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All that you are to do is to send us the list of multiplechoice objective questions which were asked in that examinationon the basis of your memory. The questions should be completewith all the alternatives for answering them. We shall includeall the entries in the contest. Participants with maximum numberof correct questions will be awarded.

PRIZES

(i) First three participants sending maximum number of correctquestions will receive Rs. 200, 100 and 60 as prize.

(ii) No prize is admissible for less than 75% questions.

C.S.V. / October / 2009 / 930

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TTHOUGHTS FOR THE MONTHTHOUGHTS FOR THE MONTH

❥ Any machine is as good or reliable as its weakest component be itso humble a part as a nut or bolt.

❥ All happy families seem alike but every unhappy family is sad in itsown way.

❥ Tomorrow is two days late for yesterday’s dead line.

❥ ‘Success’ comes before ‘work’ only in dictionary.

❥ Good is not good where better is expected.

❥ To be vanquished and yet not surrender, that is victory.

❥ The humblest citizen of all land, when clad in the armour of righteouscause, is stronger than all the hosts of error.

❥ They never die who live for others.

❥ A highbrow is a person educated beyond his intelligence.

❥ Handsome is that handsome does.

❥ He that knows least commonly assumes most.

❥ Genius is often perseverance in disguise.

❥ Our greatest evils flow from ourselves.

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

❥ Prosperity is only an instrument to be used, not a deity to beworshipped.

❥ Where true principles lack, the results are imperfect.

In This IssueIn This IssueIn This Issue

EditorMAHENDRA JAIN

Editor/Publisher is not responsible forviews, data, figures etc. expressed in thearticles by the authors.

—Editor

No part of this publication can bereproduced or transmitted in any formwithout the prior written permission fromthe publishers.

Edited, printed and published by MahendraJain for M/s. Pratiyogita Darpan, 2/11A,Swadeshi Bima Nagar, AGRA–2 andprinted by him at Pratiyogita DarpanPrinting Unit, 5 & 6, Bye pass Road, Agra.Phone : 4053333, 2531101, 2530966Fax : (0562) 4031570, 4053330E-mail : publisher@pdgroup.inWebsite : www.pdgroup.in

Branch Office :

4845, Ansari Road,Daryaganj, New Delhi–110 002Phone : 23251844/66

October 2009

Year—12 Issue—140

C.S.V. / October/ 2009 / 931 / 1A

Regulars

Editorial 933

Science and Technology 934

Latest General Knowledge 937Inspiring Young Talents—

(i) Topper : Uttarakhand, PMT 2009 (3rd Position)

—Pavitra Saxena 941(ii) Topper : U.P. CPMT 2009 (Rank-13)—Satyendra Singh 943

Science Tips 945

Physics

Sound-II : Interference and Beats 947

Nuclear Physics-II : Structure of Nucleus 953Typical Model Paper 957

Typical Model Paper 962

Chemistry

Alcohols 967

Typical Model Paper 978Typical Model Paper 984

Zoology

Communicable Diseases and Control of Microbes 989Anatomy of Liver and its Disorders 995Human Population 999Typical Model Paper 1002Typical Model Paper 1004Typical Model Paper 1007

Botany

Funaria : A Terrestrial Moss 1010Reproduction and Development in Angiosperms 1014Typical Model Paper 1024Typical Model Paper 1026Typical Model Paper 1028Typical Model Paper 1031

Other Other Other Other Other FeaturesFeaturesFeaturesFeaturesFeatures

Assertion and Reason Type Questions 1035True or False 1038Do You Know ? 1041Correct Solution and Prize Winners of CSV Quiz No. 134 1045CSV Quiz Contest No. 137 1046Mental Ability Test 1049General Awareness 1053

To Our ReadersDear Readers,

It gives us great pleasure and satisfaction in presenting to you the OctoberIssue of your favourite and frontline magazine ‘Competition Science Vision’. Ourreaders generally agree that each issue of the magazine is unique in itself. Thepresent issue is improved in extent and quality of subject matter to make it moreuseful and examination-oriented. The ever changing pattern of PMT examinationshas found our special attention.

CSV gives all that you want to be successful in any such competition. The vastmajority of our readers claim it second to none. It fully covers all medical entrancetests held throughout the country at present.

It is hardwork under proper guidance that makes one successful in these tests.In giving you the best subject matter and guidance CSV has no parallel. Hence, weadvise you to make CSV as the base of your preparations.

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 and a bright future.Sincerely yours,

Mahendra Jain (Editor)

C.S.V. / October / 2009 / 932

2009Chhattisgarh State Service (Mains) Exami-nation, 2008 (Sept. 12 & 13)SBI Management Executives Exam. (Sept. 13)U.P.S.C. Combined Defence Services Exami-nation (II), 2009 (Sept. 13)U.P. Trained Graduate Teachers Examination (Sept. 13 & 20)U.P. Lecturer Examination (Sept. 14)PNB Clerical Cadre Exam. (Sept. 20)Delhi SSSB Trained Graduate Teachers (English,Mathematics, Natural Science) Exam. (Sept. 27)Bihar Telecom Technical Assistant Exam., 2008 (Oct. 4)Delhi SSSB Trained Graduate Teacher (Social Science)Exam. (Oct. 10)UPSC CPF Assistant Commandants Exam., 2009 (Oct. 11)Reserve Bank of India Officers Grade ‘B’ Examination(I Stage) (Oct. 11)Delhi SSSB Trained Graduate Teachers (Sanskrit,Hindi, Urdu, Punjabi) Exam. (Oct. 11)The New India Assurance Company Ltd. AdministrativeOfficer Examination (Oct. 25)Gurgaon Gramin Bank Officer (Scale-I) CadreExamination (Oct. 25)Shreyas Gramin Bank Specialist Officers Exam. (Nov. 1)Uttaranchal Gramin Bank Clerk-cum-Cashiers Exam. (Nov. 1)Gurgaon Gramin Bank Clerk-cum-Cashier CadreExamination (Nov. 1)

Shreyas Gramin Bank Clerk-cum-Cashiers Exam. (Nov. 8)Madhya Pradesh Civil Judge (Main) Examination (Nov. 8)National Talent Search Exam., 2009 (For VIII ClassStudying Students) (Nov. 8)Madhya Pradesh National Means-cum-Merit ScholarshipExamination, 2009-10 (Nov. 8)S.B.I. Clerical Cadre Exam., 2009 (Nov. 8 & 15)(Online Closing Date : 15 Sept., 2009)S.S.C. Auditors and Accountants Exam. (Indian Auditand Account Deptt.) (Nov. 15)Madhya Pradesh State Service (Mains) Exam., 2008 (Nov. 16)Indian Economic Service/Indian Statistical ServiceExamination, 2009 (Nov. 21)Corporation Bank Probationary Officers Exam. (Nov. 22)(Online Closing Date : 18 Sept., 2009)Rajasthan State Eligibility Test (SET), 2009 (Nov. 22)Rajasthan Gramin Bank Officer (Scale-I) CadreExamination (Nov. 22)Rajasthan Gramin Bank Clerk-cum-Cashier CadreExamination (Nov. 29)Corporation Bank Clerical Cadre Exam. (Nov. 29)(Online Closing Date : 18 Sept., 2009)Indian Air Force Airman Selection Test [Group ‘X’(Technical) Trades] (Nov.)Rashtriya Military Schools Common Entrance Test(Class VI) (Dec. 20)(Closing Date : 10 Sept., 2009)CSIR-UGC National Eligibility Test, 2009 (Dec. 20)(Closing Date : 23 Sept., 2009)

FORTHCOMING COMPETITIVE EXAMS.

C.S.V. / October / 2009 / 935 / 2

also revealed, for the first time, thechanges in rock and mineral com-position.’’

Researchers say the presence ofmagnesium, calcium, iron and siliconin lunar rocks could help to confirmwhether the Moon was covered withmolten magma ocean early in itshistory.

Brain Surgery withUltrasound

Neurosurgeons might soon beable to say goodbye to the scalpel. Anew technique uses ultrasound wavesto remove parts of the brain. High-intensity ultrasound—a different typethan what is used in prenatal screen-ing—heats up parts of the brain,thereby killing sections of tissues thatare damaged.

Focussed ultrasound surgery hasnow been performed successfully onnine human patients, according to apreliminary study done. The groundbreaking finding here is that you canmake lesions deep in the brain—through the intact skull and skin—withextreme precision, accuracy andsafety.

The traditional treatment involvesdestroying a small part of thethalamus, a structure that relays mes-sages between different brain areas.In the past, this has been accom-plished with radio frequency ablation,in which a probe is inserted into theskull or with radio surgery whichfocuses radiation on the area.Surgeons believe that the new tech-nique will be faster—acting and moreprecise than the current methods.

In this procedure ultrasoundbeams are focussed on a specificpoint in the brain; the exact locationdepends on the condition being

treated. The small portion of braintissue at the focus (about the size ofthe rice grain) absorbs the energy andconverts it to heat; the temperature inthis area rises to about 130 degreeFahrenheit, killing the cells. The entiresystem is integrated with a magneticresonance scanner, which allowsneurosurgeons to make sure theytarget the correct piece of brain tissue.

Ultrasound could also potentiallybe used to treat other brain disorders,such as parkinson disease.

India’s Third Base inAntarctica

India’s first permanent researchstation, Dakshin Gangotri, in SouthPolar region set up in 1984, wasburied in ice and had to be abandonedin 1990, a year after India set upMaitri, the second polar station.Twenty-five years after India esta-blished Dakshin Gangotri in SouthPolar region, it is all set to build thethird such centre in Antartica to takecutting-edge research in variousfields. The new station, tentativelynamed Bharti, is scheduled to beoperational in 2012, making India amember of an elite group of ninenations that have multiple stations inthe region.

Maitri station in Antarctica

The National Centre for Antarcticand Ocean Research (NCAOR), Goa,will set up the new station onLarsmann Hill, 3000 km fromSchirmacher Oasis, where Maitristands. While Maitri is more than 100km from Antaractic Sea, Bharti willbe on promontory by the sea.

Dr. R. Ravindran, Director ofNCAOR, said, ‘‘Bharti will enable us totake up rare research on marineecology of polar region. Antarctica isa continent spread across 13 millionsq km and we should not confineourselves in just one area.’’ Bharti,like Maitri, will also conduct researchon seismic activities, climate changeand medicine.

The station will be a compactstructure of 30 × 50 metre, accom-modating 25 scientists. While living inAntarctica, where temperatures rang-ing from – 89 degree celsius in winterto – 25 degree celsius in summer, canbe tough and constructing a perma-nent structure is a huge challenge.The station will use wind power andsolar power as sources of energy andleave minimum carbon footprint, whileensuring optimum heating and otherfacilities for scientists.

Experiments in extreme coldclimates, as in the polar region, havecontributed immensely to scientificdevelopments.

India was admitted to the Scienti-fic Committee on Antarctic Research(SCAR), an international body thatcoordinates scientific activities in theregion. On October 1, 1984 India heldthe Vice-Chairman’s post in theCommittee. Russia, U.S.A., U.K.,France, China, Chile, France,Australia and Argentina have multiplestations in Antarctica

‘Copernicium (CP)’—ANew Element in Periodic

Table

Discovered 13 years ago, andofficially added to the periodic tablerecently, element 112 finally has aname. It will be called ‘Copernicium’,with the symbol ‘CP’, in honour of theastronomer Nicolaus Copernicus. Theteam of scientists who discovered theelement chose the name to honourthe man who changed our view. TheInternational Union of Pure andApplied Chemistry (IUPAC) willofficially endorse the new element’sname.

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C.S.V. / October / 2009 / 937

AWARDS/HONOURS

Miss Universe 2009

Eighteen-year-old Stefania Fer-nandez of Venezuela was crownedMiss Universe 2009 on August 24,2009 in Los Angeles by incumbentcrown holder compatriot DayanaMendoza. She is the sixth MissUniverse from a South Americancountry, known for its obsession withbeauty.

Stefania Fernandez

The first runner-up title is sharedby Miss Australia Rachael Finch andMiss Puerto Rico Mayra Matos Perezwith the title of second runner-up goingto Miss Kosovo Gona Dragusha.

Magsaysay Awards 2009

Names of the winners of Mag-saysay Awards 2009 were announcedin Manila on August 3, 2009. India’ssocial activist, Deep Joshi, who hasdone pioneering work for develop-ment of rural communities, wasnamed for Ramon Magsaysay Award.

Ms. Ka Hsaw Wa of Myanmar,cofounder of Earth Rights Inter-national, was chosen for the Awardfor dauntlessly pursuin non-violent yeteffective channels for redress, expo-sure and education for the defence ofhuman rights, environment and demo-cracy.

Other winners are—Yu Xiaogangand Ma Jun of China, Antonio OposaJr. of Philippines and Ms. KrisanaKraisintu of Thailand.

Gallantry Awards—On theoccasion of Independence Day, Presi-dent Pratibha Patil approved 148gallantry medals, including 2 AshokChakra, 4 Kirti Chakra, 26 ShauryaChakra, 2 Bar to Sena Medals, 100Sena Medals, 4 Nao Sena Medals, 5Vayu Sena Medals and 3 TatrakshakMedals. Ashok Chakra is the country’shighest peace time gallantry award.Major Mohit Sharma (posthumously)from 1 Para Special Force and MajorD. Shree Ram Kumar from 30 AssamRifles were awarded Ashok Chakra.

Kirti Chakra Awardees are—Major Amit Oskar Fernandes of 7Maratha Light Infantry.

Major Deepak Tiwari of Electro-nics and Mechanical EngineeringCorps.

Shabir Ahmed Malik (posthu-mously) of para Brigade. NaikRishikesh Gujjar of 10 Rashtriya Rifles(Rajput).

Nooyi, Sonia, Kochhar andMazumdar Shaw in Forbes List—PepsiCo Chief Executive Indira Nooyi,Congress President Sonia Gandhi,ICICI Bank Chief Chanda Kochharand Kiran Majumdar Shaw, Chair-person of Biocon have been listedamong the world’s 100 most powerfulwomen by Forbes magazine. Nooyihas been ranked third, Sonia Gandhi13th, Kochhar 20th and Mazumdar91th.

Reliance names Yusuf Pathanand Jadeja as Brand Ambas-sadors—Two prominent cricketersYusuf Pathan and Ravindra Jadejaare picked up by the RelianceIndustries Ltd. (RIL) as its BrandAmbassadors on a long term contract.Their main job will be to identify bud-ding sportspersons, particularly in thefield of cricket, which may ultimatelylead to the RIL forming its own team.

BOOKS

Jinnah : India Partition Inde-pendence—Jaswant Singh (Theauthor is the former Union Minister

and a senior leader of BJP. Now he isexpelled from BJP. In this book,Jaswant Singh has highly appreciatedMohammad Ali Jinnah and bitterlycriticised great Indian leaders likeSardar Vallabhbhai Patel and others.The book is controversial).

Midnight Diaspora—Ed. DanielHerwitz and Ashutosh Varshney(Various reading of Pakistan as acountry in the light of its portrayal inRushdie’s novels).

Why People Protest—SubhashSharma (In this book, the authoranalyses the theory and practice ofecological movements in general. Itpresents a study of six such move-ments in the third world from anecological and socio-economic standpoint).

DAYS

September 2—Coconut DaySeptember 5—Teachers’ Day,

Sanskrit DaySeptember 8—International

Literacy Day (UNESCO)September 14—Hindi DivasSeptember 15—Engineers’ DaySeptember 16—World Ozone

DaySeptember 21—Alzheimer’s Day,

Day for Peace and Non-violence(U.N.O.)

September 22—Rose Day (Wel-fare of Cancer Patients)

September 26—Day of the DeafSeptember 27—World Tourism

Day

Gayatri Devi—Former QueenMother of the erstwhile Jaipur Stateand once ranked among the ten mostbeautiful women in the world. Maha-rani Gayatri Devi, passed away inJaipur after brief illness. She was 90and is survived by her grand childrenand step sons, who include theformer Maharaja of Jaipur, BhawaniSingh.

C.S.V. / October / 2009 / 938

She was the former princess ofCooch Bihar in West Bengal. Shestudied in Shantiniketan. She was thethird wife of the former ruler of Jaipur,Man Singh-II. Her husband passedaway in 1970. Among the first royalsto join the democratic process ofelections, Ms. Gayatri Devi won theLok Sabha elections in 1962, 1967and 1971 contesting on SwatantraParty ticket. During emergency, sheunderwent a jail term in Tihar.

She was given a State honourand her step son Bhawani Singh litthe pyre. Just after departure ofRajmata, royal battle for property haserupted.

Leela Naidu—Hindi actressLeela Naidu passed away in Mumbaiafter prolonged illness. She was 69.She made her debut in 1960 withBalraj Sahni in Hrishikesh Mukherjee’s‘Anuradha’. However, she achievedfame only after starring with in a non-orthodox role with Sunil Dutt in ‘‘YehRaste Hain Pyar Ke’’ (1963). Shemarried Bikki Oberoi, owner of Oberoichain of hotels with whom she had twindaughters. She divorced Oberoi andmarried a poet and journalist and gotsettled in Hong Kong.

Devendra Nath Dwivedi—Constitutional expert Devendra NathDwivedi, who was recently appointedGovernor of Gujarat, passed away inNew Delhi due to liver problem, evenbefore he could take up his assign-ment. He was 74.

Subhash Chakraborty—Seniorleader of Communist Party of India(Marxist) and West Bengal’s Ministerfor Transport, Sports and YouthServices, Subhash Chakraborty

passed away on August 3, 2009 inKolkata. He was very popular in WestBengal.

Beithullah Mehsud—Pakistan’smost wanted man, Beithullah Mehsud,was killed in a United States missilestrike on August 7, 2009 in SouthWaziristan tribal area. The 35-year oldMehsud was the face of the PakistaniTaliban and he was involved inseveral dreadful terrorist attacks.Earlier this year, Govt. of Pakistanannounced a 50 million bounty forhim. Before that, the U.S. Govt. hadannounced a $ 5 million head moneyfor him.

Nandita Das (New Chief, CFSI)—Film actress Nandita Das took overas the Chairperson of Children’s FilmSociety of India (CFSI). She will havea tenure of three years. The actress,who created waves with her roles infilms like ‘Fire’ and ‘Earth’ has baggedseveral national and internationalawards.

Her predecessor, Nafisa Ali, hadresigned.

Jagannath Pahadia (NewGovernor, Haryana)—Former ChiefMinister of Rajasthan, JagannathPahadia, took over as the newGovernor of Haryana. He succeededA. R. Kidwai. Mr. Pahadia (77) wasborn in Bhusawar village in Bharatpurdistrict of Rajasthan. He served UnionMinistry as Deputy Minister for severalyears and Chief Minister of Rajasthansince June 6, 1980 till July 14, 1981.He had also been the Governor ofBihar.

Margaret Alva (New Governor,Uttarakhand)—Senior Congressleader, Margaret Alva took over asthe Governor of Uttarakhand. She isthe first woman Governor of this hillstate.

Justice J. Bhalla (New CJ,Rajasthan)—Justice Jagdish Bhallawas sworn in on August 8, 2009 asthe Chief Justice of Rajasthan HighCourt. He has been transferred toRajasthan from Himachal Pradesh.Born on November 1, 1948, JusticeBhalla started legal practice inAllahabad High Court in 1971. Hewas elevated as Judge of AllahabadHigh Court in 1995 and Chief Justiceof Himachal Pradesh High Court inFebruary 2008.

Ahmad Tariq Karim (New HighCommissioner of Bangladesh)—Aformer career diplomat, Mr. AhmadTariq Karim, presented his credentialsto the President, Pratibha Patil, as thenew High Commissioner of Bangla-desh to India. During 1980s, Mr.Karim had served as the Deputy HighCommissioner in New Delhi. Mr.Karim will head Bangladesh missionhere at a time when both the countriesare attempting to forge a newrelationship based on consultationand trust.

Jaswant Singh—The BhartiyaJanata Party on August 19, 2009expelled Jaswant Singh its veteranleader and the M.P. A meeting of theParliamentary Board in Shimlaunanimously decided to remove the

Jaswant Singh

former Union Minister from the primarymembership. The decision is beingread in the party as a stern messageof zero tolerance to ideological devia-tion and indiscipline. In the meeting,the tempers were high among thesenior leaders. They viewed his praiseof Pakistan’s founder Mohammad AliJinnah, and his adverse commentson India’s first Deputy Prime Minister,Sardar Vallabhbhai Patel, as ideologi-cal heresy. Mr. Singh did his commentin his recently launched book, Jinnah :India-Pakistan-Independence (Rupa &Co. New Delhi, 2009, 669 pages).

Mr. Jaswant Singh’s Jinnah is animpressive, personally attractive,intellectually brilliant freedom-loving,politically iron-willed, tactically unstop-pable figure.

Deep Joshi (Magsaysay AwardWinner 2009)—Joshi (60) has beenhonoured with Ramon MagsaysayAward 2009 for his extraordinaryachievement in seeking to transformthe lives of 68,000 families in several

C.S.V. / October / 2009 / 939

backward states, mainly the naxal-affected belt Jharkhand and Bankuraand Purlia in West Bengal. He isrecognised for bringing professiona-lism to the NGD movement in India.

Making an Impact : Joshi

Mr. Joshi took his B. Tech. degreefrom NIT Allahabad and M. Tech. andMBA degrees from MassachusettsInstitute of Technology, U.S.A. Healso worked as a Ford FoundationProgramme Officer.

In 1983, he co-founded the non-profit organization : ‘‘ProfessionalAssistance for Development Action(PRADAN)’’ that recruits collegegraduates to do community works. In2006 Joshi had received HarmonySilver Award for his contributions tosociety.

SPORTS

CricketIndian Team for Champions

Trophy and Tri-Series—Rahul Dravidis a prominent member of the India’sODI squad for the tri-series in SriLanka and the ICC ChampionshipTrophy. The National Selection Panelwhich met in Chennai on August 16,2009, picked the same 15 for both thetournaments. Dravid’s return will addthe quality to the Indian top-order. Thekey batsman Virender Sehwag,recovering from a shoulder surgery,has not been considered by the selec-tion panel headed by KrishnamachariSrikanth.

In the pace bowling section,Praveen Kumar, with his swingchange of pace and yorkers has beenpreferred over Munaf Patel. Praveen’sability to swing the new ball andreverse the old could have clinchedthe argument in his favour.

The tri-series in Sri Lanka, sinceSeptember 8 till 14, involves the host,India and New Zealand. The Cham-pions Trophy will be held in SouthAfrica since September 22 till Oct. 5.

The squads (for Sri Lankan tri-series and the Champions Trophy)—M.S. Dhoni (Captain), Yuvraj Singh(Vice Captain), Sachin Tendulkar,Gautam Gambhir, Rahul Dravid,Suresh Raina, Yusuf Pathan,Harbhajan Singh, Aashish Nehra, R.P.Singh, Ishant Sharma, Dinesh Karthik,Praveen Kumar, Amit Mishra andAbhishek Nayar.

Sri Lanka-Pakistan ODI Series—Sri Lanka clinched the ODI series3-2 against Pakistan. On August 9,2009 in Colombo, Pakistan beat SriLanka by 132 runs in the fifth and thefinal match. Sri Lanka won first threematches of the series while Pakistanclinched the last two matches. ThilanTushara of Sri Lanka was adjustedthe Man of the Series.

Sri Lanka had already clinchedthe Test Series 2-0 against Pakistan.

Wrestling

World Wrestling Champion-ship : Indian Squad—Seventeenwrestlers have been selected torepresent India in the World WrestlingChampionship to be held in Herning(Denmark) since September 20 till 27.Rajiv Gandhi Khel Ratna awardeeand Olympic bronze medalist, SushilKumar will lead the India squad.

The squad : Men (Freestyle)—Balaraj Singh (55 kg), Hardeep Singh(60 kg), Sushil Kumar (66 kg),Ramesh Kumar (74 kg), Anil Mann(96 kg) and Joginder Kumar (120 kg).

Greco-Roman—Rajender Kumar(55 kg), Ravinder Singh (60 kg), SunilKumar (66 kg), Anil Kumar (96 kg)and Dharmender Dalal (120 kg).

Women—Nirmal Devi (48 kg),Babita (51 kg), Alka Tomar (59 kg),Suman Kundu (63 kg), Geetika Jakhar(67 kg) and Gursharanpreet Kaur (72kg).

Chess

Asian Youth Meet—Asian Youthconcluded on August 8, 2009 in NewDelhi. Living up to the expectationsIndians came up with a strong finishin the most sections to reassert theirsupremacy in the Asian Youth ChessChampionship at the Tivoli GardenResort in New Delhi. India won 23 outof 36 medals at stake and its haulincluded 8 gold, 7 silver and 8 bronzemedals. This was an improvement

over last year’s performance of 5gold, 5 silver and 8 bronze, out of 30medals.

Medalists—1. Guliskhan (Kaza-khstan), 2. Krithika, 3. Bhakti.

Medals tally (read as country,gold, silver, bronze, total).

India 8-7-23; Vietnam 2-1-1-4;Iran 1-1-2-4, UAE 1-0-0-1; Kaza-khstan 1-0-0-1; Uzbekistan 0-2-1-3.

Asian Zonal Chess Champion-ship—International Master ShriramJha emerged as the men’s championwith 10 points out of a possible 12 inthe Asian Zonal Chess Championshipon August 17, 2009 at the AirportsAuthority of India Club in New Delhi.He beat Asharaf Ahmed of Maldivesin the 13th and final round.

Drought in 177 Districts—South-West monsoon has proved tobe unpredictable, variable and uncer-tain this year—with the officialannouncement that 177 districts sufferfrom either drought or drought-likeconditions indicating the magnitude ofthe crisis.

Little or no rain, late rain andheavy rain have all been features ofmonsoon behaviour so far in differentparts of India. Officially, the monsoonends on September 30, 2009 and it ispossible that September will witnessheavy rain at some places, leading tofloods and damage to crops. Foragriculture, what matters is not totalrainfall, but its distribution.

It is crucial for the Union andState governments to formulate planson the understanding that drought-like crises hit the poor, especially agri-cultural labourers and land-poorpeasants, the socially underprivilegedsections and women the hardest.Women are badly affected, becausethey do not have equal access to non-farm employment opportunities andare forced to take up jobs involvinghigh drudgery but low wages.

The first priority for the ‘NationalCrisis Management Committee’ ,chaired by Finance Minister PranabMukherjee, will be to ensure that thedrought relief and rehabilitation pro-grammes are pro-poor, pro-sociallyunderprivileged and pro-women.

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C.S.V. / October / 2009 / 941

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

‘‘Hardwork, a strong will to succeed, encouragement of my parents andteachers and above all blessings of the Almighty are the main elements

of my success.’’—Pavitra Saxena

Topper—Uttarakhand, PMT 2009 (3rd Position)

[‘Competition Science Vision’ arranged an exclusive interview with Miss Pavitra Saxena who has thecredit of securing a high position on the list of successful candidates of Uttarakhand PMT, 2009. In addition,she has also cleared other pre-medical tests with equivalent ranks viz., AFMC, CBSE 09, AIPVT and UPTU.For her brilliant success she deserves all praise and our heartiest congratulations. This important interviewis presented here in its original form.]

CSV—Congratulations on yourbrilliant success.

Pavitra—I am very thankful toyou.

CSV—Before knowing your resultwhat did you think about those whoachieve top positions ?

Pavitra—I thought that they wereapart from the general mass like thatof us.

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

Pavitra—Actually I did my paperswell, but I never thought that I wouldattain such a meritable position insuch a renowned examination.

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

Pavitra—First of all I would liketo thank the Almighty, my parents, myteachers for my success. Actually it isthe hardwork and the will of a personwhich can make him lead the world.The same is with me.

CSV—In how many attempts didyou get this success ?

Pavitra—Frankly speaking in twoyears I was able to grasp the exam.

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

Pavitra—Earlier I never botheredabout the theory of Chemistry andPhysics. But now with the help of myteachers and with my more efforts,paying more attention on classlectures I was able to cope up withmy shortcomings.

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

— It is a very helpful maga-zine containing conceptual andimportant questions related tovarious pre-medical tests. Thismagazine has been highly usefulto me in grasping these examina-tions so easily.

—Pavitra Saxena

Pavitra—From my father who isChief Pharmacist in S. N. MedicalCollege, Agra.

CSV—From when did you startthe preparation for it ?

Pavitra—After 12th examination.

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

Pavitra—I actually planned verylittle for that. My schedule was not somuch arranged whenever I thought tostudy a subject I picked it up.

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

Pavitra—Frankly, I didn’t plantime individually for these subjectsbut roughly around 2-21/2 hours. Iused to give separately for Physics,Chemistry, Zoology and Botany.

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

Pavitra—In my views havingcommand in Biology is like coveringhalf of the exam paper which canmake a path of success. Secondly,commanding Chemistry and Physicsis the most important.

Bio-Data

Name—Pavitra SaxenaFather’s Name—Mr. Pravin Kumar

SaxenaMother’s Name—Mrs. Neeta

SaxenaEducational Qualifications—H.S./Std. X—86% (St. Francis Sec.

School, Agra), 2005.Inter/Std. XII—84% (St. Patrick’s

Junior College, Agra), 2007.Special Ahievements—

● Stood 3rd in UPMT-09● Obtained position in merit list of

AFMC-09● Selected in CBSE (Mains)-09● Selected in AIPVT (120 rank)● 17th position in UPTU

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

Pavitra—Initially I made thoroughstudy of each and every chapter ofeach subject but finally around thetime of exam. I made selective topicsto study.

CSV—How did you give finaltouches to your preparation ?

Pavitra—By solving those ques-tions which were of utmost impor-tance and were conceptual and byreading the underlined parts of mynotes.

CSV—Did you prepare notes ?Pavitra—Yes, actually notes

should be like that in which each andevery important thing can be summedup and these notes can help to give afinal touch to our preparation.

C.S.V. / October / 2009 / 942

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

Pavitra—I used to solve thequestions myself whether I was ableor not then only used to see itsanswer and solution and then used torepeat the way of solving the samequestion. They are of utmost impor-tance counting a lot in exam.

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

Pavitra—About two years time issufficient for the preparation of theseexaminations.

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

Pavitra—From 11th a studentshould start preparing for it and shouldpay attention both to his medicaltarget and his school studies along-with his/her recreation time.

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

Pavitra—First of all I preferMBBS. Secondly, it is BDS andnothing other than these.

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

Pavitra—Science magazinesgive a child new and foremost infor-mations which are very helpful incompeting these competitive exami-nations.

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

Pavitra—A magazine containingnew and important informations whichcould help in competing these examswould be most preferred.

CSV—What is your opinion aboutour ‘Competition Science Vision’ ?How much helpful and useful do youfind it ?

Pavitra— It is a very helpfulmagazine which a student can optcontaining the conceptual andimportant questions related tocompetitive examinations. I amvery thankful to this because thismagazine helped me to grasp theexam easily.

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

Pavitra—Although it contains anumber of useful facts but more andmore new informations could beadded to it which can make a child tolead the world.

Personal Qualities

Hobbies—Seeing T.V., hearingmusic, playing

Strong Point—My family

Weak Point—My attitude towardstheoretical studies

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

Pavitra—In UPMT-09, I secured3rd rank and obtained 163 marks with42 in Physics, 40-40 in Botany andZoology, 41 Chemistry. Also came inmerit list of AFMC, selected in CBSE(mains), UPTU and AIPVT.

(Continued on Page 944 )

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Awareness & Marketing❖ Descriptive English — Letter Writing — Precis Writing — Essays

C.S.V. / October / 2009 / 943 / 3

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

‘‘Regular daily study, confidence in my study, hardwork and faith inGod’s grace are the secrets of my success.’’

—Satyendra SinghTopper—U.P. CPMT 2009 (Rank-13)

[‘Competition Science Vision’ arranged an exclusive interview with Mr. Satyendra Singh who has thecredit of being successful in U.P. CPMT–2009 with high rank. For his brilliant success he deserves all praiseand our heartiest congratulations. This important interview is presented here in its original form.]

CSV—Congratulations on yourbrilliant success.

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

what did you think about those whoachieve top positions ?

Satyendra—I thought that thoseare extra brilliant persons, but now Irealised regularity and hardwork isthe key of success.

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

Satyendra—I was confident ofgood rank but not such a top rank.

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

Satyendra—As above regularityand hardwork is the key of success.

CSV—In how many attempts didyou get this success ?

Satyendra—I had tried for fouryears.

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

Satyendra—In earlier attempts, Icould not be able to revise the wholesyllabus in last month, but this time Ihad started revision before 2 months.

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

Satyendra—My father is a doctorand many poor people come to himfor treatment. To see the happinesson the faces of these patients, I chosea medical career.

CSV—From when did you startthe preparation for it ?

Satyendra—I had started pre-parations since 2005.

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

Satyendra—Physics was myweak point at the time of preparation.So firstly, I decided to do hardwork inPhysics. I solved MCQs of GRBobjective and before 2 months fromexam I had started complete revision.

—‘Competition Science Vision’is a unique magazine for pre-medical competitions. It givesfull study material in all the foursubjects. Its good quality ques-tions and facts sharpen the brainand thinking power of thereaders.

—Satyendra Singh

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

Satyendra—Daily I devoted 7-8hours for study, but I focussed mainlyon Physics and it took 3-4 hours.

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

Satyendra—I gave more atten-tion on my Physics because I thoughtit requires more and more practice.

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

Satyendra—I studied all thetopics completely and before exam. Irevised my syllabus.

CSV—How did you give finaltouches to your preparation ?

Satyendra—By revising mytopics frequently and solving a lot ofMCQs.

CSV—Did you prepare notes ?Satyendra—Yes, I prepared my

own notes. These helped me duringrevision at the time of exam.

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

Bio-DataName—Satyendra Singh YadavFather’s Name—Shri Bhoop

Narayan SinghMother’s Name—Smt. Madhu

YadavEducational Qualifications—H.S./Std. X—65% (Subhash Smark

Inter College, Kanpur), 2003.Inter/Std. XII—68% (Subhash

Smark Inter College, Kanpur), 2005.Special Achievements—

● 13th rank in U.P. CPMT 2009(OBC-7)

● 9th rank in AIPVT 2009

Satyendra—I always tried toclear the concepts of Physics forquestion solving and did to properrevision of Biology and Chemistry.

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

Satyendra—If anyone studies7-8 hours regularly in a day then2 years are sufficient for preparing forthis examination.

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

Satyendra—According to mestudent should start preparation for itafter 10th standard.

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

Satyendra—MBBS, BDS, BAMS,BHMS.

C.S.V. / October / 2009 / 944

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

Satyendra—CSV objective forPhysics, Ramesh Gupta’s ModernZoology, R. K. Pillai for Zoology,Modern Botany of M.P. Kaushik andO.P. Tandon for Chemistry. I usedCSV for all subjects.

CSV—Did you take coaching inyour preparation ?

Satyendra—Yes, I had takencoaching in New Light Institute,Kanpur and this year I also joinedNew Tech Education for Physics.

Personal QualitiesHobbies—Listening to music, see-

ing films.

Ideal Person—My mother and myelder brother.

Strong Point—Hardwork .

Weak Point—Silly mistakes duringsolving MCQs.

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

Satyendra—I used sciencemagazines CSV for solving MCQsand Learning Science Tips.

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

Satyendra—I selected a maga-zine for revision which contains impor-tant points and formulae which arehelpful to me during revision in lastmonth.

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

Satyendra—It helped me to pre-pare my own notes. I collected impor-tant points, formulae and importantMCQs on my notes. These helpedme during last month revision.

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

Satyendra—By adding NCERTbased topics. Many new points hadbeen added in NCERT. So I thinkCSV should contain a separateNCERT corner containing extra pointsfor revision.

CSV—Please mention your posi-tion in the merit list as well as themarks obtained in different subjects.

What was your aggregate percentageor marks ?

Satyendra—13th position inCPMT–2009

Physics—43/50Chemistry—47/50Zoology—47/50Botany—49/50Total—186/200 = 93%.CSV—What books/magazines/

newspapers did you read for G.K.preparations ?

Satyendra—I had read onlynewspaper (Dainik Jagaran). In CPMTexam. G.K. is not asked.

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

Satyendra—I would like to givethe credit of my success to God, myparents and my elder brother. Theywere always with me during my pre-paration. I also thankful to myrespected teachers.

CSV—Please tell us somethingabout your family.

Satyendra—My father is aDoctor. My mother is a housewife. Ihave one brother and two sisters.They are all elder to me. My brotheris doing M. Tech. from I.I.T., Kanpur,my one sister is doing BPT fromKanpur University and the other isdoing B.Ed.

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

Satyendra—My biggest mistakewas that after giving the test in mycoaching I lost 2-3 days. I could notstudied 4-5 hours continuously.

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

Satyendra—Always maintainregularity, study 6-7 hours regularlydaily. To be confident on your studyand faith on God. Best wishes to allreaders of CSV. ●●●

(Continued from Page 942 )

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

Pavitra—Daily reading any news-paper is my hobby which can givenew and new informations. Generalknowledge books from Upkar are thebest ones.

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

Pavitra—Firstly, the Almighty, myparents, my relatives and teachers.

CSV—Please tell us somethingabout your family.

Pavitra—Actually mine is a jointfamily. I have an elder brotherPavneet Saxena obtaining engineer-ing education, my father Pravin KumarSaxena, mother Neeta Saxena.

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

Pavitra—Frankly speaking Inever paid more attention on theoryof Chemistry and Physics which wasa mess.

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

Pavitra—Never loose confidence,always try to do better than what youhave done earlier. Always pay atten-tion and think on each and everyquestion which you read.

●●●

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C.S.V. / October / 2009 / 945

Physics

1. What is the value of the nuclear radius ?

➠ 1·3 ×××× 10–15 metre2. Given vectors are neither perpendicular nor parallel

when➠ Neither their dot product nor their cross

product is zero3. Name some body centred lattices (bcc).

➠ Na, K, Rb, Cr, Mo4. Slope of a time-velocity graph is increasing. It repre-

sents➠ Motion with increasing acceleration

5. What type of semiconductor is the silicon doped withboron ?

➠ p-type semiconductor6. The area under the velocity-time graph gives

➠ The distance covered by the particle

7. What are the very small stars having diameter 150 th

that of sun are called ?➠ White dwarfs

8. Automobiles are provided with spring system to mini-mise the damage of the automobile by jerks. Whatdoes the spring do ?

➠ It increases the time of jerk9. What is the unit of self-inductance ?

➠ Henry10. The relation between millibar and Pascal unit of pres-

sure is➠ 1 milli bar = 100 Pa

11. Which of these two has a greater permeability–softiron or steel ?

➠ Soft iron12. 105 N m– 2 is called

➠ 1 bar

13. What is the order of drift velocity ?➠ 10–4 metre/second

14. The energy per unit volume of a stretched wire is

➠ 12 (Stress ×××× Strain)

15. What is the axial chromatic aberration of a lens ofdispersive power 0·02 and mean focal length 15 cm ?

➠ 0·3 cm

16. From phase considerations the output voltage and theinput voltage of diode are

➠ In phase

17. What is the Sabine’s formula for reverberation time ?

➠ T = KVAs

18. Thermoelectric power is defined as the➠ Rate of change of thermo emf with temperature

19. What is the use of Doppler effect ?➠ In determining the velocity of an aeroplane in air

and of submarine moving under water20. For a system of particles under central force field, the

total angular momentum is conserved because➠ Torque on such a system is zero

Chemistry

21. Hydrated carbonate of lead, used in paintmaking isknown as

➠ White lead

22. What is the chemical name and formula of war gas,lewisite ?

➠ ββββ-chlorovinyldichloroarsine, ClCH—— CH–AsCl2

23. A soluble silicate of sodium or potassium is known as➠ Water glass

24. What is the blasting gelatin ?➠ Mixture of gun-cotton and nitroglycerine

25. The friction of a liquid against its carrier is known as➠ Viscosity

26. What is the Brinell test used for ?➠ A test for hardness of metals

27. The property by which some compositions becomesolid at rest and liquefy again on agitation is knownas

➠ Thixotropy

28. What is the name of an instrument used for measure-ment of viscosity ?

➠ Ostwald viscometer

29. The resolution of a compound into its parts is knownas

➠ Analysis

30. What is an instrument used for measuring density andexpansivity of a liquid, known as ?

➠ Pyknometer

31. A knifelike instrument with flexible blade used formixing very small amounts of chemicals is known as

➠ Spatula

32. What is a purple pigment, consisting of colloidal goldand tin (IV) acid, known as ?

➠ Purple of cassius

33. A material which passes into solution when mixedwith a solvent, is known as

➠ Solute

C.S.V. / October / 2009 / 946

34. What are the alloys which emit sparks when scrapedor struck, known as ?

➠ Pyrophoric alloys35. Potassium carbonate (K2CO3.H2O) is also known as

➠ Salt of tartar

36. What is the common name of an intensely poisonoussolution of hydrocyanic acid ?

➠ Prussic acid

37. A rare metal used for pen-point alloys and for harden-ing platinum is

➠ Ruthenium

38. What is a powder of finely divided silver and proteinknown as ?

➠ Protargol

39. A ferric oxide, produced by heating copperas(FeSO4), used as a pigment and for polishing glass,metal or gems, is known as

➠ Rouge

40. What is the absorption of a gas into the bulk of a solidis known as ? ➠ Occlusion

Zoology

41. What is the effect of interspecific competition onniches ?

➠ Make niches smaller

42. By which kind of control system our body temperatureis controlled ?

➠ Negative feedback

43. How old is the human embryo when it becomesfoetus ?

➠ 2 months

44. Which cell organelles in human beings produce lyso-somes ?

➠ Golgi apparatus

45. Which ion must be present for binding of the crossbridges which connect two molecules of a fibril dur-ing muscle contraction ?

➠ Calcium ions

46. Which male reproductive paired glands open into theurethra at the base of the penis and release a buffer-ing and lubricating fluid ?

➠ Bulbourethral glands

47. What acts as a shock absorber to cushion the tibiaand femur ?

➠ Cartilage

48. Which enzyme catalyzes the synthesis of a new strandfor a DNA molecule, by linking nucleotides to thedeveloping strand ?

➠ DNA polymerase

49. Which fat derivative synthesizes all the hormones ofadrenal cortex ?

➠ Cholesterol

50. Which branch of medicine deals with the study ofsigns and symptoms ?

➠ Semeiology

51. What kind of receptors are present in lateral linesystem of fishes ?

➠ Mechanoreceptors52. What is called an assemblage of an ecological

community ?➠ Interacting population

53. Where are located the cell bodies of neurons bringingafferent informations into the spinal cord ?

➠ Dorsal root ganglia54. Name the hormones which function both as hormones

and neurotransmitters ?➠ Epinephrine and norepinephrine

55. What are called the lymphocytes that inhibit the deve-lopment and proliferation of T and B cells ?

➠ Suppressor T cells56. The release of which hormone is inhibited when the

stomach acidity reaches 2 pH ?➠ Gastrin

57. How a camel is able to tolerate the heat of a desert ?➠ By allowing its body temperature to drop at night

58. Which enzyme is involved in light production in certaininsects ?

➠ Luciferase59. Which part of the body acts as ‘thermostat’ in a

mammal ?➠ Hypothalamus

60. What acts as a shock absorber to cushion the tibiaand femur ?

➠ Cartilage

Botany

61. Which enzyme catalyses the joining of the ends oftwo chains ?

➠ DNA ligase

62. Who noticed a colour substance, chromatin, whilestudying salamander cells ?

➠ Walter Flemming

63. During which pathway electrons more from waterthrough photosystem-II to photosystem-I and then toNADP+ ?

➠ Non-cyclic electron pathway

64. What are the two important classes of higher fungi ?➠ Ascomycetes and Basidiomycetes

65. What is called alcoholic distillate obtained from fer-mented sugar ?

➠ Rum

66. What type of gland is found in Utricularia plant ?➠ Digestive gland

67. What is the central theme of evolution ?➠ Descent with modification

(Continued on Page 988 )

C.S.V. / October / 2009 / 947

Principle of Superposition

Two or more progressive waves can travel simul-taneously in a medium without affecting the motion of oneanother. In such a situation the displacement of eachparticle of medium at any instant is equal to the vectorsum of the displacements each wave would produce inabsence of the other wave or waves.

Interference

Interference is the characteristic property of all wavemotion. Since sound is wave motion, it exhibits thephenomenon of interference. Thus when two wavestravelling in the same direction in the medium, it should bepossible under suitable conditions to have minimum andmaximum intensities at definite points of the medium andthe resultant intensities in the region of superposition is ingeneral different from the sum of intensities due toindividual waves acting separately. This modification inthe distribution of intensity due to the superpositionof waves is called interference.

If the resultant intensity exceeds the sum ofintensities due to individual waves, the phenomenon iscalled constructive interference and if the resultantintensity is less than this sum, the phenomenon is calleddestructive interference. If the resultant intensity is zero,the phenomenon is termed completely destructiveinterference.

Analytical Treatment of InterferenceLet the amplitudes of the two waves be A1 and A2

and the two waves differ in phase by an angle φ. So

y1 = A1 sin(kx – ωt)

and y2 = A2 sin(kx – ωt + φ)

The resultant wave is given byy = y1 + y 2

= A1 sin(kx – ωt) + A2 sin(kx – ωt + φ)= A1 sin(kx – ωt) + A2 sin(kx – ωt) cos φ

+ A2 cos(kx – ωt) sin φ

= sin(kx – ωt) (A1 + A2 cos φ)

+ cos(kx – ωt)A2 sin φ

Let A1 + A2 cos φ = A cos ε …(i)

and A2 sin φ = A sin ε …(ii)

∴ y = A[sin(kx – ωt) cos ε

+ cos(kx – ωt) sin ε]

= A sin(kx – ωt + ε)

Thus, the resultant is indeed a sine wave ofamplitude A with a phase difference ε with the first wave.By (i) and (ii),

A2 cos2 ε + A2 sin2 ε = A2 (cos2 ε + sin2 ε) = A2

= (A1 + A2 cos φ)2 + A22 sin2 φ

= A12 + A2

2 + 2A1 A2 cos φ

or A = A12 + A2

2 + 2A1A2 cos φ

Thus for maximum A i.e., constructive interference,

cos φ = 1 or φ = 2n π

or in terms of path difference

δ = n λ

where n = 0, 1, 2, 3, …

Thus for constructive inter-ference

δ = 0, λ, 2λ, 3λ… nλ

S1

S2

- - - - - - - - - ••

O

P

Qδ

- - - -

= 0, 2λ2 ,

4λ2 ,

6λ2 … 2n

λ2

i.e. the path difference should be even multiples of λλλλ2

and for minimum A i.e., destructive interference,cos φ = –1

or φ = (2n + 1)πor in terms of path difference for destructive interference

δ = (2n + 1) λ2

=λ2 ,

3λ2 ,

5λ2 , … (2n + 1)

λ2

i.e., for destructive interference the path difference

should be odd multiples of λλλλ2.

Condition for Interference

For sustained interference between two sound waves,the following conditions must be fulfilled—

(1) The phase difference between the waves mustremain constant—If the phase difference betweeninterferring waves changes with time, the intensity ofsound at any point will also change with time and theinterference will not be sustained. Hence for observing theinterference both the waves should originate from thesame source.

(2) The amplitudes of the waves should be nearlyequal—If the difference in amplitudes is large, the intensityat points of destructive interference will be quite large.Hence, the interference will not be clear. For best inter-ference the amplitudes of the waves must be exactlyequal.

(3) The displacements produced by the two wavesmust be along the same straight line—If it is not so, theintensity at the point where the waves meet in oppositephases will not be zero.

C.S.V. / October / 2009 / 948

Beats

When two sounding bodies of nearly equal frequen-cies are sounded together, the resultant sound comprisesof alternate maxima and minima of sound. The pheno-menon of alternate waxing and waning of sound at regularintervals is called beats.

Beat frequency—The number of beats heard persecond is called beat frequency. It is equal to thedifference in frequencies of the two sounding bodies.

Beats are heard only when the difference infrequencies is not more than ten.

Beat period—The time from each loud sound to thenext loud sound is called one beat-period.

Beat—One loud sound plus the subsequent faintsound constitute one beat.

Analytical Treatment of Beats

Let there be two waves of slightly differing fre-quencies n1 and n2. Let a be the amplitude of each wave.There displacements y1 and y2 are given as

y 1 = a cos 2πn1t

y 2 = a cos 2πn2t

If they superimpose, the resultant displacement is

y = y 1 + y 2 = a cos 2πn 1t + a cos 2πn 2t

= a ⎣⎢⎢⎡

⎦⎥⎥⎤

2 cos 2π(n 1 + n 2)t

2 · cos 2π(n 1 – n 2)t

2

= 2a cos π(n 1 + n 2)t · cos π(n 1 – n 2)t

= A cos π(n 1 + n 2)t

where A = 2a cos π(n 1 – n 2)t

A is the amplitude of the resultant wave.For maximum A,cos π(n 1 – n 2)t = 1

or π(n 1 – n 2)t = nπ ; n = 0, 1, 2, …

or t =n

(n 1 – n 2)

= 0, 1

(n 1 – n 2) ,

2(n 1 – n 2)

, 3

(n 1 – n 2)Hence time interval between successive maxima

=1

n 1 – n 2or the frequency of beats

= (n 1 – n 2)Similarly for minimum A,

cos π(n 1 – n 2)t = 0

or π(n 1 – n 2)t =

n + 12

. π

t =

n + 12

(n 1 – n 2)

=1/2

(n 1 – n 2) , 3/2

(n 1 – n 2) , 5/2

(n 1 – n 2)

So the time inerval between two successive minima

=1

n 1 – n 2

∴ The frequency of beats = (n 1 – n 2)

In 1 second the intensity of sound will be (n 1 – n 2)times maximum and (n 1 – n 2) times minimum i.e.,(n 1 – n 2) beats will be heard in 1 second.

∴ No. of beats per sec (beat frequency)

= n 1 – n 2

= Difference of the frequencies of the twosound sources.

Graphical Representation of Beats

Formation of beats is shown in the figure.

Uses of Beats

(i) Determination of unknown frequency—Thetuning fork of unknown frequency N′ is sounded with astandard tuning fork of known frequency N and thenumber of beats per second ‘x ’ is heard. Then N′ is(N + x) or (N – x). Now one of the prongs of the tuning forkof unknown frequency is loaded with wax. This slightlydecreases its frequency.

Both the tuning forks are again sounded together.Now if the beat frequency is found to be greater than ‘x ’,N´ is (N – x); but if the beat frequency is found to be lessthan ‘x’ N´ is (N + x).

(ii) Tuning the musical instruments—The tensionin the string of the two instruments is altered till beats areheard. Adjusting is continued till beats disappear. Theinstruments get tuned.

(iii) In electronics—Electronic beat frequencyoscillators are commonly used to generate a beat fre-quency which is audible. Also in modern radio receivers,ultrasonic beats are generated and radio reception isobtained.

(iv) In mines—The presence of dangerous gases inmines may be detected by the use of beats. Theapparatus consists of two identical pipes; one blown withthe air from a reservoir and the other with the air from themine. If both the airs are the same, no beats will be heard.But if the air in the mine is polluted, beats will be heard.Thus, the method serves as an early warning system tosafeguard workers against possible dangerous explo-sions.

C.S.V. / October / 2009 / 949

SOME IMPORTANT SOLVED EXAMPLES

Example 1. Two sound sources are vibrating inthe same phase with a frequency of 100 sec–1 each.Speed of sound is 340 m/s. For what value of pathdifference x one would hear the successive maximumand minimum sounds ?

Solution :

Example 2. A tuning fork is in unison with anothertuning fork of frequency 260 c.p.s. When waxed, itproduces 4 beats per second. What is its frequencywhen waxed ?

Solution :

Example 3. A tuning fork that is in unison withanother tuning fork of frequency 400 c.p.s. is filed.Now it produces 3 beats per second with the sametuning fork. What is its new frequency after filing ?

Solution :

Example 4. A tuning fork of frequency 250 is usedto tune a piano. As the string of piano is tightened,the number of beats decreases till it reduces to 4beats per second. What is the new frequency of thepiano string ?

Solution :

Example 5. Calculate the frequency of the beatsheard by stationary observer when a source of soundof frequency 100 Hz moves directly away from himwith a speed of 10·0 ms–1 towards a vertical wall.

(speed of sound in air = 340 ms–1)

Solution :

Example 6. Two waves having intensity in theratio 9 : 1 produce superposition. What is the ratio ofmaximum to minimum intensity ?

Solution :

Example 7. When beats are produced by twoprogressive waves of the same amplitude and ofnearly the same frequency, what is the ratio ofmaximum loudness to the loudness of one of thewaves ?

Solution :

Example 8. A fork of unknown frequency produces4 beats with a fork of 256 Hz frequency. When the forkof unknown frequency is loaded with wax, it againproduces 4 beats per second with the other fork. Whatis unknown frequency of the fork ?

Solution :

C.S.V. / October / 2009 / 950

OBJECTIVE QUESTIONS

1. Which of the following representsthe superposition of two progres-sive waves ?(A) y = f (x – vt)(B) y = ym sin k(x + vt)(C) y = ym log(x – vt)

(D) y = f (x 2 – v

2t 2)

2. Two waves represented byy 1 = 10 sin (2000 πt ) and

y 2 = 10 sin 2000 πt + π2

. are

superposed at a particular ins-tant. The resultant amplitude is—(A) 10 units (B) 20 units(C) 14·1 units (D) Zero

3. An accurate oscillator is used tostandardise a tuning fork markedas 512 Hz. When the oscillatorreading is 514, two beats areheard per second. When theoscillator reading is 510, the beatfrequency is 6. The frequency ofthe tuning fork is—(A) 506 (B) 510(C) 516 (D) 518

4. An earth satellite has a velocitycomponent of 7 km/s towards anearth observer. It emits a signalof frequency 100 MHz. This iscombined with a signal of samefrequency produced by a localoscillator. The approximate beatfrequency will be—

(A) 1200 (B) 2400(C) 3600 (D) 4800

5. 56 tuning forks are so arrangedin series that each fork gives 4beats per second with theprevious one. The frequency ofthe last fork is the octave of thefirst. The frequency of the firstfork is—(A) 220 Hz (B) 224 Hz

(C)2207 Hz (D) 110 Hz

6. When two simple harmonicmotions of same periods, sameamplitudes, having phase diffe-

rence of 3π2 and at right angles to

each other are superimposed.The resultant wave form is a—(A) Circle(B) Parabola

(C) Ellipse(D) Figure of eight

7. A wave of frequency 500 Hz hasa speed of 360 ms– 1. Thedistance between two points 60°out of phase will be—(A) 72 cm (B) 36 cm(C) 12 cm (D) 1·8 cm

8. A, B and C are three tuningforks. The frequency of A is350 sec–1. A and B produce5 beats/sec, while B and Cproduce 4 beats/sec. When A isloaded with some wax, itproduces 2 beats/sec with B, and6 beats/sec with C. Frequenciesof B and C respectively are—(A) 345, 347(B) 341, 345(C) 345, 341(D) None of these

9. When two simple harmonicvibrations of the same frequencyand amplitude are combinedwhile acting on a particle at rightangles. The resulting motion ofthe particle is circular when thephase difference between themis—

(A) π (B)π4

(C) 2π (D)π2

10. A tuning fork of frequency 480Hz produces ten beats perseconds when sounded with avibrating sonometer. What shouldbe the frequency of the sono-meter wire if a slight increase intension produces less number ofbeats ?(A) 460 (B) 480(C) 470 (D) 490

11. Two sound producing bodiesproduce progressive wavesgiven by y 1 = 4 sin 400πt, y 2 = 3sin 404 πt. A person standingnearby will hear—(A) 2 beats per second of inten-

sity ratio 4 : 3(B) 2 beats per second of inten-

sity ratio 49 : 1(C) 4 beats per second of inten-

sity ratio 7 : 2(D) 4 beats per second of inten-

sity ratio 4 : 3

12. A man facing a wall holds atuning fork of frequency 256between himself and a verticalwall. He moves the tuning forktowards the wall with a velocity1

100 of the velocity of sound. The

number of beats heard perminute would be nearly—(A) 75 (B) 150(C) 300 (D) Zero

13. A tuning fork of frequency 256Hz is excited and held at themouth of a resonance column offrequency 254 Hz. Pick up thecorrect statement—(A) 2 beats per second will be

heard(B) 4 beats per second will be

heard(C) 1 beat per second will be

heard(D) No beats will be heard

14. The displacement of a particleexecuting periodic motion isgiven by

y = 4 cos2

.

12

. t sin (1000 t)

This expression may be con-sidered to be a result of thesuperposition of—(A) Two waves(B) Three waves(C) Four waves(D) Five waves

15. What is the speed of sound in agas in which two waves ofwavelengths 1·00 m and 1·01 mproduce 3 beats in 10 seconds ?(A) 303 m/s(B) 340 m/s(C) 320 m/s(D) None of these

16. Two waves passing through aregion are represented by

y1 = (1·0 cm) sin[(3·14 cm–1)x

– (157 s–1)t ]y 2 = (1·5 cm) sin[(1·57 cm–1) x

– (314 s–1)t ]Find the displacement of theparticle at x = 4·5 cm at timet = 5·0 ms—(A) 0·5 cm (B) 2·5 cm(C) 0·35 cm (D) – 0·35 cm

C.S.V. / October / 2009 / 951

17. The equations of the two wavesare

y 1 = 10 sin 3πt + π3

.

and y 2 = 5(sin 3πt + 3 cos 3πt)

What is the ratio of the ampli-tudes ?(A) 1 : 1 (B) 2 : 1(C) 1 : 2 (D) 2 : 3

18. Following is the resultant waveproduced by—

A B

C D

(A) Two sources of slightly diffe-rent frequencies sounded atthe same time

(B) Two sources of exactly equalfrequencies sounded at thesame time

(C) A single sound source(D) Two coherent light sources

19. In order for two sound waves toproduce beats it is most impor-tant that the two waves—(A) Have the same frequency(B) Have slightly different ampli-

tudes(C) Have slightly different fre-

quencies(D) Have the same number of

overtones

20. When two sound waves ofexactly equal frequencies movingin same directions superpose,then—(A) Beats are produced(B) Interference of sound takes

place(C) Stationary waves are pro-

duced(D) None of the above

21. Interference of sound can bedemonstrated by—(A) Kundt’s tube(B) Quineke’s tube(C) Resonance tube(D) Hebb’s method

22. For constructive interference,phase difference between twowaves should be (in radian)—

(A)π2 (B) π

(C)π4 (D) 2π

23. For destructive interferencephase difference between twowaves should be (in radians)—

(A)π2 (B) π

(C)π4 (D) Zero

24. For constructive interference pathdifference between two wavesshould be—

(A)λ4 (B)

λ2

(C) λ (D)3λ2

25. For destructive interference, pathdifference between two wavesshould be—

(A)λ4 (B)

λ2

(C) λ (D) Zero

26. Two waves of I1 and I2 intensitiesinterfere. The intensity of theresultant wave will be—

(A) I1 + I2 + 2 I1I2 cos φ

(B) I12 + I22 + 2I1I2 cos φ

(C) I1 + I2 + 2I1I2 cos φ

(D) I12 + I22 + 2I1I2 cos φ

27. In interference intensity changes—(A) With time(B) With distance(C) Both with time and distance

(D) Neither with time nor withdistance

28. In interference total energy—(A) Increases(B) Decreases(C) Remains constant and is not

redistributed(D) Remains constant and is

redistributed

29. Can beats be observed by twolight sources ?(A) Yes (B) Never(C) Seldom (D) Indiscernible

30. The composition of two S.H.M. ofequal periods at right angles toeach other and with a phase diffe-rence of π results in the displace-ment of the particles along a—(A) Straight cone(B) Circle(C) Ellipse(D) Figure of eight

ANSWERS WITH HINTS

C.S.V. / October / 2009 / 953

Important Points to Remember

Nucleus

1. The nucleus is very small part which exists at thecentre of the atom.

2. Nucleus was discovered by Rutherford through hisα-scattering experiments.

3. The whole positive charge and almost the wholemass of an atom resides inside the nucleus.

4. The charge on the nucleus is (+ Ze). It is due toprotons present in the nucleus.

5. The nucleus is highly dense. Its density is of the orderof 1014 gm/c.c.

6. The radius of the nucleus is of the order of 10–15 to10–14 m.

7. If the nucleus is presumed to be spherical its radiusr = r0 A1/3 where r0 = 1·2 × 10–15 m and A is atomicmass number.

8. The constituents of nucleus are neutrons and pro-tons. In an atom electrons, equal in number to pro-tons, revolve round the nucleus.

9. In lighter nuclei the proton number equals the neutronnumber (N = Z) e.g. 7N14, 9Fe18 etc.

10. In heavier nuclei the number of neutrons is greaterthan the number of protons (N > Z) e.g. 90Th232,92U238 etc.

11. For all stable nuclei NZ = 1 to 1·5.

12. The neutrons and protons present inside the nucleustaken together are known as nucleons.

Formation of Nucleus

1. When neutrons and protons combine to form anucleus, then the mass of the nucleus is some-what less than the sum of the masses of itsconstituent particles.

2. The decrease in mass in the process of formation ofnucleus is called mass defect.

3. The mass defectΔm = Total mass of neutrons + Total mass of

protons – Mass of the nucleus.

= N × mn + Z × mp – mZA

where N = number of neutrons present in nucleus.

4. When neutrons and protons combine to form a nuc-leus, then its Δm mass transforms into energy i.e.,

ΔE = Δm × c2

where c is the speed of light.

Binding Energy

1. The minimum energy required to keep the neutronsand protons bound inside the nucleus is defined asbinding energy.

2. The minimum energy required to dissociate the nuc-leus into its constituent particles is also equal to itsbinding energy.

3. Binding energyΔE = Δmc2

= [ ]Nmn + Zmp – mZA × c2

4. Binding energy represents the stability of nucleus.

5. B.E. per nucleon = Total binding energy

No. of nucleons

=ΔEA

=[ ]Nmn + Zmp – mZA c 2

Awhere A is mass number and is equal to the sum ofneutrons and protons present in the nucleus.

6. The binding energy per nucleon represents the stabi-lity of nucleus. Higher the binding energy per nuc-leon, more stable is the nucleus.

Binding Energy Curve

A graph of the binding energy per nucleon and themass number of nuclei is called the binding energy curve.It is shown in the figure.

We have following important informations from thebinding energy curve.

1. The nuclei having mass number A ≈ 60 (e.g. Fe withA = 56) have maximum binding-energy per nucleon(≈ 8·7 MeV). So these nuclei are most stable.

0 20 40 60 80 100 120 140 160 180 200 220 240

9.0

8.0

7.0

6.0

5.0

4.0

3.0

2.0

1.0

0.0

Bin

ding

ene

rgy

per n

ucle

on (M

eV)

Mass number (A)

H2

U238

O16

C12

N14F18

Fe56

He4

Li7

2. For nuclei having mass number above about 60, thebinding energy per nucleon gradually decreases. Forexample, for uranium (A = 238) it is about 7·6 MeV.

3. For nuclei having mass number below 56 also, thebinding energy per nucleon decreases and below 20,

C.S.V. / October / 2009 / 954

it decreases very rapidly. For example, for heavyhydrogen (A = 2) it is only 1 MeV. This means thatnuclei having mass number below 20 are relativelyless stable.

4. The special positions for He4, C12 and O16 on thecurve show that these nuclei are more stable than thenuclei in their vicinity.

5. The binding energy of every nucleus is positive whichrepresents that there exist attractive forces amongthe nucleons due to which these remain stable.These forces are known as nuclear forces.

Packing Fraction (P)

1. Packing fraction is defined as mass defect pernucleon. Thus

P =Mass defect

A = M – A

Awhere M is actual mass of the nucleus and A is themass number.

2. Packing fraction may be positive or negative.3. P also represents the stability of nucleus.4. The nuclei having negative value of packing fraction

are more stable.5. In general, the smaller the value of packing fraction,

the more stable are the nuclei.

Nuclear Forces

The forces which keep protons and neutrons boundinside the nucleus are known as nuclear forces. Theyhave following characteristics :1. They are strongly attractive, otherwise the nucleus

would be disrupted under electrostatic repulsion bet-ween protons.

2. They are non-electric. If they were electric, the pro-tons would repel one another. Thus leading to disrup-tion rather than to stability.

3. They are non-gravitational. The gravitational forcesbetween the nucleons are found to be about 10–40

times the attractive forces demanded. Thus nuclearforces cannot be gravitational in nature.

4. They are extremely short range forces. Thenuclear forces do not obey inverse square law andare effective for only very short distances ≈ 10–15 m.

5. They are charge-independent. Nuclear forces donot at all depend on charge. Thus the nuclear forcebetween all nucleons (e.g. between p-p, n-n or n-p)is the same.

6. They are non-central forces. The force between twonucleons does not act along the line joining theircentres and is, therefore, called a non-central force.

7. They are exchange forces. According to Yukawathe nuclear force between two nucleons is the resultof exchange of π mesons (π0, π+, π–) between them.

Classification of Nuclei

The nuclei have been classified on the basis of thenumber of protons (atomic number Z) or the total numberof nucleons (mass number A) in them.

1. Isotopes—The atoms of an element having samenumber of protons (Z) but different number of nucleons

(A) i.e. different number of neutrons are called isotopes,e.g.

Hydrogen : 1H1, 1H2, 1H3

Oxygen : 8O16, 8O17, 8O18

Special FeaturesIsotopes :

1. All isotopes of an element have same number of protons(i.e. same number of electrons). Hence their chemicalproperties of different isotopes of an element are same.Hence they cannot be separated by any chemicalprocess.

2. The mass number (i.e. number of nucleons) of all isotopesof an element are different. Hence their physical proper-ties are not the same to separate them atomic massdependent physical properties, like gaseous diffusion, areused.

3. Among isotopes of the same element, some may bestable and some radioactive. This is due to difference intheir nuclear structure. For example 6C12 is stable while

6C14 is radioactive similarly 11N23 is stable while 11Na24

is radioactive.

Isobars :1. Their atomic numbers (Z) are different, hence they occupy

different places in periodic table.2. They differ in chemical properties.3. Isobars differ in physical properties also.

4. Nuclei of isobars belong to different elements.5. The daughter nucleus remaining after emission of β-

particles is an isobar of the parent nucleus.

2. Isobars—The nuclei having same number ofnucleons (A) but different number of protons (Z) are calledisobars. They also have different number of neutrons. Forexample :

(a) 1H3 and 2He3

(b) 6C14 and 7N14

(c) 8O17 and 9F17

are isobars.3. Isotones—The nuclei having equal number of

neutrons are called isotones. For them both Z and A aredifferent but (A – Z) is same. For example :

(a) 3Li7 and 4Be8, A – Z = 4(b) 1H3 and 2He4, A – Z = 2(c) 11Na23 and 12Mg24, A – Z = 12

are isotones.

Proton

1. Proton is a fundamental particle and was discoveredby Rutherford in 1919 in artificial nuclear disinte-gration.

2. It has a positive charge (+ 1·6 × 10–19 coulomb) andmass 1·673 × 10–27 kg.

3. It is represented by 1H1.

4. In free state proton is a stable particle.5. The number of protons in the nucleus of the atoms of

an element gives the atomic number (Z) of thatelement.

6. Elements can be disintegrated artificially by bombar-ding them with protons

C.S.V. / October / 2009 / 955

Neutron1. Neutron was discovered by James Chadwick.

2. For this important discovery, Chadwick was honouredby Nobel Prize in Physics in 1935.

3. Neutron is a fundamental particle of matter and itresides in the nucleus along with proton.

4. Its mass is 1·675 × 10–27 kg, which is slightly greaterthan the mass of proton.

5. It is uncharged particle.6. It is represented by 0n1 because its atomic number

(charge) is zero and atomic mass number is 1.

7. It is not deflected by electric and magnetic fields.

8. Being chargeless, it does not ionise gases and doesnot produce a track in Wilson Cloud Chamber.

9. Its penetrating power is very high. It penetratesthrough thick sheets of lead.

10. On striking an atom, it easily enters its nucleusbecause it being chargeless is not deflected by thepositive charge of the nucleus. Hence neutron hasproved most useful for nuclear disintegration andfission.

11. A free neutron outside the nucleus is unstable and isconverted into a proton by emitting β-particle (elec-tron) and an antineutrino.

0n1 ⎯→ 1H1 + –1β0 + –υNeutron Proton Electron Antineutrino

Positron

1. Positron is a fundamental particle and was dis-covered by an American scientist Anderson in 1932.Anderson was honoured by Nobel Prize in physics in1936.

2. Positron is a positively charged particle whose massand charge are exactly equal to the mass and chargeof electron.

3. Positron is anti-particle of electron.

Pair-production

When an energetic γ-ray photon falls on a heavysubstance, it is absorbed by some nucleus of the subs-tance and an electron and a positron are produced. Thisphenomenon is called ‘pair-production’. It may be repre-sented as

hυ = 1β0 + –1β0

Photon Positron Electron

Rest-mass energy of each of the electron and theproton is

E0 = m0c2

= (9·1 × 10–31) × (3·0 × 108)2

= 8·2 × 10–14 joule

= 0·51 MeV

Hence for pair-production, the energy of γ-photonmust be at least 2 × 0·51 = 1·02 MeV.

Pair-annihilationIt is converse to pair-production. When an electron

and a positron come very close to each other they annihi-late each other by combining together and two γ-photonsare produced.

+1β0 + –1β0 = hυ + hυPositron Electron γ-photon γ-photon

SOME IMPORTANT SOLVED EXAMPLES

Example 1. How many electrons, protons andneutrons are there in 14 gram of 6C14 ? Avogadro’snumber N = 6 ×××× 1023.

Solution :

Example 2. How many αααα and ββββ particles are emit-ted when 92U238 changes into 82Pb206 ?

Solution :

C.S.V. / October / 2009 / 956

Example 3. An electron-positron pair is producedby the materialisation of a γγγγ-ray photon of 2·26 MeV.How much kinetic energy is imparted to each of thecharged particle ?

Solution :

OBJECTIVE QUESTIONS

1. The atoms of an element differ-ing in mass though possessingthe same chemical properties arecalled—(A) Isobars (B) Isotopes(C) Isotones (D) Isomers

2. The daughter nucleus remainingafter emission of β-particle is an……… of the parent nucleus.(A) Isotope (B) Isobar(C) Isotone (D) Isomer

3. The equation

3Li6 + 1H2 ⎯→ 4Be7 + ……

will be completed by—

(A) α-particle (B) 2β-particles

(C) Neutron (D) None of these

4. Isotopes have—

(A) Same number of protons

(B) Same number of neutrons

(C) Same number of nucleons

(D) None of these

5. Which of the following particles isunstable ?(A) Electron (B) Proton

(C) α-particle (D) Neutron

6. An electron and a positron areformed by a photon of energy2·62 MeV. The total kineticenergy of both these particles willbe—

(A) 2·6 MeV (B) 1·6 MeV

(C) 1·0 MeV (D) 3·6 MeV

7. Which of the following is correct ?Positron is called antiparticle ofelectron because—(A) It has opposite charge(B) Its mass is equal(C) It collides with electron(D) It is destroyed by combining

with electron

8. Which one of the following is nota fundamental particle ?(A) Proton(B) Meson

(C) Neutrino

(D) α-particle

9. When α-particles are bombardedon 4Be9, then 6C12 is formed.The particle emitted is—

(A) –1β0

(B) +1β0

(C) 1H1

(D) 0n1

10. Which particle is X in thefollowing nuclear reaction ?

2He4 + 7N14 ⎯→ ZXA + 1H1

(A) Oxygen with mass 16

(B) Nitrogen with mass 17

(C) Oxygen with mass 17

(D) Nitrogen with mass 16

ANSWERS WITH HINTS

●●●

(Continued from Page 952 )

●●●

C.S.V. / October / 2009 / 957

1. What is the number of significantfigures in 0·002305 × 10–27 kg ?

(A) 7 (B) 6(C) 4 (D) 3

2. The given figure is a graph of—

⎯⎯⎯⎯⎯⎯t →

v↑

(A) Speed time of a body pro-jected under gravity

(B) Acceleration-time graph of abody projected under gravity

(C) Velocity-time graph of abody projected under gravity

(D) None of these

3. Which of the following curvesdoes not represent motion in onedimension ?

4. Two springs A and B are alikeand W is the work done instretching the spring. A is stifferthan B(KA > KB). If they areelongated through the same dis-tance, then—(A) WA > WB (B) WA < WB

(C) WA = WB (D) None of these

5. A pendulum bob of weight 2N ispulled to the right by a horizontal

F

30°

force F until the string makes anangle of 30° to the vertical. Which

are the force for and tension inthe string needed to sit at 30° ?

(A) 2 newton, 1 newton(B) 4 newton, 3 newton(C) 3 newton, 3·5 newton(D) 1·2 newton, 2·3 newton

6. The amplitude and the timeperiod of a simple pendulum are0·05m and 2 second respectively.The maximum velocity of thependulum is—

(A) 0·157 m/s (B) 1·57 m/s

(C) 3·14 m/s (D) 6·28 m/s

7. A block of mass 2 kg rests on arough inclined plane making anangle of 30° with the horizontal.The coefficient of static frictionbetween the block and the planeis 0·7. The frictional force on theblock is—

(A) 9·8 N

(B) 0·7 × 9·8 × 3 N

(C) 9·8 × 3 N

(D) 0·7 × 9·8 N

8. The displacement of a particledwelling in some wave motion isy = 5 sin 2πnt. If the phasedifference between two particlesbe 30° within a time-interval of 4second and 6 second, then nshould be—

(A) 1·2 (B)112

(C)124 (D) 24

9. A man is at rest in the middle ofa pond on perfectly smooth ice.He can get himself to the shoreby making use of Newton’s—

(A) First law (B) Second law

(C) Third law (D) All the laws

10. A mass M of a certain gas isenclosed in a vessel at tempera-ture T and pressure P. A mass2M of the same gas is filled inthe same vessel until the finalvolume and temperature become

V9 and

T3 respectively. The new

pressure will be—(A) 3P (B) 9P(C) 27P (D) None of these

11. A body of 5 kg moves on africtionless horizontal surfacewith a speed 3 m/s. It com-presses a spring put along itsway and stops. What is the com-pression in spring ? The forceconstant of spring =10 kg wt permetre.(A) 0·68 m (B) 1 m(C) 0·2 m (D) None of these

12. The number of degrees of free-dom for oxygen molecule is—

(A) 3 (B) 5

(C) 6 (D) 7

13. If the change in the value of ‘g ’at a height h above the surfaceof the earth is the same as at adepth x below its, surface then—

(Both x and h being muchsmaller than the radius of the

earth)(A) x = h (B) x = 2h

(C) x = h2 (D) x = h2

14. The liquid surface towards theupper end of the capillary will beconcave when—

(A) The liquid is denser thanwater

(B) The surface tension of theliquid is zero

(C) The liquid moistens thecapillary

(D) The nearby medium is air

15. 5 gm air is heated from 4°C to6°C. If the specific heat of air atconstant volume be 0·172 cal/gm-°C, the increase in the inter-nal energy of air will be—

(A) 7·2 erg (B) 17·2 joule

(C) 7·2 calorie (D) 1·72 calorie

16. The dispersive power of a prismdepends on—

(A) Angle of the prism(B) Material of the prism

(C) Angle of the prism andmaterial of the prism both

(D) None of the angle of prismand material of prism

C.S.V. / October / 2009 / 958

17. The temperature of the filamentof a lamp is 2100 K and itssurface area is 4 × 10–4 m2. If theemissivity of the filament is 0·453then the power of lamp is—

(A) 100 watt (B) 200 watt

(C) 400 watt (D) 0 watt

18. Fluorescence is caused—

(A) In all substances due toultraviolet light

(B) In some specific substancesby any a kind of light

(C) In some specific substancesby only a specific light

(D) By only long wavelengthlight

19. In Meldis experiment the stringvibrates in 7 segments undertension of 9 gm-wt. If the string isto be vibrated in 3 segments, thetension required will be—

(A) 1·4 gm-wt (B) 49 gm-wt

(C) 61 gm-wt (D) 13 gm-wt

20. The phenomenon of bending oflight at opaque edges is called—

(A) Refraction (B) Reflection

(C) Diffraction (D) Interference

21. Two small spheres each havingthe charge + Q are suspendedby insulating threads of length Lfrom a hook. This arrangement istaken in space where there is nogravitational effect, then theangle between the two suspen-sions and the tension in eachthread will be—

(A) 180°, 1

4πε0

Q2

(2L)2

(B) 90°, 1

4πε0 Q2

L2

(C) 180°, 1

4πε0 Q2

2L2

(D)1

180° ,

14πε0

·Q2

L2

22. The wavelength of first line ofBalmer series in hydrogen spec-

trum is 6561 °A. What is the wave-

length of the second line ofBalmer series ?

(A) 4860 °A (B) 5400 °A

(C) 3600 °A (D) 6000 °A

23. The resistance of a wire is 10 Ω.Its length is increased by 10% bystretching. The new resistancewill now be—(A) 12 Ω (B) 1·2 Ω

(C) 13 Ω (D) 11 Ω

24. The plane of a tangent galvano-meter is lying in a plane perpen-dicular to the magnetic meridian.A neutral point is obtained atthe centre when a current ispassed through the coil. If theplane of the coil is rotated through90°, the deflection in the needlewill be—(A) 0° (B) 45°

(C) 60° (D) 30°

25. If two bulbs of wattage 25 and100 respectively each rated at220 volt are connected in serieswith the supply of 440 volt, thenwhich of the bulbs will fuse ?(A) 100 watt bulb(B) 25 watt bulb(C) None of these(D) Both (A) and (B)

26. An oil drop floats between theparallel plates of a condenser.The plates are horizontal and thelower plate has got + Q charge.Area of each plate is A and thedistance between them is D.Mass of oil drop is M. Charge onthe drop in CGS units will be—

(g is the acceleration due togravity)

(A) ( )AQ ( )g

M (B)(Mg A)(4π Q)

(C) – ( )g AD Q (D) –

(Mg A)4πQ

27. The magnetic flux densityapplied in a cyclotron is 3·5 tesla.The frequency of the electric fieldthat must be applied between thedees in order to accelerateprotons, will be—(A) 6·53 × 107 Hz

(B) 3·55 × 107 Hz

(C) 5·34 × 107 Hz

(D) None of these

28. The distance between the platesof a parallel plate condenser is5 cm. It is filled with two media ofdielectric constants 3 and 2.Media and the plates have equalarea. Thickness of the media are

3 and 2 cm respectively. Thedistance between the plates ofan equivalent air condenser is—(A) 5 cm (B) 3 cm(C) 2 cm (D) 2·5 cm

29. A circular disc of area

(4 i∧

+ 5 j∧

) × 10–3 m2 is placed ina uniform magnetic field of inten-

sity (0·2 i∧

+ 0·3 j∧

) tesla. The fluxcrossing the disc will be—(A) 23 weber(B) 23 × 10–2 weber

(C) 23 × 10–3 weber

(D) 23 × 10–4 weber

30. A galvanometer of 600 Ω isconnected in series with a 300Ω resistance and a battery of 18volt. If the galvanometer is shun-ted with a 200 Ω resistance, thepotential difference across thegalvanometer will be—(A) 4·5 volt (B) 6 volt(C) 9 volt (D) 12 volt

31. If applied voltage on a motor is200 volt and back e.m.f. is 160volt. The efficiency of the motoris—(A) 100% (B) 80%(C) 50% (D) 25%

32. Kilowatt hour is the unit of—(A) Energy(B) Power(C) Electric charge(D) Electric current

33. An electromagnetic radiation hasan energy 14·4 keV. To whichregion of electromagnetic spec-trum does it belong ?(A) Infra red region(B) Visible region(C) X-rays region(D) γ-ray region

34. Hot wire ammeter is used tomeasure—(A) Ionisation currents(B) Direct current only(C) Alternating current only(D) Direct current and alterna-

ting current both

35. The power of lens in the spec-tacles of a person is + 2D. Theperson suffers from—(A) Hypermetropia(B) Myopia

C.S.V. / October / 2009 / 959 / 4

(C) Colour blindness(D) Presbyopia

36. Voltameter is a device—(A) To measure the potential

difference between twopoints

(B) To measure electrochemicalequivalent by electrolysis

(C) To compute electric power(D) Made from voltaic pile of

volta

37. Electrons move at right angles toa magnetic field of 0·03 T andenter with a velocity 9 × 107 m/s.

The value of em will be—

(Given radius of circularpath = 1·764 cm)

(A) 1·7 × 1011 C kg–1

(B) 2 × 1011 C kg–1

(C) 2·5 × 1011 C kg–1

(D) None of these

38. An electric charge moving withuniform speed creates—(A) Only electric field(B) Only magnetic field(C) Both the electric and mag-

netic fields(D) No such a field

39. The activity of sea water isabout—(A) 5 Bq per c.c.(B) 7 Bq per m3

(C) 3 × 1010 Bq(D) 11 Bq per litre

40. When the current flowing througha conductor suddenly breaks,then—(A) No current is induced in the

conductor(B) A current is induced in the

direction of primary current

(C) A current opposite to theprimary current is induced

(D) None of the above

41. For a common base transistor ifthe values of emitter current andcollector current are respectively103 μA and 0·96 mA, then thevalue of base current will be—

(A) 0·04 mA (B) 4 mA(C) 0·4 mA (D) 0·004 mA

42. The average power consumedon connecting a 200 volt 50 hertzA.C. source on 50 ohm resis-tance will be—

(A) 0 watt (B) 200 watt

(C) 400 watt (D) 800 watt

43. To form a p-type semiconductorthe germanium crystal must bedoped with an impurity ofvalency—

(A) 6 (B) 5

(C) 4 (D) 3

44. In state of saturation the dynamicresistance of diode is—

(A) Zero(B) 103 kilo ohm

(C) In between 1 kΩ at 10 kΩ

(D) Infinite

45. A jocker can throw balls upto amaximum height of 20m. Hethrows up four balls in succes-sion with the same initial velocityof 20 m/s and keeps them in air.While one ball is passing throughhis hand, find the position ofother three balls from his hand—

(g = 10 m/s2)(A) 10 m, 15 m, 20 m(B) 20 m, 15 m, 10 m(C) 15 m, 15 m, 20 m(D) 15 m, 20 m, 15 m

46. When a tuning fork of frequency341 Hz is vibrated with anothertuning fork of unknown frequency,6 beats per second are heard.When the tuning fork of unknownfrequency is waxed, the numberof beats heard per secondbecomes 2. What is the funda-mental frequency of the secondtuning fork ?(A) 335 Hz (B) 339 Hz(C) 343 Hz (D) 347 Hz

47. Two parallel wires are carryingelectric currents of equal magni-tude and in the same direction.They exert—(A) An attractive force on each

other(B) A repulsive force on each

other

(C) No force on each other

(D) A rotational torque on eachother

48. An astronomical telescope has anobjective of focal length 125 cmand an eyepiece of focal length25 cm. The diameter of theobjective is 4 cm. Find the dia-meter of the image—(A) 5 cm (B) 50 cm(C) 20 cm (D) 0·8 cm

49. The temperature at and abovewhich a ferromagnetic materialbecomes paramagnetic iscalled—(A) Critical temperature(B) Temperature of inversion(C) Curie temperature(D) Debye temperature

50. What is the angle made byvector i – k with y-axis ?(A) 30° (B) 45°

(C) 60° (D) 90°

ANSWERS WITH HINTS

C.S.V. / October / 2009 / 962

1. The displacement in the n thsecond, of a uniformly accele-rated motion is given by

sn = u + a2 (2n – 1)

This equation is dimensionally—(A) Correct(B) Not correct(C) Can be made correct by

multiplying the right handside of the equation by n

(D) Can be made correct bydividing the left hand side ofthe equation by n

2. An inductor L has resistance R. Itis connected to an alternatingvoltage source of variable frequ-ency. The current in the circuit—(A) Decreases with increase of

frequency(B) Does not change with

change in frequency(C) Varies linearly with fre-

quency(D) It is inversely proportional to

the square of the frequency

3. A particle thrown upwardsreturns to the earth after 4s. Att = 3 second it is at a height of—

(A) 14·7 m (B) 19·6 m

(C) 29·4 m (D) 39·2 m

4. The moment of inertia of acollapsing star changes to one-third of its initial value. The ratioof the new rotational kineticenergy to the initial rotationalkinetic energy is—(A) 3 : 1 (B) 1 : 3(C) 9 : 1 (D) 1 : 9

5. Two springs A and B of springconstants k1 and k2 are hangingfrom a ceiling in series with amass m . The effective springconstant of the combined systemof two springs described willbe—(A) k1 + k2

(B) k1 – k2

(C) k2 – k1 > 1(D) k1 k2/k1 + k2

6. The vertical escape velocity of abody from earth’s surface is 11·2km/sec. If the body is projectedat an angle of 45° from the verti-cal, its escape velocity will be—

(A) 11·2 × 2 km/s

(B)11·2

2 km/s

(C) 11·2 × 2 km/s

(D) 11·2 km/s

7. A stone of mass m 1 moving witha uniform speed v suddenly exp-lodes into two fragments. If thefragment of mass m 2 is at rest,the speed of the other fragmentis—(A) m 1v/(m 1 – m 2)(B) m 2v/(m 1 – m 2)(C) m 1v (m 1 + m 2)(D) m 1/m 2 ·v

8. The focal length of a convex lensis f. When it is divided in twoparts by a plane parallel to theprincipal axis, focal length ofeach part will be—

(A) f (B)f2

(C) 2f (D) Zero

9. Acceleration due to gravity varieswith height (h ) at which g i smeasured, as proportional to—(A) 1/h (B) 1/h

2

(C) g ·h (D) g /h 2

10. The maximum intensity in theinterference pattern of two equaland parallel slits is I. If one of theslits is closed, the intensity at thesame point is I0. Then—(A) I = I0(B) I = 2I0(C) I = 4I0(D) There is no relation between

I and I0

11. The semi major axes of the orbitsof Mercury and Mars are respec-tively 0·387 and 1·524 in astro-nomical units. If the period ofMercury is 0·241 year, what isthe period of Mars ?(A) 1·9 year (B) 9·1 year(C) 19 year (D) 91 year

12. The current amplification ofcommon base N-P-N transistor is0·96. What will be the currentgain if it is used as commonemitter amplifier ?(A) 16 (B) 24(C) 20 (D) 32

13. A bottle weighing 200 gm and ofarea of cross section 50 cm2 andheight 4 cm oscillates on thesurface of water in verticle posi-tion. Its frequency of oscillation is(Hz)—(A) 1·5 Hz (B) 2·5 Hz(C) 3·5 Hz (D) 4·5 Hz

14. Amplification factor of a triode is20 and its plate resistance is 20kΩ. Its mutual conductance willbe—

(A) 2 × 105 mho

(B) 2 × 104 mho

(C) 500 mho

(D) 2 × 10–3 mho

15. A person measure the time periodof simple pendulum inside astationary lift and finds it to be T.If the lift starts accelerating up-wards with an acceleration 2gthe time period of pendulum willbe—

(A) 3T (B) 3/2 T

(C) T/ 3 (D) T/3

16. If p is the pressure of a gas and ρis its density, then dimension ofvelocity is given by—

(A) p1/2 ρ–1/2 (B) p1/2 ρ1/2

(C) p–1/2 ρ1/2 (D) p–1/2 ρ–1/2

17. By leaving the door of a smallstandard domestic refrigeratoropen it is not possible to cool aroom, because it violates the—(A) First law of thermodynamics(B) Second law of thermodyna-

mic(C) Law of conservation of mo-

mentum(D) Law of conservation of

energy

18. If the horizontal range of a pro-jectile is equal to the maximumheight reached, then the corres-ponding angle of projection is—

(A) tan–1 1 (B) tan–1 3

(C) tan–1 4 (D) tan–1 12

C.S.V. / October / 2009 / 963

19. Speed of sound in mercury at acertain temperature is 1450 m/s.Given the density of mercury as13·6 × 103 kg/m3. Bulk modulusfor mercury is—(A) 2·86 × 1010 N/m2

(B) 3·86 × 1010 N/m2

(C) 4·86 × 1010 N/m2

(D) 5·86 × 1010 N/m2

20. The acceleration of a particleperforming S.H.M. is 12 cm/s2 ata displacement of 3 cm from themean position. Its time periodis—(A) 6·28 s (B) 3·14 s(C) 10·0 s (D) 5·0 s

21. Sound travelling at 340 m/senters water where the speed ofsound becomes 1480 m/s. Criticalangle for total reflection is—(A) 10·3° (B) 13·3°

(C) 86·7° (D) 89·7°

22. An electron of mass 9 × 10–31 kgrevolves in a circle of radius

0·53 A° around the nucleus ofhydrogen atom with a velocity of2·2 × 106 ms–1. What is the ang-ular momentum of the electron ?

(A)h2π (B)

2h3π

(C)hπ (D)

h2π

23. In Doppler effect, if the sourcemoves towards the observer, thespectral line is shifted towardsthe—(A) Violet end of the spectrum(B) Green end of the spectrum(C) Red end of the spectrum(D) Blue end of the spectrum

24. According to Rutherford model ofatom the atom consists of—(A) Positively charged nucleus

surrounded by a cloud ofnegative charge

(B) Electrons orbiting a posi-tively charged nucleus indefinite orbits

(C) Same as (B) with electronsspinning

(D) A rigid sphere only

25. In minimum deviation conditionsa light ray passing through anequilateral prism travels—(A) Parallel to the base (non-

refracting side) of the prism(B) Perpendicular to the base

(C) Perpendicular to the firstrefracting surface

(D) Perpendicular to the secondrefracting surface

26. Along with β-particle emissionfrom a radioactive nucleus onemore particle with zero charge isemitted to conserve the energyand momentum. This particle iscalled—(A) Meson (B) Positron(C) Antineutrino (D) Neutron

27. On a rainy day, if there is an oildrop on tar road, coloured ringsare seen around this drop. Thisis because of—(A) Interface pattern produced

due to thin oil film(B) Diffraction pattern(C) Polarization(D) Total internal reflection of

light

28. A doubly ionised lithium atom ishydrogen like with atomicnumber Z = 3. The wavelength ofradiation required to excite theelectron in Li2+ from first to thirdBohr orbit will be—(Ionisation energy of hydrogenatom is 13·6 eV)

(A) 72·53 A° (B) 113·74 A°

(C) 212·52 A° (D) 17·72 A°

29. Conductivity is the reciprocalof—(A) Drift velocity (B) Resistivity(C) Inductance (D) Permittivity

30. A current carrying coil is freelysuspended in a uniform magneticfield. The coil tends to set itsplane—(A) Parallel to the magnetic field(B) Perpendicular to the magne-

tic field(C) Inclined to the magnetic field(D) Continuously rotating

31. The magnetic field due to a verylong wire carrying current variesaccording to—(A) Square of the distance from

the wire(B) Inverse of the distance from

wire(C) Square root of the distance

from wire(D) Linearly as the distance from

wire

32. The figure shows the viewthrough the eyepiece of a prismspectrometer with its slit illumina-ted by a source of light emittingwavelengths corresponding toyellow (Y), green (G) and anunknown colour (X). The colourX may be—

X G Y

(A) Red (B) Orange(C) Pink (D) None of these

33. The induced electromagneticfield in a coil is proportional to—(A) Magnetic flux through the

coil(B) Area of the coil(C) Rate of change of magnetic

flux through the coil(D) Product of magnetic flux and

area of the coil

34. SI unit of Stefan’s constant is—(A) Nm–2K–4 (B) Jm–1K–4

(C) Jm–2K–4 (D) Wm–2K–4

35. The hard ferromagnetic materialis characterised by—(A) Narrow hysterisis loop(B) Fat hysterisis loop(C) High mechanical hardness,

all over(D) Mechanically hard surface

36. The radioactive constant ofradium is 4·28 × 10–4 per year,its half period is approximately—

(A) 2000 year (B) 1240 year

(C) 1620 year (D) 2440 year

37. The total energy of the electron inthe hydrogen atom in the groundstate is –13·6 eV. The kineticenergy of this electron is—

(A) – 13·6 eV (B) 0

(C) 6·8 eV (D) 13·6 eV

38. Energy equivalent to 1 kg ofmatter is about—

(A) 1011 joule (B) 1016 joule(C) 1017 joule (D) 1020 joule

39. N0 is the number of radioactiveatoms at any instant and N thenumber of the radioactive atomsremaining undecayed after time

C.S.V. / October / 2009 / 964

‘t’. The graph drawn with loge N,where e is the base of naturallogarithm along Y-axis and ‘t ’along the X-axis will be a straightline with slope—

(A) λ (B) –λ

(C) 1/λ (D) –1/λ

40. If the elements with principalquantum number n > 4 were notexhisted in nature, the number ofpossible element would be—(A) 60 (B) 32(C) 4 (D) 64

41. Express 1 BeV in joule—

(A) 1·6 × 1010 J

(B) 1·6 × 10–9 J

(C) 1·6 × 10–10 J

(D) None of these

42. An electron moving with uniformvelocity enters a uniform electricfield perpendicular to its directionof motion. The path of the elec-tron will be—(A) Circular (B) Parabolic(C) Straight line (D) Helical

43. In Thomson’s method of determi-ning e/m of cathode rays, mag-netic field (B) and electric field(E) are parallel, a parabola is notobtained on the screen. Thereason is—(A) Cathode rays consist of

electrons which carry nega-tive charge

(B) Electron is a very lightparticle

(C) There is no velocity distribu-tion in cathode rays

(D) Parallel fields B and E donot interact with electrons

44. A body is projected verticallyupward from point A, the top of atower. It reaches the ground in t 1sec. If it is projected verticallydownwards from A with the samevelocity, it reaches the ground int 2 sec. If it falls freely from A, itwould reach the ground in—

(A)t 1 + t 2

2 sec (B)t 1 – t 2

2 sec

(C) t 1 t 2 sec (D) t 1t 2 sec

45. The acceleration of a particleperforming S.H.M. is 12 cm/sec2

at a distance of 3 cm from themean position. Its time period is—

(A) 2 sec (B) 4 sec

(C) 1·54 sec (D) 3·14 sec

46. Which is the correct relation bet-ween inter-atomic force-constant,Young’s modulus and the normaldistance a0 between the atoms ofa wire ?

(A) Y= k × r 0 (B) k = Y × r0

(C) k = Y × r02 (D) k = Y × r0

47. To decrease the magnifyingpower of an astronomical tele-scope—

(A) Focal length of objectiveshould be increased

(B) Focal length of eyepieceshould be decreased

(C) Focal length of eyepieceshould be increased

(D) Focal length of objectiveshould be increased andthat of eyepiece should bedecreased

48. The aperture of the objective of atelescope is 0·1 m and wave

length of light is 6000 A° . Theresolving limit of the telescopewill approximately be—(A) 6 × 10–5 rad(B) 6 × 10–4 rad(C) 6 × 10–3 rad(D) 6 × 10–6 rad

49. A spot light S rotates in ahorizontal plane with constantangular velocity of 0·1 rad/sec.The spot of light P moves alongthe wall at a distance of 3 m fromS. The velocity of the spot P,where θ = 45° is—

S

AP θ

˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙

(A) 0·3 m/s (B) 3·0 m/s(C) 0·6 m/s (D) 6·0 m/s

50. A stationary wave y = 2a sin kxcos ωt in a closed organ pipe isthe result of the superposition ofy = a sin (ωt – kx) and—(A) a sin (ωt + kx)(B) – a sin (ωt – kx)(C) – a sin (ωt + kx)(D) a sin (ωt + kx)

ANSWERS WITH HINTS

C.S.V. / October / 2009 / 967

Introduction

The word alcohol is derived from the Arabic Kuhl(also Kohl or Kohol). It originally was used to mean a ‘veryfine powder’ but gradually came to cannote ‘essence’.Later the term was applied to wine spirits, which werereferred to as alcool vini, and eventually simply alcohol.The compounds derived from hydrocarbons by replace-ment of one or more hydrogens by hydroxyl groups areknown as organic hydroxy compounds. The parent hydro-carbons may be acyclic aliphatic saturated or unsaturated.The hydroxyl compounds which are derived from acyclicand alicyclic hydrocarbons are called alcohols. Thoseobtained by replacement of nuclear aromatic hydrogensare called Phenols. Since phenols differ from alcohols inmany respects, they are treated separately from alcohols.Aryl substituted alcohols (C6H5CH2OH) are also known asaromatic alcohols.

General Methods of Preparation

1. Hydrolysis of halides—Alkyl halides, when boiledwith aqueous solution of an alkali hydroxide, give alcoholsthrough nucleophilic substitution mechanism.

R—X + KOH ⎯→ R—OH + KXThis general procedure produces primary and secon-

dary alcohols. Glycerol can be synthesized from propy-lene by a series of reactions including the hydrolysis of ahalide as one step in the process.● Unsaturated alcohols can be prepared by high tem-

perature chlorination followed by hydrolysis of interme-diate halide of an alkene. An example is the productionof alkyl alcohol from propylene.

CH2 —— CHCH3 + Cl2 ⎯⎯→ CH2——CHCH2Cl + HCl

Propylene Allyl chloride

CH2 —— CHCH2Cl + NaOH ⎯⎯→CH2 —— CHCH2OH + NaClAllyl alcohol

Note : There is a serious limitation of this method.Elimination is a serious competing reactionspecially with sec. and tert-halides. However, ifa weaker nucleophile like silver acetate(AgOAc) replaces alkali hydroxide, betterresults are obtained even with tert. halides.For example :(CH3)3CCl + AgOCOCH3 ⎯⎯→tert-butyl chloride

(CH3)3C—O—COCH3 + AgCltert-butyl acetate

(CH3)3—C—O—COCH3 H2O

⎯⎯→(CH3)3C—OH + CH3COOHtert-butyl alcohol

● As alkyl halides themselves are prepared from alco-hols, this method will be of interest only when alkylhalides are readily available and are very cheap.

2. Hydration of alkenes—Direct hydration takesplace by adding water in presence of catalyst.

CH3—

CH3

|C——CH2 + H2O

H⊕⎯⎯→

2-methylpropene

(CH3)3C—O⊕

H2 –H⊕

⎯⎯→ (CH3)3—C—OH2-methylpropane-2-ol

● Indirect hydration is achieved by addition of sulphuricacid to alkene followed by hydrolysis of the alkylhydrogen sulphate.

CH2——CH2 + H2SO4 ⎯→Ethene

CH3CH2OSO2OH H2O

⎯⎯→Δ

CH3CH2OH

EthanolCH3CH——CH2 + H2SO4 ⎯→

Propene

CH3CH(OSO2OH)CH3 H2O

⎯⎯→Δ

CH3CH(OH)CH3

2-propanol

3. Hydroformylation of alkenes—Lower molecularweight olefins react with carbon monoxide and hydrogenin the presence of a catalyst in a reaction called hydro-formylation or the oxo reaction.

C —— C + CO + H2 Cobalt

⎯⎯→ CH —|C|

—CO H

Alkene Aldehyde

The resulting aldehyde is, subsequently, hydrogenatedto form an alcohol.

CH—|C|

—CO H

+ H2 ⎯⎯→ CH —|C|

—CH2OH

Alcohol

4. Hydroboration-oxidation of alkenes—Alkeneswhen treated with diborane (BH3)2, give alkylboranes,R3B. Alkylboranes on oxidation with alkaline hydrogenperoxide give alcohols.

6CH2 —— CH2 (BH3)2⎯⎯⎯→ 2(CH3CH2)3B

H2O2/OH–⎯⎯⎯⎯→

Ethene Triethyl borane

6 CH3CH2OH

Ethanol

6CH3CH —— CH2 (BH3)2 ⎯⎯→ 2(CH3CH2CH2)3B

Propene Tri-n-propylborane

H2O2/OH–⎯⎯⎯→ 6CH3CH2CH2OH

1-propanol

Note : It is significant to note that this methodalways leads to the anti-Markovnikov’saddition of water to alkenes.

C.S.V. / October / 2009 / 968

5. Grignard synthesis—All the three types of alco-hols (primary, secondary, tertiary) can be prepared fromthe Grignard reagents by interaction with suitable carbonylcompounds.

R—MgX + R'R" C —— O →

Grignard Ketonereagent

R'RR"

C—OMgX H2O⎯→

H+ R'RR"

C—OH + Mg(OH)X

Alcohol

● The reaction of Grignard reagent with formaldehydeleads to primary alcohols, that with aldehydes otherthan formaldehyde yield secondary alcohols and thatwith ketones give tertiary alcohols.

● Mechanism of above reaction is illustrated as :

(i) R—:: MgX → R

– : + M

+gX

(ii) R : + |C|

O—— → R—|C|

—–O

+MgX

+ M+

gX

(iii) R–|C|

—–O—

+MgX + HOH → R—

|C|—OH + Mg(OH)X

(iv) Mg(OH)X + H+ → Mg2+ + X– + H2O

6. Reduction of carbonyl compounds—Carbonylcompounds i.e., aldehydes and ketones etc. give alcoholson reduction. This reaction can be effected by :

(i) Catalytic hydrogenation

(ii) The use of metal-solvent combination such assodium or potassium in alcohol.

(iii) The use of complex metal hydrides.

For example :

● CH3—

O

||C—CH3

2-propanone

+ H2 PtO2⎯⎯→ CH3

OH

|CH—CH3

2-propanol

● CH3CHOEthanal

+ H2 Ni

⎯⎯⎯→or LiAlH4

CH3CH2OHEthanol

●

CH3 CH3

C —— O + H2

Ni⎯⎯⎯⎯→

or LiAlH4 CH3

OH|CH—CH3

Propanone 2-propanol

● CH3—

O

||C—CH2CH2CH3

Pentanone-2

+ H2

Na⎯⎯→

EtOH

CH3CHOHCH2CH2CH32-pentanol

● C2H5COOC2H5Ethyl propanoate

+ 2H2

Na⎯⎯→

EtOH C2H5CH2OH

1-propanol + C2H5OH

Ethanol

Note : Aldehydes and ketones, in presence ofmetals like magnesium undergo bimolecularreduction to form symmetrical glycols(Pinacols). For example :

CH3—

CH3

|C ||O

+

CH3

|C—CH3

||O

Mg

⎯→ CH3—

CH3

|C—–

|O

CH3

|C—C

|O

H3

Propanone Mg

H2O⎯⎯→ CH3—

CH3

|C—–

|OH

—

CH3

|C—C

|OH

H3

2, 3-dimethyl butane2, 3-diol (Pinacol)

More about the Preparation of Alcohols

● Oxymercuration—Demercuration of alkenes leadsto the formation of alcohols. It involves the reaction ofalkene with mercuric acetate in presence of tetra-hydrofuran (THF). This step is known as Oxymer-curation as :

R2—

R

|C ==

R3

|C —R4

Hg(OCOCH3)2⎯⎯⎯⎯⎯→THF-H2O

R2—

R1

|C—–

|OH

R3

|C—R4

|HgOCOCH3

This is followed by the reduction of intermediatehydroxy mercuryl compounds known as demercura-tion.

R2—

R1

|C—–

|OH

R3

|C—R4

|HgOCOCH3

NaBH4⎯⎯→ R2—

R1

|C—–

|OH

R3

|C—R4

|H

Hydroxy mercuryl Alcoholcompound

Alcohols obtained are those which will be formed bythe Markovnikov’s addition of H2O to carbon-carbondouble bond.

● Grignard reagents also react with ethylene oxide toform primary alcohols containing two carbon atomsmore than the Grignard reagent.

CH3MgBr +Grignard reagent

CH 2—CH 2

O

H2O/H+⎯⎯→ CH3CH2CH2OH

Propanol-1Ethylene oxide

Mechanism of this reaction can be illustrated asbelow :

CH3—MgX ⎯⎯→ ··CH3 +

⊕MgX

: CH3 + CH 2 —CH 2

O⎯⎯→ CH3CH2CH2 O⎯

⎯→ H+

CH3CH2CH2OHPropanol-1

Note : This reaction sequence is useful in ascendingthe series of organic compounds.

C.S.V. / October / 2009 / 969

● Hydroxylation of alkenes leads to the formation of1, 2-glycols. For example :

OH

|CH2—

OH

|CH2

Ethane-1‚ 2-diol

KMnO4←⎯⎯⎯ CH2 —— CH2

Ethene

C6H5CO2OH⎯⎯⎯⎯→

H2O/H+

OH

|CH2—

OH

|CH2

Ethane-1‚ 2-diolNote : KMnO4 leads to syn-hydroxylation or cis-

hydroxylation and per acids to anti-hydroxy-lation or trans-hydroxylation of alkenes,where structure permits.

● All esters, except that of formic acid yield tertiaryalcohols on reaction with Grignard reagent followedby acidic hydrolysis. The esters of formic acid givesecondary alcohols.

O

||CH3 — C—OC2 Ethyl ethanoate

H5 CH3MgBr

⎯⎯⎯⎯→

⎣⎢⎢⎡

⎦⎥⎥⎤

CH3—

OMgBr

|C—OC2H5

|CH3

– C2H5OMgBr⎯⎯⎯⎯→ CH3—

O

||C—CH3

CH3MgBr⎯⎯⎯⎯→

CH3—

OMgBr

|C—CH3

|CH3

H2O/H+

⎯⎯⎯→ CH3—

OH

|C—C

|CH3

H3 + MgBr OH

General Physical Properties of AlcoholsThe lower members of alcohols are colourless, vola-

tile liquids with a characteristic alcoholic smell and burningtaste, whereas higher alcohols are odourless and taste-less. Alcohols having 12 or more than 12 carbon atomsare solids. Branched chain alcohols with much fewer car-bon atoms are, however, solids.

Boiling points—Boiling points of alcohols are muchhigher than those of alkanes, haloalkanes or ethers ofcomparable molecular mass. This is due to the inter-molecular hydrogen bonding in alcohols as :

R|O

HO|R

H

R|O

H HO|R

- - - - -

- -

- - - -- - - -- - -

-- - - - -

- -

- - - -

● For isomeric alcohols, the boiling points follow theorder as : primary alcohol > secondary alcohol >tertiary alcohol.

● Lower alcohols are found to form solid derivativeswith CaCl2 and MgCl2.

CaCl2 + 4CH3OH ⎯→ CaCl2·4CH3OHMgCl2 + 6C2H5OH ⎯→ MgCl2·6C2H5OH

It is because of this reason that alcohols cannot bedried with anhydrous CaCl2 and MgCl2.

● Alcohols are known to have intoxicating effects.Methanol is poisonous and is not good for drinkingpurposes. It causes blindness. Ethanol on the otherhand is used for drinking purposes.

● Additional hydroxyl groups in an alcohol enhances itssweetness. For example ethanol is not sweet; pro-pylene glycol, C3H6(OH)2 is slightly sweet; glycerol,C3H5(OH)3 is quite sweet; and mannitol C6H8(OH)6is so sweet that it is known as sweet alcohol.

Chemical Properties of AlcoholsIn alcohols –OH group is functional group, therefore,

the chemical properties of alcohols generally involve thereactions of –OH group. These undergo substitution andelimination reactions. The chemical reactivity of alcoholsalso depends upon the carbon chain attached to –OHgroup. The reactions of alcohols can be classified intothree types :

(i) Reactions involving cleavage of O—H bond(ii) Reactions involving cleavage of C—OH bond(iii) Reactions involving both the alkyl and hydroxyl

groups.

Reactions Involving Cleavage of O :——

: H

Bond(a) Acidic nature of alcohols—Since the oxygen

attached to hydrogen in alcohols is highly electronegative,it facilitates the separation of the relatively positive hydro-gen as H+. In other words alcohols behave as acids as isevident from the following reactions :● Reactions with metals—When treated with metals like

Na and K these liberate hydrogen with the formation ofalkoxides.

2CH3CH2O:

—:

H + 2Na ⎯→ 2CH3CH2 –O Na+ + H2 ↑

Ethanol Sod. ethoxide

2 (CH3)3CO :

—:

H + 2K ⎯→ 2 [(CH3)3 C–O] K+ + H2 ↑

tert-butyl alcohol Pot. tert-butoxide

6 (CH3)2CHO :

—:

H + 2Al ⎯→Isopropyl alcohol

2[(CH3)2CH–O]3 Al3+ + 3H2 ↑

Aluminium isopropoxide● Reactions with metal hydrides—Formation of

alkoxide with evolution of hydrogen takes place.

R—O :

—:

H + MH ⎯→ R—–O

+M + H2 ↑

● Alcohols are very weak acids (Ka = 10–16 — 10–18)even feeble than water (Ka = 1·78 × 10–16). They donot turn blue litmus red. Thus alcohols are weakeracids than water but stronger than acetylene as isevident from the following reactions :

R –O N

+a + H — OH ⎯→ N

+a

–OH + RO — H

Strongerbase

Strongeracid

Weakerbase

Weakeracid

HC ≡ –C Na+ + ROH ⎯→ RO– Na+ + HC ≡ CH

Strongerbase

Strongeracid

Weakerbase

Weakeracid

● Thus the decreasing order of acid strength isH2O > ROH > HC ≡ CH

● The decreasing order of basic strength of the corres-ponding anions is as :

HC ≡ –C : > R

–O : > H

–O:

C.S.V. / October / 2009 / 970

● Relative acid strength—Acid strength of substancedepends on how well the resulting anion can accommo-date the negative charge. An alkyl group being electronreleasing, intensifies the negative charge on alkoxideion and consequently the anion is rendered less stable.Thus electron releasing inductive effect of alkylgroups makes alcohols weaker acid than water.

R →⎯ |C|

— OH R →⎯ |C|

—–O + H+

This inductive effect will be greatest for tertiaryalcohols, less for secondary, still less for primary and leastfor methanol. The decreasing order of acid strength ofalcohols is as :

CH3OH > Primary > Secondary > Tertiary alcohols.

R →⎯

H|C|H

→⎯ OH > R →⎯

R

C→⎯OH|H

> R →⎯

R

C→⎯ OH

RPrimaryalcohol

Secondaryalcohol

Tertiaryalcohol

(b) Reaction with Grignard reagent—Alcoholsreact with Grignard reagents to form alkanes. In thesereactions the alkane is obtained from alkyl part of Grig-nard reagent.

RO:

—:

H + CH3:

—:

MgX ⎯→ CH4 ↑ + Mg X OR

Strongeracid

Weakeracid

● This reaction can be considered as displacement ofweaker acid (R – H) from its salt (Grignard reagent) bystronger acid alcohol (R – OH).

● This reaction makes basis of the Zerevitinov methodfor the estimation of the number of –OH groups in anunknown compound. The volume of evolved methaneis measured.(c) Reaction with carboxylic acids—In presence of

an acid (H2SO4 or HCl gas) ester is formed. This reactionis known as esterification.

CH3

O||C

:—:

OH + H:

—:

OC2H5

H+

CH3

O||COC2H5 + H2O

Ethanoic acid Ethanol Ethyl ethanoate

● Esterification is a reversible reaction. It can be pushedforward by using any of the reactants in large excessor by removing any of the products as soon as it isformed.

● This reaction shows considerable steric hindrance.The bulkier the acid or alcohol, the slower the rate ofesterification. For example :CH3OH > CH3CH2OH > (CH3)2 CHOH > (CH3)3C–OHHCOOH > CH3COOH > (CH3)2 CHCOOH > (CH3)3–

CCOOH● Isotopic tracer technique shows that the esterification

involves the cleavage of the O—H bond of alcohol andC—OH bond of acid :

R—

O||C—OH + H— O* R'

H+

R—

O||C—

*OR' + H2O

*O is the radio isotope of oxygen.

(d) Reaction with acid halides or acid anhy-drides—When treated with acid chloride or acid anhy-dride in the presence of bases like pyridine or dimethylaniline (as catalyst) alcohols form esters. This reaction iscalled acylation.

R—

O||C—Cl + H — OR'

Base⎯→ R—

O||C—OR' + HCl

Acid chloride Alcohol Ester

R—

O||C—O—

O||C—R + H—OR' ⎯⎯→ R—

O||C—OR'

Acid anhydride Alcohol Ester

+ R—

O||C—OH

Acid

Reactions Involving Cleavage of C :—: OHBond :

(a) Reactions with hydrogen halides—The hydro-gen halides react with alcohols to form alkyl halides.Various reactions are summarised below :

R—OH

HCl/ZnR—Cl + H2O

48% HBr

or NaBr + Conc. H2 SO4R—Br + H2O

58% HI

or KI + H3PO4R—I + H2O

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

● The order of reactivity of various alcohols is as :

tertiary > secondary > primary alcohols.● The order of reacting hydrogen halides is as :

HI > HBr > HCl.

A Closer Look● Secondary and tertiary alcohols react with hydrogen halide

through SN 1 reaction mechanism as :

(a) R—OH + HX RO⊕H2 + : X

(b) R—⊕OH2 R

⊕ + H2O

(c) R⊕

+ : X → R—X

● In step (a) : The alcohol accepts hydrogen ion to formprotonated alcohol which dissociates into carbocation andwater in step (b). The carbocation then combines withhalide ion to form alkyl halide. In this sort of mechanismalkyl group of alcohol may undergo rearrangement due torearrangement in the intermediate carbocation.

CH3—

CH3|C——|H

H|C——|OH

CH3 HCl⎯→ CH3—

CH3|C—|Cl

H|C—|H

CH3

● Primary alcohols react through SN 2 mechanism as :

(a) R—OH + HX R—⊕OH2 + :X

(b) R—⊕OH2 + :X ⎯→ [ ]δ

–

X - - - R - - - δ +

O H 2

⎯→ X—R + H2O

C.S.V. / October / 2009 / 971

(b) Reactions with phosphorus halides—Alkyl hali-des are formed as :

R–OH + PCl5 ⎯→ R–Cl + POCl3 + HCl ↑Alcohol Alkyl halide

C2H5OH + PCl5 ⎯→ C2H5Cl + POCl3 + HCl ↑Ethanol Chloro ethane

3ROH + PX3 ⎯→ 3RX + H3PO3

3C2H5OH + PCl3 ⎯→ 3C2H5Cl + H3PO3● PBr3 and PI3 are generally prepared in situ by reaction

between phosphorus bromine and iodine respectively.

3C2H5OH + PI3 (P4 + I2)

⎯⎯⎯→ 3C2H5I + H3PO3.

(c) Reaction with thionyl chloride (SOCl2)—Alco-hols react with SOCl2 in the presence of pyridine to formchloroalkanes.

R–OH + SOCl2 Pyridine⎯⎯→ RCl + SO2 ↑ + HCl ↑

C2H5OH + SOCl2 Pyridine⎯⎯→ C2H5Cl + SO2 ↑ + HCl ↑

Ethanol ThionylChloride Chloroethane

● The order of reactivity of various alcohols towards thistype of reaction is :

tertiary > secondary > primary alcohols.● This can be explained in terms of electron releasing

inductive effect of alkyl groups. The alkyl groups bytheir electron releasing effect tend to increase theelectron displacement towards oxygen.

R→⎯

R

C→⎯OH

R

> R→⎯

R

C→⎯OH|H

> R→⎯

H|C|H

→⎯OH

● In other words, the polarity of C—O bond increasesand this makes the breaking of the bond betweencarbon and oxygen easier. Therefore, the alcohols withgreater number of alkyl groups attached to carbon willbe more effective. This justifies the above order ofreactivity of alcohols in this type of reactions.

Reactions Involving both Alkyl as well asHydroxyl Groups :

The important reactions of this type are :(a) Acidic Dehydration, (b) Oxidation, (c) Dehydro-

genation(a) Acidic dehydration—When heated with concen-

trated H2SO4, phosphoric acid or boric acid, alcoholsundergo dehydration to form alkenes. The reaction withconcentrated H2SO4 is carried out at 443 K, whereasphosphoric acid and boric acid react at higher tempera-ture.

H—

H|C|H

—

OH|C—H|H

˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙

˙˙˙˙

conc. H2SO4

⎯⎯⎯⎯⎯→443 K

H—C|H

——C|H

—H + H2O

Ethanol Ethene

H—

H|C|H

—

H|C|H

—

OH|C—H|H

˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙˙

˙˙˙˙

conc. H2SO4⎯⎯⎯⎯⎯→

443 K CH3—CH——CH2 +

Propene H2O

1-propanol

A Closer Look

● The mechanism of dehydration of alcohol is as :

(a) H—

H|C|H

—

H|C|H

—OH Protonation

⎯⎯⎯⎯→+ H+

H—

H|C|H

—

H|C|H

—⊕OH2

Oxonium

(b) H—

H|C|H

—

H|C|H

—⊕OH2

Loss of H2O⎯⎯⎯⎯→ H—

H|C|H

—

H|C|H

⊕ + H2O

(c) H—

H|C|H

—

H|C|H

⊕ Loss of H+

⎯⎯⎯⎯→–H+

H—C

|H

——C|H

—H

● The relative ease of dehydration of alcohols—Step (b)involving the formation of carbocation is slowest and rate-determining step. The more easily a carbocation is formed,the more easily that alcohol would dehydrate. Further, thegreater the stability of a carbocation, the greater the easewith which it would be formed. Since the decreasing orderof stability of carbocations is

Tertiary > Secondary > Primarythe decreasing order of dehydration of alcohols should be :

Tertiary > Secondary > Primary alcohols.● Formation of unexpected products—Sometimes the

alkenes containing double bonds at position different fromthose anticipated from the original position of the –OHgroup are formed in predominating amounts. For example.

CH3CH2CH ——CH21-butene (20%)

(expected)

H+

←⎯⎯–H2O

CH3CH2CH2CH2OH ⎯⎯

⎯→

H – H O2

CH3CH = CH CH32-butene (80%)(unexpected)

Such migration of double bond is in agreement with themechanism proposed for dehydration of alcohols. Thecarbocations always rearrange whenever, possible, to astabler carbocation by 1, 2-hydride or alkyl shifts. Theprimary, n-butyl cation will, therefore, rearrange to a morestable secondary butyl cation in above example and 2-butene rather than 1-butene is major product.

CH3CH2CH—CH2

H

⊕

1‚ 2 hydride⎯⎯⎯→

Shift CH3CH2

⊕CHCH3

sec-butyl cation(more stable)

n-butyl cation (less stable)

● Dehydration of ethanol under different experimentalconditions gives different products.At 383 K, ethyl hydrogen sulphate is obtained.

C2H5OH + H2SO4 383 K

⎯⎯→ C2H5HSO4 + H2ODistillation under reduced pressure gives diethyl sul-phate.2C2H5OH + H2SO4

Heat⎯⎯→ (C2H5)2SO4 + 2H2O

At 413 K, if alcohol is used in excess, the loss of H2Omolecule takes place from two different molecules ofalcohol and ether is obtained as the product.

2C2H5OH Conc. H2SO4⎯⎯⎯⎯→

413 K C2H5OC2H5 + H2O

C.S.V. / October / 2009 / 972

(b) Oxidation of alcohols—Alcohols undergo oxida-tion with oxidising agents (in neutral or acidic or alkalinemedium) such as chromium trioxide (CrO3), potassiumdichromate (K2Cr2O7), potassium permanganate (KMnO4)and nitric acid.● Primary alcohols—They are oxidised to carboxylic

acids by potassium permanganate.

RCH2OH + KMnO4 ⎯→ RCOOK + MnO2 ↓ + KOH

RCOOK + H+ ⎯→ RCOOH + K+

An acidic solution of potassium dichromate can oxi-dise them to aldehydes, provided the products are dis-tilled away as soon as they are formed. If the aldehydeformed continues to be available to the oxidant, carboxylicacids are formed ultimately.

RCH2OH + Cr2O72– ⎯→ R—

H|C——O

Cr2O72–⎯⎯→ R—

OH|C ⎯⎯ O

Thus, primary alcohols are oxidised to aldehydesand carboxylic acids containing the same number ofcarbon atoms as the original alcohol.

● Secondary alcohols—They are oxidised to ketonesby chromic acid.

K2Cr2O7 + H2SO4 or Cr2O3 + CH3COOH or CrO3 in(Pyridine)

R2CHOHA secondary

Cr2O72–⎯⎯→ R2C ⎯⎯ O

Ketonealcohol

Ketones resist the further oxidation, but under vigo-rous conditions they are oxidised to a mixture of carboxylicacids. For example.

CH3—

OH|CH—CH3 ⎯→ CH3COOH + HCOOH

Isopropyl alcohol Acetic acid Formic acid

Thus, the ketones, the first stage oxidation pro-ducts of sec. alcohols, contain same number of car-bon atoms as original alcohol, but carboxylic acidscontain fewer carbon atoms than the parent alcohol.● Tertiary alcohols—They are not oxidised under neutral

or alkaline conditions, but acidic oxidising agentsoxidise them, presumably, through the alkene formedunder the acidic conditions, to a mixture of aldehydes,ketones and acids. For example :

CH3—

CH3|C—OH |CH3

H+

⎯→ CH3—

CH3|C——CH2

2-methylpropane-2-ol 2-methylpropene⎯⎯

→ O

CH3—

CH3|C ⎯⎯ O + H—

O||C—H

Propanone Methanal

O O

⎯→

⎯→

CH3COOH + HCOOH HCOOH

● A variation of the above oxidation reaction is observedwhen vapours of an alcohol are passed over reducedcopper at high temperature. The primary, secondaryand tertiary alcohols give different products.

● Primary alcohols—They give aldehydes as :

R—

H|C|H

—OH Cu/575 K

⎯⎯⎯→ R—

H|C ⎯⎯ O + H2 ↑

● Secondary alcohols—They give ketones as :

R—

R|C|H

—OH Cu/575 K

⎯⎯⎯→ R—

R|C ⎯⎯ O + H2 ↑

● Tertiary alcohols—They do not undergo this type ofreaction due to absence of α-hydrogen. However, itgets dehydrated to form an alkene.

CH3—

CH3|C—OH|CH3

Cu/575 K

⎯⎯⎯→ CH3—

CH3|C ⎯⎯ CH2Isobutylene

+ H2O

tert-butyl alcoholSince this oxidation reaction literally involves loss of

hydrogen from alcohol, it is known as catalytic dehydro-genation. On the basis of products of oxidation, the dis-tinction between primary, secondary and tertiary alcoholscan be made.

More about the Distinction of Primary, Secondaryand Tertiary Alcohols

● Lucas test—On treating with Lucas reagent (a mixture ofconc. HCl and ZnCl2), alcohols give cloudy appearancedue to formation of alkyl chlorides. A tertiary alcoholreacts very fast, a secondary alcohol reacts within fiveminutes and a primary alcohol does not react appreciablyat ordinary temperature.

● Victor Meyer’s test—The alcohol is subjected to thereaction sequence given below and the colours obtainedare noted.

Primaryalcohol

Secondaryalcohol

Tertiaryalcohol

RCH2OH R2CHOH R3C–OH

P4 + I2

⎯→ P4 + I2

⎯→ P4 + I2

⎯→

RCH2I R2CHI R3CI

AgNO2

⎯→ AgNO2

⎯→ AgNO2

⎯→

RCH2NO2 R2CHNO2 R3C–NO2

HONO

⎯→ HONO

⎯→ HONO

⎯→

R—C—NO2||N–OH

R

R C—NO2 |NO

No reaction

⎯→ KOH

Nitrolic acid Pseudo nitrol

⎯→ KOH

⎯→ KOH

Red colour Blue colour Colourless

C.S.V. / October / 2009 / 973

Some Important Alcohols

(1) Methyl alcohol or methanol, CH3OH—It is manu-factured by following methods :

(i) From water gas—

C + H2O (steam) → CO + H214243

Water gas

CO + 2H2 CuO/ZnO/Cr2O3

⎯⎯⎯⎯⎯⎯⎯→300–400°C/200 atm

CH3OH

The crude methanol is fractionally distilled. Thismethod gives methanol of excellent purity and inexcellent yield.

(ii) From natural gas—Controlled air oxidation ofmarsh gas by passing its mixture with oxygen (9 : 1)through a copper tube at 200°C, under a pressure of100 atm. gives methanol.

CH4 + 12 O2

Cu tube ⎯⎯⎯⎯⎯→200°C/100 atm

CH3OH

(iii) From pyroligneous acid—This is also referred toas destructive distillation of wood.

Wood

Destructive Distillation at 400°°°°C

↓ ↓Volatile Gases Non-volatile residue

Passed into condenser (Wood charcoal)

↓ ↓Uncondensed gases Distillate

(Wood gases used as fuel) allowed to settle(CO, CH4, C2H4, C2H6, H2 etc.)

↓ ↓Upper aqueous layer Lower layer

Pyroligneous acid Wood Tar(Acetic acid, Methanol, Acetone) (A mixture of cresols,

10% 2-4% 0·5% used for preservation of timberunder the name of creosote)

The aqueous layer is distilled and the vapour is passedover milk of lime. Acetic acid is retained as calciumacetate, but methanol, acetone and other volatile com-pounds pass over as wood spirit. The wood spirit is thenfractionally distilled to give a low boiling fraction, a crudemethanol fraction and a higher boiling fraction containinga mixture of alcohols and ketones. The crude methanolfraction is then carefully refractionated to give puremethanol.

Purification—Methanol obtained from pyroligneousacid is treated with anhydrous CaCl2, when crystallinederivative, CaCl2·4CH3OH is formed, leaving acetoneunaffected. The crystals are filtered and boiled with waterwhen alcohol is regenerated. This is distilled. To removelast trace of water, the distillate is dried over quick lime.

In another method, impure methanol is treated withoxalic acid when solid methyl oxalate is obtained. This is

filtered off, boiled with calculated amount of caustic potashand distilled. Pure methyl alcohol distills over at 65°C. It isdried over quick lime.

COOH|COOH

+ 2CH3OH → COOCH3|COOCH3

2KOH⎯→

COOK|COOK

+ 2CH3OH

Oxalic acid Methyl oxalate

Uses of methyl alcohol—It has following importantuses :

(i) Methanol is an important source material for theproduction of formaldehyde. Formaldehyde is araw material for plastic industry.

(ii) Important chemicals like methylaniline, dimethyl-aniline, methyl chloride, dimethyl sulphate,methyl salicylate, terylene, polyvinyl alcohol etc.are manufactured from methanol.

(iii) Methanol is used for denaturing alcohol rende-ring it unfit for drinking.

(iv) Methanol finds application in antifreeze composi-tion for automobile and aeroplane radiators.

(v) Methanol finds important place as a solvent inmany industrial processes, in certain adhesivecompositions and in wood stains.

(vi) It is also used as component of motor spiritblends.

(2) Ethyl alcohol or ethanol, C2H5OH—It is manu-factured by following methods :

Fermentation of carbohydrates—Molasses andstarchy materials are two important raw materials for largescale preparation of ethyl alcohol. Fermentation is actuallydecomposition of organic compounds into simpler com-pounds through the agency of enzymes, the bio-catalysts.In commerce the ethyl alcohol is known as spirit of wineor grain alcohol.

(a) From molasses—Molasses, the mother liquor leftafter the crystallisation of canesugar from sugarcane juice,is diluted with water to reduce the concentration of sugarto about 10 per cent, sterilised by heating with live steamfor a short time and acidified with sulphuric acid to pH = 4.This checks the growth of any undesirable bacteria. Suit-able quantities of ammonium sulphate and ammoniumphosphate may be added which act as supplementaryfood for yeast. The liquid (wort) so obtained, is placed in alarge fermentation tank, maintained at the temperature ofabout 35°C. In the presence of yeast culture, fermentationstarts accompanied by the following reactions :

(i) The enzyme, invertase present in yeast convertssucrose into glucose and fructose.

C12H22O11Sucrose

+ H2O Invertase

⎯⎯⎯⎯→(Yeast)

C6H12O6Glucose

+ C6H12O6Fructose

(ii) The enzyme Zymase, further converts the glucoseand fructose into ethyl alcohol and carbon dioxide.

C6H12O6 Zymase

⎯⎯⎯→(Yeast)

2C2H5OH + 2CO2 ↑

C.S.V. / October / 2009 / 974

When the alcohol content of fermented liquor rises toabout 15%, the yeast cells are killed and the process offermentation stops. The liquor thus obtained is subjectedto fractional distillation to get following fractions :

First runnings—These consist of acetaldehyde andare used as an important source of acetaldehyde.

Rectified spirit or Industrial alcohol—This consistsof 93 – 95 per cent of ethyl alcohol.

Final runnings or Fusel oil—This is obtained bet-ween temperature range 125 – 140°C and is a mixture of :

n-propyl, n-butyl, isobutyl, n-amyl, isoamyl and activeamyl alcohols.

It must be noted that these alcohols are not producedby the fermentation of sugar but are formed by the actionof yeast on certain amino acids obtained from the proteinspresent in raw materials.

(b) From starch—The starchy materials like pota-toes, maize, barley, rice etc. are used. The production ofethanol from starchy materials can be outlined as follows :

(i) Malting—Barley moistened with water and spreadin a room in layers about five inches thick, is allowed togerminate in the dark at about 15°C. After suitable timethe germination is stopped by heating the barley to about60°C. This germinated product is known as malt.

(ii) Liberation of starch—The malt is crushed andtreated with steam at 140 – 150°C under pressure whenthe starch present in the malt is brought into solution. Thissolution is known as mash.

(iii) Saccharification—Malt is added to the mash andis kept at 50°C. The enzyme, diastase present in maltconverts the starch into maltose.

(C6H10O5)n Starch

+ n/2 H2O Diastase

⎯⎯⎯→(Malt)

n/2 C12H22O11Maltose

In an alternative method, starch may be convertedinto glucose by heating with dilute sulphuric acid orhydrochloric acid, and the excess of acid is neutralised byadding lime.

(C6H10O5)n Starch

+ n H2O ⎯→ n C6H12O6Glucose

(iv) Fermentation—The maltose solution, obtained instep (iii) is cooled to about 30°C and fermented as usualby yeast for 3 days, when the following reaction occurs :

C12H22O11Maltose

+ H2OMaltase

⎯⎯⎯→(Yeast)

2C6H12O6Glucose

C6H12O6Glucose

Zymase⎯⎯⎯→

(Yeast)2C2H5OH

Ethyl alcohol

+ 2CO2 ↑

The fermented liquor contains about 10 per centalcohol and is subjected to fractional distillation toindustrial alcohol.

By-products of alcoholic fermentation—Followingare important by-products of alcoholic fermentation :

(i) Acetaldehyde

(ii) Fusel oil

(iii) Carbon dioxide—Compressed in iron cylindersused in aerated water industry or as a dry ice.

(iv) Spent wash—It is the solid mass left after distilla-tion of various fractions, and is used as a Cattle Feed.

Absolute alcohol—Ethyl alcohol forms a constantboiling mixture i.e., azeotrope, containing 95·87% byweight of alcohol with water. Since this mixture boils at78·15°C, a temperature slightly lower than the boiling pointof pure ethanol (78·3°C), it is not possible to effect acomplete separation of ethyl alcohol from water byfractional distillation alone. Alcohol containing only 0·8%water (lime alcohol) can be prepared by distilling rectifiedspirit repeatedly over fresh quick lime. The last trace ofwater is removed by distilling the lime-alcohol over arequisite amount of metallic sodium or magnesium orcalcium.

On large scale the absolute alcohol is prepared bythe azeotropic distillation of industrial alcohol. In thismethod the advantage is taken of the fact that alcoholforms a ternary constant boiling mixture with water andbenzene.

Water = 7·5%Alcohol = 18·5%Benzene = 74·0% by w/W

This mixture boils at 64·9°C. The industrial alcoholwhich contains 4·13% water is mixed with benzene,sufficient enough to form a ternary constant boilingmixture with almost the entire amount of water presentand then distilled. The ternary azeotrope distils at 64·9°Ccarrying entire water present and absolute alcohol is leftbehind which is totally free from water and benzene.

Absolute alcohol blended with petrol in the ratio of20%, is used as a motor fuel, hence it is named as poweralcohol also.

Uses of ethyl alcohol—Following are important usesof ethyl alcohol :

(i) Ethyl alcohol is important component of alcoholicBeverages.

For example :Beer contains 3 – 5% of ethyl alcoholCider contains 2 – 4% of ethyl alcoholGin contains 35 – 40% of ethyl alcoholBrandy contains 35 – 40% of ethyl alcoholWhisky contains 35 – 40% of ethyl alcoholRum contains 35 – 40% of ethyl alcohol(ii) Ethanol is used as a solvent for gums, resins,

paints, varnishes, stains, pharmaceutical products, per-fumes, flavourings etc.

(iii) Ethyl alcohol is used for the preparation ofacetaldehyde, acetic acid, acetic anhydride, esters,chloral, chloroform etc.

(iv) Biological specimen are preserved in ethanol.(v) Ethanol is used as a low freezing and mobile

liquid in scientific equipments like thermometers, spiritlevels etc.

(vi) It is used as a component of fuels (power alcohol)for the internal combustion engines in many countries.

(vii) It is also used as ethylating agent in the manu-facture of dye intermediates, drugs etc.

C.S.V. / October / 2009 / 975 / 5

Points to Remember

● The strength of an alcohol preparation is expressed in the terms of proof spirit. Proof spirit is the aqueous ethyl alcoholcontaining 57·1% by volume of ethyl alcohol. The sample is referred to as over-proof or under-proof according as it is strongeror weaker than proof-spirit.

A 20° under-proof sample means that 100 volumes of the sample contain as much alcohol as 80 volumes of proof-spirit.Similarly 20° over-proof sample is one whose 100 volumes contain as much alcohol as 120 volumes of proof-spirit.

● The terms distilled and undistilled are used in describing alcoholic beverages. The undistilled beverages, which have lowalcoholic content, are generally prepared by fermentation of fruit juices. The distilled beverages containing much higheralcoholic contents are made by distillation of fermented liquors.

● Since ethanol can be used for drinking purposes, it is heavily taxed. But the ethanol used for industrial purposes is dutyfree. In order to make industrial alcohol unfit for drinking, it is denatured by mixing poisonous substances like methanol,acetone, rubber thinner, and pyridine or bone oil. Such a denatured alcohol is known as methylated spirit.

● Mineralised methylated spirit is coloured and is made by adding 0·5 part by volume of crude pyridine and 9·5 parts by volumeof methanol to 90 parts by volume of rectified spirit (95·5% ethanol) and adding to every 100 gallons of resulting mixture notless than 3/8 of a gallon of mineral naphtha and not less than 1/40 oz. of the dye methyl violet.

● In 1860, the pioneer worker Pasteur suggested that fermentation is purely a physiological process carried out by living micro-organisms. However, Liebig considered it to be purely a chemical reaction. Buchner (1897) showed that the presence of livingcells is not necessary for this reaction. Thus the truth regarding the nature of fermentation lies between the views of Pasteurand Liebig.

The fermentation is slow decomposition of complex organic compounds by the activity of non-living complex nitrogenoussubstances (enzymes) produced in living organisms.

OBJECTIVE QUESTIONS

1. The number of alcohol isomersarrived at from molecular formulaC4H10O is—(A) 2 (B) 3(C) 4 (D) 5

2. When equimolar quantities ofethanol and methanol are mixedand heated with conc. H2SO4,the product formed is—(A) C2H5OC2H5(B) CH3OCH3(C) C2H5OCH3(D) All of these

3. The alcohol which reacts fastestwith Luca’s reagent at normaltemperature is—(A) 2-methyl propane-1-ol(B) 2-methyl propane-2-ol(C) Butane-1-ol(D) Butane-2-ol

4. Which of the following com-pounds can be used for thepreparation of chloroform ?(A) CH3CH2COC3H7(B) CH3COC2H5(C) CH3CH2COCH2CH3(D) All of these

5. Ethanol when heated with conc.H2SO4 may give—(A) Diethyl sulphate only(B) Diethyl ether only(C) Ethylene only(D) All of these

6. The compound which is not iso-meric with diethyl ether is—(A) Butane-1-ol(B) n-propylmethyl ether

(C) 2-methyl propane-2-ol(D) None of these

7. Denatured spirit is mainly usedas a—(A) Medicine(B) Good fuel(C) Solvent(D) Component of beverages

8. An organic compound whenpassed over heated copper at575 K, gives an alkene, the com-pound is—(A) Alkane(B) Akyne(C) Secondary alcohol(D) Tertiary alcohol

9. Which of the following com-pounds has highest boiling point ?(A) Ethanol(B) Methoxymethane(C) Chloromethane(D) Propane

10. Which of the following bonds ofan alcohol is cleaved when itreacts with carboxylic acids ?(A) C—H (B) C—O(C) O—H (D) All of these

11. Which of the following alcoholscannot be dehydrogenated ?(A) CH3CH2CH2CH2OH

(B) CH3

OH|CH—CH2CH3

(C) (CH3)3COH

(D) (CH3)2CHOH

12. In the reaction sequence

X HBr

⎯→ CH3—CH|Br

—CH3 alc.KOH⎯⎯→

Y HBr

⎯⎯⎯→Peroxide

Z

X, Y and Z are respectively—(A) 2-propanol, Propene,

1-bromopropane(B) Propene, 2-bromopropane,

Propene(C) 2-propanol, Propyne,

2-bromopropane(D) 1-propanol, Propene,

1-bromopropene

13. An organic compound (A) givespositive Lucas test in 5 minutes.When 6·0 gm of (A) is treatedwith sodium metal, 1120 ml ofhydrogen is evolved at STP. Theorganic compound is—(A) CH3CH2—CH

|OH

—CH3

(B) CH3—CH|OH

—CH3

(C) CH3—

CH3|C—OH|CH3

(D) CH3CH2OH

14. Which of the following alcoholscan be obtained from HCHO ?(A) CH3OH(B) C2H5OH

C.S.V. / October / 2009 / 976

(C) CH3CH2CH2OH

(D) All of these

15. Phenol can be distinguished fromethyl alcohol by all reagentsexcept—(A) NaOH (B) FeCl3(C) Br2/H2O (D) Na

16. Alcohol can be obtained by allmethods except—(A) Hydroboration–oxidation(B) Oxymercuration–demercura-

tion(C) Reduction of aldehydes with

Zn-Hg/HCl(D) By fermentation of starch

17. Which of the following alcohols isleast soluble in water ?(A) n-butyl alcohol(B) Iso-butyl alcohol(C) Tert-butyl alcohol(D) Sec-butyl alcohol

18. Which of the following com-pounds is isomeric with 1-pro-panol ?(A) Ethanol(B) 2-methyl-2-propanol(C) 1-butanol(D) Ethyl-methyl ether

19. Which of the following com-pounds would yield carboxylicacid as the product on oxidationwith acidified K2Cr2O7 ?

(A) 1-butanol(B) 1-propanol(C) Both (A) and (B)(D) None of these

20. Which one of the following canconvert 2-propanol to acetone ?(A) K2Cr2O7/H+

(B) Cu/575 K(C) Both (A) and (B)(D) None of these

21. The enzyme which convertsglucose and fructose into ethyl-alcohol is—(A) Diastase (B) Invertase(C) Zymase (D) Maltase

22. A solution of ethyl alcohol—(A) Decolourises the litmus

paper(B) Changes red litmus blue(C) Changes blue litmus red(D) Does not affect litmus paper

23. On industrial scale ethanol ismanufactured by the fermen-tation of—(A) C6H12O6 (B) CH3COOH(C) Molasses (D) C12H22O11

24. The percentage of ethyl alcoholin rectified spirit is—(A) 75·00 (B) 85·5(C) 95·6 (D) 100·0

25. Which of the following alcoholswill be most acidic ?(A) CH3OH (B) R–CH2OH(C) R2CHOH (D) R3COH

26. Dehydration of ethanol cannotgive—(A) C2H5OC2H5 (B) C2H5HSO4(C) C2H4 (D) C2H2

27. An alcohol on oxidation givesCH3COOH and CH3CH2COOH,the alcohol is—(A) CH3CH(OH) CH2CH2CH3(B) CH3(CH2)2CHOH(C) (CH3)2 C(OH) CH2CH3(D) CH3CH2CH2OH

28. Which of the following com-pounds gives a positive iodoformtest ?(A) 3-pentanol(B) 2-phenyl-ethanol(C) 1-phenyl ethanol(D) Pentanal

29. Determination of percentage ofalcohol in wine is called—(A) Iodometry(B) Iodimetry(C) Alcoholometry(D) Acidometry

30. Formation of 2-butene as majorproduct by dehydration of 2-butanol is according to—(A) Saytzeff rule(B) Peroxide effect(C) Markownikoff’s rule(D) Anti-Markownikoff’s rule

31. Primary, secondary and tertiaryalcohols are distinguished by—(A) Oxidation(B) Lucas reagent(C) Victor Meyer method(D) All of these

32. Which will respond to iodoformtest ?(A) CH3OH(B) (CH3)3C·CHO(C) (CH3)2CHOH(D) CH3CH2CH2OH

33. The percentage of ethanol byweight in proof spirit is—(A) 90 (B) 10(C) 48 (D) 4·5

34. Fermentation of starch solutionto ethanol does not require—(A) Maltase (B) Diastase(C) Invertase (D) Zymase

35. The correct order of boiling pointof alcohols having comparablemolar mass is—

(A) 1° < 2° < 3°

(B) 3° < 2° < 1°

(C) 2° < 1° < 3°

(D) None is correct

36. The —OH group of CH3OH cannot be replaced by the chlorineby the action of—(A) HCl (B) PCl3(C) PCl5 (D) Cl2

37. Which of the following com-pounds is known as wood spirit ?(A) Wood tar (B) Methanol(C) Ethanol (D) 95% ethanol

38. Cyclohexanol is a—(A) Phenol(B) Primary alcohol(C) Secondary alcohol(D) Tertiary alcohol

39. During the dehydration of alco-hols, the ease of formation ofcarbocation follows the order—(A) 1° > 2° > 3° (B) 3° > 2° > 1°

(C) 3° > 1° > 2° (D) 2° > 1° > 3°

40. C2H5OH can be distinguishedfrom CH3OH—

(A) By the action of HCl(B) By the action of NH3

(C) By determining the solubilityin water

(D) By iodoform test

41. Which of the following products isformed when ter-butanol reactswith C6H5MgBr ?

(A) Methane (B) Methanal(C) Benzene (D) None of these

42. Propanol-1 and propanol-2 canbe best distinguished by—(A) Oxidation with K2Cr2O7 fol-

lowed by reaction with CuO

(B) Oxidation by heating with Cufollowed by reaction withFehling solution

C.S.V. / October / 2009 / 977

(C) Oxidation with KMnO4 fol-lowed by reaction withFehling solution

(D) Oxidation with conc. H2SO4followed by reaction withcopper

43. Cyclohexanol can be convertedinto cyclohexene on reactingwith—(A) conc. HBr (B) conc. H3PO4

(C) HCl + ZnCl2 (D) conc. HCl

44. When 0·037 gm of an alcohol(ROH) reacts with excess ofCH3MgBr, 11·2 cm3 of gas wasevolved at STP. What will be themolecular weight of alcohol ?(A) 33·00 (B) 74·00(C) 66·00 (D) 47·00

45. What is the correct IUPAC name

of

H OH

?

(A) Cyclohexanol(B) Cyclohexenol(C) Cyclohexenol-3(D) 3-cyclohexene-1-ol

46. Which one of the following isused as rubbing alcohol ?(A) CH3OH

(B) C2H5OH

(C) CH3CH2OH

(D) (CH3)2CHOH

47. Which of the following is the leastsuitable solvent for dissolving anionic compound ?

(A) H—OH(B) CH3—OH

(C) CH3CH2—OH

(D) CH3(CH2)3CH2OH

48. What is the correct decreasingorder of the boiling point ofpentanol-1(A), 2-methylbutanol-2(B) and 3-methylbutanol-2(C) ?(A) A > B > C (B) A > C > B(C) B < A < C (D) B > C > A

49. Which of the following productsis expected from the reaction of

== O with LiAlH4 ?

(A) HOH

(B) HOH

(C)OH

OH

(D) CH3–CH2–CH2–CH2–CH2–CH2OH

50. Which of the following products isformed from catalytic hydrogena-tion of CH3CH== CHCH2CHO ?(A) Unsaturated alcohol(B) Saturated alcohol(C) Saturated aldehyde

(D) None of these

ANSWERS

●●●

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METRO RAIL CORPORATION

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

DELHI

Code No. 971

Price : Rs. 299/-

(Technical Cadre)

● Station Controller ● Train Operator● Section Engineer ● Junior

Engineer ● Junior Station Controller

IncludingPrevious Years’Solved Papers

For–

By : Dr. Lal & Jain

C.S.V. / October / 2009 / 978

1. The mass of carbon anode con-sumed (giving only carbondi-oxide) in the production of 270 kgof aluminium metal from bauxiteby the Hall process is—

(Atomic mass : Al = 27)

(A) 270 kg (B) 540 kg

(C) 90 kg (D) 180 kg

2. If the electron is present in then = 5 level, how many lineswould be observed, in case ofH-atom ?

(A) 10 (B) 5

(C) 7 (D) 4

3. A nuclide of an alkaline earthmetal undergoes radioactivedecay by emission of the α-parti-cles in succession. The group ofthe periodic table to which theresulting daughter element wouldbelong is—

(A) Gr. 4 (B) Gr. 6

(C) Gr. 14 (D) Gr. 16

4. Δ = XA – XB = 2. What is the per-cent ionic character for a cova-lent molecule A – B ?

(A) 46 (B) 50

(C) 20 (D) 30

5. The correct sequence of increas-ing covalent character is repre-sented by—

(A) LiCl < NaCl < BeCl2(B) BeCl2 < LiCl < NaCl

(C) NaCl < LiCl < BeCl2(D) BeCl2 < NaCl < LiCl

6. Three fold symmetrical axis isassociated with the molecule—

(A) NH3 (B) C2H4

(C) CO2 (D) SO2

7. Which one of the following oxidesis expected to exhibit paramag-netic behaviour ?(A) CO2 (B) SiO2

(C) SO2 (D) ClO2

8. A substance whose initial con-centration is ‘a’ reacts according

to zero order kinetics. The timerequired for completion will be—

(A)ak0

(B)k0a

(C)a

2k0(D) 2kaa

9. Which of the following would havea permanent dipole moment ?(A) SiF4 (B) SF4

(C) XeF4 (D) BF3

10. For the reaction A + B C + D,the initial concentration of A andB were taken as 0·9 mole dm–3.At equilibrium the concentrationof D was found to be 0·6 moledm–3. The value of equilibriumconstant will be—(A) 9 (B) 3(C) 8 (D) 4

11. In face centred cubic lattice, aunit cell is shared equally by howmany unit cells ?(A) 2 (B) 4(C) 6 (D) 8

12. Bronsted Lowry acid is—(A) BF3

(B) C2O42–

(C) [N(Me)4]+

(D)—CO—CO

NH

13. The vapour pressure of twoliquids ‘P’ and ‘Q’ are 80 and 60torr, respectively. The total vapourpressure of solution obtained bymixing 3 mole of P and 2 mole ofQ would be—(A) 72 torr (B) 140 torr(C) 68 torr (D) 20 torr

14. One equivalent of Na2CO3 and0·5 equivalent of MgCO3 aretreated with excess of HCl in twoseparate vessels. The volume ofCO2 evolve at STP will be res-pectively—(A) 22·4 lit., 11·2 lit.(B) 11.2 lit., 5·6 lit.(C) 11.2 lit., 11·2 lit.(D) 22.4 lit., 22·4 lit.

15. What is the correct relationshipbetween the pH of isomolarsolutions of sodium oxide (pH1),sodium sulphide (pH2), sodiumselenide (pH3) and sodium tellu-ride (pH4) ?(A) pH1 > pH2 > pH3 > pH4

(B) pH1 > pH2 ≈ pH3 > pH4

(C) pH1 < pH2 < pH3 < pH4

(D) pH1 < pH2 < pH3 ≈ pH4

16. The number of atoms in 200 g ofan f.c.c. crystal with a densityρ = 10 g cm–3 and cell edge 200pm is equal to—

(A) 3 × 10 (B) 4 × 1025

(C) 1 × 1025 (D) 2 × 1025

17. The absolute enthalpy of neutrali-sation of the reaction :

MgO(s) + 2HCl(aq) → MgCl2(aq)

+ H2O(l)

will be—(A) – 57·33 kJ mol– 1

(B) Greater than – 57·33 kJmol– 1

(C) Less than – 57·33 kJ mol– 1

(D) 57·33 kJ mol– 1

18. Which of the following electrolyteis least effective in coagulatingferric hydroxide sol ?(A) K3Fe(CN)6 (B) K2CrO4(C) KBr (D) K2SO4

19. The rate of reaction between tworeactants A and B decreases by afactor of 4 if the concentration ofreactant B is doubled. The orderof this reaction with respect toreactant B is—(A) 2 (B) – 2(C) 1 (D) – 1

20. The density of the gas is equalto—(A) P/RT (B) nP

(C)MPRT (D) M/V

21. Which of the following pairs for achemical reaction is certain toresult in a spontaneous reaction ?(A) Exothermic and increasing

disorder(B) Exothermic and decreasing

disorder(C) Endothermic and increasing

disorder(D) Endothermic and decreasing

disorder

C.S.V. / October / 2009 / 979

22. Solubility of sparingly solublesalt, S, specific conductance k,

and equivalent conductance ^°

are related as—

(A) S = 1000 ^°

k

(B) S = k ^°

(C) S = k

1000 ^°

(D) S = 1000 k

^°

23. 4·5 g of aluminium (at mass 27amu) is deposited at cathodefrom Al3+ solution by a certainquantity of electric charge. Thevolume of hydrogen produced atSTP from H+ ions in solution bythe same quantity of electriccharge will be—

(A) 44·8 L (B) 22·4 L

(C) 11·2 L (D) 5·6 L

24. The metal which is found in freeand combined state in nature—

(A) Na (B) Mn

(C) Cu (D) Au

25. The mole fraction of the solute inone molal aqueous solution is—

(A) 0·009 (B) 0·018

(C) 0·027 (D) 0·036

26. The ore associated with ruby,amethyst, sapphire, topaz etc.is—(A) CuFeS2 (B) Na3AlF6

(C) Cu2O (D) Al2O3

27. The surface tension of which ofthe following liquid is maximum ?(A) C2H5OH (B) CH3OH(C) H2O (D) C6H6

28. Sodium peroxide in contact withmoist air turns white due to theformation of—(A) Na2CO3 (B) NaHCO3

(C) NaNO2 (D) NaOH

29. The chirality of the compoundBr

C

H3C

|

Cl

H

is—

(A) R (B) S

(C) E (D) Z

30. Which species are present inacidic and basic media respec-tively ?(A) CrO4

2– in basic and Cr2O72–

in acidic.(B) Cr2O7

2– in basic and CrO42–

in acidic.(C) CrO4

2– is present in acidicand basic both.

(D) Cr2O72– is present in acidic

and basic both.

31. Which of the following undergoesnucleophilic substitution exclusi-vely by SN1 mechanism ?(A) Ethyl chloride(B) Isopropyl chloride(C) Chlorobenzene(D) Benzyl chloride

32. CsAuCl3 is diamagnetic com-pound. It should contain—(A) Au(II) ions(B) Au(III) ions(C) Au(I) ion only(D) Equal number of Au(I) and

Au(III) atoms

33. Which one of the followingalkenes will react faster with H2under catalytic hydrogenationconditions ?

(R = Alkyl substituent)

(A)R H

R H

(B)R R

R R

(C)R R

R H

(D)R H

R R

34. The chemical composition of‘Slag’ formed during the smeltingprocess in extraction of copperis—(A) CuO + FeS (B) FeSiO3(C) CuFeS2 (D) CuS + FeO

35. Aniline in a set of reactionsyielded a product D.

2NHNaNO2⎯⎯⎯→

HCl A

CuCN⎯⎯→ B

H2⎯→Ni

C HNO2⎯⎯→ D

The structure of the product Dwould be—(A) C6H5NHOH(B) C6H5NHCH2CH3

(C) C6H5CH2NH2

(D) C6H5CH2OH

36. Different layers of graphite areheld together by—(A) Ionic bonding(B) Metallic bonding(C) Covalent bonding(D) van der Waals forces

37. Which one of the following com-pounds is most acidic ?(A) Cl—CH2—CH2—OH

(B)

OH

(C)

OH

NO2

(D)

OH

CH3

38. HO

has the IUPAC name—(A) 3, 4-dimethyl-1-penten-3-ol(B) Isopropyl-3-methyl vinyl

carbinol(C) 2, 3-dimethyl-4-penten-3-ol(D) None of the above

39. Electrolytic reduction of nitro-benzene in weakly acidic mediumgives—(A) N-phenylhydroxylamine(B) Nitrosobenzene(C) Aniline(D) p-hydroxyaniline

40. Optical active compound is—(A) n-butanol(B) n-propanol(C) 2-chlorobutane(D) 4-hydroxy heptane

41. In a set of reactions acetic acidyielded a product D.

CH3COOH SOCl2⎯⎯→ A

Benzene⎯⎯⎯⎯→

Anhy. AlCl3 B

HCN⎯⎯→ C

H . OH⎯⎯→ D

The structure of D would be—

(A)CH2—

COOH

|C—CH3

|OH

(B)

CN

|C—CH3

|OH

C.S.V. / October / 2009 / 980

(C)CH2—

OH

|C—CH3

|CN

(D)

OH

|C—COOH

|CH3

42. What is the theoretical yield ofethane in the following reaction(in volume) ?

CH4 ⎯⎯→Cl2 CH3Cl

ether⎯⎯⎯→

Na C2H6

(A) 1 lit (B) 2 lit(C) 1·5 lit (D) 0·5 lit

43. The cell membranes are mainlycomposed of—(A) Fats(B) Proteins(C) Phospholipids(D) Carbohydrates

44. Secondary gem halide of pro-pane on heating with zincgives—(A) Sym. dimethyl ethylene(B) 2-butene(C) 2, 3-dimethyl-2-butene

(D) β-butylene

45. Which one of the followingarrangements represents thecorrect order of electron gain

enthalpy (with negative sign) ofthe given atomic species ?(A) S < O < Cl < F(B) Cl < F < S < O(C) F < Cl < O < S(D) O < S < F < Cl

46. All the three hydrocarbons X, Yand Z of molecular formula C5H8decolourises Br2 in CCl4 solution.All the three are soluble in coldconc. H2SO4. In presence ofmetal catalyst X and Y on hydro-genation yield n-pentane. X giveswhite precipitate with ammonia-cal silver nitrate solution while Yand Z donot. What are X, Y andZ respectively ?(A) 1-pentyne, 2 pentyne, cyclo-

pentene(B) Ethylmethylacetylene, cyclo-

pentene, 2-pentyne(C) 3-methyl-1-butyne, 2-pen-

tyne, 1-pentyne(D) 1-pentyne, 2-pentyne, 3-

methyl -1-butyne

47. Four successive members of thefirst row transition elements arelisted below with their atomicnumbers. Which one of them isexpected to have the highest thirdionization enthalpy ?(A) Vanadium (Z = 23)(B) Chromium (Z = 24)

(C) Manganese (Z = 25)(D) Iron (Z = 26)

48. Which of the following reactioncomes within the frame work ofelimination ?

(A) (CH3)2CHCl + SH(B) (CH3)3C–Br + Ethanolic KCN

(C) CH3CH2CH2Cl + I(D) (CH3)2 CHBr + aq KOH

49. Which one of the following isexpected to exhibit optical iso-merism ? (en = ethylenediamine)(A) cis-[Pt(NH3)2 Cl2](B) trans-[Pt(NH3)2Cl2](C) cis-[Co(en)2Cl2]+

(D) trans-[Co(en)2Cl2]+

50. Alkaline hydrolysis of C4H8Cl2gives a compound which onheating with NaOH and I2 pro-duces a yellow precipitate ofCHI3. The compound shouldbe—

(A) CH3CH2CH2CHO

(B) CH3—CH2— C||O

—CH3

(C) CH3—CH— C|O

H|H

— C|O

H2|

H

(D) CH3— C|O

H|H

— C|O

H|H

—CH3

ANSWERS WITH HINTS

,

C.S.V. / October / 2009 / 984

1. The increasing order of energy ofelectromagnetic radiation can berepresented as—(A) Microwave < infrared < visi-

ble < X-ray(B) X-ray < visible < infrared <

microwave(C) Microwave < infrared < visi-

ble < radiowaves(D) X-ray < infrared < visible <

microwave

2. The work function of a metal is

1eV. If 3300 °A wavelength lightis incident, the value of stoppingvoltage is—(A) 1V (B) 3·75 V(C) 3·2 V (D) 0·75 V

3. For a real gas, PV is a constantover a small range of pressures,at—(A) Boyle’s temperature(B) Critical temperature(C) Inversion temperature(D) Ordinary temperature

4. A monoatomic ideal gas iscompressed to its 1/8 volumeadiabatically at 17°C. Tempera-ture after compression will be—(A) None (B) 17°C

(C) 136°C (D) 887°C

5. The electronegativities of ele-ments P, As, Cl and S decreasein the order—(A) S > Cl > As > P(B) As > P > Cl > S(C) P > Cl > As > S(D) Cl > S > P > As

6. Which formula is incorrect forroot mean square velocity ?

(A)3RTM (B)

3PVM

(C)3dP (D)

2KEM

7. The oxidation number of oxygenin Cl2O and H2O2 are respecti-vely—(A) – 2 and + 1 (B) + 2 and + 1(C) – 2 and – 1 (D) + 2 and – 1

8. A liquid boils at that temperatureat which the pressure of satu-rated vapour is—

(A) More than atmosphericpressure

(B) Double to atmospheric pres-sure

(C) Equal to atmospheric pres-sure

(D) Less than atmospheric pres-sure

9. The depression in the freezingpoint of water (ΔT) caused by0·5M HCl, 0·5 M glucose and 0·5M MgCl2 are in the order—

(A) ΔTglucose > ΔTMgCl2 >ΔTHCl

(B) ΔTMgCl2 > ΔTglucose >ΔTHCl

(C) ΔTHCl > ΔTMgCl2 > ΔTglucose

(D) ΔTMgCl2 > ΔTHCl > ΔTglucose

10. If a proton totally converts intoenergy, the value of energy willbe—

(A) 190 MeV (B) 931 MeV

(C) 93·1 MeV (D) 931 joule

11. The plot of log K versus 1T is linear

with a slope of—

(A)EaR (B)

– EaR

(C)Ea

2·303R (D)– Ea

2·303R

12. In the following nuclear reaction

2He3 + zXa → z + 1Ya + 2 + Q

What is Q ?

(A) Neutron (B) Proton

(C) Positron (D) Electron

13. The pH of 10–3 M solution ofCa(OH)2 is—

(A) 8·0 (B) 11·3(C) 10·5 (D) 5·0

14. Work function of photoelectricmetal is 3·13 eV. Thresholdfrequency is—

(A) 4 × 1011 Hz (B) 5 × 1011 Hz

(C) 8 × 1014 Hz (D) 8 × 1010 Hz

15. The most efficient packing ofsimilar spheres is obtained in—(A) The simple cubic system and

the body centred cubicsystem

(B) The simple cubic system andthe hexagonal close packedsystem

(C) The face centred cubicsystem and the hexagonalclose packed system

(D) The body centred cubic sys-tem and the face centredcubic system

16. In which compound the numberof 3° carbon is maximum ?(A) 2, 5 dimethyl hexane(B) 2, 3, 4 trimethyl pentane(C) 2, 2, 4, 4 tetramethyl pen-

tane(D) 2, 2, 3 trimethyl pentane

17. The chemical composition of‘slag’ formed during the smeltingprocess in the extraction ofcopper is—(A) Cu2O + FeS(B) FeSiO3(C) CuFeS2(D) Cu2S + FeO

18. Dichlorocarbene is—(A) A neutral diamagnetic(B) A carbonation(C) A carbanion(D) A free radical

19. Na2CO3 + Fe2O3 → A + CO2 ;A is—(A) NaFeO2 (B) Na3FeO3

(C) Fe3O4 (D) Na2FeO2

20. Which test is not ideal todistinguish 2-butanol and 1-propanol ?(A) Hydrogenation(B) Iodoform test(C) Lucas test(D) Oxidation test

21. Thallium shows different oxida-tion states because—(A) Of its high reactivity(B) Of inert pair effect(C) Of its amphoteric nature(D) It is a transition metal

22. The hydrocarbon formed by elec-trolysis of sodium propionate—(A) CH3—CH = CH2(B) CH3—CH2—CH2—CH3

C.S.V. / October / 2009 / 985

(C) CH3—CH2—CH3

(D) CH3—CH3

23. Dental amalgam is composedof—(A) Hg + Ag + Cd + Au + Fe(B) Cu + Sn + Hg + Ag + Zn(C) Cd + Cu + Ni + Au + Fe(D) Cu + Sn + Au + Hg + CO

24. The product of the reaction bet-ween ethylene and ozone, whenhydrolysed in presence of Zn, thenew product formed is—(A) Alcohol(B) Ethylene oxide(C) HCHO(D) CH2OHCH2OH

25. For the molecule PF4Cl3, which ofthe following structures is themost stable, considering thatCH3— is more electropositivethan F ?

(A)F

F

F|P—CH3|F

(B)F

F

CH3|P—F|F

(C)F

CH3|P

F

F

F

(D)

FF

F|P

CH3F

26. The chemical formula of potashalum is K2SO4·Al2(SO4)3 XH2O.Here X is—

(A) 7 (B) 12

(C) 6 (D) 24

27. In the following groups,—OAc—OMe—OSO2Me

I II III—OSO2CF3

IV

the order of leaving group abilityis—

(A) I > II > III > IV

(B) IV > III > I > II(C) III > II > I > IV(D) II > III > IV > I

28. The electronic configuration ofFe26 is [Ar]—

(A) 3d 8 4s

2 (B) 3d 7 4s

2

(C) 3d 6 4s

2 (D) 3d 5 4s

2

29. In Kjeldahl’s method, nitrogenpresent is estimated as—(A) N2 (B) NH3

(C) NO2 (D) None of these

30. The element with electronicconfiguration 1s

2, 2s 2p

6, 3s 2p

6,4s

2 shows same property as—(A) Mo (B) Rb(C) Ca (D) Sr

31. The following compound is usedas—

O—

O||C—CH3

COOH(A) An anti-inflammatory com-

pound(B) Analgesic(C) Hypnotic(D) Antiseptic

32. The formula of acetaldehydesemicarbazone—(A) CH3—CH = N NHCONHCH3

(B) CH3—CH = N—OH

(C) CH3CH = N—NHCONH2

(D) CH3—CH = N—NHCONH

—CONH2

33. Which of the following cannot bediazotised ?

(A)2 HN

(B)

CH3

2NH

(C)

CH3

2NH

(D) C6H5CH2NH2

34. Phenol, chloroform and causticpotash are heated. The com-pound formed is—(A) Salicylic acid(B) p-hydroxybenzaldehyde(C) m-hydroxybenzaldehyde(D) Salicylaldehyde

35. Nessler’s reagent is used todetect the presence of—

(A) CrO2 –4 (B) PO3 –

4

(C) MnO–4 (D) NH+

4

36. The compound obtained by thereaction of acetic anhydride andammonia is—(A) CH3COONH4

(B) CH3CN

(C) CH3CONHCH3

(D) CH3CONH2

37. The precipitate of CaF2 (Ksp =1·7 × 10–10) is obtained whenequal volumes of the followingare mixed—

(A) 10 – 4 M Ca2 + + 10– 4 M F–

(B) 10– 2 M Ca2 + + 10– 3 M F–

(C) 10– 5 M Ca2+ + 10– 3 M F–

(D) 10– 3 M Ca2+ + 10– 5 M F–

38. When aniline reacts with aceticanhydride the product formedis—

(A) p-aminobenzoic acid

(B) m-aminobenzoic acid

(C) Acetanilide

(D) o-aminobenzoic acid

39. The cyanide ion, CN– and N2 areisoelectronic but in contrast toCN–, N2 is chemically inert, beca-use of—

(A) Low bond energy

(B) Absence of bond polarity

(C) Unsymmetrical electron dis-tribution

(D) Presence of more number ofelectrons in bonding orbitals

40. The element with highest ioni-sation potential is—

(A) N (B) S

(C) C (D) Be

41. The order of reactivities of thefollowing alkyl halides for a SN2reaction is—

(A) RF > RCl > RBr > RI

(B) RF > RBr > RCl > RI

(C) RCl > RBr >RF > RI

(D) RI > RBr > RCl > RF

42. Maximum melting point is of—(A) MgCl2 (B) BaCl2(C) CaCl2 (D) BeCl2

C.S.V. / October / 2009 / 986

43. The colour of the colloidal parti-cles of gold obtained by differentmethods differ because of—(A) Variable valency of gold(B) Different concentration of

gold particles(C) Different types of impurities(D) Different diameters of colloi-

dal particles

44. The process requiring the absor-ption of energy is—(A) F → F– (B) Cl → Cl–

(C) O → O2– (D) H → H–

45. Which statement, regarding themole fraction x of a component insolution is correct ?

(A) – 2 ≤ x ≤ 2(B) 0 < x ≤ 1(C) x ≤ 1(D) x is always non-negative

46. A piece of magnesium ribbonwas heated to redness in anatmosphere of nitrogen and oncooling water was added. Thegas evolved was—

(A) Ammonia (B) Hydrogen

(C) Nitrogen (D) Oxygen

47. The end product of 4n series is—(A) 82Pb208 (B) 82Pb207

(C) 82Pb209 (D) 83Pb205

48. Mixture used in Holmes’s signalis—(A) CaC2 and CaCl2(B) CaCl2 and Ca3F2

(C) CaC2 and Ca3N2 (D) CaC2 and Ca3P2

49. Percentage of gold in 21·6 caratgold is—(A) 21·6 (B) 90(C) 10 (D) 70

50. Chlorine acts as a bleachingagent only in the presence of—

(A) Dry air (B) Sunlight

(C) Moisture (D) Pure oxygen

ANSWERS WITH HINTS

C.S.V. / October / 2009 / 988

●●●

(Continued from Page 946 )

68. What are amino acids in a protein called ?➠ αααα-amino acids

69. What is the basis of modern classification of plants ?➠ Phylogeny

70. Who gave the term ‘plasmid’ for the first time ?➠ Lederberg

71. What are called those blue-green algae which live inprotozoans ? ➠ Cyanellae

72. In what type of solution the plant cells neither lose norgain water ? ➠ Isotonic solution

73. What is called the spatial arrangement of atoms in aprotein ? ➠ Conformation

74. How many polypeptides of a protein is/are specifiedfor each gene ? ➠ One

75. What are the three simple sugars of monosac-charide ? ➠ Glucose, Fructose, and Galactose

76. What are the two important components of fluid-mosaic model of plasma membrane ?

➠ Lipids and proteins77. What is called a group of structural and regulating

genes that functions as a single unit ? ➠ Operon78. Which is the earliest era in the geological record ?

➠ Precambrian79. What measures the gel electrophoresis ?➠ Measures the charge and size of proteins and DNA

fragments80. What is generally called the lower portion of the meso-

phyll in the leaf cell ? ➠ Spongy parenchyma●●●

(Continued from Page 983 )

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C.S.V. / October / 2009 / 989

Communicable diseases are caused by pathogenicmicrobes and readily transmitted from infected to healthypersons. Therefore, the diseases transmitted directly orindirectly between hosts are called communicablediseases. The pathogenic microbe may be bacteria,viruses, protozoa, fungi. Microbes are capable of growingin human tissues and cause diseases.

Pathogens vary in the severity of the diseases theycause. Some produce mild to moderate discomfort(common cold and flu) and are quickly controlled by thebody’s defenses, if affected person is healthy. Otherpathogens, such as these causing rabies, cholera, andplague, overwhelm the body defenses of every healthypeople and may prove fatal.

Some microbes infect and cause disease only if anopportunity arises. They are appropriately called oppor-

tunistic pathogens and produce opportunistic infections.These microbes often infect a host whose health ornatural defenses have been diminished in some way.Pneumocystic pneumonia is an opportunistic infectionof lung found in AIDS patients with severely impairedimmune defenses.

Even without illness, some beneficial microbes of thenormal flora can become opportunistic pathogens if theygain access to a different body location. For example,when beneficial bacteria in the digestive tract calledEscherichia coli accidently gain access to the uretheraand bladder, they can cause a urinary tract infection.

Methods of Pathogens Entry into the Body

Before pathogens can grow and cause disease, theymay first gain access, they must first gain access to

Selected Microbial Diseases, the Scientific Names of Microbes and the Areas of the Body Infected

Disease Microbe Primary Location of Infection

Bacteria

Botulism Clostridium botulinum Neuromuscular junction

Cholera Vibrio cholerae Intestine

Food poisoning Staphylococcus aureus Intestine

Gas gangrene Clostridium perfringens Infected tissues

Gonorrhea Neisseria gonorrhoeae Reproductive tract

Syphilis Treponema pallidum Reproductive tract

Chlamydia Chlamydia trachomatis Reproductive tract

Infantile and traveler’s diarrhea Escherichia coli (E. coli) Intestine

Salmonellosis Salmonella enteritidis Intestine

Tetanus Clostridium tetani Nerves

Tuberculosis Mycobacterium tuberculosis LungsTyphoid Salmonella typhi Intestine

Protozoa

Amoebic dysentery Entamoeba histolytica IntestineGiardiasis (beaver fever) Giardia lamblia IntestineMalaria Plasmodium malariae Bloodstream

Fungi

Aspergillosis Aspergillus fumigatus LungsAthlete’s foot Trichophyton SkinHistoplasmosis Histoplasma capsulatum LungsCandidiasis (yeast infection) Candida albicans Mouth, intestine, vagina

Viruses

Influenza Influenza virus Upper respiratory tractChicken pox Varicella virus SkinRabies Rabies virus Brain, spinal cordHepatitis B Hepatitis B virus LiverHepatitis A Hepatitis A virus Intestine, liverFever blisters, genital herpes Herpes simplex SkinGerman measles Rubella SkinPolio Poliovirus Spinal cord (paralytic polio)AIDS HIV Immune system

C.S.V. / October / 2009 / 990

tissues. The sites through which pathogens enter thebody are called portals of entry. Portals of entry arebreaks in the skin or the exposed mucous membranes ofthe eye or the respiratory, digestive, reproductive, andurinary tracts. Some pathogens have specific protals ofentry and others do not. For example, cold viruses enterthrough the mucous membranes of the eyes and upperrespiratory tract but not the mucous membranes of thereproductive and urinary tracts. By contrast, the bacteriathat cause the sexually transmitted disease gonorrheacan enter through the mucous membranes of the eye,upper respiratory tract, reproductive tract and anus.

Transmission of PathogensMany pathogens are transmitted by passing directly

from an infected host, either human or nonhuman, to anuninfected host. Some are transmitted indirectly bycontact with contaminated objects. Microbes may notactually grow on these objects but may remain alive longenough to be passed to others upon contact. There aremany possible ways in which pathogens are transmitted.It is important to realize that not all pathogens are trans-mitted in the same way. The viruses for the common coldand hepatitis can be transmitted after exposure to dryconditions on non-living objects, but this is not the casefor other viruses. For example, the virus responsible forAIDS (HIV) is easily inactivated by exposure to air,dryness, and soap and water. For transmission, it requiresa moist medium—a body fluid such as blood or semen.This fact explains why the disease profile for AIDS is verydifferent from the disease profile for the common cold(Prologue).

Actions of PathogensIf pathogens are not killed by the host’s initial defen-

ses, the process of infection begins. During an infection, amicrobe enters a tissue area, multiplies in number, andbegins to cause tissue damage. The accumulated tissuedamage eventually alters body functions and producesthe symptoms characteristic of the disease. Each type ofpathogen has particular processes which allow it toinvade, multiply, and damage a specific tissue or group oftissues.

Actions of Pathogenic BacteriaBacteria damage tissues in many ways. Usually they

produce and release enzymes and chemical toxins thathelp them invade and destroy tissues. As bacteria destroycells, nutrients are, liberated which the bacteria use fortheir growth, making more bacteria. More bacteria lead tomore tissue damage. Most types of bacteria proliferate inthe space outside cells and do their damage from theoutside. However, a few bacteria penetrate into cells,destroying them.

Tissue Damage by Secreted Enzymes

Enzymes are proteins that speed up specific chemi-cal reactions. Through the action of enzymes secreted bybacteria, the structure and chemical composition oftissues can be changed rapidly to benefit the microbe anddamage the host.

The bacteria that cause gas gangrene secrete anenzyme that allows them to disrupt the molecular ‘glue’

that binds cells together in a tissue. This permits thebacteria to pass between the cells and spread throughoutthe tissue. These bacteria are often introduced by woundsthat penetrate deeply and cause extensive tissuedamage. A variety of other enzymes are secreted bydifferent bacteria for tissue penetration and destruction.Some of these enzymes destroy molecules in the plasmamembranes of the host cell, causing its membrane toburst open and resulting in the host cell’s death. Othersecreted enzymes destroy white blood cells and allowcaptured bacteria to escape phagocytosis.

Tissue Damage by Bacterial Toxins

Poisonous chemicals that damage host tissue arecalled toxins. There are two categories of bacterialtoxins—

(i) Those produced and released by living, growingbacteria and

(ii) Those released from dead, disintegrating bacteria.The first category consists of toxins produced either

by pathogenic bacteria infecting the body (cholera andtetanus) or by bacteria contaminating the food (staph foodpoisoning).

Cholera bacteria enter the digestive tract with conta-minated food and water. As these bacteria grow in theintestines, they release a toxin that irritates the intestinallining causing severe diarrhea. Such diarrhea results inrapid dehydration and electrolyte loss.

Tetanus bacteria typically enter the body through awound. As they grow in the tissues, tetanus bacteriarelease a toxin which is transported by the blood andcauses damage elsewhere. The toxins affects the nervesthat stimulate skeletal muscles. As a consequence,muscles do not relax properly and spasms result.

Staph food poisoning, once known as ptomainepoisoning, results from toxins produced by staphyloccusbacteria growing on food. These bacteria can be part ofthe normal flora of the nasal passages but can also causeboils, pimples, and other skin infections. Therefore,sneezing and coughing on food and preparing food withunprotected hands can lead to the deposition of millions ofstaph bacteria. If such contaminated food remains at roomtemperature for a few hours, the bacteria will have time togrow, producing and releasing harmful quantities of toxin.When ingested with food, the toxin irritates the gastroin-testinal tract, causing vomiting, nausea, and diarrhea.

The second category of toxins are released from thedisintegrating cell wall of dead bacteria. Such toxins areassociated with the bacteria that cause salmonellosis, acommon contaminant of poultry. When these bacteria areingested, stomach acid and other digestive secretions killmany of them. As the bacteria disintegrate, the toxin isreleased. These toxins cause fever, weakness, intestinalbleeding, and even shock.

Actions of Pathogenic Viruses

Viruses reproduce by entering a living host cell. Oncethey are inside, viral nucleic acid (RNA or DNA) repro-grammes cell operations to serve the goals of viral repli-cation. The specific symptoms of a viral disease are dueto the type of cell infected, damaged and killed.

C.S.V. / October / 2009 / 991

Viral infection of a host cell is traditionally divided intofour sequential phases : attachment, penetration, biosyn-thesis and assembly, the release of viruses from hostcells occur in several different ways, depending on thetype of virus. Viruses lacking an envelope are often allreleased rapidly as the dead host cell disintegrates.Viruses with an envelope are released slowly over aperiod of time without immediate disintegration of the hostcell. As single viral particles are released, they acquire alittle bit of the host’s plasma membrane, and thisconstitutes their envelope. Of course, the host celleventually disintegrates as a result of the accumulateddamage.

Envelope

Protein of envelope

Nucleic acid

Capsid

Receptorsites

PlasmaMembrane

Copies ofnucleic acid

Newcapsidprotein

Newlyassembledvirus

Protein of virus envelope binds to protein receptors on host cell membrane.

Attachment

Copies of viral nucleic acid are replicated. Viral nucleic acid is used to synthesize new capsid proteins by employing organelles of host cell.

Biosynthesis

Newly synthesized capsid proteins and viral nucleic acid molecules are assembled into new viruses.

Assembly

Viruses with envelopes acquire their envelopes as they are relea-sed from host cell.

Release

Virus envelope fuses with plasma membrane, and virus capsid and nucleic acid enter host cell. Capsid disintegrates and nucleic acid is released.

Penetration

m-RNA

Fig. : Stages and associated events of viral multiplication. The virus shown here has an envelope.

In addition to rapid release and slow release, there isa third possibility. In this case, new viruses do not kill thehost cell and are not immediately released. Instead, theybecome lifelong residents of the host cell. During thisresidency, their periodic release may cause painful symp-toms. The herpes simplex virus that causes fever blistersand genital herpes is an example of this possibility.

Initially, the herpes virus infects and damages epithe-lial tissue. It then spreads to nerve cells and becomesinactive. During the inactive period, the viruses reside innerves near the skin. Physical and emotional stress may

reactivate the viruses. Upon reactivation, they leave thenerve cells (without damaging them) and infect epithelialcells of the skin, where damage occurs. These active andinactive episodes can be periodically repeated throughouta lifetime.

Chicken pox is another lifelong viral disease. After theusual childhood case of chicken pox, this virus remainsinactive in a person’s nerve cells. If reactivated later in life,it causes the painful skin condition known as shingles.

There is yet another strategy for viral survival, repro-duction, and damage to host cells. After penetrating a cellsome viruses do not immediately follow the sequence of

events outlined above. Instead, before it produces viru-ses, the viral nucleic acid is inserted into and becomespart of the host cell’s DNA, a process called viralintegration. The viral nucleic will remain integrated withinthe host cell’s DNA for the life of the cell. Every time thehost cell’s DNA is replicated before mitosis, the viralnucleic acid will also be replicated. When the host celldivides after mitosis, the viral nucleic acid will be passedwith the cellular DNA to the two progeny cells.

Viral integration may persist for months or yearswithout damage to the host cells. However, all is not well.

C.S.V. / October / 2009 / 992

Eventually the viral nucleic acid may become activated bychemicals or other microbial infections. Upon activation,the integrated viral nucleic acid produces nucleic acid andviral proteins that are assembled into viruses. This pro-cess damages and eventually destroys the host cell. Forgood reason, such viruses are getting much attentiontoday. The Human Immunodeficiency Virus (HIV) thatcauses Acquired Immune Deficiency Syndrome (AIDS) isan RNA virus, and its integration into the host’s DNAchromosomes requires that its RNA be copied as DNA.

Actions of Pathogenic Fungi and ProtozoaFungi have mechanisms similar to those of bacteria

for invading and killing human tissues. By releasingenzymes and chemical toxins, pathogenic fungi invadeand digest the tissues, using the breakdown products asnutrients for their growth. Pathogenic fungi have two mainportals of entry : the respiratory system and the skin.Fungal spores may be inhaled with the air.

A common respiratory system disease caused by afungus is histoplasmosis. This fungus grows in soils richin nutrients from birds congregate are often places wherehistoplasmosis infections are prevalent.

The most common fungal infections of the skin affectits upper layer and associated structures. Ringworm is thecollective name of such fungal skin diseases, whichinclude athlete’s foot. These fungi release an enzymewhich breaks down the keratin of the skin’s upper layer,hair, and nails. Yeasts are fungi, and some yeast infec-tions are caused by a pathogenic yeast named Candida.These fungi can infect the upper skin surface as well asthe mucous membranes of the mouth, the intestinal tract,and the vagina.

Pathogenic protozoa infect a range of organs, includ-ing the intestinal tract, brain, liver, and blood. Intestinaltract infections are contracted from food and liquidscontaminated with the pathogen. Amoebic dysenteryand Giardiasis are two common intestinal tract infectionscaused by protozoa. In these two diseases, the activeprotozoa cells are destroyed by stomach acid beforeentering the intestines. However, their cysts are resistantto stomach acid and they enter the intestinal tract, wherethey develop into active cells. It is these cells that invadethe intestinal lining, causing violent attacks of diarrhea,abdominal cramps and nausea.

Malaria is the world’s most widespread diseasecaused by protozoa. It is transmitted by between humansby a mosquito that feeds on human blood. In malaria, theprotozoa enter the bloodstream, mature in the liver andthen reenter the blood to infect red blood cells (RBCs).Over a period of weeks, RBCs are destroyed, leading toanaemia, capillary obstruction, and general tissuedamage. However, with appropriate care, most patientsrecover.

Control of MicrobesMicrobes are everywhere in the environment. There

are medical and economic situations in which microbesmust be greatly reduced. Sterilization destroys all themicrobes present, while various other processes, suchas pasteurization, greatly reduce their numbers. Theseprocesses and methods are of direct personal impor-

tance in the prevention of infections disease and foodspoilage.

Physical methods of controlling microbes—Tosurvive and grow, all microbes require certain minimumconditions of moisture, pH, temperature, and O2 or CO2.These physical conditions vary among bacteria, fungi, andprotozoa and among the microbes in each group. Suchvariation means that some microbes are more suscep-tible to physical damage than are others. Methods suchas heating, cooling, drying, and radiation are used todestroy microbes, reduce their numbers, or slow theirgrowth.

Chemical methods of controlling microbes—Avariety of chemical substances either destroy microbes orinhibit their growth. Chemicals that do this on nonlivingsurfaces such as kitchen equipment and bathroom facili-ties are known as disinfectants. Milder chemicals calledantiseptics do the same thing on body surfaces withoutcausing irritation.

Disinfectants and antiseptics act by disrupting plasmamembranes, cell walls, nucleic acid molecules, or pro-teins, including enzymes. If applied at appropriate concen-trations, chemicals such as alcohols, chlorine, and iodineare effective in destroying microbes. Chemicals are alsoused to purify drinking water. For example, drinking watercan be purified by adding 2 drops of household bleach(such as chlorox or purex) to a litre of water and waiting30 minutes before drinking.

Use of Antibiotics

Antibiotics are natural chemical substances producesby a variety of bacteria and fungi. Antibiotics are effectiveagainst bacteria because they interfere with cellularprocesses and structures unique to bacteria. Protozoa,viruses, fungi, and human cells which lack these struc-tures are not harmed. Antibiotics kill or inhibit bacteria byinhibiting cell wall synthesis and protein synthesis or bydamaging the plasma membrane.

Antibiotics such as penicillin and bacitracin interferewith the production of bacterial cell walls. A weakened cellwall decreases the bacteria’s ability to withstand theosmotic pressure of the surrounding tissue fluid, and thebacterial cell ruptures and dies.

Other antibiotics, such as tetracycline and erythromy-cin, inhibit bacterial protein synthesis without affectingprotein synthesis in eukaryotic cells. This fortunate distinc-tion occurs because bacterial ribosomes, the site ofprotein synthesis, are slightly different from eukaryoticribosomes.

Bacterial plasma membranes have the same generalmolecular structure as the plasma membranes of euka-ryotic cells. However, bacterial membranes are composedof different lipids. Antibiotics such as colistin and poly-myxin B react with these lipids and damage the mem-brane. The damaged membrane leaks, and as a result,the bacteria die.

In general, antibiotics have been wonder drugsagainst specific pathogenic microbes. Many of us arealive today because of them. New antibiotics that have

C.S.V. / October / 2009 / 993

been discovered in nature or synthesized in the laboratoryhave greatly increased our ability to successfully treatinfections diseases. However, more and more, physiciansand scientists are finding pathegenic bacteria that areresistant to antibiotics.

Antibiotic Resistance

Antibiotics kill susceptible cells, but there are alwaysa few cells whose genes make them resistant toantibiotics. Before the widespread use of antibiotics, therewere so few resistant cells that they did not present muchof a problem. However, with widespread indiscriminateuse of antibiotics over 50 years, these resistant cells havesurvived and flourished. Today, resistant microbes are amajor medical concern. Tuberculosis, gonorrhea, syphilis,and pneumonia are examples of bacterial diseases whosepathogens have developed some degree of antibioticsresistance.

A recent and serious example of antibiotic resistanceis provided by the bacteria that cause tuberculosis (TB).Tuberculosis is a serious infection of the lungs. It istransmitted through the air and is extremely contagious.

Before antibiotics, TB was difficult to treat, debilitating,deadly. TB has been in decline for the last four decades.With antibiotics, patients with TB could be effectivelytreated at home without risk to family members, friends,and the general population. However, since 1985, thenumber of reported TB cases has increased an alarming20 per cent.

This resurgence of TB has occurred for severalreasons. Difficult economic times have reduced themoney available for medical diagnosis, care, and antibio-tics. At greatest risk are the impoverished, the homeless,and drug adicts. A lack of adequate health care amongthese populations had made it difficult to initiate andcomplete antibiotic therapy for TB. To successfully treatTB, antibiotics must be taken continuously for 6 to 9months. If treatment is discontinued before completion,the surviving tuberculosis bacteria become most resis-tant to be killed by antibiotics. As a consequence, anti-biotic-resistant forms of the bacteria proliferate and arespread to others. Because the TB pathogen is spreadthrough the air, everyone is potentially susceptible toinfection.

A Summary of Selected Methods Used to Kill Microbes or Inhibit their Growth

Method ofControl

How to Perform Control Method Suggested Uses Action on Microbes

Heat

Incineration Burn by fire or pass materialthrough open flame

Flammable objects (paper);inflammable objects (knife)

Destroys living microbes, endospores,and cysts

Dry heat Oven temperatures : 2 hours at160°C (320°F)

Glassware, metals, soil Destroys living microbes, endospores,and cysts

Moist heat Boiling water or steam for 30minutes

Glassware, metal, food Destroys living microbes and cysts; doesnot destroy endospores, some viruses

Autoclave or pressure cooker (15minutes at 121°C (250°F)

Glassware, metal, food Destroys living microbes, endospores,cysts, and viruses

Pasteurization 15 seconds at 72°C(162°F)

Foods such as milk and beer Reduces microbe numbers; destroyspathogens

Cold

Refrigeration Temperature at 4°C (40°F) Foods and other perishables Slows growth of microbes

Freezing Temperature below 0°C (32°F) Foods and other perishables Destroys some microbes; stops growthof others

Drying

Elimination of water from product Foods Destroys some microbes; stops growthof others

Radiation

Ultraviolet Surfaces of materials Radiation destroys living microbes,

Gamma rays Surface and interior of materials endospores, cysts and viruses

Chemicals

Isopropyl alcohol

Apply at 70–90% concentration Minor cuts and abrasions; non-living surface structures

Destroys living microbes ineffective forendospores, cysts and some viruses

Chlorine Household bleach

(5–10% concentration)

Nonliving surface structures Destroys living microbes and viruses;slowly destroys endospores and cysts

Household bleach

(2 drops per litre)

Treat questionable drinking waterfor 30 minutes

Destroys living microbes but ineffectivefor endospores, cysts, and many viruses

Iodine Tincture of iodine

(2% dissolved in alcohol)

Minor cuts and abrasions Destroys living microbes, endospores,cysts, and many viruses

C.S.V. / October / 2009 / 994

Review at a Glance� Bacteria are prokaryotes. Bacterial shapes are maintained by a rigid cell wall.� Fungi are eukaryotes and have a cell wall chemically different from that of bacteria. Fungi are either unicellular (yeast) or multi-

cellular filaments (molds).� Protozoa are eukaryotes and lack a cell wall. Some protozoa develop protective cysts that are resistant to environmental extre-

mes.� Viruses must infect a host cell to produce new viruses. A virus consists of nucleic acid (DNA or RNA) surrounded by a capsid of

protein. Some viruses have an outer envelope surrounding the capsid.� Pathogens are microbes that cause disease.� Opportunistic pathogens usually infect only when there is an opportunity for their growth.� Communicable diseases are transmitted between individuals either directly or indirectly.� Bacteria produce disease by releasing either enzymes or toxins that damage tissue. Enzymes and many toxins are secreted by

living bacteria, some toxins are released from the cell wall after bacteria die.� Virus infection of a host cell occurs in four stages : attachment, penetration, synthesis and assembly, and release.� Viral release can happen in four ways : rapidly killing the host cell; slowly over time without immediately killing the host cell;

periodically without killing the host cell, and slowly after integration into host DNA.� Fungi and protozoa release enzymes and toxins for invading and destroying tissues.� Physical methods for controlling or killing microbes include heating, cooling, drying, and radiation.� Chemical methods for killing or inhibiting microbes include disinfectants for use on nonliving surfaces and antiseptics for use on

body surfaces.� Antibiotics are natural chemicals produced by certain bacteria and fungi.� Antibiotics resistance is a worldwide concern.� Tuberculosis is a disease caused by bacteria that are now resistant to many antibiotics.

OBJECTIVE QUESTIONS

1. The disease commonly called‘Lock Jaw’ is caused by—(A) Bacillus pertusis(B) Mycobacterium(C) Clostridium tetani(D) None of the above

2. Food poisoning is caused by—(A) Entamoeba coli(B) Salmonella(C) Giardia(D) Shigella

3. A dreaded viral disease whichhas been almost completely era-dicated all over the world—(A) Chicken pox(B) Polio(C) Measles(D) Small pox

4. Syphilis is a veneral diseasewhich is caused by—(A) Treponema pallidum(B) Neisseria(C) Vibrio(D) Cornybacterium

5. Which one of the following is nota communicable disease ?(A) Tuberculosis(B) Diphtheria(C) Cholera(D) Cancer

6. Meningitis is—(A) Viral disease(B) Bacterial disease

(C) Fungal disease(D) None of the above

7. The toxins produced by tetanusmicrobes affects—(A) Voluntary muscles(B) Involuntary muscles(C) Both voluntary and involun-

tary muscles(D) None of the above

8. Mumps is a viral disease causeddue to inflammation of—(A) Submaxillary gland(B) Parotid gland(C) Sublingual gland(D) Infraorbital gland

9. All the following diseases arerelated with respiratory systemexcept—(A) Asthma(B) Bronchitis(C) Encephalitis(D) Pneumonia

10. Which of the following disease iscaused by protozoa ?(A) Amoebic dysentery(B) Tuberculosis(C) Taeniasis(D) Typhoid

11. Which one of the following is acommunicable disease ?(A) Diabetes(B) Hypertension

(C) Kwashiorkor(D) Malaria

12. Which of the following set inclu-des bacterial diseases ?(A) Cholera, typhoid, mumps(B) Tetanus, tuberculosis, mea-

sles(C) Malaria, mumps, poliomye-

litis(D) Diphtheria, leprosy, plague

13. Diarrhea causes—

(A) Typhoid

(B) Pneumonia

(C) Dehydration

(D) Whooping cough

14. Viral disease Trachoma is rela-ted with—(A) Eyes (B) Skin(C) Liver (D) Muscles

15. Bacillary dysentery is causedby—(A) Shigella(B) Salmonella(C) Entamoeba(D) Proteus

ANSWERS

●●●

C.S.V. / October / 2009 / 995

Anatomy

Liver is largest gland, usuallyendodermal in origin and arising as adiverticulum of gut. It is situated onright side beneath the diaphragm;occupies the right hypochondrium,epigastrium and part of left hypo-chondrium; level with bottom ofsternum; undersurface concave;covers stomach, duodenum; hepaticflexure of colon, right kidney andsuprarenal capsule. Liver consists ofa continuous parenchymal massarranged to form a system of wallsthrough which venous blood emana-ting from the gut must pass. This stra-tegic localization between nutrient-laden capillary beds and the generalcirculation is associated with hepaticregulation of metabolite levels in theblood through storage and mobili-zation mechanisms.

Ligaments supportingliverDiaphragm

Right lobe

Left lobe

Falciformligament

Ligamentum teres

Gall bladder

Liver of Man (Anterior view)

The human liver is a massivewedge-shaped organ divided into alarge right lobe and a smaller leftlobe. It is completely covered by atough fibrous sheath, Glisson’scapsule, which is thickest at thetransverse fissure. At this point thecapsule carries blood vessels andhepatic duct, which enter the organ atthe hilus. Strands of connective tissueoriginating from the capsule enter theliver parenchyma and form thesupporting network of the organ andseparate the functional units of theliver, the hepatic lobules. Structuralunit lobule is roughly hexagonal blockof cuboidal cells (hepatocytes).

The many intrahepatic bilepassages converge and anastomose,finally leading into the hepatic duct,the excretory channel of the liver.This structure receives the cysticduct, on the end of which is situatedthe gall bladder. The union of thecystic and the hepatic ducts forms thecommon bile duct or the ductuscholedochus, which enters theduodenum at the papilla of vater. Aring of smooth muscle at the terminalportion of the choledochus, theSphincter of Oddi, permits thepassage of bile into the duodenum byrelaxing. The bile leaving the liverenters the gall bladder, where itundergoes concentration principallythrough loss of fluids absorbed by thegall bladder mucosa. When bile isneeded in the small intestine fordigestive purposes, the gall bladdercontracts and the sphincter relaxes,thus permitting escape of the viscidgall bladder bile.

Usually, within the wall ofduodenum the first portion of thesmall intestine, the common bile ductand the main channel for pancreaticjuice come together to form thechannel called the ampulla of vater,which then empties into theduodenum. The portion of thecommon duct within the duodenalwall is encircled by muscle fibrescalled the Sphincter of Oddi.

Gross Structure

The liver has been described asa collection of units called lobules,each of which contains in its centre, abranch of the hepatic vein and in itsouter areas, a complex known as aportal tract. The portal tract includes abile duct, a small branch of the portalvein and a branch of the hepaticartery. Between the tributary of thehepatic vein and the portal tract arecolumns of hepatic cells and bloodchannels called sinusoids.

The walls of the sinusoids aremade up of two types of cells,

endothelial or lining cells and Kupffercells. The Kupffer cells are capable ofphagocytosis—the ingestion of othercells and foreign particles. They alsohave important functions in theproduction of antibodies. Ligamentsand pressures exerted by the mus-cles of the abominal wall hold theliver in position. The hepatic cells, thefunctioning cells that make up about60 per cent of the liver are polyhedral.Each cell usually has one nucleusand the cells multiply by mitosis.Between the hepatic cells and thewalls of the sinusoids there is tissuefluid, which flows outward into thelymphatic vessels. The lymph chan-nels are located in connective tissuearound the portal vein.

Liver cells are metabolically veryactive cells, having abundantorganelles, particularly the mitochon-dria and also abundant glycogengranules, fat globules and vacuolesfilled with enzymes (urease, peroxi-dase etc.) and iron—containing com-pounds (ferritin, haemosiderin etc.).

Intracellular Structure

The hepatic cell, also called thehepatocyle, has more metabolicfunctions than any other cell of the

Bilecapillary

Intra lobularvein

Inter lobularvein

Bile ductVein

Artery

Kupffer'scells

Kupffer'scells

Glisson'scapsule

Sinusoid

Hepaticcells

Hepatic cord

Hepatic cells

Sinusoid

Lymph vessel

Intracellular Structure of Liver

C.S.V. / October / 2009 / 996

body and is the producer of bile. In itscytoplasm (substance outside thenucleus) the net like structure knownas the smooth endoplasmic reticulumis the site at which the bile pigmentbilirubin is metabolized and manyenzymes, such as those necessary tomake glucose available to the blood,are synthesized. The bile, needed inthe digestion of fats, are formed atthis point and drugs are detoxicated.

The rough endoplasmic reticulumproduces certain proteins, such asthe albumin and clotting factors of theblood. The mitochondria have manyfunctions, including the production ofenzymes that play a role in thesynthesis of glycogen and enzymesinvolved in the metabolism of fats.

Other structures of the hepaticcell cytoplasm—the network of finetubes known as the Golgi apparatusand the minute bodies and channelsknown, respectively, as the lysosomesand canaliculi—act as the cellsexcretory apparatus. The canaliculiand the nearby tissues are especiallyinvolved in bile excretion.

Blood Supply

Venous blood from the intestine(carrying digested nutrients), and to alesser extent from spleen andstomach, converges upon a shortbroad vessel, called the hepatic portalvein which enters the liver through adepression in the dorsocaudal surfacetermed the porta hepatis. There thehepatic portal vein divides into a shortright branch and a longer left branch.These vessels then ramify into smallbranches which actually penetrate thefunctional parenchymal mass as theinner tubes of the portal canals.

The hepatic artery (a branch ofceliac artery) also enters at the portalhepatis and ramifies into smallerbranches, which flank the portalvenules within the portal canals. Thebranches of the portal vein andhepatic artery then empty intosinusoids, which are major regions ofhepatovascular exchange. Theycommunicate with small branches ofthe hepatic veins and through thehepatic vein, the blood is returned tothe heart by way of the vena cava.

Gall Bladder

It is a hollow muscular organ pre-sent in humans and most vertebrates

Key Concepts● Liver is also known as HEPAR.

● Liver is characterised by presenceof Glisson’s capsule.

● Glisson’s capsules are polygonaland formed by connective tissue.

● Liver is largest gland of body.● Kupffer’s cells are found in liver.● Liver is endodermal in origin.

● Liver of man is bilobed, right lobe ismuch larger than left lobe.

● Liver of Frog is trilobed.

● Liver of Rabbit is 5—lobed :1st lobe — Caudate2nd lobe — Right central and islargest lobe.3rd lobe — Left central4th lobe — Left lateral5th lobe—Spigelian and is smallest.

● Gall bladder is situated with rightlobe.

● Gall bladder is absent in horses.

● Bile capillaries unite to form hepaticduct.

● Bile duct is also known ascholedochus duct.

● In Rabbit bile duct opens intoduodenum separately.

● In man bile duct first opens intoAmpulla of vater.

● Ampulla of vater also receivespancreatic duct in Man.

● Opening of Ampulla of vater issurrounded by Sphinctor of Oddi.

● Pancreatic duct is also known asWirsung duct.

● Sinusoids are present betweenhepatic cords.

● Liver is attached with the wall ofcoelom by septum transversum.

(absent in horses), which receivesdilute bile from the liver, store andconcentrates it, discharges it into theduodenum. Although not a vitalorgan, it is of great importance inhumans because it stores bile andregulates binary tract pressures.

In humans, evacuation of the gallbladder is accomplished by a triggermechanism which is set off by thepresence of fatty foods, meat andhydragogue cathartics in the duode-num and upper jejunum. Absorptionof these substances by the mucousmembrane results in the release ofcholecystokinin, a hormone whichrapidly circulates in the bloodstreamand simultaneously produces con-traction of the gall bladder andrelaxation of the sphincter of Oddi.The most effective food is egg yolkwhich contains certain l -amino acids.

Gallstones and CholecystitisGallstones are round, oval or

faceted concretions formed within thegall bladder from the salts and pigmentof bile. Such stones may also beformed in any of the bile ducts within oroutside the liver but the incidence thereis low compared to the numberoriginating in the gall bladder.Gallstone may be composed solely ofcalcium, cholesterol or bilirubin, but themost common type is the cholesterol—containing stone gallstones form whenthe bile contains more cholesterol thancan be kept in solution. The mostfrequent complication of gallstones ischolecystitis.

Cholecystitis is inflammation ofthe gall bladder, a common disease inhumans. It is nearly always associatedwith gallstones and is particularlycommon in obese middle-agedwomen.

Functions of LiverThe liver receives blood from the

portal vein and thus in the first organto receive blood from intestines,where the blood has absorbed thefinal products of digestion and decom-position products. From this blood theliver removes glucose, from which itsynthesizes glycogen, which it stores.It deaminizes amino acids with theresultant formation of ammonia, whichis converted into urea. Hippuric acidand uric acid are synthesized in theliver. The liver incorporates suchproteins as albumin, prothrombincomponent, fibrinogen, transferrin andglycoprotein. The liver is important inthe biotransformation (i.e., so calleddetoxification) of such substances asindole and skatole, which may beabsorbed into the blood from theintestine.

The liver excretes bile pigmentsbilirubin and biliverdin, formed in thecells of the reticuloendothelial systemin various parts of the body fromhaemoglobin derived from effete(exhausted and no longer functioning)red corpuscles. It synthesizes fibrino-gen and prothrombin, blood consti-tuents essential for clotting. Liver isthe source of heparin, an anticoagu-lant and of red blood cells in thefoetus and is the main site for theproduction of plasma proteins.Reticuloendothelial cells (Kupffercells) present in the lining of thesinusoids, act to filter out and destroybacteria present in the blood stream.

C.S.V. / October / 2009 / 997

The liver also performs thefollowing functions. It is a storageplace for vitamin B12 (the anti-pernicious anaemia factor) and thefat-soluble vitamins A, D, E and K. Itplays a role in the regulation of bloodvolume and is one of the mainsources of body heat. It is importantin lipid metabolism. Cholesterol, whichis found in most body cells and is amajor constituent of bile, is manu-factured mainly in the liver.

BileBile is secretion of liver. It is a

thick, viscous fluid with a bitter taste.The bile from liver is straw coloured,while that from the gall bladder variesfrom yellow to brown and green. Liversecretes 800 to 1000 ml of bile in 24hrs. The rate of bile secretion isgreatly increased during digestion.Bile is slightly alkaline having pH 7·8to 8·6. The secretion of bile by liver iscalled choleresis.

Bile Acids

Bile acid is a steroid acid produ-ced in the liver. The bile acids lowersurface tension and promote emulsi-fication of fat to aid in digestion andabsorption of fats from the intestine.The bile acids have a five-carbon sidechain and occur in the liver incombination, through peptide linkage,with glycine and taurine, formingglycoholic acid and taurocholic acidrespectively. Other bile acids arecholic, deoxycholic, lithocholic andchenodeoxycholic acid.

Bile Salts

Bile salts are alkali salts of bilesodium glycocholate and sodiumtaurocholate.

Bile Pigments

Bile pigments are principally bili-rubin and biliverdin. In addition, bilecontains cholesterol, lecithin, mucinand other organic and inorganic subs-tances.

Bilirubin is the predominantorange pigment of bile. It is the majormetabolic break down product ofhaeme, the prosthetic group ofhaemoglobin in red blood cells andother chromoproteins such as myo-globin, cytochrome and catalase. Inmammalian bile essentially all of thebilirubin is present as a glucuronideconjugate. Bacterial flora of intestinefurther reduces the bilirubin to colour-

less urobilinogen. Most of the urobili-nogen is either reduced t o sterco-bilinogen or oxidized to urobilin.These two compounds are then con-verted to stercobilin, which is ex-creted in the faeces and gives thestool its brown colour.

Pathological TestsGmelin’s Test—A test for bile in urine.Hay’s Test—A test for bile acids inurine.Van den Bergh’s Test—A test todetect the type of bilirubin.Iodine Test—A test for bile in urine.

Disorders of Liver

JaundiceJaundice is the yellowness of the

eyes and skin that comes fromexcess amounts of bile pigment—bilirubin in the blood, may arise in anyof a number of ways. (1) The hepaticcells may form more bilirubin thanusual, because of an abnormally highlevel of red blood cell destruction.This type is called haemolytic jaun-dice. (2) Impaired uptake or transportof bilirubin by the hepatic cells mayoccur. This congenital disorder isknown as Gilbert’s syndrome. (3)A defect within the hepatic cell itselfor an obstruction in the bile ductsystem may prevent the excretion ofbilirubin glucuronide into the bile. (4)A number of these defects may occurat one time. In hepatocellular jaun-dice, for example, there may beabnormalities in the transport of biliru-bin, in its combining with glucuronicacid to form bilirubin glucuronide andits excretion in the bile.

HepatitisIt is an inflammation of the liver

caused by a number of etiologicalagents, including viruses, bacteriafungi, parasites, drugs and chemi-cals. All types of hepatitis are charac-terized by distortion of the normalhepatic lobular architecture due tovarying degrees of necrosis of livercells, inflammation and Kupffer cellenlargement and proliferation.

The most common infectioushepatitis is of a viral etiology. The twowell-recognized forms are infectioushepatitis (hepatitis-A) and SerumHepatitis (hepatitis-B). Hepatitis-A isorally acquired, has an incubationperiod of 15 to 50 days. Hepatitis-B isparenterally transmitted, has anincubation period of 60 to 90 days.

The incubation of delta agent hepa-titis is 14 to 70 days.

Hepatitis-A and delta agenthepatitis are spread mostly by personto person via the faecal-oral route butmay occur by water or food contami-nated by the virus. Hepatitis-B isspread by blood and serum—derivedfluids.

Cirrhosis

Cirrhosis is a chronic disease ofthe liver characterized by formation ofdense perilobular connective tissue,degenerative changes in parenchy-mal cells, alteration in structure of thecords of liver lobules, fatty andcellular infiltration and sometimesdevelopment of areas of regeneration.In addition to clinical signs andsymptoms inherent in the cause ofthe cirrhosis, those due to cirrhosisare the result of loss of functioningliver cells and increased resistance toflow of blood through the liver (portalhypertension). When severe enough,this leads to ammonia toxicity. Mostcommon form is that of nutritional(also known as alcoholic;Laennec’s or portal) cirrhosis.

Wilson’s Disease

Copper deposits in the liver andin other tissues are characteristic of arare form of cirrhosis known asWilson’s disease or hereditary, hepa-tolenticular degeneration. Deposits ofcopper in the brain may cause tremorand other abnormalities related to thenervous system. A brownish-greencolouration of the cornea of the eye,called Kayser—Fleischer rings, isalso caused by the copper deposits.

Alcoholism is clearly associatedwith liver cirrhosis. An alcoholic withcirrhosis shows striking improvementin condition when he has rested inbed and stopped consumption ofalcohol.

Weil’s Disease

Weil’s disease is a form ofhepatitis caused by infection withLeptospira icterohaemorrhagiaefrom the urine of infected rats. Itoccurs most often in men engaged insuch pursuits as farm work or coalmining. The disease results in deepjaundice.

Other diseases of liver arecarcinoma of liver cells, syphilis ofliver, pyrogenic liver abscess andtuberculosis of liver.

C.S.V. / October / 2009 / 998

OBJECTIVE QUESTIONS

1. Site of heat production in thebody is—(A) Liver (B) Lung(C) Kidney (D) Spleen

2. Liver is also excretory organbecause—(A) Urea is formed these(B) Deamination takes place(C) Eliminates of bile pigments(D) None of the above

3. Choleresis is related with—(A) Pancreatic juice secretion(B) Bile secretion(C) Bacterial infection of liver(D) None of the above

4. Bile pigments are—(A) Helpful in digestion(B) Toxic in nature(C) Helpful in emulsification(D) None of the above

5. Cystic duct arises from—(A) Kidney(B) Pancreas(C) Gall bladder(D) Liver

6. Liver is—(A) Ectodermal in origin(B) Endodermal in origin(C) Mesodermal in origin(D) None of the above

7. Liver of frog is—(A) Bilobed (B) Trilobed(C) Five-lobed (D) Single-lobed

8. The site at which bilirubin ismetabolized—(A) Smooth endoplasmic reticu-

lum(B) Rough endoplasmic reticu-

lum(C) Both of the above(D) None of the above

9. In liver, albumin and blood clot-ting factors are produced by—(A) Smooth endoplasmic reticu-

lum(B) Rough endoplasmic reticu-

lum(C) Both of the above(D) None of the above

10. Liver is characterised by thepresence of—(A) Glisson’s capsule(B) Kupffer’s cells

(C) Both of the above(D) None of the above

11. Bile duct in Man is known as—

(A) Whartons’s duct

(B) Wirsung duct

(C) Choledochus duct

(D) Cisternae

12. Bilirubin is produced from haemo-globin by—

(A) Liver

(B) Reticuloendothelial cells inbone marrow

(C) Gall bladder

(D) All of the above

13. Biliverdin is formed by—(A) Oxidation of bilirubin(B) Liver(C) Urea(D) None of the above

14. Smallest lobe of liver of Rabbitis—

(A) Caudate lobe

(B) Right central lobe

(C) Spigelian

(D) None of the above

15. In man Ampulla of vaterreceives—

(A) Pancreatic duct

(B) Bile duct

(C) Both of the above

(D) None of the above

16. Bile acids is/are—

(A) Glycoholic acid

(B) Taurocholic acid

(C) Lilhocholic acid

(D) All of the above

17. Bile salts are alkali salts of—

(A) Sodium glycocholate

(B) Sodium taurocholate

(C) Both of the above

(D) None of the above

18. Emulsification of fat is broughtabout by—

(A) Bile pigments

(B) Bile salts

(C) Pancreatic juices

(D) HCL

19. Bile pigments are—(A) Secretory product(B) Excretory product(C) Digestive product(D) All of the above

20. Cholecystitis refers to—(A) Stone in gall bladder(B) Appendix pain(C) Stomach pain(D) None of the above

21. Gilbert’s syndrome is relatedwith—(A) Jaundice(B) Hepatitis(C) Cirrhosis(D) All of the above

22. Which of the following is trans-mitted by blood and serum-derived fluids ?(A) Hepatitis-A(B) Hepatitis-B(C) Both of the above(D) None of the above

23. Laennec’s disorder is relatedwith—(A) Cirrhosis(B) Hepatitis(C) Jaundice(D) All of the above

24. Liver cirrhosis caused by copperdeposition is known as—

(A) Weil’s disease

(B) Wilson’s disease

(C) Laennec’s disorder

(D) None of the above

25. Hepatitis caused by Leptospiraicterohaemorrhagiae is knownas—

(A) Weil’s disease

(B) Wilson’s disease

(C) Both of the above

(D) None of the above

ANSWERS

●●●

C.S.V. / October / 2009 / 999

The ever-growing human population is overexploitingnatural ecosystems to satisfy the variety of needs, whichreflect the increasingly energy-intensive lifestyle. Thisoverexploitation is disturbing the natural balance.

Modern humans appeared around fifty thousand yearsago. Initially the human population was small. Therefore,human interference with nature was minimal. Humanpopulation reached the one billion mark around 1850. Itincreased to 2 billion by 1930, and reached 6·1 billion by2000.

Exponential Growth and Human PopulationExplosion

In 1700 A.D., human population was around 6·6billion. At the beginning of the twentieth century, it reachedto 1·6 billion and by the end of the century, the humanpopulation stood at 6·1 billion. This dramatic increase inpopulation size over a relatively short period is calledpopulation explosion. In the 150 years from 1700 A.D.,human population doubled from 0·6 billion to 1·2 billion. Incontrast, it increased five-fold during the next 150 years.

There is a limit to the maximum population size thatcan be supported with a given space and resource base.The maximum population size that can supported by theenvironment is called the maximum carrying capacity.

Environment has the following three major compo-nents :

1. The first component consists of productive systems,such as croplands, orchards, etc. and provides foodand fibre.

2. The second component comprises protective sys-tems, such as climax forests, oceans etc. It buffers airand water cycles, moderates extremes in temperatureetc.

3. The final component has waste assimilative systems,such as water ways, wetlands, etc. that assimilate thewastes produced by human activities.

The first two of these components constitute the life-supporting capacity and third makes up the waste-assimilative capacity of the environment. The maximumcarrying capacity of the environment depends on theabove two capacities. It is understandable that the popu-lation size should not exceed the maximum carryingcapacity and the utilisation of resources should be suchthat lasting damage to the environment does not occur.

The carrying capacity of the human environment hasbeen increased many times by clever application ofscience and technology, particularly to the productive sys-tems of the environment. As a result, human populationhas been able to maintain exponential growth during thepast 100 years.

When the population increase is nearly a fixed pro-portion of its own size during any period of time, the

growth is said to be exponential (J-shaped growth curve).When population grows exponentially, utilisation ofresources and generation of wastes also grow exponen-tially. However, the exponential growth in resource useand waste generation cannot continue indefinitely.

Environment and Human Population Pressure

The increased levels of environmental degradationexperienced today arise from the following :

Firstly, the world population has increased dramati-cally and secondly, population densities within differentparts of the world are markedly different.

About half of the 6·1 billion people live in poverty andat least one-fifth are severly undernourished or malnour-ished. It is estimated that it takes 12–20 times as muchresource to raise a child in a developing nation. An impor-tant related fact is that 15 per cent of the world’s popu-lation controls about 85 per cent of the resources.

This imbalance is connected to the demographictransition in developed countries. This process has led toaggressive natural resource acquisition and colonialisa-tion. Developing nations, like India, contain the majority ofthe world’s population. They also have a large rural popu-lation, which is shifting to thickly populated urban areas insearch of material wealth.

Typically, urban areas produce little food, consumemore natural resources and generate more waste productsper capita than rural areas. In most cases, urban wastesare hazardous and contaminate the environment withcompounds that are foreign to natural ecosystem and areless subject to natural degradation. Increased urbanisationalso puts pressure on agriculture to produce more foodless land, leading, to increased pollution by intensiveagriculture practice. Thus, the huge human populationpressurises and degrades the environment physically,chemically, biologically and even ethically.

Development and Environment

The extent of resource exploitation is determined bythe size of human population, socio-economic structureand technological advancement of a country. Techno-logical advancement causes an increasing detachment ofthe society from nature, and interferes with the physio-chemical and biological interactions of resources in theform of excessive deforestation, intensive agriculture,indiscriminate mining operations and thoughtless use offossil fuel, etc., deplete in the environment.

Rapidly growing population demands more resourcesand is seldom concerned about the consequences ofresource acquisition and use. The issues of developmentversus environment have led to the concept of sustainabledevelopment. The most widely quoted definition of sus-tainable development is the ‘development that meets theneeds of the present without compromising the ability offuture generations to meet their own needs.’

C.S.V. / October / 2009 / 1000

Sustainable development encourages a process ofchange in which the exploitation of resources, the directionof investments, the orientation of technological develop-ment and institutional changes are all in harmony. Suchdevelopment enhances both present and future potentialto meet human needs and aspirations. Developmentshould not endanger the natural systems that support life.

Human Population Growth

Human population growth rate is measured as theannual average growth rate which can be calculated asfollows :

Average annual growth rate (in per cent)

= ⎣⎢⎢⎡

⎦⎥⎥⎤P2 – P1

P1 × N × 100

where P1 is the population size in the previous census; P2is population size in the present census and N is numberof years between the two census.

Census ascertains the number of individuals presentin a given region at a given time. The time required for apopulation to double itself is called the doubling time.Annual average growth rate and doubling time are the twoimportant indicators of the pace of population growth.Growth rate depends on several factors such as rates ofbirth, death and migration and age-sex ratio.

Growth CurvesTwo basic forms of growth curves can be identified,

the J-shaped growth curve and the S-shaped or sigmoidgrowth curve.

The S-shaped or sigmoid growth curve describes asituation in which a new environment, the population den-sity of an organism increases slowly initially, as it adaptsto new conditions and establishes itself, then increasesrapidly, approaching an exponential growth rate. It thenshows a declining rate of increase until a zero populationgrowth rate is achieved where rate of reproduction(natality) equals rate of death (mortality). This type ofpopulation growth is said to be density-dependent since,for a given set of resources, growth rate depends on thenumbers present in the population. The point of stabilisa-tion or zero growth rate is the maximum carrying capacityof the given environment for the organism concerned.

Time

J-shaped

No.

of i

ndiv

idua

ls

(a)

The J-shaped growth curve describes a situation inwhich, after the initial establishment phase (lag phase),population growth continues in an exponential form until

No.

of i

ndiv

idua

ls

K

S-shaped

Time(b)

Fig. : Population growth forms : (a) J-shaped, (b) S-shaped.K represents the carrying capacity.

stopped abruptly, as environmental resistance becomessuddenly effective. Growth is said to be density-independent since regulation of growth rate is not tied tothe population density until the final crash.

FertilityFertility is the deteminant of the current growth of

population. Birth rate is the number of babies producedper thousand individuals. It is distinct from the populationgrowth rate as it can never be negative, while the lattercan be negative.

Total fertility rate (TFR) is the average number ofchildren that would be born to a woman during her lifetime,assuming the age-specific birth rate of a given year.

Replacement level (RL) is the number of children, acouple must produce to replace themselves. Fertility islargely controlled by economics and by human aspirations.

MortalityMortality is the death rate per thousand individuals. In

most countries, the death rate has dropped almost conti-nuously since the industrial revolution, mainly due toimproved personal hygiene, sanitation, and modern medi-cine. A decrease in death rate would result in increasedpopulation growth rate.

Demographers usually employ crude birth rate andcrude death rate, which are the numbers of live births anddeaths per thousand persons, respectively in the middle ofa given year, i.e., on July 07. The difference between thenumber of births and that of deaths is the rate of naturalincrease. If birth and death rates were equal, a zero popu-lation growth rate would result, which is called demo-graphic transition.

MigrationMigration is the movement of individuals into (immi-

gration) or out of (emigration) a place or country. Migra-tion may also occur within a country, from one region toanother. But migration only between country, influences anation’s population. Only the net immigration, i.e., immi-gration minus the emigration, is added to the populationgrowth by birth. The net immigration may be positive, zeroor even negative.

Age and Sex StructuresThe age structure of a given population refers to the

proportion of individuals of different ages within thatpopulation. This aspect is important because many

C.S.V. / October / 2009 / 1001

functional aspects of the individuals are related to age.For example, infants below one year of age and the olderpeople have higher mortality rate than individuals of otherages. In addition the proportion of reproductively active

males and females in a population influences the popu-lation growth. The number of female individuals in activereproductive age influences the birth rate within apopulation.

OBJECTIVE QUESTIONS

1. Which of the following does notaffect the biotic potential ?(A) Carrying capacity of environ-

ment(B) Female’s age(C) Both (A) and (B)(D) None of these

2. Carrying capacity of environmentis determined by—(A) Birth rate(B) Death rate(C) Limiting resources(D) Population growth rate

3. Exponential growth is associatedwith—(A) J-shaped growth curve(B) S-shaped growth curve(C) Sigmoid growth curve(D) All the above

4. The population is said to be den-sity-dependent when the growthhas—(A) Sigmoid growth curve(B) J-shaped growth curve

(C) Both (A) and (B)(D) None of these

5. Population growth is said to bedensity-independent when thereis—(A) J-shaped growth curve(B) S-shaped growth curve(C) Both (A) and (B)(D) None of these

6. Essay of population was pub-lished by—(A) Darwin(B) Malthus(C) Hugo de Vries(D) None of these

7. The physiological capacity to pro-duce offspring is called—(A) Population growth(B) Birth rate(C) Biotic potential(D) Population explosion

8. The number of individuals inpopulation are added by—(A) Immigration

(B) Birth(C) Emigration(D) Both (A) and (B)

9. The main factor for populationgrowth in India is—(A) More birth rate(B) Less death rate(C) Lack of education(D) All the above

10. Population explosion being wit-nessed at present is mainly dueto—(A) Increase in agricultural pro-

duction(B) Better job facilities(C) Better health care(D) None of these

ANSWERS

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C.S.V. / October / 2009 / 1002

1. Logistic growth occurs whenthere is—(A) Asexual reproduction only(B) Sexual reproduction only(C) No inhibition from crowding(D) A fixed carrying capacity

2. Which of these were first formedduring origin of life ?(A) Coaservates (B) Cells(C) Eobionts (D) Genes

3. Which of the following snake isviviparous ?(A) Krait (B) Viper(C) Cobra (D) Hydrophis

4. The term anaerobic means—(A) Presence of oxygen(B) Without oxygen(C) With glucose(D) Without glucose

5. The extra-embryonic membranesof mammalian embryo arederived from—

(A) Formative cells

(B) Trophoblast

(C) Follicle cells

(D) Inner mass cells

6. Nucleus pulposus is found in—

(A) Medulla oblongata

(B) Femur

(C) Intervertebral disks

(D) Testes

7. Basic unit of protein is—(A) Amide (B) Proton(C) Amino acid (D) Peptide

8. The primates are—(A) Gregarious(B) Plantigrades(C) Omnivorous(D) All of the above

9. Intermediate host of Liver Flukeis—(A) Pila and Unio(B) Pila and Lymnaea

(C) All the above

(D) None of these

10. Pelvic inflammatory disease (PID)is related with—(A) Gonorrhea(B) Chlamydiosis(C) Both (A) and (B)(D) None of the above

11. Antigenic determinant sites bindto which portion of an antibodymolecule ?(A) Heavy chains(B) Light chains(C) Both (A) and (B)(D) None of these

12. Recombination nodules are asso-ciated with—(A) Axons(B) Synaptonemal complex(C) Both (A) and (B)(D) Synovial fluid

13. Liver cirrhosis caused by copperdeposition is known as—

(A) Weil’s disease

(B) Laennec’s disease

(C) Wilson’s disease

(D) None of the above

14. Each redia larva of Fasciolahepatica produces—

(A) 14 to 20 cercaria larvae

(B) 1 to 5 cercaria larvae(C) 14 to 20 metacercaria larvae(D) 100 cercaria larvae

15. Parasympathetic effect—(A) Lowers blood pressure(B) Slows heart rate(C) Promotes digestion(D) All the above

16. In earthworm (Pheretima), thearrangement of blood vessels is—

(A) Different in middle 13 seg-ments

(B) Different in last 15 segments(C) Different in first 13 segments(D) Same throughout

17. Yawning is due to more—(A) CO2 concentration in blood

(B) O2 concentration in blood

(C) Both (A) and (B)(D) None of these

18. Which of the following dissolvesblood clots during fibrinolysis ?

(A) Fibrin

(B) Platelet vector VIII

(C) Plasmin

(D) Thrombin

19. Ratio of oxyhaemoglobin andhaemoglobin in the blood isbased upon—

(A) Oxygen tension

(B) Carbon dioxide tension

(C) Carbonate tension

(D) Bicarbonate tension

20. The genetic material in ‘HIV’ is—

(A) Single-stranded DNA

(B) Double-stranded DNA

(C) Single-stranded RNA

(D) Double-stranded RNA

21. Sum of constructive processes inbody cells is called—(A) Catabolism(B) Anabolism(C) BMR(D) All the above

22. The scientific (zoological) nameof ‘pork tapeworm’ is—(A) Taenia solium(B) Fasciola hepatica(C) Rhabditis citaria(D) Hystrichis engenia

23. The reaction in which glucoseand fructose combine to formsucrose and water is—(A) Exergonic(B) Endergonic(C) Spontaneous(D) Both (A) and (B)

24. Which of these are stellate cells ?(A) Astrocytes(B) Kupffer’s cells(C) Both (A) and (B)(D) None of the above

25. In human embryo and foetus, thebypasses which limit the bloodflow to the liver and lungs is—(A) Ductus venosus(B) Foramen ovale(C) Ductus arteriosus(D) All the above

C.S.V. / October / 2009 / 1003

26. Which form of Trypanosoma gam-biense contains reduced fla-gellum ?(A) Leptomonad(B) Crithidial(C) Trypanosomid(D) Leishmanial

27. Which of the layer of heart-wallconsists cardiac muscles ?(A) Endocardium(B) Myocardium(C) Epicardium(D) All the above

28. Epithelium that contains fatglobules is known as—(A) Nepiepithelium(B) Duraepithelium(C) Pioepithelium(D) Seismoepithelium

29. Septicemia is—(A) Food poisoning(B) Blood poisoning(C) Mental disorder(D) None of these

30. Which of the following animals hasno need of a gallbladder ?(A) Human (B) Horse(C) Dog (D) Lion

31. Cardiac output is determinedby—(A) Heart rate(B) Stroke volume(C) Both (A) and (B)(D) None of these

32. Which of the following is causedin human liver by anaerobic bac-teria ?(A) Hobnail fever(B) Liver spots(C) Foamy liver(D) None of the above

33. Which particular fatty acid is notsynthesized in the human body ?(A) Linoleic acid(B) Glycerol(C) Cholesterol(D) All the above

34. Which of the following is calledliving fossil ?(A) Latimarice(B) Sphenodon(C) Heloderma

(D) Both (A) and (B)

35. Catabolism takes place in—(A) Golgi bodies(B) Nucleus(C) Mitochondria(D) All the above

36. Ligaments join—

(A) Muscle to bone

(B) Muscle to muscle

(C) Bone to bone

(D) Cartilage to bone

37. Tumour in blood vessels iscalled—

(A) Arteriosclerosis

(B) Varicose vein

(C) Angioma

(D) Phlebitis

38. Which of the following body partsis the major cholesterol producingsite in the human ?

(A) Gall bladder

(B) Liver

(C) Pancreas

(D) All of the above

39. Fishes migrating from ocean tofresh water for breeding arecalled—(A) Catadromous(B) Anadromous(C) Amphidromous(D) Bathydromous

40. Which of these factors has littleeffect on blood flow in arteries ?(A) Skeletal muscle contraction(B) Heart beat(C) Total cross-sectional area of

vessels(D) Blood pressure

41. Which of the following animal isan example of class mammalia ?(A) Planorbis (B) Manis(C) Hydrophis (D) Psittacula

42. Interstitial fluid closely resem-bles—(A) Lake water (B) Rain water(C) Sea water (D) Pond water

43. Most appropriate term to describethe life-cycle of Obelia is—(A) Neoteny(B) Metagenesis

(C) Metamorphosis

(D) None of these

44. A working combination of anapoenzyme and a coenzyme istermed as—(A) Enzyme-substrate complex(B) Prosthetic group(C) Enzyme-product complex(D) Holoenzyme

45. Which of the following arthropodsis a chelicerate ?(A) Horseshoe crab(B) Lobster(C) Millipede(D) Grasshopper

46. Pressure filtration is associatedwith—(A) Distal convoluted tubule(B) Collecting duct(C) Glomerular capsule(D) None of the above

47. Meckel’s discs are example of—(A) Tangoreceptors(B) Phonoreceptors(C) Algesireceptors(D) Proprioreceptors

48. A virus that can reproducewithout killing its host is called—(A) Temperate virus(B) Lytic virus(C) Virion(D) Retroactive virus

49. Angulo-splenial bone is foundin—(A) Lower jaw of rabbit(B) Lower jaw of frog(C) Pectoral girdle of frog(D) Pelvic girdle of frog

50. A human population has a higherthan usual percentage of indi-viduals are expected to make upa larger proportion of the nextgeneration. The most likely expla-nation is—(A) Genetic drift(B) Gene flow(C) Natural selection(D) None of the above

ANSWERS WITH HINTS

(Continued on Page 1006 )

C.S.V. / October / 2009 / 1004

1. Czermak's space is found in—(A) Vertebral column(B) Tooth(C) Sternum(D) Synovial joint

2. Capsule of Tenon is associatedwith—

(A) Brain (B) Eyeball

(C) Skin (D) Kidneys

3. The inherited cystic fibrosisdisease is associated with—

(A) Exocrine glands

(B) Apocrine glands

(C) Both (A) and (B)

(D) None of these

4. Which type of brain wave patternis observed in persons solvingmathematical problems ?(A) Beta waves

(B) Alpha waves

(C) Theta waves

(D) Delta waves

5. Curschmann's spiral are relatedwith—

(A) Asthma patients

(B) Tuberculosis patients

(C) Neuritis patients

(D) Congenital heart patients

6. Heart murmur produced inaneurysm, is called—

(A) Hemic murmur

(B) Bruit

(C) Both (A) and (B)

(D) None of these

7. In which of the following organsketone bodies are produced ?(A) Kidneys (B) Spleen(C) Liver (D) Brain

8. The arrangement of numeroussetae in a ring in each segmentof earthworm is known as—(A) Lumbricine(B) Oligochaetine(C) Otochaetine(D) Perichaetine

9. Scyphozoan medusae are morecommonly known as—(A) Sea anemones (B) Hydra(C) Jellyfish (D) Corals

10. Which of the following is calledtissue membrane ?(A) Serous membrane(B) Cutaneous membrane(C) Mucous membrane(D) All the above

11. Diaderm is composed of—(A) Ectoderm(B) Endoderm(C) Both (A) and (B)(D) None of these

12. Accumulation of free radicals inhuman body, can cause—

(A) Cancer

(B) Hardening of the arteries

(C) Cataracts

(D) All the above

13. Ginglymus is—

(A) Hinge joint

(B) Fixed joint

(C) Synovial joint

(D) None of these

14. ‘Tree line’ is a biological indicatorof environmental conditions. Treeline is—

(A) Line of similar trees in aforest

(B) Last line of trees beyondwhich no trees occur

(C) A line of trees of uniformheight and productivity

(D) Trees growing in shade thatsurvive acid rain

15. Which of the following is warm-blooded animal ?

(A) Bat (B) Shark(C) Snake (D) Lizard

16. Which developmental structure inmammalian embryo is regardedas phyletic ?(A) Gill pouches (B) Lungs(C) Heart (D) Kidneys

17. Insertion of viral nucleic acid intoDNA of host cell, is called—(A) Integration(B) Biosynthesis(C) Assembly(D) None of these

18. Clisere is related with—(A) Serial community(B) Xerosere community(C) Climax community(D) Pioneer community

19. In humans, removal parathyroidgland leads to—(A) Tetany(B) Acromegaly(C) Polyuria(D) Diabetes insipidus

20. Muscles associated with hair-roots are responsible for goose-flesh are—(A) Smooth muscles(B) Antagonistic muscles(C) Striated muscles(D) None of these

21. Scotopic vision is—(A) Night vision(B) Daytime vision(C) Defective vision(D) None of these

22. New systematics based ongenetic inter-relationship is knownas—(A) Cytotaxonomy(B) Numerical taxonomy(C) Experimental taxonomy(D) Chemotaxonomy

23. When a person is alert with eyesopen, and actively trying to solvecomplex problems, the EEGshows—(A) Alpha waves(B) Beta waves(C) Delta waves(D) Gamma waves

24. Which tumour is capsulated ?(A) Benign(B) Malignant(C) Both (A) and (B)(D) None of these

25. Endoneurium is also called—(A) Henle's ligament(B) Henle's sheath

C.S.V. / October / 2009 / 1005

(C) Henle's tubule(D) Henle's layer

26. Enzyme that perform condensa-tion reactions involving ATPcleavage is—(A) Ligases (B) Isomerases(C) Lyases (D) Transferases

27. Involuntary discharge of urine inchildren below the age of five, iscalled—(A) Entrain (B) Enuresis(C) Entotic (D) Entropy

28. Thebesian valve occurs at theentrance of—(A) Coronary sinus into right

auricle(B) Left auricle into left ventricle(C) Sinus venosus into right

auricle(D) None of these

29. The membranous lining of thespinal canal is—(A) Endorrhachis(B) Endosalpingia(C) Endosalpinx(D) None of these

30. The reaction in which glucoseand fructose combine to formsucrose and water, is—(A) Exergonic(B) Endergonic(C) Both (A) and (B)(D) None of these

31. The cartilage forming the lowerjaw of cartilaginous fishes istermed—

(A) Dentary

(B) Mentomeckelian

(C) Meckel's cartilage

(D) Angulosplenial

32. Which of the following is ‘inbornerror of metabolism’ ?

(A) Colour blindness

(B) Haemophilia

(C) Hurler syndrome

(D) None of these

33. Sympathetic nerves in humansarise from—

(A) Sacral region(B) Cervical region(C) Thoraco-lumbar region(D) None of the above

34. Which of the following helpsin consolidation of long-termmemory ?(A) Medulla and pons(B) Hippocampus(C) Both (A) and (B)(D) None of these

35. Osteoarthritis disease is usuallyassociated with aging and iscalled a—(A) Communicable disease(B) Degenerative disease(C) Deficiency disease(D) Allergy

36. Septicemia is—(A) Food poisoning(B) Blood poisoning(C) Mental disorder(D) None of these

37. Which brain wave pattern iscommon in young children andsleeping adults ?(A) Beta waves(B) Alpha waves(C) Theta waves(D) Delta waves

38. Which particular fatty acid is notsynthesized in the human body ?(A) Linoleic acid(B) Glycerol(C) Cholesterol(D) None of these

39. The shock caused by bee andwasp sting bite is called—(A) Anaphylatic shock(B) Hyperglycemic shock(C) Hypovolemic shock(D) None of these

40. The posterior part of Amoeba iscalled—(A) Amniton (B) Basal disc(C) Uroid (D) None of these

41. Which of the following enzyme isactivated by peptide hormone toconvert ATP to cyclic AMP ?(A) Pepsin(B) Adenylate cyclase(C) Trypsin(D) All the above

42. The phenomenon of fertilizationwas first perceived by—(A) Hertwig(B) Leuwenhoek

(C) Weismann

(D) None of these

43. Bells palsy is caused due toswelling of which cranial nerve ?

(A) Seventh (B) Tenth

(C) Fourth (D) Fifth

44. Schlemm’s canal is located in—

(A) Eye (B) Cochlea

(C) Spinal cord (D) Vertebra

45. The haploid complement of chro-mosomes of an organism cons-titute its—

(A) Genome

(B) Genotype

(C) Phenotype

(D) Genetic system

46. Inherited Rh factor gene is foundin—

(A) Rh+ individuals

(B) Rh– individuals

(C) AB blood group individuals

(D) O blood group individuals

47. Which of the following dance isperformed by worker honeybee,when food is far away ?

(A) Round dance

(B) Waggle dance

(C) Both (A) and (B)

(D) None of these

48. The prominence in the neckregion known as Adam’s apple isformed by—

(A) Cartilage hyaline

(B) Cartilage semilunar

(C) Cartilage yellow

(D) Cartilage thyroid

49. The protein which maintains themuscular storage of oxygen is—

(A) Myosin

(B) Myoglobin

(C) Actomyosin

(D) All the above

50. Which type of tissue forms theinner lining of a blood vessel ?

(A) Connective

(B) Epithelial

(C) Muscular

(D) None of these

C.S.V. / October / 2009 / 1007

1. Hemoccult test is done to spot anearly stage of cancer in—(A) Colon (B) Breast(C) Blood (D) Skin

2. Hellin’s law is related with—

(A) Eye defect

(B) Twins

(C) Helminth infection

(D) Hearing defect

3. In mitochondria, cristae act assites for—

(A) Breakdown of macromolecu-les

(B) Protein synthesis

(C) Oxidation-reduction reaction(D) Phosphorylation of flavopro-

teins

4. Bowman’s membrane is foundin—(A) Lungs (B) Kidneys(C) Liver (D) Eyes

5. A cell-coded protein, that isformed in response to viral infec-tion, is—(A) Antigen (B) Interferon(C) Histone (D) Antibody

6. An abnormal decrease in totalnumber of WBCs is termed—(A) Leucocytopenia(B) Leucocytopiania(C) Leucocytosis(D) None of these

7. Hassall’s corpuscles are presentin—(A) Thymus (B) Thyroid(C) Kidney (D) Liver

8. Which of these viruses has diffe-rent and varied associations indifferent geographical regions ?(A) Kapos’s virus(B) Epstein-Barr virus(C) Papova virus(D) All the above

9. Hu..rthle cells are present in—

(A) Thyroid gland (B) Liver(C) Spleen (D) Lymph

10. When the individual’s genotypeis XXX, the person is affectedby—(A) Metafemale syndrome(B) Turner’s syndrome(C) Klinefelter’s syndrome(D) Down’s syndrome

11. The transgenic animals are thosewhich have—(A) Foreign DNA is some of its

cells(B) Foreign DNA in all its cells(C) Foreign RNA in all its cells(D) Both (B) and (C)

12. Which of the following is a clam-worm ?(A) Earthworm(B) Hook worm(C) Nereis(D) Filarial worm

13. The concept that ‘populationtends to increase geometricallywhile food supply increases arith-metically’ was put forward by—(A) Stuart Mill(B) Adam Smith(C) Thomas Malthus(D) Charles Darwin

14. Menstrual cycle in mammals isinfluenced by—(A) Estrogen only(B) FSH, LH and estrogen(C) Progesterone only(D) FSH and LH only

15. Although much CO2 is carried inblood, yet blood does not becomeacidic, because—(A) CO2 combines with water to

form H2CO3, which is neutra-lized by NaCO3

(B) CO2 continuously diffusesthrough the tissues

(C) Buffer system of blood playsan important role

(D) All the above

16. Sensation of stomach pain wouldbe carried by—(A) Exteroreceptors(B) Interoreceptors

(C) Proprioreceptors(D) Teloreceptors

17. In protein synthesis, the polyme-rization of amino acids involvesthree steps. Which one of thefollowing is not involved ?(A) Elongation(B) Transcription(C) Termination(D) Initiation

18. Ring chromosomes can pro-duce—(A) Cat-eye syndrome(B) Cri-du-chat syndrome(C) Jacob’s syndrome(D) Edward’s syndrome

19. Which of the following evidencesdoes not favour the Lamarckianconcept of inheritance of acquiredchracters ?

(A) Asence of limbs in snakes

(B) Presence of webbed toes inaquatic birds

(C) Lack of pigment in cave-dwelling animals

(D) Melanization in pepperedmoth

20. The dark band of a muscle is—(A) Isotropic band(B) Anisotropic band(C) Henson band(D) None of these

21. The heart sound ‘dup’ is produ-ced when—(A) Tricuspid valve is opened(B) Mitral valve is opened(C) Mitral valve is closed(D) Semi-lunar valves at the

base of aorta get closed

22. Community health servicesinvolve—(A) Control of communicable

diseases(B) School and health education(C) Awareness of clean environ-

ment(D) All of the above

23. Which of the following cell notfound in connective tissues ?(A) Chondroblasts(B) Osteoblasts(C) Myoblasts(D) Fibroblasts

C.S.V. / October / 2009 / 1008

24. Why cholera patients are pro-vided with saline drips ?(A) NaCl is important contituent

of the blood which maintainsthe RBCs

(B) Na+ ions help to retain waterin body

(C) Cl– ions help in formation ofHCl

(D) Cl– ions are essential com-ponent of blood plasma

25. Epidermal layer consisting ofdividing cells, is—(A) Stratum lucidum(B) Stratum Malpighii(C) Stratum granulosum(D) Stratum corneum

26. Which of the following controlsthe peristaltic movements of thegut ?(A) Brachial plexus(B) Auerbach’s plexus(C) Saral plexus(D) None of these

27. Which of the following carries pro-tein and lipid to other parts of thecell ?(A) Rough endoplasmic reticu-

lum(B) Smooth endoplasmic reticu-

lum(C) Both (A) and (B)(D) None of the above

28. A localised tumour covered byconnective tissue is called—(A) Metastasis(B) Neoplasm(C) Benign tumour(D) Malignant tumour

29. Carbohydrate is a prostheticgroup in—(A) Glycoprotein(B) Chromoprotein(C) Nucleoprotein(D) Lycoprotein

30. Graft between isogenic indivi-duals is known as—(A) Syngraft(B) Allograft(C) Xenograft(D) None of these

31. Which of the following carriesabsorbed product from digestivetract ?(A) Hepatic artery(B) Hepatic portal vein

(C) Pulmonary vein(D) None of the above

32. The primates are—

(A) Plantigrade

(B) Omnivorous

(C) Gregarious

(D) All the above

33. Which of the following is regardedas an unit of nervous tissue ?

(A) Axon (B) Dendrite

(C) Neuron (D) Myelin sheath

34. The presence of gill slits in theembryos of all vertebrates sup-port the theory of—

(A) Origin of evolution

(B) Recapitulation

(C) Metamorphosis

(D) Biogenesis

35. In the fertile human female, app-roximately on which day of theovulation takes place ?

(A) 1st day (B) 8th day

(C) 14th day (D) 18th day

36. Earliest fossil form in phylogenyof horse was—

(A) Merychippus

(B) Eohippus

(C) Equus

(D) Mesohippus

37. Typhus disease in humans iscaused by—

(A) Virus

(B) Protozoans(C) Rickettsiae(D) None of the above

38. Enzymes are polymers of—(A) Fatty acids(B) Amino acids(C) Hexose sugar(D) Inorganic phosphate

39. At high altitude, the RBCs in thehuman blood will—(A) Increase in size(B) Decrease in size(C) Increase in number(D) Decrease in number

40. Silk contains a protein, knownas—(A) Fibrin (B) Fibroin(C) Casein (D) None of these

41. The most striking example ofpoint mutation is found in adisease, called—(A) Thalassemia(B) Night blindness(C) Down’s syndrome(D) Sickle-cell anaemia

42. African sleeping sickness iscaused by—(A) Entamoeba(B) Trypanosoma(C) Trichomonas(D) Leishmania

43. Pneumatic bones are found in—(A) Pigeon (B) Whale(C) Shark (D) Rana

44. SARS virus affects—(A) Heart (B) Brain(C) Lungs (D) All of these

45. The maximum formation of m -RNA occurs in—(A) Cytoplasm(B) Nucleolus(C) Ribosome(D) Nucleoplasm

46. Atocia is—(A) Female sterlity(B) Nulliparity(C) Both (A) and (B)(D) None of these

47. In human beings, the eggs are—(A) Microlecithal(B) Macrolecithal(C) Mesolecithal(D) Alecithal

48. Auer’s bodies are found in thecytoplasm of—(A) Myeloblasts(B) Myelocytes(C) Monoblasts(D) All the above

49. Which cranial nerve has the hig-hest number of branches ?(A) Trigeminal(B) Facial nerve(C) Vagus nerve(D) All the above

50. SARS virus is—(A) Corona virus(B) Picoma virus(C) Retro virus(D) None of these

C.S.V. / October / 2009 / 1010

Habit and HabitatFunaria is a terrestrial moss plant of class Bryopsida.

It chiefly grows in dense patches of bright, green colour;usually in moist shady places (damp soil). It may alsofound on the walls and trunks of trees.

Funaria comprises more than 117 species including15 species reported from Indian soil. Of these, Funariahygrometrica is cosmopolitan in distribution. Southernreported that F. hygometrica is the characteristic moss ofthe initial phase recolonization of burnt areas. Steeropined that there is an increase in the level of soil nutrientsdue to burning. It results in high concentration of solubleorganic matter and inorganic nutrients. For rapid growth ofFunaria, according to Southern, the concentration ofcalcium, nitrogen, potassium and phosphorus must all behigh in the soil.

Gametophyte PhaseGametophyte phase of Funaria consists of two growth

stages—(i) juvenile stage represented by primary pro-tonema and (ii) the leafy stage represented by leafygametophore.

Calyptra

Capsule

Seta

Leaf

Stem

Rhizoids

Malebranch Female

branch

Fig. : Funaria hygrometrica : (A) A female branch ofleafy gametophore with sporophyte, (B) A leafygametophore with a male branch and a femalebranch with young sporophyte.

Juvenile stage results from the germinatingmeiospore. When fully grown it consists of a slender, butgreen branching system of filaments called the proto-nema. The protonema stage in Funaria is only vegetativeand transitory.

Leafy stage starts as a lateral bud on the protonema.The buds may develop into an erect leafy stem. The latterbears numerous rhizoids at its base which anchor it to thesubstratum.

Gametophyte plant is erect and measures about aninch in height with an erect radial stem and spirallyarranged simple leaves. The leaves are more crowdednear the apex where they appear like a rosette thoughactually arranged in three rows corresponding to the threecutting faces when young. This arrangement disap-pears in the mature parts. Branches arise on the mainstem from extra-axillary parts. At its base, the gameto-phytic plant bears a much strong and branched rhizoidswhich become brown and cable like when mature. Becauseof their apparent leaves, stem and rootlike rhizoids theplant is likely to be mistaken for any small flowering plant(phanerogams). These structures, however, are neitherhomologous nor structurally similar to the leaves, stem androots of a phanerogams. But they perform the same func-tion, and are said to be analogous. Koch called stem ascanloid and leaves phylloids.

Leaf Anatomy

In transverse section (T. S.) the leaf is seen asseveral layered thick in the midrib but single-layered inother regions (i.e., lamina is only one cell in thickness).Mid rib bears a small central strand of narrow, thin-walledcells surrounded by a sheath of narrower cells with thickerwalls. The leaf cells are rich in chloroplasts which arecomparatively larger and prominent. The marginal cells arespecialized. They are narrow and thick-walled in contrastto the rest and afford to the frail moss leaves. The struc-tures characteristics of the leaves of the vascular plantsare absent in Funaria.

Mid rib

Mid rib

A

B

WingWing

Chloroplast

Fig. : Moss : (A) One ‘leaf’, (B) T. S. of ‘leaf’.

Stem AnatomyT. S. of the stem of Funaria reveals a simple internal

structure. The cells are arranged in three distinct zones—(i) the central cylinder (strand), (ii) the cortex and (iii) theepidermis.

C.S.V. / October / 2009 / 1011

The epidermis usually consists of a single surfacelayer of cells. The cells are small and contain chloroplastsin the younger portions of the stem. In the mature portionsof stem the epidermal cells become thick-walled andwithout chloroplasts.

The cortex surrounds the central cylinder and is com-posed of undifferentiated, large, thin-walled, parenchyma-tous cells. Near the periphery of the cortex may be seensmall, isolated patches of thin-walled parenchyma cellsconstituting the ‘false leaf traces’.

The central cylinder, which forms the core of thestem, consists of vertically elongated thin-walled, com-pactly arranged cells without protoplasm. These cells withnon-lignified walls are now commonly known as hydroids.The water reaches the base of stem more rapidly byexternal capillary channels.

Epidermis

Cortex

Thickwalled cortex

Thinwalled cortex

Conducting strand

A BFig. : Moss (A) T. S. of young ‘stem’, (B) T. S. of old ‘stem’.

Reproduction

The leafy gametophore of Funaria reproduces vege-tatively as well as sexually.

Vegetative reproduction—Funaria reproduces vege-tatively by means of multiplication of the protonemal stage,secondary protonema, apospory, bulbils and by bud-likestructures, the gemmae.

The primary protonema breaks into short fragmentsby the death of cells at intervals. Each detached fragmentgrows into a new protonema which gives rise to newindividual moss plant.

When surrounded by moist air, filamentous proto-nema may also be formed from any cell of a detachedinjured portion of the leaf, stem and rhizoid. Now, it isknown as secondary protonema from which arises theleafy gametophores as lateral buds.

Gametophore

Secondary protonema

Bud

BulbilRhizoids

Fig. : Funaria hygrometrica : Lower part of the gameto-phore with some of the rhizoids coming above thesurface of the substratum to form secondary proto-nema which bears buds. Note also the formation ofa bulbil (tuber) on the rhizoid. (After Luerssen).

Green protonemal filaments may arise from theunspecialized cells of the sporogonium. These protonemalfilaments bear lateral buds each of which develops into aleafy gametophyte. The production of leafy gametophytedirectly from the vegetative cell of sporogonium without theintervention of spore is known as apospory.

Bulbils are resting buds found on rhizoids. Undersuitable conditions they produce a protonema which bearsa crop of gametophores as lateral buds.

Gemmae, which are available in ‘gemma cup’,commonly develop from the cells of protonemal branches.The cells constituting the gemmae are thin walled andcontain chloroplasts. Gemmae, after detachment, undersuitable conditions directly develop into new gameto-phores.

Sexual reproduction : Funaria reproduces sexuallyby means of male and female reproductive structuresknown as antheridia and archegonia respectively.

Male branch (Antheridiophore) expands to form aslightly convex male receptacle with the antheridia closelypacked on it to form a cluster (red spot). The antheridialcluster with the surrounding perigonial leaves is known asperigonium. The perigonium surrounds a number ofstalked club-shaped antheridia. Associated with theantheridia in the cluster are green, hairlike filaments calledparaphyses. Each antheridium emerges from a singlesuperficial embryonic cell called the antheridial initial.Antheridia at various stages of development occur in asingle male receptacle which is thus capable of releasingsperms in succession. The mature antheridium of F.hygrometrica may be 0·25 mm in length.

ParaphysesAntheridium

Capcells

Involucral leaves

Androgonialmass

Androcyte

Jacket

Stalk

StemA B

BO

DY

Fig. : Moss : (A) L. S. of antheridial head, (B) L. S. ofone antheridium (young stage).

The archegonia are borne in cluster on the archego-nial shoot (female branch). Perigonium in this case is notdifferentiated. The archegonia are typically flask-shaped,each consisting of two usual parts—venter and neck. Theventer consists of a double layer of sterile cells enclosinga cavity known as the venter cavity. In the venter cavitylies the ventral canal cell (above) and the egg cell (below).The long neck consists of six rows of neck cells. Each

C.S.V. / October / 2009 / 1012

archegonium develops from a single archegonial initial.The cover cell in Funaria contributes to the formation ofthe archegonial neck and axial row of neck canal cells.The archegonial cluster with the perichaetial leaves consti-tutes the perichaetium.

ArchegoniaParaphyses

Cover cells

Neck

Neckcanalcells

Ventralcanalcell

Oosphere

Venter

A B

Stem

Fig. : Moss : (A) L. S. of archegonial head, (B) Onearchegonium.

Fertilization

Fertilization in Funaria occurs when the plants are wetwith rain or dew as in other bryophytes. In the maturearchegonium ready for fertilization, the axial row of cells(neck canal cells and the ventral canal cells) except theegg, degenerates to form a slimy fluid. The overlappingperigonial bracts surrounding the terminal antheridialcluster form a shallow, cup-like structure known as splashcup. The process of fertilization also stimulates cells of theventer wall that divide to form a protective covering calledthe calyptra which covers the capsule till maturity.Calyptra also acts as transpiration shield around theimmature capsule. Muggoch and Walton suggested thatthe spreading of sperm vesicle in a thin film of watersurface in the cup presents an excellent opportunity forantherozoid dispersal by rain drop splash which strikes thearchegonial cluster. Reaching the archegonial cluster thesperms swim to the archegonia. Only one sperm uniteswith the egg of archegonium to accomplish fertilizationafter which a zygote develops in a sporophyte.

Sporophyte Phase

The diploid sporophyte is formed from the zygote andis usually called sporogonium. Its sole function includesthe formation of meiospores and their dispersal. Thesporogonium of Funaria is differentiated into foot, seta andcapsule.

Foot forms the basal portion of the sporogonium. It isembedded in the tissue of the tip of the leafy femalebranch. It functions both as anchorage and an absorbingorgan. The wall of foot consists of transfer cells which

absorb water and nutrients for the better development ofsporophyte from the parent plant. The foot is thus similarin function to the root of the fern plant.

Seta is a long, reddish brown, stalk-like structure. It isdifferentiated into a central strand of small, thin-walledcells surrounded by thick-walled cuticularized epidermisand cortex. The cortex gives mechanical strength to bearthe weight of capsule. The central strand acts as conduct-ing strand. Bopp suggested that seta in Funaria plays animportant role in the photosynthesis of sporophyte.

EpidermisPhotosynthetic

tissue

Substomatalcavity

Cortex Conductingstrand

(medulla) Stoma

Stoma

Stomatalaperture

Sub-stomatalcavity

Ring-shapedguard cell

AB

C

D

E

Fig. : (A–E). Funaria hygrometrica : (A) Transverse sec-tion seta, (B) Transverse section through apophy-sis, (C) A portion of B magnified to show stoma,substomatal cavity and photosynthetic tissue, (D)Young stoma in surface view, (E) Mature stoma insurface view.

Capsule is pear-shaped, variously coloured, highlyorganized structure. Its upper portion is covered by aconical cap, the calyptra. Externally, capsule shows threedistinct regions—apophysis, operculum and theca.

OperculumPeristome

Annulus

Rim

Columella

Air space

Seta

Apophysis

Conductingstrand

Outer wallof spore sac

Inner wallof spore sac

Spores

Hypodermallayer

Spongylayer

Trabeculae

Fig. : Moss : L. S. of mature capsule.

C.S.V. / October / 2009 / 1013

Apophysis is the solid, somewhat swollen, basalgreen photosynthetic part of the capsule. The lower partthe axis of the apophysis shows a central strand of thin-walled elongated cells connected with the similar tissue ofthe seta below. On the above side, the axial strand dis-appears into columella. This axis develops from theendothecium and it is surrounded by a green spongytissue formed from amphithecium. The green spongytissue is surrounded by a single compact layer ofepidermis provided with stomata which are connectedwith air spaces. Thus, this green spongy tissue acts asthe site of photosynthesis.

The air space within the capsule wall is traversed bystrands of green, elongated cells which are known astrabeculae. The trabeculae connect the innermost layer ofthe capsule wall with the outer wall of the spore sac.Between the inner and outer walls of spore sac is a cavitywhich contains numerous spores. A solid cylinder of pith-like columella is situated in the centre of theca region.

The terminal obliquely placed cap-like portion of thecapsule is the lid or operculum. The operculum at first iscontinuous with the theca region, eventually it becomesdelimited by the appearance of a narrow, circular constric-tion. Below the constriction radially elongated, thick-walledcells form a circular rim or diaphragm. The latter consti-tutes the upper end of the open theca. Above the rim ofthe theca, forming the broadest lower part of the opercu-lum is the annulus consisting of 4-5 layers of cells. Theupper 2 or 3 layers constitute a special ring of modifiedcells forming the edge of the detached lid.

Peristome, which lies immediately underneath theoperculum, consists of two sets of long, conical teeth, onewithin the other. There are 16 teeth in each set. The peris-tome teeth close the opening of the spore sac. At maturitythe distended cell walls of annulus and the loosened oper-culum are shed leaving the peristome teeth exposed. Thetips of the peristome teeth are attached to a small circle ofa few cells. This attachment soon disintegrates and theouter peristome teeth (exostome), which are hygroscopic,display jerky movements with changes in humidity. Thiscondition is favourable for spore dissemination. Thespores, when they are most likely to be carried by the aircurrents, are thus liberated in dry weather.

Formation of New GametophyteThe moss spores are haploid reproductive units with

genetic potential for the production of moss gametophyte.The spores are formed from the Spore Mother Cells

(SMCs) in tetrad. The diploid nucleus of SMC undegoesmeiotic division to form haploid spores. They are the firstcells of the next (new) gametophyte generation. Underfavourable conditions of moisture, light and temperature,the liberated spores germinate. The first step of germina-tion of spores involves imbibition of water. The sequentialsteps of spore germinations are : → swelling phase →spore wall rupture phase → protrusion phase → sporedistension phase.

Each spore of Funaria consists of four layered walls,the inner one is termed the intine. Following protrusion ofaperture region intine, the spore protoplast adjacent to itfollows suit. The protruding protoplast is covered and thusprotected by protruded intine of the aperture region. Thistubular outgrowth is known as green germ tube. Thisoutgrowth grows in length and undergoes septation toform filaments of green cells which grow over the suitablesubstratum (soil) and branch leafy. The branched, green-algae like filament thus formed is the primary protonema.They are positively phototrophic. Sironwal recognized twodistinct growth phases in the development of primaryprotonema; these are chloronema and caulonemastages. Buds develop on the caulonema give rise to leafygametophyte. The caulonema filaments are negativelyphototrophic.

AB C DSpore

E

F

Bud

RhizoidsProtonema

Fig. : Moss : (A) Spore, (B-C) Germinating spore, (D-E)Developing protonema, (F) Mature protonema withgemma and bud.

Systematic PositionDivision — BryophytaClass — BryopsidaSubclass — BryidaeOrder — FunarialesFamily — FunariaceaeType — Funaria

OBJECTIVE QUESTIONS

1. In Funaria the spores on germi-nation give rise to—(A) Primary protonema(B) Secondary protonema(C) Sporophyte(D) All of the above

2. In bryophytes the embryonicdevelopment of zygote takesplace in—(A) Protonema

(B) Columella

(C) Archesporium

(D) Antheridium

3. The sperms of Funaria move withthe help of—

(A) A single flagellum

(B) Two flagella

(C) Cilia

(D) Pseudopodia

4. The leaves on the stem ofFunaria are arranged—(A) Spirally (B) Alternately(C) Irregularly (D) Circinately

5. The leaf-like and root-like struc-tures of a moss plant and leavesand roots of a vascular plantare—(A) Analogous structures(B) Homologous structures

(Continued on Page 1023 )

C.S.V. / October / 2009 / 1014

Reproduction

All organisms born on this earth show a characteristiclife cycle involving birth, growth maturation, reproductionand death. Of the processes occurring during such a cyclereproduction is necessary for continuation of species fromone generation to another. Reproduction is a unique pro-perty of the living beings which can produce young oneslike themselves after having attained maturity, and can,thus maintain continuity of life. Reproduction are of twotypes— (1) Asexual and (2) Sexual.

1. Asexual ReproductionAsexual reproduction is of following types—(i) Agamospermy—In these methods the seed is

formed but without gametic union.(ii) Vegetative reproduction—In such type of methods

only the vegetative parts of the body are involved formultiplication.

Agamospermy—It is in the form of vegetative repro-duction by plants in which seeds are formed without fusionof gametes. It also includes processes like apogamy andapospory.

In agamospermy, the seeds possess an embryodeveloped not from a diploid zygote but an abnormallyformed diploid egg. In these cases not only the egg butthe entire embryo sac is diploid because it developswithout meiosis from—

(i) a megaspore mother cell (the phenomenon iscalled diplospory)

(ii) any cell of the nucellus (this is called apospory)Another type Parthenogenesis (Gr Parthenos = vir-

gin, genesis = origin) may be defined as the developmentof female gamete into a new individual without fertilization.It is a type of apomixis in which the megaspore mothercell undergoes the usual meiotic division to form haploidegg and the embryo develops from egg without fertiliza-tion.

Beside this, embryos are also produced directly fromthe cells of the nucellus. This is called Adventive polyembryony. It results into the formation of many embryosin each seed. The commonest example is citrus seedwhich contains 2–40 embryos.

Vegetative reproduction—Vegetative propagationincludes reproduction by means of bulbs, bulbils, runners,suckers and so on. There propagules are formed by thesporophyte only. Gustafsson (1946) has distinguishedthree types of vegetative reproduction in higher plants.

(i) The propagules are formed outside the floralregions, and the plants are sexually sterile. Fritillariaunperialis and lilium bulbiferum are typical representativeof this group. They propagate by means of bulbils andbulblets.

(ii) The propagules are formed on the floral brancheseither in addition to the flowers or in place of stem. The

phenomenon is commonly described as vivipary. Sincethis term is also used to refer to situation where sexuallyformed seeds germinate on the mother plant (as in Mang-roves), in the present contest the vegetative vivipary maybe used.

Vegetative vivipary is quite common in grasses(Deschampasia, Festuca, Poa) and Allium.

2. Sexual Reproduction

The fundamental steps involved in the sexual repro-duction in all the plants are as follows—

(a) Formation of spores by meiosis in the sporo-phyte

(b) Formation of gametophytes from spores(c) Production of gametes in the gametophytes(d) Fusion of gametes (fertilization) and the restora-

tion of diploid condition(e) Formation of zygote which may produces embryo,

seed and the new plant.Important events and structures involved in the

sexual reproduction in angiosperms—(i) Flowers are the sex organs in angiosperms.(ii) The androecium and gynoecium are the male

and female reproductive structures respectively.(iii) As in the plants the sexual reproduction involves

the fusion of male and female gametes.(iv) The male gametes are produced in pollen grain

and the female in the ovule.(v) The fusion of male and female gemetes results

in the formation of zygote.(vi) Zygote develops into embryo within the seed.(vii) During germination of seed the embryo grows

into a new plant.(viii) The Endosperm developing within the seed

serve as a food tissue for the developing em-bryo.

Do You Know● Rudolph Camerarius (1694) first to describe sexual

reproduction in plants.● Flowers develop on the peduncle in the axil of bracts.● The arrangement of flowers on the peduncle is called

anthotaxy.● N. Grew (1682) was first to point out that flowers are

reproductive organs of plants.● Stylopodium is the swollen base of style.

Parts of flower—A typical flower is made of calyx,corolla, androecium and gynoecium. Out of these, calyxand corolla only help the process of sexual reproductionbut do not directly take part in this process.

Androecium is the male reproductive whorl and itconsists of stamens. Each stamen is made of filamentand anther.

C.S.V. / October / 2009 / 1015

Androeciumcollection of

stamens

Antherdithecousintrorse

Stigmabilobedshort

Filamentlong stalk

Calyxcollection of sepals,green, polysepalous

Pedicelstalk of flower

OvaryBicarpellary, syncarpous,bilocular, superioraxile placentationReceptacleBase of flower on whichfloral organs are arranged,also called thalamus

Corollacollection of petals,bright yellow, expandedabove, free orpolypetalous

Style

Gyn

oec

ium

Fig. : L. S. of a typical flower

Gynoecium is the female reproductive whorl and ismade of ovary, style and stigma.

Anther, Microsporogenesis and theMicrospore

Structure of anther—A mature anther is made of thewall and the pollen chamber.

1. Wall of the anther—The anther wall consists offour layers. These are epidermis, endothecium, middlelayers and tapetum.

(a) Epidermis—This is the outermost layer. It is onlyone cell in thickness.

(b) Endothecium—This layer is situated just belowthe epidermis. It is a single layer of radially elongatedcells. These cells are characterised by fibrous thickeningswhich help in the dehiscence of anther. In between thesecells, a few cells without thickenings are also present.These thin walled cells collectively form the stomium.

(c) Middle layers—Three to four layers of thin walledcells situated just below the endothecium are known asmiddle layers. In mature anther, these layers generallydisintegrate.

Connective

Epidermis

EndotheciumMiddle layers

Tapetum

Stomium

Pollen grains

Fig. : T. S. of a mature dithecous anther

(d) Tapetum—This is the innermost layer of the wall.The cells are multinucleate. These provide nutrition to the

developing microspores. Tapetum, therefore, secretesboth enzymes and hormones. Tapetum is of two types :

(i) Secretory or glandular—The cells of this type oftapetum remain in contact with the anther wall throughout.

(ii) Amoeboid or plasmodial tapetum—The cells of thistype of tapetum separate from the wall and move freely inthe pollen chamber.

2. The pollen chamber—This forms the central cavityof the anther lobe. The process of microsporogenesistakes place in this region.

Development of AntherThe development of an anther is eusporangiate

(develops from a group of cells). A young anther consistsof a homogenous mass of meristematic cells surroundedby an epidermis. As an anther grows, it becomes slightlyfour–lobed. In each lobe, commonly a row of hypoder-mal archesporial cells is differentiated which form thearchesporium of the anther.

Each archesporial initial divides into an outer primaryparietal cell and an inner primary sporogenous cell.

The primary parietal cell divides to form 3–5 walllayers, i.e., endothecium, middle layers and tapetum.The primary sporogenous cells divide to produce a massof sporogenous cells or microsporocytes. The sporoge-nous cells undergo a few mitotic divisions simultaneouslywith the growth of anther. These derivatives function asmicrospore mother cells.

Epidermis

Archesporialinitial

Wall layersSporogenous cell

Primary parietal cell

Primary sporogenous cellMicrospore mother cell

EpidermisEndothelium

Middle layers Tapetum

A

B

C

D

E

Fig. : (A-E) Different stages of development of anther

MicrosporogenesisThe microspore mother cells (MMC) also called as

pollen mother cells (PMC) which undergoes meiosis togive rise to tetrads of four haploid pollen grains. Wallformation (Cytokinesis) during microsporogenesis issuccessive or simultaneous.

(i) Successive type—Each nuclear division duringmicrosporogenesis is followed by wall formation resultingin the formation of isobilateral tetrads as found in mono-cotyledons.

C.S.V. / October / 2009 / 1016

(ii) Simultaneous type—Walls are laid down simul-taneously after both the division of microspore mother cellnucleus are over. This results in the formation of tetrahed-ral tetrads commonly found in dicots. In some cases thetetrads may be linear, T shaped or decussate type.

Microspore mother cell

Simulta

neous

type

Successivetype

Fig. : Microsporogenesis and two types of cytokinesis

The microspores soon separate out but in somecases adhere in tetrads to form the compound pollengrains, e.g., in Drinys Drosera etc.

In Asclepiadaceae all microspores (pollen grain) of asporangium cohere in a single mass called pollinium.

Tetrahedral Isobilateral Decussate T-Shaped Linear

Fig. : Microscopic Tetrads

Window

● In Aristolochia elegans all the five types of tetrads arepresent.

● In some plants e.g., Mimosa, 6–8 pollen grains sticktogether to form compound pollen grains.

● In plants e.g., Malva and Althea, 10–14 pollen tubesarise from the same pollen grain, but only one is functio-nal.

● The pollen grains of many plants including Partheniumhysterophorus, Amaranthus spinosus, Sorghum vulgare,Chenopodium and Prosopis juliflora etc. are allergic.

● Pollen of myosotis alpestris are smallest (2·5–3·5 μ) butthat of mirabilisjalapa are largest (250 μ).

● About 80% pollen are yellow in colour due to thepresence of flavonoids and carotenoids.

Structure of MicrosporeThe wall of microspore is comprised of outer exine

and inner intine. The exine is made of sporopolleninwhile intine is pectocellulosic.

Sporopollenin is highly resistant to microbes and che-mical decomposition. Therefore pollen grains are wellpreserved during fossilization.

In the exine there are 1-3 germ pores in dicots andgerminal furrows in monocots.

The wall of pollen grain surrounds an uninucleatemass of protoplasm.

Pollen Kit

It is found on the outer side of the mature pollengrains of many insect pollinated flowers. It mainly consistsof lipids and carotinoids, the latter being responsible foryellow or orange colour of the pollen.

The pollen kit acts as insect attractant and adherentto the insect body. It helps in sporophytic incompatibilityand protects the pollen against the damaging effects ofUV-radiations.

Microgametogenesis

The haploid pollen grain represents the first stage ofmale gametophyte. A pollen grain may be referred to asmale gametophyte or partially developed male gameto-phyte or male gamete. Study of pollen grains is called aPalynology.

Exine has one or more weak spots known as germpores through which intine comes out in the form of wallof pollen tube.

The nucleus of the pollen grain divides mitotically toform a vegetative nucleus (tube nucleus) and a generativenucleus. The latter gets surrounded by cytoplasm tobecome generative cell. At this stage, pollen grain is 2-celled-a large vegetative cell and a small lenticulargenerative cell. The pollen grain may be discharged fromthe anther at this 2-celled stage.

However, in some plants generative cell dividesfurther to give rise to two male gametes before the pollengrains are shed. These pollen grains are, thus, 3-celled atthe time of shedding.

Exine Intine

NucleusVacuole

Cytoplasm

Vegetativecell

Generativecell

Pollentube

Male gametes

Malegametes

Vegetativenucleus

A B C

D

Fig. : Different stages of microgametogenesis

In most of the plants, 3-celled condition is formed afterpollen grain is shed. The liberated pollen grain germinateson the stigma and produces a pollen tube. The generativecell divides inside the pollen tube to form two male game-tes.

PollinationPollination is the transfer of pollen grains from

anthers to the stigma (within the same flower or fromone flower to another).

Types of Pollination(a) Self-Pollination (autogamy)—Transfer of pollen

grains within the same flower.

C.S.V. / October / 2009 / 1017

(b) Cross-Pollination (allogamy)—Transfer of pollengrains from one flower to another.

(i) Xenogamy—Transfer of pollen grains from oneplant to another of the same species.

(ii) Geitonogamy—Transfer of pollen grains from oneflower to another on the same plant. It is also called gene-tically self pollination.

(iii) Hybridism—Transfer of pollen grains from oneplant to another related allied species.

Advantages of Cross Pollination

(a) Progeny of offsprings are healthier.(b) Seeds are more in number and viable (living

embryo germinates quickly).(c) Adaptability is better.(d) New varieties may be produced.

Adaptations for Self Pollination

(a) Homogamy—Anthers and stigmas of a flowermature at the same time.

(b) Cleistogamy—Flowers remain closed. They neveropen throughout their life span e.g., groundnut.

Cross pollination is always brought about by externalagents like insects, wind, water and animals.

Entomophily is pollination by insects. The pollengrains are rough and sticky. Insects are attracted by nec-tar / scent / colour. Some common insects which help inpollination are Bees (commonest), butterflies and moths.

Anemophily—Pollination by wind; Flowers are notcoloured; Pollen grains are produced in large amounts.Style and stigma are hairy (feathery) e.g., Grasses,wheat, maize, rice, barley and palms.

Hydrophily—Pollination by water currents e.g.,Vallisneria, Hydrilla, Ceratophyllum. Pollen grains aresticky.

Zoophily—Pollination by animals. e.g. in Bombax(silk cotton) by birds (ornithophily); in Anthocephalous bybats (Cheiropterophily); in Arisaema by Snakes—(Ophiophily); in Colocasia by snails (Malacophily).

Adaptations for Cross Pollination

(a) Dicliny (unisexuality)—e.g. Palms, cucumber,pumpkin.

(b) Self Sterility—e.g. Tea, Malva, Orchids.

(c) Dichogamy—Anthers and stigma of a flowermature at different times.

(i) Protandry—Anthers mature earlier than thestigmas of the same flower. e.g. Cotton.

(ii) Protogyny—Stigma matures earlier than theanthers of the same flowers. e.g. Ficus.

(d) Heterostyly—Stamens, style and stigma are diffe-rent heights.

Plants are dimorphic. e.g. Oxalis, Primula, Linum,Oldenlandia.

(e) Herkogamy—A natural barrier develops betweenanthers and stigma to check self-pollination. e.g. Calot-ropis.

Pollination is essential for fertilization and for fruit andseed formation.

Commonly, the pollen tube enters the ovule throughmicropyle (Porogamy). Sometimes, it enters through cha-laza (Chalazogamy). e.g. Causarina; or through base ofovule or through funicle or integuments (mesogamy).

Irrespective of the place of entry of pollen tube intothe ovule, the tube invariably enters the embryo sac atmicropylar end in between the egg cell and the syner-gids. Entry of pollen tube in the embryo sac is chemo-tropic. Micropyle and synergids secrete some chemicals.

Ovule, Megasporogenesis and theMegaspore

The Ovule— The process of megasporogenesis, i.e.,the formation of megaspores from megaspore mother celltakes place in ovule. Ovule is considered to be an integu-mented megasporangium. The ovules are situated insidethe ovary, attached to a special tissue called placenta.

1. Structure of ovule—A fully mature ovule consistsof the stalk and the body. The stalk is called funicle. Oneend of the funicle is attached to placenta and the otherend to the body of the ovule. The point of attachement offunicle with the body is called hilum. Funicle sometimesextends up to the base of ovule (i.e., chalaza). The ridgethus formed is called raphe.

Raphe

Hilum

Chalaza

Embryo sac

Integuments

Micropyle

Funicle

Fig. : An Organised Ovule

The body of the ovule shows two ends—the basalend, often called the chalaza and the upper end is calledthe micropylar end. The main body of the ovule is cove-red with one or two envelope called integuments. Theseleave an opening at the top of the ovule called micropyle.The integuments enclose a large parenchymatous tissueknown as nucellus. In the centre of the nucellus issituated a female gametophyte known as embryo sac.

Following are some of the conditions seen in ovule inrelation to integuments—

(a) Unitegmic—Ovule with a single integument; e.g.,sympetalous or gamopetalous dicotyledons.

(b) Bitegmic—Ovule with two integuments as in poly-petalous (archichlamydeae) dicotyledons and monocoty-ledons.

(c) Aril—This is a collar-like outgrowth from the baseof the ovule and forms third integument. Aril is found inlitchi, nutmeg, etc.

C.S.V. / October / 2009 / 1018

In angiosperms ovules are of following types—Orthrotropous—The micropyle, chalaza and funicle

are in straight line. This is most primitive type, e.g., Pipe-raceae.

Anatropous—The ovule turns at 180° angle. It iscalled resupination. Thus, in this type micropyle andfunicle are closer, e.g., 82% of angiospermic families.

Hemianatropous—The ovule turns at 90° angle uponthe funicle, e.g., Ranunculus.

Amphitropous—The body of ovule itself rotates sothat nucellus becomes horse shoe shaped, e.g., Lemna.

Campylotropous—Ovule is bend so micropyle andchalaza do not lie in one line, e.g., some crucifers, gram,etc.

Circinotropous—Ovule turns at more than 360°angle, e.g., Opuntia.

Orthotropous Anatropous Campylotropous

Hemiantropous AmphitropousCircinotropous

A B C

D EF

Fig. : Different types of Ovule

● In Asphodelus and Litchi the third integument is presentin the form of aril.

● In castor (Riccinus communis) the integument any out-growth is present at micropylar end called caruncle. Ithelps in seed germination and seed dispersal.

● Ategmic ovules are without any integument e.g., loran-thaceae and santalaceae.

● Chloroplast are present in the cells of the outer integu-ment in several monocotyledons such as Gladiolus,Lilium, Amaryllis Moringa.

● In Nerine, Nelumbium, Aquilegia, Canna etc. stomatahave been reported in the outer integument, whichperhaps helps in exchange of gases.

MegasporogenesisThe process begins with the differentiation of arches-

porial initial in the nucellar hypodermis. Archesporialinitial either acts directly as a megaspore mother cell ordivides periclinally into an outer primary parietal cell andthe inner primary sporogenous cell. Accordingly, follow-ing two conditions are found—

1. Tenuinucellate ovule—In this condition, the arc-hesporial initial directly behaves as megaspore mother

Nucellar epidermis

Megasporemother cell

Parietaltissue

A B

Fig. : (A-B) Nucellus and the ovule : (A) Tenuinucellate ovule,(B) Crassinucellate ovule

cell. Hence, in this case primary parietal cell is not formedand as such there is little or no nucellar tissue betweenthe megaspore mother cell and nucellar epidermis. Theseovules are called tenuinucellate and are found in gamope-talous dicotyledons with unitegmic ovules.

2. Crassinucellate ovule—The division of the arche-sporial initial into primary parietal cell and primary sporo-genous cell results in the formation of crassinucellateovule. The primary parietal cell, thus formed, divides togive rise to a large amount of nucellar tissue betweenprimary sporogenous cell and nucellar epidermis. Suchovules occur in polypetalous dicotyledons with bitegmicovules.

The megaspore mother cell (megasporocyte) dividesmeiotically to form a linear tetrad of four haploid mega-spores. Occasionally, T-shaped or inverted T-shaped (⊥)tetrads are also formed. Megaspore is the first cell offemale gametophyte.

Of the linear tetrad, three megaspores towards themicropyle degenerate. The lowermost, i.e., the chalazalmegaspore enlarges and remains functional. It later pro-duces an embryo sac.

Archesporialinitial

Primary parietal cell

Primarysporogenous

cell

Nucellus

Parietaltissue

Megasporemother

cell

Degeneratingmegaspores

Functionalmegaspore

A B C

D E F

Fig. : (A-F) Different stages of megasporogenesis

C.S.V. / October / 2009 / 1019

Worth to Remember

● Polygonum type of embryo sac is most simple, mostprimitive and normal type of embryo sac.

● In an embryo sac, all the cells are haploid except thesecondary nucleus (2n).

● In majority of angiosperms, the chalazal megaspore isfunctional. In casuarina, all the four megaspores arefunctional.

● Synergids, are also known as helpers.

● In polygonum type of embryo sac three antipodals but inZea mays 20, Sasa paniculata 300 antipodals have re-ported.

● Antipodals are altogether absent in Oenothera type ofembryo sac.

Megagametogenesis and FemaleGametophyte

The process of development of female gametophyteor embryo sac from megaspore is called megagame-togenesis.

Development of Female Gametophyte

Megaspore is the first cell of female gametophyte.Female gametophyte may be monosporic, bisporic ortetrasporic depending upon whether respectively one,two or all the four megaspores formed during megaspo-rogenesis take part in their development.

Out of the three, the monosporic type of female game-tophytes (embryo sacs) are very common and are reportedin 70% of the angiosperms. It was studied for the first timeby Strassberger (1879) in a plant called polygonum dive-ricatum hence it is also called as Polygonum type.

Development of Embryo Sac (Female Gameto-phyte)

1. Monosporic type—Polygonum is the best exampleof monosporic type embryo, which develops from thelower most megaspores (chalazal megaspore) of themegaspore tetrads formed during megasporogenesis.

The other three non-functional megaspores degene-rate. The nucleus of functional megaspore divides bythree successive mitotic divisions to form eight nuclei ofwhich four are at the micropylar end and the other four atthe chalazal end.

Degenerating megaspores

Functionalmegaspore

A

B C D

Synergids

Beak

Polar nuclei

Egg

AntipodalsE F

Fig. : Development of monosporic embryo sac (femalegametophyte) : (A-E) Successive stages of develop-ment, (F) Organised embryo sac

Out of 4 nuclei at each pole one migrates to thecenter and functions as polar nuclei which fuse togetherjust before fertilization to form a secondary nucleus.Three nuclei towards the micropylar end form and eggapparatus (central egg cell and two lateral synergids) andthe other three towards the chalazal end form threeantipodal cells which are vestigial.

Mature embryo sac—The embryo sac is 7 celled and8 nucleated i.e.—

(a) 3 cells at micropylar end form egg apparatus.One is egg cell and two are synergids or coope-rative cells or potential egg.

(b) 3 cells at chalazal end form antipodals or vege-tative cells of gametophyte.

(c) Two nuclei (one from each pole) in the centreform central cell. These two nuclei are calledpolar nuclei.

Vacuole

Nucleus

Filiform apparatus

Beak

Nucleus

Vacuole

Synergid

Egg

Fig. : Detailed structure of egg apparatus

As the embryo sac matures these polar nuclei getfused to form a secondary nucleus.

Rarely embryo sac develops haustorium which ab-sorbs food from nucellus, integument and placental tissue,e.g., Loranthus. etc.

2. Bisporic embryo sac—In this type two mega-spore nuclei take part in embryo sac formation.

3. Tetrasporic embryo sac—This embryo sac deve-lops from four megaspore nuclei.

C.S.V. / October / 2009 / 1020

Fertilization

The union of male and female gametes during sexualreproduction is called fertilization. In angiosperms malegametes are formed in the pollen tube and female gameteis formed in the embryosac. Fertilization is a very lengthyprocess involving many step-wise changes.

The Growth of Pollen TubePollen grains germinate to form pollen tubes when

they fall on compatible stigma. Pollen tubes emergethrough the apertures of the sporoderm.

Exine

Intine

Nucleus

Fig. : Germination of Pollen Grain

After germination of pollen grains pollen tubes growalong the surface of the stigmatic papillae or through thecellulose-peptic layer of their walls. After reaching thebase of the papillae pollen tubes grow through the inter-cellular spaces of the stigmatic tissues. Pollen tubes thenpass to the style. Styles are of two main types, viz., hollowand solid.

Pollen grain

Stigma

Pollen tube

Style

Ovary

Ovule

Fig. : Growth of Pollen Tube in Lantana camara

In case of hollow style the pollen tubes grow alongthe surface of the stylar canal. In case of solid style pollentubes grow through specialised conducting tissues. Cellslining the stylar canal are secretory in nature.

These cells secrete proteins, carbohydrates andlipids (Tilton and Horner, 1980).

Pollen tube Pollen grain

Canal

Secretoryzone

Stylartissue

Papilla

Hollow style

Pollentube

Stylartissue

Solid style

Pollen tubes always grow in the direction of the ovaryand according to Strasburger (1887) the path of pollentubes is guided by a secretion of the ovule.

Entry of Pollen Tube into the OvuleThe pollen tube carrying the two male gametes

reaches the ovule. It may enter the ovule in any one of thefollowing three ways—

(1) Porogamy—Pollen tube enters the ovule throughmicropyle. It is the commonest type, e.g., Ottelia.

Micropyle

Pollen tube

(2) Chalazogamy—Pollen tube enters the ovulethrough chalaza, e.g., Casuarina.

Chalazalend

(3) Mesogamy—Pollen tube enters the ovule laterallythrough the integuments or through funicle, e.g., Cucurbita.

Integuments

C.S.V. / October / 2009 / 1021

The entry of pollen tubes into the ovule is guided by achemotropic stimulus. The filiform apparatus (these arefinger-like projections of wall in the cytoplasm of syner-gids; each projection is provided with packed microfibrils,a non-fibrillar sheath and is rich in polysaccharides) isconsidered as the source of the secretion. But recentinvestigations have revealed the absence of chemotropicstimulus. The following points have been raised againstthe suggestion of filiform apparatus as the source of thestimulus—

(i) Pollen tubes may even enter such ovules whichhave aborted before the formation of embryosac,

(ii) Pollen tubes may enter such ovules in whichembryo sac has been fertilized,

(iii) Pollen tubes may also enter such ovules in whichembryo sac lacks a synergid.

Chao (1972) has carried out a detailed cytologicalinvestigation of Paspalum orbiculare and demonstratedthat the distal part of the integuments, by dissolution of itscells secretes a mucilagenous substance into themicropyle which provides a way of least resistance for thepollen tube and guides it towards its ultimate destination.The mucilagenous secretion is largely a water solublecarbohydrate and it aids the pollen tube growth bothmechanically as well as chemotropically.

Again a special structure which facilitates the entry ofpollen tube into the ovule is the obturator. Although someworkers are of the opinion that it may have associatedwith the nutrition of pollen tubes. It forms a sort of bridgefor the pollen tube to reach the ovule. There are severaltypes of obturators. In Acanthaceae, Anacardiaceae andLamiaceae obturator is formed by the swelling of funicle.

Obturators

Fig. : Ovule of Tetragonia

In Tetragonia glandular epidermal hairs at the base of thefunicle represent obturator. After fertilization the obturatorshrinks and disappears.

In Loranthaceae the proper ovule-like structure isabsent. Here the embryosac undergoes remarkableelongation and meets the pollen tube at some point in thestylar canal.

Entry of Pollen Tube into Embryosac

Pollen tube finally enters the embryosac and it entersonly through the micropylar end. The following threemodes of pollen tube entry are known to occur—

(i) Entry of pollen tube between the egg and one ofthe synergids,

(ii) Entry of pollen tube in between the walls of theembryosac and one of the synergids,

(iii) Entry of pollen tube directly into one of thesynergids.

Electron microscopic studies have conclusivelyproved that pollen tubes enter directly into one of thesynergids through the filiform apparatus. The penetratedsynergid starts disintegration and the other synergidremains unaffected.

The Act of Fertilization

There are two fertilizations in the embryosac ofangiosperms. These are—

(a) Syngamy or zygotic fertilization—The contentsof the pollen tube are discharged into the penetratedsynergid. Usually pollen tubes do not grow beyond thesynergids but in some plants (e.g., Plumbago) pollentube penetrates the egg wall. The pollen tube dischargecontains two sperms, single vegetative nucleus and a fairamount of cytoplasm. Due to disorganisation of thepenetrated synergid, sperms are released from thesynergid. During this period two darkly stained bodies areobserved in the embryosac. These are called X-bodiesand are interpreted as the remains of the synergidnucleus and the vegetative nucleus of the pollen tube(Jensen, 1972; Russell, 1982).

The mode of sperm transfer is a matter of specula-tion. According to Jensen (1972), one of the sperms

Fig. : (A–C) The Double Fertilization

comes in contact to the plasma membrane of the egg cell.Due to disintegration of the plasma membrane, nucleus ofthe sperm enters the egg. The next event is syngamy,i.e., the union of sperm and egg nuclei. The resultingdiploid nucleus produces the embryo.

Three different types of syngamy are reported inangiosperms. These are as follows—

Type 1 : Premitotic—The sperm nucleus fusesimmediately on coming in contact withthe egg nucleus, and the zygotenucleus divides subsequently, e.g.,Poaceae, Asteraceae.

C.S.V. / October / 2009 / 1022

Type 2 : Post mitotic—The sperm nucleus andthe egg nucleus remain in contact for awhile and fuse only after both the nucleihave entered into divisions, e.g., Lilium.

Type 3 : Intermediate—The sperm nucleus fuseswith the egg nucleus after completing itsprevious mitosis, e.g., Impatiens.

(b) Vegetative fertilization—The second malegamete fuses with two haploid polar nuclei or diploidsecondary nucleus (if two polar nuclei fuse to form

FIRST MALE GAMETE(n)

SECOND MALE GAMETE(n)

PRIMARY ENDOSPERMNUCLEUS (3n)

Egg (n)

POLAR NUCLEI (n, n)

ZYGOTE (2n)×

×

SYNGAMY

TRIPLE FUSION

Schematic Representation of the Process of Fertilization

secondary nucleus). As a result, a triploid primary endo-sperm nucleus is produced from which the endospermis developed.

Since, there are two fertilizations in the sameembryosac, it is described as double fertilization. S. G.

Nawaschin (1898) for the first time reported thisphenomenon and it is a characteristic feature of angio-sperms.

However, in angiosperms the double fertilization maysometimes occur in a more specialised way. When anembryosac receives two pollen tubes instead of one, thesingle sperm nucleus from one tube may fuse with theegg nucleus and the polar nucleus may receive spermnucleus from another pollen tube. This phenomenon istermed as heterofertilization.

The ultimate aim of fertilization in plants is to formseeds which are essential for their survival on motherearth. Although in some higher plants the traditional ferti-lization processes are lacking. In these plants the procrea-tion takes place through a process called apomixis.

OBJECTIVE QUESTIONS

1. A plant species that has, on eachindividual plant, flowers with car-pel is—

(A) Perfect(B) Imperfect(C) Monoecious(D) Dioecious

2. An anther produces—

(A) Haploid gametes(B) Diploid gametes(C) Haploid spores(D) Diploid spores

3. Author of an introduction to theembryology of angiosperm is—(A) P. Maheshwari(B) S. R. Kashyap(C) T. S. Sadasivan(D) K. C. Mehta

4. Anther is typically—(A) Tetrasporangiate(B) Bisporangiate(C) Trisporangiate(D) Monosporangiate

5. Tapetal cells are—(A) Uninucleate(B) Binucleate(C) Multinucleate(D) Enucleate

6. Compound pollen grains occurin—(A) Calotropis or Asclepias(B) Orchids(C) Juncus or Cryptostegia(D) Brassica

7. A pollinium consists of—

(A) A bag of pollen grain formedin a microsporangium

(B) A cluster of pollen grains be-longing to a chamber of mi-crosporangium

(C) Group of four pollen grainsderived from a single mothercell

(D) Two pollen tetrads attachedby a small stalk

8. Abundant occurrence of fossili-sed pollen grains is due to resis-tant—

(A) Ligno cellulose

(B) Sporopollenin

(C) Pectocellulose

(D) Pectolignin

9. Polar nuclei are located in—(A) Pollen tube(B) Embryo sac

C.S.V. / October / 2009 / 1023

(C) Ovule(D) Thalamus

10. Ubisch bodies are connected withthe formation of—(A) Sporopollenin(B) Intine and pollen kit(C) Exine(D) Pollen kit and Pollinia

11. Process of fusion between maleand egg nuclei is—(A) Syngamy(B) Double fertilization(C) Conjugation(D) Triple fusion

12. Female gamete of angiospermsis represented by—(A) Oospore (B) Carpel(C) Egg (D) Pollen grain

13. A typical embryo sac possesses–(A) Egg, synergids and secon-

dary cell(B) Egg, synergid, central cell

and polar nuclei(C) Egg, synergids, polar nuclei

and antipodals(D) Egg, synergids and secon-

dary wall

14. Meiosis is best seen in—(A) Gamete(B) Microsporangium(C) Pollen grain(D) Anther wall

15. Fertilization involving carring ofmale gametes by pollen tube is—(A) Porogamy(B) Siphonogamy(C) Chalazogamy(D) Syngonogamy

16. Pollen tube was discovered in1824 by an Italian mathematiciannamed—(A) Robert Brown(B) A. Takhtajan(C) G. B. Amlci(D) T. S. Eliot

17. Parthenogenesis is defined in thedevelopment of—(A) Egg without fertilization(B) Synergid without fertilization(C) Fruit without fertilization(D) Fruit without pollination

18. Anthesis is—(A) Development of anther(B) Period of opening of the flo-

wer buds

(C) Stigma receptors(D) Meiotic division of the pri-

mary mother cell19. The transfer of pollen grains from

anthers to the stigma is knownas pollination; however, when amale gamete fuses with a femalegamete the process is termedas—(A) Cross-pollination(B) Autogamy(C) Dichogamy(D) Fertilization

20. On fertilization secondary nucleusforms—(A) Endosperm (B) Seed(C) Embryo (D) Cotyledons

21. When the stigma matures beforethe anthers, it is known as—(A) Protandry (B) Heterostyly(C) Protogyny (D) Dicliny

22. The sperm nucleus and the eggnucleus remain in contact for awhile and fuse only after both thenuclei have entered into divi-sions. This type of syngamy istermed as—(A) Premitotic(B) Postmitotic(C) Intermediate(D) None of the above

23. Secondary nucleus formed by thefusion of two polar nuclei is alsocalled—(A) Coenocyte(B) Vegetative nucleus(C) Tube nucleus(D) Definitive nucleus

24. The process of double fertiliza-tion was discovered by—(A) Nawaschin(B) Robert Hook(C) Robert Koch(D) Darlington

25. The main embryo develops by thefusion of one of male gameteswith—(A) Egg(B) One of the synergids(C) Two polar nuclei of the

embryo sac(D) One of the nuclei of the

secondary nuclei

26. The point where the funicle isattached to the body of the ovuleis called—(A) Micropyle (B) Chalaza(C) Hilum (D) Node

27. Which one of the following actsas a passage for a pollen tube toreach ovary to fertilize the egg ?(A) Placenta (B) Ovule(C) Style (D) Stigma

28. The obturator is an outgrowth ofplacenta or funiculus. It helpsin—(A) Formation of ovule(B) Entry of water into seed(C) Pollination(D) Entry of pollen tube into

ovule

ANSWERS

●●●

(Continued from Page 1013 )

(C) Both (A) and (B)(D) Vestigial organs

6. The antheridia in Funaria formedat the apex of male shoot are in—(A) Groups of many(B) Groups of two(C) Solitary form(D) Shaped like open book

7. The middle sterile part of thecapsule of Funaria is—(A) Protocorn(B) Protonema(C) Columella(D) All of the above

8. The antherozoids of Funaria areusually—(A) Nonciliated (B) Biciliated(C) Multiciliated (D) Uniciliated

9. Buds in Funaria give rise to—(A) Sporophyte(B) Leafy gametophyte(C) Spore tetrad(D) All of the above

10. The first step in germination ofFunaria spore is—(A) Protrusion phase(B) Spore wall rupture phase(C) Spore distension phase(D) Swelling phase (Imbibition)

ANSWERS

●●●

C.S.V. / October / 2009 / 1024

1. Chlorophyll-e is found in—(A) All brown algae(B) Diatoms(C) Vaucheria(D) None of the above

2. The stroma of chloroplasts con-tains—(A) DNA(B) Ribosomes(C) Enzymes(D) All of the above

3. Natural selection is defined asoccurring when the environmentcauses—

(A) Differential mortality

(B) Assortative mating

(C) The bottleneck effect

(D) Rapid gene flow

4. The embryo of Helianthus annuusbears—

(A) Many cotyledons

(B) Three cotyledons

(C) Two cotyledons

(D) One cotyledon

5. The most complex cellular struc-tures are found in—

(A) Protozoa (B) Bacteria

(C) Fungi (D) Algae

6. How many bases of nucleotidetriphosphate are joined to formDNA polynucleotide chain ?

(A) Three (B) Two

(C) One (D) None of these

7. Which of these types of cells ismost likely to divide ?

(A) Epidermis

(B) Parenchyma

(C) Xylem

(D) Meristem

8. Which of the following enzymesjoins DNA fragments ?(A) Topoisomerase(B) DNA ligase(C) DNA polymerase(D) DNA gyrase

9. Cleavage polyembryony is foundin—(A) Cycas(B) Rhizophora(C) Pinus(D) All of the above

10. Viral genes are made up of—(A) DNA only(B) RNA only(C) Either DNA or RNA(D) Either proteins or nucleic

acids

11. The ‘genic balance’ theory wasproposed by—(A) Sutton and Boveri(B) Bateson and Punnett(C) Watson and Crick(D) Bridges

12. Genes not located within thenucleus are almost always foundin the—(A) Cell membrane(B) Cytoskeleton(C) Cytosol(D) Organelles

13. The root of the plants—(A) Goes down(B) Forms intricate system in the

soil(C) Extensive system in the soil(D) Can perform all of the above

14. Sucrose, a common table sugar,is composed of—(A) Fructose + galactose(B) Glucose + fructose(C) Glucose + galactose

(D) Fructose + fructose

15. Role of mutation in evolution is—(A) Genetic drift(B) Reproductive isolation(C) Genetic variation(D) None of the above

16. In recombinant DNA (r DNA) tech-nology, a plasmid vector must becleaved by—(A) Modified DNA ligase(B) The same enzyme that

cleaves the donor gene

(C) Four separate enzymes(D) A heated alkaline solution

17. Bud scales are found in—(A) Jack fruit(B) Ficus(C) Magnolia(D) All of the above

18. The sporangia of a conifer arelocated on the—(A) Tips of the needle(B) Axils of the branches(C) Scales of the cones(D) Bases of the needles

19. The phenomenon of male (O→)sterility in plants is found to becontrolled—(A) By plasmagenes only(B) By nuclear genes only(C) Either by plasmagenes or by

nuclear genes(D) None of the above

20. Chromosomes with mediancentromere and equal arms arecalled—(A) Metacentric(B) Telocentric(C) Acrocentric(D) Submetacentric

21. Pollination by bat is called—(A) Ornithophily(B) Malcophily(C) Chiropteriphily(D) Entomophily

22. In all microbial cells the end-products of biosynthesis is—(A) Proteins(B) Nucleic acids(C) Both (A) and (B)(D) None of the above

23. The pollinia are found in—(A) Vallisneria, China-rose and

Guava(B) Asclepias(C) Calotropis procera(D) Both (B) and (C)

24. The mean annual temperature fortropical rain forest is—

(A) 23—27°C (B) 80—90°C

(C) 0—5°C (D) 5—15°C

25. The movement of substancewithin a plant, i.e., from one partto another, is mainly by—(A) Turgidity(B) Plasmolysis

C.S.V. / October / 2009 / 1025

(C) Diffusion pressure deficit(D) Diffusion

26. Heteropolysaccharide contains—(A) Only two monosaccharide

units(B) No any monosaccharide unit(C) Only one monosaccharide

unit(D) Two or more different mono-

saccharide units

27. NO2 is converted into NO3 withthe help of bacteria—(A) Nitrobacter(B) Nitrosomonas(C) Nitrosococcus(D) All of the above

28. Artocarpus integrifolia belongs tofruit—(A) Berry (B) Pome(C) Sorosis (D) Syconus

29. Dermatogen is a tissue formed bythe apical meristem which deve-lops into—(A) Endodermis (B) Pericycle(C) Epidermis (D) Stele

30. The term ‘ATP’ was coined by—(A) F. Lipmann(B) F. A. Janssen(C) H. J. Muller(D) J. D. Watson

31. Scales leaves are also called—(A) Sporophylls (B) Prophylls(C) Cataphylls (D) Hypsophylls

32. If at the end of meiosis, the fourdaughter cells have four chromo-somes, how many chromosomeswere in the mother cell ?(A) 8 (B) 16(C) 2 (D) 32

33. Which of the following is correctregarding photosynthesis ?(A) It produces oxygen(B) Provides food either directly

or indirectly for most livingthings

(C) Provided energy to createtoday’s fossil fuels

(D) All of the above

34. Which of the following generaexhibits an alternation of genera-tions with haploid and diploidmulticellular phases ?(A) Volvox(B) Ulva

(C) Ulothrix(D) Chlamydomonas

35. The causal agent of ‘red rot ofsugarcane’ is—(A) Xanthomonas citri(B) Colletotrichum falcatum(C) Ustilago maydis(D) All of the above

36. Water acts as excellent solventand helps in the uptake and dis-tribution of mineral requirementsand other solutes for—(A) Growth(B) Development(C) Both (A) and (B)(D) None of the above

37. In rainy season, the surface of theearth becomes slipping due togrowth of—(A) Fungi (B) Algae(C) Mosses (D) Bacteria

38. Most fossils are found in—(A) Granite(B) Sedimentary rocks(C) Lava flows(D) Black soil

39. The effect of interspecific com-petition on niches is to makethem—(A) Smaller(B) Change location(C) Larger(D) More triangular

40. Each molecule of FADH pro-duces—(A) Six molecules of ATP(B) Four molecules of ATP(C) Three molecules of ATP(D) Two molecules of ATP

41. Plants are killed at low tempera-ture because—(A) Of cell protein precipitation(B) Desiccation occurs due to

the withdrawal of water fromvacuolated protoplasm

(C) Cell bursts due to mechani-cal pressure of ice or coldwater

(D) Of all of the above reasons

42. Which of the following kinds ofplant fixes carbon dioxide by wayof crassulacean acid meta-bolism ?(A) Red alga (B) Cactus(C) Oak tree (D) Grass

43. The sporophyta of fern consistsof—(A) Leaf (B) Stem(C) Root (D) All of these

44. If a new allele suddenly becomesvery abundant in a population,most likely it is—(A) Flowing with emigrants(B) A product of assortative

mating(C) Mutating rapidly(D) Strongly selected for

45. Spoilage of oil can be detected bywhich fatty acid ?(A) Oleic acid(B) Linoleic acid(C) Linolenic acid(D) Erucic acid

46. In ecosystem, energy entersthrough—(A) Herbivores(B) Carnivores(C) Omnivores(D) None of the above

47. What is the another term foradaptive evolution ?(A) Biopoiesis(B) Microevolution(C) Population(D) Geographic barrier

48. Anemophily, wind-pollination, isfound in—(A) Poplar (B) Willow(C) Oak (D) All of these

49. Compound capitulum is foundin—(A) Echinops echinata(B) Cassia(C) Pyrus torminalis(D) All of the above

50. Apospory is found in—(A) Taraxacum albidum(B) Eupatorium glandolusum(C) Both (A) and (B)(D) None of the above

ANSWERS WITH HINTS

(Continued on Page 1034 )

C.S.V. / October / 2009 / 1026

1. RNA processing is—

(A) The same as transcription

(B) The rejection of old, wornoutRNA

(C) An event that occurs afterRNA is transcribed

(D) Both (B) and (C) are correct

2. The energy storage at consumerlevel is—

(A) Net productivity

(B) Net primary productivity

(C) Secondary productivity

(D) Gross primary productivity

3. Floral leaves are also called—(A) Sporophylls (B) Prophylls(C) Cataphylls (D) Hypsophylls

4. The rate of photosynthesis ismore in—(A) Orange light(B) Green light(C) Red light(D) Yellow light

5. In Pinus the resin secreting cellsform one or two peripheral layersthat surround a schizogenouslydeveloped canal or duct in the—

(A) Leaves

(B) Stem

(C) Both (A) and (B)

(D) Root

6. The proteins associated withDNA in eukaryotic chromosomesare synthesized during—

(A) S-stage (B) M-stage(C) G2-stage (D) G1-stage

7. When the stigma matures beforethe anthers, this condition isknown as—

(A) Heterostyly (B) Dicliny

(C) Protogyny (D) Protandry

8. Stem climber is found in—

(A) Piper betle

(B) Asparagus

(C) Vitis

(D) All of the above

9. Retting is a process by whichbacteria bring about the—(A) Nitrogen fixation(B) Separation of fibres of flax

and coconut(C) Vitamin B6 synthesis

(D) None of the above

10. The small size of cells is bestcorrelated with—(A) An adequate surface area for

exchange of materials(B) The fact they are self-repro-

ducing(C) Their prokaryotic versus

eukaryotic nature(D) All of the above are correct

11. Reininke described the connec-tion between fungus and algaas—

(A) Consortium

(B) Helotism

(C) Symbiosis

(D) Heterothallism

12. The sporophytic phase of Ricciacomprises—

(A) Zygote

(B) Embryo

(C) Sporogonium

(D) All of the above

13. The common name of Funaria, bywhich it is generally known, is—(A) Bog moss (B) Green moss(C) Peat moss (D) Pond moss

14. Which of the following is the mostcommon monomer of carbo-hydrate molecules ?(A) Maltose (B) Amino acid(C) Glucose (D) Phospholipid

15. Biciliate antherozoids are foundin—(A) Rhizopus (B) Selaginella(C) Spirogyra (D) Dryopteris

16. Biologists who study thesequences of organisms in thefossil record are—(A) Palaeobiologists

(B) Systematists

(C) Taxonomists(D) Embryologists

17. Which of the following disease iscaused in stem by the boron, atrace element, deficiency ?(A) Black necrosis of stem(B) Stem rot of Grapes(C) Red rot of stem(D) All of the above

18. Which of the following plantsdoes not belong to the familyPapilionaceae ?(A) Phaseolus mungo(B) Vigna sinensis(C) Crotolaria juncea(D) Bauhinia variegata

19. The interconnected food chainsare called—(A) Food web(B) Epigenesis(C) Tropic level(D) Pyramid of energy

20. The middle sterile part of mosscapsule is called—(A) Protonema (B) Elaters(C) Columella (D) Foot

21. Adaxial outgrowth from the baseof the leaf of Selaginella is—(A) Trabecula (B) Velum(C) Stipule (D) Ligule

22. Each sorus has a protectivecovering called—(A) Scutellum(B) Tonoplast(C) Indusium(D) Glossopodium

23. The transformation from a singlecell into an adult individual withmany different kinds of cells iscalled—(A) Development(B) Adaptation(C) Inheritance(D) Evolution

24. The type of leaf of Mimosapudica is—(A) Decompound(B) Simple pinnate(C) Simple palmate(D) Compound bipinnate

25. The first community to inhabit anarea is known as—(A) Pioneer community(B) Climax stage

C.S.V. / October / 2009 / 1027

(C) Consumer level(D) Trophic level

26. Allopatric but not sympatricspeciation requires—(A) Spontaneous differences in

males and females(B) Prior hybridization(C) Reproductive isolation(D) Geographic isolation

27. Calvin cycle includes—(A) CO2 fixation(B) CO2 reduction

(C) Regeneration of ribulosebiphosphate

(D) All of the above

28. The function of amyloplast is to—(A) Store fat(B) Absorb water(C) Store starch(D) Absorb light

29. Unconnected complex units ofGolgi apparatus occurring ineukaryotic plant cells are termedas—(A) Dictyosome (B) Pyrenoid(C) Microsome (D) Plastid

30. Continuity of cytoplasm from cellto cell is maintained through—(A) Middle lamella(B) Plasmodesmata(C) Nuclear pore(D) Nucleolus

31. Wood fibres are sclerenchy-matous fibres found in—(A) Phloem(B) Xylem(C) Both (A) and (B)(D) Cortex

32. Substances obtained from micro-organisms and used to inhibit thegrowth of other microorganismsare—(A) Antibodies (B) Antigens(C) Antiseptics (D) Antibiotics

33. A fruit has got a wall or pericarpwhich was the wall of the—(A) Ovary (B) Stigma(C) Style (D) Stamen

34. Which of the following is/aretimber yielding plant(s) ?(A) Dalbergia sissoo(B) Desmodium gangetium(C) Vigna sinensis(D) Erythrina variegata

35. The activity of repressor dependson whether—(A) The repressor is positioned

next to the operon(B) There is enough RNA poly-

merase present(C) The repressor is positioned

next to the promoter(D) A key substance in the meta-

bolic pathway is present

36. Conjoint, collateral, closed andscattered vascular bundles withsclerenchymatous sheath arecharacteristic of—

(A) Monocot root

(B) Monocot stem

(C) Dicot root

(D) Dicot stem

37. Biologists who study thesequence of organisms in thefossil record are—

(A) Systematists

(B) Taxonomists

(C) Paleobiologists

(D) Radiologists

38. When two mutations are locatedin the same functional unit or indifferent functional units is con-firmed by—

(A) Back cross

(B) Test cross

(C) Reciprocal cross

(D) Complementation test

39. Each chemical reaction in cellularrespiration requires—

(A) A molecule of NAD

(B) A molecule of FAD

(C) A molecule of ATP

(D) A specific enzyme

40. Retting is a process by whichbacteria bring about the—

(A) Nitrogen fixation

(B) Curdling of milk(C) Synthesis of Vitamin B6

(D) Separation of fibres ofcoconut, husk and flax

41. Monocot leaves grow by meansof—(A) Intercalary meristem(B) Apical meristem(C) Lateral meristem(D) All of the above

42. When a plant is not reproducing,most of its cytokinins are repro-duced in its—(A) Lateral buds (B) Roots(C) Shoot apex (D) Leaves

43. Which of the following is/are thewater pollutant ?(A) Aeroplanes(B) Smoke(C) 2, 4-D and pesticides(D) Automobile exhaust

44. Passage cells are found in—(A) Pricycle(B) Exodermis, pith and peri-

cycle(C) Endodermis(D) Both (B) and (C)

45. The botanical name of ‘coralwood’ is—(A) Adenanthera pavorina(B) Aegle marmelos(C) Amaranthus spinosus(D) Amaranthus tricolor

46. A cell or an organism whichshows the effects of a mutation iscalled a—(A) Mutant(B) Disinfectant(C) Mutation(D) Transformation

47. The antibiotic ‘Erythromycin’ isobtained from—(A) Penicillium notatum(B) Penicillium chrysogenum(C) Streptomyces griseus(D) Streptomyces erytheraaeus

48. In older trees, the inner annualrings are called—(A) Phloem(B) Heartwood(C) Sapwood(D) All of the above

49. Mendelian recombinations aredue to—(A) Mitosis(B) Amitosis(C) Meiosis II(D) Crossing-over

50. The point at which no more CO2can be taken up in C4 plants is—(A) 450–950 ppm CO2

(B) 40–60 ppm CO2

(C) 30–50 ppm CO2

(D) None of the above

(Continued on Page 1030 )

C.S.V. / October / 2009 / 1028

1. Interphase—(A) Includes stages G1, S and G2

(B) Rarely occurs(C) Is the same as prophase,

metaphase, anaphase andtelophase

(D) Requires the use of polarfibres and kinetochore fibres

2. Mitosis is the process by whicheukaryotic cells—(A) Expose the genes for protein

synthesis(B) Grow(C) Become specialized in struc-

ture and function(D) Multiply

3. What are the functions of micro-tubules ?(A) Movement of cilia and

flagella(B) Formation of spindle fibres(C) Both (A) and (B)(D) None of the above

4. Which of the following arearchaebacteria ?

(A) Pseudomonads

(B) Chlamydias

(C) Green sulphur

(D) Methanogens

5. Hydrophytes generally lack—

(A) Root caps

(B) Leaves(C) Stem(D) All of the above

6. Which of the following zonesis/are shown by a root tip ?(A) Cell division zone(B) Maturation zone(C) Elongation zone(D) All of the above

7. Blue-green algae which live inprotozoan association arereferred to as—(A) Cyanellae(B) Phycobionts(C) Mycobionts(D) Linger groups

8. The ventral canal cell is presentin—(A) Ginkgo(B) Pinus(C) Cycas(D) All of the above

9. Which of the following is not anabiotic factor ?

(A) Water(B) Blue-green alga(C) Soil(D) Air

10. Which of the following tissuesconducts water and minerals upfrom the soil in vascular plants ?(A) Phloem(B) Xylem(C) Both xylem and phloem(D) None of the above

11. The offspring of better adaptedindividuals are expected to makea larger portion of the next gene-ration. Which of the followingprocesses explains it well ?

(A) Microevolution

(B) Genetic drift

(C) Gene population

(D) Natural selection

12. The study of the distribution ofplants in the different parts of theworld and the factors responsiblefor it, is known as—

(A) Climatology

(B) Taxonomy

(C) Phytogeography

(D) Morphology

13. Which of the following is notincluded in kingdom-protista ?

(A) Water molds (B) Slime molds

(C) Rhizopus (D) Algae

14. Which of the following groups ofplants bear flowers ?

(A) Thallophyta

(B) Bryophyta

(C) Pteridophyta

(D) Spermatophyta

15. The condition in which there isone to many or one to few chro-mosomes are less than diploidset is called—(A) Monoploidy(B) Aneuploidy(C) Polyploidy(D) Polytene

16. Mutations can be induced by—(A) Temperature(B) Radiation(C) Chemicals(D) All of the above

17. The hypothesis that the earlyatmosphere, combined with anenergy source, produced organicmonomers was tested in 1953by—(A) Fox and Pauling(B) Oparin and Haldane(C) Curie and Pasteur(D) Miller and Urey

18. A group of isodiametric cells withintercellular spaces must be—(A) Collenchyma(B) Parenchyma(C) Sclerenchyma(D) Prosenchyma

19. ‘Emerson’s effect’ is associatedwith—(A) Green house effect(B) Photosynthesis(C) Plant growth hormone(D) Tropism

20. The clockwise shell coiling inLimnaea is called—(A) Sinistral(B) Kappa(C) Dextral(D) Sinistral cleavage

21. Phelloderm consists of—(A) Collenchyma cells(B) Living parenchyma cells(C) Dead parenchyma cells(D) Both (B) and (C)

22. When the anthers mature beforethe stigma, it is said to be—(A) Heterostyly (B) Protogyny(C) Dicliny (D) Protandry

23. In terms of cells which of thefollowing factors usually deter-mines water potential ?(A) Cell wall(B) Solute concentration across

a membrane

C.S.V. / October / 2009 / 1029

(C) Water pressure across amembrane

(D) Both (B) and (C)

24. Lateral meristems are presentalong the lateral sides of—(A) Stem(B) Root(C) Root cap(D) Both (A) and (B)

25. The plants requiring longer expo-sure to light than their criticalperiod, are called—(A) Day-neutral plants(B) Long day plants(C) Short day plants(D) Both (B) and (C)

26. Differentially permeable mem-branes allow the diffusion of—(A) Solvent particles(B) Solute particles(C) Both solvent and solute

particles(D) None of the above

27. The trophic structure of anecosystem can be summarized inthe form of—(A) Ecological pyramid(B) Trophic level(C) ATP formation(D) Sub-trophic level

28. The conducting elements of thephloem are collectively knownas—(A) Sieve cells(B) Phloem parenchyma(C) Sclereids(D) Sieve elements

29. The flower part that containsovule is the—(A) Stamen (B) Sepal(C) Carpel (D) Petal

30. Which of the following plants givesblackwood ?(A) Dalbergia(B) Albizia(C) Acacia(D) None of the above

31. ‘Tricarboxylic acid cycle’ was dis-covered by—(A) Krebs (B) Bent(C) Houseleit (D) Blackman

32. Phyllode is present in which ofthe following plants ?(A) Parkinsonia(B) Citrus

(C) Asparagus(D) All of the above

33. When the change from sol intogel is reversible, the colloid iscalled—(A) Reversible colloid(B) Gelation(C) Solation(D) Suspension

34. The transverse section (T.S.) ofBoerhaavia stem shows—(A) More or less circular outline(B) Uniseriate epidermis having

stomata(C) The stele which includes vas-

cular bundles in three rings(D) All of the above

35. Circular DNA is found in—(A) Mitochondria(B) Bacteria(C) Chloroplasts(D) All of the above

36. Offset is found in—(A) Mint(B) Potato(C) Eichhornia(D) All of the above

37. Which of the following do nothave double bonds between theircarbons ?(A) Unsaturated fatty acids(B) Saturated fatty acids(C) Both (A) and (B)(D) None of these

38. In DNA replication, the helixis unwind by which type ofenzyme ?(A) Helicase(B) Topoisomerase(C) Primase(D) DNA polymerase

39. Transgenic bacteria perform ser-vices in the field of—

(A) Mineral processing

(B) Chemical production(C) Bioremediation(D) All of the above

40. Division of centromere occurs in—(A) Metaphase I(B) Anaphase I(C) Anaphase II(D) Telophase I

41. Both the palisade and spongymesophylls of a leaf are madeof—(A) Sclerenchyma cells(B) Parenchyma cells(C) Dermal cells(D) Collenchyma cells

42. Myrmecophily is a beneficialassociation between floweringplants and—(A) Ants(B) Fungi(C) Mycoplasma(D) Bacteria and Viruses

43. Which of the following sources ofenergy is virtually inexhaustible ?(A) Coal(B) Natural gas(C) Nuclear fusion(D) Oil

44. Which of the following plants iscommonly known as spike moss ?(A) Equisetum (B) Dryopteris(C) Funaria (D) Selaginella

45. Which of the following algae is/are not flagellated ?(A) Volvax and Spirogyra(B) Dinoflagellates and Spirogyra(C) Chlamydomonas and Volvox(D) Spirogyra

46. A common phycobiont in lichenis—(A) Microcystis(B) Trebouxia(C) Cetraria(D) All of the above

47. ‘Perforation theory’ for continuityof life was proposed by—(A) Swammerdan and Malpighi(B) Callan and Tomlin(C) Beadle and Tatum(D) Zinder and Lederberg

48. In terms of cells which of thefollowing factors usually deter-mine water potential ?(A) Water pressure across a

membrane(B) Solute concentration across

a membrane(C) Both (A) and (B)(D) Cell wall

49. The taxonomy of lichens is basedmainly upon the character ofthe—(A) Fungal symbiont(B) Algal symbiont

C.S.V. / October / 2009 / 1030

(C) Cyanophycean symbiont(D) None of the above

50. Environmental resistance—(A) Helps determine the carrying

capacity(B) Prevents exponential growth

from occurring(C) Includes density in depen-

dent effects(D) Both (A) and (B) are correct

ANSWERS WITH HINTS

●●●

(Continued from Page 1027 )

ANSWERS WITH HINTS●●●

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

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

Useful for Various Competitive Exams.

Website : www.upkar.in

C.S.V. / October / 2009 / 1031

1. Centromere separate at whichstage of meiosis—

(A) Anaphase-I

(B) Anaphase-II

(C) Metaphase-I

(D) Prophase-I

2. Sandwich model of plasmamembrane was proposed by—

(A) Singer and Nicolson

(B) Robertson

(C) Gorter and Grendel

(D) None of the above

3. Which one of the following is con-cerned with photo respiration ?

(A) Endoplasmic reticulum

(B) Dictysomes

(C) Peroxisomes

(D) Glyoxisomes

4. Which of these gives a possiblesequence of organic chemicalsprior to the protocell ?

(A) Inorganic gases, nucleo-tides, nucleic acids, genes

(B) Inorganic gases, aminoacids, polypeptide, micro-sphere

(C) Both (A) and (B)

(D) Water, salts, protein, oxygen

5. Root hairs occur in the—

(A) Root cap(B) Region of cell elongation(C) Apical meristem(D) Region of maturation

6. Which of these best shows thatplant cells are totipotent ?

(A) Protoplasmic culture for thepurpose of genetic enginee-ring of plants

(B) Leaf, stem, and root culturefor the purpose of cell sus-pension cultures

(C) A shoot culture for the pur-pose of micropropagation

(D) Flower meristem culture forthe purpose of somatic em-bryos

7. Leaf tip tendrils are present in—(A) Smilax (B) Lathyrus(C) Pisum (D) Gloriosa

8. Water is drawn into root cellsbecause—(A) They have the higher water

potential(B) They have the lower water

potential(C) They have both essential

and beneficial inorganicnutrients

(D) All of the above

9. Gyanobasic style is found in—

(A) Labiatae (B) Gramineae

(C) Liliaceae (D) Compositae

10. Plant growth toward or awayfrom a directional stimulus iscalled—

(A) Tropism

(B) Thigmotropism

(C) Phototropism

(D) None of the above

11. In an embryo sac of a typicalangiosperm there are—

(A) Egg, synergids and anti-podals

(B) Egg, synergids, polar nucleiand antipodals

(C) Egg, synergids, central celland polar nuclei

(D) Egg, synergids and secon-dary wall

12. Which of the following is/arefossil fuel ?

(A) Oil

(B) Coal

(C) Natural gas

(D) All of the above are fossilfuels

13. The special tissue include—

(A) Scleroids

(B) Sclerenchyma

(C) Secretory tissues

(D) Collenchyma

14. How many amino acids areconverted into α-ketoglutarate ?

(A) Three (B) Two(C) Five (D) Six

15. Stilt roots are found in—(A) Rice (B) Sugarcane(C) Groundnut (D) Gram

16. The term synteny refers to—

(A) Genes that are located onthe same chromosome,whether or not they showrecombination

(B) Genes that are located onthe different chromosome,whether or not they showrecombination

(C) Genetic loci that have beenshown by recombinationstudies to be in the samechromosome

(D) All of the above

17. Young sporophyte of fern drawsnourishment from prothallusthrough its—(A) Foot (B) Haustorium(C) Root (D) Rhizoids

18. All demographic research beginswith a statistical treatment ofquantitative data on the popula-tion, this is called—(A) Mortality(B) Fertility(C) Both (A) and (B)(D) Demographic analysis

19. DNA elements which can beswitch their position are called—(A) Cistrons (B) Exons(C) Transposons (D) Introns

20. Which one of the following plantsis called Shepherd’s purse ?

(A) Capsella bursa-pastoris

(B) Selaginella krausiana

(C) Hypericum uralum

(D) Neottia

21. The chemical linking betweenglycolysis and Kreb's cycle is—(A) Pyruvic acid(B) Citric acid(C) Acetyl coenzyme A(D) Succinic acid

22. Who among the following dis-covered a simple procedure forplacing bacteria into either a

C.S.V. / October / 2009 / 1032

gram-negative class or a gram-positive class ?(A) Carl Weigert(B) Joseph Lister(C) Spallanzani(D) None of the above

23. Which one of the following areinitiator codons ?(A) UUU, UUC (B) UAA, UAG(C) UGA, UAG (D) AUG, GUG

24. Opposite decussate phyllotaxy isfound in—(A) Quisqualis(B) Calotropis(C) Mangifera indica(D) Hibiscus rosa-sinensis

25. Richmond lang effect is shownby—(A) Auxins (B) Gibberallins(C) Kinetin (D) Sugars

26. In which of the following forms,sugar is transported within thebody of a plant ?(A) Sucrose (B) Lactose(C) Glucose (D) Maltose

27. Which one of the following is afungicide ?

(A) 2, 4–D

(B) DDT

(C) BHC

(D) Bordeaux mixture

28. The element essential for IAAsynthesis is—

(A) Iron (B) Calcium

(C) Sodium (D) Zinc

29. The correct definition of bio-sphere is—

(A) The earth and its atmos-phere which inhabit livingorganism

(B) All the living organism on theearth

(C) All the plants on earth

(D) All the animals on earth

30. Any substance or mixture ofsubstances which prevents,repels, destroys or mitigates anypest is known as—(A) Attenuation(B) Epinasty(C) Pesticide(D) Growth hormone

31. Sunken stomata are found in theleaves of—

(A) Nelumbium (B) Neem

(C) Maize (D) Nerium

32. Which of the following organsfunction as absorbing andattaching organs in bryophytes ?

(A) Root hairs (B) Columella

(C) Thallus (D) Rhizoids

33. Which one of the following isobtained from algae ?

(A) Chocolate (B) Wax(C) Carragenin (D) Butter

34. Carrageenin is obtained from—

(A) Red algae

(B) Blue-green algae

(C) Green algae

(D) Brown algae

35. A specialised multicellular struc-ture in leaves which excreteswater droplets is—

(A) Lenticel

(B) Stomata

(C) Hydathode

(D) Bordered pit

36. Caryophyllales include—

(A) Wide variety of dicotyle-donous plants

(B) Wide variety of monocoty-ledonous plants

(C) Both (A) and (B)

(D) None of the above

37. To initiate cell plasmolysis, thesalt solution should be—

(A) Isotonic

(B) Hypertonic

(C) Hypotonic

(D) None of these

38. Cell division requires that thegenetic material be able to—

(A) Undergo rare mutations

(B) Be replicated

(C) Store information(D) All of the above

39. In an ecosystem bacteria areconsidered as—(A) Micro consumers(B) Macro consumers(C) Primary consumers(D) Secondary consumers

40. Between the bark and the woodin a woody stem, there is a layerof meristem called—(A) Apical meristem(B) The zone of cell division(C) Cork cambium(D) Vascular cambium

41. The microbial conversion ofammonia to nitrate is called as—(A) Ammonification(B) Nitrification(C) Denitrification(D) Nitrogen fixation

42. The outermost primary meristemgives rise to—(A) Epidermis(B) Ground meristem(C) Procambium(D) All of the above

43. Vascular strands in rhizome ofPteridium are—(A) Collateral open(B) Bicollateral open(C) Concentric amphivasal(D) Concentric amphicribral

44. Available form of nitrogen toplants is—(A) Ammonium (NH4

+)

(B) Nitrate (NO–3)

(C) Both (A) and (B)(D) Atmospheric nitrogen

45. Petaloid staminode is seen in—(A) Cassia (B) Solanum(C) Caesalpinia (D) Canna

46. Which one of the following isusually absent in cortical cells ?(A) Nucleolus(B) Nucleus(C) Nuclear membrane(D) Chloroplast

47. Which one of the following pro-vide food rich in carbohydrates ?(A) Cruciferae(B) Leguminosae(C) Graminae(D) Palmae

48. Which one of the following is thechief reason to consider that theprotocell was probably a fermen-ter ?(A) Fermentation provides the

most amount of energy(B) It did not have any enzyme(C) The atmosphere did not

have any oxygen(D) All of the above are correct

C.S.V. / October / 2009 / 1033

49. For each molecule of glucoserespired, TCA cycle must rotate—(A) Twice (B) Thrice(C) Four times (D) Five times

50. Green algae occur in the formof—(A) Unicellular(B) Multicellular(C) Colonial(D) All of the above

ANSWERS WITH HINTS

-

-

-

C.S.V. / October / 2009 / 1035

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) : [M1L– 1T– 1] isthe dimensional formula of coeffi-cient of viscosity.

Reason (R) : Coefficient ofviscosity is force acting per unitarea per unit velocity gradient.

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

2. Assertion (A) : Light can forminterference pattern.

Reason (R) : Light is a wavemotion.

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

3. Assertion (A) : Graph betweenpotential energy of a spring ver-sus the extension or compression(x ) of the spring is a straight line.Reason (R) : This is becausepotential energy is directly pro-portional to x.(A) (B) (C) (D) (E)

4. Assertion (A) : Red light pro-duces fringes of greater separa-tion in comparison to blue light inYoung’s double slit experiment .

Reason (R) : Smaller slit separa-tion produces greater fringeseparation.

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

5. Assertion (A) : The shape of aliquid drop is spherical.Reason (R) : The pressure insidethe drop is greater than thatoutside.(A) (B) (C) (D) (E)

6. Assertion (A) : In a nuclearreactor graphite is used to cap-ture neutrons.Reason (R) : Successive colli-sions of neutrons with the gra-phite nuclei result in loss ofenergy which slows the neutronsdown.(A) (B) (C) (D) (E)

7. Assertion (A) : The energystored in a coil of 50 mH onpassing 2A current is 0·2 J.

Reason (R) : E = 12 LI2

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

8. Assertion (A) : Energy levels ofan atom must have negativevalues.Reason (R) : When detachedfrom atom, an electron is at anenergy level of zero. When atta-ched, energy is given off and sothe energy of electron is belowzero and is, therefore, negative.(A) (B) (C) (D) (E)

9. Assertion (A) : Bulk modulus ofelasticity (K) represents incom-pressibility of the material.

Reason (R) : K = ΔP

ΔV/V , where

symbols have their usual mean-ing.(A) (B) (C) (D) (E)

10. Assertion (A) : Room heatersand refrigerators lose most oftheir heat by convection.Reason (R) : A hot surface heatsthe air next to it. The hot airrises, to be replaced by cooler airwhich then heats, and so on.(A) (B) (C) (D) (E)

CHEMISTRY

11. Assertion (A) : Poly-hydroxy-butyrate-co— β-Hydroxyvalerate(PHBV) is generally used in con-trolled drug release in the stom-ach of patient.Reason (R) : When drug is put ina capsule of PHBV, it is releasedonly after the polymer is de-graded because PHBV is a bio-degradable polymer.(A) (B) (C) (D) (E)

12. Assertion (A) : Acetate ion ismore basic than formate ion.Reason (R) : The + I effect ofmethyl group in acetate ion inten-sifies electron density on oxygenatom.(A) (B) (C) (D) (E)

13. Assertion (A) : All the fourquantum numbers viz. n, l, ml andms are derived from the solutionof Schrodinger wave equationfor hydrogen atom.Reason (R) : All the four quantumnumbers viz. n, l, ml and ms areessential to designate each orbitalin an atom. These four quantumnumbers also help to designatethe electron present in an orbital.(A) (B) (C) (D) (E)

14. Assertion (A) : When CO2 gas ispassed through a solution ofCa(OH)2 in water, the white pre-cipitate, initially formed, dis-solves.

Reason (R) : Calcium com-pounds are always soluble inbasic solution.

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

15. Assertion (A) : The phenome-non of reverse osmosis is usedas a technique for desalination ofsea water.

Reason (R) : Cellulose acetate ispermeable to water and imperme-able to impurities and ions pre-sent in sea water.

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

16. Assertion (A) : Phenoxide ion isgreatly stabilised by resonanceas compared to ethoxide ion.Reason (R) : Ethoxide is a stron-ger base than the phenoxide ion.(A) (B) (C) (D) (E)

17. Assertion (A) : A catalyst speedsup the forward reaction and re-tards the reverse reaction therebyincreasing the value of Keq.

Reason (R) : A catalyst functionsby lowering the energy of activa-tion which in turn makes the rateconstant larger.(A) (B) (C) (D) (E)

C.S.V. / October / 2009 / 1036

18. Assertion (A) : All the C == Cbonds in organic molecules arecapable of showing geometricalisomerism.Reason (R) : The rotationaround C —— C bond is restricted.(A) (B) (C) (D) (E)

19. Assertion (A) : Magnesium hy-droxide, magnesium carbonate,magnesium trisilicate, aluminiumhydroxide gel, sodium bicarbo-nate and aluminium phosphateare commonly used as antacids.Reason (R) : Acid Gastritis is oneof the commonest ailments asso-ciated with digestion and it iscaused by excess of hydrochloricacid in the gastric juice.(A) (B) (C) (D) (E)

20. Assertion (A) : The copper rodturns the colourless solution ofzinc sulphate to light blue.Reason (R) : Zinc reducescopper(II) ions to metallic copper.(A) (B) (C) (D) (E)

ZOOLOGY

21. Assertion (A) : Coenzymes is anon-protein group without whichcertain enzymes are incompleteor inactive.

Reason (R) : Coenzymes notonly provide a point attachmentfor the chemical group beingtransformed but also influencesproperties of the group.

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

22. Assertion (A) : Organic nutrientsincrease plant growth and lead toeutrophication, has tening con-version of wetlands to dry land,as part of primary succession.

Reason (R) : Inorganic nutrientsincrease bacterial activity inwater, which reduces oxygenconcentration and thus the abilityof water to support other life.

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

23. Assertion (A) : DNA code iscopied in the synthesis of transferRNA.

Reason (R) : Transfer RNAmoves out of the nucleus andafter attaching on the ribosomes,forms the template.

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

24. Assertion (A) : Restrictionenzymes are necessary for thepreparation of recombinant DNA.Reason (R) : Because restrictionenzymes are not used to cleaveplasmid DNA.(A) (B) (C) (D) (E)

25. Assertion (A) : Proteins and RNAare synthesized in G2 phase.

Reason (R) : Z-DNA has righthanded coiling.

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

26. Assertion (A) : Morphogenesisis the change in shape of theembryo.

Reason (R) : Differentiation isthe specialization of cell structureand function as some genes areturned on and others off.

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

27. Assertion (A) : The pitch of thesound is perceived by the positionof the phonoreceptors in thecochlear duct.

Reason (R) : The loudness of thesound is encoded in frequency ofaction potential and the numberof phenoreceptors stimulated.

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

28. Assertion (A) : Temperatures inthe scrotum are about 2° to 3° Ccooler than those in abdomen,and this is important for the survi-val of the temperature-sensitivesperm.

Reason (R) : In the wall of scro-tum, there is a layer of smoothmuscle, the dartos muscle, whichresponds to changing tempera-tures. In the cold, this musclecontracts and the testes aredrawn closer to the abdomen forwarmth, while heat causes themuscle to relax and the testesdescend further for cooling.

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

29. Assertion (A) : Carotid body issmall neurovascular structurenear branch of internal and exter-nal carotids and supplied byvagus and glossopharyngealnerves.

Reason (R) : They are sensitiveto oxygen content of blood andassisting in homeostatic reflex.

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

30. Assertion (A) : During proteincatabolism, proteins are brokendown to amino acids which losetheir amino group.Reason (R) : Protein anabolismconsists of protein synthesis andthe synthesis of non-essentialamino acids.

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

BOTANY

31. Assertion (A) : Green algae livein the ocean but are more likelyfound in fresh water.

Reason (R) : Green algae arebelieved to be closely related tothe first plants because both ofthese groups have a cell wall thatcontains cellulose.

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

32. Assertion (A) : A few moleculesof enzyme may provide reactingsurface to a large number of subs-trate molecules.

Reason (R) : The size of enzymemolecules is far larger than thesubstrate molecules.

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

33. Assertion (A) : The evolution ofmolecular oxygen is associatedwith photosystem-II during photo-synthesis.

Reason (R) : Photosystem-I isnot involved in evolution of mole-cular oxygen.

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

34. Assertion (A) : Viruses thatinfect bacteria (bacteriophages)are another type of vector.

Reason (R) : Scientists canmanipulate them so that theytransport genetic material but donot cause diseases.

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

35. Assertion (A) : The venus’s fly-trap leaves do not interlock.Reason (R) : Because sensitivehairs are not found in this plant.(A) (B) (C) (D) (E)

36. Assertion (A) : Plasmids areextrachromosomal single-stran-ded DNA.Reason (R) : Only eukaryoticcells possess plasmids.(A) (B) (C) (D) (E)

C.S.V. / October / 2009 / 1037

37. Assertion (A) : Fermentationdifferentiates green variety of teafrom black variety of tea.Reason (R) : In tea, the tannin ispartly oxidised and the leafchanges colour and turns brightcoppery red. This happens duringfermentation.(A) (B) (C) (D) (E)

38. Assertion (A) : When the chro-mosomes are highly coiled andcondensed during cell division, itis impossible to count them.

Reason (R) : Each species has acharacteristic chromosome num-bers.(A) (B) (C) (D) (E)

39. Assertion (A) : A non-over-lapping code means that a basein an mRNA is not used fordifferent codons.Reason (R) : In translating mRNAmolecules the codons do notoverlap but are read sequentially.(A) (B) (C) (D) (E)

40. Assertion (A) : The two hydro-gen atoms and one oxygen atomof water (H2O) are held togetherby polar covalent bonds.

Reason (R) : Because theoxygen attracts the negativelycharged hydrogen electronsmore strongly than the hydrogennuclei do.

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

ANSWERS WITH HINTS

●●●

C.S.V. / October / 2009 / 1038

Physics

1. The ratio of maximum acceleration to the maximumdisplacement of a particle performing S.H.M. is equalto angular velocity.

—T/F

2. The earth moving around the sun in a circular orbit isacted upon by a force and hence work must be doneto earth by this force.

—T/F

3. Mercury thermometers can be used to measuretemperatures upto 500°C. (given B.P. of mercury= 367°C)

—T/F

4. Coefficient of frictional force is expressed in newton.

—T/F

5. The average kinetic energy of 1 gm of all ideal gasesat the same temperature is the same.

—T/F

6. A man is sitting in a boat which is floating on a pond.If the man drinks some water from the pond, the levelof water in the pond decreases.

—T/F

7. On a winter night we feel warmer when clouds coverthe sky than when the sky is clear.

—T/F

8. In Young double slit experiment performed with asource of white light, only black and white fringes areobserved.

—T/F

9. A diver in a lake wants to signal to a person lying onthe edge of the pool. He should beam his water proofflash light vertically upwards.

—T/F10. The root mean square speeds of the molecules of

different ideal gases, maintained at the same tem-perature, are same.

—T/F

11. Sun glasses which have a curved surfaces have nopower.

—T/F

12. The γ-decay generally occurs after the α or the βdecay.

—T/F13. It is possible to eliminate dispersion by combining two

prisms of same refracting angles and of differentmaterials.

—T/F

14. No net force acts on a rectangular coil carrying asteady current when suspended freely in a uniformmagnetic field.

—T/F15. A thick rope can not be stretched perfectly horizontal.

—T/F

Chemistry

16. When pressure (P) and temperature (T) of a gas are

constant, its volume, V ∝ 1n.

—T/F17. In the van der Waals equation, constant ‘a ’ reflects

actual volume of one mole of gaseous molecules.—T/F

18. Density of methane gas (CH4) at 20°C and 6 atmpressure will be equal to 3·99 g/L.

—T/F

19. In a given electrical field beta particles (β) aredeflected more than alpha (α) particles, though αparticles have larger charge.

—T/F20. The sum of the formal charges on all atoms of a mole-

cule or ion equals the charge on the species.—T/F

21. The dependence of electrode potential for anelectrode Mn +/M with a concentration under STP isexpressed as

E = E° + 0·0591

n log [Mn+]

—T/F22. Sulphur dioxide (SO2) is a polar molecule while car-

bondioxide (CO2) is non-polar in nature.

—T/F23. The addition of water to acetylene is catalysed by

Ni/H2.

—T/F24. sp

2 hybridized orbitals form stronger bond than sphybridized orbitals.

—T/F25. The heat content of reactants is less than the heat

contents of products in an endothermic reaction.—T/F

26. The pH of 10 ml 5 × 10– 4 M H2SO4 solution changesfrom 3 to 5 when it is diluted to 1 litre.

—T/F27. The species CH3

+ has isoelectronic structure withH3O+.

—T/F

C.S.V. / October / 2009 / 1039

28. Amino methane on treatment with sodium nitrile andHCl produces a diazonium salt at 0°C.

—T/F29. The dipole moment corresponding to 100% ionic

character of KCl will have different value whencompared with value of actual dipole moment of KCl.

—T/F30. The product obtained by heating phenol with zinc

dust, when treated with CH3Cl/anhydrous AlCl3, orthoand para hydroxy toluene are formed.

—T/F

Zoology

31. Lactose is monosaccharide that contains galactoseand fructose.

—T/F

32. Opsonin is a kind of antigen.

—T/F

33. The first menstrual cycle is called menarche and theend of menstrual cycle is called menopause.

—T/F

34. Spontaneous mutations in germ cells alter allele fre-quencies and reintroduce harmful alleles into popula-tion.

—T/F

35. A male human who suffers from sex-linked trait ishomozygous.

—T/F

36. The initiating hormones for the menstrual cycle arisein the hypothalamus.

—T/F

37. Plasmagene is a gene contained in a self-reflectingcytoplasmic particle and inheritance of the charactersare controlled by such genes is Mendelian.

—T/F

38. Peyer’s patches are aggregation of lymph nodesfound chiefly in the ileum near junction with the colon.

—T/F

39. Intestinal microorganisms are capable of synthesizingconsiderable amounts of phylloquinone andmetaquinone vitamins.

—T/F

40. The coxal glands of scorpion are homologous with thegreen glands of crustaceans.

—T/F

41. Planula is solid free-swimming ciliated larva of mostcnidaria and a few of the ctenophores.

—T/F

42. Enterogastrone in hibits the secretion of enterocrinin.

—T/F

43. Planarians are free-living Turbellarians.

—T/F

44. Nematodes are pseudocoelomates.

—T/F

45. Cyclic AMP is a second messenger within cells.

—T/F

Botany

46. The central core of an axis is called a cambium.

—T/F

47. Transpiration of a plant increases with its total leafsurface area.

—T/F

48. Cynellae is the term used for algae present in proto-zoa.

—T/F

49. Lesion most commonly induced by ultravioletradiation is called a spot.

—T/F

50. Green pigment for capturing of sunlight is located onthe cell wall.

—T/F

51. Lichens imperfecti is a class of the lichens containingspecies with no known method of sexual reproduction.

—T/F

52. A plant cell has the potential to develop into entireplant body.

—T/F

53. The two subunits of 80S ribosomes are 50S and 30S.

—T/F

54. The common form of bacteria occurring in groups butnot more than two is called Streptobacillus.

—T/F

55. Whittaker suggested a five-kingdom system that hasbecome widely accepted.

—T/F

56. Phospholipid is more abundant in cell membrane.

—T/F

57. Ribosomes are large particles composed of proteinsand carbohydrates but no r RNA.

—T/F

58. Evidence suggests that there is a ‘clock’ runninginside the cells or organs, determining how long theylive.

—T/F

59. Darwin’s geological observations were consistent withthose of Hutton and Lyell.

—T/F

60. Germplasm is the sum total of all the alleles of thegenes.

—T/F

C.S.V. / October / 2009 / 1041

Physics

Q. What are the rules forcounting the significant figures ?

☞ Rule I : All non zero digits aresignificant.

Rule II : All zeros occurring bet-ween the non zero digits are signifi-cant. For example 230089 containssix significant figures.

Rule III : All zeros to the left ofnon zero digits are not significant. Forexample, 0·0023 contains two signifi-cant figures.

Rule IV : All zeros to the right ofnon zero digits are significant. Forexample, 23·000 as well as 23000contain five significant figures.

Q. What is polygon law ofvector addition ?

☞ (i) If a number of vectorsacting at a point be represented bothin magnitude and direction by thesides of a closed polygon taken inorder, then the resultant of thesevector is a null vector.

(ii) If n vectors are arranged suchthat each one is making an angle2π/n with the preceding vector, thenthe resultant of these vectors is equalto zero.

Q. What is unit force ?☞ It is defined as the force which

changes the momentum of a body byunity in unit time. According to this—

→F =

d→P

dt = ddt (m

→v )

= m.d

→v

dt + →v .

dmdt

If the mass of the system is finiteand remains constant w.r.t. time, then

( )dmdt = 0 and

→F = m

d→v

dt = m→a

=(→P2 –

→P1)

tQ. What are inertial and gravi-

tational masses ?☞ Inertial mass—It is defined

as the ratio of the magnitude of

external force applied on the body tothe magnitude of acceleration pro-

duced in it, i.e., a = ( )Fm

Properties of inertial mass :(i) It is proportional to the

quantity of matter present in the body.(ii) It is independent of shape,

size and state of the body.(iii) It increases as the speed of

body increases : If m 0 be the restmass of the body and c be the speedof light then

m =m 0

1–( )v 2

c 2

Gravitational mass—Mass ofthe material of the body which isdetermined by gravitational pull actingon it, is called as gravitational mass,i.e.,

m =FR2

GMInertial and gravitational masses

are equivalent. However the defini-tions are independent of each otherand two masses differ in the methodsof their measurement.

Q. What is law of equipartitionof energy ?

☞ According to the law of Equi-partition of energy, the total kineticenergy of a dynamical system con-sisting of a large number of particlesis equally distributed among itsdegrees of freedom. We know thataverage K.E. of translation of a mole-

cule of perfect gas is ( )32.k T and the

molecule has three degrees of free-dom. Hence, the average kineticenergy associated with each degree

of freedom is ( )12.kT . Here k is

Boltzmann constant, k = RN = 1·38 ×

10–23 JK– 1 per molecule R = Gasconstant and N = Avogadro’s number.

Q. What is total internal reflec-tion ?

☞ The phenomenon of totalinternal reflection occurs when lighttravels from a medium of higher

refractive index to a medium of lowerrefractive index. At the critical angle(θc), the refrected ray just grazes theboundary between the two media.Using Snell’s law we get

μ1 sin θc = μ2 sin 90°

θc = sin–1 μ2

μ1

For an angle of incidence greaterthan θc the light is totally reflectedback into the medium of higherrefractive index. This phenomenon iscalled total internal reflection.

Q. In a transistor, the emitter isalways forward-biased but thecollector is reverse biased. Why ?

☞ The forward bias of the emitterof transistor forces the charge carriersto cross the junction. So the currentflows in the transistor. If the emitter isreverse biased, there will be practi-cally no current because the majoritycharge carriers will be prevented fromcrossing the emitter junction.

The reverse bias of the collectorhelps in fast collection of the chargecarriers from the base region. Thus,the reverse bias of the collector helpsto reduce the recombination rate inthe bias region.

Q. Why a common emitter tran-sistor amplifier is preferred to acommon base transistor amplifier ?

☞ This is because the currentgain of common emitter transistoramplifier is much higher as comparedto that of a common base transistoramplifier.

Q. The magnetic field at thecentre of a current-carrying cubemade of twelve wires is zero. Doyou know why ?

☞ A current-carrying cube of 12wires can be regarded as a set of sixcurrent-carrying pairs. The contribu-

C.S.V. / October / 2009 / 1042

tion of each pair is zero. So the netmagnetic field induction at the centreof the cube is zero (see figure).

Q. Doppler effect in light issymmetrical but the same effect insound is asymmetrical. Explain.

☞ Whether source of light isreceding from the observer or theobserver is receding from the source,both situations are physically identicaland show exactly the same Dopplerfrequency. So the Doppler effect inlight is symmetrical. On the otherhand, the source of sound recedingfrom the observer or the observer isreceding from the source of sound aredifferent physical situations. So theywould show different Doppler fre-quency. Thus, Doppler effect in soundis asymmetrical.

Q. What do you mean by‘atomicity of electricity’ ?

☞ This is just another name forquantisation of charge. The chargecarried by a body is always an integralmultiple of the fundamental charge of1·6 × 10–19 C.

Q. Why is the value of specificcharge not constant for positiverays ?

☞ The positive rays consist ofions of different isotopes of the gas inthe discharge tube. The ions mayhave the same electric charge. How-ever, masses of different isotopes aredifferent. Consequently, the value of

charge to mass ratio, i.e., em is not

constant.Q. What is the basic principle

of electron microscope ?☞ It is based on de-Broglie

hypothesis. A beam of acceleratedelectrons behaves like a wave. Thiswave can be handled by electric andmagnetic fields in exactly the sameway as electromagentic waves can behandled by lenses.

Q. A hydrogen atom containsonly one electron, but the spectrumof hydrogen atom has many lines.Do you know why ?

☞ In the ground state of hydro-gen atom, an electron occupies thelowest energy level. When the atom isexcited, the electron jumps to somehigher energy level. But within 10– 8

second the electron jumps to thelowest energy level. However, it mayso happen that the electron jumpsfrom higher energy level to somelower energy level and then to thelowest energy level. When an electronjumps from higher energy level tolower energy level, a spectral line isemitted. Since a large number ofenergy levels are available, therefore,a large number of electron-transitionsare possible. This explains why thespectrum of hydrogen atom contains alarge number of lines.

Chemistry

Q. What is doctor solution ?☞ An alkaline solution of sodium

plumbite containing sulphur is knownas doctor solution. It is used inpreparation of disulphides from thiols.

2RSH + Na2PbO2 + S →144243

R–S–S–R + PbS + 2NaOHDiallyl disulphide (CH2––CH–CH2–)2S2,occurs in garlic.

Q. What are important charac-teristics of resonating structures ofa molecule ?

☞ Resonance structures of amolecule must have an identicalarrangement of atoms, nearly sameenergy content, same number of lonepairs of electrons. However, they donot have identical bonding in them.

Q. What is the charge on onemole of electrons ?

☞ The charge on one electron is1·6021892 × 10–19 coulomb. We knowthat one mole of electrons = 6·022045× 1023 electrons. Hence, the chargeon one mole of electrons is1·6021892 × 10–19 × 6·022045 × 1023

= 96,484·56 coulomb.The value 96,484·56 C/mol is knownas Faraday’s constant.

Q. Why the product of fre-quency and wavelength of infraredand ultraviolet radiations is alwayssame ?

☞ The electromagnetic spectrumincludes radio and TV waves. micro-waves, infrared, visible light, ultra-violet, X-rays, γ-rays and cosmic rays.

They travel at the speed of light (c).They differ however, in their frequen-cies and wavelengths. The product offrequency and wavelength of electro-magnetic radiation is always equal tothe speed of light (c ).

νλ = c

As a result, electromagneticradiation that has long wavelengthhas low frequency and a radiationwith high frequency has short wave-length.

Q. Fahrenheit and Celsiusscales of temperature converge at–40°°°°.

☞ The temperature at which twoscales reach zero is not the same;zero on Celsius scale is equivalent to32°F. To convert from Celsius toFahrenheit we, therefore, have tomultiply by 9/5 and then add 32°.

T°F =95 T°C + 32

This equation can be rearrangeto convert from Fahrenheit to Celsius

T°C =59 [T°F – 32]

It does not matter in which direc-tion we try to do the conversion, from°C to °F or vice-versa, – 40°F is equalto – 40°C.

Q. What is the mass of asingle atom of 12C in grams andthe number of grams per amu ?

☞ We know the fact that a moleof 12C has mass of exactly 12 gram.We then construct a unit factor basedon the fact that a mole of any subs-tance contains 6·022 × 1023 atom.

12·00 g 12C1 mol C ×

1 mol C6·022 × 1023 atom

= 1·993 × 10–23 g/atom

The number of grams per amu

= 1·993 × 10–23 g

12·00 amu = 1·661 × 10–24 g

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 ?

☞ According to ideal gas law,the product of pressure and volumeof an ideal gas divided by the productof amount of gas (moles) and abso-lute temperature is a constant

PVnTK = R

C.S.V. / October / 2009 / 1043

We can calculate the value of Rfor any set of units of P, V, n and T bysimply substituting the known values

1·0 atom × 22·414 L

1·0 mol × 273·15 K

= 0·082057 L-atm/mol-KQ. Would H2O2 behave as

oxidant or reductant with respect tocouple Fe3+/Fe2+ at standard con-centration ?

☞ We know that H2O2 can actboth as oxidant and reductant. Follow-ing reactions occur with respect toFe3+/Fe2+ couple.

Fe3+ + e – → Fe2+ ;

E° = 0·771 V

2Fe3+ + H2O2 → 2Fe2+ + O2 + 2H+;

E° = 0·02 VFe2+ → Fe3+ + e

–;E° = – 0·771 V

2Fe2+ + H2O2 + 2H+ → 2Fe3+ + 2H2O;

E° = 1·00 VSince both potentials are positive,

H2O2 will act as an oxidant and reduc-tant.

In fact, iron(II) or iron(III) saltscatalyse self-oxidation-reduction ofH2O2.

Q. Why do cyanic acid and iso-cyanic acid have the same conju-gate base ?

☞ N ≡ C—OH is cyanic acid andHN = C —— O is isocyanic acid.

The conjugate bases of each acidare contributing structures of sameresonance hybrid, i.e., the resonancestructures of both conjugate bases ofboth acids have same hybrid struc-ture.

H : . .N —— C —— O

– H+⎯→

: –. .N —— C —— O ←→

. .N ≡ C—

. .O. .:–

H : . .O. .—C ≡ N

– H+⎯→ :

–. .O. .—C ≡ N ←→ O —— C =

. .N:–

Both have hybrid structure

: Nδ–

˙˙—— C˙˙—δ –O. . :

Hence conjugate base is same inboth cases.

Q. What is the caramel ?☞ A brown substance obtained

by heating cane sugar or other carbo-hydrate materials. Its chemical natureis not exactly known. Its reactions vary

with its method of preparation. It issoluble in water and is used as acolouring material for foodstuffs anddrinks.

Q. What is the difference bet-ween nucleophilicity and basicity ?

☞ The nucleophilicity is definedas a reaction of a nucleophile (: Nu–)with electrophilic carbon atom and itinfluences the rate of reaction asreflected by rate constant kr.

: Nu– +|C : X

kr⎯→ Nu—|C

+ + X–

The basicity of a base (: B–) isthe ability to remove H+ from an acidas is represented by equilibriumconstant kb.

: B– + H : A k eq

B : H + : A–

The basicity of a nucleophiledetermines the equilibrium of a reac-tion while nucleophilicity the kinetics.Remember that :

(a) Bases are better nucleophilesthan their conjugate acids :

H. .O. .:– >> H2

. .O:

and H2

. .N:– >> : NH3

(b) From left to right in a periodboth basicity and nucleophilicitydecrease

H3C:– > H2

. .N:– > H

. .O. .:– > :

. .F. .:–

(c) In going down the groupnucleophility increases and basicitydecreases.

: . .F. .:– < :

. .Cl. .:– < :

. .Br. . :– < :

. .I. .:–

(Nucleophility)

: . .F. .:– > :

. .Cl. .:– > :

. .Br. . :– > :

. .I. .:–

(Basicity)Q. Why is the following general

reaction,R—Cl + R′′′′–ONa →→→→ R—O—R′′′′ + NaClcatalysed by a trace amount ofNaI ?

☞ In the presence of NaI, theoverall reaction occurs in two steps,each of which is faster than, uncata-lysed reaction.(I) R—Cl + I– → R—I + Cl–

This step is faster because I– is asoft base and has more nucleo-philicity than OR–, a hard base.

(II) R—I + R′O– : → R–O–R′ + I–

This step is faster because I–

is a better leaving group than Cl–.

Zoology

Q. What are good and badcholesterol ?

☞ In human blood, cholesterol iscarried by special proteins, the lipo-proteins, which are manufactured inthe liver. Two classes of these chole-sterol–containing lipoproteins aresignificant for the heart and bloodvessels. Low density lipoproteins(LDL) contain triglyceride and chole-sterol. If cells are in need of chole-sterol for plasma membranes orsteroid hormone synthesis, they takeup LDL. If this process does notoccur, the LDL loaded with chole-sterol remain in the plasma anddeposition of cholesterol as plaquesbecomes possible. High densitylipoproteins (HDL) are the secondgroup of lipoproteins. HDL picks upcholesterol from cells and transportsit to the liver for disposal. The liverreleases the excess cholesterol withthe bile into the small intestine. It hasbeen found that higher HDL levelsare associated with a fewer risk ofdeveloping atherosclerosis, whilehigh LDL levels are associated with ahigher risk. That is why HDL hasbeen called ‘good’ cholesterol andLDL ‘bad’ cholesterol. LDL levelsbelow 130 milligrams per decilitre andHDL levels above 40 mg/dl are consi-dered desirable.

Q. How do the digestive tractsof carnivores differ from those ofhervivores ?

☞ Animal meat requires morestorage and less processing thanplant materials. The teeth of a carni-vore are pointed and sharp for killingits prey and tearing it into piecessmall enough to swallow. Animalmeat does not require chewing. Theteeth of a herbivore, by contrast, areflat for crushing and grinding plantmaterials. A mammal cannot digestcellulose and so derives no nutrientsfrom plant materials unless the cellwalls are ruptured by chewing. Acarnivore has a large stomach forfood storage, since it eats large andinfrequent meals. An herbivore has asmaller stomach because it eatssmaller amounts of food more fre-quently. Most digestion and allabsorption of food take place in smallintestine and a carnivore, whose foodrequires less processing has ashorter small intestine than an herbi-vore, whose food requires extensiveprocessing, has a longer small intes-tine.

C.S.V. / October / 2009 / 1044

Q. What is sodium pump ?☞ Sodium pump is active trans-

port mechanism present in plasmamembranes of most animal cells, con-suming an estimated third of a cell’sATP production in pumping sodiumions (Na+) out of the cell and potas-sium ions (K+) into it in the ratio 3 : 2.As a result of its (electrogenic) role inthe establishment of a cell’s restingpotential this pump serves to regulatecell volume by casting out Na+ whichwould tend to enter along its electro-chemical gradient, adding to thenegative osmotic potential of the celland drawing water in. The cell’sinternal electrical negativity preventsions (Cl–) from entering and havingthe same effect. The pump is blockedby external ouabain, and animal cellmay, therefore, swell or burst if this orother inhibitors of ATP synthesis orhydrolysis are added. It is indirectlyresponsible for glucose and aminoacid uptake by cells since it creates asodium gradient necessary for Na+-based symports.

Q. What is Transduction ?☞ Transduction is process in

which usually a bacteriophage picksup DNA from one bacterial cell andcarries it to another, when the DNAfragment may become incorporatedinto the bacterial host’s genome. Twobasic types : (a) generalized trans-duction, where the phage DNA-packaging mechanism picks up ‘bymistake’ any phagesized fragment ofchromosomal DNA, which can beintegrated by homologous recombi-nation into the recipient genome afterinjection into the cell by the phageapparatus, (b) specialized tranductionin which on induction, integrated phageDNA genome is imprecisely excisedfrom the chromosome, carrying adja-cent chromosomal DNA with it; sincethe phage is generally integrated at aspecific site in the chromosome, onlya few bacterial genes can be trans-duced this way.

Q. What is Staph Food Poison-ing ?

☞ Staph food poisoning, onceknown as ptomaine poisoning, resultsfrom toxins produced by staphylococ-cus bacteria growing on food. Thesebacteria can be part of the normalflora of the nasal passage but canalso cause boils, pimples and otherskin infections. Therefore, sneezingand coughing on food and preparingfood with unprotected hands can leadto the deposition of millions of staphbacteria. If such contaminated foodremains at room temperature for few

hours, the bacteria will have time togrow, producing and releasing harmfulquantities of toxin. When ingested withfood, the toxin irritates the gastroin-testinal tract, causing vomiting,nausea and diarrhoea. This causesstaph food poisoning.

BotanyQ. How are monosaccharides

formed from polysaccharides ?☞ Polysaccharides are broken

down into their monosaccharide com-ponents by hydrolysis. This processis essentially reverse of dehydrationsynthesis : each bond betweenmonomers is broken and a moleculeof water is added to form the originalmonosaccharides. This process isessential for use of sugars by cells;starch and glycogen must be brokendown into monomers (glucose) beforetheir energy is available for cellularwork.

Q. Why is CO2 commonlyknown as green house gas ?

☞ Since CO2 is confined exclu-sively to the troposphere, its higherconcentration may act as a seriouspollutant. Although CO2 is a minorcomponent (about 0·0314%) of theatmosphere, it plays a significant roleby absorbing radiant heat, catchingmuch like the glass of a green house.That’s why we often term CO2 andother heat-trapping gases as greenhouse gases. And we call the warm-ing caused by these gases the greenhouse effect.

Q. What do you mean byEnterobacter ?

☞ Enterobacter is a genus ofbacteria, formerly called Aerobacter.Its members are motile, ferment lac-tose, form gas from glucose, decar-boxylate ornithine and utilize citrate,but do not produce indole or hydro-gen sulphide. The two recognizedspecies, Enterobacter cloacae and E.aerogenes, are widely distributed innature (sewage, soil and watersources) and are also found in humanand animal feces. A few speciesEnterobacter agglomerans has beenproposed for strains formerly assig-ned to the genus Erwinia.

Q. What is parsnip ? What areits importance ?

☞ Parsnip (Pastinaca sativa) is ahardy biennial plant of mediterraneanorigin. It belongs to plant order—Umbellales. The parsnip is grown forits thickened taproot and is usedprimarily as a cooked vegetable.Exposure of mature roots to low tem-

peratures, not necessarily freezingimproves the quantity of the root byfavouring the conversion of starch tosugar.

Q. What do you mean by rhizo-sphere ? What are its effects onplants ?

☞ Rhizosphere is the soil regionsubject to the influence of plant roots.It is characterized by a zone ofincreased microbiological activity andis an example of the relationship ofsoil microbes to higher plants.

At the roof surface the rhizo-sphere effect is most intense, fallingoff sharply with increasing distance.In the rhizosphere there are moremicroorganisms than in soil distantfrom the plant. The rhizosphere effectis seen in seedling plants; it increaseswith the age of the plant and usuallyreaches a maximum at the stage ofgreatest vegetative growth.

Q. What do you mean by intra-specific competition ?

☞ Intraspecific competition is thecompetition occurring between mem-bers of the same species. It is likely tobe intense because individuals willtend to share requirements for thesame resources. Although there maybe age differences in resource require-ments or sex differences. By depress-ing the fitness of individuals in crowdedpopulations, it influences both pro-cesses like mortality and fecundity,and hence regulates population size,leading to behavioural adaptations toovercome the competition, such asterritoriality and dispersal.

Q. What is oregano or marjo-ram ? What are its important chrac-teristics ?

☞ Oregano (marjoram) is a herb.The dried leaves of several aromaticplants are known as oregano. Thusoregano is a common name for ageneral flavour and aroma rather thanthe name of a specific plant.

An European oregano (origanumvulgare) and Greek oregano (Origa-num hervacleoticum) are both in themint family Laminaceae. Euporopeanoregano can be distinguished by itsstrong piquant character and tallgrowth with dark, broad leaves; it is aperennial erect herb, 2-3 feet tall withpubescent stem, ovate, dark greenleaves and white or purple flowers.Mexican oregano is obtained fromplants of Lippia graveolens.

Dried oregano leaves are used asa culinary herb in meat and sausageproducts, soup and salads. Theessential oil of oregano is used in foodproducts, cosmetics and liqueurs.

●●●

C.S.V. / October / 2009 / 1045

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 PrizeVijay KumarNew MedicareL. B. Palace Kadamkuan, PatnaBihar–800 003

Second PrizeGagandeep SinghC/o Dayaram VermaL-971 Shastri Nagar, MeerutU.P.–250 004

Third Prize

Ravi JaiswalC/o Gaurav JaiswalRoom No. 88, A. N. Jha Hostel,University of Allahabad, AllahabadU.P.–211 002

ANSWERS WITH HINTS

●●●

C.S.V. / October / 2009 / 1046

1. A block of mass m = 2 kg is keptat the floor of an elevator. Theelevator is suddenly pulled downwith an acceleration of 14·7 m/s2.The distance moved by the blockin 0·3 sec from start is—(A) 0·6615 m downwards(B) 0·4410 m downwards(C) 0·2205 m downwards(D) 0·6615 m upwards

2. Light is a transverse electro-magnetic wave carrying energythrough any cross-section at therate of— (in terms of electric and

magnetic field vector)

(A)ε0 | E2 |2

2 (B)| B |2

2 μ0

(C)| E × B |

μ0(D)

Ec B

3. A lump of copper wire (mass m,density d and resistivity ρ ) isdrawn in a fine wire of length land radius r. The wire is foldedinto a circular loop of diameter Dand then placed in a time varyingmagnetic field varying at a rated Bdt = 100 gauss/sec. The induced

electric current in the circular loopis—(A) Proportional to the length of

the wire(B) Proportional to the diameter

of the wire(C) Proportional to the radius of

the loop(D) Proportional to the rate d B/dt

4. A steel rod [density ρ = 7·8 × 103

kg/m3 and Young’s modulus(Y = 2 × 1011 N/m2)] of length 50cm is clamped at its mid point.The number of natural longitudi-nal oscillations of the rod withinaudible frequency range will be—(A) 2 (B) 3(C) 4 (D) None of these

5. A charge Q is placed at each ofthe opposite corners of a square.A charge q is placed at each ofthe other two corners. If the netelectrical force on Q is zero, thenQ/q equals—

(A) – 12

(B) – 2 2

(C) – 1 (D) 1

6. The reagent not used in thechloromethylation of benzene is—(A) Cl2 (B) HCHO(C) ZnCl2 (D) HCl

7. The most acidic of the followingis—(A) CH3CH2COOH(B) C6H5COOH(C) ClCH2COOH(D) All have the same acidic

value

8. If one litre of a gas X at 600 mmand half litre of gas Y at 800 mmare taken in a 2 litre tube, theresulting pressure is—(A) 500 mm (B) 1000 mm(C) 5 mm (D) 2000 mm

9. The enzyme carbonic anhydrasecatalyses the change—(A) Maltose to glucose(B) Lactose to glucose and

galactose(C) Carbonic acid to CO2 and

H2O(D) All of the above

10. The stage in a titration when theindicator, in its widest sense,undergoes maximum change incolour for a small amount ofadded titrant, is called—(A) End point(B) Filter press(C) Fischer-Hepp rearrange-

ment(D) Feedback inhibition

11. A receptor—(A) Initiates nerve impulses(B) Responds to only one type

of stimulus(C) Is the first part of a reflex arc(D) Performs all of the above

12. In amphibians and most reptiles,the heart bears—(A) No any atrium(B) Two atria(C) One atrium(D) Three atria and one ventricle

13. A cross section at the mid pointof the middle piece of a healthyhuman sperm will show—(A) 9 + 2 arrangement of micro-

tubules only(B) Centriole, mitochondria and

9 + 2 arrangement of micro-tubules

(C) Mitochondria and 9 + 2arrangement of microtubules

(D) Centriole and mitochondriawith three flagella

14. Lakes that are deficient innutrients and consequently low inproductivity are called—(A) Oligotrophic(B) Metatrophic(C) Mesotrophic(D) Eutrophic

15. Which part of the adult humancolon joins the rectum ?(A) Sigmoid colon(B) Transverse colon(C) Ascending colon(D) Descending colon

16. The feature of flowers of Stapeliagigantea includes—(A) Flowers are star-shaped(B) Flowers have an elaborate

circular fleshy disc in thecentre

(C) Flowers have fringes of softwhite hairs on the petals

(D) All of the above

17. Amino acids derived from thebreakdown of well-stored seedproteins also yield precursorsfor—(A) Gluconeogenesis(B) Diuresis(C) Atherosclerosis(D) Amoebiasis

18. Which of the following is/arecorrect regarding Cyanophyta ?(A) No plastids(B) Pyrenoid absent(C) Cyanophycean starch found

in the chromatoplasm(D) All of the above

19. Theory of spontaneous genera-tion was prevalent in—(A) Chinese(B) Egyptian(C) Babylonian(D) All of the above

20. The spores of Funaria aftergermination give rise to—(A) Archegonia(B) Antheridia(C) Protonema(D) All of the above

●●●

C.S.V. / October / 2009 / 1047

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☞ General Principles of Taxonomy☞ Transgenic Organisms☞ Photosynthesis : Part-I☞ Soil Erosion and Conservation☞ Many Typical Model Papers and other

Regular Features

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Heat Engines☞ Atomic Physics-VII : Black Body Radia-

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April 2009☞ Atomic Physics-IV : Semiconductor

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Directions—(Q. 1–3) Thesequestions are based on the followinginformation :

‘A + B’ means ‘A is father of B’.‘A × B’ means ‘A is wife of B’.‘A – B’ means ‘A is sister of B’.‘A ÷ B’ means ‘A is brother of B’.

1. Which of the following expres-sions represents ‘J is daughter ofD’ ?(A) D × K + J ÷ H

(B) D × K + H – J

(C) D × K + J – H(D) D + K ÷ J(E) None of these

2. In M ÷ L + T × R, how is T relatedto M ?(A) Nephew(B) Niece(C) Nephew or Niece(D) Daughter(E) Cannot be determined

3. Which of the following expres-sions represents ‘V is mother ofL’ ?(A) V ÷ F + J – L(B) F × V + J – L(C) F ÷ V + J – L(D) V × F + J – L(E) None of these

Directions—(Q. 4 and 5) Thesequestions are based on following setof numbers :

349 483 766 598 6744. If in each of the numbers the

positions of the first two digits areinterchanged and then thenumbers are arranged in ascend-ing order which number will be atthe second position ?(A) 349 (B) 483(C) 766 (D) 598(E) 674

5. If in each of the numbers thepositions of the first and the thirddigits are interchanged and thenthe numbers are arranged indescending order which numberwill be at the fourth position ?(A) 349 (B) 483(C) 766 (D) 598(E) 674

6. In a certain code CONCISE iswritten as FTJBBNM. How isFISHERY written in that code ?(A) ZSFIGJT(B) ZSFGIHR(C) ZSFGEHR(D) ZSFEHGR(E) None of these

7. Four of the following five arealike in a certain way and soform a group. Which is the onethat does not belong to thegroup ?(A) Snake (B) Crocodile(C) Frog (D) Lizard(E) Fish

8. How many meaningful Englishwords can be made from theletters ADER, using each letteronly once in each word ?(A) None (B) One(C) Two (D) Three(E) More than three

9. How many such pairs of lettersare there in the word PRELI-MINARY each of which have asmany letters between them in theword, as they have in the Englishalphabet ?(A) None (B) One(C) Two (D) Three(E) More than three

10. How many such digits are therein the number 57683421, each ofwhich is as far away from thebeginning of the number, as theywill be when arranged in descen-ding order within the number ?(A) None (B) One(C) Two (D) Three(E) More than three

11. If in the word CALIBRE, theprevious letter in the Englishalphabet replaces each conso-nant and each vowel is replacedby the next letter and then theorder of letters is reversed, whichletter will be third from the rightend ?(A) A (B) C(C) B (D) K(E) None of these

12. In a row of children facing North,Bharat is eleventh from the rightend and is third to the right ofSamir who is fifteenth from theleft end. Total how manychildren are there in the row ?(A) 29 (B) 28(C) 30 (D) 27(E) None of these

13. In a certain code READ is writtenas #5%6 and PAID is written as$%46. How is RIPE written inthat code ?(A) #4$5 (B) #6$5(C) $4#5 (D) $4#6(E) None of these

14. If it is possible to make only onemeaningful word from the first,the fourth, the fifth and the ninthletters of the word VERSATILEusing each letter only oncesecond letter of that word is youranswer. If more than one suchword can be formed your answeris M and if no such word can beformed your answer is N.

(A) A (B) S

(C) E (D) N

(E) M

15. What will come next in the seriesgiven below ?1 1 2 1 2 3 1 2 3 4 1 2 3 4 5 1 23 4 5 6 1 2 3 4 5 6 7 1 2 3 4 5(A) 1 (B) 5(C) 6 (D) 2(E) None of these

Directions—(Q. 16–23) In thesequestions the symbols @, #, ★, $ and© are used with different meanings asfollows :

‘A @ B’ means ‘A is not greaterthan B’.

‘A # B’ means ‘A is neithergreater than nor equal to B’.

‘A ★ B’ means ‘A is not smallerthan B’.

‘A $ B’ means ‘A is neithersmaller than nor equal to B’.

‘A © B’ means ‘A is neithergreater than nor smaller than B’.

C.S.V. / October / 2009 / 1050

Now in each of the followingquestions assuming the given state-ments to be true, find out which of theconclusions I, II, III given below themis/are definitely true and mark youranswer accordingly.

16. Statements :H $ K, K © R, R @ J, J # FConclusions : I. R # H

II. F $ RIII. H $ J

(A) Only I and II are true(B) Only I is true(C) Only II is true(D) Only II and III are true(E) All are true

17. Statements :L @ M, M ★ P, M # D, D $ FConclusions : I. L @ P

II. P @ DIII. M © F

(A) Only I is true(B) Only III is true(C) Only either I or III is true(D) Only I and II are true(E) None is true

18. Statements :T @ V, V # Q, Q © L, L ★ MConclusions : I. M @ Q

II. T @ LIII. T # L

(A) Only I is true(B) Only II is true(C) Only III is true(D) Only I and III are true(E) None of these

19. Statements :J ★ E, D @ E, E $ K, K © T

Conclusions : I. J $ DII. J ★ D

III. E $ T

(A) None is true(B) Only II and III are true(C) Only I and III are true(D) All are true(E) None of these

20. Statements :H @ I, I # L, L ★ A, A $ Q

Conclusions : I. H # LII. H ★ L

III. Q # H(A) Only I is true(B) Only I and II are true

(C) Only III and either I or II aretrue

(D) Only either I or II is true(E) None is true

21. Statements :V # W, W ★ T, T © K, K @ FConclusions : I. T @ V

II. T $ VIII. F ★ T

(A) Only either I or II is true(B) Only III is true(C) Only I and II are true(D) Only III and either I or II are

true(E) None of these

22. Statements :F ★ E, E © H, H @ I, I $ WConclusions : I. W # H

II. F $ HIII. E # I

(A) None is true(B) Only I and II are true(C) Only III is true(D) Only either I or III is true(E) None of these

23. Statements :L @ R, R # M, N $ M, N # KConclusions : I. L # N

II. K $ MIII. R # N

(A) None is true(B) Only I and II are true(C) Only I and III are true(D) Only II and III are true(E) All are true

Directions—(Q. 24–28) In eachof these questions, a group of lettersis given followed by four combinationsof digits/symbols lettered (A), (B), (C)and (D). The letters are to be codedas per the scheme and conditionsgiven below. You have to find outwhich of the four digit/symbol combi-nations correctly represents the groupof letters. The serial letter of thatcombination is your answer. If noneof the combinations is correct, youranswer is (E) i.e. ‘None of these’.Letters :

P E Q A R T J L F I H U B D GDigit/Symbol :

3 7 # 9 6 2 $ 1 © 4 5 @ % 8 ★

Conditions : (i) If the first letteris a consonant and the last letter is avowel their codes are to be inter-changed.

(ii) If the first as well as the lastletter is a vowel, both are to be codedas £.

(iii) If the first letter is a vowel andthe last letter is a consonant both areto be coded by the code for con-sonant.24. PTLAFI—

(A) 3219©4 (B) 4219©3(C) 4219©4 (D) 3219©3(E) None of these

25. FQUEJL—(A) %#@7$% (B) 1#@7$©(C) 1#@7$1 (D) ©#@7$1(E) None of these

26. AJQTHI—(A) 9$#259 (B) 9$#254(C) £$#25£ (D) 4$#259(E) None of these

27. EBGLRQ—(A) #%★16# (B) 7%★L6#(C) #%★167 (D) 7%★167(E) None of these

28. DUARFE—(A) 8@96©7 (B) 8@96©8(C) 7@96©7 (D) %@96©%(E) None of these

Directions—(Q. 29–33) Studythe following information carefully toanswer these questions.

J, K, H, R, F, L, N and Q aresitting around a circular table facingthe centre. H is third to the left of Land is to the immediate right of K. Ris third to the left of N but is not aneighbour of H or L. J is second tothe right of Q.29. Who is second to the left of N ?

(A) Q (B) K(C) J (D) F or J(E) None of these

30. Which of the following groups ofpersons has the first personsitting between the next two ?(A) LKN (B) QFL(C) JHR (D) JHF(E) None of these

31. Who is to the immediate left ofR ?(A) Q (B) K(C) F (D) N(E) None of these

32. Which of the following is correctposition of J with respect to K ?(A) Third to the left(B) Third to the right(C) Second to the left(D) Second to the right(E) Fourth to the right

C.S.V. / October / 2009 / 1052

33. Four of the following five are alikein a certain way on the basis oftheir seating positions and soform a group. Which is the onethat does not belong to thegroup ?(A) RQ (B) LK(C) HJ (D) JR(E) FN

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C.S.V. / October / 2009 / 1053

1. If a new State of the Indian Unionis to be created which one of thefollowing Schedules of the Cons-titution must be amended ?(A) First (B) Second(C) Third (D) Fifth

2. When water is heated from 0°Cto 10°C its volume—(A) Decreases(B) Does not change(C) Increase(D) First decreases and then

increases

3. Which one of the following is notradioactive ?(A) Tritium (B) Astatine(C) Francium (D) Zirconium

4. Centre for DNA finger printingand diagnostics is situated at—(A) Chandigarh (B) Hyderabad(C) New Delhi (D) Lucknow

5. Which one of the following ani-mals was not represented on theseals and terracotta art of Harap-pan culture ?(A) Cow (B) Rhinoceros(C) Tiger (D) Elephant

6. The range of Agni-II missile isaround—(A) 5000 km (B) 2000 km(C) 500 km (D) 3500 km

7. The approximate age of theAravalli range is—(A) 370 million years(B) 670 million years(C) 470 million years(D) 570 million years

8. In reference to the provisions asenshrined in the Constitution ofIndia which of the followingstatements is incorrect ?(A) Article 56 provides for the

President of India(B) In Article 75 is enshrined the

appointment of the PrimeMinister and Council ofMinisters

(C) Article 155 mentions theappointment of the Governorof a state

(D) Article 164 provides for theappointment of Chief Ministerand Council of Ministers of aState

9. World Ozone Day is observedon—(A) March 20(B) June 5(C) September 16(D) October 3

10. Under the Permanent Settlementof 1793, the Zamindars wererequired to issue pattas to thefarmers which were not issuedby many of the Zamindars. Thereason was—(A) It was the responsibility of

the British government(B) There was no official check

upon the Zamindars(C) The farmers were not inte-

rested in getting pattas(D) The Zamindars were trusted

by the farmers

11. The Mongols under ChenghisKhan invaded India during thereign of—(A) Feroz Tughlaq(B) Muhammad Bin Tughlaq(C) Iltutmish(D) Balban

12. The most appropriate measure ofa country’s economic growth isits—(A) Net National Product(B) Gross Domestic Product(C) Net Domestic Product(D) Per Capita Real Income

13. The Supreme Court of Indiatenders advice to the Presidenton a matter of law or fact—(A) Only if the issue poses a

threat to the unity andintegrity of the country

(B) Only if he seeks such advice(C) On its own initiative(D) Only if the matter relates to

the Fundamental Rights ofCitizens

14. Which of the following year wasobserved as the Internationalyear of Coral Reefs—(A) 2007 (B) 2006(C) 2008 (D) None of these

15. Mangroves in India account forabout—(A) 9% of world’s mangrove

vegetation(B) 5% of world’s mangrove

vegetation(C) 8% of world’s mangrove

vegetation(D) 3% of world’s mangrove

vegetation

16. Which of the following was notthe major achievement of LordHastings ?(A) Subsidiary Alliance(B) Suppression of Pindaris(C) Treaty of Sagauli (1816)(D) End of the Maratha Con-

federacy

17. When Sindh was conquered in1843 by the Britishers, theGovernor-General was—(A) Lord Ellenborough(B) Lord Auckland(C) Lord Dalhousie(D) Lord Hastings

18. Which of the following was notthe party to the Tripartite Treatysigned in 1839 ?(A) Maharaja Ranjit Singh(B) Shah Shuja of Afghanistan(C) Mughal Emperor Shah

Alam II(D) The English

19. Ogaden region has been asource of conflict between whichtwo countries ?(A) Ethiopia and Somalia(B) Morocco and Algeria(C) Nigeria and Cameroon(D) Angola and Zambia

20. During the time of which MughalEmperor did the English EastIndia Company establish its firstfactory in India ?(A) Akbar (B) Jahangir(C) Shahjahan (D) Aurangzeb

21. Which one of the following UnionMinistries is implementing theBiodiesel Mission (as NodalMinistry) ?(A) Ministry of Agriculture(B) Ministry of Science and

Technology

C.S.V. / October / 2009 / 1054

(C) Ministry of New and Renew-able Energy

(D) Ministry of Rural Develop-ment

22. Who among the Gandhian fol-lowers was a teacher by profes-sion ?(A) A. N. Sinha(B) Braj Kishore Prasad(C) J. B. Kriplani(D) Rajendra Prasad

23. Which one of the following lasertypes is used in a laser typeprinter ?(A) Dye laser(B) Gas laser(C) Semiconductor laser(D) Excimer laser

24. Among the following which onehas the minimum population onthe basis of data of Census ofIndia, 2001 ?(A) Chandigarh (B) Mizoram(C) Puducherry (D) Sikkim

25. What is the name of the schemewhich provides training andskills to women in traditional andnon-traditional trades ?(A) Kishori Shakti Yojana(B) Rashtriya Mahila Kosh(C) Swayamsiddha(D) Swawlamban

26. Pedology is a science/study of—(A) Vegetation (B) Mountains(C) Soil (D) Continent

27. Which one of the followingfigures represents the age of theearth ?(A) 4·6 million year(B) 4·6 billion year(C) 13·7 billion year(D) 13·7 trillion year

28. Which one of the following hasthe longest duration ?(A) Eons (B) Era(C) Period (D) Epoch

29. The prices at which the govern-ment purchases foodgrains formaintaining the public distribu-tion system and for building upbufferstocks is known as—(A) Ceiling prices(B) Procurement prices(C) Minimum support prices(D) Issue prices

30. The first biosphere reserve ofIndia is—(A) Gulf of Mannar biosphere

reserve(B) Nilgiri biosphere reserve(C) Nanda Devi(D) Sunderbans

31. The Cartagena Protocol on Bio-safety came into force on—(A) September 11, 2003(B) June 5, 2002(C) May 12, 2005(D) February 5, 2005

32. Who among the following was thePresident of the All India State’sPeople Conference in 1939 ?(A) Sardar Vallabbhai Patel(B) Jawahar Lal Nehru(C) Jaya Prakash Narayan(D) Sheikh Abdullah

33. The Government of India hasnotified as Minorities—(A) Five Communities(B) Six Communities(C) Seven Communities(D) Four Communities

34. The earlier name of WTO was—(A) OECD (B) GATT

(C) UNCTAD (D) UNIDO

35. Volcanic eruptions do not occurin the—

(A) Baltic Sea

(B) Caspian Sea

(C) Black Sea(D) Caribbean Sea

36. Athletes Foot is a disease causedby—(A) Nematode (B) Fungus(C) Bacteria (D) Protozoa

37. Who among the following wasthe first Bhakti saint to use Hindifor the propagation of his mes-sage ?(A) Dadu (B) Kabir(C) Ramanand (D) Tulsidas

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