Do not allow the quest for perfection to ruin your life because whatever you do you will always feel that you could have done better
Volume - 5 Issue - 3 September, 2009 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009
Editorial
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Dear Students,
Pramod Maheshwari [B.Tech. IIT-Delhi] Analyst & Correspondent Mr. Ajay Jain [B.E] Cover Design & Layout Niranjan Jain Om Gocher, Govind Saini Circulation & Advertisement Ankesh Jain, Praveen Chandna Ph (0744)- 3040007, 9001799502 Subscription Sudha Jaisingh Ph. 0744-2500492, 2500692 © Strictly reserved with the publishers • No Portion of the magazine can be published/ reproduced without the written permission of the publisher • All disputes are subject to the exclusive jurisdiction of the Kota Courts only. Every effort has been made to avoid errors or omission in this publication. In spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from.
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The difference between success and failure is your attitude towards success and the strategies that you employ to achieve it. The difference between success and failure is only a few minutes or a few hours everyday. You have to keep on striving for success at every conceivable opportunity. Never postpone your happiness and zest for life and work. You should make it a habit to enjoy your profession and your job all the time. Never be a quitter because a quitter can never be a winner. You should always remember that People live not by the reason of any care they have for themselves but by the love for them that is in other people. Have only those people for friends and companions who do their best to bring out the best in you. They will be of unlimited worth to you. Such persons understand what life means to you and your goal. They feel for you as you feel for yourselves. They are the ones who are bound to you in triumph and disaster. They provide a purpose to live and break the spell of loneliness. A true friend is worth befriending as he will always stand by you. But before you expect others to be the right person to be your friend you must also become one. Be always committed to your cause. Be so engrossed in your work that you have hardly any time to think of anything else. The great secret of success is to do whatever you are to do and do it wholeheartedly. Make yourself the star of your workplace. For this you must have clear and precise objectives to be achieved within a definite time-frame. Always respect and value time. Be result-oriented and keep track of the hours. Respect the time of others as well as your own. Be always organized and write down everything you want to accomplish. Always make an assessment of yesterday's "To Do" list to crosscheck how realistic it has turned out to be today. This will help you to avoid or rectify mistakes, if any, in your planning. Keep on visualizing your goals and lists of the task to be done. Forever presenting positive ideas to your success. Yours truly
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Editor : Pramod Maheshwari XtraEdge for IIT-JEE
Pramod Maheshwari, B.Tech., IIT Delhi
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SEPTEMBER 2009
Volume-5 Issue-3 September, 2009 (Monthly Magazine) NEXT MONTHS ATTRACTIONS
CONTENTS INDEX
Regulars ..........
Key Concepts & Problem Solving strategy for IIT-JEE. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics, Chemistry & Maths
NEWS ARTICLE
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IITian ON THE PATH OF SUCCESS
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KNOW IIT-JEE
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IIT-Kanpur students made Nano-satellite IIT-K all set to start FM radio station
Much more IIT-JEE News. Xtra Edge Test Series for JEE-2010 & 2011
PAGE
Mr. Kannan M. Modgalya
Previous IIT-JEE Question
Study Time........ DYNAMIC PHYSICS S
• "True success is overcoming the fear of being unsuccessful."
8-Challenging Problems [Set# 5] Students’ Forum Physics Fundamentals Current Electricity Circular Motion, Rotational Motion
• "Get up one time more than you're knocked down."
CATALYST CHEMISTRY
Success Tips for the Months
• "When your physical environment is in alignment with your aspiration, success becomes the norm." • "The most important single ingredient in the formula of success is knowing how to get along with people." • "Dictionary is the only place that success comes before work. Hard work is the price we must pay for success. I think you can accomplish anything if you're willing to pay the price."
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Key Concept Aliphatic hydrocarbon Oxygen Family & Hydrogen Family Understanding : Inorganic Chemistry
• "Most people who succeed in the face of seemingly impossible conditions are people who simply don't know how to quit." • "The truth is that all of us attain the greatest success and happiness possible in this life whenever we use our native capacities to their greatest extent."
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DICEY MATHS
42
Mathematical Challenges Students’ Forum Key Concept Probability Binomial Theorem
Test Time .......... XTRAEDGE TEST SERIES
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Class XII – IIT-JEE 2010 Paper Class XII – IIT-JEE 2011 Paper
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SEPTEMBER 2009
IIT-Kanpur students made Nano-satellite
Kanpur: The Indian Space Research Organization (ISRO) will launch Nano-satellite "Jugnu" made by students of the Indian Institute of Technology, Kanpur (IIT-K) in December 2009. "A team of 20 students from our institute has made the nanosatellite," said Prof. S.G. Dhande, Director of IIT-Kanpur. The weight of the satellite is less than 10 kilograms. The satellite is made worth of 2 ½ Crores. The satellite will give information related with drought, flood, agriculture and forestry. In addition to this IIT-K will also celebrate its golden jubilee from August 2009 to December 2010. The inauguration is from August 8th-9th 2009. Inaugural address will be delivered on August 8. Mr. N. R. Narayana Murthy, the non-executive chairman and chief mentor of Infosys, will join the function as Chief Guest. He is a distinguished alumnus of IIT-K.
IIT-K all set to start FM radio station
Kanpur: The Indian Institute of Technology, Kanpur (IIT-K) will soon start an FM radio station in the campus. This is for the first time an FM radio station is going to be operational from IIT campus. The FM station will broadcast scientific programmes and classical music.
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IIT-K has already received green signal for the project from the Ministry of Information and Broadcasting. The FM station in IIT-K will operational following the model of America's National Public Radio (NPR). Cultural programme will also air by the upcoming FM radio station at IIT-K. "We have decided the land for the FM station and instruments worth Rs.22 lakh have been purchased. All formalities have also been completed to start the FM station," said Prof. Dhande, Director IIT-K. A four-member committee headed by Prof. Sanjay Kasalkar, Registrar of the institute has been constituted to put the whole project into operation. The most important feature of the FM station will be the phone-in programme in which students from across the city can put their queries relating to science to the professors of the college.
IIT Delhi to hone marketing skills at tourism ministry NEW DELHI: The tourism ministry has, for the first time, organised a three-day training programme for its overseas officials by experts at the Indian Institute of Technology (IIT) Delhi in an effort to sharpen their marketing skills. Talking about the programme, Tourism Minister Kumari Selja said that the ministry's officials will be trained to promote India better. 'Our overseas officials who promote India as an ideal travelling destination will get this training on better and more effective marketing and promotion 3
technique from IIT Delhi's management wing. It will be a three-day training programme,' Selja said at the inaugural function of the ministry's two-day overseas marketing meet here. 'After this, we are planning a similar training programme for our domestic officials as well,' Selja added. Twenty tourism officials will attend the first programme. 'This meet is very important, especially in the light of the global meltdown and its drastic effects on tourism. From January to June this year our foreign tourist arrival fell by 9.3 percent as compared to last year,' Selja said. 'However, our efforts to bring India back on the tourists' map, the several roadshows in different countries and promotion of niche tourism products has ensured that in June this year there was a positive growth of 0.2 percent in the foreign tourist arrival to India,' she added. Selja further said that the meet, which will see overseas tourism officials and the private sector here interacting, is bound to throw up innovative ideas to attract more travellers to the country.
More than 500 candidates refused to get into IITs Mumbai: To get into Indian Institutes of Technology (IITs) is considered as "one of the hardest nuts to crack", but this year more than 500 students have refused to get into IITs after qualifying for the Joint Entrance Test, conducted by IITs. All students have their own reason to refuse to study in IITs. Some of them have no confidence SEPTEMBER 2009
in the new IITs, while some of them did not get their own choice of streams. An IIT official said, "This experience might force HRD minister Kapil Sibal to do a rethink on his expansion plans for the IITs." In addition to existing seven IITs, eight IITs that are more new have been added during the eleventh five-year plan in the country. After not getting sufficient number of qualified student for the reserve categories, IITs have transferred 1100 reserved-category seats to the one-year preparatory course. The one-year 'prep course' trains quota students to bring them up to the mark. But this year more than 500 candidates who have refused admission to IITs are from open category.
IT BHU soon to become IIT New Delhi: The Institute of Technology, Banaras Hindu University (IT BHU) will soon join the league of Indian Institute of Technology (IIT). The Ministry of Human Resource Development (MHRD) has decided to convert IT BHU into an IIT. The decision was conveyed in the Lok Sabha on Monday by D. Purandeswari, Minister of State for HRD. The decision was conveyed when she replied to a question on whether the government has taken any steps in developing technology institutions across the country on the lines of IITs. For a long time, IT BHU had been in a process of upgrading itself into an IIT. The admission process for students is already being done through the IIT Joint Entrance Examination (IIT JEE). As per S.N. Upadhyaya, Director, IT BHU, "The IIT status will not only improve the infrastructure and academics of the institute but
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will also get better students and attract bright faculties". "With more academic freedom, the institute will be able to introduce inter-disciplinary courses that were not possible earlier as we were tied up with the rules and regulations of the university, to go through various councils to get a new academic programme passed. But the status of IIT will enable us to introduce new courses without these limitations" he added. "Appointing faculties will also become much easier," he avers. Seven other institutes comprising of Bengal Engineering College; Howrah, Cochin University of Science and Technology; Kochi, Engineering and Technology Department of Jadavpur University and Zakir Hussain College of Engineering and Technology; Aligarh Muslim University apart from IT BHU are being considered to become a part of the IIT brigade.
IIT-D to conduct annual convocation in Aug New Delhi: The Indian Institute of Technology, Delhi (IIT-D) will award 180 PhD degrees to students,the highest ever. Scholars will be awarded their degree at the convocation taking place next month on August 8 and 9 in the campus. Last year 147 students were awarded PhD. IIT-D has taken several steps to make the institute as a reputed research centre in the country and attract students towards research programme. Undergraduate and Postgraduate students will also be awarded their degrees during the convocation
IIT-Delhi to spend Rs.1 crore on tightening security New Delhi: Trespassing is passé at the Indian Institute of Technology, Delhi (IIT-D) as the institute is all geared up to tighten their security levels. 4
IIT-D is planning to shell out Rs.1 crore on enhancing the level of security within next few months. This will include installation of close circuit television (CCTV) cameras, introducing electronic access system in hostels, academic areas such as library, reading room and canteen. "The raising number of population in the campus made us take this step. This being an educational campus it was practically not possible to keep a vigil on all those entering or leaving the campus. The new measures will help us in safeguarding the campus," said Surendra Prasad, Director, IIT-D. The surveillance at all the five entry gates of the campus by 17 CCTV cameras, will ensure that the 325 acres of the campus which is pretty permeable to the people from the neighbouring villages, cannot be easily breached. With the new security measures, the institute hopes to overcome the menace of trespassers and petty thefts at the campus. Nearly 5,500 students of IIT-D will be issued new identification cards before the end of this year. "The new cards will act as a single admit card to the library, hostel and reading room. Instead of the current cards that act as their identity cards as well as the library card. This will limit the entry and exit of the outsiders in the academic areas," said Captain B.N. Yadav, Security Officer, IIT-D.
Three new MTech programmes at IIT this year CHENNAI: Three new M.Tech programmes in Catalysis Technology, Nuclear Engineering and Petroleum Engineering will be offered at IIT-Madras in the 2009-10 academic year, director M S Ananth announced on Friday Delivering the director's report at the 46th convocation of the institute here on Friday, he said a five-year integrated dual degree programme would also be introduced in these disciplines. SEPTEMBER 2009
"IIT-Madras is establishing a research park adjacent to its campus in Chennai, the first of its kind in the country. For this purpose, IIT-M has obtained 11.42 acres of land from the state government," he added. Pointing out that the facility was likely to be inaugurated in September, professor Ananth said they hoped to have the prime minister, the Tamil Nadu chief minister and Union Minister for Human Resource Development (MHRD) Kapil Sibal at the inaugural event. A total of 1,439 students received degrees 162 were awarded Ph.D, 121 M.S., 354 M.Tech, 83 M.Sc, 62 MBA, 313 B.Tech awards and 172 got dual degrees. Professor Jagdish N Bhagwati of the department of economics at Columbia University, who was conferred the degree of Doctor of Science (Honoris Causa) by the institute on the occasion, delivered the convocation address. "You should have learnt from your IIT education that great universities teach you two things: creation of knowledge and the practice of virtue. As scientists, you must imbibe knowledge and seek to extend its frontiers. As part of humanity, you must also learn to put it to good use," he said. In addition to instructing students on the need for them to periodically adjust and revise their knowledge, professor Bhagwati reminded his audience that each of them owed his/her success equally to their parents. "Great opportunities are open now to put your knowledge, and your ingenuity, at the service of India and her poor," he said. Touching on the problem of limited availability of qualified faculty in the higher education system today, R Chidambaram, principal scientific advisor to Government of India, and chairman, board of governors, IITMadras, said online teaching could be one way of ensuring quality
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technical education. "An alternative to the required growth in brick-and-mortar through expansion and addition is a massive online education system. A country like India therefore has no choice. The National Programme on Technology Enhanced Learning (NPTEL), liberally funded by the MHRD, is an opportunity to provide quality online engineering education," he said.
Bill Clinton to address IIT summit in Chicago Former American President Bill Clinton will be among many top leaders to address the seventh Pan-IIT Global Conference being held in Chicago in October. Kapil Sibal, Indian minister for human resource development; Sam Pitroda, Indian Knowledge Commission chairman; Aneesh Chopra, America's chief technology officer; and Meera Shankar, Indian ambassador in the US, will be among other keynote speakers at the Pan-IIT Global Conference to be held in Chicago from Oct 9-11. According to conference chairman Ray Mehra, 'Entrepreneurship and Innovation in a Global Economy' is the theme of this year's techie summit. Over 3,000 IITians from around the world will attend the annual gathering to be opened by Sibal. Chopra will deliver the keynote address. "We have invited President Clinton since the goals of his William J. Clinton Foundation and the Pan-IIT conference are the same. We have common areas like energy, climate change, health care and education to work on,'' Mehra told IANS. "The president and other global leaders in their fields will discuss how we can transform ideas into action on both sides of the ocean (in the US and India),'' he said. Mehra said the Pan-IIT summit will take a holistic approach to
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problems in areas like health and energy in India. "We will discuss how the public health system (PHS) in India can be steered with inputs from the PHS in the US which is under massive changes now,'' he said. Mehra said the conference will also discuss its proposal called Panch Ratnas submitted to the Indian government to revolutionize higher education in the country. "We presented a white paper titled PanIIT Panch Ratnas to President Pratibha Patil last month, proposing a five-point action plan to make India the global hub for knowledge creation and talent development by 2022. We will debate this in detail in Chicago,'' he said. Pan-IIT's Panch Ratnas include implementation of wholesale policy reforms in education, quality control and increase capacity, and 'quantum improvement in faculty service conditions, deployment of technology for teaching and collaborative research, and the establishment of an industryacademia interface, according to Mehra. Other prominent speakers at the summit include James Owens, chairman and CEO of Caterpillar Inc., Sharon Oster, dean of the Yale School of Management, Tulsi Tanti, chairman of Suzlon Energy, Carl Shramm, president and CEO of the Kauffman Foundation, and Prof Raghuram G Rajan of the University of Chicago and former chief economist of the IMF. There are said to be an estimated 35,000 IITians in the US.
Science learning materials on Net soon By the end of this year, elementary school teachers and students can find an interactive, experiential science learning programme developed in association with IIT-Madras, freely available on the Internet. The SEPTEMBER 2009
Kuruvila Jacob Initiative for promoting excellence in school education, which was set up in memory of the headmaster of the Madras Christian College High School, has been running a programme to develop multimedia-enhanced science learning materials for Classes VI to IX, in partnership with IIT-Madras since February 2007. At the sixth annual function of the Initiative, launched on Kuruvila Jacob’s birth centenary, IIT-M Director M.S. Ananth launched the latest DVDs of Physics and Chemistry material, and announced that the whole project would be ready for a Web launch by the end of 2009. “From the very beginning, it was decided to make it available in the public domain,” said Dr. Ananth, explaining how the modules were developed using the same equipment and team which are putting IIT’s engineering courses on YouTube. About 80 teachers, from 22 schools in the city, were involved in developing the materials. “An experiment showing how the electron moves is worth three lectures in quantum mechanics,” said Dr. Ananth, discussing the experiential and demonstrative mode of teaching that the project has attempted to use in each module. He believes that distance education can help fill the gaps in the Indian school education system, just as it is starting to do in higher education as well. “The ratio of teachers to students is 1:100 or worse,” he pointed out. “We are moving from a gurukulam education at the home of the guru to what I call sishyakulam at the homes of the sishyas, no matter how scattered they may be,” he said, explaining that the Internet version of these science modules would allow students to study from the comfort of their own homes. Earlier, historian and writer Ramachandra Guha delivered the
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keynote address at the function, on the topic ‘Why India is the most interesting country in the world.’ He explained that the country was undergoing simultaneous revolutions in at least five areas – industrial, urban, national, democratic and social – in a way that was unprecedented in international history.
IIT-K disappointed over rejection of setting up of n-reactor In a disappointment to the officials of IIT Kanpur, the Department of Atomic Energy has rejected the college's proposal of setting up of a small nuclear reactor in its campus citing security reasons. However, the officials of IIT Kanpur said they were trying to clear the doubts and will try to convince the Department of Atomic Energy at the 'International Nuclear Meet' in November. Director of IIT Kanpur Prof Sanjay Govind Dhande said that the college asked for a permission from the Department of Atomic Energy, Mumbai to set up a nuclear reactor for M.Tech students, who are doing Nuclear Engineering. "We have asked for the permission from the Department of Atomic Energy, Mumbai, for a small reactor in the campus. But due to security reasons as cited by the Department it has been rejected," he said. He said the college was hopeful of a positive reply after India signed the 123 agreement with the US. According to the college officials, since the last 35 years the college has been providing Nuclear engineering course to 15 students each year but they were able to give only theory classes. "The college has been giving only theory classes to the students. Due to the lack of a nuclear reactor facility in the college, the students are devoid of any practical training. 6
He added the Department has not only cited security problems but has also mentioned about the lack of nuclear fuel in the country.
IIT-D faculty gets National Award for Atmospheric Science and Technology
New Delhi: Prof. Shishir Kumar Dube, former Director of Indian Institute of Technology, Kharagpur and currently Professor at the Centre for Atmospheric Sciences, IIT, Delhi was awarded with National Award in Atmospheric Sciences for the year 2009.The award has been given to Prof. Dube by Ministry of Earth Sciences in recognition of his outstanding contributions in the field. He was born in Kalpi District of Uttar Pradesh on 4th October 1947. He worked at the India Meteorological Department from 1972-78 and then joined the faculty of IIT, Delhi. Research interests of Prof. Dube include Numerical Storm Surge Prediction, Ocean Wave Modeling, Coastal Marine Hazards and Regional Ocean State Forecasting Models. He is internationally recognized for his pioneering contributions in the field of storm surge prediction in the Bay of Bengal and the Arabian Sea. Prof. Dube is responsible for the development of real time operational surge prediction systems which under the auspices of World Meteorological Organisation have been transferred to the National Weather Services of Bangladesh, Maldives, Myanmar, Oman, Pakistan, Sri Lanka and Thailand for their operational use. Prof. Dube has received several honors and awards including the 12th MAUSAM Award for the year 1982-83.
SEPTEMBER 2009
Success Story This article contains story of a person who get succeed after graduation from different IIT's
Mr. Kannan M. Modgalya B.Tech, IIT Madras Master of Electrical Engineering, Doctorate (Ph.D) in Chemical Engineering, Rice University
Mr. Kannan M. Modgalaya completed his B.Tech in Chemical Engineering, IIT Madras, June 1980 and after that he was awarded by mastered degree as Master of Electrical Engineering from Rice University in May 1985 then achived Doctorate (Ph.D) in Chemical Engineering, Rice University, May 1985
of performance deterioration through network delays is being studied. Simulation
Presently he is related to various reseach work and awarded with with various awards and fellowship that are described below:
He has worked on the topic of simulation methodologies and simulation environments through modern computing tools. This approach has been applied to simulation of neuro transmission in muscle cells. Automatic model derivation from first principles and data driven model generation are some focus areas.
Research Areas
Awards and Affiliations
Dynamical Systems, Discontinuity Detection, Discontinuity Sticking, Differential Algebraic Equations, Process Control, Digital Control, Reactor Control, Grade Transition.
1.
Best Paper Presentation Award for "Control of a high index DAE system through a linear control law" by P. Vora,
Current Research
2.
K. Moudgalya and A. K. Pani, American Contro Conference, Anchorage, 8 May 2002.
3.
Lovraj Kumar Industry-Academia Exchange Fellowship, April 1997
4.
Best Poster Award for the paper ``An Integrated Simulation Environment'' by S. H. Rao, K. Moudgalya, K. V. Nori.
5.
G. Sivakumar, International Conference on Advances in Chemical Engineering, 11-13 Dec. 1996, IIT Madras.
Dynamics He is studying discontinuous dynamical systems that exhibit the property of sliding. The concept of equivalent dynamics has been shown to speed up sliding DAE systems by 10,000 times. Control He is looking at the control of chemical processes, especially reactors. He also looked at control of systems, such as, an inverted pendulum through internet. Handling
Dare to dream, dare to try, dare to fail, dare to succeed XtraEdge for IIT-JEE
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SEPTEMBER 2009
KNOW IIT-JEE By Previous Exam Questions
Q = mc ∆T ...(iii) From (i) and (iii) Since U = Q Therefore
PHYSICS
1 M 2g 2l 2 πr 2 Y 1 M 2g 2l ∴ ∆T = 2 πr 2 Ycm Here m = mass of string = density × volume of string = ρ × πr2l
∴ mc∆T =
1.
A transverse harmonic disturbance is produced in a string. The maximum transverse velocity is 3 m/s and maximum transverse acceleration is 90 m/s2. If the wave velocity is 20 m/s then find the waveform. [IIT-2005] Sol. The wave form of a transverse harmonic disturbance y = a sin (ωt ± kx ± φ) ...(i) Given vmax = aω = 3 m/s ....(ii) Amax = aω2 = 90 m/s2 Velocity of wave v = 20 m/s ...(iii) Dividing (ii) by (i)
∴ ∆T =
1 M 2g 2 2 (πr 2 ) 2 Ycρ
1 (100 × 10) 2 × 2 (3.14 × 2 × 10 −3 ) 2 × 2.1× 1011 × 420 × 7860 = 0.00457ºC
=
aω2 90 = ⇒ ω = 30 rad/s ...(iv) aω 3 Substituting the value of ω in (i) we get 3 a= = 0.1 m ...(v) 30 Now 3 2π 2π 2πv ω 30 k= = = = = = ...(vi) 2 λ v/v v v 20 From (iv), (v) and (vi) the wave form is 3 y = 0.1 sin 30t ± x ± φ 2
3.
A 5m long cylindrical steel wire with radius 2 × 10–3 m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring radiation losses. (For the steel wire : Young's modulus = 2.1 × 1011 Pa; Density = 7860 kg/m3; Specific heat = 420 J/kg-K). [IIT-2001] Sol. When the mass of 100 kg is attached, the string is under tension and hence in the deformed state. Therefore it has potential energy (U) which is given by the formula. 1 × stress × stain × volume U= 2 1 (Stress) 2 × × πr2l = 2 Y 1 M 2g 2l 1 (Mg / πr 2 ) 2 × πr2l = ...(i) = 2 Y 2 πr 2 Y This energy is released in the form of heat, thereby raising the temperature of the wire
An unknown resistance X is to be determined using resistance R1, R2 or R3. Their corresponding null points are A, B and C. Find which of the above will give the most accurate reading and why ? [IIT-2005] R = R1 or R2 or R3 Sol. All null point, the wheat stone bridge will be balanced r X R = ⇒X=R 1 ∴ r1 r2 r2
2.
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R
X G r1
r2
M A C N B where R is a constant r1 and r2 are variable. The maximum fraction error is ∆r ∆r ∆X = 1 + 2 X r1 r2 Here ∆r1 = ∆r2 = y (say) then ∆X For to be minimum r1 × r2 should be max X [Q r1 + r2 = c (Constt.] Let E = r1 × r2 ⇒ E = r1 × (r1 – c) dE = (r1 – c) + r1 = 0 ∴ dr1
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SEPTEMBER 2009
⇒ τ = MB sin θ where θ is the angle between M and B nhe nhe B ⇒ τ= × B sin 30º = 4πm 8πm The direction of torque can be found by right hand thumb rule. The direction of torque is perpendicular to the plane
c c ⇒ r2 = ⇒ r1 = r2 2 2 ⇒ R2 gives the most accurate value.
⇒ r1 =
4.
An electron in the ground state of hydrogen atom is revolving in anticlock-wise direction in a circular orbit of radius R. r B
r n
→
containing nˆ and B as shown.
30º
5.
(i) Obtain an expression for the orbital magnetic dipole moment of the electron. (ii) The atom is placed in a uniform magnetic r induction B such that the plane-normal of the electron-orbit makes an angle of 30º with the magnetic induction. Find the torque experienced by the orbiting electron. Sol. (i) Orbital magnetic dipole moment M = IA where I is the current due to orbital motion of e ⇒ M = × πR2 T electron and A is the area of the loop made by electron. 1 eω ⇒ M= × πR2 ⇒ M = eωR2 2 2π
e–
→
induction B is applied perpendicular and into the plane of rotation as shown in the figure below. An inductor L and an external resistance R are connected through a switch S between the point O and a point C on the ring to form an electrical circuit. Neglect the resistance of the ring and the rod. Initially, the switch is open. [IIT-1995] Y
R
r B
30º
→
τ
∫
(ii) We know that torque →
→
τ = M × B
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× × × ×
ω θ O
A
× × × × C
× × X × ×
r
x2 Br 2 ω e = B xω dx = Bω = 0 2 2 0 (b) i. The above diagram can be reconstructed as the adjacent figure e is a constant. O will accumulate positive charge and A negative when the switch S is
e– →
× × × ×
L (a) What is the induced emf across the terminals of the switch ? (b) The switch S is closed at time t = 0. (i) Obtain an expression for the current as a function of time. (ii) In the steady state, obtain the time dependence of the torque required to maintain the constant angular speed, given that the rod OA was along the positive X-axis at t = 0. Sol. (a) Let us consider a small length of metal rod dx at a distance x from the origin. Small amount of emf (de) induced in this small length (due to metallic rod cutting magnetic lines of force) is de = B(dx)v ...(i) where v is the velocity of small length dx v = xω ...(ii) From (i) and (ii) de = B(dx)xω ∴ The total emf across the whole metallic rod OA is
But according to Bohr's postulate nh nh ⇒ Rω2 = mRω2 = 2π 2πm e nh nhe ⇒ M= × = 2 2πm 4πm The direction of magnetic momentum is same as the direction of area vector, i.e. perpendicular to the plane of orbital motion.
r n
× × × ×
S
R
ω
A metal rod OA of mass 'm' and length 'r' is kept rotating with a constant angular speed ω in a vertical plane about a horizontal axis at the end O. The free end A is arranged to slide without friction along a fixed conducting circular ring in the same plane as that of rotation. A uniform and constant magnetic
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r
SEPTEMBER 2009
closed, transient current at any time t, when current I is flowing in the circuit, S
i
CHEMISTRY A sample of hard water contains 96 ppm of SO42– and 183 ppm of HCO3–, with Ca2+ as the only cation. How many moles of CaO will be required to remove HCO3– from 1000 kg of this water ? If 1000 kg of this water is treated with the amount of CaO calculated above, what will be concentration (in ppm) of residual Ca2+ ions? (Assume CaCO3 to be completely insoluble in water.) If the Ca2+ ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH ? (One ppm means one part of the substance in one million part of water, [IIT-1997] mass/mass. ) Sol. In 106g (= 1000 kg) of the given hard water, we will have Mass of SO42– ions = 96 g Mass of HCO3– ions = 183 g 96 g = 1 mol Thus Amount of SO42– ions = 96 g mol −1
O A
R
6.
L
I = I0(1 – e t / τb )
e Bωr 2 L = and TL = R 2R R Bωr 2 Therefore, I = [1 – e–(R/L)t ] 2R (ii) In steady state
Here
I0 =
In steady state I =
Bωr 2 2R R − t
[Q t has a large value and e L → 0] When current flows in the circuit in steady state, there is a power loss through the resistor. Also since the rod is rotating in a vertical plane, work needs to be done to keep it at constant angular speed. Power loss due to current I will be
Amount of HCO3– ions =
= 3 mol 61g mol −1 These ions are present as CaSO4 and Ca(HCO3)2. Hence, Amount of Ca+ ions = (1 + 1.5) mol = 2.5 mol. The addition of CaO causes the following reaction : CaO + Ca(HCO3)2 → 2CaCO3 + H2O To remove 1.5 mol of Ca(HCO3)2, 1.5 mol of CaO will be required in the treated water. After this, the solution contains only CaSO4. Thus, 1 mol of Ca2+ ions will be present in 106 g of water. Hence its concentration will be 40 ppm. Molarity of Ca2+ ions in the treated water will be 10–3 mol L–1. If the Ca2+ ions are exchanged by H+ ions, then Molarity of H+ in the treated water = 2 × 10–3 M. Thus, pH = – log (2 × 10–3) = 2.7
2
Br 2 ω R P = I2R = 2R ⇒
P=
B 2 r 4 ω2 4R t=t r/2
θ r/2 cosθ mg
The equilibrium constant Kp of the reaction 2SO3(g) is 900 atm–1 at 800 2SO2(g) + O2(g) K. A mixture containing SO3 and O2 having initial partial pressures of 1 atm and 2 atm, respectively, is heated at constant volume to equilibrate. Calculate the pressure of each gas at 800 K. [IIT- 1989] Sol. Since to start with SO2 is not present, it is expected that some of SO3 will decompose to give SO2 and O2 at equilibrium. If 2x is the partial pressure of SO3 that is decreased at equilibrium, we would have 2SO2(g) + O2(g) 2SO3(g) t=0 0 2 atm 1 atm teq 2x 2 atm + x 1 atm – 2x 7.
The torque required for this power ρ = τ1ω 2 4
B r ω 4R Torque required to move the rod in circular motion against gravitation field r τ2 = mg × cos θ 2 The total torque τ = τ1 + τ2 (clock wise) ⇒ τ1 =
B2r 4ω mgr + cos ωt 4R 2 The required torque will be of same magnitude and in anticlockwise direction. The second term will change signs as the value of cos θ can be positive as well as negative. τ=
XtraEdge for IIT-JEE
183 g
Hence, Kp =
(p SO3 ) 2 (p SO 2 ) 2 ( p O 2 )
=
(1 atm − 2 x ) 2 (2 x ) 2 (2 atm + x )
= 900 atm–1 10
SEPTEMBER 2009
Assuming x fa (b) If µm > µg, then the nature of the lens changes. The focal length may increase or decrease depending µg − µ m as compared to (µg – 1) on the value µm
µg − µ m fm increase if µm
1 2 3 (A) 70 cm divergent (C) 72 cm divergent
Passage : II (No. 18 to 20) When an incompressible and non-viscous fluid flows in streamline motion from one place to another, then at every point in its path, the total energy per unit volume (pressure energy + kinetic energy + potential energy) is constant. According to Bernoulli's theorem 1 P+ ρv2 + ρ g h = constant 2 1 1 i.e., P1 + ρv12 + ρ g h1 = P2 + ρv22 + ρgh2 2 2 where P refers to pressure, v for speed and h for height of the fluid 1 and 2 are used for different points in the system. The equation of continuity expresses the law of conservation of mass in a fluid dynamics. If a1, v1 and ρ be the area of cross section of the tube, velocity of flow of the fluid particles and density of fluid respectively at one point and a2, v2 and ρ the similar quantities at another point, then according to continuity equation a1v1 = a2v2 therefore, the equation of continuity states that as the area of cross-section of the tube of flow becomes larger, the liquid speed becomes smaller and vice versa.
> (µg – 1)
µg − µ m fm decreases if µm
< (µg – 1)
µg − µm = (µg – 1) and fm = fa if µm (c) If µm = µg, the lens is invisible and behaves as a plane glass.
15. As shown in fig. a spherical air lens of radii R1 = R2 = 10 cm is cut in glass µ = 1.5 cylinder.
µ = 1.5 µ = 1.5
µ=1 The focal length of the lens is (A) 15 cm divergent (B) 20 cm divergent (C) 15 cm convergent (D) 20 cm convergent
XtraEdge for IIT-JEE
(B) 70 cm convergent (D) 72 cm convergent
55
SEPTEMBER 2009
18. A tank is filled with water upto a height H. Water is allowed to come out of a hole P in one of the walls at a depth h below the surface of water (see fig.) Express the horizontal distance x in terms of H and h h
P
speed of the cylinder about an axis passing through P is given by ω = vcm/R. The velocity of a point at Q is therefore given by ω . 2R = vcm at that instant. The point P is instantaneously at rest. So from the point of view of the pure rotation about P, the situation is shown in fig. (1) Q 2vcm
v
H x (A) x =
h (H − h )
(B) x =
(C) x = 2 h (H − h )
C
[h (H − h ) / 2]
P
(D) x = 4 h (H − h )
Fig. (1) The rolling of a cylinder as a combination of the centre of mass and rotation of centre of mass about its axis is shown in fig. (2) Q vcm = ωR Q Q vcm 2vcm
19. As shown in the fig. water squirts horizontally out of two small holes in the side of the cylinder and the two streams strike the ground at the same point. If the hole Q is at a height h above the ground and the level of water stands at height H above the ground, then the height of P above ground level is
C v cm
P H
P
O
(B) H/h
A a
(C) H – h
(2gh )
(C)
[2gh (H1 − H 2 )]
(D)
A g [ H1 – a 2
(D) H/2
P
Pure rotation
Combined motion
B
C
v C
(A) v,v and v
(B) 2v,
2 v and zero
(C) 2v, v and zero
(D) 2v,
2 v and
2v
22. A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let vp, vQ and vC be the magnitude of velocities of points P, Q and C respectively, then
H2 ]
Passage : III (No. 21 to 23) The combined effect of translation of centre of mass ans rotation about an axis through the centre of mass are equivalent to pure rotation with the same angular speed about an axis through the point of contact of rolling body. To illustrate the above result, let us represent the instantaneous velocities acting at various points on the cylinder when it is rolling. Let vcm be the speed of centre of mass as observed by an observer fixed with respect to surface. Then the instantaneous angular
XtraEdge for IIT-JEE
C v cm
21. A solid disc rolls clockwise without slipping over a horizontal path with a constant speed v. Then the magnitude of the velocities of points A, B and C (see fig.) with respect to standing observer are respectively A
2 [ H1 – H 2 ] g
(B)
C P –ωR = –vcm Fig (2)
20. The vessel of area of cross-section A has liquid to a height H. There is a hole at the botom of vessel having area of cross-section a. The time taken to decrease the level from H1 to H2 will be (A)
vcm
=
+
Pure translation
h
(A) 2h
vcm
Q P
C
(A) vQ > vC > vP
(B) vQ < vC < vP 1 (C) vQ = vP, vC = vP (D) vQ < vC > vP 2 56
SEPTEMBER 2009
23. A cylinder is pulled by a force F acting at a point above the centre of mass of the cylinder as shown in fig. The direction of friction force (f) acting on the cylinder pushed on a rough surface will be represented by F
(A) 600 K (C) 200 K
10 moles of Fe3O4 is treated with excess of KI solution in presence of dilute H2SO4, the products are 2+ Fe and I2 (g). What volume of 0.1 (M) Na2S2O3 will be needed to reduce the liberated I2 (g)? (A) 50 ml (B) 100 ml (C) 200 ml (D) 400 ml
(A)
5.
An alkene (A) C16H16 on ozonolysis gives only one product (B) (C8H8O). Compound (B) on reaction with NH2OH/H2SO4,∆ gives N-methyl benzamide the compound 'A' is H C = C (A) CH3 CH3 H
(B)
C
f
F C f
F (D) Cannot be interpreted
(C)
f=0
CHEMISTRY
(B)
C=C CH3
Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
1.
–2
4
C F
(C)
CH3
CH2–CH=CH–CH2
Most stable free radical is CH3 (A)
(B) 400 K (D) 120 K
CH3 (D)
(B)
CH=CH CH3
(D)
(C)
6.
CH3
2.
(CH3)3COΘ
CH2 – CH – CH – CH3 t-BuOH OTs pre dominant product A is (A)
A,
CH2 – CH2 – C = CH2
Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.
CH3 (B)
CH = CH – CH – CH3
7.
CH3 (C)
When one mole of monoatomic ideal gas at T K undergoes adiabatic change under a constant external pressure of 1 atm changes volume from 1 litre to 2 litre. The final temperature in Kelvin would be T 2 (B) T + × 0.0821 (A) ( 2 / 3) 3 2 2 (C) T (D) T – × 0.0821 3
CH2 – CH = C – CH3
Refer to the figure given : Which of the following statements is wrong ?
CH3
gas C
(D) None of these
3.
Z 1
A gas is present in a cylinder fitted with movable piston. Above and below of the piston there is equal number of moles of gas. The volume above is two times the volume below at a temperature of 300K. At what temperature will the volume above be four times the volume below-
XtraEdge for IIT-JEE
gas A Ideal gas gas B
P 57
SEPTEMBER 2009
(A) For gas A, a = 0 and Z will linearly depend on pressure (B) For gas B, b = 0 and Z will linearly depend on pressure (C) Gas C is a real gas and we can find 'a' and 'b' if intersection data is given (D) All van der Waal gases will behave like gas C and give positive slope at high pressure.
8.
(A) If both (A) and (R) are true, and (R) is the correct explanation of (A). (B) If both (A) and (R) are true but (R) is not the correct explanation of (A). (C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. 11. Assertion (A) : The heat absorbed during the isothermal expansion of an ideal gas against vacuum is zero. Reason (R) : The volume occupied by the molecules of an ideal gas is zero.
When nitrobenzene is treated with Br2 in presence of FeBr3, the major product formed is mbromonitrobenzene. Statements which are related to obtain the m-isomer are (A) The electron density on meta carbon is more than that on ortho and para positions (B) The intermediates carbonium ion formed after initial attack of Br+ at the meta position is least destabilished (C) Loss of aromaticity when Br+ attacks at the ortho and para positions and not at meta position (D) Easier loss of H+ to regain aromaticity from the meta position than from ortho and para positions
12. Assertion : The value of van der Waals constant 'a' is larger for ammonia than for nitrogen. Reason : Hydrogen bonding is present in ammonia. 13. Assertion : 3-hydroxy - butan-2-one on treatment with [Ag(NH3)2]⊕ cause precipitation of silver. Reason : [Ag(NH3)2] ⊕ oxidises 3-hydroxy butan-2one to butan-2-3-dione 14. Assertion : HBr adds to 1,4-pentadiene at a faster rate than to 1,3-pentadiene Reason : 1,4-pentadiene is less stable than 1,3-pentadiene.
HBr / Boiling OH → product.
9.
Which of the following are possible products (in significant amount) (A)
(C)
(B)
Br
Br
This section contains 3 paragraphs, each has 3 multiple choice questions. (Question 15 to 23) Each questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
(D)
Passage : I (No. 15 to 17) 10. Which of the following are possible products from aldol condensation of 6-oxoheptanal ? CH3 O O H C C (B)
(A)
O (C)
(D)
The temperature dependence of the dissociation constant for the formic acid in aqueous is given by 1400 + 5 – 0.01 T log Ka = – T The enthalpy of neutralisation between strong acid and strong base at 27°C is equal to –56 kJ/equal.
CH3
CH3
O
15. What is the enthalpy of neutralisation of HCOOH against NaOH at 27°C ? (A) –56 kJ/mol (B) + 56 kJ/mol (C) –46.427 kJ/mol (D) –9.573 kJ/mol
This section contains 4 questions numbered 11 to 14, (Reason and Assertion type question). Each question contains Assertion and Reason. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct The following questions given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer.
XtraEdge for IIT-JEE
16. What is the standard entropy change for the dissociation of HCOOH at 300 K ? (A) –19.147 J K–1 mol–1 (B) +19.147 kJ K–1 mol–1 (C) –73 J K–1 mol–1 (D) – 40 JK–1 mol–1
58
SEPTEMBER 2009
17. At what approximate temperature the dissociation for formic acid is maximum ? (A) 300 K (B) 374 K (C) 474 K (D) There is no maximum temperature for dissociation
reactivity is significant. The more common reaction of ether is cleavage of C–O bond by strong acids. This may occure by SN1 of E1 mechanism for 3º alkyl groups or by SN2 mechanism for 1º alkyl groups.
21. Which of the following reaction is not assumed to proceed-
Passage : II (No. 18 to 20) OH
CH3
P HCN/NH4Cl H3O+/∆ (R) α-amino acid CHCl3 + NaOH (zwitter ion) O
(A) CH3–CH2–O–CH2–CH3 PCl 5 → – MgX (B) CH3–CH2–O–CH2–CH3 CH 3 → O2 (C) CH3–CH2–O–CH2–CH3 → hν
H⊕ Isomerisation
Al2 O 3 / ∆
(D) CH3–CH2–O–CH2–CH3 →
CH3 CHCl2 OHΘ Q
18. Product P is OH
22. If 2º alkyl group is supposed to prefer SN2 path of cleavage. Which of the following is correct order of reactivity towards conc. HI ? CH3 CH3 CH3 | | | CH3–CH–O–C–CH3 CH3–O–CH–CH3 | (II) CH3 (I)
OH
(A)
CHO
(B)
CHO
OH
OH
OHC
CHO
(C)
CH3 | CH3–CH2–O–CH–CH3
CHO (D)
(III)
CH3
CH3
(A) IV > II >I > III (C) II > I > IV > III
19. Product R is OH OH CH–CH2–COOΘ CH–COOΘ (B) (A) ⊕ ⊕ NH3 NH3 CH3 CH3 OΘ
OH
CH–COOH (C)
⊕
CH3
(D)
CHO
+
CHO
CHCl2
A. Product molecule A is -
(C) CH3–CH=C | CH3
(D)
CH3 CHO
CH3 | (D) CH3–CH2–C | OSO3H
Passage : III (No. 21 to 23) Ethers are widely used as solvents due to their relatively unreactive nature. In the acidic medium the
XtraEdge for IIT-JEE
∆
OH | (B) CH3–CH2–C | CH3
O
CH3
Conc.H SO
2 4 →
CH3 | (A) HO–CH–CH | CH3
CH3
CH3 OH (C)
CH–COOΘ
OH (B)
(B) I > III > IV > II (D) IV > II > III > I
OH
NH3
20. Product Q is OH (A)
23.
⊕
NH3
CH3 | O–CH–CH | CH3
CH3 | CH3–CH2–O–C–CH3 | (IV) CH3
59
SEPTEMBER 2009
Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.
MATHEMATICS Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.
2.
7.
∫x
dx 4
1 1 1 (A) 4 a2 + x2 − 3 a x 3x (B)
8.
, then I equals
a2 + x2
(
1 1 1 a2 + x2 − 3 a2 + x2 4 a x 3x
(
2
)
1 1 2 1 a + x2 − a2 + x2 4 x a 2 x (D) None of these
(C)
9.
a +x +C 2
3/ 2
+C
)
( x − a )( x − b) will assume x−c
all real values provided (A) a > b > c (B) a < b < c (C) a > c > b (D) a < c < b
Reflection of the line az + az = 0 in the real axis is (A) az + az = 0 z a = (B) z a (C) (a + a ) (z + z ) = 0 (D) None of these If I =
For real x, the expression
3/ 2
+C
cos 8x − cos 7 x dx is expressed as 1 + 2 cos 5x K sin 3x + M sin 2x + C then (A) K = – 1/3 (B) K = 1/3 (C) M = – 1/2 (D) M = 1/2 If
∫
α
dx A = + B (a ≠ 0). Then 1 − cos α cos x sin α possible values of A and B are π π π (A) A = , B = 0 (B) A = , B = 2 4 4 sin α π π π (D) A = π, B = (C) A = , B = 6 sin α sin α If
∫
0
10. The solution of y1(x2y3 + xy) = 1 is 2
(A) 1/x = 2 – y2 + C e − y / 2 (B) the solution of an equation which is reducible to linear equation (C) 2/x = 1 – y2 + e–y/2
2
3.
x + 20
If I =
∫ (x sin x + 5 cos x)
(A) –
x + tan x + C cos x ( x sin x + 5 cos x )
(B)
2
dx, then I equals
(D) e y
x + cot x + C sin x ( x sin x + 5 cos x )
If I =
∫
3
1/ 3
1 1 sin − x dx, then I equals x x (B) π + 3 / 2 (D) None of these
(A) 3 / 2 (C) 0
5.
The orthogonal trajectories of the family of curve y = cxk are given by (A) x2 + cy2 = const. (B) x2 + ky2 = const. (C) kx2 + y2 = const. (D) x2 – ky2 = const.
6.
The sum
∑ ∑(
10
C j ) (jCi) is equal to
11
0≤ i < j≤10
10
(A) 2 – 1 (C) 310 – 1
XtraEdge for IIT-JEE
/ 2 1 − 2x
x
+ y2 = C
This section contains 4 questions numbered 11 to 14, (Reason and Assertion type question). Each question contains Assertion and Reason. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct The following questions given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer. (A) If both (A) and (R) are true, and (R) is the correct explanation of (A). (B) If both (A) and (R) are true but (R) is not the correct explanation of (A). (C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true.
(C) (x sin x – 5 cos x)–1 sin x + 7x + C (D) None of these
4.
2
10
(B) 2 (D) 310
60
Assertion (A) : The number 1000 is not divisible 500 by 11.
SEPTEMBER 2009
Reason (R) : If p is a prime, the exponent of p in n! n n n is + 2 + 3 + .... p p p where [x] denote the greatest integer ≤ x. 12. Assertion (A) : F(x) =
∫
log t
x
1
1+ t + t 2
(A) 0 (C) –1
Passage : II (No. 18 to 20) The average value of a function f(x) over the interval, [a, b] is the number b 1 µ= f ( x ) dx b−a a
dt then
∫
F(x) = –F(1/x).
∫
x
13. Assertion (A) : The derivative of F(x) =
∫
x
1/ x
18. The average ordinate of y = sin x over the interval [0, π] is (A) 1/π (B) 2/π (C) 4/π2 (D) 2/π2
3 cos t 2 dt (x > 0) at x = 1 is cos 1 2
Reason (R) :
d dx
∫
φ( x )
ψ(x)
f ( t ) dt = f(φ(x)) – f(ψ(x))
19. The average value of pressure varying from 2 to 10 atm if the pressure p and the volume v are related by pv3/2 = 160 is 10 20 (A) (B) 3 3 3 3 10 + 3 2 20 10 + 2
14. Assertion (A) : The solution of the equation 1 dy x + 6y = 3xy4/3 is y(x) = dx ( x + Cx 2 ) 3
(
Reason (R) : The solution of a linear equation is obtained by multiplying with its integrating factor.
(C)
This section contains 3 paragraphs, each has 3 multiple choice questions. (Question 15 to 23) Each questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
r
r
ar (nCr) = k (nC[n/3])
where [x] denotes the greatest integer ≤ x, then k equals (A) 1 (B) 0 (C) 3 (D) –1
17. If a =
∑(
n
20 ( 10 + 2 ) 3
3
(C) 6/π
cos 2 x sin 2 x + 4 cos 2 x
on
(D) 1/6
23. Solution set of the inequality
13 x − 5 ≤
2k
C 2 k )( C k ) where [x] denotes the
2(13 x + 12) –
(A) (– ∞, –5) (C) [log13 5, 1]
k =0
greatest integer ≤ x, then an – a equals.
XtraEdge for IIT-JEE
160
3
22. Solution set of the inequality 2(25)x – 5(10x) + 2(4x) ≥ 0 is (A) (–1, ∞) (B) (0, ∞) (C) (2, ∞) (D) None of these
r =0
[ n / 2]
(D)
21. Solution set of the inequality 3x(0.333 ....)x–3 ≤ (1/27)x is (A) [3/2, 5] (B) (– ∞, 3/2] (D) None of these (C) (0, ∞)
16. If n is not a multiple of 3, and r
3
3
identity xy = a y log a x where a > 0, a ≠ 1.
15. Which of the following is true ? (A) ar = an – r (B) a2r = an – r (C) ar = a2n – r (D) None of these
n
20 ( 10 + 2 )
Passage : III (No. 21 to 23) To solve equation or inequality involving exponential expression f(x)g(x), we may use logarithm or the
r =0
∑ (−1)
40
3
[0, π/2] is (A) π/6 (B) 4/π
2n
∑a x
)
20. The average value of f(x) =
Passage : I (No. 15 to 17) For n ∈ N, we put (1 + x + x2)n =
1/ 2
b 1 The square root [f ( x )]2 dx is called the a b a − root mean square of f on [a, b]. The average value µ is attained if f is continuous on [a, b]
log t Reason (R) : If F(x) = dt then 1 t +1 F(x) + F(1/x) = (1/2) (log x)2
∫
(B) 1 (D) 2
61
13 x + 5 is
(B) [5, ∞) (D) [0, log13 5]
SEPTEMBER 2009
Based on New Pattern
IIT-JEE 2011 XtraEdge Test Series # 5
Time : 3 Hours Syllabus : Physics : Laws of motion, Friction, Work Power Energy, Gravitation, S.H.M., Laws of Conservations of Momentum, Rotational Motion (Rigid Body), Elasticity, Fluid Mechanics, Surface Tension, Viscosity. Chemistry : Gaseous state, Chemical Energetics, Oxidation-Reduction, Equivalent Concept, Volumetric Analysis. Mathematics : Logarithm & Modulus Function, Quadratic Equation, Progressions, Binomial Theorem, Permutation & Combination, Complex Number.
Instructions : Section - I • Question 1 to 6 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 7 to 10 are multiple choice questions with multiple correct answer. +4 marks and no negative marking for wrong answer. • Question 11 to 14 are Reason and Assertion type questions with only one correct answer in each. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 15 to 23 are passage based single correct type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer.
PHYSICS
3.
A stick is thrown in the air and lands on the ground at some distance from the thrower. The centre of mass of the stick will move along a parabolic path : (A) in all cases (B) only if the stick is uniform (C) only if the stick has linear motion but no rotational motion (D) only if the stick has a shape such that its centre of mass is located at some point on it and not outside it
4.
A simple pendulum of length 1m is attached to the ceiling of an elevator which is accelerating upward at the rate of 1 m/s2. Its frequency is approximately : (A) 2 Hz (B) 1.5 Hz (C) 5 Hz (D) 0.5 Hz
5.
The volume of an air bubble is double as it rises from the bottom of a lake to its surface. If the atmospheric pressure is H m of mercury and the density of mercury is n times that of lake water, the depth of the lake is : (A) nH (B) 2 n H (C) n/H (D) H/n
6.
A bowl made of stainless steel is floating in water, contained in a pan. There is small hole in bottom of the bowl. Water leaks into the bowl through the hole. Level of free water surface in the pan :
Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.
A large rectangular box falls vertically with an acceleration 'a', A toy gun fixed at A and aimed towards C fires a particle P. C B
S u
P D A (A) P will hit C if a = g (B) P will hit the roof BC if a > g (C) P will hit the wall CD if a < g (D) May be either (a), (b) or (c), depending on the speed of projection of P
2.
A spacecraft of mass m describes a circular orbit of radius r1 around the earth of mass M. Calculate the additional energy to be imparted to the spacecraft to transfer it to a circular orbit of larger radius r2 : GMm(r2 − r1 ) GMm(r1 − r2 ) (A) (B) 2r1r2 2r1r2 (C)
GMmr1 r2 r2 − r1
XtraEdge for IIT-JEE
(D)
GMmr1r2 (r2 + r1 ) 62
SEPTEMBER 2009
10. A liquid flows through a horizontal tube. The velocities of the liquid in the two section, which have areas of cross-section A1 and A2, are v1 and v2 respectively. The difference in the levels of the liquid in the two vertical tubes is h :
(A) always remains constant (B) remain constant for some times then falls suddenly and then becomes constant (C) falls gradually for some time and then becomes constant (D) rises gradually for some time and then becomes constant
h
Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 7.
8.
Four identical rods, each of mass m and length l, are joined to form a rigid square frame. The frame lies in the xy plane, with its centre at the origin and the sides parallel to the x and y axes. Its moment of inertia about : 2 (A) the x-axis is ml2 3 4 (B) the z-axis is ml2 3 (C) an axis parallel to the z-axis and passing through 10 m l2 a corner is 3 5 (D) one side is ml2 2
P
(B) v2 – v1 = 2
This section contains 4 questions numbered 11 to 14, (Reason and Assertion type question). Each question contains Assertion and Reason. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct The following questions given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer. (A) If both (A) and (R) are true, and (R) is the correct explanation of (A). (B) If both (A) and (R) are true but (R) is not the correct explanation of (A). (C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. 11
U Liquid C
Q Liquid B
R
T S
13. Assertion (A) : As the radius of earth reduces by 50% with out any change in mass, length of a day reduces. Reason (R) : Angular momentum conservation provides drop in time as I decreases to 25% of the original.
When a body is weighed in a liquid, the loss in its weight depends upon : (A) Volume of the body (B) Mass of the body (C) Shape of the body (D) CG of the body
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Assertion (A) : Two blocks of masses m1 and m2 are at rest. They are moving towards each other under the mutual internal force. The velocity of centre of mass is zero. Reason (R) : If no external force act on the system, then velocity of centre of mass unchanged but can never be zero.
12. Assertion (A) : For stable equilibrium force has to be zero and potential energy should be minimum. Reason (R) : For equilibrium, it is not necessary that the force is not zero.
(A) Height of column of liquid A is greater than that of C (B) Density of liquid A is less than that of liquid C (C) Density of liquid B is maximum among these three liquids (D) Pressure at R is greater than that of S
9.
2gh
2
(C) v2 – v1 = 2gh (D) the energy per unit mass of the liquid is the same in both sections of the tube
Air Liquid A
A2 v2
(A) the volume of the liquid flowing through the tube in unit time is A1v1
A U-tube is held in a vertical plane such that its two limbs are vertical with middle portion horizontal as shown in fig. Three liquids A, B and C are poured into it and in steady state their interface are as shown in the figure. Which of the following statements (s) is/are correct ?
Air
v1
A1
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14. Assertion (A) : Cross-sectional area of the water pouring out of a tap decreases as the height from the ground decreases. Reason (R) : Work done by gravity reduces the cross-sectional area.
F
Passage : I (No. 15 to 17)
19. Which of the blocks will have received the greatest impulse during 2 m push (A) block A (B) block B (C) block C (D) all will have the same impulse
The energy of a body is defined as the capacity of doing work. The energy possessed by a body by virtue of its motion is called as kinetic energy and the energy possessed by a body by virtue of its position or configuration in some field is defined as potential energy. When work is done on a body, its kinetic or potential energy increases. On the other hand, when the work is done by the body, its potential or kinetic energy decreases. According to work energy theorem, the workdone is equal to change in energy, i.e., W = ∆E.
20. Which of the blocks would have greatest kinetic energy at the end of 2 m push (A) block A (B) block B (C) block C (D) all will have the same kinetic energy
15. Choose the correct statement related to validity of work-energy theorem (A) it is valid in all inertial frames only (B) it is valid in non-inertial frames only (C) it is valid in both frames (D) None of these
Passage : III (No. 21 to 23) A tube of very small bore is called capillary. If a glass capillary tube is dipped into a liquid such as water, which wets the glass, the liquid rises in the tube. However, if the glass capillary tube is dipped in a liquid such as mercury which does not wet the glass, the liquid falls in the tube. This phenomenon is known as capillarity.
16. Consider the following two statements (i) Linear momentum of a system of particles is zero (ii) Kinetic energy of a system of particles is zero. Then (A) (i) implies (ii) and (ii) implies (i) (B) (i) does not imply (ii) and (ii) does not imply (i) (C) (i) implies (ii) but (ii) does not imply (i) (D) (i) does not imply (ii) but (ii) implies (i)
21. A liquid rises in the capillary tube more than water does. Which one of the following in the cause of it (A) temperature of liquid is higher than water (B) surface tension of water is less than the liquid (C) density of liquid is more than water (D) viscosity of the liquid is more than water
17. Two protons are brought towards each other. The potential energy of the system will (A) increase (B) decrease (C) remains same (D) None of these
22. Water rises to a height of 1.25 cm in a capillary tube. If the hight of the tube is 1 cm, then water will (A) stay at the top of the tube (B) continuously flow out of the tube (C) be depressed a little below the upper end of the tube (D) no rise in the tube
Passage : II (No. 18 to 20) Three blocks (A, B and C) are each pushed by equal forces F, frictionlessly across a horizontal surface for a distance of 2 m as shown in fig. Here the masses are such that MA > MB > MC. 2m F A
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C
18. Which of the block will be travelling faster after two metre push ? (A) block A (B) block B (C) block C (D) all moving with same speed
This section contains 3 paragraphs, each has 3 multiple choice questions. (Question 15 to 23) Each questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
F
2m
23. In gravity free space, the liquid in a capillary tube will rise to (A) same height as on earth (B) less height as on earth (C) slightly more height than that on earth (D) infinite height
2m B 64
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Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.
CHEMISTRY Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.
2.
Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy of a helium atom is (A) two times that of a hydrogen molecule (B) same as that of a hydrogen molecule (C) four times that of a hydrogen molecule (D) half that of hydrogen molecule
(C) RT
4.
10 volume H2O2 solution is present, then it means (A) 10 ml of H2O2 solution liberates 1 ml of oxygen at STP (B) 1 ml of H2O2 solution liberates 10 ml of oxygen at STP (C) 0.0303 g of H2O2 in 10 ml of solution liberates 10 ml O2 at STP (D) 0.0303 g of H2O2 in 1 ml of the solution liberates 10 ml O2 at STP
8.
Silver metal in ore is dissolved by potassium cyanide solution in the presence of air by the reaction 4Ag + 8KCN + O2 + 2H2O → 4KAg (CN)2 + 4KOH (A) The amount of KCN required to dissolve 100 g of pure Ag is 120 g. (B) The amount of oxygen used in this process is 0.742 g (C) The amount of oxygen used in this process is 7.40g (D) The volume of oxygen used at STP is 5.20 litres.
9.
Identify the intensive properties among the following: (A) Enthalpy (B) Temperature (C) Volume (D) Refractive index
In van der Waals equation of state for a nonideal gas the term that accounts for intermolecular forces is (A) (V – b)
3.
7.
a (B) p + 2 V (D) (RT)–1
White phosphorus reacts with caustic soda. The products are PH3 and NaH2PO2. This reaction is an example of (A) oxidation (B) reduction (C) oxidation and reduction (D) neutralization
10. During the Joule Thomson effect (A) A gas is allowed to expand adiabatically from a high pressure region to a low pressure region (B) A gas is allowed to expand adiabatically at constant pressure (C) ∆Η = 0 for ideal gas (D) ∆U = 0 for ideal gas
Which of the following reactions is a redox reaction? (A) Cr2O3 + 6HCl → 2CrCl3 + 3H2O (B) CrO3 + 2NaOH → Na2CrO4 + H2O Cr2O72– + OH– (C) 2CrO42– + H+ (D) Cr2O72– + 6I– + 14H+
2Cr3+ + 3I2 + 7H2O
5.
The reaction of cyanamide, NH2CN(s), with oxygen was run in a bomb calorimeter and ∆U at 300 K was found to be –743 kJ mol–1. The value of ∆H at 300K for the combustion reaction NH2CN(s) + (3/2) O2(g) → N2(g) + CO2(g) + H2O (1) would be (A) – 741.75 kJ mol–1 (B) –743 kJ mol–1 (C) – 744.25 kJ mol–1 (D) – 740.5 kJ mol–1
6.
The combustion reaction occurring in an automobile is 2C8H18(s) + 5O2(g) → 16CO2(g) + 18H2O(1). This reaction is accompanied with (A) ∆H = –ve, ∆S = + ve, ∆G = + ve (B) ∆H = + ve, ∆S = –ve, ∆G = + ve (C) ∆H = – ve, ∆S = +ve, ∆G = – ve (D) ∆Η = +ve, ∆S = +ve, ∆G = – ve
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This section contains 4 questions numbered 11 to 14, (Reason and Assertion type question). Each question contains Assertion and Reason. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct The following questions given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer. (A) If both (A) and (R) are true, and (R) is the correct explanation of (A). (B) If both (A) and (R) are true but (R) is not the correct explanation of (A). (C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. 11. Assertion (A) : At zero degree Kelvin the volume occupied by a gas is negligible. Reason (R) : All molecular motion ceases at 0 K. 65
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12. Assertion (A) : Compressibility factor for hydrogen varies with pressure with positive slope at all pressures. Reason (R) : Even at low pressures, repulsive forces dominate hydrogen gas.
easily show larger deviation. Further, it is found that higher the speed of the gas molecules, less are the deviations. However, for every gas, there is a particular temperature above which they show ideal behavior over an appreciable range of pressure. This temperature is called Boyle temperature. The plots of compressibility factor versus pressure for a few gases and for the same gas at different temperatures are given below in figs (a) and (b) respectively. The ideal gas equation has, therefore, been modified and for real gases, we apply van der Wasl's equation, a P + 2 (V – b) = RT for 1 mole of the gas. V IV T3 T 2 III 1.4 II T1 1.2 Z I Z 1.0 1.0 1.5 1.6 P P (a) (b)
13. Assertion (A) : Enthalpy of graphite is lower than that of diamond. Reason (R) : Entropy of graphite is greater than that of diamond. 14. Assertion (A) : The temperature of a gas change when it undergoes an adiabatic expansion. Reason (R) : During an adiabatic expansion of a real gas, the internal energy of the gas remains constant. This section contains 3 paragraphs, each has 3 multiple choice questions. (Question 15 to 23) Each questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 15 to 17) Redox reactions are those in which oxidation and reduction take place simultaneously. Oxidising agent can gain electron whereas reducing agent can lose electron easily. The oxidation state of any element can never be in fraction. If oxidation number of any element comes out be in fraction, it is average oxidation number of that element which is present in different oxidation states.
18. If Vo is the observed volume of a gas and Vi is the ideal gas volume, then the compressibility factor (Z) for the gas is V V (A) o (B) i Vi Vo (C) Vo + Vi
19. In fig (b), the correct order of temperatures is (A) T1 > T2 > T3 (B) T3 > T2 > T1 (C) T2 > T1 > T3 (D) T2 > T3 > T1
1
15.
N N
3
N–H In this compound HN3 (hydrazoic acid),
2
20. The gas which can be liquefied most easily is (A) I (B) II (C) III (D) IV
oxidation state of N1 N2 and N3 are (A) 0, 0, 3 (B) 0, 0, –1 (C) 1, 1, –3 (D) –3, –3, –3
Passage : III (No. 21 to 23)
16. Equivalant weight of chlorine molecule in the equation 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O (A) 42.6 (B) 35.5 (C) 59.1 (D) 71
A student wanted to apply his knowledge of thermodynamics to obtain practical applications. He performed a number of experiments. In one experiment, he took a closed vessel in which he dissolved 28 g of iron in hydrochloric acid at 27ºC. In another experiment, he dissolved the same amount of iron but in an open vessel. However, he compressed the gas to 10 atm pressure at 27ºC and then again allowed to expand it isothermally and reversibly until the pressure fell down to 1 atm. In one case, he used the work obtained to lift a heavy body of mass 20 kg till the pressure fell down to atmospheric pressure and in another case, he used the work obtained to heat up 1 litre of water (Assume that hydrogen behaves like an ideal gas).
17. The oxidation number of sulphur in K2S2O8 is (A) + 2 (B) + 4 (C) + 7 (D) + 6 Passage : II (No. 18 to 20) Real gases show deviations from ideal behaviour. Consequently, the observed molar volume of a gas is found to be different from theoretically calculated volume from ideal gas equation. The extent of deviations is measured in terms of compressibility factor, Z. It is found that gases which can be liquefied
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(D) Vo – Vi
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21. In the second experiment, the work done by the system, if the gas were not condensed, would have been nearly (A) 1246 J (B) zero (C) 2492 J (D) 1145 J
Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 7.
If z1, z2, z3, z4 are the vertices of a square in that order, then (A) z1 + z3 = z2 + z4 (B) |z1 – z2| = |z2 – z3| = |z3 – z4| = |z4 – z1| (C) |z1 – z3| = |z2 – z4| (D) (z1 – z3)/(z2 – z4) is purely imaginary
8.
Let a and k be two natural numbers greater than 1. If ak – 1 is prime then (A) a = 2 (B) k is prime (C) k = 2m for some m ∈ N (D) None of these
9.
If x satisfies log3(2x + 1) < log35 then x contains the interval.
22. The work done by the compressed gas would be nearly (A) 1247 J (B) 2500 J (C) 4000 J (D) 5000 J 23. The rise in temperature of the water would be (A) 1.25º (B) 2.50º (C) 2.0º (D) 3.0º
MATHEMATICS Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.
1 (A) − , 0 2 (C) [1, 2)
The equation
x + 3 − 4 x − 1 + x + 8 − 6 x − 1 = 1 has (A) no solution (B) only one solution (C) only two solution (D) more than two solutions 2.
If log30 3 = c, log305 = d then the value of log308 (A) 2(1 – c – d) (B) 3(1 + c + d) (C) 3(1 + c – d) (D) 3(1 – c – d)
4.
Solution set of the inequality log3(x + 2) (x + 4) + log1/3(x + 2) < (A) (–2, –1) (C) (–1, 3) n
5.
If an =
∑ r =0
Cr
n
, then
n an 2 (C) nan (A)
6.
n
3
This section contains 4 questions numbered 11 to 14, (Reason and Assertion type question). Each question contains Assertion and Reason. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct The following questions given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer. (A) If both (A) and (R) are true, and (R) is the correct explanation of (A). (B) If both (A) and (R) are true but (R) is not the correct explanation of (A). (C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true.
7 is
(B) (–2, 3) (D) (3, ∞)
1 n
1 log 2
∑ r =0
r n
Cr
equals
11. Assertion (A) : If a, b, c ∈ R and 2a + 3b + 6c = 0, then the equation ax2 + bx + c = 0 has at least one root in [0, 1]. Reason (R) : If a continuous function f defined on R assumes both positive and negative values, then it vanishes at least once.
n an 4 (D) (n – 1)an (B)
The coefficient of xk in the expansion of E = 1 + (1 + x) + (1 + x)2 + ... + (1 + x)n is (A) nCk (B) n+1Ck n+1 (C) Ck+1 (D) None of these
XtraEdge for IIT-JEE
(D) (2, 3)
1 10. Let an = 1 + . Then for each n ∈ N n (A) an ≥ 2 (B) an < 3 (D) an < 2 (C) an < 4
The positive integer n for which 2 × 22 + 3 × 23 + 4 × 24 + .... + n × 2n = 2n + 10 is (A) 510 (B) 511 (C) 512 (D) 513
3.
(B) [0, 2)
12. Assertion (A) : There exists no A.P. whose three terms are 67
3,
5 and
7. SEPTEMBER 2009
Reason (R) : If tp, tq and tr are three distinct terms of tr − tp is a rational number. an A.P., then tq − tp
Passage : II (No. 18 to 20) If n is a natural number define polynomial Pn(x) of degree n as follows : cos nθ = Pn (cos θ) For example, P2(x) = 2x2 – 1 and P3(x) = 4x3 – 3x.
13. Assertion (A) :
n (n + 1) 12 22 n2 + + .... + = (1)(3) (3)(5) (2n − 1)( 2n + 1) 2(2n + 1)
18.
Reason (R) : 1 1 1 1 + + ... + = (1)(3) (3)(5) (2n − 1)(2n + 1) 2n + 1
19. (x + x 2 − 1 )n + (x – x 2 − 1 )n equals (A) Pn(x) (B) Pn+1 + Pn–1(x) (C) 2Pn(x) (D) None of these
14. Assertion (A) : The set of all x satisfying the equation 2
x log5 x + log5 x −12 = 1/x4 is {1, 25, 1/125, 1/625} Reason (R) : A polynomial equation of degree n can have at most n real roots.
20. P6(x) equals (A) 36x6 – 45x4 + 18x2 – 8 (B) 32x6 – 48x4 + 18x2 – 1 (C) 36x6 – 48x4 + 18x2 – 5 (D) None of these
This section contains 3 paragraphs, each has 3 multiple choice questions. (Question 15 to 23) Each questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
Passage : III (No. 21 to 23) Let a, b ∈ I and n ∈ N. We write a ≡ b(mod n) if and only if n|(a – b) If a ≡ b (mod n) and c ≡ d (mod n), then a ± c ≡ b ± d (mod n), ac = bd(mod n) and ak ≡ bk (mod n) ∀ k ∈ N.
Passage : I (No. 15 to 17) Let x0, x1 and x2 be three distinct real numbers. Define three polynomials l0(x), l1(x), l2(x) and l(x) as follows : ( x − x 1 )( x − x 2 ) l0(x) = ( x 0 − x 1 )( x 0 − x 2 )
and
l1(x) =
( x − x 0 )( x − x 2 ) ( x 1 − x 0 )( x 1 − x 2 )
l2(x) =
( x − x 0 )( x − x 1 ) ( x 2 − x 0 )( x 2 − x 1 )
21. The remainder when 10073 is divided by 7 is (A) 1 (B) 2 (C) 3 (D) 4 22. The number of solutions of the system of congruences x ≡ –3 (mod 49) and x ≡ 2 (mod 11) is (A) 0 (B) 1 (C) 2 (D) infinite
l(x) = (x – x0)(x – x1) (x – x2) 15. p(x) = l0(x) + l1(x) + l2(x) equals (A) 1 (B) x (D) 1 + x + x2 (C) x2
23. The number of values of x ∈ I for which 2x ≡ 1 (mod 2006) is (A) 0 (B) 1 (C) 2 (D) infinite
16. p(x) = (x1 + x2)l0(x) + (x2 + x0) l1(x) + (x0 + x1)l2(x) then p(x) equals 1 (A) x (B) x – (x0 + x1 + x2) 2 (D) None of these (C) x0 + x1 + x2 – x 17.
Behavior •
l( x ) l( x ) l( x ) + + ( x − x 0 )l´(x 0 ) ( x − x 1 )l´(x 1 ) ( x − x 2 )l´(x 2 ) equals (A) 1 (B) x (C) x2 + x (D) None of these
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1 [Pn+1(x) + Pn–1(x)] equals 2x (A) Pn+2 (x) (B) Pn–1 (x) + Pn(x) (D) Pn+1(x) – Pn(x) (C) Pn(x)
• •
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Behavior is a mirror in which everyone displays his image. Behavior is what a man does, not what he thinks, feels, or believes. Behave the way you'd like to be and soon you'll be the way you behave.
SEPTEMBER 2009
XtraEdge Test Series ANSWER KEY IIT- JEE 2010 (September issue) PHYSICS Ques Ans Ques Ans Ques Ans
1 A 11 A 21 C
2 B 12 C 22 A
3 A 13 C 23 D
4 A 14 A
5 A 15 A
6 A 16 C
7 A ,B ,C , D 17 C
8 A ,B , C , D 18 C
9 A 19 C
10 A , B ,C 20 D
C H E MI S T R Y Ques Ans Ques Ans Ques Ans
1 C 11 C 21 A
2 B 12 A 22 A
3 D 13 A 23 C
4 C 14 D
5 B 15 C
6 A 16 A
7 B 17 B
8 A ,D 18 D
9 B,C 19 B
10 A ,C 20 C
7 C,D 17 B
8 B,C 18 B
9 A ,B 19 C
10 A ,B , D 20 D
MATHEMATICS Ques Ans Ques Ans Ques Ans
1 A 11 A 21 B
2 B 12 D 22 D
3 A 13 C 23 C
4 C 14 B
5 B 15 C
6 C 16 B
IIT- JEE 2011 (September issue) PHYSICS Ques Ans Ques Ans Ques Ans
1 B 11 C 21 B
2 B 12 C 22 D
3 A 13 D 23 A
4 D 14 C
5 A 15 C
6 B 16 D
7 A ,B ,C , D 17 A
8 B,C 18 C
9 A 19 A
8 A,C,D 18 A
9 B, D 19 A
10 A ,C , D 20 D
C H E MI S T R Y Ques Ans Ques Ans Ques Ans
1 B 11 C 21 A
2 B 12 A 22 A
3 C 13 B 23 A
4 D 14 C
5 A 15 B
6 C 16 A
7 B, D 17 D
10 A ,C , D 20 C
MATHEMATICS Ques Ans Ques Ans Ques Ans
1 D 11 B 21 B
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2 D 12 A 22 D
3 D 13 C 23 A
4 B 14 D
5 A 15 A
69
6 C 16 C
7 A ,B ,C , D 17 A
8 A ,B 18 C
9 A ,B ,C 19 C
10 A,B,C 20 B
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