main reservoir of nitrogen is present inatmosphere ?

➠ Atmosphere79. What is zeatin ?

➠A cytokinin-like substance isolated from milkyendosperm of corn

80. What is that physiological process called where ionsmove by mass flow and diffusion through theapoplast ?

➠ Passive uptake●●●

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C.S.V. / August / 2009 / 682

de-Broglie HypothesisThe wave theory of light was capable of explaining

the phenomena like reflection, refraction, interference,diffraction and polarisation but it failed to explain photo-electric effect and Compton effect. Quantum theory couldexplain these. Quantum theory says that a beam of lightof frequency ν consists of small packets each of energy

hν, called quanta or photons. These photons behave likeparticles, thus, light sometimes behaves like a wave andsometimes like a corpuscle.

In 1924 de-Broglie proposed that a material particlesuch as an electron, proton, atom etc. might have a dualnature as light does. According to de-Broglie hypothesis amoving particle, whatever its nature, has waveproperties associated with it. de-Broglie proposed thatthe wavelength λ associated with any moving materialparticle of momentum p is

λ = hp ,

where h is Planck’s constant.The waves associated with a material particle are

called matter-waves or de-Broglie waves and the wave-

length given by λ = hp is called de-Broglie wavelength.

If m is the mass and v the velocity of the materialparticle, then

p = mv

or λ =h

mvIf E is the kinetic energy of the material particle, then,

E =12 mv

2 = p

2

2m

or p = 2m E

Therefore, the de-Broglie wavelength is given by

λ =h

2m E

Diffraction effects have been obtained with streams ofprotons, neutrons and alpha particles, but it is evident fromde-Broglie’s equation that the greater the mass of themoving particle, the smaller is the associated wavelengthand so the more difficult detection becomes. Ordinary ob-jects have extremely small wavelengths.

If a charged particle carrying charge q is acceleratedthrough a potential difference of V volts, then

Kinetic energy E = q VIn that case,

λ =h

2mq V

When the material particles like neutrons are inthermal equilibrium at absolute temperature T, then theypossess Maxwellian distribution of velocities and so theiraverage kinetic energy is given by

E =12 mv

2rms

=32 k T

where k is Boltzmann’s constant whose value is1·38 × 10–23 joule/K. So that

λ =h

3mk T

Davisson and Germer Experiment—This experi-ment performed by Davisson and Germer is the firstexperimental proof of the wave nature of material parti-cles. The de-Broglie wavelength for an electron is lessthan 1

°A. A crystal lattice in which the atomic distancesbetween layers are of the order of 1

°A is ideal for thepurpose of studying the diffraction of electron waves. Anickel crystal C is taken. Electrons are made incident on itas a narrow beam. The incident electrons are producedby an electron gun G. The gun consists of a tungstenfilament F which is connected to a low tension battery.Electrons emitted from the filament F are made to passthrough pin holes under suitable accelerating potentials.The accelerating potential is provided with the help of ahigh-tension battery. Ordinarily, we get an electron beamof energy 50 eV from the gun.

The resolving power of any microscope increases as thewavelength used to illuminate the object decreases. Thus,the electron waves have a smaller wavelength than lightwaves and so an electron microscope reveals much moredetail. The field ion microscope gives even greater resolutionbecause the waves associated with the helium ions usedhave an even shorter wavelength.

The electron beam is made to fall normally on thecrystal. The beam is diffracted by the crystal and receivedat an angle θ by a detector D. The intensity of the diffrac-ted electrons is measured by the detector as a function ofangle θ.

In the original experiment, the detector was set at anangle of 50° to the direction of the incident beam. Thescattered electron current for different values of voltage Vwas noted. The graph between the voltage and thedetector current is as shown in figure on next page. Theexistence of peak in graph can be explained as due toconstructive interference of waves scattered from atomsin different planes of the crystal. The peak occurs at 54 V.The observed phenomenon is similar to the diffraction ofX-rays. So the experiment establishes the wave nature ofelectrons.

C.S.V. / August / 2009 / 683

The detector current is maximum when

2d sin θ = n λ,

where n is the order of diffraction, d is the atomic spacingbetween successive crystal planes and θ is the angle atwhich strong reflection takes place. The value of d asdetermined by X-ray reflection by nickel crystal, comesout to be 0·91 °A. The value of θ comes out to be 65°.

For n = 1, λ = 2d sin θ = 2 × 0·91 × sin 65° = 1·65 °A.We know that

λ =h

2me V

In the given experiment

V = 54 volt

λ =6·63 × 10–34

(2 × 9 × 10–31 × 1·6 × 10–19 × 54)1/2

= 1·67 °A.

So, the wavelength determined by two differentmethods comes out to be the same. This confirms thatelectrons are diffracted in the same way as the de-Brogliewaves.

F

G

θ

D

C 54 VVoltage V

Det

ecto

r cu

rren

t

Fig. : Davisson and Fig. : Variation of detectorGermer experiment current with voltage

Calculation shows that electrons accelerated through ap. d. of about 100V should be associated with de-Brogliewaves having a wavelength of the order of 10–10 m. This isabout the same as for X-rays and it was suggested that theconditions required to reveal the wave nature of X-rays mightalso lead to the detection of electron waves.

Thomson’s Experiment

In this experiment a beam of electrons obtained froman electron gun is made to fall normally on a thin platinumfoil. The foil is nearly 10–8 m thick. The foil can also be ofaluminium or gold. A photographic plate is placed behindthe foil. Diffraction pattern in the form of concentric rings isobtained on the photographic plate. These rings areclearly due to the diffraction of waves associated withelectrons. This is because of the randomly orientedcrystals in the foil. From the geometry of the apparatusthe voltage used to accelerate the electrons and thediameter of the electron diffraction rings. The wavelengthassociated with the beam of electrons can be determined.It is observed that the experimental value of wavelength isin close conformity with the theoretical wavelength. Thus,the de-Broglie hypothesis is verified experimentally.

Electron gun Thinplatinum

foil Photographicplate

Electrondiffractionrings

Electron and neutron diffraction phenomenon is nowcommonly used for studying crystal structure like X-rays.The electron beams can be conveniently produced andcontrolled as compared to X-rays.

Study of Crystal Structure—The electrons interactwith the surface atoms much more intimately. Therefore,electron diffraction is superior to X-ray diffraction to studythe surface structure of crystals. The great advantage ofneutron diffraction over X-ray diffraction lies in the factthat the neutron diffraction gives better information, forneutrons being neutral particles can penetrate deeper intothe nucleus. Moreover, neutron diffraction can giveinformation regarding lattice vibrations also and neutronbeam can be produced with very much smaller energythan is needed for the production of X-rays.

Q. What is the de-Broglie wavelength of a neutronwhose energy is 2 electron-volt ?

Mass of neutron = 1·676 ×××× 10–27 kg

Planck’s constant = 6·62 ×××× 10–34 joule-sec

Solution :

λ =h

2m E

=6·62 × 10–34

2 × 1·676 × 10–27 × 2 × 1·6 × 10–19

= 0·202 × 10–10 m

= 0·2 °A

Electron LensIt is a device used to focus an electron beam. It is

analogous to an optical lens but instead of using arefracting material, such as glass, it uses a coil or coils toproduce a magnetic field or an arrangement of electrodesbetween which an electric field is created. Electron lensesare used in electron microscopes.

Source of electrons

Condenser lens

Object

Objective lens

Intermediate image

Projector lens

Final image

Fig. : Principle of Transmission electron microscope

C.S.V. / August / 2009 / 684

de-Broglie wavelength associated withcharged particles

(1) For electrons (me = 9·1 × 10–31 kg)

λ =h

2mqv

=6·62 × 10–34

2 × 9·1 × 10–31 × 1·6 × 10–19 V m

=12·27

V °A

Note—The potential difference required to have an

electron of wavelength λ °A is

V =150·6

λ2 (From above equation)

(2) For protons (mp = 1·67 × 10–27 kg)

λ =0·286

V °A

(3) For deuterons (md = 2 × 1·67 × 10–27 kg)

λ =0·202

V °A

(4) For αααα-particles (mα = 4 × 1·67 × 10–27 kg)

λ =0·101

V °A

Electron Microscope

It is a form of microscope that uses a beam ofelectrons instead of a beam of light (as in the opticalmicroscope) to form a large image of a very small object.In optical microscopes the resolution is limited by thewavelength of light. High-energy electrons, however, canbe associated with a considerably shorter wavelength

than light; for example, electrons accelerated to an energyof 105 electron volt have a wavelength of 0·04 nanometre(de-Broglie wavelength) enabling a resolution of 0·2–0·5nm to be achieved. The Transmission electron micro-scope has an electron beam, sharply focussed by elec-tron lenses, passing through a very thin metallised speci-men (less than 50 nanometre thick) onto a fluorescentscreen, where a visual image is formed. This image canbe photographed.

de-Broglie wavelength associated withuncharged particles

(1) For neutrons (mn = 1·67 × 10–27 kg)

λ =h

2m E =

6·62 × 10–34

2 × 1·67 × 10

=0·286

E (eV) °A

–27 E

(2) For thermal neutrons at ordinary temperatures

E = k T

∴ λ =h

2mk T

=30·835

T °A

(3) For gas molecules

λ =h

m × Crms

⇒ For gas molecules at T K

E =32 k T

∴ λ =h

3mk T

SOME TYPICAL SOLVED EXAMPLES

Example 1. Calculate the de-Broglie wavelengthassociated with a proton moving with a velocity equal

to 120 th of the velocity of light.

Solution :

Example 2. Show that the electrons acceleratedthrough a potential difference of V volts will have a

wave of wavelength 12·27

√⎯ V °°°°A associated with them.

Solution :

Example 3. What voltage must be applied to anelectron microscope to produce electrons of wave-

length 0·5 °°°°A ?

Solution :

Example 4. Calculate the de-Broglie wavelengthof an αααα-particle accelerated through a potential diffe-rences of 4000 volt.

Given Planck’s constant

h = 6·62 ×××× 10–34 joule-sec.

Mass of proton mp = 1·67 ×××× 10–27 kg

Electronic charge e = 1·6 ×××× 10–19 coulomb

Solution :

C.S.V. / August / 2009 / 685

Example 5. Find the energy of the neutron in unitsof electron volt whose de-Broglie wavelength is

1 °°°°A.

(Given : Mass of the neutron = 1·674 ×××× 10–27 kg

Planck’s constant h = 6·60 ×××× 10–34 joule-sec)

Solution :

Example 6. Energy of a particle at absolutetemperature T is of the order of k T. Calculate the

wavelength of thermal neutrons at 27°°°°C. (Given : Massof the neutron = 1·67 ×××× 10–27 kg. Planck’s constanth = 6·60 ×××× 10–34 joule-sec, and Boltzmann’s constantk = 8·6 ×××× 10–5 eV deg–1.)

Solution :

Example 7. What is the energy of gamma photon

having a wavelength of 1 °°°°A ?

(Given : Planck’s constant h = 6·62 ×××× 10–34 joule-sec and speed of light c = 3 ×××× 108 m/sec.)

Solution :

Example 8. For a moving electron with massm = 2m 0, calculate the de-Broglie wavelength in termsof rest mass m 0 and velocity of light c.

Solution :

Example 9. Calculate the de-Broglie wavelengthof an electron which has kinetic energy equal to15 eV.

Solution :

C.S.V. / August / 2009 / 686

OBJECTIVE QUESTIONS

1. Matter waves—(A) Are electromagnetic waves(B) Are transverse waves(C) Are longitudinal waves(D) Exhibit diffraction

2. Neglecting variation of mass withvelocity, the wavelength asso-ciated with an electron havingthe kinetic energy E is proportio-nal to—

(A) E1/2 (B) E–1/2

(C) E (D) E–2

3. A proton when acceleratedthrough a potential difference ofV volts has a wavelength λ asso-ciated with it. An α -particle inorder to have the same λ mustbe accelerated through a poten-tial difference of—

(A) V volt (B) 4V volt

(C) 2V volt (D)V8 volt

4. An electron accelerated througha potential difference of V voltshas a wavelength λ associatedwith it. Mass of proton is nearly2000 times that of an electron. Inorder to have the same λ forproton, it must be acceleratedthrough a potential differenceof—

(A) V volt (B) 2000 V volt

(C)V

2000 volt (D) 2000 V volt

5. In an electron microscope if thepotential is increased from 20 kVto 80 kV, the resolving power Rof the microscope will become—(A) R (B) 2R(C) 4R (D) R/2

6. A proton and an α-particle areaccelerated through the samepotential difference. The ratio oftheir de-Broglie wavelengthsλp/λα is—(A) 1 (B) 2

(C) 8 (D)18

7. One can study crystal structureby electron diffraction as well asby neutron diffraction. In order tohave the same wavelength λ forthe electron (mass me) and

neutron (mass mn ), their velo-cities should be in the ratio—

(electron velocity/neutron velocity)

(A) One (B)memn

(C) me × mn (D)mnme

8. Which of the following figuresrepresents the variation of parti-cle momentum and associatedde-Broglie wavelength ?

(A)P

(B)P

(C)

P

(D)

P

9. A beam of monoenergetic neu-trons corresponding to 27°C isallowed to fall on a crystal. A firstorder reflection is observed at aglancing angle 30°, calculate theinterplanar spacing of the crys-tal—(Given : Planck’s constant h= 6·62 × 10–34 joule-sec, Mass ofneutron mn = 1·67 × 10–27 kg,Boltzmann’s constant, k = 1·38× 10–23 joule/degree)

(A) 1·78 °A (B) 2·18 °A

(C) 8·12 °A (D) 3·12 °A

10. Calculate the de-Broglie wave-length of neutron of kineticenergy 54 eV— (Given : Mass ofneutron = 1·67 × 10–27 kg)

(A) 0·4 °A (B) 4·0 °A

(C) 0·04 °A (D) 0·004 °A

11. Calculate the de-Broglie wave-length associated with a protonmoving with velocity 3 × 107 m/s—

(A) 1·32 × 10–14 m

(B) 1·14 × 10–14 m

(C) 1·14 × 10–32 m

(D) 4·14 × 10–10 m

12. Calculate the de-Broglie wave-length associated with an elec-tron of 50 eV energy—

(A) 17·3 °A (B) 1·73 °A

(C) 7·13 °A (D) 3·17 °A

13. What is the momentum of an

electron if its wavelength is 2 °A ?(Given : h = 6·63 × 1 0–34 joule-sec.)(A) 3·22 × 10–20 kg ms–1

(B) 23·2 × 10–21 kg ms–1

(C) 2·32 × 10–23 kg ms–1

(D) 3·32 × 10–24 kg ms–1

14. Calculate the de-Broglie wave-length of an α-particle accelera-ted through a potential differenceof 2000 volt—

(A) 3·2 × 10–1 °A

(B) 5·0 × 10–2 °A

(C) 2·3 × 10–3 °A

(D) 1·1 °A

15. The de-Broglie wavelength of an

electron is 1·224 °A. The energyof electron is—(A) 1 eV (B) 10 eV(C) 100 eV (D) 1224 eV

16. The de-Broglie wavelength of150 eV electron will be—

(A) 1 °A (B) 12·27 °A

(C) 0·5 °A (D) 1·5 °A

17. Obtain the de-Broglie wave-length of a neutron of kineticenergy 150 eV—(Given : Mass of neutron = 1·675× 10–27 kg)(A) 3·351 × 10–12 m(B) 2·331 × 10–10 m(C) 2·335 × 10–12 m(D) 5·332 × 10–12 m

18. Calculate the de-Broglie wave-length for electrons if their speedis 105 ms–1—

(Given : h = 6·63 × 10–34 joule-sec)(A) 7·3 × 10–9 m(B) 3·7 × 10–9 m(C) 5·3 × 10–9 m(D) 5·5 × 10–9 m

19. Calculate the de-Broglie wave-length for protons if their speedis 105 ms–1—(Given : h = 6·63 × 10–34 joule-sec)(A) 39·71 × 10–12 m(B) 3·97 × 10–12 m(C) 9·37 × 10–12 m

(D) 7·39 × 10–12 m

C.S.V. / August / 2009 / 687

20. Electron microscope works onthe principle of—(A) Particle nature of electrons(B) Wave nature of light(C) Quantum nature of light(D) Wave nature of moving ele-

ctrons

ANSWERS WITH HINTS

(Continued on Page 747 )

C.S.V. / August / 2009 / 688

Isothermal Process

It is a thermodynamic process in which the pressureand volume of system change but temperature remainsconstant. An isothermal process is carried out either bysupplying heat to the substance or by extracting heat fromit. A process has to be extremely slow to be isothermal.

Examples of Isothermal Process

(1) The temperature of a substance remains constantduring melting. So, the melting process is an isothermalprocess.

(2) When a substance boils, its temperature remainsconstant. So, boiling is an isothermal process.

(3) Consider an ideal gas enclosed in a conductingcylinder fitted with a conducting piston. Let the gas beallowed to expand very slowly. This will cause a very slowcooling of the gas. But heat will be conducted into thecylinder from the surroundings. Thus, the temperature ofthe gas remains constant.

If the gas is compressed very slowly, heat will beproduced. But this heat will be conducted to thesurroundings. So, the temperature of the gas remainsconstant.

Adiabatic Process

It is the thermodynamic process in which pressure,volume and temperature of the system change but thereis no exchange of heat between the system and thesurroundings. A process has to be sudden and quick to beadiabatic.

Examples of Adiabatic Process

(1) Consider a gas enclosed in a thermally insulatedcylinder fitted with a non-conducting piston. If the gas iscompressed suddenly by moving the piston downwards,some heat is produced. This heat cannot escape thecylinder. Consequently, there will be an increase in thetemperature of the gas.

(2) If the above gas is suddenly expanded by movingthe piston outwards, there will be a decrease in thetemperature of the gas.

(3) Bursting of a cycle tube.(4) Propagation of sound waves in a gas.(5) Expansion of gases in internal combustion engine.(6) Expansion of steam in the cylinder of a steam

engine.

Thermodynamical system is an assembly of anextremely large number of particles (atoms or molecules) sothat the assembly has a certain value of pressure, volumeand temperature.

Indicator diagrams : Pressure-volume graphs arecalled indicator diagrams. They can be used to show thecycle of changes taking place in an engine. The diagrambelow shows in simplified form, what happens in a

cylinder of a petrol engine, where there is a compressionand expansion of a gas as a piston goes up and down.

P

C

B

D

A

V

networkdone

Thermodynamic variables or parameters are thequantities like pressure, volume and temperature which helpus to study the behaviour of a thermodynamic system.

A to B—Gas (air-petrol mixture) is compressedadiabatically by the raising piston. This causes a rise intemperature.

B to C—Ignited by a spark, the mixture explodes.The further rise in temperature causes a further rise inpressure.

C to D—The hot high pressure gas pushes down thepiston as it expands adiabatically and the temperaturefalls.

Thermodynamic process is said to take place if somechange occurs in the state of a thermodynamic system i.e.the thermodynamic variable of the system change with time.

D to E—The warm waste gas is removed andreplaced by cooler, fresh gas mixture, ready for the nextcycle.

Note—From A to B, work is done on the gas. From Cto D work is done by the gas. The shaded area representsthe net work done during the cycle.

● Cyclic process is that thermodynamic process in whichthe system returns to its initial stage after undergoing aseries of changes.

● Non-cyclic process is that process in which the systemdoes not return to its initial stage.

● Isolated system is that system which is completelyisolated from its surroundings.

First law of thermodynamics : If some quantity ofheat is supplied to a system capable of doing externalwork. Then the quantity of heat (d Q) absorbed by thesystem is equal to the sum of the increase in the internalenergy (d U) of the system and the external work (d W)done by the system, i.e.

d Q = d U + d W

Equation of isothermal process : The perfect gasequation is

PV = RT,

C.S.V. / August / 2009 / 689

where R is gas constant in an isothermal process, T isconstant. Therefore,

PV = constant (∴

R is constant)

That is, the product of the pressure (P) and volume(V) of a given mass of a perfect gas remains constant inan isothermal process. So, Boyle’s law is obeyed in anisothermal process.

Isotherm : A graph between pressure and volume ofa given mass of a gas at constant temperature is knownas isotherm or isothermal of a gas. In the figure below,two isotherms for a given gas at two different tempera-tures T1 and T2 are shown.

P

V

T2 > T1

T2

T1

Fig. : Isotherms of a gas

Application of first law of thermodynamics toisothermal process : The first law of thermodynamicsstates

d Q = d U + d WThe internal energy of an ideal gas depends only on

temperature. In an isothermal process, temperatureremains constant. Therefore

d U = 0

and d Q = d W

When an ideal gas expands isothermally, it doesmechanical work d W and absorbs an equivalent amountof heat d Q from the surroundings. Similarly, when anideal gas is compressed isothermally by doing amechanical work d W on it, it rejects an equivalent quantityof heat d Q to the surroundings.

Q. Calculate the work done by an expanding gas.Solution :

Above, a gas at pressure P exerts a force PA on thepiston and moves it a short distance Δx. If the expansion ofthe gas is so small that the pressure does not change :

Work done by the gas = Force × displacement

= PAΔx

But, AΔx = ΔV, the increase in volume

So, Work done by gas = PΔV

Work done in an isothermal process : Considerone mole of an ideal gas enclosed in a cylinder havingperfectly non-conducting wallsand a perfectly conductingbottom. Let the cylinder be fittedwith a frictionless and insulatingpiston of cross-sectional area A.Let d W be the work done by thegas when the piston moves upthrough an elementary distancedx. Let P be the pressure of thegas. Then,

d W = P × A × dx(work = force × distance)

Fig. : Work done inisothermal process

or d W = PdV

where d V is the infinitesimally small increase in thevolume of the gas.

Let Wiso be the total work done by the gas when thegas expands isothermally from an initial volume V1 to thefinal volume V2.

Then Wiso = ⌡⎮⎮⌠

V2

V1

Pd V

But, PV = RT or P = RTV

where T is the constant temperature at which isothermalexpansion takes place.

∴ Wiso = ⌡⎮⎮⌠

V2

V1

RTV d V

= RT ⌡⎮⎮⌠

V2

V1

1Vd V

or Wiso = RT [ ]1oge V V2V1

= RT [loge V2 – loge V1]

or Wiso = RT loge V2V1

= 2·3026 RT log10 V2V1

The graph below left shows the expansion of a gas at

constant pressure. The area under the graph gives the work

done by the gas (PΔV). The same principle applies when the

pressure is not constant, as shown below right :

P

O V

P

O V

area =ea =workworkdonedone

area =workdone areaea

= work done= work donearea= work done

Let P1 and P2 be the pressures corresponding tothe volumes V1 and V2 respectively.

C.S.V. / August / 2009 / 690

Then, P2V2 = P1V1 or V2V1

= P1P2

∴ Wiso = RT loge P1P2

= 2·3026 RT log10 P1P2

For μ moles of an ideal gas

Wiso = μRT loge V2V1

= 2·3026 μRT log10 V2V1

If we consider one gram of an ideal gas,

Wiso =RTM loge

V2V1

where M is the molecular weight of gas

or Wiso = r T loge V2V1

= 2·3026 r T log10 V2V1

where r is the principal gas constant, i.e., gas constantfor one gram of gas.

Equations of Adiabatic Process

(i) Adiabatic relation between P and V for idealgas : Consider one mole of a gas contained in a perfectlynon-conducting cylinder fitted with a non-conductingpiston. Let P, V and T be thepressure, volume and tempera-ture respectively of the gas. Letthe gas be compressed adiabati-cally so that the piston movesinwards through a distance dx.Let A be the cross-sectionalarea of a piston.

Force acting on the piston

= P × AFig. : Work done in

adiabatic process

Work done,d W = Force × distance

= PAdx = Pd V

where d V is the decrease in the volume of the gas.

Q. A certain volume of gas suffers an expansion of

0·25 m3 at a constant pressure of 103 Nm–2. Calculate

the work done.

Solution : The work doneW = PΔV

where P = Pressure

ΔV = Change in volume

∴ W = 103 × 0·25 = 250 joule

The heat generated due to compression causes arise of temperature d T. This heat energy is equal toCvdT, where Cv is gram molecular specific heat at

constant volume. This is equal to change in the internalenergy of the gas.

d U = Cv d T

But, d Q = d U + d W

In an adiabatic process, no exchange of heatbetween the system and the surrounding is allowed.

∴ d Q = 0

So, Cv d T + Pd V = 0

For an ideal gas, PV = RT

Differentiating, Pd V + Vd P = Rd T

or d T =Pd V + Vd P

R

But, Cvd T + Pd V = 0

∴ Cv [ ]Pd V + Vd PR + Pd V = 0

or Cv Pd V + Cv Vd P + RPd V = 0

or (Cv + R) Pd V + Cv Vd P = 0

But, Cp –Cv = R

or Cp = Cv + R

∴ Cp Pd V + Cv Vd P = 0

Dividing both sides by Cv PV, we get

Cp Pd VCv PV +

Cv Vd PCv PV = 0

or γ dVV +

dpP = 0 [ ]∴ Cp

Cv = γ

Integrating, ∫ γdVV + ∫

dpP = constant

or γ ∫

1V d V + ∫

1Pd P = constant

or γ log V + log P = constant

or log Vγ + log P = constant

or log PVγ = constant

or PVγ = antilog (constant)

= another constant K

∴ PVγ = K

which is the required relation between the pressure andvolume of a gas.

If P1V1 be the initial and P2V2 be the final pressureand volume respectively of the gas for an adiabaticchange, then

P1V1γ = P2V2

γ

(ii) Adiabatic relation between volume andtemperature : For a perfect gas

PV = RT

or P =RTV

But, PVγ = K

∴RTV Vγ = K

C.S.V. / August / 2009 / 691

or RT Vγ – 1 = K

or TVγ – 1 =KR = K1 (say)

∴ TVγ – 1 = constant

(iii) Adiabatic relation between pressure andtemperature : For a perfect gas,

PV = RT

or V =RTP

But PVγ = K

∴ P [ ]RTP

γ

= K

or P RγTγ

Pγ = K

or P1 – γ · Rγ Tγ = K

or P1 – γ Tγ =KRγ = K2 (say)

∴ P1 – γ Tγ = constant

An adiabatic is steeper than an isotherm : For anisothermal process, PV = constant

Differentiating, Pd V + Vd P = 0

ord PdV = –

PV

But d Pd V represents the slope of isotherm

∴ Slope of isotherm,( )d Pd V iso

= – PV

For an adiabatic process, PVγ = constant

Differentiating,

P·γ Vγ – 1dV + VγdP = 0

or Vγd P = – γ PVγ – 1d V

ord Pd V = –

γ PVγ – 1

Vγ

= – γ PV

Hence, slope of adiabatic,d Pd V = – γ

PV

= γ ( )– PV

= γ ( )d Pd V iso

,

(From above)

But, γ > 1

Therefore, the slope of adiabatic is greater than theslope of isotherm

or an adiabatic is steeper than an isotherm.Work done in an adiabatic process : From first law

of thermodynamics.d Q = d U + d W

In an adiabatic process, no exchange of heat isallowed between the system and the surroundings.

∴ d Q = 0

∴ d U + d W = 0

or d W = – d UBut d U = Cv d T

where Cv is the molar specific heat of gas at constantvolume.

∴ d W = – Cv d T

Let Wadia be the work done when the gas expandsadiabatically from temperature T1 to temperature T2.

Then, Wadia = – ∫T2

T1Cv d T

= – Cv ∫T2

T1d T

= – Cv [ T ]

T2T1

= – Cv [T2 – T1]

or Wadia = Cv [T1 – T2]

which is the expression for the work done for one mole ofan ideal gas during adiabatic process.

Now, Cp – Cv = Rwhere Cp is the molar specific heat at constant pressureand R is the molar gas constant.

Dividing both sides by Cv, we get

CpCv

– CvCv

=RCv

or γ – 1 =RCv

or Cv =R

γ – 1

So, Wadia =R

γ – 1 (T1 – T2)

which is another expression for the work done duringadiabatic process.

If we consider of one gram of ideal gas.

Then Wadia =r

– 1 (T1 – T2)

where r is the principal gas constant, i.e., gas constant for1 gram of gas.

The work done by an ideal gas during adiabatic expansion(or compression) is proportional to the fall (or rise) in thetemperature of the gas.

Note :● If the gas expands adiabatically, work is done by the

gas. So, Wadia is positive.

∴ T1 > T2

● So, the gas cools during adiabatic expansion.

● If the gas is compressed adiabatically, work is doneon the gas. So, Wadia is negative.

∴ T1 < T2

So, the gas heats up during adiabatic compres-sion.

C.S.V. / August / 2009 / 692

Comparison of Amounts of Work Done duringIsothermal and Adiabatic Process

In the case of expansion, the work done in anisothermal process is morethan the work done in anadiabatic process as isshown in fig. below. AB isthe isotherm and AC, thecorresponding adiabatic.The shaded area gives theexcess of work done inisothermal expansion overthe work done in corre-sponding adiabatic expan-sion.

P

A

V

B

C

Isotherm

Adiabatic

Fig. : Works done in isother-mal and adiabatic expansion

Isothermal lies above the abiabatic in the case ofexpansion. But in case of compression, the adiabatic liesabove the isothermal.

In the case of compression, the work done in anadiabatic process is morethan the work done in anisothermal process asshown in figure below. ABis the isotherm while AC isthe corresponding adiabatic.The shaded area gives theexcess of work done inadiabatic compression overthe work done in corre-sponding isothermal com-pression.

PA

V

B

C

Isotherm

Adiabatic

Fig. : Works done in isother-mal and adiabatic com-

pressions

First law of thermodynamics and melting process :When a substance melts, the change in volume (dV) isvery small and is negligible. The temperature remainsconstant during the melting process.

Let a mass m of a substance be melted.Heat absorbed during melting process d Q = m L

where L is the latent heat of fusion of substance.

Also d W = Pd V = P × 0 = 0

According to first law of thermodynamics

d Q = d U + d W

So, m L = d U

That is, the internal energy increases by mL duringthe melting process.

First law of thermodynamics and boiling process :On boiling a liquid, it changes into vapour at constanttemperature called boiling point.

Let a liquid of mass m vaporise. Let Vl and Vv be thevolumes of the liquid and vapours respectively.

The work done in expanding at constant temperatureand pressure P.

d W = Pd V = P (Vv – Vl)

Heat absorbed during boiling process,

d Q = m L

where L is the latent heat of vaporisation.

Change in internal energy,

d U = Uv – U l

where Uv and Ul are the internal energies of the liquid andvapours respectively.

First law of thermodynamics states

d Q = d U + d W

∴ m L = (Uv – U l ) + P (Vv – Vl )

or Uv – U l = m L – P(Vv – Vl )

● The latent heat of fusion of ice is 80 cal g– 1 or k cal kg– 1

at normal atmospheric pressure.

● The latent heat of vaporisation of water is 540 k cal kg– 1

or 540 cal g– 1 under normal atmospheric pressure.

From which the gain in internal energy can becomputed.

SOME TYPICAL SOLVED EXAMPLES

Example 1. Calculate the work done in compress-ing 3 moles of a gas from 4 litre to 1 litre at constanttemperature. (R = 8·3 J mol–1, K–1)

Solution :

Example 2. Keeping the temperature constant at27°°°°C one mole of a perfect gas is allowed to expandfrom 4 atmospheric pressure to 1 atmospheric pres-sure. Calculate the work done by the gas.

(R = 8·314 JK– 1 mol– 1)

C.S.V. / August / 2009 / 693

Solution :

Example 3. Having suddenly compressed thevolume of dry air is reduced to one quarter atatmospheric pressure. What will be its pressure ?

(γγγγ = 1·5)

Solution :

Example 4. A gas is compressed adiabatically toone quarter of its initial volume at 17°°°°C. Calculate theresulting temperature. (γγγγ for gas = 1·5)

Solution :

Example 5. The density of a gas at 27°°°°C and105 N-m– 2 pressure is 1·775 kg m– 3 and its specificthermal capacity at constant pressure is 846 J kg– 1

K– 1. Determine the ratio of specific thermal capacityat constant pressure to that at constant volume.

Solution :

Example 6. The volume of 1 mole of oxygen atconstant pressure and temperature is 22·4 litre.Calculate the two specific heats of oxygen.

Solution :

Example 7. Dry air at 15°°°°C and 10 atmosphericpressure is suddenly released at atmospheric pres-sure. Find out the temperature of air. (γγγγ for air = 1·41)

Solution :

C.S.V. / August / 2009 / 694

Example 8. At N.T.P. certain gas expands adia-batically from 1 litre to 2 litre. Change the newtemperature of the gas. (γγγγ = 1·5)

Solution :

Example 9. A certain mass of air is expandedadiabatically at 0°°°°C so that its volume gets threefold.How much does its temperature fall ? (γγγγ = 1·4)

Solution :

Example 10. A cylinder fitted with a pistoncontains 0·2 mole of air at temperature 27°°°°C. Thepiston is pushed so slowly that the air within thecylinder remains in thermal equilibrium with thesurroundings. Find the work done by the system if thefinal volume is twice the initial volume.

Solution :

OBJECTIVE QUESTIONS

1. The first law of thermodynamicsis essentially a restatement ofthe law of—

(A) Conservation of momentum

(B) Conservation of charge

(C) Conservation of spin

(D) Conservation of energy

2. A certain gas at atmosphericpressure is compressed adiabati-cally so that its volume becomeshalf of its original volume.Calculate the resulting pressurein dyne cm– 2. (Given : γ = 1·4)

(A) 4·0 × 103 dyne cm– 2

(B) 26·7 × 106 dyne cm– 2

(C) 2·67 × 106 dyne cm– 2

(D) 2·67 × 108 dyne cm– 2

3. Two samples of a gas initially atsame temperature and pressureare compressed from a volume V

to V2 · One sample is compressed

isothermally and the other adia-

batically. Now which of the follo-wing is true ?(A) Padia = Piso

(B) Padia > Piso

(C) Padia < Piso

(D) Padia << Piso

4. A motor tyre pumped to apressure of 3 atmosphere sud-denly bursts. What is the fall intemperature due to adiabatic ex-pansion ? The temperature of airbefore expansion is 27°C.

(Given γ = 1·4)(A) 219·2 K (B) 300 K(C) 75 K (D) 80·8 K

5. A gram molecule of a gas at127°C expands isothermally untilits volume is doubled. Find theamount of work done and heatabsorbed.

(A) 2·301 × 1010 erg, 500 k cal

(B) 2·301 × 1010 J, 547·9 cal

(C) 2·301 × 1010 erg, 547·9 cal

(D) 2·301 J, 600 cal

6. Calculate the work done tocompress isothermally 1 g ofhydrogen gas at NTP to half ofits initial volume. Gas constantR = 8·31 J mol–1 K–1. Also findthe amount of heat evolved andthe change in internal energy.

(A) – 786·3 J, 187·21 cal, 100 J

(B) 786·3 J, 187·21 cal, 1000 J

(C) – 786·3 J, 187·21 cal, zero

(D) 786·3 erg, 187·21 cal, 500erg

7. Ten moles of hydrogen at NTPare compressed adiabatically sothat its temperature becomes400°C. How much work is doneon the gas ? Also calculate theincrease in the internal energy ofthe gas.

(Given R = 8·4 J mol–1 K–1, γ = 1·4)

(A) – 8·4 × 104 J, 8·4 × 104 J

(B) 8·4 × 104 J, – 8·4 × 104 J

(C) 8·4 × 104 J, 8·4 × 104 J

(D) – 8·4 × 104, – 8·4 × 104 J

C.S.V. / August / 2009 / 695 / 4

8. A certain volume of dry air at NTPis allowed to expand four timesof its original volume under iso-thermal conditions. Calculate thefinal pressure and temperature.

(Given : γ = 1·4)

(A) 19 dyne cm– 2, 273°C

(B) 19 dyne cm– 2, 0°C

(C) 19 Nm– 2, 273 K(D) 19 cm of Hg, 273 K

9. A certain volume of dry air at NTPis allowed to expand four timesof its orginal volume under adia-batic conditions. Calculate thefinal pressure and temperature.

(Given γ = 1·4)(A) 10·91 dyne cm– 2, 0 K(B) 10·91 cm of Hg, 156·8 K(C) 10·91 Nm– 2, 273 K(D) 10·91 cm of Hg, 200°C

10. A litre of hydrogen at 27°C and106 dyne cm– 2 pressure expandsisothermally until its volume isdoubled. Find the final tempera-ture, pressure and work done.

(Given γ = 1·4)

(A) 27°C, 5 × 1010 dyne cm–2,6·9 × 105 erg

(B) 27°C, 5 × 10 5 dyne cm– 2,

6·9 × 108 erg(C) – 45·6°C, 1·895 × 105 dyne

cm– 2, 6·06 × 108 erg

(D) – 40°C, 2 × 104 dyne cm– 2,7 × 103 erg

11. The equation of a certain gascan be written as

( )T7

P2

1/ 5 = constant

The specific heat at constantvolume of the gas is—

(in J/mol K)(A) 0·5 R (B) 1·5 R(C) 2 R (D) 2·5 R

12. Consider a hypothetical gaswhose temperature increases to

2 times when compressed

adiabatically to half the volume.Its equation can be written as—

(A) PV5/ 3 = constant

(B) PV7/ 5 = constant

(C) PV3/ 2 = constant

(D) PV4/ 3 = constant

13. If internal energy of a gasdecreases by an amount equal tothe external work, the gas isundergoing—

(A) Adiabatic expansion

(B) Adiabatic compression

(C) Isothermal expansion

(D) Isobaric expansion

14. The pressure and volume of agiven mass of a gas at a giventemperature are P and Vrespectively. Keeping tempera-ture constant, the pressure isincreased by 10% and thendecreased by 10%. The volumenow will be—

(A) Less than V

(B) More than V

(C) Equal to V

(D) Less than V for diatomic andmore than V for monatomic

15. 1 mole of oxygen is mixed with 1mole of argon. The external workdone when the mixture is heatedat constant pressure through 1°Cis—

(A) R (B) 2 R

(C) 3 (D) 4 R

16. When a gas is isothermallyexpanded so that its volumebecomes twice, the rms velo-city—(A) Becomes twice

(B) Becomes 2 times

(C) Remains same(D) Becomes 0·7 times

17. Four processes are shown in fig.The one representing adiabaticcompression is—

P

V

A

B

B

C

(A) A (B) B(C) C (D) D

18. Four curves A, B, C and D aredrawn in the fig. for a givenamount of gas. The curves whichrepresent adiabatic and isother-mal changes are—

P

V

A

B

DC

(A) C and D respectively(B) D and C respectively(C) A and B respectively(D) B and A respectively

19. Check the correct statement—(A) Heat is a path function, while

internal energy is not(B) Internal energy is a path

function, while heat is not(C) Both heat and internal

energy are path functions(D) Both heat and internal

energy are not path func-tions

20. When a gas expands adiabati-cally—(A) No energy is required for

expansion(B) Energy is required and it

comes from the wall of thecontainer of the gas

(C) Internal energy of the gas isused in doing work

(D) Law of conservation ofenergy does not hold

ANSWERS WITH HINTS

C.S.V. / August / 2009 / 698

1. Four smooth steel balls of equalmass at rest are free to movealong a straight line without fric-tion The first ball is given a velo-city of 0·4 m/s. It collides head onwith the second elastically, thesecond one similarly with thethird and so on. The velocity ofthe last ball is—

(A) 0·4 m/s (B) 0·2 m/s

(C) 0·1 m/s (D) 0·05 m/s

2. A toy gun uses a spring of forceconstant k. When charged beforebeing triggered in the upwarddirection, the spring is compres-sed by x. If the mass of a shot ism , on being triggered it will goupto a height of—

(A)kx 2

2mg (B)kx 2

mg

(C)(kx)2

mg (D)x 2

kmg

3. If the density of a small planet isthe same as that of earth, whilethe radius of the planet is 0·2times that of the earth, thegravitational acceleration on thesurface of that planet is—

(A) 0·2 g (B) 0·4 g

(C) 2 g (D) 4 g

4. A small hollowsphere, which hasa small hole in it, is immersed inwater to a depth of 0·5 m beforeany drop penetrates into it. If thesurface tension of water is 0·073N/m, the radius of the hole is—

(A) 0·03 mm (B) 0·06 mm

(C) 0·09 mm (D) 0·15 mm

5. A satellite moves round the earthin a circular orbit of radius Rmaking one revolution per day. Asecond satellite, also moving in acircular orbit, moves round theearth once in 8 days. The radiusof the orbit of the second satelliteis—

(A) 8R (B) 4R

(C) 2R (D) R

6. If M = mass, L = length, T = timeand I = electric current, then thedimensional formula for resis-tance R will be—(A) [R] = [M1L2T–3I–2](B) [R] = [M1L2T–3I2](C) [R] = [M1L3T–2I–2](D) [R] = [M1L–2T–3I2]

7. A wire of length L and crosssectional area A is made of amaterial of Young’s modulus Y. Ifthe wire is stretched by anamount x, the work done is—

(A)YAL x (B)

YAL x2

(C)YA2L x (D)

YA2L x2

8. A convex mirror of focal length fproduces an image 1/n of thesize of the object. The distanceof the object from the mirror is—(A) f/(1–n) (B) (1 – n)f

(C) nf – 1 (D)(1 – n)f

n9. The total energy radiated from a

black body source is collected forone minute and is used to heat aquantity of water. The tempera-ture of water is found to increasefrom 20°C to 20·5°C. If the abso-lute temperature of the blackbody was doubled and the expe-riment repeated with the samequantity of water at 20°C, thetemperature of water will be—(A) 21°C (B) 22°C(C) 24°C (D) 28°C

10. An ideal monoatomic gas istaken round the cycle ABCD asshown in the figure. The workdone during the cycle is—

(P,V) (P, 2 V)

(2 P, 2 V)(2 P, V)

A D

CB

V →

P↑

(A)12 PV (B) PV

(C) 2 PV (D) Zero

11. In circuit shown below, the resis-tances are given in ohms and thebattery is assumed ideal withe.m.f. equal to 3 volt. The voltageacross the resistance R4 is—

(A) 0·4 V (B) 0·6 V(C) 1·2 V (D) 1·5 V

12. The internal energy U of a gas, ingeneral, consists of—

(A) Only translational K.E. of allits molecules

(B) Only translational and vibra-tional K.E. of all its mole-cules

(C) Only translational, vibratio-nal and rotational K. E. of allits molecules

(D) Translational, vibrational androtational K.E. plus P.E. cor-responding to molecularforces, of all its molecules

13. A bi-convex lens made of glass(refractive index 1·5) is put in aliquid of refractive index 1·7. Itsfocal length will—

(A) Decrease and change sign

(B) Increase and change sign

(C) Decrease and remain of thesame sign

(D) Increase and remain of thesame sign

14. The concept of temperature tomeasure hotness or coldness ofa body is a consequence of—

(A) Joule’s law

(B) First law of thermodynamics

(C) Newton’s law of cooling

(D) Zeroth law of thermodyna-mics

15. The displacement x (in metre) ofa particle in simple harmonicmotion is related to time t (inseconds) as x = 0·01 cos

( )πt + π4 . The frequency of the

motion will be—

(A) 0·5 Hz (B) 1·0 Hz

(C)π2 Hz (D) π Hz

C.S.V. / August / 2009 / 699

16. In the phase diagram shown, thepoint O corresponding to thetriple point of water. The regionsI, II and III respectively corre-spond to—

T→

P↑

…………

……

…

273·16 K

4·50 mmof Hg

OIII

III

(A) Liquid, Solid, Vapour(B) Solid, Liquid, Vapour(C) Liquid, Vapour, Solid(D) Solid, Vapour, Liquid

17. If the energy of the photon isincreased by a factor of 4, thenits momentum—(A) Does not change(B) Decreases by a factor of 4(C) Increases by a factor of 4(D) Decreases by a factor of 2

18. The natural frequency of a vibrat-ing body depends upon—(1) Its length(2) Mass distribution(3) Elasticity of the body(4) Boundary conditions of the

body(A) 1 and 2 only(B) 1, 2 and 3 only(C) 1 and 3 only(D) All the four

19. A piece of copper and another ofgermanium are cooled from roomtemperature to 80 K. The resis-tance of—(A) Each of them increases(B) Each of them decreases(C) Copper increases and that

of germanium decreases(D) Copper decreases and that

of germanium increases

20. A biprism experiment is set upwith edge of biprism verticalwhen half the area on the righthand side of the biprism is cove-red with an opaque material,then we shall see—(A) Fringes only in the left half of

the field of view(B) Fringes only in the right half

of the field of view(C) Intensity of the bright bands

will be halved(D) Uniform illumination

21. A 4 μF condenser is charged to400 V and then its plates arejoined through a resistance. Theheat produced in the resistanceis—(A) 0·16 J (B) 0·32 J(C) 0·64 J (D) 1·28 J

22. A charge Q is circulating withconstant speed v in a semi-circular loop of wire of radius R.The magnetic moment of thisloop is—

(A)Qv R × π

π + 2(B)

12 Qv R

(C)Qv R × π

2(π + 2)(D) Qv R

23. If F is the force required to keepa train moving at a constantspeed v, the power required is—

(A)12 Fv2 (B) Fv2

(C)12 Fv (D) Fv

24. A potential difference of 220 V ismaintained across a 12,000 Ωrheostat, as shown in the figure.The voltmeter has a resistance of6000 Ω and point C is at one-fourth of the distance a to b.Therefore, the reading of thevoltmeter will be—

V

220 V c

a

b

⎯⎯

→

←⎯⎯⎯→

(A) 32 V (B) 36 V(C) 40 V (D) 42 V

25. In an elliptic orbit under gravita-tional force, in general—(A) Tangential velocity is cons-

tant(B) Angular velocity is constant(C) Radial velocity is constant(D) Areal velocity is constant

26. A charge Q is fixed at each oftwo opposite corners of a square.A charge q is placed at each ofthe other two corners. If theresultant electric force on Q iszero, then Q and q are relatedas—

(A) Q = 2 q

(B) Q = – 2 2 q

(C) Q = – 2 q

(D) Q = 2 2 q

27. An electron beam in an X-raytube is accelerated through apotential of 40 × 103 volt. Theseare then made to fall on aTungsten target. The shortestwavelength of X-rays emitted bythe tube is—

(A) 0·31 °A (B) 3·1 °A(C) 0·31 cm (D) 3·1 cm

28. A conductor of length l is movingperpendicular to a uniform mag-netic field of induction B and per-pendicular to its length with auniform velocity v. As a result, ane.m.f. e = B l v is induced bet-ween its ends. Under this condi-tion—(A) Free-electrons in the con-

ductor drift from higherpotential end to lower poten-tial end

(B) Free-electrons in the con-ductor drift from lower poten-tial end to higher potentialend

(C) Free electrons in the con-ductor do not drift along itslength, since there is noelectric field inside the con-ductor

(D) Free electrons in the con-ductor do not drift along itslength, since the force oneach electron due to theelectric field is balanced bythe force produced on it bythe magnetic field.

29. In the spectrum of hydrogenatom, the ratio of the longestwavelength in Lyman series tothe longest wavelength in theBalmer series is—(A) 5/27 (B) 1/3(C) 4/9 (D) 3/2

30. Assertion (A) : It is convenientto define two specific heats Cpand Cv in case of gases. How-ever, it is not generally neces-sary to define two specific heatsin case of solid or liquid.Reason (R) : For a given tempe-rature rise, the expansion of asolid or liquid is negligible ascompared to a gas.(A) Only A is true(B) Only R is true

C.S.V. / August / 2009 / 700

(C) Both A and R are true and Ris the cause of A.

(D) Both A and R are true, butthey are not related to eachother.

31. In order to prepare a p-type semi-conductor, pure silicon can bedoped with—(A) Phosphorous(B) Aluminium(C) Antimony(D) Germanium

32. Plane polarised light is passedthrough a polaroid. On viewingthrough the polaroid we find thatwhen the polaroid is given onecomplete rotation about thedirection of light, one of thefollowing is observed—(A) Intensity of light gradually

decreases and remains atzero

(B) Intensity of light graduallydecreases but does notbecome zero

(C) There is no change in inten-sity

(D) The intensity of light is twicemaximum and twice zero

33. The dimension of e2/4πε0hc,where e, ε0, h and c are electro-nic charge, electric permittivity,Planck’s constant and velocity oflight in vacuum respectively, is—(A) M0L0T0 (B) M1L0T0

(C) M0L1T0 (D) M0L0T1

34. A wire of length l, radius r, den-sity d has a tension T. The esti-mated graphs for frequency n ofwire are drawn below. Whichgraph is correct ?

l →

n↑

T →

n↑

(A) (B)

r →

n↑

d →

n↑

(C) (D)

35. A pendulum bob of mass 30·7 ×10–6 kg and carrying a charge2 × 10–8 C is at rest in a hori-zontal uniform electric field of20000 V/m. The tension in the

thread of the pendulum is g = 9·8m/s2).(A) 3 × 10–4 N (B) 4 × 10–4 N(C) 5 × 10–4 N (D) 6 × 10–4 N

36. The stationary wave

y = 2a sin kx cos ωt

in a closed organ pipe is theresult of superposition of

y = a sin (ωt – kx) and—

(A) y = a sin (ωt + kx)

(B) y = – a sin (ωt – kx)

(C) y = – a sin (ωt + kx)

(D) None of these

37. If a long spring is stretched by0·02 m, its potential energy is U.If the spring is stretched by 0·1m, then its potential energy willbe—

(A)U5 (B) U

(C) 5U (D) 25 U

38. In Young’s experiment mono-chromatic light is used to illu-minate two slits A and B. Interfe-rence fringes are observed on ascreen placed in front of the slits.Now a thin foil of mica is placenormally in the path of the beamcoming from the slit A. Then—

(A) The fringes will disappear

(B) The fringe width willdecrease

(C) The fringe width willincrease

(D) The fringe width will remainunchanged

39. A police car horn emits a soundat a frequency 240 Hz when thecar is at rest. If the speed ofsound is 330 m/s, the frequencyheard by an observer who isapproaching the car at a speedof 11 m/s is—

(A) 248 Hz (B) 244 Hz

(C) 240 Hz (D) 230 Hz

40. Air is filled at 60°C in a vessel ofopen mouth. Upto what tempera-ture should the vessel be heated

so that 14 part the air may

escape ?

(A) 273°C (B) 212°C

(C) 80°C (D) 171°C

41. A charge Q is placed at the cen-tre of a cube of side ‘a’. The fluxthrough any one of the sides is—(A) Q/ε0 (B) Q/6ε0

(C) Q/4πε0 (D) Q/24ε0

42. Two point charges of + 3coulomb and + 9 coulomb repeleach other with a force of 27newton. Charges of –3 coulombare given to each of thesecharges, then the force of inte-raction is—

(A) 27 N (B) 18 N

(C) 9 N (D) Zero

43. Consider elastic collision of aparticle of mass m moving with avelocity u with another particle ofthe same mass at rest. After thecollision the projectile and thestruck particle move in directionsmaking angles θ1 and θ2 respec-tively with the initial direction ofmotion. The sum of the angles,θ1 + θ2, is—

(A) 45° (B) 90°(C) 135° (D) 180°

44. Identify the gate for which thetruth table is given below—

A B Y

1 1 01 0 10 1 10 0 1

(A) NAND gate (B) NOR gate(C) OR gate (D) AND gate

45. An ideal spring of force constantk is broken into two identicalpieces. The force constant ofeach of the smaller springs is—

(A) k/2 (B) k

(C) 2k (D) 4k

46. With usual notations, the relationbetween current gain α and βis—

(A) β = α

1 – α(B) β = α

(C) β = 1 – α

α(D) β =

α

1 + α

47. A horizontal pipe of cross sectio-nal diameter 5 cm carries waterat a velocity of 4 m/s. The pipe isconnected to a smaller pipe witha cross sectional diameter 4 cm.

C.S.V. / August / 2009 / 701

The velocity of water through thesmaller pipe is—(A) 6·25 m/s (B) 5·0 m/s(C) 3·2 m/s (D) 2·56 m/s

48. Energy of an electron in an exci-ted hydrogen atom is – 3·4 eV.The angular momentum of theelectron according to Bohr’stheory is—(A) 1·05 × 10–34 J-s

(B) 2·1 × 10–34 J-s

(C) 6·3 × 10–34 J-s(D) None of the above

49. A Carnot engine absorbs anamount Q of heat from a reser-

voir at an absolute temperature Tand rejects heat to a sink at atemperature of T/3. The amountof heat rejected is—(A) Q/4 (B) Q/3(C) Q/2 (D) 2Q/3

50. The frequency and intensity of abeam of light falling on the sur-face of a photoelectric materialare increased by a factor of 2.This will—(A) Increases the maximum

kinetic energy of the photo-electrons as well as thephotoelectric current by afactor of two

(B) Increase the maximum kine-tic energy of the photo-elec-trons and would increase thephotoelectric current by afactor of two

(C) Increase the maximum kine-tic energy of the photo-elec-trons by a factor of two andwill have no effect on themagnitude of the photoelec-tric current produced

(D) Not produce any effect onthe kinetic energy of theemitted electrons and wouldincrease the photoelectriccurrent by a factor of two.

ANSWERS WITH HINTS

C.S.V. / August / 2009 / 703

1. When a coil carrying a steadycurrent is short-circuited, thecurrent in it decreases n times intime t0. The time constant of thecircuit is—

(A) t0 log n (B)t0

log n

(C)t0

n (D)t0

n – 1

2. Which of the following is used indiesel engine ?(i) Cylinder (ii) Piston (iii) Spark-ing plug(A) (i) and (ii) only(B) (i) and (iii) only(C) All (i), (ii) and (iii)(D) None of the above

3. Which of the following state-ments is correct ?(A) A charged particle can be

accelerated by a magneticfield

(B) A charged particle can notbe accelerated by a mag-netic field

(C) The speed of a chargedparticle can be increased bya uniform magnetic field

(D) The speed of a chargedparticle can be increased bya non-uniform magnetic field

4. Mirage is formed due to—(A) Reflection(B) Refraction(C) Total internal reflection(D) Change in refractive index of

air with change in tempe-rature

5. The activity of a sample ofradioactive material is A1 at timet1 and A2 at time t2 (t2 > t1). Itsmean life is T—(A) A1t1 = A2t2

(B)A1 – A2t2 – t1

= constant

(C) A2 = A1 e( )t1 –

t2T

(D) A2 = A1 e( )t1Tt2

6. ALGOL resembles—

(A) COBOL

(B) BASIC

(C) FORTRAN

(D) All of the above

7. When a hydrogen atom emits aphoton in going from n = 5 ton = 1, its recoil speed is almost—

(A) 10–4 m /s

(B) 2 × 10–2 m /s

(C) 4 m /s

(D) 8 × 102 m /s

8. In an N-P-N transistor circuit, thecollector current is 10 mA. If 90%of the electrons emitted reachthe collector—

(A) The emitter current will be 9mA

(B) The emitter current will be11 mA

(C) The base current will be 1mA

(D) The base current will be –1mA

9. The emitter current in a transis-tor circuit is 4 mA. The transistorused has α = 0·96. What are thevalues of base current and collec-tor current ?

(A) 160 μA , 3·84 mA

(B) 80 μA, 1·92 mA

(C) 60 μA, 1·24 mA

(D) 40 μA, 0·96 mA

10. In the Bohr model of hydrogenatom, the ratio of the kineticenergy to the total energy of theelectron in n th quantum stateis—

(A) – 1 (B) + 1

(C)12 (D) 2

11. A thin prism P of angle 4° andmade of glass of refractive index1·54 is combined with anotherthin prism P′ of refractive index1·72 for dispersion without devia-

tion. The angle of prism of P′ willbe—

(A) 2·6° (B) 4°

(C) 5·33° (D) 3°

12. In figure below are given the fourlower energy levels of hydrogenatom. The probable number oftransitions is—

= 1n

= 2n

n = 3n = 4

(A) 3 (B) 4(C) 5 (D) 6

13. The illuminance of a surface is10 lux. If the total area of thesurface is 30 cm2, then theluminous flux incident on it willbe—

(A) 3 × 10 –4 lm (B) 3 × 10

–3 lm

(C) 3 × 10 –2 lm (D) 3 × 10

–1 lm

14. A ray of light passes through aglass slab of thickness t andrefractive index μ. If the speed oflight in air be ‘c ’, the time takenby the ray to cross through theplate is—

(A)t

μc(B)

tcμ

(C)μtc (D)

μct

15. A beam of unpolarised light ispassed first through a tourmalinecrystal A and then through ano-ther tourmaline crystal B orientedso that its principal plane isparallel to that of A. The intensityof final emergent light is I. Thevalue of I is—

(A)I02 (B)

I04

(C)I08 (D) None of these

16. The mass of a proton is 1840times that of an electron. Anelectron and a proton, with equalkinetic energies, enter perpendi-cularly uniform magnetic field.Now—(A) The path of proton will be

more curved than that of theelectron

(B) The path of proton will beless curved than that ofelectron

C.S.V. / August / 2009 / 704

(C) The paths of both protonand electron will be equallycurved

(D) The paths of both will bestraight

17. In an interference experimentthird bright fringe is obtained at apoint on the screen with a light

source of wavelength 7000 A° .What should be the wavelengthof light source in order to obtainfifth bright fringe at the samepoint ?

(A) 5000 A° (B) 4200 A°

(C) 6300 A° (D) 7500 A°

18. Ice skating can be used todemonstrate that when ice isunder pressure, its—(A) Melting point is lowered(B) Melting point is raised(C) Melting point remains un-

changed(D) Coefficient of friction with

metal is reduced

19. The intensity of light due to asource of light in a room full ofsmoke particles—(A) Is uniform at all distances(B) Obeys inverse square law(C) Will increase more rapidly

with distance than inversesquare law

(D) Will decrease more rapidlywith distance than inversesquare law

20. The three primary colours usedin a colour television are—(A) Red-blue-green(B) Green-yellow-red(C) Yellow-blue-black(D) Yellow-blue-red

21. The wavelength of Hα line inhydrogen spectrum was found to

be 6563 A° in laboratory. If thewavelength of same line in thespectrum of a milky way is

observed to be 6586 A° , then therecessional velocity of the milkyway will be—(A) 105 m/s

(B) 1·05 × 106 m /s

(C) 10·5 × 106 m /s

(D) 0·105 × 106 m /s

22. A magnet of magnetic momentM is freely suspended in a uni-form magnetic field of strength B.The work done in rotating themagnet through an angle θ isgiven by—(A) B M

(B) M B sin θ

(C) M B cos θ

(D) M B (1 – cos θ)

23. When an air column at 15°C anda tuning fork are soundedtogether, 4 beats per second areproduced. The frequency of thefork is less than that of the aircolumn. When the temperaturefalls to 10°C, the beat frequencydecreases by one. The fre-quency of the fork will be—(A) 110 Hz (B) 114 Hz(C) 113 Hz (D) 106 Hz

24. When the number of turns in acoil is doubled without anychange in the length of coil, itsself-inductance becomes—(A) Four times (B) Doubled(C) Halved (D) Squared

25. A source of A.C. voltageV = 180 sin ωt is connected inseries to a resistance and tworeactances XL = 10 Ω and XC =15 Ω. Calculate the active powerif the current in the circuit is 3 A—(A) 100 watt (B) 200 watt(C) 380·3 watt (D) 250 watt

26. The system shown in the figure,when slightly displaced andreleased oscillates with a periodT. If only one spring is used, theperiod of oscillation will be—

M

K

K

(A) T (B) T/2

(C) T/ 2 (D) 2 T

27. A radio can tune over thefrequency range of a portion ofMW broadcast band (800 kHz to1200 kHz). If its LC circuit has an

effective inductance of 200 μH.What must be the range of itsvariable condenser ?

(A) 88–198 pf

(B) 50 – 100 pf

(C) 25 – 75 pf

(D) None of these

28. The units nanometer, fermi, ang-strom and attometer, arranged indecreasing order will read as—

(A) Angstrom, nanometer, fermi,attometer

(B) Fermi, attometer, angstrom,nanometer

(C) Nanometer, angstrom, fermi,attometer

(D) Attometer, angstrom, fermi,nanometer

29. The power radiated by a blackbody is P, and it radiatesmaximum energy around thewavelength λ0. If the tempera-ture of the black body is nowchanged so that it radiatesmaximum energy around wave-

length 3λ04 , the power radiated by

it will increase by a factor of—

(A) 4/3 (B) 16/9

(C)6427 (D)

25681

30. The vector →B = 5

^i + 2

^j – s

^k is

perpendicular to the vector

→A = 3

^i +

^j + 2

^k for s =

(A) 1 (B) 4·7

(C) 6·3 (D) 8·5

31. Let ω be the angular velocity ofthe earth's rotation about its axis.Assume that the accelerationdue to gravity on the earth’s sur-face has the same value at theequator and the poles. An objectweighed by a spring balancegives the same reading at theequator as at a height h abovethe poles (h << R). The value ofh is—

(A)ω2 R2

g (B)ω2 R2

2g

(C)2ω2 R2

g (D)Rgω

C.S.V. / August / 2009 / 705

32. When a given amount of water isheated from 2°C to 8°C, itsvolume varies with temperatureaccording to the curve—

(A) V

t( C)°

(B)

V

t

(C) V

t

(D)

V

t

33. Two soap bubbles with radii r1and r2 (r1 > r2) come in contact.Their common surface has aradius of curvature r —

(A) r = r1 + r2

2 (B) r = r1r2

r1 – r2

(C) r = r1r2

r1 + r2(D) r = r1r2

34. The focal length of a convex lenshaving a magnifying power of12·5 X is—(A) 8 m (B) 2 cm(C) 12·5 cm (D) 8 cm

35. A metal wire of length L, area ofcross-section A and Youngmodulus Y behaves as a springof spring constant k—

(A) k = YAL (B) k =

2YAL

(C) k = YA2L (D) k =

YLA

36. The critical angle will be maxi-mum when light travels from—(A) Glass to air(B) Water to air(C) Glass to water(D) Water to glass

37. The escape velocity for a planetis ve. A particle starts from rest ata large distance from the planet,reaches the planet only undergravitational attraction, andpasses through a smooth tunnelthrough its centre. Its speed atthe centre of the planet will be—(A) ve (B) 1·5 ve

(C) 1·5 ve (D) 2 ve

38. When an electron jumps fromthe second allowed orbit to thefirst allowed orbit in hydrogen, itsangular momentum changesby—

(A) 6.6 × 10–34 kg m2/s

(B) 6.6 × 10–27 kg m2/s

(C) 1.06 × 10–34 kg m2/s

(D) 3·24 × 10–34 kg m2/s

39. A coin is placed on a horizontalplatform which undergoes simpleharmonic motion about a meanposition O. The angular frequ-ency of the simple harmonicmotion is ω. The coefficient offriction between the coin and theplatform is μ. The amplitude ofoscillation is gradually increa-sed. The coin will begin to slip onthe platform for the first time—

(A) For an amplitude of gω2

(B) At the mean position

(C) For an amplitude of μg

ω2

(D) For an amplitude of g

μ ω2

40. In the reaction

92U235 + 0n1 ⎯→ 54X142 + 36Kr89

+ 0n1 + α-particle

The number of α-particles emit-ted is—(A) Four (B) Three(C) Two (D) One

41. A sphere of mass m is suspen-ded from a string of length l fromthe point O. The sphere rotatesin a circular path in a horizontalplane. The string makes an angleα with the vertical. Find the timeperiod of rotation—

(A) T = 2π h3g

(B) T = 12π

hg

(C) T = 2π hg

(D) T = π2

hg

42. The fermi energy for a subs-tance—(A) Is directly proportional to T

(B) Is proportional to T(C) Is independent of T(D) Varies as T2

(T is the temperature in kelvin)

43. A double-star is a system of twostars moving around the centreof inertia of the system due togravitation. Find the distance

between the components of thedouble star, if the total massequals M and the period ofrevolution is T—

(A) ( )GMT2

4π21/3

(B) ( )GMT4π

1/3

(C) ( )GT4π2M

1/3

(D) ( )MT2

4π2G1/2

44. An aircraft is going at a speed ofMach 2. Its speed is nearly……… km/hr.

(A) 660 (B) 1080

(C) 1440 (D) 2380

45. A uniform wire is bent in the formof a circle of radius r = 9·8 cm. Itis initially at rest and its diameterOB is horizontal. It is swingingabout O in the vertical plane.Calculate its angular velocity ωwhen its diameter occupies theposition OB'—

O

B

G

G

B

(A) 2 rad/s (B) 15 rad/s(C) 10 rad/s (D) 20 rad/s

46. If the intensity of sound isdoubled, the sound level willincreease by nearly—(A) A factor of 2(B) 2 db(C) 3 db(D) 4 db

47. A man swimming at a rate of 5km/h wants to cross a 120 m wideriver in a direction perpendicularto the stream. If the stream speedis 4 km/h, find the direction inwhich the man swims, and thetime he takes to cross the river—

(A) cos–1 ( )– 45 , 144s

(B) sin–1 ( )– 45 , 100s

C.S.V. / August / 2009 / 706

(C) tan–1 ( )– 45 , 200s

(D) cot–1 ( )– 45 , 50s

48. Which of the following combina-tions is in order of increasingresistance ?(A) Galvanometer, voltmeter,

ammeter

(B) Ammeter, galvanometer,voltmeter

(C) Ammeter, voltmeter, gal-vanometer

(D) Voltmeter, ammeter, gal-vanometer

49. If y = x – 12 x2 is the equation of a

trajectory, find the time of flight—

(A)2g

(B)2

g

(C)2g (D)

12g

50. The unit of e.m.f. is—(A) joule(B) joule-coulomb(C) volt-coulomb(D) joule/coulomb

ANSWERS WITH HINTS

C.S.V. / August / 2009 / 709

Acidity of X–H Bond in Non-metalHydrides

1. The Polarity of the X–H Bond● When all other factors are kept constant, acids become

stronger as the X–H bond becomes more polar. Thesecond row non-metal hydrides, for example, becomemore acidic as the difference between electronegativity(ΔEN) of X and H atoms increases.

H—F : Ka = 7·2 × 10– 4 ΔEN = 1·8

H2O : Ka = 1·8 × 10– 16 ΔEN = 1·2

NH3 : Ka = 1 × 10– 33 ΔEN = 0·8

CH4 : Ka = 1 × 10– 49 ΔEN = 0·4

● HF is strongest and CH4 is the weakest acid amongabove four compounds.

● 0·1 M solution of HF is moderately acidic, H2O is muchless acidic and the acidity of NH3 is so small that thechemistry of aqueous solution of NH3 is dominated byits ability to act as base.

HF (0·1 M) ; pH = 2·1H2O (0·1 M) ; pH = 7

NH3 (0·1 M) ; pH = 11·1

● Stronger the acid weaker will be its conjugate base orvice-versa. The increasing order of basic nature ofconjugate bases is as :

F– < OH– < NH2– < CH3

–

2. Size of Atom, X

● The Ka data for HF, HCl, HBr and HI reflect the factthat X—H bond dissociation enthalpy (BDE) becomessmaller as the X atom becomes larger

HF : Ka = 7·2 × 10– 4 BDE = 569 kJ/mole

HCl : Ka = 1 × 106 BDE = 431 kJ/mole

HBr : Ka = 1 × 109 BDE = 370 kJ/mole

HI : Ka = 3 × 109 BDE = 300 kJ/mole● The increasing order of acid strength is

HF < HCl < HBr < HISimilarly H2O < H2S < H2Se

● The presence of lone pair of electrons on the centralatoms of NH3 and PH3, makes these hydrides Lewisbases.

MH3 + H+ ⎯→ MH4+

Base Conjugate acid

● As the size of central atom (N, P) increases, thestability of conjugate acid decreases, i.e., M—H bondbecomes weaker as size of central atom increases andhence basic nature decreases as :

NH3 > PH3 > AsH3 ………

● The acid strength of conjugate acids is in the order asNH4

+ < PH4+ < AsH4

+

3. Charge on Acid or Base

● The charge on molecule or ion can influence its abilityto act as an acid or a base. This is clearly shown whenthe pH of 0·1 M solutions of H3PO4 and H2PO4

–,HPO4

2– and PO43– ions are compared

H3PO4 : pH = 1·5

H2PO4– : pH = 4·4

HPO42– : pH = 9·3

PO43– : pH = 12·0

● A compound or ion becomes less acidic and morebasic as negative charge increases

Acidity : H3PO4 > H2PO4– > HPO4

2–

Basicity : H2PO4– < HPO4

2– < PO43–

● Similarly for H2SO4

Acidity : H2SO4 > HSO4–

Basicity : HSO4– > H2SO4

4. Oxidation State of Central Atom

● There is no difference in polarity, size or charge whenwe compare oxyacids of the same element such asH2SO4 and H2SO3 or HNO3 and HNO2

H2SO4 ; Ka = 1 × 103 HNO3; Ka = 28

H2SO3 ; Ka = 1·7 × 10– 2 HNO2; Ka = 5·1 × 10– 4

Acidity of these oxyacids increases significantly as theoxidation number of central atom becomes larger.H2SO4 is much stronger than H2SO3 and similarlyHNO3 is much stronger than HNO2.

● This trend is easiest to be seen in four oxyacids ofchlorine.

Acid KaOxidation number of

Cl atom

HOCl 2·9 × 10– 8 + 1

HOClO 1·1 × 10– 2 + 3

HOClO2 5·0 × 102 + 5

HOClO3 1 × 103 + 7

HOCl is the weakest and HOClO3 is the strongest acid.

● As the oxidation number of Cl atom increases, theatom becomes more electronegative. This tends todraw electrons away from oxygen atom whichsurround the Cl atom, thereby making oxygen atommore electronegative as well. As a result O—H bondbecomes more polar and compounds becomes moreacidic.

C.S.V. / August / 2009 / 710

POINTS TO REMEMBER

● O and N atoms are about the same size, yet H2O is muchstronger acid as compared to NH3. This is only on account ofO is more electronegative and O—H bond is more polar.

● NH4+ ion is a stronger acid than NH3 molecule because it is

easier to remove H+ ion from NH4+ than from NH3 molecule.

● PH3 is a stronger acid than NH3 because the compoundsbecome more acidic as the size of central atom holdinghydrogen atom increases and X—H bond becomes weaker.

● OH– ion is a conjugate base of H2O and NH2– ion is the

conjugate base of NH3. H2O is a stronger acid than NH3 andhence OH– ion is a weaker base than NH2

– ion.

● PH3 is a stronger acid than NH3, which means the PH 2– ion

must be a weaker base than NH2– ion.

● The relative strength of Bronsted bases can be predictedfrom relative strengths of their conjugate acids, combined withthe general rule that the stronger of a pair of acids always hasweaker conjugate base.

● A weaker acid is known to be displaced from its salt by astronger acid, e.g., acetylene displaces NH3 from its saltsodamide.

HC ≡ CH + NaNH2 HNH2 + HC ≡ C NaStronger Stronger Weaker Weaker

acid base acid base

Water displaces acetylene from its salt.

H–OH + HC ≡ CNa HC ≡ C–H + NaOHStronger Stronger Weaker Weaker

acid base acid base

Hence acetylene is stronger acid than NH3, but weaker acidthan H2O.

H2O > HC ≡ CH > NH3

● The decreasing order of acidity of hydrocarbons is as

HC ≡ CH > CH2 = CH2 > CH3–CH3

Conversely the decreasing order of basicities of anionsresulting from these hydrocarbons is as :

: C2H5 > H2C —— CH : > CH ≡ C :

● The concentration of an acid solution is determined by howmany mole of acid is dissolved per litre and its strength isdetermined by how completely it ionizes.

● Liquid ammonia is an amphoteric solvent like water becausethe auto-ionization of NH3 is as

2 NH3 NH4+ + NH2

–

It is just like waterH2O H+ + OH–

● H2SO4, a stronger acid, gives proton to HNO3, a weaker acidthan H2SO4.

H—OSO2—OH + H—O–NO2 → H—

H|O+

—NO2 + –OSO2OH

Acid Base Acid Base

↓H2O + NO2

+

Nitronium ion

HNO3 is acting as a base.

● H2SO4 is a weaker acid than HClO4 and acts as a base toaccept proton from HClO4.

H—OSO2–OH + H—OClO3 → –OClO3 + H—

H|O+

—SO2OH

Base Acid Base Acid

● A strong acid must be stronger than H3O+ and strong basestronger than OH– ion.

● Feeble acids and bases are weaker than H2O. Exampleacetylene (HC ≡ CH)

● Weak acids are weaker than H3O+ but stronger than H2O.Weak bases are weaker than OH– but stronger than H2O.Most of the organic acids and bases are weak.

CH3COOH + H2O H3O+ + CH3COO–

Acid Base Acid Base(Weak) (Weak) (Strong) (Strong)

CH3NH2 + H2O OH– + CH3NH3+

Base Acid Base Acid(Weak) Weak (Strong) (Strong)

● The substance which can act both as acid and base is said tobe ampholytic or amphoteric. For example : Liquid NH3, H2O,HF etc.

NH3 + NH3 NH4+ + NH2

–

Acid Base Acid Base

H2O + H2O H3O+ + OH–

Acid Base Acid Base

HF + HF H2F+ + F–

Acid Base Acid Base

OBJECTIVE QUESTIONS

1. Which of the following subs-tances is not amphoteric ?(A) HCO3

– (B) H2O(C) NH3 (D) NH4

+

2. Which is the strongest acidamong the following ?(A) Acetylene (B) Water(C) Ammonia (D) Ethylene

3. Weak acids are—(A) Weaker than H2O(B) Stronger than H3O+

(C) Stronger than H2O

(D) Always amphoteric

4. Which of the following acids isstrongest ?(A) HF (B) HCl(C) HBr (D) HI

5. Which is the weakest Bronstedbase ?(A) F– (B) Cl–

(C) Br– (D) I–

6. Which of the following sub-stances has the highest pH for0·1 M solution ?(A) NaH2PO4 (B) Na2HPO4(C) Na3PO4 (D) H3PO4

7. Which of the following acids isleast ionized in 0·1 M solution ?(A) HCN (B) HF(C) H2SO3 (D) H2CO3

8. Which is the weakest Bronstedacid among the following ?(A) HF (B) H2O(C) NH3 (D) CH4

9. For the reactionZn2+ + X– ZnX+

The keq is greatest when X– is—(A) NO3

– (B) I–

(C) ClO3– (D) F–

C.S.V. / August / 2009 / 711 / 5

10. The correct decreasing order ofbasic character is—(A) CH3

– > OH– > NH2– > F–

(B) CH3– > – NH2

– > OH– > F–

(C) F– > NH2– > OH– > CH3

–

(D) NH2– > CH3

– > OH– > F–

11. Among HS–, I–, R–NH2 and NH3,the proton accepting tendencywill be maximum and lowestrespectively in—(A) R–NH2 and I–

(B) NH3 and HS–

(C) I– and HS–

(D) HS– and I–

12. The electronegativity of chlorineatom will be highest in—(A) HOCl (B) HOClO(C) HOClO2 (D) HOClO3

13. Which among the following is theweakest base ?(A) PO4

3–

(B) H2PO4–

(C) HPO42–

(D) All are equally basic

14. Which of the following factorsexplains the higher acid strengthof HI than that of HF ?(A) Polarity in H–X bond(B) Charge on the molecule(C) Size of atom X(D) Oxidation state

15. Which of the following isincorrect statement ?(A) PH3 is a stronger acid than

NH3

(B) NO3– is a weaker base than

NO2–

(C) PH2– is a weaker base than

NH2–

(D) H2O and liquid NH3 are notamphiprotic solvents

16. The strongest Bronsted baseamong the following is—(A) ClO– (B) ClO2

–

(C) ClO3– (D) ClO4

–

17. Which is the weakest Lewisbase ?(A) H– (B) OH–

(C) Cl– (D) HCO3–

18. HNO3 in liquid HF behaves as—

(A) An acid(B) A base(C) Neither a base nor an acid(D) As a base as well as an acid

19. The conjugate base of ammo-nium ion is—(A) NH2

– (B) NH4+

(C) NH3 (D) OH–

20. Which is the correct order ofbasic nature ?(A) H2O > NH3 > PH3

(B) NH3 > PH3 > H2O(C) NH3 > H2O > PH3

(D) PH3 > NH3 > H2O

21. NH3 gas dissolves in H2O to giveNH4OH. H2O acts as—

(A) An acid(B) A base(C) A conjugate base(D) Amphoteric solvent

22. In the reactionHClO4 + H2O H3O+ + ClO4

–

(A) HClO4 is a conjugate acid ofH2O

(B) H2O is a conjugate acid ofH3O+

(C) H3O+ is a conjugate base ofH2O

(D) ClO4– is the conjugate base

of HClO4

23. An aqueous solution of aceticacid contains—(A) CH3COO– and H+

(B) CH3COO–, H3O+ and H+

(C) CH3COO–, H3O+ andCH3COOH

(D) CH3COOH, CH3COOH andH+

24. Which is the correct decreasingorder of basic strength ?

(A) CH3–CH2– > NH2

– > HC ≡ C–

> OH–

(B) HC ≡ C– > CH3–CH2– > NH2

–

> OH–

(C) OH– > NH2– > HC ≡ C–

> CH3–CH2–

(D) NH2– > HC ≡ C– > OH–

> CH3–CH2–

25. The conjugate base of HPO42–

is—(A) PO4

3– (B) H2PO4–

(C) H3PO4 (D) H4PO3

26. Correct decreasing order of acidstrength is—(A) H2SO4 > HClO > H3PO4

(B) H2SO4 > H3PO4 > HClO

(C) HClO > H3PO4 > H2SO4

(D) H3PO4 > HClO > H2SO4

27. Which can act both as Bronstedacid and base ?(A) Cl–

(B) HCO3–

(C) H3O+

(D) Both (B) and (C)

28. Correct increasing order of aci-dity is—(A) H3PO4 < HCl < H2CO3 < HI(B) H3PO4 < H2CO3 < HCl < HI(C) H2CO3 < H3PO4 < HCl < HI

(D) None is correct

29. The conjugate acid of azide ionis—(A) NH3 (B) HN3

(C) NH2– (D) N2

–

30. Strongest conjugate base resultsfrom—(A) Formic acid (B) Acetic acid(C) Ethylene (D) Acetylene

31. Liquid ammonia, like water is anamphiprotic solvent. Which is theappropriate auto-ionization equa-tion for liquid NH3 ?(A) NH3 NH2

– + H+

(B) NH3 + H+ NH4+

(C) 2 NH3 NH4+ + NH2

–

(D) All of these

32. Which of the following ions inaqueous solution gives a neutralsolution ?(A) SO3

2– (B) NH4+

(C) Na+ (D) F–

33. Which is the correct representa-tion of acidic nature of AlCl3 inwater ?(A) AlCl3 + 3 H2O Al(OH)3

+ 3 HCl(B) [Al (H2O)6]3+

[Al (H2O)5]3+ + H2O

(C) [Al (H2O)6]3+

[Al (H2O)5 OH]2+ + OH–

(D) [Al (H2O)6]3+ + H2O

[Al (H2O)5 OH]2+ + H3O+

34. Amino acid, glycine exists pre-dominantly in the form of+NH3CH2COO–. Which is theconjugate acid of glycine ?(A) NH2CH2COOH

(B) NH2CH2COO–

C.S.V. / August / 2009 / 712

(C)+NH3 CH2COOH

(D)+NH3 CH2COO–

35. Which is the Lewis acid-basereaction ?(A) Ca + S → Ca2+ + S2–

(B) NH3 + HCl → NH4+ + Cl–

(C) NH3 + BF3 → H3N : BF3(D) None of these

36. According to Lowry and Bronstedconcept, Cl– ion in aqueous solu-tion is a—(A) Weak base (B) Strong base(C) Weak acid (D) Strong acid

37. Ionic dissociation of CH3COOHis represented as—CH3COOH + H2O

H3O+ + CH3COO–

According to Lowry and Brons-ted, in this equation we have—(A) One acid and three bases(B) One acid and one base(C) Two acids and two bases(D) Three acids and one base

38. An aqueous solution of Na+

HCO3– is alkaline because—

(A) Bicarbonate ion is alkaline(B) It is an example of cationic

hydrolysis

(C) It is an example of anionichydrolysis

(D) Bicarbonate ion forms ano-ther anion

39. Which of the following reactionswill occurs when sodium hydrideis dissolved in water ?

(A) H–(aq) + H2O → H3O–

(B) H+(aq) + H2O → H3O+

(C) H–(aq) + H2O → OH– + H2

(D) None of these

40. Water can act as an acid in pre-sence of—(A) HCl (B) NH3

(C) C6H6 (D) H2SO4

41. Which among the following is theweakest base ?(A) C2H5O– (B) F–

(C) NO3– (D) CH3COO–

42. Which is the correct statement ?(A) BCl3 and AlCl3 are both

Lewis acids and BCl3 isstronger than AlCl3

(B) Both BCl3 and AlCl3 are notLewis acids

(C) BCl3 and AlCl3 are bothequally strong Lewis acids

(D) BCl3 and AlCl3 are bothLewis acids and AlCl3 isstronger than BCl3

43. The strongest Lewis base amongthe following is—(A) CH3

– (B) F–

(C) NH2– (D) OH–

44. H+ is a—(A) Lewis acid(B) Lewis base(C) Bronsted base(D) None of these

45. Cl– is a conjugate base of—(A) HOCl (B) HOClO(C) HCl (D) None of these

ANSWERS

●●●

Exam. Date6 Sept., 2009

● E-mail : publisher@upkar.in ● Website : www.upkar.in

By : Dr. Lal & Jain

Code No. 307 Rs. 135/-

UPKAR PRAKASHAN, AGRA-2

HINDI EDITION Code No. 1129 Rs. 140/-

(Including Previous Years’ Solved Papers)(Including Previous Years’ Solved Papers)(Including Previous Years’ Solved Papers)(Including Previous Years’ Solved Papers)(Including Previous Years’ Solved Papers)

� Test of Reasoning Ability

� Test of English Language

� Test of Numerical Ability

� Officework Aptitude

Main Features :Main Features :Main Features :Main Features :Main Features :

Upkar Prakashan, AGRA-2

By : Dr. Lal & Jain Code No. 1619 Rs. 270/-

HINDI EDITION Code No. 1119 Rs. 230/-

E-mail : publisher@upkar.in Website : www.upkar.in

HighlightsHighlights

(Including Previous Years’ Solved Papers)

● General Awareness ● Educa-tional Awareness ● General English Comprehension ● Logical and Analytical Reasoning ● Teaching- Learning and the School.

Subject Competence ● Science ● Mathematics ● Social Studies ● English

Exam. Date16 Aug., 2009

C.S.V. / August / 2009 / 713

Introduction

● Aromatic hydroxy compounds are of two kinds (i)Phenols : in which hydroxy group (groups) is attachedto aromatic nucleus (benzene) (ii) aromatic alcohols :in which hydroxy group is present in side chain e.g.,benzyl alcohol, C6H5CH2OH.

● Phenols are of following kinds :

(i) Monohydric phenols—

Phenol

O H

,

OH

C H3

o-cresol

,

p -cresolOH

CH3

,

OH

C H 3

m-cresol

,

OH

, OH

α-naphthol β-naphthol

(ii) Dihydric phenols—

O H

OH

Catechol(o-dihydroxy benzene)

,

OH

O HResorcinol

(m-dihydroxy benzene)

,

OH

OH

Hydroquinone or quinol(p-dihydroxy benzene)

(iii) Trihydric phenols—O H

O H

Pyrogallol

O H ,

Hydroxy quinol

OH

OH

OH ,

O H

PhloroglucinolOH OH

Methods of Preparation

● From diazonium compounds—When diazoniumsulphates are boiled with water or with solution ofCuSO4 or steam distilled, phenols are formed

N+

2HSO4–

H2O‚ H+⎯⎯→

Δ OH + N2↑ + H2SO4

Benzene diazonium Phenolhydrogen sulphate

● From sulphonic acids—Fusion of sodium salt of sul-phonic acid with NaOH followed by acidification,gives phenols

SO3Na + 2NaOH Fuse⎯→ ONa + Na2SO3

Sod. sulphonate Sod. phenoxide

H2SO4 ⎯⎯→ OH

Phenol● From aryl halides (Dow’s method)—(i) This method

has limited application in laboratory preparation ofsimple phenols.

Cl + NaOH

Chlorobenzene

300 – 350°C‚ High pressure

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→Cu. Salt

OH + NaCl

Phenol

(ii) Aryl halides are first converted into their magne-sium compound (Grignard reagent) in presence ofether, which soon changes to phenol by oxidationand subsequent hydrolysis.

C6H5Br Mg

⎯⎯→Ether

C6H5MgBr [O]

⎯→ C6H5OMgBr

H2O

⎯⎯→ C6H5OH + HOMgBrPhenol

● Higher homologues from lower ones—This isachieved by heating lower homologue with alcohol inpresence of anhydrous zinc chloride.

C6H5OH Phenol

+ CH3OH anhy. ZnCl2⎯⎯⎯⎯→ C6H4

CH3

OH + H2O

o- and p-cresols

● Raschig process—

C6H5H + HCl + O(Air)Cu/Fe

⎯⎯→ C6H5Cl + H2O

C6H5Cl + H2O (Steam)425°C

⎯⎯→ C6H5OH + HClPhenol

● From decarboxylation of salicylic acid with sodalime—

OH

COOH NaOH‚ CaO

⎯⎯⎯⎯→360 K

ONa

HCl⎯→

O H

Salicylic acid Sod. phenoxide Phenol

● From cumene—Petroleum is recent commercialsource of manufacture of phenol. Benzene and pro-pene obtained from petroleum are made to undergoFriedel-Crafts reaction to give cumene. This onoxidation in presence of a metal catalyst followed bytreatment with acid gives phenol.

C.S.V. / August / 2009 / 714

C6H6 + CH3CH——CH2 Anhy. AlCl3 ⎯⎯⎯⎯→ C6H5 —

CH3 |CH—CH3

Cumene

Oxidation⎯⎯⎯→Aerial

C6H5 —

CH3 |C— |CH3

O—OHDil. H2SO4 ⎯⎯⎯→

Cumene hydroperoxide

CH3COCH3Acetone

+ C6H5OHPhenol

Properties of Phenols● Phenols usually have high boiling points due to inter-

molecular hydrogen bonding. For example b.p. ofphenol (C6H5OH), mol. wt. = 94, is 453K while that oftoluene (C6H5CH3), mol. wt. 92, is 384 K.

● o-nitrophenol (intramolecular hydrogen bonding) hasmuch lower b.p. (Steam volatile) than m-and p-nitro-phenols (intermolecular hydrogen bonding.)

● Phenol on account of corrosive action, producesblisters when comes in contact with skin. It is poi-sonous and is good antiseptic (0·2%) and disinfectant(1%). Dilute solution of phenol is used for causteriz-ing wounds caused by mad dog bite.

● Phenol is stronger acid than alcohols since phen-oxide ion is stabilised by resonance while alkoxideion is not. Electron withdrawing groups increase theacidity while electron releasing groups decrease theacidity of phenol. The increasing order of acidicnature is as—

OH

C H 3

<

OH

<

OH

N O 2p-cresol Phenol p-nitrophenol

<

OH

N O 2

N O 2

<

OH

NO2

N O 2NO 2

2, 4-dinitrophenol Picric acid● Phenol itself is weaker acid than carbonic acid

(H2CO3) and hence does not decompose NaHCO3 togive CO2. Instead phenols are recovered fromaqueous solution of phenoxides by bubbling CO2.C6H5ONa + CO2 + H2O ⎯→ C6H5OH + NaHCO3

● Acid ionization constants (Ka) for some phenols arelisted below. Higher the value of Ka higher is theacidic nature.

Name Ka Name Ka

Phenol 1·1 × 10–10

o-cresol 0·63 × 10–10

m-cresol 0·98 × 10–10

p-cresol 0·67 × 10–10

o-fluorophenol 15 × 10–10

m-fluorophenol 5·2 × 10–10

p-fluorophenol 1·1 × 10–10

o-chlorophenol 77 × 10–10

m-chlorophenol 16 × 10–10

p-chlorophenol 6·3 × 10–10

o-aminophenol 2·0 × 10–10

m-aminophenol 69 × 10–10

o-nitrophenol 600 × 10–10

m-nitrophenol 50 × 10–10

p-nitrophenol 690 × 10–10

{2‚ 4-dinitrophenol2‚ 4‚ 6-trinitrophenol

1000‚000 × 10–10

Very large

● Properties of the nitrophenols

Name B.P. °°°°C at70 mm

Solubility ing/100 gm H2O

Remark

o-nitrophenol 100 0·2 Steam volatile dueto intramolecularH-bonding.

m-nitrophenol 194 1·35 Non-volatile insteam

p-nitrophenol High (Dec.) 1·69 Non-volatile insteam

(A) Properties in which Phenols Resemble Alco-hols (Reactions due to —OH Group)

● Reaction with Na—Sodium phenoxide is formed.

2C6H5OH + 2Na → 2C6H5ONa + H2↑ Phenol Sod. phenoxide

● Esterification—With acids in presence of poly-phosphoric acid or toluene sulphonic acid, phenylesters are formed. However, the yield of ester is poorthan that obtained from alcohols.

ArOH + R—COOH TsOH⎯⎯→ ArOCOR + H2O

Phenyl esters are better prepared by action of acidchlorides or anhydrides on phenol.

C6H5OH + C6H5COCl :–OH⎯→ C6H5OCOC6H5 + HCl

Phenol Benzoyl chloride Phenyl benzoate

This is Benzoylation or Schotten-Baumann reac-tion.

C6H5OH + (CH3CO)2O :–OH⎯→

Phenol Acetic anhydride

C6H5OCOCH3 + CH3COOHPhenyl acetate

This is acylation reaction.

Key Point

● Phenyl esters, when heated with anhy. AlCl3‚ undergoFries rearrangement. The acyl group migrates to o- or p-position to form phenolic ketones.

OCOCH3

Anhy. AlCl3⎯⎯⎯⎯→

OH

COCH3

Phenyl acetate o-hydroxyacetophenone

+

OH

COCH3 p-hydroxyaceto

phenone

● Formation of Ethers (Alkylation)

C6H5ONa + CH3—I Δ⎯→ C6H5OCH3 + NaIMethoxy benzene

(Anisole)

C6H5ONa + C2H5—I Δ⎯→ C6H5OC2H5Ethoxy benzene

(Phenetole)This reaction is known as Williamson's synthesis. Itis a nucleophilic substitution reaction.

C.S.V. / August / 2009 / 715

Key Points

● Aryl halides (ArX) are too inert to react with phenol to giveesters. However, at high temperature and in presence offinely divided copper, the reaction is achieved.

C6H5ONa + BrC6H5 High temperature⎯⎯⎯⎯⎯⎯→

CuC6H5OC6H5 + NaBr (Ullmann reaction)

Diphenyl ether

● The reaction of halogen substituted acids and halogensubstituted phenol is important because the products areWeedicides used in agriculture.

C l

OHCl + ClCH2COOH

2, 4-dichlorophenol Chloroacetic acid

⎯→

C l

C l O

C H2C

O OH

2, 4-dichlorophenoxy acetic acid (2, 4-D), a weedicide

● Bucherer reaction—Phenolic group is replaced byamino group by heating phenol with ammoniumsulphite or bisulphite.

C6H5OH (NH4)2SO3/NH3 ⎯⎯⎯⎯⎯→

150°C C6H5NH2

Replacement of phenolic group by amino group canalso be achieved by heating phenol with double com-pound of zinc chloride and ammonia at 300°C.

C6H5OH ZnCl2/NH3⎯⎯⎯⎯→

300°C C6H5NH2 + H2O

Note : This reaction is not used for preparation of ani-line but finds extensive use in commercial pre-paration of naphthyl amines from correspondingnaphthols.

● Reaction with PCl5 —Replacement of —OH group ofphenols by halogen atom is rather difficult. Whentreated with PCl5 or PBr5 halo benzenes are obtainedin small amount and main product being triphenylphosphate.

(B) Reactions in which Phenols Differ from Alco-hols

● Reaction with ferric chloride—Aqueous solution ofphenol on reacting with aqueous solution of FeCl3gives violet coloured solution probably due to forma-tion of hexa-coordinated complex.6C6H5OH + FeCl3 → [Fe(OC6H5)6]–3 + 3HCl + 3H+

This reaction is used as qualitative test of phenolicgroup.

● Reaction with zinc—Phenol on distillation with zincdust gives benzene.

C6H5OH + Zn → ZnO + C6H6

● Reduction of phenol—When reduced with H2 in pre-sence of Ni (200 – 250°C) cyclohexanol is formedwhich is used as a solvent in rubber industry.

O H

Phenol

+ 3H2 Ni⎯⎯⎯⎯→

200 – 250°C

HH O HHHHH

H HHHH

Cyclohexanol

The reduction in presence of molybdenum oxidegives benzene.

C6H5OH + H2 MoO⎯⎯→ C6H6 + H2O

● Reaction with phenyl isocyanate—Phenyl urethaneis obtained as a colourless crystalline compoundwhich is used for characterisation of phenols.

C6H5OH + C6H5N——C——O → C6H5NH·COOC6H5

Phenol Phenyl isocyanate Phenyl urethane

● Reactions of aromatic ring—The —OH group inphenol is o- and p-directing because it increaseselectron density at o- and p-positions due to reso-nance. Thus phenol undergoes electrophilic substitu-tion reactions as—

➠ Halogenation—Like —NH2 group, —OH group is somuch activating that it is rather difficult to prevent polysubstitution.

O H

Phenol

3Br2‚ H2O⎯⎯⎯→

O H

Br

B r B r

2, 4, 6-tribromo phenol

O H

O H

Resorcinol

3Br2

water⎯⎯⎯⎯→

O H

Br

B r B rOH

2, 4, 6-tribromo resorcinol

Note : If it is required to arrest the reaction at mono

substitution stage, the reaction should be carriedout in non-polar solvents like CCl4 and CS2 andat lower temperature.

O H

Phenol

Br2/CCl4 ⎯⎯⎯→or CS2

O H

Br

+

O H

B r

➠ Sulphonation—This reaction is temperature depen-dent. Low temperature favours o-sulphonation whilehigh temperature p-sulphonation.

O H(o-phenol sulphonic

acid)

(p-phenol sulphonic acid)

O H

O H

H2SO4

100

° C

S O 3H

SO3H

15

° C

➠ Nitration—(a) With dil. HNO3 at 293 K, a mixture of2-nitrophenol (major) and 4-nitrophenol (minor) isformed.

C.S.V. / August / 2009 / 716

O H

20

O H

N O2

(10%)

O H

N O2

Dil. HNO3

° C

(50%) and steam volatile

(b) Nitration of phenol with mixture of conc. HNO3and H2SO4 gives a poor yield of 2, 4, 6-trinitro-phenol (Picric acid). Poor yield is due to theexcessive oxidation side reaction.

O H

Phenol

HNO3-H2SO4⎯⎯⎯⎯→

OH

NO2

N O 2N O 2

2, 4, 6-trinitrophenol(Picric acid)

O H

O H

Resorcinol

HNO3-H2SO4 ⎯⎯⎯⎯→

OH

NO2

N O 2N O 2OH

Styphinic acid

(c) The picric acid in good yield can be prepared byfollowing reaction :

O H

Phenol

Conc. H2SO4 ⎯⎯⎯⎯→373 K

O H

S O 3H +

O H

SO3H

Conc. HNO3 ⎯⎯⎯⎯→

OH

NO2

N O 2N O 2

Picric acid

➠ Friedel-Crafts alkylation reaction—When phenolsare treated with alkyl halides in presence of anhy.AlCl3, alkyl phenols are formed.

O H

+ CH3Cl Anhy. AlCl3⎯⎯⎯⎯→

O H

C H 3

o-cresol

+

O H

C H 3p-cresol

➠ Kolbe-Schmidt reaction—

O Na

+ CO2 150°C⎯⎯→5-7 atm

O H

C OO N a

Sod. phenoxide Sod. salicylate

Dil. HCl⎯⎯⎯→—NaCl

O H

COOH

Salicylic acid

Key Point● Salicylic acid obtained by Kolbe’s-Schmidt reaction is a

very useful starting chemical for preparation of a numberof medicinal compounds.

OH

C O O H

Salicylicacid

SO

2-Acetoxyacid (aspirin) analgesic and antipyretic

Phenol salicylate (Salol)intestinal antiseptic

(CH 3C O)2O

Conc. H 2SO 4

C6H5 O HPOCl3

CH3OHConc. H2 4

benzoic

Methyl salicylate (oil of wintergreen)Use in perfumery and flavouring agent in food,drinks and cosmetics. Analgesic in rheumatic and sciatica pains

O C O C H 3

C O O H

O HC O O C 6H 5

O H

CO O CH3

➠ Reimer-Tiemann reaction—(a) The treatment ofphenol with chloroform (CHCl3) in presence of NaOHat 340K followed by hydrolysis gives o-hydroxy ben-zaldehyde (salicylaldehyde) with small amount of p-hydroxy benzaldehyde.

O H

CHCl3/NaOH⎯⎯⎯⎯→

70°C/HCl

O H

C H O

Salicylaldehyde(Main)

+

O H

C H Op-hydro-

xybenzaldehyde(Minor)

Note : o-isomer, due to chelation (intramolecular H-bonding) is more volatile than p-isomer (inter-molecular H-bonding) and can be separated bysteam distillation.

(b) By using carbon tetrachloride in place of chloro-form a mixture of o-hydroxy and p-hydroxybenzoic acids is formed.

O H

CCl4/NaOH⎯⎯⎯→

70°C

O Na

C O ONa Dil. HCl⎯⎯→—NaCl

O H

C O OH

Phenol Salicylic acid

➠ Reaction with formaldehyde—Under different con-ditions, different products are formed.(i) Phenol when treated with formalin in presence of

a dil. acid or alkali at low temperature, p-hydroxy benzyl alcohol along with small quantityof o-isomer is formed. This is known as LedererManasse reaction.

O H

+ HCHO NaOH⎯⎯→5 days

O H

C H 2 O H

+

O H

C H 2 O H

(Main product)

C.S.V. / August / 2009 / 717

(ii) Phenol on condensation with insufficientamount of HCHO in presence of an acid catalyst,gives a linear polymer called novolak.

O H

HCHO⎯⎯→

O H

C H 2 O H

C6H5OH ⎯⎯→

O H

C H 2

O H

HCHO⎯⎯→

O H

C H 2

O H

C H 2 O H

(iii) In presence of excess of formaldehyde and usinga basic catalyst like ammonia or strong aqueousalkali at higher temperature, a different kind ofcondensation occurs. Two kinds of products arefirst formed

(a) 2

O H

+ 2HCHO ⎯→

O H

C H 2 O H

C H 2 O H +

HOCH2

O H

C H2O H

Bis (hydroxymethyl) phenols

(b) 2

O H

+ 2HCHO

⎯→ HO CH2 OH

p, p′-dihydroxydiphenyl methane

These products undergo slow polymerisation to givea resin called Bakelite, a three dimensional polymer.

CH

C H 2

O H

C H 22O H

C H 2

C H2

C H2

C H 2H O

➠ Liebermann’s reaction—When a crystal of phenol isadded to a mixture of conc. H2SO4 and sodium nitriteand reaction mixture is warmed, a characteristic blueor green colour is obtained. On dilution with watercolour becomes red but again turns deep blue onadding excess of caustic soda. The chemistry of thisreaction is of diagnostic value—

HO NaNO2/H2SO4⎯⎯⎯⎯⎯→

HO N——O Tautomerises

O—— ——NOH

p-nitrosophenol Quinonemonoxime

C6H5OHConc.

2SO4H

⎣⎢⎢⎡

⎦⎥⎥⎤HO N —— ——OH+ HSO4

Phenol indophenol hydrogen sulphate(Deep blue)

2OH

HO N —— ——O

Phenol indophenol(Red)

NaOH

⎣⎢⎢⎡

⎦⎥⎥⎤–O N—— ——O Na+

Sodium salt of phenol indophenol(Deep blue)

➠ Phthalein reaction—When phenols are heated withphthalic anhydride in presence of conc. H2SO4 oranhydrous zinc chloride, phthaleins, with characteri-stic colours, depending on the pH, are formed.

Phenol Phenol

C

OH OH

H HO

O

C

O

Anhy. ZnCl2⎯⎯⎯⎯→

–H2O

Phthalic anhydride

C

OH OH

O

C

OPhenolphthalein

Phthalein reaction is a diagnostic test for phenols.Phenolphthalein is used an important indicator inacid-base titrations.When treated with alkali, it undergoes an interestingcolour change. It turns red first but on addition ofexcess of alkali it turns colourless again. Thesechanges are shown as—

C.S.V. / August / 2009 / 718

C

OH OH

O

C

O

NaOH⎯⎯→

Phenolphthalein(Colourless)

C

OH ONa

O

C

O

+–

Colourless

C

OH

ONa

C

O

+–

O

NaOH

C

O

OH

COONa+–

:

(Red colour)

NaOH

NaO+ –

–

C

OH

COONa+–

Trisodium salt (Colourless)

NaO+ – ONa

+–

Similarly resorcinol and phthalic anhydride give fluo-rescein, which dissolves in dilute alkali to produceintense green fluorescence.

➠ Oxidation—(a) Phenols are much susceptible toatmospheric oxidation than alcohols, but the nature of

products is uncertain. However, the followingsequence of reactions is predicted.

O H

O2⎯→

O

O

||

||

+ H2O

Quinone

Quinone forms brilliant red addition product withphenol.

C6H5OH + O—— ——O

⎯→ OH… O—— ——O… HO

Phenoquinone (Red)

(b) With potassium permanganate :

C6H5OHPhenol

KMnO4⎯⎯⎯→

[O] CH(OH)COOH |CH(OH)·COOH

+ 2CO2

Mesotartaricacid

(c) With a mixture of KClO3 and conc. HCl.

O H

Phenol

KClO3⎯⎯→

HCl

O

C lC l

O

C lC l

Chloranilor Tetrachloroquinone

(d) With alkaline solution of potassium persulphate.O H

K2S2O8⎯⎯⎯→Alkaline

O H

OHHydroquinone or quinol

➠ Gattermann’s aldehyde synthesis :O H

+ HCN + HClGas

Anhy. AlCl3 ⎯⎯⎯⎯→

OH

CH NH——

H2O ⎯→

OH

CHO + NH3

o-hydroxy benzaldehyde

Small quantity of p-hydroxy benzaldehyde is also for-med.

Points to Remember● Phenol is also known as carbolic acid. It is used as a disin-

fectant in carbolic soaps and lotions and as a preservativein inks.

● Bakelite is a phenol-formaldehyde thermosetting polymerwhich is generally used as an insulator for electrical work.

● 2, 4, 6-Trinitrophenol (Picric acid) is used in dyeing silk andwool and also used in preparation of explosives. It is effectivein treatment of burns.

● Phenol finds extensive use in preparation of drugs like,salicylic acid, aspirin, salol, phenacetin etc.

● The reaction of phenol and phthalic anhydride leads to theformation of phenolphthalein which is used as an indicatorin volumetric analysis and as purgative in medicines.

C.S.V. / August / 2009 / 719

● Phenyl salicylate (salol) is used in toothpastes because it isa good antiseptic.

● Methyl salicylate is main constituent of oil of wintergreen.It is obtained by direct esterification of salicylic acid withCH3OH (Fischer-Speter’s method). It is used in preparation ofiodex.

● Certain phenols like eugenol (clove oil), isoeugenol (nutmegoil), thymol (thyme and mint oil), vanillin (vanila beans) arefound in nature.

● Decreasing order of acidic nature of some derivatives ofphenol is as :(i) 2, 4-dinitrophenol > p-nitrophenol > p-cresol > m-amino-

phenol.

(ii) Phenol > p-cresol > m-cresol > o-cresol● 3, 4, 5-trihydroxy benzoic acid (gallic acid) is present in tea

and roots of pomegranate. In combined state it is present ingall nuts, oak bark and acacia bark. At m.p. it gives pyro-gallol.

HO

H O C OO H

HO

Δ⎯→

H O

HO

H O+ CO2

Gallic acid Pyrogallol

Gallic acid on exposure to air turns brown to black, thus usedin the manufacture of blue-black ink. Bismuth gallate, underthe name of dermatol is used in treatment of skin infection.

● o-hydroxy benzoic acid (salicylic acid) is used in the treat-ment of eczema and other skin diseases and rheumaticpains.

● Claisen-rearrangement—When allylphenyl ether is heatedat 573 K, it undergoes rearrangement as

573 K⎯⎯→

O–C H 2—CH = C H 2 O H

C H —CH = C H 22

Allylphenyl ether o-allylphenol

● Elbs-persulphate oxidation :

HO

Phenol

K2S2O8‚ KOH⎯⎯⎯⎯→

H O

HO

+

H O

HO

Catechol (Minor)

Hydroquinone or quinol(Major)

● Hydroxy toluenes (cresols) are used in preservation of timberunder the name of creosote oil. Lysol is an emulsion ofcrude cresols in soap solution.

● Ehrlich, the founder of chemotherapy discovered Salvarsanor 606. It is an effective drug for treatment of trypano-soniasis. Hydrochloride of salvarsan is used in treatment ofsyphilis and it is also an effective antimalarial. It is preparedfrom phenol.

N H2 HO

N H 2

A s A s

OH

=

Salvarsan (606)

● Neosalvarsan has been found more useful as it is readilywater soluble. It can be used for injection purpose.

H O

N H2

A s A s

OH

=

N H C H 2OSO2 Na

Neosalvarsan

● 1, 2, 3-trihydroxy benzene (Pyrogallol) is used in hair dye toconvert grey hair to black. In gas analysis it is also used forabsorbing unpleasant gases.

● Levodopa, 3-(3, 4-dihydroxy phenyl)-L-alanine is a drugused in treatment of perkinsonism.

LevodopaH O

HO

CH2—

NH2 |CH—COO H

● Paracetamidophenol (para-acetamol) is actually hydroxyderivative of acetanilide and is a main antipyretic.

H O N HC OCH3

Paracetamol

● Phenacetin (acetophenetidine) is ethyl ether of paracetamol.

NHCOCH3C2H5O

Phenacetin

It is widely used as an analgesic and antipyretic, usually incombination with aspirin, caffeine and codeine.

● Acid strength of phenol (C6H5OH) with respect to othercompounds is in following decreasing order :

R—COOH > H2CO3 > C6H5OH > H2O > R—OH

OBJECTIVE QUESTIONS

1. Phenol is acidic in nature and itcan react with—(A) Sodium bicarbonate(B) Potassium carbonate(C) Sodium hydroxide(D) All of these

2. Which of the following com-pounds does not contain acarboxylic group ?

(A) Anisole

(B) Picric acid(C) Carbolic acid(D) All of these

3. The fusion of sodium benzenesulphonate with NaOH, followedby acid hydrolysis, gives—(A) Benzene(B) Benzoic acid(C) Phenol(D) Sodium salicylate

4. Which does not react with phenol(C6H5OH) ?

(A) Sodium

(B) Caustic soda(C) Caustic potash(D) Washing soda

5. Salicylic acid is—(A) m-hydroxybenzoic acid(B) p-hydroxybenzoic acid(C) o-hydroxy benzoic acid(D) All of these

6. Intramolecular hydrogen bondingis present in—(A) Phenol(B) p-nitrophenol(C) o-nitrophenol(D) m-nitrophenol

7. Which of the following com-pounds is the most acidic ?

C.S.V. / August / 2009 / 720

(A) p-nitrophenol(B) p-chlorophenol(C) p-aminophenol(D) o-cresol

8. Which of the following com-pounds is formed when phenolreacts with PCl5 ?

(A) Chlorophenol(B) Chlorobenzene(C) Triphenyl phosphate(D) Both (B) and (C)

9. Which of the following reactionsdoes not give phenol or sodiumphenoxide ?

(A) C6H5COCl NaOH

⎯⎯→

(B) C6H5N2ClAlc. KOH⎯⎯→

(C) C6H5N2ClAqu. KOH⎯⎯→

(D) C6H5NNCl H2O

⎯⎯→Δ

10. Which of the following reactionsconverts phenol into salicylicacid ?(A) Kolbe’s reaction(B) Etard’s reaction(C) Reimer-Tiemann reaction(D) Dow’s reaction

11. A compound which is useful intreatment of burns is—

(A) Creosote oil(B) Picric acid(C) 2, 4, 6-trinitro resorcinol(D) Acetyl salicylic acid

12. In the reaction sequence

PhenolCCl4⎯⎯→NaOH

(A) HCl

⎯⎯→ (B)

Sodalime⎯⎯⎯→ (C)

Product (C) is—

(A) Sodium salicylate

(B) Benzoic acid

(C) Phenol

(D) Salicylic acid

13. Aspirin is an acetylation productof—(A) m-hydroxybenzoic acid(B) p-hydroxybenzoic acid(C) o-hydroxybenzoic acid(D) Phenol

14. Cumene on oxidation in pre-sence of metal catalyst followedby treatment with acid givesphenol. The cumene is—(A) o-cresol

(B) Phenyl-n-propane(C) Isopropyl benzene(D) 2, 4-dimethylbenzene

15. Which of the following com-pounds is attacked by an electro-phile most easily ?(A) Toluene(B) Phenol(C) Benzene

(D) Chlorobenzene

16. When phenol is distilled with zincdust, the product is—

(A) Toluene

(B) Benzene

(C) Benzoic acid

(D) Zinc phenoxide

17. Which compound is formed whensodium phenoxide is heated withethyl iodide ?

(A) Ethyl phenyl alcohol

(B) Phenetole

(C) Phenol(D) None of these

18. Which of the following com-pounds undergoes nitration mostreadily ?(A) Benzoic acid(B) Toluene

(C) Nitrobenzene

(D) Phenol

19. Cyclohexanol is a—

(A) Primary alcohol

(B) Secondary alcohol

(C) Phenol

(D) Tertiary alcohol

20. When phenol is treated withexcess of bromine water, whichof the following compounds isformed ?(A) m-bromophenol(B) o- and p-bromophenol(C) 2, 4, 6-tribromophenol(D) All of these

21. In the reaction sequence

Phenol Zn

⎯⎯⎯→Distillation

(X)

Conc. H2SO4⎯⎯⎯⎯⎯→

conc. HNO3‚ 60°C (Y)

Zn⎯⎯⎯→

NaOH (Z)

The product (X), (Y) and (Z)respectively are—(A) Benzene, nitrobenzene, ani-

line

(B) Benzene, dinitrobenzene, m-nitroaniline

(C) Benzene, nitrobenzene,hydrazobenzene

(D) Toluene, m -nitrobenzene,m-toluedine

22. Phenol is less acidic than—

(A) p-methoxyphenol(B) p-nitrophenol(C) Carbonic acid(D) Both (B) and (C)

23. Reimer-Tiemann reaction invol-ves—

(A) Carbonium ion intermediate

(B) Carbanion intermediate

(C) Carbene intermediate

(D) Free radical intermediate

24. C6H5OH + ClCOCH3aq. NaOH⎯⎯⎯→ C6H5OCOCH3

is called—

(A) Kolbe reaction

(B) Reimer-Tiemann reaction

(C) Schotten-Baumann reaction

(D) Dow’s reaction

25. Sodium phenoxide when heatedwith CO2 under pressure, gives—

(A) Sodium benzoate

(B) Benzoic acid

(C) Phenol

(D) o-hydroxy benzoic acid

26. Which product is formed whenphenol reacts with benzenediazonium chloride ?

(A) Phenyl hydroxylamine

(B) Para aminoazobenzene

(C) Phenyl hydrazine

(D) Para hydroxyazobenzene

27. Phenol on reacting with dil.HNO3 at room temperaturegives—

(A) Picric acid

(B) m-nitrophenol

(C) o - and p-nitrophenol

(D) All of these

28. Phenol and ethanol can be dis-tinguished by using—

(A) Caustic soda(B) Anhydrous AlCl3(C) FeCl3(D) All of these

C.S.V. / August / 2009 / 721

29. A diazonium salt reacts withphenol to give an azo dye. Thisreaction is termed as—(A) Diazotization(B) Coupling(C) Both (A) and (B)(D) None of these

30. The Dow process is used forconversion of chlorobenzeneinto—(A) Benzene(B) Nitrophenol(C) Phenol(D) Chlorophenol

31. Phenol on reacting with HNO3 orH2SO4—

(A) Forms ester(B) Gives ether(C) Does not give ester(D) Forms toluene

32. Carbolic acid is—(A) C6H5OH(B) H2CO3

(C) HCOOH(D) None of these

33. Main component of oil of wintergreen is—(A) Salicylic acid(B) Methyl salicylate(C) Phenyl salicylate(D) Ethyl salicylate

34. Phenol on heating at 60°C with amixture of chloroform and excessof NaOH and on subsequentacidification gives—

(A) 2-hydroxybenzoic acid(B) 2-hydroxy benzaldehyde(C) 3-hydroxybenzoic acid(D) 3-hydroxy benzaldehyde

35. Which of the following com-pounds gives violet colour withneutral solution of FeCl3 ?

(A) Benzoic acid(B) Formic acid(C) Salicylic acid(D) Acetic acid

36. When phenol reacts with CHCl3in presence of NaOH, salicyl-aldehyde is formed, if we usepyrene in place of CHCl3, theproduct is—(A) Salicylaldehyde(B) Phenolphthalein

(C) Salicylic acid(D) Cyclohexanol

37. Which statement regardingC6H5OH is correct ?

(A) It is a carboxylic acid(B) It is insoluble in water(C) It has higher b.p. than

toluene(D) None of these

38. Phenol is a—(A) Carboxylic acid(B) Base(C) Alcohol(D) None of these

39. Which compound gives purplecolour with neutral ferric chloridesolution ?(A) Boric acid(B) Phenol(C) Aniline(D) Benzoic acid

40. Phenol is most easily solublein—(A) NaHCO3 solution

(B) NaOH solution(C) Dil. HCl(D) None of these

41. The reaction

HO

HC O

H2O2 /OH–⎯⎯⎯→

H O

H O

is known as—(A) Reimer-Tiemann reaction(B) Liebermann’s nitroso reac-

tion(C) Dakin reaction(D) Lederer-Manasse reaction

42. Which will undergo Friedel-Craftsalkylation reaction ?

O H C H3

N O 2

COOH

(I) (II) (III)C H 2 C H 3

(IV)(A) I and IV (B) I, II, III(C) III and II (D) II and IV

43. Correct increasing order of acidicnature is—

(A) H2O < H2C2 < H2CO3< C6H5OH

(B) C2H2 < H2O < C6H5OH< H2CO3

(C) C6H5OH < C2H2 < H 2CO3

< H2O

(D) H2C2 < H 2O < H 2CO3

< C6H5OH

44. Which of the following will notform phenol or phenoxide ?(A) C6H5SO3Na(B) C6H5N2Cl(C) C6H5COOH(D) C6H5Cl

45. Phenol dissolves in aqueousNaOH to give sodium phenate,but on passing CO2 into solution,phenol is thrown out. Thisindicates that—(A) Phenol is weakly acidic, but

stronger than H2CO3

(B) Phenol is basic(C) Phenol is amphoteric(D) Phenol is weakly acidic

weaker than H2CO3

46. If salicylic acid is heated withzinc dust, the product formed willbe—(A) Phenol(B) Benzene(C) Benzoic acid(D) Zinc benzoate

47. When salicylic acid is treatedwith bromine water, the productformed is—(A) 2, 3, 4-tribromophenol(B) 3, 4, 5-tribromosalicylic acid(C) 3, 6-dibromosalicylic acid(D) 2, 4, 6-tribromophenol

48. Chemically salol is—

(A) Sodium salicylate

(B) Acetyl salicylic acid

(C) Phenyl salicylate(D) Methyl salicylate

49. Which of the following propertiesof benzoic acid and phenol aresimilar ?(A) Both are weak acids and do

not react with NaHCO3

(B) Both are equally acidic innature

C.S.V. / August / 2009 / 722

(C) Both liberate hydrogen onreacting with metallic sodium

(D) Both form esters

50. The electrophilic substitution inphenol takes place at—

(A) Orthoposition

(B) Metaposition

(C) Paraposition

(D) Both ortho and para posi-tions

51. Anisole is the reaction product ofphenol and dimethyl sulphate.This reaction is termed as—

(A) Coupling

(B) Esterification

(C) Etherification

(D) None of these

52. Which of the following reactionsis given by phenol ?

(A) Williamson's reaction

(B) Coupling reaction

(C) Lederer-Manasse reaction

(D) All of these

53. The reactionO H

+ HCHO OH– or⎯⎯→H+

O H

C H 2 O H +

O H

C H 2 O His called—

(A) Reimer-Tiemann reaction

(B) Sand Meyer reaction

(C) Lederer-Manasse reaction

(D) Kolbe reaction

54. A compound ‘A’ when treatedwith CH3OH and a few drops ofH2SO4, gave a smell of wintergreen. The compound ‘A’ is—

(A) Succinic acid

(B) Phenol(C) Salicylic acid(D) Salicylaldehyde

55. Schotten-Baumann reaction is—

(A) Phenol + CHCl3 NaOH

⎯⎯→Salicylaldehyde

(B) Phenol + Benzoyl chlorideNaOH

⎯⎯→ Phenyl benzoate

(C) Sodium phenate + CH3I⎯→

–HI Anisol

(D) None of these

56. Phenol on reacting with brominedissolved in carbon disulphide(CS2), gives—

(A) m-bromophenol only(B) o-bromophenol(C) Both o- and p-bromophenol(D) 2, 4, 6-tribromophenol

57. When phenetole reacts with HI,which is formed ?(A) C6H5OH and C2H5I(B) C2H5OH and C6H5OH(C) C2H5OH and iodobenzene

(D) Ethane and benzene

58. Which is not true statement ?(A) Paracetamol is parahydroxy

derivative of acetanilide(B) Salol is phenyl salicylate(C) Phenyl esters on heating

with AlCl3 give cresols

(D) Benzoylation of phenol givesphenyl benzoate

59. Phenol gives novolak on react-ing with—(A) CH3CHO in presence of acid

at low temp.(B) HCHO in presence of acid at

low temp.(C) HCOOH in presence of

NaOH at low temp.(D) Formalin in presence of

alkali at high temp.

60. The amount of bromine requiredto convert 1·0 g of phenol into2, 4, 6-tribromophenol, is—(A) 2·5 g (B) 5·1 g(C) 10·22 g (D) 15·3 g

ANSWERS

●●●

(Continued from Page 708 )

●●●

C.S.V. / August / 2009 / 723

1. 60 gm of a compound on ana-lysis produce 24 gm carbon, 4 gmhydrogen and 32 gm oxygen. Theempirical formula of the com-pound is—(A) CH2O2 (B) CH2O(C) H2 (D) CH4

2. The energy of second Bohr orbitof the hydrogen atom is – 328 kJmol– 1; hence the energy of fourthBohr orbit would be—(A) – 41 kJ mol– 1

(B) – 82 kJ mol– 1

(C) – 164 kJ mol– 1

(D) – 1312 kJ mol– 1

3. The number of neutrons in theelement 4Be9 is—(A) 3 (B) 5(C) 7 (D) 9

4. Which of the following is theelectron deficient molecule ?(A) C2H6 (B) B2H6(C) SiH4 (D) PH3

5. 8·2 litre of an ideal gas weight9·0 gm at 300 K and 1 atm pres-sure. The molecular mass of thegas is—(A) 9 (B) 18(C) 25 (D) 36

6. The correct order in which theO–O bond length increases in thefollowing is—(A) O2 < H2O2 < O3

(B) O3 < H2O2 < O2

(C) H2O2 < O2 < O3

(D) O2 < O3 < H2O2

7. Both ionic and covalent bond ispresent in—(A) KCN (B) KCl(C) H2 (D) CH4

8. Which of the following moleculeshas trigonal planar geometry ?(A) BF3 (B) NH3

(C) PCl3 (D) IF3

9. The pH of pure water at 80° willbe—(A) > 7 (B) < 7(C) = 7 (D) None of these

10. The number of moles of KMnO4reduced by one mole of KI inalkaline medium is—(A) One (B) Two

(C) Five (D) One fifth

11. 2 Ag + 2 H2SO4 → Ag2SO4

+ 2 H2O + SO2

In the above reaction H2SO4 actsas a—

(A) Catalyst

(B) Oxidising agent

(C) Reducing agent

(D) Acid as well as oxidant

12. A solution has a 1 : 4 mole ratioof pentane to hexane. The vapourpressures of the pure hydro-carbons at 20°C are 440 mm Hgfor pentane and 120 mm Hg forhexane. The mole fraction ofpentane in the vapour phasewould be—

(A) 0·200 (B) 0·549

(C) 0·786 (D) 0·478

13. N2(g) + 3H2(g) → 2 NH3(g)

For the above reaction thecorrect statement is—

(A) ΔH < ΔE (B) ΔH = ΔE

(C) ΔH > ΔE (D) None of these

14. Equilibrium constants K1 and K2for the following equilibria :

NO(g) + 12 O2

K1 NO2(g)

and 2NO2(g) K2 2NO(g) + O2(g)

are related as—

(A) K2 = 1K2

1(B) K2 = K2

1

(C) K2 = 1K1

(D) K2 = K12

15. Variation of heat of reaction withtemperature is known as—

(A) Kirchhoff’s equation

(B) van’t Hoff’s isotherm

(C) van’t Hoff’s isochore

(D) None of the above

16. At 25°C, the dissociation constantof a base, BOH, is 1 × 10–12. Theconcentration of hydroxyl ions in0·01 M aqueous solution of thebase would be—

(A) 1·0 × 10– 5 mol L– 1

(B) 1·0 × 10– 6 mol L– 1

(C) 2·0 × 10– 6 mol L– 1

(D) 1·0 × 10– 7 mol L– 1

17. The function of moderator in anuclear reactor is—

(A) To produce more neutrons

(B) To stop nuclear reaction

(C) To increase speed of neu-tron

(D) To slow down speed of neu-tron

18. For a first order reaction A → Bthe reaction rate at reactant con-centration of 0·01 M is found tobe 2·0 × 10– 5 mol L– 1 s– 1. Thehalf-life period of the reaction is—

(A) 30 s (B) 220 s

(C) 300 s (D) 347 s

19. A colloidal solution is subjectedto an electric field. The colloidalparticles move towards anode.The coagulation of the samesolution studied using NaCl,BaCl2 and AlCl3 solutions. Thecoagulation power is in the orderof—(A) BaCl2 > NaCl > AlCl3(B) NaCl > BaCl2 > AlCl3(C) AlCl3 > BaCl2 > NaCl

(D) NaCl > AlCl3 > BaCl2

20. A reaction occurs spontaneouslyif—(A) TΔS < ΔH and both ΔH and

ΔS are + ve(B) TΔS > ΔH and ΔH is + ve and

ΔS is – ve(C) TΔS > ΔH and both ΔH and

ΔS are + ve(D) TΔS = ΔH and both ΔH and

ΔS are + ve

21. The pH of the solution obtainedby mixing 40 ml of 0·10 M HClwith 10 ml of 0·45 M of NaOHis—(A) 4 (B) 8(C) 12 (D) 14

C.S.V. / August / 2009 / 724

22. The correct order of acid strengthis—(A) HClO4 < HClO3 < HClO2

< HClO(B) HClO < HClO2 < HClO3

< HClO4

(C) HClO4 < HClO < HClO2

< HClO3

(D) HClO2 < HClO3 < HClO4

< HClO

23. 50 ml of H2SO4 require 10 gmCaCO3 for complete decomposi-tion. The normality of acid is—(A) 2 (B) 0·30(C) 4 (D) 0·20

24. A solution of urea (mol. mass 56g mol– 1) boils at 100·18 °C at theatmospheric pressure. If Kf andKb for water are 1·86 and 0·512 Kkg mol– 1 respectively, the abovesolution will freeze at—(A) 0·654°C (B) – 0·654°C

(C) 6·54°C (D) – 6·54°C

25. The isomerism represented byethyl acetoacetate is—(A) Keto-enol isomerism(B) Geometrical isomerism(C) Enantiomerism(D) Diastereoisomerism

26. Which one of the following formsmicelles in aqueous solutionabove certain concentration ?(A) Dodecyl trimethyl ammo-

nium chloride(B) Glucose(C) Urea(D) Pyridinium chloride

27. The compound known as oil ofwintergreen is—(A) Phenyl acetate(B) Phenyl salicylate(C) Methyl salicylate(D) Methyl acetate

28. Names of some compounds aregiven. Which one is not thecorrect IUPAC name ?(A) CH3—CH2—CH2

—CH—

CH3

|CH—

|CH2CH3

CH2CH3

3-methyl-4-ethyl heptane

(B) CH3—CH

|OH

—CH

|CH3

—CH3

3-methyl-2-butanol

(C) CH3—CH2—C⎯ ||CH2

CH—

|CH3

CH3

2-ethyl-3-methyl-but-1-ene

(D) CH3—C ≡ C—CH—(CH3)24-methyl-2-pentyne

29. C3H9N cannot represent—

(A) 1° amine

(B) 2° amine

(C) 3° amine

(D) Quaternary ammonium salt

30. Which one of the following pairsrepresents stereoisomerism ?

(A) Structural isomerism andgeometrical isomerism

(B) Optical isomerism and geo-metrical isomerism

(C) Chain isomerism and rota-tional isomerism

(D) Linkage isomerism and geo-metrical isomerism

31. The main product obtained fromphenol with PCl5 is—

(A) BHC(B) Hexachlorobenzene(C) Chlorobenzene(D) Triphenyl phosphate

32. Which amongst the following isthe most stable carbocation ?

(A)+CH3 (B) CH3

+CH2

(C) CH3—+CH

|CH3

(D) CH3—

CH3

|C+

|CH3

33. Hydrolysis of sucrose is—(A) Inversion(B) Hydration(C) Saponification(D) Inhibition

34. Products of the following reac-tion—

CH3–C ≡ C–CH2–CH3 (i) O3

⎯⎯⎯⎯⎯→(ii) Hydrolysis

are—(A) CH3COOH + CO2

(B) CH3COOH + HOOC–CH2CH3

(C) CH3CHO + CH3CH2CHO(D) CH3COOH + CH3COCH3

35. The pairs of base in DNA areheld together by—(A) Ionic bonds(B) Hydrogen bond

(C) Phosphate group(D) Deoxyribose group

36. The best method for the separa-tion of naphthalene and benzoicacid from their mixture is—(A) Distillation(B) Sublimation(C) Chromatography(D) Crystallisation

37. The first ionisation potential ofNa, Mg, Al and Si are in theorder—(A) Na < Mg < Al < Si(B) Na < Al < Mg < Si(C) Na > Mg > Al > Si(D) Na < Si < Al < Mg

38. The major organic product formedfrom the following reaction :

O

(i) CH3NH2(ii) LiAlH4⎯⎯⎯→(iii) H2O

…

is—

(A)OH

HNCH3

(B)OH

HNCH3

(C) O—NHCH3

(D)

HNCH3

39. Which is not present in chloro-phyll ?(A) Carbon (B) Calcium(C) Magnesium (D) Hydrogen

40. Which functional group partici-pates in disulphide bond forma-tion in proteins ?(A) Thioester (B) Thioether(C) Thiol (D) Thioacetone

41. The molecular formula of cryoliteis—(A) Na2F.AlF6 (B) Na3AlF6(C) Na3AlF5 (D) Na2AlF3

42. The monomer of the polymer;

CH2—

CH3

|C—CH2

|CH3

—C⊕CH3 CH3

is—

(A) H2C = CCH3 CH3

(B) CH3CH = CHCH3

C.S.V. / August / 2009 / 725

(C) CH3CH = CH2

(D) (CH3)2C = C(CH3)2

43. Aluminium appears like goldwhen mixed with—

(A) 90% Cu (B) 75% Ni

(C) 80% Sn (D) 80% Co

44. The aqueous solution containingwhich one of the following ionswill be colourless ?

(Atomic No. : Sc = 21, Fe = 26,Ti = 22, Mn = 25)

(A) Sc3+ (B) Fe2+

(C) Ti3+ (D) Mn2+

45. Lithophone is a mixture of—(A) CuSO4 + ZnS

(B) CaSO4 + ZnS

(C) BaSO4 + CaSO4

(D) BaSO4 + ZnS

46. The main reason for larger num-ber of oxidation states exhibited

by the actinoids than the corres-ponding lanthanoids, is—(A) More energy difference bet-

ween 5f and 6d orbitals thanbetween 4f and 5d orbitals

(B) Lesser energy difference bet-ween 5f and 6d orbitals thanbetween 4f and 5d orbitals

(C) Larger atomic size of acti-noids than the lanthanoids

(D) Greater reactive nature of theactinoids than the lantha-noids

47. By passing air over red hot cokethe gas obtained is—(A) Coal gas (B) Water gas(C) Oil gas (D) Producer gas

48. Which one of the following is aninner orbital complex as well asdiamagnetic in behaviour ?

(Atomic number : Zn = 30,Cr = 24, Co = 27, Ni = 28)

(A) [Zn (NH3)6]2+

(B) [Cr (NH3)6]3+

(C) [Co(NH3)6]3+

(D) [Ni (NH3)6]2+

49. The most acidic oxide is—(A) P2O5 (B) N2O5

(C) Sb2O5 (D) As2O5

50. H2S gas when passed through asolution of cations containing HClprecipitates the cations of secondgroup of qualitative analysis butnot those belonging to the fourthgroup. It is because—

(A) Presence of HCl decreasesthe sulphide ion concentra-tion

(B) Solubility product of group IIsulphides is more than that ofgroup IV sulphides

(C) Presence of HCl increasesthe sulphide ion concentra-tion

(D) Sulphides of group IV cationsare unstable in HCl

ANSWERS WITH HINTS

C.S.V. / August / 2009 / 727

1. Two solutions (a) containingFeCl3 (aq) and (b) containingK4[Fe (CN)6] are separated bysemipermeable membrane asshown below. If FeCl3 on reactionwith K4 [Fe (CN)6] produces bluecolour of—

(a) (b)

FeCl 3 K4 [Fe(CN)6]

5 pm

………………………………

……

……

Fe4[Fe(CN)6], the blue colour willbe noticed in—(A) (a)(B) (b)(C) In both (A) and (B)(D) Neither in (A) nor in (B)

2. Volume of a mixture of 6·02 × 1023

oxygen atoms and 3·01 × 1023

hydrogen molecules at STP is—(A) 28·0 litre (B) 33·6 litre(C) 11·2 litre (D) 22·4 litre

3. The molal freezing point constantof water is 1·86 K molality– 1. If342 g of cane-sugar (C12H22O11)are dissolved in 1000 g of water,the solution will freeze at—(A) – 1·86°C (B) 1·86°C(C) – 3·92°C (D) 2·42°C

4. The speed of electron in the firstorbit of hydrogen atom in theground state is—

[c = velocity of light]

(A)c

1·37 (B)c

1370

(C)c

13·7 (D)c

137

5. ZSM –5 is used to convert—(A) Alcohol to petrol(B) Benzene to toluene(C) Toluene to benzene(D) Heptane to toluene

6. Which among the following iswrong about isodiapheres ?(A) They have the same diffe-

rence of neutrons andprotons or same isotopicnumber

(B) Nucleide and its decay pro-duct after α emission areisodiapheres

(C) mZA → m – 4

Z – 2B + 42He

A and B are isodiapheres(D) All correct

7. Which curve represents zeroorder reaction ?

(A) [A]

↑

→t

(B) [A]↑

→t

(C) [A]↑

→t

(D) [A]

↑

→t

8. If one mole of monoatomic gas

( )γ = 53 is mixed with one mole

of diatomic gas ( )γ = 75 the

value of γ for the mixture is—

(A) 1·40 (B) 1·50(C) 1·53 (D) 3·07

9. A certain weak acid has a disso-ciation constant 1·0 × 10– 4. Theequilibrium constant for its reac-tion with a strong base is—

(A) 1·0 × 10– 4 (B) 1·0 × 10–10

(C) 1 × 1010 (D) 1·0 × 10–14

10. If glycerol and methanol weresold at same price in the market,which would be cheaper for pre-paring an artifreezer solution forthe radiator of an automobile ?(A) Glycerol(B) Methanol(C) Both equal(D) None of these

11. When a solid melts, there is—(A) An increase in enthalpy(B) No change in enthalpy

(C) A decrease in enthalpy(D) A decrease in internal

energy

12. Fog is a colloidal system inwhich—(A) Liquid particles dispersed in

gas(B) Gaseous particles disper-

sed in liquid(C) Solid particles dispersed in

liquid(D) Solid particles dispersed in

gas

13. Which arrangement of electronslead to anti-ferromagnetism ?

(A) ↑↑↑↑

(B) ↑↓↑↓

(C) Both (A) and (B)(D) None of these

14. A particular reaction at 27°C forwhich ΔH > 0 and ΔS > 0 is foundto be nonspontaneous. The reac-tion may proceed spontaneouslyif—(A) Temperature is decreased(B) Temperature is kept constant(C) Temperature is raised(D) It is carried out in open

vessel at 27°C

15. The correct order of relative sta-bility of half filled and completelyfilled shells is—(A) p 3 < d 5 < d10 < p6

(B) d 5 < p3 < d10 < p6

(C) d 5 < p3 < d10 < p6

(D) p3 < d10 < d 5 < p6

16. For the reaction A + B → C + Dthe variation of the concentrationof product is given by the curve—

X

Y

W

Z

Time →

Con

cent

ratio

n↑

(A) X (B) Y(C) Z (D) W

17. Water glass is—(A) Another name for sodium

silicate(B) A special form of glass to

store water only

C.S.V. / August / 2009 / 728

(C) Hydrated form of glass(D) Hydrated silica

18. Which one is correct representa-tion for

2 SO2 (g) + O2 (g) 2 SO3 (g) ?

(A) Kp = [PSO3

]2

[PSO2]2 [PO2

]

(B) Kc = [SO3]2

[SO2]2 [O2]

(C) Kp = [nSO3

]2

[nSO2]2 [nO2

]

× [ ]PTotal moles

– 1

(D) All correct

19. Super halogen is—(A) F2 (B) Cl2(C) Br2 (D) I2

20. In the following reaction[Cu (H2O)3 (OH)]+ + [Al (H2O)6]3+

(a) (b)

→ [Cu(H2O)4]2+ + [Al (H2O)5 (OH)]2+

(c) (d)

(A) (a) is an acid and (b) is abase

(B) (a) is a base and (b) is anacid

(C) (c) is conjugate acid of (a)and (d) is conjugate base of(b)

(D) (c) is conjugate base of (a)and (d) is conjugate acid of(b)

21. A complex shown below canexhibit—

(A) Optical isomerism only(B) Geometrical isomerism only(C) Both optical and geometrical

isomerism(D) None of these

22. In the chemical reaction

Ag2O + H2O + 2e → 2Ag + 2OH–

(A) Water is oxidised(B) Electrons are reduced(C) Silver is reduced(D) Silver is oxidised

23. 0·2 g of an organic compoundcontaining C, H and O, on com-bustion yielded 0·147 g CO2 and0·12 g water. The percentage ofoxygen in it is—(A) 73·3% (B) 78·45%(C) 83·23% (D) 89·50%

24. At 25°C, the solubility of iodine inwater is 0·35 g/l. If distributioncoefficient of I2 between CS2 andwater is 600, the solubility of I2 inCS2 in g/l will be about—

(A) 1714 (B) 210(C) 569·6 (D) 857

25. The reaction described below is—

CH3 (CH2)5

C—Br OH–

⎯→H3C

H(CH2)5 CH3

OH—C CH3

H(A) SE1 (B) SN2(C) SN1 (D) SE2

26. Chloride ion and potassium ionare isoelectronic—(A) Their sizes are same(B) Cl– ion is bigger than K+ ion(C) K+ ion is relatively bigger(D) Their sizes depend on other

cation and anion

27. The polymer H2C

CH2

O

CH2

O

O

is obtained when HCHO isallowed to stand. It is a whitesolid. The polymer is—(A) Trioxane(B) Formose(C) Para formaldehyde(D) Metaldehyde

28. The function of flux during thesmelting of the ore is—(A) To make the ore porous(B) To remove gangue(C) To facilitate reduction(D) To facilitate oxidation

29. RMgX on heating with cyanogenchloride gives—(A) R—NC (B) R—Cl(C) R—CN (D) None of these

30. An orange coloured when solu-tion acidified with H2SO4 treatedwith a substance X gives a bluecoloured solution of CrO5. Thesubstance X is—(A) H2O (B) dil. HCl(C) H2O2 (D) Conc. HCl

31. In the reaction, C6H5CH3Oxidation⎯⎯⎯⎯→ A NaOH⎯⎯→ B

Soda lime⎯⎯⎯⎯→Δ

C, the product C is—

(A) C6H5OH (B) C6H6

(C) C6H5COONa (D) C6H5ONa

32. In the purification of bauxite byBayer’s process, the chemicalused is—(A) Na2CO3

(B) Cryolite(C) NaOH(D) A mixture of NaOH and

Na2CO3

33. Toilet soap is—(A) A mixture of calcium and

sodium salts of higher fattyacids

(B) A mixture of potassiumstearate and glycerol

(C) A mixture of sodium salts ofhigher fatty acids

(D) A mixture of potassium saltsof higher fatty acids

34. When concentrated nitric acid isheated it decomposes to give—(A) Oxygen and nitrogen(B) Nitric oxide(C) Oxygen(D) Nitrogen dioxide and oxygen

35. Which of the following is anexample of zwitter ion ?(A) Urea(B) Glycine hydrochloride(C) Ammonium acetate

(D) α-alanine

36. When SO2 is passed throughcupric chloride solution—

(A) The solution becomescolourless and a white pre-cipitate of Cu2Cl2 is obtained

(B) A white precipitate is ob-tained

(C) The solution becomes colour-less

(D) No visible change takesplace

C.S.V. / August / 2009 / 729

37. Indigo belongs to the class of—(A) Mordant dye(B) Vat dye(C) Direct dye(D) Disperse dye

38. 0·5 g of an organic compoundwas Kjeldahlised and the NH3evolved was absorbed in certainvolume of 1N H2SO4. The resi-dual acid required 60 cm3 of N/2NaOH. If the percentage of nitro-gen is 56 then the volume of 1NH2SO4 taken was—

(A) 30 ml (B) 40 ml

(C) 50 ml (D) 60 ml

39. Equal weights of methane andoxygen are mixed in an emptycontainer at 25°C. The fraction ofthe total pressure exerted by oxy-gen is—

(A) 1/3 (B) 1/2

(C) 2/3 (D)13 ×

273298

40. Which of the following is the leaststable carbanion ?(A) HC ≡ C– (B) (C6H5)3C–

(C) (CH3)3C– (D) CH3–

41. The values of vander Waalsconstant ‘a’ for the gases O2, N2,NH3 and CH4 are 1·360, 1·390,4·170 and 2·253 L2 atm mol– 2

respectively. The gas which canmost easily be liquefied is—(A) O2 (B) N2(C) NH3 (D) CH4

42. In presence of peroxide, hydro-gen chloride and hydrogen iodidedo not give anti-Markownikov’saddition to alkenes because—(A) Both are highly ionic(B) One is oxidising and the

other is reducing(C) One of the steps in both the

reactions is endothermic(D) All the steps are exothermic

in both the cases

43. The number of P—O—P bondsin cyclic metaphosphoric acid is—(A) Zero (B) Two(C) Three (D) Four

44. Which among MeX, RCH2X,R2CHX, R3CX is most reactivetowards SN2 reaction ?(A) MeX (B) RCH2X(C) R2CHX (D) R3CX

45. Which one has the pyramidalshape ?(A) PF3 (B) CO2 –

3(C) SO3 (D) BF3

46. Which is not a moth repellent ?

(A) Cl Cl

(B) Perchloroethane

(C) CHCl3

(D)

47. Grignard’s reagents add to—(A) C—— O (B) —C ——— N(C) C—— S (D) All of these

48. Which of the following is mostacidic ?

(A) Phenol

(B) m-chlorophenol

(C) Benzyl alcohol

(D) Cyclohexanol

49. Number of σ and π-bonds inbenzaldehyde is—

(A) 4π-bonds and 13 σ-bonds

(B) 4π-bonds and 8 σ-bonds

(C) 4π-bonds and 14 σ-bonds

(D) 8π-bonds and 10 σ-bonds

50. An organic compound X, ontreatment with acidified K2Cr2O7gives compound Y which reactswith I2 and sodium carbonate toform triiodomethane. The com-pound X can be—(A) CH3OH(B) CH3COCH3

(C) CH3CHO(D) CH3CHOHCH3

ANSWERS WITH HINTS

Blood vessels, lymphaticvessels and nerves pass

alongside the vas deferensto supply the testis.

Vas deferens

Epididymis

A tubule network connectsseminiferous tubules in

compartments tothe epididymis.

Compartment

Tubule network

Seminiferoustubules

Lumen ofseminiferous

tubule

Sperm

Spermatid

Secondary spermatocyte

Primary spermatocyte

Spermatogonium

Sertoli cells nourish andprotect sperm during

their development

Interstitial cellsproduce testosterone

C.S.V. / August / 2009 / 733

Introduction● The formation of gametes usually takes place in

specially restricted parts of the body which are knownas gonads (i.e., Ovary in female and testis in male).

● The lining of sex organs (gonads) is termed as germi-nal epithelium which possesses power of division andproduces the cells which undergo reduction to formthe gametes.

● The process by which gametes are produced in thegonads is known as gametogenesis. The process offormation of male sex cells or male gametes is calledspermatogenesis and the products are known asspermatids. The maturing of ova from the cells ofovary is termed oogenesis and the products areknown as ova or ootids.

Spermatogenesis● The process of maturation of reproductive cells in the

testis of male so as to form male gametes or sperm isknown as spermatogenesis.

Fig. : Sectioned view of testis showing seminiferous tubules and epididymis.

● The process is much complicated and can be studiedunder two heads :(a) Formation of spermatids from the cells of ger-

minal epithellium produced as a result of meioticdivision.

(b) Metamorphosis of spermatids into sperm, thespermiogenesis or spermatoleosis.

Formation of Spermatids

● The formation of spermatids starts from the primarygerm cells or primordial germ cells of the testis and isdifferentiated into three phases :

(i) Multiplication phase—The testes are com-posed of seminiferous tubules which are lined bygerminal epithelium. Some of the cells of ger-minal epithelium get differentiated as primarygerm cells or primordial germ cells. These be-come comparatively bigger in size and possess aprominent nucleus. During multiplication phase,the primordial germ cells multiply by mitosis andthe ultimate products are known as sperma-togonia. Each spermatogonium represents adiploid cell consisting of 2n-number of chromo-somes.

(ii) Growth phase—The spermatogonia thus formedafter repeated divisions from primary germ cellsdo not undergo further division by mitosis butprepare themselves for maturation division.These increase in size by the accumulation of

C.S.V. / August / 2009 / 734

nourishing material obtained from germinal cells.The enlarged cells are known as primary sper-matocytes. The nucleus of each primary sper-matocyte is of ordinary size but soon enlargesconsiderably so that in a mature primary sperma-tocyte the nucleus is much larger than that of thespermatogonial cell.

(iii) Maturation phase—Each diploid primary sper-matocyte goes through the first meiotic division(reduction division) with its long drawn outprophase. The pairing and splitting of homolo-gous chromosomes takes place and leads to theformation of tetrad. Crossing over exchange ofhomologous. Chromosomes also takes place. Asa result, halving of the chromosome numbertakes place in the first maturation division and thetwo cells formed are haploid having n-chromo-somes. These are known as secondary sper-matocytes.

Fig. : The stages of spermatogenesis

● The secondary spermatocytes undergo 2nd matura-tion division which is simple mitotic division. As aresult, each secondary spermatocyte is divided intoequal cells which are known as spermatids.

● Each spermatid consists only one set of homologouspairs and is, therefore, haploid. It undergoes meta-morphosis to form the sperm. As a result of these twoconsecutive divisions, four haploid spermatids areformed from each spermatocyte.

Spermiogenesis or Spermatoleosis

● Spermiogenesis is the gradual differentiation ofstationary rounded spermatid into an active motile

sperm cell. The spermatid contains mitochondria,Golgi body and centriole, but with one haploid set ofchromosomes. In this form it is not capable of func-tioning as a male gamete because the male gameteis motile and has to reach the ovum.

● During spermiogenesis, the nucleus shrinks by losingwater from the nuclear sap and the chromosomesbecome closely packed.

● The nucleus changes from its usual spherical formand becomes elongated and narrow.

● The centrosome of the spermatid consists of twocentrioles. During spermiogenesis, these move andcome to lie behind the nucleus. One of them entersthe depression developed in the posterior part of thenucleus and is called the proximal centriole.

● The other one is called the distal centriole. It occupiesa position behind the proximal one with its axiscoinciding with the longitudinal axis of the spermato-zoon. This distal centriole gives rise to the axial fila-ment of the flagellum and act as basal granule.

● Mitochondria from different parts of the spermatidaccumulate around the proximal part of axial filamentand distal centriole in the middle piece of the sper-matozoon.

● The Golgi complex of the spermatid forms the acro-some of the spermatozoon and forms a cap aroundthe nucleus. It helps the sperm in penetration into theovum.

Structure of Spermatozoon

● A typical spermatozoon consists of head, middlepiece and tail.

● Head is the anterior most part of the spermatozoon. Itis often conical but its shape varies considerably. Theanterior tip of the head is formed of acrosome whichenables the spermatozoon to penetrate through theegg membrane.

● Acrosome contains hyaluronidase enzymes andhydrolases, released when the sperm cell-membranefuses at several points with the acrosome duringacrosome reaction, dissolving the jelly around theegg so that sperm can penetrate it.

● The remaining portion of the head is occupied by thenucleus, whose posterior margin is depressed toaccommodate the proximal centriole, which initiatescell-division in the fertilized egg.

● Middle piece of the spermatozoon encloses distalcentriole and mitochondria, which carry oxidativeenzymes which provides energy for the propulsion ofthe spermatozoon.

● Tail or flagellum is the longest but most narrow partof the spermatozoon and its movements propel thespermatozoon. It consists of an axial filament, a thinlayer of cytoplasm and an outer sheath of plasma-lemma.

C.S.V. / August / 2009 / 735

C.S.V. / August / 2009 / 736

Oogenesis

Germinalepithelium

Immature ovum(primary oocyte)in primary follicle

Secondaryfollicle Mature

graafian follicle

Immature ovum(secondary oocyte)

Fluid-filled cavity

Rupturing of follicleduring ovulation

Ovulated immature ovum(secondary oocyte)

Developingcorpus luteum

Maturecorpus luteum

Degeneratingcorpus luteum

Blood vessels, nervesand lymphatic vesselsservice the ovary.

The oocyte completesmeiosis while the follicleenlarges and matures.

Fig. : Sectioned view showing primary secondary and graafian follicles in which oocytes form through meiosis.

● Oogenesis is the process of development and matu-ration of ova from the primary germ cell produced bythe division of cells of the germinal epithelium in theovary of female organism. The process includes threesteps or phases—Multiplication phase, Growth phaseand Maturation phase.

● Multiplication phase is the first phase and similar tothat found in spermatogenesis. The cells of germinalepithelium divide and produce oogonia. Theseundergo repeated mitotic divisions and the final pro-

Stages Cells Nuclear and cell divisions

Meiosis I

Meiosis II

Chromosomesin each cell

Primaryoocytesin primaryfollicles

Enlargedprimaryoocyte

Polarbody

Secondary oocytein secondary andgraafian follicles

Oogonium

Polar body

Polar bodiesdisintegrate.

Matureovum

DiploidHaploid

46

46

23

23

46

After puberty, one primary oocyte matures and is ovulated about once a month until menopause.

Proliferation of cells by mito-sis before birth

The secondary oocyte is released at ovulation and does not complete meio-sis II unless it is ferti-lized by a sperm.

Fig. : The stages of oogenesis

ducts are known as oocytes. These are diploid cellscontaining the same number of chromosomes as inthe parent body. The oocytes cease to divide andenter the growth phase.

● Growth phase of female gametes is much prolonged.The growth in size of the oocyte is not only due toincrease in the amount of its cytoplasm but mostlydue to accumulation of nutritive material. The mostcommon form of food storage consists of granules ofyolk. Yolk consists of proteins, phospholipids andneural fat. The follicle cells assist the growth ofoocyte by secreting substances which are taken upby the oocyte.

● In the primary oocyte, large amount of fat and proteinsbecome accumulated in the form of yolk and due toits heavy weight, it is usually concentrated towardsthe lower portion of egg, forming the vegetal pole.The portion of the cytoplasm containing the eggpronucleus remains separated from the yolk andoccurs towards the upper side of egg forming theanimal pole. During the growth phase, tremendouschanges occur in the nucleus of the primary oocyte.

Nuclear Changes

● Simultaneously with the growth of oocyte, its nucleusenters into the prophase of meiosis. The homologouschromosomes pair but further steps are postponeduntil the end of growth period.

● The nucleus increases in size by the formation ofnuclear sap. It is called the germinal vesicle. Thenucleoli are greatly increased to compensate thelarge amount of RNA required for the increasedmetabolism of growing oocyte.

C.S.V. / August / 2009 / 737

Differences betweenSpermatogenesis and Oogenesis

Spermatogenesis Oogenesis

1. It occurs in the testes. 1. It occurs in the ovaries.

2. Growth phase is short sothat spermatocytes areonly twice the size ofspermatogonia.

2. Growth phase is very longso that oocytes are muchlarger than oogonia.

3. Spermatocytes have cyto-plasm and nucleus withnormal contents.

3 . Oocytes have cytoplasmrich in RNA, ATP, enzy-mes and yolk; and nucleuswith giant chromosomesand large nucleoli.

4. A primary spermatocytedivides equally to formtwo similar secondaryspermatocytes.

4. A primary oocyte dividesunequally to form onelarge secondary oocyteand one minute polarbody.

5. A secondary spermato-cyte also divides equally,forming two similar sper-matids.

5. A secondary oocyte alsodivides unequally, formingone large ootid and oneminute polar body.

6. Upto the formation ofspermatids, the cells oftenremain interconnected.

6. Oogonia are separate andsurrounded by folliclecells.

7. A spermatogonium pro-duces four functional sper-matozoa.

7. An oogonium producesone functional ovum andthree non-functional polarbodies.

8. Spermatozoa are minute,streamlined, yolkless andmotile.

8. Ova are much larger,rounded, often with yolkand non-motile.

9. Spermatogenesis is oftena continuous process.

9. Oogenesis is not necessa-rily a continuous process.

10. Spermatogenesis isusually completed in thetestes so that maturesperms are released.

10. Oogenesis is often comp-leted in the reproductivetract of the female or evenin water as oocytes leavethe ovaries.

● Maturation phase—A full grown primary oocyteundergoes first meiotic division and produces twodaughter cells. The division of its cytoplasm is un-equal. One daughter cell is extremely small, whereasthe other is almost as large as the primary oocyte

itself. This large cell is called secondary oocyte. Itreceives almost whole of the cytoplasm of the primaryoocyte. The small cell is known as the first polarbody. It consists almost a bare nucleus. But both thecells have haploid (n) number of chromosomes.

● The haploid secondary oocyte and first polar bodypass through the second maturation division. Due tosecond maturation division, the secondary oocyteforms a mature egg and a second polar body. By thesecond maturation division, the first polar body alsodivides into two secondary polar bodies. These polarbodies ooze out from the egg and degenerate, whilethe haploid egg cell or secondary oocyte is releasedfrom the ovary for fertilization.

● When it is released from the ovary, the secondaryoocyte is surrounded by a protective sphere of cellscalled the corona radiata which is derived from thefollicle in the ovary. The size of the corona radiataand its irregular surface makes it easier for the entireassembly to be trapped by the uterine tube.

Key Points

● Amount of yolk in the ooplasm varies from species tospecies.

● The eggs with very little amount of yolk are known as themicrolecithal eggs, e.g., Amphioxus, Eutherian mammals.

● The eggs containing moderate amount of yolk are calledmesolecithal eggs, e.g., Petromyzon, Dipnoi amphibia.

● The eggs with large amount of the yolk are known asmacrolecithal eggs, e.g., Myxine, Cartilaginous andbonyfishes, reptiles, birds and monotremata.

● The eggs with evenly distributed yolk contents in theooplasm are known as Homolecithal eggs, e.g., eggs ofEchinoderms.

● The eggs in which the yolk is not evenly distributed in theooplasm are known as Heterolecithal eggs.

● When in heterolecithal eggs, the amount of yolk is con-centrated in the one half of the egg to form the vegetativepole of the egg; then this condition is known as teloleci-thal. Telolecithal eggs may be moderately telolecithal(e.g., Amphibians) or highly telolecithal (e.g., Hen).

● In macrolecithal and highly telolecithal eggs, the amountof yolk is very large and it occupies the largest portionexcept a small disc-shaped portion of the cytoplasm. Thecytoplasm contains the zygote nucleus and is known asthe germinal disc, e.g., eggs of fishes, reptiles and birds.

● In the centrolecithal eggs, the yolk accumulates in thecentre of the ooplasm, e.g., Insects.

OBJECTIVE QUESTIONS

1. In spermiogenesis, the sperma-tozoa are produced from—(A) Spermatids(B) Primary spermatocytes(C) Spermatogonia(D) Secondary spermatocytes

2. Between the spermatogonia arethe—(A) Epithelial cells

(B) Cells of Sertoli(C) Lymph spaces

(D) Capillaries

3. The minute cells which are sepa-rated from the developing overduring their maturation phase, arecalled—(A) Primary oogonia(B) Secondary oogonia

(C) Polar bodies(D) None of these

4. Spermatogenesis requires thepresence of—(A) Fructose(B) Progesterone(C) Testosterone(D) Thyroxine

(Continued on Page 765 )

C.S.V. / August / 2009 / 738

Introduction

● Chordates are bilateral and deuterostomial eucoelo-mate eumetazoa, basically possessing, in theembryo or throughout life a flexible but firm support-ing skeletal rod, called notochord.

● Phylum Chordate is the largest of the deuterostomephyla. It is the highest and the most important phylumcomprising a vast majority of living and extinct ani-mals including humans.

● Most of the living chordates are well known familiarvertebrate animals such as the fishes, amphibians,reptiles, birds and mammals. Besides, they include anumber of marine forms such as the tunicates andlancelets(amphioxus).

● The chordates are probably the most conspicuousand the best-known group in the entire animalkingdom, partly because of their large size and partlybecause of the important role they perform in theirecosystems. They are of primary interest becausehumans himself are member of the group.

● From a purely biological viewpoint, chordates areinteresting because they illustrate well the broadbiological principles of evolution, development andrelationship.

The Fundamental Chordate Characters

● All the chordates possess three outstanding uniquecharacteristics at some stage in their life cycle. Thesethree fundamental morphological features include:

1. A dorsal hollow or tubular nerve cord.

2. A longitudinal supporting rod-like notochord.

3. A series of pharyngeal gill slits.

1. Dorsal hollow nerve cord—The central nervoussystem of the chordates is present dorsally in thebody. It is in the form of a longitudinal, hollow ortubular nerve cord lying just above the notochord andextending lengthwise in the body.

● The nerve cord or neural tube is derived from thedorsal ectodermal neural plate of the embryo andencloses a cavity or a canal, called neurocoel.

● Nerve cord persists throughout life in the lower chor-dates, but in the higher chordates, it is surrounded orreplaced partly or fully, in adults by a jointed vertebralcolumn.

● In vertebrates, the anterior region of nerve cord isspecialized to form a cerebral vesicle or brain whichis enclosed by a protective bony or cartilaginouscranium. The posterior part of nerve cord becomesthe spinal cord and protected within the vertebralcolumn.

Pharyngeal gill slits Dorsal hollow nerve cord

Notochord

Mouth

Pharynx Trunk Hepatic caecumAnus Tail

Digestive tube

Fig. : Diagrammatic side view of a chordate showing thethree fundamental chordate characters.

2. Notochord or chorda dorsalis—The notochord isan enlarged rod-like flexible structure extending thelength of the body. It is present immediately beneaththe nerve cord and just above the alimentary canal.

● Notochord originates from the endodermal roof of theembryonic archenteron.

● Structurally, it is composed of large vacuolated noto-chordal cells containing a gelatinous matrix and sur-rounded by an outer fibrous and an inner elasticsheath.

● The notochord is the prime diagnostic feature of thephylum chordata which derives its name from it.

● Notochord is present at some stage in all chordates.In most vertebrates, it occurs complete only inembryo, but remnants may persist between the verte-brae, which obliterate it.

● The notochord may persist even in adult cephalo-chordates.

3. Pharyngeal gill slits—In all the chordates, at somestage of their life history, a series of paired lateral gillslits perforate through the pharyngeal wall of the gutbehind the mouth. These are also termed as pharyn-geal pouches.

● They are seen only during embryonic development inmost vertebrates. In the lower chordates, fishes andamphibian larvae, the pharyngeal pouches becomefunctional gills.

● In terrestrial vertebrates, the pouches are modified forvarious purposes.

● In humans, the first pair of pouches become theeustachian tubes. The second pair become thetonsils, while the third and fourth pairs become thethymus gland and parathyroid.

● The above common features appear during earlyembryonic life of all the chordates, but all the featuresrarely persist in the adult. Often they are modified oreven lost in adult stages of higher chordates.

● The notochord disappears during development inmost vertebrates, the nerve cord remain in adults.

C.S.V. / August / 2009 / 739

Advancement of Chordates over other Phyla

● Phylum chordata has some advantages over otherphyla due to certain characters :

1. Living endoskeleton—Only chordates possess aliving endoskeleton. It grows in size with the rest ofthe body. This living endoskeleton permits greaterfreedom of movement.

2. Efficient respiration—The gills in aquatic chordatesand the lungs in terrestrial forms, form efficientorgans of respiration.

3. Efficient circulation—The circulatory system of thechordates is well developed and the blood flowsfreely in the respiratory organs ensuring rapid ex-change of gases.

4. Centralized nervous system—A growing tendencyof centralization of nervous system is found in chor-dates and the sensory system is modified accord-ingly. The advancement of nervous and sensoryorgans explains the great power of the chordates foradapting themselves most successfully to a variety ofenvironments.

Major Subdivisions of Chordata

● Chordata is divided into three subphyla : Urochor-data, Cephalochordata and Vertebrata.

● The first two subphyla are groups of primitive chor-dates without vertebral column and are commonlycalled lower chordates. They are grouped together asProtochordates.

Group A : Acraniata(Protochordata) or Lower

Chordata

Group B : Craniata(Euchordata) or Higher

Chordata

1. Exclusively marine, small-sized chordates.

1. Aquatic or terrestrial,mostly large sizedvertebrates.

2. No appendages, cephali-zation and exoskeleton.

2. Usually 2 pairs ofappendages, well-developed head andexoskeleton present.

3. Coelom enterocoelic,budding off from embryo-nic archenteron.

3. Coelom schizocoelic,arising by splitting ofmesoderm.

4. Notochord persistent. Noskull, cranium and verte-bral column.

4. Notochord covered orreplaced by a vertebralcolumn. Skull and cra-nium well developed.

5. Pharynx with permanentgill clefts. Endostyle pre-sent.

5. Pharyngeal gill cleftspersist or disappearendostyle absent.

6. Heart chamberless whenpresent. No red bloodcorpuscles in blood.

6. Heart made of 2, 3 or 4chambers. Blood con-tains R.B.C.

7. Kidneys protonephridia. 7. Kidneys meso- ormetanephridia.

8. Sexes separate or unitedreproduction asexual aswell as sexual. Gono-ducts usually absent.

8. Sexes separate. Onlysexual reproducing,Gonoducts always pre-sent.

9. Development indirectwith a free-swimminglarval stage.

9. Development indirector direct, with or with-out a larval stage.

● The vertebrates are commonly called higher chor-dates.

● The protochordates lack a head and a cranium, sothat they are known as Acraniata.

● Vertebrates have a distinct head and cranium, there-fore, called Craniata.

● The vertebrates are further subdivided into Agnathaand Gnathostomata.

● Agnathans lack true jaws and paired appendages,include a small number of primitive but highly specia-lized fish-like forms, the extinct Ostracoderms andthe modern Cyclostomes.

● All other vertebrates have true jaws and pairedappendages, called Gnathostomata.

● A basic division of Gnathostomata recognises twosuper classes Pisces and Tetrapoda. The super-class Pisces includes all the fishes which are strictlyaquatic forms with fins. The superclass Tetrapoda isformed by four-legged vertebrates including amphi-bians, reptiles, birds and mammals.

General Characteristic of Phylum Chordata

1. Aquatic, aerial or terrestial.2. Body bilaterally symmetrical and metamerically seg-

mented.3. A postanal tail usually project beyond the anus at

some stage and may or may not persist in the adults.4. Exoskeleton often present in some forms.5. A cartilaginous or bony living jointed endoskeleton

present in the majority of vertebrates.6. Body triploblastic with three germinal layers.7. Coelomates, having a true coelom.8. A skeletal rod, the notochord present at some stage

in life cycle.9. Pharyngeal gill slits present at some stage.

10. Digestive system complete with digestive glands.

Classification of Chordates

● Phylum Chordata is separated into three primarysubdivisions, called subphyla, based on the characterof notochord. These are :1. Subphylum—Urochordata2. Subphylum—Cephalochordata3. Subphylum—Vertebrata

Subphylum—Urochordata● Characters—The urochordates are marine, mostly

sessile, filter-feeders. They have the followingcharacters :1. The notochord occurs only in the tail of the larva

and disappears in the adult.2. The nerve cord is present in the larva, but is

replaced by a single dorsal ganglion in the adult.3. The gill slits are numerous, persist in the adult and

open into an ectoderm-lined cavity, the atrium,instead of to the exterior. There are no gills.

4. The tail does not persist throughout life.Examples—Herdmania, Doliolum, Salpa.

C.S.V. / August / 2009 / 740

Subphylum—Cephalochordata● Characters—The cephalochordates are also marine

and filter-feeders. They have the following charac-ters :1. The notochord extends upto the tip of snout and

persists throughout life.2. The nerve cord persists throughout life, but no

brain is formed.3. The gill slits are numerous and persist in the adult.

They open in the atrium. There are no gills.4. The body wall consists of myotomes.5. Tail persists throughout life.

Subphylum VertebrataThe term vertebrate was introduced by Lamarck.Vertebrates show the following characters :1. Body regions—Vertebrate body typically con-

sists of 4 regions—head, neck, trunk and tail.The neck and tail may be lacking in some cases.Head is prominent and sense organs are welldeveloped.

2. Segmentation—There is some degree of seg-mentation. It is indicated by muscles, vertebrae,ribs, paired blood vesssels and nerves.

3. Appendages—There are as a rule 2 pairs ofappendages, which may be fins or limbs. One orboth pairs are absent in certain forms.

Outline Classification of Phylum Chordata

PhylumChordata

GroupAcrania orProtochordata(Lower chordata)

GroupCraniataor Euchordata(Higher chordata)

SubphylumUrochordata SubphylumCephalochordata

{SubphylumVertebrata

DivisionAgnatha

DivisionGnathostomata

SuperclassPisces

SuperclassTetrapoda

Classes1. Ascidiacea2. Thaliacea3. Larvacea1. Leptocardii

1. Ostracodermi2. Cyclostomata

{3. Placodermi4. Chondrichthyes5. Ostelchthyes

⎩⎪⎨⎪⎧6. Amphibia

7. Reptilia8. Aves9. Mammalia

Anamniota(Lowervertebrata)

Amniota(Highervertebrata)

4. Integument—Skin consists of epidermis of manylayers of cells and dermis of connective tissue.

5. Exoskeleton—Epidermis often produces anexoskeleton of keratinized cells. It may consist ofscales, feather, hair.

6. Muscles—Three types of muscles–striped, un-striped and cardiac are present.

7. Coelom—There is a spacious true coelom thatcontains viscera loosely suspended in it bymesenteries.

8. Endoskeleton—An internal framework of carti-lage or bone or both is always present. It growswith the body. The notochord is replaced partlyor fully by a jointed vertebral column in the adult,hence called vertebrata.

9. Circulatory system—Heart is ventral and con-sists of 2 to 4 chambers. Blood contains redcorpuscles that have haemoglobin. There isclosed circulatory system consisting of bloodvascular and lymphatic system. Hepatic portalsystem is present. In many vertebrates, renalportal system is also present.

10. Respiratory organs—Respiratory organs maybe gills, skin in aquatic vertebrates. Higher terre-strial vertebrates have lungs for respiration.

11. Nervous system—Nerve cord is differentiatedinto anterior brain and posterior spinal cord. Askull develops to provide cranium for protectingthe brain.

12. Sense organ—Sense organs are eyes, ears,tongue, nasal chambers and skin.

13. Cranial nerves—Cranial nerves may be 8, 10 or12 pairs.

14. Digestive tract—Digestive tract is complete andis ventral to the central nervous system.

15. Endocrine system—All vertebrates have duct-less endocrine glands that secrete hormones formetabolic regulation.

16. Excretory system—Vertebrates have a pair ofkidneys that require a high blood pressure forworking.

17. Reproductive system—The sexes are sepa-rate. They have gonoducts. Fertilization mayinternal or external. Asexual reproduction is lack-ing.

C.S.V. / August / 2009 / 741

Geological Representation

● Cambrian and Ordovician periods—The first fossilsof vertebrates were found in the rocks of the Ordovi-cian period in the form of Ostracoderms. These weresmall jawless bony fishlike forms related to cyclo-stomes. This shows that their chordate ancestorsmust have existed much before in the late Cambrian.

● Silurian and Devonian periods—Some fossil fisheswere found in the Silurian period and far more werepresent in the succeeding Devonian period which isknown as 'Age of Fishes'.

● Ostracoderms were jawless fishes, but during Devo-nian, the first jawed fish—the Placoderm arose. Theplacoderms became extinct without leaving any livingrepresentative. It is likely that early placoderms wereancestors of cartilaginous and bony fishes both,which had true jaws.

● Carboniferous periods—In late Devonian to earlyCarboniferous period, the lobed-finned fishes arrived.They were the first vertebrates to walk on land andbecame the primitive stem to amphibia.

● The amphibians became abundant and mutated inmany directions during carboniferous, which isusually known as the ‘Age of Amphibians’.

● Mesozoic era—In the early carboniferous, the veryprimitive amphibians also gave origin to primitivereptiles. Reptiles reached their peak during Mesozoicera, which is aptly known as the ‘Age of Reptiles’.They included dianosaurs.

● The ancestral mammals were derived from theprimitive reptiles during Triassic period. The first birdalso appeared in the late Jurassic period and one ofthe fossil Archaeopteryx had both reptilian as wellas avian characters.

● Cenozoic era—Following the decline of the reptilesduring the late Mesozoic era, both birds and mam-mals started flourishing.

● The mammals became more diversified of all the ani-mals during Cenozoic era, which is called the ‘Age ofMammals’.

Classification of Subphylum Vertebrata

● Subphylum vertebrata is divided into two sections :Agnatha and Gnathostomata.

● Agnatha—These are jawless vertebrates and aremost primitive. The mouth does not possess jawshence called Agnatha. They do not have exoskeletonand paired appendages, and have single nostril.

● Section Agnatha is further divided into followingclasses :1. Class Ostracodermi—These were group of

extinct agnatha, and were primitive, heavilyarmoured. World's first vertebrates collectivelycalled Ostracoderms.

2. Class Cyclostomata—These have body withoutscales, jaws and lateral fins. Mouth rounded andsuctorial, gills 5–16 pairs. They are parasites andscavengers.

Examples—Lampreys (Petromyzon) and Hagfishes(Myxine).

● Gnathostomata—These are jawed vertebrates, hav-ing true jaws and paired limbs. Embryonic notochordis usually replaced in adult by a vertebral column.Mouth has jaws hence called gnathostomata. Pairednostril are present.

● Gnathostomata is divided into two super classes :Pisces and Tetrapoda.

● Super class Pisces—It includes true fishes. All areaquatic. The body bears paired as well as medianfins. Respiration occurs typically by gills. They arecold-blooded. Each eye has a well developed nictita-ting membrane and possess scaly skin.

Super class Pisces is divided into following threeclasses :

1. Class Placodermi—It includes the earliest (Paleo-zoic) fossil fishes which lived in fresh water. Body hadan external protective armour of bony scales orplates. Primitive jaws with teeth were present. Skele-ton was bony and fins were mostly formed of largespines.Examples—Climatius, Dinichthys.

2. Class Chondrichthyes—Mostly marine fishes hav-ing cartilaginous endoskeleton and skin with placoidscales. Gill slits not covered by operculum. Malefishes have pelvic claspers.Examples—Scoliodon (Dogfish) and Chimera (Rat-fish).

3. Class Osteichthyes—Fresh water as well as marinefishes. Endoskeleton mostly bony. Skin havingvarious types of scales (cycloid, clenoid) other thanplacoid. Gill slits are covered by operculum. Malefishes without claspers.Examples—Labeo, Hippocampus, Protopterus.

● Super class Tetrapoda—Land vertebrates with twopair of pentadactyle limbs. They have cornified skinand lungs. Sensory organs are adapted for vision,hearing and smelling etc. Tetrapoda is divided into 4classes :

1. Class amphibia—Adults are amphibious and respireby lungs. Larval stages usually aquatic and breatheby gills. Skin moist, glandular and with no externalscales. Heart 3-chambered. Cold blooded.Example—Rana (frog), Bufo (toad).

2. Class Reptilia—Terrestrial, skin dry and covered byectodermal horny scales or bony plates. Heart incom-pletely 4-chambered. Cold blooded.Example—Hemidactylus (lizard), snakes, crocodiles,turtles.

3. Class Aves—Body covered with feathers. Forelimbsmodified into wings. No teeth in beak. Heart 4-chambered. Warm blooded.Example—Columba (Pigeon).

4. Class Mammalia—Body covered by hairs. Skin glan-dular. Mammary glands present. Heart 4-chambered.Warm-blooded.Example—Homo (humans), whale, elephant.

C.S.V. / August / 2009 / 742

Review at a Glance

● Phylum chordata name was established by Balfour.

● Early chordates evolved during Ordovician period of Paleozoic era.

● According to Garstrong, chordates have evolved from echinoderm larva by paedogenesis.

● All vertebrates are chordates but all chordates are not vertebrates.

● Presence of notochord, pharyngeal gill slits and dorsal tubular nerve cord are three characteristic features of chordates.

● In vertebrates notochord is replaced by vertebral column.

● Notochord is derived from mesoderm.

● In urochordata, notochord is found only in tail region of larval stage.

● In cepalochordata, notochord is present throughout the life.

● In mammals, remanent of notochord is present among invertebral disks as nucleus pulposus.

● Dorsal tubular nerve cord is ectodermal in origin.

● All chordates are deuterostomes, i.e., blastopore forms anus.

● All chordates are enterocoelic.

● Members of urochordata show retrogressive metamorphosis.

● The first fossile of vertebrates were found in rocks of the Ordovician period in the form of Ostracoderms.

● Devonian period is known as age of fishes.

● Carboniferous period usually known as age of amphibians.

● Mesozoic era is known as age of reptiles.

● Cenozoic era is known as age of mammals.

● World's first vertebrates collectively called Ostracoderms.

OBJECTIVE QUESTIONS

1. Vertebrates skin is covered by—(A) Scales(B) Feathers(C) Hairs(D) All the above

2. The vertebrates are the mem-bers of—(A) Cephalochordata(B) Agnatha(C) Gnathostomata(D) Urochordata

3. The study of fishes is called—(A) Ichthyology(B) Herpatology(C) Saurology(D) Ornithology

4. Cartilaginous fishes do nothave—(A) Gill slits (B) Operculum(C) Scales (D) Pelvic fin

5. Urinary bladder is absent in—(A) Reptiles (B) Aves(C) Mammals (D) Amphibians

6. Which of the following bird hasteeth in beak ?(A) Ostrich(B) Kiwi(C) Pelican(D) Archaeopteryx

7. A common characteristic amongall mammals is—(A) They are carnivorous(B) They have ventral nerve

cord(C) They do not moult(D) They have seven cervical

vertebrae

8. Which of the following is notfound in the amphibian skin ?(A) Epidermis(B) Mucous glands(C) Scales(D) Chromatophores

9. The presence of true placenta ischaracteristic feature of sub-class—(A) Eutheria(B) Metatheria(C) Prototheria(D) All the above

10. Which of the following era isknown as age of mammals ?(A) Devonian(B) Carboniferous(C) Mesozoic(D) Cenozoic

ANSWERS

●●●

(Continued from Page 702 )

●●●

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Introduction

● Muscles are in use during every waking and sleepingmoment. Some muscles are under conscious voluntarycontrol, while others operate without conscious control.

● All muscle tissues have three characteristics incommon :

1. Muscles can contract and shorten in length.

2. After contraction, muscles relax and return to theirformer length.

3. Muscles are excitable, responding to electrical orchemical signals (stimuli) from the nervous andendocrine systems.

● Muscle is the only tissue in the body that has all theseproperties.

● A muscle is composed of many muscle cells that arebundled into fascicles.

Antagonistic Muscles

● A muscle can pull a part of the body by its contraction(shortening). It cannot push that part by relaxation(elongation). Hence, the muscles are typically arran-ged in antagonistic (opposing) pairs, one of whichmoves a body part in one direction and the othermoves that part in the opposite direction by its contrac-tion. Of course, when one muscle contracts, its oppo-sing muscle relaxes.

Muscle

Fascicle

Blood vessels

Tendon

Bone

Connective tissuesurrounding fascicle

Muscle cell

Connective tissuesurrounding eachmuscle cell

Fasciasurroundingmuscle

Fig. : Relationship of a skeletal muscle with its muscle cellsand connective tissue wrappings.

Skeletal Muscles Structure

● Skeletal muscles attach to bones and move the skele-ton. They also attach to other muscles and to the facialskin.

● Some skeletal muscles are very large, such as thegastrocnemius, the main muscle forming the calf in

the lower leg. Others are very small, such as themuscle of the eyelid.

● Muscle shape vary. Some muscles are triangular(deltoid) while others are rectangular (rectus abdo-minis) or trapezoidal (trapezius).

● Many are spindle-shaped (biceps brachii).

● Each skeletal muscle is composed of parallel bundlesof individual muscle cells. Each muscle cell is sur-rounded by delicate connective tissue. Bundles ofthese cells, called fascicles are wrapped in anotherthin layer of connective tissue.

● The connective tissue wrappings are actually conti-nuous with each other at the end of the muscle, wherethey unite to form an extremely tough fibrous connec-tive tissue material called a tendon. Tendons attachmuscle to bone.

● Individual skeletal muscle cells have a tubular shape,contain many nuclei per cell and appear striated orstriped. These muscle cells are relatively large com-pared with other human cells.

● Muscle cells contain the same cellular components ororganelles found in most other human cells, but somecomponents have specialized functions or are unique.

● The smooth endoplasmic reticulum forms into parallelsaclike compartments called the sarcoplasmic reticu-lum or SR.

● On the outer surface of the muscle cell’s plasma mem-brane are the openings of tiny tubes called transversetubules or T-tubules. These tubules are extensions ofthe plasma membrane that run deep into the cell andlie close to the SR. They relay the electrical signalsthat trigger contraction.

● Within each muscle cell and running its entire lengthare numerous protein fibres called myofibrils. Myofi-brils are the structures responsible for contraction.

● Each myofibril consists of a series of units calledsarcomeres. The orderly arrangement of sarcomeresgives a striated appearance to skeletal muscle cells.

● Each sarcomere is made of a rather complex arrange-ment of several kinds of protein. Two of these proteinsare actin and myosin.

● Thousands of actin molecules link together to form along threadlike macromolecules, called actin fila-ments. Similarly, thousands of myosin molecules formthe somewhat thicker myosin filaments. The mole-cular interaction between actin and myosin filamentscontributes directly to muscle contraction.

C.S.V. / August / 2009 / 744

C.S.V. / August / 2009 / 745

Skeletal Muscle Contraction

● For muscle contraction, a stimulus is needed. Before amuscle cell can contract, it must be stimulated. Stimu-lation is provided by nerve impulses that pass alongnerve cells.

● A nerve cell stimulates a muscle cell by secreting achemical substance called acetylcholine (ACh). Acetyl-choline permits a nerve cell to communicate chemi-cally with a muscle cell at a site called the neuromus-cular junction.

● Upon its release, acetylcholine diffuses across thejunction and binds to specific receptors on the plasmamembrane of the muscle cell. This binding initiates anelectrical impulse which travels down the T-tubulesinto the interior of the cell.

● When the impulse reaches the SR, Ca2+ stored in theSR is released. The released Ca2+ forms bond with thetroponin-tropomyosin protein complex which isassociated with the actin filaments.

● When Ca2+ forms this bond, the troponin-tropomyosincomplex shifts position, exposing sites on the actinfilaments for attachment to the head portions of themyosin filaments.

● After exposing sites for attachment, a series of eventscause the shortening or contraction of a muscle cell.Each myosin head projecting from a myosin filamentbonds with an actin filament. After forming thesebonds, the myosin heads change shape and pull theactin filaments toward the centre of sarcomere.

Spinal cord

Nerve cell

Nerve cells carryimpulses fromspinal cord tomuscle cells.

Neuromuscularjunction End of

nerve cell

Plasma membraneof muscle cell Acetylcholine

(ACh)

ACh receptors

Musclecells

ACh released fromnerve cell attaches toreceptors on musclecell plasma membrane.

Fig. : Pathway of an impulse from the spinal cord to theneuromuscular junction.

● This reaction requires energy, which comes from theATP attached to each myosin head.

● Myosin acts as an enzyme (when Ca2+ is present) andbreaks the ATP into ADP + P + energy. This energy isused to change the shape of the myosin head, whichcauses the actin filaments to slide a little towards thecentre of sarcomere.

● After sliding, a new ATP attaches to the myosin head,breaking the myosin-actin bond. When this bond isbroken, the myosin head is released and returns to itsoriginal shape. Immediately, the myosin head bonds toanother site on the actin filament. Then it changesshape again and the filament slide a little past oneanother.

● The sequence of myosin-actin bonding, sliding andrelease is repeated rapidly over and over until themuscle has shortened sufficiently. This is the slidingfilament mechanisms for muscle contraction, and itconsumes a great deal of ATP.

● When an individual muscle is stimulated, the actin fila-ment in every sarcomere unit slide toward the middleof the sarcomere. As a result, the myofibrils shorten. Amuscle cell shortens (contracts) then all its myofibrilsshorten. An entire muscle contract when many of itsmuscle cells contract simultaneously.

● Contraction ends when the muscle cell ceases to bestimulated by the nerve cell. If an impulse no longertravels down the T-tubules to the cell’s interior, Ca2+ isactively transported back into the SR. Without Ca2+,the troponin-tropomyosin protein complex shifts backinto its original position, masking the sites on the actinfilaments and preventing further bonding by the myosinheads. When this happens, the muscle relaxes andreturn to its original shape.

● In muscle contraction, oxygen is used, carbon dioxideis produced, glycogen is consumed and heat is gene-rated. Only a part of energy released in the process isused. In muscle contraction, the rest is converted intoheat.

● Muscle Tonus (Tone)—All apparently relaxed skeletalmuscles always remain in partial contraction as longas their nerves are intact. This state of sustainedpartial contraction is called muscle tonus or tone. It isa sort of mild tetanus.

● The tone is maintained by a constant flow of nerveimpulses to muscle fibres. A muscle under slight ten-sion can react more rapidly and can contract morestrongly than one which is completely relaxed. Almostall human daily activities are carried out by tetaniccontractions of muscles. Tetanus is necessary tomaintain posture and form of body.

● Muscle Fatigue—The duration for which a muscle canremain contracted depends on its ability to supplyATPs to the contractile proteins (myosin and actin).The reduction in force of contraction of a muscle afterprolonged stimulation is called muscle fatigue. Amuscle can contract in the absence of oxygen, but itgets fatigued sooner. This is due to the fact that lacticacid is not disposed off without oxygen and collects inthe muscle fibres.

C.S.V. / August / 2009 / 746

● Fatigue is thus primarily caused by an excess of lacticacid. A completely fatigued muscle refuses to respondto nervous stimuli.

Types of Muscles According to Type ofMotion

1. Flexors—The flexors bend one part of a limb onanother at a joint.Example—Biceps. It brings the forearm towards theupper arm.

2. Extensors—The extensors extend or straighten a limb.Example—Triceps. It extends the forearm.

3. Abductors—The abductors pull a limb away from themiddle of the body.Example—Deltoideus. It draws the entire forelimb tothe side.

4. Adductors—The adductors bring a limb toward themidline of the body.Example—Latissimus dorsi. It presses the entireforelimb against the side.

5. Depressors—The depressors lower a part.Example—Depressor mandibulae. It lowers down thelower jaw to open the mouth.

6. Elevators—The elevators raise a part.Example—Masseter. It lifts up the lower jaw to closethe mouth.

7. Rotators—The rotators rotate a part.Example— Pyriformis. It raises and rotates thefemur.

8. Pronators—These rotate the forearms to turn thepalms downward or backward.

9. Supinators—These rotate the forearm to turn thepalms upward or forward.

10. Sphincters—These decrease the size of the aper-tures to close them.

11. Dilators—These widen the apertures.The adductor and abductor; elevator and depressor;

pronator and supinator; and sphincters and dilators are allantagonistic muscles.

● The single isolated contraction of a muscle fibrecaused by a single nerve impulse or artificial stimulusis called muscle twitch. Immediately after a twitch, themuscle fibre relaxes.

● Fundamentally, skeletal muscles function by twitching,and most graceful movements of body are the productof twitches.

● The contraction period is the time during which themuscles actually shortens. The relaxation period is thetime it takes the muscle to relax and return to itsoriginal length.

● A muscle twitch can be altered by increasing the fre-quency of stimulation (the number of impulses persecond). Muscle contractions are known as summa-tion, and are produced when the frequency of stimula-tion is so rapid that a muscle does not have time torelax completely after each impulse. Summation pro-duces a greater strength of contraction by addingtogether many separate contractions.

● If the frequency of stimulation is further increased sothat there is no relaxation, the muscle will remain in acompletely contracted state called tetanus.

● A muscle contracts only when the stimulus it receivesis of sufficient strength. The smallest strength of sti-mulus that causes contraction is called the thresholdstimulus.

● When a muscle cell receives a threshold stimulus, itcontracts with maximum strength. Even if a muscle cellreceives a stimulus greater than the threshold, it willcontract with the same maximum contraction strength.This is called the all-or-none law, and it states that amuscle cell will contract maximally or not at all.

● The all-or-none law applies to individual muscle cells,but not to an entire muscle. This explains that musclecells in a muscle do not all have the same thresholdvalue.

● Isotonic and isometric muscle contractions are usedin certain exercise activities.

● Isotonic means ‘same strength’ (constant tension, andin an isotonic contraction, the strength of the contrac-tion remains the same but the muscle shortens. Exa-mple—Weight lifting.

● Isometric means ‘same length’. In an isometric con-traction, the strength of the contraction increases, butthere is practically no shortening of the muscle. Thistype of contraction does not result in body movement.Sitting upright in a chair while contracting abdominalmuscles is an example of an isometric contraction.

● Tonic (Slow, Red) Muscle Fibres—These are thin,dark red and slow contracting muscle fibres. Theycontain a high content of haem-protein pigment calledmyoglobin, abundant mitochondria, low glycogen con-tent and poorly formed sarcoplasmic reticulum (SR).

● Myoglobin imparts them dark colour and stores oxygenas oxyhaemoglobin. Its oxygen by anaerobic oxidationin mitochondria provides energy for muscle contrac-tion. Less lactic acid accumulates in this process. Thisenables the red muscle fibres to carry on slow andsustained contractions for long periods without fatigue.

● The tonic muscle fibres are innervated by thin, slowconducting nerve fibres.

● The body muscles meant for sustained work at a slowrate for a prolonged duration are composed mostly ofred muscle fibres. The extenser muscles of the back inhumans remain in sustained contraction to maintainerrect posture against gravity, and are rich in redmuscle fibres.Twitch (Fast, White) Muscle Fibres—These aremuch thicker, lighter in colour and fast-contractingmuscle fibres. They have a low content of myoglobin,few mitochondria, abundant glycogen granules andwell formed SR.

● They derive energy for their fast contraction mainly byanaerobic oxidation, accumulate lactic acid during stre-nuous work and soon get fatigued.

● These muscle fibres are innervated by thick fast con-ducting nerve fibres.

● The body muscles, which are meant for fast and stre-nuous work for short durations, are composed mostlyof white muscle fibres.

● The muscles that move eyeballs are very rich in whitemuscle fibres.

C.S.V. / August / 2009 / 747

Differences between Red and White Skeletal Muscle Fibres

Red (Slow) Muscle Fibres White (Fast) Muscle Fibres

1. They are thin. 1. They are much thicker.

2. They contain abundant mitochondria, low glycogen contentand poorly formed sarcoplasmic reticulum.

2. They are poor in mitochondria, and have abundant glycogengranules and well formed sarcoplasmic reticulum.

3. They are dark red as they contain abundant pigmentmyoglobin.

3. They are light in colour as they have very little myoglobin.

4. Their myoglobin stores O2 as oxymyoglobin that relea-ses O2 for oxidation during muscle contraction.

4. They have little or no store of oxygen.

5. They get energy for contraction by aerobic respiration. 5. They get energy for contraction mainly by anaerobicrespiration.

6. They accumulate little lactic acid. 6. They accumulate lactic acid during strenuous work.

7. They undergo slow sustained contractions for long periods. 7. They undergo fast contractions for short periods.

8. They are not fatigued with work. 8. They soon get fatigued with work.

9. They are innervated by thin, slow-conducting nerve fibres. 9. They are innervated by thick fast-conducting nerve fibres.

Example : extensor muscles of the back in man. Example : eyeball muscles.

OBJECTIVE QUESTIONS

1. Myoglobin is found in—(A) Slow muscle fibres(B) Fast muscle fibres(C) Blood(D) Lymph

2. Which of these plays a role inmuscle contraction ?(A) K+ (B) Na+

(C) Mg2+ (D) Ca2+

3. In skeletal muscle, the T-tubules—(A) Secrete acetylcholine(B) Give mechanical support(C) Relay electrical signals(D) Store calcium

4. As compare to the slow muscle,fast muscle has—(A) More myoglobin(B) More mitochondria(C) More sarcoplasmic reticulum(D) All the above

5. Which of the following is the con-tractile protein of a muscle ?(A) Tubulin(B) Myosin(C) Tropomyosin(D) All the above

6. In muscle, the contraction occursdue to—(A) Myosin(B) Actin(C) Both (A) and (B)(D) None of these

7. The contraction and relaxationphases of a muscle constitute—(A) Beat (B) Twitch(C) Condition (D) Stimulus

8. The sprain is caused due toexcessive pulling of—(A) Muscles (B) Nerves(C) Tendons (D) Ligaments

9. Sarcoplasmic reticulum plays amajor role during—

(A) Muscle contraction

(B) Muscle excitement

(C) Muscle relaxation

(D) All the above

10. When a muscle bends one partupon other, it is called—

(A) Extensor (B) Flexor(C) Abductor (D) Regulator

ANSWERS

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(Continued from Page 687 )

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C.S.V. / August / 2009 / 748

1. A type of dentition in which theteeth are replaced throughout thelifetime, is called—

(A) Monophyodont

(B) Polyphyodont

(C) Diastemic

(D) None of the above

2. Which of the following lipids donot have fatty acids ?

(A) Steroids

(B) Cholesterol

(C) Phospholipids

(D) Both (A) and (B)

3. Which of the following hormoneinitiates the preparation of theuterus for implantation of theovum ?

(A) Oxytocin

(B) Vasopressin

(C) Progesterone

(D) HCG.

4. Which brain wave pattern iscommon in young children andsleeping adults ?

(A) Alpha waves

(B) Beta waves

(C) Theta waves

(D) Delta waves

5. In birds, a comb like structure ispresent projecting into the cavityof eye, is called—

(A) Keel (B) Pecten

(C) Baleen (D) Ossicles

6. German measles is caused by—

(A) Rubella virus

(B) Varicella virus(C) Papilloma virus(D) None of the above

7. Quill feathers present on wingsare also called—(A) Down feathers(B) Barbules(C) Coverts(D) Remiges

8. Sensory cells that occur in thepits or canals of the lateral linesystem of fishes ?(A) Neuromast(B) Neurochrome(C) Neuroglia(D) None of the above

09. Adipocytes are—

(A) Bones

(B) Cartilage

(C) Specialized connectivetissues

(D) Nerves

10. Formation m-RNA from DNA iscalled—

(A) Translocation

(B) Transcription

(C) Transduction

(D) Duplication

11. Diapedesis is—

(A) Formation of WBC

(B) Bursting of WBC

(C) Formation of Pus

(D) Passage of WBC

12. The chief function of utriculus inhumans is—

(A) To perceive sound vibrations

(B) To maintain body equilibrium

(C) To perceive pressure

(D) To act as shock absorber

13. Housefly has—

(A) Chewing mouth parts

(B) Piercing-sucking mouth parts

(C) Sponging mouth parts

(D) Siphoning mouth parts

14. Which of the following conditionsmeet the Hardy-Weinberg equili-brium ?

(A) The homozygous recessivephenotype is non-viable

(B) The dominant allele mutatesto a recessive allele

(C) The alleles are sex-linked

(D) Mating is random

15. What is the name of cycle whichproduces CO2, ATP, NADH andFADH2?

(A) Oxidative carboxylation(B) Oxidative phosphorylation(C) Citric acid cycle(D) Glycolysis

16. The cartilage making the lowerjaw of cartilaginous fishes, iscalled—(A) Dentary(B) Mentomeckelian(C) Meckel’s cartilage(D) Angulosplenial

17. In a DNA molecule, the—(A) Bases are covalently bonded

to the sugars(B) Sugars are covalently bon-

ded to the phosphates(C) Bases are hydrogen bonded

to one another(D) All of these are correct

18. Ecotone is characterised by—(A) Transitional zone between

two diverse communities(B) Terrestrial ecosystem(C) Zone of transition between

water and land(D) Forest ecosystem

19. DNA nucleotide differences bet-ween organisms—(A) Indicate how closely related

organisms are(B) Indicate that evolution occurs(C) Explain why there are phe-

notypic differences(D) All of these are correct

20. Process of urea formation inhumans is also known as—(A) Krebs cycle(B) Hans Krebs cycle(C) Nitrogen cycle(D) Transamination

21. Ends of long bones are coveredwith—(A) Cartilage (B) Muscles(C) Ligaments (D) Blood cells

22. In some animals, allantois is alsorelated with—(A) Storage of nitrogenous

wastes(B) Blood formation(C) Digestion(D) All the above

C.S.V. / August / 2009 / 749

23. Territoriality occurs as a result ofpredation—

(A) Predation(B) Parasitism(C) Competition(D) Cooperation

24. During respiration, failure ofventilation leads to—(A) Decreased oxygen tension(B) Decreased carbon dioxide

tension(C) Carbonate tension(D) Bicarbonate tension

25. Forest destruction leads to den-trimental effects on—(A) Loss of a CO2 sink(B) Loss of biodiversity(C) Loss of possible medicinal

plants(D) All of the above

26. The association of sea anemoneand hermit crab is an exampleof—(A) Mutualism(B) Commensalism(C) Parasitism(D) None of the above

27. Simple two neuron reflex arcinvolves—(A) Sensory neuron(B) Spinal cord(C) Effector neuron(D) All of the above

28. Tube feet are locomotory organsof—(A) Jelly fish (B) Carp fish(C) Silver fish (D) Star fish

29. Down syndrome—(A) Is always caused by non-

disjunction of chromosome21

(B) Shows no overt abnormali-ties

(C) Is more often seem inchildren of elderly mothersafter the age of thirty fiveand above

(D) Both (A) and (C) are correct

30. In the development of the humanembryo, the ectoderm is respon-sible for the development of—(A) Nervous system(B) Lens of eyes(C) Sweat glands(D) All the above

31. The protein which polymerizes toform microtubules is—(A) Actin (B) Chitin(C) Tubulin (D) Ferritin

32. The centrum of a typical vertebraof a frog is—

(A) Procoelus

(B) Heterocoelus

(C) Opisthocoelus

(D) Acoelus

33. Which vitamin does not act as acoenzyme ?

(A) Riboflavin

(B) Pantothenic acid

(C) Biotin

(D) Folic acid

34. Fovea centralis in eye, lacks—

(A) Rod cells

(B) Blood vessels

(C) Nerve fibres

(D) All the above

35. Balbiani rings are the charac-teristics of—

(A) Polytene chromosomes

(B) Lampbrush chromosomes

(C) Sex chromosomes

(D) Ring chromosomes

36. The stage of animal embryonicdevelopment at which the gutcavity and germ layers firstappear, is called—(A) Involution(B) Epiboly(C) Gastrulation(D) Formative movement

37. Pebrine is a disease of—(A) Honey bee (B) Fish(C) Silk worm (D) Lac insect

38. Most primitive living mammalswhich provide an evidence oforganic evolution from geogra-phical distribution are found in—(A) China (B) India(C) Africa (D) Australia

39. Which one of the following pro-cesses make direct use ofoxygen ?(A) Glycolysis(B) Fermentation(C) Krebs cycle(D) Electron transport

40. Waste product of adenine andguanine metabolism is—(A) Ammonia (B) Uric acid(C) Urea (D) All these

41. The skin of certain sharks andrays containing small knobs, iscalled—(A) Shagreen (B) Pulp(C) Dentine (D) Ganoin

42. Sympathetic nerves in mammalsarise from—(A) Sacral region(B) Cervical region(C) Thoraco-lumbar region(D) None of the above

43. Which one of the following iscorrect about malaria ?(A) Gametocytes fuse in mos-

quito's stomach to form eggcells.

(B) Mosquito pick up gametocy-tes when sucking blood froman infected person

(C) Merozoites invade red bloodcells and become tropho-zoites

(D) All the above are correct

44. Osteoarthritis disease usuallyassociated with aging and iscalled—

(A) Communicable disease

(B) Degenerative disease

(C) Deficiency disease

(D) Allergy

45. Medullary cavity of which follo-wing bone has yellow bonemarrow ?

(A) Long bones(B) Short bones(C) Spongy bones(D) All of the above

46. Stomach pain is felt through—(A) Interoceptors(B) Proprioceptors(C) Teloceptors(D) None of the above

47. Camel's hump conserve waterfor longer time because it iscomposed of—(A) Muscular tissue(B) Skeletal tissue(C) Areolar tissue(D) Adipose tissue

C.S.V. / August / 2009 / 750

48. Black-water fever is caused by—(A) Plasmodium falciparum(B) Leishmania donovani(C) Plasmodium ovale(D) Plasmodium malariae

49. The transfer of genetic materialof one bacterium to another byvirus is called—(A) Transduction(B) Translation(C) Transcription(D) Replication

50. Charcot-Leydon crystals arefound in faeces of man duringinfection of—(A) Entamoeba histolytica(B) Trypanosoma gambiense(C) Ascaris(D) All the above

ANSWERS WITH HINTS

(Continued on Page 790 )

C.S.V. / August / 2009 / 751

1. Which of these provides phos-pholipids dual properties ?

(A) Fatty acid non-polar tails

(B) Charged phosphate head

(C) Fatty acid polar tails

(D) Both (A) and (B)

2. Which of the following is presentin microsomes of liver cells andplays a part in detoxification ?

(A) Cytochrome P-450

(B) Adenylate kinase

(C) Cytochrome C-reductase

(D) Nucleoside diphosphokinase

3. When atoms share one or moreelectron pairs, the bond is—

(A) Covalent bond

(B) Ionic bond

(C) Hydrogen bond

(D) All the above

4. Gooseflesh caused by erectionof skin papillae, as from cold,shock, fright or fear, is called—

(A) Cutis anserina

(B) Cutis aurantiasis

(C) Cutis hyperelastica

(D) Cutis marmorata

5. Which of these viruses becomelife-long residents of host cells ?

(A) Herpes virus

(B) Varicella virus

(C) Rubella virus

(D) Both (A) and (B)

6. Cuboid bone is associated with—

(A) Carpels

(B) Metacarpels

(C) Metatarsals

(D) None of these

7. Which of these are a group ofviruses important in investigatingviral carcinogenesis ?

(A) Papova viruses

(B) Parvo viruses

(C) Adeno viruses

(D) Pox viruses

8. The wishbone of a bird, whichsupport the flight muscles, formedfrom—(A) Fusion of the collar bones(B) Ventral extensions of the

vertebrae(C) Anterior ribs(D) Posterior extensions of the

larynx

9. Synthesis of m-RNA and t-RNAoccurs in—(A) G1 phase (B) G2 phase(C) S-phase (D) All these

10. Diapedesis is associated with—(A) Leucocytes(B) Urinary system(C) Nervous system(D) Peristalsis

11. Absence of lack of ‘intrinsic fac-tor’ in diet causes—(A) Pernicious anaemia(B) Cooley’s anaemia(C) Aplastic anaemia(D) None of the above

12. Which of the following is dinu-cleotide ?(A) FAD(B) NAD(C) Nucleic acid(D) Both (A) and (B)

13. Food energy passed at eachtrophic level is approximately—(A) 10% (B) 30%(C) 50% (D) 90%

14. Which of the following is viraldisease ?(A) Impetigo(B) Warts(C) Both (A) and (B)(D) None of these

15. A non-essential hereditary factorthat may exist either free within acell or in a state in which it isintegrated in a chromosome, isknown as—(A) Episome(B) Cosmid(C) Both (A) and (B)(D) None of the above

16. Ticks and mites are—(A) Myriapods(B) Arachnids(C) Insects(D) None of these

17. The malleus, incus and stapes ofear ossicles of mammals are res-pectively modified bones of—(A) Articular, hyomandibular and

quadrate(B) Quadrate, articular and hyo-

mandibular(C) Articular, quadrate and hyo-

mandibular(D) Quadrate, hyomandibular

and articular

18. Hashimoto disease is causedwhen—(A) Adrenal gland is destroyed

by autoimmunity(B) Thyroid gland is destroyed by

autoimmunity(C) Kidney is destroyed(D) Pancreas is destroyed

19. Long chain molecules of fattyacids are formed by—(A) Polymerisation of 2 carbon

compounds(B) Decomposition of fats(C) Polymerisation of glycogen(D) None of the above

20. Volkmann's canals are found in—(A) Bones of frog(B) Bones of fowl(C) Bones of rabbit(D) Cartilage of rabbit

21. Bedbugs can survive during longtime of starvation because—(A) It stores glycogen(B) It convert uric acid to amino

acid and thus it can use itsexcretory materials

(C) Its life span is very long(D) It can minimise its require-

ments

22. Which of the following is involvedin DNA synthesis and cell divi-sion ?(A) Vitamin K (B) Folic acid(C) Vitamin E (D) Vitamin D

23. Oxygen toxicity is related with—(A) Failure of ventilation of lungs(B) Collapse of alveolar walls(C) Blood poisoning(D) None of the above

C.S.V. / August / 2009 / 752

24. For most people, dreamingoccurs during which stage ofsleep ?(A) Deep sleep(B) REM sleep(C) Alpha sleep(D) Slow wave sleep

25. Latissimus dorsi muscles inhumans—(A) Draws arm downward and

backward(B) Draws legs forward(C) Moves head(D) Moves ankles

26. Trochanter is associated with—(A) Femur (B) Humerus(C) Ulna (D) Radius

27. Heparin is produced by—(A) Liver cells(B) Nervous cells(C) Kidney cells(D) Spleen

28. Podocytes are found in—(A) Spleen (B) Pancreas(C) Kidneys (D) Heart

29. Which of the following stimulateserythropoiesis ?(A) Fe++ (B) Mg++

(C) Ca++ (D) Cu++

30. Fovea centralis in eye lacks—(A) Rod cells(B) Blood vessels(C) Nerve fibres(D) All the above

31. Ichthyophis is a member of—(A) Mollusca (B) Pisces(C) Amphibia (D) Annelida

32. Waste product of adenine andguanine metabolism is—(A) Uric acid (B) Ammonia(C) Urea (D) All these

33. Cis-trans test is related with—(A) Crossing over(B) Heredity(C) Mutation(D) All the above

34. Parathion, malathion and femi-trothion insecticides belong thegroup of—(A) Carbamates(B) Pyretheroids(C) Organophosphates(D) Triazines

35. Which one of the following is nota Ketone body ?(A) Acetoacetic acid(B) Succinic acid

(C) β-hydroxybutyrate

(D) Acetone

36. German measles is caused by—(A) Rubella virus(B) Varicella virus(C) Papiloma virus(D) None of these

37. Antihistamine drugs relieve—

(A) Allergy

(B) Angina pectoris

(C) Stroke

(D) Nephrites

38. When diminished blood volumeis present in human body, it istermed—(A) Hypovolemia(B) Oligemia(C) Oligohaemia(D) All the above

39. Middle part of human sperm con-tains—(A) Nucleus(B) Mitochondria(C) Centriole(D) Nucleus and mitochondria

40. Another term for adaptive evo-lution is—(A) Clinal change(B) Microevolution(C) Macroevolution(D) All the above

41. Eggs having yolk in the centreand cytoplasm in the peripheralarea, are called—(A) Isolecithal(B) Microlecithal(C) Centrolecithal(D) Telolecithal

42. Which of the following belongs tophylum Arthropoda ?(A) Gold fish (B) Silver fish(C) Cuttle fish (D) Star fish

43. Parathion, Malathion and Femi-trothion insecticides belong tothe group of—(A) Carbamates(B) Pyretheroids(C) Organophosphates(D) Triazines

44. Exotoxins are related with—(A) Tetanus(B) Diphtheria(C) Both (A) and (B)(D) None of these

45. Degradative reactions—

(A) Are oxidation reactions

(B) Are synthetic reactions

(C) Requires a supply of NADPHmolecules

(D) Both (A) and (C)

46. Which of the following is anexample of bile acids ?(A) Glycocholic acid(B) Taurocholic acid(C) Both (A) and (B)(D) None of these

47. Which one of the following is notrelated with the path of soundvibrations ?(A) Auditory canal(B) Tympanic membrane(C) Semicircular canals(D) Cochlea

48. Wolffian body is known as—(A) Pronephros(B) Mesonephros(C) Metanephros(D) None of these

49. Chaga’s disease is caused by—(A) Trypanosoma gambiense(B) Trypanosoma cruzi(C) Trypanosoma bruci(D) None of the above

50. Inheritance of which of the follow-ing is not Mendelian ?(A) Plasmagene(B) Oncogene(C) Polygene(D) All the above

ANSWERS WITH HINTS

C.S.V. / August / 2009 / 754

1. Which of the following preventsthe mother’s immune response tothe developing embryo ?(A) Trophoblast(B) Blastocyst(C) Umblical vein(D) Umblical artery

2. Auer’s bodies are present in thecytoplasm of—(A) Myeloblasts(B) Myelocytes(C) Monoblasts(D) All the above

3. Milkman’s syndrome is charac-terized by—(A) Failure of reabsorption of

phosphates by renal tubules(B) Failure of absorption of lac-

tose(C) Failure of absorption of

amino acids(D) None of these

4. Certain aquatic mammals, suchas seals, sealions, and walrusesare included in the order—(A) Pinnipedia (B) Fissipedia(C) Hyracoidea (D) Cetacea

5. Failure of an organ or part todevelop or grow, is called—(A) Agenesia (B) Agenitalism(C) Ageism (D) Agerasia

6. Group of related species formingfertile hybrids are called—(A) Coenospecies(B) Subspecies(C) Sibling species(D) None of these

7. Olfactory organ of a snake is—(A) Jacobson’s organ(B) Johnston’s organ(C) Organ of Bojanus(D) None of these

8. Which of the following is sourceof energy to an ecosystem ?(A) Heat liberated during respi-

ration(B) ATP

(C) NADPH2

(D) Solar energy

9. What is the function of porphyrinin earthworm ?(A) To protect against harmful

germs(B) To help in respiration(C) To help in excretion(D) To protect from harmful ultra-

violet rays

10. Which of the following is not awater soluble vitamin ?(A) Tocopherol(B) Niacin(C) Folic acid(D) Ascorbic acid

11. Which division of PNS is oftencalled ‘voluntary nervous sys-tem’ ?(A) Somatic(B) Autonomic(C) Both (A) and (B)(D) None of these

12. FAD is derived from—(A) Thiamin (B) Riboflavin(C) Pyridoxine (D) Folic acid

13. The final hormonal stimulus lead-ing to ovulation in human femaleis provided by—(A) Estrogen(B) Progesterone(C) Follicle stimulating hormone(D) Luteinizing hormone

14. Autoimmune thyroiditis is asso-ciated with—(A) Hashimoto’s disease(B) Grave’s disease(C) Goitre(D) Acromegaly

15. Convulsions in infants is causeddue to deficiency of—(A) Iodine (B) Vitamin B6

(C) Vitamin D (D) Vitamin C

16. A cell-coded protein, that isformed in response to viral infec-tion, is—(A) Antigen (B) Interferon(C) Histone (D) Antibody

17. Which of the following is aninactive enzyme precursor ?(A) Polyglycoids(B) Cholenzymes(C) Activases(D) Zymogens

18. Restriction fragment length poly-morphisms (RFLPs)—

(A) Identify individuals geneti-cally

(B) Are the basis for DNAfingerprints

(C) Can be subjected to gelelectrophoresis

(D) All the above

19. Notochord is usually regardedas—

(A) Ectodermal

(B) Mesodermal

(C) Endodermal

(D) Blastodermal

20. In cladistics—(A) A clad must contain the

common ancestor plus all itsdescendents

(B) Derived characters helpconstruct cladograms

(C) Data for the cladogram ispresented

(D) All the above

21. Bones enlarge by—(A) Auxentic growth(B) Multiplicable growth(C) Accretionary growth(D) Appositional growth

22. Free-living Planaria is a—(A) Roundworm(B) Flatworm(C) Tapeworm(D) Fluke

23. Bilharziasis is a parasitic disease,caused by—(A) Schistosoma(B) Taenia solium(C) Fasciola hepatica(D) Enterobius vermicularis

24. Guillain-Barre syndrome is rela-ted with—(A) Acute polyneuritis(B) Infectious polyneuritis(C) Landry’s paralysis(D) All the above

C.S.V. / August / 2009 / 755

25. Which of the following is the onlyamino acid metabolized by thebrain ?(A) Alanine(B) Glutamic acid(C) Histidine(D) Glutanine

26. Electron micrographs followingfreeze-fracture of the plasmamembrane indicate that—(A) The membrane is a phos-

pholipid bilayer(B) Some proteins span the

membrane(C) Protein is found only on the

surface of the membrane(D) Glycolipids and glucoproteins

are antigenic

27. Enzymes which catalyse reac-tions involving electron transfer,called—(A) Transferases(B) Hydrolases(C) Ligases(D) Oxidoreductase

28. Arytenoid cartilages are foundin—

(A) Sternum (B) Hyoid

(C) Nose (D) Larynx

29. Which of the following animals aredevoid of respiratory, excretoryand circulatory organs ?

(A) Tapeworms

(B) Sponges

(C) Liver flukes

(D) None of these

30. Active transport—

(A) Requires a carrier

(B) Moves a molecule againstits concentration gradient

(C) Requires supply of energy

(D) All the above are correct

31. Parasympathetic effect—

(A) Lowers blood pressure(B) Slows heart rate(C) Promotes digestion(D) All the above

32. Enzyme enterokinase converts—(A) Proteins into peptides(B) Caseinogen into casein(C) Trypsinogen into trypsin(D) Pepsinogen into pepsin

33. Sum of constructive processes inbody cells is called—(A) Catabolism(B) Anabolism(C) BMR(D) All the above

34. Which of these is present inhuman buccal cavity ?(A) Trypsin (B) Ptyalin(C) Lipase (D) Pepsin

35. Which layer of the heart wallconsists cardiac muscles ?(A) Endocardium(B) Myocardium(C) Epicardium(D) All of these

36. Human nerve cells originate fromthe embryonic—(A) Ectoderm(B) Ectoderm and mesoderm(C) Endoderm(D) Mesoderm

37. Which of the following is calledincomplete proteins ?(A) Most animal proteins(B) Most plant proteins(C) Both (A) and (B)(D) None of these

38. Which of the following syndromeis associated with easily dislo-cation of body joints ?

(A) Ehlers-Danlos syndrome

(B) Tay-Sachs syndrome

(C) Marfan’s syndrome

(D) None of these

39. Reticulocytes are—(A) WBCs(B) Immature RBCs(C) Blood platelets(D) Lymphocytes

40. Which of the following division ofperipheral nervous system isoften called ‘voluntary nervoussystem ?(A) Somatic(B) Autonomic(C) Both (A) and (B)(D) None of these

41. Meroblastic cleavage refers towhich type of division of eggs ?(A) Incomplete (B) Spiral(C) Complete (D) Horizontal

42. Which part of human brain isassociated with integration ofsympathetic and parasympathe-tic activities ?(A) Hypothalamus(B) Cerebrum(C) Medulla oblongata(D) Neopallium

43. Iter or Aqueduct of sylvius islocated between—(A) Lateral ventricles(B) Optocoels(C) III and IV ventricles(D) None of these

44. Autologous blood tranfusion is—(A) Use of patient’s own blood(B) Use of patient’s parental

blood(C) Use of patient’s children

blood(D) All the above

45. Which type of tissue forms thethin surface for gas exchange inlungs ?(A) Epithelial (B) Connective(C) Nervous (D) Muscular

46. Unmyelinated axons in spinalcord are present in—(A) White matter(B) Gray matter(C) Both (A) and (B)(D) None of these

47. Carcinoid syndrome is associatedwith—(A) Serotonin(B) Bradykinin(C) Prostaglandins(D) All of these

48. Which of the mitosis stage provi-des best opportunity for prepar-ing human karyotype ?(A) Anaphase (B) Metaphase(C) Prophase (D) Telophase

49. Castle’s intrinsio factor helps in—(A) Absorption of cyanoco-

balamin(B) Nervous stimulation(C) Cardiac reflex(D) All the above

50. Bright’s disease is associatedwith—(A) Kidneys (B) Liver(C) Lung (D) Bone

C.S.V. / August / 2009 / 757

Introduction

Edaphic factors take in account the structure andcomposition of soil with its physical and chemical charac-teristics. The study of soil (Latin = Solum) is called pedo-logy or edaphology, which is important for geology,mineralogy, paleobotany, paleozoology and petrology.Soil is a natural habitat for plants, animals and micro-organisms. It may be defined as “any part of earth'scrust in which plants root grow.” However, soil is notmerely mineral matter, but a complex of several othertypes of components. Thus biologically the soil may bedefined as “the weathered superficial layer of theearth's crust in which the living organisms grow andalso release the products of their activities, death anddecay.”

Composition of Soil

From the above classical definition of soil it is evidentthat soil is, thus, not merely a mineral particles, but it hasalso a biological system of living organisms as well assome other components. Therefore, it is preferred to call ita soil complex, which includes the following categories ofcomponents—

(i) Mineral matter—The matrix of mineral particlesderived by varying degrees of break down of the parentmaterial–rock occur in the soil. The minerals representabout 90% of the total weight of the soil.

(ii) Soil atmosphere—Soil atmosphere occupiesthe pore space between soil particles, which is not water–filled. The soil atmosphere differs from the above groundatmosphere as it is normally lower in oxygen and higher incarbon dioxide content.

(iii) Soil organic matter or humus—Soil organicmatter or humus are partially decayed organic compo-nents derived from long and short-term addition of mate-rial from organisms growing above and below the groundthat is microorganisms, plants and animals.

Dokuchayev’s View

● According to Dokuchayev, a famous Russian pedologist,“the soil is a result of the actions and reciprocal influencesof parent rocks, climate, topography, plants, animals andage of the land.

● Hence, soil can be represented by the following formula :

S = (g.e.b) Δt

where, S = Soil

g = Geology

e = Environment

b = Biological influences

t = Time

(iv) Soil water or solution—Soil water is held bycapillary and absorptive forces between and around thesurface of soil particles. Actually soil water is a dilute solu-tion of various organic and inorganic compounds, whichbecome available to plants as mineral nutrients. Water inthe soil comes mainly through infiltration of precipitatedwater (rain, sleet, snow and hail) and irrigation, whereas itis lost from the soil chiefly through evaporation, percola-tion and transpiration.

Soil Formation (Pedogenesis)

C. F. Marbut states that “a mature soil is one that hasassumed the profile features characteristic of predominantsoils on the smooth uplands within the general climaticand botanic regions in which it is found.

Soil is a stratified mixture of organic and inorganicmaterials, both of which are decomposition products. Thesoil forming rocks are the parent material from whichmineral constituents of soil are derived by fragmentationor weathering. Organic components of soil are formedeither by decomposition (or transformation) of deadremains of plants or animals or through metabolic acti-vities of living organisms present in the soil.

Soil-forming rocks—There are two basic kinds ofsoil-forming rocks as given below—

(i) Sedimentary rocks—These are formed bydeposition of weathered minerals which are derived fromigneous rocks.

(ii) Igneous rocks—These are formed due to coolingof molten magma or lava.

Chemical Nature of Minerals of Soil-formingRocks

Rocks are the chemical mixture of various kinds ofminerals. The chemical nature of most common andabundant minerals of soil-forming rocks are given in thefollowing table—

Table : Chemical Composition of Some Common SoilMinerals

Minerals Chemical Constituents

A. Sand and Silt minerals

1. Quartz or silica SiO2

2. Feldspars

(a) Orthoclase

(b) Calcium feldspar

(c) Plagioclase

K2Al2Si6O16

CaAl2Si2O8

NaAlSi3O8

3. Amphibole (Mg, Fe)7 (Si4O11)2 (OH)2

4. Calcite; magnesite;and dolomite

CaCO3, MgCO3 and(CaCO3, MgCO3)

C.S.V. / August / 2009 / 758

5. Micas

(a) Biotite

(b) Muscovite

K, Mg, Fe, Al silicate

K (OH)2 Al2 (AlSi3) O10

6. Iron oxides

(a) Limonite

(b) Haematite

(c) Magnetite

FeO (OH), X H2O

Fe2O3

Fe2O4

7. Pyroxene

8. Olivine and serpentine

(Mg, Fe) SiO4

(Mg, Fe)2 SiO4

B. Clay minerals

1. Montmorillonite (Ca, MgO) Al2O3,

5 SiO3, 5 H2O

2. Kaolin Al2O3, 2 SiO2, 2 H2O

Process of Soil Formation

(1) Weathering of Soil Forming Rocks—Soils arederived from parent rocks by the process called weather-ing. Formation of soil is initiated by disintegration orweathering of parent rocks by certain physical, chemicaland biological agents as a result, soil-forming rocks arebroken down into small particles called regoliths, whichfinally develop into mature soil by pedogenesis weatheringis accomplished by three types of forces—physical,chemical and biological.

(A) Physical weathering—It takes place due to thefollowing processes—

(i) Wetting-drying—Wetting-drying causes the dis-ruption of layer or lattice structure of the rock, whichswells upon wetting and contract on drying leading todisintegration of rock, which provides more aeration.

(ii) Heating-cooling—It causes the disruption ofheterogenous crystalline rocks in which inclusions havedifferential coefficients of thermal expansion.

(iii) Sand blast—In arid and desert conditions therocks are disrupted by physical action of wind and sand.

(iv) Freezing—It causes the disruption of porous,lamellar or vesicular rocks by frost shatter due to expan-sion of water.

(v) Glaciation—At mountain tops, ice formation takesplace mostly in the winter season. When the summerapproaches, ice starts melting and glaciers (huge slidingmasses of ice) move downwardly on the slops. In theglacier movement, the rocks are corroded and finallybroken into sand particles.

(B) Chemical weathering—Chemical weathering,which brings about disappearance of original rock mine-rals either completely or partly, includes the following pro-cesses—

(i) Solution—Certain mobile components of rocks,such as calcium sulphates and chlorides are simply remo-ved by agents like water (solution) making the rock porousand hence liable to further disintegration.

(ii) Hydrolysis—During hydrolysis, components likealumino silicates of rocks break down and potassium orsilicon are washed out which give rise to simpler mineralmatter like clay alumino-silicates.

(iii) Chelation—Certain chemical exudates, producedthrough biochemical activities of microorganisms like

bacteria, cyanobacteria and lichens dissolve the mineralcomponents of the rocks. These metals dissolved withorganic products of microbial activities are known aschelates.

(iv) Hydration—Reversible changes occur due towater absorption by haematite to limonite, (FeO3 Fe2O3 . 3 H2O), as a result the rocks swell up and causesdisruption.

(v) Carbonation—Various chemicals produced inatmosphere and by metabolism of microorganisms bringabout carbonation. For example, reversible change ofCaCO3 to Ca (HCO3)2 leads to solution loss of limestoneor disruption of calcium carbonate cemented rocks ashydrogen carbonate is more soluble than the carbonate.

(vi) Oxidation-reduction—Certain chemicals bringabout oxidation-reduction reactions, such as reversiblechange of Fe3+ to Fe2+, cause disruption of rocks,because Fe3+ is less soluble than Fe2+.

(C) Biological weathering—According to Jacks(1965-66), a number of microorganisms (e.g., bacteria,fungi, nematodes, protozoans, etc.) lichens and mossesare early colonizers which transform the rock into adynamic system, storing energy and synthesizing organicmaterial. Their activities alter the chemical compositionand physical structure of the rock. For example, lichensare present in the initial stages of biological successionand their growth may cause cracking or flaking, exposinggreater area of rock to further weathering. Lichens as wellas mosses extract mineral nutrients such as Ca, Mg, S, P,Al and Si from the rock. These elements are combinedwith organic complexes and eventually return to develop-ing soil when vegetation decomposes. Joffe states thatthere is no biogeochemical weathering. According to him,it is either chemical or physical weathering by biologicalagencies.

Products of Weathering

● The surface rocks of the earth are weathered and as aresult of weathering, small particles of parental materialsare formed.

● The soil which is formed by weathering of rocks is calledembryonic or primary soil.

● The soil which develops in situ above parent bedrocks isknown as sedimentary and residual soil.

● In a few places soil material comes from accumulatedorganic matter as peat.

● Soil material transported from one area to another is calledeolian or loess.

● Transported or secondary soils are those which arecarried to other places by carriers like gravity, glacier,water and wind.

(2) Mineralization and Humification—Break downof organic debris into humus is accompanied by decom-position and finally by mineralization. Higher organisms inthe soil consume fresh material and leave partiallydecomposed products in their excreta. This is furtherdecomposed by bacteria and fungi into various com-pounds of carbohydrates, proteins, fats, lignin and resins.These compounds are broken down into simpler productssuch as carbohydrate, water salts and minerals. This

C.S.V. / August / 2009 / 759

process is called mineralization. The residual amorphousincompletely decomposed black coloured organic matterwhich undergoes mineralization is called humus. Theprocess of humus formation is called humification. Thesoils of deserts in which vegetation is scanty and newsoils freshly formed from rocks, apparently contain verylittle humus, while mature soil with vegetation, the most.

(3) Organo-mineral Complex Formation—In thefinal stage of pedogenesis, colloidal particles, which areformed as a result of weathering, humification and mine-ralization, accumulate and may aggregate into crumbsor concretions. About 60–70% colloidal humus particlesbecome associated with mineral particles to form organo-mineral complexes. The crumbs of organo-mineral com-plexes, according to Wallwork (1970), are formed by twomechanisms—electrochemical bonding and cement-ing.

Aggregation of negatively charged colloidal clayand/or humus particles of water molecules and metallicions particularly calcium takes place in electrochemicalbonding method of crumb formation. The cementingmechanism involves the action of substances adsorbed onthe surface of soil particles which effectively make themcompact. The crumbs increase the total pore space in thesoil, providing good aeration and drainage. Some solubleorganic compounds are removed from the top soil bywater which percolates downwardly through humus mixedsoil.

OBJECTIVE QUESTIONS1. Eroded soils are—

(A) Richer in plant nutrients

(B) Devoid of plant nutrients(C) Fit for agriculture(D) None of the above

2. Soil is composed of—(A) Air + water + minerals(B) Water + organic matter + minerals(C) Air + water + organic matter + minerals(D) Water + organic matter

3. The chemical weathering involves the—(A) Solution (B) Hydrolysis(C) Oxidation (D) All of these

4. Humus is an example of—(A) Fertilizer(B) Crystalloid(C) Component of soil structure(D) Organic colloids

5. Sequence of humification and mineralization is—(A) Dead organic matter → litter → duff → humus(B) Minerals → humus → litter → duff(C) Dead organic matter → duff → litter → minerals

→ humus(D) Humus → minerals → litter → duff

ANSWERS

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C.S.V. / August / 2009 / 760

IntroductionBiochemistry has traditionally been considered

‘organic’ and in fact much of the biochemistry involvesorganic or carbon chemistry. It has been known for a longtime, however, that in plants many other elements in theform of macro or micronutrients are also present. Mate-rials are added to the soil, or applied directly to cropfoliage, to supply elements needed for plant nutrition.These materials may be in the form of solids, semisolids,slurry suspensions, pure liquids, aqueous solutions orgases.

FertilizersCrop requirements of fertilizer components could be

satisfied by the spreading of individual materials for eachelement deficient in the soil. However, economy favoursthe single application of a balanced mixture that satisfiesall nutritional needs of a crop. Many commercial fertilizers,therefore, contain more than one of the primary fertilizerelements. The chemical elements, such as nitrogen (N),phosphorus (P) and potassium (K) are the macronutrients,or primary fertilizer elements, which are required ingreatest quantity.

Sulphur (S), calcium (Ca) and magnesium (Mg),called secondary elements, are also necessary to thehealth and growth of vegetation, but they are required inlesser amounts compared to the macronutrients. Theother elements of agronomic importance, called macro-nutrients and provided for plant ingestion in small (trace)amounts, include boron, zinc, molybdenum, cobalt,copper and iron. All these fertilizer elements along withother chemical elements, occur naturally in agriculturalsoils in varying concentrations and mineral compositionswhich may or may not be in forms readily accessible toroot systems of plants.

Many commercial fertilizers contain more than one ofthe primary fertilizer elements. The composition of ferti-lizer mixtures, in terms of the primary fertilizer elements,are identified by an N-P-K code : N denotes elementalnitrogen; P denotes the anhydride of phosphoric acid(P2O5); K denotes the oxide of potassium (K2O). All areexpressed numerically in percentage composition, or unitsof 20 lb each per short ton (10 kg per metric ton) offinished fertilizer as packaged. Formula 8-32-16 thuscontains a mixture aggregating 8 wt% N in some form ofnitrogen compounds, 32 wt% P2O5 in some form of phos-phate and 16 wt% K2O in some form of potassium com-pounds, to give a product with a total of 56 fertilizer units.The commercial N-P-K formulas totals 100% plantnutrients because the formulas indicate only the nutrientportions of the primary-element compounds.

Growing plants can assimilate only fertilizer elementsin the combined state of inorganic compounds that areamenable to osmotic absorption. Many modern fertilizer

materials consist of compounds that are immediatelyusable by the crops to which they are applied. Others arequickly converted within the soil to forms that can beassimilated. Some fertilizer chemicals are specificallydesigned to dissolve slowly or to delay reaction within thesoil and, therefore, prolong the release of easily absorbedcompounds to provide sustained feeding over the growthcycle of the plants.

Aqueous solutions of urea, ammonia and ammoniumnitrate (UAN solutions) are used directly by the farmersas well as in the preparation of granular N-P-K productsby mixing with other materials, such as normal superphos-phate and triple superphosphate. UAN solutions are alsospread directly by field application or used to preparecomplete N-P-K fertilizer solutions or suspensions.

Fertilizing

● Addition of elements or other materials to the soil toincrease or maintain plant yields is known as fertilizing.

● Fertilizers may be organic or inorganic.

● Organic fertilizers are usually manures and waste mate-rials which in addition to providing small amounts of growthelements also serve as conditioners for the soil.

● Methods of applying fertilizers vary widely and depend onsuch factors as kind of crop and stage of growth, applica-tion rates, physical and chemical properties of the ferti-lizers and soil type.

● Two basic application methods of fertilizers used are bulkspreading and precision placement.

● Liquid fertilizer of the high pressure type (e.g., anhydrousammonia) is usually regulated by valves or positive dis-placement pumps.

● Non-volatile fertilizer solutions are often pumped into thesupply lines of irrigation systems to allow simultaneousfertilization and irrigation.

BiofertilizersBiofertilizers are the organisms which enrich the soil

in nutrients due to their biological activity. To combat theill effects of chemical fertilizers, biofertilizers have nowbeen introduced for the pollution-free and better growth ofthe plants of useful aspects. The chief sources of biofer-tilizers are blue-green algae (cyanobacteria), fungi andbacteria.

1. Algae as Biofertilizers

Early in the history of agriculture in coastal Asia thevalue of sea weeds in fertilizing the soil was discovered.Long before the recognition of the potash content, seaweeds were employed as fertilizers by the farmers. Notonly as the fertilizers, but also the water-holding capacityof fragments of the algae in the soil proved effective.Furthermore, the yield of paddy is increased substantially

C.S.V. / August / 2009 / 761

when paddy field is inoculated with nitrogen-fixing blue-green algae, such as Anabaena oryzae, Tolypothrixtenius, Calothrix, Cylindrospermum bengalense, andNostoc commune.

Blue-green algae (cyanobacteria) secrete growth pro-moting substances like IBA, NAA, IAA and various pro-teins and vitamins. They add sufficient amounts of organicmatter in the soil. They can grow and multiply under widepH range of 6·5–8·5. Therefore, they can be used as thepossible tool to reclaim saline or alkaline soil because oftheir ameliorating effect on the physio-chemical propertiesof soil.

2. Bacteria as Biofertilizers

Rhizobium is one of the best fertilizers of leguminousplants. Rhizobium is a bacterium living in the root nodulesof leguminous plants and making symbiotic associationwith them. The root cells of leguminous plants possess apurple coloured pigment known as leghaemoglobin inwhich Rhizobium floats to fix atmospheric nitrogen.

Clostridium, Azotobacter, Aerobacter and Methano-bacterium are free-living bacteria which fix atmosphericnitrogen. Rhizobial biofertilizer can fix 50–150 kg/Na/ha/yr.

3. Pteriodophytes as Biofertilizers

In the cavities of leaves of certain aquatic pterido-phytes, such as Azolla, a large number of plants ofAnabaena (Anabaena azollae, a blue-green alga) are pre-sent which have the capacity to fix atmospheric nitrogenand make it available to Azolla (an aquatic pteridophyte).Azolla supplies nitrogen, increases organic matter andfertility in soil and shows tolerance against heavy metals.Dr. P. K. Singh has done an outstanding work on masscultivation of Azolla and its uses as biofertilizer in rice andother crop fields.

4. Micorrhiza as Biofertilizer

Mycorrhiza is the symbiotic relationship between afungus and the roots of higher plant. Mycorrhiza performsthe following functions :

● It enhances phosphate nutrition in plant.● It enhances water and various other nutrient up-

take.● It enhances greater plant vigour (heterosis),

growth and yield.Mycorrhiza is of two types—ectomycorrhiza and

endomycorrhiza. Ectomycorrhiza increases water andnutrient intake by plants. It occurs in plants like Eucalyp-tus, Ficus, Oak, Pine, etc. It absorbs and stores nitrogen,potassium, phosphorus and calcium in fungus. It alsoconverts complex organic molecules into simpler andeasily absorbable form. Endomycorrhiza is important inphosphate nutrition of plants. It is also called VesicularArbascular Mycorrhiza (VAM). VAM has capacity topenetrate even the cortical cells.

Other Benefits of Mycorrhiza to Plants● VAM fungi enhance water uptake in plants.● They play a key role for selective absorption of immobile

(Cu, Zn and P) and mobile (Fe, Mn, Cl, N, Br, S, Ca andK) elements to plants. These are available to plants in lessamount.

● They increase resistance in plants and with their presencethe effects of pathogens and pests on plant health isreduced.

● Some of the trees like pines cannot grow in new areasunless soil has mycorrhizal inocula because of limited orcoarse root hairs.

● They increase the longevity of feeder roots, surface areasof roots by forming mantle and spreading mycelia into soiland, in turn, the rate of absorption of major and minornutrients from soil resulting in enhanced plant growth.

OBJECTIVE QUESTIONS

1. An aquatic fern used as bioferti-lizer is—(A) Azolla(B) Marsilea

(C) Salvia

(D) None of the above

2. Manure containing a mixture ofcattle dung and crop residues isknown as—

(A) Green manure

(B) Farm yard manure

(C) Organic manure

(D) None of the above

3. Materials of biological origin thatare applied commonly to main-tain and improve soil fertility aretermed—(A) Nitrogenous fertilizers(B) Biofertilizers

(C) Manures(D) All of the above

4. A legume having symbiotic asso-ciation with Rhizobium and Aero-rhizobium is—(A) Sebania rostrata(B) Sebanea aculeata(C) Crotolaria juncea(D) None of the above

5. Fertilizers applied to crop plantspollute—(A) Chiefly water resources(B) Chiefly atmosphere(C) Soil resources(D) Soil resources and water

resources

6. A green manure—(A) Protects soil against erosion

and leaching(B) Supplies additional nitrogen

(C) Supplies organic matter(D) None of the above

7. Leaves of Azolla possess thecolonies of—(A) Anabaena(B) Rhizobium(C) Both (A) and (B)(D) Azotobacter

8. Biofertilizer present in the rootsof legume plants is—

(A) Azospirillum

(B) Anabaena(C) Rhizobium(D) All of the above

ANSWERS

●●●

C.S.V. / August / 2009 / 762

Introduction and Occurrence● Club fungi (division—Basidiomycota), which have

septate hyphae, include the familiar mushroomsgrowing on lawns and the shelf or bracket fungi foundon dead trees.

● Less but well known club fungi are puffballs, bird’snest fungi and stinkhorns.

● This big group of fungi containing about 16000 spe-cies includes both saprophytic and parasitic species.

● Basidiocarps, also called fruiting bodies, contain thebasidia, club-shaped structures which produce basi-diospores and from which this division takes thisname (club-fungi).

Mycelium● The mycelium generally is a weft of interlacing and

anastomosing hyphae. In a few genera, however, themycelial hyphae run parallel to one another and getbundled together to form definite and conspicuousthick cords called rhizomorphs.

● Most of the Basidiomycetes are heterothallic. Itmeans primary or homokaryotic mycelium in them isof two distinct strains which are called plus (+) andminus (–) strains.

● The well-developed filamentous mycelium consistingof a mass of branched, septate hyphae generallyspreads in a fan-shaped manner.

● The mycelium of Basidiomycetes passes throughthree distinct stages—primary, secondary and tertiarybefore the fungus completes its life cycle.

Clamp Connection● The dikaryotic cell divides repeatedly by conjugate

divisions to give rise to a secondary or dikaryoticmycelium. During nuclear divisions of the dikaryoticcell special structures called clamp connections areformed.

Diplodisation (Dikaryotisation)

The process by which the primary mycelium is convertedinto secondary mycelium or dikaryotic mycelium is calleddiplodisation or dikaryotisation. It may take place by thefollowing methods—

(i) By hyphal fusion—In this case, fusion occurs betweenthe vegetative cells of two neighbouring hyphae.

(ii) By the fusion a germinating basidiospore and a diploidcell of the basidium.

(iii) By the fusion between the two haploid cells of oppositestrains of the basidium.

(iv) By the fusion between germinating odium of onestrains of the basidium.

(v) By conjugation of basidiospores—In this case twobasidiospores of opposite strains meet and conjugate.

● The clamp connections are usually formed on theterminal cells of the hyphae of the secondary myce-lium.

● The clamp connections by some mycologists are con-sidered to be homologous to the hooks of ascoge-nous hyphae of the Ascomycetes.

● Presence of hooklike clamp connection is a safecriterion for distinguishing a secondary or dikaryoticmycelium from the primary or monokaryotic myce-lium.

A B C D E F

Fig. (A–F) Basidiomycetes. Diagram illustrating theformation of clamp connections.

Sexual Reproduction● Club fungi usually reproduce sexually. However,

development of sex organs, the male antheria andfemale ascogonia are universally absent throughoutthe class.

● The rudimentary differences in sex, shown at the timeof sexual fusion, are designated as plus (+) andminus (–) signs. These signs are called sexualstrains. Either of these mycelia, if cultured artificially,remains sterile. They form no fructifications.

● Fructifications are formed only if two mycelia of oppo-site strains come in contact. The sexual processbeing extremely simplified, consists of three funda-mental processes such as karyogamy, sexual fusionor plasmogamy and meiosis.

(i) Karyogamy

● The terminal binucleate or dikaryotic cells of thehyphae of the secondary mycelium develop into basi-dia.

● The two nuclei in the dikaryotic cell fuse. This fusionis called karyogamy. The resultant diploid fusionnucleus is called a synkaryon.

● The young basidium containing the synkaryon iscalled probasidium which represents the transitorydiplophase.

C.S.V. / August / 2009 / 763

Brand spores

Dikaryotic Diploid

A

BC

Youngbasidium

Basidio-spore

Brandspore

D E

F

Fig. : (A–E) Basidiomycetes. Ustilago sp. (A) Dikaryotichypha forming brand spores; (B–C) brand spores(dikaryotic and diploid); (D) Germinated diploidbrand spore to form an epibasidium with fourhaploid nuclei arranged in a row; (E) mature four-celled epibasidium bearing basidiospores; (F)Brand spore of Ustilago budding in a nutrientsolution (after Brefeld).

(ii) Plasmogamy

● Plasmogamy means the union of two protoplastswhereby the sexual nuclei of opposite strains comeclose together in a pair within the same cell.

● In Basidiomycetes, plasmogamy is achieved either bysomatogamy or by spermatization.

● The two somatic hyphae of the primary mycelia ofopposite strains come in contact and lie side by sidein the fusion cell. This sexual union or plasmogamyby fusion of somatic cell is called somatogamy.

● In the homothallic species plasmogamy occurs by theformation of tubular connections between the somaticcells of the same mycelium.

● Plasmogamy by the union of a spermatium with areceptive hyphae (female organ) is known as sper-matization.

● Plasmogamy by spermatization exclusively occurs inthe rusts which produce numerous tiny, uninucleate,nonmotile spore-like bodies called spermatia. Theyare formed in flask-shaped organs, the spermato-gonia.

(iii) Meiosis :

● The synkaryon in the probasidium undergoes twosuccessive nuclear divisions which constitute meio-sis.

● Meiosis restores the haploid condition in the lifecycle.

● Both karyogamy and meiosis take place in the basi-dium at different stages of development.

Asexual Reproduction

(i) By Conidia :

● The production of conidia is not so common occur-rence in the Basidiomycetes. They are formed in therusts, smuts and some other Basidiomycetes.

● In smuts, conidia are budded off from the basidio-spores and the mycelium.

● The conidia in Basidiomycetes are produced by thedikaryotic mycelium. They serve to propagate thedikaryophase in the life cycle.

(ii) By Oidia :

● Oidia are hyaline, small and thin-walled unicellularsections or fragments of the mycelium.

● Oidia may be uni—or binucleate as whether they areproduced by the breaking up of the primary or secon-dary mycelium.

● Oidia serve a double function; they may either germi-nate to form primary mycelia or bring about diplodi-sation. In the latter case the germinating odium actsas a spermatium and fuses with the somatic hyphalwall of opposite strain.

● In some species, the oidia are segmented from spe-cial, short lateral hyphal branches called the oidio-phores.

(iii) By Fragmentation

● Any part detached from parent cell containing coni-diophores become able to form a new mycelium(plant body).

Basidiocarp and Basidium

● In higher Basidiomycetes (class = Homobasidiomycetidae)the secondary mycelium develops fruiting bodies calledbasidiocarps.

● The basidiocarps are usually massive aerial sporophoreswhich bear basidia.

● The basidia, which are characteristic reproductive struc-tures of Basidiomycetes, are of two types in general, theholobasidium and phragmobasidium.

● Holobasidia are aseptate and thus unicellular, whereasphragmobasidia are septate basidia.

● Holobasidia are characteristic of most of the Basid-iomycetes particularly the gilled (gill-containing) or fleshyfungi.

● Basidia are developed in a palisade-like layer on thebasidiocarp. This fertile layer is called hymenium.

● Basidia produce basidiospores, which are often wind-blown, when they germinate, the new haploid mycelia areformed.

● The septate basidium or phragmobasidium is typical ofrusts and smuts which usually does not form any fructifica-tion or basidiocarp.

C.S.V. / August / 2009 / 764

Epiba-sidium

A

BC

Hyp

obas

idiu

m

Epi

basi

dium

Hyp

obas

idiu

m

BasidiosporesSporidia

Migratingnucleus

Sterigma

Epibasidium

Brand spore

DE

Fig. : (A–E) Basidiomycetes. Different types of basidia. (A)Stichobasidial type; (B) Chiastobasidial type; (C) Tuning-forktype; (D) Holobasidium; (E) Stichobasidial type with a terminalcluster of septate, sickle-shaped sporidia.

Pileus

GillsAnnulus

Pores

Stipe

Volva

MyceliumAB

C

D

Pileus

Fruit body

G Log of wood

PeridioleVolva

E F

Fig. : (A–G) Basidiomycetes. Common types of basidiocarps.(A) basidiocarp of Agaricus with gills on the underside ofpileus; (B) basidiocarp of Boletus with pores on the undersideof pileus; (C) basidiocarp of Lycoperdon; (D) basidiocarp ofGeaster; (E) fruit bodies of Cyathus; (F) mature fruit body ofPhallus; (G) fruit bodies of Fomes.

Rusts and Smuts are Parasitic Club Fungi● Rusts and smuts are club fungi that parasitize cereal

crops, such as corn, wheat, rye and oats.● They do not form the basidiocarps.● Some smuts enter seeds of the host and exist inside

the host plant, becoming visible only near maturity.● In corn smuts, the mycelium grows between the corn

kernel and secretes substances that cause the deve-lopment of tumors on the ears of corn.

● The life cycle of rusts, which may be particularly com-plex, often requires two different plant host species tocomplete the life cycle.

● Black stem rust of wheat uses barberry leaf as analternate host.

Differences Between Rusts and Smuts

Rusts Smuts

1. Rusts may be heteroe-cious and autoecious.For example, wheatrusts are heteroeciousand others are autoe-cious.

1. All smuts are autoe-cious.

2. The rusts are intercellu-lar and obtain theirnutrition by means ofhaustoria.

2. The smuts may beintercellular or intracel-lular (Ustilago maydis).Haustoria are absent.

3. Clamp connections onthe secondary myce-lium are rare.

3. Clamp connections arecommon.

4. The dikaryotic mycel-ium produces threekinds of spores :uredospores and tele-utospores on the pri-mary host and aecio-spores on the alternatehost.

4. It produces only onekind of binucleatespores called the smutspores which are com-parable to the teleuto-spores of rust fungi.

5. Teleutospores are two-celled, stalked; andeach cell is binucleate.

5. The brand spores(teleutospores) are uni-cellular and binucleate.

6. The teleutospores aredeveloped from theterminal cells of themycelium.

6. Smut spores are for-med from the inter-calary cells.

7. Each cell of the 2-celled teleutosporesproduces an epibasi-dium which bears fourbasidiospores. Theyare borne on sterig-mata and are dis-charged violently by thewaterdrop method.

7. The single-celled, brandspore germinates toproduce a single epiba-sidium which bears avariable number ofbasidiospores. Theyare not borne on sterig-mata nor are they dis-charged violently.

Economic Importance● Several members of Basidiomycetes are of great

economic importance because of their beneficial aswell as harmful nature.

C.S.V. / August / 2009 / 765

● Some of them are causative agents of most destruc-tive diseases of cereal crops such as ‘smut disease’of oats, wheat, corn, barley as well as the wheat rust.

● Stem rust of wheat is caused by Puccinia graministritici.

● Some of the higher Basidiomycetes, such as sporefungi, are the common wood rotters; they destroylumbar and timber.

● Mushrooms which also belong to this group are ofgreat economic value as food.

● The young fleshy sporophores of many species of‘puff balls’, e.g., Lycoperdon and Clavatia are alsoedible.

● Clavatia has medicinal value as it contains anticancersubstance calvacin.

● Many members of this class form ectomorphic mycor-rhizal associations with the roots of forest trees.

Important Smut Diseases of CerealsCereals Loose smut

(Causal organisms)Covered smut

(Causal organisms)

Wheat Ustilago nuda (race ofU. tritici)

Tilletia caries

Oat Ustilago avenae Ustilago horde; ( arace of U. Kolleri )

Corn Ustilago maydis ——

Barley Ustilago nuda Ustilago hordei

OBJECTIVE QUESTIONS

1. In certain members of Basidio-mycetes, the basidium bears fourspores exogenously, eachusually at the tip of a minute stalkknown as—(A) Sterigmata(B) Basidiophore(C) Both (A) and (B)(D) Conidiophore

2. ‘Loose smut’ of corn is causedby—(A) Ustilago maydis(B) Ustilago avenae(C) Ustilago nuda(D) All of the above

3. Which division of fungi is com-monly known as club-fungi ?(A) Zygomycetes(B) Oomycetes(C) Deuteromycetes(D) Basidiomycetes

4. Clamp connection is characteris-tic of a certain group of—(A) Algae and cyanobacteria(B) Fungi(C) Bryophytes(D) None of the above

5. Puccinia graminis tritici causes—(A) Loose smut of oat(B) Black stem rust of wheat(C) Loose smut of barley(D) Covered smut of oat

6. Anticancer substance ‘calvacin’is obtained from—(A) Club fungi(B) Red algae(C) Myxophycean cell(D) None of the above

7. Conidiospores are formed—(A) When nutrients are in short

supply(B) During sexual reproduction(C) By sporangia(D) By sac, club and imperfect

fungi

8. Which of the following group(s)is/are comprised of club fungi ?

(A) Mushrooms

(B) Puff balls

(C) Truffles

(D) All of the above

9. In which of the following thedikaryotic stage is longer-lasting ?

(A) Sac fungi

(B) Club fungi

(C) Imperfect fungi

(D) Zygospore fungi

10. The part of mushroom that isvisible above the ground is—(A) Ascogonium (B) Ascocarp(C) Zygospore (D) Basidiocarp

ANSWERS

●●●

(Continued from Page 737 )

5. During oogenesis, each haploidcell produces—(A) Four functional eggs(B) Two functional eggs(C) One functional egg and two

polar bodies(D) Four functional polar bodies

6. During spermatogenesis, thereare three phases in which theprocess is completed. The correctsequence of phases is—

(A) Growth → Multiplication →Maturation

(B) Multiplication → Growth →Maturation

(C) Maturation → Multiplication→ Growth

(D) Multiplication → Maturation→ Growth

7. Growth phase prepares oogoniaor spermatogonia for first—

(A) Mitotic division

(B) Meiotic division

(C) Binary fission

(D) Multiple fission

8. The sperm head is covered by amembrane—enclosed structurecalled—

(A) Head (B) Neck

(C) Middle piece (D) Acrosome

9. A pair of centrioles found in thesperm, are located in the region—

(A) Head (B) Neck

(C) Middle piece (D) Tail

10. Which of the following is thegenetic part of the sperm ?

(A) Head (B) Neck

(C) Middle piece (D) Tail

ANSWERS

●●●

C.S.V. / August / 2009 / 766

1. What happens during photo-respiration when stomates areclosed ?(A) The concentration of CO2

increases in leaves(B) The concentration of CO2

decreases in the leaves(C) Oxygen, a by-product of

photosynthesis, increases(D) Both (B) and (C)

2. In which of the following waysthe RNA differs from DNA ?

(A) The pentose sugar is ribose,not deoxyribose

(B) The base uracil replacesthymine

(C) RNA is single stranded

(D) All of the above

3. The fibres which are used in themanufacture of brooms areknown as—

(A) Brush fibres

(B) Rough weaving fibres

(C) Soft or bast fibres

(D) None of the above

4. The genetic code is a triplet codeand each codon consists of—

(A) One base

(B) Two bases

(C) Three bases

(D) Four bases

5. Tendrils of Smilax are homolo-gous to—

(A) Axillary buds

(B) Stipules

(C) Leaf apices

(D) Leaflets

6. The internal mechanism bywhich a biological rhythm ismaintained in the absence ofappropriate environmental stimuliis termed as a—

(A) Circardian rhythm

(B) Sleep movement

(C) Thigmotropism

(D) Biological clock

7. The first-formed elements ofphloem are called—(A) Protophloem(B) Metaphloem(C) Medullary ray(D) Both (B) and (C)

8. Which part of a leaf carries onmost of the photosynthesis of aplant ?(A) Guard cells(B) Epidermis(C) Epidermal layer(D) Mesophyll

9. Which of the following is comm-only known as ‘thorn apple’ ?(A) Aegle marmelos(B) Ananas comosus(C) Datura stramonium(D) Garcinia mangostana

10. A change in the amount ofenergy in the form of heat libe-rated or absorbed by the systemduring physical or changes is ter-med as—

(A) Enthalpy

(B) Entropy

(C) Ecological energetics

(D) Pyramid of energy

11. Amino acids are converted intoammonia by a group of bacteriacalled—

(A) Ammonifying bacteria(B) Denitrifying bacteria(C) Nitrifying bacteria(D) All of the above

12. During the noncyclic electronpathway, electrons move from—(A) Water through PS-II to PS-I

and then to NADP+

(B) Water through PS-I to PS-IIand then to NADP+

(C) PS-I to PS-II(D) ADP to ATP

13. The mode of reproduction inphycomycetes is—(A) Asexual(B) Sexual

(C) Both (A) and (B)(D) None of the above

14. The botanical name of ‘prayerplant’ is—(A) Ipomoea(B) Lantana Camara(C) Osimum sanctum(D) None of the above

15. The chloroplast envelope en-closes a liquid protenaceoussubstance called—(A) Protoplasm (B) Cytoplasm(C) Granum (D) Stroma

16. An agent, such as radiation or achemical, that brings about amutation is called—(A) Mutagen (B) Transcription(C) Polysome (D) Intron

17. In Utricularia, leaves are modi-fied into—(A) Petiole (B) Bladder(C) Tendril (D) Pitcher

18. Which of the following scientistsproved experimentally that DNAand not protein is the geneticmaterial of the T2 bacterio-phage ?(A) Watson and Crick(B) Hershey and Chase(C) Jacob and Monod(D) Miescher and Griffith

19. In shade leaves the ratio ofchlorophyll to carotenoids appro-aches—(A) 5 : 1 (B) 2 : 3(C) 3 : 2 (D) 1 : 1

20. Plasmodesmata are formedaround the elements of—(A) Golgi bodies(B) Chloroplast(C) Nucleus(D) None of the above

21. Corymb inflorescence has—(A) Flattened peduncle(B) Long peduncle(C) Short peduncle(D) No peduncle

22. Which of the following is correctregarding aerobic respiration ?(A) It is common in all higher

plants(B) Energy is liberated in the

form of ATP

C.S.V. / August / 2009 / 767

(C) End products are H2O andCO2

(D) All of the above

23. Cyathium and hypanthodiumtypes of inflorescence are similarin having—(A) Unisexual flowers(B) Petaloid bracts(C) Apical pore(D) Nectar glands

24. Which of the following RNAcarries a sequence of codons tothe ribosomes ?(A) t-RNA (B) m-RNA(C) r-RNA (D) All of these

25. The structure which can showthe effect of traits brought by themale gamete immediately afterits formation is—(A) Plumule (B) Cotyledon(C) Endosperm (D) Embryo

26. A microbial metabolite excretedor released by lysed cells whichin very low concentration is dire-ctly toxic to cells of the suscept,is defined as—(A) Pathogenesis (B) Taxis(C) Toxin (D) Etiology

27. Guttation is the process ofelimination of water from plantsthrough—(A) Lenticels (B) Stomata(C) Wounds (D) Hydathodes

28. The core of the axis whichincludes the vascular system, theinterfascicular portion, the pith inthe vicinity of vascular bundles(pericycle) is called—(A) Stele (B) Cortex(C) Tracheid (D) Endodermis

29. Which of the following factors ismost important in regulatingtranspiration ?(A) Wind (B) Temperature(C) Light (D) Humidity

30. Hira (HD 1941) and Moti (HD1949) are the varieties of—(A) Wheat (B) Rice(C) Maize (D) Pulse

31. Sugarcane, maize and someother tropical plants have highefficiency of CO2 fixation beca-use they operate—(A) TCA cycle(B) Calvin cycle

(C) Hatch-Slack cycle(D) PP-pathway

32. An increase in the dry weight of aplant is dependent upon the pre-sence of—(A) Oxygen(B) Iron(C) Nitrogen(D) Carbon dioxide

33. RNA retroviruses have a special(particular) type of enzyme res-ponsible for—(A) Transcribing viral RNA to a

DNA(B) Polymerizing host DNA(C) Disintegrating host DNA(D) Translating host DNA

34. Which of the following gives apossible sequence of organicchemicals prior to the protocell ?(A) Inorganic gases, nucleo-

tides, nucleic acids, genes.(B) Inorganic gases, amino

acids, polypeptide, micro-sphere

(C) Both (A) and (B)(D) Water, salt, protein, oxygen

35. Nitrifying bacteria are keptunder—(A) Vibrio group(B) Bacillus group(C) Coccus group(D) None of the above

36. Which of the following xylem ele-ments is living ?(A) Fibres (B) Vessels(C) Parenchyma (D) Tracheids

37. A rootless aquatic plant in whicha portion of leaf is modified toform a bladder for catching smallaquatic animals is—(A) Utricularia (B) Drosera(C) Nepenthes (D) Dionaea

38. Golgi bodies are absent in—(A) Higher plants(B) Bacteria(C) Blue green algae(D) Both (B) and (C)

39. Herbaceous angiosperms flouri-shed in—(A) Holocene epoch(B) Pleistocene epoch(C) Pliocene epoch(D) Miocene epoch

40. How many primary types of his-tone molecules are known ?(A) One (B) Two(C) Four (D) Five

41. Who among the following coinedthe term cryptovegetarian ?(A) Draghetti(B) Spalanzani(C) Lamarck(D) Darwin

42. For each of the twenty aminoacids found in proteins, thereshould be atleast—(A) One t-RNA molecule(B) Two t-RNA molecules(C) Three t-RNA molecules(D) Twenty t-RNA molecules

43. Cinnamon is obtained from whichpart of the plant body ?(A) Bark (B) Root(C) Leaf (D) Fruit

44. The edible part of tomato is—

(A) Mesocarp only

(B) Epicarp

(C) Thalamus

(D) Placenta and pericarp

45. A substance or mixture of subs-tances which prevents, miti-gates, destroys or repels anypest is commonly called—(A) Epinasty(B) Growth hormone(C) Antitransparent(D) Pesticide

46. Apple and pear are—(A) Schizocarpic and pome fruits(B) Schizocarpic and pepo fruits(C) Succulent and pome fruits(D) True and schizocarpic fruits

47. In which of the following plants,the seed germinates while stillattached to the parent (main)plant ?(A) Rhizophora(B) Screwpine(C) Mango(D) All of the above

48. Polysome is a group of—(A) Nucleolus(B) Endoplasmic reticulum(C) Spherosomes(D) None of the above

C.S.V. / August / 2009 / 768

49. Which of the following plants iscommonly called Swallow wart ?

(A) Calotropis procera(B) Eucalyptus globulus(C) Jatropha curcas(D) Bryophyllum pinnatum

50. An allosteric site on an enzymeis—(A) Often involved in feedback

inhibition(B) Where ATP attaches and

gives up its energy(C) The same as the active site(D) All of the above

ANSWERS WITH HINTS

(Continued on Page 790 )

C.S.V. / August / 2009 / 769

1. Match Column A (Different typesof fruits) with Column B (Differentexamples) then select the correctanswer from the options givenbelow—

Column A Column B

(a) Capsule 1. Acer(b) Lomentum 2. Iris(c) Samara 3. Genarium(d) Regma 4. Acacia

(a) (b) (c) (d)

(A) 2 4 3 1

(B) 3 2 1 4

(C) 2 4 1 3

(D) 3 4 2 1

2. Proteins and polar lipids accountfor almost all of the mass ofbiological membranes; the smallamount of carbohydrate presentis generally part of—

(A) Glycoproteins

(B) Glycolipids

(C) Both (A) and (B)

(D) None of the above

3. Mobilization of stored food in ger-minating seed is triggered by—

(A) Gibberellin (B) Auxin

(C) Abscisic acid (D) Cytokinin

4. Match Column ‘A’ (Commonname) with Column ‘B’ (Botani-cal name) then select the correctanswer from the options givenbelow—

Column ‘A’ Column ‘B’(a) Palmyra

palm1. Erianthus

ravennae(b) Toddy palm 2. Borassus

flabellifer(c) Plum grass 3. Typha

angustata(d) Elephant grass

4. Caryotaurens

(a) (b) (c) (d)

(A) 4 3 2 1

(B) 4 3 1 2

(C) 2 4 1 3

(D) 2 4 3 1

5. The oil is extracted from theseeds of—(A) Brassica nigra(B) Brassica campestris(C) Both (A) and (B)(D) None of the above

6. Which of the following is themain internal ground tissue ?(A) Parenchyma (B) Epidermis(C) Endodermis (D) Pith

7. The small size of cells is bestcorrelated with—(A) The fact that they are self-

reproducing(B) An adequate surface area for

exchange of materials(C) Their prokaryotic versus

eukaryotic nature(D) All of the above are correct

8. Which of the following mem-branes prohibits the entry of bothsolvent and solute particles ?(A) Permeable membrane(B) Semipermeable membrane(C) Both (A) and (B)(D) Impermeable membrane

9. Which of the following micro-organisms charges protein intoammonia ?(A) Nitrobacter(B) Azotobacter(C) Rhizobium(D) Bacillus mycoides

10. Synthesis of quinine was achie-ved by—(A) Alexander Fleming(B) Woodward and Doering(C) M.S. Swaminathan(D) U. Brahmchari

11. ‘Safed Lerma’ is a new varietyof—(A) Onion (B) Grape(C) Garlic (D) Wheat

12. Which of the following algae iscommonly known as ‘frog-spawn’ ?(A) Ulothrix(B) Volvox

(C) Chlamydomonas(D) Batrachospermum

13. The water held tightly by the soilparticle around them is knownas—(A) Capillary water(B) Field capacity(C) Hygroscopic(D) None of the above

14. Which of the following state-ments is not correct regardingCycas ?(A) Cycas plant is arboreal(B) It looks like a palm tree(C) The stem is columnar nor-

mally branched with deci-duous leaf bases

(D) Both (A) and (C)

15. The sporophyte of Riccia isrepresented by—(A) Spore sac only(B) Foot, seta and capsule(C) Foot and capsule(D) Spores and elaters

16. The process by which DNA ofnucleus passes information toRNA is called—(A) Transcription(B) Translation(C) Transformation(D) Transduction

17. Chief means of perennation inliverworts is—(A) Tuber formation(B) Persistent apices(C) Both (A) and (B)(D) None of the above

18. Dry rot of sugarbeet and internalbrowning of cauliflower diseasesare due to the deficiency of—(A) Zinc (B) Boron(C) Copper (D) Manganese

19. Membrane often contains steroidwhich is a type of lipid that con-tains—(A) No carbon rings(B) Four carbon rings(C) Only one carbon rings(D) None of the above

20. A red pigment found in the fruitof tomato is called—(A) Lycopene(B) Xanthophyll

C.S.V. / August / 2009 / 770

(C) Myxoxanthin(D) Rhodopsin

21. Who coined the term ‘metabolism’for all chemical processes carriedon in cells ?(A) Schwann (B) Nawashchin(C) Twort (D) Strasburger

22. Inflorescence having a flattenedaxis, sessile flowers and a whorlof involucral bracts is—(A) Corymb (B) Head(C) Raceme (D) Umbel

23. For the purpose of gene therapya retrovirus should be equippedwith—(A) Recombinant RNA(B) t RNA(C) m RNA(D) r RNA

24. Non transfer of pollen fromanther to stigma of the sameflower due to mechanical barrieris—(A) Heterostyly(B) Herkogamy(C) Dichogamy(D) Cleistogamy

25. Regulation of gene activity inprokaryotes usually occurs at thelevel of—(A) Translation(B) Transcription(C) Both (A) and (B)(D) None of the above

26. Genotypically the pollen grainsproduced by an anther belongto—(A) One type(B) Many types(C) Two types(D) All of the above

27. Cross-pollination within a speciesis called—(A) Explant(B) Xenogamy(C) Allopolyploidy(D) Autopolyploidy

28. At constant temperature, the rateof transpiration will be higherat—(A) 1 km below sea level(B) 1 km above sea level(C) 2 km above sea level(D) Sea level

29. The cytoplasmic portion of theprotoplast is referred to as—(A) Endoplasm (B) Cytoplasm(C) Cytosome (D) Ectoplasm

30. Herbicide DCMU kills plants dueto spoilage of—(A) Photophosphorylation(B) Electron transport(C) O2-evolution

(D) Rubisco activity

31. Trisomy is a type of—(A) Euploidy (B) Polyploidy(C) Hyperploidy (D) Hypoploidy

32. Balance between CO2 and O2 ismaintained by—(A) Photosynthesis(B) C4-pathway

(C) Transpiration(D) Photorespiration

33. The potential energy of water isreferred to as—(A) Protoplasmic streaming(B) Thermodynamics(C) Water relation(D) Water potential

34. From photosystem-I ‘high en-ergy’ electrons pass to NADPwhere they combine with hydro-gen ions to form—(A) NAD (B) ADP(C) NADPH2 (D) FAD

35. The noncyclic electron pathwaygenerates—(A) No ATP(B) ATP only(C) NADPH only(D) Both ATP and NADPH

36. The first person to associatespecific gene with a specificchromosome is/was—(A) Morgan (B) Swaminathan(C) Correns (D) Maheswari

37. The baloonlike outgrowth ofparenchyma into the lumen of thevessels is known as—(A) Tunica (B) Tyloses(C) Histogen (D) Phellogen

38. Drosera catches insects bymeans of—(A) Pitcher(B) Adhesive disc(C) Bladder(D) Tentacles secreting shining

liquid

39. Which of the following plantsdoes not belong to family Papil-ionaceae ?(A) Phaseolus mungo(B) Bauhinia variegata(C) Crotolaria juncea(D) All of the above

40. Which of the following is asaprophytic angiosperm ?(A) Neottia (B) Agaricus(C) Cuscuta (D) Eucalyptus

41. Autopolyploidy arised by increasein the number of chromosomesets of the same species iscalled—(A) Intraspecific polyploidy(B) Interspecific polyploidy(C) Both (A) and (B)(D) Explant

42. Which of the following is a truehemp ?(A) Cannabis sativa(B) Boehmeria nivea(C) Girardinia heterophylla(D) Agave sisalana

43. Quadrifoliate leaves are foundin—(A) Paris quadrifoliata(B) Bombax ceiba(C) Hardwickia(D) All of the above

44. The meristematic layer betweenbark and the wood in a woodystem is called—(A) Cork cambium(B) Zone of cell division(C) Vascular cambium(D) None of the above

45. Tribulus fruit is dispersed bymeans of—(A) Water (B) Animals(C) Wind (D) Explosion

46. In bryophytes, absorbing andattaching organs are—(A) Columella(B) Rhizoids(C) Root hairs(D) Whole thallus

47. The vertical temperature gradientover earth’s surface is called—(A) Phenology(B) Littoral zone(C) Sigmoid curve(D) Lapse rate

C.S.V. / August / 2009 / 771

48. If concentration of CO2 in theexternal atmosphere is increa-sed, the stomata are closed in—(A) Dark only(B) Light only(C) Both in light and dark(D) None of the above

49. Leaf primordia are produced bythe—(A) Vascular bundle(B) Cambia(C) Cortex(D) Apical meristem

50. Carrageenin is used as emulsi-fying and stabilizing agent in—(A) Chocolates(B) Ice creams(C) Toothpastes(D) All of the above

ANSWERS WITH HINTS

●●●

C.S.V. / August / 2009 / 772

In each of the following ques-tions, a statement of Assertion (A)is given and a corresponding state-ment of Reason (R) is given justbelow it. Of the statements, markthe correct answer as—

(A) If both A and R are trueand R is the correct expla-nation of A

(B) If both A and R are true butR is not the correct expla-nation of A

(C) If A is true but R is false(D) If both A and R are false(E) If A is false but R is true

PHYSICS

1. Assertion (A) : Light is diffractedaround the edges of obstaclesand the bending is so slight thatit is not easily observed.Reason (R) : The wavelength oflight is very small.(A) (B) (C) (D) (E)

2. Assertion (A) : Mass and energyare not conserved separately, butare conserved as a single entitycalled ‘mass energy’.Reason (R) : This is because onecan be obtained at the cost ofother as per Einstein’s equationE = mc

2. (A) (B) (C) (D) (E)

3. Assertion (A) : Coherent sourcesgenerate waves of identical fre-quencies.

Reason (R) : Two coherentsources of circular wave patternscan be produced by passing aplane wave through narrow slits.

(A) (B) (C) (D) (E)

4. Assertion (A) : The force experi-enced by a charged particlemoving in a magnetic field cannot do any work.

Reason (R) : The force does notdisplace the charged particle.

(A) (B) (C) (D) (E)

5. Assertion (A) : With a white lightsource the central fringe is darkin Young’s double slit experi-ment.

Reason (R) : Light is formed ofmaterial particles.(A) (B) (C) (D) (E)

6. Assertion (A) : If there existsCoulomb attraction between twocharged bodies, both of themmay not be charged.Reason (R) : They will be oppo-sitely charged.(A) (B) (C) (D) (E)

7. Assertion (A) : White light con-tains the range of colours in lightfrom violet with a wavelength of4 × 10– 7 m to red light with awavelength of 7 × 10– 7 m.Reason (R) : When Young’sdouble slit experiment is carriedout with light, multicolouredfringes are formed.(A) (B) (C) (D) (E)

8. Assertion (A) : To float, a bodymust displace liquid whose weightis greater than the actual weightof the body.Reason (R) : A floating body willexperience no net downwardforce.(A) (B) (C) (D) (E)

9. Assertion (A) : Dark lines areobserved in the spectrum of lightproduced from the sun rays.Reason (R) : Dark lines are dueto absorption of certain radiationsby gases in the outer atmos-phere of the sun.(A) (B) (C) (D) (E)

10. Assertion (A) : Two sources Aand B each carrying a sound of400 Hz are standing a few metresapart. When A moves towards B,both persons hear the samenumber of beats per second.Reason (R) : Doppler shift infrequency of sound is the samewhether the observer approachesthe source or the sourceapproaches the observer withsame speed.(A) (B) (C) (D) (E)

CHEMISTRY

11. Assertion (A) : Iodine dissolvesmore appreciably in potassium

iodide solution as compared topure water.

Reason (R) : Potassium iodidesolution absorbs iodine formingcompound of potassium in whichK+ is oxidised to K3 +.

(A) (B) (C) (D) (E)

12. Assertion (A) : At isoelectricpoint of an amino acid, it does notmigrate under the influence of anelectric field.Reason (R) : At isoelectric pointan amino acid is totally ionized.(A) (B) (C) (D) (E)

13. Assertion (A) : Mercury metalstarts sticking to the side of theglass tube on bubbling ozonethrough it.Reason (R) : In the presence ofozone, mercury reacts with glassto form mercury (II) silicide.(A) (B) (C) (D) (E)

14. Assertion (A) : Methanoic acidchanges mercuric chloride tomercurous chloride on heatingbut acetic acid does not do sounder similar conditions.Reason (R) : Methanoic acid is astronger acid than ethanoic acid.(A) (B) (C) (D) (E)

15. Assertion (A) : Sometimes ayellow turbidity appears whilepassing H2S gas even in theabsence of radicals of 2nd group.Reason (R) : Group IV radicalsare precipitated as their sul-phides.(A) (B) (C) (D) (E)

16. Assertion (A) : In the electrolysis,the quantity of electricity neededto deposit one mole atom of silveris different from that required fordepositing one mole atom ofcopper.Reason (R) : Atomic weights ofsilver and copper are different.(A) (B) (C) (D) (E)

17. Assertion (A) : The gelatinouswhite precipitate of aluminiumhydroxide dissolves both in HClas well as in concentrated NaOHsolution.

C.S.V. / August / 2009 / 773

Reason (R) : Aluminium hydro-xide is a strongly ionic as well ascovalent in nature.(A) (B) (C) (D) (E)

18. Assertion (A) : Nitrogen obtainedby fractional distillation of liquidair is heavier than the nitrogenobtained by decomposing nitro-genous compounds.Reason (R) : Nitrogen obtainedfrom fractional distillation of liquidair is not completely pure.(A) (B) (C) (D) (E)

19. Assertion (A) : The methylcarbanion is pyramidal in shapelike the structure of ammoniamolecule.Reason (R) : The carbon atom ofmethyl carbanion is sp

2 hybridi-sed.(A) (B) (C) (D) (E)

20. Assertion (A) : Most of the endo-thermic reactions are not sponta-neous under ordinary conditionsbut become so at elevated tem-perature.Reason (R) : Entropy of thesystem increases with increase intemperature.(A) (B) (C) (D) (E)

ZOOLOGY

21. Assertion (A) : Dramatic popula-tion growth has occurred in thelast 300 years, with more growthin less developed countries(LDCs) than in more developedcountries (MDCs).Reason (R) : Total fertility rate(TFR) is lower in MDCs, due tofamily planning and access tocontraceptives. Infant mortality islower in MDCs than LDCs.(A) (B) (C) (D) (E)

22. Assertion (A) : Capsule of tendonis associated with brain.Reason (R) : Inherited Rh factorgene is found in Rh– individuals.(A) (B) (C) (D) (E)

23. Assertion (A) : Baroreceptor isthe receptor for hydrostatic pres-sure of blood.Reason (R) : Receptors of atriumstimulate the Cardio-acceleratorycentre, helping to regulate bloodpressure.(A) (B) (C) (D) (E)

24. Assertion (A) : Coenzyme is anon-protein group without whichcertain enzymes are incompleteor inactive.Reason (R) : Coenzymes not onlyprovide a point of attachment forthe chemical groups being trans-formed but also influence the pro-perties of the group.(A) (B) (C) (D) (E)

25. Assertion (A) : It is the brain, notthe sense organs, that interpretsthe stimulus.Reason (R) : Sense organs aretransducers, they transform theenergy of stimulus to the energyof nerve impulses.(A) (B) (C) (D) (E)

26. Assertion (A) : Jelly fish is not atrue fish which is a vertebrateanimal with backbone.Reason (R) : The name jelly fishis given to invertebrate coelente-rate animal, Aurelia, because it ismade up of jelly-like substances.(A) (B) (C) (D) (E)

27. Assertion (A) : Girls have alltheir primary oocytes at birth,oogenesis begins at puberty.Reason (R) : The first menstrualcycle is called menopause, andthe end of menstrual cycles iscalled menarche.(A) (B) (C) (D) (E)

28. Assertion (A) : Morphogenesisinvolves change in the shape ofthe embryo.Reason (R) : Differentiation is thespecialization of cell structure andfunction as some genes areturned on and others off.(A) (B) (C) (D) (E)

29. Assertion (A) : The sympatheticdivision acts to mobilize the bodyfor emergency or quick action—the ‘fight-or-flight’ response.Reason (R) : It decreases theheart rate, respiration rate anddigestive activity; dilates bloodvessels of the skeletal muscles,and constricts those of the skin.(A) (B) (C) (D) (E)

30. Assertion (A) : Neurosecretion ispivotal in the vertebrate endocrinesystem.Reason (R) : Hormones are pro-duced by endocrine glands.(A) (B) (C) (D) (E)

BOTANY

31. Assertion (A) : In Mucor mucedoall of the sporangiophores obtai-ned from a single germ sporan-gium give rise to mycelia whichare of the same mating type. Azygospore germination of thistype is known as pure homothal-lic.Reason (R) : Sexual repro-duction in a heterothalic speciesof Mucor is not effected by copu-lation of two multinucleate iso-gametangia.(A) (B) (C) (D) (E)

32. Assertion (A) : Hershey andChase turned to blue-green algaeas their experimental material.Reason (R) : Franklin preparedan X-ray photograph of DNA thatshowed its certain dimensions.(A) (B) (C) (D) (E)

33. Assertion (A) : The sterile por-tion of the sporangium in Mucoris called pileus.Reason (R) : In Mucor, theremainder of the progametan-gium is known as the suspen-sor.(A) (B) (C) (D) (E)

34. Assertion (A) : Archegonium isthe female sex organ of liverworts,ferns and most gymnosperms.Reason (R) : Archegonium ismulticellular of which the swollenbase, i.e., venter contains egg-cell.(A) (B) (C) (D) (E)

35. Assertion (A) : Transposableelements can block gene expres-sion by inserting into a gene.Reason (R) : Transposable ele-ments can move and insert intogenes, disrupting their expres-sion.(A) (B) (C) (D) (E)

36. Assertion (A) : Daily periodicityof stomatal movement is not acharacteristic of succulent plant.Reason (R) : In succulent plantsstomata are closed at night andopened during the day.(A) (B) (C) (D) (E)

37. Assertion (A) : Scientists useseveral approaches when theydesign experiments in order tomake data as valid as possible.

C.S.V. / August / 2009 / 774

Reason (R) : Researchers mightexamine hundreds of cells,because a large sample sizehelps ensure meaningful results.(A) (B) (C) (D) (E)

38. Assertion (A) : Lecanora is agenus of lichens in which thethallus is crustose.Reason (R) : Some species ofLecanora are more resistant than

most lichens to atmosphericpollution.(A) (B) (C) (D) (E)

39. Assertion (A) : The energy cap-turing portion of photosynthesistakes place in thylakoid mem-branes and can not proceedwithout solar energy.Reason (R) : The synthesis partof photosynthesis occurs in chlo-

roplast and does not directlyrequire solar energy.(A) (B) (C) (D) (E)

40. Assertion (A) : Cellular respira-tion includes glycolysis and fer-mentation.Reason (R) : Both glycolysis andfermentation are only aerobicrespiration.(A) (B) (C) (D) (E)

ANSWERS WITH HINTS

(Continued on Page 790 )

C.S.V. / August / 2009 / 775

Physics

1. A convex lens is made of two different materials. Apoint is placed on the principal axis. The number ofimages formed by the lens will be two.

—T/F2. No net force acts on a rectangular coil carrying a

steady current when suspended freely in a uniformmagnetic field.

—T/F3. In an electric field, the electron moves away from

higher potential to the lower potential.—T/F

4. Birds fly off a high tension wire when the current isswitched on.

—T/F5. Fuse wire must have high resistance and low melting

point.—T/F

6. In a perfectly elastic collision, the kinetic energy maynot remain constant.

—T/F7. The sensitivity of a moving coil galvanometer can be

increased by using a suspension wire of shorterlength.

—T/F8. When a monocycle turns, along the curve, remaining

vertical at a constant speed, the angular velocity ofthe wheel does not change.

—T/F9. Cathode rays constitute a stream of negatively

charged particles, called electrons.—T/F

10. A piece of red glass is heated till it glows in dark. Thecolour of the glowing piece will be somewhat orange.

—T/F11. A body, whatever its motion, is always at rest in a

frame of reference which is fixed to the body itself.—T/F

12. An electron revolves round a nucleus of charges Ze.In order to excite the electron from the state n = 2 ton = 3, the energy required is 47·2 eV. Z is equal to 5.

—T/F13. On reducing the volume of a gas at constant tem-

perature, the pressure of the gas increases.—T/F

14. X-rays cannot be diffracted by means of grating.—T/F

15. Thermal conductivity of air being less than that of felt,even then we prefer felt to air for thermal insulation.

—T/F

Chemistry

16. The structural-pair geometry of xenon difluoride(XeF2) is linear.

—T/F17. The heat of hydrogenation of trans-2-butene is more

than that of cis-2-butene.—T/F

18. A balloon is inflated with He to a volume of 4·5 L atroom temperature. If balloon is taken to the placehaving temperature of – 10°C, the volume of theballoon remains unchanged.

—T/F19. The electronic configuration of fluorine atom is

1s 2, 2s

2 2px 2 2py

2 2pz 1, and according to Hund’s rule

the last electron enters to 2pz orbital with clockwisespin.

—T/F20. Phenol is a weaker acid than carbonic acid.

—T/F21. A redox reaction in which reactants containing same

element in different oxidation states react to give aspecies in which the element is in same oxidationstate is called a comproportionation reaction.

—T/F22. The oxidation of sucrose with conc. HNO3 in

presence of V2O5 leads to the formation of 1, 6-hexanedioic acid.

—T/F23. When one mole electron is passed separately through

each of solutions of CuSO4 and AgNO3, the mole ratioof Cu and Ag deposited will be 1 : 1.

—T/F24. Colour is shown by a transition metal ion which

contains all (n – 1) d vacant orbitals.—T/F

25. The vapour density of a substance has no unit and isindependent of temperature and pressure.

—T/F26. Alkaline KMnO4 solution is more readily decolourised

by benzene than by acetylene.—T/F

27. A gas has vapour density of 17, 10 g of this gas at17°C and 1 atm will occupy a volume of 7 mL.

—T/F28. Gatta percha and natural rubber are actually,

stereoisomers.—T/F

29. SnCl2 is a crystalline solid while SnCl4 is a volatileliquid.

—T/F30. Methyl cyanide is soluble in water while methyl

isocyanide is insoluble.—T/F

C.S.V. / August / 2009 / 776

Zoology

31. IgE is not fastened by its C-terminal end.—T/F

32. The rate of oxygen consumption of an organism ortissue is called oxygen quotient.

—T/F33. Osteocytes maintain bone, osteoblasts form new bone

and osteoclasts attack and destroy bones.—T/F

34. Pentose phosphate pathway is alternative toglycolysis.

—T/F35. Twelve pairs of cranial nerves and thirty one pairs of

spinal nerves comprise the somatic peripheral nervesin human nervous system.

—T/F36. Some developmental structures or processes such as

gill pouches in mammalian embryos are regarded asphyletic.

—T/F37. The circulatory system of cockroach, Periplaneta

americana, is of open type.—T/F

38. The occurrence of different morphological stagesduring the life of an organism is called pleiomorphism.

—T/F39. Buffers limit pH changes and protect living cells from

injury by strong acids and bases.—T/F

40. Prosimian is a group of primates that include apesand humans.

—T/F41. Restriction enzymes are not used to cleave the plas-

mid DNA for genetic engineering.—T/F

42. Planarians are free-living turbellarians.—T/F

43. Epitope is an antibody determinant.—T/F

44. Cyclic AMP is a second messenger within cells.—T/F

45. Organic evolution is a change in the frequency ofalleles in a population.

—T/F

Botany

46. Mitochondria supply most of the necessary biologicalenergy by breaking down of protein, sugar andcarbohydrates.

—T/F47. The highest point in the reaction coordinate diagram

represents the idiogram.—T/F

48. Genes which confer antibiotic resistance on bacteriaare located on plasmid.

—T/F

49. Some plants of dry and arid habitats shed their leavesto reduce water loss.

—T/F

50. Colour-blindness is not sex-linked inheritance.

—T/F

51. The edible part of Oryza sativa is epicarp andthalamus.

—T/F

52. Ergotism is caused by Claviceps purpurea.

—T/F

53. Plant tissue culture cannot produce haploid plants.

—T/F

54. Xylem transports water from the root to the leaf.

—T/F55. Stomata arise from the protoderm cells.

—T/F

56. Stomata in angiosperms open and close owing tochange in turgor pressure in guard cells.

—T/F

57. The pine stem is characterized by the presence ofconspicuous resin ducts, which are distributedthroughout the stem.

—T/F

58. The hormone cytokinin is used for early ripening offruits.

—T/F

59. Permeable membranes are those which allow diffusionof both solvent and solute molecules through them.

—T/F60. Mustard shows hypogeal germination.

—T/F

ANSWERS WITH HINTS

C.S.V. / August / 2009 / 778

Physics

Q. Can a body have zero velo-city and constant acceleration ?

☞

Q. What is meant by band-width ?

☞

Q. What is the difference bet-ween the interference and diffrac-tion of light ?

☞

Q. A transistor is a tempera-ture sensitive device. Why ?

☞

Q. What is a solenoid ?Comment on the magnetic fieldaround a solenoid ?

☞

Q. A hydrogen atom containsone electron. But the spectrum ofhydrogen atom has many lines.Why ?

☞

Q. What is axial or pseudovector ?

☞

Q. What is Lami’s theorem ?☞

Q. What is the radius of gyra-tion ?

☞

C.S.V. / August / 2009 / 779

Q. What is law of equipartitionof energy ?

☞

Q. What is total internal reflec-tion ?

☞

Chemistry

Q. How the process of osmo-sis helps developing adema andpreserving meat and fruits ?

☞

Q. What are gemstones ?☞

Q. What are mustard oils ?

☞

Q. What unbreakable plastic-crockery is made of ?

☞

Q. Hydrogen peroxide findsdifferent uses in different concen-trations.

☞

Q. How calcium cyanamide-carbon mixture acts as a fertilizer ?

☞

Q. What is the differencebetween mass and weight of anobject ?

☞

C.S.V. / August / 2009 / 780

Q. Why the product of fre-quency and wavelength of infraredand ultraviolet radiations is alwayssame ?

☞

Q. What is the mass of asingle atom of 12C in grams andthe number of grams per amu ?

☞

Q. What will be the value ofideal gas constant, R, if exactly1 mol of an ideal gas occupies avolume of 22·414 litre at 0°°°°C and1 atom pressure ?

☞

Zoology

Q. What is perception ofpitch ?

☞

Q. What is teleology ?☞

Q. What are good and badcholesterol ?

☞

Q. How do the digestive tractsof carnivores differ from those ofhervivores ?

☞

C.S.V. / August / 2009 / 781

Botany

Q. What do you mean by pureculture ?

☞

Q. Why does coconut oil freezeduring winter ?

☞

Q. What is forest engineeringand its significance ?

☞

Q. How are monosaccharidesformed from polysaccharides ?

☞

Q. Why is CO2 commonlyknown as green house gas ?

☞

Q. What do you mean byEnterobacter ?

☞

Q. What is parsnip ? What areits importance ?

☞

●●●

(Continued from Page 777 )

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C.S.V. / August / 2009 / 782

1. A police van moving on a highway with a speed 30 km h–1 firesa bullet at a thief’s car speedingaway in the same direction with aspeed of 192 km h–1. If the muzzlespeed of the bullet is 150 ms–1,with what speed does the bullethit the thief’s car ?(A) 100 ms–1 (B) 105 ms–1

(C) 95 ms–1 (D) None of these

2. A 4 kg block A is placed on thetop of 8 kg block B which restson a smooth table. A just slips onB when a force of 12N is appliedon A. What is maximum horizon-tal force F required to make bothA and B move together ?

(A) 16N (B) 24N(C) 36N (D) 72N

3. Point P, Q and R are in a verticalline such that PQ = QR. A ball atP is allowed to fall freely. Find theratio of time of descent throughPQ and QR—

(A)1

2 – 1(B)

12 – 1

(C)2 – 11 (D)

32

4. An ideal gas confined to aninsulated chamber is allowed toenter into an evacuated insulatedchamber. If Q, W and ΔEint havethe usual meanings, then—(A) Q = 0, W ≠ 0(B) W = 0, Q ≠ 0(C) ΔEint = 0, Q ≠ 0

(D) Q = W = ΔFint = 0

5. A hydrogen atom and a doublyionised lithium atom are both inthe second excited state. If LHand LLi respectively representtheir electronic angular momentaand EH and ELi their energies,then—(A) LH > LLi and | EH | > | ELi |(B) LH = LLi and | EH | < | ELi |

(C) LH = LLi and | EH | > | ELi |(D) LH < LLi and | EH | < | ELi |

6. Which of the following is notcorrectly matched ?(A) Hexadecane—C6H14(B) Heptadecane—C17H36(C) Eicosane—C20H42(D) Docosane—C22H46

7. Alkenes may be hydrated toalcohols by absorption in—(A) Dilute sulphuric acid(B) Dilute hydrochloric acid(C) Concentrated sulphuric acid(D) Concentrated hydrochloric

acid

8. ‘Cement fondu’ is a/an—(A) Hydraulic cement composed

chiefly of calcium aluminate(B) Hydraulic cement composed

chiefly of sodium silicate(C) Portland cement(D) Cement made from granu-

lated blast furnace slag offairly low alumina

9. Which of the following is/are saidto be diamagnetic species ?(A) Carbanion(B) Carbocation(C) Singlet carbene(D) All of the above

10. Phenols are more acidic thanalcohols because—(A) Alcohols do not lose H+ ions(B) Phenoxide ion is stabilized

by resonance(C) Phenoxide ion does not

exhibit resonance(D) Phenols are more soluble in

polar solvents

11. Medrysone is a/an—(A) Newly discovered mutagen

from various insects(B) Adrenal corticosteroid drug(C) Pill for contraceptive treat-

ment(D) Endonuclease enzyme

12. In which of the following orga-nisms, the head is prolonged intoa tubular rostrum ?(A) Sea horse (B) Varanus(C) Heloderma (D) Ophisaurus

13. Which of the following is/arecorrect about interclavicular ofpigeon ?(A) It is a median(B) It is unpaired(C) Connected to the secondary

bronchi of both lungs(D) All of the above

14. How many nucleotide pairs arereported in β-globin gene ofmouse by Leder et al ?(A) 116 (B) 216(C) 96 (D) 106

15. The induction of regionalanesthesia by preventing sensorynerve impulse from reachingcentres of consciousness, isknown as—(A) Nerve back(B) Nerve entrapment syndrome(C) Nerve growth factor(D) Nephrydrosis

16. Soil is chiefly composed of—(A) Only water and organic

matter(B) Air + water + organic matter

+ minerals(C) Only organic colloids(D) None of the above

17. In balausta type of fruit, the ediblepart is—(A) Aril(B) Fleshy thalamus(C) Succulent testa(D) Succulent sepal

18. From which one of the followingflowers is the insecticide‘Pyrethrum’ derived ?(A) Rosa(B) Nelumbo(C) Iberis(D) Chrysanthemum

19. The mutagen proflavin is a/an—(A) Base analog(B) Hydroxylating agent(C) Acridine dye(D) Alkalylating agent

20. Bud scales are found in—(A) Jack fruit(B) Ficus(C) Magnolia(D) All of the above

●●●

C.S.V. / August / 2009 / 783

Age.................. Academic Qualification........................

Competition examination for which preparing

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................................... (Signature)

Solution to Quiz No. 135 Competition Science Vision

Last date for sending 28th August, 2009

Name Mr./Miss/Mrs. ...........................….........................

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RESULT

No. of questions attempted..........................................

No. of correct answers.................................................

No. of wrong answers...................................................

Marks obtained.............................................................

ANSWER FORM

Q. No. A B C D Q. No. A B C D

1. 11.

2. 12.

3. 13.

4. 14.

5. 15.

6. 16.

7. 17.

8. 18.

9. 19.

10. 20.

Rules for taking part in Quiz Contestof Competition Science Vision

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2. Candidates taking part in quiz contest will necessarilyhave to send their entries by a fixed date. Entries areto be sent by ordinary post. Please mark yourenvelope 'Quiz–Competition Science Vision' onthe top left hand side.

3. Answers given only on the form of the magazine willbe admissible.

4. In the form there are four squares against eachquestion number. Contestants should put a cross (×)in the square for the answer they think is correct.Giving more than one answer to a question willdisqualify it.

5. Contestants should essentially write the number ofquestions they have solved.

6. Marks will be deducted for wrong answers.7. The candidate sending the maximum number of

correct answers will be given Rs. 600 as first prize.Next two candidates after that will get Rs. 400 andRs. 300 as second and third prize respectively. Ifthere are more than one candidate eligible for aprize, the amount will be equally distributed amongthem.

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C.S.V. / August / 2009 / 785

According to the rules of the CSVQuiz, all entry forms were examined.As a result, the following participantshave qualified for various prizes. CSVsends them greetings and goodwishes for their bright future. It alsoplaces on record its appreciation fortheir inquisitive nature and expressesobligation for their co-operation.

PRIZE WINNERS

First PrizeRavi Jaiswal C/o Gaurav JaiswalRoom No. 88, A. N. Jha Hostel,University of Allahabad, AllahabadU.P.–211 002

Second PrizeVishal TiwariIII-B-231Vidyut Vihar ColonyShaktinagar, SonebhadraU.P.–231 222

Third Prize1. Gagandeep Singh

C/o Dayaram VermaL-971, Shastri Nagar, MeerutU.P.–250 004

2. Sarvdaman SharmaLane No. 23Greater Kailash, JammuJ & K–180 011

Note : Amount of third prize has beendistributed among two third prizewinners.

C.S.V. / August / 2009 / 786

1. For which one of the followingcountries, is Spanish not anofficial language ?(A) Chile(B) Colombia(C) Republic of Congo(D) Cuba

2. For which one of the following, isTirupur well known as a hugeexporter to many parts of theworld ?(A) Gems and Jewellery(B) Leather goods(C) Knitted garments(D) Handicrafts

3. The great Asian river Mekongdoes not pass through—(A) China (B) Malaysia(C) Cambodia (D) Loas

4. Which one of the following doesnot border Panama ?(A) Costa Rica(B) Pacific Ocean(C) Colombia(D) Venezuela

5. The waterfall Victoria is asso-ciated with the river—(A) Amazon (B) Missouri(C) St. Laurence (D) Zambezi

6. Israel has common borderswith—(A) Lebanon, Syria, Jordan and

Egypt(B) Lebanon, Syria, Turkey and

Jordan(C) Cyprus, Turkey, Jordan and

Egypt(D) Turkey, Syria Iraq and

Yemen

7. What is the correct sequence ofthe rivers Godavari, Mahanadi,Narmada and Tapti in the des-cending order of their lengths ?(A) Godavari – Mahanadi –

Narmada – Tapti(B) Godavari – Narmada –

Mahanadi – Tapti(C) Narmada – Godavari – Tapti

– Mahanadi(D) Narmada – Tapti – Godavari

– Mahanadi

8. Which one of the following state-ments is correct ?(A) Liquid Sodium is employed

as a coolant in nuclearreactors

(B) Calcium Carbonate is aningredient of toothpaste

(C) Bordeaux mixture consistsof Sodium Sulphate and lime

(D) Zinc amalgams are used asa dental filling

9. Diffusion of light in the atmos-phere takes place due to—(A) Carbon dioxide(B) Methane(C) Helium(D) Water Vapours and Dust

Particle

10. In which country is the com-mittee which selects winners forNobel Peace Prize located ?(A) Norway (B) Sweden(C) Finland (D) Denmark

11. Consider the following state-ments—1. Caffeine a constituent of tea

and coffee is a diuretic.2. Citric acid is used in soft

drinks.3. Ascorbic acid is essential for

the formation of bones andteeth.

4. Citric acid is a good substi-tution for ascorbic acid in ournutrition

Which of the statements givenabove are correct ?

(A) 1 and 2 only(B) 1, 2 and 3 only(C) 3 and 4 only(D) 1, 2, 3 and 4 only

12. Which one of the followingsubjects is under the Union list inthe Seventh Schedule of theConstitution of India ?(A) Regulation of labour and

safety in mines and oil fields(B) Agriculture(C) Fisheries(D) Public Health

13. Other than India and Chinawhich of the following groups ofcountries border Myanmar ?(A) Bangladesh, Thailand and

Vietnam(B) Cambodia, Laos and

Malaysia(C) Thailand, Vietnam and

Malaysia(D) Thailand, Laos and

Bangladesh

14. Through which one of thefollowing groups of countriesdoes the Equator pass ?(A) Brazil, Zambia and Malaysia(B) Colombia, Kenya and

Indonesia(C) Brazil, Sudan and Malaysia(D) Venezuela, Ethiopia and

Indonesia

15. Which one of the following pairsis not correctly matched ?

Railway Zone Headquarters

(A) NorthEasternRailway

Gorakhpur

(B) SouthEasternRailway

Bhubaneshwar

(C) EasternRailway

Kolkata

(D) South EastCentralRailway

Bilaspur

16. Huangpu River flows throughwhich one of the following cities ?(A) Beijing(B) Ho Chi Minh City(C) Shanghai(D) Manila

17. Consider the following state-ments—1. Dengue is a protozoan

disease transmitted by mos-quitoes

2. Retro-orbital pain is not asymptom of Dengue.

3. Skin rash and bleeding fromnose and gums are some ofsymptoms of Dengue.

C.S.V. / August / 2009 / 787

Which of the statements givenabove is/are correct ?(A) 1 and 2 (B) 3 only(C) 2 only (D) 1 and 3

18. Which one of the following pairsis not correctly matched ?(A) Southern Air Command

—Thiruvananthapuram(B) Eastern Naval Command

—Vishakhapatnam(C) Armoured Corps Centre and

School —Jabalpur(D) Army Medical Corps Centre

and School —Lucknow

19. Itaipu Dam built on the riverParana is one of the largest damsin the world. Which of the follow-ing two countries have this as ajoint project ?(A) Brazil and Peru(B) Paraguay and Equador(C) Brazil and Paraguay(D) Colombia and Paraguay

20. Consider the following state-ments—1. There are 25 High Courts in

India.2. Punjab, Haryana and the

Union Territory of Chandi-garh have a common HighCourt.

3. National Capital Territory ofDelhi has a High Court of itsown.

Which of the statements givenabove is/are correct ?(A) 2 and 3 (B) 1 and 2(C) 1, 2 and 3 (D) 3 only

21. Which one of the following majorIndian cities is most eastwardlocated ?(A) Hyderabad(B) Bhopal(C) Lucknow(D) Bengaluru (Bangalore)

22. Out of the four southern states :Andhra Pradesh, Karnataka,Kerala and Tamil Nadu whichone shares boundaries with themaximum number of Indianstates ?(A) Andhra Pradesh only(B) Karnataka(C) Andhra Pradesh and

Karnataka both(D) Tamil Nadu and Kerala both

23. Which one of the following typesis used by computerised tomo-graphy employed for visualisa-tion of the internal structure ofhuman body ?(A) X-rays(B) Sound waves(C) Magnetic resonance(D) Radioisotopes

24. Production of which one of thefollowing is a function of theliver ?(A) Lipase(B) Urea(C) Mucus(D) Hydrochloric acid

25. Which one of the following is nota digestive enzyme in the humansystem ?(A) Trypsin (B) Gastrin(C) Ptyalin (D) Pepsin

26. Which one of the following isprinted on a commonly usedflourescent tubelight ?(A) 220 K (B) 273 K(C) 6500 K (D) 9000 K

27. What does the 104th ConstitutionAmendment Bill relate to ?(A) Abolition of Legislative

Councils in certain states(B) Introduction of dual citizen-

ship for persons of Indianorigin living outside India

(C) Providing quota to sociallyand educationally backwardclasses in private educationalinstitutions

(D) Providing quota for religiousminorities in the servicesunder the Central Govern-ment

28. From North towards South, whichone of the following is the correctsequence of the given rivers inIndia ?(A) Shyok—Spiti—Zaskar—

Satluj(B) Shyok—Zaskar—Spiti—

Satluj(C) Zaskar—Shyok—Satluj—

Spiti(D) Zaskar—Satluj—Shyok—

Spiti

29. In which state is the Rajiv GandhiNational Institute of YouthDevelopment located ?(A) Tamil Nadu(B) Karnataka(C) Himachal Pradesh(D) Uttarakhand

30. What is the Universal ProductCode (UPC) adopted for ?(A) Fire safety code in buildings(B) Earthquake resistant build-

ing code(C) Bar code(D) Against adulteration in

eatables

C.S.V. / August / 2009 / 788

Directions—(Q. 1–7) In eachquestion below is given a group ofletters followed by four combinationsof digits/symbols lettered (A), (B), (C)and (D). You have to find out which ofthe combinations correctly representsthe group of letters based on thecoding system and mark the letter ofthat combination as your answer. Ifnone of the combinations correctlyrepresents the group of letters, mark(E) i .e. ‘None of these’ as youranswer.Letter :

M A T W R E K I H F U B N PDigit/Symbol Code :

4 @ 3 7 © 2 8 % 1 # $ 6 9 5Conditions :(i) If both the first and the last letters

are consonants, both are to becoded as the code for the firstletter.

(ii) If both the first and the last lettersare vowels, both are to be codedas the code for the last letter.

(iii) If the first letter is a vowel andthe last letter is a consonant, thecodes are to be interchanged.

(iv) If the first letter is a consonantand the last letter is a vowel,both are to be coded as ★.

1. AEBRMH(A) @26©41 (B) 126©4@(C) 126©41 (D) @26©4@(E) None of these

2. HBEAFU(A) 162@#$ (B) ★62@#★

(C) 162@#1 (D) $62@#1(E) None of these

3. BKNIRT(A) 689%©6 (B) 689%©3(C) 698%©3 (D) 389%©3(E) None of these

4. KFUBNA(A) ★#$69★ (B) 8#$69@(C) 8#$698 (D) @#$69@(E) None of these

5. IRFAME(A) 2©#@4% (B) 2©#@42(C) %©#@4% (D) ★©#@4★

(E) None of these

6. MRTPFW(A) 4©35#7 (B) 7©35#4(C) 7©35#7 (D) 4©35#4(E) None of these

7. ENTHWR(A) 29317© (B) 293172(C) ©9317© (D) ★9317★

(E) None of these

Directions—(Q. 8–12) Below ineach question are given two state-ments (A) and (B). These statementsmay be either independent causes ormay be effects of independent causesor a common cause. One of thesestatements may be the effect of theother statement. Read both the state-ments and decide which of the follow-ing answer choice correctly depictsthe relationship between these twostatements. Give answer—

(A) if statement (A) is the causeand statement (B) is itseffect.

(B) if statement (B) is the causeand statement (A) is itseffect.

(C) if both the statements (A)and (B) are independentcauses.

(D) if both the statements (A)and (B) are effects of inde-pendent causes.

(E) if both the statements (A)and (B) are effects of somecommon cause.

8. (A) The prices of vegetableshave increased substantiallyduring past few weeks.

(B) Consumer price index at theend of previous week wasincreased by 2 per cent.

9. (A) Many employees of thecompany proceeded on aday’s leave on Friday.

(B) Both Thursday and Saturdaywere declared holiday bythe company.

10. (A) Many anti-social elementsare caught by the police fromthe locality.

(B) Many people in the localityare detained by the police forquestioning.

11. (A) Many pilgrims used Govt.transport to travel to the holyshrine.

(B) The cost of travel by privatetransport is very high.

12. (A) There has been heavy rainsin the catchment area of thelakes supplying drinkingwater to the city.

(B) The municipal authority hassuspended the proposed cutin water supply to the city.

13. Among P, Q, R, S and T, eachhaving different weight, R isheavier than only P. S is lighterthan Q and heavier than T. Whoamong them is the heaviest ?(A) Q(B) P(C) S(D) Data inadequate(E) None of these

14. What should come next in thefollowing number series ?

8 7 6 5 4 3 2 1 7 6 5 4 3 2 1 6 5 4 3 2 1

(A) 6 (B) 4(C) 7 (D) 5(E) None of these

15. In a certain code language, ‘comenow’ is written as ‘ha na’; ‘nowand then’ is written as ‘pa da na’and ‘go then’ is written as ‘sa pa’.How is ‘and’ written in that codelanguage ?(A) sa(B) pa(C) na(D) Cannot be determined(E) None of these

16. Each consonant in the wordDISTEMPER is replaced by thenext letter in the English alphabetand each vowel in the word isreplaced by the previous letter inthe English alphabet, which ofthe following will be the fourthletter from the right end after thereplacement ?

(A) T (B) M(C) S (D) P(E) None of these

C.S.V. / August / 2009 / 789

17. How many meaningful Englishwords can be made with theletters NDOE using each letteronly once in each word ?(A) None (B) One(C) Two (D) Three

(E) More than three

18. Four of the following five are alikein a certain way and so form agroup. Which is the one thatdoes not belong to that group ?(A) BD (B) NQ(C) RP (D) MK(E) FH

19. In a certain code GEAR is writtenas ‘5934’ and RIPE is written as‘4869’. How is PAGE written inthat code ?(A) 6359 (B) 6539(C) 4359 (D) 6459(E) None of these

20. How many such digits are therein the number 64382179 each ofwhich is as far away from thebeginning of the number as whenthe digits are arranged in ascend-ing order within the number ?(A) None (B) One(C) Two (D) Three(E) More than three

21. Four of the following five are alikein a certain way and so form agroup. Which is the one thatdoes not belong to that group ?(A) 39 (B) 91(C) 78 (D) 52(E) 45

22. How many such pairs of lettersare there in the word ORDINALeach of which has as many lettersbetween them in the word as inthe English alphabet ?(A) None (B) One(C) Two (D) Three(E) More than three

Directions—(Q. 23 and 24) Readthe following information carefully andanswer the question which follow—

(i) ‘P × Q’ means ‘P is brotherof Q’.

(ii) ‘P ÷ Q’ means ‘P is sister ofQ’.

(iii) ‘P – Q’ means ‘P is motherof Q’.

(iv) ‘P + Q’ means ‘P is father ofQ’

23. Which of the following means ‘Dis maternal uncle of K’ ?(A) D ÷ N – K (B) D ÷ N + K(C) D × N – K (D) D × N + K(E) None of these

24. Which of the following means ‘Mis nephew of R’ ?(A) M × T + R

(B) R ÷ J + M × T(C) R ÷ J + M(D) R × J + M(E) None of these

Directions—(Q. 25–27) Followingquestions are based on the five three-digit numbers given below—

519 378 436 624 893

25. If the positions of the first and thethird digits within each numberare interchanged, which of thefollowing will be the secondsmallest number ?(A) 519 (B) 378(C) 436 (D) 624(E) 893

26. If ‘1’ is subtracted from the firstdigit in each number and ‘1’ isadded to the second digit in eachnumber, which of the followingwill be the third digit of thesecond highest number ?(A) 9 (B) 8(C) 6 (D) 4(E) 3

27. If the positions of the first and thesecond digits within each numberare interchanged, which of thefollowing will be the highestnumber ?(A) 519 (B) 378(C) 436 (D) 624(E) 893

Directions—(Q. 28–32) Studythe following information carefully andanswer the questions given below :

M, D, P, K, R, T and W are sittingaround a circle facing at the center. Dis second to the right of P who is thirdto the right of K. T is third to the rightof W who is not an immediateneighbour of D. M is third to the left ofR.28. Who is third to the left of D ?

(A) W(B) P(C) K(D) Data inadequate(E) None of these

29. Who is to the immediate left ofM ?(A) K (B) W(C) D (D) T(E) None of these

30. In which of the following pairs isthe second person sitting to theimmediate right of the firstperson ?(A) DT (B) TP(C) PR (D) KW(E) None of these

31. Who is second to the right of T ?(A) D(B) K(C) M(D) Data inadequate(E) None of these

32. Who is to the immediate left ofR?(A) W (B) P(C) K (D) T(E) None of these

Directions—(Q. 33–35) Studythe following arrangement carefullyand answer the questions givenbelow —

R % E 5 D 2 # 9 A F B @ J 3 I KM 4 1 W U 8 V © N ★ G Z δ 6 733. How many such symbols are

there in the above arrangement,each of which is immediatelyfollowed by a consonant andimmediately preceded by anumber ?(A) None (B) One(C) Two (D) Three(E) More than three

34. Which of the following is the fifthto the left of the sixteenth fromthe left end of the abovearrangement ?(A) B (B) U(C) W (D) N(E) None of these

35. Which of the following is theseventh to the right of the four-teenth from the right end ?(A) B (B) #(C) N (D) U(E) None of these