XtraEdge for IIT-JEE

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MARCH 2012

"Faliure is Success if we learn from it"

Volume - 7 Issue - 9 March, 2012 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) Tel. : 0744-2500492, 2500692, 3040000 e-mail : xtraedge@gmail.com

Editorial

Editor : Pramod Maheshwari [B.Tech. IIT-Delhi]

Cover Design Govind Saini

Layout Rajaram Gocher

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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. Inr 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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Editor : Pramod Maheshwari XtraEdge for IIT-JEE

Dear Students, Find a mentor who can be your role model and your friend ! A mentor is someone you admire and under whom you can study. Throughout history, the mentor-protege relationship has proven quite fruituful. Socrates was one of the early mentors. Plato and Aristotle studied under him and later emerged as great philosophers in their own right. Some basic rules to know mentors : • The best mentors are successful people in their own field. Their behaviors are directly translatable to your life and will have more meaning to you. • Be suspicious of any mentors who seek to make you dependent on them. It is better to have them teach you how to fish than to have them catch the fish for you. That way, you will remain in control. • Turn your mentors into role models by examining their positive traits. Write down their virtues. without identifying to whom they belong. When you are with these mentors, look for even more behavior that reflect their success. Use these virtues as guidelines for achieving excellence in your field. Be cautious while searching for a mentor : • Select people to be your mentors who have the highest ethical standards and a genuine willingness to help others. • Choose mentors who have and will share superb personal development habits with you and will encourage you to follow suit. • Incorporate activities into your mentor relationship that will enable your mentor to introduce you to people of influence or helpfulness. • Insist that your mentor be diligent about monitoring your progress with accountability functions. • Encourage your mentor to make you an independent, competent, fully functioning, productive individual. (In other words, give them full permission to be brutally honest about what you need to change.) Getting benefited from a role-mode : Acquiring good habits from others will accelerate you towards achieving your goals. Ask yourself these questions to get the most out of your role model/mentors : • What would they do in my situation? • What do they do every day to encourage growth and to move closer to a goal ? • How do they think in general ? in specific situations ? • Do they have other facts of life in balance ? What effect does that have on their well-being ? • How do their traits apply to me ? • Which traits are worth working on first ? Later ? A final word : Under the right circumstances mentors make excellent role models. The one-to-one setting is highly conducive to learning as well as to friendship. But the same cautions hold true here as for any role model. It is better to adapt their philosophies to your life than to adopt them . Presenting forever positive ideas to your success.

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Volume-7 Issue-9 March, 2012 (Monthly Magazine) NEXT MONTHS ATTRACTIONS

CONTENTS INDEX

PAGE

Regulars ..........

Much more IIT-JEE News. Know IIT-JEE With 15 Best Questions of IIT-JEE

NEWS ARTICLE

Challenging Problems in Physics,, Chemistry & Maths

• Wavelet & Fractal • IITian bags job with Rs 73 lc salary

Key Concepts & Problem Solving strategy for IIT-JEE. IIT-JEE Mock Test Paper with Solution AIEEE & BIT-SAT Mock Test Paper with Solution

3

IITian ON THE PATH OF SUCCESS

5

Mr. Pramod Maheshwari

ALL ABOUT ISEET

7

Frequently Asked Questions

KNOW IIT-JEE

9

Previous IIT-JEE Question

Study Time........ S

DYNAMIC PHYSICS

Success Tips for the Months • "All of us are born for a reason, but all of us don't discover why. Success in life has nothing to do with what you gain in life or accomplish for yourself. It's what you do for others."

8-Challenging Problems [Set # 11] Students’ Forum Physics Fundamentals Calorimetry, K.T.G., Heat transfer Atomic Structure, X-Ray & Radio Activity

CATALYSE CHEMISTRY

• "Success does not consist in never making mistakes but in never making the same one a second time." • "A strong, positive self-image is the best possible preparation for success." • "Failure is success if we learn from it." • "The first step toward success is taken when you refuse to be a captive of the environment in which you first find yourself."

XtraEdge for IIT-JEE

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Key Concept Purification of Organic Compounds Boron Family & Carbon Family Metallurgy Understanding : Inorganic Chemistry

• "Don't confuse fame with success. Madonna is one; Helen Keller is the other." • "Success is not the result of spontaneous combustion. You must first set yourself on fire."

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DICEY MATHS

39

Mathematical Challenges Students’ Forum Key Concept Definite Integrals & Area under curves Probability

Test Time .......... XTRAEDGE TEST SERIES

50

Mock Test IIT-JEE Paper-1 & Paper-2 Mock Test AIEEE Mock Test BIT SAT

SOLUTIONS

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Wavelet & Fractal The subject of wavelet and fractal analyses is fast developing and has drawn a great deal of attention of scientists in varied disciplines of science and engineering. The wavelet transformation is a localized transformation of signals in spacetime and time-frequency domains. This property can be effectively utilized to extract information from signals that is not possible with the conventional signal processing tools. Over the past decade, wavelets, multiresolution and multifractal analyses have been formalized into a thorough mathematical framework and have found a variety of applications with significant impact in the analyses of several geophysical processes such as geomagnetism, atmospheric turbulence, space-time rainfall, ocean wind waves, fluid dynamics, seafloor bathymetry, welllogging and climate change studies among others. It is likely that there will be a variety of applications of wavelets and fractals in geophysics in the years to come. This workshop aims to create a platform to discuss the developments in wavelet-based and fractal-based data analysis techniques and their applications in various processes of Earth, Ocean and Atmospheric sciences. Papers related to but not limited to the following themes are welcome. •

Construction of wavelets

•

Discrete and Continuous wavelet transforms in geophysics

•

Multiresolution and Multifractal Analysis in geophysics

•

Wavelet-based optimization

•

Wavelet-based techniques

•

Wavelets and Fractals in Earth, Ocean and Atmospheric Sciences

data

XtraEdge for IIT-JEE

processing

IIT-ian bags job with Rs 73 lc salary Bagging a job in US company, Pocket Gems with a magnificient salary (per annum) of $137,000 (over Rs 73 lakh), an IIT student from Kanpur, Karan Narain made a record. With these huge amount of salary, Narain became one of the first IIT-ian and one of the Indians who have offered such a huge huge salary during their campus selection. Speaking about his selection in Pocket Gems, Narain said, "Pocket Gems is into mobile application development for Android and iOS. The interviews held in Kanpur were completely technical." The IIT student gave all credit to his schooling in different cities (Delhi, Chennai, Hyderabad and Agra) in India and claimed that the schools, where he has done his schooling, explored a new world of knowledge to Narain who is pursuing M Tech in the IIT-Kanpur.

affiliated to the WRCBB for the research projects. The Department of Biosciences & Bioengineering is relatively a young department but has already achieved recognition for its research excellence in multiple areas. These include -- cancer cell biology, signalling mechanisms in immune cells, computational biology, computational neurobiology, bionanotechnology, biosensors and drug delivery systems, among others. In addition, WRCBB will focus on understanding cell motility and cancer invasion as its research area with the ultimate goal to build a better knowledge base in this field.

IIT to design early earthquake warning system

Ex-IITian NRI's 5 mn dollar gift to cancer esearch project Living up to his promise made nearly four years ago, a US-based Indian and IT entrepreneur came to his almamater Indian Institute of Technology (IIT) here to launch a research centre for biosciences and bio-engineering with the focus on cancer. An alumnus of the 1969 batch of IIT, Romesh Wadhwani, founder of Symphony Technology Group and chairman of Wadhwani Foundation, along with his wife Kathy and IIT Bombay director Devang Khakhar, inaugurated the Wadhwani Research Centre in Biosciences & Bioengineering (WRCBB) at the IITBombay campus The WRCBB follows a generous gift of US$ 5 million announced by Wadhwani in New York, towards the project, designed to focus on understanding cell and cancer invasion. The IIT-B's Faculty of Department of Biosciences & Bioengineering will be 3

Indian Institute of Technology, Gandhinagar, Ahmedabad has taken up a project to design a 'slightly early' earthquake warning system. The IITGN researchers have partnered with California Institute of Technology (CIT) , Pasadena , USA, to develop the system which will have a strong network of low cost motion sensor and the ability to detect earthquake's early seismic activity. "The goal of our project is to put in place a network of small devices called accelerometers near an active fault line which can pick up earth vibrations. They can be standalone devices transmitting data to a central server," said Prof Girish Singhal, project in-charge at IIT-GN. Singhal clamied that "A mesh of very low-cost sensors in that area MARCH 2012

shall be able to pick up velocity of shock waves, issuing slight early alerts of an earthquake". According to experts, the alert from the system will help to close down big machines, which are major sources of secondary losses during an earthquake. Meanwhile, the project is expected to be of significant help to first responders during a natural calamity.

IIT KGP to develop technology for Coal India Ltd. Coal India Limited (CIL) has approached IIT Kharagpur for developing technology to improve safety standards at mines. CIL had searched for appropriate technology at Indian and overseas institutes and universities without much success. Currently CIL has adopted the same practices used in Australian mines, where safety standards are considered to be the best in the world.The other idigenous partner in this project along with IIT KGP is Jadavpur University. The outgoing Chairman of CIL commented that there has been progress in this regard. The organiszation is ready to spend considerable amount for developing the new technology.

IIT honors Digvijay Singh for inventing call records eraser technology Kanpur : Mighty impressed with his ability to erase call records from telecom companies’ database with help of some secret technology, IIT here has decided to confer an honorary degree on Digvijay Singh, senior Congress leader and former CM of Madhya Pradesh. The secret technology came to public knowledge when it was found that call records didn’t confirm the claims of Digvijay Singh about him talking to Hemant Karkare hours before he (Mr. Karkare, unfortunately) was killed fighting the terrorists during the Mumbai terror attacks in November 2008. Digvijay Singh apparently has a special phone with his secretly developed technology that seamlessly erases the call records as soon as the conversation is over XtraEdge for IIT-JEE

Faking News tried to call up Digvijay Singh many times to get his reactions on this development, but all our calls were unanswered. Strangely, none of our mobile phones showed Digvijay Singh’s number in its outgoing call records list afterwards “Of course he erased the call records; he can’t be lying,” Professor P K Dhutt of IIT Kanpur said. “It’s amazing,” Professor Dhutt added, “it surpasses the hacking abilities of both Mark Zuckerberg and Julian Assange, and since he (Digvijay Singh) is from a party that idolizes Mahatma Gandhi, we have no option but to conclude that it was a case of ethical hacking.” If Diggy Raja, as Digvijay Singh is respectfully called, accepts this offer from IIT Kanpur, he’d be awarded PhD in Electronics and Communications Engineering. In fact, IIT administration has suggested that Diggy Raja should file an application for patent over the technology that he had developed. “Need for such a technology is being felt strongly all over the world,” Professor Dhutt explained, “imagine if all those who have been caught on tapes talking to Niira Radia could erase all the call records and claim that the released tapes were work of some mimicry artists.” Professor Dhutt further informed that the institute was willing to spend on research and development and “enrich” the technology developed by Digvijay Singh so that in could be applied in other areas too. “It could help us in curbing student suicides as well; badly performing students could simply erase their bad grades from our database with this technology and choose to start a new life,” Prof Dhutt wondered about the possibilities. Meanwhile sources at IIT Kanpur reveal that Diggy Raja had almost lost this honorary degree to some unidentified “capable people” in BJP, whom he thought could tamper with WikiLeaks, rendering the whistleblower website to some kind of wiki site. But the latest revelations of missing called records clinched the deal for Diggy Raja. 4

IITs will set papers for ISEET: Kapil Sibal Kapil Sibal in a conference with state ministers said that the prestigious IITs (Indian Institutes of Technology) of the country will prepare questions papers for the ISEET (Indian Science Engineering Eligibility Test) for next year. This is the first time when the government has decided to hold a common entrance test for all IITs, NITs of the country rather than conducting individual tests for them. Central Board of Secondary Education (CBSE) will take care of management and also conduct the entrance test. HRD Ministry want to replace the existing exam AIEEE and IITJEE by ISEET. Both the exams are conducted by CBSE for engineering college admission.JEE exam is conducted by IITs for UG and Integrated PG engineering courses for admission in IITs only. This aspiring exam is going to replace all state level entrance exams too for engineering courses. The ISEET 2013 will be held in two phases: Main exam and advance examination. IITs and other Central Educational Institutions were also proposing to accept a weightage of 40% for state Board marks. Some states are not happy with this change and not willing to accept this exam. They said that states conduct the exam in regional language, but this ISEET exam will be held only in Hindi or in English. This will create problem for those students who do not know both the languages. The States of Tamil Nadu, Himachal Pradesh, Odisha, Puducherry and West Bengal asked for some more time to study the proposal in detail. However, the ministry want to reach up to some conclusion over this proposal within two months. MARCH 2012

Success Story This article contains story/interview of persons who succeed after graduation from different IITs

“Pursue Excellence and all else shall follow….” Mr. Pramod Maheshwari B-Tech, IIT-Delhi CMD & CEO, Career Point Infosystems Ltd., Kota offering quality academic courses is contingent upon recruiting and retaining experienced faculty members, providing updated educational content and effective academic administration and control on content delivery. We retain faculty and instructors with relevant industry experience and appropriate academic credentials. Our Research and Development Cell helps in reassessing and updating our tutorial courses on a regular basis which also helps us in designing new academic courses. Regular feedbacks from our students are also an inbuilt standard procedure for our content delivery.

He holds a B.Tech Degree from IIT Delhi. He is a first generation entrepreneur and the key founder member of the company. Recently he has been awarded as the 'Star CEO' of the country. Pramod Maheshwari has over 15 years of experience in developing and implementing training methodologies. He plays a major role in providing thought, leadership and strategic guidance, in addition to supervising the functional heads. He is responsible for the overall operation and growth of the company. Career Point was established in 1993, to impart quality education to students preparing for various competitive examinations. With the sky-high ideals and commitment to excellence, now Career Point has taken a shape of vibrant, dynamic and responsible institute of the country. Today, Career Point stands apart and well above the rest on a distinguished platform, as an epitome of success. Since beginning Career Point’s objective is to enable each aspirant to achieve success in different competitive examinations. In the pursuit of which, Career Point has to its credit a team of outstanding faculty members including IITians, NITians and Doctors, added with the complete & finest study material, excellent coaching methodology and a stimulating academic environment. Career Point believes that effective guidance is the primary need of every student, which would create motivation and instill courage and confidence to face all challenges. And that is exactly what Career Point imparts in all its coaching programmes. For every course, Career Point has a strategic & a well charted programme, which aims at skills in a well organized manner so that it leads them like a self guided missile to unfailingly hit the target in the bull’s eye.

Strong brands and geographic presence We believe that our training centres have established a competitive position and brand recognition among students. We currently have presence across 13 states and providing our expert services to them. In our Kota centre students from across the country and even from Singapore and the Middle East are also there, which in turn reinforces the brand equity and our geographical reach. Qualified faculty team We believe that our qualified and experienced faculty members contribute to our success. Our faculty members are graduates in engineering and science from Indian Institute of Technology, National Institute of Technology and other colleges in India. Our faculty members are well equipped with subject knowledge guiding and tutoring students. We also have an ongoing in-house faculty training facility which ensures that all our faculty members undergo training on our teaching methodologies and skills and subject matter of relevant courses and to keep them abreast of the changes in competitive entrance examination trends and changing student needs.

Sir, can you tell us what the major competitive strength of Career Point are? We believe the following competitive strengths contribute to our success and differentiate us from our competitors:

Experience management team Our senior management team, comprising of senior vice presidents and above, has collective experience of over 65 years and over 13 years of average experience in the education industry. We believe our management led by our Promoters, some of whom have extensive tutorial

Commitment to offering quality courses and student success We offer quality tutorial courses, and intend to improve the learning experience for our students. We believe XtraEdge for IIT-JEE

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MARCH 2012

experience, have deep understanding of the education industry, which enables us to successfully manage our operations and facilitate our growth.

DAY 1 Class of ABC

Quality teaching methodology We have over a period of time developed a scientific teaching methodology and system of teaching, which we believe is essential for success in any competitive entrance examination. We understand that in order to achieve success, one needs knowledge which should be acquired through a comprehensive systematic approach, rigorous practice, time management and confidence.

Revision of ABC

Our focus is to train our students by developing necessary conceptual knowledge base, enhance speed and accuracy levels, infuse confidence and build the right temperament to face the competitive entrance examination. In such competitive examinations, we believe our teaching methodology plays a key role in enhancing studentsâ€™ overall performance.

Questions of ABC

Sir, recently Career Point has penetrated in capital market through IPO? Can you tell us what your future plans are? The IPO proceeds would be utilised for constructing and developing an integrated campus facility; secondly for the expansion of classroom infrastructure and office facility; thirdly, for acquisitions and other strategic initiatives; and to meet expenses towards general corporate purposes.

Question and Pending work of ABC

DAY 3 Question and Pending work of PQR

Quick Reviev for the class of PQR

Quick Review for the class of ABC

Class of PQR

Class of ABC

Revision of PQR

Revision of ABC

Questions of PQR

Questions of ABC

Next Class

DAY1 On day 1, suppose topic ABC is being taught in the class, now after the class student should revise the ABC at home (shown as ABCRevision) in the diagram. After revision of ABC student should do the questions of ABC (shown as Questions of ABC) to make the ABC topic perfect.

The entire requirement of funds set forth will be met from the proceeds of the issue. We intend to set-up an integrated campus facility at Kota, for 3,000 students which in addition to providing tutoring facility will also provide facilities such as accommodation for students, library, guest house for visiting parents, primary health centre, auditorium, canteen, cafeteria, indoor and outdoor recreation, staff quarters, provision for utilities like departmental store, bank facilities etc.

DAY2 Now on day 2, student should complete the questions and pending work of ABC (Questions and Pending work of ABC). Before the class of PQR on this day student should study the PQR which is going to be taught in the class (shown as Quick Review for the class of PQR). After the class of PQR student will do the revision and problems of PQR (shown as Revision of PQR and Questions of PQR).

Sir, brief us about your recent foray into Education Consultancy and Management Services and formal education? We plan to participate in formal education through ECAMS, we believe there is a huge potential for ECAMS in the K-12 and Higher Education segment in India. We will explore opportunities to provide ECAMS to a number of privately and/or Government-run schools, colleges and universities. We also intend to enter into partnerships with the Governments under the PPP model to manage schools, colleges and universities in rural and/or urban areas.

What is your take on the new education bill? The initiatives taken by the government are applaudable. In my opinion, the government should introduce more education reforms. We would be happy to take part in anything which is for the benefit of Indian education system.

Sir, can you suggest some study pattern to the students at home after taking class in coaching? Ans. This study plan is meant for the students to save their time and decide the strategy to study the current day topic and go through the next day topic. Regular revision is very important to understand the topic and the subject in depth.

XtraEdge for IIT-JEE

DAY 2

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ALL ABOUT ISEET Frequently Asked Questions What is the pattern of the ISEET 2013? • ISEET 2013 will be conducted twice in 2013, first in the month of April or May and second in November or December and the ultimate aim is to conduct it thrice or four times in a year • Results will be valid for two years while multiple attempts are allowed for students to improve scores by getting multiple attempts • ISEET 2013 will be a single day exam in two sessions o Morning Session – 10:00 AM to 01:00 PM (3 Hours) o Afternoon Session – 02:00 PM to 05:00 PM (3 Hours) • ISEET 2013 will have two papers o ISEET Main – Objective type aptitude test to be held in the morning session to access the abilities of comprehension, critical thinking, logical reasoning of students o ISEET Advance– Objective type test to be held in the afternoon session to access the problem solving abilities of a student in basic science subjects i.e. Physics, Chemistry & Mathematics

What is ISEET - Indian Science-Engineering Eligibility Test and when it will be in operation? • ISEET (Indian Science-Engineering eligibility Test) is the proposed Single National level entrance test. • It is proposed to be conducted from year 2013 for all students who seek admissions in central engineering institutions i.e. IITs, NITs, IIITs and IISERs • As of now ISEET 2013 will replace IIT-JEE & AIEEE but it is proposed that ISEET will gradually replace all State level exams like RPET, MH-CET, Karnataka CET, etc. Why has the pattern changed? • ISEET 2013 has been proposed by the Ministry of Human Resource Development, Government of India to reduce the burden on aspirants due to multiple entrance examinations conducted across India each year. • The burden in terms of time, money (examination & Transit fee) and the stress caused in scheduling and preparing for each examination syllabus. Is this change beneficial for students in general? • Yes, it is definitely beneficial for students. Earlier they had to prepare for many engineering entrance tests. • ISEET implementation will reduce mental and financial burden on the student/parent and save time as well. • Students can concentrate on the single test. It will improve the chances of getting admission in a good college. • As ISEET will be conducted twice or more during a year and the scores will be valid for two years, students will have multiple chances to improve their scores

What will be the admission process from 2013 for IITs? • Board Percentage of Class 12th will be given a minimum of 40% weightage • Different State Boards and Central Board results normalized on the basis of percentile formula. • ISEET will replace IITJEE and will come into place from academic session 2012-13 • ISEET Main and Advance will be compulsory. • Main will have 30% to 60% weightage and Advance exam will have 0% to 30% weightage. The actual weightage within this limit will be decided by the board of directors if IITs. • Merit list for the admissions will be prepared by the admission committee for IITs • All other institutes who took admissions based on JEE will also follow the same pattern

What major changes come with the new pattern? • With emphasis on School Board marks a minimum 40% weightage will be mandatorily given to School Board marks in the admission process across India • ISEET Main (Objective Aptitude Test comprehension, critical thinking, logical reasoning) made a mandatory part of admission process with a minimum of 30% weightage • ISEET Advance (Advanced to School Board curriculum & in-between AIEEE and IIT-JEE pattern basic science subject knowledge objective test) with a maximum of 30% weightage in admission process made a mandatory part of for taking admissions into IITs and NITs while it is optional for other central and state level institutions XtraEdge for IIT-JEE

What will be that admission process from 2013 for institutes currently using AIEEE rank? • Board Percentage of class 12th will be given a minimum 40% weightage. • Different State Boards and Central Board results normalized on the basis of percentile formula. • ISEET will replace AIEEE and will come into place from academic session 2012-13 • ISEET Main and Advance will be compulsory. 7

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•

• •

•

Main will have 30% to 60% weightage and Advance exam will have 0% to 30% weightage. The actual weightage within this limit will be decided by the admission committee for NITs.. Merit list for the admissions will be prepared by the common admission committee for NITs. All other institutes who took admissions based on AIEEE will also follow the same pattern

•

What is the selection process for admissions to other institutes accepting ISEET 2013 scores? • The final cut-off list for admissions with ISEET score will be generated in three steps. Each state government or institute will be able to decide the specific weightage for Board, Main and Advance exam scores. • Following are guidelines mentioned below: o Class XII Board Score: weightage not less than 40% and can go up to 100% of the total score o ISEET Main - weightage not less than 30% and can go up to 60% of the total score o ISEET Advance - weightage not more than 30% of the total score. o Combined Score- of ISEET Main & Advance does not exceed weightage 60% of the total score • It would be up to each institution/ groups of institutions/State agencies to carry out the task of counseling and finally the admission in a coordinated manner

• •

Will I have to take coaching for the same pattern? Yes, you will require coaching to score good marks in boards, ISEET main (aptitude test) and advance. Now onwards ISEET will be the only exam. To take admission in a good engineering colleges like IITs, NITs good preparation is mandatory. “Early Start” to the preparation will help the student. I am an intelligent student, will this change be beneficial for me? Yes, this change will increase your chances to get IITs instead of NITs because selection criteria for both the institutes will be same in general.

What is ISEET Main? • ISEET Main is an Objective Aptitude Test like CAT for IIMs, SAT in the USA, partly BITSA. • Main will test the inherent intelligence of the student

I am an average/ below average student, how it will be beneficial for me? It will be even beneficial for you since you will be free from tension of preparing different test syllabus for different colleges and can concentrate for only one test. Your chances of getting good college will increase.

What is ISEET Advance? • ISEET Advance is expected to be Advance to School Board curriculum & its difficulty level will be some what in between AIEEE and IIT-JEE. • This is a mandatory part of for taking admissions into IITs and NITs while it is optional for other central and state level institutions • Since HRD Ministry of India has indicated to pay more emphasis on board education, ISEET Advance exam level is likely to be closer to AIEEE which has the difficulty level advanced to board curriculum

Will Career Point be able to prepare me according to that examination pattern? • Yes, Career point has already completed its preparation for the change. • We have already included 12th Board (CBSE/ State Board) in our academic curriculum in 2007 and we are providing the complete study material for board examinations. In fact we have adopted these patterns far before than other institutes in Kota. • Our students have been performing tremendously good in board examination nation wide along with NTSE, Olympiads & other reputed national level scholarship tests. • We have been conducting National Science Proficiency Test (NSPT) every year at national level where we have tested more that 4.5 lacs students for their of comprehension, critical thinking, logical reasoning of students • Secondly, the pattern of ISEET will be more close to AIEEE in which Career Point has proved its leadership in the very first year of AIEEE Examination in year 2001-02 and maintaining the leading position since then.

Who will conduct ISEET? • CBSE in collaboration with State Boards will physically conduct and manage the examinations across India • CBSE is the body which currently conducts AIEEE • Question paper will be set by IITs. Will ISEET replace all State level engineering and science entrance exams? • All states have accepted the new pattern of common entrance exam except Tamil Nadu, Himachal Pradesh, Odisha, Puducherry and West Bengal which will take the decision by the end of April 2012 XtraEdge for IIT-JEE

Further, It is proposed that ISEET will gradually replace all entrance exams within a couple years and aspirants will have to prepare for only one national entrance exams to take admissions into all science and engineering colleges in India vis a vis IITs, NITs, IIITs, IISc, IISERs, other Technical and Deemed Universities as well as all Private colleges States which will base its admissions on ISEET as of now are Delhi, Haryana, Chandigarh and Uttarakhand The decision to implement ISEET by the states of Gujarat, Madhya Pradesh and Chhattisgarh is final stages The state of Tamil Nadu has rejected ISEET and admissions to state government colleges will be done with 100% weightage for Tamil Nadu State Board results

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KNOW IIT-JEE By Previous Exam Questions

Now as loop goes down, magnetic flux linked with it increases, hence induced current flows in such a direction so a to reduce the magnetic flux linked with it. Hence induced current flows in anticlockwise direction. (B) Each side of the cube will experience a force as shown (since a current carrying segment in a magnetic field experience a force).

PHYSICS 1.

^

A magnetic field B = B0(y/a) K is into the paper in the +z direction. B0 and a are positive constants. A square loop EFGH of side a, mass m and resistance R, in x – y plane, starts falling under the influence of gravity see figure) Note the direction of x and y axes in figure [IIT-1999] O x ⊗ ⊗ ⊗

E H

⊗ ⊗ ⊗

F G

⊗ ⊗

→ → → ^ ^ B y^ F1 = i( l × B) = i – a i× 0 k = B0yi j ; a → ^ ^ B (y + a) ^ F3 = i + a i× 0 k = – iB0(y + a) j a

g

⊗

→

Find (a) the induced current in the loop and indicate its direction. (b) the total Lorentz force acting on the loop and indicate its direction, and (c) an expression for the speed of the loop, v(t) and its terminal value. Sol. Suppose at t = 0, y = 0 and t = t, y = y

→

∴

→

m

B 02 a 2 v( t ) ^ j ; R

B 2 a 2 v( t ) dv = mg – 0 R dt v

Integrating it, we get,

∫ 0

t

dv B 2 a 2 v( t ) g– 0 mR

=

∫ dt 0

( v) t

B 2 a 2 v( t ) log g – 0 mR 0 – B 02 a 2 mR

dφ dy Net emf., e = – = – B0a = – B0av(t) dt dt As total resistance = R B av( t ) |e| = 0 |i|= R R X → F1 y

or

or

1–

or

1 – e – B0 a

→

F2

=t

B 02 a 2 v( t ) g – 2 2 mR = – B 0 a t log g mR

y+a

2 2 B 02 a 2 v( t ) = e – B0a t mR mgR 2 2

t

=

B 02 a 2 v( t ) ; mgR

2 2 mgR e – B0 a t ∴ v(t) = 2 2 1 – mR B0 a

→

F3

Y

XtraEdge for IIT-JEE

→

^

B y ∴ φ = 0 .a2 = B0ya a

F4

→

= – iB0a j = –

→

→

→

F = F1 + F2 + F3 + F4

(A) Total magnetic flux = AB → → ^ B y^ Where A = a 2 k and B = 0 k a

∴

→

Please note that F2 = – F4 and hence will cancel out each other.

y

9

MARCH 2012

For the point P, x = 0 ∴ Path difference = t (µ2 – µ1) = t(1.7 – 1.4) = 0.3 t ...(i) th But the point P lies between the 5 maximum and 6th minimum (given). Therefore the path difference = 5λ + ∆ ...(ii) Equating equations (i) and (ii), we get ...(iii) 0.3t = 5λ + ∆ The path difference ∆ can be determined from the I 3 = . given intensity at P, which is I0 4

When terminal velocity is attained, V(t) does not depend on t mgr ∴ V(t) = 2 2 B0 a

In Young's experiment, the upper slit is covered by a thin glass plate of refractive index 1.4 while the lower slit it covered by another glass plate, having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength 5400 Å. It is found that the point P on the screen where the central maximum (n = 0) fell before the glass plates were inserted now has 3/4 the original intensity. It is further observed that what used to be the fifth maximum earlier, lies below the point P while the sixth minimum lies above P. Calculate the thickness of the glass plate. (Absorption of light by [IIT-1997] glass plate may be neglected.) Sol. The time taken by the ray to reach P' from S1 d plate d = air + Vplate Vair 2.

=

The expression of I/I0 in terms of ∆ is I π∆ = cos 2 I0 λ 3 π∆ For I/I0 = 3/4 , we get cos = 2 λ π∆ π λ = or ∆ = λ 6 6 Hence, the thickness of the glass plates (Eq. 3) is λ 0.3t = 5λ + λ/6 or ∆ = 6 Hence, the thickness of the glass plates (Eq. 2) is

or

S1P'– t t + c c / µ1

S1P'– t + tµ1 c Effective path travelled = S1P' – t + tµ1 where c is the speed of light in air. P' =

t S1

1 31 0.3t = 5λ + λ/6 or t = λ 0.3 6 1 31 4 –6 = × 5400 Å = 9.3 × 10 Å = 9.3×10 m 0.3 6

x

µ1

d

A 3.6 m long vertical pipe resonates with a source of frequency 212.5 Hz when water level is at certain height in the pipe. Find the height of water level (from the bottom of the pipe) at which resonance occurs. Neglect end correction. Now, the pipe is filled to a height H(≈ 3.6m). A small hole is drilled very close to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of water level in the pipe as a function of H. If the radii of the pipe and the hole are 2 × 10–2 m and 1 × 10–3 m respectively, calculate the time interval between the occurance of first two resonances. Speed of sound in [IIT-2000] air is 340 m/s and g = 10 m/s2. Sol. Speed of sound, V = 340 m/s. Let l0 be the length of air column corresponding to the fundamental frequency. Then V = 212.5 4l 0 3.

P S2

µ2 D

Similarly the time taken by the ray to reach P' from S2 S P'– t + tµ 2 = 2 c Effective path travelled = S2P' – t + tµ2 ∴ Path difference = S2P' – t + tµ2 – S1P' + t – tµ1 = (S2P' – S1P') + t(µ2 – µ1) Also when there ware no plates infront of the slits. xd = S2P' – S1P' = D xd S2P' – S1P' = D xd ∴ Path difference = + t(µ2 – µ1) D XtraEdge for IIT-JEE

or

10

l0 =

V 340 = = 0.4 m. 4(212.5) 4(212.5)

MARCH 2012

In close pipe only odd harmonics are obtained. Now, let l1, l2, l3, l4 etc. be the lengths corresponding to the 3rd harmonic, 5th harmonic, 7th harmonic etc. Then

3.14 × 10 –6 – dH = dt 1.26 × 10 – 3

⇒

V = 212.5 ⇒ l1 = 1.2 m ; 3 4l 1

= (1.11 × 10–2

H

H

Between first two resonances, the water level falls from 3.2 m to 2.4 m.

V = 212.5 ⇒ l2 = 2.0 m 5 4l 2

∴

V = 212.5 ⇒ l3 = 2.8 m; 7 4l 3

⇒

V 9 4l 4

– dH

2 × 10 × H

dH 2.4

= 212.5 ⇒ l4 = 3.6 m

∫

3.2

⇒

= – 1.11 × 10–2 dt

H

dH H

1

= – (1.11 × 10–2)

[

∫ dt 0

]

2 2.4 – 3.2 = – (1.11 × 10–2) . t

⇒ t ≈ 43 second 0.4m

3.2m

4.

1.2m

3.4m

Three particles A, B and C, each of mass m, are connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side l. This body is placed on a horizontal frictioness table (x-y plane) and is hinged to it at the point A so that it can move without friction about the vertical axis through A (see figure). The body is set into rotational motion on the table about A with a [IIT-2002] constant angular velocity ω. y A

2.0m 2.8m

x ω

1.6m 0.8m

→ F C B l (a) Find the magnitude of the horizontal force exerted by the hinge on the body. (b) At time T, when the side BC is parallel to the x-axis, a force F is applied on B along BC (as shown). Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time T. Sol. The mass B is moving in a circular path centred at

or heights of water level are (3.6 – 0.4) m, (3.6 – 1.2) m (3.6 – 2.0) m and (3.6 – 2.8) m. Therefore heights of water level are 3.2 m, 2.4 m, 2.4 m, 1.6 m and 0.8 m. Let A and a be the are of cross-sections of the pipe and hole respectively. Then A = π (2 × 10–2)2 = 1.26 × 10–3 m–2. and a = π(10–3)2 = 3.14 × 10–6 m2 Velocity of efflux, V =

A. The centripetal force (mlω2) required for this circular motion is provided by F′. Therefore a force

2gH

Continuity equation at 1 and 2 gives, – dH a 2gH = A dt

F′ acts on A (the hinge) which is equal to mlω2. The same is the case for mass C. Therefore the net force on the hinge is

Therefore rate of fall of water level in the pipe,

Fnet = F'2 + F'2 +2F' F' cos 60º

a – dH 2gH = A dt Substituting the values, we get

XtraEdge for IIT-JEE

Fnet = 2F' 2 +2F' 2 ×

11

1 = 3 F′ = 2

3 mlω2

MARCH 2012

Sol. Let the piston be displaced by a distance x.

Y A F′ l

Fnet

Mg γ Mg + p ( v 0 − Ax ) γ Then p 0 + V0 = p 0 + A A

X

60º F′

Q Initial pressure on the gas P1 = p0 +

l

F′

P2 = p0 +

Final pressure on the gas

F′

Mg A Mg +p A

P0

C B l (b) The force F acting on B will provide a torque to the system. This torque is

A V0

l 3 = Iα 2 3 F 3l F× = (2ml2)α ⇒ α = 4 ml 2 The total force acting on the system along x-direction is F + (Fnet)x This force is responsible for giving an acceleration ax to the system.

x

F×

c.m l

where p is the extra pressure due to which the compression x takes place. Final volume of the gas V2 = V0 – Ax The above equation can be rearranged as Mg Ax γ p + p (V0 − Ax) γ p0 + 1 + 1− A Mg V0 = = + p Mg 0 γ A p0 + V A 0

3 2

⇒ F Therefore F + (Fnet)x = 3m(ax) c.m. 3 F l F F = 3m Q ax = αr = × = 4 ml 4m 3 4 3F = 4 F ∴ (Fnet)x = 4 (Fnet)y remains the same as before = 3 mlω2.

5.

γAx V 0

Negligible as p and x are small ∴

An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M. The piston and the cylinder have equal cross-sectional area A. Atmospheric pressure is P0, and when the piston is in equilibrium, the volume of the gas is V0. The piston is now displaced slightly from its equilibrium position. Assuming that the system is completely isolated from its surroundings, show that the piston executes simple harmonic motion and find the frequency of oscillation. [IIT-1981]

XtraEdge for IIT-JEE

p γAx 1=1+ – + Mg Mg V0 p 0 + p0 + A A p

p Mg p0 + A

=

γAx V0

∴

Mg γAx p = p0 + A V0

⇒

F Mg γAx = p0 + A A V0

⇒

Mg γA 2 x F = p0 + A V0

⇒

Mg γA 2 x Ma = p 0 + A V0

⇒

Mg γA 2 x a = p0 + A V0 M

Comparing it with a = ω2x we get Mg γA 2 x ω2 = p 0 + A V0 M

12

MARCH 2012

Mg γA 2 x ω = p0 + A V0 M

∴

If

H3N

Mg is small as compared to p0 then A

∴

f=

H 3N

p0γ V0 M

A 2π

3+

NH3

H3N

or

NH3

NH3

NH3

Co3+

H3N

NH3

NH3

In [Ni(CN)42– nickel is present as Ni2+ ion and its coordination numbers is four Ni28 =1s2, 2s22p6, 3s23p63d8, 4s2 Ni2+ ion = 1s2, 2s22p6, 3s23p63d8 3d 4s 4p Ni2+ ion =

CHEMISTRY 6.

NH3

Co

p 0 γA 2 = 2πf V0 M

ω=

NH3

(a) Write the chemical reaction associated with the "brown ring test".

3d

4s

4p

Ni2+ion in Complex ion

(b) Draw the structures of [Co(NH3)6]3+, [Ni(CN)4]2– and [Ni(CO)4]. Write the hybridization of atomic orbital of the transition metal in each case.

dsp2 hybridization

Hence structure of [Ni(CN)4]2– is

(c) An aqueous blue coloured solution of a transition metal sulphate reacts with H2S in acidic medium to give a black precipitate A, which is insoluble in warm aqueous solution of KOH. The blue solution on treatment with KI in weakly acidic medium, turns yellow and produces a white precipitate B. Identify the transition metal ion. Write the chemical reaction involved in the [IIT-2000] formation of A and B.

C≡N

N≡C Ni2+ N≡C

C≡N

In [Ni(CO)4, nickel is present as Ni atom i.e. its oxidation number is zero and coordination number is four. 3d 4s 4p Ni in Complex

Sol. (a) NaNO3 + H2SO4 → NaHSO4 + HNO3

2HNO3 + 6FeSO4 + 3H2SO4 → 3Fe2(SO4)3 + 2NO + 4H2O

sp3 hybridization

[Fe(H2O)6]SO4.H2O + NO

Its structure is as follows :

Ferrous Sulphate → [Fe(H2O)5NO] SO4 + 2H2O

CO

(Brown ring) (b) In [Co(NH3)6]3+ cobalt is present as Co3+ and its coordination number is six. 1

2

6

2

6

7

2

6

Ni OC

2

Co27 = 1s , 2s 2p , 3s 3p 3d , 4s 3+

2

2

6

Co ion = 1s , 2s 2p , 3s 3p 3d 3d

4s

4p

3d

4s

4p

CO (c) The transition metal is Cu2+. The compound is CuSO4.5H2O

Hence

CuSO4 + H2S Acidic medium → CuS ↓ + H2SO4 Black ppt

3+

Co ion in Complex ion

2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4 (B) white I2 + I– → I3– (yellow solution)

d2sp3 hybridization

XtraEdge for IIT-JEE

CO

6

13

MARCH 2012

Hence structure of [Ni(CN)4]2– is

(a) Write the chemical reaction associated with the "brown ring test". (b) Draw the structures of [Co(NH3)6]3+, [Ni(CN)4]2– and [Ni(CO)4]. Write the hybridization of atomic orbital of the transition metal in each case. (c) An aqueous blue coloured solution of a transition metal sulphate reacts with H2S in acidic medium to give a black precipitate A, which is insoluble in warm aqueous solution of KOH. The blue solution on treatment with KI in weakly acidic medium, turns yellow and produces a white precipitate B. Identify the transition metal ion. Write the chemical reaction involved in the [IIT-2000] formation of A and B. Sol. (a) NaNO3 + H2SO4 → NaHSO4 + HNO3 2HNO3 + 6FeSO4 + 3H2SO4 → 3Fe2(SO4)3 + 2NO + 4H2O [Fe(H2O)6]SO4.H2O + NO Ferrous Sulphate → [Fe(H2O)5NO] SO4 + 2H2O (Brown ring) (b) In [Co(NH3)6]3+ cobalt is present as Co3+ and its coordination number is six. Co27 = 1s1, 2s22p6, 3s23p63d7, 4s2 Co3+ion = 1s2, 2s22p6, 3s23p63d6 3d 4s 4p Hence 7.

3d

4s

C≡N

N≡C Ni2+ N≡C

C≡N

In [Ni(CO)4, nickel is present as Ni atom i.e. its oxidation number is zero and coordination number is four. 3d 4s 4p Ni in Complex sp3 hybridization

Its structure is as follows : CO

Ni OC

CO

CO (c) The transition metal is Cu2+. The compound is CuSO4.5H2O

CuSO4 + H2S Acidic medium → CuS ↓ + H2SO4 Black ppt

2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4 (B) white – – I2 + I → I3 (yellow solution)

4p

3+

Co ion in Complex ion

(a) A white solid is either Na2O or Na2O2. A piece of red litmus paper turns white when it is dipped into a freshly made aqueous solution of the white solid. (i) Identify the substance and explain with balanced equation. (ii) Explain what would happen to the red litmus if the white solid were the other compound. (b) A, B and C are three complexes of chromium (III) with the empirical formula H12O6Cl3Cr. All the three complexes have water and chloride ion as ligands. Complex A does not react with concentrated H2SO4, whereas complexes B and C lose, 6.75% and 13.5% of their original mass, respectively, an treatment with conc. H2SO4. [IIT-1999] Identity A, B and C. Sol. (a) The substance is Na2O2 (i) Na2O2 + 2H2O → 2NaOH + H2O2 (strong base) (Weak acid) H2O2 + red litmus → White H2O2 → H2O + [O] Nascent oxygen bleaches the red litumus. (ii) Na2O + H2O → 2NaOH 8.

d2sp3 hybridization

H3N

NH3

NH3

3+

or

Co H 3N

NH3

H3N

NH3

NH3

NH3

Co3+

H3N

NH3

NH3

In [Ni(CN)42– nickel is present as Ni2+ ion and its coordination numbers is four Ni28 =1s2, 2s22p6, 3s23p63d8, 4s2 Ni2+ ion = 1s2, 2s22p6, 3s23p63d8 3d 4s 4p Ni2+ ion = 3d

4s

4p

Ni2+ion in Complex ion dsp2 hybridization

XtraEdge for IIT-JEE

14

MARCH 2012

NaOH solution turns colour of red litmus paper into blue due to stronger alkaline nature. (b) A = [Cr(H2O)6]Cl3. It has no reaction with conc. H2SO4 as its all water molecular are present in coordination sphere. B = [Cr(H2O)5Cl]Cl2.H2O Conc. H2SO4 removes its one mol of H2O as it is outside the coordination sphere. Molecular Weight of complex = 266.5 18 × 100 = 6.75% % loss = 266.5 C = [Cr(H2O)4Cl]Cl2.2H2O Conc. H2SO4 removes its 2H2O which are outside of the coordination sphere. 18 × 100 = 13.5 % % loss = 2 × 266.5 Hence complexes A = [Cr(H2O)6]Cl3 B = [Cr(H2O)5Cl]Cl2.H2O C = [Cr(H2O)4Cl2]Cl.2H2O

2 =5 0.4 (b) According to adiabatic gas equation, P1V1γ = P2V2γ Here, P1 = P V1 = V V2 = 5.66 V Hence, PVγ = P2 × (5.66V)γ = P2 ×(5.66)γ × Vγ P P P = = [using eq.(1)] or P2 = (5.66) γ (5.66)1.4 11.32

or

Hence, work done by the gas during adiabatic expansion P PV − × 5.66V P1V1 − P2 V2 11 .32 = = γ –1 1.4 – 1 PV 2 = PV = 1.25 PV 0.4 2 × 0.4

PV −

=

An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V, while its temperature falls to T/2. (a) How many degrees of freedom do the gas molecules have ? (b) Obtain the work done by the gas during the expansion as a function of the initial pressure P and [IIT-1990] volume V. Sol. (a) According to adiabatic gas equation, TVγ–1 = constant or T1V1γ–1 = T2V2γ–1 Here, T1 = T ; T2 = T/2 V1 = V and V2 = 5.66 V T × (5.66V)γ–1 Hence, TVγ–1 = 2 T × (5.66)γ–1 × Vγ–1 = 2 or (5.66)γ–1 = 2 ...(1) Taking log, (γ – 1)log 5.66 = log 2 0.3010 log 2 or γ–1= = 0.4 = 0.7528 log 5.66 9.

10. An organic compound X, on analysis gives 24.24 percent carbon and 4.04 percent hydrogen. Further, sodium extract of 1.0 g of X gives 2.90 g of silver chloride with acidified silver nitrate solution. The compound X may be represented by two isomeric structures, Y and Z. Y on treatment with aqueous potassium hydroxide solution gives a dihydroxy compound, while Z on similar treatement gives ethanal. Find out the molecular formula of X and [IIT-1989] give the structures of Y and Z. Sol. C = 24.24%, H = 4.04% 35.5 2.90 g Percentage of Cl = × × 100 = 71.74% 143.5 1g 24.24 = 2.02 12 4.04 = 4.04 Relative number of H atoms = 1 71.74 Relative number of Cl atoms = = 2.02 35.5 Atomic ratio = C : H : Cl = 2.02 : 4.04 : 2.02 = 1 : 2: 1 Empirical formula of X = CH2Cl It is given that isomer of Y of the compound X gives a dihydroxy compound on treatment with aqueous potassium hydroxide. Therefore, the given compound should contain two Cl atoms. Thus molecular formula of X = C2H4Cl2 The two isomers of X are : H H | | Y = ClCH2CH2Cl → Cl − C − C − Cl | | H H

Relative number of C atoms =

or γ = 1.4 If f, be the number of degrees of freedom, then 2 2 or 1.4 = 1 + γ=1+ f f 2 = 1.4 – 1 = 0.4 or f XtraEdge for IIT-JEE

f=

15

MARCH 2012

1 2 4π (a – b2) sin 2θ + ...(2) 2 3 Equation of normal to the ellipse at R is ax sin(θ + 4π/3) – by cos(θ + 4π/3) 1 = (a2 – b2) sin (2θ + 8π/3) ...(3) 2 But sin (θ + 4π/3) = sin(2π + θ – 2π/3) = sin(θ – 2π/3) and cos (θ + 4π/3) = cos (2π + θ – 2π/3) and sin (2θ + 8π/3) = sin (4π + 2θ – 4π/3) = sin (2θ – 4π/3) Now, (3) can be written as ax sin (θ – 2π/3) – by cos (θ – 2π/3) 1 = (a2 – b2) sin (2θ – 4π/3) ...(4) 2 For the lines (1), (2) and (4) to be concurrent, we must have the determinant

H H | | H − C − C − Cl Z = CH3CHCl2 → | | H Cl

=

The relevant reactions are : ( aq ) (i) ClCH2–CH2Cl KOH → HO–CH2–CH2–OH ( aq ) → CH3CHO (ii) CH3CHCl2 KOH

MATHEMATICS 11. Let ABC be an equilateral triangle inscribed in the circle x2 + y2 = a2. Suppose perpendiculars from A, x2 y2 B, C to the major axis of the ellipse 2 + 2 = 1, a b (a > b) meets the ellipse respectively at P, Q, R so that P, Q, R lie on the same side of the major axis as A, B, C respectively. Prove that the normals to the ellipse drawn at the points P, Q and R are concurrent. [IIT-2000] Sol. Let A ≡ (a cosθ, b sinθ) so the coordinates of B ≡ {a cos(θ + 2π/3), a sin (θ + 2π/3)} and C ≡ {a cos(θ + 4π/3), a sin (θ + 4π/3)}. y

∆1 =

P

Q R

x Ellipse

C

According to the given condition, coordinates of P are (a cosθ, b sinθ), and that of Q are (a cos(θ + 2π/3), b sin(θ + 2π/3)) and that of R are (a cos(θ + 4π/3), b sin (θ + 4π/3)) (It is given that P, Q, R are on the same side of x-axis as A, B and C) Equation of the normal to the ellipse at P is ax by – = a2 – b2 cos θ sin θ 1 or ax sin θ – by cos θ = (a2 – b2) sin 2θ ...(1) 2 Equation of normal to the ellipse at Q is 2π 2π ax sin θ + – by cos θ + 3 3 XtraEdge for IIT-JEE

=0

12. A right circular cone with radius R and height H contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = k > 0). Find the time [IIT-2003] after which the cone is empty. Sol. Given : liquid evaporates at a rate proportional to its surface area dv ⇒ ∝–S ...(1) dt 1 We know, volume of cone = πr2h and 3 3 surface area = πr (of liquid in contact with air) 1 ...(2) or V = πr2h and S = πr2 3 r R and = tan θ ...(3) where tan θ = h H from (2) and (3) 1 S = πr2 ...(4) V = πr3 cotθ and 3 R

Circle

O

1 2 (a − b 2 ) sin 2θ 2 1 2 4π (a − b 2 ) sin 2θ + 2 3 1 2 4π ( a − b 2 ) sin 2θ − 2 3

Thus line (1), (2) and (4) are concurrent.

A(a cos θ, b sinθ)

B

a sin θ − b cos θ 2π 2π a sin θ + − b cos θ + 3 3 2π 2π a sin θ − − b θ − 3 3

r h

θ

H

Substituting (4) in (1), we get 1 dr cotθ . 3r2 . = – kπr2 3 dt 16

MARCH 2012

⇒ cot θ

∫

0

R

dr = – k

∫

T

0

⇒ sin (2A + 60º) = sin (C – A) = – sin(60º + 2C) =

dt

⇒ 2A + 60º = 30º, 150º {neglecting 30º, as not possible}

⇒ cot θ(0 – R) = – k(T – 0) ⇒ R cot θ = kT ⇒ H = kT (using (3)) H ⇒T= k ∴ required time after which the cone is empty H =T= k

⇒

⇒

∴ C = 75º Hence, A = 45º, B = 60º, C = 75º 15. Find the centre and radius of the circle formed by all the points represented by z = x + iy satisfying the z−α =k(k ≠ 1), where α and β are relation z −β

constant complex numbers given by α = α1 + iα2, β = β1 + iβ2. [IIT-2004] Sol. As we know; |z|2 = z. z

2

⇒ 1 ≤ x ≤ 2 the curve y = 2x lies above 2 as compared to y = logex Hence, the required area

In the region

=

∫

2

1/ 2

60º + 2C = 210º, 330º

⇒ C = 75º or 135º Also from (1) sin (C – A) = 1/2 C – A = 30º, 150º, 195º for A = 45º, C = 75º and C = 195º (not possible)

y = logex 1

2A + 60º = 150º

⇒ A = 45º again from (1), sin (60º + 2c) = –1/2

13. Sketch the curves and identify the region bounded by x = 1/2, x = 2, y = ln x and y = 2x. Find the area [IIT-1991] of this region. Sol. The required area is the shaded portion in following figure. y y = 2x

O 1/2

⇒

| z − α |2 | z − β |2

= k2

(z – α)( z – α ) = k2(z – β)( z – β ) |z|2 – α z – α z + |α|2 = k2(|z|2 – β z – β z+ |β|2)

or |z|2 (1 – k2) – (α – k2β) z – ( α – β k2) z

(2 x − log x) dx

+ (|α|2 – k2|β|2) = 0 2

2x = − ( x log x − x) log 2 1/ 2

⇒

|z|2 –

(α − k 2β) (1 − k 2 )

4− 2 5 3 = – log 2 + log 2 2 2

+

z–

(α − β k 2 ) (1 − k 2 )

| α |2 −k 2 | β |2 (1 − k 2 )

z

=0

...(i)

On comparing with equation of circle, |z|2 + a z + α z + b = 0

14. ABC is a triangle such that

whose centre is (– a) and radius = | a |2 −b

1 2 If A, B and C are in Arithmetic Progression, determine the values of A, B and C. [IIT-1990] Sol. Given that in ∆ABC, A, B and C are in A.P. A + C = 2B also A + B + C = 180º ⇒ B = 60º Also given that, sin (2A + B) = sin (C – A) = – sin (B + 2C) = 1/2 ...(1)

sin(2A + B) = sin(C – A) = –sin(B + 2C) =

XtraEdge for IIT-JEE

1 2

∴ centre for (i) =

α − k 2β 1− k 2

and radius

α − k 2β α − k 2 β α α − k 2β β = 1 − k 2 1 − k 2 − 1 − k 2

radius =

17

k (α − β) 1− k 2

MARCH 2012

Physics Challenging Problems

Set # 11

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics, that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants. By : Dev Sharma Director Academics, Jodhpur Branch

So lutions will b e p ub lished in nex t issue 1.

(C) the speed of the missile when it was detected gx is 2

A metallic conductor of irregular cross section is as shown in figure A constant potential difference is applied across the ends (A) and (B). Then A

Q

P

4.

(A) Electric current at cross section P is equal to that of cross section Q (B) Electric field intensity at P is less than that at Q (C) The number of electrons crossing per unit area per unit time at cross section P is less than that at Q (D) The rate of heat generating per unit time at Q is greater than that of P 2.

(C) has a maximum value Bvl 2 (D) has a maximum value 2Bvl

A circular ring of radius R with uniform positive charge density λ per unit length is located in the y-z plane with its centre at the origin O. A particle of

5.

charge –q0 is released from x = 3R on x-axis at t = 0 then kinetic energy of particle when it passes through origin, is

3.

(A)

λq 0 2 ∈0

(B)

(C)

q 0λ ∈0

(D)

q 0λ 3 ∈0 q 0λ 4 ∈0

6.

(A) the velocity with which the missile was

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Figure shows a parabolic reflector in x-y plane given by y2 = 8x. A ray of light traveling along the line y = a is incident on the reflector. Find where the ray intersects the x-axis after reflection. y-axis y2=8x P(0,a)

gx

(B) you have a warning time of

The variation of pressure versus volume is shown in the figure. The gas is diatomic and the molar specific heat capacity for the process is found to be xR. Find the value of x. P

V

Missile is fired for maximum range at your town from a place in the enemy country at a distance ‘x’ from your town. The missile is first detected at its half-way point. Then projected is

x 4 Figure shows a square loop being pulled out with a constant speed out of region of uniform magnetic field. The induced emf in the loop B × × × × × l × l × × × × v × × × × × × × l × × × × (A) first increases, then decreases (B) first decreases, then increases

(D) the maximum height attained by the missile is

B

line y = a incident ray x-axis

x 2g

18

MARCH 2012

7.

A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine the value of Z. (Ionisation energy of hydrogen atom is 13.6 eV)

8.

Consider the circuit showing in figure. There are three switches S1, S2, S3. Match the columns.

Column-I

Column-II

(A) If S2 and S3 are opened and S1 is

(P) CV/4

Puzzle : Marble Mix Up

•

Years ago, to puzzle his friends, a scientist gave one of four containers containing blue and/or yellow marbles to each of the friends; Tom, Dick, Harry, and Sally.

•

There were 3 marbles in each container, and the number of blue marbles was different in each one. There was a piece of paper in each container telling which color marbles were in that container, but the papers had been mixed up and were ALL in the wrong containers.

•

He then told all of his friends to take 2 marbles out of their container, read the label, and then tell him the color of the third marble.

•

So Tom took two blue marbles out of his container and looked at the label. He was able to tell the color of the third marble immediately.

•

Dick took 1 blue marble and 1 yellow marble from his container. After looking at his label he was able to tell the color of his remaining marble.

•

Harry took 2 yellow marbles from his container. He looked at the label in his container, but could not tell what color the remaining marble was.

•

Sally, without even looking at her marbles or her label, was able to tell the scientist what color her marbles were. Can you tell what color marbles Sally had? Can you also tell what color marbles the others had, and what label was in each of their containers?

closed then in steady state, charge on the capacitor is (B) If switch S2 only is closed then

(Q) 2CV/5

maximum charge on the capacitor is (C) If switch S3 only is closed then

(R) CV/3

maximum charge on the capacitor is (D) If all the switches are closed then (S) CV maximum charge on the capacitor is (T) zero

Cartoon Law of Physics As speed increases, objects can be in several places at once. This is particularly true of tooth-and-claw fights, in which a character's head may be glimpsed emerging from the cloud of altercation at several places simultaneously. This effect is common as well among bodies that are spinning or being throttled. A `wacky' character has the option of self- replication only at manic high speeds and may ricochet off walls to achieve the velocity required.

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19

MARCH 2012

1.

2.

8

Solution

Physics Challenging Problems Qu e s tio ns we r e Pub lis he d in Feb rua ry I ssu e

Ans. Remain Same Hint: KE = QU Magnetic moment = i × Area Q = × πR 2 T 2mKE 2MU 2πm QT = = R= qB qB qB

Magnetic moment =

Q2 × B 2m × U × π× 2 2 2πm Q B

Magnetic moment =

U B

Ans.

Set # 10

4.

Ans. None Hint:

Both dF get cancel out net force on the loop is zero. 5.

Ans. 2q Ea Hint: Work done by field = – (UB – UA) = −q[VB − VA ] = +q[E × d] = qE × 2a = 2qEa

6.

Ans.

B0 v λ

Hint:

2B 0 πqR 2 3

2π × B 0 (p + qt ) 3 dφ 2π e= = R 2 × × B0q dt 3

Hint: φ = R 2 ×

3 B 0 lv 3λl B0 v i= λ Ans. Zero if both wires slide in opposite direction, 0.2mA if both wires move towards left Hint: i=

3.

7.

Ans. 4 Hint:

φ = 3B 0 t × π12 − B 0 t (π 22 − π12 )

φ = 2B 0 t × π12 − B 0 t × πr22 dφ = 2B 0 π12 − B 0 πr22 dt dφ E × 2πr = dt As E = 0 dφ =0 Q 4r12 = r22 dt

When both are moving in same direction BLv i= = 0.2mA (9 + 1)

r1 r2

When both are moving in opposite direction equation emf of battery = 0 ∴i = 0 XtraEdge for IIT-JEE

8.

20

2

1 = 4

Ans. Α→P,Q,R; B→P,Q,S; C→Q,T; D→P,Q,S MARCH 2012

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21

MARCH 2012

Students Forum Expert’s Solution for Question asked by IIT-JEE Aspirants

PHYSICS

For vertical forces on cylinder,

A triangular prism of mass M = 1.12 kg having base angle 37º is placed on a smooth horizontal floor. A solid cylinder of radius R = 20 cm and mass m = 4 kg is placed over the inclined surface of the prism. If sufficient friction exists between the cylinder surface and the prism, so that cylinder does not slip, calculate acceleration of prism when the system is released. Calculate also, force of friction existing between the cylinder and the prism. (g = 10 ms–2) m

1.

mg – N cos37º – F sin37º = m (0.2α . sin37º) ...(iii) Taking moments (about O) of forces acting on the cylinder, FR = Iα where

...(iv) I=

mR 2

From above equation

2

= 0.08 kg m2 N = 23 newton

α = 30 rad/sec2 M

37º

Sol. Let angular acceleration of cylinder be α clockwise and acceleration of prism be a leftwards.

a = 3.75 ms–2

Ans.

F = 12 newton

Ans.

A cylindrical tank of base area A has a small orifice of area a at the bottom. At time t = 0, a tap starts to supply water into the tank at a constant rate Q m3 s–1. Calculate relation between height h of water in the tank and time t. Sol. When water supply is started, water starts to accumulate in the tank. But leakage of water through orifice at bottom also start simultaneously. 2.

Acceleration of cylinder axis (relative to prism) is Rα = 0.2 α down the plane. Its horizontal and vertical components are (rightwards) and 0.2αsin37º 0.2α .cos37º (downwards) respectively. But resultant acceleration of cylinder axis is vector sum of Rα and a, therefore horizontal and vertical components of resultant acceleration of cylinder axis become (0.2α cos37º – α) rightwards and (0.2α sin 37º) downwards respectively.

Let at instant t, height of water in the tank be y as shown in figure

Considering free body diagrams (fig. (a) and Fig. (b)) Mg

F

N

Iα

y

m (0.2 α cos 37º – a) O

M.g

37º

N

F

m (0.2 a sin 37º)

Then flow velocity through orifice,

M.a

v Fig. (a)

v= Fig. (b)

∴ Volume flowing out per second through orifice,

For horizontal forces acting on prism, N sin 37º – F cos 37º = Ma

q = a 2gy

...(i)

But rate of supply to the tank is Q. Therefore net of increase of volume in tank = (Q – q) m3s–1. Since,

For horizontal forces acting on cylinder, N sin 37º – F cos 37º = m (0.2α cos 37 – a) XtraEdge for IIT-JEE

2gy

...(ii) 22

MARCH 2012

area of tank base is A, therefore, net rate of increase of height of water in tank,

Gravitational potential energy of body at A = that due to larger star + that due to smaller star.

Q – 2gy dy (Q – q ) = = dt A A

∴ U1 = –

integrating above equation with limits, at t = 0, y = 0 and at t, y = h,

Similarly, gravitational potential energy at P,

h

∴

∫Q– 0

t=

U2 = –

t

dy 2gy

∫

= A . dt =–

0

Q – 2gh A Q – 2h + log e Q g g

Distance between centres of two stars is 10 α. Mass of these stars in M and 16 M and their radii are a and 2a respectively. A body of mass m is fired straight from the surface of larger star directly towards the smaller star. Calculate minimum initial speed of the body so that it can reach the surface of smaller star. Obtain the expression in terms of G, M, and a. Sol. Since the body is projected from surface of large star towards smaller star, therefore, the body follows a straight line path AB, as shown in figure (a). Near point A, magnitude of gravitational force exerted by larger star on the body is greater than that exerted by smaller star. Therefore, near point A, the body experiences a resultant force directed towards larger star. Hence, the body retards till this resultant force becomes zero. It means velocity of body is minimum at that point where magnitudes of gravitational force exerted by two stars are equal. If initial velocity of star is such that it crosses this point, then it will reach the smaller star. 3.

B

2a

G (16M)m GMm – 8a a 5 GMm a 2

Minimum kinetic energy required at A = Increase in potential energy from A to P

∴

A

G (16M )m G (M)m 65 GMm – =– 2a 8a 8 a

1 mv 02 = U2 – U1 2

∴ v0 =

4.

45GM 4a

Ans.

A non-conducting piston of mass m and area S divides a non-conducting, closed cylinder into two parts as shown in figure. Piston is connected with left wall of cylinder by a spring of force constant K. Left part is evacuated and right part contains an ideal gas at pressure P. Adiabatic constant of the gas is γ and in equilibrium length of each part is l. Calculate angular frequency of small oscillations of the piston. K

Pressure P

Sol. If the piston is slightly displaced leftwards from its equilibrium position, spring is further compressed and gas expands. Due to expansion of gas, its pressure decreases. Piston is restored due to both the reasons, i.e., increase in compression and decrease in pressure.

a

10.a Fig.(a)

Let the piston be displaced through dx. Then increase in compression in spring = Kdx

16M

P

2a

Increase in volume of gas is dV = Sdx

M a

Since piston and cylinder both are non-conducting, therefore, gas undergoes an adiabatic expansion.

(10a–x)

x

Hence, it obeys the law PVγ = constant.

Fig.(b)

Taking log, log P + γ. log V = constant.

Let distance of this point P from centre of larger star be x. Then, or

G (16M )m x2

=

Differentiating the above equation. dV γP dP + γ = 0 or dP = – dV V V P

GMm (10a – x ) 2

or dP = –

x = 8a

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23

γP γP (S.dx) = – dx (Sl) l

MARCH 2012

Restoring force, F = K.dx + S. |dP | = K.dx + or F =

passing through a section in one second in n = 6.25 × 1011

γPS dx l

∴ Number of α-particles in unit length of the n beam = = 3.125 × 105 per m. v

Kl + γPS dx l

∴ Number of α-particle in length l of the beam

F Kl + γPS ∴ Restoring acceleration = = dx m ml

=

...(i)

= 6.25 × 104

Since acceleration of piston is restoring and is directly proportional to displacement dx, therefore, it performs SHM.

5.

ω=

Kl + γPS ml

Ans.(ii)

Let potential difference of the source be V volt. Kinetic energy of α-particle accelerated by this source,

Comparing equation (i) with Restoring acceleration = ω2 . (displacement) Angular frequency,

n l v

E = qV or V =

Ans.

E = 41.75 kV q

Power supplied by the source to accelerate α-particles, P = VI = 8.35 × 10–3 watt

A steady beam of α-particles travelling with kinetic energy E = 83.5 ke V carries a current of I = 0.2 µA. (i) If this beam strikes a plane surface at an angle θ = 30º with normal to the surface, how many α-particles strike the surface in t = 4 second ?

Ans.(iii)

GLOBAL WARMING IS REAL

(ii) How many α-particles are there in length l = 20 cm of the beam? (iii) Calculate power of the source used to accelerate these α-particles from rest. (Mass of α-particle = 6.68 × 10–27 kg) Sol. Since, current is rate of flow of charge through a section, therefore, a current I = 0.2µA means that a charge 0.2 µC is flowing per second. Charge of an α-particle is q = 2e = 3.2 × 10–19 C

∴ Rate of flow of α-particles, n =

The arctic ice is receding and global warming is no longer a theory but a reality. Scientists predict that by the year 2100, the average surface temperature will jump up by 6 degrees Fahrenheit. Nighttime temperatures will be higher and there will be hotter days.

I = 6.25 × 1011 per second q

∴ Number of α-particles striking against a surface in t = 4 second = n × t = 6.25 × 1011 × 4 = 2.5 × 1012

Ans.(i)

Since air temperature is a powerful component of climate, there will be unavoidable climate changes in the future. Some climate changes involve extreme weather disturbances such as more severe hurricanes and longer droughts. There will be an increased precipitation of snow and rain during winter. The faster melting of snow during the spring will result in flooding. All these climate changes are predicted based on the assumption that changes will be relatively gradual.

(Not : these is no significance of angle θ for calculation of number of α-particles striking the surface.) Kinetic energy of each α-particle is E = 83.5 Ke V or E = (83.5 × 103) (1.6 × 10–19) J But E =

1 mv2 2

where m = 6.68 × 10–27kg

∴ Velocity of α-particles is v = 2 × 106 ms–1. It means a beam of length v = 2 × 106 m crosses a section in one second. But number of α-particles XtraEdge for IIT-JEE

24

MARCH 2012

P HYSICS F UNDAMENTAL F OR IIT-J EE

Calorimetry, K.T.G., Heat transfer KEY CONCEPTS & PROBLEM SOLVING STRATEGY Calorimetry : The specific heat capacity of a material is the amount of heat required to raise the temperature of 1 kg of it by 1 K. This leads to the relation

where C´ is a constant that depends on the nature and extent of the surface exposed. Simplifying dθ C´ = constant = –C(θ – θ0) where C = dt ms

Q = ms θ

Kinetic theory of gases :

where Q = heat supplied, m = mass, θ = rise in temperature. The relative specific heat capacity of a material is the ratio of its specific heat capacity to the specific heat capacity of water (4200 J kg–1K–1). Heat capacity or thermal capacity of a body is the amount of heat required to raise its temperature by 1 K. [Unit : J K–1] Thus

heat capacity = Q/θ = ms

Also

dθ 1 dQ = × dt ms dt

The pressure of an ideal gas is given by p =

where µ = mass of each molecule, n = number of molecules per unit volume and C is the root square speed of molecules. p=

pV =

1 mC2 3

This is defined as C=

m×s sw

C12 + C 22 + C 32 + ... + C 2N N

where N = total number of molecules. It can be obtained through these relations C=

3p = ρ

3RT M

Total Energy of an ideal gas (E) :

This is equal to the sum of the kinetic energies of all the molecules. It is assumed that the molecules do not have any potential energy. This follows from the assumption that these molecules do not exert any force on each other.

Σmsθ Σ sθ , for equal masses θ = Σms´ Σs

E=

Newton's law of cooling : The rate of loss of heat from a body in an environment of constant temperature is proportional to the difference between its temperature and that of the surroundings.

1 3 m 3 mC2 = RT = pV 2 2 M 2

Thus, the energy per unit mass of gas = The energy per unit volume =

If θ = temperature of the surroundings then

The energy per mole =

dθ – ms = C´(θ – θ0) dt

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or

Root Mean Square Speed of Molecules :

where m = mass of body, s = specific heat capacity of the body, sw = specific heat capacity of water. Principle of Calorimetry : The heat lost by one system = the heat gained by another system. Or, the net heat lost or gainsed by an isolated system is zero. It system with masses m1, m2, ...., specific heat capacities s1, s2, ...., and initial temperatures θ1, θ2, .... are mixed and attain an equilibrium temperature θ then

θ=

1 2 ρC 3

where ρ is the density of the gas and m = mass of the gas.

i.e., the rate of heating (or cooling) of a body depends inversely on its heat capacity. The water equivalent of a body is that mass of water which has the same heat capacity as the body itself. [Unit : g or kg] This is given by W=

1 µnC2 3

25

1 2 C 2

3 p 2

3 3 pV = RT 2 2

MARCH 2012

velocities between c and c + dc per unit volume is

Perfect gas equation : From the kinetic theory of gases the equation of an ideal gas is pV = RT for a mole m RT for any mass m and pV = M Avogadro number (N) and Boltzmann constant (k) : The number of entities in a mole of a substance is called the Avogadro number. Its value is 6.023 × 1023 mol–1. The value of the universal gas constant per molecular is called Boltzmann constant (k). Its value is 1.38 × 10–23 J K–1. Degrees of Freedom : Principle of equipartition of energy : The number of ways in which energy may be stored by a system is called its degrees of freedom. Principle of Equipartition of Energy : This principle states that the total energy of a gas in thermal equilibrium is divided equally among its degrees of freedom and that the energy per degree of freedom is kT/2 where T is the temperature of the gas. For a monoatomic atom the number of degrees of freedom is 3, for a diatomic atom it is 5, for a polyatomic atom it is 6. Hence the energy of a mole of a monoatomic gas is

2

dn = 4πna3 e − bc c2 dc where b=

m 2πkT

and the number of molecules with the velocity c per unit volume is 2

nc = 4πna3 e − bc c2 The plot of nc and c is shown in the figure. The velocity possessed by the maximum number of molecules is called the most probable velocity

α c Crms

α=

2kT / m

The mean velocity 8kT / mπ and

c =

3kT / mπ

vrms =

Conduction : The transfer of heat through solids occurs mainly by conduction, in which each particle passes on thermal energy to the neighboring particle but does not move from its position. Very little conduction occurs in liquids and gases. θ2 θ1

3 1 µ = N 3 × kT = RT 2 2 Which is the same as that given by the kinetic theory. For a mole of diatomic gas µ

Q

5 1 = N 5 × kT = RT 2 2 For a mole of polyatomic gas µ

Q d A Consider a slab of area A and thickness d, whose opposite faces are at temperature θ1 and θ2 (θ1 > θ2). Let Q heat be conducted through the slab in time t.

1 = N 6 × kT = 3RT 2 When the irrational degrees of freedom are also taken into account, the number of degrees of freedom = 6n – 6 for non-linear molecules = 6n – 5 for linear molecules where n = number of atoms in a molecule. Kinetic Temperature : The kinetic temperature of a moving particle is the temperature of an ideal gas in thermal equilibrium whose rms velocity equals the velocity of the given particle. Maxwellian distribution of velocities : In a perfect gas all the molecules do not have the same velocity, rather velocities are distributed among them. Maxwell enunciated a law of distribution of velocities among the molecules of a perfect gas. According to this law, the number of molecules with XtraEdge for IIT-JEE

m and a = 2kT

Then

θ − θ2 Q = λA 1 t d

where λ = thermal conductivity of the material. This has a fixed value for a particular material, being large for good conductors (e.g., Cu, Ag) and low for insulators (e.g., glass, wood). Heat Current : The quantity Q/t gives the heat flow per unit time, and is called the heat current. In the steady state, the heat current must be the same across every cross-section. This is a very useful principle, and can be applied also to layers or slabs in contact. θ − θ2 Q dθ dθ = – λA where the quantity = 1 is t dx dx d called the temperature gradient. 26

MARCH 2012

Unit of λ : Different units are used, e.g., cal cm s ºC–1, cal m–1 s–1 ºC–1, jm´1 s–1 ºC–1. Convection : It is a process by which heat is conveyed by the actual movement of particles. Particles closest to the source receive heat by conduction through the wall of the vessel. They rise up-wards and are replaced by colder particles from the sides. Thus, a circulation of particles is set up – hot particles constitute the upward current and cold particles, the side and downward current. The transfer of heat by convection occurs only in fluids, and is the main mode of heat transfer in them. Most fluids are very poor conductors. Radiation : Thermal Radiation : Thermal radiations are electromagnetic waves of long wavelengths. Black Body : Bodies which absorb the whole of the incident radiation and emit radiations of all wavelengths are called black bodies. It is difficult to realize a perfect black body in practice. However, a cavity whose interior walls are dull black does behave like a black body. Absorption : Every surface absorbs a part or all of the radiation falling on it. The degree of absorption depends on the nature and colour of the surface. Dull, black surfaces are the best absorbers. Polished, white surfaces absorb the least. The coefficient of absorption for a surface is

aλ =

Stefan-Boltzmann Law : If a black body at an absolute temperature T be surrounded by another black body at an absolute temperature T0, the rate of loss of radiant energy per unit area is

E = σ(T4 – T04) where σ is a constant called Stefan constant and its value is 5.6697 × 10–8 W m–2 K–4 The total energy radiated by a black body at an absolute temperature T is given by E = σT4 × surface area × time Note : Remember that rate of generation of heat by

electricity is given by H = I2 R or

Solved Examples 1.

An earthenware vessel loses 1 g of water per second due to evaporation. The water equivalent of the vessel is 0.5 kg and the vessel contains 9.5 kg of water. Find the time required for the water in the vessel to cool to 28ºC from 30ºC. Neglect radiation losses. Latent heat of vaporization of water in this range of temperature is 540 cal g–1.

Sol. Here water at the surface is evaporated at the cost of the water in the vessel losing heat.

Heat lost by the water in the vessel

radiation absorbed radiation incident

= (9.5 + 0.5) × 1000 × (30 – 20) = 105 cal Let t be the required time in seconds.

The suffix λ denotes the wavelength of the radiation being considered, Clearly, aλ = 1 for a black body, for all values of λ. Emission : Each surface emits radiation (radiates) continuously. The emissive power (eλ) is defined as the radiation emitted normally per second per unit solid angle per unit area, in the wave-length range λ and λ + dλ. Clearly, the emissive power of a black body (denoted by Eλ) is the maximum. Kirchhoff's Law : According to this law, for the same conditions of temperature and wavelength, the ratio eλ /aλ is the same for all surfaces and is equal to Eλ. This simply means that good absorbers are good emitters. Hence, a black body is the best emitter, and a polished white body, the poorest emitter. Prevost's Theory of Exchanges : All bodies emit radiations irrespective of their temperatures. They emit radiations to their environments and receive radiations from their environments simultaneously. In the equilibrium state the exchange between a body and the environment of energy continues in equal amounts. XtraEdge for IIT-JEE

V2 or VI Js–1 or W. R

Heat gained by the water at the surface = (t × 10–3) × 540 × 103 (Q L = 540 cal g–1 = 540 × 103 cal kg–1)

∴ 105 = 540t or t = 185 s = 3 min 5s 2.

15 gm of nitrogen is enclosed in a vessel at temperature T = 300 K. Find the amount of heat required to double the root mean square velocity of these molecules.

Sol. The kinetic energy of each molecule with mass m is given by 1 3 m v 2rms = kT 2 2

...(1)

If we want to increase the r.m.s. speed to η times, then the temperature has to be raised to T´. Then,

27

1 3 1 3 mv 2rms = kT´ or mη2 v 2rms = kT´ 2 2 2 2

...(2)

From eqs. (1) and (2), T´ = η2T

...(3) MARCH 2012

The internal energy of n molecules at temperature T is given by 5 U = nRT 2 5 Similarly, U´ = nRT´ 2 5 ∴ Change in internal energy ∆U = nR[T´ – T] 2 5 ∆U = nRT[η2 – 1] or 2 =

5 m 2 RT[η – 1] 2 M

=

5 15 4 (8.31) (300) [4 – 1] = 10 J 2 28

Sol. The quantity of heat Q passing across the stone is given by KA(T1 − T2 ) t Q= d Here A = 3600 sq. cm = 0.36 m2 d = 10 cm = 0.10 m, (T1 – T2) = 100 – 0 = 100ºC and t = 1 hour = 3600 sec. K × 0.36 × 100 × 3600 ∴ Q= kilo-calories ...(1) 0.10 Now heat gained by the ice in one hour = mass of the ice × latent heat of ice = 4.8 × 80 kilo calories ...(2) From eqs. (1) and (2) K × 0.36 × 100 × 3600 4.8 × 80 = 0.10 4.8 × 80 × 0.10 0.36 × 100 × 3600 = 3 × 10–4 kilo cal m–1(ºC)–1s–1

or K =

10 gm of oxygen at a pressure 3 × 105 N/m2 and temperature 10ºC is heated at constant pressure and after heating it occupies a volume of 10 litres (a) find the amount of heat received by the gas and (b) the energy of thermal motion of gas molecules before heating. Sol. (a) The states of the gas before and after heating are M M PV1 = RT1 and PV2 = RT2 µ µ 3.

A flat bottomed metal tank of water is dragged along a horizontal floor at the rate of 20m/sec. The tank is of mass 20 kg and contains 1000 kg of water and all the heat produced in the dragging is conducted to the water through the bottom plate of the tank. If the bottom plate has an effective area of conduction 1 m2 and the thickness 5 cm and the temperature of water in the tank remains constant at 50ºC, calculate the temperature of the bottom surface of the tank, given the coefficient of friction between the tank and the floor is 0.343 and K for the material of the tank is 25 cal m–1 s–1 K–1. Sol. Frictional force = µ m g = 0.343 × (1000 + 20) × 9.81 = 3432 N The rate of dragging, i.e., the distance travelled in one second = 20 m. ∴ Work done per second = (3432 × 20) Nm/sec. This work done appears as heat at the bottom plate of the tank. Hence 3432 × 20 cal/sec H= 4.18 5.

Solving these equations for T2, we have T2 =

µV2 P 32 × (10 × 10 −3 )(3 × 10 5 ) = = 1156 K MR (10 × 10 −3 )(8.31× 10 3 )

Now T2 – T1 = 1156 – 283 = 873 K The amount of heat received by the gas is given by M ∆Q = CP(T2 – T1) µ (10 × 10 −3 )29.08 × 10 3 × 873 32 = 7.9 × 103 J (b) The energy of the gas before heating M i × × RT1 E1 = µ 2

=

where i = number of degrees of freedom = 5 (for oxygen) (10 × 10 −3 )5 × (8.31× 10 −3 )(283) = 2 × 32 3 = 1.8 × 10 J

4.

H=

Now

25 × 1× (T1 − T2 ) 3432 × 20 = 4.18 0.05

(Q t = 1 sec)

3432 × 20 × 0.05 = 32.84 4.18 × 25 × 1 Temp. of bottom surface T1 = 50 + 32.84 = 82.84ºC

∴ T1 – T2 =

A slab of stone of area 3600 sq cm and thickness 10 cm is exposed on the lower surface of steam 100ºC. A block of ice at 0ºC rests on upper surface of the slab. In one hour 4800 gm of ice is melted. Calculate the thermal conductivity of the stone.

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KA(T1 − T2 ) d

But

28

MARCH 2012

P HYSICS F UNDAMENTAL F OR IIT-J EE

Atomic Structure, X-Ray & Radio Activity KEY CONCEPTS & PROBLEM SOLVING STRATEGY Atomic Structure : According to Neil Bohr's hypothesis is the angular momentum of an electron is quantised.

The maximum number of electrons that can be accommodated in an orbit is 2n2. X-rays :

h h mvr = n or L = n 2π 2π

When fast moving electron strikes a hard metal, X-rays are produced. When the number of electrons striking the target metal increases, the intensity of Xrays increases. When the accelerating voltage/kinetic energy of electron increases λmin decreases. X-rays have the following properties :

2πr = nλ h z –1 c = × ms 2πmr 137 n

h2 rn = 2 2 4π mke ke 2 fn = hr

K.E. = T.E. =

n2 n2 1 = 0.529 Å where k = Z Z 4 πε 0

(a) Radiations of short wavelength (0.01 Å – 10Å); high pentrating power; having a speed of 3 × 108 m/s in vacuum.

15 × 1 = 6.58 × 10 Hz n n

Intensity

vn = Zn

1 ke 2 Z ke 2 − ke 2 ; P.E. = × Z; T.E. = – ×Z 2 r r 2r − 13.6 Z 2

ev/atom where –13.6 n2 = Ionisation energy + P.E. ⇒ +T.E. = = – K.E. 2 Note : If dielectric medium is present then εr has to be taken into consideration.

λmin

(c)

1 1 = R(Z – b)2 1 − 2 λ n

(d) Moseley law

1 p mv 1 = RZ 2 − 2 = = h h n1 n 2

n=2

hc hc 12400 = = Å eV K.E V

b = 1 for k-line transfer of electron

2

n=3

λ

(b) λmin =

me 4 z 2 1 1 v 1 − 2 = =v= 2 3 2 c λ 8ε 0 h c n1 n 2

n=∞ n=7 n=6 n=5 n=4

Continuous spectrum (Varies & depends on accelerating voltage) Characteristic spectrum Kα (fixed for a target material) Kβ Lγ Lβ Lα

R = R0A

1/3

ν = a(z – b)

where R0 = 1.2 × 10–15 m

R = radius of nucleus of mass number A. * Nucleus density is of the order of 1017 kg/m3

Kδ

Lγ Lβ

Kγ Kβ

Lα

Paschen (I.R.)

Isomers are nuclides which have identical atomic number and mass number but differ in their energy states.

–0.85 eV Pfund Brackett (I.R.) (I.R.) –1.5 eV

Nuclear binding energy ∆mc 2 = A Nucleon

–3.4 eV

Balmer (Visible) Limiting line of Lyman series

n=1

where ∆m = mass defect –13.6 eV

=

Lyman Series (U.V. rays)

XtraEdge for IIT-JEE

29

[ Zm p + (A − Z)m n − M ]c 2 A

MARCH 2012

(f) In a nuclear fusion reaction small nuclei fuse to give big nuclei whereas in a nuclear fusion reaction a big nuclei breaks down.

* The binding energy per nucleon is small for small nuclei. * For 2 < A < 20, there are well defined maxima which indicate that these nuclei are more stable.

Thermal neutrons produce fission in fissile nuclei. Fast moving neutrons, when collide with atoms of comparable masses, transfer their kinetic energy to colliding particle and slow down.

* For 30 < A < 120 the average B.E./A is 8.5 MeV / nucleon with a peak value of 8.8 MeV for Iron. * For A > 120, there is a gradual decreases in B.E./nucleon. * More the B.E./A, more is the stability. Radioactivity :

η=

(a) N = N0e–λt

out put In put

dN −dN = activity level = λN where dt dt n

Solved Examples

t 1 T1 / 2

1 (c) N = N0 = N0 2 2

1.

n

1 ⇒ A = A0 where A = activity level 2

(d) T1/2 = (e) τ =

=

The energy of an excited hydrogen atom is –3.4 eV. Calculate the angular momentum of the electron according to Bohr theory.

Sol. The energy of the electron in the nth orbit is

0.693 λ

En = –

1 λ

Here,

(f) τ = 1.4 T1/2 (g) t =

E nhν nhc = = t t λt

Power, P =

β particles are electrons emitted from the nucleus. (n → p + β)

(b)

–

or

N A 2.303 2.303 log10 0 = log10 0 λ N λ A

13.6 n2

13.6 n2

eV

= –3.4

n=2

Angular momentum =

2 × 6.63 × 10 −34 nh = 2π 2 × 3.14

= 2.11 × 10–34 Js.

m 2.303 log 0 λ m

2.

(h) If a radioactive element decays by simultaneous −dN emission of two particle then = λ1N + λ2N dt The following parameters remain conserved during a nuclear reaction

The wavelength of the first member of the Balmer series in the hydrogen spectrum is 6563 Å. Calculate the wavelength of the first member of the Lyman series.

Sol. For the first member of the Balmer series 5R 1 1 1 = R 2 − 2 = λ 36 2 3

(a) linear momentum (b) Angular momentum

...(1)

For the first member of the Lyman series

(c) Number of nucleons

1 1 3R 1 = R 2 − 2 = λ´ 4 2 1

(d) Charge (e) The energy released in a nuclear reaction

...(2)

Dividing Eq. (1) by Eq. (2)

X+P→Y+Z+Q

λ´ 5× 4 5 = = λ 36 × 3 27

Q = [mx + mp) – (my + mz)]c2 = ∆m × c2 Q = ∆m × 931 MeV or XtraEdge for IIT-JEE

∆λ v = λ c

According to Doppler's effect of light

30

λ´ =

5 5 λ= × 6563 = 1215 Å 27 27

MARCH 2012

3.

Hydrogen atom in its ground state is excited by means of a monochromatic radiation of wavelength 970.6 Å. How many different wavelengths are possible in the resulting emission spectrum ? Find the longest wavelength amongst these.

Sol. Initial kinetic energy of the electron = 50.0 keV

Energy of the photon produced in the first collision, E1 = 50.0 – 25.0 = 25.0 keV Wavelength of this photon

Sol. Energy the radiation quantum

E = hv =

λ1 =

6.6 × 10 −34 × 3 × 108 hc = λ 970.6 × 10 −10 × 1.6 × 10 −19

= 0.99 × 10–10 m = 0.99 Å

= 12.75 eV

Kinetic energy of the electron after third collision = 0

Energy of the excited sate

Energy of the photon produced in the third collision ,

En = – 13.6 + 12.75 = – 0.85 eV Now, we know that En = – or n2 = –

6.6 × 10 −34 × 3 × 108 hc = E1 1.6 × 10 −19 × 12.5 × 10 3

E3 = 12.5 – 0 = 12.5 keV

13.6

This is same as E2. Therefore, wavelength of this photon, λ3 = λ2 = 0.99 Å

n2

13.6 −13.6 = = 16 En − 0.85

5.

or n = 4 The number of possible transition in going to the ground state and hence the number of different wavelengths in the spectrum will be six as shown in the figure. n 4 3

In an experiment on two radioactive isotopes of an elements (which do not decay into each other), their mass ratio at a given instant was found to be 3. The rapidly decaying isotopes has larger mass and an activity of 1.0 µCi initially. The half lives of the two isotopes are known to be 12 hours and 16 hours. What would be the activity of each isotope and their mass ratio after two days ?

Sol. We have, after two days, i.e., 48 hours, 4

1 N1 = N10 = N10 /16 2

2

3

1 N2 = N 02 = N 02 /8 2 1

Mass ratio =

The longest wavelength corresponds to minimum energy difference, i.e., for the transition 4 → 3. Now

E3 = –

13.6 32

Now,

= – 1.51 eV

λmax =

A1 = λ1N1 = λ1 N10 /16 = A10 /16 = (1/16)µCi A2 = λ2N2 = λ2 N 02 /8

6.6 × 10 −34 × 3 × 108 (1.51 − 0.85) × 1.6 × 10 −19

= 18.75 × 10–7m = 18750 Å 4.

X-rays are produced in an X-ray tube by electrons accelerated through a potential difference of 50.0 kV. An electron makes three collisions in the target before coming to rest and loses half its kinetic energy in each of the first two collisions. Determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms.

XtraEdge for IIT-JEE

A10 = λ1 N10 = 1.0 µCi

After two days,

hc = E4 – E3 λ max

or

N0 8 N1 3 3× 8 = = 10 . = N2 162 2 N 2 16

But

T λ2 12 3 = 1 = = 16 λ1 T2 4

or

λ2 =

3 λ1 4

1 3 1 A2 = λ1 × N10 × 8 3 4

=

1 1 A10 λ1 N10 = 32 32

= (1/32) µCi

31

MARCH 2012

KEY CONCEPT

PURIFICATION OF ORGANIC CHEMISTRY

Organic Chemistry Fundamentals

Qualitative Analysis : Qualitative analysis of an organic compound involves the detection of various elements present in it. The elements commonly present in organic compounds are carbon, hydrogen, oxygen, nitrogen, halogens, sulphur and sometimes phosphorus. Detection of carbon and Hydrogen : Principle. Carbon and hydrogen are detected by strongly heating the organic compound with cupric oxide, (CuO). The carbon present in the organic compound is oxidised to carbon dioxide and hydrogen is oxidised to water. Carbon dioxide is tested by lime water test, whereas water is tested by anhydrous copper sulphate test. Mixture of orgainc compoud and dry copper oxide (CuO)

Preparation of Lassaigne's Extract (or sodium extract): A small piece of sodium is gently heated in an ignition tube till it melts. The ignition tube is removed from the flame, about 50–60 mg of the organic compound added and the tube heated strongly for 2–3 minutes to fuse the material inside it. After cooling , the tube is carefully broken in a china dish containing about 20–30 mL of distilled water. The fused material along with the pieces of ignition tube are crushed with the help of a glass rod and the contents of the china dish are boiled for a few minutes. The sodium salts formed in the above reactions (i.e., NaCN, Na2S, NaX or NaSCN) dissolve in water. Excess of sodium, if any, reacts with water to give sodium hydroxide. This alkaline solution is called Lassaigne's extract or sodium extract. The solution is then filtered to remove the insoluble materials and the filtrate is used for making the tests for nitrogen, sulphur and halogens. Reactions : An organic compound containing C, H, N, S, halogens when fused with sodium metal gives the following reactions.

Guard tube containing sodalime

Cotton plug

Anhydrous copper sulphate (white)

Lime water

C

Reactions :

C + 2CuO in the compound

→ CO2 + 2Cu

2H + CuO in the compound

→ H2O + Cu

CO2 + Ca(OH)2

limewater

N + Na

fusion →

in organic compound

NaCN

sodium cyanide fusion

X(Cl, Br, I) + Na→ NaX(X=Cl,Br, I) from organic compound

sodium halide

S + 2Na fusion → Na2S

→ CaCO3 + H2O

from organic compound sodium sulphide If nitrogen and sulphur both are present in any organic compound, sodium thiocyanate (NaSCN) is formed during fusion which in the presence of excess sodium, forms sodium cyanide and sodium sulphide.

milky

5H2O + CuSO4 (anhyd) → CuSO4.5H2O white blue Process : The given organic compound is mixed with dry cupric oxide (CuO) and heated in a hard glass tube. The products of the reaction are passed over (white) anhydrous copper sulphate and then bubbled through limewater. The copper sulphate turns blue (due to the formation of CuSO4.5H2O) by water vapour, showing that the compound contains hydrogen. The limewater is turned milky by CO2, showing that the compound contains carbon. Detection of Nitrogen, Sulphur and Halogens : Nitrogen, sulphur and halogens in any organic compound are detected by Lassaigne's test.

XtraEdge for IIT-JEE

+

Na + C

+

N +

→ NaCNS S fusion

in organic compound sodium thiocyanate Detection of Nitrogen : Take a small quantity of the sodium extract in a test tube. If not alkaline, make it alkaline by adding 2–3 drops of sodium hydroxide (NaOH) solution. To this solution, add 1 mL of freshly prepared solution of ferrous sulphate. Heat the mixture of the two solutions to boiling and then acidify it with dilute sulphuric acid. The appearance of prussion blue or green colouration or precipitate confirms the presence of nitrogen in the given organic compound. 32

MARCH 2012

Chemistry of the test : The following reactions describe the chemistry of the tests of nitrogen. The carbon and nitrogen present in the organic compound on fusion with sodium metal give sodium cyanide (NaCN). NaCN being ionic salt dissolves in water. So, the sodium extract contains sodium cyanide. Sodium cyanide on reaction with ferrous sulphate gives sodium ferrocyanide. On heating, some of the ferrous salt is oxidised to the ferric salt and this reacts with sodium ferrocyanide to form ferric-ferrocyanide.

sulphide (if present) to hydrogen cyanide and hydrogen sulphide gases, respectively. This solution is cooled and silver nitrate solution added. A white precipitate soluble in ammonia shows chlorine, a yellowish precipitate sparingly soluble in ammonia indicates bromine, and a yellow precipitate insoluble in ammonia shows the presence of iodine in the given organic compound. NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq) white precipitate (soluble in ammonia)

6 NaCN + FeSO4 → Na4[Fe(CN)6] + Na2SO4 sodium ferrocyanide 3Na4[Fe(CN)6] + 2Fe2(SO4)3 formed during boiling of the solution

NaBr(aq) + AgNO3(aq) → AgBr(s) + NaNO3(aq) light yellow ppt. (sparingly soluble in ammonia)

→ Fe4[Fe(CN)6]3 + 6Na2SO4 prussian blue When nitrogen and sulphur both are present in any organic compound, sodium thiocyanate is formed during fusion. When extracted with water sodium thiocynate goes into the sodium extract and gives blood red colouration with ferric ions due to the formation of ferric thiocyanate NaCNS Na + C + N + S Sod. thiocyanate

from organic 3+

3NaCNS + Fe

NaI(aq) + AgNO3(aq) → AgI(s) + NaNO3(aq) yellow precipitate (insoluble in ammonia) (b) CS2 layer test for detecting bromine and iodine : Boil a small quantity of sodium extract with dilute HNO3 for 1–2 min and cool the solution. To this solution, add a few drops of carbon disulphide (CS2) and 1–2 mL fresh chlorine water, and shake. Appearance of orange colour in the CS2 layer confirms the presence of bromine, whereas that of a violet/purple colouration confirms the presence of iodine in the compound. 2NaBr(aq) + Cl2 in sodium extract

→ Fe(CNS)3 + 3Na+ ferric thiocyanate (blood red)

Detection of Sulphur : The presence of sulphur in any organic compound is detected by using sodium extract as follows : (a) Lead acetate test : Acidify a small portion of sodium extract with acetic acid and add lead acetate solution to it. A black precipitate of lead sulphide indicates the presence of sulphur.

CS 2 → 2NaCl(aq) + Br2 dissolves in CS2 to give orange colour. 2NaI(aq) + Cl2 in sodium extract CS

2 → 2NaCl(aq) + I2 dissolves in CS2 to give purple/violet colour Detection of Phosphorus : The organic compound is fused with sodium peroxide. The fused mass is then extracted with water. The aqueous solution so obtained is boiled with concentrated nitric acid, and ammonium molybedate solution is added to it. A yellow solution or precipitate indicates the presence of phosphorus in the organic compound. The yellow precipitate is of ammonium or phosphomolybedate (NH4)3[PMo12O40] (NH4)3PO4.12MoO3.

+

(CH3COO)2Pb + Na2S H→ PbS + 2CH3COONa lead acetate

in sodium black ppt extract (b) Sodium nitroprusside test : To a small quantity of sodium extract taken in a test tube, add 2-3 drops of sodium nitroprusside solution. A violet colour indicates the presence of sulphur. This colour fades away slowly on standing. Na2S + Na2[Fe(CN)5NO] → Na4[Fe(CN)5NOS] sodium nitroprusside violet or purple colour Detection of Halogens : The presence of halogens in any organic compound is detected by using sodium extract (Lassaigne's extract) by silver nitrate test. (a) Silver nitrate test : Sodium extract (or Lassaigne's extract) is boiled with dilute nitric acid to decompose sodium cyanide or sodium XtraEdge for IIT-JEE

33

MARCH 2012

KEY CONCEPT

BORON FAMILY & CARBON FAMILY

Inorganic Chemistry Fundamentals

Now, the tendency for back donation is maximum in the case of fluorine due to its small size and more interelectronic repulsions, therefore, it is the least acidic. The tendency of back bonding falls as we move from BF3 to BCl3 and BCl3 to BBr3 due to increase in the size of halogen atoms consequently, the acidic character increase accordingly. π

Boron Trihalides :

The trihalides of boron are electron deficient compounds having a planar structure as shown. They act as Lewis acids because of incomplete octet. X 120º B

F

BF3

Lewis acid

BF3 Lewis acid

+ : NH 3 → Lewis base

+

: F−

Lewis base

→

F π

F3 B ← NH3

Empty 2p-orbital 2p-orbital with lone pair pπ-pπ back bonding

Addition product

BF4− Fluoroborate ion

Acidic nature of H3BO3 or B(OH)3 : Since B(OH)3 only partially reacts with water to form H3O+ and [B(OH)4]–, it behaves as a weak acid. Thus H3BO3 or (B(OH)3) cannot be titrated satisfactorily with NaOH, as a sharp end point is not obtained. If certain organic polyhydroxy compounds such as glycerol, mannitol or sugars are added to the titration mixture, then B(OH)3 behaves as a strong monobasic acid. It can now be titrated with NaOH, and the end point is detected using phenolphthalein as indicator (indicator changes at pH 8.3 – 10.0). 2B(OH)3 + 2NaOH

The acid strength of trihalides decreases as : BF3 < BCl3 < BBr3 < BI3 Explanation :

This order of acid strength is reverse of what may normally be expected on the basis of electronegativity of halogens. Since F is most electronegative, hence BF3 should be most electron deficient and thus should be strongest acid. The anomalous behaviour is explained on the basis of tendency of halogen atom to back-donate its electrons to boron atom. For example, in BF3 one of the 2p-orbital of F atom having lone pair overlaps sidewise with the empty 2p-orbital of boron atom to form pπ-pπ back bonding. This is also known as back donation. Further, due to back-π donation of three surrounding fluorine atoms. BF3 can be represented as a resonance hybrid of following three structures. F F

–

+

B =F

+ F F

B– – F

F + F

B– — F ≡

F F

Na[B(OH)4] + NaBO 2 + 2H 2 O sodium metaborate

The added compound must be a cis-diol, to enhance the acidic properties in this way. (This means that it has OH groups on adjacent carbon atoms in the cis configuration.) The cis-diol forms very stable complexes with the [B(OH)4]– formed by the forward reaction above, thus effectively removing it from solution. The reaction is reversible. Thus removal of one of the products at the right hand side of the equation upsets the balance, and the reaction proceeds completely to the right. Thus all the B(OH)3 reacts with NaOH : in effect it acts as a strong acid in the presence of the cis-diol.

B– — F

Probable hybrid structure As a result of this back donation, the electron deficiency of boron gets compensated and its Lewis acid character decreases. Resonating forms of BF3

XtraEdge for IIT-JEE

F

B

X Planar structure of X boron trihalides

34

MARCH 2012

– C – OH HO + – C – OH HO

OH

–

B OH

– OH HO – C –

–C–O

–2H2O

B –C–O

beyond Me4B2H2 without breaking the molecule into BMe3. The terminal B – H distance are the same as the bond lengths measured in non-electron-deficient compounds. These are assumed to be normal covalent bonds, with two electrons shared between two atoms. We can describe these bonds as twocentre two-electron bonds (2c-2e).

+ OH HO – C –

–C–O –2H2O

O–C–

–

Thus the electron deficiency must be associated with the bridge groups. The nature of the bonds in the hydrogen bridges is now will established. Obviously they are abnormal bonds as the two bridges involve only one electron from each boron atom and one from each hydrogen atom, making a total of four electrons. An sp3 hybrid orbital from each boron atom overlaps with the 1s orbital of the hydrogen. This gives a delocalized molecular orbital covering all three nuclei, containing one pair of electrons and making up one of the bridges. This is a three centre two-electron bond (3c-2e). A second three-centre bond is also formed.

B –C–O

O–C–

Borax : The most common metaborate is borax Na2[B4O5(OH)4] . 8H2O. It is a useful primary standard for titrating against acids.

(Na2[B4O5(OH)4] . 8H2O) + 2HCl → 2NaCl + 4H3BO3 + 5H2O One of the products H3BO3 is itself a weak acid. Thus the indicator used to detect the end point of this reaction must be one that is unaffected by H3BO3. Methyl orange is normally used, which changes in the pH range 3.1 – 4.4. One mole of borax reacts with two moles of acid. This is because when borax is dissolved in water both B(OH)3 and [B(OH)4]– are formed, but only the [B(OH)4]– reacts with HCl. [B4O5(OH)4]2– + 5H2O 2B(OH)3 + 2[B(OH)4]–

B

H

H

H

H H

H

B H

2[B(OH)4]– + 2H3O+ → 2B(OH)3 + 4H2O The last reaction will titrate at pH 9.2, so the indicator must have pKa < 8. Borax is also used as a buffer since its aqueous solution contains equal amounts of weak acid and its salt. Structures of the Boranes : The bonding and structures of the boranes are of great interest. They are different from all other hydrides. There are not enough valency electrons to form conventional two-electron bonds between all of the adjacent pair of atoms, and so these compounds are termed as electron dificient. In diborane there are 12 valency electrons, three from each B atom and six from the H atoms. Electron diffraction results indicate the structure shown in fig. H H 1.33Å H 1.19Å B B 1.19Å H 1.33Å H H

B

H

B H

H

Overlap of approximately sp2 hybrid orbitals from B with an s orbital from H to give a banana-shaped three-centre two-electron bond.

The higher boranes have an open cage structure. Both normal and multi-centre bonds are required to explain these structures. Terminal B–H bonds. These are normal covalent bonds, that is two centre two-electron (2c-2e) bonds.

The two bridging H atoms are in a plane perpendicular to the rest of the molecules and prevent rotation between the two B atoms. Specific heat measurements confirm that rotation is hindered. Four of the H atoms are in a different environment from the other two. This is confirmed by Raman spectra and by the fact that diborane cannot be methylated XtraEdge for IIT-JEE

H

H

B – B bonds. These are also normal 2c-2e bonds. Three-centre bridge bonds including B ... H ... B as in diborane. These are 3c-2e bonds. Three-centre bridge bonds including B....B.....B, similar to the hydrogen bridge. These are called 35

MARCH 2012

'open boron bridge bonds' and are of the type 3c-2e.

R

R

R

(iii) HO – Si – O – Si – O – Si – OH

Closed 3c-2e bonds between three B atoms.

OH HO

B

OH HO

OH HO

–3H2O

HO – Si – O – Si – O – Si – OH R

R

B B Silicones : These are organosilicon polymers containing Si – O – Si linkages. These are formed by the hydrolysis of alkyl or aryl substituded chlorosilanes and their subsequent polymerisation. The alkyl or aryl substitued chlorosilanes are prepared by the reaction of Grignard reagent and silicon tetrachloride.

RMgCl

R R R Cross linked silicone Cyclic (ring) silicones are formed when water is eliminated from the terminal –OH group of linear silicones. R R

Si R

O

Si

Si

R

• A porcelain funnel used for filtration by suction is ® Bucher Funnel known as • What is diazomethane ?

HO – Si – O – Si – OH R R Polymerisation continues on both the ends and thus chain increases in length. RSiCl3 on hydrolysis gives a cross linked silicone. The formation can be explained in three steps : Cl OH 3H2O (i) R – Si – Cl R – Si – OH –3HCl OH Cl R R R

+

• A drying chamber, containing chemicals such as concentrated sulphuric acid or silica gel is known as ® Desiccator • Reforming of a gasoline fraction to increase branching in presence of AlCl3 is known as ® Isomerization • A condenser consisting of glass tube surrounded by another glass tube through which cooling water flows is known as ® Liebig condenser

OH

HO – Si – O – Si – O – Si – OH

• What is • ® Hot wire ammeter

OH

• What quantity has the

OH

R

–

® [CH 2 = N = N or CH 2 N 2 ]

(ii) HO – Si – OH + H O – Si – OH + H O – Si – OH R

R

SCIENCE TIPS

HO – Si – OH + H O – Si – OH

XtraEdge for IIT-JEE

O

R O R3SiCl on hydrolysis forms only a dimer R3Si – O – Si R3 R3Si OH + OH Si R3 R

R OH Si R OH Dialky silandiol Polymerisation of dialkyl silandiol yields linear thermoplastic polymer. R R R Cl H OH –2HCl Si + R Cl H OH

OH

O

– O – Si – O – Si – O – Si – O –

2RMgCl + SiCl4 → R2SiCl2 + 2MgCl2

OH

O

O

3RMgCl + SiCl4 → R3SiCl + 3MgCl2 R stands for – CH3, –C2H5 or –C6H5 groups Hydrolysis of substituted chlorosilanes yield corresponding silanols which udergo polymerisation.

R

R

– O – Si – O – Si – O – Si – O –

+ SiCl4 → R – SiCl3 + MgCl2

R

R

R

Grignard reagent

R

R

R OH

36

MARCH 2012

UNDERSTANDING

Inorganic Chemistry

(c) 2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 + 2H2O

(a) When Mn(OH)2 is made by adding an alkali to a solution containing Mn2+ ions, the precipitate quickly darkens, and eventually goes black. What might be the chemical giving the black colour, and how is it made ? (b) Dimercury (I) iodide, Hg2I2 is a greenish colour and is precipitated if iodide ions are added to a solution of dimercury (I) sulphate, Hg2SO4. Likewise the red mercury (II) iodide, HgI2, is precipitated from a solution of mercury (II) sulphate, HgSO4. However, both precipitates dissolve in excess iodide solution. What might be the reason for this ? Sol. (a) The black colour is due to the manganese (IV) oxide, MnO2. It is made by the Mn(OH)2 being oxidised by oxygen in the air : Mn(OH)2 → MnO + H2O MnO + ½O2 → MnO2 air black (b) It is due to formation of HgI42– (a soluble complex) in both the cases with HgI2 : HgI2 + 2I– → HgI42– But in Hg2I2, first there is oxidation of Hg(I) to Hg(II) and then complex formation takes place; it is by following disproportionation reaction : 1.

Hg 22 + +1

–

+ 4I →

HgI 24+ +2

2 mol NH4Cl ≡ 1 mol Ca(OH)2 1 mol NH4Cl ≡ 0.5 mol Ca(OH)2 (d) 2NaHCO3 + Ca(OH)2 → Na2CO3 + CaCO3 + 2H2O 2 mol NaHCO3 ≡ 1 mol Ca(OH)2 1 mol NaHCO3 ≡ 0.5 mol Ca(OH)2 A colourless salt (A), soluble in water, gives a mixture of three gases (B), (C) and (D) along with water vapours. Gas (B) is blue towards litmus paper, gas (C) red and gas (D) is neutral. Gas (B) is also obtained when (A) is heated with NaOH and gives brown ppt. with K2HgI4. Solution thus obtained gives white ppt. (E) with CaCl2 solution in presence of CH3COOH. Precipatete (E) decolorises MnO −4 /H+. Gas (C) turns lime water milky while gas (D) burns with blue flame and is fatal when inhaled. Identify (A) to (D) and explain chemical reactions. Sol. Gas (B) gives brown ppt. with K2HgI4 3.

⇒ gas (B) is NH3 ⇒ gas (A) has NH4+ (C) turns lime water milky ⇒ gas (C) can be SO2 or CO2 Gas (D) is also obtained along with (C). Gas (D) burns with blue flame and is fatal when inhaled

+ Hg 0

⇒ gas (D) is CO

Calculate mol of Ca(OH)2 required to carry out following conversion taking one mol in each case : COO COOH into Ca (a) COO COOH (b) H3PO4 into CaHPO4 (c) NH4Cl into NH3 (d) NaHCO3 into CaCO3 COOH Sol. (a) is a dibasic acid COOH COOH COO + Ca(OH)2 Ca COOH COO 1 mo l 1 mo l Ca(OH)2 required = 1 mol (b) H3PO4 + Ca(OH)2 → CaHPO4 + 2H2O 1 mol of H3PO4 ≡ 2H+ neutralised by 1 mol of Ca(OH)2 Ca(OH)2 required = 1 mol 2.

XtraEdge for IIT-JEE

⇒ gas (C) is CO2

2–

⇒ (A) has C2O4 It is confirmed by the fact that CaCl2 gives white ppt. CaC2O4(E) which decolourises MnO4–/H+ ⇒ (A) is (NH4)2C2O4 Explanation : ∆ (NH4)2 C2O4 → 2NH3 + CO2 + CO + H2O

(A) (B) (C) (B) is blue towards litmus (basic) (C) is red toward litmus (acidic) (D) is neutral

(D)

∆ Na2C2O4+ 2 NH 3 +2H2O (NH4)2C2O4+2NaOH → (B)

Na2C2O4+ CaCl2 →

37

CaC 2 O 4 ↓ + 2NaCl White ppt. (E )

MARCH 2012

Sol. (i) Acetic acid on heating with C2H5OH gives original compound (A).

Hg

NH3 + K2HgI4 →

O Hg

NH2I

4 CH3COOH + C2H5OH H2SO → CH 3COOC 2 H 5

brown ppt

∆

(Iodide of Million’s base)

+ H2O (ii) CH3COOC2H5 (A) on heating with C2H5ONa undergoes Claisen condensation to give (B), which is aceto acetic ester.

2MnO4– +16H++5C2O42– → 10CO2+2Mn2++ H2O violet

colourless

A solution of a salt (A) when treated with calculated quantity of sodium hydroxide gave a green coloured ppt (B), which dissolve in excess of NaOH. (B) acts as a weak base and loses water on heating to give a green powder (C). The green powder is used as refractory material. When (C) is fused with an alkali in presence of air or oxidising agent, a yellow coloured solution (D) is obtained. Identify the compounds from (A) to (D) Sol. The compound (A) is chromic salt. The chemical reactions are as under (i) With calculated quantity of sodium hydroxide 4.

CH3CO OC2H5 + H CH2COOC2H5 (A)

Reflux

+ C2H5OH + CH3COCH2COOC2H5

(iii) (B) on heating in acidic solution gives (C) and ethyl alcohol. +

CH 3COCH 2 COOC 2 H 5 + HOH H→ ( B)

CH 3COCH 2 COOH + C2H5OH (C)

(iv) (C) on decarboxylation gives acetone (D).

green ppt (B)

∆ CH 3COCH 2 COOH → CH 3COCH 3

(ii) In excess of sodium hydroxide, soluble NaCrO2 is formed

(C)

− CO 2

( D)

(v) (D) reacts with NaNH2 to form sodium salt (E), which on heating with CH3I gives butanone (F).

Cr(OH)3 + NaOH → NaCrO2 + 2H2O (sod. chromite)

∆ CH 3COCH 3 + NaNH2 → CH 3COCH 2 Na

(iii) Since Cr(OH)3 contains -OH group, so it will act as a base. On heating it will lose water to give Cr2O3 powder (C)

− NH 3

( D)

(E)

CH I

3 → CH 3COCH 2 CH 3 – NaI

2Cr(OH)3 → Cr2O3 + 3H2O (C) (iv) On fusing Cr2O3 with an alkali in presence of oxygen or oxidising agent, a yellow soluble chromate will be formed -

( F)

∆ (vi) CH 3COCH 2 CH 3 + 3I2 + 4NaOH → ( F)

CHI3 + CH3CH2COONa + 3NaI + 3H2O ∆ (vii) CH 3COCH 3 + 3I2 + 4NaOH →

2Cr2O3 + 8NaOH + 3O2 → 4Na2CrO4 + 4H2O

( D)

yellow soln. (D)

CHI3 + CH3COONa + 3NaI + 3H2O → CH3COOH + NaCl CH3COONa HCl

Two moles of an anhydrous ester (A) are condensed in presence of sodium ethoxide to give a β-keto ester (B) and ethanol. On heating in an acidic solution compound (B) gives ethanol and a β-keto acid (C). (C) on decarboxylation gives (D) of molecular formula C3H6O. Compound (D) reacts with sodamide to give a sodium salt (E), which on heating with CH3I gives (F), C4H8O, which reacts with phenyl hydrazine but not with Fehling reagent. (F) on heating with I2 and NaOH gives yellow precipitate of CHI3 and sodium propionate. Compound (D) also gives iodoform, but sodium salt of acetic acid. The sodium salt of acetic acid on acidification gives acetic acid which on heating with C2H5OH in presence of conc. H2SO4 gives the original ester (A). What are (A) to (F) ?

XtraEdge for IIT-JEE

C2H5ONa

(B)

CrCl3 + 3NaOH → Cr(OH)3 + 3NaCl

5.

(A)

Thus, (A) CH3COOC2H5 (B) CH3COCH2COOC2H5 (C) CH3COCH2COOH (D) CH3COCH3 (E) CH3COCH2Na (F) CH3COCH2CH3

38

MARCH 2012

XtraEdge for IIT-JEE

39

MARCH 2012

Set

`tà{xÅtà|vtÄ V{tÄÄxÇzxá 11 This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants. By : Shailendra Maheshwari So lu t ion s wi l l b e p ub lished in nex t issue Joint Director Academics, Career Point, Kota 1.

For complex numbers z1 = x1 + iy1 and z2 = x2 + iy2

7.

we write z1 ∩ z2, if x1 ≤ x2 and y1 ≤ y2. The for all complex numbers z with 1 ∩ z, we have

1− z ∩ 0, 1+ z

Justify the result. 2.

AP and BQ are fixed parallel tangents to a circle, and a tangent at any point C cuts them at P and Q respectively. Show that CP.CQ is independent of the position of C on the circle.

3.

Let f(x) = ax2 + bx + c & g(x) = cx2 + bx + a, such

Passage : Let Z denotes the set of integers. Let p be a prime number and let z1 ≡ {0, 1}. Let f : z → z and g : z → z1 are two functions defined as follows : f(n) = pn; if n ∈ z and g(n) = 1; if n is a perfect square = 0, otherwise.

that |f(0)| ≤ 1, |f(1)| ≤ 1 and |f(–1)| ≤ 1. Prove that

8.

g(f(x)) is (A) many one into (B) many one onto (C) one one onto (D) one one into

9.

f(g(x)) = p has (A) no real root (B) at least one real root (C) infinity many roots (D) exactly one real root

|f(x)| ≤ 5/4 and |g(x)| ≤ 2 4.

A straight line is drawn throguh the origin and parallel to the tangent to the curve x + a2 − y2 a

a + a2 − y2 at an arbitrary = ln y

point M. Show that the locus of the points P of intersection of this straight line and the straight line parallel to the x-axis and passing through the point M is a circle. n

5.

Show that

∑ (−2) r =0

6.

Let In =

1

∫x 0

n

r

n

Cr

r +2

Cr

2 + f ( x) + f ( y ) x+ y If f = 3 3 for all real x and y. If f ´(2) = 2, then f(2) is -

10. g(f(x)) is – (A) non periodic function (B) odd function (C) even function (D) None of these

1 , If n is even = n +1 1 , If n is odd n + 2

tan −1 x dx , then expression In in terms

of In–2.

XtraEdge for IIT-JEE

40

MARCH 2012

MATHEMATICAL CHALLENGES SOLUTION FOR FEBRUARY ISSUE (SET # 10)

1.

g(x) = sin x 1 sin2 x/2 lim g(x) = lim x →π 2−

x →π 2+

; 0 ≤ x < π/2 π/2 ≤ x ≤ π ; ; π<x g(x) = g(π/2) = 1

A

lim g(x) = lim g(x) = g(π) = 1

x→π−

2.

3.

4.

x→π+

g´(π–) = g´(π +) = 0 g´(π/2–) = g´(π/2+) = 0 and Hence g(x) is continuous and differentiable in (0,∞) sin(sin x) sin x < x sin x sin θ ; 0 < θ < π/2 Let f(θ) = θ θ cos θ − sin θ f ´(θ) = θ2 cos θ . (θ − tan θ) = < 0 as tan θ > θ θ2 so f(θ) ↓ so f(x) < f(sinx) as sin x < x 6.5 (i) 6C4 = = 15 2 (ii) coeff. of x4 in (1 – x)–6 9.8.7.6 = 4 + 6 – 1C6 – 1 = 9C5 = = 126 4.3.2 (iii) select 3 different flavours : 6C3 ways choose (at least one from each) 4 cones : 4–1 C3 – 1 = 3C2 = 3 ways 6.5.4 so required ways = 6C3 × 3 = × 3 = 60 3.2 (iv) Select 2 different flavours : 6C2 ways choose (at least one from each) 4 cones ; 4–1 C2 – 1 = 3C1 = 3 so required ways (either 2 or 3 different flavours) 6.5 = 60 + 6C2 3 = 60 + × 3 = 105 2 Let A at origin & P.V. of B & C are b & c. b c + So line AD ⇒ r = t |b| |c|

E

C

D

t =1+s ...(1) |b| t = –s ...(2) |c| | b || c | t t so =1– ⇒ t= |b| |c| |b|+|c|

use it in line AD . b c b | c | +c | b | | b || c | = . pt D : + | b | + | c | | b | | c | |b |+|c | which divides BC in ratio of |c| : |b| similary use eq. of external angle bisector line AE b c ⇒ r = p − |b | | c | solve it with BC to find pt. E. 5.

Consider eix(1 + eix)n = eix [1 + nC1eix + nC2ei2x + .... + nCneinx] n+2 x i 2

e

. 2cosn

x = eix + 2

n

C1ei2x +

n

C2ei3x +...

....+ nCnei(n+ 1)x Compare real parts & get (a) Compare imaginary. parts & get (b) 6.

& line BC ⇒ r = b + ∆ ( b – c ) solve them together to find pt. D b c = b + s ( b – c ) + t |b| |c| XtraEdge for IIT-JEE

B

Let Ei = the event that originator will not receive a letter in the ith stage. Originator sands letters to two persons so in 1st stage he will not get letter. Prob. that letter sent by 1st received is not received n−2 C (n − 2)(n − 3) n − 3 by originator is n −1 2 = = n −1 (n − 1)(n − 2) C1 similarly prob. that letter sent by 2nd receipiant is not n−3 received by originator is n −1 so P(E2) = prob. that originator not received letter in 2

n−3 2nd stage is = . n −1

41

MARCH 2012

similarly P(E3) = prob. that originator not receive letter sent by the four person getting letters from two recipients is 4

n −3 n −3 n −3 n −3 n − 3 n −3 . . . = = n −1 n −1 n −1 n −1 n −1 n −1 8

n −3 n −3 P(E4) = = n −1 n −1

Similarly,

π/ 2 1 1 1 – + – sin 2θ sec θdθ 0 7 5 3 29 181 + 2(cos θ) 0π / 2 – 0 dθ = – = 105 105

∫

∫

22

23

9.

2k –1

n − 3 Similarly, P(Ek) = n −1 So the required prob. is P(E) = prob. the originator not receive letter in 1st k stages = P(E1) . P(E2) . ........ P(Ek)

n−3 = n −1 n−3 = n −1

7.

y = f(x) = y´ =

x

∫

=–

0

x

∫

0

2+ 22 + 23 +....2k −1

2.

2k −1 −1 2−1

n −3 = n −1

2

e zx− z dz =

∫

x

0

2

e zx .e − z dz + 1 = –

1 z 2 zx x (e .e ) 0 − 2

∫

x

0

( 2k − 2 )

2

3x − 4 y + l 4x + 3 y + m = a 5 5 (3x – 4y + l)2 – 5a(4x + 3y + m) = 0 9x2 – 24xy + 16y2 + (6l – 20a)x + (–8l – 15a)y + (l2 – 5am) = 0 comp. it with given equation. 6l – 20a = –18 ⇒ 24l – 80a = –72 ...(1) –8l –15a = –101 ⇒ –24l – 45a = –303 ...(2) From (1) & (2) ⇒ 125a = –375 ⇒ a=3

2

e zx .e − z dz

1 2

∫

x

0

2

e zx (−2 ze − z ) dz + 1

2 1 xe − z .e zx dz + 1 = xy + 1 2

dy 1 – xy = 1 dx 2 2 − x / 2 dx = e−x / 4 I.F. = e ∫

Sol is y . e − x y = ex 8.

2

/4

∫

x

0

9x2 – 24xy + 16y2 – 18x – 101y + 19 = 0 (3x – 4y)2 = 18x + 101y – 19. Let the vertex of the parabola be A(α, β). Shift origin to A and y-axis along the tangent at vertex (3x – 4y + l). So the axis of parabola be 4x + 3y + m = 0 (along x axis) If L.R. of parabola be a then it’s equation is

2

e−z

/4

2

/4

=

∫

e− x

2

/4

∫

dx =

x

0

e−z

2

/4

10. circle : (x – 1)2 + (y – 1)2 = 1 ⇒ x2 + y2 – 2x – 2y + 1 = 0

dz

dz.

∫ sin n θ sec θ dθ = ∫ sin (n –1 + 1) θ sec θ dθ = ∫ sin (n – 1)θ + cos (n – 1)θ sin θ sec θ ) dθ = ∫ sin (n – 1)θ + [ sin (n – 1)θ cos θ – sin (n – 2)θ sec θ ] dθ = ∫ (2 sin (n – 1)θ – sin (n – 2)θ sec θ ) dθ 2 cos(n − 1)θ =– – ∫ sin (n – 2)θ secθ dθ n −1 1 π 2 sin 8θ − sin 2θ = dθ 2 0 cos θ π/2 π2 1 2 = sin 6θ sec θdθ − − cos 7θ 0 2 7 0

(0,1)B D A(1,0)

Let the line be y = mx 1 Altitude of ∆ = 1 + m2 For DE length : solve line with circle. x2 + m2x2 – 2x – 2mx + 1 = 0 (1 + m2)x2 – 2(1 + m)x + 1 = 0

∫

∫

−

1 2 2 2 − − (cos 3θ) 0π / 2 − 2 7 5 3

∫

π/2

0

π2

0

sin 2θ sec θdθ

|x1 – x2| = =

sin 2θ sec θdθ

−

XtraEdge for IIT-JEE

∫ ∫

π2

0

E

sin 2θ sec θdθ

|DE| =

42

( x1 + x 2 ) 2 − 4 x1 x 2

4(1 + m) 2 2 2

(1 + m )

−4

1 1+ m

2

x12 + 1 |x1 – x2| = 2

=

2 1 + m2

2m

2m 1 + m2

MARCH 2012

Students' Forum Expert’s Solution for Question asked by IIT-JEE Aspirants

MATHS

Given a point P on the circumference of the circle |z| = 1, and vertices A1, A2, ......, An of an inscribed regular polygon of n sides. Prove using complex numbers that (PA1)2 + (PA2)2 + ......... + (PAn)2 is a constant. Sol. Without loss of generality we can take P as 1 + 0i. i.e., P ≡ C is 0

Find the point inside a triangle from which the sum of the squares of distance to the three side is minimum. Find also the minimum value of the sum of squares of distance. Sol. If a, b, c are the lengths of the sides of the ∆ and x, y, z are length of perpendicular from the points on the sides BC, CA and AB respectively, we have to minimise : ∆ = x2 + y2 + z2 1 1 1 we have, ax + by + cz = ∆ 2 2 2 ⇒ ax + by + cz = 2∆ A

1.

2.

A2

A3

θ2

A1 θ1

P

θn

z

y

An x B C where ∆ is the area of ∆ABC. We have the identity : ⇒ (x2 + y2 + z2) (a2 + b2 + c2) – (ax + by + cz)2 = (ax – by)2 + (by – cz)2 + (cz – ax)2 ⇒ (x2 + y2 + z2)(a2 + b2 + c2) ≥ (ax + by + cz)2 ⇒ (x2 + y2 + z2) (a2 + b2 + c2 ≥ 4∆2 4∆2 ⇒ x2 + y2 + z2 ≥ 2 a + b2 + c2 Equality holds only when ax + by + cz 2∆ x y z = = = 2 = 2 2 2 a b c a +b +c a + b2 + c2 ∴ The minimum value of ∆ is ; 4( s − a)( s − b)( s − c) s 4∆2 = 2 2 2 a2 + b2 + c2 a +b +c

Let Ar ≡ C is θr, r = 1, 2, ......, n. PAr = |Cis θr – Cis 0| = |(cosθr – 1) + i(sinθr)| PAr2 = (cos θr – 1)2 + (sinθr)2 = 2 – 2cos θr n

⇒

∑

( PAr ) 2 = 2n – 2

r =1

∑ cos θ

r

r =1

n

Now,

n

∑ cos θ

r

r =1

n = Re Cisθ r r =1

∑

= Re [e iθ1 + e iθ 2 + ....... + e iθ n ] n 2π e iθ1 1 − e i n = Re 2π i 1− e n

Q θ2 – θ1 = θ3 – θ2 = ..... = θn – θn–1 =

Let a1, a2, ......, an be real constant, x be a real variable 1 1 and f (x) = cos(a1 + x) + cos(a2 + x) + cos(a3 + x) 2 4 1 +...... + n −1 cos(an + x). Given that f (x1) = f (x2) = 0, 2 prove that (x2 – x1) = mπ for integer m. Sol. f (x) may be written as, 3.

2π n

iθ1 e (1 − 1) =0 = Re 2π i 1 − e n n

Hence,

∑ ( PA ) r

2

n

f (x) =

= 2n = constant.

k =1

r =1

XtraEdge for IIT-JEE

∑2

43

1 k −1

cos(ak + x)

MARCH 2012

n

∑2

=

k =1

1 k −1

n cos ak = cos x . k −1 k =1 2

∑

n sin a k – sin x k −1 k =1 2

∑

n

= A cos x – B sin x, where A =

∑ k =1

n

B=

2 (x – y) = 0 r ⇒ Given family of line passes through fixed point (7/2, 7/2).. equation of line in normal form x cos α + y sin α = p 7 7 passes through , 2 2 2d ⇒ cos α + sin α = 7

(7 – 2x) +

{cosak. cos x – sin ak . sin x}

cos a k 2 k −1

and

sin ak

∑2 k =1

but – 2 ≤ cos α + sin α ≤ 7 2d ≤ 2 ⇒ |d| ≤ 7 2

k −1

since f (x1) = f (x2) = 0 ⇒ A cos x1 – B sin x1 = 0 and A cos x2 – B sin x2 = 0 A ⇒ tan x1 = B A ⇒ tan x2 = B ⇒ tan x1 = tan x2 ⇒ (x2 – x1) = mπ 4.

There are n straight lines in a plane such that n1 of them are parallel in one direction, n2 are parallel in different direction and so on, nk are parallel in another direction such that n1 + n2 + ... + nk = n. Also no three of the given lines meet at a point. Prove that the total number of points of intersection is k 1 2 nr2 n – 2 r =1 Sol. If no two of n given lines are parallel and no three of them meet at a point, then the total number of points of intersection is nC2. But it is given that there are k sets of n1, n2, n3, ... , nk parallel lines such that no line in one set is parallel to a line in any other set. Also, lines of one set do not intersect with each other. Therefore, lines of one set do not provide any point of intersection. Hence, Total number of points of intersection 6.

∑

If (a, b, c) is a point on the plane 3x + 2y + z = 7, then find the least value of a2 + b2 + c2, using vector methods. →

Sol. Let A = a iˆ + b ˆj + c kˆ →

⇒ B = 3 iˆ + 2 ˆj + kˆ → →

→

→

⇒ (A . B) 2 ≤ | A |2 | B |2 a2 + b2 + c2

3 a + 2b + c ≤

= nC2 –

(

n1

C 2 + n2 C 2 + ... + nk C 2

)

n (n – 1) n(n – 1) n1 (n1 – 1) n2 (n2 – 1) – + + ... + k k 2 2 2 2 n(n – 1) 1 – {(n12 + n22 + ... + nk2) – (n1 + n2 + ... + nk)} = 2 2 n(n – 1) 1 – {(n12 + n22 + ... + nk2) – n} = 2 2 n2 1 = – (n12 + n22 + ... + nk2) 2 2 k 1 2 = nr2 n – 2 r =1

14

=

(7)2 ≤ (a2 + b2 + c2) (14) {Q 3a + 2b + c = 7, point lies on the plane}

a2 + b2 + c2 ≥

2

49 7 = 14 2

If parameters p, r, q are in H.P. and d be the length of perpendicular from origin to any member of family of lines xr(p + q – 2pq) – 2pq(y – 5r) – 3pqr = 0 then 7 . show that |d| ≤ 2 Sol. Given family of line is xr(p + q – 2pq) – 2pq(y – 5r) – 3pqr = 0 dividing by pqr, we get 1 1 y x + − 2 – 2 − 5 – 3 = 0 q p r 5.

∑

2 y x − 2 – 2 − 5 – 3 = 0 r r XtraEdge for IIT-JEE

44

MARCH 2012

MATHS

DEFINITE INTEGRALS & AREA UNDER CURVES Mathematics Fundamentals

Properties 1 :

Every continuous function defined on [a, b] is integrable over [a, b]. Every monotonic function defined on [a, b] is integrable over [a, b] If f(x) is a continuous function defined on [a, b], then there exists c ∈ (a, b)such that

∫ f ( x) dx = F(x), then

If

∫

b a

f ( x) dx = F(b) – F(a), b ≥ a

Where F(x) is one of the antiderivatives of the function f(x), i.e. F´(x) = f(x) (a ≤ x ≤ b). Remark : When evaluating integrals with the help of the above formula, the students should keep in mind the condition for its legitimate use. This formula is used to compute the definite integral of a function continuous on the interval [a, b] only when the equality F´(x) = f(x) is fulfilled in the whole interval [a, b], where F(x) is antiderivative of the function f(x). In particular, the antiderivative must be a function continuous on the whole interval [a, b]. A discontinuous function used as an antiderivative will lead to wrong result. If F(x) =

∫

x

a

∫

∫

b a

∫

a

∫

b

f ( x) dx =

∫

a

∫

b

∫

a

f ( x) dx =

f ( x) dx =

0

or

a

f ( x ) dx =

−a

∫

2a

0

a

defined by g(x) =

c

∫

a

∫

b

∫

a

m(b – a) ≤

b

a

f ( x) dx ≥ 0

d dx

∫ ∫

c

b

a

2 f ( x) dx = 0

∫

XtraEdge for IIT-JEE

b

a

x

a

f (t ) dt for x ∈ [a, b] is derivable

b a

f ( x) dx ≤ M(b – a)

∫

ψ( x)

φ( x )

f (t ) dt = f(ψ(x)) ψ´(x) – f(φ(x)) φ´(x) b

∫ | f ( x) | dx

f ( x) dx ≤

a

2

If f (x) and g (x) are integrable on [a, b], then

f ( x) dx, a < c < b

∫

b

a

f ( x) g ( x) dx ≤

∫

b

a

1/ 2

f 2 ( x) dx

∫

1/ 2

g 2 ( x) dx a b

Change of variables : If the function f(x) is continuous on [a, b] and the function x = φ(t) is continuously differentiable on the interval [t1, t2] and a = φ(t1), b = φ(t2), then

f (a + b − x) dx

∫

∫

2

f (a − x) dx

2 f ( x) dx = 0

b

a

b

∫

If the function φ(x) and ψ(x) are defined on [a, b] and differentiable at a point x ∈ (a, b) and f(t) is continuous for φ(a) ≤ t ≤ ψ(b), then

f ( x) dx

f ( x) dx +

a

0

a

∫

on [a, b] and g´(x) = f(x) for all x ∈ [a, b]. If m and M are the smallest and greatest values of a function f(x) on an interval [a, b], then

f (t ) dt

∫

f ( x) dx = –

b

a

b

∫

∫

f ( x) dx = f(c) . (b – a)

b 1 f ( x) dx is called the (b − a) a mean value of the function f(x) on the interval [a, b]. If f is continous on [a, b], then the integral function g

Properties of Definite Integrals :

If f(x) ≥ 0 on the interval [a, b], then

a

The number f(c) =

f (t ) dt, t ≥ a, then F´(x) = f(x)

b

b

f ( x ) dx if f(–x) = f(x)

∫

if f(–x) = – f ( x )

b a

f ( x) dx =

∫

t2

t1

f (φ(t )) φ´(t) dt

Let a function f(x, α) be continuous for a ≤ x ≤ b and c ≤ α ≤ d. Then for any α ∈ [c, d], if

f ( x ) dx if f(2a – x) = f(x) if f(2a – x) = – f ( x )

I(α) = 45

∫

b

a

f ( x, α) dx, then I´(α) =

∫

b a

f ´(x, α) dx, MARCH 2012

Where I´(α) is the derivative of I(α) w.r.t. α and f ´(x, α) is the derivative of f(x, α) w.r.t. α, kepping x constant.

2 n −1 n − 3 n − 5 n . n − 2 . n − 4 ..... 3 In = 1 π n −1 n − 3 n − 5 . . ..... . 2 2 n n−2 n−4

Integrals with Infinite Limits :

If a function f(x) is continuous for a ≤ x < ∞, then by definition

∫

∞

f ( x) dx = lim

b

∫

f ( x) dx

b→∞ a

a

If In =

∫ ∫

∞

−∞

−∞

∫

b

f ( x) dx =

∫

a

−∞

d dx

f ( x) dx and

a →−∞ a

f ( x) dx +

∫

∞

a

a

0

and

∫

∫

0

π/2

0

1 a 2

f ( x) dx

∫

a

0

∫

∞

0

log sin x dx =

∫

e −ax cos bx dx = e −ax sin bx dx = e −ax xndx =

If In =

1

0

r =0

f ( x) dx

∑

π/2

0

1 1 cosα + (n − 1)β sin nβ 2 2 cos(α + rβ) = 1 r =0 sin β 2

log cos x dx

n −1

∑

π π 1 log 2 = log 2 2 2

1

π

∫

π/2

0

–

12 1

+

2

1

+

22 1

+

2

1 32 1 2

– .... =

π2 12

+ .... =

π2 6

1 2 3 Area under Curves : Area bounded by the curve y = f(x), the x-axis and the ordinates x = a, x = b

Reduction Formulae of some Define Integrals :

0

r 1

∑ f n . n = ∫

n −1

∫

∫

d d v(x) – f{u(x) u(x) dx dx

1 1 sin α + (n − 1)β sin nβ 2 2 sin(α + rβ) = 1 r =0 sin β 2

x f ( x) if f(a – x) = f(x)

m +1 n +1 Γ Γ π/ 2 2 2 sin m x cos n x dx = 0 m+n+2 2Γ 2

∞

u ( x)

n →∞

If m and n are non-negative integers, then

0

f (t ) dt = f{v(x)}

n −1

1 Γ(n + 1) = n Γ (n), Γ(1) = 1, Γ = 2

∞

v( x)

∫

lim

a f ( x) dx = 2 f ( x) + f (a − x)

=–

∫

( when n is even)

Some Important Results :

x f ( x) dx = a

( when n is odd )

Summation of Series by Integration :

properties :

∫

cos n x dx , then

Leibnitz's Rule : If f(x) is continuous and u(x), v(x) are differentiable functions in the interval [a, b], then

Geometrically, the improper integral (i) for f(x) > 0, is the area of the figure bounded by the graph of the function y = f(x), the straight line x = a and the x-axis. Similarly, f ( x) dx = lim

0

( when n is even)

2 n −1 n − 3 n − 5 n . n − 2 . n − 4 ..... 3 Im = 1 π n −1 n − 3 n − 5 . . ..... . 2 2 n n−2 n−4

....(i)

If there exists a finite limit on the right hand side of (i), then the improper integrals is said to be convergent; otherwise it is divergent.

b

π/ 2

∫

( when n is odd )

=

a

∫

b

a

y dx =

∫

b

a

f ( x) dx

Y

2

a + b2

y = f (x)

b a2 + b2

y

x=b

n! a n +1

O

sin n x dx , then

XtraEdge for IIT-JEE

δx

X

Area bounded by the curve x = f(y), the y-axis and the abscissae y = a, y = b 46

MARCH 2012

∫

=

b

a

x dy =

∫

b

Black Holes-The Most Efficient Engines in the Universe

f ( y ) dy

a

Y

y=b x

δy

The scientists have just found the most energyefficient engines in the universe. Black holes, whirling super dense centres of galaxies that suck in nearly everything. Jets of energy spurting out of older ultra-efficient black holes also seem to be playing a crucial role as zoning police in large galaxies preventing to many stars from sprouting. This explains why there are fewer burgeoning galaxies chock full of stars than previously expected.

x = f (y)

y=a O

X The area of the region bounded by y1 = f1(x), y2 = f2(x) and the ordinates x = a and x = b is given by

=

b

∫

a

b

∫

f 2 ( x) dx –

a

f1 ( x) dx

For the first time, the scientists have measured both the mas of hot gas that is being sucked into nine older black holes and the unseen super speedy jets of high energy particles split out, which essentially form a cosmic engine. Then they determined a rate of how efficient these older black hole engines are and were awe-struck. These black holes are 25 times more efficient than anything man has built, with nuclear power being the most efficient of man-made efforts, said the research's lead author, Professor Steve Allen of Stanford University.

Y B

x=a

x=b

A

O

X where f2(x) is y2 of the upper curve and f1(x) is y1 of the lower curve, i.e. the required area

=

b

∫

a

[ f 2 ( x) − f1 ( x)] dx =

b

∫ (y a

The galaxies in which these black holes live are bigger than the Milky way, which is the Earth's galaxy and are 50 million to 400 million light-years away.

− y1 ) dx

2

f(x) ≤ 0 for all x in a ≤ x ≤ b, then area bounded by x-axis, the curve y = f(x) and the ordinates x = a, x = b is given by

=–

∫

b

a

f ( x) dx Y

C X

D

O

B A

If f(x) ≥ 0 for a ≤ x ≤ c and f(x) ≤ 0 for c ≤ x ≤ b, then area bounded by y = f(x), x-axis and the ordinates x = a, x = b is given by

∫

c a

f ( x) dx +

∫

b

c

− f ( x) dx =

x=a

A f (x)≥0

O

M

∫

C

f (x)≤0

c a

f ( x) dx –

∫

b

c

The results were surprising because the types of black holes studied were older, less powerful and generally considered boring, scientists said. But they ended up being more efficient than originally thought, possibly as efficient as their younger, brighter and more potent black hole siblings called quasars. One way the scientists measured the efficiency of black holes was by looking at the jets of high energy spewed out.

f ( x ) dx

N x=b

=

Black holes are the most fuel-efficient engines in the universe.

B

XtraEdge for IIT-JEE

47

MARCH 2012

MATHS

PROBABILITY Mathematics Fundamentals Probability : In a random experiment, let S be the sample space and E ⊆ S, then E is an event. The probability of occurrence of event E is defined as

Some Definitions : Experiment : A operation which can produce some well defined outcomes is known as an experiment. Random experiment : If in each trail of an experiment conducted under identical conditions, the outcome is not unique, then such an experiment is called a random experiment. Sample space : The set of all possible outcomes in an experiment is called a sample space. For example, in a throw of dice, the sample space is {1, 2, 3, 4, 5, 6}. Each element of a sample space is called a sample point. Event : An event is a subset of a sample space. Simple event : An event containing only a single sample point is called an elementary or simple event. Events other than elementary are called composite or compound or mixed events. For example, in a single toss of coin, the event of getting a head is a simple event. Here S = {H, T} and E = {H} In a simultaneous toss of two coins, the event of getting at least one head is a compound event. Here S = {HH, HT, TH, TT} and E = {HH, HT, TH} Equally likely events : The given events are said to be equally likely, if none of them is expected to occur in preference to the other. Mutually exclusive events : If two or more events have no point in common, the events are said to be mutually exclusive. Thus E1 and E2 are mutually exclusive in E1 ∩ E2 = φ. The events which are not mutually exclusive are known as compatible events. Exhaustive events : A set of events is said to be totally exhaustive (simply exhaustive), if no event out side this set occurs and at least one of these event must happen as a result of an experiment. Independent and dependent events : If there are events in which the occurrence of one does not depend upon the occurrence of the other, such events are known as independent events. On the other hand, if occurrence of one depend upon other, such events are known as dependent events.

XtraEdge for IIT-JEE

P(E) = =

number of distinct elements in E n(E) = number of distinct element in S n(S)

number of outocomes favourable to occurrence of E number of all possible outcomes

Notations : Let A and B be two events, then

A ∪ B or A + B stands for the occurrence of at least one of A and B. A ∩ B or AB stands for the simultaneous occurrence of A and B. A´ ∩ B´ stands for the non-occurrence of both A and B. A ⊆ B stands for "the occurrence of A implies occurrence of B". Random variable : A random variable is a real valued function whose domain is the sample space of a random experiment. Bay’s rule : Let (Hj) be mutually exclusive events such that n

P(Hj) > 0 for j = 1, 2, ..... n and S =

U

H j . Let A

j =1

be an events with P(A) > 0, then for j = 1, 2, .... , n P( H j ) P(A / H j ) Hj = P n A ∑ P(H k ) P( A / H k ) k =1

Binomial Distribution : If the probability of happening of an event in a single trial of an experiment be p, then the probability of happening of that event r times in n trials will be n Cr pr (1 – p)n – r. Some important results : (A)

P(A) = =

48

Number of cases favourable to event A Total number of cases n(A) n(S)

MARCH 2012

P(A) =

(i) Probability of happening none of them

Number of cases not favourable to event A Total number of cases

=

= (1 – p1) (1 – p2) ........ (1 – pn) (ii) Probability of happening at least one of them

n(A) n(S)

= 1 – (1 – p1) (1 – p2) ....... (1 – pn) (iii) Probability of happening of first event and not happening of the remaining

(B) Odd in favour and odds against an event : As a result of an experiment if “a” of the outcomes are favourable to an event E and b of the outcomes are against it, then we say that odds are a to b in favour of E or odds are b to a against E. Thus odds in favour of an event E

=

= p1(1 – p2) (1 – p3) ....... (1 – pn) If A and B are any two events, then B P(A ∩ B) = P(A) . P or A

a Number of favourable cases = Number of unfavourable cases b

B P(AB) = P(A) . P A

Similarly, odds against an event E =

b Number of unfavourable cases = Number of favorable cases a

B Where P is known as conditional probability A means probability of B when A has occurred.

Note : If odds in favour of an event are a : b, then the probability of the occurrence of that event is a and the probability of non-occurrence of a+b b a that event is . a +b a+b

Difference between mutually exclusiveness and independence : Mutually exclusiveness is used when the events are taken from the same experiment and independence is used when the events are taken from the same experiments. (E)

If odds against an event are a : b, then the probability of the occurrence of that event is b and the probability of non-occurrence of a+b a that event is . a+b (C)

P(AB) + P( AB ) = 1 P( A B) = P(B) – P(AB) P(A B ) = P(A) – P(AB) P(A + B) = P(A B ) + P( A B) + P(AB) Some important remark about coins, dice and playing cards :

P(A) + P( A ) = 1 0 ≤ P(A) ≤ 1

(D)

Coins : A coin has a head side and a tail side. If an experiment consists of more than a coin, then coins are considered to be distinct if not otherwise stated.

P(φ) = 0 P(S) = 1 If S = {A1, A2, ..... An}, then P(A1) + P(A2) + .... + P(An) = 1 If the probability of happening of an event in one trial be p, then the probability of successive happening of that event in r trials is pr. If A and B are mutually exclusive events, then P(A ∪ B) = P(A) + P(B) or P(A + B) = P(A) + P(B) If A and B are any two events, then

Dice : A die (cubical) has six faces marked 1, 2, 3, 4, 5, 6. We may have tetrahedral (having four faces 1, 2, 3, 4,) or pentagonal (having five faces 1, 2, 3, 4, 5) die. As in the case of coins, If we have more than one die, then all dice are considered to be distinct if not otherwise stated. Playing cards : A pack of playing cards usually has 52 cards. There are 4 suits (Spade, Heart, Diamond and Club) each having 13 cards. There are two colours red (Heart and Diamond) and black (Spade and Club) each having 26 cards.

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) or P(A + B) = P(A) + P(B) – P(AB) If A and B are two independent events, then

In thirteen cards of each suit, there are 3 face cards or coart card namely king, queen and jack. So there are in all 12 face cards (4 kings, 4 queens and 4 jacks). Also there are 16 honour cards, 4 of each suit namely ace, king, queen and jack.

P(A ∩ B) = P(A) . P(B) or P(AB) = P(A) . P(B) If the probabilities of happening of n independent events be p1, p2, ...... , pn respectively, then XtraEdge for IIT-JEE

P(A A ) = 0

49

MARCH 2012

MOCK TEST FOR IIT-JEE PAPER - I

Time : 3 Hours Instructions :

Total Marks : 240

• This question paper contains 69 questions in Chemistry (23), Mathematics (23) & Physics (23). • In section -I (7 Ques. SCQ Type) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer. • In section -II (4 Ques. MCQ Type) of each paper +4 marks will be given for correct answer no negative marking for wrong answer. • In section -III contains 2 groups of questions [Pass. 1 (2 Ques.) + Pass. 2 (3 Ques.) = 5 Ques.] of each paper +3 marks will be given for each correct answer & –1 mark for wrong answer. • In section -IV contain (7 Ques. of Numerical Response with single-digit Ans.) of each paper +4 marks will be given for correct answer & No Negative marking for wrong answer.

OH

CHEMISTRY SECTION – I

3.

Questions 1 to 7 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. 1. Arrange the following compounds according to decreasing order of combustion CH3

Assign double bond configurations to the following– COOH C NC CH2OH C=C CN NH2–CH2 (A) E

4.

II

O

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

(P)

major

tautomer

HO OH

CH3

(C)

5.

(enol

OH (B)

50

CH3 OH

(D) OH

form) among the following is -

OH

(B) OH CH3

⊕

H

CH2–OH

(A)

(B) I > IV > III > II (D) IV > I > III > II

(D) Z, Z

⊕

H / H2O

CH3 CH3

IV

III

XtraEdge for IIT-JEE

(C) E.E.

OH

I

(A)

(B) Z

Product (P) in the following reaction is -

CH3

2.

(D)

(C)

Straight Objective Type

O

OH

CH3

O CH3 CH3

V1 mL of NaOH of normality X and V2 mL of Ba(OH)2 of normality Y are mixed together. the mixture is completely neutralised by 100 mL of 0.1 1 X N HCl. If V1 / V2 = and = 4, what fraction of 4 Y the acid is neutralized by Ba(OH)2 : (A) 0.5 (B) 0.25 (C) 0.33 (D) 0.67 MARCH 2012

6.

Equal moles of CO, B2H6, H2 and CH4 are placed in a container. If a hole was made in container, after 5 minute partial pressure of gases in container would be : (at wt. of C, O, B and H are 12, 16, 11 and 1 respectively)

11.

SECTION – III

(A) PCO > PB2 H 6 > PH 2 > PCH 4

Comprehension Type

(B) PCO = PB2 H 6 > PCH 4 > PH 2

This section contains 2 paragraphs; passage- I has 2 multiple choice questions (No. 12 & 13) and passage- II has 3 multiple (No. 14 to 16). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and –1 mark for each wrong answer. Paragraph # 1 (Ques. 12 to 13)

(C) PCO > PB2 H 6 = PH 2 > PCH 4 (D) PB2 H 6 > PH 2 > PCH 4 > PCO 7.

The dipole moment of HBr is 2.6 × 10–30 c–m and the interatomic spacing is 1.41Å. The percentage of ionic character in HBr is : (A) 10.5 (B) 11.5 (C) 12.5 (D) 13.5

De-Broglie proposed dual nature for electron by

h . Later on mu Heisenburg proposed uncertainty principle as ∆p.∆x ³ h h h = . On the contrary particle nature of 2 2π electron was established on the basis of photoelectric effect. When a photon strikes the metal surface,, it gives up its energy to the electron. Part of this energy (say W) is used by the electrons of escape from the metal and the remaining imparts the kinetic energy (½ mu2) to the photoelectron. The potential applied on the surface to reduce the velocity of photoelectron to zero is known as stopping potential. putting his famous equation λ =

SECTION – II Multiple Correct Answers Type Questions 8 to 11 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and no negative marks. O

8.

The compound C H 2 5

may be named as -

CH3 (A) 3– ethyl, 2-methyl oxirane (B) 1, 2 – epoxy – 2 – methyl butane (C) 1, 2 – oxa pentane (D) 2 – methyl –1, 2– butoxide

9.

12. With what velocity must an electron travel so that its momentum is equal to that of photon of wavelength of λ = 5200 Å : (A) 800 m s–1 (B) 1400 m s–1 (C) 400 m s–1 (D) 200 m s–1

In which of the following reactions 3º alcohol will be obtained as a product -

13. The wavelength of helium atom whose speed is equal to its rms speed at 27ºC : (A) 7.29 × 10–11 m (B) 4.28 × 10–10 m –11 (C) 5.31 × 10 m (D) 6.28 × 10–11 m

⊕

(A)

H

MgBr + HCOCl →

(excess) O ⊕

H (B) Ph–Mg–Br + CH3—C—Cl → → (excess)

Paragraph # 2 (Ques. 14 to 16)

From following sets of compounds give answer of following question

⊕

H

(C) CH3–Mg–Br + (CH3 CO2)O → → (excess)

Set – A

⊕

(D) CH3–Mg – Br + Cl–C–OC2H5 → → (excess) O Which of the following species has same number of unpaired electron ? (A) Cr3+ (B) Mn 2+ (C) Fe3+ (D) Cu2+

XtraEdge for IIT-JEE

O

O

H

10.

Which of the following metals become passive when dropped into conc. HNO3 ? (A) Cu (B) Fe (C) Cr (D) Al

O

Set – B

51

I

I

O

O

O

II

II

III

III

O

O

IV

IV

MARCH 2012

O

O

O

CH3

O

CH3

O

Set – C

O I

(vii) D

II

O

O

18.

O

III IV 14. Correct statement regarding set A (A) I is stronger acid than III (B) II is stronger acid then I (C) III & IV are equal acidic strength (D) IV is weaker acidic then I

(ii)

D

MATHEMATICS SECTION – I Straight Objective Type Questions 1 to 7 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

N

C2H5

CH2 =CH–CH2

O

H

CH3O

(iv) CH3

S CH3

1. H

O H

Si

D

XtraEdge for IIT-JEE

CH3

D

x 1 1 Let f (x) = sin πx 2 x 2 1 , f (x) be an odd function x3 3x 4 1 and its odd value is equal to g(x), then f (1) g(1) is (A) –1 (B) – 4 (C) – 5 (D) 1

(vi)

T

(viii) – SO3H

23. How many P —O—P bonds are present in P4O8 ?

⊕

Ph

condition

22. How many π-bonds are present in Marhall's acid?

3

T

C (mild-basic) D

20. How many compound which is given below is isomer of D-glucose. D-Mannose, D – Fructose, D-Idose, D-Galactose, D-Arabinose, D-Ribose 21. What is oxidation state of sulphur in Caro's acid?

17. How many of the following compound contain chiral atom. CH3 CH ⊕

(1) Sn / HCl

A + B (2)NaNO / HCl 2

O (ix) —N = O (x) —N = NH

This section contains 7 questions (Q.17 to 23). +4 marks will be awarded for each correct answer and no negative marking for wrong answer. The answer to each question is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the OMR.

P

OH

O (vii) —C—NH—Me

Numerical Response Type

(v)

H

19. How many groups are o/p director in electrophilic aromatic substitution reaction – (ii) — CHO (i) — NH2 (iii) — COOH (iv) —OMe (iv) — Et (v) —O—C—Me

SECTION – IV

(iii)

Ph

More Less volatile volatile

16. Correct statement regarding set C (A) II is stronger acid than I (B) I is stronger acid than IV (C) II and III are equal acidic (D) In IV acid anion is not stablises by resonance

N

(viii)

CH3 C2H5 14 In the following reaction double bond equivalent of D is –

dil.HNO3

15. Correct statement regarding set B (A) I & II compound show both resonance and hyperconjugation (B) III compound show five hyperconjugation structure (C) IV is more stable than II (D) II is more stable than I

H

CH3

OH

O

(i)

C

Ge

H

52

MARCH 2012

2.

3.

sin 3α < 0 if α lies in cos 2α (A) (13π/48, 14π/48) (B) (14π/48, 18π/48) (C) (18π/48, 23π/48) (D) any of these intervals

(C) f (x) is an increasing function in the interval – ∞ < x ≤ 0 and decreasing in the interval 0≤x<∞ (D) f (x) is a decreasing function in the interval – ∞ < x ≤ 0 and increasing in the interval 0≤x<∞

If x1, x2, x3, x4 are roots of the equation x4 – x3 sin 2β + x2 cos 2β – x cos β – sin β = 0, then 4

∑ tan

−1

11. If u =

a 2 sin 2 θ + b 2 cos 2 θ , then

xi is equal to

(A) max. u2 =

i =1

(A) π – β (C) π/2 – β 4.

π 2

SECTION – III

(A) Does not exist (C) e

(B) 1 (D) 4

5.

If f (x) = logx (ln (x)), then f '(x) at x = e is (A) 0 (B) 1 (C) e (D) 1/e

6.

If the function f (x) = |x2 + a |x| + b| has exactly three points of non-differentiability, then which of the following may hold (A) b = 0, a < 0 (B) b < 0, a ∈ R (C) b > 0, a ∈ R (D) b < 0, a ∈ R–

7.

Comprehension Type This section contains 2 paragraphs; passage- I has 3 multiple choice questions (No. 12 to 14) and passage- II has 2 multiple (No. 15 & 16). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and –1 mark for each wrong answer. Passage # 1 (Ques. 12 to 14)

f (x) = sin {cot–1 (x + 1)} – cos (tan–1 x) a = cos tan–1 sin cot–1 x b = cos (2 cos–1 x + sin–1 x)

The inequality log2(x) < sin–1 (sin(5)) holds if x ∈ (A) (0, 25–2π) (B) (25–2π, ∞) 2π–5 (C) (2 , ∞) (D) (0, 22π–5)

SECTION – II

12. The value of x for which f (x) = 0 is (A) – 1/2 (B) 0 (C) 1/2 (D) 1

Multiple Correct Answers Type Questions 8 to 11 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and no negative marks. 8.

a2 + b2

(B) max. u2 = 2 a 2 + b 2 (C) min. u2 = 2(a + b)2 (D) min u2 = (a + b)2

(B) π – 2β (D) π/2 – 2β

Lim− [1 + (cos x)cos x]2 is equal to

x→

a 2 cos 2 θ + b 2 sin 2 θ +

2 tan–1(–3) is equal to (A) – cos–1(–4/5) (C) – π/2 + tan–1(–4/3)

13. If f (x) = 0, then a2 is equal to (A) 1/2 (B) 2/3 (C) 5/9 (D) 9/5 14. If a2 = 26/51, then b2 is equal to (A) 1/25 (B) 24/25 (C) 25/26 (D) 50/51

(B) – π + cos–1 (4/5) (D) cot–1 (4/3)

Let [x] denotes the greatest integer less than or equal to x. If f (x) = [x sin πx], then f (x) is (A) continuous at x = 0 (B) continuous in (–1, 0) (C) differentiable at x = 1 (D) differentiable at (–1, 1) x2 , then 10. Let f be the function f (x) = cos x – 1 − 2 (A) f (x) is an increasing function in (0, π/2) (B) f (x) is a decreasing function in (–∞, ∞) 9.

XtraEdge for IIT-JEE

Passage # 2 (Ques. 15 & 16)

If y =

∫

v( x) u ( x)

f (t ) dt, let us define

dy in different dx

dy = v'(x) f (v(x)) – u'(x) f (u(x)) and the dx equation of tangent at (a, b) is. dy (x – a ) (y – b ) = dx ( a ,b )

manner

53

MARCH 2012

15. If y =

∫

x2 x

If f (x) =

x

1

et

2

/2

at x = 1 is (A) 0 (C) 2

(1 – t2)dt, then

→

→

→

→

→

→

→

→

→

SECTION – I Straight Objective Type Questions 1 to 7 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

SECTION – IV This section contains 7 questions (Q.17 to 23). +4 marks will be awarded for each correct answer and no negative marking for wrong answer. The answer to each question is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the OMR.

1.

17. The number of elements in the range of

2 f (x) = [x] + [2x] + x + [3x] + [4x] + [5x] 3 for 0 ≤ x < 3 is m + 21. Then the value of m is -

A block of mass 'm1' attached to the free end of a spring of force constant 'k' is mounted on a smooth horizontal surface as shown in figure. The block execute S.H.M. with amplitude A and frequency 'f '. If an object of mass 'm2' is put on it, when the block was passing through mean position and both move together, then the new amplitude of oscillation is -

m2 m1 smooth

h S

= e , then

2.

19. The number of polynomials of the form x3 + ax2 + bx + c which are divisible by x2 + 1, where a, b, c ∈ {1, 2, 3, ….. 10} must be 2k. Value of k will be.

d2y at x = 0 is e–λ, then dx 2 numerical quantity λ should be equal to……….

(A) A

m1 m2

(B) A

(C) A

m1 (m1 + m 2 )

(D) A

v B

21. If a circle S (x, y) = 0 touches at the point (2, 3) of the line x + y = 5 and S (1, 2) = 0, then

(A) v cos θ (1 − e) v cos θ (C) 2

( 2 × Radius) of such circle is.

x2 y 2 + =1 16 9 at A and D meets the x-axis and y-axis at B and C respectively, so that PA.PD = PB.PC, then | λ | is greater than or equal to ……

22. A line through P(λ, 3) meets the ellipse

3.

54

m2 (m1 + m 2 )

Two smooth, identical billiard balls A and B collide with B at rest and A in motion as shown. If speed of A is 'v' before collision and coefficient of restitution for the collision is 'e', the speed of B after collision will be-

If ey + xy = e, then

XtraEdge for IIT-JEE

→

PHYSICS

Numerical Response Type

20.

→

d f (x), dx

(B) 1 (D) – 1

1 1 (h + 1) h + h + 2 2 2 2 18. Lim h − h h →∞ ..................... h + 1 2 h −1 find the value of S.

→

( c + a ), c ⊥ ( a + b ), then | a + b + c | is……

(B) x + y = 1 (D) y = x

∫

→

23. If | a | = 3, | b | = 4, | c | = 5 and a ⊥ ( b + c ), b ⊥ →

x = 1 is (A) y = x + 1 (C) y = x – 1 16

→

t 2 dt , then equation of the tangent at

A

θ

(B) ev cosθ (1 + e) v cos θ (D) 2

The moment of inertia of a flat annular ring is having mass M, inner radius 'a' and outer radius 'b' about the perpendicular axis through the centre is(A) M (b2 – a2) (B) M/2 (b2 – a2) 2 2 (C) M/2 (b + a ) (D) M/2 (b – a) MARCH 2012

4.

A solid sphere of mass 2 kg is pulled by a constant force acting at its centre on a rough surface having co-efficient of friction 0.5. The maximum value of F so that the sphere rolls without slipping is-

9.

F

(A) 70 N (C) 40 N 5.

A

(B) 25 N (D) 35 N

3GM 4R

(B)

2GM 3R

(C)

GM 2R

(D)

2GM R

10.

6.

The work done to break a spherical drop of radius R in n drops of equal size is proportional to1 1 (A) 2 / 3 − 1 (B) 1/ 3 − 1 n n (C) n1/3 –1 (D) n4/3 – 1

7.

A rod of length L kept on a smooth horizontal surface is pulled along its length by a force F. The area of cross-section is A and Young's modulus is Y. The extension in the rod isF (A)

FL AY

(B)

2FL AY

(C)

FL 2AY

11.

(D) zero

Multiple Correct Answers Type Questions 8 to 11 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and no negative marks.

The horizontal distance x and the vertical height y of a projectile at time t are given by x = at and y = bt2 + ct, where a, b and c are constant. Then(A) the speed of the projectile 1 second after it is fired is (a2 + b2 + c2)1/2 (B) the angle with the horizontal at which the c projectile is fired is tan–1 a is

A block having mass m and charge q is connected by spring of force constant k. The block lies on a frictionless horizontal track and a uniform electric field E acts on system as shown. The block is released from rest when spring is unstretched (at x = 0). Then-

E q,m (A) maximum elongation in the spring is

Under the action of a force, 2 kg body moves such that its position x as a function of time is given by x t3 = , x is in meter and t in second3 (A) body is acted by constant force (B) body is acted by a force which is proportional to time (C) body is acted by a force which is proportional to square of time (D) work done by the force in first 2 second is 16 J

XtraEdge for IIT-JEE

B

Q

(C) the acceleration due to gravity is –2b (D) the initial speed of the projectile (a2 + c2)1/2

SECTION – II

8.

P

(A) Electric current at cross section P is equal to that of cross section Q (B) Electric field intensity at P is less than that at Q (C) The number of electrons crossing per unit area per unit time at cross section P is less than that at Q (D) The rate of heat generating per unit time at Q is greater than that of P

A tunnel is dug across the diameter of Earth. A ball is released from the surface of Earth into the tunnel. The velocity of ball when it is at a distance R/2 from centre of Earth is (where R = radius of earth and M = mass of Earth) (A)

A metallic conductor of irregular cross section is as shown in figure A constant potential difference is applied across the ends (A) and (B). Then-

2qE k

(B) at equilibrium position, elongation in the spring is

qE k

qE k 2qE (D) amplitude of oscillation of block is k (C) amplitude of oscillation of block is

55

MARCH 2012

SECTION – III

17.

Comprehension Type This section contains 2 paragraphs; passage- I has 3 multiple choice questions (No. 12 & 14) and passage- II has 2 multiple (No. 15 to 16). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and –1 mark for each wrong answer.

18.

19.

Passage # 1 (Ques. 12 & 14) Consider a circuit shown in figure, then 7µF 23µF A

1µF 12µF

10µF

Two simple pendulum have lengths l and 25l. At t = 0 they are in same phase, after how many oscillations of smaller pendulum will they be again in phase for first time ? In the resonance tube experiment, the first and second states of resonance are observed at 20 cm and 66 cm. Find the value of end correction (in cm). The variation of pressure versus volume is shown in the figure. The gas is diatomic and the molar specific heat capacity for the process is found to be xR. Find the value of x.

P B

5µF

1µF

V 12V

12.

13.

14.

20.

Equivalent capacitance of the circuit between A and B is(A) 10 µF (B) 7.5 µF (C) 5 µF (D) 14 µF

C

A

Energy stored by 10 µF capacitor is(A) 20 µJ (B) 40 µC (C) 400 µJ (D) 245 µF

Total mechanical energy is(A) 16 J (B) 26 J (C) 6 J

(D) 5 J

Minimum potential energy is(A) 5 J (B) 10 J (C) 6 J

(D) 9 J

21.

A rod of length 20 cm is placed along the optical axis of a concave mirror of focal length 30 cm. One end of the rod is at the centre of curvature and other end lies between F and C. Calculate the linear magnification of the rod.

22.

Figure shows a parabolic reflector in x-y plane given by y2 = 8x. A ray of light traveling along the line y = a is incident on the reflector. Find where the ray intersects the x-axis after reflection. y-axis y2 =8x

P(0,a)

23.

SECTION – IV Numerical Response Type This section contains 7 questions (Q.17 to 23). +4 marks will be awarded for each correct answer and no negative marking for wrong answer. The answer to each question is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the OMR.

XtraEdge for IIT-JEE

O 0ºC

Potential energy of particle is given by U = 5 + (x – 1)2. Particle have kinetic energy at x = 2m is 10 J. Then

16.

20ºC

100ºC

Charge stored by 5 µF capacitor is(A) 20 µC (B) 70 µC (C) 10 µC (D) zero

Passage # 2 (Ques. 15 & 16)

15.

Three identical rods are joined at point O as shown in the figure. In the steady state, find the ratio of thermal current through rod AO and OC.

56

line y = a incident ray x-axis

A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine the value of Z. (Ionisation energy of hydrogen atom is 13.6 eV). MARCH 2012

XtraEdge for IIT-JEE

57

MARCH 2012

MOCK TEST FOR IIT-JEE PAPER - II

Time : 3 Hours Instructions :

Total Marks : 240

• This question paper contains 60 questions in Chemistry (20,) Mathematics (20) & Physics (20). • In section -I (8 Ques. SCQ Type) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer. • In section -II (4 Ques. MCQ Type) of each paper +4 marks will be given for correct answer no negative marking for wrong answer. • In section -III (2 Ques. Column Matching Type) of each paper +8(2×4) marks will be given for correct answer. No Negative marking for wrong answer. • In section -IV contain (6 Ques. of Numerical Response with single-digit Ans.) of each paper +4 marks will be given for correct answer & No Negative marking for wrong answer. 4.

CHEMISTRY SECTION – I Straight Objective Type Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. 1. In the enolisation of the given molecule the, H-atom involved is : O Ha

H

2.

(B) Hb

(C) Hc

The no. of position are CH3—CH—CH2—C≡CH CH3 including this structure) : (A) 1 (B) 2 (C) 0

3.

5.

XtraEdge for IIT-JEE

H Br

The metals present in insulin, haemoglobin and vitamin B12 are respectively : (A) Zn, Hg, Cr (B) Co, Fe, Zn (C) Mg, Fe, Co (D) Zn, Fe, Co

6. (A) Fused with Na2CO3 Green solid

(D) Hd

(B) (H2SO4 + H2O) (C) Yellow solution

Evaporisation

Orange

(CH3COO)2Pb

isomers for (give answer

(D) Yellow Here, A, B, C and D are respectively : A B C D (A) FeSO4 FeCO3 Fe(OH)3 (B) Cr2O3 Na2CrO4Na2Cr2O7 (C) FeCl2 FeSO4 PbSO4 (D) FeSO4 FeCl3 Fe(OH)3

(D) 3

Rank the following alkyl halides in order of increasing SN2 reactivity: Br Br Br I II (A) I < III < II (B) III < I < II (C) II < I < III (D) I < II < III

D

(A) Identical (B) Diastereomers (C) Optically inactive (D) enantiomers

H Hcc

(A) Ha

Cl I

Hb

Ha

The two compounds given below are : D Cl H Br I H

7.

III

58

PbCO3 PbCrO4 Fe(OH)3 PbCl2

The negative charge on As2S3 sol is due to absorption of : (A) H¯ (B) OH¯ (C) O2– (D) S2–

MARCH 2012

8.

pH of a mixture of 1M benzoic acid (pKa = 4.20) and 1M sodium benzoate is 4.5 In 300 mL buffer, benzoic acid is (log 2 = 0.3) : (A) 200 mL (B) 150 mL (C) 100 mL (D) 50 mL

SECTION – III Matrix Match Type This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column-I and five statements (P, Q, R, S and T) in Column-II. Any given statement in Column–I can have correct matching with One or More statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in Q and R, then for the particular question, against statement B, darken the bubbles corresponding to Q and R in the OMR. +8 marks will be given for each correct answer (i.e. +2 marks for each correct row) and no negative marking for each wrong answer.

SECTION – II Multiple Correct Answers Type Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) may be correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and no negative marks for wrong answer. 9.

13. Match the column:

Which of the following pair have same IUPAC naming : Br COOCH3 COOH (A) & Br COOH OCOCH3 I Br Br Cl F (B) & Cl I I I F ` (C) &

Column -I (A) Electro chamical cell st

(B) I law of faraday (C) Electrolytic cell (D) lead acid storage cell

F H2N

OH Cl

Column -I (A) lithium (B) Barium (C) calcium (D) magnesium

H2N &

Cl

OH

10. Benzene can be nitration by : (A) NO2 BF4 (B) NO2ClO4 (C) conc. HNO3 + H2SO4 (D) Cl2 / AlCl3

(Q) ∆G = –ve (R) w = z . i . t. (S) salt bridge (T) rechargable cell

Column-II (P) violet (Q) Brick red (R) Apple green (S) Crimson red (T) No flame colour

SECTION – IV Numerical Response Type This section contains 6 questions (Q.15 to 20). The answer to each of the questions is a Single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the OMR. +4 marks will be given for each correct answer and no negative marking for each wrong answer.

11. In which of the following salt bridge is not needed : (A) Pb(s) |PbSO4(s)| H2SO4|PbO2(s)|Pb (s) (B) Zn(s)|ZnSO4|CuSO4|Cu(s) (C) Cd(s)|CdO(s)|KOH(aq)|NiO2(s)|Ni(s) (D) Fe(s) | FeO(s)|KOH(aq)|Ni2O3(s) |Ni(s)

15. How many acidic group is present in given compound :

12. Which of the following is (are) true : (A) H2S + H2O H3O+ + HS¯ ; Kc acidity constant of H2S (B) AgCl + 2NH3 Ag(NH3)2Cl; Kc is stability constant for Ag(NH3)2Cl (C) H2O H+ + OH¯ ; Kc is equilibrium constant for dissociation of water

(D) RNH2 + H2O

(P) ∆G = +ve

14. Match the column:

F (D)

Column-II

⊕

NH3—CH—CH2—CH2—COOH COO Θ

16. How many isomers of ‘X’ C8H10 when reacts with hot KMnO4 give only aromatic dicarboxylic acid ? How many isomers of 'Y' C4H8 when reacts with hot alkaline KMnO4 give carbondioxide sum of X + Y = ?

RNH 3+ + OH¯;

KC is basicity constant for RNH2 XtraEdge for IIT-JEE

59

MARCH 2012

CH3

17.

CH3—CH–CH—CH3

C2H5OH

If f (θ) = cos2 θ + sin4 θ. Then minimum value of f (θ) is (A) 1/4 (B) 3/4 (C) 1 (D) 1/2

4.

Let f (x) =

‘X’ (SN1 + E1)

Consider all products

Br

products Find total no. of products (including stereoisomer). 18.

3.

4

equal to

N exhaustive methylation to remove nitrogen from given compound

5.

Alc.KOH

‘Y’ is total number of possible ∆ Br product (including stereoisomers) sum of X+ Y = ?

dx = K[H+]n. If the pH of reaction dt medium changes from two to one, the rate becomes 100 times of the value of at pH = 2. What is the order of reaction -

MATHEMATICS

7.

Let f (x) = 2x3 + ax2 + bx – 3cos2x is an increasing function for all x ∈ R, then (B) a2 – 6b + 18 < 0 (A) a2 – 6b – 18 > 0 (C) a2 – 3b – 6 < 0 (D) a > 0, b > 0

8.

The function y = f(x) is represented parametrically by x = t5 – 5t3 – 20t + 7 and y = 4t3 – 3t2 – 18t + 3, (–2 < t < 2). The minimum of y = f(x) occurs at (B) t = 0 (A) t = – 1 (D) t = 3/2 (C) t = 1/2

SECTION – I Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

r1 r2 r3 + + is equal to bc ca ab

1 1 − 2R r

(B) 2R – r

(A)

(C) r – 2R

XtraEdge for IIT-JEE

(D)

; x ≤1 ; x >1

2

The angle between the tangents at any point P and the line joining P to origin O, where P is a point on the curve ln (x2 + y2) = c tan–1 y/x, c is a constant, is (A) constant (B) varies as tan–1 (x) (D) None of these (C) varies as tan–1 (y)

Straight Objective Type

In a triangle ABC,

x 2 e 2( x −1) Let f (x) = 2 a cos(2 x − 2) + bx

3

6

20. For a reaction

2.

2

(D)

2

(B)

f (x) will be differentiable at x = 1, if (B) a = 1, b = – 2 (A) a = – 1, b = 2 (D) None of these (C) a = 1, b = 2

19. If the density of Fe2O3 and Al are 5.2 g/mL and 2.7 g / mL respectively. Calculate the fuel value in kcal mL–1 of mixture according to thermite reaction (If ∆Hf Al2O3 = – 399 Kcal / mol & ∆HfFe2O3 = – 199 Kcal / mol)

In a triangle, if the sum of two sides is x and their product is y such that (x + z) (x – z) = y, where z is the third side of the triangle, then the triangle is (A) equilateral (B) right angled (C) obtuse angled (D) none of these

1

(C) 1

(A)

‘X ’ is total number of hofmann

1.

2 2 − (cos x + sin x) 3 , then Lim f (x) is π 1 − sin(2 x) x→

SECTION – II Multiple Correct Answers Type Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) may be correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and no negative marks for wrong answer. 9.

If in a triangle ABC, θ is the angle determined by cos θ = (a – b)/c, then (A)

1 1 − r 2R

(B)

60

(a + b) sin θ 2 ab (a + b) sin θ 2 ab

= cos

A− B 2

= cos

A+ B 2 MARCH 2012

(C) (D)

c sin θ 2 ab

c sin θ 2 ab

= cos

A− B 2

= cos

A+ B 2

13. Match the column : Column –I

sin 2 x (A) f (x) = x , x ≠ 0 0, x=0 x 3 − 2 x, (B) f (x) = 2 x − 2 sin( x),

10. Which of the following function are defined for all x

(A) sin[x] + cos [x] ([x] denotes greatest integer ≤ x) (B) sec–1 (1 + sin2 x) (C) tan (log x) (D)

9 + cos x + cos 2 x 8

11. If x + |y| = 2y, then y as a function of x is (A) defined for all real x (B) continuous at x = 0 (C) differentiable for all x (D) such that dy/dx = 1/3 for x < 0

c 2 + 2ab

(B) minimum value of PA + PB is

c 2 + 4ab

(C) minimum value of PA + PB is

c 2 + ab

4 − x 2 , (C) f (x) = 2 5− x ,

x<0 x≥0

(R) Differentiable

3 − 2 x , (D) f (x) = 7 − 6 x,

x <1 x ≥1

(S)Non- differentiable (T) continuous & differentiable

Column –I (A) If x2 + y2 = 1, then minimum value of x + y is (B) If maximum value of 1 y = acos(x) – cos (3x) 3

occurs, when x =

Column-II (P) – 3

(Q) − 2

π , then 6

value of 'a' is (C) If f (x) = x – 2sin(x), 0 ≤ x ≤ 2π is increasing in interval (aπ, bπ) then, (a + b) is (D) If equation of tangent to the curve y = – e–x/2, where it crosses the y-axis is x/p + y/q = 1, then p – q is

(D) The required location must lie on the imaginary line joining village A and image of village B in the river.

(R) 3

(S) 2 (T) 1

SECTION – IV Numerical Response Type This section contains 6 questions (Q.15 to 20). The answer to each of the questions is a Single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the OMR. +4 marks will be given for each correct answer and no negative marking for each wrong answer. 15. An urn containing '14' green and '6' pink ball. K (< 14, 6) balls are drawn and laid a side, their colour being ignored. Then one more ball is drawn. Let P(E) be the probability that it is a green ball, then 10 P(E) = ..............

SECTION – III Matrix Match Type This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column-I and five statements (P, Q, R, S and T) in Column-II. Any given statement in Column–I can have correct matching with One or More statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in Q and R, then for the particular question, against statement B, darken the bubbles corresponding to Q and R in the OMR. +8 marks will be given for each correct answer (i.e. +2 marks for each correct row) and no negative marking for each wrong answer.

XtraEdge for IIT-JEE

(Q) Discontinuous

14. Match the column

12. Two villages A and B are on the same side of a straight river. A pump set is to be installed by the river side at a point P. Then if the villages are situated at a distance c, then (where a =distance of village A from river side, b = distance of village B from river side, c = distance between villages.)

(A) minimum value of PA + PB is

x<0 x≥0

Column-II (P) continuous

16. Two lines zi – z i + 2 = 0 and z(1+i) + z (1–i) + 2 = 0 intersect at a point P. Find the sum of minimum and maximum modulus of complex number of a point on second line which is at a distance of 2 units from point P.

61

MARCH 2012

3 − 2 3 17. If A = 2 1 −1 . Solve the system of equations 4 − 3 2 3 0 3 x 8 2 y 2 1 0 y = 1 + z , then find the value of 4 0 2 z 4 3 y y z x+ + 2 3

2.

5 gm of water of 30ºC and 5 gm of ice at –20ºC are mixed together in a calorimeter. Then temperature of the mixture will be (water equivalent of calorimeter is negligible, specific heat of ice is 0.5 cal/gmºC and latent heat of ice is 80 cal/gm) (A) 10ºC (B) 5ºC (C) zero (D) 25ºC

3.

Six resistors each of 5Ω are connected as shown in figure. Reading of ideal ammeter is-

20V

18. If the planes x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0 pass through a straight line, then find the value of a2 + b2 + c2 + 2abc.

A

(A) 4A

19. If

the solution of differential equation dy d y x2 2 + 2x = 12y is y = Axm + Bx–n then find the dx dx value of m + n, if m & n ∈ N.

4.

2

20. If f(x) = x +

f(x) =

∫

1

5.

Ax 2 + Bx A+ B ⇒ then the value of is 260 119

SECTION – I

A wooden plank of length 0.8 m and uniform crosssection is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water upto a height of 0.4 m. The specific gravity of the plank is 0.5. Find the angle θ that the plank makes with the horizontal in the equilibrium position-

h (A)

π 3

(B)

XtraEdge for IIT-JEE

6.

(C)

π 4

(D)

u2 2

B

2r

C

A r

(A)

q 02 3 log e 2π ∈0 2

(B)

(C)

q 0λ 2 log e 2π ∈0 3

(D)

q 0λ 3 log e 2π ∈0 2 q 0λ 2 π ∈0

A circular ring of radius R with uniform positive charge density λ per unit length is located in the y-z plane with its centre at the origin O. A particle of charge –q0 is released from x = 3 R on x-axis at t = 0 then kinetic energy of particle when it passes through origin, is-

θ π 6

(D)

+ + λ + + + + +

Straight Objective Type

1.

(D) 6A

The arc AB with centre C and infinitely long wire have linear charge density λ, are kept in the same plane as shown. The minimum amount of work to be done to move a point charge q0 from point A to B through a circular path AB radius r is equal to-

PHYSICS Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

(C) 8A

A particle is moving in parabolic path y = x2 with constant speed 'u'. Then acceleration of the particle when it crosses origin, is(A) u2 (B) 2u2 (C) zero

(xy2 + x2y) f (y) dy, then

0

(B) 2A

π 5

62

(A)

λq 0 2 ∈0

(B)

q 0λ 3 ∈0

(C)

q 0λ ∈0

(D)

q 0λ 4 ∈0 MARCH 2012

7.

A long current carrying wire is bent as shown. The magnetic field at O is-

11. Figure shows a square loop being pulled out with a constant speed out of region of uniform magnetic field. The induced emf in the loopB × × × × × × × l× × l× × × v × × × × l × × × l× (A) first increases, then decreases (B) first decreases, then increases

90º

i L (A) zero 8.

i

O (B)

L

2 2µ 0 i πL

(C)

µ0i πL

(D)

2µ 0 i πL

In a given series RLC circuit, average power dissipated in the circuit is-

(C) has a maximum value Bv l 2 (D) has a maximum value 2Bvl

50Ω 500mH 100µC

12. The current a certain circuit varies with time as shown. The peak value of current is i0. If iv and im represent the virtual (rms) and mean value of current for a complete cycle respectively. Theni i0

~

π V = 200 sin 100 t + 4

(A) 200 W (C) 400 W

(B) 800 W (D) 100 W

O T/2 T

SECTION – II

t

Multiple Correct Answers Type Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) may be correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and no negative marks for wrong answer. 9. Missile is fired for maximum range at your town from a place in the enemy country at a distance 'x' from your town. The missile is first detected at its half-way point. Then(A) the velocity with which the missile was projected

is

(A) im = (C) irms =

gx 2

(D) the maximum height attained by the missile is

i0 2

(D) irms =

3

i0 2

SECTION – III

x 2g

(C) the speed of the missile when it was detected is

(B) im =

Matrix Match Type This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column-I and five statements (P, Q, R, S and T) in Column-II. Any given statement in Column–I can have correct matching with One or More statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in Q and R, then for the particular question, against statement B, darken the bubbles corresponding to Q and R in the OMR. +8 marks will be given for each correct answer (i.e. +2 marks for each correct row) and no negative marking for each wrong answer. 13. Consider the circuit showing in figure. There are three switches S1, S2, S3. Match the columns.

gx

(B) you have a warning time of

2i 0 π i0

x 4

C

10. A particle A of mass 'm' and charge Q moves directly towards a fixed particle B, which has charge Q. The speed of A is 'v' when it is far away from B. The minimum separation between the particles is proportional to(A) Q2 (B) 1/v2 (C) 1/v (D) 1/m

3r

S1 r S3

S2

2r

V

XtraEdge for IIT-JEE

63

MARCH 2012

Column -I (A) If S2 and S3 are opened and S1 is closed then in steady state, charge on the capacitor is (B) If switch S2 only is closed then maximum charge on the capacitor is (C) If switch S3 only is closed then maximum charge on the capacitor is (D) If all the switches are closed then maximum charge on the capacitor is-

17. Two blocks A and B are connected by a massless string as shown in figure. Friction coefficient of the inclined plane is 0.5. The mass of block A is 5 kg. If minimum and maximum values of mass of the block B for which the block A remains in equilibrium are m1 and m2 then find the value of (m2 – m1) in kg.

Column-II (P) CV/4

(Q) 2CV/5

M=5kg

(R) CV/3

A 37º

(S) CV

18. A split lens has two parts separated by 4 cm and focal length is f. An object is placed at a distance 3f/2 from C as shown in figure. The distance the images formed by the two halves (in cm) is.

(T) zero 14. A particle is projected from ground with velocity 40 m/s at an angle θ = 60º with vertical. Match the quantities given in Column I to the results in Column II (g = 10 m/s2) Column -I (A) Time to flight

O

Column-II (P) 40 m (Q) 80 meter when θ = 0º

(B) Range (C) Maximum height

SECTION – IV

10cm

30º

10cm

10 m/s

20. The ratio of respective acceleration due to gravity at the surface of two planets having masses in the ratio 'x' and density in the ratio 'y' is (xy2)1/n then find value of n.

30º

10cm

Cartoon Law of Physics

then electric field in the space is n × 102 N/C. Find value of n.

Any body passing through solid matter will leave a perforation conforming to its perimeter.

16. Two electric bulbs of power rating (200 V, 40 W) and (200 V, 50 W) are connected in series with the main supply as shown in figure. The voltage (V0) of the main supply, so that 40 W bulb glows with full intensity is 60 n volt. Find value of n.

40 W

XtraEdge for IIT-JEE

Also called the silhouette of passage, this phenomenon is the specialty of victims of directedpressure explosions and of reckless cowards who are so eager to escape that they exit directly through the wall of a house, leaving a cookie-cutout-perfect hole. The threat of skunks or matrimony often catalyzes this reaction.

50 W

~

f

20 m/s

15. In a space, equipotential surfaces are shown below. 80V 60V 40V 20V

30º

4 cm

19. A solid cylinder is rolling without slipping on a plank which also moving with speed 10 m/s on a horizontal surface as shown. The speed of centre of cylinder is 20 m/s. The mass of cylinder is 2 kg. The kinetic energy of the cylinder when observed from ground is 90 nJ. Then find value of n.

3 meter (S) 4 second (T) 17 second

Numerical Response Type This section contains 6 questions (Q.15 to 20). The answer to each of the questions is a Single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the OMR. +4 marks will be given for each correct answer and no negative marking for each wrong answer.

30º

C 3f/2

(R) 80

(D) Maximum possible height

B

V0 64

MARCH 2012

MOCK TEST - AIEEE Time : 3 Hours

Total Marks : 360

Instructions : • There are three parts in question paper A, B, C consisting of chemistry, Physics & Mathematics having 30 questions in each part of equal weightage. Each question is allotted four marks for each correct response.

•

1/4 (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

CHEMISTRY (Part-A) 1.

2.

3.

4.

5.

6.

Consider the following statements : I. Atomic hydrogen is obtained by passing hydrogen through an electric arc. II. Hydrogen gas will not reduce heated aluminium oxide III. Finely divided palladium adsorbs large volume of hydrogen gas IV. Pure nascent hydrogen is best obtained by reacting Na with C2H5OH Which of the above statements is/are correct ? (1) I alone (2) II alone (3) I, II and III (4) II, III and IV When ethanal reacts with CH3MgBr and C2H5OH/dry HCl, the product formed are (1) ethyl alcohol and 2-Propanol (2) ethane and hemi-acetal (3) 2-propanol and acetal (4) propane and ethyl acetate

A drug that is antipyretic as well as analgesic is(1) Chloroquin (2) Penicillin (3) Paracetamol (4) Chloropromazine hydrochloride

8.

Which of the following can possibly be used as analgesic without causing addiction and modifications ? (1) Morphine (2) N-Acetyl-para-aminophenol (3) Diazepam (4) Tetrahydrocatenol

9.

Na2HPO4 is used to test (1) Ca+2 (3) Ni+2

(2) Ba+2 (4) Mg+2

10. 1 mole of N2 & 3 mole of H2 filled in one litre bulb were allowed to reaction when the reaction attained equilibrium, two third of N2 converted to NH3. If a hole is then made in the bulb, the mole ratio of the gases N2, H2 & NH3 effusing out initially would be respectively -

How many moles of CH3I will react with one mole of the ethylamine to form a quarternary salt ? (1) 2 (2) 3 (3) 4 (4) 5

(1) 1 : 3 : 4 (3)

26.8 g of Na2SO4⋅nH2O contains 12.6 g of water . The value of ‘n’ s (1) 1 (2) 10 (3) 6 (4) 7

1 28

:

(2) 1 2

:

1 17

(4)

28 :

2 : 17

1

3

28

:

2

:

4 17

11. Which of the following is not a state function (1) q + w (2) Q/T (3) E + PV (4) Q/W

The concentration of oxalic acid is‘x’ mol litre–1 40 mL of this solution reacts with 16 mL of 0.05 M acidified KMnO4. What is the pH of 'x' M oxalic acid solution (Assume that oxalic acid dissociates completely) (1) 1.3 (2) 1.699 (3) 0.05 (4) 2 Which pair of species given below produce bakelite? (1) phenol, methanol (2) phenol, NaOH (3) phenol, urea (4) phenol, formaldehyde

XtraEdge for IIT-JEE

7.

12. The relative rate of acid catalysed dehydration of following alcohols would be CH3 {P} Ph – CH – CH – CH3 OH CH3 {Q} Ph – CH – CH2 – CH2 – OH

65

MARCH 2012

20. Which of the following contains minimum number of lone pairs around Xe atom ? (1) XeF4 (2) XeF6 (3) XeOF2 (4) XeF2

CH3 {R} Ph – C – CH2 – CH3 OH CH3

21. The two isomers X and Y with the formula Cr(H2O)5CIBr2 were taken for experiment on depression in freezing point. It was found that one mole of X gave depression corresponding to 2 moles of particles and one mole of Y gave depression due to 3 moles of particles. The structural formulae of X and Y respectively are (1) [Cr(H2O)5Cl]Br2; [Cr(H2O)4Br2]Cl⋅H2O (2) [Cr(H2O)5Cl]Br2; [Cr(H2O)3CIBr2] ⋅2 H2O (3) [Cr(H2O)5Br]BrCl; [Cr(H2O)4CIBr] Br ⋅H2O (4) [Cr(H2O)4Br2]Cl⋅H2O; [Cr(H2O)5Cl]Br2

{S} Ph – C – CH2 – OH CH3 (1) R > P > Q > S (2) P > R > S > Q (3) R > S > P > Q (4) R > S > Q > P

O CH2

13. The ether

when treated with

HI produces – [I]

CH2 I

[III]

I

[IV]

(1) I and III (3) I and IV

[II]

CH2 – OH

OH

22

(2) Only II (4) Only III

H2C=CH–C–H (Q) CH3

H2C=CH2 (P)

14. 40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) 2NH3 (g) at constant temperature and + 3H2 (g) pressure. Then the ratio of the final volume to the initial volume of gases are : (1) 4 : 5 (2) 5 : 4 (3) 7 : 10 (4) 8 : 5

CH3–CH2–HC=CH2 (R)

CH3–C=CH2 (S)

(1) S > R > P > Q (3) P > Q > R > S

15. ∆Gº for the reaction, X + Y Z is – 4.606 kcal. The equilibrium constant for the reaction at 227ºC is (1) 100 (2) 10 (3) 2 (4) 0.01 23.

16. According to Bronsted Lowry concept, the correct order of strength of bases follows the order : (1) CH3COO– > OH– > Cl– (2) OH– > CH3COO– > Cl– (3) CH3COO– > Cl– > OH– (4) OH– > Cl– > CH3COO–

(2) S > P > Q > R (4) P > Q > S > R

(i) O3 (i) Na/ NH3(l) (A) (ii) H2O-Zn (ii) C2H5OH

Product

Product will be (1) CHO–CHO (2) CHO–CH2–CHO O O (3) CH3–C–C–CH3

17. The element having the highest ionization energy has the outer shell configuration as (1) ns2 np3 (2) ns2 np6 2 (3) ns (4) ns2 np5 18. Which of the following processes is exothermic ? (1) EA of N (2) IE of O– (3) EA of Cl (4) IE of Cl

(4) CHO–CHO & CHO–CH2–CHO 24. In the reaction Br CH3–CH–CH2–Br

(i) X mole NaNH2 (ii) C2H5Br

The value of [X] is (1) One (2) Two

19. Which of the following is not correct ? (1) XeO3 has pyramidal shape (2) The hybrid state of Xe in XeF4 is sp3 d2 (3) In calcium carbide, between carbon atoms one sigma and two π-bonds are present (4) In silica(SiO2), one Si atom is attached with two oxygen atoms

XtraEdge for IIT-JEE

Rates of addition of Cl2/H2O of the following alkenes are O

CH3–C≡C–C2H5

(3) Three (4) Four

25. Relate the following compounds F F Br F C C C Br R Cl R Cl R Cl Br

66

F C

Cl Br S

MARCH 2012

(1) Identical (3) Diastereomers

(2) Enantiomer (4) Meso

∞

33. If Sλ =

∑λ r =0

OH

r

n

, then

∑ (λ − 1)S

(1)

CH3

λ

=

λ =1

n(n − 1) 2 n(n + 2) (3) 2

26. The correct order of acidic strength of following acid is COOH COOH COOH COOH OH HO

1

(2)

n(n + 1) 2

(4) none of these k

OH (IV) (2) I > II > III > IV (4) II > IV > I > III

(II)

(I)

2π 2π cos − sin 7 7 = 1 0 , Then the least 34. If 0 1 sin 2π cos 2π 7 7 positive integral value of k is (1) 6 (2) 7 (3) 3 (4) 4

(III)

(1) II > I > III > IV (3) II > I > IV > III

27. In the phenomenon of osmosis, the semipermeable membrane allows the passage of (1) solute particles (2) solvent molecules only (3) both solute and solvent (4) none

35. The number of 2 digit numbers , which are of the form xy with y < x are given by (1) 45 (2) 55 (3) 17 (4) None 36. Let A, B, C be three independent events such that 1 1 1 P(A) = , P(B) = , P(C) = . Then probability of 3 2 4 exactly two events occuring out of three events is (1) 1/2 (2) 1/3 (3) 1/4 (4) none of these

28. What is the contribution of the atom present at the edge centre to the cubic unit cell ? (1) 1/2 (2) 1/4 (3) 1/8 (4) 1 29. In the cell, Zn | Zn2+ | | Cu2+ | Cu, the negative terminal is (1) Cu (2) Cu2+ (3) Zn (4) Zn2+

37. If a variable x takes values xi such that a ≤ xi ≤ b, for i = 1, 2, ---- n, then (1) a ≤ Var (x) ≤ b (2) a2 ≤ Var (x) ≤ b2

(3)

30. Assign double bond configuration of the following COOH CH2OH NC H2N–H2C

(2) Z, Z (4) Z, E

MATHEMATICS (Part-B) 31. f(x) = log

x2

39. If it is possible to draw a line which belongs to all the given family of lines y – 2x + 1 + λ1 (2y – x – 1) = 0, 3y – x– 6 + λ2 (y – 3x + 6) = 0 and ax + y – 2 + λ3 (6x + ay – a) = 0 , then (1) a = 4 (2) a = 3 (3) a = – 2 (4) a = 2

25 and g(x) = logx 5 then f(x) = g(x)

holds for x belonging to (1) R (2) (0, 1)∪ (1, ∞) (4) none (3) φ

40. If (α, 0) is an interior point of ∆ABC formed by the lines x – y = 0, 4x + 3y – 12 = 0 and y + 2 = 0 then integral values of α are (1) 1 (2) 2 (3) 4 (4) 8

32. The area enclosed by the parabola y2 = 4ax between the ordinates x = a and x = 9a is -

(1) 8a2

(2) 108

XtraEdge for IIT-JEE

a2 3

(3) 208

a2 3

(4) (b – a)2 ≥ Var (x)

38. A variable line has it's intercepts on the coordinate e e' , are eccentricies of axes e, e' where 2 2 hyperbola and it's conjugate hyperbola then the line always touches the circle x2 + y2 = r2, where r = ? (1) 1 (2) 2 (3) 3 (4) cannot be decided

CN

(1) E, Z (3) E, E

a2 ≤ Var (x) 4

(4) a2

67

MARCH 2012

48. If A = {1, 2, 3} and B = { 4, 5, 6} then which of the following sets are relation from A to B ? (i) R1 = {(4, 2) (2, 6) (5, 1) (2, 4)} (ii) R2 = {(1, 4) (1, 5) (3, 6) (2, 6) (3, 4)} (iii) R3 = {(1, 5) (2, 4) (3, 6)} (iv) R4 = {(1, 4) (1, 5) (1, 6)} (1) R1, R2, R3 (2) R1, R3, R4 (3) R2, R3, R4 (4) R1, R2, R3, R4

41. ABCD is a square of unit area. A circle is tangent to two sides of ABCD and passes through exactly one of its vertices. The radius of the circle is -

(1) 2 – (3)

(2)

2

1 2

(4)

2 –1 1 2

42. If A, B, C, D are four points in space satisfying

AB . CD = k[| AD |2 + | BC |2 –| AC |2 –| BD |2] then the value of k is (1) 2 (2) 1/3 (3) 1/2 (4) 1 r r r 43. Unit vectors a , b and c are coplanar. A unit vector r d is perpendicular to them . If r r r r 1 1 1 (a × b ) × (c × d ) = iˆ − ˆj + kˆ 6 3 3 r r r and the angle between a and b is 30º, then c is (iˆ − 2 ˆj + 2kˆ) (2iˆ + ˆj − kˆ) (1) (2) 3 3 ˆ (−iˆ + 2 ˆj − 2kˆ) (−2iˆ − 2 ˆj + k ) (3) (4) 3 3

49. If the area of the triangle on the complex plane formed by the complex numbers z, ωz, z + ωz is 3 square units then |z + ωz| equals 100

(1) 5 50. f (x) = x

222

(2) – 20 ±

(3) – 20 ±

224

(4) none of these

π 2

221

52.

4ax − x 2 , (a > 0) then f (x) is -

sin 2 x

∫ cos

dx is a x (1) polynomial of degree 5 in sin x (2) polynomial of degree 4 in tan x (3) polynomial of degree 5 in tan x (4) polynomial of degree 5 in cos x 6

53. The real value of m for which the substitution, y = um will transform the differential equation, dy 2x4y + y2 = 4x6 into a homogeneous equation is dx (1) m = 0 (2) m = 1 (3) m = 3/2 (4) no value of m

(4) 0

51/ x ; x < 0 46. Let f(x) = ; λ ∈ R then at x = 0 λ[ x] ; x ≥ 0

(1) f is discontinuous (2) f is continuous only if λ = 0 (3) f is continuous whatever λ may be (4) None

54. The mirror image of the parabola y2 = 4x in the tangent to the parabola at the point (1, 2) is (1) (x – 1)2 = 4(y + 1) (2) (x + 1)2 = 4(y + 1) (3) (x + 1)2 = 4(y – 1) (4) (x – 1)2 = 4(y – 1)

47. If R = {(x, y) : x, y ∈ Z} , x2 + y2 ≤ 4 is relation in Z, then DR is (1) {–2, –1, 0, 1, 2} (2) {–2, – 1, 0} (3) {0, 1, 2} (4) None of these XtraEdge for IIT-JEE

(4) |ωz|

51. Let f(x) be a twice differentiable function for all real values of x & satisfies f (1) = 1, f (2) = 4, f (3) = 9. then which of the following is true (1) f "(x) = 2 for ∀ x ∈ (1, 3) (2) f "(x) = f ' (x) = 5 for some x ∈ (2, 3) (3) f "(x) = 3 ∀ x ∈ (2, 3) (4) f "(x) = 2 for some x ∈ (1, 3)

45. The direction ratios l, m, n of two lines are connected by the relations l + m + n = 0 and lm = 0 then angle between them is π π (1) (2) 3 4

(3)

(3) |z|

(1) Increasing in (0, 3a), decreasing in (– ∞, 0) ∪ (3a, ∞) (2) Increasing in (a, 4a) decreasing in (5a, ∞) (3) Increasing in (0, 4a), decreasing in (– ∞, 0) (4) None of these

44. If the shortest distance between the lines z −3 x −1 y z x +1 y L1 : = = and L2 : = = is λ 1 −1 2 2 2 unity then λ is equal to -

(1) – 20 ±

(2) 1/5

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MARCH 2012

55. Let f(x) = minimum ({x + 1}, {x – 1}) ∀ x ∈ R,

PHYSICS (Part-C)

4

where{.} denotes the fractional part, then

∫

f (x) d(x) =

These questions of two statements each, printed as statement-1 & statement-2. While answering these Questions you are required to choose any one of the following four responses. (1) If both statement-1 and statement-2 are true but statement-2 is not a correct explanation of the statement-1. (2) If both statement-1 & statement-2 are true & the statement-2 is a correct explanation of the statement-1. (3) If statement-1 is true and statement-2 is false. (4) If statement-1 is false and statement-2 is true.

−5

(1) 9/2

(2) –

1 2

(3) 9

(4) None

Statement based questions : (Q. No. 56 to 60) Each of these questions contains two statements. Statement-I and Statement-II. Each of these has four alternatives choices. You have to select the correct choice. (1) If both statement-I and statement-II are true but statement-II is not the correct explanation of statement-I. (2) If both statement-I and statement- II are true, and statement-II is correct explanation of Statement-I. (3) If statement-I is true but statement-II is false. (4) If statement-I is false but statement-II is true

61. Statement–1 : Average energy in the interference pattern is the same as it would be if there were no interference. Statement–2 : Interference is the only rare phenomenon in which law of conservation of energy does not hold good.

56. Statement I : Let h(x) =xm/n for x ∈ R, where m & n are odd No. & 0 < m < n then y = h (x) has no Extreme. Statement II : If h' (x) does not change sign in neighborhood of x = a then x = a is not an Extreme pt.

62. A light ray is incident upon a prism in minimum deviation position and suffers a deviation of 34º. If the shaded half of the prism is knocked of, the ray will –

57. Statement I : If a2 x4 + b2 y4 = c6 then maximum c3 value of xy is . 2ab

(1) suffer a deviation of 34º (2) suffer a deviation of 68º (3) suffer a deviation of 17º (4) not come out of the prism

Statement II : For any + ve f(x), AM ≥ GM 58. Let A and B are two independent events in a sample space. Statement I : If P(A) = 0.3, P(B) = 0.4, Then P (A∩ B ) = 0.18.

63. A father (60 kg) and his daughter (20 kg) are both at rest on a frictionless ice pond. The father lifts a 1 kg ball and throws it to his daughter with horizontal speed 5 ms–1 ; the daughter catches it. The speeds of father and daughter are (in ms–1) is (1) 1/12, 5/21 (2) 5/59, 1/4 (3) 5/61, 1/4 (4) 5/59, 5/21

Statement II : P(A∩ B ) = P(A) – P(A) P(B) 59. Statement I : If p and q two statements then contra positive of conditional statement ~ ( p ∧ q) → q is (∧ q →~ p ∧ ~ q ) . Statement II : If ~ (p ∧ q) = ~ p∨ ~ q.

64. A large sheet carries uniform surface charge density σ. A rod of length 2l has a linear charge density λ on one half and −λ on the other half. The rod is hinged at mid-point O and makes angle θ with the normal to the sheet. The torque experienced by the rod is +λ

60. Let C1 be the circle with centre O1(0, 0) and radius 1 and C2 be the circle with centre O2 (t, t2 + 1), t ∈ R and radius 2 Statement I : Circle C1 and C2 always have at least one common tangent for any value of t. Statement II : For the two circles O1O2 ≥ |r1 – r2| where r1 and r2 are their radii for any value of t.

O

θ

+σ –λ

XtraEdge for IIT-JEE

69

MARCH 2012

(1)

σλl 2 cos θ 2ε 0

(2)

σλl cos2θ ε0

(3)

σλl 2 sin θ 2ε 0

(4)

σλl sin2θ ε0

Passage : (Q. No. 70 to 72) Bulb 2

Bulb 1

65. A body cools from 60ºC to 50ºC in 10 minutes. If the room temperature is 25ºC and assuming Newton's law of cooling to hold good the temperature of the body at the end of the next 10 minutes will be (1) 38.5ºC (2) 40ºC (3) 42.85ºC (4) 45ºC

+

12 volt

66. A passenger is at a distance of x from a bus when the bus begins to move with constant acceleration a. What is the minimum velocity with which the passenger should run towards the bus so as to reach it –

(1)

(2) 2ax

2ax

(3)

ax

A 12 volt battery is connected in two light bulbs, as shown in figure. Light bulb 1 has resistance 3Ω while light bulb 2 has resistance 6Ω. The battery has essentially no internal resistance and all the wires are essentially resistanceless too. When a light bulb is unscrewed, no current flows through that branch of the circuit. For instance, if light bulb 2 is unscrewed, current flows only around the lower loop of the circuit, which consists of the battery and light bulb 1. When two resistance are joined in series, their equivalent resistances Req. = R1 + R2 but when two resistances are wired in parallel. Their net resistance is given by: 1 1 1 = + Req. R1 R2

(4) ax

67. A bus is moving with a speed of 10 ms–1 on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase the bus ? (2) 25 ms–1 (1) 40 ms–1 (4) 20 ms–1 (3) 10 ms–1

6'8. For a equilateral glass prism the angle of minimum deviation is 30°, then refractive index of prism is (1)

1

(2)

2

(3)

2

1 2

(4)

3 2

70. When bulb 1 is screwed in, but bulb 2 is unscrewed, the power generated in bulb 1 is (1) 4 watt (2) 12 watt (3) 36 watt (4) 48 watt

69. A zener diode is to be used as voltage regulator. Identify the correct set up -

(1)

⊕

RS

71. Bulb 2 is now screwed in, as a result, bulb 1 (1) turns off (2) becomes dimmer (3) stays about the same brightness (4) becomes brighter

RL Θ

(2)

⊕

RS RL

Θ

⊕

72. When both light bulbs are screwed in, the current through the battery is (1) 1.2 ampere (2) 2 ampere (3) 4 ampere (4) 6 ampere

RS

(3)

RL Θ

⊕

73. String 1 has twice the length, twice the radius, twice the tension and twice the density of another string 2. The relation between the fundamental frequencies of 1 and 2 is : (2) f1 = 4f2 (1) f1 = 2f2 (4) f1 = f2 (3) f2 = 4f1

RS

(4)

RL Θ

XtraEdge for IIT-JEE

–

70

MARCH 2012

74. When a source of sound of frequency f crosses a stationary observer with a speed vs (<< speed of sound v), the apparent change in frequency ∆f is given by 2 fv s (1) (2) 2 fv vs v 2f v (3) vs

82. A man running has half the kinetic energy of a boy of half his mass. The man speeds up by 1 ms–1 and then has KE as that of the boy. What were the original speeds of man and the boy ?

(2) ( 2 – 1) ms–1; 2( 2 – 1) ms–1

f vs (4) v

(3) ( 2 + 1) ms–1; 2( 2 + 1) ms–1 (4) None of the above

75. N atoms of a radioactive element, emit n, α particles per second. Half-ilfe of the element is (1) n/N (2) N/n (3) 0.693 N/n (4) 0.693 n/N

Passage Based Questions : (Q.83 to Q.84)

The centre of mass of a body is a point at which the whole mass of the body is supposed to be concentrated. If the body consists of two particles of

76. If Ma is the mass of an oxygen isotope 8O17, Mp and Mn are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is (1) (Ma – 8Mp)c2 (2) (M0 – 8Mp – 9Mn)c2 (4) (M0 – 17 Mn)c2 (3) M0c2

→

→

masses m1, m2 with r1 and r2 as their position vectors, then the position vector of the centre of mass is. →

→

……. rCM =

77. The photoelectric threshold of Tungsten is 2300 Å. the energy of the electrons ejected from the surface by ultraviolet light of wavelength 1800 Å is : (h = 6.6 × 10–34 J-s) (1) 0.15 eV (2) 1.5 eV (3) 15 eV (4) 150 eV

→

m1 r2 + m2 r2 m1 + m2

Similarly, velocity of centre of mass of two particles →

→

moving with velocities v1 and v2 is →

→

m v + m2 v 2 = 1 1 , and m1 + m2

→ vCM

Acceleration of centre of mass of two particles

78. In a transistor amplifier, β = 62, RL = 5000Ω and internal resistance of the transistor is 500 Ω. The voltage amplification of the amplifier will be (1) 500 (2) 620 (3) 780 (4) 950

→

→

having acceleration a1 and a2 is → aCM

79. A planet is moving in an elliptical orbit. If T, V, E and L represent respectively the kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of the orbit, then which of the following is correct – (1) T is conserved (2) V is always positive (3) E is always negative

→

→

m a + m2 a 2 = 1 1 m1 + m2

For an isolated system, where no external force acts, →

→

aCM = 0 and vCM = constant.

Read the above paragraph and answer the following questions. 83. Two bodies of masses 2 kg and 4 kg are moving towards each other with equal velocity 5 m/s. The velocity of centre of mass is 5 5 (1) – ms–1 (2) ms–1 3 3

→

(4) L is conserved but the direction of L is continuously changing 80. The motion of particle executing SHM is given by x = 0.01 sin (100 π t + 0.05), where x is in metre and time t is in second. The time period is – (1) 0.2 s (2) 0.1 s (3) 0.02 s (4) 0.01 s

(3)

3 –1 ms 5

(4) –

3 ms–1 5

84. An electron and a proton move towards each other with velocities v1 and v2 due to mutual attraction. The velocity of their centre of mass is v +v v –v (1) 1 2 (2) 1 2 2 2 (4) zero (3) v2

81. If the surface tension of water is 0.07 N m–1, the weight of water supported by surface tension in a capillary tube of radius 0.1 mm is – (2) 22 µN (1) 11 µN (3) 33 µN (4) 44 µN XtraEdge for IIT-JEE

2 ms–1;2 2 – 1 ms–1

(1)

71

MARCH 2012

85. If the space between the lenses in the lens combination shown were filled with water, what would happen to the focal length and power of the lens combination ?

MOTIVATION •

Pull the string, and it will follow wherever you wish. Push it, and it will go nowhere at all.

•

Be the change that you want to see in the world.

•

Efficiency is doing things right; effectiveness is doing the right things.

•

Formula for success: under promise and over deliver.

•

A life spent making mistakes is not only more honorable, but more useful than a life spent doing nothing.

•

Discovery consists of seeing what everybody has seen and thinking what nobody else has thought.

•

The best way to teach people is by telling a story.

88. The work done in turning a magnet of magnetic moment M by an angle of 90º from the meridian is n times the corresponding work done to turn it through an angle of 60º. The value of n is given by – (1) 1 (2) 1/4 (3) 4 (4) 2

•

If you'll not settle for anything less than your best, you will be amazed at what you can accomplish in your lives.

•

I had to pick myself up and get on with it, do it all over again, only even better this time.

89. As compared to ordinary diode an zener diode is (1) also connected in F.B (2) also connected in F.B as well as RB (3) always connected in RB (4) exactly same

•

Improvement begins with I.

•

Success depends upon previous preparation, and without such preparation there is sure to be failure.

•

The man of virtue makes the difficulty to be overcome his first business, and success only a subsequent consideration.

•

As a general rule the most successful man in life is the man who has the best information.

•

The secret of success is constancy to purpose.

•

One secret of success in life is for a man to be ready for his opportunity when it comes.

Focal length Power (1) Decreased increased (2) Decreased unchanged (3) Increased unchanged (4) Increased decreased 86. If two coils have self-inductances L1 and L2, the coefficient of mutual induction will be –

(1) M ∝

L1 L2

(3) M ∝

L2 L1

(2) M ∝

L1 L2

(4) None of these

87. Vapour pressure at any temperature is equal to saturated vapour pressure : (1) At that temperature (2) At dew point (3) At boiling point (4) At freezing point

90. There are two identical concentric coils X and Y with their planes at right angles to each other. The coil X lies in the horizontal plane while coil Y lies in the vertical plane. If the coil X carries a current of 1 A then what value of current in coil Y be passed so that the resultant field at the centre of the coils just balances the earth's magnetic field of 10–5 tesla inclined at 30º with the vertical ?

(1) 1 A

(2)

(3) 2 A

(4) (1/ 3 )A

XtraEdge for IIT-JEE

3A

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MARCH 2012

XtraEdge for IIT-JEE

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MARCH 2012

MOCK TEST – BIT-SAT Time : 3 Hours

Total Marks : 450

Instructions :

•

This question paper contains 150 questions in Physics (40) Chemistry (40), Mathematics (45), Logical Reasoning (10) & English (15). There is Negative Marking

•

Each question has four option & out of them, ONLY ONE is the correct answer. There is – ve marking.

•

+3 Marks for each correct & – 1 Mark for the incorrect answer.

PHYSICS 1.

2.

3.

A particle moving along x-axis has acceleration f, at t time t, given by f = f0 1 – , Where f0 and T are T constants. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle's velocity (vx) is 1 (A) f0 T 2 (B) f0 T 2 2 1 (D) f0 T (C) f0 T 2 Two forces P and Q acting at a point are such that if P is reversed, the direction of the resultant is turned through 90°. Then (A) P = Q (B) P = 2Q Q (C) P = 2 (D) No relation between P and Q Three particles A, B and C of equal mass, move with equal speed v along the medians of an equilateral triangle as shown in Fig. They collide at the centroid G of the triangle. After collision, A comes to rest and B retraces its path with speed v. What is the speed of C after collision ? A

4.

A man throws bricks to a height of 12 m where they reach with a speed of 12 ms–1. If he throws the bricks such that they just reach that height, what percentage of energy will be saved ? (A) 9 % (B) 19 % (C) 38 % (D) 46 %

5.

If the Earth shrinks to half of the present radius, without any change in mass, then the duration of day and night becomes (A) 24 hours (B) 12 hours (C) 6 hours (D) 3 hours

6.

Two white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. What is the maximum distance at which these dots can be resolved by the eye ? (λ =500 nm) (A) 1 m (B) 5 m (C) 3 m (D) 6 m

7.

An object is placed at a distance of 10 cm from a concave mirror of radius of curvature 0.6 m. Which of the following statements is incorrect ? (A) The image is formed at a distance of 15 cm from the mirror (B) The image formed is real (C) The image is 1.5 times the size of the object (D) The image formed is virtual and erect

8.

Two bodies of masses 1 kg and 3 kg have position vectors iˆ + 2 ˆj + kˆ and – 3iˆ – 2 ˆj + kˆ , respectively. The centre of mass of this system has a position vector (B) – 2iˆ + 2kˆ (A) – iˆ + ˆj + kˆ (C) – 2iˆ – ˆj + kˆ

G

9. C

B

(A) 0 (C) v XtraEdge for IIT-JEE

v 2 (D) 2v

(B)

74

(D) 2iˆ – ˆj – 2kˆ

A perfect gas at 27°C is heated at constant pressure so as to triple its volume. The temperature of the gas will be (A) 81°C (B) 900°C (C) 627°C (D) 450°C MARCH 2012

10. An ideal gas of mass m in a state A goes to another state B via three different processes as shown in figure. If Q1, Q2 and Q3 denote the heat absorbed by the gas along the three paths, then P A 1 2

(A) Q1 < Q2 < Q3 (C) Q1 = Q2 > Q3

15. In a galvanometer 5% of the total current in the circuit passes through it. If the resistance of the galvanometer is G, the shunt resistance S connected to the galvanometer is G G (A) 19 G (B) (C) 20 G (D) 19 20

3

16

B V (B) Q1 < Q2 = Q3 (D) Q1 > Q2 > Q3

11. Two sound waves (expressed in CGS units) given by 2π 2π (vt – x) and y2 = 0.4 sin (vt – x + y1 = 0.3 sin λ λ θ) interfere. The resultant amplitude at a place where phase difference is π/2 will be (A) 0.7 cm (B) 0.1 cm 1 (C) 0.5 cm (D) 7 cm 10

17. The number of turns in primary and secondary coils of a transformer is 50 and 200 respectively. If the current in the primary coil is 4 A, then the current in the secondary coil is (A) 1 A (B) 2 A (C) 4 A (D) 5 A

12. A tuning fork of frequency 100 when sounded together with another tuning fork of unknown frequency produces 2 beats per second. On loading the tuning fork whose frequency is not known and sounded together with a tuning fork of frequency 100 produces one beat, then the frequency of the other tuning fork is (A) 102 (B) 98 (C) 99 (D) 101

18. Which of the following statements is not correct when a junction diode is in forward bias ? (A) The width of depletion region decreases (B) Free electrons on n-side will move towards the junction. (C) Holes on p-side move towards the junction. (D) Electron on n-side and holes on p-side will move away from junction.

13. A current of 2 A flows in an electric circuit as shown in figure. The potential difference (VR – VS), in volts (VR and VS are potentials at R and S respectively) is R 3Ω

19. The displacement of a charge Q in the electric field r E = e1iˆ + e2 ˆj + e3 kˆ is rr = aiˆ + bˆj . The work done is (A) Q (ae1 + be2)

7Ω

P

(B) Q (ae1 ) 2 + (be2 ) 2

Q

2A

(C) Q(e1 + e2 ) a 2 + b 2

2A 7Ω

3Ω

(D) Q( e12 + e22 )(a + b)

S

(A) – 4

(B) + 2

(C) + 4

20. An electric line of force in the xy plane is given by equation x2 + y2 = 1. A particle with unit positive charge, initially at rest at the point x = 1, y = 0 in the xy plane (A) not move at all (B) will move along straight line (C) will move along the circular line of force (D) Information is insufficient to draw any conclusion

(D) – 2

14. When a battery connected across a resistor of 16 Ω, the voltage across the resister is 12 V. When the same battery is connected across a resistor of 10 Ω, voltage across it is 11V. The internal resistance of the battery (in ohm) is 10 20 25 30 (B) (C) (D) (A) 7 7 7 7 XtraEdge for IIT-JEE

Two concentric coils of 10 turns each are placed in the same plane. Their radii are 20 cm and 40 cm and carry 0.2 A and 0.3 A current respectively in opposite directions. The magnetic induction (in tesla) at the centre is 3 5 (A) µ0 (B) µ0 4 4 7 9 (C) µ0 (D) µ0 4 4

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MARCH 2012

21. The relation between voltage sensitivity (σv) and current sensitivity (σi) of a moving coil galvanometer is (resistance of galvanometer is G). σ σ (A) i = σv (B) v = σi G G

(C)

G = σi σv

(D)

28. A dielectric is introduced in a charged and isolated parallel plate capacitor, which of the following remains unchanged ? (A) Energy (B) Charge (C) Electric field (D) Potential difference

G = σv σi

29. Positively charged particles are projected into a magnetic field. If the direction of the magnetic field is along the direction of motion of the charge particles, the particles get : (A) Accelerated (B) Decelerated (C) Deflected (D) no changed in velocity

22. An inductor of 2H and a resistance of 10Ω are connected in series with a battery of 5V. The initial rate of change of current is (A) 0.5 A/s (B) 2.0 A/s (C) 2.5 A/s (D) 0.25 A/s 23. A and B are two radioactive substances whose halflives are 1 and 2 years respectively. Initially 10 g of A and 1 g of B is taken. The time (approximate) after which they will have same quantity remaining is (A) 6.62 year (B) 5 year (C) 3.2 year (D) 7 year

30. Fusion reaction takes place at high temperature because : (A) KE is high enough to overcome repulsion between nuclei (B) Nuclei are most stable at this temperature (C) Nuclei are unstable at this temperature (D) None of the above 31. Among the following properties describing diamagnetism identify the property that is wrongly stated : (A) Diamagnetic material do not have permanent magnetic moment (B) Diamagnetism is explained in terms of electromagnetism induction (C) Diamagnetic materials have a small positive susceptibility (D) The magnetic moment of individual electrons neutralize each other.

24. In the circuit, the potential difference across PQ will be nearest to 100Ω

48V 100Ω

80Ω

Q

20Ω P

(A) 9.6 V (C) 4.8 V

(B) 6.6 V (D) 3.2 V

25. Radiations of intensity 0.5 W/m2 are striking a metal plate. The pressure on the plate is (A) 0.166 × 10–8 N/m2 (B) 0.332 × 10–8 N/m2 (C) 0.111 × 10–8 N/m2 (D) 0.083 × 10–8 N/m2

32. Electron of mass m and change q is travelling with a speed v along a circular path of radius r at right angles to a uniform magnetic field of intensity B. If the speed of the electron is doubled and the magnetic field is halved the resulting path would have a radius. (A) 2r (B) 4r (C) r/4 (D) r/2

26. A cell of constant emf first connected to a resistance R1 and then connected to a resistance R2. If power delivered in both cases is same then the internal resistance of the cell is :

(A)

R1R2

(B)

(C)

R1 − R2 2

(D)

33. The density of water at 0°C is 0.998 g/cc. While at 4°C it is 1 g/cc. The average coefficient of volume expansion of water in the temperature range 0°C to 4°C is (A) 5 × 10– 4/°C (B) – 5 × 10– 4/°C –4 (D) – 6 × 10– 4/°C (C) 6 × 10 /°C

R1 R2 R1 + R2 2

27. In a magnetic field of 0.05 T area of coil changes from 101 cm2 to 100 cm2 without changing the resistance which is 2Ω. The amount of charge that flow during this period is (A) 2.5 × 10–6 C (B) 2 × 10–6 C –6 (D) 8 × 10–6 C (C) 10 C XtraEdge for IIT-JEE

34. A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10 cm/ s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the snap-shot of the wave is shown in the figure. The velocity of point P when its displacement is 5 cm, is -

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MARCH 2012

y P

3π ˆ (A) j m/s 50

(C)

ML2 K KL2 (C) 2M

(A) x

(D) –

(C)

R V

(B)

R V

(D)

→

3π ˆ i m/s 50

365 days 2

R

The mass of CaCO3 formed by passing CO2 gas through 50 mL of 0.5 M Ca(OH)2 solution is (A) 10 g (B) 2 g (C) 2.5 (D) 5 g

2.

2g of hydrogen diffuse out from a container in 10 min. How many gram of chlorine will diffuse out from the same container under similar conditions ?

V

3.

2 days

(D) 365 × 4 days

ML2 3

(B)

ML2 6

(C)

ML2 9

(D)

ML2 12

(A)

2 × 71

(B)

2 71

(C)

71 2

(D)

71

Total volume of atoms present in face-centred cubic unit cell of a metal is (r = atomic radius) 20 3 24 3 πr (B) πr (A) 3 3 (C)

38. The moment of inertia of a thin uniform rod of length L and mass M about an axis passing through a point L from one of its ends and at a distance of 3 perpendicular to the rod is -

(A)

→

1.

V

37. If Earth describes an orbit round the Sun of double its present radius, the year on Earth will be of -

(C)

→

CHEMISTRY

R

(B) 365 × 2 ×

→

between B and C is (A) 0° (B) 90° (C) 180° (D) 270°

36. A diver is 10 m below the surface of water. The approximate pressure experienced by the diver is (A) 105 Pa (B) 2 × 105 Pa (D) 4 × 105 Pa (C) 3 × 105 Pa

(A) 365 days

→

MK L

40. Given : A . B = 0 and A × C = 0. The angle

35. Which of the following diagrams is a correct presentation of deviation and dispersion of light by prism ?

(A)

(D) →

3π ˆ (B) – j m/s 50

3π ˆ i m/s 50

(B) zero

39. The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is -

12 3 πr 3

(D)

16 3 πr 3

4.

A ball of 100 g mass is thrown with a velocity of 100 ms-1. The wavelength of the de Broglie wave associated with the ball is about (B) 6.63 × 10–30 m (A) 6.63 × 10–35 m (D) 6.63 × 10–33 m (C) 6.63 × 10–35 cm

5.

The unicertainty in the position of an electron moving with a velocity of 1.0 × 104 cm s–1 (accurate up to 0.011%) will be (A) 1.92 cm (B) 7.68 cm (C) 0.528 cm (D) 3.8 cm

6.

∆Hº for a reaction, F2 + 2HCl —→ 2HF + Cl2

is given to be –352.18 kJ. If ∆H 0f for HF is

M

–

268.3 kJ mol–1 the ∆H 0f of HCl would be (A) – 22 kJ mol–1 (C) – 91.9 kJ mol–1 XtraEdge for IIT-JEE

77

(B) 88.0 kJ mol–1 (D) – 183.8 kJ mol–1 MARCH 2012

7.

If CH3COOH(aq) + OH¯ (aq) —→ CH3COO¯ + H2O + q1, H+ + OH¯ —→ H2O (l) + q2 Then enthalpy change for the reaction, CH3COOH —→ CH3COO¯ + H+ is (B) q1 + q2 (A) q1 – q2 (D) q1 / q2 (C) q2 – q1

8.

In which of the following reaction Kp > Kc 2NH3 (A) N2 + 3H2 2HI (B) H2 + I2 (C) PCl3 + Cl2 PCl5 O2 + 2SO2 (D) 2SO3

9.

If the equilibrium constant for the reaction – A2 + B2 is 49, what is the value of 2AB equilibrium constant for 1 1 AB A2 + B2 2 2 (A) 49 (B) 2401 (C) 7 (D) 0.02

14. The rise in boiling point of a solution containing 1.8 g of glucose in 100 g of solvent is 0.1°C. The molal elevation constant of the liquid is (A) 0.01 K/m (B) 0.1 K/m (C) 1 K/m (D) 10 K/m. 15. The value of E ° Zn / Zn 2+ = 0.76 V and that of E ° Fe 2+ / Fe = – 0.41 V. The E°

16. The hydrogen electrode can exhibit electrode potential > 0 if (A) H2 is bubbled through the solution at 2 atm. pressure (B) concentration of H+ ion in solution is increased (C) concentration of H+ ions in solution is decreased

(D) concentration of H+ ions is decreased and simultaneously pressure of H2 gas is increased. +

17.

Rate

Rate

(B)

Br (C) both (a) and (b)

Br (D) none of these

OsO

H 2O2

(B) racemic diol (D) none of the above

19. Following compound is treated with NBS CH2CH=CH2+ NBS → A

Compound formed A is -

(A)2

(A) 13. For an elementary process 2X + Y —® Z + W (A) 2 (B) 1 (C) 3 (D) unpredictable.

XtraEdge for IIT-JEE

have

4 → A, A is -

Rate

Rate (A)2

will

Br

(A) meso diol (C) both (a) and (b) (A)2

(D)

A

(A)

18.

(C)

CCl 4 → A,

Br

12. Which of the following graph is for a second order reaction ?

(A)2

Br2

configuration :

11. What is the solubility of Al(OH)3, Ksp=1 × 10–33, in a solution having pH = 4 ? (B) 10–6 M (A) 10–3 M –4 (D) 10–10 M (C) 10 M

(B)

of the cell with net

cell reaction Zn + Fe2+ → Zn2+ + Fe is (A) – 0.35 V (B) – 1.17 V (C) + 1.17 V (D) + 0.35 V.

10. An equimolar solution of CH3COOH and CH3COONa has a pH of 6. The value of Kα for acetic acid is (B) 1 × 10–6 (A) 106 –6 (D) cannot be predicted (C) 2 × 10

(A)

cell

CHCH=CH2 Br

(B)

78

CH = CHCH2Br

MARCH 2012

27. Pinacol is : (A) 3-methylbutane-2-ol (B) 2,3-dimethyl-2,3-butanediol (C) 2,3-dimethyl-2-propanone (D) none of the above

CH2CH=CH2

(C)

Br

Br

CH2CH = CH2

(D)

28. Aldol condensation will not take place in : (A) HCHO (B) CH3CH2CHO (D) CH3COCH3 (C) CH3CHO

Br

20. Bicyclo (1, 1, 0) butane is (B) (A)

(C)

29. IUPAC name of the following compound is :

(D) H3C CH3 (A) 3,5-dimethylcyclohexene (B) 3,5-dimethyl -1-cyclohexene (C) 1,5-dimethyl-5-cyclohexene (D) 1,3-dimethyl-5-cyclohexene

21. Grignard reagent reacts with HCHO to produce: (A) secondary alcohol (B) anhydride (C) and acid (D) primary alcohol

30. The ions O2–, F–, Na+, Mg2+ and Al3+ are isoelectronic. Their ionic radii show. (A) A decrease from O2– to F– and then increase from Na+ to Al3+ (B) A significant increase from O2– to Al3+ (C) A significant decrease from O2– to Al3+ (D) An increase from O2– to F– and then decrease from Na+ to Al3+

22. Cyanohydrin of which of the following forms lactic acid : (A) HCHO (B) CH3CHO (D) CH3COCH3 (C) CH3CH2CHO 23. Isopropyl bromide on Wurtz reaction gives : (A) hexane (B) propane (C) 2, 3-dimethylbutane (D) neo-hexane

31. Which of the following values in electron volt per atom represent the first ionisation energies of oxygen and nitrogen atom, respectively ? (A) 14.6, 13.6 (B) 13.6, 14.6 (C) 13.6, 13.6 (D) 14.6, 14.6

24. On heating with oxalic acid at 110°C, glycerine gives: (A) glyceryl trioxalate (B) formic acid (C) glyceryl dioxalate (D) none of the above 25. The compound, whose stereo-chemical formula is written below, exhibits x geometrical isomers and y optical isomers. H CH3 OH C=C H CH2–CH2–C–CH3

32. Which of the following is not involved in any diagonal relationship ? (A) C (B) B (C) Al (D) Si 33. A metal M readily forms water soluble sulphate MSO4, water insoluble hydroxide M(OH)2 and oxide MO which becomes insert on heating. The hydroxide is soluble in NaOH. The M is (B) Mg (A) Be (D) Sr (C) Ca

H The values of x and y are : (A) 4 and 4 (B) 2 and 2 (C) 2 and 4 (D) 4 and 2

34. Plaster of paris is (A) CaSO4 (C) 2CaSO4.H2O

26. Which one of the following product is formed when calcium salt of adipic acid is heated ? CH2–CH2 (A) O CH2–CH2 CH2–CH2 C=O (B) CH2–CH2 CH2CH2CO (C) C=O CH2CH2CO CH2CH2COOH (D) CH2CH2COOH XtraEdge for IIT-JEE

(B) CaSO4.H2O (D) CaSO4.2H2O

35. In diborane (A) 4-bridged hydrogens and two terminal hydrogens are present (B) 2-bridged hydrogens and four terminal hydrogens are present (C) 3-bridged hydrogens and three terminal hydrogens are present (D) None of these

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36

Which of the following statements is correct ? (A) BCl3 and AlCl3 are both Lewis acids and BCl3 is stronger than AlCl3 (B) BCl3 and AlCl3 are both Lewis acids and AlCl3 is stronger than BCl3 (C) BCl3 and AlCl3 are both equally strong Lewis acids (D) Both BCl3 and AlCl3 are not Lewis acids

37. When sodium thiosulphate solution is made to react with copper sulphate solution, a complex compound is formed. What is that ? (A) Na4[Cu3(S2O3)5] (B) Na2[Cu6(S2O3)4] (C) Na4[Cu6(S2O3)5] (D) Na4[Cu3(S2O3)4]

4.

The centres of a set of circles, each of radius 3, lie on the circle x2 + y2= 25. The locus of any point in the set is (B) x2 + y2 ≤ 25 (A) 4 ≤ x2 + y2 ≤ 64 (D) 3 ≤ x2 + y2 ≤ 9 (C) x2 + y2 ≥ 25

5.

The pairs of straight lines x2 – 3xy + 2y2 = 0 and x2 – 3xy + 2y2 + x – 2 = 0 form a (A) square but not rhombus (B) rhombus (C) parallelogram (D) rectangle but not a square

6.

The distance between the foci of the hyperbola x2 – 3y2 – 4x – 6y – 11 = 0 is (A) 4 (B) 6 (C) 8 (D) 10

7.

If θ is the acute angle of intersection at a real point of intersection of circle x2 + y2 = 5 and the parabola y2 = 4x, then tan θ is equal to -

38. Which one of the following does not contain zinc ? (A) Brass (B) German silver (C) Gun metal (D) Bell metal 39. Stability constants for some copper complexes are given below : Cu+2, + 4NH3 [Cu(NH3)4]+2 K = 4.5 × 1011 +2 – [Cu(CN)4]–2 K = 2.0 × 1027 Cu , + 4CN +2 [Cu(en)2]+2 K = 9.5 × 1015 Cu , + 2en +2 [Cu(H2O)4]+2 K = 9.5 × 108 Cu , + 4H2O Which is the strongest ligand ? (B) CN– (A) NH3 (D) H2O (C) en 40. The pair [Co(NH3)5NO3]SO4 and [Co(NH3)5 SO4] NO3 will exhibit (A) Hydrate isomerism (B) Linkage isomerism (C) Ionisation isomerism (D) Coordinate isomerism

The transformed equation of 3x2 + 3y2 + 2xy = 2 when the coordinate axes are rotated through an angle of 45º, is (B) 2x2 + y2 = 1 (A) x2 + 2y2 = 1 2 2 (D) x2 + 3y2 = 1 (C) x + y = 1

3.

The length of the common chord of the ellipse ( x – 1) 2 ( y – 2) 2 + = 1 and the circle (x – 1)2 + (y–2 )2 = 1 9 4 is (A) 0

(B)

(C) 4

(D) 5

XtraEdge for IIT-JEE

(C) 3

(D) 1 / 3

3

8.

The tangents from a point (2 2 , 1) to the hyperbola 16x2 – 25y2 = 400 include an angle equal to (B) π/4 (C) π (D) π/3 (A) π/2

9.

The slope of a common tangent to the ellipse x2 y2 + = 1 and a concentric circle of radius ‘r’ a2 b2 is -

(C)

The point on the line 3x + 4y = 5 which is equidistant from (1, 2) and (3, 4) is (A) (7, – 4) (B) (15, – 10) (C) (1/7, 8/7) (D) (0, 5/4)

2.

(B)

(A) tan–1

MATHEMATICS 1.

(A) 1

r 2 – b2 a2 – r 2

r 2 – b2 a2 – r 2

(B)

r 2 – b2 a2 – r 2

(D)

a2 – r 2 r 2 – b2

10. If f : R → R and g : R → R are defined by f(x) = | x | and g (x) = [x – 3] for x ∈ R, then 8 8 g ( f ( x)) : – < x < is equal to 5 5 (A) {0, 1} (B) {1, 2} (C) {–3, –2} (D) {2, 3}

11. If f : R → R is defined by cos 3x – cos x , for x ≠ 0 f (x) = x2 λ , for x = 0

and if f is continuous at x = 0, then λ is equal to (A) – 2 (B) – 4 (C) – 6 (D) – 8

3

80

MARCH 2012

21. The length of the subtangent at (2, 2) to the curve x5 = 2y4 is 5 8 2 5 (A) (B) (C) (D) 2 5 5 8

12. The solution of the differential equation xy2dy – (x3 + y3) dx = 0 is 3 3 (B) y3 = 3x3 log (cx) (A) y = 3x + c (C) y3 = 3x3 + log (cx) (D) y3 + 3x3 = log (cx) 13. If

∫

θ 22. If x = a cos θ + log tan and y = a sin θ, then 2 dy is equal to dx (A) cot θ (B) tan θ (D) cos θ (C) sin θ

e x (1 + x).sec2 (xex) dx = f (x) + constant, then

f (x) is equal to (A) cos (xex) (C) 2 tan–1 (x)

(B) sin (xex) (D) tan (x ex)

14. If f : R → R is defined by f (x) = [x – 3] + | x – 4 | for x ∈ R, then lim– f (x) is equal to where [.] is G.I.F. -

23. The value of the expression 1 1 1 1 2 1 + 1 + 2 + 3 2 + 2 + 2 + ω ω ω ω

x→3

(A) – 2 15.

16.

(B) – 1

(C) 0

(D) 1

1 1 1 1 4 3 + 3 + 2 + ..... + (n + 1) n + n + 2 , ω ω ω ω

2x – 1 ∈ R equals x ∈ R : 3 x + 4 x 2 + 3x

(A) R – {0}

(B) R – {0, 1, 3}

where ω is an imaginary cube root of unity, is

(C) R – {0, –1, – 3}

1 (D) R – 0, – 1, – 3, + 2

(A)

n(n 2 + 2) 3

(B)

(C)

n 2 (n + 1) 2 + 4n 4

(D) None of these

d dx

1 x – 1 –1 a tan x + b log x + 1 = 4 x – 1

⇒ a – 2b is equal to (A) 1 (B) – 1 (C) 0

a –a a2 a3 a +a = 2 3 = 3 2 3 , then a1, a2, a3, a4 a1a4 a1 + a4 a1 – a4 are in (A) A.P. (B) G.P. (C) H.P. (D) None of these

24. If

(D) 2

17. The function f (x) = x3 + ax2 + bx + c, a2 ≤ 3b has (A) one maximum value (B) one minimum value (C) no extreme value (D) one maximum and one minimum value

x2 – 1 for every real number x then the x2 +1 minimum value of f (A) does not exist because f is unbounded (B) is not attained even though f is bounded (C) is equal to 1 (D) is equal to – 1

25. If f (x) =

18. Area of the region satisfying x ≤ 2, y ≤ | x | and x ≥ 0 is (A) 4 sq. unit (B) 1 sq. unit (C) 2 sq. unit (D) None of these

∫

19. If I1 =

I4 =

∫

2 1

1 0

x2

2 dx , I2 =

∫

1 0

x3

2 dx , I3 =

∫

2 1

26. If x + y and y + 3x are two factors of the expression λx3 – µx2y + xy2 + y3, then the third factor is (B) y – 3x (A) y + 3x (D) None of these (C) y – x

x2

2 dx and

3

2 x dx , then

(A) I3 > I4 (C) I1 > I2

27. The number of ways in which 30 marks can be allotted to 8 questions if each question carries at least 2 marks, is (A) 115280 (B) 117280 (C) 116280 (D) None of these

(B) I3 = I4 (D) I2 > I1

20. Let A = [–1, 1] and f : A → A be defined as f (x) = x | x | for all x ∈ A, then f (x) is (A) many-one into function (B) one-one into function (C) many-one onto function (D) one-one onto function XtraEdge for IIT-JEE

n(n 2 – 2) 3

28. The numerically largest term in the binomial 2 expansion of (4 – 3x)7, when x = is 3 (A) 46016 (B) 66016 (C) 86016 (D) None of these

81

MARCH 2012

r r 36. Let a and b be two r r r r r u = a – ( a . b ) b and to r (A) | u | r (C) 2| v |

2 4n 29. The fractional part of is 15 1 2 (B) (A) 15 15 4 (D) none of these (C) 15

–1 –1 (B) –1 –1

2 2 (C) 2 2

(D) None of these

5 + sin 2 x 2

31. If f(x) =

cos 2 x

4 sin 2 x

2

sin x

5 + cos x

4 sin 2 x

sin 2 x

cos 2 x

5 + 4 sin 2 x

(C)

(C)

1 36

5 6

5 (B) 6

40. {x ∈ R : cos 2x + 2 cos2 x = 2} is equal to π π (A) 2nπ + : n ∈ Z (B) nπ ± : n ∈ Z 3 6 π (C) nπ + : n ∈ Z 3

π (D) 2nπ – : n ∈ Z 3

3 4 π 41. If sin–1 + sin–1 = , then x is equal to x x 2 (A) 3 (B) 5 (C) 7 (D) 11

4

2

(D) None of these

42. In ∆ABC, if

1 1 3 + = , then C is b+c c+a a+b+c

equal to (A) 90º (B) 60º

34. Three persons A, B and C are to speak at a function along with five others. If they all speak in random order, the probability that A speak before B and B speaks before C, is 3 1 (B) (A) 8 6 3 (D) None of these (C) 5

(C) 45º

(D) 30º

43. From the top of a hill h metres high the angles of depressions of the top and the bottom of a pillar are α and β respectively. The height (in metres) of the pillar is h(tan β – tan α) h(tan α – tan β) (B) (A) tan α tan β

35. The group of 10 items has arithmetic mean 6. If the arithmetic mean of 4 of these items is 7.5, then the mean of the remaining items is (A) 6.5 (B) 5.5 (C) 4.5 (D) 5.0 XtraEdge for IIT-JEE

iˆ – 4 ˆj + kˆ 3

39. The image of the point P (1, 3, 4) in the plane 2x – y + z + 3 = 0 is (A) (3, 5, – 2) (B) (–3, 5, 2) (C) (3, –5, 2) (D) (3, 5, 2)

33. The probability of getting a sum of 12 in four throws of an ordinary dice, is 3

(D)

find that P = AC + BD is collinear with AD . If r P = µ AD , then (A) µ = λ + 1 (B) λ = µ + 1 (C) λ + µ = 1 (D) µ = 2 + λ

32. If R be a relation < from A = {1, 2, 3, 4} to B = {1, 3, 5} i.e. (a, b) ∈ R iff a < b, then RoR–1 is (A) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} (B) {(3, 1), (5, 1), (3, 2), (5, 2), (5,3), (5, 4)} (C) {(3, 3), (3, 5), (5, 3), (5, 5)} (D) {(3, 3), (3, 4), (4, 5)}

5 6

1 ˆ (–i + 4 ˆj – kˆ) 3

38. In a trapezoid the vector BC = λ AD we will then

, then

(A) domain of function f (x) ∈ (0, ∞) (B) range of function f (x) ∈ (0, ∞) (C) period of function f (x) is 2π f ( x) – 150 (D) lim = 200 x →0 x

1 (A) 6

r r r (B) | u | + | v . a | r r r r (D) | u | + u . ( a + b )

r r r 37. The vectors a . b , and c are equal in length and taken pairwise, they make equal angles. If r r r a = iˆ + ˆj , b = ˆj + kˆ and c makes an obtuse angle r with x-axis then c is equal to (B) iˆ + kˆ (A) – iˆ + 4 ˆj – kˆ

1 2 1+ x 30. If A = , then f (A) is and f (x) = 2 1 1– x 1 1 (A) 1 1

non-collinear unit vectors if r r r r v = a × b , then | v | is equal

(C)

82

h(tan β + tan α) tan β

(D)

h(tan β + tan α ) tan α

MARCH 2012

44. In a ∆ABC if the sides are a = 3, b = 5 and c = 4, then B B + cos is equal to sin 2 2

(A)

2

(B)

3 +1 (C) 2

7.

Directions : In following question, find out which of the answer figures (A), (B), (C) and (D) completes the figure – matrix ?

3 –1 (D) 1 2

45. If sin θ + cosec θ = 2 then sin11θ + cosec21 θ = (A) 2 (B) 221 (C) 232 (D) 1

?

(A)

(B)

(C)

(D)

LOGICAL REASONING 8. 1.

Fill in the blank spaces 6, 13, 28, . ?. . . (A) 56 (B) 57 (C) 58

(D) 59

2.

Choose the best alternative Car : Petrol : : T.V. : ? (A) Electricity (B) Transmission (C) Entertainment (D) Antenna

3.

Pick the odd one out – (A) Titan (B) Mercury (C) Earth (D) Jupiter

4.

Direction : In questions, find out which of the figures (A), (B), (C) and (D) can be formed from the pieces given in (x).

Directions : The questions that follow contain a set of three figure X, Y and Z showing a sequence of folding of piece of paper. Fig. (Z) shows the manner in which the folded paper has been cut. These three figure are followed by four answer figure from which you have to choose a figure which would most closely resemble the unfolded form of figure. (Z)

X

(B)

(A) A

9.

5.

(B)

(D) C

D

Direction : In following questions, complete the missing portion of the given pattern by selecting from the given alternatives (A), (B), (C) and (D). ?

(C)

(D) (X)

Directions : In question, choose the set of figures which follows the given rule. Rule : Closed figures become more and more open and open figures more and more closed.

(A)

(B)

(C)

(D)

6.

Z

(C) B

(x)

(A)

Y

(B)

(A)

(C)

(D)

10. Directions : In question below, you are given a figure (x) followed by four figures (A), (B), (C) and (D) such that (X) is embedded in one of them. Trace out the correct alternative.

Directions : In question below, you are given a figure (X) followed by four figures (A), (B), (C) and (D) such that (X) is embedded in one of them. Trace out the correct alternative.

(X)

(A)

B)

(C)

(D)

(x)

(A)

(B)

XtraEdge for IIT-JEE

(C)

(D)

83

MARCH 2012

10. Which one of the two sentences given below is wrong on the basis of the underlined words : 1. He is a very "ingenuous" businessman. 2. I like him for his "Ingenious" nature. (A) Sentence 1 is correct (B) Sentence 2 is correct (C) Both the sentences can be made correct by interchanging the underlined words. (D) Both the sentences can not be interchanged hence, both are wrong

ENGLISH 1.

Find the correctly spelt word â€“ (A) Geraff (B) Giraffe (C) Giraf (D) Gerraffe

2.

Find out that word where the spelling is wrong â€“ (A) Puncture (B) Puntuation (C) Pudding (D) Pungent

3.

Pick up the correct synonym for the following words Plush : (A) Luxurious (B) Delicious (C) Comforting (D) Tasty

4.

Choose the alternative which can replace the word printed in underline without changing the meaning of the sentence. When he returned, he was accompanied by 'sprightly' young girl. (A) Lively (B) Beautiful (C) Sportive (D) Intelligent

5.

Choose one alternative which is opposite in meaning to the given word : Astute : (A) Wicked (B) Impolite (C) Cowardly (D) Foolish

6.

7.

8.

9.

11. Choose from the given words below the two sentences, that word which has the same meaning and can be used in the same context as the part given underlined in both the sentences : 1. His "aloof" behaviour is an indication of his arrogance. 2. During our field visits we visited "remote" parts of Rajasthan. (A) Far-off (B) Introvert (C) Distant (D) Depressed 12. Find out which part of the sentence has an error. If there is no mistake, the answer is 'No error'. " Meatless days" / have been made / into a film / No Error (a )

Choose the word which is closest to the 'opposite' in meaning of the underlined word Many snakes are 'innocuous' : (A) Deadly (B) Ferocious (C) Poisonous (D) Harmful

(c)

(d)

(B) have been made (D) No Error

13. Which part of the following sentence has an error ? If the sentence is correct, the answer will be 'No Error". Looking forward / to / meet you here / No Error (a )

(b)

(A) looking forward (C) meet you here

Choose the one which can be substituted for the given words/sentences : Giving undue favours to one's kith and kin' (A) Corruption (B) Worldliness (C) Favouritism (D) Nepotism

( c)

(d )

(B) to (D) No error

14. Choose the one which best expresses the meaning of the given Idiom/Proverb : The 'pros and cons' (A) Good and Evil (B) Former and Latter (C) For and Against a thing (D) Foul and Fair

Find out which one of the words given below the sentence can most appropriately replace the group of words underlined in the sentence : The bus has to "go back and forth" every six hours. (A) Cross (B) Shuttle (C) Travel (D) Run

15. Replace the underlined word with one of the given options : The Second World War started in 1939. (A) Broke out (B) Set out (C) Took out (D) Went out

Read both the sentences carefully and decide on their correctness on the basis of the underlined words : 1. I am out of practise these days 2. I practice law (A) Only 1 is correct (B) Only 2 is correct (C) Both the sentences 1 & 2 are correct (D) Both the sentences 1 & 2 are incorrect

XtraEdge for IIT-JEE

( b)

(A) Meatless days (C) into a film

84

MARCH 2012

SOLUTION FOR MOCK TEST IIT-JEE (PAPER - I) CHEMISTRY 1.

CH3–Mg–Br + (CH3.CO)2O → CH3—C—CH3 (excess)

[A]

1 HOC ∝ stability So II > IV > I > III 2.

3.

4.

H

CH3–Mg – Br + Cl–C–OC2H5 → (excess)

O

6.

7.

CH3—C—CH3 CH3

10. [B,C] Cr3+ → 3, Mn2+ → 5, Fe3+ → 5, Cu2+ → 1

CH2–OH ⊕

H / H2O

[B]

11. [B,C,D] Fe, Cr, Al have protected film HO

5.

OH

Both double are E, E isomer O

CH3

h 6.626 × 10 −34 ⇒ 9.1 × 10–31 u = λ 5200 × 10 −10 ⇒ u = 1400 m/sec

12. [B]mu =

[A] Meq of NaOH + Meq of Ba(OH)2 = Meq of HCl XV1 + YV2 = 0.1 × 100 V Also 4Y 2 + YV2 = 10 ⇒ YV2 = 5 4 ∴ V1X = 5 5 ∴ Fraction of acid by Ba(OH)2 = = 0.5 10 [B]

3 × 8.314 × 300 4 × 10 −3 = 1367.8 M/sec. 6.626 × 10 −34 −3 λ .8 = 7.29 × 10–11 m λ= = 4 × 10 × 1367 23 mu 6.023 × 10

13. [A]

r ∝ M. CO = 28, B2H6 = 28, H2 = 2, CH4 = 16

[B]%ionic =

2.6 × 10 −30 1.41 × 10 −10 × 1.6 × 10 −19 = 11.5%

[B]

× 100

17. [6]

18. [9]

19. [6]

20. [4]

21. [6]

HO – S – O – O – H

1, 2-epoxy-2-methyl butane

+6

O

[B,C,D] O ⊕

22. [4]

OH

O

H Ph–Mg–Br + CH3—C—Cl → Ph—C—Ph

XtraEdge for IIT-JEE

15. [A,B,D]

O

C2H5

(excess)

3RT = M

16. [A,B]

CH3

9.

u=

14. [A,B,D]

O

8.

CH3 ⊕

[B] In acidic médium enol form depend on stability of alkene [C]

OH

⊕

H

O

HO – S – O – O – S – OH

CH3

O 85

O MARCH 2012

23. [6]

4.

P O

[D] Lim− [1 + (cos x)cosx]2 x→

O

O

y = Lim− (cos x)cos x

O←P–O

x→

P→O

O

π 2

π 2

log(y) = Lim− (cos x) log cos x

O

x→

P

x

3

2x2 3x

x2 + 2

4

1

x

3

x3 sin(πx) + 6x6 + 1

g(1) f (1) = – 3 5

3x

4

3.

π 2

π 2

y = e0 = 1 Now limit is (1 + 1)2 = 22 = 4

1

4x 4 + 1

[D] f(x) =

6.

[A] b = 0 , a < 0 f(x) = x2 + ax + b is quadratic polynomial cut x axis at x = 0 and x = – a f(x) = x2 + ax + b

x6 + 9x4 + 1

5 7 =–4

sin 3α < 0 if sin 3α > 0 and cos 2α < 0 cos 2α or sin 3α < 0 and cos 2α > 0 i.e. if 3α∈ (0, π) and 2α∈ (π/2, 3π/2) or 3α∈ (π, 2π) and 2α∈ (–π/2, π/2) i.e. if α∈ (0, π/3) and α∈ (π/4, 3π/4) or α∈ (π/3, 2π/3) and α∈ (–π/4, π/4) α∈ (π/4, π/3) i.e. if since (13π/48, 14π/48) ⊂ (π/4, π/3) (A) is correct

ln(ln( x)) ln ( x ) 1 1 × × ln( x) − ln(ln( x)) ×1 / x f '(x) = lnx x [ln( x)]2 1 −0 e f '(x) = = 1/e 1

5.

5 7 11 2.

sin x 1 ×– cos x sec x tan x

x→

= – x sin(πx) + 2x2 + 1 sin2 (πx) + 4x4 + 1 x3 sin(πx) + 6x6 + 1

3 3

log(y) = Lim−

= Lim− – cos x = 0

x 1 1 1 1 sin πx 2 x 2 1

x sin(πx) + 2x2 + 1

4x4 + 1

log(cos x) (∞/∞) L'hospital sec( x)

x→

[B] f(x) is odd function g(x) = f(–x) = – f(x) x 1

g(x).f(x) = – sin πx

log(y) = Lim− π x→ 2

MATHEMATICS 1.

π 2

[A]

[C] We have S1 = Σ x1 = sin 2β S2 = Σ x1x2 = cos 2β S3 = Σ x1x2x3 = cos β S4 = x1x2x3x4 = – sin β 4

so that

∑ tan

−1

xi = tan–1

i =1

f (| x |) = x2 + a |x| + b

S1 − S3 1 − S2 + S4

| f |x|| = | x2 + a |x| + b|

sin 2β − cos β 1 − cos 2β − sin β cos β(2 sin β − 1) = tan–1 sin β(2 sin β − 1) = tan cot β = tan–1 (tan (π/2 – β)) = π/2 – β = tan–1

XtraEdge for IIT-JEE

Exactly at three points function is not differentiable.

86

MARCH 2012

7.

[A] Graph of sin–1 sin(x) = f(x) –π

π

0

5

11. [B,D] u2 = a2 + b2

2π 3π

+2

f(x) = sin–1 (sinx) = x – 2π 3π/2 ≤ x ≤ 5π/2 f(5) = sin–1 (sin 5) = 5 – 2π log2(x) < 5 – 2π x>0 x < 25–2π So, (0, 25–2π) 8.

a2 + b2 max. u2 = a2 + b2 + 2 2 2 2 2 min. u = a + b +2

[A,B,C] Let tan–1 (–3) = α ⇒ tan α = – 3 π <α<0 and – 2 ⇒ – π < 2α < 0 ⇒ cos (– 2α) = cos 2α

=

1 − tan 2 α 2

1 + tan α

=

⇒

4 ⇒ – 2α = cos − 5 –1 ⇒ 2 tan (–3) = – cos–1 (– 4/5) Again –π < 2α < 0 ⇒ 0 < 2α + π < π 4 So cos (π + 2α) = – cos 2α = 5 4 ⇒ π + 2α = cos–1 5 4 ⇒ 2 tan–1 (–3) = – π + cos–1 5 Finally – π < 2 α < 0 ⇒ –π/2 < 2α + π/2 < π/2

2

= (a + b)2

1 2

1

=

⇒ 1 + x2 = 2 + x2 + 2x

2

1 + ( x + 1) 1+ x ⇒ x = – 1/2, so f (x) = 0 for x = – 1/2

13. [C] a = cos tan–1 sin cot–1 x = cos tan–1 sin α, where x = cot α 1 = cos β where = cos tan–1 1 + x2

tan β =

⇒ a2 =

1 1 + x2

1

=

1+

1 1 + x2

=

x2 + 1 x2 + 2

x2 + 1 5 = for x = – 1/2 x2 + 2 9

14. [B] Now a2 = 26/51 ⇒ x = ± 1/5 and b = cos (2 cos–1 x + sin–1 x) = cos (cos–1x + π/2)

−1 tan 2α

⇒ b = – sin (cos–1 x) = – 1 − x 2 ⇒ b2 = 1 – x2 = 24/25 for x = ± 1/5

− 1 + tan 2 α − 4 = 2 tan α 3 –1 ∴ π/2 + 2α = tan (–4/3) ⇒ 2 tan–1 (–3) = – π/2 + tan–1(–4/3)

=

Passage # 2 (Q.15 to Q.16)

dy = 2x (x4) – 1 (x2) dx dy =2–1=1 dx x =1

[A,B,D] f (x) = [x sin πx] = 0 – 1 ≤ x ≤ 1. So, A, B, D are true

15. [C]

10. [A,D]

1

x2 f (x) = cos x – 1 − 2 f '(x) = – sin x + x sin x < x if x > 0 and sin x > x if x < 0 (A) follow from (D) since (0, π/2) is a subset of (0, ∞) XtraEdge for IIT-JEE

2

a2 − b2 − 2

12. [A] f (x) = 0 ⇒ sin {cot–1 (x + 1)} = cos (tan–1 x) 1 1 ⇒ sin sin–1 = cos cos–1 2 1 + ( x + 1) 1+ x 2

–1

9.

a2 + b2 2

= 2(a2 + b2)

Passage # 1 (Q.12 to Q.14)

1− 9 4 =− 1+ 9 5

So tan (π/2 + 2α) = – cot 2α =

2

2

(a 2 − b 2 ) cos 2 2θ − 2

a2 + b2 2

at x = 1

y=

∫t

2

dt = 0

1

(y – 0) = 1 (x – 1) ⇒y=x–1

87

MARCH 2012

2

16. [A] f '(x) = e x / 2 (1 – x2) = 0 [f '(x)]x = 1 = e(1 – 1) = 0

h

1 1 = Lim 1 + . 1 + h →∞ h 2h

1 2 h −1 h ......... 1 + n −1 2 h 2 h −1

2 17. [9] f (x) = [x] + [2x] + x + [3x] +[4x] + [5x] [kx] 3 changes its value of every integral multiple of 1/k [x] will change at every integral multiple of 1 [2x] will change at every integral multiple of 1/2 [3x] will change at every integral multiple of 1/3 [4x] will change at every integral multiple of 1/4 [5x] will change at every integral multiple of 1/5 2 and x will change at every integral multiple of 3 3/2 They would change all together at every multiple of LCM of {1, 1/2, 1/3, 1/4, 1/5, 3/2} = 3 No. of total points at which f (x) will changes its value in the interval [0, 3] will depend on the total number of different terms in the following cases – [x] = 0, 1, 2 [2x] = 0, 1/2, 2/2, 3/2, 4/2, 5/2 [3x] = 0, 1/3, 2/3, 3/3, 4/3, 5/3, 6/3, 7/3, 8/3 [4x] = 0, 1/4, 2/4, 3/4, 4/4, 5/4, 6/4, 7/4, 8/4, 9/4, 10/4 [5x] = 0, 1/5, 2/5, 3/5, 4/5, 5/5, 6/5, 7/5, 8/5, 9/5, 10/5, 11/5, 12/5, 13/5, 14/5 2 3 x = 0, 3/2, 6/2

1

= e1. e1/2. e1/4........ e 2

1 Lim 1 + h →∞ ah e

n −1

.......∞

ah

= ea

1 1 1+ + +.......∞ 2 4 1

= e 1−1 / 2 = e2 = eS ⇒ S = 2 ⇒ 0002 19. [5] x2 + 1 = (x + i) (x – i) The cubic polynomial must vanish for x = i , x = – i – i – a + bi + c = 0, i – a – bi + c = 0 –1+b=0 – a + c = 0, 1–b=0 – a + c = 0, b = 1, a = c ⇒ b is fixed Now a can be chosen in 10 ways and c = a, c can be chosen in 1 way only ⇒ Number of ways of choosing a, b, c = 10 ⇒ 10 = 2k ⇒ k = 5 20. [ 2] ey + xy = e on putting x = 0, we get ey = e y = 1 when x = 0 on differentiating the relation (i) we get dy dy + 1.y + x. =0 dx dx on putting x = 0, y = 1 we get

∴ f (x) will change its values in the intervals 14 ≤ 5

0 ≤ x < 1/5, 1/5 ≤ x < 1/4, 1/4 ≤ x < 1/3,...........

2h 2

x<3 Total no. of different terms in above equation = 30. So, no. of terms in the range of f (x) for 0 ≤ x < 3 is 30 ⇒ m +21 = 30 ⇒ m =9

dy −1 dy ey + 1= 0 ⇒ = dx e dx on differentiating solution (ii) we get 2

18. [2] Lim h h →∞

= Lim h →∞

1 1 (h + 1) h + ..... h + h −1 2 2

h

1 1 (h + 1) h + ....... h + h −1 2 2 hh

h

−h2

h

h +1 = Lim h →∞ h

h

XtraEdge for IIT-JEE

1 1 h + h −1 h+ 2 ......... 2 h h

d2y d2y dy dy dy e + ey + + + x =0 dx dx dx 2 dx 2 dx −1 dy = we get on putting x = 0, y = 1, dx e y

d2y 1 = 2 = e–2 ⇒ λ = 2 dx 2 e 21. h

88

[1] Circle (x – 2)2 + (y – 3)2 + λ (x + y – 5) = 0 it passes through (1, 2) 1 + 1 + λ (1 + 2 – 5) = 0 ⇒ λ = 1 x2 – 4x + 4 + y2 – 6y + 9 + x + y – 5 = 0 (x – 3/2)2 + (y – 5/2)2 = 1/2 So, number of circle is 1. MARCH 2012

22.

[8] Let the equation of line through P(λ, 3) be x−λ y −3 = r ⇒ x = λ + r cos θ =

2.

[D]

ucosθ A B θ usinθ

cos θ sin θ and y = 3 + r sin θ

v1

x2 y2 Line meets the ellipse + =1 16 9 Such that 9x2 + 16y2 = 144 at A and D ⇒ 9(λ + r cos θ)2 + 16(3 + r sin θ)2 = 144 ⇒ 9(λ2 + r2 cos2θ + 2λr cosθ) + 16 (9 + r2 sin2θ + 6r sin θ) = 144 2 ⇒ (9cos θ + 16 sin2 θ) r2 + (18λ cos θ + 96 sin θ)r + 9λ2 = 0

9λ2 … (i) 9 cos 2 θ + 16 sin 2 θ Since line meet the axes at B and C 3λ So, PB.PC = …. (ii) sin θ cos θ from (i) & (ii) λ≥8 23.

[7]

→

→

→

→

3.

[C]

r

b

dI =

→ →

⇒ a . b + a . c = 0 and two similar results → →

→ →

→

→

→

→

→

→

I= →

→

Now | a + b + c |2 = ( a + b + c ). ( a + b + c ) →

→

→

→ →

→ →

4.

→ →

= | a |2 + | b |2 + | c |2 + 2( a . b + b . c + c . a ) = 9 + 16 + 25 + 0 = 50 →

→

and dm =

M (2πrdr ) π(b 2 − a 2 )

M 2 (b + a 2 ) 2 α

→

F aCM

f F – f = maCM ….(i) fR = Iα I a CM ….(ii) or f = R2 from (i) & (ii) F and aCM = I m + 2 R

PHYSICS [C] When block m1 passes through mean position, its speed is maximum. Let v1 and v2 be the speed of blocks m1 and (m1 + m2) respectively at equilibrium position. Then m1v1 = (m1 + m2) v2 at mean position v = Aω = A2πf hence

v1 A f = 1 × 1 and f1 = 2π v2 A2 f2 f2 = 2π A2 = A

2

[D]

∴ | a + b + c | =5 2

1.

∫ dm r a

→ →

adding, 2 ( a . b + b . c + c . a ) = 0 →

a

b

→

a ⊥(b +c )⇒a .(b +c )=0

→ →

usinθ

before collis ion after collision Apply conservation of momentum mv1 + mv2 = mvcosθ or v1 + v2 = v cosθ ….(i) …(ii) and v1 – v2 = evcosθ from (i) + (ii) (1 + e) v cos θ v1 = 2

∴ PA.PD =

→

v2

f=

m1 k

I µmg m + 2 R Fmax = I 2 R

m1 + m 2 k

m1 m1 + m 2

XtraEdge for IIT-JEE

I F ≤ µmg 2 I R m + 2 R

89

MARCH 2012

5.

[A] We have gravitational potential at internal point. GM V= − (3R 2 − r 2 ) 2R 3 at surface 3 GM V1 = − 2 R

x at r =

R 2

V2 = −

GM 2 R 2 3R − 4 2R 3

8.

t3 3 dx v= = t2 dt x=

d2x = 2t dt 2 F = ma F = 4t or F ∝ t Apply work – Energy theorem 1 W = ∆k = mv2 2 at t = 0, u = 0 at t = 2, v = 4 m/s during 0 – 2 sec 1 W= × 2 (4)2 2 W = 16 J a=

11GM 8R we have 1 mv2 = m | V2 – V1 | 2

V2 = −

6.

7.

[C] W = surface tension (T) × change in surface area (∆A)

9.

Current density,

F m Consider an element of length dx at a distance 'x' from one end of the rod. x F a=

10.

dx stress developed at a distance x m (l − x ) F stress = Al m

at

F x 1 − m l elongation in the dx length stress dl = dx Y

∫

∆l =

a 2 + ( 2 b + c) 2

at t = 0 vx = a, vy = c c and | u | = tanθ = a

F x 1 − dx l YA 0

∫

a 2 + c2

d2x d2y = 0 and = 2b dt dt 2 i.e. ay = 2b or g = –2b

Fl 2YA

XtraEdge for IIT-JEE

v 2x + v 2y

t=1

|v|=

l

dl =

JP < JQ (Q J =

[B,C,D] Here x = at and y = bt2 + ct dx dy vx = = a and vy = = 2bt + c dt dt

|v|=

stress =

or

[A,B,C,D] Here iP = iQ

i ) A IP < EQ (Q J = σE) ρdx ) Resistance, RP < Rq (Q R = A or (i2R)P < (i2R)Q

Acceleration of the rod (a) F r a

[C]

[A,B,D]

90

MARCH 2012

11.

[A,B,C] At equilibrium

Passage : II Sol. for Q.No. 15.[A] & 16.[A] U = 5 + (x – 1)2 at x = 2, U=6J KE = 10 J M.E. = 10 + 6 = 16 J

qE k for maximum elongation, apply Work – Energy theorem 1 (qE)x0 – kx02 = 0 2 2qE x0 = k the block perform oscillation about mean position qE ) (i.e. x = k 2qE qE – amplitude (A) = k k qE A= k kx = qE or x =

dU =0 dx 2(x – 1) = 0 x=1

for minima of U,

Umin = 5 J 17.

Sol. for Q.No. 12.[B], 13.[D] & 14.[A] The given circuit can be simplified as 23µF 7µF

10µF

1µF

19.

1µF

λ 4

…(i)

3λ 4

…(ii)

[3] For a polytropic process PVn = constant

R n −1 5 R = R− 2 −1−1 (Q P ∝ V or PV–1 = constant) C = 3R

7µF B

10µF

Here l1 e =

C = CV –

OR A

[3]

from (i) & (ii) l − 3l 1 e= 2 2 e=3

12V

35µF

25l g

l2 + e =

B 5µF

l = T and g

T2 = 5T at t = 5T, both pendulum will again be in phase for 1st time during that time, 5 oscillations are made by smaller pendulum. 18.

12µF

T1 = 2π

T2 = 2π

Passage : I

A

[5]

12µF

20.

[3] Temperature of junction

100 + 20 + 0 = 40ºC 3 i1 100 − 40 = =3 40 − 20 i2

TC =

12V

35 × 7 20 = 7.5 µF Ceq = + 35 + 7 12 Charge on 5 µF = 0 20 = 20 µC Charge on 10 µF = 12 × 12 20 × 20 = 20 µJ U12µF = 2 × 10

21.

[3]

20cm A C

B f

O f = 30cm

60cm XtraEdge for IIT-JEE

91

MARCH 2012

image one end A is formed at same point (i.e. at C) for image of other end 1 1 1 + = v − 40 − 30 v = –120 cm length of image l = 120 – 60 = 60 cm 60 =3 linear magnification ml = 20 22.

Chemistry Facts At 0 degress Celsius and 1 atmospheric pressure, one mole of any gas occupies approximately 22.4 liters.

•

Atomic weight is the mass of an atom relative to the mass of an atom of carbon-12 which has an atomic weight of exactly 12.00000 amu.

•

If the atom were the size of a pixel (or the size of a period), humans would be a thousand miles tall.

•

It would require about 100 million (100,000,000) atoms to form a straight line one centimeter long.

•

The weight (or mass) of a proton is 1,836.1526675 times heavier than the weight (or mass) of an electron.

•

The electron was first discovered before the proton and neutron, in 1897 from English physicist John Joseph Thomson.

•

The neutron was discovered after the proton in 1932 from British physicist James Chadwick, which proved an important discovery in the development of nuclear reactors.

•

Carbon dioxide was discovered by Scottish chemist Joseph Black.

•

When silver nitrate is exposed to light, it results in a blackening effect. (Discovered by Scheele, which became an important discovery for the development of photography).

[2]

P θ θ 2θ θ α N Q R

a O N = a /8 2

Normal x0

y2 = 8x

In ∆PNQ tan 2θ =

a a2 x0 − 8

…(i)

y2 = 8x dy 4 = dx y at P y=a dy 4 = = tan(90º – θ) dx a

a 4 eq.(i) and (ii) x0 = 2

or tan θ =

23.

•

…(ii)

[3] We have

1 1 E = 13.6 Z2 2 − 2 (in eV) n1 n 2

1 1 10.2 + 17 = 13.6 Z2 2 − 2 n 2 and 1 1 4.25 + 5.95 = 13.6 Z2 2 − 2 n 3 from (i) & (ii) n = 6 and Z = 3

XtraEdge for IIT-JEE

…(i)

…(ii)

92

MARCH 2012

SOLUTION FOR MOCK TEST PAPER - II - II) IIT-JEE (PAPER

CHEMISTRY O

D

1.

O Θ

OH

[C]

[A,B,C] Na2BF3, Na2ClO4, conc HNO3 + H2SO4 used for nitration of benzene.

11.

[A,C,D] Salt bridge is used to eliminate liquid junction potential arised due to different speed of ions persent in cathodic & anodic compartment

12.

[A,B,C,D]

Θ

HC O

OH

(A)

– + OH

H– OH →

2.

10.

(B) (C)

[B]

(D)

CH3–CH–CH2–C≡CH CH3–CH–C≡C–CH3 CH3

3.

[B]

4.

[D]

CH3

,

Br I Cl D

H H Br

I Both are enantiomers.

Insulin → Zinc (Zn); Haemoglobin → Fe [D] B12 → Co

6.

[B]

7. 8.

9.

[D]

14.

[A → S; B → R; C → Q; D → T;]

15.

[2] ⊕

5.

Green

[A → Q,S; B → R; C → P; D → T; ] Electrochenical cell → ∆G < 0, Salt bridge Ist law of faraday → W = Zit Electrolytic cell → ∆G > 0 Lead acid cell → rechargeable

Cl

H H

Cr2O3

13.

III < I < II

D

∆ Na2CO3

[ H 3O + ][ HS ¯] [H 2 S ] Higher the Kc high is stability Fact [ RNH 3+ ][OH ¯] Kb = [ RNH 2 ] Ka =

NH3–CH–CH2–CH2–COOH COO Θ Two group present in above compound. 16.

[5]

17.

[6]

PbCrO4

18.

[6]

Yellow

19.

[4] 2Al + Fe2O3 —→ 2Fe + Al2O3 ∆H = – 399 – (– 199) = – 200 Kcal / ml

Na2CrO4

H2O + H2SO4

(CH3COO)2 Pb

Na2CrO7 Orange

As2S3 sol particles absorb S2– as common ion

54 160 + = 50.7 ml 2.7 5.2 ∆H 200 ∴ = = 4 Kcal / ml ml 50.7 vol. =

(300 – V ) × 1 pH = pKa + log [C] 1× V (300 – V ) 4.5 = 4.2 + log ; V = 100 ml V

20.

[2]

rλ = 100 r1 ; C2 = 10C1 ⇒ n = 2

[C,D] Both C and D pair are same IUPAC name.

XtraEdge for IIT-JEE

93

MARCH 2012

MATHEMATICS 1.

= lim x→

[C] Let ABC be the triangle with b + c = x and bc = y, then a = z, and from the given relations we have (b + c + a) (b + c – a) = bc ⇒ b2 + c2 – a2 = – bc b2 + c2 − a2 1 ⇒ =− 2bc 2

= lim

π x→ 4

[A] Function is diff. at x = 1 it means function is continuous at x = 1, diff. at x = 1 R.H.L. at x = 1 = L.H.L. at x = 1, R.H.D. = L.H.D. a cos (0) + b = 1 a+b=1 ....(1) (–2 a sin (2x – 2) + 2bx)x = 1 = (2x2 e2(x – 1) + 2xe2(x – 1))x = 1 0 + 2b = 2 + 2 2b = 4 b=2 so a = 1 – b = 1 – 2 = – 1

6.

[A] P(x, y) be a point on the curve ln (x2 + y2) = c tan–1 y/x differentiating both side with respect to x 2 x + 2 yy ' c( xy '− y )

1 = cos 120º 2 ⇒ A = 120º and the triangle is obtuse angled. ⇒ A is an obtuse angle. r1 4 R sin( A / 2) cos( B / 2) cos(C / 2) = bc 2 R sin B.2 R sin C sin( A / 2) = 4 R sin( B / 2) sin(C / 2)

[D]

=

sin 2 ( A / 2) 4 R sin( A / 2) sin( B / 2) sin(C / 2)

sin 2 ( A / 2) r r1 r2 r3 So that + + bc ca ab 1 = [(sin2 (A/2) + sin2 (B/2)+ sin2 (C/2)] r 1 (1 – cos A + 1 – cos B + 1 – cos C) = 2r 1 = [3 – (cos A + cos B + cos C)] 2r 1 [3 – (1 + 4 sin (A/2) sin (B/2) sin (C/2))] = 2r r 1 1 1 2− = − = 2r R r 2R =

3.

[B] f (θ) = cos2θ + sin4 θ f (θ) = cos2θ + (1 – cos2 θ)2 f (θ) = (cos2 θ – 1/2)2 + 3/4 (f (θ))min = 3/4.

4.

[D]

(x 2 + y 2 ) ⇒y'=

π 4

3(cos x + sin x) 2 (− sin x + cos x) − 2 cos 2 x

XtraEdge for IIT-JEE

m1 − m2 1 + m1m2

2 x + cy y − cx − 2 y x = = 2/c 2 xy + cy 2 1+ 2 cx − 2 xy

θ = tan–1 (2/c) which is independent of x and y

π 4

x→

x2 + y2

2 x + cy = m1 cx − 2 y

so tan θ =

2 2 − (cos x + sin x) 3 (0/0) 1 − sin 2 x L'Hospital Rule lim f (x) = lim –

=

slope of OP = y/x = m2

f (x) =

x→

3 3 2 3 × (cos x + sin x) = × = 2 2 2 2

5.

⇒ cos A = –

2.

π 4

3(cos x + sin x) 2 (cos x − sin x) 2(cos x − sin x) (cos x + sin x)

94

7.

[B] f (x) = 2x3 + ax2 + bx – 3cos2x f '(x) = 6x2 + 2ax + b + 3 sin 2x > 0 as sin 2x ≥ – 1 6x2 + 2ax + b – 3 > 0 ∴ 4a2 – 4b(b – 3) < 0 ⇒ a2 – 6b + 18 < 0

8.

[D] x = φ(t) = t5 – 5t3 – 20t + 7 dx = φ '(t) = 5t4 – 15 t2 – 20 = 5(t2 – 4) (t2 + 1) ≠ 0 dt If – 2 < t < 2 y = ψ(t) = 4t3 – 3t2 – 18t + 3 dy = ψ'(t) = 12 t2 – 6t – 18 dt MARCH 2012

dy = 0 ⇒ t = – 1 or 3/2 dt

9 (D) Range of function cosx + cos 2x is − ,2 8 put cos x = t, t∈[–1, 1] therefore 9 + cos x + cos 2x > 0 for all x 8

d2y = ψ"(t) = 24t – 6 ⇒ ψ"(–1) = – 30 dt 2 and ψ"(3/2) = 30 y = f (x) is minimum at t = 3/2 11. 9.

[A,D]

a − b sin A − sin B = c sin C [by the law of sine] 2 cos[(1 / 2) ( A + B)] sin[(1 / 2) ( A − B)] = 2 sin[(1 / 2) C ] cos[(1 / 2)C ] cos θ =

=

2 sin[(1 / 2)C ] sin[(1 / 2) ( A − B)] 2 sin[(1 / 2)C ] sin[(1 / 2)( A + B)]

=

sin[(1 / 2) ( A − B)] sin[(1 / 2)( A + B)]

= ⇒

=

y = x, y ≥ 0 y = x/3

y = 1/3 x y<0 y=x y>0 (A) domain and range of function is set of real numbers so (A) is true (B) f (0) = L.H.L = R.H.L so (B) is true (D) L.H.D. = 1/3 and R.H.D. = 1 so (D) is true

sin 2 [(1 / 2) ( A + B )] − sin 2[(1 / 2) ( A − B )] sin[(1 / 2) ( A + B )]

⇒ sin θ =

[A,B,D] x + |y| = 2y 3y = x if y < 0 y = x is y ≥ 0

sin A sin B sin[(1 / 2) ( A + B)] (a + b) sin θ 12.

2 ab sin A + sin B

sin A sin B 2 sin A sin B sin[(1 / 2) ( A + B)]

2 sin(1 / 2)( A + B ) cos(1 / 2)( A − B ) 1 2 sin ( A + B ) 2 = cos [(1/2) (A – B)] c sin θ sin C = and 2 ab 2 sin A sin B

[B,D] Take river as x axis, village A as x axis

line joining origin and

B(b, k)

=

P (–a, 0)

10.

BA' =

2 sin(C / 2) cos(C / 2) = cos ((A + B)/2) 2 sin[( A + B) / 2]

= 13.

[A,B,D] (A) sin[x] + cos[x] is defined for all x, since [x], sinθ, cosθ are always defined. (B) sin x is always defined and 1 + sin2 x ≥ 1 ⇒ sec–1(1+ sin2 x) is defined for all x π (C) tan(log x) is not defined if logx = (2k + 1) 2

XtraEdge for IIT-JEE

A(a, 0)

k2 = c2 – (b – a)2 image of A in the river, this is A' and BA' must be minimum value of PA + PB

sin A sin B sin[(1 / 2) ( A + B)] =

c

A'

(b + a ) 2 + k 2 (b + a ) 2 + c 2 − (b − a ) 2 =

c 2 + 4ab

A → P,R,T; B → P,R,T; C → Q,S; D → P,S

(A) Lim f(x) = x →0

sin 2 ( x) = Lim sin(x) = 0 = f(0) x →0 x

⇒ continuous

sin 2 x −0 x f '(0) = Lim =1 x →0 x 95

MARCH 2012

⇒ differentiable (B) f(0–) = Lim (x3 – 2x) = 0

6.

x →0

f(0) = f(0+) = Lim (x2 – 2sin (x)) = 0 x →0

⇒ continuous f '(0) = Lim (3x2 – 2) = – 2 x →0

+

f '(0 ) = Lim+ (2π – 2cos(x)) = – 2 x →0

14.

⇒ differentiable (C) f (0–) = 4 & f (0+) = 5 ⇒ discontinuous ⇒ non diff. (D) At x = 1, f (x) = 3 – 2x, which is polynomial ⇒ continuous & nondifferentiable

3i

A → Q, B → S, C → S, D → R (A) x = sin θ & y = cos θ x + y = sin θ + cos θ

–i

i

|z|max = |3i| = 3 |z|min = | –i| = 1 sum = 4

2 sin (θ + π/4) ⇒ min. value = – 2 y = a cos x – 1/3 cos (3x) 9 y'(π/6) = 0 ⇒ – a sin π/6 + sin π/2 = – +1=0 2 ⇒a=2 (C) f '(x) = 1 – 2 cosx f '(x) > 0 ⇒ cos(x) < 1/2 ⇒ x ∈ (π/3, 5π/3) ⇒ a = 1/3, b = 5/3 ⇒ a + b = 2 (D) at x = 0, y = – e0 = – 1 y' = 1/2 e–x/2 = y' (0) = 1/2 Equation of tangent y + 1 = 1/2 (x – 0) x y ⇒ + ⇒ p = 2, q = – 1 2 −1 So, p – q = 3 =

(B)

15.

[4] zi – z i + 2 = 0 ⇒ (z – z ) = 2i ⇒ Im (z) = 1 z (1 + i) + z (1 –i) + 2 = 0 ⇒ (z + z ) + i (z – z ) + 2 = 0 ⇒z=–z ⇒z=i ⇓ Re(z) = 0 Let the point on the line be z so |z –i| = 2

17.

− 1 − 5 − 1 adj A = − 8 − 6 9 & |A| = – 17 − 10 1 7 3 0 3 x 8 2 y ⇒ 2 1 0 y = 1 + z is equivalent to 4 0 2 z 4 3 y 3 − 2 3 x 8 2 1 − 1 y = 1 4 − 3 2 z 4 ∴ solution is given by − 1 − 5 − 1 8 − 17 x y = 1 − 8 − 6 9 1 = − 1 − 34 − 17 17 − 10 1 − 51 z 7 4

[7] Let Ei denote the event that out of the first k balls drawn, i balls are green. Let A denote the event that (k + 1)th ball drawn is also green. 14

and

P(Ei) =

Ci × 6 C k −i ; 0≤i≤k 20 Ck

P(A/Ei) =

14 − i 20 − k

k

Now P(A) =

∑

14

j =0

∴x = 1 , y = 2, z = 3 6

C j × Ck −i 20

Ck

+

14 − j 20 − k

18.

Also (1 + x)14 – 1 (1 + x)6 = (14–1C0 + 14–1C1x +.......+ 14 – 1C14 – 1 x14–1) (6C0 + 6C1x + .......+ 6C6x6) k

⇒

∑(

14−1

C j + 6 Ck − j ) = co-efficient of xk

j=0

∴ P(E) =

14 14 = ⇒ 10P(Ε) = 7 6 + 14 20

XtraEdge for IIT-JEE

[3]

3 − 2 3 A = 2 1 − 1 4 − 3 2

96

[1] Given planes are x – cy – bz = 0 ... (i) cx – y + az = 0 ... (ii) bx + ay – z = 0 ... (iii) equation of plane passing through the line of intersection of plane (i) and (ii) may be taken as (x – cy –bz) + λ (cx – y + az) = 0 ⇒ (1 + λc) x – y (c + λ) + z (aλ – b) = 0 ....(iv) If plane (iii) and (iv) are same then MARCH 2012

1 + λ c − ( c + λ ) aλ − b = = b a −1 a + bc ab + c ⇒λ=– =– ac + b 1 − a2 ⇒ a – a3 + bc – a2 bc = a2 bc + ac2 + ab2 + bc ⇒ a2 + b2 + c2 + 2abc = 1 19.

Ax 2 + Bx 180 2 80 x + x= 119 119 119 A = 180 B = 80 A+ B A + B = 260 ⇒ =1 260 f (x) =

y = Axm + Bx–n dy ⇒ = Amxm–1 – nBx–n–1 dx

[7]

⇒

from (2) and (3) 180 80 a= and b = 119 119

d2y = Am (m –1) xm–2 + n (n +1) Bx–n –2 dx 2

PHYSICS

d2y dy + 2x = 12y 2 dx dx We have = m (m + 1) Axm + n (n –1) Bx–n

1.

Putting these values in x2

[C]

= 12 (Axm + Bx–n) ⇒ m (m +1) = 12 or n (n –1) = 12 ⇒ m = 3, –4 or n = 4, –3 1

20.

[1] f (x) = x + x

0.4 m θ

y2 f (y) dy + x2

0

∫

y f (y) dy

0

1 1 2 2 = x 1 + y f ( y )dy + x yf ( y )dy 0 0 f (x) is quadratic expression f (x) = ax + bx2 or f (y) = ay + by2

∫

∫

…(1)

(xA0 ρWg)

1

a=1+

∫

2

y f (y) dy

0

x2 = l2

1

=1+

∫

y2 (ay + by2) dy

x=

0

5

a b a=1+ + 4 5 20a = 20 + 5a + 4b 15a – 4b = 20 1

b=

∫

…(2) 2.

1

y f (y) dy =

0

∫

y (ay + by)2 dy

0

1

ay 3 by 4 a b ⇒b= + = + 3 4 3 4 0 12 b = 4a + 3b 9b – 4a = 0 XtraEdge for IIT-JEE

mg

ρp ρw

x l = lA0ρp g 2 2 = (0.8)2 × 0.5

0 .8

2 from (i) and (ii) π 1 or θ = sin θ = 4 2

1

ay by =1+ + 5 0 4 4

A

O Let x length of plank is inside the water (i.e. AO = x) x sinθ = 0.4 …(i) buoyant force, FB = xA0ρWg for rotational equilibrium about O x l cos θ = mg cos θ FB 2 2

1

∫

FB

…(3) 97

….(ii)

[C] When water cools down to 0ºC then heat released ∆Q = 5 × 1 × 30 = 150 cal. while heat required to convert ice into water at 0ºC (ice)–20ºC → (ice)0ºC → (water)0ºC Q1 = 5 × 0.5 × 20 = 50 cal Q2 = 5 × 80 = 400 ∆Q′ = 450 cal here 50 < ∆Q < 450 i.e final temperature is 0ºC with some ice melt.

MARCH 2012

3.

Electric potential at A 1 Q 1 Q VA = = 4π ∈0 (2R ) 4π ∈0 R 2 + 3R 2

[B] The given circuit can be simplified as

i1

A

at origin 1 Q V0 = 4π ∈0 R We have W = q(V0 – VA) = ∆K

20V reading of ammeter, i1 = 4.

20 = 2A 10

7.

[C]

[B]

2

y

1 i

ay O

dv x = 2ux dt

µ i B2 = B3 = 0 2πL r r r r r B net = B1 + B 2 + B3 + B 4 r µ i B net = 0 π L

we have y = x2 dy dx = 2x =0 dt dt

d2y dx = 2u 2 dt dt ay = 2u2

8.

a 2x + a 2y

a = 2u2 along y-axis 5.

[B] + + + + +

r

XL = ωL = 50 Ω 1 XC = = 100 Ω ωC

[A]

Z=

R 2 + (X C − X L ) 2 = 50 2 Ω

irms =

Vrms = 2A Z

Pav = i 2rms R = 200 W

λ E= 2π ∈0 r

9.

We have

[A,B,C,D] For maximum range, θ = 45º

2r

r r VB – VA = dV = − E . d r

∫

i.e. x =

3r

W = q dV λq 0 3 ln W= 2π ∈0 2 6.

u2 ⇒u= g

A

x = 3R

t=

u g 2

=

at max. height, v = u cos 45º =

x

max. height H =

Q=λ2πR XtraEdge for IIT-JEE

gx

T 2

warning time t =

[D] + + + + R+ + + + + +

L

2 Here B1 = B4 = 0 µ i B2 = B3 = 0 × (2 sin 45°) 2π (L / 2 )

vx = u

a=

L r=

x

u

i 4

O

L

at origin, ax =

90º 3

r

98

x 2g gx 2

x u2 u2 (sin 45º)2 = = 2g 4g 4 MARCH 2012

10.

[A,B,D] Q fixed

VAB =

v

⇓

Q

V at steady state. 4

Hence charge on capacitor, q1 =

r0

When only S2 → closed then

fixed v′=0 Apply conservation of mechanical energy

C

3r A

1 1 Q2 mv2 = 2 4π ∈0 r0

11.

VAB =

1 1 ,r∝ 2 m v

3r

[A,C] We have

dA dt

|ε|=B here

2r

2 V at steady state 5 2 hence max. charge, q2 = CV 5 When only S3 → closed then

Q 2π ∈0 mv 2

r0 ∝ Q2, r0 ∝

B

V

2

r0 =

CV 4

C

V

q3 = CV When all switches are closed then

dA increases first then decreases. dt

C

3r

εmax = Bvl 2 12.

[B,C] We have

no charge appear on capacitor.

T

1 iav = T im =

V

1 i dt = (Area of i – t curve) T 0

∫

14.

i0 2 T

1 2 i dt = T0

∫

irms =

1 ×2 T

T/2

∫i

2

dt

R=

0

we have

H= 15.

A→P; B→Q;C→S; D→T When only S1 → closed then C

3r A

40 × 40 3 u 2 sin 2α = × g 10 2

R = 80 3 m

T 2i i = 0 t for 0 ≤ t ≤ 2 T We get i irms = 0 3 13.

A → S ; B → R ; C → P; D → Q α = 90º – 60º = 30º 2 × 40 10 2u sin α × = = 4 sec T= 10 2 g

40 × 40 u 2 sin 2 α = = 40 m 2 × 10 × 4 2g

[4] Electric field lines are perpendicular to equipotential surfaces and electric potential decreases along electric field.

80V

90º

B

30º

r

B 120º

V

XtraEdge for IIT-JEE

60V

99

40V

20V

30º r E MARCH 2012

From A → B | dV | = –E dr cos 120º 20 = E × (10 × 10–2) ×

18.

[6]

1 2

O

E = 400 N/C 16.

[6]

3f/2

V2 V2 or R = R P 50 W

V0 200 × 200 R40W = = 1000 Ω 40 200 × 200 = 800 Ω R50W = 50 Let max. voltage of main supply is V0 then 4 × V0 = 200 V40W = 9 V0 = 450 V and 5 V50W = V0 = 200 = 360 V 9 hence for safety of both bulb, V0 should be 360 V and for this main supply voltage V40W = 200V hence it glows with full intensity.

hi = 2 × 0.5 = 1 cm i.e. image is formed at a height of 3 cm (i.e. 2 + 1) from main principle axis. (above principle axis) Similarly for lower part, image is formed 3 cm below main principle axis. Hence distance between image = 3 + 3 = 6 cm. 19.

[5] Velocity of centre of mass of cylinder w.r.t. plank, vcp = 20 – 10 = 10 m/s 10 We have vcp = Rω ⇒ ω = R

Kinetic energy KE = =

17.

[4]

5kg A

f

For upper part of lens 3f , h0 = 2 cm u= − 2 h f = 0.5 m= i = h0 − 3f f + 2

We have P =

40 W

2 cm 2 cm

C

1 1 2 mv cm + Iω2 2 2

R2 1 1 × 2 × 400 + ×2× × 2 2 2

100 R2

B

= 450 J i.e. n = 5

37º 4 = 20 N 5 mg sin 37 º +f lim max. mass of block B is = g

flim = µmgcosθ = 0.5 × 50 ×

GM 4 πR 3ρ = G × 3 R2 R2 4 g ∝ Rρ and m = πR 3ρ 3

g=

30 + 20 50 = = 5 kg 10 10 min. mass of block B mg sin 30º −f lim m1 = g

m2 =

R ρ g1 = 1 1 g2 R 2 ρ2

m1 = 1 kg

m and 1 m2

m2 – m1 = 4

XtraEdge for IIT-JEE

We have

20. [3]

100

ρ2 R1 × = ρ1 R 2

3

MARCH 2012

SOLUTION FOR MOCK TEST PAPER AIEEE- II

or NH3 formed is 0.16 mole or 2a = 0.16 ∴ a = 0.08 ∴ Initial mole = 0.8 Final mole = (0.2 – 0.08) + (0.6 – 0.24) + 0.16 = 0.12 + 0.36 + 0.16 = 0.64 0.64 4 ∴Ratio of final to initial mole = = 0.8 = 0.8 5

CHEMISTRY 1. 2.

[3] [3]

4.

OH CH3–Mg–Br CH3–CH–CH3 CH3–C–H H3O⊕ (Propan-2-ol) C2H5OH CH3–CH(OC2H5)2 H⊕ (Acetal) [2] CH3 .. ⊕ C2H5–N–CH3 I 3CH3–I + C2H5NH2 CH3 [4] 5. [3]

6.

[4] Laderer monassey reaction

18.

[3] Cl(g) + e– → Cl–(g) + EA ; ∆H = – EA

7.

[3] Paracetamol

19.

[4] In silica, one Si atom is attached with four oxygen atoms.

8.

[2] Fact

20.

9.

[4] Na2HPO4 is group reagent of VIth group mol [4] r ∝ M .M

[2] XeF4 contains two, XeF6 one, XeOF2 two and XeF2 three lone pairs of electrons.

21.

[4]

22.

[1] Addition of Cl2/H2O is an electrophilic addition reaction and Rate µ stability of carbocation formed.

23.

[2]

O

3.

10. 11.

[4]

12.

[1] Rate of dehydration ∝ stability of carbocation of alcohol

15.

[1] ∆Gº = – 2.303 RT logK – 4.606 × 103 = – 2.303 × 2 × 500 log K K = 100

16.

[2]Stronger is acid, weaker is its conjugate base.

17.

[3] He has highest ionization energy.

Na-liq.NH3 C2H5OH

13.

[3]

24.

2 CH2

[3] Br 2 NaNH2

OH + I – CH2

CH3–CH–CH2–Br –2HBr

N2(g) + 3H2 (g) 2NH3 (g) Initially at eq. 0.2 0.6 0 (0.2 – a) (0.6 – 3a) 2a Total mixture is 0.8; 40% of it reacts, i.e., 0.8 × 40 0.8 × 40 1 reacts to give × mole of NH3 100 100 2 [1]

XtraEdge for IIT-JEE

(ii) H2O-Zn

CHO

O CH2 HI

14.

CHO

(i) O3

H3C–C≡CH –NH3 NaNH2 ⊕

CH3–C≡C Na C2H5Br

CH3–C≡C–C2H5

101

MARCH 2012

F

F

25.

[2]

Br

C1–––––C2

Cl

MATHEMATICS 31.

Cl Br (3) F Cl (2)

Cw C––––(2) R-form

(4)

equality hold only for x > 0, x ≠1 or

At C2 : → F (3) Br(C1)FC–––C (4) Cl (2)

Br (1)

(3)––C

2

ACw S-form

1

F

32.

F

[3] The given curve is symmetric about the X-axis as shown below y

Cl

C–––––C

Cl

x'

Br

At C1 : → (3) F Br (1) Cl (2)

(0, 1) ∪ (1, ∞) 2 log 5 f(x) = = log x5 2 log x

4

Excharge Method

(4) C1––CF(Cl)Br

Exchange

(2)

O

y'

Cw C1––(3) R-form

9a

∴

The required area = 2

(1)

4ax dx

a

9a

(4) Br(Cl)FC––C

∫

[

8 a x x 27 a a − a a = 4 a = 3 3 / 2 a F (3) Cl (2)

1 Exchange

(2)––C

Br (1)

4

=

ACw S-form

33.

3

[1] Due to formation of Intramolecular H.B Conjugated base if I and II is more stable. Due to ortho effect III is more acidic than IV

[2] Edge-centre atom is shared in 4 cubic unit cells.

29.

[3] In electrochemical cells, anode = –ve

XtraEdge for IIT-JEE

λ ⇒ λ = (λ – 1 )Sλ λ −1

∑

n

(λ − 1) S λ =

λ =1

∑λ λ =1

=

[2] SPM allows only solvent molecules to pass through

28.

1 1 + + ............ ∞ λ λ2

n

∴

]

208a 2 3

[2] Sλ = 1 +

=

II > I > III > IV

27.

y2= 4ax x

(4)

At C2 : →

26.

log 5 log x

Domain x > 0, x ≠1

(3)

Br (1)

Br

, g(x) =

Domain x ≠ 0 x≠0 and x≠1

(1) Exchange Method

log x

2

2

AT C1 : → (4) C–– CF(Br)Cl

log 25

[2] f(x) =

n(n + 1) 2 7

34.

102

2π 2π cos − sin 7 7 = 1 0 [2] 0 1 π 2π 2 sin cos 7 7 here k = 7

MARCH 2012

35.

[1] We have 10 digits {0, 1, 2, ---, 9} Select any 2 and write in descending order 10.9 n = 10C2 .1 = = 45 2

36.

[3] P (Exactly two of A, B and C occur) = P(B ∩ C) + P(C ∩ A) + P(A ∩ B) –3 P(A ∩ B ∩ C) = P(B). P(C) + P(C) . P(A) + P(A). P(B) –3 P(A). P(B). P(C) 1 1 1 1 1 1 1 1 1 1 = = × + × + × – 3× × × 2 4 4 3 2 3 3 2 4 4

37.

[4] We have a ≤ xi ≤ b ; i = 1, 2, 3.........n n

∑

n

i =1

xi ≤

i =1

n

∑

na ≤

i =1

⇒ ⇒ ⇒ ⇒

y

40.

...(1)

∑

∑

12 12 , 7 7

(0,0) (1,0)(2,0)(3,0)

i =1

1 xi ≤ nb ⇒ a ≤ n

y=x

A

[2]

b

x 4x + 3y = 12

(–2,–2) n

C

B

∑x ≤ b

41.

[1]

y +2 = 0

9 ,−2 2

i

i =1

a≤ x≤b⇒–b≤– x≤–a ...(2) By (1) and (2), we get (a – b) ≤ (xi– x ) ≤ (b –a) , i = 1, 2 , ........ n |xi– x | ≤ (b –a) , i = 1, 2 , ........ n (xi – x )2 ≤ (b – a)2 ; i = 1, 2, ......n n

ax + y – 2 = 0, & 6x + ay – a = 0 a a 2 − 12 ,y= 2 solving x = 2 a −6 a −6 2 a a − 12 = 2 a2 − 6 a −6 a2 – a – 12 = 0 (a – 4) (a + 3) = 0 a = 4, a = – 3

n

∑

a≤

⇒

C (1, 1)

B (0, 1) (r,r)

( xi − x ) 2 ≤ n(b – a)2

i =1

A

var (x) ≤ (b – a)2

D (1, 0)

(r − 1) 2 + (r − 1) 2 = r

38.

[2]

⇒ ⇒ ⇒

e' e

x y + =1 ....(1) e e' Hyperbola & it's conjugate hyperbola 1 1 + 2 1 2 e1 e2 4 2

e 1

2

+ +

4

e' 2 1 2

e e' x2 + y2 = r2 p=r | 0 + 0 −1 | 1 1 + 2 2 e e'

39.

42.

1 4

4 ± 16 − 8 2

r=

4±2 2 2

[3]

2

r r r r LHS → AB . CD = (b − a ) . (d − c ) r r r r r r r r = b .d – b .c – a .d +a .c

RHS → k{| AD |2 + | BC |2– | AC |2 – | BD |2} r r r r r r r r = k { | d − a |2 + | c − b |2 – | c − a |2 – | d − b |2 } rr rr rr rr = k {– 2 (a.d + c .b ) + 2[ (a.c ) + (d .b ) ]} rr r r rr r r ...(2) = 2k {a.c + b .d − b .c − a.d }

= r ⇒ r =2

By (1) and (2) ⇒ k =

[1] First two family of lines passes through (1, 1) and (3, 3) respectively. ⇒ Point of intersection of lines belonging to third family will lie on y = x

XtraEdge for IIT-JEE

r=

r=2±

=1 =

2(r – 12) =r2 2(r2 – 2r + 1) = r2 r2 – 4r + 2 = 0

103

1 2

MARCH 2012

43.

[1] r r r r 1 1 ˆ 1 ˆ j+ k Given (a × b ) × (c × d ) = iˆ – 6 3 3 r r r r 3 Q a . b = | a || b | cos30º = 2 r Q ar, b , cr are coplaner vectors r rr ∴ [a b c ] = 0 r r r r ( a × b ) × (c × d ) r r r r r r r r = { (a × b ).d } c –{ (a × b ).c} d r r r r 1 ˆ 1ˆ 1ˆ i – j + k {By (1)} = [a b d ]c = 3 6 3

=

⇒ cosθ =

...(1) ...(2) 46.

45.

[3]

f(0) = λ [0] = 0

+ c22

=

1 2

h→0

47.

x2 + y2 ≤ 4 [1] R = {(–2, 0) , (–1, 0), (–1, 1) , (–1, –1), (0, 0), (0, 1), (0, 2), (0, –1), (0, –2), (0,0), (1, 1), (1, –1), (2, 0)} ⇒ DR = {–2, –1, 0 , 1, 2}

48.

[3]

(i) R1 is not a relation Q 4 ∉ A (ii) R2 is subset of A × B, ∴ it is a relation (iii) R3 is subset of A × B, ∴ it is relation (iv) R4 is subset of A × B, ∴ it is relation 49.

λ=?

[2] Area of ∆ formed by z, ωz, z + ωz 1 = |z|2. sin 120º 2

But it is given ...(1) ...(2)

By (1) l = – m – n (3) Case (I) If m = 0 By (3) ⇒ l = – n ∴ DR's of line (1) is

|z|2 =

1 25

|z | =

1 5

3 1 2 (z) sin (120º) = 100 2

| z + ωz| = |z| | 1 +ω| =

50.

a1, b1, c1 − n, 0, n − 1, 0, 1

51.

Case (II) If l = 0 By (3) ⇒ m = – n DR's of line (2) is a 2, b2, c2 l, m, n ⇒ 0, − n, n 0, − 1, 1 XtraEdge for IIT-JEE

π 3

+ b22

R.H.L. lim λ [0 + h] = 0 ∀ λ ∈ R

x −1 y z = = ...(1) 1 −1 2 z −3 x +1 y L2 : = = ...(2) λ 2 2 Shortest distance between the lines (1) & (2) is r r r r (a1 − a 2 ).(b1 × b2 ) r r ⇒ =1 | b1 × b2 |

⇒

⇒θ=

a22

h→0

[3] L1 :

[1] l+m+n=0 and lm = 0 ⇒ l = 0 or m = 0

+ c12

1

1 ˆ (i − 2 ˆj + 2kˆ) 6

⇒

+ b12

L.H.L. lim 5 0 − h = 0

r iˆ − 2 ˆj + 2kˆ = c= 3 44.

a1a 2 + b1b2 + c1c2 a12

104

1 1 1 | –ω2| = × 1 = 5 5 5

(Q 1 + ω = –ω2) [1] do your self [4] Let g(x) = f(x) – x2 ⇒ [g(1) = 0 , g(2) = 0, g(3) = 0 as f (1) = 1, f (2) = 4, f (3) = 9] From RT on g(x), g' (c1) = 0 for at least x ∈ (1, 2) ⇒ c1 ∈ (1, 2) RT on g (x), g' (c2) = 0 for at least x ∈ (2, 3) ⇒ c2 ∈ (2, 3) ∴ Now g' (c1) = g' (c2) = 0 ⇒ so between x ∈ [c1, c2], g" (x) = 0 ⇒ f" (x) –2 = 0 ⇒ f" (x) = 2 MARCH 2012

sin 2 x

∫ cos

52. [3]

6

x

dx =

∫

tan 2 x . sec4 x dx

∫

= tan 2 x (1 + tan2x). sec2 xdx

57.

[2] Statement I and Statement II both correct and statement II is correct explanation of Statement I.

58.

[2] P(A∩ B ) = P(A) P( B ) = P(A) ⋅ (1–P(B))

59.

[4] Contra positive of compound statement ~ (p ∧ q) → q ≡ ~ q → ~ (~( p ∧ q)) ~ (p ∧ q) → q ≡ ~ q → (p ∧ q) so, statement I is wrong. but statement II is correct.

60.

[2] Both statements are correct and statement 2 is correct explanation of statement 1

tan x = t dy du = mum–1. dx dx du Hence, 2x4.um.m um–1. + u4m = 4x6 dx

53. [3] y = um ⇒

4 x 6 − u 4m du 3 = ⇒ 4m = 6 ⇒ m = and dx 2mx 4u 2 m −1 2 3 2m – 1 = 2 ⇒ m = 2

PHYSICS

(1,2) (t2, 2t)

54. [3]

61.

[3] Statement 1 is true, statement 2 is false.

62.

[3] A halved, δ halved. 5 [1] Vfather = 60 5 For the daughter V = , after catches M = 21 21

63.

Equation of tangent at (1, 2) y . 2 = 2 (x + 1) x–y+1=0 image pt of (t2, 2t) about line x – y + 1 = 0

64.

+λ

y − 2t x −t2 − 2(t 2 − 2t + 1) = = 1 −1 2 x = t2 – t2 + 2t – 1, y = 2t + t2 – 2t + 1 x = 2t – 1 y = t2 + 1 x +1 =t y – 1 = t2 2 Eliminating

y–1= 55.

–λ

σ (2x sin θ) = λ(dx) 2ε 0 σλx sin θ dx = ε0

[1] f (x) = min ({x + 1}, {x –1}) = min ({x}, {x}) = {x}

so 56.

+

( x + 1) 2 ⇒ (x + 1)2 = 4 (y – 1) 4

4

1

−5

0

τ=

∫ f ( x)dx = 9. ∫ x − [ x]dx

65.

[2] Statement I and Statement II both correct and statement II is correct explanation of Statement I. m−n

Even

− m m n as h' (x) = x = .(x) odd n n as h' (x) is undefined at x = 0 so h' (x) does not change. sign in neighbour hood ⇒ No extreme

XtraEdge for IIT-JEE

[3] dτ = (dq)E(2x sin θ)

σλl 2 sin θ 2ε 0

[3] According to Newton law of cooling rate of loss of heat (T – T0), where T is the average temperature in the given time interval hence. 60 + 50 60 − 50 – 25 mC ∝ 10 2 (50 – T ) 50 + T ∝ 625 10 2 Solving we get : T = 42.85ºC

and mC

105

MARCH 2012

66.

1 arel.t2 2

[1] xrel = (v0)rel t +

1 2 at [arel = aman – abus = 0 – a] 2 at2 – 2v0 t + 2x = 0

or

x = v0t –

2v 0 ±

t=

4v02

f1 = f2

∴

74.

[1] vs

s

∆f = f1 – f2 v = f v − vs

a v0 + v02 – 2ax

v02 – 4ax = 0

67.

68. 69.

71.

[3]

P=

72.

76.

[2]

⇐

i=

[3]

12V

77.

∝

T / ρs l

∝

T / ρr 2 l

f ∝

3Ω

12V

T /m l

(m = mass per unit length = ρs)

106

hc e

1 1 – λ λ (in eV) 0

6.6 × 10 –34 × 3 × 108 1010 1010 – = 1.5 eV 1800 2300 1.6 × 10 –19

78.

[2] Voltage gain = β(Resistance gain)

79.

[3] E is always negative

80.

[3] ω = 100 π =

81.

[4] F = 2πrT 22 × 0.1 × 10–3 × 0.07 F =2 × 7 1 [3] (K.E.)max = – (K.E.) of boy 2 1 1 1 (2m) u2 = × mu'2 2 2 2 (K.E.)man = (K.E.) of boy 1 1 (2m) (u + 1)2 = mu'2 2 2

82.

T /ρ rl

XtraEdge for IIT-JEE

[2] Ek = =

12 = 6A 2 v f ∝ ∝ l

Binding energy

= (MO – 8Mp – 9Mn)c2

6Ω

[4] i

73.

[3] From –

BE = (Mnucleus – Mnucleuon)c2

V 2 12 2 = = 48 W R 3

2Ω

dN = λN dt n n = λN or λ = N

75.

[1] zener diode is used in parallel to load resistance is connected in R.B. [4]

v v 2 f vs = f 1 + s − 1 − s = v v ν

2ax

1km [4] Scooter ← → Bus Relative distance = (time) (relative velocity) (In uniform motion) 1000 = (100) vrel. ∴ 10 = (vs – vb) vs = 20 m/s [3] Use, δm = (µ − 1) A

70.

v – f v+v s

v −1 v −1 = f 1 − s − 1 + s v ν

a For time to be minimum

v0 =

vs

O

s

v0 ± v02 – 4ax

or t =

1 1 2 4

f2 = 4f1

– 8ax

2a

t=

ρ 2 r2 l2 = ( 2) ρ1 r1l1

T1 T2

2π T

MARCH 2012

m1v1 + m2 v2 5 = – m/s 3 m1 + m2

83.

[1] Vcm =

84.

[4] As no external force is applied ∴ (vcm = constant) = 0

85.

1 1 [4] P = (µ – 1) – R1 R2

CHEMISTRY JOKES If you didn't get the joke, you probably didn't understand the science behind it. If this is the case, it's a chance for you to learn a little chemistry. Chemistry Joke 1: Q: Why do chemists call helium, curium and barium the medical elements? A: Because if you can't helium or curium, you barium!

µ decreases, P decreases, f increases.

Chemistry Joke 2: Q: What is the name of the molecule CH2O? A: Seawater

86.

[1] M ∝ k L1 L2

87.

[2]

88.

[4] W1 = MB (cos 0º– cos 90º) W2 = MB (cos 0º – cos 60º) W1 = nW2 n=2

89.

90.

Chemistry Joke 3: Q: What do you call a joke that is based on cobalt, radon, and yttrium? A: CoRnY. Chemistry Joke 4: Q: If a mole of moles were digging a mole of holes, what would you see? A: A mole of molasses.

[3] Zener diode is always connected in RB and it act as voltage regulator.

Chemistry Joke 5: Q: What does a teary-eyed, joyful Santa say about chemistry? A: HOH, HOH, HOH!

[4] y i2

B1

Chemistry Joke 6: Susan was in chemistry. Susan is no more, for what she thought was H2O was H2SO4.

BR 30º

⇐

x i1

Chemistry Joke 7: Q: Why is potassium a racist element? A: Because, when you put three of them together, you get KKK.

B2 Be

tan 30º = B1 =

B2 B1

Chemistry Joke 8: An electron sitting in a prison asked a second electron cellmate, "What are you in for?" To which the latter replied, "For attempting a forbidden transition."

3 B2

µ0i1 µi = 3 02 2R 2R i1 1 i2 = = Amp. 3 3

Chemistry Joke 9: Q: What is the dullest element? A: Bohrium Chemistry Joke 10: At the end of the semester, a 10th-grade chemistry teacher asked her students what was the most important thing that they learned in lab. A student promptly raised his hand and said, "Never lick the spoon."

XtraEdge for IIT-JEE

107

MARCH 2012

SOLUTION FOR MOCK TEST PAPER - II BIT-SAT

or Q2 sin2θ = P2 – Q2 cos2 θ or Q2(sin2θ + cos2θ) = P2 But sin2θ + cos2θ = 1 ∴ Q2 = P2 or Q = P No need for negative sign.

PHYSICS 1.

t f = f0 1 – T At t = 0, v = 0 t = 1 or t = T When f = 0, then T

[C]

dv t = f0 1 – dt T

or dv = f0 dt – f0 vx

∫ 0

dt – f 0

0

vx = f0T – 2.

∫

[C] In order to conserve momentum, C should move with speed v in a direction opposite to that of B.

4.

[C] Percentage energy saved 1 2 mv v2 2 = × 100 = 2 × 100 1 2 v + 2 gh mv + mgh 2

t dt T

T

dv = f 0

3.

1 T

T

∫ t dt 0

=

f0 T 2 1 1 = f0T – f0T = f0T T 2 2 2

[A]

Q sin θ tan β = P + Q cos θ

Q

R' Q

5.

[C] Iω = constant, 2 T 2π MR2 × = constant or 2 = constant 5 T R 4T ' 24 T' T = 2 or 2 = 2 or T ' = 6 hours or 2 R /4 R R R

6.

[B] Limit of resolution of eye = θ =

……. (i)

R

β'

θ

Q β

P Q sin(180º– θ) ……. (ii) tan β' = P + Q cos(180º– θ)

180–θ

or tan β' =

β'

θ

90°

β

10 –3 1 mm = = 2.03 × 10–4 x x

or x =

7.

Multiplying (i) and (iii),

10 –3 2.03 × 10 – 4

m ≈ 5m

0.6 m = – 0.3 m = – 30 cm 2 1 1 1 + = v – 10 – 30

[B] (a) f = –

1 1 1 = – v 10 30

=1

XtraEdge for IIT-JEE

1.22 × 5 × 10 –7

θ=

But β' + 90º + β = 180° or β' = 90° – β tan β' = tan(90° – β) = cot β Q sin θ ∴ cot β = ……… (iii) P – Q cos θ

P 2 – Q 2 cos 2 θ

1.22λ D

= 2.03 × 10–4 rad 3 × 10 – 3 If the maximum distance at which dots are resolved is x, then =

Q sin θ P – Q cos θ

Q 2 sin 2 θ

12 × 12 × 100 ≈ 38 12 × 12 + 2 × 9.8 × 12

108

MARCH 2012

1 3 –1 = v 30 30 or v = cm = 15 cm 2 v 15 (c) m = – =– = 1.5 u – 10 (d) Object lies between principal focus and pole. So, the image is virtual and erect. →

8.

[C] R CM

[C] V∝ T ⇒

11.

12.

13.

14.

[C] Resultant amplitude =

18.

[D] In forward biasing both electrons and protons move towards the junction and hence the width of depletion region decreases. r r r r r r [A] W = Q ( E ∆r ) = F . ∆r F = Q E

i1 i2 − r1 r2

NP IS = NS IP

21.

[C] Change will move along the circular line of force because x2 + y2 = 1 is the each of circle. θ θ [A] σi = = . G = σ vG i iG σ ⇒ i = σv G

22.

Rt − [C] i = i0 1 − e L

Rt

a12 + a 22 + 2a1a 2 cos φ

0.32 + 0.4 2 + 2 × 0.3 × 0.4 × cos

π = 0.5 cm 2

[A] Suppose nA = known frequency = 100 Hz. nB = ? x = 2 = Beat frequency, which is decreasing after loading (i.e.x ↓) Unknown tuning fork is loaded so nB↓ Hence nA – nB ↓ = x↓ …… (i) → Wrong nB ↓ – nA = x ↓ …….(ii) → Correct ⇒ nB = nA + x = 100 + 2 = 102 Hz.

23.

di i0 R − L = e dt L initially t = 0 di i0 R = dt L 1 N 1 [A] = = t /T n N0 2 H 2

NA =

10

2t / 1 NA = NB 10 = 2t/2

1 + NB = 1 2

⇒ log1010 =

[C] Current through each arm PRQ & PSQ = 1 A VP – VR = 3V VP – VS = 7V VR – VS = 4V

XtraEdge for IIT-JEE

[A]

20.

V1 T V (273 + 27) = 1 ⇒ = 3V V2 T2 T2

[B] V < E 12 r E = 12 + 6 11 & E = 11 + r 10 20 On solving r = 7

17.

µ0 N 2

W = Q(e1a + e2b)

[A] Initial and final states are same in all the process. Hence ∆U = 0; in each case By FLOT; ∆Q = ∆W = Area enclosed by curve with volume axis. Q (Area)1 < (Area)2 < (Area)3 ⇒ Q1 < Q2 < Q3

=

[B] Bnet =

W = Q[e1 iˆ + e2 ˆj + e3 kˆ ] . (a iˆ + b ˆj )

⇒ T2 = 900 K → 627°C [Q T(K) = 273 + t°C] 10.

16.

19.

1(iˆ + 2 ˆj + kˆ) + 3(–3iˆ – 2 ˆj + kˆ) = 1+ 3 ˆ = – 2iˆ – ˆj + k

9.

[B] S =

→

m r + m2 r 2 = 1 1 m1 + m2

IgG I − Ig

15.

t/2

t log102 2

t = 6.62 yr 24.

[D] Potential difference across the resistance 20Ω. Which is V = i × 20 48 i= 100 + 100 + 80 + 20

25.

[A] Intensity = Power per limit area. P = pv P 0.5 p= = v 3 × 108

_____(i) _____(ii)

109

MARCH 2012

26.

[A] I =

37.

E R+r

P1 = I2R1 P2 = I2R2 Power delivered is same in both cases. 2

factor of [2]3/2 or

2

38.

E E R1 = R2 R1 + r R2 + r

R1

( R1 + r ) 2

=

( R2 + r ) 2

B∆A [A] q = R

28.

[B] F = qvB sinθ

39.

29.

[D] θ = 0 F = 0 Hence no change in the velocity.

40.

30. 31. 32.

[A] [C] Diamagnetic susceptibility mV [D] r = qB

[B] ρ =

have

negative

36.

[D]

1 P2 KL2 = or p = L 2 2M

[B] A . B = 0 ⇒ A ⊥ B

→

→

→

→

→

→

MK

→

→

∴ θ = 90°

1.

[C] Ca(OH)2 (aq.) + CO2(g) —→ CaCO3(s) 1 mol 100 g given: 0.05 × 0.5 mol ? 0.05 × 0.5 × 100 = 2.5 g = 1

2.

[A]

r=

⇒ w∝

∂y ∂x

3.

[D]

n ∝ t

1 m

5

[B] 1 atmosphere ≈ 10 Pa Also, p = hρg = 10 × 1000 × 10 Pa = 105 Pa So, total pressure is nearly 2 × 105 Pa

110

1

w ∝ M

No. of atoms per unit cell =

M

8 16 + =4 8 2

4 3 16 3 πr = πr 3 3

h mv

4.

[C]

λ=

5.

[A]

∆x. ∆v =

Here

⇒

m

Vol. of 4 atoms = 4 ×

[C] Deviation should take place at each face. Dispersion takes place at first face only.

XtraEdge for IIT-JEE

ML2 ML2 4 ML2 ML2 + = = 36 36 9 12

CHEMISTRY

∂y is – and V is along +ve At location of P, ∂x x-axis. So, VP is along +ve x-axis

35.

I=

∴ B⊥ C

ρ0 1 + γ∆T

[A] From VP = – V ×

ML2 L + M 12 6

A × C = 0 ⇒ A || C

or 1(1 + γ × 4) = 0.998 ∴ γ = – 5 × 10–4/°C Negative sign tells that for 0 – 4°C, water contracts on heating. 34.

2

I=

→

r1 V1 B2 = . r2 V2 B1

33.

So, the new time period is 365 × 2 2 days. L/3 L/6 [C]

→

materials

8 or 2 2

L/2 Using theorem of parallel axes,

R2

⇒ r = R1R2 27.

[B] T ∝ (r)3/2 Since r is doubled therefore T is increased by a

h 4πm

∆v = 104 ×

0.011 100

MARCH 2012

∆rHº =

∑ AgH º ( p) − ∑ AgH º ( R)

6.

[D]

7.

[C]

8.

[C]

∆Hg > 0

9.

[C]

1 K2 = K1

10.

[B]

pH = pKa + log

20.

[C]

21.

[D]

Bicylo[1,1,0]

δ–

1/ 2

O

Θ ⊕

Θ ⊕

R/Mgx + H–C–H → R–CH2–O/Mg x δ+ HOH hydrolysis

[CH 3COO − ] [CH 3COOH ]

R–CH2–OH+ Mg(OH)x

6 = pKa + log 1 ⇒ Ka = 10–6 11.

12.

3+

Al (aq) + 3OH¯(aq) S 3S Given pH = 4 ∴ pm = 10 ∴ [m–] = 10–10 ksp = [Al3+][OH¯]3 ⇒ 10–33 = S × (10–10)3 ⇒ S = 10–3 M

[A]

[B]

13.

[C]

14.

[C]

Al(OH)3(s)

[C]

24.

16.

[B] 2H+ + 2e– → H2

E = E° +

[D]

OH (i) H/CN CH3– C–H CH3– C–COOH (NAR) ⊕ (ii) H3O H Lactic acid

CH– Br + 2Na + Br – CH

25.

Br + Br2 CCl 4 → (Anti-addn reaction) [Syn-addn]

Br

Ether

CH3 CH3 CH–CH

CH3

CH2–OH

[B]

H

H

C=C

OH | CH2–CH2–*C–CH3 | H Chiral centre (optical activity is shown)

two geometrical isomers will be formed due to double bond. two optical isomers with one chiral centre

O O

S 4 → meso product

H 2O2

CH3

CH3 2,3-dimethylbutane H3C

CH3

E can be greater than E° if [H+] is increased.

cis-alkene

⊕ Θ

°C CH–OH + COOH 110 → HCOOH formic acid CH2–OH COOH Glycerol

0.059 [ H + ]2 log 2 [H 2 ]

[A]

..

O

CH3

[D] E°cell = E°red (c) + E°oxi(A) = – 0.41 + 0.76 = 0.35 V

18.

23.

CH3

∆Tb = Kb × m 1.8 1000 0.1 = Kb × × 180 100 Kb = 1K/m

[B]

[B]

r = k[A]2

15.

17.

22.

C2H5 OH

H

OH C2H5

19.

[B] CH2–CH=CH2 NBS →

CH–CH=CH2 | Br

[NBS. substitutes bromine at allylic position] XtraEdge for IIT-JEE

111

MARCH 2012

26.

[B]

COOH CH2

34.

[C]

CH2

35.

[B]

POP ⇒ CaSO4.

CH2 H

Ca(OH)2

CH2

H

COOH Adipic acid O || C–O

Ca

C–O || O

Cyclopentanone

27.

[B]

29.

H–C–H, do not show aldol condensation αH = 0 1 2 6 [A] 3 CH3 H3C 5 4

–2 L O

F–

Na+ Mg+2 Al+3

[C] Fact

38.

[D] Brass ⇒ Cu + Zn non metal ⇒ Cu + Sn + Zn German silver ⇒ Cu + Zn + Ni

39.

[B] CN– is a strongest ligand

40.

[C] Ionisation Isomerism

1.

[B] Let the point is (x1, y1) then … (1) 3x1 + 4y1 = 5 2 2 2 Also, (x1 – 1) + (y1 – 2) = (x1 – 3) + (y1–4)2 ⇒ 4x1 + 4y1 = 20 … (2) Solving these we get, x1 = 15, y1 = – 10

2.

[B] x = X cos 45º – Y sin 45º =

y = X cos 45º + Y cos 45º =

R

O<N 13.6 ,14.6

32.

[A]

33.

[A] Order of solubility of sulphate BeSO4 > MgSO4 > CaSO4 > SrSO4

2

X +Y

2

2

2

X 2 –Y2 + 2 2 2 2 2 2 3(x – y) + 3(x + y) + 2(x – y ) = 4 ⇒ 8x2 + 4y2 = 4 ⇒ 2x2 + y2 = 1 X –Y 3 2

∴ Correct order O–2 > F– > Na+ > Mg+2 > Al+3 [B]

X –Y

Hence equation be

Size↓ ENC ↑

31.

H

MATHEMATICS

CH3 CH3 2,3-dimethyl–2,3-butanediol [A] O

[C]

H

Bridge H ⇒ 2 Terminal H ⇒ 4

H

37.

(A) 3, 5–dimethyl cyclohexene 30.

B

[B] Order of Lewis Acid BCl3 < AlCl3

Pinacol → (vic-diols) OH OH

H3C—C—C—CH3

28.

B

36.

–CaCO3 dry distillation

O

H

1 H2O 2

3.

X +Y + 3 2

=2

[A]

(Li – Mg), (Be – Al), (B – Si) ⇒ show Diagonal Relationship

XtraEdge for IIT-JEE

112

MARCH 2012

4.

[A]

10.

[C] f (x) = | x |, g (x) = [x – 3] –8 8 <x< ⇒ 0 ≤ f (x) 5 5 Now for 0 ≤ f (x) < 1 =–3 [Q – 3 ≤ f (x) – 3 < – 2] again for 1 ≤ f(x) < 1.6 g (f (x)) = – 2 [Q – 2 ≤ f (x) – 3 < 1.4] required get = {–3, – 2}

11.

[B]

12.

[B]

3

5.

6.

7.

[C] x2 – 3xy + 2y2 = 0 ⇒ (x – 2y) (x – y) = 0 ⇒ x – 2y = 0 and x – y = 0 Also x2 – 3xy + 2y2 + x – 2 = 0 x – 2y + 2 = 0 and x – y – 1 = 0 Clearly it is parallelogram ( x – 2) 2 ( y + 1) 2 – =1 12 4 a2e2 = a2 + b2 ⇒ ae = 4 ⇒ 2ae = 8

[C]

1 + v3 dy = dx v2 After solving y3 = 3x3 log cx

13.

9.

14.

[C] Put x = 3 – h lim [– h] + [– 1 – h] = –1 + 1 = 0

x3 + 4x2 + 3x = x (x2 + 4x + 3) = x (x + 1) (x + 3) x ≠ 0, – 1, – 3 A = R – {0, – 1, – 3}

[C]

16.

[B] Integrate both sides a tan–1 x + b [log (x – 1) – ln (x + 1)] dx = 2 ( x – 1)( x 2 + 1)

∫

a tan–1 x + b [log (n – 1) – ln (x + 1) dx 1 dx = 2 – 2 x –1 x 2 + 1

∫

a 2m 2 + b 2

=

As it touch circle x2 + y2 = r2 then a2m2 + b2 = r2 + r2m2

∫

1 1 x –1 –1 – tan x ln 2 2 x +1

Compare a = –1/2, b = + 1/4

r 2 – b2 a2 – r 2

XtraEdge for IIT-JEE

tan ( xe x ) +c f ( x)

15.

=3

[B] Eqn of tangent having slope 'm' to ellipse

⇒m=

∫

sec 2 t dt = tan t + c =

h →0

[A] As the point (2 2 , 1) satisfies the eqn of director circle x2 + y2 = 25 – 16 ⇒ x2 + y2 = 9

y = mx +

[D] Put x ex = t (x + 1) ex dx = dt

∴

1+

8.

1 + ( y / x )3 dy = dx ( y / x) 2 put y = vx

v+x

[C] x2 + 4x – 5 = 0 ⇒ x2 + 5x – x – 5 = 0 ⇒ x(x + 5) – (x + 5) = 0 ⇒ x = –5, 1 Hence points are (1, 2) and (1, –2) At (1, 2) tangent on circle be x + 2y = 5 its slope is (–1/2) Also at (1, 2) tangent on parabola be 2y = 2(x + 1) Whose slope = 1 1 2 than tan θ = 1 1– 2

f (0) = f (0 +) cos 3 h – cos h λ = lim h →0 h2 2 sin 2 h. sin h =λ=–4 = lim – h →0 ( 2h) ( h)

a – 2b = –

113

1 – 2 (+ 1/4) = – 1 2

MARCH 2012

17.

[C] f ′(x) = 3x2 + 2ax + b A=3>0

dy (cos θ).sin θ = dθ cos h

D = 4 (a2 – 3b)

Q a2 – 3b < 0

∴D<0

23.

∴ f ′(x) > 0 ⇒ f is strictly increasing ⇒ no maxi./min. lie. 18.

[C]

y = –x

∴ Sn =

∑ r =1

O A= 19.

∫

2 0

2

∫

y dx =

2 0

24.

x dx = 2

so,

⇒ x > x3 2

∫ 20.

1 0

2

3

2 x dx >

∫

1 0

3

2 x dx

1 1 1 1 , , , are in A.P. or a1, a2, a3, a4 are a1 a2 a3 a4 in H.P.

25.

One-one function Range = [–1, 1] = co domain

x2 +1 – 2 2 =1– 2 ; 2 x +1 x +1 2 f (x) is minimum when 2 is maximum or (x2 x +1 + 1) is minimum i.e. x2 + 1 ≥ 1 for all x 2 =2 Hence 2 x + 1 max . 2 2 is minimum, when 2 is or f (x) = 1 – 2 x +1 x +1 maximum or f (x) = 1 – 2 = – 1

x5 – 2y4 = 0

dy 5 x 4 = dx 8 y 3 5 dy 80 =– =– 4 dx 2, 2 8×8

y 2 8 = = y1 5/ 4 5

[B] dy dy / dθ = = dx dx / dθ

XtraEdge for IIT-JEE

[D]

f (x) =

⇓ onto function

22.

… (ii)

so,

1

length =

1 1 1 1 1 1 1 1 + = + or – = – a4 a1 a3 a2 a4 a3 a2 a1

1 1 1 1 So, 3 – = – a4 a1 a3 a2 Clearly, (i) and (ii) 1 1 1 1 1 1 ⇒ – = – = – ; a2 a1 a3 a2 a4 a3

⇒ I1> I2

[D]

[B]

r =1

n 2 (n + 1) 2 +n 4

3(a2 – a3 ) a1 – a4 Also, = ; a2 a3 a1a4

–1

21.

∑

(r 3 + 1) =

… (i)

2

2x > 2x

n

tr =

a +a a1 + a4 = 2 3; a1a4 a2 a3

[C]

0<x<1

[C]

dy = tan θ dx

1 1 [C] tn = (n + 1) n + n + 2 ω ω 1 1 1 1 = n3 + n2 2 + + 1 + n 1 + 2 + + 1 ω ω ω ω 3 2 2 2 = n + n (ω + ω + 1) + n (ω + ω + 1) + 1 = n3 + 1 n

y=x

⇒

26. a cos θ

1 a – sin θ + 2 sin θ / 2 cos θ / 2

114

[B] As it is a third-degree homogeneous expression in x, y, we have y3 + y2x – µ yx2 + λ + x3 =(y + x) (y + 3x)(y + mx) = y3 + (m + 3 + 1) y2x + (3 + m + 3m) yx2 + 3mx3 ⇒ 1 = m + 4, – µ = 3 + 4m, λ = 3m, ⇒m=–3 ∴ y – 3x is third factor MARCH 2012

27.

[C] Required number = coeff. of x30 in (x2 + x3 + …..+ x16)8 = coeff. of x30 in x16 (1 + x + …..+ x14)8 1 – x15 = coeff. of x14 in 1– x

= 25

r≥ 2

if 8 – r ≥ 2r or

[C] We have R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} ∴ R = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} hence R oR–1 = {(3, 3), (3, 5), (5, 3), (5, 5)}

8 ≥r 3

33.

[A] n (S) = 6 × 6 × 6 × 6 n(E) = the number of integral solutions of x1 + x2 + x3 + x4 = 12, where 1≤ x1 ≤ 6, ….,1≤ x4 ≤ 6 = coefficient of x12 in (x + x2 + …. x6)4

2 3

∴ T[r] + 1 is greatest term ∴ Numerically largest term = | t3 | = 7C2.45.(3x)2

4

1 – x6 = coefficient of x in 1– x = coefficient of x8 in (1 – x6)4 . (3C0 + 4C1 x + 5C2x2 + ….) = 11C8 – 4. 5C2 = 125 .125 ∴ P (E) = 6× 6× 6× 6 8

2

=

29.

7! 5 2 .4 . 3. = 86016. 2! 5! 3

[A]

2 4 n 16 n (1 + 15) n = = 15 15 15

=

1 + n C115 + n C2152 + .... + n Cn15n 15

=

1 + 15k , where k ∈ N, 15

=

1 +k 15

2 4 n ∴ = 15

30.

[B]

34.

1 1 + k = 15 15

Here f (x) =

8

Hence, required probability =

–1

C3 × 5! 1 = 8! 6

10

∑x

i

35.

– 1/ 2 2 2 0 = – 1/ 2 2 2 0

Since,

[D]

i =1

10

=6

(Q A.M. = 6) 10

–1 –1 = –1 –1

∑

∑x

i

10

⇒

xi = 60

and

xi = 30,

∴

i =1

4

i =1 4

Applying R1 → R1 – R2 and R2 → R2 – R3 5 –5 0 0 5 –5 f (x) = 2 2 sin x cos x 5 + 4 sin 2 x

⇒

[D]

XtraEdge for IIT-JEE

[B] The total number of ways in which 8 persons can speak is 8P8 = 8!. The number of ways in which A, B and C can be arranged in the specified speaking order is 8C3. There are 5! ways in which the other five can speak. So, favourable number of ways is 8C3 × 5!

1+ x ⇒ f (A) = (I + A) (I – A)–1 1– x

2 2 0 –2 = 2 2 –2 0

31.

–1

32.

8–r 3 2 8–r . . = r 4 3 2r

∴ |tr + 1| ≥ | tr |

0

1

⇒ f (x) = 150 + 100 sin 2x Clearly (a) domain (– ∞, ∞) (b) range [50,250] f ( x) – 150 (c) period π (d) lim = 200 x →0 x

[C] Here |tr + 1| = 7Cr.47–r. (3x)r and | tr | = 7Cr–1 48–r (3x)r–1 |t | 7! (r – 1)!(8 – r )! 1 ∴ r +1 = × . .3 x 4 | tr | r !(7 – r )! 7!

=

–1

0

sin 2 x cos 2 x 5 + 4 sin 2 x

8

= coeff. of x14 in (1 – x2)–8 =21C14 = 116280 28.

1

∑ i =1

10

∑x

i

= 60 – 30 = 30

i =5

⇒ Mean of remaining items =

115

= 7.5

30 =5 6

MARCH 2012

36.

37.

r [A] v = sin θ nˆ r | v | = sin θ r r r r r | u |2 = | a |2 + cos2 θ | b |2 – 2 cos θ a . b = 1 + cos2 θ – 2 cos2 θ r 2 | u | = sin2θ r r r | u | = sin θ |v | = |u | r [C] Let c = x iˆ + yˆj + zkˆ r r r | a |2 = | b |2 = | c |2 = x2 + y2+ z2 = 2 r c . iˆ < 0

41.

39.

… (1)

3 x 2 – 16 = x x 2 ⇒ x – 16 = 9 ⇒x=5

1 x+ y y+z = = 2 2 2 x = z, y=1–z Put x = z, y = 1 – z in eqn (i) we get z = 1, – 1/3 1 x=z= – 3 r 1 c= (– iˆ + 4 ˆj – kˆ ) 3 r [A] P = AC + BD = AC + BC + CD r P = AC + λ AD + CD r P = λ AD + AD r P = (λ +1) AD µ = λ +1

42.

[B]

43.

1 1 3 + = b+c c+a a+b+c c+a+b+c 3 ⇒ = (b + c)(c + a ) (a + b + c )

[B]

⇒ (2c + a + b) (a + b + c) = 3 (b +c) (c + a) ⇒ 2ac + 2bc + 2c2 + a2 + ab + ac + ab + b2 + bc = 3bc + 3ab + 3c2 + 3ac ⇒ a2 + b2 = ab + c2 a 2 + b –2 – c 2 a2 + b2 – c2 1 =1⇒ = ab 2ab 2 1 ⇒ cos c = ∠c = 60º 2

⇒

C D

x – x1 y – y1 z – z1 = = a b c –2(ax1 + by1 + cz1 + d ) = a 2 + b2 + c2 x – 1 y – 3 z – 4 –2(2 – 3 + 4 + 3) = = = 2 –1 1 4 +1+1 x = – 3, y =5, z = 2 Image of point (–3, 5, 2)

[A]

α

Eh

x

β A B Let height of piller be x m. In ∆ABC h tan β = AB AB = h cot β … (1) In ∆CDE h–x tan α = AB AB = (h – x) cot θ … (ii) from (1) & (2) h cot β = (h – x) cot α ⇒ h cot β = h cot α – x cot α h (tan β – tan α) h (cot α – cot β) ⇒x= ⇒x= cot α tan β

[B] cos 2x + 2 cos2 x = 2 ⇒ 2 cos2 x –1 + 2 cos2 x = 2 ⇒ 4 cos2x = 3 3 ⇒ cos2x = 4 π ⇒ cos2x = cos2 6 π ⇒ x = nπ ± ,n∈z 6

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x 2 – 16 3 = sin–1 x x

⇒

For image of point

40.

sin–1

⇒ sin–1

x < 0s rr rr rr a.b b .c a.c r r = r r = r r | a |.| b | | a | | c | | b | | c |

38.

3 4 π + sin–1 = x x 2 3 π 4 ⇒ sin–1 = – sin–1 x 2 x 3 4 ⇒ sin–1 = cos1 x x

[B]

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44.

[A]

a = 3, b = 5, a = 4

ENGLISH

c2 + a 2 – b2 2ac 16 + 9 – 25 ⇒ cos B = =0 2(3) (4)

cos B =

1.

[B] Geraff : Incorrect spelling. • 'e' should be replaced with 'i' • The word should end with 'e' after 'ff' Giraffe : Correct spelling. Giraf : 'fe' is to be added in the end. Gerraffe : • 'Ge' is to be replaced with 'Gi' to make the correct spelling.

2.

[B] Puncture : No error. It makes the tyre flat. Puntuation : Error of spelling Correct spelling is 'Punctuation' Hence 'c' is missing. Pudding : No error It is used as 'Dessert' Pungent : No Error It is some what 'sharp' and 'shrill'.

3.

[A] Luxurious : (Plush) Something full of all 'amenities' making life 'cozy' and 'snug'. Delicious : Irrelevant as it means 'something very tasty.' Comforting : 'Irrelevant' as it means 'giving necessary comforts', whereas 'Plush' means more than comforts. Tasty : (Irrelevant) It means 'delicious'

4.

[A] Lively : Correct synonym to 'sprightly' as both means, 'someone dashing/energetic/enthusiastic'. Beautiful : (Irrelevant) Sportive : (Irrelevant) Intelligent : (Irrelevant)

5.

[D] Wicked : It is almost a synonym to 'Astute' Impolite : Irrelevant because it is the antonym of 'polite'. Cowardly : Irrelevant as it is the opposite of 'bravely'. Foolish : (It's the correct antonym of 'Astute' which itself means 'clever, shrewd'.

6.

[D] Deadly : It means 'Fatal'. Hence, this is not a proper antonym to 'innocuous'. Ferocious : It means 'horrible' Hence, irrelevant to the opposite of 'innocuous'.

⇒ ∠B = 90º β β ∴ sin + cos = sin 45º + cos 45º 2 2 1 1 + = 2 = 2 2 45.

sin θ + cosec θ = 2 1 ⇒ sin θ + =2 sin θ ⇒ sin2θ – 2 sin θ + 1 = 0 ⇒ (sin θ – 1)2 = 0 ∴ sin11θ + cosec21θ = (1)11 + (1)21 = 2

[A]

LOGICAL REASONING 1.

[D] The pattern is x2 +1, x2 + 2, . . . . Missing number = 28 × 2 + 3 = 59

2.

[A]A car runs on petrol and a television works by electricity.

3.

[A] All except Titans are planets of the solar system.

4.

[C];

6.

[D]

7.

[B] The third figure in each row comprises of parts which are not common to the first two figure.

8.

[A]

9.

[C]

10.

[A]

5. [B]

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MARCH 2012

Poisonous : It means 'venomous'. Hence, an irrelevant 'antonym'. Harmful : It is a perfect antonym of innocuous which itself means 'harmless'. 7.

8.

9.

10.

[D] Corruption : Irrelevant Worldliness : Irrelevant Favouritism : Irrelevant Nepotism : (Correct Answer) because It's a kind of corruption in which the authority in power takes the advantage of giving opportunity to their relatives in their self interest. [B] Cross : (to pass by, to intersect) It means different Hence, irrelevant. Shuttle : (Proper answer) It's a kind of "regular beats" of an air flight or bus service between the two stations. Travel : It means to journey. Hence, irrelavent. Run : (to move regularly) Hence, irrelevant. [D] Only 1 is correct : Inappropriate answer because sentence 1 can't be correct using 'practise' as it is a verb, whereas the required word should be a noun. Only 2 is correct : Sentence 2 is also wrong because the word 'practice' is wrongly used as a verb. It should be a verb like 'practise'. Hence, incorrect answer. Both the sentences 1 and 2 are correct. This is not relevant. Both the sentences 1 and 2 are not correct. Correct option, if both the words, i.e. 'practice' and 'practise' are interchanged respectively, it really makes a meaningful sentence. [C] Sentence 1 is correct : This option is wrong because the word 'ingenuous' means 'frank and simple' which is inappropriate. Sentence 2 is correct : This option is also wrong because the word 'ingenious' means 'clever or prudent' and this is inappropriate. Both the words, i.e. 'ingenuous' and 'ingenious' if interchanged together respectively, it really makes both the sentences meaningful. Hence, appropriate option. Both the sentences can't be interchanged. This is an incorrect option because words have been misinterpreted together. Incorrect option.

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11.

[C] Far off : It can't be used in place of 'aloof' as far off' means long-long ago. Hence, incorrect alternative . Introvert : It means 'self-centred', Hence, It is an incorrect alternative. distance : This is an appropriate word because one of the meaning of 'aloof' is distant also while keeping distance between two nouns. Depressed : (it means 'hopeless') Hence, quite irrelevant.

12.

[A] "Meatless days" This is the name of a novel. Hence, no error is there. Have been made : (Erroneous) Because 'have' should be replaced with 'has' because 'meatless days' is a singular noun. Into a film : No error in this part of the sentence. No error : Incorrect option because there is an error in the sentence.

13.

[C] Looking forward : (No error) This is a phrase. 'to' (no error) This is a preposition. 'Meet you here' (erroneous) Because 'meet will be replaced with 'meeting' Phrase 'looking forward to' is followed by present participle (V. I + ing) form of the Verb. No error : (incorrect option) Part 'C' is erroneous.

14.

[C] Good and Evil This is a wrong interpretation. Former and Latter : Wrong interpretation. For and against a thing. Appropriate option as it really suits the Idiom ins and outs. Foul and Fair : (by hook or by crook) This is an inappropriate option.

15.

[A] Broke out : (to start suddenly) 'Correct and relevant' option because it is used for 'wars' and 'diseases' e.g. cholera broke out in Surat in 1985. Set out : (to start) it is different because it is used when one leaves for somewhere e.g. He set out on his long voyage to Achilese. took out : (incorrect use) Because it means differently. e.g. He took out a one rupee coin to give to the beggar. Went out : (Incorrect use) Because meaning is different e.g. : The light went out when I was preparing for my Board Exams. Hence, inappropriate option. MARCH 2012

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Xtraedge March 2012

Published on Mar 15, 2012

Xtraedge for IIT-JEE is a Monthly Magzine for Engineering Aspirants.

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