To Our Readers

To Our Readers Dear Readers, It gives us great pleasure and satisfaction to present to you the August issue of your favourite and frontline magazine ‘Competition Science Vision’. It is generally claimed by the toppers and high ranking successful candidates of all pre-medical examinations that each issue of this magazine is highly useful, unique and unbeaten in matter of its contents and way of presentation. Traditionally, we always try to improve the extent and quality of the subject matter and make it more and more examination-oriented keeping in view the changes introduced in the examination pattern. CSV meets fully your requirements in all the four subjects. It has been marked as second to none in its field by its readers. It covers all pre-medical tests held throughout the country at present. Hardwork under proper guidance, constant practice and revision have been widely claimed by successful candidates as the core elements of their success. In matters of guidance CSV stands matchless in the worthy estimation of our wise readers. Read CSV regularly and intelligently. It gives you the power to master your career and shape your destiny. With best wishes for your all-round success. Sincerely yours,

Mahendra Jain (Editor)

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There is always a section of students which wants that the examination dates be postponed for sometime so that they may make a thorough revision of the whole course or complete the chapters which have been left unfinished. Behind this demand there is only one argument that they did not have enough time for proper preparations. On the other side, there is another section of examinees which is always ready to take their examination and wants that the examination may finish at the earliest. It does not look nice that the examinations may be postponed once a date has been fixed for them. In our daily life also we find two sections of people—one section of the people seems to be having no time for anything, while the other section of the people has time for every useful thing. The people of this section never complain of having no time for anything. These people have time to complete all assignments such as, complete their studies, attend games, go to important meetings and also join important functions of their friends and relations etc. There are some people in this section who have to earn their living as well. As opposed to this the people of the section one have only one complaint to make that they had no time. They even make their business a cause of great excuse and make it convenient to forget many things under their cover. If proper account is kept of work, they have done during the twenty four hours, it will be seen that they have hardly devoted two to three hours to serious work. It is no surprise if such persons are not successful in anything in their lives. A student may lag behind in his preparations on account of his limitations or circumstances but in his heart of hearts he also wants to be one of those who are always ready to welcome the examinations. The

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examinee who is always ready to take the examination is apparently full of self confidence and as such is always full of hope and optimism. We believe that you also would like to make the best of your time and complete your course within the definite period of time and keep always ready to take the examination.

then sit for study for the period as fixed in your time table. This would do you two benefits—there would remain a continuity in your study and secondly you will be free from the tension which is the result of irregular study. One who makes regular studies keeps also some hours for extra study which makes him full of enthusiasm. Regular studies will do you one more good that you will get pleasure in doing constructive work and will never feel depressed. The continuous study creates boredom which needs to be removed. For this the candidate must make a change. For example, after reading for sometime if one sits to write something he will see that boredom is released to a great extent. For recreation it seems to be more proper that we develop the habit of reducing to writing what we read. This makes what you read a part of your personality and you become exact in your expression. As has been rightly said, “Reading makes a man perfect and writing makes a man exact.”

To make the best of time means that we do not waste our time in useless things and gossipping. Every examinee should first of all draw out his time table as to for how many hours he has to devote to his studies and also the timings of his studies. He should be clear in his mind that particular hours he must devote to his books and do nothing else during that period. Such students may have to face criticism of many types such as he does not talk to anybody, he tries to show himself, a studious man, he is very proud etc. but he has to ignore such things and follow his time table with full determination. Then and then alone will he be able to make himself ready to take the examination. Develop the habit of leaving your bed in the morning at a particular time and

We want to emphasize that if we keep a constant eye on our time table we shall see that all our things are done in their own way and without resistance and then we will have no complaint of having no time. If we work regularly and with full devotion we will always be ready to take the examination and will never think of its postponment. This is a known fact that without proper tests no one gets perfect knowledge. So you should make up your mind to welcome examinations and as such be always ready for it. Also, do things systematically. Hotch-potch working is the result of hotch-potch of ideas. Systematise your ideas so as to acquire a system in your working and then your success is assured. That would give you confidence to go ahead and open the gates for progress. ●●●

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Climate Change—Immediate and Biggest Threat to Health

walking and cycling, it would tend to lower stress levels, reduce obesity, lessen heart and lung diseases and stroke risks.

Europe, who inhabited the ancient forests of the Carpathian Mountains in what is the present day Romania, about 35,000 years ago.

Top experts of medical sciences of U.K. have published a report warning that climate change is the biggest threat to global health of the 21st century. Rising global temperatures would have a catastrophic effect on human health and patterns of infection would change with insectborne diseases such as malaria and dengue fever spreading more easily.

Indian Space Shuttle will be a Milestone

The reconstruction of the face that could be male or female is based on the skull and jaw-bone found in a cave where bears were known to hibernate.

Insect Invasion : Patterns of infection would change, with insect-borne diseases such as malaria and dengue fever spreading more easily.

The report says that the poorest people in the world will be worst affected. Although the carbon footprint of the poorest billion people is about 3 per cent of the world’s total footprint, loss of life is expected to be 500 times greater in Africa than in the wealthy countries. The impact of heat waves, flooding and food shortages will be felt globally. Climate change is an immediate danger. It is going to affect you and it will inevitably affect your children. The report says the evidence of greenhouse gas emissions, temperature and sea-level rises, the melting of icesheets, ocean acidification and extreme climatic events suggest that the forecasts by the Intergovernmental panel on Climate Change might be too conservative. There is an awful lot we can do. Reducing carbon emissions would encourage people to cut use of vehicles, and if that led to more

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An Indian space shuttle will be test-fired from the spaceport at Sriharikota within a year. Reusable Launch Vehicle-Technology Demonstrator (RLV-TD), as it is called, will be a rocket-aircraft combination : the aircraft with a winged body, which is the RLV, will sit vertically on the rocket. The engineering model of the aircraft is ready at the Vikram Sarabhai Space Centre (VSSC) in Thiruvananthapuram. The first stage of the Satellite Launch Vehicle-3 will form the booster rocket. Weighing nine tonnes, it is called S-9. After it takes off like a rocket, the booster will release the unmanned aircraft, which will go into space. At the end of the mission, the aircraft will land on the sea. Next year, the prototype of the RLV-TD will be ready for flight-testing. This will be a milestone for ISRO. The RLV will open a new dimension in the launch vehicle technology and transportation system of ISRO. Ground testing of the booster rocket is done recently at Sriharikota. The aircraft will stand over the rocket, nosetip up, and its tail will be interfaced with the rocket. In other words, the entire RLV will stand vertically on top of the booster. The booster rocket will take the RLV to a specific altitude, release the RLV and fall into the sea. On re-entry into the Earth’s atmosphere, the RLV will land in the sea, to be recovered.

First European Face Unveiled A scientist of U.K. succeeded to reconstruct the face of the first anatomically modern human to live in

Face of an ancient European

The facial features indicate the close affinity of these early Europeans to their immediate African ancestors, although it was not still possible to determine the person’s sex. Professor Richard Neave, who reconstructed this face, based his assessment on a careful and minute measurement of the bone fragments and his deep understanding of how the soft tissues of the face are built around the bones of the skull. The reconstruction is intended to record, how human origins and evolution from our birth place in Africa to the long migratory routes, led us to populate even the most distant parts of the globe. It is impossible from the bones to determine the skin colour of the individual, although scientists speculate it was probably darker than modern day Europeans, reflecting a more recent African origin. Taken together, the material is the first that securely documents what modern humans looked like when they spread into Europe.

Extended Sleep Improves Performance Wondering how star athletes win world-level championships as recently Roger Federer won the French Open Tennis Championship. The secret may lie in increased sleep. In accordance with a study, by the researchers of Stanford University in California, the athletes, who extended their nightsleep, reported improvements in various drills conducted after every regular practice. The research stated that many of the athletes, who participated in this study, for the first time realized the importance of sleep and how it affects their performance during competitions. While most athletes and coaching staff may believe that sleep is an important contributing factor in sports and peak performance can only occur when an athlete’s sleep and sleep habits are optimal. Five healthy students between the age-group of 18 to 21 participated in the study. A record of their sleep/ wake pattern for a three-week-period was recorded. Athletic’s performance, including sprinting and hitting drills, was recorded after every practice. Athletes extended their sleep to 10 hours a night for six weeks. Mood and daytime sleepiness were recorded. Furthermore, daily sleep/wake activities were monitored using sleep actigraphy. Results suggested that sleep extension in athletes was associated with a faster sprinting drill, increased hitting depth in drill and increased hitting accuracy. According to the research team the findings of this study could be relevant in all walks of life. The results of this study were presented at the 23rd Annual Meeting of Associated Professional Sleep Societies.

As Alaska Glaciers Melt, Sea Level Falls Global warming conjures images of rising seas that threaten coastal areas, but in Juneau, Alaska, as almost nowhere else in the world, climate change is having the opposite effect. As the glaciers melt there, the land is rising, causing the sea to retreat.

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The geology is complex, but it boils down to this : Relieved of billions of tons of glacial weight, the land has risen much as a cushion. The land is ascending so fast that the rising seas—a ubiquitous byproduct of global warming—cannot keep pace. As a result, the relative sea-level is falling, at a rate ‘among the highest ever recorded’, according to a report by a panel of experts.

The Ape Laughs Like Humans What happens if you tickle a gorilla ? The ape laughs like humans, according to a research, which suggests that the origins of laughter can be traced back more than 16 million years. Findings of a research team of the University of Portsmouth (Britain) indicate that humans have inherited ability to laugh from the last common ancestor from which they and great apes evolved. This investigation is the first phylogenetic test of the evolutionary continuity of a human emotional expression. It supports the idea that there is laughter in apes.

Humour in Evolution : Origin of laughter can be traced back to more than 16m years.

Professor Marina Davila Ross led the research team. The experts who carried out the research on gorillas, chimpanzees and orangutans, believe that it started out as a ‘grunt like’ noise with our distant ancestors and gradually turned into the more sophisticated chuckles and guffaws, we know today. The research proves that laughter evolved gradually over the last 10 to 16 million years of primate evolutionary history.

food and nutrition, not only helps tackle obesity but cuts cancer risk also. Professor Soo-Yeun Lee from the University of Illinois has come up with a cinnamon-flavoured soy cereal that can cut the risk of prostate and breast cancer.

Giant Space Tornadoes Power the Northern Lights Different cultures have attributed their spectacular light show to firebreathing dragons, dancing gods and ghostly clans at war. Now research has found that Northern Lights or aurora borealis, are powered by giant, electrical tornadoes spinning at more than a million miles an hour and stretching thousands of miles into space. Scientists used a set of five satellites designed to measure the Earth’s magnetic field to generate the first images of the whirling vertices. They show how vast quantities of charged particles emitted by the Sun first pile up in huge clouds about 40,000 miles above the night side of Earth. Then as the energy they hold becomes too great, the particles explode downwards towards Earth, spinning as they go. ‘‘When these space tornadoes reach the upper atmosphere, their enormous energy heats the air so strongly that it starts glowing. That is what generates the aurorae’’, said Professor Karl Heinz Glassmeier (Germany), the leader of research team. The Northern lights and their Southern equivalent, the aurora australis, create spectacular moving displays of different shades and colours of light. ●●●

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AWARDS/HONOURS 10th IIFA Awards, 2009 The 10th IIFA (International Indian Film Academy) Awards ceremony for 2009 was concluded in Macau, China. Major names from Bollywood transcended on this beautiful island of China. The function took place in the Venetian Macao Resort Hotel, Macau. The event continued for three days. Bollywood director Ashutosh Govarikar’s historical love epic Jodhaa Akbar swept the 10th IIFA awards, taking home trophies for the best picture, best director and best actor. Awardees are— Best Film—Jodhaa Akbar Best Actor in a Leading Role (Male)—Hrithik Roshan—Jodhaa Akbar Best Actor in a Leading Role (Female)—Priyanka Chopra— Fashion Best Actor in Supporting Role (Male)—Arjun Rampal—Rock On Best Actor in a Supporting Role (Female)—Kangana Ranaut— Fashion Best Actor in a Comic Role— Abhishek Bachchan—Dostana Best Actor in a Negative Role—Akshaye Khanna—Race Best Debutant Star (Male)— Farhan Akhtar Best Debutant Star (Female)— Asin Best Director—Ashutosh Gowariker—Jodhaa Akbar Best Story—Neeraj Pandey—A Wednesday Best Music—A. R. Rahman— Jodhaa Akbar Best Lyrics—Javed Akhtar— Jashn-E-Bahara (Jodhaa Akbar) Best Playback Singer (Male)— Javed Ali—Jashn-E-Bahara (Jodhaa Akbar)

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Best Playback Singer (Female) —Shreya Ghoshal—Teri Ore (Singh is King) Outstanding Achievement by an Indian in International Cinema— Aishwarya Rai Bachchan Lifetime Achievement Award— Rajesh Khanna Best Dialogue—Manu Rishi— Oye Lucky ! Lucky Oye ! Best Screenplay—Neeraj Pandey—A Wednesday ! Sound Editing—Ghajini Sound Recording—Ghajini Best Action—Ghajini Special Effects—Ghajini Best Sound Recording—Rock On !! Best Editing—Jodhaa Akbar Best Make-up—Jodhaa Akbar Art Direction—Jodhaa Akbar Best Costume Award—Jodhaa Akbar Best Background Score—A. R. Rahman Star of the Decade (Male)— Shahrukh Khan Star of the Decade (Female)— Aishwarya Rai Bachchan Music Director of the Decade— A. R. Rahman

Man Booker Award, 2009

International

Canadian short story writer, Alice Munro, beat Mahasweta Devi and host of other literary heavyweights, including Nobel Laureate V. S. Naipaul, to win the £ 60,000 Man Booker International Prize. Ms. Munro (78) is regarded as one of Canada’s most celebrated writers. She was amazed and delighted to win the prize. The prize, different from the annual Booker Prize for Fiction, is awarded once every two years to a living author for his or her lifetime achievements.

BOOKS Twilight of the Tigers— G. H. Peiris (The book deals with what would be the next for Tamils in Sri Lanka. The dismissal of devolution of power as a solution has dark foreboding. After three decades holding a gun to Sri Lanka’s head, the Liberation Tigers of Tamil Eelam has finally been defeated militarily and its leader Velupillai Prabhakaran is dead. The decisive battles between the LTTE and the Sri Lankan military were fought since January 2009 till the middle of May 2009. The simple point is that in Sri Lanka, Sinhalese are 73 per cent and Tamils are 12 per cent. Sri Lankan Tamils are liberated from the LTTE, but are afraid of not having a strong voice to speak up for them. The book deals with these facts). World of Work— Pub. by International Institute for Labour Studies, Geneva (The book gives the details that globalisation has resulted in the widening of income–inequality across and within the countries.) India’s Energy Security—Ed. by Ligia Noronha and Anant Sudarshan (The book draws the attention to the need for formulating holistic policies on energy.) Democracy and Human Development in India— Naresh Gupta (The book explains how the democracy and human development are interrelated.)

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

Ustad Akbar Ali Khan—Sarod maestro Ustad Akbar Ali Khan (88) passed away in Francisco (U.S.A.) on June 19, 2009 after a prolonged kidney ailment. He is survived by his

wife Mary, three sons and a daughter. He was born in 1922 in Comilla district, now in Bangladesh. A recipient of Padma Bhushan and Padma Vibhushan, the Ustad was a colossus in the world of Indian classical music for the last five decades. He was admired by both eastern as well as western musicians for his brilliant compositions and his mastery of the 25 string instrument. He also composed music for Indian films of Chetan Anand, Satyajit, Tapan Sinha and others. Habib Tanvir—Noted playwright Habib Tanvir passed away on June 8, 2009. He was one of the greatest stalwarts of the Indian stage, known for blending theatre, folk art and poetry in his works. He left an indelible mark on the minds of the viewers. He was born on September 1, 1923 in Raipur. He took M.A. degree from Aligarh Muslim University and joined All India Radio, Bombay as Director in 1945. His full name was Habib Ahmed Khan and he had adopted the pen name ‘Tanvir’. He also wrote songs for Hindi films. He won Sangeet Natak Akademi award in 1969. He was honoured with Bhushan. He had been a Member of Rajya Sabha (1972–1978). Rajeev Motwani—Rajiv Motwani (47) was talented mathematician whose contribution to the world of computer science influenced the development of algorithm-based search technology. He passed away in a freak drowning accident in his swimming pool on June 7, 2009. In Atherton, California. He was professor at Stanford University. He is survived by his wife Asha Jadeja and the daughters Naitri and Anya. Google’s founders Sergey Brin and Laurence page, were mentored by Dr. Motwani as he conducted research to launch what would become an iconic service. His contribution to frame the ‘world wide web (www)’ is incomparable.

G. E. Vahanvati ( New Attorney Gen. )—Goolam E. Vahanvati (60) till recently the Solicitor-General of India, is appointed as the next AttorneyGeneral of India for a period of three years. He will succeed Milon K.

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Banerjee. Prior to being appointed the Solicitor-General, he was AdvocateGeneral of Maharashtra till June 2004. Mr. Vahanvati is the first Muslim to occupy the top law officer’s post in the country in the last six decades. Pradeep Vasant Naik (New Chief of the Air Staff)—Air Chief Marshal Pradeep Vasant Naik (60) took over as the 19th Chief of the Indian Air Force succeeding Air Chief Marshal Fali Homi Major. With more than four decades of distinguished services, Air Chief Marshal Naik took serious part in Indo-Pak war in 1971. He was honoured with the Vishisht Seva Medal. He was commissioned into the IAF in 1969 as a fighter pilot. So far he has clocked more than 3000 hours of flying. P. K. Barbora ( New Vice-Chief of Air Staff )—Air Marshal P. K. Barbora took over as the new ViceChief of the Air Staff. Nirmal Kumar Verma (New Navy Chief)—Vice-Admiral, Nirmal Kumar Verma, will be the next Chief of the Naval Staff. He will take charge from Admiral Sureesh Mehta who will retire on August 31, 2009. ViceAdmiral Verma, currently Flag Officer Commanding-in-Chief, Eastern Naval Command, commanded aircraft carrier INS Viraat. He was decorated with the Param Vishisht Seva Medal. R. Khullar (New Commerce Sec.)—Union Government, on June 5, 2009, appointed Disinvestment Secretary, Rahul Khullar as Commerce Secretary. He will replace G. K. Pillai a 1975 batch IAS Officer, Mr. Khullar was shifted to the Department of Disinvestment from the Commerce Ministry. Father’s Name Mother’s Name Date of Birth Place of Birth Marital Status Date of Marriage Spouse’s Name No. of Son No. of Daughters Educational Qualifications

Profession Permanent Address

A. K. Kembhavi (New Director, IUCAA )—Ajit Keshav Kembhavi has been appointed as the Director of Inter-University Centre for Astronomy and Astrophysics (IUCAA), Pune. IUCAA is one of the five interuniversity research centres of the University Grants Commission. Prof. Khembhavi will take charge on September 1, 2009 from Professor Naresh Dadhich. Professor Kembhavi took his Ph.D. degree in 1979 under the guidance of Professor Jayant Narlikar. His research interests included quasars and other active galaxies, intermediate red-shift galaxies, galaxy morphology, tidal capture binary stars in globular clusters, astronomical data bases and the virtual observatory (VO). P. Varghese (New Australian Envoy)—Australia has named Peter Varghese as the next High Commissioner to India with concurrent acreditation to Bhutan. He succeeds John McCarthy. Mr. Varghese will take charge in August 2009. Mr. Varghese served as the High Commissioner in Malaysia and had been also a part of Australian missions Vienna, Washington and Tokyo.

Meira Kumar—History was made on June 3, 2009 when diplomatturned-politician Meira Kumar became the first woman Speaker of the Lok Sabha, with the otherwise fractious House setting aside its differences to elect the Dalit leader unanimously. The Meira Kumar

Late Shri Jagjivan Ram Smt. Indrani Devi 31.03.1945 Patna (Bihar) Married 29 Nov., 1968 Shri Manjul Kumar 1 2 M.A., LL.B., Advanced Diploma in Spanish Educated at Indraprastha College and Miranda House, Delhi (Delhi) Social Worker, Advocate, Civil Servant D-1029, New Friends Colony, New Delhi–110 065 (011) 26910618, 26910639, 9810630165 (M) Fax—91-11-26910618

name of the 64-year-old Congress leader from Bihar was proposed by party President Sonia Gandhi and seconded by the Prime Minister Manmohan Singh and BJP leader L. K. Advani together led her to the podium and later paid tribunes to a woman.

Iran’s Political Turmoil

This is Ms. Meira Kumar’s fifth term in Lok Sabha. At present she represents Sasaram (Bihar). Karia Munda—Seven-time Member of Parliament, Karia Munda, was unanimously elected Deputy Speaker of the Lok Sabha. A dozen of nominations were filed in Mr. Munda’s favour. The motion was first moved by the opposition leader L. K. Advani and seconded by BJP President Karia Munda Rajnath Singh. Subsequently, leaders representing other parties, including Congress, moved similar motions. A 32-year-old unbroken tradition of having the Deputy Speaker from the opposition—which began in 1977, the very year Mr. Munda entered the Lok Sabha—has been preserved with the unanimous election of Mr. Munda. This sentiment was echoed by the entire house. Mr. Munda is elected to this 15th Lok Sabha from Khunti in Jharkhand on BJP ticket. Mr. Karia Munda was born on April 20, 1936 in Ranchi district of Jharkhand. He took his M.A. degree from Ranchi University. He is a senior leader of Bhartiya Janta Party and had been a Cabinet Minister in Vajpayee’s Government.

‘IRAN’ on the Boil—Thirty years after momentous events brought Ayatollah Ruhollah Khomeini to the fore as Iran’s man of destiny, his Islamic revolution has skidded into uncharted territory. The official declaration of incumbent President Mohamoud Ahmadinejad as a run away winner in a hotly disputed and possibly rigged presidential election has generated a white heat of popular anger that has no precedent in Iran’s post-revolution history.

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The purported victor’s triumphalism at a huge victory rally was immediately challenged on the streets by close to a million supporters of Mir Hosain Mausavi, the principal opponent. They braved a ban order and choked a nine-kilometre stretch leading to Azadi (freedom) Square, a prominent Tehran landmark. The unrest has now spread to other cities, including Tabriz, Shiraj and others. Only a fair, free and credibly monitored fresh election can heal a nation that is treading a thin line between fear of theocratic authoritarianism and hope of genuine reform that is not in conflict with revolution’s fundamentals. Lalgarh (West Bengal)—Social unrest is known to manifest itself as problems of law and order. But the reverse can also be true sometimes, politically inspired violence seeks the cover of socio-economic grievance. Maoist, outlaws who went on the rampage in the Lalgarh of West Midnapore district of West Bengal, carrying out murderous attacks on the ruling Communist Party of India (Marxist) workers and destroying their houses and party offices, were looking to settle political scores in the guise of protesting ‘police atrocities’. Using the neighbouring State of Jharkhand as the base, they established reign of terror and drove out security

personnel and CPI(M) workers and sympathisers. With tribal folk as human shield, they have now shought to create ‘liberated zones’ in the district. The offensive, time to take advantage of the electoral debacle of CPI(M) in the recently concluded Lok Sabha election, would not have been possible without the support of the main opposition party, the Trinamool Congress. Either directly or indirectly, the Police Santrosh Birodhi Janashadharaner Committee or the People’s Committee against police atrocities, which spread heads the agitation, has drawn substance from the opportunism of the Trinamool Congress and its partner the Congress. Union Government is reluctant to extend full assistance to West Bengal Government. The situation in West Midnapore is too serious to allow such crass politicking. Manmohan Singh Government must not lose any more time in coming to the aid of West Bengal’s Left Front Government to tackling the Maoists and their surrogates. Else there will be a heavier cost to pay.

SPORTS Badminton Indonesian Open—Ace Indian shuttler, Saina Nehwal, on June 21, 2009 in Jakarta, scripted history by

becoming the first Indian to win a Super Series tournament after she clinched the Indonesian Open title with a stunning victory over the higher rankeded Chinese Lin Wang. Saina beat Wang 12-21, 21-18, 21-9 in a thriller that lasted for 49 minutes.

Federer achieved this victory without his having to combat the formidable rivals such as Rafael Nadal, Andy Murrray and Novak Djokovic, all of whom suffered shock defeats.

French Open 2009

Big Moment : Saina Nehwal scripted a remarkable victory over her higherranked opponent China’s Wang Lin in the Indonesian Open.

In the final, World No. 8 Saina came from behind to outsmart World No. 3 Wang and avenge her last week’s Singapore Open Super Series loss to her. Hailing Saina’s victory as an important milestone, Badminton Association of India announced a cash prize of Rs. 2 lakh as a recognition of here feat. The results (finals) : Women : Saina Nehwal (Ind) bt Lin Wang (Chn) 12-21, 21-18, 21-9. Doubles : Eei Hui Chin & Pei Tty Wong (Mas) bt Shu Cheng & Yunlie Zhao (Chn) 21-16, 21-16.

Tennis French Open Tennis Tournament 2009—Roger Federer gloriously equalled Pete Sampras’ record of 14 Grand Slam title, when he won his maiden French Open title in Paris on June 7, 2009.

Top Five Grand Slam Winners 14-Pete Sampras (U.S.) and Roger Federer (Switzerland), 12-Roy Emerson (Australia), 11-Bjorn Borg (Sweden) and Rod Laver (Australia), 10-Bill Tilden (U.S.), 8-Andre Agassi (U.S.), Jimmy Connors (U.S.), Ivan Lendl (Czechoslovakia), Fred Perry (Britain) and Ken Rosewall (Australia).

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

Men’s Singles—Switzerland’s Roger Federer beat Robin Soderling of Sweden 6-1, 7-6, 6-4 in the men’s singles final of the French Open 2009 in Paris (France) on June 7, 2009. This victory took him level with great friend Pete Sampras of USA as the holder of 14 Grand Slam titles. With this title Federer has joined the elite club of five others—Fred Perry (Britain), Don Budge (USA), Rod Laver (Australia), Roy Emerson (Australia) and Andre Agassi (USA)—who have won all four Grand Slams. Women’s Singles—Russia’s Svetlana Kuznetsova claimed the women’s singles title by defeating her compatriot Dinara Safina 6-4, 6-2 on June 6, 2009. By defeating her she claimed her second Grand Slam title. Men’s Doubles—India’s Leander Paes and the Czech Republic’s Lukas Dlouhy won the men’s doubles title outclassing South Africa’s Wesley Moodie and Belgium’s Dick Norman. Women’s Doubles—Anabel Medina Garrigues and Virginia Ruano Pascual of Spain clinched the women’s doubles crown by defeating Victoria Azarenka of Belarus and Elena Vesnina of Russia. Mixed Doubles—In the mixed doubles, Liezel Huber and Bob Bryan of USA beat Vania King of USA and Marcelo Melo of Brazil.

Cricket ICC World Twenty–20 Cricket Championship—Final of this tournament was played between Pakistan and Sri Lanka on June 21, 2009 at Lord’s in London. Pakistan won the final by eight wickets. Man of the match—Shahid Afridi Man of the series—Tillkaratne Dilshan

New Union Government’s Agenda—While addressing the joint

session of Parliament on June 4, 2009, the President Pratibha Patil unveiled the new Union Government’s agenda—

Manmohan’s Agenda ● Internal security and preservation of communal harmony. ● Economic growth in agriculture manufacturing and services. ● Consolidation of the existing flagship programmes. ● Concerted action for the welfare of women, youth, children, other backward classes, SCs, STs, minorities, differently-abled and elderly. ● Governance reform. ● Creation and modernisation of infrastructure and capacity addition. ● Prudent fiscal management. ● Energy security and environment protection. ● Constructive and creative engagement with the world. ● Promotion of a culture of enterprise and innovation. ‘‘The yearning of our young people for inclusiveness—economic, social and cultural—and the rejection of the forces of divisiveness…continues as both its (govt’s) inspiring vision and unfinished business. —President, Pratibha Patil

New Apex Body for Higher Education—The creation of a National Commission for Higher Education and Research that will subsume as many as 13 existing profession councils and regulatory agencies, including the University Grants Commission, Medical Council of India and the AICTE, is a key recommendation of a committee headed by well-known educationist, Professor Yashpal. A draft legislation and constitutional amendment are recommended. The proposal autonomous statutory body will comprise six members and a chairman appointed by the President. State Higher Education Councils, along the lines of those existing in West Bengal, Kerala and Andhra Pradesh, will form the second tier of the system. ●●●

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

‘‘My hard working, teachers’ guidance, support of family members and belief in God are the elements of my success.’’ —Rashmi Singh

Topper—U.P.-CPMT 2009 (1st Position) [‘Competition Science Vision’ arranged an exclusive interview with Miss Rashmi Singh who has the credit of standing first on the list of successful candidates in U.P.-CPMT, 2009. In addition she has also cleared AIPMT (CBSE) and BHU Pre-medical Test with high ranks. For her brilliant success, she deserves all praise and our heartiest congratulations. This important interview is presented here in its original form.] CSV—Congratulations on your brilliant success. Rashmi—Thank you. CSV—Before knowing your result what did you think about those who achieve top positions ? Rashmi—I used to think that God has given some special feature in them. But now I am convinced that nothing is special in them. CSV—Achieving top position has come as surprise to you or were you confident of achieving it ? Rashmi—It was really surprising to me. I had never thought of becoming topper. But after giving examination of U.P.-CPMT and calculating my marks, I was expecting my position in top ten and not of a topper. CSV—What do you think is the secret of your success ? Rashmi—My hard working, teachers’ guidance, my brother family members support as well belief in God. CSV—In how many attempts did you get this success ? Rashmi—This was my 2nd attempt. CSV—What were the shortcomings in your preparation for earlier attempts ? How did you make up for them this time ? Rashmi—In earlier attempts, I did not prepare well any subject, I just gave the exam for knowing pattern etc. This time I had joined coaching and from the starting of session I started preparing for competitive exam. CSV—From where did you get the inspiration of choosing a medical career ? Rashmi—I belong to rural area. I used to see that many poor people were not getting proper treatment. So, for them I thought that I have to be a doctor.

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

Rashmi—To both Botany and Zoology I have given more weightage because they together make 50% of our question paper. These are scorable subjects but Physics and Chemistry gave rank in exam. CSV—Did you make complete study of all topics or of some selective topics ? Rashmi—I studied all the topics.

— CSV is really very helpful in PMT examinations. It helps to understand basic concepts. Its science tips, multiple choice questions and assertion-reason type questions are really wonderful for these tests.

—Rashmi Singh CSV—What planning did you make for preparation ? Please tell something in detail. Rashmi—First, I used to make list of work. I have to do in a day. I used to use my full effort to complete those pieces of work. I planned to read and solve problems of NCERT text book at least for 3 times. Apart from this I used to solve objective after reading text book. CSV—How much time did you devote daily and regularly for Physics, Chemistry, Zoology and Botany ? Rashmi—For me it was not possible to read all the four subjects daily. But I used to read three subjects atleast. I devoted 3 hours for Botany, 3 hours for Zoology, 3 hours for Botany. If time remained then 1 or 2 hours for Physics. CSV—Out of the above four subjects, to which subject did you give more weightage and why ?

CSV—How did you give final touches to your preparation ? Rashmi—Last time I prepared the topics but I used to forget what I read.

Bio-Data Name—Rashmi Singh Father’s Name—Sri Bhishma Pitamah Singh Mother’s Name—Smt. Chanda Singh Educational Qualifications— H.S./Std. X—85·4% (Raj English School, Varanasi), 2006. Inter/Std. XII—87% (Raj English School, Varanasi), 2008. Special achievements— ● 1st rank in U.P.-CPMT, 2009. ● AIPMT (CBSE) Main—AIR-110 (OBC-18) ● BHU Screening—AIR 26th (OBC-5)

CSV—Did you prepare notes ? Rashmi—Yes, In coaching my teachers helped me to prepare notes. CSV—What was your attitude for solving numerical questions ? What weightage did you give them ? Rashmi—2 or 3 hours before commencement of any exam I used to solve numerical questions randomly of any chapter. Due to this step and tricks of numerical striked immediately in mind. I gave much importance to them because they helped me in becoming topper.

CSV—How much time is sufficient for preparing for this examination ? Rashmi—Time is not an important factor. I did planning for a day. Whatever the topics I had to complete them, I used to complete in specified time whether it took 2 hours or 10 hours.

Personal Qualities Hobby—Watching cartoon Ideal Person—Dr. A.P.J. Abdul Kalam Strong Point—Hard work Weak Point—Nervousness

CSV—From what level of education should an aspirant begin preparing for it ? Rashmi—You must know basic concepts of all subject. You should try to feel the subjects. CSV—What was your order of preference for various branches for which this test is held ? Rashmi—MBBS, BDS, BAMS, BHMS. CSV—Please mention various books in each subject and magazines on which you based your preparation. Rashmi—NCERT Text Book and Competition Science Vision magazine. CSV—Did you take coaching in your preparation ? Rashmi—Yes, JRS Tutorial, Varanasi. I was very much impressed with the director of JRS Tutorial Mr. A. K. Jha who taught me Physics in very scientific manner. I was also impressed with Professor of Botany Mr. Diwedi. CSV—What help do the science magazines render in the preparations for this examination ? Rashmi—They helped me understanding basic concept in many topics. CSV—What will be your criterion for selecting a magazine for these examination ? Rashmi—Select those magazine which has contents of your syllabus and latest G.K. CSV—What is your opinion about our Competition Science Vision ? How much helpful and useful do you find it ?

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

Rashmi— It was really very helpful to me. It helped me understanding basic concepts. Science Tips, multiple choice questions as well as assertion-reason were very wonderful for the medical students. CSV—Please suggest in what way CSV can be made more useful for medical aspirants. Rashmi—Latest discovery in various medical areas should be published. For AIIMS something should be specially published. CSV—Please mention your position in the merit list as well as the marks obtained in different subjects. What was your aggregate percentage of marks ? Rashmi—Ist Position. Physics—48/50 Chemistry—48/50 Zoology—46/50 Botany—50/50 Total—192/200, i.e., 96%. CSV—What books/magazines/ newspapers did you read for G.K. preparations ? Rashmi—Competition Science Vision. CSV—Whom would you like to give credit for your success ? Rashmi—I would like to give credit of my success to my family, relatives as well as my teachers. CSV—Please tell us something about your family. Rashmi—I have mummy, papa, elder brother and sister. My father name is Mr. Bhishma Pitamah Singh who is engineer in DRDA (Sonbhadra) and my brother is doing MBBS, Allahabad and sister also doing MBBS from Jhansi, my mother is housewife. CSV—What in your frank opinion has been the biggest mistake in your preparation for this test ? Rashmi—After giving any competitive exam. I used to rest for atleast 3 days and I did not solve Physics problem much. CSV—What message would you like to give for our readers of CSV ? Rashmi—Please read CSV very seriously for easy selection in PMT as it is very useful for medical students because all the four subjects are given together in this magazine. ●●●

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

‘‘Continuous hard work, Long practice and faith & belief in God are the elements of my success.’’ —Anjali Singh

Topper—Uttarakhand, PMT–2009 (1st Position) [‘Competition Science Vision’ arranged an exclusive interview with Miss Anjali Singh who has the credit of standing first on the list of successful candidates in Uttarakhand PMT, 2009 in the very first attempt. In addition, she has also cleared AFMC in the written examination. For this brilliant success she deserves all praise and our heartiest congratulations. This important interview is presented here in its original form.] CSV—Congratulations on your brilliant success. Anjali—Thank you, sir. CSV—Before knowing your result what did you think about those who achieve top positions ? Anjali—Same as now others think of me. As brilliant and very hardworking personalities. CSV—Achieving top position has come as surprise to you or were you confident of achieving it ? Anjali—As a surprise. But I was confident of achieving a good rank in this PMT. CSV—What do you think is the secret of your success ? Anjali—Continuous hardwork, long practice and faith & belief in God. CSV—In how many attempts did you get this success ? Anjali—In Uttarakhand PMT, I had appeared 1st time but after 10 + 2 (2005), I had prepared for medical entrance exam for four years. CSV—What were the shortcomings in your preparation for earlier attempts ? How did you make up for them this time ? Anjali—Earlier, I had not prepared some topics (Like Animal taxonomy, families of angiosperms etc.). Also, I was lazy at the last time after a lot of hardwork in the beginning. But this year from the very starting. I prepared these topics well and continued my studies until the last exam. CSV—From where did you get the inspiration of choosing a medical career ? Anjali—It was a dream for me since my childhood. But when my cousins and friends were selected, it became an aim for me, which I had to achieve anyway. In childhood I liked the clinic of paediatricians very much.

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

—It is a very good magazine and has influenced me much. It contains good MCQs as well as sound and brief material for revision. Interviews of toppers and high ranking candidates are very inspiring. —Anjali Singh CSV—From when did you start the preparation for it ? Anjali—I started my preparation of medical entrance exam after intermediate. In XI and XII Std., I was not so sincere and passed intermediate with both Maths and Bio. as an average student. CSV—What planning did you make for preparation ? Please tell something in detail. Anjali—First I prepared my weak topics in all subjects and had a command on them. I did a lot of practice by solving MCQs of all subjects. I gave a lot of attention to Botany and Physics. At the time of exams, instead of solving MCQs, I concentrated on the revision of topics, thoroughly. CSV—How much time did you devote daily and regularly for Physics, Chemistry, Zoology and Botany ? Anjali—My study hours were not fixed. I studied all the time but when I felt exhausted. I liked to take rest or a

sleep. Atleast 2 hours are required for each subject and extra time is needed for solving MCQs. CSV—Out of the above four subjects, to which subject did you give more weightage and why ? Anjali—I payed more attention to Physics and Botany. I gave maximum time to Biology as a whole as I think I am a little weak in Biology (Botany especially). I had a good command on Physics, but maintaining the same, it requires a lot of practice and a lot of time.

Bio-Data Name—Anjali Singh Father’s Name—Mr. Desh Raj Singh Mother’s Name—Mrs. Vinay Singh Educational Qualifications— H.S./Std. X—75% (St. Francis Sec. School, Agra), 2003 Inter/Std. XII—77% (Holy Public School, Agra), 2005. Special achievements— ● 1st rank in state PMT ● I had been selected in AFMC (written) also called for an interview.

CSV—Did you make complete study of all topics or of some selective topics ? Anjali—I studied all the topics completely in the starting and solved all MCQs but at the time of exams, I left few topics of low weightage (Like, Aging, Virus etc.) CSV—How did you give final touches to your preparation ? Anjali—By solving more and more multiple choice questions and revising my topics frequently. CSV—Did you prepare notes ? Anjali—I had taken coaching. So I already had notes. I wrote all the formulae and important points in

Personal Qualities

Anjali—I had not given much importance to this aspect. But I admit that for quick revision, magazines are helpful.

Hobbies—Listening music (filmy songs), cooking Ideal Person—My mother and all the selected students Strong Point—My continuous hardwork when I feel it is a requirement. Weak Point—My sleep and silly mistakes while solving MCQs.

CSV—What was your attitude for solving numerical questions ? What weightage did you give them ? Anjali—I had Maths in XII, so I can do well in numerical questions. I gave them as much time as required by using formulae and practising MCQs. CSV—How much time is sufficient for preparing for this examination ? Anjali—Two years study during XI and XII is sufficient if we study properly and seriously. CSV—From what level of education should an aspirant begin preparing for it ? Anjali—According to me, a student preparing for medical entrance exam should begin the preparation from XI Std. CSV—What was your order of preference for various branches for which this test is held ? Anjali—Only MBBS. CSV—Please mention various books in each subject and magazines on which you based your preparation. Anjali—Elementary Biology, Pradeep Physics and Comprehensive Chemistry for subjective study. For preparing MCQs, I used objective Dinesh for Biology, CSV for all subjects and Pradeep for Chemistry. CSV—Did you take coaching in your preparation ? Anjali—Yes, I had taken coaching for first two years (after XII) and then for one year at Agra. CSV—What help do the science magazines render in the preparations for this examination ?

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

CSV—What will be your criterion for selecting a magazine for these examination ?

CSV—What is your opinion about our Competition Science Vision ? How much helpful and useful do you find it ? Anjali—It contains good MCQs as well as a sound and brief material for revision. It contain interviews of top rankers, which is very much inspiring for the students preparing for the same. It influenced me very much. CSV—Please suggest in what way CSV can be made more useful for medical aspirants. Anjali—By adding NCERT based topics. Many new points had been added in NCERT. So, I think, CSV should contain a separate NCERT corner containing extra points for revision. CSV—Please mention your position in the merit list as well as the marks obtained in different subjects. What was your aggregate percentage of marks ? Anjali—I got 1st rank in Uttarakhand PMT (general) with 44 marks in Physics, 43 in Chemistry, 41 each in Zoology and Botany. All are out of 50 on aggregate 169 marks out of 200, i.e., 84·5%. CSV—What books/magazines/ newspapers did you read for G. K. preparations ? Anjali—I had read only daily newspaper (Amar Ujala) as only 20 questions in G. K. are asked in AIIMS. So, I payed only a negligible importance to G. K. CSV—Whom would you like to give credit for your success ? Anjali—I would like to give my regards to God and my parents, who had been always with me during my preparation. I am also very much thankful to all my respected teachers. CSV—Please tell us something about your family.

Anjali—My father is a Chemistry lecturer. My mother is housewife. I have one sister and two brothers, all are younger to me. My sister (Shilpi Singh) had also qualified in CBSE this year. CSV—What in your frank opinion has been the biggest mistake in your preparation for this test ? Anjali—My unlimited sleep and laziness. When I lost my interest in studies in between and found it difficult to regain it. When I got good ranks in tests (in watching), I became quite loose. CSV—What message would you like to give for our readers of CSV ? Anjali—First make your concepts clear. Read the theory deeply, solve MCQs, complete a book wholly, instead of solving few topics from different books. Nothing is impossible if we have patience and strong will power. Impossible word itself says ‘‘I am possible’’. Best of luck to all. ●●●

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Anjali—I used science magazines CSV for solving MCQs and last years exam questions.

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Physics and Physical Chemistry. But did not prepared any notes in other subjects, because biology, organic and inorganic chemistry don’t have any certainty from where questions may be asked.

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Physics 1. Smaller is the potential gradient along a potentiometer wire, the more is the ➠ Sensitivity of potentiometer 2. The standard equation of trajectory of a projectile in terms of range (R) is x ➠ y = x 1 – R tan α

( )

3. Can Kirchhoff's laws be applied to both the direct and alternating currents ? ➠ Yes 4. What are the limitations of the Newton’s laws ?

→ → ➠ (a) The relation F m a would not hold good in case m does not remain constant.

12. The magnifying power of a telescope for image at infinity is f ➠ M = – fo e 13. The three characteristics of a heating wire are ➠ High resistivity, high melting point, heat resistant 14. Intensity of electric field due to a point charge ‘q ’ at a distance r → 1 q ^ ➠ E = . 2r πε 0 r 4π 15. Which of the two has more resistance : 100W tungsten bulb or 1000W heater, both marked for 220 V. ➠ 100W tungsten bulb 16. What is the one basic difference between BiotSavart’s law and Coulomb’s law ? ➠ The charge element appearing in Coulomb’s law is a SCALAR but the current element (idl ) appearing in Biot-Savart’s law is a VECTOR

(b) The first two laws do not hold good in each and every frame of reference. It is only in a very special frame of reference that the two laws of motion hold good. Such a frame of reference is called an inertial frame of reference.

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

5. What limits the energy that can be provided by a cell ? ➠ Amount of reactants

18. The force between two magnetic poles of pole strengths m and m′ separated by distance r is given by

→

(chemicals used in the cell) 6. Excess pressure inside a liquid drop; inside a liquid bubble, inside an air bubble in a liquid are 2T 4T 2T ➠ R ; R ; R 7. Why is it necessary to keep the concentration of CuSO 4 solution constant in Daniell cell ?

➠F=

μ0 π 4π

( ) mm ′ r2

19. Planets are natural satellites of the

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

➠ This helps to get a steady e.m.f. 8. What is the equation of a plane progressive simple harmonic wave ?

➠ y = a sin

π 2π λ

π (vt – x ) = a sin 2π

( ) t x – T λ

With usual meaning of notations 9. How are the coils wound in a resistance box ? ➠ The resistance coil is doubly wound on a bobbin to avoid electromagnetic induction 10. The equation of a stationary wave when the wave is reflected from a rigid boundary is

➠ y = 2a cos

πx 2π λ

cos

πt 2π T

11. Board of trade unit is defined as the amount of work done when a power of one kilowatt is consumed for ➠ One hour

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

Chemistry 21. The true use of chemistry is not to make gold, but to prepare medicines. Who said this ? ➠ Paracelsus (1493–1541) 22. The first super-conductor of family 1-2-3, discovered in 1987 is ➠ Y Ba2Cu3O7 23. First modern text book of chemistry was written by : ➠ Berzelius 24. The reactions in which a single reagent undergoes both oxidation and reduction are known as ➠ Disproportionation reactions 25. Meson theory of nuclear stability was given by : ➠ Yukawa

26. Isothermal gel-sol transformation brought about by shaking is termed as ➠ Thixothropy

43. An union or joining of 2 or more arteries, veins or other vessels ➠ Anastamosis

27. The quantitative aspect of dealing with the mass and volume relation among the reactants and products is known as : ➠ Stoichiometry

44. What is called a unit of recombination ?

28. Fuel value of hydrogen is

➠ 142 kJ/gm 29. A mole is Chemists unit for counting atoms, molecules, ions and other microscopic species. A mole means :

➠ Recon 45. Phenomenon in which synthesis of fatty acids and keto acids takes place from amino acids ➠ Ketogenesis 46. What type of RNA carries a sequence of codons to ribosomes ? ➠ m RNA

➠ A collection of 6·022 × 1023 particles 47. Structures in which a bone fits into a socket of the 30. An instrument used to detect and measure radiation by the fluorescence is known as ➠ Scintillation counter 31. The ratio of charge (e) of the electron to its mass was found to be 1·76 × 108 coulomb/gm (5·28× 1017 esu/gm). It was measured for first time by : ➠ Sir J. J. Thomson (1887) 32. Two liquids which mix in all proportions are called ➠ Miscible 33. The amount of radiant energy in a photon is proportional to the frequency of the radiation. This was the observation of : ➠ Max Planck (1900) 34. Bonding in which the bonding electrons are relatively free to move throughout the three-dimensional structure, is called ➠ Metallic-bonding 35. The theory of quantum mechanics was independently and simultaneously formulated by ➠ Werner Heisenberg and Erwin Schrodinger (1925-26) 36. A substance capable of behaving as either an acid or a base is called ➠ Amphoteric 37. Cyclic aliphatic hydrocarbons are known as : ➠ Alicyclic hydrocarbons 38. A substance which acts as a proton donor is called ➠ Bronsted-Lowry acid

other.

➠ Gomphosis 48. Which serous membrane lines the body cavity and covers the organs therein in many animals ? ➠ Peritoneum 49. A hormone that stimulates growth of seminiferous tubules and spermatogenesis in men ➠ Follicle stimulating Hormone 50. What is called the duct which derives from mesoderm and conveys gametes or excretory products from the codon to the exterior ? ➠ Coelomoduct 51. Honey bees repair, filling and cementing cracks and crevices called ➠ Propolis 52. What is called an animal that warms his body mainly by the heat of cellular respiration ? ➠ An Ednotherm 53. Abiotically polymerized amino acids that are joined in a preferred manner ➠ Proteinoid 54. What is called the product made from swim bladder of fishes containing 90% gelatin ? ➠ Isinglass 55. A group of structural and regulating genes that functions as a single unit ➠ Operon

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

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

40. A cell used to obtain sodium metal by electrolysis of molten NaCl, is known as ➠ Downs cell

57. A blood substance that produces vaso contraction ➠ Angiotensin

Zoology 41. DNA that contains genes from more than one source ➠ Recombinant DNA 42. Who first discovered the lethal genes ?

➠ Cuenot C.S.V. / August / 2009 / 680

58. Which form of coelom is formed by the splitting of embryonic mesoderm ? ➠ Schizocoel 59. A pelagia free swimming, colonial luminiscent thaliacean is called ➠ Pyrosoma 60. Which portion of internal ear is coiled and contains organ of hearing ? ➠ Cochlea

Botany 61. What are bacteriophages attacking E. coli called ? ➠ Coliphages 62. What often glucose metabolism provides ? ➠ Energy in the form of ATP 63. How many capsomeres are present in a virion of φ ×174 ? ➠ Twelve 64. What causes separation of the complementary chains of a DNA molecule ? ➠ Denaturation 65. What is the time between infection of host and assembly of new phages called ? ➠ Eclipse period 66. What is the chief means of achieving the genetic variation ? ➠ Mutation 67. Who gave the concept of pure line ? ➠ W. L. Johanssen 68. Which factor causes most mutation in nature ? ➠ Radiation 69. Which acid was first produced by fermentation ? ➠ Lactic acid 70. Who gave the concept of pure line ? ➠ W. L. Johanssen 71. What is Triticale ? ➠ A man-made cereal 72. What term is applied for specific sequence of development of a community relating to particular set of physical and chemical conditions ? ➠ Sere 73. What happens when a cell is placed in strong salt solution ? ➠ Water comes out by exosmosis, as a result the cell is shrinked 74. CO2, CFC, NO2 and CH4 causes global warming of atmosphere which is called ➠ Green house effect 75. What type of inflorescence is found in Ocimum sanctum ? ➠ Verticillaster 76. What are climbers growing on large trees and becoming woody due to secondary growth called ? ➠ Lianas 77. What refers an allelomorph ? ➠ Pairs of determiners 78. Where the main reservoir of nitrogen is present in atmosphere ? ➠ Atmosphere 79. What is zeatin ? ➠ A cytokinin-like substance isolated from milky endosperm of corn 80. What is that physiological process called where ions move by mass flow and diffusion through the apoplast ? ➠ Passive uptake ●●●

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

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UPKAR PRAKASHAN, AGRA-2 ● E-mail

: [email protected] ● Website : www.upkar.in

de-Broglie Hypothesis The wave theory of light was capable of explaining the phenomena like reflection, refraction, interference, diffraction and polarisation but it failed to explain photoelectric effect and Compton effect. Quantum theory could explain these. Quantum theory says that a beam of light of frequency ν consists of small packets each of energy hν, called quanta or photons. These photons behave like particles, thus, light sometimes behaves like a wave and sometimes like a corpuscle. In 1924 de-Broglie proposed that a material particle such as an electron, proton, atom etc. might have a dual nature as light does. According to de-Broglie hypothesis a moving particle, whatever its nature, has wave properties associated with it. de-Broglie proposed that the wavelength λ associated with any moving material particle of momentum p is h λ= , p where h is Planck’s constant. The waves associated with a material particle are called matter-waves or de-Broglie waves and the waveh length given by λ = is called de-Broglie wavelength. p If m is the mass and v the velocity of the material particle, then p = mv h or λ = mv If E is the kinetic energy of the material particle, then, 1 p2 E = mv 2 = 2 2m

p =

or

2m E

Therefore, the de-Broglie wavelength is given by h λ = 2m E Diffraction effects have been obtained with streams of protons, neutrons and alpha particles, but it is evident from de-Broglie’s equation that the greater the mass of the moving particle, the smaller is the associated wavelength and so the more difficult detection becomes. Ordinary objects have extremely small wavelengths.

If a charged particle carrying charge q is accelerated through a potential difference of V volts, then Kinetic energy E = q V In that case, h λ = 2mq V

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

When the material particles like neutrons are in thermal equilibrium at absolute temperature T, then they possess Maxwellian distribution of velocities and so their average kinetic energy is given by 1 E = mv 2rms 2 3 = kT 2 where k is Boltzmann’s constant whose value is 1·38 × 10–23 joule/K. So that h λ = 3mk T Davisson and Germer Experiment—This experiment performed by Davisson and Germer is the first experimental proof of the wave nature of material particles. The de-Broglie wavelength for an electron is less ° A crystal lattice in which the atomic distances than 1 A. ° is ideal for the between layers are of the order of 1 A purpose of studying the diffraction of electron waves. A nickel crystal C is taken. Electrons are made incident on it as a narrow beam. The incident electrons are produced by an electron gun G. The gun consists of a tungsten filament F which is connected to a low tension battery. Electrons emitted from the filament F are made to pass through pin holes under suitable accelerating potentials. The accelerating potential is provided with the help of a high-tension battery. Ordinarily, we get an electron beam of energy 50 eV from the gun. The resolving power of any microscope increases as the wavelength used to illuminate the object decreases. Thus, the electron waves have a smaller wavelength than light waves and so an electron microscope reveals much more detail. The field ion microscope gives even greater resolution because the waves associated with the helium ions used have an even shorter wavelength.

The electron beam is made to fall normally on the crystal. The beam is diffracted by the crystal and received at an angle θ by a detector D. The intensity of the diffracted electrons is measured by the detector as a function of angle θ. In the original experiment, the detector was set at an angle of 50° to the direction of the incident beam. The scattered electron current for different values of voltage V was noted. The graph between the voltage and the detector current is as shown in figure on next page. The existence of peak in graph can be explained as due to constructive interference of waves scattered from atoms in different planes of the crystal. The peak occurs at 54 V. The observed phenomenon is similar to the diffraction of X-rays. So the experiment establishes the wave nature of electrons.

The detector current is maximum when

Electron diffraction rings

2d sin θ = n λ, where n is the order of diffraction, d is the atomic spacing between successive crystal planes and θ is the angle at which strong reflection takes place. The value of d as determined by X-ray reflection by nickel crystal, comes ° The value of θ comes out to be 65°. out to be 0·91 A. ° For n = 1, λ = 2d sin θ = 2 × 0·91 × sin 65° = 1·65 A. We know that h λ = 2me V In the given experiment V = 54 volt λ =

6·63 × 10–34 (2 × 9 ×

10 –31

× 1·6 × 10–19 × 54)1/2

° = 1·67 A. So, the wavelength determined by two different methods comes out to be the same. This confirms that electrons are diffracted in the same way as the de-Broglie waves. F

D θ

C Fig. : Davisson and Germer experiment

Detector current

G

Fig. : Variation of detector current with voltage

Thomson’s Experiment In this experiment a beam of electrons obtained from an electron gun is made to fall normally on a thin platinum foil. The foil is nearly 10 –8 m thick. The foil can also be of aluminium or gold. A photographic plate is placed behind the foil. Diffraction pattern in the form of concentric rings is obtained on the photographic plate. These rings are clearly due to the diffraction of waves associated with electrons. This is because of the randomly oriented crystals in the foil. From the geometry of the apparatus the voltage used to accelerate the electrons and the diameter of the electron diffraction rings. The wavelength associated with the beam of electrons can be determined. It is observed that the experimental value of wavelength is in close conformity with the theoretical wavelength. Thus, the de-Broglie hypothesis is verified experimentally.

Thin platinum foil

Photographic plate

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

Study of Crystal Structure—The electrons interact with the surface atoms much more intimately. Therefore, electron diffraction is superior to X-ray diffraction to study the surface structure of crystals. The great advantage of neutron diffraction over X-ray diffraction lies in the fact that the neutron diffraction gives better information, for neutrons being neutral particles can penetrate deeper into the nucleus. Moreover, neutron diffraction can give information regarding lattice vibrations also and neutron beam can be produced with very much smaller energy than is needed for the production of X-rays. Q. What is the de-Broglie wavelength of a neutron whose energy is 2 electron-volt ? Mass of neutron = 1·676 × 10–27 kg Planck’s constant = 6·62 × 10–34 joule-sec Solution : h λ = 2m E 6·62 × 10–34 = 2 × 1·676 × 10–27 × 2 × 1·6 × 10–19 = 0·202 × 10–10 m ° = 0·2 A

54 V Voltage V

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

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Electron gun

Electron Lens It is a device used to focus an electron beam. It is analogous to an optical lens but instead of using a refracting material, such as glass, it uses a coil or coils to produce a magnetic field or an arrangement of electrodes between which an electric field is created. Electron lenses are used in electron microscopes. Source of electrons

Condenser lens Object Objective lens

Intermediate image Projector lens

Final image Fig. : Principle of Transmission electron microscope

de-Broglie wavelength associated with charged particles (1) For electrons ( me = 9·1 × 10–31 kg) h λ = 2mqv 6·62 × 10–34 = m 2 × 9·1 × 10–31 × 1·6 × 10–19 V 12·27 ° = A V Note—The potential difference required to have an ° electron of wavelength λ A is 150·6 V = (From above equation) λ2 (2) For protons (mp = 1·67 × 10–27 kg) 0·286 ° λ = A V (3) For deuterons (md = 2 × 1·67 × 10–27 kg) 0·202 ° λ = A V (4) For α-particles (mα = 4 × 1·67 × 10–27 kg) 0·101 ° λ = A V

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

than light; for example, electrons accelerated to an energy of 105 electron volt have a wavelength of 0·04 nanometre (de-Broglie wavelength) enabling a resolution of 0·2–0·5 nm to be achieved. The Transmission electron microscope has an electron beam, sharply focussed by electron lenses, passing through a very thin metallised specimen (less than 50 nanometre thick) onto a fluorescent screen, where a visual image is formed. This image can be photographed.

de-Broglie wavelength associated with uncharged particles (1) For neutrons (m n = 1·67 × 10–27 kg) h 6·62 × 10–34 λ = = 2m E 2 × 1·67 × 10 –27 E 0·286 ° = A E (eV) (2) For thermal neutrons at ordinary temperatures E = kT h ∴ λ = 2mk T 30·835 ° = A T (3) For gas molecules h λ = m×C rms ⇒ For gas molecules at T K 3 E = 2kT h ∴ λ = 3mk T

SOME TYPICAL SOLVED EXAMPLES Example 1. Calculate the de-Broglie wavelength associated with a proton moving with a velocity equal 1 to th of the velocity of light. 20 Solution :

Example 2. Show that the electrons accelerated through a potential difference of V volts will have a 12·27 ° A associated with them. wave of wavelength ⎯V √

Example 3. What voltage must be applied to an electron microscope to produce electrons of wave° length 0·5 A ? Solution :

Example 4. Calculate the de-Broglie wavelength of an α -particle accelerated through a potential differences of 4000 volt. Given Planck’s constant

h = 6·62 × 10–34 joule-sec.

Solution :

Mass of proton mp = 1·67 × 10 –27 kg Electronic charge e = 1·6 × 10 –19 coulomb Solution :

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

Example 7. What is the energy of gamma photon ° ? having a wavelength of 1 A (Given : Planck’s constant h = 6·62 × 10 –34 joulesec and speed of light c = 3 × 108 m/sec.) Solution :

Example 5. Find the energy of the neutron in units of electron volt whose de-Broglie wavelength is ° 1 A. (Given : Mass of the neutron = 1·674 × 10–27 kg Planck’s constant h = 6·60 × 10–34 joule-sec) Solution :

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

Example 6. Energy of a particle at absolute temperature T is of the order of k T. Calculate the wavelength of thermal neutrons at 27°C. (Given : Mass of the neutron = 1·67 × 10–27 kg. Planck’s constant h = 6·60 × 10–34 joule-sec, and Boltzmann’s constant k = 8·6 × 10 –5 eV deg–1.) Solution :

Example 9. Calculate the de-Broglie wavelength of an electron which has kinetic energy equal to 15 eV. Solution :

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

OBJECTIVE QUESTIONS 1. Matter waves— (A) Are electromagnetic waves (B) Are transverse waves (C) Are longitudinal waves (D) Exhibit diffraction 2. Neglecting variation of mass with velocity, the wavelength associated with an electron having the kinetic energy E is proportional to— (A) E1/2

(B) E–1/2

(C) E

(D) E–2

3. A proton when accelerated through a potential difference of V volts has a wavelength λ associated with it. An α -particle in order to have the same λ must be accelerated through a potential difference of— (A) V volt (C) 2V volt

(B) 4V volt V (D) volt 8

4. An electron accelerated through a potential difference of V volts has a wavelength λ associated with it. Mass of proton is nearly 2000 times that of an electron. In order to have the same λ for proton, it must be accelerated through a potential difference of— (A) V volt V (C) volt 2000

(B) 2000 V volt (D)

2000 V volt

5. In an electron microscope if the potential is increased from 20 kV to 80 kV, the resolving power R of the microscope will become— (A) R (B) 2R (C) 4R (D) R/2 6. A proton and an α-particle are accelerated through the same potential difference. The ratio of their de-Broglie wavelengths λp/λα is— (A) 1 (B) 2 1 (C) 8 (D) 8 7. One can study crystal structure by electron diffraction as well as by neutron diffraction. In order to have the same wavelength λ for the electron (mass me) and

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

neutron (mass m n ), their velocities should be in the ratio— (electron velocity/neutron velocity) me (A) One (B) mn mn (C) me × mn (D) me 8. Which of the following figures represents the variation of particle momentum and associated de-Broglie wavelength ? P

(A)

P

(B) P

P

(C)

(D)

9. A beam of monoenergetic neutrons corresponding to 27°C is allowed to fall on a crystal. A first order reflection is observed at a glancing angle 30°, calculate the interplanar spacing of the crystal—(Given : Planck’s constant h = 6·62 × 10–34 joule-sec, Mass of neutron m n = 1·67 × 10 –27 kg, Boltzmann’s constant, k = 1·38 × 10–23 joule/degree) ° (A) 1·78 A

° (B) 2·18 A

° (C) 8·12 A

° (D) 3·12 A

10. Calculate the de-Broglie wavelength of neutron of kinetic energy 54 eV— (Given : Mass of neutron = 1·67 × 10–27 kg) ° (A) 0·4 A

° (B) 4·0 A

° (C) 0·04 A

° (D) 0·004 A

11. Calculate the de-Broglie wavelength associated with a proton moving with velocity 3 × 107 m/s— (A) 1·32 × 10–14 m (B) 1·14 × 10–14 m (C) 1·14 × 10–32 m (D) 4·14 × 10–10 m 12. Calculate the de-Broglie wavelength associated with an electron of 50 eV energy— ° (A) 17·3 A

° (B) 1·73 A

° (C) 7·13 A

° (D) 3·17 A

13. What is the momentum of an ° electron if its wavelength is 2 A ? (Given : h = 6·63 × 1 0–34 joulesec.) (A) 3·22 × 10–20 kg ms–1 (B) 23·2 × 10–21 kg ms–1 (C) 2·32 × 10–23 kg ms–1 (D) 3·32 × 10–24 kg ms–1 14. Calculate the de-Broglie wavelength of an α-particle accelerated through a potential difference of 2000 volt— ° (A) 3·2 × 10–1 A ° (B) 5·0 × 10–2 A ° (C) 2·3 × 10–3 A ° (D) 1·1 A 15. The de-Broglie wavelength of an ° The energy electron is 1·224 A. of electron is— (A) 1 eV (B) 10 eV (C) 100 eV (D) 1224 eV 16. The de-Broglie wavelength of 150 eV electron will be— ° ° (A) 1 A (B) 12·27 A ° (C) 0·5 A

° (D) 1·5 A

17. Obtain the de-Broglie wavelength of a neutron of kinetic energy 150 eV— (Given : Mass of neutron = 1·675 × 10–27 kg) (A) 3·351 × 10–12 m (B) 2·331 × 10–10 m (C) 2·335 × 10–12 m (D) 5·332 × 10–12 m 18. Calculate the de-Broglie wavelength for electrons if their speed is 105 ms–1— (Given : h = 6·63 × 10–34 joulesec) (A) 7·3 × 10–9 m (B) 3·7 × 10–9 m (C) 5·3 × 10–9 m (D) 5·5 × 10–9 m 19. Calculate the de-Broglie wavelength for protons if their speed is 105 ms–1— (Given : h = 6·63 × 10–34 joulesec) (A) 39·71 × 10–12 m (B) 3·97 × 10–12 m (C) 9·37 × 10–12 m (D) 7·39 × 10–12 m

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

ANSWERS WITH HINTS

(Continued on Page 747 ) C.S.V. / August / 2009 / 687

Isothermal Process It is a thermodynamic process in which the pressure and volume of system change but temperature remains constant. An isothermal process is carried out either by supplying heat to the substance or by extracting heat from it. A process has to be extremely slow to be isothermal.

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

P B

Examples of Isothermal Process (1) The temperature of a substance remains constant during melting. So, the melting process is an isothermal process. (2) When a substance boils, its temperature remains constant. So, boiling is an isothermal process. (3) Consider an ideal gas enclosed in a conducting cylinder fitted with a conducting piston. Let the gas be allowed to expand very slowly. This will cause a very slow cooling of the gas. But heat will be conducted into the cylinder from the surroundings. Thus, the temperature of the gas remains constant. If the gas is compressed very slowly, heat will be produced. But this heat will be conducted to the surroundings. So, the temperature of the gas remains constant.

Adiabatic Process It is the thermodynamic process in which pressure, volume and temperature of the system change but there is no exchange of heat between the system and the surroundings. A process has to be sudden and quick to be adiabatic.

Examples of Adiabatic Process (1) Consider a gas enclosed in a thermally insulated cylinder fitted with a non-conducting piston. If the gas is compressed suddenly by moving the piston downwards, some heat is produced. This heat cannot escape the cylinder. Consequently, there will be an increase in the temperature of the gas. (2) If the above gas is suddenly expanded by moving the piston outwards, there will be a decrease in the temperature of the gas. (3) Bursting of a cycle tube. (4) Propagation of sound waves in a gas. (5) Expansion of gases in internal combustion engine. (6) Expansion of steam in the cylinder of a steam engine. Thermodynamical system is an assembly of an extremely large number of particles (atoms or molecules) so that the assembly has a certain value of pressure, volume and temperature.

Indicator diagrams : Pressure-volume graphs are called indicator diagrams. They can be used to show the cycle of changes taking place in an engine. The diagram below shows in simplified form, what happens in a

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

net work done

D A

V Thermodynamic variables or parameters are the quantities like pressure, volume and temperature which help us to study the behaviour of a thermodynamic system.

A to B—Gas (air-petrol mixture) is compressed adiabatically by the raising piston. This causes a rise in temperature. B to C—Ignited by a spark, the mixture explodes. The further rise in temperature causes a further rise in pressure. C to D—The hot high pressure gas pushes down the piston as it expands adiabatically and the temperature falls. Thermodynamic process is said to take place if some change occurs in the state of a thermodynamic system i.e. the thermodynamic variable of the system change with time.

D to E—The warm waste gas is removed and replaced by cooler, fresh gas mixture, ready for the next cycle. Note—From A to B, work is done on the gas. From C to D work is done by the gas. The shaded area represents the net work done during the cycle. ●

● ●

Cyclic process is that thermodynamic process in which the system returns to its initial stage after undergoing a series of changes. Non-cyclic process is that process in which the system does not return to its initial stage. Isolated system is that system which is completely isolated from its surroundings.

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

dQ = dU+dW Equation of isothermal process : The perfect gas equation is PV = RT,

where R is gas constant in an isothermal process, T is constant. Therefore, ∴ PV = constant ( R is constant) That is, the product of the pressure (P) and volume (V) of a given mass of a perfect gas remains constant in an isothermal process. So, Boyle’s law is obeyed in an isothermal process. Isotherm : A graph between pressure and volume of a given mass of a gas at constant temperature is known as isotherm or isothermal of a gas. In the figure below, two isotherms for a given gas at two different temperatures T1 and T 2 are shown.

Work done in an isothermal process : Consider one mole of an ideal gas enclosed in a cylinder having perfectly non-conducting walls and a perfectly conducting bottom. Let the cylinder be fitted with a frictionless and insulating piston of cross-sectional area A. Let d W be the work done by the gas when the piston moves up through an elementary distance d x . Let P be the pressure of the gas. Then, d W = P × A × dx Fig. : Work done in (work = force × distance) isothermal process

d W = Pd V

or T2 > T1

P

where d V is the infinitesimally small increase in the volume of the gas. T2 T1

Let Wiso be the total work done by the gas when the gas expands isothermally from an initial volume V1 to the final volume V 2.

V

Then

Fig. : Isotherms of a gas

Application of first law of thermodynamics to isothermal process : The first law of thermodynamics states

dQ = dU + dW The internal energy of an ideal gas depends only on temperature. In an isothermal process, temperature remains constant. Therefore dU = 0 and

But,

V2 ⌠ ⎮ Pd V ⎮ ⌡ V1

Wiso =

where T is the constant temperature at which isothermal expansion takes place. ∴

V2 ⌠ ⎮ RT d V ⎮ ⌡ V1 V

Wiso =

dQ = dW

When an ideal gas expands isothermally, it does mechanical work d W and absorbs an equivalent amount of heat d Q from the surroundings. Similarly, when an ideal gas is compressed isothermally by doing a mechanical work d W on it, it rejects an equivalent quantity of heat d Q to the surroundings.

RT V

PV = RT or P =

V2 ⌠ ⎮ 1d V ⌡ V1 V

= RT ⎮

Wiso = RT [1oge V ]

or

V2 V1

= RT [loge V 2 – loge V 1] Q. Calculate the work done by an expanding gas. Solution :

or

Wiso = RT loge

V2 V2 = 2·3026 RT log10 V1 V1

The graph below left shows the expansion of a gas at constant pressure. The area under the graph gives the work done by the gas (PΔV). The same principle applies when the pressure is not constant, as shown below right :

Above, a gas at pressure P exerts a force PA on the piston and moves it a short distance Δx. If the expansion of the gas is so small that the pressure does not change : Work done by the gas = Force × displacement = PAΔx But, AΔx = ΔV, the increase in volume So, Work done by gas = PΔV

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

P

P area ea = work done

O

V

O

area ea = work done V

Let P 1 and P2 be the pressures corresponding to the volumes V1 and V2 respectively.

Then, ∴

P2V2 = P1V1 or Wiso = RT loge

V2 P1 = V1 P2

P1 P2

= 2·3026 RT log 10

constant volume. This is equal to change in the internal energy of the gas. d U = Cv d T

P1 P2

In an adiabatic process, no exchange of heat between the system and the surrounding is allowed. ∴

For μ moles of an ideal gas V2 V1

= 2·3026 μRT log 10

For an ideal gas, V2 V1

Differentiating,

Wiso = r T loge

V2 V1

V2 = 2·3026 r T log 10 V1

∴

Cv d T + Pd V = 0 Cv

[

= P×A

Cv P d V + Cv V d P + RPd V = 0

or

(Cv + R) Pd V + C v V d P = 0

= PAdx = Pd V where d V is the decrease in the volume of the gas. Q. A certain volume of gas suffers an expansion of 0·25 m3 at a constant pressure of 10 3 Nm–2. Calculate the work done. Solution : The work done W = PΔV where

P = Pressure

C p = Cv + R ∴

W = 103 × 0·25 = 250 joule

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

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

Cp P d V + Cv V d P = 0

Dividing both sides by C v PV, we get Cp P d V Cv V d P + = 0 Cv PV Cv PV γ

or

∫

Integrating, γ

or

[

d V dp + = 0 V P

γd V + V

∴ Cp =γ Cv

∫ dpP

= constant

∫ 1V d V + ∫ 1Pd P

= constant

or

γ log V + log P = constant

or

log V + log P = constant

or

log PV = constant

]

γ

γ

γ

PV = antilog (constant)

or

= another constant K ∴

γ

PV = K

which is the required relation between the pressure and volume of a gas. If P1V1 be the initial and P2V2 be the final pressure and volume respectively of the gas for an adiabatic change, then γ

P 1V1 = P2V2

γ

(ii) Adiabatic relation between volume and temperature : For a perfect gas PV = RT RT P = V

ΔV = Change in volume ∴

Cp –C v = R

But, or

Fig. : Work done in adiabatic process

Work done, d W = Force × distance

]

Pd V + Vd P + Pd V = 0 R

Equations of Adiabatic Process

Force acting on the piston

Pd V + V d P R

or

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

(i) Adiabatic relation between P and V for ideal gas : Consider one mole of a gas contained in a perfectly non-conducting cylinder fitted with a non-conducting piston. Let P, V and T be the pressure, volume and temperature respectively of the gas. Let the gas be compressed adiabatically so that the piston moves inwards through a distance dx. Let A be the cross-sectional area of a piston.

P d V + V d P = Rd T

dT = But,

V2 RT = log e M V1

where M is the molecular weight of gas

PV = RT

or

If we consider one gram of an ideal gas,

or

dQ = 0 Cv d T + Pd V = 0

So,

Wiso = μRT log e

Wiso

dQ = dU+dW

But,

or But, ∴

γ

PV = K RT γ V = K V

γ–1

or

RT V

or

TV ∴

TV

γ–1 γ–1

= K K = = K1 (say) R = constant

(iii) Adiabatic relation between pressure and temperature : For a perfect gas, PV = RT RT or V = P γ But PV = K RT γ ∴ P = K P

∴ or But

∴

d W = – Cv d T

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

Then,

Wadia = –

∫

= – Cv

P

or or

γ

∴

P1 – T

γ

= constant

An adiabatic is steeper than an isotherm : For an isothermal process, PV = constant Pd V + V d P = 0 P dP = – V dV

Differentiating, or

( )

iso γ

For an adiabatic process, PV

or

γ–1

or

γ–1 =

= constant

or

Cv =

γ

So,

γ

γ– 1

dV

PVγ – 1

γ dP = – dV Vγ

or

= γ

R (T1 – T 2) γ–1

Wadia =

r (T – T 2) –1 1

where r is the principal gas constant, i.e., gas constant for 1 gram of gas.

P – V

dP dV

R γ–1

which is another expression for the work done during adiabatic process.

Then

( ) ( )

Wadia =

R Cv

If we consider of one gram of ideal gas.

γP = – V P dP Hence, slope of adiabatic, = –γ V dV = γ

= – Cv [T2 – T 1]

Cp Cv R – = Cv Cv Cv

P = – V

V d P = – γ PV

T2 T1

which is the expression for the work done for one mole of an ideal gas during adiabatic process. Now, Cp – C v = R where Cp is the molar specific heat at constant pressure and R is the molar gas constant. Dividing both sides by C v , we get

dV + V dP = 0

or

dT

T1

Wadia = Cv [T1 – T 2]

Differentiating, P·γ V

∫

= – Cv [ T ]

dP But represents the slope of isotherm dV dP ∴ Slope of isotherm, dV

Cv d T

T1

T2

γ γ

RT = K γ P γ γ γ P1 – · R T = K K γ γ P1 – T = γ = K2 (say) R

dW = –dU d U = Cv d T

where Cv is the molar specific heat of gas at constant volume.

[ ]

or

dU+dW = 0

, iso

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

(From above) γ > 1

But,

Therefore, the slope of adiabatic is greater than the slope of isotherm an adiabatic is steeper than an isotherm. Work done in an adiabatic process : From first law of thermodynamics. dQ = dU+dW

Note : ● If the gas expands adiabatically, work is done by the gas. So, Wadia is positive. ∴

or

In an adiabatic process, no exchange of heat is allowed between the system and the surroundings. ∴

dQ = 0

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

T1 > T2

●

So, the gas cools during adiabatic expansion.

●

If the gas is compressed adiabatically, work is done on the gas. So, W adia is negative. ∴

T1 < T2

So, the gas heats up during adiabatic compression.

Comparison of Amounts of Work Done during Isothermal and Adiabatic Process In the case of expansion, the work done in an isothermal process is more than the work done in an Isotherm A adiabatic process as is shown in fig. below. AB is P B the isotherm and AC, the corresponding adiabatic. C Adiabatic The shaded area gives the excess of work done in isothermal expansion over the work done in correV sponding adiabatic expan- Fig. : Works done in isothersion. mal and adiabatic expansion

Also

dQ = dU+dW So,

mL = dU

That is, the internal energy increases by m L during the melting process. First law of thermodynamics and boiling process : On boiling a liquid, it changes into vapour at constant temperature called boiling point. Let a liquid of mass m vaporise. Let Vl and Vv be the volumes of the liquid and vapours respectively. The work done in expanding at constant temperature and pressure P.

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

In the case of compression, the work done in an adiabatic process is more than the work done in an C Adiabatic isothermal process as P shown in figure below. AB B A is the isotherm while AC is Isotherm the corresponding adiabatic. The shaded area gives the excess of work done in adiabatic compression over V the work done in correFig. : Works done in isothersponding isothermal commal and adiabatic compressions pression. First law of thermodynamics and melting process : When a substance melts, the change in volume ( d V) is very small and is negligible. The temperature remains constant during the melting process. Let a mass m of a substance be melted. Heat absorbed during melting process d Q = m L where L is the latent heat of fusion of substance.

d W = Pd V = P × 0 = 0

According to first law of thermodynamics

d W = Pd V = P (Vv – Vl ) Heat absorbed during boiling process,

dQ = mL where L is the latent heat of vaporisation. Change in internal energy,

d U = Uv – U l where U v and U l are the internal energies of the liquid and vapours respectively. First law of thermodynamics states

dQ = dU+dW ∴ or

m L = (Uv – U l ) + P (Vv – Vl ) Uv – U l = m L – P(Vv – Vl )

●

The latent heat of fusion of ice is 80 cal g– 1 or k cal kg– 1 at normal atmospheric pressure.

●

The latent heat of vaporisation of water is 540 k cal kg– 1 or 540 cal g– 1 under normal atmospheric pressure.

From which the gain in internal energy can be computed.

SOME TYPICAL SOLVED EXAMPLES Example 1. Calculate the work done in compressing 3 moles of a gas from 4 litre to 1 litre at constant temperature. (R = 8·3 J mol–1, K–1) Solution :

Example 2. Keeping the temperature constant at 27°° C one mole of a perfect gas is allowed to expand from 4 atmospheric pressure to 1 atmospheric pressure. Calculate the work done by the gas. (R = 8·314 JK– 1 mol– 1)

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

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

Solution :

Solution :

Example 3. Having suddenly compressed the volume of dry air is reduced to one quarter at atmospheric pressure. What will be its pressure ? (γγ = 1·5) Solution :

Example 4. A gas is compressed adiabatically to one quarter of its initial volume at 17°° C. Calculate the resulting temperature. (γγ for gas = 1·5) Solution :

Example 7. Dry air at 15°° C and 10 atmospheric pressure is suddenly released at atmospheric pressure. Find out the temperature of air. (γγ for air = 1·41) Solution :

Example 5. The density of a gas at 27°° C and 105 N-m– 2 pressure is 1·775 kg m– 3 and its specific thermal capacity at constant pressure is 846 J kg– 1 K – 1. Determine the ratio of specific thermal capacity at constant pressure to that at constant volume. Solution :

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

Example 8. At N.T.P. certain gas expands adiabatically from 1 litre to 2 litre. Change the new temperature of the gas. (γγ = 1·5) Solution :

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

Example 10. A cylinder fitted with a piston contains 0·2 mole of air at temperature 27°° C. The piston is pushed so slowly that the air within the cylinder remains in thermal equilibrium with the surroundings. Find the work done by the system if the final volume is twice the initial volume. Solution :

OBJECTIVE QUESTIONS 1. The first law of thermodynamics is essentially a restatement of the law of— (A) Conservation of momentum (B) Conservation of charge (C) Conservation of spin (D) Conservation of energy 2. A certain gas at atmospheric pressure is compressed adiabatically so that its volume becomes half of its original volume. Calculate the resulting pressure in dyne cm – 2. (Given : γ = 1·4) (A) 4·0 × 103 dyne cm – 2 (B) 26·7 × 106 dyne cm – 2 (C) 2·67 ×

106 dyne

cm – 2

(D) 2·67 × 108 dyne cm – 2 3. Two samples of a gas initially at same temperature and pressure are compressed from a volume V V to · One sample is compressed 2 isothermally and the other adia-

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

batically. Now which of the following is true ? (A) Padia = Piso (B) Padia > Piso (C) Padia < Piso (D) Padia t1). Its mean life is T— (A) A1t 1 = A2t 2 A1 – A2 (B) = constant t2 – t1 (C) A2 = A1 e

( ) ( ) t1 –

t2 T

t1 T (D) A2 = A1 e t2

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

6. ALGOL resembles— (A) COBOL

(A) 2·6°

(B) 4°

(C) 5·33°

(D) 3°

12. In figure below are given the four lower energy levels of hydrogen atom. The probable number of transitions is—

(B) BASIC (C) FORTRAN

n=4 n =3

(D) All of the above

n=2

7. When a hydrogen atom emits a photon in going from n = 5 to n = 1, its recoil speed is almost— (A) 10–4 m /s (B) 2 × 10–2 m /s (C) 4 m /s (D) 8 × 102 m /s 8. In an N-P-N transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector— (A) The emitter current will be 9 mA (B) The emitter current will be 11 mA (C) The base current will be 1 mA (D) The base current will be –1 mA 9. The emitter current in a transistor circuit is 4 mA. The transistor used has α = 0·96. What are the values of base current and collector current ? (A) 160 μA , 3·84 mA (B) 80 μA, 1·92 mA (C) 60 μA, 1·24 mA (D) 40 μA, 0·96 mA 10. In the Bohr model of hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in n th quantum state is— (A) – 1 1 (C) 2

(B) + 1 (D) 2

11. A thin prism P of angle 4° and made of glass of refractive index 1·54 is combined with another thin prism P′ of refractive index 1·72 for dispersion without devia-

n=1

(A) 3 (C) 5

(B) 4 (D) 6

13. The illuminance of a surface is 10 lux. If the total area of the surface is 30 cm2, then the luminous flux incident on it will be— (A) 3 × 10 –4 lm (B) 3 × 10 –3 lm (C) 3 × 10 –2 lm (D) 3 × 10 –1 lm 14. A ray of light passes through a glass slab of thickness t and refractive index μ. If the speed of light in air be ‘c ’, the time taken by the ray to cross through the plate is— t tc (A) (B) μc μ μt μc (C) (D) c t 15. A beam of unpolarised light is passed first through a tourmaline crystal A and then through another tourmaline crystal B oriented so that its principal plane is parallel to that of A. The intensity of final emergent light is I. The value of I is— I0 I0 (A) (B) 2 4 I0 (C) (D) None of these 8 16. The mass of a proton is 1840 times that of an electron. An electron and a proton, with equal kinetic energies, enter perpendicularly uniform magnetic field. Now— (A) The path of proton will be more curved than that of the electron (B) The path of proton will be less curved than that of electron

(C) The paths of both proton and electron will be equally curved (D) The paths of both will be straight 17. In an interference experiment third bright fringe is obtained at a point on the screen with a light ° source of wavelength 7000 A. What should be the wavelength of light source in order to obtain fifth bright fringe at the same point ? ° ° (A) 5000 A (B) 4200 A ° (C) 6300 A

° (D) 7500 A

18. Ice skating can be used to demonstrate that when ice is under pressure, its— (A) Melting point is lowered (B) Melting point is raised (C) Melting point remains unchanged (D) Coefficient of friction with metal is reduced 19. The intensity of light due to a source of light in a room full of smoke particles— (A) Is uniform at all distances (B) Obeys inverse square law (C) Will increase more rapidly with distance than inverse square law (D) Will decrease more rapidly with distance than inverse square law 20. The three primary colours used in a colour television are— (A) Red-blue-green (B) Green-yellow-red (C) Yellow-blue-black (D) Yellow-blue-red 21. The wavelength of H α line in hydrogen spectrum was found to ° be 6563 A in laboratory. If the wavelength of same line in the spectrum of a milky way is ° observed to be 6586 A, then the recessional velocity of the milky way will be— (A) 105 m/s (B) 1·05 × 106 m /s (C) 10·5 × 106 m /s (D) 0·105 ×

106 m /s

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

22. A magnet of magnetic moment M is freely suspended in a uniform magnetic field of strength B. The work done in rotating the magnet through an angle θ is given by— (A) B M

effective inductance of 200 μH. What must be the range of its variable condenser ? (A) 88–198 pf (B) 50 – 100 pf (C) 25 – 75 pf (D) None of these

(B) M B sin θ (C) M B cos θ (D) M B (1 – cos θ) 23. When an air column at 15°C and a tuning fork are sounded together, 4 beats per second are produced. The frequency of the fork is less than that of the air column. When the temperature falls to 10°C, the beat frequency decreases by one. The frequency of the fork will be— (A) 110 Hz (B) 114 Hz (C) 113 Hz (D) 106 Hz 24. When the number of turns in a coil is doubled without any change in the length of coil, its self-inductance becomes— (A) Four times (B) Doubled (C) Halved (D) Squared 25. A source of A.C. voltage V = 180 sin ω t is connected in series to a resistance and two reactances X L = 10 Ω and XC = 15 Ω. Calculate the active power if the current in the circuit is 3 A— (A) 100 watt (B) 200 watt (C) 380·3 watt (D) 250 watt 26. The system shown in the figure, when slightly displaced and released oscillates with a period T. If only one spring is used, the period of oscillation will be—

K

K M

(A) T

(B) T/2

(C) T/ 2

(D) 2 T

27. A radio can tune over the frequency range of a portion of MW broadcast band (800 kHz to 1200 kHz). If its LC circuit has an

28. The units nanometer, fermi, angstrom and attometer, arranged in decreasing order will read as— (A) Angstrom, nanometer, fermi, attometer (B) Fermi, attometer, angstrom, nanometer (C) Nanometer, angstrom, fermi, attometer (D) Attometer, angstrom, fermi, nanometer 29. The power radiated by a black body is P, and it radiates maximum energy around the wavelength λ0. If the temperature of the black body is now changed so that it radiates maximum energy around wave3λ0 length , the power radiated by 4 it will increase by a factor of— (A) 4/3 64 (C) 27

(B) 16/9 256 (D) 81

→ ^ ^ ^ 30. The vector B = 5 i + 2 j – s k is perpendicular to the vector → ^ ^ ^ A = 3 i + j + 2k for s = (A) 1

(B) 4·7

(C) 6·3

(D) 8·5

31. Let ω be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth’s surface has the same value at the equator and the poles. An object weighed by a spring balance gives the same reading at the equator as at a height h above the poles ( h r2) come in contact. Their common surface has a radius of curvature r — r1 + r2 r1 r2 (A) r = (B) r = 2 r1 – r2 r1 r2 (C) r = (D) r = r1r2 r1 + r2 34. The focal length of a convex lens having a magnifying power of 12·5 X is— (A) 8 m (B) 2 cm (C) 12·5 cm (D) 8 cm 35. A metal wire of length L, area of cross-section A and Young modulus Y behaves as a spring of spring constant k— YA 2YA (B) k = (A) k = L L YA YL (C) k = (D) k = 2L A 36. The critical angle will be maximum when light travels from— (A) Glass to air (B) Water to air (C) Glass to water (D) Water to glass 37. The escape velocity for a planet is v e. A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be— (A) v e (B) 1·5 v e (C)

1·5 ve

(D) 2 ve

38. When an electron jumps from the second allowed orbit to the first allowed orbit in hydrogen, its angular momentum changes by— (A) 6.6 × 10–34 kg m2/s (B) 6.6 × 10–27 kg m2/s

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

(C) 1.06 × 10–34 kg m2/s (D) 3·24 × 10–34 kg m2/s 39. A coin is placed on a horizontal platform which undergoes simple harmonic motion about a mean position O. The angular frequency of the simple harmonic motion is ω. The coefficient of friction between the coin and the platform is μ. The amplitude of oscillation is gradually increased. The coin will begin to slip on the platform for the first time— g (A) For an amplitude of 2 ω (B) At the mean position (C) For an amplitude of

μg

ω2 g (D) For an amplitude of μ ω2 40. In the reaction 92 U

235

+ 0n1 ⎯→ 54X142 + 36Kr89 + 0n1 + α-particle

The number of α-particles emitted is— (A) Four (B) Three (C) Two (D) One 41. A sphere of mass m is suspended from a string of length l from the point O. The sphere rotates in a circular path in a horizontal plane. The string makes an angle α with the vertical. Find the time period of rotation— (A) T = 2π

h 3g

1 (B) T = 2π

h g

(C) T = 2π

h g

(D) T =

π 2

h g

42. The fermi energy for a substance— (A) Is directly proportional to T (B) Is proportional to T (C) Is independent of T (D) Varies as T 2 (T is the temperature in kelvin) 43. A double-star is a system of two stars moving around the centre of inertia of the system due to gravitation. Find the distance

between the components of the double star, if the total mass equals M and the period of revolution is T— GMT2 1/3 (A) 4π2 (B) (C) (D)

( ) ( ) ( ) ( ) GMT 4π

1/3

GT 4π2M

1/3

MT2 4π2G

1/2

44. An aircraft is going at a speed of Mach 2. Its speed is nearly ……… km/hr. (A) 660

(B) 1080

(C) 1440

(D) 2380

45. A uniform wire is bent in the form of a circle of radius r = 9·8 cm. It is initially at rest and its diameter OB is horizontal. It is swinging about O in the vertical plane. Calculate its angular velocity ω when its diameter occupies the position OB'—

G

O

B

G

B

(A) 2 rad/s (C) 10 rad/s

(B) 15 rad/s (D) 20 rad/s

46. If the intensity of sound is doubled, the sound level will increease by nearly— (A) A factor of 2 (B) 2 db (C) 3 db (D) 4 db 47. A man swimming at a rate of 5 km/h wants to cross a 120 m wide river in a direction perpendicular to the stream. If the stream speed is 4 km/h, find the direction in which the man swims, and the time he takes to cross the river— (A) cos–1 (B) sin –1

( ) ( ) –

–

4 , 144s 5

4 , 100s 5

(C) tan –1 (D) cot–1

( ) ( ) –

4 , 200s 5

–

4 , 50s 5

48. Which of the following combinations is in order of increasing resistance ? (A) Galvanometer, voltmeter, ammeter

(B) Ammeter, galvanometer, voltmeter (C) Ammeter, voltmeter, galvanometer (D) Voltmeter, ammeter, galvanometer 1 2 x is the equation of a 2 trajectory, find the time of flight—

49. If y = x –

ANSWERS WITH HINTS

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

2 g 2 (C) g (A)

(B)

2 g

(D)

1 2g

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

Acidity of X–H Bond in Non-metal Hydrides

● The acid strength of conjugate acids is in the order as NH4+ < PH4+ < AsH4+

1. The Polarity of the X–H Bond

3. Charge on Acid or Base

● When all other factors are kept constant, acids become stronger as the X–H bond becomes more polar. The second row non-metal hydrides, for example, become more acidic as the difference between electronegativity (ΔEN) of X and H atoms increases.

● The charge on molecule or ion can influence its ability to act as an acid or a base. This is clearly shown when the pH of 0·1 M solutions of H3 PO4 and H2 PO4–, HPO 42– and PO43– ions are compared

H—F :

Ka = 7·2 × 10– 4

ΔEN = 1·8

H2O :

Ka = 1·8 × 10– 16

ΔEN = 1·2

NH3 :

Ka = 1 × 10– 33

ΔEN = 0·8

CH4 :

Ka = 1 × 10– 49

ΔEN = 0·4

● HF is strongest and CH4 is the weakest acid among above four compounds. ● 0·1 M solution of HF is moderately acidic, H2O is much less acidic and the acidity of NH3 is so small that the chemistry of aqueous solution of NH3 is dominated by its ability to act as base. HF (0·1 M) ; H2O (0·1 M) ;

pH = 2·1 pH = 7

NH3 (0·1 M) ;

pH = 11·1

● Stronger the acid weaker will be its conjugate base or vice-versa. The increasing order of basic nature of conjugate bases is as : F– < OH– < NH2– < CH3–

H3PO4

:

pH = 1·5

–

:

pH = 4·4

2–

:

pH = 9·3

:

pH = 12·0

H2PO4 HPO 4 PO4

3–

● A compound or ion becomes less acidic and more basic as negative charge increases Acidity :

H3PO4 > H2PO4– > HPO42–

Basicity :

H2PO4– < HPO42– < PO43–

● Similarly for H2SO4 Acidity :

H2SO4 > HSO4–

Basicity :

HSO 4– > H2SO4

4. Oxidation State of Central Atom ● There is no difference in polarity, size or charge when we compare oxyacids of the same element such as H2SO4 and H 2SO3 or HNO3 and HNO2 H2SO4 ; Ka = 1 × 103

HNO3; Ka = 28

2. Size of Atom, X

H2SO3 ; Ka = 1·7 ×

HNO2; Ka = 5·1 × 10 – 4

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

Acidity of these oxyacids increases significantly as the oxidation number of central atom becomes larger. H2SO4 is much stronger than H2 SO3 and similarly HNO3 is much stronger than HNO 2.

Ka = 7·2 × 10– 4

BDE = 569 kJ/mole

HCl :

Ka = 1 ×

106

BDE = 431 kJ/mole

HBr :

Ka = 1 × 109

BDE = 370 kJ/mole

HF

:

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

HF < HCl < HBr < HI Similarly H2O < H2S < H2Se ● The presence of lone pair of electrons on the central atoms of NH3 and PH 3, makes these hydrides Lewis bases. MH3 + H+ ⎯→ MH4+ Base

Conjugate acid

● As the size of central atom (N, P) increases, the stability of conjugate acid decreases, i.e., M—H bond becomes weaker as size of central atom increases and hence basic nature decreases as : NH3 > PH3 > AsH3 ………

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

10– 2

● This trend is easiest to be seen in four oxyacids of chlorine. Acid

Ka

Oxidation number of Cl atom

HOCl

2·9 × 10– 8

+1

HOClO

1·1 × 10– 2

+3

HOClO2

5·0 ×

+5

HOClO3

1 × 103

102

+7

HOCl is the weakest and HOClO 3 is the strongest acid. ● As the oxidation number of Cl atom increases, the atom becomes more electronegative. This tends to draw electrons away from oxygen atom which surround the Cl atom, thereby making oxygen atom more electronegative as well. As a result O—H bond becomes more polar and compounds becomes more acidic.

POINTS TO REMEMBER ● O and N atoms are about the same size, yet H 2O is much stronger acid as compared to NH 3. This is only on account of O is more electronegative and O—H bond is more polar. ● NH 4+ ion is a stronger acid than NH3 molecule because it is easier to remove H + ion from NH4+ than from NH3 molecule. ● PH 3 is a stronger acid than NH3 because the compounds become more acidic as the size of central atom holding hydrogen atom increases and X—H bond becomes weaker. ● OH – ion is a conjugate base of H2 O and NH2 – ion is the conjugate base of NH3. H2O is a stronger acid than NH3 and hence OH– ion is a weaker base than NH2– ion.

● Liquid ammonia is an amphoteric solvent like water because the auto-ionization of NH3 is as 2 NH3 NH 4+ + NH 2– It is just like water H 2O H+ + OH– ● H2SO4, a stronger acid, gives proton to HNO3, a weaker acid than H 2SO4. H | H—OSO 2—OH + H—O–NO2 → H—O—NO2 + –OSO2OH + Acid

Base

Acid

H2O + NO2+

● PH3 is a stronger acid than NH3, which means the PH 2– ion must be a weaker base than NH2– ion. ● The relative strength of Bronsted bases can be predicted from relative strengths of their conjugate acids, combined with the general rule that the stronger of a pair of acids always has weaker conjugate base. ● A weaker acid is known to be displaced from its salt by a stronger acid, e.g., acetylene displaces NH3 from its salt sodamide. HC ≡ CH + NaNH 2 Stronger acid

Stronger base

HNH2 + HC ≡ C Na Weaker acid

Weaker base

Water displaces acetylene from its salt. H–OH + HC ≡ CNa Stronger acid

Stronger base

HC ≡ C–H + NaOH Weaker acid

Weaker base

Hence acetylene is stronger acid than NH3, but weaker acid than H 2O. H2O > HC ≡ CH > NH3 ● The decreasing order of acidity of hydrocarbons is as HC ≡ CH > CH2 = CH2 > CH3–CH3 Conversely the decreasing order of basicities of anions resulting from these hydrocarbons is as : : C2H5 > H2C — — CH : > CH ≡ C : ● The concentration of an acid solution is determined by how many mole of acid is dissolved per litre and its strength is determined by how completely it ionizes.

Base

↓ Nitronium ion

HNO3 is acting as a base. ● H2SO4 is a weaker acid than HClO4 and acts as a base to accept proton from HClO4. H | H—OSO 2–OH + H—OClO3 → –OClO3 + H— O—SO 2OH +

Base

Acid

Base

Acid

● A strong acid must be stronger than H 3O+ and strong base stronger than OH– ion. ● Feeble acids and bases are weaker than H 2O. Example acetylene (HC ≡ CH) ● Weak acids are weaker than H 3O+ but stronger than H2O. Weak bases are weaker than OH– but stronger than H2 O. Most of the organic acids and bases are weak. CH 3COOH + H2O H3O+ + CH3COO– Acid (Weak)

Base (Weak)

Acid (Strong)

OH– + CH3NH3+

CH 3NH2 + H2O Base (Weak)

Acid Weak

Base (Strong)

Base (Strong)

Acid (Strong)

● The substance which can act both as acid and base is said to be ampholytic or amphoteric. For example : Liquid NH3, H2O, HF etc. NH 3 + NH3

NH 4+ + NH2–

Acid

Acid

Base

H 2O + H 2O Acid

Base

HF + HF Acid

Base

Base

H3O+ + OH – Acid

Base

H2F+ + F– Acid

Base

OBJECTIVE QUESTIONS 1. Which of the following substances is not amphoteric ? (A) HCO3– (B) H2O (C) NH3 (D) NH4+

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

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

2. Which is the strongest acid among the following ? (A) Acetylene (B) Water (C) Ammonia (D) Ethylene 3. Weak acids are— (A) Weaker than H2O (B) Stronger than H 3O+ (C) Stronger than H 2O

5. Which is the weakest Bronsted base ? (A) F– (B) Cl– – (C) Br (D) I –

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

6. Which of the following substances has the highest pH for 0·1 M solution ? (A) NaH2PO4 (B) Na 2HPO 4 (C) Na 3PO4 (D) H3PO4

9. For the reaction Zn 2+ + X– ZnX + The k eq is greatest when X – is— (A) NO3– (B) I – – (C) ClO3 (D) F–

(D) Always amphoteric

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

10. The correct decreasing order of basic character is— (A) CH3– > OH– > NH2– > F– (B) CH3– > – NH2– > OH– > F– (C) F– > NH2– > OH– > CH3– (D) NH2– > CH3– > OH– > F– 11. Among HS –, I –, R–NH 2 and NH3, the proton accepting tendency will be maximum and lowest respectively in— (A) R–NH2 and I – (B) NH3 and HS – (C) I – and HS – (D) HS – and I – 12. The electronegativity of chlorine atom will be highest in— (A) HOCl (B) HOClO (C) HOClO2 (D) HOClO3 13. Which among the following is the weakest base ? (A) PO43– (B) H2PO4– (C) HPO 42– (D) All are equally basic 14. Which of the following factors explains the higher acid strength of HI than that of HF ? (A) Polarity in H–X bond (B) Charge on the molecule (C) Size of atom X (D) Oxidation state 15. Which of the following is incorrect statement ? (A) PH 3 is a stronger acid than NH3 (B) NO3– is a weaker base than NO2– (C) PH 2– is a weaker base than NH2– (D) H2O and liquid NH3 are not amphiprotic solvents 16. The strongest Bronsted base among the following is— (A) ClO– (B) ClO2– (C) ClO3– (D) ClO4– 17. Which is the weakest Lewis base ? (A) H– (B) OH– – (C) Cl (D) HCO3– 18. HNO3 in liquid HF behaves as— (A) (B) (C) (D)

An acid A base Neither a base nor an acid As a base as well as an acid

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

19. The conjugate base of ammonium ion is— (A) NH2– (B) NH4+ (C) NH3 (D) OH– 20. Which is the correct order of basic nature ? (A) H2O > NH3 > PH3 (B) NH3 > PH3 > H2O (C) NH3 > H2O > PH3 (D) PH 3 > NH3 > H2O 21. NH3 gas dissolves in H2O to give NH4OH. H2O acts as— (A) (B) (C) (D)

An acid A base A conjugate base Amphoteric solvent

22. In the reaction HClO4 + H2O

(C) HClO > H3PO4 > H2SO4 (D) H3PO4 > HClO > H2SO4 27. Which can act both as Bronsted acid and base ? (A) Cl– (B) HCO3– (C) H3O+ (D) Both (B) and (C) 28. Correct increasing order of acidity is— (A) H3PO4 < HCl < H2CO3 < HI (B) H3PO4 < H2CO3 < HCl < HI (C) H2CO3 < H3PO4 < HCl < HI (D) None is correct

H3O+ + ClO4–

(A) HClO4 is a conjugate acid of H2O (B) H2O is a conjugate acid of H3O+ (C) H3O+ is a conjugate base of H2O (D) ClO4– is the conjugate base of HClO4 23. An aqueous solution of acetic acid contains— (A) CH3COO– and H + (B) CH3COO–, H3O+ and H + (C) CH3COO–, H3O+ and CH3COOH (D) CH3COOH, CH3COOH and H+ 24. Which is the correct decreasing order of basic strength ? (A) CH3–CH2– > NH2– > HC ≡ C– > OH– (B) HC ≡ C– > CH3–CH2– > NH2–

29. The conjugate acid of azide ion is— (B) HN3 (A) NH3 (C) NH2– (D) N2– 30. Strongest conjugate base results from— (A) Formic acid (B) Acetic acid (C) Ethylene (D) Acetylene 31. Liquid ammonia, like water is an amphiprotic solvent. Which is the appropriate auto-ionization equation for liquid NH 3 ? (A) NH3 NH2– + H+ NH4+

(B) NH3 + H+

NH4+ + NH2–

(C) 2 NH3 (D) All of these

32. Which of the following ions in aqueous solution gives a neutral solution ? (A) SO32– (B) NH4+ (C) Na + (D) F– 33. Which is the correct representation of acidic nature of AlCl3 in water ? (A) AlCl3 + 3 H2O Al(OH)3

> OH– (C)

OH–

>

NH2–

> HC ≡ > CH3–CH2– C–

(D) NH2– > HC ≡ C– > OH– > CH3–CH2– 25. The conjugate base of HPO42– is— (A) PO43– (B) H2PO4– (C) H3PO4

(D) H4PO3

26. Correct decreasing order of acid strength is— (A) H2SO4 > HClO > H3PO4 (B) H2SO4 > H3PO4 > HClO

+ 3 HCl (B) [Al (H2O)6] 3+ [Al (H2O)5] 3+ + H2O (C) [Al (H2O)6] 3+ [Al (H2O)5 OH]2+ + OH– (D) [Al (H2O)6] 3+ + H2O [Al (H2O)5 OH]2+ + H3O+ 34. Amino acid, glycine exists predominantly in the form of + NH3CH2COO–. Which is the conjugate acid of glycine ? (A) NH2CH2COOH (B) NH2CH2COO–

+

(C) BCl 3 and AlCl3 are both equally strong Lewis acids (D) BCl3 and AlCl3 are both Lewis acids and AlCl3 is stronger than BCl3

(C) It is an example of anionic hydrolysis

(C) NH3 CH2COOH +

(D) NH3 CH2COO– 35. Which is the Lewis acid-base reaction ? (A) Ca + S → Ca2+ + S2– (B) NH3 + HCl → NH4+ + Cl– (C) NH3 + BF3 → H3N : BF3 (D) None of these 36. According to Lowry and Bronsted concept, Cl– ion in aqueous solution is a— (A) Weak base (B) Strong base (C) Weak acid (D) Strong acid 37. Ionic dissociation of CH 3COOH is represented as— CH3COOH + H2O H3O+ + CH3COO– According to Lowry and Bronsted, in this equation we have— (A) One acid and three bases (B) One acid and one base (C) Two acids and two bases (D) Three acids and one base 38. An aqueous solution of Na + HCO3– is alkaline because— (A) Bicarbonate ion is alkaline (B) It is an example of cationic hydrolysis

(D) Bicarbonate ion forms another anion 39. Which of the following reactions will occurs when sodium hydride is dissolved in water ? (A) H–(aq) + H2O → H3O– (B) H+(aq) + H2O → H3O+ (C) H–(aq) + H2O → OH – + H2 (D) None of these 40. Water can act as an acid in presence of— (A) HCl (B) NH3 (C) C6H6

(D) H2SO4

43. The strongest Lewis base among the following is— (A) CH3– (B) F– – (C) NH2 (D) OH– 44. H+ is a— (A) Lewis acid (B) Lewis base (C) Bronsted base (D) None of these 45. Cl– is a conjugate base of— (A) HOCl (B) HOClO (C) HCl (D) None of these

ANSWERS

41. Which among the following is the weakest base ? (A) C2H5O– (B) F– (C) NO3–

(D) CH3COO–

42. Which is the correct statement ? (A) BCl3 and AlCl 3 are both Lewis acids and BCl3 is stronger than AlCl3 (B) Both BCl3 and AlCl3 are not Lewis acids

●●●

Exam. Date 6 Sept., 2009

Exam. Date 16 Aug., 2009

(Including Previous Years’ Solved Papers)

(Including Previous Years’ Solved Papers)

Main Features :  Test of Reasoning Ability  Test of English Language  Test of Numerical Ability  Officework Aptitude By : Dr. Lal & Jain Code No. 307 HINDI EDITION

Code No. 1129

Rs. 135/Rs. 140/-

UPKAR PRAKASHAN, AGRA-2 ● E-mail : [email protected]

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

● Website : www.upkar.in

By : Dr. Lal & Jain

Code No. 1619 Rs. 270/-

Highlights ● General Awareness ● Educational Awareness ● General English Comprehension ● Logical and Analytical Reasoning ● TeachingLearning and the School.

Subject Competence ● Science ● Mathematics ● Social Studies ● English HINDI EDITION

Code No. 1119

Rs. 230/-

Upkar Prakashan, AGRA-2 E-mail : [email protected]

Website : www.upkar.in

●

Introduction ●

●

Aromatic hydroxy compounds are of two kinds (i) Phenols : in which hydroxy group (groups) is attached to aromatic nucleus (benzene) (ii) aromatic alcohols : in which hydroxy group is present in side chain e.g., benzyl alcohol, C 6H5CH2OH.

From sulphonic acids—Fusion of sodium salt of sulphonic acid with NaOH followed by acidification, gives phenols Fuse

SO3Na + 2NaOH ⎯→ Sod. sulphonate

ONa + Na2SO3

Sod. phenoxide H2SO 4

Phenols are of following kinds :

⎯⎯→

OH

(i) Monohydric phenols— CH3

OH

OH

,

,

,

o-cresol

Phenol

Phenol

●

CH 3

CH3

OH

OH

,

Cl + NaOH

m -cresol

p -cresol

Chlorobenzene

OH

α-naphthol

300 – 350°C‚ High pressure Cu. Salt

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

OH

,

OH + NaCl Phenol

(ii) Aryl halides are first converted into their magnesium compound (Grignard reagent) in presence of ether, which soon changes to phenol by oxidation and subsequent hydrolysis.

β-naphthol

(ii) Dihydric phenols— OH

OH

From aryl halides (Dow’s method)—(i) This method has limited application in laboratory preparation of simple phenols.

Mg Ether

[O]

C6H5Br ⎯⎯→ C6H5MgBr ⎯→ C6H5OMgBr

OH ,

OH

H O

2 ⎯⎯→ C6H5OH + HOMgBr

,

Catechol Resorcinol ( o-dihydroxy benzene) ( m-dihydroxy benzene)

Phenol

● OH

Higher homologues from lower ones—This is achieved by heating lower homologue with alcohol in presence of anhydrous zinc chloride. anhy. ZnCl

2 C6H5OH + CH3OH ⎯⎯⎯⎯→ C6H4

Hydroquinone or quinol ( p-dihydroxy benzene)

o- and p-cresols

●

Cu/Fe

OH

425°C

C6H5Cl + H2O (Steam) ⎯⎯→ C6H5OH + HCl Phenol

OH , ● HO

OH

OH

Phloroglucinol

Hydroxy quinol

From diazonium compounds—When diazonium sulphates are boiled with water or with solution of CuSO 4 or steam distilled, phenols are formed +

H O‚ H+ Δ

2 N2HSO 4– ⎯⎯→

Benzene diazonium hydrogen sulphate

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

From decarboxylation of salicylic acid with soda lime— OH OH ONa NaOH‚ CaO COOH ⎯⎯⎯⎯→

Methods of Preparation ●

Raschig process— C6H5H + HCl + O(Air) ⎯⎯→ C6H5Cl + H2O

(iii) Trihydric phenols— OH OH

Pyrogallol

+ H 2O OH

Phenol

OH

OH , OH

CH3

OH + N2↑ + H2SO4 Phenol

360 K

Salicylic acid

●

HCl ⎯→

Sod. phenoxide

Phenol

From cumene—Petroleum is recent commercial source of manufacture of phenol. Benzene and propene obtained from petroleum are made to undergo Friedel-Crafts reaction to give cumene. This on oxidation in presence of a metal catalyst followed by treatment with acid gives phenol.

CH3 | Anhy. AlCl 3 ⎯⎯⎯⎯→ C6H5 —CH—CH3

C6H6 + CH3CH — — CH2

Cumene

CH3 | Dil. H 2SO 4 Oxidation ⎯⎯⎯→ C6H5 —C— O—OH ⎯⎯⎯→ Aerial | CH3

●

Properties of the nitrophenols

o-nitrophenol

B.P. °C at 70 mm 100

m -nitrophenol

194

p-nitrophenol

High (Dec.)

Name

Cumene hydroperoxide

CH3COCH3 + C6H5OH Acetone

Phenol

Properties of Phenols ●

●

●

●

(A) Properties in which Phenols Resemble Alcohols (Reactions due to —OH Group) ●

Phenols usually have high boiling points due to intermolecular hydrogen bonding. For example b.p. of phenol (C6H5OH), mol. wt. = 94, is 453K while that of toluene (C 6H5CH3), mol. wt. 92, is 384 K. o-nitrophenol (intramolecular hydrogen bonding) has much lower b.p. (Steam volatile) than m -and p-nitrophenols (intermolecular hydrogen bonding.) Phenol on account of corrosive action, produces blisters when comes in contact with skin. It is poisonous and is good antiseptic (0·2%) and disinfectant (1%). Dilute solution of phenol is used for causterizing wounds caused by mad dog bite. Phenol is stronger acid than alcohols since phenoxide ion is stabilised by resonance while alkoxide ion is not. Electron withdrawing groups increase the acidity while electron releasing groups decrease the acidity of phenol. The increasing order of acidic nature is as— OH OH OH

Solubility in Remark g/100 gm H 2O 0·2 Steam volatile due to intramolecular H-bonding. 1·35 Non-volatile in steam 1·69 Non-volatile in steam

Reaction with Na—Sodium phenoxide is formed. 2C 6H5OH + 2Na → 2C6H5ONa + H2↑ Phenol

●

Sod. phenoxide

Esterification—With acids in presence of polyphosphoric acid or toluene sulphonic acid, phenyl esters are formed. However, the yield of ester is poor than that obtained from alcohols. TsOH

ArOH + R—COOH ⎯⎯→ ArOCOR + H2O Phenyl esters are better prepared by action of acid chlorides or anhydrides on phenol. – :OH

C6H5OH + C6H5COCl ⎯→ C6H5OCOC6H5 + HCl Phenol

Benzoyl chloride

Phenyl benzoate

This is Benzoylation or Schotten-Baumann reaction. :

Phenol

–

OH ⎯→

C6H5OH + (CH3CO)2O Acetic anhydride

C6H5OCOCH3 + CH3COOH Phenyl acetate

< CH3

p-cresol

This is acylation reaction.

<

Key Point

NO2 p-nitrophenol

Phenol

OH

OH

NO2 <

O2N

NO2

3

< NO2

2, 4-dinitrophenol

● Phenyl esters, when heated with anhy. AlCl 3‚ undergo Fries rearrangement. The acyl group migrates to o- or pposition to form phenolic ketones. OH OCOCH Anhy. AlCl 3

NO2

Picric acid

Phenyl acetate

●

Phenol itself is weaker acid than carbonic acid (H2CO3) and hence does not decompose NaHCO 3 to give CO2. Instead phenols are recovered from aqueous solution of phenoxides by bubbling CO2. C6H5ONa + CO2 + H2O ⎯→ C6H5OH + NaHCO3 ● Acid ionization constants (Ka) for some phenols are listed below. Higher the value of K a higher is the acidic nature. Name

Ka

Phenol 1·1 × 10–10 o-cresol 0·63 × 10–10 m-cresol 0·98 × 10–10 p-cresol 0·67 × 10–10 o-fluorophenol 15 × 10–10 m-fluorophenol 5·2 × 10–10 p-fluorophenol 1·1 × 10–10

o-chlorophenol m-chlorophenol p-chlorophenol

77 × 10–10 16 × 10–10 6·3 × 10–10

Name

Ka

o-aminophenol m-aminophenol

2·0 × 10–10 69 × 10–10

o-nitrophenol m-nitrophenol p-nitrophenol

600 × 10–10 50 × 10–10 690 × 10–10

{

C.S.V. / August / 2009 / 714

COCH 3

⎯⎯⎯⎯→

2‚ 4-dinitrophenol 1000‚000 × 10–10 2‚ 4‚ 6-trinitrophenol Very large

o-hydroxyaceto phenone

OH + COCH 3 p-hydroxyaceto phenone

●

Formation of Ethers (Alkylation) Δ

C6H5ONa + CH3—I ⎯→ C6H5OCH3 + NaI Δ

Methoxy benzene (Anisole)

C6H5ONa + C2H5 —I ⎯→ C6H5OC2H5 Ethoxy benzene (Phenetole)

This reaction is known as Williamson's synthesis. It is a nucleophilic substitution reaction.

The reduction in presence of molybdenum oxide gives benzene.

Key Points ●

Aryl halides (ArX) are too inert to react with phenol to give esters. However, at high temperature and in presence of finely divided copper, the reaction is achieved.

MoO

C6H5OH + H2 ⎯⎯→ C6H6 + H2O ●

Reaction with phenyl isocyanate—Phenyl urethane is obtained as a colourless crystalline compound which is used for characterisation of phenols. — C— — O → C6H5NH·COOC6H5 C6H5OH + C6H5N —

●

Reactions of aromatic ring—The —OH group in phenol is o - and p -directing because it increases electron density at o - and p -positions due to resonance. Thus phenol undergoes electrophilic substitution reactions as— Halogenation—Like —NH 2 group, —OH group is so much activating that it is rather difficult to prevent poly substitution. OH OH

High temperature

C6H5ONa + BrC6H5 ⎯⎯⎯⎯⎯⎯→ Cu

C6H5OC6H5 + NaBr (Ullmann reaction) Diphenyl ether

●

The reaction of halogen substituted acids and halogen substituted phenol is important because the products are Weedicides used in agriculture. Cl Cl

OH + ClCH2COOH

2, 4-dichlorophenol

Chloroacetic acid

Cl

⎯→ Cl

➠ O CH2CO OH

2, 4-dichlorophenoxy acetic acid (2, 4-D), a weedicide

●

Phenol

3Br ‚ H O Br

Br

Phenol

(NH4) 2SO 3/NH3 ⎯⎯⎯⎯⎯→ C6H5NH2 150°C

Replacement of phenolic group by amino group can also be achieved by heating phenol with double compound of zinc chloride and ammonia at 300°C.

Phenyl urethane

2 2 ⎯⎯⎯→

Bucherer reaction—Phenolic group is replaced by amino group by heating phenol with ammonium sulphite or bisulphite. C6H5OH

Phenyl isocyanate

Br 2, 4, 6-tribromo phenol

OH

OH 3Br water

2 ⎯⎯⎯⎯→

ZnCl2/NH3

C6H5OH ⎯⎯⎯⎯→ C6H5NH2 + H2O

Br OH

Br

OH

300°C

●

●

Br /CCl

2 4 ⎯⎯⎯→

+

Br

or CS2

Phenol

Br

➠ Sulphonation—This reaction is temperature dependent. Low temperature favours o-sulphonation while high temperature p-sulphonation. OH SO3 H OH

( o-phenol sulphonic acid)

°

H2SO 4 °

OH

C

C.S.V. / August / 2009 / 715

substitution stage, the reaction should be carried out in non-polar solvents like CCl4 and CS 2 and at lower temperature. OH OH OH

0

Cyclohexanol

Note : If it is required to arrest the reaction at mono

10

●

Reaction with ferric chloride—Aqueous solution of phenol on reacting with aqueous solution of FeCl 3 gives violet coloured solution probably due to formation of hexa-coordinated complex. 6C 6H5OH + FeCl3 → [Fe(OC6H5)6] –3 + 3HCl + 3H+ This reaction is used as qualitative test of phenolic group. Reaction with zinc—Phenol on distillation with zinc dust gives benzene. C6H5OH + Zn → ZnO + C6H6 Reduction of phenol—When reduced with H2 in presence of Ni (200 – 250°C) cyclohexanol is formed which is used as a solvent in rubber industry. OH HO H H H Ni H + 3H2 ⎯⎯⎯⎯→ H H 200 – 250°C H H H Phenol H H

2, 4, 6-tribromo resorcinol

C

(B) Reactions in which Phenols Differ from Alcohols

Br

Resorcinol

15

Note : This reaction is not used for preparation of aniline but finds extensive use in commercial preparation of naphthyl amines from corresponding naphthols. ● Reaction with PCl5 —Replacement of —OH group of phenols by halogen atom is rather difficult. When treated with PCl 5 or PBr5 halo benzenes are obtained in small amount and main product being triphenyl phosphate.

( p-phenol sulphonic acid)

SO3 H

➠ Nitration—(a) With dil. HNO 3 at 293 K, a mixture of 2-nitrophenol (major) and 4-nitrophenol (minor) is formed.

OH

Key Point ● Salicylic acid obtained by Kolbe’s-Schmidt reaction is a very useful starting chemical for preparation of a number of medicinal compounds. O CO CH3

(10%)

OH NO 2

Dil. HNO 3 20 °C

OH N O2 (50%) and steam volatile

(b) Nitration of phenol with mixture of conc. HNO 3 and H2SO4 gives a poor yield of 2, 4, 6-trinitrophenol (Picric acid). Poor yield is due to the excessive oxidation side reaction. OH OH HNO3-H2SO 4

OH CO O H Salicylic acid

OH CH 3OH Conc. H2SO4

NO2

⎯⎯⎯⎯→ NO2 Phenol

COO H (CH3CO)2O Conc. H2SO 4 2-Acetoxy benzoic acid (aspirin) analgesic and antipyretic OH C 6H 5 O H C O O C 6H 5 POCl3 Phenol salicylate (Salol) intestinal antiseptic

Methyl salicylate (oil of wintergreen) Use in perfumery and flavouring agent in food, drinks and cosmetics. Analgesic in rheumatic and sciatica pains

NO2

2, 4, 6-trinitrophenol (Picric acid)

OH

OH NO2

HNO3-H2SO 4

OH

⎯⎯⎯⎯→ NO2

➠ Reimer-Tiemann reaction—(a) The treatment of

OH

Resorcinol

NO2

Styphinic acid

(c) The picric acid in good yield can be prepared by following reaction : OH OH OH Conc. H2SO 4

SO3 H

⎯⎯⎯⎯→ 373 K

COOCH3

phenol with chloroform (CHCl 3) in presence of NaOH at 340K followed by hydrolysis gives o-hydroxy benzaldehyde (salicylaldehyde) with small amount of phydroxy benzaldehyde. OH OH OH CHCl /NaOH

CHO

3 ⎯⎯⎯⎯→ 70°C/HCl

+

+

Salicylaldehyde (Main)

Phenol

CHO p-hydroxybenzaldehyde (Minor)

SO3 H OH Conc. HNO3

⎯⎯⎯⎯→

NO2 Picric acid

➠ Friedel-Crafts alkylation reaction—When phenols are treated with alkyl halides in presence of anhy. AlCl3, alkyl phenols are formed. OH OH OH Anhy. AlCl3

CH3 p-cresol

➠ Kolbe-Schmidt reaction— OH 150°C 5-7 atm

Sod. phenoxide

COONa

Sod. salicylate

OH Dil. HCl ⎯⎯⎯→

⎯⎯→

COOH

Salicylic acid

➠ Reaction with formaldehyde—Under different conditions, different products are formed. (i) Phenol when treated with formalin in presence of a dil. acid or alkali at low temperature, p hydroxy benzyl alcohol along with small quantity of o-isomer is formed. This is known as Lederer Manasse reaction. OH OH OH NaOH 5 days

+ C H 2OH

Salicylic acid

COOH

—NaCl

+ HCHO ⎯⎯→

—NaCl

C.S.V. / August / 2009 / 716

COO Na Dil. HCl

Phenol

o-cresol

+ CO2 ⎯⎯→

CCl4/NaOH ⎯⎯⎯→ 70°C

CH3 +

+ CH3Cl ⎯⎯⎯⎯→

ONa

: o -isomer, due to chelation (intramolecular Hbonding) is more volatile than p -isomer (intermolecular H-bonding) and can be separated by steam distillation. (b) By using carbon tetrachloride in place of chloroform a mixture of o-hydroxy and p-hydroxy benzoic acids is formed. OH ONa OH

Note

NO2

NO2

(Main product)

C H 2OH

(ii) Phenol on condensation with insufficient amount of HCHO in presence of an acid catalyst, gives a linear polymer called novolak. OH

OH CH 2OH

HCHO ⎯⎯→

2 2 4 ⎯⎯⎯⎯⎯→

HO

N— —O

Tautomerises

p-nitrosophenol

OH

⎡⎢ ⎢⎣ HO

CH2

⎯⎯→

H 2O

(iii) In presence of excess of formaldehyde and using a basic catalyst like ammonia or strong aqueous alkali at higher temperature, a different kind of condensation occurs. Two kinds of products are first formed OH

— N—

HO

— —O

Phenol indophenol (Red) NaOH

OH C H 2OH

+ 2HCHO ⎯→

(a) 2

⎤ ⎦

Phenol indophenol hydrogen sulphate (Deep blue)

C H 2OH

CH2

— — NOH

— — OH+ ⎥ ⎥ HSO 4

— N—

OH

OH

— O—

Quinonemonoxime Conc. H 2SO 4 C6H5OH

OH

C6H5OH

HCHO ⎯⎯→

NaNO /H SO

HO

⎡⎢ ⎢⎣ –O

+

Sodium salt of phenol indophenol (Deep blue)

C H 2OH OH HOCH2

⎤ ⎦

— — O⎥ ⎥ Na+

N— —

➠ Phthalein reaction—When phenols are heated with CH2OH

Bis (hydroxymethyl) phenols

phthalic anhydride in presence of conc. H2SO4 or anhydrous zinc chloride, phthaleins, with characteristic colours, depending on the pH, are formed. HO OH

OH (b) 2

+ 2HCHO

Phenol

⎯→ HO

CH2

OH

Anhy. ZnCl

2 ⎯⎯⎯⎯→ –H O

C

p, p′-dihydroxydiphenyl methane

These products undergo slow polymerisation to give a resin called Bakelite, a three dimensional polymer.

Phenol

H H O

2

O C O

OH

C

CH2

CH2

OH

HO OH

CH2

CH2

Phthalic anhydride

CH2

HO

O

CH2 C CH2

O Phenolphthalein

➠ Liebermann’s reaction—When a crystal of phenol is added to a mixture of conc. H 2SO4 and sodium nitrite and reaction mixture is warmed, a characteristic blue or green colour is obtained. On dilution with water colour becomes red but again turns deep blue on adding excess of caustic soda. The chemistry of this reaction is of diagnostic value—

C.S.V. / August / 2009 / 717

Phthalein reaction is a diagnostic test for phenols. Phenolphthalein is used an important indicator in acid-base titrations. When treated with alkali, it undergoes an interesting colour change. It turns red first but on addition of excess of alkali it turns colourless again. These changes are shown as—

products is uncertain. However, sequence of reactions is predicted. O OH ||

OH

HO

C

O

2 ⎯→

NaOH ⎯⎯→

O

the following

+ H 2O

|| O

C

Quinone

O

Quinone forms brilliant red addition product with phenol.

Phenolphthalein (Colourless)

C6H5OH + O — —

– +

O Na

HO

⎯→

— —O — OH… O —

C

Phenoquinone (Red)

(b) With potassium permanganate : CH(OH)COOH KMnO 4 | + 2CO2 C6H5OH ⎯⎯⎯→ [O] CH(OH)·COOH Phenol

O C O

Mesotartaric acid

Colourless

(c) With a mixture of KClO3 and conc. HCl. O OH

O

HO

KClO

3 ⎯⎯→ HCl

Cl Cl

Cl Cl

Phenol

C

O Chloranil or Tetrachloroquinone

– +

O Na

(d) With alkaline solution of potassium persulphate. OH OH

C O

K S O

2 2 8 ⎯⎯⎯→ Alkaline

NaOH +

— — O… HO

–

NaO

OH

O

Hydroquinone or quinol

➠ Gattermann’s aldehyde synthesis : OH C

Anhy. AlCl

–

3 + HCN + HCl ⎯⎯⎯⎯→

: OH

Gas

– +

COONa

OH

(Red colour)

CH — — NH

NaOH +

Points to Remember

C

●

OH – +

➠

COONa

●

Trisodium salt (Colourless)

●

Similarly resorcinol and phthalic anhydride give fluorescein, which dissolves in dilute alkali to produce intense green fluorescence. Oxidation—(a) Phenols are much susceptible to atmospheric oxidation than alcohols, but the nature of

C.S.V. / August / 2009 / 718

+ NH3 CHO

Small quantity of p-hydroxy benzaldehyde is also formed.

ONa

NaO

⎯→

o-hydroxy benzaldehyde

– +

–

OH H 2O

● ●

Phenol is also known as carbolic acid. It is used as a disinfectant in carbolic soaps and lotions and as a preservative in inks. Bakelite is a phenol-formaldehyde thermosetting polymer which is generally used as an insulator for electrical work. 2, 4, 6-Trinitrophenol (Picric acid) is used in dyeing silk and wool and also used in preparation of explosives. It is effective in treatment of burns. Phenol finds extensive use in preparation of drugs like, salicylic acid, aspirin, salol, phenacetin etc. The reaction of phenol and phthalic anhydride leads to the formation of phenolphthalein which is used as an indicator in volumetric analysis and as purgative in medicines.

● ●

●

●

●

Phenyl salicylate (salol) is used in toothpastes because it is a good antiseptic. Methyl salicylate is main constituent of oil of wintergreen. It is obtained by direct esterification of salicylic acid with CH 3OH (Fischer-Speter’s method). It is used in preparation of iodex. Certain phenols like eugenol (clove oil), isoeugenol (nutmeg oil), thymol (thyme and mint oil), vanillin (vanila beans) are found in nature. Decreasing order of acidic nature of some derivatives of phenol is as : (i) 2, 4-dinitrophenol > p -nitrophenol > p-cresol > m-aminophenol.

HO HO

Δ

COOH

⎯→

NH 2

OH As = As Salvarsan (606)

●

Neosalvarsan has been found more useful as it is readily water soluble. It can be used for injection purpose. NH2

NHCH2 OSO 2 Na

HO

OH As = As Neosalvarsan

+ CO2 ●

Pyrogallol

Gallic acid on exposure to air turns brown to black, thus used in the manufacture of blue-black ink. Bismuth gallate, under the name of dermatol is used in treatment of skin infection. o -hydroxy benzoic acid (salicylic acid) is used in the treatment of eczema and other skin diseases and rheumatic pains. Claisen-rearrangement—When allylphenyl ether is heated at 573 K, it undergoes rearrangement as

1, 2, 3-trihydroxy benzene (Pyrogallol) is used in hair dye to convert grey hair to black. In gas analysis it is also used for absorbing unpleasant gases. Levodopa, 3-(3, 4-dihydroxy phenyl)-L-alanine is a drug used in treatment of perkinsonism. NH2 | CH2— CH—COO H Levodopa

HO

●

HO Paracetamidophenol (para-acetamol) is actually hydroxy derivative of acetanilide and is a main antipyretic.

OH

573 K ⎯⎯→

NH2

HO

●

HO HO

O–CH2—CH = C H2

NHCOCH3

HO CH2—CH = C H2

Paracetamol ●

Phenacetin (acetophenetidine) is ethyl ether of paracetamol.

o-allylphenol

Allylphenyl ether ●

Hydroxy toluenes (cresols) are used in preservation of timber under the name of creosote oil. Lysol is an emulsion of crude cresols in soap solution. Ehrlich, the founder of chemotherapy discovered Salvarsan or 606. It is an effective drug for treatment of trypanosoniasis. Hydrochloride of salvarsan is used in treatment of syphilis and it is also an effective antimalarial. It is prepared from phenol.

OH

Gallic acid

●

●

(ii) Phenol > p-cresol > m-cresol > o-cresol 3, 4, 5-trihydroxy benzoic acid (gallic acid) is present in tea and roots of pomegranate. In combined state it is present in gall nuts, oak bark and acacia bark. At m.p. it gives pyrogallol. OH

●

●

Elbs-persulphate oxidation : OH

OH K 2S2O8‚ KOH

⎯⎯⎯⎯→ Phenol

OH

Phenacetin

OH

+ OH

NHCOCH3

C2 H5 O

Catechol (Minor)

Hydroquinone or quinol (Major)

●

It is widely used as an analgesic and antipyretic, usually in combination with aspirin, caffeine and codeine. Acid strength of phenol (C6 H 5 OH) with respect to other compounds is in following decreasing order : R—COOH > H 2CO3 > C6H5OH > H2O > R—OH

OBJECTIVE QUESTIONS 1. Phenol is acidic in nature and it can react with— (A) Sodium bicarbonate (B) Potassium carbonate (C) Sodium hydroxide (D) All of these 2. Which of the following compounds does not contain a carboxylic group ?

3. The fusion of sodium benzene sulphonate with NaOH, followed by acid hydrolysis, gives— (A) Benzene (B) Benzoic acid (C) Phenol (D) Sodium salicylate 4. Which does not react with phenol (C6H5OH) ?

(A) Anisole

(A) Sodium

(B) Picric acid (C) Carbolic acid (D) All of these

(B) Caustic soda (C) Caustic potash (D) Washing soda

C.S.V. / August / 2009 / 719

5. Salicylic acid is— (A) m-hydroxybenzoic acid (B) p-hydroxybenzoic acid (C) o-hydroxy benzoic acid (D) All of these 6. Intramolecular hydrogen bonding is present in— (A) Phenol (B) p-nitrophenol (C) o-nitrophenol (D) m-nitrophenol 7. Which of the following compounds is the most acidic ?

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

p-nitrophenol p-chlorophenol p-aminophenol o-cresol

(B) Phenyl- n-propane (C) Isopropyl benzene (D) 2, 4-dimethylbenzene

8. Which of the following compounds is formed when phenol reacts with PCl5 ? (A) (B) (C) (D)

Chlorophenol Chlorobenzene Triphenyl phosphate Both (B) and (C)

9. Which of the following reactions does not give phenol or sodium phenoxide ? NaOH

(A) C6H5COCl ⎯⎯→ Alc. KOH

(B) C6H5N2Cl ⎯⎯→

Aqu. KOH

(C) C6H5N2Cl ⎯⎯→ H O Δ

2 (D) C6H5NNCl ⎯⎯→

10. Which of the following reactions converts phenol into salicylic acid ? (A) Kolbe’s reaction (B) Etard’s reaction (C) Reimer-Tiemann reaction (D) Dow’s reaction 11. A compound which is useful in treatment of burns is— (A) (B) (C) (D)

Creosote oil Picric acid 2, 4, 6-trinitro resorcinol Acetyl salicylic acid

12. In the reaction sequence CCl NaOH Sodalime

HCl

4 Phenol⎯⎯→ (A) ⎯⎯→ (B)

⎯⎯⎯→ (C) Product (C) is— (A) Sodium salicylate (B) Benzoic acid (C) Phenol (D) Salicylic acid 13. Aspirin is an acetylation product of— (A) m-hydroxybenzoic acid (B) p-hydroxybenzoic acid (C) o-hydroxybenzoic acid (D) Phenol 14. Cumene on oxidation in presence of metal catalyst followed by treatment with acid gives phenol. The cumene is— (A) o-cresol

C.S.V. / August / 2009 / 720

15. Which of the following compounds is attacked by an electrophile most easily ? (A) Toluene (B) Phenol (C) Benzene (D) Chlorobenzene 16. When phenol is distilled with zinc dust, the product is— (A) Toluene

(B) Benzene, dinitrobenzene, m nitroaniline (C) Benzene, nitrobenzene, hydrazobenzene (D) Toluene, m -nitrobenzene, m-toluedine 22. Phenol is less acidic than— (A) (B) (C) (D)

p-methoxyphenol p-nitrophenol Carbonic acid Both (B) and (C)

23. Reimer-Tiemann reaction involves—

(B) Benzene

(A) Carbonium ion intermediate

(C) Benzoic acid

(B) Carbanion intermediate

(D) Zinc phenoxide

(C) Carbene intermediate

17. Which compound is formed when sodium phenoxide is heated with ethyl iodide ?

(D) Free radical intermediate 24. C6H5OH + ClCOCH3 aq. NaOH

⎯⎯⎯→ C6H5OCOCH3

(A) Ethyl phenyl alcohol

is called—

(B) Phenetole

(A) Kolbe reaction

(C) Phenol (D) None of these

(B) Reimer-Tiemann reaction

18. Which of the following compounds undergoes nitration most readily ? (A) Benzoic acid (B) Toluene (C) Nitrobenzene (D) Phenol 19. Cyclohexanol is a— (A) Primary alcohol (B) Secondary alcohol

(C) Schotten-Baumann reaction (D) Dow’s reaction 25. Sodium phenoxide when heated with CO 2 under pressure, gives— (A) Sodium benzoate (B) Benzoic acid (C) Phenol (D) o-hydroxy benzoic acid 26. Which product is formed when phenol reacts with benzene diazonium chloride ?

(C) Phenol

(A) Phenyl hydroxylamine

(D) Tertiary alcohol

(B) Para aminoazobenzene

20. When phenol is treated with excess of bromine water, which of the following compounds is formed ? (A) m-bromophenol (B) o- and p-bromophenol (C) 2, 4, 6-tribromophenol (D) All of these 21. In the reaction sequence Zn Distillation Conc. H2SO 4 Zn ⎯⎯⎯⎯⎯→ (Y) ⎯⎯⎯→ (Z) conc. HNO3‚ 60°C NaOH

Phenol ⎯⎯⎯→ (X)

The product (X), (Y) and (Z) respectively are— (A) Benzene, nitrobenzene, aniline

(C) Phenyl hydrazine (D) Para hydroxyazobenzene 27. Phenol on reacting with dil. HNO3 at room temperature gives— (A) Picric acid (B) m-nitrophenol (C) o - and p-nitrophenol (D) All of these 28. Phenol and ethanol can be distinguished by using— (A) Caustic soda (B) Anhydrous AlCl3 (C) FeCl3 (D) All of these

29. A diazonium salt reacts with phenol to give an azo dye. This reaction is termed as— (A) Diazotization (B) Coupling (C) Both (A) and (B) (D) None of these 30. The Dow process is used for conversion of chlorobenzene into— (A) Benzene (B) Nitrophenol (C) Phenol (D) Chlorophenol 31. Phenol on reacting with HNO 3 or H2SO4— (A) (B) (C) (D)

Forms ester Gives ether Does not give ester Forms toluene

32. Carbolic acid is— (A) C6H5OH (B) H2CO3 (C) HCOOH (D) None of these 33. Main component of oil of winter green is— (A) Salicylic acid (B) Methyl salicylate (C) Phenyl salicylate (D) Ethyl salicylate 34. Phenol on heating at 60°C with a mixture of chloroform and excess of NaOH and on subsequent acidification gives— (A) (B) (C) (D)

2-hydroxybenzoic acid 2-hydroxy benzaldehyde 3-hydroxybenzoic acid 3-hydroxy benzaldehyde

35. Which of the following compounds gives violet colour with neutral solution of FeCl 3 ? (A) (B) (C) (D)

Benzoic acid Formic acid Salicylic acid Acetic acid

36. When phenol reacts with CHCl 3 in presence of NaOH, salicylaldehyde is formed, if we use pyrene in place of CHCl3, the product is— (A) Salicylaldehyde (B) Phenolphthalein

C.S.V. / August / 2009 / 721

(C) Salicylic acid (D) Cyclohexanol

43. Correct increasing order of acidic nature is—

37. Which statement C6H5OH is correct ?

regarding

(A) It is a carboxylic acid (B) It is insoluble in water (C) It has higher b.p. than toluene (D) None of these 38. Phenol is a— (A) Carboxylic acid (B) Base (C) Alcohol (D) None of these 39. Which compound gives purple colour with neutral ferric chloride solution ? (A) Boric acid (B) Phenol (C) Aniline (D) Benzoic acid 40. Phenol is most easily soluble in— (A) NaHCO3 solution (B) NaOH solution (C) Dil. HCl (D) None of these 41. The reaction OH CHO H2O2 / OH–

OH

⎯⎯⎯→

OH

is known as— (A) Reimer-Tiemann reaction (B) Liebermann’s nitroso reaction (C) Dakin reaction (D) Lederer-Manasse reaction 42. Which will undergo Friedel-Crafts alkylation reaction ? CH3 COOH OH

(A) H2O < H2C2 < H2CO3 < C6H5OH (B) C2H2 < H2O < C6H5OH < H2CO3 (C) C6H5OH < C2H2 < H 2CO3 < H2O (D) H2C2 < H 2O < H 2CO3 < C6H5OH 44. Which of the following will not form phenol or phenoxide ? (A) C6H5SO3Na (B) C6H5N2Cl (C) C6H5COOH (D) C6H5Cl 45. Phenol dissolves in aqueous NaOH to give sodium phenate, but on passing CO2 into solution, phenol is thrown out. This indicates that— (A) Phenol is weakly acidic, but stronger than H2CO3 (B) Phenol is basic (C) Phenol is amphoteric (D) Phenol is weakly acidic weaker than H2CO3 46. If salicylic acid is heated with zinc dust, the product formed will be— (A) Phenol (B) Benzene (C) Benzoic acid (D) Zinc benzoate 47. When salicylic acid is treated with bromine water, the product formed is— (A) 2, 3, 4-tribromophenol (B) 3, 4, 5-tribromosalicylic acid (C) 3, 6-dibromosalicylic acid (D) 2, 4, 6-tribromophenol 48. Chemically salol is— (A) Sodium salicylate (B) Acetyl salicylic acid

(I)

N O2 (II)

(C) Phenyl salicylate (D) Methyl salicylate

(III) C H 2C H 3

(A) I and IV (C) III and II

(IV) (B) I, II, III (D) II and IV

49. Which of the following properties of benzoic acid and phenol are similar ? (A) Both are weak acids and do not react with NaHCO 3 (B) Both are equally acidic in nature

(C) Both liberate hydrogen on reacting with metallic sodium

(C) Sodium phenate ⎯→ Anisol

(D) Both form esters

(D) None of these

50. The electrophilic substitution in phenol takes place at— (A) Orthoposition (B) Metaposition (C) Paraposition (D) Both ortho and para positions 51. Anisole is the reaction product of phenol and dimethyl sulphate. This reaction is termed as— (A) Coupling (B) Esterification

+ CH3I

(Continued from Page 708 )

–HI

56. Phenol on reacting with bromine dissolved in carbon disulphide (CS2), gives— (A) (B) (C) (D)

m-bromophenol only o-bromophenol Both o- and p-bromophenol 2, 4, 6-tribromophenol

57. When phenetole reacts with HI, which is formed ? (A) C6H5OH and C2H5I (B) C2H5OH and C6H5OH (C) C2H5OH and iodobenzene (D) Ethane and benzene

(C) Etherification (D) None of these 52. Which of the following reactions is given by phenol ? (A) Williamson's reaction (B) Coupling reaction (C) Lederer-Manasse reaction (D) All of these 53. The reaction OH OH– or H

+ HCHO ⎯⎯→ + OH

OH C H 2O H +

C H 2O H is called— (A) Reimer-Tiemann reaction (B) Sand Meyer reaction (C) Lederer-Manasse reaction

58. Which is not true statement ? (A) Paracetamol is parahydroxy derivative of acetanilide (B) Salol is phenyl salicylate (C) Phenyl esters on heating with AlCl3 give cresols (D) Benzoylation of phenol gives phenyl benzoate 59. Phenol gives novolak on reacting with— (A) CH3CHO in presence of acid at low temp. (B) HCHO in presence of acid at low temp. (C) HCOOH in presence of NaOH at low temp. (D) Formalin in presence of alkali at high temp. 60. The amount of bromine required to convert 1·0 g of phenol into 2, 4, 6-tribromophenol, is— (A) 2·5 g (B) 5·1 g (C) 10·22 g (D) 15·3 g

(D) Kolbe reaction 54. A compound ‘A’ when treated with CH 3OH and a few drops of H2SO4, gave a smell of winter green. The compound ‘A’ is—

ANSWERS

(A) Succinic acid (B) Phenol (C) Salicylic acid (D) Salicylaldehyde 55. Schotten-Baumann reaction is— NaOH

(A) Phenol + CHCl3 ⎯⎯→ Salicylaldehyde (B) Phenol + Benzoyl chloride NaOH

⎯⎯→ Phenyl benzoate C.S.V. / August / 2009 / 722

●●●

●●●

16. At 25°C, the dissociation constant of a base, BOH, is 1 × 10–12. The concentration of hydroxyl ions in 0·01 M aqueous solution of the base would be— (A) 1·0 × 10– 5 mol L– 1 1. 60 gm of a compound on analysis produce 24 gm carbon, 4 gm hydrogen and 32 gm oxygen. The empirical formula of the compound is— (A) CH2O2 (B) CH2O (C) H2 (D) CH4 2. The energy of second Bohr orbit of the hydrogen atom is – 328 kJ mol– 1; hence the energy of fourth Bohr orbit would be— (A) – 41 kJ mol– 1 (B) – 82 kJ mol– 1 (C) – 164 kJ mol– 1 (D) – 1312 kJ mol– 1 3. The number of neutrons in the element 4Be9 is— (A) 3 (B) 5 (C) 7 (D) 9 4. Which of the following is the electron deficient molecule ? (A) C2H6 (B) B2H6 (C) SiH4 (D) PH 3 5. 8·2 litre of an ideal gas weight 9·0 gm at 300 K and 1 atm pressure. The molecular mass of the gas is— (A) 9 (B) 18 (C) 25 (D) 36 6. The correct order in which the O–O bond length increases in the following is— (A) O2 < H2O2 < O3 (B) O3 < H2O2 < O2 (C) H2O2 < O2 < O3 (D) O2 < O3 < H2O2 7. Both ionic and covalent bond is present in— (A) KCN (B) KCl (C) H2 (D) CH4 8. Which of the following molecules has trigonal planar geometry ? (A) BF 3 (B) NH3 (C) PCl3

(D) IF3

9. The pH of pure water at 80° will be— (A) > 7 (B) < 7 (C) = 7 (D) None of these

C.S.V. / August / 2009 / 723

10. The number of moles of KMnO 4 reduced by one mole of KI in alkaline medium is— (A) One (B) Two (C) Five

(D) One fifth

11. 2 Ag + 2 H2SO4 → Ag2SO4 + 2 H2O + SO2 In the above reaction H2SO4 acts as a— (A) Catalyst

(B) 1·0 × 10– 6 mol L– 1 (C) 2·0 × 10– 6 mol L– 1 (D) 1·0 × 10– 7 mol L– 1 17. The function of moderator in a nuclear reactor is— (A) To produce more neutrons (B) To stop nuclear reaction (C) To increase speed of neutron (D) To slow down speed of neutron

(B) Oxidising agent (C) Reducing agent (D) Acid as well as oxidant 12. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20°C are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be— (A) 0·200

(B) 0·549

(C) 0·786

(D) 0·478

13. N2(g) + 3H2(g) → 2 NH3(g) For the above reaction the correct statement is— (A) ΔH < ΔE (B) ΔH = ΔE

18. For a first order reaction A → B the reaction rate at reactant concentration of 0·01 M is found to be 2·0 × 10 – 5 mol L– 1 s– 1. The half -life period of the reaction is— (A) 30 s

(B) 220 s

(C) 300 s

(D) 347 s

19. A colloidal solution is subjected to an electric field. The colloidal particles move towards anode. The coagulation of the same solution studied using NaCl, BaCl 2 and AlCl 3 solutions. The coagulation power is in the order of— (A) BaCl 2 > NaCl > AlCl3

(C) ΔH > ΔE (D) None of these

(B) NaCl > BaCl2 > AlCl3

14. Equilibrium constants K1 and K2 for the following equilibria :

(C) AlCl3 > BaCl2 > NaCl

K1 1 NO(g) + O 2 2

and 2NO 2(g)

K2

NO2(g) 2NO(g) + O2(g)

are related as— 1 (A) K2 = 2 (B) K2 = K12 K1 (C) K2 =

1 K1

(D) K2 =

K1 2

15. Variation of heat of reaction with temperature is known as— (A) Kirchhoff’s equation (B) van’t Hoff’s isotherm (C) van’t Hoff’s isochore (D) None of the above

(D) NaCl > AlCl3 > BaCl2 20. A reaction occurs spontaneously if— (A) TΔS < ΔH and both ΔH and ΔS are + ve (B) TΔS > ΔH and ΔH is + ve and ΔS is – ve (C) TΔS > ΔH and both ΔH and ΔS are + ve (D) TΔS = ΔH and both ΔH and ΔS are + ve 21. The pH of the solution obtained by mixing 40 ml of 0·10 M HCl with 10 ml of 0·45 M of NaOH is— (A) 4 (B) 8 (C) 12 (D) 14

22. The correct order of acid strength is— (A) HClO4 < HClO3 < HClO2 < HClO (B) HClO < HClO2 < HClO3 < HClO4 (C) HClO4 < HClO < HClO2 < HClO3 (D) HClO2 < HClO3 < HClO4 < HClO 23. 50 ml of H2 SO4 require 10 gm CaCO3 for complete decomposition. The normality of acid is— (A) 2 (B) 0·30 (C) 4 (D) 0·20 24. A solution of urea (mol. mass 56 g mol– 1 ) boils at 100·18 °C at the atmospheric pressure. If Kf and Kb for water are 1·86 and 0·512 K kg mol– 1 respectively, the above solution will freeze at— (A) 0·654°C (B) – 0·654°C (C) 6·54°C

(D) – 6·54°C

25. The isomerism represented by ethyl acetoacetate is— (A) Keto-enol isomerism (B) Geometrical isomerism (C) Enantiomerism (D) Diastereoisomerism 26. Which one of the following forms micelles in aqueous solution above certain concentration ? (A) Dodecyl trimethyl ammonium chloride (B) Glucose (C) Urea (D) Pyridinium chloride 27. The compound known as oil of wintergreen is— (A) Phenyl acetate (B) Phenyl salicylate (C) Methyl salicylate (D) Methyl acetate 28. Names of some compounds are given. Which one is not the correct IUPAC name ? (A) CH3—CH2—CH2 CH3 | —CH—CH—CH2CH3 | CH2CH3 3-methyl-4-ethyl heptane

(B) CH3—CH —CH —CH3 | | OH CH3 3-methyl-2-butanol

C.S.V. / August / 2009 / 724

(C) CH3—CH2—C⎯ CH—CH3 || | CH2 CH3 2-ethyl-3-methyl-but-1-ene

(D) CH3—C ≡ C—CH—(CH3)2 4-methyl-2-pentyne

29. C3H9N cannot represent— (A) 1° amine (B) 2° amine (C) 3° amine (D) Quaternary ammonium salt 30. Which one of the following pairs represents stereoisomerism ? (A) Structural isomerism and geometrical isomerism (B) Optical isomerism and geometrical isomerism (C) Chain isomerism and rotational isomerism (D) Linkage isomerism and geometrical isomerism

(C) Phosphate group (D) Deoxyribose group 36. The best method for the separation of naphthalene and benzoic acid from their mixture is— (A) Distillation (B) Sublimation (C) Chromatography (D) Crystallisation 37. The first ionisation potential of Na, Mg, Al and Si are in the order— (A) Na < Mg < Al < Si (B) Na < Al < Mg < Si (C) Na > Mg > Al > Si (D) Na < Si < Al < Mg 38. The major organic product formed from the following reaction : O

(iii) H 2O

(A)

BHC Hexachlorobenzene Chlorobenzene Triphenyl phosphate

(B) (C)

32. Which amongst the following is the most stable carbocation ? +

+

(A) CH3 +

(C) CH3—CH | CH3

(B) CH3CH2 CH3 | (D) CH3—C+ | CH3

33. Hydrolysis of sucrose is— (A) Inversion (B) Hydration (C) Saponification (D) Inhibition

H NCH3 OH H NCH3 OH O—NHCH3

(D) NCH 3 H 39. Which is not present in chlorophyll ? (A) Carbon (B) Calcium (C) Magnesium (D) Hydrogen 40. Which functional group participates in disulphide bond formation in proteins ? (A) Thioester (B) Thioether (C) Thiol (D) Thioacetone

41. 34. Products of the following reaction— (i) O 3 CH3–C ≡ C–CH 2–CH3 ⎯⎯⎯⎯⎯→ (ii) Hydrolysis 42. are— (A) CH3COOH + CO2 (B) CH3COOH + HOOC–CH2CH3 (C) CH3CHO + CH3CH2CHO (D) CH3COOH + CH3COCH3 35. The pairs of base in DNA are held together by— (A) Ionic bonds (B) Hydrogen bond

(ii) LiAlH

is—

31. The main product obtained from phenol with PCl 5 is— (A) (B) (C) (D)

(i) CH3NH2 4 ⎯⎯⎯→ …

The molecular formula of cryolite is— (A) Na 2F.AlF6 (B) Na 3AlF6 (C) Na 3AlF5 (D) Na 2AlF3 The monomer of the polymer; CH3 CH3 | CH2—C—CH2—C⊕ CH3 | CH3 is— CH3 (A) H2C = C CH3 (B) CH3CH = CHCH3

(C) CH3CH = CH2 (D) (CH3)2C = C(CH3)2 43. Aluminium appears when mixed with—

like gold

(A) 90% Cu

(B) 75% Ni

(C) 80% Sn

(D) 80% Co

44. The aqueous solution containing which one of the following ions will be colourless ? (Atomic No. : Sc = 21, Fe = 26, Ti = 22, Mn = 25) (A) Sc 3+ (B) Fe 2+ 3+ (C) Ti (D) Mn2+ 45. Lithophone is a mixture of— (A) CuSO 4 + ZnS (B) CaSO 4 + ZnS (C) BaSO4 + CaSO4 (D) BaSO4 + ZnS 46. The main reason for larger number of oxidation states exhibited

by the actinoids than the corresponding lanthanoids, is— (A) More energy difference between 5f and 6d orbitals than between 4 f and 5d orbitals (B) Lesser energy difference between 5f and 6d orbitals than between 4 f and 5d orbitals (C) Larger atomic size of actinoids than the lanthanoids (D) Greater reactive nature of the actinoids than the lanthanoids 47. By passing air over red hot coke the gas obtained is— (A) Coal gas (B) Water gas (C) Oil gas (D) Producer gas 48. Which one of the following is an inner orbital complex as well as diamagnetic in behaviour ? (Atomic number : Zn = 30, Cr = 24, Co = 27, Ni = 28) (A) [Zn (NH3)6] 2+ (B) [Cr (NH3)6] 3+

ANSWERS WITH HINTS

C.S.V. / August / 2009 / 725

(C) [Co(NH3)6] 3+ (D) [Ni (NH3)6] 2+ 49. The most acidic oxide is— (A) P2O5 (B) N2O5 (C) Sb2O5 (D) As 2O5 50. H2S gas when passed through a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because— (A) Presence of HCl decreases the sulphide ion concentration (B) Solubility product of group II sulphides is more than that of group IV sulphides (C) Presence of HCl increases the sulphide ion concentration (D) Sulphides of group IV cations are unstable in HCl

(C) A decrease in enthalpy (D) A decrease in internal energy

…………

(a)

(b)

……………………………… FeCl 3

(C)

m A Z

→

m–4 B Z–2

+ 24He

A and B are isodiapheres (D) All correct 7. Which curve represents zero order reaction ?

(A)

↑

13. Which arrangement of electrons lead to anti-ferromagnetism ?

[A]

K 4 [Fe(CN) 6]

t →

(A) ↑↑↑↑

5 pm

Fe 4[Fe(CN)6], the blue colour will be noticed in— (A) (a) (B) (b) (C) In both (A) and (B) (D) Neither in (A) nor in (B) 2. Volume of a mixture of 6·02 × 1023 oxygen atoms and 3·01 × 10 23 hydrogen molecules at STP is— (A) 28·0 litre (B) 33·6 litre (C) 11·2 litre (D) 22·4 litre 3. The molal freezing point constant of water is 1·86 K molality– 1. If 342 g of cane-sugar (C 12H22O11) are dissolved in 1000 g of water, the solution will freeze at— (A) – 1·86°C (B) 1·86°C (C) – 3·92°C (D) 2·42°C 4. The speed of electron in the first orbit of hydrogen atom in the ground state is— [ c = velocity of light] c c (B) (A) 1·37 1370 c c (C) (D) 13·7 137 5. ZSM –5 is used to convert— (A) Alcohol to petrol (B) Benzene to toluene (C) Toluene to benzene (D) Heptane to toluene 6. Which among the following is wrong about isodiapheres ? (A) They have the same difference of neutrons and protons or same isotopic number (B) Nucleide and its decay product after α emission are isodiapheres

C.S.V. / August / 2009 / 727

(B)

↑

(B) ↑↓↑↓

[A]

(C) Both (A) and (B) (D) None of these

t →

↑ (C) [A] t →

↑ (D) [A] t →

8. If one mole of monoatomic gas 5 γ= is mixed with one mole 3 7 γ= the of diatomic gas 5

( )

( )

value of γ for the mixture is— (A) 1·40 (C) 1·53

(B) 1·50 (D) 3·07

9. A certain weak acid has a dissociation constant 1·0 × 10 – 4 . The equilibrium constant for its reaction with a strong base is— (A) 1·0 × 10– 4 (B) 1·0 × 10–10 (C) 1 × 1010

(D) 1·0 × 10–14

10. If glycerol and methanol were sold at same price in the market, which would be cheaper for preparing an artifreezer solution for the radiator of an automobile ? (A) Glycerol (B) Methanol (C) Both equal (D) None of these 11. When a solid melts, there is— (A) An increase in enthalpy (B) No change in enthalpy

14. A particular reaction at 27°C for which ΔH > 0 and ΔS > 0 is found to be nonspontaneous. The reaction may proceed spontaneously if— (A) Temperature is decreased (B) Temperature is kept constant (C) Temperature is raised (D) It is carried out in open vessel at 27°C 15. The correct order of relative stability of half filled and completely filled shells is— (A) p 3 < d 5 < d10 < p6 (B) d 5 < p3 < d10 < p6 (C) d 5 < p3 < d10 < p6 (D) p3 < d10 < d 5 < p6 16. For the reaction A + B → C + D the variation of the concentration of product is given by the curve— Y

↑

Z

Concentration

1. Two solutions (a) containing FeCl3 (aq) and (b) containing K4[Fe (CN)6] are separated by semipermeable membrane as shown below. If FeCl3 on reaction with K4 [Fe (CN) 6] produces blue colour of—

12. Fog is a colloidal system in which— (A) Liquid particles dispersed in gas (B) Gaseous particles dispersed in liquid (C) Solid particles dispersed in liquid (D) Solid particles dispersed in gas

W

X Time →

(A) X (C) Z

(B) Y (D) W

17. Water glass is— (A) Another name for sodium silicate (B) A special form of glass to store water only

(C) Hydrated form of glass (D) Hydrated silica 18. Which one is correct representation for 2 SO 2 (g) + O2 (g) 2 SO3 (g) ? [P SO3]2 (A) Kp = [P SO2]2 [PO2] (B) Kc (C) Kp

[SO 3] 2 = [SO 2] 2 [O2] [ nSO3]2 = [ nSO2]2 [nO2] ×

[

]

P Total moles

–1

(D) All correct

30. An orange coloured when solution acidified with H2SO4 treated with a substance X gives a blue coloured solution of CrO5. The substance X is— (A) H2O (B) dil. HCl (C) H2O2 (D) Conc. HCl

24. At 25°C, the solubility of iodine in water is 0·35 g/l. If distribution coefficient of I 2 between CS 2 and water is 600, the solubility of I2 in CS 2 in g/l will be about—

31. In the reaction, C 6H5CH3

(A) 1714 (C) 569·6

(B) 210 (D) 857

25. The reaction described below is—

19. Super halogen is— (A) F2 (B) Cl2 (C) Br2 (D) I 2 20. In the following reaction [Cu (H2O)3 (OH)]+ + [Al (H2O)6] 3+ (a)

CH 3 (CH 2)5

OH– C—Br ⎯→

H3C H

(CH 2)5 CH 3 OH—C

CH 3

(b)

→ [Cu(H2O)4] 2+ + [Al (H2O)5 (OH)]2+ (c)

23. 0·2 g of an organic compound containing C, H and O, on combustion yielded 0·147 g CO 2 and 0·12 g water. The percentage of oxygen in it is— (A) 73·3% (B) 78·45% (C) 83·23% (D) 89·50%

(d)

(A) (a) is an acid and (b) is a base (B) (a) is a base and (b) is an acid (C) (c) is conjugate acid of (a) and (d) is conjugate base of (b) (D) (c) is conjugate base of (a) and (d) is conjugate acid of (b) 21. A complex shown below can exhibit—

H (B) SN2 (D) SE 2

(A) SE 1 (C) SN1

26. Chloride ion and potassium ion are isoelectronic— (A) Their sizes are same (B) Cl– ion is bigger than K + ion (C) K+ ion is relatively bigger (D) Their sizes depend on other cation and anion O 27. The polymer H2C

CH 2

O

O CH 2

(A) Optical isomerism only (B) Geometrical isomerism only (C) Both optical and geometrical isomerism (D) None of these 22. In the chemical reaction Ag2O + H2O + 2e → 2Ag + 2OH– (A) (B) (C) (D)

Water is oxidised Electrons are reduced Silver is reduced Silver is oxidised

C.S.V. / August / 2009 / 728

is obtained when HCHO is allowed to stand. It is a white solid. The polymer is— (A) Trioxane (B) Formose (C) Para formaldehyde (D) Metaldehyde 28. The function of flux during the smelting of the ore is— (A) To make the ore porous (B) To remove gangue (C) To facilitate reduction (D) To facilitate oxidation 29. RMgX on heating with cyanogen chloride gives— (A) R—NC (B) R—Cl (C) R—CN (D) None of these

Oxidation NaOH ⎯⎯⎯⎯→ A ⎯⎯→ B Soda lime C, the product C is— ⎯⎯⎯⎯→

Δ (A) C6H5OH (B) C6H6 (C) C6H5COONa (D) C6H5ONa

32. In the purification of bauxite by Bayer’s process, the chemical used is— (A) Na 2CO3 (B) Cryolite (C) NaOH (D) A mixture of NaOH and Na 2CO3 33. Toilet soap is— (A) A mixture of calcium and sodium salts of higher fatty acids (B) A mixture of potassium stearate and glycerol (C) A mixture of sodium salts of higher fatty acids (D) A mixture of potassium salts of higher fatty acids 34. When concentrated nitric acid is heated it decomposes to give— (A) Oxygen and nitrogen (B) Nitric oxide (C) Oxygen (D) Nitrogen dioxide and oxygen 35. Which of the following is an example of zwitter ion ? (A) Urea (B) Glycine hydrochloride (C) Ammonium acetate (D) α-alanine 36. When SO2 is passed through cupric chloride solution— (A) The solution becomes colourless and a white precipitate of Cu 2Cl2 is obtained (B) A white precipitate is obtained (C) The solution becomes colourless (D) No visible place

change takes

37. Indigo belongs to the class of— (A) Mordant dye (B) Vat dye (C) Direct dye (D) Disperse dye 38. 0·5 g of an organic compound was Kjeldahlised and the NH 3 evolved was absorbed in certain volume of 1N H 2SO4. The residual acid required 60 cm3 of N/2 NaOH. If the percentage of nitrogen is 56 then the volume of 1N H2SO4 taken was— (A) 30 ml

(B) 40 ml

(C) 50 ml

(D) 60 ml

39. Equal weights of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is— (A) 1/3 (C) 2/3

(B) 1/2 1 273 (D) × 3 298

40. Which of the following is the least stable carbanion ? (A) HC ≡ C– (B) (C6H5)3C– – (C) (CH3)3C (D) CH3– 41. The values of vander Waals constant ‘ a’ for the gases O2, N 2, NH3 and CH4 are 1·360, 1·390, 4·170 and 2·253 L2 atm mol– 2

respectively. The gas which can most easily be liquefied is— (A) O2 (B) N2 (C) NH3 (D) CH4

(D)

42. In presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markownikov’s addition to alkenes because— (A) Both are highly ionic (B) One is oxidising and the other is reducing (C) One of the steps in both the reactions is endothermic (D) All the steps are exothermic in both the cases

47. Grignard’s reagents add to— — (A) C — —O (B) —C — —N (C) C — —S (D) All of these

43. The number of P—O—P bonds in cyclic metaphosphoric acid is— (A) Zero (B) Two (C) Three (D) Four

49. Number of σ and π-bonds in benzaldehyde is—

44. Which among MeX, RCH2X, R2CHX, R3 CX is most reactive towards SN2 reaction ? (A) MeX (B) RCH2X (C) R2CHX (D) R3CX

(B) 4π-bonds and 8 σ-bonds

45. Which one has the pyramidal shape ? (A) PF 3 (B) CO32 – (C) SO3 (D) BF 3 46. Which is not a moth repellent ? (A)

Cl

Cl

(B) Perchloroethane

ANSWERS WITH HINTS

C.S.V. / August / 2009 / 729

(C) CHCl3

48. Which of the following is most acidic ? (A) Phenol (B) m-chlorophenol (C) Benzyl alcohol (D) Cyclohexanol

(A) 4π-bonds and 13 σ-bonds (C) 4π-bonds and 14 σ-bonds (D) 8π-bonds and 10 σ-bonds 50. An organic compound X, on treatment with acidified K2Cr2O7 gives compound Y which reacts with I2 and sodium carbonate to form triiodomethane. The compound X can be— (A) CH3OH (B) CH3COCH3 (C) CH3CHO (D) CH3CHOHCH3

Introduction ●

●

●

The formation of gametes usually takes place in specially restricted parts of the body which are known as gonads (i.e., Ovary in female and testis in male). The lining of sex organs (gonads) is termed as germinal epithelium which possesses power of division and produces the cells which undergo reduction to form the gametes. The process by which gametes are produced in the gonads is known as gametogenesis. The process of formation of male sex cells or male gametes is called spermatogenesis and the products are known as spermatids. The maturing of ova from the cells of ovary is termed oogenesis and the products are known as ova or ootids.

(b) Metamorphosis of spermatids into sperm, the spermiogenesis or spermatoleosis.

Formation of Spermatids ●

(i)

Spermatogenesis ●

The process of maturation of reproductive cells in the testis of male so as to form male gametes or sperm is known as spermatogenesis. Blood vessels, lymphatic vessels and nerves pass alongside the vas deferens to supply the testis.

The formation of spermatids starts from the primary germ cells or primordial germ cells of the testis and is differentiated into three phases : Multiplication phase—The testes are composed of seminiferous tubules which are lined by germinal epithelium. Some of the cells of germinal epithelium get differentiated as primary germ cells or primordial germ cells. These become comparatively bigger in size and possess a prominent nucleus. During multiplication phase, the primordial germ cells multiply by mitosis and the ultimate products are known as spermatogonia. Each spermatogonium represents a diploid cell consisting of 2n-number of chromosomes.

Seminiferous tubules

Vas deferens Epididymis Interstitial cells produce testosterone

Tubule network Compartment

Lumen of seminiferous tubule

Sperm Spermatid Secondary spermatocyte Primary spermatocyte Spermatogonium A tubule network connects seminiferous tubules in compartments to the epididymis.

Sertoli cells nourish and protect sperm during their development

Fig. : Sectioned view of testis showing seminiferous tubules and epididymis. ●

The process is much complicated and can be studied under two heads : (a) Formation of spermatids from the cells of germinal epithellium produced as a result of meiotic division.

C.S.V. / August / 2009 / 733

(ii) Growth phase—The spermatogonia thus formed after repeated divisions from primary germ cells do not undergo further division by mitosis but prepare themselves for maturation division. These increase in size by the accumulation of

nourishing material obtained from germinal cells. The enlarged cells are known as primary spermatocytes. The nucleus of each primary spermatocyte is of ordinary size but soon enlarges considerably so that in a mature primary spermatocyte the nucleus is much larger than that of the spermatogonial cell. (iii) Maturation phase—Each diploid primary spermatocyte goes through the first meiotic division (reduction division) with its long drawn out prophase. The pairing and splitting of homologous chromosomes takes place and leads to the formation of tetrad. Crossing over exchange of homologous. Chromosomes also takes place. As a result, halving of the chromosome number takes place in the first maturation division and the two cells formed are haploid having n-chromosomes. These are known as secondary spermatocytes.

sperm cell. The spermatid contains mitochondria, Golgi body and centriole, but with one haploid set of chromosomes. In this form it is not capable of functioning as a male gamete because the male gamete is motile and has to reach the ovum. ●

During spermiogenesis, the nucleus shrinks by losing water from the nuclear sap and the chromosomes become closely packed.

●

The nucleus changes from its usual spherical form and becomes elongated and narrow.

●

The centrosome of the spermatid consists of two centrioles. During spermiogenesis, these move and come to lie behind the nucleus. One of them enters the depression developed in the posterior part of the nucleus and is called the proximal centriole.

●

The other one is called the distal centriole. It occupies a position behind the proximal one with its axis coinciding with the longitudinal axis of the spermatozoon. This distal centriole gives rise to the axial filament of the flagellum and act as basal granule.

●

Mitochondria from different parts of the spermatid accumulate around the proximal part of axial filament and distal centriole in the middle piece of the spermatozoon.

●

The Golgi complex of the spermatid forms the acrosome of the spermatozoon and forms a cap around the nucleus. It helps the sperm in penetration into the ovum.

Structure of Spermatozoon

Fig. : The stages of spermatogenesis ●

●

The secondary spermatocytes undergo 2nd maturation division which is simple mitotic division. As a result, each secondary spermatocyte is divided into equal cells which are known as spermatids. Each spermatid consists only one set of homologous pairs and is, therefore, haploid. It undergoes metamorphosis to form the sperm. As a result of these two consecutive divisions, four haploid spermatids are formed from each spermatocyte.

Spermiogenesis or Spermatoleosis ●

Spermiogenesis is the gradual differentiation of stationary rounded spermatid into an active motile

C.S.V. / August / 2009 / 734

●

A typical spermatozoon consists of head, middle piece and tail.

●

Head is the anterior most part of the spermatozoon. It is often conical but its shape varies considerably. The anterior tip of the head is formed of acrosome which enables the spermatozoon to penetrate through the egg membrane.

●

Acrosome contains hyaluronidase enzymes and hydrolases, released when the sperm cell-membrane fuses at several points with the acrosome during acrosome reaction, dissolving the jelly around the egg so that sperm can penetrate it.

●

The remaining portion of the head is occupied by the nucleus, whose posterior margin is depressed to accommodate the proximal centriole, which initiates cell-division in the fertilized egg.

●

Middle piece of the spermatozoon encloses distal centriole and mitochondria, which carry oxidative enzymes which provides energy for the propulsion of the spermatozoon.

●

Tail or flagellum is the longest but most narrow part of the spermatozoon and its movements propel the spermatozoon. It consists of an axial filament, a thin layer of cytoplasm and an outer sheath of plasmalemma.

C.S.V. / August / 2009 / 735

Oogenesis Immature ovum (primary oocyte) in primary follicle

The oocyte completes meiosis while the follicle enlarges and matures.

Secondary follicle

Mature graafian follicle Immature ovum (secondary oocyte)

Germinal epithelium

Fluid-filled cavity

Rupturing of follicle during ovulation

Ovulated immature ovum (secondary oocyte)

Degenerating corpus luteum Blood vessels, nerves and lymphatic vessels service the ovary.

Mature corpus luteum

Developing corpus luteum

Fig. : Sectioned view showing primary secondary and graafian follicles in which oocytes form through meiosis. ●

●

Oogenesis is the process of development and maturation of ova from the primary germ cell produced by the division of cells of the germinal epithelium in the ovary of female organism. The process includes three steps or phases—Multiplication phase, Growth phase and Maturation phase.

ducts are known as oocytes. These are diploid cells containing the same number of chromosomes as in the parent body. The oocytes cease to divide and enter the growth phase. ●

Growth phase of female gametes is much prolonged. The growth in size of the oocyte is not only due to increase in the amount of its cytoplasm but mostly due to accumulation of nutritive material. The most common form of food storage consists of granules of yolk. Yolk consists of proteins, phospholipids and neural fat. The follicle cells assist the growth of oocyte by secreting substances which are taken up by the oocyte.

●

In the primary oocyte, large amount of fat and proteins become accumulated in the form of yolk and due to its heavy weight, it is usually concentrated towards the lower portion of egg, forming the vegetal pole. The portion of the cytoplasm containing the egg pronucleus remains separated from the yolk and occurs towards the upper side of egg forming the animal pole. During the growth phase, tremendous changes occur in the nucleus of the primary oocyte.

Multiplication phase is the first phase and similar to that found in spermatogenesis. The cells of germinal epithelium divide and produce oogonia. These undergo repeated mitotic divisions and the final proStages

Cells

Nuclear and cell divisions Chromosomes in each cell

Oogonium

46

Proliferation of cells by mitosis before birth Primary oocytes in primary follicles

46 After puberty, one primary oocyte matures and is ovulated about once a month until menopause.

Enlarged primary oocyte Meiosis I Polar body

Secondary oocyte in secondary and graafian follicles

Meiosis II

23 Mature ovum

Fig. : The stages of oogenesis

C.S.V. / August / 2009 / 736

Nuclear Changes

The secondary oocyte is released at 23 ovulation and does not complete meiosis II unless it is fertilized by a sperm.

Polar body Polar bodies disintegrate.

46 Diploid Haploid

●

Simultaneously with the growth of oocyte, its nucleus enters into the prophase of meiosis. The homologous chromosomes pair but further steps are postponed until the end of growth period.

●

The nucleus increases in size by the formation of nuclear sap. It is called the germinal vesicle. The nucleoli are greatly increased to compensate the large amount of RNA required for the increased metabolism of growing oocyte.

Differences between Spermatogenesis and Oogenesis Spermatogenesis

Oogenesis

1. It occurs in the testes.

1.

It occurs in the ovaries.

2. Growth phase is short so that spermatocytes are only twice the size of spermatogonia.

2.

Growth phase is very long so that oocytes are much larger than oogonia.

3. Spermatocytes have cyto- 3 . Oocytes have cytoplasm plasm and nucleus with rich in RNA, ATP, enzynormal contents. mes and yolk; and nucleus with giant chromosomes and large nucleoli. 4. A primary spermatocyte divides equally to form two similar secondary spermatocytes.

4.

A primary oocyte divides unequally to form one large secondary oocyte and one minute polar body.

5. A secondary spermatocyte also divides equally, forming two similar spermatids.

5.

A secondary oocyte also divides unequally, forming one large ootid and one minute polar body.

6. Upto the formation of spermatids, the cells often remain interconnected.

6.

Oogonia are separate and surrounded by follicle cells.

7. A spermatogonium pro- 7. duces four functional spermatozoa.

An oogonium produces one functional ovum and three non-functional polar bodies.

8. Spermatozoa are minute, streamlined, yolkless and motile.

8.

Ova are much larger, rounded, often with yolk and non-motile.

9. Spermatogenesis is often a continuous process.

9.

Oogenesis is not necessarily a continuous process.

10. Spermatogenesis is usually completed in the testes so that mature sperms are released.

●

10. Oogenesis is often completed in the reproductive tract of the female or even in water as oocytes leave the ovaries.

Maturation phase—A full grown primary oocyte undergoes first meiotic division and produces two daughter cells. The division of its cytoplasm is unequal. One daughter cell is extremely small, whereas the other is almost as large as the primary oocyte

●

●

itself. This large cell is called secondary oocyte. It receives almost whole of the cytoplasm of the primary oocyte. The small cell is known as the first polar body. It consists almost a bare nucleus. But both the cells have haploid (n) number of chromosomes. The haploid secondary oocyte and first polar body pass through the second maturation division. Due to second maturation division, the secondary oocyte forms a mature egg and a second polar body. By the second maturation division, the first polar body also divides into two secondary polar bodies. These polar bodies ooze out from the egg and degenerate, while the haploid egg cell or secondary oocyte is released from the ovary for fertilization. When it is released from the ovary, the secondary oocyte is surrounded by a protective sphere of cells called the corona radiata which is derived from the follicle in the ovary. The size of the corona radiata and its irregular surface makes it easier for the entire assembly to be trapped by the uterine tube.

Key Points ● ● ● ●

●

● ●

●

●

Amount of yolk in the ooplasm varies from species to species. The eggs with very little amount of yolk are known as the microlecithal eggs, e.g. , Amphioxus, Eutherian mammals. The eggs containing moderate amount of yolk are called mesolecithal eggs, e.g., Petromyzon, Dipnoi amphibia. The eggs with large amount of the yolk are known as macrolecithal eggs, e.g. , Myxine, Cartilaginous and bonyfishes, reptiles, birds and monotremata. The eggs with evenly distributed yolk contents in the ooplasm are known as Homolecithal eggs, e.g., eggs of Echinoderms. The eggs in which the yolk is not evenly distributed in the ooplasm are known as Heterolecithal eggs. When in heterolecithal eggs, the amount of yolk is concentrated in the one half of the egg to form the vegetative pole of the egg; then this condition is known as telolecithal. Telolecithal eggs may be moderately telolecithal ( e.g., Amphibians) or highly telolecithal (e.g., Hen). In macrolecithal and highly telolecithal eggs, the amount of yolk is very large and it occupies the largest portion except a small disc-shaped portion of the cytoplasm. The cytoplasm contains the zygote nucleus and is known as the germinal disc, e.g., eggs of fishes, reptiles and birds. In the centrolecithal eggs, the yolk accumulates in the centre of the ooplasm, e.g. , Insects.

OBJECTIVE QUESTIONS 1. In spermiogenesis, the spermatozoa are produced from— (A) Spermatids (B) Primary spermatocytes (C) Spermatogonia (D) Secondary spermatocytes 2. Between the spermatogonia are the— (A) Epithelial cells

C.S.V. / August / 2009 / 737

(B) Cells of Sertoli (C) Lymph spaces (D) Capillaries 3. The minute cells which are separated from the developing over during their maturation phase, are called— (A) Primary oogonia (B) Secondary oogonia

(C) Polar bodies (D) None of these 4. Spermatogenesis requires the presence of— (A) Fructose (B) Progesterone (C) Testosterone (D) Thyroxine

(Continued on Page 765 )

Introduction ●

Chordates are bilateral and deuterostomial eucoelomate eumetazoa, basically possessing, in the embryo or throughout life a flexible but firm supporting skeletal rod, called notochord.

●

Phylum Chordate is the largest of the deuterostome phyla. It is the highest and the most important phylum comprising a vast majority of living and extinct animals including humans.

●

●

●

Most of the living chordates are well known familiar vertebrate animals such as the fishes, amphibians, reptiles, birds and mammals. Besides, they include a number of marine forms such as the tunicates and lancelets(amphioxus). The chordates are probably the most conspicuous and the best-known group in the entire animal kingdom, partly because of their large size and partly because of the important role they perform in their ecosystems. They are of primary interest because humans himself are member of the group. From a purely biological viewpoint, chordates are interesting because they illustrate well the broad biological principles of evolution, development and relationship.

Pharyngeal gill slits

Notochord

Mouth Pharynx

All the chordates possess three outstanding unique characteristics at some stage in their life cycle. These three fundamental morphological features include: 1. A dorsal hollow or tubular nerve cord. 2. A longitudinal supporting rod-like notochord. 3. A series of pharyngeal gill slits.

1. Dorsal hollow nerve cord—The central nervous system of the chordates is present dorsally in the body. It is in the form of a longitudinal, hollow or tubular nerve cord lying just above the notochord and extending lengthwise in the body. ●

The nerve cord or neural tube is derived from the dorsal ectodermal neural plate of the embryo and encloses a cavity or a canal, called neurocoel.

●

Nerve cord persists throughout life in the lower chordates, but in the higher chordates, it is surrounded or replaced partly or fully, in adults by a jointed vertebral column.

●

In vertebrates, the anterior region of nerve cord is specialized to form a cerebral vesicle or brain which is enclosed by a protective bony or cartilaginous cranium. The posterior part of nerve cord becomes the spinal cord and protected within the vertebral column.

C.S.V. / August / 2009 / 738

Anus Trunk

Hepatic caecum

Tail

Digestive tube

Fig. : Diagrammatic side view of a chordate showing the three fundamental chordate characters.

2. Notochord or chorda dorsalis—The notochord is an enlarged rod-like flexible structure extending the length of the body. It is present immediately beneath the nerve cord and just above the alimentary canal. ●

Notochord originates from the endodermal roof of the embryonic archenteron.

●

Structurally, it is composed of large vacuolated notochordal cells containing a gelatinous matrix and surrounded by an outer fibrous and an inner elastic sheath.

●

The notochord is the prime diagnostic feature of the phylum chordata which derives its name from it.

●

Notochord is present at some stage in all chordates. In most vertebrates, it occurs complete only in embryo, but remnants may persist between the vertebrae, which obliterate it.

●

The notochord may persist even in adult cephalochordates.

The Fundamental Chordate Characters ●

Dorsal hollow nerve cord

3. Pharyngeal gill slits—In all the chordates, at some stage of their life history, a series of paired lateral gill slits perforate through the pharyngeal wall of the gut behind the mouth. These are also termed as pharyngeal pouches. ●

They are seen only during embryonic development in most vertebrates. In the lower chordates, fishes and amphibian larvae, the pharyngeal pouches become functional gills.

●

In terrestrial vertebrates, the pouches are modified for various purposes.

●

In humans, the first pair of pouches become the eustachian tubes. The second pair become the tonsils, while the third and fourth pairs become the thymus gland and parathyroid.

●

The above common features appear during early embryonic life of all the chordates, but all the features rarely persist in the adult. Often they are modified or even lost in adult stages of higher chordates.

●

The notochord disappears during development in most vertebrates, the nerve cord remain in adults.

Advancement of Chordates over other Phyla

●

● Phylum chordata has some advantages over other

phyla due to certain characters : 1. Living endoskeleton—Only chordates possess a living endoskeleton. It grows in size with the rest of the body. This living endoskeleton permits greater freedom of movement. 2. Efficient respiration—The gills in aquatic chordates and the lungs in terrestrial forms, form efficient organs of respiration. 3. Efficient circulation—The circulatory system of the chordates is well developed and the blood flows freely in the respiratory organs ensuring rapid exchange of gases. 4. Centralized nervous system—A growing tendency of centralization of nervous system is found in chordates and the sensory system is modified accordingly. The advancement of nervous and sensory organs explains the great power of the chordates for adapting themselves most successfully to a variety of environments.

Major Subdivisions of Chordata ● ●

Chordata is divided into three subphyla : Urochordata, Cephalochordata and Vertebrata. The first two subphyla are groups of primitive chordates without vertebral column and are commonly called lower chordates. They are grouped together as Protochordates. Group A : Acraniata (Protochordata) or Lower Chordata 1. Exclusively marine, smallsized chordates.

Group B : Craniata (Euchordata) or Higher Chordata

1. Aquatic or terrestrial, mostly large sized vertebrates. 2. No appendages, cephali- 2. Usually 2 pairs of zation and exoskeleton. appendages, welldeveloped head and exoskeleton present. 3. Coelom enterocoelic, 3. Coelom schizocoelic, budding off from embryoarising by splitting of nic archenteron. mesoderm. 4. Notochord persistent. No 4. Notochord covered or skull, cranium and vertereplaced by a vertebral bral column. column. Skull and cranium well developed. 5. Pharynx with permanent 5. Pharyngeal gill clefts gill clefts. Endostyle prepersist or disappear sent. endostyle absent. 6. Heart chamberless when 6. Heart made of 2, 3 or 4 present. No red blood chambers. Blood concorpuscles in blood. tains R.B.C. 7. Kidneys protonephridia. 7. Kidneys meso- or metanephridia. 8. Sexes separate or united 8. Sexes separate. Only reproduction asexual as sexual reproducing, well as sexual. GonoGonoducts always preducts usually absent. sent. 9. Development indirect 9. Development indirect with a free-swimming or direct, with or withlarval stage. out a larval stage.

C.S.V. / August / 2009 / 739

● ● ● ●

● ●

The vertebrates are commonly called higher chordates. The protochordates lack a head and a cranium, so that they are known as Acraniata. Vertebrates have a distinct head and cranium, therefore, called Craniata. The vertebrates are further subdivided into Agnatha and Gnathostomata. Agnathans lack true jaws and paired appendages, include a small number of primitive but highly specialized fish-like forms, the extinct Ostracoderms and the modern Cyclostomes. All other vertebrates have true jaws and paired appendages, called Gnathostomata. A basic division of Gnathostomata recognises two super classes Pisces and Tetrapoda. The superclass Pisces includes all the fishes which are strictly aquatic forms with fins. The superclass Tetrapoda is formed by four-legged vertebrates including amphibians, reptiles, birds and mammals.

General Characteristic of Phylum Chordata 1. Aquatic, aerial or terrestial. 2. Body bilaterally symmetrical and metamerically segmented. 3. A postanal tail usually project beyond the anus at some stage and may or may not persist in the adults. 4. Exoskeleton often present in some forms. 5. A cartilaginous or bony living jointed endoskeleton present in the majority of vertebrates. 6. Body triploblastic with three germinal layers. 7. Coelomates, having a true coelom. 8. A skeletal rod, the notochord present at some stage in life cycle. 9. Pharyngeal gill slits present at some stage. 10. Digestive system complete with digestive glands.

Classification of Chordates ●

Phylum Chordata is separated into three primary subdivisions, called subphyla, based on the character of notochord. These are : 1. Subphylum—Urochordata 2. Subphylum—Cephalochordata 3. Subphylum—Vertebrata

Subphylum—Urochordata ●

Characters—The urochordates are marine, mostly sessile, filter-feeders. They have the following characters : 1. The notochord occurs only in the tail of the larva and disappears in the adult. 2. The nerve cord is present in the larva, but is replaced by a single dorsal ganglion in the adult. 3. The gill slits are numerous, persist in the adult and open into an ectoderm-lined cavity, the atrium, instead of to the exterior. There are no gills. 4. The tail does not persist throughout life. Examples—Herdmania, Doliolum, Salpa.

Subphylum—Cephalochordata ●

Characters—The cephalochordates are also marine and filter-feeders. They have the following characters : 1. The notochord extends upto the tip of snout and persists throughout life. 2. The nerve cord persists throughout life, but no brain is formed. 3. The gill slits are numerous and persist in the adult. They open in the atrium. There are no gills. 4. The body wall consists of myotomes. 5. Tail persists throughout life.

Subphylum Vertebrata The term vertebrate was introduced by Lamarck. Vertebrates show the following characters : 1. Body regions—Vertebrate body typically consists of 4 regions—head, neck, trunk and tail. The neck and tail may be lacking in some cases. Head is prominent and sense organs are well developed. 2. Segmentation—There is some degree of segmentation. It is indicated by muscles, vertebrae, ribs, paired blood vesssels and nerves. 3. Appendages—There are as a rule 2 pairs of appendages, which may be fins or limbs. One or both pairs are absent in certain forms.

8. Endoskeleton—An internal framework of cartilage or bone or both is always present. It grows with the body. The notochord is replaced partly or fully by a jointed vertebral column in the adult, hence called vertebrata. 9. Circulatory system—Heart is ventral and consists of 2 to 4 chambers. Blood contains red corpuscles that have haemoglobin. There is closed circulatory system consisting of blood vascular and lymphatic system. Hepatic portal system is present. In many vertebrates, renal portal system is also present. 10. Respiratory organs—Respiratory organs may be gills, skin in aquatic vertebrates. Higher terrestrial vertebrates have lungs for respiration. 11. Nervous system—Nerve cord is differentiated into anterior brain and posterior spinal cord. A skull develops to provide cranium for protecting the brain. 12. Sense organ—Sense organs are eyes, ears, tongue, nasal chambers and skin. 13. Cranial nerves—Cranial nerves may be 8, 10 or 12 pairs. 14. Digestive tract—Digestive tract is complete and is ventral to the central nervous system.

Outline Classification of Phylum Chordata Group Acrania or Protochordata (Lower chordata)

Classes 1. Ascidiacea 2. Thaliacea 3. Larvacea 1. Leptocardii

Subphylum Urochordata Subphylum Cephalochordata

Phylum Chordata Division Agnatha Group Craniata or Euchordata (Higher chordata)

1. Ostracodermi 2. Cyclostomata

{

Subphylum Vertebrata

Superclass Pisces

{

Anamniota (Lower vertebrata)

Superclass Tetrapoda

⎧⎪6. Amphibia Reptilia ⎨⎪7. 8. Aves ⎩9. Mammalia

Amniota (Higher vertebrata)

Division Gnathostomata

4. Integument—Skin consists of epidermis of many layers of cells and dermis of connective tissue. 5. Exoskeleton—Epidermis often produces an exoskeleton of keratinized cells. It may consist of scales, feather, hair. 6. Muscles—Three types of muscles–striped, unstriped and cardiac are present. 7. Coelom—There is a spacious true coelom that contains viscera loosely suspended in it by mesenteries.

C.S.V. / August / 2009 / 740

3. Placodermi 4. Chondrichthyes 5. Ostelchthyes

15. Endocrine system—All vertebrates have ductless endocrine glands that secrete hormones for metabolic regulation. 16. Excretory system—Vertebrates have a pair of kidneys that require a high blood pressure for working. 17. Reproductive system—The sexes are separate. They have gonoducts. Fertilization may internal or external. Asexual reproduction is lacking.

Geological Representation ●

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●

Cambrian and Ordovician periods—The first fossils of vertebrates were found in the rocks of the Ordovician period in the form of Ostracoderms. These were small jawless bony fishlike forms related to cyclostomes. This shows that their chordate ancestors must have existed much before in the late Cambrian. Silurian and Devonian periods—Some fossil fishes were found in the Silurian period and far more were present in the succeeding Devonian period which is known as 'Age of Fishes'. Ostracoderms were jawless fishes, but during Devonian, the first jawed fish—the Placoderm arose. The placoderms became extinct without leaving any living representative. It is likely that early placoderms were ancestors of cartilaginous and bony fishes both, which had true jaws.

●

Carboniferous periods—In late Devonian to early Carboniferous period, the lobed-finned fishes arrived. They were the first vertebrates to walk on land and became the primitive stem to amphibia.

●

The amphibians became abundant and mutated in many directions during carboniferous, which is usually known as the ‘Age of Amphibians’.

●

Mesozoic era—In the early carboniferous, the very primitive amphibians also gave origin to primitive reptiles. Reptiles reached their peak during Mesozoic era, which is aptly known as the ‘Age of Reptiles’. They included dianosaurs.

●

The ancestral mammals were derived from the primitive reptiles during Triassic period. The first bird also appeared in the late Jurassic period and one of the fossil Archaeopteryx had both reptilian as well as avian characters.

●

Cenozoic era—Following the decline of the reptiles during the late Mesozoic era, both birds and mammals started flourishing.

●

The mammals became more diversified of all the animals during Cenozoic era, which is called the ‘Age of Mammals’.

Classification of Subphylum Vertebrata ●

Subphylum vertebrata is divided into two sections : Agnatha and Gnathostomata.

●

Agnatha—These are jawless vertebrates and are most primitive. The mouth does not possess jaws hence called Agnatha. They do not have exoskeleton and paired appendages, and have single nostril.

●

Section Agnatha is further divided into following classes : 1. Class Ostracodermi—These were group of extinct agnatha, and were primitive, heavily armoured. World's first vertebrates collectively called Ostracoderms. 2. Class Cyclostomata—These have body without scales, jaws and lateral fins. Mouth rounded and suctorial, gills 5–16 pairs. They are parasites and scavengers.

C.S.V. / August / 2009 / 741

●

● ●

Examples—Lampreys (Petromyzon) and Hagfishes (Myxine). Gnathostomata—These are jawed vertebrates, having true jaws and paired limbs. Embryonic notochord is usually replaced in adult by a vertebral column. Mouth has jaws hence called gnathostomata. Paired nostril are present. Gnathostomata is divided into two super classes : Pisces and Tetrapoda. Super class Pisces—It includes true fishes. All are aquatic. The body bears paired as well as median fins. Respiration occurs typically by gills. They are cold-blooded. Each eye has a well developed nictitating membrane and possess scaly skin.

Super class Pisces is divided into following three classes : 1. Class Placodermi—It includes the earliest (Paleozoic) fossil fishes which lived in fresh water. Body had an external protective armour of bony scales or plates. Primitive jaws with teeth were present. Skeleton was bony and fins were mostly formed of large spines. Examples—Climatius, Dinichthys. 2. Class Chondrichthyes—Mostly marine fishes having cartilaginous endoskeleton and skin with placoid scales. Gill slits not covered by operculum. Male fishes have pelvic claspers. Examples—Scoliodon (Dogfish) and Chimera (Ratfish). 3. Class Osteichthyes—Fresh water as well as marine fishes. Endoskeleton mostly bony. Skin having various types of scales (cycloid, clenoid) other than placoid. Gill slits are covered by operculum. Male fishes without claspers. Examples—Labeo, Hippocampus, Protopterus. ● Super class Tetrapoda—Land vertebrates with two pair of pentadactyle limbs. They have cornified skin and lungs. Sensory organs are adapted for vision, hearing and smelling etc. Tetrapoda is divided into 4 classes : 1. Class amphibia—Adults are amphibious and respire by lungs. Larval stages usually aquatic and breathe by gills. Skin moist, glandular and with no external scales. Heart 3-chambered. Cold blooded. Example—Rana (frog), Bufo (toad). 2. Class Reptilia—Terrestrial, skin dry and covered by ectodermal horny scales or bony plates. Heart incompletely 4-chambered. Cold blooded. Example—Hemidactylus (lizard), snakes, crocodiles, turtles. 3. Class Aves—Body covered with feathers. Forelimbs modified into wings. No teeth in beak. Heart 4chambered. Warm blooded. Example—Columba (Pigeon). 4. Class Mammalia—Body covered by hairs. Skin glandular. Mammary glands present. Heart 4-chambered. Warm-blooded. Example—Homo (humans), whale, elephant.

Review at a Glance ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●

Phylum chordata name was established by Balfour. Early chordates evolved during Ordovician period of Paleozoic era. According to Garstrong, chordates have evolved from echinoderm larva by paedogenesis. All vertebrates are chordates but all chordates are not vertebrates. Presence of notochord, pharyngeal gill slits and dorsal tubular nerve cord are three characteristic features of chordates. In vertebrates notochord is replaced by vertebral column. Notochord is derived from mesoderm. In urochordata, notochord is found only in tail region of larval stage. In cepalochordata, notochord is present throughout the life. In mammals, remanent of notochord is present among invertebral disks as nucleus pulposus. Dorsal tubular nerve cord is ectodermal in origin. All chordates are deuterostomes, i.e., blastopore forms anus. All chordates are enterocoelic. Members of urochordata show retrogressive metamorphosis. The first fossile of vertebrates were found in rocks of the Ordovician period in the form of Ostracoderms. Devonian period is known as age of fishes. Carboniferous period usually known as age of amphibians. Mesozoic era is known as age of reptiles. Cenozoic era is known as age of mammals. World's first vertebrates collectively called Ostracoderms.

OBJECTIVE QUESTIONS 1. Vertebrates skin is covered by— (A) Scales (B) Feathers (C) Hairs (D) All the above 2. The vertebrates are the members of— (A) Cephalochordata (B) Agnatha (C) Gnathostomata (D) Urochordata 3. The study of fishes is called— (A) Ichthyology (B) Herpatology (C) Saurology (D) Ornithology 4. Cartilaginous have— (A) Gill slits (C) Scales

fishes

do

not

(B) Operculum (D) Pelvic fin

5. Urinary bladder is absent in— (A) Reptiles (B) Aves (C) Mammals (D) Amphibians 6. Which of the following bird has teeth in beak ? (A) Ostrich (B) Kiwi (C) Pelican (D) Archaeopteryx

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7. A common characteristic among all mammals is— (A) They are carnivorous (B) They have ventral nerve cord (C) They do not moult (D) They have seven cervical vertebrae

(Continued from Page 702 )

8. Which of the following is not found in the amphibian skin ? (A) Epidermis (B) Mucous glands (C) Scales (D) Chromatophores 9. The presence of true placenta is characteristic feature of subclass— (A) Eutheria (B) Metatheria (C) Prototheria (D) All the above 10. Which of the following era is known as age of mammals ? (A) Devonian (B) Carboniferous (C) Mesozoic (D) Cenozoic

ANSWERS

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Introduction ● Muscles are in use during every waking and sleeping moment. Some muscles are under conscious voluntary control, while others operate without conscious control. ● All muscle tissues have three characteristics in common : 1. Muscles can contract and shorten in length. 2. After contraction, muscles relax and return to their former length. 3. Muscles are excitable, responding to electrical or chemical signals (stimuli) from the nervous and endocrine systems. ● Muscle is the only tissue in the body that has all these properties. ● A muscle is composed of many muscle cells that are bundled into fascicles.

Antagonistic Muscles ● A muscle can pull a part of the body by its contraction (shortening). It cannot push that part by relaxation (elongation). Hence, the muscles are typically arranged in antagonistic (opposing) pairs, one of which moves a body part in one direction and the other moves that part in the opposite direction by its contraction. Of course, when one muscle contracts, its opposing muscle relaxes. Muscle Blood vessels

Fascia surrounding muscle

Connective tissue surrounding fascicle Fascicle

Tendon Connective tissue surrounding each muscle cell Muscle cell Bone

Fig. : Relationship of a skeletal muscle with its muscle cells and connective tissue wrappings.

Skeletal Muscles Structure ● Skeletal muscles attach to bones and move the skeleton. They also attach to other muscles and to the facial skin. ● Some skeletal muscles are very large, such as the gastrocnemius, the main muscle forming the calf in

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the lower leg. Others are very small, such as the muscle of the eyelid. ● Muscle shape vary. Some muscles are triangular (deltoid) while others are rectangular (rectus abdominis) or trapezoidal (trapezius). ● Many are spindle-shaped (biceps brachii). ● Each skeletal muscle is composed of parallel bundles of individual muscle cells. Each muscle cell is surrounded by delicate connective tissue. Bundles of these cells, called fascicles are wrapped in another thin layer of connective tissue. ● The connective tissue wrappings are actually continuous with each other at the end of the muscle, where they unite to form an extremely tough fibrous connective tissue material called a tendon. Tendons attach muscle to bone. ● Individual skeletal muscle cells have a tubular shape, contain many nuclei per cell and appear striated or striped. These muscle cells are relatively large compared with other human cells. ● Muscle cells contain the same cellular components or organelles found in most other human cells, but some components have specialized functions or are unique. ● The smooth endoplasmic reticulum forms into parallel saclike compartments called the sarcoplasmic reticulum or SR. ● On the outer surface of the muscle cell’s plasma membrane are the openings of tiny tubes called transverse tubules or T-tubules. These tubules are extensions of the plasma membrane that run deep into the cell and lie close to the SR. They relay the electrical signals that trigger contraction. ● Within each muscle cell and running its entire length are numerous protein fibres called myofibrils. Myofibrils are the structures responsible for contraction. ● Each myofibril consists of a series of units called sarcomeres. The orderly arrangement of sarcomeres gives a striated appearance to skeletal muscle cells. ● Each sarcomere is made of a rather complex arrangement of several kinds of protein. Two of these proteins are actin and myosin. ● Thousands of actin molecules link together to form a long threadlike macromolecules, called actin filaments. Similarly, thousands of myosin molecules form the somewhat thicker myosin filaments. The molecular interaction between actin and myosin filaments contributes directly to muscle contraction.

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Skeletal Muscle Contraction ● For muscle contraction, a stimulus is needed. Before a muscle cell can contract, it must be stimulated. Stimulation is provided by nerve impulses that pass along nerve cells. ● A nerve cell stimulates a muscle cell by secreting a chemical substance called acetylcholine (ACh). Acetylcholine permits a nerve cell to communicate chemically with a muscle cell at a site called the neuromuscular junction. ● Upon its release, acetylcholine diffuses across the junction and binds to specific receptors on the plasma membrane of the muscle cell. This binding initiates an electrical impulse which travels down the T-tubules into the interior of the cell. ● When the impulse reaches the SR, Ca 2+ stored in the SR is released. The released Ca 2+ forms bond with the troponin-tropomyosin protein complex which is associated with the actin filaments. ● When Ca2+ forms this bond, the troponin-tropomyosin complex shifts position, exposing sites on the actin filaments for attachment to the head portions of the myosin filaments. ● After exposing sites for attachment, a series of events cause the shortening or contraction of a muscle cell. Each myosin head projecting from a myosin filament bonds with an actin filament. After forming these bonds, the myosin heads change shape and pull the actin filaments toward the centre of sarcomere. Spinal cord Nerve cell

Nerve cells carry impulses from spinal cord to muscle cells.

Muscle cells

Neuromuscular junction

Plasma membrane Acetylcholine of muscle cell (ACh) ACh receptors

End of nerve cell

ACh released from nerve cell attaches to receptors on muscle cell plasma membrane.

Fig. : Pathway of an impulse from the spinal cord to the neuromuscular junction.

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● This reaction requires energy, which comes from the ATP attached to each myosin head. ● Myosin acts as an enzyme (when Ca2+ is present) and breaks the ATP into ADP + P + energy. This energy is used to change the shape of the myosin head, which causes the actin filaments to slide a little towards the centre of sarcomere. ● After sliding, a new ATP attaches to the myosin head, breaking the myosin-actin bond. When this bond is broken, the myosin head is released and returns to its original shape. Immediately, the myosin head bonds to another site on the actin filament. Then it changes shape again and the filament slide a little past one another. ● The sequence of myosin-actin bonding, sliding and release is repeated rapidly over and over until the muscle has shortened sufficiently. This is the sliding filament mechanisms for muscle contraction, and it consumes a great deal of ATP. ● When an individual muscle is stimulated, the actin filament in every sarcomere unit slide toward the middle of the sarcomere. As a result, the myofibrils shorten. A muscle cell shortens (contracts) then all its myofibrils shorten. An entire muscle contract when many of its muscle cells contract simultaneously. ● Contraction ends when the muscle cell ceases to be stimulated by the nerve cell. If an impulse no longer travels down the T-tubules to the cell’s interior, Ca2+ is actively transported back into the SR. Without Ca2+, the troponin-tropomyosin protein complex shifts back into its original position, masking the sites on the actin filaments and preventing further bonding by the myosin heads. When this happens, the muscle relaxes and return to its original shape. ● In muscle contraction, oxygen is used, carbon dioxide is produced, glycogen is consumed and heat is generated. Only a part of energy released in the process is used. In muscle contraction, the rest is converted into heat. ● Muscle Tonus (Tone)—All apparently relaxed skeletal muscles always remain in partial contraction as long as their nerves are intact. This state of sustained partial contraction is called muscle tonus or tone. It is a sort of mild tetanus. ● The tone is maintained by a constant flow of nerve impulses to muscle fibres. A muscle under slight tension can react more rapidly and can contract more strongly than one which is completely relaxed. Almost all human daily activities are carried out by tetanic contractions of muscles. Tetanus is necessary to maintain posture and form of body. ● Muscle Fatigue—The duration for which a muscle can remain contracted depends on its ability to supply ATPs to the contractile proteins (myosin and actin). The reduction in force of contraction of a muscle after prolonged stimulation is called muscle fatigue. A muscle can contract in the absence of oxygen, but it gets fatigued sooner. This is due to the fact that lactic acid is not disposed off without oxygen and collects in the muscle fibres.

● Fatigue is thus primarily caused by an excess of lactic acid. A completely fatigued muscle refuses to respond to nervous stimuli.

Types of Muscles According to Type of Motion 1. Flexors—The flexors bend one part of a limb on another at a joint. Example— Biceps . It brings the forearm towards the upper arm. 2. Extensors—The extensors extend or straighten a limb. Example—Triceps. It extends the forearm. 3. Abductors—The abductors pull a limb away from the middle of the body. Example—Deltoideus. It draws the entire forelimb to the side. 4. Adductors—The adductors bring a limb toward the midline of the body. Example— Latissimus dorsi. It presses the entire forelimb against the side. 5. Depressors—The depressors lower a part. Example— Depressor mandibulae. It lowers down the lower jaw to open the mouth. 6. Elevators—The elevators raise a part. Example—Masseter. It lifts up the lower jaw to close the mouth. 7. Rotators—The rotators rotate a part. Example— Pyriformis . It raises and rotates the femur. 8. Pronators—These rotate the forearms to turn the palms downward or backward. 9. Supinators—These rotate the forearm to turn the palms upward or forward. 10. Sphincters—These decrease the size of the apertures to close them. 11. Dilators—These widen the apertures. The adductor and abductor; elevator and depressor; pronator and supinator; and sphincters and dilators are all antagonistic muscles.

● The single isolated contraction of a muscle fibre caused by a single nerve impulse or artificial stimulus is called muscle twitch. Immediately after a twitch, the muscle fibre relaxes. ● Fundamentally, skeletal muscles function by twitching, and most graceful movements of body are the product of twitches. ● The contraction period is the time during which the muscles actually shortens. The relaxation period is the time it takes the muscle to relax and return to its original length. ● A muscle twitch can be altered by increasing the frequency of stimulation (the number of impulses per second). Muscle contractions are known as summation, and are produced when the frequency of stimulation is so rapid that a muscle does not have time to relax completely after each impulse. Summation produces a greater strength of contraction by adding together many separate contractions. ● If the frequency of stimulation is further increased so that there is no relaxation, the muscle will remain in a completely contracted state called tetanus.

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● A muscle contracts only when the stimulus it receives is of sufficient strength. The smallest strength of stimulus that causes contraction is called the threshold stimulus. ● When a muscle cell receives a threshold stimulus, it contracts with maximum strength. Even if a muscle cell receives a stimulus greater than the threshold, it will contract with the same maximum contraction strength. This is called the all-or-none law, and it states that a muscle cell will contract maximally or not at all. ● The all-or-none law applies to individual muscle cells, but not to an entire muscle. This explains that muscle cells in a muscle do not all have the same threshold value. ● Isotonic and isometric muscle contractions are used in certain exercise activities. ● Isotonic means ‘same strength’ (constant tension, and in an isotonic contraction, the strength of the contraction remains the same but the muscle shortens. Example—Weight lifting. ● Isometric means ‘same length’. In an isometric contraction, the strength of the contraction increases, but there is practically no shortening of the muscle. This type of contraction does not result in body movement. Sitting upright in a chair while contracting abdominal muscles is an example of an isometric contraction. ● Tonic (Slow, Red) Muscle Fibres—These are thin, dark red and slow contracting muscle fibres. They contain a high content of haem-protein pigment called myoglobin, abundant mitochondria, low glycogen content and poorly formed sarcoplasmic reticulum (SR). ● Myoglobin imparts them dark colour and stores oxygen as oxyhaemoglobin. Its oxygen by anaerobic oxidation in mitochondria provides energy for muscle contraction. Less lactic acid accumulates in this process. This enables the red muscle fibres to carry on slow and sustained contractions for long periods without fatigue. ● The tonic muscle fibres are innervated by thin, slow conducting nerve fibres. ● The body muscles meant for sustained work at a slow rate for a prolonged duration are composed mostly of red muscle fibres. The extenser muscles of the back in humans remain in sustained contraction to maintain errect posture against gravity, and are rich in red muscle fibres. Twitch (Fast, White) Muscle Fibres—These are much thicker, lighter in colour and fast-contracting muscle fibres. They have a low content of myoglobin, few mitochondria, abundant glycogen granules and well formed SR. ● They derive energy for their fast contraction mainly by anaerobic oxidation, accumulate lactic acid during strenuous work and soon get fatigued. ● These muscle fibres are innervated by thick fast conducting nerve fibres. ● The body muscles, which are meant for fast and strenuous work for short durations, are composed mostly of white muscle fibres. ● The muscles that move eyeballs are very rich in white muscle fibres.

Differences between Red and White Skeletal Muscle Fibres Red (Slow) Muscle Fibres

White (Fast) Muscle Fibres

1. They are thin.

1. They are much thicker.

2. They contain abundant mitochondria, low glycogen content and poorly formed sarcoplasmic reticulum.

2. They are poor in mitochondria, and have abundant glycogen granules and well formed sarcoplasmic reticulum.

3. They are dark red as they contain myoglobin.

3. They are light in colour as they have very little myoglobin.

abundant

pigment

4. Their myoglobin stores O2 as oxymyoglobin that releases O2 for oxidation during muscle contraction.

4. They have little or no store of oxygen.

5. They get energy for contraction by aerobic respiration.

5. They get energy for contraction mainly by anaerobic respiration.

6. They accumulate little lactic acid.

6. They accumulate lactic acid during strenuous work.

7. They undergo slow sustained contractions for long periods. 7. They undergo fast contractions for short periods. 8. They are not fatigued with work.

8. They soon get fatigued with work.

9. They are innervated by thin, slow-conducting nerve fibres.

9. They are innervated by thick fast-conducting nerve fibres.

Example : extensor muscles of the back in man.

Example : eyeball muscles.

OBJECTIVE QUESTIONS 1. Myoglobin is found in— (A) Slow muscle fibres (B) Fast muscle fibres (C) Blood (D) Lymph 2. Which of these plays a role in muscle contraction ? (A) K+ (B) Na + (C) Mg2+ (D) Ca 2+ 3. In skeletal muscle, the Ttubules— (A) Secrete acetylcholine (B) Give mechanical support (C) Relay electrical signals (D) Store calcium 4. As compare to the slow muscle, fast muscle has— (A) More myoglobin (B) More mitochondria (C) More sarcoplasmic reticulum (D) All the above 5. Which of the following is the contractile protein of a muscle ? (A) Tubulin (B) Myosin (C) Tropomyosin (D) All the above 6. In muscle, the contraction occurs due to— (A) Myosin (B) Actin (C) Both (A) and (B) (D) None of these

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7. The contraction and relaxation phases of a muscle constitute— (A) Beat (B) Twitch (C) Condition (D) Stimulus 8. The sprain is caused due to excessive pulling of— (A) Muscles (B) Nerves (C) Tendons (D) Ligaments 9. Sarcoplasmic reticulum plays a major role during— (A) Muscle contraction (B) Muscle excitement (C) Muscle relaxation (D) All the above 10. When a muscle bends one part upon other, it is called— (A) Extensor (C) Abductor

(B) Flexor (D) Regulator

ANSWERS

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(Continued from Page 687 )

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15. What is the name of cycle which produces CO2, ATP, NADH and FADH2?

1. A type of dentition in which the teeth are replaced throughout the lifetime, is called— (A) Monophyodont (B) Polyphyodont (C) Diastemic (D) None of the above 2. Which of the following lipids do not have fatty acids ?

8. Sensory cells that occur in the pits or canals of the lateral line system of fishes ? (A) Neuromast (B) Neurochrome (C) Neuroglia (D) None of the above 09. Adipocytes are— (A) Bones

(A) Steroids

(B) Cartilage

(B) Cholesterol

(C) Specialized connective tissues

(C) Phospholipids (D) Both (A) and (B) 3. Which of the following hormone initiates the preparation of the uterus for implantation of the ovum ? (A) Oxytocin

(D) Nerves 10. Formation m -RNA from DNA is called— (A) Translocation (B) Transcription (C) Transduction (D) Duplication

(B) Vasopressin (C) Progesterone (D) HCG.

11. Diapedesis is— (A) Formation of WBC (B) Bursting of WBC

4. Which brain wave pattern is common in young children and sleeping adults ? (A) Alpha waves (B) Beta waves (C) Theta waves (D) Delta waves 5. In birds, a comb like structure is present projecting into the cavity of eye, is called—

(C) Formation of Pus (D) Passage of WBC 12. The chief function of utriculus in humans is— (A) To perceive sound vibrations (B) To maintain body equilibrium (C) To perceive pressure (D) To act as shock absorber 13. Housefly has—

(A) Keel

(B) Pecten

(A) Chewing mouth parts

(C) Baleen

(D) Ossicles

(B) Piercing-sucking mouth parts

6. German measles is caused by— (A) Rubella virus (B) Varicella virus (C) Papilloma virus (D) None of the above 7. Quill feathers present on wings are also called— (A) Down feathers (B) Barbules (C) Coverts (D) Remiges

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(C) Sponging mouth parts (D) Siphoning mouth parts 14. Which of the following conditions meet the Hardy-Weinberg equilibrium ? (A) The homozygous recessive phenotype is non-viable (B) The dominant allele mutates to a recessive allele (C) The alleles are sex-linked (D) Mating is random

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

Oxidative carboxylation Oxidative phosphorylation Citric acid cycle Glycolysis

16. The cartilage making the lower jaw of cartilaginous fishes, is called— (A) Dentary (B) Mentomeckelian (C) Meckel’s cartilage (D) Angulosplenial 17. In a DNA molecule, the— (A) Bases are covalently bonded to the sugars (B) Sugars are covalently bonded to the phosphates (C) Bases are hydrogen bonded to one another (D) All of these are correct 18. Ecotone is characterised by— (A) Transitional zone between two diverse communities (B) Terrestrial ecosystem (C) Zone of transition between water and land (D) Forest ecosystem 19. DNA nucleotide differences between organisms— (A) Indicate how closely related organisms are (B) Indicate that evolution occurs (C) Explain why there are phenotypic differences (D) All of these are correct 20. Process of urea formation in humans is also known as— (A) Krebs cycle (B) Hans Krebs cycle (C) Nitrogen cycle (D) Transamination 21. Ends of long bones are covered with— (A) Cartilage (B) Muscles (C) Ligaments (D) Blood cells 22. In some animals, allantois is also related with— (A) Storage of nitrogenous wastes (B) Blood formation (C) Digestion (D) All the above

23. Territoriality occurs as a result of predation— (A) (B) (C) (D)

Predation Parasitism Competition Cooperation

24. During respiration, failure of ventilation leads to— (A) Decreased oxygen tension (B) Decreased carbon dioxide tension (C) Carbonate tension (D) Bicarbonate tension 25. Forest destruction leads to dentrimental effects on— (A) Loss of a CO 2 sink (B) Loss of biodiversity (C) Loss of possible medicinal plants (D) All of the above 26. The association of sea anemone and hermit crab is an example of— (A) Mutualism (B) Commensalism (C) Parasitism (D) None of the above 27. Simple two neuron reflex arc involves— (A) Sensory neuron (B) Spinal cord (C) Effector neuron (D) All of the above 28. Tube feet are locomotory organs of— (A) Jelly fish (B) Carp fish (C) Silver fish (D) Star fish 29. Down syndrome— (A) Is always caused by nondisjunction of chromosome 21 (B) Shows no overt abnormalities (C) Is more often seem in children of elderly mothers after the age of thirty five and above (D) Both (A) and (C) are correct 30. In the development of the human embryo, the ectoderm is responsible for the development of— (A) Nervous system (B) Lens of eyes (C) Sweat glands (D) All the above

C.S.V. / August / 2009 / 749

31. The protein which polymerizes to form microtubules is— (A) Actin (B) Chitin (C) Tubulin (D) Ferritin

40. Waste product of adenine and guanine metabolism is— (A) Ammonia (B) Uric acid (C) Urea (D) All these

32. The centrum of a typical vertebra of a frog is—

41. The skin of certain sharks and rays containing small knobs, is called— (A) Shagreen (B) Pulp (C) Dentine (D) Ganoin

(A) Procoelus (B) Heterocoelus (C) Opisthocoelus (D) Acoelus 33. Which vitamin does not act as a coenzyme ? (A) Riboflavin (B) Pantothenic acid (C) Biotin (D) Folic acid 34. Fovea centralis in eye, lacks— (A) Rod cells (B) Blood vessels (C) Nerve fibres (D) All the above 35. Balbiani rings are the characteristics of— (A) Polytene chromosomes (B) Lampbrush chromosomes (C) Sex chromosomes (D) Ring chromosomes 36. The stage of animal embryonic development at which the gut cavity and germ layers first appear, is called— (A) Involution (B) Epiboly (C) Gastrulation (D) Formative movement 37. Pebrine is a disease of— (A) Honey bee (B) Fish (C) Silk worm (D) Lac insect

42. Sympathetic nerves in mammals arise from— (A) Sacral region (B) Cervical region (C) Thoraco-lumbar region (D) None of the above 43. Which one of the following is correct about malaria ? (A) Gametocytes fuse in mosquito's stomach to form egg cells. (B) Mosquito pick up gametocytes when sucking blood from an infected person (C) Merozoites invade red blood cells and become trophozoites (D) All the above are correct 44. Osteoarthritis disease usually associated with aging and is called— (A) Communicable disease (B) Degenerative disease (C) Deficiency disease (D) Allergy 45. Medullary cavity of which following bone has yellow bone marrow ? (A) (B) (C) (D)

Long bones Short bones Spongy bones All of the above

38. Most primitive living mammals which provide an evidence of organic evolution from geographical distribution are found in— (A) China (B) India (C) Africa (D) Australia

46. Stomach pain is felt through— (A) Interoceptors (B) Proprioceptors (C) Teloceptors (D) None of the above

39. Which one of the following processes make direct use of oxygen ? (A) Glycolysis (B) Fermentation (C) Krebs cycle (D) Electron transport

47. Camel's hump conserve water for longer time because it is composed of— (A) Muscular tissue (B) Skeletal tissue (C) Areolar tissue (D) Adipose tissue

48. Black-water fever is caused by— (A) Plasmodium falciparum (B) Leishmania donovani (C) Plasmodium ovale (D) Plasmodium malariae 49. The transfer of genetic material of one bacterium to another by virus is called— (A) Transduction (B) Translation (C) Transcription (D) Replication 50. Charcot-Leydon crystals are found in faeces of man during infection of— (A) Entamoeba histolytica (B) Trypanosoma gambiense (C) Ascaris (D) All the above

ANSWERS WITH HINTS

(Continued on Page 790 ) C.S.V. / August / 2009 / 750

16. Ticks and mites are— (A) Myriapods (B) Arachnids (C) Insects (D) None of these 1. Which of these provides phospholipids dual properties ? (A) Fatty acid non-polar tails (B) Charged phosphate head (C) Fatty acid polar tails (D) Both (A) and (B) 2. Which of the following is present in microsomes of liver cells and plays a part in detoxification ? (A) Cytochrome P-450 (B) Adenylate kinase (C) Cytochrome C-reductase (D) Nucleoside diphosphokinase 3. When atoms share one or more electron pairs, the bond is— (A) Covalent bond (B) Ionic bond (C) Hydrogen bond (D) All the above 4. Gooseflesh caused by erection of skin papillae, as from cold, shock, fright or fear, is called— (A) Cutis anserina (B) Cutis aurantiasis (C) Cutis hyperelastica (D) Cutis marmorata 5. Which of these viruses become life-long residents of host cells ? (A) Herpes virus (B) Varicella virus (C) Rubella virus (D) Both (A) and (B) 6. Cuboid bone is associated with— (A) Carpels (B) Metacarpels (C) Metatarsals (D) None of these 7. Which of these are a group of viruses important in investigating viral carcinogenesis ? (A) Papova viruses (B) Parvo viruses (C) Adeno viruses (D) Pox viruses

C.S.V. / August / 2009 / 751

8. The wishbone of a bird, which support the flight muscles, formed from— (A) Fusion of the collar bones (B) Ventral extensions of the vertebrae (C) Anterior ribs (D) Posterior extensions of the larynx 9. Synthesis of m -RNA and t-RNA occurs in— (A) G1 phase (B) G2 phase (C) S-phase (D) All these 10. Diapedesis is associated with— (A) Leucocytes (B) Urinary system (C) Nervous system (D) Peristalsis 11. Absence of lack of ‘intrinsic factor’ in diet causes— (A) Pernicious anaemia (B) Cooley’s anaemia (C) Aplastic anaemia (D) None of the above 12. Which of the following is dinucleotide ? (A) FAD (B) NAD (C) Nucleic acid (D) Both (A) and (B) 13. Food energy passed at each trophic level is approximately— (A) 10% (B) 30% (C) 50% (D) 90% 14. Which of the following is viral disease ? (A) Impetigo (B) Warts (C) Both (A) and (B) (D) None of these 15. A non-essential hereditary factor that may exist either free within a cell or in a state in which it is integrated in a chromosome, is known as— (A) Episome (B) Cosmid (C) Both (A) and (B) (D) None of the above

17. The malleus, incus and stapes of ear ossicles of mammals are respectively modified bones of— (A) Articular, hyomandibular and quadrate (B) Quadrate, articular and hyomandibular (C) Articular, quadrate and hyomandibular (D) Quadrate, hyomandibular and articular 18. Hashimoto disease is caused when— (A) Adrenal gland is destroyed by autoimmunity (B) Thyroid gland is destroyed by autoimmunity (C) Kidney is destroyed (D) Pancreas is destroyed 19. Long chain molecules of fatty acids are formed by— (A) Polymerisation of 2 carbon compounds (B) Decomposition of fats (C) Polymerisation of glycogen (D) None of the above 20. Volkmann's canals are found in— (A) Bones of frog (B) Bones of fowl (C) Bones of rabbit (D) Cartilage of rabbit 21. Bedbugs can survive during long time of starvation because— (A) It stores glycogen (B) It convert uric acid to amino acid and thus it can use its excretory materials (C) Its life span is very long (D) It can minimise its requirements 22. Which of the following is involved in DNA synthesis and cell division ? (A) Vitamin K (B) Folic acid (C) Vitamin E (D) Vitamin D 23. Oxygen toxicity is related with— (A) Failure of ventilation of lungs (B) Collapse of alveolar walls (C) Blood poisoning (D) None of the above

24. For most people, dreaming occurs during which stage of sleep ? (A) Deep sleep (B) REM sleep (C) Alpha sleep (D) Slow wave sleep 25. Latissimus dorsi muscles in humans— (A) Draws arm downward and backward (B) Draws legs forward (C) Moves head (D) Moves ankles 26. Trochanter is associated with— (A) Femur (B) Humerus (C) Ulna (D) Radius 27. Heparin is produced by— (A) Liver cells (B) Nervous cells (C) Kidney cells (D) Spleen 28. Podocytes are found in— (A) Spleen (B) Pancreas (C) Kidneys (D) Heart 29. Which of the following stimulates erythropoiesis ? (A) Fe ++ (B) Mg++ (C) Ca ++ (D) Cu ++ 30. Fovea centralis in eye lacks— (A) Rod cells (B) Blood vessels (C) Nerve fibres (D) All the above 31. Ichthyophis is a member of— (A) Mollusca (B) Pisces (C) Amphibia (D) Annelida 32. Waste product of adenine and guanine metabolism is— (A) Uric acid (B) Ammonia (C) Urea (D) All these 33. Cis-trans test is related with— (A) Crossing over (B) Heredity (C) Mutation (D) All the above 34. Parathion, malathion and femitrothion insecticides belong the group of— (A) Carbamates (B) Pyretheroids (C) Organophosphates (D) Triazines

C.S.V. / August / 2009 / 752

35. Which one of the following is not a Ketone body ? (A) Acetoacetic acid (B) Succinic acid (C) β-hydroxybutyrate (D) Acetone

44. Exotoxins are related with— (A) Tetanus (B) Diphtheria (C) Both (A) and (B) (D) None of these

36. German measles is caused by— (A) Rubella virus (B) Varicella virus (C) Papiloma virus (D) None of these

45. Degradative reactions— (A) Are oxidation reactions (B) Are synthetic reactions (C) Requires a supply of NADPH molecules (D) Both (A) and (C)

37. Antihistamine drugs relieve— (A) Allergy (B) Angina pectoris (C) Stroke (D) Nephrites

46. Which of the following is an example of bile acids ? (A) Glycocholic acid (B) Taurocholic acid (C) Both (A) and (B) (D) None of these

38. When diminished blood volume is present in human body, it is termed— (A) Hypovolemia (B) Oligemia (C) Oligohaemia (D) All the above

47. Which one of the following is not related with the path of sound vibrations ? (A) Auditory canal (B) Tympanic membrane (C) Semicircular canals (D) Cochlea

39. Middle part of human sperm contains— (A) Nucleus (B) Mitochondria (C) Centriole (D) Nucleus and mitochondria

48. Wolffian body is known as— (A) Pronephros (B) Mesonephros (C) Metanephros (D) None of these

40. Another term for adaptive evolution is— (A) Clinal change (B) Microevolution (C) Macroevolution (D) All the above

49. Chaga’s disease is caused by— (A) Trypanosoma gambiense (B) Trypanosoma cruzi (C) Trypanosoma bruci (D) None of the above

41. Eggs having yolk in the centre and cytoplasm in the peripheral area, are called— (A) Isolecithal (B) Microlecithal (C) Centrolecithal (D) Telolecithal 42. Which of the following belongs to phylum Arthropoda ? (A) Gold fish (B) Silver fish (C) Cuttle fish (D) Star fish 43. Parathion, Malathion and Femitrothion insecticides belong to the group of— (A) Carbamates (B) Pyretheroids (C) Organophosphates (D) Triazines

50. Inheritance of which of the following is not Mendelian ? (A) Plasmagene (B) Oncogene (C) Polygene (D) All the above

ANSWERS WITH HINTS

17. Which of the following is an inactive enzyme precursor ? (A) Polyglycoids (B) Cholenzymes (C) Activases (D) Zymogens 1. Which of the following prevents the mother’s immune response to the developing embryo ? (A) Trophoblast (B) Blastocyst (C) Umblical vein (D) Umblical artery 2. Auer’s bodies are present in the cytoplasm of— (A) Myeloblasts (B) Myelocytes (C) Monoblasts (D) All the above 3. Milkman’s syndrome is characterized by— (A) Failure of reabsorption of phosphates by renal tubules (B) Failure of absorption of lactose (C) Failure of absorption of amino acids (D) None of these 4. Certain aquatic mammals, such as seals, sealions, and walruses are included in the order— (A) Pinnipedia (B) Fissipedia (C) Hyracoidea (D) Cetacea 5. Failure of an organ or part to develop or grow, is called— (A) Agenesia (B) Agenitalism (C) Ageism (D) Agerasia 6. Group of related species forming fertile hybrids are called— (A) Coenospecies (B) Subspecies (C) Sibling species (D) None of these 7. Olfactory organ of a snake is— (A) Jacobson’s organ (B) Johnston’s organ (C) Organ of Bojanus (D) None of these 8. Which of the following is source of energy to an ecosystem ? (A) Heat liberated during respiration (B) ATP

C.S.V. / August / 2009 / 754

(C) NADPH2

18. Restriction fragment length polymorphisms (RFLPs)—

(D) Solar energy 9. What is the function of porphyrin in earthworm ? (A) To protect against harmful germs (B) To help in respiration (C) To help in excretion (D) To protect from harmful ultraviolet rays 10. Which of the following is not a water soluble vitamin ? (A) Tocopherol (B) Niacin (C) Folic acid (D) Ascorbic acid 11. Which division of PNS is often called ‘voluntary nervous system’ ? (A) Somatic (B) Autonomic (C) Both (A) and (B) (D) None of these 12. FAD is derived from— (A) Thiamin (B) Riboflavin (C) Pyridoxine (D) Folic acid 13. The final hormonal stimulus leading to ovulation in human female is provided by— (A) Estrogen (B) Progesterone (C) Follicle stimulating hormone (D) Luteinizing hormone 14. Autoimmune thyroiditis is associated with— (A) Hashimoto’s disease (B) Grave’s disease (C) Goitre (D) Acromegaly 15. Convulsions in infants is caused due to deficiency of— (A) Iodine (B) Vitamin B 6 (C) Vitamin D

(D) Vitamin C

16. A cell-coded protein, that is formed in response to viral infection, is— (A) Antigen (B) Interferon (C) Histone (D) Antibody

(A) Identify cally

individuals geneti-

(B) Are the basis fingerprints

for DNA

(C) Can be subjected to gel electrophoresis (D) All the above 19. Notochord is usually regarded as— (A) Ectodermal (B) Mesodermal (C) Endodermal (D) Blastodermal 20. In cladistics— (A) A clad must contain the common ancestor plus all its descendents (B) Derived characters help construct cladograms (C) Data for the cladogram is presented (D) All the above 21. Bones enlarge by— (A) Auxentic growth (B) Multiplicable growth (C) Accretionary growth (D) Appositional growth 22. Free-living Planaria is a— (A) Roundworm (B) Flatworm (C) Tapeworm (D) Fluke 23. Bilharziasis is a parasitic disease, caused by— (A) Schistosoma (B) Taenia solium (C) Fasciola hepatica (D) Enterobius vermicularis 24. Guillain-Barre syndrome is related with— (A) Acute polyneuritis (B) Infectious polyneuritis (C) Landry’s paralysis (D) All the above

25. Which of the following is the only amino acid metabolized by the brain ? (A) Alanine (B) Glutamic acid (C) Histidine (D) Glutanine 26. Electron micrographs following freeze-fracture of the plasma membrane indicate that— (A) The membrane is a phospholipid bilayer (B) Some proteins span the membrane (C) Protein is found only on the surface of the membrane (D) Glycolipids and glucoproteins are antigenic 27. Enzymes which catalyse reactions involving electron transfer, called— (A) Transferases (B) Hydrolases (C) Ligases (D) Oxidoreductase 28. Arytenoid cartilages are found in— (A) Sternum

(B) Hyoid

(C) Nose

(D) Larynx

29. Which of the following animals are devoid of respiratory, excretory and circulatory organs ?

33. Sum of constructive processes in body cells is called— (A) Catabolism (B) Anabolism (C) BMR (D) All the above 34. Which of these is present in human buccal cavity ? (A) Trypsin (B) Ptyalin (C) Lipase (D) Pepsin 35. Which layer of the heart wall consists cardiac muscles ? (A) Endocardium (B) Myocardium (C) Epicardium (D) All of these 36. Human nerve cells originate from the embryonic— (A) Ectoderm (B) Ectoderm and mesoderm (C) Endoderm (D) Mesoderm 37. Which of the following is called incomplete proteins ? (A) Most animal proteins (B) Most plant proteins (C) Both (A) and (B) (D) None of these 38. Which of the following syndrome is associated with easily dislocation of body joints ?

(A) Tapeworms

(A) Ehlers-Danlos syndrome

(B) Sponges

(B) Tay-Sachs syndrome

(C) Liver flukes

(C) Marfan’s syndrome

(D) None of these

(D) None of these

30. Active transport— (A) Requires a carrier (B) Moves a molecule against its concentration gradient (C) Requires supply of energy (D) All the above are correct 31. Parasympathetic effect— (A) (B) (C) (D)

Lowers blood pressure Slows heart rate Promotes digestion All the above

32. Enzyme enterokinase converts— (A) Proteins into peptides (B) Caseinogen into casein (C) Trypsinogen into trypsin (D) Pepsinogen into pepsin

C.S.V. / August / 2009 / 755

39. Reticulocytes are— (A) WBCs (B) Immature RBCs (C) Blood platelets (D) Lymphocytes 40. Which of the following division of peripheral nervous system is often called ‘voluntary nervous system ? (A) Somatic (B) Autonomic (C) Both (A) and (B) (D) None of these 41. Meroblastic cleavage refers to which type of division of eggs ? (A) Incomplete (B) Spiral (C) Complete (D) Horizontal

42. Which part of human brain is associated with integration of sympathetic and parasympathetic activities ? (A) Hypothalamus (B) Cerebrum (C) Medulla oblongata (D) Neopallium 43. Iter or Aqueduct of sylvius is located between— (A) Lateral ventricles (B) Optocoels (C) III and IV ventricles (D) None of these 44. Autologous blood tranfusion is— (A) Use of patient’s own blood (B) Use of patient’s parental blood (C) Use of patient’s children blood (D) All the above 45. Which type of tissue forms the thin surface for gas exchange in lungs ? (A) Epithelial (B) Connective (C) Nervous (D) Muscular 46. Unmyelinated axons in spinal cord are present in— (A) White matter (B) Gray matter (C) Both (A) and (B) (D) None of these 47. Carcinoid syndrome is associated with— (A) Serotonin (B) Bradykinin (C) Prostaglandins (D) All of these 48. Which of the mitosis stage provides best opportunity for preparing human karyotype ? (A) Anaphase (B) Metaphase (C) Prophase (D) Telophase 49. Castle’s intrinsio factor helps in— (A) Absorption of cyanocobalamin (B) Nervous stimulation (C) Cardiac reflex (D) All the above 50. Bright’s disease is associated with— (A) Kidneys (B) Liver (C) Lung (D) Bone

Introduction Edaphic factors take in account the structure and composition of soil with its physical and chemical characteristics. The study of soil (Latin = Solum) is called pedology or edaphology, which is important for geology, mineralogy, paleobotany, paleozoology and petrology. Soil is a natural habitat for plants, animals and microorganisms. It may be defined as “any part of earth's crust in which plants root grow.” However, soil is not merely mineral matter, but a complex of several other types of components. Thus biologically the soil may be defined as “the weathered superficial layer of the earth's crust in which the living organisms grow and also release the products of their activities, death and decay.”

Composition of Soil From the above classical definition of soil it is evident that soil is, thus, not merely a mineral particles, but it has also a biological system of living organisms as well as some other components. Therefore, it is preferred to call it a soil complex, which includes the following categories of components— (i) Mineral matter—The matrix of mineral particles derived by varying degrees of break down of the parent material–rock occur in the soil. The minerals represent about 90% of the total weight of the soil. (ii) Soil atmosphere—Soil atmosphere occupies the pore space between soil particles, which is not water– filled. The soil atmosphere differs from the above ground atmosphere as it is normally lower in oxygen and higher in carbon dioxide content. (iii) Soil organic matter or humus—Soil organic matter or humus are partially decayed organic components derived from long and short-term addition of material from organisms growing above and below the ground that is microorganisms, plants and animals.

Dokuchayev’s View ● According to Dokuchayev, a famous Russian pedologist, “the soil is a result of the actions and reciprocal influences of parent rocks, climate, topography, plants, animals and age of the land. ● Hence, soil can be represented by the following formula : S = (g.e.b) Δt where,

S = Soil

g = Geology e = Environment b = Biological influences t = Time

C.S.V. / August / 2009 / 757

(iv) Soil water or solution—Soil water is held by capillary and absorptive forces between and around the surface of soil particles. Actually soil water is a dilute solution of various organic and inorganic compounds, which become available to plants as mineral nutrients. Water in the soil comes mainly through infiltration of precipitated water (rain, sleet, snow and hail) and irrigation, whereas it is lost from the soil chiefly through evaporation, percolation and transpiration.

Soil Formation (Pedogenesis) C. F. Marbut states that “a mature soil is one that has assumed the profile features characteristic of predominant soils on the smooth uplands within the general climatic and botanic regions in which it is found. Soil is a stratified mixture of organic and inorganic materials, both of which are decomposition products. The soil forming rocks are the parent material from which mineral constituents of soil are derived by fragmentation or weathering. Organic components of soil are formed either by decomposition (or transformation) of dead remains of plants or animals or through metabolic activities of living organisms present in the soil. Soil-forming rocks—There are two basic kinds of soil-forming rocks as given below— (i) Sedimentary rocks—These are formed by deposition of weathered minerals which are derived from igneous rocks. (ii) Igneous rocks—These are formed due to cooling of molten magma or lava.

Chemical Nature of Minerals of Soil-forming Rocks Rocks are the chemical mixture of various kinds of minerals. The chemical nature of most common and abundant minerals of soil-forming rocks are given in the following table— Table : Chemical Composition of Some Common Soil Minerals Minerals A.

Chemical Constituents

Sand and Silt minerals 1. Quartz or silica

SiO2

2. Feldspars (a) Orthoclase

K2Al2Si6O16

(b) Calcium feldspar

CaAl2Si2O8

(c) Plagioclase

NaAlSi3O8

3. Amphibole

(Mg, Fe)7 (Si4O11) 2 (OH)2

4. Calcite; magnesite; and dolomite

CaCO3, MgCO3 (CaCO3, MgCO3)

and

B.

5. Micas (a) Biotite (b) Muscovite 6. Iron oxides (a) Limonite (b) Haematite (c) Magnetite 7. Pyroxene 8. Olivine and serpentine Clay minerals 1. Montmorillonite 2. Kaolin

K, Mg, Fe, Al silicate K (OH)2 Al2 (AlSi3) O10 FeO (OH), X H2O Fe2O3 Fe2O4 (Mg, Fe) SiO4 (Mg, Fe)2 SiO4 (Ca, MgO) Al 2O3, 5 SiO3, 5 H2O Al2O3, 2 SiO2, 2 H 2O

Process of Soil Formation (1) Weathering of Soil Forming Rocks—Soils are derived from parent rocks by the process called weathering. Formation of soil is initiated by disintegration or weathering of parent rocks by certain physical, chemical and biological agents as a result, soil-forming rocks are broken down into small particles called regoliths, which finally develop into mature soil by pedogenesis weathering is accomplished by three types of forces—physical, chemical and biological. (A) Physical weathering—It takes place due to the following processes— (i) Wetting-drying—Wetting-drying causes the disruption of layer or lattice structure of the rock, which swells upon wetting and contract on drying leading to disintegration of rock, which provides more aeration. (ii) Heating-cooling—It causes the disruption of heterogenous crystalline rocks in which inclusions have differential coefficients of thermal expansion. (iii) Sand blast—In arid and desert conditions the rocks are disrupted by physical action of wind and sand. (iv) Freezing—It causes the disruption of porous, lamellar or vesicular rocks by frost shatter due to expansion of water. (v) Glaciation—At mountain tops, ice formation takes place mostly in the winter season. When the summer approaches, ice starts melting and glaciers (huge sliding masses of ice) move downwardly on the slops. In the glacier movement, the rocks are corroded and finally broken into sand particles. (B) Chemical weathering—Chemical weathering, which brings about disappearance of original rock minerals either completely or partly, includes the following processes— (i) Solution—Certain mobile components of rocks, such as calcium sulphates and chlorides are simply removed by agents like water (solution) making the rock porous and hence liable to further disintegration. (ii) Hydrolysis—During hydrolysis, components like alumino silicates of rocks break down and potassium or silicon are washed out which give rise to simpler mineral matter like clay alumino-silicates. (iii) Chelation—Certain chemical exudates, produced through biochemical activities of microorganisms like

C.S.V. / August / 2009 / 758

bacteria, cyanobacteria and lichens dissolve the mineral components of the rocks. These metals dissolved with organic products of microbial activities are known as chelates. (iv) Hydration—Reversible changes occur due to water absorption by haematite to limonite, (FeO3 Fe 2O3 . 3 H2O), as a result the rocks swell up and causes disruption. (v) Carbonation—Various chemicals produced in atmosphere and by metabolism of microorganisms bring about carbonation. For example, reversible change of CaCO3 to Ca (HCO3)2 leads to solution loss of limestone or disruption of calcium carbonate cemented rocks as hydrogen carbonate is more soluble than the carbonate. (vi) Oxidation-reduction—Certain chemicals bring about oxidation-reduction reactions, such as reversible change of Fe3+ to Fe2+, cause disruption of rocks, because Fe3+ is less soluble than Fe 2+. (C) Biological weathering—According to Jacks (1965-66), a number of microorganisms ( e.g., bacteria, fungi, nematodes, protozoans, etc.) lichens and mosses are early colonizers which transform the rock into a dynamic system, storing energy and synthesizing organic material. Their activities alter the chemical composition and physical structure of the rock. For example, lichens are present in the initial stages of biological succession and their growth may cause cracking or flaking, exposing greater area of rock to further weathering. Lichens as well as mosses extract mineral nutrients such as Ca, Mg, S, P, Al and Si from the rock. These elements are combined with organic complexes and eventually return to developing soil when vegetation decomposes. Joffe states that there is no biogeochemical weathering. According to him, it is either chemical or physical weathering by biological agencies.

Products of Weathering ● The surface rocks of the earth are weathered and as a result of weathering, small particles of parental materials are formed. ● The soil which is formed by weathering of rocks is called embryonic or primary soil. ● The soil which develops in situ above parent bedrocks is known as sedimentary and residual soil. ● In a few places soil material comes from accumulated organic matter as peat. ● Soil material transported from one area to another is called eolian or loess. ● Transported or secondary soils are those which are carried to other places by carriers like gravity, glacier, water and wind.

(2) Mineralization and Humification—Break down of organic debris into humus is accompanied by decomposition and finally by mineralization. Higher organisms in the soil consume fresh material and leave partially decomposed products in their excreta. This is further decomposed by bacteria and fungi into various compounds of carbohydrates, proteins, fats, lignin and resins. These compounds are broken down into simpler products such as carbohydrate, water salts and minerals. This

process is called mineralization. The residual amorphous incompletely decomposed black coloured organic matter which undergoes mineralization is called humus. The process of humus formation is called humification. The soils of deserts in which vegetation is scanty and new soils freshly formed from rocks, apparently contain very little humus, while mature soil with vegetation, the most. (3) Organo-mineral Complex Formation—In the final stage of pedogenesis, colloidal particles, which are formed as a result of weathering, humification and mineralization, accumulate and may aggregate into crumbs or concretions. About 60–70% colloidal humus particles become associated with mineral particles to form organomineral complexes. The crumbs of organo-mineral complexes, according to Wallwork (1970), are formed by two mechanisms—electrochemical bonding and cementing. Aggregation of negatively charged colloidal clay and/or humus particles of water molecules and metallic ions particularly calcium takes place in electrochemical bonding method of crumb formation. The cementing mechanism involves the action of substances adsorbed on the surface of soil particles which effectively make them compact. The crumbs increase the total pore space in the soil, providing good aeration and drainage. Some soluble organic compounds are removed from the top soil by water which percolates downwardly through humus mixed soil.

OBJECTIVE QUESTIONS 1. Eroded soils are— (A) Richer in plant nutrients

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UPKAR PRAKASHAN, AGRA-2

C.S.V. / August / 2009 / 759

2. Soil is composed of— (A) Air + water + minerals (B) Water + organic matter + minerals (C) Air + water + organic matter + minerals (D) Water + organic matter 3. The chemical weathering involves the— (A) Solution (B) Hydrolysis (C) Oxidation (D) All of these 4. Humus is an example of— (A) Fertilizer (B) Crystalloid (C) Component of soil structure (D) Organic colloids 5. Sequence of humification and mineralization is— (A) Dead organic matter → litter → duff → humus (B) Minerals → humus → litter → duff (C) Dead organic matter → duff → litter → minerals → humus (D) Humus → minerals → litter → duff

ANSWERS ●●●

Just Released

By : Dr. Vijay Agarwal Code No. 1647

(B) Devoid of plant nutrients (C) Fit for agriculture (D) None of the above

Rs. 110/-

● E-mail : [email protected] ● Website : www.upkar.in

Introduction Biochemistry has traditionally been considered ‘organic’ and in fact much of the biochemistry involves organic or carbon chemistry. It has been known for a long time, however, that in plants many other elements in the form of macro or micronutrients are also present. Materials are added to the soil, or applied directly to crop foliage, to supply elements needed for plant nutrition. These materials may be in the form of solids, semisolids, slurry suspensions, pure liquids, aqueous solutions or gases.

Fertilizers Crop requirements of fertilizer components could be satisfied by the spreading of individual materials for each element deficient in the soil. However, economy favours the single application of a balanced mixture that satisfies all nutritional needs of a crop. Many commercial fertilizers, therefore, contain more than one of the primary fertilizer elements. The chemical elements, such as nitrogen (N), phosphorus (P) and potassium (K) are the macronutrients, or primary fertilizer elements, which are required in greatest quantity. Sulphur (S), calcium (Ca) and magnesium (Mg), called secondary elements, are also necessary to the health and growth of vegetation, but they are required in lesser amounts compared to the macronutrients. The other elements of agronomic importance, called macronutrients and provided for plant ingestion in small (trace) amounts, include boron, zinc, molybdenum, cobalt, copper and iron. All these fertilizer elements along with other chemical elements, occur naturally in agricultural soils in varying concentrations and mineral compositions which may or may not be in forms readily accessible to root systems of plants. Many commercial fertilizers contain more than one of the primary fertilizer elements. The composition of fertilizer mixtures, in terms of the primary fertilizer elements, are identified by an N-P-K code : N denotes elemental nitrogen; P denotes the anhydride of phosphoric acid (P2O5); K denotes the oxide of potassium (K2O). All are expressed numerically in percentage composition, or units of 20 lb each per short ton (10 kg per metric ton) of finished fertilizer as packaged. Formula 8-32-16 thus contains a mixture aggregating 8 wt% N in some form of nitrogen compounds, 32 wt% P 2O5 in some form of phosphate and 16 wt% K2O in some form of potassium compounds, to give a product with a total of 56 fertilizer units. The commercial N-P-K formulas totals 100% plant nutrients because the formulas indicate only the nutrient portions of the primary-element compounds. Growing plants can assimilate only fertilizer elements in the combined state of inorganic compounds that are amenable to osmotic absorption. Many modern fertilizer

C.S.V. / August / 2009 / 760

materials consist of compounds that are immediately usable by the crops to which they are applied. Others are quickly converted within the soil to forms that can be assimilated. Some fertilizer chemicals are specifically designed to dissolve slowly or to delay reaction within the soil and, therefore, prolong the release of easily absorbed compounds to provide sustained feeding over the growth cycle of the plants. Aqueous solutions of urea, ammonia and ammonium nitrate (UAN solutions) are used directly by the farmers as well as in the preparation of granular N-P-K products by mixing with other materials, such as normal superphosphate and triple superphosphate. UAN solutions are also spread directly by field application or used to prepare complete N-P-K fertilizer solutions or suspensions.

Fertilizing ● Addition of elements or other materials to the soil to increase or maintain plant yields is known as fertilizing. ● Fertilizers may be organic or inorganic. ● Organic fertilizers are usually manures and waste materials which in addition to providing small amounts of growth elements also serve as conditioners for the soil. ● Methods of applying fertilizers vary widely and depend on such factors as kind of crop and stage of growth, application rates, physical and chemical properties of the fertilizers and soil type. ● Two basic application methods of fertilizers used are bulk spreading and precision placement. ● Liquid fertilizer of the high pressure type ( e.g., anhydrous ammonia) is usually regulated by valves or positive displacement pumps. ● Non-volatile fertilizer solutions are often pumped into the supply lines of irrigation systems to allow simultaneous fertilization and irrigation.

Biofertilizers Biofertilizers are the organisms which enrich the soil in nutrients due to their biological activity. To combat the ill effects of chemical fertilizers, biofertilizers have now been introduced for the pollution-free and better growth of the plants of useful aspects. The chief sources of biofertilizers are blue-green algae (cyanobacteria), fungi and bacteria.

1. Algae as Biofertilizers Early in the history of agriculture in coastal Asia the value of sea weeds in fertilizing the soil was discovered. Long before the recognition of the potash content, sea weeds were employed as fertilizers by the farmers. Not only as the fertilizers, but also the water-holding capacity of fragments of the algae in the soil proved effective. Furthermore, the yield of paddy is increased substantially

when paddy field is inoculated with nitrogen-fixing bluegreen algae, such as Anabaena oryzae , Tolypothrix tenius, Calothrix, Cylindrospermum bengalense, and Nostoc commune. Blue-green algae (cyanobacteria) secrete growth promoting substances like IBA, NAA, IAA and various proteins and vitamins. They add sufficient amounts of organic matter in the soil. They can grow and multiply under wide pH range of 6·5–8·5. Therefore, they can be used as the possible tool to reclaim saline or alkaline soil because of their ameliorating effect on the physio-chemical properties of soil.

2. Bacteria as Biofertilizers Rhizobium is one of the best fertilizers of leguminous plants. Rhizobium is a bacterium living in the root nodules of leguminous plants and making symbiotic association with them. The root cells of leguminous plants possess a purple coloured pigment known as leghaemoglobin in which Rhizobium floats to fix atmospheric nitrogen. Clostridium, Azotobacter, Aerobacter and Methanobacterium are free-living bacteria which fix atmospheric nitrogen. Rhizobial biofertilizer can fix 50–150 kg/Na/ ha/yr.

3. Pteriodophytes as Biofertilizers In the cavities of leaves of certain aquatic pteridophytes, such as Azolla , a large number of plants of Anabaena ( Anabaena azollae, a blue-green alga) are present which have the capacity to fix atmospheric nitrogen and make it available to Azolla (an aquatic pteridophyte). Azolla supplies nitrogen, increases organic matter and fertility in soil and shows tolerance against heavy metals. Dr. P. K. Singh has done an outstanding work on mass cultivation of Azolla and its uses as biofertilizer in rice and other crop fields.

4. Micorrhiza as Biofertilizer Mycorrhiza is the symbiotic relationship between a fungus and the roots of higher plant. Mycorrhiza performs the following functions : ● It enhances phosphate nutrition in plant. ● It enhances water and various other nutrient uptake. ● It enhances greater plant vigour (heterosis), growth and yield. Mycorrhiza is of two types—ectomycorrhiza and endomycorrhiza. Ectomycorrhiza increases water and nutrient intake by plants. It occurs in plants like Eucalyptus, Ficus, Oak, Pine, etc. It absorbs and stores nitrogen, potassium, phosphorus and calcium in fungus. It also converts complex organic molecules into simpler and easily absorbable form. Endomycorrhiza is important in phosphate nutrition of plants. It is also called Vesicular Arbascular Mycorrhiza (VAM). VAM has capacity to penetrate even the cortical cells.

Other Benefits of Mycorrhiza to Plants ● VAM fungi enhance water uptake in plants. ● They play a key role for selective absorption of immobile (Cu, Zn and P) and mobile (Fe, Mn, Cl, N, Br, S, Ca and K) elements to plants. These are available to plants in less amount. ● They increase resistance in plants and with their presence the effects of pathogens and pests on plant health is reduced. ● Some of the trees like pines cannot grow in new areas unless soil has mycorrhizal inocula because of limited or coarse root hairs. ● They increase the longevity of feeder roots, surface areas of roots by forming mantle and spreading mycelia into soil and, in turn, the rate of absorption of major and minor nutrients from soil resulting in enhanced plant growth.

OBJECTIVE QUESTIONS 1. An aquatic fern used as biofertilizer is— (A) Azolla (B) Marsilea (C) Salvia (D) None of the above 2. Manure containing a mixture of cattle dung and crop residues is known as— (A) Green manure (B) Farm yard manure (C) Organic manure (D) None of the above 3. Materials of biological origin that are applied commonly to maintain and improve soil fertility are termed— (A) Nitrogenous fertilizers (B) Biofertilizers

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(C) Manures (D) All of the above

(C) Supplies organic matter (D) None of the above

4. A legume having symbiotic association with Rhizobium and Aerorhizobium is— (A) Sebania rostrata (B) Sebanea aculeata (C) Crotolaria juncea (D) None of the above

7. Leaves of Azolla possess the colonies of— (A) Anabaena (B) Rhizobium (C) Both (A) and (B) (D) Azotobacter

5. Fertilizers applied to crop plants pollute— (A) Chiefly water resources (B) Chiefly atmosphere (C) Soil resources (D) Soil resources and water resources

8. Biofertilizer present in the roots of legume plants is—

6. A green manure— (A) Protects soil against erosion and leaching (B) Supplies additional nitrogen

ANSWERS

(A) Azospirillum (B) Anabaena (C) Rhizobium (D) All of the above

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Introduction and Occurrence ●

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Club fungi (division—Basidiomycota), which have septate hyphae, include the familiar mushrooms growing on lawns and the shelf or bracket fungi found on dead trees. Less but well known club fungi are puffballs, bird’s nest fungi and stinkhorns. This big group of fungi containing about 16000 species includes both saprophytic and parasitic species. Basidiocarps, also called fruiting bodies, contain the basidia, club-shaped structures which produce basidiospores and from which this division takes this name (club-fungi).

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The clamp connections are usually formed on the terminal cells of the hyphae of the secondary mycelium.

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The clamp connections by some mycologists are considered to be homologous to the hooks of ascogenous hyphae of the Ascomycetes.

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Presence of hooklike clamp connection is a safe criterion for distinguishing a secondary or dikaryotic mycelium from the primary or monokaryotic mycelium.

Mycelium ●

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The mycelium generally is a weft of interlacing and anastomosing hyphae. In a few genera, however, the mycelial hyphae run parallel to one another and get bundled together to form definite and conspicuous thick cords called rhizomorphs. Most of the Basidiomycetes are heterothallic. It means primary or homokaryotic mycelium in them is of two distinct strains which are called plus (+) and minus (–) strains. The well-developed filamentous mycelium consisting of a mass of branched, septate hyphae generally spreads in a fan-shaped manner. The mycelium of Basidiomycetes passes through three distinct stages—primary, secondary and tertiary before the fungus completes its life cycle.

Clamp Connection ●

The dikaryotic cell divides repeatedly by conjugate divisions to give rise to a secondary or dikaryotic mycelium. During nuclear divisions of the dikaryotic cell special structures called clamp connections are formed.

A

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C

D

E

F

Fig. (A–F) Basidiomycetes. Diagram illustrating the formation of clamp connections.

Sexual Reproduction ●

Club fungi usually reproduce sexually. However, development of sex organs, the male antheria and female ascogonia are universally absent throughout the class.

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The rudimentary differences in sex, shown at the time of sexual fusion, are designated as plus (+) and minus (–) signs. These signs are called sexual strains. Either of these mycelia, if cultured artificially, remains sterile. They form no fructifications.

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Fructifications are formed only if two mycelia of opposite strains come in contact. The sexual process being extremely simplified, consists of three fundamental processes such as karyogamy, sexual fusion or plasmogamy and meiosis.

Diplodisation (Dikaryotisation) The process by which the primary mycelium is converted into secondary mycelium or dikaryotic mycelium is called diplodisation or dikaryotisation. It may take place by the following methods— (i) By hyphal fusion—In this case, fusion occurs between the vegetative cells of two neighbouring hyphae. (ii) By the fusion a germinating basidiospore and a diploid cell of the basidium. (iii) By the fusion between the two haploid cells of opposite strains of the basidium. (iv) By the fusion between germinating odium of one strains of the basidium. (v) By conjugation of basidiospores—In this case two basidiospores of opposite strains meet and conjugate.

B

(i) Karyogamy ●

The terminal binucleate or dikaryotic cells of the hyphae of the secondary mycelium develop into basidia.

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The two nuclei in the dikaryotic cell fuse. This fusion is called karyogamy. The resultant diploid fusion nucleus is called a synkaryon. The young basidium containing the synkaryon is called probasidium which represents the transitory diplophase.

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Asexual Reproduction

Brand spores

(i) By Conidia : Dikaryotic

Diploid

C

B A

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The production of conidia is not so common occurrence in the Basidiomycetes. They are formed in the rusts, smuts and some other Basidiomycetes.

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In smuts, conidia are budded off from the basidiospores and the mycelium.

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The conidia in Basidiomycetes are produced by the dikaryotic mycelium. They serve to propagate the dikaryophase in the life cycle.

Young basidium

(ii) By Oidia : Basidiospore

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Oidia are hyaline, small and thin-walled unicellular sections or fragments of the mycelium.

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Oidia may be uni—or binucleate as whether they are produced by the breaking up of the primary or secondary mycelium.

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Oidia serve a double function; they may either germinate to form primary mycelia or bring about diplodisation. In the latter case the germinating odium acts as a spermatium and fuses with the somatic hyphal wall of opposite strain.

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In some species, the oidia are segmented from special, short lateral hyphal branches called the oidiophores.

Brand spore

F D

E

Fig. : (A–E) Basidiomycetes. Ustilago sp. (A) Dikaryotic hypha forming brand spores; (B–C) brand spores (dikaryotic and diploid); (D) Germinated diploid brand spore to form an epibasidium with four haploid nuclei arranged in a row; (E) mature fourcelled epibasidium bearing basidiospores; (F) Brand spore of Ustilago budding in a nutrient solution (after Brefeld).

(ii) Plasmogamy ●

Plasmogamy means the union of two protoplasts whereby the sexual nuclei of opposite strains come close together in a pair within the same cell.

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In Basidiomycetes, plasmogamy is achieved either by somatogamy or by spermatization.

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The two somatic hyphae of the primary mycelia of opposite strains come in contact and lie side by side in the fusion cell. This sexual union or plasmogamy by fusion of somatic cell is called somatogamy.

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In the homothallic species plasmogamy occurs by the formation of tubular connections between the somatic cells of the same mycelium. Plasmogamy by the union of a spermatium with a receptive hyphae (female organ) is known as spermatization. Plasmogamy by spermatization exclusively occurs in the rusts which produce numerous tiny, uninucleate, nonmotile spore-like bodies called spermatia. They are formed in flask-shaped organs, the spermatogonia. (iii) Meiosis :

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The synkaryon in the probasidium undergoes two successive nuclear divisions which constitute meiosis.

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Meiosis restores the haploid condition in the life cycle.

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Both karyogamy and meiosis take place in the basidium at different stages of development.

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(iii) By Fragmentation ●

Any part detached from parent cell containing conidiophores become able to form a new mycelium (plant body).

Basidiocarp and Basidium ● In higher Basidiomycetes (class = Homobasidiomycetidae) the secondary mycelium develops fruiting bodies called basidiocarps. ● The basidiocarps are usually massive aerial sporophores which bear basidia. ● The basidia, which are characteristic reproductive structures of Basidiomycetes, are of two types in general, the holobasidium and phragmobasidium. ● Holobasidia are aseptate and thus unicellular, whereas phragmobasidia are septate basidia. ● Holobasidia are characteristic of most of the Basidiomycetes particularly the gilled (gill-containing) or fleshy fungi. ● Basidia are developed in a palisade-like layer on the basidiocarp. This fertile layer is called hymenium. ● Basidia produce basidiospores, which are often windblown, when they germinate, the new haploid mycelia are formed. ● The septate basidium or phragmobasidium is typical of rusts and smuts which usually does not form any fructification or basidiocarp.

Rusts and Smuts are Parasitic Club Fungi Epibasidium

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● Hypobasidium

Hypobasidium

Epibasidium

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A C Basidiospores

B

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Sporidia

Rusts and smuts are club fungi that parasitize cereal crops, such as corn, wheat, rye and oats. They do not form the basidiocarps. Some smuts enter seeds of the host and exist inside the host plant, becoming visible only near maturity. In corn smuts, the mycelium grows between the corn kernel and secretes substances that cause the development of tumors on the ears of corn. The life cycle of rusts, which may be particularly complex, often requires two different plant host species to complete the life cycle. Black stem rust of wheat uses barberry leaf as an alternate host. Differences Between Rusts and Smuts Rusts

Migrating nucleus Sterigma

Rusts may be heteroe- 1. All smuts are autoecious and autoecious. cious. For example, wheat rusts are heteroecious and others are autoecious.

2.

The rusts are intercellular and obtain their nutrition by means of haustoria.

2. The smuts may be intercellular or intracellular (Ustilago maydis ). Haustoria are absent.

3.

Clamp connections on the secondary mycelium are rare.

3. Clamp connections are common.

4.

The dikaryotic mycelium produces three kinds of spores : uredospores and teleutospores on the primary host and aeciospores on the alternate host.

4. It produces only one kind of binucleate spores called the smut spores which are comparable to the teleutospores of rust fungi.

5.

Teleutospores are twocelled, stalked; and each cell is binucleate.

5. The brand spores (teleutospores) are unicellular and binucleate.

6.

The teleutospores are developed from the terminal cells of the mycelium.

6. Smut spores are formed from the intercalary cells.

7.

Each cell of the 2celled teleutospores produces an epibasidium which bears four basidiospores. They are borne on sterigmata and are discharged violently by the waterdrop method.

7. The single-celled, brand spore germinates to produce a single epibasidium which bears a variable number of basidiospores. They are not borne on sterigmata nor are they discharged violently.

Epibasidium Brand spore D E

Fig. : (A–E) Basidiomycetes. Different types of basidia. (A) Stichobasidial type; (B) Chiastobasidial type; (C) Tuning-fork type; (D) Holobasidium; (E) Stichobasidial type with a terminal cluster of septate, sickle-shaped sporidia. Pileus

Gills Annulus

Pores

Stipe Volva A

Mycelium B

D Pileus

C

Fruit body

G Log of wood Peridiole Volva E

F

Fig. : (A–G) Basidiomycetes. Common types of basidiocarps. (A) basidiocarp of Agaricus with gills on the underside of pileus; (B) basidiocarp of Boletus with pores on the underside of pileus; (C) basidiocarp of Lycoperdon; (D) basidiocarp of Geaster; (E) fruit bodies of Cyathus; (F) mature fruit body of Phallus; (G) fruit bodies of Fomes.

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Smuts

1.

Economic Importance ●

Several members of Basidiomycetes are of great economic importance because of their beneficial as well as harmful nature.

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Some of them are causative agents of most destructive diseases of cereal crops such as ‘smut disease’ of oats, wheat, corn, barley as well as the wheat rust.

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Stem rust of wheat is caused by Puccinia graminis tritici. Some of the higher Basidiomycetes, such as spore fungi, are the common wood rotters; they destroy lumbar and timber. Mushrooms which also belong to this group are of great economic value as food. The young fleshy sporophores of many species of ‘puff balls’, e.g., Lycoperdon and Clavatia are also edible.

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Clavatia has medicinal value as it contains anticancer substance calvacin. Many members of this class form ectomorphic mycorrhizal associations with the roots of forest trees. Important Smut Diseases of Cereals Cereals

Loose smut (Causal organisms)

Covered smut (Causal organisms)

Wheat

Tilletia caries

Oat

Ustilago nuda (race of U. tritici ) Ustilago avenae

Corn Barley

Ustilago maydis Ustilago nuda

Ustilago horde; ( a race of U. Kolleri ) —— Ustilago hordei

OBJECTIVE QUESTIONS 1. In certain members of Basidiomycetes, the basidium bears four spores exogenously, each usually at the tip of a minute stalk known as— (A) Sterigmata (B) Basidiophore (C) Both (A) and (B) (D) Conidiophore 2. ‘Loose smut’ of corn is caused by— (A) Ustilago maydis (B) Ustilago avenae (C) Ustilago nuda (D) All of the above 3. Which division of fungi is commonly known as club-fungi ? (A) Zygomycetes (B) Oomycetes (C) Deuteromycetes (D) Basidiomycetes 4. Clamp connection is characteristic of a certain group of— (A) Algae and cyanobacteria (B) Fungi (C) Bryophytes (D) None of the above 5. Puccinia graminis tritici causes— (A) Loose smut of oat (B) Black stem rust of wheat (C) Loose smut of barley (D) Covered smut of oat 6. Anticancer substance ‘calvacin’ is obtained from— (A) Club fungi (B) Red algae (C) Myxophycean cell (D) None of the above

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7. Conidiospores are formed— (A) When nutrients are in short supply (B) During sexual reproduction (C) By sporangia (D) By sac, club and imperfect fungi 8. Which of the following group(s) is/are comprised of club fungi ? (A) Mushrooms

6. During spermatogenesis, there are three phases in which the process is completed. The correct sequence of phases is— (A) Growth → Multiplication → Maturation (B) Multiplication → Growth → Maturation (C) Maturation → Multiplication → Growth (D) Multiplication → Maturation → Growth

(B) Puff balls (C) Truffles (D) All of the above 9. In which of the following the dikaryotic stage is longerlasting ?

7. Growth phase prepares oogonia or spermatogonia for first— (A) Mitotic division (B) Meiotic division

(A) Sac fungi

(C) Binary fission

(B) Club fungi

(D) Multiple fission

(C) Imperfect fungi (D) Zygospore fungi 10. The part of mushroom that is visible above the ground is— (A) Ascogonium (B) Ascocarp (C) Zygospore (D) Basidiocarp

ANSWERS

8. The sperm head is covered by a membrane—enclosed structure called— (A) Head

(B) Neck

(C) Middle piece (D) Acrosome 9. A pair of centrioles found in the sperm, are located in the region— (A) Head

(B) Neck

(C) Middle piece (D) Tail ●●●

(Continued from Page 737 ) 5. During oogenesis, each haploid cell produces— (A) Four functional eggs (B) Two functional eggs (C) One functional egg and two polar bodies (D) Four functional polar bodies

10. Which of the following is the genetic part of the sperm ? (A) Head

(B) Neck

(C) Middle piece (D) Tail

ANSWERS

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(C) Both (A) and (B) (D) None of the above

1. What happens during photorespiration when stomates are closed ? (A) The concentration of CO 2 increases in leaves (B) The concentration of CO 2 decreases in the leaves (C) Oxygen, a by-product of photosynthesis, increases (D) Both (B) and (C) 2. In which of the following ways the RNA differs from DNA ? (A) The pentose sugar is ribose, not deoxyribose (B) The base uracil replaces thymine (C) RNA is single stranded (D) All of the above 3. The fibres which are used in the manufacture of brooms are known as— (A) Brush fibres (B) Rough weaving fibres (C) Soft or bast fibres (D) None of the above 4. The genetic code is a triplet code and each codon consists of— (A) One base (B) Two bases (C) Three bases (D) Four bases 5. Tendrils of Smilax are homologous to— (A) Axillary buds (B) Stipules (C) Leaf apices (D) Leaflets 6. The internal mechanism by which a biological rhythm is maintained in the absence of appropriate environmental stimuli is termed as a— (A) Circardian rhythm (B) Sleep movement (C) Thigmotropism (D) Biological clock

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7. The first-formed elements of phloem are called— (A) Protophloem (B) Metaphloem (C) Medullary ray (D) Both (B) and (C) 8. Which part of a leaf carries on most of the photosynthesis of a plant ? (A) Guard cells (B) Epidermis (C) Epidermal layer (D) Mesophyll 9. Which of the following is commonly known as ‘thorn apple’ ? (A) Aegle marmelos (B) Ananas comosus (C) Datura stramonium (D) Garcinia mangostana 10. A change in the amount of energy in the form of heat liberated or absorbed by the system during physical or changes is termed as— (A) Enthalpy (B) Entropy (C) Ecological energetics (D) Pyramid of energy 11. Amino acids are converted into ammonia by a group of bacteria called— (A) (B) (C) (D)

Ammonifying bacteria Denitrifying bacteria Nitrifying bacteria All of the above

12. During the noncyclic electron pathway, electrons move from— (A) Water through PS-II to PS-I and then to NADP + (B) Water through PS-I to PS-II and then to NADP + (C) PS-I to PS-II (D) ADP to ATP 13. The mode of reproduction in phycomycetes is— (A) Asexual (B) Sexual

14. The botanical name of ‘prayer plant’ is— (A) Ipomoea (B) Lantana Camara (C) Osimum sanctum (D) None of the above 15. The chloroplast envelope encloses a liquid protenaceous substance called— (A) Protoplasm (B) Cytoplasm (C) Granum (D) Stroma 16. An agent, such as radiation or a chemical, that brings about a mutation is called— (A) Mutagen (B) Transcription (C) Polysome (D) Intron 17. In Utricularia, leaves are modified into— (A) Petiole (B) Bladder (C) Tendril (D) Pitcher 18. Which of the following scientists proved experimentally that DNA and not protein is the genetic material of the T2 bacteriophage ? (A) Watson and Crick (B) Hershey and Chase (C) Jacob and Monod (D) Miescher and Griffith 19. In shade leaves the ratio of chlorophyll to carotenoids approaches— (A) 5 : 1 (B) 2 : 3 (C) 3 : 2 (D) 1 : 1 20. Plasmodesmata are formed around the elements of— (A) Golgi bodies (B) Chloroplast (C) Nucleus (D) None of the above 21. Corymb inflorescence has— (A) Flattened peduncle (B) Long peduncle (C) Short peduncle (D) No peduncle 22. Which of the following is correct regarding aerobic respiration ? (A) It is common in all higher plants (B) Energy is liberated in the form of ATP

(C) End products are H2O and CO2 (D) All of the above 23. Cyathium and hypanthodium types of inflorescence are similar in having— (A) Unisexual flowers (B) Petaloid bracts (C) Apical pore (D) Nectar glands 24. Which of the following RNA carries a sequence of codons to the ribosomes ? (A) t-RNA (B) m-RNA (C) r-RNA (D) All of these 25. The structure which can show the effect of traits brought by the male gamete immediately after its formation is— (A) Plumule (B) Cotyledon (C) Endosperm (D) Embryo 26. A microbial metabolite excreted or released by lysed cells which in very low concentration is directly toxic to cells of the suscept, is defined as— (A) Pathogenesis (B) Taxis (C) Toxin (D) Etiology 27. Guttation is elimination of through— (A) Lenticels (C) Wounds

the process of water from plants (B) Stomata (D) Hydathodes

28. The core of the axis which includes the vascular system, the interfascicular portion, the pith in the vicinity of vascular bundles (pericycle) is called— (A) Stele (B) Cortex (C) Tracheid (D) Endodermis 29. Which of the following factors is most important in regulating transpiration ? (A) Wind (B) Temperature (C) Light (D) Humidity

(C) Hatch-Slack cycle (D) PP-pathway 32. An increase in the dry weight of a plant is dependent upon the presence of— (A) Oxygen (B) Iron (C) Nitrogen (D) Carbon dioxide 33. RNA retroviruses have a special (particular) type of enzyme responsible for— (A) Transcribing viral RNA to a DNA (B) Polymerizing host DNA (C) Disintegrating host DNA (D) Translating host DNA 34. Which of the following gives a possible sequence of organic chemicals prior to the protocell ? (A) Inorganic gases, nucleotides, nucleic acids, genes. (B) Inorganic gases, amino acids, polypeptide, microsphere (C) Both (A) and (B) (D) Water, salt, protein, oxygen 35. Nitrifying bacteria are under— (A) Vibrio group (B) Bacillus group (C) Coccus group (D) None of the above

kept

36. Which of the following xylem elements is living ? (A) Fibres (B) Vessels (C) Parenchyma (D) Tracheids 37. A rootless aquatic plant in which a portion of leaf is modified to form a bladder for catching small aquatic animals is— (A) Utricularia (B) Drosera (C) Nepenthes (D) Dionaea

30. Hira (HD 1941) and Moti (HD 1949) are the varieties of— (A) Wheat (B) Rice (C) Maize (D) Pulse

38. Golgi bodies are absent in— (A) Higher plants (B) Bacteria (C) Blue green algae (D) Both (B) and (C)

31. Sugarcane, maize and some other tropical plants have high efficiency of CO2 fixation because they operate— (A) TCA cycle (B) Calvin cycle

39. Herbaceous angiosperms flourished in— (A) Holocene epoch (B) Pleistocene epoch (C) Pliocene epoch (D) Miocene epoch

C.S.V. / August / 2009 / 767

40. How many primary types of histone molecules are known ? (A) One (B) Two (C) Four (D) Five 41. Who among the following coined the term cryptovegetarian ? (A) Draghetti (B) Spalanzani (C) Lamarck (D) Darwin 42. For each of the twenty amino acids found in proteins, there should be atleast— (A) One t-RNA molecule (B) Two t-RNA molecules (C) Three t-RNA molecules (D) Twenty t-RNA molecules 43. Cinnamon is obtained from which part of the plant body ? (A) Bark (B) Root (C) Leaf (D) Fruit 44. The edible part of tomato is— (A) Mesocarp only (B) Epicarp (C) Thalamus (D) Placenta and pericarp 45. A substance or mixture of substances which prevents, mitigates, destroys or repels any pest is commonly called— (A) Epinasty (B) Growth hormone (C) Antitransparent (D) Pesticide 46. Apple and pear are— (A) Schizocarpic and pome fruits (B) Schizocarpic and pepo fruits (C) Succulent and pome fruits (D) True and schizocarpic fruits 47. In which of the following plants, the seed germinates while still attached to the parent (main) plant ? (A) Rhizophora (B) Screwpine (C) Mango (D) All of the above 48. Polysome is a group of— (A) Nucleolus (B) Endoplasmic reticulum (C) Spherosomes (D) None of the above

49. Which of the following plants is commonly called Swallow wart ? (A) (B) (C) (D)

Calotropis procera Eucalyptus globulus Jatropha curcas Bryophyllum pinnatum

50. An allosteric site on an enzyme is— (A) Often involved in feedback inhibition (B) Where ATP attaches and gives up its energy (C) The same as the active site (D) All of the above

ANSWERS WITH HINTS

(Continued on Page 790 ) C.S.V. / August / 2009 / 768

(C) Chlamydomonas (D) Batrachospermum

1. Match Column A (Different types of fruits) with Column B (Different examples) then select the correct answer from the options given below— Column A (a) (b) (c) (d)

Column B

Capsule Lomentum Samara Regma (a)

1. 2. 3. 4.

Acer Iris Genarium Acacia

(b)

(c)

(d)

(A) 2

4

3

1

(B) 3

2

1

4

(C) 2

4

1

3

(D) 3

4

2

1

2. Proteins and polar lipids account for almost all of the mass of biological membranes; the small amount of carbohydrate present is generally part of— (A) Glycoproteins (B) Glycolipids (C) Both (A) and (B) (D) None of the above 3. Mobilization of stored food in germinating seed is triggered by— (A) Gibberellin

(B) Auxin

(C) Abscisic acid (D) Cytokinin 4. Match Column ‘A’ (Common name) with Column ‘B’ (Botanical name) then select the correct answer from the options given below— Column ‘A’ Column ‘B’ (a) Palmyra 1. Erianthus ravennae palm (b) Toddy palm 2. Borassus flabellifer (c) Plum grass 3. Typha angustata (d) Elephant 4. Caryota urens grass (a)

(b)

(c)

(d)

(A) 4

3

2

1

(B) 4

3

1

2

(C) 2

4

1

3

(D) 2

4

3

1

C.S.V. / August / 2009 / 769

5. The oil is extracted from the seeds of— (A) Brassica nigra (B) Brassica campestris (C) Both (A) and (B) (D) None of the above 6. Which of the following is the main internal ground tissue ? (A) Parenchyma (B) Epidermis (C) Endodermis (D) Pith 7. The small size of cells is best correlated with— (A) The fact that they are selfreproducing (B) An adequate surface area for exchange of materials (C) Their prokaryotic versus eukaryotic nature (D) All of the above are correct 8. Which of the following membranes prohibits the entry of both solvent and solute particles ? (A) Permeable membrane (B) Semipermeable membrane (C) Both (A) and (B) (D) Impermeable membrane 9. Which of the following microorganisms charges protein into ammonia ? (A) Nitrobacter (B) Azotobacter (C) Rhizobium (D) Bacillus mycoides 10. Synthesis of quinine was achieved by— (A) Alexander Fleming (B) Woodward and Doering (C) M.S. Swaminathan (D) U. Brahmchari 11. ‘Safed Lerma’ is a new variety of— (A) Onion (B) Grape (C) Garlic (D) Wheat 12. Which of the following algae is commonly known as ‘frogspawn’ ? (A) Ulothrix (B) Volvox

13. The water held tightly by the soil particle around them is known as— (A) Capillary water (B) Field capacity (C) Hygroscopic (D) None of the above 14. Which of the following statements is not correct regarding Cycas ? (A) Cycas plant is arboreal (B) It looks like a palm tree (C) The stem is columnar normally branched with deciduous leaf bases (D) Both (A) and (C) 15. The sporophyte of Riccia is represented by— (A) Spore sac only (B) Foot, seta and capsule (C) Foot and capsule (D) Spores and elaters 16. The process by which DNA of nucleus passes information to RNA is called— (A) Transcription (B) Translation (C) Transformation (D) Transduction 17. Chief means of perennation in liverworts is— (A) Tuber formation (B) Persistent apices (C) Both (A) and (B) (D) None of the above 18. Dry rot of sugarbeet and internal browning of cauliflower diseases are due to the deficiency of— (A) Zinc (B) Boron (C) Copper (D) Manganese 19. Membrane often contains steroid which is a type of lipid that contains— (A) No carbon rings (B) Four carbon rings (C) Only one carbon rings (D) None of the above 20. A red pigment found in the fruit of tomato is called— (A) Lycopene (B) Xanthophyll

(C) Myxoxanthin (D) Rhodopsin 21. Who coined the term ‘metabolism’ for all chemical processes carried on in cells ? (A) Schwann (B) Nawashchin (C) Twort (D) Strasburger 22. Inflorescence having a flattened axis, sessile flowers and a whorl of involucral bracts is— (A) Corymb (B) Head (C) Raceme (D) Umbel 23. For the purpose of gene therapy a retrovirus should be equipped with— (A) Recombinant RNA (B) t RNA (C) m RNA (D) r RNA 24. Non transfer of pollen from anther to stigma of the same flower due to mechanical barrier is— (A) Heterostyly (B) Herkogamy (C) Dichogamy (D) Cleistogamy 25. Regulation of gene activity in prokaryotes usually occurs at the level of— (A) Translation (B) Transcription (C) Both (A) and (B) (D) None of the above 26. Genotypically the pollen grains produced by an anther belong to— (A) One type (B) Many types (C) Two types (D) All of the above 27. Cross-pollination within a species is called— (A) Explant (B) Xenogamy (C) Allopolyploidy (D) Autopolyploidy 28. At constant temperature, the rate of transpiration will be higher at— (A) 1 km below sea level (B) 1 km above sea level (C) 2 km above sea level (D) Sea level

C.S.V. / August / 2009 / 770

29. The cytoplasmic portion of the protoplast is referred to as— (A) Endoplasm (B) Cytoplasm (C) Cytosome (D) Ectoplasm 30. Herbicide DCMU kills plants due to spoilage of— (A) Photophosphorylation (B) Electron transport (C) O2-evolution (D) Rubisco activity 31. Trisomy is a type of— (A) Euploidy (B) Polyploidy (C) Hyperploidy (D) Hypoploidy 32. Balance between CO2 and O2 is maintained by— (A) Photosynthesis (B) C4-pathway (C) Transpiration (D) Photorespiration 33. The potential energy of water is referred to as— (A) Protoplasmic streaming (B) Thermodynamics (C) Water relation (D) Water potential 34. From photosystem-I ‘high energy’ electrons pass to NADP where they combine with hydrogen ions to form— (A) NAD (B) ADP (C) NADPH2 (D) FAD 35. The noncyclic electron pathway generates— (A) No ATP (B) ATP only (C) NADPH only (D) Both ATP and NADPH 36. The first person to associate specific gene with a specific chromosome is/was— (A) Morgan (B) Swaminathan (C) Correns (D) Maheswari 37. The baloonlike outgrowth of parenchyma into the lumen of the vessels is known as— (A) Tunica (B) Tyloses (C) Histogen (D) Phellogen 38. Drosera catches insects by means of— (A) Pitcher (B) Adhesive disc (C) Bladder (D) Tentacles secreting shining liquid

39. Which of the following plants does not belong to family Papilionaceae ? (A) Phaseolus mungo (B) Bauhinia variegata (C) Crotolaria juncea (D) All of the above 40. Which of the following is a saprophytic angiosperm ? (A) Neottia (B) Agaricus (C) Cuscuta (D) Eucalyptus 41. Autopolyploidy arised by increase in the number of chromosome sets of the same species is called— (A) Intraspecific polyploidy (B) Interspecific polyploidy (C) Both (A) and (B) (D) Explant 42. Which of the following is a true hemp ? (A) Cannabis sativa (B) Boehmeria nivea (C) Girardinia heterophylla (D) Agave sisalana 43. Quadrifoliate leaves are found in— (A) Paris quadrifoliata (B) Bombax ceiba (C) Hardwickia (D) All of the above 44. The meristematic layer between bark and the wood in a woody stem is called— (A) Cork cambium (B) Zone of cell division (C) Vascular cambium (D) None of the above 45. Tribulus fruit is dispersed by means of— (A) Water (B) Animals (C) Wind (D) Explosion 46. In bryophytes, absorbing and attaching organs are— (A) Columella (B) Rhizoids (C) Root hairs (D) Whole thallus 47. The vertical temperature gradient over earth’s surface is called— (A) Phenology (B) Littoral zone (C) Sigmoid curve (D) Lapse rate

48. If concentration of CO2 in the external atmosphere is increased, the stomata are closed in— (A) Dark only (B) Light only (C) Both in light and dark (D) None of the above 49. Leaf primordia are produced by the— (A) Vascular bundle (B) Cambia (C) Cortex (D) Apical meristem 50. Carrageenin is used as emulsifying and stabilizing agent in— (A) Chocolates (B) Ice creams (C) Toothpastes (D) All of the above

ANSWERS WITH HINTS

●●●

C.S.V. / August / 2009 / 771

In each of the following questions, a statement of Assertion (A) is given and a corresponding statement of Reason (R) is given just below it. Of the statements, mark the correct answer as— (A) If both A and R are true and R is the correct explanation of A (B) If both A and R are true but R is not the correct explanation of A (C) If A is true but R is false (D) If both A and R are false (E) If A is false but R is true

PHYSICS 1. Assertion (A) : Light is diffracted around the edges of obstacles and the bending is so slight that it is not easily observed. Reason (R) : The wavelength of light is very small. (A) (B) (C) (D) (E) 2. Assertion (A) : Mass and energy are not conserved separately, but are conserved as a single entity called ‘mass energy’. Reason (R) : This is because one can be obtained at the cost of other as per Einstein’s equation E = mc 2. (A) (B) (C) (D) (E) 3. Assertion (A) : Coherent sources generate waves of identical frequencies. Reason (R) : Two coherent sources of circular wave patterns can be produced by passing a plane wave through narrow slits. (A)

(B)

(C)

(D)

(E)

4. Assertion (A) : The force experienced by a charged particle moving in a magnetic field can not do any work. Reason (R) : The force does not displace the charged particle. (A)

(B)

(C)

(D)

(E)

5. Assertion (A) : With a white light source the central fringe is dark in Young’s double slit experiment.

C.S.V. / August / 2009 / 772

Reason (R) : Light is formed of material particles. (A) (B) (C) (D) (E) 6. Assertion (A) : If there exists Coulomb attraction between two charged bodies, both of them may not be charged. Reason (R) : They will be oppositely charged. (A) (B) (C) (D) (E) 7. Assertion (A) : White light contains the range of colours in light from violet with a wavelength of 4 × 10– 7 m to red light with a wavelength of 7 × 10– 7 m. Reason (R) : When Young’s double slit experiment is carried out with light, multicoloured fringes are formed. (A) (B) (C) (D) (E) 8. Assertion (A) : To float, a body must displace liquid whose weight is greater than the actual weight of the body. Reason (R) : A floating body will experience no net downward force. (A) (B) (C) (D) (E) 9. Assertion (A) : Dark lines are observed in the spectrum of light produced from the sun rays. Reason (R) : Dark lines are due to absorption of certain radiations by gases in the outer atmosphere of the sun. (A) (B) (C) (D) (E) 10. Assertion (A) : Two sources A and B each carrying a sound of 400 Hz are standing a few metres apart. When A moves towards B, both persons hear the same number of beats per second. Reason (R) : Doppler shift in frequency of sound is the same whether the observer approaches the source or the source approaches the observer with same speed. (A) (B) (C) (D) (E)

CHEMISTRY 11. Assertion (A) : Iodine dissolves more appreciably in potassium

iodide solution as compared to pure water. Reason (R) : Potassium iodide solution absorbs iodine forming compound of potassium in which K+ is oxidised to K3 +. (A)

(B)

(C)

(D)

(E)

12. Assertion (A) : At isoelectric point of an amino acid, it does not migrate under the influence of an electric field. Reason (R) : At isoelectric point an amino acid is totally ionized. (A) (B) (C) (D) (E) 13. Assertion (A) : Mercury metal starts sticking to the side of the glass tube on bubbling ozone through it. Reason (R) : In the presence of ozone, mercury reacts with glass to form mercury (II) silicide. (A) (B) (C) (D) (E) 14. Assertion (A) : Methanoic acid changes mercuric chloride to mercurous chloride on heating but acetic acid does not do so under similar conditions. Reason (R) : Methanoic acid is a stronger acid than ethanoic acid. (A) (B) (C) (D) (E) 15. Assertion (A) : Sometimes a yellow turbidity appears while passing H2S gas even in the absence of radicals of 2nd group. Reason (R) : Group IV radicals are precipitated as their sulphides. (A) (B) (C) (D) (E) 16. Assertion (A) : In the electrolysis, the quantity of electricity needed to deposit one mole atom of silver is different from that required for depositing one mole atom of copper. Reason (R) : Atomic weights of silver and copper are different. (A) (B) (C) (D) (E) 17. Assertion (A) : The gelatinous white precipitate of aluminium hydroxide dissolves both in HCl as well as in concentrated NaOH solution.

Reason (R) : Aluminium hydroxide is a strongly ionic as well as covalent in nature. (A) (B) (C) (D) (E) 18. Assertion (A) : Nitrogen obtained by fractional distillation of liquid air is heavier than the nitrogen obtained by decomposing nitrogenous compounds. Reason (R) : Nitrogen obtained from fractional distillation of liquid air is not completely pure. (A) (B) (C) (D) (E) 19. Assertion (A) : The methyl carbanion is pyramidal in shape like the structure of ammonia molecule. Reason (R) : The carbon atom of methyl carbanion is sp 2 hybridised. (A) (B) (C) (D) (E) 20. Assertion (A) : Most of the endothermic reactions are not spontaneous under ordinary conditions but become so at elevated temperature. Reason (R) : Entropy of the system increases with increase in temperature. (A) (B) (C) (D) (E)

ZOOLOGY 21. Assertion (A) : Dramatic population growth has occurred in the last 300 years, with more growth in less developed countries (LDCs) than in more developed countries (MDCs). Reason (R) : Total fertility rate (TFR) is lower in MDCs, due to family planning and access to contraceptives. Infant mortality is lower in MDCs than LDCs. (A) (B) (C) (D) (E) 22. Assertion (A) : Capsule of tendon is associated with brain. Reason (R) : Inherited Rh factor gene is found in Rh – individuals. (A) (B) (C) (D) (E) 23. Assertion (A) : Baroreceptor is the receptor for hydrostatic pressure of blood. Reason (R) : Receptors of atrium stimulate the Cardio-acceleratory centre, helping to regulate blood pressure. (A) (B) (C) (D) (E)

C.S.V. / August / 2009 / 773

24. Assertion (A) : Coenzyme is a non-protein group without which certain enzymes are incomplete or inactive. Reason (R) : Coenzymes not only provide a point of attachment for the chemical groups being transformed but also influence the properties of the group. (A) (B) (C) (D) (E) 25. Assertion (A) : It is the brain, not the sense organs, that interprets the stimulus. Reason (R) : Sense organs are transducers, they transform the energy of stimulus to the energy of nerve impulses. (A) (B) (C) (D) (E) 26. Assertion (A) : Jelly fish is not a true fish which is a vertebrate animal with backbone. Reason (R) : The name jelly fish is given to invertebrate coelenterate animal, Aurelia, because it is made up of jelly-like substances. (A) (B) (C) (D) (E) 27. Assertion (A) : Girls have all their primary oocytes at birth, oogenesis begins at puberty. Reason (R) : The first menstrual cycle is called menopause, and the end of menstrual cycles is called menarche. (A) (B) (C) (D) (E) 28. Assertion (A) : Morphogenesis involves change in the shape of the embryo. Reason (R) : Differentiation is the specialization of cell structure and function as some genes are turned on and others off. (A) (B) (C) (D) (E) 29. Assertion (A) : The sympathetic division acts to mobilize the body for emergency or quick action— the ‘fight-or-flight’ response. Reason (R) : It decreases the heart rate, respiration rate and digestive activity; dilates blood vessels of the skeletal muscles, and constricts those of the skin. (A) (B) (C) (D) (E) 30. Assertion (A) : Neurosecretion is pivotal in the vertebrate endocrine system. Reason (R) : Hormones are produced by endocrine glands. (A) (B) (C) (D) (E)

BOTANY 31. Assertion (A) : In Mucor mucedo all of the sporangiophores obtained from a single germ sporangium give rise to mycelia which are of the same mating type. A zygospore germination of this type is known as pure homothallic. Reason (R) : Sexual reproduction in a heterothalic species of Mucor is not effected by copulation of two multinucleate isogametangia. (A) (B) (C) (D) (E) 32. Assertion (A) : Hershey and Chase turned to blue-green algae as their experimental material. Reason (R) : Franklin prepared an X-ray photograph of DNA that showed its certain dimensions. (A) (B) (C) (D) (E) 33. Assertion (A) : The sterile portion of the sporangium in Mucor is called pileus. Reason (R) : In Mucor, the remainder of the progametangium is known as the suspensor. (A) (B) (C) (D) (E) 34. Assertion (A) : Archegonium is the female sex organ of liverworts, ferns and most gymnosperms. Reason (R) : Archegonium is multicellular of which the swollen base, i.e., venter contains eggcell. (A) (B) (C) (D) (E) 35. Assertion (A) : Transposable elements can block gene expression by inserting into a gene. Reason (R) : Transposable elements can move and insert into genes, disrupting their expression. (A) (B) (C) (D) (E) 36. Assertion (A) : Daily periodicity of stomatal movement is not a characteristic of succulent plant. Reason (R) : In succulent plants stomata are closed at night and opened during the day. (A) (B) (C) (D) (E) 37. Assertion (A) : Scientists use several approaches when they design experiments in order to make data as valid as possible.

Reason (R) : Researchers might examine hundreds of cells, because a large sample size helps ensure meaningful results. (A) (B) (C) (D) (E) 38. Assertion (A) : Lecanora is a genus of lichens in which the thallus is crustose. Reason (R) : Some species of Lecanora are more resistant than

most lichens to atmospheric pollution. (A) (B) (C) (D) (E)

roplast and does not directly require solar energy. (A) (B) (C) (D) (E)

39. Assertion (A) : The energy capturing portion of photosynthesis takes place in thylakoid membranes and can not proceed without solar energy. Reason (R) : The synthesis part of photosynthesis occurs in chlo-

40. Assertion (A) : Cellular respiration includes glycolysis and fermentation. Reason (R) : Both glycolysis and fermentation are only aerobic respiration. (A) (B) (C) (D) (E)

ANSWERS WITH HINTS

(Continued on Page 790 ) C.S.V. / August / 2009 / 774

Physics

Chemistry

1. A convex lens is made of two different materials. A point is placed on the principal axis. The number of images formed by the lens will be two. —T/F 2. No net force acts on a rectangular coil carrying a steady current when suspended freely in a uniform magnetic field. —T/F 3. In an electric field, the electron moves away from higher potential to the lower potential. —T/F 4. Birds fly off a high tension wire when the current is switched on. —T/F 5. Fuse wire must have high resistance and low melting point. —T/F 6. In a perfectly elastic collision, the kinetic energy may not remain constant. —T/F 7. The sensitivity of a moving coil galvanometer can be increased by using a suspension wire of shorter length. —T/F 8. When a monocycle turns, along the curve, remaining vertical at a constant speed, the angular velocity of the wheel does not change. —T/F 9. Cathode rays constitute a stream of negatively charged particles, called electrons. —T/F 10. A piece of red glass is heated till it glows in dark. The colour of the glowing piece will be somewhat orange. —T/F 11. A body, whatever its motion, is always at rest in a frame of reference which is fixed to the body itself. —T/F 12. An electron revolves round a nucleus of charges Ze. In order to excite the electron from the state n = 2 to n = 3, the energy required is 47·2 eV. Z is equal to 5. —T/F 13. On reducing the volume of a gas at constant temperature, the pressure of the gas increases. —T/F 14. X-rays cannot be diffracted by means of grating. —T/F 15. Thermal conductivity of air being less than that of felt, even then we prefer felt to air for thermal insulation. —T/F

16. The structural-pair geometry of xenon difluoride (XeF2) is linear. —T/F 17. The heat of hydrogenation of trans-2-butene is more than that of cis-2-butene. —T/F 18. A balloon is inflated with He to a volume of 4·5 L at room temperature. If balloon is taken to the place having temperature of – 10°C, the volume of the balloon remains unchanged. —T/F 19. The electronic configuration of fluorine atom is 1s 2, 2s 2 2px 2 2py 2 2pz 1, and according to Hund’s rule the last electron enters to 2 pz orbital with clockwise spin. —T/F 20. Phenol is a weaker acid than carbonic acid. —T/F 21. A redox reaction in which reactants containing same element in different oxidation states react to give a species in which the element is in same oxidation state is called a comproportionation reaction. —T/F 22. The oxidation of sucrose with conc. HNO3 in presence of V2O5 leads to the formation of 1, 6hexanedioic acid. —T/F 23. When one mole electron is passed separately through each of solutions of CuSO4 and AgNO3, the mole ratio of Cu and Ag deposited will be 1 : 1. —T/F 24. Colour is shown by a transition metal ion which contains all ( n – 1) d vacant orbitals. —T/F 25. The vapour density of a substance has no unit and is independent of temperature and pressure. —T/F 26. Alkaline KMnO4 solution is more readily decolourised by benzene than by acetylene. —T/F 27. A gas has vapour density of 17, 10 g of this gas at 17°C and 1 atm will occupy a volume of 7 mL. —T/F 28. Gatta percha and natural rubber are actually, stereoisomers. —T/F 29. SnCl 2 is a crystalline solid while SnCl4 is a volatile liquid. —T/F 30. Methyl cyanide is soluble in water while methyl isocyanide is insoluble. —T/F

C.S.V. / August / 2009 / 775

Botany

Zoology 31. IgE is not fastened by its C-terminal end. 32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42. 43. 44. 45.

—T/F The rate of oxygen consumption of an organism or tissue is called oxygen quotient. —T/F Osteocytes maintain bone, osteoblasts form new bone and osteoclasts attack and destroy bones. —T/F Pentose phosphate pathway is alternative to glycolysis. —T/F Twelve pairs of cranial nerves and thirty one pairs of spinal nerves comprise the somatic peripheral nerves in human nervous system. —T/F Some developmental structures or processes such as gill pouches in mammalian embryos are regarded as phyletic. —T/F The circulatory system of cockroach, Periplaneta americana, is of open type. —T/F The occurrence of different morphological stages during the life of an organism is called pleiomorphism. —T/F Buffers limit pH changes and protect living cells from injury by strong acids and bases. —T/F Prosimian is a group of primates that include apes and humans. —T/F Restriction enzymes are not used to cleave the plasmid DNA for genetic engineering. —T/F Planarians are free-living turbellarians. —T/F Epitope is an antibody determinant. —T/F Cyclic AMP is a second messenger within cells. —T/F Organic evolution is a change in the frequency of alleles in a population. —T/F

46. Mitochondria supply most of the necessary biological energy by breaking down of protein, sugar and carbohydrates. —T/F 47. The highest point in the reaction coordinate diagram represents the idiogram. —T/F 48. Genes which confer antibiotic resistance on bacteria are located on plasmid. —T/F 49. Some plants of dry and arid habitats shed their leaves to reduce water loss. —T/F 50. Colour-blindness is not sex-linked inheritance. —T/F 51. The edible part of Oryza sativa is epicarp and thalamus. —T/F 52. Ergotism is caused by Claviceps purpurea. —T/F 53. Plant tissue culture cannot produce haploid plants. —T/F 54. Xylem transports water from the root to the leaf. —T/F 55. Stomata arise from the protoderm cells. —T/F 56. Stomata in angiosperms open and close owing to change in turgor pressure in guard cells. —T/F 57. The pine stem is characterized by the presence of conspicuous resin ducts, which are distributed throughout the stem. —T/F 58. The hormone cytokinin is used for early ripening of fruits. —T/F 59. Permeable membranes are those which allow diffusion of both solvent and solute molecules through them. —T/F 60. Mustard shows hypogeal germination.

ANSWERS WITH HINTS

C.S.V. / August / 2009 / 776

—T/F

Physics Q. Can a body have zero velocity and constant acceleration ?

☞

Q. What is a solenoid ? Comment on the magnetic field around a solenoid ? Q. What is meant by bandwidth ?

☞

☞

Q. A hydrogen atom contains one electron. But the spectrum of hydrogen atom has many lines. Why ? Q. What is the difference between the interference and diffraction of light ?

☞

☞

Q. What is axial or pseudo vector ?

☞ Q. A transistor is a temperature sensitive device. Why ?

☞ Q. What is Lami’s theorem ?

☞ C.S.V. / August / 2009 / 778

Q. What is the radius of gyration ?

☞

Q. What is law of equipartition of energy ?

☞ Q. What are gemstones ?

☞

Q. Hydrogen peroxide finds different uses in different concentrations.

☞

Q. What is total internal reflection ?

☞ Q. What are mustard oils ?

☞

Q. How calcium cyanamidecarbon mixture acts as a fertilizer ?

☞

Q. What unbreakable plasticcrockery is made of ?

☞

Q. What is the difference between mass and weight of an object ?

☞

Chemistry Q. How the process of osmosis helps developing adema and preserving meat and fruits ?

☞

C.S.V. / August / 2009 / 779

Q. Why the product of frequency and wavelength of infrared and ultraviolet radiations is always same ?

☞

Zoology Q. What pitch ?

is

perception

of

☞

Q. What are good and bad cholesterol ?

☞

Q. What is the mass of a single atom of 12C in grams and the number of grams per amu ?

☞

Q. What will be the value of ideal gas constant, R, if exactly 1 mol of an ideal gas occupies a volume of 22·414 litre at 0°° C and 1 atom pressure ?

Q. What is teleology ?

☞

☞ Q. How do the digestive tracts of carnivores differ from those of hervivores ?

☞

C.S.V. / August / 2009 / 780

Q. What is parsnip ? What are its importance ?

☞

Botany Q. What do you mean by pure culture ?

☞

●●● Q. How are monosaccharides formed from polysaccharides ?

(Continued from Page 777 )

☞

Q. Why does coconut oil freeze during winter ?

☞ Q. Why is CO2 commonly known as green house gas ?

☞

Q. What do you Enterobacter ?

mean by

☞ Q. What is forest engineering and its significance ?

☞

●●●

C.S.V. / August / 2009 / 781

1. A police van moving on a high way with a speed 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 ms–1, with what speed does the bullet hit the thief’s car ? (A) 100 ms –1 (B) 105 ms –1 (C) 95 ms –1 (D) None of these 2. A 4 kg block A is placed on the top of 8 kg block B which rests on a smooth table. A just slips on B when a force of 12N is applied on A. What is maximum horizontal force F required to make both A and B move together ?

(A) 16N (C) 36N

(B) 24N (D) 72N

3. Point P, Q and R are in a vertical line such that PQ = QR. A ball at P is allowed to fall freely. Find the ratio of time of descent through PQ and QR— 1 1 (A) (B) 2–1 2–1 (C)

2–1 1

(D)

3 2

4. An ideal gas confined to an insulated chamber is allowed to enter into an evacuated insulated chamber. If Q, W and ΔE int have the usual meanings, then— (A) Q = 0, W ≠ 0 (B) W = 0, Q ≠ 0 (C) ΔEint = 0, Q ≠ 0 (D) Q = W = ΔFint = 0 5. A hydrogen atom and a doubly ionised lithium atom are both in the second excited state. If LH and LLi respectively represent their electronic angular momenta and EH and ELi their energies, then— (A) LH > LLi and | EH | > | ELi | (B) LH = LLi and | EH | < | ELi |

C.S.V. / August / 2009 / 782

(C) LH = LLi and | EH | > | ELi | (D) LH < LLi and | EH | < | ELi | 6. Which of the following is not correctly matched ? (A) Hexadecane—C6H14 (B) Heptadecane—C17H36 (C) Eicosane—C 20H42 (D) Docosane—C22H46 7. Alkenes may be hydrated to alcohols by absorption in— (A) Dilute sulphuric acid (B) Dilute hydrochloric acid (C) Concentrated sulphuric acid (D) Concentrated hydrochloric acid 8. ‘Cement fondu’ is a/an— (A) Hydraulic cement composed chiefly of calcium aluminate (B) Hydraulic cement composed chiefly of sodium silicate (C) Portland cement (D) Cement made from granulated blast furnace slag of fairly low alumina 9. Which of the following is/are said to be diamagnetic species ? (A) Carbanion (B) Carbocation (C) Singlet carbene (D) All of the above 10. Phenols are more acidic than alcohols because— (A) Alcohols do not lose H + ions (B) Phenoxide ion is stabilized by resonance (C) Phenoxide ion does not exhibit resonance (D) Phenols are more soluble in polar solvents 11. Medrysone is a/an— (A) Newly discovered mutagen from various insects (B) Adrenal corticosteroid drug (C) Pill for contraceptive treatment (D) Endonuclease enzyme 12. In which of the following organisms, the head is prolonged into a tubular rostrum ? (A) Sea horse (B) Varanus (C) Heloderma (D) Ophisaurus

13. Which of the following is/are correct about interclavicular of pigeon ? (A) It is a median (B) It is unpaired (C) Connected to the secondary bronchi of both lungs (D) All of the above 14. How many nucleotide pairs are reported in β-globin gene of mouse by Leder et al ? (A) 116 (B) 216 (C) 96 (D) 106 15. The induction of regional anesthesia by preventing sensory nerve impulse from reaching centres of consciousness, is known as— (A) Nerve back (B) Nerve entrapment syndrome (C) Nerve growth factor (D) Nephrydrosis 16. Soil is chiefly composed of— (A) Only water and organic matter (B) Air + water + organic matter + minerals (C) Only organic colloids (D) None of the above 17. In balausta type of fruit, the edible part is— (A) Aril (B) Fleshy thalamus (C) Succulent testa (D) Succulent sepal 18. From which one of the following flowers is the insecticide ‘Pyrethrum’ derived ? (A) Rosa (B) Nelumbo (C) Iberis (D) Chrysanthemum 19. The mutagen proflavin is a/an— (A) Base analog (B) Hydroxylating agent (C) Acridine dye (D) Alkalylating agent 20. Bud scales are found in— (A) Jack fruit (B) Ficus (C) Magnolia (D) All of the above ●●●

Rules for taking part in Quiz Contest of Competition Science Vision 1. 2.

3. 4.

5. 6. 7.

8.

All students or those appearing in competitive examinations can take part in this contest. Candidates taking part in quiz contest will necessarily have to send their entries by a fixed date. Entries are to be sent by ordinary post. Please mark your envelope 'Quiz–Competition Science Vision' on the top left hand side. Answers given only on the form of the magazine will be admissible. In the form there are four squares against each question number. Contestants should put a cross (×) in the square for the answer they think is correct. Giving more than one answer to a question will disqualify it. Contestants should essentially write the number of questions they have solved. Marks will be deducted for wrong answers. The candidate sending the maximum number of correct answers will be given Rs. 600 as first prize. Next two candidates after that will get Rs. 400 and Rs. 300 as second and third prize respectively. If there are more than one candidate eligible for a prize, the amount will be equally distributed among them. The decision of the editor will be final and binding in all cases, and will not be a matter for consideration of any court.

Solution to Quiz No. 135 Competition Science Vision Last date for sending 28th August, 2009 Name Mr./Miss/Mrs. ...........................…......................... Full Address ...................................….......................... .............................................................………………… ............................................................………………… State .......................Pin Code No. Age.................. Academic Qualification........................ Competition examination for which preparing .................................................................................... I have read and understood the rules of quiz contest of Competition Science Vision issued by Pratiyogita Darpan and agree to abide by them. ................................... (Signature) RESULT No. of questions attempted.......................................... No. of correct answers................................................. No. of wrong answers................................................... Marks obtained............................................................. ANSWER FORM Q. No. A

C.S.V. / August / 2009 / 783

B

C

D

Q. No. A

1.

11.

2.

12.

3.

13.

4.

14.

5.

15.

6.

16.

7.

17.

8.

18.

9.

19.

10.

20.

B

C

D

According to the rules of the CSV Quiz, all entry forms were examined. As a result, the following participants have qualified for various prizes. CSV sends them greetings and good wishes for their bright future. It also places on record its appreciation for their inquisitive nature and expresses obligation for their co-operation.

PRIZE WINNERS First Prize Ravi Jaiswal C/o Gaurav Jaiswal Room No. 88, A. N. Jha Hostel, University of Allahabad, Allahabad U.P.–211 002 Second Prize Vishal Tiwari III-B-231 Vidyut Vihar Colony Shaktinagar, Sonebhadra U.P.–231 222 Third Prize 1. Gagandeep Singh C/o Dayaram Verma L-971, Shastri Nagar, Meerut U.P.–250 004 2. Sarvdaman Sharma Lane No. 23 Greater Kailash, Jammu J & K–180 011 Note : Amount of third prize has been distributed among two third prize winners.

C.S.V. / August / 2009 / 785

1. For which one of the following countries, is Spanish not an official language ? (A) Chile (B) Colombia (C) Republic of Congo (D) Cuba 2. For which one of the following, is Tirupur well known as a huge exporter to many parts of the world ? (A) Gems and Jewellery (B) Leather goods (C) Knitted garments (D) Handicrafts 3. The great Asian river Mekong does not pass through— (A) China (B) Malaysia (C) Cambodia (D) Loas 4. Which one of the following does not border Panama ? (A) Costa Rica (B) Pacific Ocean (C) Colombia (D) Venezuela 5. The waterfall Victoria is associated with the river— (A) Amazon (B) Missouri (C) St. Laurence (D) Zambezi 6. Israel has common borders with— (A) Lebanon, Syria, Jordan and Egypt (B) Lebanon, Syria, Turkey and Jordan (C) Cyprus, Turkey, Jordan and Egypt (D) Turkey, Syria Iraq and Yemen 7. What is the correct sequence of the rivers Godavari, Mahanadi, Narmada and Tapti in the descending order of their lengths ? (A) Godavari – Mahanadi – Narmada – Tapti (B) Godavari – Narmada – Mahanadi – Tapti (C) Narmada – Godavari – Tapti – Mahanadi (D) Narmada – Tapti – Godavari – Mahanadi

C.S.V. / August / 2009 / 786

8. Which one of the following statements is correct ? (A) Liquid Sodium is employed as a coolant in nuclear reactors (B) Calcium Carbonate is an ingredient of toothpaste (C) Bordeaux mixture consists of Sodium Sulphate and lime (D) Zinc amalgams are used as a dental filling

13. Other than India and China which of the following groups of countries border Myanmar ? (A) Bangladesh, Thailand and Vietnam (B) Cambodia, Laos and Malaysia (C) Thailand, Vietnam and Malaysia (D) Thailand, Laos and Bangladesh

9. Diffusion of light in the atmosphere takes place due to— (A) Carbon dioxide (B) Methane (C) Helium (D) Water Vapours and Dust Particle

14. Through which one of the following groups of countries does the Equator pass ? (A) Brazil, Zambia and Malaysia (B) Colombia, Kenya and Indonesia (C) Brazil, Sudan and Malaysia (D) Venezuela, Ethiopia and Indonesia

10. In which country is the committee which selects winners for Nobel Peace Prize located ? (A) Norway (B) Sweden (C) Finland (D) Denmark 11. Consider the following statements— 1. Caffeine a constituent of tea and coffee is a diuretic. 2. Citric acid is used in soft drinks. 3. Ascorbic acid is essential for the formation of bones and teeth. 4. Citric acid is a good substitution for ascorbic acid in our nutrition Which of the statements given above are correct ? (A) (B) (C) (D)

1 and 2 only 1, 2 and 3 only 3 and 4 only 1, 2, 3 and 4 only

12. Which one of the following subjects is under the Union list in the Seventh Schedule of the Constitution of India ? (A) Regulation of labour and safety in mines and oil fields (B) Agriculture (C) Fisheries (D) Public Health

15. Which one of the following pairs is not correctly matched ? Railway Zone

Headquarters

(A) North Eastern Railway

Gorakhpur

(B) South Eastern Railway

Bhubaneshwar

(C) Eastern Railway

Kolkata

(D) South East Bilaspur Central Railway 16. Huangpu River flows through which one of the following cities ? (A) Beijing (B) Ho Chi Minh City (C) Shanghai (D) Manila 17. Consider the following statements— 1. Dengue is a protozoan disease transmitted by mosquitoes 2. Retro-orbital pain is not a symptom of Dengue. 3. Skin rash and bleeding from nose and gums are some of symptoms of Dengue.

Which of the statements given above is/are correct ? (A) 1 and 2 (B) 3 only (C) 2 only (D) 1 and 3 18. Which one of the following pairs is not correctly matched ? (A) Southern Air Command —Thiruvananthapuram (B) Eastern Naval Command —Vishakhapatnam (C) Armoured Corps Centre and School —Jabalpur (D) Army Medical Corps Centre and School —Lucknow 19. Itaipu Dam built on the river Parana is one of the largest dams in the world. Which of the following two countries have this as a joint project ? (A) Brazil and Peru (B) Paraguay and Equador (C) Brazil and Paraguay (D) Colombia and Paraguay 20. Consider the following statements— 1. There are 25 High Courts in India. 2. Punjab, Haryana and the Union Territory of Chandigarh have a common High Court. 3. National Capital Territory of Delhi has a High Court of its own. Which of the statements given above is/are correct ? (A) 2 and 3 (B) 1 and 2 (C) 1, 2 and 3 (D) 3 only 21. Which one of the following major Indian cities is most eastward located ? (A) Hyderabad (B) Bhopal (C) Lucknow (D) Bengaluru (Bangalore) 22. Out of the four southern states : Andhra Pradesh, Karnataka, Kerala and Tamil Nadu which one shares boundaries with the maximum number of Indian states ? (A) Andhra Pradesh only (B) Karnataka (C) Andhra Pradesh and Karnataka both (D) Tamil Nadu and Kerala both

C.S.V. / August / 2009 / 787

23. Which one of the following types is used by computerised tomography employed for visualisation of the internal structure of human body ? (A) X-rays (B) Sound waves (C) Magnetic resonance (D) Radioisotopes 24. Production of which one of the following is a function of the liver ? (A) Lipase (B) Urea (C) Mucus (D) Hydrochloric acid 25. Which one of the following is not a digestive enzyme in the human system ? (A) Trypsin (B) Gastrin (C) Ptyalin (D) Pepsin 26. Which one of the following is printed on a commonly used flourescent tubelight ? (A) 220 K (B) 273 K (C) 6500 K (D) 9000 K 27. What does the 104th Constitution Amendment Bill relate to ? (A) Abolition of Legislative Councils in certain states (B) Introduction of dual citizenship for persons of Indian origin living outside India (C) Providing quota to socially and educationally backward classes in private educational institutions (D) Providing quota for religious minorities in the services under the Central Government 28. From North towards South, which one of the following is the correct sequence of the given rivers in India ? (A) Shyok—Spiti—Zaskar— Satluj (B) Shyok—Zaskar—Spiti— Satluj (C) Zaskar—Shyok—Satluj— Spiti (D) Zaskar—Satluj—Shyok— Spiti 29. In which state is the Rajiv Gandhi National Institute of Youth Development located ? (A) Tamil Nadu (B) Karnataka (C) Himachal Pradesh (D) Uttarakhand

30. What is the Universal Product Code (UPC) adopted for ? (A) Fire safety code in buildings (B) Earthquake resistant building code (C) Bar code (D) Against adulteration in eatables

Directions—(Q. 1–7) In each question below is given a group of letters followed by four combinations of digits/symbols lettered (A), (B), (C) and (D). You have to find out which of the combinations correctly represents the group of letters based on the coding system and mark the letter of that combination as your answer. If none of the combinations correctly represents the group of letters, mark (E) i.e. ‘None of these’ as your answer. Letter : MATWREKIHFUBNP Digit/Symbol Code : 4@ 3 7 © 2 8%1 # $ 6 9 5 Conditions : (i) If both the first and the last letters are consonants, both are to be coded as the code for the first letter. (ii) If both the first and the last letters are vowels, both are to be coded as the code for the last letter. (iii) If the first letter is a vowel and the last letter is a consonant, the codes are to be interchanged. (iv) If the first letter is a consonant and the last letter is a vowel, both are to be coded as ★. 1. AEBRMH (A) @26©41 (B) 126©4@ (C) 126©41 (D) @26©4@ (E) None of these 2. HBEAFU (A) 162@#$ (B) ★62@#★ (C) 162@#1 (D) $62@#1 (E) None of these 3. BKNIRT (A) 689%©6 (B) 689%©3 (C) 698%©3 (D) 389%©3 (E) None of these 4. KFUBNA (A) ★#$69★ (B) 8#$69@ (C) 8#$698 (D) @#$69@ (E) None of these 5. IRFAME (A) 2©#@4% (B) 2©#@42 (C) %©#@4% (D) ★©#@4★ (E) None of these

C.S.V. / August / 2009 / 788

6. MRTPFW (A) 4©35#7 (B) 7©35#4 (C) 7©35#7 (D) 4©35#4 (E) None of these 7. ENTHWR (A) 29317© (B) 293172 (C) ©9317© (D) ★9317★ (E) None of these Directions—(Q. 8–12) Below in each question are given two statements (A) and (B). These statements may be either independent causes or may be effects of independent causes or a common cause. One of these statements may be the effect of the other statement. Read both the statements and decide which of the following answer choice correctly depicts the relationship between these two statements. Give answer— (A) if statement (A) is the cause and statement (B) is its effect. (B) if statement (B) is the cause and statement (A) is its effect. (C) if both the statements (A) and (B) are independent causes. (D) if both the statements (A) and (B) are effects of independent causes. (E) if both the statements (A) and (B) are effects of some common cause. 8. (A) The prices of vegetables have increased substantially during past few weeks. (B) Consumer price index at the end of previous week was increased by 2 per cent. 9. (A) Many employees of the company proceeded on a day’s leave on Friday. (B) Both Thursday and Saturday were declared holiday by the company. 10. (A) Many anti-social elements are caught by the police from the locality. (B) Many people in the locality are detained by the police for questioning.

11. (A) Many pilgrims used Govt. transport to travel to the holy shrine. (B) The cost of travel by private transport is very high. 12. (A) There has been heavy rains in the catchment area of the lakes supplying drinking water to the city. (B) The municipal authority has suspended the proposed cut in water supply to the city. 13. Among P, Q, R, S and T, each having different weight, R is heavier than only P. S is lighter than Q and heavier than T. Who among them is the heaviest ? (A) Q (B) P (C) S (D) Data inadequate (E) None of these 14. What should come next in the following number series ? 87 65432 1 765 4 3 2 1 6 5 4 3 2 1

(A) 6 (B) 4 (C) 7 (D) 5 (E) None of these 15. In a certain code language, ‘come now’ is written as ‘ha na’; ‘now and then’ is written as ‘pa da na’ and ‘go then’ is written as ‘sa pa’. How is ‘and’ written in that code language ? (A) sa (B) pa (C) na (D) Cannot be determined (E) None of these 16. Each consonant in the word DISTEMPER is replaced by the next letter in the English alphabet and each vowel in the word is replaced by the previous letter in the English alphabet, which of the following will be the fourth letter from the right end after the replacement ? (A) T (B) M (C) S (D) P (E) None of these

17. How many meaningful English words can be made with the letters NDOE using each letter only once in each word ? (A) None (B) One (C) Two (D) Three (E) More than three 18. Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group ? (A) BD (B) NQ (C) RP (D) MK (E) FH 19. In a certain code GEAR is written as ‘5934’ and RIPE is written as ‘4869’. How is PAGE written in that code ? (A) 6359 (B) 6539 (C) 4359 (D) 6459 (E) None of these 20. How many such digits are there in the number 64382179 each of which is as far away from the beginning of the number as when the digits are arranged in ascending order within the number ? (A) None (B) One (C) Two (D) Three (E) More than three 21. Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group ? (A) 39 (B) 91 (C) 78 (D) 52 (E) 45 22. How many such pairs of letters are there in the word ORDINAL each of which has as many letters between them in the word as in the English alphabet ? (A) None (B) One (C) Two (D) Three (E) More than three Directions—(Q. 23 and 24) Read the following information carefully and answer the question which follow— ‘P × Q’ means ‘P is brother of Q’. (ii) ‘P ÷ Q’ means ‘P is sister of Q’. (iii) ‘P – Q’ means ‘P is mother of Q’. (iv) ‘P + Q’ means ‘P is father of Q’ (i)

C.S.V. / August / 2009 / 789

23. Which of the following means ‘D is maternal uncle of K’ ? (A) D ÷ N – K (B) D ÷ N + K (C) D × N – K (D) D × N + K (E) None of these

29. Who is to the immediate left of M? (A) K (B) W (C) D (D) T (E) None of these

24. Which of the following means ‘M is nephew of R’ ? (A) M × T + R

30. In which of the following pairs is the second person sitting to the immediate right of the first person ? (A) DT (B) TP (C) PR (D) KW (E) None of these

(B) (C) (D) (E)

R÷J+M×T R÷J+M R×J+M None of these

Directions—(Q. 25–27) Following questions are based on the five threedigit numbers given below— 519 378 436 624 893 25. If the positions of the first and the third digits within each number are interchanged, which of the following will be the second smallest number ? (A) 519 (B) 378 (C) 436 (D) 624 (E) 893 26. If ‘1’ is subtracted from the first digit in each number and ‘1’ is added to the second digit in each number, which of the following will be the third digit of the second highest number ? (A) 9 (B) 8 (C) 6 (D) 4 (E) 3 27. If the positions of the first and the second digits within each number are interchanged, which of the following will be the highest number ? (A) 519 (B) 378 (C) 436 (D) 624 (E) 893 Directions—(Q. 28–32) Study the following information carefully and answer the questions given below : M, D, P, K, R, T and W are sitting around a circle facing at the center. D is second to the right of P who is third to the right of K. T is third to the right of W who is not an immediate neighbour of D. M is third to the left of R. 28. Who is third to the left of D ? (A) W (B) P (C) K (D) Data inadequate (E) None of these

31. Who is second to the right of T ? (A) D (B) K (C) M (D) Data inadequate (E) None of these 32. Who is to the immediate left of R? (A) W (B) P (C) K (D) T (E) None of these Directions—(Q. 33–35) Study the following arrangement carefully and answer the questions given below — R%E5D2#9AFB@J3IK M41WU8V©N★GZδ67 33. How many such symbols are there in the above arrangement, each of which is immediately followed by a consonant and immediately preceded by a number ? (A) None (B) One (C) Two (D) Three (E) More than three 34. Which of the following is the fifth to the left of the sixteenth from the left end of the above arrangement ? (A) B (B) U (C) W (D) N (E) None of these 35. Which of the following is the seventh to the right of the fourteenth from the right end ? (A) B (B) # (C) N (D) U (E) None of these