In This IssueRegulars Editorial Science and Technology Latest General Knowledge Inspiring Young Talent—Topper BHU (Med.) Rank-22, 2007—Lokesh Kumar Garg Science Tips 1547 1548 1551 1555 1557 February 2008 Year—10 Issue—120 Physics Sound-III Nuclear Physics-III Typical Model Paper Typical Model Paper 1560 1566 1572 1578 Editor MAHENDRA JAIN Chemistry Chemistry of Alkali Metals Chemical Thermodynamics Typical Model Paper Typical Model Paper 1583 1589 1598 1604 Zoology Advertisement ATUL KAPOOR (Business Manager) 4840/24, Govind Lane, Ansari Road, Daryaganj, New Delhi–110 002 Phone : 23251844, 23251866 Connective Tissue Lamarckism and Darwinism Typical Model Paper Typical Model Paper 1608 1612 1618 1621 Botany Fat (Lipids) Metabolism Editor/Publisher is not responsible for views, data, figures etc. expressed in the articles by the authors. —Editor No part of this publication can be reproduced or transmitted in any form without the prior written permission from the publishers. 1624 1627 1630 1634 1637 1640 1643 Different Forms of Green Algae General Aspects of Basidiomycetes (Club Fungi) Typical Model Paper Typical Model Paper Typical Model Paper Solved Paper : CBSE Medical Entrance (Mains) Examination, 2007 Edited, printed and published by Mahendra Jain for M/s. Pratiyogita Darpan, 2/11A, Swadeshi Bima Nagar, AGRA–2 and printed by him at Pratiyogita Darpan Printing Unit, 5 & 6, Bye pass Road, Agra. Phone : 2531101, 2530966, 3208693 Fax : (0562) 2531940 E-mail : pratiyogita_darpan@sancharnet . in Website : www.competitionsciencevision.com Other Features Assertion and Reason Type Questions True or False Do You Know ? CSV Crossword No. 20 CSV Quiz Contest No. 117 Correct Solution and Prize Winners of CSV Quiz No. 114 General Awareness 1659 1662 1665 1669 1670 1673 1676 C.S.V. / February/ 2008 / 1545 To Our Readers Dear Readers, The February issue of your favourite magazine Competition Science Vision is in your hands. Like all previous issues, this issue is also full of examination-oriented reading material for your pre-medical tests. The issue has been designed to pay you rich dividends in your examinations. It focusses on improving your performance and giving you an extra-edge over other competitors. If you want to get the best out of a book you must look for the best that is in it. If you have the positive attitude, you will highly be benefited by CSV and its unparalleled guidance. Read CSV regularly and intelligently. It gives you the power to master your career and shape your destiny. We take this opportunity to send our warmest new year greetings to you. May it be a peaceful, prosperous and very successful year for you. With best wishes for your bright future. Sincerely yours. Mahendra Jain (Editor) FORTHCOMING COMPETITIVE EXAMS. 2008 Recruitment of Religion Teachers in Indian Army (Jan. 13) Delhi Police Sub-Inspector (Executive) Examination (Jan. 13) Haryana Naib Tehsildar Recruitment Exam. (Jan. 20) Union Bank of India Probationary Officers Examination (Feb. 3) Navodaya Vidyalaya Entrance Exam. (Class VI) (Feb. 10) Recruitment of Clerical Staff (Customer Relations Associates) in State Bank of India (Feb. 10) Combined Defence Services Exam. (I), 2008 (Feb. 17) Uttar Pradesh Sainik School Entrance Exam., 2008-09 (Class VII and IX) (Feb. 17) BPSC Combined Preliminary Exam., 2008 (Feb. 24) Jharkhand P.C.S. Preliminary Exam., 2007 (Feb. 24) U.P. Naib Tehsildar/Assistant Jailor Exam., 2006 (Feb. 24) Bihar Newly Appointed Primary Teacher Exam. (Feb. 28) Vijaya Bank Probationary Assistant Manager Exam. (March 2) United Bank of India Clerk-cum-Cashier Exam. (March 9) Uttarakhand Combined State Civil/Lower Subordinate (Pre.) Exam., 2008 (March 9) Vijaya Bank Probationary Clerk Exam. (March 16) (Closing Date : 21 Jan., 2008) S.S.C. (Matric Level) Combined Preliminary Exam., 2008 (March 30) Education Worker Recruitment Examination (Category 1, 2 and 3) in Chhattisgarh (March 30) CBSE All India Pre-Medical/Pre-Dental Entrance Examination, 2008 (April 6) I.I.T. Joint Entrance Exam., 2008 (April 13) RPSC Hostel Warden Competitive Exam., 2007 (April 13) (Closing Date : 24 Jan., 2008) National Defence Academy & Naval Academy Examination (I), 2008 (April 20) Mahatma Gandhi Institute, Vardha M.B.B.S. Courses Entrance Exam. (April 20) (Closing Date : 29 Feb., 2008) All India Engineering Entrance Exam., 2008 (April 27) U.P. Polytechnic Combined Entrance Exam., 2008 (April 27-29) (Closing Date : 15 Jan., 2008) Uttarakhand Polytechnic Combined Entrance Exam., 2008 (May 4-6) (Closing Date : 31 Jan., 2008) BHU MBBS Screening Test, 2008 (May 7) (Closing Date : 8 March, 2008) National Talent Search Exam. (Second Stage) (May 11) CBSE All India Pre-Medical/Pre-Dental Main Exam., 2008 (May 11) Civil Services (P) Exam., 2008 (May 18) (Closing Date : 28 Jan., 2008) BHU PMT (Mains) (June 15) C.S.V. / February / 2008 / 1546 We generally meet two classes of people. Some people appear to be happy and gay always, while others present sorrowful and funereal looks. The former are always full of hope and optimism, while the latter are always dejected and pessimist. They always think that they will meet success nowhere and their no work will be done. The optimistic person is always full of hope and enthusiasm for his work, he always feels light in his life. The dejected person does not want to take up any work. According to him his work will not be completed. He always worships darkness and hopelessness in life. The optimistic person sees a ray of light and hope even in darkness. He knows that hope is immortal. His labours never go waste. The person who starts work with hope and determination, gets means and ways to complete it and new horizons get opened before him and the path of progress is always open before him. If some difficulties and obstructions come in the way, he gives them a fight with full force because hope continuously inspires him to reach the goal. Even when he does not meet success, he looks at it as a passing phase and does not leave hope. He knows that only he falls who rides a horse. He gets up and again and rides the horse. For fear of falling if the riding of the horse is abandoned and for fear of death if life is not lived then what would be the plight of the human society, which can be easily imagined. In Mahabharat it is said that all the top warriors like Bhishm, Dron and Karn were killed. Even then Duryodhan was hopeful of winning the battle because he thought that Shalya might deliver the goods and win the war. and continued to fight to the end. He lost the war and the empire but no one would say that he was a coward. After sending all his friends and relations into the jaws of death, he did not beg for life and lived like a dejected man. In the modern time the fight for Indian freedom is a glaring example of undying hope and getting victory in the end. The fight for Indian freedom was suppressed many times and often appeared to be a failure. But the freedom fighters always remained devoted to their cause and were always hopeful of getting to the goal. In the end victory smiled on them and Bharat became a free country. Dejection is a great sin. By losing hope we destroy our good actions and earn sin. Dejection is like that devil who is always in search of doing our destruction. Those who are optimistic have courage to face failure every now and then and they have also to suffer pangs of dejection, but their dejection is momentary; their dejection does not deter them from the path of hope. In the words of the great novelist Prem Chand, dejection shows them the right path and works as a stare of the blind man. There is a group of such unsuccessful persons as considered themselves incapable and powerless. In short they were always seized of inferiority complex. Such persons are generally those people who have been discouraged at every step in their childhood. Inferiority complex gives rise to the feeling of pity. A man of inferiority complex is always afraid of taking up any great work. He always considers himself helpless and powerless and he considers that the society will give him recognition on his face value without looking at his actions. We should remember these words of Mahaprabhu Shankaracharya that “So long as you consider yourself to be an object of pity, you will always remain a beggar and will not be able to do anything worthy of achievement. Actionlessness makes a man victim of inferiority complex and dejection makes the man actionless. The combination of dejection and actionlessness do not allow a person to proceed on the path of progress. In short, dejection does not allow a person to fight the difficulties and face the path of progress. Those who want to progress and achieve something should not suffer from dejection and inferiority complex. They should be always prepared to face difficulties and remove the impediments that come on the road of success. ●●● Hope makes a man full of enthusiasm for work while a dejected man becomes almost a dead person and suffers from inferiority complex. Those who want to do progress must get rid of these maladies and be prepared to face difficulties and remove the speed breakers on the road to success. The man without hope creates an atmosphere of darkness all about him and becomes like a dead log, with the result that he becomes totally inactive. He neither hopes to get success nor he becomes successful. So far as his ledger of success is concerned it remains totally blank. Pointing to the persons without hope the great thinker Swet Martin has said that the life without hope destroys all the valuable elements of life and there is no place in civilization for a person who is dejected and without hope. Civilization is the collective tressure of achievements. One who is without hope is afraid of proceeding towards achievements. Dejection makes him so weak and coward that he becomes indifferent to achivements and progress. Hope is both of cause and effect of good action. ·grs HhÆes grs nzks. d`.s Z ok f(fnoaxrs vk'kk c¥orh jktuΩ 'kY;ks tksIpfr ik≥MokuΩ‚\ Even after the departure of all his companions he did not lose courage C.S.V. / February / 2008 / 1547 / 2 Violent Black Holes Radiating High-Energy Cosmic Rays Pierre Auger Observatory, located in Argentina, is the largest cosmic ray observatory in the world. A team of researchers from 17 countries used the Auger Observatory and found that the sources of the highest-energy particles that reach the Earth are not distributed uniformly across the sky. Instead, these results link the origins of these mysterious particles to the locations of nearby galaxies that have active nuclei in their centres. These nuclei called Active Galactic Nuclei (AGN), have long been considered sites where high-energy particles-production might take place. They swallow gas, dust and other matter from their host galaxies and spew out particles and energy. The research team announced that Active Galactic Nuclei—thought to be powered by super massive black holes that devour large amounts of matter—are the most likely candidate for the source of the highest energy cosmic rays that hit the Earth. While most galaxies have black holes at their centres, only a fraction of all galaxies have an AGN according to this international research group. The exact mechanism of how AGNs can accelerate particles, to energies 100 million times higher than the most powerful particle acceleration on Earth, is still a mystery. Cosmic rays are comprised of protons and atomic nuclei, which travel across the universe at close to the speed of light. When these particles smash into the upper atmosphere of our planet, they create a cascade of secondary particles called an ‘air glower’ that can spread across 40 or more square kilometres as they reach the Earth’s surface. “We find the southern hemisphere sky as observed in ultra-highenergy cosmic rays is non-uniform. This is a fundamental discovery. In the next few years our data will permit us to identify the exact source of these cosmic rays and how they accelerate these particles. These enormously energetic particles are very rare, but they pack a real punch. We have taken a big step forward in solving the mystery of nature and origin of the highest-energy cosmic rays”, says Professor Cronin, the leader of the research group. This discovery is a major step towards understanding some of the most extreme processes in the universe. Interceptor Missile Defence : A Great Success In a test, carried out on December 6, 2007, a new, fully solid interceptor missile developed by the Defence Research and Development Organisation (DRDO) successfully destroyed a Prithvi missile fired five minutes earlier. The interceptor named, Advanced Air Defence (AAD02), homed on to the target missile, though the latter climbed five kilometres higher than the expected altitude. Interception was just like hitting a bullet with a bullet. The interceptor scored a ‘direct hit’ and destroyed the target missile over the Bay of Bengal. The interceptor crossed the target missile at the correct point. The target missile went into fragments thereafter. A milestone : The trajectories of the target missile and the interceptor missile and the interception point on display on the consoles of the Mission Control Centre at Wheeler Island, off the Orissa coast. (Right) : The interceptor missile roaring off from Wheeler Island. It is a single-stage missile, powered by solid propellants. It is 7·5 metres tall and weighs around 1·2 tonnes. It had a diameter of less than 0·5 metre. Data showed that the radiofrequency seeker, used for the first time against ballistic missile, had acquired the target in real time, while the homing guidance too worked perfectly. The new technologies used in AAD-02 included electro-mechanical acurators, homing guidance, the inertial navigation system, unique rocket motors with high specific impulse and specially developed jet vanes for control of the interceptor during the boost phase. This success has boosted the confidence of the scientists in networking the array radars, optics, command, control and communication systems to track an incoming missile in real time, validate all the software computation and send the command to the seeker to home in on the target. The mission signified the DRDO’s capability to network massive software with hardware actuation. The target missile is a modified single-stage Prithvi missile, fuelled by liquid propellants. It is 11 metres tall and weighs 5 tonnes. Its diameter is one metre. In April 2008, the DRDO would launch two interceptor missiles to intercept a single incoming target missile in both exo-atmosphere (over 40 km altitude) and endo-atmosphere (below 30 km altitude). Our this interceptor missile (AAD02) is better than the PAC-3 (Patriod Advanced Capability) of the U.S. in terms of range and altitude. The direct hit compared very well with the PAC3 in terms of accuracy. Neighbouring Pakistan has in its armoury a range of missiles that can carry nuclear warheads. With a 700 kg. payload, it is estimated that the Ghaznavi could have a range of about 350 km, the Shaheen-1 of about 495 km, the Ghauri of over 920 km and the Shaheen-2 of more than 1100 km. The longer the range of missile, the faster the warhead is propelled and the more difficult become the problems of interception. India’s unease with so many dangerous missiles right next door, and that too in a country showing alarming signs of instability, explains its eagerness to C.S.V. / February / 2008 / 1548 have some form of defence against these missiles. So an impenetrable missile shield against a nucleararmed neighbour is a very tall order indeed. Rover Probes ‘Mars was Habitable’ NASA’s Mars rover, ‘Spirit’, has proved an invaluable science tool, turning up evidence of a once habitable environment. Meanwhile, images from Mars Reconnaissance orbiter have largely unravelled the mystery of geological patterns called ‘Spiders’ that appear around each spring of south pole. now, the rocks that Spirit has examined have largely been volacanic Rapid Acceleration in Human Evolution Humans are evolving more quickly than at any time in history. In the past 5,000 years, humans have evolved up to 100 times more quickly than any time since the split with the ancestors of modern chimpanzees 6 million years ago. This research is carried out by a team of researcher of University of Wisconsin led by Professor Hawks. In fact, people today are genetically more different from people living 5,000 years ago. The genetic changes have related to numerous different human characteristics. The research also suggests that human races in different parts of the world are becoming more genetically distinct, although this is likely to reverse in future as populations become mixed. The changes have been driven by the colossal growth in the human population—from a few million to 6·5 billion in the past 10,000 years—with people mixing into new environments to which they needed to adapt. The researchers analysed data from the international haplotype map of the human genome, and analysed genetic markers in 270 people from four groups Han Chinese, Japanese, Africa’s Yoruba and northern Europeans. They found that at least 7 per cent of the human genes undergone recent evolution. The changes include lighter skin and blue eyes in northern Europe and partial resistance to disease such as malaria among some African populations, according to the research published in the Proceedings of the National Academy of Sciences, Washington. Many Chinese and African adults cannot digest lactose in milk, but across Europe a lactose-tolerance gene is now widespread. The surge in global population has also led to faster evolution since more mutations occur. The central finding is that human evolution is happening very fast– faster than any of us thought. Most of the acceleration is in the last 10,000 years, basically corresponding to population growth after agriculture is invented. ‘Spirit’ rover is cracking open the rock basalt. Whichever conditions produced it, this concentration of silica is probably the most significant discovery by Spirit rover for revealing a habitable niche that existed on Mars in the past. Stem Cells Developed from Skin Cells New Find : Images from the Mars orbiter help unravel mystery of patterns called ‘spiders’ around its south pole In November 2007, the scientists noticed a bright spot in the trail of overturned dirt. They turned ‘Spirit’ around for a closer look, finding high levels of silica, the main ingredient of window glass. Then they aimed the rover at a nearby rock, wanting to break it apart to determine if the silica was just a surface coating or if the rock was silica all the way through. As per instruction by the scientists, a rock was cracked open by the Spirit’s charge. The interior of that rock, which the scientists in formally named ‘Innocent Bystander’, turned out to be rich in silica. On Earth, such high concentrations of silica can form in only two places : a hot spring where the silica is dissolved away and deposited elsewhere or a fumarole, an environment often near a volcano, where acidic stream rises through cracks. The acids dissolve other minerals, leaving mostly silica. On Earth both environments team with life. Spirit’s twin ‘opportunity’, which has been exploring a spot on the other side of Mars, has found the evidence of an environment once steeped in acidic ground water. The discovery of silica for the first time is the evidence that Spirit has been signs of widespread water in its surroundings, a 90 milewide impact crater known as Gusev Crater. Gusev was chosen as landing site because it looks as if it were once a lake with what appears to be river channels flowing away from it. But, till Stem cells are the versatile cells that have the ability to grow into any kind of tissue—skin, heart, liver, any organ. These are precursor cells that can give rise to multiple tissue types. Stem cells have a remarkable potential to develop into different cells types within the body. Serving as a sort of repair system, they can divide without limit to replenish other cells for as long as a person or animal is alive. Pluripotent : A scientific team from the University of Wisconsin-Madison created genetic modifications in skin cells, pictured here, to induce the cells into what scientists call a pluripotent state— a condition that is essentially the same as that of embryonic stem cells. Researchers reported on November 20, 2007, that they have transformed ordinary human skin cells into batches of cells that look and act like embryonic stem cells—but without using cloning technology and without making embryos. Stem cells have two important characteristics that distinguish them from other cells. When a stem cell divides, each ‘daughter’ cell can either (Continued on Page 1556 ) C.S.V. / February / 2008 / 1550 AWARDS Nano National Award—C.N.R. Rao, Chairman, Science Advisory Council to the Prime Minister and Honorary President of Jawaharlal Centre for Advanced Scientific Research (JNCASR) was conferred the first Nano National Award for his outstanding achievements and sustained works in the field of nanotechnology. Business Leader Award, 2007 —CNBC-TV 18 has announced the winners of the India Business Leader Awards 2007. Business Leader Award winners are—Mukesh Ambani (Reliance Group), Ratan Tata (Tata Group) and Indra Nooyi (PepsiCo). CNBC-TV also celebrated the spirit of enterprise of leaders in fields such as politics and entertainment by recognising Pranab Mukherjee and Shah Rukh Khan. Asian Television Award— Television journalist Karan Thapar is awarded the Asian Television Award for the ‘Best Current Affairs Presenter’ in Asia in Singapore. Mr. Thapar is the only South Asian anchor to have ever won the award. He won this award for his ‘Devil’s Advocate’ series on CNN-IBN. International Children’s Peace Prize, 2007—A 16-year-old girl of Zambia won the 2007 International Children’s Peace Prize for her efforts to help educate children in Zambia. The Award includes Euro (€) 1,00,000 grant. The winner, Thandiwe Chama, is from Lusaka. When she was eight, she began lobbying for better school facilities after her school was closed because of shortage of teachers. Atmaram Prize—President Pratibha Patil gave away the prestigious Atmaram Prize for developing scientifc technical literature in Hindi to Mahendra Madhup of Jaipur. Dr. Madhup has specialisation in agriculture and edits a monthly journal ‘Sharad Krishi’. FIFA World Player of the Year Award 2007—Brazil’s midfielder, ‘Kaka’ (25), completed a virtual sweep, on December 18, 2007 in Zurich, by winning FIFA’s World Player of the Year Award, 2007. For Kaka, it is third accolade this year; he also won European football’s Golden Ball and World Soccer magazine’s Player of the Year Award. BCCI’s Award 2007—India’s top cricketers—men and women—figured in the Board of Control for Cricket in India’s Awards 2007 function held in Mumbai on December 16, 2007. Nari Contractor was honoured with Lifetime Achievement Award. The awardees are— Lifetime achievement award— Nari Contractor, Rs. 15 Lakh. Polly Umrigar award for outstanding performance in international cricket—Sachin Tendulkar, Rs. 5 lakh. Best performance in domestic cricket—Mumbai Cricket Association (Trophy). Felicitation award to Arjuna Award winners—Anju Jain and Anjum Chopra. ICC Woman cricketer of 2007— Jhulan Goswami, Rs. 1 lakh. Felicitation awards—Sachin Tendulkar (for crossing 11,000 runs in Test cricket and 15,000 runs in ODIs); Rahul Dravid (for crossing 10,000 runs in ODIs); Anil Kumble (for crossing 550 wickets in Tests and being appointed captain of the Indian team). Madhavrao Scindia award for the highest run-getter in Ranji Trophy 2007—Robin Uthappa, Rs. 1 lakh. Madhavrao Scindia award for the highest wicket-taker in Ranji Trophy 2007—Ranadeb Bose, Rs. 1 lakh. M. A. Chidambaram Trophy for the best under-15 player—Mandeep Singh (Punjab), Rs. 50,000. M. A. Chidambaram Trophy for the best under-17 player—K. Ajay Rana (Himachal Pradesh), Rs. 50,000. M. A. Chidambaram Trophy for the best under-19 player—Ajinkya Rahane (Mumbai), Rs. 50,000. M. A. Chidambaram Trophy for the best under-22 player—A. S. Bali (Delhi), Rs. 50,000. M. A. Chidambaram Trophy for the best woman cricketer—Jaya Sharma, Rs. 50,000—Special Correspondent. Anand and Dola honoured— Government of India awarded a cash amount of Rs. 10 lakh each to Vishwanathan Anand and Dola Banerjee in appreciation of their outstanding achievements. Anand won the World Chess Championship in Mexico City, while Dola won Archery World Cup Finals in women’s recurve division in Dubai. BOOKS Indian Clerk—David Leavitt (The book is packed with informations, insights and historical facts about Ramanujan. For the story of Ramanujan, the ordinary clerk from Madras in colonial India, whose extraordinary mathematical genius would take him to Cambridge and catapult him to immortality in a white men’s world, is no less intriguing that the academic politics of the time). India’s New Middle Class— Leela Fernandes (The book is a courageous attempt at researching a subject, the new Indian Middle Class, which is intrinsically problematic and intractable, at both the conceptual and empirical levels). From Jinnah to Jihad—Arvin Bahi (Pakistan’s Kashmir quest and the limits of realism). Globalization and Development—Ashwini Deshpande (A handbook of new perspectives). C.S.V. / February / 2008 / 1551 DAYS January 12—National Youth Day January 21—Army Day January 26—Republic Day, International Customs Day January 30—Martyr’s Day president. He is the first person of Indian origin to scale such stratopheric heights in the financial world. PLACES IN THE NEWS Malaysia—Indians were brought to Malaysia by Britons as indentured labour to work in rubber plantations. At present Indian minority constitutes 8% of Malaysia’s total population. 50 years ago, Britain granted independence to Malaysia. Most of Indian Malaysians are Tamilians. Bilateral relations between India and Malaysia have been excellent. But India and Indians were well within their rights to communicate their quiet over Malaysia’s harsh handling of political demonstrations by Malaysian Indians with legitimate grievances. The demonstrators, were met with tear gas shells and water cannons and are now facing serious criminal charges, were protesting the failure of colonial Britain, at the time of Malaysian independence, to protect the rights of ethnic Indians. Organised under the umbrella of the Hindu Rights Action Force (Hindraf) they were trying to link their ‘marginalisation’ in Malaysia directly to that failure. There is little doubt about Indian Tamils being a disadvantaged ethnic group or minority in Malaysia. They have their longstanding grievances. These include the absence of equal opportunities for minorities, affirmative action in favour of majority Malay, a bigoted unofficial policy of temple demolition and the increasing adoption of laws based on Sharia. Malaysia, a dynamic economic performer, needs seriously to address and resolve issues of equality, discrimination and minority rights. PERSONS IN THE NEWS Dmitry Medvedev (Next Russian President)—President Vladimir Putin of Russia has made his choice of successor to lead Russia after he steps down in March 2008 : his trusted ally and close associate of 17 years, First Deputy Prime Minister Dmitry Medvedev, will run for President in the March 2, 2008 election. The United Russian Party formally named his as its candidate. The fact that Mr. Medvedev is Mr. Putin’s chosen successor renders the presidential election a one-horse race. Mr. Putin is leaving to his successor a resurgent country with a roaring economy, a restored global role and a stable political system—a solid basis for the further rejuvenation of the Russian Federation. Challenging Job : Vikram Pandit, new CEO of Citigroup. Citigroup has operations in more than 100 countries, with 3,00,000 employees and $ 2 trillion in assets. Lal Krishna Advani—Bhartiya Janata Party on December 10, 2007 announced that senior leader and Leader of the Opposition in the Lok Sabha, L. K. Advani, will be its Prime Ministerial candidate in the next Lok Sabha election. This brings to an end of a perceived leadership tussle in the party. 2007 has been the year of caution and political correctness for the Prime Minister in waiting. He might be all of 80 years old–five years older than the Congress incumbent— but he is fit octogenarian with plenty of fight in him. It will be costly if not fatal, for the BJP’s political opponents to underestimate the strategy, tactics and mobilisational capabilities of the Shadow Prime Minister. Mr. Advani’s nomination is a significant move on nation’s political chessboard. Putting their Heads Together : Russian President Vladimir Putin (Right) with First Deputy Prime Minister Dmitry Medvedev. Indians living in South-East Asia Myanmar: 25,00,000 Thailand: 90,000 Vietnam: 299 Philippines: 50,000 Vikram Pandit (New CEO of Citigroup—World’s Largest Bank)— Nagpur born NRI, Vikram Pandit (50) may have taken over as the CEO of the Citigroup in the middle of its worst crisis and many see it as a rescue mission. He had joined Citigroup only a few months ago. Vikram Pandit replaces Charles ‘O’ Chuck Prince-III, who was forced out of the bank in November 2007. Charles had reported bank’s loss of $ 17 billion amid a massive financial crisis. Pandit is an ex-Morgan Stanley India Cambodia: 300 Ocean Malaysia: 20,50,000 Singapore : 4,00,000 Indian Ocean Indonesia: 82,000 Source: Lok Sabha Unstarred Question # 5740 KBK C.S.V. / February / 2008 / 1552 Dantewada (Chhattisgarh)—The manner, in which 300 prisoners detained in Dantewada jail escaped, points to an enormous security lapse—one caused by dangerous mix of negligence and incompetence. It is shocking that so many detenus—over 100 of them either naxalites or their sympathisers—could have fled in a operation that lasted just 15 minutes. It is specially outrageous that such an incident could have occurred in a maximum-security prison such as Dantewada, that it was pulled off without any apparent help from the outside. The pre-meditated strike took place around 4·35 p.m. when a naxalite commander Sujit Kumar over-powered a guard and snatched his weapon. Then he fired and injured three guards. The naxal inmates snatched six rifles and a wireless set before fleeing. Protecting the Climate at Bali—The conference of the UN Framework Convention on Climate Change (UNFCCC) held in Bali (Indonesia) overcame considerable wrangling and produced the Bali Action Plan, a basis but promising road map to 2009, when major economies must decide on new actions to reduce greenhouse gas emissions. All countries will need bold initiatives on emission cuts beyond 2012, when existing Kyoto Protocol commitments expire. Bali resolution accepts the scientific evidence and emphasises the ‘urgency’ for combating climate change. Developed nations as per Bali plan, adopt ‘‘measurable, reportable and verifiable emission limits and reductions’’, while developing countries can emphasise mitigation rather than emission reduction. The U.S., which emits the most Green-house Gases (GHGs) and i s a Kyoto Sceptic, obviously felt compelled to endorse the Bali plan under global moral pressure. UNFCCC has agreed, as part of the road map, to help, protect and expand forests through special funding. This provision can fund forestry schemes and generate income for rural and tribal communities. 29th SAARC Ministers Meeting—SAARC Council of Ministers signed the declaration of the meeting in New Delhi on December 7, 2007. The meeting resolved to C.S.V. / February / 2008 / 1553 operationalise the SAARC Development Fund, initially on the basis of India’s unilateral commitment of $ 100 million. The fund was envisaged to have a corpus of $ 300 million and India had made the offer even before the fund was conceptualised. The amount would be earmarked for social projects. The meeting also saw a SAARC declaration on climate change. The declaration said the way forward must include provision of adequate resources to tackle climate change without detracting from the funds for development effective access and funding assistance for transfer of environment-friendly technologies and adoption of binding GHG (Greenhouse Gas emission) reduction commitment by developed countries with effective timeframes. Australia as a consultant for a couple of weeks probably after the second test match. Jaysuriya quits Test Cricket— Sri Lanka’s highest scorer in both forms of cricket, Sanath Jaisuriya, declared to retire from Test cricket. Jaisuriya’s 6,973 runs from 110 Tests and 12,207 runs in 403 one-dayers are the highest by any Sri Lankan batsman. This 38 year old left-hander batsman has also claimed 97 Test and 307 one-day wickets with his left arm spin. Murali is the highest wickettaker—Sri Lankan spinner Muttiah Muralitharan broke Shane Warne’s all-time test wickets record of 708 during the first Test against England in Kandy on December 3, 2007. India-Pakistan Test series— India triumphed in a home Test series against Pakistan after 27 years with a 1-0 verdict in the three—Test Indian Oil series. Bad light helped Pakistan hold on for a draw as Indian captain Anil Kumble threatened to pull his side to victory with a burst of wickets on the last day of third and final Test in Bangalore on December 12, 2007. Leg-spinner Kumble took five for 60 off 14 overs after setting Pakistan an unlikely victory target of 374 from 48 overs, but the light intervened with the tourists clinging on at 167 for seven. India-Australia series—Indian cricket team is now in Australia. Skipper Anil Kumble and Harbhajan Singh are the lone spinners in Indian side which lacks a left-arm spinner. Pathan is the only all-rounder. Dinesh Karthik and Sehwag are the openers. M. S. Dhoni and Karthik are the batsman wicket keepers. victory in the A1GP at the Zhuhai International Circuit in China on December 16, 2007. He staved off a stiff challenge from New Zealand’s Jonny Reid to post his maiden A1GP Proud Moment : Narain Karthikeyan SPORTS 1,160 Flats in Commonwealth Games Villages—The Delhi Development Authority has declared that it will build a residential complex of 1,160 flats in the Commonwealth Games Village to accommodate athletes coming over to participate in the events in 2010. The Games Village is being built on the Yamuna river bed near the mammoth Akshardham Temple in East Delhi. feature race victory. It was Narain’s first podium finish after a gap of nearly three-and-a-half years in any form of racing. Narain comes from Coimbatore. Final standings (Top five) : Feature race—(1) Narain Karthikeyan (India) 1 : 0830·759, (2) Jonny Reid (New Zealand) 1 : 0831·261, (3) Adrian Zaugg (South Africa), (4) Michael Ammermüller (Germany), (5) Oliver Jarvis (Great Britain). ●●● Cricket Cricket is included in Olympic Games after 100 years—The International Cricket Council (ICC) has welcomed the decision by the International Olympic Committee (IOC) to return cricket into the Olympic Movement after a gap of more than 100 years. This will lead to closer collaboration between the ICC and IOC. Cricket is already a part of Asian Games programme and will be seen in the 2010 Guangzhou Games. Gary Kirsten is India’s new coach—BCCI on December 5, 2007 confirmed the appointment of former South African cricketer, Gary Kirsten, as India’s new coach. His tenure will be for two years starting since March 1, 2008. Gary will be with Indian team in Gary Kirsten UPKAR’S By Major P. N. Joshi (Retd.) Anil Kumble (Capt.), M. S. Dhoni (Vice-Capt.), Waseem Jaffer, Dinesh Karthik, Virendra Sehwag, Rahul Dravid, Sachin Tendulkar, Saurabh Ganguly, V.V.S. Laxman, Yuvraj Singh, Irfan Pathan, Harbhajan Singh, Zaheer Khan, R. P. Singh, Ishant Sharma and Pankaj Singh. It Includes ❖ Intelligence Tests ❖ Psychological Tests ❖ GTO’s Tests ❖ The Interviews Techniques ❖ Pilot’s Aptitude Tests UPKAR PRAKASHAN, AGRA–2 Motor Race A1GP Motor Race—Narain Karthikeyan (31) drove Team India to C.S.V. / February / 2008 / 1554 Code 916 The Squad Rs. 105/- Inspiring Young Talent “Hardwork, better understanding of exam. pattern, Blessings of God and family members are the main elements of my success.” —Lokesh Kumar Garg Topper BHU (Med.) Rank 22, 2007 [‘Competition Science Vision’ arranged an exclusive interview with Sri Lokesh Kumar Garg who has the credit of being selected in BHU (Med.) and in nearly half a dozen other institutes. For his brilliant success, he deserves all praise and our heartiest congratulations. This important interview is presented here in its original form.] CSV—Congratulations on your brilliant success. Lokesh—Thank you, sir. CSV—Before knowing your result what did you think about those who achieve top positions ? Lokesh—My opinion about toppers has always been that they are sincere, hard-working, determined and keep the faith in God and themselves. CSV—Achieving top position has come as a surprise to you or were you confident of achieving it ? Lokesh—I was hopeful about my success but I was never sure to be selected in so many exams simultaneously. CSV—What do you think is the secret of your success ? Lokesh—Blessings of God and family, hardwork, never have die attitude and better understanding about pattern of exams. CSV—In how many attempts did you get this success ? Lokesh—It was my first attempt in BHU-PMT and AFMC and second attempt in IIT-JEE, AIEEE, CBSEPMT and MP-PMT. CSV—What were the shortcomings in your preparation in the earlier attempts ? How did you make up for them this time ? Lokesh—In earlier attempts, I was not prepared for facing competitive exams. This attempt was quite to the point. CSV—From where did you get the inspiration of choosing a medical career ? Lokesh—From my father and my friends. CSV—From when did you start the preparation for it ? Lokesh—I had started my preparation two months after the board exam of 12th class. Lokesh—I put more weightage on Physics as it is the subject in which you can make difference with other candidates specially in medical entrance exams. CSV—Did you make complete study of all the topics or of some selective topics ? Lokesh—Complete study of all the topics is always better. But I focussed more on topics having more weightage in exams. Furthermore, paper of CBSE-PMT mains require selective intensive study. CSV—How did you give final touches to your preparation ? Lokesh—At the completion of syllabus, I gave many mock tests and revised formulae for Chemistry and Physics. For Biology, I read important topics many times from NCERT book. CSV—Did you prepare notes ? Lokesh—Yes, I prepared short notes by writing all concepts and formulae of topic at the end of chapter. ……Competition Science Vision is a standard magazine and a ‘must read’ for medical aspirants. Its each issue is prepared according to the latest pattern of exams. —Lokesh CSV—What planning did you make for preparation ? Please tell something in detail. Lokesh—Initially I tried to understand the basic concepts of the subject and completed topics one by one with patience. I didn’t get nervous and bettered my position gradually. CSV—How much time did you devote daily and regularly for Physics, Chemistry, Zoology and Botany ? Lokesh—I never stick to the strict time division according to subjects. One should read according to the necessity of topic and practice rotation of subjects to avoid boredom. Apart from PCB, I also gave some time for Mathematics. CSV—Out of the above four subjects, to which subject did you give more weightage and why ? Bio-Data Name—Lokesh Kumar Garg Father’s Name—Sri P.L. Garg Mother’s Name—Smt. Uma Garg Educational Qualifications— H.S./Std. X—89·6% (J.N.V. Hatta, Madhya Pradesh), 2004. Inter/Std. XII—75·6% (J.N.V. Hatta, Madhya Pradesh), 2006. Special Achievements— ● BHU-PMT (Mains)—AIR 22 ● Selected in AFMC (written, interview, physical) ● MP-PMT—Rank 73 ● CBSE-PMT—AIR 2352 ● IIT-JEE—AIR 3396 ● AIEEE—AIR 7100 ● State Rank—354 C.S.V. / February / 2008 / 1555 CSV—What was your attitude for solving numerical questions ? What weightage did you give them ? Lokesh—I used to proceed step by step for such questions. Making small diagram is also helpful. You should remember proper formula and apply it with proper units. CSV—How much time is sufficient for preparing for this examination ? Lokesh—I think, one year of dedicated study is enough to clear any medical entrance exam. short-tricks and memorizing techniques in it. 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 ? Lokesh— BHU-PMT (Mains) Physics—78 Chemistry—64 Zoology—64 Botany—65 Total Marks—250 Percentage—67·75% AIR—22 IIT-JEE Score—247 AIR—3396 AIEEE Score—226 AIR—7100 State Rank—354 MP-PMT Rank—73 Physics + Chemistry = 90·82 Biology = 90·00 180·82 Percentage = 90·41% CSV—What books/magazines/ newspapers did you read for G.K. preparations ? Lokesh—I read newspapers (Times of India) regularly. CSV—To whom would you like to give the credit for your success ? Lokesh—I give full credit to my loving parents and teachers who gave me support and courage to crack the exams. CSV—Please tell us something about your family. Lokesh—My father is a lecturer and mother is a housewife. CSV—What in your frank opinion has been the biggest mistake in your preparation for this test ? Lokesh—As I joined IIT Roorkee, I think I should have chosen only one stream, either Maths or Bio. CSV—What message would you like to give for our readers of CSV ? Lokesh—Read only standard text. Make your study circle. Improve your receptive power. Be honest toward yourself. Your family and the God. Always put your cent-per cent efforts. ●●● (Continued from Page 1550 ) remain as an unspecialized cell that can renew itself for long periods through cell division, or under certain physiological or experimental conditions can be induced to become specialised cell like the beating cell of a heart, muscle or an insulin-producing cell of the pancreas, a red blood cell or even a brain cell. Scientists primarily work with two kinds of stem cells from animals and human : embryonic and adult stem cells. For treating incurable and genetic diseases, there is extreme shortage of stem cells. Personal Qualities Hobbies—Cricket, Philately Ideal Person—Dr. A.P.J. Abdul Kalam Strong Point—My emotions Weak Point —I am too introvert Solar System is not Round but Dented NASA’s Voyager-2 spacecraft has found that our solar system is not round but is ‘dented’ by the local intersteller magnetic field of deep space. The data was gathered by the spacecraft on its 30 year journey into the edge of the solar system when it crossed into a sweeping region called the Termination shock. It showed that the southern hemisphere of the solar system’s heliosphere is being pushed in or ‘dented’. CSV—From what level of education should an aspirant begin preparing for it ? Lokesh—In ideal conditions, one should start his preparation immediately after 10th class. CSV—Please mention various books in each subject and magazines on which you based your preparation ? Lokesh—I mainly relied on coaching notes and Pradeep’s books for preparation. In magazines, I read CSV regularly. CSV—Did you take coaching in your preparation ? Lokesh—Yes, I took coaching at ALLEN Career Institute, Kota. CSV—What help do the science magazines render in the preparation for this examination ? Lokesh—These magazines provide vital guidelines and useful facts for aspirants. CSV—What will be your criterion for selecting a magazine for these examinations ? Lokesh—Magazine should be state-of-art to the latest pattern of exams. CSV—What is your opinion about our Competition Science Vision ? How much helpful and useful do you find it ? Lokesh—CSV is standard magazine and a ‘must read’ for medical aspirants. I read interviews of toppers in CSV and found these so much energetic and helpful. CSV—Please suggest in what way CSV can be made more useful for medical aspirants. Lokesh—I think, CSV should be made more useful by publishing more Brought to Light : The southern hemisphere of the solar system’s heliosphere is ‘dented’ Voyager-2 is the second spacecraft to enter this region of the solar system, behind Voyager-1, which entered the northern region of helioseath. The Termination shock is turbulent area far beyond the Pluto’s orbit where the solar winds emanating from the sun are significantly slowed as they run up against the thin gas of intersteller space. Solar winds blow in all directions from our sun, and shape what was once thought to be a bubble around the solar system called the heliosphere. ‘Voyager-2’ entered the Termination Sock almost one billion miles closer within the southern hemisphere of the heliosphere of the solar system than Voyager-1 previously had. ●●● C.S.V. / February / 2008 / 1556 Physics 1. Zero vector or null vector is a vector which has zero magnitude and an arbitrary direction. It is represented as 14. What is the electronic configuration of sodium ? ➠ Na (Atomic number = 11) 1s2 2s2 2p 6 3s1 15. At any point on the equatorial line, the dipole field E is ➠ In the direction opposite to that of dipole moment p → → ➠ 0 → 2. What is the forbidden energy gap for a diamond crystal ? ➠ 6 eV 3. The equilibrant and the resultant vectors are ➠ Equal in magnitude but opposite in direction 4. NOR gate is a combination of 16. Latent heat of ice is 3·34 × 105 J/kg. Express it in cal/gm ➠ L = 3·34 × 105 J/kg = = 3·34 × 105 103 × 4·2 3·34 × 105 J/gm 103 cal/gm = 80 cal/gm ➠ OR and NOT gates 5. If the horizontal range of a projectile for two angles α and β is the same, then α + β should be equal to ° ➠ 90° 6. What is 1 coulomb equal to in electrostatic units ? 17. The amount of work required to reverse an electric dipole from the direction of electric field is ➠ 2 pE 18. The kinetic energy of an ideal gas at absolute zero is ➠ Zero 19. Assuming earth to be a spherical conductor of radius 6400 km, its capacitance will be ➠ 1 coulomb = 3 × 109 e.s.u. 7. If the greatest height to which you can throw a ball is h, then what is the maximum horizontal distance to which you throw the ball ? v2 v2 ➠ 2h Note : h = 2g ‚ Rmax = g = 2h ➠ 711 µF 20. What does F × v represent ? Distance Work ➠ F × v = Force × Time = Time = Power [ ] 8. What is International Ampere ? ➠ The International Ampere is that unvarying current which will deposit 0·0011183 grams of silver per second from a solution of silver nitrate of a specified nature in cell of given specifications. 9. Two balls are thrown with same initial velocity at angles 25° and 45° with the horizontal. Which ball will come to the ground earlier ? ° ➠ The ball thrown at 25° [Note : Time of flight is proportional to sin θ when v and g are constant] 10. What is a low frequency choke ? ➠ It is an iron core inductor and is applicable for frequency 50 Hz. It is used in the filter circuit of the rectifier. 11. The net charge of an electric dipole is ➠ Zero 12. Calculate the diameter of the K orbit of He. n2 1 ° ➠ r = 0·529 × Z = 0·529 × 2 ⇒ 2r = 0·529 A 13. The S. I. unit of electric dipole moment p is ➠ Cm (coulomb-metre) [Note : It is not mC which represents milli coulomb] Chemistry 21. The property of O 2 molecules which is almost identical to that of Xe atom is ➠ Ionization energy 22. The relation between standard free energy change and equilibrium constant is as ° ➠ ∆G° = – 2·303 RT log Kc 23. The gas which diffuses fastest through glass and plastic materials is ➠ Helium 24. The function of a system which takes both enthalpy and entropy into account is called ➠ Free energy 25. The sorption of hydrogen by metals is termed as ➠ Occlusion 26. First organic compound (Carbamide) was prepared by heating a mixture of two inorganic salts ➠ Potassium cyanate + Ammonium sulphate 27. The balance which is used for magnetic measurements of compounds is known as ➠ Gouy balance C.S.V. / February / 2008 / 1557 28. A vitamin whose deficiency causes lengthening the time of blood clotting is ➠ Vitamin K 29. The movement of ions from a sol across the membrane can be expedited by applying electrical field and this process is called ➠ Electrodialysis 30. When chlorobenzene is heated with chloral in presence of conc. H2SO4, the product formed is— 44. What kind of RNA is involved in protein synthesis ? ➠ Messenger RNA 45. Which enzyme is involved in light production in certain insects ? ➠ Luciferase 46. By which process, glucose diffuses rapidly across the plasma membrane ? ➠ Facilitated diffusion 47. What is called a square grid that shows all the possible genotypes of the progeny, resulting from a mating between organisms with known genotypes ? ➠ Punnet square 48. Which kind of base sequences of eukaryotic genes play role in protein synthesis ? ➠ Exons 49. What is called a double wrapping of DNA around histone proteins to form a packing unit in the chromosome ? ➠ Nucleosome 50. By which process, the diffusion of water takes place across a membrane ? ➠ Osmosis 51. Which lymphatic capillary in a villus of the small intestine, receives fats, cholesterol and fat-soluble vitamins ? ➠ Lacteal 52. Which process is involved in the formation of m-RNA ? ➠ Transcription 53. Which term is used for the capacity of the human body to maintain stable internal condition ? ➠ Homeostasis 54. What is called the three-base sequence in m-RNA ? ➠ Codon 55. What is called an embryological connection between the hepatic portal vein and inferior vena cava that allows blood to bypass the foetal liver ? ➠ Ductus venosus 56. Which kind of cartilage is found in human external pinna ? ➠ Elastic cartilage 57. Which extra embryonic membrane directly surrounds the developing human embryo and forming a fluidfilled protective covering ? ➠ Amnion 58. Which cells produce the matrix of cartilage ? ➠ Chondroblasts 59. Which paired male reproductive glands open into the urethra at the base of the penis and release a buffering and lubricating fluid ? ➠ Bulbourethral glands 60. What is called the noncellular layer below the epithelial cells ? ➠ Basement membrane ➠ pp ′ dichlorodiphenyl trichloroethane or D.D.T. 31. A salt whose aqueous solution is used in marking linen, is ➠ Silver nitrate (AgNO3) 32. When methanol is heated with salicylic acid, in presence of conc. H2SO4, the product is known as ➠ Methyl salicylate or oil of winter green 33. A solid substance in which certain physical properties are different in different directions is termed as ➠ Anisotropic 34. The process which is used to make iron soft and ductile is known as ➠ Annealing 35. An apparatus used for measuring the change in volume of gases during chemical reactions is known as ➠ Eudiometer 36. For isoelectronic species as the positive charge increases, the size ➠ Decreases 37. The process of removing an adsorbed material from an adsorbent by washing it in a liquid is known as ➠ Elution 38. Organic liquids which have very high boiling point or those which decompose at or below their boiling points can be purified by ➠ Vacuum distillation 39. A trade name of an aluminium-based alloy of high reflectivity for light and UV radiation which contains 1-2% Cu and 5–30% Mg, is ➠ Magnelium 40. A synthetic rubber obtained by co-polymerisation of butadiene and styrene in presence of Na is known as ➠ Buna-S Zoology 41. What is called an artery in mammals that arise from an arch of the aorta and divides to form the right carotid and right subclavian arteries ? ➠ Innominate artery 42. Which term is designated to the maintenance of internal conditions ? ➠ Homeostasis 43. What is called a movement response to air or water current ? ➠ Rheotaxis C.S.V. / February / 2008 / 1558 Botany 61. What helps fragmentation of hypha in fungi ? ➠ Helps in multiplication (where each fragment grows into new mycelium) 62. How many ascospores are contained in a single ascus ? ➠ 4 or 8 63. What is called an outgrowth of micropylar end of castor seed ? ➠ Caruncle 64. Who coined the term ‘metabolism’ for all chemical processes carried on in cells ? ➠ Schwann 65. Why is mercury used in anaerobic respiration experiment ? ➠ Because it does not react with CO2 66. What causes antiparallel nature of DNA strands ? ➠ H-bonds 67. During which phase of cell cycle protein synthesis occurs ? ➠ Interphase 68. What type of phyllotaxy is found in Calotropis procera ? ➠ Opposite phyllotaxy 69. Which chromosomal fragments are most often prone to deletion ? ➠ Chromosome without centromere 70. What is called the basal part of the ligule of Selaginella ? ➠ Glossopodium 71. What measures a ‘potometer’ ? ➠ Transpiration 72. What causes Synchitrium endobioticum in potato ? ➠ Wart disease of potato 73. To which category of hydrophytes does Vallisneria belong ? ➠ Submerged hydrophyte 74. What is the botanical name of ‘Little blue stem’ ? ➠ Schyzacharium scoparium 75. Which trace element causes ‘Khaira’ disease of rice ? ➠ Zinc (Zn) 76. Which plant of pteridophyte is commonly called ‘resurrection plant’ ? ➠ Selaginella 77. Which cell organelle produces lysosomes ? ➠ Golgi apparatus 78. Which type of meristem is responsible for growth in thickness of plant body ? ➠ Lateral meristem 79. Which kind of base sequences of eukaryotic genes play role in protein synthesis ? ➠ Exons 80. What happens as the electrons move down the electron transport chain (ETS) ? ➠ Energy is released ●●● C.S.V. / February / 2008 / 1559 Stationary Waves (i) When two progressive waves of equal wavelength and amplitude moving in opposite directions in a medium with the same speed superpose on each other, a new wave is formed that does not seem to proceed in any direction. Such a wave is called the stationary wave. (ii) A stationary wave does not transmit energy in the medium. t Graphical Representation of Stationary Waves II I t II I Important Let the wave moving in positive x-direction be represented by 2π y1 = a sin λ (vt –x) and the wave moving in negative x-direction be given by 2π y2 = a sin λ (vt + x) Then the resultant wave due to superposition is given by y = y1 + y2 2πx 2πvt = 2a cos λ sin λ 2πvt = A sin λ t I II I t II t (iii) The amplitude of the stationary wave is given by A = 2a cos 2πx λ Characteristics of Stationary Waves (i) Some points of the medium are always stationary i.e., their amplitude is zero. These are called nodes. (ii) In between the nodes there are points with maximum displacements. These are called antinodes. (iii) The distance between two consecutive nodes or nearest antinodes is λ/2· The distance between a node and the nearest antinode is λ/4· (iv) Excluding nodes every point of the medium executes simple harmonic motion about its mean position. (v) The amplitudes of all the particles of the medium are not same. Amplitude at the nodes is zero and at antinodes maximum. (vi) All the points occurring between two nodes vibrate in the same phase. They reach their respective positions of maximum displacement simultaneously and also pass through their mean position at the same time. (vii) Like progressive waves the stationary wave do not proceed forward. (viii) The variations of pressure and density are maximum at nodes but are minimum at antinodes. (iv) The points where the amplitude of the resultant wave is maximum are called the antinodes. For antinodes x = 0, and λ 3λ , λ, , …… 2 2 A = ± 2a. (v) The points where the amplitude of the resultant wave is minimum are called the nodes. For nodes x = and λ 3λ 5λ , , , …… 4 4 4 A = 0. (vi) The distance between two consecutive nodes or antinodes is λ/2. (vii) Stationary waves can be produced by the superposition of both the longitudinal and transverse waves. (viii) Longitudinal stationary waves are formed in flutes and air columns. (ix) Transverse stationary waves are formed in sonometer, sitar, gitar and strings. C.S.V. / February / 2008 / 1560 Comparison of Progressive and Stationary Waves Progressive Waves 1. These waves move in medium with a definite speed. Stationary Waves 1. These waves do not move in any direction and remain confined within two boundaries. (iii) In fig. (c), a pulse passes from a heavy spring on the left to a light spring. Partial reflection and transmission again occur but the reflected pulse is not turned upside down. (iv) In fig. (d), the left-hand of the long narrow spring is fastened to a length of a thin string and is in effect free. Here almost the whole of the incident pulse is reflected the right way up i.e., a crest is reflected as a crest and no phase change occurs. 2. All the medium particles in 2. All the medium particles in these waves vibrate with these waves, except equal amplitude and timenodes, vibrates with equal period. time-periods but different amplitudes. 3. The phase of each particle changes continuously. 3. All the particles between two consecutive nodes are in same phase but the particles on the opposite sides of the node are in opposite phase. Phase Changes in Longitudinal Waves (i) Phase changes also occur when longitudinal waves are reflected, as can be shown by sending pulses along a slinky spring to ‘denser’ and ‘less dense’ boundaries, i.e., to fixed and free ends. At a fixed end a compression is reflected as a compression, at a free it is reflected as a rarefaction. (ii) Similar effects are obtained when sound waves are reflected in pipes with closed and open ends; a compression is reflected as a compression at a closed end and as a rarefaction at an open end. 4. None of the particles is 4. Node particles are permapermanently stationary. In nently stationary; the rest the position of maximum are momentarily stationary displacement every partiin the position of maximum cle is momentarily statiodisplacement. nary. 5. The compressions and rarefactions move onward with a definite speed. 6. Same changes of pressure and density occur at all points of medium. 7. These waves transmit every in the medium. 5. Compressions and rarefactions produce alternately at definite places. 6. Changes of pressure and density are maximum at nodes but minimum (zero) at antinodes. 7. These waves do not transmit energy in the medium. Important Points ● When a transverse wave on a spring is reflected at a ‘denser’ medium, there is a phase change of 180°. ● The phase change occurs in the case of the spring with one end fixed for example, because there can be no displacement of the fixed end, it must be a node. The incident and reflected waves, therefore, cause equal and opposite displacements at the fixed end so that they superpose to give resultant zero displacement as shown in the figure below— Incident wave Fixed end Reflection and Phase Changes in Mechanical Waves The behaviour of a wave at a boundary can be studied by sending pulses along a long narrow spring as shown in figure below : 1 Fixed end 2 (a) 1 Heavy spring Light spring 2 (b) (d) Light thread Spring (c) Heavy spring Light spring Reflected wave N N N N N Vibrations of Air Columns (i) An organ pipe is a pipe that sets in vibration the air enclosed in it when the air is blown into it. As a result sound is produced in it. (ii) Organ pipes are of two types—closed end organ pipe and open end organ pipe. (iii) A closed end organ pipe has one of its ends closed and the other open. (iv) An open end organ pipe has both its ends open. (v) In a closed end pipe a node is always formed at the closed end and an antinode is formed at the open end. (vi) Longitudinal stationary waves are formed in an organ pipe. (i) In fig. (a), the left-hand end of the spring is fixed and a transverse upward pulse travelling towards it is ° reflected as a trough. A phase change of 180° or π rad has occurred and there is a phase difference of half a wavelength (λ/2) between the incident and reflected pulse. (ii) In fig. (b), the left-hand end of the spring is attached to a heavier spring and at the boundary the pulse is partly transmitted and partly reflected, the reflected pulse being inverted. C.S.V. / February / 2008 / 1561 (vii) Various stages of resonance in a closed end organ pipe are represented in the following diagrams : A A A N N t λ1 4 3λ2 4 5λ3 4 This frequency is called the fundamental frequency or the fundamental note or the first harmonic. v v (xvi) If l = λ2 then n2 = = = 2n1. λ2 l This frequency is called the second harmonic or first overtone. (xvii) If l = 3λ3 2l v 3v then λ3 = ∴ n3 = = = 3n1 3 2 λ3 2l N N N N This frequency is called the third harmonic or second overtone. (xviii) Both the odd and even harmonics are produced in an open end organ pipe. That is, n1 : n2 : n3 : …… : : 1 : 2 : 3 : …… (xix) The sound emitted by an open end organ pipe is musical. (viii) If the length of the pipe λ1 l = then λ1 = 4l 4 v v ∴ Frequency, n1 = = λ1 4l This frequency is called fundamental frequency or fundamental note or first harmonic. 3λ2 4l (ix) If l = then λ2 = 3 4 v 3v ∴ Frequency, n2 = = = 3 n1 λ2 4l This frequency is called third harmonic or first overtone. 5λ3 4l (x) If l = then λ3 = 5 4 v 5v ∴ Frequency, n3 = = = 5n1. λ3 4l This frequency is called fifth harmonic or second overtone. (xi) Only odd harmonics can be produced in a closed end organ pipe. That is, n1 : n2 : n3 ! …… : : 1 : 3 : 5 : …… (xii) Longitudinal stationary waves are formed in an open end organ pipe too. (xiii) The antinodes are formed at both the ends of an open pipe. (xiv) Various stages of resonance in an open organ pipe have been represented in the following diagrams : A A A N N A l N λ1 2 A λ2 A N N A A A N 3λ3 2 End Correction (i) In an organ pipe antinode is formed a bit above the open end as shown below by distance e. e A e A l λ 4 l N λ 2 N e A (xv) If ∴ Frequency, l = n1 λ1 then λ1 = 2l 2 v v = = λ1 2l (ii) The length of the vibrating air column is a little greater than the length of the pipe. (iii) The distance from the free end of the pipe to the antinode is called the end-correction represented by e. (iv) Length of air column in closed end pipe = l + e. Length of air column in open end pipe = l + 2e. (v) If r be the radius of pipe then the end-correction e = 0·6r. (vi) Frequency of the fundamental note of closed end pipe v n1 = 4(l + 0·6r ) and the frequency of the fundamental note of the open end pipe v n2 = · 2(l + 1·2r ) (vii) The frequency of the fundamental note of open end pipe is not exactly the double of that of the closed end pipe but is a bit less. Resonance Tube (i) Resonance tube is a closed organ pipe with an air column of variable length. C.S.V. / February / 2008 / 1562 (ii) By resonance tube we can determine the speed of sound in air and the frequency of tuning fork. (iii) When the frequency of the air column of resonance tube becomes equal to that of the tuning fork, resonance occurs and the amplitude of vibrations of air column becomes very large. A loud sound is heard in such a case. A l N λ 4 l2 A N 3λ 4 N (v) If a string of length l vibrates in one loops then λ1 = 2l and frequency n1 = v 1 = 2l 2l T · m This frequency is called the fundamental note or first harmonic. (vi) If the string vibrates in two loops, then v λ = l and n2 = = 2n1. l This frequency is called the first overtone or second harmonic. (vii) If the string vibrates in three loops, then 3λ3 2l = l ⇒ λ3 = 3 2 3v ∴ n3 = = 3n1. 2l This frequency is called the second overtone or third harmonic. (viii) Both the odd and even harmonics are emitted from a stretched string. That is, n1 : n2 : n3 : …… : : 1 : 2 : 3 : …… (iv) If the lengths of air column at first and second resonance be l1 and l2 respectively, then l1 + e = and l2 + e = ∴ Speed of sound and end-correction λ 4 3λ 4 λ = 2(l2 – l1) v = 2n(l2 – l1) l2 – 3l1 e = 2 Laws of Transverse Vibrations in a Stretched String (i) Law of length : n ∝ 1 ; if T and m are constant. l 1 ; if T and l are constant. m (v) Speed of sound at 0°C is given by v 0 = v t – 0·61t where v t is the speed of sound at t °C. Vibrations of Stretched String (i) The vibrations of a thin, long and perfectly elastic string are transverse stationary. (ii) On both the ends of string there are nodes, and an antinode is there in the middle. (iii) The speed of transverse wave in a stretched string is given by T m where T and m are respectively the tension and mass per unit length of the string. (iv) Modes of vibration in a stretched string are as under : v = l N A N λ1 2 A N λ 2 N A N λ 2 λ 2 A N λ 2 3λ3 2 λ 2 A = λ2 N N (ii) Law of tension : n ∝ T ; if l and m are constant. (iii) Law of mass : n ∝ (iv) Law of radius : n ∝ 1 ; if T, l and the density d r of the string are constant; r being the radius of string. 1 (v) Law of density : n ∝ ; if T, l and r are cond stant. Melde’s Experiment (i) Melde’s experiment is a simple and beautiful example for the demonstration of stationary waves and the harmonics of their transverse vibrations. (ii) Melde’s experiment is performed by two methods. (iii) Transverse arrangement of vibrations : (a) In this arrangement the tuning fork is held such that the direction of its vibration is perpendicular to the direction of length of string. (b) In this arrangement the frequencies of the tuning fork and the string are equal. (c) If the tension T in the string is adjusted such that p loops are produced in the string of length l, then p 2T = 4n 2l 2m = Constant, where m is the mass per unit length of the string. This is Melde’s law. λ 2 N A C.S.V. / February / 2008 / 1563 / 3 (iv) Longitudinal arrangement of vibrations : (a) In this arrangement the tuning fork is held such that the direction of its vibration is along the length of the string. (b) In this arrangement p 2T = n 2l 2m = Constants. (c) Here the frequency of string is half the frequency of tuning fork. SOME IMPORTANT SOLVED EXAMPLES Example 1. An open end organ pipe emits a note of frequency 256 Hz which its fundamental. What would be the smallest frequency produced by a closed end pipe of the same length ? Solution : For open end organ pipe v n = 2l v 256 = 2l ⇒ v = 512l For closed end organ pipe v 512l = = 128 Hz 4l 4l Example 2. An air column with a tuning fork of frequency 256 Hz gives resonance at column lengths 33·4 cm and 101·8 cm. Deduce (i) the end correction, and (ii) the speed of sound in air. l2 – 3l1 Solution : (i) e = 2 101·8 – 3 × 33·4 = 2 101·8 – 100·2 = 2 1·6 = = 0·8 cm 2 (ii) λ = 2 (l2 – l1)2 = 2 (101·8 – 33·4) = 136·8 cm = 1·368 m v = nλ = 256 × 1·368 = 350·2 ms–1 Example 3. A pipe 30 cm long is open at both ends. Which harmonic mode of the pipe is resonantly excited by a 1·1 kHz source ? (Given speed of sound = 330 ms–1) Solution : Frequency of the nth harmonic for an open nv end pipe is νn = where n = 1, 2, 3, …… v = 330 ms–1, 2l l = 0·3 m n × 330 = 550 ns–1 ∴ νn = 2 × 0·3 Now, 550n = 1100 ⇒ n = 2 (second harmonic) Example 4. A wire is under tension of 32N and length between the two bridges is 1 m. A 10 m length of the sample of the wire has mass of 2g. Deduce the speed of transverse waves on the wire and frequency of the fundamental. Solution : T = 32 N, 2 × 10–3 m = kg m–1 10 = 0·0002 kg m –1 n = l = 1m v = 1 2l T = m T m 32 = 400 ms–1 0·0002 Also, n = = 1 × 400 = 200 Hz 1×1 Example 5. The length of the sonometer wire between two fixed ends is 100 cm. Where should the two bridges be placed to divide the wire into three segments whose frequencies are in the ratio of 1 : 2 : 3 ? Solution : In a sonometer n ∝ Since n1 : n2 : n3 = 1 : 2 : 3 Hence, l1 : l2 : l3 = ∴ l1 = l2 = l3 = 1 1 1 : : =6:3:2 1 2 3 6 × 100 = 54·54 cm 11 3 × 100 = 27·27 cm 11 2 × 100 = 18·18 cm 11 1 l Example 6. A tuning fork and an air column at ° 51° C produce 4 beats in one second when sounded together. The same tuning produces 1 beat per second when the temperature of the air column is reduced to ° 16° C. Determine the frequency of the tuning fork. Solution : When temperature decreases, speed of v , hence n sound in air also decreases. Since ν = λ decreases. Since the number of beats per second is less at lower temperature, therefore, we conclude that the frequency of the air column is higher than the frequency of the tuning fork. Let ν be the frequency of the tuning fork. Then at 51°C, frequency of air column = (n + 4) where n is the frequency of the tuning fork. At 16°C, the frequency of air column = (n + 1) ∴ v 51 (n + 4) λ = v 16 (n + 1) λ n+4 273 + 51 = n+1 273 + 16 ⇒ n+4 = n+1 324 18 = 289 17 ⇒ 18n + 18 = 17n + 68 n = 50 Hz C.S.V. / February / 2008 / 1564 OBJECTIVE QUESTIONS 1. The length of an organ pipe open at both ends is 0·5 m. Calculate the fundamental frequency of the pipe if the speed of sound in air is 350 m/s. If one end of the pipe is closed, then what will be its fundamental frequency ? (A) 175 Hz, 350 Hz (B) 300 Hz, 150 Hz (C) 350 Hz, 175 Hz (D) 150 Hz, 350 Hz 2. A resonance tube is resonated with a tuning fork of frequency 512 s –1. Two successive lengths of the resonated air-column are 16·0 cm and 51·0 cm. The experiment is performed at room temperature 40°C. Calculate the speed of sound and end-correction at 0°C— (A) 334 ms –1, 1·5 cm (B) 334 ms –1, 2·1 cm (C) 350 ms –1, 1·5 cm (D) 350 ms –1, 2·1 cm 3. A wire of length 1·5 m under tension emits a fundamental note of frequency 120 Hz. (a) What would be its fundamental frequency if the length is increased by half its length under the same tension (b) By how much should the length be shortened so that the frequency is increased threefold ? (A) 90 Hz, 1·0 m (B) 80 Hz, 1·0 m (C) 80 Hz, 0·5 m (D) 100 Hz, 0·5 m 4. A string A has thrice the length, thrice the diameter, thrice the tension and thrice the density of another wire B. Which overtone of A will have the same frequency as the fundamental of B ? (A) 9th (C) 6th (B) 8th (D) 10th the string is 0·01 kg/m. Find the tension in the string in kg-wt— (Speed of sound in air = 330 ms–1) (A) 100 kg-wt (B) 102 kg-wt (C) 15·0 kg-wt (D) 10·2 kg-wt 8. Two perfectly identical wires are in unison. When the tension in one wire is increased by 1%, then on sounding together, 3 beats are heard in 2 second. What is the frequency of each wire ? (A) 350 Hz (C) 300 Hz (B) 256 Hz (D) None of these 5. A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2·53 kHz. What is the speed of sound in steel ? (A) 2·53 km/s (B) 253 km/s (C) 5·06 km/s (D) None of these 6. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 425 Hz source ? Will this same source be in resonance with the pipe if both ends are open ? (Speed of sound = 340 ms–1) (A) First harmonic, No (B) Second harmonic, No (C) Third harmonic, No (D) None of these 7. The pitch of the fundamental tone of an open organ pipe 66 cm long is the same as that of a stretched string 20 cm long vibrating transversely. If the mass per unit of 9. Two tuning forks A and B when sounded together give 4 beats/s. A is unison with the note emitted by a 0·96 m length of a sonometer wire under a certain tension. B is in unison with 0·97 m length of the same wire under the same tension. Calculate the frequency of the fork— (A) 388 Hz (C) 256 Hz (B) 512 Hz (D) 540 Hz 10. Two tuning forks A and B when sounded together give 8 beats/ sec. Fork A resonates with a closed column of air 16 cm long and B with an open column 32·5 cm long. Calculate their frequencies— (A) 520 Hz, 512 Hz (B) 512 Hz, 504 Hz (C) 256 Hz, 248 Hz (D) None of these ANSWERS WITH HINTS 1. (C) 3. (B) 2. (A) v = 1 2l T m Also, 3n1 × l3 = n1l1 3l3 = 1·5 ⇒ l3 = 0·5 m The wire should be shortened by 1·5 – 0·5 = 1·0 m 4. (B) n = = = nB 1 2l 1 2lr 1 lD T πr 2ρ 1 πρ T πρ T πρ nA = Since p 3l × 3D 3T π×3ρ nA = nB , we get Since T and m are constant, hence n1l1 = n2l2 = n3l3 n2 p = 9 i.e., 8th overtone. 5. (C) ⇒ l = λ = 1m 2 ( ) l1 + l1 2 = n1 l1 λ = 2m 3 n = n1 2 2 2 n2 = × 120 3 = 80 Hz A N A 1 = l×D (Continued on Page 1674 ) C.S.V. / February / 2008 / 1565 Important Concepts Einstein’s mass energy equivalence relation— Einstein on the basis of his relativity theory has proved that if a substance loses an amount ∆m of its mass, an equivalent amount ∆E of energy is produced where ∆E = (∆m) × c2 where c is the speed of light. This is called Einsteins mass-energy relation. In view of this the laws of conservation of mass and the conservation of energy have been unified into a single law which states that the total (mass + energy) of the universe is conserved. Atomic mass unit (amu)—The unit chosen to express extremely small masses of atoms, nuclei and fundamental particles (electrons, protons, neutrons etc.) is called ‘atomic mass unit’ (amu). 1 amu is defined as one-twelfth part of the mass of carbon (6C12) atom. This definition makes the mass of 12 6C atom exactly equal to 12·00000…amu. The mass of 1 gm-atom of carbon is 12 gm and it contains N = 6·02 × 1023 atoms. 1 ∴ 1 amu = (mass of 1 carbon atom) 12 1 12 × = 12 N 1 = gram N 1 = 6·02 × 1023 = 1·66 × 10–24 gram = 1·66 × 10–27 kg. Unit electron volt—Electron volt is the unit of energy. 1 electron volt is the energy which an electron acquires when it is accelerated through a potential difference of 1 volt. 1 eV = (1·6 × 10–19 coulomb) × 1 volt = 1·6 × 10–19 joule Larger units are kilo electron volt (keV), million electron volt (MeV) and billion electron volt (BeV). 1 keV = 103 eV = 1·6 × 10–16 joule 1 MeV = 106 eV = 1·6 × 10–13 joule 1 BeV = 109 eV = 1·6 × 10–10 joule But 1·6 × 10–19 joule = 1 electron volt (eV) ∴ ∆E = 1·49 × 10–10 1·6 × 10–19 eV = 0·931 × 109 eV = 931 MeV (million electron volt) ∴ 1 amu = 931 MeV Mass defect—The mass of the atomic nucleus is slightly less than the sum of the masses of the nucleons (i.e. protons and neutrons) present in the nucleus. This mass difference is called mass-defect (∆m). Thus ∆m = (Mass of protons + Mass of neutrons) – Mass of the nucleus For an atom ZXA ∆m = [Zmp + (A – Z)mn] – mN where mN is the mass of the nucleus. The significance of mass defect is that when protons and neutrons combine to form the nucleus, an amount of mass ∆m disappears and an equivalent amount of energy (∆m) × c2 is liberated. It is due to this energy that protons and neutrons remain bound in the nucleus. Binding energy—The nucleons are bound together in a nucleus and energy must be supplied to the nucleus to separate the constituent nucleons to large distances. The amount of energy needed to do this is called the binding energy of the nucleus. Thus, the binding energy of a nucleus is the energy required to take its nucleons away from one another. Physical Constants Mass of proton, mp = 1·673 × 10–27 kg = 1·00728 amu Mass of neutron, mn = 1·675 × 10–27 kg = 1·00867 amu 1 amu = 931 MeV Mass defect for α-particle (2He4) = 0·03040 amu Binding energy of α-particle = 28·3 MeV = 7·07 MeV per nucleon Binding energy of deuteron (nucleus of deuterium 1H2) = 2·25 MeV = 1·12 MeV per nucleon Energy equivalent to 1 amu mass— ∆E = (∆m) × c2 = (1·66 × 10–27) × (3·0 × 108)2 = 1·49 × 10–10 joule C.S.V. / February / 2008 / 1566 The binding energy of a nucleus is generally expressed as binding energy per nucleon. It is a measure of the stability of nucleus. Higher the binding energy per nucleon, more stable is the nucleus. Binding energy curve—A graph between the binding energy per nucleon and the mass number of nuclei is called the binding energy curve. This is shown in the figure below : Binding energy per nucleon (MeV) Important Features of Nuclear Fission 1. The mass of the compound nucleus must be greater than the sum of masses of fission products. 2. The binding energy per nucleon for compound nucleus must be less than that of the fission products. 3. Energy liberated is equivalent to difference in masses of the nuclei before and after fission. 4. The energy release in the fission of U235 is estimated to be about 200 MeV per fission (or about 0·9 MeV per nucleon). 5. The energy obtained by fission of 1 gm of uranium is about 5 × 10 23 MeV. So much energy is obtained from explosion of 20 tones of T.N.T. From this energy about 2 × 10 4 kWh electrical energy can be produced. 6. 92U238 is fissionable only by fast neutrons (1·2 MeV energy) whereas 92U235 is fissionable by slow neutrons (1 eV) or thermal neutrons (0·025 eV) as well as fast neutrons. 7. Each of the three neutrons carries an energy of about 2 MeV. The fast neutrons will escape and will not cause the fission of U235 nuclei. In order to utilise them to cause fission of three other nuclei of U235, these neutrons have to be slowed down. 8. On an average 2·5 neutrons are emitted per fission. 9. Energy is released in the form of kinetic energy of fission fragments. Some of the energy is also released in the form of γ-rays, heat energy, sound energy and light energy. 10. The pressure and temperature are very high in the fission process. 11. The elements formed lie nearly in the middle of the periodic table. 12. Fission fragments are radioactive and they decay to stable products by emitting α-, β- and γ-rays. 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 F18 He N14 C12 4 O16 Fe 56 U238 Li7 H2 0 20 40 60 80 100 120 140 160 180 200 220 240 Mass number (A) Important Conclusions 1. The nuclei having mass number around 60 (for example Fe with A = 56) have maximum binding energy per nucleon (≈ 8·7 MeV). These nuclei are most stable. 2. For heavier nuclei, the binding energy per nucleon gradually decreases. For uranium A = 238 it is about 7·6 MeV. 3. For nuclei having mass number below 56 also, the binding energy decreases and below A = 20, it decreases very rapidly. For example, for heavy hydrogen (A = 2) it is only 1 MeV. It shows that nuclei of A < 20 are relatively less stable. 4. The special positions of He4, C12 and O16 on the curve indicate that these nuclei are more stable than their neighbouring nuclei. 5. The binding energy per nucleon of very light and very heavy nuclei is generally less than that of the nuclei in the middle. Thus if a very heavy nucleus (e.g., uranium) is broken into comparatively lighter nuclei, then the binding energy per nucleon will increase. Hence a large quantity of energy will be liberated in this process. This phenomenon is called ‘nuclear fission’. 6. Similarly, if two or more very light nuclei (e.g., nuclei of 1H2) are combined into a relatively heavier nucleus (e.g., 2He 4), then also the binding-energy per nucleon will increase. In this process also energy is liberated. The phenomenon is called ‘nuclear fusion’. Nuclear fission—It is the process in which a heavy unstable nucleus breaks into two nuclei of nearly same mass with the liberation of energy. Nuclear fission was discovered by two German scientists Otto-Hahn and Strassman. 92 U 235 + 0n1 → (92U236) → 56Ba141 + 36Kr92 + 3(0n1) It is not necessary that the fission products are Ba and Kr. Other nuclei may also be produced. Chain reaction in nuclear fission—In the nuclear fission of each uranium nucleus 2 or 3 fresh neutrons are liberated which under favourable conditions fission other uranium nuclei. This establishes a chain of nuclear fission which continues until the whole uranium is consumed. In the chain reaction, the number of nuclei undergoing fission increases very fast producing tremendous amount of energy. Difficulties in chain reaction—(i) The major part of natural uranium is the isotope U238 the isotope U235 is very little (only 0·7%). U 238 can be fissioned only by fastneutrons (energy more than 1 MeV). The neutrons of energy less than this are absorbed by U 238 . U235 can be fissioned with slow as well as fast neutrons. Though the low-energy neutrons can fission U235, but the probability of their absorption by U238 is much more. Thus the fresh neutrons released in the fission of ordinary uranium are not able to continue the chain reaction. (ii) The second difficulty in the maintenance of chain reaction is that the fast neutrons liberated by the fission of U235 nucleus travel a distance of about 10 cm in the substance before they are slowed down and fission other nuclei. If the uranium block to be fissioned is of small size, the most of the neutrons will escape before fissioning any nucleus, and the chain reaction will stop. Therefore to continue the chain-reaction, the size of the fissionable substance should be bigger than a certain critical size. C.S.V. / February / 2008 / 1567 Removal of diffculties—There are two ways to overcome the above difficulties : (i) The first is to separate the lighter isotope U235 from the ordinary uranium by diffusion method and to carry out the fission of U235 . The fission of U235 is possible by neutrons of any energy (very high or very low). In this case the chain reaction will continue. But this process is very expensive and cumbersome. (ii) The second method is to slow down the neutrons so that their energy remains about 0·03 eV. Then the probability of their absorption by U238 becomes very low while the probability of their fissioning U 235 becomes high. This slowing down of the neutrons is achieved by use of moderators. (iii) The second difficulty mentioned above is removed by taking the critical size of the fissionable substance. Critical size—Suppose, r = radius of the fissionable piece N = Number of neutrons produced by primary fission. A = Number of neutrons absorbed by the substance without causing fission. L = Number of neutrons escaping the substance. Then obviously N ∝ r 3 = k 1r 3, A ∝ r 3 = k 2r 3, L ∝ r 2 = k 3r 2 Case I. If N < A + L or N–A <1 L a very short time as a violent explosion. This happens in a nuclear bomb. In controlled chain-reaction, the fission is so controlled by artificial means that only one of the neutrons produced in each fission is able to cause further fission. The rate of reaction remains constant. Thus the rate of fission is kept constant. Therefore, this process is slow and the energy release is steady which can be utilised for useful purposes. Nuclear reactors are based on this process. Parts of a Modern Nuclear Reactor Fuel—It is the substance used for fission. U 235 or Pu 239 is used for this purpose. Moderator—It is used to slow down neutrons. Heavy water, graphite or beryllium-oxide is used for this purpose. Heavy water is best moderator. Coolant—Heat energy released in reactor is removed by coolant. For this purpose air, water or CO2 is flown in the reactor. Heat removed is utilised in producing steam which is used to drive turbines to produce electricity. Controller—It controls the rate of fission in a reactor. Cadmium rods are used for this purpose. These rods are fixed in reactor-walls. When they are pushed into the reactor, the fission rate decreases and when they are pulled out, the fission grows. Cadmium is a very good absorber of neutrons. Shield—To protect the workers from the injurious radiation emitted in the reactor, thick concrete walls are erected around the reactor. There will be no fission and the chain reaction will stop. N–A Case II. If N > A + L or > 1. L Then fission will continue and the chain reaction will be maintained. k 1 r3 – k 2 r3 N–A Now, = L k 3 r2 = k1 – k2 × r = kr k3 Breeder reactors—The reactors in which energy is produced by fission of U235 by slow neutrons are called ‘thermal reactors’. Since the major part in ordinary uranium is of U238 (U235 is only 0·7%), therefore, the fission of U235 is very costly. This will also lead to an early depletion of uranium reserves. It is known that besides U235, Pu 239 is also a fissionable substance. But Pu239 is not a naturally occurring isotope. It is produced from U 238 . 92 U 238 where k is a new constant. Hence for chain reaction to continue. N–A > 1 L or kr > 1 1 ⇒ r > k Hence for the chain-reaction to continue, the size (r ) of the substance to be fissioned should be larger than a 1 1 critical value . The value is called the ‘critical k k size’ of the substance. If the size of the substance is even slightly less than the critical size, the chain-reaction shall stop. Uncontrolled and controlled chain-reactions—In uncontrolled chain-reaction, more than one of the neutrons produced in fission cause further fissions so that the number of fissions increases very rapidly. Thus this is a very fast reaction and the whole substance is fissioned in a few moments liberating a huge quantity of energy within β + n → 92U239 → 93 Np β 239 → 94 Pu 239 If more than one neutron can be absorbed by U238 per fission, then we produce more fuel than what we consume. Thus apart from nuclear energy these reactors give us fresh nuclear fuel which often exceeds the nuclear fuel used. Hence they are called ‘Breeder reactors’. Note—In these reactors, in addition to Pu239 , U233 (fissionable fuel) is also produced from Th232. Nuclear fusion—When two lighter nuclei moving at very high speeds fuse together to form a single heavier nucleus, then this phenomenon is called nuclear fusion. The mass of the nucleus obtained after fusion is less than the sum of the masses of the nuclei which are fused together. The lost mass is obtained in the form of energy. Hydrogen bomb is based on phenomenon of nuclear fusion. () () C.S.V. / February / 2008 / 1568 As an example, in the fusion of two nuclei of deuterium (heavy hydrogen), following reactions take place 1H 2 At a Glance Nuclear Fission 1. Neutrons are required for it. 2. It is possible at normal temperature and pressure. + 1H2 → 1H3 + 1H1 + 4·0 MeV (energy) Nuclear Fusion Protons are required for it. It is possible at extremely high temperature and pressure. The nucleus of tritium (1H3) so formed can again fuse with a deuterium nucleus : 1H 3 + 1H2 → 2He 4 + 0n1 + 17·6 MeV (energy) The net result of these two reactions is that 3 deuterium nuclei fuse together to form a helium nucleus and liberate 21·6 MeV energy. Source of solar energy—The sun is continuously emitting huge amount of energy since millions of years. Emission of such a large amount of energy by chemical reactions is not possible. The source of huge solar energy is the fusion of lighter nuclei. At present, it is believed that proton-porton cycle is more probable in the sun (instead of carbon cycle). In this cycle (also), hydrogen nuclei fuse together to form of helium nucleus through the following reactions : 2[ 1H1 + 1H1 → 2[ 1H1 + 1H2 → 2 1H 2 3. For this the energy relea- For this the energy released sed per nucleon per nucleon 200 ∆E ∆E 27 = 235 ≈ 0·8 MeV = 4 = 6·75 MeV A A 4. Fissionable materials are expensive. The materials used in it are cheap (e.g., hydrogen) Nuclear Fission 1. Fission is not a spontaneous process. It is produced by bombarding the nuclei of fissionable material by neutrons. 2. In this process a heavy unstable nucleus breaks into two nuclei of almost same mass. Energy is obtained in the form of kinetic energy of these fragments, heat, light and sound. 3. Tremendous amount of energy is obtained in fission process. Radioactive Disintegration It is a spontaneous process. It cannot be started, stopped, accelerated or retarded by any chemical or physical process. In radioactive decay the unstable nuclei spontaneously emit light particle (α- and β-) and energy is obtained in the form of γ-rays. + +1β0 + υ + 0·4 MeV] 3 2He 2 + 5·5 MeV] + 2(1H1) + 12·9 MeV He 3 + 3 2He 4 → He 4 Adding we get 4(1H1) → 2He + 2(+1β0) + 2υ + 24·7 MeV In full cycle 24·7 MeV energy is liberated. At a Glance Uses of Nuclear Reactor 1. To generate electricity. 2. To produce Pu239. 3. To produce a neutron beam of high energy for neutron bombardment. 4. To produce artificially radioactive isotopes for medical, industrial and biological uses. 5. To generate power for driving the engines and the propulsion of ships, submarines and air crafts, thus replacing steam, coal and petrol. Energy obtained in radioactive disintegration is very less. Fast Breeder Test Reactor 1. In this liquid sodium is used as coolant. 2. The fuel used is a mixture of plutonium and natural uranium. 3. 60–70% fraction of natural uranium is used. 4. The chain reaction is maintained by fast neutrons. 5. Number of neutrons produced per fission is more. Thermal Reactor In this water is used as coolant. The fuel used is natural uranium. Only 1·2% fraction of natural uranium is used. The chain reaction is maintained by slow neutrons. The number of neutrons produced per fission is less. Research Atomic Reactors 1. Apsara—This is 1MW reactor situated in Mumbai. An alloy of uranium and aluminium is used as fuel in it. This is also known as swimming pool type reactor. 2. Cirus—It is a 40MW reactor, made in collaboration with Canada and is used to produce radioactive isotopes. 3. Zerlina—0 MW reactor. 4. Purnima—0 MW reactor. Atom Bomb 1. It is based on fission process. 2. In it critical size is important. Hydrogen Bomb It is based on fusion process. There is no limit to size. Power Reactors 1. Tarapur (Maharashtra) 400 MW 2. Rana Pratap Sagar (Rajasthan) 400 MW 3. Kalpakkam (Tamil Nadu) 220 MW 4. Narora (U.P.) 200 MW 5. Kaiga (Karnataka) 200 MW 6. Kakarapar (Gujarat) 200 MW 3. In this, explosion is possi- In this extremely high tempeble at normal temperature rature and pressure is requiand pressure. red to explode it. 4. In this, harmful radiations are produced. In this harmful radiations are not produced. C.S.V. / February / 2008 / 1569 SOME TYPICAL SOLVED EXAMPLES Example 1. Find the binding energy per nucleon for 3Li7 if mass of 3Li7 is 7·01653 amu [Given : mp = 1·00759 amu, mn = 1·00898 amu] Solution : The binding energy per nucleon ∆E E = A ∆m × 931 = MeV A ∆m = (3 mp + 4 mn) – Mass of 3Li7 = (3 × 1·00759 + 4 × 1·00898) – 7·01653 = 0·04216 amu E = 0·04216 × 931 7 39·25 = = 5·6 MeV per nucleon 7 Given the binding energy per nucleon of 1H2 and is 1·125 MeV and 7·2 MeV respectively. Solution : B.E. per nucleon of ∆E = 1·125 MeV ∴ E = A × ∆E = 2 × 1·125 MeV = 2·25 MeV ∴ B.E. of two 1 = 2 × 2·25 MeV Ed = 4·5 MeV H2 2 He 4 B.E. of α-particle = 4 × 7·2 MeV Eα = 28·8 MeV ∴ Energy released = Eα – Ed = 28·8 – 4·5 = 24·3 MeV Example 5. If 200 MeV energy is released in the fission of a single nucleus of 92U235, how many fission must occur per second to produce a power of 1 kW ? Solution : Energy released per second (power) = 1 kW = 103 Js –1 Energy released per fission = 200 MeV = 200 × 1·6 × 10–13 J ∴ No. of fissions per second 103 = 200 × 1·6 × 10–13 = 3·125 × 1013 Example 6. The energy supplied to a city by state electricity board is 40 million kilowatt-hour. If this energy could be obtained by the conversion of matter, how much mass would have to be annihilated ? Solution : Energy supplied to the city = 40 × 106 kWh = 40 × 106 × 36 × 10 5 joule Now E = ∆m.c2 ∆m = 4 × 36 × 10 12 E 2 = c (3 × 10 8)2 Example 2. If the mass defect in the formation of helium from hydrogen is 0·5%, then find the energy obtained in kWh, in forming helium from 1 kg of hydrogen. Solution : ∆E = ∆mc2 0·5 ∆m = × 1 = 0·005 kg 100 ∆E = 0·005 × (3 × 108) 4·5 × 1014 60 × 60 2 = 4·5 × 1014 joule or watt-sec = = 1·25 × 1011 watt-hour = 1·25 × 108 kWh Example 3. The mass of helium nucleus is less than that of its constituent particles by 0·03 amu. Find the binding energy per nucleon of 2He 4 nucleus. Solution : ∆m = 0·03 amu No. of nucleons A = 4 0·03 × 931 ∆E = = 7 MeV 4 Example 4. How much energy is released in the following reaction ? 1H 2 + 1H2 = 2He 4 = 16 × 10–4 kg = 1·6 gm OBJECTIVE QUESTIONS 1. m, mn and mp are the masses of A ZX nucleus, neutron and proton respectively. If the nucleus is broken into its constituents, then— (A) m > [(A – Z)mn + Zmp] (B) m < [(A – Z)mn + Zmp] (C) m = [(A – Z)mn + Zmp] (D) m = [Zmn + (A – Z)mp] 2. An element A is converted to C through following reactions : A → B + 2He 4 B → C + 2e– Then— (A) A and C are isobars (B) A and C are isotopes (C) A and B are isobars (D) A and B are isotopes 3. The nucleus with maximum binding energy per nucleon out of the following is— (A) 92U238 (B) 2He 4 16 (C) 8O (D) 26Fe 56 4. The following nuclear reaction represents 14 1 15 7N + 1H → 8 O + 7·3 MeV (A) Nuclear fusion (B) Nuclear fission C.S.V. / February / 2008 / 1570 (C) Scattering of particles (D) Element transformation 5. The binding energies per nucleon of deuterium and helium are 1·1 MeV and 7 MeV respectively. When two deuterons fuse to form a helium nucleus, the amount of energy released will be— (A) 23·6 MeV (B) 7 MeV (C) 6 MeV (D) 200 MeV 6. The most suitable material for moderator in a nuclear reactor is— (A) D2O (B) Cd (C) B (D) 92U235 7. The energy of thermal neutrons is nearly— (A) 0·25 MeV (B) 0·025 MeV (C) 200 MeV (D) 0·025 joule 8. The first atomic reactor was made by— (A) Hahn (B) Strassma (C) Fermi (D) Bethe 9. The fissionable material used in the bomb dropped at the city of Nagasaki of Japan was— (A) Plutonium (B) Uranium (C) Thorium (D) Neptunium 10. When the number of nucleons in the nucleus increases, the binding energy per nucleon— (A) Decreases continuously with A (B) Increases continuously with A (C) Remains constant with A (D) First increases with A and then decreases 11. The source of soloar energy is— (A) Fission of helium (B) Chemical reaction (C) Burning of carbon (D) Fusion of hydrogen nuclei 12. Average binding energy of nucleons is— (A) 8 eV (B) 8 MeV (C) 8 BeV (D) 8 joule 13. The energy equivalent to 1 kg of matter is about— (A) 1011 joule (B) 1014 joule (C) 1017 joule (D) 1020 joule 14. The controller rods in the nuclear reactor are made of— (A) Cadmium (B) Uranium (C) Graphite (D) Plutonium 15. Which of the following is the main source of energy emission in the stars ? (A) Chemical reaction (B) Fusion of heavy nuclei (C) Fission of heavy nuclei (D) Fusion of lighter nuclei 16. In a nuclear reactor— (A) Moderator is used to control the number of neutrons (B) Moderator is used to slow down the speed of neutrons (C) Controller rods are used to slow down the speed of neutrons (D) Coolant is used to control the number of neutrons 17. The equation 4(1H1) → 2He 4 + 2e– + 26 MeV represents— (A) β-decay (C) Fusion (B) γ-decay (D) Fission 18. The mass defect in a nuclear fusion reaction is 0·3 per cent. What amount of energy will be liberated in one kg fusion reaction ? (A) 5·2 × 1014 joule (B) 2·7 × 1014 joule (C) 2·7 × 1014 kWh (D) None of these 19. Enriched uranium is better fuel for a reactor because it has greater proportion of— (A) Slow neutrons (B) Fast neutrons (C) 92U235 (D) 92U238 20. In a nuclear reactor— (A) Rate of reaction may be controlled by boron steel rods (B) Fast neutrons are slowed down by cadmium rods (C) Plutonium is used as coolant (D) Hydrogen is used as fuel ANSWERS WITH HINTS 1. (B) The mass of nucleus is less than the sum of the masses of its constituent particles. 2. (B) In the first reaction Z becomes (Z – 2) and A becomes (A – 4). In the second reaction (Z – 2) becomes Z again and (A – 4) remains unchanged. Thus for A and C, Z is the same and A differs i.e., they are isotopes. 3. (D) The nuclei having mass number ≈ 60 (for example, Fe whose mass no. is 56) have maximum binding energy per nucleon (about 8·7 MeV). 4. (A) 5. (A) 1H2 + 1H2 → 2He 4 Net initial binding energy = 4 × 1·1 = 4·4 MeV Net final binding energy = 4 × 7 = 28 MeV ∴Energy released = 28 – 4·4 = 23·6 MeV (A) Heavy water (D2O) is the best moderator. (B) 8. (C) 9. (A) 10. (D) 11. (D) (B) (C) ∆E = ∆m × c2 = 1 × (3 × 10 8)2 = 9 × 1016 joule ≈ 1017 joule 14. (A) 15. (D) 16. (B) 17. (C) 0·3 18. (B) ∆m = × 1 = 0·003 kg 100 ∆E = ∆mc2 = 0·003 × (3 × 108)2 = 2·7 × 1014 joule 19. (C) Ordinary uranium is U238. The isotope U235 is only 0·7% in it. In enriched uranium the isotope U235 is increased from 0·7% to 2·3%. 20. (A) ●●● 6. 7. 12. 13. C.S.V. / February / 2008 / 1571 Model Paper for Various Medical Entrance Examinations PHYSICS 1. For a glass prism ( µ = 3 ) , the angle of minimum deviation is equal to the angle of the prism. The angle of prism is— (A) 30° (B) 45° (C) 60° (D) 90° 7. An electric heater, assumed to be a black body has a temperature of 727°C. If its temperature is raised to 1727°C, the amount of energy radiated per unit time now as compared with that in the first case will be— (A) Twice (B) Four times (C) Sixteen times (D) Hundred times 8. A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to (– k /r 2) where k is a constant. The total energy of the particle is— k k (A) (B) – 2r r k (C) – (D) None of these 2r Directions—In each of the following questions (From Q. 9 to 16), 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 A is false but R is true (E) If both A and R are wrong 9. Assertion (A) : If the image formed by a mirror is virtual, erect and enlarged, then it is definitely a concave mirror only. Reason (R) : The convex mirror can never produce an enlarged virtual image. (A) (B) (C) (D) (E) 10. Assertion (A) : A black-hole is an astronomical entity that cannot be seen by telescope. Reason (R) : The gravitational field on a black hole is so strong that it does not allow even light to escape. (A) (B) (C) (D) (E) 11. Assertion (A) : In the process of refraction of light, the speed, the wavelength and the intensity of light change. Reason (R) : The change of the speed of light results in the change of its frequency. (A) (B) (C) (D) (E) 2. A stone is just released from the window of a train moving along a horizontal straight track. The stone will hit the ground following a— (A) Parabolic path (B) Hyperbolic path (C) Straight line path (D) Circular path 3. In Ingen Housz’s experiment, the wax melts upto 4 cm and 10 cm on the iron and the silver rods respectively. If K for silver is taken as 1, then its value for iron is— (A) 0·4 (B) 0·16 (C) 2·5 (D) 6·25 4. The magnitude of displacement of a point of a wheel (of radius r) initially in contact with the ground when the wheel rolls forward half-revolution along a straight line is— (A) (C) 2r 2 πr (B) r π2 + 4 (D) r π2 + 1 12. Assertion (A) : When some metals are cooled towards absolute zero, a transition temperature is reached at which the resistance suddenly falls to zero. Reason (R) : Some specially developed metal compounds have transition temperatures above 100 K. (A) (B) (C) (D) (E) 13. Assertion (A) : With rise of temperature the resistance of a metal increases. Reason (R) : When the temperature of the metal rises, the atoms of lattice vibrate more vigorously. Free electrons collide with the lattice more frequently. (A) (B) (C) (D) (E) 14. Assertion (A) : A monoatomic gas has only 3 degrees of freedom. Reason (R) : Monoatomic gas cannot have rotational degrees of freedom at all temperatures. (A) (B) (C) (D) (E) 5. A refrigerator with its power on and door open, is kept in a closed room. The temperature of the room— (A) Rises (B) Falls (C) Remains constant (D) Rises or falls depending on the area of room 6. A man moves on a cycle with a velocity of 4 km/hr. The rain appears to fall to him with a velocity of 3 km/hr vertically. The actual velocity of rain is— 4 (A) 7 km hr–1 (B) km hr–1 3 3 –1 (D) 5 km hr–1 (C) km hr 4 15. Assertion (A) : Time taken for a hot container to cool from 50°C to 40°C is more than that to cool from 60°C to 50°C. Reason (R) : Rate of cooling depends only on the temperature of the surroundings. (A) (B) (C) (D) (E) 16. Assertion (A) : In adiabatic compression the internal energy and temperature of the system gets decreased. Reason (R) : The adiabatic compression is a slow process. (A) (B) (C) (D) (E) C.S.V. / February / 2008 / 1572 17. A planet is revolving round the sun in an elliptical orbit. The maximum and the minimum distances of the planet from the sun are 3 × 10 12 m and 2 × 10 10 m respectively. The speed of the planet when it is nearest to the sun is 2 × 10 7 m/s. What is the speed of the planet when it is farthest from the sun ? (A) 1·33 × 105 m/s (B) 3 × 105 m/s (C) 1·5 × 107 m/s (D) 2·66 × 105 m/s 18. An earth satellite is moved from one circular orbit to another orbit of larger radius. The quantity which increases for the satellite as a result of this change is— (A) Gravitational energy (B) Angular speed (C) Linear speed (D) Centripetal acceleration 19. Two glass plates, one upon the other with a little water in between them can not be separated easily because of— (A) Viscosity (B) External pressure (C) Surface tension (D) None of these 20. The initial pressure and volume of a gas are P and V respectively. First its volume is expanded to 4V by isothermal process and then its volume is restored to V by adiabatic process. The final pressure of the gas is— (γ = 1·5) (A) 8P (C) P (B) 4P (D) 2P potential (C) The molecular agitation increases at constant pressure (D) The molecular agitation decreases at constant volume 22. An air column in an organ pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 264 Hz, if the length of the column is— (For sound v = 330 m/sec) (A) 31·25 cm (B) 62·25 cm (C) 93·75 cm (D) 125·75 cm 23. The velocities of sound in an ideal gas at temperatures T1 and T2 are v 1 and v 2 respectively. If the rms speeds of the same gas at the same temperatures T1 and T2 are v 1′ and v 2′ respectively, then— v2 (A) v 2′ = v1′ v1 v1 (B) v 2′ = v1′ v2 (C) v 2′ = v 1′ (D) v 2′ = v 1′ v2 v1 v1 v2 27. An A. C. voltage of V = 220 2 sin (ωt + π/2 ) will be read as by a hot wire voltmeter— (A) 220 (C) 220 V 2V (B) 2V (D) 440 V 28. In Young’s double slit experiment the fringe-width is 0·4 mm. If the whole apparatus be immersed in a liquid of refractive index 4 , then the new fringe-width will 3 become— (A) 0·3 mm (C) 0·53 mm (B) 0·4 mm (D) 450 micron 29. Three identical resistances each of 10Ω are connected as shown in figure. The maximum power that can be consumed by individual resistor is 20 watt. Then the maximum power consumed by the combination is— 10Ω 10Ω A 10Ω B (A) 60 watt (C) 30 watt (B) 15 watt (D) 40 watt 24. What will be the angle between two plane mirrors when a ray of light, incident on one and parallel to the other, emerges parallel to the first after reflection at the second ? (A) 30° (B) 45° (C) 60° (D) 90° 25. A Carnot’s engine works as a refrigerator between 250 K and 300 K. If it receives 750 calories of heat from the reservoir at lower temperature, the amount of heat rejected at the higher temperature is— (A) 900 cal (B) 625 cal (C) 750 cal (D) 1000 cal 26. A real image of a distant object is formed by a plano-convex lens on its optical axis. Spherical aberration is— (A) Absent (B) Smaller if the curved surface of the lens faces the object (C) Smaller if the plane surface of the lens faces the object (D) The same whichever side of the lens faces the object 30. A star emitting light of wave° length 6000 A is moving towards the earth with a speed of 3·0 × 106 m/s. The apparent wavelength of emitted light is— (c = 3 × 10 8 m/s) ° (A) 6060 A ° (C) 5500 A ° (B) 5940 A ° (D) 6500 A 31. Consider the following statements in case of specific heat of gases— 1. Specific heat of a gas is zero when it undergoes an adiabatic change. 2. Specific heat of a gas is infinite when it undergoes isothermal change. 3. The difference between principal specific heats is the same for all gases. 4. The ratio of principal specific heats is the same for all gases. Of these correct statements are— (A) 1, 2, 3 (C) 1, 2 (B) 2, 3, 4 (D) 1, 2, 3, 4 21. The specific heat of a gas at constant pressure is greater than its specific heat at constant volume because— (A) At constant pressure work is done in expanding the gas against constant external pressure (B) At constant volume work is done when pressure increases C.S.V. / February / 2008 / 1573 → → 32. A conductor PQ with PQ = r → moves with a velocity v in a uniform magnetic field of induction → B . The e.m.f. in the conductor is— → → → (A) ( v × B )· r → → → (B) v ·( r × B ) → → → (C) B ·( r × v ) → → → (D) r × ( v × B ) 33. At an instant, the ratio of the amounts of radioactive substances is 2 : 1. If their half lives be respectively 12 and 16 hours, then after 2 days, the amounts of the substances left over will be in the ratio— (A) 1 : 1 (C) 1 : 2 (B) 2 : 1 (D) 1 : 4 principal quantum number n as— Z4 Z2 (B) 4 (A) 2 n n Z (C) (D) None of these n 37. If a very short pulse of white light were to travel through a piece of glass, the colour which would exit the glass first is— (A) Green (C) Yellow (B) Red (D) Violet 41. Pressure (P) volume (V) plots for two gases during adiabatic process are shown. Plot 1 and 2 correspond to— P 1 2 V 38. In the circuit shown in figure, power developed across 1Ω, 2Ω and 3Ω resistance is in the ratio of— 1Ω i 3Ω 2Ω i (A) He and O 2 (B) O2 and He (C) He and Ar (D) O2 and N 2 42. When separation between two charges is increased, the electric potential energy of the charges— (A) Increases (B) Decreases (C) Remains the same (D) May increase or decrease 43. The moment of inertia of a flat annular ring of mass M, inner radius r and outer radius R about a perpendicular axis through its centre is— 1 M (R – r )2 (A) 2 1 (B) M (R2 + r 2) 2 (C) M (R2 – r 2) 1 (D) M (R2 – r 2) 2 44. The drift velocity of the free electrons in a conducting wire carrying a current is v. If in a wire of the same metal, but of double the radius, the current be 2i, then the drift velocity of the electrons will be— v v (A) (B) 4 2 (C) v (D) 4v (A) 1 : 2 : 3 (C) 6 : 4 : 9 (B) 4 : 2 : 27 (D) 2 : 1 : 27 34. If an electron enters into a space between the plates of a parallel plate capacitor at an angle α with the plates and leaves at an angle β to the plates. The ratio of its K.E. while entering the capacitor to that while leaving it will be— (A) 39. A ray of light is incident at an angle i on a surface of a prism of small angle A and emerges normally from the opposite face. If the refractive index of the material of the prism is µ, the angle of incidence i is nearly— (A) A/µ (C) µA (B) A/2µ (D) µA/2 (C) cos α 2 cos β sin α 2 sin β (B) (D) cos β 2 cos α sin β 2 sin α 35. An AC voltage V is applied across a R–C series combination. At the instant during the cycle when the current in the circuit becomes zero— (A) The voltage across the resistor is equal and opposite to the voltage across the capacitor 1 (B) Energy equal to CV 2 is 2 stored in the capacitor (C) Energy equal to CV2 is stored in the capacitor (D) Energy stored in the capacitor is zero 36. In hydrogen and hydrogen-like atoms the ratio of difference of energies E4n – E2n and E2n – En varies with atomic number Z and 40. Figure shows an equilateral triangular loop PQR of side l and carrying a current i. A uniform magnetic field B exists in direction parallel to PQ. Which statement is incorrect about the magnitude of forces acting on the sides of the triangle ? R B i i 45. In the circuit shown in the figure, switch is turned on at t = 0, then— Q + 12V − 2 kΩ R2 R3 2kΩ S 2kΩ R1 10µF P i (A) On PQ, F = 0 3 (B) On QR, F = ilB 2 3 ilB (C) On RP, F = 2 (D) None of these 1. At t = 0, current supplied by battery is 4 mA. C.S.V. / February / 2008 / 1574 2. At t = 0, current in R3 is 2 mA. 3. In the steady state current supplied by battery is 3 mA. 4. In the steady state current in R3 is zero. Of these statements correct are— (A) Only 1, 4 (B) Only 2, 3 (C) All the statements 1, 2, 3 and 4 (D) None of the statements 46. In a radioactive material, after a time interval equal to the reciprocal of decay constant, the undecayed atoms are— (A) 63·2% of their initial number (B) 50% of their initial number (C) 36·8% of their initial number (D) None of these 47. The 5Ω resistor in the given circuit produces 10 calories per second. The heat per second in 6Ω resistor will be— 4Ω A 5Ω B 6Ω (A) 6 cal (C) 3 cal (B) 2 cal (D) 10 cal 49. A convex lens and a convex mirror are placed co-axially at a distance of 10 cm from each other. The focal length of the lens is 15 cm. If an object is placed on the axis at a distance of 20 cm from the lens, it is found that the image coincides with the object. The radius of curvature of the mirror is— (A) 60 cm (B) 55 cm (C) 50 cm (D) 45 cm 50. The given truth table belongs to— A 0 1 Y 1 0 48. Which of the following processes is not related to radioactive decay ? (A) Positron emission (B) Electron capture (C) Nuclear fusion (D) α-decay (A) XOR gate (B) NOT gate (C) NOR gate (D) NO gate ANSWERS WITH HINTS sin 1. (A) Since µ = A+δ 2 sin A/2 ( ) 6. (D) δ = A, we get 3 = sin A sin A/2 2 sin A/2 cos A/2 = sin A/2 3 = cos 30° 2 ∴ ⇒ 7. (C) T1 T2 H2 H1 → ^ v RM = 3(– j ) → → = vR – vM → ^ ^ vR = – 3 j + 4 i vR = 32 + 42 = 5 km hr–1 = 727 + 273 = 1000 K = 1727 + 273 = 2000 K T2 4 = T1 = 9. (B) 10. (A) 11. (C) 12. (B) 13. (A) 14. (A) 15. (C) 16. (E) 17. (A) In planetary motion angular momentum remains constant, i.e., mv1r1 = mv2r2 v 1 r1 v2 = r2 = 2 × 107 × 2 × 10 10 3 × 1012 A cos = 2 A/2 = 30° ∴ 2. (A) 3. (B) Iron l1 = 4 cm, k1 = ? Silver l2 = 10 cm, k2 = 1 In Ingen Housz experiment l1 2 k1 = 2 k2 l2 k1 = (4)2 (10)2 × 1 = 0·16 A = 60° 2000 4 = 16 1000 2 k mv = 2 8. (C) r r 1 k K.E. = mv2 = …(i) 2 2r Force acting on the particle is conservative in nature. Hence du F = – dr or du = – Fdr U = – = ⇒ () ( ) = 1·33 × 105 m/s 18. (A) 19. (C) The thin layer of water wets the plates over an almost circular area which is concave outwards as shown in figure. r d ∫ r ∞ r The excess pressure p inside the liquid film over the outside atmospheric pressure is given by p = T Since r >> d, Fdr ∞ 4. (B) Point has two displacements πr and 2r in two mutually perpendicular directions. The resultant displacement = (πr)2 + (2r)2 ∫ k dr r2 ( ) 1 2 – r d 2T d U = – k r 1 can be neglected r in comparison to 2/d. p = – Total energy k k – K.E. + P.E. = 2r r k = – 2r Hence = r π2 + 4 5. (A) Thus, the excess pressure inside the film is less than the outside atmospheric pressure (as is clear C.S.V. / February / 2008 / 1575 from –ve sign) by 2T/d . The excess pressure of atmosphere on the two plates pushes them closer together. Smaller the value of d, greater will be the force required to pull them apart. 20. (D) For isothermal process P1V1 = P2V2 or PV = P2 (4V) P ⇒ P2 = 4 For adiabatic process P2V2γ = P3V3γ P (4V)γ = P3(V)γ 4 ⇒ P3 = P(4)γ – 1 = P(4)0·5 = 2P 21. (A) 22. (A) For an organ pipe closed at one end, in fundamental mode of vibration l = ⇒ λ 4 25. (A) In Carnot’s engine Q1 T1 = T2 Q2 750 250 = Q2 300 ⇒ Q2 = 900 cal 26. (B) Spherical aberration is small when the deviation is shared by the two faces of the lens. s Fm Fp 30. (B) Since the star is moving towards earth, frequency of light emitted by the star will increase or its wavelength will decrease. v ·λ Now, ∆λ = c = 3·0 × 106 3 × 108 ° × 6000A ° = 60A Therefore, apparent wavelength λ′ = λ – ∆λ = 6000 – 60 ° = 5940A 31. (C) 32. (A) Imagine a unit charge inside the conductor. As the conductor → moves with velocity v , the charge inside it will experience a → → → force F = v × B . If this unit charge is moved from one end of the conductor to its other end, its → displacement is r . Hence, the work done on it is →→ W = F· r → → → = ( v × B )· r This by definition is the e.m.f. across the conductor. 33. (A) Let the substances be A and B For A, mass = 2m No. of half lives = 4 1 4 NA = 2m × 2 m = 8 For B, mass = m No. of half lives = 3 1 3 NB = m × 2 m = 8 NA ∴ = 1 NB 34. (B) Let u and v be the velocity of the electron when entering and when leaving the capacitor respectively. Since component of velocity parallel to the plates remains unchanged, we have u cos α = v cos β s′ Fm Fp Longitudinal axial spherical aberration s is smaller than s′. 27. (C) ∴ 28. (A) ⇒ V0 = 220 2 Vrms = 220 2 = 220 V 2 λa λl 1 3 = µ 4 D λ d λ = 4l v n = λ = 330 = 264 4×l µ = λl λa = ⇒ 330 l = 4 × 264 = 0·3125 m = 31·25 cm Now fringe width β = β ∝ λ for a given arrangement βl βa ⇒ = λl λa 23. (A) For sound v2 = v1 For rms speeds T2 T1 v 1′ v2 v 2′ = v 1′· v1 v 2′ = T2 T1 ⇒ 1 µ 3 = 0·4 × 4 = 0·3 mm 29. (C) Maximum current than be sent through resistances is βl = βa × 20 = I 2 × 10 ⇒ I = 2A Clearly, therefore, max. current through each bulb combined in parallel is 1 1 = 2A = A 2 2 Therefore, max. power consumed () () 24. (C) Conditions of the problem are shown in the figure from which it is clear that M2 A C θ θ O θ θ D θ B θ M1 = ( ) 1 2 2 × 10 + × 10 + ( ) 1 2 2 θ + θ + θ = 180° ⇒ θ = 60° ( 2)2 × 10 = 5 + 5 + 20 = 30 W ⇒ cos β u = v cos α C.S.V. / February / 2008 / 1576 Hence, the required ratio 1 mu2 u 2 cos β 2 2 = 1 2 = v mv cos α 2 () 42. (D) 43. (B) Face area of annular ring = π (R2 – r 2) Mass per unit area of the disc M = π (R2 – r 2) At t = 0, Total resistance 2×2 = 3 kΩ 2+2 12 = 4 mA ∴ I = 3 × 103 At t = 0. This current divides into two equal parts. Hence current in R3 is 2 mA. In the steady state, current supplied by battery 12 = 3 mA = 4 × 103 In the steady state, condenser blocks the current. Hence in this state current in R 3 is zero. 46. (C) N = N0e – λt 1 Putting t = λ N = N0e–1 N0 = e 1 = N 2·718 0 = 0·368 N 0 i.e., N = 36·8% of N 0. 2+ 47. (C) Heat produced in 5Ω resistor = 10 cal/sec = 42 J/s V2 ∴ = 42 5 ⇒ V = 42 × 5 35. (C) When current is zero, potential drop across resistance is zero. Hence all the voltage appears across the capacitor. V is RMS voltage, so that its peak voltage is 2 V . Hence energy stored in the capacitor 1 2 = C ( 2 V ) = CV 2 2 E1 E1 – E4n – E2n 16n2 4n2 36. (D) = E2n – En E1 E1 – 4n2 n2 1 = = Constant 4 37. (B) Speed of light in glass is maximum for red colour as seen from the relation. c v = µ 38. (B) At point A current i devides itself in the inverse ratio of resistance. 1Ω i A i1 = 2i /3 i2 = i /3 2Ω i B r O R Thus i1 = 2i i and i2 = 3 3 H ∝ i 2R ∴ H1 : H2 : H3 2i 2 i = ×1: 3 3 () () 2 × 2 : i2 × 3 Imagine the disc to be made up a number of thin circular rings and consider one such ring of radius x and thickness dx. Mass of this element M = 2πx dx π (R2 – r 2) 2 Mx dx = (R2 – r 2) Its moment of inertia about an axis passing through O and perpendicular to its plane 2 Mx = . dx . x 2 (R2 – r 2) 2 Mx 3 = dx (R2 – r 2) Hence moment of inertia of the annular disc R 2M = = x3 dx (R2 – r 2) r 2M x4 R = 2 R – r2 4 r ∫ = = = 4 : 2 : 27 39. (C) A = r1 + r2 = r + 0 ⇒ r = A i µ = r when angles are small i = µr = µA F = i B l sin 120° = On RP, 3 iB l 2 2M R2 – r 2 [ ] [ ] R4 – 4 r4 This is the potential difference between points A and B. Heat produced in 6Ω resistor = I2 × R = 2 ⇒ 40. (D) On PQ, F = i B l sin 0 = 0 On QR, F = i B l sin 120° = 3 iB l 2 41. (B) For monoatomic gas γ (= Cp/Cv ) is larger than that for a diatomic gas. 2M 1 . . [(R2 + r 2) R2 – r 2 2 (R2 – r 2)] 1 = M (R2 + r 2) 2 44. (B) Drift velocity of free electrons i vd = ne A i = ne π r 2 i vd ∝ 2 r when i and r are both doubled v vd = 2 45. (C) At t = 0 the capacitor acts as a short whereas in the steady state it acts as open circuit. 42 × 5 × 6 cal 10 = 12·6 J 12·6 = = 3 cal 4·2 48. (C) 49. (C) The rays emerging from the lens are directed towards the centre of curvature of the mirror, i.e., these are normally incident on the mirror and retrace their path to form the image coincident with the object L O 20 cm M 60 cm C (Continued on Page 1633 ) C.S.V. / February / 2008 / 1577 Model Paper for Various Medical Entrance Examinations PHYSICS 1. The range of a particle when launched at an angle of 15° with the horizontal is 1·5 km. What is the range of the projectile when launched at an angle of 45° to the horizontal ? (A) 6·0 km (B) 0·75 km (C) 1·5 km (D) 3·0 km 2. A large tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both the holes are the same. Then the radius R is equal to— (A) L 2π (C) L (D) L 2π (B) 2πL 7. Assertion (A) : A cyclist always bends inwards while negotiating a curve. Reason (R) : By bending he lowers his centre of gravity. Of these statements— (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false (D) A is false but R is true 8. Brakes of very small contact area are not used although friction is independent of area, because friction— (A) Resists motion (B) Causes wear and tear (C) Depends on the nature of materials (D) Sliding friction operates in this case → ^ 9. Work done when a force F = i + ^ ^ 2 j + 3k acting on a particle takes → ^ ^ ^ it from the point r1 = ( i + j + k) → ^ ^ ^ to the point r2 = ( i – j + 2k) is— (A) – 3J (C) Zero (B) – 1 J (D) 2J 11. A particle executing SHM along a straight line has a velocity 4 cm s–1 when at a distance of 3 cm from its mean position and 3 cm s–1 when at a distance of 4 cm from it. The time taken by the particle to travel 2·5 cm from equilibrium position is— (A) 0·25 sec (B) 0·5 sec (C) 0·75 sec (D) 1·0 sec 12. The network shown in the figure is a part of a circuit. If at a certain instant, the current 5A and is decreasing at a rate of 103 A/s, then VB – VA is— 1Ω A + 5 mH B 15 V (A) 20V (C) 15V (B) 10V (D) 5V 13. A satellite goes along an elliptical path around earth. The rate of change of area swept by the line joining earth and satellite is proportional to— (A) r (B) r 2 (C) r 1/2 (D) r 3/2 14. An ideal fluid is— (A) One which obeys Newton’s law of viscosity (B) One which satisfies continuity equation (C) One which flows through pipes with least friction (D) Frictionless and incompressible 15. A sphere is at a temperature 600K. Its cooling rate is R in an external environment of 200K. When its temperature falls to 400K, its cooling rate R′ will be— (A) 3 R 16 1 (C) R 3 (B) 16 R 3 3. A spherical ball rolls on a table without slipping. The fraction of its total energy associated with rotation is— 3 3 (A) (B) 5 7 2 2 (C) (D) 5 7 4. In which of the following processes, does the internal energy of an ideal gas remain constant ? (A) Adiabatic (C) Isobaric (B) Isochoric (D) Isothermal 10. In the circuit shown in the figure B is a battery of e.m.f. E and K1 B 6Ω K2 V 12 Ω 5. The unit of ‘Poynting vector’ is— (A) Watt per second (B) Watt per metre square (C) Watt (square metre–1)(sec)–1 (D) Watt/metre cube 6. At 0 K intrinsic semiconductor behaves as— (A) A perfect insulator (B) A semiconductor (C) A super conductor (D) A perfect conductor (D) None of these internal resistance r. When key K1 is closed and K2 is open, the ideal voltmeter shows a reading 18V. When key K 2 is closed and K1 is open, the reading of the voltmeter is 24V. When K1 and K2 are both open, the voltmeter reading is— (A) 12V (B) 36V (C) 24V (D) 8V 16. A transformer transforms— (A) Energy (C) Power (B) Frequency (D) Voltage 17. At 0K which of the following properties of a gas will be zero ? (A) Kinetic energy (B) Potential energy (C) Vibrational energy (D) Density C.S.V. / February / 2008 / 1578 18. Cores are laminated to— (A) Reduce eddy currents (B) Reduce hysteresis (C) Better assembly (D) Increase current drawn 19. A string is cut into three parts, having fundamental frequencies n1, n2 and n3 respectively. The original fundamental frequency n is related to these frequencies as— 1 1 1 1 (A) = + + n n1 n2 n3 (B) n = n1 + n2 + n3 n1 + n 2 + n 3 (C) n = 3 (D) n = n1 × n2 × n3 20. The switches that must be closed for obtaining resistance of 2Ω are— S1 1 23. A microscope is focussed on a mark at the bottom of a beaker. The microscope is then raised through 1·5 cm. To what height must water be poured into the beaker to bring the mark again into focus ? (A) 4·5 cm (B) 6·0 cm (C) 8·0 cm (D) 12 cm 24. Which phaser is correct for a series RC circuit ? (A) I I V (B) I V (D) I V V 31. The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is x. If the current and radius both are doubled the new ratio will become— (A) 2x (B) 4x x x (C) (D) 4 8 32. What will be the wave velocity, if the radar gives 54 waves per minute and wavelength of given wave is 10 m ? (A) 4 ms– 1 (B) 6 ms– 1 (C) 9 ms– 1 (D) None of these 33. Which one of the following graphs represents the variation of voltage V across the inductor L with time t in a series L-R circuit ? (A) V O t (C) S2 4Ω 5Ω 10 Ω 10 Ω 4Ω S3 S4 25. A ray of light is inclined to one face of a prism at an angle of 50°. If the angle of the prism is 60° and the ray is deviated through an angle of 42°, then the angle which the emergent ray makes with second face of the prism is— (A) 30° (B) 62° (C) 50° (D) 28° (B) V O t (C) V O t (D) V O t A B (A) S1 and S4 (C) S2 and S3 (B) S2 and S4 (D) S1 and S2 26. A sonometer wire 65 cm long is in resonance with a tuning fork of frequency N. If the length of the wire is decreased by 1 cm and it is vibrated with same tuning fork, 8 beats are heard per second. The value of N is— (A) 256 Hz (B) 384 Hz (C) 480 Hz (D) 512 Hz 27. Two lenses of powers + 4D and – 2D are placed 10 cm apart. The focal length of the combination is— (A) – 75 cm (B) – 35·7 cm (C) 35·7 cm (D) 75 cm 28. An aeroplane flying 490 m above the ground level at 100 m/s releases a block. How far on the ground will it strike ? (A) 0·1 km (B) 1 km (C) 2 km (D) None of these 29. Sound waves cannot exhibit— (A) Interference (B) Diffraction (C) Reflection (D) Polarisation 30. A diode is used as a— (A) Rectifier (B) Amplifier (C) Detector (D) Modulator 34. GaAs is— (A) Element semiconductor (B) Alloy semiconductor (C) Bad conductor (D) Metallic semiconductor 35. The figure shows the energy levels of a certain atom. When the system moves from 2E level to E, a photon of wavelength λ is emitted. The wavelength of the photon produced during its 4E transition from level to E is— 3 2E 4E 3 E 21. The equation of a simple harmonic progressive wave is y = 8 sin (0·628 x – 12·56 t) where y and x are in cm and t in second. The phase difference between two particles at a distance of 2·0 cm apart at any instant is— (A) 90° (C) 0° (B) 72° (D) 270° O 22. In a double slit experiment the distance between the slits is d. The screen is at a distance D from the slits. If a bright fringe is formed opposite to a slit, find its order— d D (A) (B) 2Dλ 2d λ d2 (C) 2Dλ dD (D) 2λ (A) (C) λ 3 4λ 3 (B) 3λ 4 (D) 3λ 36. When a load of 5 kg is hung on a wire, then extension of 0·3 m takes place. Work done is— (A) 7·5 J (B) 6·0 J (C) 15·0 J (D) 5·0 J C.S.V. / February / 2008 / 1579 / 4 37. The K α X-ray emission of tungsten occurs at λ = 0·021 nm. The energy difference between K and L levels in this atom is about— (A) 0·51 MeV (B) 1·2 MeV (C) 59 keV (D) 13·6 eV 38. A gas behaves as an ideal gas at— (A) Low pressure and high temperature (B) High pressure and low temperature (C) High pressure and high temperature (D) Low pressure and low temperature 39. An electrolysis experiment is stopped and the battery terminals are reversed— (A) The electrolysis will stop (B) The rate of liberation of material at the electrodes will increase (C) The rate of liberation of material will remain the same (D) Heat will be produced at a greater rate 40. A person is standing on a rotating stool spreading his arms. He suddenly contracts his arms. His— (A) Moment of inertia increases (B) Angular momentum decreases (C) Angular momentum increases (D) Angular velocity increases 41. A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It— (A) Will stay in north-south direction (B) Will stay in east-west direction (C) Will become rigid showing no movement (D) Will stay in any position 42. Longest wavelength of Balmer ° series is 6563 A. What will be the shortest series? wavelength of this ted from an unstable atom whose atomic number Z remains unchanged. (B) γ-radioactivity is the process in which the daughter nucleus has atomic-number 1 unit more than that of the parent nucleus. (C) α-radioactivity is the process in which an unstable atom emits the nucleus of a helium atom (D) α-radioactivity is the process in which a heavy atom emits electromagnetic radiation of very high frequency. 48. A large drop is made from two similar smaller drops each charged to 100 volt. The voltage on the bigger drop is— (A) 100 × 2– 2/3 volt (B) 100 × 22/3 volt (C) 100 × 21/3 volt (D) 100 volt 49. A diode is connected to 200 V a.c. mains in series with a capacitor as shown in the figure. The final voltage V across the capacitor is— ° (A) 912 A ° (C) 3646 A ° (B) 4861 A ° (D) 4341 A 43. A proton and an electron are placed in a uniform electric field— (A) The electric forces acting on them will be equal (B) The magnitudes of the forces will be equal (C) The accelerations will be equal (D) The magnitudes of their accelerations will be equal 44. Which of the prisms is used to study infra red spectrum ? (A) Crown glass (B) Flint glass (C) Rock salt (D) Nicol 45. Which of the following is not the property of ionic solids ? (A) They are good conductors of electricity (B) They are strong but nondirectional (C) They have high melting points (D) They are transparent to visible light 46. In an scattering experiment number of scattered particles at 90° is 56. What will be their number when scattered at 60° ? (A) 256 (B) 224 (C) 95 (D) 102 47. In the four statements given below the only one correct statement is— (A) β-radioactivity is the process in which an electron is emit- 200 V a.c. V (A) 200 V (B) 200 V 2 (C) 200 2 V (D) None of these 50. On increasing the height of a satellite, its time period will— (A) Increase (B) Decrease (C) Remain unchanged (D) Increase or decrease depends on height ANSWERS WITH HINTS 1. (D) Range R = ∴ u 2 sin 2α g sin 2α 2 2. (A) The volume of water flowing out per second = Velocity × Area of crosssection of the hole = vA According to problem v 1A1 = v 2A2 ⇒ 2gy × L 2 = R = 2g × 4y . πR 2 L 2π 1 mv2 2 1 2 = Iω 2 R2 = R1 sin 2α 1 sin 90° R2 = 1·5 . sin 30° = 1·5 × 2 = 3·0 km 3. (D) (K.E.)linear = (K.E.)rot. C.S.V. / February / 2008 / 1580 4. 5. 6. 9. 10. S1 d/2 S D S2 ( ) 1 GM ·r 2 r 1 = GMr 2 1/2 Rate ∝ r = 14. (D) 15. (A) According to Stefan’s law R′ (400)4 – (200)4 = R (600)4 – (200)4 3 ⇒ R′ = R 16 16. (D) 17. (A) 18. (A) 1 19. (A) For a string n ∝ since T and l m are constant. L = l1 + l2 + l3 1 1 1 1 ∴ = + + n n1 n 2 n 3 20. (A) Closing S1 and S4 will put in circuit two resistances of 4Ω each connected in parallel and exclude 23. (B) nβ = d 2 d n = 2β d d d2 = × = 2 Dλ 2Dλ µ = = R A R R – 1·5 ⇒ Now r = 6Ω E = 18 6+r 6 6+6 = 18 = 36V 6 11. (B) v 1 = ω a2 – x 12 , v 2 = ω a2 – x 22 v1 = v2 a2 – x 12 a2 – x 22 –9 a2 – 16 a2 A R 1.5 Cm cm O O′ O' or ⇒ 4 = 3 R 4 = R – 1·5 3 ⇒ R = 6·0 cm a = 5 cm C.S.V. / February / 2008 / 1581 ↑ 1 2 v2 · mr 2 × 2 2 5 r 1 = mv2 5 1 1 Total energy = mv2 + mv2 2 5 7 = mv2 10 1 mv2 (K.E.) rot. 5 2 = = 7 Total K.E. 7 mv2 10 (D) In isothermal change, temperature remains constant. (B) Poynting vector → → = E ×B volt amp-turn = × metre metre = watt/m2 (A) 7. (B) 8. (B) → → → (B) W = F ·( r2 – r1 ) → → ^ ^ ^ ^ ^ ^ r2 – r1 = ( i – j + 2k) – ( i + j + k) ^ ^ = – 2j + k → → → ∴ W = F ·( r2 – r1 ) ^ ^ ^ ^ ^ = ( i + 2 j + 3k)·(– 2 j + k) = –4+3=–1J (B) When both keys are open, the ideal voltmeter will read the e.m.f. E of the battery. V = E – Ir r E = E– R+r ER = R+r 6E Hence 18 = 6+r 12E and 24 = 12 + r 18 6E 12 + r Dividing = × 24 6 + r 12E = Also, ⇒ Now, 4 = ω 25 – 9 ω = 1 rad/sec. y = a sin ωt 2·5 = 5 sin ωt π ⇒ ωt = 6 π ∴ t = sec 6 3·14 = ≈ 0·5 sec. 6 12. (C) Proceeding in the direction of current dI VA – VB = IR – E + L dt all other resistance. Hence equivalent resistance in the circuit will be 2Ω. 21. (B) y = 8 sin (0·628 x – 12·56 t) Comparing it with standard wave equation 2πx 2πt y = a sin – T λ ( ) () = 1 × 5 – 15 + 5 × 10– 3 (– 10)3 = – 15V ∴ VB – VA = 15V 1 13. (C) Area EPQ = EP × PQ 2 1 r. r d θ = 2 Q dθ E P Rate at which area is swept 1 2dθ = r 2 dt 1 2 = r ω 2 1 2 v = r · 2 r 1 = vr 2 We get 2π = 0·628 λ 0·628 ⇒ λ = 2 × 3·14 = 0·1 m = 10 cm Phase difference 2π ∆φ = × ∆x λ 2π × 2·0 = 10 2 ×π = 5 2 = × 180° = 72° 5 22. (C) Fringe width Dλ β = d If nth bright fringe is opposite S 1 on the screen, then ↑ 24. (C) In C–R circuit current I leads the applied voltage V by a phase angle φ given by VC XC tan φ = = VR R = 25. (D) 1/ωC R 29. (D) 30. (A) 31. (D) B x = M = ⇒ x ∝ Note E = hν= µ0i 2r 1 iπr2 hc λ ( )( ) = = (4·14 × 10–15 eV-s) × (3 × 108 m/s) λ(in nm) 12·42 × 10–7 eV-m δ = (i – r) + (i′ – r ′ ) = i + i′ – (r + r′ ) = i + i′ – A A 60 D r E r' A o δ R i θ S C 1 r3 Hence when radius and current both are doubled this ratio x becomes · 8 32. (C) Frequency 54 – 1 n = s , 60 λ = 10 m v = nλ 54 = × 10 = 9 ms– 1 60 di 33. (D) VL = L. dt ∴ = L. = 0 λ(in nm) 1242 eV - nm = λ(in nm) 38. (A) A low pressure and high temperature molecular attraction becomes negligible and the gas behaves as an ideal gas. 39. (C) 40. (D) On contracting (i.e. pulling inside) the arms, the moment of inertia of the man decreases. Therefore by conservation of angular momentum, the angular velocity increases since J = Iω 41. (D) 42. (C) λL for Balmer series corresponds to transition of electron from n = 3 to n = 2 and is given by 1 1 1 = R – 22 3 2 λL 36 ⇒ λL = 5R Shortest wavelength λS corresponds to transition from n = ∞ to n = 2. 1 1 1 = R – 22 ∞ 2 λS 4 ⇒ λS = R λS 4 5R 5 ∴ = × = R 36 9 λL 5 ⇒ λS = × 6563 9 ° = 3646 A i P B 50 o Q 42 = (90° – 50°) + i ′ – 60° ⇒ ∴ i′ = 62° θ = 90° – 62° = 28° 26. (D) For a sonometer nl = constant Decreasing length increases frequency, so the frequency of sonometer wire is N for l1 = 65 cm and N + 8 for l2 = 64 cm n1 l2 = n2 l1 N 64 ⇒ = N+8 65 ⇒ N = 512 Hz 27. (C) For two lenses separated by a distance, power is given by P = P1 + P2 – d × P1P2 = + 4 – 2 – 0·1 × 4 × (– 2) = 4 – 2 + 0·8 = + 2·8 D 100 F = cm 2·8 = 35·7 cm 1 28. (B) s = ut + αt 2 2 1 = 0 + gt 2 2 t = = 2s g = At t = 0, V L is maximum and falls off exponentially to zero (as the current approaches its maximum steady value). 34. (B) Gallium Arsenide (GaAs) semiconductor is used in lightemitting-diodes (LEDs) and semiconductor lasers. hc 35. (D) ∆E = 2E – E = E = λ 4E E hc ∆E′ = –E = = 3 3 λ′ hc hc or = 3λ λ′ ⇒ λ′ = 3λ 36. (A) Work done 1. = Force × Extension 2 1 = × 5 × 10 × 0·3 2 = 7·5 joule hc 37. (C) ∆E = λ 1242 eV - nm = λ(in nm) 1242 eV - nm ∴ EK – EL = 0·021 nm 1242 = eV 0·021 = 59 × 103 eV = 59 k eV () () i Re () Ee – R i0 e L – R t L – R t L R t L ( ) [ ] 2 × 490 9·8 = 10 sec Horizontal distance covered in 10 sec = 100 × 10 m = 1 km 43. (B) Electric force F = qE Acceleration qE α = m 44. (C) 45. (A) Reasons— (A) Ionic solids do not have free electrons. Hence they are poor conductors of electricity. On dissolving in water, ions become free and conduct electricity. (Continued on Page 1674 ) C.S.V. / February / 2008 / 1582 CHEMISTRY OF ALKALI METALS Introduction ● Group I consists of the elements : Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Caesium (Cs) and Francium (Fr). ● Francium (Fr) is radioactive, its longest-lived isotope 223 has a half-life of only 21 minutes. 87 Fr ● Atomic and ionic size—Alkali metal atoms have largest sizes in a particular period in periodic table. With increase in atomic number atomic size gradually increases down the group. The monovalent ions (M+) are smaller than the parent atoms. ● Ionization enthalpy—The ionization enthalpies of these elements are considerably low and decrease down the group from Li to Cs. This is due to the large atomic size which out do the increasing nuclear charge. (B) Physical Properties ● All the alkali metals are silvery white, soft and light elements. Due to large size, these elements have low density which increases down the group from Li to Cs. However, potassium is lighter than sodium. Lithium is the lightest metal known. ● Photoelectric effect—When exposed to sun-light, the higher alkali metals (K and Cs) eject electrons due to their low ionization potential. Hence K and Cs are used in photoelectric cells. ● Flame-colouration—Alkali metals and their salts impart characteristic colours to an oxidising flame. Metal → Colour → λ (nm) → Li Crimson 670·8 Na Yellow 589·2 K Violet 766·5 Rb Red violet (Purple) 780·0 Cs Blue 455·5 → 88 + – 1 (beta decay) 87 ● All these elements are collectively known as alkalimetals after the Arabic word al-qis meaning plant ashes since ashes of plants are particularly rich in carbonates of sodium and potassium. ● Alkali metals belong to s-block of long form of periodic table as all of these have valence shell electronic configuration, ns 1. ● The elements are all metals, excellent conductors of electricity and typically soft and highly reactive. ● Sodium and its compounds are by far the most important and most familiar of all alkali metals and their compounds. ● Commercial production of sodium is carried out with the help of Downs cell. Occurrence and Abundance ● Lithium (Li) is the thirty-fifth most abundant element by weight and is mainly found as silicate minerals, spodumene LiAl (SiO3)2 and Lepidolite, (Li, Na, K)2 Al2 (SiO3)3 (F.OH)2. ● Sodium and potassium are the seventh and eighth most abundant elements respectively by weight in the earth’s crust. ■ NaCl and KCl occur in large amounts in sea water. The largest source of Na is the rock salt (NaCl). Sea water contains nearly 3% NaCl. ■ Various salts including, Na2B4O7.10H2O (borax), Na 2CO3.NaHCO3.2H2O (trona), NaNO3 (Chile salt petre) and Na2SO4.10H2O (mirabilite) are found as natural deposits. ■ Potassium occurs mainly as deposit of KCl (sylvine), a mixture of KCl and NaCl (sylvinite) and the double salt KCl.MgCl2.6 H 2O (carnallite). ● There is no convenient source of rubidium and only one of caesium and hence these elements are obtained as by-products from lithium processing. ● Francium does not occur appreciably in nature due to its transitory existence. Fr223 Ra 223 e0 ● Mobility and electrical conductivity of ions in aqueous solution : ■ Alkali metal ions exist as hydrated ions M+ (H2O)x in the aqueous solution. ■ The degree of hydration decreases with increase in ionic size from Li+ to Cs+, i.e., Li+ ion is the mostly hydrated [Li (H2O)6] +. ■ The mobility of ions and electrical conductivity of ions in aqueous solution are inversely proportional to the size of hydrated ions. Hence among alkali metal ions, the lithium ion has lowest mobility and electrical conductivity. The mobility of ions and electrical conductivity decrease in the order Cs+ > Rb+ > K+ > Na+ > Li+ ● Reducing character—The increasing order of reducing character in gaseous state depends upon ionization energy (IE) of atoms and hence expected order is as : Li < Na < K < Rb < Cs IE (kJ mol– 1) 520 496 419 403 376 ■ The reducing character of any metal in solution is best expressed in terms of its Standard electrode potential (E°) which depends on its (I) sublimation (II) ionization and (III) hydration enthalpies. General Characteristics of Alkali Metals (A) Atomic Properties ● Electronic configuration—All the elements have one valence electron (ns 1) over the electronic configuration of inert gases. The outermost s-electron is very well screened from the nuclear charge in these elements. C.S.V. / February / 2008 / 1583 E° (V) at 298 K for M+(aq) + e – → M (s) } Li Na K Rb Cs – 3·050 – 2·714 – 2·925 – 2·99 – 3·02 ■ Hence in aqueous solution the increasing order of reducing tendency is as : Na < K < Rb < Cs < Li With the smallest size of its ion lithium has highest hydration enthalpy which accounts for its highest negative value of E° and its highest reducing power. (C) Chemical Properties ● Alkali metals are highly reactive metals and the reactivity gradually increases down the group. ● Reaction with water—Alkali metals react with water liberating hydrogen. The reaction becomes increasingly violent on descending the group. Na and K always catch fire 2M + H2O → 2 MOH + H2↑ ■ Hydrides of alkali metals are salt-like and react with proton donors like alcohols, gaseous ammonia and alkynes. Decreasing order of stability of these hydrides is as LiH > NaH > KH > RbH > CsH ■ Hydrides react with water liberating hydrogen. LiH is used as a source of hydrogen for military purposes and for filling meteorological balloons. LiH + H2O → Li (OH) + H2↑ Some complex hydrides like Li [AlH 4], Na [BH 4] and K [BH 4] are also known. These are powerful reducing agents and are widely used in organic chemistry. ● Reaction with liquid ammonia—Alkali metals dissolve in liquid NH3, the solubility is as high as 5M. These solutions are coloured and metastable. ■ The dilute solutions are blue but the colour changes to bronze with increasing concentration. The blue colour of dilute solution is attributed to the presence of solvated electrons. M + (x + y ) NH3 → [M (NH3)x ] + + [e (NH3)y ] – ■ Solvated electron ■ In concentrated solutions, ammoniated metal ions The standard electrode potential, E° and Gibbs free energy, ∆G are related by equation ∆G = – n FE° (n = No. of electrons removed) (F = Faraday constant) ■ The reaction Li+ + e – → Li has the largest negative E° value, and hence the largest positive ∆G value. Thus, this reaction does not occur spontaneously. However, the reverse reaction Li → Li + + e – has largest negative value of ∆G, so Li liberates more energy than other metals when it reacts with water. In view of above fact, it is rather surprising that Li reacts with water less violently as compared to Na and K. The explanation lies in the kinetics, i.e., the rate at which reaction proceeds rather than in the thermodynamics, i.e., the total energy liberated. Potassium has low melting point (63°C) as compared to Li (181°C) and the heat of reaction is just sufficient to make it melt or vaporise. The molten metal spreads out and exposes a large surface to the water, so it reacts even faster, gets even hotter and catches fire. Same explanation holds good for vigorous reaction of water with Rb and Cs. are bound by free unpaired electrons. If the blue solution is allowed to stand, the colour slowly fades until it disappears due to the formation of metal amide. 1 M+(am) + e –(am) + NH3(l) → MNH2(am) + H2(g) 2 (am denotes solution in ammonia) ■ Blue solutions are paramagnetic, whereas bronze ■ solutions are diamagnetic. ■ At concentration above 3M, solutions are copper- ■ ● Reaction with air—All alkali metals tarnish in air due to the formation of oxide or hydroxide on the surface. When heated in excess of air they form oxides Li2O, Na 2O2, KO2, RbO 2 and CsO2 respectively. ■ bronze coloured and have metallic lustre (expanded metals) because metal ion clusters are formed. ■ These solutions of metals in liquid ammonia conduct electricity better than any salt in any liquid and their conductivity is similar to that of pure metals. The conduction is mainly due to the presence of solvated electrons. ■ These solutions of alkali metals in liquid ammonia act as powerful reducing agents for many compounds and coordination complexes and they may even reduce an aromatic ring. Abnormal Behaviour of Lithium ● The anomalous behaviour of Lithium (Li) is mainly due to the exceptionally small size of its atom and ion (Li+). Its ionic potential is greatest of all alkali metal ions. Charge Ionic potential = Radius ● High polarising power of Li+ results in increased covalent nature of its compounds which in turn is responsible for their solubility in non-polar (organic) solvents. ● The main points of difference between lithium and rest of alkali metals are summarised as follows : ■ Li is much harder, its m.p. and b.p. are higher than other alkali metal. ■ Li is least reactive but the strongest reducing agent in aqueous solution among all alkali metals. On Lithium shows exceptional behaviour in reacting with air forming a mixture of oxide and nitride. 4 Li + O2 → 2 Li 2O 6 Li + N2 → 2 Li 3N ■ Li3N is an ionic compound and is a ruby red solid. It reacts with water giving ammonia. Li3N + 3 H2O → 3 LiOH + NH3↑ ● Reaction with hydrogen—When heated with hydrogen, they form the hydrides of the formula M+H– (MH). They are ionic solids with high melting point. C.S.V. / February / 2008 / 1584 combustion in air it forms mainly monoxide (Li 2O) and nitride (Li 3N), unlike other alkali metals. ■ Li2CO3 is the least stable of all alkali metal carbonates and hence it decomposes on heating unlike other carbonates. ■ ■ General Characteristics of Some Important Compounds of Alkali Metals (a) Oxides ● The monoxides are ionic, e.g., (2 Li+ and O2–). Li2O and Na2O are pure white solids, but surprisingly K2O is pale yellow, Rb2O is bright yellow and Cs 2O is orange. ● The crystal structures of Li2O, Na 2O, K2O and Rb 2O are anti-fluorite structures. Cs2O has anti-CdCl2 layer structure. ● Sodium forms mainly peroxide (Na2O2) with some Na 2O and all others form mainly superoxides (MO2) with traces of monoxides. ● The peroxides contain [—O—O—]2– ion and hence they are all diamagnetic and strong oxidising agents. ● Na 2O2 is pale yellow and is used industrially for bleaching wool, pulp, paper and fabric. Because it reacts with CO 2 in the air it has been used to purify air in submarines and confined spaces, as it removes CO 2 and produces O2. 2 Na2O2 + 2 CO2 → 2 Na2CO3 + O2 ● The superoxides (MO2) contain [O2] – ion, which has an unpaired electron. Hence they are paramagnetic and are all coloured. LiO2 and NaO 2 are yellow, KO2 is orange, RbO 2 is brown and CsO2 is orange. ● Superoxides are even stronger oxidising agents as compared to peroxides and give both H2O2 and O 2 with either water or acids. 1 KO2 + 2 H2O → KOH + H2O2 + O 2 2 ● KO2 is used in space capsules, submarines and breathing masks because it both produces dioxygen and removes CO2. 4 KO 2 + 2 CO2 → 2 K2CO3 + 3 O2 ● The stability of peroxides and superoxides increases with increase in metal ion size. Na 2O2 < K2O2 < Rb2O2 < Cs2O2 NaO 2 < KO2 < RbO2 < CsO2 (b) Halides ● All the halides, MX have negative enthalpies of formation which indicates that thermodynamically it is possible to form these halides from elements directly. ● The most negative enthalpies of formation occurs with fluorides. For any given metal, the value decreases in the order Fluoride > Chloride > Bromide > Iodide NaF > NaCl > NaBr > NaI Thus fluorides are most stable and iodides, the least stable. ● The enthalpies of formation of chlorides, bromides and iodides become more negative on descending the group. This trend is observed in most of the salts. The stability increases in the order : NaCl < KCl < RbCl < CsCl NaBr < KBr < RbBr < CsBr NaI < KI < RbI < CsI Note : Opposite trend is found in case of fluorides. LiF is the most stable and CsF is the least stable. Li2CO3 → Li2O + CO2↑ LiHCO3 does not exist as a solid while other bicarbonates exist as solid. LiCl is deliquescent and crystallises as a hydrate, LiCl.2H2O. Other alkali metal chlorides do not form such hydrates. ∆ ■ Lithium unlike other alkali metals forms no acetylide on reacting with acetylene. ■ LiNO3 on heating decomposes to give NO2 and O2 while nitrates of other alkali metals form nitrites and O2 under similar conditions. 4 LiNO3 → 2 Li 2O + 4 NO2 + O2 2 NaNO 3 → 2 NaNO 2 + O2 ■ ∆ ∆ Fluoride, oxalate and phosphate salts of lithium are sparingly soluble in water while corresponding salts of all other alkali metals are soluble. Lithium shows diagonal relationship with magnesium (Mg), the second element of second group. Lithium resembles more with Mg as compared to the elements of its own group because both Li and Mg have almost same ionic potential. ■ Fajans’ Rules ● Partial covalent characters in ionic compounds are developed due to the polarization of the anion by cation. As a result of this effect the electron density between two nuclei increases. ● The covalent characters in an ionic compound are governed by following factors, which are collectively known as Fajans’ rules. ■ Large anion—Larger the anion, greater is its polarization and more are the covalent characters. For example, among KF, KCl, KBr and KI, KI has highest covalent character and KF has the least. Small cation—Smaller the cation, greater is its polarising power for anion and more are the covalent characters. For example : among LiCl, NaCl, KCl and RbCl; LiCl has highest covalent character and RbCl has least. Pseudo inert gas configuration of cation—Of the two cations, having same charge and size, one with the pseudo inert gas configuration (mostly transition metal ions) has more polarizing power than the cation having inert gas configuration. For example : ° Ag + (Ionic radius = 1·26 A) forms more covalent ° AgCl as compared to KCl (K + radius = 1·33 A). ■ ■ C.S.V. / February / 2008 / 1585 ● The melting points and boiling points always follow the trend. Fluoride > Chloride > Bromide > Iodide NaF > NaCl > NaBr > NaI ● All halides are soluble in water. The lowest solubility of LiF is due to its highest lattice energy whereas the low solubility of CsI is due to small hydration energy of its two ions. (c) Salts of oxo-acids ● Since alkali metals are the most electropositive, they form salts with all oxo-acids. They are generally soluble in water and thermally stable. ● Carbonates and bicarbonates are highly stable. As the electropositive character increases down the group, the stability of carbonates (M2CO3) and bicarbonates (MHCO3) increases down the group. Lithium is exceptional in so far as its carbonate is not stable and bicarbonate does not exist as a solid. Thermal stability and solubility in water follow the following trend— Li2CO3 < Na2CO3 < K2CO3 < Rb2CO3 < Cs2CO3 NaHCO3 < KHCO3 < RbHCO3 < CsHCO3. ● NaHCO3 is not very soluble in water and, therefore, it makes a basis for the Solvay process of manufacture of sodium carbonate (Na2CO3.10H2O). K2CO3 cannot be prepared by Solvay process because KHCO3 is fairly soluble in water as compared to NaHCO 3. ● Except lithium bicarbonate, all other bicarbonates form carbonates on heating. 2 NaHCO3 → Na2CO3 + CO2 + H2O Uses of Alkali Metals and their Compounds ● Lithium stearate, C 17H35COOLi, is used in making automobile grease. ● Lithium is alloyed with Pb to make white metal which is used to make bearings for motor engines. ● Armour plates are made from Li-Mg alloy. ● Large amounts of Na 2SO4 are used to make soaps, detergents, paper, textiles and glass. ● NaOCl is used as a bleach and a disinfectant. NaHCO3 is used in baking powder. ● The largest use of sodium is to make Na.Pb alloy needed to make PbEt4 and PbMe4. These organolead compounds are used as anti-knock additives to petrol. ● KOH is used to make potassium phosphates and also soft soaps (potassium stearate), both of which are used in liquid detergents. ● KNO 3 is used in most of the explosives. Roughly 95% of potassium salts are used as fertilizers for plants. ● KBr finds an extensive use in photography. ● Sodium vapour is used to fill street lamps, which emits a characteristic yellowish-orange glow. ● Caesium is used in vapour form in the atomic clock, the vibration of its atoms acting as a regulator. ● High thermal conductivity of Na makes it a very effective heat-transfer medium. Points to Remember ● Lithium can not be stored in kerosene oil because it is very light and hence floats over surface. Hence it is kept wrapped in paraffin wax. Other alkali metals can be stored in kerosene oil. ● Microcosmic salt, Na (NH4) HPO4.4 H2O, is some times used in place of borax for performing bead test for coloured basic radicals. Hence, while dilute solutions are paramagnetic, the concentrated solutions become diamagnetic. ● A granular material prepared from quick-lime and NaOH solution followed by heating to dryness, is known as sodalime. It is used as an absorbent for CO2 and in decarboxylation organic reactions. ● All alkali metals except Li, dissolve in mercury forming ● Following are some important compounds of alkali metals which are most often known by common and commercial amalgams with evolution of heat. These amalgams are names : used as reducing agents in organic chemistry. ● Polyhalide formation—Alkali metal halides react with halo(i) Na2CO3, Soda ash gens and interhalogens forming ionic polyhalides. (ii) K2CO3, Pearl ash KI + I2 → KI 3 (iii) A mixture of Na O and dil. HCl, Oxone or Soda-bleach KBr + ICl KF3 + BrF3 K [BrF 4] ● All alkali metal sulphates form alums (except Li2SO4).Li2SO4 is not isomorphous with other sulphates. ● Sodium-lead (Na-Pb) alloy is used in the preparation of antiknock compound, Pb (C2H5)4. ● Liquid sodium is used as a coolant in nuclear reactors. ● Lithium metal is used as scavenger in the metallurgical operations to remove oxygen and nitrogen; because Li reacts with both O2 and N2. ● When the concentration of alkali metals in liquid ammonia is increased beyond 3M, the solvated electron undergoes a pairing process. [2e (NH3)x]– [e2 (NH3)x]2– → → 2 2 K [BrICl] (iv) (v) (vi) Na2CO3.NaHCO3.2 H2O, Sodium sesquicarbonate, used for wool washing NaCl, Rock salt Na2SO4.10 H2O, Glauber’s salt (vii) NaNO3, Chile salt petre (viii) Na2B4O7.10 H2O, Borax (ix) Na3[AlF6], Cryolite ● Some complexes of alkali metals with crown ethers and cryptands are known. Complexes with cryptands are known as cryptates. The term cryptate comes from the fact that metal ion remains hidden in the structure. ● Base consisting of a soluble metal hydroxide, is known as alkali. Alkali metals form caustic alkalies. C.S.V. / February / 2008 / 1586 OBJECTIVE QUESTIONS 1. Which of the following is the lightest metal known in whole of periodic table ? (A) Aluminium (B) Tin (C) Lithium (D) Beryllium 2. Alkali metals in each period of the periodic table have— (A) Smallest atomic radii (B) Highest electronegativity (C) Highest melting point (D) Lowest ionization potential 3. The strongest metal-metal bond exists in— (A) Lithium (B) Sodium (C) Potassium (D) Caesium 4. Which of the following compounds does not exist in the solid state ? (A) Na 2CO3 (B) NaHCO3 (C) Li2CO3 (D) LiHCO3 5. Which of the following is the lightest element ? (A) Potassium (B) Sodium (C) Rubidium (D) Caesium 6. When metallic sodium is heated in presence of sufficient oxygen, which of the following compounds is mainly formed ? (A) Na 2O (B) Na 2O2 (C) NaO (D) NaO 2 7. Which of the following compounds is better used as a source of oxygen in space capsules, submarines and in breathing masks ? (A) Na 2O2 (B) K2O (C) Li2O (D) KO2 8. Which of the following hydrides of alkali metals possesses the highest enthalpy of formation ? (A) LiH (B) NaH (C) PbH (D) CsH 9. Which of the following ions has lowest ionic mobility in aqueous solution ? (A) Cs+ (B) Rb + + (C) K (D) Li+ 10. The metal which acts as strongest reducing agent in aqueous solution, is— (A) Lithium (B) Sodium (C) Potassium (D) Rubidium 11. Which of the following compounds is the most deliquescent in nature ? (A) LiCl (B) NaCl (C) KCl (D) RbCl 12. Which of the following compounds has highest covalent characters ? (A) NaF (B) NaCl (C) NaBr (D) NaI 13. A colourless salt imparts yellow colour to Bunsen flame and its aqueous solution does not change the colour of litmus paper. The salt is— (A) K2CO3 (B) Na 2CO3 (C) NaCl (D) CH3COONa 14. Which of the following compounds has highest electrovalent characters ? (A) LiCl (B) BeCl 2 (C) MgCl2 (D) KCl 15. Which of the following halides has the highest lattice energy and highest stability ? (A) LiF (B) LiCl (C) LiBr (D) LiI 16. Which of the following ions has the highest conductivity ? (A) Li+ (B) Na + (C) K+ (D) Rb + 17. Which of the following halides has the highest melting point ? (A) NaF (B) NaCl (C) NaBr (D) NaI 18. Ionic potential of Li+ is almost equal to that of— (A) Mg2+ ion (B) Rb + ion (C) Cs+ ion (D) Both (B) and (C) 19. The heat of hydration of alkali metal ions decreases in order : Li+ > Na+ > K+ > Rb+ Which metal will react with water at the highest rate ? (A) Li (B) Na (C) K (D) Rb 20. Which among the following compounds is paramagnetic in nature ? (A) K2O2 (B) BaO2 (C) KO2 (D) None of these 21. The ionization potentials of alkalimetals are in order Cs < Rb < K < Na < Li but Li is the strongest reducing agent. The strongest reducing nature of Li is attributed to— (A) Small size and high ionization energy (B) Small size and high heat of hydration (C) Small size and low heat of hydration (D) None of these 22. Which of the following nuclear reactions gives francium ? (A) Alpha decay of actinium (B) Beta decay of actinium (C) Gamma decay of actinium (D) Alpha decay of uranium 23. Which of the following compounds is extensively used in explosives ? (A) K2SO4 (B) Na 2SO4 (C) KNO 3 (D) K2CO3 24. Sodium metal can be stored in— (A) Kerosene oil (B) Benzene (C) Toluene (D) All of these 25. Which of the following is the incorrect statement ? (A) Lithium has highest ionization potential among alkali metals (B) Lithium is the lightest metal known (C) Lithium is the strongest reducing agent among alkali metals (D) Lithium chloride is totally insoluble in water 26. A dilute solution of sodium metal in liquid ammonia is strongly reducing on account of— (A) Sodium amide (B) Solvated cation (C) Solvated electron (D) Both (B) and (C) C.S.V. / February / 2008 / 1587 27. Which of the following sodium salts is used for performing bead test in qualitative analysis ? (A) Borax (B) Microcosmic salt (C) Potash alum (D) Both (A) and (B) 28. Which of the following alkali metal ions is the best conductor of electricity ? (A) Li+ (B) Na + + (C) K (D) Cs+ 29. Which of the following ions has the biggest size in aqueous solution ? (A) Li+ (B) Na + (C) Rb + (D) Cs+ 30. A ruby red compound of an alkali metal when reacts with water gives ammonia. The compound is— (A) NaN3 (B) LiNO2 (C) Li3N (D) LiN3 31. Which of the following will be diamagnetic in nature ? (A) A dilute solution of Na in liquid ammonia (B) A concentrated solution of Na in liquid ammonia (C) Sodium metal (D) None of these 32. Anhydrous form of sodium carbonate is commercially known as— (A) Washing soda (B) Salt soda (C) Pearl ash (D) Soda ash 33. Squashes are preserved adding— (A) Sodium sulphate (B) Sodium metabisulphite (C) Microcosmic salt (D) None of these by 36. Which of the following is the byproduct during the manufacture of Na 2CO3 by Solvay process ? (A) (B) (C) (D) Lime stone Calcium chloride Ammonia Carbon dioxide 42. What is the percentage of NaCl in sea water ? (A) 10 per cent (B) 5 per cent (C) 3 per cent (D) 15 per cent 43. Molten sodium is used in nuclear reactors to— (A) Absorb neutrons (B) Slow down the neutrons (C) Absorb the heat (D) Absorb radioisotopes 44. The fire caused by combustion of alkali metals can be extinguished by— (A) Water (B) Kerosene (C) Carbon tetrachloride (D) None of these 45. Which of the following substances is formed when CO is passed over solid NaOH heated at 200°C ? (A) Sodium carbonate (B) Sodium bicarbonate (C) Sodium acetate (D) Sodium formate 46. In the disproportionation of chlorine with NaOH, the main reaction products are— (A) NaCl and NaClO4 (B) NaCl and NaOCl (C) NaCl and NaClO2 (D) None of these 47. When sodium chloride is dissolved in water— (A) Na + ion is oxidised (B) Na + ion is hydrolysed (C) Cl– ion is hydrolysed (D) Both Na + and Cl– ions are hydrated 48. Which of the following is the correct increasing order of solubility in water ? (A) CaCO3 < NaHCO3 < KHCO3 (B) KHCO3 < NaHCO3 < CaCO3 (C) NaHCO3 < KHCO3 < CaCO3 (D) CaCO3 < KHCO3 < NaHCO3 49. Soda is— (A) NaHCO3.Na2CO3 (B) NaOH (C) Na 2CO3.10 H 2O (D) Na 2CO3 37. Which of the following properties of lithium is more similar to that of other alkali metals as compared to that of magnesium ? (A) Oxidation state (B) Electronegativity (C) Solubility of hydroxide (D) All of these 38. In which of the following properties lithium resembles more with magnesium as compared to other alkali metals ? (A) Direct reaction with atmospheric nitrogen (B) Greater solubility of its bicarbonate than carbonate (C) Absence of peroxide in the product of reaction with excess oxygen (D) All of these 39. Which of the following is the wrong statement regarding francium (Fr) ? (A) It has bcc type of lattice (B) The main product of combustion of Fr metal in air is FrO2 (C) Melting point of Fr is highest of all alkali metals (D) Melting point of FrCl is lowest of all chlorides of other alkali metals 40. The compound Cs BrCl2 contains— (A) Cs3+ and Cl – and Br– ions (B) Cs+ and BrCl 2– ions (C) Cs2+ and 2 Cl– and Br2 (D) None of these 41. Which is the correct statement of the following ? (A) E° red of Li is more positive than that of Rb (B) E° of Li is more negative ox than that of Rb (C) Rb is the stronger reducing agent than lithium (D) None of these 34. Dry powder fire extinguishers contain— (A) Soda ash and sand (B) Baking soda and sand (C) Pearl ash and sand (D) Sand only 35. The reactivity of sodium with water is made used in— (A) Drying alcohol (B) Drying ammonia (C) Drying benzene (D) None of these (Continued on Page 1603 ) C.S.V. / February / 2008 / 1588 CHEMICAL THERMODYNAMICS Introduction Thermodynamics is the branch of science which deals with the energy changes accompanying all types of physical and chemical processes. The word ‘thermodynamics’ implies flow of heat. ● Importance of thermodynamics—It is a fundamental subject of great importance in physical chemistry. Most of the important laws of physical chemistry like van’t Hoff law of dilute solutions, distribution law, Raoult’s law, law of chemical equilibrium, the phase rule and the laws of thermochemistry can be deduced from the laws of thermodynamics. It also helps to lay down the criteria for predicting feasibility or spontaneity of a process under a given set of conditions. It also helps to predict the yields of products. ● Limitations—Laws of thermodynamics apply only to matter in bulk and to no individual atoms or molecules. In other words, the laws of thermodynamics apply to behaviour of assemblages of a vast number of molecules and not to individual molecules. It also does not tell anything about the rate at which a given process may proceed and gives no information about the mechanism or path taken by a process. free energy, heat capacity etc. Properties which do not depend on the quantity of matter are called intensive properties. Examples : pressure, temperature, surface tension, viscosity, specific heat etc. 5. State of a System The conditions for existence of a particular system when its macroscopic properties have definite values. A system to be in a state must have definite values assigned to its macroscopic properties such as temperature, pressure, volume and composition. These four properties are called thermodynamic properties or parameters or state variables. 6. State Functions The thermodynamic parameters which depend only upon the initial and final states of the system and are independent of how the change is accomplished are called state functions. Examples : internal energy, enthalpy, entropy, pressure, temperature etc. Thermodynamic Processes and their Types The operation by which a system changes from one state to another is called a process. The following types of processes are known— (a) Isothermal—Temperature of the system remains constant during each stage of the process. (b) Adiabatic—Process in which no heat enters or leaves the system during any step of the process. (c) Isobaric—Process in which the pressure of the system remains constant during each step of the process. (d) Isochoric—Those processes in which the volume remains constant. (e) Cyclic—When a process in a given state goes through a number of different steps and finally returns to its initial state. Terminology of Thermodynamics 1. System A system is that part of the universe which is under thermodynamic study and the rest of universe is known as surroundings. A system can be of three types. (a) Isolated system—A system which can exchange neither energy nor matter with its surroundings. (b) Closed system—A system which can exchange energy but not matter with its surroundings. (c) Open system—A system which can exchange matter as well as energy with its surroundings. 2. Macroscopic Properties The properties associated with a macroscopic system (i.e., consisting of large number of particles) are called macroscopic properties. These properties are pressure, volume, temperature, density, surface tension etc. Internal Energy (E) and Change in Internal ∆ Energy (∆ E) The total energy stored in a substance by virtue of its chemical nature is called its internal energy (E) or intrinsic energy. E is a state function, i.e., its value depends upon state of the substance but does not depend upon how the state is achieved. The various energies which contribute towards internal energy are translational energy of the molecule (Et), rotational energy of the molecule (Er), vibrational energy of the molecule (Ev ), electronic energy (Ee), nuclear energy (En ) and interaction energy of molecule (E i ) E = Et + Er + Ev + Ee + En + Ei The absolute value of internal energy cannot be determined but the change in internal energy, which occurs during chemical reaction, can be determined by ∆E = ΣEproducts – ΣEreactants 3. Thermodynamic Equilibrium A system in which the macroscopic properties do not undergo any change with time is said to be in thermodynamic equilibrium. It implies the existence of three kinds of equilibria in the system (a) Thermal equilibrium, (b) Mechanical equilibrium and (c) Chemical equilibrium. 4. Extensive and Intensive Properties Properties of a substance which depend on the quantity of matter (or amount of the substance) are called extensive properties. Examples : mass, volume, enthalpy, C.S.V. / February / 2008 / 1589 For exothermic reactions ∆E is negative because ΣEreactants > ΣEproducts For endothermic reactions ∆E is positive because ΣEproducts > ΣEreactants ∆ Enthalpy (H) and Enthalpy Change (∆H) Enthalpy is the heat content of a system at constant temperature and pressure. It is the sum of internal energy and volume energy of a system. Mathematically, H = E + PV where PV = volume energy of a system. If pressure is kept constant ∴ ∆H = ∆E + P∆V The change in enthalpy may be expressed as— ∆H = H(products) – H (reactants) = Hp – H r The enthalpy change of a reaction is equal to the heat absorbed or evolved during a reaction at constant temperature and constant pressure. From first law of thermodynamics, we know that for expansion work ∆E = q – P∆V ∴ ∴ ∆H = q – P∆V + P∆V ∆H = q(p) (Pressure being constant) Zeroth Law of Thermodynamics The law states that if two bodies X and Y are separately in thermal equilibrium with a third body, they all will be in thermal equilibrium with each other. This law introduces the concept of temperature. It is the underlying principle in the measurement of temperature. First Law of Thermodynamics This law was given by Meyer and Helmholtz in 1840. Energy can neither be created nor destroyed, however, it can be transformed from one form to another. This is also known as the law of conservation of energy. The law is based on the cumulative experience of ages that it is impossible to construct a perpetual motion machine, i.e., a machine which can produce energy without expenditure of energy. For the mathematical expression, let us assume that a system having internal energy E1 absorbs a certain amount of heat energy (q). ∴ Its internal energy = q + E1 Let an amount of work w be done on it, so that its internal energy changes to E 2 ∴ or E2 = E1 + q + w E2 – E1 = q + w ∆E = q + w …(1) A Closer Look ● For exothermic reactions ∆E or ∆H = – ve ● For endothermic reactions— ∆E or ∆H = + ve ● A process which is carried out so slowly that the system remains in equilibrium at every stage is called reversible process. ● Mechanical work is said to be done when there is a change in volume of the system at a constant temperature and pressure. ● The relation of unit mechanical work to thermal unit is called mechanical equivalent of heat. Its numerical value is 4·185 × 10 7 erg = 4·185 joule. Thus, for expenditure of 4·185 × 107 erg = 4·185 joule of mechanical energy, one calorie of heat is produced. Hence, 1 calorie > 1 joule > 1 erg ● Heat change taking place at constant volume and constant temperature is represented by ∆E i.e., change in internal energy. ● Heat change taking place at constant pressure and constant temperature is represented by ∆H, i.e. change in enthalpy. ∴ (Change in internal energy) = (Heat absorbed) + (Work done on the system) When work is done by the system ∆E = q – w Similarly q will be positive sign when heat is absorbed and negative sign when heat is evolved from the system. For a cyclic process, where initial and final states are same, E2 = E1 Hence, and ∆E = 0 q = w For the change in volume (∆V) during the process at constant pressure P, the work is referred to as pressure– volume work and its expression is given as (– P∆V) Thus ∴ or w = – P∆V ∆E = q – P∆V q = ∆E + P∆V …(2) If during the change, the volume remains constant, then no work is done i.e., Hence, P∆V = 0 qv = ∆E Thus, enthalpy change (∆H) represents, the heat change taking place during the process occurring at constant temperature and constant pressure. Thermochemistry—A study of heat or energy changes accompanying a chemical reaction is termed as thermochemistry. An equation which indicates the amount of heat evolved or absorbed in the reaction process is called a thermochemical equation. It must essentially (a) be balanced, (b) give the value of ∆E or ∆H (c) mention the physical state of the reactants and products. Heat of reaction—It is defined as “the amount of heat evolved or absorbed when the number of gram molecules of the substance as specified by the equation have completely reacted.” For example— C(s) + O2 (g) Thus, change in internal energy represents the heat change taking place during the process occurring at constant volume and constant temperature. → CO2 (g) + 393·5 kJ ∆H = – 393·5 kJ C.S.V. / February / 2008 / 1590 According to the equation when 1 gram atom (12 g) of solid carbon reacts with 1 mole (32 g) of oxygen gas to give 1 mole (44 g) of CO2 at 25°C and at atmospheric pressure, 393·5 kJ (94380 calories) of heat is evolved. This is thus the heat of above reaction. Heat of reaction at constant volume—Let us consider a chemical reaction taking place at constant temperature and constant volume. In such case w = 0 and hence, from first law (viz., ∆E = q – w) ∆E = qV Suppose E r and Ep are the internal energies of reactants and products respectively, then— ∆E = Ep – Er ∆E = Ep – Er = qv = Heat of reaction at constant volume Thus the heat of reaction at constant volume and at a given temperature is given by the difference in the internal energies of the products and the reactants. Heat of reaction at constant pressure—For the reaction at constant pressure and temperature ∆H = qp If Hp and Hr are the enthalpies of the products and reactants respectively, then ∆H = Hp – H r ∆H = Hp – H r = qp = Heat of reaction at constant pressure Thus, the heat of reaction at constant pressure and at a given temperature is given by the difference in the enthalpies of the products and reactants. Relation between heat of reaction at constant ∆ ∆ volume (∆ E) and constant pressure (∆ H) ∆H = ∆E + P∆V = ∆E + ∆nRT ∆n = Difference in number of moles of gaseous products and gaseous reactants. R = Gas constant, T = Temperature in kelvin. If If If ∆n = 0 ∆n = Positive ∆n = Negative ∆H = ∆E ∆H > ∆E ∆H < ∆E A Closer Look ● ● ● If Ep > Er , reaction will be endothermic, ∆E will be positive. If E p < Er, reactions will be exothermic, ∆E will be negative. The value of ∆V is negligible in the reactions involving solids and liquids. Hence, for such reactions P∆V = 0 and ∆H = ∆E That is heat of reaction at constant pressure is equal to that at constant volume d∆H = ∆Cp dt . p d∆E and = ∆Cv dt . v are known as Kirchhoff’s equations which show variation of heat of reaction at constant pressure and constant volume respectively. Kirchhoff’s equation shows that change in heat of reaction at constant pressure or volume per degree change of temperature is shown by differences in heat capacities of products and reactants at constant pressure or volume. If molar heat capacities of products and reactants are same, then the heat of reaction is independent of temperature. The molar heat capacities of solids are almost equal to each other and hence heat of reactions involving solids are almost independent of temperature. ● ( ) ( ) ● ● ● Types of Heat of Reactions (1) Heat of formation—The quantity of heat evolved or absorbed when one mole of a compound is formed from its elements. It is expressed as ∆Hf ·e.g., H2(g) + 1 → H2O(l) , O 2 2(g) ∆Hf = – 68 k cal (2) Heat of combustion—The quantity of heat evolved when one mole of a substance is completely oxidised e.g., CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆H = – 213 k cal The heat of combustion is very useful in— (a) Calculating the heat of formation which is otherwise not possible in many cases. (b) Calculating the calorific value of fuels. (c) Elucidating the structure of organic compounds. (3) Heat of solution—The quantity of heat evolved or absorbed when one mole of a solute is dissolved completely in a large excess of water, so that the further dilution does not produce any heat change e.g., HCl(g) + nH2O → HCl(aq), ∆H = –39·3 k cal (4) Heat of neutralisation—The quantity of heat evolved when one equivalent (equivalent mass) of an acid is completely neutralised by one equivalent (equivalent mass) of a base in dilute solution e.g., HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –13·68 k cal Variation of heat of reaction with temperature— The variation of heat of reaction with temperature is given by Kirchhoff’s equation ∆E2 – ∆E1 = ∆Cv (T2 – T 1) and ∆H2 – ∆H1 = ∆Cp (T2 – T 1) where ∆Cv is difference in heat capacities of products and reactants at constant volume and ∆Cp is difference in heat capacities of products and reactants at constant pressure. Heat of reaction depends on— (i) Physical state and the amount of substance (ii) Temperature (iii) In the case of reactions involving gases whether the reaction is taking place at constant volume or at constant pressure. C.S.V. / February / 2008 / 1591 The heat of neutralisation of a strong acid and a strong base is always – 13·7 k cal. On the basis of electrolytic dissociation theory, it has been clearly explained that this heat of neutralisation is merely the heat of formation of water from H + of an acid and OH– of a base. H+(aq) + OH–(aq) → H2O(l), ∆H = –13·7 k cal The heat of neutralisation in case of a weak acid or a weak base is somewhat less than – 13·7 k cal because some of the energy is used up in dissociating these weak electrolytes. The difference in the value gives the dissociation energy of the weak acid or a weak base. The heat of neutralisation of CH3COOH by NaOH is found to be –13·2 k cal. The decrease in heat of neutralisation is due to the fact that some heat is utilised in dissociation of weak acid (CH3COOH). Thus, the heat of dissociation (x) of acetic acid is— – 13·7 + x = – 13·2 or x = 0·5 k cal. Similarly, the heat of dissociation of weak base (NH4OH) in its neutralisation by HCl is 13·7 – 12·3 = 1·4 k cal. (The heat of neutralisation of NH4OH by HCl is – 12·3) (5) Heat of dissociation or Ionisation—The quantity of heat absorbed when one mole of a substance completely dissociates into its ions. e.g., H2O(l) → H+ + OH– , ∆H = 13·7 k cal (2) Hess’s law of constant heat summation—It states that “the amount of heat evolved or absorbed in a chemical change is the same whether the process takes place in one step or in several steps”. For e.g., A system changes from state ‘A’ to state ‘B’ in one step. A → B+Q Now suppose the system changes from state A to B in three steps C A D B A → C + q1 C → D + q2 D → B + q3 Then according to Hess’s law Q = q1 + q2 + q3 Applications—(1) By using Hess’s law, thermochemical equations may be multiplied, added, divided or subtracted just like ordinary algebraic equations. (2) Hess’s law helps us to calculate the heat of formation, heat of combustion of compounds. (3) Hess’s law is used in calculating the heat of transition, i.e., conversion of one allotropic form into another. Bond energy—It may be defined as the energy required to break a bond between two atoms, in one mole of a gaseous substance, forming gaseous atoms. For a diatomic molecule like H 2 that has only one bond, the dissociation energy of the molecule is identical to the bond energy. It may be noted that in molecules involving more than one bonds of the same type one takes an average value. e.g., in CH4 there are 4 C—H bonds. The breaking of each of these C—H bonds is associated with a different enthalpy change. Hence C—H bond energy is taken as the average value. e.g., CH4(g) → C(g) + 4H(g) ; ∆H = 398 k cal According to definition, the average bond energy of 398 = 99·5 k cal. C—H bonds 4 Bond energies are used in the determination of enthalpies of reactions, enthalpies of formation of compounds and determination of resonance energy. (6) Heat of dilution—The quantity of heat evolved or absorbed when solution containing one mole of a solute is diluted from one concentration to another. KCl(s) + 20H2O → KCl (20H2O), ∆H1 = +3·8 k cal KCl(s) + 200H2O → KCl (200H2O), ∆H2 = +4·44 k cal ∴ Heat of dilution = ∆H2 – ∆H1 = 0·64 k cal (7) Heat of precipitation—The quantity of heat evolved in the precipitation of one mole of a sparingly soluble substance on mixing dilute solutions of suitable electrolytes e.g., Ba2+(aq) + SO42–(aq) → BaSO4(s), ∆H = 4·66 k cal Laws of Thermochemistry (1) Lavoisier and Laplace law—This law states that the amount of heat required to decompose a compound into its elements is always equal to the heat of formation of that compounds from its elements. In other words the heat evolved in a particular reaction is equal to heat absorbed when reaction is reversed. This thermochemical equation can be reversed only by changing the sign of heat evolved or absorbed. For example— C(s) + O2(g) → CO2(g); CO2(g) → C(s) + O2(g); ∆H = –94·05 k cal ∆H = +94·05 k cal Limitations of First Law : Need for the Second Law of Thermodynamics (1) The first law states that during a process one form of energy can be changed into another form but the total amount of energy remains the same. However, it does not predict whether the process in question can occur spontaneously or not and if so in which direction. (2) The first law states that energy of one form can be converted into an equivalent amount of energy of another form. However, in experience it has been observed that although various forms of energy can be completely transformed into other; but heat energy cannot be completely converted into equivalent amount of work without producing some changes elsewhere. C.S.V. / February / 2008 / 1592 Second Law of Thermodynamics Second law can be stated in various ways as— (a) It is impossible to construct a machine working in cycles which can convert heat completely into an equivalent amount of work without producing any additional changes elsewhere (Planck’s definition). (b) It is impossible to construct a heat engine operating in cycles which can perform work at the expense of heat obtained from a thermal reservoir (Kelvin’s statement) (c) It is impossible for a self acting machine unaided by any external agency to transform heat from a colder body to a hotter body (Clausius statement). Although second law has been stated in a number of ways but all statements are modifications of the same fundamental concept “work can always be converted into heat but the conversion of heat into work does not always takes place under all conditions.” A Closer Look ∆ ● Units of entropy change (∆S) q ∆S = , thus T In C.G.S. units, ∆S = In S.I. units, ∆S = calorie = cal deg–1 degree joules degrees in kelvin = J-K–1 ● An irreversible spontaneous reaction is accompanied by an increase in the value of ∆S i.e., for such a reaction, ∆S has a positive value. q ∆S > T or ∆S > 0 ● A non spontaneous reaction is accompanied by a decrease in the value of ∆S i.e., for such a reaction ∆S has a negative value. ● For a reversible spontaneous reaction ∆S is equal to zero i.e., qrev ∆S = T = 0 ● When a solid melts ∆Hfusion ∆Sfusion = Melting temperature ∆Hvaporisation ∆Svaporisation = Temperature of vaporisation ● In general ∆S for any phase transition ∆H = Transitiontransition temperature ● All periodic machines working reversibly between the same temperatures of source and sink have same efficiency. ● If an isothermal change is carried out reversibly, it leads to maximum work of expansion. Therefore, reversible machine will give maximum efficiency. ● The fraction of heat absorbed by a machine which it can convert into work or other energy is called the efficiency of the machine. ● First law— The total amount of energy in nature is constant. ● Second law—The total amount of entropy in nature is increasing. ● Entropy always increases in the direction of spontaneous change. ● Entropy change in isothermal expansion of an ideal gas is given by V2 ∆S = nR ln V1 Carnot Cycle or Carnot Engine In 1924, S. Carnot (a French engineer) observed that maximum conversion of heat into work occurs in a cyclic process in which all the necessary steps are carried out reversibly. Such a reversible cycle is called Carnot cycle and the engine working on the basis of this cycle is called Carnot engine or heat engine. Efficiency of a heat engine is defined as the fraction of heat absorbed by the engine which it can convert into work. It is expressed as q 2 – q1 η = q2 T2 – T 1 = T2 w = q2 η is efficiency, q2 is heat absorbed by the engine at higher temperature equal to T2 and q1 is the heat evolved by the engine at lower temperature T1. Since, (T2 – T1)/T2 is always less than unity, the efficiency of a heat engine is always less than unity. In fact no heat engine has so far been constructed whose efficiency is equal to unity. Entropy (S)—The concept of entropy was introduced by Clausius in 1854. It is a thermodynamic state quantity i.e., It is a measure of the randomness or disorder of the molecules of the system. The change in entropy, ∆S for any process is given by the equation ∆S = Sfinal – Sinitial When Sfinal > Sinitial, ∆S is positive. “A change in a system which is accompanied by an increase in entropy tends to be spontaneous.” When a system is allowed to change from one state to another state, the heat q, is absorbed or evolved during the change depending upon the way, the change is carried out. ( ) ● Entropy of mixing of ideal gases is given by ∆Smix = –(n 1R ln N1 + n2R ln N2 ……) where N1 and N2 are the mole fractions of gases and n1 and n2 are number of moles of gases. We know that ∆E = q – w; ∆E = q – P∆V q = ∆E + P∆V q (∆E + P∆V) ∆S = T = T C.S.V. / February / 2008 / 1593 Giving +ve sign to heat absorbed (q1) and –ve sign to heat evolved (q2) by a system, then for a reversible cycle q1 q2 = – T1 T2 q1 q2 + = 0 T1 T2 . . + When a reversible process occurs the entropy remains constant, ∆Suniverse = 0. Since, the entire universe is undergoing spontaneous change, the second law can be most concisely stated as “the entropy of the system is constantly increasing”. or Free Energy (G) and Work Function (A) Entropy change is considered in terms of other functions which can be determined more easily. Two such state functions are free energy and work function, represented by G and A respectively. These are defined by the equations, A = E – TS …(1) and G = H – TS …(2) where A → Helmholtz work function G → Gibbs free energy E → Internal energy H → Enthalpy S → Entropy T → Temperature Since, E, H and S depend upon the state of a system, the functions A and G also depend upon the state of the system, ∴ and ∆A = ∆E – T∆S ∆G = ∆H – T∆S …(3) …(4) Thus, for reversible cycle the summation of q /T terms is equal to zero. Thus q Σ = 0 T For infinitesimal changes, equation is dq Σ = 0 T Now, we know that dq and T are thermodynamic functions whose change measured by dq /T is independent of path of change of system. This function is called entropy (S). If SA is the entropy in the initial state A and SB is the entropy in the final state B, then the change in entropy, ∆S, is given by the equation ∆S = SB – SA = B A ∫ dq T For each infinitesimally small change dq dS = T At constant temperature for a finite change, dS becomes ∆S and dq becomes q q ∴ ∆S = T Like ∆H and ∆E, the entropy change (∆S) is a definite quantity and it depends upon initial and final state of a system. For an adiabatic change, q = 0 hence, ∆S = 0. Such type of change is called isoentropic change. Entropy is an extensive property and its value depends upon the amount of the substance involved. Standard entropy (S°)—Standard entropy is the entropy of 1 mole of a substance in its pure state at 1 atm. pressure and 25° C. If the study is carried out reversibly at constant temperature T, and that the heat absorbed is equal to qrev. the increase in entropy qrev ∆S = T Substituting it in equation (3), we get ∆A = ∆E – qrev …(5) From first law of thermodynamics (∆E = q – w) we have –w rev = ∆E – qrev Combining equations (5) and (6), we get –∆A = wrev …(7) As the process is carried out reversibly, w represents the maximum work. Thus decrease in the function (–∆A) gives the maximum work obtainable from a system during a given change. Thus, A is known as work function. Free energy—At constant pressure, we know ∆H = ∆E + P∆V ∆G = ∆H – T∆S ∆G = ∆E + P∆V – T∆S or or or ∆G = ∆A + P∆V –∆G = wrev – P∆V [from equation (3)] …(6) Statement of Second Law of Thermodynamics in Terms of Entropy Change It states that whenever a spontaneous process takes place, it is accompanied by an increase in the total energy of the universe, thus ∆Suniverse = ∆Ssystem + ∆Ssurroundings The second law as stated above tells us that when an irreversible spontaneous process occurs, the entropy of the system and the surroundings increases. In other words ∆Suniverse > 0. ∆G = –w rev + P∆V [from equation (7)] Thus, –∆G gives the maximum work obtainable from a system other than that due to change of volume at C.S.V. / February / 2008 / 1594 constant temperature and pressure. The work other than that due to change of volume is called the net work. Thus, net work = wrev – P∆V = –∆G ∆H (1) Negative ∆S Positive ∆ ∆G = ∆H – T∆S Negative Remarks Reaction is spontaneous at all temperatures. Reaction is nonspontaneous at all temperatures. So, –∆G is a measure of the maximum net work that can be obtained from a system at constant temperature and pressure. The function G, is due to Gibbs and is known as Gibbs free energy or simply free energy. The free energy change (∆G) for a chemical reaction is given by ∆G = Σ(∆Gproducts) – Σ(∆Greactants) (2) Positive Negative Positive (3) Negative Negative Negative at low Spontaneous at temperature low temperatures. positive at high Non-spontaneous at temperature high temperatures. Positive at low Non-spontaneous at temperature low temperatures. Negative at high Spontaneous at high temperature temperatures. (4) Positive Positive A Closer Look Work function (A) and free energy (G) are extensive properties. For a n isothermal process, these can be written as ∆A = ∆E – T∆S ∆G = ∆H – T∆S Subtraction (i) from (iii) ∆G – ∆A = ∆H – ∆E If ∆H is measured at constant pressure Then ∆H – ∆E = P∆V Therefore, ∆G = ∆A + P∆V ‘A’ is known as Helmholtz function and ‘G’ is known as Gibbs function. Gibbs and Helmholtz equations δ(∆A) (1) ∆A – ∆E = T δT (2) ∆G – ∆H = …(i) …(ii) Relation between the standard free energy change ∆ ° (∆ G°) and equilibrium constant (K) of a reaction ∆G° = – RT l n K or ∆G° = – 2·303 RT log K If R is taken as 8·314 JK –1 mol–1 ∆G° = – 2·303 × 8·314 T log K = – 19·147 T log K Third law of thermodynamics—This law helps in determining the absolute values of entropy of various substances. For a crystalline substance at 0 K, there is perfect order, consequently, the disorder or entropy of substance at absolute zero is zero. This law was propounded by Nernst in 1906. It states as— “At absolute zero temperature, the entropy of perfectly crystalline substance is zero.” It may be stated in a number of other ways— “At any pressure, the entropy of any crystalline solid in thermodynamic equilibrium at absolute zero is always zero.” “It is rather impossible to reduce the temperature of any system to absolute zero by any process.” As the absolute temperature reaches zero, increase in entropy for isothermal process in crystalline state approaches zero as the limit. ( ) δ(∆G) T ( δT ) V p Gibbs equation ∆G = ∆H – T∆S For a system at equilibrium, (∆G)T, P = 0. All spontaneous processes taking place at constant temperature and pressure are accompanied by a decrease in free energy i.e., ∆G is negative. A process is thermodynamically not feasible when ∆G is positive. Thus, depending on the sign of ∆H and T∆S, the following four cases may be studied. SOME IMPORTANT EXAMPLES WITH SOLUTIONS Example 1. In a certain process, 678 J of heat is absorbed by a system while 290 J of work is done on the system. What is the change in internal energy for the process ? Solution : q = + 678 J Since, work is done on the system, w has a positive value w = + 290 J ∴ ∆E = q + w = 678 + 290 = 968 J Example 2. Given the following information A+B → C+D, C + D → E, ° ∆ H° = – 10·0 k cal ° ∆ H° = 15·0 k cal ° Calculate ∆H° for each of the following reactions : (a) C + D → A + B (b) 2C + 2D → 2A + 2B (c) A + B → E C.S.V. / February / 2008 / 1595 / 5 Solution : (a) Since, the reaction is the reverse of the first given so the sign of ∆H° becomes positive + 10·0 k cal. (b) Since there are twice the number of moles of each reactant, the value of ∆ H° will also be doubled. ∴ ∆H° = 20·0 k cal A+B → C+D +C+D → E A+B → E ∆H° = –10 + 15 = + 5·0 k cal Example 3. The reaction of cyanamide NH2CN(s) with oxygen was run in a bomb calorimeter and ∆E was found to be –742·7 kJ/mol of NH2 CN(s) at 298 K. Calculate ∆H298 for the reaction. 3 NH2CN(s) + O 2(g) → N2(g) + CO2(g) + H2O(l) 2 Solution : 3 The number of moles of gaseous reactant (O2) is , 2 the number of moles of gaseous products (N 2 + CO2) is 2. ∆H = ∆E + P∆V = ∆E + ∆nRT 3 1 ∆n = 2 – = 2 2 ∆H = – 742·7 kJ + (0·500 mol) (8·314 J/mol K) (298 K) = – 742·7 kJ + 1240 J (˙.˙ 1240 J = 1·240 kJ) = – 741·46 kJ Example 4. What is the enthalpy change when ° 1·00 g of water is frozen at 0° C ? ∆ (∆ Hfusion = 1·435 k cal/mol) Solution : Freezing is just the reverse of melting, hence, the enthalpy change must be – 1·435 k cal/mol ∆H = – 1435 cal/mol 18·0 g/mol and (c) Adding the two equations given Adding two equations HF → H+ + F– ∆H° = 13·4 + (–16·4) = – 3·0 k cal. Example 6. For the reaction at 298 K, 2A + B → C ∆ H = 100 k cal ∆ S = 0·050 k cal/K Assuming ∆ H and ∆ S to be constant over the temperature range, at what temperature will the reaction become spontaneous ? Solution : The reaction will just be spontaneous when ∆G = 0 ∆G = ∆H – T∆S = 0 ∆H = T∆S ∆S 100 k cal = 0·050 k cal/K = 2000 K Example 7. What is the standard free energy ° change for the melting of 3·0 mol of water at 0° C ? Determine the entropy change for this process. Is the entropy greater for the liquid than that of the solid ? ∆ (∆ Hfusion = 1·435 k cal/mol) Solution : ∆G° is 0 for this reversible process ∆G = ∆H – T∆S = 0 ∆S = ∆H T + 1435 cal/mol = 273 K = 5·26 cal/mol K (5·26 cal/mol K) (3·0 mol) = 15·78 cal/K. The entropy is greater for the liquid as seen by the positive sign of ∆ S. The melting of a solid into liquid results into increase in randomness or increase in entropy. ° Example 8. ∆ G° f for the formation of HI(g) from its gaseous elements is –10·10 kJ/mol at 500 K when the partial pressure of HI is 10·0 atm, and that of I 2 is 0·001 atm. What must the partial pressure of hydrogen be at this temperature to reduce the magnitude of ∆G for the reaction to 0 ? Solution : 1 1 H + I → HI, ∆G° = –10·10 kJ at 500 K 2 2 2 2 When ∆G = 0, the reaction is at equilibrium – ∆G° = RTln K = 2·303 RT log K log K = –∆G° 2·303 RT ∴ T = ∆H = –79·7 cal/g Therefore, 79·7 cal of heat is liberated when 1·00 g of water is frozen at 0°C and 1·00 atm. pressure. Example 5. The heat released on neutralisation of CsOH with all strong acids is 13·4 k cal/mole. The heat released on neutralization of CsOH with HF(weak acid) ° is 16·4 k cal/mol. Calculate ∆ H° of ionization of HF in water. Solution : (1) (2) H2O OH– + H+, ∆H° = + 13·4 k cal OH– + HF → H2O + F–, ∆H° = –16·4 k cal C.S.V. / February / 2008 / 1596 = + 10·10 × 103 J (2·303) (8·31 J/K) (500 K) P(HI) [P (H2)]1/2 [P(I2)]1/2 = 1·055 K = 11·36 = = P(H2) = 10·0 (10–3)1/2 [P(H2)]1/2 100 (10–3) (11·36)2 = 775 atm. Example 9. Calculate the entropy change when 1 mole of an ideal gas expands reversibly from an initial ° volume of 2 litres to a final volume of 20 litres at 25° C. Solution : We know that ∆S = n R ln V2 V1 Example 10. An athlete is given 100 g of glucose (C6H12O6) of energy equivalent to 1560 kJ. He utilises 50 per cent of this gained energy in the event. In order to avoid storage of energy in the body, calculate the mass of water he would have to perspire. The enthalpy of evaporation of water is 44 kJ/mole. Solution : Net amount of energy given to athlete = 1560 kJ 1560 × 50 Energy lost in an event = 100 = 780 kJ Energy left out = 1560 – 780 kJ = 780 kJ Now, consider the evaporation of water H2O(l) → H2O(g); ∆H = 44 kJ mole–1 Thus, for consumption of 44 kJ of energy the amount of water evaporated = 18 g. ∴ For consumption of 780 kJ of energy the amount of water to be evaporated 18 = × 780 g 44 = 319·09 g 20 = 2·303 n R log 2 = 2·303 × 1 × 8·314 × 1 = 19·15 J/mol K OBJECTIVE QUESTIONS 1. The heat change taking place during the reaction at a particular temperature which does not involve any change of volume is represented by— (A) ∆G (B) ∆H (C) ∆E (D) ∆n 2. Which of the following proceeds with the positive value of ∆S ? 1 (A) SO2(g) + O 2(g) → SO3(g) 2 (B) HCl(g) → H(g) + Cl(g) (C) Freezing of ice-cream (D) None of these 3. In which of the following reactions heat of neutralisation has highest numerical value ? (A) NH4OH and HCl (B) CH3COOH and NaOH (C) NH4OH and CH 3COOH (D) NaOH and HCl 4. At constant T and P, which one of the following statements is correct for the reaction ? 1 CO(g) + O 2(g) → CO2(g) 2 (A) ∆H = ∆E (B) ∆H < ∆E (C) ∆H > ∆E (D) ∆H is independent of physical state of reactants 5. The enthalpies of the elements in their standard states are assumed to be— (A) Zero at 298 K (B) Unity at 298 K (C) Zero at all temperatures (D) Zero at 273 K 6. 2C (s) + 2O2(g) → 2CO2(g) ∆H = –188 k cal The heat of combustion of carbon (s) is— (A) – 188·0 k cal (B) + 188 k cal (C) – 94·0 k cal (D) + 94·0 k cal 7. The occurrence of a reaction is impossible when— (A) ∆H is +ve and ∆S is +ve (B) ∆H is –ve and ∆S is +ve (C) ∆H is +ve and ∆S is –ve (D) ∆H is –ve and ∆S is –ve 8. Hess’s law deals with— (A) Change in heat of reaction (B) Rate of reactions (C) Equilibrium constants (D) Influence of pressure on volume of a gas 9. For a reversible reaction at equilibrium— (A) ∆H = T∆S (B) ∆H > T∆S (C) ∆H < T∆S (D) ∆H = Zero 10. The property which can be classified as an intensive property is— (A) Mass (B) Temperature (C) Volume (D) Heat capacity 11. If for H2(g) = 2H(g) ; ∆H = 104 k cal the heat of atomisation of hydrogen is— (A) 52 k cal (B) 104 k cal (C) 208 k cal (D) None of these 12. 2·1 g of Fe combines with S evolving 3·77 kJ. The heat of formation of FeS in kJ/mole is— (A) –3·77 (B) –1·79 (C) –100·5 (D) None of these 13. The temperature of 5 ml of a strong acid increases by 5°C when 5 ml of a strong base are added to it. If 10 ml of each are mixed, temperature should increase by— (A) 5° (B) 10° (C) 15° (D) Cannot be known (Continued on Page 1603 ) C.S.V. / February / 2008 / 1597 Model Paper for Various Medical Entrance Examinations (C) Hunsdiecker reaction (D) Liebermann’s reaction 13. Formal change at chlorine in ClF3— (A) 0 1 (C) + 3 (B) + 1 (D) + 3 CHEMISTRY 1. Number of electrons present in 1·6g of methane— (A) 6·02 × (C) 6·02 × 1022 1021 (B) 6·02 × 1023 (D) 3·01 × 1022 The formula of the gas would be— (B) C2H4 (A) CH2 (C) C3H6 (D) C4H8 7. 214·2g sugar syrup contains 34·2g of sugar. Mole fraction of sugar in the syrup is— (A) 0·0099 (B) 0·0990 (C) 0·9000 (D) 0·1000 2 2 — 8. 2HC — CH → X — 2. Which of the following are aromatic compound ? CHO | (a) N N O (c) O The correct answer is— (A) a, c (C) a, b, d (B) a, b, c (D) c, d (d) (b) 14. For a given value of l, a single line in the normal spectrum will appear as …… lines if a magnetic field is applied. (A) 2l (B) 2l + 1 (C) 2l – 1 (D) l + 1 15. A finely divided state of the catalyst is more efficient because— (A) It preferentially decreases the surface area (B) More surface area is available (C) More energy is stored in the catalyst (D) Positive charge is acquired 16. Formula to evaluate effective nuclear charge is— (A) Z – S (B) Z + S (C) ZS –1 (D) Z · S [Here Z = Atomic number and S = Shielding constant] 17. The oxidation state of Cr in CrO 5 is— (A) + 3 (B) + 10 (C) + 6 (D) + 2·5 18. For the reaction N2 + 3H2 2NH3 Cu Cl NH 4Cl .. S X + HCl → Y Compounds X and Y are respectively— (A) 1, 3-butadiene, 2-chloro-1buten-3-yne (B) 1-butene-3-yne, chloroprene (C) Butenyne, chloropicrine (D) Vinyl acetylene, chloroprene 9. A system is supplied 500 calorie of heat energy and system does 350 calorie of work on the surroundings. What is the energy change of the system ? (A) 150 cal (B) 850 cal (C) 350 cal (D) 500 cal 10. Which of the following is a cyclic ether ? (A) Ethyl ether (B) Vinyl ether (C) Phenyl ether (D) Tetrahydrofuran 11. N2O4 is 25% dissociated at 30°C and 1 atm pressure. The equilibrium constant K p at this temperature is— (A) 0·267 atm (B) 2·67 atm (C) 2·67 × 10–2 atm (D) 2·67 × 10–3 atm O || 12. For the elimination of —C — group of amide following reaction is used— (A) Hofmann hypobromite reaction (B) Kolbe reaction 3. The wavelength associated with an electron moving with a velocity of 1010 cm/sec— (h = 6·62 × 10 –27 ergs-sec) (A) 7·27 × 10–10 cm (B) 7·27 × 10–9 cm (C) 2·57 × 10–10 cm (D) 3·86 × 10–9 cm 4. Select the positive ion which comes under the purview of Bronsted acid but not considered as Lewis acid— (A) NH+ (B) Al3+ 4 (C) Cu 2+ (D) Ag+ 5. A 1000 ml sample of a gas at – 73°C and 2 atm is heated to 123°C and pressure is reduced to 0·5 atm. The final volume of the gas is— (A) 2000 ml (C) 6000 ml (B) 4000 ml (D) 8000 ml at 500°C the value of Kp is 1·44 × 10–5. The value of Kc will be— (A) (1·44 × 10–5) × (0·082 × 773)2 (B) (0·082 × 10–5) (1·44 × 10–5) (C) (1·44 × 10–5) (0·082 × 773)–2 (D) (1·44 × 10–5) (0·082 × 500)–2 19. The electronic configuration of element A is 1s2, 2s22p6, 3s2 while of element B is 1s2, 2s22p5. The formula of the compound containing A and B will be— (A) AB (B) A2B (C) AB2 (D) A2B6 20. In pure water ionic product of water is correctly evaluated by using— (A) [H3O+] 2 (B) [H+] + [OH–] 6. It was found from the chemical analysis of a gas that its one carbon atom is linked to two hydrogen atoms. At S.T.P. its density is 1·25 grams per litre. C.S.V. / February / 2008 / 1598 (C) [H3O+] [OH–] –1 (D) [OH–] [H+] –1 21. Pair of amphiprotic species is— (A) H2CO3 and HCO3– (B) H2CO3 and H 2O (C) CO32– and H 3O+ (D) HCO3– and H 2O 22. pH of 0·2 F HClO4 is— (A) 0·70 (C) 13·3 (B) 1·0 (D) None of these 29. Which of the following is correct composition of water gas ? (A) CO + Cl2 (B) CO + H2 (C) CO + N2 (D) CO + H2 + N2 30. The van der Waals equation at low pressure may be written as : a P + 2 V = RT V (C) Roasting (D) Concentration 37. The Belistein test is a rapid test used for organic compounds to detect— (A) Phosphorous (B) Sulphur (C) Halogens (D) Nitrogen 38. Match List-I with List-II and choose the correct answer from the codes given below : (i) (ii) (iii) (iv) List I NaNO3 Na(NH4)HPO4 NaHCO3 Na 2CO3·10H2O ( ) 23. The incorrect statement is— (A) Calamine and siderite are carbonates (B) Argentite and cuprite are oxides (C) Zinc blende and iron pyrities are sulphides (D) Malachite and azurite are ores of copper 24. On adding NH4Cl to NH4OH solution the pH of the solution would— (A) Increase (B) Decrease (C) Remain unchanged (D) None of these 25. Which of the following has lowest melting point ? (A) Li (C) K (B) Na (D) Cs The compressibility factor would be— a RTV (A) 1– (B) 1 – RTV a (C) ( ( 1+ a RTV ) ( ) ( (D) 1+ RTV a ) ) 31. Number of atoms in a molecule of sulphur is— (A) 3 (B) 4 (C) 8 (D) Infinite 32. The specific conductivity of 0·01 mole/dm3 aqueous acetic acid at 300 K is 19·5 × 10–5 ohm–1 cm–1 and the limiting molecular conductivity of acetic acid at the same temperature is 390 ohm–1 cm2 mole–1. The degree of dissociation of acetic acid is— (A) 0·5 (B) 0·05 –5 (C) 5 × 10 (D) 5 × 10–7 33. In a given sample of bleaching powder the percentage of available chlorine is 49. The volume of chlorine obtained if 10 g of sample is treated with HCl at STP is approximately— (A) 1·5 litre (B) 3·0 litre (C) 15·0 litre (D) 150 litre 34. pH of a buffer made by HA and NaA is 4. The concentration of HA is 1·0 M. What is the concentration of NaA ? [Given K a = 10–5] (A) 1·0 M (B) 0·01 M (C) 0·1 M (D) 1·5 M List II (a) Baking soda (b) Chile salt petre (c) Microcosmic salt (d) Washing soda Code (i) (ii) (iii) (A) a b c (B) b c a (C) c a b (D) d a b (iv) d d d c 26. The position of Cs+ ion in CsCl structure will be— (A) At the corners of the cube (B) At the centre of each face of the cube (C) At the body centre of the cube (D) At the edge centre of the cube 27. Sodalime is used extensively in decarboxylation reaction to obtain alkanes. Sodalime is— (A) NaOH (B) NaOH and CaO (C) CaO (D) Na 2CO3 28. Solid alcohol is the colloidal solution of— (A) Calcium nitrate (B) Calcium chloride (C) Calcium carbonate (D) Calcium acetate 39. In which of the following the bond length between carbon and carbon is equal— (A) 2-butene (B) Benzene (C) 1-butene (D) Propyne 40. The metallic lustre exhibited by sodium is explained by— (A) Diffusion of sodium ions (B) Oscillation of loose electrons (C) Excitation of free protons (D) Existence of body centered cubic lattice 41. In an industrial process, coke is heated with quick lime in an electric furnace and the cooled product is then treated with water to 35. K4Fe(CN)6 is named as— produce— (A) Potassium hexacyanoferrate(II) (A) Acetylene (B) Ethylene (B) Potassium ferricyanate (C) Ethane (D) Methane (C) Potassium ferricyanide 42. A substance on heating gives (D) Prussian blue oxygen, turns acidified KI solution violet and reduces acidified 36. The process of converting KMnO 4 solution. Hence, the subhydrated alumina to anhydrous stance is— alumina is called— (A) SO3 (B) KNO 3 (A) Calcination (C) H2O2 (D) All of these (B) Smelting C.S.V. / February / 2008 / 1599 43. Acetylene when treated with dil. HCl at 60°C in presence of HgCl 2 produces— (A) Methyl chloride (B) Vinyl chloride (C) Acetaldehyde (D) Formaldehyde 44. The chemical composition of glass is— (A) K2SiO3 CaSiO4·4SiO 2 (B) Na 2SiO3 CaSiO3·4SiO 2 (C) Na 2SiO3 MgSiO3·4SiO 2 (D) Na 2SiO3 K 2SiO3·4SiO 2 45. Which of the following will not give iodoform test ? (A) Ethanol (B) Ethanal (C) Isopropyl alcohol (D) Benzyl alcohol 46. Which one of the following is calcium superphosphate ? (A) Ca(H2PO4)2 (B) Ca 3(PO4)2 + 2CaSO4 (C) CaSO 4 + CaO (D) Ca(H2PO4)2H2O + CaSO4 47. Formaldehyde reacts with ammonia to give— (A) Hexamethylene tetramine (B) Formaldehyde ammonia (C) Formalin (D) Hydrobenzamide 48. On heating ozone its volume— (A) Increases to 1·5 times (B) Decreases to half (C) Remains unchanged (D) Becomes double 49. The pairs of bases in DNA are held together by— (A) Hydrogen bonds (B) Ionic bonds (C) Phosphate groups (D) Oxygen linkage 50. For Ar and CO2 the ratio of the molar heat capacities C p /Cv would be respectively— (A) 1·66, 1·66 (B) 1·66, 1·33 (C) 1·40, 1·66 (D) 1·40, 1·40 ANSWERS WITH HINTS 1. (B) Number of moles in 1·6g CH4 = 1·6 = 0·1 16 At S.T.P. 1 mole of a gas occupies 22·4 litre ∴ Molar mass = 22·4 × 1·25 = 28 g Molecular weight = 28 Formula of the gas = (CH2)n ∴ (12 + 2)n = 28 or n = 2 Therefore, gas is C 2H4. and 7. (A) Wt. of sugar syrup = 214·2 g Wt. of sugar = 34·2 g ∴ Wt. of water = 214·2 – 34·2 = 180 g 34·2 = 0·1 Moles of sugar = 342 180 = 10 Moles of water = 18 Moles of sugar Mole fraction of sugar = Total number of moles 0·1 = = 0·0099 10 + 0·1 Cu2Cl 2 — — 8. (B) 2 HC — CH → CH2 — CH—C — CH — — — Acetylene (two molecules) NH 4Cl Monovinyl acetylene (X) Number of molecules of CH 4 = 0·1 × 6·02 × 1023 Each CH4 molecule contains 6 + 4 = 10 electrons Number of electrons present in 1·6g CH 4 = 0·1 × 6·02 × 10 23 × 10 = 6·02 × 1023 CHO 2. (C) Compounds N .. S N are aromatic as they contain 6π electrons, i.e., conjugate systems of three double bonds. Velocity of electron = 1010 cm/sec h = 6·62 × 10–27 erg.sec. λ = 6·62 × 10–27 h = mv 9·1 × 10–28 × 1010 ; 3. (A) Mass of electron = 9·1 × 10–28g = 7·27 × 10–10 cm 4. (A) Bronsted acid is proton donor in solution (whereas Lewis acids are electrons pair acceptor NH4+ + H2O NH3 + H3O+ Hence NH 4+ is Bronsted acid. 5. (D) P1 = 2 atm P2 = 0·5 atm V1 = 1000 ml V2 = ? T1 = – 73°C = 200 K P1V1 P2V2 = T1 T2 V2 = T2 = 123°C = 400 K ∴ 2 × 1000 0·5 × V2 = 200 400 = 8000 ml — — — — CH2 — CH—C — CH + HCl → CH2 — CH—CCl — CH2 — (X) (Y) 2-chloro-1, 3-butadiene (chloroprene) 2 × 1000 × 400 200 × 0·5 6. (B) One carbon atom is linked with two carbon atoms. The empirical formula is CH2. Density at S.T.P. = 1·25 g/lit 9. (A) Heat supplied to the system q = + 500 calorie Work done (by the system) on the surroundings W = 350 calorie Energy change = 500 – 350 = 150 calorie Note : Heat supplied to the system q = + ve Work done by the system on surroundings W = – ve C.S.V. / February / 2008 / 1600 10. (D) Tetrahydrofuran 11. (A) N2O4 O 2NO 2 0·5 a is a cyclic ether. (a – 0·25a) at ( ) 17. (C) CrO5 is oxodiperoxochromium(VI). It has two peroxide bonds (4 oxygen atoms have oxidation state-1 each) O.N. of Cr = x (Say) O O Cr O O x + 4(–1) + 1(–2) = 0 x = +6 N2 + 3H2 2NH3 Kp = 1·44 × 10–5 Kp = Kc (RT)∆n ∆n = 2 – (1 + 3) = – 2 T = 500 + 273 = 773 K R = 0·082 litre atm/K –1 mole–1 Kp 1·44 × 10–5 Kc = = (RT)∆n (0·082 × 773)–2 O Total moles at ( ) = a – 0·25a + 0·5a = 1·25a Total pressure = 1 atm 0·75a × 1 atm PN2O4 = 1·25a 0·5a × 1 atm PNO 2 = 1·25a 2 P NO 2 (0·5a)2 1·25a = × Kp = PN2O4 (1·25a)2 0·75a = (0·5)2 1·25 × 0·75 = 0·267 atm 18. (A) 12. (A) Hofmann hypobromite reaction is used to eliminate O || Kc = (1·44 × 10–5)(0·082 × 773)2. —C— group of amide. 19. (C) Electronic Outer Valency O configuration electrons || CH3—C—NH2 + Br2 + KOH → CH3CONHBr + KBr 2 2 A 1s 2, 2s 2 2p6, 3s 2 + H 2O B 1s 2 2s2 2p5 7 1 O A B Valency || 2 1 CH3—C—NHBr + KOH → CH3NCO + KBr + H2O Formula of compound is AB2. CH3NCO + 2 KOH → CH3NH2 + K2CO3 20. (A) In pure waters which is neutral [H3O+] = [OH–] …(1) CH3CONH2 + Br2 + 4 KOH → CH3NH2 + 2 KBr Ionic product of water + K2CO3 + 2 H2O Kw = [H3O+] [OH–] …(2) 1 From equations (1) and (2) 13. (A) Formal charge = V – N – B 2 Kw = [H3O+] 2 V = Valence electrons of the free atom 21. (D) Amphiprotic species are those which can accept N = Number of non-bonding electrons or donate a proton B = Number of bonding electrons –H+ + H+ CO32– ← HCO3– → H2CO3 . . .. . F. + + H+ –H OH– ← H2O → H3O+ . .. .. . In . .F . .Cl. .F.. formal charge at chlorine. . . 22. (A) 0·2 F HClO4 = 0·2 M HClO4 1 (Since the given formula is molecular formula of per= 7–4– ×6=0 2 chloric acid) HClO4 H+ + ClO4– Here V = 7, N = 4, B = 6. 0·2 M 0·2 M 14. (B) For a given value of l, m l can have values – l to +l HClO4 being strong acid ionises completely inclusive zero, i.e., total values of ml = 2l + 1. Thus in [H+] = 0·2 M = 2 × 10 –1 M magnetic field for a given value of l, a single line will appear as 2l + 1 lines. pH = – log 2 × 10–1 15. (B) The catalytic activity of a heterogeneous catalyst = 1 – 0·3010 ≈ 0·7. depends on its free valencies. In finely divided state 23. (B) the surface area is increased to a large extent and Calamine ZnCO3 Carbonate Option A is thereby free valencies are also increased correct FeCO3 Siderite ores Cu 2O Oxide ore Cuprite Option B is Ni Ni Ni Ni Subdivision + Ag2S incorrect Argentite Sulphide ore Ni Ni Ni Ni (Not an oxide ore) 16. (A) Effective nuclear charge Zinc blende ZnS Sulphide Option C is FeS 2 Iron pyrites correct Zeff = Z–S (or σ) ores Malachite CuCO3·Cu(OH)2 Z = Atomic number Ores Option D is Cu(OH)2·2CuCO 3 of Cu correct Azurite S (or σ) = Screening constant } } } C.S.V. / February / 2008 / 1601 24. (B) NH4OH NH4Cl NH4+ + OH– NH4+ + Cl– NH4OH being weak base ionises feebly. In presence of NH4 Cl its dissociation is further supressed. In solution [OH –] is decreased. pH is also decreased. 25. (D) Energy binding the atoms of alkali metals is relatively low on account of a single electron in valence shell. The melting points decrease in moving down the group from Li to Cs as the cohesive force decreases Li Na K Cs 181 98 63 28·5 m.p. (°C) 26. (C) CsCl crystal has body centered cubic lattice structure in which Cl– ions are at the corners of the cube and Cs + is present at the centre of the cube. —Cs + —Cl – 33. (A) In bleaching powder (CaOCl2 ) the available chlorine is 49%, i.e., 100g bleaching powder give 49 g Cl2 10 g sample of bleaching powder yields 4·9 g Cl2 4·9 Mole of Cl2 = 71 4·9 Volume of Cl 2 at STP = × 22·4 71 = 1·5 lit (Approx.) 34. (C) pH of acidic buffer is given by Henderson eqn. [Salt] pH = p Ka + log [Acid] pH = 4, p Ka = – log Ka = – log 10–5 = 5, [Acid] = 1·0 M ∴ 4 = 5 + log log [Salt] = – 1 log [Salt] = log 1 10 [Salt] 1·0 27. (B) Sodalime is the mixture of NaOH and CaO. 28. (D) Colloidal solution of calcium acetate in alcohol sets as jelly and is termed as solid alcohol. 29. (B) Water gas is the mixture of CO and H2 in 1 : 1 ratio. 30. (A) van der Waals equation at low pressure is a P + 2 V = RT V a = RT PV + V a PV + = 1 RT RTV a PV = 1– RTV RT PV = z (Compressibility factor) RT a 1– ∴ z = RTV 31. (C) Molecular formula of sulphur molecule is S8. It has staggered ring structure ∴ [Salt] = 0·1 M 35. (A) IUPAC name of K4 Fe(CN)6 is potassium hexacyanoferrate(II). 36. (A) Heating of the ore below its fusion temperature in absence of air is called calcination. This step expels organic matter and moisture from the ore. Al2O3·2H2O → Al2O3 + 2H2O Hydrated alumina Anhydrous alumina ( ) 37. (C) The Balestein test is performed to detect halogens in organic compound. A copper wire is heated in the Bunsen flame till it does not impart any green colour to flame. The heated end is dipped in the organic compound and heated again. Formation of a green or bluish green flame due to formation of volatile cupric halides indicates the presence of some halogen in organic compound. 38. (B) NaNO3 Chile salt petre Na(NH4)HPO4 Microcosmic salt NaHCO3 Baking soda Na 2CO3·10H2O Washing soda 39. (B) In benzene each carbon atom is sp 2 hybridized and further due to resonance C—C bond lengths are equal ( ( ) ) S S S S S S 32. (B) S S k V = 19·5 × 10–5 ohm–1 cm–1 ←→ 40. (B) In metallic sodium outer shell electron is loosely attached to the Kernel of the atom. To and fro oscillation of these loose electrons cause lustre of sodium metal. 2 41. (A) CaO → CaC2 → C2H2 C = 0·01 mole/dm3, V = 100 × 1000 cm3 mole–1 µV = k V × V = 19·5 × 10 –5 × 105 = 19·5 ohm –1 cm2 mole–1 µ∞ = 390 ohm–1 cm2 mole–1. Acetic acid being monobasic acid its equivalent conductance is equal to molar conductance. (λV = µV and λ∞ = µV ). The degree of dissociation 19·5 α = = = 0·05 λ∞ 390 λV C(Coke) H O Calcium carbide Acetylene ∆ 42. (C) H2O2 + 2H 2O2 → 2KI → (Acidified) 2H 2O + O2 2KOH + I2 (Violet) C.S.V. / February / 2008 / 1602 2KMnO 4 + 3H2SO4 → [H2O2 + O → K2SO4 + 2MnSO4 + 3H2O + 5O H2O + O2] × 5 (Continued from Page 1588 ) 50. Which of the following has the highest pH in water ? (A) NaCl (B) NaHCO3 (C) Na 2CO3 (D) KCl 51. The correct order of solubility of lithium halides in nonpolar solvents is— (A) LiF > LiI > LiBr > LiCl (B) LiI > LiBr > LiCl > LiF (C) LiCl > LiF > LiI > LiBr (D) LiBr > LiCl > LiF > LiI 2KMnO 4 + 3H2SO4 + 5H2O2 → K2SO4 + 2MnSO4 + 8H2O + 5O2 CH HgCl2 CH2 43. (B) ||| + HCl → || 333K CH CHCl Acetylene Vinyl chloride ANSWERS 1. (C) 8. (A) 15. (A) 22. (A) 29. (A) 36. (B) 43. (C) 50. (C) 2. (D) 9. (D) 16. (D) 23. (C) 30. (C) 37. (A) 44. (D) 51. (B) 3. (A) 10. (A) 17. (A) 24. (D) 31. (B) 38. (D) 45. (D) 4. (D) 11. (A) 18. (A) 25. (D) 32. (D) 39. (C) 46. (B) 5. (A) 12. (D) 19. (D) 26. (C) 33. (B) 40. (B) 47. (D) 6. (B) 13. (C) 20. (C) 27. (D) 34. (B) 41. (D) 48. (A) 7. (D) 14. (D) 21. (B) 28. (D) 35. (C) 42. (C) 49. (C) ●●● 44. (B) Ordinary glass is a mixture of sodium and calcium silicates Na2SiO3·CaSiO3·4SiO 2. 45. (D) Ethanol (C2H5OH), Ethanal (CH3CHO) and isopropyl alcohol (CH 3 CHOHCH3) contain either CH3CHOH or CH 3CO group and hence give iodoform test. Benzyl alcohol (C 6H5CH2OH) does not process this grouping. Its iodoform test is negative. 46. (D) It is a mixture of primary (or mono) calcium phosphate and gypsum Ca(H2PO4)2·H2O + CaSO 4·2H2O. It is used as phosphate fertilizer. 47. (A) Formaldehyde reacts with ammonia to yield urotropine (hexamethylene tetramine) 6HCHO + 4NH3 → (CH2)6N4 + 6H2O Hexamethylene tetramine (Continued from Page 1597 ) 14. In which case of mixing of a strong acid and a base each of 1N concentration, temperature increase is the highest ? (A) 20 ml acid and 30 ml alkali (B) 10 ml acid and 40 ml alkali (C) 25 ml acid and 25 ml alkali (D) 35 ml acid and 15 ml alkali 15. The dissociation energy of CH4 and C2H6 are 360 and 620 k cal/mole respectively. The bond energy of C—C is— (A) 260 k cal/mole (B) 180 k cal/mole (C) 130 k cal/mole (D) 80 k cal/mole 16. In the process of ice melting at –15°C— (A) ∆G < 0 (C) ∆G = 0 (B) ∆G > 0 (D) It cannot be predicted 48. (A) 2O 3 → 3O2 2 Vol. 3 Vol. Vol. of O2 3 = = 1·5. Vol. of O3 2 N H2C CH2 N N CH2 CH2 CH2 N CH2 17. For an adiabatic process which of the following is correct ? (A) q = 0 (B) ∆E = q (C) q = +w (D) P∆V = 0 18. In a closed vessel 2 moles of CO and 1 mole O 2 are ignited to get CO2. If ∆ H and ∆E are the change in enthalpy and change in internal energy respectively, then— (A) ∆H < ∆E (B) ∆H > ∆E (C) ∆H = ∆E (D) Not definite 19. The relationship which describes variation of vapour pressure with temperature is called— (A) Hess’s law (B) Arrhenius equation (C) Clausius-Clapeyron equation (D) Kirchhoff’s equation 20. Two moles of an ideal gas expand spontaneously into a vacuum. The work done is— (A) Zero (B) 2 J (C) 4 J (D) 8 J On heating ozone is decomposed to oxygen and volume is increased to 1·5 times. 49. (A) In DNA pairs of bases are held together with Hbonding. 50. (B) For gases 5 x = 0 for monoatomic Cp = R+x x = R for diatomic 2 3 3 x = R for polyatomic gases R+x Cv = 2 2 For Ar (monoatomic gas) 5 R Cp 2 = = 1·66 Cv 3 R 2 For CO2 (polyatomic gas) 3 5 R+ R Cp 2 2 4 = = = 1·33. 3 3 Cv 3 R+ R 2 2 ●●● ANSWERS 1. (C) 2. (B) 3. (D) 4. (B) 5. (A) 6. (C) 7. (C) 8. (A) 9. (A) 10. (B) 11. (A) 12. (C) 13. (A) 14. (C) 15. (D) 16. (B) 17. (A) 18. (A) 19. (C) 20. (A) ●●● C.S.V. / February / 2008 / 1603 Model Paper for Various Medical Entrance Examinations CHEMISTRY 1. The molecular mass of CO2 is 44 a.m.u. and Avogadro number is 6·02 × 10 23. Mass of one molecule of CO 2 is— (A) 7·31 × 10–23 g (B) 3·65 × (C) 1·01 × 10–23 10–23 g g 9. A gas is filled in a balloon of 80 litres at 10 atm. If the whole gas is filled in small balloons of 3 litres at 1·11 atm and at same temperature, find the number of small balloons required— (A) 150 (C) 100 (B) 240 (D) 250 of the acid and water layer 1·843 g of the acid. The distribution coefficient between water and ether is— (A) 7·25 (C) 0·725 (B) 72·5 (D) 725 16. Find the oxidation state of sulphur in H 2SO4— (A) + 6 (C) + 2 (B) + 4 (D) + 3 (D) 2·01 × 10–23 g 2. A compound contains 38·8% C, 16% H and 45·2% N. The empirical formula of the compound will be— (A) CH5N (B) C2H5N (C) CHN (D) CH10N2 3. The number of σ and π bonds in 1-butene-3 yne are— (A) 5σ + 5π (B) 7σ + 3π (C) 8σ + 2π (D) 6σ + 4π 4. Find the temperature of hydrogen gas which has the same velocity as that of oxygen at a temperature of 0°C— 273 (A) K (B) 273 × 8K 16 273 (C) K (D) 273 × 4K 32 5. Which is the oxidant in the following reaction ? 5KI + KIO3 + 6HCl (A) I 2 (C) KI → 3I2 + 6KCl + 3H 2O (B) KIO 3 (D) None of these 17. If edge length of a bcc crystal of an element is a cm. M is the atomic mass and N 0 is the Avogadro number. Then the density of the crystal is— (A) (C) 2×M a 3N0 a3 M 2N 0 (B) (D) 4×M a 3N0 2N 0 Ma 3 10. What is Bond order in case of O2+ ? (A) 2·5 (C) 1·5 (B) 2 (D) 3 11. The reaction between A and B is first order with respect to A and second order with respect to B. If the concentration of A is halved and the concentration of B is doubled, the rate of reaction will be— (A) Same as the initial value (B) Three times the initial value (C) Double the initial value (D) Half the initial value 12. In the reaction 2C (s) + O2(g) 2CO (g) 18. Which of the following is correct – β– for 60Co → ? (A) (C) 19. A (IUPAC)— (A) Vinyl acetyline (B) 1-butene-3-yne (C) 1-butyne-3-ene (D) Both (B) and (C) are true 20. 100 ml 0·1 (N) HCl is mixed with 100 ml of 0·2 (N) NaOH, then the resulting concentration of the solution is— (A) 0·05 (N) (C) 2 (N) (B) 5 (N) (D) 0·1 (N) 61 Co 59 Ni (B) (D) is 60 Ni 60 Cu named as the partial pressure of CO and O2 is 8 atm. and 4 atm. respectively, then find its equilibrium constant— (A) 16 (C) 8 (B) 24 (D) 2 6. Which is the correct electronic 13. When 48250 C of electricity is passed through an aqueous soluconfiguration of Cr (chromium) ? tion of NiI2 (Ni = 58·8) the mass (A) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 of Ni metal deposited would be— (B) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4 (C) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 (A) 7·3 g (B) 14·7 g (D) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 8 (C) 22·0 g (D) 29·4 g 7. CuSO 4 is not stored in aluminium 14. 1M solution of a weak monobasic bottles because— acid has a pH value of 6, then its (A) Cu gets oxidised dissociation constant will be— (B) Cu 2+ gets reduced (B) 400 (A) 10–4 (C) Al gets reduced (D) CuSO 4 is decomposed (D) None of these (C) 10–8 8. Which of the following is not linear ? (A) BeCl 2 (B) HCN (C) ZnCl2 (D) H2O 15. When succinic acid was shaken with a mixture of 100 ml water and 50 ml ether, it was found that ether layer consisted of 0·127 g 21. C4H4O4 can be— (A) A cyclic ester (B) cis-and trans-diabasic acid (C) gem-α, β unsaturated diabasic acid (D) All correct 22. For detection of sulphur in an organic compound, sodium nitroprusside is added to the lassaigne’s filtrate. The ppt. obtained is— (A) Purple colour (B) Black colour (C) Blood-red colour (D) White colour C.S.V. / February / 2008 / 1604 23. Chlorine in vinyl chloride is less reactive because— (A) sp 2 hybridized carbon has more acidic character than sp 3 hybridized carbon (B) C—Cl bond develops partial double bond character (C) Of resonance (D) All are correct 24. Which of the following compound is optically active ? (A) CH3CH2COOH (B) CH3CHOHCOOH (C) HOOC·CH2·COOH (D) CH3·CO·COOH 25. Which of the following can act as a nucleophile as well as electrophile ? (A) BF 3 (B) FeCl3 (C) ZnCl2 (D) C2H5MgBr 26. Which of the following free radical is most stable ? (A) Primary (B) Methyl (C) Secondary (D) Tertiary 27. When butene-1 is mixed with HBr, the major reaction product is— (A) 1, 2-dibromobutane (B) 1-bromobutane (C) 2-bromobutane (D) None of these 28. Number of oxygen molecules requires in the combustion of C4H10 is— 13 (A) O (B) 8O2 2 2 15 11 (C) O (D) O 2 2 2 2 29. n-propyl bromide reacts with ethanoic KOH to form— (A) Propane (B) Propene (C) Propyne (D) Propyl alcohol 30. Lindlar’s catalyst contains— (A) Pd supported over CaCO3 (B) Hg supported over PbSO 4 (C) Ni supported over CuSO4 (D) Ni supported over CdSO4 31. The end product of the reaction CH3OH → X → B (A) (B) (C) (D) Alkane Carboxylic acid Sodium salt of carboxylic acid Ketone Cu 300°C NaOH 32. Product obtained by heating (C) Aqueous solution electroC2H5Br in presence of dry silver lysis oxide is— (D) Thermite reduction (A) C2H5OC2H5 (B) C2H5COOH 40. The boiling point of phenols are (C) C2H5OH (D) C2H5CHO higher than the hydrocarbons of comparable masses. It is due 33. Carboxylic acid reacts with diato— zomethane to yield— (A) More polarising power (A) Amines (B) Alcohols (B) Presence of hydrogen bond(C) Esters (D) Amides ing 34. The reduction of aldehyde and (C) Due to resonance stabilised ketones to the corresponding (D) Due to acidic character hydrocarbons with amalgamated zinc and concentrated HCl is called— (A) Wolff-Kishner reduction (B) Clemmensen reduction (C) Coupling reduction (D) Cross-Cannizzaro reaction 35. After partial roasting, the sulphide of copper is reduced by— (A) Cyanide process (B) Electrolysis (C) Reduction with carbon (D) Self reduction 36. Write the from— product obtained 41. Mineral of aluminium that does not contain oxygen is— (A) Corundum (B) Diaspore (C) Bauxite (D) Cryolite 42. In the manufacturing of polymer, neoprene called artificial rubber, the monomer present is— (A) Butene (B) Propene (C) Styrene (D) 2-chloro-1, 3-butadiene 43. Graphite conducts because of— electricity 2 CH3CH2CONH2 → (A) CH3CH2NH2 (B) CH3CH2CH2NH2 (C) CH3NH2 (D) CH3CH2OH Br + NaOH (A) Weak vander Waals forces between layers (B) Covalent bonding between carbon atoms of layers (C) Delocalized electrons in each layer (D) sp 2 hybridization of carbon atoms in each layer 44. Which of the following is correct for electron affinity ? (A) F > Cl > Br > I (B) Cl > F > Br > I (C) Cl > Br > I > F (D) None of these 45. In P4O6 the number of oxygen atoms bonded to each P atom is— (A) 1·5 (B) 2 (C) 3 (D) 4 46. Which one require lowest bond dissociation energy ? (A) H—F (B) H—Br (C) H—Cl (D) H—I 47. Most of the elementry gases are obtained by chemical reaction of their compounds. For example, Cl2 is obtained by allowing KMnO4 to react with hydrochloric acid. 37. Tailing of mercury is a laboratory test for— (A) O3 (B) Hg (C) Cl2 (D) O2 38. The major product obtained when benzene reacts with CH3CH2CH2Cl in presence of anhydrous AlCl3 is— CH2CH2C H3 (A) CH3 | C—CH 3 | CH3 C H (CH 3)2 (B) (C) (D) None of these 39. Which of the following process is used in the extractive metallurgy of magnesium ? (A) Fused salt electrolysis (B) Self reduction C.S.V. / February / 2008 / 1605 fluorine, however, can be obtained only by the electrolysis of a fluoride. This is because— (A) Fluorine is highly reactive gas (B) Fluorine is strongest chemical oxidising agent (C) Fluorine is highly poisonous (D) It is easy to electrolyse a fluoride 48. Concentration of ores is related to— (A) Froth floatation (B) Roasting (C) Calcination (D) Thermal decomposition 49. Cu 2+ ion would be reduced to cuprous ion if their solutions are treated with an aqueous— (A) KI solution (B) KCl solution (C) K2CO3 solution (D) K2SO4 solution 50. Coordination number and oxidation number of Cr in K3Cr(C2O4)3 are respectively— (A) 6 and + 3 (C) 3 and 0 (B) 4 and –2 (D) 3 and + 3 ANSWERS WITH HINTS 1. (A) Mass of 1 mole (= 6·02 × 1023 molecules) of CO2 = 44g 44 g ∴ Mass of one molecule = 6·02 × 1023 = 7·31 × 10–23 g 2. (A) 3. (B) H H | | H—C ≡ C—C = C—H 4 3 2 1 Concentration of A is halved and B is doubled the rate 1 r′ = k [A] {2[B]} 2 …(2) 2 { } 1-butene-3-yne No. of σ bonds = 7 No. of π bonds = 3 4. (A) 5. (B) In oxidising agent O.N. of key atom is decreased. In the reaction 5KI + KIO3 + 6HCl → 3I2 + 6KCl + 3H 2O +5 0 12. 13. O.N. of iodine atom in KIO3 decreases from + 5 to zero. KIO3 is oxidising agent. 6. (A) 7. (B) Standard reduction potential of copper (E° Cu2+/Cu = + 0·34 V) is much higher than aluminium (E°Al3+/Al = – 1·66 V) Cu 2+ ions are reduced when placed its solution in Al container. 8. (D) 9. (B) Gas law equation is PV = nRT PV ∴ n = RT Moles of gas in larger balloon n = 10 × 80 RT 14. 15. Moles of gas in smaller balloon 3 × 1·11 n′ = RT No. of smaller balloons filled = 10 × 80 3 × 1·11 = 240·24 16. 17. ≈ 240 10. (A) 11. (C) The rate law equation for the reaction r = k[A] [B]2 …(1) Equation (2) is divided by (1) 1 [A] {2[B]} 2 k 2 r′ = r k[A] [B]2 1 = × (2)2 = 2 2 r ′ = 2r (A) (B) w = ZIt Q = It = 48250 C E 58·8 Z = = 96500 2 × 96500 Ionic wt. of Ni2+ E= 2 58·8 w = × 48250 = 14·7 g 2 × 96500 (D) (A) Concentration of succinic acid in water layer 1·843 = g/ml 100 Concentration of succinic acid in ether layer 0·127 g/ml = 50 Concentration in Distribution coefficient (K) water layer between water and ether = Concentration in ether layer 1·843 50 K = × 100 0·127 = 7·25 (A) Mass (A) Density = Volume Unit cell content (Z) in bcc = 2 Molar mass = M g 2M Mass of unit cell = g N0 Edge length of unit cell = a Volume of unit cell = a 3 2M ∴ Density = N0a 3 { } ( ) C.S.V. / February / 2008 / 1606 18. (B) 19. (B) 1 2 CH — CH 2— 3 C ≡ CH 33. (C) Carboxylic acid reacts with diazomethane to yield methyl esters. RCOOH + CH2N2 → RCOOCH3 + N2 Methyl ester According to IUPAC double bond carbon is given lowest number. IUPAC name is 1-butene-3-yne. 20. (A) 21. (D) (i) Glycol oxalate O CH2 CH2 O H (ii) H cis CO (cyclic ester) CO H C COOH COOH C and C COOH COOH C HOOC trans H — (iii) CH2 — C COOH ( + gem-α‚ β unsaturated dibasic acid ) 22. (A) 23. (C) Less reactivity of chlorine in vinyl chloride is due to resonance in the compound. → — CH2 — CH—Cl ← CH2—CH — Cl– — 24. (B) 25. (D) Grignard reagent can act as nucleophile as well as an electrophile δ– δ– R—MgX + A—B → R—A + MgX δ– δ+ H OH + – C — O + R—MgX → C— O MgX — | R 34. (B) 35. (D) By partial roasting sulphide ore of copper is partially oxidised to Cu 2O. Now self reduction of sulphide occurs as follows : Cu 2S + 2 Cu2O → 6 Cu + SO2 36. (A) 37. (A) Mercury loses its meniscus in contact with ozone. This is termed as tailing of mercury. This is because of oxidation of Hg in presence of O 3 to suboxide. 2 Hg + O3 → Hg2O + O2 It is the laboratory test for O3. 38. (C) 39. (A) Magnesium is highly electropositive having a high value of oxidation potential E° Mg|Mg2+ = 2·36 V. It can be reduced by electrolysis of its fused salt. Mg2+ + 2e → Mg (Cathode) 2 Cl– → Cl2 + 2e (Anode) 40. (B) 41. (D) Corundum (Al2O3) Diaspore (Al2O3.H2O) Bauxite (Al2O3.2H2O) Cryolite (Na3AlF6) Thus, ore cryolite does not contain oxygen. 42. (D) 43. (C) Graphite has hexagonal planar structure. Each carbon atom is sp 2 hybridized. The fourth valence electron present in unhybridized orbital is delocalised, which is responsible for conduction of electricity. 44. (B) 45. (C) P4O6 has the structure shown in fig. It is evident from the fig. that each phosphorus atom is linked with three oxygen atoms. P :O : :O : ↓ OH C—OH + Mg X | R 26. (D) 27. (C) CH3 CH2CH — CH2 + HBr — 2-bromobutane (According to Markownikoff’s rule) P :O: P : :O :O: :O : P → CH3 CH2 CHBr CH3 46. (D) 47. (B) Fluorine has highest reduction potential (E°1/2F2|F– = + 2·87 V). It can only be oxidised from F– → F2 by electrolysis of its fused salt (fluorides). 48. (A) 49. (A) Cu2+ ion is reduced to cuprous ion by KI. [CuSO4 + 2KI → CuI2 + K2SO4] × 2 2 CuI 2 → 2 CuI + I2 2 CuSO4 + 4KI → 2CuI + I2 + 2K2SO4 50. (A) ●●● 28. (A) 29. (B) When n-propyl bromide reacts with ethanoic KOH elemination reaction takes place. alc. KOH CH3CH2CH2Br → CH3—CH — CH2 — – HBr Propene 30. (A) 31. (C) CH3OH → HCHO Cu 300°C NaOH → HCOONa + CH3OH (Cannizzaro reaction) 32. (A) C.S.V. / February / 2008 / 1607 Introduction ● Connective tissues have great variety in terms of their structure, composition and function and are abundantly distributed in the body. A characteristic cell type is the fibroblast, producing fibres of the protein collagen and elastin, providing tensile strength and elasticity respectively. Another protein, reticulin, is associated with polysaccharides in the basement membranes, and surrounding fat cells in adipose tissue. Loose connective tissue (Areolar tissue) binds many other tissues together. Collagen fibres align along the direction of tension, as in tendons and ligaments. The viscosity of many connective tissues is due to ‘space-filling’ hyaluronic acid. The Peritoneum, Pleura and Pericardium are modified connective tissues as are the bone and cartilage. Besides support, connective tissue is defensive, due to largely presence of histiocytes (macrophages), which may be as numerous as fibroblasts. Connective tissues are frequently well vascularized and permeated by tissue fluid. ● ● connective tissue. It fixes the skin with the muscles, attaches blood vessels and nerves with the surrounding tissues. It forms the dermis of the skin and submucosa in the wall of the alimentary canal. ● ● ● The areolar tissue consists of a transparent jelly-like matrix containing numerous fibres, cells and mucin. The white fibres occur in bundles called fascia and are formed of collagen. Collagen is a rope-like protein that can bend but not stretch. These fibres are abundant in tendons, which join muscles to bones, and ligaments join muscles together. In genetic disorder called ‘Ehlers–Danlos Syndrome’, weakened collagen fibres in ligaments cause the bones of body joints to dislocate easily. The yellow fibres are fewer and thicker than the white fibres. They are straight and occur singly. Yellow fibres are flexible, elastic and branched, the branches join with one another to form an irregular network. They are formed of a protein called elastin, which is resistant to boiling. Yellow fibres provide elasticity to the tissue. Collagen fibres formed at the injured site help in tissue repair. ● ● ● ● ● ● ● ● ● ● Classification of Connective Tissues Connective Tissues Proper Skeletal Tissues Bone Vascular Tissues Lymph Cartilage Blood Compact Bone Hyaline Cartilage Fibrous Cartilage Calcified Cartilage Spongy Bone White Fibrocartilage Elastic Cartilage Areolar Tissue Adipose Tissue White Fibrous Tissue Yellow Elastic Tissue Reticular Tissue Connective Tissue Proper ● The connective tissue proper has a soft gel-like matrix composed of proteoglycans, which are complex carbohydrates linked to proteins. They are of five major types : Areolar, Adipose, white fibrous, yellow elastic and reticular. Areolar tissue is the most widely distributed connective tissue in the body. They are also called loose Cells Present in Connective Tissue Matrix ● Fibroblasts are the principal cells of the areolar tissues. They have abundant euchromatin, nuclei, rough endoplasmic reticulum and Golgi apparatus. When the tissue matures and stops growing, the cells become less active and are called Fibrocytes. They are small and contain acidophilic cytoplasm. ● ● C.S.V. / February / 2008 / 1608 ● ● ● ● ● ● ● ● ● Macrophages or histiocytes are almost as numerous as the fibroblasts. They are large, long-lived amoeboid cells. They are capable of phagocytosis. Phagocytosis permits them to engulf and digest dead human cells and combat bacteria, viruses and other foreign pathogens that may cause disease. Mast cells secrete substances involved in defense against foreign material or cells. One of these substances—heparin, prevents blood clotting in the blood vessels. Another secreted substance—histamine, causes blood vessels to enlarge in diameter and become leaky. Both of these substances encourage white blood cells to move out of the blood vessels into the tissue to fight infection. Adipose tissue is connective tissue and functions as storage. It differs from fibrous connective tissues in that it has very little ground substance and few fibres. It contains cells called adipocytes which are specialized to store fat. Brown adipose tissue (brown fat) comprises cells whose granular cytoplasm is due to high concentration of cytochromes and whose function appears to release heat in the neonatal mammals. Distributed around neck and between scapulae in hibernating mammals. White adipose tissue is distributed widely in the body. The stored fat is largely composed of triglyceride. White fibrous tissue is specially rich in white collagen fibres. These are similar but thicker than those found in the areolar tissue. This tissue is very tough and inelastic. It has two forms—cords and sheets. White fibres run parallel to form cords, called tendons. They are arranged in bundles bound together by areolar tissue. The white fibres lie criscross in one plane to form sheets. It occurs in the pericardium of heart, duramater of the brain and spinal cord, sclerotic coat and cornea of the eye ball, capsule of the kidney, perichondrium of cartilage and periosteum of bone. ● Yellow elastic tissue consists mainly of a loose network of yellow elastic fibres. The fibres are much thicker but similar to those of areolar tissue. The fibroblasts are irregularly scattered. Elastic connective tissue has numerous bundles of elastic fibres, which impart their characteristics to the entire tissue, allowing it to stretch without deforming permanently. Elastic tissue is found in the walls of body organs that regularly change shape, such as the stomach, the lungs, the blood vessels and even the heart. Elastic tissue are also found in vocal cords and in the ligaments that connect the bones of the spine. ● ● Differences between Tendons and Ligaments Tendons 1. Are composed of white fibrous tissue. 2. Fibroblasts lie in almost continuous rows. 3. Are tough and inelastic. 4. Connect skeletal muscles with bones. Ligaments 1. Are composed of yellow elastic tissue. 2. Fibroblasts lie scattered. 3. Are strong but elastic. 4. Connect bones with one another. Also connect other parts together or support an organ. ● Reticular connective tissue has a finely branched network of reticular fibres. It provides an internal framework of the cells that perform the functions of soft organs, for example, the liver, spleen, lymphnodes and tonsils. It also forms the lamina propria of the gut wall. The reticular cells are phagocytic and form defense mechanism of the body. Skeletal tissues have a tough matrix. They form a rigid framework, which supports the body, protects the more vital organs, provides hand surface for the attachment of the tendons of the muscles and helps in locomotion in collaboration with the muscles. There are two types of skeletal tissue : Cartilage and bone. Cartilage has a high concentration of collagen fibres like dense connective tissue. However, the ground substance of cartilage has a unique chemical composition and special properties that permit it to absorb large amounts of bound water. This bound water makes cartilage resistant to compression. Cartilage lacks blood vessels, nerves and lymphatic vessels in its extracellular matrix. Due to absence of blood vessels, the injuries to cartilage are slow to heal. Chondroblast cells of cartilage become chondrocytes when surrounded by ground substance within the lacunae. The amorphous matrix (chondrin) contains glycoproteins, basophilic chondroitin and fine collagen fibres, varying proportions of which determine whether it is hyaline, elastic, or fibrocartilage. The surface of cartilage is surrounded by irregular connective tissue forming the perichondrium. ● Differences between White and Yellow Fibres White Fibres 1. 2. 3. Are abundant, very fine and wavy. Occur in bundles called fascia. Are inelastic and unbranched. Formed of a protein called collagen. Digested by pepsin. Provide tenacity to the tissues. Composed of microfibrils having alternating light and dark bands. 1. 2. 3. Yellow Fibres Are fewer, thicker and straight. Occur singly. Are elastic and branched, branches join to form a network. Formed of a protein called elastin. Digested by trypsin. Provide elasticity to the tissues. Not composed of microfibrils and lack striations. ● ● ● 4. 5. 6. 7. 4. 5. 6. 7. ● ● C.S.V. / February / 2008 / 1609 ● The blood vessels that nourish cartilage are all located in perichondrium. There are three types of cartilage present in the body, i.e., hyaline cartilage, fibrocartilage and elastic cartilage. Hyaline cartilage is both flexible and strong. Front part of nose is made of hyaline cartilage. It also covers the ends of many bones, providing a slippery surface for the movement of joints. It joins the ribs to the breast bone (sternum). Fibrocartilage is tougher and less flexible than hyaline cartilage. Its collagen fibres are thicker and are arranged in dense bundles. Fibrocartilage joins bones together in areas where considerable stress may occur. Pads of fibrocartilage form the intervertebral discs between vertebral, where it acts as cushions. Elastic cartilages is more elastic and flexible than any other type of cartilage. It readily recovers its shape after distortion. This cartilage is found in pinna (external ear) and external auditory canal of the ear, epiglottis, eustachian tubes. Bone is a living connective tissue composed of a hard matrix material and bone cells. If a bone is kept in an acid for some time, its inorganic part is dissolved and organic part is left behind. Such a bone is said to be decalcified. Small bones are solid, while long bones have a cavity, the bone marrow cavity. ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● Differences between Bone and Cartilage Bone 1. Matrix is composed of a tough, inflexible material, the ossein. Matrix is always impregnated with calcium salts. Bone cells lie in lacunae singly. 1. Cartilage Matrix is composed of a firm, but flexible material, the chondrin. Matrix may be free of or impregnated with calcium salts. Cartilage cells lie in lacunae singly or in groups of two or four. Chondroblasts are oval and devoid of processes. Lacunae lack canaliculi. There are no special cartilage-forming cells. Cartilage grows by division of all chondroblasts. ● 2. 2. ● ● ● 3. 3. 4. Osteocytes are irregu- 4. lar and give off branching processes in the developing bone. Lacunae give off canaliculi. There are outer and inner layers of special bone-forming cells, the osteoblasts, that produce new osteocytes, which secrete new lamellae of matrix. Matrix occurs largely in concentric lamellae. 5. 6. ● 5. 6. ● 7. 8. 9. 7. Matrix occurs in a homogenous mass. Cartilage is nonvascular. No such tissue is present. ● Bone is highly vas- 8. cular. Bone may have bone marrow at the centre. 9. The substance of bone is distinguishable into 3 regions—periosteum, matrix and endosteum. Marrow cavity contains bone marrow. Periosteum covers the bone externally and is a tough sheath composed of white fibrous tissue. Tendons of the muscles are attached to this sheath. Bundles of Collagen fibres, called Sharpey-Schafer fibres, from the periosteum penetrates the bone to provide a firm connection between the two and as a firm base for tendon insertions. The periosteum contains blood vessels, which send branches into the bone for providing nourishment. It also contains active bone-forming cells, the Osteoblasts, which produce new bone material. Endosteum lines the marrow cavity. It is composed of white fibrous tissue and contains the bone-forming cells, which produce new bone material. The long bones thus grow in thickness from two sides. Matrix of bone is dense and hard. It is formed of a protein called Ossein. Ossein is the collagen of bone and forms a strengthening framework in the matrix. The hardness and rigidity of bone comes from the minerals primarily calcium and phosphate, which precipitate out of body fluids coating the collagen fibres. The main minerals (salts) include calcium phosphate, calcium carbonate, sodium chloride and magnesium phosphate. Bones thus contains both inorganic and organic matter. The matrix occurs as layers, the lamellae, which are largely arranged in concentric rings around narrow longitudinal cavities—the Haversian Canals. In the lamellae are found small fluid-filled spaces, the bone lacunae, which give off fine radiating channels called Canaliculi. A Haversian canal together with its lamellae, lacunae and canaliculi forms a Haversian system or Osteon. The Haversian canals are interconnected by transverse channels, the Volkmann’s canals. In compact bone, the numerous bone cells or osteocytes are embedded in a solid non-living matrix of calcium phosphate that is deposited on collagen fibres. The osteocytes closer to the blood vessels obtain nutrients by diffusion and pass them to more distant osteocytes by means of their slender cellular extensions. Spongy bones also contain osteocytes, but they are not arranged around Haversian canals. The bony bars and plates of spongy bone are not wider than a few cells, therefore, each osteocyte has intimate access to red marrow for the exchange of nutrient and wastes. Bone removing cells are called osteoclasts. Osteoclasts are located on the surface of bone and remove bone by secreting enzymes and dissolve the solid calcium phosphate. As a result calcium and phosphate ions are released in the blood when insufficient calcium and phosphate dietary is consumed. C.S.V. / February / 2008 / 1610 Types of Bone ● In compact bone, osteocytes are located in lacunae that are arranged in concentric circles within Haversian system. Compact bone forms outer cylinder of shafts of long bones of limbs and is typified by Haversian system. Its cavity contains yellow bone marrow. Spongy bone is present at the end of long bones and has entirely different structure. ● Spongy bones form vertebrae, flat skull bones, and ends of long bones (epiphyses), typified by the presence of traberculae. ● Types of Bones According to their Source of Formation ● The process of bone formation is called ossification. Bones are of the following types according to their source of formation— (1) Cartilaginous or replacing bones—These bones develop from the pre-existing cartilage and particularly replace the cartilage. They are also called endochondrial bones. Examples— Humerus, Femur. (2) Investing or dermal or membranous bones— These bones develop in the dermis of the integument as thin plate and sink to get attached over the original cartilaginous endoskeleton. Example —Frontal, nasal, vomers and parietal of the skull. (3) Sesamoid bones—These bones are formed in the tendons at the joints. Example—Patella (Knee-cap). (4) Visceral bones—These bones are formed in organs dissociated from rest of the skeleton. Examples—in the heart of some ruminants and called os cordis. Bone also develops as os penis in the copulatory organs of most bats, insectivores and rodents. Small bone also develops as os palpebae in the eye lids of crocodile. ● Blood is a liquid connective tissue. Blood contains approximately 22% solids and 78% water. ● Differences between Compact and Spongy Bones Compact Bone 1. Forms shaft (diaphysis) 1. of long bones. Spongy Bone Forms epiphyses of long bones, besides forming skull bones, vertebrae and ribs. Lamellae arranged as interlacing network. Small spaces occur between lamellae, hence spongy bone. Marrow cavity lacking. Spaces between lamellae contain red marrow. Marrow produces red corpuscles and granular white corpuscles. 2. 3. 4. 5. Lamellae arranged in regular Haversian systems. No gaps between lamellae, hence compact bone. Marrow cavity present. Marrow cavity has yellow marrow. Marrow stores fat. 2. 3. 4. 5. 6. 6. ● Spongy bone contains numerous bony bars and plates separated by irregular spaces. These spaces contain red bone marrow. OBJECTIVE QUESTIONS 11. Protein chondrin is found in— (A) Blood (B) Areolar tissue (C) Bone (D) Cartilage 12. Which of the following proteins, is not found in the connective tissue ? (A) Collagen (B) Elastin (C) Ossein (D) Actin 13. Ligaments join— (A) Muscle to bone (B) Bone to bone (C) Skin to muscles (D) Nerves to muscles 14. Osteoclasts are associated with— (A) Bone formation (B) Bone breakdown (C) Muscle formation (D) Bone regeneration 15. The mast cells are found in— (A) Adipose tissue (B) Yellow fibrous tissue (C) Areolar tissue (D) White fibrous tissue 16. The camel’s hump is composed of a tissue which provides water when oxidized. It is— (A) Areolar tissue (B) Adipose tissue (C) Muscular tissue (D) None of these 17. The formation of cartilage is known as— (A) Chondrogenesis (B) Diapendesis (C) Haemopoisis (D) Ossification 18. A bone which is formed by the transformation of the connective tissue, called— (A) Replacing bone (B) Investing bone (C) Cartilage bone (D) Reticular bone 19. The bone of mammals contains longitudinal Haversian canals, which are connected by transverse canals, called— (A) Inguinal canal (B) Bidder’s canal (C) Volkmann’s canal (D) None of these 10. The pericardium of heart is composed of— (A) Areolar tissue (B) Reticular tissue (C) White fibrous tissue (D) Yellow fibrous tissue ANSWERS 1. (D) 2. (D) 3. (B) 4. (B) 5. (C) 6. (B) 7. (A) 8. (B) 9. (C) 10. (C) ●●● C.S.V. / February / 2008 / 1611 Lamarckism Lamarckism is also known as the theory of inheritance of acquired characters. Lamarckism is the first theory of evolution, and was proposed by Jean Baptiste de Lamarck (1744–1829), a French biologist. Although the outline of the theory was brought to notice in 1801, but his famous book 'Philosophic Zoologique' was published in 1809, in which he discussed his theory in detail. Lamarck coined the terms 'invertebrates' and 'Annelida'. The term 'Biology' was given by Lamarck and Treviranus. Lamarck also introduced the term 'Vertebrates'. development, while disuse for a long time make them undeveloped and ultimately lead to their degeneration. (4) Competition—There is a sort of competition in nature to avoid overcrowding on earth. The stronger organisms try to destroy the weaker ones. The smaller multiply faster, while larger ones slowly, thus a balance is maintained. (5) Inheritance of acquired characters—The favourably structural changes gained by an individual due to use or disuse during its life time is preserved and are passed on to its offsprings. These changes become more and more pronounced if they are exposed to similar stress of the environment as was faced by their ancestors or parents. Such cumulative effects will finally result in the appearance of new species. Cross breeding—These various peculiarities produced in the organisms will always appear in successive generations (provided breeding is confined to such unions only). As a result, crosses between individuals not acquiring these peculiarities, result in the disappearance of such characters acquired in particular circumstances. Isolation—The separation of various generations brings about their different features, which in the course of evolution, become specialized into particular species. Lamarckian Postulates or Laws The complete theory of evolution given by Lamarck (Lamarckism) is popularly known as the inheritance of acquired characters in organisms. It is in the form of four postulates or laws, as follows : 01. The internal forces of life tend to increase the size of an organism, i.e. the whole body and also the different parts upto a limit determined by life itself. 02. The formation of a new organ or a part in the body is the result of a new need or new want, which has arisen and continues to be left by the organism. 03. The development of an organ and its power of action is directly proportional to its use, continuous and constant use strengthening the organ, while disuse results in its degeneration. 04. All the changes, which organisms acquire during their life time are transmitted to their offsprings by the process of inheritance. It means changes are cumulative over a period of time. The fourth principle that is acquired characters are inherited, is the most attractive and controversial law, now known as 'Lamarckian doctrine'. A more precise account of Lamarck’s theory may be given by the following factors, which according to him, played an essential role in evolution : (1) Role of environmental factors—Lamarck believed that various factors like soil, food, temperature etc. causing changes in environment act directly in case of plants, while indirectly in case of animals (since they possess nervous system). The environment influence leads to change in their habits which results in unusual activity of an organ or structure. Lamarck, thus assumed that living organisms react to external conditions and become modified. (2) Effects of needs or physical wants—Lamarck thought that change of habits may bring about the modification of existing organs or may initiate the formation of new organ. (3) Use and disuse—The continuous and constant use of organs make them efficient and lead to their better Analysis of Lamarck’s Theory First law—The first law of Lamarck is merely the growth process of the organism with which most scientists agree. The increase in size in living beings is common due to metabolic activities, which are controlled by vital forces of life. Second law—This law of development of an organ due to its need is unacceptable to the modern evolutionists. Nobody can believe that any organ can develop due to the need of its presence. Third law—The principle of use and disuse of organs were based on direct observation of nature. Some of the examples are as follows— (1) Ancestors of modern horse lived in soft ground in forests and were plantigrade. These forests were later replaced by dry grass plains, and these had to graze on hard grass and to walk on dry land. These changes in habit were followed by changes in premolars and molars, reduction in number of digits and lengthening of legs. Thus the foot was gradually changed to unguligrade, suited for swift running over hard ground. (2) The continuous stretching of neck for several generations by giraffe to catch high located leaves and fruits of tall trees, caused the lengthening of its neck. This was due to continuous use of particular organ (neck) that enabled them to eat the food. (3) The muscles on the upper part of the body of a blacksmith are well developed due to constant use. C.S.V. / February / 2008 / 1612 (4) The whales lost their hindlimbs as the consequences of the inherited effect of disuse. (5) Snakes have elongated body accompanied by loss of limbs. The continuous creeping through holes and crevices made limbs continuously useless, hence finally degenerated. Besides these, there are other examples, like vestigial organs in living animals due to disuse; claws in carnivores, sensitive skin and tactile points on the ventral side of the body; callosities of palm in hard workers; webbed feet in swimming birds, etc., exemplifying Lamarckian theory. The transmission of acquired characters is the important principle usually termed as the ‘Lamarckian doctrine’. The acquired characters use hereditary or not, i.e., whether any structural change induced, in the body by use or disuse or by a change in surrounding environment can affect the germplasm in such a way that offspring will acquire these structural modifications or not, is debatable. and every generation. This clearly proved that acquired characters are not inherited. (5) Experiment by Castle and Phillips—Recently Castle and Phillips performed transplantation experiments to show the environment has no effect on heredity. They took a black female guinea pig, its ovary was removed and transplanted into the body of white female guinea pig. The white female (with the ovary from black female) was mated with a white male guinea pig. He found that all the offsprings of this pair were black. This clearly showed that environment has no effect on heredity. Hence, this again disapproves Lamarckism. However some later workers have found that in certain cases environment has some effect on heredity. Likewise, there are numerous cases non-hereditary nature of acquired characters. Hence, it is obvious that variations due to mutilations and diseases are not inherited; otherwise, non of us would have existed without some trace of hereditary crippling. Significance of Lamarckism Lamarckian theory was simple and it had some appeal, as it provided a way in which changes in organisms could come about. It was the first completely comprehensive mechanistic theory that was offered. Furthermore, it was the theory that lent itself to predictions and, therefore, to testing. Thus, Lamarckian theory enjoyed popular acceptance for near about 70 years, because it was exemplified by many common examples. Most persons know that exercise results in larger muscles. Neo-Lamarckism The evolutionist who support the Lamarckian doctrine of inheritance of acquired characters come under the heading of Neo-Lamarckism. Among the notable supporters of New-Lamarckism are Cope, Giard, Packard, Spencer and McBride. They tried to modify Lamarckism in order to make it acceptable. These Neo-Lamarckians considered that adaptation is universal. It arises as a result of casual relationship of structure, function and environment. Changed environmental conditions alter habits of organisms, hence, in response to new habits, organisms acquire new structures in place of old structures. Consequently, variations among animals have become distinct. These variations have finally become engrained in the heredity of the race. The kind of argument is, in fact, a modified version of Lamarckian principles, because it has omitted Lamarck’s view that of general perfecting tendency in evolution. Thus modern modified form of Lamarckism is known as Neo-Lamarckism. Weismann’s theory is also not very satisfactory. In some cases the characters are also transmitted through somatic cells, e.g., regeneration in some animals like earthworms and vegetative propagation in plants. It means that somatic cells also contain all the characters. Environment can influence the form, size colour etc., of the organism, which may be inherited to the next generation. Tower exposed the young ones of potato beetles to abnormal conditions of temperature and moisture and allowed them to breed. He observed marked changes in the next generation of beetles. Morgan exposed normal fruit flies (Drosophila) to X-rays in a particular surrounding and found some remarkable changes in the offspring, which were also transmitted to the next generation. Agar reared water fleas in a culture of green flagellates and found that some abnormalities were developed in their structures. The parthenogentic eggs of such individuals when kept in ordinary water and allowed to hatch, produced individuals with the same abnormalities. Guyer and Smith broke the lens of eye a male rabbit by means of a needle. The surrounding blood capillaries penetrated the damaged area. The made rabbit Criticism of Lamarckism The Lamarckian doctrine has been criticised much. Cuvier and Weisman were the great critics of Lamarckism. The following are the main objections against the inheritance of acquired characters— (1) Experiment by Weismann—August Weismann, a German biologist was the main opposer of the inheritance of acquired characters. He put forward the theory of continuity of germplasm. According to Weismann, the characters influencing the germ cells are only inherited. There is a continuity germplasm (Protoplasm of germ cells) but the somatoplasm (protoplasm of somatic cells) is not transmitted to the next generation, hence it does not carry characters to next generation. Weismann cut off tails of rats for about 80 generations and allowed the mutilated parents to breed, but tailless rats were never born. It means that acquired character was not inherited. (2) Experiment by Loeb—Loeb produced artificial parthenogenesis in the sea urchin’s egg with the help of chemical stimuli. Similar eggs, were produced in corresponding environments in brine shrimp Artemia salina but none of their generation showed acquired characters. (3) Indian traditions—Boring of ears and nostrils in Indians has been continued from centuries among human beings, but their offsprings do not show any trace of holes in ears and nostrils. (4) Experiment by Pavlov—Pavlov was a Russian scientist, who wanted to show the inheritance of learning. He trained the mice to come for food on hearing a bell. But it was found that this training was necessary for each C.S.V. / February / 2008 / 1613 was allowed to mate a normal female who gave birth to seven young ones, four of which had defective eyes. It is clear that the antibodies induce mutation in genes controlling eye development. It indicates that the acquired characters can be inherited, if the body changes produce antibodies which in turn induce changes in the genes concerned with the character. The radioactive rays, certain chemicals and hormones can influence the chromosomes and genes of the germ cells and can bring about the changes in the next generation. It is also believed that somatic changes may be affecting the germ cells through some means. Thus Neo-Lamarckism proposes that— 1. Environment influences the organism and change its heredity. 2. Some of the variations acquired by an individual can be transmitted to the next generation. 3. Internal vital force and appentency do not play any role in evolution. 4. Only those variations are transferred to the next generation which also affect germ cells or where somatic cells give rise to germ cells. Darwinism Darwin’s Theory of Natural Selection Charles Darwin was an English naturalist. In 1831, Darwin got an opportunity to travel on H.M.S. Beagle (a ship in which Charles Darwin sailed around the world) for a voyage of world exploration. The voyage lasted for five years. During that period Darwin explored the fauna and flora of a number of continents and islands. Later Beagle was sailed to the Galapagos Islands, where Darwin observed great variations among the organisms that lived on these islands. The common birds of Galapagos Islands, the finches were markedly different from the finches of mainland. In fact Darwin took idea from the finches found on the Galapagos Islands for his theory of natural selection. Alfred Russel Wallace, another English naturalist, who travelled widely, and studied the fauna and flora of South America and South East Asia. Evolutionary ideas similar to those of Darwin developed in Wallace’s mind. He expressed these ideas in an essay titled ‘On the Tendency of Varieties of Depart Indefinitely from the Original Type’. Wallace sent his essay to Darwin. By the time Darwin received Wallace’s essay, he had developed a clear idea about organic evolution. The thinking of both Darwin and Wallace in respect of organic evolution was similar. Finally in 1859, Darwin published his observations and conclusions under the name ‘Origin of Species’. Darwin’s this publication became very popular and changed people’s thinking about organic evolution. Conclusion The above objections and evidences on the inheritance of acquired characters are unconvincing and unsatisfactory. The main theories in this regard are ‘Mneme theory of Semon’ and ‘Centro-epigenesis theory of Rignano’. Semon holds that every condition of life or functional activity of organism leaves a permanent record of itself in the form of ‘engramme’. If these conditions are continued for long period, these engrammes affect heredity and become inheritable; but if they are of short duration, they do not inherit. It is clearly established that only those characters, which can affect germplasm (germinal or blastogenic) can be inherited, while other characters (somatic or somatogenic) affecting body cells do not transmit to their offsprings. The Main Features of the Theory of Natural Selection are as Follows (1) Overproduction (Prodigality or Fecundity)—All organisms possess enormous fertility. The productivity of all living organisms is far beyond the ultimate numbers, which can possibly survive. The space and the available food supply remain the same, while their quantity increases enormously, i.e., in geometric ratio. It is the innate desire of all living beings to produce their own progeny for the continuity of race. For example; insects lay hundreds of eggs. Plants produce thousands of seeds. A female rabbit gives birth to six young one in one time. If all the rabbits survived and multiply at this rate, their number would be very large after sometime. It is assumed that elephant is the slowest breeder, which matures at the age of 30 years and lives for about 90 years. Each female gives rise to about six offsprings and if all survived, in 750 years a single pair would produce about 19 million elephants. Thus some organisms produce more offsprings and others produce fewer offsprings. This is called differential reproduction. All this indicates that every species of animals and plants has an extraordinary power of production, and for controlling this over productivity, these should be some efficient check. Darwin and Wallace both recognized the natural check and called it the struggle for existence. (2) Struggle for existence—As a result of geometrical multiplication of individuals, the food and space remain almost constant, with the result that a struggle Differences between Lamarckism and NeoLamarckism Lamarckism Neo-Lamarckism 1. It is the original theory given 1. It is a modification of the by Lamarck. original theory of Lamarck in order to make it more suitable to modern knowledge. 2. The theory lays stress on 2. Neo-Lamarckism does not internal vital force, appegive any importance to tency and use and disuse these factors. of organs. 3. It believes that change in 3. The theory stresses on the environment brings about a direct effect of changed enconscious reaction in anivironment on the orgamals. nisms. 4. According to Lamarckism 4. Normally only those modithe acquired characters are fications are transferred to passed on to the next the next generation which generation. influence germ cells or where somatic cells give rise to germ cells. C.S.V. / February / 2008 / 1614 inevitably follows for existence. This struggle for existence may be between the individuals of the same species (intraspecific struggle) or between different species (interspecific struggle) or due to environmental factors (struggle with the environment of inanimate nature). Consequently it is of three types— (i) Intraspecific struggle—It is the struggle between the individuals of the same species or in closely related forms to gain upper hand. It is the most severe check, because their requirements like food, shelter, breeding places are similar. Many human wars are the examples of intraspecific struggle. Cannibalism (eating the individuals of its own species) is another example of this type of struggle. (ii) Interspecific struggle—This struggle is more frequent occurring between the individuals of different species living together. Individuals of the species compete with individuals of other species, normally for food and shelter. (iii) Environmental struggle—It is the struggle between the organisms and the environmental factors, such as drought, heavy rains, extreme heat or cold, earthquake, disease etc. Thus climate and other natural factors also help in restricting the number of individuals of particular species. (3) Variations—The fact that no two organisms or parts of the organisms are exactly alike, no matter how closely related, is a commonly observed phenomenon. These differences are called variations. It is the basic prerequisite and progressive factor for evolution, because without variations, no change could occur and evolution would be impossible. But all the variations are not significant from evolutionary point of view. Some of them are changes occurred temporarily in the soma of the organisms and are not inherited to the offsprings. Only those variations which can be inherited can take part in the evolution of species. These variations are called heritable variations. Changes occurred in the genes, or the chromosomes of the germplasm are the only heritable variations. Darwin observed the various useful variations which are selected by individuals and thus evolution results. Darwin assumed variations as axiomatic without describing their real nature and origin in plants and animals. Thus according to Darwin, the variations are continuous and those which are helpful in the adaptations of an organism towards its surroundings would be passed on to the next generation, while the others disappear. (4) Survival of the fittest and natural selection— The organisms which are provided with favourable variations would survive, because they are the fittest to face their surroundings, while others are destroyed. Originally it was an idea of Herbert Spencer, who used the phrase ‘the survival of the fittest’ first time. While Darwin named it as ‘Natural Selection’. The survival of the fittest is the result of natural selection, which enforces adaptations. In the course of long periods those best fitted and suitable individuals survive and become adjusted to nature. Nature being the super power selects only those adapted organism which have accumulated variations. Darwin illustra- ted the survival of the fittest by natural selection by taking example of Lamarck’s giraffe. Giraffe showed great variations in the length of their neck and legs. Due to scarcity of grass on the land, these had to eat the leaves of tall trees. Naturally giraffe with long neck and longer legs had an advantage over those with shorter necks and legs, since these could get enough food easily, had better chances of survival and others having shorter necks and legs gradually became extinct. Due to variations and inheritance, the successive generations gradually became better adapted to their environment. These adaptations are preserved and accumulated in the organisms of a species and ultimately lead to the origin of new species. The environment is ever changing due to which further changes occur and new adaptations appear in the organisms. The descendants after several generations become quite distinct from their ancestors. In this way new species appear. Some of the variations exhibited by living things make it easier for them to survive, others are handicaps which bring about the elimination of their possessors. The idea of ‘the survival of the fittest’ is the core of the theory of natural selection. Evidences in Favour of Natural Selection (1) Rate of reproduction—Rate of reproduction is many times higher than the rate of survival in all organisms. (2) Limitation of resources—Food, space and other resources are limited. (3) Struggle for existence—Competition or struggle for existence is seen in all organisms. (4) Abundance of variations—Variations are so abundant in nature that no two individuals of a species are similar, not even the monozygotic twins. (5) Production of new varieties of plants and animals by sexual selection—When man can produce various new varieties of plants and animals in a short period, nature with its vast resources and long time at its disposal can easily produce new species by selection. (6) Mimicry and protective colouration—They are found in certain animals and are products of natural selection. (7) Pedigrees of some animals—Pedigrees of horses, camels and elephants also support the Natural Selection Theory. Objections to the Theory of Natural Selection (Darwinism) Some of the objections to the theory of natural selection, which Darwin explained vaguely are the following— (1) If species have descended as a result of gradations, there should be innumerable transitional stages and the species should not be so well defined as they are. Darwin’s theory stresses upon small fluctuating variations which are to a large extent noninheritable and can play no part in evolution. C.S.V. / February / 2008 / 1615 (2) His theory does not explain the effect of use and disuse and the presence of vestigial organs. (3) He could not explain whether the instincts are acquired and modified through natural selection or not. (4) He did not differentiate between somatic and germinal variations and considered all variations as heritable. (5) Natural selection cannot account for degeneracy. (6) One of the classical objections to natural selection is that new variations would be lost by ‘dilution’ as the individuals possessing them breed with others without them. Darwin indirectly accepted the Lamarckian idea of inheritance of acquired characters in the form of ‘pangenesis hypothesis’, which cannot be accepted in the light of knowledge of genetics. (1) Mutations—Alteration in the chemistry of gene (DNA molecule) is able to change its effect is called gene mutation. They are discontinuous variations which develop due to permanent changes in genotype. Mutation can produce drastic changes or can remain insignificant. There are equal chances of a gene to mutate back to normal. Most of the mutations are harmful or deleterious but not all. Most of the mutants are recessive to normal gene and these are able to express only in homozygous state. Thus mutations produce variations in the offsprings. (2) Variations and heredity—The nature of genetic variations caused by reshuffling of genes during sexual reproduction was very little known at the time of Darwin. The phenomenon of meiosis causes random assortment of genes during synapsis and rearrangement of maternal and paternal chromosomes in both kinds of gametes. Such a reassortment of genes in a large population with large gene pool is the basis of appearance of new organisms. Crossing-over of genes during meiosis also adds to the variations and chromosomal aberrations like inversion, translocation and polyploidy also result in the origin of new species. Heredity is the transmission of characteristics or variations from parent to offsprings, is an important mechanism of evolution. Organisms possessing hereditary characteristics that are helpful, either in the animal's native environment or in some other environment that is open to it, are favoured in the struggle for existence. Consequently, the offsprings are able to benefit from the advantageous characteristics of their parents. (3) Genetic drift—It is the elimination of the genes of certain characters when a section of population migrates or dies of natural calamity. It changes gene frequency of remaining population. (4) Natural selection—If differential reproduction (i.e., some individuals produce abundant offspring, some only a few and still others none) continues for many generations, genes of the individuals which produce more offspring will become predominant in the gene pool of the population. Thus natural selection occurs through differential reproduction in successive generations. Thus the population becomes diverged either from the parent population or from sister populations. (5) Isolation—Isolation is a segregation or separation of populations by some barriers which prevent interbreeding between related organisms. The reproductive isolation between the populations due to certain barriers leads to the formation of new species. (6) Origin of new species—The populations of a species present in different environments and are segregated by geographical and physiological barriers, accumulate different genetic differences due to mutations, recombination, hybridization, genetic drift and natural selection. The populations thus become different from each other morphologically and genetically, and they become reproductively segregated, forming new species. Thus, new species arise by the establishment of some reproductive barrier or isolation which checks the free gene flow among populations of different environments. Darwin’s Pangenesis Theory To explain the inheritance of characters from parents to the offspring, Darwin proposed the theory of pangenesis. It is now totally discarded. The following are the main points of pangenesis hypothesis : (1) All the somatic cells of the body not only multiply by cell division but also give off minute particles throughout their life called as pangenes or gemmules, which wander throughout the body. (2) There gemmules later on become concentrated in the germ cells in both the sexes. (3) Each germ cell is a minute replica of the parent’s body and is capable of developing into the same kind of body even in minute details. (4) Each gemmule in the developing individual regulates the development of the organ from which it originated from the parent. (5) These gemmules are continuously produced at all stages of development. (6) Sometimes, certain gemmules might lie dormant for several generations and then develop. It results in the appearance of characteristics in the offsprings which were born by their ancestors (atavism) and are not visible in the parents. (7) The weakest link in Darwin’s theory of natural selection was his ignorance of the mechanism of heredity. Galton—Made pangenesis hypothesis untenable by presenting several experimental proofs. According to pangenesis theory, every somatic cell is a germ cell producing gemmules, while the actual germ cells are the collection place for the gemmules coming from different somatic cells. Thus this theory has no basis at all and has been replaced by modern theory of germplasm put forward by Weismann. Neo-Darwinism The modern theory of origin of species (evolution) is called modern synthetic theory of evolution. It is combination of mutations, variations, heredity, isolation and natural selection. It is the modified form of Darwin’s theory of natural selection. C.S.V. / February / 2008 / 1616 Differences between Darwinism and NeoDarwinism Darwinism (Natural Selection) 1. It is the original theory given by Charles Darwin (1859) to explain the origin of new species. 2. According to this theory accumulation of continuous variations causes change in individuals to form new species. 3. It believes in the selection of individuals on the basis of accumulation of variations. 4. Darwinism does not believe in isolation. 5. It can explain the origin of new characters. 6. Darwinism cannot explain the persistence of certain forms in the unchanged condition. Neo-Darwinism 1. Neo-Darwinism is a modification of the original theory of Darwin to remove its shortcomings. 2. Instead of continuous variations, mutations are believed to help form new species. 3. Variations accumulate in the gene pool and not in the individuals. 4. Neo-Darwinism incorporates isolation as an essential component of evolution. 5. The theory cannot explain the origin of new characters. 6. The theory can explain the occurrence of unchanged forms over millions of years. The modern synthetic theory of evolution is the result of the work of a number of workers, namely R. A. Dobzhansky, R. A. Fisher, J. B. S. Haldane, Sewall Wright, Ernst Mayr and G. L. Stebbins. Mutations and natural selection both are important for organic evolution. Darwin and DeVries in their theory of organic evolution did not consider the other factors which were essential beside natural selection and mutation for organic evolution. Stebbins in his book ‘Process of Organic Evolution’ discussed the synthetic theory and other factors which were responsible for evolution. Significant Facts of Evolution ● ● ● Alpheus S. Packard Jr. was the first to use the term ‘Neo-Lamarckism’. G. J. Romanes coined the term ‘Neo-Darwinism’. Dollo’s rule state that evolution is irreversible. The generalization that evolution does not proceed back along its own path, or repeat routes. Cope’s rule states that there is a tendency for animals to increase in size during the long course of evolution. Bergmann’s rule states that in geographically variable species of Homoiothermic animals, body size tends to larger in cooler regions of a species range. Allen’s rule states that the extremities (tail, ears, feet, bill) of endothermic animals tend to the relatively smaller in cooler regions of a species range. Gause’s rule or the competitive exclusion rule states that two species having the same ecological requirements cannot continue to occupy indefinitely the same habitat. Gloger’s rule states that among warm blooded individuals living in warm and wet climates, develop more melanin pigment, and are darker than those living in cold climates. Jordan’s rule states that fishes inhabiting water of low temperature tend to have more vertebrae. The sum total of all genes in a breeding population is called gene pool. ● ● Difference between Neo-Darwinism and NeoLamarckism Neo-Darwinism Neo-Lamarckism 1. The theory explains that the 1. It is silent about high biotic number of organisms of potential and geometrical different species remain the increase in population. same despite their high biotic potential and ability to increase by geometrical ratio. 2. It stresses the role of 2. The theory does not touch struggle for existence and these aspects of evolunatural selection in face of tionary forces. limited resources. 3. It explains the role of varia- 3. The theory considers that tions, their origin and accuthe change in environment mulation in the formation of produces modifications new species. directly due to its effect on germ cells or rarely indirectly through somatic cells. ● ● ● ● ● OBJECTIVE QUESTIONS 1. The key point in Lamarck’s view of organic evolution is— (A) Origin of species (B) Inheritance of acquired characters (C) Overpopulation (D) Natural selection 2. Co-discoverer was— (A) Ruskin (C) Malthus of Darwinism 4. Who published the book ‘Origin of Species by Natural Selection in 1859 ? (A) Lamarck (B) Oparin (C) Wallace (D) Darwin 5. The ultimate source of organic evolution is— (A) Mutation (B) Sexual reproduction (C) Natural selection (D) Hormonal action 6. Phenomenon of ‘Industrial Melanism’ demonstrates— (A) Natural selection (B) Induced mutation (C) Geographical isolation (D) Reproductive isolation 7. The unit of natural selection is the— (A) Individual (B) Family (C) Population (D) Species 8. The Darwinian fitness of an organism is a measure of— (A) Its ability, relative to others in the population, to pass its genes to the next generation (B) The number of offspring it produces (B) Buffon (D) Wallace 3. Galapagos islands are associated with the name of— (A) Wallace (B) Malthus (C) Darwin (D) Lamarck (Continued on Page 1623 ) C.S.V. / February / 2008 / 1617 Model Paper for Various Medical Entrance Examinations ZOOLOGY 1. Parapatric speciation is called— (A) Demes (B) Cline (C) Clisere (D) None of these 2. Which particular fatty acid is not synthesized in the human body ? (A) Linoleic acid (B) Glycerol (C) Cholesterol (D) None of the above 3. Outermost sheath of connective tissue that surrounds a skeletal muscle is— (A) Epimer (B) Epimere (C) Epimerite (D) Epimysium 4. The ‘soft spot’ on the top of an infant’s skull is called— (A) Suture (B) Fontanel (C) Ligament (D) Fascia 5. Dohle’s bodies are associated with— (A) Neutrophils (B) Erythrocytes (C) Platelets (D) None of these 6. People who regularly run, jog or perform other aerobic exercises often develop a painful condition called— (A) Tendinitis (B) Shinsplints (C) Sprain (D) All of these 7. Which of these displays immune tolerance ? (A) B-cells (B) T-cells (C) Both (A) and (B) (D) None of these 8. Cardiac output is determined by— (A) Heart rate (B) Stroke volume (C) Both (A) and (B) (D) None of the above 9. The cycle of reactions that include glycogen conversion to lactic acid in muscles and lactic acid conversion to glycogen in liver is called— (A) Glucosamine pathway (B) Cori cycle (C) Glyoxylate cycle (D) Glycolysis 10. Septicemia is— (A) Food poisoning (B) Blood poisoning (C) Mental disorder (D) All the above 11. When enzyme molecule transforms substrate molecules per unit time under maximum activity, it is called— (A) Biological oxidation (B) Specific activity (C) Turnover number (D) Oxidation number 12. Large arteries have more elastic connective tissue and less smooth muscle, such as— (A) Aorta (B) Arterioles (C) Capillaries (D) All the above 13. The disease caused by the deficiency of parathormone hormone is— (A) Cretinism (B) Tetany (C) Hypercalcemia (D) Myxoedema 14. Staph food poisoning is related with— (A) Salmonella bacteria (B) Clostridium bacteria (C) Staphylococcus bacteria (D) None of the above 15. Manas biosphere reserve is famous for— (A) Lions (B) Elephants (C) Rhino (D) Wild buffalo 16. Which of the following helps in consolidation of long-term memory ? (A) Medulla and Pons (B) Hippocampus (C) Both (A) and (B) (D) None of the above 17. Which of the following is liverproduced growth-promoting anabolic peptides of vertebrate blood serum ? (A) Somatostatin (B) Somatomedins (C) Somatotrophin (D) None of these 18. In females, which of the following is released during ovulation ? (A) Primary oocyte (B) Secondary oocyte (C) Oogonium (D) All the above 19. Chelicerate arthropods include— (A) Merostomata (B) Arachnida (C) Both (A) and (B) (D) None of these 20. The skin of human foetus is covered with thin silky hairs, called— (A) Fontanel (B) Lanugo (C) Primary skin (D) None of the above 21. The disease ‘itai-itai’ is related with— (A) Fragile bones (B) Minamata (C) Respiratory disorder (D) None of these 22. Polygenic inheritance occurs when— (A) Several separate genes control a single phenotypic trait (B) A single gene controls a single phenotypic trait (C) Many genes control many phenotypic traits (D) None of the above 23. L-arabinose is— (A) Restrictive enzyme (B) Pesticide (C) Monosaccharide (D) None of these 24. The reaction in which glucose and fructose combine to form sucrose and water is— (A) Exergonic (B) Endergonic (C) Spontaneous (D) Both (A) and (B) C.S.V. / February / 2008 / 1618 25. The release of which hormone is inhibited when the stomach acidity reaches pH 2 ? (A) Secretin (B) Gastrin (C) Cholecystokinin (D) All the above 26. Sum of constructive processes in body cells is called— (A) Catabolism (B) Anabolism (C) BMR (D) All the above 27. Most of the chemical digestion and absorption takes place in— (A) Jejunum (B) Ileum (C) Colon (D) Both (A) and (B) 28. Ratio of oxyhaemoglobin and haemoglobin in the blood is based upon— (A) Oxygen tension (B) Carbon dioxide tension (C) Carbonate tension (D) Bicarbonate tension 29. Which part of the brain regulates food intake ? (A) Pons varoli (B) Hypothalamus (C) Medulla oblongata (D) Crura cerebri 30. Dissociation curve is shifted to right side when there is— (A) Rise of pH (B) Fall of pH (C) Decreased CO2 concentration (D) None of the above 31. Which of the following secretes HCG hormone ? (A) Placenta (B) Pituitary gland (C) Ovary (D) Adrenal gland 32. β-oxidation is the process, which occurs during— (A) Fatty acid oxidation (B) Protein digestion (C) Carbohydrate oxidation (D) Nucleic acid oxidation 33. Camouflage of Chameleon is associated with— (A) Chromomere (B) Chromoplast (C) Chromatophore (D) Chromosome 34. The free energy in the breaking down of glucose into CO2 and H2O is— (A) (B) (C) (D) Negative and exergonic Positive and endergonic Only endergonic None of the above 43. Achondroplasia is a disease related with the defect in the formation of— (A) Membrane (B) Mucosa (C) Cartilage (D) All of these 44. Fabeliae bones are associated with— (A) Angular joints (B) Elbow joints (C) Knee joints (D) Neck region 45. Amoeba ingests food by— (A) Circumfluence (B) Circumvallation (C) Invagination (D) All the above 46. If 30% of an organism’s DNA is thymine, then— (A) 70% is purine (B) 20% is guanine (C) 30% is adenine (D) Both (B) and (C) are correct 47. Which of the following epidermal layer cells die and worn off in humans ? (A) Stratum malpighi (B) Stratum lucidum (C) Stratum granulosum (D) Stratum corneum 48. The extra embryonic membranes of mammalian embryo are derived from— (A) Formative cells (B) Trophoblast (C) Follicle cells (D) Inner mass cells 49. Which of the following gland is present in arms pit and pubic region in humans ? (A) Ecrine (B) Apocrine (C) Merocrine (D) None of the above 50. Which of the snake is viviparous ? (A) Krait (B) Viper (C) Cobra (D) Hydrophis 35. Synovial fluid is found in— (A) Freely moveable joints (B) Around the brain (C) Intercellular spaces (D) Internal ear 36. Parasympathetic effect is/are— (A) Lowers blood pressure (B) Slows heart rate (C) Promotes digestion (D) All the above 37. Pashmina wool is obtained from— (A) Sheep (B) Goat (C) Rabbit (D) Deer 38. Ecotone is characterised by— (A) Transition zone between two vegetational types (B) Terrestrial ecosystem (C) Zone of transition between water and land (D) Forest ecosystem 39. Desmosomes are related with— (A) Cell excretion (B) Cell adherence (C) Cell division (D) Cytolysis 40. Laennec’s disorder is related with— (A) Liver cirrhosis (B) Hepatitis (C) Jaundice (D) All the above 41. Action potential on outer surface of plasma membrane is— (A) Negative (B) Positive (C) Neutral (D) Variable 42. Heart pumps only impure blood in case of— (A) Shark fish (B) Whale (C) Frog (D) Rat ANSWERS WITH HINTS 1. (A) Parapatric speciation is that speciation which occurs in small, local populations, called demes. 2. (A) Fat can be synthesized in the body from dietary carbohydrates and proteins, but a particular fatty acid called linoleic acid C.S.V. / February / 2008 / 1619 must be obtained from the food, because it cannot be synthesized by the body. 3. (D) Epimysium is the outermost sheath of connective tissue that surrounds a skeletal muscle. It consists of irregularly distributed collagenous, reticular and elastic fibres, connective tissue cells and fat cells. 4. (B) In a newborn infant, the bones of the skull are still developing and there are relatively wide spaces between them. The spaces contain tough sheets of connective tissue that connect the bones and are called fontanels. 5. (A) Dohle’s bodies are leukocyte inclusions in the periphery of neutrophils. They are present in association with burns, infections, trauma and neoplastic diseases. 6. (B) 7. (C) Immune tolerance is acquired inability to react to particular selfor non-self antigens. Both B-cells and T-cells display tolerance generally to their specific antigen classes. 8. (C) Cardiac output is the volume of blood pumped by each ventricle in 1 minute. It is calculated by multiplying the heart rate by the stroke volume. 9. (B) Cori cycle refers to carbohydrate metabolism, the breakdown of muscle glycogen, with formation of lactic acid which enters the bloodstream, and is converted to liver glycogen, which in turn breaks down into glucose. This glucose is carried to muscle where it is converted muscle glycogen. 10. (B) Blood poisoning is a common name for an infection of blood that is also called septicemia or toxemia. 11. (C) Number of substrate molecules catalysed per minute by a single enzyme molecule is called turnover number (molecular activity). 12. (A) Large arteries, such as the aorta, have more elastic connective tissue and less smooth muscle. This arrangement allows them to expand and receive the surge of blood pumped with each heart-beat. 13. (B) Hormone parathormone is secreted by parathyroid glands, its deficiency causes tetany and muscle cramps. 14. (C) Staph food poisoning, once known as ptomaine poisoning, results from toxins produced by staphylococcus bacteria growing on food. 15. (B) 16. (B) All informations first enters the brain as short-term memories. Then in a process called consolidation, some short-term memories are transformed to long-term storage. The hippocampus region of the limbic system is associated with consolidation. 17. (B) Somatomedins are liver produced growth promoting (anabolic) peptides of vertebrate blood serum whose synthesis and activity depends upon the presence of growth hormone and whose activity the mediate. 18. (B) As oogenesis continues, a diploid primary oocyte undergoes meiosis I, producing a functional cell called a secondary oocyte and a small nonfunctional polar body. The polar body degenerates immediately. The secondary oocyte begins meiosis II but stops midway in the process at metaphase II. It is this secondary oocyte that is the immature ovum released from the ovary at ovulation. 19. (C) Chelicerates are natural assemblage containing those arthropods with chelicerate and includes Merostomata and Arachnida. 20. (B) 21. (A) The disease ‘itai-itai’ is caused by cadmium poisoning and causes fragile bones. 22. (A) 23. (C) 24. (B) In the reaction of glucose and fructose to form sucrose and water requires absorption of energy from the surroundings, therefore, called endergonic ? 25. (B) Normally gastrin release is inhibited when stomach acidity reaches pH 2, as a result the further release of HCl and pepsinogen is reduced or stopped. 26. (B) 27. (D) 28. (A) Haemoglobins are adapted for maximal loading and unloading of oxygen within the oxygen tension ranges. 29. (B) Food intake is regulated by the hunger centre and the satiety centre present in the hypothalamus. 30. (B) When pCO 2 and acidity increases due to fall in pH, oxyhaemoglobin dissociates into oxygen and haemoglobin and the sigmoid curve shifts to right. 31. (A) Placenta is also an endocrine organ, secreting HCG, progesterone, and estrogen. HCG sustains progesterone production by the corpus luteum in the mother’s ovary. 32. (A) 33. (C) Chromatophores lie in the skin with permanent radiating processes containing pigments that can be concentrated or dispersed with in the cell under nervous and/or hormonal stimulation, affecting colour changes. They often result in camouflage in Chameleon. 34. (A) 35. (A) 36. (D) The endings of parasympathetic nerves release acetylcholine, which slows heart rate, lowers blood pressure and promotes digestion. 37. (B) Pashmina wool is obtained from the mountain goat. This animal is found in Laddakh and Tibet. 38. (A) Ecotone is the transition between two or more diverse communities, as between forest and grassland. 39. (B) 40. (A) Cirrhosis is a chronic disease of liver. When severe enough, this leads to ammonia toxicity. Most common form is that of nutritional, also known as alcoholic, Laennec’s or portal cirrhosis. 41. (A) 42. (A) 43. (C) Achondroplasia is a defect in the formation of cartilage at the epiphyses of long bones, producing a form of dwarfism. 44. (C) 45. (D) 46. (D) Within each species, however, DNA has the constancy required of the genetic material. (Continued on Page 1642 ) C.S.V. / February / 2008 / 1620 Model Paper for Various Medical Entrance Examinations (C) Both (A) and (B) (D) None of these 16. Which of the following attacks and destroys the large endoparasites, which are too large to be engulfed by phagocytes ? (A) Neutrophils (B) Eosinophils (C) Lymphocytes (D) Monocytes 17. Colustrum is secretion from mammary glands during the first few days after the birth of a child, is the important source of— (A) Passive immunity (B) Active immunity (C) Both (A) and (B) (D) None of these 18. Which of the following cellular layer undergoes breakdown and regeneration ? (A) Corneal layer of eyes (B) Endometrium of uterus (C) Both (A) and (B) (D) None of these 19. Which of these is bile salt ? (A) Glycocholate (B) Sodium taurocholate (C) Both (A) and (B) (D) Bilirubin 20. Atocia is— (A) Female sterility (B) Nulliparity (C) Both (A) and (B) (D) None of these 21. Mammalian cervical vertebrae can be identified by the presence of— (A) Odontoid process (B) Transverse process (C) Amphiplatyon centrum (D) Large neural canal 22. Mesothelioma is— (A) Liver cancer (B) Skin cancer (C) Lung cancer (D) Breast cancer 23. Perforated and plate-like mitochondrial cristae are found in— (A) Paramecium (B) Flight muscles of certain insects (C) Brown fat cells of rats (D) Liver cells in rats ZOOLOGY 1. Mitosome is associated with— (A) Spermatozoon (B) Liver (C) Brain (D) Pancreas 2. The application of genetic principles for the improvement of mankind is— (A) Genetic engineering (B) Biotechnology (C) Eugenics (D) Anthropology 3. Cavities unguium is also known as— (A) Leukonychia (B) Leukopathia (C) Leukopedesis (D) Leukonecrosis 4. The common name of octopus mollusc is— (A) Devil fish (B) Cuttlefish (C) Squid (D) None of these is often 8. Viral disease Trachoma is associated with— (A) Skin (B) Liver (C) Eyes (D) Muscles 9. Which organelle in the cell, other than nucleus, contains DNA ? (A) Centriole (B) Golgi apparatus (C) Lysosome (D) Mitochondria 10. The element phosphorus needed to produce— (A) RNA and DNA (B) ATP and GTP (C) NAD and FAD (D) All the above is 11. Blood vessels dilate and become more permeable in response to which of the following ? (A) Histamine (B) Interferons (C) Pyrogens (D) Antibodies 12. The study of all the living and relevant non-living components of a natural community and their relationship with each other, is called— (A) Synecology (B) Autecology (C) Paraecology (D) Synergism 13. Introduction of Gambusia fish in a pond to control mosquitoes, is an example of— (A) Mechanical control (B) Biological control (C) Systemic control (D) None of these 14. The toxins produced by tetanus microbes affect— (A) Voluntary muscles (B) Involuntary muscles (C) Both (A) and (B) (D) None of these 15. Excess of which of the following in human body is excreted in urine ? (A) Fat-soluble vitamins (B) Water-soluble vitamins 5. Kaposi’s sarcoma associated with— (A) AIDS (B) Herpes (C) Hepatitis-B (D) All the above 6. Which of these derives most of its water requirement from the metabolism and does not drink water ? (A) Kangaroo rat (B) Marine fishes (C) Kangaroo (D) Penguins 7. A form of bioregulation in which one cell type of tissue influences the activity of an adjacent cell type, is called— (A) Endocrine control (B) Exocrine control (C) Paracrine control (D) None of these C.S.V. / February / 2008 / 1621 24. The first organ receiving nutrient rich messenteric blood from intestine after absorption of chyle is— (A) Liver (B) Heart (C) Brain (D) Lungs 25. Which of the following ecosystems has the highest gross primary productivity ? (A) Coral reef ecosystem (B) Mangroves ecosystem (C) Rain forest ecosystem (D) Grassland ecosystem 26. Lacteals absorb— (A) Lactic acid (B) Lactose (C) Amino acids (D) Fatty acids 27. Which of the following only amino acid is metabolized by brain ? (A) Alanine (B) Glutamic acid (C) Glutamine (D) Histidine 28. Drones in a colony of honeybees originate by— (A) Cyclic parthenogenesis (B) Arrhenotoky (C) Thelotoky (D) Diploid parthenogenesis 29. Which of the following snakes makes nest ? (A) Viper (B) Krait (C) Coral snake (D) King cobra 30. ‘A tube within a tube’ body plan is met within— (A) Hydra (B) Planaria (C) Ascaris (D) Fasciola 31. Infection of Trypanosoma gambiense is initiated in a healthy person by— (A) Metacyclic form (B) Stumpy form (C) Long slender form (D) Intermediate form 32. Foetal sex can be determined by examining cells from amniotic fluid, looking for— (A) Chiasmata (B) Barr body (C) Sex chromosome (D) Kinetochores 33. The euphoric condition achieved by joggers called ‘runner’s high’ and is thought to be produced by high concentration in blood of— (A) Endorphins (B) Ach (C) Norepinephrine (D) Dopamine 34. Tectum is dominant brain centre in— (A) Fishes (B) Amphibians (C) Mammals (D) Both (A) and (B) 35. Which of the following mollusc is cultivated for producing pearls ? (A) Pinctada (B) Haliotis (C) Anodonta (D) Mytilus 36. Which of these do not exist free in human body ? (A) Triglycerides (B) Fatty acids (C) Chloesterol (D) All the above 37. The end of spinal cord in humans is called— (A) Cauda equina (B) Conus terminalis (C) Filum terminale (D) Funiculus 38. Heart lacks sinus venosus in— (A) Amphibians (B) Fishes (C) Mammals (D) None of these 39. Hay’s test is conducted detect— (A) Bile pigments (B) Bile acids (C) Cholesterol (D) Cholecystokinin to 42. Canities unguium is also known as— (A) Leukonychia (B) Leukonecrosis (C) Leupathia (D) Leukopedesis 43. Which of the following represents class Mastigophora ? (A) Monocystis (B) Paramecium (C) Trypanosoma (D) Amoeba 44. Which of the following is/are ketone bodies ? (A) Acetone (B) Beta-hydroxy butyric acid (C) Acetoacetic acid (D) All the above 45. Goldschmidt has classified evolution into micro-, macro-, and mega-evolution. Which one of the following is referred to as micro-evolution ? (A) Evolution at variety level (B) Evolution at subspecies level (C) Evolution at species and genus level (D) Evolution at family level 46. Loss of ability to speak due to defect or paralysis of the vocal organs is called— (A) Alalia (B) Alaorina (C) Alate (D) Albidus 47. Inherited Rh gene is found in— (A) Rh + individuals (B) Rh – individuals (C) AB blood group individuals (D) O blood group individuals 48. Which of the following lacks blood vessels ? (A) Connective tissues (B) Epithelial tissues (C) Muscular tissues (D) None of these 49. Which constituent of certain flavoproteins function as coenzyme in cellular oxidation ? (A) Thiamine (B) Riboflavin (C) Cyanocobalamin (D) Tocopherol 50. Inducible enzyme is found in— (A) Certain vertebrates (B) Microorganisms (C) Mammals (D) None of these 40. Dane particles are associated with— (A) Hepatitis A (B) Hepatitis B (C) Both (A) and (B) (D) None of these 41. Which of the following insect is vector for Bubonic plague ? (A) Xenopsylla (B) Cimex (C) Pediculus (D) Phlebotomus C.S.V. / February / 2008 / 1622 ANSWERS WITH HINTS 1. (A) Mitosome is a body giving rise to the middle piece of the spermatozoon. 2. (C) 3. (A) White spots or streaks on the nails is called leukonychia, also known as cavities unguium. 4. (A) 5. (A) Kaposi’s sarcoma is multiple area of cell proliferation initially in the skin and eventually in other body sites. These lesions may eventually become sarcomatous. It is felt to be related to the immunocompromised state that accompanies AIDS. 6. (A) Different animals use different combinations of strategies to conserve water. A kangaroo rat does not drink water and derives most of its water requirement from metabolism. 7. (C) Paracrine control is a general form of bioregulation in which one cell type in a tissue selectively influences the activity of an adjacent cell type by secreting chemicals that diffuse into the tissue and act specifically on cells of that area. 8. (C) 9. (D) Mitochondria are cytoplasmic organelles of all eukaryotic cells. Mitochondrion contains DNA. Mitochondrial genetic code differs from both nuclear and bacterial codes. 10. (D) Phosphorus enters into the composition of complex molecules such as RNA, DNA, ATP, GTP, NAD, FAD etc. 11. (A) Histamine is potent vasodilator, formed by decarboxylation of the amino acid histadine and released by mast cells. 12. (A) 13. (B) Mosquito larvae in ponds are destroyed by introduction of larvivorous Gambusia fish. This method of control is called biological control. 14. (A) 15. (B) 16. (B) One function of eosinophils is to attack large endoparasites, which cannot be engulfed by phagocytes. Numerous eosinophils cluster around a large endoparasite and secrete powerful digestive enzymes to destroy it. 17. (A) 18. (B) 19. (C) Bile contains bile salts, such as glycocholate and sodium taurocholate. Bile also contains pigments such as bilirubin. 20. (C) 21. (D) Cervical vertebrae are present in neck region, which form a flexible portion of the vertebral column in mammals and are usually seven in number. They are all characterized by relatively large neural canal. 22. (C) 23. (B) 24. (A) 25. (A) Due to availability of large number of autotrophs per unit area, coral reef ecosystem has highest gross primary productivity. 26. (D) 27. (B) Glutamic acid is an amino acid formed in the hydrolysis of proteins. It is the only amino acid metabolized by brain. 28. (B) 29. (D) 30. (C) 31. (A) The crithidial forms of Trypanosoma gambiense multiply in the lumen of salivary glands of Tse-tse fly and transform into metacyclic forms. When this fly bites a healthy person, it transfers the metacyclic forms along with saliva into blood where they initiate infection. 32. (B) 33. (A) 34. (D) 35. (A) Pearl oysters are sedentary bivalve mollusc. The Indian pearl oyster is Pinctada vulgaris that is known for producing precious pearls. 36. (B) 37. (C) Filum terminale is a long, slender filament at the end of spinal cord. 38. (C) 39. (B) Hay’s test is conducted to detect bile acids in urine. 40. (B) Dane particles are present in the serum of patients with hepatitis B. They contain hepatitis B surface (HBa Ag) and core (HBc Ag) antigens. 41. (A) Xenopsylla is a genus of fleas belonging to the family pulicidae, order siphonoptera. Xenopsylla cheopsis is a vector of bubonic plague. 42. (A) 43. (C) Trypanosoma is flagellate genus and belonging to subclass Zoomastigophora of class Mastigophora. 44. (D) 45. (C) 46. (A) 47. (A) 48. (B) Epitheliel tissues lack blood vessels and obtain nutrition by diffusion from neighbouring tissues. 49. (B) Riboflavin is vitamin B2 and is a constituent of flavoproteins that function as coenzymes in cellular oxidation and is essential for tissue repair. 50. (B) Inducible enzyme is also called adaptive enzyme. The enzyme is produced by an organism only in the presence of its substrate. Adaptive enzymes are known only in microorganisms. ●●● (Continued from Page 1617 ) (C) Its life span (D) Its physical vigour 9. The unit of evolution is now known to be the— (A) Individual (B) Family (C) Population (D) Species 10. The idea of common descent was first suggested to Darwin by his observations on— (A) Geographic distribution of species (B) Human pedigree (C) Comparative embryology (D) All the above 11. According to Darwin, two different areas within a continent have different species because they have different— (A) Evolutionary mechanisms (B) Ancestors (C) Environments (D) Evolutionary times 12. The struggle for existence is a consequence of— (A) Each organism leaving more offspring than needed to replace itself (B) Innate competitive tendencies (C) The inevitable difficulty of coping with climatic conditions (D) Territories and dominance hierarchies ANSWERS 1. (B) 2. (D) 6. (A) 7. (A) 11. (C) 12. (A) 3. (C) 4. (D) 5. (A) 8. (A) 9. (C) 10. (A) ●●● C.S.V. / February / 2008 / 1623 Topic on Botany FAT (LIPIDS) METABOLISM —Kumar Pushkar Introduction ● Fats, or lipids are best defined as a diverse group of organic compounds found in plants, animals and micro-organisms. Fats are produced in all actively metabolising plant cells and serve a number of indispensable roles in plants, particularly as reserve food substance. Fats and certain other compounds, more or less closely related, are often called lipids. Characteristically, they are greasy to touch and insoluble in water but soluble in alcohol, ether, acetone and other organic solvents. The proportion of lipids in food-stuffs varies from 0·2% in white potatoes to 70% in some nut kernels. Fats, such as olive oil, contain a mixture of fatty substances called triglycerides. Lipids made by cells are important not only because they serve as an energy source but also because they form the structural components. ● ● It may be of the following types— (i) Phospholipids Phospholipids are fats containing phosphoric acid. Lecithin, the best known among phospholipids in plant cells, is an essential structural material for living cell membrane, where it maintains continuity between the water and lipid phases inside and outside the cell. The function of certain enzymes depends upon their attachment to lipids such as lecithin. (ii) Galactolipids Galactolipids are the major lipid constituents of greenleaf tissue. Monogalactosyl diglyceride and diglactosyl diglyceride are commonly occurring galactolipids in chloroplasts. ● ● ● ● ● ● ● ● ● Distribution of Fats in Plants ● Fats are widely distributed throughout the plant kingdom. ● Fats are especially found in abundance in reproductive tissues (e.g., seeds and fruits) of certain higher plants where they form important reserve food material such as in cotyledons of sunflower, rape (Brassica napus), peanuts (Arachis hypogaea) almonds (Prunus amygdalus) and mesocarp of avocado pear (Persea americana). ● Certain cereal seeds such as wheat and barley (Hordeum distichon) which store starch as chief reserve food in their endosperm, have rich fat content in their aleurone cell layers. ● Carnauba wax is found in the leaves of carnauba palm (Copernicia prunifera). Simple Classification of Lipids in Plants 1. Simple Lipids ● Simple lipids are esters of fatty acids with alcohol. These are of following types— (i) Fats ● Fats are esters of fatty acids with glycerol, i.e., fatty acids are esterified with glycerol to form triglycerides. ● These are true fats and are indispensable as reserve food material and as a source of energy in living cells. (ii) Waxes ● Waxes are esters of higher aliphatic fatty acids with long chain of alcohols other than glycerols. ● Sometimes waxes are important components of cuticle of epidermal cell walls. ● Waxes must : be solid at 68°F (20°C); be crystalline; melt above 140° F (40°C) without decomposition; have relatively low viscosity above the melting point; have consistency and solubility properties that are strongly dependent upon temperature. ● Waxes have been isolated from the outer layer of bacteria, the roots, stems, leaves, fruit and flowers of plants. ● Carnauba wax is extracted from an exudate on the leaves of the carnuba palm (Copernicia prunifera ). ● Candelilla wax is obtained from a coating on the stem of Euphorbia antisyphilitica, a leafless desert shrub. 2. Compound Lipids ● Compound lipids are esters of fatty acids containing groups in addition to alcohol and fatty acid radicals. Fatty Acids ● The fatty acids of naturally occurring lipids have an even number of carbon (C) atoms because they are synthesized from acetyl groups, each of which contains two carbon atoms. ● Fatty acids with 16 (palmitic acid) and 18 (stearic acid) carbon atoms are most commonly found in nature. ● Two typical fatty acids are palmitic and oleic fatty acids. Saturated Fatty Acids ● Typical naturally occurring saturated fatty acids are chainlike (non-branched) compound with an even number of carbon atoms. ● Several important short-chain fatty acids include butyric acid (4C–atoms and caproic acid (6C– atoms), octanoic acid (8C–atoms) and decanoic acid (10C–atoms), which are present in palm oil. Major Saturated Plant Fatty Acids Common Name Lauric acid Myristic acid Palmitic acid Stearic acid Arachidic acid Behenic acid Number of C–atoms 12 14 16 18 20 22 C.S.V. / February / 2008 / 1624 Unsaturated Fatty Acids ● Fatty acid molecules containing one or more double — bonds (—CH — CH—) are called unsatured fatty acids. Fatty acids with one double bond are called monounsaturated fatty acids; those with two or more are called polysaturated ones. Monounsaturated fatty acids comprise the largest group of unsaturated fatty acids. Major Unsaturated Fatty Acids of Plants Common Name Palmitic acid Oleic acid Linoleic acid α-Linoleic acid Erucic acid Number of Carbon Atoms 16 18 18 18 22 ● ● Because β-carbon (i.e., carbon atom number 3) of the fatty acid is oxidised, it is called β -oxidation which involves the following sequential steps— (i) The first step involves the activation of fatty acid in presence of an enzyme thiokinase and ATP. CoASH is consumed and CoA derivative of fatty acid is produced. Fatty acid + CoASH → ATP + Mg2+ Thiokinase ● Fatty acyl-CoA + AMP + PPi The AMP (Adenosine monophosphate), thus produced, reacts with another ATP molecule to form ZADP molecule, in presence of an enzyme adenylate kinase. Adenylate kinase AMP + ATP 2ADP Breakdown of Fatty Acids The long chain fatty acids are broken down by the process of α-oxidation and β-oxidation. The latter produces acetyl CoA (2-carbon unit). α -oxidation ● ● The long chain of fatty acid by this process is broken down until it is reduced to 12 C–atoms. Fatty acids with less than 13 C–atoms are not affected by this process. α-oxidation occurs in the following sequential series— (i) In presence of an enzyme fatty acid peroxidase and hydrogen peroxide (H2O2), the fatty acid is oxidatively decarboxylated to form an aldehyde. In this process CO2 comes from carboxylic group and oxidation occurs at α-carbon atom which is converted into aldehyde group. (ii) In presence of an enzyme fatty aldehyde dehydrogenase, the aldehyde is further oxidised to form the new fatty acid containing one less carbon atom than in the original fatty acid. NAD+ is reduced in this reaction. Aldehyde → NAD + H2O Aldehyde dehydrogenase (ii) In this step, two hydrogen atoms are removed between α- and β-carbons, as a result α-, βunsaturated fatty acyl CoA is formed in presence of FAD–containing enzyme acyl–CoA dehydrogenase. (iii) This step involves the addition of a water molecule across the double bond to form corresponding β-hydroxyacyl–CoA in presence of enzyme enol hydrase. (iv) In presence of NAD—specific β-hydroxyacyl– CoA dehydrogenase enzyme, β-hydroxyacyl– CoA is dehydrogenated. Two H–atoms are removed from the β-carbon atom (β-oxidation) which now bears a carbonyl function and β -keto fatty acyl–CoA is formed. (v) This is the final step which involves the thioclastic cleavage of β-keto fatty acyl—CoA in presence of an enzyme β-ketoacyl thiolase to form an active 2C unit acetyl–CoA and a fatty acyl– CoA molecule. The fatty acyl–CoA so produced again re-enters the β-oxidation losing further 2 carbon unit. This sequence continues until whole molecule is degraded. New fatty acid + NADH + H + The new fatty acid will be oxidised again and again till it consists of 12 C–atoms. β-oxidation ● ● β-oxidation is the chief method of fatty acid degradation in plants. It takes place in mitochondria (and also in glyoxysomes) and involves sequential removal of 2C in the form of acetyl CoA molecules from the carboxyl end of fatty acid. Worth Remembering ● ● ● Each turn of β-oxidation generates one FADH2 [step (ii)], one NADH + H+ [step (iv)] and one acetyl–CoA [step (v)]. In the last turn of β-oxidation spiral, two acetyl–CoA molecules are produced. Each turn of β-oxidation generates 5 ATPs. However, in the first turn there is consumption of 2 ATP molecules [in the step (i)] hence, in this turn there will be a net gain of only 3 ATPs. A huge amount of energy is generated in the form of ATP by mitochondrial oxidation of fatty acids through βoxidation spiral and TCA cycle. For example, one molecule of palmitic acid (16–C atom) on complete oxidation will produce 129 ATP molecules. ● C.S.V. / February / 2008 / 1625 Fatty Acid Synthesis ● Although the reactions of β-oxidation of fatty acids are reversible, the fatty acids are not formed by the reverse reactions of β-oxidation. Long chain fatty acids are synthesized in plants from active two carbon units, the acetyl–CoA. Synthesis of fatty acids from acetyl–CoA occurs in sequential steps and takes place in the cytosol at the endoplasmic reticulum. In each step the fatty acid chain is increased by 2C–atoms. Each step involves the following two reaction— (i) In the first reaction, acetyl–CoA combines with CO2 in presence of enzyme acetyl–CoA Carboxylase to form malonyl–CoA. Mn2 + and biotin are required as cofactors, while ATP provides energy during this reaction. Acetyl–CoA + CO2 + ATP → 2+ Biotin‚ Mn Acetyl–CoA carboxylase Malonyl–CoA + Acetyl–CoA → 2 NADPH 2 Fatty acid synthetase Butyryl–CoA + CO2 + H2O + 2NADP Butyryl–CoA, in the next step, will combine with malonyl–CoA to form CoA derivative of fatty acid containing 6C–atoms. This process is repeated till coenzyme–A derivative of long chain fatty acid (which may contain upto 16–18 carbon atoms) is formed. The reaction (ii) described above summarises a number of reactions, involved in the synthesis of fatty acid from acetyl–CoA and malonyl–CoA, which can be grouped in the following heads— (i) Initiation reaction—In this case acetyl–CoA transfers its acetyl group to one of the —SH group of multienzyme complex, i.e., fatty acid synthetase. ● ● (ii) Elongation and termination reactions— ● When the fatty acid residue has attained a desired length the chain elongation stops and the cycle is not repeated. The acyl group instead of being transferred to the —SH of the enzyme is transferred to —SH group of Co-enzyme–A (CoASH). Thus, CoA derivative of fatty acid is produced. Malonyl–CoA + ADP + iP (ii) In presence of an enzyme fatty acid synthetase and coenzyme NADPH2, malonyl–CoA reacts with another molecule of acetyl–CoA to form butyryl–CoA (Coenzyme–A derivative of butric acid). ● OBJECTIVE QUESTIONS 1. Lipids or fats are a diverse group of organic compounds found in— (A) Plants (B) Animals (C) Micro-organisms (D) All of the above 2. In which of the following organic or inorganic compounds, lipids are soluble ? (A) Water (B) Alcohol (C) Ether (D) In both (B) and (C) 3. Triglycerides called— are sometimes house of 5. Which one of the following is a polyunsaturated fatty acid ? (A) Linoleic acid (B) Oleic acid (C) Erucic acid (D) All of the above 6. Palmitoleic fatty acid is obtained from— (A) Olive oil (B) Marine algae (C) Pine oil (D) Both (B) and (C) 7. Triolein is a simple triglyceride since it contains— (A) One type of fatty acid (B) Two types of fatty acids (C) Three types of fatty acids (D) Four types of fatty acids 8. How many fatty acids are found in each molecule of triglyceride ? (A) Two (B) Three (C) One (D) Four 9. The energy in fatty acid molecule is transformed into ATP by a process known as— (A) β-oxidation (B) α-reduction (C) Both (A) and (B) (D) Fatty acid reduction 10. Fatty acid molecules containing one or more double bonds (—CH — —CH—) are called— (A) Saturated fatty acids (B) Unsaturated fatty acids (C) Both (A) and (B) (D) Stearic acid ANSWERS 1. (D) 2. (D) 6. (D) 7. (A) 3. (A) 4. (D) 5. (A) 8. (B) 9. (A) 10. (B) ●●● (A) Nature’s store energy (B) Enzymes (C) Co-factors (D) All of the above 4. Which one of the following is a saturated fatty acid ? (A) Lignoceric acid (B) Palmitic acid (C) Myristic acid (D) All of the above C.S.V. / February / 2008 / 1626 There is no taxonomic category called algae, this term has long been used in biology as a matter of convenience to mean aquatic organisms that carry on photosynthesis. In the ocean and fresh water lakes and ponds, algae are a part of the phytoplankton, which photosynthesize and produce the food that maintains an entire community of organisms. Green algae (Phylum–Chlorophyta) live in the ocean but are more likely found in fresh water and can even be found on land, especially if moisture is available. Some, however, have modifications that allow them on tree trunks even in bright sun. Zygote (2n) Zygospore (2n) Isogametes pairing (n) Fertilization Haploid ( (n) Sexual Reproduction (n) Diploid (2n) Meiosis Green Algae are Most Plant-like Green algae are believed to be closely related to the first plants because both of these groups : (i) (ii) Have a cell wall that contains cellulose. Possess chlorophyll-a and chlorophyll-b Gamete formation Eyespot Nucleus Flagellum Chloroplast Starch granule Pyrenoid Zoospores (n) (iii) Store reserve food as starch inside the chloroplast. (Other types of algae store reserve food outside the chloroplast). Flagellated Green Algae Chlamydomonas is a unicellular green alga usually less than 25 µm long that has been studied in detail. It has a definite cell wall and single, large cup-shaped chloroplast that contains a pyrenoid, a dense body where starch is synthesized. The chloroplast also contains a redpigmented eyespot (stigma), which is sensitive to light where photosynthesis can occur. Two long whiplash flagella project from the anterior end of this alga and curl backward. When growth conditions are favourable, Chlamydomonas reproduces asexually. The adult divides, forming zygospores (flagellated spores) that resemble the parent cell. A spore is a haploid that develops into a mature adult under favourable conditions. Under unfavourable conditions of growth, Chlamydomonas reproduces sexually. Two different mating types of cells, morphologically similar but physiologically different acting as gametes come into contact and unite to form a zygote. Later on, zygote is surrounded by a thick wall and becomes a resistant zygospore. When a zygospore germinates, it produces four zoospores by meiosis. In most species gametes are isogamous (identical). In other species, there may be anisogamy, where gametes of different sizes are fused to form a zygote. Still in other species a nonmotile egg specialized for storing food and the motile sperms are specialized for seeking out an egg. This condition is termed oogamy and the gametes as heterogametes. Asexual Reproduction Zoospore (n) Zoospore formation Fig. : The structure and life cycle of Chlamydomonas, a motile green alga. During asexual reproduction, all structures are haploid; during sexual reproduction, meiosis follows the zygote stage, which is the diploid part of the cycle. Colonial Green Algae A number of colonial forms occur among the flagellated green algae. A Volvox colony is a hollow sphere with thousands of cells arranged in a single layer surrounding the watery interior. Each cell of a Volvox colony resembles a Chlamydomonas cell. In Volvox, the cells cooperate in that the flagella beat in a coordinated fashion. Some cells are specialized for reproduction and each of these can divide asexually to form a new daughter colony. This daughter colony resides for a time within the parental colony, but then it leaves by releasing an enzyme that dissolves away a portion of the parental colony, allowing it to escape. Among these algae sexual reproduction occurs by means of oogamy. C.S.V. / February / 2008 / 1627 Daughter Colonies Fig. : Volvox sp. Parent colony with a number of daughter colonies. Daughter colony Zygote Egg 200 µm Sperm Fig. : Volvox, a colonial green alga. The adult Volvox colony often contains daughter colonies, which are asexually produced by special cells. During sexual reproduction, colonies produce a definite sperm and egg. Filamentous Green Algae Filaments are end-to-end chains of cells that formed after cell divisions in only one plane. Spirogyra a filamentous green alga, is found in green masses on the surfaces of ponds and streams. Being characteristically slimy, due to mucilaginous covering, it is called water silk or pond scum. It possesses ribbonlike chloroplasts arranged as spiral bands within the cell. Conjugation, the temporary union of two individuals during which there is an exchange of genetic material, occurs during sexual reproduction. The two filaments line up next to each other, and the cell contents of one filament move into the cells of other filament, forming diploid zygote. These zygote survive the winter and in the spring they undergo meiosis to produce new haploid filaments (new individuals). In Spirogyra, Cell wall sexual conjugation may be of two types; scalariform and lateral conjugation. Scalariform is of common Nucleus occurrence and takes Pyrenoid place between the opposite cells of the two neighbouring filaments. Lateral Chloroplast conjugation is rarely found and occurs between two adjacent cells of the same filament. Ulothrix is a common, fresh water, thread-like alga Fig. : Spirogyra, a filamentous green alga, in found in rather cold, flowwhich each cell has a ing water. The thallus is an ribbonlike chloroplast. unbranched filament conDuring conjugation the cell contents of sisting of short, cylindric or one filament enter the quadrate uninucleate cells. cells of another filaThe haploid filament reproment. Zygote formaduces both sexually as well tion follows. asexually. Asexual reproduction takes place by vegetative method of fragmentation and sporulation. The spores are chiefly zoospores sometimes aplanospores, both serve to reduplicate the haploid Ulothrix filament. The simple life history of Ulothrix consisting of a haploid plant body (filament) with the zygote as the haploid stage is called haplontic. It is characterized by zygotic meiosis. Is Ulothrix a Colony or a Multicellular Plant ? Nearly all colonies of single-celled algae are spherical or disc-like or irregular. In several species, successive cell divisions occur in parallel planes so that a chain of cells is formed. Such colonies are filamentous. The best example of this type of colony is represented by Spirogyra. In Ulothrix, the basal cell may be colourless, somewhat modified in shape and attached to the substratum. It simply anchors the plant to its substrate, whereas the others are capable of photosynthesis, cell division and form either spores or gametes. Because Rhizoidal of this slight differentiation betcell ween cells, a filament of Ulothrix represents a transition between filamentous colony of single-celled plant and a very simply constructed multicellular plant. Fig. : Ulothrix filament. C.S.V. / February / 2008 / 1628 Oogamy does occur among the filamentous green algae. The genus Oedogonium contains filamentous algae in which the cells are cylindrical with netlike chloroplasts; during sexual reproduction there is a definite egg and sperm. Cap cell formation is also reported in Oedogonium. Antheridia marine cosmopolitan genus. A few species are found in saline or fresh water but grow in profusion in waters polluted by organic matter or sewage. Ulva has an alternation of generations like that of plants except that both generations look exactly alike, the gametes look alike (isogamy), and the spores are flagellated. In plants one generation is typically dominant over (lasts longer than) the other, egg and sperm are produced (oogamy). Spores are without any flagellum. (A) Sperms and eggs are produced (B) Generations look alike (C) Both (A) and (B) (D) Alternation of generation 5. Both lateral and scalariform conjugation occur in— (A) Spirogyra (B) Ulva (C) Volvox (D) Nostoc 6. Ribbon-like chloroplast is found in— Nucleus Vegetative cell Oogonium Egg Suffultory cell Apical caps Gametes Fertilization Zygote (A) Ulva (B) Oedogonium and Ulva (C) Spirogyra Meiosis Diploid (2n) Haploid (n) Plus (+) gametophyte (D) Volvox 7. Which of the following algae shows colonial form ? (A) Chlamydomonas (B) Volvox (C) Spirogyra Spores Minus (–) Gametophyte (D) All of the above 8. Which of the following algae can change direction by regulating the strength ? (A) Volvox (B) Nostoc (C) Spirogyra (D) Chlamydomonas 9. The pyrenoid of Spirogyra is meant for— (A) Protein synthesis (B) Starch synthesis (C) Amino acid synthesis (D) All of the above 10. Which of the following algae is commonly known as sea lettuce ? (A) Chlamydomonas (B) Ulva (C) Spirogyra (D) Volvox Reticulate chloroplast Cell wall Pyrenoid Cap cell Fig. : Ulva Life Cycle. OBJECTIVE QUESTIONS 1. In Chlamydomonas— (A) Sexual reproduction occurs (B) The adult is haploid (C) Spores survive times of stress (D) All of the above are correct 2. Which of these algae are not flagellated ? Basal holdfast Flattened disc Fig. : Oedogonium. A monoecious filament. (A) Chlamydomonas (B) Volvox (C) Spirogyra (D) Dinoflagellates 3. Which of these is not a green alga ? (A) (B) (C) (D) Chlamydomonas Volvox Spirogyra Fucus Multicellular Green Algae Multicellular form of green algae is best represented by Ulva. Multicel lular Ulva is commonly called sea lettuce because of its leafy appearance. The thallus is two cells thick and can be a metre long. Ulva is a ANSWERS 1. (D) 2. (C) 6. (C) 7. (B) 3. (D) 4. (B) 5. (A) 8. (D) 9. (B) 10. (B) ●●● 4. Which of the following characteristic parts is found in Ulva but not in plants ? C.S.V. / February / 2008 / 1629 Occurrence ● Club fungi (division = Basidiomycota 16,000 species), which have septate hyphae, include the familiar mushrooms growing on lawns and the shelf or bracket fungi found on dead trees. Less well known are puffballs, bird’s nest fungi, and stinkhorns. All the above structures are fruiting bodies called basidiocarps. Basidiocarps contain the basidia, club-shaped structures which produce basidiospores and from which this division takes its name. This big group of fungi includes both saprophytic and parasitic species. ● The clamp connections are usually formed on the terminal cells of the hyphae of the secondary mycelium. The presence of a hook-like clamp connection is a safe criterion for distinguishing a secondary or dikaryotic mycelium from the primary or monokaryotic mycelium. The clamp connections by some mycologists are considered homologous to the hooks of ascogenous hyphae of the ascomycetes. ● ● ● ● ● ● Diploidisation (Dikaryotisation) ● The process by which the primary mycelium is converted into secondary mycelium or dikaryotic mycelium is called diploidisation or dikaryotisation. Diploidisation may take place by the following methods— (i) By hyphal fusion : In this case fusion occurs between the vegetative cells of two neighbouring hyphae. Mycelium ● The well-developed filamentous mycelium consists of a mass of branched, septate hyphae generally spreading in a fan-shaped manner. The mycelium generally is a weft of interlacing and anastomasing 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 of macroscopic size which are called rhizomorphs. The mycelium of Basidiomycetes passes through three distinct stages—primary, secondary and tertiary before the fungus completes its life cycle. 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 (–) strain. ● ● (ii) By conjugation of basidiospores : In this case two basidiospores of opposite strains meet and conjugate. (iii) By the fusion between a germinating basidiospore and a diploid cell of the basidium. Secondary mycelium Clamp connection ● 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. Dikaryon Primary mycelium A B C D E F Fig. : (A–F). Basidiomycetes. Diagram illustrating the formation of clamp connections. Fig. : Basidiomycetes. Sketch showing the formation of a secondary mycelium from a dikaryotised cell produced by somatogamous copulation between two uninucleate cells of primary mycelia of opposite strains. C.S.V. / February / 2008 / 1630 (iv) By the fusion between germinating oidium of one strain with a cell of primary mycelium of the opposite strain. (v) By the fusion between two haploid cells of opposite strains of the basidium. (ii) Plasmogamy : ● Plasmogamy is achieved by the union of two protoplasts whereby the sexual nuclei of opposite strains come close together in a pair within the same cell. Plasmogamy in Basidiomycetes is achieved either by somatogamy or by spermatisation. 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 cells is called somatogamy. 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 spermatisation. Plasmogamy by spermatisation exclusively takes place in the rusts which produce numerous tiny, uninucleate, nonmotile spore-like bodies called spermatia. (iii) Meiosis : ● ● The synkaryon in the probasidium undergoes two nuclear divisions which constitute meiosis. Meiosis restores the haploid condition in the life cycle. ● ● Sexual Reproduction ● Club fungi usually reproduce sexually. But the development of sex organs, antheridia and ascogonia, are universally absent throughout the class. The rudimentary differences in sex, shown at the time of sexual fusion, are designated by plus (+) and minus (–) signs. These signs are called sexual strains. Fructifications are formed only if two mycelia of opposite strains come into contact. The sexual process, being extremely simplified, consists of three fundamental processes such as karyogamy, sexual fusion or plasmogamy and meiosis. (i) Karyogamy : ● The terminal binucleate or dikaryotic cells of the hyphae of the secondary mycelium develop into basidia. 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. Brand spores Dikaryotic Diploid ● ● ● ● ● ● ● Asexual Reproduction (i) By Conidia : ● Conidia are not of so common occurrence in Basidiomycetes. They are produced in the rusts, smuts and some other Basidiomycetes. In smuts, conidia are budded off from the basidiospores and the mycelium. Conidia are produced by the dikaryotic mycelium and serve to propagate the dikaryophase in the life cycle. ● ● B A Young basidium C (ii) By Oidia : ● ● Basidiospore F D Brand spore E Oidia are small, hyaline, thin-walled unicellular fragments of the mycelium. Oidia may be uni or binucleate as they are produced respectively by the breakage of primary or secondary mycelium. Oidia serve a double function. They may either germinate to form primary mycelia or bring about dipodisation. In the latter case the germinating oidium acts as a spermatium and fuses with the somatic hyphal wall of opposite strain. ● Fig. : (A–F). 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). (iii) By Fragmentation : ● Any part detached from parent cell containing conidiophores becomes able to form a new mycelium (plant body). C.S.V. / February / 2008 / 1631 Basidiocarp and Basidia ● ● ● ● ● ● ● In the higher Basidiomycetes 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, while phragmobasidium is a septate basidium. Holobasidia are characteristic of most of the Basidiomycetes particularly the gilled or fleshy fungi. Basidia are developed in a palisade-like layer on the basidiocarps. This fertile layer is called hymenium. Basidia produce basidiospores which are often wind blown, when they germinate, a new haploid mycelium is formed. Pileus Epibasidium Stipe Gills Annulus Pores Volva Epibasidium A Mycelium B Hypobasidium Hypobasidium A C Basidiospores B Sporidia D C Pileus Migrating nucleus Sterigma Fruit body G Log of wood Epibasidium Brand spore D E E Peridiole Volva F 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. 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. Rusts and Smuts are Parasitic Club Fungi ● ● ● ● Rusts and Smuts are club fungi that parasitize cereal crops such as corn, wheat, oats and rye. Rusts and smuts do not form basidiocarps. They spores produce, resembling soot. Some smuts enter seeds and exist inside the plant, becoming visible only near maturity. In corn smuts, the mycelium grows between the corn kernels and secrets 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 cycle. ● Black stem rust of wheat uses barberry leaf (bushes) as an alternate host. Important Smut Diseases of Cereals Name of Cereals Wheat Oat Barley Corn Loose Smut Ustilago nuda (race of Ustilago tritici ) Ustilago avenae Ustilago nuda Ustilago maydis Covered Smut Tilletia caries Ustilago hordei (a race of Ustilago kolleri ) Ustilago hordei ● C.S.V. / February / 2008 / 1632 Economic Importance ● Several members of basidiomycetes are of great economic importance because of their beneficial as well as harmful nature. Some of them are causative agents of most destructive diseases of our cereal crops such as smut diseases of corn, wheat, oats, barley as well as the wheat rust. Stem rust of wheat is caused by Puccinia graminis tritici. Some of the higher basidiomycetes such as pore fungi are the common wood rotters ; they destroy lumbar and timber. Mushrooms, a good source of proteins, are of great economic value as food. The young fleshy sporophores of many species of puffballs (Lycoperdon and Clavatia) are also edible. Clavatia contains an anticancer substance Calvacin. Differences between Rusts and Smuts Rusts 1. Rusts are intercellular and obtain their nutrition by means of haustoria. Clamp connection on the secondary mycelium are rare. 2. Wheat rusts are heteroecious and others are autoecious. 3. The dikaryotic mycelium produces three kinds of spores–uredospores and teleutospores on primary host and aeciospores on the alternate host. 4. The teleutospores are stalked, 2-celled and each cell is binucleate. 5. Teleutospores are developed from the terminal cells of the mycelium. 6. Each cell of the 2-celled teleutospores produces an epibasidium which bears 4 basidiospores. They are borne on sterigmata. Smuts 1. The smuts may be intercellular or intracellular (Ustilago maydis). Haustoria are present. Clamp connections are common. 2. All smuts are autoecious. 3. Produces only one kind of binucleate spores called smut spores which may be comparable to the teleutospores of rusts. 4. The brand spores (teleutospores) are unicellular and binucleate. 5. Smut spores are formed from the intercalary cells. 6. The single-celled, brand spore, equivalent to teleutospore, germinates to form single epibasidium which bears a variable numbers of basidiospores. ● ● ● ● ● ● OBJECTIVE QUESTIONS 1. Which division of fungi is commonly known as club fungi ? (A) Deuteromycetes (B) Oomycota (C) Basidiomycota (D) Zygomycetes 2. Which of the following groups is/ are comprised of club fungi ? (A) Truffles (B) Puffballs (C) Mushrooms (D) All of the above 3. The part of mushroom that is visible above the ground is a— (A) Ascocarp (B) Basidiocarp (C) Ascogonium (D) Zygospore 4. In which fungus is the dikaryotic stage longer lasting ? (A) Club fungi (B) Imperfect fungi (C) Sac fungi (D) Zygospore fungi 5. Conidiospores are formed— (A) By sporangia (B) By sac, club and imperfect fungi (C) When nutrients are in short supply (D) During sexual reproduction 6. Clamp connection is characteristic of certain group of— (A) Fungi (B) Algae (C) Bryophytes (D) All of the above 7. Anticancer substance calvacin is obtained from— (A) Sac fungi (B) Club fungi (C) Red algae (D) Myxophycean cells 8. In certain members of Basidiomycetes, the basidium bears four spores exogenously, each usually at the tip of a minute stalk known as— (A) Conidiophore (B) Basidiophore (C) Sterigmata (D) Both (B) and (C) 9. Puccinia graminis tritici causes— (A) Black stem rust of wheat (B) Loose smut of oat (C) Covered smut of oat (D) Loose smut of barley 10. 'Loose smut of corn' is caused by— (A) Ustilago nuda (B) Ustilago maydis (C) Ustilago avenae (D) All of the above ANSWERS 1. (C) 2. (D) 6. (A) 7. (B) 3. (B) 4. (A) 5. (B) 8. (C) 9. (A) 10. (B) ●●● = – 20 cm, = 15 cm, = ? 1 = f 1 = 15 (Continued from Page 1577 ) For lens u f v 1 1 – v u 1 1 + 20 v ⇒ v = 60 cm Radius of curvature of the mirror r = 60 – 10 = 50 cm 50. (B) There is one input (A) and one output (Y) connected according to Boolean expression – A = Y read as ‘NOT A equals Y’ meaning that Y is negation (or inversion) of A. Since there are only two digits 0 and 1 in binary system, we have, Y = 0 if A = 1 and Y = 1 if A = 0. The logic symbol of NOT gate is A Y ●●● C.S.V. / February / 2008 / 1633 Model Paper for Various Medical Entrance Examinations BOTANY 1. Match column-A (Name of diseases) with column-B (Causal agents), then select the correct answer from the options given below— Column-A (Name of diseases) (a) Powdery mildew of wheat (b) Flag smut of wheat (c) Hill bunt of wheat (d) Karnal bunt of wheat Column-B (Causal agents) 1. Urocystis tritici 2. Neovossia indica 3. Erysiphe graminis tritici 4. Tilletia foetida (a) (b) (c) (d) (A) 3 2 4 1 (B) 3 4 2 1 (C) 3 1 4 2 (D) 3 4 1 2 2. An activator is often the substrate itself, and regulatory enzymes for which substrate and modulator are identical are called— (A) Heterotropic (B) Homotropic (C) Feed-back inhibition (D) None of the above 3. The casparian strip is found— (A) Between xylem and phloem (B) On four sides of endodermal cells (C) Within the secondary wall of parenchyma (D) Between pericycle and endodermis 4. The plant of Cycas, which belongs to sporophytic generation, is differentiated into— (A) Root, stem and leaves (B) Male and female cones (C) Leaves only (D) Sometimes leaves and sometimes modified stems 5. Which of the following cell organelles produces lysosomes ? (A) ER (B) Peroxisomes (C) Golgi apparatus (D) None of the above 6. Epidermis and Epiblema are produced from— (A) Phellogen (B) Protoderm (C) Procambium (D) Calyptrogen 7. Which of the following is the characteristic of water storage tissue in xerophytes ? (A) Presence of large sized cells (B) Presence of thin cell walls (C) Presence of mucilage (D) Presence of vacuole 8. What happens when lactose is present ? (A) The repressor becomes able to bind to the operator (B) Transcription of lac-Y, lac-Z, and lac-a genes occurs (C) The repressor becomes unable to bind to the operator (D) Both (B) and (C) 9. Which of the following organelles is usually absent in chloroplast of algae ? (A) Lamellae (B) Grana (C) Pigments (D) Quantasomes 10. Abiogenesis theory states that— (A) Life originated due to spontaneous generation (B) Origin of life is due to preexisting organisms (C) Origin of life occurred from blue-green algae like microorganisms (D) Origin of life is due to organic evolution owing to chemical reactions in presence of light 11. Which of the process(es) is/are referred to as translation ? (A) Decoding of the triplet codons of mRNA by t RNA (B) Decoding of amino acids to proteins (C) Decoding of the triplet codons by t RNA to m RNA (D) All of the above 12. Leafless stem of onion which is produced to bear flowers is called— (A) Scape (B) Torus (C) Thalamus (D) Pedicel 13. Feathery stigma and versatile anthers are found in— (A) Compositae (B) Graminae (C) Leguminosae (D) All of the above 14. Genotoxicity refers to the detection of agents that will damage— (A) Vitamins (B) Enzymes (C) DNAs (D) Proteins 15. Which one of the following families is commonly known as nightshade-family ? (A) Compositae (B) Malvaceae (C) Both (A) and (B) (D) Solanaceae 16. The phospholipids are broken down to their residues by the action of enzymes called— (A) Triose phosphate isomerase (B) Cytochrome oxidase (C) Peroxidase (D) Malic dehydrogenase 17. E. coli have the full complement of enzymes for the glyoxylate and citric acid cycles in the— (A) Mesosomes (B) Mitochondria (C) Cytosol (D) Polysomes 18. At what phase of meiosis are there two cells, each with separated sister chromatids that have been moved to opposite spindle poles ? (A) Anaphase-I (B) Anaphase-II (C) Metaphase-I (D) Metaphase-II C.S.V. / February / 2008 / 1634 19. Plants that retain their evergreen foliage throughout the year are called— (A) Draugh escaping plants (B) Evergreen plants (C) Xerophytic plants (D) All of the above 20. The term ‘allelomorphic’ means— (A) Sex-linked characters (B) Sex determining chromosomes (C) A pair of contrasting characters (D) Representatives of the same gene locus 21. The microbial metabolites released or excreted by lysed cells which in very low concentration is directly toxic to cells of suscept is defined as— (A) Susceptibility (B) Pathogen (C) Detoxification (D) Toxin 22. The soil characteristic that determines the movement of water through pore spaces is referred to as— (A) Soil water (B) Soil permeability (C) Soil temperature (D) Porosity 23. Mycorrhiza establishes symbiosis between— (A) Fungi and bryophytes (B) Phaeophyceae and rhodophyceae (C) Fungi and algae (D) None of the above 24. Each ovule is attached on the placenta by a small stalk called— (A) Funiculus (B) Nucellus (C) Raphe (D) Hilum 25. When two nuclei lie side by side after plasmogamy, this stage is termed as— (A) Haplophase (B) Karyogamy (C) Dikaryophase (D) Syngamy 26. Besides their utility as simple fuels, fossil fuels also give rise to by-products such as— (A) Certain synthetic chemicals (B) Drugs (C) Plastic (D) All of the above 27. Consider the following fruit— 1. Apricot 2. Guava 3. Tomato 4. Papaya Which of these are berries ? (A) 1 and 3 (B) 2, 3 and 4 (C) 1, 3 and 4 (D) 1, 2, 3 and 4 28. Which of the following plants is/are more efficient at photosynthesis in hot, dry conditions ? (A) C3 plants (B) C4 plants (C) Both (A) and (B) (D) Xerophytes 29. What is the source of energy that flows through the living world ? (A) Sun (B) Green plants (C) Photosynthesis (D) Chemical bonds 30. An accumulation of CO2, ozone and other pollution-related gases in the lower atmosphere, trapping heat near the earth’s surface leading to global warming is known as— (A) Green house effect (B) Biomass (C) Energy pyramid (D) Autecology 31. In protein synthesis of prokaryotes, the initiation of polypeptide chain is induced by— (A) Lysine (B) Sesine (C) Leucine (D) Methionine 32. How many carbon atoms (Catoms) are there in a molecule of ribulose biphosphate (RuBP) ? (A) Two (B) Three (C) Four (D) Five 33. POP is abbreviated for— (A) Pak occupied people (B) Persistent organic pollutant (C) Plant organising principle (D) Plants of pacific province 34. Which of the following phenomena is associated with the burning of fossil fuel ? (A) Acid rain (B) Green house effect (C) Both (A) and (B) (D) None of the above 35. Haploid plants can be obtained by culturing— (A) Pollen grain (B) Endosperm (C) Root apices (D) None of the above 36. During the fermentation of grapes for wine preparation, a reddish brown crystalline mass of acid potassium salt is obtained, which is known as— (A) Rochelle salt (B) Argol (C) Pyruvic acid (D) Glyceraldehyde 37. Members of a biological species are potentially able to— (A) Express all the same genes (B) Introgress (C) Complete (D) Interbreed 38. Which of the following is the earliest land plant ? (A) Rhynia (C) Riccia (B) Cordaites (D) Pinus 39. Exergonic reactions are usually— (A) Exothermic (B) Endothermic (C) Both of the above (D) Neither of the above 40. A measure of the frequency of crossing-over between two linked chromosome marker loci in recombining meiotic product is called— (A) Map distance (B) Loop distance (C) Chromosomal aberration (D) Chromosome distance 41. In the phosphorus cycle weathering makes phosphate available first to— (A) Producers (B) Consumers (C) Tertiary consumers (D) Decomposers 42. The shape of polypeptide is— (A) Important to its function (B) Dependent upon the primary structure C.S.V. / February / 2008 / 1635 (C) Maintained by bonding between parts of the polypeptide (D) All of the above 43. In a certain dicotyledonous stem, the fascicular cambium is a meristematic tissue referred to as— (A) Apical (B) Intercalary (C) Lateral (D) Secondary 44. Nucleolus of the cell is a special region of chromatin where— (A) m-RNA is produced (B) t-RNA is produced (C) r-RNA is produced (D) All of the above are produced 45. Which of the following plants is commonly called ‘royal fern’ ? (A) Campyloneurum scolopendrium (B) Adiantum pedatum (C) Both (A) and (B) (D) Osmunda regalis 46. The most important feature of all living systems from the viewpoint of their continuity is their capacity to— (A) Utilize oxygen to generate energy (B) Utilize solar energy for metabolic activities (C) Replicate the genetic information (D) Produce gametes 47. The C4 molecules, formed after the CO 2 fixation by CAM plants, are stored in— (A) Vacuoles of mesophyll cells (B) Cortex (C) Endodermal casparian stripe (D) All of the above 48. Transcription requires— (A) Enzyme RNA polymerase (B) Ribonucleoside triphosphate (C) DNA template (D) All of the above 49. At each trophic level of a food chain (pyramid) the energy not used or passed along is given off as— (A) Heat (B) Water (C) Free energy (D) Matter 50. ‘Chemical cycling’ through an ecosystem involves— (A) Physical environments (B) Biotic components (C) Both (A) and (B) (D) None of the above 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 16. 17. 18. 19. (B) Plants bearing evergreen foliage throughout the year are called evergreen plants. Popu(A) larly, needle-leaved trees (Pine, (B) The modulators for allosteric Juniper, etc.) are called everenzymes may be either inhibitory green plants. or stimulatory. An activator is 20. (C) 21. (D) 22. (B) often the substrate itself, and regulatory enzymes for which 23. (D) Symbiosis is also established by mycorrhiza, but in this case substrate and modulator are symbiosis is always established identical and are called homobetween fungus and the roots of tropic. higher plants. (B) The endodermal cells fit snugly together and are bordered 24. (A) on four sides by a layer of imper- 25. (C) In certain classes of fungi meable lignin and suberin known such as Ascomycetes and Basias casparian strip. diomycetes, as a result of plas(A) mogamy (fusion of the proto(C) Lysosomes are membraneplasm of two compatible gametes) bound vesicles produced by a the nuclei of the opposite strains Golgi apparatus that contain get themselves arranged in pairs hydrolytic digestive enzymes. but do not fuse. This phase in life (B) The outermost meristematic cycle is termed as dikaryophase layer of the young growing region and the process is called dikais known as protoderm, which ryotization. develops into epidermis and 26. (D) epiblema. 27. (B) Apricot (Prunus persica) is a (B) In xerophytes mostly water kind of drupe fruit. storage tissues possess thin- 28. (B) C plants are more efficient at 4 walled cells having a few interphotosynthesis than C 3 plants in cellular spaces, e.g., Opuntia, hot, dry conditions. Euphorbia etc. 29. (A) 30. (A) (D) 31. (D) In prokaryotic protein syn(B) In the algal plastids (chlorothesis, the initiation of polypeptide plasts) the thylakoids are only of chain is always brought about by one kind and restricted to the an amino acid, methionine, stack itself. They are not closely which is coded by the codon packed or fused as in the grana AUG. of higher plants. 32. (D) (A) Abiogenesis theory of origin of life stated that different types of 33. (B) POP is an abbreviation of ‘Persistent organic pollutants’. organisms are formed automatiThe WHO estimates an annual cally due to chemical reactions in death toll of around 20000 due to presence of high energy. The poisoning from POPs and other spontaneous generation or abiopesticides. genesis of life was visualized as beginning with either inorganic 34. (C) materials or with putrefying orga- 35. (A) Haploid plants possess only one set of chromosomes. Haploid nic matter. produced from microspores or (A) 12. (A) 13. (B) 14. (C) 15. (D) pollen grains (anthers) are called (A) androgenic haploids. Mahesh(C) Some bacteria, including E. wari and Guha (1964, 65) succoli, have the full complement of cessfully reported anther culenzymes for the glyoxylate and tured plants from Datura innoxia. citric acid cycles in the cytosol. 36. (B) 37. (D) E. coli can, therefore, grow with acetate as its sole source of 38. (A) Rhynia, which belongs to carbon and energy. division Psilophyta, was discovered in 1917 by Kidston and (B) During anaphase-II of Lang from Rhynie chert bed meiosis-II, the centromeres divide of Middle Devonian. and the daughter chromosomes move towards the spindle poles. (Continued on Page 1639 ) ANSWERS WITH HINTS C.S.V. / February / 2008 / 1636 Model Paper for Various Medical Entrance Examinations BOTANY 1. Match column A (Types of inflorescence) with column B (Examples) then select the correct answer from the options given below : Column A (a) (b) (c) (d) 1. 2. 3. 4. (A) (B) (C) (D) Cymose umbel Thyrsus Cymose corymb Mixed panicle Column B Grape vine Allium cepa Ligustrum Ixora (a) (b) 1 2 2 1 1 3 2 3 (C) Roots of Digitaria in association with Spirillum notatum (D) All of the above 6. Which of the following is an artificial ecosystem ? (A) Grassland (B) Rice field (C) Lake (D) Forest 7. Which of the following insectivorous plants are partial stem parasite ? (A) Viscum album (B) Santalum album (C) Both (A) and (B) (D) Botrichium (d) 4 3 2 4 8. Edible part of pomegranate is— (A) Fleshy thalamus (B) Fleshy aril (C) Succulent testa (D) Accrescent sepals 9. With which of the following tissues/cells, companion cells are associated ? (A) Collenchyma (B) Medullary parenchyma (C) Sieve tube (D) None of the above 10. In older trees, the inner annual rings are called— (A) Heartwood (B) Sapwood (C) Phloem (D) Dermatogen 11. Soil showing permanently wilted plants has— (A) Capillary water (B) Hygroscopic water (C) Gravitational water (D) Ground water 12. Parasitic fungi take their nutrition from their host with the help of— (A) Paraphyses (B) Haustoria (C) Soredia (D) None of the above 14. Root apical meristem cells are distinct from other root cells, because of their— (A) Thick cell walls (B) Triploid nuclei (C) Enlarged water vacuoles (D) Smaller size 15. Protein that forms chromatin with DNA is— (A) Myosin (B) Actin (C) Troponin (D) Histone 16. Ethnobotany is a branch of botanical science which deals with the study of— (A) Shrubs and trees (B) Plants used by people, particularly by primitive people (C) Plants of economic value (D) All of the above 17. Those plants which grow luxuriantly in rainy season and shed all the leaves in summer are called— (A) Xerophytes (B) Lianus (C) Trophophytes (D) None of the above 18. ‘SCP’ stands for— (A) Single cell protein (B) Special cell protein (C) Sweetener cell protein (D) Special cell for photosynthesis 19. The transition between meiosis-I and meiosis-II is known as— (A) Interphase (B) Cell cycle (C) Interkinesis (D) Cytokinesis 20. Epithem tissue is associated with— (A) Guttation (B) Photorespiration (C) Exudation (D) None of the above 21. The exchange of genetic material between nonsister chromatids of a bivalent during meiosis-I is known as— (A) Diakinesis (B) Hybridization (C) Both (A) and (B) (D) Crossing-over (c) 3 4 4 1 2. In certain algae, the hapteron is a/an— (A) Attaching organ (B) Respiratory organ (C) Reproduction type in which sexual spores are involved (D) Apical zone 3. Some plant tissues, that are modified for the storage of starch, derive most of their energy from— (A) Photosynthesis (B) Glycolysis (C) Glyoxylate metabolism (D) Fermentation 4. Plasmids that can integrate into the bacterial DNA are called— (A) Plasmogenes (B) Neurogenes (C) Episomes (D) Hydrogenes 5. Which of the following associations fixes nitrogen through nonnodulation ? (A) Roots of maize associated with spirillum notatum (B) Coralloid root of Cycas in association with Anabaena 13. In which of the following plant groups, the root is feebly deve- 22. Entropy is a state or condition of— loped and stem is soft with a (A) Energy large number of air cavities ? (B) Matter (A) Xerophytes (B) Hydrophytes (C) Both (A) and (B) (C) Mesophytes (D) Helophytes (D) None of the above C.S.V. / February / 2008 / 1637 23. Mushrooms are placed in division— (A) Ascomycota (B) Deuteromycota (C) Basidiomycota (D) Zygomycota 24. Flavoproteins contain flavin nucleotide, which is/are— (A) FMN (B) FAD (C) Both (A) and (B) (D) None of the above 25. The term meiosis-I and meiosis-II were used by— (A) Flemming (B) Strasburger (C) Gregoire (D) Mendel 26. Laticiferous vessels are found in— (A) Phloem tissue (B) Cortex (C) Xylem tissue (D) All of the above 27. Ozone is produced naturally by high energy solar radiation and the activity of methane (CH 4) in the presence of— (A) Nitrogen oxide (B) Sulphur dioxide (C) Carbon monoxide (D) Carbon dioxide 28. Shifting cultivation requires— (A) Excessive soil erosion for crop production (B) Alternate crop pattern on a particular area (C) Long time for the regeneration of soil (D) Huge amount of commercial fertilizer 29. The chemical formula of carotene obtained from carrot is— (A) C40H56 (B) C55H72N4 Mg (C) C55H27N2 Mg (D) C55H72 N6 Mg 30. The primary cell wall may contain— (A) Cellulose (B) Pectic compounds (C) Certain polysaccharides (D) All of the above 31. The fertilizer named ‘Nitrolim’ is prepared by the use of— (A) N + P + K (B) CaC2 + P (C) CaC2 + N2 (D) None of the above 32. The average process of nitrogen fixation, the reduction of NO3– in the soil into molecular nitrogen is called— (A) Chlorination (B) Ammonification (C) Denitrification (D) Nitrification 33. Who coined the term ‘telomere’ ? (A) Steffensen (B) Waldeyer (C) Altman (D) Muller 34. Soil particles, particularly clay and organic matter in soil, contain negative charges and attract positively charged ions, such as— (A) Mg2+ (B) Ca 2+ + (C) K (D) All of these 35. The acid present in tomatoes is mainly— (A) Citric acid (B) Oxalic acid (C) Tartaric acid (D) Lactic acid 36. Genes not located within the nucleus are almost always found in the— (A) Cytoskeleton (B) Plastids (C) Cytosol (D) All of the above 37. Cladosporium is a/an— (A) Fungus (B) Alga (C) Cyanobacterium (D) PPLO 38. Divergent palmate venation is found in— (A) China-rose (B) Cinnamon (C) Zizyphus jujuba (D) All of the above 39. Genes that are similar in phenotypic effect when present separately but together interact to form a different phenotypes in the ratio of 9 : 7. Such genes are known as— (A) Epistasis genes (B) Lethal genes (C) Dominant epistatis genes (D) Complementary genes 40. The basic monomers used in DNA replication are— (A) Glucose (B) Amino acids (C) DNA nucleotides (D) RNA nucleotides 41. Who among the following proposed the signal hypothesis for selective translocation of m RNAs ? (A) T. H. Morgan (B) Blobel and Dobberstein (C) Bateson and Punnett (D) Watson and Crick 42. The rate of growth in plant is not uniform throughout its length in the apical meristems of— (A) Roots (B) Stems (C) Both (A) and (B) (D) None of the above 43. The DHU arm of transfer RNA contains— (A) Unusual pyrimidine nucleotide (B) Dihydrouracil (C) Both (A) and (B) (D) None of the above 44. The term mutation was introduced by— (A) Hugo de Vries (B) Lamarck (C) Darwin (D) Morgan 45. Which of the following diseases of stem is caused by the deficiency of Boron (B) ? (A) Black necrosis (B) Die-back disease (C) Whip-tail disease (D) None of the above 46. Several ribosomes are often attached to and translating the same m-RNA. The entire complex is called— (A) Clover leaf (B) Anticodon (C) Codon (D) Polysome 47. Which of the following properties makes plasmids suitable vectors for gene cloning ? (A) Plasmids are small circular DNA molecules with their own replication origin site (B) Plasmids often carry antibiotic resistance gene C.S.V. / February / 2008 / 1638 (C) Plasmids are small circular DNA molecules that can integrate with host chromosomal DNA (D) Plasmids can shuttle between prokaryotic and eukaryotic cells 48. A micromutation is— (A) Change in gene (B) Polyploidy (C) Deletion of chromosome (D) Addition of chromosome 49. The approximate size of molecules that can pass through plasma membrane is— ° (A) 220—250 A ° (B) 170—150 A ° (C) 150—125 A ° (D) 8—10 A 50. Chlorophyll-c is found in— (A) Brown algae (B) Some flagellates (C) Diatoms (D) All of the above ANSWERS WITH HINTS 1. (B) 2. (A) In certain algae, like Fucus, hapteron is a special attaching organ. It is also known as holdfast. 3. (B) Glycolysis is an almost universal central pathway of glucose catabolism. Some plant tissues that are modified for the storage of starch (such as potato tubers) derive most of their energy from glycolysis. 4. (C) 5. (D) A large number of examples are there where root nodules are not formed but symbiotic nitrogen fixation occurs. Other examples (other than given example in the options) are—Anthoceros associated with Nostoc (a blue-green alga). 6. (B) 7. (A) The partial stem parasite is represented well by Viscum album which parasitizes a number of shrubs and trees. 8. (C) 9. (C) Companion cells extended through the whole length of the sieve tube but sometimes there is a vertical file of two or more companion cells next to each tube cell. 10. (A) 11. (A) The amount of water present around the soil particles and held by surface tension and attraction force of water molecules is called capillary water. 12. (B) 13. (B) Due to availability of water in plenty, roots, the principal organs of water absorption, in such (hydrophytes) plants become of secondary importance, and less significant. The overall development is usually very poor and insignificant in most of the hydrophytes. 14. (D) 15. (D) DNA and histone (protein) together comprise chromatin, forming the bulk of the eukaryotic chromosome. Histones are of five major types—H1, H 2A, H 2B, H3 and H 4. 16. (B) All those plants which are used by primitive people for various purposes, such as worship are studied in ethnobotany. 17. (C) 18. (A) 19. (C) 20. (A) 21. (D) As the lattice breaks down, homologues are temporarily held together by chiasmata, regions where the non-sister chromatids are attached due to crossingover. Then, homologues separate and are distributed to different daughter cells. 22. (C) Entropy is a state or condition not only of energy but also of matter. 23. (C) 24. (C) Flavoproteins contain a highly, sometimes covalently bound flavin nucleotide, either FMN and FAD. 25. (C) Gregoire (1904) called the first division as meiosis-I and second division as meiosis-II. Because during meiosis nucleus divides twice but chromosomes only once. 26. (A) 27. (A) 28. (C) 29. (A) 30. (D) Primary cell wall consists mainly of cellulose and pectic substance and may also contain other polysaccharides. 31. (C) 32. (C) 33. (D) The chromosome extremities or terminal regions on either side are called telomeres, the term coined by Muller in 1938. 34. (D) 35. (B) 36. (B) 37. (B) 38. (A) 39. (D) 40. (C) 41. (B) 42. (C) 43. (C) The cloverleaf model of transfer RNA (t-RNA) composed of a series of stem-loop structures, known as arms. The D or DHU arm is a stem-loop structure containing dihydrouracil, an unusual pyrimidine nucleotide. 44. (A) 45. (A) Boron (B) is a trace element (microelement). Its deficiency causes black necrosis of stem. 46. (D) 47. (A) Plasmids are small accessory rings of DNA found in some bacteria that carry genes not present in bacterial chromosome. Plasmids that are used as vectors have been removed from bacteria and have had a foreign gene inserted into them. 48. (A) 49. (D) 50. (D) In brown algae, diatoms and some flagellates, a chlorophyll component has been isolated and termed chlorophyll-c. ●●● (Continued from Page 1636 ) 39. (A) Exothermic refers to the release of heat (∆H is negative). Exothermic reactions are generally, but not always, exergonic. 40. (A) 41. (A) 42. (D) 43. (C) Lateral meristems are present along the lateral sides of stem and roots. Interstelar cambium ring formed by intrafascicular and interfascicular cambium and cork cambium are examples of lateral meristem. 44. (C) Chromatin (Chromosome) has a special region called nucleolus where ribosomal RNA (r-RNA) is produced. 45. (D) 46. (C) 47. (A) CAM plants use PEPCase to fix CO 2 at night, forming a C4 molecule, which is stored in large vacuoles in their mesophyll cells until the next day. 48. (D) The process of formation of m-RNA from DNA is called transcription, which involves promotors (P), binding to the promotors site, RNA chain initiation and elongation, and RNA chain termination. 49. (A) 50. (C) ‘Chemical cycling’ through an ecosystem involves not only the biotic components of an ecosystem but also the physical environment. ●●● C.S.V. / February / 2008 / 1639 Model Paper for Various Medical Entrance Examinations BOTANY 1. Match Column A (Types of Fruit) with Column B (Common english name of fruits) then select the correct answer from the options given below— Column A Column B (a) Drupe 1. Strawberry (b) Pome 2. Plum (c) Etaerio of 3. Brinjal achnes (d) Berry 4. Pear (a) (b) (c) (d) (A) 2 4 3 1 (B) 2 4 1 3 (C) 2 1 4 3 (D) 2 3 4 1 2. Ethnobotany is a branch of Botanical science which deals with— (A) The study of plants used by people (particularly by primitive people) (B) The study of plants of economic value (C) The study of shrubs and trees (D) None of these 3. Most of the history of life concerns the evolution of— (A) Plants and animals (B) Prokaryotes (C) Eukaryotes (D) Photosynthesizers 4. Birbal Sahni Institute of Palaeobotany is situated at— (A) Lucknow (B) Delhi (C) Kolkata (D) Kanpur 5. An octamer of four histones complexed with DNA is called— (A) Nucleosome (B) Telomere (C) Satellite (D) Karyotype 6. Protoplasm term was used by Purkinje (1839) and Von Mohl (1846). It was called as physical basis of life by— (A) Julian Huxley (B) Hugo de Vries (C) Dujardin (D) Linnaeus 7. The least number of t RNA molecules for each of the twenty amino acids found in protein is— (A) Four (B) Three (C) One (D) Two 8. Quantasome discovered by Park and Biggins (1964) has— (A) 100 chlorophyll molecules (B) 300 chlorophyll molecules (C) 230 chlorophyll molecules (D) 230 chlorophyll and about 50 carotenoid molecules 9. In pteridophytes, xylem consists of— (A) Tracheids (B) Xylem parenchyma (C) Both (A) and (B) (D) None of the above 10. Lock and Key theory (Template theory) of enzyme action was given by— (A) Buchner (B) Fischer (C) Koshland (D) Umberger 11. The skeleton of the shoot is formed by the— (A) Root (B) Stem (C) Leaf (D) Bud 12. Coenzyme is— (A) Always a protein (B) Non-protein organic molecule (C) A non-protein inorganic compound (D) Often a metal 13. True bulbs are found— (A) Only among monocotyledons (B) Only among dicotyledons (C) Equally both in monocotyledons and in dicotyledons (D) Largely in dicotyledons and rarely in monocotyledons 14. Glycolysis is found in cytoplasm of virtually all types of aerobic/ anerobic cells. In this process, glucose is converted into a compound which is— (A) Citric acid (B) Pyruvic acid (C) Acetyl CoA (D) PEP 15. The causal organism of ‘foot rot of papaya’ is— (A) Armillaria mellea (B) Phymatorichum omnivorum (C) Ophiobolus graminis (D) Pythium aphanidermatum 16. RQ is measured by— (A) Auxanometer (B) Potometer (C) Ganong’s Respirometer (D) Porometer 17. Gynobasic style is chiefly found in family— (A) Solanaceae (B) Cucurbitaceae (C) Malvaceae (D) Labiatae 18. Fermentation is— (A) Anaerobic respiration (B) Aerobic respiration (C) Incomplete oxidation of carbohydrates (D) Anaerobic complete oxidation of carbohydrates 19. All the appendages which are epidermal in origin are referred to as— (A) Trichomes (B) Ground tissues (C) Mesophylls (D) None of the above 20. Net gain of ATP in glycolysis from one molecule of glucose is— (A) 8 (C) 2 (B) 4 (D) 0 21. The ultimate biological unit which controls heredity is called— (A) Gene (B) Genotype (C) Genome (D) Chromosome 22. When yeast ferments glucose, the products are— (A) C2H5OH + Energy (B) C2H5OH + CO2 + Energy (C) CO2 + H2O + Energy (D) CH3OH + H2O + Energy 23. The nucleus of a bacterium is— (A) Absent (B) Well-defined (C) Incipient (D) Found always near the pili C.S.V. / February / 2008 / 1640 24. Which intermediate compound acts as connecting link between glycolysis and Krebs cycle/link between carbohydrate and fat metabolism ? (A) Acetyl CoA (B) Cytochrome (C) OAA (D) Pyruvic acid 25. Gametophyte is the main plant body in— (A) Angiosperms (B) Gymnosperms (C) Pteridophytes (D) Bryophytes 26. Number of meiotic divisions required to produce 100 macrospores in angiosperm is— (A) 25 (B) 100 (C) 50 (D) 125 27. How many nucleotides are present in one turn of DNA helix ? (A) Ten (B) Thirteen (C) Two (D) Four 28. A diploid somatic cell can divide by— (A) Meiosis but not mitosis (B) Mitosis but not meiosis (C) Meiosis or Mitosis (D) Amitosis only 29. Intrafascicular cambium is situated— (A) (B) (C) (D) Outside the vascular bundles In between vascular bundles Inside the vascular bundles In the pith 33. Transgenic bacteria perform services in the field of— (A) Bioremediation (B) Protection and enhancement of plants (C) Chemical production (D) All of the above 34. Gram stain is— (A) Technique developed in black gram (B) A stain obtained from gram seeds (C) A chemical for staining bacteria (D) A trade mark 35. The fruit of cereals is mostly referred to as— (A) Pome (B) Berry (C) Caryopsis (D) Drupe 36. Bacteria differ from viruses in the presence of— (A) Cytoplasm (B) True nucleus (C) Causing disease (D) All of these 37. Point mutation involves a change in— (A) A single nucleotide (B) A specific codon (C) Both (A) and (B) (D) None of the above 38. Natural system of classification is good for practical purpose, who proposed natural classification of plants ? (A) Bentham and Hooker (B) Engler and Prantle (C) Linnaeus (D) Oswald Tippo 39. Pear fruit contains a lot of— (A) Stone cells (B) Collenchyma (C) Parenchyma (D) Both (B) and (C) 40. Phenomenon of heterothallism was first observed by— (A) Alexopoulas (B) K. C. Mehta (C) Blakeslee (D) None of these 41. Stilt roots are found in— (A) Sugarcane (B) Rice (C) Gram (D) Groundnut 42. Tannin is obtained from which part of Acacia nilotica— (A) Leaves (B) Bark (C) Heart wood (D) Pods 43. Between the bark and the wood in a woody stem, there is a layer of meristem called— (A) Cork cambium (B) Vascular cambium (C) Apical meristem (D) Cell division zone 44. ‘Chipko Movement’ is the world’s most known Eco-development programme, started by S. L. Bahuguna in 1973. It is related with— (A) Conservation of forest (B) Afforestation (C) Planting (D) Population 45. Which of the following is a homopolysaccharide ? (A) Dextran (B) Glycogen (C) Both (A) and (B) (D) Hyaluronic acid 46. Difference between angiosperm and gymnosperm is that in gymnosperm— (A) Ovules are absent (B) Seeds are naked (C) Companion cells are absent (D) Sieve elements are absent 47. Osmotic pressure of pure water is— (A) Zero (B) 4 (C) 40 (D) 100 48. Velamen in orchids helps in— (A) Absorption of moisture from air (B) Absorption of water from soil (C) Guttation (D) Clinging the weak plant 49. Phytoplanktons are more likely to be found in which zone of the ocean ? (A) Epipelagic zone (B) Bathypelagic zone (C) Continental zone (D) Mesopelagic zone 50. The cord like tendrils in Smilax are— (A) Stem tendrils (B) Leaf tendrils (C) Leaflet tendrils (D) Stipular tendrils 30. In which stage chromosomes go towards the opposite poles ? (A) Telophase (B) Metaphase (C) Prophase (D) Anaphase 31. Plants with little or no secondary growth are— (A) Deciduous (B) Evergreen (C) Dicot (D) Herbaceous 32. Importance of meiosis lies in— (A) Bringing discontinuous variations (B) Maintaining constancy in the number of chromosomes in the next generation (C) Addition in number of chromosomes (D) None of the above C.S.V. / February / 2008 / 1641 ANSWERS WITH HINTS 1. (B) 2. (A) All those plants which are used by primitive people for various purposes, such as worship are studied in Ethnobotany. 3. (B) Prokaryotes existed alone on the surface of the earth for atleast 1·5 billion years. During this time all the metabolic pathways found in modern cells developed. 4. (A) 5. (A) 6. (A) Protoplasm is a polyphasic crystallo-colloidal solution and includes cytoplasm and nucleoplasm. Since in living organism all processes of life occur in protoplasm and hence Huxley called it as physical basis of life. 7. (C) 8. (D) Inner membrane of thylakoid bears quantasomes (functional unit of chloroplast). These are called photosynthetic units (PSU) where primary act of photosynthesis (i.e. release of electron) occurs. A quantasome has 230 chlorophyll molecules (160 chl. a + 70 chl. b) and about 50 carotenoid molecules. 9. (C) 10. (B) 11. (B) 12. (B) Coenzymes are non protein molecules that assist enzymes in performing their reactions. 13. (A) 14. (B) In glycolysis one molecule of hexose sugar (glucose/fructose, a 6C compound) is splitted to form two molecules of a 3C compound, Pyruvic acid (CH 3CO. COOH). 15. (D) 16. (C) R. Q. (Respiratory Quotient) or respiratory ratio is the ratio of CO2 evolved and O2 used during respiration. It is measured by Ganong’s Respirometer. 17. (D) 18. (C) Fermentation is incomplete oxidation of carbohydrates. It can be aerobic or anaerobic but is more rapid in anaerobic conditions. 19. (A) Outgrowths of diverse forms, structures and functions develop from the epidermis are referred to as trichomes. 20. (C) In glycolysis 4 ATP molecules are produced and 2 are consumed, therefore, net gain is 2 ATP. 21. (A) 22. (B) Reaction of fermentation is— C6H12O6 36. (A) In bacteria and Viruses true nucleus is absent. However, cytoplasm is found in bacteria, while absent in viruses. 37. (C) Point mutations involve a change in a single nucleotide and, therefore, a change in a specific codon. When one base is substituted for another, the result can be variable. 38. (A) 39. (A) 40. (C) 41. (A) 42. (C) Tannin is a secretion of heart wood in Acacia nilotica, commonly called Babool. 43. (B) 44. (A) 45. (C) 46. (B) In gymnosperms the seeds are naked, whereas in angiosperms these are enclosed in ovary (fruits). 47. (A) 48. (A) Velamen is composed of layers of spongy tissues, which help in the absorption of moisture from air. 49. (A) 50. (D) In Smilax, stipules are modified into tendrils. ●●● → Yeast 2 CO 2 + 2 C2H5OH + 56 k cal 23. (C) In bacterial cell, a well-defined membrane bound nucleus is not found, i.e., nuclear membrane, nucleolus and chromosomes, like a typical eukaryotic cell, are absent. Such ill-defined nucleus is termed incipient nucleus. 24. (A) Acetyl CoA is the connecting link between glycolysis, Krebs cycle and oxidation of fat. 25. (D) 26. (B) Macrospore : one meiosis produces one macrospore (egg). 27. (A) The distance between two ° base pairs is 3·4 A, and since each turn of DNA helix measures ° 34 A long, there are ten base pairs at each turn. 28. (B) Mitosis occurs in somatic cells. It is also called somatic division. 29. (C) 30. (D) At anaphase stage of cell division chromosomes move towards opposite poles due to tractile fibres arising from Kinetochore. Acentric chromosomes (without centromere) do not move as they do not have centromere. 31. (D) 32. (B) Meiosis maintains the fixed number of chromosomes in sexually reproducing organisms. It produces haploid gametes by reducing the chromosome number to half. These gametes on fertilization restore diploidy. 33. (D) Transgenic bacteria perform services in protecting plants and the environments to increase industrial production. 34. (C) Gram stain is a chemical which is used for staining bacteria. All bacteria take crystal violet stain. Gram stain was discovered by Gram in 1884. 35. (C) (Continued from Page 1620 ) Further, the percentage of ‘A’ equals the percentage of ‘T’ and the percentage of ‘G’ equals the percentage of ‘C’. The percentage of A + G equals 50% and the percentage of T + C equals 50%. These relationships are called Chargaff’s rules. 47. (D) Stratum corneum is the outer layer of epidermis of vertebrate skin; cells undergo cornification and die, becoming worn off. 48. (B) Trophoblast is epithelium surrounding the mammalian blastocyst, forming outer layer and becoming part of extra embryonic membranes. 49. (B) Apocrine glands are located in the axillae and pubic region that open into the hair follicles rather than directly on the surface of the skin. 50. (D) Hydrophis is marine aquatic snake and is viviparous. ●●● C.S.V. / February / 2008 / 1642 Solved Paper CBSE MEDICAL ENTRANCE (MAINS) EXAMINATION, 2007 (Based on Memory) Paper I Max. Marks : 100] [Time : 2 Hours Physics 1. (a) How many photons of wavelength 439 nm should strike on a perfectly reflecting surface in 1 second so that it may exert a force of 10 N ? (b) Can water be boiled without heating ? 2. (a) Equation for two waves is given as y 1 = a sin (ωt + φ1), y2 = a sin (ωt + φ2). If amplitude and time period of resultant wave does not change then calculate (φ 1 – φ 2). (b) A solid sphere of radius a having charge q is placed inside spherical shell of inner radius r, outer radius R. Find potential at distance x, where r < x < R. 3. (a) Prove that for a monoatomic gas ratio of specific heat γ = 5/3. (b) Give the truth table of the following : A A Y B B (b) There are two wires each produces frequency of 500 Hz. By what percentage tension in one wire is increased so that 5 beats per second can be heard ? 6. (a) Find the force on conductor carrying current i as shown in the figure. b i I a x I (b) A conducting cone is given charge q. How will the charge density and electric potential varies at different points of cone ? 7. (a) When 4 amp current flows through battery from positive to negative terminal potential difference is 12 V obtained, when 2 amp current passes from negative to positive terminal of the battery potential difference 9 V is obtained, calculate emf and internal resistance of the battery. (b) A small pulley of radius 20 cm and moment of inertia 0·32 kg m2 is used to hang a 2 kg mass with the help of a massless string. If this load is released then calculate 2kg acceleration of the block. 8. (a) Capacitance of 6 µF is charged by 6 V battery. Now it is connected with inductor of 5 mH. Find current in inductor when 1/3rd of total energy is magnetic. (b) An object is thrown vertically upward with some speed. It crosses two points p, q which are separated by h metre. If t p is the time between p and highest point and coming back and tq is the time between q and highest point and coming back, relate acceleration due to gravity t p, t q and h. 9. (a) Two coils m and n having 10 turns and 15 turns respectively are placed close to each other. When 2A current is passing through coil m, 4. (a) Write the difference between nuclear force and coulombic force. (b) An airplane is moving horizontally with speed of 100 m/sec at height of 2000 m from ground. A small object is detached from it and strikes the ground. Calculate the angle from vertical with which it strikes the ground. (c) Which of the following quantities have same dimensional formula ? Angular momentum, impulse, energy, torque, force and moment of inertia. 5. (a) From a table, a rod is hinged as shown in the figure. When the support is withdrawn, calculate the acceleration of centre of mass. I C.S.V. / February / 2008 / 1643 then flux linked in coil n is 1·8 × 10–4 weber per turn. If 3A current is passed through coil n then, calculate the flux linked per turn of coil m. (b) A string having tension 360 N and mass/length = 4 × 10–3 kg/m. It produces two consecutive resonant frequencies with a tuning fork, which are 375 Hz and 450 Hz. Find mass of string. 10. (a) In photoelectric effect a photon of wavelength 3300 Å is incident on metal surface of work function 2·5 eV. Now emitted electrons enter in a transverse magnetic field 6·7 × 10 –6 T and turn in a circular path of radius 50 cm. Calculate charge of electron from the given data. (b) If temperature and magnetic field applied across paramagnetic substance are tripled, how many times intensity of magnetization of substance will change ? (b) Identify which of the following given compounds is optically active ? (i) 2-chloro 3-methyl pent-1, 4-diene (ii) 3-methyl 3-hydroxy pentan 1, 5-diol (iii) 2-chloro 2-methyl butane (c) Convert : (i) Ph—C ≡ CH → Ph—C—CH3 || O (ii) Ar—NH2 → ArNC An alkene C4H8 reacts with HBr both in the presence and absence of peroxides to give the same product. Identify the alkene. C4H10O is produced on reaction of an alkane with H2 O/H2SO4, which is not resolvable into optical isomers. Identify the compound. Make two possible dispeptides from the amino acids given below : NH2—(CH2)5—CH—CHOOH and | NH2 NH2—CH—(CH2)3—COOH | COOH The amino acid alanine when kept in a solution with pH less than its isoelectric point it coagulates at the cathode and if pH is greater than isoelectric point it coagulates at anode. Explain this phenomenon. Which out of 1-butene and 2-butene react easily with Br2 in CS 2 and why ? Why 1-butyne gives sodium salt with NaNH2 but 2-butyne does not ? Draw the structures for DNA purines. Ka2 = 8·3 × 10 – 5. Find the concentrations of H+, HSO 3– and SO32– . N2O4 dissociates with a degree of dissociation as 0·4. Establish Kχ. Relation between Kχ and Kp and the value of Kp. Given total pressure = 1 atm and T = 315 K. 1 mole of nitrogen and 4 mole of hydrogen react to form ammonia in a 20 litre vessel. 10 litres of water are added and the vessel properly shaken. What will be the pressure of the residual gases ? Why is F2 more reactive than Cl2 ? Why is CrO42– more oxidizing than MoO 42– ? Out of (SiH3)2O and (CH3)2O which is more basic and why ? The empirical formula of an insoluble compound is PtCl2.(NH3)2. On churning this compound with AgNO3 we get [Pt(NH3)4](NO3)2 and Ag2[PtCl4]. What will be the molecular formula of the compound will be ? Out of trimethylamine and triethyl phosphine, which one has higher dipole moment ? (d) 5. (a) Chemistry 1. (a) Two silver rods are dipped in 1M HCl and 1M HNO3. In which of the two acids will the silver rods dissolve under standard conditions ? Given : EAg/Ag = – 0·79 volt, ENO 2/NO = + 0·96 volt. (b) A 0·1M acetic acid solution ionizes to 1·2%. What is its Ka ? 2. (a) Why NH3 is more soluble than PH3 in water ? (b) Why BH3 dimerizes but BF3 does not ? (c) The complex K[PtCl3C2H4] has 3 chlorine atoms bonded to platinum. Why does the chlorine atom lying opposite to ethene have higher bond length ? 3. (a) An electron in which orbit of lithium ion will have same energy as an electron in the second orbit of hydrogen. (b) I 2 + CH3COCH3 → CH3COCH2I + HI For the reaction, Rate = K[CH3COCH3] 1[H+] 1. What is the order of the reaction with respect to I 2 ? Also give the total order. [I2] mol/lit [CH3COCH3] mol/lit [H+] mol/lit 3 × 10– 3 2·5 × 10– 2 1·5 × 10– 3 Rate mol/lit sec 2 × 10– 2 Also find the value and unit of the rate constant from the data given above. (c) For a photoelectron, the frequency is given by 1 1 the expression ν = 3·3 × 1015 – · If 22 n2 ° the wavelength of the photoelectron is 6600 A, H+ 0 0 (b) (c) (d) 6. (a) (b) 7. (a) For 0·5 M H2SO3 solution K a1= 1·8 × 10– 2 and (b) (c) ( ) 8. (a) (b) (c) 9. (a) what will be the value of ‘n’ ? 4. (a) Complete the reaction given below : NOH → A → Nylon 6 Rearrangement Beckmann’s Polymerize (b) Cyclohexanoxime C.S.V. / February / 2008 / 1644 10. (a) Why PO 43– ions exist but NO43– ions don’t ? (b) Why B 2 is paramagnetic but C2 is not ? (c) For an octahedral field splitting ∆0 > P when the pairing energy is less and ∆ 0 < P when pairing energy is higher. Explain the spin magnetic moments acquired by d 5 and d 6 configurations of metal ions in this field. ∴ V3 = potential at P due to the outer Kq surface of the shell = R Kq K(– q) Kq Kq V = + + = x R R x 2 , n 3. (a) The ratio of specific heats, γ is given by γ = 1+ ANSWERS Physics 1. (a) Let n photons strike on the surface per second. Then the force exerted by the photons, → pi where ∆p = Change in momentum per second → pf → → → → → ∆ p = pi – (– pf ) = pi + pf → → → h (˙.˙ | pi | = | pf |) | ∆ p | = 2pi = 2 × λ Fλ 2h or n = 2h λ Here, F = 10 N, λ = 439 nm = 439 × 10–9 m, h = 6·625 × 10– 34 Js Putting all these values, we get 10 × 439 × 10 – 9 n = = 3·313 × 10 27. 2 × 6·625 × 10– 34 (b) Water can be boiled without supplying heat if it is kept inside a closed vessel and increasing the pressure on its surface such as water in a pressure cooker and also it can be boiled if pressure is reduced in a vessel containing water. 2. (a) Here, A 1 = a, A2 = a and A = a We have, A = A12 + A22 + 2A1A2 cos (φ 1 – φ 2) ∴ F = n× or or or a2 = a2 + a2 + 2a.a cos (φ1 – φ 2) a2 = 2a2 + 2a2 cos (φ 1 – φ 2) F = n (∆p) where n is the degree of freedom. We know that the degree of freedom of a molecule is the number of independent ways in which it can have energy. Also, a monoatomic molecule can move linearly but can’t rotate, so it can have energy along three directions viz. x, y and z axes only. ∴ For a monoatomic gas, n = 3 2 2 Hence, γ = 1+ =1+ n 3 5 3 (b) From the figure the Boolean expression is – – Y = (A. B ) or γ = After simplification, we get Y = A+B=A+B – – (˙.˙ A.B = A + B by absorptive law of Boolean algebra) ∴ The truth table is as given below : Inputs A 0 0 1 1 B 0 1 0 1 Output Y 0 1 1 1 = = 4. (a) The differences between nuclear force and coulombic force are tabulated below : Nuclear Force 1. The force acting between the constituent particles of an atom is called nuclear force. a2 = 2a2 [1 + cos (φ1 – φ 2)] 1 or = 1 + cos (φ1– φ2) 2 1 or cos (φ1 – φ 2) = – 2 ⇒ φ 1 – φ 2 = 120° (b) Let the point at a distance x be P. Then the potential at P is given by V = V1 + V2 + V3 a where, V1 = potential q at P due to the solid Kq sphere = x Coulombic Force 1. The force acting between two charged particles is called coulombic force. 2. It is charge indepen- 2. It is charge dependent. dent. 3. It is a short-range force. 3. It is a short as well as a long range force. x r R P V2 = potential at P due to the inner K(– q) surface of the shell = x 4. It does not obey the 4. It obeys the inverse inverse square law square law of disbut it obeys inverse tance. law of r n where n is a very large number. 5. It is stronger than coulombic force. 5. It is weaker than nuclear force. C.S.V. / February / 2008 / 1645 (b) Let θ be the required angle. Then tan θ = where, v x vx · vy = horizontal component of velocity when the object strikes the ground at B = ux + ax t = ux (˙.˙ ax = 0) = 100 m/s ux = 100 m/s uy = 0 Torque acting on the rod about the axis of rotation l = mg 2 () A Also, torque = Iα, where I = moment of inertia of the rod about the axis of rotation ml 2 = 3 If aCM be the acceleration of the centre of mass, aCM then α = (l /2) mgl ml 2 aCM l ∴ mg = Iα or = × 2 2 3 l /2 () h = 2000m B O vy v vx v y = vertical component of velocity at B = uy + ayt = gt (˙.˙ uy = 0, ay = + g) But ∴ h = uy t + vy = g × = ∴ ∴ tan θ = 1 1 a t 2 = gt 2 ⇒ t = 2 y 2 2h = 2gh g 2h g 2ml 2.aCM mgl = 2 3l 3gl 2 3g ⇒ aCM = = ms– 2 4ml 2 4m 3 × 10 = ms– 2 4m 15 ∴ aCM = ms– 2 2m where m is the mass of the rod. (b) We know that or n ∝ T where, n = frequency and T = tension in the wire. n = KT 1/2 where, K is the constant of proportionality. ∆n ∆T = 2 , n T where, ∆T is change in tension and ∆n change in frequency. ∆T % change in tension = × 100 T ∆n × 100 = 2 n 2×5 × 100 = 2 = 500 6. (a) Let us consider a current element d y at a distance y from the wire carrying current I. b The magnetic field acting on the current element due to infinii tely long conductor, l 2 × 10 × 2000 = 200 ms– 1 v x 100 1 = = v y 200 2 ( ) θ = tan – 1 () 1 2 = = = = = = [ML2T– 1] [MLT– 1] [ML2T–2] [ML2T– 2] [MLT– 2] [ML2] (c) Clearly, [Angular momentum] [Impulse] [Energy] [Torque] [Force] [Moment of Inertia] ( ) Hence, the dimensional formulae of energy and torque are same. 5. (a) When the support is withdrawn, the rod will rotate about the axis passing through the hinged point. l B = µ0 2πy y dy a x I ∴ Force acting on dy length, d F = idy B Now, total force acting on the conductor of length (l ), l l /2 l /2 C.M. mg (x + l ) (x + l ) F = ∫ x (x + l ) idyB = ∫ x µ0I 2πy idy = µ0Ii 2π = ∫ x () 1 y dy = µ0Ii 2π = [lny ]x (x + l ) C.S.V. / February / 2008 / 1646 = = or F = µ0Ii 2π µ0Ii 2π µ0Ii = [ln(x +l ) – ln(x)] = or a = mg [ (x + l ) ln x 1+ l x 2π (b) We know that, charge density q σ = or q = (4πr 2)σ 4πr2 Also, the potential on the surface of a conductor of any configuration is constant, q i.e., = constant 4πε0r or or (4πr 2)σ 1 · = constant r 4πε0 = ln ] ( ) ( [ m+ I r2 ) ] Putting the values of m, g, I and r, we get a = 2 × 10 20 = ms– 2 2+8 0·32 2+ (0·2)2 ∴ 2ms– 2 = 20 ms– 2 10 8. (a) Energy of capacitor, 1 1 EC = × CV2 = × (6 × 10 – 6) × (6) 2 2 2 = 108 × 10 – 6 J When the capacitor is connected to the inductor, the magnetic energy, 1 2 Li EB = 2 1 or EB = × (5 × 10 – 3) i 2 3 = (2·5 × 10– 3) i 2 According to question, 1 EB = × EC or (2·5 × 10– 3)i 2 2 1 = × 108 × 10 – 6 3 or i2 = 108 × 10 – 6 3 × 2·5 × 10– 3 = 43·2 × 10– 3 3 r.σ = constant constant 1 or σ = ⇒σ∝ r r Hence, the charge density varies inversely with distance but potential remains constant. 7. (a) First case : In this case, the battery is being i = 4A charged. ∴ V = E + ir where E = emf of battery and r = internal resistance of the battery or 12 = E + 4r Second case : In this case, the battery is being discharged. ∴ …(1) i = 2A = 14·40 × 10– 3 = 144 × 10– 4 ∴ or i = 144 × 10 – 4 A V = E – ir or 9 = E – 2r …(2) Subtracting (2) from (1), we get 3 1 3 = 6r ⇒ r = = = 0·5 Ω 6 2 Putting this value of r in (1), we get 12 = E + (0·554) or E = 10 V ∴ E = 10 V and r = 0·5 Ω (b) Considering the rotational motion of the pulley, we get a Tr = Iα or Tr = I r where, a is the linear acceleration of the pulley in downward direction. a Ia T ∴ T = 2 …(1) r T Now, considering the m = 2kg motion of the mass m (= 2 kg), we get mg i = 12 × 10 – 2 A () (b) Let A be the highest point and h′ the distance between the point p and A. Here, A t p = time taken in going from p to A and again from A to p q h′ ∴ Time taken in coming from h tp A to p = 2 p Similarly, time taken in falling from tq A to q = · O 2 Now, and h′ = h′ – h = 1 g 2 1 g 2 () () tp 2 tq 2 2 = 2 gtp2 8 gtq2 8 …(1) …(2) = or or or a ( mg – T Ia mg – 2 r Ia ma + 2 r I m+ 2 r = ma = ma [from (1)] = mg = mg Subtracting (2) from (1), we get gt p2 gt q2 g 2 h = – = (tp – t q2) 8 8 8 8h or g = t p2 – t q2 which is the required relation. 9. (a) For the coil n, we have 15 × φ n = MIm or 15 × 1·8 × 10– 4 = M × 2 ∴ M = 13·5 × 10– 4 H ) C.S.V. / February / 2008 / 1647 For the coil m, we have 10 × φ m = MIn or φm MIn 13·5 × 10– 4 × 3 = = 10 10 = 40·5 × 10– 5 Wb. (b) We know that the velocity of transverse wave in a string of mass per unit length µ and tension T is given by v= T = µ 360 = 3 × 102 m/s = 300 m/s 4 × 10 – 3 Putting the values of m, R, B and Tmax, we get q = = 2 × 9·1 × 10– 31 × 2·02 × 14– 19 50 × 10 – 2 × 6·7 × 10– 6 6·04 × 10– 25 3·35 × 10– 6 coulomb coulomb = 1·8 × 10– 19 C (approx.) (b) We have, from Curie’s law, B I ∝ T where, I = intensity of magnetization B = magnetic field T = absolute temperature when B and T are tripled, the new intensity of magnetization, 3B B I′ ∝ ∝ 3T T So, intensity of magnetization will not change. ˙. ˙ The string produces two consecutive resonant frequencies 375 Hz and 450 Hz with a tuning fork. v We have n = 375 …(1) 2l v = 450 …(2) and (n + 1) 2l where n is an integer. Subtracting (1) from (2), we get v = 75 (n + 1 – n). 2l 300 v = 75 or = 75 or 2l 2l ⇒ l = 2 –3 Chemistry 1. (a) Less reactive metals like Ag are oxidised more readily in HNO3 than HCl because HNO3 behaves as an oxidising acid while HCl as nonoxidising acid. Ag + 2HNO3(dil) → AgNO3 + NO2 + H2O Further if we calculate E° for the reaction, we cell ˙. ˙ Mass per unit length (µ) = 4 × 10 m or = 4 × 10 – 3 l or ∴ m = 8 × 10 – 3 kg. get ° Ecell = Ecathode – Eanode = + 0·96 – (– 0·79) = + 1·75 volt From the relation ∆G = – nFE0, we find ∆G to be negative. Since the E° is +ve, the reaction would be cell m = 4 × 10 – 3 × l = 4 × 10– 3 × 2 10. (a) Here, wavelength (λ) of photon ° = 3300 A = 3300 × 10– 10 = 3·3 × 10– 7 m Work function (φ) of the surface = 2·5 eV = 2·5 × 1·6 × = 4·0 × 10– 19 J Energy of incident photon = hc 6·625 × 10– 34 × 3 × 10 8 = J λ 3·3 × 10– 7 10– 19 J spontaneous under standard conditions. (∆G = – nFE) (b) Given α = 1·2 = 0·012 and C = 0·1 M. 100 Ka = α 2C 1–α = 6·02 × 10– 19 J By Einstein’s equation of photoelectric effect, we have E = φ + Tmax or Tmax = E – φ = 6·02 × 10–19 – 4·0 × 10–19 J = 2·02 × 10– 19 J Now, as the emitted electrons enter a magnetic field and turn in a circular path, the radius of the path is given by R = 2m (T max) ⇒ q= qB 2m (T max) RB 2. (a) Intermolecular hydrogen bonding is found in NH3 which increases its solubility whereas hydrogen bonding does not exist in PH3 which makes it less soluble. (b) BF 3 is a monomeric covalent species. It does not dimerize like BH3 because the lone pairs on the halogens interact with the vacant p orbital on boron due to which the electron deficiency of the central boron is satisfied whereas this property is not seen in BH 3. (c) [PtCl3(C2H4)]– is essentially square planar — — CH 2 CH 2 Cl Pt Cl Cl C.S.V. / February / 2008 / 1648 A dative bond in which electron pair in the filled π orbital on ethene overlaps with and empty hybrid orbital on the metal, giving a σ bond. Overlap also occurs between a filled metal d orbital and an empty antibonding orbital on ethene. This results in back bonding. Due to the building up of high electron density between Pt and ethene, the bonded electron Pt—Cl bond opposite to Pt-ethene, experiences more repulsion and hence Pt–Cl bond is rather longer than other Pt—Cl bonds. 3. (a) For H-like particles energy of electron in nth shell. 1312 Z2 En = – kJ mol– 1 n2 ∴ Energy in 2nd orbit of hydrogen 1312 × 12 kJ mol– 1 22 1312 = – kJ mol– 1 4 1312 × 32 and for Li ++, En = – 62 – 1312 = kJ mol– 1 4 Therefore, if the electron is in sixth orbit of Li++, energy would be same. (b) As the rate law expression does not involve the concentration of I2, therefore, order of reaction w.r.t. I2 is 0. As given in the expression that Rate = k[CH3COCH3][H+]. The total order of reaction is 2. Now, substitute the values in the expression. 2 × 10 – 2 = k[2·5 × 10– 2][1·5 × 10– 3] = – ∴ k = 2 × 10 – 2 mol lit– 1 s–1 2·5 × 10– 2 × 1·5 × 10– 3 mol2 lit–2 10– 3 carboxylic amino acid containing six carbon atoms. This polymerises to give nylon-6. H N O → H2N—(CH2)5—COOH Heat Amino caproic acid Caprolactum (A) H 2O O || ∆ → —NH—(CH2)5—C—n [ ] – H2O Nylon-6 (b) (i) 2-chloro 3-methyl pent-1, 4-diene is optically active as it contains chiral carbon atom. Cl | * H2C = C—CH—CH = CH2 | CH3 Note : A chiral carbon atom is attached to four different atoms or groups. Ph—C≡ CH3 → OH | Ph— C—CH2 Hg2+‚ dil H2SO 4 (c) (i) O || Ph—C—CH3 Acetophenone (ii) Carbylamine reaction—When a primary amine reacts with chloroform and alcoholic KOH give isocyanides or carbylamines which have unpleasant smell. Ar—NH2 + CHCl3 + 3KOH (alc.) → → Ar—N==C + 3HCl + 3H2O Aryl carbylamine ∆ → Tautomerises = 0·535 × (c) Given = 5·33 × 10– 4 mol– 1 lit sec– 1 or L mol–1 sec –1 1 1 ν = 3·3 × 1015 – 22 n 2 ( ) and ∴ 3 × 108 ° λ = 6600A = 6600 × 10–10 metre c ν = λ c 1 1 = 3·3 × 1015 – 22 n 2 λ = 3·3 × 1015 6·6 × 10–7 ⇒ ( ) ( ) 1 1 – 4 n2 1 1 – 4 n2 1 1 138 1000 – 552 448 ⇒ = – = = n2 4 1000 4000 4000 4000 63·24 ⇒ n2 = ⇒n= = 2·988 ≈ 3 448 21·16 4. (a) Caprolactum on heating with traces of water gives ε-amino caproic acid which is mono0·138 = (d) As the given alkene is buten-2 which is a symmetrical alkene thus, addition of HBr to it does not follow Markovnikoff’s rule hence the product formed would be same in both cases i.e., in presence of peroxide and in absence of peroxide. 5. (a) The formula suggests that the compound is CH3 | CH3—C—OH | CH3 i.e., a tertiary alcohol, therefore, alkane from which it is formed must be CH3 CH3 | | H2O/H2SO 4 CH3—CH → CH3—CH—OH | | CH3 CH3 Isobutane t-butanol C.S.V. / February / 2008 / 1649 (b) Dipeptides are formed when two molecules of αamino acids are joined together by peptide bonds. The possible dipeptide formed from these amino acids would be H2N—(CH2)5— CH—COOH | NH2 + H2N— CH—(CH2)3—COOH | COOH H2N–(CH2)5– CH–CO–NH–CH–(CH2)3–COOH | | NH2 COOH COOH | and H2N—(CH2)5—CH—NH2 NH2 | + HOOC—CH—(CH2)3—COOH → NH2 COOH | | H2N–(CH2)5– CH–NH–CO–CH–(CH2)3–COOH (c) At isoelectric point amino acid exists as Zwitter ion (a dipolar ion) R | + NH3—CH—COO– Zwitter ion C—H bond are attracted more towards carbon than towards H atom. As a result H-atom is being less tightly held by carbon and hence can easily be removed as a proton. CH3—CH2—C ≡ C—H + NaNH2 → CH3—CH2—C ≡ C—Na + NH3 Sodium butylide (b) Following are the structures of DNA purines : NH2 O NH N N Adenine NH HN N H2N N Guanine N 7. (a) Given : H2SO3 HSO 3– H2SO3 at t = 0 at teq 0·5 0·5 – α H+ + HSO3– Ka1 = 1·8 × 10 – 2 H+ + SO32– Ka2 = 8·3 × 10 – 5 H+ + HSO3– α α Zwitter ions are neutral but in presence of acid dissociation favours formation of cation. R R | + | Acid + H3N—CH—COO– + H+ → NH3—CH—COOH Zwitter ion Cation where α = degree of dissociation. [H+] [HSO3–] Now, Ka 1 = [H2SO3] = 1·8 × 10 – 2 (0·5 – α) or α2 = 0·9 × 10– 2 – 1·8 × 10– 2 α 2 + 1·8 × 10 – 2 α – 0·9 × 10– 2 = 0 or α ∴ Ka1 = On solving the value of α, we get α = 0·08625 mol/L [H+] = [HSO3–] = 0·08625 mol/L [SO 3– –][H+] and Ka 2 = [HSO3–] Now, substitute of the value of [H+] and [HSO 3–] in above equation, we get [SO 3– –][0·08625] 8·3 × 10– 5 = [0·08625] [SO 3– –] = 8·3 × 10– 5 mol/L ∴ (b) t=0 teq α2 Hence it will coagulate at cathode (– ve electrode). In presence of alkali (pH > 7) dissociation of Zwitter ion favours formation of an anion R + | H3N—CH—COO– + OH– Zwitter ion R | → H2N—CH—COO– + H2O Base Anion Hence it will coagulate at anode (+ ve electrode) (d) The structures of 1-butene and 2-butene are as CH3—CH2—CH = CH2 CH3—CH = CH—CH3 1-butene 2-butene N2O4 1 1 – 0·4 2NO 2 0 0·4 × 2 = 0·8 More substituted alkenes are more stable than less substituted alkenes. The more is the stability, the lesser is the reactivity. Note : More substituted alkenes contain more no. of alkyl group. Hence, 2-butene is more stable and less reactive to Br2 than 1-butene. 6. (a) In 1-butyne the terminal hydrogen is weakly acidic whereas in 2-butyne, there is no such hydrogen is available. The reason is that carbon atoms of the triple bond are more electronegative due to sp hybridization. Due to greater electronegativity, the shared pair of electron of Total no. of moles = 1 – 0·4 + 0·8 = 1·4 x NO 2 Now, Kx = xN O [ [ 2 4 ] ] 2 where ∴ x NO 2 0·8 0·6 = and x N O = 1·4 2 4 1·4 Kx = ( ) 0·8 1·4 0·6 = 0·76 1·4 As Kp = Kx (Ptotal)∆ng = 0·76 × (1) = 0·76 atm (˙.˙ ∆ng = 2 – 1 = 1) C.S.V. / February / 2008 / 1650 (c) N2 1 mole + 3H 2 3 mole 2NH3 2 mole (b) Following are the electronegativities of elements involved in bonding C = 2·55; N = 3·00; P = 2·2 Structures of (CH3)3N and (C2H5)3P can be represented as : Both are pyramidal. ·· N i.e., 1 mole of N2 react with 3 mole of H2 to form 2 mole of NH3, but the amount of H2 taken is 4 moles thus the H2 will be the residual gas and hence PH × V = nRT 2 ·· P CH3 PH × 10 = 1 × 0·0821 × 298 2 CH3 C 2H 5 C2H 5 C2H 5 PH 2 = 2·45 atm. CH2 Note : The effective volume of the vessel occupied by H2 gas is 20 – 10 = 10 litre. 8. (a) Fluorine (F 2) is more reactive than chlorine (Cl2). This is on account of lower bond dissociation energy (158·8 kJ/mol) of F2 molecule as compared to Cl 2 molecule (242·6 kJ/mol). Lower bond dissociation energy of F2 is due to the greater interelectronic repulsion which is consequently due to very small size of F atoms. 2– 2– (b) CrO4 is more oxidising than MoO4 . CrO2– ion 4 The dipole moment due P—C bonds in (C2H5)3P are in opposite direction of the dipole moment of the lone pair on P atom, while the dipole moments of N—C bonds in (CH3)3N are in same direction as the direction of dipole moment of lone pair on N atom. Hence, the dipole moment of (CH3)3N is definitely higher than that of (C2H5)3P. 10. (a) P(15) = 1s22s 22p63s 23p33d 0, N(7) = 1s22s 22p 3 As P contains vacant d-orbit, thus it can exceed its octet and utilize d-orbitals for multiple bonding, whereas N cannot exceed its octet due to the absence of d-orbital and hence cannot form NO43– ion. (b) The molecular orbital configuration of B 2 is B2 KK σ (2s)2 σ* (2s)2 π(2px )1 π (2py )1. As it contains two unpaired e–, therefore, it is paramagnetic and the molecular orbital configuration of C2 is C2 = KK σ (2s)2 σ* (2s)2 π (2px )2 π (2py )2. At it does not contain any unpaired e– thus it is diamagnetic. (c) When ∆0 < P eg ∆0 ∆0 eg has very high standard reduction potential as 2– 2– compared to MoO4 and hence CrO4 is strong oxidising agent. Moreover, the most stable oxidation state of Mo is + 6 and hence its reduction to lower oxidation state is rather difficult. (c) Two substances can be represented as : ·· ·· SiH3—O—SiH3 and CH3—O—CH3 ·· ·· In case of (SiH3)2O, Si—O bond shows pπ-dπ bonding due to availability of vacant d orbitals on Si atom. As a result of this lone pair of electrons on O atom are engaged in π bond formation and are not available for donation. In case of (CH3)2O, CH3—O bond cannot form such π bonding as C atom does not possess vacant d orbitals. The lone pairs of oxygen atom in this case are available for donation. Hence (SiH3)2O is a weaker base than (CH 3)2O. 9. (a) The empirical formula of the compound is ·· PtCl2(NH3)2. Since chlorine is not precipitated by AgNO3 and hence all Cl atoms are in coordination sphere. Since the product of churning with AgNO3 are [Pt(NH3)4](NO3)2 and Ag2[PtCl4], the compound should be a dimer of PtCl2(NH3)2 i.e., a coordination isomer of PtCl2(NH3)2 [Pt(NH3)4] 2+ [PtCl4] 2– T2g d5 (5 unpaired electrons) d6 T2g (4 unpaired electrons) µ = (5 + 2) 5BM µ = (6 + 2) 6BM When ∆0 > P eg µ = √1(1 + 2) ∆0 ∆0 eg µ = √0(0 + 2) [Pt(NH3)4] (NO3)2 + Ag2[PtCl4] C.S.V. / February / 2008 / 1651 → AgNO3 T2g d5 (Unpaired electron) d6 T2g (Zero unpaired electron) Paper II Max. Marks : 100] Biology [Time : 2 Hours 1. Write the location and function of the following : (a) Cytoskeleton (b) Synergid (c) Phytol chain (d) Sieve tube element (e) Centromere 2. Refer the following diagrams (a) to (e) : P (a) Mucilagenous sheath Q (c) Which type of ovule is shown in diagram (c) ? Give one reason. (d) What is the type of flower called given in diagram (d) ? Give examples of such flower. (e) Which type of aestivation is shown in (e) ? Give examples. 3. Write down the types of placentation, inflorescence and fruit of the following : (a) Pea (b) Coriander (c) Wheat (d) Sunflower (e) Mustard 4. Differentiate between the following [Give one important difference] : (a) Culm and Caudex (b) Slime moulds and Fungi (c) Biological magnification and eutrophication (d) White rust and Brown rust 5. Match the Column I and Column II : Column II (i) Neurospora (ii) Operon (iii) Reverse transcriptase (iv) Okazaki fragments (v) Transformation (vi) Transduction (vii) Conjugation 6. Fill in the blanks with suitable words given in boxes : Marsilea Coralloid root Monoecious Usnea Pneumato- Geitonophores gamy Vivipary Eichhornia Cycas Head P Sheath (b) Column I (a) Jacob and Monod (b) One gene one enzyme hypothesis (c) Griffith (d) Temin and Baltimore (e) DNA polymerase Q Chalaza Raphe Nucellus Integuments Antipodal cells Embryo sac Secondary nucleus Egg cell Synergids Micropyle Hilum Funicle Pinus Dioecious Xenogamy Oscillatoria (c) (d) Ovary Thalamus (e) (a) What is shown in diagram (a) ? What are (P) and (Q) ? (b) Tell whether the diagram (b) is T1-phage or T2phage. What are (P) and (Q) ? (a) Rhizophora has both……… and ………… (b) Aquatic plants are ………… (Pteridophyte) and …………… (Angiosperm). (c) Plant bearing both male and female flowers is called ………… (and the pollination between different flowers of the same plant is called ……… (d) …………… has …………… 7. (a) Give any two conditions for seed habit. (b) Among the pairs which one shows the tendency of seed habit ? (i) Lycopodium and Equisetum (ii) Selaginella and Pteris (iii) Dryopteris and Pteridium. (c) Suggest three ways to break seed dormancy. 8. Fill in the blanks : (a) Mutation can be defined as………… variation. (b) Mutation leads to evolution of ………… of a gene. (c) Genes expressed only in homozygous state are ………… C.S.V. / February / 2008 / 1652 9. 10. (a) kept on complete diet (b) fed basal diet without nutrient to be tested (c) basal diet + carotene by feeding carrots. 14. Out of following animals—answer the questions : Whale, Earthworm, Bat, Starfish, Scorpion, Honey bee, Peafowl. (a) Which animal is different from rest of animals ? (b) How many of these are poisonous, which class they belong to ? (c) Which animals belong to same class ? 11. (d) Which animals are only representatives of their phylum ? (e) Which of the following have 3 ear ossicles ? 12. Name the ossicles in the order of sound transmission. 15. The following diagrams represent Age-Sex pyramid of (a) developed (b) developing nation. On the basis of this graph give the answer of the following questions : (a) What types of phases are shown by the graphs in (a) and (b) ? (b) What is the position of job opportunity in graph (a) and is it possibility of migration of population from developing nation to developed nation ? Males Age 75 70 60 50 40 30 20 10 0 120 80 40 0 40 80 Females (d) When both alleles of a gene are expressed it is………… (e) Gene, which is most frequent is ………type. (a) Who discovered photoperiodism ? (b) Select one SDP and one LDP from following plants : Chrysanthemum, Rice, Spinach, Barley, Radish. (c) Who gave the term ‘Phytochrome’ ? Give one specific feature of phytochrome. Match the Column I and Column II : Column I Column II (a) Indian Rhinoceros (i) Extinct (b) Acid rain (ii) Endangered (c) Somaclonal (iii) Gir forest variations (iv) Corbett National (d) Protoplasm fusion Park (e) Biopiracy (v) CFC (vi) Nitrogen oxide (vii) Sulphur oxide (viii) PEG (ix) Mutation (x) Turmeric Differentiate between : (a) Habitat and Niche (b) Flora and Vegetation Answer the following questions : (a) Apart from CO 2, name other green house gases. (b) What is the type of movement of twiner and opening of flower ? 13. Three groups of baby mice : a Wt c b Days 320 280 240 200 160 120 160 200 240 280 Population (a) Age Males 75 70 60 50 40 30 20 10 0 120 80 40 0 40 80 Females 320 280 240 200 160 120 160 200 240 280 Population (b) C.S.V. / February / 2008 / 1653 16. Refer the following diagram : Muscle spindle Efferent pathway (receptor) Dorsal root ganglion Gray matter White matter Stimulus Response Interneuron Motor endplate Afferent pathway (effector) Motor neuron late orientation and distribution of cell organelles, intracellular transport and movement of cells. They are of three types–microfilaments, intermediate filaments and microtubules. (b) Synergid—Synergids are the part of egg apparatus which is located inside the embryo sac. Synergids nourish the ovum. Chalazal end Antipodals (a) (b) (c) What kind of reflex is it ? Where is it striking ? List the errors in the above diagram. Polar nuclei Central cell Egg Synergids Filiform apparatus 17. (a) Nictitating membrane and Vermiform appendix are examples of which type of organs. (b) Define vestigial organs. (c) Which of the following four can be categorised in the same category ? Out of coccyx, mammary glands muscles of external ear and opacity of eye due to cataract are these 4 in same categories. 18. Refer the following diagrams (a) and (b) : Micropylar end (c) Phytol chain—Phytol chain tail in association with porphyrin (ring head) through ester linkage forms a chlorophyll molecule. X CH3 in Chl a CHO in Chl b CH3 H 3C Porphyrin ring (Head) H N H 3C III I N Mg N IV CH3 N H II 1 (a) 2 1 Ester linkage O { O O O O CH3 H3C CH3 CH3 CH3 2 (b) Write the functions of (1) and (2) in each diagram. 19. In photorespiration, RuBP is oxidized to form glycolate and glycerate. Glycolate enters the PCO cycle to regenerate glycerate. Draw a flow diagram of photorespiration to show different intermediates and the cell organelles involved. Phytol (Tail) H 3C H3C H3C H3C Chlorophyll a and b CH3 ANSWERS 1. (a) Cytoskeleton—They are extremely minute, fibrous and tubular structures which form the structural frame work inside the cell. Cytoskeletal structures occur only in eukaryotic cells. They were discovered with the help of fluorescence microscopy. Cytoskeletal structures maintain shape of the cell and its extensions, regu- (d) Sieve tube element—Sieve tubes are elongated tubular conducting channels of phloem. Each sieve tube is formed of several cells called sieve tube elements or members, sieve tube cells or sieve elements. Sieve tube members are placed end to end. The endwalls are generally bulged out. They may be transverse or oblique. They have many small pores or sieve pits. Each sieve pore is lined by a layer of callose. Its function is translocation of solute. C.S.V. / February / 2008 / 1654 (e) Centromere—The two chromatids of chromosomes are attached to each other by a narrow area which is also known as primary constriction. It is the site at which the spindle attaches during cell division which is also concerned with anaphasic movement of chromosome. On the basis of position of centromere chromosomes shape is also determined like telocentric, acrocentric, submetacentric and metacentric. 2. (a) Nostoc is a filamentous blue-green algae. The region marked as (P), in the figure is Heterocyst which is large-sized, pale coloured thick-walled cell which occurs in terminal intercalary or lateral position. It is specialised to perform nitrogen fixation. The region marked (Q) is Akinete which helps in asexual reproduction. (b) It is T2 -phase of bacteriophage which comes under T-even phage and is also known as coliphages. The region marked as (P) in the figure is collar and the region marked as (Q) is tail fibre. (c) This ovule is anatropous because the ovule is inverted at 180° angle which is also called resupination. In this case, micropyle and funicle are closer. (d) Epigynous flower (inferior ovary)—The thalamus grows further upwards thus enclosing the ovary and finally fusing with it. (e) This is vaxillary aestivation which is seen in the whorl of five petals where the posterior one is the largest and almost covers the two lateral petals and the latter in turn nearly overlap two anterior petals, e.g., Papilionaceae (soyabean, gram, etc.) 3. (a) Pea—In this case, placenta develops along the Junction of the two margins of the carpels in one-chambered ovary which is known as marginal placentation. Inflorescence may be solitary axillary or terminal, raceme. Fruit type is pod or legume, where fruit is formed from superior unilocular ovary of a monocarpellary pistil. (b) Coriander—In this case, the ovary is multichambered and placenta bearing ovules develop from the central axis which is known as axile placentation. Inflorescence is compound umbel because peduncle or main axis gives branches repeatedly once or twice in umbel of racemose manner. Fruit is of cremocarp type where fruit develops from anterior, bilocular ovary of a bicarpellary syncarpous pistils with persistant stylopod or stylopodium. (c) Wheat—Since the ovary is unilocular and placenta develops directly on the thalamus and bears a single ovule at the base of the ovary. So, the placentation is of basal type. The inflorescence is spikelet which bears two empty glumes (bracts) and also bears IIIrd bract (flowering glumes) or lemma or lower palea. The fruit is of caryopsis type. (d) Sunflower—In this case, placentation is again of basal type because the ovary is unilocular and placenta develops directly on the thalamus and bears single ovule at the base of ovary. The inflorescence is capitulum or head because its receptacle is flattened, bear sessile flowers and florets in a centripetal manner. Fruit type is of cypsela–in this case fruit develops from a monocarpellary pistil with superior, unilocular and uniovuled ovary. (e) Mustard—Parietal placentation is present in mustard where ovary is one chambered and placenta bears the ovules which develops on the inner-wall of the ovary and their position corresponds to the confluent margins of the carpels and the number of placenta is equal to the number of carpel. The fruit is of siliqua type. 4. (a) Culm—It is an erect stem in members of poaceae. The stem with solid nodes and hollow internodes is called culm, e.g., Bambusa is branched cylindrical. Caudex—The caudex is an enlarged, woody base of the stem or trunk (located just below the ground) on some plants–it is used for water storage. Many desert plants have a caudex, an adaptation to dry conditions. Some palms, cycads, and succulents have a caudex, e.g., Date palm. (b) Slime moulds—They are vegetative cells without cell wall while spores are with cellulosic cell wall Protistans that may represent a transition between protistans and fungi are called slime moulds. Phylum of fungus like organisms within the kingdom Protista, commonly known as true slime moulds. They exhibit characteristics of both protozoans (one-celled microorganisms) and fungi. Distributed worldwide, they usually occur in decaying plant material. Fungi—They are vegetative cell with cell wall which is made up of either fungal cellulose or chitin or both. Saprophytic and parasitic organisms that lack chlorophyll and include molds, rusts, mildews, smuts, mushrooms and yeast; singular, fungus. They are aerobic, multicellular, nonphoto-synthetic, heterotrophic microorganisms. There is gradual and progressive simplification and ultimate elimination of the sexual apparatus from the lower to higher forms. (c) Biological magnification—Refers to the process whereby certain substances such as pesticides or heavy metals move up the food chain, work their way into rivers or lakes, and are eaten by aquatic organisms such as fish, which in turn are eaten by large birds, animals or humans. The substances become concentrated in tissues or internal organs as they move up the chain. Eutrophication—Over-enrichment of a water body with nutrients, resulting in excessive growth of organisms and depletion of oxygen C.S.V. / February / 2008 / 1655 concentration. The process by which a body of water becomes either naturally or by pollution rich in dissolved nutrients (as phosphates) and often shallow with a seasonal deficiency in dissolved oxygen. e.g., disease of crucifers caused by Albuginaceae (Peronosporales: Oomycota). Zinc oxide; the powdery product of corrosion of zinc or zinc-coated surfaces. Fungus causing a disease characterized by a white powdery mass of conidia. (d) White rust—It is a fungal disease caused by Cystopus candidus or Albugo candida on vegetative parts of crucifers. Brown rust—It is a fungal disease caused by Puccinia recondita on vegetative parts of wheat plant. 5. (a) Jacob and Monod gave the ‘operon’ model to explain the regulation of gene expression in the bacterium E. coli in 1961. (b) Beadle and Tatum proposed the ‘one gene one enzyme’ theory. One gene codes for the production of one protein. ‘One gene one enzyme’ has since been modified to ‘one gene one polypeptide’ since many proteins (such as hemoglobin) are made of more than one polypeptide. (c) An Okazaki fragment is a relatively short fragment of DNA (with an RNA primer at the 5′ terminus) created on the lagging strand during DNA replication. These fragments are processed by the replication machinery to produce a continuous strand of DNA and hence a complete daugher DNA helix. It was given by Griffith. (d) Reverse transcriptase is the replication enzyme of retroviruses which was discovered by Temin and Baltimore. 6. (a) Pneumatophores, Vivipary (b) Marsilea, Eichhornia (c) Monoecious, Geitonogamy (d) Cycas and coralloid root 7. (a) Heterospory (which is a condition in which an organism produces two different types and sizes of spores, viz., microspores and megaspores) and formation of female gametophyte inside megasporangium. (b) Selaginella—The advent of reproduction by seeds was one of the most essential evolutionary steps in plant history : the vast majority of living plants are seed plants (spermatophytes). The seed habit includes the following set of defining characteristics : (1) heterospory, (2) occurrence of a single megaspore that germinates within an indehiscent megasporangium (nucellus) retained on the sporophyte), (3) enclosure of the megasporangium in an integument, and (4) capture of pollen before seed dispersal. Contrasting hypotheses about the single/multiple, saltational/gradual origin of the seed habit. Pteris also possesses the seed habit. (c) Mechanical Scarification, Stratification, Impaction. Seeds of almost all plants growing in areas with marked seasonal temperature variations require a period of cold treatment prior to germination, just as buds do. This requirement is usually satisfied by our winter temperatures. There are several ways to break the dormancy of seeds. For example, for some species, if moist seed is exposed to a low temperature for many days, dormancy may be broken and the seed germinates. The average temperature and time are 5°C for 100 days). This horticultural procedure is called Stratification. Mechanical abrasion or breaking the seed coat, which is termed Scarification with a knife, file or sandpaper may allow the hard seed condition or inhibitor to be removed or metabolic activity requisite to germination to be initiated. Impaction or the third condition is to keep the seed under certain condition which breaks the seed dormancy. 8. (a) Sudden or inheritable—It arises due to change in base pair of the genetic materials. (b) New Alleles—It is the term used by Mendel to define genes which is also known as factor. It is the particular form of gene. (c) Recessive—If the gene is in homozygous state then both alleles are expressed. If it is in heterozygous then one is expressed (dominant). (d) Codominance—Both genes of allelomorphic pair express themselves equally in F 1-hybrids. (e) Wild type gene—Wild type refers to the most common phenotype in the natural population. The phenotype can be dominant or recessive. In biology it relates specifically to the difference between a naturally occurring organism and one that has been deliberately mutated. Mutant is an antonym of wild type. 9. (a) Photoperiodism was first studied by Garner and Allard in 1920. (b) SDP (short day plants) or long night plantsChrysanthemum LDP (long day plants-spinach. (c) Borthwick and Hendricks in 1952 reported phytochrome pigment. Induction of flowering. 10. (a) Indian Rhinoceros are now in the list of endangered species because its populations have decreased or habitats have been reduced to the levels that pose immediate danger of extinction. (b) Acid rain refers to precipitation with a pH of less than 5. It is, in fact, a mixture of H2SO4 and HNO3. (c) Somaclonal variations is a type of mutation which is spontaneous in origin. The variations observed among plants regenerated from tissue culture is termed as somaclonal variation. (d) Protoplasm fusion–PEG. (e) Biopiracy–turmeric. C.S.V. / February / 2008 / 1656 11. (a) Habitat and Niche—Habitat means to dwell. It is a specific place where a species or population normally lives in nature, which is a physical area, some particular part of earth’s surface, air, soil or water. It includes both living and nonliving objects. Niche—The word niche literally means a specific place. However, the ecologists use it for the habitat along with the role a species or population plays in its ecosystem. In other words, niche means the total interaction of a species with its environment or its functional position or status in an ecosystem. Xerophytic vegetation is found in desert. (b) Flora—All species of plants that are found in a particular region, period, or special environment. Six floral kingdoms are commonly distinguished : Boreal (Holarctic), Paleotropical, Neotropical, South African (Capensic), Australian and Antarctic. These kingdoms are further broken down into subkingdoms and regions, over which there is some dispute. Vegetation—Types of plant species found in an area constitute the vegetation of the area. It is also the process of growth in plants. An abnormal growth or excrescence (especially a warty excrescence on the valves of the heart) is also known as vegetation. 12. (a) CFCs = 14%, CH 4 = 20%, N2 O = 6% and the rest 60% are CO2. N 2O 6% CFC s 14% 13. (a) According to the given conditions and graph a shows complete growth, b shows retarded growth and c shows moderate growth. (b) Vitamin i.e., Retinol. (c) Hyper-concentration of retinol leads to hypervitaminosis A which is characterised by anorexia, headache, irritability, hepatosplenomegaly, dermatitis, bone pain, loss of hair in patch etc. Overdoses of phyuridoxine can cause peripheral neuropathy. Overdose of ascorbie acid is less likely to cause any problem because it can be rapidly cleared from the body. 14. (a) Earthworm is different from rest of animals because earthworm is monoecious (or) hermaphrodite while rest are dioecious. (b) Scorpion and Honey bee carry poison to defend themselves from the enemy. Scorpion belongs to class–Arachnida of phylum–Arthropoda. Honey bee belongs class Insecta of phylum– Arthropoda. (c) Bat and whale belong to class Mammalia. The bats are the only mammals which have wings and can really fly. Whale is the largest animal in existence and is the inhabitants of the open ocean, strandings and offshore sightings are rare. (d) Earthworm which is also known as Pheretima is the representative of phylum–Annelida. Starfish is found in rocky area floor which is carnivorous, which possess greater power of regeneration and shows autotomy represents the phylum Echinodermata. 20% Methane 60% Carbon dioxide (b) Type of movement of twiner is thigmotropic movement e.g., cucurbit plant. Certain flowers, e.g., Tulip, show thermonasty by opening at high temperature. Support (e) Bat and whales which are the members of class Mammalia, like human have 3 ear oscicles in the order Maleus ↓ Hammer shaped → incus ↓ anvil shaped → stapes ↓ stirrup shaped 3 ear oscicles are the characteristics of mammalia. More growth on upper surface Tendrill Opening Epinasty 15. (a) In the graph (a) population remains stable that means it will be statronary phase or plateau phase, while in the graph (b) it will be the lag phase or exponential phase when the population will increase tremendously. (b) Graph (a) shows the stable growth so the resources available will be more like getting jobs ultimately more professional job opportunity will be in the developed nation which will force to migrate the population from developing nation to the developed nation. Branch C.S.V. / February / 2008 / 1657 16. Muscle spindle (receptor) Afferent pathway Dorsal root ganglion White matter Gray matter Stimulus Response Motor endplate Efferent pathway (effector) 19. CHLOROPLAST O - RuBP PHOSPHOGLYCOLATE-PGA ADP ATP GLYOXYLATE GLYCERATE Interneuron Motor neuron PEROXISOME Fig. : Diagrammatic presentation of reflex action (showing knee jerk reflex) GLYOXYLATE O2 NAD GLYOXYLATE NADH2 (a) It is knee jerk reflex. (b) In case of knee jerk reflex tendon of patella is tapped. Patella is a lens-shaped sesamoid bone situated in front of the knee. (c) Gray matter is replaced by white matter. Efferent Pathway should be afferent pathway and afferent pathway should be efferent pathway because afferent end transmits the impulse via a dorsal nerve root in the CNS (at the level of spinal cord) and the efferent neuron then carries signals from CNS to the effector. 17. (a) These two are the examples of vestigial organs which were functional in ancestors but functionless in modern humans. (b) The organs which occur in reduced form and are useless to the possessor, but are homologous to the fully developed, functional organs of related animals are called vestigial organs. The existence of vestigial or useless organs is satisfactorily explained by the doctrine of organic evolution. These organs were fully developed, functional and necessary in the ancestral forms, due to change in their mode of life. (c) Coccyx, mammary glands in males, muscles of external ear in human (both male and female) belong to the same category of vestigial organs. 18. The given figures (a) and (b) are chloroplast and mitochondria respectively. Function of thylakoids which is marked as (1) figure (a)— In thylakoids the primary process of energy transduction occurs i.e., light energy is converted into chemical energy. The reaction centres photosystem I and II are present in thylakoids. It also increases the surface area of inner membrane for photosynthesis. Function of stroma which is marked as (2) in figure (a) stroma is the site of dark reaction of photosynthesis. It has ATP synthetase and e– carriers. It has enzymes for amino acid synthesis, calvin cycle, fatty acid synthetase, nucleic acid and protein synthesis sulphate reduction. In the diagram (b) the— (1) represents outer membrane of mitochondria, and (2) represents the matrix. Outermembrane is an outer protective covering. Matrix contains all the enzymes of Krebs cycle. The matrix alongwith inner mitochondrial membrane constitutes the respiratory unit. GLYCINE SERINE MITOCHONDRIA GLYCINE O2 CO 2 SERINE SERINE Fig. : Photorespiratory Pathway ●●● UNT DISCO h Wit TS GIF FREE 31st Chennai Book Fair From 4th to 17th Jan., 2008 At St. George Anglo Indian Hr. Secondary School CHENNAI National Book Fair From 11th to 20th Jan., 2008 At Kasturchand Park NAGPUR Kolkata Book Fair From 30th Jan. to 10th Feb., 2008 At KOLKATA UPKAR’S CAREER BOOKS India's Largest Selling Competition Books C.S.V. / February / 2008 / 1658 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 is equal to the potential at a point on the surface of the conductor. (A) (B) (C) (D) (E) 7. Assertion (A) : If a convex lens is kept in water, its converging power decreases. Reason (R) : Focal length of convex lens in water increases. (A) (B) (C) (D) (E) 8. Assertion (A) : An inductance of 1H is connected to a 200 V, 50 cps supply to have a maximum current of 0·9 A flowing through the circuit. Reason (R) : In an inductive E0 circuit I0 = ωL (A) (B) (C) (D) (E) 9. Assertion (A) : A cricket player lowers his hands to catch a falling ball. Reason (R) : The rate of change of momentum of the ball is reduced in the process. (A) (B) (C) (D) (E) 10. Assertion (A) : Heavy water is preferred over ordinary water as a moderator in reactors. Reason (R) : Mass of the nucleus of heavy water is nearly the same as the mass of a neutron so that the exchange of momentum is more. (A) (B) (C) (D) (E) 13. Assertion (A) : White phosphorus is always kept in kerosene oil whereas metallic sodium can be kept in any anhydrous medium. Reason (R) : White phosphorus reacts with atmospheric oxygen while metallic sodium readily reacts with atmospheric moisture. (A) (B) (C) (D) (E) 14. Assertion (A) : Xenon trioxide on reacting with aqueous alkali gives hydrogen xenate ion which slowly decomposes to give xenon and perxenate ion. Reason (R) : Conversion of XeO3 to hydrogen xenate ion is oxidation and conversion of hydrogen xenate to perxenate is displacement reaction. (A) (B) (C) (D) (E) 15. Assertion (A) : Addition of Br2 to butene-1 gives two optical isomers. Reason (R) : The product of addition of bromine to 1-butene contains one asymmetrical carbon atom. (A) (B) (C) (D) (E) 16. Assertion H I H CH3 Br Br represents total eight PHYSICS 1. Assertion (A) : ‘Light Year’ and ‘Wavelength’ both measure distance. Reason (R) : Both have dimensions of time. (A) (B) (C) (D) (E) 2. Assertion (A) : Light waves are electromagnetic in nature. Reason (R) : Polarisation of light waves is possible. (A) (B) (C) (D) (E) 3. Assertion (A) : If current is flowing through a machine of iron, eddy currents are produced. Reason (R) : Change in magnetic flux through an area causes eddy currents. (A) (B) (C) (D) (E) 4. Assertion (A) : Mercury is used in thermometers. Reason (R) : Mercury is good conductor of electricity. (A) (B) (C) (D) (E) 5. Assertion (A) : For depth of camera, the aperture should be reduced. Reason (R) : Smaller the aperture larger is its power. (A) (B) (C) (D) (E) 6. Assertion (A) : The electric field inside a charged spherical conductor is zero. Reason (R) : Electric potential at all points inside a charged spherical conductor is same and (A) : Molecule, CHEMISTRY 11. Assertion (A) : When carbondioxide is continuously bubbled through lime-water a precipitate is formed which dissolves in due course. Reason (R) : Lime water reacts with CO2 to initially form Ca(HCO3)2 which reacts with excess of CO 2 to give CaCO3 in due course. (A) (B) (C) (D) (E) 12. Assertion (A) : A positive catalyst always lowers the energy of activation of a particular reaction. Reason (R) : A positive catalyst does not affect the value of equilibrium constant. (A) (B) (C) (D) (E) stereoisomers and all are optically active. Reason (R) : Each molecule is the enantiomer of one of the other seven structures. (A) (B) (C) (D) (E) 17. Assertion (A) : CH 3MgI should be prepared under strictly anhydrous conditions. Reason (R) : Grignard reagents reacts very rapidly with water. (A) (B) (C) (D) (E) 18. Assertion (A) : The introduction of acetyl group (CH 3CO–) in alcohols is known as acetylation reaction and when this reaction is carried out with acid chloride, pyridine is essentially required. C.S.V. / February / 2008 / 1659 Reason (R) : Pyridine acts as a specific catalyst for acetylation of alcohols with acetyl chloride. (A) (B) (C) (D) (E) smooth muscle contraction or increase inflammation, causing pain and fever. (A) (B) (C) (D) (E) 24. Assertion (A) : Bacteria produce disease by releasing their enzymes or toxins that damage tissue. Reason (R) : Enzymes and many toxins are secreted only by living bacteria. (A) (B) (C) (D) (E) 25. Assertion (A) : Cataract results from reduced lens flexibility. Reason (R) : Presbyopia results from reduced lens transparency. (A) (B) (C) (D) (E) 26. Assertion (A) : Retrovirus are DNA animal viruses that have a RNA stage. Reason (R) : The DNA is called cDNA because it is DNA copy of viral genome in this case. (A) (B) (C) (D) (E) 27. Assertion (A) : Non-specific defences decline as the skin ages and external glandular activity decreases. Reason (R) : The thymus, which is large during childhood shrinks during adulthood reducing cell mediated immunity and antibody mediated immunity. (A) (B) (C) (D) (E) 28. Assertion (A) : Brain stem includes the area of the brain between the thalamus and the spinal cord. Reason (R) : Hypothalamus nestles at the base of the thalamus. (A) (B) (C) (D) (E) 29. Assertion (A) : The production of a dilute urine requires reabsorbing more water and less ions. Reason (R) : The production of a concentrated urine requires an osmotic gradient in the kidney medulla. (A) (B) (C) (D) (E) 30. Assertion (A) : The driving force for passive absorption is nonmetabolic in origin. Reason (R) : The driving force for active absorption needs energy, derived from metabolic process. (A) (B) (C) (D) (E) BOTANY 31. Assertion (A) : In tropical region no any deciduous forest is reported. Reason (R) : The broad-leaved trees do not lose their leaves in this region. (A) (B) (C) (D) (E) 32. Assertion (A) : The available water to the plants, due to excessive use of fertilizers, becomes hypotonic in relation to cell sap. Reason (R) : The water molecules, as a result, diffuse out of the cells due to endosmosis. (A) (B) (C) (D) (E) 33. Assertion (A) : Sedimentation, a process that has been going on since the earth was formed, cannot occur on land or in waterbody. Reason (R) : Erosion of rocks and weathering produce an accumulation of particles that vary in size and nature, and are called sediment. (A) (B) (C) (D) (E) 34. Assertion (A) : Different chromosomes stain in different patterns. Reason (R) : Dark staining genetic material (heterochromatin) is more highly coiled than the lighter-staining (euchromatin). (A) (B) (C) (D) (E) 35. Assertion (A) : Many globular proteins also undergo small conformational changes in the course of their biological function. Reason (R) : Such changes are associated with the binding of a ligand. (A) (B) (C) (D) (E) 36. Assertion (A) : The genetic material of a prokaryote is a single circle of DNA. Reason (R) : The site or part of prokaryotic cell where DNA is located is termed as nucleosome which is circular in shape. (A) (B) (C) (D) (E) 37. Assertion (A) : Notch when present at the apex of the leaf becomes deep and divides it into two lobes is known as truncate apex. Reason (R) : Syncarpous is the condition of free carpels in a flower. (A) (B) (C) (D) (E) 19. Assertion (A) : On adding a few drops of dilute hydrochloric acid to freshly precipitated ferric hydroxide, a red coloured colloidal solution is obtained. Reason (R) : A freshly prepared salt can be peptised by addition of a small amount of an electrolyte containing an ion common with material to be dispersed. (A) (B) (C) (D) (E) 20. Assertion (A) : Kelvin scale of temperature is called absolute scale and it is also known as thermodynamic temperature scale; while the celsius scale is relative. Reason (R) : Thermodynamic scale has its zero point representing the lowest possible temperature. In celsius scale zero point is the arbitrary creation. (A) (B) (C) (D) (E) ZOOLOGY 21. Assertion (A) : Alleles are different forms of the same gene, creating different phenotypic versions of a trait. Reason (R) : Dominant alleles are expressed in the homozygous or heterozygous condition, while recessive alleles are expressed in the heterozygous condition. (A) (B) (C) (D) (E) 22. Assertion (A) : In the human body, changes in the pH of fluids can be dangerous and life threatening. Reason (R) : If the pH of blood falls below 7·0 or rises above 7·8, nerves do not function properly and a coma or convulsions may occurs. (A) (B) (C) (D) (E) 23. Assertion (A) : The penal gland in the brain secretes the hormone melatonin, which influences the daily rhythms of sleep, hunger and body temperature. Reason (R) : The prostaglandins are a family of hormone like substances that either increase C.S.V. / February / 2008 / 1660 38. Assertion (A) : For usual course of transpiration light is one of the important external factors. Reason (R) : Because light induces opening of stomata and the rate of transpiration increases in presence of light. (A) (B) (C) (D) (E) 39. Assertion (A) : Pteridophytes and gymnosperms are placed under archegoniatae. Reason (R) : Because both of these plant groups have archegonia as female reproductive organ. (A) (B) (C) (D) (E) 40. Assertion (A) : Oils and fats are trigylcerides of long chain of fatty acids. Reason (R) : Fatty acids containing one or more double bonds (C — C) are said to be unsatu— rated. (A) (B) (C) (D) (E) ANSWERS WITH HINTS 1. (C) 2. (B) Light waves are transverse. Polarisation is the characteristic of transverse waves only. 3. (A) 4. (B) 5. (C) dV 6. (A) E = – = 0 dr 7. (A) 8. (A) ⇒ I0 = L = E0 ωL = 2E 2πn L Oxidation state of + 6 changes to + 8 and O and hence it is a disproportionation reaction. 15. (A) Addition of bromine (Br2) to butene-1 or 1-butene gives a product which contains one asymmetric carbon atom as— H H H H H H H H | | | | | | | | H—C—C—C = C—H + Br2 → H— C—C—C*— C —H | | | | | | H H H H Br Br * Shows asymmetrical carbon atom. 16. (D) Molecule has two chiral centres and hence total stereoisomers would be 2n 2 = 4. H I H CH3 Br Br H 1·414 × 200 2E = 2πn I0 2 × 3·14 × 50 × 0·1 282·8 = = 1H 282·6 9. (A) 10. (A) In a reactor the fast moving neutrons are slowed down by colliding them with the nuclei of a moderator. We know that in head on elastic collision of two bodies the exchange of momentum is maximum when the masses of the bodies are equal. The mass of heavy water nucleus is nearly the same as the mass of neutron. So exchange of momentum between neutron and heavy water is maximum. Thus neutrons are slowed down more efficiently by heavy water. 11. (C) First of all a precipitate of CaCO3 is formed which dissolves when excess of CO 2 is passed due to the formation of soluble calcium bicarbonate, Ca(HCO3)2. 12. (B) A positive catalyst always lowers the energy of activation (Ea) of a reaction by providing a different reaction mechanism, i.e., reaction mechanism changes in presence of catalyst. The equilibrium constant is not affected by a catalyst because it speeds up both forward and backward reactions to the same extent. 13. (E) White P is soluble in organic solvents and hence cannot be kept in kerosene oil, however it is kept in water as it does not react with water at all. Sodium is kept in kerosene but cannot be kept in all anhydrous solvents because it reacts with them’ as— 2C 2H5OH + 2Na → 2C2 H5ONa + H2 14. (C) Reactions are as : XeO3 + OH– → HXeO4– Since oxidation state of Xe in both XeO3 and HXeO 4– is + 6 and hence it is not oxidation. HXeO 4– → XeO64– + Xe + O 2 ←→ Br Br I H CH3 (Enantiomers) Diastereomers H I Br CH3 Br H H ←→ Br H I Br CH3 (Enantiomers) 17. (A) CH 3MgI is a Grignard reagent and it reacts with water very rapidly as— CH3MgI + HOH → CH3H(CH4) + Mg(OH)I Stronger acid Weaker acid 18. (C) Acylation of alcohols with acetyl chloride is termed as acetylation R—OH + CH3COCl ROCOCH3 + HCl The reaction is reversible and, therefore, in order to remove HCl, pyridine is used as base. It helps in shifting the equilibrium to right hand side. 19. (A) 20. (A) 21. (C) Dominant alleles are expressed in the homozygous or heterozygous condition, while recessive alleles are expressed in the homozygous condition. 22. (A) 23. (B) 24. (C) Enzymes and toxins are secreted by living bacteria as well as some toxins are also released from the cell wall of dead bacteria. 25. (D) Cataract results from reduced lens transparency, while presbyopia results from reduced lens flexibility. 26. (E) Retroviruses are RNA animal viruses that have a DNA stage. The viral DNA remains in the host genome and is replicated when host DNA is replicated. (Continued on Page 1664 ) C.S.V. / February / 2008 / 1661 Physics 1. When a particle is in motion, its acceleration may be in any direction. —T/F 2. Understanding boolean data types may require some extra thought. It may help to think of a boolean variable as a switch. —T/F 3. The air resistance increases the time of rise as well as the time of fall. —T/F 4. Substance, which expand in volume on melting, have their melting points raised due to the application of pressure. —T/F 5. The work done can be zero even when the body has accelerated motion. —T/F 6. Ampere’s circuital theorem states that the line integral of the magnetic induction around any closed path containing the current is equal to the permeability (free space) times the current enclosed by the path. —T/F 7. The continuity equation A1V1 = A2V2 is valid only in the steady state condition. —T/F 8. An emf can be induced between the two ends of a straight copper wire when it is moved through a uniform magnetic field. —T/F 9. Displacement node is a pressure antinode. —T/F 10. No net force acts on a rectangular coil carrying a steady current when suspended freely in a uniform magnetic field. —T/F 11. Kirchhoff’s current law is applicable at any junction or node in the circuit. —T/F 12. Heating of water under atmospheric pressure is an isothermal process. —T/F 13. Magnetic field due to a straight wire varies in inverse square proportion with distance. —T/F 14. When a soap bubble is given an electric charge its size increases. —T/F 15. Only Balmer series of hydrogen atom lies in the visible range. —T/F Chemistry 16. Equal volumes of two gases which do not react together are enclosed in two separate vessels and their pressures are 100 mm and 400 mm respectively. If two vessels are joined together the pressure of mixture will be 500 mm. —T/F 17. Complex [CO (NO2)3 (NH3)3] does not show geometrical isomerism. —T/F 18. The equivalent weight of copper in Cu2O is 63·6 while in CuO it is 31·8. —T/F 19. Setting of the plaster of Paris is on account of process of dehydration. —T/F 20. In the solid state, the conduction of electricity in the ionic compound is due to the presence of imperfaction in crystals. —T/F 21. Benzene undergoes substitution reactions more easily than addition reactions due to resonance. —T/F 22. There is no difference between atomic mass and mass number of an element. —T/F 23. The rate of substitution in phenol is faster than that in benzene. —T/F 24. The favourable conditions for the formation of ionic solid are high ionization energy of metal atom, low electron affinity of non-metal atom. —T/F 25. Mesotartaric acid is optically inactive due to the presence of plane of symmetry. —T/F 26. The lowering of vapour pressure is equal to the mole fraction of the solute in the solution. —T/F 27. The kinetic energy of a gas molecule becomes zero at 0°C. —T/F 28. The group C6H5CH2O— is called benzoyloxy. —T/F 29. The equivalent weight of crystalline oxalic acid is equal to 90/2. —T/F 30. Carbon-carbon bond in benzene is longer than carbon-carbon bond in ethylene. —T/F C.S.V. / February / 2008 / 1662 Zoology 31. Epitope is antibody determinant. —T/F 32. Nematodes are pseudocoelomates. —T/F 33. A conserved DNA sequence of 180 base pairs encodes a protein domain in many proteins. —T/F 34. Enterogastrone inhibits the secretion of enterocrinin. —T/F 35. Planula is solid free-swimming ciliated larva of most cnidaria and a few of the ctenophora. —T/F 36. The coral glands of scorpion are homologous with the green glands of crustaceans. —T/F 37. Intestinal micro-organisms are capable of synthesizing considerable amounts of phylloquinone and menaquinone vitamins. —T/F 38. Peyer’s patches are aggregation of lymph nodes found chiefly in the ileum near the junction with the colon. —T/F 39. Plasmagene is a gene contained in a self-replicating cytoplasmic particle and inheritance of the characters controlled by such genes is Mendelian. —T/F 40. The initiating hormones for the menstrual cycle arise in the hypothalamus. —T/F 41. Perissodactyla is the order of mammals that contains odd-toed ungulates, such as horses and rhinoceros. —T/F 42. Spontaneous mutations in germ cells alter allele frequencies and reintroduce harmful alleles into population. —T/F 43. Natural parabiosis occurs in Siamese twins. —T/F 44. Opsonin is a kind of antigen. —T/F 45. Gingiva is the gum tissue that surrounds the neck of the teeth and covers the alveolar processes of the maxilla and mandible. —T/F Botany 46. When doing genetic problems, first decide on the appropriate key and then determine the genotype and gametes for both parents. —T/F 47. Axillary buds are not served by strands of vascular tissue. —T/F 48. Plants growing on sandy-soil are called lithophytes. —T/F 49. The growth of the vascular plants depends upon the activity of meristems, which are, in sense, always embryonic. —T/F 50. Dionea is commonly known as butterwort. —T/F 51. If phytoplanktons are killed in the pond, no effect will take place in food chain. —T/F 52. Sieve tube bears multinuclei. —T/F 53. Dendrochronology determines age of the tree by radiocarbon dating. —T/F 54. In old trees or wood plants transpiration occurs by lenticels. —T/F 55. Conjugated proteins containing carbohydrates as prosthetic groups are termed glycoproteins. —T/F 56. The closely related genus Anabaena differs from Nostoc in that firm colony is formed. —T/F 57. In hypogynous type of flowers all floral parts arise above the gynoecium. —T/F 58. In C 4 plants, as opposed to C3 plants, the enzymes PEPCase fixes CO 2 to phosphoenol pyruvate to form oxaloacetate. —T/F 59. Nucleolus is the chief site for the synthesis of DNA. —T/F 60. The zygote, which is always haploid, contains heterologous chromosomes. —T/F ANSWERS WITH HINTS 1. (True) The direction of acceleration is the direction of change in velocity and the velocity may change in any direction. 2. (True) 3. (False) Air resistance decreases the time of rise as it increases the decelerating rate. On the other hand, it increases the time of fall because it decreases the accelerating rate. 4. (True) 5. (True) Work done along a closed path is zero, when only conservative forces are acting. 6. (True) It is symbolically stated as φ B dl = µ0i 7. (True) 8. (True) 9. (True) Maximum pressure variation occurs at the displacement node. →→ C.S.V. / February / 2008 / 1663 10. 12. 13. 14. 16. (True) 11. (True) (False) It is an isobaric process. (True) Bio-Savart law. (True) 15. (True) (False) After mixing the volume is doubled, so the pressures are halved. Hence pressure after mixing = 50 + 200 = 250 mm. 17. (False) Complex [CO (NO2)3 (NH3)3] shows following geometrical isomers NH3 NO2 CO NO2 NO2 Facial-form NH3 NH2 NO2 CO NH3 NO2 NH3 Meridional-form NH3 NO2 50. (False) Dionea is commonly called ‘Venus fly trap’. 51. (False) If phytoplanktons are killed in the pond, then it will effect the food chain because they are chief producers of pond ecosystem. 52. (False) Sieve tube bears no any nucleus. 53. (True) 54. (True) 55. (True) 56. (False) Anabaena differs from Nostoc in that no firm colony is formed. 57. (False) In hypogynous type of flowers all floral parts arise below the gynoecium. 58. (True) 59. (False) Nucleolus is the chief site for the synthesis of ribosomal RNA. 60. (False) The zygote, which is always diploid, contains homologous chromosomes. ●●● (Continued from Page 1661 ) 27. (A) 28. (B) 29. (E) The production of a dilute urine requires reabsorbing more ions and less water. 30. (B) 31. (D) In tropical region, some forests are deciduous–the broad-leaved trees lose their leaves because of a dry season. 32. (D) Available water to the plants becomes hypertonic in relation to cell sap of the plants due to excessive use of fertilizers. As a result, water molecules diffuse out of the cells due to exosmosis. 33. (E) Sedimentation can occur on land or in bodies of water. 34. (A) 35. (B) 36. (C) The part of eukaryotic cell where the DNA is located is called nucleoid, which sometimes under microscope appears fibrous. 37. (D) 38. (A) 39. (A) 40. (B) Triglycerides may be solid or liquid at ordinary room temperature and are termed as fats and oils respectively. ●●● 18. (True) 19. (False) 20. (True) 21. (True) 22. (False) The mass number is a whole number and atomic mass is a fractional number. Atomic masses are fractional due to isotopes which have different masses. 23. (True) 24. (False) Low ionization energy of metallic-atom and high electron affinity of non-metallic atom favours the formation of ionic bond. 25. (True) 26. (False) The relative lowering of vapour pressure is equal to the mole fraction of the solute in the solution. 27. (False) All the molecular motions of a gas molecule becomes zero at absolute zero temp., i.e., at – 273°C or at 273 K. 28. (False) The group C 6H5CH2O– is called benzyloxy. 29. (False) Molecular weight of oxalic acid in crystalline state (H2C2O4.2 H2O) is 126 and hence equivalent 126 weight is equal to = 63. 2 30. (True) 31. (False) Epitope is an antigenic determinant. 32. (True) 33. (True) 34. (False) Enterogastrone inhibits the secretion of HCl by gastric glands. 35. (True) 36. (True) 37. (True) 38. (True) 39. (False) Plasmagene is found in cytoplasmic particles such as mitochondria, plastids and centrioles. Inheritance controlled by such genes is not Mendelian, because appreciable amounts of cytoplasm are passed only by the female gamete. 40. (True) 41. (True) 42. (True) 43. (True) 44. (False) Opsonin is a kind of antibody that is absorbed on the surface of bacteria. 45. (True) 46. (True) 47. (False) Axillary buds are also served by strands of vascular tissue departing from the vascular system of the stem. 48. (False) Plants growing on sandy-oil are called psammophytes. 49. (True) Just Released Journalism & Mass Communication By : Hena Naqvi Code No. 1567 Rs. 130/- This book is very useful for any professional course based on Journalism and Mass Communication. Hindi Edition Code No. 1332 Rs. 90/- UPKAR PRAKASHAN, AGRA-282 002 C.S.V. / February / 2008 / 1664 Physics Q. What is a periodic wave function ? ☞ A wave function y (x, t) which satisfies the periodicity conditions of position and time is called a periodic wave function, i.e. (i) y (x + m λ; t ) = y (x, t ) (ii) y (x, t + nt) = y (x, t ) where λ = wavelength of wave, T = period of the wave n and m are integers. Q. Heating system based on circulation of steam are more efficient in warming a building than those based on circulation of hot water. Explain why. ☞K = 2 Hr µ0n Also R = ρ L Area where H = Horizontal component of earths magnetic field in tesla. r = Radius of circular coil n = Number of turns in the coil Q. What is the function of grid in a triode valve ? ☞ The grid eliminates space charge and it controls the number of electrons reaching the plate and hence it is known as control grid. Q. Explain why sun appears red at sun rise or sun set. where ρ is resistivity R × Area Length 1 Conductance = Resistivity Length = R × area ρ = Therefore, Conductance = [Length] [Resistance] [Area] [L] = [ML2T–3A–2] [L 2] = [M–1L–3T3A2] Q. Why is it easier to start a car engine on a warm day than on a chilly day ? ☞ At sunrise or sunset, the sun is near the horizon. Sunlight has to travel much greater distance than at noon. So, a much larger portion of the blue component of sunlight gets scattered away. Therefore, the light reaching the observer has a larger proportion of the remaining colours specially red. Hence, the sun appears red at sun rise or sun set. Q. What does a welder protect against when he wears a mask ? ☞ Steam at 100°C possesses more heat than the same mass of water at 100°C. One gram of steam at 100°C possesses 540 calories of heat more than that possessed by 1 gm of water at 100°C. That is why heating systems based on circulation of steam are more efficient than those based on circulation of hot water. Q. What is the relation between dielectric constant (K) and electric χ susceptibility (χe) ? ☞ The internal resistance of a car battery decreases with increase in temperature. Hence more current is drawn on a warm day and it becomes easier to start the engine. Q. Why electrolytes have lower conductivity than metallic conductors ? ☞ This is because of the following two reasons : (i) The number of ions in electrolyte is small compared to the number of free electrons in metallic conductors. (ii) The mobility of ions in an electrolyte is small compared to the mobility of free electrons in a metallic conductor. ☞ The mask has a filter which absorbs the ultra-violet radiation produced by the welding arc. This radiation is dangerous for the eyes. Q. Does the relation V = i R apply to non-ohmic resistors ? ☞ Relation between dielectric constant K and electric susceptibility χe is K = 1 + χe. Q. Why is electric power transmitted through A.C. and not D.C. ? ☞ Transformer is a device which can increase or decrease the A.C. voltage. We know that we transmit power at high voltages to reduce transmission losses. Therefore, by using step-up transformers, we increases the voltage at generating station and step-down the voltage to 230 V at the distribution. No such equivalent device of transformers is available in case of D.C. Therefore, we transmit power through AC and not D.C. Q. What is the value of reduction factor K of tangent galvanometer ? ☞ Relation V = i R applies to non-ohmic resistors as well. For nonohmic resistors, ‘R’ is not constant. It may noted that V = i R is not the statement of Ohm’s law. It is merely the definition of the resistance of a conductor whether it obeys Ohm’s law or not. Q. Derive dimensional formula for electrical resistance and electrical conductance. [V] [Work done] ☞ [R] = = [I] [Charge] [I] [ML2T– 2] = [AT] [A] = [ML2T–3A–2] Chemistry Q. How are colloidal impurities removed during water purification ? ☞ The neutralisation of surface charge is used to precipitate colloidal impurities in water purification. When water is taken into municipal purification plants, it contains dispersed material such as colloidal clay. Clay particles are negatively charged, which keeps particles apart and dispersed in water. To precipitate clay, aluminium salts such as Al2(SO4)3 are added. An aqueous solution, Al3+ ions C.S.V. / February / 2008 / 1665 are hydrolysed to give [Al (H2O)6] 3+ and other larger ions such as : OH | O (H2O)4 Al O | H Al (H 2O)4 4+ These highly charged ions are adsorbed onto the surface of the negatively charged colloidal particles, thereby neutralising the surface charge and allowing the colloidal particles to come together, and the clay precipitates from solution. Q. Why does sulphur (density 2·1 g cm 3), when carefully placed on top of water, float while the same sinks in water when a few drops of detergent are placed on water ? Q. What is the chemistry of usefulness of mortar ? ☞ Mortar consists of one part of lime to three parts of sand, with water added to make a thick paste. The very first reaction which occurs, is the hydrolysis or ‘slaking’ of the lime. When mortar is placed between bricks or stone blocks, it slowly absorbs CO 2 from air, and the slaked lime converts to calcium carbonate. Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(l) Although the sand in the mortar is chemically inert, the grains are bound together by the particles of CaCO 3 and the hard material results. Q. What is the thyroxine ? ☞ It is now known that the thyroid gland produces a growth regulating hormone, thyroxine which contains iodine. Thyroxine has following structure : I H | | C— C HO—C C— C — | | I H I H | | — C —C — C— C | | I H C—O ☞ Suppose an electron exists inside the nucleus, then its uncertainty in position will be of the same order as the radius of the nucleus. The radius of the nucleus is of the order of 10–13 cm. Then the uncertainty in its velocity h ∆v = 4πm × ∆x = 6·62 × 10– 27 4 × 3·14 × 9·1 × 10 – 28 × 10–13 ≈ 5 × 1012 cm/sec The velocity of electron is greater than the velocity of light when it is present inside the nucleus. This condition is impossible and hence electron cannot exist in nucleus. Q. What will be the average life of radium (Ra) if its t1/2 is 1690 years ? ☞ The reciprocal of radioactive decay constant, k is called the average life of a radioactive substance. 1 Average life (T) = k 0·693 We know that t 1/2 = k 0·693 1690 years = k 0·693 ∴ k = 1690 Hence average life, 1690 T = 0·693 = 2433·6 years or the value of 1 = 1·44 0·693 ∴ T = 1·44 × t1/2 = 1·44 × 1690 = 2433·6 years Q. What is the basic difference between a sol and gel ? ☞ Surface tension of water is affected by adding detergent. The high surface tension of the water keeps more dense sulphur float on the water surface. When some drops of detergent are placed on the surface of water, the surface tension of water is reduced and the sulphur sinks to the bottom of the pot containing water. Q. Why does even painstakingly purified water conducts —C electricity to a very small extent ? C—C H2CH(NH2) COOH. ☞ The existence of so-called autoionization of water was proved many years ago by Friedrich Kohlrauch. He found that even after water is painstakingly purified, it still conducts electricity to a very small extent, since autoionization produces very low concentration of H3O + and OH– ions even in the purest water. 2 H2O(l) H3O+(aq) + OH– (aq) Most of the table salt sold in the market contains 0·01% NaI added to provide the necessary iodine in the diet. Q. Cryolite is found in only small amounts in nature but large amount of cryolite is needed in production of Al. How is demand met with ? ☞ Most of the hydrogen fluoride (HF) produced by the reaction of fluorspar (CaF 2) and H 2SO4, is used to make cryolite. CaF2(s) + H2SO4 Q. What biological function is performed by mineral apatite (calcium and phosphorus containing mineral) ? ☞ Hydroxy-apatite, Ca (OH)2.3 Ca 3(PO4)2, is the main component of tooth enamel. Cavities in our teeth are formed when acids decompose the weakly basic apatite coating. This decay can be prevented by converting hydroxy-apatite to much more acidresistant coating, fluorapatite, CaF2, 3 Ca3 (PO 4)2, by adding a source of fluoride ion. C.S.V. / February / 2008 / 1666 ☞ A colloidal solution in which dispersion medium is liquid and dispersed phase is solid, then such a system is called sol. For example gold sol and sulphur sol. When the dispersion medium is solid and dispersed phase is liquid then the colloidal system is called gel. For example, boot polish, butter, gellies etc. Q. What would be the vapour density (VD) of PCl5 if it is 90% ° dissociated at 250°C ? → CaSO 4(s) + 2HF(g) 6 HF(aq) + Al(OH)3(s) + 3 NaOH → Na 3AlF6(s) + 6 H2O(l) Cryolite Q. Why cannot an electron exist inside the nucleus of an atom ? ☞ As PCl5 is 90% dissociated at 250°C and hence the degree of dissociation (α) of PCl5 will be 90/100 = 0·9. The molecular mass of PCl5 is 180·5 and hence the theoretical V.D. of PCl5 will be 104·25. We know that α = DT – Do Do 104·25 – Do Do 104·25 ≈ 54·8 1·9 Q. Using diazomethane as one of the reagents, how is acetic acid converted into propanoic acid ? What is this synthesis known as ? ☞ The acid chloride of acetic acid on reacting with diazomethane, an intermediate, diazoketone is formed. Diazoketone on losing nitrogen gives a ketene. When ketene is passed into water propanoic acid is formed. 5 CH3COOH → CH3COCl (Do observed V.D. of PCl 5) Hence ∴ 0·9 = Do = PCl Acetic acid O || → CH3—C—CH—N ≡ N CH 2N2 Diazoketone Hence, the vapour density of PCl 5 at 250°C will be 54·8. Q. What is the relation between equivalent mass and electrochemical equivalent of a substance ? O || CH3—C—CH—N ≡ N → CH3CH = C = O – N2 Ketene Propanoic acid (qualitative) characters are those exemplified throughout the phenotypic range, tending to be determined by polygenes. Discontinuous (quantitative) characters are those unrepresented in all parts of the phenotypic range tending to be polymorphic, determined by genetic ‘switchmechanisms’. Controversy over the relative influence of heredity and environment in producing phenotypic differences fuels the Nature—Nature debate. Problems arise in obtaining acceptable control populations to test hypothesis. By starting with genetically uniform material (e.g., by cloning or repeated inbreeding) it is often possible to compare phenotypes produced under different environmental regimes and to estimate heritability. Q. What is Brain stem ? ☞ The equivalent mass of a substance is the mass of substance deposited by passage of one mole of electrons or one Faraday (96,500 C) of electricity. The mass of substance deposited by one coulomb is called electrochemical equivalent. Hence equivalent mass = 96,500 × electrochemical equivalent Q. What will be the % composition of a mixture of CO and CO 2 which on passing over red hot coke registered a 25% increase in its volume ? ☞ We are aware of the following facts : CO2 + C → 2 CO CO + C → No reaction Suppose the gaseous mixture contains x ml of CO2 and (100 – x ) ml of CO gas. From equation we can conclude that x ml CO2 produces 2x ml CO gas after passing on red hot coke. Hence, the total volume of CO after reaction can be expressed as 100 – x + 2x = (100 + x ) ml Given volume is increased by 25% and hence the final volume is 125 ml. Hence 100 + x = 125 ml x = 25 ml Thus, the mixture contains 25% CO2 and 75% CO gases. and → CH3CH2COOH This reaction is known as Arnt Eistert synthesis. H 2O ☞ Brain stem is the stem like part of the brain that connects cerebral hemispheres with the spinal cord and comprises the medulla oblongata, the pons and the midbrain. Thus, the medulla oblongata, the pons and the midbrain lie in a portion of the brain known as the brain stem. The medulla oblongata lies between the spinal cord and the pons and is anterior to the midbrain. It contains a number of vital centres for regulating heartbeat, breathing and vasoconstriction. It also contains the reflex centres for vomiting, coughing, sneezing, hiccups and swallowing. The pons contains bundles of axons travelling between the cerebellum and the rest of the central nervous system. In addition, the pons functions with the medulla to regulate the breathing rate and has reflex centres concerned with head movements in response to visual and auditory stimuli. Aside from acting as a relay station for tracts passing between the cerebrum and the spinal cord or cerebellum, the midbrain has reflex centres for visual, auditory and tactile responses. Q. What are good and bad cholesterols ? ☞ In human blood, cholesterol is carried by special proteins, the lipoproteins, which are manufactured in the liver. Two classes of these cholesterol containing lipoproteins are significant for the heart and blood vessels. Low density lipoproteins (LDL) contain triglyceride and cholesterol. If cells are in need of cholesterol for plasma Zoology Q. What is Trophic Level ? ☞ Trophic level is theoretical term in ecology. One of a succession of steps in the transfer of matter and energy through a community, as may be brought about by such events as grazing, predation, parasitism, decomposition. For theoretical and heuristic purposes, organisms are often treated as occupying the same trophic level when the matter and energy they contain have passed through the same number of steps since their fixation in photosynthesis. Primary producers, herbivores, primary, secondary and tertiary carnivores and decomposers all commonly figure as trophic levels in the analysis of ecosystems. Different developmental stages and/or sexes within a species may occur in more than one trophic level. The number of trophic levels in a community is thought to be limited by inefficiency in energy flow from one trophic level to the next; however, food chains are no longer in tropical communities, where energy input is high, than they are in Arctic communities, where energy input is low. Q. What is Variation ? ☞ Variation is phenotypic and/or genotypic differences between individuals of a population. Continuous C.S.V. / February / 2008 / 1667 membranes or steroid hormone synthesis, they take up LDL. If this process does not occur, the LDL loaded with cholesterol remain in the plasma and deposition of cholesterol as plagues becomes possible. High density lipoproteins (HDL) are the second group of lipoproteins. HDL picks up cholesterol from cells and transports it to the liver for disposal. The liver releases the excess cholesterol with the bile into the small intestine. It has been found that the higher HDL levels are associated with a fewer risk of developing atherosclerosis, while high LDL levels are associated with a higher risk. That is why HDL has been called ‘good’ cholesterol and LDL ‘bad’ cholesterol. LDL levels below 130 milligram per decilitre and HDL levels above 40 mg/dl are considered desirable. cell. The genetic material (DNA) is transferred form male bacterium to female bacterium along the groove of pili. Fig. : Bacterial cell showing the distribution of pili. Q. What do you mean by essential oil ? Botany Q. What do you mean by trace fossils ? ☞ Trace fossils are tracks, traits and burrows made by animals and found in ancient Sediments such as Sandstones, Shales or limestones. Very different biological activities of animals produce these biological structures. Study of trace fossils is known as ichnology or palichnology and is a part of the Science of Palaeontology. In the early years of Palaeontology, most of the ramified burrows creeping trails or other trace fossils were considered to be remains of marine algae. This is proven by many names ending ‘Phycus’ given to trace fossils. Generally palaeontology use for trace fossils the same bionomial nomenclature, with Latin names, as in naming animals and plants scientifically. Q. What are fimbrie (pili) in certain bacteria ? ☞ Many gram negative bacteria have minute, hair like, straight and non-flagellar appendages called fimbrie or pili. They are given off from the cell surface and thinner than flagellum. Pili are composed entirely of protein known as Pilin. Their function is attachment of bacteria to any plant or animal. During bacterial conjugation, one or two sex pili from the male cells form bridges to the female ☞ The essential oils are odoriferous, oily products of plant origin which are distillable. They occur in leaves, twig, fruit, blossoms, root and trunk of plants. The principal constituents of essential oils are the terpenes. Benzenoid and aliphatic compounds may also be present. Most of the constituents are hydrocarbons and oxygenated derivatives of hydrocarbons. A few contain sulphur and nitrogen. For example, oil of mustard contains organic isothiocyanate; garlic and onion oils contain organic sulphides. Anthranilates, indole and skatole sometimes occur in small amounts. Most essential oils are exceedingly complex mixtures of non-terpene and terpene ingredients. Q. What do you mean by rusts and smuts fungi ? histones. In which type of histone the largest amount of arginine is found ? ☞ The DNA in the chromatin is very highly associated with proteins called histones, which package and order the DNA into structural units called nucleosomes. Histones have molecular weight of between 11,000 and 21,000 and are very rich in the basic amino acids arginine and lysine. The H3 histones are nearly identical in amino acid sequence in all eukaryotes, as are the H4 histones, suggesting strict conservation of their functions. Histone H4 bears the largest amount of argine (about 14%). Q. What is cytochrome and its application ? ☞ Rusts and smuts are club fungi. They parasitize cereal crops such as rye, wheat, corn and oat. Rusts and smuts do not form basidiocarps and their spores are small and numerous, resembling shoot. Some smuts enter seeds and exist inside the plant, becoming visible only near maturity. Some other smuts externally infect plants. In corn smuts, the mycelium grows between the corn kernels and secretes substances that cause the development of tumours on the ears of corn. The life cycle of rusts often requires two different plant host species to complete the cycle. Black stem rust of wheat uses barberry plant leaves as an alternate host and blister rust of white pine uses current and gooseberry bushes. The black rust of wheat is caused by Puccinia graminis-tritici, which is a heteroecious parasite. Q. What are histones ? Give the molecular weight of H3 and H 4 ☞ Cytochrome is a complex protein occurring within cells of a wide variety of animals and plants. The integral part of a cytochrome is a heme (iron tetrapyrrole) moiety. The heme imparts generally red colour to the cytochrome, as well as characteristic absorption bands which are used for the spectroscopic observation and identification of these intracellular hemoproteins. The iron atom of the heme may be reduced and oxidized by appropriate substances in the cell, thus constituting the function of the cytochrome as electron carriers. Q. What does a cell face challenges while it undergoes a great increase in size ? ☞ Under such conditions, many portion of the cell may well suffer deprivation of vital fuel materials or the essential molecules as a result of enlargement. The cell must exchange materials with the environment across the surface membrane. An increase cell size will result in a greater increase in volume and mass than in surface area, so that the cell wall lose effective exchange capacity. This will impose restriction on the amount of food and oxygen that can move across the membrane to service the metabolic needs of the increased living mass in the interior. The distribution of materials by diffusion will also take longer as the cell grows larger. ●●● C.S.V. / February / 2008 / 1668 CSV Crossword–20 1 2 3 4 5 6 7 8 9 10 11 13 13. Not ‘analog’, but ‘………’ (7). 15. We select an infinitesimal ……… before performing the integration (7). 16. Rank and ……… are important ideas in linear vector spaces (7). 19. Prefix for 100 (5). 21. Delay (3). Note : Its solution will be published in the next issue. 13 14 15 16 CSV Crossword–19 ANSWERS Across : (1) KAON (3) COULOMBS (9) LATTICE (10) MORSE (11) MOLAR (12) ION (13) ON (15) LAURENT SERIES (18) BALMER (20) EARTH (23) CHI (24) ACETATE (25) DIMINISH (26) BTUS 17 18 19 20 21 22 23 24 Across : A long thin solenoid turned into a ring (6). A mathematical entity, an array (6). Unit of 1/(Capacitance) (5). Geiger and ……… bombarded thin gold foils with Alpha particles, on suggestions of Ernest Rutherford (7). 10. His mass spectrograph was very important (5). 11. Essentially oxygen, molecular weight = 48 (5). 14. A supplementary unit (9). 17. Excess pressure above the atmosphere is known as the ……… pressure (5). 18. A type of diffraction grating, where very high orders of diffraction are observed (7). 20. Methyl radical attached to benzene ring (7). 22. An alkali halide of the most electro positive element (Formula) (3). 23. Something supreme : the rational basis of human activity and conduct (3). 24. According to the Aristotlean view of nature, all material objects fall down. The ………--……… falls slower (7, 4). Down : 1. A cheap Indian liquor derived from palm trees (5). 2. What a student does, when he does not understand a topic in the first reading. He ………… it (7). 3. In ‘Field Effect Transistors’, in contrast to ‘MOS’ type, the other type is the ………… (Acronym) (5). 5. From cause to effect, from a general law to a particular instance (7). 6. A wireless tuned receiver (5). 9. A hard–brittle, grayish–white, element metal, used chiefly in alloying steel for strength (9). 12. Gases generally cool during ……… (9). 1. 4. 8. 9. D o w n : ( 1 ) KILOMOLE (2) OCTAL (4) OPEN INTERVALS (5) LUMEN (6) MARCONI (7) SKEW (8) PIERRE (14) ISOHYETS (16) URANIUM (17) ELEVEN (19) M BORN (21) REACT (22) ACID. 1 2 3 4 5 6 7 K I 9 A O C N 8 C P O P U L U 10 O M A B S K L O 11 A T A T I E C E N 12 M E O N 17 O R C 13 S E W M O O L 16 A R R I O N 14 N N T E S E L 20 N R I 21 I E S O 15 L E A 18 U R R E 19 B 22 A N L M B O R E R V 24 E V C E N A R E T H Y A 23 C I 25 H I U A L I S T 26 A C T E T D I M I N H B T U S ●●● New Release WIT AND HUMOUR By : Radharaman Agarwal Code No. 1523 Rs. 30/- Published by : UPKAR PRAKASHAN, AGRA–2 E-mail :
[email protected] Website : www.upkarprakashan.com C.S.V. / February / 2008 / 1669 01. Two identical steel wires are under tension and are in unison. When the tension in one of the wires is increased by 1 per cent, 4 beats/second are heard. Find the original frequency of wires— (A) 512 Hz (B) 256 Hz (C) 800 Hz (D) 400 Hz 02. A body executing linear simple harmonic motion has a velocity of 0·03 ms–1. When its displacement is 0·04 m and a velocity of 0·04 ms–1 when its displacement is 0·03 m. If the mass of the body is 50 × 10–3 kg, calculate the total energy of oscillation— (A) 6·25 × 10–4 J (B) 6·25 × 10–3 J (C) 6·25 × 10–5 J (D) 6·25 × 10–2 J 03. A brass boiler has a base area of 0·15 m2 and thickness 1·0 cm. It boils water at the rate of 6·0 kg/ min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s–1 m–1 °C–1— (Heat of vaporisation of water = 2256 × 10 3 J kg–1) (A) 327·98°C (B) 237·98°C (C) 400°C (D) 373·98°C same range of temperature at constant volume is— (A) 384 J (B) 144 J (C) 276 J (D) 452 J 06. Aldehydes that have no α-hydrogen atoms— (A) Undergo the Cannizzaro reaction (B) Do not undergo the Cannizzaro reaction (C) Both (A) and (B) are correct (D) Do not undergo any type of reaction 07. Who among the following first measured the charge/mass ratio (e/m) of the electron shortly after 1900 ? (A) Bohr (B) Schrodinger (C) Thomson (D) Millikan 08. Valence electrons not used in bond formation are called— (A) Unshared electrons only (B) Lone pair electrons only (C) Unshared or lone pair electrons (D) None of the above 09. ‘VSEPR’ stands for— (A) Valence Spin Electron Pair Repulsion (B) Variable Shell Electron Pair Repulsion (C) Valence Shell Electron Pair Repulsion (D) Variable Sigma Energy Production Rate 10. The pentaphenyl phosphorus is— (A) Ph3P (B) PH 5P (C) Ph5P (D) Ph.P 5 11. The kingdom Monera gave rise to the Protista about— (A) 3·5 billion years ago (B) 2·5 billion years ago (C) 1·5 billion years ago (D) 2 million years ago 12. Members of the phylum microspora are commonly called— (A) Microsporidia (B) Microzooflagellata (C) Macrosporidia (D) None of the above 13. Plasmodium vivax causes malaria in which the paroxyms recur every— (A) 8 hours (B) 48 hours (C) 12 hours (D) 24 hours 14. The immune system of vertebrate animals protects them from invasion by— (A) Pathogenic microorganisms (B) Foreign substances (C) Both (A) and (B) (D) None of the above 15. The class of antibody is determined by— (A) Its heavy chain constant region (B) Its light chain constant region (C) Both (A) and (B) (D) B lymphocytes 16. Recombination of genes on the same chromosome is accomplished by— (A) Hybridization (B) Crossing-over (C) Mitosis (D) Self-pollination 17. Which of the following is a total stem parasite ? (A) Twiner dodder (B) Orobanche (C) Monotropa (D) Sarcodes 18. A very common example of root stock is found in— (A) Guava (B) Mango (C) Jack fruit (D) Banana 19. Distichous phyllotaxy is/are found in— (A) (B) (C) (D) Cynodon Rice Both (A) and (B) None of the above 04. Calculate the heat produced between C and D in the circuit shown in the figure— 10Ω A 15Ω B 20Ω 25Ω C D (A) 4 J (C) 25 J (B) 8 J (D) None of these 05. 310 J of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 25°C to 35°C. The amount of heat required to raise the temperature of the gas through the 20. Leaf base is also known as— (A) Epipodium (B) Mesopodium (C) Hypopodium (D) Phyllopodium ●●● C.S.V. / February / 2008 / 1670 Rules for taking part in Quiz Contest of Competition Science Vision 1. 2. All students or those appearing in competitive examinations can take part in this contest. 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PRATIYOGITA DARPAN C.S.V. / February / 2008 / 1672 2/11A, Swadeshi Bima Nagar, AGRA-2 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. 2. (B) When the particle is dropped from a height h and spring is compressed by x, then loss in potential energy of the particle = mg (h + x) h a Total resistance in circuit 100R = r + RA + 100 + R 4 100R = 2+ + 3 100 + R 10 100R = + 3 100 + R 1000 + 310R = 3 (100 + R) V PRIZE WINNERS First Prize Ram Krishna Mishra S/o Shiv Narayan Mishra Behind 33/11 K.V. Hydel Colony, Vivekanand Nagar, Sultanpur U.P.–228 001 Second Prize Rahul Kumar Singh S/o Shobha Sinha Noorani Bagh Colony, 5A P.O.–Gulzarbagh, Patna Bihar–800 007 Third Prize Vijay Kumar New Medicare L. B. Palace Kadam Kuan, Patna Bihar–800 003 k 100 Ω The gain in the elastic potential energy of the spring 1 2 = kx 2 1 2 ∴ mg (h + x) = kx 2 1 0·1 × g (0·24 + 0·01) = k (0·01)2 2 …(i) Similarly in the second case 1 0·1 × g (h ′ + 0·04) = k (0·04)2 2 …(ii) Dividing (ii) by (i) h ′ + 0·04 = 0·24 + 0·01 ⇒ A RA = 4 Ω 3 E = 1·4V r =2Ω Current circuit 1·4 1000 + 310 R 3 (100 + R) Solving R = 200 Ω (ii) The equivalent resistance of the voltmeter and 100Ω resistance is given by 200 × 100 Req = 200 + 100 200 = Ω 3 ∴ p.d. across the voltmeter = i × Req 200 = 0·02 × = 1·33V 3 ∴ Error in voltmeter reading = 1·33 – 1·10 = 0·23V 5. (D) The energy in the nth state of a hydrogen like atom of atomic number Z is given by 13·6 En = – Z2 2 …(i) n The first excited state, n = 2 and energy is given by 13·6 …(ii) E2 = – Z2. 2 (2) Since the excited atom can make a transition to the first excited state by successively emitting two photons of energies 10·20 eV and 17·00 eV, the energy difference between these two states will be En – E2 = 10·20 + 17·00 = 27·20 eV …(iii) i = 0·02 = ( ) 0·04 0·01 2 h′ = 3·96 m ANSWERS WITH HINTS 1. (C) Let m be the instantaneous mass of the ascending rocket plus fuel and v r the relative velocity of the gas ejecting from the rocket. Suppose the fuel is burnt at the rate ∆m/∆t to provide the rocket an acceleration α. Then, α = Here, v r ∆m –g m ∆t 3. (D) Length of the fish = 6 cm Real depth of the fish’s head = 18 cm aµw = 4/3 ∴ Apparent depth of fish’s head = 18 = 13·5 cm 4/3 Also real depth of fish’s tail = 18 + 6 = 24 cm ∴Apparent depth of fish’s tail 24 = = 18 cm 4/3 Hence, the length of the fish = 18 – 13·5 = 4·5 cm 4. (C) (i) Let R be the resistance of the voltmeter. The voltmeter is connected to 100 Ω resistance in parallel. Hence, their equivalent 100R resistance is . 100 + R v r = 250 ms –1, m = 500 kg, g = 10 ms –2, α = 20 ms –2 ∆m ∆t = = m (α + g) vr 500 (20 + 10) 250 = 60 kg s–1 C.S.V. / February / 2008 / 1673 From equations (i), (ii) and (iii) 1 1 – 13·6Z 2 – = 27·20 n2 2 2 – Z2 6. (A) ( ( 1 1 – n2 2 2 ) ) = 27·20 13·6 = 2·00 …(iv) Similarly for transition to the second excited state (n = 3) from the same excited state, we have 4·25 + 5·95 1 1 – Z2 – = 13·6 n2 32 ( ) = 0·75 …(v) Subtracting (v) from (iv), we get 1 1 Z2 – = 2·0 – 0·75 4 9 ( ) l = 20 cm = 0·2 m, v = 340 ms –1 v 340 n = = = 425 Hz 4l 4 × 0·2 Since it is equal to the given frequency, therefore, the first harmonic mode will be excited. Frequency of open organ pipe v 340 n = = = 850 Hz 2l 2 × 0·2 So the same source will not be in resonance with the pipe if its both ends are open. v 330 7. (D) n = = 2l 2 × 0·66 = 250 Hz Now, 250 = ⇒ 1 T 0·01 2 × 0·2 T = (250 × 2 × 0·2)2 × 0·01 N 100 kg-wt = 100 N = 9·8 = 10·2 kg-wt 1 2l T m (Continued from Page 1582 ) (B) Ionic bond is very strong bond being of electrostatic nature but ionic solids are non-directional because the filled shells of the ions are spherically symmetric. (C) Ionic solids have high melting point because large energy (k T) is required to break bonds. (D) Ionic solids are transparent to visible light because electron excitation in ions require energy much higher than visible-photon energy which, therefore, pass right through. 46. (B) In scattering 1 n ∝ sin 4 (θ/2) where θ is angle of scattering ∴ n2 sin 4 (θ1/2) = n1 sin 4 (θ2/2) = sin 4 (45°) sin 4 (30°) 1 4 2 = = 4 1 4 2 ⇒ Z = 3 Putting Z = 3 in eq. (iv), we get 1 1 – 32 – = 2 n2 4 ( ) 6. 7. 9. 10. 11. 12. 15. 16. 17. 18. 19. ⇒ n = 6 (D) All measurements contain atleast three elements—a number that indicates the size of quantity being measured, units that provide a basis for comparing this quantity with a standard reference, and some uncertainty or error. (A) 8. (D) (C) Lipowitz’ alloy is a fusible alloy consisting of 50% bismuth, 27% lead, 13% tin and 10% cadmium. (A) (A) All viruses have atleast two parts—an outer capsid composed of protein subunits and an inner core of nucleic acid, either DNA or RNA but not both. (B) 13. (D) 14. (B) (D) If father is Rh+ and mother is Rh –, a baby born to this couple is likely to suffer from hemolytic disease. (D) (C) Polymer is a macromolecule consisting of covalently bonded monomers, for example, a protein is a polymer of monomers called amino acids. (A) During oxidative phosphorylation ten NADH molecules and two FADH2 molecules take electrons to the electron transport system. (D) 20. (B) ●●● 8. (C) n = 3 1 = 2 2l Dividing, n+ n + 1·5 = n 101 100 ( 1 T+ T 100 m ) ⇒ 47. (C) 48. (B) ( ) () (Continued from Page 1565 ) v = nλ = 2·53 × 103 × 2 = 5·06 km/s = 1·01 = 1·005 1·5 = 1·005 1+ n 1·5 = 0·005 n 1·5 n = = 300 Hz 0·005 9. (A) A is in unison with a small length of the wire as compared to B. Hence, A has higher frequency as compared to B. nA lB = nB lA Let n be the frequency of A. Then 0·97 97 n = = 0·96 96 n–4 ⇒ n = 388 Hz 10. (A) nA – nB = 8, nA 2 × 32·5 65 = = 64 nB 4 × 16 65 nA = n 64 B 65 ∴ n – nB = 8 64 B ⇒ nB = 512 Hz ∴ nA = 512 + 8 = 520 Hz ●●● n2 = 4 × n1 = 4 × 56 = 224 C = C1 = C2 = 4πε0r 4 4 πR3 = 2 × πr 3 3 3 ⇒ R = 21/3 × r For bigger drop C′ = 4πε0R = 21/3 C Total charge V = Total capacity 4πε0r (100 + 100) = 21/3 . 4πε0r = 200 × 2–1/3 = 100 × 22/3 volt 49. (C) The capacitor will be charged upto the peak value of the applied voltage, i.e., 2 × 200 volt. 50. (A) For a satellite T = 2π (Re + h)3 g R e2 This shows greater is the distance of the satellite above the earth’s surface, greater is its period of revolution. ●●● C.S.V. / February / 2008 / 1674 1. The ‘Miss Earth-Air Title at the Miss Earth pageant in Philippines has been won by— (A) Jessica Nicole Trisko (B) Silvana Santaella Arellano (C) Pooja Chitgopekar (D) Angela Gomez 2. As per the World’s Economic Forum’s report, 2007 Gender Gap Index has been topped by— (A) Norway (B) Finland (C) Sweden (D) New Zealand 3. ‘Chang’e-1’ is— (A) A spacecraft launched by Japan (B) China’s first lunar probe (C) An artificial satellite launched by Russia (D) None of these 4. The 2006 Indira Gandhi Award for National Integration has been given to— (A) Ram Puniyani and J. S. Bandukwalla (B) Javed Akhtar (C) Shyam Benegal (D) Acharya Mahaprajna 5. The caretaker Prime Minister of Pakistan is— (A) Mohammedmian Soomro (B) Benazir Bhutto (C) Sahbaz Sharif (D) Nawaz Sharif 6. The Chairman of the 13th Finance Commission is— (A) Indira Rajaraman (B) Vijay L. Kelkar (C) Atul Sharma (D) Abusaleh Shariff 7. The 4th International Conference on Federalism was hosted by— (A) Canada (B) Switzerland (C) Belgium (D) India 8. As per the study of World’s Power Stations, which among the following countries has been found to be the world’s worst polluter in terms of carbon dioxide emission ? (A) China (B) India (C) U.S. (D) Australia 9. ‘Skynet 5-B’ is a new military communication satellite launched by— (A) China (B) Britain (C) India (D) Russia 10. Which among the following nations would host 2014 Football World Cup ? (A) South Africa (B) Brazil (C) Germany (D) South Korea and Japan 11. The theme of the 13th ASEAN Summit was— (A) One caring and sharing community (B) One vision, one identity, one community (C) Advancing a secure and dynamic ASEAN family through greater solidarity, economic integration and social progress (D) One ASEAN at the Heart of Dynamic Asia 12. The 2007 Durand Cup has been won by— (A) Churchill brothers (B) Central Railways (C) Mahindra United (D) Air India 13. Which among the following States has received the Telecom Excellence Award, 2007 ? (A) Andhra Pradesh (B) Kerala (C) Haryana (D) Uttar Pradesh 14. Vigilance Awareness Week was from— (A) November 12-16 (B) November 14-18 (C) November 1-7 (D) November 7-14 15. As per the World Bank study, the rank of India in global trade Logistics Performance Index is— (A) 2nd (B) 1st (C) 39th (D) 14th 16. Which among the following books has been written by M.S. Swaminathan ? (A) Freedom on Trial (B) Agriculture cannot wait : New Horizons in Indian Agriculture (C) Agriculture Technology (D) Conserving Agriculture 17. Who among the following won the WTA Championship Title ? (A) Maria Sharapova (B) Justine Henin (C) Ana Ivanovic (D) Sunita Williams 18. As per the Forbes’ India’s 40 Richest list for 2007, the richest Indian is— (A) L. N. Mittal (B) Mukesh Ambani (C) Anil Ambani (D) Kushal Pal Singh 19. Chandrayaan-2 project was signed on November 12, 2007 between which among the following nations ? (A) India and Russia (B) India and China (C) India and Japan (D) India and USA 20. Which among the following statements regarding the supercomputer ‘EKA’ is incorrect ? (A) It has been developed by IBM systems (B) It has been named as Asia’s fastest supercomputer (C) It is world’s fourth fastest supercomputer (D) None of these 21. ‘Abuja Declaration’ has been signed between which among the following nations ? (A) India and Nigeria (B) India and Sri Lanka C.S.V. / February / 2008 / 1676 (C) India and Nepal (D) India and China 22. UN General Assembly has launched year 2008 as— (A) International Year of Rice (B) International Year of Potato (C) International Year of Water (D) None of these 23. Who among the following is the new member of the Asia-Pacific partnership on clean development and climate ? (A) Canada (B) China (C) Japan (D) India 24. Which among the following countries would host the 3rd IBSA Summit in 2008 ? (A) South Africa (B) Brazil (C) India (D) None of these 25. Which among the following players has recently taken retirement from International Test Cricket ? (A) Javed Miandad (B) Inzamam-ul-Haque (C) Shoaib Akhtar (D) Anil Kumble 26. Nobel Prize in Physics for 2007 has been won by— (A) Gerhard Estl (B) Albert Arnold (Al) Gore Jr. (C) Doris Lessing (D) Albert Fert and Peter Grunberg 27. As per the Report of Law Commission on Dowry Deaths, in case of dowry deaths— (A) The minimum sentence should be increased from seven to ten years (B) It deserves a death penalty (C) It deserves a life time imprisonment (D) None of these 28. The Right Livelihood Prize for 2007, an annual alternative to the Nobel Prizes has not been awarded to— (A) Ibrahim Abdi (B) Christopher Weera Mantry (C) Ruth Manorama (D) Percy and Louise Schmeiser 29. The theme of the 12th World Lake Conference was— (A) Conserving Lakes and Wetlands for Future (B) Balancing Agriculture and Lake Protection (C) Global threats to large lakes (D) None of these 30. ‘OCTOPUS’ is— (A) Indian Navy’s Trainingship (B) Organisation to Counter Terrorist Operations (C) India-Bangladesh naval exercise (D) None of these 31. Japan Open Tennis Tournament 2007 has been won by— (A) Venus Williams (B) Virginie Razzano (C) James Blake (D) None of these C.S.V. / February / 2008 / 1677 32. The 5th Military World Games 2011 will be held at— (A) Brazil (B) Italy (C) India (D) Croatia 33. Who among the following has become the richest person of the world ? (A) Bill Gates (B) Carlos Slim Helu (C) Mukesh Ambani (D) Laxmi Mittal 34. The 2nd meeting of SAARC Interior/Home Ministers was held in New Delhi. Which among the following attended the meeting for the first time ? (A) Sri Lanka (B) Afghanistan (C) Nepal (D) India 35. The Man Booker Prize 2007 has been awarded to— (A) Nicola Barker (B) Mohsin Hamid (C) Anne Enright (D) Lloyd Jones ANSWERS WITH HINTS 1. (C) Jessica Nicole Trisko from Canada is Miss Earth 2007, Silvana Santaella Arellano from Venezuela is Miss Earth-Water, Pooja Chitgopekar from India is Miss Earth-Air and Angela Gomez from Spain is Miss EarthFire. 2. (C) 2007 Gender Gap Index has been topped by Sweden with a gender equality of 81·5%, followed by Norway, Finland, Ice-land and New Zealand. 3. (B) Chang’e-1 entered its working orbit on November 7, 2007 after completing its two million km journey to the moon successfully. It entered the moon’s orbit on November 5, 2007. 4. (A) The 2006 Indira Gandhi Award for National Integration was given to Dr. J. S. Bandukwalla and Ram Puniyani on October 31, 2007 for furthering the cause of communal harmony by personal harmony. 5. (A) Mohammedmian Soomro, Chairman of the Pakistan Senate, was sworn in by President Pervez Musharraf on November 16, 2007 as the caretaker Prime Minister of Pakistan. 6. (B) Indira Rajaraman, Atul Sharma and Abusaleh Shariff are the other full time members of the Commission. 7. (D) The first International Conference on Federalism was hosted by Canada in 1999, 2nd was hosted by Switzerland in 2002, third was hosted by Belgium in 2005 and the fourth Conference was hosted by India. It was held from November 5 to 7, 2007 in New Delhi. 8. (D) CO 2 emissions per capita of Australia are 10·0 tonnes, US 8·2 tonnes, UK 3·2 tonnes, China 1·8 tonnes, India 0·5 tonnes. 9. (B) Skynet 5-B was launched on November 14, 2007 from Europe’s Kourou spaceport, in French Guiana. 10. (B) South Africa will host the FIFA World Cup in 2010, Germany hosted in 2006 and South Korea and Japan hosted in 2002. 11. (D) The 13th ASEAN Summit was held from November 18 to 22, 2007 in Singapore. 12. (A) Churchill brothers won the Durand Cup Football Title on November 7, 2007 defeating Mahindra United. 13. (C) 14. (A) 15. (C) Singapore is at 1st position followed by Netherlands at 2nd and Germany at 3rd. The US is at 14th position. 16. (B) 17. (B) 18. (A) L. N. Mittal is the richest Indian with a wealth of $ 51 billion, Mukesh Ambani is at second position with net worth of $ 49 billion, Anil Ambani is at third position with net worth of $ 45 billion and Kushal Pal Singh is at 4th position with net worth of $ 35 billion. 19. (A) Chandrayaan-2 will be a joint plan which will include a lander and a rover to walk around the moon’s surface and collect samples and data for analysis. 20. (A) It has been developed by Tata Group’s Pune based Computational Research Laboratory (CRL). 21. (A) ‘Abuja Declaration’ was signed between India and Nigeria on October 15, 2007 at Abuja. 22. (B) The General Assembly of United Nations declared 2008 as the International Year of Potato on October 18, 2007 to “increase awareness of the importance of potato as a food in the developing nations.” 23. (A) Canada joined as the 7th member of Asia-Pacific partnership on clean development and climate, which held its 2nd ministerial meeting in New Delhi on October 15, 2007. 24. (C) India would host the 3rd IBSA Summit in 2008. It was decided at the 2nd IBSA Summit at Tshwane in Pretoria, South Africa. 25. (B) Inzamam-ul-Haque, the former Pakistani Captain took retirement from International Test Cricket on October 12, 2007. 26. (D) Albert Fert and Peter Grunberg have won the Nobel Prize in Physics 2007 for the discovery of Giant Magneto resistance. 27. (A) 28. (C) Ruth Manorama from India received the prize in 2006. Grameen Shakti, a company in Bangladesh was also the recipient of the prize for 2007. 29. (A) 30. (B) ‘OCTOPUS is an organisation to fight terrorism, which was established by Andhra Pradesh Government in Hyderabad in September 2007. 31. (B) Virginie Razzano of France won Japan Open Tennis Tournament in Tokyo on October 6, 2007 defeating Venus Williams. 32. (A) 33. (C) With sensex touching the mark of 20,000, Mukesh Ambani became the richest person of the world leaving behind Carlos Slim and Bill Gates. His total assets now stand at 63·2 billion dollars. 34. (B) 35. (C) Anne Enright was on October 16, 2007 named the winner of Man Booker Prize 2007 for fiction for her novel ‘The Gathering’. ●●● C.S.V. / February / 2008 / 1678