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Volume - 5 Issue - 6 December, 2009 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009

Editorial

Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected] Editor : Pramod Maheshwari [B.Tech. IIT-Delhi] Analyst & Correspondent Mr. Ajay Jain Cover Design & Layout Mohammed Rafiq Om Gocher, Govind Saini Circulation & Advertisement Ankesh Jain, Praveen Chandna Ph (0744)- 3040007, 9001799502 Subscription Sudha Jaisingh Ph. 0744-2500492, 2500692 © Strictly reserved with the publishers • No Portion of the magazine can be published/ reproduced without the written permission of the publisher • All disputes are subject to the exclusive jurisdiction of the Kota Courts only. Every effort has been made to avoid errors or omission in this publication. In spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from.

Unit Price Rs. 20/Special Subscription Rates 6 issues : Rs. 100 /- [One issue free ] 12 issues : Rs. 200 /- [Two issues free] 24 issues : Rs. 400 /- [Four issues free]

Dear Students, All of us want to get organized. The first thing in getting organized is to find our obstacles and conquer them. Make a beginning by conquering obstacles for starting in right earnest. Sometimes the quest for perfectionism holds us back. We occasionally feel that we should start when we have enough time to do a job thoroughly. One way to tackle this kind of mindset is to choose smaller projects or parts of projects that can be completed within 15 minutes to one hour or less. It is important to keep yourself motivated. Approach your projects as something which are going to give you pleasure and fun. Reward yourself for all that you accomplish no matter how small they may be. Never hesitate to ask for help from a friend. Make all efforts to keep your motivational level high. You might feel overwhelmed because you are focusing on every trivial thing that needs to be got done. How do you act or react to your life? When you are merely reacting to events in your life, you are putting yourself in a weak position. You are only waiting for things to happen in order to take the next step in your life. On the other hand when you are enthusiastic about your happiness you facilitate great things to happen. It is always better to act from a position of power. Never be a passive victim of life. Be someone who steers his life in exactly the direction he wants it to go. It is all upto you now. If you do what you have always done and in the way you have done it you shall get only such results which you have always got. Getting organized requires that not doing things that cause clutter, waste of time and hurt your chances adversely of realizing your goals. You should concentrate only on doing things that eliminate clutter, waste of times and hurt your chances adversely of realizing your goals. You should concentrate only on doing things that eliminate clutter, increase your productivity and provide the best chance for achieving your goals. The first step should be to stop leaving papers and other things on tables, desks, counter tops and in all kind of odd place. The more things you leave around in places other than rightful places the quicker the clutter will accumulate. Keep things in their assigned places after you have finished using them. It does not take along to put something away. If you leave things lying around they will build into a mountain of clutter. It could take hours if not weeks or months to trace them and declutter the atmosphere. Presenting forever positive ideas to your success. Yours truly

Owned & Published by Pramod Maheshwari, 112, Shakti Nagar, Dadabari, Kota & Printed by Naval Maheshwari, Published & Printed at 112, Shakti Nagar, Dadabari, Kota.

Editor : Pramod Maheshwari XtraEdge for IIT-JEE

Pramod Maheshwari, B.Tech., IIT Delhi

1

DECEMBER 2009

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DECEMBER 2009

Volume-5 Issue-6 December, 2009 (Monthly Magazine) NEXT MONTHS ATTRACTIONS

CONTENTS INDEX

Regulars ..........

Key Concepts & Problem Solving strategy for IIT-JEE. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics, Chemistry & Maths Much more IIT-JEE News. Xtra Edge Test Series for JEE-2010 & 2011

PAGE

NEWS ARTICLE

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IITian ON THE PATH OF SUCCESS

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KNOW IIT-JEE

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IIT-Jodhpur to begin functioning from February - March 2010 IIT-K students develop autonomous vehicle Dr. Rajeewa Arya

Previous IIT-JEE Question

Study Time........ S Success Tips for the Month • The greatest adventure is what lies ahead. • Fixers believe they can fix. Complainers believe they can complain. They are both right. • The tire model for motivation: People work best at the right pressure.

DYNAMIC PHYSICS 8-Challenging Problems [Set# 8] Students’ Forum Physics Fundamentals Reflection at plane & curved surfaces Fluid Mechanics

CATALYST CHEMISTRY

• People who expect to fail are usually right. • The path to success is paved with mistakes. • You've got to cross that lonesome valley. You've got to cross it by yourself.

DICEY MATHS

Test Time ..........

• Learn to mock the woe-mongers.

XTRAEDGE TEST SERIES

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Mathematical Challenges Students’ Forum Key Concept Monotonicity, Maxima & Minima Function

• Appreciate what your brain does. In case nobody else does. • Be confident. Even if you are not, pretend to be. No one can tell the difference.

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Key Concept Carboxylic Acid Chemical Kinetics Understanding : Inorganic Chemistry

• Trust the force, Luke. • Use your feelings or your feelings will use you.

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Class XII – IIT-JEE 2010 Paper Class XII – IIT-JEE 2011 Paper Mock Test CBSE Pattern Paper -1 [Class # XII]

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DECEMBER 2009

IIT-Jodhpur to begin functioning from February - March 2010 Indian Institute of Technology in Rajasthan is all set to begin functioning from the MBM Engineering College, here from next academic session commencing in February-March 2010. The decision to this effect was made after the visit of a central team headed by additional secretary, ministry of Human Resource Development, Ashok Thakur during recent visit to different sites, which included the sites proposed for the construction of the IIT-Jodhpur and the existing engineering college of the city. So, if everything goes well, the next session of IIT-J, which is presently being run from the IITKanpur campus, will start functioning from the campus of this college here. Thakur, who himself approved the MBM college during a visit on Saturday, said, "The final decision is to be taken by the ministry, to whom we will submit the report in a 2-3 days." Agarwal, who was quite ecstatic following the visit of the team, expressed hope that the next session of the IIT will start here from February–March 2010.

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He added that the new building of the IIT will take atleast 2-3 years, but owing to repeated reminders by IIT-Kanpur citing its inability to continue running the session of IIT-J from its own campus, there is a growing pressure to shift it to Jodhpur.

IIT-K students develop autonomous vehicle True to its reputation of being ahead in technological developments, the Indian Institute of Technology, Kanpur, in collaboration with the Boeing Company is all set to unveil a new autonomous vehicle `Abhyast' as part of its Golden Jubilee celebrations. Prof Shantanu Bhattacharya, faculty in mechanical engineering department of IIT-K and coordinator of the project, while talking to TOI said, "The vehicle will set new landmarks in terms of applied robotics research. Besides being unique, it will probably be one of its kind in India. Such vehicles play important role in defence applications and disaster management plans." He added that quite a few modules of such vehicles have already been developed by the United States and some other countries. But, India has so far not produced vehicles of this nature. Further, Prof Bhattacharya informed, `Abhyast' will serve as a 4

first running prototype for such vehicles in the country. "It has a very small footprint -- 30x30x15 cm -- and thus it can be easily carried by soldiers and relief workers for disaster management operations. This technology was widely used by the US in investigating the World Trade Centre attack as well as in the war zones of Afghanistan and Iraq," he added. Earlier this year, as a part of its University relations programme, Boeing decided to actively collaborate with IIT-K to foster research and innovation among undergraduate students. As part of this endeavour, eight students of IIT-K -- Abhilash Jindal, Ankur Jain, Faiz Ahmed, Gaurav Dhama, Palash Soni, Shishir Pandya and Sriram Ganesan -- were selected by Prof Bhattacharya to work on the project. "My role was mainly confined to that of a mentor. `Abhyast' in fact the result of students' labour and skills and should serve as an inspiration for other students," said a beaming Prof Bhattacharya. `Abhyast' is a fully autonomous vehicle capable of navigating in unstructured and unknown environments. The user needs to specify only the end coordinates where he wants the vehicle to reach, and the task of reaching there would be taken by the vehicle itself, requiring no intervention by the user. The DECEMBER 2009

vehicle is equipped with high end sensors like GPS, IMU and SONARS to navigate and avoid obstacles in its vicinity. The vehicle has a tank like chassis design that allows it to move even on uneven and slippery terrain, thus making it a robust vehicle for warlike and other disastrous situations.

IIT forecast system for Oman, Maldives Ocean State Forecasting System (OSFS), a technology developed by Indian Institute of Technology, Kharagpur, will now be adopted by Oman and Maldives. The World Meteorological Organisation (WMO) feels that OSFS should be immediately adopted by other nations that have sea coasts. The system, that will continuously measure height, direction and period of waves, will help shipping and navigation. IIT-Kgp was given the prestigious project jointly by the union ministry of ocean development and the department of science and technology. It was headed by the institute's former director, S K Dubey. The model was developed by the department of ocean engineering and naval architecture. It was completed about a year ago and handed over to the India Meteorological Department by the institute for adoption. Dubey is presently in IIT Delhi to coordinate the project. "WMO was so impressed that it immediately referred the technology to all the coastal nations. But it cannot be transferred without proper training to end users, so we will

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be training meteorologists from other countries in batches. This will start with Oman and Maldives. Their scientists will be on the campus for an intensive training," Dubey said.

IITs come up with their RTI 'shield' Stung by the exposure of admission anomalies in recent years, the IIT system has come up with an innovative method of blocking transparency even as it agreed to give data under RTI on the marks obtained by the four lakh candidates in this year’s joint entrance examination (JEE). It insisted on giving the data only in the hard copy running into hundreds of thousands of pages rather than in the more convenient form of a CD.

In his first mail to CIC on October 2, Barua said that as there were a number of RTI applications seeking the CD, “we are apprehensive that this request for electronic data is to profit from it by using it for IIT JEE coaching purposes (planning, targeting particular cities, population segments, etc).” The reference to the coaching institutes is reminiscent of the recent controversy over the move to raise the bar on 12th class marks to be eligible for IIT selection. Asserting that IITs had “nothing to hide regarding the results”, Barua said, “We are ready to show the running of the software with the original data to the CIC, if it so desires.”

The information seeker, Rajeev Kumar, a computer science professor in IIT Kharagpur, is crying foul. For, the hard copy would not only result in a steep increase in the cost of information (running into six figures) but also make it almost impossible for him to detect irregularities in the latest JEE as he did in the three previous ones by analyzing the electronic data that had then by given to him under RTI.

As a corollary, Barua made an issue of the fact that Kumar “has not asked to see the data, but he wants an electronic version delivered to him. Why is this so?” Kumar responded to that by pointing out that the irregularities he had uncovered in the JEE of the previous three years was on the basis of “compute intensive scientific calculations and analysis, which could not have been done just by looking at the data.”

As a result of this change in the strategy of the IIT system, central information commissioner Shailesh Gandhi fixed a hearing for November 6 specially to resolve this soft vs hard debate. The hearing follows the unusual reasons given by Gautam Barua, director of IIT Guwahati and overall in-charge of JEE 2009, for his failure to comply with the CIC’s disclosure direction passed on July 30.

Barua’s explanation in his subsequent mail on October 3 is: “By seeing, I meant that the appellant could come to IIT Guwahati and view the data, see the software being run, etc.” He added that if this option was unacceptable to CIC, “we will wish to provide the data in hard copy form, the costs of printing having to be borne by the appellant.”

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DECEMBER 2009

If Kumar is pressing that the data be given to him “in the form in which it is originally available”, it is because the access to the electronic data of the previous three years helped him unearth, for instance, the shocking fact that general category candidates got into IITs after scoring in JEE as little as little as 5% in Physics and 6% in Mathematics.

Kerala seeks second IIT report on record Kerala has filed an application in the apex court urging it to take on record the second part of the report titled Seismic Stability of Mullaiperiyar Composite Dam, submitted by D K Paul, Professor of earthquake engineering department, Indian Institute of Technology, Roorkee. The first part of the report Structural Stability of Mullaiperiyar Dam Considering Seismic Effects — Part I — Seismic Hazard Assessment was submitted by IITRoorkee in May 2008. The said report is already before the apex court, it averred. Kerala would like to file the second part of the report in support of its argument that Mullaiperiyar dam is not safe for storage, the application stated. The report stated that the earthquake safety of old concrete or masonry gravity dams under moderate to strong ground motions is of great concern. Although there is no evidence of catastrophic failure of gravity dams, yet the possibility of tensile cracking is never ruled out. The finite element analysis of dam subjected to static and seismic

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loading shows tensile stresses at the heel of dam-foundation interface, the report indicated. “The Mullaiperiyar dam is a composite gravity dam built during 1887-1895. The front and rear faces of the dam were built with un-coursed rubble masonry in lime surkhi mortar. The hearting is constructed of lime surkhi concrete. It lies in seismic zone III as per seismic zoning map of India. The 176 feet-high composite gravity dam is now over 114 years old,” the report went on to say.

IITs told to candidate details

reveal

The Central Information Commission has ordered the IITs to disclose most details of candidates who sat the 2009 entrance examination, rejecting the institutes’ argument that revealing candidates’ names would be a breach of their privacy.

IITs want to be accredited by statutory body

India’s apex watchdog for the Right to Information Act has ordered the IITs to reveal the names, addresses, pin codes and marks of all students who appeared in the Joint Entrance Examination this year.

As the government wants to make accreditation mandatory for all institutions, the IITs have said they would like to be accredited by a statutory body and not by the National Board of Accreditation (NBA).

In its November 6 order, the commission asked IIT Guwahati, the chief organiser among the IITs of the 2009 examination, to disclose by November 25 the information sought by the appellant.

The IIT directors have told the government that they have no objection to accreditation of the institutes, but insisted that the accreditation agency should be a statutory and autonomous organisation.

The order follows efforts by the IIT to withhold information on candidates who appeared in the 2009 JEE despite earlier orders mandating the release of similar data on IIT candidates over the past three years.

They expressed these views at the meeting of the IIT Council, the highest decision making body for the IITs, held here last month. HRD Minister Kapil Sibal, who is the chairman of the council, told them that the government would set up an accreditation agency by introducing a bill in the Parliament soon. "The directors said the accreditation should be conducted by a statutory body

The order is significant because a similar disclosure in 2006 revealed discrepancies between cutoff marks used by the IITs that year and the cutoffs arrived at by using the formula the institutes claimed to have used.

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At least 994 students, who cleared the cutoffs arrived at by using the formula the IITs claimed to have used, were denied admission because the institutes used different cutoffs. DECEMBER 2009

The IITs have till now been unable to explain how they arrived at the cutoffs — by using the formula they claim to have used. The appellant in the 2009 case is the parent of a student who appeared in the controversial 2006 examination and is trying to use the RTI Act to ensure that the IITs do not repeat their errors. The IITs, in the 2009 case, argued that the release of candidate details sought by the appellant could compromise the privacy of these candidates. The appellant had sought the registration numbers, names, gender, parents’ names, pin codes and marks in physics, chemistry and mathematics of all students who appeared in the 2009 examination. The appellant has expressed concern that the IITs may be admitting the children of institute administrators or certain faculty members despite poor marks, and has argued that details he has sought would help clarify his doubts. The commission, in its order, has argued that while providing email addresses and mobile phone numbers of candidates would constitute a violation of privacy, merely disclosing their names and addresses would not.

Bombay (IIT-B), highlighted this on Monday. Moudgalya, who heads the Centre for Distance Engineering Education Programme (CDEEP) at the IIT, was delivering the keynote address at the launch of kPoint, a software solution for interactive learning and training. kPoint, developed by city-based Great Software Laboratory (GSL), was launched by noted computer expert Vijay Bhatkar, creator of India's Param series of supercomputers. Heads and professionals from leading IT companies as well as principals of engineering institutions were present at the occasion. Open source software refers to computer software provided under a license that is in the public domain. "Open source software has a distinct cost advantage over the expensive commercial software packages. However, a considerable marketing effort is required to secure a greater and wider audience of students for courses transmitted live using ICT tools based on open source software," Moudgalya said.

Open source software needs marketing

"Open source software is often sufficient in most distance education programmes, except for some niche academic segments. However, academic institutions don't train students in using good open source software," he further stated.

PUNE: There is a need for greater promotion of the use of open source software for information and communication technology (ICT)-based teaching and learning. Professor Kannan M Moudgalya of the Indian Institute of Technology,

Moudgalya gave an overview of the CDEEP's involvement in the Talk to teacher' programme, which is funded by the Union human resource development ministry and aims to train students as well as teachers in higher

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7

education. IIT-B started disseminating its courses live on the internet nearly a decade ago. For the last two years, he stated, the CDEEP has been using the education satellite Edusat, provided by the Indian Space Research Organisation, and has raised a network of 75 centres for transmission of live courses. In his brief address, Bhatkar made out a strong case for Indian ICT professionals acknowledging and adopting technologies and innovations developed indigenously. Sunil Gaitonde, chief executive of GSL, said, "Technology must bridge the gap between the growing number of learners and lesser number of teachers. The prevailing knowledge economy needs highly skilled workers and the existing faculty crunch can be tackled only through apt use of technology." Gaitonde said: "Factors like grassroots videos, collaborations, mobile broadband, data mash-ups, collective intelligence and social operating systems are bound to make a sea change in the way education is delivered." College of Engineering, Pune (CoEP) principal Anil Sahasrabudhe, Vishwakarma Institute of Technology principal Hemant Abhyankar, Persistent Systems chief Anand Deshpande and founderCEO of music education web portal ShadjaMadhyam, Nandu Kulkarni, were among those present at the event.

DECEMBER 2009

Success Story This article contains story of a person who get succeed after graduation from different IIT's

Dr. Rajeewa Arya B.E., M.E.(IIT – Kanpur) Chief Executive Officer, Moser Bear (I), Ltd.

Dr. Rajeewa (Rajiv) Arya, M. Tech., Ph. D is presently

includes product design, process scale-up, process

the Chief Executive Officer l at Moserbaer Photovoltaic

transfer, piloting and start-up of a thin-film solar module

(MBPV) in New Delhi, India. He was previously the COO

plant. He has maintained a professional interest in many

& CTO for the Thin Film Vertical. He joined MBPV as a

aspects of renewable energy components and systems.

Senior Vice-President & CTO (Thin Film) in September,

Dr. Arya holds a Masters of Science degree in Solid-State

2007. and Electrical Engineers (IEEE) and the Materials

Physics from Jadavpur University, Calcutta, India and a

Research Society (MRS). Prior to that Dr. Arya was a

Master of Technology degree in Material Science from the

founder, Vice-President and CTO at Optisolar (previously

Indian Institute of Technology, Kanpur, India. He obtained

called Gen3Solar) in Hayward, California. Before founding

his Ph.D. in Engineering from Brown University, Rhode

Gen3solar, he was the Director of Oregon Renewable

Island, in 1983. He has extensive management training in

Energy Center (OREC), an academic/research center at

Total Quality Management, Finance, Project management,

the Oregon Institute of Technology (OIT).

and Technology Innovation Management.

Dr. Arya launched Arya International, Inc., a Solar

Dr. Arya has authored and co-authored over 100

Technology and Business Consulting firm, in 2003. Prior

technical papers and holds 6 U.S. Patents. He is the

to that Dr. Arya worked at Solarex/ BPSolar for 19 years

recipient of the “Outstanding Paper Award” at the 7th

in various capacities, from Scientist to Executive Director,

PVSEC, the “Team of the Year” award from Solarex

thin-film technology.

Quality Process, and his group received an R&D 100

He has over 25 years experience in thin-film solar cells

award for the Power view Product. He chaired the

and modules. His R&D activities have centered on

Program Committee for the 29th IEEE Photovoltaic

material and device aspects of three types of thin-film

Specialists Conference in 2002. He is a member of the

solar cells and modules – amorphous silicon, copper-

Institute of Electronics.

indium-diselenide, and cadmium telluride. His work

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8

DECEMBER 2009

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9

DECEMBER 2009

KNOW IIT-JEE By Previous Exam Questions

FB=πR2hσg

PHYSICS 1.

A circular disc with a groove along its diameter is placed horizontally on a rough surface. A block of mass 1 kg is placed as shown. The co-efficient of friction in contact is µ = 2/5. The disc has an acceleration of 25 m/s2 towards left. Find the acceleration of the block with respect to disc. Given cos θ = 4/5, sin θ = 3/5. [IIT-2006]

θ h/2

mg

l . Let OG = y 2 For vertical equilibrium FG = FB ⇒ (M + m)g = F B ⇒ πR 2Lρg + mg = πR 2 l σ g

25 m/s2 θ

OB =

l=

πR 2 Lρ + m

...(1) πR 2 σ Now using the concept of centre of mass to find y. Then My1 + my 2 y= M+n Since mass m is at O the origin, therefore y2 = 0 M(L / 2) + m × O ML ∴ y= = M+m 2(M + m)

Sol. Applying pseudo force ma and resolving it. Applying Fnet = ma x for x-direction ma cos θ – (f1 + f2+ = ma x ma cos θ – µN1 – µN2 = ma x ma cos θ – µma sin θ – µmg = ma x ⇒ a x = a cos θ – µa sin θ – µg 4 2 3 2 – × 25 × – × 10 = 10 m/s2 5 5 5 5

=

(πR 2 Lρ)L

2(πR 2 Lρ + m) Therefore for stable equilibrium l >y 2 πR 2 Lρ + m (πR 2 Lρ)L ∴ > 2(πR 2 σ) 2(πR 2 Lρ + m)

A wooden stick of length L, radius R and density ρ has a small metal piece of mass m (of negligible volume) attached to its one end. Find the minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in equilibrium in liquid of density σ(>p). [IIT-1999] Sol. For the wooden stick-mass system to be in stable equilibrium the centre of gravity of stick-mass system should be lower than the centre of buoyancy. Also in equilibrium the centre of gravity (G) and the centre of buoyancy (B) lie in the same vertical axis. The above condition 1 will be satisfied if the mass is towards the lower side of the stick as shown in the figure. The two forces will create a torque which will bring the stick-mass system in the vertical position of the stable equilibrium Let l be the length of the stick immersed in the liquid. 2.

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L/2 (πR2Lρ)g

θ

Then

= 25 ×

C

⇒ ∴ 3.

10

...(2)

m ≥ π R2L ( ρσ – ρ) minimum value of m is πr2L ( ρσ – ρ)

A gaseous mixture enclosed in a vessel of volume V consists of one mole of a gas A with λ (=Cp/Cv) = 5/3 and another gas B with λ = 7/5 at a certain temperature T. The relative molar masses of the gases A and B are 4 and 32, respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation PV19/13 = constant, in adiabatic processes. [IIT-1995] DECEMBER 2009

(a) Find the number of moles of the gas B in the gaseous mixture. (b) Compute the speed of sound in the gaseous mixture at T = 300 K. (c) If T is raised by 1 K from 300 K, find the % change in the speed of sound in the gaseous mixture. (d) The mixture is compressed adiabatically to 1/5 of its initial volume V. Find the change in its adiabatic compressibility in terms of the given quantities. Sol. (a) The ratio of specific heat of mixture of gases (C p ) m γm = * m stands for mixture. (C v ) m n A C pA + n B C pB Also (Cp)m = nA + nB

Mass of the gas, m = nA MA + nBMB = 1 × 4 + 2 × 32 = 68 g/mol mol = 0.068 kg/mol 19(1 + 2) × 8.314 × 300 = 400.03 ms–1 13 × 0.068 (c) We know that the velocity of sound ∴ v=

v=

(Cv)m =

R 3R = 5 / 3 −1 2 3R 5R ∴ (Cp)A = R + (Cv)A = R + = 2 2 R 5R For Gas B (Cv)B = = 7 / 5 −1 2 5R 7R ∴ (Cp)B = R + = 2 2 (C p ) m n A C pA + n B C pB γm = = (C v ) m n A C vA + n B C vB For Gas A (Cv)A =

⇒ ⇒ ∴ ∴

1× (5R / 2) + n B (7 R / 2) 5 + 7n B = 1 × (3R / 2) + n B (5R / 2) 3 + 5n B According to the relationship PV19/13 = constant we get γ m = 19/13 5 + 7n B 19 ∴ = ⇒ nB = 2 mol. 3 + 5n B 13 Alternatively we may use the following formula n1 n1 n2 = + γ m −1 γ1 − 1 γ 2 −1 where γm = Ratio of specific heats of mixture (b) We know that velocity of sound in air is given by the relationship

∴

=

v= Also, ⇒ ∴

γ (n A + n B )RT = m V× V

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γR (T + ∆T ) M

1/ 2

v + ∆v = v

dP γPV γ −1 = = γPVγ–1 dV dV Vγ −dP = –γ P dV / V Bulk Modulus B = γP 1 1 Compressibility K = = B γP 1 1 K1 = and K2 = γP1 γP2 ∆K = K2 – K1 =

1 1 1 1 1 – = − γP2 γP1 γ P2 P1 γ

γ

∴ Since the process is adiabatic, P2V 2 = P1V 1 V ∴ P2 = P 1 1 V2

γ

= P1

γ

γ V1 = P1V 1 V1 / 5

1 1 1 1 1 − = − 1 γ γ P1 5 P1 γP1 5 γ ( n A + n B ) RT (1 + 2) × 8.31× T 24.93 T P1 = = = V V V 1 1 ⇒ ∆K = − 1 19 T 519 /13 × 24.93 × 13 V V = – 0.025 Pa–1 T This unit Pa–1 is valid when V, T are taken in S.I. units. ∴ ∆Κ =

γP m where d = density = d v PV = (nA + nB)RT (n A + n B) PV = RT V v=

γRT and v + ∆v = M

T + ∆T 1 + ∆T = T T ∆v 1 ∆T ⇒ 1+ =1+ v 2 T ∆T when ∆T 2d It is observed that 1st bright fringe is observed in front of one of the slit.

For the given circuit

3.

a

R R

4R 8R

4.

What should be the value of R so that equivalent

(A)

5.

Relation of the maximum wavelength missing with wave length of Blue light is (B) 3λb

(C) 4λb

(D) 5λb

b

12R

2 Ω 5

(C) 2Ω

(A) 2λb

R

R

resistance between terminals a and b is 1Ω -

2 λ max. = .λ b 3 ( missing )

(B)

5 Ω 2

(D) 15Ω

If current passing through the circuit is 1 amp then(A) Potential difference across 4R and 8R is in the ratio of 1 : 2 (B) Potential difference across ab is 1 volt when measured by an ideal voltmeter if R =

With blue light if the slit widths are made unequal then (A) Position of 1st bright fringe will not be in front of one of the slit (B) Dark fringe which was of black colour earlier with same slit widths now is of blue colour

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2R 3R

If monochromatic light source of blue color is replaced by the white colored light source then maximum wavelength which is missing in front of one of the slit is (A) Never of indigo and violet colors (B) It is always of less than blue color (C) Missing wave lengths can have wave length more or less than blue color (D)

2.

R

2 Ω 5

(C) Maximum potential difference will appear across 12R resistance (D) Maximum potential difference will appear across 3R resistance

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DECEMBER 2009

Passage # 3 (Ques. 6 to 8)

SCIENCE TIPS

Behaviour of capacitor in electric circuits is very typical because of it's energy storing nature. Capacitor behaves in just opposite manner to

• What is the expression for growing current, in LR R − t circuit ? ® I = I0 1 − e L

inductor, Inductor 'L' which is measured in Henary in SI system stores the energy in magnetic field instead of capacitor which stores in electric field

• What is the range of infrared spectrum ? ® This covers wavelengths from 10–3 m down to 7.8 × 10–7 m

Inductor opposes the change in current and capacitor opposes change in voltage. Behaviour of inductor: t=0

open switch

t→∞

• What is the nature of graph between electric field and potential energy (U) ? ® The nature of the graph will be parabola having symmetry about U-axis

L/R is known as time constant of R-L series circuit which is measured in ohm

• Why no beats can be heard if the frequencies of the two interfering waves differ by more than ten ? ® this is due to persistence of hearing

Closed switch

Steady State

For the electric circuit shown ε1

R1

R

C

ε2

6.

• Why heating systems based on steam are more efficient than those based on circulation of hot water ? ® This is because steam has more heat than water a the same temperature

R2

• Can the specific heat of a gas be infinity ? ® Yes

If capacitance C varies even after that energy stored

• What is the liquid ascent formula for a capillary ? 2T cos θ r ®h= – γpg 3 where h is the height through which a liquid of density ρ and surface tension T rises in a capillary tube of radius r

in capacitor is zero at steady state then R 1 ε1 = R 2 ε2 (C) ε1 + ε2 = 0

(A)

7.

• What is the expression for total time of flight (T) 2u sin θ for oblique projection ? ®T= g • The space charge limited current iP in the diode value is given by ® iP = k Vp3/2

Time constant for the circuit (A) RC (C) R2C if ε1 < ε2 where εeq = Req =

8.

R1 ε 2 = R 2 ε1 (D) ε1R1 + ε2R2 = 0 (B)

(B) R1C if ε1 > ε2 R R (D) R + 1 2 C R1 + R 2

ε1 / R 1 − ε1 / R 2 1 / R1 + 1 / R 2

• What is an ideal gas ?

R 1R 2 R1 + R 2

• Can a rough sea be calmed by pouring oil on its surface ? ® Yes

Maximum current passing through resistance R ε eq ε eq (A) (B) R eq R + R eq (C)

ε eq R

XtraEdge for IIT-JEE

(D)

® An ideal gas is one in which intermolecular forces are absent

• What is the expression for fringe width (β) in Young's double slit experiment? ® β =Dλ /d where D is the distance between the source and screen and d is distance between two slits

| ε1 − ε 2 | R

19

DECEMBER 2009

1.

8

Solution

Physics Challenging Problems Que s t i ons we r e Publ is he d i n Nove m be r I s s ue

As the resistances of voltmeters in upper branch are R, R/2, R/4 ...................... the equivalent circuit is as shown below R

R/2

R/4

3.

current in upper and lower branch are in the ratio of 1 : 2. R

i b

V

From current division formula we can conclude that

upper Branch ...................

a Lower Branch

2i

R′ = R voltmeter V

Reading of voltmeter V is (2i.)R So V = 2V1

1 =R = 2R 1 −1 / 2 further the equivalent circuit is R

x.

upper branch

a

A

V

b. l.

Lower Branch

l = length of rod = b – a

the resistance of voltmeter V should be 2R so that current in upper and lower branch is same.

charge on element of length d x is dq dq = λ dx

Entire upper branch is having the resistance of 2R

Equivalent current due to element of length d x

can conclude that equivalent resistance of all the

di = ω .dq =

voltmeters in upper branch except V1 is R and the upper branch is as follows:

a

i

V2 R

V3 R

C V 1=X

λ = 3x

as

dq = 3xd x

and voltmeter V1 is having the resistance of R so we

.....up to infinite

B dx.

a.

4.

b

V1

b

Reading of voltmeter V1 is i.R

1 1 = R 1 + + + ..... 2 4

a

R C

a

the resistance of upper branch is = R + R/2 + R/4 + ............. up to infinite

2.

ω (3xd x) 2π

∫

Total equivalent current i = d i =

b

b

3ω = 2π

V 2=Y

As reading of voltmeter V1 is X = i.R

b

x 2 3ω = 2 a 2π

Except V1 in upper branch

Option A is correct

So,

(B) Equivalent current

20

ω

∫ 2π (3xd

x)

a

3ω 2 2 (b – a ) 4π

=

X=Y

b

b2 − a 2 3 ω 2 2 2 = 2 . 2π (b – a )

sum of the readings of voltmeters is Y = i.R

XtraEdge for IIT-JEE

Set # 7

DECEMBER 2009

=

3ω 2 2 3ω (b – a ) = (b – a)(b + a) 4π 4π

Option B is correct As charge on cone C3 ≠ charge on cone C 4

3ω 3 = (b + a)(b – a) = ω . (b + a).l 4π 4π

Option C correct Part-A and part-B will have different charges so

As ω = 4π/3 So,

option D incorrect

3 4π Equivalent current = . . (b + a).l 4π 3

Ans. B, C

= (b + a).l = const.l

6.

i∝l

The circuit is as follows 10Ω

CT

Option B is correct. b

b x2 q = d q = 3xd x = 3. 2 a a

∫

=

∫

a

ig =

Option D is correct

5.

using R =

For part B >

open cone 7.

for part A q

=

q

closed cone

Using R =

ω .q 2π

ω .q , 2π cone - C1 (closed cone) i=

ω = .q 2π (closed cone)

R2 = 2000 – 1000 = 1000Ω

ω .q 2π cone-C3 (closed cone)

R2 = 1000Ω

i=

8.

ω i= .(σ) 2π (Surface area of closed cone)

(coneC1 )

i

(cone - C2) (ConeC1 )

=

(cone C1)

XtraEdge for IIT-JEE

and

i

(coneC2 )

=

ω . q 2π (ConeC2 )

V 5 × 10 −3

– 10

⇒

990 + 1000 + 3000 =

⇒

5000 =

⇒

V = 25 volt

q

(cone C1) q

V –G ig

R1 + R 2 + R3 =

Option A incorrect

ω . 2π

Range between CT and c is V so Using R =

their currents will be different. =

V 10 – G ⇒ R1 + R 2 = – 10 ig 5 ×10 −3

990 + R2 = 2000 – 10

If σ varies then charge on cone C1 differs from C3 So

=

Range between CT and b is 10 volt so,

open cone

Equivalent current i =

i

V 5 – G ⇒ R1 = – 10 = 990Ω ig 5 × 10 −3

R1 = 990Ω

q

closed cone

q

50m = 5mA 10Ω

For terminals CT and a range is 5V so

A, B, D

q

c

b

Full scale deflection current for galvanometer is

3 2 2 (b – a ) 2

Ans.

R3

R2

R1

(D) Charge on rod

V 5 × 10 −3

– 10

V 5 × 10 −3

So range between CT and C is 25 volts.

i (cone C2) 21

DECEMBER 2009

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

PHYSICS 1.

A homogeneous sphere of radius r rolls without slipping with constant angular speed ω' over bigger sphere, of radius R, which is in pure rotation with constant angular speed ω about its centre O. (Fig.) Find the time taken. r

ω

(a) If t is the direction of common tangent, then from eqn. (2) vp'o (t) = vpc (t) + vco (t) dθ ω R = – ω′R + (R + r) dt

ω′

dθ ωR + ω' r = dt R +r R+r dt = dθ (ωR + ω' r )

or,

C

so, O

R

for the centre C of the rolling sphere to return to its initial position (with respect to O), and (ii) for the point of contact of the rolling sphere to make one full revolution over the bigger rotating sphere. (b)(i)Determine the acceleration of the contact point of the rolling sphere, and (ii) the point of greatest acceleration of the rolling sphere (both w.r.t. the centre O) Sol. Suppose that at t = 0, the contact points of lie along the fixed line (reference line) OX. Let at time t the line OC makes the angle θ with OX (Fig.) C

ω O

θ P′

dθ with angular speed – ω to satisfy the condition dt of the problem. dθ dt

ω

Hence the sought time t′ (say) is given by 2π 2π(R + r) = [using Eq. (4)] r (ω'− ω) dθ −ω dt (b) From apo = apc + aco = apc (tangential) + apc (normal) + aco In our case apc (tangential) = 0, because ω′ = constant Hence, apo = apc (normal) + aco (5) t′ =

ω′

P X

It is better to express the velocity and acceleration of any point including contact point P (say) at an arbitrary instant of time t as, v po = v co and apo = aco = apc + apc + aco (1) We know that in the case of pure rolling the velocity of contact point of the rolling body has zero velocity and zero tangential acceleration relative to the contact point of the surface on which it rolls. So if P′ be the contact point of rotating sphere at time t, then vp′o = vpo So, vp'o = vpc + vco (2)

XtraEdge for IIT-JEE

(4)

(ii) It is simple to observer if rotating sphere were at rest, the CM of rolling sphere C will turn by the angle 2 π,

(i)

^t

(3)

2 dθ 2 apo = − ω' r + (R + r) dt

dθ ωR + ω' r = from equation (3) or part (a)) dt R +r (ii) From eqn. (5) it is obvious that the maximum value |apo |max = |apc (normal) | + | aco | (using

2

dθ = – ω′2r + (R + r) dt where

22

dθ will be substituted from Eqn. (3). dt

DECEMBER 2009

2.

Show that the temperature of a planet varies inversely as the square root of its distance from the Sun.

Volume of compressed air when the sphere is in position C = [ πr2x – (2/3)πr3] Atmospheric pressure = h cm of water Let the pressure of compressed air be p cm of water.

Sol. Let Rs be the radius of the Sun. Consider the Sun as a black body. Energy emitted per sec by it equals 4 π R s2 σ Ts4

A

This energy falls uniformly on the inner surface of spheres centred on the Sun. If d is the distance of the planet from the Sun, then energy falling on unit area of the sphere of radius d is : 4πR 2s σTs4

=

4πd 2

L

x d

σR 2s Ts4

According to Boyle's Law (assuming no change in the temperature of compressed air) p[πr2x – (2/3)πr3] = h[πr 2L – (2/3)πr3] or p[x – (2r/3) = h[L – (2r/3)] ...(1) In the equilibrium position C, the weight of the sphere is balanced by the difference of vertical thrust on either side due to atmospheric and compressed air.

d2

This energy received by the planet is given by Q = πr2 =

σR 2s Ts4

where r is the radius of planet.

d2

If T is the temperature of the planet, then energy lost by it per sec is 4 π r2 σ T4

Hence, πr2(p – h)g = mg = (4/3)πr3sg or p – h = (4/3)rs From equation (1) and (2), we get

In the steady state the rate of reception of energy is equal to the loss of energy

3.

Hence,

4 π r2 σ T4 =

Thus,

T∝

πr 2 σR s2 Ts4

x=

d2

1 d

A sphere of specific gravity s just fits into a vertical cylinder with lower end closed. The sphere is allowed to drop slowly until it is held in equilibrium by the thrust of the compressed air. There is no leakage of air. If the diameter of the sphere is d, the length of the cylinder is L and the height of the water barometer is h, then what will be the position of sphere ?

4.

...(2)

h (3L − 2r ) 2r + 3p 3

or x =

h (3L − 2r ) 2r 9 hL + 8r 2s + = 3h + 4 rs 3 3(3h + 4rs)

or x =

9hL + 2d 2s 9h + 6ds

Given the position of the object O and the image I as shown in the figure. Find (a) the position of the convex lens (b) its focal length and the magnification of the image. Verify graphically. Y

Sol. Initially, the cylinder contained air at atmospheric pressure. When the sphere comes down into the cylinder by the action of its own weight, it presses the air downwards. Suppose the sphere comes to position C which is at height x above the closed end. Let the sphere remain in equilibrium in this position.

2 1 0 –1 –2

O

1 2 3 4

5

6

7

8

9 10 11 12 13 14

X

I

Sol. A ray of light from the object passes undeviated through the optic centre of the lens (C) and also the image I. So join OI. So it cuts the principal axis XY and C. So AB is the position of the lens. A ray parallel to the principal axis from the object after refraction meets the principal axis at F. F is the focus.

Volume of sphere = (4/3)πr3 Weight of the sphere = (4/3)πr3sg Volume of cylinder = πr2L Volume of air inside the cylinder when the sphere is in position A = πr2L – (2/3)πr3

XtraEdge for IIT-JEE

C

23

DECEMBER 2009

A

dB =

X

8

F

C

4

14

10.4

B

X

I

F is at a distance 2.4 cm from C.

x2

B=

∴ Focal length of lens = 2.4 × 10 = 24 cm By calculation

µ i dB = 0 4π

1 1 100 1 – = = 60 − 40 2400 24

v 60 3 Magnification = = = u 40 2 =

length of image length of object

=

3 mm 3 = 2 mm 2

µ iR B= 0 4π

∴

i

x2

∫ (x

x1

dx 2

+ R2

...(ii)

)

3/ 2

At

x x = x1 φ1 = tan–1 1 and x = x2, R x φ2 = tan–1 2 R

D

distance R from a straight conductor of length l. x2

r

dx x1

Consider a typical element dx. The magnitude of the contribution dB of this element to the magnetic field at P as found from Biot-Savart's law is 24

R sec 2 φ

µ 0 iR 4π

=

µ 0 iR 4π

=

µ 0 iR −1 x 2 x − sin tan −1 1 ...(iii) sin tan 4π R R

∫

x tan −1 2 R x tan −1 1 R

∫

R 3 sec 2 φ cos φ dφ

Let,

x z = tan–1 , R

Thus,

tan z =

⇒

R

x tan −1 2 R x tan −1 1 R

B=

Sol. First of all, we determine the expression of B at a

XtraEdge for IIT-JEE

r2

x1

x = R tan φ, such that dx = R sec2 φ dφ

i

φ

sin θ dx

Also, x2 + R 2 = R 2 tan2φ + R2 = R2 sec2 φ Hence, equation (ii) becomes

i

x

∫

Let

Calculate the magnetic field B at the point P shown in the figure. Assume that i = 10 A and a = 8.0 cm. a/4 i B C a 4 P

A

x2

R and r = (x2 + R2)1/2 r

where, sin θ =

∴ f = 24 cm

5.

∫

x1

1 1 1 = – f v u =

µ 0 i dx sin θ

...(i) 4πr 2 Since, the direction of the contribution dB at the point P for all elements are identical viz, at right angle into the plane of the figure, the resultant field is obtained by simply integration equation (i), which gives

O

x R

sin z

=

1 − sin 2 z

x R

⇒

R2sin2z = x2(1 – sin2z)

⇒

sin2z(R2 + x2) = x2

⇒

sin z =

⇒

x z = sin–1 2 2 x +R

x 2

x + R2

DECEMBER 2009

Hence, equation (iii) becomes

MEMORABLE POINTS

µ0 i x2 x1 ...(iv) − 2 2 4πR x 2 + R 2 x R + 1 2 Applying the equation (iv) to the given problem For AB, we get B=

• Can a current be measured by a voltameter ? ® Yes, direct current can be measured by a voltameter • The equivalent resistance of n resistances each equal to r and connected in parallel is given by ® r/n

a a 3 x1 = – a, x2 = and R = 9 4 4

a µ 0i 4 Hence, B1 = − a a2 a2 4π + 4 16 16 µ i 1 3 + = 0 πa 2 10

• Repeated use of which digital gate or gates can produce all the three basic gates (OR, AND and NOT) ® NAND gate and NOR gate

2 2 9a a + 16 16 −3 a 4

• What is Turnbull's blue ?

• An hypothesis tested by experiments is known as ® Theory

• What is magnesia alba ? ® Mg(OH)2.MgCO3.3 H2O

For BC :

• The humidity of air is measured by

−a 3a a x1 = , x2 = and R = 4 4 4 This also gives B2 =

µ 0i 1 3 + πa 2 10

• What is oleum ?

=

• An amino acid which does not contain a chiral centre. ® Glycine [NH2–CH2–COOH] • Scientist who perfected the technique for converting pig iron into steel. ® Hynry Besemer (1856)

• When the pH of the blood is lower than the normal value, this condition is known as ® Acidosis • The electrolytic method of obtaining aluminium from bauxite was first developed by

µ 0i 1 1 + 3π a 10 2

® Charles Hall (1886)

• Which compound possesses characteristic smell like that of mustard oil ? ® Ethyl isothiocyanate [C2H5N = C = S]

For DA : x1 =

3a −a 3a , x2 = and R = 4 4 4

This also gives

B4 =

• First solar battery was developed in the ® Bell Telephone Laboratory (1954)

µ0 i 1 1 + 3πa 10 2

• What is Wilkinson's catalyst ? ® tris (triphenylphosphine) chlororhodlum (I)

Total magnitude at magnetic field, B = B1 + B 2 + B 3 + B 4 =

• In 1836 the galvanised iron was introduced first in ® France

µ 0i 1 3 1 3 1 1 1 1 + + + + + + + πa 2 10 2 10 3 10 3 2 3 10 3 2

• What is caro's acid ? ® Permonosulphuric acid [H2SO5]

2µ0 i = ( 10 + 2 2 ) 3πa = 2.0 × 104 T

XtraEdge for IIT-JEE

® H2S 2O7

• Vulcanised rubber was invented by ® Charies Goodyear (1839)

a −3a −3a , x2 = and R = 4 4 4

a − 3a µ0 i 4 4 ∴ Β3 = − 2 − 3a 9a 2 9a 2 a 9a 2 4π + + 4 16 16 16 16

® Hygrometer

Also known as fuming sulphuric acid

For CD : x1 =

® Fe4[Fe(CN)6]3

25

DECEMBER 2009

P HYSICS F U NDAMENTAL F OR IIT-J EE

Reflection at plane & curved surfaces KEY CONCEPTS & PROBLEM SOLVING STRATEGY For solving the problem, the reference frame is chosen in which optical instrument (mirror, lens, etc.) is in rest. The formation of image and size of image is independent of size of mirror. Visual region and intensity of image depend on size of mirror. P P'

Key Concepts : (a) Due to reflection, none of frequency, wavelength and speed of light change. (b) Law of reflection : Incident ray, reflected ray and normal on incident point are coplanar. The angle of incidence is equal to angle of reflection Incident n Reflected Ray Ray θ θ

n

n Tangent at point P

θ θ

α α

P Convex surface

Plane surface

θ θ

n αα Convex surface

If the plane mirror is rotated through an angle θ, the reflected ray and image is rotated through an angle 2θ in the same sense. If mirror is cut into a number of pieces, then the focal length does not change. The minimum height of mirror required to see the full image of a man of height h is h/2.

A

Tangent at point P

Some important points : In case of plane mirror For real object, image is virtual. For virtual object, image is real. The converging point of incident beam behaves as a object. If incident beam on optical instrument (mirror, lens etc) is converging in nature, object is virtual. If incident beam on optical instrument is diverging in nature, the object is real. The converging point of reflected or refracted beam from an optical instrument behaves as image. If reflected beam or refracted beam from an optical instrument is converging in nature, image is real.

Rest

vsinθ Object

P

P P Virtual Object

Real Object

n n

Real Object

n αα

α Real n α Object

XtraEdge for IIT-JEE

v θ

Image

Rest

vcosθ

vsinθ

vcosθ Image

P Virual Object

Object v

If reflected beam or refracted beam from an optical instrument is diverging in nature, image is virtual.

P

v

Object v

P'

Object In rest

Virual Object

26

vm Image

vm Image

2vm–v

2vm

DECEMBER 2009

v

Object

vm Image

These formulae are only applicable for paraxial rays. All distances are measured from optical centre. It means optical centre is taken as origin. The sign conventions are only applicable in given values. The transverse magnification is

2vm+v

(c) Number of images formed by combination of two plane mirrors : The images formed by combination of two plane mirror are lying on a circle whose centre is at the meeting points of mirrors. Also, object is lying on that circle. Here, n =

β=

360º θ

1. If object and image both are real, β is negative. 2. If object and image both are virtual, β is negative. 3. If object is real but image is virtual, β is positive.

where θ = angle between mirrors. 360º is even number, the number of images is θ n – 1. If

4. If object is virtual but image is real, β is positive. 5. Image of star; moon or distant object is formed at focus of mirror. If y = the distance of sun or moon from earth. D = diameter of moon or sun's disc f = focal length of the mirror d = diameter of the image

360º is odd number and object is placed on θ bisector of angle between mirror, then number of images is n – 1. If

360º is odd and object is not situated on θ bisector of angle between mirrors, then the number of images is equal to n. (d) Law of reflection in vector form : Let eˆ1 = unit vector along incident ray. If

θ = the angle subtended by sun or moon's disc Then tan θ = θ =

Sun D

nˆ = unit vector along normal on point of Incidence Then eˆ 2 = eˆ1 − 2(eˆ1.nˆ ) nˆ

(e) Spherical mirrors : It easy to solve the problems in geometrical optics by the help of co-ordinate sign convention. y y y x

x

x' y'

y

y x

x'

The mirror formula is

y'

x

x'

y'

x

x'

θ

θ

Problem solving strategy : Image formation by mirrors Step 1: Identify the relevant concepts : There are two different and complementary ways to solve problems involving image formation by mirrors. One approach uses equations, while the other involves drawing a principle-ray diagram. A successful problem solution uses both approaches. Step 2: Set up the problem : Determine the target variables. The three key quantities are the focal length, object distance, and image distance; typically you'll be given two of these and will have to determine the third. Step 3: Execute the solution as follows : The principal-ray diagram is to geometric optics what the free-body diagram is to mechanics. In any problem involving image formation by a mirror, always draw a principal-ray diagram first if you have enough information. (The same

eˆ 2

y'

F d

n nˆ

eˆ1

D d = y f

Here, θ is in radian.

eˆ 2 = unit vector along reflected ray

x'

image size −v = object size u

y'

1 1 1 + = v u f

Also, R = 2f

XtraEdge for IIT-JEE

27

DECEMBER 2009

advice should be followed when dealing with lenses in the following sections.) It is usually best to orient your diagrams consistently with the incoming rays traveling from left to right. Don't draw a lot of other rays at random ; stick with the principal rays, the ones you know something about. Use a ruler and measure distance carefully ! A freehand sketch will not give good results. If your principal rays don't converge at a real image point, you may have to extend them straight backward to locate a virtual image point, as figure (b). We recommend drawing the extensions with broken lines. Another useful aid is to color-code the different principal rays, as is done in figure(a) & (b). Q I 3

P

4 C P' F

2 2 4

Q'

Solved Examples 1.

Sol. B θ i1

i1

2 4

3

v 3

Rays of light strike a horizontal plane mirror at an angle of 45º. At what angle should a second plane mirror be placed in order that the reflected ray finally be reflected horizontally from the second mirror. Sol. The situation is shown in figure G C D A S θ θ

P' F

C

4 (b)

45º

1 1 1 + = and the s s' f y' s' magnification equation m = = − . The y s results you find using this equation must be consistent with your principal-ray diagram; if not, double-check both your calculation and your diagram. Pay careful attention to signs on object and image distances, radii or curvature, and object and image heights. A negative sign on any of these quantities always has significance. Use the equations and the sign rules carefully and consistently, and they will tell you the truth ! Note that the same sign rules (given in section) work for all four cases in this chapter : reflection and refraction from plane and spherical surfaces. Step 4: Evaluate your answer : You've already checked your results by using both diagrams and equations. But it always helps to take a look back and ask yourself. "Do these results make sense ?".

P

N 45º

Q B The ray AB strikes the first plane mirror PQ at an angle of 45º. Now, we suppose that the second mirror SG is arranged such that the ray BC after reflection from this mirror is horizontal. From the figure we see that emergent ray CD is parallel to PQ and BC is a line intersecting these parallel lines. So, ∠DCE = ∠CBQ = 180º ∠DCN + ∠NCB + ∠CBQ = 180º θ + θ + 45º = 180º ∴ θ = 67.5º As ∠NCS = 90º, therefore the second mirror should be inclined to the horizontal at an angle 22.5º.

Check your results using Eq.

XtraEdge for IIT-JEE

C

2.

Q' v

P

i2

The total deviation of the ray is given by δ =180 – 2i1 + 180 – 2i2 = 360 – 2(i1 + i2) For the resultant ray to be parallel, δ should be 180º ∴ 360 – 2(i1 + i2) = 180 i.e., i1 + i2 = 90º From the geometry of the figure i1 + i2 = θ θ Angle between the mirrors should be 90º.

1

2

i2

A

(a) 1

Q

P

1

Q

How will you arrange the two mirrors so that whatever may be the angle of incidence, the incident ray and the reflected ray from the two mirrors will be parallel to each other.

3.

28

An object is placed exactly midway between a concave mirror of radius of curvature 40 cm and a convex mirror of radius of curvature 30 cm. The mirrors face each other and are 50 cm apart. Determine the nature and position of the image formed by the successive reflections, first at the concave mirror and then at the convex mirror. DECEMBER 2009

Sol. The image formation is shown in figure.

The image will be virtual. This is formed at I1 behind the convex mirror at a distance of 10 cm. The image I1 acts as an object for convex mirror. (ii) For concave mirror, u2 = P2I1 = 60 + 10 = 70 cm, f2 = +15 cm and v 2 = ? 1 1 1 ∴ = + 15 70 v2 Solving we get, v2 = (210/11) cm. As v2 is positive, the image I2 is formed in front of concave mirror at a distance of (210/11) cm. Magnification m1 for first reflection v 10 1 = 1 = = u1 30 3 Magnification m2 for second reflection v ( 210 / 11) 3 = 2 = = u2 70 11

50cm

I2

P2 C

F

P1

25cm

r = 30 cm

r = 40 cm

I1 (i) For concave mirror, u1 = 25 cm, f1 = 20 cm and v1 = ? 1 1 1 Now = + f1 u1 v1

1 1 1 = + 20 25 v1 v1 = 100 cm. As v1 is positive, hence the image is real. In the absence of convex mirror, the rays after reflection from concave mirror would have formed a real image I1 at distance 100 cm from the mirror. Due to the presence of convex mirror, the rays are reflected and appear to come from I2. (ii) For convex mirror, In this case, I1 acts as virtual object and I2 is the virtual image. The distance of the virtual object from the convex mirror is 100 – 50 = 50 cm. Hence u2 = –50 cm. As focal length of convex mirror is negative and hence f2 = –30/2 = –15 cm. Here we shall calculate the value of v 2. Using the mirror formula, we have 1 1 1 − = − + 15 50 v 2 or v2 = –21.42 cm As v2 is negative, image is virtual. So image is formed behind the convex mirror at a distance of 21.43 cm. or

1 3 1 × = 3 11 11 1 5 ∴ Size of the image = 5 × = 11 11 5. An object is placed infront of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm it is found that there is no parallex between the images formed by the two mirrors. What is the radius of curvature of the convex mirror ? Sol. Let O be the object placed infront of a convex mirror MM' at a distance of 50 cm as shown in figure. The distance of the plane mirror NN' from the object is 30. We know that in a plane mirror the image is formed behind the mirror at the same distance as the object infront of it. It is also given that there is no parallax between the images formed by the two mirrors, i.e., the image is formed at a distance of 30 cm behind the plane mirror. For convex mirror, u = 50 cm, v = 10 cm [Q QP = QN – PN] as v is negative in convex mirror. M 50cm Final magnification = m1 × m2 =

4.

A convex and a concave mirror each 30 cm in radius are placed opposite to each other 60 cm apart on the same axis. An object 5 cm in height is placed midway between them. Find the position and size of the image formed by reflection, first at convex and then at the concave mirror. Sol. The image formation is shown in figure.

N Q

P

O

N' 20cm 30cm P2 I2

O

P1

M'

I1

1 1 1 = + , we have f u v 1 1 1 4 50 = − = − , ∴f= − f 50 10 50 4 50 × 2 Now v = 2f = − = –25 m 4 The radius of curvature of convex mirror is 25 cm. Using the mirror formula

r = 30 cm r = 30 cm (i) For convex mirror, u1 = +30 m, f1 = –15 cm and v 1 = ? 1 1 1 ∴− = + 15 50 v1 or v1 = –10 cm

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DECEMBER 2009

P HYSICS F U NDAMENTAL F OR IIT-J EE

Fluid Mechanics KEY CONCEPTS & PROBLEM SOLVING STRATEGY Hydrodynamics :

Hydrostatics : Pressure at a point inside a Liquid : p = p0 + ρ gh where p0 is the atmospheric pressure, ρ is the density of the liquid and h is the depth of the point below the free surface. p0 h

1 2 p v + gh + = a constant 2 ρ for a streamline flow of a fluid (liquid or gas). Here, v is the velocity of the fluid, h is its height above some horizontal level, p is the pressure and ρ is the density. p1 Bernoulli's Theorem :

p

v1

ρ

h1 v2 h2

Pressure is a Scalar : The unit of pressure may be atmosphere or cm of mercury. These are derived units. The absolute unit of pressure is Nm–2. Normal atmospheric pressure, i.e, 76 cm of mercury, is approximately equal to 10 5 Nm– 2. Thrust : Thrust = pressure × area. Thrust has the unit of force. Laws of liquid pressure (a) A liquid at rest exerts pressure equally in all directions. (b) Pressure at two points on the same horizontal line in a liquid at rest is the same. (c) Pressure exerted at a point in a confined liquid at rest is transmitted equally in all directions and acts normally on the wall of the containing vessel. This is called Pascal's law. A hydraulic press works on this principle of transmission of pressure. The principle of floating bodies (law of flotation) is that W = W´, that is, weight of body = weight of displaced liquid or buoyant force. The weight of the displaced liquid is also called buoyancy or upthrust. Hydrometers work on the principle of floating bodies. This principle may also be applied to gases (e.g., a balloon). Liquids and gases are together called fluids. The important difference between them is that liquids cannot be compressed, while gases can be compressed. Hence, the density of a liquid is the same everywhere and does not depend on its pressure. In the case of a gas, however, the density is proportional to the pressure.

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p2

v2 > v1

p2 < p1

According to this principle, the greater the velocity, the lower is the pressure in a fluid flow. It would be useful to remember that in liquid flow, the volume of liquid flowing past any point per second is the same for every point. Therefore, when the cross-section of the tube decreases, the velocity increases. Note : Density = relative density or specific gravity × 1000 kg m– 3. Surface tension and surface energy : Surface Tension : The property due to which a liquid surface tends to contract and occupy the minimum area is called the surface tension of the liquid. It is caused by forces of attraction between the molecules of the liquid. A molecule on the free surface of a liquid experiences a net resultant force which tends to draw it into the liquid. Surface tension is actually a manifestation of the forces experienced by the surface molecules. If an imaginary line is drawn on a liquid surface then the force acting per unit length of this line is defined as the surface tension. Its unit is, therefore, newton / metre. This force acts along the liquid surface. For curved surfaces, the force is tangent to the liquid surface at every point. Surface Energy : A liquid surface possesses potential energy due to surface tension. This energy per unit area of the surface is called the surface energy of the liquid. Its units is joule per square metre. The surface energy of a liquid has the same numerical values as the surface tension. The surface 30

DECEMBER 2009

tension of a liquid depends on temperature. It decreases with rise in temperature. Excess of Pressure : Inside a soap bubble or a gas bubble inside a liquid, there must be pressure in excess of the outside pressure to balance the tendency of the liquid surface to contract due to surface tension.

The upward force by which a liquid surface is pulled up in a capillary tube is 2πrTcos θ, and the downward force due to the gravitational pull on the mass of liquid in the tube is ( πr2h + v)ρ g, where v is the volume above the liquid meniscus. If θ = 0º, the meniscus is hemispherical in shape. Then v = difference between the volume of the cylinder of radius r and height r and the volume of the hemisphere of radius r

1 1 p(excess of pressure) = T + in general r1 r2

= πr3 –

where T is surface tension of the liquid, and r1 and r2 are the principal radii of curvature of the bubble in two mutually perpendicular directions. For a spherical soap bubble, r1 = r2 = r and there are two free surfaces of the liquid.

When θ ≠ 0, we cannot calculate v which is generally very small and so it may be neglected. For equilibrium (πr2h + v) ρ g = 2 πrT cos θ When a glass capillary tube is dipper in mercury, the meniscus is convex, since the angle of contact is obtuse. The surface tension forces now acquire a downward component, and the level of mercury inside the tube the falls below the level outside it. the relation 2T cos θ = hρgr may be used to obtain the fall in the mercury level. Problem Solving Strategy Bernoulli's Equations : Bernoulli's equation is derived from the work-energy theorem, so it is not surprising that much of the problem-solving strategy suggested in W.E.P. also applicable here. Step 1: Identify the relevant concepts : First ensure that the fluid flow is steady and that fluid is incompressible and has no internal friction. This case is an idealization, but it hold up surprisingly well for fluids flowing through sufficiently large pipes and for flows within bulk fluids (e.g., air flowing around an airplane or water flowing around a fish). Step 2: Set up the problem using the following steps Always begin by identifying clearly the points 1 and 2 referred to in Bernoulli's equation. Define your coordinate system, particular the level at which y = 0. Make lists of the unknown and known quantities 1 1 in Eq. p1 + ρ gy1 + ρ v12 = p2 + ρ gy2 + ρ v22 2 2 (Bernoulli's equation) The variables are p1, p2, v 1, v 2, y1 and y2, and the constants are ρ and g. Decide which unknowns are your target variables. Step 3: Execute the solutions as follows : Write Bernoulli's equation and solve for the unknowns. In some problems you will need to use the continuity equation, Eq. A1v1 = A2v 2 (continuity equation, incompressible fluid), to get a relation between the two speeds in terms of cross-sectional areas of pipes

4T ∴ p= r For a gas bubble inside a liquid, r1 = r2 = r and there is only one surface. 2T r For a cylindrical surface r1 = r and r2 = ∞ and there are two surfaces.

∴

p=

2T r Angle of Contact : The angle made by the surface of a liquid with the solid surface inside of a liquid at the point of contact is called the angle of contact. It is at this angle that the surface tension acts on the wall of the container.

∴

p=

The angle of contact θ depends on the natures of the liquid and solid in contact. If the liquid wets the solid (e.g., water and glass), the angle of contact is zero. In most cases, θ is acute (figure i). In the special case of mercury on glass, θ is obtuse (figure ii). θ

θ fig. (i)

fig. (ii)

Rise of Liquid in a Capillary Tube : In a thin (capacity) tube, the free surface of the liquid becomes curved. The forces of surface tension at the edges of the liquid surface then acquire a vertical component. meniscus θ T θ T θ

θ

h

r

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2 3 1 3 πr = πr 3 3

31

DECEMBER 2009

or containers. Or perhaps you will know both speeds and need to determine one of the areas. You may also dV need to use Eq. = Av (volume flow rate) to find dt the volume flow rate. Step 4: Evaluate your answer : As always, verify that the results make physical sense. Double-check that you have used consistent units. In SI units, pressure is in pascals, density in kilograms per cubic meter, and speed in meters per second. Also note that the pressures must be either all absolute pressure or all gauge pressures.

•1

H

p0 1 2 p 1 + v 1 + gH = + v 22 + g (h – x) ρ 2 ρ 2 p 1 = 0 + v 23 + 0 ρ 2 By continuity equation v 1A1 = A2v 2 = A2v 3 Since A1 >> A2,v1 is negligible and v2 = v 3 = n (say). p0 p 1 ∴ + gH = 2 + v2 + g (h – x) ρ ρ 2 p0 1 2 = + v ρ 2

A vertical U-tube of uniform cross-section contains mercury in both arms. A glycerine (relative density 1.3) column of length 10 cm is introduced into one of the arms. Oil of density 800 kg m–3 is poured into the other arm until the upper surface of the oil and glycerine are at the same horizontal level. Find the length of the oil column. Density of mercury is 13.6 × 103 kg m–3. Sol. Draw a horizontal line through the mercury-glycerine surface. This is a horizontal line in the same liquid at rest namely, mercury. Therefore, pressure at the points A and B must be the same.

∴

v = 2gH

(i)

p0 p + gH = 2 + gH + g (h – x) ρ ρ ⇒ p0 + p2 + ρ g (h – x) ⇒ p2 = p0 – ρ g (h – x) (ii) Thus pressure varies with distance from the upper end of the pipe according to equation (ii) and velocity is a constant and is given by (i). and

10 cm

h (1 – h)

h

•3

1.

3.

A rod of length 6m has a mass of 12 kg. It is hinged at one end at a distance of 3 m below the water surface. (i) What weight must be attached to the other end so that 5 m of the rod is submerged ? (ii) Find the magnitude and direction of the force exerted by the hinge on the rod. The specific gravity of the material of the rod is 0.5 Sol. Mass per unit length of the rod is 2 kg. Therefore, mass of the submerged portion of the rod is 10 kg and 10 3 its volume = m (using the simple formula, 500 mass volume = and density = specific gravity × density 1000 kg m– 3). B Fb W θ C O 3m O´ 12 kgf H A N

B

Pressure at B = p0 + 0.1 × (1.3 × 1000) × g Pressure at A = p0 + h × 800 × g + (0.1 – h) × 13.6 × 1000g ∴ p0 + 0.1 × 1300 × g = p0 + 800gh + 1360g – 13600 × g × h ⇒ 130 = 800h + 1360 – 13600h 1230 ⇒ h= = 0.096 m = 9.6 cm 12800 2.

A liquid flows out of a broad vessel through a narrow vertical pipe. How are the pressure and the velocity of the liquid in the pipe distributed when the height of the liquid level in the vessel is H from the lower end of the length of the pipe is h ? Sol. Let us consider three points 1, 2, 3 in the flow of water. The positions of the points are as shown in the figure. Applying Bernoulli's theorem to points 1, 2 and 3

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•2

2

Solved Examples

A

x

32

DECEMBER 2009

Therefore, buoyant force 10 Fb = × 1000 = 20 kgf 500 Let N and H be the vertical downward and horizontal reactions of the hinge on the rod. Considering horizontal and vertical translational equilibrium of the rod, N + 12 + W = 20 (where W is the weight to be attached and H = 0). N+ W=8 and H=0 ..(i) Considering the rotational equilibrium of the rod about A –20 × g × 2.5 cos θ + 12 × g × 3 cos θ + W × 6 cos θ = 0 ⇒ 6W = 50g – 36g = 14g 7 7 ⇒ W = g = kgf 3 3 7 17 Ν=8– = kgf 3 3

1 1 ∴ pressure difference = 2T cos θ − r1 r2 B

Let this pressure difference correspond to h units of the liquid. 1 1 Then 2T cos θ − = ρ gh r1 r2

4.

The end of a capillary tube with a radius r is immersed in water. Is mechanical energy conserved when the water rises in the tube ? The tube is sufficiently long. If not, calculate the energy change. Sol. In the equilibrium position (θ = 0º for pure water and glass) 2πrT cos 0º = πr2hρ g 2T or h= ρgr

• • • •

2πT 2 ρg Thus it is seen that the mechanical energy is not conserved. 4πT 2 2πT 2 ∴ mechanical energy loss = – ρg ρg

•

2πT 2 ρg This energy is converted into heat.

•

U=

• •

=

•

5.

Calculate the difference in water levels in two communicating tubes of diameter d = 1 mm and d = 1.5 mm. Surface tension of water = 0.07 Nm–1 and angle of contact between glass and water = 0º. 2T cos θ Sol. Pressure at A = p0 – r2

• • •

(Q pressure inside a curved surface is greater than that outside) 2T cos θ Pressure at B = p0 – r1

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⇒ h=

2T cos θ 1 1 − ρg r1 r2

∴ h=

2 × 0.07 1 1 − = 4.76 mm − 3 − 3 1000 × 9.8 1 ×10 1.5 ×10

Interesting Science Facts

4πT 2 Work done by surface tension = (2 πrT) × h = ρg U, potential energy of water in the tube = (πr2hρ)gh/2; it is multiple by h/2 because the cg of the water in the capillary tube is at a height h/2.

⇒

A

•

33

The dinosaurs became extinct before the Rockies or the Alps were formed. Female black widow spiders eat their males after mating. When a flea jumps, the rate of acceleration is 20 times that of the space shuttle during launch. The earliest wine makers lived in Egypt around 2300 BC. If our Sun were just inch in diameter, the nearest star would be 445 miles away. The Australian billy goat plum contains 100 times more vitamin C than an orange. Astronauts cannot belch - there is no gravity to separate liquid from gas in their stomachs. The air at the summit of Mount Everest, 29,029 feet is only a third as thick as the air at sea level. One million, million, million, million, millionth of a second after the Big Bang the Universe was the size of a …pea. DNA was first discovered in 1869 by Swiss Friedrich Mieschler. The molecular structure of DNA was first determined by Watson and Crick in 1953. The thermometer was invented in 1607 by Galileo. Englishman Roger Bacon invented the magnifying glass in 1250.

DECEMBER 2009

KEY CONCEPT

Organic Chemistry Fundamentals

CARBOXYLIC ACID

Acidity of carboxylic acids. Fatty acids are weak acids as compared to inorganic acids. The acidic character of fatty acids decreases with increase in molecular weight. Formic acid is the strongest of all fatty acids. The acidic character of carboxylic acids is due to resonance in the acidic group which imparts electron deficiency (positive charge) on the oxygen atom of the hydroxyl group (structure II). O– O R

C

O

R

H

I Non-equivalent structures

C

+

O

(1.27Å) which is nearly intermediate between C O and C—O bond length values. This proves resonance in carboxylate anion. O H

O+H

–

C O

Resonance hybrid of carboxylate ion

Due to equivalent resonating structures, resonance in carboxylate anion is more important than in the parent carboxylic acid. Hence carboxylate anion is more stabilised than the acid itself and hence the equilibrium of the ionisation of carboxylic acids to the right hand side. RCOOH RCOO– + H+ The existence of resonance in carboxylate ion is supported by bond lengths. For example, in formic acid, there is one C=O double bond (1.23 Å) and one C—O single bond (1.36Å), while in sodium formate both of the carbon-oxygen bond lengths are identical

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O

Alkoxide ion (No resonance)

Relative acidic character of carboxylic acids with common species not having —COOH group. RCOOH > Ar—OH > HOH > ROH > HC CH > NH3 > RH Effect of Substituents on acidity. The carboxylic acids are acidic in nature because of stabilisation (i.e., dispersal of negative charge) of carboxylate ion. So any factor which can enhance the dispersal of negative charge of the carboxylate ion will increase the acidity of the carboxylic acid and vice versa. Thus electron-withdrawing substitutents (like halogens, —NO2, —C6H5, etc.) would disperse the negative charge and hence stabilise the carboxylate ion and thus increase acidity of the parent acid. On the other hand, electron-releasing substituents would increase the negative charge, destabilise the carboxylate ion and thus decrease acidity of the parent acid.

Resonating forms of carboxylate ion (Equivalent structures) (Resonance more important)

R

Na+

Sodium formate

Alcohol (No resonance)

+

The positive charge (electron deficiency) on oxygen atom causes a displacement of electron pair of the O—H bond towards the oxygen atom with the result the hydrogen atom of the O—H group is eliminated as proton and a carboxylate ion is formed. Once the carboxylate ion is formed, it is stabilised by means of resonance. O O– R C R C O– O

O

OH

C

–

It is important to note that although carboxylic acids and alcohols both contain –OH group, the latter are not acidic in nature. It is due to the absence of resonance (factor responsible for acidic character of –COOH) in both the alcohols as well as in their corresponding ions (alkoxide ions). R—O—H R—O– + H+

H

C

H

Formic acid

II (Resonance less important) O– R

C

O

O X

C

–

O

The substituent X withdraws electrons, disperses negative charge, stabilises the ion and hence increases acidity –

O Y

C

O

The substituent Y releases electrons, intensifies negative charge, destabilises the ion and hence decreases acidity

34

DECEMBER 2009

Now, since alkyl groups are electron-releasing, their presence in the molecule will decrease the acidity. In general, greater the length of the alkyl chain, lower shall be the acidity of the acid. Thus, formic acid (HCOOH), having no alkyl group, is about 10 times stronger than acetic acid (CH3COOH) which in turn is stronger than propanoic acid (CH3CH2COOH) and so on. Similarly, following order is observed in chloro acids. Cl Cl Cl pKa

C

CO2H > Cl

C

Comparison of nucleophilic substitution (e.g., hydrolysis) in acid derivatives. Let us first study the mechanism of such reaction. O R

C

R C Nu + Z (where Z= —Cl, —OCOR, —OR, —NH2 and Nu = A nucleophile) Nucleophilic substitution in acid derivatives O O OH

R

H CO2H > H

C

δ–

Cl

δ+ Acid chloride

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δ+

R

R'

Nu

R

C

Nu

H+

R

C

Nu

R'

(where R' = H or alkyl group) Nucleophilic addition on aldehydes and ketones The (i) step is similar to that of nucleophilic addition in aldehydes and ketones and favoured by the presence of electron withdrawing group (which would stabilise the intermediate by developing negative charge) and hindered by electron-releasing group. The (ii) step (elimination of the leaving group Z) depends upon the ability of Z to accommodate electron pair, i.e., on the basicity of the leaving group. Weaker bases are good leaving groups, hence weaker a base, the more easily it is removed. Among the four leaving groups (Cl–, –OCOR, –OR, and –NH2) of the four acid derivatives, Cl– being the weakest base is eliminated most readily. The relative order of the basic nature of the four groups is – NH2 > –OR > –O.COR > Cl– Hence acid chlorides are most reactive and acid amides are the least reactive towards nucleophilic acyl substitution. Thus, the relative reactivity of acid derivatives (acyl compounds) towards nucleophilic substitution reactions is ROCl > RCO.O.COR > RCOOR > RCONH2 Acid Acid Esters Acid chlorides anhydrides amides OH– being stronger base than Cl–, carboxylic acids (RCOOH) undergo nucleophilic substitution (esterfication) less readily than acid chlorides.

δ–

R C

C

R'

CO2H

H H pKa 2.86 4.76 Decreasing order of aliphatic acids (i) O2NCH2COOH > FCH2COOH > ClCH2COOH > BrCH2COOH (ii) HCOOH > CH3COOH > (CH3)2 CHCOOH > (CH3)3CCOOH (iii) CH3CH2CCl2COOH > CH3CHCl.CHCl.COOH > ClCH2CHClCH2COOH (iv) F3CCOOH > Cl3CCOOH > Br3CCOOH Benzoic acid is somewhat stronger than simple aliphatic acids. Here the carboxylate group is attached to a more electronegative carbon (sp2 hybridised) than in aliphatic acids (sp3 hybridised). HCOOH > C6H5COOH > CH3COOH. Nucleophilic substitution at acyl carbon : It is important to note that nucleophilic substitution (e.g., hydrolysis, reaction with NH3, C 2H5OH, etc.) in acid derivatives (acid chlorides, anhydrides, esters and amides) takes place at acyl carbon atom (difference from nucleophilic substitution in alkyl halides where substitution takes place at alkyl carbon atom). Nucleophilic substitution in acyl halides is faster than in alkyl halides. This is due to the presence of > CO group in acid chlorides which facilitate the release of halogen as halide ion. O

Nu

C O

1.48

> Cl

R

(ii) Elimination step

CO2H

H

Z + Nu

Z

H

Cl 0.70

C

O (i) Addition step

δ–

Cl

Alkyl chloride

35

DECEMBER 2009

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KEY CONCEPT

Physical Chemistry Fundamentals

CHEMICAL KINETICS

The temperature dependence of reaction rates : The rate constants of most reactions increase as the temperature is raised. Many reactions in solution fall somewhere in the range spanned by the hydrolysis of methyl ethanoate (where the rate constant at 35ºC is 1.82 times that at 25ºC) and the hydrolysis of sucrose (where the factor is 4.13). (a) The Arrhenius parameters : It is found experimentally for many reactions that a plot of ln k against 1/T gives a straight line. This behaviour is normally expressed mathematically by introducing two parameters, one representing the intercept and the other the slope of the straight line, and writing the Arrhenius equaion.

behaviour is a signal that the reaction has a complex mechanism. The temperature dependence of some reactions is non-Arrhenius, in the sense that a straight line is not obtained when ln k is plotted against 1/T. However, it is still possible to define an activation energy at any temperature as

dln k Ea = RT2 .......(ii) dT This definition reduces to the earlier one (as the slope of a straight line) for a temperature-independent activation energy. However, the definition in eqn.(ii) is more general than eqn.(i), because it allows Ea to be obtained from the slope (at the temperature of interest) of a plot of ln k against 1/T even if the Arrhenius plot is not a straight line. Non-Arrhenius behaviour is sometimes a sign that quantum mechanical tunnelling is playing a significant role in the reaction. (b) The interpretation of the parameters : We shall regard the Arrhenius parameters as purely empirical quantities that enable us to discuss the variation of rate constants with temperature; however, it is useful to have an interpretation in mind and write eqn.(i) as

Ea ......(i) RT The parameter A, which corresponds to the intercept of the line at 1/T = 0(at infinite temperature, shown in figure), is called the pre-exponential factor or the 'frequency factor'. The parameter Ea, which is obtained from the slope of the line (–Ea/R), is called the activation energy. Collectively the two quantities are called the Arrhenius parameters. ln k = ln A –

ln A

k = Ae − Ea / RT .......(iii) To interpret Ea we consider how the molecular potential energy changes in the course of a chemical reaction that begins with a collision between molecules of A and molecules of B(shown in figure).

ln k

Slope = –Ea/R

Potential energy

1/T A plot of ln k against 1/T is a straight line when the reaction follows the behaviour described by the Arrhenius equation. The slope gives –E a/R and the intercept at 1/T = 0 gives ln A.

The fact that Ea is given by the slope of the plot of ln k against 1/T means that, the higher the activation energy, the stronger the temperature dependence of the rate constant (that is, the steeper the slope). A high activation energy signifies that the rate constant depends strongly on temperature. If a reaction has zero activation energy, its rate is independent of temperature. In some cases the activation energy is negative, which indicates that the rate decreases as the temperature is raised. We shall see that such

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Ea Reactants

Products Progress of reaction

A potential energy profile for an exothermic reaction. The height of the barrier between the reactants and products is the activation energy of the reaction

40

DECEMBER 2009

As the reaction event proceeds, A and B come into contact, distort, and begin to exchange or discard atoms. The reaction coordinate is the collection of motions, such as changes in interatomic distances and bond angles, that are directly involved in the formation of products from reactants. (The reaction coordinate is essentially a geometrical concept and quite distinct from the extent of reaction.) The potential energy rises to a maximum and the cluster of atoms that corresponds to the region close to the maximum is called the activated complex. After the maximum, the potential energy falls as the atoms rearrange in the cluster and reaches a value characteristic of the products. The climax of the reaction is at the peak of the potential energy, which corresponds to the activation energy Ea. Here two reactant molecules have come to such a degree of closeness and distortion that a small further distortion will send them in the direction of products. This crucial configuration is called the transition state of the reaction. Although some molecules entering the transition state might revert to reactants, if they pass through this configuration then it is inevitable that products will emerge from the encounter. We also conclude from the preceding discussion that, for a reaction involving the collision of two molecules, the activation energy is the minimum kinetic energy that reactants must have in order to form products. For example, in a gas-phase reaction there are numerous collisions each second, but only a tiny proportion are sufficiently energetic to lead to reaction. The fraction of collisions with a kinetic energy in excess of an energy Ea is given by

ratio of the two rates, and therefore of the two rate constants : [P2 ] k = 2 [ P1 ] k1 This ratio represents the kinetic control over the proportions of products, and is a common feature of the reactions encountered in organic chemistry where reactants are chosen that facilitate pathways favouring the formation of a desired product. If a reaction is allowed to reach equilibrium, then the proportion of products is determined by thermodynamic rather than kinetic considerations, and the ratio of concentration is controlled by considerations of the standard Gibbs energies of all the reactants and products. The kinetic isotope effect The postulation of a plausible mechanism requires careful analysis of many experiments designed to determine the fate of atoms during the formation of products. Observation of the kinetic isotope effect, a decrease in the rate of a chemical reaction upon replacement of one atom in a reactant by a heavier isotope, facilitates the identification of bond-breaking events in the rate-determining step. A primary kinetic isotope effect is observed when the ratedetermining step requires the scission of a bond involving the isotope. A secondary isotope effect is the reduction in reaction rate even though the bond involving the isotope is not broken to form product. In both cases, the effect arises from the change in activation energy that accompanies the replacement of an atom by a heavier isotope on account of changes in the zero-point vibrational energies. First, we consider the origin of the primary kinetic isotope effect in a reaction in which the ratedetermining step is the scission of a C–H bond. The reaction coordinate corresponds to the stretching of the C–H bond and the potential energy profile is shown in figure. On deuteration, the dominant change is the reduction of the zero-point energy of the bond (because the deuterium atom is heavier). The whole reaction profile is not lowered, however, because the relevant vibration in the activated complex has a very low force constant, so there is little zero-point energy associated with the reaction coordinate in either isotopomeric form of the activated complex.

the Boltzmann distribution as e −E a / RT . Hence, we can interpret the exponential factor in eqn(iii) as the fraction of collision that have enough kinetic energy to lead to reaction. The pre-exponential factor is a measure of the rate at which collisions occur irrespective of their energy. Hence, the product of A and the exponential factor,

Potential energy

e −E a / RT , gives the rate of successful collisions. Kinetic and thermodynamic control of reactions : In some cases reactants can give rise to a variety of products, as in nitrations of mono-substituted benzene, when various proportions of the ortho-, meta-, and para- substituted products are obtained, depending on the directing power of the original substituent. Suppose two products, P 1 and P2, are produced by the following competing reactions : A + B → P1 Rate of formation of P 1 = k1[A][B] A + B → P2 Rate of formation of P 2 = k2[A][B] The relative proportion in which the two products have been produced at a given state of the reaction (before it has reached equilibrium) is given by the

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C–H C–D

Ea(C–H) Ea (C–D)

Reaction coordinate

41

DECEMBER 2009

UNDERSTANDING

Inorganic Chemistry

1.

A black coloured compound (A) on reaction with dil. H2SO4 gives a gas (B) which on passing in a solution of an acid (C) gives a white turbidity (D). Gas (B) when passed through an acidified solution of a compound (E), gives ppt.(F) which is soluble in dilute nitric acid. After boiling this solution an excess of NH4OH is added, a blue coloured compound (G) is produced. To this solution, on addition of CH3COOH and aqueous K4[Fe(CN) 6], a chocolate ppt. (H) is produced. On addition of an aqueous solution of BaCl2 to aqueous solution of (E), a white ppt. insoluble in HNO3 is obtained. Identify compounds (A) to (H). Sol. From the data on compounds (G) and (H), it may be inferred that the compound (E) contains cupric ions (Cu2+), i.e., (E) is a salt of copper. Since the addition of BaCl2 to (E) gives a white ppt. insoluble in HNO3, it may be said that the anion in the salt is sulphate ion (SO42–). Hence, (E) is CuSO4. The gas (B) which is obtained by adding dil. H2SO4 to a black coloured compound (A), may be H2S since it can cause precipitation of Cu2+ ions in acidic medium. The black coloured compound (A) may be ferrous sulphide (iron pyrite). Hence, the given observation may be explained from the following equations. Fe S + H2 SO4 → FeSO4 + H2S (A) Dil. (B) H2S + 2HNO3 → 2NO2 + 2H2O + S (D) (C) White turbidity CuSO4 + H2S → CuS ↓ + H2SO4 (E) (B) (F) Black ppt. 3CuS + 8HNO3 → Dil. 3Cu(NO3)2 + 2NO + 3S + 4H2O ++ Cu + 4NH3 → [Cu(NH3) 4]2+ (G) Blue colour 2+ [Cu(NH3)4] + 4CH3COOH → Cu2+ + 4CH3COONH4 2+ 4– Cu + [Fe(CN)6] → Cu2[Fe(CN)6] (H) Chocolate colour CuSO4(aq) + BaCl2(aq) → BaSO4 ↓ + CuCl2 (E) White ppt. Insuluble in HNO3 Hence, (A) is FeS, (B) is H2S, (C) is HNO3, (D) is S, (E) is CuSO4, (F) is CuS, (G) is [Cu(NH3)4]SO4 and (H) is Cu2[Fe(CN)6]

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2.

An unknown inorganic compound (X) gives the following observations : (i) When added to CuSO4 solution it liberates iodine and a white ppt. (Y) is formed. The liberated I2 reacts with Na 2S2O3 solution to give NaI and a colourless compound (Z). (ii) When CHCl3 and Cl2 water is added to aqueous solution of (X), a violet layer of chloroform is formed. (iii)(X) gives a violet colour flame when heated in Bunsen burner flame. (iv)When aqueous solution of (X) is added to aqueous lead nitrate, a yellow ppt. (M) is formed. (v) Addition of (X) to HgCl2 gives a red ppt. which dissolves in excess of (X) to give Nessler's reagent. (vi) On heating (X) with dil. HCl and KNO2, violet vapours of a compound (N) are formed which condenses on the wall of test tube. What are (X), (Y), (Z), (M) and (N) ? Explain the reactions. Sol. Observation (iii) indicates that the compound (X) contains K+, it is because it gives a violet coloured flame. On the other hand set (ii) confirmed that (X) contains I– ions, thus (X) is KI. Now the different reactions may be formulated as follows. 2KI + Cl2 CHCl 3 → I2 + 2KCl Violet layer (i) When CuSO4 reacts with KI, I2 is liberated and a ppt. of cuprous iodide (Y) is formed. [CuSO4 + 2KI → K2SO4 + CuI2] × 2 2CuI2 → Cu2I2 + I2 Unstable On adding, 2CuSO4 + 4KI → 2K2SO4 + Cu2I2 + I2 (Y) white ppt. The librated iodine is titrated against standard hypo solution when a colourless sodium tetrathionate (Z) is formed. 2Na2S2O3 + I2 → Na2S4O6 + 2NaI (Z) (colourless) (iv) Pb(NO3)2 + 2KI → PbI2 + 2KNO3 (M) yellow ppt (v) HgCl2 + 2KI → HgI2 ↓ + 2KCl Red ppt. HgI2 + 2KI → K2HgI4 (Soluble) K2HgI4 + NaOH → Nessler's reagent (vi) HCl + KNO2 → HNO2 + KCl HCl + KI → KCI + HI 42

DECEMBER 2009

2HNO2 + 2HI → I2 + 2H2O + 2NO (N) Hence, (X) is KI, (Y) is Cu2I2, (Z) is Na2S4O6, (M) is PbI2 and (N) is I2.

∆ 2Bi + 6HCl → 2BiCl3 + 3H2 (G) (A) Hence, (A) is BiCl3, (B) is Bi2S3, (C) is H2S, (D) is Bi(NO3)2, (E) is Bi(OH)3, (F) is BiOCl and (G) is Bi

3.

An inorganic halide (A) gives the following reactions. (i) The cation of (A) on raction with H2S in HCl medium, gives a black ppt. of (B). (A) neither gives ppt. with HCl nor blue colour with K4Fe(CN) 6. (ii) (B) on heating with dil.HCl gives back compound(A) and a gas (C) which gives a black ppt. with lead acetate solution. (iii) The anion of (A) gives chromyl chloride test. (iv) (B) dissolves in hot dil. HNO3 to give a solution, (D). (D) gives ring test. (v) When NH4OH solution is added to (D), a white precipitate (E) is formed. (E) dissolves in minimum amount of dil. HCl to give a solution of (A). Aqueous solution of (A) on addition of water gives a whitish turbidity (F). (vi) Aqueous solution of (A) on warming with alkaline sodium stannite gives a black precipitate of a metal (G) and sodium stannate. The metal (G) dissolves in hydrochloride acid to give solution of (A). Identify (A) to (G) and give balanced chemical equations of reactions. Sol. Observation of (i) indicates that cation (A) is Bi3+ because it does not give ppt. with HCl nor blue colour with K4Fe(CN)6, hence it is neither Pb2+ nor Cu2+. Since anion of (A) gives chromyl chloride test, hence it contains Cl– ions. Thus, (A) is BiCl3. Its different reactions are given below : (i) 2BiCl3 + 3H2S → Bi2S3 + 6HCl (A) (B) (ii) Bi2S3 + 6HCl → 3H2S + 2 BiCl3 (B) (C) (A)

4.

(i) An inorganic compound (A) is formed on passing a gas (B) through a concentrated liquor containing Na2S and Na 2SO3. (ii) On adding (A) into a dilute solution of AgNO3, a white ppt. appears which quickly changes into black coloured compound (C). (iii) On adding two or three drops of FeCl3 into excess of solution of (A), a violet coloured compound (D) is formed. This colour disappears quickly. (iv) On adding a solution of (A) into the solution of CuCl2, a white ppt. is first formed which dissolves on adding excess of (A) forming a compound (E). Identify (A) to (E) and give chemical equations for the reactions at steps (i) to (iv) Sol. (i) The compound (A) appears to be Na2S 2O3 from its method of preparation given in the problem. Na2S + Na2SO3 + I2 → 2NaI + Na2S 2O3 (B) (A) or Na2SO3 + 3Na 2S + 3SO2 → 3Na 2S2O3 (B) (A) (ii) White ppt. of Ag2S2O3 is formed which is hydrolysed to black Ag2S Na2S2O3 + 2AgNO3 → 2NaNO3 + Ag2S 2O3 ↓ White ppt Ag2S 2O3 + H2O → Ag2S + H2SO4 (C) (iii) A violet ferric salt is formed. 3Na2S2O3 + 2FeCl3 → Fe2(S 2O3)3 + 6NaCl (D)(violet) (iv) 2CuCl2 + 2Na2S2O3 → 2CuCl + Na2S4O6 + 2NaCl White ppt. 2CuCl + Na2S 2O3 → Cu2S2O3 + 2NaCl 3Cu2S2O3 + 2Na 2S2O3 → Na 4[Cu6(S2O3) 5] (E) or 6CuCl + 5Na2S2O3 → Na4[Cu6(S2O3)5] + 6NaCl (E) Hence, (A) is Na2S2O3, (B) is I2 or SO2, (C) is Ag2S, (D) is Fe2(S2O3)3 and (E) is Na4[Cu6(S2O3) 5]

∆ (iii) Bi2S3 + 8HNO3 → 2Bi(NO3)3 + 2NO (B) (D) + 3S + 4H2O (iv) Bi(N O3)3 + 3NH4OH → (D) Bi(OH)3 ↓ + 3NH4NO3 (E) White ppt. ∆ Bi(OH)3 + 3HCl → BiCl3 + 3H2O Dil. (A) BiCl3 + H2O → BiOCl + 2HCl (A) (F) Bismuth oxychloride (White turbidity) (v) BiCl3 + 2NaOH +Na2[SnO2] → (A) Bi + NaSnO3 + H2O + 3NaCl (G) Black ppt.

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5.

43

A colourless solid (A) on heating gives a white solid (B) and a colourless gas (C). (B) gives off reddishbrown fumes on treating with H2SO4. On treating with NH4Cl, (B) gives a colourless gas (D) and a DECEMBER 2009

residue (E). The compound (A) on heating with (NH4)2SO4 gives a colourless gas (F) and white residue (G). Both (E) and (G) impart bright yellow colour to Bunsen flame. The gas (C) forms white powder with strongly heated Mg metal which on hydrolysis produces Mg(OH) 2. The gas (D) on heating with Ca gives a compound which on hydrolysis produces NH3. Identify compounds (A) to (G) giving chemical equations involved. Sol. The given information is as follows :

TRUE OR FALSE

(i) A Heat → B + C Colourless Solid Colourless Solid gas ∆ (ii) B + H2SO4 → Reddish brown gas ∆ (iii) B + NH4Cl → D + E Colourless gas ∆ (iv) A + (NH4)2SO4 → F + G olourless gas White Residue (v) E and G imparts yellow colour to the flame.

(vi) C + Mg Heat → White powder 2O H → Mg(OH)2 2O (vii) D + Ca Heat → Compound H → NH3 Information of (v) indicates that (E) and (G) and also (A) are the salts of sodium because Na+ ions give yellow coloured flame. Observations of (ii) indicate that the anion associated with Na+ in (A) may be NO3–. Thus, the compound (A) is NaNO3. The reactions involved are as follows :

1.

The magnitude of charge on one gram of electrons is 1.60 × 10–19 coulomb.

2.

Chromyl chloride test of Cl– radical is not given by HgCl2.

3.

The energy levels in a hydrogen atom can be compared with the steps of a ladder placed at equal distance.

4.

In SN1 mechanism, the leaving group in the molecule, leaves the molecule, well before joining of an attacking group.

5.

Metamerism is special type of isomerism where isomers exist simultaneously in dynamic equilibrium.

6.

Addition of HCN with formaldehyde is an example of electrophilic addition reaction.

7.

Ligroin is essentially petroleum ether containing aliphatic hydrocarbons and is generally used in dry cleaning clothes.

Sol.

∆ (i) 2NaNO3 → 2NaNO2 + O2 ↑ (A) (B) (C) (ii) 2NaNO2 + H2SO4 → Na2SO4 + 2HNO2 (B) Dil. 3HNO2 → HNO3 + H2O + 2NO↑ 2NO + O2 → 2NO2 ↑ Reddish brown Fumes (iii) NaNO2 + NH4Cl → NaCl + N2 ↑ + 2H2O (B) (E) (D)

1.

[False] Thomson through his experiment determined the charge to mass ratio of an electron and the value of e/m is equal to 1.76 × 108 coulomb/gm. Hence one gm of electrons have charge 1.76 × 10 8 C.

2.

[True]

3.

[False]

4.

[True] SN1 reaction mechanism takes place in two steps as : R—X Slow → R+ + X– R+ + OH– Fast → ROH

5.

∆ (iv) 2NaNO3 + (NH4)2SO4 → Na2SO4 + 2NH3 (A) (G) (F) 2HNO3

For example :

∆ 2O (v) O2 + 2Mg → 2MgO H → Mg(OH)2 (C)

CH3CH2–O–CH2CH3 and CH3–OCH2CH2CH3

∆

Conditions mentioned in the statement are associated with phenomenon of trautomerism.

(vi) N2 + 3Ca → Ca3N2 (D) Ca3N2 + 6H2O → 3Ca(OH)2 + 2NH3 ↑ Hence, (A) is NaNO3, (B) is NaNO2, (C) is O2, (D) is N2, (E) is NaCl, (F) is NH3 and (G) is Na2SO4.

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[False] In metamerism isomers differ in structure due to difference in distribution of carbon atoms about the functional group.

6.

[False] H H – C = O + H+ CN–

H H – C = OH CN

7. 44

[True] DECEMBER 2009

Set

8

`tà{xÅtà|vtÄ V{tÄÄxÇzxá

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 Sol ut i ons w il l be publ i s he d i n ne xt is s ue Joint Director Academics, Career Point, Kota 1.

Show that the six planes through the middle point of each edge of a tetrahedron perpendicular to the opposite edge meet in a point.

2.

Prove that if the graph of the function y = f (x), defined throughout the number scale, is symmetrical about two lines x = a and x = b, (a < b), then this function is a periodic one.

3.

Show that an equilateral triangle is a triangle of maximum area for a given perimeter and a triangle of minimum perimeter for a given area.

4.

Let az2 + bz + c be a polynomial with complex coefficients such that a and b are non zero. Prove that the zeros of this polynomial lie in the region b c |z|≤ + a b

5.

9.

10. ABC is a triangle inscribed in a circle. Two of its sides are parallel to two given straight lines. Show that the locus of foot of the perpendicular from the centre of the circle on to the third side is also a circle, concentric to the given circle.

Dimensional Formulae of Some Physical Quantities

x2

+

y2

= 1 is

a2 b2 circumscribed with all the three sides touching the ellipse. Find the least possible area of the triangle. 6.

7.

2

If one of the straight lines given by the equation ax + 2hxy + by2 = 0 coincides with one of those given by a′x2 + 2h′ xy + b′ y2 = 0 and the other lines represented ha´b´ h´ab´ by them be perpendicular, show that = b´− a´ b−a Prove that n m n m + 1 n m + 2 + + + ....... to 0 n 1 n 2 n (n + 1) terms n m n m n m = + 2 + 22 + ..... to (n + 0 0 1 1 2 2 1) terms 1

8.

∫

Torque (τ)

[ML2T–2]

Moment of Inertia (I) Coefficient of viscosity (η) Gravitational constant (G)

[ML2] [ML–1T–1]

Specific heat (S)

[L2T–2θ–1]

Coeficient of thermal conductivity (K)

[MLT–3θ–1]

Universal gas constant (R) Potential (V)

[ML2T–2θ–1] [ML2T–3A–1]

Intensity of electric field (E)

[MLT–3A–1]

Permittivity of free space (ε0)

[M–1L–3T4A2]

Specific resistance (ρ) Magnetic Induction (B)

[ML3T–3A2]

Planck's constant (h)

[ML2T–1]

Boltzmann's constant (k)

[ML2T–2θ–1]

Entropy (S)

[ML2T–2θ–1] [T–1]

Decay constant (λ ) Bohr magnetic (µB)

If n ≥ 2 and In = (1 − x 2 ) n cos mx dx, then show that −1

m 2I n = 2n(2n – 1) I n–1 – 4n(n – 1) I n–2.

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Work (W) Stress

Dime nsional Formulae [ML2T–2] [ML–1T–2]

Physical Quantity

An isosceles triangle with its base parallel to the major axis of the ellipse

Find the sum to infinite terms of the series 3 5 7 9 11 + + + + + ........ ∞ 4 36 144 400 900

Thermmionic current density (J) 45

[M–1L3T–2]

[MT–2A–1]

[L2A] [AL–2] DECEMBER 2009

MATHEMATICAL CHALLENGES SOLUTION FOR NOVEMBER ISSUE (SET # 7)

1.

Let the line be y = 2x + c

9 − c 9 + 2c Point A , 3 6

=

AD DC + BD . BC BD.CD

2c − 3 + c − 6 Point B , −3 −3

=

AD AD AD 1 = = . 2 BD.CD AD AD AD

so it is vector along AB with magnitude

c + 6 5c + 12 Point C , 3 3 mid point of B & C is

1 2

|a+b |=

2c − 3 c + 6 + . , 3 −3

3.

1 − c + 6 5c + 12 9 − c 2c + 9 + = , 2 + 3 3 6 3

1 AD

The line PQ always passes through (α, β) so it is y –β = m(x – α) Let the circle be x2 + y2 – 2hx – 2ky = 0

which is point A, so AB and AC are equal.

Joint equation of OP and OQ. x2 + y2 – 2 (hx + ky)

2. A

( y − mx ) =0 β − mα P

b

a

O B

=

1 AB

2

1 AB2

Q

AB +

1 AC2

h − mk 2k 2 1 − y – 2 xy + β − mα β − mα

AC

(AD + DB) +

1 AC2

2hn 2 1 + x = 0 β − mα

It must represent y2 – x2 = 0 so

(AD + DC)

1 DB DC 1 + = + AD + 2 2 BD.DC CD.CB AC AB

and

h − mk = 0 ⇒ m = h/k β − mα 1–

...(1)

2k 2hm = –1 – β − mα β − mα

⇒ β – mα – 2k = –β + mα – 2hm

DB DC 1 1 1 + = AD + BD + CD BC AB2 AC 2

⇒ –β + mα + k – hm = 0 ⇒ –β + k + h/k(α – h) = 0

1 1 + = AD . BD.DC CD.CB =

(h,k)

C

D

1 AB 1 AC a+b = . + . AB AB AC AC =

1 . AD

(using (1) in it)

⇒ k2 – βy + αh – h2 = 0 so required locus is x2 – y2 – α x + β y = 0

AD 1 1 + . CD BD CD

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46

DECEMBER 2009

4.

π π As |f(x)| ≤ |tan x| for ∀ x ∈ − , 2 2

(–1,log23)

so f(0) = 0 so |f(x) – f(0)| ≤ |tan x| divides both sides by |x|

⇒ lim

x →0

f ( x ) − f (0) tan x ≤ lim x →0 x x

1

=

ai

∑i

∫

0

log 2 (2 − x ) dx +

−1

= – log2

≤1

∫ (2 − 2

y

) dy +

−1

= log 2 3 −

1 1 1 ⇒ a 1 + a 2 + a 3 + ..... + a n ≤ 1 2 3 n

⇒

(3/2,–1)

(0,–1)

⇒ |f´(0)| ≤ 1

n

(1,0)

(–1,0)

f ( x ) − f (0) tan x ≤ ⇒ x x

1 π 4

2 1 π + 2 log 2 3 + 2 – + ln 2 2ln 2 4

e2 e π +2+ sq units 27 4

i =1

7. 5.

A

Let the number is xyz, here x < y and z < y. Let y = n, then x can be filled in (n – 1) ways. (i.e. from 1 to (n – 1)) and z can be filled in n ways (i.e. from 0 to (n – 1))

F

here 2 ≤ n ≤ 9

B

so total no. of 3 digit numbers with largest middle digit =

=

9

9

n =2

n =2

∑ n(n −1) = ∑ n – ∑ n

so tan A =

n =2

9.10.19 9.10 – 6 2

so

= 285 – 45 = 240 240 required probability = 9 × 10 ×10

so

a2 r12 a2 r12

BD BC BC a = = = MD 2MD 4 r1 4r1

= tan2A +

b2 r22

+

c2 r32

= 16 (tan2A + tan2B + tan2C)

8 = 30 =

C

D

∠BMC = 2∠BAC = 2∠BMD

9

2

E M

...(1)

Now as tan A + tan B + tan C ≥ 3 (tan A . tan B . tan C)1/3 and for a triangle tan A + tan B + tan C = tan A . tan B . tan C

4 15

so (tan A . tan B . tan C)2/3 ≥ 3 6.

The region bounded by the curve y = log2(2 – x) and

⇒ tan A . tan B . tan C ≥ 3 3

the inequality (x – |x|)2 + (y – |y|)2 ≤ 4 is required area

⇒ tan2A + tan2B + tan2C ≥ 3(tan A. tan B tan C)2/3 ≥ 3.3

is

so from (1),

XtraEdge for IIT-JEE

47

a2 r12

+

b2 r22

+

c2 r32

≥ 144. DECEMBER 2009

8.

= a + (1 – r2)

Z1, Z2, Z3 are centroids of equilateral triangles ACX, ABY and BCZ respectively.

∫ (1+ r

Z1 − Z A iπ/6 e ZC − ZA

Z1 – ZA = (ZC – ZA)

a

=a+ C

B Z3

1 3 i − 3 2 2

...(1)

...(2)

Now and

–

Tr = 2

=

∫ 0

=

2

−1 1 + r tan t 1− r

tan a / 2

0

lim Tr = a –

2(1 + r )(r − 1) π = a–π (1 + r )(r − 1) 2

lim+ Tr = a +

2(1 − r)( r + 1) π = a+π (1 + r )(r − 1) 2

r→1

∫ du = a r→1

so x 2 + ax 2 + bx + c = (x – γ) (x 2 + (a + γ) x + (γ 2 + aγ + b)) where – γ (γ2 + aγ + b) = c, as γ is the root of given equation, so x2 + (a + γ) x + (γ2 + aγ + b) = 0 must have two roots i.e. α and β. So its discriminant is non negative, thus (γ + a)2 – 4(γ2 + aγ + b) ≥ 0

1 − r cos u

2

du.

1 − 2r cos u + r 2 − r 2 + 1 1 − 2r cos u + r 2

...(1)

3γ2 + 2aγ – a2 + 4b ≤ 0 so γ ≤

du

− a + 2 a 2 − 3b 3

so greatest root is also less than or equal to

1− r 1 + 1 − 2r cos u + r 2 du 0

XtraEdge for IIT-JEE

1− r 1+ r

10. Let α, β, γ be the three real roots of the equation without loss of generality, it can be assumed that α ≤ β ≤ γ.

1 3 (ZA + ZC – 2ZB)) + i 2 2 2 3

∫

t2 +

0

difference π.

i

a

2 dt

∫

2

r→1+

r→1

0

a

(1 + r)

tan a / 2

Hence lim+ Tr, T1, lim− Tr form an A.P. with common

1 (ZC – ZA) 2

∫ 1 − 2r cos u + r

(1 − r) 2 (1 + r) 2

0

(ZA + ZC – 2ZB) ..(4)

= Z1 – Z2

9.

tan 2 u / 2 +

a

1 i = (ZC – ZB) + (ZC + ZB – 2ZA) 2 2 3

a

1− r 2

and (from (1)) T1 =

To prove ∆xyz as equilateral triangle, we prove that (Z3 – Z2)eiπ/3 = Z1 – Z2 So, (Z3 – Z2)eiπ/ 3 = (

0

tan 2 u / 2 + (1 − r ) 2

sec2 u / 2 du

2(1 − r 2 ) 1 + r =a+ (1 + r) 2 1 − r

1 (ZA – ZC) 2 2 3

∫

(1 + r) 2

so, Tr = a +

...(3)

i

a

1− r 2

ZC

1 i (ZC – ZB) + (ZC + ZB – 2ZA) 2 2 3

+

2

Let tan u/2 = t

1 3 i + Z1 – ZA = (ZC – ZA) 3 2 2 similarly,

similarly Z2 – Z3 =

∫ (1+ r )

sec 2 u / 2

0

z

So, Z1 – Z2 =

)(1 + tan 2 u / 2) − 2r(1 − tan 2 u / 2)

Z1

A

Z2 – ZA = (ZB – ZA)

2

= a + (1 – r2)

Z2 ZB

sec 2 u / 2

0

x

ZA y

a

2

− a + 2 a 2 − 3b . 3 48

DECEMBER 2009

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

MATHS 1.

Suppose f(x) = x3 + ax2 + bx + c, where a, b, c are

3.

chosen respectively by throwing a die three times.

4) = g(x + 2) + g(x + 6), then prove that

Find the probability that f(x) is an increasing

∫

x +8 x

g (t ) dt

is a constant function.

function. Sol. f´(x) = 3x2 + 2ax + b

Sol. given that g(x) + g(x + 4) = g(x + 2) + g(x + 6) ...(1) putting x = x + 2 in (1) ........

y = f(x) is strictly increasing

⇒ f´(x) > 0 ∀ x

g(x + 2) + g(x + 6) = g(x + 4) + g(x + 8)

...(2)

from (1) & (2)

⇒ (2a)2 – 4.3.b < 0

g(x) = g(x + B)

This is true for exactly 15 ordered pairs (a, b); 1 ≤ a, 15 5 b ≤ 6, so probability = = 36 12 2.

Let g be a real valued function satisfying g(x) + g(x +

Now,

f(x) =

∫

x +8 x

g (t ) dt

f´(x) = g(x + 8) – g(x) = 0

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

⇒ g is constant function

methods. 4.

→

Sol. Let A = a ˆi + b ˆj + c kˆ

If exactly three distinct chords from (h, 0) point to the circle x2 + y2 = a2 are bisected by the parabola y2 = 4ax, a > 0, then find the range of 'h' parameter.

→

⇒ B = 3 ˆi + 2 ˆj + kˆ → →

→

Sol. Let M(at2, 2at) is mid-point of chord AB, then chord AB = T = S1

→

⇒ ( A . B) 2 ≤ | A |2 | B |2 a 2 + b2 + c 2

3a + 2b + c ≤

B

14

M A

(7) ≤ (a + b + c ) (14) 2

2

2

2

{Q 3a + 2b + c = 7, point lies on the plane} a2 + b2 + c 2 ≥

49 7 = 14 2 AB :

x.at2 + y.2at = a2t4 + 4a2t2

since AB chord passes through (h, 0)

XtraEdge for IIT-JEE

49

DECEMBER 2009

so, h.at2 = a 2t4 + 4a2t2

Sol. Let P be (x1, y1),

at2 [at2 + (4a – h)] = 0

Q

A P

⇒ 4a – h < 0

If a > 0

⇒ h > 4a

...(i) S

Now point (at2,2at) must lie inside the circle, on solving B

we get, h < a ( 5 + 2)

...(ii)

Points of intersection of tangent and normal at P

from (i) & (ii)

b2 a2y points with y-axis and A 0, , B 0, y1 − 2 1 , b y1

4a < h < a ( 5 + 2) 5.

S : (ae, 0) slope (SA). slope (SB) = –1

Find the sum of the terms of G.P. a + ar + ar 2 + ..... + ∞

⇒ ∠ASB = 90º (PA and PB are tangent and normal) P must lie on the circle with AB as diameter. Hence the point of intersection of the ellipse and the

where a is the value of x for which the function 7 + 2x loge25 – 5x – 1 – 52– x has the greatest value and r is the Lt

∫

x →0 0

Sol. S =

t 2dt

x

circle is P. Due to symmetry the angles made by AB

x 2 tan(π + x )

at P,Q,M, N are all 90º.

a , 1− r

Do you know

To get the greatest value f´(x) = 2log e25 – 5 x – 1 log 5 + 52– x log 5

•

100 years ago: The first virus was found in both plants and animals.

•

90 years ago: The Grand Canyon became a national monument & Cellophane is invented.

•

80 years ago: The food mixer and the domestic refrigerator were invented.

•

70 years ago: The teletype and PVC (polyvinylchloride) were invented.

1 π

•

60 years ago: Otto Hahn discovered nuclear fission by splitting uranium, Teflon was invented.

1 2π ⇒ sum of G.P. = π π −1

• •

50 years ago: Velcro was invented.

•

30 years ago: The computer mouse was invented.

= 1 meets the y-axis at A and B

•

respectively. Find the angle subtended by AB at the

20 years ago: First test-tube baby born in England, Pluto’s moon, Charon, discovered.

•

10 years ago: First patent for a geneticallyengineered mouse was issued to Harvard Medical School.

f´(x) = 4 loge5 – 5

x– 1

loge5 + 5.5

1– x

loge5

⇒ f´(x) = 0 put 5 x – 1 t(> 0) t2 – 4t – 5 = 0 ⇒ t = 5 ⇒ 5 x –1 = 5 ⇒ x = 2 to evaluate r : r = Lt

∫

t 2dt

x

x →0 0

2

x tan(π + x )

since a = 2, r = 6.

M

N

a2t4 + 4a2t2 – a 2 < 0

=

The tangent and normal at a point P on the ellipse x2 a2

+

y2 b2

points of intersection of the circle (through A,S,B) and the ellipse. S being one of the foci.

XtraEdge for IIT-JEE

50

40 years ago: An all-female population of lizards was discovered in Armenia.

DECEMBER 2009

MATH

MONOTONICITY, MAXIMA & MINIMA Mathematics Fundamentals

Monotonic Functions : A function f(x) defined in a domain D is said to be (i) Monotonic increasing :

according as f(x) is monotonic increasing or decreasing at x = a. So at x = a, function f(x) is monotonic increasing ⇔ f´(a) > 0

x < x 2 ⇒ f (x 1 ) ≤ f (x 2 ) ⇔ 1 ∀ x1, x2 ∈ D x 1 > x 2 ⇒ f (x 1 ) ≥ f (x 2 ) y y

O

x

O

monotonic decreasing ⇔ f´(a) < 0 (ii) In an interval : In [a, b], f(x) is monotonic increasing ⇔ f´(x) ≥ 0 monotonic decreasing ⇔ f´(x) ≤ 0 ∀ x ∈ (a, b) constant ⇔ f´(x) = 0 x

Note : (i) In above results f´(x) should not be zero for all values of x, otherwise f(x) will be a constant function. (ii) If in [a, b], f´(x) < 0 at least for one value of x and f´(x) > 0 for at least one value of x, then f(x) will not be monotonic in [a, b]. Examples of monotonic function : If a functions is monotonic increasing (decreasing ) at every point of its domain, then it is said to be monotonic increasing (decreasing) function. In the following table we have example of some monotonic/not monotonic functions Monotonic Monotonic Not increasing decreasing monotonic x3 1/x, x > 0 x2 x|x| 1 – 2x |x| x –x e e ex + e– x log x log2x sin x sin h x cosec h x, x > 0 cos h x [x] cot hx, x > 0 sec h x

x < x 2 ⇒ f (x 1 ) >/ f ( x 2 ) i.e., ⇔ 1 ∀ x1, x2 ∈ D x 1 > x 2 ⇒ f (x 1 ) x 2 ⇒ f (x 1 ) ≤ f ( x 2 ) y

y

O

x

O

x

x < x 2 ⇒ f (x 1 ) x 2 ⇒ f (x 1 ) /> f ( x 2 ) A function is said to be monotonic function in a domain if it is either monotonic increasing or monotonic decreasing in that domain. Note : If x1 < x2 ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ D, then f(x) is called strictly increasing in domain D and similarly decreasing in D. Method of testing monotonicity : (i) At a point : A function f(x) is said to be monotonic increasing (decreasing) at a point x = a of its domain if it is monotonic increasing (decreasing) in the interval (a – h, a + h) where h is a small positive number. Hence we may observer that if f(x) is monotonic increasing at x = a then at this point tangent to its graph will make an acute angle with xaxis where as if the function is monotonic decreasing there then tangent will make an obtuse angle with xaxis. Consequently f´(a) will be positive or negative

XtraEdge for IIT-JEE

Properties of monotonic functions : If f(x) is strictly increasing in some interval, then in that interval, f–1 exists and that is also strictly increasing function. If f(x) is continuous in [a, b] and differentiable in (a, b), then f´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f(x) is monotonic increasing in [a, b] f´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f(x) is monotonic decreasing in [a, b] 51

DECEMBER 2009

If both f(x) and g(x) are increasing (or decreasing) in [a, b] and gof is defined in [a, b], then gof is increasing. If f(x) and g(x) are two monotonic functions in [a, b] such that one is increasing and other is decreasing then gof, it is defined, is decreasing function. Maximum and Minimum Points : The value of a function f(x) is said to be maximum at x = a if there exists a small positive number δ such that f(a) > f(x) y

O

( ) a

( ) b

( ) c

where x = c is a point such that f´(c) = 0. If a continuous function has only one maximum (minimum) point, then at this point function has its greatest (least) value. Monotonic functions do not have extreme points. Conditions for maxima and minima of a function Necessary condition : A point x = a is an extreme point of a function f(x) if f´(a) = 0, provided f´(a) exists. Thus if f´(a) exists, then x = a is an extreme point ⇒ f´(a) = 0

f´(a) ≠ 0 ⇒ x = a is not an extreme point But its converse is not true i.e. f´(a) = 0 ⇒ / x = a is an extreme point. For example if f(x) = x3, then f´(0) = 0 but x = 0 is not an extreme point. Sufficient condition : For a given function f(x), a point x = a is a maximum point if f´(a) = 0 and f´´(a) < 0 a minimum point if f´(a) = 0 and f´´(a) > 0 not an extreme point if f´(a) = 0 = f´´(a) and f´´´(a) ≠ 0. Note : If f´(a) = 0, f´´(a) = 0, f´´´(a) = 0 then the sign of f(4)(a) will determine the maximum or minimum point as above. Working Method : Find f´(x) and f´´(x). Solve f´(x) = 0. Let its roots be a, b, c, ... Determine the sign of f´´(x) at x = a, b, c, .... and decide the nature of the point as mentioned above. Properties of maxima and minima : If f(x) is continuous function, then Between two equal values of f(x), there lie atleast one maxima or minima. Maxima and minima occur alternately. For example if x = –1, 2, 5 are extreme points of a continuous function and if x = –1 is a maximum point then x = 2 will be a minimum point and x = 5 will be a maximum point. When x passes a maximum point, the sign of dy/dx changes from + ve to – ve, where as when x passes through a minimum point, the sign of f´(x) changes from –ve to + ve. If there is no change in the sign of dy/dx on two sides of a point, then such a point is not an extreme point. If f(x) is maximum (minimum) at a point x = a, then 1/f(x), [f(x) ≠ 0] will be minimum (maximum) at that point. If f(x) is maximum (minimum) at a point x = a, then for any λ ∈ R, λ + f(x), log f(x) and for any k > 0, k f(x), [f(x)]k are also maxmimum (minimum) at that point.

x

Also then the point x = a is called a maximum point for the function f(x). Similarly the value of f(x) is said to be minimum at x = b if there exists a small positive number δ such that f(b) < f(x) ∀ x ∈ (b – δ, b + δ) Also then the point x = b is called a minimum point for f(x) Hence we find that : (i) x = a is a maximum point of f(x)

f (a ) – f(a + h) > 0 ⇔ f(a) – f(a – h) > 0 (ii) x = b is a minimum point of f(x) f (b ) – f(b + h) < 0 ⇔ f(b) – f(b – h) > 0 (iii) x = c is neither a maximum point nor a minimum point f (c) – f (c + h ) ⇔ and have opposite signs. f (c) − f (c − h ) Where h is a very small positive number. Note : The maximum and minimum points are also known as extreme points. A function may have more than one maximum and minimum points. A maximum value of a function f(x) in an interval [a, b] is not necessarily its greatest value in that interval. Similarly a minimum value may not be the least value of the function. A minimum value may be greater than some maximum value for a function. The greatest and least values of a function f(x) in an interval [a, b] may be determined as follows : Greatest value = max. {f(a), f(b), f(c)} Least value = min. {f(a), f(b), f(c)}

XtraEdge for IIT-JEE

or

52

DECEMBER 2009

MATH

FUNCTION Mathematics Fundamentals If f and g are two functions then their sum, difference, product, quotient and composite are denoted by

Definition of a Function : Let A and B be two sets and f be a rule under which every element of A is associated to a unique element of B. Then such a rule f is called a function from A to B and symbolically it is expressed as

or

f + g, f – g, fg, f/g, fog and they are defined as follows :

f:A→B

(f + g) (x) = f(x) + g(x)

f A → B

(f – g) (x) = f(x) – g(x) (fg) (x) = f(x) f(g)

Function as a Set of Ordered Pairs Every function f : A → B can be considered as a set of ordered pairs in which first element is an element of A and second is the image of the first element. Thus

(f/g) (x) = f(x)/g(x) (fog) (x) = f[g(x)]

Formulae for domain of functions :

f = {a, f(a) /a ∈ A, f(a) ∈ B}.

Df ± g = Df ∩ Dg

Domain, Codomain and Range of a Function :

Dfg = Df ∩ Dg

If f : A → B is a function, then A is called domain of f and B is called codomain of f. Also the set of all images of elements of A is called the range of f and it is expressed by f(A). Thus

Df/g = Df ∩ Dg ∩ {x |g(x) ≠ 0} Dgof = {x ∈ Df | f(x) ∈ Dg} D

f(A) = {f(a) |a ∈ A} obviously

f(A) ⊂ B.

f

= Df ∩ {x |f(x) ≥ 0}

Classification of Functions

Note : Generally we denote domain of a function f by Df and its range by R f.

1.

Algebraic and Transcendental Functions : Algebraic functions : If the rule of the function consists of sum, difference, product, power or roots of a variable, then it is called an algebraic function. Transcendental Functions : Those functions which are not algebraic are named as transcendental or non algebraic functions.

Equal Functions : Two functions f and g are said to be equal functions if domain of f = domain of g codomain of f = codomain of g 2.

f(x) = g(x) ∀ x. Algebra of Functions :

XtraEdge for IIT-JEE

(g(x) ≠ 0)

53

Even and Odd Functions : Even functions : If by replacing x by –x in f(x) there in no change in the rule then f(x) is called an even function. Thus f(x) is even ⇔ f(–x) = f(x) DECEMBER 2009

Odd function : If by replacing x by –x in f(x) there is only change of sign of f(x) then f(x) is called an odd function. Thus

Period of f(x) = T

⇒ Period of f(nx + a) = T/n Periods of some functions :

f(x) is odd ⇔ f(–x) = – f(x) 3.

Function

Explicit and Implicit Functions : Explicit function : A function is said to be explicit if its rule is directly expressed (or can be expressed( in terms of the independent variable. Such a function is generally written as

sin x, cos x, sec x, cosec x,

2π

tan x, cot x

π

sinnx, cosn x, secn x, cosecn x

2π if n is odd

π if n is even

y = f(x), x = g(y) etc.

4.

Implicit function : A function is said to be implicit if its rule cannot be expressed directly in terms of the independent variable. Symbolically we write such a function as

tann x, cotn x

π∀ n∈ N

|sin x|, |cos x|, |sec x|, |cosec x|

π

|tan x|, |cot x|,

π

f(x, y) = 0, φ(x, y) = 0 etc.

|sin x| + |cos x|, sin4x + cos 4x

π 2

|sec x| + |cosec x|

Continuous and Discontinuous Functions : Continuous functions : A functions is said to be continuous if its graph is continuous i.e. there is no gap or break or jump in the graph. Discontinuous Functions : A function is said to be discontinuous if it has a gap or break in its graph atleast at one point. Thus a function which is not continuous is named as discontinuous.

5.

Period

x – [x]

1

Period of f1(x) = T1, period fo f2(x) = T 2

⇒ period of a f1(x) + bf2(x) ≤ LCM {T1, T 2}

Increasing Functions : A function f(x) is said to be increasing function if for any x1, x2 of its domain

Kinds of Functions : One-one/ May one Functions :

x1 < x2 ⇒ f(x1) ≤ f(x2)

A function f : A → B is said to be one-one if different elements of A have their different images in B.

or x1 > x2 ⇒ f(x1) ≥ f(x2) Decreasing Functions : A function f(x) is said to be decreasing function if for any x1, x2 of its domain

Thus

⇒ f (a ) ≠ f (b) a≠b f is one-one ⇔ or f (a ) = f (b) ⇒ a =b

x1 < x2 ⇒ f(x1) ≥ f(x2) or x1 > x2 ⇒ f(x1) ≤ f(x2)

A function which is not one-one is called many one. Thus if f is many one then atleast two different elements have same f-image.

Periodic Functions : A functions f(x) is called a periodic function if there exists a positive real number T such that

Onto/Into Functions : A function f : A → B is said to be onto if range of f = codomain of f

∀x

Also then the least value of T is called the period of the function f(x).

XtraEdge for IIT-JEE

π 2

Period of f(x) = T ⇒ period of f(ax + b) = T/|a|

Increasing and Decreasing Functions :

f(x + T) = f(x).

|tan x| + |cot x|

Thus f is onto ⇔ f(A) = B 54

DECEMBER 2009

Hence f : A → B is onto if every element of B (co-domain) has its f–preimage in A (domain).

Domain and Range of some standard functions : Function

A function which is not onto is named as into function. Thus f : A → B is into if f(A) ≠ B. i.e., if there exists atleast one element in codomain of f which has no preimage in domain. Note : Total number of functions : If A and B are finite sets containing m and n elements respectively, then total number of functions which can be defined from A to B = nm. total number of one-one functions from A to B

n p = m 0

if if

m≤n m>n

total number of onto functions from A to B (if m ≥ n) = total number of different n groups of m elements. Composite of Functions :

Domain

Range

Polynomial function

R

R

Identity function x

R

R

Constant function c

R

{c}

Reciprocal function 1/x

R0

R0

x2, |x|

R

R+ ∪ {0}

x3, x |x|

R

R

Signum function

R

{–1, 0, 1}

x + |x|

R

R+ ∪ {0}

x – |x|

R

R– ∪ {0}

[x]

R

Z

x – [x]

R

[0, 1)

x

[0, ∞)

[0, ∞)

ax

R

R+

log x

R+

R

sin x

R

[–1, 1]

cos x

R

[–1, 7]

tan x

R – {± π/2, ± 3π/2, ...}

R

cot x

R – {0, ± π. ± 2 π, .....

R

The following properties of composite functions can easily be established.

sec x

R – (± π/2, ± 3π/2, .....

R – (–1, 1)

cosec x

R – {0, ±π, ± 2 π, ......}

R –(–1, 1)

Composite of functions is not commutative i.e.,

sinh x

R

R

cosh x

R

[1, ∞)

tanh x

R

(–1, 1)

coth x

R0

R –[1, –1]

sech x

R

(0, 1]

cosech x

R0

R0

Let f : A → B and g : B → C be two functions, then the composite of the functions f and g denoted by gof, is a function from A to C given by gof : A → C, (gof) (x) = g[f(x)]. Properties of Composite Function :

fog ≠ gof Composite of functions is associative i.e. (fog)oh = fo(goh) Composite of two bijections is also a bijection. Inverse Function :

–1

sin x

If f : A → B is one-one onto, then the inverse of f i.e., f–1 is a function from B to A under which every b ∈ B is associated to that a ∈ A for which f(a) = b. Thus

–1

cos x

XtraEdge for IIT-JEE

[0, π]

R

(–π/2, π/2}

R

(0, π)

–1

sec x –1

cosec x

55

[–1, 1]

–1

cot x

f–1(b) = a ⇔ f(a) = b.

[–π/2, π/2]

–1

tan x

f–1 : B → A,

[–1, 1]

R –(–1, 1)

[0, π] – {π/2}

R – (–1, 1)

(– π/2, π/2] – {0}

DECEMBER 2009

Based on New Pattern

IIT-JEE 2010 XtraEdge Test Series # 8

Time : 3 Hours Syllabus :

Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus

Instructions :

Section - I • Question 1 to 4 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 5 to 9 are multiple choice questions with multiple correct answer. +4 marks and -1 mark for wrong answer. Section - II • Question 10 to 11 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer and No Negative marks for wrong answer. However, +2 marks will be given for a correctly

marked answer in any row.

Section - III • Question 12 to 19 are numerical response type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer.

PHYSICS

4.

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

A particle starts from rest and travels a distance x with uniform acceleration, then moves uniformly a distance 2x and finally comes at rest after moving further 5x distance with uniform retardation. The ratio of maximum speed to average speed is -

1.

(A)

5 2

(B)

5 3

(C)

7 4

(D)

7 5

A deuterium plasma is a neutral mixture of negatively charged electrons and positively charged deuterium charged electrons and positively charged deuterium nuclei. Temperature required to produced fusion in deuterium plasma is : (Take the range of nuclear forces 2 fermi) (A) 835 K (C) 8.35 × 107 K

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

7

(B) 835 × 10 K (D) 273 K

5. 2.

3.

As observed in the laboratory system, a 6 MeV proton is incident on a stationary 12C target velocity of center of mass of the system is : (Take mass of proton to be 1 amu) (A) 2.6 × 106 m/s (B) 6.2 × 106 m/s (C) 10 × 106 m/s (D) 10 m/s

(A) v =

Find the de Broglie wavelength of Earth. Mass of Earth is 6 × 1024 kg. Mean orbital radius of Earth around Sun is 150 × 106 km (A) 3.7 m (C) 3.7 × 1063 m

XtraEdge for IIT-JEE

A body moves in a circular path of radius R with deceleration so that at any moment of time its tangential and normal accelerations are equal in magnitude. At the initial moment t = 0, the velocity of body is v0 then the velocity of body will bev0 at time t v0 t 1+ R

(B) v = v 0 e −S / R after it has moved S meter (C) v = v0e–SR after it has moved S meter (D) None of these

–63

(B) 3.7 × 10 m (D) 3.7 × 10–63 cm 56

DECEMBER 2009

6.

Three identical rods of same material are joined to form a triangular shape ABC as shown. Angles at edge A and C are respectively θ1 and θ2 as shown. When this triangular shape is heated then A

9.

(B) the indices of refraction of the two media are same (C) the boundary is not visible (D) angle of incidence is lesser than angle of

θ1

µ refraction but greater then sin–1 R µD

θ2 B

7.

In passing through a boundary refraction will not take place if (A) light is incident normally on the boundary

C

(A) θ1 decreases and θ2 increases (B) θ1 increases and θ2 decreases (C) θ1 increases (D) θ2 increases Direction of current in coil (2) is opposite to direction of current in coil (1) and coil (3). All three coils are coaxial and equidistant coil (1) and coil (3) are fixed while coil (2) is suspended thus able to move freely. Then -

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T

R

R

1

R

A P B P C P D P

3

2

(A) coil (2) is in equilibrium (B) equilibrium state of coil (2) is stable equilibrium along axial direction (C) equilibrium state of coil (2) is unstable equilibrium along axial direction (D) if direction of current in coil (2) is same as that of coil (1) and coil (3) then state of equilibrium of coil (2) along axial direction is unstable 8.

10.

Which of the following statements are correct about the circuits shown in the figure where 1 Ω and 0.5 Ω are internal resistances of the 6 V and 12 V batteries respectively – 11.

P

4Ω R

(A) The potential at point P is 6 V (B) The potential at point Q is – 0.5 V (C) If a voltmeter is connected across the 6 V battery, it will read 7 V (D) If a voltmeter is connected across the 6 V battery, it will read 5 V

XtraEdge for IIT-JEE

(A) Interference

(P) Non-mechanical waves

(B) Diffraction

(Q) Electromagnetic waves (R) Visible light

(D) Reflection

(S) Sound waves (T) None

Column-I

Column-II

(A) 5000 Å (B) 1 Å

(P) De-Broglie wavelength of electron in x-ray tube (Q) Photoelectric threshold

(C) 0.1 Å

wavelength (R) x-ray wavelength

0.5Ω S

Q R S T Q R S T

Match column I with column II in the light of possibility of occurrence of phenomena listed in column I using the systems in column II Column-I Column-II

(C) Polarisation

6V, 1Ω 12V, 0.5Ω Q

Q R S T Q R S T

(D) 10 Å

(S) De-Broglie wavelength of most energetic photoelectron emitted from metal surface (T) None

57

DECEMBER 2009

This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

between the particles in process of their motion in meters is (g = 10 m/s2).

u = 2 m/s A 45°

22m

X Y Z W 0 1 2 3 4 5 6 7 8 9

12.

13.

14.

15.

0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9

17.

2mH S

18.

The minimum speed in m/s with which a projectile must be thrown from origin at ground so that it is able to pass through a point P (30 m, 40 m) is : (g = 10 m/s 2) (Ans. in .............. × 10)

9Ω

9Ω 9Ω

2mH

A rectangular plate of mass 20 kg is suspended from points A and B as shown. If the pin B is suddenly removed then the angular acceleration in rad/sec2 of the plate is : (g = 10 m/s2). B

A

In U238 ore containing Uranium the ratio of U234 to Pb206 nuclei is 3. Assuming that all the lead present in the ore is final stable product of U238. Half life of U238 to be 4.5 × 10 9 years and find the age of ore. (in 109 years)

b =0.15m

l =0.2m

Under standard conditions the gas density is 1.3 mg/cm3 and the velocity of sound propagation in it is 330 m/s, then the number of degrees of freedom of gas is.

19.

If the temperature of a gas is raised by 1 K from 27ºC. Find the percentage change in speed of sound. (Speed = 300 ms–1) (Ans. in .............. × 102)

A single conservative force acts on a body of mass 1 kg that moves along the x-axis. The potential energy U(x) is given by U (x) = 20 + (x – 2)2, where x is in meters. At x = 5.0 m the particle has a kinetic energy of 20 J, then the maximum kinetic energy of body in J is.

CHEMISTRY Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Two particles are simultaneously thrown from top of two towers with making angle 45º with horizontal. Their velocities are 2 m/s and 14 m/s. Horizontal and vertical separation between these particles are 22 m and 9 m respectively. Then the minimum separation

XtraEdge for IIT-JEE

Consider the circuit shown in figure. What is the current through the battery just after the switch is closed. 18V

(Ans. in .............. × 10) 16.

v = 14 m/s 45° B

9m

1.

58

In the precipitation of sulphides of second group basic radicals. H2S is passed into acidified solution with dilute HCl. If the solution is not acidified, then which of the following is correct ?

DECEMBER 2009

(A) Only the sulphides of second precipitated

group

get

(B) Only the sulphides precipitated

group

get

of

fourth

5.

(C) Neither of the sulphides of second and fourth groups get pricipitated

(C) they are nuclear spin isomer, where the spins of protons are same in para and different in ortho isomer (D) they are nuclear spin isomers, where the spins of protons are opposite in para but same in ortho isomer

(D) Sulphides of both the groups second and fourth get precipitated 2.

The difference between ortho and para hydrogen is/are (A) they are electron spin isomer, where the spins of electrons are opposite (B) ortho hydrogen is more stable at lower temperature

The geometrical shapes of XeF5+, XeF6 and XeF82– respectively are (A) trigonal bipyramidal, octahedral and square planar

6.

(B) square based pyramidal, distorted octahedral and octahedral

NGP assistance support for S N2 reaction will be seen in OH

(C) planar pentagonal, octahedral and square anti prismatic

(A)

(B) CH3–S–CH2–CH2–Cl

(D) square based pyramidal, distorted octahedral and square anti prismatic

Cl

OH

OH

3.

Regarding graphite the following informations are available :

(C)

(D)

Cl

Θ

COO

Top view

7.

When the compound called isoborneol is heated with 50% sulfuric acid the product of the reaction is/are ?

3.35Å

HO Isoborneol

The density of graphite = 2.25 gm/cm3. What is C–C bond distance in graphite ?

4.

(A) 1.68Å

(B) 1.545Å

(C) 2.852 Å

(D) 1.426Å

The molecular formula of a non-stoichiometric tin oxide containing Sn(II) and Sn (IV) ions is Sn4.44 O8. Therefore, the molar ratio of Sn(II) to Sn(IV) is approximately (A) 1 : 8

(B) 1 : 6

(C) 1 : 4

(D) 1 : 1

8.

(B)

(C)

(D)

Regarding the radial probability distribution (4nr2R 2nl) vs r plot which of the following is/are correct ? (A) The number of maxima is (n-l) (B) The number of nodal points is (n-l-1) (C) The radius at which the radial probability density reaches to maxima is 3s < 3p < 3d

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

XtraEdge for IIT-JEE

(A)

(D) The number of angular nodes is l

59

DECEMBER 2009

9.

H2C 2O4 and NaHC2O4 behave as acids as well as reducing agents which is/are correct statement (s) ? (A) Equivalent wt. of H2C 2O4 and NaHC 2O4 are equal to their molecular weights when behaving as reducing agent (B) 100 mL of 1N solution of each is neutralised by equal volumes of 1M Ca(OH)2 (C) 100 mL of 1N solution of each is neutralised by equal volumes of 1N Ca(OH)2 (D) 100 mL of 1M solution of each is oxidised by equal volumes of 1M KMnO4

Cl

Cl

Br

Me Br

Me Me

Me

Me Br

Me

Cl

Column- II (P) Optically active (Q) Cis compound (R) Trans compound (S) Optically inactive (T) Chiral axis (element of chirality) This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

at high temperature (T) PV/RT = 1–a/VRT 11. Column- I Cl

12.

Me

XtraEdge for IIT-JEE

Me Me

Me Me

(C) At STP (for real gas) (R) PV = RT + Pb (D) At low pressure and (S) PV = RT

Me Me

Me Me

(D) Me

gas molecules be negligible

(A) Me

Me Br

Cl

Me

Cl

a (P) P + 2 (V–b)= RT V

Me Me

Me Me

(C) Me

among the gas molecules be negligible (B) If the volume of the (Q) PV = RT –a/V

Me Me

Me Me Me

Column- II

(A) If force of attraction

Me Me

(B)

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T 10. Column- I

Me Me

Cl

What is the number of lone pair on the molecule XeO2F 2 ?

Me Me

60

DECEMBER 2009

13.

15g of a solute in 100g water makes a solution freeze at –1ºC. What will be the depression in freezing point if 30g of a solute is dissolved in 100g of water ?

14.

The half lives of decomposition of gaseous CH3CHO at constant temperature but at initial pressure of 364mm and 170mm Hg were 410 second and 880 second respectively. Hence what is the order of reaction.

4.

x x and g(x) = where 0 < x ≤ 1, sin x tan x then in this interval (A) both f(x) and g(x) are increasing functions If f(x) =

(B) both f(x) and g(x) are decreasing functions (C) f(x) is an increasing function (D) g(x) is an increasing function

15.

The number of S–S bonds in sulphur trioxide trimer (S3O9) is ....... .

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

16.

The number of peroxo linkages present in the [H4B2O8] 2– is ..... .

5.

17.

The maximum number of structural isomers (acyclic and cyclic) possible for C4H8 are ....... .

(A) a = b = c

(B) a, b, –

18.

What is the no. of lone pair of electrons present on N in Trisilylamine ?

(C) a, b, c are in H.P.

(D) –

19.

A small amount of solution containing 24Na with activity 2 × 103 dps was administered into the blood of patient in a hospital. After 5 hour a sample of the blood drawn out from the patient shared an activity of 16 dps per cm3. (t1/2 of 24Na = 15 hrs.) Find the volume (in L) of blood in patient.

6.

(C) infinite number of solutions if a + b + c = 0 (D) none of these

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. The largest term in the expansion of (3 + 2x)50, where x = 1/5, is (A) 5th (B) 6th

3.

If

1 sin–1 2

8.

(C) (1/3) tan θ

(D) 3 cot θ

The integer n for which lim

x

n

XtraEdge for IIT-JEE

(B) [0, ∞)

(C) (–∞, 0)

(D) (– ∞,∞)

x 4 cos 2 x − x sin x + cos x dx If l = e x sin x +cos x x 2 cos 2 x

∫

sec x x − (C) ex sin x + cos x +C tan x x

is

a finite nonzero number is (A) 1 (C) 3

(A) (0, ∞)

cos x (B) ex sin x + cos x x sin x − x

(cos x − 1)(cos x − e x )

x →0

x is differentiable on 1+ | x |

sec x (A) ex sin x + cos x x − +C x

3 sin 2θ –1 5 + 4 cos 2θ = tan x, then x = (B) 3 tan θ

The function f(x) =

then l equals -

(D) 9th

(A) tan 3 θ

The system of equations –2x + y + z = a

(B) unique solution if a + b + c = 0

7.

2.

1 a, b, c are in G.P. 2

(A) no solution if a + b + c ≠ 0

MATHEMATICS

(C) 8th

1 c are in G.P. 2

x – 2y + z = b x = y – 2z = c has

[Given : log 1.2598 = 0.1003]

1.

If a, b, c are in A.P., and a 2, b2, c 2 are in H.P., then

cos x − x sin x (D) xe x sin x+cos x – e x sin x +cos x 1 − dx x 2 cos 2 x

∫

(B) 2 (D) 4 61

DECEMBER 2009

9.

If for the differential equation y′ = general solution is y =

This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

x y + φ the x y

x then f (x / y) is log | Cx |

given by (A) – x2 / y2 (C) x2 / y2

(B) y2 / x2 (D) – y2 / x2

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T

12.

Suppose X follows a binomial distribution with parameters n = 6 and p. If 9P(X = 4) = P(X = 2), find 4p.

13.

If Q is the foot of the perpendicular from the point x −5 y + 2 z− 6 P(4, –5,3) on the line = = then 3 −4 5 100 (PQ)2 is equal to 457

14.

If a = (0, 1, –1) and c = (1, 1, 1) are given vectors, then |b|2 where b satisfies a × b + c = 0 and a . b = 3 is equal to

15.

ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to BC. If the triangle ABC has perimeter P and area ∆ ∆ then lim 512r 3 is equal to h →0 p

16.

If f(x) = sin x, x ≠ n π, =0 and g(x) = x2 + 1, =4 =5 then lim g(f(x)) is .....

10. The domain of the functions Column- I Column- II –1 (A) sin (x/2 – 1) (P) (3 – 2π, 3 – π) ∪ (3,4] + log (x – [x]) (B) ex + 5sin π 2 / 16 − x 2 (Q) (0, 4) – {1, 2, 3} (R) [– π/4, π/4]

(C) log10sin (x – 3) + 16 − x 2 (D) cos–1

1 − 2x 4

(S) [– 3/2, 5/2] (T) None

11. Column- I I denotes an integral (A)

∫

π 0 ∞

(B)

∫

0

(C)

∫

0

(D)

∫

x log sin x dx log (x+x–1)

π/4

π 0

Column- II

dx 1+ x 2

(P) I = ( π/8) log 2 (Q) I =

− π2 log 2 2

x →0

log (1+ tan x)dx (R) I = π log 2

17.

(S) I = π log 2

18.

log (1 – cos x)dx

If y = (1 + 1/x) x then

2 y 2 ( 2) + 1 / 8 (log 3 / 2 − 1 / 3)

is equal to

If the greatest value of y = x/log x on [e, e3] is u then e3/u is equal to 19. If z ≠ 0 and 2 + cos θ + i sin θ = 3/z, then find the value of 2(z + z ) – |z|2.

(T) None

XtraEdge for IIT-JEE

n = 0, ± 1, ± 2, ..... otherwise x ≠ 0, 2 x=0 x=2

62

DECEMBER 2009

Based on New Pattern

IIT-JEE 2011 XtraEdge Test Series # 8

Time : 3 Hours Syllabus :

Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus

Instructions :

Section - I • Question 1 to 4 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 5 to 9 are multiple choice questions with multiple correct answer. +4 marks and -1 mark for wrong answer. Section - II • Question 10 to 11 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer and No Negative marks for wrong answer. However, +2 marks will be given for a correctly

marked answer in any row.

Section - III • Question 12 to 19 are numerical response type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer.

PHYSICS Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

Velocity of block 2 shown in figure is u

(A) (C)

60º

4. 2

u

(A) 2.

3.

3u 2

(B)

3u 2

1

(C)

3 3u (D) 2u 2

(B)

MR 2 ω 2∆t

(D) Zero

2 ∆t

Select the incorrect statement (A) The velocity of the centre of mass of an isolated system must stay constant (B) Only a net external force can change the velocity of the center of mass of a system (C) A system have non-zero kinetic energy but zero linear momentum →

(D) Fext

Which of the following restoring force can give rise to S.H.M (A) F = 2x (B) F = 2 – 4x (C) F = – 2x2 (D) None of these

→

d v → dm =m +v is true for all situation dt dt

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 5.

A disk of mass M and radius R is rotating about its axis with angular velocity ω . Axis of disk is rotated by 90º in time ∆t. Average torque acting on disk is -

XtraEdge for IIT-JEE

2 Mω2 ∆t MR 2 ω

63

Velocity of a particle moving on straight line varies 1 as th power of displacement. Then 2 (A) K.E. ∝ S (B) P ∝ S1/2 (C) a = constant (D) S ∝ t2

DECEMBER 2009

6.

7.

A block of density ρ is floating in a liquid X kept in container. A liquid Y of density ρ′ (< ρ) is slowly poured into container – (A) The block will move up if liquid X and Y are immiscible (B) The block will sink more if liquid X and Y are immiscible (C) The block will sink more if liquid X and Y are miscible (D) The block will not move if liquid X and Y are miscible

P Q R S T A P B P C P D P

10.

A

B

(A) A

Q

Q R S T Q R S T

There are four identical rod having thermal resistance 10 Ω, each column I contains various arrangement of rod. Column II contains current flowing across point C when a temperature difference of 100ºC is maintained across A & B. Match them. Column-I Column-II

Six identical rod are connected as shown in figure and temperature difference of 100ºC is maintained across P and Q –

P

Q R S T Q R S T

C

B

(P) 2 J/sec

C

(A) (B) (C) (D) 8.

9.

C Temperature of point 'A' is 50ºC 200 Temperature of point A is ºC 3 Thermal current passing through B is zero Thermal current passing through A is twice of that through C

(B) A

(C) A

C

B

(Q) 4 J/sec

(R) 6 J/sec

A

A solid iron cylinder A rolls down a ramp and an identical iron cylinder B slides down the same ramp without friction – (A) B reaches the bottom first (B) A and B have the same kinetic energy (C) B has greater translational kinetic energy than that of A (D) Linear speed of centre of mass of B is greater than that of A

(D)

C

B

(S)

20 J/sec 7

(T) None 11.

A car of mass 500 kg is moving in a circular road of radius 35 / 3 . Angle of banking of road is 30º. Coefficient of friction between road and tyres 1 is µ = . Match the following: 2 3 Column-I Column-II

A mass and spring system oscillates with amplitude A and angular frequency ω – (A) The average speed during one complete cycle 2Aω of oscillation is π (B) Maximum speed is ω A (C) Average velocity of particle during one complete cycle of oscillation is zero (D) Average acceleration of particle during one complete cycle of oscillation is zero

(A) Maximum speed (in m/s) of car for safe turning (B) Minimum speed (in m/s) of car for safe turning

(P) 5 2 (Q) 12.50

(C) Speed (in m/s) at which friction (R) force between tyres and road is zero

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

XtraEdge for IIT-JEE

B

(D) Friction force (in 102 Newton)

(S)

210

350 3

between tyres and road if speed is

350 m/s 6 (T) None

64

DECEMBER 2009

This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 12.

14.

A longitudinal wave of frequency 220 Hz travels down a copper rod of radius 8.00 mm. The average power in the wave is 6.50 µW. The amplitude of the wave is n × 10–8 m. Find n.

15.

A piston-cylinder device with air at an initial temperature of 30ºC undergoes an expansion process for which pressure and volume are related as given below P (kPa) 100 25 6.25 3 V (m ) 0.1 0.2 0.4 The work done by the system is n × 103 J. Find n.

16.

The block connected with spring is pushed to compress the spring by 10 cm and then released. All surfaces are frictionless and collision are elastic. Time period of the motion in sec (mass of block = 9 kg and spring constant 4 π2 N/m).

Pulley 'P' shown in figure is pulled upward with F = 2t N, where t is time in sec. Velocity of block of mass 1 at the time block 2 is about to lift is (in cm/sec). (Ans. in ................. × 10 1) F = 2t

5 cm

17.

300 × 381 m/s.

RMS velocity of gas at 27ºC is

RMS velocity (in m/s) when temperature is increased four times is. (Ans. in ................. × 10 2)

2

1

18.

A block of mass 2 kg is placed on a wedge of mass 10 kg kept on a horizontal surface. Coefficient of friction between all surfaces is µ = 0.2. If block is slipping down the wedge with constant speed then friction force on wedge due to horizontal surface is (in Newton) :

19.

A particular quantity 'y' varies as 'x' as shown in figure. RMS value of y with respect to x for large values of 'x' is.

m 1 = 0.5 kg m 2 = 1 kg

13.

The upper edge of a gate in a dam runs along the water surface. The gate is 2.00 m high and 4.00 m wide and is hinged along a horizontal line through its center. The torque about the hinge arising from the force due to the water is (n × 104 Nm). Find value of n.

y

2m

60º

60º 1

XtraEdge for IIT-JEE

65

2

3

4 x

DECEMBER 2009

(C) The first ionization energies of elements in a period do not increase with the increase in atomic numbers (D) For transition elements the d-subshells are filled with electrons monotonically with the increase in atomic number

CHEMISTRY Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

An ideal gaseous mixture of ethane (C2H6) and ethene (C 2H4) occupies 28 litre at STP. The mixture reacts completely with 128 gm O2 to produce CO2 and H2O. Mole fraction of C2H4 in the mixture is (A) 0.6 (B) 0.4 (C) 0.5 (D) 0.8

2.

A bulb of constant volume is attached to a manometer tube open at other end as shown in figure. The manometer is filled with a liquid of density (1/3rd) that of mercury. Initially h was 228 cm.

Gas

h

When 300 mL of 0.2 M HCl is added to 200 mL of 0.1 M NaOH. Resultant solution require how many equivalent of Ba(OH)2 ? (A) 0.06 (B) 0.12 (C) 0.3 (D) 0.04

4.

The dipole moment of HCl is 1.03D, if H–Cl bond distance is 1.26Å, what is the percentage of ionic character in the H–Cl bond ? (A) 60% (B) 29% (C) 17% (D) 39%

6.

Consider the following carbides CaC2, BeC2, MgC2 and SrC 2 which of the given carbides on hydrolysis yield same product (A) CaC2 (B) Be2C (C) MgC2 (D) SrC2

Which of the following is/are state function ? (A) q (B) q – w (C) q + w (D) q / w

9.

The IUPAC name of the following compound is OH

10. Column- I (A) 5.4 g of Al (B) 1.2 g of Mg2+ (C) Exact atomic weight of mixture of oxygen isotopes (D) 0.9 mL of H2O

Which of the following is/are correct regarding the periodic classification of elements ? (A) The properties of elements are the periodic function of their atomic number (B) Non metals are lesser in number than metals

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8.

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 5.

Which of the following statements regarding hydrogen peroxide is/are correct ? (A) Hydrogen peroxide is a pale blue viscous liquid (B) Hydrogen peroxide can act as oxidising as well as reducing agent (C) The two hydroxyl groups in hydrogen peroxide lie in the same plane (D) In the crystalline phase, H2O2 is paramagnetic

Br CN (A) 3-Bromo-3-cyano phenol (B) 3-Bromo-5-hydroxy benzonitrile (C) 3-Cyano-3-hydroxybromo benzene (D) 5-Bromo-3-hydroxy benzonitrile

Through a small hole in the bulb gas leaked assuming dp pressure decreases as = – kP. dt If value of h is 114 cm after 14 minutes. What is the value of k (in hour–1) ? [Use : ln(4/3) = 0.28 and density of Hg = 13.6 g/mL] (A) 0.6 (B) 1.2 (C) 2.4 (D) None of these 3.

7.

66

Column- II (P) 0.5 NA electrons (Q) 15.9994 amu (R) 0.2 mole atoms

(S) 0.05 moles (T) 3.1 × 1023 electrons

DECEMBER 2009

11. Column- I (Ionic species) (A) XeF5+ (B) SiF5– (C) AsF4+ (D) ICl4–

Column- II (Shapes) (P) Tetrahedral (Q) Square planar (R) Trigonal bipyramidal (S) Square pyramidal (T) Octahedral

13.

14.

The equivalent weight of a metal is 4.5 and the molecular weight of its chloride is 80. The atomic weight of the metal is.

19.

No. of π bond in the compound H2CSF 4 is.

MATHEMATICS

This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 12.

18.

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

If 0 < r < s ≤ n and nPr = nPs , then value of r + s is (A) 2n – 2 (B) 2n – 1 (C) 2 (D) 1

2.

If sin x + sin2 x + sin3 x = 1, then cos6 x – 4 cos4 x + 8 cos2 x is equal to (A) 0 (B) 2 (C) 4 (D) 8

3.

If x is real, and k=

then x2 + x +1 (A) 1/3 ≤ k ≤ 3 (C) k ≤ 0 4.

How much volume (in mL) 0.001 M HCl should we add to 10 cm3 of 0.001 M NaOH to change its pH by one unit ?

5.

An acid type indicator, HIn differs in colour from its conjugate base (In– ). The human eye is sensitive to colour differences only when the ratio [In– ] / [HIn] is greater than 10 or smaller than 0.1. What should be the minimum change in the pH of the solution to osberve a complete colour change ? (Ka = 1.5 × 10–5) What is the sum of total electron pairs (b.p. + l.p.) present in XeF6 molecule ?

16.

The number of geometrical isomers of CH3CH=CH–CH=CH–CH=CHCl is.

17.

At 200ºC, the velocity of hydrogen molecule is 2.0 × 105 cm/sec. In this case the de-Broglie wavelength (in Å) is about.

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(B) k ≥ 5 (D) none of these

A flagstaff stands in the centre of a rectangular field whose diagonal is 1200 m, and subtends angles 15º and 45º at the mid points of the sides of the field. The height of the flagstaff is (A) 200 m

(B) 300 2 + 3 m

(C) 300 2 − 3 m

(D) 400 m

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

The stopcock, connecting the two bulbs of volumes 5 litres and 10 litres containing an ideal gas at 9 atm and 6 atm respectively, is opened. What is the final presure (in atm) in the two bulbs if the temperature remained the same ?

15.

x2 − x +1

If a, b, c are the sides of the ∆ABC and a2, b 2, c2 are the roots of x3 – px2 + qx – k = 0, then cos A cos B cos C p (A) + + = a b c 2 k (B) a cos A + b cos B + c cos C = (C) a sin A + b sin B + c sin C =

4q − p 2

2 k 2p∆ k

3

(D) sin A sin B sin C = 6.

67

8∆ k

The coordinates of the feet of the perpendiculars from the vertices of a triangle on the opposite sides are (20, 25), (8, 16) and (8, 9). The coordinates of a vertex of the triangle are (A) (5, 10) (B) (50, –5) (C) (15, 30) (D) (10, 15)

DECEMBER 2009

7.

(D) 3 x–1 + 3 x–2 + 3 x–3 + ... (S) 7 1 1 = 2 5 2 + 5 + 1 + + 2 + ... 5 5 (T) None

1 1 1 2 Let E = + + + + ... upto 50 terms, then 3 50 3 50 (A) E is divisible by exactly 2 primes (B) E is prime (C) E ≥ 30 (D) E ≤ 35

8.

If m is a positive integer, then [( 3 + 1) 2m ] + 1, where [x] denotes greatest integer ≤ n, is divisible by(A) 2m (B) 2m+1 (C) 2m+2 (D) 22m

9.

If A and B are acute angles such that sin A = sin2 B, 2 cos2 A = 3 cos2 B; then (A) A = π/6 (B) A = π/2 (C) B = π/4 (D) B = π/3

This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T

12.

Fifteen persons, among whom are A and B, sit down at random at a round table. if p is The probability that there are exactly 4 persons between A and B find 14 p.

13.

If l is the length of the intercept made by a common tangent to the circle x2 + y2 = 16 and the ellipse x2/25 + y2/4 = 1, on the coordinate axes, then

10. For the circle x2 + y2 + 4x + 6y – 19 = 0 Column- I Column- II (A) Length of the tangent (P)

81l 2 + 3 is equal to 1059

72 226 113

14.

from (6, 4) to the circle (B) Length of the chord (Q) of contact from (6, 4) to the circle

113

parabola on this normal; then

(C) Distance of (6, 4) (R) 113 – 32 from the centre of the circle (D) Shortest distance of (S) 9 (6, 4) from the circle (T) None 11. Value of x when Column- I Column- II (A) 52 54 5 6 ... 52x = (0.04)–28 (P) 3 log3 5 2

(B) x = (0.2) (C)

log

1 + 1 + 1 +... 4 8 16

5

1 1 1 log 2.5 + 2 + 3 +... 3 3 3 x = (0.16)

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If x + y = k is a normal to the parabola y2 = 12x, p is the length of the perpendicular from the focus of the

15.

The volume of the tetrahedron whose vertices are (0, 1, 2) (3, 0, 1) (4, 3, 6) (2, 3, 2) is equal to

16.

Let a1 =

17.

(Q) 4 (R) 2 68

3k 3 + 2p 2 is equal to 741

1 , ak+1 = ak2 + ak ∀ k ≥ 1 and 2 1 1 1 + + ... + xn = a 1 +1 a 2 + 1 a n +1 Find [x100] where [x] denotes the greatest integer ≤ x. Find the value of x which satisfy the equation log2 (x2 – 3) – log2 (6x – 10) + 1 = 0

18.

Find the coefficient of x2009 in the expansion of (1 – x)2008 (1 + x + x2)2007

19.

Find the value of x satisfying 4

1 / log2 x

= 2.

DECEMBER 2009

MOCK TEST PAPER-1 CBSE BOARD PATTERN

CLASS # XII SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS Sol ut i ons w il l be publ i s he d i n ne xt is s ue General Instructions : Physics & Che mistry • Time given for each subject paper is 3 hrs and Max. marks 70 for each. • All questions are compulsory. • Marks for each question are indicated against it. • Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each. • Question numbers 9 to 18 are short-answer questions, and carry 2 marks each. • Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each. • Question numbers 28 to 30 are long-answer questions and carry 5 marks each. • Use of calculators is not permitted. General Instructions : Mathematics • Time given to solve this subject paper is 3 hrs and Max. marks 100. • All questions are compulsory. • The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each. Section B comprises of 12 questions of four marks each. Section C comprises of 7 questions of six marks each. • All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. • There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2 question of six marks each. You have to attempt only one of the alternatives in all such questions. • Use of calculators is not permitted.

PHYSICS 1.

q =

q0

7.

If nuclear density d ∝ An, where A is the atomic number then write the value of n.

8.

Why standard resistors are made of alloys.

9.

Name the quantities whose SI units are given below : 1. V – m 2. C-m out of the two also name the vector quantity.

is valid or not, q is the charge on

v2 c2 particle when it is moving with velocity v, q0 is the rest charge and c is the velocity of light . 1−

2.

A concave mirror is dipped inside the liquid of absolute refractive index 1.25. What will be the percentage change in its focal length.

3.

Write the name of a compound semiconductor.

4.

If one of the slit get closed in Young's Double slit experiment then fringe pattern will be observed or not on the screen.

5.

Write one of the use of Zener Diode.

6.

Name the experiment which proves the Dual Nature of electron.

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10. A transistor is shown in figure.

+2V

.

+1V +3V (i) Name the type of transistor (ii) Is the transistor is properly biased.

11. A time variant current is given i(t) = 1 + 3 2 sin (314 t + 30º) Find its root mean square value. 69

DECEMBER 2009

12. A metallic conductor of non-uniform cross-sectional area is shown in figure.

18. (i) Write the order of colors in secondary rainbow. (ii) Why sun appears reddish at the time of sunrise and sunset.

i p1

p2

19. The charges on the capacitors are Qa, Qb and Qc then calculate 1µF

p3 b

a

Qa

(i) Out of p 1, p 2, and p3 at which point the drift speed of the electron is maximum. (ii) Out of p 1, p2 and p3 at which point the current density is minimum.

12µF Qc

13. Stopping potential versus freq. of the incident light graph is shown in figure for metal-1 and metal-2 metal-1 metal-2 V0 θ1

θ2

30V

3µF

key-K

(i) Ratio of energy stored in 1 µF and 3 µF capacitor. (ii) Potential difference across 12 µF capacitor (iii) Energy supplied by the battery.

υ

20. (i) A uniform magnetic field of 0.5 T exist in the given solenoid. If an electron is projected along the axis of solenoid from a towards b with the speed of 3 × 10 2 m/s then find the Lorentz force working on electron.

(i) Which metal will have the higher value of threshold wavelength. (ii) Is θ1 = θ2 if yes then why ? 14. Draw the circuit diagram for finding the internal resistance of the cell using potentiometer.

a

15. (i) Define angle of Dip. (ii) What is the value of angle of Dip at a place on earth where Horizontal component and vertical component of earth magnetic field are equal.

i

b

(ii) When the current flows through the metallic spring why it get shrinked. 21. (i) For the given circuit diagram find the position of the null point. 5Ω 10 Ω

16. (i) State Kirchhoff's current law. (ii) Find the potential difference across the 2 ohm resistance for the given circuit diagram. 2Ω 4A 1Ω

A

3Ω

B G

1A

17. (i) State Lenz's law. (ii) If the current i increases then what will be the direction of induced current in the circular coil for the given figure.

i

Qb

Vbb

100 cm

Key-k Rh Rheostat

(ii) For the given figure draw truth table. A

b

B

Y

a a >> b

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70

DECEMBER 2009

22. (i) State Brewster's law for polarization . (ii) If the Brewster's angle for the given pair of medium is ip then express the critical angle for the same pair of media in terms of ip. (iii) If the lens is made of medium-2 and placed in medium-1. If the absolute refractive Index of medium-2 is µ2 and of medium-1 is µ1 then µ2 > µ1. Is it correct or not.

(iii) Where to put the proton that it will have maximum force on it. 27. For the given configuration find the –2q 2a

2a

Incident rays Refracted rays

+q

Medium -2

+q

2a (i) Dipole moment of the system. (ii) Electric potential energy of the system.

Medium-1 Medium-1

23. What is the need of Modulation in communication system. Draw the shapes of signal, carrier wave and amplitude Modulated wave.

28. Draw the circuit diagram for common emitter transistor amplifier. Explain its working. What is the phase difference between input and output voltage in case of common emitter transistor amplifier.

24. How a Galvanometer can be converted into Ammeter explain it by drawing the circuit diagram.

29. Explain construction and working of cyclotron. Why cyclotron can not be used to accelerate the electrons.

ig G Prove that S = i − ig G = Resistance of Galvanometer coil S = value of shunt ig = Full scale deflection current for Galvanometer i = Range of Ammeter.

30. State Ampere's circuital law. Using Ampere's circuital law find the magnetic field at the axis of the long solenoid.

CHEMISTRY

25. (i) Find the equivalent resistance between A and B for given fig.

1.

Give the IUPAC name of the organic compound (CH3)2 C = CH – C – CH3 . || O

2.

Complete the following reaction : CH3 – CH2 – CH = CH2 + HCl → …..

3.

State a use for the enzyme streptokinase in medicine.

4.

What is copolymerization ?

5.

Which type of a metal can be used in cathodic protection of iron against rusting ?

6.

Why is the bond dissociation energy of fluorine molecule less than that of chlorine molecule ?

7.

What is meant by inversion of sugar ?

8.

What are the types of lattice imperfections found in crystals ?

R R

A

R

R R

B

R

R

(ii) α is the symbol for the temperature coefficient of resistivity for the given material. If α → 0 then the material will be copper or constantan ? 26. (i) What is the angle between electric field line and equi-potential surface. (ii) If Ea, Eb and Ec are the electric field intensities at points a, b and c respectively, where to put a proton that it will have the maximum electric potential energy. c b a

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71

DECEMBER 2009

9.

Describe the mechanism of the formation of diethyl ether from ethanol in the presence of concentrated sulphuric acid.

21. Identify the substances A and B in each of the following sequences of reactions : alc.KOH

Br

2 (i) C2H5 Br → A → B

10. Predict, giving reasons, the order of basicity of the following compounds in (i) gaseous phase and (ii) in aqueous solutions (CH3)3 N, (CH3)2 NH, CH3NH2, NH3. 11. Write names of monomer/s of the following polymers and classify them as addition or condensation polymers. (i) Teflon (ii) Bakelite (iii) Natural Rubber

NaNO + HCl

Cu ( CN )

(ii)

2 2 → A2 → B NH 2 0 ºC

(iii)

Heat NH 2 → A → B

H2SO 4

22. Give the electronic configuration of the (a) d-orbitals of Ti in [Ti (H2O)6]3– ion in an octahedral crystal field. (b) Why is this complex coloured ? Explain on the basis of distribution of electrons in the d-orbitals.

12. Give an example for each of the following reactions : (i) Kolbe’s reaction. (ii) Reimer-Tiemann reaction.

23. State reasons for the following : (a) Rusting of iron is said to be an electrochemical phenomenon. (b) For a weak electrolyte, its molar conductance in dilute solutions increases sharply as its concentration in solution is decreased.

13. Write one distinction test each for : (i) Ethyl alcohol and 2–propanol (ii) Acetaldeyde and acetone 14. Distinguish between multimolecular and macromolecular colloids. Give one example of each type.

24. Using the valence bond approach predict the shape and magnetic character of [Co (NH3) 6] 3+. (Atomic number of Co is 27).

15. Draw the structure of pyrophosphoric acid and how it is prepared.

25. Explain the following : (a) F-centre (b) Schottky & Frenkel defect

16. If E° for copper electrode is +0.34 V how will you calculate its emf value when the solution in contact with it is 0.1 M in copper ions ? How does emf for copper electrode change when concentration of Cu2+ ions in the solution is decreased ? 17. Physical and chemical adsorptions respond differently to a rise in temperature. What is this difference and why is it so ?

26. Account for the following : (i) Ferric hydroxide sol is positively charged. (ii) The extent of physical adsorption decreases with rise in temperature. (iii) A delta is formed at the point where the river enters the sea.

18. Using the valence bond approach, predict the shape and magnetic character of [Ni (CO)4]. (At No. of Ni = 28).

27. Taking two examples of heterogeneous catalytic reactions, explain how a heterogeneous catalyst helps in the reaction.

19. Describe the following giving a chemical equation for each : (i) Markownikoff’s rule (ii) Hofmann Bromide Reaction

28. (a) An organic compound ‘A’ with molecular formula C5H8O2 is reduced to n-pentane on treatment with Zn-Hg/HCl. ‘A’ forms a dioxime with hydroxylamine and gives a positive lodoform test and Tollen’s test. Identify the compound A and deduce its structure. (b) Write the chemical equations for the following conversions : (not more than 2 steps) (i) Ethyl benzene to benzene (ii) Acetaldehyde to butane – 1, 3–diol (iii) Acetone to propene

20

(a) Write chemical equations and reaction conditions for the conversion of : (i) Ethene to ethanol (ii) Phenol to phenyl ethanoate (iii) Ethanal to 2-propanol

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72

DECEMBER 2009

29. Describe how potassium dichromate is made from chromite ore and give the equations for the chemical reactions involved. Write balanced ionic equations for reacting ions to represent the action of acidified potassium dichromate solution on : (i) Potassium iodide solution (ii) Acidified ferrous sulphate solution Write two uses of potassium dichromate.

8.

→

9.

3.

4. 5.

6.

7.

3 and 2 respectively.

11. Consider f : N → N, g : N → N and h : N → R defined as f (x) = 2x, g (y) = 3y + 4 and h (z) = sin z ∀ x, y and z in N. Show the ho(gof) = (hog)of.

1− x 12. Differentiate cot–1 w.r.t. x. 1+ x 13. Solve the differential equations : (1 + e2x) dy + e x (1 + y2) dx = 0. Give that y = 1, when x = 0. or dy Solve the differential equation : x − y − 2x 3 = 0 dx π/ 4

14. Evaluate :

∫ sin 2x sin 3x dx. 0

x. tan −1 x dx.

or π/ 4

Evaluate :

Find the differential equation of the family of curves given by- x2 + y2 = 2ax.

∫ log(1 + tan x)dx 0

Find the principle value of tan–1 (–1).

15. Evaluate :

Find a matrix C such that 2A – B + C = 0 3 1 − 2 1 Where A = and B = 0 2 0 3

∫ 2x

3x + 1 2

− 2x + 3

dx .

16. If x = a (θ – sin θ) and y = a (1 – cos θ), find

If A is a square matrix of order 3 such that | adj A | = 64, find | A |.

d2y dx

2

at θ =

π . 2

17. Find the value of K so that the Kx + 1, if x ≤ π f ( x) = cos x , if x > π or

4 3 5 Find the value of x if the matrix A = 3 − 2 7 is 10 − 1 x singular.

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→

Section B

Show that the relation R in the set {1, 2, 3} is given by R = {(1, 2), (2, 1)} is symmetric.

∫

→

10. Find the direction cosines of a line which make equal angles with the co ordinate axes.

Section A

Evaluate :

→

What is the angle between vector a and b with magnitude

MATHEMATICS

2.

→

in the direction of a + b + c .

30. Give appropriate reasons for each of the following observations : (i) Sulphur vapour exhibits some paramagnetic behaviour. (ii) Silicon has no allotropic form analogous to graphite. (iii) Of the noble gases only xenon is known to form real chemical compounds. (iv) Nitrogen shows only a little tendency for catenation, whereas phosphorus shows a clear tendency for catenation.

1.

→ → → If a = ˆi + ˆj ; b = ˆj + kˆ ; c = kˆ + ˆi find a unit vector

function,

x 3 + 3, if x ≠ 0 Show that the function f ( x ) = 1 , if x = 0

73

DECEMBER 2009

24. Using integration, find the area of the circle x2 +y2 = 16, which is exterior to the parabola y2 = 6x. or Find the area of the smaller region bounded by the

x −1 x +1 π 18. Solve for x : tan −1 + tan −1 = ; x 2 − x+2 4 x < 1.

4 − 5 − 11 19. If A = 1 − 3 1 find A–1 2 3 − 7 or Using the properties of determinates prove that a +x y z x a+y z = a 2 (a + x + y + z ) x y a+z

ellipse

y2 b

2

= 1 and the line

x y + = 1. a b

1 − 1 1 26. For A = 2 1 − 3 , find A–1 and hence solve the 1 1 1 system of equations. x + 2y + z = 4 –x+ y+z=0 x – 3y + z = 2

→

21. If a and b are unit vectors and θ is the angle between them, then prove that cos

27. A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope just two consecutive letters TA are visible. What is the probability that the letter has come from(i) TataNagar (ii) Calcutta or Find the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement from a bag containing 4 white and 6 red balls. Also find mean and variance of the distribution.

θ 1→ → = a+ b 2 2

22. A football match may be either won, drawn or lost by the host country team. So there are three ways of forecasting the result of any one match, one correct and two incorrect. Find the probability of forecasting at least three correct results for four matches. or A candidate has to reach the examination centre in time. Probabilities of him going by bus or scooter or 3 1 3 by other means of transport are , , 10 10 5 respectively. The probability that he will be late is 1 1 and respectively, if he travels by bus or scooter. 4 3 But he reaches in time if he uses any other mode of transport. He reached late at the center. Find the probability that he traveled by bus.

28. Find the distance of the point (3, 4, 5) from the plane x + y + z = 2 measured parallel to the line 2x = y = z. 29. Every gram of wheat provides 0.1 gm of proteins and 0.25 gm of carbohydrates. The corresponding values for rice are 0.05 gm and 0.5 gm respectively. Wheat costs Rs. 4 per kg and rice Rs. 6 per kg. The minimum daily requirements of protein and carbohydrates for an average child are 50 gms and 200 gms respectively. In what quantities should wheat and rice be mixed in the daily diet to provide minimum daily requirements of proteins and carbohydrates at minimum cost. Frame an L.P.P. and solve it graphically.

Section C π/ 2

23. Evaluate :

a

2

+

25. Show that a right circular cylinder which is open at the top, and has a given surface area, will have the greatest volume if its height is equal to the radius of its base.

20. Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0, measured parallel to the line x +3 y −2 z = = 3 6 2 →

x2

∫ 0

cos x dx (1 + sin x )(2 + sin x )

XtraEdge for IIT-JEE

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DECEMBER 2009

XtraEdge Test Series ANSWER KEY IIT- JEE 2010 (December issue) PHYSICS Ques Ans Ques Ques Ques Ans

1 B 10 11 12 3

2 A A → P,Q,R,S A→ Q 13 2

3 B

4 5 C A, B B → Q,P,Q,R B→R 15 16 2 6

14 5

6 7 C, D A, B, D C → P,Q,R,S C→P 17 18 2 1

8 9 B, C A, B, C, D D → P,Q,R,S D→S 19 1

C H EM I STR Y Ques Ans Ques Ques Ques Ans

1 D 10 11 12 1

2 D A→ R A → Q,S 13 2

3 D

14 2

4 C B → Q,T B → R,S 15 0

5 D

16 2

6 A, B, C C→P C → P,Q,T 17 9

7 B

18 0

8 A, B, D D→S D → R,S 19 6

9 B, C, D

MATHEMATICS Ques Ans Ques Ques Ques Ans

1 B 10 11 12 1

2 C A→ Q A→ Q 13 4

3 C

14 6

4 C B→R B→S 15 4

5 A, B, D

16 1

6 A, C C→P C→P 17 3

7 A, B, C, D

18 3

8 A, D D→S D→R 19 3

9 D

IIT- JEE 2011 (December issue) PHYSICS Ques Ans Ques Ques Ques Ans

1 D 10 11 12 2

2 B A→ Q A→ R 13 3

3 C

14 3

4 D B→S B→P 15 5

5 A, B, C, D

16 2

6 A, C C→R C→S 17 3

7 A, C, D

6 A, B, D C→Q C→P 17 1

7 A, B, D

18 0

8 A, B, C, D D→P D→Q 19 1

9 A, B, C, D

8 C D → P,T D→Q 19 1

9 D

8 A, B D→R D→P 19 4

9 A, C

C H EM I STR Y Ques Ans Ques Ques Ques Ans

1 A 10 11 12 1

2 B A→ R A→ S 13 7

3 D

14 2

4 5 C A, C, D B → P,S,T B→R 15 16 7 8

18 9

MATHEMATICS Ques Ans Ques Ques Ques Ans

1 B 10 11 12 2

XtraEdge for IIT-JEE

2 C A→ S A→ S 13 5

3 A

14 3

4 C B→P B→R 15 6

5 A, B, C, D

16 1

75

6 A, B, C C→Q C→Q 17 2

7 B, D

18 0

DECEMBER 2009

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DECEMBER 2009

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Volume - 5 Issue - 6 December, 2009 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009

Editorial

Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected] Editor : Pramod Maheshwari [B.Tech. IIT-Delhi] Analyst & Correspondent Mr. Ajay Jain Cover Design & Layout Mohammed Rafiq Om Gocher, Govind Saini Circulation & Advertisement Ankesh Jain, Praveen Chandna Ph (0744)- 3040007, 9001799502 Subscription Sudha Jaisingh Ph. 0744-2500492, 2500692 © Strictly reserved with the publishers • No Portion of the magazine can be published/ reproduced without the written permission of the publisher • All disputes are subject to the exclusive jurisdiction of the Kota Courts only. Every effort has been made to avoid errors or omission in this publication. In spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from.

Unit Price Rs. 20/Special Subscription Rates 6 issues : Rs. 100 /- [One issue free ] 12 issues : Rs. 200 /- [Two issues free] 24 issues : Rs. 400 /- [Four issues free]

Dear Students, All of us want to get organized. The first thing in getting organized is to find our obstacles and conquer them. Make a beginning by conquering obstacles for starting in right earnest. Sometimes the quest for perfectionism holds us back. We occasionally feel that we should start when we have enough time to do a job thoroughly. One way to tackle this kind of mindset is to choose smaller projects or parts of projects that can be completed within 15 minutes to one hour or less. It is important to keep yourself motivated. Approach your projects as something which are going to give you pleasure and fun. Reward yourself for all that you accomplish no matter how small they may be. Never hesitate to ask for help from a friend. Make all efforts to keep your motivational level high. You might feel overwhelmed because you are focusing on every trivial thing that needs to be got done. How do you act or react to your life? When you are merely reacting to events in your life, you are putting yourself in a weak position. You are only waiting for things to happen in order to take the next step in your life. On the other hand when you are enthusiastic about your happiness you facilitate great things to happen. It is always better to act from a position of power. Never be a passive victim of life. Be someone who steers his life in exactly the direction he wants it to go. It is all upto you now. If you do what you have always done and in the way you have done it you shall get only such results which you have always got. Getting organized requires that not doing things that cause clutter, waste of time and hurt your chances adversely of realizing your goals. You should concentrate only on doing things that eliminate clutter, waste of times and hurt your chances adversely of realizing your goals. You should concentrate only on doing things that eliminate clutter, increase your productivity and provide the best chance for achieving your goals. The first step should be to stop leaving papers and other things on tables, desks, counter tops and in all kind of odd place. The more things you leave around in places other than rightful places the quicker the clutter will accumulate. Keep things in their assigned places after you have finished using them. It does not take along to put something away. If you leave things lying around they will build into a mountain of clutter. It could take hours if not weeks or months to trace them and declutter the atmosphere. Presenting forever positive ideas to your success. Yours truly

Owned & Published by Pramod Maheshwari, 112, Shakti Nagar, Dadabari, Kota & Printed by Naval Maheshwari, Published & Printed at 112, Shakti Nagar, Dadabari, Kota.

Editor : Pramod Maheshwari XtraEdge for IIT-JEE

Pramod Maheshwari, B.Tech., IIT Delhi

1

DECEMBER 2009

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2

DECEMBER 2009

Volume-5 Issue-6 December, 2009 (Monthly Magazine) NEXT MONTHS ATTRACTIONS

CONTENTS INDEX

Regulars ..........

Key Concepts & Problem Solving strategy for IIT-JEE. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics, Chemistry & Maths Much more IIT-JEE News. Xtra Edge Test Series for JEE-2010 & 2011

PAGE

NEWS ARTICLE

4

IITian ON THE PATH OF SUCCESS

8

KNOW IIT-JEE

10

IIT-Jodhpur to begin functioning from February - March 2010 IIT-K students develop autonomous vehicle Dr. Rajeewa Arya

Previous IIT-JEE Question

Study Time........ S Success Tips for the Month • The greatest adventure is what lies ahead. • Fixers believe they can fix. Complainers believe they can complain. They are both right. • The tire model for motivation: People work best at the right pressure.

DYNAMIC PHYSICS 8-Challenging Problems [Set# 8] Students’ Forum Physics Fundamentals Reflection at plane & curved surfaces Fluid Mechanics

CATALYST CHEMISTRY

• People who expect to fail are usually right. • The path to success is paved with mistakes. • You've got to cross that lonesome valley. You've got to cross it by yourself.

DICEY MATHS

Test Time ..........

• Learn to mock the woe-mongers.

XTRAEDGE TEST SERIES

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46

Mathematical Challenges Students’ Forum Key Concept Monotonicity, Maxima & Minima Function

• Appreciate what your brain does. In case nobody else does. • Be confident. Even if you are not, pretend to be. No one can tell the difference.

35

Key Concept Carboxylic Acid Chemical Kinetics Understanding : Inorganic Chemistry

• Trust the force, Luke. • Use your feelings or your feelings will use you.

18

58

Class XII – IIT-JEE 2010 Paper Class XII – IIT-JEE 2011 Paper Mock Test CBSE Pattern Paper -1 [Class # XII]

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DECEMBER 2009

IIT-Jodhpur to begin functioning from February - March 2010 Indian Institute of Technology in Rajasthan is all set to begin functioning from the MBM Engineering College, here from next academic session commencing in February-March 2010. The decision to this effect was made after the visit of a central team headed by additional secretary, ministry of Human Resource Development, Ashok Thakur during recent visit to different sites, which included the sites proposed for the construction of the IIT-Jodhpur and the existing engineering college of the city. So, if everything goes well, the next session of IIT-J, which is presently being run from the IITKanpur campus, will start functioning from the campus of this college here. Thakur, who himself approved the MBM college during a visit on Saturday, said, "The final decision is to be taken by the ministry, to whom we will submit the report in a 2-3 days." Agarwal, who was quite ecstatic following the visit of the team, expressed hope that the next session of the IIT will start here from February–March 2010.

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He added that the new building of the IIT will take atleast 2-3 years, but owing to repeated reminders by IIT-Kanpur citing its inability to continue running the session of IIT-J from its own campus, there is a growing pressure to shift it to Jodhpur.

IIT-K students develop autonomous vehicle True to its reputation of being ahead in technological developments, the Indian Institute of Technology, Kanpur, in collaboration with the Boeing Company is all set to unveil a new autonomous vehicle `Abhyast' as part of its Golden Jubilee celebrations. Prof Shantanu Bhattacharya, faculty in mechanical engineering department of IIT-K and coordinator of the project, while talking to TOI said, "The vehicle will set new landmarks in terms of applied robotics research. Besides being unique, it will probably be one of its kind in India. Such vehicles play important role in defence applications and disaster management plans." He added that quite a few modules of such vehicles have already been developed by the United States and some other countries. But, India has so far not produced vehicles of this nature. Further, Prof Bhattacharya informed, `Abhyast' will serve as a 4

first running prototype for such vehicles in the country. "It has a very small footprint -- 30x30x15 cm -- and thus it can be easily carried by soldiers and relief workers for disaster management operations. This technology was widely used by the US in investigating the World Trade Centre attack as well as in the war zones of Afghanistan and Iraq," he added. Earlier this year, as a part of its University relations programme, Boeing decided to actively collaborate with IIT-K to foster research and innovation among undergraduate students. As part of this endeavour, eight students of IIT-K -- Abhilash Jindal, Ankur Jain, Faiz Ahmed, Gaurav Dhama, Palash Soni, Shishir Pandya and Sriram Ganesan -- were selected by Prof Bhattacharya to work on the project. "My role was mainly confined to that of a mentor. `Abhyast' in fact the result of students' labour and skills and should serve as an inspiration for other students," said a beaming Prof Bhattacharya. `Abhyast' is a fully autonomous vehicle capable of navigating in unstructured and unknown environments. The user needs to specify only the end coordinates where he wants the vehicle to reach, and the task of reaching there would be taken by the vehicle itself, requiring no intervention by the user. The DECEMBER 2009

vehicle is equipped with high end sensors like GPS, IMU and SONARS to navigate and avoid obstacles in its vicinity. The vehicle has a tank like chassis design that allows it to move even on uneven and slippery terrain, thus making it a robust vehicle for warlike and other disastrous situations.

IIT forecast system for Oman, Maldives Ocean State Forecasting System (OSFS), a technology developed by Indian Institute of Technology, Kharagpur, will now be adopted by Oman and Maldives. The World Meteorological Organisation (WMO) feels that OSFS should be immediately adopted by other nations that have sea coasts. The system, that will continuously measure height, direction and period of waves, will help shipping and navigation. IIT-Kgp was given the prestigious project jointly by the union ministry of ocean development and the department of science and technology. It was headed by the institute's former director, S K Dubey. The model was developed by the department of ocean engineering and naval architecture. It was completed about a year ago and handed over to the India Meteorological Department by the institute for adoption. Dubey is presently in IIT Delhi to coordinate the project. "WMO was so impressed that it immediately referred the technology to all the coastal nations. But it cannot be transferred without proper training to end users, so we will

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be training meteorologists from other countries in batches. This will start with Oman and Maldives. Their scientists will be on the campus for an intensive training," Dubey said.

IITs come up with their RTI 'shield' Stung by the exposure of admission anomalies in recent years, the IIT system has come up with an innovative method of blocking transparency even as it agreed to give data under RTI on the marks obtained by the four lakh candidates in this year’s joint entrance examination (JEE). It insisted on giving the data only in the hard copy running into hundreds of thousands of pages rather than in the more convenient form of a CD.

In his first mail to CIC on October 2, Barua said that as there were a number of RTI applications seeking the CD, “we are apprehensive that this request for electronic data is to profit from it by using it for IIT JEE coaching purposes (planning, targeting particular cities, population segments, etc).” The reference to the coaching institutes is reminiscent of the recent controversy over the move to raise the bar on 12th class marks to be eligible for IIT selection. Asserting that IITs had “nothing to hide regarding the results”, Barua said, “We are ready to show the running of the software with the original data to the CIC, if it so desires.”

The information seeker, Rajeev Kumar, a computer science professor in IIT Kharagpur, is crying foul. For, the hard copy would not only result in a steep increase in the cost of information (running into six figures) but also make it almost impossible for him to detect irregularities in the latest JEE as he did in the three previous ones by analyzing the electronic data that had then by given to him under RTI.

As a corollary, Barua made an issue of the fact that Kumar “has not asked to see the data, but he wants an electronic version delivered to him. Why is this so?” Kumar responded to that by pointing out that the irregularities he had uncovered in the JEE of the previous three years was on the basis of “compute intensive scientific calculations and analysis, which could not have been done just by looking at the data.”

As a result of this change in the strategy of the IIT system, central information commissioner Shailesh Gandhi fixed a hearing for November 6 specially to resolve this soft vs hard debate. The hearing follows the unusual reasons given by Gautam Barua, director of IIT Guwahati and overall in-charge of JEE 2009, for his failure to comply with the CIC’s disclosure direction passed on July 30.

Barua’s explanation in his subsequent mail on October 3 is: “By seeing, I meant that the appellant could come to IIT Guwahati and view the data, see the software being run, etc.” He added that if this option was unacceptable to CIC, “we will wish to provide the data in hard copy form, the costs of printing having to be borne by the appellant.”

5

DECEMBER 2009

If Kumar is pressing that the data be given to him “in the form in which it is originally available”, it is because the access to the electronic data of the previous three years helped him unearth, for instance, the shocking fact that general category candidates got into IITs after scoring in JEE as little as little as 5% in Physics and 6% in Mathematics.

Kerala seeks second IIT report on record Kerala has filed an application in the apex court urging it to take on record the second part of the report titled Seismic Stability of Mullaiperiyar Composite Dam, submitted by D K Paul, Professor of earthquake engineering department, Indian Institute of Technology, Roorkee. The first part of the report Structural Stability of Mullaiperiyar Dam Considering Seismic Effects — Part I — Seismic Hazard Assessment was submitted by IITRoorkee in May 2008. The said report is already before the apex court, it averred. Kerala would like to file the second part of the report in support of its argument that Mullaiperiyar dam is not safe for storage, the application stated. The report stated that the earthquake safety of old concrete or masonry gravity dams under moderate to strong ground motions is of great concern. Although there is no evidence of catastrophic failure of gravity dams, yet the possibility of tensile cracking is never ruled out. The finite element analysis of dam subjected to static and seismic

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loading shows tensile stresses at the heel of dam-foundation interface, the report indicated. “The Mullaiperiyar dam is a composite gravity dam built during 1887-1895. The front and rear faces of the dam were built with un-coursed rubble masonry in lime surkhi mortar. The hearting is constructed of lime surkhi concrete. It lies in seismic zone III as per seismic zoning map of India. The 176 feet-high composite gravity dam is now over 114 years old,” the report went on to say.

IITs told to candidate details

reveal

The Central Information Commission has ordered the IITs to disclose most details of candidates who sat the 2009 entrance examination, rejecting the institutes’ argument that revealing candidates’ names would be a breach of their privacy.

IITs want to be accredited by statutory body

India’s apex watchdog for the Right to Information Act has ordered the IITs to reveal the names, addresses, pin codes and marks of all students who appeared in the Joint Entrance Examination this year.

As the government wants to make accreditation mandatory for all institutions, the IITs have said they would like to be accredited by a statutory body and not by the National Board of Accreditation (NBA).

In its November 6 order, the commission asked IIT Guwahati, the chief organiser among the IITs of the 2009 examination, to disclose by November 25 the information sought by the appellant.

The IIT directors have told the government that they have no objection to accreditation of the institutes, but insisted that the accreditation agency should be a statutory and autonomous organisation.

The order follows efforts by the IIT to withhold information on candidates who appeared in the 2009 JEE despite earlier orders mandating the release of similar data on IIT candidates over the past three years.

They expressed these views at the meeting of the IIT Council, the highest decision making body for the IITs, held here last month. HRD Minister Kapil Sibal, who is the chairman of the council, told them that the government would set up an accreditation agency by introducing a bill in the Parliament soon. "The directors said the accreditation should be conducted by a statutory body

The order is significant because a similar disclosure in 2006 revealed discrepancies between cutoff marks used by the IITs that year and the cutoffs arrived at by using the formula the institutes claimed to have used.

6

At least 994 students, who cleared the cutoffs arrived at by using the formula the IITs claimed to have used, were denied admission because the institutes used different cutoffs. DECEMBER 2009

The IITs have till now been unable to explain how they arrived at the cutoffs — by using the formula they claim to have used. The appellant in the 2009 case is the parent of a student who appeared in the controversial 2006 examination and is trying to use the RTI Act to ensure that the IITs do not repeat their errors. The IITs, in the 2009 case, argued that the release of candidate details sought by the appellant could compromise the privacy of these candidates. The appellant had sought the registration numbers, names, gender, parents’ names, pin codes and marks in physics, chemistry and mathematics of all students who appeared in the 2009 examination. The appellant has expressed concern that the IITs may be admitting the children of institute administrators or certain faculty members despite poor marks, and has argued that details he has sought would help clarify his doubts. The commission, in its order, has argued that while providing email addresses and mobile phone numbers of candidates would constitute a violation of privacy, merely disclosing their names and addresses would not.

Bombay (IIT-B), highlighted this on Monday. Moudgalya, who heads the Centre for Distance Engineering Education Programme (CDEEP) at the IIT, was delivering the keynote address at the launch of kPoint, a software solution for interactive learning and training. kPoint, developed by city-based Great Software Laboratory (GSL), was launched by noted computer expert Vijay Bhatkar, creator of India's Param series of supercomputers. Heads and professionals from leading IT companies as well as principals of engineering institutions were present at the occasion. Open source software refers to computer software provided under a license that is in the public domain. "Open source software has a distinct cost advantage over the expensive commercial software packages. However, a considerable marketing effort is required to secure a greater and wider audience of students for courses transmitted live using ICT tools based on open source software," Moudgalya said.

Open source software needs marketing

"Open source software is often sufficient in most distance education programmes, except for some niche academic segments. However, academic institutions don't train students in using good open source software," he further stated.

PUNE: There is a need for greater promotion of the use of open source software for information and communication technology (ICT)-based teaching and learning. Professor Kannan M Moudgalya of the Indian Institute of Technology,

Moudgalya gave an overview of the CDEEP's involvement in the Talk to teacher' programme, which is funded by the Union human resource development ministry and aims to train students as well as teachers in higher

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7

education. IIT-B started disseminating its courses live on the internet nearly a decade ago. For the last two years, he stated, the CDEEP has been using the education satellite Edusat, provided by the Indian Space Research Organisation, and has raised a network of 75 centres for transmission of live courses. In his brief address, Bhatkar made out a strong case for Indian ICT professionals acknowledging and adopting technologies and innovations developed indigenously. Sunil Gaitonde, chief executive of GSL, said, "Technology must bridge the gap between the growing number of learners and lesser number of teachers. The prevailing knowledge economy needs highly skilled workers and the existing faculty crunch can be tackled only through apt use of technology." Gaitonde said: "Factors like grassroots videos, collaborations, mobile broadband, data mash-ups, collective intelligence and social operating systems are bound to make a sea change in the way education is delivered." College of Engineering, Pune (CoEP) principal Anil Sahasrabudhe, Vishwakarma Institute of Technology principal Hemant Abhyankar, Persistent Systems chief Anand Deshpande and founderCEO of music education web portal ShadjaMadhyam, Nandu Kulkarni, were among those present at the event.

DECEMBER 2009

Success Story This article contains story of a person who get succeed after graduation from different IIT's

Dr. Rajeewa Arya B.E., M.E.(IIT – Kanpur) Chief Executive Officer, Moser Bear (I), Ltd.

Dr. Rajeewa (Rajiv) Arya, M. Tech., Ph. D is presently

includes product design, process scale-up, process

the Chief Executive Officer l at Moserbaer Photovoltaic

transfer, piloting and start-up of a thin-film solar module

(MBPV) in New Delhi, India. He was previously the COO

plant. He has maintained a professional interest in many

& CTO for the Thin Film Vertical. He joined MBPV as a

aspects of renewable energy components and systems.

Senior Vice-President & CTO (Thin Film) in September,

Dr. Arya holds a Masters of Science degree in Solid-State

2007. and Electrical Engineers (IEEE) and the Materials

Physics from Jadavpur University, Calcutta, India and a

Research Society (MRS). Prior to that Dr. Arya was a

Master of Technology degree in Material Science from the

founder, Vice-President and CTO at Optisolar (previously

Indian Institute of Technology, Kanpur, India. He obtained

called Gen3Solar) in Hayward, California. Before founding

his Ph.D. in Engineering from Brown University, Rhode

Gen3solar, he was the Director of Oregon Renewable

Island, in 1983. He has extensive management training in

Energy Center (OREC), an academic/research center at

Total Quality Management, Finance, Project management,

the Oregon Institute of Technology (OIT).

and Technology Innovation Management.

Dr. Arya launched Arya International, Inc., a Solar

Dr. Arya has authored and co-authored over 100

Technology and Business Consulting firm, in 2003. Prior

technical papers and holds 6 U.S. Patents. He is the

to that Dr. Arya worked at Solarex/ BPSolar for 19 years

recipient of the “Outstanding Paper Award” at the 7th

in various capacities, from Scientist to Executive Director,

PVSEC, the “Team of the Year” award from Solarex

thin-film technology.

Quality Process, and his group received an R&D 100

He has over 25 years experience in thin-film solar cells

award for the Power view Product. He chaired the

and modules. His R&D activities have centered on

Program Committee for the 29th IEEE Photovoltaic

material and device aspects of three types of thin-film

Specialists Conference in 2002. He is a member of the

solar cells and modules – amorphous silicon, copper-

Institute of Electronics.

indium-diselenide, and cadmium telluride. His work

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8

DECEMBER 2009

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9

DECEMBER 2009

KNOW IIT-JEE By Previous Exam Questions

FB=πR2hσg

PHYSICS 1.

A circular disc with a groove along its diameter is placed horizontally on a rough surface. A block of mass 1 kg is placed as shown. The co-efficient of friction in contact is µ = 2/5. The disc has an acceleration of 25 m/s2 towards left. Find the acceleration of the block with respect to disc. Given cos θ = 4/5, sin θ = 3/5. [IIT-2006]

θ h/2

mg

l . Let OG = y 2 For vertical equilibrium FG = FB ⇒ (M + m)g = F B ⇒ πR 2Lρg + mg = πR 2 l σ g

25 m/s2 θ

OB =

l=

πR 2 Lρ + m

...(1) πR 2 σ Now using the concept of centre of mass to find y. Then My1 + my 2 y= M+n Since mass m is at O the origin, therefore y2 = 0 M(L / 2) + m × O ML ∴ y= = M+m 2(M + m)

Sol. Applying pseudo force ma and resolving it. Applying Fnet = ma x for x-direction ma cos θ – (f1 + f2+ = ma x ma cos θ – µN1 – µN2 = ma x ma cos θ – µma sin θ – µmg = ma x ⇒ a x = a cos θ – µa sin θ – µg 4 2 3 2 – × 25 × – × 10 = 10 m/s2 5 5 5 5

=

(πR 2 Lρ)L

2(πR 2 Lρ + m) Therefore for stable equilibrium l >y 2 πR 2 Lρ + m (πR 2 Lρ)L ∴ > 2(πR 2 σ) 2(πR 2 Lρ + m)

A wooden stick of length L, radius R and density ρ has a small metal piece of mass m (of negligible volume) attached to its one end. Find the minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in equilibrium in liquid of density σ(>p). [IIT-1999] Sol. For the wooden stick-mass system to be in stable equilibrium the centre of gravity of stick-mass system should be lower than the centre of buoyancy. Also in equilibrium the centre of gravity (G) and the centre of buoyancy (B) lie in the same vertical axis. The above condition 1 will be satisfied if the mass is towards the lower side of the stick as shown in the figure. The two forces will create a torque which will bring the stick-mass system in the vertical position of the stable equilibrium Let l be the length of the stick immersed in the liquid. 2.

XtraEdge for IIT-JEE

L/2 (πR2Lρ)g

θ

Then

= 25 ×

C

⇒ ∴ 3.

10

...(2)

m ≥ π R2L ( ρσ – ρ) minimum value of m is πr2L ( ρσ – ρ)

A gaseous mixture enclosed in a vessel of volume V consists of one mole of a gas A with λ (=Cp/Cv) = 5/3 and another gas B with λ = 7/5 at a certain temperature T. The relative molar masses of the gases A and B are 4 and 32, respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation PV19/13 = constant, in adiabatic processes. [IIT-1995] DECEMBER 2009

(a) Find the number of moles of the gas B in the gaseous mixture. (b) Compute the speed of sound in the gaseous mixture at T = 300 K. (c) If T is raised by 1 K from 300 K, find the % change in the speed of sound in the gaseous mixture. (d) The mixture is compressed adiabatically to 1/5 of its initial volume V. Find the change in its adiabatic compressibility in terms of the given quantities. Sol. (a) The ratio of specific heat of mixture of gases (C p ) m γm = * m stands for mixture. (C v ) m n A C pA + n B C pB Also (Cp)m = nA + nB

Mass of the gas, m = nA MA + nBMB = 1 × 4 + 2 × 32 = 68 g/mol mol = 0.068 kg/mol 19(1 + 2) × 8.314 × 300 = 400.03 ms–1 13 × 0.068 (c) We know that the velocity of sound ∴ v=

v=

(Cv)m =

R 3R = 5 / 3 −1 2 3R 5R ∴ (Cp)A = R + (Cv)A = R + = 2 2 R 5R For Gas B (Cv)B = = 7 / 5 −1 2 5R 7R ∴ (Cp)B = R + = 2 2 (C p ) m n A C pA + n B C pB γm = = (C v ) m n A C vA + n B C vB For Gas A (Cv)A =

⇒ ⇒ ∴ ∴

1× (5R / 2) + n B (7 R / 2) 5 + 7n B = 1 × (3R / 2) + n B (5R / 2) 3 + 5n B According to the relationship PV19/13 = constant we get γ m = 19/13 5 + 7n B 19 ∴ = ⇒ nB = 2 mol. 3 + 5n B 13 Alternatively we may use the following formula n1 n1 n2 = + γ m −1 γ1 − 1 γ 2 −1 where γm = Ratio of specific heats of mixture (b) We know that velocity of sound in air is given by the relationship

∴

=

v= Also, ⇒ ∴

γ (n A + n B )RT = m V× V

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γR (T + ∆T ) M

1/ 2

v + ∆v = v

dP γPV γ −1 = = γPVγ–1 dV dV Vγ −dP = –γ P dV / V Bulk Modulus B = γP 1 1 Compressibility K = = B γP 1 1 K1 = and K2 = γP1 γP2 ∆K = K2 – K1 =

1 1 1 1 1 – = − γP2 γP1 γ P2 P1 γ

γ

∴ Since the process is adiabatic, P2V 2 = P1V 1 V ∴ P2 = P 1 1 V2

γ

= P1

γ

γ V1 = P1V 1 V1 / 5

1 1 1 1 1 − = − 1 γ γ P1 5 P1 γP1 5 γ ( n A + n B ) RT (1 + 2) × 8.31× T 24.93 T P1 = = = V V V 1 1 ⇒ ∆K = − 1 19 T 519 /13 × 24.93 × 13 V V = – 0.025 Pa–1 T This unit Pa–1 is valid when V, T are taken in S.I. units. ∴ ∆Κ =

γP m where d = density = d v PV = (nA + nB)RT (n A + n B) PV = RT V v=

γRT and v + ∆v = M

T + ∆T 1 + ∆T = T T ∆v 1 ∆T ⇒ 1+ =1+ v 2 T ∆T when ∆T 2d It is observed that 1st bright fringe is observed in front of one of the slit.

For the given circuit

3.

a

R R

4R 8R

4.

What should be the value of R so that equivalent

(A)

5.

Relation of the maximum wavelength missing with wave length of Blue light is (B) 3λb

(C) 4λb

(D) 5λb

b

12R

2 Ω 5

(C) 2Ω

(A) 2λb

R

R

resistance between terminals a and b is 1Ω -

2 λ max. = .λ b 3 ( missing )

(B)

5 Ω 2

(D) 15Ω

If current passing through the circuit is 1 amp then(A) Potential difference across 4R and 8R is in the ratio of 1 : 2 (B) Potential difference across ab is 1 volt when measured by an ideal voltmeter if R =

With blue light if the slit widths are made unequal then (A) Position of 1st bright fringe will not be in front of one of the slit (B) Dark fringe which was of black colour earlier with same slit widths now is of blue colour

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2R 3R

If monochromatic light source of blue color is replaced by the white colored light source then maximum wavelength which is missing in front of one of the slit is (A) Never of indigo and violet colors (B) It is always of less than blue color (C) Missing wave lengths can have wave length more or less than blue color (D)

2.

R

2 Ω 5

(C) Maximum potential difference will appear across 12R resistance (D) Maximum potential difference will appear across 3R resistance

18

DECEMBER 2009

Passage # 3 (Ques. 6 to 8)

SCIENCE TIPS

Behaviour of capacitor in electric circuits is very typical because of it's energy storing nature. Capacitor behaves in just opposite manner to

• What is the expression for growing current, in LR R − t circuit ? ® I = I0 1 − e L

inductor, Inductor 'L' which is measured in Henary in SI system stores the energy in magnetic field instead of capacitor which stores in electric field

• What is the range of infrared spectrum ? ® This covers wavelengths from 10–3 m down to 7.8 × 10–7 m

Inductor opposes the change in current and capacitor opposes change in voltage. Behaviour of inductor: t=0

open switch

t→∞

• What is the nature of graph between electric field and potential energy (U) ? ® The nature of the graph will be parabola having symmetry about U-axis

L/R is known as time constant of R-L series circuit which is measured in ohm

• Why no beats can be heard if the frequencies of the two interfering waves differ by more than ten ? ® this is due to persistence of hearing

Closed switch

Steady State

For the electric circuit shown ε1

R1

R

C

ε2

6.

• Why heating systems based on steam are more efficient than those based on circulation of hot water ? ® This is because steam has more heat than water a the same temperature

R2

• Can the specific heat of a gas be infinity ? ® Yes

If capacitance C varies even after that energy stored

• What is the liquid ascent formula for a capillary ? 2T cos θ r ®h= – γpg 3 where h is the height through which a liquid of density ρ and surface tension T rises in a capillary tube of radius r

in capacitor is zero at steady state then R 1 ε1 = R 2 ε2 (C) ε1 + ε2 = 0

(A)

7.

• What is the expression for total time of flight (T) 2u sin θ for oblique projection ? ®T= g • The space charge limited current iP in the diode value is given by ® iP = k Vp3/2

Time constant for the circuit (A) RC (C) R2C if ε1 < ε2 where εeq = Req =

8.

R1 ε 2 = R 2 ε1 (D) ε1R1 + ε2R2 = 0 (B)

(B) R1C if ε1 > ε2 R R (D) R + 1 2 C R1 + R 2

ε1 / R 1 − ε1 / R 2 1 / R1 + 1 / R 2

• What is an ideal gas ?

R 1R 2 R1 + R 2

• Can a rough sea be calmed by pouring oil on its surface ? ® Yes

Maximum current passing through resistance R ε eq ε eq (A) (B) R eq R + R eq (C)

ε eq R

XtraEdge for IIT-JEE

(D)

® An ideal gas is one in which intermolecular forces are absent

• What is the expression for fringe width (β) in Young's double slit experiment? ® β =Dλ /d where D is the distance between the source and screen and d is distance between two slits

| ε1 − ε 2 | R

19

DECEMBER 2009

1.

8

Solution

Physics Challenging Problems Que s t i ons we r e Publ is he d i n Nove m be r I s s ue

As the resistances of voltmeters in upper branch are R, R/2, R/4 ...................... the equivalent circuit is as shown below R

R/2

R/4

3.

current in upper and lower branch are in the ratio of 1 : 2. R

i b

V

From current division formula we can conclude that

upper Branch ...................

a Lower Branch

2i

R′ = R voltmeter V

Reading of voltmeter V is (2i.)R So V = 2V1

1 =R = 2R 1 −1 / 2 further the equivalent circuit is R

x.

upper branch

a

A

V

b. l.

Lower Branch

l = length of rod = b – a

the resistance of voltmeter V should be 2R so that current in upper and lower branch is same.

charge on element of length d x is dq dq = λ dx

Entire upper branch is having the resistance of 2R

Equivalent current due to element of length d x

can conclude that equivalent resistance of all the

di = ω .dq =

voltmeters in upper branch except V1 is R and the upper branch is as follows:

a

i

V2 R

V3 R

C V 1=X

λ = 3x

as

dq = 3xd x

and voltmeter V1 is having the resistance of R so we

.....up to infinite

B dx.

a.

4.

b

V1

b

Reading of voltmeter V1 is i.R

1 1 = R 1 + + + ..... 2 4

a

R C

a

the resistance of upper branch is = R + R/2 + R/4 + ............. up to infinite

2.

ω (3xd x) 2π

∫

Total equivalent current i = d i =

b

b

3ω = 2π

V 2=Y

As reading of voltmeter V1 is X = i.R

b

x 2 3ω = 2 a 2π

Except V1 in upper branch

Option A is correct

So,

(B) Equivalent current

20

ω

∫ 2π (3xd

x)

a

3ω 2 2 (b – a ) 4π

=

X=Y

b

b2 − a 2 3 ω 2 2 2 = 2 . 2π (b – a )

sum of the readings of voltmeters is Y = i.R

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Set # 7

DECEMBER 2009

=

3ω 2 2 3ω (b – a ) = (b – a)(b + a) 4π 4π

Option B is correct As charge on cone C3 ≠ charge on cone C 4

3ω 3 = (b + a)(b – a) = ω . (b + a).l 4π 4π

Option C correct Part-A and part-B will have different charges so

As ω = 4π/3 So,

option D incorrect

3 4π Equivalent current = . . (b + a).l 4π 3

Ans. B, C

= (b + a).l = const.l

6.

i∝l

The circuit is as follows 10Ω

CT

Option B is correct. b

b x2 q = d q = 3xd x = 3. 2 a a

∫

=

∫

a

ig =

Option D is correct

5.

using R =

For part B >

open cone 7.

for part A q

=

q

closed cone

Using R =

ω .q 2π

ω .q , 2π cone - C1 (closed cone) i=

ω = .q 2π (closed cone)

R2 = 2000 – 1000 = 1000Ω

ω .q 2π cone-C3 (closed cone)

R2 = 1000Ω

i=

8.

ω i= .(σ) 2π (Surface area of closed cone)

(coneC1 )

i

(cone - C2) (ConeC1 )

=

(cone C1)

XtraEdge for IIT-JEE

and

i

(coneC2 )

=

ω . q 2π (ConeC2 )

V 5 × 10 −3

– 10

⇒

990 + 1000 + 3000 =

⇒

5000 =

⇒

V = 25 volt

q

(cone C1) q

V –G ig

R1 + R 2 + R3 =

Option A incorrect

ω . 2π

Range between CT and c is V so Using R =

their currents will be different. =

V 10 – G ⇒ R1 + R 2 = – 10 ig 5 ×10 −3

990 + R2 = 2000 – 10

If σ varies then charge on cone C1 differs from C3 So

=

Range between CT and b is 10 volt so,

open cone

Equivalent current i =

i

V 5 – G ⇒ R1 = – 10 = 990Ω ig 5 × 10 −3

R1 = 990Ω

q

closed cone

q

50m = 5mA 10Ω

For terminals CT and a range is 5V so

A, B, D

q

c

b

Full scale deflection current for galvanometer is

3 2 2 (b – a ) 2

Ans.

R3

R2

R1

(D) Charge on rod

V 5 × 10 −3

– 10

V 5 × 10 −3

So range between CT and C is 25 volts.

i (cone C2) 21

DECEMBER 2009

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

PHYSICS 1.

A homogeneous sphere of radius r rolls without slipping with constant angular speed ω' over bigger sphere, of radius R, which is in pure rotation with constant angular speed ω about its centre O. (Fig.) Find the time taken. r

ω

(a) If t is the direction of common tangent, then from eqn. (2) vp'o (t) = vpc (t) + vco (t) dθ ω R = – ω′R + (R + r) dt

ω′

dθ ωR + ω' r = dt R +r R+r dt = dθ (ωR + ω' r )

or,

C

so, O

R

for the centre C of the rolling sphere to return to its initial position (with respect to O), and (ii) for the point of contact of the rolling sphere to make one full revolution over the bigger rotating sphere. (b)(i)Determine the acceleration of the contact point of the rolling sphere, and (ii) the point of greatest acceleration of the rolling sphere (both w.r.t. the centre O) Sol. Suppose that at t = 0, the contact points of lie along the fixed line (reference line) OX. Let at time t the line OC makes the angle θ with OX (Fig.) C

ω O

θ P′

dθ with angular speed – ω to satisfy the condition dt of the problem. dθ dt

ω

Hence the sought time t′ (say) is given by 2π 2π(R + r) = [using Eq. (4)] r (ω'− ω) dθ −ω dt (b) From apo = apc + aco = apc (tangential) + apc (normal) + aco In our case apc (tangential) = 0, because ω′ = constant Hence, apo = apc (normal) + aco (5) t′ =

ω′

P X

It is better to express the velocity and acceleration of any point including contact point P (say) at an arbitrary instant of time t as, v po = v co and apo = aco = apc + apc + aco (1) We know that in the case of pure rolling the velocity of contact point of the rolling body has zero velocity and zero tangential acceleration relative to the contact point of the surface on which it rolls. So if P′ be the contact point of rotating sphere at time t, then vp′o = vpo So, vp'o = vpc + vco (2)

XtraEdge for IIT-JEE

(4)

(ii) It is simple to observer if rotating sphere were at rest, the CM of rolling sphere C will turn by the angle 2 π,

(i)

^t

(3)

2 dθ 2 apo = − ω' r + (R + r) dt

dθ ωR + ω' r = from equation (3) or part (a)) dt R +r (ii) From eqn. (5) it is obvious that the maximum value |apo |max = |apc (normal) | + | aco | (using

2

dθ = – ω′2r + (R + r) dt where

22

dθ will be substituted from Eqn. (3). dt

DECEMBER 2009

2.

Show that the temperature of a planet varies inversely as the square root of its distance from the Sun.

Volume of compressed air when the sphere is in position C = [ πr2x – (2/3)πr3] Atmospheric pressure = h cm of water Let the pressure of compressed air be p cm of water.

Sol. Let Rs be the radius of the Sun. Consider the Sun as a black body. Energy emitted per sec by it equals 4 π R s2 σ Ts4

A

This energy falls uniformly on the inner surface of spheres centred on the Sun. If d is the distance of the planet from the Sun, then energy falling on unit area of the sphere of radius d is : 4πR 2s σTs4

=

4πd 2

L

x d

σR 2s Ts4

According to Boyle's Law (assuming no change in the temperature of compressed air) p[πr2x – (2/3)πr3] = h[πr 2L – (2/3)πr3] or p[x – (2r/3) = h[L – (2r/3)] ...(1) In the equilibrium position C, the weight of the sphere is balanced by the difference of vertical thrust on either side due to atmospheric and compressed air.

d2

This energy received by the planet is given by Q = πr2 =

σR 2s Ts4

where r is the radius of planet.

d2

If T is the temperature of the planet, then energy lost by it per sec is 4 π r2 σ T4

Hence, πr2(p – h)g = mg = (4/3)πr3sg or p – h = (4/3)rs From equation (1) and (2), we get

In the steady state the rate of reception of energy is equal to the loss of energy

3.

Hence,

4 π r2 σ T4 =

Thus,

T∝

πr 2 σR s2 Ts4

x=

d2

1 d

A sphere of specific gravity s just fits into a vertical cylinder with lower end closed. The sphere is allowed to drop slowly until it is held in equilibrium by the thrust of the compressed air. There is no leakage of air. If the diameter of the sphere is d, the length of the cylinder is L and the height of the water barometer is h, then what will be the position of sphere ?

4.

...(2)

h (3L − 2r ) 2r + 3p 3

or x =

h (3L − 2r ) 2r 9 hL + 8r 2s + = 3h + 4 rs 3 3(3h + 4rs)

or x =

9hL + 2d 2s 9h + 6ds

Given the position of the object O and the image I as shown in the figure. Find (a) the position of the convex lens (b) its focal length and the magnification of the image. Verify graphically. Y

Sol. Initially, the cylinder contained air at atmospheric pressure. When the sphere comes down into the cylinder by the action of its own weight, it presses the air downwards. Suppose the sphere comes to position C which is at height x above the closed end. Let the sphere remain in equilibrium in this position.

2 1 0 –1 –2

O

1 2 3 4

5

6

7

8

9 10 11 12 13 14

X

I

Sol. A ray of light from the object passes undeviated through the optic centre of the lens (C) and also the image I. So join OI. So it cuts the principal axis XY and C. So AB is the position of the lens. A ray parallel to the principal axis from the object after refraction meets the principal axis at F. F is the focus.

Volume of sphere = (4/3)πr3 Weight of the sphere = (4/3)πr3sg Volume of cylinder = πr2L Volume of air inside the cylinder when the sphere is in position A = πr2L – (2/3)πr3

XtraEdge for IIT-JEE

C

23

DECEMBER 2009

A

dB =

X

8

F

C

4

14

10.4

B

X

I

F is at a distance 2.4 cm from C.

x2

B=

∴ Focal length of lens = 2.4 × 10 = 24 cm By calculation

µ i dB = 0 4π

1 1 100 1 – = = 60 − 40 2400 24

v 60 3 Magnification = = = u 40 2 =

length of image length of object

=

3 mm 3 = 2 mm 2

µ iR B= 0 4π

∴

i

x2

∫ (x

x1

dx 2

+ R2

...(ii)

)

3/ 2

At

x x = x1 φ1 = tan–1 1 and x = x2, R x φ2 = tan–1 2 R

D

distance R from a straight conductor of length l. x2

r

dx x1

Consider a typical element dx. The magnitude of the contribution dB of this element to the magnetic field at P as found from Biot-Savart's law is 24

R sec 2 φ

µ 0 iR 4π

=

µ 0 iR 4π

=

µ 0 iR −1 x 2 x − sin tan −1 1 ...(iii) sin tan 4π R R

∫

x tan −1 2 R x tan −1 1 R

∫

R 3 sec 2 φ cos φ dφ

Let,

x z = tan–1 , R

Thus,

tan z =

⇒

R

x tan −1 2 R x tan −1 1 R

B=

Sol. First of all, we determine the expression of B at a

XtraEdge for IIT-JEE

r2

x1

x = R tan φ, such that dx = R sec2 φ dφ

i

φ

sin θ dx

Also, x2 + R 2 = R 2 tan2φ + R2 = R2 sec2 φ Hence, equation (ii) becomes

i

x

∫

Let

Calculate the magnetic field B at the point P shown in the figure. Assume that i = 10 A and a = 8.0 cm. a/4 i B C a 4 P

A

x2

R and r = (x2 + R2)1/2 r

where, sin θ =

∴ f = 24 cm

5.

∫

x1

1 1 1 = – f v u =

µ 0 i dx sin θ

...(i) 4πr 2 Since, the direction of the contribution dB at the point P for all elements are identical viz, at right angle into the plane of the figure, the resultant field is obtained by simply integration equation (i), which gives

O

x R

sin z

=

1 − sin 2 z

x R

⇒

R2sin2z = x2(1 – sin2z)

⇒

sin2z(R2 + x2) = x2

⇒

sin z =

⇒

x z = sin–1 2 2 x +R

x 2

x + R2

DECEMBER 2009

Hence, equation (iii) becomes

MEMORABLE POINTS

µ0 i x2 x1 ...(iv) − 2 2 4πR x 2 + R 2 x R + 1 2 Applying the equation (iv) to the given problem For AB, we get B=

• Can a current be measured by a voltameter ? ® Yes, direct current can be measured by a voltameter • The equivalent resistance of n resistances each equal to r and connected in parallel is given by ® r/n

a a 3 x1 = – a, x2 = and R = 9 4 4

a µ 0i 4 Hence, B1 = − a a2 a2 4π + 4 16 16 µ i 1 3 + = 0 πa 2 10

• Repeated use of which digital gate or gates can produce all the three basic gates (OR, AND and NOT) ® NAND gate and NOR gate

2 2 9a a + 16 16 −3 a 4

• What is Turnbull's blue ?

• An hypothesis tested by experiments is known as ® Theory

• What is magnesia alba ? ® Mg(OH)2.MgCO3.3 H2O

For BC :

• The humidity of air is measured by

−a 3a a x1 = , x2 = and R = 4 4 4 This also gives B2 =

µ 0i 1 3 + πa 2 10

• What is oleum ?

=

• An amino acid which does not contain a chiral centre. ® Glycine [NH2–CH2–COOH] • Scientist who perfected the technique for converting pig iron into steel. ® Hynry Besemer (1856)

• When the pH of the blood is lower than the normal value, this condition is known as ® Acidosis • The electrolytic method of obtaining aluminium from bauxite was first developed by

µ 0i 1 1 + 3π a 10 2

® Charles Hall (1886)

• Which compound possesses characteristic smell like that of mustard oil ? ® Ethyl isothiocyanate [C2H5N = C = S]

For DA : x1 =

3a −a 3a , x2 = and R = 4 4 4

This also gives

B4 =

• First solar battery was developed in the ® Bell Telephone Laboratory (1954)

µ0 i 1 1 + 3πa 10 2

• What is Wilkinson's catalyst ? ® tris (triphenylphosphine) chlororhodlum (I)

Total magnitude at magnetic field, B = B1 + B 2 + B 3 + B 4 =

• In 1836 the galvanised iron was introduced first in ® France

µ 0i 1 3 1 3 1 1 1 1 + + + + + + + πa 2 10 2 10 3 10 3 2 3 10 3 2

• What is caro's acid ? ® Permonosulphuric acid [H2SO5]

2µ0 i = ( 10 + 2 2 ) 3πa = 2.0 × 104 T

XtraEdge for IIT-JEE

® H2S 2O7

• Vulcanised rubber was invented by ® Charies Goodyear (1839)

a −3a −3a , x2 = and R = 4 4 4

a − 3a µ0 i 4 4 ∴ Β3 = − 2 − 3a 9a 2 9a 2 a 9a 2 4π + + 4 16 16 16 16

® Hygrometer

Also known as fuming sulphuric acid

For CD : x1 =

® Fe4[Fe(CN)6]3

25

DECEMBER 2009

P HYSICS F U NDAMENTAL F OR IIT-J EE

Reflection at plane & curved surfaces KEY CONCEPTS & PROBLEM SOLVING STRATEGY For solving the problem, the reference frame is chosen in which optical instrument (mirror, lens, etc.) is in rest. The formation of image and size of image is independent of size of mirror. Visual region and intensity of image depend on size of mirror. P P'

Key Concepts : (a) Due to reflection, none of frequency, wavelength and speed of light change. (b) Law of reflection : Incident ray, reflected ray and normal on incident point are coplanar. The angle of incidence is equal to angle of reflection Incident n Reflected Ray Ray θ θ

n

n Tangent at point P

θ θ

α α

P Convex surface

Plane surface

θ θ

n αα Convex surface

If the plane mirror is rotated through an angle θ, the reflected ray and image is rotated through an angle 2θ in the same sense. If mirror is cut into a number of pieces, then the focal length does not change. The minimum height of mirror required to see the full image of a man of height h is h/2.

A

Tangent at point P

Some important points : In case of plane mirror For real object, image is virtual. For virtual object, image is real. The converging point of incident beam behaves as a object. If incident beam on optical instrument (mirror, lens etc) is converging in nature, object is virtual. If incident beam on optical instrument is diverging in nature, the object is real. The converging point of reflected or refracted beam from an optical instrument behaves as image. If reflected beam or refracted beam from an optical instrument is converging in nature, image is real.

Rest

vsinθ Object

P

P P Virtual Object

Real Object

n n

Real Object

n αα

α Real n α Object

XtraEdge for IIT-JEE

v θ

Image

Rest

vcosθ

vsinθ

vcosθ Image

P Virual Object

Object v

If reflected beam or refracted beam from an optical instrument is diverging in nature, image is virtual.

P

v

Object v

P'

Object In rest

Virual Object

26

vm Image

vm Image

2vm–v

2vm

DECEMBER 2009

v

Object

vm Image

These formulae are only applicable for paraxial rays. All distances are measured from optical centre. It means optical centre is taken as origin. The sign conventions are only applicable in given values. The transverse magnification is

2vm+v

(c) Number of images formed by combination of two plane mirrors : The images formed by combination of two plane mirror are lying on a circle whose centre is at the meeting points of mirrors. Also, object is lying on that circle. Here, n =

β=

360º θ

1. If object and image both are real, β is negative. 2. If object and image both are virtual, β is negative. 3. If object is real but image is virtual, β is positive.

where θ = angle between mirrors. 360º is even number, the number of images is θ n – 1. If

4. If object is virtual but image is real, β is positive. 5. Image of star; moon or distant object is formed at focus of mirror. If y = the distance of sun or moon from earth. D = diameter of moon or sun's disc f = focal length of the mirror d = diameter of the image

360º is odd number and object is placed on θ bisector of angle between mirror, then number of images is n – 1. If

360º is odd and object is not situated on θ bisector of angle between mirrors, then the number of images is equal to n. (d) Law of reflection in vector form : Let eˆ1 = unit vector along incident ray. If

θ = the angle subtended by sun or moon's disc Then tan θ = θ =

Sun D

nˆ = unit vector along normal on point of Incidence Then eˆ 2 = eˆ1 − 2(eˆ1.nˆ ) nˆ

(e) Spherical mirrors : It easy to solve the problems in geometrical optics by the help of co-ordinate sign convention. y y y x

x

x' y'

y

y x

x'

The mirror formula is

y'

x

x'

y'

x

x'

θ

θ

Problem solving strategy : Image formation by mirrors Step 1: Identify the relevant concepts : There are two different and complementary ways to solve problems involving image formation by mirrors. One approach uses equations, while the other involves drawing a principle-ray diagram. A successful problem solution uses both approaches. Step 2: Set up the problem : Determine the target variables. The three key quantities are the focal length, object distance, and image distance; typically you'll be given two of these and will have to determine the third. Step 3: Execute the solution as follows : The principal-ray diagram is to geometric optics what the free-body diagram is to mechanics. In any problem involving image formation by a mirror, always draw a principal-ray diagram first if you have enough information. (The same

eˆ 2

y'

F d

n nˆ

eˆ1

D d = y f

Here, θ is in radian.

eˆ 2 = unit vector along reflected ray

x'

image size −v = object size u

y'

1 1 1 + = v u f

Also, R = 2f

XtraEdge for IIT-JEE

27

DECEMBER 2009

advice should be followed when dealing with lenses in the following sections.) It is usually best to orient your diagrams consistently with the incoming rays traveling from left to right. Don't draw a lot of other rays at random ; stick with the principal rays, the ones you know something about. Use a ruler and measure distance carefully ! A freehand sketch will not give good results. If your principal rays don't converge at a real image point, you may have to extend them straight backward to locate a virtual image point, as figure (b). We recommend drawing the extensions with broken lines. Another useful aid is to color-code the different principal rays, as is done in figure(a) & (b). Q I 3

P

4 C P' F

2 2 4

Q'

Solved Examples 1.

Sol. B θ i1

i1

2 4

3

v 3

Rays of light strike a horizontal plane mirror at an angle of 45º. At what angle should a second plane mirror be placed in order that the reflected ray finally be reflected horizontally from the second mirror. Sol. The situation is shown in figure G C D A S θ θ

P' F

C

4 (b)

45º

1 1 1 + = and the s s' f y' s' magnification equation m = = − . The y s results you find using this equation must be consistent with your principal-ray diagram; if not, double-check both your calculation and your diagram. Pay careful attention to signs on object and image distances, radii or curvature, and object and image heights. A negative sign on any of these quantities always has significance. Use the equations and the sign rules carefully and consistently, and they will tell you the truth ! Note that the same sign rules (given in section) work for all four cases in this chapter : reflection and refraction from plane and spherical surfaces. Step 4: Evaluate your answer : You've already checked your results by using both diagrams and equations. But it always helps to take a look back and ask yourself. "Do these results make sense ?".

P

N 45º

Q B The ray AB strikes the first plane mirror PQ at an angle of 45º. Now, we suppose that the second mirror SG is arranged such that the ray BC after reflection from this mirror is horizontal. From the figure we see that emergent ray CD is parallel to PQ and BC is a line intersecting these parallel lines. So, ∠DCE = ∠CBQ = 180º ∠DCN + ∠NCB + ∠CBQ = 180º θ + θ + 45º = 180º ∴ θ = 67.5º As ∠NCS = 90º, therefore the second mirror should be inclined to the horizontal at an angle 22.5º.

Check your results using Eq.

XtraEdge for IIT-JEE

C

2.

Q' v

P

i2

The total deviation of the ray is given by δ =180 – 2i1 + 180 – 2i2 = 360 – 2(i1 + i2) For the resultant ray to be parallel, δ should be 180º ∴ 360 – 2(i1 + i2) = 180 i.e., i1 + i2 = 90º From the geometry of the figure i1 + i2 = θ θ Angle between the mirrors should be 90º.

1

2

i2

A

(a) 1

Q

P

1

Q

How will you arrange the two mirrors so that whatever may be the angle of incidence, the incident ray and the reflected ray from the two mirrors will be parallel to each other.

3.

28

An object is placed exactly midway between a concave mirror of radius of curvature 40 cm and a convex mirror of radius of curvature 30 cm. The mirrors face each other and are 50 cm apart. Determine the nature and position of the image formed by the successive reflections, first at the concave mirror and then at the convex mirror. DECEMBER 2009

Sol. The image formation is shown in figure.

The image will be virtual. This is formed at I1 behind the convex mirror at a distance of 10 cm. The image I1 acts as an object for convex mirror. (ii) For concave mirror, u2 = P2I1 = 60 + 10 = 70 cm, f2 = +15 cm and v 2 = ? 1 1 1 ∴ = + 15 70 v2 Solving we get, v2 = (210/11) cm. As v2 is positive, the image I2 is formed in front of concave mirror at a distance of (210/11) cm. Magnification m1 for first reflection v 10 1 = 1 = = u1 30 3 Magnification m2 for second reflection v ( 210 / 11) 3 = 2 = = u2 70 11

50cm

I2

P2 C

F

P1

25cm

r = 30 cm

r = 40 cm

I1 (i) For concave mirror, u1 = 25 cm, f1 = 20 cm and v1 = ? 1 1 1 Now = + f1 u1 v1

1 1 1 = + 20 25 v1 v1 = 100 cm. As v1 is positive, hence the image is real. In the absence of convex mirror, the rays after reflection from concave mirror would have formed a real image I1 at distance 100 cm from the mirror. Due to the presence of convex mirror, the rays are reflected and appear to come from I2. (ii) For convex mirror, In this case, I1 acts as virtual object and I2 is the virtual image. The distance of the virtual object from the convex mirror is 100 – 50 = 50 cm. Hence u2 = –50 cm. As focal length of convex mirror is negative and hence f2 = –30/2 = –15 cm. Here we shall calculate the value of v 2. Using the mirror formula, we have 1 1 1 − = − + 15 50 v 2 or v2 = –21.42 cm As v2 is negative, image is virtual. So image is formed behind the convex mirror at a distance of 21.43 cm. or

1 3 1 × = 3 11 11 1 5 ∴ Size of the image = 5 × = 11 11 5. An object is placed infront of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm it is found that there is no parallex between the images formed by the two mirrors. What is the radius of curvature of the convex mirror ? Sol. Let O be the object placed infront of a convex mirror MM' at a distance of 50 cm as shown in figure. The distance of the plane mirror NN' from the object is 30. We know that in a plane mirror the image is formed behind the mirror at the same distance as the object infront of it. It is also given that there is no parallax between the images formed by the two mirrors, i.e., the image is formed at a distance of 30 cm behind the plane mirror. For convex mirror, u = 50 cm, v = 10 cm [Q QP = QN – PN] as v is negative in convex mirror. M 50cm Final magnification = m1 × m2 =

4.

A convex and a concave mirror each 30 cm in radius are placed opposite to each other 60 cm apart on the same axis. An object 5 cm in height is placed midway between them. Find the position and size of the image formed by reflection, first at convex and then at the concave mirror. Sol. The image formation is shown in figure.

N Q

P

O

N' 20cm 30cm P2 I2

O

P1

M'

I1

1 1 1 = + , we have f u v 1 1 1 4 50 = − = − , ∴f= − f 50 10 50 4 50 × 2 Now v = 2f = − = –25 m 4 The radius of curvature of convex mirror is 25 cm. Using the mirror formula

r = 30 cm r = 30 cm (i) For convex mirror, u1 = +30 m, f1 = –15 cm and v 1 = ? 1 1 1 ∴− = + 15 50 v1 or v1 = –10 cm

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DECEMBER 2009

P HYSICS F U NDAMENTAL F OR IIT-J EE

Fluid Mechanics KEY CONCEPTS & PROBLEM SOLVING STRATEGY Hydrodynamics :

Hydrostatics : Pressure at a point inside a Liquid : p = p0 + ρ gh where p0 is the atmospheric pressure, ρ is the density of the liquid and h is the depth of the point below the free surface. p0 h

1 2 p v + gh + = a constant 2 ρ for a streamline flow of a fluid (liquid or gas). Here, v is the velocity of the fluid, h is its height above some horizontal level, p is the pressure and ρ is the density. p1 Bernoulli's Theorem :

p

v1

ρ

h1 v2 h2

Pressure is a Scalar : The unit of pressure may be atmosphere or cm of mercury. These are derived units. The absolute unit of pressure is Nm–2. Normal atmospheric pressure, i.e, 76 cm of mercury, is approximately equal to 10 5 Nm– 2. Thrust : Thrust = pressure × area. Thrust has the unit of force. Laws of liquid pressure (a) A liquid at rest exerts pressure equally in all directions. (b) Pressure at two points on the same horizontal line in a liquid at rest is the same. (c) Pressure exerted at a point in a confined liquid at rest is transmitted equally in all directions and acts normally on the wall of the containing vessel. This is called Pascal's law. A hydraulic press works on this principle of transmission of pressure. The principle of floating bodies (law of flotation) is that W = W´, that is, weight of body = weight of displaced liquid or buoyant force. The weight of the displaced liquid is also called buoyancy or upthrust. Hydrometers work on the principle of floating bodies. This principle may also be applied to gases (e.g., a balloon). Liquids and gases are together called fluids. The important difference between them is that liquids cannot be compressed, while gases can be compressed. Hence, the density of a liquid is the same everywhere and does not depend on its pressure. In the case of a gas, however, the density is proportional to the pressure.

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p2

v2 > v1

p2 < p1

According to this principle, the greater the velocity, the lower is the pressure in a fluid flow. It would be useful to remember that in liquid flow, the volume of liquid flowing past any point per second is the same for every point. Therefore, when the cross-section of the tube decreases, the velocity increases. Note : Density = relative density or specific gravity × 1000 kg m– 3. Surface tension and surface energy : Surface Tension : The property due to which a liquid surface tends to contract and occupy the minimum area is called the surface tension of the liquid. It is caused by forces of attraction between the molecules of the liquid. A molecule on the free surface of a liquid experiences a net resultant force which tends to draw it into the liquid. Surface tension is actually a manifestation of the forces experienced by the surface molecules. If an imaginary line is drawn on a liquid surface then the force acting per unit length of this line is defined as the surface tension. Its unit is, therefore, newton / metre. This force acts along the liquid surface. For curved surfaces, the force is tangent to the liquid surface at every point. Surface Energy : A liquid surface possesses potential energy due to surface tension. This energy per unit area of the surface is called the surface energy of the liquid. Its units is joule per square metre. The surface energy of a liquid has the same numerical values as the surface tension. The surface 30

DECEMBER 2009

tension of a liquid depends on temperature. It decreases with rise in temperature. Excess of Pressure : Inside a soap bubble or a gas bubble inside a liquid, there must be pressure in excess of the outside pressure to balance the tendency of the liquid surface to contract due to surface tension.

The upward force by which a liquid surface is pulled up in a capillary tube is 2πrTcos θ, and the downward force due to the gravitational pull on the mass of liquid in the tube is ( πr2h + v)ρ g, where v is the volume above the liquid meniscus. If θ = 0º, the meniscus is hemispherical in shape. Then v = difference between the volume of the cylinder of radius r and height r and the volume of the hemisphere of radius r

1 1 p(excess of pressure) = T + in general r1 r2

= πr3 –

where T is surface tension of the liquid, and r1 and r2 are the principal radii of curvature of the bubble in two mutually perpendicular directions. For a spherical soap bubble, r1 = r2 = r and there are two free surfaces of the liquid.

When θ ≠ 0, we cannot calculate v which is generally very small and so it may be neglected. For equilibrium (πr2h + v) ρ g = 2 πrT cos θ When a glass capillary tube is dipper in mercury, the meniscus is convex, since the angle of contact is obtuse. The surface tension forces now acquire a downward component, and the level of mercury inside the tube the falls below the level outside it. the relation 2T cos θ = hρgr may be used to obtain the fall in the mercury level. Problem Solving Strategy Bernoulli's Equations : Bernoulli's equation is derived from the work-energy theorem, so it is not surprising that much of the problem-solving strategy suggested in W.E.P. also applicable here. Step 1: Identify the relevant concepts : First ensure that the fluid flow is steady and that fluid is incompressible and has no internal friction. This case is an idealization, but it hold up surprisingly well for fluids flowing through sufficiently large pipes and for flows within bulk fluids (e.g., air flowing around an airplane or water flowing around a fish). Step 2: Set up the problem using the following steps Always begin by identifying clearly the points 1 and 2 referred to in Bernoulli's equation. Define your coordinate system, particular the level at which y = 0. Make lists of the unknown and known quantities 1 1 in Eq. p1 + ρ gy1 + ρ v12 = p2 + ρ gy2 + ρ v22 2 2 (Bernoulli's equation) The variables are p1, p2, v 1, v 2, y1 and y2, and the constants are ρ and g. Decide which unknowns are your target variables. Step 3: Execute the solutions as follows : Write Bernoulli's equation and solve for the unknowns. In some problems you will need to use the continuity equation, Eq. A1v1 = A2v 2 (continuity equation, incompressible fluid), to get a relation between the two speeds in terms of cross-sectional areas of pipes

4T ∴ p= r For a gas bubble inside a liquid, r1 = r2 = r and there is only one surface. 2T r For a cylindrical surface r1 = r and r2 = ∞ and there are two surfaces.

∴

p=

2T r Angle of Contact : The angle made by the surface of a liquid with the solid surface inside of a liquid at the point of contact is called the angle of contact. It is at this angle that the surface tension acts on the wall of the container.

∴

p=

The angle of contact θ depends on the natures of the liquid and solid in contact. If the liquid wets the solid (e.g., water and glass), the angle of contact is zero. In most cases, θ is acute (figure i). In the special case of mercury on glass, θ is obtuse (figure ii). θ

θ fig. (i)

fig. (ii)

Rise of Liquid in a Capillary Tube : In a thin (capacity) tube, the free surface of the liquid becomes curved. The forces of surface tension at the edges of the liquid surface then acquire a vertical component. meniscus θ T θ T θ

θ

h

r

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2 3 1 3 πr = πr 3 3

31

DECEMBER 2009

or containers. Or perhaps you will know both speeds and need to determine one of the areas. You may also dV need to use Eq. = Av (volume flow rate) to find dt the volume flow rate. Step 4: Evaluate your answer : As always, verify that the results make physical sense. Double-check that you have used consistent units. In SI units, pressure is in pascals, density in kilograms per cubic meter, and speed in meters per second. Also note that the pressures must be either all absolute pressure or all gauge pressures.

•1

H

p0 1 2 p 1 + v 1 + gH = + v 22 + g (h – x) ρ 2 ρ 2 p 1 = 0 + v 23 + 0 ρ 2 By continuity equation v 1A1 = A2v 2 = A2v 3 Since A1 >> A2,v1 is negligible and v2 = v 3 = n (say). p0 p 1 ∴ + gH = 2 + v2 + g (h – x) ρ ρ 2 p0 1 2 = + v ρ 2

A vertical U-tube of uniform cross-section contains mercury in both arms. A glycerine (relative density 1.3) column of length 10 cm is introduced into one of the arms. Oil of density 800 kg m–3 is poured into the other arm until the upper surface of the oil and glycerine are at the same horizontal level. Find the length of the oil column. Density of mercury is 13.6 × 103 kg m–3. Sol. Draw a horizontal line through the mercury-glycerine surface. This is a horizontal line in the same liquid at rest namely, mercury. Therefore, pressure at the points A and B must be the same.

∴

v = 2gH

(i)

p0 p + gH = 2 + gH + g (h – x) ρ ρ ⇒ p0 + p2 + ρ g (h – x) ⇒ p2 = p0 – ρ g (h – x) (ii) Thus pressure varies with distance from the upper end of the pipe according to equation (ii) and velocity is a constant and is given by (i). and

10 cm

h (1 – h)

h

•3

1.

3.

A rod of length 6m has a mass of 12 kg. It is hinged at one end at a distance of 3 m below the water surface. (i) What weight must be attached to the other end so that 5 m of the rod is submerged ? (ii) Find the magnitude and direction of the force exerted by the hinge on the rod. The specific gravity of the material of the rod is 0.5 Sol. Mass per unit length of the rod is 2 kg. Therefore, mass of the submerged portion of the rod is 10 kg and 10 3 its volume = m (using the simple formula, 500 mass volume = and density = specific gravity × density 1000 kg m– 3). B Fb W θ C O 3m O´ 12 kgf H A N

B

Pressure at B = p0 + 0.1 × (1.3 × 1000) × g Pressure at A = p0 + h × 800 × g + (0.1 – h) × 13.6 × 1000g ∴ p0 + 0.1 × 1300 × g = p0 + 800gh + 1360g – 13600 × g × h ⇒ 130 = 800h + 1360 – 13600h 1230 ⇒ h= = 0.096 m = 9.6 cm 12800 2.

A liquid flows out of a broad vessel through a narrow vertical pipe. How are the pressure and the velocity of the liquid in the pipe distributed when the height of the liquid level in the vessel is H from the lower end of the length of the pipe is h ? Sol. Let us consider three points 1, 2, 3 in the flow of water. The positions of the points are as shown in the figure. Applying Bernoulli's theorem to points 1, 2 and 3

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•2

2

Solved Examples

A

x

32

DECEMBER 2009

Therefore, buoyant force 10 Fb = × 1000 = 20 kgf 500 Let N and H be the vertical downward and horizontal reactions of the hinge on the rod. Considering horizontal and vertical translational equilibrium of the rod, N + 12 + W = 20 (where W is the weight to be attached and H = 0). N+ W=8 and H=0 ..(i) Considering the rotational equilibrium of the rod about A –20 × g × 2.5 cos θ + 12 × g × 3 cos θ + W × 6 cos θ = 0 ⇒ 6W = 50g – 36g = 14g 7 7 ⇒ W = g = kgf 3 3 7 17 Ν=8– = kgf 3 3

1 1 ∴ pressure difference = 2T cos θ − r1 r2 B

Let this pressure difference correspond to h units of the liquid. 1 1 Then 2T cos θ − = ρ gh r1 r2

4.

The end of a capillary tube with a radius r is immersed in water. Is mechanical energy conserved when the water rises in the tube ? The tube is sufficiently long. If not, calculate the energy change. Sol. In the equilibrium position (θ = 0º for pure water and glass) 2πrT cos 0º = πr2hρ g 2T or h= ρgr

• • • •

2πT 2 ρg Thus it is seen that the mechanical energy is not conserved. 4πT 2 2πT 2 ∴ mechanical energy loss = – ρg ρg

•

2πT 2 ρg This energy is converted into heat.

•

U=

• •

=

•

5.

Calculate the difference in water levels in two communicating tubes of diameter d = 1 mm and d = 1.5 mm. Surface tension of water = 0.07 Nm–1 and angle of contact between glass and water = 0º. 2T cos θ Sol. Pressure at A = p0 – r2

• • •

(Q pressure inside a curved surface is greater than that outside) 2T cos θ Pressure at B = p0 – r1

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⇒ h=

2T cos θ 1 1 − ρg r1 r2

∴ h=

2 × 0.07 1 1 − = 4.76 mm − 3 − 3 1000 × 9.8 1 ×10 1.5 ×10

Interesting Science Facts

4πT 2 Work done by surface tension = (2 πrT) × h = ρg U, potential energy of water in the tube = (πr2hρ)gh/2; it is multiple by h/2 because the cg of the water in the capillary tube is at a height h/2.

⇒

A

•

33

The dinosaurs became extinct before the Rockies or the Alps were formed. Female black widow spiders eat their males after mating. When a flea jumps, the rate of acceleration is 20 times that of the space shuttle during launch. The earliest wine makers lived in Egypt around 2300 BC. If our Sun were just inch in diameter, the nearest star would be 445 miles away. The Australian billy goat plum contains 100 times more vitamin C than an orange. Astronauts cannot belch - there is no gravity to separate liquid from gas in their stomachs. The air at the summit of Mount Everest, 29,029 feet is only a third as thick as the air at sea level. One million, million, million, million, millionth of a second after the Big Bang the Universe was the size of a …pea. DNA was first discovered in 1869 by Swiss Friedrich Mieschler. The molecular structure of DNA was first determined by Watson and Crick in 1953. The thermometer was invented in 1607 by Galileo. Englishman Roger Bacon invented the magnifying glass in 1250.

DECEMBER 2009

KEY CONCEPT

Organic Chemistry Fundamentals

CARBOXYLIC ACID

Acidity of carboxylic acids. Fatty acids are weak acids as compared to inorganic acids. The acidic character of fatty acids decreases with increase in molecular weight. Formic acid is the strongest of all fatty acids. The acidic character of carboxylic acids is due to resonance in the acidic group which imparts electron deficiency (positive charge) on the oxygen atom of the hydroxyl group (structure II). O– O R

C

O

R

H

I Non-equivalent structures

C

+

O

(1.27Å) which is nearly intermediate between C O and C—O bond length values. This proves resonance in carboxylate anion. O H

O+H

–

C O

Resonance hybrid of carboxylate ion

Due to equivalent resonating structures, resonance in carboxylate anion is more important than in the parent carboxylic acid. Hence carboxylate anion is more stabilised than the acid itself and hence the equilibrium of the ionisation of carboxylic acids to the right hand side. RCOOH RCOO– + H+ The existence of resonance in carboxylate ion is supported by bond lengths. For example, in formic acid, there is one C=O double bond (1.23 Å) and one C—O single bond (1.36Å), while in sodium formate both of the carbon-oxygen bond lengths are identical

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O

Alkoxide ion (No resonance)

Relative acidic character of carboxylic acids with common species not having —COOH group. RCOOH > Ar—OH > HOH > ROH > HC CH > NH3 > RH Effect of Substituents on acidity. The carboxylic acids are acidic in nature because of stabilisation (i.e., dispersal of negative charge) of carboxylate ion. So any factor which can enhance the dispersal of negative charge of the carboxylate ion will increase the acidity of the carboxylic acid and vice versa. Thus electron-withdrawing substitutents (like halogens, —NO2, —C6H5, etc.) would disperse the negative charge and hence stabilise the carboxylate ion and thus increase acidity of the parent acid. On the other hand, electron-releasing substituents would increase the negative charge, destabilise the carboxylate ion and thus decrease acidity of the parent acid.

Resonating forms of carboxylate ion (Equivalent structures) (Resonance more important)

R

Na+

Sodium formate

Alcohol (No resonance)

+

The positive charge (electron deficiency) on oxygen atom causes a displacement of electron pair of the O—H bond towards the oxygen atom with the result the hydrogen atom of the O—H group is eliminated as proton and a carboxylate ion is formed. Once the carboxylate ion is formed, it is stabilised by means of resonance. O O– R C R C O– O

O

OH

C

–

It is important to note that although carboxylic acids and alcohols both contain –OH group, the latter are not acidic in nature. It is due to the absence of resonance (factor responsible for acidic character of –COOH) in both the alcohols as well as in their corresponding ions (alkoxide ions). R—O—H R—O– + H+

H

C

H

Formic acid

II (Resonance less important) O– R

C

O

O X

C

–

O

The substituent X withdraws electrons, disperses negative charge, stabilises the ion and hence increases acidity –

O Y

C

O

The substituent Y releases electrons, intensifies negative charge, destabilises the ion and hence decreases acidity

34

DECEMBER 2009

Now, since alkyl groups are electron-releasing, their presence in the molecule will decrease the acidity. In general, greater the length of the alkyl chain, lower shall be the acidity of the acid. Thus, formic acid (HCOOH), having no alkyl group, is about 10 times stronger than acetic acid (CH3COOH) which in turn is stronger than propanoic acid (CH3CH2COOH) and so on. Similarly, following order is observed in chloro acids. Cl Cl Cl pKa

C

CO2H > Cl

C

Comparison of nucleophilic substitution (e.g., hydrolysis) in acid derivatives. Let us first study the mechanism of such reaction. O R

C

R C Nu + Z (where Z= —Cl, —OCOR, —OR, —NH2 and Nu = A nucleophile) Nucleophilic substitution in acid derivatives O O OH

R

H CO2H > H

C

δ–

Cl

δ+ Acid chloride

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δ+

R

R'

Nu

R

C

Nu

H+

R

C

Nu

R'

(where R' = H or alkyl group) Nucleophilic addition on aldehydes and ketones The (i) step is similar to that of nucleophilic addition in aldehydes and ketones and favoured by the presence of electron withdrawing group (which would stabilise the intermediate by developing negative charge) and hindered by electron-releasing group. The (ii) step (elimination of the leaving group Z) depends upon the ability of Z to accommodate electron pair, i.e., on the basicity of the leaving group. Weaker bases are good leaving groups, hence weaker a base, the more easily it is removed. Among the four leaving groups (Cl–, –OCOR, –OR, and –NH2) of the four acid derivatives, Cl– being the weakest base is eliminated most readily. The relative order of the basic nature of the four groups is – NH2 > –OR > –O.COR > Cl– Hence acid chlorides are most reactive and acid amides are the least reactive towards nucleophilic acyl substitution. Thus, the relative reactivity of acid derivatives (acyl compounds) towards nucleophilic substitution reactions is ROCl > RCO.O.COR > RCOOR > RCONH2 Acid Acid Esters Acid chlorides anhydrides amides OH– being stronger base than Cl–, carboxylic acids (RCOOH) undergo nucleophilic substitution (esterfication) less readily than acid chlorides.

δ–

R C

C

R'

CO2H

H H pKa 2.86 4.76 Decreasing order of aliphatic acids (i) O2NCH2COOH > FCH2COOH > ClCH2COOH > BrCH2COOH (ii) HCOOH > CH3COOH > (CH3)2 CHCOOH > (CH3)3CCOOH (iii) CH3CH2CCl2COOH > CH3CHCl.CHCl.COOH > ClCH2CHClCH2COOH (iv) F3CCOOH > Cl3CCOOH > Br3CCOOH Benzoic acid is somewhat stronger than simple aliphatic acids. Here the carboxylate group is attached to a more electronegative carbon (sp2 hybridised) than in aliphatic acids (sp3 hybridised). HCOOH > C6H5COOH > CH3COOH. Nucleophilic substitution at acyl carbon : It is important to note that nucleophilic substitution (e.g., hydrolysis, reaction with NH3, C 2H5OH, etc.) in acid derivatives (acid chlorides, anhydrides, esters and amides) takes place at acyl carbon atom (difference from nucleophilic substitution in alkyl halides where substitution takes place at alkyl carbon atom). Nucleophilic substitution in acyl halides is faster than in alkyl halides. This is due to the presence of > CO group in acid chlorides which facilitate the release of halogen as halide ion. O

Nu

C O

1.48

> Cl

R

(ii) Elimination step

CO2H

H

Z + Nu

Z

H

Cl 0.70

C

O (i) Addition step

δ–

Cl

Alkyl chloride

35

DECEMBER 2009

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KEY CONCEPT

Physical Chemistry Fundamentals

CHEMICAL KINETICS

The temperature dependence of reaction rates : The rate constants of most reactions increase as the temperature is raised. Many reactions in solution fall somewhere in the range spanned by the hydrolysis of methyl ethanoate (where the rate constant at 35ºC is 1.82 times that at 25ºC) and the hydrolysis of sucrose (where the factor is 4.13). (a) The Arrhenius parameters : It is found experimentally for many reactions that a plot of ln k against 1/T gives a straight line. This behaviour is normally expressed mathematically by introducing two parameters, one representing the intercept and the other the slope of the straight line, and writing the Arrhenius equaion.

behaviour is a signal that the reaction has a complex mechanism. The temperature dependence of some reactions is non-Arrhenius, in the sense that a straight line is not obtained when ln k is plotted against 1/T. However, it is still possible to define an activation energy at any temperature as

dln k Ea = RT2 .......(ii) dT This definition reduces to the earlier one (as the slope of a straight line) for a temperature-independent activation energy. However, the definition in eqn.(ii) is more general than eqn.(i), because it allows Ea to be obtained from the slope (at the temperature of interest) of a plot of ln k against 1/T even if the Arrhenius plot is not a straight line. Non-Arrhenius behaviour is sometimes a sign that quantum mechanical tunnelling is playing a significant role in the reaction. (b) The interpretation of the parameters : We shall regard the Arrhenius parameters as purely empirical quantities that enable us to discuss the variation of rate constants with temperature; however, it is useful to have an interpretation in mind and write eqn.(i) as

Ea ......(i) RT The parameter A, which corresponds to the intercept of the line at 1/T = 0(at infinite temperature, shown in figure), is called the pre-exponential factor or the 'frequency factor'. The parameter Ea, which is obtained from the slope of the line (–Ea/R), is called the activation energy. Collectively the two quantities are called the Arrhenius parameters. ln k = ln A –

ln A

k = Ae − Ea / RT .......(iii) To interpret Ea we consider how the molecular potential energy changes in the course of a chemical reaction that begins with a collision between molecules of A and molecules of B(shown in figure).

ln k

Slope = –Ea/R

Potential energy

1/T A plot of ln k against 1/T is a straight line when the reaction follows the behaviour described by the Arrhenius equation. The slope gives –E a/R and the intercept at 1/T = 0 gives ln A.

The fact that Ea is given by the slope of the plot of ln k against 1/T means that, the higher the activation energy, the stronger the temperature dependence of the rate constant (that is, the steeper the slope). A high activation energy signifies that the rate constant depends strongly on temperature. If a reaction has zero activation energy, its rate is independent of temperature. In some cases the activation energy is negative, which indicates that the rate decreases as the temperature is raised. We shall see that such

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Ea Reactants

Products Progress of reaction

A potential energy profile for an exothermic reaction. The height of the barrier between the reactants and products is the activation energy of the reaction

40

DECEMBER 2009

As the reaction event proceeds, A and B come into contact, distort, and begin to exchange or discard atoms. The reaction coordinate is the collection of motions, such as changes in interatomic distances and bond angles, that are directly involved in the formation of products from reactants. (The reaction coordinate is essentially a geometrical concept and quite distinct from the extent of reaction.) The potential energy rises to a maximum and the cluster of atoms that corresponds to the region close to the maximum is called the activated complex. After the maximum, the potential energy falls as the atoms rearrange in the cluster and reaches a value characteristic of the products. The climax of the reaction is at the peak of the potential energy, which corresponds to the activation energy Ea. Here two reactant molecules have come to such a degree of closeness and distortion that a small further distortion will send them in the direction of products. This crucial configuration is called the transition state of the reaction. Although some molecules entering the transition state might revert to reactants, if they pass through this configuration then it is inevitable that products will emerge from the encounter. We also conclude from the preceding discussion that, for a reaction involving the collision of two molecules, the activation energy is the minimum kinetic energy that reactants must have in order to form products. For example, in a gas-phase reaction there are numerous collisions each second, but only a tiny proportion are sufficiently energetic to lead to reaction. The fraction of collisions with a kinetic energy in excess of an energy Ea is given by

ratio of the two rates, and therefore of the two rate constants : [P2 ] k = 2 [ P1 ] k1 This ratio represents the kinetic control over the proportions of products, and is a common feature of the reactions encountered in organic chemistry where reactants are chosen that facilitate pathways favouring the formation of a desired product. If a reaction is allowed to reach equilibrium, then the proportion of products is determined by thermodynamic rather than kinetic considerations, and the ratio of concentration is controlled by considerations of the standard Gibbs energies of all the reactants and products. The kinetic isotope effect The postulation of a plausible mechanism requires careful analysis of many experiments designed to determine the fate of atoms during the formation of products. Observation of the kinetic isotope effect, a decrease in the rate of a chemical reaction upon replacement of one atom in a reactant by a heavier isotope, facilitates the identification of bond-breaking events in the rate-determining step. A primary kinetic isotope effect is observed when the ratedetermining step requires the scission of a bond involving the isotope. A secondary isotope effect is the reduction in reaction rate even though the bond involving the isotope is not broken to form product. In both cases, the effect arises from the change in activation energy that accompanies the replacement of an atom by a heavier isotope on account of changes in the zero-point vibrational energies. First, we consider the origin of the primary kinetic isotope effect in a reaction in which the ratedetermining step is the scission of a C–H bond. The reaction coordinate corresponds to the stretching of the C–H bond and the potential energy profile is shown in figure. On deuteration, the dominant change is the reduction of the zero-point energy of the bond (because the deuterium atom is heavier). The whole reaction profile is not lowered, however, because the relevant vibration in the activated complex has a very low force constant, so there is little zero-point energy associated with the reaction coordinate in either isotopomeric form of the activated complex.

the Boltzmann distribution as e −E a / RT . Hence, we can interpret the exponential factor in eqn(iii) as the fraction of collision that have enough kinetic energy to lead to reaction. The pre-exponential factor is a measure of the rate at which collisions occur irrespective of their energy. Hence, the product of A and the exponential factor,

Potential energy

e −E a / RT , gives the rate of successful collisions. Kinetic and thermodynamic control of reactions : In some cases reactants can give rise to a variety of products, as in nitrations of mono-substituted benzene, when various proportions of the ortho-, meta-, and para- substituted products are obtained, depending on the directing power of the original substituent. Suppose two products, P 1 and P2, are produced by the following competing reactions : A + B → P1 Rate of formation of P 1 = k1[A][B] A + B → P2 Rate of formation of P 2 = k2[A][B] The relative proportion in which the two products have been produced at a given state of the reaction (before it has reached equilibrium) is given by the

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C–H C–D

Ea(C–H) Ea (C–D)

Reaction coordinate

41

DECEMBER 2009

UNDERSTANDING

Inorganic Chemistry

1.

A black coloured compound (A) on reaction with dil. H2SO4 gives a gas (B) which on passing in a solution of an acid (C) gives a white turbidity (D). Gas (B) when passed through an acidified solution of a compound (E), gives ppt.(F) which is soluble in dilute nitric acid. After boiling this solution an excess of NH4OH is added, a blue coloured compound (G) is produced. To this solution, on addition of CH3COOH and aqueous K4[Fe(CN) 6], a chocolate ppt. (H) is produced. On addition of an aqueous solution of BaCl2 to aqueous solution of (E), a white ppt. insoluble in HNO3 is obtained. Identify compounds (A) to (H). Sol. From the data on compounds (G) and (H), it may be inferred that the compound (E) contains cupric ions (Cu2+), i.e., (E) is a salt of copper. Since the addition of BaCl2 to (E) gives a white ppt. insoluble in HNO3, it may be said that the anion in the salt is sulphate ion (SO42–). Hence, (E) is CuSO4. The gas (B) which is obtained by adding dil. H2SO4 to a black coloured compound (A), may be H2S since it can cause precipitation of Cu2+ ions in acidic medium. The black coloured compound (A) may be ferrous sulphide (iron pyrite). Hence, the given observation may be explained from the following equations. Fe S + H2 SO4 → FeSO4 + H2S (A) Dil. (B) H2S + 2HNO3 → 2NO2 + 2H2O + S (D) (C) White turbidity CuSO4 + H2S → CuS ↓ + H2SO4 (E) (B) (F) Black ppt. 3CuS + 8HNO3 → Dil. 3Cu(NO3)2 + 2NO + 3S + 4H2O ++ Cu + 4NH3 → [Cu(NH3) 4]2+ (G) Blue colour 2+ [Cu(NH3)4] + 4CH3COOH → Cu2+ + 4CH3COONH4 2+ 4– Cu + [Fe(CN)6] → Cu2[Fe(CN)6] (H) Chocolate colour CuSO4(aq) + BaCl2(aq) → BaSO4 ↓ + CuCl2 (E) White ppt. Insuluble in HNO3 Hence, (A) is FeS, (B) is H2S, (C) is HNO3, (D) is S, (E) is CuSO4, (F) is CuS, (G) is [Cu(NH3)4]SO4 and (H) is Cu2[Fe(CN)6]

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2.

An unknown inorganic compound (X) gives the following observations : (i) When added to CuSO4 solution it liberates iodine and a white ppt. (Y) is formed. The liberated I2 reacts with Na 2S2O3 solution to give NaI and a colourless compound (Z). (ii) When CHCl3 and Cl2 water is added to aqueous solution of (X), a violet layer of chloroform is formed. (iii)(X) gives a violet colour flame when heated in Bunsen burner flame. (iv)When aqueous solution of (X) is added to aqueous lead nitrate, a yellow ppt. (M) is formed. (v) Addition of (X) to HgCl2 gives a red ppt. which dissolves in excess of (X) to give Nessler's reagent. (vi) On heating (X) with dil. HCl and KNO2, violet vapours of a compound (N) are formed which condenses on the wall of test tube. What are (X), (Y), (Z), (M) and (N) ? Explain the reactions. Sol. Observation (iii) indicates that the compound (X) contains K+, it is because it gives a violet coloured flame. On the other hand set (ii) confirmed that (X) contains I– ions, thus (X) is KI. Now the different reactions may be formulated as follows. 2KI + Cl2 CHCl 3 → I2 + 2KCl Violet layer (i) When CuSO4 reacts with KI, I2 is liberated and a ppt. of cuprous iodide (Y) is formed. [CuSO4 + 2KI → K2SO4 + CuI2] × 2 2CuI2 → Cu2I2 + I2 Unstable On adding, 2CuSO4 + 4KI → 2K2SO4 + Cu2I2 + I2 (Y) white ppt. The librated iodine is titrated against standard hypo solution when a colourless sodium tetrathionate (Z) is formed. 2Na2S2O3 + I2 → Na2S4O6 + 2NaI (Z) (colourless) (iv) Pb(NO3)2 + 2KI → PbI2 + 2KNO3 (M) yellow ppt (v) HgCl2 + 2KI → HgI2 ↓ + 2KCl Red ppt. HgI2 + 2KI → K2HgI4 (Soluble) K2HgI4 + NaOH → Nessler's reagent (vi) HCl + KNO2 → HNO2 + KCl HCl + KI → KCI + HI 42

DECEMBER 2009

2HNO2 + 2HI → I2 + 2H2O + 2NO (N) Hence, (X) is KI, (Y) is Cu2I2, (Z) is Na2S4O6, (M) is PbI2 and (N) is I2.

∆ 2Bi + 6HCl → 2BiCl3 + 3H2 (G) (A) Hence, (A) is BiCl3, (B) is Bi2S3, (C) is H2S, (D) is Bi(NO3)2, (E) is Bi(OH)3, (F) is BiOCl and (G) is Bi

3.

An inorganic halide (A) gives the following reactions. (i) The cation of (A) on raction with H2S in HCl medium, gives a black ppt. of (B). (A) neither gives ppt. with HCl nor blue colour with K4Fe(CN) 6. (ii) (B) on heating with dil.HCl gives back compound(A) and a gas (C) which gives a black ppt. with lead acetate solution. (iii) The anion of (A) gives chromyl chloride test. (iv) (B) dissolves in hot dil. HNO3 to give a solution, (D). (D) gives ring test. (v) When NH4OH solution is added to (D), a white precipitate (E) is formed. (E) dissolves in minimum amount of dil. HCl to give a solution of (A). Aqueous solution of (A) on addition of water gives a whitish turbidity (F). (vi) Aqueous solution of (A) on warming with alkaline sodium stannite gives a black precipitate of a metal (G) and sodium stannate. The metal (G) dissolves in hydrochloride acid to give solution of (A). Identify (A) to (G) and give balanced chemical equations of reactions. Sol. Observation of (i) indicates that cation (A) is Bi3+ because it does not give ppt. with HCl nor blue colour with K4Fe(CN)6, hence it is neither Pb2+ nor Cu2+. Since anion of (A) gives chromyl chloride test, hence it contains Cl– ions. Thus, (A) is BiCl3. Its different reactions are given below : (i) 2BiCl3 + 3H2S → Bi2S3 + 6HCl (A) (B) (ii) Bi2S3 + 6HCl → 3H2S + 2 BiCl3 (B) (C) (A)

4.

(i) An inorganic compound (A) is formed on passing a gas (B) through a concentrated liquor containing Na2S and Na 2SO3. (ii) On adding (A) into a dilute solution of AgNO3, a white ppt. appears which quickly changes into black coloured compound (C). (iii) On adding two or three drops of FeCl3 into excess of solution of (A), a violet coloured compound (D) is formed. This colour disappears quickly. (iv) On adding a solution of (A) into the solution of CuCl2, a white ppt. is first formed which dissolves on adding excess of (A) forming a compound (E). Identify (A) to (E) and give chemical equations for the reactions at steps (i) to (iv) Sol. (i) The compound (A) appears to be Na2S 2O3 from its method of preparation given in the problem. Na2S + Na2SO3 + I2 → 2NaI + Na2S 2O3 (B) (A) or Na2SO3 + 3Na 2S + 3SO2 → 3Na 2S2O3 (B) (A) (ii) White ppt. of Ag2S2O3 is formed which is hydrolysed to black Ag2S Na2S2O3 + 2AgNO3 → 2NaNO3 + Ag2S 2O3 ↓ White ppt Ag2S 2O3 + H2O → Ag2S + H2SO4 (C) (iii) A violet ferric salt is formed. 3Na2S2O3 + 2FeCl3 → Fe2(S 2O3)3 + 6NaCl (D)(violet) (iv) 2CuCl2 + 2Na2S2O3 → 2CuCl + Na2S4O6 + 2NaCl White ppt. 2CuCl + Na2S 2O3 → Cu2S2O3 + 2NaCl 3Cu2S2O3 + 2Na 2S2O3 → Na 4[Cu6(S2O3) 5] (E) or 6CuCl + 5Na2S2O3 → Na4[Cu6(S2O3)5] + 6NaCl (E) Hence, (A) is Na2S2O3, (B) is I2 or SO2, (C) is Ag2S, (D) is Fe2(S2O3)3 and (E) is Na4[Cu6(S2O3) 5]

∆ (iii) Bi2S3 + 8HNO3 → 2Bi(NO3)3 + 2NO (B) (D) + 3S + 4H2O (iv) Bi(N O3)3 + 3NH4OH → (D) Bi(OH)3 ↓ + 3NH4NO3 (E) White ppt. ∆ Bi(OH)3 + 3HCl → BiCl3 + 3H2O Dil. (A) BiCl3 + H2O → BiOCl + 2HCl (A) (F) Bismuth oxychloride (White turbidity) (v) BiCl3 + 2NaOH +Na2[SnO2] → (A) Bi + NaSnO3 + H2O + 3NaCl (G) Black ppt.

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5.

43

A colourless solid (A) on heating gives a white solid (B) and a colourless gas (C). (B) gives off reddishbrown fumes on treating with H2SO4. On treating with NH4Cl, (B) gives a colourless gas (D) and a DECEMBER 2009

residue (E). The compound (A) on heating with (NH4)2SO4 gives a colourless gas (F) and white residue (G). Both (E) and (G) impart bright yellow colour to Bunsen flame. The gas (C) forms white powder with strongly heated Mg metal which on hydrolysis produces Mg(OH) 2. The gas (D) on heating with Ca gives a compound which on hydrolysis produces NH3. Identify compounds (A) to (G) giving chemical equations involved. Sol. The given information is as follows :

TRUE OR FALSE

(i) A Heat → B + C Colourless Solid Colourless Solid gas ∆ (ii) B + H2SO4 → Reddish brown gas ∆ (iii) B + NH4Cl → D + E Colourless gas ∆ (iv) A + (NH4)2SO4 → F + G olourless gas White Residue (v) E and G imparts yellow colour to the flame.

(vi) C + Mg Heat → White powder 2O H → Mg(OH)2 2O (vii) D + Ca Heat → Compound H → NH3 Information of (v) indicates that (E) and (G) and also (A) are the salts of sodium because Na+ ions give yellow coloured flame. Observations of (ii) indicate that the anion associated with Na+ in (A) may be NO3–. Thus, the compound (A) is NaNO3. The reactions involved are as follows :

1.

The magnitude of charge on one gram of electrons is 1.60 × 10–19 coulomb.

2.

Chromyl chloride test of Cl– radical is not given by HgCl2.

3.

The energy levels in a hydrogen atom can be compared with the steps of a ladder placed at equal distance.

4.

In SN1 mechanism, the leaving group in the molecule, leaves the molecule, well before joining of an attacking group.

5.

Metamerism is special type of isomerism where isomers exist simultaneously in dynamic equilibrium.

6.

Addition of HCN with formaldehyde is an example of electrophilic addition reaction.

7.

Ligroin is essentially petroleum ether containing aliphatic hydrocarbons and is generally used in dry cleaning clothes.

Sol.

∆ (i) 2NaNO3 → 2NaNO2 + O2 ↑ (A) (B) (C) (ii) 2NaNO2 + H2SO4 → Na2SO4 + 2HNO2 (B) Dil. 3HNO2 → HNO3 + H2O + 2NO↑ 2NO + O2 → 2NO2 ↑ Reddish brown Fumes (iii) NaNO2 + NH4Cl → NaCl + N2 ↑ + 2H2O (B) (E) (D)

1.

[False] Thomson through his experiment determined the charge to mass ratio of an electron and the value of e/m is equal to 1.76 × 108 coulomb/gm. Hence one gm of electrons have charge 1.76 × 10 8 C.

2.

[True]

3.

[False]

4.

[True] SN1 reaction mechanism takes place in two steps as : R—X Slow → R+ + X– R+ + OH– Fast → ROH

5.

∆ (iv) 2NaNO3 + (NH4)2SO4 → Na2SO4 + 2NH3 (A) (G) (F) 2HNO3

For example :

∆ 2O (v) O2 + 2Mg → 2MgO H → Mg(OH)2 (C)

CH3CH2–O–CH2CH3 and CH3–OCH2CH2CH3

∆

Conditions mentioned in the statement are associated with phenomenon of trautomerism.

(vi) N2 + 3Ca → Ca3N2 (D) Ca3N2 + 6H2O → 3Ca(OH)2 + 2NH3 ↑ Hence, (A) is NaNO3, (B) is NaNO2, (C) is O2, (D) is N2, (E) is NaCl, (F) is NH3 and (G) is Na2SO4.

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[False] In metamerism isomers differ in structure due to difference in distribution of carbon atoms about the functional group.

6.

[False] H H – C = O + H+ CN–

H H – C = OH CN

7. 44

[True] DECEMBER 2009

Set

8

`tà{xÅtà|vtÄ V{tÄÄxÇzxá

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 Sol ut i ons w il l be publ i s he d i n ne xt is s ue Joint Director Academics, Career Point, Kota 1.

Show that the six planes through the middle point of each edge of a tetrahedron perpendicular to the opposite edge meet in a point.

2.

Prove that if the graph of the function y = f (x), defined throughout the number scale, is symmetrical about two lines x = a and x = b, (a < b), then this function is a periodic one.

3.

Show that an equilateral triangle is a triangle of maximum area for a given perimeter and a triangle of minimum perimeter for a given area.

4.

Let az2 + bz + c be a polynomial with complex coefficients such that a and b are non zero. Prove that the zeros of this polynomial lie in the region b c |z|≤ + a b

5.

9.

10. ABC is a triangle inscribed in a circle. Two of its sides are parallel to two given straight lines. Show that the locus of foot of the perpendicular from the centre of the circle on to the third side is also a circle, concentric to the given circle.

Dimensional Formulae of Some Physical Quantities

x2

+

y2

= 1 is

a2 b2 circumscribed with all the three sides touching the ellipse. Find the least possible area of the triangle. 6.

7.

2

If one of the straight lines given by the equation ax + 2hxy + by2 = 0 coincides with one of those given by a′x2 + 2h′ xy + b′ y2 = 0 and the other lines represented ha´b´ h´ab´ by them be perpendicular, show that = b´− a´ b−a Prove that n m n m + 1 n m + 2 + + + ....... to 0 n 1 n 2 n (n + 1) terms n m n m n m = + 2 + 22 + ..... to (n + 0 0 1 1 2 2 1) terms 1

8.

∫

Torque (τ)

[ML2T–2]

Moment of Inertia (I) Coefficient of viscosity (η) Gravitational constant (G)

[ML2] [ML–1T–1]

Specific heat (S)

[L2T–2θ–1]

Coeficient of thermal conductivity (K)

[MLT–3θ–1]

Universal gas constant (R) Potential (V)

[ML2T–2θ–1] [ML2T–3A–1]

Intensity of electric field (E)

[MLT–3A–1]

Permittivity of free space (ε0)

[M–1L–3T4A2]

Specific resistance (ρ) Magnetic Induction (B)

[ML3T–3A2]

Planck's constant (h)

[ML2T–1]

Boltzmann's constant (k)

[ML2T–2θ–1]

Entropy (S)

[ML2T–2θ–1] [T–1]

Decay constant (λ ) Bohr magnetic (µB)

If n ≥ 2 and In = (1 − x 2 ) n cos mx dx, then show that −1

m 2I n = 2n(2n – 1) I n–1 – 4n(n – 1) I n–2.

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Work (W) Stress

Dime nsional Formulae [ML2T–2] [ML–1T–2]

Physical Quantity

An isosceles triangle with its base parallel to the major axis of the ellipse

Find the sum to infinite terms of the series 3 5 7 9 11 + + + + + ........ ∞ 4 36 144 400 900

Thermmionic current density (J) 45

[M–1L3T–2]

[MT–2A–1]

[L2A] [AL–2] DECEMBER 2009

MATHEMATICAL CHALLENGES SOLUTION FOR NOVEMBER ISSUE (SET # 7)

1.

Let the line be y = 2x + c

9 − c 9 + 2c Point A , 3 6

=

AD DC + BD . BC BD.CD

2c − 3 + c − 6 Point B , −3 −3

=

AD AD AD 1 = = . 2 BD.CD AD AD AD

so it is vector along AB with magnitude

c + 6 5c + 12 Point C , 3 3 mid point of B & C is

1 2

|a+b |=

2c − 3 c + 6 + . , 3 −3

3.

1 − c + 6 5c + 12 9 − c 2c + 9 + = , 2 + 3 3 6 3

1 AD

The line PQ always passes through (α, β) so it is y –β = m(x – α) Let the circle be x2 + y2 – 2hx – 2ky = 0

which is point A, so AB and AC are equal.

Joint equation of OP and OQ. x2 + y2 – 2 (hx + ky)

2. A

( y − mx ) =0 β − mα P

b

a

O B

=

1 AB

2

1 AB2

Q

AB +

1 AC2

h − mk 2k 2 1 − y – 2 xy + β − mα β − mα

AC

(AD + DB) +

1 AC2

2hn 2 1 + x = 0 β − mα

It must represent y2 – x2 = 0 so

(AD + DC)

1 DB DC 1 + = + AD + 2 2 BD.DC CD.CB AC AB

and

h − mk = 0 ⇒ m = h/k β − mα 1–

...(1)

2k 2hm = –1 – β − mα β − mα

⇒ β – mα – 2k = –β + mα – 2hm

DB DC 1 1 1 + = AD + BD + CD BC AB2 AC 2

⇒ –β + mα + k – hm = 0 ⇒ –β + k + h/k(α – h) = 0

1 1 + = AD . BD.DC CD.CB =

(h,k)

C

D

1 AB 1 AC a+b = . + . AB AB AC AC =

1 . AD

(using (1) in it)

⇒ k2 – βy + αh – h2 = 0 so required locus is x2 – y2 – α x + β y = 0

AD 1 1 + . CD BD CD

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46

DECEMBER 2009

4.

π π As |f(x)| ≤ |tan x| for ∀ x ∈ − , 2 2

(–1,log23)

so f(0) = 0 so |f(x) – f(0)| ≤ |tan x| divides both sides by |x|

⇒ lim

x →0

f ( x ) − f (0) tan x ≤ lim x →0 x x

1

=

ai

∑i

∫

0

log 2 (2 − x ) dx +

−1

= – log2

≤1

∫ (2 − 2

y

) dy +

−1

= log 2 3 −

1 1 1 ⇒ a 1 + a 2 + a 3 + ..... + a n ≤ 1 2 3 n

⇒

(3/2,–1)

(0,–1)

⇒ |f´(0)| ≤ 1

n

(1,0)

(–1,0)

f ( x ) − f (0) tan x ≤ ⇒ x x

1 π 4

2 1 π + 2 log 2 3 + 2 – + ln 2 2ln 2 4

e2 e π +2+ sq units 27 4

i =1

7. 5.

A

Let the number is xyz, here x < y and z < y. Let y = n, then x can be filled in (n – 1) ways. (i.e. from 1 to (n – 1)) and z can be filled in n ways (i.e. from 0 to (n – 1))

F

here 2 ≤ n ≤ 9

B

so total no. of 3 digit numbers with largest middle digit =

=

9

9

n =2

n =2

∑ n(n −1) = ∑ n – ∑ n

so tan A =

n =2

9.10.19 9.10 – 6 2

so

= 285 – 45 = 240 240 required probability = 9 × 10 ×10

so

a2 r12 a2 r12

BD BC BC a = = = MD 2MD 4 r1 4r1

= tan2A +

b2 r22

+

c2 r32

= 16 (tan2A + tan2B + tan2C)

8 = 30 =

C

D

∠BMC = 2∠BAC = 2∠BMD

9

2

E M

...(1)

Now as tan A + tan B + tan C ≥ 3 (tan A . tan B . tan C)1/3 and for a triangle tan A + tan B + tan C = tan A . tan B . tan C

4 15

so (tan A . tan B . tan C)2/3 ≥ 3 6.

The region bounded by the curve y = log2(2 – x) and

⇒ tan A . tan B . tan C ≥ 3 3

the inequality (x – |x|)2 + (y – |y|)2 ≤ 4 is required area

⇒ tan2A + tan2B + tan2C ≥ 3(tan A. tan B tan C)2/3 ≥ 3.3

is

so from (1),

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47

a2 r12

+

b2 r22

+

c2 r32

≥ 144. DECEMBER 2009

8.

= a + (1 – r2)

Z1, Z2, Z3 are centroids of equilateral triangles ACX, ABY and BCZ respectively.

∫ (1+ r

Z1 − Z A iπ/6 e ZC − ZA

Z1 – ZA = (ZC – ZA)

a

=a+ C

B Z3

1 3 i − 3 2 2

...(1)

...(2)

Now and

–

Tr = 2

=

∫ 0

=

2

−1 1 + r tan t 1− r

tan a / 2

0

lim Tr = a –

2(1 + r )(r − 1) π = a–π (1 + r )(r − 1) 2

lim+ Tr = a +

2(1 − r)( r + 1) π = a+π (1 + r )(r − 1) 2

r→1

∫ du = a r→1

so x 2 + ax 2 + bx + c = (x – γ) (x 2 + (a + γ) x + (γ 2 + aγ + b)) where – γ (γ2 + aγ + b) = c, as γ is the root of given equation, so x2 + (a + γ) x + (γ2 + aγ + b) = 0 must have two roots i.e. α and β. So its discriminant is non negative, thus (γ + a)2 – 4(γ2 + aγ + b) ≥ 0

1 − r cos u

2

du.

1 − 2r cos u + r 2 − r 2 + 1 1 − 2r cos u + r 2

...(1)

3γ2 + 2aγ – a2 + 4b ≤ 0 so γ ≤

du

− a + 2 a 2 − 3b 3

so greatest root is also less than or equal to

1− r 1 + 1 − 2r cos u + r 2 du 0

XtraEdge for IIT-JEE

1− r 1+ r

10. Let α, β, γ be the three real roots of the equation without loss of generality, it can be assumed that α ≤ β ≤ γ.

1 3 (ZA + ZC – 2ZB)) + i 2 2 2 3

∫

t2 +

0

difference π.

i

a

2 dt

∫

2

r→1+

r→1

0

a

(1 + r)

tan a / 2

Hence lim+ Tr, T1, lim− Tr form an A.P. with common

1 (ZC – ZA) 2

∫ 1 − 2r cos u + r

(1 − r) 2 (1 + r) 2

0

(ZA + ZC – 2ZB) ..(4)

= Z1 – Z2

9.

tan 2 u / 2 +

a

1 i = (ZC – ZB) + (ZC + ZB – 2ZA) 2 2 3

a

1− r 2

and (from (1)) T1 =

To prove ∆xyz as equilateral triangle, we prove that (Z3 – Z2)eiπ/3 = Z1 – Z2 So, (Z3 – Z2)eiπ/ 3 = (

0

tan 2 u / 2 + (1 − r ) 2

sec2 u / 2 du

2(1 − r 2 ) 1 + r =a+ (1 + r) 2 1 − r

1 (ZA – ZC) 2 2 3

∫

(1 + r) 2

so, Tr = a +

...(3)

i

a

1− r 2

ZC

1 i (ZC – ZB) + (ZC + ZB – 2ZA) 2 2 3

+

2

Let tan u/2 = t

1 3 i + Z1 – ZA = (ZC – ZA) 3 2 2 similarly,

similarly Z2 – Z3 =

∫ (1+ r )

sec 2 u / 2

0

z

So, Z1 – Z2 =

)(1 + tan 2 u / 2) − 2r(1 − tan 2 u / 2)

Z1

A

Z2 – ZA = (ZB – ZA)

2

= a + (1 – r2)

Z2 ZB

sec 2 u / 2

0

x

ZA y

a

2

− a + 2 a 2 − 3b . 3 48

DECEMBER 2009

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

MATHS 1.

Suppose f(x) = x3 + ax2 + bx + c, where a, b, c are

3.

chosen respectively by throwing a die three times.

4) = g(x + 2) + g(x + 6), then prove that

Find the probability that f(x) is an increasing

∫

x +8 x

g (t ) dt

is a constant function.

function. Sol. f´(x) = 3x2 + 2ax + b

Sol. given that g(x) + g(x + 4) = g(x + 2) + g(x + 6) ...(1) putting x = x + 2 in (1) ........

y = f(x) is strictly increasing

⇒ f´(x) > 0 ∀ x

g(x + 2) + g(x + 6) = g(x + 4) + g(x + 8)

...(2)

from (1) & (2)

⇒ (2a)2 – 4.3.b < 0

g(x) = g(x + B)

This is true for exactly 15 ordered pairs (a, b); 1 ≤ a, 15 5 b ≤ 6, so probability = = 36 12 2.

Let g be a real valued function satisfying g(x) + g(x +

Now,

f(x) =

∫

x +8 x

g (t ) dt

f´(x) = g(x + 8) – g(x) = 0

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

⇒ g is constant function

methods. 4.

→

Sol. Let A = a ˆi + b ˆj + c kˆ

If exactly three distinct chords from (h, 0) point to the circle x2 + y2 = a2 are bisected by the parabola y2 = 4ax, a > 0, then find the range of 'h' parameter.

→

⇒ B = 3 ˆi + 2 ˆj + kˆ → →

→

Sol. Let M(at2, 2at) is mid-point of chord AB, then chord AB = T = S1

→

⇒ ( A . B) 2 ≤ | A |2 | B |2 a 2 + b2 + c 2

3a + 2b + c ≤

B

14

M A

(7) ≤ (a + b + c ) (14) 2

2

2

2

{Q 3a + 2b + c = 7, point lies on the plane} a2 + b2 + c 2 ≥

49 7 = 14 2 AB :

x.at2 + y.2at = a2t4 + 4a2t2

since AB chord passes through (h, 0)

XtraEdge for IIT-JEE

49

DECEMBER 2009

so, h.at2 = a 2t4 + 4a2t2

Sol. Let P be (x1, y1),

at2 [at2 + (4a – h)] = 0

Q

A P

⇒ 4a – h < 0

If a > 0

⇒ h > 4a

...(i) S

Now point (at2,2at) must lie inside the circle, on solving B

we get, h < a ( 5 + 2)

...(ii)

Points of intersection of tangent and normal at P

from (i) & (ii)

b2 a2y points with y-axis and A 0, , B 0, y1 − 2 1 , b y1

4a < h < a ( 5 + 2) 5.

S : (ae, 0) slope (SA). slope (SB) = –1

Find the sum of the terms of G.P. a + ar + ar 2 + ..... + ∞

⇒ ∠ASB = 90º (PA and PB are tangent and normal) P must lie on the circle with AB as diameter. Hence the point of intersection of the ellipse and the

where a is the value of x for which the function 7 + 2x loge25 – 5x – 1 – 52– x has the greatest value and r is the Lt

∫

x →0 0

Sol. S =

t 2dt

x

circle is P. Due to symmetry the angles made by AB

x 2 tan(π + x )

at P,Q,M, N are all 90º.

a , 1− r

Do you know

To get the greatest value f´(x) = 2log e25 – 5 x – 1 log 5 + 52– x log 5

•

100 years ago: The first virus was found in both plants and animals.

•

90 years ago: The Grand Canyon became a national monument & Cellophane is invented.

•

80 years ago: The food mixer and the domestic refrigerator were invented.

•

70 years ago: The teletype and PVC (polyvinylchloride) were invented.

1 π

•

60 years ago: Otto Hahn discovered nuclear fission by splitting uranium, Teflon was invented.

1 2π ⇒ sum of G.P. = π π −1

• •

50 years ago: Velcro was invented.

•

30 years ago: The computer mouse was invented.

= 1 meets the y-axis at A and B

•

respectively. Find the angle subtended by AB at the

20 years ago: First test-tube baby born in England, Pluto’s moon, Charon, discovered.

•

10 years ago: First patent for a geneticallyengineered mouse was issued to Harvard Medical School.

f´(x) = 4 loge5 – 5

x– 1

loge5 + 5.5

1– x

loge5

⇒ f´(x) = 0 put 5 x – 1 t(> 0) t2 – 4t – 5 = 0 ⇒ t = 5 ⇒ 5 x –1 = 5 ⇒ x = 2 to evaluate r : r = Lt

∫

t 2dt

x

x →0 0

2

x tan(π + x )

since a = 2, r = 6.

M

N

a2t4 + 4a2t2 – a 2 < 0

=

The tangent and normal at a point P on the ellipse x2 a2

+

y2 b2

points of intersection of the circle (through A,S,B) and the ellipse. S being one of the foci.

XtraEdge for IIT-JEE

50

40 years ago: An all-female population of lizards was discovered in Armenia.

DECEMBER 2009

MATH

MONOTONICITY, MAXIMA & MINIMA Mathematics Fundamentals

Monotonic Functions : A function f(x) defined in a domain D is said to be (i) Monotonic increasing :

according as f(x) is monotonic increasing or decreasing at x = a. So at x = a, function f(x) is monotonic increasing ⇔ f´(a) > 0

x < x 2 ⇒ f (x 1 ) ≤ f (x 2 ) ⇔ 1 ∀ x1, x2 ∈ D x 1 > x 2 ⇒ f (x 1 ) ≥ f (x 2 ) y y

O

x

O

monotonic decreasing ⇔ f´(a) < 0 (ii) In an interval : In [a, b], f(x) is monotonic increasing ⇔ f´(x) ≥ 0 monotonic decreasing ⇔ f´(x) ≤ 0 ∀ x ∈ (a, b) constant ⇔ f´(x) = 0 x

Note : (i) In above results f´(x) should not be zero for all values of x, otherwise f(x) will be a constant function. (ii) If in [a, b], f´(x) < 0 at least for one value of x and f´(x) > 0 for at least one value of x, then f(x) will not be monotonic in [a, b]. Examples of monotonic function : If a functions is monotonic increasing (decreasing ) at every point of its domain, then it is said to be monotonic increasing (decreasing) function. In the following table we have example of some monotonic/not monotonic functions Monotonic Monotonic Not increasing decreasing monotonic x3 1/x, x > 0 x2 x|x| 1 – 2x |x| x –x e e ex + e– x log x log2x sin x sin h x cosec h x, x > 0 cos h x [x] cot hx, x > 0 sec h x

x < x 2 ⇒ f (x 1 ) >/ f ( x 2 ) i.e., ⇔ 1 ∀ x1, x2 ∈ D x 1 > x 2 ⇒ f (x 1 ) x 2 ⇒ f (x 1 ) ≤ f ( x 2 ) y

y

O

x

O

x

x < x 2 ⇒ f (x 1 ) x 2 ⇒ f (x 1 ) /> f ( x 2 ) A function is said to be monotonic function in a domain if it is either monotonic increasing or monotonic decreasing in that domain. Note : If x1 < x2 ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ D, then f(x) is called strictly increasing in domain D and similarly decreasing in D. Method of testing monotonicity : (i) At a point : A function f(x) is said to be monotonic increasing (decreasing) at a point x = a of its domain if it is monotonic increasing (decreasing) in the interval (a – h, a + h) where h is a small positive number. Hence we may observer that if f(x) is monotonic increasing at x = a then at this point tangent to its graph will make an acute angle with xaxis where as if the function is monotonic decreasing there then tangent will make an obtuse angle with xaxis. Consequently f´(a) will be positive or negative

XtraEdge for IIT-JEE

Properties of monotonic functions : If f(x) is strictly increasing in some interval, then in that interval, f–1 exists and that is also strictly increasing function. If f(x) is continuous in [a, b] and differentiable in (a, b), then f´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f(x) is monotonic increasing in [a, b] f´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f(x) is monotonic decreasing in [a, b] 51

DECEMBER 2009

If both f(x) and g(x) are increasing (or decreasing) in [a, b] and gof is defined in [a, b], then gof is increasing. If f(x) and g(x) are two monotonic functions in [a, b] such that one is increasing and other is decreasing then gof, it is defined, is decreasing function. Maximum and Minimum Points : The value of a function f(x) is said to be maximum at x = a if there exists a small positive number δ such that f(a) > f(x) y

O

( ) a

( ) b

( ) c

where x = c is a point such that f´(c) = 0. If a continuous function has only one maximum (minimum) point, then at this point function has its greatest (least) value. Monotonic functions do not have extreme points. Conditions for maxima and minima of a function Necessary condition : A point x = a is an extreme point of a function f(x) if f´(a) = 0, provided f´(a) exists. Thus if f´(a) exists, then x = a is an extreme point ⇒ f´(a) = 0

f´(a) ≠ 0 ⇒ x = a is not an extreme point But its converse is not true i.e. f´(a) = 0 ⇒ / x = a is an extreme point. For example if f(x) = x3, then f´(0) = 0 but x = 0 is not an extreme point. Sufficient condition : For a given function f(x), a point x = a is a maximum point if f´(a) = 0 and f´´(a) < 0 a minimum point if f´(a) = 0 and f´´(a) > 0 not an extreme point if f´(a) = 0 = f´´(a) and f´´´(a) ≠ 0. Note : If f´(a) = 0, f´´(a) = 0, f´´´(a) = 0 then the sign of f(4)(a) will determine the maximum or minimum point as above. Working Method : Find f´(x) and f´´(x). Solve f´(x) = 0. Let its roots be a, b, c, ... Determine the sign of f´´(x) at x = a, b, c, .... and decide the nature of the point as mentioned above. Properties of maxima and minima : If f(x) is continuous function, then Between two equal values of f(x), there lie atleast one maxima or minima. Maxima and minima occur alternately. For example if x = –1, 2, 5 are extreme points of a continuous function and if x = –1 is a maximum point then x = 2 will be a minimum point and x = 5 will be a maximum point. When x passes a maximum point, the sign of dy/dx changes from + ve to – ve, where as when x passes through a minimum point, the sign of f´(x) changes from –ve to + ve. If there is no change in the sign of dy/dx on two sides of a point, then such a point is not an extreme point. If f(x) is maximum (minimum) at a point x = a, then 1/f(x), [f(x) ≠ 0] will be minimum (maximum) at that point. If f(x) is maximum (minimum) at a point x = a, then for any λ ∈ R, λ + f(x), log f(x) and for any k > 0, k f(x), [f(x)]k are also maxmimum (minimum) at that point.

x

Also then the point x = a is called a maximum point for the function f(x). Similarly the value of f(x) is said to be minimum at x = b if there exists a small positive number δ such that f(b) < f(x) ∀ x ∈ (b – δ, b + δ) Also then the point x = b is called a minimum point for f(x) Hence we find that : (i) x = a is a maximum point of f(x)

f (a ) – f(a + h) > 0 ⇔ f(a) – f(a – h) > 0 (ii) x = b is a minimum point of f(x) f (b ) – f(b + h) < 0 ⇔ f(b) – f(b – h) > 0 (iii) x = c is neither a maximum point nor a minimum point f (c) – f (c + h ) ⇔ and have opposite signs. f (c) − f (c − h ) Where h is a very small positive number. Note : The maximum and minimum points are also known as extreme points. A function may have more than one maximum and minimum points. A maximum value of a function f(x) in an interval [a, b] is not necessarily its greatest value in that interval. Similarly a minimum value may not be the least value of the function. A minimum value may be greater than some maximum value for a function. The greatest and least values of a function f(x) in an interval [a, b] may be determined as follows : Greatest value = max. {f(a), f(b), f(c)} Least value = min. {f(a), f(b), f(c)}

XtraEdge for IIT-JEE

or

52

DECEMBER 2009

MATH

FUNCTION Mathematics Fundamentals If f and g are two functions then their sum, difference, product, quotient and composite are denoted by

Definition of a Function : Let A and B be two sets and f be a rule under which every element of A is associated to a unique element of B. Then such a rule f is called a function from A to B and symbolically it is expressed as

or

f + g, f – g, fg, f/g, fog and they are defined as follows :

f:A→B

(f + g) (x) = f(x) + g(x)

f A → B

(f – g) (x) = f(x) – g(x) (fg) (x) = f(x) f(g)

Function as a Set of Ordered Pairs Every function f : A → B can be considered as a set of ordered pairs in which first element is an element of A and second is the image of the first element. Thus

(f/g) (x) = f(x)/g(x) (fog) (x) = f[g(x)]

Formulae for domain of functions :

f = {a, f(a) /a ∈ A, f(a) ∈ B}.

Df ± g = Df ∩ Dg

Domain, Codomain and Range of a Function :

Dfg = Df ∩ Dg

If f : A → B is a function, then A is called domain of f and B is called codomain of f. Also the set of all images of elements of A is called the range of f and it is expressed by f(A). Thus

Df/g = Df ∩ Dg ∩ {x |g(x) ≠ 0} Dgof = {x ∈ Df | f(x) ∈ Dg} D

f(A) = {f(a) |a ∈ A} obviously

f(A) ⊂ B.

f

= Df ∩ {x |f(x) ≥ 0}

Classification of Functions

Note : Generally we denote domain of a function f by Df and its range by R f.

1.

Algebraic and Transcendental Functions : Algebraic functions : If the rule of the function consists of sum, difference, product, power or roots of a variable, then it is called an algebraic function. Transcendental Functions : Those functions which are not algebraic are named as transcendental or non algebraic functions.

Equal Functions : Two functions f and g are said to be equal functions if domain of f = domain of g codomain of f = codomain of g 2.

f(x) = g(x) ∀ x. Algebra of Functions :

XtraEdge for IIT-JEE

(g(x) ≠ 0)

53

Even and Odd Functions : Even functions : If by replacing x by –x in f(x) there in no change in the rule then f(x) is called an even function. Thus f(x) is even ⇔ f(–x) = f(x) DECEMBER 2009

Odd function : If by replacing x by –x in f(x) there is only change of sign of f(x) then f(x) is called an odd function. Thus

Period of f(x) = T

⇒ Period of f(nx + a) = T/n Periods of some functions :

f(x) is odd ⇔ f(–x) = – f(x) 3.

Function

Explicit and Implicit Functions : Explicit function : A function is said to be explicit if its rule is directly expressed (or can be expressed( in terms of the independent variable. Such a function is generally written as

sin x, cos x, sec x, cosec x,

2π

tan x, cot x

π

sinnx, cosn x, secn x, cosecn x

2π if n is odd

π if n is even

y = f(x), x = g(y) etc.

4.

Implicit function : A function is said to be implicit if its rule cannot be expressed directly in terms of the independent variable. Symbolically we write such a function as

tann x, cotn x

π∀ n∈ N

|sin x|, |cos x|, |sec x|, |cosec x|

π

|tan x|, |cot x|,

π

f(x, y) = 0, φ(x, y) = 0 etc.

|sin x| + |cos x|, sin4x + cos 4x

π 2

|sec x| + |cosec x|

Continuous and Discontinuous Functions : Continuous functions : A functions is said to be continuous if its graph is continuous i.e. there is no gap or break or jump in the graph. Discontinuous Functions : A function is said to be discontinuous if it has a gap or break in its graph atleast at one point. Thus a function which is not continuous is named as discontinuous.

5.

Period

x – [x]

1

Period of f1(x) = T1, period fo f2(x) = T 2

⇒ period of a f1(x) + bf2(x) ≤ LCM {T1, T 2}

Increasing Functions : A function f(x) is said to be increasing function if for any x1, x2 of its domain

Kinds of Functions : One-one/ May one Functions :

x1 < x2 ⇒ f(x1) ≤ f(x2)

A function f : A → B is said to be one-one if different elements of A have their different images in B.

or x1 > x2 ⇒ f(x1) ≥ f(x2) Decreasing Functions : A function f(x) is said to be decreasing function if for any x1, x2 of its domain

Thus

⇒ f (a ) ≠ f (b) a≠b f is one-one ⇔ or f (a ) = f (b) ⇒ a =b

x1 < x2 ⇒ f(x1) ≥ f(x2) or x1 > x2 ⇒ f(x1) ≤ f(x2)

A function which is not one-one is called many one. Thus if f is many one then atleast two different elements have same f-image.

Periodic Functions : A functions f(x) is called a periodic function if there exists a positive real number T such that

Onto/Into Functions : A function f : A → B is said to be onto if range of f = codomain of f

∀x

Also then the least value of T is called the period of the function f(x).

XtraEdge for IIT-JEE

π 2

Period of f(x) = T ⇒ period of f(ax + b) = T/|a|

Increasing and Decreasing Functions :

f(x + T) = f(x).

|tan x| + |cot x|

Thus f is onto ⇔ f(A) = B 54

DECEMBER 2009

Hence f : A → B is onto if every element of B (co-domain) has its f–preimage in A (domain).

Domain and Range of some standard functions : Function

A function which is not onto is named as into function. Thus f : A → B is into if f(A) ≠ B. i.e., if there exists atleast one element in codomain of f which has no preimage in domain. Note : Total number of functions : If A and B are finite sets containing m and n elements respectively, then total number of functions which can be defined from A to B = nm. total number of one-one functions from A to B

n p = m 0

if if

m≤n m>n

total number of onto functions from A to B (if m ≥ n) = total number of different n groups of m elements. Composite of Functions :

Domain

Range

Polynomial function

R

R

Identity function x

R

R

Constant function c

R

{c}

Reciprocal function 1/x

R0

R0

x2, |x|

R

R+ ∪ {0}

x3, x |x|

R

R

Signum function

R

{–1, 0, 1}

x + |x|

R

R+ ∪ {0}

x – |x|

R

R– ∪ {0}

[x]

R

Z

x – [x]

R

[0, 1)

x

[0, ∞)

[0, ∞)

ax

R

R+

log x

R+

R

sin x

R

[–1, 1]

cos x

R

[–1, 7]

tan x

R – {± π/2, ± 3π/2, ...}

R

cot x

R – {0, ± π. ± 2 π, .....

R

The following properties of composite functions can easily be established.

sec x

R – (± π/2, ± 3π/2, .....

R – (–1, 1)

cosec x

R – {0, ±π, ± 2 π, ......}

R –(–1, 1)

Composite of functions is not commutative i.e.,

sinh x

R

R

cosh x

R

[1, ∞)

tanh x

R

(–1, 1)

coth x

R0

R –[1, –1]

sech x

R

(0, 1]

cosech x

R0

R0

Let f : A → B and g : B → C be two functions, then the composite of the functions f and g denoted by gof, is a function from A to C given by gof : A → C, (gof) (x) = g[f(x)]. Properties of Composite Function :

fog ≠ gof Composite of functions is associative i.e. (fog)oh = fo(goh) Composite of two bijections is also a bijection. Inverse Function :

–1

sin x

If f : A → B is one-one onto, then the inverse of f i.e., f–1 is a function from B to A under which every b ∈ B is associated to that a ∈ A for which f(a) = b. Thus

–1

cos x

XtraEdge for IIT-JEE

[0, π]

R

(–π/2, π/2}

R

(0, π)

–1

sec x –1

cosec x

55

[–1, 1]

–1

cot x

f–1(b) = a ⇔ f(a) = b.

[–π/2, π/2]

–1

tan x

f–1 : B → A,

[–1, 1]

R –(–1, 1)

[0, π] – {π/2}

R – (–1, 1)

(– π/2, π/2] – {0}

DECEMBER 2009

Based on New Pattern

IIT-JEE 2010 XtraEdge Test Series # 8

Time : 3 Hours Syllabus :

Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus

Instructions :

Section - I • Question 1 to 4 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 5 to 9 are multiple choice questions with multiple correct answer. +4 marks and -1 mark for wrong answer. Section - II • Question 10 to 11 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer and No Negative marks for wrong answer. However, +2 marks will be given for a correctly

marked answer in any row.

Section - III • Question 12 to 19 are numerical response type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer.

PHYSICS

4.

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

A particle starts from rest and travels a distance x with uniform acceleration, then moves uniformly a distance 2x and finally comes at rest after moving further 5x distance with uniform retardation. The ratio of maximum speed to average speed is -

1.

(A)

5 2

(B)

5 3

(C)

7 4

(D)

7 5

A deuterium plasma is a neutral mixture of negatively charged electrons and positively charged deuterium charged electrons and positively charged deuterium nuclei. Temperature required to produced fusion in deuterium plasma is : (Take the range of nuclear forces 2 fermi) (A) 835 K (C) 8.35 × 107 K

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

7

(B) 835 × 10 K (D) 273 K

5. 2.

3.

As observed in the laboratory system, a 6 MeV proton is incident on a stationary 12C target velocity of center of mass of the system is : (Take mass of proton to be 1 amu) (A) 2.6 × 106 m/s (B) 6.2 × 106 m/s (C) 10 × 106 m/s (D) 10 m/s

(A) v =

Find the de Broglie wavelength of Earth. Mass of Earth is 6 × 1024 kg. Mean orbital radius of Earth around Sun is 150 × 106 km (A) 3.7 m (C) 3.7 × 1063 m

XtraEdge for IIT-JEE

A body moves in a circular path of radius R with deceleration so that at any moment of time its tangential and normal accelerations are equal in magnitude. At the initial moment t = 0, the velocity of body is v0 then the velocity of body will bev0 at time t v0 t 1+ R

(B) v = v 0 e −S / R after it has moved S meter (C) v = v0e–SR after it has moved S meter (D) None of these

–63

(B) 3.7 × 10 m (D) 3.7 × 10–63 cm 56

DECEMBER 2009

6.

Three identical rods of same material are joined to form a triangular shape ABC as shown. Angles at edge A and C are respectively θ1 and θ2 as shown. When this triangular shape is heated then A

9.

(B) the indices of refraction of the two media are same (C) the boundary is not visible (D) angle of incidence is lesser than angle of

θ1

µ refraction but greater then sin–1 R µD

θ2 B

7.

In passing through a boundary refraction will not take place if (A) light is incident normally on the boundary

C

(A) θ1 decreases and θ2 increases (B) θ1 increases and θ2 decreases (C) θ1 increases (D) θ2 increases Direction of current in coil (2) is opposite to direction of current in coil (1) and coil (3). All three coils are coaxial and equidistant coil (1) and coil (3) are fixed while coil (2) is suspended thus able to move freely. Then -

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T

R

R

1

R

A P B P C P D P

3

2

(A) coil (2) is in equilibrium (B) equilibrium state of coil (2) is stable equilibrium along axial direction (C) equilibrium state of coil (2) is unstable equilibrium along axial direction (D) if direction of current in coil (2) is same as that of coil (1) and coil (3) then state of equilibrium of coil (2) along axial direction is unstable 8.

10.

Which of the following statements are correct about the circuits shown in the figure where 1 Ω and 0.5 Ω are internal resistances of the 6 V and 12 V batteries respectively – 11.

P

4Ω R

(A) The potential at point P is 6 V (B) The potential at point Q is – 0.5 V (C) If a voltmeter is connected across the 6 V battery, it will read 7 V (D) If a voltmeter is connected across the 6 V battery, it will read 5 V

XtraEdge for IIT-JEE

(A) Interference

(P) Non-mechanical waves

(B) Diffraction

(Q) Electromagnetic waves (R) Visible light

(D) Reflection

(S) Sound waves (T) None

Column-I

Column-II

(A) 5000 Å (B) 1 Å

(P) De-Broglie wavelength of electron in x-ray tube (Q) Photoelectric threshold

(C) 0.1 Å

wavelength (R) x-ray wavelength

0.5Ω S

Q R S T Q R S T

Match column I with column II in the light of possibility of occurrence of phenomena listed in column I using the systems in column II Column-I Column-II

(C) Polarisation

6V, 1Ω 12V, 0.5Ω Q

Q R S T Q R S T

(D) 10 Å

(S) De-Broglie wavelength of most energetic photoelectron emitted from metal surface (T) None

57

DECEMBER 2009

This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

between the particles in process of their motion in meters is (g = 10 m/s2).

u = 2 m/s A 45°

22m

X Y Z W 0 1 2 3 4 5 6 7 8 9

12.

13.

14.

15.

0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9

17.

2mH S

18.

The minimum speed in m/s with which a projectile must be thrown from origin at ground so that it is able to pass through a point P (30 m, 40 m) is : (g = 10 m/s 2) (Ans. in .............. × 10)

9Ω

9Ω 9Ω

2mH

A rectangular plate of mass 20 kg is suspended from points A and B as shown. If the pin B is suddenly removed then the angular acceleration in rad/sec2 of the plate is : (g = 10 m/s2). B

A

In U238 ore containing Uranium the ratio of U234 to Pb206 nuclei is 3. Assuming that all the lead present in the ore is final stable product of U238. Half life of U238 to be 4.5 × 10 9 years and find the age of ore. (in 109 years)

b =0.15m

l =0.2m

Under standard conditions the gas density is 1.3 mg/cm3 and the velocity of sound propagation in it is 330 m/s, then the number of degrees of freedom of gas is.

19.

If the temperature of a gas is raised by 1 K from 27ºC. Find the percentage change in speed of sound. (Speed = 300 ms–1) (Ans. in .............. × 102)

A single conservative force acts on a body of mass 1 kg that moves along the x-axis. The potential energy U(x) is given by U (x) = 20 + (x – 2)2, where x is in meters. At x = 5.0 m the particle has a kinetic energy of 20 J, then the maximum kinetic energy of body in J is.

CHEMISTRY Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Two particles are simultaneously thrown from top of two towers with making angle 45º with horizontal. Their velocities are 2 m/s and 14 m/s. Horizontal and vertical separation between these particles are 22 m and 9 m respectively. Then the minimum separation

XtraEdge for IIT-JEE

Consider the circuit shown in figure. What is the current through the battery just after the switch is closed. 18V

(Ans. in .............. × 10) 16.

v = 14 m/s 45° B

9m

1.

58

In the precipitation of sulphides of second group basic radicals. H2S is passed into acidified solution with dilute HCl. If the solution is not acidified, then which of the following is correct ?

DECEMBER 2009

(A) Only the sulphides of second precipitated

group

get

(B) Only the sulphides precipitated

group

get

of

fourth

5.

(C) Neither of the sulphides of second and fourth groups get pricipitated

(C) they are nuclear spin isomer, where the spins of protons are same in para and different in ortho isomer (D) they are nuclear spin isomers, where the spins of protons are opposite in para but same in ortho isomer

(D) Sulphides of both the groups second and fourth get precipitated 2.

The difference between ortho and para hydrogen is/are (A) they are electron spin isomer, where the spins of electrons are opposite (B) ortho hydrogen is more stable at lower temperature

The geometrical shapes of XeF5+, XeF6 and XeF82– respectively are (A) trigonal bipyramidal, octahedral and square planar

6.

(B) square based pyramidal, distorted octahedral and octahedral

NGP assistance support for S N2 reaction will be seen in OH

(C) planar pentagonal, octahedral and square anti prismatic

(A)

(B) CH3–S–CH2–CH2–Cl

(D) square based pyramidal, distorted octahedral and square anti prismatic

Cl

OH

OH

3.

Regarding graphite the following informations are available :

(C)

(D)

Cl

Θ

COO

Top view

7.

When the compound called isoborneol is heated with 50% sulfuric acid the product of the reaction is/are ?

3.35Å

HO Isoborneol

The density of graphite = 2.25 gm/cm3. What is C–C bond distance in graphite ?

4.

(A) 1.68Å

(B) 1.545Å

(C) 2.852 Å

(D) 1.426Å

The molecular formula of a non-stoichiometric tin oxide containing Sn(II) and Sn (IV) ions is Sn4.44 O8. Therefore, the molar ratio of Sn(II) to Sn(IV) is approximately (A) 1 : 8

(B) 1 : 6

(C) 1 : 4

(D) 1 : 1

8.

(B)

(C)

(D)

Regarding the radial probability distribution (4nr2R 2nl) vs r plot which of the following is/are correct ? (A) The number of maxima is (n-l) (B) The number of nodal points is (n-l-1) (C) The radius at which the radial probability density reaches to maxima is 3s < 3p < 3d

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

XtraEdge for IIT-JEE

(A)

(D) The number of angular nodes is l

59

DECEMBER 2009

9.

H2C 2O4 and NaHC2O4 behave as acids as well as reducing agents which is/are correct statement (s) ? (A) Equivalent wt. of H2C 2O4 and NaHC 2O4 are equal to their molecular weights when behaving as reducing agent (B) 100 mL of 1N solution of each is neutralised by equal volumes of 1M Ca(OH)2 (C) 100 mL of 1N solution of each is neutralised by equal volumes of 1N Ca(OH)2 (D) 100 mL of 1M solution of each is oxidised by equal volumes of 1M KMnO4

Cl

Cl

Br

Me Br

Me Me

Me

Me Br

Me

Cl

Column- II (P) Optically active (Q) Cis compound (R) Trans compound (S) Optically inactive (T) Chiral axis (element of chirality) This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

at high temperature (T) PV/RT = 1–a/VRT 11. Column- I Cl

12.

Me

XtraEdge for IIT-JEE

Me Me

Me Me

(C) At STP (for real gas) (R) PV = RT + Pb (D) At low pressure and (S) PV = RT

Me Me

Me Me

(D) Me

gas molecules be negligible

(A) Me

Me Br

Cl

Me

Cl

a (P) P + 2 (V–b)= RT V

Me Me

Me Me

(C) Me

among the gas molecules be negligible (B) If the volume of the (Q) PV = RT –a/V

Me Me

Me Me Me

Column- II

(A) If force of attraction

Me Me

(B)

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T 10. Column- I

Me Me

Cl

What is the number of lone pair on the molecule XeO2F 2 ?

Me Me

60

DECEMBER 2009

13.

15g of a solute in 100g water makes a solution freeze at –1ºC. What will be the depression in freezing point if 30g of a solute is dissolved in 100g of water ?

14.

The half lives of decomposition of gaseous CH3CHO at constant temperature but at initial pressure of 364mm and 170mm Hg were 410 second and 880 second respectively. Hence what is the order of reaction.

4.

x x and g(x) = where 0 < x ≤ 1, sin x tan x then in this interval (A) both f(x) and g(x) are increasing functions If f(x) =

(B) both f(x) and g(x) are decreasing functions (C) f(x) is an increasing function (D) g(x) is an increasing function

15.

The number of S–S bonds in sulphur trioxide trimer (S3O9) is ....... .

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

16.

The number of peroxo linkages present in the [H4B2O8] 2– is ..... .

5.

17.

The maximum number of structural isomers (acyclic and cyclic) possible for C4H8 are ....... .

(A) a = b = c

(B) a, b, –

18.

What is the no. of lone pair of electrons present on N in Trisilylamine ?

(C) a, b, c are in H.P.

(D) –

19.

A small amount of solution containing 24Na with activity 2 × 103 dps was administered into the blood of patient in a hospital. After 5 hour a sample of the blood drawn out from the patient shared an activity of 16 dps per cm3. (t1/2 of 24Na = 15 hrs.) Find the volume (in L) of blood in patient.

6.

(C) infinite number of solutions if a + b + c = 0 (D) none of these

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. The largest term in the expansion of (3 + 2x)50, where x = 1/5, is (A) 5th (B) 6th

3.

If

1 sin–1 2

8.

(C) (1/3) tan θ

(D) 3 cot θ

The integer n for which lim

x

n

XtraEdge for IIT-JEE

(B) [0, ∞)

(C) (–∞, 0)

(D) (– ∞,∞)

x 4 cos 2 x − x sin x + cos x dx If l = e x sin x +cos x x 2 cos 2 x

∫

sec x x − (C) ex sin x + cos x +C tan x x

is

a finite nonzero number is (A) 1 (C) 3

(A) (0, ∞)

cos x (B) ex sin x + cos x x sin x − x

(cos x − 1)(cos x − e x )

x →0

x is differentiable on 1+ | x |

sec x (A) ex sin x + cos x x − +C x

3 sin 2θ –1 5 + 4 cos 2θ = tan x, then x = (B) 3 tan θ

The function f(x) =

then l equals -

(D) 9th

(A) tan 3 θ

The system of equations –2x + y + z = a

(B) unique solution if a + b + c = 0

7.

2.

1 a, b, c are in G.P. 2

(A) no solution if a + b + c ≠ 0

MATHEMATICS

(C) 8th

1 c are in G.P. 2

x – 2y + z = b x = y – 2z = c has

[Given : log 1.2598 = 0.1003]

1.

If a, b, c are in A.P., and a 2, b2, c 2 are in H.P., then

cos x − x sin x (D) xe x sin x+cos x – e x sin x +cos x 1 − dx x 2 cos 2 x

∫

(B) 2 (D) 4 61

DECEMBER 2009

9.

If for the differential equation y′ = general solution is y =

This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

x y + φ the x y

x then f (x / y) is log | Cx |

given by (A) – x2 / y2 (C) x2 / y2

(B) y2 / x2 (D) – y2 / x2

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T

12.

Suppose X follows a binomial distribution with parameters n = 6 and p. If 9P(X = 4) = P(X = 2), find 4p.

13.

If Q is the foot of the perpendicular from the point x −5 y + 2 z− 6 P(4, –5,3) on the line = = then 3 −4 5 100 (PQ)2 is equal to 457

14.

If a = (0, 1, –1) and c = (1, 1, 1) are given vectors, then |b|2 where b satisfies a × b + c = 0 and a . b = 3 is equal to

15.

ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to BC. If the triangle ABC has perimeter P and area ∆ ∆ then lim 512r 3 is equal to h →0 p

16.

If f(x) = sin x, x ≠ n π, =0 and g(x) = x2 + 1, =4 =5 then lim g(f(x)) is .....

10. The domain of the functions Column- I Column- II –1 (A) sin (x/2 – 1) (P) (3 – 2π, 3 – π) ∪ (3,4] + log (x – [x]) (B) ex + 5sin π 2 / 16 − x 2 (Q) (0, 4) – {1, 2, 3} (R) [– π/4, π/4]

(C) log10sin (x – 3) + 16 − x 2 (D) cos–1

1 − 2x 4

(S) [– 3/2, 5/2] (T) None

11. Column- I I denotes an integral (A)

∫

π 0 ∞

(B)

∫

0

(C)

∫

0

(D)

∫

x log sin x dx log (x+x–1)

π/4

π 0

Column- II

dx 1+ x 2

(P) I = ( π/8) log 2 (Q) I =

− π2 log 2 2

x →0

log (1+ tan x)dx (R) I = π log 2

17.

(S) I = π log 2

18.

log (1 – cos x)dx

If y = (1 + 1/x) x then

2 y 2 ( 2) + 1 / 8 (log 3 / 2 − 1 / 3)

is equal to

If the greatest value of y = x/log x on [e, e3] is u then e3/u is equal to 19. If z ≠ 0 and 2 + cos θ + i sin θ = 3/z, then find the value of 2(z + z ) – |z|2.

(T) None

XtraEdge for IIT-JEE

n = 0, ± 1, ± 2, ..... otherwise x ≠ 0, 2 x=0 x=2

62

DECEMBER 2009

Based on New Pattern

IIT-JEE 2011 XtraEdge Test Series # 8

Time : 3 Hours Syllabus :

Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus

Instructions :

Section - I • Question 1 to 4 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 5 to 9 are multiple choice questions with multiple correct answer. +4 marks and -1 mark for wrong answer. Section - II • Question 10 to 11 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer and No Negative marks for wrong answer. However, +2 marks will be given for a correctly

marked answer in any row.

Section - III • Question 12 to 19 are numerical response type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer.

PHYSICS Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

Velocity of block 2 shown in figure is u

(A) (C)

60º

4. 2

u

(A) 2.

3.

3u 2

(B)

3u 2

1

(C)

3 3u (D) 2u 2

(B)

MR 2 ω 2∆t

(D) Zero

2 ∆t

Select the incorrect statement (A) The velocity of the centre of mass of an isolated system must stay constant (B) Only a net external force can change the velocity of the center of mass of a system (C) A system have non-zero kinetic energy but zero linear momentum →

(D) Fext

Which of the following restoring force can give rise to S.H.M (A) F = 2x (B) F = 2 – 4x (C) F = – 2x2 (D) None of these

→

d v → dm =m +v is true for all situation dt dt

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 5.

A disk of mass M and radius R is rotating about its axis with angular velocity ω . Axis of disk is rotated by 90º in time ∆t. Average torque acting on disk is -

XtraEdge for IIT-JEE

2 Mω2 ∆t MR 2 ω

63

Velocity of a particle moving on straight line varies 1 as th power of displacement. Then 2 (A) K.E. ∝ S (B) P ∝ S1/2 (C) a = constant (D) S ∝ t2

DECEMBER 2009

6.

7.

A block of density ρ is floating in a liquid X kept in container. A liquid Y of density ρ′ (< ρ) is slowly poured into container – (A) The block will move up if liquid X and Y are immiscible (B) The block will sink more if liquid X and Y are immiscible (C) The block will sink more if liquid X and Y are miscible (D) The block will not move if liquid X and Y are miscible

P Q R S T A P B P C P D P

10.

A

B

(A) A

Q

Q R S T Q R S T

There are four identical rod having thermal resistance 10 Ω, each column I contains various arrangement of rod. Column II contains current flowing across point C when a temperature difference of 100ºC is maintained across A & B. Match them. Column-I Column-II

Six identical rod are connected as shown in figure and temperature difference of 100ºC is maintained across P and Q –

P

Q R S T Q R S T

C

B

(P) 2 J/sec

C

(A) (B) (C) (D) 8.

9.

C Temperature of point 'A' is 50ºC 200 Temperature of point A is ºC 3 Thermal current passing through B is zero Thermal current passing through A is twice of that through C

(B) A

(C) A

C

B

(Q) 4 J/sec

(R) 6 J/sec

A

A solid iron cylinder A rolls down a ramp and an identical iron cylinder B slides down the same ramp without friction – (A) B reaches the bottom first (B) A and B have the same kinetic energy (C) B has greater translational kinetic energy than that of A (D) Linear speed of centre of mass of B is greater than that of A

(D)

C

B

(S)

20 J/sec 7

(T) None 11.

A car of mass 500 kg is moving in a circular road of radius 35 / 3 . Angle of banking of road is 30º. Coefficient of friction between road and tyres 1 is µ = . Match the following: 2 3 Column-I Column-II

A mass and spring system oscillates with amplitude A and angular frequency ω – (A) The average speed during one complete cycle 2Aω of oscillation is π (B) Maximum speed is ω A (C) Average velocity of particle during one complete cycle of oscillation is zero (D) Average acceleration of particle during one complete cycle of oscillation is zero

(A) Maximum speed (in m/s) of car for safe turning (B) Minimum speed (in m/s) of car for safe turning

(P) 5 2 (Q) 12.50

(C) Speed (in m/s) at which friction (R) force between tyres and road is zero

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

XtraEdge for IIT-JEE

B

(D) Friction force (in 102 Newton)

(S)

210

350 3

between tyres and road if speed is

350 m/s 6 (T) None

64

DECEMBER 2009

This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 12.

14.

A longitudinal wave of frequency 220 Hz travels down a copper rod of radius 8.00 mm. The average power in the wave is 6.50 µW. The amplitude of the wave is n × 10–8 m. Find n.

15.

A piston-cylinder device with air at an initial temperature of 30ºC undergoes an expansion process for which pressure and volume are related as given below P (kPa) 100 25 6.25 3 V (m ) 0.1 0.2 0.4 The work done by the system is n × 103 J. Find n.

16.

The block connected with spring is pushed to compress the spring by 10 cm and then released. All surfaces are frictionless and collision are elastic. Time period of the motion in sec (mass of block = 9 kg and spring constant 4 π2 N/m).

Pulley 'P' shown in figure is pulled upward with F = 2t N, where t is time in sec. Velocity of block of mass 1 at the time block 2 is about to lift is (in cm/sec). (Ans. in ................. × 10 1) F = 2t

5 cm

17.

300 × 381 m/s.

RMS velocity of gas at 27ºC is

RMS velocity (in m/s) when temperature is increased four times is. (Ans. in ................. × 10 2)

2

1

18.

A block of mass 2 kg is placed on a wedge of mass 10 kg kept on a horizontal surface. Coefficient of friction between all surfaces is µ = 0.2. If block is slipping down the wedge with constant speed then friction force on wedge due to horizontal surface is (in Newton) :

19.

A particular quantity 'y' varies as 'x' as shown in figure. RMS value of y with respect to x for large values of 'x' is.

m 1 = 0.5 kg m 2 = 1 kg

13.

The upper edge of a gate in a dam runs along the water surface. The gate is 2.00 m high and 4.00 m wide and is hinged along a horizontal line through its center. The torque about the hinge arising from the force due to the water is (n × 104 Nm). Find value of n.

y

2m

60º

60º 1

XtraEdge for IIT-JEE

65

2

3

4 x

DECEMBER 2009

(C) The first ionization energies of elements in a period do not increase with the increase in atomic numbers (D) For transition elements the d-subshells are filled with electrons monotonically with the increase in atomic number

CHEMISTRY Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

An ideal gaseous mixture of ethane (C2H6) and ethene (C 2H4) occupies 28 litre at STP. The mixture reacts completely with 128 gm O2 to produce CO2 and H2O. Mole fraction of C2H4 in the mixture is (A) 0.6 (B) 0.4 (C) 0.5 (D) 0.8

2.

A bulb of constant volume is attached to a manometer tube open at other end as shown in figure. The manometer is filled with a liquid of density (1/3rd) that of mercury. Initially h was 228 cm.

Gas

h

When 300 mL of 0.2 M HCl is added to 200 mL of 0.1 M NaOH. Resultant solution require how many equivalent of Ba(OH)2 ? (A) 0.06 (B) 0.12 (C) 0.3 (D) 0.04

4.

The dipole moment of HCl is 1.03D, if H–Cl bond distance is 1.26Å, what is the percentage of ionic character in the H–Cl bond ? (A) 60% (B) 29% (C) 17% (D) 39%

6.

Consider the following carbides CaC2, BeC2, MgC2 and SrC 2 which of the given carbides on hydrolysis yield same product (A) CaC2 (B) Be2C (C) MgC2 (D) SrC2

Which of the following is/are state function ? (A) q (B) q – w (C) q + w (D) q / w

9.

The IUPAC name of the following compound is OH

10. Column- I (A) 5.4 g of Al (B) 1.2 g of Mg2+ (C) Exact atomic weight of mixture of oxygen isotopes (D) 0.9 mL of H2O

Which of the following is/are correct regarding the periodic classification of elements ? (A) The properties of elements are the periodic function of their atomic number (B) Non metals are lesser in number than metals

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8.

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 5.

Which of the following statements regarding hydrogen peroxide is/are correct ? (A) Hydrogen peroxide is a pale blue viscous liquid (B) Hydrogen peroxide can act as oxidising as well as reducing agent (C) The two hydroxyl groups in hydrogen peroxide lie in the same plane (D) In the crystalline phase, H2O2 is paramagnetic

Br CN (A) 3-Bromo-3-cyano phenol (B) 3-Bromo-5-hydroxy benzonitrile (C) 3-Cyano-3-hydroxybromo benzene (D) 5-Bromo-3-hydroxy benzonitrile

Through a small hole in the bulb gas leaked assuming dp pressure decreases as = – kP. dt If value of h is 114 cm after 14 minutes. What is the value of k (in hour–1) ? [Use : ln(4/3) = 0.28 and density of Hg = 13.6 g/mL] (A) 0.6 (B) 1.2 (C) 2.4 (D) None of these 3.

7.

66

Column- II (P) 0.5 NA electrons (Q) 15.9994 amu (R) 0.2 mole atoms

(S) 0.05 moles (T) 3.1 × 1023 electrons

DECEMBER 2009

11. Column- I (Ionic species) (A) XeF5+ (B) SiF5– (C) AsF4+ (D) ICl4–

Column- II (Shapes) (P) Tetrahedral (Q) Square planar (R) Trigonal bipyramidal (S) Square pyramidal (T) Octahedral

13.

14.

The equivalent weight of a metal is 4.5 and the molecular weight of its chloride is 80. The atomic weight of the metal is.

19.

No. of π bond in the compound H2CSF 4 is.

MATHEMATICS

This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 12.

18.

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

If 0 < r < s ≤ n and nPr = nPs , then value of r + s is (A) 2n – 2 (B) 2n – 1 (C) 2 (D) 1

2.

If sin x + sin2 x + sin3 x = 1, then cos6 x – 4 cos4 x + 8 cos2 x is equal to (A) 0 (B) 2 (C) 4 (D) 8

3.

If x is real, and k=

then x2 + x +1 (A) 1/3 ≤ k ≤ 3 (C) k ≤ 0 4.

How much volume (in mL) 0.001 M HCl should we add to 10 cm3 of 0.001 M NaOH to change its pH by one unit ?

5.

An acid type indicator, HIn differs in colour from its conjugate base (In– ). The human eye is sensitive to colour differences only when the ratio [In– ] / [HIn] is greater than 10 or smaller than 0.1. What should be the minimum change in the pH of the solution to osberve a complete colour change ? (Ka = 1.5 × 10–5) What is the sum of total electron pairs (b.p. + l.p.) present in XeF6 molecule ?

16.

The number of geometrical isomers of CH3CH=CH–CH=CH–CH=CHCl is.

17.

At 200ºC, the velocity of hydrogen molecule is 2.0 × 105 cm/sec. In this case the de-Broglie wavelength (in Å) is about.

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(B) k ≥ 5 (D) none of these

A flagstaff stands in the centre of a rectangular field whose diagonal is 1200 m, and subtends angles 15º and 45º at the mid points of the sides of the field. The height of the flagstaff is (A) 200 m

(B) 300 2 + 3 m

(C) 300 2 − 3 m

(D) 400 m

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

The stopcock, connecting the two bulbs of volumes 5 litres and 10 litres containing an ideal gas at 9 atm and 6 atm respectively, is opened. What is the final presure (in atm) in the two bulbs if the temperature remained the same ?

15.

x2 − x +1

If a, b, c are the sides of the ∆ABC and a2, b 2, c2 are the roots of x3 – px2 + qx – k = 0, then cos A cos B cos C p (A) + + = a b c 2 k (B) a cos A + b cos B + c cos C = (C) a sin A + b sin B + c sin C =

4q − p 2

2 k 2p∆ k

3

(D) sin A sin B sin C = 6.

67

8∆ k

The coordinates of the feet of the perpendiculars from the vertices of a triangle on the opposite sides are (20, 25), (8, 16) and (8, 9). The coordinates of a vertex of the triangle are (A) (5, 10) (B) (50, –5) (C) (15, 30) (D) (10, 15)

DECEMBER 2009

7.

(D) 3 x–1 + 3 x–2 + 3 x–3 + ... (S) 7 1 1 = 2 5 2 + 5 + 1 + + 2 + ... 5 5 (T) None

1 1 1 2 Let E = + + + + ... upto 50 terms, then 3 50 3 50 (A) E is divisible by exactly 2 primes (B) E is prime (C) E ≥ 30 (D) E ≤ 35

8.

If m is a positive integer, then [( 3 + 1) 2m ] + 1, where [x] denotes greatest integer ≤ n, is divisible by(A) 2m (B) 2m+1 (C) 2m+2 (D) 22m

9.

If A and B are acute angles such that sin A = sin2 B, 2 cos2 A = 3 cos2 B; then (A) A = π/6 (B) A = π/2 (C) B = π/4 (D) B = π/3

This section contains 8 questions (Questions 12 to 19). The answer to each of the questions is a SINGLEDIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbe rs in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

This section contains 2 questions (Questions 10 to 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T

12.

Fifteen persons, among whom are A and B, sit down at random at a round table. if p is The probability that there are exactly 4 persons between A and B find 14 p.

13.

If l is the length of the intercept made by a common tangent to the circle x2 + y2 = 16 and the ellipse x2/25 + y2/4 = 1, on the coordinate axes, then

10. For the circle x2 + y2 + 4x + 6y – 19 = 0 Column- I Column- II (A) Length of the tangent (P)

81l 2 + 3 is equal to 1059

72 226 113

14.

from (6, 4) to the circle (B) Length of the chord (Q) of contact from (6, 4) to the circle

113

parabola on this normal; then

(C) Distance of (6, 4) (R) 113 – 32 from the centre of the circle (D) Shortest distance of (S) 9 (6, 4) from the circle (T) None 11. Value of x when Column- I Column- II (A) 52 54 5 6 ... 52x = (0.04)–28 (P) 3 log3 5 2

(B) x = (0.2) (C)

log

1 + 1 + 1 +... 4 8 16

5

1 1 1 log 2.5 + 2 + 3 +... 3 3 3 x = (0.16)

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If x + y = k is a normal to the parabola y2 = 12x, p is the length of the perpendicular from the focus of the

15.

The volume of the tetrahedron whose vertices are (0, 1, 2) (3, 0, 1) (4, 3, 6) (2, 3, 2) is equal to

16.

Let a1 =

17.

(Q) 4 (R) 2 68

3k 3 + 2p 2 is equal to 741

1 , ak+1 = ak2 + ak ∀ k ≥ 1 and 2 1 1 1 + + ... + xn = a 1 +1 a 2 + 1 a n +1 Find [x100] where [x] denotes the greatest integer ≤ x. Find the value of x which satisfy the equation log2 (x2 – 3) – log2 (6x – 10) + 1 = 0

18.

Find the coefficient of x2009 in the expansion of (1 – x)2008 (1 + x + x2)2007

19.

Find the value of x satisfying 4

1 / log2 x

= 2.

DECEMBER 2009

MOCK TEST PAPER-1 CBSE BOARD PATTERN

CLASS # XII SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS Sol ut i ons w il l be publ i s he d i n ne xt is s ue General Instructions : Physics & Che mistry • Time given for each subject paper is 3 hrs and Max. marks 70 for each. • All questions are compulsory. • Marks for each question are indicated against it. • Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each. • Question numbers 9 to 18 are short-answer questions, and carry 2 marks each. • Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each. • Question numbers 28 to 30 are long-answer questions and carry 5 marks each. • Use of calculators is not permitted. General Instructions : Mathematics • Time given to solve this subject paper is 3 hrs and Max. marks 100. • All questions are compulsory. • The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each. Section B comprises of 12 questions of four marks each. Section C comprises of 7 questions of six marks each. • All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. • There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2 question of six marks each. You have to attempt only one of the alternatives in all such questions. • Use of calculators is not permitted.

PHYSICS 1.

q =

q0

7.

If nuclear density d ∝ An, where A is the atomic number then write the value of n.

8.

Why standard resistors are made of alloys.

9.

Name the quantities whose SI units are given below : 1. V – m 2. C-m out of the two also name the vector quantity.

is valid or not, q is the charge on

v2 c2 particle when it is moving with velocity v, q0 is the rest charge and c is the velocity of light . 1−

2.

A concave mirror is dipped inside the liquid of absolute refractive index 1.25. What will be the percentage change in its focal length.

3.

Write the name of a compound semiconductor.

4.

If one of the slit get closed in Young's Double slit experiment then fringe pattern will be observed or not on the screen.

5.

Write one of the use of Zener Diode.

6.

Name the experiment which proves the Dual Nature of electron.

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10. A transistor is shown in figure.

+2V

.

+1V +3V (i) Name the type of transistor (ii) Is the transistor is properly biased.

11. A time variant current is given i(t) = 1 + 3 2 sin (314 t + 30º) Find its root mean square value. 69

DECEMBER 2009

12. A metallic conductor of non-uniform cross-sectional area is shown in figure.

18. (i) Write the order of colors in secondary rainbow. (ii) Why sun appears reddish at the time of sunrise and sunset.

i p1

p2

19. The charges on the capacitors are Qa, Qb and Qc then calculate 1µF

p3 b

a

Qa

(i) Out of p 1, p 2, and p3 at which point the drift speed of the electron is maximum. (ii) Out of p 1, p2 and p3 at which point the current density is minimum.

12µF Qc

13. Stopping potential versus freq. of the incident light graph is shown in figure for metal-1 and metal-2 metal-1 metal-2 V0 θ1

θ2

30V

3µF

key-K

(i) Ratio of energy stored in 1 µF and 3 µF capacitor. (ii) Potential difference across 12 µF capacitor (iii) Energy supplied by the battery.

υ

20. (i) A uniform magnetic field of 0.5 T exist in the given solenoid. If an electron is projected along the axis of solenoid from a towards b with the speed of 3 × 10 2 m/s then find the Lorentz force working on electron.

(i) Which metal will have the higher value of threshold wavelength. (ii) Is θ1 = θ2 if yes then why ? 14. Draw the circuit diagram for finding the internal resistance of the cell using potentiometer.

a

15. (i) Define angle of Dip. (ii) What is the value of angle of Dip at a place on earth where Horizontal component and vertical component of earth magnetic field are equal.

i

b

(ii) When the current flows through the metallic spring why it get shrinked. 21. (i) For the given circuit diagram find the position of the null point. 5Ω 10 Ω

16. (i) State Kirchhoff's current law. (ii) Find the potential difference across the 2 ohm resistance for the given circuit diagram. 2Ω 4A 1Ω

A

3Ω

B G

1A

17. (i) State Lenz's law. (ii) If the current i increases then what will be the direction of induced current in the circular coil for the given figure.

i

Qb

Vbb

100 cm

Key-k Rh Rheostat

(ii) For the given figure draw truth table. A

b

B

Y

a a >> b

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70

DECEMBER 2009

22. (i) State Brewster's law for polarization . (ii) If the Brewster's angle for the given pair of medium is ip then express the critical angle for the same pair of media in terms of ip. (iii) If the lens is made of medium-2 and placed in medium-1. If the absolute refractive Index of medium-2 is µ2 and of medium-1 is µ1 then µ2 > µ1. Is it correct or not.

(iii) Where to put the proton that it will have maximum force on it. 27. For the given configuration find the –2q 2a

2a

Incident rays Refracted rays

+q

Medium -2

+q

2a (i) Dipole moment of the system. (ii) Electric potential energy of the system.

Medium-1 Medium-1

23. What is the need of Modulation in communication system. Draw the shapes of signal, carrier wave and amplitude Modulated wave.

28. Draw the circuit diagram for common emitter transistor amplifier. Explain its working. What is the phase difference between input and output voltage in case of common emitter transistor amplifier.

24. How a Galvanometer can be converted into Ammeter explain it by drawing the circuit diagram.

29. Explain construction and working of cyclotron. Why cyclotron can not be used to accelerate the electrons.

ig G Prove that S = i − ig G = Resistance of Galvanometer coil S = value of shunt ig = Full scale deflection current for Galvanometer i = Range of Ammeter.

30. State Ampere's circuital law. Using Ampere's circuital law find the magnetic field at the axis of the long solenoid.

CHEMISTRY

25. (i) Find the equivalent resistance between A and B for given fig.

1.

Give the IUPAC name of the organic compound (CH3)2 C = CH – C – CH3 . || O

2.

Complete the following reaction : CH3 – CH2 – CH = CH2 + HCl → …..

3.

State a use for the enzyme streptokinase in medicine.

4.

What is copolymerization ?

5.

Which type of a metal can be used in cathodic protection of iron against rusting ?

6.

Why is the bond dissociation energy of fluorine molecule less than that of chlorine molecule ?

7.

What is meant by inversion of sugar ?

8.

What are the types of lattice imperfections found in crystals ?

R R

A

R

R R

B

R

R

(ii) α is the symbol for the temperature coefficient of resistivity for the given material. If α → 0 then the material will be copper or constantan ? 26. (i) What is the angle between electric field line and equi-potential surface. (ii) If Ea, Eb and Ec are the electric field intensities at points a, b and c respectively, where to put a proton that it will have the maximum electric potential energy. c b a

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71

DECEMBER 2009

9.

Describe the mechanism of the formation of diethyl ether from ethanol in the presence of concentrated sulphuric acid.

21. Identify the substances A and B in each of the following sequences of reactions : alc.KOH

Br

2 (i) C2H5 Br → A → B

10. Predict, giving reasons, the order of basicity of the following compounds in (i) gaseous phase and (ii) in aqueous solutions (CH3)3 N, (CH3)2 NH, CH3NH2, NH3. 11. Write names of monomer/s of the following polymers and classify them as addition or condensation polymers. (i) Teflon (ii) Bakelite (iii) Natural Rubber

NaNO + HCl

Cu ( CN )

(ii)

2 2 → A2 → B NH 2 0 ºC

(iii)

Heat NH 2 → A → B

H2SO 4

22. Give the electronic configuration of the (a) d-orbitals of Ti in [Ti (H2O)6]3– ion in an octahedral crystal field. (b) Why is this complex coloured ? Explain on the basis of distribution of electrons in the d-orbitals.

12. Give an example for each of the following reactions : (i) Kolbe’s reaction. (ii) Reimer-Tiemann reaction.

23. State reasons for the following : (a) Rusting of iron is said to be an electrochemical phenomenon. (b) For a weak electrolyte, its molar conductance in dilute solutions increases sharply as its concentration in solution is decreased.

13. Write one distinction test each for : (i) Ethyl alcohol and 2–propanol (ii) Acetaldeyde and acetone 14. Distinguish between multimolecular and macromolecular colloids. Give one example of each type.

24. Using the valence bond approach predict the shape and magnetic character of [Co (NH3) 6] 3+. (Atomic number of Co is 27).

15. Draw the structure of pyrophosphoric acid and how it is prepared.

25. Explain the following : (a) F-centre (b) Schottky & Frenkel defect

16. If E° for copper electrode is +0.34 V how will you calculate its emf value when the solution in contact with it is 0.1 M in copper ions ? How does emf for copper electrode change when concentration of Cu2+ ions in the solution is decreased ? 17. Physical and chemical adsorptions respond differently to a rise in temperature. What is this difference and why is it so ?

26. Account for the following : (i) Ferric hydroxide sol is positively charged. (ii) The extent of physical adsorption decreases with rise in temperature. (iii) A delta is formed at the point where the river enters the sea.

18. Using the valence bond approach, predict the shape and magnetic character of [Ni (CO)4]. (At No. of Ni = 28).

27. Taking two examples of heterogeneous catalytic reactions, explain how a heterogeneous catalyst helps in the reaction.

19. Describe the following giving a chemical equation for each : (i) Markownikoff’s rule (ii) Hofmann Bromide Reaction

28. (a) An organic compound ‘A’ with molecular formula C5H8O2 is reduced to n-pentane on treatment with Zn-Hg/HCl. ‘A’ forms a dioxime with hydroxylamine and gives a positive lodoform test and Tollen’s test. Identify the compound A and deduce its structure. (b) Write the chemical equations for the following conversions : (not more than 2 steps) (i) Ethyl benzene to benzene (ii) Acetaldehyde to butane – 1, 3–diol (iii) Acetone to propene

20

(a) Write chemical equations and reaction conditions for the conversion of : (i) Ethene to ethanol (ii) Phenol to phenyl ethanoate (iii) Ethanal to 2-propanol

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72

DECEMBER 2009

29. Describe how potassium dichromate is made from chromite ore and give the equations for the chemical reactions involved. Write balanced ionic equations for reacting ions to represent the action of acidified potassium dichromate solution on : (i) Potassium iodide solution (ii) Acidified ferrous sulphate solution Write two uses of potassium dichromate.

8.

→

9.

3.

4. 5.

6.

7.

3 and 2 respectively.

11. Consider f : N → N, g : N → N and h : N → R defined as f (x) = 2x, g (y) = 3y + 4 and h (z) = sin z ∀ x, y and z in N. Show the ho(gof) = (hog)of.

1− x 12. Differentiate cot–1 w.r.t. x. 1+ x 13. Solve the differential equations : (1 + e2x) dy + e x (1 + y2) dx = 0. Give that y = 1, when x = 0. or dy Solve the differential equation : x − y − 2x 3 = 0 dx π/ 4

14. Evaluate :

∫ sin 2x sin 3x dx. 0

x. tan −1 x dx.

or π/ 4

Evaluate :

Find the differential equation of the family of curves given by- x2 + y2 = 2ax.

∫ log(1 + tan x)dx 0

Find the principle value of tan–1 (–1).

15. Evaluate :

Find a matrix C such that 2A – B + C = 0 3 1 − 2 1 Where A = and B = 0 2 0 3

∫ 2x

3x + 1 2

− 2x + 3

dx .

16. If x = a (θ – sin θ) and y = a (1 – cos θ), find

If A is a square matrix of order 3 such that | adj A | = 64, find | A |.

d2y dx

2

at θ =

π . 2

17. Find the value of K so that the Kx + 1, if x ≤ π f ( x) = cos x , if x > π or

4 3 5 Find the value of x if the matrix A = 3 − 2 7 is 10 − 1 x singular.

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→

Section B

Show that the relation R in the set {1, 2, 3} is given by R = {(1, 2), (2, 1)} is symmetric.

∫

→

10. Find the direction cosines of a line which make equal angles with the co ordinate axes.

Section A

Evaluate :

→

What is the angle between vector a and b with magnitude

MATHEMATICS

2.

→

in the direction of a + b + c .

30. Give appropriate reasons for each of the following observations : (i) Sulphur vapour exhibits some paramagnetic behaviour. (ii) Silicon has no allotropic form analogous to graphite. (iii) Of the noble gases only xenon is known to form real chemical compounds. (iv) Nitrogen shows only a little tendency for catenation, whereas phosphorus shows a clear tendency for catenation.

1.

→ → → If a = ˆi + ˆj ; b = ˆj + kˆ ; c = kˆ + ˆi find a unit vector

function,

x 3 + 3, if x ≠ 0 Show that the function f ( x ) = 1 , if x = 0

73

DECEMBER 2009

24. Using integration, find the area of the circle x2 +y2 = 16, which is exterior to the parabola y2 = 6x. or Find the area of the smaller region bounded by the

x −1 x +1 π 18. Solve for x : tan −1 + tan −1 = ; x 2 − x+2 4 x < 1.

4 − 5 − 11 19. If A = 1 − 3 1 find A–1 2 3 − 7 or Using the properties of determinates prove that a +x y z x a+y z = a 2 (a + x + y + z ) x y a+z

ellipse

y2 b

2

= 1 and the line

x y + = 1. a b

1 − 1 1 26. For A = 2 1 − 3 , find A–1 and hence solve the 1 1 1 system of equations. x + 2y + z = 4 –x+ y+z=0 x – 3y + z = 2

→

21. If a and b are unit vectors and θ is the angle between them, then prove that cos

27. A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope just two consecutive letters TA are visible. What is the probability that the letter has come from(i) TataNagar (ii) Calcutta or Find the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement from a bag containing 4 white and 6 red balls. Also find mean and variance of the distribution.

θ 1→ → = a+ b 2 2

22. A football match may be either won, drawn or lost by the host country team. So there are three ways of forecasting the result of any one match, one correct and two incorrect. Find the probability of forecasting at least three correct results for four matches. or A candidate has to reach the examination centre in time. Probabilities of him going by bus or scooter or 3 1 3 by other means of transport are , , 10 10 5 respectively. The probability that he will be late is 1 1 and respectively, if he travels by bus or scooter. 4 3 But he reaches in time if he uses any other mode of transport. He reached late at the center. Find the probability that he traveled by bus.

28. Find the distance of the point (3, 4, 5) from the plane x + y + z = 2 measured parallel to the line 2x = y = z. 29. Every gram of wheat provides 0.1 gm of proteins and 0.25 gm of carbohydrates. The corresponding values for rice are 0.05 gm and 0.5 gm respectively. Wheat costs Rs. 4 per kg and rice Rs. 6 per kg. The minimum daily requirements of protein and carbohydrates for an average child are 50 gms and 200 gms respectively. In what quantities should wheat and rice be mixed in the daily diet to provide minimum daily requirements of proteins and carbohydrates at minimum cost. Frame an L.P.P. and solve it graphically.

Section C π/ 2

23. Evaluate :

a

2

+

25. Show that a right circular cylinder which is open at the top, and has a given surface area, will have the greatest volume if its height is equal to the radius of its base.

20. Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0, measured parallel to the line x +3 y −2 z = = 3 6 2 →

x2

∫ 0

cos x dx (1 + sin x )(2 + sin x )

XtraEdge for IIT-JEE

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DECEMBER 2009

XtraEdge Test Series ANSWER KEY IIT- JEE 2010 (December issue) PHYSICS Ques Ans Ques Ques Ques Ans

1 B 10 11 12 3

2 A A → P,Q,R,S A→ Q 13 2

3 B

4 5 C A, B B → Q,P,Q,R B→R 15 16 2 6

14 5

6 7 C, D A, B, D C → P,Q,R,S C→P 17 18 2 1

8 9 B, C A, B, C, D D → P,Q,R,S D→S 19 1

C H EM I STR Y Ques Ans Ques Ques Ques Ans

1 D 10 11 12 1

2 D A→ R A → Q,S 13 2

3 D

14 2

4 C B → Q,T B → R,S 15 0

5 D

16 2

6 A, B, C C→P C → P,Q,T 17 9

7 B

18 0

8 A, B, D D→S D → R,S 19 6

9 B, C, D

MATHEMATICS Ques Ans Ques Ques Ques Ans

1 B 10 11 12 1

2 C A→ Q A→ Q 13 4

3 C

14 6

4 C B→R B→S 15 4

5 A, B, D

16 1

6 A, C C→P C→P 17 3

7 A, B, C, D

18 3

8 A, D D→S D→R 19 3

9 D

IIT- JEE 2011 (December issue) PHYSICS Ques Ans Ques Ques Ques Ans

1 D 10 11 12 2

2 B A→ Q A→ R 13 3

3 C

14 3

4 D B→S B→P 15 5

5 A, B, C, D

16 2

6 A, C C→R C→S 17 3

7 A, C, D

6 A, B, D C→Q C→P 17 1

7 A, B, D

18 0

8 A, B, C, D D→P D→Q 19 1

9 A, B, C, D

8 C D → P,T D→Q 19 1

9 D

8 A, B D→R D→P 19 4

9 A, C

C H EM I STR Y Ques Ans Ques Ques Ques Ans

1 A 10 11 12 1

2 B A→ R A→ S 13 7

3 D

14 2

4 5 C A, C, D B → P,S,T B→R 15 16 7 8

18 9

MATHEMATICS Ques Ans Ques Ques Ques Ans

1 B 10 11 12 2

XtraEdge for IIT-JEE

2 C A→ S A→ S 13 5

3 A

14 3

4 C B→P B→R 15 6

5 A, B, C, D

16 1

75

6 A, B, C C→Q C→Q 17 2

7 B, D

18 0

DECEMBER 2009

XtraEdge for IIT-JEE

76

DECEMBER 2009

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