May 2011 - Career Point

Volume - 6 Issue - 11May, 2011 (Monthly Magazine)Editorial / Mailing Office :112-B, Shakti Nagar, Kota (Raj.)Tel. : 0744-2500492, 2500692, 3040000e-mail : xtraedge@gmail.comEditor :Pramod Maheshwari[B.Tech. IIT-Delhi]Cover DesignOm Gocher, Govind SainiLayoutRajaram GocherCirculation & AdvertisementPraveen ChandnaPh 0744-3040000, 9672977502SubscriptionSudha Jaisingh Ph. 0744-2500492, 2500692© Strictly reserved with the publishers• No Portion of the magazine can bepublished/ reproduced without thewritten permission of the publisher• All disputes are subject to theexclusive jurisdiction of the KotaCourts only.Every effort has been made to avoid errors oromission in this publication. Inr spite of this,errors are possible. Any mistake, error ordiscrepancy noted may be brought to ournotice which shall be taken care of in theforthcoming edition, hence any suggestion iswelcome. It is notified that neither thepublisher nor the author or seller will beresponsible for any damage or loss of action toany one, of any kind, in any manner, there from.Unit Price ` 20/-Special Subscription Rates6 issues : ` 100 /- [One issue free ]12 issues : ` 200 /- [Two issues free]24 issues : ` 400 /- [Four issues free]Owned & Published by PramodMaheshwari, 112, Shakti Nagar,Dadabari, Kota & Printed by NavalMaheshwari, Published & Printed at 112,Shakti Nagar, Dadabari, Kota.Editor : Pramod MaheshwariDear Students,All of us live and work within fixed patterns. These patterns and habitsdetermine the quality of our life and the choices we make in life. Thereare a few vital things to know about ourselves. We should becomeaware of how much we influence others, how productive we are andwhat can help us to achieve our goals. It is important to create anenvironment which will promote our success. We should consciouslycreate a system that would enable us to achieve our goals. Most of uslive in systems which have come our way by an accident, circumstancesor people we have met over a period of time. We are surrounded by ourcolleagues or subordinates who happened to be there by the fact ofsheer recruitment earlier or later by the management. Our daily routinesand schedules have been formed on the basis of convenience,coincidence, and the expectations of society and sometimes due tosuperstitions. The trick for success is to have an environment that helpsin attaining our goals. Control your life. Make an effort to launch yourday with a great start. A law of physics says that an object set in motiontends to remain in motion. It is the same thing with daily routine. Tohave a good start each morning will keep you upbeat during the day. Ifyou begin the day stressed, you will tend to remain so that way. Thebest is to create a course of action or conditions where you are nothassled for being late for a meeting, worried about household affairs ordistracted by happenings in the world.Aim to be highly successful. Control the direction of your life. Not onlyshould you start the day on a cheerful note but also continue to do soduring the day. Keep yourself stimulated and invigorated during theentire day. Start your day with a purpose. Have a daily direction andtrajectory of action. It will keep you on your course all day long.Throughout the day reinforce your positive values and your choices.Anything that helps you in maintaining your highest values and yourmost important priorities should be welcome. Be in control of your lifeand work. Create and sustain a wonderful environment filled withbeauty, peace, inspiration and hope.Plan your day in such a way that suits your plans objectives and makesyou feel just right with the right amount of encouragement during theentire day. You should give a direction to your day and timing.Presenting forever positive ideas to your success.Yours trulyPramod Maheshwari,B.Tech., IIT DelhiWorry is a misuse of imagination.EditorialXtraEdge for IIT-JEE 1 MAY 2011

XtraEdge for IIT-JEE 2 MAY 2011

Volume-6 Issue-11May, 2011 (Monthly Magazine)NEXT MONTHS ATTRACTIONS Much more IIT-JEE News. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics,, Chemistry & Maths Key Concepts & Problem Solving strategy for IIT-JEE. Xtra Edge Test Series for JEE- 2012 & 2013 AIEEE 2011 Examination Paper & SolutionINDEXCONTENTSRegulars ..........PAGENEWS ARTICLE 4• Pollen, Nanoparticles, and Asthma :The Aerosol Science of Allergic Disease• Seminar on IIT olympiad CoursesIITian ON THE PATH OF SUCCESS 6Mr. NarayanaswamyKNOW IIT-JEE 7Previous IIT-JEE QuestionStudy Time........DYNAMIC PHYSICS 14SSuccess Tips for the Months• If you can't make a mistake, you can't makeanything.• Sometimes a big step is safer; you can'tcross a ditch in small jumps• Self-confidence grows not from what youcan do, but what you know you can do.• Children focus on what they can’t do.Adults focus on what they can do.• The secret of confidence is to know yourresources.• You never need to feel fear if you don'twant to do anything.• You got to know when to hold ‘em andknow when to fold ‘em…• An ounce of success is worth a pound ofpositive thinking.• To understand motivation, know the powerof the Hunter.• Defeat is advance payment for victory.8-Challenging Problems [Set # 1]Students’ ForumPhysics Fundamentals• Electrostatics-I• 1-D Motion, Projectile MotionCATALYSE CHEMISTRY 30Key Concept• Gaseous State• General Organic ChemistryUnderstanding : Physical ChemistryDICEY MATHS 38Mathematical ChallengesStudents’ ForumKey Concept• Complex Number• Matrices & DeterminantsTest Time ..........XTRAEDGE TEST SERIES 46Class XII – IIT-JEE 2012 PaperClass XI – IIT-JEE 2013 PaperIIT - 2011 Examination Paper with Solution 62XtraEdge for IIT-JEE 3 MAY 2011

"Pollen, Nanoparticles, andAsthma: The Aerosol Science ofAllergic Disease"Pollen particles are so large thatinhaled pollen deposits primarily inthe nose and throat, yet allergists warnasthmatics that pollen may triggerairway inflammation deep in thelungs, leading to bronchoconstriction.At the same time, epidemiologicalstudies yield conflicting results on thelink between pollen and asthma, butdo show links between exposures tofine particles and asthma. What beganas an effort to unravel the physicsbehind this seemingly contradictoryadvice stimulated studies of windpollinatingplants, pollen grains, andrespirable allergens. Botanists hadobserved that live pollen grains willrupture when immersed in water, andspeculated that the cytoplasmicmaterial was the source of respirableallergen that could trigger asthma.Initial experiments confirmed ourspeculation that adhesion forces at themicrometer scale are too strong for themechanism to explain the link, at leastas it was originally stated.This seminar will examine how thestructures of the rarely noticed flowersenable pollen antigen to be released inrespirable particles. Along the way,we will examine some of theremarkable methods that windpollinatingplants use to release theirpollen. The new understanding of theways that pollen antigen enters the air,and the wide range of sizes ofparticles in which it is found posesnew challenges for the measurementof these airborne allergens. Presentmeasurements are limited to manualmicroscopic analysis of samplescollected from the air, a tedious andlabor intensive undertaking. As aresult, the number of locations wherehistorical pollen counts have beenrecorded is very limited, challengingefforts both to quantify the associationbetween pollen and asthma and tounderstand how climate change mayaffect allergic disease in the comingdecades. Measurement of the respirableallergen have only been reported for afew localized studies. A computervision system has been developed toaddress the former problem. Anultrasensitive, label-free antibody assaybased upon high-Q whispering gallerymode optical resonators is underdevelopment to address the latter.Seminar on IIT OlympiadCoursesRavindra Bharathi IIT OlympiadSchool conducted an IIT Olympiadawareness seminar here on Saturday atRavindra Bharathi School in Kakinada.The motto of the seminar is to createawareness about IIT Olympiad courseson students and parents. Speaking onthis, Chief guest Ravindra Bharathi IITOlympiad School Principal Mr. M.RamNaresh explained about the importanceof IIT Olympiad courses for thestudents and parents presided in theseminar.IIT-Roorkee new campusinauguratedIIT-Roorkee today inaugurated itsGreater Noida Campus at theKnowledge Park-II area here. It wasinaugurated by Ashok Bhatnagar,Chairman, Board of Governors, IITRoorkee in the presence of Director S CSaxena.IITs in world’s top 100 universitieson technical subjectsThe Indian Institutes of Technology(IITs) India’s premier technicalinstitutes have made it to the world’stop 100 universities as per the recentlyreleasedsubject-wise rankings byQuacquarelli Symonds (QS). In theengineering and technology rankings,IIT Bombay has bagged the 47th spotwhile IIT Delhi is at 52nd position.IIT Kanpur follows at 63rd and IITMadras at 68th.HCL FOUNDER S HIV N ADARTO HEAD IIT-KHARAGPUR’ SBOARDShiv Nadar, founder of softwaremajor HCL, was appointed thechairman of the board of governorsof the Indian Institute of Technology(IIT), Kharagpur, a statement fromthe company said Thursday.IIT-Kharagpur is the onlyengineering institution from Indialisted among top 500 universitiesworldwide in the Shanghai Jiao TongUniversity’s ‘Academic Ranking ofWorld Universities’.‘It is an honour to be entrusted theresponsibility to further advance theglobal stature this great institutionhas built already,’ said Nadar.Former chairpersons of IIT-Kharagpur include former WestBengal chief minister B.C. Roy,former chairman and managingdirector of Tata Steel Russi Mody,chairman emeritus of BallarpurIndustries Ltd. L.M. Thapar, andchairman emeritus of RPG GroupR.P. Goenka.IIT Delhi's golden jubileelecture by Professor SusanGreenfieldIIT Delhi's Golden Jubilee Lectureby Professor Susan Greenfield, titled‘The 21st century mind: How willtechnology change it?’Susan Greenfield is a scientist,XtraEdge for IIT-JEE 4 MAY 2011

writer, broadcaster and member of theHouse of Lords. She is also Directorof the Institute for the Future of theMind. She has been awarded 30Honorary Degrees from British andForeign universities and heads amulti-disciplinary research groupexploring novel brain mechanismslinked to neurodegenerative diseasessuch as Alzheimer’s and Parkinson’s.In addition she has developed aninterest in the impact of 21st Centurytechnologies on how young peoplethink and feel, as discussed in herbook ID.IIT student wins award forrevolutionary researchMumbai: Prashanthi Kovur, a PhDstudent from the Centre for Excellencein Nanoelectronics, EE Department,IIT Bombay has been awarded theoutstanding student researcher in thefield of Physics, Chemistry ofMaterial for nano-scale devices by theworld largest semiconductor companyTSMC.Taiwan Semiconductor ManufacturingCompany (TSMC) is one of theworld's largest semi conductorfoundry, which provides leadingtechnologies for the semiconductormanufacturing industry. The purposeof the TSMC Outstanding StudentResearch Award is to recogniseexceptional semiconductor relatedresearch carried out by graduatestudents. The competition for thisaward attracted hundreds ofapplications from students fromvarious universities all over the world.Prashanthi's research is based on thedevelopment of a novel multiferroicsystem for MEMS applications.Magnetism and ferroelectricity areessential to many forms of currenttechnology, and the quest formultiferroic materials, where thesetwo phenomena are intimatelycoupled, is of great technological andfundamental importance.It has been shown for the first timethat room temperature multiferrocitywith significant coupling could beachieved in Dy modified BiFeO3system. The current work focuses onthe fabrication of RF MEMS switchusing this novel multiferroic basedactuator.This work was done under the guidanceof Prof Vaijayanti R Palkar from EEDepartment, IIT Bombay. Prof Palkarwho was invited as an honourable guestfor the award ceremony delivered alecture on 'Semiconductors in India andFunctional Nano-oxides'.IIT Bombay, Applied Materialslaunch State-of-the-Art LaboratoryIIT Bombay, Applied MaterialsCollaboration and the inauguration ofthe "Chemistry Laboratory for Energyand Nanoelectronics" (CLEAN)Applied Materials, Inc. the world'sleading supplier of manufacturingsolutions for the semiconductor, displayand solar industries, and IIT Bombay,India's highest-rated niversity and aleader in education and research, todayannounced the state-of-the-art "AppliedMaterials Chemistry Laboratory forEnergy and Nanoelectronics" (CLEAN)at the IIT Bombay (IITB) campus. Thisnew laboratory expands the scope ofcollaboration between IIT Bombay andApplied Materials to include thedevelopment of new materials that canpotentially be used in a variety ofelectronic and renewable energyfocusedapplications, including thefabrication of next-generation solarcells.The lab was launched at a ceremonyattended by Mike Splinter, chairman ofthe board, president and CEO ofApplied Materials, OmkaramNalamasu, chief technology officer ofApplied Materials, and Prof. DevangKhakhar, director, IIT Bombay."This is a great example of the kind ofuniversity and corporate collaborationthat is helping to advance technologyby enabling world-class research,innovation and workforcedevelopment," said Mike Splinter. "Ourgoal is to serve as a catalyst fordeveloping the critical technologyneeded to solve the many challenges ofnext-generation electronic and solardevice manufacturing.""Applied Materials has grown tobecome IIT Bombay's mostimportant industry collaborator interms of the scale of researchcollaboration," said ProfessorDevang Khakhar. "We welcome theestablishment of the AppliedMaterials CLEAN laboratorythrough Applied's generous support.This will begin a new phase of thecollaboration in areas related torenewable energy, which are a focusof IIT Bombay's research."The event also celebrates asuccessful five-year relationshipbetween IIT Bombay and AppliedMaterials for nanoelectronics andsolar photovoltaic technologyresearch. During the course of thisspecial alliance, Applied Materialshas endowed IIT Bombay with over$12 million for the followingprojects:• The establishment of theAppliedMaterialsNanomanufacturing Laboratory,India's first 200mmsemiconductor fabricationfacility and one of the fewuniversity-based 200mmfacilities worldwide.• Collaborative research onnanoelectronics and solar PVtechnology.• Applied's donation of threestate-of-the-art physical vapourdeposition and chemical vapourdeposition process chambers tothe National Centre forPhotovoltaic Research andEducation (NCPRE) fordepositing thin films for solarcell applications.• A solar PV and LED lightingsystem that lights up the mainavenue at IITB's campus• The establishment of TheApplied Materials ChemistryLaboratory for Energy andNanoelectronicsXtraEdge for IIT-JEE 5 MAY 2011

Success StoryThis article contains story/interviews of persons who succeed after graduation from different IITsMr. NarayanaswamyIIT Chennai, IAS officer.(Secretary General Education, kerala)On passing out from IIT Chennai Mr. Narayanaswamywas offered scholarship by the prestigious MassachusettsInstitute of Technology, USA. He who came from amiddle class family believed that he had a moral obligationto give something in return for the lakhs of rupees thegovernment spent on him as an IIT student. He had theintelligence and conviction to realize that this money camealso from the poorest of the poor - who pay up the exciseduty on textiles when they buy cloth, who pay up customs,excise and sales tax on diesel when they travel in a bus,and in numerous other ways indirectly pay thegovernment. So he decided to join IAS hoping he could dosomething for the people of this country. How many youngmen have the will power to resist such an offer from USA?Narayanaswamy did never look at IAS as a black moneyspinner as his later life bears testimony to this fact.After a decade of meritorious service in IAS, today,Narayanaswamy is being forced out of the IAS profession.Do you know why?A real estate agent wanted to fill up a paddy field which isbanned under law. An application came up beforeNarayanaswamy who was sub collector the, for anexemption from this rule for this plot of land. Uponvisiting the site he found that the complaint from 60 poorfamilies that they will face water logging due to the wastewater from a nearby Government Medical College if thispaddy field was filled up was correct. Narayanswamycame under intense political pressure but he did what wasright - refused permission for filling up the paddy field.That was his first confrontation with politicians.Soon after his marriage his father-in-law closed down apublic road to build compound wall for his plot of land.People approached Narayanaswamy with complaint. Whentalking with his own father-in-law did not help, heremoved the obstructing wall with police help. The result,his marriage broke up.As district Collector he raided the house of a liquor baronwho had defaulted Rupees 11 crores payment togovernment and carried out revenue recovery. A Ministerdirectly telephoned him and ordered to return the forfeitedarticles to the house of the liquor baron. Narayanswamypolitely replied that it is difficult. The minister replied thatNarayanaswamy will suffer.In his district it was a practice to collect crores of rupeesfor earthen bunds meant for poor farmers, but which werenever constructed. A bill for rupees 8 crores came upbefore Narayanaswamy. He inspected the bund. He foundit very weak and said that he will pass the bill after therainy season to ensure that the bund served the purpose. Asexpected the earthen bund was too weak to stand the rainand it disappeared in the rain. But he created a lot ofenemies for saving 8 crores public money.The net result of all such unholy activities was that he wasasked to go on leave by the government. Later such anillustrious officer was posted as "State Co-Ordinator,Quality Improvement Programme for Schools". This iswhat the politician will do to a honest officer withbackbone - post him in the most powerless position toteach him a lesson.Since he found that nothing can be achieved for the peopleif he continued with the State Service he opted for centralservice. But that too was denied on some technical ground.What will you do when you have a brilliant computercareer anywhere in the world you choose with the backingof several advanced technical papers too published ininternational journals to your credit? When you arepowerless to do anything for the people, why should youwaste your life as the Co-Ordinator for a SchoolsProgramme?Mr. Narayanaswamy is on the verge of leaving IAS to goto Paris to take up a well paid United Nations assignment.The politicians can laugh thinking another obstacle hasbeen removed. But it is the helpless people of this countrywho will lose - not Narayanaswamy. But you have thepower to support capable and honest bureaucrats likeNarayaswamy, G.R.Khairnar and Alphons Kannamthanamwho have suffered a lot under self seeking politicians whorule us. You have even the power to replace suchpoliticians with these kind of people dedicated to thecountry. The question is will you do the little you can doNOW? At least a vote or word in support of suchpersonalities?XtraEdge for IIT-JEE 6 MAY 2011

KNOW IIT-JEEBy Previous Exam QuestionsPHYSICS1. A non-conducting disc of radius a and uniform positivesurface charge density σ is placed on the ground, withits axis vertical. A particle of mass m and positivecharge q is dropped, along the axis of the disc, from aheight H with zero initial velocity, the particle hasq/m = ε 0 g/σ.[IIT-1990](a) Find the value of H if the particle just reaches thedisc(b) Sketch the potential energy of the particle as afunction of its height and find its equilibriumposition.Sol. (A) Given that : a = radius of disc, σ = surfacecharge density, q/m = 4ε 0 g/sThe K.E. of the particle, which it react reaches thedisc can be taken as zero.Potential due to a charge disc at any axial pointsituated at a distance x from 0.σ 2 2V(x) = [ a + x – x]2ε0σ 2 2Hence, V(H) = [ a + H – H]2ε0σaand V(O) =2ε0According to law of conservation energy, Loss ofgravitation potential energy = gain in electricpotential energy.H(m,q)or H = 2[a + H – (a2 2+ H ) ]2 +or H = 2a + 2H – 2 (a H )or 2 (a2+ H ) = H + 2aor 4a 2 + 4H 2 = H 2 + 4a 2 + 4aHor 3H 2 = 4aH or4aH =3[Q H = O is not valid](B) Total potential energy of the particle at height hU(x) = mgx + qV(x) = mgx + σ 2( a + x– x)]2ε0= mgx + 2mg [ (a2+ x ) – x]= mg [2 (a2 2+ x ) – x]...(ii)dUFrom equilibrium : = 0 dxaThis gives : x =3From equation (ii) graph between U(x) and x is andshown aboveU2 mga3 mgaO a/ 3 H = 4a/3X2HO amgH = q∆H = q[V(0) – V(H)]mgH = q[a – { (a2 2 σ+ H ) – H }]2ε 0σqFrom the given relation : = 2 mg (given)2ε0Putting this is equation (i), we get,mgH = 2mg [a – { (a2 2+ H ) – H }]...(i)2. Light is incident at an angle α on one planar end of atransparent cylindrical rod of refractive index.Determine the least value of n so that the light enteringthe rod does not emerge from the curved surface of rodirrespective of the value α [IIT- 1992]αβ90 – βSol. The light entering the rod does not emerge from thecurved surface of the rod when the angle 90 – r isgreater than the critical angle.nXtraEdge for IIT-JEE 7 MAY 2011

k 21[Where L is the relative distance between A and B]i.e., µ ≤ where C is the critical angle.sin Cor t 2 2L 2L= =Here C = 90 – ra A / B a A – a B1L = 2m ⇒ L = aA/B t 2 = 0.785 + 0.5 + 1.05 + 0.52 = 2.83 (app.)⇒11Putting values we get, t 2 = 4 or t = 2s.]µ ≤⇒ µ ≤sin(90 – r)cos rDistance moved by B during that time is given by11As a limiting case µ =...(i)S = 2Bcos r2t1 0.79 2× 0.7Applying Snell's law at A= × × 4 =2sin αsin α2 2µ =⇒ sin r = ...(ii)sin rµSimilarly for A = 8 2 m.× 10 = 7 2 mThe smallest angle of incident on the curved surface 4. Two light springs of force constants k 1 and k 2 and aπ block of mass m are in one line AB on a smoothis when α = . This can be taken as a limiting case2 horizontal table such that one end of each spring isfor angle of incidence on plane surfacefixed on rigid on rigid supports and the other end is freesin π / 2 1shown in the figure. The distance CD between the freeFrom (ii) sin r = ⇒ µ =...(iii)µ sin rends of the springs is 60 cms. If the block moves alongFrom (i) and (ii) sin r = cos rAB with a velocity 120 cm/sec in between the springs,⇒ r = 45ºcalculate the period of oscillation of the block.⇒1 1(k 1 = 1.8 N/m, k 2 = 3.2 N/m. m = 200 gm)µ = =cos 45º 1/ 2[IIT-1985]60 cm⇒ µ = 2k 1This is the least value of the refractive index of rodVfor light entering the rod and not leaving it from thecurved surface.A C m D B3. Two block A and B of equal masses are placed onrough inclined plane as shown in figure. When andwhere will the two blocks come on the same line on theinclined plane if they are released simultaneously?Initially the block A is 2 m behind the block B.Sol. The mass will strike the right spring, compress it.The K.E. of the mass will convert into P.E. of thespring. Again the spring will return to its natural sizethereby converting its P.E. to K.E. of the block. TheTtime taken for this process will be . where 2Co-efficient of kinetic friction for the blocks A andmT = 2π .B are 0.2 and 0.3 respectively (g = 10 m/s 2 ).k2 m[IIT-2005]T m 0.2∴ t 1 = = π = π 2 K 2 1. 8= 0.785AThe block will move from A to B without anyacceleration. The time taken will beB A60t 2 = = 0.5 120BNow the block will compress the left spring and then45ºthe spring again attains ite natural length. The timemgsinθ – µ kmg cosθtaken will beSol. a =mm 0.2∴ a A = g sin θ – µ kA g cos θ ...(i)t 3 = π = π = 1.05K 1 1. 8and a B = g sin θ – m kB g cos θ ...(ii)Putting values we getAgain the block moves from B to A, completing oneoscillation. the time taken for doing so0.89 0.79a A = and α B =6022t 4 = = 0.5 120a AB is relative acceleration of A' w.r.t.∴ The complete time of oscillation will beB = a A – a B= t 1 + t 2 + t 3 + t 4XtraEdge for IIT-JEE 8 MAY 2011

60 cmK 1 K 2VB5. A solid sphere of copper of radius R and a hollowsphere of the same material of inner radius r and outerradius A are heated to the same temperature andallowed to cool in the same environment. Which ofthem starts cooling faster ?[IIT-1982]Sol. Since the temperature and surface area is same,therefore the Energy emitted per second by bothspheres is same.We know that Q = mc∆TSince Q is same and c is same (both copper)1∴ m ∝∆TMass of hollow sphere is less∴ Temperature change will be more.∴ Hollow sphere will cool faster.mCHEMISTRY6. The molar volume of liquid benzene(density = 0.877 g ml –1 ) increases by a factor of 2750as it vaporizes at 20ºC and that of liquid toluene(density = 0.867 g ml –1 ) increases by a factor of 7720at 20ºC. A solution of benzene and toluene at 20ºChas a vapour pressure of 46.0 torr. Find the molefraction of benzene in vapour above the solution.[IIT-1996]Sol. Given that,Density of benzene = 0.877 g ml –1Molecular mass of benzene (C 6 H 6 )= 6 × 12 + 6 × 1 = 7878∴ Molar volume of benzene in liquid form = ml 0.87778 1= × L = 244.58 L0.877 1000And molar volume of benzene in vapour phse78 2750= × L = 244.58 L0.877 1000Density of toluene = 0.867 g ml –1Molecular mass of toluene (C 6 H 5 CH 3 )= 6 × 12 + 5 × 1 + 1 × 12 + 3 × 1 = 92∴ Molar volume of toluene in liquid form92 92 1= ml = × L0.867 0. 867 1000And molar volume of toluene in vapour phase92 7720= × L = 819.19 L0.867 1000Using the ideal gas equation,PV = nRTAAt T = 20ºC = 293 K0 nRTFor benzene, P = P B =V1 × 0.082×293== 0.098 atm244.58= 74.48 torr (Q 1 atm = 760 torr)Similarly, for toluene,0 nRTP = P T =V1 × 0.082×293== 0.029 atm819.19= 22.04 torr (Q 1 atm = 760 torr)According to Raoult's law,0P B = PBx B = 74.48 x B0P T = PTx T = 22.04 (1 – x B )0 0And P M = PBx B + PTx Tor 46.0 = 74.48 x B + 22.04 (1 – x B )Solving, x B = 0.457According to Dalton's law,'P B = P M x B (in vapour phase)or mole fraction of benzene in vapour form,' PB74 .48×0.457x B = == 0.74P 46.0M7. 0.15 mol of CO taken in a 2.5 L flask is maintained at705 K along with a catalyst so that the followingreaction takes placeCO(g) + 2H 2 (g) CH 3 OH(g)Hydrogen is introduced until the total pressure of thesystem is 8.5 atm at equilibrium and 0.08 mol ofmethanol is formed. Calculate (a) K p and K c and (b)the final pressure if the same amount of CO and H 2 asbefore are used, but with no catalyst so that thereaction does not take place. [IIT-1993]Sol. We haveCO(g) + 2H 2 (g) CH 3 OH(g)t = 0 0.15 molt eq 0.15 mol – x ( n H 2) 0 – 2x xIt is given that 0.08 mol of CH 3 OH is formed atequilibrium. Hencen CH 3 OH = x = 0.08 moland n CO = 0.15 mol – x = 0.07 molFrom the total pressure of 8.5 atm equilibrium, wecalculate the total amount of gases, i.e. CO, H 2 andCH 3 OH at equilibrium.pV (0.08mol / 2.5L)n total = = RT– 1 −1(0.082atm L K mol )(705K)= 0.3676 molNow, the amount of H 2 at equilibrium is given asn H 2= n total – n CO – n CH 3 OH= (0.367 – 0.07 – 0.08) mol = 0.2176 molXtraEdge for IIT-JEE 9 MAY 2011

Hence, K C =[CH OH]3[CO][H2]2(8.5atm)(2.5L)=2(0.07 mol / 2.5L )(0.2176mol / 2.5L)= 150.85 (mol L –1 ) –2Now K p = K c (RT) ∆v g= (150.85 mol –2 L 2 ){(0.082 L atm K –1 mol –1 )(705 K)} –2= 0.04513 atm –2Since n = ( n ) 0 – 2x, we have( n ) 0 =H 2H 2n H 2H 2+ 2x = (0.2176 + 2 × 0.08)mol= 0.3776 molTotal amount of CO and H 2 in the reacting systembefore the reaction sets in is given asn 0 = (n CO ) 0 + ( n H 2) 0= (0.15 + 0.3776)mol = 0.5276 molnHence, p 0 = 0 RTV=(0.5276mol)(0.082L atm K(2.5L)= 12.20 atm−1 −1mol)(705K)8. An ester A(C 4 H 8 O 2 ), on treatment with excess methylmagnesium chloride followed by acidification, givesan alcohol B as the sole organic product. Alcohol B,on oxidation with NaOCl followed by acidification,gives acetic acid. Deduce the structures of A and B.Show the reactions involved. [IIT-1998]Sol. The reactions of an ester with methyl magnesiumchloride are as follows.OOMgClOCHR–C–OR´ 3MgClHR–C–OR´+R–C–CH 3 + R´OH–HOMgCl(A)CH 3CH 3 MgClOHOMgClHR–C–CH 3–HOMgClR–C–CH 3CH 3CH 3(B)Since the given ester (C 4 H 8 O 2 ) produces only onealcohol B, it follows that RC(CH 3 ) 2 OH and R´OHmust be identical. Thus, the alkyl group R´ must beRC(CH 3 ) 2 – and the given ester A isO CH 3R – C – O – C – CH 3(molecular formula R 2 C 4 H 6 O 2 )RFrom the molecular formula of A, we conclude that Rmust be H atom. Hence, the given ester isOH – C – O – CH – CH 3CH 3The alcohol B is a secondary alcohol.Isopropyl formateCH 3 – CH – CH 3Isopropyl alcoholOHThe oxidation of alcohol B with NaOCl will give aketone which further undergoes a haloform reaction.CH 3 – CH – CH 3 + NaOClOHCH 3 – C – CH 3 + 3NaOClOCH 3 – C – CCl 3 + NaOHOCH 3 – C – CH 3 + NaCl + H 2 OOCH 3 – C – CCl 3 + 3NaOHOThe acidification of sodium acetate will produceacetic acid.OCH 3 – C – O – Na + + CHCl 39. (a) Write the chemical reaction associated with the"brown ring test".(b) Draw the structures of [Co(NH 3 ) 6 ] 3+ , [Ni(CN) 4 ] 2–and [Ni(CO) 4 ]. Write the hybridization of atomicorbital of the transition metal in each case.(c) An aqueous blue coloured solution of a transitionmetal sulphate reacts with H 2 S in acidic medium togive a black precipitate A, which is insoluble inwarm aqueous solution of KOH. The blue solution ontreatment with KI in weakly acidic medium, turnsyellow and produces a white precipitate B. Identifythe transition metal ion. Write the chemical reactioninvolved in the formation of A and B. [IIT-2000]Sol. (a) NaNO 3 + H 2 SO 4 → NaHSO 4 + HNO 32HNO 3 + 6FeSO 4 + 3H 2 SO 4 →3Fe 2 (SO 4 ) 3 + 2NO + 4H 2 O[Fe(H 2 O) 6 ]SO 4 .H 2 O + NOFerrous Sulphate⎯→ [Fe(H 2 O) 5 NO] SO 4 + 2H 2 O(Brown ring)(b) In [Co(NH 3 ) 6 ] 3+ cobalt is present as Co 3+ and itscoordination number is six.Co 27 = 1s 1 , 2s 2 2p 6 , 3s 2 3p 6 3d 7 , 4s 2Co 3+ ion = 1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 6HenceCo 3+ ion inComplex ion3d 4s 4p3d 4s 4pd 2 sp 3 hybridizationXtraEdge for IIT-JEE 10 MAY 2011

∴ cos A ==cos B ==cos C ==b22+ c − a2bc236λa2+ 25λ2222(30) λ2+ c − b2ac225λa2249λ222+ 49λ270λ+ b − c2ab22+ 36λ22− 49λ2− 36λ2− 25λ= 5119= 35= 7584λ1 19 5∴ cos A : cos B : cos C = : : = 7 : 19 : 255 35 712. Complex numbers z 1 , z 2 , z 3 are the vertices A, B, Crespectively of an isosceles right angled triangle withright angle at C. show that(z 1 – z 2 ) 2 = 2(z 1 – z 3 ) (z 3 – z 2 ) [IIT-1986]Sol. Since ∆ is right angled isosceles ∆.∴ Rotating z 2 about z 3 in anticlock wise directionthrough an angle of π/2, we getA(z 1 )B(z 3 )C(z 2 )z2− z3| z2− z3|= e iπ/2z1− z3| z1− z3|where, |z 2 – z 3 | = |z 1 – z 3 | ⇒ (z 2 – z 3 ) = i(z 1 – z 3 )squarring both sides we get,(z 2 – z 3 ) 2 = – (z 1 – z 2 ) 22 2⇒ z 2 + z 3 – 2z 2 z 3 = –z 2 1 – z 2 3 + 2z 1 z 32 2⇒ z 1 + z 2 – 2z 1 z 2 = 2z 1 z 3 + 2z 2 z 3 – 2 z 2 3 – 2z 1 z 2⇒ (z 1 – z 2 ) 2 2= 2{(z 1 z 3 – z 3 ) + (z 2 z 3 – z 1 z 2 )}⇒ (z 1 – z 2 ) 2 = 2(z 1 – z 3 )(z 3 – z 2 )13. Let n be a positive integer and(1 + x + x 2 ) n = a 0 + a 1 x + .... + a 2n x 2nShow that a 2 0 – a 2 1 + .... + a 2 2n = a n [IIT-1994]Sol. (1 + x + x 2 ) n = a 0 + a 1 x + ... + a 2n x 2n ...(1)Replacing x by – x1 , we obtain⎛ 1 1 ⎞⎜1 − + ⎟ = a⎝ x2 0 –x ⎠na 1 a +2x2x–a 3 a3 + ... +2n2nx x...(2)2Now, a 0 – a 2 1 – a 2 3 + ... + a 2 2n = coefficient of theterm independent of x in[a 0 + a 1 x + a 2 x 2 + ... + a 2n x 2n ⎡ a1a2a2n⎤] ⎢a0 − + −...+n ⎥⎣ x2 2x x ⎦= coefficient of the term independent of x inn(1 + x + x 2 ) n ⎛ 1 1 ⎞⎜1− + ⎟⎝ x2x ⎠But R.H.S. = (1 + x + x 2 ) n ⎛ 1⎜1− +⎝ x x=2n( 1+x + x ) ( x − x + 1)2x2n422nn12n⎞⎟⎠2[( x + 1) − x ]=2nx( 1+2x+ x − x ) ( 1+x + x )==2n2nxxThus, a 2 0 – a 2 1 + a 2 22 + ....a 2n= coefficient of the term independent of x in1n (1 + x 2 + x 4 ) nx 2= coefficient of x 2n in (1 + x 2 + x 4 ) n= coefficient of t n in (1 + t + t 2 ) n = a n14. A rectangle PQRS has its side PQ parallel to the liney = mx and vertices P, Q and S lie on the lines y = a,x = b and x = –b, respectively. Find the locus of thevertex R.[IIT-1996]Sol. Let the coordinates of R be (h, k). It is given that Plies on y = a. So, let the coordinates of P be (x 1 , a).Since PQ is parallel to the line y = mx. Therefore,Slope of PQ = (Slope of y = mx) = mAnd, Slope of PS = –= – m1[∴ PS ⊥ PQ]Now, equation of PQ isy – a = m(x – x 1 )x = –bx´ S(0, – b)(0, a)R1(Slope of y = mx)Oyy = 0y´P24x = bQ(0, b)nx22n...(i)XtraEdge for IIT-JEE 12 MAY 2011

It is given that Q lies on x = b. So, Q is the point ofintersection if (i) and x = b.Putting x = b in (i), we gety = a + m(b – x 1 )So, coordinates of Q are (b, a + m(b – x 1 )).Since PS passes through P(x 1 , a) and has slope – m1 .So, Equation of PS isy – a = – m1 (x – x1 )...(ii)It is given that S lies on x = – b. So, S is the point ofintersection of (ii) and x = –b.Solving (ii) and x = – b, we gety = a + m1 (b + x1 )⎛ 1 ⎞So, coordinates of S are ⎜− b , a + ( b + x1)⎟⎝ m ⎠1k − a − ( b + x1)Now, Slope of RS =m= mh + bBut RS is parallel to PQ.1k − a − ( b + x1)∴m= mh + b⇒ b + x 1 = m(k – a) – m 2 (h + b) ...(iii)Similarly,k − a − m( b − x1)Slope of RQ =h − bBut, RQ is perpendicular to PQ whose slope is m.∴k − a − m( b − x1 ) 1= –h − b m1 1⇒ b – x 1 = (k – a) + (h – a) ...(iv)m2mWe have only one variable x 1 . To eliminate x 1 , add(iii) and (iv) to obtain⎛ 1 ⎞2b = (k – a) ⎜m + ⎟ – m 2 (h + b) +⎝ m ⎠12m(h – b)⎛2⎞⇒ 2b = (k – a) ⎜m + 1 ⎛4⎞⎟– h ⎜m + 1 ⎛4⎞⎟⎝ m2– b ⎜m + 1⎟⎠ ⎝ m2⎠ ⎝ m ⎠⎛2⎞⇒ (k – a) ⎜m + 1⎟–⎝ m ⎠2h ( m −1)(m + 1)2m–22b ( m + 1)2m2= 02 −2 +h ( m 1)b ( m 1)⇒ (k – a) – – = 0m m⇒ m(k – a) – h(m 2 – 1) – b(m 2 + 1) = 0Hence, the locus of R(h, k) ism(y – a) – x(m 2 – 1) – b(m 2 + 1) = 015. Let f [(x + y)/2] = {f (x) + f (y)} / 2 for all real x and y,If f ´(0) exists and equals –1 and f (0) = 1, find f (2).[IIT- 1995]⎛ x + y ⎞ f ( x)+ f ( y)Sol. f ⎜ ⎟ =∀ x, y ∈ R (given)⎝ 2 ⎠ 2Putting y = 0, we get⎛ x ⎞ f ( x)+ f (0) 1f ⎜ ⎟ == [1 + f (x)] [Q f (0) = 1]⎝ 2 ⎠ 2 2⇒ 2f (x/2) = f (x) + 1⇒ f (x) = 2f (x/2) – 1 ∀x, y ∈ R ...(1)Since f´(0) = –1, we getf (0 + h)− f (0)f ( h)−1lim = – 1 ⇒ lim = –1h→0 hh→0 hNow, let x ∈ R then applying formula ofdifferentiability.⎛ 2x+ 2h⎞f ⎜ ⎟ − f ( x)f ( x + h)− f ( x)2f ´(x) = lim = lim⎝ ⎠h→0 hh→0 hf (2x)+ f (2h)− f ( x)= lim2h→0h1 ⎧ ⎛ 2x⎞ ⎛ 2h⎞ ⎫⎨2f ⎜ ⎟ −1+2 f ⎜ ⎟ −1⎬− f ( x)2 22= lim⎩ ⎝ ⎠ ⎝ ⎠ ⎭h→0h[Using equations (1)1{2 f ( x)−1+2 f ( h)−1}− f ( x)= lim2h→0hf ( x)+ f ( h)−1−f ( x)= limh→0hf ( h)−1= lim = –1h→0 hTherefore f ´(x) = – 1 ∀ x ∈ R⇒∫f ´( x)dx =∫ −1dx⇒ f (x) = –x + k where k is a constant.But f (0) = 1, therefore f (0) = – 0 + k⇒ f (x) = 1 – x ∀ x ∈ R ⇒ f (2) = – 1XtraEdge for IIT-JEE 13 MAY 2011

Physics Challenging ProblemsSet # 1This section is designed to give IIT JEE aspirants a thorough grinding & exposure to varietyof possible twists and turns of problems in physics that would be very helpful in facing IITJEE. Each and every problem is well thought of in order to strengthen the concepts and wehope that this section would prove a rich resource for practicing challenging problems andenhancing the preparation level of IIT JEE aspirants.By : Dev SharmaSolutions available in next issueDirector Academics, Jodhpur Branch1. A thin rod of length 2L is placed horizontally in adown ward uniform electric field E. Half of lengthis charged positively and other half with linearcharge density λ . The rod of is hinged at its middlepoint and free to rotate in vertical plane, A weight wis hanged at a distance x from hinge to keep ithorizontal. The value of x isEλλ– – – – – – + + + + + + +O(A)(C)wλEEλL2wwx2LEλL2(B)2w(D)22EλLw2. S is a cross section of a solenoid where magneticfields is increasing at a constant positive rate. V 1 andV 2 are two ideal voltmeters, then V 1 /V 2 isV 1 1Ω× ×× ×× × ×2Ω V 2S(A) 1 (B) 0.5(C) 2 (D) 0.25Passage # 1 (Q. No. 3 to Q. No. 4)In the given circuits three inductors L 1 , L 2 and L 3and a resistance R are connected to a V volt batteryneglecting mutual inductance. At t = 0 switch isclosed and L 3 > L 2 > L 1 .L 2L 3 L 13. Find current through L 1 will beV L2(A)R L1+ L2+ L3V L1(B)R L1+ L2+ L3V L3(C)R L1+ L2+ L3(D)Can’t be determined4. Time constant of the circuit is :(A)(C)L1 + L2+ L3R⎛R⎜⎝1L1+L1+2 1 L3⎞⎟⎠V(B)⎛⎜⎝1LR1R+L1+3 1 L2(D) None of thesePassage # 2 (Q. No. 5 to Q. No. 6)Hydrogen gas in the atomic state is excited to aenergy level such that the electrostatic potentialenergy of H-atom becomes -1.7eV. Now aphotoelectric plate having work function W = 2.3eVis exposed to the emission spectra of this gas.Assuming all transitions to be possible. Answer thefollowing questions.⎞⎟⎠XtraEdge for IIT-JEE 14 MAY 2011

5. How many electric transition may result in anejected photoelectron(A) 3 (B) 4(C) 5 (D) 26. The minimum de-broglie wavelength of the ejectedphoto electrons is(A) 13.43Å(B) 3.43 Å(C) 2Å(D) None of thesePassage # 3(Q. No. 7 to Q. No. 8)A charged dielectric sphere of negligible mass isconnected to a conducting square shaped loop oftotal resistance R and side 'l', by means of a lightnon conducting thread, which passes over two fixedpulleys as shown in the figure. The sphere carries acharge q and kept in uniform downward electricfield E. The loop has mass ‘m’ (m < qE) and kept ina horizontal magnetic field B as shown. Answer thefollowing.2 qE2Take g = 10m / s , = 20m / s andm22B lmRE= 2N − s / mq2× B× × × × ×× × × × × ×× × × × × ×× × × × × ×7. Choose the incorrect (wrong) statements(A) The initial acceleration of loop is 10m/s 2(B) Loop’s one edge leaves magnetic field after 0.5sec, its acceleration drops to zero(C) The magnetic force on loop is acting upwards(D) When loop is completely off the magnetic fieldits acceleration is 10m/s 28. If after 2 sec, one sides of the loop emerges from themagnetic field, the speed of the loop as a function oftime is−2t−( 2t−4)(A) 5[1 + e ](B) 5[1 + e ]−2t(C) 5[1 − e ](D) None of theseBlack Holes-The Most EfficientEngines in the UniverseThe scientists have just found the most energyefficientengines in the universe. Black holes,whirling super dense centres of galaxies that suck innearly everything. Jets of energy spurting out ofolder ultra-efficient black holes also seem to beplaying a crucial role as zoning police in largegalaxies preventing to many stars from sprouting.This explains why there are fewer burgeoninggalaxies chock full of stars than previouslyexpected.For the first time, the scientists have measured boththe mas of hot gas that is being sucked into nineolder black holes and the unseen super speedy jetsof high energy particles split out, which essentiallyform a cosmic engine. Then they determined a rateof how efficient these older black hole engines areand were awe-struck. These black holes are 25times more efficient than anything man has built,with nuclear power being the most efficient of manmadeefforts, said the research's lead author,Professor Steve Allen of Stanford University.The galaxies in which these black holes live arebigger than the Milky way, which is the Earth'sgalaxy and are 50 million to 400 million light-yearsaway.Black holes are the most fuel-efficient engines inthe universe.The results were surprising because the types ofblack holes studied were older, less powerful andgenerally considered boring, scientists said. Butthey ended up being more efficient than originallythought, possibly as efficient as their younger,brighter and more potent black hole siblings calledquasars. One way the scientists measured theefficiency of black holes was by looking at the jetsof high energy spewed out.XtraEdge for IIT-JEE 15 MAY 2011

PHYSICSStudents'ForumExpert’s Solution for Question asked by IIT-JEE Aspirants1. Two circular rings A and B, each of radius a = 30 cmare placed coaxially with their axes horizontal in auniform electric field E = 10 5 NC –1 directed verticallyupwards as shown in figure. Distance betweencentres of these rings A and B is h = 40 cm. Ring Ahas a positive charge q 1 = 10 µC while ring B has anegative charge of magnitude q 2 = 20 µC. A particleof mass m = 100 gm and carrying a positive chargeq = 10 µC is released from rest at the centre of thering A. Calculate its velocity when it has moved adistance of 40 cm.B E AahSol. Weight of the particle is W = mg = 1 newtonForce on particle, due to electric field E is F = qE = 1newton (upwards)It means weight of the particle is balanced by theforce F. Hence, net force on the particle is due tocharge on rings.Since particle is at centre of ring A therefore initiallyno force is exerted due to charge of this ring. But ringB is negatively charged. Therefore particleexperiences a resultant force towards centre of ringB.Separation between centres of rings is h = 40 cm anddistance moved the particle is also 40 cm. It meansvelocity of particle is to be calculated when it reachesthe centre of ring B.According to law of conservation of energy, kinetic⎛ 1 2 ⎞energy ⎜ mv ⎟ of particle at centre of B = Loss of⎝ 2 ⎠its electrical potential energy. Potential energy ofparticle at centre of A is1 qq11 q(–q 2)U 1 = U 0 + +4πε 0 a 4πε2 20 a + hWhere U 0 is potential energy due to electric field E.Potential energy at centre of B is1 qq1U 2 = U 0 ++1 q(–q ).24πε2 20 a + h 4πε 0 aaBut1mv22= U 1 – U 2∴ v = 6 2 ms –12. Two thin similar convex glass pieces are joinedtogether, front to front, with rear portion silvered andthe combination of glass pieces is placed at a distancea = 60 cm from a screen. A point object is placed onoptical axis of the combination such that its m = 2times the magnified image formed on the screen. Ifair between the glass pieces is replaced by water (µ =4/3), calculate the distance through which the objectmust be displaced so that a sharp image is againformed on the screen.Sol. A thin convex glass piece has both the surfacesequally curved, therefore, it works as a glass plate.When two thin similar convex glass pieces are joinedfront to front, though they form shape of an equiconvexlens as shown in figure(1), but its opticalpower is zero because medium inside this lens is air.AirFig. (1)When its rear surface is silvered, it works as aconcave mirror. This concave mirror has focal lengthf m = – r/2 where r is the radius of curvature of eachglass piece. Since, this concave mirror forms animage on screen, therefore the image is real. Hencescreen and object both are on same side of the mirroras shown figure (2).ScreenObjectaAirFig. (2)Since, image is m = 2 times magnified, therefore,distance of object from the mirror is equal toa/m = 30 cmXtraEdge for IIT-JEE 16 MAY 2011

for this mirror,v = –a = – 60 cm, u = – 30 cm,Using mirror formula,1 1 1 + = , fm = – 20 cm or r = 40v u fwhen water is filled in space between glass pieces, anequi-convex lens of water is formed whose onesurface is silvered. This silvered convex glass may beassumed as a combination of a convex lens and aconcave mirror. Effective focal length of the1 2combination is given by = – where fl isF f1m flfocal length of the equiconvex lens of water.For the equiconvex lens, R 1 + r , R 2 = – r, µ = 4/31 ⎛ 1 1Using, = (µ–1) f ⎟ ⎞⎜ – , f = f 1 = 60 cm⎝ R1 R 2 ⎠Hence, effective focal length F is given by1 1 2= –F (–20)60or F = – 12 cmsince, a sharp image is again formed on the screentherefore, fore this effective mirror,F = – 12 cm, v = – a = – 60 cm, u = ?1 1 1Using, mirror formula, = + , u = – 15 cmF v uInitially the object was at a distance (–a/2) = – 30 cmfrom the mirror but now object must be at a distance(–15 cm). Hence, the object must be moved through[– 15 – (– 30)] cm = 15 cm towards the mirror(combination).3. A uniform rod of length l = 75 cm is hinged at one ofits ends and is free to rotate in vertical plane. It is areleased from rest when the rod is horizontal. Whenrod becomes vertical, it is broken at mid point andlower part now moves freely. Calculate distance ofthe centre of lower part from hinge, when it againbecomes vertical for the first time. (g = 10 ms –2 )Sol. When rod is released, it rotates about upper end itsgravitational potential energy converts into rotationalkinetic energy. Initially centre of mass of the rod is inlevel of O while its depth from O is l/2 when rodbecomes vertical as shown in figure. (1). If mass ofrod is m, then loss of gravitational potentialenergy = mg 2l .OFig. (1)ml2Moment of inertia of rod about O is I =3Let angular velocity or rod be ω when it becomesvertical. According to law of conservation energy,1 2 lI ω = mg2 2or ω =3gl= 40 rad/secNow the rod gets broken at its mid point as shown infigure (2). At this instant velocity of centre of lowerpart is v 0 = 43l ω (horizontally leftwards) and itsangular velocity is ω (clockwise).ωv 0Fig. (2)Now this part moves freely under gravity. Therefore,its angular velocity ω remains constant. This partagain becomes vertical for the first time aftercompleting half rotation. Therefore, time taken by theπ πlower part during this rotation is t = = sec.ω 40During this time. horizontal displacement centre of3πllower part is x = v 0 t = = 1.77 m4and vertical displacement = 21 gt2= 1.23 mBut initial depth of centre of lower part from hinge3lwas , therefore vertical depth of its centre from O4⎛ 3l 1 2 ⎞at this instant is y = ⎜ + gt ⎟ = 1.80 m.⎝ 4 2 ⎠Hence, distance of its centre from O isr =22x + y = 2.52 m4. Length of a horizontal arm of a U-tube is l = 21 cmand ends of both of the vertical arms are open tosurroundings of pressure 10500 Nm –2 . A liquid ofdensity ρ = 10 3 kg m –3 is poured into the tube suchthat liquid just fills horizontal part of the tube. Now,one of the open ends is sealed and the tube is thenrotated about a vertical axis passing through the othervertical arm with angular velocity ω 0 = 10radians/sec. If length of each vertical arm be a = 6cm. calculate the length of air column in the sealedarm.ωXtraEdge for IIT-JEE 17 MAY 2011

Q 1 = 1200 JEnginel = 21 cma = 6 cmSol. When tube is rotated, liquid starts to flow radiallyoutward and air in sealed arm is compressed. Let theshift of liquid be x as shown in figure.xA(l – x)Ba–xLet cross-sectional area of tube be S.Initial volume of air, V 0 = Sa and initial pressureP 0 = 10500 Nm –2Final volume, V = S(a – x)P V∴ Final pressure P = 0 0 P= 0 .a.V (a – x)Por Pressure at B, P 2 = P + xρg = 0 a+ xρg(a – x)Centripetal force required for circular motion ofvertical column of height x of liquid is provided byreaction of the tube while that to horizontal length(l – x) is provided by excess pressure at B.Force exerted by pressure difference is⎡ PF 1 = (P B – P A ) S = (P 2 – P 0 ) S = 0 x ⎤⎢ + xρg⎥S⎣(a– x) ⎦Mass of horizontal arm AB of liquid is,m = S (l – x)ρRadius of circular path traced by its centre of mass isl – x ⎛ l + x ⎞r = x + = ⎜ ⎟2 ⎝ 2 ⎠∴ Centripetal force, F 2 = mω 2 0 r⎧ l + x ⎫ 2But F 2 = F 1 or {S ρ (l – x)} ⎨ ⎬ ω 0⎩ 2 ⎭⎧ P= 0 x⎨ + xρgS⎩(a– x) ⎭ ⎬⎫or x = .01 m = 1cm∴ Length of air column in sealed arm = (a – x)= 5cm5. A reversible Carnot engine is coupled with three heatreservoirs A,B and C as shown in Figure.Temperature of these reservoirs isxABQ 2Q 3CT 1 = 400 K T 2 = 300 K T 3 = 200 KT 1 = 400 K, T 2 = 300 K and T 3 = 200K, respectively.During an integral number of complete cycles, theengine absorbs Q 1 = 1200 joule of heat energy fromreservoir A and performs W = 200 joule ofmechanical work. Calculate quantities Q 1 and Q 2 ofheat energy exchanged with the other two reservoirsB and C respectively. State whether the reservoirsabsorb or lose heat energy.Sol. Given engine can be assumed to work in two ways ;(i) engine working between a source of temperatureT 1 and sink of temperature T 2 and(ii) an engine working between source of temperatureT 1 and sink of temperature T 3 . Hence, efficiency ofthese two isT1– T21 Tη 1 = = and1 – T31η2 = = respectivelyT14 T14If an engine having efficiency η absorbs heat Q fromsource then heat energy converted into mechanicalwork by the engine will be equal to ηQ andremaining part (Q – ηQ) = (1 – η)Q should berejected to the sink.Since, heat rejected to sink by first engine, havingefficiency η 1 , is Q 2 , therefore, heat absorbed by itQ24Qfrom source is equal to = 2(1– η1)3and work done by this engine isQ2η1QW 1 = = 2(1– η1)3Similarly, theat absorbed by the other engine fromQ3source is equal to = 2Q 3(1– η2)Q3.η2and work done by this engine is W 2 = = Q 3(1– η2)But W 1 + W 2 = 200 J1∴ Q2 + Q 3 = 200 J ...(1)3Total heat extracted from source A is⎛ 4 ⎞ ⎛ 4 ⎞Q 1 = ⎜ Q2 + 2Q3⎟ or ⎜ Q2 + 2Q3⎟ = 1200 J⎝ 3 ⎠ ⎝ 3 ⎠...(2)Solving equation (1) and (2),Q 2 = 1200 J and Q 3 = – 200 JIt means reservoir B absorbs 1200 J of heat andreservoir C loses 200 J of heat.In fact engine works as an engine between reservoirsA and B and as heat pump between reservoirs B andC.XtraEdge for IIT-JEE 18 MAY 2011

XtraEdge for IIT-JEE 19 MAY 2011

PHYSICS FUNDAMENTAL FOR IIT-JEEElectrostatics-IKEY CONCEPTS & PROBLEM SOLVING STRATEGY• Coulomb's Law :1 q1q2F 0 =(in vacuum)24πε0 rVectorially → 1 q1q2F = rˆ24πε0 r1 q1q2In any material medium F =24πε0εrrwhere ε r is a constant of the material medium calledits relative permittivity, and ε 0 is a universal constant,called the permittivity of free space.ε 0 = 8.85 × 10 –12 1or4πε = 9 × 1090The unit of ε 0 is C 2 N m –2 or farad per metre.1 q1q2Also F =24πε0εrrWhere ε is called the absolute permittivity of themedium.Obviously, ε r = F 0 /F. Remember ε r = ∞ forconductors.Conductors and insulators Each body containsenormous amounts of equal and opposite charges. A'charged' body contains an excess of either positive ornegative charge.In a conductor, some of the negative charges are freeto move around. In an insulator (also called adielectric), the charges cannot move. They can onlyundergto small localized displacements, causingpolarization.Induction When a charged body A is brought nearanother body B, unlike charges are induced on thenear surface of B (called bound charges) and likecharges appear on the far surface of B (called freecharges) If B is a conductor, the free charges can beremoved by earthing B, e.g., by touching it. If B is aninsulator, separation of like and unlike charges willstill occur due to induction. However, the likecharges cannot then be removed by earthing B.• Electric Field And PotentialElectric Field An electric field of strength E is saidto exist at a point if a test charge ∆q at that pointexperiences a force given by→→ →→∆ F∆ F = ∆qF or E =∆qThe unit of electric field is Newton per coulomb orvolt per metre. The electric field strength at adistance r from a point charge q in a medium ofpermittivity ε is given byE =14πεq2rVectorially → 1 qE = 4πε2 rˆrWith reference to any origin→E =q4πε→R−r→→3→R−rWhere R → is the position vector of the field point and→r , the position vector of q.Due to a number of discrete chargesi=N→→→q E =1 R−ri∑34πε→ →i=1R−riElectric Potential The electric potential at a point isthe work done by an external agent in bringing a unitpositive charge from infinity up to that point alongany arbitrary path.∆W∞→(by an external agent)V P = Pvolt(V) or JC –1∆qThe potential difference between two points P and Qis given by∆W Q → P (by agent)V P – V Q =volt (V)∆qThe potential at a distance r from a point charge q ina medium of permittivity ε isϕ or V =14πεq 1 =r 4πεq→ →R− rwith reference to any arbitrary origin.Due to a number of chargesi=N1 q1ϕ or V = ∑4πε→ →i= 1 R−riXtraEdge for IIT-JEE 20 MAY 2011

In a conductor, all points have the same potential.If charge q (coulomb) is placed at a point where thepotential is V (volt), the potential energy of thesystem is qV (joule). It follows that if charges q 1 , q 2are separated by distance r, the mutual potentialqenergy of the system is 1 q 24πε r.• Relation Between Field (E) And Potential (V)The negative of the rate of change of potential alonga given direction is equal to the component of thefield that direction.∂VE r = – along r∂ r∂V and E ⊥ = perpendicular to rr ∂θWhen two points have different potentials, an electricfield will exist between them, directed from thehigher to the lower potential.• Lines of ForceA line of force in an electric field is such a curve thatthe tangent to it at any point gives the direction of thefield at that point. Lines of force cannot intersecteach other because it is physically impossible for anelectric field to have two directions simultaneously.• Equipotential SurfacesThe locus of points of equal potential is called anequipotential surface. Equipotential surfaces lie atright angles to the electric field. Like lines of force,they can never intersect.Note: For solving problems involving electrostaticunits, remember the following conversion factors:3 × 10 9 esu of charge = 1 C1 esu of potential = 300 V• Electric FluxThe electric flux over a surface is the product of itssurface area and the normal component of the electricfield strength on that surface. Thus,dϕ = (E cos θ) ds = E n ds = → E . → dsdsOThe total electric flux over a surface is obtained bysumming :→ →→ →ϕ E = ∑ E . ∆ s or∫ E .d sGauss's Theorem The total electric flux across a1closed surface is equal to times the total chargeε0inside the surface.Mathematically ∑ → E . ∆ →s = q/ε 0where q is the total charge enclosed by the surface.ENProblems in electrostatics can be greatly simplifiedby the use of Gaussian surfaces. These are imaginarysurfaces in which the electric intensity is eitherparallel to or perpendicular to the surface everywhere.There are no restrictions in constructing aGaussian surface.The following results follow from Gauss's law1. In a charged conductor, the entire charge residesonly on the outer surface. (It must always beremembered that the electric field is zero inside aconductor.)2. Near a large plane conductor with a chargedensity σ (i.e., charge per unit area), the electricintensity isE = σ/ε 0 along the normal to the plane3. Near an infinite plane sheet of charge with acharge density σ, the electric intensity isE = σ/2ε 0 along the normal to the plane4. The electric intensity at a distance r from the axisof a long cylinder with λ charge per unit length(called the linear density of charge), is1 λ→E = along r2πεr0Problem solving strategy: Coulomb's Law :Step 1 : The relevant concepts : Coulomb's lawcomes into play whenever you need to know theelectric force acting between charged particles.Step 2 : The problem using the following steps :Make a drawing showing the locations of thecharged particles and label each particle with itscharge. This step is particularly important if morethan two charged particles are present.If three or more charges are present and they donot all lie on the same line, set up an xycoordinatesystem.Often you will need to find the electric force onjust one particle. If so, identify that particle.Step 3 : The solution as follows :For each particle that exerts a force on the particleof interest, calculate the magnitude of that force1 | q1q2 |using equation F =24πε0 rSketch the electric force vectors acting on theparticle(s) of interest due to each of the otherparticles (that is, make a free-body diagram).Remember that the force exerted by particle 1 onparticle 2 points from particle 2 toward particle 1if the two charges have opposite signs, but pointsfrom particle 2 directly away from particle 1 if thecharges have the same sign.Calculate the total electric force on the particle(s)of interest. Remember that the electric force, likeany force, is a vector. When the forces acting on acharge are caused by two or more other charges,the total force on the charge is the vector sum ofXtraEdge for IIT-JEE 21 MAY 2011

the indivual forces. It's often helpful to usecomponents in an xy-coordinate system. Be sureto use correct vector notation; if a symbolrepresents a vector quantity, put an arrow over it.If you get sloppy with your notation, you will alsoget sloppy with your thinking.As always, using consistent units is essential.With the value of k = 1/4πε 0 given above,distances must be in meters, charge in coulombs,and force in newtons. If you are given distance incentimeters, inches, or furlongs, donot forget toconvert ! When a charge is given inmicrocoulombs (µC) or nanocoulombs (nC),remember that 1µC = 10 –6 C and 1nC = 10 –9 C.Some example and problems in this and laterchapters involve a continuous distribution ofcharge along a line or over a surface. In thesecases the vector sum described in Step 3 becomesa vector integral, usually carried out by use ofcomponents. We divide the total chargedistribution into infinitesimal pieces, useCoulomb's law for each piece, and then integrateto find the vector sum. Sometimes this processcan be done without explicit use of integration.In many situations the charge distribution will besymmetrical. For example, you might be asked tofind the force on a charge Q in the presence oftwo other identical charges q, one above and tothe left of Q and the other below and to the left ofQ. If the distance from Q to each of the othercharges are the same, the force on Q from eachcharge has the same magnitude; if each forcevector makes the same angle with the horizontalaxis, adding these vectors to find the net force isparticularly easy. Whenever possible, exploit anysymmetries to simplify the problem-solvingprocess.Step 4 : your answer : Check whether your numericalresults are reasonable, and confirm that the likecharges repel opposite charges attract.Problem solving strategy : Electric-field calculationsStep 1: the relevant concepts : Use the principle ofsuperposition whenever you need to calculate theelectric field due to a charge distribution (two ormore point charges, a distribution over a line, surface,or volume or a combination of these).Step 2: The problem using the following steps :Make a drawing that clearly shows the locationsof the charges and your choice of coordinate axes.On your drawing, indicate the position of the fieldpoint (the point at which you want to calculate theelectric field E r ). Sometimes the field point willbe at some arbitrary position along a line. Forexample, you may be asked to find E r at point onthe x-axis.Step 3 : The solution as follows :Be sure to use a consistent set of units. Distancesmust be in meters and charge must be incoulombs. If you are given centimeters ornanocoulombs, do not forget to convert.When adding up the electric fields caused bydifferent parts of the charge distribution,remember that electric field is a vector, so youmust use vector addition. Don't simply addtogether the magnitude of the individual fields:the directions are important, too.Take advantage of any symmetries in the chargedistribution. For example, if a positive charge anda negative charge of equal magnitude are placedsymmetrically with respect to the field point, theyproduce electric fields of the same magnitude butwith mirror-image directions. Exploiting thesesymmetries will simplify your calculations.Must often you will use components to computevector sums. Use proper vector notation;distinguish carefully between scalars, vectors, andcomponents of vectors. Be certain thecomponents are consistent with your choice ofcoordinate axes.In working out the directions of E rvectors, becareful to distinguish between the source pointand the field point. The field produced by a pointcharge always points from source point to fieldpoint if the charge is positive; it points in theopposite direction if the charge is negative.In some situations you will have a continuousdistribution of charge along a line, over a surface,or through a volume. Then you must define asmall element of charge that can be considered asa point, finds of all charge elements. Usually it iseasiest to do this for each component of E rseparately, and often you will need to evaluateone or more integrals. Make certain the limits onyour integrals are correct; especially when thesituation has symmetry, make sure you don'tcount the charge twice.Step 4 : your answer : Check that the direction of E ris reason able. If your result for the electric-fieldmagnitude E is a function of position (say, thecoordinate x), check your result in any limits forwhich you know what the magnitude should be.When possible, check your answer by calculating itin a different way.Problem solving strategy : Gauss's LawStep 1 : Identify the relevant concepts : Gauss's lawis most useful in situations where the chargedistribution has spherical or cylindrical symmetry oris distributed uniform over a plane. In these situationswe determine the direction of E r from the symmetryof the charge distribution. If we are given the chargedistribution. we can use Gauss's law to find the themagnitude of E r . Alternatively, if we are given thefield, we can use Gauss's law to determine the detailsXtraEdge for IIT-JEE 22 MAY 2011

of the charge distribution. In either case, begin youranalysis by asking the question, "What is thesymmetry ?"Step 2 : Set up the problem using the following stepsSelect the surface that you will use with Gauss'slaw. We often call it a Gaussian surface. If youare trying to find the field at a particular point,then that point must lie on your Gaussian surface.The Gaussian surface does not have to be a realphysical surface, such as a surface of a solidbody. Often the appropriate surface is animaginary geometric surface; it may be in emptyspace, embedded in a solid body, or both.Usually you can evaluate the integral in Gauss'slaw (without using a computer) only if theGaussian surface and the charge distribution havesome symmetry property. If the chargedistribution has cylindrical or sphericalsymmetry, choose the Gaussian surface to be acoaxial cylinder or a concentric sphere,respectively.Step 3 : Execute the solution as follows :Carry out the integral in Eq.ΦE =∫E cosφ dA =∫E dA =∫Er .dAr Q= enclε0(various forms of Gauss's law)This may look like a daunting task, but thesymmetry of the charge distribution and yourcareful choice of a Gaussian surface makes itstraightforward.Often you can think of the closed surface as beingmade up of several separate surfaces, such as theside and ends of a cylinder. The integral∫ E dAover the entire closed surface is always equal tothe sum of the integrals over all the separatesurfaces. Some of these integrals may be zero, asin points 4 and 5 below.If E r is perpendicular (normal) at every point to asurface with area A, if points outward from theinterior of the surface, and if it equal to EA. Ifinstead E r is perpendicular and inward, then E ⊥ =– E and∫ E ⊥dA = – EA.If E r is tangent to a surface at every point, then E ⊥= 0 and the integral over that surface is zero.If E r = 0 at every point on a surface, the integralis zero.In the integral∫ E ⊥dA , E ⊥ is always theperpendicular component of the total electric fieldat each point on the closed Gaussian surface. Ingeneral, this field may be caused partly bycharges within the surface and partly by chargesoutside it. Even when there is no charge withinthe surface, the field at points on the Gaussiansurface is not necessarily zero. In that case,however, the integral over the Gaussian surface –is always zero.Once you have evaluated the integral, use eq. tosolve for your target variable.Step 4 : Evaluate your answer : Often your result willbe a function that describes how the magnitude of theelectric field varies with position. Examine this functionwith a critical eye to see whether it make sense.1. Supposing that the earth has a charge surface densityof 1 electron/metre 2 , calculate (i) earth's potential, (ii)electric field just outside earths surface. Theelectronic charge is – 1.6 × 10 –19 coulomb and earth'sradius is 6.4×10 6 metre (ε 0 = 8.9 × 10 –12 coul 2 /nt–m 2 ).Sol. Let R and σ be the radius and charge surface densityof earth respectively. The total charge, q on the earthsurface is given byq = 4 p R 2 σ(i) The potential V at a point on earth's surface is sameas if the entire charge q were concentrated at itscentre. Thus,1 qV = .4πε 0 R1 4πR2 σ R. σ= . =4πε0R ε0Substituting the given values−6−192(6.4×10 metre) × ( −1.6×10 coul / metre )V =−122 2(8.9×10 coul / nt − m )(ii) E =Solved Examplesnt − m= – 0.115 coul1 q 12 =4πε0R 4πε . 4πRσ20 R−19joule= – 0.115 = – 0.115 volt.coul−1.6×10 coul / metre== – 1.8 × 10 –8 nt/coul.−122 28.9×10 coul / nt − mThe negative sign shows that E is radially inward.2. Determine the electric field strength vector if thepotential of this field depends on x, y co-ordinates as(a) V = a(x 2 – y 2 ) and (b) V = axy.Sol. (a) V = a(x 2 – y 2 )∂V∂VHence, E x = – = – 2ax, E y = – = + 2ay∂ x ∂ y∴ E = – 2axi + 2ayjor E = – 2a(xi – yj)(b) V = a x y22=σε 0XtraEdge for IIT-JEE 23 MAY 2011

Hence, E x = –∴∂V∂ xE = – ayi – axj= – a[yi + xj]∂V= –ay, E y = –∂ y= – ax3. A charge Q is distributed over two concentric hollowspheres of radii r and R (> r) such the surfacedensities are equal. Find the potential at the commoncentre.q′ qORSol. Let q and q′ be the charges on inner and outer sphere.Thenq + q′ = Q…(1)As the surface densities are equal, henceq q'=2 24πr4πR(∴ Surface density = charge/area)∴ q R 2 = q′ r 2…(2)From eq. (1) q′ = (Q – q), henceq R 2 = (Q – q)r 2q(R 2 + r 2 ) = Q r 222Q rQ R∴ q = and q′ = Q – q =2 22 2R + rR + rNow potential at O is given by1 q 1 q'V = +4πε0r 4πε0r==14πε002Q r 1+2 2(R + r ) r 4πεQ (r + R)4πε (R2 2+ r )r0(RQ r222+ r ) r4. S 1 and S 2 are two parallel concentric spheres enclosingcharges q and 2q respectively as shown in fig.(a) What is the ratio of electric flux through S 1 and S 2 ?(b) How will the electric flux through the sphere S 1change, if a medium of dielectric constant 5 isintroduced in the space inside S 1 in place of air ?S 2Sol. (a) Let Φ 1 and Φ 2 be the electric flux through spheresS 1 and S 2 respectively.q2qS 1∴qΦ 1 =ε0q / εΦ 1 =0Φ 2 3q / ε0and Φ 2 == 31q + 2q 3q=ε ε(b) Let E be the electric field intensity on the surface ofsphere S 1 due to charge q placed inside the sphere.When dielectric medium of dielectric constant K isintroduced inside sphere S 1 , then electric fieldintensity E′ is given byE′ = E/KNow the flux Φ′ through S 1 becomes' 1 qΦ′ =∫E .dS =∫E.dS =K Kε∴ Φ′ =q5ε05. A charge of 4 × 10 –8 C is distributed uniformly on thesurface of a sphere of radius 1 cm. It is covered by aconcentric, hollow conducting sphere of a radius5 cm. (a) Find the electric field at a point 2 cm awayfrom the centre. (b) A charge of 6 × 10 –8 C is placedon the hollow sphere. Find the surface charge densityon the outer surface of the hollow sphere.Sol. (a) See fig. (a) Let P be a point where we have tocalculate the electric field. We draw a Gaussiansurface (shown dotted) through point P. The fluxthrough this surface isq = 6 × 10 –8 C5cmΦ =2cmPFig. (a) Fig. (b)∫∫E.dS = E dS = 4π(2×10 ) E000−22According to Gauss's law, Φ = q/ε 0∴ 4π × (2 × 10 –2 ) 2 E = q/ε 09−8q (9×10 ) × (4×10 )or E ==−22−44πε0 × (2×10 ) 4×10= 9 × 10 5 N/C(b) See fig. (b) We draw a Gaussian surface (showndotted) through the material of hollow sphere. Weknow that the electric field in a conducting material iszero, therefore the flux through this Gaussian surfaceis zero. Using Gauss's law, the total charge enclosedmust be zero. So, the charge on the inner surface ofhollow sphere is 6 × 10 –8 C. So, the charge on theouter surface will be 10 × 10 –8 C.XtraEdge for IIT-JEE 24 MAY 2011

PHYSICS FUNDAMENTAL FOR IIT-JEE1-D Motion, Projectile MotionKEY CONCEPTS & PROBLEM SOLVING STRATEGYKinematics :Velocity (in a particular direction)=Displacement(in that direction)Time taken( V r AB ) x = V r Ax – V r Bx and ( V r AB ) x =dxtWhere dx is the displacement in the x direction intime t.Swimmer crossing a riverdv sθv s sinθv s cosθdTime taken to cross the river =VscosθFor minimum time, θ should be zero.xv sv rin this case resultant velocity2 2 dV R = V s + vrand t = .v sAlso x = v r × tFor reaching a point just opposite the horizontalcomponent of velocity should be zero.v sinθ = V r| Displacement ||Average Velocity| =timea r vr – ur r r=t= v + (–u)t22v + u – 2uvcosθ⇒ |a| =tWhere θ is the angle between v and u.The direction of acceleration is along the resultant ofv r and (– u r ).v rv RGraphsDuring analysis of a graph, the first thing is see thephysical quantities drawn along x-axis and y-axis.If y = mx, the graph is a straight line passing throughthe origin with slope = m. [see fig. (a)]Ym = tanθθ is acute andm is positiveYm = tanθθ is abtuse andm is positiveθθXX(i) fig.(a) (ii)if y = mx + c, the graph is a straight line not passingthrough the origin and having an intercept c whichmay be positive or negative [see fig. (b,) (c)]]YYc is negativem is '+' vem is '+' vec(i)YXfig(b)c(ii)c is positivem is negative(ii) Xfig(c)For y = kx 2 , where k is a constant, we get parabola[see fig (d)]YParabolaFig.(d)x 2 + y 2 = r 2 is equation of a circle with centre atorigin and radius r.XXXtraEdge for IIT-JEE 25 MAY 2011

For (x – a) 2 + (y – b) 2 = r 2 , the motion is in a circularpath with centre at (a, b) and radius r22x y2 +2 = 1 is equation of an ellipsea bx × y = constant give a rectangular hyperbola.Note : To decide the path of motion of a body, arelationship between x and y is required.Area under-t graph represents change in velocity.Calculus method is used for all types of motion (a = 0or a = constt or a = variable)s = f(t)Differentiate w.r.ttimev = f(t)integrate w.r.t.timeDifferentiate w.r.ttimea = f(t)integrate w.r.t.timeS stand for displacement2dv dv d sa = v = = ds dt 2dtdx dyAlso v x = ⇒ vy = dt dt2dva x = x d x dv 2 y d y= and ay = =dt 2dtdt 2dtThe same concept can be applied for z-co-ordintae.Projectile motion :Pucosθu ucosθusinθ ucosθθucosθucosθQ θuusinθgg FFgFgFProjectile motion is a uniformly accelerated motion.For a projectile motion, the horizontal component ofvelocity does not change during the path becausethere is no force in the horizontal direction. Thevertical component of velocity goes on decreasingwith time from O to P. At he highest point it becomeszero. From P to Q again. the vertical component ofvelocity increases but in downwards direction.Therefore the minimum velocity is at the topmostpoint and it is u cos θ directed in the horizontaldirection.The mechanical energy of a projectile remainconstant throughout the path.the following approach should be adopted for solvingproblems in two-dimensional motion :Resolve the 2-D motion in two 1-D motions in twomutually perpendicular directions (x and y direction)Resolve the vector quantitative along thesedirections. Now use equations of motion separatelyfor x-direction and y-directions.If you do not resolve a 2-D motions in two 1-Dmotions in two 1-D motion then use equations ofmotion in vector formv r = u r + at ; s r = ut + 21 ar t2; v r . v r – u r . u = 2 a r s rs = 21 ( ur + vr )tWhen y = f(x) and we are interested to find(a) The values of x for which y is maximum forminimum(b) The maximum/minimum values of y then we mayuse the concept of maxima and minima.Problem solving strategy :Motion with constant Acceleration :Step 1: Identify the relevant concepts : In moststraight-line motion problems, you can use theconstant-acceleration equations. Occasionally,however, you will encounter a situation in which theacceleration isn't constant. In such a case, you'll needa different approacha x =dυxdtd = dt2⎛ dx ⎞ d x⎜ ⎟ =2⎝ dt ⎠ dtStep 2: Set up the problem using the following steps:You must decide at the beginning of a problemwhere the origin of coordinates are usually amatter of convenience. If is often easiest to placethe particle at the origin at time t = 0; then x 0 = 0.It is always helpful to make a motion diagramshowing these choices and some later positions ofthe particle.Remember that your choice of the positive axisdirection automatically determines the positivedirections for velocity and acceleration. If x ispositive to the right of the origin, the v x and a x arealso positive toward the right.Restate the problem in words first, and thentranslate this description into symbols andequations. When does the particle arrive at acertain point (that is, what is the value of t)?where is the particle when its velocity has aspecified value (that is, what is the value of xwhen v x has the specified value)? "where is themotorcyclist when his velocity is 25m/s?"XtraEdge for IIT-JEE 26 MAY 2011

Translated into symbols, this becomes "What isthe value of x when v x = 25 m/s?"Make a list of quantities such as x, x 0 ,v x ,v 0x ,a x andt. In general, some of the them will be knownquantities, and decide which of the unknowns arethe target variables. Be on the lookout for implicitinformation. For example. "A are sits at astoplight" Usually means v 0x = 0.Step 3 : Execute the solution :Choose an equation from Equation v x = v 0x + a x tx = x 0 + v 0x t + 21ax t 22v x =(constant acceleration only)2v 0x+ 2a x (x – x 0 ) (constant accelerations only)⎛ v0x+ vx⎞x – x 0 = ⎜ ⎟ t (constant acceleration only)⎝ 2 ⎠that contains only one of the target variables. Solvethis equation for the equation for the target variable,using symbols only. then substitute the known valuesand compute the value of the target variable.sometimes you will have to solve two simultaneousequations for two unknown quantities.Step 4 : Evaluate your answer : Take a herd look atyour results to see whether they make sense. Arethey within the general range of values youexpected?Problem solving strategy :Projectile Motion :Step 1 : Identify the relevant concepts : The keyconcept to remember is the throughout projectilemotion, the acceleration is downward and has aconstant magnitude g. Be on the lookout for aspectsof the problem that do not involve projectile motion.For example, the projectile-motion equations don'tapply to throwing a ball, because during the throwthe ball is acted on by both the thrower's hand andgravity. These equations come into play only afterthe ball leaves the thrower's hand.Step 2 : Set up the problem using the following stepsDefine your coordinate system and make a sketchshowing axes. Usually it's easiest to place theorigin to place the origin at the initial (t = 0)position of the projectile. (If the projectile is athrown ball or a dart shot from a gun, thethrower's hand or exits the muzzle of the gun.)Also, it's usually best to take the x-axis as beinghorizontal and the y-axis as being upward. Thenthe initial position is x 0 = 0 and y 0 = 0, and thecomponents of the (constant) acceleration are a x = 0,a y = – g.List the unknown and known quantities, anddecide which unknowns are your target variables.In some problems you'll be given the initialvelocity (either in terms of components or interms of magnitude and direction) and asked tofind the coordinates and velocity components assome later time. In other problems you might begiven two points on the trajectory and asked tofind the initial velocity. In any case, you'll beusing equationsx = (v 0 cosα 0 )t (projectile motion) through ...(1)v y = v 0 sin α 0 – gt (projectile motion) ...(2)make sure that you have as many equations asthere are target variables to be found.It often helps to state the problem in words andthen translate those words into symbols. Forexample, when does the particle arrive at a certainpoint ? (That is at what value of t?) Where is theparticle when its velocity has a certain value?(That is, what are the values of x and y when v x orv y has the specified value ?) At the highest pointin a trajectory, v y = 0. so the question "When doesthe particle reach its highest points ?" translatesinto "When does the projectile return to its initialelevation?" translates into "What is the value of twhen y = y 0 ?"Step 3 : Execute the solution use equation (1) & (2)to find the target variables. As you do so, resist thetemptation to break the trajectory into segments andanalyze each segment separately. You don't have tostart all over, with a new axis and a new time scale,when the projectile reaches its highest point ! It'salmost always easier to set up equation (1) & (2)at the starts and continue to use the same axes andtime scale throughout the problem.Step 4 : Evaluate your answer : As always, look atyour results to see whether they make sense andwhether the numerical values seem reasonable.Relative Velocity :Step 1 : Identify the relevant concepts : Wheneveryou see the phrase "velocity relative to" or "velocitywith respect to", it's likely that the concepts ofrelative will be helpful.Step 2 : Set up the problem : Label each frame ofreference in the problem. Each moving object has itsown frame of reference; in addition, you'll almostalways have to include the frame of reference of theearth's surface. (Statements such as "The car istraveling north at 90 km/h" implicitly refer to thecar's velocity relative to the surface of the earth.) Usethe labels to help identify the target variable. Forexample, if you want to find the velocity of a car (C)with respect to a bus (B), your target variable is v C/B .Step 3 : Execute the solution : Solve for the targetvariable using equationv P/A = v P/B + v B/A (relative velocity along a line) ...(1)XtraEdge for IIT-JEE 27 MAY 2011

(If the velocities are not along the same direction, (As the ball returns to its initial position, the change→ → →→| s1+s2| | snet| (a) velocity of the particle after 2 sec.average velocity = VaV= =s1s2t(b) angle between initial velocity and the velocity after 2 sec.+netV1V(c) the maximum height reached by the projectile2(d) horizontal range of the projectile→ →Since s 1 = s 2 = d and s net = s + s | = 0you'll need to use the vector from of this equations,derived later in this section.) It's important to note thein position, the change in position vector of the ball,that is the net displacement will be zero).order of the double subscripts in equation (1) v A/B→always means "velocity of A relative to B." These ∴ | VaV| = 0.subscripts obey an interesting kind of algebra, asequation (1) shown. If regard each one as a fraction,2. A long belt is moving horizontally with a speed of 4then the fraction on the left side is the product of theKm/hour. A child runs on this belt to and fro with afractions on the right sides : P/A = (P/B) (B/A). Thisspeed of 9 Km/hour (with respect to the belt) betweenis a handy rule you can use when applying Equationhis father and mother located 50 m apart on the(1) to any number of frames of reference. Formoving belt. For an observer on a stationary platformexample, if there are three different frames ofoutside, what is thereference A, B, and C, we can write immediately. (a) speed of the child running in the direction of motionv P/A = v P/C + v C/B + V B/Aof the belt,Step 4 : Evaluate your answer : Be on the lookout forstray minus signs in your answer. If the target(b) speed of the child running opposite to the direction ofmotion of the belt andvariable is the velocity of a car relative to a bus(v V/B ), make sure that you haven't accidentallycalculated the velocity of the bus relative of the car(c) time taken by the child in case (a) and (b) ?Which of the answers change, if motion is viewed byone of the parents ?(v B/C ). If you have made this mistake, you canrecover using equation.v A/B = – v B/ASol. Let us consider positive direction of x-axis from leftto right(a) Here, v B = + 4 Km/hourSpeed of child w.r.t. belt, v C = = 9 Km/hourSolved Examples∴ Speed of child w.r.t. stationary observer,v C ′ = v C + v B or v C ′ = 9 + 4 = 13 Km/hour1. A small glass ball is pushed with a speed V from A.It moves on a smooth surface and collides with thewall at B. If it loses half of its speed during thecollision, find the distance, average speed andvelocity of the ball till it reaches at its initial position.(b) Here, v B = + 4 Km/hour, v C = – 9 Km/hour∴ Speed of child w.r.t. stationary observer,v C ′ = v C + v B or v C ′ = – 9 + 4 = –5 Km/hourThe negative sign shows that the child appears to runin a direction opposite to the direction of motion ofthe belt.A V 0.5V B(c) Distance between the parents, s = 50 m = 0.05 KmSince parents and child are located on the same belt,the speed of the child as observe by stationarydSol. The ball moves from A to B with a constant speed V.Since it loses half of its speed on collision, it returnsobserver in either direction (either father to mother orfrom mother to father) will be 9 Km/hour.Time taken by the child in case (a) and (b),from B to A with a constant speed V/2.0.50 km∴ V 1 = V and V 2 = V/2t = = 20 sec.9 km / hourd1+ d2Using the formula, V aV =If the motion is observed by one of parents, answer to(d1/ V 1)+ (d2/ V2)case (a) case (b) gets altred. It is because the speedPutting d 1 = d 2 = d; V 1 = V and V 2 = V/2of the child w.r.t. either of mother or father isd1+ d22V9 Km/hour.We obtain, V aV ==(dV) + (d / 0.5V) 33. A particle is projected with velocity v 0 = 100 m/s atFrom the formula,an angle θ = 30º with the horizontal. Find :| 1 2XtraEdge for IIT-JEE 28 MAY 2011

Sol. (a)(b)(c)→vt→= vxt→î + vytĵwhere î and ĵ are the unit vectors along +ve x and+ve y-axis respectively→tv =(u x + a x t) î + (u y + a y t) ĵHere, u x = v 0 cos θ = 50 3 m/s, a x = 0u y = v 0 sin θ = 50 m/s, a y = – g(Q g acts downwards)→tv = 503 î + (50 – 10 × 2) ĵ=[50 3 î + 30 ĵ ] m/s∴ | → 2 2v 2 | = (vx + v y ) = 2 2( 50 3) + (30)∴→0→2→0v = 50v = 503 î + 50 ĵ3 î + 30 ĵv . v → 2 = 7500 + 1500 = 9000If α is the angle between v → 0 and v→ 2→v0.v29000Then, cos α = =→ →100×91.65| v0| × | v2|α = cos –1 (0.98) = 10.8ºv 2 y – u 2 y = 2a y yAt y = y max , v y = 0∴ 0 – v 2 0 sin 2 θ = 2 (–g)y max2 2v0 sin θ∴ y max = = 125 m2g→u 2 sin 2θ (d) R = = 1732 mg4. A ball starts falling with zero initial velocity on asmooth inclined plane forming an angle α with thehorizontal. Having fallen the distance 'h', the ballrebounds elastically off the inclined plane. At whatdistance from the impact point will the ball reboundfor the second time ?αSol. Just before impact magnitude of velocity of the ball,v = ( 2gh)ααAs the ball collides elastically and the inclined planeis fixed, the ball follows the law of reflection.Now along the incline, velocity component afterimpact is v sin α and acceleration is g sin α.Perpendicular to the incline, velocity component isvcos α and acceleration (– g cos α). Hence, if wemeasure x and y-coordinate along the incline andperpendicular to the incline, thenx = (v sin α) t + ½ (g sin α)t 2and y = (v cos α) t – ½ (g cos α)t 2When the ball hits the plane for a second time,y = 0, (v cos α)t – ½(g cos α)t 2 or t = (2v/g)Putting this value of t in x,4v2 sin αx = = 8h sin αg5. A batsman hits a ball at a height of 1.22m above theground so that ball leaves the bat at an angle 45º withthe horizontal. A 7.31 m high wall is situated at adistance of 97.53 m from the position of the batsman.Will the ball clear the wall if its range is 106.68 m.Take g = 10 m/s 2v02 sin 2θSol. R(range) =gor,1.22m2 Rgv 0 = = Rg as θ = 45ºsin 2θA45ºv 0106.68mor, v 0 = ( Rg)…(1)Equation of trajectoryor, y = x –y = x tan 45º –2vgx 22Rg.½20gx2cos2gx 2= x – RgB45ºPutting x = 97.53, we get210×(97.53)y = 97.53 –= 8.35 cm106.68×10Hence, height of the ball from the ground level ish = 8.35 + 1.22 = 9.577 mAs height of the wall is 7.31 m so the ball will clearthe wall.XtraEdge for IIT-JEE 29 MAY 2011

KEY CONCEPTPhysicalChemistryFundamentalsGASEOUS STATEReal Gases :Deviation from Ideal Behaviour :Real gases do not obey the ideal gas laws exactlyunder all conditions of temperature and pressure.Experiments show that at low pressures andmoderately high temperatures, gases obey the laws ofBoyle, Charles and Avogadro approximately, but asthe pressure is increased or the temperature isdecreased, a marked departure from ideal behaviouris observed.Ideal gaspVPlot of p versus V of hydrogen, ascompared to that of an ideal gasThe curve for the real gas has a tendency to coincidewith that of an ideal gas at low pressures when thevolume is large. At higher pressures, however,deviations are observed.Compressibility Factor :The deviations can be displayed more clearly, byplotting the ratio of the observed molar volume V m tothe ideal molar volume V m,ideal (= RT/p) as a functionof pressure at constant temperature. This ratio iscalled the compressibility factor Z and can beexpressed asZ =VVmm,idealp= RTVmPlots of Compressibility Factor versus Pressure :For an ideal gas Z = 1 and is independent of pressureand temperature. For a real gas, Z = f(T, p), afunction of both temperature and pressure.A graph between Z and p for some gases at 273.15 K,the pressure range in this graph is very large. It canbe noted that:(1) Z is always greater than 1 for H 2 .(2) For N 2 , Z < 1 in the lower pressure range and isgreater than 1 at higher pressures. It decreases withincrease of pressure in the lower pressure region,passes through a minimum at some pressure and thenH 2increases continuously with pressure in the higherpressure region.(3) For CO 2 , there is a large dip in the beginning. Infact, for gases which are easily liquefied, Z dipssharply below the ideal line in the low pressureregion.1.0t = 0ºCH 2N 2CH 4ideal gasCO 2Z0 100 200 300p/101.325 barPlots of Z versus p of a few gasesThis graph gives an impression that the nature of thedeviations depend upon the nature of the gas. In fact,it is not so. The determining factor is the temperaturerelative to the critical temperature of the particulargas; near the critical temperature, the pV curves arelike those for CO 2 , but when far away, the curves arelike those for H 2 (below fig.)Z1.0T 1 >T 2 >T 3 >T 4ideal gas0 200 400 600p/101.325 kPaT 4T 3T 2Plots of Z versus p of a single gasat various temperaturesProvided the pressure is of the order of 1 bar or less,and the temperature is not too near the point ofliquefaction, the observed deviations from the idealgas laws are not more than a few percent. Underthese conditions, therefore, the equation pV = nRTand related expressions may be used.Van der Waals Equation of state for a Real gasCauses of Deviations from Ideal Behaviour :The ideal gas laws can be derived from the kinetictheory of gases which is based on the following twoimportant assumptions:T 1XtraEdge for IIT-JEE 30 MAY 2011

(i) The volume occupied by the molecules isnegligible in comparison to the total volume ofthe gas.(ii) The molecules exert no forces of attraction uponone another.Derivation of van der Waals Equation :Van der Waals was the first to introducesystematically the correction terms due to the abovetwo invalid assumptions in the ideal gas equationp i V i = nRT. His corrections are given below.Correction for volume :V i in the ideal gas equation represents an idealvolume where the molecules can move freely. In realgases, a part of the total volume is, however,occupied by the molecules of the gas. Hence, the freevolume V i is the total volume V minus the volumeoccupied by the molecules. If b represents theeffective volume occupied by the molecules of 1mole of a gas, then for the amount n of the gas V i isgiven byV i = V – nb ...(1)Where b is called the excluded volume or co-volume.The numerical value of b is four times the actualvolume occupied by the gas molecules. This can beshown as follows.If we consider only bimolecular collisions, then thevolume occupied by the sphere of radius 2rrepresents the excluded volume per pair ofmolecules as shown in below Fig.excludedvolume2rExcluded volume per pair of moleculesThus, excluded volume per pair of molecules4= π(2r) 3 ⎛ 4 ⎞= 8 ⎜ πr3 ⎟ 3 ⎝ 3 ⎠Excluded volume per molecule1 ⎡ ⎛ 4 ⎞⎤= ⎢ ⎜ π 3 ⎛ 4 ⎞8 r ⎟⎥ = 4 ⎜ πr3 ⎟2 ⎣ ⎝ 3 ⎠ ⎦ ⎝ 3 ⎠= 4 (volume occupied by a molecule)Since b represents excluded volume per mole of thegas, it is obvious that⎡ ⎛ 4 ⎞⎤b = N A ⎢4⎜πr3 ⎟⎥ ⎣ ⎝ 3 ⎠ ⎦Correction for Forces of Attraction :Consider a molecule A in the bulk of a vessel asshown in Fig. This molecule is surrounded by othermolecules in a symmetrical manner, with the resultthat this molecule on the whole experiences no netforce of attraction.AArrangement of molecules within andnear the surface of a vesselNow, consider a molecule B near the side of thevessel, which is about to strike one of its sides, thuscontributing towards the total pressure of the gas.There are molecules only on one side of the vessel,i.e. towards its centre, with the result that thismolecule experiences a net force of attractiontowards the centre of the vessel. This results indecreasing the velocity of the molecule, and hence itsmomentum. Thus, the molecule does not contributeas much force as it would have, had there been noforce of attraction. Thus, the pressure of a real gaswould be smaller than the corresponding pressure ofan ideal gas, i.e.p i = p + correction term ...(2)This correction term depends upon two factors:(i) The number of molecules per unit volume of thevessel Large this number, larger will be the net forceof attraction with which the molecule B is draggedbehind. This results in a greater decrease in thevelocity of the molecule B and hence a greaterdecrease in the rate of change of momentum.Consequently, the correction term also has a largevalue. If n is the amount of the gas present in thevolume V of the container, the number of moleculesper unit volume of the container is given asnNN' = A nor N' ∝VVThus, the correction term is given as :Correction term ∝ n/V ...( 2a)(ii) The number of molecules striking the side of thevessel per unit time Larger this number, larger willbe the decrease in the rate of change of momentum.Consequently, the correction term also has a largervalue,. Now, the number of molecules striking theside of vessel in a unit time also depends upon thenumber of molecules present in unit volume of thecontainer, and hence in the present case:Correction term ∝ n / V...(2b)Taking both these factors together, we haveBXtraEdge for IIT-JEE 31 MAY 2011

⎛ n ⎞ ⎛ n ⎞Correction term ∝ ⎜ ⎟ ⎜ ⎟⎝ V ⎠ ⎝ V ⎠2nor Correction term = a...( 3)2VWhere a is the proportionality constant and is ameasure of the forces of attraction between themolecules. Thus2np i = p + a...(4)2VThe unit of the term an 2 /V 2 will be the same as that ofthe pressure. Thus, the SI unit of a will be Pa m 6 mol –2 .It may be conveniently expressed in kPa dm 6 mol –2 .When the expressions as given by Eqs (1) and (4) aresubstituted in the ideal gas equation p i V i = nRT, weget⎛2⎞⎜n ap + ⎟ (V – nb) = nRT ...(5)2⎝ V ⎠This equation is applicable to real gases and is knownas the van der Waals equation.Values of van der Waals Constants :The constants a and b in van der Waals equation arecalled van der Waals constants and their valuesdepend upon the nature of the gas. TheyVan Der Waals ConstantsaGas 6 2kPa dm mol –H 2HeN 2O 2Cl 2NONO 2H 2 OCH 4C 2 H 6C 3 H 8C 4 H 10 (n)C 4 H 10 (iso)C 5 H 12 (n)COCO 221.7643.457140.842137.802657.903135.776535.401553.639228.285556.173877.8801466.1731304.0531926.188150.468363.959bdm3 mol –10.026 610.023 700.039 130.031 830.056 220.027 890.044 240.030 490.042 780.063 800.084 450.122 60.114 20.146 00.039 850.042 67are characteristics of the gas. The values of theseconstants are determined by the critical constants ofthe gas. Actually, the so-called constant vary to someextent with temperature and this shows that the vander Waals equation is not a complete solution of thebehaviour of real gases.Applicability of the Van Der Waals Equation :Since the van der Waals equation is applicable to realgases, it is worth considering how far this equationcan explain the experimental behaviours of realgases. The van der Waals equation for 1 mole of agas is⎛ ⎞⎜ap + ⎟ (V 2m – b) = RT ..(i)⎝ V m ⎠At low pressure When pressure is low, the volume issufficiently large and b can be ignored in comparisonto V m in Eq. (i). Thus, we have⎛ ⎞⎜a⎟ap + V 2m = RT or pV m + =RT⎝ V m ⎠V maor Z = 1 –...(ii)VmRTFrom the above equation it is clear that in the lowpressure region, Z is less than 1. On increasing thepressure in this region, the value of the term(a/V m RT) increase as V is inversely proportional to p.Consequently, Z decreases with increase of p.At high pressure When p is large , V m will be smalland one cannot ignore b in comparison to V m .2However, the term a / V m may be considerednegligible in comparison to p in Eq. (i) Thus,pbp(V m – b) = RT or Z = 1 + ...(iiii)RTHere Z is greater than 1 and increases linearly withpressure. This explains the nature of the graph in thehigh pressure region.A high temperature and low pressure Iftemperature is high, V m will also be sufficiently large2and thus the term a / V m will be negligibly small. Atthis stage, b may also be negligible in comparison toV m . Under these conditions, Eq. (i) reduces to anideal gas equation of state:pV m = RTHydrogen and helium The value of a is extremelysmall for these gases as they are difficult to liquefy.Thus, we have the equation of state as p(V m – b) = RT,obtained from the van der Waals equation by2ignoring the term a / V m . Hence, Z is always greaterthan 1 and it increases with increase of p.The van dar Waals equation is a distinctimprovement over the ideal gas law in that it givesqualitative reasons for the deviations from idealbehaviour. However, the generality of the equation islost as it contains two constants, the values of whichdepend upon the nature of the gas.XtraEdge for IIT-JEE 32 MAY 2011

KEY CONCEPTOrganicChemistryFundamentalsGENERAL ORGANICCHEMISTRYStability of different types of carbocations indecreasing order :⊕C⊕> CH⊕⊕ ⊕(Ph) 3 C > (Ph)2 CH > Ph – C H2 ≥⊕ ⊕ ⊕CH 2 = CH – CH 2 ≥ R – C – R > R – CH – R>R⊕ ⊕> R – CH 2 > CH 2 = CHA special stability is associated with cycloproylmethyl cations and this stability increases with everyadditional cyclopropyl group.This is undoubtedly because of conjugation betweenthe bent orbitals of the cyclopropyl ring and thevacant p-orbital of the cation carbon.HHHHCyclopropyl methyl cation orbital representation conjugationwith the p-like orbital of the ringNucleophilicity versus basicity :If the nucleophilic atoms are from the same period ofthe periodic table, strength as a nucleophile parallelsstrength as a base. For example :H 2 O < NH 3CH 3 OH ≈ H 2 O < CH 3 CO Θ 2 < CH 3 O Θ ≈ OH Θ⇒Increasing base strengthIncreasing nucleophile strengthNucleophile strength increases down a column of theperiodic table (in solvents that can have hydrogenbond, such as water and alcohols). For example :ΘR O ΘF < Θ Cl < Θ Br < Θ Iincreasing nucleophilic strengthdecreasing base strength⇒ Steric bulk decreases nucleophilicity. Forexample :CH 3ΘH 3 C – C – O < HO ΘCH 3weaker nucleophileStronger basestronger nucleophileweaker baseLeaving Groups :A good leaving groups is the one which becomes astable ion after its departure. As most leaving groupsleave as a negative ion, the good leaving groups arethose ions which stabilize this negative charge mosteffectively. The weak bases do this best, thus the bestgroups are weak bases. If a group is a weak base i.e.,the conjugate base of a strong acid, it will generallybe a good leaving group. In an S N 2 reaction theleaving group begins to gain negative charge as thetransition state is reached. The more the negativecharge is stabilized, the lower is the energy of thetransition state; this lowers the energy of activationand thereby increases the rate of reaction.The acids HCl, HBr, HI and H 2 SO 4 are all strongacids since the anions Cl – , Br – , I – and HSO – 4 arestable anions these anions (weak bases) are also goodleaving groups in S N 2 reactions. Of the halogens, aniodide ion is the best leaving group and the fluorideion is the poorest :I – > Br – > Cl – > F –The order of basicity is opposite : F – > Cl – > Br – > I – ,the reason that alkyl fluorides are ineffectivesubstrates in S N 2 reactions is related, to the relativelylow acidity of HF (pK a = 3). Sulfonic acids, RSO 2 OH are similar to sulfuric acid in acidity and thesulfonate ion RSO – 3 is a very good leaving group.Alky benzenesulfonates, alkyl p-toluenesulfonatesare therefore, very good substrates in S N 2 reactions.The triflate ion (CF 3 SO – 3 ) is one of the best leavinggroups known, it is the anion of CF 3 SO 3 H which is astrong acid much stronger than sulfuric acid.XtraEdge for IIT-JEE 33 MAY 2011

Kinetic Isotope Effects :The kinetic isotope effect is a change of rate thatoccurs upon isotopic substitution and is generallyexpressed as a ratio of the rate constants,k light/k heavy. A normal isotope effect is one wherethe ratio of k light to k heavy is greater than 1. In aninverse isotope effect, the ratio is less than 1. Aprimary isotope effect is one which results from themaking or breaking of a bond to an isotopicallysubstituted atom and this must occur in the ratedetermining step. A secondary isotope effect isattributable to isotopic substitution of an atom notinvolved in bond making or breaking in the ratedetermining step. Thus when a hydrogen in asubstrate is replaced by deuterium, there is often achange in the rate. Such changes are known asdeuterium, isotope effects and are expressed by theratio k H /k D , the typical value for this ratio is 7. Theground state vibrational energy (the zero-pointvibrational energy) of a bond depends on the mass ofthe atoms and is lower when the reduced mass ishigher. Consequently, D – C, D – O, D – N bonds,etc., have lower energies in the ground state than thecorresponding H – C, H – O, H – N bonds, etc. Thus,complete dissociation of deuterium bond wouldrequire more energy than that for a correspondinghydrogen bond in the same environment. In case a H– C, H –O, or H – N bond is not broken at all in areaction or is broken in a non-rate-determining step,substitution of deuterium for hydrogen generally doesnot lead to a change in the rate, however, if the bondis broken in the rate-determining step, the rate mustbe lowered by the substitution. This helps indetermination of mechanism. In the bromination ofacetone, the rate determining step is thetautomerization of acetone which involves cleavageof a C–H bond. In case this mechanistic assignmentis correct, one should observe a substantial isotopeffect on the bromination of deuterated acetone.Indeed k H /k D was found to be around 7.CH 3 COCH 3 + Br 2 ⎯→ CH 3 COCH 2 Brrate-determining stepBromoacetoneOHCH 3 COCH 3 CH 3 C = CH 2Several mechanisms get support from kinetic isotopeeffect. Some of these are, oxidation of alcohols withchromic acid and electrophilic aromatic substitution.An example of a secondary isotope effect, where it issure that the C – H bond does not break at all in thereaction. Secondary isotope effects for k H /k D aregenerally between 0.6 and 2.0.(CZ 3 ) 2 CHBr + H 2 O → (CZ 3 ) 2 CHOH + HBrthe solvolysis of isopropyl bromide where Z = H or D, k H /k D is1.34 Secondary isotope effect.The substitution of tritium for hydrogen gives isotopeeffects which are numerically larger (k H /k T = 16).E2 elimination like S N 2 process takes place in onestep (without the formation of any intermediates). Asthe attacking base begins to abstract a proton from acarbon next to the leaving group, the C – H bondbegins to break, a new carbon-carbon double bondbegins to form and leaving group begins to depart. Inconfirmation with this mechanism, the base inducedelimination of HBr from (I) proceeds 7.11 timesfaster than the elimination of DBr from (II). ThusC–H or C – D bond is broken in the rate determiningstep. If it was not so there would not have been anyrate difference.HBase– C – CH 2 Br – CH = CH 2H1-Bromo-2-phenylethane (I)DFaster reactionBase– C – CH 2 Br – CD = CH 2DSlower reaction1-Bromo-2,2-dideuterio-2-phenylethane (II)No deuterium isotope effect is found in E1 reactionssince the rupture of C – H (or C – D) bond occursafter the rate determing step, rather than during it.Thus no rate difference can be measured between adeuterated and a non deuterated substrate.Mechanism Review : Substitution versus EliminationS N 2Primary substrateBack-side attack of Nu : withrespect to LGStrong/polarizable unhinderednucleophileBimolecular in ratedeterminingstepConcerted bond forming/bondbreakingInverse of stereochemistryFavored by polar aproticsolvent.S N 2 and E2Secondary or primary substrateStrong unhinderedbase/nucleophile leads to S N 2Strong hinderedbase/nucleophile leads to E2Low temperatrue (S N 2)/hightemperature (E2)S N 1 and E 1Tertiary substrateCarbocation intermediateWeak nucleophile/base (e.g.,solvent)Unimolecular in ratedeterminingstepRacemization if S N 1Removal of β-hydrogen if E1Protic solvent assists ionizationof LGLow temperature (S N 1)/hightemperature (E2)E2Tertiary or secondary substrateConcerted anti-coplanar TSBimolecular in ratedeterminingstepStrong hindered baseHigh temperatureXtraEdge for IIT-JEE 34 MAY 2011

UNDERSTANDINGPhysical Chemistry1. The freezing point of an aqueous solution of KCNcontaining 0.189 mol kg –1 was – 0.704 ºC. On adding0.095 mol of Hg(CN) 2 , the freezing point of thesolution became –0.530ºC. Assuming that thecomplex is formed according to the equationHg(CN) 2 + x CN – x –→ Hg (CN) x + 2Find the formula of the complex.Sol. Molality of the solution containing only KCN is(–∆Tf) (0.704 K)m = == 0.379 mol kg –1K–1f (1.86 K kg mol )This is just double of the given molality( = 0.189 mol kg –1 ) of KCN, indicating completedissociation of KCN. Molality of the solution afterthe formation of the complexm =(–∆TKff)=(0.530K)(1.86K kg mol–1)= 0.285 mol kg –1If it be assumed that the whole of Hg(CN) 2 isconverted into complex, the amounts of variousspecies in 1 kg of solvent after the formation of thecomplex will ben(K + ) = 0.189 mol,n(CN – ) = (0.189 – x) molx–n(Hg(CN) x+ 2 ) = 0.095 molTotal amount of species in 1 kg solvent becomesn total = [0.189 + (0.189 – x) + 0.095] mol= (0.473 – x) mol Equating this to 0.285 mol,we get(0.473 – x) mol = 0.285 moli.e. x = (0.473 – 0.285) = 0.188Number of CN – units combined =Thus, the formula of the complex is0.188mol0.095mol= 22–Hg (CN) 4 .2. The equilibrium constant K p of the reaction2SO 2 (g) + O 2 (g) 2SO 3 (g) is 900 atm –1 at 800K. A mixture containing SO 3 and O 2 having initialpartial pressures of 1 atm and 2 atm, respectively, isheated at constant volume to equilibrate. Calculatethe pressure of each gas at 800 K. [IIT- 1989]Sol. Since to start with SO 2 is not present, it is expectedthat some of SO 3 will decompose to give SO 2 and O 2at equilibrium. If 2x is the partial pressure of SO 3 thatis decreased at equilibrium, we would have2SO 2 (g) + O 2 (g) 2SO 3 (g)t = 0 0 2 atm 1 atmt eq 2x 2 atm + x 1 atm – 2x2(p )2SO3(1 atm − 2x)Hence, K p ==22(pSO) (p )2 O (2x) (2 atm + x)2= 900 atm –1Assuming x

(m / 32) 1Moles fraction of O 2 ==(m / 32) + (m /16) 3According to Dalton's law of partial pressure,P ´O 2= P × mole fraction of O 2∴ Fraction of total pressure exerted by O 2 ,P´O= 21= Mole fraction of O2 =P3(b) Given that,Mass of O 2 = Mass of CH 4 = 32 g32No. of moles of O 2 = = 1 3232No. of moles of CH 4 = = 2 16According to Dalton's law of partial pressure,P ´O 2= P × mole fraction of O 2nRT= × Mole fraction of O2V3 × 0.0821×300 1=× = 24.63 atm1 3Similarly,P ´CH 4= P × mole fraction of CH 4nRT= × Mole fraction of CH4V3 × 0.0821×300 2=× = 49.26 atm1 3Total pressure, P = 24.63 + 49.26 = 73.89 atm4. A solution contains Na 2 CO 3 and NaHCO 3 .10 ml ofthis requires 2.0 ml of 0.1 M H 2 SO 4 for neutralizationusing phenolphthalein as indicator. Methyl orange isthen added when a further 2.5 ml of 0.2 M H 2 SO 4was required. Calculate the strength of Na 2 CO 3 andNaHCO 3 in solution.[IIT-1978]Sol. Step 1.Molecular massEquivalent mass of Na 2 CO 3 =2106= = 532mMeq. of Na 2 CO 3 in solution = 1× 1000 53Step 2.Molecular massEquivalent mass of NaHCO 3 =1= 84mMeq. of NaHCO 3 in solution = 2× 1000 84Step 3.Meq. of H 2 SO 4 used with phenolphthalein= Valency factor × Molarity × Volume (ml)= 2 × 0.1 × 2.0 = 0.42Na 2 CO 3 + H 2 SO 4 → 2NaHCO 3 + Na 2 SO 4Meq. of H 2 SO 4 used with phenolphthalein= 21 Meq. of Na2 CO 3 ∴ 21 Meq. of Na2 CO 3 = 0.4Step 4.Meq. of H 2 SO 4 used with methyl orange= Valency factor × molarity × volume(ml)= 2 × 0.2 × 2.5 = 1Meq. of H 2 SO 4 used with methyl orange= Meq. of NaHCO 3 + 21 Meq. of Na2 CO 3∴ Meq. of NaHCO 3 + 21 Meq. of Na2 CO 3 = 1∴ Meq. of NaHCO 3 = 1 – 0.4 = 0.6and Meq. of Na 2 CO 3 = 2 × 0.4 = 0.8Step 5.m 1 0 .8×53× 1000 = 0.8 or m 1 = = 0.04245310000 .0424×1000∴ Strength of Na 2 CO 3 solution =10= 4.24 g L –1Step 6.m 2× 1000 = 0.6 or m 2 =84∴ Strength of NaHCO 3 solution == 5.04 g L –10 .6×84= 0.050410000 .0504×1000105. Using the data given below, calculate the bondenthalpy of C–C and C–H bonds.∆ C Hº(ethane) = –1556.5 kJ mol –1∆ C Hº (propane) = –2117.5 kJ mol –1C(graphite) → C(g); ∆H = 719.7 kJ mol –1Bond enthalpy of H–H = 435.1 kJ mol –1∆ f Hº(H 2 O, 1) = –284.5 kJ mol –1∆ f Hº(CO 2 , g) = –393.3 kJ mol –1 [IIT-1990]Sol. From the enthalpy of combustion of ethane andpropane, we writeXtraEdge for IIT-JEE 36 MAY 2011

(1) C 2 H 6 (g) + 27O2 (g) → 2CO 2 (g) + 3H 2 O(1) :∆ C H = 3∆ f H(H 2 O, 1) + 2∆ f H(CO 2 , g) – ∆ f H(C 2 H 6 , g)Thus,∆ f H(C 2 H 6 ,g) = – ∆ C H + 3∆ f H(H 2 O, 1)+ 2∆ f H(CO 2 , g)= (1556.5 – 3 × 284.5 – 2 × 393.3) kJ mol –1= – 83.6 kJ mol –1(2) C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(1)∆ C H = 3∆ f H(CO 2 , g)+ 4∆ f H(H 2 O), 1) – ∆ f H(C 3 H 8 , g)Thus∆ f H(C 3 H 8 , g) = –∆ C H + 3∆ f H(CO 2 , g) + 4∆ f H(H 2 O, 1)= (2217.5 – 3 × 393.5 – 4 × 284.5) kJ mol –1= –101.0 kJ mol –1To calculate the ε C–H and ε C–C , we carry out thefollowing manipulations.(i) 2C(graphite) + 3H 2 (g) → C 2 H 6 (g)∆H = – 83.6 kJ mol –12C(g) → 2C (graphite)∆H = –2 × 719.7 kJ mol –16H(g) → 3H 2 (g)∆H = –3 × 435.1 kJ mol –1Add2C(g) + 6H(g) → C 2 H 6 (g)∆H (i) = (–83.6 – 2 × 719.7 – 3 × 435.1) kJ mol –1= – 2828.3 kJ mol –1(ii) 3C(graphite) + 4H 2 (g) → C 3 H 8 (g)∆H = –101.0 kJ mol –13C(g) → 3C (graphite)∆H = –3 × 719.7 kJ mol –18H(g) → 4H 2 (g)∆H = – 4 × 435.1 kJ mol –1Add3C(g) + 8H(g) → C 3 H 8 (g)∆H (ii) = (– 101 – 3 × 719.7 – 4 × 435.1) kJ mol –1= – 4000.5 kJ mol –1Now,∆H (i) = ε C–C – 6ε C–H= –2828.3 kJ mol –1∆H (ii) = –2ε C–C – 8ε C–H= –4000.5 kJ mol –1Solving for ε C–C and ε C–H , we getε C–H = 414.0 kJ mol –1and ε C–H = 344.3 kJ mol –1GalenaGalena is the natural mineral form of lead sulfide. Itis the most important lead ore mineral.Galena is one of the most abundant and widelydistributed sulfide minerals. It crystallizes in thecubic crystal system often showing octahedralforms. It is often associated with the mineralssphalerite, calcite and fluorite.Galena deposits often contain significant amountsof silver as included silver sulfide mineral phases oras limited solid solution within the galena structure.These argentiferous galenas have long been themost important ore of silver in mining. In additionzinc, cadmium, antimony, arsenic and bismuth alsooccur in variable amounts in lead ores. Seleniumsubstitutes for sulfur in the structure constituting asolid solution series. The lead telluride mineralaltaite has the same crystal structure as galena.Within the weathering or oxidation zone galenaalters to anglesite (lead sulfate) or cerussite (leadcarbonate). Galena exposed to acid mine drainagecan be oxidized to anglesite by naturally occurringbacteria and archaea, in a process similar tobioleaching [3]Galena uses :One of the earliest uses of galena was as kohl,which in Ancient Egypt, was applied around theeyes to reduce the glare of the desert sun and torepel flies, which were a potential source ofdisease.[4]Galena is a semiconductor with a small bandgap ofabout 0.4 eV which found use in early wirelesscommunication systems. For example, it was usedas the crystal in crystal radio sets, in which it wasused as a point-contact diode to detect the radiosignals. The galena crystal was used with a safetypin or similar sharp wire, which was known as a"cat's whisker". Making such wireless sets was apopular home hobby in the North of England duringthe 1930s. Derbyshire was one of the main areaswhere Galena was mined. Scientists that werelinked to this application are Karl Ferdinand Braunand Sir Jagdish Bose. In modern wirelesscommunication systems, galena detectors have beenreplaced by more reliable semiconductor devices,though silicon point-contact microwave detectorsstill exist in the market.XtraEdge for IIT-JEE 37 MAY 2011

`tà{xÅtà|vtÄ V{tÄÄxÇzxá1 SetThis section is designed to give IIT JEE aspirants a thorough grinding & exposure to varietyof possible twists and turns of problems in mathematics that would be very helpful in facingIIT JEE. Each and every problem is well thought of in order to strengthen the concepts andwe hope that this section would prove a rich resource for practicing challenging problems andenhancing the preparation level of IIT JEE aspirants.By : Shailendra MaheshwariSolutions will be published in next issueJoint Director Academics, Career Point, KotaPassage :A bag contains ‘n’ cards marked 1, 2, 3, ......, n. ‘X’draws a card from the bag and the card is put backinto the bag. Then ‘Y’ draws a card. The probabilitythat ‘X’ draws.1. The same card as ‘Y’ is –11(A) (B) n 2n12(C)(D)2nn2. a higher card than ‘Y’ is –n −1n −1(A)(B)n2nn −1n −1(C)(D)22n2n3. a lower card than ‘Y’ is –n −1n −1(A)(B)n2nn −1n − 1(C)(d)22n2n4. Evaluate : 228∫∫1010xx2524(1 − x)(1 − x)5049dxdx= ?5. Find the minimum value of⎛2(x 1 – x 2 ) 2 1+ (17 2 )( 2 13)20 ⎟ ⎞⎜x− − x x −⎝⎠where x 1 ∈ R + and x 2 ∈ (13, 17).26. Let f (x) = a 1 tan x + a 2 tan 2x + a3 tan 3x + ................... + a n tan nx , where a1 , a 2 , a 3 , ... a n ∈ R andn ∈ N. If | f (x) | ≤ | tan x | for ∀ x ∈ ⎛ π π ⎞⎜−, ⎟ , Prove⎝ 2 2 ⎠nthat ∑i=1aii≤ 17. Let az 2 + bz + c be a polynomial with complexcoefficients such that a and b are non zero. Prove thatthe zeros of this polynomial lie in the region.b c| z | ≤ + + a b8. Find the fifth degree polynomial which leavesremainder 1 when divided by (x – 1) 3 and remainder–1 when divided by (x + 1) 3 .9. A quadrilateral ABCD is inscribed in a circle ofradius R such that AB 2 + CD 2 = 4R 2 . Using vectormethod prove that its diagonals are at right angle.10. Through a focus of an ellipse two chords are drawnand a conic is described to pass through theirextremities, and also through the centre of the ellipse.Prove that it cuts the major axis in another fixedpoint.Honesty• To be persuasive, You must be believable.To be believable, You must be credible.To be credible, You must be truthful.• An honest man is the noblest work of God.• If I am honesty in all my dealings, I can neverexperience fear.• Prefer a loss to a dishonest gain; one brings painfor the moment, the other for all time.XtraEdge for IIT-JEE 38 MAY 2011

XtraEdge for IIT-JEE 39 MAY 2011

Students' ForumExpert’s Solution for Question asked by IIT-JEE AspirantsMATHS1. Let S be the coefficients of x 49 in given expressionf (x) and if P be product of roots of the equationf (x) = 0, then find the value of PS , given that :f (x) = (x – 1) 2 ⎛⎜⎝x2⎞ ⎛ 1 ⎞ ⎛− 2⎟ ⎜ x − ⎟ ⎜⎠ ⎝ 2 ⎠ ⎝x3⎞ ⎛ 1 ⎞− 3⎟ ⎜ x − ⎟ ,⎠ ⎝ 3 ⎠⎛ x ⎞ ⎛......... ⎜ − 25⎟ ⎜ x⎝ 25 ⎠ ⎝Sol. Here we can write f(x) as :⎧ ⎛ x ⎞⎛x ⎞ ⎛ x ⎞⎫f (x) = ⎨(x −1)⎜ − 2⎟⎜− 3⎟...⎜ − 25⎟⎬⎩ ⎝ 2 ⎠⎝3 ⎠ ⎝ 25 ⎠⎭1− 25⎧ ⎛ 1 ⎞⎛1 ⎞ ⎛ 1 ⎞⎫× ⎨( x −1)⎜ x − ⎟⎜x − ⎟...⎜ x − ⎟⎬⎩ ⎝ 2 ⎠⎝3 ⎠ ⎝ 25 ⎠⎭Now roots of f (x) = 0 are;1 2 , 2 2 , 3 2 , ..... , 25 2 1 1 1and 1, , , ....., 2 3 25Now f (x) is the polynomial of degree 50,So coefficient of x 49 will be :S = – (sum of roots)= – (1 2 + 2 2 + ... + 25 2 ⎛ 1 1 1 ⎞) – ⎜1+ + + .... + ⎟⎝ 2 3 25 ⎠⎧25×26×51 ⎫= – ⎨ + K⎬where, K =⎩ 6 ⎭ ∑n=⇒ S = –(K + 5525).Product of roots :1 2 . 2 2 . 3 2 .... 25 2 1 1 1. 1 . . .... = 1 . 2 . 3 ...252 3 25∴ P = 25 !S −( K + 5525)Hence = P 25!2511n25, where K = ∑n=2. From an external point P(α, 2) a variable line is2 2x ydrawn to meet the ellipse + = 1 at the points9 4A and D. Same line meets the x-axis and y-axis at thepoints B and C respectively. Find the range of valuesof 'α' such that PA. PD = PB.PC.11n⎞⎟⎠Sol. We have been given,yOCDP(α,2)APA.PD = PB.PCEquation of any line through point 'P' is :x − α y − 2= = rcosθsin θor x = α + r cos θ, y = 2 + r sin θPutting this point in the equation of given ellipse, we get4(r cos θ + α) 2 + 9(2 + r sin θ) 2 = 36⇒ r 2 (4 cos 2 θ + 9 sin 2 θ) + 4r (9 sin θ + 2 α cos θ)+ 4 α 2 = 0Since PA and PD are the roots of this quadratic in r,we get24αPA.PD =...(i)2 2(4cos θ + 9sin θ)Similarly, putting x = r cos θ + α, y = r sin θ + 2 inthe equation of coordinate axis i.e. xy = 0(r cos θ + α). (r sin θ + 2) = 0⇒ r 2 sin θ cos θ + r (2 cos θ + α sin θ) + 2α = 0Since PB and PC and the roots of this quadratic in 'r',2α4αwe get, PB.PC = =...(ii)sin θcosθsin 2αThus, we get24α4α={from (i) and (ii)}2 2sin 2θ4cos θ + 9sin θ4α8α2⇒ =sin 2θ4(1 + cos 2θ)+ 9(1 − cos 2θ)B1 2α⇒ =sin 2θ 13 − 5cos 2θ⇒ 13 = 5 cos 2θ + 2α sin 2θwhere; 5 cos 2θ + 2α sin 2θ ≤∴ 4α 2 + 25 ≥ 13 ⇒ α 2 ≥⇒ α ∈ (–∞, – 6] ∪ [6, ∞)25 + 4α169 − 254x2= 36XtraEdge for IIT-JEE 40 MAY 2011

3. Find the sum of the terms of G.P. a + ar + ar 2 + ..... + ∞where a is the value of x for which the function7 + 2x log e 25 – 5 x – 1 – 5 2–x has the greatest value andx t dtr is the Ltx→∫ 02x tan( π + x20 )aSol. S =1− r,To get the greatest value f´(x) = 2log e 25 – 5 x – 1 log 5+ 5 2–x log 5f ´(x) = 4 log e 5 – 5 x–1 log e 5 + 5.5 1 – x log e 5⇒ f ´(x) = 0 put 5 x – 1 t(> 0)t 2 – 4t – 5 = 0 ⇒ t = 5 ⇒ 5 x –1 = 5 ⇒ x = 2to evaluate r :x 2t dt 1r = Ltx→∫=0 02x tan( π + x)π1 2πsince a = 2, r = ⇒ sum of G.P. = π π −14. Let [x] stands for the greatest integer function find23 x + sin xthe derivative of f (x) = ( x + [ x + 1]) , where itexists in (1, 1.5). Indicate the point(s) where it doesnot exist. Give reason(s) for your conclusion.Sol. The greatest integer [x 3 + 1] takes jump from 2 to 3 at3 2 and again from 3 to 4 at 3 3 in [1, 1.5] andtherefore it is discontinuous at these two points. As aresult the given function is discontinuous at 3 2 andhence not differentiable.To find the derivative at other points we write :in (1, 3 2 ), f (x) = ( x + 2)2x + sin x−12x + sin x⇒ f ´(x) = ( x + 2){x 2 + sin x + (x + 2) (2x + cos x) log (x + 2)}2x + sin xin ( 3 2 , 3 3 ), f (x) = ( x + 3) ,2x + sin x−1f ´(x) = ( x + 3) {x 2 + sin x+ (2x + cos x) (x + 3) × log e (x + 3)}2x + sin xin ( 3 5 , 1.5), f (x) = ( x + 4) ,2x + sin x−1f ´(x) = ( x + 4) , {x 2 + sin x + (2x + cos x)(x + 4) × log e (x + 4)}5. For three unit vectors â , bˆ and ĉ not all collineargiven that aˆ× cˆ= cˆ× bˆand b ˆ× aˆ= aˆ×c ˆ . Show thatcosα + cos β + cos γ = –3/2, where α, β and γ are theangles between â and bˆ , bˆ and ĉ and ĉ and ârespectively.Sol. â × ĉ = cˆ× bˆ⇒ ( â + bˆ ) × ĉ = → 0⇒ ĉ is collinear with â + bˆ ⇒ â + bˆ = λ ĉ forsame λ ∈ RSimilarly bˆ + ĉ = µ â for some scalar uNow â + bˆ = λ ĉ ⇒ â + bˆ + ĉ = (λ + 1) → cSimilarly ⇒ â + bˆ + ĉ = (µ + 1) âHence (λ + 1) ĉ = (µ + 1) â ,either λ + 1 = µ + 1 = 0 or ĉ is collinear with â .But ĉ can not be collinear to â other wise cˆ × aˆ= 0⇒ cˆ × bˆ= 0⇒ bˆ is collinear to with ĉ⇒ â bˆ and ĉ are collinear.Hence ĉ is not collinear to â⇒ λ + 1 = µ + 1 = 0⇒ λ ± µ = –1Hence bˆ + ĉ = µ â⇒ â + bˆ + ĉ = → 0⇒ ( â + bˆ + ĉ ) . ( â + bˆ + ĉ ) = 0⇒ 1 + 1 + 1 + 2 ( â . bˆ + bˆ . ĉ + ĉ . â ) = 03⇒ â . bˆ + bˆ . ĉ + ĉ . â = – 2⇒ cos α + cos β + cos γ = – 236. If a, b, c and n are positive integers such thata + b + c = n, show that(a a b b c c ) 1/n + (a b b c c a ) 1/n + (a c b a c b ) 1/n ≤ n.Sol. Since a, b, c are integers, from A.M. – G.M.inequality we can write( a + a + .... atimes)+ ( b + b + ... btimes)+ ( c + c + ... ctimes)a + b + c1/a +b+c≥ [(a.a....a times)(b.b....b times)(c.c...c times)]1a.a + b.b + c.c a b c⇒≥ ( a b c ) a+b+ca + b + c1c.a + a.b + b.c c a cSimilarly,≥ ( a b c ) a+b+cc + a + b1b.a + c.b + a.c b c aand≥ ( a b c ) b+c+ab + c + aAdding these three inequalities, we get2 2 2a + b + c + 2ab+ 2bc+ 2ca≥ (a a b b c c ) 1/na + b + c+ (a c b a c b ) 1/n + (a b b c c a ) 1/n2( a + b + c)where LHS =a + b + c= a + b + c Hence provedXtraEdge for IIT-JEE 41 MAY 2011

MATHSCOMPLEX NUMBERMathematics Fundamentals− 1 is denoted by ‘i’ and is pronounced as ‘iota’.i = − 1 ⇒ i 2 = –1, i 3 = –i, i 4 = 1.If a, b ∈ R and i = − 1 then a + ib is called acomplex number. The complex number a + ib is alsodenoted by the ordered pair (a, b)If z = a + ib is a complex number, then :(i) a is called the real part of z and we writeRe (z) = a.(ii) b is called the imaginary part of z and we writeIm (z) = bTwo complex numbers z 1 and z 2 are said to be equalcomplex numbers if Re (z 1 ) = Re (z 2 ) and Im(z 1 ) = Im (z 2 ).If z = x + iy is a non zero complex number, then 1/z iscalled the multiplicative inverse of z.If x + iy is a complex number, then the complexnumber x – iy is called the conjugate of the complexnumber x + iy and we write x + iy = x – iy.Algebra of Complex Numbers(i) Addition : (a + ib) + (c + id) = (a + c) + i(b + d)(ii) Subtraction :(a + ib) – (c + id) = (a – c) + i(b – d)(iii) Multiplication :(a + ib) + (c + id) = (ac – bd) + i(ab + bc)(iv) Division by a non-zero complex number :a + ib ac + bd bc − ad= + i , (c + id) ≠ 02 2 2 2c + id c + d c + dProperties : If z 1 , z 2 are complex numbers, then(i) ( z 1 ) = z 1(ii) z + z = 2 Re (z)(iii) z – z = 2i Im (z)(iv) z = z iff z is purely real(v) z = z iff z is purely imaginary(vi) z 1 + z2= z1+ z2(vii) z 1 + z2= z1+ z2(viii) z 1 – z2= z1– z2⎛ z(ix)⎜⎝ z12⎞ z⎟ =⎠ z12provided z 2 ≠ 0If x + iy is a complex number, then the non-negative2 2ral number x + y is called the modulus of thecomplex number x + iy and write| x + iy| =2x + y2Properties : If z 1 , z 2 are complex numbers, then(i) | z 1 | = 0 iff z 1 = 0(ii) | z 1 | = | z 1 | = | – z 1 |(iii) – | z 1 | ≤ Re (z 1 ) ≤ | z 1 |(iv) – | z 1 | ≤ Im (z 1 ) ≤ | z 1 |(v) | z 1 z 1 | = | z 1 | 2(vi) | z 1 + z 2 | ≤ | z 1 | + | z 2 |(vii) | z 1 – z 2 | ≥ | z 1 | – | z 2 |(viii) | z 1 z 2 | = | z 1 | | z 2 |(ix)zz12| z1|= , provided z 2 ≠ 0| z |2(x) | z 1 + z 2 | 2 = | z 1 | 2 + | z 2 | 2 + 2 Re (z 1 z 2 )(xi) | z 1 – z 2 | 2 = | z 1 | 2 + | z 2 | 2 – 2 Re (z 1 z 2 )(xi) | z 1 + z 2 | 2 + | z 1 – z 2 | 2 = 2 [| z 1 | 2 + | z 2 | 2 ].De Moivre’s Theorem(i) If n is any integer (positive or negative), then(cos θ + i sin θ) n = cos nθ + i sin nθ(ii) If n is a rational number, then the value or one ofthe values of (cos θ + i sin θ) n is cos nθ + i sin nθEuler’s Formulae iθ = cos θ + i sin θ and e –iθ = cos θ – i sin θSquare root of complex numberSquare root of z = a + ib are given by⎡ ⎛ | z | + a ⎞ ⎛ | z | −a⎞ ⎤± ⎢ ⎜ ⎟ + i ⎜ ⎟ ⎥ for b > 0 and⎢⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦⎡ ⎛ | z | + a ⎞ ⎛ | z | −a⎞ ⎤± ⎢ ⎜ ⎟ – i ⎜ ⎟ ⎥ for b < 0.⎢⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦XtraEdge for IIT-JEE 42 MAY 2011

−1+i 3If ω = , then the cube roots of unity are 1,2ω and ω 2 . We have:(i) 1 + ω + ω 2 = 0 (ii) ω 3 = 1Let z = x + iy be any complex number.Let z = r (cos θ + i sin θ) where r > 0.∴ x = r cos θ and y = r sin θ∴ x 2 + y 2 = r 2⇒ r =22x + y (Q r > 0)∴ cos θ =x2 2x + yand sin θ =y2 2x + yThe value of θ is found by solving these equations. θis called the argument (or amplitude) of z.If – p < θ ≤ π, then θ is called the principal argumentof z.Identification of θ –x y arg(z) Interval of θ⎛ π ⎞+ + θ ⎜0< θ < ⎟⎝ 2 ⎠⎛ – π ⎞+ – –θ ⎜ < θ < 0⎟ ⎝ 2 ⎠⎛ π ⎞– + (π – θ) ⎜ < θ < π⎟ ⎝ 2 ⎠⎛ – π ⎞– – –(π – θ) ⎜ – π < θ < ⎟⎝ 2 ⎠If z 1 and z 2 are two complex numbers then(i) | z 1 – z 2 | is the distance between the points withaffixes z 1 and z 2 .mz(ii) 2 + nz 1 is the affix of the point dividing them + nline joining the points with affixes z 1 and z 2 in theratio m : n internally.mz2 – nz1(iii)is the affix of the point dividing them – nline joining the points with affixes z 1 and z 2 in theratio m : n externally where m ≠ n.(iv) If z 1 , z 2 , z 3 are the affixes of the vertices of az 1 + z2+ z3triangle then the affix of its centroid is.3(v) z = tz 1 + (1 – t)z 2 is the equation of the line joiningpoints with affixes z 1 and z 2 . Here ‘t’ is a parameter.z − z1z − z1(vi) = is the equation of the linez2− z1z2− z1joining points with affixes z 1 and z 2 .Three points with affixes z 1 , z 2 , z 3 are collinear ifzzz123zzz123111= 0.The general equation of a straight line isa z + az + b = 0 , where b is any real number.(i) | z – z 1 | < r represents the circle with centre z 1and radius r.(ii) | z – z 1 | < r represents the interior of the circlewith centre z 1 and radius r.z − z1= k represents a circle line which is thez − z1perpendicular bisector of the line segment joiningpoints with affixes z 1 and z 2 .(z – z 1 ) ( z − z2)+ ( z − z1)+ (z – z 2 ) = 0 represents thecircle with line joining points with affixes z 1 and z 2 asa diameter.| z – z 1 | + | z – z 2 | = 2k, k ∈ R + represents the ellipsewith foci at points with affixes z 1 and z 2 .If z 1 , z 2 , z 3 be the affixes of the points A, B, Crespectively, then the angle between AB and AC is⎛ z ⎞given by arg⎜3 − z1⎟ .⎝ z2− z1⎠If z 1 , z 2 , z 3 , z 4 are the affixes of the points A, B, C, Drespectively, then the angle between AB and CD is⎛ z ⎞given by arg⎜2 − z1⎟ .⎝ z4− z3⎠nth roots of a complex numberLet z = r (cos θ + i sin θ), r > 0 be any complexnumber. n th root o z = z 1/n= r 1/n ⎛ 2 π + θ kπ + θ ⎞⎜cos k + isin2 ⎟ ,⎝ nn ⎠where k = 0, 1, 2, ………, n – 1.There are n distinct values and sum of all thesevalues is 0.Logarithm of a complex numberLet z = re iθ be any complex number.Then log z = log re iθ = log r + log e iθ= log r + iθ log e = log r + iθ.∴ log z = log | z | + i amp (z).XtraEdge for IIT-JEE 43 MAY 2011

MATHSMATRICES &DETERMINANTSMathematics FundamentalsMatrices :An m × n matrix is a rectangular array of mn numbers(real or complex) arranged in an ordered set of mhorizontal lines called rows and n vertical lines calledcolumns enclosed in parentheses. An m × n matrix Ais usually written as :⎡a11⎢⎢a21⎢ MA = ⎢⎢ ai1⎢ M⎢⎢⎣am1aaaa1222Mi2Mm2............aaa1 j2 jaijmj............a1n⎤a⎥2n⎥⎥⎥ain⎥⎥⎥amn⎥⎦Where 1 ≤ i ≤ m and 1 ≤ j ≤ nand is written in compact form as A = [a ij ] m× nA matrix A = [a ij ] m × n is called(i) a rectangular matrix if m ≠ n(ii) a square matrix if m = n(iii) a row matrix or row vector if m = 1(iv) a column matrix or column vector if n = 1(v) a null matrix if a ij = 0 for all i, j and is denoted byO m× n(vi) a diagonal matrix if a ij = 0 for i ≠ j(vii) a scalar matrix if a ij = 0 for i ≠ j and all diagonalelements a ii are equalTwo matrices can be added only when thye are of sameorder. If A = [a ij ] m × n and B = [b ij ] m × n , then sum of Aand B is denoted by A + B and is a matrix [a ij + b ij ] m × nThe product of two matrices A and B, written as AB,is defined in this very order of matrices if number ofcolumns of A (pre factor) is equal to the number ofrows of B (post factor). If AB is defined , we say thatA and B are conformable for multiplication in theorder AB.If A = [a ij ] m × n and B = [b ij ] n × p , then their product ABis a matrix C = [c ij ] m × p whereC ij = sum of the products of elements of ith row of Awith the corresponding elements of j th column of B.Types of matrices :(i) Idempotent if A 2 = A(ii) Periodic if A k+1 = A for some positive integer k.The least value of k is called the period of A.(iii) Nilpotent if A k = O when k is a positive integer.Least value of k is called the index of thenilpotent matrix.(iv) Involutary if A 2 = I.The matrix obtained from a matrix A = [a ij ] m × n bychanging its rows into columns and columns of A intorows is called the transpose of A and is denoted by A′.A square matrix a = [a ij ] n × n is said to be(i) Symmetric if a ij = a ji for all i and j i.e. if A′ = A.(ii) Skew-symmetric ifa ij = – a ji for all i and j i.e., if A′ = –A.Every square matrix A can be uniquely written as sumof a symmetric and a skew-symmetric matrix.A = 21 (A + A′) + 21 (A – A′) where 21 (A + A′) issymmetric and 21 (A – A′) is skew-symmetric.Let A = [a ij ] m × n be a given matrix. Then the matrixobtained from A by replacing all the elements by theirconjugate complex is called the conjugate of the matrixA and is denoted by A = a ] .Properties :(i) ( A ) = A(ii) ( A + B)= A + B[ ij(iii) (λ A)= λ A , where λ is a scalar(iv) ( A B) = A B .Determinant :Consider the set of linear equations a 1 x + b 1 y = 0 anda 2 x + b 2 y = 0, where on eliminating x and y we getthe eliminant a 1 b 2 – a 2 b 1 = 0; or symbolically, wewrite in the determinant notationa1b1≡ a 1 b 2 – a 2 b 1 = 0a2b2Here the scalar a 1 b 2 – a 2 b 1 is said to be the expansiona1b1of the 2 × 2 order determinant having 2a2b2rows and 2 columns.Similarly, a determinant of 3 × 3 order can beexpanded as :XtraEdge for IIT-JEE 44 MAY 2011

aaa123bbb123ccc123b2c2= a 1b3c3a2c2a2b2– b 1 + c 1a3c3a3b3= a 1 (b 2 c 3 – b 3 c 2 ) – b 1 (a 2 c 3 – a 3 c 2 ) + c 1 (a 2 b 3 – a 3 b 2 )= a 1 (b 2 c 3 – b 3 c 2 ) – a 2 (b 1 c 3 – b 3 c 1 ) + a 3 (b 1 c 2 – b 2 c 1 )= Σ(± a i b j c k )To every square matrix A = [a ij ] m × n is associated anumber of function called the determinant of A and isdenoted by | A | or det A.Thus, | A | =aaa1121Mn1aaa1222Mn2.........aaa1n2nIf A = [a ij ] n × n , then the matrix obtained from A afterdeleting ith row and jth column is called a submatrixof A. The determinant of this submatrix is called aminor or a ij .Sum of products of elements of a row (or column) ina det with their corresponding cofactors is equal tothe value of the determinant.ni.e., ∑i=1a ijC ij = | A | and ∑j=n1Mnna ijC ij = | A |.(i) If all the elements of any two rows or two columnsof a determinant ate either identical orproportional, then the determinant is zero.(ii) If A is a square matrix of order n, then| kA | = k n | A |.(iii) If ∆ is determinant of order n and ∆′ is thedeterminant obtained from ∆ by replacing theelements by the corresponding cofactors, then∆′ = ∆ n–1(iv) Determinant of a skew-symmetric matrix of oddorder is always zero.The determinant of a square matrix can be evaluatedby expanding from any row or column.If A = [a ij ] n × n is a square matrix and C ij is thecofactor of a ij in A, then the transpose of the matrixobtained from A after replacing each element by thecorresponding cofactor is called the adjoint of A andis denoted by adj. A.Thus, adj. A = [C ij ]′.Properties of adjoint of a square matrix(i) If A is a square matrix of order n, thenA . (adj. A) = (adj . A) A = | A | I n .(ii) If | A | = 0, then A (adj. A) = (adj. A) A = O.(iii) | adj . A | = | A | n –1 if | A | ≠ 0(iv) adj. (AB) = (adj. B) (adj. A).(v) adj. (adj. A) = | A | n – 2 A.Let A be a square matrix of order n. Then the inverse ofA is given by A –1 1= adj. A.| A |Reversal law : If A, B, C are invertible matrices of sameorder, then(i) (AB) –1 = B –1 A –1(ii) (ABC) –1 = C –1 B –1 A –1Criterion of consistency of a system of linear equations(i) The non-homogeneous system AX = B, B ≠ 0 hasunique solution if | A | ≠ 0 and the unique solution isgiven by X = A –1 B.(ii) Cramer’s Rule : If | A | ≠ 0 and X = (x 1 , x 2 ,..., x n )′| Athen for each i =1, 2, 3, …, n ; x i =i |where| A |Ai is the matrix obtained from A by replacing theith column with B.(iii) If | A | = 0 and (adj. A) B = O, then the systemAX = B is consistent and has infinitely manysolutions.(iv) If | A | = 0 and (adj. A) B ≠ O, then the systemAX = B is inconsistent.(v) If | A | ≠ 0 then the homogeneous system AX = Ohas only null solution or trivial solution(i.e., x 1 = 0, x 2 = 0, …. x n = 0)(vi) If | A | = 0, then the system AX = O has non-nullsolution.(i) Area of a triangle having vertices at (x 1 , y 1 ), (x 2 , y 2 )and (x 3 , y 3 ) is given by12(ii) Three points A(x 1 , y 1 ), B(x 2 , y 2 ) and C(x 3 , y 3 ) arecollinear iff area of ∆ABC = 0.A square matrix A is called an orthogonal matrix ifAA′ = AA′ = I.A square matrix A is called unitary if AA θ = A θ A = I(i) The determinant of a unitary matrix is of modulusunity.(ii) If A is a unitary matrix then A′, A , A θ , A –1 areunitary.(iii) Product of two unitary matrices is unitary.Differentiation of Determinants :Let A = | C 1 C 2 C 3 | is a determinant thendA = | C′1 C 2 C 3 | + | C 1 C′ 2 C 3 | + | C 1 C 2 C′ 3 |dxSame process we have for row.Thus, to differentiate a determinant, we differentiate onecolumn (or row) at a time, keeping others unchanged.xxx123yyy123111XtraEdge for IIT-JEE 45 MAY 2011

aBased on New PatternIIT-JEE 2012XtraEdge Test Series # 1Time : 3 HoursSyllabus : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectilemotion, Circular motion, Electrostatics & Gauss's Law, Capacitance, Current electricity, Alternating Current,Magnetic Field, E.M.I. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table, ChemicalKinetics, Electro Chemistry, Solid state, Solutions, Surface Chemistry, Nuclear Chemistry. Mathematics:Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii ofCircle, Function, Limit, Continiuty, Differentiation, Application of Differentiation (Tangent & Normal,Monotonicity, Maxima & Minima)Instructions : [Each subject contain]Section – I : Question 1 to 7 are multiple choice questions with only one correct answer. +3 marks will beawarded for correct answer and -1 mark for wrong answer.Section – II : Question 8 to 11 are multiple choice questions with multiple correct answer. +4 marks will beawarded for correct answer and No Negative marks for wrong answer.Section – III : Question 12 to 16 are passage based single correct type questions. +3 marks will be awarded forcorrect answer and -1 mark for wrong answer.Section – IV : Question 17 to 23 are Numerical Response Question (single digit Ans. type) +4 marks will beawarded for correct answer and No Negative marks for wrong answer.PHYSICSQuestion 1 to 7 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. A rectangular piece of aluminum is 5.10 ± 0.01 cmlong and 1.90 ± 0.01 cm wide. The area of therectangle is written as -(A) (9.69 ± 0.03) cm 2 (B) (9.69 ± 0.01) cm 2(C) (9.69 ± 0.02) cm 2 (D) (9.69 ± 0.07) cm 22. One body fall freely from a point P at a heightH + h as shown in figure whilst another body isprojected upwards with an initial velocity v 0 . frompoint R at the same time as the first body begins tofall and second body meet at a point Q at a height h.The maximum height attained by of the secondbody for the given initial velocity is -P(A) H + h(C)(H + h)4H2HhQR(B)(D)(H + h)4h2(H + h)h23. The velocity of a car is given byv = (10 m/s 2 )t – (5 m/s 2 ) t 2Initially car at x = 0 at t = 0. Time taken by the carto reach its maximum positive x-coordinates is -(A) 0 s(B) 1 s(C) 2 s(D) 1.5 s4. For the circuit shown here keys k 1 and k 3 are closedfor 1 second. Key k 2 is closed at the instant k 1 andk 3 are opened. Maximum charge on the capacitorafter key k 2 is closed is –2V k 1⎛ 1 ⎞(A) 2⎜1–⎟Cb⎝ e ⎠⎛ 1 ⎞(C) 8⎜1–⎟Cb⎝ e ⎠2F4 V2 H2 Ω0.5 Ωk 2k 3⎛ 1 ⎞(B) 4 2⎜1–⎟Cb⎝ e ⎠(D) zeroXtraEdge for IIT-JEE 46 MAY 2011

5. The circuit shown in the diagram extends to theright into infinity. Each battery has emf ε(unknown)and internal resistance r. Each resistor has theresistance 4r. The reading of ideal ammeter shownin the diagram is I. Find the value of ε is terms of Iand r4 r 4 r 4 rAε,r(A) 4.82 I r (B) 6.82 I r(C) 1.82 I r(D) 10.82 I r6. The metal (hollow) sphere of radius R, 2R and 3Rare placed into each other such that their centres areat the same point. The inner sphere is given acharge of Q. the middle one is charged to 2Q andthe outer one is charged to 3Q. Find the potentials,measured from the common centre of the circles, ata distance 4R, if the potential at the centre is takento be zero -(A)(C)3 KQ2 RKQRε,r(B) –(D) –3 KQ2 RKQ2R7. Which of these is not correct regarding eddycurrent?(A) Eddy currents result due to motion of ametallic plate in magnetic field(B) Eddy currents are minimised in transformer byusing laminated core with metal laminations(C) In induction furnace, eddy currents raisetemperature of metal upto melting point(D) Eddy currents are named so, as they propagatesimilar to swirling eddies in waterQuestions 8 to 11 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.8. Acceleration-time graph of particle moving alongx-axis is given bya(m/s 2 )42ε,rt(s)2 4 6Initially particle is at rest and at origin -(A) Velocity of the particle at t = 6 s is 14 m/s(B) Position of the particle at t = 2 s is x = 4 m(C) Velocity of the particle at t = 4 s is 4 m/s(D) Position of the particle at t = 2 s is x = 8 m9. A particle is projected with a speed v and an angleθ with horizontal. Choose the correct statements -(A) Speed of the particle is zero at highest point(B) Speed of the particle is minimum at highestpoint(C) Acceleration of particle through out he motionis constant(D) Acceleration of particle varies with time10.A20 ΩB100 Ω30 Ω150 ΩC(A) V B – V C is greatest (B) V A = V C(C) V B – V D is greatest (D) V A = V D11. Two charged particles M and N enters a space ofuniform magnetic field with velocitiesperpendiculars to the magnetic field. The paths areas shown in the figure. The possible reasons fordifferent paths may be -× × × × × × × × × × ×× × × × × × × × × × ×× × × × × × × × × × ×× × × × × × × × × × ×× × × × × × × × × × ×× × × × × × × × × × ×× × × × × × × × × × ×× × × × × × M × × N × × ×× × × × × × × × × × ×(A) The charge of M is greater than that of N.(B) The momentum of M is greater than that of N(C) Specific charge of M is greater than that of N(D) The speed of M is less than that of NThis section contains 2 paragraphs, one has 3multiple choice questions and other has 2 multiplechoice questions (Question 12 to 16). Each questionshas 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.Passage : I (Q. No. 12 to 14)Velocity of the river with respect to ground isgiven v 0 with of the river is d. A swimmer swims(with respect to water) perpendicular to the currentwith acceleration a = t (where t is time) startingfrom rest from the origin O at t = 0DXtraEdge for IIT-JEE 47 MAY 2011

dxO12. The time of crossing the river is -(A) (d) 1/3 (B) (6d) 1/3(C) (2d) 1/3(D) None of these13. The drift of the swimmer is -(A) v 0 (d) 1/3 (B) v 0 (2d) 1/3(C) v 0 (6d) 1/3 (D) None of thesey14. The equation of trajectory of the path followed bythe swimmer -2xx(A) y =(B) y =v202v(C) y =x6v330v 0(D) None of thesePassage : II (Q. No. 15 to 16)A metal ring having three metallic spokes of lengthr = 0.2 m is in vertical plane and can spin around afixed horizontal axis in a homogeneous magneticfield of a magnetic induction B = 0.5 T. The linesof magnetic field are perpendicular to the plane ofmetal ring. Between the axis of the metal ring andits perimeter we connect a consumer of resistanceof 0.15 Ω with the help of two sliding contact. Wefix a thread of negligible mass to the rim of the ringand wind it several times around the ring and to itsend we fix a body of a mass of 20 g. At a givenmoment we release the body of mass m. the frictionis negligible everywhere, the resistance of the ring,the spokes and the connected wiring is alsonegligible.15. What is the torque exerted on the ring with spokesby the magnetic force when the body of mass m ismoving with a constant velocity -0B16. What current is flowing through the consumerwhen the velocity of the body of mass m is 3m/s -(A) 1 Amp(B) 3 Amp(C) 2 Amp(D) 4 AmpThis section contains 7 questions (Q.17 to 23).+4 marks will be given for each correct answer andno negative marking. The answer to each of thequestions is a SINGLE-DIGIT INTEGER, rangingfrom 0 to 9. The appropriate bubbles below therespective question numbers in the OMR have to bedarkened. For example, if the correct answers toquestion numbers X, Y, Z and W (say) are 6, 0, 9 and2, respectively, then the correct darkening of bubbleswill look like the following :X Y Z W0123456789012345678917. The speed of a motor launch with respect to theflow of water in stream was v = 7 m/s, the speed ofthe stream was u = 3 m/s. When the launch begantraveling upstream, a float was dropped from it.The launch traveled l = 4.2 km upstream, turnedabout and caught up with the float. How long wasit before the launch reached the float (Answer in................× 10 1 minutes)18. A man is running with a speed 8 m/s constant inmagnitude and direction passes under a lanternhanging at a height 10 m above the ground. Findthe velocity which the edge of the shadow of theman's head moves over the ground with if hasheight is 2 m. (Answer in ............ × 10 1 m/sec)19. Two motor vehicles run at constant speeds 5 m/seach along highways intersecting at an angle 60º.In what time after they meet at the intersection willthe distance between the vehicles be 10 3 m.01234567890123456789(A) 0.02 N-m(C) 0.06 N-mm(B) 0.04 N-m(D) 0.08 N-m20. A boy is standing on an open truck. Truck ismoving with an acceleration 2m/s 2 on horizontalroad. When speed of truck is 10 m/s, boy projecteda ball with velocity 10 m/s in vertical upwarddirection relative to himself (take g = 10 m/s 2 ).After how many seconds of projection of ball, ballis moving backward horizontally as seen by boy.XtraEdge for IIT-JEE 48 MAY 2011

21. A current carrying loop is placed in a uniformmagnetic field pointing in negative z direction.Branch PQRS is a three quarter circle, whilebranch PS is straight. If force on branch PS is2 F. Force on branch PQR is given a × F. Thenvalue of a is.yPQR22. One of the sides of the frame shown in the figurecan move. The frame is in uniform magnetic fieldof B 0 . Which is perpendicular to the plane of theframe. At the time t= 0. the magnetic field beginsB 0to decrease as B(t) = . Side of the frame is1+ktmoving with velocity v in order not to induceelectric current in it. If the value of k is 3 sec –1 ,then v = ........... × a.at = 0bBSxBv23. Two circular rings of identical radii and resistanceof 36 Ω each are placed in such a way that theycross each other's centre C 1 and C 2 a shown infigure. Conducting joints are made at intersectionpoints A and B of the rings. An ideal cell of emf 20V is connected across AB. The power delivered bycell is ........ × 25 watt.AC 1BC 2CHEMISTRYQuestion 1 to 7 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. For a general nth order process A → P with initialconcentration of reactant "a" and rate constant k,the expression for time for 75% completion ofreaction is -(A)1 ⎛⎜2n – 1⎝ an–1– 2 ⎞⎟⎠n–11k(B)1 ⎛⎜2n – 1⎝2n–2an–1– 2 ⎞⎟⎠1k(C)1 ⎛⎜2n – 1⎝ a2n–2n–1– 1⎞⎟⎠1k(D)1 ⎛⎜2n – 2⎝ a2n–22n–1– 1⎞⎟⎠2. An important application of radioactive decay is inthe dating of rocks, fossils, and ancient objects.Naturally occurring radioactive uranium -23892decays in a series of 238 U steps that finally lead tothe stable isotope 206 Pb.A sample of rock is found to contain 23.2 mg ofuranium -238 and 1.42 mg of lead -206. If the halflifeof 238 U 92 is 4.51 × 10 9 years, what is the age ofrock ?(A) 1.8 × 10 10 years (B) 4.4 × 10 8 years(C) 1.7 × 10 9 years (D) 2.8 × 10 7 years3. For a reaction A → Product, half-life measured fortwo different values of initial concentrations 5 ×10 –3 M and 25 × 10 –4 M are 1.0 and 8.0 hrsrespectively. If initial concentration is adjusted to1.25 × 10 –3 M, the new half-life would be :(A) 16 hrs(B) 32 hrs(C) 64 hrs(D) 256 hrs4. Consider the following standard reductionpotentials :Half reaction(V)Ni 2+ (aq) + 2e – Ni (s) Eº = – 0.23 VFe 2+ (aq) + 2e – Fe (s) Eº = – 0.41 VMn 2+ (aq) + 2e – Mn (s) Eº = – 1.03 VCo 2+ (aq) + 2e – Co (s) Eº = – 0.28 VCr 2+ (aq) + 2e – Cr(s) Eº = – 0.74 VWhich of the following metals could be usedsuccessfully to galvanize steel?(A) Ni only (B) Ni and Co(C) Fe only (D) Mn and Cr5. A beaker contains a small amount of gold dust(Au(s)). Which of the following aqueous solution,when added to the beaker, would dissolve the golddust (ie, convert Au (s) to + Au 3+ (aq))?Half reactionEº (at 25ºC)Zn 2+ + 2e – ⎯→ Zn – 0.76Al 3+ + 3e – ⎯→ Al –1.66O 2 + 2H + + 2e – ⎯→ H 2 O 2 0.702–Cr 2 O 7 + 6e – + H – ⎯→ 2Cr 3+ 1.23Au 3+ + 3e – ⎯→ Au 1.50O 2 + 4H + + 4e – ⎯→ 2Η 2 Ο 1.30(A) Cr 2 O 7 2– (acidic solution)(B) H 2 O 2 (acidic solution)(C) Al 3+(D) Zn 2+1kXtraEdge for IIT-JEE 49 MAY 2011

6. Choose the INCORRECT statement from thefollowing(A) Hexagonal closest packed (HCP) and cubicclosest packed (CCP) structures havecoordination numbers of 12 and 7, respectively(B) HCP and CCP structures involve ABABABand ABCABCABC stacking, respectively(C) HCP and CCP structures involve the same netpacking density(D) Metals can crystallize in either HCP or CCPstructures, depending on the element7. C 60 (buckyball) is cubic closest packed (facecentredcubic) in its crystalline form. If you insertpotassium atoms into all the tetrahedral andoctahedral holes of the C 60 structure, the formulawould become K x C 60 . What is the value of x ?(A) 1 (B) 2 (C) 3 (D) NoneQuestions 8 to 11 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.8. Which one of the following statements is (are) in aconcentration cell made of Fe in 0.010 M Fe 2+ (aq)and Fe in 0.10 M Fe 2+ (aq)?(A)The Eº cell = 0(B) ∆G = 0(C) The cell reaction is Fe 2+ (aq 0.010 M) →Fe 2+ (aq, 0.10 M)(D) At equilibrium, the [Fe 2+ ] anode = [Fe 2+ ] cathode9. The emf of a galvanic cell depends upon(A) surface area of electrodes(B) concentration of electrolyte(C) volume of electrolyte(D) temperature10. To 20 mL of 0.1 M barium chloride solution 0.2mL of 3 M chromium sulphate is added BaSO 4formed is completely insoluble and all the salts arecompletely dissociated (Change in volume isignored). Then,(A) boiling point is raised(B) boiling point is lowered(C) freezing point does not change(D) freezing point is increased11. An atomic solid has hexagonal arrangement of unitcell with height of hexagonal (in close packedarrangement) as "h" The radius of atom in terms ofheight is2(A) 4 h(B)3(C)h232(D)h4h42332This section contains 2 paragraphs, one has 3multiple choice questions and other has 2 multiplechoice questions (Question 12 to 16). Each questionshas 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.Passage : I (Q. No. 12 to 14)The structure of unit cell of perovskite – a salt oflanthanum(La), manganese (Mn) and oxygen, hasMn 2+ at the each corner, oxide on every edgecentre and a lanthanum ion at the body centre.12. What is the empirical formula of the salt?(A) LaMnO 2 (B) LaMn 2 O 3(C) LaMnO 3 (D) Both (B) and (C)13. What are the coordination ion numbers of Mn; Laand O, respectively?(A) 6, 2, 8 (B) 6, 4, 12(C) 8, 4, 8 (D) 6, 12, 214. Considering the closed packed arrangement of ions,which of the following statement regarding theionic radii is most the appropriate?(A) r 4+> r 2La O–(B) r 4+> r 2La Mn+(C) Edgelength of unit cell = r 2++ r 2Mn O–(D) None of the abovePassage : II (Q. No. 15 to 16)Properties such as vapour pressure, boiling point,freezing point of pure solvent changes when solutemolecules are added in order to preparehomogeneous solution. These are called colligativeproperties. Application of colligative properties arevery useful in day to day life. One of its examplesis the use of ethylene glycol and water mixture asanti-freezing agent in the radiator of automobiles. Asolution M is prepared by mixing ethanol andwater. The mole fraction of ethanol in the mixtureis 0.90. Also given are the following information :K f of water = 1.86 K kg mol –1 ; K f of ethanol = 2.00K kg mol –1 ; K b of water = 0.52 K kg mol –1 , k b ofethanol = 1.20 K kg mol –1 ; Standard freezing pointof water = 273 K; Standard freezing point ofethanol = 155.7 KStandard boiling point of water = 373 K; Standardboiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm of Hg;vapour pressure of pure ethanol = 40 mm of Hg. Inanswering the following questions, consider thesolutions to be ideal dilute solutions and solute tobe non-volatile, non-electrolytic.15. The freezing point of solution M is(A) 268.7 K (B) 268.5 K(C) 234.2 K (D) 150.9 KXtraEdge for IIT-JEE 50 MAY 2011

16. The vapour pressure of solution M is(A) 39.3 mm of Hg (B) 36.0 mm of Hg(C) 29.5 mm of Hg (D) 28.8 mm of HgThis section contains 7 questions (Q.17 to 23).+3 marks will be given for each correct answer andno negative marking. The answer to each of thequestions is a SINGLE-DIGIT INTEGER, rangingfrom 0 to 9. The appropriate bubbles below therespective question numbers in the OMR have to bedarkened. For example, if the correct answers toquestion numbers X, Y, Z and W (say) are 6, 0, 9 and2, respectively, then the correct darkening of bubbleswill look like the following :X Y Z W0123456789012345678917. A body-centred cubic lattice is made-up of twotypes of atoms, A and B in which A occupy thebody-centre, while B is the corners. Due to someimperfections in the solid, one corner is leftunoccupied per unit cell and formula of the solid isA x B y . Here x is.18. In HCP arrangement of atoms, coordinationnumber of atoms in the same layer is,19. If an aqueous NaCl solution is electrolysed using acurrent of 5 A for 200 min, Volume of Cl 2 (g) inlitres, produced under STP condition is(F = 96000 C).20. Equivalent conductance of 0.2 aqueous solution ofa weak monobasic acid (HA) is 10 S cm 2 equiv –1and that at infinite dilution is 200 S cm 2 equiv –1 .Hence, pH of this solution is.21. How many Faradays of electricity is required forelectrolysis of 72 g of acidified water?22. A 0.4 molal aqueous solution of Na x A has freezingpoint of –3.27º C. If K f of water is 1.86 K kg mol –1 ,the value of x is : (Assume complete of ionizationof salt Na x A).23. Solubility product of PbCl 2 at 25ºC is 4 × 10 –6 . K fof water is 2K kg mol –1 . When PbCl 2 is dissolvedin water its saturated solution, the freezing point⎛ m ⎞ºdecreases by ⎜ ⎟⎠ . The value of "m" is -⎝10001234567890123456789MATHEMATICSQuestion 1 to 7 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. If tan 2 ⎛ πx⎞⎜ ⎟ f (x) = 1 + cos 2πx+ | f (x)|, then⎝ 9 ⎠f (3) is equal to -11(A)(B) –22 2(C)–12(D)2. If y = x 3 – c and y = x 2 + ax + b touch each other at| a | + | b | + | c |the point (1, 2), thenis equal to -| a + b – c |(A) 1 (B) 0(C) 2(D) None of these3. If the domain of the functionf (x) =1is R then number of2ln ( x + 2ax+ 9 – 3a)integral values of a is -(A) 8 (B) 10 (C) 5 (D) 64. If the vertical distance between the graphs3 52 x + 3 x + 5 xy =& y = k approaches to zero33x– 5 + 2x– 3as x approaches to infinity, then the value of 'k' is -23(A) (B) (C) 1 (D) 03 25. The length of the sub tangent to the curve32x ( x + 1)y = 3at x = 1 is -55 – x81(A) 2020(B) 8121227(C) 2020(D) 276. If (f (x)) (n) represents n th order derivative of thefunction y = f (x) & g(x) = (ln (x)) (5) , then g(x) isalways -(A) increasing and concave up(B) increasing & concave down(C) decreasing and concave up(D) decreasing & concave downXtraEdge for IIT-JEE 51 MAY 2011

7. If the equation 3x 4 – 16x 3 + 30x 2 – 24x + 12 = 12ahas exactly 2 solution then the complete set ofvalue of 'a' is -⎛ 1 ⎞(A) ⎜ ,∞⎟ (B) (3, ∞)⎝ 3 ⎠(C) (– 1, ∞) (D) (1, ∞)Questions 8 to 11 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.| |{ a x sgn x}| |[ a x sgn x]8. If f (x) = a ; g(x) = a for a > 1 andx ∈ R 0 , where {.} & [.] denotes the fractional partand integral part functions respectively, then whichof the following statements holds good for thefunction h(x), where (lna) h(x) = (ln f (x) + lng(x))(A) h is even (B) h is odd(C) h is decreasing (D) h is increasing9. If f (x) = max (sin –1 x, cos –1 x) and⎛ 1g(x) = f ⎟ ⎞ π ⎜ x + – , then identify correct⎝ 2 ⎠ 4statements⎡ 3π⎤(A) range of g(x) is ⎢0, ⎥⎣ 4 ⎦(B) g(x) is not differentiable at x = 0⎛ π 3π⎞(C) g(x) = k have only one solution is k ∈ ⎜ , ⎟⎝ 4 4 ⎠(D) f (cos 5) + f (sin 5) = 7π – 1510. Identify surjective functions -⎛ π ⎞(A) f : ⎜0 , ⎟ → [12, ∞), f (x) = 4 tan 2 x + 9 cot 2 x⎝ 2 ⎠(B) f : [0, 2π] → [0, 50], f (x) = 24cosx – 7sinx +25(C) f : R → R, f (x) = (2x + 5) (3x + 7) (4x + 9)| x |(D) f : R → [0, 1), f (x) =| x | + 111. If f is a function defined from R to R such thatf(f (f (x))) – (f (f (x))) 3 = f (f (x)), then -(A) f (x) has exactly two real roots(B) f (x) is continuous everywhere(C) f (x) is surjective function(D) f (x) = f –1 (x) at x = 0This section contains 2 paragraphs, one has 3multiple choice questions and other has 2 multiplechoice questions (Question 12 to 16). Each questionshas 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.Passage : I (Q. No. 12 to 14)Consider a function f (x) satisfying a functionalrule f (x) + 2f (1 – x) = x 2 + 1, ∀ x ∈ R. Let g(x) beanother function such that g'(x) = 3 f (x) ∀ x ∈ Rand g(0) = 0. x-axis touches the graph of y = g(x) at(a, 0).12. The range of the function f (sin x) is -(A) [0, 8] (B) [1, 8]⎡18⎤⎡ 8⎤(C) ⎢ , ⎥ (D)⎣33⎢0, ⎥ ⎦ ⎣ 3 ⎦13. If the number of solutions of the equation g(|x|) = kbe 2, then the range of values of k, are -⎛ 4 ⎞⎛ 4⎤(A) ⎜0 , ⎟ (B) ⎜0,⎝ 3⎥ ⎠ ⎝ 3 ⎦⎛ 4 ⎞(C) ⎜ ,∞⎟ ⎝ 3 ⎠(D)⎡4⎞⎢ ,∞⎟ ⎣3⎠14. If the area of triangle formed by the points(k, 0), (k, g(k)) & (a, 0) is maximum, wherek ∈ (0, 3) then the value of k is345(A) (B) 1 (C) (D) 4 3 2Passage : II (Q. No. 15 to 16)Let the function f and g be defined by⎧ x⎪f (x) = ⎨ 2 – x⎪ f ( x + 2)⎩for x ∈[0,1)for x ∈[1,2)for all xand g(x) = 4f (3x) + 1 for all x.15. Sum of all the solutions of the equation f (x) = 0.6for 0 ≤ x ≤ 7 is(A) 23.64 (B) 13.44(C) 31.04(D) None of these⎛ –1 ⎛ 5 ⎞16. ⎟ ⎞tan⎜2 tan f ⎜ ⎟ is equal to -⎝ ⎝ 3 ⎠⎠(A) 43(B) 31(C) 32(D) 34This section contains 7 questions (Q.17 to 23).+4 marks will be given for each correct answer andno negative marking. The answer to each of thequestions is a SINGLE-DIGIT INTEGER, rangingfrom 0 to 9. The appropriate bubbles below therespective question numbers in the OMR have to bedarkened. For example, if the correct answers toquestion numbers X, Y, Z and W (say) are 6, 0, 9 and2, respectively, then the correct darkening of bubbleswill look like the following :XtraEdge for IIT-JEE 52 MAY 2011

X Y Z W01234567890123456789⎧2[x] –{ x}, If x < 017. Let f (x) = ⎨, where {x} and⎩[x]+ 3{ x}, If x ≥ 0[x] are the fractional part and greatest integer of xrespectively. The number of solutions of theequation f (x) – x = {x} ∀ x ∈ [– 8, 8] are.18. If f (x) = lim h ⎛ sin( x + 1/ h)⎟ ⎞πl n⎜; 0 < x < ,h→∞⎝ sin x ⎠2then number of points where f (x) is discontinuousis.19. If set of value of 'p' for which the equation–1 2xsin + px = 0 has exactly three solutions is21+x(l, m) then 8m – l is.20. If the equation 2|x – 2| – |x + 1| + x = k, has infinitesolutions then the value of k is.⎡⎤⎢ n⎥21. The value of ⎢2 1lim ⎥⎢∑, where [.]n→∞n ⎛ 2r⎞⎥r=1⎢ 1+cos⎜⎟⎥⎣⎝ n ⎠⎦represents greatest integer function, is.0123456789012345678922. The greatest value of the functionlog(4 +10 )23. From a point on the curve(cos 2 θ – 6 sinθ cosθ + 3 sin 2 θ + 2) is.y = sin –1 x + cos –1 – x a tangent is drawn to thecurve g(x) = x + cos x + 2π – 1. Then sum of itsintercepts on coordinate axes is.WHAT ARE EARTHQUAKES?Earthquakes like hurricanes are not only superdestructive forces but continue to remain a mysteryin terms of how to predict and anticipate them. Tounderstand the level of destruction associated withearthquakes you really need to look at someexamples of the past.If we go back to the 27th July 1976 in Tangshan,China, a huge earthquake racked up an officialdeath toll of 255,000 people. In addition to this anestimated 690,000 were also injured, wholefamilies, industries and areas were wiped out in theblink of a second. The scale of destruction is hard toimagine but earthquakes of all scales continue tohappen all the time.So what exactly are they ? Well the earths outerlayer is made up of a thin crust divided into anumber of plates. The edges of these plates arereferred to as boundaries and it’s at theseboundaries that the plates collide, slide and rubagainst each other. Over time when the pressure atthe plate edges gets too much, something has togive which results in the sudden and often violenttremblings we know as earthquakes.The strength of an earthquake is measured using amachine called a seismograph. It records thetrembling of the ground and scientists are able tomeasure the exact power of the quake via a scaleknown as the richter scale. The numbers range from1-10 with 1 being a minor earthquake (happenmultiple times per day and in most case we don’teven feel them) and 7-10 being the stronger quakes(happen around once every 10-20 years). There’s alot to learn about earthquakes so hopefully we’llrelease some more cool facts in the coming months.XtraEdge for IIT-JEE 53 MAY 2011

Based on New PatternIIT-JEE 2013XtraEdge Test Series # 1Time : 3 HoursSyllabus : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion,Circular motion. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table. Mathematics:Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii ofCircleInstructions : [Each subject contain]Section – I : Question 1 to 7 are multiple choice questions with only one correct answer. +3 marks will be awardedfor correct answer and -1 mark for wrong answer.Section – II : Question 8 to 11 are multiple choice questions with multiple correct answer. +4 marks will be awardedfor correct answer and No Negative marks for wrong answer.Section – III : Question 12 to 16 are passage based single correct type questions. +3 marks will be awarded for correctanswer and -1 mark for wrong answer.Section – IV : Question 17 to 23 are Numerical Response Question (single digit Ans. type) +4 marks will be awardedfor correct answer and No Negative marks for wrong answer.PHYSICS4. Correct graph of 3x + 4y + 1 = 0 is -YYQuestion 1 to 7 are multiple choice questions. Each11/4questions has four choices (A), (B), (C) and (D), out of(A)X (B)Xwhich ONLY ONE is correct.d1. tan xYYdx(A) 2 sec 2 x (tan x) –1/2 1(B) sec 2 x (tan x) –1/2(C)X (D)X2 –1/4–1/41(C) (tan x)–1/2(D) 2 (tan x ) –1/225. Correct graph of y – 1 = x 2 is -2YYd (x + 1)2.dx x + 1(A)X (B)X22x + 2x –1x – 2x + 1(A)(B)22(x + 1)(x + 1)YYx 2 2+ 2x –1x + 2x + 1(C)(D)2(C)x + 1X (D)X(x + 1)3. If | A → → → →→ →× B| = 3 (A.B), then the value of (A + B)is -6. If → r = bt 2 ^i + ct 3 j ^, where b and c are positive1/ 2constant, the time at which velocity vector makes(A) (A 2 + B 2 + AB) 1/2 ⎛ 2 2 AB ⎞(B)⎜A+ B +3⎟an angle θ = 60º with positive y-axis is -⎝⎠c2b(A) (B)b(C) A + B (D) )A 2 + B 2 + 3 AB) 1/2 3 3c2cb(C)(D)3b23 cXtraEdge for IIT-JEE 54 MAY 2011

7. A man throws a stone in vertical upward directionfrom ground and reach maximum height H. Duringascent, he travels a distance of H/5 in the lastsecond. Then H is equal to : (Take g = 10 m/s 2 )(A) 25 m(B) 20 m(C) 10 m(D) data is insufficientQuestions 8 to 11 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.8. Which of the following graph(s) is/are not possible?(A)(C)DistanceODistanceTimeO t 1Time(B)(D)DistanceODistance9. If the velocity of a body is constant -(A) |Velocity | = speed(B) |Average velocity| = speed(C) Velocity = average velocity(D) Speed = average speedOTimeTime10. The two vectors → A and → B are drawn from acommon point and → C = A → + → B , then anglebetween A → →and B is -(A) 90º if C 2 = A 2 + B 2(B) greater than 90º if C 2 < A 2 + B 2(C) greater than 90º if C 2 > A 2 + B 2(D) less than 90º if C 2 > A 2 + B 211. Which of the following quantities are independentof the choice of orientation of the co-ordinate axes?(A) → a + → b(C) | → a + → b – → c |(B) a x + b y(D) Angle between → a and → bThis section contains 2 paragraphs, one has 3multiple choice questions and other has 2 multiplechoice questions (Question 12 to 16). Each questionshas 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.Passage : I (No. 12 to 14)An object is rotating on a horizontal circular trackwith constant speed 'u 0 ' in clock-wise direction.Initially object was moving in east direction.12. Change in velocity as object rotates by an angle 90ºis -(A) zero (B) 2u 0(C) 2 u 0 (D) 3 u 013. Direction of change in velocity in above questionis-(A) south-east (B) south-west(C) south(D) west14. Charge in velocity as objects rotates by an angle180º is -(A) zero (B) 2u 0 towards west(C) 2 u 0 towards west (D) 2 u 0 towards southPassage : II (No. 15 to 16)A particle is projected from ground with a velocityu and an angle 53º with the horizontal. Duringrising after t= 1 s angle made by the velocity vectorwith the horizontal is 37º. (Take g = 10 m/s 2 )15. Value of u is100 200(A) m/s (B) m/s77300(C) m/s (D) date is insufficient716. Speed of particle at the moment when angle is37º -150(A) 100 m/s (B) m/s7100(C) m/s (D) date is insufficient7This section contains 7 questions (Q.17 to 23).+4 marks will be given for each correct answer andno negative marking. The answer to each of thequestions is a SINGLE-DIGIT INTEGER, rangingfrom 0 to 9. The appropriate bubbles below therespective question numbers in the OMR have to bedarkened. For example, if the correct answers toquestion numbers X, Y, Z and W (say) are 6, 0, 9 and2, respectively, then the correct darkening of bubbleswill look like the following :X Y Z W0123456789012345678901234567890123456789XtraEdge for IIT-JEE 55 MAY 2011

17. A stone is projected from level ground. At a heightof 0.4 m above ground its velocity is found to be→v = ( 6^i + 2^j ) m/s. x-axis is along horizontaland y-axis vertically upwards). The angle ofprojection with the vertical is found to be (10 m)degree. Find the m.18. A particle starts from the origin goes along thex-axis to the point (30 m, 0 m) and then returnsalong the same line to the point (–30 m, 0 m). Thedistance travelled by the particle is (10 p) metre.Find the value of 'p'.19. A body initially at rest moving along x-axis in sucha way so that its acceleration Vs displacement plotis as shown in figure. what will be the maximumvelocity of particle in m/sec.1 m/s 2 a0.51m20. If a cone of radius R and height R is taken out of asolid hemi sphere of radius R as shown in figure.Find the volume of remaining shaded part.15(given that R = 3 )π21. If y = 4x 2 – 4x + 7. Find the minimum value of y.22. The area of a rectangle of size 1.25 cm × 1.55 cm is1.9 y, where y is single digit numbers. Find y.23. A 2m wide truck is moving with a speed of 5 5m/s along a straight horizontal road. A man startscrossing the road with a uniform speed v when thetruck is 4m away from him The minimum value ofv (in m/s) to cross the truck safely is -RSCHEMISTRYQuestion 1 to 7 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. Which of the following have identical geometrywith same number of lone pair (s)?(I) PCl 5 (II) PF 3 Br 2(III) BrF 52–(IV) SbF 5(A) I and II (B) II and III(C) III and IV (D) III, II and IV2. Polyethylene can be produced from calcium carbideaccording to the following sequence of reactionsCaC 2 + H 2 O → CaO + HC ≡ CHnHC ≡ CH + nH 2 → —CH 2–CH 2 —nThe mass of polyethylene which can be producedfrom 40.0 kg of CaC 2 is -(A) 6.75 kg (B) 17.5 kg(C) 8.75 kg (D) 9.75 kg3. sp 3 hybridisation is in -––(A) AlH 4 (B) CH 3(C) Be 4 (CH 3 COO) 6 O (D) All of these4. What will be the maximum spin multiplicity of 3dorbital?(A) 4 (B) 6(C) 5 (D) 105. Photons having energy equivalent to bindingenergy of 2 nd state of Li + ion are used at metalsurface of work function 10.6 eV. If the ejectedelectrons are further accelerated through thepotential difference of 5 V then the minimum valueof de-Broglie wavelength associated with theelectron is -(A) 2.45 Å(B) 9.15 Å(C) 5 Å(D) 11 Å6. An impure sample of Ba(OH) 2 (mol. wt. = 171) ofmass 1 gram allowed to react with 80 ml of 0.20 MHCl (aq). When excess acid was titrated withNaOH, 20 ml of NaOH (aq) was required. 10 ml ofthe same NaOH (aq) required 30 ml of the 0.1 MHCl (aq.) in a separate titration percentage purity ofBa(OH) 2 sample is -(A) 4.275 (B) 42.75(C) 85.5(D) None of theseXtraEdge for IIT-JEE 56 MAY 2011

7. What may be the correct orbital notation if thewave function is –Ψ =1 ⎛⎜81 6π⎝r/a 0 and a 0 =(A) 4s(C) 3P yh1 ⎞⎟⎠a 02∈πme3/ 202σ 2 e –σ/3 (3cos 2 θ – 1); Here σ =(B)2 P x2(D) 3d zQuestions 8 to 11 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.8. Which of the following statements is/are true forP 4 S 3 molecule -(A) It contains six P–S bonds and three P-P-bonds(B) It contains six P-S bonds and ten lone pairs(C) It has all atoms sp 3 -hybridised(D) It contains six P-P bonds and ten lone pairs9. Which of the following species are correctlymatched with their geometries according to theVSEPR theory -(A) ClF 2 – → linear(B) IF 4 + → see – saw(C) SnCl 5 – → trigonal bipyramidal••(D) N(SiH3)3 → pyramidal10. Sodium borohydride (NaBH 4 ) reacts with iodine toform boron triodide, Sodium iodide and HI. In anexperiment, 76 gram sodium borohydride is mixedwith 300 gram of Iodine. In this reaction -(A) NaBH 4 is the excess reagent(B) Iodine is the excess reagent(C) If 39.2 gram of BI 3 are formed, its percentageyield is 40(D) The weight of the reagent remaining in excessis 64.78 gram11. KCl has a dipole moment of 10 D. The inter ionicdistance in KCl is 2.6 Å. Which of the followingstatement are true for this compound ?(A) The theoretical value of dipole moment, if thecompound were completely ionic is 12.5 D.(B) The % ionic character of the compound is 85 %(C) It is a poor conductor of electricity(D) The forces operating in this molecule arecoulombic typeThis section contains 2 paragraphs, one has 3multiple choice questions and other has 2 multiplechoice questions (Question 12 to 16). Each questionshas 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.Passage : I (Q. No. 12 to 14)In order to explain the existance of doublets in thespectra of alkali metals, Goudsmit and Uhlenbeckin 1925 proposed that electron has an intrinsicangular momentum due to spining about its ownaxis.The value of spining a angular momentum ofelectron can be described by 2 spin quantumnumber s and m s . The physical significance of sand m s is similar as of l and m l .12. The possible value of s for electron is -(A) 1/2 (B) – 1/2(C) 0 (D) 113. Relation between s and m s is :(A) s (s + 1)h2π. cos θ = ms(B) s (s + 1)cos θ = m s3h(C) = m s4π(D) None of these14. Spin angular momentum of electron has magnitudeequal to :(A)3h4π3h(B) 2πnh(C) (Here n is any positive integer)2π1(D) 2Passage : II (Q. No. 15 to 16)Ozone in the upper atomsphere absorbs ultravioletradiation which induce the following chemicalreaction in –O 3 (g) ⎯→ O 2 (g) + O(g)O 2 produced in the above photochemicaldissociation undergoes further dissociation into onenormal oxygen atom and one more energeticoxygen atom O*.O 2 (g) ⎯→ O + O*If O* has 3eV more energy than O and normaldissociation energy of O 2 is 480 KJ/mol. (Given1 eV/photon = 96 KJ/mol)15. What is maximum wavelength effective for thephotochemical dissociation of O 2 molecule -(A) 2440 Å (B) 1547 Å(C) 1000 Å (D) 155 ÅXtraEdge for IIT-JEE 57 MAY 2011

16. Half a mole of photon is used to break the O 3molecule completely according to I st reaction. O 2produced in this reaction dissociates according to2 nd reaction, then what is the total energy requiredin KJ to carry out the dissociation of O 2 .(A) 384 KJ(B) 280 KJ(C) 480 KJ(D) 300 KJThis section contains 7 questions (Q.17 to 23).+4 marks will be given for each correct answer andno negative marking. The answer to each of thequestions is a SINGLE-DIGIT INTEGER, rangingfrom 0 to 9. The appropriate bubbles below therespective question numbers in the OMR have to bedarkened. For example, if the correct answers toquestion numbers X, Y, Z and W (say) are 6, 0, 9 and2, respectively, then the correct darkening of bubbleswill look like the following :X Y Z W012345678901234567890123456789012345678921. Find number of 120º bond angles in O = CF 2 .22. Give number of H-atoms in H 2 C = SF4 which arein plane of axial F-atoms.23. For an element Z eff for its outermost electron is3.55. If atomic radius of element is 0.75 Å, thenfind out its electro negativity.MATHEMATICSQuestion 1 to 7 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1– sin 2θ + cos 2θ1. If f (θ) =, then value of2cos 2θf (11º) f (34º) equals -13(A) (B) 2 4(C) 41(D) None of these2. If 3 ≤ a < 4 then value ofsin –1 (sin [a]) + tan –1 (tan [a]) + sec –1 (sec[a])(where [.] denotes greatest integer function) is -(A) 2π (B) 2π – 3(C) 3 (D) 2π + 317. A compound exists in the gaseous phase both asmonomer (A) and dimer (A 2 ). The molecularweight of A is 60. In an experiment 240 g of thecompound was confined in a vessel of volume32.84 litre and heated to 127º C. Calculate thepressure (in atm) developed if the compound existsas dimer to the extent of 50% by weight underthese conditions.18. Li 2+ ion in its ground state absorbs a photon ofenergy 183 electron volt. Electron of Li 2+ ionstrikes the He + ion. (in Å) after being struck byelectron of Li 2+ ion.3. If sum of all solutions of equation3cot 2 kπθ + 10 cotθ + 3 = 0 in [0, 2π] is 2k ∈ I then k equals -(A) 3 (B) 6(C) 10 (D) 154. Range of k for whichk cos 2 x – k cos x + 1 ≥ 0 ∀ x ∈ R is -1(A) k < – (B) k > 42where19. First ionization energy of Li is 13.6 eV. If it isassumed that outermost electron of Li revolvesunder the influence of nucleus which is shielded byinner two electrons in first orbit of Li, then find outby what amount of charge inner two electronsshield the nucleus?20. A photon of 50 eV energy strikes a metal surface(having work function 1.64 eV). Photoelectronejected from metal with maximum kinetic energystrikes He + ion (in ground state).Find out de-Broglie wavelength of electron of He +ion finally in Å.(C) – 21 ≤ k ≤ 4(D) None of these5. If sin (sin x + cos x) = cos (cos x – sin x ) thenlargest possible value of sin x is -1(A)(B) 12(C)16 – π42(D) 4πXtraEdge for IIT-JEE 58 MAY 2011

6. If ∆ is area of ∆ABC and length of two sides are 3& 5 respectively, if third side is c, then22c + 16c + 64 c + 16c + 54(A) ∆ ≤(B) ∆ =12 38(C) ∆ >c2+ 16c + 7443(D) None of these7. A point Q is selected at random inside theequilateral triangle. If sum of lengths ofperpendicular dropped on sides from Q is P. Thenaltitude of triangle is -PP(A) (B) 2 3(C) P(D) None of theseQuestions 8 to 11 are multiple choice questions. Eachquestions has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.8. If cotθ + tanθ = x and secθ – cosθ = y, then -(A) x sinθ . cosθ = 1(B) sin 2 θ = y cosθ(C) (x 2 y) 1/3 + (xy 2 ) 1/3 = 1(D) (x 2 y) 2/3 – (xy 2 ) 2/3 = 1θ9. If 0 ≤ θ ≤ π and sin = 1 + sin θ – 1 – sin θ2then possible values of tanθ is/are -43(A) (B) – 3 4(C) – 34(D) 0⎛ 2 1 ⎞10. If ⎜cos x + ⎟ (1 + tan 2 2y) (3 + sin 3z) = 42⎝ cos x ⎠then -(A) x is a multiple of π(B) x is a multiple of 2π(C) y is a multiple of π/2(D) None of these11. In a ∆ABC, if r = 1, R = 3, s = 5 which of thefollowing is/are correct (where a, b, c representsides of triangle) -(A) ar ∆ABC = 5(B) Product of sides of ∆ABC is 60(C) a 2 + b 2 + c 2 = 24(D) Sum of exradii of ∆ABC is 13This section contains 2 paragraphs, one has 3multiple choice questions and other has 2 multiplechoice questions (Question 12 to 16). Each questionshas 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.Passage : I (Q. No. 12 to 14)Let α ± β is not an odd multiple of π.α + βIf cos α + cos β = b, sin α + sin β = a, θ =2na(a – b)and sin 2θ + cos 2θ = 1 + where n ∈ I2 2a + bthen12. Value of n is -(A) 0 (B) 1(C) – 2(D) None of these13. If cosec n x = A then sin 3A. sin A is polynomial inx, whose degree is equal to -(A) 5 (B) 4 (C) 2 (D) 314. If degree of polynomial obtained in above questionis p then max. value of (p + 1) sin x + (p + 2) cos xis -(A) p + 2 (B) p + 3(C) p + 1(D) None of thesePassage : II (Q. No. 15 to 16)Let sides of a triangle are a = n + 1, b = n + 2 &c = n with sin C = 54 , then answer the following.15. Area of ∆ABC is -(A) 84 (B) 72(C) 60(D) None o these16. Largest exradius of circle escribing ∆ABC -(A) 12 (B) 14(C) 16(D) None of theseThis section contains 7 questions (Q.17 to 25).+4 marks will be given for each correct answer andno negative marking. The answer to each of thequestions is a SINGLE-DIGIT INTEGER, rangingfrom 0 to 9. The appropriate bubbles below therespective question numbers in the OMR have to bedarkened. For example, if the correct answers toquestion numbers X, Y, Z and W (say) are 6, 0, 9 and2, respectively, then the correct darkening of bubbleswill look like the following :X Y Z W0123456789012345678901234567890123456789XtraEdge for IIT-JEE 59 MAY 2011

17. Ifsin 3θcos 2θ2–147137= 0Then no. of values of θ in [0, 2π].18. Minimum value ofy = (sin x + cosec x) 2 + (cos x + sec x) 2 ∀ x ∈ R is.19. If sin –1 (sin 5) > x 2 – 4x then number of possibleintegral values of x will be.20. Find no. of solution of x for e x cot x = 1 wherex ∈ (0, 2π).21. In ∆ABC if angles A, B, C are in A. P. & ∠Aexceeds lowest angle by 30º & D divides BC∠BAD1internally in 1 : 3 then = . Find k∠CADk22. The number of solution ofcos(2sin –1 (cot(tan –1 (sec(6cosec –1 x))))) + 1 = 0where x > 0 is23. Circumcentre is at origin & a ≤ sin A. P(x, y) lie1inside the circumcircle & k = , then least8 | xy |integer value of k can be.PREFIXESPrefix Symbol Powers of 10deciSub multipliesd 10 –1centi c 10 –2milli m 10 –3micro µ 10 –6nano n 10 –9pico p 10 –12femto f 10 –15atto a 10 –18zepto z 10 –21yocto y 10 –24Multiplesdeca da 10 1hecto h 10 2kilo k 10 3mega (or million) M 10 6giga (or billion) G 10 9tera (or trillion) T 10 12peta P 10 15exa E 10 18zetta Z 10 21Yotta Y 10 24Facts on the SunThe Milky Way galaxy is home to Planet Earth andis energized by the enormous ball of energy, theSun. This big star is much closer to us than themillions of stars that twinkle in the night sky and itsupports life with the heat generated.The quest to know the space or void that surroundsthe planet and the celestial bodies that intrigue manby their shimmer and movements have led scientiststo delve deep into the study of the universe, with thesupport of harnessed technology. Research revealsthat the Sun is average in size and features,compared to the million others that just appearsmall in the night sky due to their proximity toearth. The Sun is the center of our galaxy and ismade up of hot gases that comprise elements likecalcium, hydrogen, helium, sodium, iron andmagnesium.The Sun is a big star that simply looks small in thesky because it is approximately 93 million miles or150 million km away. While the Earth is about 13thousand kilometers across, the sun is 1.4 millionkilometers and the difference in size between theEarth and the Sun is better understood with thecalculation that it would take more than a millionEarths to fill the Sun, if possible! Considering that abeam of sunlight takes approximately 8 minutes toreach us, scientists have calculated that it wouldtake man about a hundred and seventy six years toreach the sun; this is in conjunction to the speeds weallow ourselves on the planet.Although actually weighing the Sun with a scale isimpossible, scientists compute its weight with theunderstanding that it contains the same materialmass that is weighable in our solar system. Certaincomparisons and the updated study that the Sun isover 300,000 times heavier than Earth, defines anapproximate gravitational pull and the weightincrease factor. The Sun has no solid surface! TheSun is estimated to be more than four billion yearsold. In comparison to the time frame within whichearliest evidence of life forms on the planet havebeen recorded, the Big Bang seemed to have takenplace much, much earlier. And, the Sun is notexpected to change in form or size at least another 5billion years.Once the hydrogen exhausts, or burns out, scientistsexpect this Center of the Galaxy to enter a newphase of existence, burning out the heliumcomponent next. The burning of helium will resultin the expansion of the star to nearly 100 times itscurrent size! The prediction and study reveals thatthe resultant Red Giant will then collapse into asmaller White Dwarf!XtraEdge for IIT-JEE 60 MAY 2011

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IIT-JEE 2011PAPER-I (PAPER & SOLUTION)Time : 3 Hours Total Marks : 240Instructions : [Each subject contain]Section – I :Section – II :Multiple choice questions with only one correct answer. +3 marks will be awarded for correct answerand -1 mark for wrong answer. [No. of Ques. : 7]Multiple choice questions with multiple correct answer. +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 4]Section – III : Passage based single correct type questions. +3 marks will be awarded for correct answer and -1 markfor wrong answer. [2 passage, No. of Ques. : 5]Section – IV : Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 7]CHEMISTRYSECTION – ISingle Correct Choice TypeThis section contains 7 multiple choice questions. Eachquestion has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. Geometrical shapes of the complexes formed bythe reaction of Ni 2+ with Cl – , CN – and H 2 O,respectively, are(A) octahedral, tetrahedral and square planar(B) tetrahedral, square planar and octahedral(C) square planar, tetrahedral and octahedral(D) octahedral, square planar and octahedralAns. [B]Sol. Complexes are : [NiCl 4 ] –2 , [Ni(CN) 4 ] –2 &[Ni(H 2 O) 6 ] +2Ni +2 = 3d 8 4s 0[NiCl 4 ] –2 : Now Since, Cl – is a weak legand so nopairing of electron take place and geometry istetrahedral[Ni(CN) 4 ] –2 : Since, CN – is a strong legand sopairing of electron will take place & geometry issquare planar.[Ni(H 2 O) 6 ] +2 : It will formed octahedral complexsince C.N. = 62. AgNO 3 (aq.) was added to an aqueous KClsolution gradually and the conductivity of thesolution was measured. The plot of conductance( Λ ) versus the volume of AgNO 3 is -Ans.Sol.ΛΛvolume(P)volume(R)ΛΛvolume(Q)volume(S)(A) (P) (B) (Q) (C) (R) (D) (S)[D]Because in the beginning of the reaction no ofions remain constant so conductivity remainsconstant but after complete precipitation of Cl –the no. of ions increases in solution. Soconductivity increases.3. Bombardment of aluminum by α-particle leads toits artificial disintegration in two ways, (i) and (ii)as shown. Products X, Y and Z respectively are -Ans.(i)273013 Al15 P + Y3014 Si +X(ii)3014 Si +(A) proton, neutron, positron(B) neutron, positron, proton(C) proton, positron, neutron(D) positron, proton, neutron[A]ZXtraEdge for IIT-JEE 62 MAY 2011

Sol.27134He30152Al ⎯⎯→ P +42 He3014 Si + H1 301 Si0 114 + + e4. Extra pure N 2 can be obtained by heating -(A) NH 3 with CuO (B) NH 4 NO 3(C) (NH 4 ) 2 Cr 2 O 7 (D) Ba(N 3 ) 2Ans. [D]Sol. Ba(N 3 ) 2 ⎯⎯→∆ Ba + 3N 2 ↑5. Among the following compounds, the most acidicis -(A) p-nitrophenol(B) p-hydroxybenzoic acid(C) o-hydroxybenzoic acid(D) p-toluic acidAns. [C]Sol. o-hydroxy benzoic acid is stronger acid due toortho effect6. The major product of the following reaction is -OC(i) KOHNHC(ii) Br CH 2 ClOO(A)(B)(C)(D)CN–CH 2COOCNCOOCNO–CH 2OCNO10nCH 2 ClBrCH 2 ClBrSol.OCNHCOOC Θ ⊕NKCOCl–CH 2–KOH– BrOCN – CH 2 –C7. Dissolving 120 g of urea (mol. wt. 60) in 1000 gof water gave a solution of density 1.15 g/mL.The molarity of the solution is -(A) 1.78 M(B) 2.00 M(C) 2.05 M(D) 2.22 MAns.[C]Sol. M =x × d × 10 10 .7 × 1.15×10== 2.05 Mmol wt 60x = percentage by weightO120x = × 100120 + 1000SECTION – IIMultiple Correct Choice TypeThis section contains 4 multiple choice questions. Eachquestions has 4 choices (A), (B), (C) and (D), out ofwhich ONE OR MORE is/are correct.8. Extraction of metal from the ore cassiteriteinvolves -(A) carbon reduction of an oxide ore(B) self-reduction of a sulphide ore(C) removal of copper impurity(D) removal of iron impurityAns. [A,C,D]Sol. Cassiterite is SnO 2 .To reduce SnO 2 into Sn, carbon reduction processis used.Sn has iron impurity.SnO 2 + C → Sn + CO 2–BrAns.[A]XtraEdge for IIT-JEE 63 MAY 2011

9. Amongst the given options, the compound(s) inwhich all the atoms are in one plane in all thepossible conformations (if any), is (are)H HH(A) C–C(B) H–C≡C–CH 2 C CH 2CH 2(C) H 2 C=C=O (D) H 2 C=C=CH 2Ans. [B,C]Sol. Factual.10. The correct statement(s) pertaining to theadsorption of a gas on a solid surface is (are)(A) Adsorption is always exothermic(B) Physiosorption may transform intochemisorption at high temperature(C) Physiosorption increases with increasingtemperature but chemisorption decreaseswith increasing temperature(D) Chemisorption is more exothermic thanphysiosorption, however it is very slow dueto higher energy of activationAns. [A, B, D]Sol. Factual.11. According to kinetic theory of gases -(A) collisions are always elastic(B) heavier molecules transfer more momentumto the wall of the container(C) only a small number of molecules have veryhigh velocity(D) between collisions, the molecules move instraight lines with constant velocitiesAns. [A, D]Sol. Factual.SECTION – IIIParagraph TypeThis section contains 2 paragraphs. Based upon thefirst paragraph 3 multiple choice questions and basedupon the second paragraph 2 multiple choice questionshave to be answered. Each of these questions has fourchoices (A), (B), (C) and (D) out of which ONLY ONEis correct.Paragraph for Questions No. 12 to 14When a metal rod M is dipped into an aqueouscolourless concentrated solution of compound N,the solution turns light blue. Addition of aqueousNaCl to the blue solution gives a white precipitateO. Addition of aq. NH 3 dissolves O and gives anintense blue solution.12. The metal rod M is -(A) Fe(B) Cu(C) Ni(D) CoAns.Sol.[B]Metal rod M is Cu13. The compounds N is -(A) AgNO 3 (B) Zn(NO 3 ) 2(C) Al(NO 3 ) 3 (D) Pb(NO 3 ) 2Ans. [A]Sol. Cu + AgNO 3 (conc.) —→ Cu(NO 3 ) 2 + Aglight blue14. The final solution contains -(A) [Pb(NH 3 ) 4 ] 2+ and [CoCl 4 ] 2–(B) [Al(NH 3 ) 4 ] 3+ and [Cu(NH 3 ) 4 ] 2+(C) [Ag(NH 3 ) 2 ] + and [Cu(NH 3 ) 4 ] 2+(D) [Ag(NH 3 ) 2 ] + and [Ni(NH 3 ) 6 ] 2+Ans. [C]Sol. AgCl + NH 2 (aq) ⎯→ [Ag(NH 3 ) 2 ] +Cu +2 + NH 3 (aq) ⎯→ [Cu(NH 3 ) 4 ] +2Intense blueParagraph for Question No. 15 to 16An acyclic hydrocarbon P, having molecularformula C 6 H 10 , gave acetone as the only organicproduct through the following sequence ofreactions, in which Q is an intermediate organiccompound.(i) dil. H 2 SO 4 /HgSO 4P(ii) NaBH 4 /ethanol(C 6 H 10 ) (iii) dil. acidQ(i) conc. H 2 SO 4(catalytic amount)(–H 2 O)(ii) O 3(iii) Zn/H 2 O15. The structure of compound P is -(A) CH 3 CH 2 CH 2 CH 2 –C≡C–HAns.(B) H 3 CH 2 C–C≡C–CH 2 CH 3H 3 C(C) H–C–C≡C–CH 3H 3 CH 3 C(D) H 3 C–C–C ≡C–HH 3 C[D]O2 CH 3 C CH 3XtraEdge for IIT-JEE 64 MAY 2011

Sol.CH 3CH 3 –C–C≡C–HCH 3(P)(i) dil. H 2 SO 4 /HgSO 4(ii) NaBH 4(iii) dil. HClH 3 C–H 2 OCH 3CH 3 –C=C(i) conc. H 2 SO 4CH 3(ii) O 3 (iii) Zn/H 2 OCH 32 C=OCH 3H 3 C OHCH 3 –C–CH–CH 3CH 3(Q)Sol.O O⊕ Θ# ∗ ∗ Θ ⊕#NaO–S–S–S–S–ONaO OO.N. S * = 0O.N. S # = + 5∴ Difference = 519. The maximum number of electrons that can haveprincipal quantum number, n = 3 and spinquantum number, m s = – 1/2, is.Ans. [9]Sol. For n = 3, max e – = 2n 2 = 18Half of them can have m s = – 1/216. The structure of compound Q is -Ans.Sol.H 3 COH(A) H–C–C–CH 2 CH 3H 3 C HH 3 COH(B) H 3 C–C–C–CH 3H 3 C HH 3 C OH(C) H–C–CH 2 CHCH 3H 3 COH(D) CH 3 CH 2 CH 2 CHCH 2 CH 3[B]Factual.SECTION – IVNumerical Response TypeThis section contains. 7 questions. The answer to eachquestions is a single digit integer ranging from 0 to 9.The bubble corresponding to the correct answer is tobe darkened in the ORS.17. Reaction of Br 2 with Na 2 CO 3 in aq. solution givessodium bromide and sodium bromate withevolution of CO 2 gas. The number of sodiumbromide molecules involved in the balancedchemical equation is.Ans. [5]Sol. 3 Na 2 CO 3 + 3Br 2 → 5NaBr + NaBrO 3 + 3CO 218. The difference in the oxidation numbers of thetwo types of sulphur atoms in Na 2 S 4 O 6 is.Ans. [5]20. A decapeptide (mol. wt. 796) on completehydrolysis gives glycine (mol. wt. 75), alanineand phenylalanine. Glycine contributes 47.0% tothe total weight of the hydrolysed products. Thenumber of glycine units present in thedecapeptide is.Ans. [6]Sol.9 molecule⎯⎯⎯waterDecapeptide ⎯ → (x) glycine + (y)alanine + (z) phenylalanineMass of hydrolysed product = 796 + 18 × 947mass of glycine = 958 × = 450.26100450.26No. of glycine unit = = 67521. To an evacuated vessel with movable pistonunder external pressure of 1 atm., 0.1 mol of Heand 1.0 mol of an unknown compound (vapourpressure 0.68 atm, at 0ºC) are introduced.Considering the ideal gas behaviour, the totalvolume (in litre) of the gases at 0ºC is close to.Ans. [7]Sol.Pext = 1 atmHe +compoundsVapour pressure of compound = 0.68∴ P He = 1 – 0.68 = 0.32 ∴ By PV =nRT, for HenHe RT 0.1×0.0821×273V = =PHe0.32V ~ 7LXtraEdge for IIT-JEE 65 MAY 2011

22. The total number of alkenes possible bydehydrobromination of 3-bromo-3cyclopentylhexane using alcoholic KOH is.Ans. [5]−1⎛ 5.7 ⎞ 3= T 1 ⎜ ⎟⎝ 0.7 ⎠1no. of mole = 45= T 1 (8) 2/3 = 4T 1Sol.C–C–C–C–C–C1 2 3 4 5 6Br+ G.IC–C–C–C= C– C+ G.IC–C–C=C– C– Calc.KOH+C–C–C–C– C– CnR(T1− T2)W ==γ −11R(4T T )4× 1 − 1 9= RT12/3 825. A ball of mass (m) 0.5 kg is attached to the end ofa string having length (L) 0.5 m. The ball isrotated on a horizontal circular path about verticalaxis. The maximum tension that the string canbear is 324 N. The maximum possible value ofangular velocity of ball (in radian/s) is-23. The work function (φ) of some metals is listedbelow. The number of metals which will showphotoelectric effect when light of 300 nmwavelength falls on the metal is.Metal Li Na K Mg Cu Ag Fe Pt Wφ (eV) 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75Ans. [4]Sol.−348hc 6.62×10 × 3×10E falling = =−9−19λ 300×10 × 1.6×10= 4.137 eVThe metals having less work function will showphotoelectric effectHence Li, Na, K, MgPHYSICSSECTION – ISingle Correct Choice TypeThis section contains 7 multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct.24. 5.6 liter of helium gas at STP is adiabaticallycompressed to 0.7 liter. Taking the initialtemperature to be T 1 , the work done in the processis-93(A) RT1(B) RT182159(C) RT1(D) RT182Ans. [A]Sol. T 1 V γ–1 γ–11 = T 2 V 21⎛ V1⎞γ−T 2 = T 1 ⎜V⎟⎝ 2 ⎠Ans.Sol.m(A) 9 (B) 18(C) 27 (D) 36[D]θmgT cos θ = mgLTLθT sin θ = mω 2 L sin θT = mω 2 Lω 2 max =ω max =T maxmLT max=mL= 36 rad/s3240.5×0.5= 324× 4r26. Consider an electric field E = E0xˆ, where E 0 is aconstant. The flux through the shaded area (asshown in the figure) due to this field is-(a,0,a)xz(a,a,a)(0,0,0) (0,a,0)yXtraEdge for IIT-JEE 66 MAY 2011

Ans.Sol.(A) 2E 0 a 2(C) E 0 a 2[C]z2(B) 2 E 0 aE a20(D)2⎛ V + V ⎞ν′ =⎜0⎟ ν⎝ V − Vs′ ⎠⎛ 320 + 10 ⎞= ⎜ ⎟ 8 KHz⎝ 320 −10⎠330= × 8 = 8.51 KHz310(a,0,a)x(a,a,a)(0,0,0) (0,a,0)yzF(0,0,a) G (0,a,a)28. A meter bridge is set-up as shown, to determinean unknown resistance 'X' using a standard 10ohm resistor. The galvanometer show null pointwhen tapping-key is at 52 cm mark. The endcorrectionsare 1 cm and 2 cm respectively for theends A and B. The determine value of 'X' is-X10ΩE(a,0,a)xD(a,0,0)AH(a,a,a)C (a,a,0)flux through EHBA= flux through EHDC= E 0 a 2B(0,a,0)27. A police car with a siren of frequency 8 kHz ismoving with uniform velocity 36 km/hr towards atall building which reflects the sound waves. Thespeed of sound in air is 320 m/s. The frequencyof the siren heard by the car driver is-yA(A) 10.2 ohm(C) 10.8 ohmAns. [B]x 52 + 1Sol. =10 48 + 253×10= = 10.650B(B) 10.6 ohm(D) 11.1 ohm29. A 2 µF capacitor is charged as shown in figure.The percentage of its stored energy dissipatedafter the switch S is turned to position 2 is-1 2VS2µF 8µF(A) 8.50 kHz(B) 8.25 kHz(C) 7.75 kHz(D) 7.50 kHzAns. [A]Sol.336 × 10V s =3600m/s = 10 m/s, ν = 8 KHzV 0 = 320 m/sV 0 = V s = 10 m/sV sObserverAns.Sol.(A) 0 % (B) 20 %(C) 75 % (D) 80 %[D]2µF8µFXtraEdge for IIT-JEE 67 MAY 2011

1 2 × 8∆U = × [V – 0]22 2 + 81 16= × × V 2 =2 10U i′= 21 × 2 × V28V210= V 2∆U 8% dissipated = = × 100U i 10= 80%30. The wavelength of the first spectral line in theBalmer series of hydrogen atom is 6561 Å. Thewavelength of the second spectral line in theBalmer series of singly-ionized helium atom is-Ans.Sol.(A) 1215 Å(B) 1640 Å(C) 2430 Å(D) 4687 Å[A]hc ⎡ 1 1 ⎤= 13.66561 ⎢ 2− 2 ⎥⎣23 ⎦hc ⎡ 1 1 ⎤= 13.6 × 4λ ⎢ 2− 2 ⎥⎣24 ⎦hc 5= 13.6 ×6561 36hc 3= 13.6 × 4 ×λ16hc 5 4= ×6561 36 35λ = 27Sol.R AQSR BQAns.(A)kQ A kQ = [Final potential will be same]R R BBAQ A R =AQBR Bas R A > R B∴ Q A > Q BQ RA2AB2B4πRQ4πR=RRRA2AB2B=σ A R =BσBR Aas R A > R B∴ σ B > σ AE B > E A Ans. [D]∴Ans. [B]Ans [C]32. A metal rod of length 'L' and mass 'm' is pivotedat one end. A thin disk of mass 'M' and radius 'R'( Angularfrequency for case B(D) Angular frequency for case A < Angularfrequency for case B[A,D]33. An electron and a proton are moving on straightparallel paths with same velocity. They enter asemi-infinite region of uniform magnetic fieldperpendicular to the velocity. Which of thefollowing statement(s) is/are true ?XtraEdge for IIT-JEE 68 MAY 2011

(A) They will never come out of the magneticfield region(B) They will come out traveling along parallelpaths(C) They will come out at the same time(D) They will come out at different timesAns. [B, D]Sol.× × × ×× × × ×e – v R× p× × ×p× × ×v×× R e × × ×× × × ×2mevevB =R emeVR e =eBRp > Re2mpvevB =R pmpvR p =eBπReπmeTe= =v eB⇒ Tp > TeπmpTp=eB34. A composite book is made of slabs A, B, C, Dand E of different thermal conductivities (given interms of a constant K) and sizes (given in termsof length, L) as shown in the figure. All slabs areof same width. Heat 'Q' flow only from left toright through the blocks. Then in steady state-heat1L3L4L0 1L 5L 6LABCD3K4K5KE6K(A) heat flow through A and E slabs are same(B) heat flow through slab E is maximum(C) temperature difference across slab E issmallest(D) heat flow through C = heat flow through B +heat flow through DAns. [A, C, D]Sol.A4AH2KA2AABCD3 H16H25 H163K4K5KE4AH6KL 4L LAll the three system shown are in series hencerate of heat flow will be same through both A &E.LR A = ;8(KA)4LR D = ;5KA4LR B = ;3KALR E =24KA4LR C = ;8KAUsing parallel combination rate of heat flowacross C = rate of heat flow through B+ rate ofheat flow through D.∆ θ A = HR A =∆ θ∆ θ∆ θ∆ θ3H= R16HL8KA3H ⎛ 4L ⎞= ⎜ ⎟16 ⎝ 3KA ⎠B B=H=2( R )H ⎛ 4L ⎞= ⎜ ⎟2 ⎝ 8KA ⎠C C=5H ⎛ 4L ⎞= ⎜ ⎟16 ⎝ 5KA ⎠D =⎛ L ⎞= H⎜⎟⎝ 24KA ⎠E =HL4KAHL24KAHL4KAHL4KASECTION – IIIParagraph TypeThis section contains 2 paragraphs.. Based upon thefirst paragraph 3 multiple choice question and basedupon on the other paragraph 2 multiple choicequestions have to be answered. Each of these questionshas four choices (A), (B), (C) and (D) out of whichONLY ONE is correct.Paragraph for Questions 35 to 37Phase space diagrams are useful tools inanalyzing all kinds of dynamical problems. Theyare especially useful in studying the changes inmotion as initial position and momentum arechanged. Here we consider some simpledynamical systems in one-dimension. For suchsystems, phase space is a plane in which positionis plotted along horizontal axis and momentum isplotted along vertical axis. The phase spacediagram is x(t) vs p(t) curve in this plane. Thearrow on the curve indicates the time flow. Forexample, the phase space diagram for a particlemoving with constant velocity is a straight line asXtraEdge for IIT-JEE 69 MAY 2011

shown in the figure. We use the sign conventionin which position or momentum upwards (or toright) is positive and downwards (or to left) isnegative.22m v − m u = 2mp 2 2– p 0 = 2m 2 gxp 2 = p 2 0 + 2m 2 gx222gxMomentumPosition35. The phase space diagram for a ball thrownvertically up from ground is-Momentum36. The phase space diagram for simple harmonicmotion is a circle centered at the origin. In thefigure, the two circles represent the sameoscillator but for different initial conditions, andE 1 and E 2 are the total mechanical energiesrespectively. Then-MomentumE2a2aE 1Position(A)PositionMomentumAns.Sol.(A) E 1 = 2 E 2 (B) E 1 = 2E 2(C) E 1 = 4E 2 (D) E 1 = 16 E 2[C]K' E Max 1(B)PositionK' E max 2K.E 1 = 0Momentummaximal positionmaximal positionK.E 2 = 0(C)PositionMomentumEE1k(2a)21 2k(a)221= =2E 1 = 4E 24(D)Position37. Consider the spring-mass system, with the masssubmerged in water, as shown in figure. Thephase space diagram for one cycle of this systemis-Ans.Sol.[D]From conservation of mechanical energy1 2 1 2mv + mgx = mu22XtraEdge for IIT-JEE 70 MAY 2011

(A)MomentumMomentumPositionhas an angular frequency ω, where a part of theenergy is absorbed and a part of it is reflected. Asω approaches ω p , all the free electrons are set toresonance together and all the energy is reflected.This is the explanation of high reflectivity ofmetals.38. Taking the electronic charge as 'e' and thepermittivity as 'ε 0 ', use dimensional analysis todetermine the correct expression for ω p .(A)Nemε 0(B)mε 0Ne(B)MomentumPositionAns.Sol.(C)[C]Nemε20F = mω 2 l ≡e24πεl02(D)mεNe02ω 2 ≡e24πεl03≡⎛ e⎜⎝ m2ε 0⎞⎟⎠⎛ Nl⎜3⎝ l3⎞⎟⎠(C)Positionω =Nemε20Ans.[B](D)MomentumPositionParagraph for Questions 38 to 39A dense collection of equal number of electronsand positive ions is called neutral plasma. Certainsolids containing fixed positive ions surroundedby free electrons can be treated as neutral plasma.Let 'N' be the number density of free electrons,each of mass 'm'. When the electrons aresubjected to an electric field, they are displacedrelatively away from the heavy positive ions. Ifthe electric field becomes zero, the electronsbegin to oscillate about the positive ions with anatural angular frequency 'ω p ', which is called theplasma frequency. To sustain the oscillations, atime varying electric field needs to be applied that39. Estimate the wavelength at which plasmareflection will occur for metal having the densityof electrons N ≈ 4 × 10 27 m –3 . Take ε 0 = 10 –11and m ≈ 10 –30 , where these quantities are inproper SI unit-(A) 800 nm(B) 600 nm(C) 300 nm(D) 200 nmAns. [B]c = λf22πc Neω p = ω = =λ mελ = 2πcmεNe02=8πce0mεN2 0−302×3.14×3×10 (10 )(10=−19271.6×10 4×10= 589 × 10 –9 m ≈ 600 nm−11SECTION – IVNumerical Response TypeThis section contains 7 questions. The answer to eachquestion is a single-digit integer, ranging from 0 to 9.The bubble corresponding to the correct answer is tobe darkened in the ORS.)XtraEdge for IIT-JEE 71 MAY 2011

40. A block is moving on an inclined plane makingan angle 45º with the horizontal and thecoefficient of friction is µ. The force required tojust push it up the inclined plane is 3 times theforce required to just prevent it from slidingdown. If we define N = 10 µ, then N is.Ans. [5]Sol.45ºN=mgcosθmgsinθ + µmgcosθ = 3(mgsinθ – µmgcosθ)1 µ 3 3µ⇒ + = −2 2 2 2⇒4µµ = 212=22⎛ 1 ⎞N = 10µ = 10 ⎜ ⎟ = 5⎝ 2 ⎠41. A boy is pushing a ring of mass 2 kg and radius0.5 m with a stick as shown in the figure. Thestick applies a force of 2 N on the ring and rolls itwithout slipping with an acceleration of 0.3 m/s 2 .The coefficient of friction between the groundand the ring is large enough that rolling alwaysoccurs and the coefficient of friction between thestick and the ring is (P/10). The value of P is.StickAns. [4]Sol.mgGroundf 22NP×2 Pf 1 = µ×2 = =10 52 – f 2 = Ma cm …….(1)f 2 = 2 – 2 × 0.3 = 1 .4 N(f 2 – f 1 ) R = I cm αa(f 2 – f 1 ) R = MR2 cm×Rf 2 – f 1 = Ma cmf 1 = f 2 – ma cm = 1.4 – 2 × 0.3 = 0.8 NP0 .8 = ⇒ P = 45Note : It has been assumed that the stick applieshorizontal force of 2N (only normal reaction)42. Four point charge, each of +q, are rigidly fixed atthe four corners of a square planar soap film ofside 'a'. The surface tension of the soap film is γ.The system of charges and planar film are inequilibrium, andAns. [3]Sol.⎡2q ⎤a = k ⎢ ⎥⎦⎣ γF AC1/NF AB, where 'k' is a constant. Then N is.F +qADA+qDF AC =BC+q +qq28πεa0F AD = F AB =F R =⇒ a 3 =22q24πεa0q ⎛ 1 ⎞⎜2cos45º+ ⎟24πε0a⎝ 2 ⎠= r (2) (BD) = 2 r( 2 a)2⎛q ⎜⎝8⎛2q ⎞a = k⎜⎟r⎝ ⎠1/3⎛ 1 ⎞⎜ 2 + ⎟where k = ⎜ 2 ⎟⎜ 8 2π⎟⎝ ⎠21 ⎞2 + ⎟2 ⎠2πεr1/30⇒ N = 3XtraEdge for IIT-JEE 72 MAY 2011

43. Four solid spheres each of diameter 5 cm andmass 0.5 kg are placed with their centers at thecorners of a square of side 4 cm. The momentumof inertia of the system about the diagonal of thesquare is N × 10 –4 kg-m 2 , then N is.Ans. [9]Sol.mmmr =lm5 cm = × 1025 –2 m2m = 21 kgl = 4 × 10 –2 mUsing parallel axis theorem⎡ 2 1 5 −4 ⎤I total = ⎢4 × × × × 10 ⎥ +⎣ 5 2 4 ⎦⎡ 1 −4 ⎤⎢2× × 8×10 ⎥⎣ 2 ⎦⇒ 10 –4 + 8 × 10 –4 ⇒ 9 × 10 –4 kg m 244. The activity of a freshly prepared radioactivesample is 10 10 disintegrations per second, whosemean life is 10 9 s. The mass of an atom of thisradioisotope is 10 –25 kg. The mass (in mg) of theradioactive sample is.Ans. [1]Sol. A = λN10 10 1=9 × N10N = 10 19mass of sample = 10 19 × 10 –25 × 1 × 10 6 = 1 mg45. A long circular tube of length 10 m and radius 0.3carries a current I along its curved surface asshown. A wire-loop of resistance 0.005 ohm andof radius 0.1 m is placed inside the tube with itsaxis coinciding with the axis of the tube. Thecurrent varies as I = I 0 cos(300 t) where I 0 isconstant. If the magnetic moment of the loop is Nµ 0 I 0 sin(300 t), then 'N' is.Ans. [6]Sol. B =µ 0 0I cos300 t10µ Iφ = 0 0× 3.14 × 0.01 cos 300 t10φ = 3.14 × µ 0 I 0 cos 300 t × 10 –3dφe = − = 3.14 × 300 µ 0 I 0 sin 300t × 10 –3dtI−3e 3.14 × 300 µ I sin 300t 10i = =0 0 ×R 0. 005i = 3.14 × 60 µ 0 I 0 sin 300 t0.01Magnetic moment = 3.14 × 60 × 3.14 × × 100µ 0 I 0 sin 300 t= 5.9 µ 0 I 0 sin 300 t= 6 µ 0 I 0 sin 300 t46. Steel wire of length 'L' at 40ºC is suspended fromthe ceiling and then a mass 'm' is hung from itsfree end. The wire is cooled down from 40ºC to30ºC to region its original length 'L'. Thecoefficient of linear thermal expansion of the steelis 10 –5 /ºC, Young's modulus of steel is 10 11 N/m 2and radius of the wire is 1 mm. Assume that L >>diameter of the wire. Then the value of 'm' in kgis nearly.Ans. [3]Sol.mmgL∆L = = Lα(∆θ)AYAYα(∆θ)⇒ m ==gm = 3.14 kg ⇒ 3 kgπ × 10−611× 10 × 1010−5× 10XtraEdge for IIT-JEE 73 MAY 2011

MATHEMATICSSECTION – ISingle Correct Choice TypeThis section contains 7 multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.ln3x sin x47. The value of∫dx is22sin x + sin( ln6 – x )Ans.Sol.(A) 41 ln 23(C) ln 23[A]ln2Let x 2 = t xdx = 2dtl 31 n ln2l 31 n 2ln2sin tI =∫dt2 sin t + sin( ln6− t)sin ( ln6− t)I =∫dtsin t + sin( ln6− t)Add (1) & (2)2I =1 l n 32 ∫l n 2dt1 1 3I = (ln3 – ln2) = l n4 4 22(B) 21 ln 23(D) 61 ln 23... (1)… (2)48. Let the straight line x = b divide the area enclosed byy = (1 – x) 2 , y = 0, and x = 0 into two parts R 1 (0 ≤ xAns.Sol.≤ b) and R 2 (b ≤ x ≤ 1) such that R 1 – R 2 = 41 .Then b equals(A) 43(C) 31[B]R 1R 20 b 1(B) 21(D) 41R 1 – R 2 = 41b22 1∫( x − 1) dx –∫( x − 1) dx = 40⎡(x − 1)⎢⎣ 3( b −1)3333⎤⎥⎦b01b⎡(x − 1)– ⎢⎣ 33⎤⎥⎦1 ( b −1)+ – 0 + 3 32( b −1)1 1 −1= – =3 4 3 12(b – 1) 3 1= – 8b – 1 =1 1− ⇒ b =2 21b= 413= 4149. Let → a = î + ĵ + kˆ , → b = î – ĵ + kˆ and → c =Ans.Sol.î – ĵ – kˆ be three vectors. A vector → v in theplane of → a and → b , whose projection on → c is1 , is given by3(A) î – 3 ĵ + 3 kˆ (B) – 3 î – 3 ĵ – kˆ(C) 3 î – ĵ + 3 kˆ (D) î + 3 ĵ – 3 kˆ[C]Let → ν = x î + y ĵ + z kˆQ [ → a → b → ν ] = 011x1− 1 1 = 0y1zOn solving x = zQ projection of → ν on → c is→νc→131 . c x − y − zSo, = ⇒ =3→| |3⇒ x – y – z = 1So solving (1) & (2)y = –1 & x = z13….(1)….(2)50. Let (x 0 , y 0 ) be the solution of the followingequations(2x) ln 2 = (3y) ln 33 ln x = 2 ln yThen x 0 isXtraEdge for IIT-JEE 74 MAY 2011

(A) 61(B) 31So, m + 3 = ± 3 (1 – 3 m)(C) 21(D) 6Ans. [C]Sol. (2x) ln 2 = ( 3y) ln 3ln 2 (ln 2 + ln x) = ln 3 (ln 3 + ln y)ln 2 . ln x – ln 3 ln y = (ln 3) 2 – (ln 2) 2 .....(1)3 ln x = 2 ln yln x . ln 3 = lny . ln 2ln3ln y = ln x.....(2)ln2Solving (1) & (2)1ln x = – ln 2 ⇒ x = 251. Let α and β be the roots of x 2 – 6x – 2 = 0, withα > β. If a n = α n – β n for n ≥ 1, then the value ofa10– 2a82a9is(A) 1 (B) 2(C) 3 (D) 4Ans. [C]Sol. ∴ x 2 – 6x – 2 = 0 has roots α, βSo, α 2 – 2 = 6α & β 2 – 2 = 6βa n = α n – β n=8So,2α ( αa2( α10− 2a2a9− 2) − β9− β8)8( β92=( α10− 2) α=8− β102( α) − 2( α9− β89)8− β(6α)− β (6β)= 3.9 92( α − β )52. A straight line L through the point (3, –2) isinclined at an angle 60° to the line 3 x + y = 1.If L also intersects the x-axis, then the equation ofL is(A) y + 3 x + 2 – 3 3 = 0(B) y – 3 x + 2 + 3 3 = 0(C) 3 y – x + 3 + 2 3 = 0(D) 3 y + x – 3 + 2 3 = 0Ans. [B]Sol. Let the slope of the line is mm + 3tan 60º =1 − 3m3 =m +1 −33m8)m + 3 = 3 – 3m m + 3 = – 3 + 3mm = 0 m = 3hence linehence liney = – 2 y + 2 = 3 (x – 3)y – 3 x + 2 + 3 3 = 0as line intersects x-axisSo line will be, y – 3 x + 2 + 3 3 = 053. Let P = {θ : sin θ – cos θ = 2 cos θ} andQ = {θ : sin θ + cos θ = 2 sin θ} be two sets.Then(A) P ⊂ Q and Q – P ≠ ∅(B) Q ⊄ P(C) P ⊄ Q(D) P = QAns. [D]Sol. P : sin θ – cos θ = 2 cos θsin θ = ( 2 + 1) cos θtan θ = 2 + 1tan θ = tan 67 21 °3πθ = nπ + , n ∈ I8Q : sin θ + cos θ = 2 sin θcos θ = ( 2 – 1) sin θ1tanθ =2 − 1= 2 + 1θ = nπ +∴ P = Q3π , n ∈ I8…(1)…(2)SECTION – IIMultiple Correct Choice TypeThis section contains 4 multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONE OR MORE may be correct.54. The vector(s) which is/are coplanar with vectorsî + ĵ + 2 kˆ and î + 2 ĵ + kˆ , and perpendicularto the vector î + ĵ + kˆ is /are(A) ĵ – kˆ(B) – î + ĵ(C) î – ĵ(D) – ĵ + kˆXtraEdge for IIT-JEE 75 MAY 2011

Ans.Sol.[A,D]r = xiˆ + yj ˆ + zkˆis coplanar with the given vectorsox y z∴ 1 1 2 = 01 2 1So, 3x = y + z ...(1)→∴ r ⊥ iˆ + ˆj+ kˆ→So, r . (ˆ i + ˆj+ kˆ)= 0So, x + y + z = 0 ...(2)On solving (1) & (2)So, x = 0 ∴ y + z = 0 ∴ (A) & (D) Satisfy55. Let f : R → R be a function such thatf(x + y) = f(x) + f(y), ∀x, y ∈ RIf f(x) is differentiable at x = 0, then(A) f(x) is differentiable only in a finite intervalcontaining zero(B) f(x) is continuous ∀ x ∈ R(C) f '(x) is constant ∀ x ∈ R(D) f(x) is differentiable except at finitely manypointsAns. [B,C]Sol. f(x + y) = f(x) + f(y)By Partial differentiation with respect to xf ' (x + y) = f ' (x)f ' (y) = f '(0)f(y) = (f '(0))y + cf(y) = ky +c∴ f(y) = ky as f(0) = 0∴ f(x) = kxAlternatef ( x + h)− f ( x)f '(x) = limh→0hf ( x)+ f ( h)− f ( x)f ( h)= lim= limh→0hh →0h= λ (let)f(x) = λx + c As f(0) = 0 ⇒ c = 0f(x) = λx56. Let M and N be two 3 × 3 non-singularskew-symmetric matrices such that MN = NM.If P T denotes the transpose of P, thenM 2 N 2 (M T N) –1 (MN –1 ) T is equal to(A) M 2 (B) –N 2(C) –M 2(D) MNAns. [C]Sol.Q MN = NMM 2 N 2 = MN MN Q (M T ) –1 = (–M) –1 = –M –1Given, M 2 N 2 (M T N) –1 . (MN –1 ) T= – MN MN N –1 M –1 N –1 MI= –M NN –1 M = – M 2The most suitable answer is (C), althoughgiven information is contradictory as Skewsymmetric matrix of odd order cannot be nonsingular57.x yLet the eccentricity of the hyperbola –2 2a b= 1be reciprocal to that of the ellipse x 2 + 4y 2 = 4. Ifthe hyperbola passes through a focus of theellipse, then2 2x y(A) the equation of the hyperbola is – = 13 2(B) a focus of the hyperbola is (2, 0)(C) the eccentricity of the hyperbola is53(D) the equation of the hyperbola is x 2 – 3y 2 = 3Ans. [B,D]Sol. Let e 1 = eccentricity of hyperbolae 2 = eccentricity of ellipse1∴ e 1 =e 2so eccentricity of ellipse =3 = e22eccentricity of ellipse =2 = e13Now focus of ellipse is (± ae 2 , 0) ≡ (± 3 , 0)Hyperbola passes through it2( 3)So, – 0 = 1 ⇒ a 2 = 32aalso b 2 = a 2 (e 2 1 – 1)b 2 ⎛ 4 ⎞= 3 ⎜ – 1⎟ = 1⎝ 3 ⎠and hyperbola22x − y = 13 1also focus (± ae 1 , 0) ≡ (± 2, 0)22XtraEdge for IIT-JEE 76 MAY 2011

SECTION – IIIParagraph TypeThis section contains 2 paragraphs. Based upon one ofthe paragraphs 3 multiple choice questions and basedon the other paragraph 2 multiple choice questionshave to be answered. Each of these questions has fourchoices (A), (B), (C) and (D) out of which ONLY ONEis correct.Paragraph for Question No. 58 to 60Let a, b and c be three real numbers satisfying⎡19 7⎤⎢ ⎥[a b c]⎢8 2 7⎥= [0 0 0] …….(E)⎢⎣7 3 7⎥⎦58. If the point P(a, b, c), with reference to (E), lieson the plane 2x + y + z = 1, then the value of7a + b + c is(A) 0 (B) 12 (C) 7 (D) 6Ans. [D]⎡19 7⎤⎢ ⎥Sol. [a b c]⎢8 2 7⎥= [0 0 0]⎢⎣7 3 7⎥⎦a + 8b + 7c = 09a + 2b + 3c = 07a + 7b + 7c = 0On solving above equation⎛ λ 6λ⎞(a, b, c) ≡ ⎜− , − , λ⎟ ⎝ 7 7 ⎠∴ (a, b, c) lies on the plane 2x + y + z = 1So2λ− –76λ + λ = 17on solving λ = – 7So 7a + b + c = 659. Let ω be a solution of x 3 – 1 = 0 with Im(ω) > 0.If a = 2 with b and c satisfying (E), then the value3 1 3ofa +b +cω ω ωis equal to(A) –2 (B) 2(C) 3 (D) –3Ans. [A]Sol.⎛ λ 6λ⎞∴ (a, b, c) ≡ ⎜− , − , λ⎟ ⎝ 7 7 ⎠∴ a = 2 is given so λ = – 14So (a, b, c) ≡ (2, 12, – 14)3 1 3So + +a b cω ω ω= –260. Let b = 6, with a and c satisfying (E). If α and βare the roots of the quadratic equationAns.ax 2 + bx + c = 0, thenn∑ ∞ ⎛ 1 1 ⎞⎜ + ⎟= 0 ⎝αβ ⎠6(A) 6 (B) 7 (C) 7[B]⎛ λ 6λ⎞Sol. ∴ (a, b, c) ≡ ⎜− , − , λ⎟ ⎝ 7 7 ⎠∴ b = 6 so λ = – 7.So (a, b, c) ≡ (1, 6, –7)So the equation ax 2 + bx + c = 0x 2 + 6x – 7 = 0So α = 1, β = – 7S = ∑∞n=0⎛ 1 1 ⎞⎜ + ⎟⎝ α β ⎠n⎛11 ⎞= ∑ n⎜ − ⎟⎝17 ⎠⎞= ∑ n2⎛ 6 6 ⎛ 6 ⎞⎜ ⎟ = 1 + + ⎜ ⎟⎠ + .... ∞⎝ 7 ⎠ 7 ⎝ 7nis(D) ∞1= = 761−7Paragraph for Question Nos. 61 and 62Let U 1 and U 2 be two urns such that U 1 contains 3white and 2 red balls, and U 2 contains only 1white ball. A fair coin is tossed. If head appearsthen 1 ball is drawn at random from U 1 and putinto U 2 . However, if tail appears then 2 balls aredrawn at random from U 1 and put into U 2 . Now 1ball is drawn at random from U 261. The probability of the drawn ball from U 2 beingwhite isAns.Sol.13(A) 3019(C) 30[B]23(B) 3011(D) 303W1 W2RU 1 U 2Required probability = P(H)[P(W/H) × P(W 2 ) +⎛ both W ⎞P(R/H)P(W 2 )] + P(T) [P ⎜ ⎟ P(W 2 ) +⎝ T ⎠⎛ both R ⎞ ⎛ R1 & W1⎞P ⎜ ⎟ P(W 2 ) + P⎜⎟ P(W 2 )]⎝ T ⎠ ⎝ T ⎠XtraEdge for IIT-JEE 77 MAY 2011

⎡323 21× ⎤=⎡32 1⎤1 C2C21 C1C12⎢ × 1+× ⎥ + ⎢ × 1+× + ×555 ⎥2 ⎣55 2⎦2 ⎣ C2C23 C23 ⎦1 ⎡31⎤1 ⎡ 3 1 2⎤2 11 23= ⎢ + ⎥ + ⎢ + + ⎥ = + =2 ⎣55⎦2 ⎣1030 5 ⎦ 5 30 3062. Given that the drawn ball from U 2 is white, theprobability that head appeared on the coin is1711(A) (B) 23231512(C) (D) 2323Ans. [D]Sol. Required probability =⎡ ⎛ W1⎞ ⎛ R1⎞ ⎤P(H) ⎢P⎜⎟P(W 2)+ P⎜⎟P(W 2)⎥⎣ ⎝ H ⎠ ⎝ H ⎠ ⎦⎡ ⎛ W1⎞ ⎛ R1⎞ ⎤ ⎡ ⎛ both W ⎞P(H) ⎢P⎜⎟P(W 2)+ P⎜⎟P(W 2)⎥ + P(T) ⎢P⎜⎟P(W 2)⎣ ⎝ H ⎠ ⎝ H ⎠ ⎦ ⎣ ⎝ T ⎠⎛ both R ⎞ ⎛ R+ P⎜⎟P(W 2)+ P⎜⎝ T ⎠ ⎝=1 ⎡32 1⎤12⎢ × + ×5 5 2⎥⎣ ⎦=233012131& W1⎞ ⎤⎟P(W 2)⎥T ⎠ ⎦SECTION – IVNumerical Response TypeThis section contains 7 questions. The answer to each ofthe questions is a single-digit integer, ranging from 0 to9. The bubble corresponding to the correct answer is tobe darkened in the ORS.63. Let a 1 , a 2 , a 3 , .., a 100 be an arithmetic progressionwith a 1 = 3 and S p = ∑=pi 1a i, 1 ≤ p ≤ 100. For anyinteger n with 1 ≤ n ≤ 20, let m = 5n. Ifnot depend on n, then a 2 isAns. [9]Sol. a 1 = 35nS[2am S1 + (5n−1)d]n= 5= 2S n S nn [2a1+ ( n −1)d]25[(6 − d)+ 5nd]=(6 − d)+ ndS5nQ is independent of n so d = 6SnSo a 2 = a 1 + d = 3 + 6 = 9SSmndoes64. Consider the parabola y 2 = 8x. Let ∆ 1 be the areaof the triangle formed by the end points of its⎛ 1 ⎞latus rectum and the point P ⎜ , 2⎟ on the⎝ 2 ⎠parabola, and ∆ 2 be the area of the triangle formedby drawing tangents at P and at the end points of∆1the latus rectum. Then is∆2Ans. [2]Sol. It is a property that area of triangle formed byjoining three points lying on parabola is twice thearea of triangle formed by tangents at these pointsAlternate : y 2 = 8x⎛ 1 ⎞P ⎜ , 2⎟ ⎝ 2 ⎠(2, 4) A•P•3/2•∆ 1 = 21 |Base × Height|= 21 × 23 × 8 = 6B(2, –4)Also⎛ 1 ⎞Equation of tangent at P ⎜ , 2⎟ ⎝ 2 ⎠•• P••⎛ 1 ⎞y (2) = 4. ⎜ x + ⎟⎝ 2 ⎠y = 2x + 1 ...(1)Tangent at A : y = x + 2Tangent at B : – y = + x + 2 ⇒ y = – x – 2Point of intersectionL(–2, 0), M (1, 3), N (–1, –1)− 2 0 11∆ 2 = 1 3 12−1−11= | 21 [–2(4) + (–1 + 3)]|XtraEdge for IIT-JEE 78 MAY 2011

1= [ − 8 + 3 −1]= 32∆ 1 6So, = = 2∆ 3265. The positive integer value of n > 3 satisfying the1 1 1equation = + is⎛ π ⎞ ⎛ 2π⎞ ⎛ 3π⎞sin⎜⎟ sin⎜⎟ sin⎜⎟⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠Ans. [7]Sol.Let nπ = θ1sin θ1sin θ=–1sin 2θ1sin 3θ+=1sin 3θ1sin 2θ[sin 3θ – sin θ] sin 2θ = sin θ sin 3θ2 sin θ cos 2θ sin 2θ = sin θ sin 3θQ sin θ ≠ 02 cos 2θ sin 2θ = sin 3θsin 4θ = sin 3θso either 4θ = 3θ or 4θ = π – 3θso θ = 0 or θ = 7π so n = 766. Let f(θ) =⎛ ⎛ ⎞sin ⎜sin θ ⎞⎟tan 1⎜⎟,⎝ ⎝ cos2θ⎠⎠whereπ π d– < θ < . Then the value of 4 4 d(tan θ)(f (θ))isAns. [1]Sol.⎛Q tan –1 sin θ⎟ ⎞⎜ = sin –1 tan θ⎝ cos2θ⎠so f (θ) = sin (sin –1 tan θ) = tan θd(f ( θ))d(tanθ)Q = = 1d(tanθ)d(tanθ)67. If z is any complex number satisfying | z – 3 – 2i| ≤ 2,then the minimum value of | 2z – 6 + 5i| isAns. [5]Sol.(3, 2)BSo, Min of |2z – 6 + 5i| = PA5i5= Min 2 z – 3 + = 2 × = 52 268. The minimum value of the sum of real numbersa –5 , a –4 , 3a –3 , 1, a 8 and a 10 with a > 0 isAns. [8]Sol.A.M. ≥ G.M.–5–4–3–3a + a + a + a + a + 1+a + a8≥ (a –5 . a –4 . a –3 . a –3 . a –3 . 1. a 8 .a 10 ) 1/8a –5 + a –4 + a –3 + a –3 + a –3 + 1 + a 8 + a 10 ≥ 8so minimum value is 869. Let f : [1, ∞) → [2, ∞) be a differentiable functionxsuch that f(1) = 2. If 6∫f ( t)dt = 3x f(x) – x 3 forall x ≥ 1, then the value of f (2) isAns. [6]xSol. 6∫f ( t)dt = 3xf(x) – x 316 f(x) = 3f(x) + 3xf '(x) –3x 23f(x) = 3x f '(x) – 3x 2dy3y = 3x – 3x2dxdyx – y = x2dxdy y – = xdx xI.F. = e–1∫ dxx1= e –ln x = x11y . = x ∫x 1 y. dx ⇒ = x + c ⇒ y = x 2 + cxx xQ f(1) = 2 ⇒ c = 1y = x 2 + xf(2) = 4 + 2 = 6–3810MinvalueA⎛ 5 ⎞P ⎜3,– ⎟⎝ 2 ⎠XtraEdge for IIT-JEE 79 MAY 2011

XtraEdge for IIT-JEE 80 MAY 2011

IIT-JEE 2011PAPER-II (PAPER & SOLUTION)Time : 3 Hours Total Marks : 240Instructions : [Each subject contain]Section – I :Section – II :Multiple choice questions with only one correct answer. +3 marks will be awarded for correct answerand -1 mark for wrong answer. [No. of Ques. : 8]Multiple choice questions with multiple correct answer. +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 4]Section – III : Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 6]Section – IV : Column Matching type question +2 marks for each correctly matched row and No Negative markingfor wrong answer. [No. of Ques. : 2]CHEMISTRYSECTION – ISingle Correct Choice TypeThis section contains 8 multiple choice questions. Eachquestion has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. Oxidation states of the metal in the mineralshaematite and magnetite respectively, are(A) II, III in haematile and III in magnetite(B) II, III in haematite and II in magnetite(C) II in haematite and II, III in magnetite(D) III in haematite and II, III in magnetiteAns. [D]+ 3Sol. Haematite is F e2O3+ 2+ 3Magnetite is Fe 3 O 4 or FeO. Fe2O32. The following carbohydrate isH OHH OHOHOOHH OHH H(A) a ketohexose (B) an aldohexose(C) an α-furanose (D) an α-pyranoseAns. [B]Sol. Aldohexose3. The major product of the following reaction isRCH 2 OHH + (anhydrous)O(A) a hemiacetal (B) an acetal(C) an ether(D) an esterAns.Sol.[B]OR- CH 2 –OH⊕H (anhydrous)OOCH 2 –RAcetal4. Amongst the compounds given, the one thatwould form a brilliant colored dye on treatmentwith NaNO 2 in dill. HCl followed by addition toan alkaline solution of β -naphthol isN(CH 3 ) 2NHCH 3(A)(B)Ans.Sol.(C)[C]H 3 CNH 2(D)NaNO 2 + HCl (dil.)CH 3 NH 2 CH 30–5°CCH 3Alkaline solutionN = NHOColoured dyeCH 2 NH 2N 2 ClOH5. The freezing point (in °C) of a solution containing0.1 g of K 3 [Fe(CN) 6 ] (mol. wt. 329) in 100 g ofwater (k f = 1.86 K kg mol –1 ) is(A) –2.3 × 10 –2 (B) –5.7 × 10 –2(C) –5.7 × 10 –3 (D) –1.2 × 10 –2XtraEdge for IIT-JEE 81 MAY 2011

Ans. [A]Sol. ∆T = k f × m × i × 10000.1= 1 .86 × × 4 × 1000329 × 100= 2.26 × 10 –2 ≈ 2.3 × 10 –26. Consider the following cell reaction :+2Fe(s)+ O 2(g) + 4H (aq)2+⎯⎯→2Fe (aq) + 2H 2O(l)At [Fe 2+ ] = 10 –3 M, P(O 2 ) = P(O 2 ) = 0.1 atm andpH = 3, the cell potential at 25 °C is(A) 1.47 V(B) 1.77 V(C) 1.87 V(D) 1.57 VAns. [D]+ 20.0591 [Fe ]Sol. E cell = E° cell − log4+ 4[P ][H ]−30.591 [10 ]= 1.67 − log4 [0.1][10= 1.57 V7. Passing H 2 S gas into a mixture of Mn 2+ , Ni 2+ ,Cu 2+ and Hg 2+ ions in an acidified aqueoussolution precipitates.(A) CuS and HgS (B) MnS and CuS(C) MnS and NiS (D) NiS and HgSAns. [A]Sol. Cu + 2 , Hg +2 are group II basic radicals8. Among the following complexes (K–P)K 3 [Fe(CN) 6 ] (K), [CO(NH 3 ) 6 ]Cl 3 (L), Na 3[Co(oxalate) 3 ] (M), [Ni(H 2 O) 6 ]Cl 2 (N),K 2 [Pt(CN) 4 ] (O) and [Zn(H 2 O) 6 ] (NO 3 ) 2 (P) thediamagnetic complexes are(A) K, L, M, N (B) K, M, O, P(C) L, M, O, P (D) L, M, N, OAns. [C](L) : [Co(NH 3 ) 6 ]Cl 3(M) : Na 3 [Co(Ox) 3 ](O) : K 2 [Pt(CN) 4 ](P) : [Zn(H 2 O) 6 ] (NO 3 ) 2O2−3SECTION – IIMultiple correct Choice TypeThis section contains 4 multiple choice questions. Eachquestions has 4 choices (A), (B), (C) and (D), out ofwhich ONE OR MORE is/are correct.]249. Reduction of the metal centre in aqueouspermanganate ion involves(A) 3 electrons in neutral medium(B) 5 electrons in neutral medium(C) 3 electrons in alkaline medium(D) 5 electrons in acidic mediumAns. [A, C, D]Sol. → In alkaline solution, KMnO 4 is first reduced toE 0 mangnate and then to insoluble MnO= 1.67 V2→→ I+ 7Neutral2KMnO + H O ⎯⎯⎯→−42+−+ 42MnO + 2KOH + 3[O]acidicMnO 4 + 8H + 5e ⎯⎯⎯→Mn+ 4H 2O10. For the first order reaction : 2N 2 O 5 (g) →4NO 2 (g) + O 2 (g)(A) the concentration of the reactant decreasesexponentially with time.(B) the half-life of the reaction decreases withincreasing temperature(C) the half-life of the reaction depends on theinitial concentration of the reactant(D) the reaction proceeds to 99.6 % completion ineight half-life duration.Ans. [A, B, D]−ktSol. C = C e [A]tA A0120.41000.693 0.693= =K −Ea/A e⎛ 1 ⎞= ⎜ ⎟⎝ 2 ⎠4=04100RT40.693= eA0+ 2E / RT⎛ 4 ⎞log⎜⎟100= n =⎝ ⎠= 8⎛ 1 ⎞log⎜⎟⎝ 2 ⎠a[D][B]11. The equilibrium: 2Cu' Cu° + Cu''in aqueous medium at 25 °C shifts towards theleft in the presence of−−(A) NO 3(B) Cl(C) SCN – (D) CN –Ans. [B, C, D]Sol. Cu 2 Cl 2 , Cu 2 (CN) 2 and Cu 2 (SCN) 2 are stable12. The correct functional group X and the reagent /reaction conditions Y in the following scheme areX– (CH 2 ) 4 –X (i) Y(ii) OOC–(CH 2 ) 4 –CHOOHHeatXtraEdge for IIT-JEE 82 MAY 2011

(A) X = COOCH 3 , Y = H 2 /Ni/Heat(B) X = CONH 2 , Y = H 2 / Ni / heat(C) X = CONH 2 , Y = Br 2 / NaOH(D) X = CN, Y = H 2 / Ni / heatAns. [A, B, C, D]Sol. FactualSECTION – IIIInteger Answer TypeThis section contains 6 Question. The answer to each ofthe question is a single-digit integer, ranging from0 to 9. The total bubble corresponding answer it to bedarkened in the ORS.13. In 1 L saturated solution of AgCl [K sp (AgCl) =1.6 × 10 –10 ], 0.1 mol of CuCl [K sp (CuCl) = 1.0 ×10 –6 ]is added. The resultant concentration of Ag + inthe solution is 1.6 × 10 –x . The value of "x" isAns. [7]Sol.+ K1[Ag ] =K + KQ K 1 < < K 21∴ K1 + K 2 ≅ K 2∴ [Agx = 7+2−101.6×10] = = 1.6 × 10 –7−61.0×1014. The maximum number of isomers (includingstereoisomers) that are possible onmonochlorination of the following compound, isCH 3CCH 3 CH 2 CH 2 CH 3HAns. [8]Sol. CH 3 –CH 2 –CH –CH 2 –CH 3ororCl 2 / hνCH 3*Cl–CH 2 –CH 2 –CH–CH 2 – CH 3CH 3(2)* *CH 3 –CH –CH–CH 2 – CH 3Cl CH 3(4)ClCH 3 –CH 2 –C–CH 2 – CH 3CH 3(1)orCH 3 –CH 2 –CH–CH 2 – CH 3(1)CH 2 Cl15. Among the following the number of compoundsthan can react with PCl 5 to give POCl 3 isO 2 , CO 2 , SO 2 , H 2 O, H 2 SO 4 , P 4 O 10Ans. [4]Sol.PCl 5 + H 2 O ⎯→ POCl 3 + 2HClPCl 5 + H 2 SO 4 ⎯→ POCl 3 + H 2 O + SO 2 Cl 26PCl 5 + P 4 O 10 ⎯→ 10POCl 3PCl 5 + SO 2 ⎯→ POCl 3 + SO 2 Cl 216. The number of hexagonal faces that are present ina truncated octahedron isAns. [8]Sol.8 Hexagonal faces17. The volume (in mL) of 0.1 M AgNO 3 required forcomplete precipitation of chloride ions present in30 mL of 0.01 M solution of [Cr(H 2 O) 5 Cl]Cl 2 , assilver chloride is close toAns. [6]Sol. 0.1 V = 30 × 0.01 × 20.3×2V = = 6 ml0.118. The total number of contributing structuresshowing hyper conjugation (involving C-Hbonds) for the following carbocation isH 3 C ⊕CH 2 CH 3Ans. [6]Sol. 6 (α – H → 6)XtraEdge for IIT-JEE 83 MAY 2011

SECTION – IVMatrix match TypeThis section contain 2 questions. Each question hasfour statements (A, B, C and D) given in column I andfive statements (p, q, r, s and t) in column II. Any givenstatement in column I can have correct matching withONE or MORE statement (s) given in Column II. Forexample, if for a given question, statement B matcheswith the statements given in q and r, then for theparticular question, against statement B, darken thebubbles corresponding to q and r in the ORS19. Match the transformation in column I withappropriate options in column IIColumn-I(A) CO 2 (s) → CO 2 (g)(B) CaCO 3 (s) → CaO(s) + CO 2 (g)(C) 2H• → H 2(g)(D) P (white solid) → P (red, solid)Column-II(p) phase transition(q) allotropic change(r) ∆H is positive(s) ∆S is positive(t) ∆S is negativeAns. [A → p, r, s; B → r, s; C → t ; D → p, q, t]Sol. [A] CO 2 (s) → CO 2 (g)p, r, s[B] CaCO 3 (s) → CaO (s) + CO 2 (g)r, s[C] 2H•→ H 2(g)t[D] P white → P redp, q, t20. Match the reactions in column I with appropriatetype of steps/reactive intermediate involved inthese reactions as given in column IIColumn-I(A)H 3 CO OO(B)Oaq NaOHCH 2 CH 2 CH 2 ClCH 3 MgIOCH(C)(D)O18CH 2 CH 2 CH 2 OHCH 2 CH 2 CH 2 C(CH 3 ) 2OHH 2 SO 4H 2 SO 4H 3 C CH 3Column-II(p) Nucleophilic substitution(q) Electrophilic substitution(r) Dehydration(s) Nucleophilic addition(t) CarbanionAns. [A → r, t, s; B → p, s, t; C → r, s ; D → r, q]Sol. FactualPHYSICSSECTION – ISingle Correct Choice TypeThis section contains 8 multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.21. A light ray travelling in glass medium is incidenton glass-air interface at an angle of incidence θ.The reflected (R) and transmitted (T) intensities,both as function of θ, are plotted. The correctsketch is :(A)(B)100%Intensity100%Intensity00TRTRθ 90ºθ 90º18OXtraEdge for IIT-JEE 84 MAY 2011

Ans.(C)(C)[C]100%Intensity0100%Intensity0TRTRθ 90ºθ 90º23. The density of solid ball is to be determined in anexperiment. The diameter of the ball is measuredwith a screw gauge, whose pitch is 0.5 mm andthere are 50 divisions on the circular scale. Thereading on the main scale is 2.5 mm and that onthe circular scale is 20 division. If the measuredmass of the ball has a relative error of 2%, therelative percentage error in the density is -(A) 0.9% (B) 2.4%(C) 3.1% (D) 4.2%Ans. [C]Pitch = 0.5 mmdivisions on the = 50circular scale0. 5⇒ least count of screw gauge = = 0.01 50θθWhen θ > θ C , no ray will transmit⇒ T = 0, T + R = 100 % and R > 0Tθ ce22. A wooden block performs SHM on a frictionlesssurface with frequency, v 0 . The block carries acharge + Q on its surface. If now a uniformelectric field E r is switched-on as shown, then theSHM of the block will beAns.E r +Q(A) of the same frequency and with shifted meanposition.(B) of the same frequency and with the samemean position(C) of changed frequency and with shifted meanposition(D) of changed frequency and with the samemean position[A]In order to have net force zero, the mean positionwill be shifted towards right but the time periodwill remain unaffected.main scale, reading = 2.5 mmcircular scale reading = 20⇒ reading = 2.5 mm + (20 × 0.01) mm= 2.5 mm + 0.2 mm = 2.7 mmmρ =34π ⎡D⎤3 ⎢ 2 ⎥⎣ ⎦∆ ρρ%error =∆ m ∆D= + 3m D∆ ρ × 100ρ⎛ 0.01⎞= 2 % + 3 ⎜ ⎟ × 100 = 3.1.⎝ 2.7 ⎠24. A ball of mass 0.2 kg rests on a vertical post ofheight 5m. A bullet of mass 0.01 kg. travellingwith a velocity V m/s in a horizontal direction,hits the centre of the ball. After the collision, theball and bullet travel independently. The ball hitsthe ground at a distance of 20 m and the bullet ata distance of 100 m from the foot of the post. Theinitial velocity V of the bullet isV m/sH = 5m020 100XtraEdge for IIT-JEE 85 MAY 2011

Ans.[D]m 1 = 0.01 kgvH = 5mm 2 = 0.2 kgAns.[B](A)(C)3π2A ,4(B) A,5π3A ,6(D)A4π3A , π320 mBallBulletθ30º120ºA100 m2HT = = 1 secgLet v 1 & v 2 be velocity of bullet & ballrespectively just after collision.v 2 × 1 = 20 ⇒ v 2 = 20& v 1 = 100From conservation of momentum0.01 × v = (0.01 × 100) + (0.2 × 20)0.01 v = 1 + 4 = 55v = = 500 m/sec.−21025. Which of the field patterns given below is validfor electric field as well as for magnetic field ?(A)(B)BHere φ = π + θA cos 30º = B sin θ ⇒ B sin θ =30º + B cosθ = A ⇒ B cos θ = 2A3A2and A sinSolving above, B = A and θ = 60º = 3π .Hence φ = 240º =4π327. A long insulated copper wire is closely wound asspiral of ‘N’ turns. The spiral has inner radius ‘a’and outer radius ‘b’. The spiral lies in the X-Yplane and a steady current ‘I’ flows through thewire. The Z-component of the magnetic field atthe centre of the spiral is -y(C)(D)IabxAns.[C] Electric lines of force for induced electric field isclosed loop.26. A point mass is subjected to two simultaneoussinusoidal displacements in x-direction, x 1 (t) = A⎛ 2π⎞sin ωt and x 2 (t) = A sin ⎜ωt+ ⎟ .Adding a third⎝ 3 ⎠sinusoidal displacement x 3 (t) = B sin(ωt + φ)brings the mass to a complete rest. The values ofB and φ are :(A)(C)µ 0NI⎛ b ⎞ln⎜⎟2( b − a)⎝ a ⎠µ 0NI ⎛ b ⎞ ln ⎜ ⎟2b⎝ a ⎠(B)(D)µ 0NI⎛ b + a ⎞ln⎜⎟2( b − a)⎝ b − a ⎠µ 0NI⎛ b + a ⎞ ln ⎜ ⎟2b⎝ b − a ⎠XtraEdge for IIT-JEE 86 MAY 2011

SECTION – IIy⇒⎛ GM ⎞Ans.[C,D]K = m ⎜ e⎟ = mv 2⎝ r ⎠Multiple correct Choice TypedrThis section contains 4 multiple choice questions. EachAns.[A]question has four choices (A), (B), (C) and (D), out ofwhich ONE OR MORE may be correct.a rx29. Two solid spheres A and B of equal volumes butof different densities d A and d B are connected by astring. They are fully immersed in a fluid ofbdensity d F . They get arranged into an equilibriumstate as shown in the figure with a tension in thestring. The arrangement is possible only if –NNo. of turns per unit thickness =b − aµmagnetic field at centre due to element = 0 (dN)iA2rµdB = 0 i ⎛ N ⎞⎜ ⎟ dr2r ⎝ b − a ⎠bBµB =0iN drµ2(b − a) ∫ =0iNln 32(b − a)a(A) d A < d F(B) d B > d F(C) d A > d F(D) d A + d B = 2d F28. A satellite is moving with a constant speed ‘V’ in Ans.[A,B,D]a circular orbit about the earth. An object of mass‘m’ is ejected from the satellite such that it justVd f gescapes from the gravitational pull of the earth. AtAthe time of its ejection, the kinetic energy of theobject is -Vd A g1 2(A) mV (B) mV2Vd f g2B3 2(C) mV (D) 2mV2Vd B g2system will be in equilibrium with tension inAns.[B]string only if d f > d A and d B > d f . If both A & BV M•are considered as a system then2Vd f g = V (d A + d b )gr⇒ d A + d B = 2d f30. Which of the following statement(s) is/arecorrect?(A) If the electric field due to a point chargevaries as r –2.5 instead of r –2 , then the Gaussmv 2 GmM law will still be valid.⇒= er2(B) The Gauss law can be used to calculate therfield distribution around an electric dipole.GM⇒ r = e...(1)(C) If the electric field between two point charges2Vis zero somewhere, then the sign of the twoIf K.E. of mass m = was k then fromcharges is the same.(D) The work done by the external force inGmME = K – e moving a unit positive charge from point A at= 0rpotential V A to point B at potential V B is (V B– V A ).XtraEdge for IIT-JEE 87 MAY 2011

31. A series R-C circuit is connected to AC voltagesource. Consider two cases ; (A) when C iswithout a dielectric medium and (B) when C isfilled with dielectric of constant 4. The current I Rthrough the resistor and voltage V C across thecapacitor are compared in the two cases. Whichof the following is/are true -A RA R(A) I > I(B) I < IAns.[B,C]RACRBC(C) V > V(D) V < VRCRACRBCSECTION – IIIInteger Answer TypeThis section contains 6 questions. The answer to each ofthe questions is a single-digit integer, ranging from 0 to9. The bubble corresponding to the correct answer is tobe darkened in the ORS33. Two batteries of different emfs and differentinternal resistances are connected as shown. Thevoltage across AB in volts is -6V 1ΩABZ1=R2~⎛ 1 ⎞+ ⎜ ⎟⎝ ωC⎠R24CAns.[5]V A – V B =3V6 3+1 21 1+1 2=2Ω6 +1.5=1.57.51.5= 5V~22 ⎛ 1 ⎞Z2= R + ⎜ ⎟⎝ 4ωC⎠z 1 > z 2 ∴AA I KVC=ωC; VARI < IBCBRBI K= ;4ωCBCV < V32. A thin ring of mass 2 kg and radius 0.5 m isrolling without slipping on a horizontal plane withvelocity 1 m/s. A small ball of mass 0.1 kg,moving with velocity 20 m/s in the oppositedirection, hits the ring at a height of 0.75 m andgoes vertically up with velocity 10 m/s.Immediately after the collision.20 m/s10 m/sAC34. A series R-C combination is connected to an ACvoltage of angular frequency ω = 500 radian/s. Ifthe impendence of the R-C circuit is R 1. 25 , thetime constant (in millisecond) of the circuit is.Ans. [4]Z = R 1 . 25τ = RC2R 2 ⎛ 1 ⎞+ ⎜ ⎟⎝ 500C⎠= Z2R 2 ⎛ 1 ⎞+ ⎜ ⎟⎝ 500C ⎠= R2 × 1. 2522⎛ 1 ⎞⎜ ⎟⎝ 500C ⎠= 0.25 R2 ⇒12500= RC1500C= 0.5RAns.0.75 m1 m/s(A) the ring has pure rotation about its stationaryC.M(B) the ring comes to a complete stop(C) friction between the ring and the ground is tothe left(D) there is no friction between the ring and theground[A,C]1 = RC2500.004 sec = RCRC = 4 mill second.35. A train is moving along a straight line with aconstant acceleration ‘a’. A boy standing in thetrain throws a ball forward with a speed of 10m/s, at an angle of 60º to the horizontal. The boyhas to move forward by 1.15 m inside the train tocatch the ball back at the initial height. Theacceleration of the train, in m/s 2 , is.XtraEdge for IIT-JEE 88 MAY 2011

Ans. [5]2×10×3T == 3 sec2×10x = 10 cos 60° (T) = 5 3 mIn frame of train,5 3 = 21 × a × ( 3 ) 2 + 1.15(a : acceleration of train)a = 5 m/sec 236. Water (with refractive index = 4/3) in a tank is 18cm deep. Oil of refractive index 7/4 lies on watermaking a convex surface of radius of curvature ‘R= 6 cm’ as shown. Consider oil to act as a thinlens. An object ‘S’ is placed 24 cm above watersurface. The location of its image is at ‘x’ cmabove the bottom of the tank. Then ‘x’ is –SAns.[2]R = 6 cmµ = 1.0µ = 7/4µ = 4/3µ 3 µ 1 µ 2 − µ 1 µ 3 − µ 2− = +v u R1R27 4 7 − 1 −4 1 – = 43V − 24 6+ 3 4∞4 1 1 4 1⇒ + = ⇒ =3V 24 8 3V12⇒ V = 16 cm∴ Ans. = (18 –16) cm = 2 cm37. A block of mass 0.18 kg is attached to a spring offorce-constant 2 N/m. The coefficient of frictionbetween the block and the floor is 0.1. Initiallythe block is at rest and the spring is un-stretched.An impulse is given to the block as shown in thefigure. The block slides a distance of 0.06 m andcomes to rest for the first time. The initialvelocity of the block in m/s is V = N/10. Then NisAns.[4]m = 0.18 kgk = 2N/mµ = 0.1Using W – E theorem1 × m(u) 2 1= K (x) 2 + µmg (x)2 21 × (0.18) u 2 1= × 2 × 36 × 10 –4 + 0.1 × 0.1822× 10 × 0.06⇒ u = 0.4 m/sec.4⇒ m/sec.1038. A silver sphere of radius 1 cm and work function4.7 eV is suspended from an insulating thread infree-space. It is under continuous illumination of200 nm wavelength light. As photoelectrons areemitted, the sphere gets charged and acquires apotential. The maximum number ofphotoelectrons emitted from the sphere is A × 10 z(where 1 < A< 10). The value of ‘Z’ is.Ans. [7]1240Energy of photon ≈ eV = 6.2 eV200Maximum KE of a electron = 6.2 eV – 4.7 eVWhen potential on surface of sphere becomesequal to 1.5Vq4π∈ 0= 1.5 V ⇒ q = 1.5 × (4π ε 0 ) × rr1.5 × (4πε0)rNo. of photoelectron emitted n =−191.6 × 10= 1. 04 × 10 7SECTION – IVMatrix match TypeThis section contains 2 questions. Each question hasfour statements (A, B, C and D) given in Column I andfive statements (p, q, r, s and t) in Column II. Anygiven statement Column I can have correct matchingwith ONE or MORE statement(s) given in Column II.For example, if for a given question, statement Bmatches with the statements given in q and r, then forthe particular question, against statement B, darkenthe bubbles corresponds to q and r in the ORS.39. One mole of a monoatomic ideal gas is takenthrough a cycle ABCDA as shown in the P-Vdiagram.Column-II gives the characteristics involved inthe cycle. Match them with each of the processesgiven in Column-I.XtraEdge for IIT-JEE 89 MAY 2011

P3PBAColumn-I(A) Pipe closed at one end0 L1PCD(B) Pipe open at both ends0 L(C) Stretched wire clamped at both endsAns.0 1V 3V 9V VColumn-I Column-II(A) Process A → B (p) Internal energy decreases(B) Process B → C (q) Internal energy increases(C) Process C → D (r) Heat is lost(D) Process D → A (s) Heat is gained(t) Work is done on the gas(A) → p,r,t; (B) → p,r; (C) → q,s; (D) → r,tProcess AB : (Pressure is constant)If T A = T ⇒ T B = 3TSo ∆U = Negative [Q ∆U = nC v ∆T]∆W = nR∆T = Negative∆Q = ∆U + ∆W = NegativeProcess BC : (Volume is constant)If T B = 3T then TC = 9T∆U = nC v ∆T = Negative∆W = Zero∆Q = NegativeProcess C → D : (Pressure is constant)If T C = 9T then TD = T∆U = nC v ∆T = positive∆W = positive∆Q = positiveProcess D → A :T D = T and T A = THence process is isothermal∆U = 0∆W = negative∆Q = negative0 L(D) Stretched wire clamped at both ends and atmid-pointColumn-II0 LL/2(p) Longitudinal waves(q) Transverse waves(r) λ 1 = L(s) λ f = 2L(t) λ f = 4LAns. (A) → p,t; (B) → p,s; (C) → q,s; (D) → q,r(A)λf= L4⇒ λ f = 4L(B) Longitudinal wavesλ f = L2(C) Stretched wire clamped at both endsλ f2= L ⇒ λf = 2L40. Column-I shows four systems, each of the samelength L, for producing standing waves. Thelowest possible natural frequency of a system iscalled its fundamental frequency, whosewavelength is denoted as λ 1 . Match each systemwith statements given in Column-II describingthe nature and waves.(D)λ f λ + f = L2 2⇒ λ f = LXtraEdge for IIT-JEE 90 MAY 2011

MATHEMATICSSECTION – ISingle Correct Choice TypeThis section contains 8 multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.41. Let P(6, 3) be a point on the hyperbola2 2x y− = 1. If the normal at the point P2 b2aintersects the x-axis at (9, 0), then the eccentricityof the hyperbola is -Ans.Sol.(A)52(B)32(C) 2 (D) 3[B]Equation of the normal at (6, 3) isa 2 x b 2 y+ = a 2 + b 26 3it passes through (9, 0)9a 2so = a 2 + b 262a⇒ b 2 =2Now b 2 = a 2 (e 2 –1)∴ e 2 1–1 = 2e 2 = 23⇒ e =42. Let (x, y) be any point on the parabola y 2 = 4x. LetP be the point that divides the line segment from(0, 0) to (x, y) in the ratio 1 : 3. Then the locus ofP is(A) x 2 = y(B) y 2 = 2x(C) y 2 = x(D) x 2 = 2yAns. [C]2tSol. h = 42t, k = 4(0, 0)1323• P(h, k)(t 2 , 2t)t 2 = 4h, t = 2kso 4k 2 = 4h∴ k 2 = hhence required locus is y 2 = x43. Let f(x) = x 2 and g(x) = sin x for all x∈R. Then the setof all x satisfying (f o g o g o f) (x) = (g o g o f) (x),where (f o g) (x) = f(g(x)), is(A) ± n π, n ∈{0, 1, 2...}(B) ± n π, n ∈{1, 2...}π(C) + 2nπ,n ∈{....,− 2, −1,0, 1, 2, ....}2(D) 2nπ, n∈{…., –2, –1, 0, 1, 2, ….}Ans. [A]Sol. gof(x) = gf(x) = g(x 2 ) = sin x 2go (gof(x)) = g(sin x 2 ) = sin (sin x 2 )fo(gogof(x)) = f(sin (sin x 2 )) = (sin(sin x 2 )) 2∴ (sin (sin x 2 )) 2 = sin (sin x 2 )sin (sin x 2 ) (sin (sin x 2 ) –1) = 0sin (sin x 2 ) = 0 or sin (sin x 2 ) = 1sin x 2 = nπsin x 2 π= 2nπ + 2At n = 0 At n = 0sin x 2 = 0sin x 2 π= 2x 2 = nπx = ±n πNot possible; n ∈ {0, 1, 2, ….}44. Let f : [–1, 2] → [0, ∞) be a continuous functionsuch that f(x) = f(1 – x) for all x∈[–1, 2]. LetAns.Sol.R21 =∫x f x)−1( dx , and R 2 be the area of the regionbounded by y = f(x), x = – 1, x = 2, and the x-axis.Then(A) R 1 = 2R 2 (B) R 1 = 3R 2(C) 2R 1 = R 2 (D) 3R 1 = R 2[C]2R 1 =∫x f ( x)dx… (i)2−1R 1 =∫( 1−x ) f (1 − x)dx =∫( 1−x ) f ( x)dx ...(ii)−1(i) + (ii)22R 1 =∫−1f ( x)dx = R 22−1∴ 2R 1 = R 2XtraEdge for IIT-JEE 91 MAY 2011

21/ x45. If lim [1 + xl n(1 + b )] = 2b sin 2 θ, b > 0 andAns.Sol.x→0θ∈(– π, π], then the value of θ is -(A)(C)π± (B)4π± (D)6π±3π±2[D]lim (1 + x ln(1 + b 2 )] 1/x = 2b sin 2 θ b > 0;x →0ln(1+b )1 ⎞22 x ln(1+b ) ⎟)]x⎟⎛⎜lim [1 x n (1 b→0⎜ + l +⎝2ln(1+b )e = 2b sin 2 θ1 + b 2 = 2b sin 2 θ2 sin 2 1θ = b + b⎟⎠RHS = b + b1 ≥ 2 as b > 0But LHS = 2 sin 2 θ ≤ 2Only possibility2 sin 2 θ = 2sin 2 θ = 1θ = ± 2π2θ ∈ (–π, π)= 2b sin 2 θ46. The circle passing through the point (–1, 0) andtouching the y-axis at (0, 2) also passes throughthe point -⎛ 3 ⎞(A) ⎜ − , 0 ⎟⎝ 2 ⎠⎞(C) ⎜⎛ 3 5 − , ⎟⎝ 2 2 ⎠⎛ 5 ⎞(B) ⎜ − , 2 ⎟⎝ 2 ⎠(D) (– 4, 0)Ans. [D]Sol. ∴ (h – 0) 2 + (2 –2) 2 = (h + 1) 2 + (2 – 0) 2h 2 = h 2 + 1 + 2h + 4(h, 2)(–1, 0)5h = −2Equation of circle is(0, 2)2⎛ 5 ⎞⎜ x + ⎟⎠ + (y –2) 2 ⎛ 5 ⎞= ⎜ – − 0⎟ ⎝ 2⎝ 2 ⎠x 2 +25 + 5x + y 2 + 4 – 4y =42254x 2 + y 2 + 5x – 4y + 4 = 0from given points only point (– 4, 0) satisfies thisequation.47. Let ω ≠ 1 be a cube root of unity and S be the setof all non-singular matrices of the form⎡ 1 a b⎤⎢ ⎥⎢ω 1 c⎥, where each of a, b and c is either⎢2⎣ωω 1⎥⎦ω or ω 2 . Then the number of distinct matrices inthe set S is :(A) 2 (B) 6(C) 4 (D) 8Ans. [A]Sol.1ωω2a1ωbc1≠ 0(1 – ωc) – a (ω – ω 2 c) + b(ω 2 – ω 2 ) ≠ 01 – ωc – aω + acω 2 ≠ 0(1 – ωc) – aω (1 – ωc) ≠ 0(1 – ωc) (1 – aω) ≠ 0c ≠ ω 2 & a ≠ ω 2 & b = ω or ω 2(a, b, c) ≡ (ω, ω, ω) or (ω, ω 2 , ω)48. A value of b for which the equationsx 2 + bx – 1 = 0, x 2 + x + b = 0have one root in common is -(A) − 2(B) − i 3(C) i 5(D) 2Ans. [B]Sol. x 2 + bx –1 = 0 … (i)x 2 + x + b = 0… (ii)b + 1(i) – (ii) we get x =b – 1Put this value in (i)2⎛ b + 1⎞⎛ b + 1⎞⎜ ⎟⎠ + b ⎜ ⎟ –1 = 0⎝ b –1 ⎝ b –1 ⎠⇒ b 3 + 3b = 0⇒ b(b 2 + 3) = 0⇒ b = 0 or b = ± i 3XtraEdge for IIT-JEE 92 MAY 2011

SECTION – IIMultiple correct Choice TypeThis section contains 4 multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONE OR MORE may be correct.⎧ π π⎪−x − , x ≤ −2 2⎪π49. If f ( x)= ⎨ − cos x,− < x ≤ 0 , then⎪2⎪x −10 < x ≤1⎪⎩ln x,x > 1π(A) f(x) is continuous at x = −2(B) f(x) is not differentiable at x = 0(C) f(x) is differentiable at x = 13(D) f(x) is differentiable at x = −2Ans. [A, B, C, D]πSol. At x = −2⎛ π ⎞LHL = 0, RHL = 0, f ⎜− ⎟ = 0, So f(x) is⎝ 2 ⎠πcontinuous at x = −2At x = 0LHD = 0; RHD = 1So f(x) is not differentiable at x = 0At x = 1LHD = 1, RHD = 1So f(x) is differentiable at x = 1⎛ π ⎤in ⎜− , 0⎥ ; f(x) = – cos x⎝ 2 ⎦so f(x) is differentiable at x =3−250. Let L be a normal to the parabola y 2 = 4x. If Lpasses through the point (9, 6), then L is given by(A) y – x + 3 = 0 (B) y + 3x – 33 = 0(C) y + x – 15 = 0 (D) y – 2x + 12 = 0Ans. [A, B, D]Sol. y = mx – 2m – m 3It passes through (9, 6)6 = 9m –2m – m 3m 3 – 7m + 6 = 0(m –1) (m –2) (m + 3) = 0∴ m = –3, 1, 2Hence equations will bey = x – 3, y = 2x –12 and y = –3x + 3351. Let E and F be two independent events. Theprobability that exactly one of them occurs is11/25 and the probability of none of themoccurring is 2/25. If P(T) denotes the probabilityof occurrence of the event T, then -4 3(A) P ( E)= , P(F)=5 51 2(B) P ( E)= , P(F)=5 52 1(C) P ( E)= , P(F)=5 53 4(D) P ( E)= , P(F)=5 5Ans. [A, D]11Sol. P(E) (1 – P(F)) + (1 – P(E)) P(F) = 2511P(E) + P(F) –2P (E) P(F) = 252(1 – P(E)) (1 – P(F)) = 2521 – P(E) – P(F) + P(E) P(F) = 2523P(E) + P(F) – P(E) P(F) = 25From (1) & (2)12P(E) P(F) = 257and P(E) + P(F) = 5so eitherP(E) = 54 , P(F) = 53 and P(E) = 53 , P(F) = 54… (1)... (2)b − x52. Let f : (0, 1) → R be defined by f ( x)= , 1 − bxwhere b is a constant such that 0 < b < 1. Then(A) f is not invertible on (0, 1)(B) f ≠ f –1 1on (0, 1) & f '(b) =f '(0)(C) f = f –1 1on (0, 1) and f '(b) =f '(0)(D) f –1 is differentiable on (0, 1)Ans. [A, B]Sol. f : (0, 1) → Rb − xf(x) = ∀ b ∈ (0, 1)1−bx2b −1f ′(x) = = (–) ve2(1 − bx)XtraEdge for IIT-JEE 93 MAY 2011

So f(x) is monotonically decreasing for x ∈ (0, 1)so for x ∈ (0, 1)f(x) ∈ (f(1), f(0))f(x) ∈ (–1, b)so f(x) is not onto.so f(x) is not invertible function.SECTION – IIIInteger Answer TypeThis section contains 6 questions. The answer to each ofthe questions is a single-digit integer, ranging from 0 to9. The bubble corresponding to the correct answer is tobe darkened in the ORS53. Let y'(x) + y(x)g'(x) = g(x)g'(x), y(0) = 0, x∈R,d f ( x)where f '(x) denotes and g(x) is a givendxnon-constant differentiable function on R withg(0) = g(2) = 0. Then the value of y(2) is.Ans. [0]dySol. + y = gdgI. F. =∫ 1 . dg = gy.e g =∫ge g . dg = ge g –∫e g . dgye g = ge g – e g + cy = g –1 + ce –gQ y(0) = 0 & g(0) = 0at x = 00 = 0 –1 + Ce –0C = 1y = g –1 + e –gat x = 2y(2) = 0 – 1 + e –0 = 0r r54. Let a = −ˆ i − kˆ, b = −ˆ i + ˆrj and c = iˆ + 2 ˆj+ 3kˆbethree given vectors. If r is a vector such thatr r r r× b = c × b and r . arr= 0 , then the value of br.is.Ans. [9]Sol. a r = – î – kˆ , b r = – î + ĵ , c r = î + 2 ĵ + 3 kˆr r r r r r r r r(− c)× b = 0 ⇒− c = λb⇒= c + λbQ r ⋅ ar = 0r r r r⇒ a. c + λb.a = 0r ra.c⇒ λ = − r r = 4b.ar r r r r⇒ . b = c.b + λ | b |2 = 955. Let ω = e iπ/3 , and a, b, c, x, y, z be non-zerocomplex numbers such that a + b + c = x,a + bω + cω 2 = y, a + bω 2 + cω = z. Then theAns. [ * ]value of| x |22| a |2+ | y |2+ | b |+ | z2|2+ | c |i 2π/3Sol. wrong question if ω = e then ans is 3.i π/3If ω = e then no integral solution is possible.⎡0⎤⎡−1⎤⎢ ⎥56. Let M be a 3× 3 matrix satisfying M⎢ ⎥⎢1⎥=⎢2⎥,⎢⎣0⎥⎦⎢⎣3 ⎥⎦⎡ 1 ⎤ ⎡ 1 ⎤ ⎡1⎤⎡ 0 ⎤M⎢ ⎥=⎢ ⎥⎢−1⎥ ⎢1⎥and M⎢ ⎥ ⎢ ⎥⎢1⎥=⎢0⎥. Then the sum⎢⎣0 ⎥⎦⎢⎣−1⎥⎦⎢⎣1⎥⎦⎢⎣12⎥⎦of the diagonal entries of M is.Ans. [9]⎡a⎢Sol. Let M =⎢d⎢⎣gQ∴behc ⎤f⎥⎥i ⎥⎦is.⎡0⎤⎡−1⎤⎢ ⎥M⎢1⎢ ⎥⎥=⎢2⎥⇒ b = –1, e = 2, h = 3⎢⎣0⎥⎦⎢⎣3 ⎥⎦⎡ 1 ⎤ ⎡ 1 ⎤⎢ ⎥M⎢−1⎢ ⎥⎥=⎢1⎥⇒ a = 0, d = 3, g = 2⎢⎣0 ⎥⎦⎢⎣−1⎥⎦⎡1⎤⎡ 0 ⎤⎢ ⎥M⎢1⎢ ⎥⎥=⎢0⎥⇒ c = 1, f = –5, i = 7⎢⎣1⎥⎦⎢⎣12⎥⎦So a + e + i = 0 + 2 + 7 = 957. The number of distinct real roots ofx 4 – 4x 3 + 12x 2 + x – 1 = 0 isAns. [2]Sol. Let f(x) = x 4 – 4x 3 + 12x 2 + x – 1Let α, β, γ, δ are the root of equation.∴ αβγδ = –1 so the equation has at least tworeal roots....(i)f '(x) = 4x 3 – 12x 2 + 24x + 1f "(x) = 12x 2 – 24x + 24 = 12((x + 1) 2 + 1)so f "(x) > 0 so f '(x) = 0 has only one real rootsso f(x) = 0 has at most two real roots. ...(ii)from (i) & (ii)f(x) = 0 has exactly two real roots.XtraEdge for IIT-JEE 94 MAY 2011

58. The straight line 2x – 3y = 1 divides the circularregion x 2 + y 2 ≤ 6 into two parts. If⎧⎛3 ⎞ ⎛ 5 3 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞⎫S = ⎨⎜2, ⎟,⎜ , ⎟,⎜ , − ⎟,⎜ , ⎟⎬, then⎩⎝4 ⎠ ⎝ 2 4 ⎠ ⎝ 4 4 ⎠ ⎝ 8 4 ⎠⎭the number of point(s) in S lying inside thesmaller part is -Ans. [2]Sol.(2, 3)⎛ 1 ⎞⎜ ,0⎟⎝ 2 ⎠Pont (x 1 , y 1 ) lies inside the region if2 2x 1 + y1− 6 ≤ 0 & 2x 1 – 3y 1 – 1 ≤ 0.⎛ 3 ⎞ 9P 1 ≡ ⎜2,⎟ 4 + − 6 ≤ 0 True⎝ 4 ⎠ 1694 − −1> 0 True4⎛ 5 3 ⎞ 25 9P 2 ≡ ⎜ , ⎟ + − 6 ≤ 0 False⎝ 2 4 ⎠ 4 16⎛ 1 −1⎞1 1P 3 ≡ ⎜ , ⎟ + − 6 ≤ 0 True⎝ 4 4 ⎠ 16 162 3+ −1> 0 True4 4⎛ 1 1 ⎞ 1 1P 4 ≡ ⎜ , ⎟ + − 6 ≤ 0 True⎝ 8 4 ⎠ 64 162 3− −1> 0 False8 4So P 1 & P 3 lies in the intervalSECTION – IVMatrix match TypeThis section contains 2 questions. Each question hasfour statements (A, B, C and D) given in Column I andfive statements (p, q, r, s and t) in Column II. Anygiven statement Column I can have correct matchingwith ONE or MORE statement(s) given in Column II.For example, if for a given question, statement Bmatches with the statements given in q and r, then forthe particular question, against statement B, darkenthe bubbles corresponds to q and r in the ORS.59. Match the statements given in Column-I with thevalues given in Column-II.Column-Irr(A) If a = ˆ j + 3kˆ, b = − ˆ j + 3kˆrand c = 2 3kˆform a triangle, then the internal angle of thetriangle between a r and b r isb2 2(B) If∫( f ( x)− 3x)dx = a − b , then the valueofa⎛ π ⎞f ⎜ ⎟ is⎝ 6 ⎠5/ 6π(C) The value of∫sec(πx)dx isln327 / 6(D) The maximum value of|z| = 1, z ≠ 1 is given byColumn-I(p) π/6(q) 2π/3(r) (r) π/3(s) π(t) π/2Ans. [A → q ; B → p ; C → s ; D → s]Sol. (A)(B)c rbrθa rr r− a ⋅ b 1cos θ = r r = −| −a|| b | 2b∫a2⎛ 1 ⎞Arg⎜⎟ for⎝ 1 − z ⎠⇒( f ( x)− 3( x))dx = a − bdifferentiating w.r.t (b).f(b) – 3b = – 2bf ( b)= b⎛ π ⎞ πSo f ⎜ ⎟ =⎝ 6 ⎠ 622πθ =3XtraEdge for IIT-JEE 95 MAY 2011

(C)25 / 6πI = x dxn ∫sec πl 327 / 6π5 / 6I = ln| sec πx+ tan πx|πln37 / 6πI = ⋅ ln3= πln3(D) ∴ |z| = 1z = cos θ + i sin θ. ∀ θ∈ (– π. π] and θ ≠ 0.1Arg(1 − z)⎛ θ ⎞⎜ icot⎟⎛ 1 ⎞⎜1= Arg ⎜⎟ = Arg + 2 ⎟⎝1− cosθ − isinθ ⎠ ⎜ 2 2 ⎟⎝ ⎠=π − θ2so maximum value is π.60. Match the statements given in Column-I with theintervals/union of intervals given in Column-II.Column-I(A) The set⎧ ⎛ 2iz ⎞⎫⎪Re⎜⎟;z is a complex number,2⎪⎨ ⎝1− z ⎠⎬ is⎪⎪⎩| z | = 1, z ≠ ± 1 ⎭(B) The domain of the function⎛ x−2⎞−1⎜8(3)f ( x)= sin⎟ is2( x−1)⎝1−3 ⎠1(C) If f ( θ)= − tan θ−1tan θ1− tan θ1tan θ , then the⎧π⎫set ⎨ f ( θ) : 0 ≤ θ < ⎬ is⎩2 ⎭3/ 2(D) If f ( x)= x (3x−10),x ≥ 0 , then f(x) isincreasing inColumn-II(p) (– ∞, –1) ∪ (1, ∞)(q) (– ∞, 0) ∪ (0, ∞)(r) [2, ∞)(s) ( – ∞, –1] ∪ [1, ∞)(t) (– ∞, 0] ∪ [2, ∞)Ans. [A → p, r, s ; B → r, t ; C → r ; D → r]1Sol.(A) Let z = cos θ + i sin θ2iz2i(cos+isinθ)so == −cosecθ21−z 1−cos2θ − isin 2θ∀ θ ≠ (2n + 1) 2π⎛ 2iz ⎞so Re⎜⎟ = − cos ec θ∈(−∞,−1]∪ [1, ∞)2⎝1− z ⎠x−28×3 8×3(B) =2x−22x1−3 9 − 3Let 3 x = t⎛So f(x) = sin –1 ⎜8 3⎝ 9 − 3x⎞⎟ = sin⎠×x−12x8t−1≤ 129 − t≤ on solvingx ∈ (– ∞, 0] ∪ [2, ∞) ∪ {1}(C) f(θ) = 2 sec 2 θso f(θ) ∈[2, ∞)(D) f(x) = 3x 5/2 – 10x 3/215 xf'(x) = ( x − 2)2So f(x) is increasing for f '(x) ≥ 0x ∈ [2, ∞)Random Facts⎛ 8t⎜⎝ 9 − t• As a gas' temperature is raised to over10,000°, its molecules collide so violentlythat they are broken apart into individualatoms.• When the tsunami reaches the coast andmoves inland, the water level can risemany meters. In extreme cases, water levelhas risen to more than 15 m (50 ft) fortsunamis of distant origin and over 30 m(100 ft) for tsunami waves generated nearthe earthquake’s epicenter.• Some minerals, notably quartz, arepiezoelectric--that is, they produceelectricity when subjected to pressure orstress. This same phenomenon is probablyalso responsible for "earthquake lights,"the luminescence sometimes reported (and,on occasion, photographed) in the skyduring earthquakes.2⎟⎠⎞XtraEdge for IIT-JEE 96 MAY 2011

XtraEdge Test SeriesANSWER KEYIIT- JEE 2012 (May issue)PHYSICSQues 1 2 3 4 5 6 7 8Ans D C C B A B A A,B,CQues 9 10 11 12 13 14 15 16Ans B,C B,C A,B,C,D B C C B CNumerical Ques 17 18 19 20 21 22 23Response Ans 3 1 2 1 2 3 4CHEMISTRYQues 1 2 3 4 5 6 7 8Ans C B D D B A C B,CQues 9 10 11 12 13 14 15 16Ans B,D B,D D C D B D BNumerical Ques 17 18 19 20 21 22 23Response Ans 1 6 7 2 8 4 6MATHEMATICSQues 1 2 3 4 5 6 7 8Ans A A D A C C A B,DQues 9 10 11 12 13 14 15 16Ans A,B,C A,B,C,D B,D D C A A ANumerical Ques 17 18 19 20 21 22 23Response Ans 9 0 2 5 1 1 0IIT- JEE 2013 (May issue)PHYSICSQues 1 2 3 4 5 6 7 8Ans B A A D A B A A,B,C,DQues 9 10 11 12 13 14 15 16Ans A,B,C A,B,D A,C,D C B C B BNumerical Ques 17 18 19 20 21 22 23Response Ans 6 9 1 5 6 4 5CHEMISTRYQues 1 2 3 4 5 6 7 8Ans A B D B A C D A,B,CQues 9 10 11 12 13 14 15 16Ans A,B,C A,D A,C,D A B A B ANumerical Ques 17 18 19 20 21 22 23Response Ans 3 1 1 5 0 2 3MATHEMATICSQues 1 2 3 4 5 6 7 8Ans A C C C D A C A,B,DQues 9 10 11 12 13 14 15 16Ans C,D A,C A,B,C,D C C B A BNumerical Ques 17 18 19 20 21 22 23Response Ans 5 9 3 2 6 3 2XtraEdge for IIT-JEE 97 MAY 2011

Subscription Offer for Students'XtraEdge for IIT-JEEIIT JEE becoming more competitive examination day by day.Regular change in pattern making it more challenging.C"XtraEdge for IIT JEE" magazine makes sure you're updated & at the forefront.Every month get the XtraEdge Advantage at your door step.✓ Magazine content is prepared by highly experienced faculty members on the latest trend of IIT JEE.✓ Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice.✓ Take advantage of experts' articles on concepts development and problem solving skills✓ Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda.✓ Confidence building exercises with Self Tests and success stories of IITians✓ Elevate you to the international arena with international Olympiad/Contests problems and Challenging Questions.SUBSCRIPTION FORM FOR “EXTRAEDGE FOR IIT-JEEThe Manager-Subscription,“XtraEdge for IIT-JEE”Career Point Infosystems Ltd,4 th Floor, CP-Tower,IPIA, Kota (Raj)-324005I wish to subscribe for the monthly Magazine “XtraEdge for IIT-JEE”Half Yearly Subscription (Rs. 100/-) One Year subscription (Rs. 200/-) Two year Subscription (Rs. 400/-)I am paying R. …………………….throughMoney Order (M.O)Bank Demand Draft of No………………………..Bank………………………………………………………..Dated(Note: Demand Draft should be in favour of "Career Point Infosystems Ltd" payable at Kota.)Name:Father's Name:Address:SpecialOffer_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________City_____________________________State__________________________PIN_________________________________________Ph with STD Code __________________________Class Studying in ________________E-Mail: ________________________________________________From months: ____________________to ________________________________________________CXtraEdge for IIT-JEE 98 MAY 2011

Subscription Offer for SchoolsXtraEdge for IIT-JEEIIT JEE becoming more competitive examination day by day.Regular change in pattern making it more challenging.C"XtraEdge for IIT JEE" magazine makes sure you're updated & at the forefront.Every month get the XtraEdge Advantage at your door step.✓ Magazine content is prepared by highly experienced faculty members on the latest trend of the IIT JEE.✓ Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice.✓ Take advantage of experts' articles on concepts development and problem solving skills✓ Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda.✓ Confidence building exercises with Self Tests and success stories of IITians✓ Elevate you to the international arena with international Olympiad/ Contests problems and Challenging Questions.FREE SUBSCRIPTION FORM FOR “EXTRAEDGE FOR IIT-JEEThe Manager-Subscription,“XtraEdge for IIT-JEE”Career Point Infosystems Ltd,4 th Floor, CP-Tower,IPIA, Kota (Raj)-324005CWe wish to subscribe for the monthly Magazine “XtraEdge for IIT-JEE”Half Yearly Subscription One Year subscription Two year SubscriptionInstitution Detail:Graduate Collage Senior Secondary School Higher Secondary SchoolName of the Institute: _____________________________________________________________________________Name of the Principal: _____________________________________________________________________________Mailing Address:_______________________________________________________________________________________________City_________________________State__________________________PIN_____________________Ph with STD Code_____________________________________Fax_______________________________ E-Mail_____________________________________Board/ University: _____________________________________________________________________________________School Seal with SignatureXtraEdge for IIT-JEE 99 MAY 2011

XtraEdge for IIT-JEE 100 MAY 2011