Csv Nov09

In This Issue Regulars

November 2009 Year—12 Issue—141

Editorial Science and Technology Latest General Knowledge Inspiring Young Talents— (i) Topper : Uttarakhand, PMT 2009 (4th Rank) —Dhananjay Dwivedi (ii) CBSE (PMT) 2009 (54th Rank)—Garima Singh Science Tips

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Physics Editor MAHENDRA JAIN

Sound-III : Stationary Waves, Vibrations of Air Columns and Stretched Strings Nuclear Physics-III : Nuclear Energy Typical Model Paper

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Typical Model Paper

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Chemistry Nuclear Chemistry Typical Model Paper Typical Model Paper Typical Model Paper 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.

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 : 4053333, 2531101, 2530966 Fax : (0562) 4053330, 4031570 E-mail : [email protected] Website : www.pdgroup.in

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Zoology Lymphatic System Biology and Therapy of Cancer Typical Model Paper Typical Model Paper Typical Model Paper

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Botany Diversity of Life : Plant Groups Pollination Typical Model Paper Typical Model Paper Typical Model Paper Solved Paper : CBSE Medical Entrance Exam., 2009

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Other Features Assertion and Reason Type Questions True or False Do You Know ? CSV Quiz Contest No. 138 Correct Solution and Prize Winners of CSV Quiz No. 135 General Awareness

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To Our Readers Dear Readers, It is a matter of immense pleasure and great satisfaction to us in presenting to you the November issue of your favourite and frontline magazine ‘Competition Science Vision’. PMT aspirants generally agree that the utility of each issue of the magazine is unique and unbeaten. Its flawless presentation in all the subjects of premedical tests is its special quality rarely found in any other such magazine. Our experienced authors are trying hard to make its each issue more and more examination-oriented including in it all the changes of examination pattern. This makes our readers updated for each year’s tests. We strive to make CSV to meet all your requirements. Hard work under proper guidance is a must for success in any examination. Revision of what you have read is also necessary before you finally take up the examination. CSV is so designed that it provides you all you need for success. Read CSV regularly and intelligently. It gives you the power to master your career and shape your destiny. With best wishes for your success. Sincerely yours,

Mahendra Jain (Editor)

FORTHCOMING COMPETITIVE EXAMS. 2009 Delhi SSSB Trained Graduate Teacher (Social Science) Exam. (Oct. 10) UPSC CPF Assistant Commandants Exam., 2009 (Oct. 11) Reserve Bank of India Officers Grade ‘B’ Examination (I Stage) (Oct. 11) Delhi SSSB Trained Graduate Teachers (Sanskrit, Hindi, Urdu, Punjabi) Exam. (Oct. 11) Civil Services (Mains) Exam., 2009 (Oct. 22) The New India Assurance Company Ltd. Administrative Officer Examination (Oct. 25) Gurgaon Gramin Bank Officer Scale-I Exam. (Oct. 25) Prathama Bank Officers Scale-I Exam. (Oct. 25) Shreyas Gramin Bank Specialist Officers Exam. (Nov. 1) Uttaranchal Gramin Bank Clerk-cum-Cashiers Exam. (Nov. 1) Gurgaon Gramin Bank Clerk-cum-Cashier Cadre Examination (Nov. 1) Prathama Bank Office Assistants/Clerks Exam. (Nov. 1) Shreyas Gramin Bank Clerk-cum-Cashiers Exam. (Nov. 8) Madhya Pradesh Civil Judge (Mains) Examination (Nov. 8) National Talent Search Exam., 2009 (For VIII Class Studying Students) (Nov. 8) Madhya Pradesh National Means-cum-Merit Scholarship Examination, 2009-10 (Nov. 8) S.B.I. Clerical Cadre Exam., 2009 (Nov. 8 & 15) S.S.C. Auditors and Accountants Exam. (Indian Audit and Account Deptt.) (Nov. 15) Madhya Pradesh State Service (Mains) Exam., 2008 (Nov. 16)

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Indian Economic Service/Indian Statistical Service Examination, 2009 Corporation Bank Probationary Officers Exam. Rajasthan State Eligibility Test (SET), 2009 Rajasthan Gramin Bank Officers Scale-I Examination Triveni Khetriya Gramin Bank Officers Scale-I Exam. Rajasthan Gramin Bank Clerk-cum-Cashier Exam. Corporation Bank Clerical Cadre Exam. Triveni Khetriya Gramin Bank Office Assistant Exam. Indian Air Force Airman Selection Test [Group ‘X’ (Technical) Trades] S.S.C. Tax Assistant Examination, 2009 (Closing Date : 30 Oct., 2009) Rashtriya Military Schools Common Entrance Test (Class VI) CSIR-UGC National Eligibility Test, 2009 (Science Subjects) Vijaya Bank Probationary Clerks Exam. KVS Post Graduate Teachers Exam. (Closing Date : 12 Oct., 2009) Jaipur Thar Gramin Bank Office Assistants Exam. (Closing Date : 10 Oct., 2009) KVS Trained Graduate Teachers Exam. (Closing Date : 12 Oct., 2009) KVS Primary Teachers Exam. (Closing Date : 12 Oct., 2009)

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AWARDS/HONOURS 55th National Film Awards Director Priyadarshan’s off-beat Tamil Film ‘Kanjeevaram’ won the best film award, 2007, while southern actor Prakash Raj has been adjusted as the best actor for his performance in the same film. Renowned Malayalam film director Adoor Gopalakrishnan is selected as the best director for ‘Naalu Pennungal’ , while Kannada actress Uma Shri is declared as the best actress for ‘Gulabi Takies’. Bollywood star Aamir Khan’s directorial debut ‘Taare Zameen Par’ got the best family welfare award and Prasoon Joshi got the award for the best lyrics for ‘Maa’ a film about a dyslexic child. Shankar Mahadevan received the National Award for the best playback singer. ‘Chak De India’ won the award for the best picture providing wholesome entertainment. Sarswati Samman—Assamese writer Lakshminandan Bora was honoured with the ‘Sarswati Samman 2008’ for his significant contributions to Assamese literature. Vyas Samman—Hindi writer Mannu Bhandari was awarded the ‘Vyas Samman 2008’ for her invaluable contributions to Hindi literature. Sarswati Samman and Vyas Samman are given by K. K. Birla Foundation for promotion of Indian literature. 61st Annual Emmy Award 2009—Emmy Awards were given in Los Angeles (U.S.A.) on September 21, 2009. 30 Rock and Mad Men were the winners of every category. 30 Rock was named the best comedy. Mad Men won the best drama category. Toni Collette won the Emmy as the best actor in a comedy. Calenn Close won her second straight Emmy for best actress.

C.S.V. / November / 2009 / 1067 / 2

ICC Awards England’s Ashes winning captain Andrew Strauss and Indian opener Gautam Gambhir are among the four players who were honoured with the ‘Cricketer of the Year Award’. India’s captain Mahendra Singh Dhoni and Australian fast bowler Mitchell Johnson are other cricketers who were honoured with the Awards by International Cricket Council in Johannesburg, South Africa, on October 1, 2009. List of the awardees : Cricketer of the Year— Mahendra Singh Dhoni (Ind), Gautam Gambhir (Ind), Mitchell Johnson (Aus), Andrew Strauss (Eng). Test Player of the Year— Gautam Gambhir (Ind), Mitchell Johnson (Aus), Thilan Samaraweera (SL), Andrew Strauss (Eng). One-Day Player of the Year— Shivnarine Chanderpal (WI), Mahendra Singh Dhoni (Ind), Virender Sehwag (Ind), Yuvraj Singh (Ind). Emerging Player of the Year— Ben Hilfenhaus (Aus), Graham Onions (Eng), Jesse Ryder (NZ), Peter Siddle (Aus). Associate and Affiliate Player of the Year (non-Test nations)— Rizwan Cheema (Can), Ryan ten Doeschate (Ned), William Porterfield (Irl), Edgar Schiferli (Ned). Twenty-20 International Performance of the Year—Shahid Afridi (Pak)-51 off 34 balls and two for 16 against South Africa in the World Twenty-20 semi-final in Nottingham on June 18. Tillakaratne Dilshan (SL)-96 not out off 57 balls against the West Indies in the World Twenty-20 semi-final at the Oval on June 19. Chris Gayle (WI)-88 off 50 balls against Australia in the World Twenty20 at the Oval on June 6. Umar Gul (Pak)-5-6 against New Zealand in the World Twenty-20 at the Oval on June 13.

Women’s Cricketer on the Year—Charlotte Edwards (Eng), Shelley Nitschke (Aus), Claire Taylor (Eng). Umpire of the Year—Aleem Dar (Pak), Tony Hill (NZ), Asad Rauf (Pak), Simon Taufel (Aus). Spirit of Cricket Award—Australia, England, New Zealand, Sri Lanka.

BOOKS Civil Society and Human Security— V. R. Raghavan (The volume touches upon some very critical issues and raises many questions related to human security). Jammu and Kashmir : The Cold War and the West— D. N. Panigrahi —(In this book the author shows that the Western support to Pakistan played a key role in shaping the India-Pakistan contestation on Jammu and Kashmir. The military and diplomatic assistance to Pakistan, sustained its long-running offensive to take control of the state.) Reducing Global Poverty : Patterns of Potential Human Progress— Bary B. Hughes (This hefty work, the first in a series of annual volumes, deals with poverty reduction issues). The Corporate Greenhouse— Yda Schreuder (The book discusses the role of MNCs in shaping the implementation of climate change agreements).

DAYS October 1—International Day for Elderly. October 2—Gandhi Jayanti, Lal Bahadur Shastri’s Jayanti, World Vegetation Day. October 3—World Habitat Day. October 4—World Animal Welfare Day. October 8—Indian Air Force Day.

years later, and never lost an Assembly or Lok Sabha election thereafter. A devout Christian, he made pilgrimages to Bethlehem and Tirumala with equal piety, showing his broad mindedness. The game-changing event was his 1500 km padayatra in 2003, which brilliantly tapped into mass unrest over the agrarian crisis and captured him to the seat of power a year later. He tirelessly presented his mission as the transformation of agriculture and farmers’ lives and the ending of agrarian distress. His ‘Jilayagnam’ drive to irrigate ten million acres of land by 2014, his Agriculture Technology Mission, his vigorous implementation and expansion of the National Rural Employment Guarantee Scheme and his schemes to supply rice at Rs. 2 per kg. to the poor and provide free power to farmers, won his enormous credibility on the ground. Norman Borlaug—Dr. Norman Y. S. R. Reddy (A Talented leader)—In the heart-rending tragedy Borlaug, who passed away in Dallas, Texas on September 13, 2009 at the of Yeduguri Sandinti age of 95, remains the only recipient Rajasekhara Reddy’s of the Nobel Prize for agriculture, demise in a helispecifically the Peace prize of 1970. copter crash, Andhra Possessed of insatiable curiosity Pradesh has lost its about matters, botanical and blessed most charismatic with remarkable energy and scientific leader and the Conincisiveness, Dr. Borlaug worked in gress its most politically talented and Yeduguri Sandinti Mexico in 1950s, where he crossbred a ‘dwarf’ strain of wheat. This, when resourceful Chief Rajasekhara Reddy treated with chemical fertilizers and Minister. A medical graduate, YSR, as he was widely pesticides, produced for greater known, was a politician of Pulivendula quantities of seeds than the earlier in Kadapa district. He became an varieties had done. His discovery was MLA at the age of 29, a Minister two commercially introduced in Mexico in October 9—World Post Office Day. October 11—World Allergy Awareness Day. October 13—UN International Day for Natural Disaster Reduction. October 14—World Standards Day. October 15—World Sight Day (Second Thursday of October), World White Cane Day (Guiding the Blind). October 16—World Food Day. October 21—Police Commemoration Day. October 24—U.N. Day, World Development Information Day. October 27—Infants Day. October 30—World Thrift Day. October 31—National Integration Day.

1956, and the 1963 harvest was six times greater than that of 1956.

An Assault on Hunger : Norman E. Borlaug and his wife Margaret with the Nobel Peace Prize he was awarded in 1970

Dr. Borlaug had a special place in the hearts of all Indians as he, alongwith M. S. Swaminathan, brought the ‘Green Revolution’ in India marking a watershed in India’s agrarian economy and rural development. At a time in sixties when India was facing the spectre of severe food shortages, his introduction of high yielding varieties of (wheat) seeds set in motion a technological revolution in Indian agriculture that led eventually to the country achieving self-sufficiency in food grains. ‘Green Revolution’ lifted the spirits of Indian people and gave them new hope and confidence. He was the father of ‘Green Revolution’ in India. The ‘grateful nation’ mourns his death and owes a debt of gratitude to this greatman. This greatest hunger-fighter for all time, made a profound difference to the lives of hundreds of millions of people in developing world. Raj Singh Dungarpur—The former President of the Board of Control for Cricket in India (BCCI), Raj Singh Dungarpur (73), passed away in Mumbai on September 12, 2009 after prolonged illness. He had been suffering from alzheimer and did not recognise his acquaintances. He was the third son of the former ruler of Dungarpur, Maharawal Lakshman Singhji. He graduated from Vikram University, Ujjain. He was an excellent cricketer and played for Rajasthan in Ranji Trophy and other national tournaments for 16 years, as a medium fast bowler. He took 206 wickets and

C.S.V. / November / 2009 / 1068

scored 1292 runs in national tournaments. He worked as the national selector for two terms. He was the founder-Chairman of National Cricket Academy, Bangalore.

V. K. Sarswat (New D.G., DRDO )—Dr. V.K. Sarswat assumed office as the Director-General of the Defence Research and Development Organisation (DRDO) and Scientific Advisor to the Ministry of Defence. He would also be the Secretary of Department of Defence Research and Development in the Ministry of Defence. He succeeded Dr. M. Natarajan who retired. Earlier Dr. Sarswat had been Chief Controller, Research and Development (Missiles and Strategic Systems), DRDO. Dr. Sarswat, who has a Ph. D. degree in combustion engineering, joined DRDO in 1972. He led the development of India’s tactical and strategic missile systems and antimissile systems. K. Rosaiah (New C.M., A.P.)— Mr. K. Rosaiah (76), the seniormost member in Y.S. Rajasekhara cabinet, was sworn-in as the new Chief Minister of Andhra Pradesh. The State Governor Mr. N. D. Tiwari administered the oath of office collectively to all the 34 members of Y. S. Rajasekhara Reddy cabinet also. Mr. Rosaiah had been a member of several Congress cabinets in the past. Irina Bokova (First Woman to Head UNESCO)—In what can only be described as a stunning upset. Irina Bokova, Bulgaria’s former foreign minister and currently ambassador to Paris, became the first woman to head UNESCO, the Paris-based United Nations’ agency for education, science and culture. Ms. Bokova was elected on September 23, 2009 by the Organisation’s 58-member Executive Board winning 31 votes to defeat her rival, Egypt’s Culture Minister, Farukh Hosni, who polled 27. Mr. Bokova also became the first person from the former Soviet bloc countries to be elected to this post.

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final touches to the 4·62 km long rail bridge connecting the mainland to the Y. Hatoyama (New P. M., Japan )—Mr. Yukio Hatoyama, on September 16, 2009, was appointed as Japan’s 93rd Prime Minister following his party’s landslide victory in the August 30 general election. Mr. Hatoyama was appointed as Prime Yukio Hatoyama Minister at the plenary session of both the House of Representatives and the House of Councillors. He is the leader of Centre-left coalition. With his assumption of office, the curtain came down over a half-century of conservative rule in Japan. Mr. Hatoyama’s agenda is sparking a previously unthinkable idea that the U. S. may have to exit from the centre-stage of the next big theatre in global affairs. Somdev Devvarman (Davis Cup Hero)—Somdev Devvarman scripted a spectacular win over South Africa’s Rik-de-Voest in a high-voltage marathan five-setter to put India back

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The structure has a clearance of 7m above water and is made of 1 3 4 pillars.

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About 18,000 tonnes of cement and 50,000 metric tonnes of steel have gone into the structure.

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The initial cost was estimated to be Rs. 246 cr, which was later revised to Rs. 298 cr.

proposed international container trans-shipment terminal on Vallarpadam island across the Vembanad lake in Ernakulam district. Due to be commissioned in November 2009, it will be the longest rail bridge in India. Currently, the record is held by the Nehru Sethu near Dehri-on-Sone, located on the Kolkata-Delhi line. The Nehru Sethu is 3·065 km-long. This bridge is expected to save exporters of crores of rupees wasted in trans-shipment of their goods through other ports. The project is also set to transform the industrial environment in the State.

SPORTS Cricket

Somdev Devvarman

in the elite Davis Cup World Group after a gap of 11 years, on September 20, 2009 in Johannesburg (South Africa). The tenacious Somdev won 3–6, 6–7 (3), 7–6 (5), 6–2, 6–4 after an energy-sapping battle of four hours and 38 minutes at the Ellis Park indoor stadium. Down by two sets, Somdev made an incredible comeback as he overcame an early break of serve to break back de-voest in the fourth game. Somdev wrapped up the set in the eighth game and it all boiled down to the fifth and decisive set. This win of Davis Cup gave India an unassailable 3–1 lead over the Rainbow Nation.

Kochi (The longest rail-bridge)— Engineers in Kochi (Kerala) are giving

Squads for Irani Trophy and Champions League—Cricket matches between Ranji Champion, Mumbai and rest of India for the Irani Trophy will be held at Nagpur during October 1 till 5. Squads for the rest of India for Irani Trophy and Challenger Series are— Irani Trophy : Rest of India— Gautam Gambhir (Capt.), S. Badrinath (Vice-Capt.), M. Vijay, Virat Kohli, Manoj Tiwary, Ravindra Jadeja, Irfan Pathan, Wriddiman Saha, S. Sreesanth, Munaf Patel, Pragyan Ojha, Kedar Jadhav, Abhinav Mukund, Uday Kaul and Sudeep Tyagi. Challenger Series : India Blue — M. S. Dhoni (Capt.), Sachin Tendulkar, Wasim Jaffer, Jalaj Saxena, Suresh Kumar, Kedar Jadhav, Abhishek Nayar, Yusuf Pathan, Naman Ojha, Harbhajan Singh, S. K. Trivedi, S. Sreesanth, Ashok Dinda and A. S. Goswami. India Red—Yuvraj Singh (Capt.), M. Vijay, Sunny Sohail, Shikhar Dhawan, Ishant Jaggi, Ravindra

Jedeja, Bhuvneshwar Kumar, Wriddiman Saha, R. Ashwin, Sudeep Tyagi, Munaf Patel, S. S. Tiwary, Ishant Sharma and H. Khadiwale. India Green—Suresh Raina (Capt.), A. Rahane, Tanmay Srivastava, Ravi Inder Singh, Manoj Tiwary, S. Badrinath, Chetanya Nanda, Irfan Pathan, Parthiv Patel, S. K. Jakati, L. Balaji, Dhawal Kulkarni, Pankaj Singh and Uday Kaul. ICC Champions Trophy 2009— ICC Champions Trophy Tournament started on September 22, 2009 in South Africa and final will be played on October 5, 2009. ICC Champions Trophy 2009 Schedule Group A : Australia, India, Pakistan and West Indies. Group B : England, New Zealand, South Africa and Sri Lanka Oct. 2

Ist Semifinal (A1 Vs. B2)

Centurion

Oct. 3

2nd Semifinal (B1 Vs. A2)

Johannesburg

Oct. 5

Final

Centurion

C.S.V. / November / 2009 / 1070

India Wins Compaq Cup— Sachin Tendulkar’s century, Harbhajan Singh’s five-wicket haul and the magnitude of the target (320 in 50 overs) ensured that India won the Compaq Cup final against Sri Lanka in Colombo on September 15, 2009. Sachin Tendulkar held centre stage at the Premdasa Stadium. His 44th ODI century—his fourth at this ground and his sixth in tournament finals— helped India to score 319 for five in 50 overs in the championship match for the Compaq Cup.

Tennis U. S. Open—India’s Leander Paes and Czech Republic’s Lucas Dlouhy won the U. S. open men’s doubles title defeating India’s Mahesh Bhupathi and Mark Knowles of Bahamas 3–6, 6–3, 6–2 in New York on September 14, 2009. This was Paes’ 10th Grand Slam title, including six men’s doubles and four mixed doubles crown.

The Results Prefix denotes seeding Women Singles Final : Kim Clijsters (Bel) bt 9-Caroline Wozniacki (Den) 7–5, 6–3. Men Singles Final : 6-Juan Martin del Potro (Arg) bt 1-Roger Federer (Sui) 3–6, 7–6(5), 4–6, 7–6(4), 6–2. Women Doubles : 4-Serena Williams & Venus Williams (USA) bt 1Cara Black (Zim) and Liezel Huber (USA) 6–2, 6–2. Men Doubles Final : 4-Leander Paes (Ind) & Lukas Dlouhy (Cze) bt 3Mahesh Bhupathi (Ind) & Mark Knowles (Bah) 3–6, 6–3, 6–2. Junior Singles Boys Final : 3Bernard Tomic (Aus) bt Chase Buchanan (USA) 6–1, 6–3. Girls Final : 11-Heather Watson (GBr) bt Yana Buchina (Rus) 6–4, 6–1.

Billiards IBSF World Professional Billiards Championship—India’s Pankaj Advani is 2009 World

(Continued on Page 1184)

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

‘‘Hardwork, motivation and encouragement of family members and positive thinking are the main elements of my success in this examination.’’ —Dhananjay Dwivedi

Topper—Uttarakhand, PMT–2009 (4th Rank) [‘Competition Science Vision’ arranged an exclusive interview with Mr. Dhananjay Dwivedi who has the credit of securing a high position in the list of successful candidates of Uttarakhand PMT, 2009. 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. Dhananjay—Thanks. CSV—Before knowing your result what did you think about those who achieve top positions ? Dhananjay—They are like us but do things differently. CSV—Achieving top position has come as surprise to you or were you confident of achieving it ? Dhananjay—Not so good but I was sure that blessings of my parents and teachers will not fail so I was expecting under 50th rank. CSV—What do you think is the secret of your success ? Dhananjay—My parents and my Bhaiya-Bhabhi’s motivating behaviour which never demoralised me. CSV—In how many attempts did you get this success ? Dhananjay—It was my 3rd attempt. CSV—What were the shortcomings in your preparation for earlier attempts ? How did you make up for them this time ? Dhananjay—I was weak in Biology. But this time my teachers and friends supported me very much so I got a position in it also. CSV—From where did you get the inspiration of choosing a medical career ? Dhananjay—My father and mother. CSV—From when did you start the preparation for it ? Dhananjay—Just after 12th but I think it must be right from the 11th standard to achieve good success in less time. CSV—What planning did you make for preparation ? Please tell something in detail.

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Dhananjay—From the beginning of this year, I planned to take more and more tests, particularly of objective questions in which material of CSV, helped me a lot.

—Competition Science Vision is a landmark in the field of science magazines for pre-medical examinations. It gives all important facts with easily understandable short explanations. —Dhananjay Dwivedi CSV—How much time did you devote daily and regularly for Physics, Chemistry, Zoology and Botany ? Dhananjay—Initially, in my first year of preparation, I give almost one hour for each subject, but in my last year I always gave more time on my weak points. Subject not mattering much. CSV—Out of the above four subjects, to which subject did you give more weightage and why ? Dhananjay—Certainly Biology because firstly we are going to become doctor and the almost in every entrance exam there are more questions on Biology. But Chemistry is very scoring subject. CSV—Did you make complete study of all topics or of some selective topics ?

Dhananjay—I always stress more on those points which my teachers tell me because being selective is also the key to success. Don’t waste time in vain for the less weightage portion. CSV—How did you give final touches to your preparation ? Dhananjay—After preparing, I joined one month test series which was helpful for whole revision and discussion with the teachers and classmates was utmost important. CSV—Did you prepare notes ? Dhananjay—Yes, I prepared notes but they were very selective and short which can be studied at the night before the examination and without tension. CSV—What was your attitude for solving numerical questions ? What weightage did you give them ? Dhananjay—One must concentrate on numericals sincerely because in medical exams they are generally formula based so we can score good marks in them.

Bio-Data Name—Dhananjay Dwivedi Father’s Name—Sri Brij Bhushan Dwivedi Mother’s Name—Smt. Shakuntla Dwivedi Educational Qualifications— H. S. / Std. X—89·4% (P.P.J. Saraswati Vihar, Nainital), 2004. Inter/Std. XII—76% (P.P.J. Saraswati Vihar, Nainital), 2006.

CSV—How much time is sufficient for preparing for this examination ? Dhananjay—Actually planning is very important because upto my 12th class I had not planned my career so I got this success late but I think two years are sufficient.

(Continued on Page 1092)

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

‘‘Sustained and scheduled study with guidance of teachers is the secret of my success.’’ —Garima Singh

CBSE (PMT) 2009 [General Rank-54, OBC Rank-5] [‘Competition Science Vision’ arranged an exclusive interview with Miss Garima Singh who has the credit of being successful in CBSE (PMT) and U.P.-CPMT with high ranks. For her brilliant success, she deserves all praise and our heartiest congratulations. This important interview is presented here in its original form.] CSV—Congratulations on your brilliant success. Garima—Thank you, Sir. CSV—Before knowing your result what did you think about those who achieve top positions ? Garima—I thought they must be some extraordinary persons, much more studious than myself. CSV—Achieving top position has come as surprise to you or were you confident of achieving it ? Garima—I was confident of achieving good rank but was not sure about such high rank. CSV—What do you think is the secret of your success ? Garima—I think sustained and scheduled study is the secret of my success along with the guidance of the teachers. CSV—In how many attempts did you get this success ? Garima—It was my second attempt. CSV—What were the shortcomings in your preparation for earlier attempts ? How did you make up for them this time ? Garima—I was not so much serious about my study in the first attempt as in second attempt. CSV—From where did you get the inspiration of choosing a medical career ? Garima—My mother always told me that for girls the most respectable job is to be a doctor. CSV—From when did you start the preparation for it ? Garima—After class X. CSV—What planning did you make for preparation ? Please tell something in detail. Garima—I used to make a time table nearly daily for what to do that

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day. I used to decide subject matters to study for the day and sleep only after completing them.

—CSV is a nice magazine and is very useful for pre-medical examinations. It contains good subject matter and quality sample questions in exhaustive numbers. —Garima Singh

CSV—Did you prepare notes ? Garima—Yes, I prepared short but comprehensive notes. CSV—What was your attitude for solving numerical questions ? What weightage did you give them ? Garima—They are very important. In PMT’s, the numericals are generally net very tough and they assure easy marks. If any numerical was very long then I solved it at last. CSV—How much time is sufficient for preparing for this examination ? Garima—Two years of deep study is sufficient. CSV—From what level of education should an aspirant begin preparing for it ? Garima—From class XI.

Bio-Data CSV—How much time did you devote daily and regularly for Physics, Chemistry, Zoology and Botany ? Garima—After 6 hours of coaching, I devoted 7-8 hours daily for self study. CSV—Out of the above four subjects, to which subject did you give more weightage and why ? Garima—I gave weightage to the subjects according to the toughness of the lesson which I am going to read. If it is from Physics then more weightage for it and if from Botany then less for it. CSV—Did you make complete study of all topics or of some selective topics ? Garima—I tried to make complete study of all subjects. CSV—How did you give final touches to your preparation ? Garima—After reading the lessons, I go to the lesson content at start and used to make ticks for what I know I studied. This gave me confidence.

Name—Garima Singh Father’s Name—Sri Dharmraj Singh Mother’s Name—Smt. Renu Singh Educational Qualifications— H.S./Std. X—90% (Kendriya Vidyalaya, DLW), 2006. Inter/Std. XII—89·8% (Kendriya Vidyalaya, DLW), 2008. Special Ahievements— ● Also selected in CPMT, 2009 ● Selected in National Science Exhibition (KVS), 2007 ● Got 3rd District Rank in Maths Olympiad one time

CSV—What was your order of preference for various branches for which this test is held ? Garima—Only MBBS. CSV—Please mention various books in each subject and magazines on which you based your preparation. Garima—NCERT Books for each subject. Elementary Biology, ABC Chemistry, Pradeep Physics, H. C. Verma for Physics and CSV magazine.

CSV—Did you take coaching in your preparation ? Garima—Yes, I took JRS Tutorials, Varanasi. It was very useful. I was much influenced by its Director Sir A. K. Jha. He gave equal attention to each student and taught every topic in a playing manner. CSV—What help do the science magazines render in the preparations for this examination ? Garima—They give extra edge in preparations and provide knowledge in some currently discovered matters related to our subject topics. CSV—What will be your criterion for selecting a magazine for these examinations ? Garima—It must have correctness in its writings and should explain difficult things in easy way.

Personal Qualities Hobbies—Listening songs and playing cricket Ideal Person—My mother Strong Point—My confidence Weak Point—Being nervous at some events

C.S.V. / November / 2009 / 1073

CSV—What is your opinion about our Competition Science Vision ? How much helpful and useful do you find it ?

Garima—I hadn’t read any specific book for G.K. I got it from the Internet, Dainik Jagran, The Times of India and CSV magazine.

Garima—It is a nice magazine. It contains subject matter and sample questions in good amount.

CSV—Whom would you like to give credit for your success ?

CSV—Please suggest in what way CSV can be made more useful for medical aspirants. Garima—It is already very useful and can be made more useful by introducing more diagrams alongwith questions. CSV—Please mention your position in the merit list as well as the marks obtained in different subjects. What were your aggregate percentage of mraks ? Garima—I obtained All India rank 5 in OBC category and 54th All India General rank. In Physics and Chemistry together I got 66/100. In Botany and Zoology I got 78/100 in CBSE Mains subjective examination. CSV—What books/magazines/ newspapers did you read for G.K. preparations ?

Garima—First my parents, then to my teachers and then myself. CSV—Please tell us something about your family. Garima—My father is a Senior Section Engineer in DLW. My mother is an architect Engineer. I have a sister and a brother who are younger to me. CSV—What in your frank opinion has been the biggest mistake in your preparation for this test ? Garima—I don’t find any big mistake. CSV—What message would you like to give for our readers of CSV ? Garima—They must try to get their dreams with whole heart, mind and soul taken together they will surely be get success. ●●●

Physics 1. For wattless current what should be the value of the power factor of the circuit ? ➠ Zero 2. For which colour is the critical angle of light, passing from glass to air, minimum ? ➠ Violet 3. Give an example of application of mutual induction in any device. ➠ Transformer

16. How does the atmospheric pressure vary with height? ➠ Atmospheric pressure P decreases with height h above sea level. For an ‘ideal’ atmosphere at constant temperature P = P0 e – kh where k is a constant and P0 is the pressure at the surface 17. How is r.m.s. velocity of gas molecules related to absolute temperature of the gas ?

➠ vrms ∝ T 18. What are transducers ? ➠ Devices which change signals from one form to another (e.g. sound to electrical) are called transducers 19. Is polarisation the property of all types of waves ? ➠ No, it is property of only transverse waves

4. What is the correct sequence of the semiconductors silicon, tellunium and germanium in the increasing order of their energy gap ? ➠ Tellurium, germanium, silicon

20. What is the peculiarity of a NAND gate ? ➠ The output is HIGH if the input is NOT HIGH, and vice-versa

5. Which ammeter is used to measure alternating current ? ➠ Hot wire ammeter

Chemistry

6. What quantity has the ampere-second as its unit ? ➠ Quantity of electricity 7. By seeing a glowing electric bulb can you say if it is being fed by A.C. or D.C. ➠ No 8. What does the sudden burst of a cycle tyre represent? ➠ Adiabatic process 9. What happens to the velocity and wavelength of light when it enters a denser medium ? ➠ Both decrease 10. The skylab space station did not have a safe landing. Why ? ➠ Because its remote control system failed 11. What happens when even a small bird hits a flying aeroplane ? ➠ It causes heavy damage 12. When does the lunar eclopse occur ? ➠ It occurs when the earth comes in between the moon and the sun 13. Why is the cooling inside a refrigerator not proper when a thick layer of ice deposits on the freezer ? ➠ Because ice is a bad conductor of heat 14. Which type of computer is often found in small business and in homes and classrooms ? ➠ The micro computer. It is the smallest and the least costly type of computer 15. Out of joule, calorie, kilowatt and electron-volt which one is not the unit of energy ? ➠ kilowatt

C.S.V. / November / 2009 / 1074

21. What is Turnbull's blue ?

➠ Fe 4[Fe (CN)6]3 22. An hypothesis tested by experiments is known as ➠ Theory 23. What is magnesia alba ? ➠ Mg (OH)2.Mg CO3.3 H 2O (used as an antacid) 24. The humidity of air is measured by ➠ Hygrometer 25. What is oleum ? ➠ H2S2O7 Also known as fuming sulphuric acid 26. Vulcanised rubber was invented by ➠ Charles Goodyear (1839) 27. An amino acid which does not contain a chiral centre. ➠ Glycine [NH2—CH2—COOH] 28. Scientist who perfected the technique for converting pig iron into steel ➠ Henry Bessemer (1856) 29. When the pH of the blood is lower than the normal value, this condition is known as ➠ Acidosis 30. The electrolytic method of obtaining aluminium from bauxite was first developed by ➠ Charles Hall (1886) 31. Which compound possesses characteristic smell like that of mustard oil ? ➠ Ethyl isothiocyanate [C2H5N = C = S]

32. First solar bettery was developed in the ➠ Bell Telephone Laboratory (1954) 33. What is Wilkinson's catalyst ? ➠ tris (triphenylphosphine) chlororhodium(I) [RhCl (PPh3)3] 34. In 1836 the galvanised iron was introduced first in ➠ France 35. What is caro's acid ? ➠ Permonosulphuric acid [H2SO5] 36. A porcelain funnel used for filtration by suction is known as ➠ Buchner Funnel 37. What is diazomethane ? +

–

➠ [CH2 = N = N or CH2N2] 38. A drying chamber, containing chemicals such as concentrated sulphuric acid or silica gel is known as ➠ Desiccator 39. Reforming of a gasoline fraction to increase branching in presence of AlCl 3 is known as ➠ Isomerization 40. A condenser consisting of glass tube surrounded by another glass tube through which cooling water flows is known as ➠ Liebig condenser

52. What is called the air spaces present in some of cranial and facial bones ? ➠ Sinuses 53. The basic monomers used in DNA replication are ➠ DNA nucleotides 54. What is called the smallest strength of stimulus that causes muscle contraction ? ➠ Threshold stimulus 55. The form of endocytosis in which specific molecules bind to receptor sites on the external surface of the membrane is called ➠ Pinocytosis 56. Where the osteocytes of compact bone are organized? ➠ Around Haversian canals 57. The term humour refers to

➠ Plasma and lymph 58. What is gradually lost from bones during aging ? ➠ Calcium and phosphate 59. Most vitamins function as

➠ Coenzymes 60. Which enzyme from small intestinal mucosa digests maltose and sucrose ? ➠ Disaccharidase

Zoology 41. The earliest prokaryotes must have been ➠ Chemoheterotrophs 42. Which ions are most concentrated outside a resting neuron ? ➠ Sodium 43. The only arthropods with two pairs of antennae are the ➠ Crustaceans 44. Which type of cartilage is present in rib cage and the tip of nose ? ➠ Hyaline 45. Most fossils are found in ➠ Sedimentary rocks 46. Where the Meissner’s corpuscles and Merkel’s disks are located ? ➠ Near the papillae of dermis 47. To make a karyotype, chromosomes are photographed during ➠ Mitosis 48. Which kind of glands have possible role in communication as scent glands ? ➠ Apocrine glands 49. Viruses are obligate parasites of

➠ Cells 50. Which kind of bone marrow is present in the interior of each epiphysis ? ➠ Red bone marrow

C.S.V. / November / 2009 / 1075 / 3

51. Lack of precision with regard to the third base in codon and anticodon is the ➠ Wobble effect

Botany 61. What are nonseptate fungal hyphae ? ➠ Fungi having no cross walls in their hyphae 62. What is alternation of sexual and asexual reproduction in the life cycle of plants called ? ➠ Alternation of generation 63. Why does distinguishing species on the basis of reproductive isolation run into problems ? ➠ Because some species hybridize and reproductive isolation is very difficult to observe 64. What is a catalytically active complex made up of an apoenzyme and a coenzyme called ? ➠ Holoenzyme 65. What does retrovirus have ? ➠ Ritroviruses are RNA animal viruses that have a DNA stage 66. What is the first stable product of photosynthesis in C4 plants ? ➠ Oxaloacetic acid 67. What is the chief means of achieving genetic variation ? ➠ Mutation

(Continued on Page 1176)

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.

Graphical Representation of Stationary Waves II

I t

(ii) A stationary wave does not transmit energy in the medium.

II

I t

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

(iii) The amplitude of the stationary wave is given by A = 2a cos

2πx λ

(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.

II

I t

I

II

t

t

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.

(viii) Longitudinal stationary waves are formed in flutes and air columns.

(vii) Like progressive waves the stationary wave do not proceed forward.

(ix) Transverse stationary waves are formed in sonometer, sitar, gitar and strings.

(viii) The variations of pressure and density are maximum at nodes but are minimum at antinodes.

C.S.V. / November / 2009 / 1076

Comparison of Progressive and Stationary Waves Progressive Waves

Stationary Waves

1. These waves move in medium with a definite speed.

1. These waves do not move in any direction and remain confined within two boundaries.

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.

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.

5. Compressions and rarefactions produce alternately at definite places.

6. Same changes of pressure and density occur at all points of medium.

6. Changes of pressure and density are maximum at nodes but minimum (zero) at antinodes.

7. These waves transmit every in the medium.

7. These waves do not transmit energy in the medium.

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 :

(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.

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.

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— Fixed end

Incident wave

Reflected wave

1 Heavy spring Light spring

Fixed end 2 (a)

N

(c) 1

Heavy spring Light spring

Spring

Light thread 2

(b)

(d)

(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. / November / 2009 / 1077

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.

(vii) Various stages of resonance in a closed end organ pipe are represented in the following diagrams : A

A

A N

This frequency is called the second harmonic or first overtone.

N λ1 4

t

N

3λ2 4

N

5λ3 4

N

N

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 N N A

l

N

λ1 2

λ2

A

N A

N N A

(xv) If ∴ Frequency,

A

A

λ1 l = then λ1 = 2l 2 v v n1 = = λ1 2l

C.S.V. / November / 2009 / 1078

(xvii) If l =

3λ3 2l v 3v then λ3 = ∴ n3 = = = 3n1 3 2 λ3 2l

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 : ……

(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

A

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

3λ3 2

(xix) The sound emitted by an open end organ pipe is musical.

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

λ

l

4

N

A

N

l

e

λ 2

A

(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.

(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

A

l N

λ 4

N 3λ 4

l2

N

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.

l2 + e =

3λ 4

(viii) Both the odd and even harmonics are emitted from a stretched string. That is,

λ = 2(l2 – l1) v = 2n(l2 – l1) l2 – 3l1 e = 2

n1 : n2 : n3 : …… : : 1 : 2 : 3 : ……

Laws of Transverse Vibrations in a Stretched String

(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

2

A

A

N

1 ; if T and m are constant. l

(ii) Law of tension : n ∝ T ; if l and m are constant. (iii) Law of mass : n ∝

1 ; if T and l are constant. m

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. (iv) Law of radius : n ∝

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.

λ 2

λ 2

= λ2

A

A

N

C.S.V. / November / 2009 / 1079

λ 2

(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,

N λ 2

(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.

N

N

Law of length : n ∝

(iii) Transverse arrangement of vibrations : λ1

λ 2

A

(i)

(ii) Melde’s experiment is performed by two methods.

A

N

λ 2

T · m

λ 4

and end-correction,

N

v 1 = 2l 2l

l1 + e =

∴ Speed of sound,

N

n1 =

(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.

(iv) If the lengths of air column at first and second resonance be l1 and l2 respectively, then

and

(v) If a string of length l vibrates in one loops then λ1 = 2l and frequency

3λ3 2

where m is the mass per unit length of the string. This is Melde’s law.

(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 :

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 ? 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.

Solution :

Solution :

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 :

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 :

C.S.V. / November / 2009 / 1080

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 :

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

(B) 8th

(C) 6th

(D) 10th

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

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

(B) 256 Hz

(C) 300 Hz

(D) None of these

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

(B) 512 Hz

(C) 256 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

(Continued on Page 1098) C.S.V. / November / 2009 / 1081

Important Concepts

But

Einstein’s mass energy equivalence relation— Einstein on the basis of his relativity theory has proved that if a substance loses an amount Δm of its mass, an equivalent amount ΔE of energy is produced where

1·6 × 10–19 joule = 1 electron volt (eV) ∴

ΔE = (Δ 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 Energy equivalent to 1 amu mass— ΔE = (Δ m) × c2 = (1·66 × 10–27) × (3·0 × 108)2 = 1·49 × 10–10 joule

C.S.V. / November / 2009 / 1082

ΔE =

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 (Δ m). Thus Δm = (Mass of protons + Mass of neutrons) – Mass of the nucleus For an atom ZXA Δ 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 Δm disappears and an equivalent amount of energy (Δ 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

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 per nucleon (MeV)

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 : 9.0 8.0 7.0

O16

C12 4

F18 He N14

56

Fe

U238

6.0 7

5.0 Li 4.0 3.0 2.0 1.0 0.0

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.

C.S.V. / November / 2009 / 1083

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.

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.

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

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

where k is a new constant. Hence for chain reaction to continue. N–A > 1 L

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

or

()

()

C.S.V. / November / 2009 / 1084

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.

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

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.

As an example, in the fusion of two nuclei of deuterium (heavy hydrogen), following reactions take place 1H

2

+ 1H2 ⎯→ 1H3 + 1H1 + 4·0 MeV (energy)

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 :

At a Glance Nuclear Fission

Nuclear Fusion

1. Neutrons are required for it.

Protons are required for it.

2. It is possible at normal temperature and pressure.

It is possible at extremely high temperature and pressure.

3. For this the energy relea- For this the energy released sed per nucleon per nucleon ΔE 200 ΔE 27 A = 235 ≈ 0·8 MeV A = 4 = 6·75 MeV 4. Fissionable materials are expensive.

The materials used in it are cheap ( e.g., hydrogen)

Nuclear Fission

Radioactive Disintegration

1. Fission is not a spontaneous process. It is produced by bombarding the nuclei of fissionable material by neutrons.

It is a spontaneous process. It cannot be started, stopped, accelerated or retarded by any chemical or physical process.

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.

In radioactive decay the unstable nuclei spontaneously emit light particle (α- and β-) and energy is obtained in the form of γ-rays.

3. Tremendous amount of energy is obtained in fission process.

Energy obtained in radioactive disintegration is very less.

Fast Breeder Test Reactor

Thermal Reactor

1. In this liquid sodium is used as coolant.

In this water is used as coolant.

5. To generate power for driving the engines and the propulsion of ships, submarines and air crafts, thus replacing steam, coal and petrol.

2. The fuel used is a mixture of plutonium and natural uranium.

The fuel used is natural uranium.

Research Atomic Reactors

3. 60–70% fraction of natural uranium is used.

Only 1·2% fraction of natural uranium is used.

4. The chain reaction is maintained by fast neutrons. 5. Number of neutrons produced per fission is more.

The chain reaction is maintained by slow neutrons.

Atom Bomb

Hydrogen Bomb

1. It is based on fission process.

It is based on fusion process.

2. In it critical size is important.

There is no limit to size.

2[ 1H1 + 1H1 →

1H

2[ 1H1 + 1H2 →

2He

2

He 3

+

3 2He

→

2

2

+ +1β0 + υ + 0·4 MeV] 3

He 4

+ 5·5 MeV] + 2(1H1) + 12·9 MeV

Adding we get 4(1H1) →

2He

4

+ 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.

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.

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

C.S.V. / November / 2009 / 1085

The number of neutrons produced per fission is less.

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.

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]

2

Given the binding energy per nucleon of 1H2 and is 1·125 MeV and 7·2 MeV respectively.

He 4

Solution :

Solution :

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.

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 :

Solution :

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 :

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 :

Example 4. How much energy is released in the following reaction ? 1H

2

+ 1H2 =

2He

4

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]

C.S.V. / November / 2009 / 1086

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 (C) 8O16 (D) 26Fe 56 4. The following nuclear reaction represents 14 1 15 + 7·3 MeV 7N + 1H ⎯→ 8 O (A) Nuclear fusion (B) Nuclear fission

(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 solar 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

represents— (A) β-decay (C) Fusion

4

+ 2e– + 26 MeV

(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

●●●

Useful for Various Competitive Exams.

By : Dr. Lal, Mishra & Kumar Code No. 1624 Rs. 250/UPKAR PRAKASHAN, AGRA-2 E-mail : [email protected] Website : www.upkar.in

C.S.V. / November / 2009 / 1087

radius of the circular orbits of these particles are respectively Re , Rp , Rd and Rα. It follow that— (A) Re = Rp (B) Rp = Rd (C) Rd = Rα (D) Rp = Rα 1. Which of the following is India’s first geostationary satellite ? (A) Apple

7. Scratch pad memory is used to— (A) Support main programme

(C) Aryabhatta

(B) Store small amounts of information that can be fetched when needed

(D) Rohini

(C) Store redundant data

(B) INSAT-B

2. A body is rolling down an inclined plane. Its translational and rotational kinetic energies are equal. The body is a— (B) Hollow sphere

(B) Linear and angular momentum

(C) Solid cylinder (D) Hollow cylinder 3. Which of the following is the main source of electricity in our country ? (A) Nuclear energy (B) Hydro electricity (C) Thermal energy (D) Geothermal energy 4. A combination of two thin convex lenses of focal length 0·3 m and 0·1 m will have minimum spherical and chromatic aberrations if the distance between them is— (A) 0·1 m

(B) 0·2 m

(C) 0·3 m

(D) 0·4 m

5. A free-floating astronaut ‘P’ pushes another free-floating astronaut ‘Q’ in space. The mass of P is greater than that of Q. The force exerted by astronaut P on astronaut Q will be— (A) Equal to zero (B) Equal to force exerted by Q on P (C) Greater than the forces exerted by Q on P (D) Less than the force exerted by Q on P. 6. When 92U235 is bombarded with one neutron, fission occurs and the products are three neutron, 94 36 Kr , and— 56 Ba

141

(B)

(C)

56 Ba

139

(D)

8. In Bohr’s model of hydrogen atom, which of the following pairs of quantities are quantized ? (A) Energy and linear momentum

(A) Solid sphere

(A)

(D) Store flexible data

139 54 Xe 142 53 I

C.S.V. / November / 2009 / 1088

(C) Energy and angular momentum (D) None of the above 9. The activity of a sample of a radioactive material is A 1 at time t 1 and A2 at time t2 ; (t2 > t 1). Its mean life is T. Then— (A) A1t 1 = A2t 2 (B)

A1 – A2 = Constant t2 – t1

13. A body of mass m slides down a rough plane of inclination α. If μ be the coefficient of friction, the acceleration of the body will be— (A) g sin α (B) g μ cos α (C) g (sin α – μ cos α) (D) g (cos α – μ sin α) 14. Spherical aberration in a lens— (A) Is minimum when most of the deviation is at the first surface (B) Is minimum when most of the deviation is at the second surface (C) Is minimum when the total deviation is equally distributed over the two surfaces (D) Does not depend on the above considerations 15. In the given arrangement of springs, the time-period of vertical oscillations of the mass m is—

(C) A2 = A1 e (t1 – t2)/T (D) A2 = A1 e

t1 t2 T

10. The output of a NAND gate is 0— (A) If both inputs are 0 (B) If one input is 0 and the other input is 1 (C) If both inputs are 1 (D) Either if both inputs are 1 or if one of the inputs is 1 and the other 0 11. The maximum frequency ν of continuous X-ray is related to the applied potential difference V as— (A) ν ∝ V

(B) ν ∝ V

(C) ν ∝ V3/2

(D) ν ∝ V2

12. An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The

(A) T = 2π

m k1 + k 2

(B) T = 2π

k1 + k2 m

(C) T = 2π

m(k 1 + k 2) k 1k 2

(D) T = 2π

mg k1 + k2

16. Kepler’s second law (law of areas) is nothing but a statement of— (A) Work-energy theorem (B) Conservation of linear momentum (C) Conservation of angular momentum (D) Conservation of energy

17. The kinetic energy Ek of a photoelectron varies with the frequency ν of the incident radiation as which of the following graph ?

section equal to φ , then the magnetic field energy is— 1 (A) N i φ (B) N φ i 2 (C) N2 i φ

(D) N i 2 φ

24. A circular metal plate of radius R is rotating with a uniform angular velocity ω with its plane perpendicular to a uniform magnetic field B. Then the e.m.f. developed between the centre and the rim of the plate is—

18. The equation of a travelling wave is given by y = 0·5 sin (20x – 400 t) where x and y are in metre and t is in second. The velocity of the wave is— (A) 10 m/s (B) 20 m/s (C) 200 m/s (D) 400 m/s 19. APT is used— (A) In inventory management (B) In CAM for NC machine tools (C) In teaching of the beginners (D) In Cobol 20. The radiation energy density per unit wavelength at a temperature T has a maximum at a wavelength λ0. At temperature 2T, it will have a maximum at a wavelength— (A) 4λ0

(B) 2λ0

(C) λ0/2

(D) λ0/4

21. A magnetising field of 1600 Am –1 produces a magnetic flux of 2·4 × 10–5 Wb in an iron bar of cross-section area 0·2 m2. The succeptibility of bar is— (A) 596 (B) 1192 (C) 298 (D) 1788 22. A short bar magnet, placed with its axis at 30° with an external magnetic field of 0·16 T, experiences a torque of magnitude 0·032 J. The magnetic moment of the bar magnet is (in units of J/T)— (A) 4 (B) 2 (C) 0·5 (D) 0·4 23. An iron tore supports N turns. If a current i produces a magnetic flux across the tore’s cross-

C.S.V. / November / 2009 / 1089

(A) πωBR2

ωBR2

(B)

(C) πωBR2/2

(D) ωBR2/2

25. The magnetic field inside a toroidal solenoid of radius R is B. If the current through it is doubled and the radius is also doubled keeping the number of turns per unit length the same, the magnetic field produced by it will be— (A) 4 B B (C) 2

(B) 2 B B (D) 4

26. A water tank, open to the atmosphere, has a leak in it in the form of a circular hole, located at a height h below the open surface of water. The velocity of the water coming out of the hole is— (A)

gh/2

(B)

gh

(C)

2gh

(D) 2 gh

27. The amplitude of an oscillating simple pendulum is 10 cm and its period is 4 sec. Its speed 1 second after it passes its equilibrium position is— (A) Zero

(B) 0·157 m/s

(C) 0·212 m/s (D) 0·32 m/s 28. A wave is represented by the equation y = a cos (kx + ωt) is superposed with another wave to form a stationary wave such that the point x = 0 is a node. The equation of the other wave is— (A) a sin (kx + ωt ) (B) –a sin ( kx – ωt ) (C) –a cos (kx + ωt ) (D) – a cos (kx – ωt ) 29. A person sitting in a rotating merry-go-round will feel— (A) An inward force (B) An outward force

(C) No force (D) An inward or outward force depending upon his weight and the velocity of rotation of the merry-go-round 30. In Young’s double slit experiment, the separation between the slits is halved and the distance between the slits and screen is doubled. The fringe-width will— (A) Be halved (B) Be doubled (C) Be quadrupled (D) Remain unchanged 31. A temperature of 4K on the Fahrenheit scale will be nearly— (A) 36° F

(B) 39°F

(C) – 237°F

(D) – 452°F

32. A square of side ‘a’ has charge Q at its centre and charge ‘q’ at one of the corners. The work required to be done in moving the charge ‘q ’ from the corner to the diagonally opposite corner is— Qq (A) Zero (B) 4πε0a (C)

Qq 2 4πε0 a

(D)

Qq 2πε0 a

33. If the rms speed of the molecules of a gas at 27°C is 141·4 m/s, the rms speed at 327°C will be nearly— (A) 1697 m/s (B) 565·6 m/s (C) 282·8 m/s (D) 200 m/s 34. The magnetic susceptibility of any paramagnetic material changes with absolute temperature T as— (A) Directly proportional to T (B) Remains constant (C) Inversely proportional to T (D) Exponentially decaying with T 35. Hot incandescent liquids and solids produce a…… spectrum. (A) Line (B) Band (C) Continuous (D) None of the above

→

36. An electric dipole of moment p placed in a uniform electric field

→

E has minimum potential energy

→

when the angle between p and

→

E is—

(A) Zero (C) π

π 2 3π (D) 2 (B)

37. The intensity of different order maxima is nearly the same in— (A) Young’s double slit experiment (B) Single slit diffraction pattern (C) Double slit diffraction pattern (D) All the above 38. X-rays are produced by accelerating electrons by voltage V and letting them strike a metal of atomic number Z. The highest frequency of X-rays produced is proportional to—

range at constant volume is (Gas constant R = 2 cal/mol-K)— (A) 70 calories (B) 60 calories (C) 50 calories (D) 30 calories 41. The amount of U235 in a sample of 20 g of naturally occurring uranium will be nearly— (A) 10 g (C) 3·36 g

(B) 14 g (D) 0·14 g

42. A body is thrown vertically upwards with a velocity u . The distance travelled by it in the fifth and the sixth seconds are equal. The velocity u is given by— (g = 9·8 m/s2) (A) 24·5 m/s (B) 49·0 m/s (C) 73·5 m/s (D) 98·0 m/s 43. The ionic usually—

crystals

are not

(A) Poor conductors

(A) V

(B) Z

(B) Opaque to visible radiation

(C) (Z – 1)

(D) (Z – 1)2

(C) Of high melting points

39. Canal rays is the name given to a beam of— (A) Electrons (B) Protons (C) Neutrons (D) Positively charged ions 40. 70 calories of heat is required to raise the temperature of 2 moles of an ideal gas from 30 °C to 35°C while the pressure of the gas is kept constant. The amount of heat required to raise the temperature of the same gas through the same temperature

(D) Very stable 44. A coil of N = 100 turns carries a current I = 5A and creates a magnetic flux φ = 10– 5 Tm2 per turn. The value of its inductance L will be— (A) 0·05 mH (B) 0·10 mH (C) 0·15 mH (D) 0·20 mH 45. The amplitude of a particle in S.H.M. decreases from 20 cm to 15 cm in 2 minutes. Its energy decreases by nearly— (A) 44% (B) 25% (C) 22·5% (D) 12·5%

ANSWERS WITH HINTS

C.S.V. / November / 2009 / 1090

46. The kinetic energy of an electron with de-Broglie wavelength of 0·3 nanometer is— (A) 0·168 eV (B) 16·8 eV (C) 1·68 eV (D) 2·5 eV 47. The intensity of sound waves in a medium is directly proportional to— (A) Square root of wave speed (B) Square of the density of the medium (C) Amplitude of the wave (D) Square of the frequency 48. The approximate nuclear radius is proportional to (A is the mass number and Z the atomic number)— (A)

A

(B) A1/3

(C)

Z

(D) Z1/3

49. The brightness of a bulb will be reduced if a resistance is connected in— (A) Series with it (B) Parallel with it (C) Series or parallel with it (D) None of the above cases 50. A molecule with a dipole moment p is placed in an electric field of strength E. Initially the dipole is aligned parallel to the field. If the dipole is to be rotated to be antiparallel to the field, the work required to be done by an external agency is— (A) –2p E (B) – p E (C) p E (D) 2p E

●●●

(Continued from Page 1071) CSV—From what level of education should an aspirant begin preparing for it ? Dhananjay—After 10th standard.

Personal Qualities Hobbies—Driving, Basket ball Ideal Person—My father Strong Point—My parents supporting behaviour Weak Point—Really I don’t give sufficient time for studies

CSV—What was your order of preference for various branches for which this test is held ? Dhananjay—Firstly M.B.B.S. then B.D.S. CSV—What help do the science magazines render in the preparations for this examination ? Dhananjay—They give us the latest information and many other necessary facts to crack exams like AIIMS, MGIMS etc. CSV—What will be your criterion for selecting a magazine for these examination ? Dhananjay—A good magazine should be precise and full of easy formulae and tricks. CSV—What is your opinion about our Competition Science Vision ? How

C.S.V. / November / 2009 / 1092

much helpful and useful do you find it ? Dhananjay—CSV is a landmark in the field of science magazines. It gives all important facts with easily understandable short explanations. CSV—Please suggest in what way CSV can be made more useful for medical aspirants. Dhananjay—By providing more model papers with Assertion and Reason type questions. 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 ? Dhananjay—4th rank Physics—35/50 Chemistry—43/50 Zoology—41/50 Botany—41/50 Total—160 CSV—What books/magazines/ newspapers did you read for G. K. preparations ? Dhananjay—Newspapers and T.V. News. CSV—Whom would you like to give the credit for your success ? Dhananjay—My respected mother and father and my BhaiyaBhabhi also helped me a lot and unforgettable support of my teachers and classmates. CSV—Please tell us something about your family. Dhananjay—My father is a lecturer in Electronics Engg. Deptt. M. G. Polytechnic Hathras, my mother is a house-wife. I have only one elder brother who is sales engineer in Delhi and Bhabhi house-wife. CSV—What in your frank opinion has been the biggest mistake in your preparation for this test ? Dhananjay—I had no planning in my mind about the career which is very-very important because in medical stream time has very much importance. CSV—What message would you like to give for our readers of CSV ? Dhananjay—Belief in God and firm determination and Blessings of your parents and teachers are something which do more than the hardwork. ●●●

(C) Acceleration (D) Momentum

→ → 1. Two vectors A and B are inclined to each other at an angle θ. Which of the following is the unit vector perpendicular to both → → A and B ? → → (A × B ) (A) → → A . B

(C)

(B)

^ ^ A ×B

(D)

sin θ

→ ^ A ×B AB sin θ ^ → A×B

(A) Time (B) Inverse time (C) Square of time (D) Square of inverse time 3. A number of bullets are fired in all possible directions with the same initial velocity u. The maximum area of ground covered by bullets is—

(C) π

( ) () u2 g u g

2

(B) π

2

(D) π

u2 2g

2

u 2g

2

( ) ()

4. Two particles having mass M and m are moving in a circular path having radius R and r. If their periods are same, the ratio of angular velocity will be— R r (A) (B) R r (C) 1

R r

(D)

5. Three blocks of masses 2 kg, 3 kg and 5 kg are connected to one another with light strings and are then placed on a smooth

F

2kg

1m/sec2 T1 3kg

T2

5kg

frictionless horizontal surface. The system is pulled with a force F from the side of the lightest mass so that it moves with an acceleration of 1 ms–2 . T1 and

C.S.V. / November / 2009 / 1093

6. The kinetic energy acquired by a mass (m ) in travelling distance (s ) starting from rest under the action of a constant force is directly proportional to— 1 (A) (B) m ⎯m √ (C)

AB sin θ

2. The dimension of RC is—

(A) π

T2 denote the tensions in other strings. The value of F is— (A) 2 N (B) 3 N (C) 5 N (D) 10 N

⎯m √

(D) m0

7. A metal ball of mass 2 kg moving with a velocity of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If after the collision, the two balls move together, the loss in kinetic energy due to collision is— (A) 40 J (B) 60 J (C) 100 J (D) 140 J

12. The translational degree of freedom of an ant moving in a plane is— (A) 0 (B) 1 (C) 2 (D) 3 13. In a surface tension experiment with a capillary tube water rises upto 0·1 m. If the same experiment is repeated in an artificial satellite, which is revolving around the earth; water will rise in the capillary tube upto a height of— (A) 0·1 m (B) 0·2 m (C) 0·98 m (D) Full length of tube 14. A thermodynamic process is shown in the figure. The pressures and volumes corresponding to some points in the figure are PA = 3 × 104 N/m2, VA = 2 × 10– 3 m3 PB = 8 × 104 N/m2, VB = 5 × 10– 3 m3

8. The escape velocity from the surface of the earth is v e. The escape velocity from the surface of a planet whose mass and radius are three times those of the earth, will be— (A) v e (B) 3 ve 1 (C) 9 ve (D) ve 3 9. A disc revolves in horizontal plane at a steady rate of 3 rev/s. A coin just remains on the disc if kept at a distance of 2 cm from the axis of rotation. What is the coefficient of friction between the coin and the disc ? (A) 0·5 (B) 0·65 (C) 0·7 (D) 0·75 10. A ring of mass m and radius r rotates about an axis passing through the centre and perpendicular to its plane with angular velocity ω. Its kinetic energy is— (A) mr ω 2 (B) mr 2ω 2 1 1 (C) mr ω 2 (D) mr 2ω 2 2 2 11. The dimension of gravitational field is same as that of— (A) Force (B) Velocity

In the process AB, 600 J heat is added to the system and in the process BC, 200 J of heat is added to the system. The change in internal energy of the system in process AC would be— (A) 800 J (B) 600 J (C) 640 J (D) 560 J 15. One poise is— (A) 1 dyne sec/cm2 (B) 1/98·1 kgf sec/m2 (C) 10–1 kg/m-sec (D) Any of the above 16. A particle executes simple harmonic motion with a frequency f. The frequency with which its kinetic energy oscillates is— f (A) (B) f 2 (C) 2f (D) 4f

17. We plot the graph having temperature in °C on x -axis and in °F on y -axis. If the graph is straight line, then the correct statement is— (A) The line intercepts the positive x-axis (B) The line intercepts the positive y-axis (C) The line passes through origin (D) The line intercepts the negative axis of both x and y axis 18. Two waves having equation x 1 = a sin (ω t + φ1) and x 2 = a sin (ω t + φ2) superimpose. If in the resultant wave the frequency and amplitude remain equal to those of superimposing waves, then phase difference between them is— π 2π (A) (B) 6 3 π π (C) (D) 4 3 19. A 10 kg iron bar (specific heat 0·11 cal/gm °C) at 80°C is placed on a block of ice. How much ice melts ? (A) 1·1 kg (B) 10 kg (C) 16 kg (D) 60 kg 20. With the propagation of a longitudinal wave through a material medium the quantities transmitted are— (A) Energy (B) Energy and linear momentum (C) Energy, momentum and mass (D) Energy and mass 21. Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is— (A) 30 K (B) 18 K (C) 50 K (D) 42 K 22. Rainbow is formed due to— (A) Scattering and refraction (B) Total internal reflection and dispersion

C.S.V. / November / 2009 / 1094

(C) Reflection only (D) Diffraction and dispersion 23. The displacement of particles in a string in x -direction is represented by y. Among the following expression for y , that describing wave motion is— (A) cos kx sin ωt (B) k 2x 2 – ω2t 2 (C) cos (k 2x 2 – ω2t 2) (D) cos2 (kx + ωt) 24. Which of the following phenomena is not explained by Huygen’s construction of wavefront ? (A) Reflection (B) Origin of spectra (C) Refraction (D) Diffraction 25. In a good tuning fork— (A) Only the first overtone along with the fundamental mode is excited (B) Only the fundamental mode is excited (C) Only the first overtone is excited (D) Only the first two overtone are excited 26. A luminous object is placed at a distance of 30 cm from a convex lens of focal length 20 cm. On the other side of the lens, at what distance from the lens a convex mirror of radius of curvature 10 cm should be placed in order to have an upright image of the object coincident with it ? (A) 12 cm (B) 30 cm (C) 50 cm (D) 60 cm 27. For a converging lens forming a real image, the object distance may be stated as p = a + f and the image distance as q = f + b. The focal length of lens will be— (A) (C)

a/b (a + b)

(B)

ab

(D)

a–b

29. The intensity ratio of two coherent sources of light is P. They are interfering in some region and produce interference pattern. Then the fringe visibility is— 1+P 2 P P (C) 1+P

(A)

(B)

2 P 1+P

(D)

2P 1+P

30. Work done in carrying a charge q1 once round a circle of radius R with a charge q2 at the centre is— q1 q2 (A) (B) Zero 4πε0R2 q1 q2 (D) Infinite (C) 4πε0R 31. In the figure, the image is formed ahead of retina, the eye has—

(A) Myopia (B) Hypermetropia (C) Astigmatism (D) Colour Blindness 32. The energy and capacity of a charged parallel plate condenser (PPC) are E and C respectively. Now a dielectric slab of εr = 6 is inserted in it. Assuming that the charge on the plate remains constant, the energy and capacity become— (A) 6E, 6C

(B) E, C

(C) E/6, 6C

(D) E, 6C

→ 33. An electric dipole of moment P is placed at the origin along the x- axis. The electric field at a point P, whose position vector makes an angle θ with the xaxis, will make an angle with xaxis equal to— 1

(where tan α = 2 tan θ)

28. Eight equal charged drops are combined to form a big drop. If the potential on each drop is 10 V, then the potential of big drop will be— (A) 40 V (C) 30 V

(B) 10 V

(A) α

(B) θ

(D) 20 V

(C) θ + α

(D) 2θ + α

34. In an A. C. circuit the value of current is π i = 5 sin 100 t – 2 and A. C. potential V = 200 sin (100 t ) what will be the power consumption in the circuit ? (A) 0 W (B) 1000 W (C) 40 W (D) 0·025 W

(

)

35. In an A.C. sub circuit the resistance R = 0·2 Ω. At a certain instant (VA – VB ) = 0·5 volt, I = 0·5 amp. and (ΔI/Δ t ) = 8 A/s. Find the inductance of the coil— L A

(A) 0·01 H (C) 0·05 H

R

B

(B) 0·02 H (D) 0·5 H

36. A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes— (A) Parallel to the magnetic field (B) Perpendicular to the magnetic field (C) Inclined at 45° to the magnetic field (D) Inclined at an arbitrary angle to the magnetic field 37. A vertical ring of radius r and resistance R falls vertically. It is in contact with two vertical rails which are joined at the top. The rails are without friction and resistance. There P Q × × is a horizontal uniform magnetic field of magnitude × × B perpendicular to the plane of the ring and the × × rails. When the speed of the ring is v the current in the section PQ is— (A) Zero (B) 2 B rv/R 4 B rv Brv (C) (D) 8 R R 38. A

bar

magnet of magnetic → moment M is placed in a field of

→

strength B , the torque acting on it is—

→ →

(A) M . B

→

→

(C) M × B

→ →

(B) – M . B

→

→

(D) B × M

C.S.V. / November / 2009 / 1095

→ 39. A dipole of moment P is placed → in a uniform electric field E . The → force on the dipole is F and the → torque is τ — → → → → (A) F = 0, τ = P × E

years, if the number of radioactive nuclei changes by a factor f, then—

( ) ( )

(A) f = e × p (B) f =

–

1 2

1 2

(C) f = loge 2

→ → → (B) F = | P | E

(D) f =

→ → → → (C) F = 0, | τ | = P . E

1 loge 2

46. An element A breaks into C by a two step process A → B + 2He 4 and B → C + 2e

(D) None of these

Then—

40. Faraday’s laws of electromagnetic induction is related to— (A) Law of energy (B) Law of charge

conservation

of

(A) A and C are isotopes (B) A and C are isobars (C) A and B are isotopes (D) A and B are isobars

conservation of

(C) Newton’s third law of motion (D) Conservation momentum

of angular

41. The plate current in a triode will become zero if the amplitude of the negative voltage applied on the grid is— (A) Vp /μ

(B) μVp

(C) Vp × μ

(D) Vp / rp

47. The acceleration of the electron in first Bohr orbit in hydrogen atom is— (where a0 = Bohr’s radius; l = h ) m = mass of electron; h– 2π – – lh2 lh2m 2 (B) (A) m 2a 30 a 30 2 –l – a 3 h 0 (C) (D) None of these m2 48. Which of the following gates will have the output 1 ?

42. The total K.E. of an electron is 3·555 MeV, then its K.E., is— (A) 3·545 MeV (B) 3·045 MeV (C) 3·5 MeV (D) None of these 43. Which one among the following exhibits particle picture of light ? (A) Interference (B) Refraction (C) Photoelectric effect (D) Diffraction 44. Atomic weight of boron is 10·81 and it has two isotopes 5B10 and 11 10 to B11 in 5B . The ratio of 5B 5 nature would be— (A) 19 : 81

(B) 81 : 19

(C) 15 : 16

(D) 10 : 11

45. The half-life of a radioactive sample is 1600 years. After 800

49. A point P lies on the axis of a flat coil carrying a current. The magnetic moment of the coil is μ. The distance of P from the coil is d, which is large compared to the radius of the coil. The magnetic field at P has magnitude— (A)

μ0 2π

(C)

μ0 2π

( ) ( ) μ d3

(B)

μ0 4π

μ d2

(D)

μ0 4π

( ) ( ) μ d3 μ d2

50. Which one of the following is the weakest kind of bonding in solids ? (A) Metallic (B) Covalent (C) Ionic

(D) Vander Waals

Introduction ● The French physicist, Henri Becquerel, in 1896

observed that a photographic plate, wrapped in a thick black paper, which had been placed, by chance, in the same drawer which contained certain uranium salt, had become affected or fogged.

behave in three different ways and consequently, three kinds of radiations are postulated. (i)

These rays are made up of nuclei of helium atoms, i.e ., each α-particle is doubly positive charged helium atom.

● Becquerel concluded that uranium salts emitted

certain rays which had penetrating properties similar to X-rays. He called these rays as radioactive rays and property of giving out these rays as radioactivity. ● A year later, Marie Curie and her husband, Pierre

Curie, discovered the element, Polonium, which was many times more radioactive than uranium. Two years later they discovered another element, Radium which was found to be three million times more radioactive than uranium. ● In fact, it is now known that elements having atomic

number greater than 83 are all radioactive. The nuclei

α) rays—These rays are deflected towards Alpha (α the negative plate in the electrical field. Hence they carry positive charge.

β) rays—These rays are deflected towards (ii) Beta (β the positive plate in the electrical field. Hence these are supposed to be made up of negatively charged particles, known as β-particles. (iii) Gamma (γγ ) rays—These rays are not deflected in electrical or magnetic field. It means they are not made up of charged particles. They are electromagnetic rays of high frequency. ● The comparative properties of α, β and γ rays are

summarised as—

α-Rays

Properties

β-Rays

γ -Rays

1. Nature

Stream of doubly + vely charged helium atoms

Stream of fast moving elec- Photons, electromagnetic radiatrons ( –1 e0) tor of short wavelengths

2. Charge

+2e (+ 3·2 × 10–19 C)

Zero

3. Mass

4 times that of proton

–e(–1·6 × 10–19 C) 1 1836 times that of a proton

4. Velocity

About 107 m/sec

Upto 97% speed of light

Equal to speed of light

5. Relative ionizing power

Maximum (10‚000)

Less than α-rays (100)

Less than β-rays (1)

6. Relative penetrating power Minimum (1)

More than α-rays (100)

More than β-rays (10‚000)

7. Fluorescence

Can cause

Can cause

Can cause

of these elements are unstable and decompose or disintegrate spontaneously with the emission of radiations. ● The radioactivity is a nuclear phenomenon and is not

affected by variation in temperature, pressure, electrical and the magnetic field. It was also found that radioactivity of a material was completely unaffected by physical or chemical changes in the material.

Characteristics of Radioactive Rays ● Photographic plates are fogged by these radiations. ● They ionize the gases, if passed through gases. ● They penetrate through thin metallic plates. ● They cause fluorescence on zinc sulphide, barium

platinocyanide, cadmium tungstate etc. ● Types of radioactive rays—When radioactive rays

are passed through electrical or magnetic fields, they

C.S.V. / November / 2009 / 1099

Zero (rest mass)

Theory of Radioactive Disintegration This theory was propounded by Rutherford and Soddy. According to this theory— ● Radioactive disintegration is a nuclear phenomenon. ● Radioactive substances emit α, β and γ rays and they

undergo disintegration accordingly to produce new substances. ● α -ray decay change—It can be represented by

following mechanism— ZX

–α A ⎯→

A–4 Z – 2Y

or ZXA ⎯→ 2He 4 + Z – 2YA – 4

● Following are consequences of emission of one α-

particle from an unstable nucleus— (a) Mass number decreases by four units or atomic weight is reduced by four units. In other words the number of neutrons is decreased by two units and the number of protons is also decreased by two units.

(b) Atomic number decreases by two units. In other words the nuclear charge is decreased by two units. (c) The daughter element occupies its position in periodic table two groups left to the position of parent element. For example— 88

Ra 226

⎯→ 86

Parent element II group

Rn 222

+

4 2He

Daughter element Zero group

● Heavy nuclides having mass number greater than 210

decay by α-emission in order to achieve stability. ● β-ray decay change—It is represented by following

mechanism— A

ZX

–β

⎯→

A Z + 1Y

or ZXA ⎯→ Z + 1YA + –1e0

● Following are the consequences of emission of a β-

particle from an unstable nucleus— (a) Mass number or atomic mass remains unchanged. (b) Number of neutrons is reduced by one unit. (c) Atomic number increases by one unit or the nuclear charge increases by one unit or the number of protons increases by one unit. (d) The nuclei having excess of neutrons, i. e., having high n/p ratio tend to decrease this ratio or tend to achieve stability by emitting β-particles.

Key Points ●

● ●

● ●

●

IVA group

–β+

⎯→ Z – 1YA or Z XA ⎯→ Z – 1YA + +1 e0 The nuclei having excess protons are found to decay by positron emission. This tends to increase n/p ratio. This kind of disintegration is mostly found in artificial radioactive isotopes. Any radioactive decay with one α followed by two βparticles emission produces isotopes. In general if the ratio of α and β emissions is 1 : 2, then the obtained element will be isotope of parent element. After a β-decay the daughter nuclide is an isobar of parent nuclide. Emission of γ-rays is a secondary effect of radioactive change and after γ-decay the daughter nuclide is isomer of parent nuclide. The group displacement law was postulated by SoddyFajan. β-particle is not present in the nucleus, even then it is emitted from the nucleus. A neutron first breaks down to a proton and an electron as 1 1 0 0n ⎯→ 1p + –1e The proton is retained by nucleus while electron is emitted as β-particle. The determination of number of α- and β-particles in a nuclear reaction : b d a A ⎯→ c B (i) First write the reaction as : b d 4 0 aA ⎯→ c B + x (2α ) + y (–1 β ) (ii) Then b = d + 4x + 0 y and a = c+2x–y Solve for x and y.

C.S.V. / November / 2009 / 1100

VA group

● γ -ray decay change—They do not bring about any

change in the number of nucleons as they are chargeless and massless. Therefore, mass number and atomic number remain unchanged. ● Rate law of radioactive disintegration—Radioactive disintegration is independent of temperature, pressure or state of the element and depends upon the concentration of the radioactive element. The rate of disintegration (decay) follows the first order kinetics. ● The rate of radioactive disintegration is directly proportional to the number of atoms of radioactive element. d Nt d Nt – ∝ Nt or – = k Nt dt dt N0 d Nt – = kt or –kt = ln Nt Nt Nt = e–kt or Nt = N0 e–kt N0 ● In the above expressions :

(i)

Some nuclides also disintegrate by emission of positron (β+ or +1e0) A ZX

●

The daughter element occupies its position in periodic table one group right to the position of parent element. For example— 14 ⎯→ 14 + 0 6C 7N –1e

(ii)

dN is rate of disintegration and negative sign, dt means the rate is decreasing with increase in time. k is disintegration constant and its value is not affected by initial concentration, temperature, pressure and chemical state.

The disintegration constant (k ) is that fraction of total number of atoms which decays in one second. (iii) This expression shows that radioactive decay is exponential form. Here Nt is the number of atoms left after time ‘ t ’ and N0 is initial number of atoms of radioactive substance. dN If N = 1, then – = k dt dN If dt = 1, then = k N The decay constant ‘k ’ is given by expression.

k=

N0 2·303 2·303 a log = log t Nt t (a – x )

where ‘a ’ = Initial amount and x = Amount that has disintegrated in time ‘t ’. The unit of k is time–1. ● Following is the graphical representation of above exponential equation— Exponential decay N

Nt = N0e–kt

t

t 1/2 = 0·693 × Average-life (τ)

● Half-life—The total decay period of a radioactive

element is infinity. So it is meaningless to use total decay period. The half-life is a characteristic of a radioactive substance. Thus half-life is the time in which half of the original amount of the radioactive substance disintegrates. 0·693 Half-life period (t1/2) = ‚ k = Decay constant k

t 1/2 = 0·693 × τ

∴ τ =

t 1/2 = 1·44 × t1/2 0·693 τ > t 1/2

i.e .,

Units of Radioactivity ● The S.I. unit of radioactivity is Becquerel (Bq) which

corresponds to one disintegration per second (dps).

Key Points ●

●

●

Since ‘k ’ is the decay constant, the half-life period of a particular radioactive substance is independent of the initial concentration of the substance. Half-life period is the measure of radioactivity of a substance. Shorter the half-life period of an element, greater is the number of disintegrating atoms per unit time and greater is its radioactivity. The radioactivity left in an element can be correlated with 1 half-life of radioactive substance. For example if 16 of original activity is left then 1 1 n 16 = 2 (Here n = total number of half-lives) 1 1 4 16 = 2 Therefore, the number of half-lives elapsed during the period of disintegration is 4. Suppose half-life of substance under question is 5 years then total time 1 required to reduce to 16 th will be = 5 × 4 = 20 years. Following formulae be remembered. Total time (T) = n × t1/2 T t1/2 = n T n = t 1/2

() ()

●

1 n 1 Nt = N0 2 = N 0 2 Amount left after time ‘t ’ and

()

●

●

()

T t 1/2

Here Nt = N0 = Original amount of radioactive substance. The radioisotopes having low half-life period, have more probability of finding in nature. The intensity of radiations is also low in these cases. There is no difference in radioactive property of Na and Na+ as radioactivity depends upon the composition of nucleus.

● Average-life

period—The average-life period is inverse of disintegration constant (k ). 1 Average-life period (τ) = or k–1 k 0·693 We know that k = t 1/2 Hence average-life (τ) =

t 1/2 1 = = 1·44 × t1/2 0·693 0·693 t 1/2

● Remember following relations :

t 1/2 =

0·693 k

= 0·693 ×

C.S.V. / November / 2009 / 1101

● Curie (ci) is an earlier unit of radioactivity which is

defined as the amount of any radioactive substance which gives 3·7 × 101 0 disintegrations per second. This much activity is shown by one gram of radium. 1 ci = 3·7 × 1010 dps = 3·7 × 1010 Bq ● Milli Curie (mci)

= 1 × 10–3 ci = 3·7 × 107 dps = 3·7 × 107 Bq

μci) = 1 × 10–6 ci ● Micro Curie (μ = 3·7 × 104 dps = 3·7 × 104 Bq ● 1ci =

103

m ci =

μci

106

● Rutherford (rd) is yet another unit of radioactivity.

This is defined as the amount of radioactive substance which undergoes 10 6 dps i.e., 1 rd = 106 dps = 106 Bq

Radioactive Equilibrium ● Suppose a radioactive element disintegrates to give B

which in turn disintegrates to still another element C. A ⎯→ B ⎯→ C A stage is reached when the rate of disintegration of A to B is equal to that of B to C, with the result the amount of B remains constant. Under these conditions B is said to be in equilibrium with A. ● It is important to note that the term equilibrium is used

for reversible reactions but radioactive reactions are irreversible, hence it is preferred to say that B is in steady state rather than in equilibrium state. It is, therefore, known as secular equilibrium. ● Thus at steady state

Rate of disintegration of A = Rate of disintegration of B d NA d NB = dt dt ∴ dN k A × NA = k B × N B = kN dt

[

∴

NA k B τ A = = NB k A τ B

● In terms of half-life period

1 k

(t1/2)A NA = NB (t1/2)B

]

⎡⎢ ∴ τ = 1 ⎤⎥ ⎢ k ⎥ ⎢⎣ τ = Average life⎥⎦

● At the steady state, the amounts of different elements

● 4 n + 1 series starts from plutonium (Pu), but com-

present in the reaction series are directly proportional to their half-life and also average life or inversely proportional to disintegration constant.

monly known as neptunium series because neptunium (Np) is longest lived member of the series. The same principle of nomenclature is applied to 4 n + 3 series which actually starts from uranium (92U235) but known as actinium series.

Nuclear Isomers ● Nuclear isomers (isomeric nuclei) are the atoms of

same element having same atomic number and the same mass number but different radioactive properties. ● Nuclear isomers have same number of electrons,

protons and neutrons. The examples of nuclear isomers are uranium-X (half-life 1·4 min.) and uraniumZ (half-life 6·7 hours). ● The reason for nuclear isomerism is the different

energy states of the two isomeric nuclei. One may be in the ground state whereas the other in an excited state. The nucleus under excited state will evidently have different half-life. ● Now-a-days as many as more than 70 pairs of nuclear

isomers have been found. Some examples are— 69 Zn

69 Zn

( t1/2 = 13·8 hour)

( t1/2 = 57 min.)

80 Br

80 Br

( t1/2 = 4·4 hour)

( t1/2 = 18 min.)

Disintegration Series ● The phenomenon of natural radioactivity continues till

stable nucleus is formed. All the nuclei from the initial element to final stable element form a series, known as disintegration series. ● The mass numbers of all the elements in a series will fit into one of the following formulae— 4n, 4n + 1, 4n + 2 and 4 n + 3 Hence there can be maximum four disintegration series. ● The number indicates that in a particular series the

mass numbers of all the elements are either divisible by four (4n series) or divisible by four with remainder of 1(4n + 1), 2(4n + 2) or 3(4n + 3). ‘n ’ being an integer. Name of Series

4n (Thorium series) 4n + 1 (Neptunium series)

4n + 2 (Uranium series) 4n + 3 (Actinium series)

Starting Nuclide

Final Nuclide

232 90 Th ( n = 58)

208 82 Pb ( n = 52)

( 94 Pu241) 237 93 Np ( n = 60) 238 92 U ( n = 59)

209 83 Bi ( n = 52)

8

5

206

8

6

7

4

( 92 U235) 227 89 Ac ( n = 58)

82 Pb

Particles Lost α 6

β 4

( n = 51) 207

82 Pb ( n = 51)

● 4n + 1 series is an artificial series while rest three are

natural series. The end product in 4 n + 1 series is bismuth (83Bi209 ) while in rest three a stable isotope of lead is the end product.

C.S.V. / November / 2009 / 1102

Significance of Radioactivity ● Carbon-14 dating (Archaeological dating)—Carbon-

14 is an unstable isotope of carbon. Its half-life is about 5700 years. It is a valuable isotope in estimating the date of origin of old samples of wood. The technique was developed by Willard Libby. ● Example—A piece of wood found in an excavation has 25·6% as much as C14 as ordinary wood today has. When did the piece get burried ? (t1/2 of C 14 = 5760 years). Solution : N0 = 100, Nt = 25·6 and t1/2 of C14 = 5760 years 0·693 k= 5760 2·303 100 Also k = log 25·6 t 0·693 2·303 100 ∴ = log 5760 25·6 t 2·303 × 5760 100 ∴ t = log 0·693 25·6 = 11330 years. ● Uranium-Lead dating (Geological dating)—This is a

valuable method for estimating the ages of old geological formations, rocks and minerals. ● The age of the earth has been determined by uraniumlead dating technique. The samples of uranium ore are found to contain 82Pb206 as result of series of α- and β-decays. It is assumed that the ore sample contained no lead at the moment of its formation and if none of the lead formed from U238 is lost, then the measurement of Pb206/U238 ratio will give the value of time of the mineral. ● Example—92U238 by successive radioactive decays changes to 82Pb206 . A sample of uranium ore was analysed and found to contain 1·0 gm of uranium and 0·1 gm of 82Pb206 . Find out the age of ore. (t1/2 of U238 = 4·5 × 109 years). Solution : The amount of uranium decayed ∴ 206 gm Pb is obtained from 238 gm of uranium-238 ∴ 0·1 gm Pb will be obtained from 238 × 0·1 206 = 0·1155 gm of U 238 Hence initial amount of uranium-238 = 1·0 + 0·1155 gm of U238 0·693 We know that, k = t 1/2 0·693 = 4·5 × 109 years =

= 0·154 × 10–9 year–1

N0 2·303 log k Nt 2·303 1·1155 = log 1 0·154 × 10–9

t =

● Since α-particles, protons and deuterons are positively

Nuclear Reactions

charged, they are repelled by the nucleus and hence are not good projectiles. On other hand, neutrons which do not carry any charge are the best projectiles. ● The positively charged particles are made more effective if they are imparted with high velocity. The cyclotron is most important device for accelerating these particle. A more recent accelerating instrument called synchrotron or bevatron makes use of the induced current or force exerted on electron in a changing magnetic field.

● The nuclear reaction is one which proceeds with a

● Artificial transmutation of elements—This type of

t = 7·099 × 108 years ● Tritium dating—Tritium (1H3) is an unstable isotope

of hydrogen with half-life 12·5 years. It is used for shorter-range-dating. Such as finding the age of water and wine.

change in the composition of nucleus resulting in the formation of an atom of a new element. The new element is known as daughter element. ● In contrasts a chemical reaction, where only rear-

rangement of electrons takes place, the nuclear reactions involve the redistribution of protons and neutrons (nucleons) present in the nucleus of parent atom. However, the total number of protons and neutrons in the reactant and product side remains the same. Consequently, the sum of the mass number and atomic number on two sides of the reaction must be equal (mass is conserved + charge is conserved). ● A nuclear reaction may be expressed as— 7N

14

+

2He

4

17 8O

⎯→

α-projectile

or

+

1H

Artificial radioactivity was discovered by Frederic Joliot and Irene Curie (Son-in-law and daughter of Marie Curie) in 1934. They were awarded Nobel prize for discovery of artificial radioactivity. ● The natural radioactivity is found in isotopes of some

heavy elements such as uranium, thorium, radium etc. while artificial radioactivity is the process whereby a stable nucleus is made radioactive by bombarding it with some high speed particles like α, protons, neutrons or deuterons. + 2 He 4 ⎯→ 14Si27 + 0n 1 (Radioactive)

→ 13 Al

5

B10

+

14

+ 2 He 4 ⎯→ 8O17 + 1H 1 (Non-radioactive)

This transformation is expressed as 14 17 7 N (α, p) 8O . ● The transmutation nuclear reactions are classified

according to the nature of bombarding particle (projectile) and ejected particle. 9 4 12 1 4 Be + 2 He ⎯→ 6 C + 0n 39 1 36 + He 4 19 K + 1 H ⎯→ 18 Ar 2

( p, α)

Proton (ejected particle)

● Artificial radioactivity or induced-radioactivity—

24

7N

(α, n) 1

14 17 or [(α, p) type of nuclear reaction] 7 N (α, p)8O

12 Mg

nuclear reaction was discovered by Rutherford in 1919. The first artificial transmutation of nitrogen was brought about by Rutherford (1919) when nitrogen nucleus was bombarded with fast α-particles.

27

+ +1e 0

(Positron)

4 13 2 He ⎯→ 7 N

+

1 0n

(Radioactive)

→ 6C

13

+ +1e 0

● The radioisotope 6C14 is produced by bombardment of

nitrogen of atmosphere with neutrons (from cosmic rays). 14 1 14 1 7 N + 0 n ⎯→ 6 C + 1H (Radioactive)

→ 7N

14

+ –1e 0 (β-particle)

C.S.V. / November / 2009 / 1103

11 Na

23

+ 1 H2 ⎯→ 10 Ne 21 + 2He 4 ( d, α)

12 1 13 0 6 C + 1 H ⎯→ 7 N + 0γ

( p, γ)

● Nuclear fission—This was discovered by Otto Hahn

and Strassmann. In this process, when U235 is hit by slow moving neutrons, it splits up into two fragments of nearly equal mass with the release of large amount of energy. The cause of release of energy is the loss of mass and conversion of lost mass into energy according to Einstein’s mass-energy relation, E = mc 2. Two or three neutrons are produced by fission of each nucleus. ● The uncontrolled nuclear fission leads to the pro-

duction of atom bomb whereas controlled nuclear fission leads to the construction of nuclear reactor. ● Some of the released neutrons do not participate in

the chain reaction as these escape from the surface without being utilised. Thus in order to sustain the chain reaction, a sufficient amount of fissionable material (U235) is needed. Thus, the minimum mass of fissionable material is required to support the selfsustaining chain reaction under given set of conditions. This minimum mass of fissionable material is known as critical mass. The critical mass of U235 is 1 kg. ● Atom-bomb—The principle of atom bomb is based on

the process of nuclear chain reaction in which no secondary neutrons escape of fissionable material without striking. The size of the fissionable material (U235) or Pu239 should not be less than the critical mass.

● Nuclear-reactor—It is the device in which nuclear

●

●

●

●

fission is carried out in a controlled manner and the energy released is utilised for peaceful purposes. Its main components are— (a) Fuel rods : (U235 or Pu239 ) (b) Control rods : These absorb excess electrons. These are made of boron or cadmium steel. (c) Moderator : It slows down the fast moving neutrons. Heavy water or graphite is used as moderator. (d) Coolant and (e) Protective shield. Research-reactors of India—Five research reactors are— Purnima, Zerlina, Dhruva, Cirus, Apsara Nuclear power stations are— Tarapur Atomic Power Station, Maharashtra Rajasthan Atomic Power Station, Kota Narora Atomic Power Station, U.P. Indira Gandhi Centre of Research at Kalpakkam, Tamilnadu. Kakrapar, Gujarat Rawatbhatta, Rajasthan. Breeder-reactor—Natural uranium contains only 0·7% of fissionable U235 and rest being nonfissionable U238 . In addition to U235 , Plutonium-239 and Uranium-233 have also been found fissionable. They are produced by bombardment of naturally occurring and abundantly available U238 and Th 232 with neutrons. Such reactors in which the neutrons produced from U235 are partly used for carrying the fission of U 235 and partly to produce Pu239 are named as breeder reactors.

● For producing

Pu239

Points to Remember ●

●

●

● ●

238 ⎯→ 234 + He4 92 U 90 Th 2 No. of p = 92 n–p = 238 : No. of n = 238 – 92 = 146 146 – 92 = 54 92 U 234 90 Th

●

●

●

● ●

●

● Nuclear fusion—It is the process in which lighter

nuclei are fused together to form heavier nuclei. The heavier nuclei have slightly less mass than the total mass of fused nuclei. This decrease in mass is converted to energy. ● Since fusion reactions always occur at a very high

●

●

temperature, therefore, they are known as thermo nuclear reactions. Hydrogen bomb is based on this principle. Tritium (1

2 3 4 1 H + 1 H ⎯→ 2 He H3) is generated as : 3 Li

6

+ 0 n 1 ⎯→

1H

3+

+ 0n 1 + 17·6 MeV 2He

4

● Source of energy at sun and stars is also fusion of

hydrogen nuclei.

C.S.V. / November / 2009 / 1104

:

} No. of p = 90 n–p = No. of n = 234 – 90 = 144} 144 – 90 = 54

The first nuclear reactor was assembled by Fermi in 1942 and in India the first nuclear reactor was put into operation in 1952 in Trombay. Beryllium has been found to be the best moderator as it occupies small space and has low absorption crosssection. Electron capture or k-capture involves the capture of electrons by nucleus of one of the atoms during the nuclear reactions. This is followed by conversion of a proton to a neutron and emission of neutrino. The orbital electron that is usually captured is from k-shell. For example— 79 Au

U238

from excess neutrons obtained in fission chain reaction of U235 are utilised. Thus the materials which undergoes fission are called fissile such as U235 , Pu 239, U 233 etc. On the otherhand the materials which do not undergo fission easily but may be fissionable by bombarding with neutrons are called fertile such as U238. Thus in breeder reactors fertile fuel is made fissile fuel.

Radioisotopes—When atoms of normally stable elements are made radioactive by artificial nuclear transformation, they are called radioisotopes. They are used in almost every field of science and industries. Cobalt-60 is used in the treatment of cancerous growth and Iodine-131 is used as a medicine to determine thyroid disorders. Na-24 is used for diagnosis of restricted circulation of blood and P32 is used for locating cancer and curing blood diseases. A number of devices, such as ionization chamber, Geiger-Muller counter, Scintillation counter, Wilson cloud chamber etc. have been used to detect and measure radioactivity. The radioactivity or activity of a radioactive nucleus is inversely proportional to half-life or average-life period. Isodiapheres—Nuclei having the same difference of neutrons and protons or same isotopic number, are known as isodiapheres. For example, the nucleide and its decay product after an α-emission are isodiapheres.

●

195 EC

⎯→ 78Pt 195 + 0n 0 (Neutrino)

W.F. Libby was awarded Nobel prize for discovering the technique of carbon-14 dating or radiocarbon dating. The machines like cyclotron and linear accelerator etc. have been constructed for accelerating α-particles or protons, so that they could be given momenta large enough to counteract coulombic repulsion of atomic nuclei. Cyclotron was designed by E. O. Lawrence. Spallation nuclear reactions are those nuclear reactions in which high speed projectile chops off a fragment of the nucleus leaving behind a smaller nucleus. 63 4 37 + 14 H1 + 16 n 1 29 Cu + 2 He ⎯→ 17Cl 1 0 The half-life of a radioactive substance is independent of physical or chemical state of radioactive substance. For example t 1/2 of C1 4 is same whether it is on CO2 or cellulose or solid coal. In the process of nuclear fission of U235 , 3·5 × 10 –11J of energy is produced per U235 nucleus. Or The fission of an atom of U235 releases 211·5 MeV of energy. The energy released from 1 mole of U 235 = 20·41 × 109 kJ 20·41 × 109 and from 1 gm of U235 = 235 = 8·68 × 107 kJ

OBJECTIVE QUESTIONS 1. Which of the following does not contain material particles ? (A) Alpha rays (B) Beta rays (C) Gamma rays (D) Canal rays 2. The phenomenon of radioactivity is due to— (A) Stable electronic configuration (B) Stable nucleus (C) Unstable electronic configuration (D) Unstable nucleus 3. In the nuclear reaction, 7N

14

+ 1H1 → 8O15 + X

The ‘X’ is— (A) 0n 1 (C)

+1

(B)

–1

e0

(D) γ

e0

4. From the nuclear reaction, 84 Po

210

→ 82Pb206 + 2He 4

deduce the group of polonium in periodic table (Pb belongs to group 14)— (A) 2 (B) 14 (C) 6 (D) 16 5. A radioactive isotope decays at such a rate that after 96 min. 1 only th of the original amount 8 remains. The value of t 1/2 of this nuclide is— (A) 16·0 min. (B) 24·0 min. (C) 32·0 min. (D) 48·0 min. 6. In the nuclear reaction, 92 U

238

→ 82Pb206

the number of α and β particles emitted is— (A) 6α, 4β

(B) 4α, 6β

(C) 8α, 6β

(D) 7α, 5β

7. The reaction, 1D

2

+ 1T3 → 2He 4 + 0n 1

is an example of— (A) (B) (C) (D)

Nuclear fission Nuclear fusion Artificial radioactivity Radioactive disintegration

8. 1 mole of an alpha emitting nuclide ZXA (half-life 10 min.) was kept in a sealed container.

C.S.V. / November / 2009 / 1105

4·52 × 10 23 helium atoms will accumulate in the container in— (A) 4·52 min. (B) 940 min. (C) 10·00 min. (D) 20·0 min. 9. The equipment used to carry out nuclear reactions in a controlled manner is called— (A) Breeder reactor (B) Nuclear reactor (C) Thermo nuclear fission (D) Cyclotron 10. Which of the following substances is used as neutron absorber in the nuclear reactor ? (A) Heavy water (B) Deuterium (C) Cadmium (D) Graphite 11. The change, 30 15 P

→ 14Si30

requires the emission of— (A) α-particle (B) β-particle (C) Neutron

(D) Positron

12. An element X loses one α- and two β-particles in three successive stages. The resulting element will be— (A) An isobar of X (B) An isotope of X (C) An isotone of X (D) X itself 13. The half-life of a radioactive substance is 60 min. During 3 hours, the fraction of total number of atoms that have decayed would be— (A) 12·5% (B) 87·5% (C) 8·5% (D) 25% 14. Which of the following is used as moderator ? (A) Heavy water (B) Graphite (C) Boron (D) Both (A) and (B) 15. After two hours, a radioactive 1 substance remains th of origi16 nal amount. The half-life in minute is— (A) 15 min. (B) 30 min. (C) 60 min. (D) 120 min.

16. For the nuclear fission 235 + n 1 → Fission products 92 U 0 + neutrons + 19·293 × 109 kJ/ mole, the energy released when 1 gm of U235 undergoes fission is—

(A) 12·75 × 108 kJ (B) 18·60 × 109 kJ (C) 8·20 × 107 kJ (D) 6·55 × 106 kJ 17. One gm of a radioactive element reduces to 125 mg in 24 hours. The half-life of the isotope is— (A) 4 hours

(B) 8 hours

(C) 6 hours

(D) 12 hours

18. A radioactive element has halflife of 60 minutes. The amount left after 3 hours is— (A) 25% of original amount (B) 50% of original amount (C) 12·5% of original amount (D) 17·5% of original amount 19. Parent atom will be an isobar of daughter nuclide when parent nuclide emits— (A) One α-particle (B) One β-particle (C) One α-particle and one βparticle (D) 2α-particles and 1 β-particle 20. The activity of a radioactive nuclide, X 100 is 6·023 curie. If the disintegration constant is 3·7 × 104 sec–1, the mass of X in gm will be— (A) 5 × 10–2 gm (B) 10–6 gm (C) 10–15 gm (D) 10–3 gm 21. The half-life of a radioactive isotope is 3 hours. If the initial mass of the isotope was 300 gm, the mass which remained undecayed in 18 hours will be— (A) 2·34 gm

(B) 1·17 gm

(C) 4·68 gm

(D) 9·36 gm

22. A radioactive element X emits 3α, 1β and 1γ particles and forms 225 . The element X is— 76 Y

(A) (C)

236 81 X 237 80 X

(B) (D)

237 81 X 236 80 X

23. Which element is the end product of every natural radioactive series ? (A) Bismuth (B) Lead (C) Both (A) and (B) (D) None of these 24. Which one of the following notations does not represent the products correctly ? (A) 14Si28 (d, n ) 15P29 (B)

14

(n, p ) 6C14

B10

N 13

7N

(α, n),

(C)

5

(D)

96 Cm

242

(α, 2n)

82 Pb

206

(D)

243

97 Bk

83 Bi

209

26. The number of neutrons in a parent nucleus, X which yields 14 after two successive β7N emissions are— (A) 6

(B) 7

(C) 8

(D) 9

27. Which is not emitted by a radioactive substance ? (A) α-rays

(B) β-rays

(C) Positron

(D) Proton

28. The neutrino can be detected during the emission of— (A) α-rays

(B) β-particles

(C) Protons

(D) X-rays

29. In the decay series, –α

–β

–β

A ⎯→ B ⎯→ C ⎯→ D (A) (B) (C) (D)

ANSWERS WITH HINTS

32. The artificial radioactivity was discovered by— (A) Madame and Pierre Curie (B) Irene Curie and F. Joliot (C) Soddy-Rutherford (D) Soddy-Fajan 33. Which is used in dating archaeological objects ? (A) 92U235 (B) 6C14 (C)

92 U

238

(D) 6C12

34. Which of the following is used for estimating the age of rocks ?

25. The end product of (4 n + 2) disintegration series is— (A) 82Pb204 (B) 82Pb208 (C)

(C) Change slightly (D) Remain unchanged

A and B are isobars A and C are isobars A and D are isotopes B and C are isotopes

30. Bismuth is the end product of radioactive disintegration series known as— (A) 4n (B) 4n + 1 (C) 4n + 2 (D) 4n + 3 31. If a radioactive element is kept in a evacuated container its rate of disintegration will— (A) Increase (B) Decrease

C.S.V. / November / 2009 / 1106

(A) C14

(B) U238

(C) P32

(D) Co 60

35. The half-life of a radioactive element is t 1/2 years. The time required for its complete decay is— (A) t 1/2 years (B) 2 × t1/2 years (C)

t 1/2 years 2

(D) ∞

36. Which of the following nuclei is most unstable ? (A) 5B10 (B) 4Be10 14 (C) 7N (D) 8O16 37. Group displacement law was given by— (A) Rutherford (B) Mendeleef (C) Soddy-Fajan (D) None of these 38. By removing a positron from the nucleus of a radioactive element, the atomic number— (A) Increases by one (B) Decreases by one (C) Does not change (D) Decreases by two 39. According to nuclear reaction 4Be

+ 2He 4 → 6C12 + 0n 1

the mass number of Be atom is— (A) 4 (B) 8 (C) 6 (D) 9 40. The symbol x in the following equation is— 11 Na

23

+ 1H1 → 12Mg23 + x

(A) A positron (C) A neutron

(B) A proton (D) A deuteron

(Continued on Page 1133)

13. The oxidation number of Cr in CrO2Cl2 is— (A) + 2 (B) + 3 (C) + 4 (D) + 6

1. Velocities of four molecules in a container are 1, 2, 2, 4 units respectively. What is the value of root mean square velocity ? (A) 2·5 unit

(B) 2·25 unit

(C) 9 unit

(D) Zero

2. Mark the correct statement about given graph—

6. For the given electrolyte Ax By , the degree of dissociation ‘α’ and ‘i ’ are related as— i–1 (A) α = (x + y – 1) (B) i = (1 – α) + x α + y α 1–i (C) α = (1 – x – y ) (D) All of these are correct 7. The solubility of Ca 3(PO4)2 is ‘s ’ moles per litre. The solubility product of it is— (A) 72s 5 (B) 108s 5 (C) 9s 2 (D) 8s 3

(A) X is threshold energy level (B) Y and Z are energy of activation for forward and backward reactions respectively (C) Q is heat of reaction and reaction is exothermic (D) All of the above 3. “Electrons are filled in energy orbitals, in increasing order of energy.” This statement is related to— (A) Planck’s rule (B) Hund’s rule (C) Pauli’s rule (D) Aufbau principle 4. A substance ‘A’ is more soluble in chloroform than water. The amount of ‘A’ left unextracted in water would be maximum when the extraction is done with 100 ml of chloroform in— (A) 4 lots each of 25 ml (B) 1 lot of 100 ml (C) 2 lots each of 50 ml (D) 5 lots each of 20 ml 5. 2 moles of PCl5 is heated in one litre vessel. At equilibrium 0·4 mole Cl2 is formed. The value of equilibrium constant will be— (A) 1 × 10– 3

(B) 1 × 10– 2

(C) 2 × 10– 1

(D) 1 × 10– 1

C.S.V. / November / 2009 / 1107 / 5

8. For the hydrolysis of esters in alkaline medium rate expression is; d [Ester] – = K [Ester] [Alkali] dt In case alkali used is in excess, then the overall order of the reaction is— (A) Zero (B) First (C) Same (D) Third 9. Which of the following is buffer ? (A) NH4OH + CH3COONH4 (B) NaOH + Na2SO4 (C) NaOH + CH3COONa (D) K2SO4 + H2SO4 10. ΔG° for the reaction X + Y Z is – 4·606 k cal. The equilibrium constant for the reaction at 227°C is— (A) 100 (B) 10 (C) 2 (D) 0·01 11. At 25°C the solubility of sparingly soluble binary salt AgCl is 1·25 × 10– 5 mole/litre. What will be the value of Ksp at same temperature ? (A) 1·25 × 10– 5 (B) 1·56 × 10– 10 (C) 1 × 10– 10

14. The internal energy of a gas is— 3 (A) RT (B) 2 RT (C) (D) 2

KT 2 3 KT 2

15. Wohler used this compound with (NH4)2SO4 to get urea— (A) Potassium chloride (B) Potassium cyanide (C) Potassium sulphate (D) Potassium cyanate 16. The edge length of face centred unit cube cell is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is— (A) 288 pm (B) 398 pm (C) 144 pm (D) 618 pm 17. The number of σ and π bonds in CH2 — — CH—CH — — CH2 (A) (B) (C) (D)

8σ and 2π bonds 9σ and 1π bond 9σ and 3π bonds 9σ and 2π bonds

18. Pitchblende is the main source of— (A) U (B) Ce (C) Th (D) Mg 19. Which one of the following is freon ? (A) CCl4 (B) CF4 (C) CCl2F2 (D) CH2Cl2 20. If a person is injured by the shot of a gun and all the pellets could not be removed, it may cause poisoning by— (A) Hg (B) Pb (C) Fe (D) As 21. Which one will form primary alcohol on reaction with CH3Mgl ? (A) Ethyl acetate (B) Ethylene oxide (C) Methyl cyanide (D) Acetone 22. In the following chemical change what are X and Y ? (MnO2) + K2CO3 + Air

(D) 1·44 × 10– 10 12. An aqueous solution of 0·1 M NH4Cl will have a pH closer to— (A) 9·1 (C) 7·1

of one mole

(B) 8·1 (D) 5·1

X + Cl2

⎯→

Δ ⎯→

X (green) Y (pink)

X and Y are— (A) K2Cr2O7, KMnO 7 (B) K2MnO4, KMnO 4

(C) KMnO 4, K2MnO (D) None of these 23. The product CH3—C —CHCl2 || O is formed when HOCl reacts with— — (A) CH3—CH2—C — — CH — CH (B) CH —C — — 3

— (C) CH3—C — — C—CH3 — CH (D) CH — — 24. The reaction, 2 RC ≡ CCu

(CH3COO)2 Cu ⎯⎯⎯⎯⎯→ Pyridine

R—C ≡ C—C ≡ C—R is called— (A) Eglinton’s reaction (B) Glaser reaction (C) Gomberg-Beckmann’s reaction (D) Leuckart reaction 25. 3-hydroxybutanol is formed by the reaction of ethanol and alkali. The reaction is— (A) Aldol condensation (B) Claisen condensation (C) Polymerisation (D) Reimer-Tiemann reaction 26. The correct order of melting and boiling points of the primary (1°), secondary (2°) and tertiary (3°) alkyl halides is— (A) P > S > T (B) T > S > P (C) S > T > P (D) T > P > S 27. Phenol is condensed with phthalic anhydride. The compound formed is— (A) Cyclohexanol (B) Isophthalic acid (C) Phenolphthalein (D) Phthalic acid 28. Rectified spirit contains— (A) 75·0% alcohol (B) 85·5% alcohol (C) 95·6% alcohol (D) 100·0% alcohol 29. Which one of the following is picric acid ? (A) 2, 4, 6-trinitrophenol (B) 2, 4, 6-trinitrobenzoic acid (C) o-nitrophenol (D) 2, 4 dinitrophenol

C.S.V. / November / 2009 / 1108

30. Magenta is— (A) Alkaline phenolphthalein (B) Red litmus (C) P-rosaniline hydrochloride (D) Methyl red 31. When ethyl amine reacts with sodium metal, the gas evolved is— (A) H2 (B) C2H2 (C) N2 (D) NH3 32. The decreasing order of solubility of methanal (a), propanaldehyde (b), benzaldehyde (c) and acetophenone (d) in water is— (A) a > b > c > d (B) d > c > b > a (C) d > a > b > c (D) b > a > c > d 33. 2, 4-dinitrophenyl hydrazine does not react with— (A) CH3—C —CH3 || O (B) CH3—C —OC2H5 || O —O (C) CH3—C — | H (D) HCHO 34. Which acid forms Zwitter ions ? (A) CH3COOH (B) Salicylic acid (C) Phthalic acid (D) Sulphanilic acid 35. The outer electronic configuration of transitional element is— (A) (B) (C) (D)

(n – 1) s 2nd 1 – 5 (n + 1)s 2nd 1 – 5 (n – 1)s 2p 6nd 1 – 10, ns 2 ns 2 (n + 1) d 1 – 10

36. The given structural formula refers to— Cl

Cl (A) BHC (C) DDT

38. In a reaction involving ring substitution of C6H5Y, the major product is meta-isomer. The group Y can be— (A) —NH2 (B) —COOH (C) —CH3

(D) —Cl

39. The element of largest atomic size is— (A) Li

(B) Be

(C) B

(D) C

40. Wax used in gramophone records is— (A) Paraffin wax (B) Bees wax (C) Carnauba wax (D) None of these 41. The correct increasing order of electron affinity is— (A) Be, B, N, C (B) C, N, B, Be (C) Be, N, B, C (D) Be, B, C, N 42. During destructive distillation of bituminous coal at 1000–1400 °C, what per cent of coal-tar is left behind ? (A) 70%

(B) 17%

(C) 8–10%

(D) 4–5%

43. Alkaline dilute KMnO4 solution is called— (A) Etard reagent (B) Benedict reagent (C) Mulliken’s reagent (D) Baeyer’s reagent 44. When cyclohexane is poured on water, it floats because— (A) Cyclohexane is in ‘boat’ form (B) Cyclohexane is in ‘chair’ form (C) Cyclohexane is in ‘crown’ form (D) Cyclohexane is less dense than water 45. Cryolite is the ore of—

C—CCl3 | H (B) DNA (D) RNA

37. Which one of the following is the largest size ? (A) I – (B) I (C) Cl (D) Br

(A) Fe

(B) Cu

(C) Ag

(D) Al

46. Sequence of increasing screening effect is— (A) s < p < d < f (B) s < p > d < f (C) s > p > d > f (D) s > p > d < f

47. The silver is extracted by Parke’s process. The basis of this method is— (A) Silver is immiscible is molten Zn (B) Ag is miscible in NaCN (C) Ag is more miscible in molten Zn than in molten Pb (D) Ag is more miscible in molten Pb in comparison to molten Zn

48. Which species are aromatic in nature ? ⊕ (A) (B)

(D) Fe 2O3 50. Which is used as an anaesthetic ? (A) CH3OCH3

⊕ (C)

(B) Fe(OH)3 (C) Fe 2O3. x H2O

(D)

(B) C2H5OC2H5 (C) CH3OC2H5

49. The formula of rust is— (A) FeO.5 H2O

(D) C3H7OC2H5

ANSWERS WITH HINTS

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17. The work of expansion for a system is 500 cal. The heat given to the system is 80 cal. The change in internal energy in the process will be— 1. Isodiapheres have— (A) The same number of neutrons but different number of protons (B) The same number of protons-electrons (C) The same radius of nucleus sphere (D) The same number of neutrons-protons 2. The quantum number which is responsible for the size of electron cloud is— (A) Spin (B) Azimuthal (C) Principal (D) Magnetic 1

3.

2

3

4

5

— CH—CH3 HC ≡ C—CH — Which carbon atom will show minimum electronegativity ? (A) Fifth (B) Third (C) Second (D) First

4. The maximum values of m for l = 2 is— (A) 0, ± 1 (B) ± 1, ± 2 (C) 0, ± 1, ± 2 (D) 0, ± 2 5. In which of the following compounds chromium shows maximum radius ? (A) K2Cr2O7 (B) CrO2Cl2 (C) Cr2(SO4)3 (D) CrCl2 6. Which one is not Lewis acid ? (A) NH3 (B) Cu 2+ (C) ZnCl2 (D) BF 3 7. At 444°C HI is 20% dissociated, if initially 3 mole of HI are taken, what is the number of moles of HI at equilibrium ? (A) 0·6 (B) 4·8 (C) 2·4 (D) 5·4 8. The pH of an aqueous solution is 4·7. Then H+ concentration will be—

9.

(A) 3 × 10– 6

(B) 3 × 10– 5

(C) 2 ×

(D) 5 ×

10– 5

A– + H2O

10– 5

HA + OH–

represents the hydrolysis reaction. This indicates that the salt is made up of— (A) Strong acid weak base (B) Strong base weak acid

C.S.V. / November / 2009 / 1110

(C) Weak acid weak base (D) Strong acid strong base 10. Strongest base is— (A) RbOH (B) LiOH (C) KOH (D) NaOH 11. A solution of 1M acetic acid contains 0·1 M HCl. What would be the value of CH 3COO– ion concentration. pKa = 5 ? (A) 10– 4 (B) 10– 5 – 6 (C) 10 (D) 10– 3 12. When solid potassium cyanide is added in water then— (A) Electrical conductivity will not change (B) pH is reduced (C) pH is increased (D) pH will remain the same 13. Electrochemical equivalent of an element is— Atomic wt. × Valency (A) 96500 Atomic wt. × 96500 (B) Valency Atomic wt. (C) Valency × 96500 (D)

Valency × 96500 Atomic wt.

14. The IUPAC name of : (CH3)3C—CH = CH2 (A) Vinyl trimethyl methane (B) 2, 2-dimethyl-3-butene (C) Tertiary butyl ethene (D) 3, 3-dimethyl-1-butene 15. Osmotic pressure is 0·0821 atmosphere at temperature 27°C. Find the concentration of solution in mole/litre— (A) 0·033 (B) 0·066 (C) 0·0033 (D) 3 16. Friedel-Crafts reaction is— (A) C2H5OH + Zn → C2H6 + ZnO (B) C6H6 + CH3COCl Anhyd. AlCl 3

⎯⎯⎯→ C6H5COCH3 + HCl (C) C6H5Br + 2 Na + CH 3Br Dry ether

⎯⎯⎯→ C6H5CH3 (D) None of these

(A) 80 cal

(B) 500 cal

(C) – 420 cal

(D) 420 cal

18. Conjugated double bond is present in— (A) Butylene

(B) Butadiene

(C) Isobutylene (D) Propylene 19. Which of the following is less than zero during adsorption ? (A) ΔG

(B) ΔS

(C) ΔH

(D) All of these

20. The active species in reaction between methane and chlorine, to get chloromethane is— (A) Cl2

(B) Cl°

(C) Cl–

(D) Cl+

21. The most convenient method to protect the bottom of ship made of iron is— (A) Coating it with red lead oxide (B) White tin plating (C) Connecting it with Mg block (D) Connecting it with Pb block 22. C2H5OH and H 2SO4 at 170° C from— (A) C2H4

(B) CH3CHO

(C) CH3COOH (D) CH3COCH3 23. The metal with least density is— (A) Be

(B) B

(C) Li

(D) Al

24. If the ozonolysis products of hydrocarbon are CH3—C —CH3 || O and CH 3CHO, then the formula of parent hydrocarbon will be— — C— — CH3 CH3—C — (A) | | CH3 CH3 (B)

CH3—C — — CH—CH3 | CH3

CH3—C — — CH—CH2—CHO | CH3 — (D) CH3—C — — C—CH3

(C)

(A) Paraformaldehyde (B) Metaldehyde (C) Trioxy methylene (D) Paraldehyde 27. Mg ribbon after heating in an atmosphere of nitrogen when dipped in cold water produces a gas which is— (A) NH3 (B) H2 (C) O2

(D) N2

28. If nitrobenzene is reduced in presence of Zn + NH4 Cl then the product will be— (A) p-amino phenol (B) Phenyl hydroxylamine (C) Aniline (D) Azobenzene 29. Microcosmic salt is— (A) Na (NH4) HPO4.4 H 2O (B) Na (NH4).H2O (C) Na (NH3) HPO4.4 H 2O (D) K (NH4) HPO3.2 H 2O 30. Which enzyme is effective in the following reaction ? C6H12O6 → 2 C2H5OH + 2 CO2 (A) Invertase (C) Maltase

(B) Zymase (D) Diastase

31. An excess of AgNO3 is added to 100 ml of 0·01 M solution of dichloro tetraaquo chromium(III) chloride. The number of moles of AgCl precipitated would be— (A) 0·001 (B) 0·002 (C) 0·003 (D) 0·01 32. What is the product of the reaction of ethyl amine and HNO2 ? (A) (B) (C) (D)

Ethene Primary alcohol Ethyl nitrile Nitrosoethane

C.S.V. / November / 2009 / 1111

(C) Benzene (D) Cyclohexane 42. Which of the following is not a member of 3 d transition series ? (A) Fe (B) Co (C) Au (D) Cu

34. Nitrobenzene, on reaction with O fuming nitric acid at 90°C, the 43. CH3—CH2—N O product is— OH NO2 NO2 CH3—CH — —N NO2 O (A) (B) is the example of— (A) Geometrical isomerism NO2 (B) Optical isomerism NO2 NO2 (C) Tautomerism NO2 (D) None of these (C) (D) 44. Malachite is the ore of— (A) Cu (B) Ni NO2 NO2 (C) Au (D) Hg 35. Iodine is liberated from KI 45. The order of reactivity of alcohols solution when treated with— with respect to acid catalysed (A) ZnSO 4 (B) CuSO 4 elimination will be— (a) (CH3)3 C—OH (C) NiSO4 (D) FeSO 4 (b) CH3 CH2 CH2 CH2OH 36. The most electronegative ele(c) CH3 CHCH2CH3 ment in periodic table is— | (A) Fr (B) F OH (C) He (D) N code is— (A) a > b > c (B) a > c > b 37. The first noble gas compound (C) b > a > c (D) b > c > a obtained was— 46. The purest form of iron is— (A) XeF 2 (B) XeF 4 (A) Wrought iron (C) Xe Pt F6 (D) XeOF 4 (B) Cast iron 38. The highest magnetic moment (C) Iron turning will be shown by— (D) Stainless steel (A) Ni (B) Co 47. Suitable reagent to obtain ethan(C) Fe (D) Zn ethiol from bromoethane is—

↓

26. Gaseous formaldehyde at room temperature polymerises to—

33. By the action of conc. H 2SO4 phosphorus changes to— (A) Phosphorus acid (B) Metaphosphoric acid (C) Orthophosphoric acid (D) Pyrophosphoric acid

↓

25. Tempering of steel is done as— (A) Steel is made red hot and dipped in water (B) Steel is made red hot and cooled slowly (C) Steel is heated at much low temperature and cooled slowly (D) Steel is made red hot and dipped in oil

39. For which of the following compound, the Lassaigne’s test for N will fail— (A) NH2NH2.2 HCl (B) NH2.CO.NH2 (C) NH2CO NH NH2HCl (D) C6H5N — — N–C6H5 40. If 19 is the atomic number of an element then this element will be— (A) A metal with + 3 oxidation state (B) A metal with + 1 oxidation state (C) An inert gas (D) A metal with – 3 oxidation state 41. 3° carbon is present in the compound— (A) Cyclopropane (B) Toluene

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

Sodium hydrogen sulphide Sodium sulphide Potassium sulphide Sodium sulphate

48. Blue vitriol is— (A) MnSO 4.7 H 2O (B) CuSO 4.5 H 2O (C) FeSO 4.7 H 2O (D) NiSO4.6 H 2O 49. Both HCHO and CH3CHO give similar reactions with all the reagents except— (A) Schiff’s reagent (B) Ammoniacal AgNO 3 (C) Fehling solution (D) Ammonia 50. Tischenko reaction proceeds in presence of— (A) Zn and ZnCl 2 (B) Al powder (C) Al (OC2H5)3 and anhydrous AlCl3 (D) FeCl3

(C) m-nitrophenol (D) 2, 4-dinitrophenol

1. How many molecules are contained in 3·5 g of CO at 0°C and 760 mm pressure ?

8. In the reaction 2Ag + 2H2SO4 → Ag2SO4 + 2H2O + SO2,

(A) 3·5 × 6·02 × 10 23

sulphuric acid acts as—

(B) 28 × 6·02 × 1023

(A) An oxidizing agent

(C) 7·525 × 1022

(B) A reducing agent

(D) None of these

(C) A catalyst

2. The nucleus of an electrically neutral atom undergoes radioactive decay. It will remain neutral after the decay if the process is a— (A) β–-decay (B) β+-decay (C) γ-decay (D) Neutron decay 3. Total number of possible isomers of formula C7H7NO 2 having a benzene ring is— (A) 6

(B) 8

(C) 10

(D) 15

4. The density of nitrogen is maximum at— (A) STP (B) 273 K and 2 atm (C) 546 K and 1 atm (D) 546 K and 4 atm 5. Which metal has maximum tendency to get oxidised ? (A) Mg

(B) Zn

(C) Cu

(D) Ag

6. What would be the weight of the slaked lime required to decompose 8·0 g of ammonium chloride ? (A) 5·53 g

(B) 2·12 g

(C) 15·52 g

(D) 7·62 g

7. Which of the following is not a colligative property ? (A) Osmotic pressure (B) Vapour pressure (C) Elevation of b.p. (D) Depression of f. p.

C.S.V. / November / 2009 / 1113

17. Mercury(II) forms a complex with— (A) H2S (B) SnCl 2 (C) KI

(D) An acid as well as an oxidant 9. An anhydride of HClO 4 is— (A) Cl2O (B) ClO2 (C) Cl2O6 (D) Cl2O7 10. How many mL of 0·1 M H2SO4 must be added to 50 mL of 0·1 M NaOH to give a solution that has a concentration of 0·05 M in H2SO4 ? (A) 400 mL (B) 50 mL (C) 200 mL (D) 100 mL 11. Which of the following will reduce acidic K 2Cr2O7 solution ? (A) White vitriol (B) Mohr’s salt (C) Chile salt petre (D) Potash alum 12. In the equilibrium 1 O , 2 2(g) the extent of dissociation of water when P = 1 atm and K = 2·08 × 10– 3 is approximately— (A) 2% (B) 0·2% (C) 20% (D) 1% H2O(g)

16. Which of the following aqueous solutions has the highest boiling point ? (A) 0·1 M KNO 3 (B) 0·1 M Na 3PO4 (C) 0·1 M BaCl 2 (D) 0·1 M K2SO4

H2(g) +

13. The IUPAC name of is— (A) Trans-3-methyl pent-2-ene (B) (Z)-3-methyl pent-2-ene (C) Trans-2-ethyl but-2-ene (D) (E)-3-methyl pent-2-ene

(D) NaOH

18. During adsorption— (A) TΔS is positive (B) ΔH-TΔS is negative (C) ΔH is positive (D) TΔS and ΔG become zero 19. Oxidation of NH3 by CuO yields a molecule in which oxidation state of nitrogen is— (A) + 6 (B) + 3 (C) 0 (D) + 4 20. Ozone is powerful oxidizing agent. It is— (A) Less oxidizing than F 2 (B) More oxidizing than F2 (C) Less oxidizing than O2 (D) Less oxidizing than H 2O2 21. Epsom salt is— (A) CuSO 4.5 H 2O (B) ZnSO 4.7 H 2O (C) FeSO 4.7 H 2O (D) MgSO 4.7 H 2O 22. Which of the following statements is incorrect ? (A) Boron carbide is used as an abrasive (B) Boron is used to increase the hardenability of steel (C) Boron oxide (B2O3) is used in the manufacture of borosilicate glass (D) Orthoboric acid undergoes intramolecular hydrogen bonding

14. A given weak acid (0·01 M) has pK a = 6. The pH of the solution is— (A) 3 (B) 4 (C) 5 (D) 6

23. Which of the following gives methane on reaction with water ? (A) Al4C3 (B) Mg2C3 (C) CaC2 (D) CaH2

15. Which of the following is strongest acid ? (A) Phenol (B) o-nitrophenol

24. Which of the following molecules is not a donor ? (A) (C2H5)2NH (B) (C2H5)3N (C) (SiH3)3N (D) C2H5OH

25. Sugar of lead is— (A) PbSO4 (B) Pb (CH3COO)2.3H2O (C) PbCl 2 (D) PbCO 3

35. Acidic solution of a salt produced deep blue colour with starch and KI. The salt is— (A) Chloride (B) Nitrite (C) Acetate (D) Bromide

26. The number of S—S bonds in the cyclic trimer of sulphur trioxide is— (A) Three (B) Two (C) One (D) Zero

36. Which of the following ions disproportionates in water ? (A) Au3+ (B) [Au (CN)4] – – (C) [AuCl4] (D) Au+

27. PCl5 is prepared by the action of Cl2 on— (A) P2O3 (B) P2O5 (C) PCl3 (D) All of these 28. Which of the following noble gases is the least polarised ? (A) Radon (B) Krypton (C) Xenon (D) Helium 29. Which of the following hydrides is most stable ? (A) CH4 (B) PbH 4 (C) SiH4 (D) SnH 4 30. The least melting metal among the following is— (A) Tin (B) Carbon (C) Germanium (D) Lead 31. Select the rate law that corresponds to the data shown for the following reaction : A+B→C

Exp. 1 2 3 4 (A) (B) (C) (D)

[A]0

[B]0

0·012 0·035 0·024 0·070 0·024 0·035 0·012 0·070 Rate = K [B]3 Rate = K [B]4 Rate = K [A] [B]3 Rate = K [A]2 [B] 2

Initial Rate 0·10 0·80 0·10 0·80

32. Which vitamin contains N, S and Cl elements ? (A) Thiamine (B) Cyanocobalamin (C) Ergocalciferol (D) Adermin 33. Eight grams of a radioactive substance are reduced to 0·5 g after 1 hour. The half-life period of radioactive substance is— (A) 15 min (B) 30 min (C) 45 min (D) 10 min 34. Purple of Classius is a colloidal solution of— (A) Silver (B) Gold (C) Copper (D) Platinum

C.S.V. / November / 2009 / 1114

COOH (B) O2N

COOH

COOH (C) COOH NO2

37. Which of the following elements shows allotropy ? (A) Zn (B) Sn (C) Mg (D) Cs

COOH NO2

(D) 38. On strong heating, which of the COOH following produces a mixture SO 2 NO2 and SO 3 ? (A) Fe 2(SO4)3.9H2O 45. The maximum number of 3d electrons having spin quantum (B) (NH4)2SO4.Fe2(SO4)3.24H2O number, s = + 1/2 is— (C) FeSO 4.7H2O (D) FeS 2 (A) 10 (B) 5 39. Which of the following expression does not give root mean square velocity ? 3RT 1/2 3P 1/2 (A) (B) M DM (C)

( ) ( ) 3P D

1/2

(D)

( ) ( ) 3PV M

1/2

40. NaCl and Na2SO4 can be distinguished by— (A) Gutzeit’s test (B) Marsh’s test (C) The flame test (D) The chromyl chloride test 41. Which of the following does not adopt hcp structure ? (A) Be (B) Mg (C) Fe (D) Mo 42. Which is the most stable conformation of cyclohexane ? (A) Chair (B) Twist (C) Boat (D) Staggered 43. Which of the following contains maximum number of electrons in the antibonding molecular orbitals ? (A) O2 (B) O22 – – (C) O2 (D) O2+ 44. The major product obtained in the reaction COOH HNO H2SO 4

3 ⎯⎯→

COOH is— COOH NO2 (A) COOH

(C) 2

(D) 1

46. Which of the following will give only one monosubstituted product ? (A) o-dinitrobenzene (B) m-dinitrobenzene (C) p-dinitrobenzene (D) None of these 47. Which of the following is laevorotatory ? (A) Glucose

(B) Sucrose

(C) Fructose

(D) Galactose

48. The product of addition polymerization reaction is— (A) PVC

(B) Nylon

(C) Terylene

(D) Polyamide

49. Which of the following reagents can be safely used to convert — CHCHO → CH2 — — CHCOOH CH2 — (A) Cold alk. KMnO4 solution (B) Diamine Silver(I) Hydroxide (C) Cold dil. Cl2/NaOH solution (D) Br2 in CCl4 50. The unit of Q (reaction quotient) is— (A) (Concentration)Δn (B) (Atmosphere)Δn (C) Pressure and temperature dependent (D) Does not have a unit

Introduction

The Lymphatic Vessels

● The lymphatic system helps to maintain the proper fluid balance in the tissues and the blood, to conserve protein and to remove bacteria and other particles from the tissues. ● It consists of a network of fluid-filled vessels that open into the blood-vessel system, together with masses lymphoid tissue. ● The lymphoid tissue is, in part, intimately associated with the network of vessels, as in the lymph nodes or glands, is less intimately associated with the vessels, as in the tonsils and thymus, or has no obvious connection with the vessel network, as in the spleen. ● The fluid in the lymphatic vessels, called lymph, is as a rule clear, or faintly opalescent, except the lymph called chyle, which comes from the intestine. Chyle at times may be white and milky in appearance because of the minute globules of fat that it contains. The thin walled vessels through which chyle flows are referred to as lacteals because of their milky appearance.

● Lymphatic vessels commence as a network of closed microscopic vessels-capillaries whose walls consists essentially of a single layer of cells, the endothelium.

The Lymphocytes ● Of the variety of cells in the lymph nodes, the most numerous are lymphocytes, which are also found in the lymph, in blood and elsewhere. ● Lymphocytes vary in size, but the smallest are about the size of red blood cells and consists almost entirely of nucleus with very little cytoplasm. ● Lymphocytes are capable of active movement and can migrate through the walls of blood vessels and through the various tissues. ● Lymphocytes exist in number of places : the spleen, the lymphoepithelial tissues, the thymus and bone marrow, and large numbers of cells are scattered throughout the connective tissues of the body. ● The lymphoepithelial tissues are masses of lymphocytes in close relation to the epithelium (covering) of the alimentary canal. ● At upper end they form both the tonsils and the tissues at the back of the nose that when enlarged are called adenoids. ● The appendix is made up largely of lympho epithelial tissue; and throughout the intestine there are lymphoid nodules, arranged either singly or in groups, known as Peyer’s patches. ● The thymus is a lymphoid organ that is particularly well developed at birth and during the growth period. After puberty it undergoes progressive atrophy. ● The bone marrow contains a large population of lymphocytes scattered among its other cells and not arranged in dense nodules. ● A considerable number lymphocytes are to be found in connective tissues all over the body.

C.S.V. / November / 2009 / 1117

● The lymphatic vessels are quite extensive, most regions of the body are richly supplied with lymphatic capillaries. The construction of the larger lymphatic vessels is similar to that of cardio-vascular veins, including the presence of valves. ● The movement of lymph within these vessels is dependent upon skeletal muscle contraction. When the muscle contract, the lymph is squeezed past a valve that closes, preventing the lymph from flowing backwards.

Differences between Blood and Lymph Blood

Lymph

1. It is a red fluid tissue.

1. It is a colourless fluid tissue.

2. It starts from the heart and flowing through the arteries, capillaries and veins returns to the heart.

2. It starts from the tissuespaces and flowing through the lymphatic capillaries and vessels enters the subclavian veins.

3. It consists of plasma, erythrocytes, leucocytes and platelets.

3. It consists of plasma and leucocytes only.

4. It contains more plasma proteins and more calcium and phosphorus.

4. It contains fewer plasma proteins and less calcium and phosphorus.

5. It has haemoglobin in erythrocytes, hence, red in colour.

5. It lacks haemoglobin containing erythrocytes hence colourless.

6. Its flow is fairly rapid.

6. Its flow is very slow.

7. It mainly transports materials from one organ to another in the body.

7. It mainly conveys materials from the tissue cells into the blood.

● Lymphatic system is a one-way system that begins with lymphatic capillaries. These capillaries take-up fluid that has diffused from blood capillaries. ● The lymphatic capillaries join to form progressively larger vessels whose walls become somewhat thicker because of their small amounts of muscle and elastic tissue. ● These larger collecting vessels usually accompany veins and possess valves. The compression of valved lymphatic trunks by surrounding muscles is an important factor in the propulsion of lymph.

● At the points at which valves are attached, the walls of lymphatic trunks are thicker than elsewhere, and consequently, distended lymph trunks usually have a beaded appearance. ● The intermediate vessels join to form a series of main lymph trunks. The main lymph trunk draining the lower limbs, enters the abdomen after passing through lymph nodes in the lumbar region, known as lumbar trunk. ● Lumbar trunk enters the cirterna chyli. Cirterna chyli is a dilated sac lying at the back of the abdomen opposite the first lumbar vertebra and from its upper end the thoracic ascends through the chest. It inclines slightly to the left and ends at the base of the neck by entering the left innominate vein formed by the junction of the great veins draining the left side of the head and neck and the left arm. ● The thoracic duct is much larger duct and returns to the blood stream, the lymph from both legs, pelvis and abdomen and from greater part of the left body. ● Lymph from the right side of the body returns to the blood through the right lymph duct, which opens into the great veins at the base of the neck on the right side. ● There are valves in both the thoracic duct and the right lymph duct.

Differences between Blood Capillaries and Lymphatic Capillaries Lymphatic Capillaries

Blood Capillaries

1. Are colourless and difficult to observe.

1. Are reddish and easy to observe.

2. Are free at one end and joined to lymphatic vessels at the other end.

2. Are joined to arterioles at one end and to venules at the other end.

3. Free end is blind and expanded into a knob.

3. Have no free end.

4. Are wider than blood capillaries.

4. Are narrower than lymphatic capillaries.

5. Do not have uniform diameter.

5. Have a uniform diameter.

6. Wall formed of highly attenuated endothelium and a poorly developed basement membrane.

6. Wall formed of normal endothelium and basement membrane.

7. Carry a colourless lymph received by diffusion from tissue spaces to lymphatic vessels.

7. Carry red blood received from arterioles to the venules.

8. Have a relatively low pressure.

8. Have relatively pressure.

high

9. Mark the beginning of the lymphatic system.

9. Join arterial venous systems.

and

10. Absorb tissue fluid from intercellular spaces.

Lymph Nodes ● Lymph nodes are relatively evolutionary development and in their characteristic form are present only in mammals, though a rudimentary type of lymph node is found in birds. ● Lymph nodes are found in both superficial and deep situations. ● There is well marked node or nodes at the back of the knee, while there are a number of both superficial and deep nodes in the region of groin. There are large collection of nodes are present in the armpit and numerous nodes in the thorax and abdomen are found especially in the mesentery. ● It is distinctive of humans and primates that lymph passes through chains of several small nymph nodes instead of through one or two large nodes as in other mammals. ● The human lymph node is enclosed in a fibrous capsule, in the interior are masses of cells, mainly lymphocytes. ● The groups of lymph nodes found in the region of groin are called inguinal nodes, while in the armpits are called axillary nodes. ● Vessels bringing lymph to the lymph node are termed afferent, while draining lymph away from lymph node are termed efferent. ● Freshly formed lymph that has not passed through a node is called peripheral lymph. Lymph that has passed through only one node is called intermediate lymph. The lymph that has passed through all the nodes that is going to traverse and is on its way to the blood without further interruption is termed central lymph. ● Peripheral lymph contains a small number of lymphocytes, whereas the central lymph contains many more. ● Lymphocytes that reach the blood via the lymph stream are termed indirect-entery lymphocytes as opposed to the direct-entery lymphocytes that obtain access to the bloodstream directly by migrating through the walls of blood vessels in lymphoid tissue without first entering the lymph. Considerable numbers of lymphocytes are of the direct-entery variety. The few cells reaching the node via peripheral lymph are derived from the blood passing first from the blood capillaries into connective tissue then into peripheral lymph. The steady drift of a small number of lymphocytes through the connective tissues of the body may possess much greater importance than the numbers suggest. ● Furthermore, under abnormal conditions, the number of lymphocytes in peripheral lymph may be greatly increased.

Lymphoid Organs

10. Add tissue into intercellular spaces.

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● Spleen is the largest mass of lymphoid tissue. It is soft spongy organ lying in upper left portion of abdominal cavity just beneath the diaphragm. ● A dense connective tissue with scattered smooth muscle cells cover the surface of spleen. Outer connective tissue divides the spleen into lobules, which contain sinuses.

Fig. : The lymphoid organs. The lymphoid organs include the lymph nodes, the thymus gland, the spleen, and the bone marrow, which all contain lymphocytes.

● In the spleen, the sinuses are filled with blood instead of lymph. The blood vessels of the spleen can expand, and this enhances the carrying capacity of this organ to serve as the blood reservoir and to make blood available in times of low pressure or when the body needs extra oxygen carrying capacity. ● Internally, there are two distinct masses of tissue : the white and the red pulp. The white pulp is packed with lymphocytes which can generate a specific immune response against foreign substances in the blood. ● The red pulp filters and stores blood. As a filter, it contains phagocytic cells which remove damaged, worn out red blood cells and platelets. ● Filtered blood is stored in the red pulp as a reserve blood volume. When there is blood loss from an injury, the smooth muscle cells in the capsule contract to squeeze out the reserve blood and increase the circulating blood volume. ● Tonsils are clusters of lymphatic tissue that contain lymphocytes and are embedded in the mucous membrane that lines the throat. Five tonsils are arranged in a ring at the junction of the nasal and oral cavities where the pharynx begins. Their job is to provide lymphocytes and immune chemicals to guard against infection entering through the mouth and nose. ● Thymus gland is located along the trachea behind the sternum in the upper thoracic cavity. This gland varies in size, and is larger in children than in adults; may disappear completely in old age. ● Thymus is divided into lobules by connective tissue. The T-lymphocytes mature in these lobules. The interior of the lobule consists mostly of epithelial cells.

C.S.V. / November / 2009 / 1119

● Red bone marrow is the site of origination of all types of blood cells including the five types of white blood cells. ● Red bone marrow contains stem cells that are capable of dividing and producing cells that go on to differentiate into the various types of blood cells. ● In a child, most bones have red bone marrow but in adults it is present only in the bones of the skull, sternum, ribs, clavicle, pelvic bones and the vertebral column. ● The red bone marrow consists of a network of connective tissue fibres, called reticular fibres, which are produced by cells called reticular cells.

Composition of Lymph ● Lymph contains all the protein fractions present in the blood plasma but at a lower level. The protein content of lymph varies in different regions. ● Liver lymph contains the highest percentage of protein, whereas lymph from the skin and subcutaneous tissues usually contains the lowest. ● Antibodies are associated with serum globulins and it is the globulin content of lymph that is more reduced in relation to blood plasma than the albumin content. ● Hence as a rule, lymph contains less antibodies than blood plasma. ● Apart from proteins, lymph, especially intestinal lymph contains appreciable amounts of fat after a fatty meal. ● Electrolytes (calcium, sodium, potassium, chloride) in lymph are found in virtually the same concentration as in blood plasma. The same is basically true of glucose and urea.

Formation of Lymph ● Lymph is derived from the blood by a process that begins with filtration through the wall of the blood capillary. This primary filtrate then undergoes some modifications as it passes through the tissues that it finally leaves by entering the lymphatics. ● The amount of tissue fluid at any time represents an equilibrium between the amount of fluid leaving the blood capillary and the amount entering the lymphatic capillary. ● Compared with the blood vascular system, the lymphatic vessels can transport relatively small amounts of fluid, hence, they can deal with only a moderate increase in lymph flow. Beyond this point lymph accumulates in the tissues which become waterlogged and said to develop edema. ● Most commonly a rise in the venous pressure is reflected almost at once by a rise in the capillary pressure and a corresponding increase in lymph formation.

Functions of Lymphatic System ● This system is closely associated with cardiovascular system and has three main functions : 1. Lymphatic vessels take up excess tissue fluid and protein, return them to the blood stream. 2. Lymphatic capillaries absorb fats at the intestinal villi and transport them to the blood stream. 3. The lymphatic system helps to defend the body against diseases.

Review at a Glance ● The mammalian lymphatic system consists of lymphatic vessels and lymphoid organs. ● Lymph flows one way from a capillary to ever-larger lymphatic vessels and finally to a lymphatic duct, which enters a subclavian vein. ● Lymph nodule called Peyer’s patches are found within the intestinal wall. ● Lymph is clear fluid within lymphatic vessels and derived from tissue fluid. ● Lymph present in the vessels between the intestine after a meal is generally milky due to presence of chylomicrons. ● Lymphocytes are produced in primary lymphoid tissue (thymus, embryonic liver, adult bone marrow) and migrate to secondary lymphoid tissue (spleen, lymph nodes, unencapsulated lymphoid regions of gut submucosa, respiratory and urogenital regions). ● Lymphoma is tumour composed of lymph tissue. ● Hodgkin’s disease is a malignant lymphoma of reticuloendothelial cells in lymph nodes and other lymphoid tissues. ● Lymph contains less antibodies than blood plasma. ● Even a minor interference in the removal of tissue fluid will be noticed as tissue swelling, a condition called edema. ● Lymphatic vessels have one-way valves along their length to prevent the backflow of lymph. ● Lymph is moved by the action of skeletal muscles. ● The process of squeezing of leucocytes out of capillaries is called Diapedesis.

OBJECTIVE QUESTIONS 1. Blood and lymph differ in— (A) Blood has RBCs which are absent in lymph (B) Blood has WBCs which are absent in lymph (C) Blood has cells while lymph is without cells (D) Blood has several inorganic substances which are absent in lymph 2. Lymph flows in the lymph vessels due to— (A) Pressure of valves in their walls (B) Contraction and dilation of heart (C) Pressure exerted by the contraction of surrounding muscles (D) All the above 3. A dilated sac into which empties the intestinal, two lumbar and two descending lymphatic trunks is— (A) Cisterna chyle (B) Thoracic duct

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(C) Both (A) and (B) (D) None of these 4. The process of squeezing of leucocytes out of capillaries is called— (A) Exocytosis (B) Diapedesis (C) Exosmosis (D) Chemotaxis 5. Lymphocytes are produced in— (A) Thymus (B) Embryonic liver (C) Adult bone marrow (D) All the above 6. Largest lymphoid organ is— (A) Liver (B) Spleen (C) Tonsils (D) Peyer’s patches 7. Which organ of the body filters the blood ? (A) Bone marrow (B) Liver (C) Lymph vessels (D) Spleen

8. Pericardial fluid is basically— (A) Plasma (B) Serum (C) Lymph (D) None of the above 9. Which of the following statement is correct regarding lymph ? (A) It consists of plasma and leucocytes only (B) It lacks haemoglobin, hence colourless (C) It contains fewer blood proteins and less Ca and P (D) All the above are correct 10. Lymph consists of— (A) Plasma (B) Leucocytes (C) Erythrocytes (D) Both (A) and (B)

ANSWERS

●●●

● Cancers are a result of uncontrolled cell division. The type of nuclear division involved is mitosis. The problem is caused by mutations or abnormal activation of the genes which control cell division. ● When the genes are abnormal they are called oncogenes. About 100 of these have been discovered. A single faulty cell may divide to form a clone of identical cells. Eventually an irregular mass of relatively undifferentiated cells called a tumour is formed. ● Tumour cells can break away and spread to other parts of the body, particularly in the bloodstream or lymphatic system, causing secondary tumours or metastases. This process is called metastasis. ● Tumours that spread and eventually cause ill health and death are described as malignant. The majority of tumours, such as common warts, do not spread and are described as benign.

Primary tumour

Invasive tumour cell (c) Secondary tumour site

How Cancer Cells Differ from Normal Cells

Tumour cell adhering to capillary

● Cell division is a highly regulated process, with a balance being maintained between production of new cells and cell death in most of the tissues and organs. ● Mature and differentiated normal cells have a finite life span. They are usually replaced by new cells generated by cell division and differentiation.

(d) Fig. : Stages in development of cancer (a) to (d). Primary tumour may become metastatic and get transformed into secondary tumour.

● Normally, the production of new cells is regulated in such a manner that at any given time, the number of a given cell type remains nearly constant. Normal cells live in a complex interdependent manner, regulating one another's proliferation. ● Occasionally, some cells may arise, which do not respond to normal growth control mechanisms. These cells proliferate in an unregulated manner and give rise to clones of cells that can expand to a considerable size; this growth is called tumour or neoplasm. ● All tumours are not malignant. Noncancerous tumours, commonly called benign tumours, remain confined to their original location and are incapable of indefinite

growth, e.g., warts. However, cancerous or malignant tumours grow rapidly with infinite life span of the proliferating cells and become progressively invasive. Only the malignant tumours are properly referred to as true cancer.

Normal epithelial cells

Basement membrane (a) Invasive cancer in Situ cancer

(b)

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Properties of Cancer Cells ● Cancer cells exhibit some features which distinguish them from normal cells. They show uncontrolled proliferative ability, with a reduced requirement for extracellular growth factors. ● They acquire the ability to invade new sites, a phenomenon designated as Metastasis. Cancer cells exhibit a number of alterations on cell surface, in the cytoplasm and in their genes; these features are used for the identification of cancers. The ability of cancer cells to resist induction of cell death promotes the development of tumours. ● The nucleus of cancer cells are usually enlarged and have irregularly distributed chromatin. The interchromatin and perichromatin granules, which are formed of RNA and proteins are more abundant in the nucleus of cancer cells. The nucleolus of cancer cells displays hypertrophy. It becomes more irregular and enlarged. Sometimes several nucleoli may be present in the nucleus of cancer cells. ● The ergastoplasm and ribosomes are more abundant in cancer cells. Formation of polysomes by the union

of various ribosomes is more frequent. This is associated with rapid growth of cancer cells. ● The normal cells have a cytoskeleton of microtubules and microfilaments arranged in some regular fashion and help in coordinated cell movement. In cancer cells, the cytoskeleton undergoes depolymerization and disintegration, thus cells become independent and their movements uncoordinated. ● Normal cells exhibit stickness or adhesiveness and their is specificity in adhesion because cells of one type stick to the cells of same type. Cancer cells have decreased adhesiveness and these do not exhibit specificity in adhesion. If skin cancer cells are mixed with normal kidney cells, the aggregates formed contain both kidney and skin cells mixed together. This is why, the cell from malignant tumour can invade normal body organs. ● Because of loss of contact and decreased adhesiveness, cancer cells are able to dissociate themselves from neighbouring cells and can invade other tissues and organs. The invasion is brought about by changes in the plasma membrane or by proteases released by cancer cells.

Normal cells Controlled growth Contact inhibition One organized layer Differentiated cells

● It is believed that development of a malignant cancer cell involves several steps and is usually caused by more than one factor operating over several years rather than a single factor. More than one mutation to the genes may occur. Upto 20% of the cancer worldwide may be caused by viruses. ● Evidence that cancers are genetic in origin was provided by work with retroviruses. Retroviruses are RNA viruses which, when they invade animal cells, use the enzyme reverse transcriptase (a viral coat protein) to make DNA copies of the viral RNA. ● The DNA is inserted into the host DNA where it may stay and be replicated for generations of cells. ● Some retroviruses are harmless. However, HIV is a harmful retrovirus and other retroviruses cause cancer. These contain a gene which alters host cell division genes, switching them on and causing the cell to become malignant. The genes become oncogenes. The advantage to the virus is that the cell makes many copies of itself and, therefore, of the virus. ● DNA viruses contain DNA as their hereditary material. Some contain their own oncogenes which can cause uncontrolled cell division of host cells. In humans, papilloma viruses cause warts. ● Some papilloma viruses have been implicated in some forms of cervical cancer, making this a sexually transmitted disease. The Epstein-Barr virus may cause one form of Burkitt’s lymphoma which is common in Africa.

Hereditary Predisposition

Cancer cells Uncontrolled growth No contact inhibition Disorganized, multilayered Non-differentiated cells Abnormal nuclei Fig. : Cancer cells differ from normal cells in the ways noted.

Types of Cancer ● Cancers can be classified on the basis of the original tissue from where they arose. Most of the cancers fall into one of the following categories : 1. Carcinomas—Cancers of this type arise from epithelial tissues, such as skin or the epithelial lining of internal organs or glands. 2. Melanomas—These are cancerous growths of melanocytes (a type of skin cells). 3. Sarcomas—These are derived from tissues of mesodermal origin, e.g., bone and cartilage. 4. Leukemias and lymphomas—These are tumours of the haematopoietic cells.

Causes of Cancer ● Changes in genes are called mutations and any factor bringing about a mutation is called a mutagen. An agent which causes cancer is called a carcinogen.

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● About 5% of human cancers show a strong genetic predisposition, in other words they tend to run in families. More than 40 types of cancer, including cancer of breast, ovary and colon, come into this category. ● Genes responsible may be oncogenes, or genes which lead to failure and cannot kill cancer cells. In most cases other factors are required, but in a few cases, such as retinoblastoma, a single faulty gene is responsible. Retinoblastoma starts in the eye and spreads to the brain, causing death if untreated. It is caused by a dominant gene. ● Two breast cancer genes have been identified and named BRCA1 and BRCA2. BRCA1 was cloned in 1994 and codes for a protein involved in transcription.

Ionising Radiation ● Ionising radiation includes X-rays, γ-rays and particles from the decay of radioactive elements. Cancers were caused in workers with X-rays at the beginning of the twentieth century and factory workers painting the dials of watches with luminous paint containing radioactive radium and thorium. ● The radiation causes the formation of chemically active and damaging ions inside cells which can break DNA strands or cause mutations. The types of cancers linked with ionising radiation include skin cancer, bone marrow cancer, lung cancer and breast cancer. Medical and dental X-rays also expose patients to ionising radiation.

Ultraviolet Light ● Ultraviolet light is the most common form of carcinogenic radiation and is non-ionising. DNA absorbs ultraviolet light and the energy is used in converting the bases into more reactive forms which react with surrounding molecules. Sunlight contains ultraviolet light and prolonged exposure to it can result in skin cancers, including melanoma which is highly malignant and commonly causes death through secondary brain tumours. Depletion of the ozone layer results in a higher proportion of ultraviolet light reaching the Earth’s surface. The brown skin pigment melanin offers some protection.

Radon Gas ● Radon gas is a natural source of radiation released from certain rocks such as granite. It may accumulate in houses in areas where these rocks are found. It has been linked to the development of leukaemia (blood cancer), lung, kidney and prostate cancers, although the evidence is inconclusive.

Growth factor

Receptor

Proto-oncogenes code for a growth factor, a receptor, or a signalling protein in a regulatory network within the cell. If a proto-oncogene becomes an oncogene, the result is active cell division. Regulatory network

Signalling protein Heredity Organic chemicals Radiation Viruses Oncogene

Protooncogene Tumour-suppressor gene

Chemical Mutagens ● Many chemicals are now recognised as causing cancer. The first example was described as soot and coaltar, when chimney sweeps were discovered to develop cancer of scrotum. ● Later mineral oils were also found to be carcinogenic, when shale oils were used as a lubricant in the cottonspinning mills. The workers developed cancers of abdominal wall where their clothes had been splashed. Workers in the synthetic dye industry in the late nineteenth century developed bladder cancer. ● The list of chemical carcinogens has steadily lengthened over the last 90 years and now includes, in addition to the above, inorganic arsenic compounds which produce skin cancer and asbestos products which cause lung cancer. Some food additives (flavours, colourings and stabilisers) have been considered as possible carcinogens because they cause cancers in experimental animals. ● Tobacco smoke contains chemicals responsible for lung cancer. The most important of these are polycyclic hydrocarbons which are converted in the body to carcinogens.

Cancer and Genes ● Francis Peyton Rous reported a virus, Rous Sarcoma Virus (RSV), is capable of causing sarcoma, a type of cancer. He was awarded Nobel prize in 1966. In the meantime, it was clear that RSV is an RNA virus. After the virus infects a cell, it inserts a DNA transcript of its genome, which includes a gene known as src, into the host chromosome. The src gene is now known to be an oncogene. In its normal, nonmutated state, it is a proto-oncogene, a gene that can be transformed into an oncogene.

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These agents can bring about the activation of oncogenes and the inactivation of tumour-suppressor genes.

Tumour-suppressor genes code for a signaling protein in a regulatory network. If a tumour-suppressor gene is inactivated, the result is active cell division.

Fig. : Causes of cancer. Heredity and environmental factors cause mutations of proto-oncogenes and tumour-suppressor genes. A regulatory pathway, which includes plasma membrane receptors for growth factors, intracellular reactions and the genes, is unbalanced. This leads to uncontrolled growth and a tumour. Cancer still does not develop unless the immune system fails to respond and kill these abnormal cells.

● The oncogene most frequently involved in human cancers belong to the ras gene family. An alteration of only a single nucleotide pair is sufficient to convert a normally functioning ras proto-oncogene to an oncogene. ● The ras oncogene is found in about 25% of lung cancers, 50% of colon cancers and 90% of pancreatic cancers. The ras oncogene is associated with leukaemias (cancer of blood-forming cells) and lymphomas (cancer of lymphoid tissue) and both ras oncogenes are frequently found in thyroid cancers. ● Researchers have identified a half-dozen tumoursuppressor genes. When these genes malfunction, a tumour results. The RB tumour-suppressor gene was discovered by Alfred Knudson. When a child receives only one normal RB tumour-suppressor gene and that gene mutates, eye tumour develop in the retina by the age of three. Loss of RB tumour-suppressor gene through chromosomal deletion is particularly frequent in a type of lung cancer called small cell lung carcinoma. ● Another major tumour-suppressor gene is called p53, agene that is more frequently mutated in human

cancers that any other known gene. It has been found that the p53 protein acts as a transcription factor and as such is involved in turning on the expression of a gene called WAFI or Cipl. The product of this gene is a cell cycle inhibitor. A number of cell proteins combine with the p53 protein. When a protein combines with p53, it is hard to tell whether it is activating p53 or is being activated by p53. This intricate situation is now being unravelled.

●

Detection and Diagnosis ● Cancer diagnosis is based on the characteristic histological features of malignant cells. Blood tests for abnormal WBCs and bone marrow biopsy are also used. ● Non-invasive techniques like X-rays (using injected dyes), CT scans and MRI scans can be used to detect cancers of internal organs like kidneys and pancreas. ● Modern techniques monitor and detect the molecular changes that occur in cancer cells, this enables an early diagnosis of cancer. Monoclonal antibodies against cancer-specific antigens are coupled to appropriate radioisotopes. These antibodies are then used for detection of cancer.

Therapy of Cancer ● Various approaches have been adapted for the treatment of cancer. Therapeutic strategies vary, depen-

●

●

●

ding on the etiology of each individual cancer. Some of the common approaches include : (I) Surgery (II) Radiation therapy, (III) Chemotherapy and (IV) Immunotherapy. The therapies can be used either singly or in a suitable combination. Surgery—Surgical manipulation/excision of tumour mass is one of the easiest approaches in the treatment of cancer. However, surgery does not ensure that all the cancer cells have been removed. Moreover, not all tumours are accessible for surgical manipulation. Surgical reduction of tumour load is also considered advantageous prior to initiation of other therapeutic approaches. Radiation therapy—This approach focuses on lethally irradiating cells in a tumour mass. However, this approach also causes tremendous damage to several tissues in the vicinity of tumour mass. Chemotherapy—Several chemotherapeutic drugs are used in this strategy to kill tumour cells. Some such drugs can specifically kill tumour cells; however, majority of them have a number of side-effects. Immunotherapy—One of the recent approaches of cancer treatment involves augmentation of natural anticancer immunological defence mechanisms. Monoclonal antibodies have been used in various ways, e.g., radioimmunotherapy etc., for treatment of cancer. Research is in progress to develop cancer vaccines.

OBJECTIVE QUESTIONS 1. Which of the following might cause a proto-oncogene to become an oncogene ? (A) Exposure of cell to radiation (B) Exposure of cell to certain chemicals (C) Viral infection of the cell (D) All the above 2. A cell is cancerous. Where might you find an abnormality ? (A) Only in nucleus (B) Only in the plasma membrane receptors (C) Only in cytoplasmic reactions (D) In any part of the cell concerned with growth and cell division 3. A tumour-suppressor gene— (A) Inhibits cell division (B) Opposes oncogenes (C) Prevents cancers (D) All the above 4. Blood cancer is known as— (A) Sarcoma (B) Carcinoma (C) Leukaemia (D) None of these 5. Sarcoma is a cancer of— (A) Connective tissue (B) Skin

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(C) Mucosal membrane (D) Glandular tissues 6. Carcinoma is a cancer of— (A) Bone (B) Muscle (C) Connective tissue (D) Skin 7. Which of the following metastasizes through lymphatic vessel ? (A) Sarcoma (B) Carcinoma (C) Both (A) and (B) (D) None of these 8. Which of the following metastasizes through blood stream ? (A) Sarcoma (B) Carcinoma (C) Both (A) and (B) (D) None of these 9. Which tumour is capsulated ? (A) Benign (B) Malignant (C) Both (A) and (B) (D) None of these 10. Malignant neoplasms which develop from mesenchyme are called— (A) Sarcoma (B) Carcinoma

(C) Angionensis (D) None of these

ANSWERS ●●●

Just Released

(Multiple Choice Questions on Computer Fundamental, Application and Advance Topics)

Useful for Various Competitive Exams. By : Dr. Alok Kumar Code No. 1664 Price : Rs. 80/UPKAR PRAKASHAN, AGRA-2 ● E-mail : [email protected] ● Website : www.upkar.in

1. When DNA is exchanged via cytoplasmic bridges between two Paramecia, the process is called— (A) Transformation (B) General transduction (C) Restricted transduction (D) Conjugation 2. Amphiblastula stage is associated with— (A) Coelenterates (B) Amphibia (C) Porifera (D) Polychaetes 3. A restriction enzyme breaks bonds between the— (A) Base pairs of a DNA molecule (B) Base pairs of a DNA-RNA hybrid molecule (C) Sugar and phosphate components of a nucleic acid molecule (D) Exons and introns of a DNA molecule 4. Actinic burn is caused by excessive exposure to— (A) Thermal energy (B) Chemicals (C) Ultraviolet sun rays (D) Electricity 5. The males of ants, bees and wasps are— (A) Haploid (B) Polyploid (C) X/X (D) X/O 6. Which of the following is true ? (A) DNA is the genetic material in most viruses and bacteria (B) RNA is the genetic material in most viruses and bacteria (C) DNA is the genetic material in all the viruses (D) RNA is the genetic material in all the viruses 7. Replicon is related with— (A) Amino acids (B) Nucleic acid (C) Blood plasma (D) None of these

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8. β-oxidation is the process, occurring during— (A) (B) (C) (D)

Fatty acid oxidation Protein digestion Carbohydrate oxidation Nucleic acid oxidation

9. When a mutation is limited to the substitution of one nucleotide pair for another, it is called a— (A) Translocation (B) Point mutation (C) Sugar-phosphate deletion (D) Base inversion 10. In matured human females, which of the following is released during ovulation ? (A) Primary oocyte (B) Secondary oocyte (C) Oogonium (D) All the above 11. In chromosomal mutation, if a section of chromosome is lost during cell division, it is called— (A) Deletion (B) Duplication (C) Translocation (D) Inversion 12. Describing similarity due common ancestory, called—

to

(A) Patristic (B) Homoplasty (C) Homostyly (D) None of these 13. Which of the following muscles help urinary bladder to expel urine forcefully ? (A) Detrusor muscle (B) Latissimus dorsi muscle (C) Hamstring group of muscles (D) None of these 14. Which layer of heart wall consists cardiac muscles ? (A) (B) (C) (D)

Endocardium Epicardium Myocardium All the above

15. Which of the following forms part of complement system of blood plasma ? (A) Kinins (B) Histamine (C) Prostaglandins (D) All the above 16. Which of the following does not pass across membranes by simple diffusion ? (A) O2 (B) CO2 (C) H2O (D) H+ 17. Waterproof nature of feathers of aquatic birds is due to secretion of— (A) Sudorific glands (B) Preen glands (C) Brunner’s glands (D) None of these 18. Reticulocytes are— (A) WBCs (B) Immature RBCs (C) Blood platelets (D) Lymphocytes 19. Members of a biologic species are potentially able to— (A) Phyletic evolution (B) Divergent evolution (C) Convergent evolution (D) Allopatry 20. Laennec’s disorder is related with— (A) Liver cirrhosis (B) Hepatitis (C) Jaundice (D) All the above 21. The goblet cells present in the intestinal lining are— (A) Holocrine (B) Apocrine (C) Merocrine (D) Endocrine 22. Desmin is characteristic of— (A) Cardiac muscle cells and fibres (B) Lung tissues (C) Liver tissues (D) None of these 23. What causes Pituitary Nanism ? (A) Hyposecretion of adrenocorticotrophic hormone (B) Hypersecretion of adrenocorticotrophic hormone (C) Hyposecretion of somatotrophic hormone (D) Hypersecretion of somatotrophic hormone

24. Lining of intestine and kidneys in humans is— (A) Keratinized (B) Brush border (C) Ciliated (D) None of these 25. Irrational fear of disease is called— (A) Algophobia (B) Myophobia (C) Haematophobia (D) Pathophobia 26. Stereocilia are non-motile protoplasmic projections from free surfaces of cells of— (A) Ductus epididymes (B) Ductus deferens (C) Ductus arteriosus (D) Both (A) and (B) 27. The posterior end of Amoeba is called— (A) Plasmalemma (B) Posterior pseudopodia (C) Uroid (D) None of these 28. Which of the following disease in humans is closely linked to the dreaded Mad Cow disease ? (A) Creutzieldt-Jacob disease (B) Bovine Spongiform Encephalopathy (C) Transmissible Spongiform Encephalopathy (D) None of the above 29. The arms of a starfish will grow again when cut off, this type of regeneration is called— (A) Morphollaxis (B) Epimorphosis (C) Autotomy (D) None of these 30. Any reaction with ΔG negative is— (A) Exergonic (B) Endergonic (C) Both (A) and (B) (D) None of these 31. Which of the following is the function of nuclear organizers ? (A) Synthesis of t RNA (B) Synthesis of m RNA (C) Regeneration of nucleoli after nuclear division (D) All the above

C.S.V. / November / 2009 / 1127

32. In humans, which brain wave pattern disappears entirely during sleep ? (A) Alpha waves (B) Beta waves (C) Theta waves (D) Delta waves 33. Which of the following cell organelles is non-membranous ? (A) Glyoxysomes (B) Endoplasmic reticulum (C) Ribosomes (D) Lysosomes 34. Which of the following method determines the concentration of protein-bound hormones in blood plasma ? (A) Radio-immunoassay (B) Radiology (C) Both (A) and (B) (D) None of these 35. Micro-organisms that can live at temperature below 20°C are called— (A) Mesophilic (B) Thermophilic (C) Psychrophilic (D) None of these 36. A powerful eye irritant present in smog is— (A) Ozone (B) Nitric acid (C) Peroxyacetyl nitrate (D) Sulphur dioxide 37. Chromosome with median centromere and equal arms is called— (A) Acrocentric (B) Submetacentric (C) Telocentric (D) Metacentric 38. Which of the following shows a taxonomically close relation ? (A) Sea horse, sea anaemone, sea urchin (B) Housefly, dragonfly, butterfly (C) Silver fish, cuttle fish, star fish (D) Earthworm, ringworm, tapeworm 39. The toxins produced by tetanus microbes affects— (A) Voluntary muscles (B) Involuntary muscles

(C) Both (A) and (B) (D) None of these 40. Striations of actin and myosin filaments are not found in— (A) Smooth muscles (B) Skeletal muscles (C) Cardiac muscles (D) All the above 41. Which type of tissue forms the inner lining of a blood vessel ? (A) Epithelial (B) Connective (C) Muscle (D) None of these 42. Amphibians are— (A) Homeothermic (B) Endothermic (C) Ectothermic (D) Both (B) and (C) 43. Wharton’s duct in humans is associated with— (A) Genital organs (B) Submandibular salivary gland (C) Renal gland (D) None of these 44. Upper round portion of skull covering the brain is known as— (A) Crest (B) Calvaria (C) Olfactory capsule (D) None of these 45. Which of these controls the peristaltic movements of the intestine ? (A) Auerbach’s plexus (B) Brachial plexus (C) Sacral plexus (D) None of these 46. Aldosterone causes— (A) Kidneys to release renin (B) Kidneys to absorb sodium (C) Blood volume to increase (D) All the above 47. Cleidoic egg is characteristic of— (A) Reptiles (B) Birds (C) Insects (D) All of these 48. Which of the following is genetic part of the sperm ? (A) Head (B) Neck (C) Middle piece (D) Tail 49. Osmotically inactive stored material in human body is— (A) Protein (B) Glycogen (C) Lipid (D) Phosphogen

(Continued on Page 1130)

(C) Phenylketonuria (D) Tay-sach disease 18. Periotic bone present in— (A) Skull (B) Jaw (C) Sternum (D) None of these 1. The sex of the developing foetus can be detected by the— (A) Twelfth week (B) Fourteen week (C) Fifteen week (D) Seventeen week 2. Bacterial resistance to antibiotics is a genetic trait carried in the bacterial— (A) Intron (B) Chromosome (C) Plasmid (D) Centromere 3. In DNA, which of the following is double-ringed molecule ? (A) Adenine (B) Guanine (C) Thymine (D) Both (A) and (B) 4. Lobo’s disease is associated with— (A) Skin (B) Spleen (C) Liver (D) Eyes 5. The feedback control mechanism is related with— (A) Bile secretion (B) HCl secretion (C) Hormonal secretion (D) Herring-Breuer reflex 6. Enterohepatic circulation related with— (A) Bile acids (B) Bile pigments (C) Cholesterol (D) Cholecystokinin

is

7. The force generating response of muscle contraction, when muscle shortens, is called— (A) Isometric (B) Isotonic (C) Both (A) and (B) (D) None of the above 8. Diseases communicable from animals to humans under natural conditions, are called— (A) Zoophile (B) Zoonoses (C) Zoophilism

(D) Zoomania

C.S.V. / November / 2009 / 1128

9. Bursa is sac-like cavity of fibrous tissue which contain a thick lubricating fluid, called— (A) Synovial fluid (B) Vitreous fluid (C) Spinal fluid (D) None of these 10. Nuhn’s glands are related with— (A) Nose (B) Ear (C) Tongue (D) Hairs 11. Which of the following technologies uses radioactive glucose and isotopes of carbon to reveal metabolic activity of tissues ? (A) CAT (B) PET (C) MRI (D) None of these 12. Bowman’s glands are located in— (A) Olfactory membrane (B) Cardiac muscles (C) Seminiferous tubules (D) Pleural membrane 13. Which of the extraembryonic membranes forms blood cells in developing embryo ? (A) Chorion (B) Yolk sac (C) Allantois (D) Amnion 14. Toothless mammals are included in— (A) Xenarthra (B) Rodentia (C) Chiroptera (D) None of these 15. Individuals affected by Klinefelter’s syndrome are— (A) Males (B) Females (C) Eunuchs (D) Both (A) and (B) 16. The protein present in the fingernail is— (A) Actin (C) Keratin

(B) Myosin (D) Globin

17. Extra-nuclear inheritance occurs in— (A) Killer strain in Paramecium (B) Colour blindness

19. Interferons are synthesized in response to— (A) Bacterial infection (B) Viral infection (C) Fungal infection (D) All the above 20. Which of these form part of complement system of blood plasma ? (A) Histamine (B) Kinins (C) Prostaglandins (D) All of these 21. The cartilage of nasal septum is— (A) Hyaline cartilage (B) Elastic cartilage (C) Calcified cartilage (D) Fibrous cartilage 22. The wings of a bird and the wings of an insect are— (A) Homologous organs (B) Analogous organs (C) Vestigial organs (D) Phylogenetic structures 23. Chromosome number 2n – 1 is an example of— (A) Monosomy

(B) Trisomy

(C) Euploidy

(D) Polyploidy

24. The total collection of genes, at any one time, in a unit of evolution is called the— (A) Genotype (B) Demotype (C) Multiple-allelic group (D) Gene pool 25. Linings of human intestine is— (A) Ciliated (B) Keratinized (C) Brush border (D) None of these 26. Liquor amnii is also known as— (A) Amniotic fluid (B) Cerebrospinal fluid (C) Vitreous humour (D) None of these

27. Theory of blood clotting is related with— (A) Fuld and Spiro theory (B) Cascade theory (C) Hyman theory (D) Huxley theory 28. Stripped muscles are— (A) Syncytial (B) Uninucleate (C) Binucleate (D) None of these 29. High level of which of the following is found in blood of gout patients ? (A) Urea (B) Uric acid (C) Cholesterol (D) All the above 30. Which immunoglobin produces immunity in infants prior to birth ? (A) IgA (B) IgD (C) IgG (D) IgE 31. Skin colour in humans is believed to be controlled by atleast three genes. This is an example of— (A) Co-dominance (B) Epistasis (C) Multiple alleles (D) Quantitative inheritance 32. Which of the following essential amino acid is found in low quantity or missing in wheat flour ? (A) Methionine (B) Tryptophan (C) Lysine (D) Aginine 33. Frequency of an allele in a isolated population may change due to— (A) Genetic drift (B) Gene flow (C) Mutation (D) Natural selection 34. Which of the following is superficial calf-muscle ? (A) Gastronemius (B) Trapezius (C) Latissimus (D) Gluteus 35. Arrector pili muscle consists of— (A) Myoepithelial cells (B) Smooth fibres (C) Striated fibres (D) None of these

C.S.V. / November / 2009 / 1129

36. Which of the following bones, has a large opening called foramen magnum ? (A) Occipital bone (B) Sphenoid bone (C) Palatine bone (D) Zygomatic bone 37. Zonula adherens is a kind of— (A) Desmosome (B) Mesosome (C) Filament (D) Membrane 38. Human blood groups are an example of— (A) Gradualism (B) Cline (C) Gradient of diploidy (D) Polymorphism 39. The layer of cells that secrete enamel of teeth is— (A) Dentoblast (B) Ameloblast (C) Odontoblast (D) Osteoblast 40. Milk teeth in a child of 3-4 years do not include— (A) Canines (B) Incisors (C) Premolars (D) Molars 41. Aquatic larva of mosquito is known as— (A) Floating raft (B) Imago (C) Maggot (D) Wriggler 42. Cleidoic egg is characteristic of— (A) Aves (B) Reptiles (C) Insects (D) All of these 43. The thumb of humans move more freely than the other fingers due to the presence of— (A) Pivotal joint (B) Gliding joint (C) Hinge joint (D) Saddle joint 44. With reference to human diet, which one of the following is an essential fatty acid ? (A) Butyric acid (B) Linoleic acid (C) Oleic acid (D) Stearic acid 45. Who introduced the term ‘phylum’ in taxonomy ? (A) Cuvier (B) Huxley (C) John Ray (D) Aristotle 46. Rathke’s pouch forms— (A) Pineal gland (B) Thyroid gland

(C) Pituitary gland (D) None of these 47. Which of these is essential for normal excitability of cardiac muscles ? (A) Potassium ions (B) Calcium ions (C) Magnesium ions (D) All the above 48. Myenteron is— (A) Muscular coat of the intestine (B) Mesentry (C) Alveoli of lung (D) None of these 49. Melanin pigments give colour to which part of human brain ? (A) Substantia nigra (B) Substantia grisa (C) Substantia gelatinosa (D) Substantia propria 50. Tusks of elephant are— (A) Molar teeth (B) Canine teeth (C) Incisor teeth (D) Premolar teeth

ANSWERS WITH HINTS

●●●

(Continued from Page 1127) 50. Growth phase prepares oogonia or spermatozoa for first— (A) Mitotic division (B) Meiotic division (C) Binary fission (D) Multiple fission

ANSWERS WITH HINTS

●●●

C.S.V. / November / 2009 / 1130

1. Spermatogenesis is under the regulatory influence of— (A) Vasopressin (B) Follicle stimulating hormone (C) Luteinizing hormone (D) Luteotrophic hormone 2. The epithelium of bronchioles is— (A) Pseudostratified and columnar (B) Squamous and sensory (C) Cuboidal columnar (D) Pseudostratified and sensory 3. Which hormone stimulates contraction of uterine muscles ? (A) Luteinizing hormone (B) Oxytocin (C) Prolactin (D) All of the above 4. Which of the following represent the correct sequence of events for defence by leucocytes ? (A) Inflammation, diapedesis, chemotaxis, phagocytosis, digestion (B) Chemotaxis, inflammation, phagocytosis, digestion, diapedesis (C) Diapedesis, digestion, inflammation, phagocytosis, chemotaxis (D) Inflammation, chemotaxis, diapedesis, phagocytosis, digestion 5. Formation of red blood corpuscles from bone marrow is stimulated by—

7. Which one is not a epithelial tissue ? (A) Simple cuboidal and stratified columnar (B) Bone and cartilage (C) Stratified squamous and simple squamous (D) All the above are epithelial tissues 8. Which of the following cases results in foetal or serious situations ? (A) Rh – male marrying Rh– female (B) Rh – male marrying Rh+ female male female

marrying

Rh+

(D) Rh + male female

marrying

Rh–

(C)

Rh +

(A) Crocodile (B) Chicken (D) Fish

10. Archenteron cavity is found in— (A) Blastula

(B) Gastrula

(C) Morula

(D) Planula

11. In a population N = 1500, if the birth rate is 42 per year and the death rate is 27 per year, the rate of natural increase will be— (A) 42%

(B) 27%

(C) 1%

(D) 0·1%

12. The phenomenon of nuclear fusion of sperm and egg is known as— (A) Karyogamy

(A) Erythropoietin

(B) Vitellogenesis

(B) Secretin

(C) Oogenesis

(C) Stomatin

(D) None of these

(D) Testosterone 6. Microlecithal eggs with small amount of food reserve are found in— (A) Insects (B) Frog (C) Humans (D) Fish

C.S.V. / November / 2009 / 1131

15. Function of Bartholin’s glands in female rabbit is— (A) Secrete viscid fluid (B) Help in sperm transfer (C) Lubricates passage

the

vaginal

(D) Both (A) and (C) 16. The number of foetal membranes in humans is— (A) 2

(B) 3

(C) 4

(D) 5

17. Biometry is the study of— (A) Forests (B) Biomass (C) Characteristics of populations (D) All the above

9. In which of the following animals, the blood entering the aorta incompletely oxygenated ? (C) Monkey

14. In mammals the female secondary sexual characters are developed by the hormone— (A) Relaxin (B) Estrogens (C) Progesterone (D) Gonadotropins

13. Which enzyme is responsible for urea formation ? (A) Arginase (B) Urease (C) Aminotransferase (D) None of the above

18. Chromosome theory of sex determination was propounded by— (A) Bridges (B) Balbiani (C) Goldschmidt (D) Mendel 19. Both olfactory receptors and sound receptors have cilia, and they both— (A) Are chemoreceptors (B) Are mechanoreceptors (C) Initiate nerve impulses (D) None of the above 20. Lampbrush chromosomes are special type of chromosomes found in— (A) Salivary gland cells of dipterans (B) Gonadal cells of reptiles (C) Amphibian oocytes (D) Brain stem cells 21. The opening in the vertebrate skull through which the spinal cord passes is— (A) Foramen magnum (B) Foramen mental (C) Foramen lacerum posterius (D) Foramen incisive

22. On the basis of morphology twenty three chromosomes in human beings are arranged in— (A) 5 groups (B) 6 groups (C) 7 groups (D) 8 groups 23. Elasmobranch is the group of fishes that includes— (A) Bony fishes (B) Lung fishes (C) Cartilaginous fishes (D) All the above 24. The sex chromosomes in females are— (A) XX

(B) XY

(C) YY

(D) XXY

25. Spongin fibres from—

are

secreted

(A) Spongioblasts (B) Choanocytes (C) Amoebocytes (D) Pinacocytes 26. Vitamin C is helpful in the— (A) Formation of visual pigment (B) Growth of bones (C) Treatment anaemia

of

pernicious

(D) Wound healing 27. Vertebrae of frog are generally procoelus except— (A) Eight only (B) Ninth only (C) Eight and ninth (D) Ninth and tenth 28. Secretin hormone stimulates— (A) Gastric glands (B) Pancreas (C) Gall bladder (D) All the above 29. Glenoid cavity is found in— (A) Pelvic girdle (B) Skull (C) Pectoral girdle (D) Radio-ulna 30. When the intensity of light is low during night the light is detected by— (A) Rods (B) Cones (C) Lens (D) Both rods and cones

C.S.V. / November / 2009 / 1132

31. Role of vitamin C is to help in— (A) Formation of visual pigment (B) Growth of bones (C) Treatment of pernicious anaemia (D) Wound-healing 32. If a man goes from sea coast to Everest peak then— (A) His breathing and heart beat will increase (B) His breathing and heart beat will decrease (C) His respiratory rate will decrease (D) His heart beat will decrease 33. Schwann cells are associated with— (A) Axon (B) Dendrites (C) Ciliary body (D) Stapes 34. Blood vessels leading into Bowman’s capsule are called— (A) Renal artery (B) Renal vein (C) Efferent atriole (D) Afferent atriole 35. Which one of the following is example of spinal reflexes ? (A) Blinking (B) Sneezing (C) Coughing (D) All of these 36. During conduction of nerve impulse— (A) Na + moves into axoplasm (B) Na + moves out of axoplasm (C) K+ moves into axoplasm (D) Ca ++ moves into exoplasm 37. Placenta is— (A) A cord that connects the foetus with wall of mother’s uterus (B) A part of mother’s uterine wall through which exchange of materials occurs between mother and foetus (C) A part of foetus through which exchange of materials takes place between foetus and mother (D) Chorional villi of foetus and a part of mother’s uterine wall through which materials exchange between foetus and mother occurs 38. Renin is released from— (A) Juxtaglomerular apparatus (B) Cortical nephron

(C) Collecting duct (D) Pelvis 39. Vestigial structures of man are— (A) Wisdom teeth, appendix, ear muscles (B) Ear pinnae (C) Ileum and molars (D) Fossa ovalis and canines 40. Complete colourblindness humans is called— (A) Monochorionic (B) Monochromasy (C) Monochromatism (D) None of these

in

41. Which one represents a connecting link ? (A) Whale between fishes and mammals (B) Archaeopteryx between birds and mammals (C) Duckbill platypus between reptiles and mammals (D) Java ape man between modern man and peking man 42. The exchange of materials between blood and interstitial fluid occurs only at the— (A) Veins (B) Capillaries (C) Arteries (D) Arterioles 43. Wild life preservation act was enacted by Indian Government in— (A) 1952 (B) 1983 (C) 1972 (D) 1986 44. Which of these enzymes is produced by infants and not by adults ? (A) Lipase (B) Pepsinogen (C) Rennin (D) Trypsin 45. Physiology of blood circulation was first described by— (A) William Harvey (B) Carl Landsteiner (C) Andreas Vesalius (D) Carl Carrens 46. Which of the following dietary protein is called incomplete protein ? (A) Most animal proteins (B) Most plant proteins (C) Both (A) and (B) (D) None of these

47. Heparin is formed by— (A) Liver cells (B) Kidney cells (C) Blood cells (D) Bone marrow 48. Digestive enzyme pepsin in the gastric juice begins digestion of— (A) Carbohydrates (B) Proteins (C) Fats (D) Nucleic acids 49. Ratitae are flightless birds which have— (A) Flat breastbones (B) Weak wing muscles (C) Strong wing muscles (D) Both (A) and (B)

●●●

(Continued from Page 1106)

50. Non-specialized dentition is found in— (A) Carnivores (B) Herbivores (C) Omnivores (D) All the above

ANSWERS WITH HINTS

●●●

C.S.V. / November / 2009 / 1133

General Account and Two Kingdoms Classification (i)

The living world around us consists of diverse forms of plants and animals. This extra-ordinary diversity around us in form of varied flowering and nonflowering plants and enormous types of animals, led the early biologists to separate the organisms of the whole living world into two Kingdoms (the highest taxonomic rank)-the Plantae and the Animalia. (ii) Thus, the plantae represented the entire Plant Kingdom, while the animalia represented the entire Animal Kingdom. (iii) However, Linnaeus as early as in 1753, on the basis of absence or presence of flowers, divided the whole Plant Kingdom into two sharply contrasting major groups : (a) Plants without flowers belong to Cryptogam group; and (b) Plants bearing flowers belong to Phanerogam group. (iv) The cryptogams, being flowerless and hence seedless are further divided into— (a) Thallophyta—Plant body is an undifferentiated thallus.

(b) Plants bearing closed seeds are called Angiosperms

However, the transit from Thallophyta to Bryophyta, then to Pteridophyta and finally to spermatophyta is followed by gradual reduction of the gametophytic phase and corresponding elaboration of the sporophytic phase by repeated mitotic division in the zygote. It has resulted in the formation of a multicellular embryo. Thus, the condition in Spermatophyta is just reverse of Thallophyta i.e., (2n) phase most dominant and the ( n) phase very short lived. Coupled with the differentiation of sporophyte into root, stem and leaves, there was development of vascular tissue (xylem and phloem) in pteridophytes, which became more elaborate in spermatophyta. On the basis of above features, Tippo (1942) divided the plant kingdom as follows :

(a) Plants bearing naked seeds are called Gymnosperms; and

C.S.V. / November / 2009 / 1134

⏐

↓ Thallophyta (No embryo)

⏐ ↓

↓

Tracheophyta (With vascular tissue)

Bryophyta (No vascular tissue)

⏐ ↓

↓

Spermatophyta (With seed)

Pteridophyta (No seed)

⏐ ↓

↓

Gymnosperms (Naked seed)

Angiosperms (Closed seed)

Characteristic Features I. Thallophyta 1.

The plant body shows no differentiation into root, stem and leaves. This is called a thallus.

2.

No vascular tissue in the plant body.

3.

Asexual reproduction by mitotic spores is common.

4.

Sexual reproduction may be isogamous, anisogamous or oogamous.

Plant Kingdom

⏐

↓

↓

Cryptogams

↓

Phanerogams

↓

Thallophyta Phyco- Mycophyta phyta n

2n

↓

Bryophyta n

Pteridophyta

ytic toph e m c Ga hyti rop o p S 2n

n

2n

⎯→

(v) The Phanerogams, bearing flowers and hence also known as seed bearing plants (Spermatophyta) are divided into—

↓ Embryophyta (With embryo)

Cytologically, the Thallophyta are characterised by a dominant gametophytic ( n) phase in the life-cycle. The sporophytic (2n ) phase comes for a very short duration. It is represented by the zygote only, which immediately goes back to the ( n) phase following meiosis. There is no embryo formation.

(b) Bryophyta—Plant body thalloid or a leafy axis lacking roots; and (c) Pteridophyta—Plant body differentiated into root, stem and leaf.

Plant Kingdom

Spermatophyta n e Phas e Phas 2n Gymno- Anthophyta phyta

5.

6. 7.

If present, the sex-organs are simple and unicellular, occasionally multicellular ( e.g., globule of Chara). No embryo formation after gametic union. Mostly the plants are aquatic.

Besides Bacteria and Lichens, the Thallophyta comprises two main sub-divisions, the Algae (Phycophyta) and the Fungi (Mycophyta). Their differences are given in the following table :

III. Pteridophyta

5. Homosporous or heterosporous.

1. Mostly terrestrial, may be aquatic e.g. , Marsilea, Salvinia, Azolla etc.

6. Sexual reproduction oogamous i.e. , by antheridia and archegonia.

2. The sporophytic (2 n ) phase is dominant and independent of the gametophyte.

7. The zygote develops into embryo of the sporophyte.

3. The sporophyte well differentiated into root, stem and leaves (except the order psilotales).

8. The gametophyte is green, independent and short lived, still provided with rhizoids. IV. Spermatophyta

4. Vascular tissues present.

1. These are the seed bearing sporophytic plants, highly differentiated into root, stem and leaves and form the dominant terrestrial flora of the present day.

Algae (Phycophyta)

Fungi (Mycophyta)

(a) Presence of chlorophyll (b) Manufacture their own food (autotrophs)

(a) Absence of chlorophyll (b) Get food from external sources (heterotrophs)

(c) Thallus made up of true parenchyma.

(c) Thallus made up of false tissue (interwoven hyphae) called pseudoparenchyma.

3. Always heterosporous.

(d) Cell wall cellulose.

(d) Cell wall made up of chitin.

5. The gametophyte is non-green and represents a highly reduced phase of the life cycle.

made up

of

(e) Commonly the reserve food is starch. (f) Usually light loving. (g) Gradual complexity of sexual apparatus from simple to higher forms.

(e) Commonly the reserve food is glycogen. (f) Usually light fearing. (g) Gradual simplification of sexual apparatus from lower to higher forms.

II. Bryophyta 1.

These are the simplest and most primitive land plants, mostly terrestrial, may be aquatic.

2.

The gametophytic (n ) phase is dominant.

3.

The gametophytic plant body may be thalloid or leafy.

4.

Roots are absent and their functions are served by rhizoids.

5.

No vascular tissue (although some sort of conducting tissues may be present).

6.

Sexual reproduction is highly oogamous i.e. , by means of multicellular male and female sex organs called antheridia and archegonia.

7.

The zygote develops into embryo (sporogonium).

8.

The sporophyte is wholly or partially dependent upon the green gametophyte for its nutrition.

C.S.V. / November / 2009 / 1135

2. Vascular tissue well developed. 4. The embryo is well organized.

The spermatophyta are further divided into Gymnosperms (Gymnophyta) and Angiosperms (Anthophyta), which differ in the following respects :

Gymnosperms (Gymnophyta)

Angiosperms (Anthophyta)

(a) Commonly woody plants. (b) Vessels and companion cells absent from xylem and phloem respectively (except Gnetum, Ephedra and Welwitschia). (c) The reproductive organs form distinct male and female cones. (d) The cones (= flowers) are devoid of perianth. (e) The megasporophylls (= carpels) are not differentiated into stigma, style and ovary. (f) Ovules are exposed on the carpels, hence seeds are naked (no fruit). (g) Pollination by air only.

(a) May be woody or herbaceous. (b) Vessels and companion cells present in xylem and phloem respectively.

(h) Male gametophytes contain prothallial cells. (i) Female gemetophytes mainly monosporic (j)

Male gametes motile or nonmotile. (k) Double fertilization absent. (l) Endosperm is haploid (n ) and develops before fertilization.

(c) The reproductive organs form uni or bisexual flowers. (d) The flowers are mostly provided with perianth. (e) The carpels are differentiated into stigma, style and ovary. (f)

Ovules are enclosed within the ovary, hence seeds are closed with the fruit. (g) Pollination by air, water, wind or different animals. (h) Prothallial cells are absent from male gametophytes. (i) Female gametophytes, mono, bi or tetrasporic, without archegonia. (j) Always non-motile. (k) Double fertilization present. (l) Endosperm triploid (3n ) and developed after fertilization.

The angiosperms are further divided into dicotyledons and monocotyledons, which are as follows : Dicotyledons

Monocotyledons

(a) Embryo with two cotyledons. (b) Tap root system. (c) Leaves show reticulate venation. (exception is calophyllum having parallel venation) (d) Vascular bundles collateral and open, arranged in a ring. (e) Flowers usually pentamerous sometimes may be tetramerous.

(a) Embryo with only one cotyledon. (b) Adventitious root system. (c) Parallel as well as reticulate (rarely) (d) Vascular bundles collateral and closed, scattered. (e) Flowers trimerous.

Remarks : 01. Sometimes the two groups of organisms may, however, be found together in associations for their symbiotic or mutual benefits e.g., (a) Mycorrhizae–associations of fungi with the roots or other underground parts of vascular plant. (b) Lichens–symbiotic association of algae and fungi. 02. The archegonium is a common female reproductive organ of Bryophytes, Pteridophytes and Gymnosperm which are collectively known as archegoniate. 03. While thallophytes and bryophytes constitute the non-vascular cryptogams, the pteridophytes alone are known as vascular cryptogams. 04. Likewise, bryophytes constitute the non-vascular archegoniate, whereas pteridophytes and gymnosperms together form the group called vascular archegoniate. 05. The gymnosperms, containing ‘plants with unprotected seeds’ have rightly been called by Goeble (1930) as the ‘phanerogams without ovary’. 06. Following are some of the homologies between the reproductive organs of heterosporous pteridophytes and gymnosperms as well as the angiosperms :

(i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

Heterosporous Pteridophytes and Gymnosperms Microsporophylls Microsporangium Microspores Male gametophytes Megasporophylls Megasporangium Megaspore Female gametophyte

C.S.V. / November / 2009 / 1136

Copeland’s (1956) Four Kingdoms Classification The groups of plants described above are all cellular organisms having a well organized cytoplasm. However, in the scheme of classification described above, no mention has been made about the viruses which do not have any cytoplasm or organized cell and are called acellular. In view of the above, the revised scheme of classification stands as follows : O rganisms ⏐ ↓ ↓ Protobiota Cytobiota (Acellular organisms (All the cellular e.g., viruses) organisms) The cytobiota, on the basis of absence or presence of a well organized (true) nucleus have been further divided into two major categories namely ‘prokaryotes’ and ‘eukaryotes’ respectively as detailed below : Cytobiota ⏐

↓ Prokaryotes (e.g., bacteria and cyanobacteria)

↓ Eukaryotes (e.g., all other plants and all animals)

Angiosperms Stamens Pollen-sac Pollen grains Germinating pollen grains Carpels Nucellus of ovule Embryosac mother cell Embryosac

Prokaryotes (i) Well organized nucleus absent i.e., (a) Nuclear membrane absent (b) Nucleolus absent (c) Chromatin devoid of histone protein (d) Spindle apparatus absent (ii) Membrane bound cell-organelles like chloroplast, mitochondria, E. R. Golgi body etc. absent (iii) Diaminopimelic acid (an amino acid) present in cell wall (iv) Flagella, if present lack 9 + 2 arrangement (v) Ribosomes of 70s type (vi) Cells divide amitotically (vii) Sexual reproduction absent, if present, it does not involve meiosis and meiotically produced gametes Eukaryotes (i) Well organized nucleus present i.e., (a) Nuclear membrane present (b) Nucleolus present (c) Chromatin with histone protein (d) Spindle apparatus present (ii) Membrane bound cell organelles present (iii) Diaminopimelic acid absent (iv) Flagella, when present show 9 + 2 arrangement (v) Ribosomes of 80s type (vi) Cells divide mitotically (vii) Sexual reproduction essentially involves meiosis and meiotically produced gametes The above concept led Copeland (1956) to establish an alternate system of classification of the ‘living world as a whole’ into following 4 major taxonomic groups : 1. Monera : This term was introduced by Haeckel. It includes all the prokaryotes. 2. Protista : This includes simpler eukaryotes and may be :

(a) autotrophic protista— e.g., All algae except the blue green (cyanobacteria) (b) heterotrophic protista— e.g., Fungi, Slime moulds and Protozoa. 3.

4.

Metaphyta : It consists of multicellular plants i.e., all the archegoniate (bryophytes, pteridophytes and gymnosperms) as well as all angiosperms. Metazoa : It includes the entire animal kingdom, except protozoa.

Whittaker’s (1969) Five Kingdoms Classification In the system of classification presented by Copeland, as discussed above, a very important modification was subsequently affected by Whittaker in the year 1969. He separated the whole of the fungi from rest of the plants (Plantae) on the basis of their heterotrophic mode of nutrition, as compared to other plants which largely possessed a photosynthetic mode of nutrition. This separation of fungi resulted in the creation of a separate Kingdom-Fungi and thereby leading to the establishment of the following 5 Kingdoms in the entire living world– I.

II.

Monera : This kingdom includes all the prokaryotic forms, in which the cells are strictly devoid of a true nucleus, plastids, mitochondria and other membrane bound cell organelles. Mode of nutrition may vary from heterotrophic to photosynthetic to even chemosynthetic one. The modes of reproduction are predominantly vegetative or asexual. Even if some of them show sexual process–the act of gametic fusion and meiosis is completely lacking. This kingdom includes– Cyanobacteria (Blue-green algae), Actinomycetes and the true bacteria (Eubacteria). Protista—This kingdom consists primarily of the unicellular but also the colonial forms of eukaryotic organisms. The modes

C.S.V. / November / 2009 / 1137

of nutrition are either by photosynthesis or by ingestion. The modes of reproduction include both asexual and sexual. This kingdom includes–unicellular and colonial algae, Diatoms (Bacillariophyta) as well as slime moulds (Myxomycetes) and Protozoa.

devoid of chloroplasts. Naturally, the mode of nutrition is heterotrophic (absorptive). The tissue differentiation is almost nil or totally absent. Reproductive modes include both a sexual as well as sexual. This kingdom includes—alga fungi (Phycomycetes). Sac fungi (Ascomycetes), Club fungi (Basidiomycetes) and the imperfect fungi (Deuteromycetes). However, sexuality is totally lacking among the deuteromycetes.

III. Plantae : This kingdom consists strictly of the multicellular eukaryotic plants, having predominantly a cellulosic cell wall and photosynthetic mode of nutrition. Mode of reproduction varies from asexual to sexual. However, the sexual mode of reproduction is predominant and more elaborate. This kingdom includes multicellular algae belonging to classes– Chlorophyceae, Phaeophyceae, Rhodophyceae as well as the Bryophytes and Tracheophytes (Pteridophytes, Gymnosperms and Angiosperms). The tissue differentiation varies from simple to complex one.

V.

IV. Fungi : This kingdom consists basically of the multicellular organisms, having chitinous cellwall and the cells absolutely

Animalia : This kingdom consists of the wallless, multicellular eukaryotic organisms. The principal mode of nutrition is ingestion. The ingested food is digested internally in the intestinal cavity (holozoic nutrition). The somatic built up shows a very complex and advanced type of tissue differentiation and the predominant mode of reproduction is sexual. This kingdom includes all the multicellular animals. The general plan of evolution and divergence of the members of above five kingdoms, suggested by Whittaker has been summarized below :

3

4

5

Plantae Multicell Plants

Fungi

Animalia Multicell animals

↑

↑

Photosynthesis

Absorption

↑ Ingestion

↑

2

Protista Unicell Eukaryotes

↑ 1

Monera The Prokaryotes

↑ The First Living Object Whittaker’s concept of five kingdom classification and diversification

OBJECTIVE QUESTIONS 1. The cryptogams include— (A) Thallophyta only (B) Bryophyta only (C) Pteridophyta only (D) All of the above 2. The group of plants included under ‘phycophyta’ may alternatively be called as— (A) Heterotrophic thallophytes (B) Non-vascular archegoniate (C) Autotrophic thallophytes (D) Vascular archegoniatae 3. A fully mature plant body, undifferentiated into root, stem and leaf is technically known as— (A) Callus

(B) Thallus

(C) Annulus

(D) Herbaceous

4. Which of the following groups of plants produce seeds but have no fruits ? (A) Gymnosperms (B) Bryophytes (C) Fungi (D) Pteridophytes 5. The division of the plant kingdom, forming dominant terrestrial flora of the present day is— (A) Pteridophyta (B) Spermatophyta (C) Thallophyta (D) Bryophyta 6. The chlorophyllous thallophytes have been grouped under— (A) Schizophyta (B) Mycophyta (C) Phycophyta (D) Anthophyta 7. The chlorophyllous thallophytes have been grouped under— (A) Algae (B) Fungi (C) Bryophytes (D) Angiosperms 8. In which of the following respects, the angiosperms resemble the gymnosperms ? (A) Presence of ovule (B) Nature of endosperm (C) Presence of vessels in wood (D) Mode of fertilization

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9. In which of the following groups of plants, embryo formation is entirely absent ? (A) Thallophyta

(C) Bryophytes only (D) All of the above 17. The term ‘Protista’ signifies— (A) All protozoans

(B) Bryophyta (C) Pteridophyta (D) Anthophyta

(B) All eukaryotic algae

unicellular

(C) All slime moulds

10. Which of the following is not included within trachaeophyta ? (A) Anthophyta (B) Gymnophyta (C) Bryophyta (D) Pteridophyta 11. A pteridophyta differs from a bryophyte in having— (A) An independent gametophyte (B) An independent sporophyte (C) Archegonia

(D) All of these 18. The stamen in angiosperms is homologous to— (A) Megasporophyll (B) Microsporangium (C) Microsporophyll (D) Embryosac mother cell 19. The microsporangium corresponds to which of the following parts of angiosperms ? (A) Pollen sac (B) Male gametophyte

(D) Antheridia

(C) Pollen grain

12. Conducting tissues, other than the xylem and phloem are found in— (A) Algae

(D) Carpel 20. The embryophyta alongwith thallophyta constitute— (A) Bryophyta only

(B) Fungi

(B) Pteridophyta only

(C) Bryophytes (D) None of these

(C) Spermatophyta only (D) The entire plant kingdom

13. Which one of the following is a prokaryote ? (A) Chlorella

(B) Rivularia

(C) Gnetum

(D) Fungi

14. Which one of the following is an acellular organism ?

21. Which of the following groups of plants have vascular tissue and reproduce asexually by spores but are devoid of seeds ? (A) Gymnosperms (B) Algae

(A) Cyanobacteria

(C) Bryophytes

(B) Viruses

(D) Pteridophytes

(C) Bacteria (D) Ephedra 15. The term ‘Monera’ was introduced by— (A) Theophrastus (B) Weismann (C) Oparin (D) Haeckel 16. The ‘non-vascular cryptogams’ includes— (A) Algae only (B) Fungi only

22. The chlorophyllous thallophytes are represented by— (A) Fungi (B) Algae (C) Mosses (D) Liverworts 23. The symbiotic association of a fungus with the roots of higher plants is known as— (A) Lichens (B) Nodules (C) Mycorrhiza (D) None of these

24. Which one of the following is neither a prokaryote nor an eukaryote ? (A) Virus (B) Bacteria (C) Blue-green algae (D) Bryophytes 25. Which of the followings are known as ‘phanerogams without ovary’ ? (A) Pteridophytes (B) Gymnosperms (C) Thallophytes (D) Bryophytes 26. The plants bearing both antheridia and archegonia are said to be— (A) Autoecious (B) Heteroecious (C) Monoecious (D) Dioecious 27. In which of the following groups of plants, the gametophytic generation is dominant over the sporophytic one ? (A) Thallophytes (B) Pteridophytes (C) Bryophytes (D) More than one 28. The simplest and highly reduced form of archegonium is to be found in— (A) Bryophytes (B) Gymnosperms (C) Pteridophytes (D) None of these 29. Which of the following is an aquatic pteridophyte ? (A) Pteris (B) Lycopodium (C) Marsilea (D) Selaginella 30. Diaminopimelic acid is the characteristic amino acid confined to— (A) All the moneras (B) All the protistas (C) All the metaphytes (D) All the metazoans 31. Which one is a feature typical to the gymnosperms ? (A) Pollination by air only

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(B) Absence of double fertilization (C) Haploid nature of the endosperm (D) All of the above 32. Which of the following has the largest gametophyte ? (A) Pinus (B) Funaria (C) Selaginella (D) Cycas 33. The female gametophyte of angiosperms is represented by the— (A) Nucellus of the ovule (B) Embryosac mother cell (C) Embryosac (D) Germinating pollen grain

Codes : (a) (A) 4 (B) 3 (C) 4 (D) 2

(b) 2 2 2 4

(c) 5 1 1 1

(d) 1 5 3 5

37. Consider the following statements : (1) Prokaryotes have well organized spindle apparatus (2) Prokaryotes have chromatin devoid of histone-proteins (3) Prokaryotes include bacteria only (4) Diaminopimelic acid is present in the cell wall of prokaryotes. Of the above statements : (A) Only 1, 2 and 4 are correct (B) Only 2, 3 and 4 are correct (C) Only 2 and 4 are correct (D) Only 2 and 3 are correct

34. Which one of the following statements is correct ? (A) Vascular cryptogams are all 38. Which one of the following heterosporous nomenclatures is correct as per (B) Vascular cryptogams are all Whittaker’s 5 kingdoms classifihomosporous cation ? (C) Archegoniates are all homos(A) Protobiota, Cytobiota, porous Metaphyta, Metazoa and (D) Spermatophytes are all Plantae heterosporous (B) Metaphyta, Plantae, Animalia, Protista and Fungi 35. Vessels and companion cells are (C) Monera, Protista, Plantae, absent in— Fungi and Animalia (A) All angiosperms (D) Monera, Protista, Cytobiota, (B) All gymnosperms Fungi and Animalia (C) All angiosperms including 39. The four kingdom classification Gnetum, Ephedra and of organism was proposed by— Welwitschia (A) Copeland (B) Whittaker (D) All gymnosperms excluding (C) Linnaeus (D) Hooker Gnetum, Ephedra and Wel40. A multicellular, eukaryotic and witschia non-chlorophyllous body organi36. Match List-I with List -II and pick zation coupled with ingestion as up the correct choice from the the mode of nutrition is characunderassigned codes : teristic of— List-I (A) Fungi (B) Animalia (a) Thallophyta (C) Protista (D) Monera (b) Embryophyta ANSWERS (c) Bryophyta (d) Tracheophyta List- II (1) Embryophyta excluding tracheophyta (2) Bryophyta and tracheophyta (3) Phycophyta and mycophyta (4) Bryophyta and pteridophyta (5) Pteridophyta and spermatophyta

●●●

Definition and Occurrence

Disadvantages of Self-Pollination

Pollination is a process by which pollen grains of an anther transfer on the stigma of a gynoecium. The phenomenon of pollination is a characteristic feature of spermatophytes. It is direct in gymnosperms as the ovules are exposed and the pollen grains are deposited directly on them. In angiosperms, pollination is indirect in the sense that the pollen grains are transferred on the stigma instead of ovules directly.

Types of Pollination The transference of pollen grain during pollination, from the anther to the stigma, may be of the same flower or of a different flowers. If a stigma be pollinated by the pollen of the same flower, it is called self-pollination or autogamy. When the pollen of a flower located on a different plant, pollinates the stigma of another flower, it is called cross-pollination or allogamy. Allogamy is again divided into geitonogamy and xenogamy. Geitonogamy is the deposition of pollen on the stigma of another flower but on the same plant Xenogamy is the deposition of pollens on the stigma of flower on another plant of the same species.

1. Hybrids or varieties are not formed. 2. New characters are not incorporated into the genome of offsprings. 3. The seeds are of low quality, lower resistance level and are usually prone to environmental factors. 4. Better quality plants are not formed.

Advantages of Cross-Pollination 1. It produces good quality seeds with a better germination percentage and a greater yield. 2. Cross-pollination may give rise to variation in the population. 3. Hybrids are created by this process. 4. Cross-pollination avoids the harmful effects of inbreeding.

Disadvantages of Cross-Pollination 1. It is fully dependent on agents. 2. Wastage of pollens in huge amount takes place. 3. Low percentage of fertilization occurs after pollination. 4. Cross-pollination cannot maintain the genomic constitution of a species.

Self-Pollination

Cross-Pollination

1. The pollen of a flower pollinates the stigma of the same flower.

1. Pollen of a flower pollinates the stigma of another flower.

2. Occurs in bisexual flower.

2. Occurs in both bisexual and unisexual flowers.

3. The anther and carpel of flower mature at the same time.

3. The anther and carpel mature at different time.

4. Wastage of pollen grains is comparatively lower.

4. Wastage of pollen grains is comparatively higher.

(b) Dioecism—Male and female flowers are on different plants, e.g., Vallisnaria.

5. The rate of pollen germination is lower.

5. The rate of pollen germination is higher.

2. Monocliny/Bisexuality—Flowers are bisexual and some special devices are present for preventing self pollination.

6. New genetic characters are not formed; so, variation is not possible.

6. Variation is always possible.

(a) Dichogamy—Flowers are bisexual but anther and stigma mature at different times.

7. Quality reduces from generation to generation.

7. Quality gradually increases from generation to generation.

Advantages of Self-Pollination 1. This process is more easy and gives better results. 2. Newly born plant bears the same genetic factors. 3. Self-pollination is not dependent on any agents. 4. Resulting seeds on germination give rise to genetically similar plants.

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Contrivances for Cross-Pollination 1. Dicliny/Unisexuality—When flowers are unisexual only allogamy is possible. (a) Monoecism—Male and female flowers are on the same plant, e.g., Cocos.

(i) Protandry—In these flowers stamen matures earlier than the stigma, e.g., Clerodendron. (ii)

Protogyny—In these flowers stigma matures earlier than the stamen, e.g. , Magnolia, Michelia.

(b) Herkogamy—Flowers are bisexual but the essential organs, i.e., stamen and stigma are arranged in such a way that self-pollination becomes mechanically impossible, e.g. , Hibiscus, Gloriosa. (c) Heterostyly—Flowers are similar except in the essential organs, i.e., the androecium and gynoecium are reciprocally different in length and are so adapted that the stamens pollinate stigmas of the same length.

Contrivances for Self-Pollination 1. Cleistogamy—In this case the flowers never open and as a result self-pollination is carried out within the closed bud. Some common examples of cleistogamous flowers are Commelina benghalensis, Viola, etc. 2. Homogamy—Homogamy means that the stamens and carpels of a flower mature at the same time. So there is a greater chance of self pollination although that is not obligatory. Some homogamous flowers, however, show special mechanisms for self-pollination. In Mirabilis jalapa, mature stamens show recoiling which brings the anther near to the stigma; so that when they burst self-pollination is achieved. Somewhat similar adaptations are seen in Argemone mexicana. 3. Incomplete dichogamy—Some flowers place their anthers at the mouth of the corolla tube. As the stigma elongates from below, it pushes out through these anthers which are already ripe and are pollinated in so doing. Incomplete dichogamy is found in Ixora. Pollen

Petal

(a)

(b)

Fig. : (a-d) Incomplete dichogamy in Ixora. 4. Safety mechanism—Many flowers are normally adapted for cross-pollination. But they also have provisions for self-pollination. In Asteraceae, the flowers are protandrous and normally adapted for cross-pollination. But when cross-pollination fails, the stigmatic lobes curl back and catch up its own pollen which results self-pollination. Stigma Anthers Filament

Petal

Pappus (a)

Ovary (b) Fig. : (a-c) Pollination in Asteraceae.

Agents of Pollination Pollination involves the transference of pollen from the stamen to the stigma and several external agents are employed in pollination. The usual agents are wind, animals and water, although the chief among them are insects. (A) Anemophily When the pollinating agent is wind, it is called anemophily or anemogamy. The wind pollinated flowers are called anemophilous. Characteristics of Anemophilous Flowers 1. Plants that are grown in wind-exposed regions are generally wind pollinated, rather than those which are confined to sheltered places. 2. The flowers occur in pendulous or catkin-type of inflorescence, e.g., Coryllus avellama, Betula verrucosa or may be of pendent type, e.g., Acalypha sanguiana, Casuarina equisetifolia.

C.S.V. / November / 2009 / 1141

3. Generally the flowers are unisexual, inconspicuous, non-nectariferous, dull and destitute of colour. 4. The stamens are protruding with long filaments, e.g., Clerodendron. 5. Anthers are versatile, moving with the direction of the wind, e.g., Poa.

What is Sulphur Shower In Pinus pollination takes place about in March-April in the Eastern Himalaya. Pinus pollens are yellow in colour, winged and are carried on by wind. The amount of pollens liberated by the pine forests at this time is prolific so that the air gets saturated with them and there is a yellow deposit of pollens on the forest floor. This phenomenon is known as sulphur shower or flowers of sulphur. Naturally, most of the pollens are wasted and only a small amount reach the female cone for fertilization.

6. Pollen grains are light-weight, small, smooth, dry and powdery; pollens are produced in large number allowing for the huge wastage, e.g., Pinus .

7. Anemophilous flowers have their stigmas long and feathery, as in Poa, to catch up the pollen grains floating in the air. (B) Hydrophily When the pollinating agent is water, it is called hydrophily. The flowers pollinated with the help of water are known as hydrophilous. Hydrophily is subdivided into two categories on the basis of the occurrence of pollination. When pollination takes place completely underwater, it is called hypohydrogamy and is found in genera like Naias, Ceratophyllum, etc. On the other hand, when pollination takes place on the surface of water, it is known as epihydrogamy and is found in common water-weeds like Vallisneria, Hydrilla, Elodea, Ruppia, etc. Characteristics of Hydrophilous Flowers 1. Plants that are grown in water are generally hydrophilous. 2. Flowers are generally unisexual, inconspicuous, dull and smaller in size. 3. Pollens are heavy as these are carried by water currents. 4. Pollination occurs either underwater or on the water surface. (C) Zoophily When the pollinating agent is animal, it is called zoophily. The animal pollinated flower is known as zoophilous. Pollination by animals represents the highest stage in the pollination mechanism of angiosperms. The agents concerned primarily with zoophily are birds, bats, snails, insects and even human beings. However, insects are the chief pollinating agents in this regard. On the basis of the types of agents, zoophily can be subdivided into following categories : (i) Ornithophily—The pollinating agent here is bird. Ornithophily may be direct or indirect in that the birds may visit the flowers for the ants already harboured by the flowers, and indirectly carry out the pollination. Honey thruses and humming birds pollinate Bignonia capreolata ; honey birds pollinate Strelitzia raginae whereas crows and mynas pollinate Bombax ceiba ; the pollinating agent in Butea monosperma is parrot. (ii) Chiropteriphily—Here the pollinating agent is bat. Bats pollinate a few flowers like Bauhinia megalandra, Eperua falcata, Anthocephalus cadamba, etc.

(iii) Malacophily—When pollination takes place through slugs and snails, it is called malacophily, e.g., Lemna, Colocasia, Alocasia, etc. (iv) Entomophily—When the pollinating agent is insect, it is called entomophily. The insect pollinated flowers are called entomophilous. Insects are the chief and most important pollinating agents and the majority of angiosperms are adapted to entomophily. It is supposed to be the most advanced method and is found mainly in advanced angiospermous flowers. Characteristics of Entomophilous Flowers There are a number of ways by which a flower may attract and induce insects to visit it. Some such instances are as follows : (1) Conspicuousness and colour—A flower may be conspicuous by the bright colour of its petals or perianth. Sometimes bracts, sepals or even stamens may become petaloid and serve the same function. The conspicuousness may be due to the individual flower or to the aggregation of flowers forming such inflorescences as capitula. It is found that certain insects prefer a particular colour. Such as, bees seem to prefer blue flowers, butterflies prefer red and so on. (2) Scent—Generally flowers with scent are usually night-blooming, and colourless or white. Sweet scent or nauseous odour attracts insects and as a result they pollinate the flower. (3) Food—Flowers attract insects to offer them foods which may be nectar, edible pollen or sap. Nectar is a watery fluid containing about 25% sugar. Some insects like bees, pollinate the flower, carrying the pollen adhered to their legs or mouth parts from anthers to stigmas, during the period of collecting nectars from flower, e.g., Ixora . Bees collect a large amount of pollens for eating purposes to nourish their youngs, e.g., in Cassia marylandica the flowers are visited by the humble bees. Some flowers like Orchis morio secrete an edible sap which attract insects to carry out pollination.

Some Special Floral Mechanism for Cross Pollination 1. Piston mechanism : Sesbania (Papilionaceae) 2. Lever mechanism : Salvia (Lamiaceae) 3. Trap mechanism : Aristolochia (Aristolochiaceae) 4. Pit-fall mechanism : Arum maculatum (Araceae) 5. Clip-mechanism : Asclepias cornutii (Asclepiadaceae)

OBJECTIVE QUESTIONS 1. A pollen grain is— (A) First a haploid and then a diploid structure (B) First a diploid and then a haploid structure (C) A haploid structure (D) A diploid structure

C.S.V. / November / 2009 / 1142

2. If a stigma be pollinated by the pollen of the same flower, it is called— (A) Geitonogamy (B) Xenogamy (C) Allogamy (D) Autogamy 3. The process of transfer of pollens on the stigma of another

flower on another plant of the same species, is called— (A) Xenogamy (B) Autogamy (C) Geitonogamy (D) All of the above

(Continued on Page 1152)

(C) Reproductive isolation of two groups of organisms (D) The geographical distribution of two groups of organisms 1. Match Column A (Different theories) with Column B (Scientists/discoverers) then select the correct answer from the options given below— Column A (a) Apical cell theory (b) Histogen theory (c) Tunica corpus theory (d) Stomatal opening theory Column B 1. Schmidt 2. Fujino 3. Hofmeister 4. Hanstein (a) (b) (c) (d) (A) 3 4 2 1 (B) 4 3 2 1 (C) 3 4 1 2 (D) 4 2 3 1 2. Mechanisms that control the rate of mitosis operate at the— (A) Molecular level (B) Cellular level (C) Tissue level (D) All of the above 3. Substrate-level phosphorylation occurs in— (A) EMP-pathway and ETS (B) EMP-pahtway and Krebs cycle (C) Krebs cycle and ETS (D) ETS and transition reaction 4. Which of the following is capable of producing itself during the cell division ? (A) Protein molecule (B) Fat molecule (C) DNA molecule (D) All of the above 5. The ‘Origin of Species’ of Darwin was published in the year— (A) 1859 (B) 1959 (C) 1791 (D) 1900 6. The rate of absorption of ion is slow at high temperature. It is due to— (A) Enzyme inactivation (B) Low transpiration rate (C) Low photosynthetic rate (D) All of the above

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7. Commensalism is— (A) When both weaker partners are benefited (B) When both partners are benefited (C) When both partners are harmed (D) None of the above 8. Ornithophily refers to the pollination by— (A) Wind (B) Water (C) Insects (D) Birds 9. In tissue culture, the callus can be induced to form shoot or root by adjusting the ratio of— (A) Gibberellin to cytokinin (B) Gibberellin to ethylene (C) Auxin to cytokinin (D) Auxin to gibberellin 10. Generally root-cap is absent in— (A) Lithophytes (B) Hydrophytes (C) Mesophytes (D) Xerophytes 11. The amount by which the water potential is reduced as a result of the presence of solute, is termed as— (A) Solute potential (B) Pressure potential (C) Matric potential (D) Osmotic pressure 12. In Funaria , reduction division occurs in— (A) Zygotic cells (B) Antheridial cells (C) Spore mother cells (D) Archegonial cells 13. A fruit dehiscing by both the sutures and derived from monocarpellary ovary with marginal placenta is— (A) Legume (B) Berry (C) Follicle (D) Siliqua 14. The biological definition of a species depends on— (A) Differences in the adaptations of two groups of organisms (B) Anatomical and developmental differences between two groups of organisms

15. The transfer of genetic material from one cell to another by a phage is termed— (A) Transduction (B) Transformation (C) Hybridization (D) Conjugation 16. The production of ATP by utilizing the energy released when H+ flows through an ATP synthetase complex in mitochondria and chloroplast is called— (A) ETS (B) Chemosynthesis (C) Feedback inhibition (D) Chemiosmotic phosphorylation 17. Which of the following levels of organization emphasizes the structural aspects in terms of relation and requirements of organisms ? (A) Autoecological level (B) Synecological level (C) Both (A) and (B) (D) None of the above 18. Enlargement of a tissue or organ due to an increase in the size of its cells or fibres is termed as— (A) Incompressibility (B) Hypertrophy (C) Shearing stress (D) Cytokinesis 19. The upper cortex in heteromerous lichens is precisely made up of— (A) Plectenchyma (B) Prosoenchyma (C) Prosplectenchyma (D) Parenchyma 20. Disk-shaped region of root apex containing slowly dividing cells is called— (A) Quiscent centre (B) Meristematic zone (C) Elongation zone (D) Tent pole 21. Crop-rotation is carried out for— (A) Increasing soil fertility (B) Decreasing soil fertility (C) Increasing soil acidity (D) All of the above

22. Which of the following is present at the chalazal end of embryosac ? (A) Egg apparatus (B) Egg cell (C) Antipodal cell (D) Synergids 23. The major cause of aneuploidy production is— (A) Mitosis and meiosis (B) Homologous chromosome (C) Aberrant chromosome at meiosis (D) Aberrant chromosome at mitosis 24. Stem rot of Jute is caused by— (A) Viruses (B) Bacteria (C) Fungi (D) The deficiency of cobalt and magnesium 25. The transformation from a single cell into an adult individual with many different kinds of cells is called— (A) Inheritance (B) Development (C) Adaptation (D) Evolution 26. Meiosis is significant because— (A) It restores the original number of chromosomes (B) It occurs only in somatic cells (C) It produces identical cells (D) There is doubling of DNA content in the cell 27. Mendel started his experiments with pure strains of peas. A pure strain was developed by removing all— (A) Male plants (B) Female plants (C) Weak plants (D) A typical plants in each generation 28. The flagellum of bacterium is driven out by— (A) ATP (B) Enzyme (C) Protonmotive force (D) All of these 29. Cyanobacteria, unlike other types of bacteria that photosynthesize, do— (A) Not have a cell wall (B) Give off oxygen

C.S.V. / November / 2009 / 1144

(C) Not have chlorophyll (D) Not give off oxygen 30. What percentage of usable radiant energy entering a reaction site of photosynthesis is converted to potential energy ? (A) 10% (B) 20% (C) 35% (D) 42% 31. In flowering plant a megaspore develops into an embryo sac containing— (A) 6 cells, one of which is an egg (B) 4 cells, one of which is an egg (C) 8 cells, one of which is an egg (D) None of the above 32. Who called a cell a ‘peculiar little organism’ ? (A) Schleiden (B) Schwann (C) Leeuwenhoek (D) Robert Hook 33. A change in the relative abundance of an allele (the allelic frequency) within a population, over a succession of generations, is called— (A) Macroevolution or speciation (B) Polygenetic evolution (C) Microevolution or adaptive evolution (D) Coevolution 34. Which one of the following plants does not belong to family papillionaceae ? (A) Crotolaria juncea (B) Vigna sinensis (C) Phaseolus mungo (D) Bauhinia variegata 35. Which of the following factors directly influence the rate of transpiration from the leaves of a flowering plant ? (A) Humidity (B) Light (C) Temperature (D) All of the above 36. Chromosome appears longer during— (A) Leptotene (B) Zygotene (C) Diplotene (D) Pachytene

37. The cellular proteins called enzymes are organic catalysts that speed up reactions— (A) Within the cells (B) Outside the cells (C) Both (A) and (B) (D) None of the above 38. Which one of the following life cycles is associated with Mucor ? (A) Heteromorphic (B) Isomorphic (C) Halontic (D) Diplontic 39. Polygenic inheritance occurs when— (A) Many genes control many phenotypic traits (B) A single gene controls a single phenotypic trait (C) Both (A) and (B) (D) Several separate genes control a single phenotypic trait 40. Which one of the following vitamins does not act as a coenzyme ? (A) Tocopherol (B) Biotin (C) Folic acid (D) Riboflavin 41. Ecotone is characterised by— (A) Forest ecosystem (B) Transition zone between two vegetational types (C) Zone of transition between water and land sometimes hydroponics (D) Terrestrial ecosystem 42. Diffusion of dissolved substances through semipermeable membrane is called— (A) Osmosis (B) Active transport (C) Facilitated diffusion (D) Dialysis 43. Who among the following provided evidence for the synthesis of a small amount of DNA during zygotene and pachytene in Trillium ? (A) Moses (B) Morgan (C) Darlington (D) Stern 44. Out of the widely known systems of classification one remains less phylogenetic and more natural. That is of— (A) Bentham and Hooker (B) Rendle (C) Hutchinson (D) Engler and Prantl

45. Which is an obligate stem parasite ? (A) Monotropa (B) Orobanche (C) Cuscuta (D) None of the above 46. All algae have pigments— (A) Chlorophyll-a and carotenes (B) Chlorophyll-a and chlorophyll-b (C) Chlorophyll-b and carotenes (D) Carotenes and phycobilins 47. Molecules like amino acids which contain both an acid and a basic part are described as— (A) Peptide bond (B) Amphoteric (C) Hydrophobic interaction (D) None of the above

.

48. For a low temperature promotion of flowering in plants, who among the following used the term vernalization ? (A) G. Melchers (B) Nawaschin (C) E. Strassburger (D) T. D. Lysenko 49. A biosensor uses a biological material like— (A) A cell (B) An enzyme (C) An anitbody (D) All of the above 50. The change involving inversion or twisting of xylem strands from one type of structure to another is referred to as— (A) Transition region (B) Vascular strand (C) Transitional point (D) None of the above

ANSWERS WITH HINTS

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C.S.V. / November / 2009 / 1145

(C) Spontaneous differences in males and females (D) Geographic isolation

1. In DNA the two chains of double helix run opposite to each other. This head to tail arrangement is called— (A) Antiparallelism (B) Semiconservatism (C) Autoconservatism (D) Alternation 2. The cellulose of cell wall has empirical formula— (A) (C6H12O6)n (B) (C4H10O5)n (C) (C6H10O4)n (D) None of the above 3. Speciation depends upon— (A) Natural selection (B) Genetic drift (C) Mutation (D) All of the above 4. From which of the following algae iodine is obtained ? (A) Chara (B) Spirogyra (C) Laminaria (D) Polysiphonia 5. Algae are described to possess heteroecious habit because they bear— (A) Prostrate and projecting system developed to a variable length (B) Short and long hair (C) Short nodal and long internodal cells (D) Short and long lateral branches 6. Latex vessels are found in— (A) Papaver (B) Carica (C) Argemone (D) All of the above 7. The difference between NAD+ and NADP + is— (A) Only NAD+ production requires niacin in the diet (B) One contains high-energy phosphate bonds and the other does not (C) One carries electrons to ETS and the other carries them to synthetic reactions (D) All of the above are correct

C.S.V. / November / 2009 / 1146

8. Functional enzymes inside the cells are called— (A) Endoenzymes (B) Apoenzymes (C) Exoenzymes (D) Isoenzymes 9. Cytochromes are found in— (A) Outer wall of mitochondria (B) Lysosomes (C) Matrix of mitochondria (D) Cristae of mitochondria 10. The term ‘osmosis’ refers to the diffusion of— (A) Positive electric charge (B) Glucose (C) Water (D) Energy 11. Fruits and milk are dried for preservation at room temperature by the process of— (A) Pasteurization (B) Dehydration (C) Vernalization (D) All of the above 12. All eukaryotic genes contain two kinds of base sequences, which of the following do not play role in protein synthesis ? (A) Exons (B) Introns (C) Both (A) and (B) (D) None of the above 13. Desired improved varieties of economically useful crops are raised by— (A) Mutation (B) Biomagnification (C) Natural selection (D) Hybridization 14. Arrangement of leaves on main stem or branches is known as— (A) Inflorescence (B) Phyllotaxy (C) Phylogeny (D) Phytogeography 15. Allopatric but not sympatric speciation requires— (A) Prior hybridization (B) Reproductive isolation

16. While small to medium sized molecules cross the plasma membrane by simple diffusion or through transport proteins, many large molecules have to cross by different technique called— (A) Exocytosis (B) Endocytosis (C) Both (A) and (B) (D) None of the above 17. Which of the following statements is the chief objection to the ‘atmospheric pressure theory’ regarding ascent of sap ? (A) Pressure of the free surface at the lower end of plant is required for the atmospheric pressure to operate which is not found in roots (B) The atmospheric pressure can raise water only upto 34 feet (C) Both (A) and (B) (D) None of the above 18. Food production is most likely to be improved in the near future by the development of new— (A) Pesticides (B) Energy sources for farm (C) Plants by genetic engineering (D) Plant by cross-breeding 19. Genes which are confined to differential region of Y-chromosome only are called— (A) Autosomal (B) Mutant (C) Holandric

(D) Barr body

20. When the temperature of the soil becomes zero, then— (A) Water absorption decreases (B) The soil will loose all capillary water (C) The water absorption increases (D) The water absorption is not effected by temperature 21. ‘Geranium oil’ is extracted from— (A) Cymbopogon nardus (B) Jasminum officinale (C) Lavandula officinalis (D) None of the above

22. Kranz anatomy is found in— (A) C3 plants (B) C4 plants (C) Succulent plants (D) All of the above 23. Which of the following is not a characteristic of true roots ? (A) They do not possess root cap over the apex (B) They are not susceptible to the influence of light (C) They possess endogenous origin and branching (D) All of the above 24. ‘Dichasial cyme’ inflorescence is found in— (A) Clerodendron (B) Hamerocallis (C) Ranunculus (D) All of the above 25. Which of the following classes of fungi had for long time been known as ‘algal-fungi’ ? (A) Ascomycetes (B) Phycomycetes (C) Basidiomycetes (D) Deuteromycetes 26. Which of the following endoplasmic reticulum lacks ribosomes ? (A) Smooth endoplasmic reticulum (B) Rough endoplasmic reticulum (C) Both (A) and (B) (D) None of the above 27. Which of the following is a specialized organ of sporophyte for attachment to the gametophyte ? (A) Foot (B) Stalk (C) Apophysis (D) Root

30. The crystalline form(s) of plant globulin(s) is/are— (A) Edestin (B) Canavalin (C) Amandin (D) All of the above 31. A nucleotide consists of— (A) A phosphate (B) A nitrogenous base (C) A sugar (D) All of the above 32. A dioecious plant has— (A) Both X- and Y-chromosomes (B) Two X-chromosomes (C) Both sexes in the same individual (D) The two sexes in separate individuals 33. The part of prokaryotic cell where the DNA is located is called— (A) Plasmid (B) Nucleosome (C) Nucleoid (D) Primary constriction 34. The movement of water from one cell to another cell through the plasmodesmata is called— (A) Symplast pathway (B) Apoplast pathway (C) Both (A) and (B) (D) Transmembrane pathway 35. Name the weed belonging to family compositae, which is a burning problem in almost all the states of India— (A) Parthenium (B) Nicotiana (C) Barleria (D) Barley 36. A green manure— (A) Protects soil against erosion and leaching (B) Supplies organic matter (C) Supplies additional nitrogen (D) All of the above are correct

28. The correct name of slow viruses is— (A) Introns (B) Cistrons (C) Prions (D) Exons

37. The core particle of nucleosome is composed of how many histones ? (A) 2 (B) 3 (C) 4 (D) 5

29. Which of the following groups of algae produces the ‘carrageenin’ ? (A) Brown algae (B) Red algae (C) Blue green algae (D) Green algae

38. The continuous change of a character within an evolving lineage is termed as— (A) Evolutionary trend (B) Systematics (C) Cladistics (D) Physiognomy

C.S.V. / November / 2009 / 1147

39. According to ‘Unit membrane model’, the thickness of the cell membrane is about— (A) 7·5 nm (B) 150 nm (C) 1·0 nm (D) 2·5 nm 40. Fucus is a/an— (A) Alga (B) Bryophyte (C) Pteridophyte (D) Gymnosperm 41. Among the following microorganisms, the smallest one is— (A) Virus (B) Toxoplasma (C) Mycoplasma (D) Rickettsiae 42. A plant requires potassium for— (A) Synthesizing chlorophyll (B) Opening and closing of stomata (C) Holding its cells together (D) Synthesizing proteins 43. The membrane-bound enzyme involved in Krebs cycle in mitochondrial inner membrane is— (A) Cisaconitase (B) Malate dehydrogenase (C) Fumarase (D) Succinate dehydrogenase 44. Which of the following is not an insectivorous plant ? (A) Orobanche (B) Utricularia (C) Drosera (D) All of the above 45. The ‘trap centre’ of radiant energy in photosystem-I is— (A) P700 (B) P800 (C) P880 (D) Carotene 46. Which of the following statements show(s) resemblances of Ginkgo with the cycadales ? (A) The embryo has two cotyledons (B) The embryo is endoscopic (C) The germination is hypogeal and cotyledons remain embedded between the endoderm tissue (D) All of the above 47. The ferns, in which the entire sporangium develops from a single superficial cell of the sporophyll is known as— (A) Unisporangiate (B) Leptosporangiate

(C) Mesosporangiate (D) Eusporangiate 48. Dry climates typically occur at— (A) 30° latitude (B) 60° latitude (C) 90° latitude (D) 0° latitude 49. Which of the following statements is correct regarding the origin of long shoots (rhizomes) of Pteridium aquilinum ? (A) They grow obliquely upwards and then run horizontally a little below (B) After their origin from parent axis they grow obliquely upwards for some distance and then run horizontally (C) They arise from the parent axis and grow in the same direction, thus, penetrating deep in the soil (D) All of the above are corrects 50. One cannot age a tree by its rings if that tree is located in— (A) Tropical evergreen forest (B) Temperate evergreen forest (C) Tropical deciduous forest (D) Temperate deciduous forest

ANSWERS WITH HINTS

●●●

C.S.V. / November / 2009 / 1148

(C) Sterile soil is converted into fertile soil (D) Top fertile soil is protected against the loss

1. In flowering plants, the diploid generation is called— (A) Gametophyte (B) Sporophyte (C) Both A and B (D) Embryony 2. Match column A (common name of plant) with column B (botanical name) then select the correct answer from the options given below : Column A (a) Fever nut

Column B 1. Caesalpinia crista (b) Life plant 2. Calotropis procera (c) Swallow 3. Bryophyllum wart pinnatum (d) Indian 4. Centella pennywort asiatica (A) (B) (C) (D)

(a)

(b)

(c)

(d)

3 3 2 1

1 1 1 3

2 4 3 2

4 2 4 4

3. The central dogma of molecular biology states that— (A) Translation proceeds transcription (B) DNA is a template only for DNA replication (C) DNA is a template for all RNA production (D) All of the above 4. Which of the following is not synthesized by DNA ? (A) m-RNA synthesis (B) Another DNA synthesis (C) Protein synthesis (D) All of the above 5. When nucellus and integuments lie more or less at right angles to the funiculus, this type of ovule is known as— (A) Campylotropous (B) Hemitropous (C) Anatropous (D) Orthotropous

C.S.V. / November / 2009 / 1149

6. The crystals of CaCO3 which appear like a bunch of grapes in the epidermal cells of leaves of certain plants are termed as— (A) Raphides (B) Cystoliths (C) Sphaerophides (D) Otoliths 7. When a pollen grain matures, it contains a nonflagellated sperm cells, which travel by way of pollen tube to the megagametophyte, which is called— (A) Nucellus

(B) Endosperm

(C) Embryosac (D) Cotyledon 8. In which of the following divisions, mushrooms are placed ? (A) Basidiomycota (B) Ascomycota (C) Zygomycota (D) Deuteromycota 9. The plasma membrane is followed by the colloidal fluid called— (A) Matrix (B) Cytosol (C) Both (A) and (B) (D) Endoplasm 10. In plants enzymes are present in— (A) All living cells (B) Leaves only (C) Flowers only (D) Storage organs only 11. The rate of storage of organic matter in plant tissues in excess of the respiratory utilization by plants during measurement period is known as— (A) Net productivity (B) Net primary productivity (C) Secondary productivity (D) Gross primary productivity 12. Soil conservation is the process whereby— (A) Soil erosion is allowed (B) Soil aeration system is well defined

13. The light-independent reactions are the— (A) First stage of photosynthesis (B) Second stage of photosynthesis (C) Both (A) and (B) (D) None of the above 14. About what percentage of the solar energy received by the earth is absorbed by the pigments and is utilized in photosynthesis ? (A) 75% (B) 1% (C) 99% (D) 90% 15. We know of about seventy gibberellins that chemically differ only slightly. The most common of them is— (A) GA1 (B) GA2 (C) GA3 (D) GA4 16. The thread-like body of Nostoc filament without mucilage sheath is known as— (A) Trichome (B) Hyphae (C) Mycelium (D) Colony 17. The portions of DNA that do not bind to m RNA are called— (A) Exons (B) Introns (C) RNA polymerase (D) DNA polymerase 18. Which one of the following is commonly called fallen star ? (A) Pinus (B) Riccia (C) Nostoc (D) Spirogyra 19. Which of the following phase is known as synthesis phage ? (A) G1-phase (B) G2-phase (C) S-phase (D) M-phase 20. Which is responsible for guttation ? (A) Transpiration (B) Osmosis (C) Photorespiration (D) Root pressure 21. Vegetative propagation is accomplished by use of— (A) Apomictic seed (B) Layers and cutting (C) Grafting and budding (D) All of the above

22. Fertilization in which male gametes are carried through pollen tube is known as— (A) Chalazogamy (B) Siphonogamy (C) Syngamy (D) Porogamy 23. Splitting of one cell into two daughter cells serving as asexual form of reproduction in bacteria is termed as— (A) Binary fission (B) Transduction (C) Endogamy (D) All of the above 24. Chilgoza is a fruit obtained from a gymnosperm which is— (A) Cycas revoluta (B) Abies balasamiana (C) Pinus gerardiana (D) Pinus roxburghii 25. m-RNA synthesis in bacteria is initiated at— (A) 3' end (B) 5' end (C) Both (A) and (B) (D) None of the above 26. Zigzag development of inflorescence axis is an example of— (A) Scorpoid cyme (B) Umbel (C) Verticillaster (D) Helicoid cyme 27. Phaseo toxin is produced by— (A) Red algae (B) Brown algae (C) Phaseolus vulgaris (D) None of the above 28. Match column A (Name of fruit) with column B (edible part) then select the correct answer from the options given below— Column A (Name of fruit) (a) Rice (b) Pea

Column B (Edible parts) 1. Seed 2. Epicarp and mesocarp (c) Peach 3. Endosperm and embryo (d) Lady's finger 4. Entire fruit (a) (b) (c) (d) (A) 3 1 4 2 (B) 3 1 2 4 (C) 2 3 1 4 (D) 2 3 4 1

C.S.V. / November / 2009 / 1150

29. Match Column A (Common english name) with Column B (Botanical name), then select the correct option as your answer from the options given below— Column B (Botanical name)

Column A (Common name) (a) Hemp

1. Musa textilis

(b) Flax

2. Cannabis sativus

(c) Manila hemp

3. Crotolaria juncea

(d) Sunhemp 4. Linum usitatissimum (A)

(a)

(b)

(c)

(d)

1

2

4

3

(B)

1

2

3

4

(C)

2

4

1

3

(D)

2

4

3

1

30. The tissue of the leaf that lies between the upper and lower epidermis and between the veins in dicotyledonous leaf consists of typically thin walled parenchyma is known as— (A) Ground tissue (B) Mesophyll (C) Plasmodesmata (D) Endodermis 31. Which of the following scientists placed gymnosperms between monocots and dicots as third taxon ? (A) Englar—Prantl (B) Bentham—Hooker (C) Hutchinson (D) Copeland 32. The achene fruit develops from— (A) Monocarpellary inferior ovary (B) Monocarpellary ovary

superior

(C) Bicarpellary inferior ovary (D) Bicarpellary superior ovary 33. The RNA which possesses the capacity to combine specifically with only one amino acid in a reaction mediated by a set of amino acid specific enzymes called— (A) RNA polymerase (B) DNA polymerase

(C) Amino-acyl t-RNA synthetase (D) None of the above 34. Which of the following is used by CAM plants to carry on carbon dioxide fixation at night ? (A) RuBP carboxylase (B) Carboxydismutase (C) PEPCase (D) All of the above 35. Evolution of the DNA →RNA→ Protein system was a milestone because the protocell— (A) Needed energy to flow (B) Could now reproduce (C) Was a heterotrophic fermenter (D) All of the above 36. The period of DNA synthesis when replication occurs is termed as— (A) M-phase (B) S-phase (C) G1-phase (D) G2-phase 37. In eukaryotes, DNA is associated with— (A) Proteins (B) Carbohydrates (C) Lipids (D) Both (B) and (C) 38. Guttation occurs only through— (A) Hydathodes (B) Lenticels (C) Stomata (D) Cuticle 39. In eukaryotes, control of gene expression occurs at— (A) One level only (B) Two levels (C) Three levels (D) Four levels 40. The function of tracheid is— (A) Conduction of water (B) Mechanical support (C) Both (A) and (B) (D) To form abscission layer 41. Which of the following functions during reproduction in plant ? (A) Root (B) Stem (C) Both (A) and (B) (D) Flower 42. Enzymes having same function but different molecular configuration are called— (A) Apoenzymes (B) Coenzymes

(C) Inducible enzymes (D) Isoenzymes 43. A collection of simple fruits or fruitlets which develop from the multicarpellary, apocarpous ovary (free carpels), is known as— (A) Multiple fruit (B) Aggregate fruit (C) False fruit (D) All of the above 44. Under more distressing environmental conditions the aplanospores of algae secrete thicker walls around them which are called— (A) Hypnospores (B) Akinetes (C) Planospores (D) Pseudospores 45. Which one of following plants have curved hooks for dispersal ? (A) Guava (B) Brinjal (C) Martynia (D) Fig 46. Of the 36 ATP molecules that are produced during the complete breakdown of glucose, most are due to the action of— (A) Chemiosmotic phosphorylation (B) Electron transport system (ETS) (C) Both (A) and (B) (D) Substrate-level phosphorylation 47. Match Column A (fruit type) with Column B (examples) and select the correct answer from the options given below— Column A (Fruit type) (a) Drupe (b) Berry (c) Pepo (d) Double samara (a) (A) 1 (B) 2 (C) 3 (D) 3

Column B (Examples) 1. Brinjal 2. Cucumber 3. Coconut 4. Acer (b) 2 1 1 2

(c) 3 3 2 1

(d) 4 4 4 4

48. Cauliflory is the production of flowers— (A) In clusters (B) From epiphyllous buds

C.S.V. / November / 2009 / 1151

(C) On old stem from dormant buds (D) On young branches 49. Lysosomes are produced by— (A) Chloroplasts (B) Ribosomes (C) Golgi apparatus (D) Mitochondria 50. Photosynthesis in CAM plant is minimal because of the limited amount of— (A) CO2 fixed at night (B) CO2 fixed in daylight (C) O2 fixed at night (D) O2 fixed in daylight

ANSWERS WITH HINTS

5. Pollination occurs when a pollen grain— (A) Releases its sperm nuclei (B) Releases its sperm nuclei and they fertilize the egg and polar nuclei (C) Matures and has three nuclei (D) Lands on stigma 6. A flower with both stamens and carpels is said to be— (A) Perfect (B) Incomplete and cyclic (C) Asexual (D) Dioecious 7. The wind-pollinated flowers are called— (A) Entomophilous (B) Anemophilous (C) Malacophilous (D) Hypophilous 8. Incomplete dichogamy is found in— (A) Ixora (B) Argemone (C) Mirabilis (D) Viola

ANSWERS

●●●

Read UPKARS ReadUPKAR’S UPKAR’S

(Through English-Hindi Medium)

●●●

(Continued from Page 1142) 4. Which of the following is/are the advantage(s) of self-pollination ? (A) Newly born plants have the same genetic factors (B) It gives a better result (C) Not dependent on any agents (D) All of the above

C.S.V. / November / 2009 / 1152

By : Dr. B. B. Jain Code No. 1589 Rs. 110/-

UPKAR PRAKASHAN, AGRA-2 ● E-mail : [email protected] ● Website : www.upkar.in

(C) [Cr (NH3)6] 3 +

Solved Paper

CBSE Medical Entrance Exam., 2009 1. Which of the following molecules acts as a Lewis acid ? (A) (CH3)3 B (B) (CH3)2 O (C) (CH3)3 P

(D) (CH3)3 N

2. Which of the following reactions is an example of nucleophilic substitution reaction ? (A) RX + KOH → ROH + KX (B) 2RX + 2Na → R – R + 2NaX (C) RX + H2 → RH + HX (D) RX + Mg → RMgX 3. From the following bond energies : H—H bond energy : 431·37 kJ mol– 1 C== C bond energy : 606·10 kJ mol– 1 C—C bond energy : 336·49 kJ mol– 1 C—H bond energy : 410·50 kJ mol– 1 Enthalpy for the reaction, H H H H | | | | C== C + H – H → H—C—C—H | | | | H H H H will be— (A) 1523·6 kJ mol– 1 (B) – 243·6 kJ mol – 1 (C) – 120·0 kJ mol – 1 (D) 553·0 kJ mol– 1 4. Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states ? (A) 3d 34s 2 (B) 3d 54s 1 5 2 (C) 3d 4s (D) 3d 24s 2 5. The ionization constant of ammonium hydroxide is 1·77 × 10– 5 at 298 K. Hydrolysis constant of ammonium chloride is— (A) 5·65 × 10– 10 (B) 6·50 × 10– 12 (C) 5·65 × 10– 13 (D) 5·65 × 10– 12 6. Which of the following oxides is not expected to react with sodium hydroxide ? (A) B2O3 (B) CaO (C) SiO2 (D) BeO

C.S.V. / November / 2009 / 1153

7. Which of the following does not show optical isomerism ? (A) [CO (en)2Cl2] + (B) [CO (NH3)3Cl3] 0 (C) [CO (en)Cl2(NH3)2)]+ (D) [CO (en)3] 3 + (en = ethylenediamine) 8. Which one of the following is employed as a tranquilizer ? (A) Equanil (B) Naproxen (C) Tetracycline (D) Chlorpheninamine 9. Al2O3 is reduced by electrolysis at low potentials and high currents. If 4·0 × 104 ampere of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced ? (Assume 100% current efficiency, At. mass of Al = 27g mol– 1) (A) 9·0 × 103g (B) 8·1 × 104g (C) 2·4 × 105g (D) 1·3 × 104g 10. Trichloroacetaldehyde, CCl 3CHO reacts with chlorobenzene in presence of sulphuric acid and produces— Cl | —C— —Cl (A) Cl— | CH2Cl Cl |

(B) Cl—

(C) Cl—

(D) Cl—

| — C— | H OH | —C— | Cl — CH— | CCl3

—Cl

—Cl

—Cl

11. Which of the following complex ions is expected to absorb visible light ? (A) [Sc (H2O)3 (NH3)3] 3 + (B) [Ti (en)2 (NH3)2] 4 +

(D) [Zn (NH3)6] 2 + (At. no. Zn = 30, Sc = 21, Ti = 22, Cr = 24) 12. Half life period of a first-order reaction is 1386 second. The specific rate constant of the reaction is— (A) 5·0 × 10– 3 s– 1 (B) 0·5 × 10– 2 s– 1 (C) 0·5 × 10– 3 s– 1 (D) 5·0 × 10– 2 s– 1 13. Consider the following reaction : Zn

CH Cl

dust

Anhydrous AlCl 3

3 Y Phenol ⎯⎯→ X ⎯⎯⎯⎯⎯⎯→

Alkaline KMnO 4 ⎯⎯⎯⎯⎯⎯⎯→ Z, the product Z is— (A) Toluene (B) Benzaldehyde (C) Benzoic acid (D) Benzene 14. Copper crystallises in a facecentred cubic lattice with a unit cell length of 361 pm. What is the radius of copper atom in pm ? (A) 128 (B) 157 (C) 181 (D) 108 15. For the reaction A + B ⎯→ Products, it is observed that : (1) on doubling the initial concentration of A only, the rate of reaction is also doubled and (2) on doubling the initial concentrations of both A and B, there is a change by a factor of 8 in the rate of the reaction. The rate of this reaction is given by— (A) Rate = k [A]2 [B] (B) Rate = k [A] [B] 2 (C) Rate = k [A]2 [B] 2 (D) Rate = k [A] [B] 16. According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order ? –

2–

(A) N2 < N2 < N2 2–

–

(B) N2 < N2 < N2 2–

–

(C) N2 < N2 < N2 –

2–

(D) N2 < N2 < N2

17. The state of hybridization of C2, C3, C5 and C6 of the hydrocarbon, CH3 CH3 | | CH3—C —CH==CH—CH —C ≡ CH 6| 5 7 4 2 1 3 CH3

23. In the case of alkali metals, the covalent character decreases in the order— (A) MCl > MI > MBr > MF (B) MF > MCl > MBr > MI (C) MF > MCl > MI > MBr (D) MI > MBr > MCl > MF

30. Lithium metal crystallises in a body centred cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of the lithium will be— (A) 240·8 pm (B) 151·8 pm (C) 75·5 pm (D) 300·5 pm

is in the following sequence— (A) sp , sp3, sp 2 and sp 3 (B) sp 3, sp 2, sp 2 and sp (C) sp , sp2, sp 2 and sp 3 (D) sp , sp2, sp 3 and sp 2

24. Given : (1) Cu 2 + + 2e– → Cu; E0 = 0·337 V (2) Cu 2 + + e– → Cu+; E0 = 0·153 V Electrode potential, E0 for the reaction, Cu+ + e– ⎯→ Cu, will be— (A) 0·52 V (B) 0·90 V (C) 0·30 V (D) 0·38 V

31. The segment of DNA which acts as the instrumental manual for the synthesis of the protein is— (A) Nucleotide (B) Ribose (C) Gene (D) Nucleoside

18. Among the following which is the strongest oxidising agent ? (A) F2 (B) Br2 (C) I 2

(D) Cl2

19. The equivalent conductance of M solution of a weak monobasic 32 acid is 8·0 mhos cm2 and at infinite dilution is 400 mhos cm2. The dissociation constant of this acid is— (A) 1·25 × 10– 5 (B) 1·25 × 10– 6 (C) 6·25 ×

10– 4

(D) 1·25 × 10– 4 20. Structure of some common polymers are given. Which one is not correctly presented ? (A) Teflon –CF ( 2 – CF2 –)n (B) Neoprene –CH2– C == CH–CH2–CH2– | Cl (C) Terylene –OC– (

–COOCH2–CH2–O–)n

3–

21. Oxidation numbers of P in PO4 , 2 – of S in SO4 and that of Cr in 2– Cr2 O 7 are respectively— + 5, + 6 and + 6 + 3, + 6 and + 5 + 5, + 3 and + 6 – 3, + 6 and + 6

22. The IUPAC name of the compound having the formula CH ≡ C—CH = CH 2 is— (A) (B) (C) (D)

3-butene-1-yne 1-butyn-3-ene but-1-yne-3-ene 1-butene-3-yne

C.S.V. / November / 2009 / 1154

26. What is the [OH– ] in the final solution prepared by mixing 20·0 mL of 0·050 M HCl with 30·0 mL of 0·10 M Ba(OH) 2 ? (A) 0·10 M (C) 0·0050 M 27.

n

(D) Nylon-66 –NH(CH [ 2)6NHCO(CH2)4–CO–]2

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

25. For the reaction, N2 + 3H2 ⎯→ 2NH3 d [NH3] if = 2 × 10– 4 mol L– 1 s– 1 dt – d [H2] the value of would be— dt (A) 3 × 10– 4 mol L– 1 s– 1 (B) 4 × 10– 4 mol L– 1 s– 1 (C) 6 × 10– 4 mol L– 1 s– 1 (D) 1 × 10– 4 mol L– 1 s– 1

(B) 0·40 M (D) 0·12 M

2– 3– Out of TiF6 , COF 6 , Cu2Cl2 2– and NiCl4 (Z of Ti = 22, CO = 27,

Cu = 29, Ni = 28) the colourless species are— 2–

3–

(A) TiF6 and COF6

2–

(B) Cu 2Cl2 and NiCl4 (C) (D)

2– TiF6 and Cu 2Cl2 3– 2– COF6 and NiCl4

28. Amongst the elements with following electronic configurations, which one of them may have the highest ionization energy ? (A) Ne [3 s23p3] (B) Ne [3 s23p2] (C) Ar [3d104s 24p3] (D) Ne [3 s23p1] 29. Maximum number of electrons in a subshell of an atom is determined by the following— (A) 4l + 2 (B) 2l + 1 (C) 4l – 2 (D) 2n2

32. Predict the product— NHCH3

(A)

(B)

+ NaNO2+ HCl ⎯→ Product CH3 | N—N=O CH3 | N—NO2

NHCH3 (C)

NO +

(D)

OH | N—CH3

NHCH 3

NO

33. Which of the following compounds will exhibit cis-trans (geometrical) isomerism ? (A) 2-butene (B) Butanol (C) 2-butyne (D) 2-butenol 34. The values of ΔH and ΔS for the reaction, C(graphite) + CO2(g) → 2CO (g) are 170 kJ and 170 JK– 1, respectively. This reaction will be spontaneous at— (A) 710 K (B) 910 K (C) 1110 K (D) 510 K 35. H2COH·CH2OH on heating with periodic acid gives— (A) 2 CO2 (B) 2 HCOOH CHO H C=O (C) | (D) 2 CHO H 36. The dissociation constants for acetic acid and HCN at 25°C are 1·5 × 10– 5 and 4·5 × 10– 10, res-

pectively. The equilibrium constant for the equilibrium CN– + CH3COOH HCN +

CH3COO–

would be— (A) 3·0 × 105

(B) 3·0 × 10– 5

(C) 3·0 × 10– 4 (D) 3·0 × 104 37. Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO3 and conc. H2SO4. In the mixture, nitric acid acts as a/an— (A) Reducing agent (B) Acid (C) Base (D) Catalyst 38. In the reaction – – BrO3 (aq) + 5Br(aq) + 6H+ → 3Br2 (l) + 3H2O(l) The rate of appearance of bromine (Br2) is related to rate of disappearance of bromide ions as following— d (Br2) 3 d (Br–) (A) =– 5 dt dt d (Br2) 5 d (Br–) (B) =– 3 dt dt d (Br2) 5 d (Br–) (C) = 3 dt dt d (Br2) 3 d (Br–) (D) = 5 dt dt 39. In which of the following mole– – cules/ions BF3, NO2, NH 2 and H2O, the central atom is sp2 hybridized ? – – (A) NO2 and NH2 –

(B) NH2 and H 2O –

(C) NO2 and H 2O (D) BF 3 and

– NO2

40. Which of the following hormones contains iodine ? (A) Insulin (B) Testosterone (C) Adrenaline (D) Thyroxine 41. 10g of hydrogen and 64g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be— (A) 2 mol (B) 3 mol (C) 4 mol (D) 1 mol

C.S.V. / November / 2009 / 1155

42. The energy absorbed by each molecule (A2) of a substance is 4·4 × 10 –19 J and bond energy per molecule is 4·0 × 10–19 J. The kinetic energy of the molecule per atom will be— (A) 2·0 × 10– 20 J (B) 2·2 × 10– 19 J (C) 2·0 × 10– 19 J (D) 4·0 × 10– 20 J 43. The straight chain polymer is formed by— (A) Hydrolysis of (CH3)3 SiCl followed by condensation polymerisation (B) Hydrolysis of CH3 SiCl3 followed by condensation polymerisation (C) Hydrolysis of (CH 3)4 Si by addition polymerisation (D) Hydrolysis of (CH3)2 SiCl 2 followed by condensation polymerisation 44. The stability of + 1 oxidation state increases in the sequence— (A) Al < Ga < In < Tl (B) Tl < In < Ga < Al (C) In < Tl < Ga < Al (D) Ga < In < Al < Tl 45. Consider the following reaction, PBr

Alc.KOH

3 X ⎯⎯⎯⎯→ Y ethanol ⎯⎯→ (i) H2SO4 room temperature ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ Z; (ii) H2O, heat the product Z is— (A) CH2=CH2 (B) CH3CH2—O—CH2—CH3 (C) CH3—CH2—O—SO3H (D) CH3CH2OH

46. A 0·0020 m aqueous solution of an ionic compound Co(NH 3)5

48. Benzene reacts with CH3Cl in the presence of anhydrous AlCl 3 to form— (A) Toluene (B) Chlorobenzene (C) Benzylchloride (D) Xylene 49. Which of the following is not permissible arrangement of electrons in an atom ? 1 (A) n = 4, l = 0, m = 0, s = – 2 1 (B) n = 5, l = 3, m = 0, s = + 2 1 (C) n = 3, l = 2, m = – 3, s = – 2 1 (D) n = 3, l = 2, m = – 2, s = – 2 50. Propionic acid with Br2 | P yields a dibromo product. Its structure would be— (A) CH2Br—CHBr—COOH Br | (B) H— C —CH2COOH | Br (C) CH2Br—CH2—CO Br Br | (D) CH3— C —COOH | Br 51. The bacterium Bacillus thuringiensis is widely used in contemporary biology as— (A) Indicator of water pollution (B) Insecticide (C) Agent for production of dairy products (D) Source of industrial enzyme

(NO2)Cl freezes at – 0·00732°C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be— (kf = – 1·86°C/m)

52. Chipko movement was launched for the protection of—

(A) 2 (C) 4

53. A health disorder that results from the deficiency of thyroxine in adults and characterised by (i) a low metabolic rate, (ii) increase in body weight and (iii) tendency to retain water in tissues is—

(B) 3 (D) 1

47. What is the dominant intermolecular force or bond that must be overcome in converting liquid CH3OH to a gas ? (A) Hydrogen bonding (B) Dipole-dipole interaction (C) Covalent bonds (D) London dispersion force

(A) Grasslands (B) Forests (C) Livestock

(D) Wet lands

(A) Hypothyroidism (B) Simple goitre (C) Myxoedema (D) Cretinism

54. Elbow joint is an example of— (A) Pivot joint (B) Hinge joint (C) Gliding joint (D) Ball and socket joint 55. The correct sequence of plants in a hydrosere is— (A) Oak → Lantana → Scirpus → Pistia → Hydrilla → Volvox (B) Volvox → Hydrilla → Pistia → Scirpus → Lantana → Oak (C) Pistia → Volvox → Scirpus → Hydrilla → Oak → Lantana (D) Oak → Lantana → Volvox → Hydrilla → Pistia → Scirpus 56. Mannitol is the stored food in— (A) Chara (B) Porphyra (C) Fucus (D) Gracillaria 57. The epithelial tissue present on the inner surface of bronchioles and fallopian tubes is— (A) Cuboidal (B) Glandular (C) Ciliated (D) Squamous 58. Which one of the following pairs of food components in humans reaches the stomach totally undigested ? (A) Protein and starch (B) Starch and fat (C) Fat and cellulose (D) Starch and cellulose 59. Uric acid is the chief nitrogenous component of the excretory products of— (A) Man (B) Earthworm (C) Cockroach (D) Frog 60. Which one of the following groups of animals is bilaterally symmetrical and triploblastic ? (A) Coelenterates (Cnidarians) (B) Aschelminthes (round worms) (C) Ctenophores (D) Sponges 61. Tiger is not a resident in which one of the following national park ? (A) Ranthambhor (B) Sunderbans (C) Gir (D) Jim Corbett

C.S.V. / November / 2009 / 1156

62. A change in the amount of yolk and its distribution in the egg will affect— (A) Formation of zygote (B) Pattern of cleavage (C) Number of blastomeres produced (D) Fertilization 63. When breast feeding is replaced by less nutritive food low in proteins and calories; the infants below the age of one year are likely to suffer from— (A) Marasmus (B) Rickets (C) Kwashiorkor (D) Pellagra 64. Stroma in the chloroplasts of higher plant contains— (A) Light-independent reaction enzymes (B) Light-dependent reaction enzymes (C) Ribosomes (D) ChlorophyII 65. Which one of the following correctly describes the location of some body parts in the earthworm Pheretima ? (A) Two pairs of accessory glands in 16-18 segments. (B) Four pairs of spermathecae in 4-7 segments. (C) One pair of ovaries attached at intersegmental septum of 14th and 15th segments. (D) Two pairs of testes in 10th and 11th segments. 66. Foetal ejection reflex in human female is induced by— (A) Pressure exerted by amniotic fluid (B) Release of oxytocin from pituitary (C) Fully developed foetus and placenta (D) Differentiation of mammary glands 67. Which one of the following has haplontic life cycle ? (A) Funaria (B) Polytrichum (C) Ustilago (D) Wheat 68. Which of the following plant species you would select for the production of bioethanol ? (A) Brassica (B) Zea mays (C) Pongamia (D) Jatropha

69. Which part of human brain is concerned with the regulation of body temperature ? (A) Medulla Oblongata (B) Cerebellum (C) Cerebrum (D) Hypothalamus 70. Which one of the following is correct pairing of a body part and the kind of muscle tissue that moves it ? (A) Heart wall—Involuntary unstriated muscle (B) Biceps of upper arm —Smooth muscle fibres (C) Abdominal wall —Smooth muscle (D) Iris—Involuntary smooth muscle 71. In the case of peppered moth (Biston betularia ) the blackcoloured form became dominant over the light coloured form in England during industrial revolution. This is an example of— (A) Natural selection whereby the darker forms were selected (B) Appearance of the darker coloured individuals due to very poor sunlight (C) Protective mimicry (D) Inheritance of darker colour character acquired due to the darker environment 72. Oxygenic photosynthesis occurs in— (A) Chromatium (B) Oscillatoria (C) Rhodospirillum (D) Chlorobium 73. The genetic defect-adenosine deaminase (ADA) deficiency may be cured permanently by— (A) Periodic infusion of genetically engineered lymphocytes having functional ADA c DNA (B) Administering adenosine deaminase activators (C) Introducing bone marrow cells producing ADA into cells at early embryonic stages (D) Enzyme replacement therapy

74. The annular and spirally thickened conducting elements generally develop in the protoxylem when the root or stem is— (A) Maturing (B) Elongating (C) Widening (D) Differentiating 75. In barley stem vascular bundles are— (A) Open and scattered (B) Closed and scattered (C) Open and in a ring (D) Closed and radial 76. Sickle cell anemia is— (A) An autosomal linked dominant trait (B) Caused by substitution of valine by glutamic acid in the beta globin chain of haemoglobin (C) Caused by a change in a single base pair of DNA (D) Characterized by elongated sickle like RBCs with a nucleus 77. If a live earthworm is pricked with a needle on its outer surface without damaging its gut, the fluid that comes out is— (A) Excretory fluid (B) Coelomic fluid (C) Haemolymph (D) Slimy mucus 78. There is no DNA in— (A) An enucleated ovum (B) Mature RBCs (C) A mature spermatozoan (D) Hair root 79. Point mutation involves— (A) Insertion (B) Change in single base pair (C) Duplication (D) Deletion 80. Which one of the following has maximum genetic diversity in India ? (A) Teak (B) Mango (C) Wheat (D) Tea 81. In a standard ECG which one of the following alphabets is the correct representation of the respective activity of the human heart ? (A) R-repolarisation of ventricles (B) S-start of systole

C.S.V. / November / 2009 / 1157

(C) T-end of diastole (D) P-depolarisation of the atria 82. Study the pedigree chart given below—

(C) Membranes connecting the nucleus with plasmalemma (D) Connections between adjacent cells 88. Which of the following is a pair of viral diseases ? (A) Ringworm, AIDS (B) Common Cold, AIDS (C) Dysentery, Common Cold (D) Typhoid, Tuberculosis

What does it show ? (A) Inheritance of a sex-linked inborn error of metabolism like phenylketonuria (B) Inheritance of a condition like phenylketonuria as an autosomal recessive trait (C) The pedigree chart is wrong as this is not possible (D) Inheritance of a recessive sex-linked disease like haemophilia 83. Middle lamella is composed mainly of— (A) Hemicellulose (B) Muramic acid (C) Calcium pectate (D) Phosphoglycerides 84. Somaclones are obtained by— (A) Tissue culture (B) Plant breeding (C) Irradiation (D) Genetic engineering 85. Which one of the following plants is monoecious ? (A) Marchantia (B) Pinus (C) Cycas (D) Papaya 86. Which one of the following is the correct matching of three items and their grouping category ? Items Group (A) Malleus, incus, cochlea —Ear ossicles (B) Ilium, ischium, pubis —Coxal bones of pelvic girdle (C) Actin, myosin, rhodopsin —Muscle proteins (D) Cytosine, uracil, thiamine —Pyrimidines 87. Plasmodesmata are— (A) Lignified cemented layers between cells (B) Locomotary structures

89. Aerobic respiratory pathway is appropriately termed— (A) Catabolic (B) Parabolic (C) Amphibolic (D) Anabolic 90. Which of the following is not used as a biopesticide ? (A) Bacillus thuringiensis (B) Trichoderma harzianum (C) Nuclear Polyhedrosis Virus (NPV) (D) Xanthomonas campestris 91. Alzhimer disease in humans is associated with the deficiency of— (A) Dopamine (B) Glutamic acid (C) Acetylcholine (D) Gamma aminobutyric acid (GABA) 92. Cytoskeleton is made up of— (A) Calcium carbonate granules (B) Callose deposits (C) Cellulosic microfibrils (D) Proteinaceous filaments 93. Compared to blood our lymph has— (A) No plasma (B) Plasma without proteins (C) More WBCs and no RBCs (D) More RBCs and less WBCs 94. Which one of the following is the most likely root cause why menstruation is not taking place in regularly cycling human female ? (A) Fertilisation of the ovum (B) Maintenance of the hypertrophical endometrial lining (C) Maintenance of high concentration of sex-hormones in the bloodstream (D) Retention of well-developed corpus luteum

99. Anatomically fairly old dicotyle- 106. Synapsis occurs between— donous root is distinguished from (A) A male and a female gamete the dicotyledonous stem by— (B) mRNA and ribosomes (A) Absence of secondary xylem (C) Spindle fibres and centro(B) Absence of secondary mere phloem (D) Two homologous chromo(C) Presence of cortex somes (D) Position of protoxylem 107. Which one of the following pairs 100. Manganese is required in— of animals comprises ‘jawless fishes’ ? (A) Nucleic acid synthesis (A) Lampreys and eels (B) Plant cell wall formation (B) Mackerals and Rohu (C) Photolysis of water during photosynthesis (C) Lampreys and hag fishes Which one of the following is the (D) Chlorophyll synthesis (D) Guppies and hag fishes correct indication of the stage/ 101. Which one of the following is com- 108. Select the incorrect statement phase in the cell cycle ? from the following— monly used in transfer of foreign (A) B-Metaphase DNA into crop plants ? (A) Linkage is a exception to the (B) C-Karyokinesis principle of independent (A) Trichoderma harzianum (C) D-Synthetic phase assortment in heredity (B) Meloidogyne incognita (D) A-Cytokinesis (B) Galactosemia is an inborn (C) Agrobacterium tumefaciens error of metabolism 96. The floral formula ⊕ O K C A (D) Penicillium expansum (5) + (5) 5 (C) Small population size results G(2) is that of— in random genetic drift in a 102. Removal of introns and joining (A) Tulip (B) Soybean population the exons in a defined order in a (C) Sunnhemp (D) Tobacco (D) Baldness is a sex-limited transcription unit is called— trait (A) Splicing 97. The most popularly known blood 109. The cell junctions called tight, grouping is the ABO grouping. It (B) Tailing adhering and gap junctions are is named ABO and not ABC, be(C) Transformation found in— cause ‘O’ in it refers to having— (D) Capping (A) Muscular tissue (A) Other antigens besides A 103. Whose experiments cracked the and B on RBCs (B) Connective tissue DNA and discovered unequivo(C) Epithelial tissue (B) Overdominance of this type cally that a genetic code is a on the genes for A and B (D) Neural tissue ‘triplet’ ? types 110. Which one of the following types (A) Nirenberg and Mathaei (C) One antibody only-either of organisms occupy more than (B) Hershey and Chase anti-A or anti-B on the RBCs one trophic level in a pond eco(C) Morgan and Sturtevant (D) No antigens A and B on system ? RBCs (D) Beadle and Tatum (A) Phytoplankton (B) Fish (C) Zooplankton (D) Frog 98. Which one of the following state- 104. Seminal plasma in humans is rich ments is true regarding digestion in— 111. Use of anti-histamines and steand absorption of food in roids give a quick relief from— (A) Fructose, calcium and certain humans ? enzymes (A) Allergy (B) Nausea (A) Oxyntic cells in our stomach (C) Cough (D) Headache (B) Fructose and calcium but secrete the proenzyme has no enzymes 112. Which one is the wrong pairing pepsinogen. (C) Glucose and certain enzyfor the disease and its causal (B) Fructose and amino acids mes but has no calcium organism ? are absorbed through intesti(D) Fructose and certain enzy(A) Late blight of potato nal mucosa with the help of mes but poor in calcium carrier ions like Na+. —Alternaria solani (C) Chylomicrons are small 105. Phylogenetic system of classifi(B) Black rust of wheat lipoprotein particles that are cation is based on— —Puccinia graminis transported from intestine (C) Loose smut of wheat (A) Evolutionary relationships into blood capillaries. —Ustilago nuda (B) Morphological features (D) About 60% of starch is (C) Chemical constituents (D) Root-knot of vegetables hydrolysed by salivary (D) Floral characters — Meloidogyne sp amylase in our mouth. →

95. Given below is a schematic breakup of the phases/stages of cell cycle—

C.S.V. / November / 2009 / 1158

(C) Non-toxic to aquatic animals (D) Water soluble

113. Given below is a diagrammatic sketch of a portion of human male reproductive system. Select the correct set of the names of the parts labelled 1, 2, 3, 4—

122. Which one of the following is a vascular cryptogam ? (A) Equisetum (B) Ginkgo (C) Marchantia (D) Cedrus 123. Steps taken by the Government of India to control air pollution include— Interpretations— (A) ‘A’ is more recent and shows slight reduction in the growth rate. (B) ‘B’ is earlier pyramid and shows stabilised growth rate. (C) ‘B’ is more recent showing that population is very young. (D) ‘A’ is the earlier pyramid and no change has occurred in the growth rate.

(A) Compulsory mixing of 20% ethyl alcohol with petrol and 20% biodiesel with diesel.

(1) ureter, prostate, seminal vesicle, bulbourethral gland. (B) Compulsory PUC (Pollution (2) vas deferens, seminal Under Control) certification of vesicle, prostate, bulbouretpetrol driven vehicles which hral gland. tests for carbon monoxide and hydrocarbons. (3) vas deferens, seminal vesicle, bulbourethral gland, (C) Permission to use only pure prostate. diesel with a maximum of (4) ureter, seminal vesicle, 500 ppm sulphur as fuel for prostate, bulbourethral gland. vehicles. 118. The correct sequence of sper(A) (1) (B) (2) (D) Use of non-polluting Commatogenetic stages leading to (C) (3) (D) (4) pressed Natural Gas (CNG) the formation of sperms in a only as fuel by all buses and 114. Vegetative propagation in mint mature human testis is— trucks. occurs by— (A) Spermatocyte—spermatogo(A) Runner (B) Offset nia—spermatid—sperms 124. Montreal Protocol aims at— (C) Rhizome (D) Sucker (B) Spermatogonia—spermato(A) Reduction of ozone deplet115. What will happen if the stretch cyte—spermatid—sperms ing substances receptors of the urinary bladder (C) Spermatid—spermatocyte— wall are totally removed ? (B) Biodiversity conservation spermatogonia—sperms (A) Urine will not collect in the (C) Control of water pollution (D) Spermatogonia—spermatidbladder (D) Control of CO2 emission spermatocyte—sperms (B) Micturition will continue (C) Urine will continue to collect 119. What is true about Bt toxin ? 125. Which one of the following acids (A) The inactive protoxin gets normally in the bladder is a derivative of carotenoids ? converted into active form in (D) There will be no micturition (A) Indole butyric acid the insect gut. (B) Indole-3-acetic acid 116. One of the synthetic auxin is— (B) Bt protein exists as active (A) NAA (B) IAA (C) Gibberellic acid toxin in the Bacillus. (C) GA (D) IBA (C) The activated toxin enters (D) Abscisic acid the ovaries of the pest to 117. A country with a high rate of 126. Guard cells help in— sterilise it and thus prevent population growth took measures (A) Protection against grazing its multiplication. to reduce it. The figure below (D) The concerned Bacillus has (B) Transpiration shows age-sex pyramids of popuantitoxins. lations A and B twently years (C) Guttation apart. Select the correct inter- 120. The kind of tissue that forms the (D) Fighting against infection pretation about them— supportive structure in our pinna (external ears) is also found in— 127. Palisade parenchyma is absent in leaves of— (A) Vertebrae (A) Sorghum (B) Mustard (B) Nails (C) Ear ossicles (D) Tip of the nose

(C) Soybean

(D) Gram

128. Cotyledons and testa respectively are edible parts in— 121. DDT residues are rapidly passed (A) Groundnut and pomegranate through food chain causing bio(B) Walnut and tamarind magnification because DDT is— (A) Lipo soluble (C) French bean and coconut (B) Moderately toxic (D) Cashew nut and litchi

C.S.V. / November / 2009 / 1159

129. Which one of the following is considered important in the development of seed habit ? (A) Dependent sporophyte (B) Heterospory (C) Haplontic life cycle (D) Free-living gametophyte 130. T.O. Diener discovered a— (A) Free infectious RNA (B) Free infectious DNA (C) Infectious protein (D) Bacteriophage

136.

131. Polyethylene glycol method is used for— (A) Gene transfer without a 137. vector (B) Biodiesel production (C) Seedless fruit production (D) Energy production from sewage 132. Which one of the following pairs is wrongly matched ? (A) Detergents—lipase (B) Alcohol—nitrogenase 138. (C) Fruit juice—pectinase (D) Textile—amylase 133. A person likely to develop tetanus is immunised by administering— 139. (A) Dead germs (B) Preformed antibodies (C) Wide spectrum antibiotics (D) Weakened germs 140. 134. Biochemical Oxygen Demand (BOD) in a river water— (A) Remains unchanged when algal bloom occurs. (B) Has no relationship with concentration of oxygen in the 141. water (C) Gives a measure of salmonella in the water (D) Increases when sewage gets mixed with river water.

(C) Development of corpus 143. Which one of the following statements is correct ? luteum (A) Patients who have under—Secretory phase and incregone surgery are given ased secretion of progescannabinoids to relieve pain. terone (B) Benign tumours show the (D) Menstruation property of metastasis. —Breakdown of myometrium (C) Heroin accelerates body funand ovum not fertilised ctions Cyclic photophosphorylation re(D) Malignant tumours may exsults in the formation of— hibit metastasis (A) NADPH 144. Transgenic plants are the ones— (B) ATP and NADPH (A) Produced by a somatic em(C) ATP, NADPH and O2 bryo in artificial medium (D) ATP (B) Generated by introducing Globulins contained in human foreign DNA in to a cell and blood plasma are primarily invoregenerating a plant from lved in— that cell. (A) Defence mechanisms of (C) Produced after protoplast body fusion in artificial medium. (B) Osmotic balance of body (D) Grown in artificial medium fluids after hybridization in the (C) Oxygen transport in the blood field. (D) Clotting of blood 145. The letter T in T-lymphocyte refers to— Which of the following is a symbiotic nitrogen fixer ? (A) Thyroid (B) Thalamus (A) Glomus (B) Azotobacter (C) Tonsil (D) Thymus (C) Frankia (D) Azolla 146. Global agreement in specific An example of axile placentation control strategies to reduce the is— release of ozone depleting subs(A) Argemone (B) Dianthus tances, was adopted by— (C) Lemon (D) Marigold (A) Rio de Janeiro Conference Semiconservative replication of (B) The Montreal Protocol DNA was first demonstrated in— (C) The Koyoto Protocol (A) Drosophila melanogaster (D) The Vienna Convention (B) Escherichia coli 147. Reduction in vascular tissue, (C) Streptococcus pneumoniae mechanical tissue and cuticle is (D) Salmonella typhimurium characteristic of— A young infant may be feeding (A) Xerophytes entirely on mother’s milk which is (B) Mesophytes white in colour but the stools (C) Epiphytes which the infant passes out is (D) Hydrophytes quite yellowish. What is this

yellow colour due to ? 148. An example of a seed with endo135. Which one of the following is the sperm, perisperm and caruncle (A) Intestinal juice correct matching of the events is— (B) Bile pigments passed throoccurring during menstrual (A) Cotton (B) Coffee ugh bile juice cycle ? (C) Lily (D) Castor (C) Undigested milk protein (A) Ovulation casein 149. What is not true for genetic —LH and FSH attain peak (D) Pancreatic juice poured into code ? level and sharp fall in the duodenum (A) A codon in m RNA is read in secretion of progesterone. a non-contiguous fashion 142. A fruit developed from hypantho(B) Proliferative phase dium inflorescence is called— (B) It is nearly universal — Rapid regeneration of (A) Hesperidium (B) Sorosis (C) It is degenerate myometrium and maturation (C) Syconus (D) Caryopsis (D) It is unambiguous of Graafian follicle.

C.S.V. / November / 2009 / 1160

(C) VC ≠ VB ≠ VA 161. A body of mass 1 kg is thrown 150. Peripatus is a connnecting link upwards with a velocity 20 m/s. It between— (D) VC = VB = VA momentarily comes to rest after (A) Ctenophora and Platyhel156. A bus is moving with a speed of attaining a height of 18m. How minthis – 1 on a straight road. A 10 ms much energy is lost due to air (B) Mollusca and Echinoderscooterist wishes to overtake the friction ? (g = 10 m/s2) mata bus in 100s. If the bus is at a (A) 20 J (B) 30 J (C) Annnelida and Arthropoda distance of 1 km from the scoo(D) Coelenterata and Porifera (C) 40 J (D) 10 J terist, with what speed should the 151. An explosion blows a rock into scooterist chase the bus ? 162. The symbolic representation of three parts. Two parts go off at –1 –1 four logic gates are given below : (A) 20 ms (B) 40 ms right angles to each other. These –1 –1 (i) (C) 25 ms (D) 10 ms two are, 1 kg first part moving with a velocity of 12 ms–1 and 2 157. Under the influence of a uniform (ii) kg second part moving with a magnetic field, a charged particle (iii) velocity of 8 ms–1. If the third part moves with constant speed v in a flies off with a velocity of 4 ms– 1, circle of radius R. The time (iv) period of rotation of the particle— its mass would be— (A) 5 kg (B) 7 kg The logic symbols for OR, NOT (A) Depends on v and not on R and NAND gates are respecti(C) 17 kg (D) 3 kg (B) Depends on R and not on v vely— 152. A conducting circular loop is (C) Is independent of both V and (A) (iii), (iv), (ii) (B) (iv), (i), (iii) placed in a uniform magnetic R (C) (iv), (ii), (i) (D) (i), (iii), (Iv) field 0·04 T with its plane (D) Depends on both v and R perpendicular to the magnetic 163. A particle is displaced from posifield. The radius of the loop starts 158. A wave in a string has an ^ ^ tion 3 i + 2 j metre to another shrinking at 2 mm/s. The induced amplitude of 2 cm. The wave ^ ^ position 14 i + 13 j metre by emf in the loop when the radius travels in the + ve direction of x ^ ^ is 2 cm is— axis with a speed of 128 m/sec applying a force 4 i + j newton. and it is noted that 5 complete (A) 3·2 π μ V (B) 4·8 π μ V Work done by the force will be— waves fit in 4 m length of the (A) 110 J (B) 55 J (C) 0·8 π μ V (D) 1·6 π μ V string. The equation describing (C) 27·5 J (D) None of these 153. If the dimensions of a physical the wave is— 164. If a diamagnetic substance is quantity are given by MaLbTc, (A) y = (0·02)m sin (7·85x brought near the north or the then the physical quantity will south pole of a bar magnet, it is— + 1005t) be— (A) Repelled by both the poles (A) Pressure if (B) y = (0·02)m sin (15·7x (B) Repelled by the north pole a = 1, b = – 1, c = – 2 – 2010 t) and attracted by the south (B) Velocity if (C) y = (0·02)m sin (15·7x pole a = 1, b = 0, c = – 1 + 2010t) (C) Acceleration if (C) Attracted by the north pole a = 1, b = 1, c = – 2 and repelled by the south (D) y = (0·02)m sin (7·85x (D) Force if pole – 1005 t) a = 0, b = – 1, c = – 2 (D) Attracted by both the poles 159. A simple pendulum performs 154. In the nuclear decay given 165. The mass of a lift is 2000 kg. simple harmonic motion about below : When the tension in the supx = 0 with an amplitude a and A A–4 A–4 A porting cable is 28000 N, then its X → Y → B* → B, Z+1 Z–1 Z–1 Z time period T. The speed of the acceleration is— a the particles emitted in the sequpendulum at x = will be— (A) 30 ms – 2 downwards 2 ence are— (B) 4 ms– 2 upwards (A) β, α, γ (B) γ, β, α πa 3 πa (C) 4 ms– 2 downwards (A) (B) (C) β, γ, α (D) α, β, γ 2T T (D) 14 ms – 2 upwards 2 a 3π π a 3 155. Three concentric spherical shells 166. The number of beta particles (C) (D) T T have radii a, b and c (a < b < c ) emitted by a radioactive subsand have surface charge densitance is twice the number of 160. A p-n photodiode is fabricated ties σ, – σ and σ respectively. If alpha particles emitted by it. The from a semiconductor with a band VA , V B and VC denote the potenresulting daughter is an— gap of 2·5 eV. It can detect a tials of the three shells, then, for (A) Isobar of parent signal of wavelength— c = a + b, we have— (B) Isomer of parent ° (A) 6000 A (B) 4000 nm (A) VC = VA ≠ VB (C) Isotone of parent ° (B) VC = VB ≠ VA (D) Isotope of parent (C) 6000 nm (D) 4000 A

C.S.V. / November / 2009 / 1161

and m is the mass per unit length 167. A rectangular, a square, a circu- 171. A thin circular ring of mass M of the water jet. What is the rate lar and an elliptical loop, all in the and radius R is rotating in a at which kinetic energy is impar(x – y) plane, are moving out of a horizontal plane about an axis ted to water ? vertical to its plane with a uniform magnetic field with a 1 → constant angular velocity ω . If ^ (A) mv 3 (B) mv 3 constant velocity, v = v i . The 2 two objects each of mass m be magnetic field is directed along 1 1 attached gently to the opposite (C) mv 2 (D) m2v 2 the negative z axis direction. The 2 2 ends of a diameter of the ring, induced emf, during the passage the ring will then rotate with an 178. The driver of a car travelling with of these loops, out of the field angular velocity— speed 30 m/sec towards a hill region, will not remain constant sounds a horn of frequency 600 ω (M – 2 ωM m ) for— (A) (B) Hz. If the velocity of sound in air M + 2m M + 2m (A) The rectangular, circular and is 330 m/s, the frequency of ω(M + 2m) ωM ellliptical loops reflected sound as heard by (C) (D) M+m M (B) The circular and the elliptical driver is— loops 172. A transistor is operated in (A) 550 Hz (B) 555·5 Hz common-emitter configuration at (C) 720 Hz (D) 500 Hz (C) Only the elliptical loop Vc = 2V such that a change in (D) Any of the four loops 179. In a Rutherford scattering experithe base current from 100 μA to ment when a projectile of charge 168. A particle starts its motion from 200 μA produces a change in the Z1 and mass M1 approaches a rest under the action of a conscollector current from 5 mA to 10 target nucleus of charge Z2 and tant force. If the distance covered mA. The current gain is— mass M2, the distance of closest in first 10 second is S1 and that approach is r0. The energy of the (A) 75 (B) 100 covered in the first 20 second is projectile is— (C) 150 (D) 50 S2,then— (A) Directly proportional to 173. A black body at 227°C radiates (B) S2 = 3S1 (A) S2 = 2S1 M1 × M2 heat at the rate of 7 cal/cm2 s. At (C) S2 = 4S1 (D) S2 = S1 (B) Directly proportional to Z1Z2 a temperature of 727°C, the rate (C) Inversely proportional to Z1 169. A wire of resistance 12 ohm per of heat radiated in the same units (D) Directly proportional to mass metre is bent to form a complete will be— M1 circle of radius 10 cm. The resis(A) 60 (B) 50 tance between its two diametri180. Three capacitors each of capaci(C) 112 (D) 80 cally opposite tance C and of breakdown voltage 174. A student measures the terminal points, A and B A V are joined in series. The capaB potential difference (V) of a cell as shown in the citance and breakdown voltage (of emf E and internal resistance figure, is— of the combination will be— r) as a function of the current (I) C V V (A) 0·6 π Ω (B) 3Ω (A) , (B) 3C, flowing through it. The slope, and 3 3 3 intercept, of the graph between V (C) 6 π Ω (D) 6Ω C (C) , 3V (D) 3C, 3V and I, then, respectively, equal— 3 170. The electric field parts of an elec(A) E and – r (B) – r and E tromagnetic wave in a medium is 181. The magnetic force acting on a (C) r and – E (D) – E and r represented by— charged particle of charge – 2μc 175. Two bodies of mass 1 kg and 3 kg Ex = 0; in a magnetic field of 2T acting in ^ ^ ^ y direction, when the particle have position vectors i + 2 j + k Ey = 2·5 N/C cos [(2π × 106 rad/m) ^ ^ ^ ^ ^ velocity is (2i + 3 j ) × 10 6 ms– 1, and – 3 i – 2 j + k , rspectively. – 2 t – (π × 10 rad/s)x]; is— The centre of mass of this sysEz = 0. The wave is— (A) 8N in – z direction tem has a position vector— ^ ^ ^ ^ ^ (B) 4N in z direction (A) Moving along y direction (A) – 2 i + 2k (B) – 2 i – j + k (C) 8N in y direction with frequency 2π × 106 Hz ^ ^ ^ ^ ^ ^ (C) 2 i – j – 2k (D) – i + j + k (D) 8N in z direction and wavelength 200 m. (B) Moving along x direction 176. The internal energy change in a 182. Which one of the following equasystem that has absorbed 2 k cal tions of motion represents simple with frequency 106 Hz and of heat and done 500 J of work harmonic motion ? wavelength 100 m. is— (A) Acceleration = – k0x + k1x 2 (C) Moving along x direction with (A) 8900 J (B) 6400 J (B) Acceleration = – k(x + a) frequency 106 Hz and wave(C) 5400 J (D) 7900 J (C) Acceleration = k(x + a) length 200 m. (D) Acceleration = kx (D) Moving along – x direction 177. An engine pumps water continu6 ously through a hose. Water where k, k0, k1 and a are all posiwith frequency 10 Hz and leaves the hose with a velocity v tive. wavelength 200 m.

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183. Monochromatic light of wave- 186. The electric potential at a point 190. See the electrical circuit shown in length 667 nm is produced by a (x, y, z) is given by : this figure. Which of the following helium neon laser. The power equations is a correct equation V = – x2y – xz3 + 4 emitted is 9 mW. The number of for it ? → photons arriving per sec on the The electric field E at that point average at a target irradiated by is— this beam is— → ^ ^ (A) E = i (2xy + z3) + j x2 16 (A) 9 × 1017 (B) 3 × 10 ^ + k 3 xz2 (C) 9 × 1015 (D) 3 × 1019 → ^ ^ (B) E = i 2 xy + j (x2 + y2) 184. The figure shows a plot of photo ^ current versus anode potential + k (3xz – y2) for a photo sensitive surface for → ^ (A) E1 – (i1 + i2) R – i1r1 = 0 ^ ^ three different radiations. Which (C) E = i z3 + j xyz + k z2 (B) E2 – i2 r2 – E1 – i1 r1 = 0 one of the following is a correct → ^ ^ (C) – E 2 – (i1 + i2) R+ i2r2 = 0 (D) E = i (2xy – z 3) + j xy2 statement ? (D) E1 – (i1 + i2) R + i1r1 = 0 ^ + k 3 z2x 191. A conductor with a cross-section 187. A bar magnet having a magnetic of 10 –4 m 2 carries an electric moment of 2 × 10 4 JT– 1 is free to current of 1·2A. If the number of rotate in a horizontal plane. A free electrons be 5 × 102 8/m3, horizontal magnetic field B = 6 × then the electron drift velocity will 10 – 4 T exists in the space. The be— work done in taking the magnet (A) 1·5 × 106 ms–1 slowly from a direction parallel to (B) 3·0 × 106 ms–1 the field to a direction 60° from (A) Curves (a) and (b) represent (C) 1·5 × 10–6 ms–1 the field is— incident radiations of different (D) 3·0 × 10–6 ms–1 (A) 0·6 J (B) 12 J frequencies and different (C) 6 J (D) 2 J 192. The number of photo electrons intensities. emitted for light of a frequency υ 188. A galvanometer having a coil (B) Curves (a) and (b) represent (higher than the threshold freresistance of 60 Ω shows full incident radiations of same quency υ0) is proportional to— scale deflection when a current of frequency but of different 1·0 amp passes through it. It can (A) υ – υ0 intensities. be converted into an ammeter to (B) Threshold frequency (υ0) (C) Curves (b) and (c) represent read currents upto 5·0 amp by— incident radiations of different (C) Intensity of light (A) Putting in parallel a resis(D) Frequency of light (υ) frequencies and different tance of 240 Ω intensities. (B) Putting in series a resistance 193. Each of the two strings of length 51·6 cm and 49·1 cm are ten(D) Curves (b) and (c) represent of 15 Ω sioned separately by 20 N force. incident radiations of same (C) Putting in series a resistance Mass per unit length of both the frequencies having same of 240 Ω strings is same and equal to intensities. 1g/m. When both the strings (D) Putting in parallel a resisvibrate simultaneously the num185. Power dissipated in an LCR tance of 15 Ω ber of beats is— series circuit connected to an a.c. 189. The two ends of a rod of length (A) 5 (B) 7 source of emf E is— L and a uniform cross-sectional (C) 8 (D) 3 E2R area A are kept at two tempera(A) 2 1 tures T1 and T2 (T1 > T 2). The rate 194. The ionization energy of the R2 + Lω – electron in the hydrogen atom in Cω dQ , through the of heat transfer, its ground state is 13·6 eV. The dt 2 atoms are excited to higher 1 rod in a steady state is given E2 R2 + Lω – energy levels to emit radiations of Cω by— (B) 6 wavelengths. Maximum waveR d Q k L(T1 – T 2) length of emitted radiation corre(A) = 2 1 A dt sponds to the transition betE2 R2 + Lω – d Q k (T1 – T 2) Cω ween— (B) = (C) LA dt R (A) n = 3 to n = 2 states dQ E2R (C) = k L A (T1 – T 2) (B) n = 3 to n = 1 states dt (D) 2 1 (C) n = 2 to n = 1 states d Q k A(T1 – T 2) R2 + Lω – = (D) Cω (D) n = 4 to n = 3 states L dt

[ (

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[ ( (

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)

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195. Four identical thin rods each of 197. The figure shows elliptical orbit of 199. A block of mass M is attached to the lower end of a vertical spring. mass M and length L, form a a planet m about the sun S. The The spring is hung from a ceiling square frame. Moment of inertia shaded area SCD is twice the shaded area SAB. If t1 is the time and has force constant value k . of this frame about an axis through the centre of the square The mass is released from rest for the planet to move from C to and perpendicular to its plane D and t2 is the time to move from with the spring initially unstretcis— hed. The maximum extension A to B then— 4 produced in the length of the 2 2 2 (A) ML (B) ML 3 spring will be— 3 13 1 Mg 2Mg (C) ML2 (D) ML2 (A) (B) 3 3 k k 4M M g g 196. In thermodynamic processes (C) (D) k 2k which of the following statements in not true ? 200. A body, under the action of a (A) In an adiabatic process the → (A) t 1 > t2 (B) t 1 = 4t2 ^ ^ ^ system is insulated from the force F = 6 i – 8 j + 10k , acqui(C) t 1 = 2t2 (D) t 1 = t2 surroundings res an acceleration of 1m/s2. The (B) In an isochoric process pres- 198. Sodium has body centred packmass of this body must be— sure remains constant ing. Distance between two near(A) 2 10 kg (C) In an isothermal process the est atoms is 3·7 Å. The lattice temperature remains consparameter is— (B) 10 kg tant ° ° (C) 20 kg (A) 6·8 A (B) 4·3 A (D) In an adiabatic process PV γ ° ° (D) 10 2 kg (C) 3·0 A (D) 8·6 A = constant

ANSWERS WITH HINTS

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In each of the following questions, a statement of Assertion (A) is given and a corresponding statement of Reason (R) is given just below it. Of the statements, mark the correct answer as— (A) If both A and R are true and R is the correct explanation of A (B) If both A and R are true but R is not the correct explanation of A (C) If A is true but R is false (D) If both A and R are false (E) If A is false but R is true

PHYSICS 1. Assertion (A) : While passing round the corners the light spreads out to some extent into the region of the geometrical shadow. Reason (R) : The bending is greater for light of longer wavelengths and less for shorter wavelengths. (A)

(B)

(C)

(D)

(E)

2. Assertion (A) : A transistor is preferred to a triode when used as an amplifier. Reason (R) : No heating of filament is needed in the case of a transistor. (A)

(B)

(C)

(D)

(E)

3. Assertion (A) : An ammeter is connected in series in the electric circuit and it should have a low resistance. Reason (R) : The introduction of the ammeter must not affect the main current. (A)

(B)

(C)

(D)

6. Assertion (A) : A null vector is also called a proper vector. Reason (R) : A null vector has a definite magnitude and a definite direction. (A) (B) (C) (D) (E) 7. Assertion (A) : The density of water is maximum at 4°C. Reason (R) : The volume of water decreases while heating from 0°C to 4°C. (A) (B) (C) (D) (E) 8. Assertion (A) : The collisions between atomic and subatomic particles are considered to be perfectly elastic. Reason (R) : In collisions of tiny particles both the momentum and energy are conserved. (A) (B) (C) (D) (E) 9. Assertion (A) : Pressure cooker is useful at high mountains. Reason (R) : Due to low atmospheric pressure on high mountains the water boils at less than 100°C. (A) (B) (C) (D) (E) 10. Assertion (A) : Illuminance is the luminous flux per unit surface area, when the area is held normal to the beam of light. Reason (R) : The luminous intensity is the radiant flux per unit angle in that direction. (A) (B) (C) (D) (E)

CHEMISTRY

(E)

4. Assertion (A) : In any relation between some physical quantities, the dimensions on either side of the equation must be the same. Reason (R) : Dimensions are the powers raised to fundamental units in a dimensional equation. (A) (B) (C) (D) (E)

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5. Assertion (A) : The calorimeters are made of metal. Reason (R) : Metals have good thermal conductivity and low specific heat. (A) (B) (C) (D) (E)

11. Assertion (A) : The bond order in both H2+ and H2– is same that 1 is , but H 2– is slightly less stable 2 than H 2+. Reason (R) : Since the bond order is the same, the bond dissociation energy must strictly be the same. (A) (B) (C) (D) (E)

12. Assertion (A) : Many endothermic reactions which are nonspontaneous at room temperature become spontaneous on raising temperature. Reason (R) : Endothermic reactions become spontaneous when ΔS is positive and TΔS > ΔH. (A) (B) (C) (D) (E) 13. Assertion (A) : The number of unpaired electrons in Ni(CO)4 is zero. Reason (R) : 4 s electrons of nickel atom enter the inner d orbitals to facilitate the sp 3 hybridisation in Ni atom. (A) (B) (C) (D) (E) 14. Assertion (A) : The dipole moment of carbon tetrafluoride molecule will be greater than that of carbon tetrachloride. Reason (R) : C—F bond is more polar than C—Cl bond. (A) (B) (C) (D) (E) 15. Assertion (A) : Unlike the chloride ion, the fluoride ion cannot be quantitatively estimated with the help of AgNO3 solution. Reason (R) : Silver fluoride is considerably soluble in water. (A)

(B)

(C)

(D)

(E)

16. Assertion (A) : Molecule CH3 ≡ N: can act both as electrophile as well as nucleophile. Reason (R) : Carbon atom in the molecule is electron deficient and nitrogen atom is electron sufficient. (A) (B) (C) (D) (E) 17. Assertion (A) : A lanthanide element with oxidation number + 2 acts as a good oxidising agent. Reason (R) : The common oxidation number of lanthanide elements is + 3. (A) (B) (C) (D) (E) 18. Assertion (A) : The rate equation for a general chemical reaction, a A + b B c C + d D can be expressed as : Rate = k [A]x [B] y

Reason (R) : The exponents x and y in rate equation are necessarily equal to coefficients a and b respectively. (A) (B) (C) (D) (E) 19. Assertion (A) : Ethyl bromide and alcoholic silver nitrite react to give nitroethane as a major product. Reason (R) : Nitrite ion (NO 2–) is an ambidented nucleophile. (A) (B) (C) (D) (E) 20. Assertion (A) : The reactivity of anions of ionic compounds is greater in a polar aprotic solvent, like dimethyl sulphoxide (DMSO) than in a protic solvent like ethanol. Reason (R) : Polar aprotic solvents shield the anion effectively from cation so that ion-pairing does not occur, thus cation remains considerably free. (A) (B) (C) (D) (E)

ZOOLOGY 21. Assertion (A) : Food intake is regulated by an area of the brain called hypothalamus. Reason (R) : In the hypothalamus, a region called the hunger centre regulates the need to eat and a satiety centre indicates when to stop eating. (A) (B) (C) (D) (E) 22. Assertion (A) : Degenerative diseases are usually associated with aging, when tissues break down more rapidly than they can be repaired. Reason (R) : Osteoarthritis is a painful and chronic joint inflammation that involves a progressive degeneration of the hyaline cartilage over the bone surfaces in movable joints. (A)

(B)

(C)

(D)

(E)

23. Assertion (A) : The release of ADH from the pituitary gland is controlled by special cells in the hypothalamus of the brain that monitor water concentration of the blood. Reason (R) : Alcohol inhibits the release of ADH, and caffeine interferes with ADH action and sodium absorption, thus both

C.S.V. / November / 2009 / 1171

cause artificially dilute urine to be produced. (A) (B) (C) (D) (E) 24. Assertion (A) : Immunization is the process of rendering an animal less susceptible to infection by pathogens and toxins. Reason (R) : This involve use of vaccine to induce passive immunity through injection of appropriate antibodies. (A) (B) (C) (D) (E) 25. Assertion (A) : Most skeletal bones begin as a cartilage model and consists of hyaline cartilage. Reason (R) : This model does not change into bone, instead, it is gradually dissolved and replaced by bone. (A) (B) (C) (D) (E) 26. Assertion (A) : 5-hydroxytryptamine is tryptophan derived neurotransmitter of vertebrate brain, especially of pons, and of pineal gland. Reason (R) : It hyperpolarizes postsynaptic membranes and activates phospho fructokinase in liver, both mediated via cyclic AMP. (A) (B) (C) (D) (E) 27. Assertion (A) : An action potential (or impulse) develops when a stimulus alters the resting membrane potential. Reason (R) : The resting membrane potential of a neuron results from an uneven distribution of ions across its plasma membrane. (A) (B) (C) (D) (E) 28. Assertion (A) : Clone is group of organisms of identical genotype, produced by some kind of asexual reproduction and some sexual processes such as haploid selfing, or inbreeding a completely homozygous line. Reason (R) : Nuclear transplantation techniques introducing genetically identical nuclei into enucleated eggs can also produce clones in some animals. (A) (B) (C) (D) (E) 29. Assertion (A) : The pitch of sound is perceived by the position of the phonoreceptors in the cochlear duct.

Reason (R) : The loudness of the sound is encoded in frequency of action potentials and the number of phonoreceptors stimulated. (A) (B) (C) (D) (E) 30. Assertion (A) : DDT and aldrin are contact insecticides and their fat-solubility (often needed to penetrate waxy epicuticle) resulting in accumulation in fat reserves of animals in higher trophic levels. Reason (R) : The mode of entry of contact insecticides to the body is via the cuticle rather than the gut. (A) (B) (C) (D) (E)

BOTANY 31. Assertion (A) : Within the anther pollen sacs, spore mother cells form 4 haploid pollen grains. Reason (R) : Spore mother cells (SMCs) divide meiotically to form 4 haploid pollen grains. (A) (B) (C) (D) (E) 32. Assertion (A) : A food web helps to understand who is eating whom, and an energy pyramid to appreciate the actual amount of energy and organic matter passing from producers to consumers. Reason (R) : Cycling of nutrients and flow of energy through the various populations in a natural community produces a food web. (A) (B) (C) (D) (E) 33. Assertion (A) : Every tissue and organ has its special requirements for optimal growth. Reason (R) : Auxins, gibberellins and cytokinins can not be added in the medium. (A) (B) (C) (D) (E) 34. Assertion (A) : The energy stored in carbohydrate molecules during photosynthesis is released during cellular oxidation of carbohydrates into CO2 and H2O. This process is called respiration. Reason (R) : During respiration, the oxidation of various organic food substances like carbohydrate, fat, protein, etc., may take place. (A) (B) (C) (D) (E)

35. Assertion (A) : Chromatid is one of a pair of replicated chromosomes found during the prophase of meiosis and mitosis. Reason (R) : Chromatin is classified as euchromatin or heterochromatin on the basis of staining properties. (A) (B) (C) (D) (E) 36. Assertion (A) : The energy capturing portion of photosynthesis occurs in thylakoids membrane and can not proceed without solar energy. Reason (R) : The synthetic portion of photosynthesis occurs in stroma and does not directly require solar energy. (A) (B) (C) (D) (E)

37. Assertion (A) : Oparin reasoned that methane, ammonia, water and hydrogen gas might have been the raw materials for the formation of nucleic acids and nucleotides. Reason (R) : Oparin realized that to build large molecules from small ones requires energy. (A) (B) (C) (D) (E) 38. Assertion (A) : Welwitschias are strange gnetophytes that are found only in deserts. Reason (R) : Gnetophyte is one of a small group of plants that grow in deserts and tropical forests. (A) (B) (C) (D) (E)

39. Assertion (A) : Zoospore fungus produces spores within sporangia. Reason (R) : During sexual reproduction, a zoospore forms prior to meiosis and production of spores. (A)

(B)

(C)

(D)

(E)

40. Assertion (A) : The direction of maturation of flower in racemose inflorescence is acropetal or centripetal. Reason (R) : Sieve-tube cells contain nuclei in their cytoplasm. (A)

(B)

(C)

(D)

(E)

ANSWERS WITH HINTS

(Continued on Page 1184) C.S.V. / November / 2009 / 1172

Physics 1. When an object is placed at the focus of a convex mirror, it is imaged at infinity.

15. A lorry and a car moving with the same kinetic energy are brought to rest by application of brakes which provide equal retarding force. Both come to rest in equal distance. —T/F

—T/F 2. The ratio of de-Broglie wavelength of an electron and a proton moving with same kinetic energy is 1840 . —T/F 3. A bullet is fired from a rifle. If the rifle recoils freely, the kinetic energy of the rifle is less than that of the bullet. —T/F 4. Coloured glass appears white after being crushed into fine powder. —T/F 5. Pressure exerted by one mole of oxygen is equal to that by one mole of hydrogen when their temperature and volume are same. —T/F 6. The laser is a source of ultrasound. —T/F 7. In an electrolyte solution, the electric current is mainly due to the movement of free electrons. —T/F 8. A body can describe a vertical circle with any velocity. —T/F 9. According to Kepler’s second law, the radius vector to a planet from the sun sweeps out equal areas in equal intervals of time. This law is a consequence of conservation of linear momentum. —T/F 10. If two slits in Young’s double slit experiment are illuminated by two identical monochromatic sources of light emitting the same wavelength, no interference pattern will be observed.

Chemistry 16. A compound having composition 4KCN.Fe(CN) 2 is a double salt and gives the test of Fe 2 + and CN– ions. —T/F 17. Atoms of an element having different number of neutrons are known as isotones. —T/F 18. The equilibrium Cr2O72– acidic medium.

CrO42– will exist in

—T/F 19. The formal charge on each hydrogen atom in ammonium ion (NH4+) is equal to zero. —T/F 20. The decreasing order of acidic strength of boron halides is as BBr3 > BCl3 > BF3 —T/F 21. Ethyl alcohol (C2H5OH) is a covalent molecule and hence it is insoluble in water. —T/F → 22. R—C ≡ N and R—N — — C constitute a pair of compounds which are tautomeric to each other. —T/F 23. The coordinate covalent bond formed between two atoms is always non-polar in nature. —T/F 24. Photochemical chlorination of alkanes starts with the homolytic fission of Cl—Cl bond.

—T/F

—T/F

11. The ratio of the velocity of sound in hydrogen (γ = 7/5) to that in helium (γ = 5/3) at the same temperature is

25. The metal ions whose chlorides have high values of solubility products (Ksp) are kept in first group of qualitative analysis of basic radicals. —T/F 26. Ethyne is more acidic than propyne. —T/F 27. Methanoic acid gives silver mirror test with Tollen’s reagent and ethanoic acid does not respond to this test. —T/F 28. T-shape geometry is possible if a molecule or ion has three bond pairs and two lone pairs of electrons. —T/F

21/5. —T/F 12. X-rays are deflected by magnetic field. —T/F 13. 300 gram of water at 25°C is added to 100 gram of ice at 0°C. The final temperature of the mixture is – 1·25°C. —T/F 14. On refraction the frequency of light gets changed. —T/F

C.S.V. / November / 2009 / 1173

29. Both ethanol and ethanal undergo a haloform reaction when treated with a solution of iodine in NaOH. —T/F 30. The dipole moment of NH 3 molecule is higher than that of NF 3. —T/F

crosslinking of proteins, development of autoimmunity, and the presence of an internal biological clock. —T/F 45. In the wall of caecum and colon, the longitudinal muscular layer is thickened along three lines forming the taeniae. —T/F

Zoology 31. Granules of calcium carbonate in vertebrate macula is called otolith. —T/F 32. The right cerebral hemisphere in humans is used for language, speech and mathematical calculations.

Botany 46. Doubling of chromosomes during S-phase of interphase occurs which is accomplished by doubling of DNA. —T/F

—T/F

47. Eukaryotic cells are far larger and more complex than the prokaryotic cells.

—T/F

—T/F

34. Age-related memory loss affects short-term memory more than long-term memory.

48. The free-energy change for ATP hydrolysis is small and negative.

—T/F

—T/F

35. The statocysts in invertebrates perceive the direction of the force of gravity and function as the organ of orientation and equilibrium.

49. The antherozoid of Funaria is nonciliated, because it is nonmotile.

—T/F

50. In sequential reactions of glycolysis, there occurs four chemical transformations.

33. Tumours of muscles are called osteofibroma.

36. Restriction fragment length polymorphism (RFLP) analysis is performed on body tissue or fluid samples to match suspect and victim in a crime. —T/F 37. Sutures are the fusion lines of junctions between adjacent bones of the skull. —T/F 38. Recessive alleles are expressed in heterozygous condition, while dominant alleles are expressed in the homozygous condition. —T/F 39. Fraternal twins are developed as a result of fertilization of a single ovum. —T/F 40. The first humans were Homo habilis, appeared about 1·5 million years ago. —T/F 41. Digestion within a digestive tract of humans is extracellular. —T/F 42. Fibroblasts secrete a living matrix which contains collagen, reticular, or elastic fibres. —T/F 43. The breathing centre in the brain responds to changes in the carbon dioxide concentration of the blood. —T/F 44. Aging may result from an accumulation of genetic mutations, free radicals, or waste products as well as

C.S.V. / November / 2009 / 1174

—T/F

—T/F 51. In lysogenic cycle, the viral nucleic acid is incorporated into the host DNA. —T/F 52. During photosynthesis, oxygen comes from water was proved by Kamen and Ruben. —T/F 53. The part of the embryo of grasses that lies next to the endosperm is called coleoptile. —T/F 54. Monotropa is an obligate saprophyte. —T/F 55. The bivalents, during metaphase-I, become arranged in the plane of the equator to form equatorial plate. —T/F 56. In a hypogynous flower the ovary occupies the highest position on the thalamus. —T/F 57. Natural selection does not maintain favourable genes within a population. —T/F 58. Mycorrhiza is the symbiotic relationship between roots of higher plants and fungi. —T/F 59. The tegmens of seeds develop from inner integument. —T/F 60. Porphyrins are important part of chlorophylls and cytochromes. —T/F

70. What is naturally occurring group of individuals of the same genotype called ? ➠ Biotype 71. What causes black necrosis of stem and root ? ➠ Deficiency of boron 72. Which part of plant— Autumn crocens —is used for the extraction of colchicine ? ➠ Root 73. What does dendrochronology determine ? ➠ Age of tree by radiocarbon dating 74. What is called the protein part of an enzyme ? ➠ Apoenzyme 75. What does primary lysosome mean ? ➠ Lysosomes containing inactive enzymes 76. What is Morchel commonly called ? ➠ Sponge mushroom 77. Who proposed the ‘Operon model’ of gene regulation and organization in prokaryotes ? ➠ Jacob and Monad 78. What is the shape of chloroplast in Ulothrix ? ➠ Girdle-shaped 79. Why does plant’s life require a vascular system on land ? ➠ Because plants must transport water and minerals 80. What is the name of most edible mushroom most often seen on the tables of restaurants ? ➠ Agaricus compestris ●●●

(Continued from Page 1169)

●●●

(Continued from Page 1075) 68. What are those movements in which whole of the plant body or the cell or cytoplasm moves from one place to another called ? ➠ Movements of locomotion 69. What is the arrangement of bundles in dicot herbaceous stem ? ➠ Bundles are arranged in a distinct ring that separates the cortex from the central pith

C.S.V. / November / 2009 / 1176

●●●

Physics Q. Why radio transistors do not work satisfactorily when used inside a railway carriage ? ☞ The railway carriage is made of iron. Iron acts as a magnetic screen. So, the carriage does not permit electromagnetic waves from the radiobroadcasting station to enter the carriage. Thus, the radio transistors do not work satisfactorily when used inside a railway carriage. Q. Do you know how PNP and NPN transistors are separated symbolically ?

☞

(a) PNP-Transistor

(b) NPN-Transistor

Q. Do you know why there is no magnetic field near a conductor when no current is flowing through it ?

☞ When no current is flowing in a conductor, the free electrons in the conductor move randomly on account of their thermal energy. The motion of each electron produces a magnetic field. But since the motion is random, the magnetic fields due to individual electrons cancel out. Thus there is no net magnetic field near a conductor in which no current is flowing. Q. What do you know about the path of a charged particle projected with velocity v into a uniform magnetic field at an angle θ other than 0°° or 90°° ? ☞ The velocity v of the charged particle can be resolved into two components v cos θ (in the direction of the

C.S.V. / November / 2009 / 1177

magnetic field) and v sin θ (perpendicular to the direction of the magnetic field). Due to component v sin θ the charged particle follows a circular path. Due to component v cos θ, the charged particle follows a linear path in the direction of the magnetic field. Thus, the charged particle will follow a helical path under the combined effect of two component velocities. Q. What are laevorotatory substances ? ☞ Laevorotatory substances are those substances which rotate the plane of polarisation towards left or in the anti-clockwise direction when the observer looks at the source of light. Q. Stokes’ formula for viscous drag is not really valid for oil drops of extremely minute sizes. Explain why ? ☞ Stokes’ formula is valid for motion through a homogeneous continuous medium. The size of the drops should be much larger than the intermolecular separation in the medium for this assumption to be valid, otherwise the drop ‘sees’ inhomogeneities in the medium (concentrated mass density in molecules and holes in between them).

→

→

Q. Why E and B are kept perpendicular in Thomson’s experiment ? ☞ The force due to the electric field is in the plane of the field. In order to balance this force, the force due to magnetic field should also act in the plane of the electric field. This is possible only if the magnetic field acts perpendicular to the direction of the electric field. Q. Are matter waves electromagnetic ? ☞ No. This is because electromagnetic waves are produced by accelerated charge. On the other hand, the de-Broglie wave (matter wave) is independent of charge of a particle. Q. What is the difference between the frictional force and gravitational force ?

☞ Frictional forces are passive. It means they come into play only when there is tendency of relative motion between two solid surfaces in contact. While gravitational forces continue to act all the time. Further frictional forces are nonconservative while gravitational force is conservative in nature. Q. What factors affect the elasticity of a material ? ☞ Hammering, rolling and annealing affect the elasticity of a material. An addition of impurity in a material also affects the elasticity. On changing the temperature of a substance its elasticity changes. However, there is no change in the elasticity on changing the temperature of invar (invariable) steel. Q. What are logic gates ? Explain. ☞ The manipulation of binary information is done by logic circuits called gates. A gate is a digital circuit that follows a certain logical relationship between the input and output voltages. They are, therefore, called logic gates. Various logic gates are commonly found in digital computer system. There are three basic gates called OR, AND, NOT. Each gate is indicated by a graphic symbol and its operation is described by means of an Boolean algebra function. The input-output relationship of the binary variables for each gate is represented in tabular form in a truth table. Q. Compare Gauss’s law and Ampere’s law. ☞ Gauss’s law

⎡⎢ → → q ⎤⎥ ⎢⎣ E . ds = ε ⎥⎦ 0

involves the surface integral and helps to determine the charge q.

→ →

B . dl = μ 0I] Ampere’s law [ involves the line integral and helps to determine the current I, i.e., rate of flow of charge. Q. What is meant by ‘inferior mirage’ and ‘superior mirage’ ? ☞ The optical illusion that water is present at some distant place is called inferior mirage. This generally

occurs on very hot summer days. This is due to total internal reflection. The optical illusion of objects floating in air is called superior mirage. It is also known as looming. This occurs in very cold regions. This is also due to total internal reflection. Q. What is the rms value of an A.C.? ☞ The rms value of an alternating current (also called the effective value) is the steady direct current which converts electrical energy to other forms of energy in a given resistance at the same rate as the A.C.

Chemistry Q. What will be the solubility of AgCl in 0·2 M AgNO3 solution ? (Ksp for AgCl = 1 × 10–10)

☞ For AgCl in aqueous solution following equilibrium exists : AgCl

Ag+ + Cl– Ksp = [Ag +] [Cl–] = 1 × 10 –10

Let

[Cl–] = x

Then

[Ag +] = 0·2 + x ≈ 0·2

Hence

0·2 x = 1 ×

10– 10

x = 5 × 10– 10 M Thus, solubility of AgCl in 0·2 M AgNO 3 will be 5 × 10– 10 M. Q. Would H2O2 behave as oxidant or reductant with respect to couple Fe3+/Fe2+ at standard concentration ? ☞ We know that H2O2 can act both as oxidant and reductant. Following reactions occur with respect to Fe 3+/Fe2+ couple. Fe 3+ + e – → Fe 2+ ;

☞ N ≡ C—OH is cyanic acid and HN = C — — O is isocyanic acid. The conjugate bases of each acid are contributing structures of same resonance hybrid, i.e., the resonance structures of both conjugate bases of both acids have same hybrid structure. .. – H+ H:N — —C— — O ⎯→ .–. .. .. – :N— —C— — O ←→ N ≡ C—O . .: .. H:O . . —C ≡ N .. – H+ .–. ⎯→ : O — C = N: – . . —C ≡ N ←→ O — Both have hybrid structure δ–

δ–

˙˙˙ — C˙˙˙ —.O. : :N — Hence conjugate base is same in both cases. Q. What is the carbamic acid ?

☞ This acid is not known in free state, but the salts and esters of this acid exist. It may be represented by chemical formula H2NCOOH. Ammonium carbamate is formed by the reaction of dry CO 2 and dry NH 3. The urethanes, widely used in plastic industry, are the esters of carbamic acid. A commercial sample of ammonium carbonate, contains an appreciable amount of ammonium carbamate. Q. What are soft and hard bases ?

☞ Soft bases have larger and more polarizable basic site atoms (examples : I, Br, S, P etc.) Hard bases have smaller and weakly polarizable sites (Examples : N, O, F etc.). E° = 0·771 V Soft bases have strong nucleophilici2Fe 3+ + H2O2 → 2Fe 2+ + O2 + 2H +; ties and hard bases have diminished E° = 0·02 V nucleophilicites. .. .. .. Fe 2+ → Fe 3+ + e –; – – : .I . : –, : Br . . : , H.S. : , : PH3 etc. are soft E° = – 0·771 V base (strong nucleophiles) 2 Fe2+ + H2O2 + 2H+ .. .. : NH3 , H 2O :, : .F. : – etc. are hard → 2Fe3+ + 2H2O; E° = 1·00 V Since both potentials are positive, bases (weak nucleophiles) H2O2 will act as an oxidant and reducQ. Why is the dipole moment of tant. 1-butyne larger (0·8 D) than 1In fact, iron(II) or iron(III) salts butene (0·3 D) ? catalyse self-oxidation-reduction of ☞ Two compounds have followH2O2. ing molecular formulae :

C.S.V. / November / 2009 / 1178

H H H | 3| 2| — H—C—C—C — | | sp 2 H H

Q. Why do cyanic acid and isocyanic acid have the same conjugate base ?

H | C | H

1-butene

H H | |3 2 H—C—C—C ≡ C—H | | sp H H 1-butyne

The bond between 2 and 3 carbon atoms in 1-butyne (C—C sp) is more polarised than bond between 2 and 3 carbon atoms of 1-butene. This is because sp carbon in 1-butyne has more s -character and high electronegativity than sp 2 carbon in 1-butene. Q. What is fluoridation ?

☞ Fluoride ions present in drinking water, tooth pastes or other sources, can react with the hydroxyapatite [Ca10(PO4)6(OH)2] present in tooth enamel to form fluoroapatite [Ca10(PO4)6F2]. This mineral in which F– has replaced OH–, is much more resistant to attack of acids because F– ion is a much weaker base than OH– ions. About 80% of all tooth pastes contain fluoride compounds at the level of 0·1% fluoride by mass. The most common compounds in tooth pastes are stannous fluoride (SnF2), Sodium monofluoro phosphate (Na2PO3F) and sodium fluoride (NaF). Because fluoride ion is so effective in preventing cavities, it is added to public water supply in many places to give a concentration of 1 mg/L (1ppm). The compounds added may be NaF or NaSiF6. The latter compounds gives fluoride ions by following reaction : 2–

SiF6(aq) + 2H2O(l) –

+

→ 6F(aq) + 4H(aq) + SiO2(s) Q. How Teflon helped development of first atomic bomb ? ☞ Uranium hexafluoride (UF6), which was used to separate fissionable U 235 by gaseous diffusion, in an extremely corrosive material. Teflon was used as a gasket material in the gaseous diffusion plant. It is now used in variety of applications from non-sticks cookwares to space-suits. In 1938 a scientist at Du Pont named Roy J. Plunkett made an accidental

discovery of teflon. This was formed by addition polymerisation of tetrafluoroethylene. Q. What are the plasticisers ? ☞ Polymers like PVC, are extremely hard and brittle resins difficult to be processed. These can be made flexible and easy to handle by adding some low molecular weight chemicals called plasticisers during their formation. For example Bis 2-ethyl hexylphthalate is one of the compounds added to polyvinyl chloride as a plasticiser; which provides it a tough leathery and rubber like texture, easy to be moulded into garden pipes, rain coats, seat covers, sheets and electrical insulators. Q. Who first predicted that chlorofluorocarbons would deplete the ozone shield ? ☞ In 1974, Molina and Rowland pointed out that chlorofluorocarbons (CFCl3 and CF 2Cl2) which had been used in refrigerants and as propellants in aerosol cans were beginning to accumulate in atmosphere. In the atmosphere, at altitude of 10 to 50 km above the earth’s surface, chlorofluorocarbons decompose to form Cl and ClO free radicals. These react with ozone and oxygen atoms. O3 + Cl → ClO + O2 ClO + O → Cl + O2

Zoology Q. What is Ubiquitin ? ☞ Ubiquitin is highly conserved protein (uniform amino acid sequence), apparently present in all eukaryotic cells. Free or covalently bound to a variety of nuclear and cytoplasmic proteins, some integral to membranes. Binding to a protein may initiate the protein’s selective degradation. May act as a protein modulator, possibly of certain histones in chromatin Q. What is Pribnow Box ? ☞ Pribnow box is nucleotide sequence in prokaryote promotor regions, located about six bases upstream of a transcribed region. Contains the sequence TATAATG. Also called a (– 10) region, because of the invariant T residue at base 10 upstream from the start of the transcribed region. However, many pro-

C.S.V. / November / 2009 / 1179

motors whose expression is controlled by sigma factor other than the normal vegetative one have different consensus sequences in different positions upstream of the actual transcription start point. Q. What is lymphoid tissue ?

☞ Lymphoid tissue is vertebrate tissue in which lymphocytes develop (e.g., Bursa of Fabricius in birds, thymus in mammals). Lymphocytes are produced in primary lymphoid tissue (thymus, embryonic liver, adult bone marrow) and migrate to secondary lymphoid tissue (spleen, lymph nodes, unencapsulated lymphoid regions of gut submucosa, respiratory and urogenital regions), where antigen-presenting cells and mature Tcells and B-cells occur. The kidney is a major secondary lymphoid organ in lower vertebrates. Tissue in animals without close chordate affinities may nonetheless form regional aggregates of phagocytes which are loosely termed lymphoid. Q. How light is focussed in human eye to form images ? ☞ Light reflected from an object can be focussed by the lens of eye to form an image of that object on retina. The curvature of the cornea and the lens causes light rays from an object more than 20 feet away to converge on the retina, forming a clear image. However, if the object is closer than 20 feet, the light rays do not converge on the retina, and a clear image is not formed. To focus an image of a near object on the ratina, the lens must change shape, becoming more round. The lens shape is changed by contraction of muscles in the ciliary body. Long periods of close-up work such as reading can cause eyestrain because prolonged contraction fatigues the ciliary muscle cells. To focus on objects 20 feet or more away, the ciliary smoth muscles relax and the lens assumes a more flattened shape. Q. What are gel electrophoresis and DNA probes ? ☞ Gel electrophoresis allows the separation of different sized DNA fragments. A DNA probe has a nucleotide sequence complementary to a desired DNA fragment and allows that fragment to be extracted from a mixture.

Gel electrophoresis separates DNA fragments by causing them to migrate through a porous gel under the influence of an electrical field. Differently sized fragments of DNA contain differents amounts of negative charges. Therefore, these fragments migrate to different locations in the gel when they are placed in an electrical field. Such migration separates different fragments. When separation is complete, the gel can be bathed with dyes or fluorescent chemicals to identify DNA fragments as bands on the gel. A band containing a fragment of interest can be cut out of the gel, and the DNA can be washed from the band for use in gene cloning and recombinant DNA formation. DNA probes are single-stranded nucleotide sequences that are complementary to and thus able to bind with a desired DNA fragment or gene. DNA probes can be synthesized in a test tube by linking either DNA or RNA nucleotides in a specified sequence. Radioactive atoms are often incorporated in the probe to help locate it, after it binds to a complementary DNA fragment. After the electrophoretic separation, a DNA probe and help locate the precise fragment of interest. Q. What changes occur in epidermis and dermis due to growing age ? ☞ As humans grow older, keratinocytes in the epidermis adhere less strongly to each other. As a result, the epidermis becomes more vulnerable to penetration by disease causing microorganisms. In addition, there is a decrease in the production of new cells by the basal stratum. This slows wound healing and enhances the chance of developing an infection. After age of 30, the number of melanocytes in the epidermis decrease by about 20 per cent each decade, making painful sunburn. The melanocytes that remain become larger and gradually cluster more closely together, producing dark spots on the skin, often called ‘liver spots’. Below the epidermis, the dermis also changes with age, becoming thinner and less elastic. Less active fibroblast cells decrease the amount of collagen and elastic fibres in the

Q. What do you mean by benthic microorganisms ?

☞ The microbial inhabitants of the bottom region of a body of water reservoir are referred to as benthic microorganisms. The richest region of an aquatic system in terms of numbers and types of organisms is the benthic region. Various aquatic microorganisms inhabit the gut of marine animals. The lowermost region of a freshwater or marine profile in which the benthos resides, is known as benthic zone. Q. What do you mean by water table ? How is water table shown ? ☞ Water table is the upper surface of the zone of saturation in permeable rocks not confined by impermeable rocks. It may also be defined as the surface underground at which the water is at atmospheric pressure. Below the water table, water

C.S.V. / November / 2009 / 1180

Q. What is mosaicism and how it is caused ? ☞ The coexistence of two or more genetically distinct cell populations derived originally from a single zygote, is termed as mosaicism (mosaic). Mosaics may arise at any stage of development, from the twocell stage onward, or in any tissue which actively proliferates thereafter. Mosaicism is commonly observed in many species of animals and plants and may be caused by somatic mutation or chromosomal nondisjunction. Chromosome non-disjunction is probably the principal cause of chromosomal aberration, which in turn may lead to the development of mosaicism. Somatic crossing-over leads to the production of a recombinant mosaic where chromosome segments, with their corresponding blocks of genes, are exchanged between homologous chromosomes during mitosis. The occurrence of this process leads to mosaicism, mostly manifested as spots (clones of variant cells) on the cuticle of insects or on leaves, petals, or stamen hairs. ●●●

New Release

Mathematical Formulae (Useful for Various Competitive Examinations) Compiled by : Dr. N. K. Singh

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Botany

is free to move under the influence of gravity. The position of the water table is shown by the level at which water stands in wells penetrating an unconfined water-bearing formation. Where a well penetrates only impermeable material, there is no water table and the well is dry. But if the well passes through impermeable rocks into waterbearing material whose hydrostatic head is higher than the level of the bottom of the impermeable rock, water will rise approximately to the level it would have assumed if the whole column of rock penetrated had been permeable. This is called artesian water. Q. What is lethal gene ? When dominant and recessive lethal genes are expressed ? ☞ A gene which brings about death of the organism carrying it, is called lethal gene. Lethal genes constitute the most common class of mutations and the reflections of the fact that the fundamental function of gene is the control of processes essential to the growth and development of organisms. Dominant lethals, expressed in heterozygotes, are rapidly eliminated and thus rarely detected. Recessive zygotic lethals are retained with considerable frequency in natural populations of cross-fertilizing organisms, whereas gametic lethals are subject to stringent selection and are accordingly rare. In higher diploid forms lethals are usually recessive and expressed only in homozygotes. Q. What is mucilage ? How is it formed inside the plant ? ☞ Mucilage is a naturally occurring, high molecular weight (about 200,000), organic plant product of unknown detailed structure. Chemically, mucilage is closely allied to gums and pectins but differ in certain physical properties. In water, mucilages form slippery and aqueous colloidal dispersions. Mucilages are formed in normal plant growth within the plant by mucilage-secreting hairs, sacs and canals, but they are not found on the surface as exudates as a result of bacterial or fungus action after mechanical injury, as are gums. The chief industiral sources of mucilages are Icelandic and Irish moss.

Code No. 1642

extracellular matrix. This substantially reduces the flexibility and elasticity of skin. The nubmer of blood vessels in the dermis also declines with age, and there is a general reduction in blood flow. The remaining vessels become less permeable, and the exchange of nutrients, O2, CO2, water, and wastes are reduced. Fewer white blood cells migrate into the tissues to fight infections and remove debris. The inflammation and blood-clotting responses are slowed, and in the elderly even a slight bump produces significant bruising and discolouration. The reduced flow of blood to the dermis also reduces service to epidermis, which has no vessels and relies on diffusion to and from the dermis. Because of the reduced blood flow, older skin is paler in colour and functions less effectively in temperature regulation. Difficulty in maintaining body temperature is associated with a decrease in the thickness of the body’s layer of fatty insulation in the hypodermis. As people age, fat storage shifts from the skin to the body cavities. This reduction in fat content below the skin, together with abnormal cross-linkage between collagen proteins, contributes to skin wrinkling. In some individuals changes in fat distribution may even alter body proportions.

HINDI EDITION

Code 248 Rs. 76/Upkar Prakashan, AGRA-2 ● E-mail : [email protected] ● Website : www.upkar.in

1. A weight of 290 N and another of 200 N are suspended by a rope on either side of a frictionless pulley. The acceleration of each weight is— (A) 1·5 ms–2 (B) 1·8 ms–2 (C) 2·2 ms–2 (D) 2·5 ms–2 2. The potential energy of a particle in a certain field is given by a b U = 2 – where a and b are r r positive constants and r is the distance from the centre of the field. The distance of particle in the stable equilibrium position is— a a (A) (B) – b b 2a 2a (C) (D) – b b 3. A solid sphere (density ρ and specific heat capacity c) of radius r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. The time required (in μs) for the temperature of the sphere to drop to 100 K is— 72 r ρc 7 r ρc (A) (B) 7 σ 72 σ (C)

27 r ρc 7 σ

(D)

7 r ρc 27 σ

4. A metallic sphere having inner and outer radii a and b respectively has thermal conductivity k0 (a ≤ r ≤ b) k = 2 r The thermal resistance between inner surface and outer surface is— b 2 – a2 b–a (B) (A) 4πk0 4πk0ab 4πk0 (C) (D) None of these b–a 5. In the following circuit current I drawn from 5V source will be— 10 Ω 5Ω 10 Ω

I

20 Ω

10 Ω + 5V

C.S.V. / November / 2009 / 1182

(A) 0·33 A (C) 0·67 A

(B) 0·5 A (D) 0·17 A

6. What is the IUPAC name of (CH3)2CHCH2CH3 ? (A) 2-methyl butane (B) Dimethyl propane (C) Trimethyl ethane (D) Tetra-2-methyl ethane 7. The speed of electron in the first orbit of hydrogen atom in the ground state is about— (c = Velocity of light) c c (A) (B) 13·7 137 c c (C) (D) 1·37 1370 8. Which of the following has maximum flocculation value ? (A) Cl– (B) SO42– (C) [Fe(CN6)4–] (D) PO43– 9. Diketones are classified according to the— (A) O-atom (B) C-atom (C) —CO-atom (D) —CO2-atom

14. Stuart power factor of blood clotting is— (A) Factor XIII (B) Factor X (C) Factor IX (D) Factor III 15. Dartos tunic muscle fibres occur in— (A) Abdominal wall (B) Uterine wall (C) Urethral wall (D) Scrotal wall 16. Mutation at one base of the codon of a gene yields a nonfunctional protein; such a mutation is called— (A) Nonsense mutation (B) Missense mutation (C) Frame-shift mutation (D) Reverse mutation 17. The primitive atmosphere of the earth was— (A) Normal

(B) Neutral

(C) Oxidizing (D) Reducing

10. Which of the following is a direct method for the production of aromatic aldehydes by the oxidation of the methylated homologues of benzene by chromyl chloride, Cr2OCl2 ? (A) Grignard’s reaction (B) Green acid (C) Grothus-Draper law (D) Etard’s reaction

18. Proteins are diverse in nature because of—

11. Lamina propria is associated with— (A) Acinus of pancreas (B) Gut mucosa (C) Graffian follicle (D) Liver

19. The pyrenoid of Spirogyra, belonging to a green alga, is specially meant for— (A) Starch synthesis (B) Amino acid synthesis (C) Protein synthesis (D) None of the above

12. The arm of a starfish will grow again after being cut off, this type of regeneration is called— (A) Autotomy (B) Epimorphosis (C) Morphollaxis (D) Cyclosis 13. The males of ants, bees and wasps are— (A) X/X (B) X/O (C) Haploid (D) Polyploid

(A) Complexity of amino acids (B) Different molecular weight of amino acids (C) Different molecular nature of amino acids (D) Different arrangement of amino acids

20. Geothermal energy is represented by— (A) Non-renewable and nonconventional (B) Renewable and non-conventional (C) Non-renewable and conventional (D) All of the above ●●●

Rules for taking part in Quiz Contest of Competition Science Vision 1. 2.

3. 4.

5. 6. 7.

8.

All students or those appearing in competitive examinations can take part in this contest. Candidates taking part in quiz contest will necessarily have to send their entries by a fixed date. Entries are to be sent by ordinary post. Please mark your envelope 'Quiz–Competition Science Vision' on the top left hand side. Answers given only on the form of the magazine will be admissible. In the form there are four squares against each question number. Contestants should put a cross (×) in the square for the answer they think is correct. Giving more than one answer to a question will disqualify it. Contestants should essentially write the number of questions they have solved. Marks will be deducted for wrong answers. The candidate sending the maximum number of correct answers will be given Rs. 600 as first prize. Next two candidates after that will get Rs. 400 and Rs. 300 as second and third prize respectively. If there are more than one candidate eligible for a prize, the amount will be equally distributed among them. The decision of the editor will be final and binding in all cases, and will not be a matter for consideration of any court.

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Solution to Quiz No. 138 Competition Science Vision Last date for sending 28th November, 2009 Name Mr./Miss/Mrs. ...........................…......................... Full Address ...................................….......................... .............................................................………………… ............................................................………………… State .......................Pin Code No. Age.................. Academic Qualification........................ Competition examination for which preparing .................................................................................... I have read and understood the rules of quiz contest of Competition Science Vision issued by Pratiyogita Darpan and agree to abide by them. ................................... (Signature) RESULT

SALES/FINANCIAL EXECUTIVES TEST By : Dr. Lal & Jain Code 1666

Price : Rs. 195/-

This book contains ☞ Reasoning ☞ Numerical Ability ☞ General Knowledge Hindi Edition

Code No. 1478 Price : Rs. 160/UPKAR PRAKASHAN, AGRA-2 C.S.V. / November / 2009 / 1183

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

No. of questions attempted.......................................... No. of correct answers................................................. No. of wrong answers................................................... Marks obtained............................................................. ANSWER FORM Q. No. A

B

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D

Q. No. A

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11.

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12.

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14.

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15.

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B

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According to the rules of the CSV Quiz, all entry forms were examined. As a result, the following participants have qualified for various prizes. CSV sends them greetings and good wishes for their bright future. It also places on record its appreciation for their inquisitive nature and expresses obligation for their co-operation.

PRIZE WINNERS First Prize Gagandeep Singh C/o Dayaram Verma L–971, Shastri Nagar, Meerut U.P.–250 004 Second Prize Manita Kumari C/o Pawan Kumar Pal, Patna Bihar–800 007

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(Continued from Page 1172)

Third Prize Ravi Jaiswal C/o Gaurav Jaiswal Room No. 88, A. N. Jha Hostel, University of Allahabad, Allahabad U.P.–211 002

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(Continued from Page 1070) Professional Billiards Champion. In a surprising one-sided five-hour final in Leeds (England) on September 7, 2009, the 24-year-old outclassed the 41-year-old Mike Russell 2031–1253 to end a remarkable domination by the Englishman who now lives in and represents Qatar. Russell has won the title a record nine times and had been in the finals 17 times. Advani, who won the World amateur snooker championship in China six years ago and the World amateur billiards championship in Malta four years back, is the second after Malta’s Paul Mifsud to bag both the amateur billiards and world snooker titles. ●●●

C.S.V. / November / 2009 / 1184

1. The language spoken by the largest number of people in the world is— (A) Hindi (B) English (C) Mandarin (D) Spanish

(C) Bahadur Shah Zafar (D) Faiz Ahmad Faiz

8.

2. The Constitution of India recognises— (A) Only religious minorities (B) Only linguistic minorities (C) Religious and linguistic minorities (D) Religious linguistic and ethnic minorities 3. The Sikh Guru who was put to death by the order of Aurangzeb was— (A) Guru Angad (B) Guru Teg Bahadur (C) Guru Govind Singh (D) None of the above 4. After the death of Aurangzeb in 1707, which of the following ascended the throne by assuming the title of Bahadur Shah I ? (A) Azam (B) Kam Baksh (C) Muazzam (D) None of the above

9.

10.

14. The most short-lived of all the Britain’s Constitutional experiments in India was the— Which of the following is not (A) Indian Councils Act 1861 correctly matched ? (B) Indian Councils Act of 1892 Article Provision (C) Indian Councils Act of 1909 (A) Article 20 (D) Government of India Act of —Protection against arrest and 1919 detention in certain cases (B) Article 21(A) 15. ‘Vilnius’ is the capital of— —Right to Education of Children (A) Estonia (B) Macedonia of the age 6 to 14 years (C) Eriterea (D) Lithuania (C) Article 24 16. Malaysia’s currency is— —Prohibition of Employment of (A) Peso (B) Rupiah Children below 14 in factories (C) Ringgit (D) Zloty (D) Article 26 17. Which of the following is not the —Freedom to Manage Religious warm wind ? Affairs (A) Zonda (B) Khamsin For which one of the following is (C) Mistral (D) Samun capillarity not the only reason ? 18. The largest river of Europe Volga (A) Blotting of Ink rises/originates from— (B) Rising of underground water (A) Altai Mountains (C) Spread of water drop on a (B) Valdai Plateau cotton cloth (C) Baikal Mountains (D) Rising of water from the (D) None of the above roots of a plant to its foliage Which of the following is not 19. The great River Mekong does not pass through— correctly matched ? (A) Malaysia (B) China Substance Element (C) Thailand (D) Laos (A) Diamond Carbon (B) Marble Calcium 20. Which of the following is not correctly matched ? (C) Sand Silicon Railway ZoneHeadquarters (D) Ruby Magnesium (A) East Central The term ‘Aryan’ denotes— Railway Hajipur (A) An ethnic group (B) West Central (B) A nomadic group Railway Jabalpur (C) A superior race (C) North Central (D) A speech group Railway Allahabad In eye donation, which of the (D) South Central following parts of donor’s eye is Railway Hyderabad utilized ? 21. The territorial jurisdiction of (A) Iris (B) Lens Guwahati High Court does not (C) Cornea (D) Retina cover—

5. Which of the following is incorrectly matched ? State Founder (A) Avadh Burhanul Mulk Saadat Khan (B) Hyderabad Chin Qulich 11. Khan (C) Bengal Murshid Quli Khan (D) Rohilakhand Shuja-UdDaula 12. 6. Which Maratha Peshwa is known by the title–Nana Sahib ? (A) Balaji Baji Rao (B) Baji Rao I (C) Balaji Vishwanath 13. In which one of the following does (D) Madhav Rao low temperatures (Cryogenics) not find application ? 7. The first writer to use Urdu as the (A) Space Travel medium of poetic expression (B) Magnetic levitation was— (C) Telemetry (A) Amir Khusro (D) Surgery (B) Mirza Ghalib

C.S.V. / November / 2009 / 1185

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

Assam Manipur Sikkim Arunachal Pradesh

22. Which of the following is not one of the Fundamental Duties as enshrined in Article 51-A (Part

IV-A) of the Constitution of India ? (A) To value and preserve the rich heritage of our Composite Culture (B) To develop the scientific temper, humanism, spirit of enquiry and reform (C) To safeguard public property and to abjure violence (D) To build public opinion to root out social evils 23. Who among the following was the speaker of the Lok Sabha before he became the President of India ? (A) Shankar Dayal Sharma (B) R. Venkataraman (C) Dr. Neelam Sanjiva Reddy (D) B.D. Jatti 24. Under which Article of the Constitution of India, has it been provided that the Vice-President shall be an ex-officio Chairman of the Council of States ? (A) Article 62 (B) Article 64 (C) Article 65 (D) Article 67 25. Under the Constitution (Fortysecond Amendment) Act 1976, which of the following clauses/ provisions was not added to the Constitution of India ? (A) Ideals of socialism, secularism and the integrity of the nation were inserted (B) A new chapter on Fundamental Duties was inserted (C) Provision of Administrative and Other Tribunal was made (D) The right to property as a Fundamental Right was omitted 26. In order to win the Grand Slam in Tennis, a player must win which one of the following groups of tournaments ? (A) Australian Open, French Open, Wimbledon Open, U.S. Open (B) Wimbledon, French Open, U.S. Open (C) Wimbledon, French Open, Paegas Czech, U.S. Open (D) Davis Cup, Wimbledon, French Open

C.S.V. / November / 2009 / 1186

27. The real founder of Portugese power in India was— (A) Francisco de-Almeida (B) Alfonso-de-Albuquerque (C) D. J. Castro (D) Vasco-da-Gama 28. The first French factory in India was established in— (A) Masulipattanam (B) Kolkata (C) Surat (D) Ahmedabad 29. Arya Samaj was founded by— (A) Swami Vivekanand (B) Swami Dayanand (C) Ram Krishna Parmahansa (D) None of the above 30. In regard to revolt of 1857, which of the following is incorrectly matched ? Revolt Centre Leader (A) Bareilly Ali Mohammad Khan (B) Bihar Vir Kunwar Singh (C) Allahabad Liyakat Ali (D) Kanpur Nana Sahib 31. The river which makes the World’s longest estuary is— (A) Lena River (B) Ob River (C) Amazon River (D) Yellow River 32. Which of the following is incorrectly matched ? City River (A) Paris Danube (B) Karachi Indus (C) Stalingrad Volga (D) Yangoon Irrawdy 33. The is— (A) (B) (C) (D)

old name of ‘Burkina Faso’ Lower Volta Upper Volta Gold Coast None of the above

34. A person of mixed European and Indian blood in Latin America is called— (A) Mulatto (B) Mestizo (C) Meiji (D) Mau Mau 35. Endoscopy, a technique used to explore the stomach or other inner parts of the body is based on the phenomenon of— (A) Total internal reflection (B) Interference (C) Diffraction (D) Polarization

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