November 2011 - Career Point

Volume - 7 Issue - 5November, 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 DesignSatyanarayan SainiLayoutRajaram GocherCirculation & AdvertisementPraveen ChandnaPh 0744-3040000, 9672977502SubscriptionHimanshu Shukla Ph. 0744-3040000© 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,Everyone wants success. Some people spend their every waking moment pursuingit, to the detriment of all else. For others, attaining success seems impossible. Theyconclude that it is destined for a select few. The rest of us are to remain "contentwith such things as we have". Having it all is not "in our stars".When you strive for success with the wrong assumptions, you will never reachit. It's like traveling somewhere with the wrong map.Zig Ziglar says that, "Success is a process, not an event," "a journey, not adestination." Jim Rohn describes it as " .... a condition that must beattracted not pursued."Success is something you must work hard and long to earn, for yourself. It has aprice, sometimes a very high one. And most people aren't really and truly readyto pay that price, to do what success demands. If success has eluded you so far,perhaps you should try changing your assumptions. You need to accept that :• You must go through a growing process, which will require time andpatience, in order to achieve success. There are no short cuts. Anything elseis a temporary illusion. Success that will remain with you, and bring you joyrather than sorrow, requires a learning process, a time to grow out of oldhabits and into new ones, a time to learn what works and what doesn't. Andyou must pay your dues, in full, in advance! so don't be in a hurry.• You will need to acquire traits and skills that attract it. What does success meanto you ? Identify, in specific terms, what you regard as success. What traits orskills will you need to achieve this goal? Find 2 or 3 people who have what youwant. Write down the habits that have made them successuf and resolve tocopy them. This is called mentoring learning from others who have arrivedwhere you want to go. Once you learn to do what it takes, you qualify. Andwhen you qualify, success comes looking for you. You just can't be denied!Remember, when parents try to teach their children to crawl, what they do?They put their favorite toy in front of them and teased them forward, inch byinch. They were after the toy, which kept them motivated. When they becamegood at reaching the toy, they had learned to crawl. After that, they could reachany destination they wanted. The DESTINATION was less important. Theybecame champion crawlers in the PROCESS!When you are ready for success you attract it, with little effort. When youare not, it runs from you, no matter how hard you chase. In other words,you repel it! Most likely, this is the reason that success eludes people.Now that you know how to attract success, why not get started on the journeythat will take you where you want to go. Any one can succeed, butunfortunately not every one will. Fate does not foist it upon you. You can haveanything you want in life, if you're ready to pay the price. But if you consider theprocess too hard, too slow, or too long and lonely, you have qualified your self asa looser; painful but true.So don't short change yourself with short-cuts. Go out there today and startattracting success. It's literally yours for the taking!Presenting forever positive ideas to your success.Yours trulyPramod Maheshwari,B.Tech., IIT DelhiImpatience never commanded success.EditorialXtraEdge for IIT-JEE 1 NOVEMBER 2011

Volume-7 Issue-5November, 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 & 2013SSuccess Tips for the Months• " Always bear in mind that your ownresolution to succeed is more importantthan any other thing."• "God gave us two ends. One to sit on andone to think with. Success depends onwhich one you use; head you win -- tails,you lose."• "The ladder of success is best climbed bystepping on the rungs of opportunity."• "Success is getting what you want.Happiness is wanting what you get."• "The secret of success in life is for a man tobe ready for his opportunity when itcomes."• "I don't know the key to success, but thekey to failure is trying to pleaseeverybody."• "The secret of success is to be in harmonywith existence, to be always calm… to leteach wave of life wash us a little farther upthe shore."INDEXCONTENTSRegulars ..........PAGENEWS ARTICLE 3• Government plans to recover costs from IITstudents• HRD to hand over 10,000 low-cost laptops toIIT RajasthanIITian ON THE PATH OF SUCCESS 5Mr. Subrah S. IyerKNOW IIT-JEE 6Previous IIT-JEE QuestionStudy Time........DYNAMIC PHYSICS 148-Challenging Problems [Set # 7]Students’ ForumPhysics Fundamentals• E.M.I. & A.C.• S.H.M.CATALYSE CHEMISTRY 27Key Concept• Nitrogen Compound• Nitrogen FamilyUnderstanding :Physical ChemistryDICEY MATHS 38Mathematical ChallengesStudents’ ForumKey Concept• Differentiation• Straight Line & CircleTest Time ..........XTRAEDGE TEST SERIES 47Class XII – IIT-JEE 2012 PaperClass XI – IIT-JEE 2013 PaperXtraEdge for IIT-JEE 2 NOVEMBER 2011

Government plans to recovercosts from IIT studentsStudents of the Indian Institutes ofTechnology (IITs) are set to be undera debt burden from their first day inclass.According to a new proposal, generalcategory IIT graduates may soon have topay back the money that the governmentincurs on their education as soon as theyfind a job after passing out. The 15premier engineering institutes and thehuman resource development ministrygave their "in principle" nod to theproposal, which has been touted as aworkable alternative to hiking tuitionfees of the IITs.The reimbursed amount will go to theIIT from where a student graduates.The step is believed to be in line withthe government's efforts to give moreadministrative and financial autonomyto the IITs.The landmark decision,taken at Wednesday's meeting ofthe IIT Council, the highest decisionmakingbody of these institutes, willnot apply to students from SC, ST andOBC (noncreamy layer) categories.Nor will it apply to those going in forhigher studies at the IITs.HRD minister Kapil Sibal said theidea of giving back to the institute wasagreed upon during a discussion onthe Kakodkar Committeerecommendation which had proposedthat the IITs increase their annualtuition fee four times from Rs 50,000each year to anything between Rs 2lakh to Rs 2.5 lakh per annum."We have shot down that propositionas we do not want to burden thestudents' families. Students willcontinue to pay Rs 50,000 as theirannual fee, but they will be expectedto pay the difference between thetuition fee and the actual expenditureincurred by the institute once theystart working," he said.Sources said the payback proposal is"reasonable". IIT graduates are in greatdemand and command high salaries.Roughly half the students of an IIT arefrom the general category because 49.5per cent of the seats are reserved. Arough estimate pegs the totalexpenditure on an IIT BTech graduateover four years at Rs 6 lakh to Rs 8lakh. The student will not be liable topay the difference between the tuitionfee and the actual expenditure in casehe studies further like pursue MTech,PhD and so on.But the moment a student gets a job,irrespective of whether it is in thegovernment or the private sector, theloan meter will start ticking. It wasdecided at the meeting that studentswill only need to return the amount ininstallments."A student who eventually becomes aresearcher or joins an IIT as a facultymember will also be exempted as wewant to encourage research andstudents to become teachers. In case thestudent remains unemployed, we won'texpect him to pay," Sibal added. It is,anyway, rare for an IIT graduate toremain unemployed. Even though theministry and the IITs have agreed uponthis proposal "in principle", it can onlybe implemented if Sibal can get thefinance ministry on board. Obviouslythen, as of now, there is no deadline forimplementation of this decision.Another hurdle would be to ensure thatstudents do not shirk theirresponsibility of paying back. Sibal saidthe shift to "demat" degrees andcertificates will take care of thisproblem.Last year, the minister had announcedthe start of a process for theestablishment of a national database ofacademic qualifications (degrees orcertificates from school to graduate andpostgraduate levels, includingprofessional degrees), which will becreated and maintained in a digitalformat by an identified, registereddepository. For this purpose, theHRD ministry had constituted a taskforce under the supervision of IIT-Kanpur director Sanjay Dhande."The degree eventually goes to theemployer. Once the demat system isin place, an IIT graduate's degreewill reflect the obligation to pay theinstitute back and the money willcome via the employer," said Sibal,adding that the details will beworked out in consultation with theIITsHRD to hand over 10,000 lowcostlaptops to IIT RajasthanJaipur: The much awaited low-costlaptops in India will be introducedthrough the IIT-Rajasthan. TheUnion HRD ministry announced thatthe first batch of low-cost laptops,designed for use by students, will behanded over to IIT Rajasthan."We would deliver the first lot of10,000 laptops to IIT-Rajasthan inJune," a ministry official said. Thelaptops would come for Rs 2,200 perunit. The original price band forthese laptops was kept betweenRs1,000-1,500 per unit."One lakh laptops have been orderedfor students across the country. Theremaining 90,000 units would bedistributed in remaining states overthe next four months," the officialsaid.The HRD ministry announced itsplans of rolling out low-costcomputing devices during aconference of education ministersfrom different states in New Delhitwo days ago. The project had beenin the pipeline for six years. Theconference, presided over by UnionHRD minister Kapil Sibbal, wasattended by school educationminister Bhanwar Lal Meghwal andXtraEdge for IIT-JEE 3 NOVEMBER 2011

higher education minister JitendraSingh.Officials at the IIT Rajasthan inJodhpur said the project of low-costlaptops was underway. But, they saidthey were not aware of the price bandat which it would be available or themode of distribution that would beadopted. "It's a Union governmentdecision, an official of the institutesaid.Officials said the government wouldsubsidise 50 per cent of the cost and astudent would pay only around Rs1,000 for the device. On the basis ofthe feedback of the field trial, thecomputers will be made available fordistribution among students under theNational Mission On Educationthrough Information andCommunication Technology.The computers will be equipped withWiFi connectivity, PDF Reader,Office applications, a web browserwith remote device managementcapability and video streaming. Theywill come with 2 USB ports, built-inkeyboard, a 7-inch touch-screen and 2GB RAM.IIT-Bombay announcesGeomat12IIT-B going to conduct Geomatrix’12,an International Conference onGeospatial Technologies andApplications to be held from 26th to29th February 2012 at Indian Instituteof Technology Bombay (IITB).Geomatrix’12 is the third in a series ofconferences on geoinformatics tools,techniques and applications beingorganized by the Centre of Studies inResources Engineering (CSRE), IITBsince 2009. The first two conferenceswere national level conferences;Geomatrix’12 is envisaged to be aninternational event with strongparticipation from national andinternational researchers and institutesof repute at both organization anddelegate level.We are expecting participation ofdelegates from various academicinstitutes, research organizations andindustries to share their researchfindings and professional experienceswith fellow researchers, professionalsand students from India and abroad. Weare looking forward to your enthusiasticparticipation in Geomatrix’12.The focus of the conference would be on:(a) tools and techniques of GIS, remoteSensing, satellite image processingand GPS, and(b) applications of GIS and remotesensing to exploration, assessmentand management of natural resources(including minerals, water resources,forests, snow and glaciers, etc.),agriculture, natural hazardassessment and disaster management,environmental impact assessmentincluding climate change studies,atmospheric studies, terrain studies,land-use planning, and related fieldsof earth sciences.The specific themes include:1. Advances in Tools and techniquesof GIS2. Recent advances in microwaveremote sensing3. State-of-the-art in satellite imageprocessing4. Global positioning systems andwireless sensor networks5. Geospatial technology in mineralsystem studies and mineralexploration6. Agroinformatics – tools andapplications7. Geoinformatics for coastal, marine,and urban environments8. Geospatial technology forprediction and management ofnatural hazards and disasters9. Monitoring and managing ourglaciers and water resources usingremote sensing10. Education and educationaltechnology in geoinformaticsGet a reality check: IITs wellbehind Chinese peers, says PMIf you think that our Indian Institutes ofTechnology are producing the bestbrains of the world, think again. In arecent speech at IIT-Kharagpur, PrimeMinister Manmohan Singh said the IITswere well behind technologycounterparts in China when it cameto research and PhDs."The Kakodkar committee reportnoted the number of PhDs is verysmall in comparison to similartechnology institutions inthe USA and China," he said. It wasimportant as it emphasised thechallenge in creating an advancedresearch-based innovationecosystem, with the involvement ofindustry and national technologyprogrammes.Talking about how Kakodkar panelwas set up last year to come up with areport card of the progress of IITs inthe country, Singh said that therecommendations of the committeewill soon be considered by theCouncil of the IITs and then by theGovernment of India. Of manysuggestions given by Kakodkar panel,the most significant one is aboutgiving more autonomy to theseinstitutions.The PM also stressed on the fact thatIITs need to take on a leadership roleon innovations to stimulate long-termgrowth and development. Heemphasises the need for a secondGreen Revolution. He said, "Wehave to usher in a soft revolution inour academic business andadministrative culture…ourscientific and entrepreneurialenergies should be channeled tospark the second Green Revolution,find new pathways for sustainablegrowth and living and make greengrowth a profitable businessproposition."PM Manmohan Singh wasaddressing the 60th anniversary andthe 57th convocation of IIT-Kgp. Atthe event, 1,966 degrees wereawarded, of which 235 were PhDs,29 MS, 692 M Tech, 84 MBAs, 380B. Tech and 216 MSc, among others.Several personalities, includingBharti Airtel chairman Sunil Mittalwere awarded honourary doctorates.XtraEdge for IIT-JEE 4 NOVEMBER 2011

Success StoryThis article contains story/interviews of persons who succeed after graduation from different IITsSUBRAH S. IYERBorn : 1957, Mumbai, Maharashtra, IndiaOccupation : CEO, WebExNet worth : US$129 million (2000)Subrah S. Iyer (b.1957) is a leading technocrat,entrepreneur and Web conferencing pioneer of Indianorigin. He is the founder and CEO of WebEx which hasrecently merged with Cisco Systems.Early lifeSubrah S. Iyer was born and brought up in Mumbai. Hehad descended from Tamil immigrants who had migratedto Mumbai. He did his schooling in Mumbai and graduatedfrom the Indian Institute of Technology. On completion ofhis graduation he moved to the US in the year 1982. Heworked with Intel, Apple Inc., Quarterdeck, and TeleosResearch prior to the establishment of WebEx.Founding of WebExIn his childhood days, his father had sternly warned himagainst dreaming of becoming an entrepreneur. However,he overruled him when in 1996; he founded WebEx inpartnership with Min Zhu.The founding of the company by Subrah Iyar was fuelledby a new-found interest in Web Conferencing. Min Zhu, aStanford-trained System Engineer had been struggling todevelop a web-conferencing tool. Coincidentally, duringthis time, he befriended Subrah Iyar who was runningQuarterdeck's research lab and the two formed apartnership.Growth of WebExWebEx struggled to make a profit in its early days, lowbandwidth being one of the main reasons. Slowly, with theadvancement of technology and the shift to broadbandtechnology, WebEx began to emerge as a potentcompetitor with clients such as Hoover's Online, Oracleand Tibco Software. However, despite the belowperformance of Webex in its early days, it was generally aboom time for digital conferencing technology with theemergence of standards such as ISDN and SwitchedDigital Service. WebEx received its first funding of $25million in December 1999.Faced with a win or lose situation, the management ofWebEx accepted the challenge with a brave heart. As aresult of the new ideas propounded by Subrah Iyer, 2000became a honeymoon year for WebEx. The revenuescrossed the million mark and Subrah Iyar's own net worthrocketed from a paltry $450,000 in January 2000 to $129million in November 2000. When enquired about it in aninterview at a later stage, Subrah Iyar remarked, "It didn'tget too scary, because I knew we had done everythingbased on fundamentals. You always have a feeling ofuncertainty. But it was never a feeling of terror."In 2003, when Microsoft purchased conferencing companyPlace ware it was thought to be the end of the road forSubrah Iyar and WebEx. However, WebEx survived andcompleted a $45 million acquisition of Intranets.com in2005. As per the company website, more than 3.5 millionpeople use Cisco’s WebEx products every month tocommunicate and collaborate online.How WebEx went the Cisco routeIn Silicon Valley, being at the top of your game in a hotmarket means you can pretty much name your price. Atleast that's what seems to have happened to Webconferencing company WebEx.In March 2007, Cisco Systems said it would pay $3.2billion for the company. Cisco plans to integrate WebEx'sonline collaboration and meeting services into its unifiedcommunications business. Subrah Iyar, chairman and chiefexecutive officer, has been with WebEx since thebeginning as a co-founder. And through the years, he hasestablished the company as a leader in the Webcollaboration market, fending off tough competitors suchas Microsoft.WebEx had already been partnering with Cisco to integratevoice over IP capabilities into its Web conferencingservices. So when potential suitors came knocking onWebEx's door, it made perfect sense for the company totalk to Cisco about a deal.XtraEdge for IIT-JEE 5 NOVEMBER 2011

KNOW IIT-JEEBy Previous Exam QuestionsPHYSICS1. Two square metal plates of side 1 m are kept 0.01 mapart like a parallel plate capacitor in air in such away that one of their edges is perpendicular to an oilsurface in a tank filled with an insulating oil. Theplates are connected to a battery of emf 500 V. Theplates are then lowered vertically into the oil at aspeed of 0.001 ms –1 . Calculate the current drawnfrom the battery during the process. (Dielectricconstant of oil = 11, ε 0 = 8.85 × 10 –12 C 2 N –1 m –1 )[IIT-1994]Sol. The adjacent figure is a case of parallel platecapacitor, The combined capacitance will beV1–xx+ –1mdC = C 1 + C 2kε(x 1)= 0 × ε+ 0 [(1– x) × 1]ddε 0C = [kx + 1 – x] ...(i)dAfter time dt, the dielectric rises by dx. The newequivalent capacitance will beC + dC = C 1 ' + C 2 'ε 0ε= [(x + dx) × 1] + 0 [(1– x – dx) × 1]d dε 0= d[kx + kdx + 1 – x – dx]Change of capacitance in time dtε 0dC = [kx + kdx + 1 – x – dx – kx – 1 + x]dε0= d(k – 1)dx ...(ii)dC ε =0 dx ε (k – 1) =0 (k – 1)v ...(iii)dt d dt ddxwhere v = dtWe know that q = CVdq dV = Vdt dtε 0⇒ I = V (k – 1)v dFrom (i) and (ii)–12...(iv)500×8.85×10I =(11 – 1) × 0.0010.01= 4.425 × 10 –9 Amp. Ans.AlternativelyWe can differentiate eq. (i) w.r.t. 't', we getdC ε =0 dx (K –1) and then proceed further.dt d dt2. An electrons gun G emits electrons of energy 2keVtravelling in the positive x-direction. The electronsare required to hit the spot S where GS = 0.1m, andthe line GS make an angle of 60º with the x-axis asshown in the figure. A uniform magnetic field→B parallel to GS in the region outside the electrongun. Find → B parallel to GS exists in the regionoutside the electron gun. Find the minimum value ofB needed to make the electrons hit S. [IIT-1993]Sv→B90ºGSol. Let us resolve the velocity two rectangularcomponents V 1 (= V cos θ) and V 2 (Vsin 60º), V 1component of velocity is responsible to move thecharge particle in the direction of the magnetic fieldwhereas V 2 component is responsible for rotating thecharged particle in circular motion. The overall pathis helical. The condition for he charged particle tostrike S with minimum value of B isPitch of Helix = GS2π mT × V 1 = GS ⇒ × v cos 60º = 0.1qB1 mv 2 = E ⇒ V =2x2EmXtraEdge for IIT-JEE 6 NOVEMBER 2011

5= = 2.5 Amp. 2I∴ I = 02deviation becomes zero.The equation for growth of current in L-R circuit isS→RT2t––BI = I 0 [1 – eL] ⇒ 2.5 = 5 [1 – e10]V 1 = v cos 60ºr = 0.1 m1⇒ = 1 – e–t/52⇒ e –t/5 1= ⇒ et/5= 2 2vGV 2 = v sin 60ºt⇒ = log e 2 5⇒ t = 5 log e 2 = 2 × 2.303 × 0.3010 = 3.466 sec. Ans.2πmVcos60º⇒ B =q × 0.14. Two identical prisms of refractive index 3 are2πm2Ekept as shown in the figure. A light ray strikes the⇒ B = × × cos 60ºq × 0.1 mfirst prism at face AB. Find, [IIT-2005]2πB D= × 2 mE × cos 60ºq × 0.160º 60º2×3.14=–191.6×10 × 0.160º 60º=–31 3–192 × 9.1×10 × 2×10 × 1.6×101× 2A C E149.8(a) the angle of incidence, so that the emergent ray= × 0.316 × 10 –23 = 47.37 × 10 –4–19from the first prism has minimum deviation.10= 4.737 × 10 –3 (b) through what angle the prism DCE should beT Ans.rotated about C so that the final emergent ray also3. A solenoid has an inductance of 10 henry and aresistance or 2 ohm. It is connected to a 10 voltbattery. How long will it take for the magnetic energyhas minimum deviation.Sol. (a) For minimum deviation of emergent ray from thefirst prism MN is parallel to ACto reach ¼ of its maximum value? [IIT-1996]∴ ∠ BMN = 90ºSol. Let I 0 be the current at steady state. The magnetic⇒ ∠ r = 30ºenergy stored in the inductor at this state will beApplying Snell's law at Msin iµ =L = 10H R = 2Ωsin rsin i = µ sin rsin i =10V33 × sin 30º = 21 ⇒ i = 60º2E = LI0 ...(i)2BThis is the maximum energy stored in the inductor.The current in the circuit for one fourth of this energycan be found as1 1P× E =260ºLI0 ...(ii)i2 2rNMDividing equation (i) and (ii)Q160º2LIE 0=2I⇒ I = 0ACE / 4 1 2LI22(b) When the prism DCE is rotated about C inanticlockwise direction, as shown in the figure,V 10Also, V = I 0 R ⇒ I 0 = = = 5 Amp.then the final emergent ray SR becomes parallelR 2to the incident ray TM. Thus, the angle ofXtraEdge for IIT-JEE 7 NOVEMBER 2011

5. A neutron of kinetic energy 65eV collidesinelastically with a singly ionized helium atom atrest. It is scattered at an angle of 90º with respectof its original direction.[IIT-1993](i) Find the allowed values of the energy of theneutron and that of the atom after the collision.(ii) If the atom get de-excited subsequently byemitting radiation, find the frequencies of theemitted radiation.[Given: mass of He atom –4×(mass of neutron),Ionization energy of H atom = 13.6eV]Sol.m 4mm K 2θ4mK 1Applying conservation of linear momentum inhorizontal direction(Initial Momentum) x = (Final Momentum) xp ix = p fx⇒ 2 Km = 2(4m)K1 cosθ...(i)Now applying conservation of linear momentumin Y-directionp iy = p fy0 = 2K2 m – (4m)K sin θyx2 1⇒ 2K2 m = 2(4m)K1 sin θ ...(ii)Squaring and adding (i) and (ii)2Km + 2K 2 m = 2(4m)K 1 + 2(4m)K 1K 1 + K 2 = 4K 1 ⇒ K = 4K 1 – K 2⇒ 4K 1 – K 2 = 65 ...(iii)When collision takes place, the electron gainsenergy and jumps to higher orbit.Applying energy conservationK = K 1 + K 2 + ∆E⇒ 65 = K 1 + K 2 + ∆E ...(iv)Possible value of ∆E For He +Case (1)∆E 1 = – 13.6 – (54.4eV) = 40.8 eV⇒ K 1 + K 2 = 24.2 eV from (4)Solving with (3), we getK 2 = 6.36 eV; K 1 = 17.84 eVCase (2)∆E 2 = – 6.04 – (–54.4 eV) = 48.36 eV⇒ K 1 + K 2 = 16.64 eV from (4)Solving with (3), we getK 2 = 0.312 eV; K 1 = 16.328 eVCase (3)∆E 3 = – 3.4 – (–54.4eV) = 51.1 eV⇒ K 1 + K 2 = 14 eVSolving with (3), we getK 2 = 15.8 eV; K 1 = – 1.8 eVBut K.E. can never be negative therefore case (3)is not possible.Therefore the allowed values of kinetic energiesare only that of case (1) and case (2) and electroncan jump upto n = 3 only.–3.4eVn=4–6.04eVn=3–13.6eVn=2–54.4eVn=1For He +(ii) Thus when electron jumps back there are threepossibilitiesn 3 → n 1 or n 3 → n 2 and n 2 → n 1The frequencies will beE3 E2ν 1 =− E3 Eν 2 =− 1 E2 ν 3 =− E1hhh= 1.82×10 15 H z = 11.67×10 15 H z = 9.84×10 15 Hz Ans.CHEMISTRY6. An organic compound C x H 2y O y was burnt with twicethe amount of oxygen needed for completecombustion to CO 2 and H 2 O. The hot gases whencooled to 0 ºC and 1 atm pressure, measure 2.24 L.The water collected during cooling weighed 0.9 g.The vapour pressure of pure water at 20ºC is 17.5mm of Hg and is lowered by 0.104 mm when 50 g ofthe organic compound are dissolved in 1000 g ofwater. Give the molecular formula of the organiccompound.[IIT-1983]Sol. According to the question, an organic compoundC x H 2y O y was burnt with twice the amount of oxygen.Hence,C x H 2y O y + 2x O 2 ⎯→ xCO 2 + yH 2 O + xO 2Volume of gases after combustion = 2.24 L (given)Volume of gases left after combustion = xCO 2 + xO 2∴ x + x = 2.24or x = 1.12 L22.4 L CO 2 = 1 mol CO 21.12∴ 1.12 L CO 2 = = 0.05 mol CO 222.4and 18 g H 2 O = 1 mol H 2 OXtraEdge for IIT-JEE 8 NOVEMBER 2011

0. 9∴ 0.9 g H 2 O = 18= 0.05 mol H 2 OThus, the empirical formula of the organic compoundis CH 2 O.Empirical formula mass = 12 + 2 + 16 = 30Vapour pressure of the pure liquid,0P A = 17.5 mm of HgLowering in vapour pressure P A0 –P A =0.104mm of HgMass of organic compound = 50 gMass of water = 1000 g50 / M∴ Mole fraction of organic compound =50 1000+M 18where M is the molecular mass of the organiccompound, the molecular mass of water being 18.We know,0A A0PAP − P= Mole fraction of organic compound0.104 50 / M∴=17.5 50 1000+M 1817.5 1000×Mor= 1 +0.104 18×50Solving, M = 150.5 ≈ 150Molecular mass 150n == = 5Empirical formula mass 30∴ Molecular formula or organic compound= 5(CH 2 O) = C 5 H 10 O 57. At room temperature, the following reactions proceednearly to completion :2NO + O 2 → 2NO 2 → N 2 O 4The dimer, N 2 O 4 , solidified at 262 K. A 250 ml flaskand a 100 ml flask are separated by a stopcock. At300 K, the nitric oxide in the larger flask exerts apressure of 1.053 atm and the smaller one containsoxygen at 0.789 atm. The gases are mixed by openingthe stopcock and after the end of the reaction theflasks are cooled to 200 K. Neglecting the vapourpressure of the dimer, find out the pressure andcomposition of the gas remaining at 220 K. (Assumethe gases to behave ideally)[IIT-1992]Sol. According to the gas equation,PV = nRTPVor n = RTAt room temperature,For NO, P = 1.053 atm, V = 250 ml = 0.250 L1.053×0.250∴ Number of moles of NO =0.0821×300= 0.01069 molFor O 2 , P = 0.789 atm, V = 100 ml = 0.1L0.789×0.1∴ Number of moles of O 2 =0.0821×300= 0.00320 molAccording to the given reaction,2NO + O 2 → 2NO 2 → N 2 O 4Composition of gas after completion of reaction,Number of moles of O 2 = 01 mol of O 2 react with = 2 mol of NO∴ 0.00320 mol of O 2 react with = 2 × 0.00320= 0.0064 mol of NONumber of moles of NO left = 0.01069 – 0.0064= 0.00429 molAlso, 1 mol of O 2 yields = 1 mol of N 2 O 4∴ Number of moles of N 2 O 4 formed = 0.00320 molN 2 O 4 condenses on cooling,∴ 0.350 L (0.1 + 0.250) contains only 0.00429 molof NOAt T = 220 K,Pressure of the gas,nRT 0 .00429×0.0821×220P = = = 0.221 atmV0. 3508. One gram of an alloy of aluminium and magnesiumwhen treated with excess of dil. HCl formsmagnesium chloride, aluminium chloride andhydrogen. The evolved hydrogen collected overmercury at 0ºC has a volume of 1.20 litres at 0.92atm pressure. Calculate the composition of the alloy.[At wt. Mg = 24, Al = 27][IIT-1978]Sol. Let the alloy contains, Al = x gthen Mg = (1 – x)gStep 1.2Al + 6HCl → 2AlCl 3 + 3H 22 × 27 = 54 g 3 × 22.4 = 67.2 LAt STP, 54 g of Al with HCl yields H 2 = 67.2 L67.2∴ x g of Al with HCl yields H 2 = x L54Step 2.Mg + 2HCl → MgCl 2 + H 224 g 22.4 LAt STP, 24 g of Mg with HCl yields H 2 = 22.4 L∴ (1 – x)g of Mg with HCl yields H 222.4=24× (1 – x)LXtraEdge for IIT-JEE 9 NOVEMBER 2011

Step 3. Also given that,P 1 = 0.92 atmP 2 = 1 atmV 1 = 1.20 L V 2 = ?T 1 = 0ºC = 273 K T 2 = 273 KTotal volume V 2 of hydrogen collected over mercuryat STP is given by,P V1 1T1or V 2 ==P1VT1P2VT12×2T 2 0 .92×=273P21.20×273 = 1.104 L1Now, Volume of H 2 evolved by Al + Volume of H 2evolved by Mg = Total volume of H 2or67.2x +5422.4(1 – x) = 1.10424or 67.2 × 24x + 22.4 × 54 × 1 – 22.4 × 54 × x= 1.104 × 24 × 54or 1612.8x + 1209.6 – 1209.6x = 1430.78or 403.2 x = 21.18or x = 0.5486∴Mass of Al = 0.5486 gand Mass of Mg = 1 – 0.5486= 0.4514 gStep 4. % composition of Al in 1 g alloyMass of Al=× 100Mass of alloy0.5486= × 100 = 54.86% Al1% composition of Mg in 1 g alloyMass of Mg=× 100Mass of alloy=0.4514 × 100 = 45.14 % Mg19. An organic compound (X), C 5 H 8 O, does not reactappreciably with Lucas reagent at room temperaturesbut gives a precipitate with ammonical AgNO 3solution. With excess CH 3 MgBr; 0.42 g of (X) gives224 ml of CH 4 at STP. Treatment of (X) with H 2 inthe presence of Pt catalyst followed by boiling withexcess HI gives n-pentane. Suggest structure of (X)and write the equations involved. [IIT-1992]Sol. Lucas test sensitive test for the distinction of p, s, andt-alcohol. A t-alcohol gives cloudiness immediately,while s-alcohol within 5 minutes. A p-alcohol doesnot react with the reagent at room temperature. Thus,the present compound (X) does not react with thisreagent, hence it is a p-alcohol.(X) = C 4 H 6 .CH 2 OH(p-alcohol)Since the compound gives a ppt. with ammonicalAgNO 3 , hence it is an alkyne containing one–C≡ CH, thus (X) may be written as :HC≡C –C 2 H 4 – CH 2 OH (X)It is given that 0.42 g of the compound (which is0.005 mol) produces 22.4 ml of CH 4 at STP (which is0.01 mol) with excess of CH 3 MgBr. This shows thatthe compound (X) contains two active H atoms (Hatom attached to O, S, N and –C≡CH is calledactive). Of these, one is due to the p-alcoholic group(–CH 2 OH) and the other is due to the –C≡CH bond,since both these groups are present in (X), hence itevolves two moles of CH 4 on reaction withCH 3 MgBr.H – C≡C. C2H 4(X)– CH 2 OH + 2CH 3 MgBr →BrMgC≡C–C 2 H 4 – CH 2 OMgBr + 2CH 4Moreover, the treatment of (X) with H 2 /Pt followedby boiling with excess of HI gives n-pentane(remember that 2HI are required to convert one–CH 2 OH into CH 3 ). This shows that the compound(X) contains a straight chain of five carbon atoms.H – C≡C–C 2 H 4 – CH 2 OH ⎯2 ⎯H 2⎯/ Pt →CH 3 CH 2 .C 2 H 4 – CH 2 OH2 ⎯→ CH CH CH CH CH + H O + I 3 2 2 2 3 2 2⎯ HI∆n-pentaneOn the basis of abvoe analytical facts (X) has thestructure :5 4 3 2 1HC≡C.CH 2 CH 2 – CH 2 OH (X)4-pentyne-1-olThe different equations of (X) are :H − C ≡ C − CHZnCl2CH2CH2OH2 + ⎯ HClRoom ⎯⎯⎯temp. ⎯(X)→ No reactionAgNO 3Ag – C≡C – CH 2 CH 2 CH 2 OH + NH 4 NO 3NH 32CH 3 MgBr2H 2 /PtWhite ppt.Br MgC≡C.CH 2 CH 2 CH 2 OMgBr + 2CH 4CH 3 CH 2 CH 2 CH 2 CH 2 OHPentanol-12 HI∆, –H 2 O; –I 2CH 3 CH 2 CH 2 CH 2 CH 3n-pentaneThe production of 2 moles of CH 4 is confirmed as thereactions give 224 ml of CH 4 .Q 84 g(X) gives = 2 × 22.4 litre CH 42 × 22.4×0.42∴ 0.42 g (X) gives =84= 224 ml of CH 4XtraEdge for IIT-JEE 10 NOVEMBER 2011

10. A white amorphous powder A when heated gives acolourless gas B, which turns lime water milky andthe residue C which is yellow when hot but whitewhen cold. The residue C dissolves in dilute HCl andthe resulting solution gives a white precipitate onaddition of potassium ferrocyanide solution. Adissolves in dilute HCl with the evolution of a gaswhich is identical in all respects with B. The solutionof A as obtained above gives a white precipitate D onaddition of excess of NH 4 OH and on passing H 2 S.Another portion of this solution gives initially a whiteprecipitate E on addition of NaOH solution, whichdissolves on further addition of the base. Identify thecompound A to E.[IIT-1979]Sol. The given information is as follows.(a)Awhite powderdiluteHCl⎯ heat ⎯→(b)C ⎯⎯⎯⎯→ solution(c) Adilute HCl(i) NH 4OH(ii) H 2SDwhite precipitateB +colourless gasturnsmilky lime waterK6⎯ 4 Fe(CN )Cresidueyellow whenhot white when cold⎯⎯⎯⎯ → white precipitateSolution + B(i) NaOHEwhite precipitatedissolvesNaOHFrom part (a), we conclude that B is CO 2 as it turnslime water milky :Ca(OH 2 ) + CO 2 → CaCO 3 + H 2 Omilky dueto thisand C is ZnO as it becomes yellow on heating and iswhite in cold. Hence, the salt A must be ZnCO 3 .From part (b), it is confirmed that C is a salt of zinc(II) which dissolves in dilute HCl and whiteprecipitate obtained after adding K 4 [Fe(CN) 6 is dueto Zn 2 [Fe(CN) 6 ].From part (c), it is again confirmed that A is ZnCO 3as on adding dilute HCl, we get CO 2 and zinc (II)goes into solution. White precipitate is of ZnS whichis precipitated in ammonical medium as its solubilityproduct is not very low. White precipitate E is ofZn(OH) 2 which dissolves as zincate, in excess ofNaOH. Hence the given information is explained asfollows.(a) ZnCO(b) ZnO⎯ heat ⎯→ CO +(B)2ZnO(C)3(A)⎯ dil ⎯HClK⎯→ZnCl(C)solution 2 ⎯⎯4 Fe(CN 6⎯ ⎯⎯)(c) ZnCO 3⎯ dil ⎯HCl ⎯→Solution 2→ ZnZnCl + CO 2 + H 2 O2 [Fe(CN) 6]White precipitateZnCl 2 + S 2– → ZnS↓ + 2Cl –(D)Zn 2+ + 2OH – → Zn(OH)2(E)2-2dissolvesZn(OH) 2 + 2OH – → ZnO + 2H 2 OMATHEMATICS11. 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→0hh→0 hNow, let x ∈ R then applying formula ofdifferentiability.⎛ 2x+ 2h⎞f ⎜ ⎟ − f ( x)f ( x + h)− f ( x)2f ´(x)= lim= lim⎝ ⎠h→0hh→0hf (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 11 NOVEMBER 2011

12. Find all the solutions of :4 cos 2 x sin x – 2 sin 2 x = 3 sin x [IIT-1983]Sol. 4 cos 2 x sin x – 2 sin 2 x = 3 sin x⇒ 4(1 – sin 2 x) sin x – 2 sin 2 x – 3 sin x = 0⇒ 4 sin x – 4 sin 3 x – 2 sin 2 x – 3 sin x = 0⇒ – 4 sin 3 x – 2 sin 2 x + sin x = 0⇒ – sin x (4 sin 2 x + 2 sin x – 1) = 0⇒ sin x = 0 or 4 sin 2 x + 2 sin x – 1 = 0⇒ sin x = sin 0 or sin x =− 2 ± 4 + 162(4)−1± 5⇒ x = nπ or sin x =4π ⎛ 3π⎞x = nπ or sin x = sin or sin x = sin ⎜− ⎟⎠ 10 ⎝ 10⇒ x = nπ + (–1) n π , nπ + (–1)n ⎛ 3π⎞⎜− ⎟10⎝ 10 ⎠∴ General solution set⎧n π ⎫⇒ (x : x = nπ} ∪ ⎨x: x = nπ + ( −1)⎬⎩10⎭⎧∪ ⎨x: x = nπ + ( −1)⎩⎛ − 3π⎞⎫⎜ ⎟⎬⎝ 10 ⎠⎭13. Let C 1 and C 2 be two circles with C 2 lying inside C 1 .A circle C lying inside C 1 touches C 1 internally andC 2 externally. Identify the locus of the centre to C.[IIT-2001]Sol. Let the given circles C 1 and C 2 have centres O 1 andO 2 and radii r 1 and r 2 respectively.Let the variable circle C touching C 1 internally, C 2externally have a radius r and centre at O.nC 2 C 1O 2r 2rO 1 OCNow, OO 2 = r + r 2 and OO 1 = r 1 – r⇒ OO 1 + OO 2 = r 1 + r 2which is greater than O 1 O 2 as O 1 O 2 < r 1 + r 2(Q C 2 lies inside C 1 )⇒ locus of O is an ellipse with foci O 1 and O 2 .r 114. Find the smallest positive number p for which theequation cos (p sin x) = sin(p cos x) has a solutionx ∈ [0, 2π][IIT-1995]Sol. cos (p sin x) = sin (p cos x) (given) ∀ x ∈[0, 2π]⎛ π ⎞⇒ cos (p sin x) = cos ⎜ − p cos x⎟ ⎝ 2 ⎠⎛ π ⎞⇒ p sin x = 2nπ ± ⎜ − p cos x⎟ , n ∈ I⎝ 2 ⎠[Q cos θ = cos α ⇒ θ = 2nπ ± α, n ∈ I]⇒ p sin x + p cos x = 2nπ + π/2or p sin x – p cos x = 2nπ – π/2, n∈ I⇒ p.⎛ 1 1 ⎞2⎜ sin x + cos x⎟ = 2nπ + π/2⎝ 2 2 ⎠or p⎛ 1 12 ⎟ ⎞⎜ sin x − cos x⎝ 2 2 ⎠= 2nπ – π/2, n ∈ I⇒ p⎛ π π ⎞ π2 ⎜cos sin x + sin cos x⎟ = 2nπ +⎝ 4 4 ⎠ 2or p⎛ π π ⎞ π2 ⎜cos sin x − sin cos x⎟ = 2nπ – , n ∈ I⎝ 4 4 ⎠ 2⇒ p 2⎡ ⎛ π ⎞⎤π⎢sin⎜x + ⎟⎥ = (4n + 1) , n ∈ I⎣ ⎝ 4 ⎠ ⎦ 2or p 2⎡ ⎛ π ⎞⎤π⎢sin⎜x − ⎟⎥ = (4n – 1) , n ∈ I⎣ ⎝ 4 ⎠ ⎦ 2Now, –1 ≤ sin (x ± π/4) ≤ 1⇒ – p 2 ≤ p 2 sin (x ± π/4) ≤ p 2⇒ – p( 4n + 1). π2 ≤2≤ p 2 , n ∈ Ior – p( 4n −1)π2 ≤2≤ p 2 , n ∈ ISecond inequality is always a subset of first,therefore, we have to consider only first.It is sufficient to consider n ≥ 0, because for n > 0,the solution will be same for n ≥ 0.If n ≥ 0, – 2 p ≤ (4n + 1) π/2⇒ (4n + 1) π/2 ≤ 2 pFor p to be least, n should be least⇒ n = 0⇒2 p ≥ π/2 ⇒ p ≥Therefore least value of p =22ππ22XtraEdge for IIT-JEE 12 NOVEMBER 2011

15. Find the range of values of t for which21−2x+ 5x⎡ π π⎤2 sin t =, t ∈2⎢−, ⎥ [IIT-2005]3x− 2x−1⎣ 2 2 ⎦21−2x+ 5x⎡ π π⎤Sol. Here, 2 sin t =, t ∈2⎢−, ⎥3x− 2x−1⎣ 2 2 ⎦Put, 2sin t = y ⇒ – 2 ≤ y ≤ 2∴ (3y – 5)x 2 – 2x(y – 1) – (y + 1) = 0Since x ∈ R – {1, –1/3} {as, 3x 2 – 2x – 1 ≠ 0}∴ D ≥ 0⇒ 4(y – 1) 2 + 4(3y – 5) (y + 1) ≥ 0⇒ y 2 – y – 1 ≥ 02⎛ 1 ⎞ 5⇒ ⎜ y − ⎟⎠ – ≥ 0⎝ 2 4⎛ ⎞⇒ ⎜1 5 ⎛ ⎞y − − ⎟ ⎜1 5y − + ⎟ ≥ 0⎝ 2 2⎠ ⎝ 2 2 ⎠1− 5 1+ 5⇒ y ≤ or y ≥221− 5 1+ 5or 2 sin t ≤ or sin t ≥22⎛ π ⎞⎛ 3π⎞⇒ sin t ≤ sin ⎜− ⎟ or sin t ≥ sin ⎜ ⎟⎝ 10 ⎠ ⎝ 10 ⎠π 3π⇒ t ≤ – or t ≥ 10 10⎡ π π⎤⎡3ππ⎤Thus, Range for t ∈ ⎢−, ⎥ ∪⎣ 2 2⎢ , ⎥ ⎦ ⎣102 ⎦Physics FactsWave Phenomena1. Sound waves are longitudinal and mechanical.2. Light slows down, bends toward the normal andhas a shorter wavelength when it enters a higher(n) value medium.3. All angles in wave theory problems are measuredto the normal.4. Blue light has more energy. A shorter wavelengthand a higher frequency than red light (remember-ROYGBIV).5. The electromagnetic spectrum (radio, infrared,visible. Ultraviolet x-ray and gamma) are listedlowest energy to highest.6. A prism produces a rainbow from white light bydispersion (red bends the least because it slowsthe least).7. Light wave are transverse (they can be polarized).8. The speed of all types of electromagnetic waves is3.0 x 10 8 m/sec in a vacuum.9. The amplitude of a sound wave determines itsenergy.10. Constructive interference occurs when two wavesare zero (0) degrees out of phase or a wholenumber of wavelengths (360 degrees.) out ofphase.11. At the critical angle a wave will be refracted to 90degrees.12. According to the Doppler effect a wave sourcemoving toward you will generate waves with ashorter wavelength and higher frequency.13. Double slit diffraction works because ofdiffraction and interference.14. Single slit diffraction produces a much widercentral maximum than double slit.15. Diffuse reflection occurs from dull surfaces whileregular reflection occurs from mirror typesurfaces.16. As the frequency of a wave increases its energyincreases and its wavelength decreases.17. Transverse wave particles vibrate back and forthperpendicular to the wave direction.18. Wave behavior is proven by diffraction,interference and the polarization of light.19. Shorter waves with higher frequencies haveshorter periods.20. Radiowaves are electromagnetic and travel at thespeed of light (c).21. Monochromatic light has one frequency.22. Coherent light waves are all in phase.Geometric Optics23. Real images are always inverted.24. Virtual images are always upright.25. Diverging lens (concave) produce only smallvirtual images.26. Light rays bend away from the normal as theygain speed and a longer wavelength by entering aslower (n) medium {frequency remains constant}.27. The focal length of a converging lens (convex) isshorter with a higher (n) value lens or if blue lightreplacesred.XtraEdge for IIT-JEE 13 NOVEMBER 2011

Physics Challenging ProblemsSet # 7This 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 will be published in next issueDirector Academics, Jodhpur Branch1. A current I enters at A in a square Dloop of uniform resistance and leavesat B. The ratio of magnetic field atthe centre of square loop due to Asegments AB and due to DC is I(A) 1 (B) 2 (C) 3 (D) 42. The potential difference acrossa 2H inductor is as a functionof time is shown. At t = 0current is zero. Then(A) current at t = 2s is 5A(B) current at t = 2s is 10A(C) current versus time graph in inductor will be(D) current versus time graph in inductor will be3. A body is thrown from the surface of earth withgRvelocity (when R is radius of earth) at some2angle from the vertical. If the maximum heightreached by body is R/4 then angle of projection withvertical is⎛−(A) sin 1 ⎜⎝⎛−(C) sin 1 ⎜⎝5 ⎞⎟4⎠3 ⎞⎟2⎠⎛−(B) cos 1 ⎜⎝5 ⎞⎟4⎠(D) None of these4. There is an infinite straight chain of alternating chargesq and –q. The distance between the two neighbouringcharges is equal to d. Then the interaction energy of anycharge with all the other charges isCBI(A)(C)− 2q4π∈d− 2q0224π∈log0de2(B)2q2log4π∈0ed2(D) None of thesePassage # (Q. No. 5 to Q. No. 7)A extension of Young’s double slit interferenceexperiment its to increase the number of slits fromtwo to a large number N. A three slit grating consiststhree slits of width a and separated by spacing d.Intensity due to single slit is I 0 .5. The intensity pattern for a three-slit grating is given2πd sin θby (assume a

8Questions1. dQ = dW + dU as dQ = - dU∴ dW = −2dU= 2CdTP.dV = 2CdTRT−2C / R.dV = 2CdT ⇒ VT = constant.VNow, dQ = -dUC.dT = -C v . dT ⇒ C = −CvOption [A] is correct2. As−2C / RVT = constant−R2C⇒ T.V = constantOption [B] is correctSolutionSet # 6Physics Challenging Problemswere Published in October Issue8.12dP0 ) ⇒ = 0 ⇒ T 2T0P = α.RT(T− T=dT∴ P[2]−1/2min = α.R.2T0(2T0− T0)y =(3x[2]20.8+ 12xt + 12t20.8=+ 4) [4 + 3(x + 2t)Space Shuttle2]3.R∆TW = where PV x = constant1 − xOption [B] is correct4. A → P, R; B → Q,S; C → P, R; D → SConceptual.5. A → Q; B → R; C → T; D → P3 9VA − VD= × 6 = V3 + 7 529V − VB= × 6 = 2V∴VD− VB= 2 −2 + 45for no energy stored in capacitor V D = V B∴ VA− VD= VA− VB9 214= × 6 ⇒ Y = Ω5 2 + Y 3A =6. φ = at(T− t)dφE = = a(T − 2t)dtT 2EH =∫.dtR0Option [D] is correct7.2T = T 0 + αV2⎛ RT ⎞T = T0+ α⎜⎟ as PV = RT⎝ P ⎠⇒ T.P2= T P02+ αR2T21V5OK here is the deal with the space shuttle. It hasthree rocket engines in the back, but there'sabsolutely no room inside for all the fuel it needs tolaunch itself up into space. All of that fuel is storedoutside the shuttle, in the big brown cylinder, calledthe external tank.The tank containing all the rocket fuel weighs seventimes more than the space shuttle itself! That's a lotof really heavy fuel, and the space shuttle enginesaren't quite strong enough to push the combinedweight of the shuttle and the big bloated externaltank up off the ground.That's what the two long white solid rocket boostersstrapped onto the sides of the external tank are for.They lift the tank! Fortunately, it was not necessaryto strap an infinite series of smaller and smallerrockets to the sides of the solid rocket boosters.It is not widely known that just behind the mainflight deck of the space shuttle is a small Starbucksadapted for use in zero gravity.XtraEdge for IIT-JEE 15 NOVEMBER 2011

PHYSICSStudents'ForumExpert’s Solution for Question asked by IIT-JEE Aspirants1. One end of an ideal spring is fixed to a wall at originO and axis of spring is parallel to x-axis. A block ofmass m = 1 kg is attached to free end of the springand its is performing SHM. Equation of position ofthe block in co-ordinate system shown in figure isx = 10 + 3. sin (10.t),t is in second and x in cm.Another block of mass M = 3 kg, moving towards theorigin with velocity 30 cm/sec collides with the blockperforming SHM at t = 0 and gets stuck to itCalculate1 Kg3 KgxO(i) new amplitude of oscillations,(ii) new equation for position of the combined bodyand(iii) loss of energy during collision. Neglect friction.Sol. Natural length of the spring is 10 cm and forceconstant of the spring is K = 100 Nm –1 .Just before collision, velocities of 1 kg block and3 kg block are (+ 0.30 ms –1 ) and (– 0.30 ms –1 )respectively. Let velocity of combined body just aftercollision be v, then, according to law of conservationof momentum, (1 + 3) v = 1 (0.30) + 3 ( – 0.30)or v = – 0.15 ms –1 .Negative sign indicates that combined body starts tomove leftward. But at the instant of collision spring isin its natural length or combined body is inequilibrium position. Hence at t = 0, phase ofcombined body becomes equal to π.Now angular frequency of oscillations of combinedbody isK 100ω' = = = 5 rad sec –1 .m + M 1+3∴ New amplitude of oscillations is| v |a' =ω '= 0.15= 0.3 m or 3 cm Ans. (i)5∴ Equation for position x of combined body isgiven byx = l 0 + a' sin(ω't + π)or x = 10 + 3 sin (5t + π) cmor x = 10 – 3 sin (5t) cmAns. (ii)Kinetic energy of two blocks (Just before collision)= 21 m(0.3)3+ 21 M(0.3)2= 0.18 jouleKinetic energy of combined body (just after collision)= 21 (m + M) v 2 = 0.045 Joule∴ Loss of energy, during collision= 0.18 – 0.045 joule= 0.135 joule Ans.(iii)2. Two identical blocks A and B of mass m = 3 kg areattached with ends of an ideal spring of forceconstant K = 2000 Nm –1 and rest over a smoothhorizontal floor. Another identical block C movingwith velocity v 0 = 0.6 ms –1 as shown in fig. strikesthe block A and gets stuck to it. Calculate forsubsequent motion(i) velocity of centre of mass of the system,(ii) frequency of oscillations of the system,(iii) oscillation energy of the system, and(iv) maximum compression of the spring.mC v 0mASol. When block C collides with A and get stuck with it,combined body moves to the right, due to whichspring is compressed. Therefore, the combined bodyretards and block B accelerates. In fact, deformationof spring varies with time and the system continues tomove rightwards. In other words, centre of mass ofthe system moves rightwards and combined body andblock B oscillate about the centre of mass of thesystem.Let just after the collision velocity of combined bodyformed by blocks C and A be v. Then, according tolaw of conservation of momentum,(m + m)v = mv 0vor v = 0= 0.3 ms –12∴ Velocity of centre of mass of the system,2m×v + m×0v c == 0.2 ms –12m + mNow the system is as shown in fig.2mmmBXtraEdge for IIT-JEE 16 NOVEMBER 2011

(2m)(m)Its reduced mass, m 0 = =2m + m∴ Frequency of oscillations,f =12πKm 0=510πHz.2m3Ans.Since, just after the collision, combined body hasvelocity v, therefore, energy of the system at that1instant, E = (2m)v 2 = 0.27 joule2Due to velocity v C of centre of mass of the system,translational kinetic energy,1E t = (3m)2v c = 0.18 joule2But total energy E of the system = its translationalkinetic (E t ) + oscillation energy (E 0 )∴ E 0 = E – E t = 0.09 jouleAt the instant of maximum compression, oscillationenergy is stored in the spring in the form of its strainenergy. Let maximum compression of spring be x 0 .1 2then Kx 0 = E 02∴ x 0 = 90 × 10 –3 m or 3 10 mm Ans.3. Calculate the inductance of a closely wound solenoidof length l whose winding is made of copper wire ofmass m. The winding resistance is equal to R. Thesolenoid diametre is considerably less than its length.Given, density of copper = d and resistivity = ρ.Sol. Let the radius of solenoid be a, total number of turnsN and let cross-sectional area of wire be S.Then length of the wire of which the solenoid ismade = 2π aNBut its resistance is R⎛ 2πaN⎞∴ R = ρ⎜⎟⎝ S ⎠RSor aN =...(i)2πρMass of the wire is given as m∴ (2πaN) S d = mmor aN =...(ii)2πSdFrom equation (i) and (ii)(aN) 2 mR=...(iii)4π 2 ρd2µSelf-Inductance of a solenoid is L = 0 ANlwhere A = πa 22 2µ 0πaN µ∴ L = = 0 π mRl l 4π 2 ρdµ 0or L =mR Ans.4πlρd4. A wire frame of area A = 4 × 10 –4 m 2 and resistanceR = 20 Ω is suspended freely by a thread of lengthl = 0.40 m. A uniform horizontal magnetic field ofinduction B = 0.8 T exists in the space such thatplane of the frame is perpendicular to the magneticfield. At time t = 0, the frame is made to oscillateunder gravity by displacing it through a = 2 × 10 –2 mfrom its initial position along the direction of themagnetic field. The plane of frame is always alongthe thread and does not rotate about wire frame as afunction of time. Calculate also, maximum current inthe frame. (g = 10 ms –2 )Sol. Since, frame oscillates under gravity, therefore, itperforms SHM with angular frequencyω = g / l = 5 rad/secSince the frame is released after displacing itthrough a distance a at t = 0, therefore, initial phaseof its harmonic oscillations is π/2. Hence, at time t,displacement x of frame from its mean position isgiven by x = a sin (ωt + π/2)or x = 0.02 cos (5t) ...(i)Since, the frame always remains along the thread,therefore, at time t, its inclination with the vertical isequal that of the thread as shown in figure.BThis inclination θ is given byθxlsin θ = lxSince, x

5. A rod of mass m and having resistance R can rotatefreely about a horizontal axis O, sliding along aPhysics Factsmetallic circular ring of radius a as shown in figure.The ring is fixed in a vertical plane with axiscoinciding with the axis of rotation of the rod. A Modern Physicsuniform magnetic field of induction B exists in thespace parallel to the axis of1. The particle behavior of light is proven by the+ E –rotation. A voltage source isphotoelectric effect.connected across the ring and the2. A photon is a particle of light {wave packet}.axis. Neglecting self inductance ofthe circuit, calculate the law3. Large objects have very short wavelengths whenaccording to which emf of the ×ωtmoving and thus can not be observed behaving asisource must vary so that roda wave. (DeBroglie Waves)Orotates with constant angular4. All electromagnetic waves originate fromvelocity ω.accelerating charged particles.Sol. When the rod rotates, it cuts magnetic lines of flux.5. The frequency of a light wave determines itsHence, an emf is induced in the rod. Magnitude of thisenergy (E = hf).induced emf is equal to flux cut per second by the rod.∴ Induced emf, e = B × area traced per second by 6. The lowest energy state of a atom is called the1 ground state.the rod = Bωa22 7. Increasing light frequency increases the kineticAccording to Fleming's right hand rule, the inducedenergy of the emitted photo-electrons.emf tries to flow a radially outward current through 8. As the threshold frequency increase for a photocell(photo emissive material) the work functionthe rod. It means that induced emf is in opposition toemf of the source.also increases.Assuming tht the magnitude of induced emf is greaterthan that of the source, net emf of the circuit becomes 9. Increasing light intensity increases the number ofequal to (e – E).emitted photo-electrons but not their KE.e – E∴ Current through the rod is i = .RInternal Energy(radially outward) 10. Internal energy is the sum of temperature (ke)Due to the current, force experienced by the rod isand phase (pe) conditions.equal to F = Bia. Direction of this force is such that itproduces a retarding moment as shown in figure11. Steam and liquid water molecules at 100 degrees+ Ehave equal kinetic energies.–12. Degrees Kelvin (absolute temp.) Is equal to zero(0) degrees Celsius.F13. Temperature measures the average kinetic energy×ωtof the molecules.iO mg14. Phase changes are due to potential energychanges.a15. Internal energy always flows from an object atThis retarding moment, τ 1 = F higher temperature to one of lower temperature.2Ba 2or τ 1 = (e – E)2RGeneralWeight mg of the rod produces an accelerating moment 16. The most important formulas in the physicsa regents are:τ 2 = mg sin (ωt) 2Since, angular velocity ω of the rod remains constant,therefore, the resultant torque on the rod must beequal to zero.17. Physics is fun. (Honest!)Hence, τ 1 = τ 21or E = Bωa 2 mgR– sin (ωt) Ans. 2 BaXtraEdge for IIT-JEE 18 NOVEMBER 2011

PHYSICS FUNDAMENTAL FOR IIT-JEEElectromagnetic Induction & A.C.KEY CONCEPTS & PROBLEM SOLVING STRATEGYElectromagnetic Induction (E.M.I.)Faraday's law states that the induced emf in a closedloop equals the negative of time rate of change ofmagnetic flux through the loop. This relation is validwhether the flux change is caused by a changingmagnetic field, motion of the loop, or both.dΦε = – BdtBφALenz's law states that an induced current or emfalways tends to oppose or cancel out the change thatcaused it. Lenz's law can be derived from Faraday'slaw, and is often easier to use.εChange in BB (increasing)When an emf is induced by a changing magnetic fluxthrough a stationary conductor, there is an inducedelectric field E r of non-electrostatic origin. This fieldis non conservative and cannot be associated with apotential.∫r r dΦE . dl = – BdtEIGBrEWhen a bulk piece of conducting material, such as ametal, is in a changing magnetic field or movesthrough a field, currents called eddy currents areinduced in the volume of the material.EεII´I 0B 0B´B inducedIf a conductor moves in a magnetic field, a motionalemf is induced.ε = vBL(conductor with length L moves in uniform B r field,L r and v r both perpendicular to B r and to each other)rε =∫( v×B).dr rl(all or part of a closed loop moves in a B r field)× B × × a × × ×+ a× × × × × ×F=qvB× × × × × ×× ×L q v× × × ×× × × × × ×F = qE× × × × × ×–× × × × × ×A time-varying electric field generates adisplacement current i D , which acts as a source ofmagnetic field in exactly the same way as conductioncurrent.dΦi D = ε E (displacement current)dtAlternating Current (A.C.)An alternator or ac source produces an emf variessinusoidally with time. A sinusoidal voltage orcurrent can be represented of the by a phasor, avector that rotates counterclockwise with constantangular velocity ω equals to the angular frequency ofthe sinusoidal quantity. Its projection on thehorizontal axis at any instant represents theinstantaneous value of the quantity.XtraEdge for IIT-JEE 19 NOVEMBER 2011

ωtIO i=I cos ωtFor a sinusoidal current, the rectified average and rms(root-mean-square) currents are proportional to thecurrent amplitude I. Similarly, the rms value of asinusoidal voltage is proportional to the voltageamplitude V.I rav = π2 I = 0.637 IIVI rms = ; V rms =22In general, the instantaneous voltage between twopoints in an ac circuit is not in phase with theinstantaneous current passing through points. Thequantity φ is called the phase angle of the voltagerelative to the current.i = I cos ωtv = V cos(ωt + φ)IVφ V cosφωtOThe voltage across a resistor R is in phase with thecurrent. The voltage across an inductor L leads thecurrent by 90º (φ = + 90º), while the voltage across acapacitor C lags the current by 90º(φ = –90º). Thevoltage amplitude across each type of device isproportional to the current amplitude I. An inductorhas inductive reactance X L = ωL and a capacitor hascapacitive reactance X C = 1/ωC.V R = IR; V L = IX L ; V C = IX CResistor connected to Inductor connected toac sourceac sourceia R b a L bCapacitor connected toac sourceiq –q ia C bIn a general ac circuit, the voltage and currentamplitude are related by the circuit impedance Z. Inan L-R-C series circuit, the values of L, R, C, and theωiangular frequency ω determine the impedance andthe phase angle φ of the voltage relative to the current.V = IZ22 22Z = R + (XL− XC) = R + [ ωL− (1/ ωC)]ωL−1/ωCtan φ =RV L = IX LV = IZV L – V COφIV R = IRV C = IX CThe average power input P av to an ac circuit dependon the voltage and current amplitudes (or,equivalently, their rms values) and the phase angle φof the voltage relative to the current. The quantitycos φ is called the power factor.P av = 21 VI cos φ = Vrms I rms cos φv, i, pφωP av = ½ VIcosφppiIn an L-R-C series circuit, the current becomesmaximum and the impedance becomes minimum atan angular frequency ω 0 = 1/ LC called theresonance angular frequency. This phenomenon iscalled resonance. At resonance the voltage andcurrent are in phase, and the impedance Z is equal tothe resistance R.I(A)0.5 200 Ω0.40.3 500 Ω0.22000 Ω0.10ω (rad/s)1000 2000A transformer is used to transform the voltage andcurrent levels in an ac circuit. In an ideal transformerwith no energy losses, if the primary winding has N 1turns and the secondary winding has N 2 turns, theamplitudes (or rms values) of the two voltages arerelated by Eq. The amplitudes (or rms values) of theprimary and secondary voltages and currents arerelated by Eq.tωtXtraEdge for IIT-JEE 20 NOVEMBER 2011

V 2 =2V1N1N ; V1 I 1 = V 2 I 2Problem Solving Strategy (P.S.S.) :Faraday' LawStep 1: Identify the relevant concepts: Faraday's lawapplies when there is a changing magnetic flux. Touse the law, make sure you can identify an areathrough which there is a flux of magnetic field. Thiswill usually be the area enclosed by a loop, usuallymade of a conducting material. As always, identifythe target variable(s).Step 2: Set up the problems using the following stepsFaraday's law relates the induced emf to the rateof change of magnetic flux. To calculate this rateof change, you first have to understand what ismaking the flux change. Is the conductor moving?Is it changing orientation? Is the magnetic fieldchanging? Remember that it's not the flux itselfthat counts, but its rate of change.Choose a direction for the area vector A r or dAr .The direction must always be perpendicular to theplane of the area. Note that you always have twochoice of direction. For instance, if the plane ofthe area is horizontal, A r could point straight upor straight down. It's like choosing whichdirection is the positive one in a probleminvolving motion in a straight line; it doesn'tmatter which direction you choose, just so youuse it consistently throughout the problem.Step 3: Execute the solution as follows :Calculate the magnetic flux using Eq.φ B = B r. Ar= BA cos φ if B r is uniform over thearea of the loop or eq. φ B =∫Br.dArif it is not uniform, being mindful of the directionyou chose for the area vector.=∫B dA cos φdφCalculate the induced emf using Eq. ε = – Bdtdφ (Faraday's law of induction) or ε = –N B . dtIf your conductor has N turns in a coil, do notforget multiply by N. Remember the sign rule forthe positive direction of emf and use itconsistently.If the circuit resistance is known, you cancalculate the magnitude of the induced current Iusing ε = IR.Step 4: Evaluate your answer : Check your results forthe proper units, and double-check that you haveproperly implemented the sign rules for calculatingmagnetic flux and induced emf.P.S.S. :: Inductors in Circuits :Step 1: Identify the relevant concepts : An inductor isjust another circuit element, like a source of emf, aresistor, or a capacitor. One key difference is thatwhen an inductor is included in a circuit, all thevoltages, currents, and capacitor charges are ingeneral functions of time, not constants as they havebeen in most of our previous circuit analysis. ButKirchhoff's rules, which we studied in section, arestill valid. When the voltages and currents vary withtime, Kirchhoff's rules hold at each instant of time.Step 2: Set up the problem using the following stepsDraw a large circuit diagram and label allquantities known and unknown. Apply thejunction rule immediately at any junction.Determine which quantities are the targetvariables.Step 3: Execute the solution as follows :Apply Kirchhoff's loop rule to each loop in thecircuit.As in all circuit analysis, getting the correct signfor each potential difference is essential. To getthe correct sign for the potential differencebetween the terminals of an inductor, rememberLenz's law and the sign rule described in sectiondiin conjunction with eq. ε = –L (self-induceddtemf) and fig. In Kirchhoff's loop rule, when wego through an inductor in the same direction asthe assumed current, we encounter a voltage dropequal to L di/dt, so the corresponding term in theloop equation is –L di/dt. When we go through aninductor in the opposite direction from theassumed current, the potential difference isreversed and the term to use in the loop equationis + L di/dt.aibdiV ab = L dtInductor with current i following from a to b:If di/dt > 0 : potential drops from a to bIf di/dt < 0: potential increases from a to bIf i is constant (di/dt = 0): no potential differenceAs always, solve for the target variables.Step 4: Evaluate your answer : Check whether youranswer is consistent with the way that inductorsbehave. If the current through an inductor ischanging, your result should indicate that thepotential difference across the inductor opposes theLXtraEdge for IIT-JEE 21 NOVEMBER 2011

change. If not, you probably used an incorrect signsomewhere in your calculation.P.S.S. :: Alternating –Current Circuits :Step 1: Identify the relevant concepts: All of theconcepts that we used to analyze direct-currentcircuits also apply to alternating current circuits.However, we must be careful to distinguish betweenthe amplitudes of alternating currents and voltagesand their instantaneous values. We must also keep inmind the distinctions between resistance (forresistors), reactance (for inductors or capacitors), andimpedance (for composite circuits).Step 2: Set up the problem using the following stepsDraw a diagram of the circuit and label all knownand unknown quantities.Determine the target variables.Step 3: Execute the solution as follows :In ac circuit problem it is nearly always easiest towork with angular frequency ω. If you are giventhe ordinary frequency f, expressed in Hz, convertit using the relation ω = 2πf.Keep in mind a few basic facts about phaserelationships. For a resistor, voltage and currentare always in phase, and the two correspondingphasor in a diagram always have the samedirection. For an inductor, the voltage alwaysleads the current by 90º (i.e., φ = + 90º), and thevoltage phasor is always turned 90ºcounterclockwise from the current phasor. For acapacitor, the voltage always lags the current by90º (i.e., φ = –90º), and the voltage phasor isalways turned 90º clockwise from the currentphasor.Remember that with ac circuits, all voltages andcurrents are sinusoidal functions of time insteadof being constant, but Kirchhoff's rules holdnonetheless at each instant. Thus, in a seriescircuit, the instantaneous current is the same in allcircuit elements; in a parallel circuit, theinstantaneous potential difference is the sameacross all circuit elements.Inductive reactance, capacitive reactance, andimpedance are analogous to resistance; eachrepresents the ratio of voltage amplitude V tocurrent amplitude I in a circuit element orcombination of elements. Keep in mind, however,that phase relations play an essential role. Theeffect of resistance and reactance have to becombined by vector addition of the correspondingvoltage phasors, as in fig(i) & (ii). When youhave several circuit elements in series, forexample, you can't just add all the numericalvalues of resistance and reactance to get theimpedance; that would ignore the phase relations.V L = IX LV L – V COV = IZφIV R =IRV C = IX CPhasor diagram for thecase X L > X CFig. (i)ωtV R = IRV L = IX LOωtV = IZIV L –V CV C =IX CPhasor diagram for thecase X L < X CFig. (ii)Evaluate your answer : When working with a seriesL-R-C circuit, you can check your results bycomparing the values of the inductive reactance X Land the capacitive reactance X C . If X L > X C , then thevoltage amplitude across the inductor is greater thanthat across the capacitor and the phase angle φ ispositive (between 0 and 90º). If X L < X C , then thevoltage amplitude across the inductor is less than thatacross the capacitor and the phase angle φ is negativebetween (0 and –90º).1. A coil of 160 turns of cross-sectional area 250 cm 2rotates at an angular velocity of 300 rad/sec. about anaxis parallel to the plane of the coil in a uniformmagnetic field of 0.6 weber/metre 2 . What is themaximum e.m.f. induced in the coil. If the coil isconnected to a resistance of 2 ohm, what is themaximum torque that has to be delivered to maintainits motion ?Sol. We know that, e max = NABω= 160 × 0.6 × (250 × 10 –4 ) × 300= 720 volt.Now i max =Solved Examplese max=R720 = 360 amp2τ(torque) = NiBA sin θτ max = NiBA= 160 × 360 × 0.6 × (250 × 10 –4 )= 864 newton metreThis torque opposes the rotation of the coil (Lenz'sLaw). Hence to maintain the rotation of the coil, anequal torque must be applied in opposite direction. Sothe torque required is = 864 newton metre.2. A closed coil having 50 turns, area 300 cm 2 , isrotated from a position where it plane makes an angleof 45º with a magnetic field of flux density 2.0weber/metre 2 to a position perpendicular to the fieldin a time 0.1 sec. What is the average e.m.f. inducedin the coil ?φXtraEdge for IIT-JEE 22 NOVEMBER 2011

Sol. The flux linked initially with each turn of the coil isΦ = B.A = BA cos θ = BA cos 45ºSubstituting the values, we get⎛ weber ⎞Φ = 2.0 ⎜ ⎟ × (300 × 10 4 metre –2 )×(0.7071)2⎝ metre ⎠= 4.24 × 10 –2 weberThe final flux linked with each turn of the coilΦ´ = BA cos 0º = BA= 2.0 × (300 × 10 –4 )= 6.0 × 10 –2 weberChange in flux = Φ´ – Φ= (6.0 × 10 –2 ) – 4.24 × 10 –2= 1.76 × 10 –2 WeberThis change is carried out in 0.1 sec. The magnitudeof the e.m.f. induced in the coil is given byd(Φ´−Φ)e = Ndt1.76×10= 50 ×0.1−5= 8.8 volt.3. A vertical copper disc of diameter 20 cm makes10 revolution per second about a horizontal axispassing through it centre. A uniform magnetic field10 –2 weber/m 2 acts perpendicular to the plane of thedisc. Calculate the potential difference between itscentre and rim in volt.Sol. The magnetic flux Φ linked with the disc is given byΦ = BAThe induced e.m.f. (potential difference) between rimand centredΦ d dA∴ e = – = – (BA) = B (numerically)dt dt dtdAwhere is the area swept out by the disc in unit time.dt∴∴dA = πr 2 × number of revolutions per seconddt= 3.14 × (0.1) 2 × 10= 0.314e = (10 –2 weber/m 2 ) × (0.314 m 2 /sec)= 3.14 × 10 –3 volt.4. A resistance R and inductance L and a capacitor C allare connected in series with an A.C. supply. Theresistance of R is 16 ohm and for a given frequency,the inductive reactance of L is 24 ohm and capacitivereactance of C is 12 ohm, If the current in the circuitis 5 amp., find(a) the potential difference across R, L and C(b) the impedance of the circuit(c) the voltage of A.C. supply(d) phase angleSol. (a) Potential difference across resistanceV R = iR = 5 × 16 = 80 voltPotential difference across inductanceV L = i × (ωL) = 5 × 24 = 120 voltPotential difference across condenserV C = i × (1/ωC) = 5 × 12 = 60 volt(b) Z =⎡⎢R⎢⎣21C⎛+ ⎜ωL−⎝ ω22⎞ ⎤⎟ ⎥⎠ ⎥⎦= [(16) + (24 −12)] = 20 ohm(c) The voltage of A.C. supply is given byE = iZ = 5 × 20 = 100 volt(d) Phase angleφ = tan –1 ⎡ωL− (1/ ωC)⎤⎢⎥⎣ R ⎦= tan –1 ⎡ 24 −12⎤⎢ ⎥⎣ 16 ⎦= tan –1 (0.75) = 36º46´5. A 100 volt A.C. source of frequency 500 hertz isconnected to a L-C-R circuit with L = 8.1 millinery,C = 12.5 microfarad and R = 10 ohm, all connectedin series. Find the potential difference across theresistance.Sol. The impedance of L-C-R circuit is given by2Z = [R + (X − X ) ]Lwhere X L = ωL = 2πfL= 2 × 3.14 × 500 × (8.1 × 10 –3 )= 25.4 ohm1and X C =ω C= 12πfC1=2×3.14×500×(12.5×10= 25.4 ohm2∴ Z = [(10) + (25.4 − 25.4) ] = 10 ohmEr.m.s 100 volt∴ i r.m.s. = = = 10 amp.Z 10 ohmPotential difference across resistanceV R = i r.m.s. × R = 10 amp × 10 ohm = 100 volt.C222−6)XtraEdge for IIT-JEE 23 NOVEMBER 2011

PHYSICS FUNDAMENTAL FOR IIT-JEESimple Harmonic MotionKEY CONCEPTS & PROBLEM SOLVING STRATEGYPeriodic motion is motion that repeats itself in adefinite cycle. It occurs whenever a body has a stableequilibrium position and a restoring force that actswhen it is displaced from equilibrium. Period T is thetime for one cycle. Frequency f is the number ofcycles per unit time. Angular frequency ω is 2π timesthe frequency.f = T1 or T = f1 ; ω = 2πf = T2πnmgaFyOxynO mgxya nxO F mgIf the net force is a restoring force F that is directlyproportional to the displacement x, the motion iscalled simple harmonic motion (SHM). In manycases this condition is satisfied if the displacementfrom equilibrium is small.F x = –kx;a x = mF x = – mk xRestoring force F xx < 0 F x = – kxF x > 00Displacement xx > 0F x < 0The circle of reference construction uses a rotatingvector called a phasor, having a length equal to theamplitude of the motion. Its projection on thehorizontal axis represents the actual motion of a bodyin simple harmonic motion.yDisplacementOAQPxx= A cos θIn SHM, the displacement, velocity, and accelerationare sinusoidal functions of time. The angularfrequency is ω = k / m ; the amplitude A and phaseangle φ are determined by the initial position andvelocity of the body.x = A cos(ωt + φ)xAO–AT2TEnergy is conserved in SHM. The total energy can beexpressed in terms of the force constant k andamplitude A.E = 21 mvx 2 + 21 kx 2 = 21 kA 2 = constantUEnergyE = K+UKx–A O AIn angular simple harmonic motion, the frequencyand angular frequency are related to the moment ofinertia I and the torsion constant k.ω =k and f =I12πkIA simple pendulum consists of a point mass m at theend of a massless string of length L. Its motion isapproximately simple harmonic for sufficiently smallamplitude; the angular frequency, frequency, andperiod depend only on g and L, not on the mass oramplitude.tThe angular frequency, frequency, and period inSHM do not depend on the amplitude, but only on themass m and force constant k.ω =k ; f =mω2π=12πkm; T = f1 = 2π kmθ TLxmg sinθmθmg cosθmgXtraEdge for IIT-JEE 24 NOVEMBER 2011

ω =2πT = ωg ω ; f = =L 2π= f1 = 2π gL12πA physical pendulum is a body suspended from anaxis of rotation a distance d from its center of gravity.If the moment of inertia about the axis of rotation is I,the angular frequency and period for small-amplitudeoscillations are independent of amplitude.mgd Iω = ; T = 2πImgdd sinθmg sinθOθmgzdcggLmg cosθProblem Solving Strategy :Simple Harmonic Motion I :Step 1: Identify the relevant concepts : An oscillatingsystem under goes simple harmonic motion (SHM)only if the restoring force is directly proportional tothe displacement. Be certain that this is the case forthe problem at hand before attempting to use any ofthe results of this section. As always, identify thetarget variables.Step 2: Set up the problem using the following stepsIdentify the known and unknown quantities, anddetermine which are the target variables.It's useful to distinguish between two kinds ofquantities. Basic properties of the system includethe mass m and force constant k. (In someproblems, m, k, or both can be determined fromother information.) They also include quantitiesderived from m and k, such as the period T,frequency f, and angular frequency ω. Propertiesof the motion describe how the system behaveswhen it is set into motion in a particular way.They include the amplitude A, maximum velocityv max , and phase angle φ, as well as the values ofx,v x , and a x at the particular time.If necessary, define an x-axis as.Step 3: Execute the solution as follows :Use the equations T = 1/f and ω = 2πf = 2π/T tosolve for the target variables.If you need to calculate the phase angle, becertain to express it in radians. The quantity ωt inEq. F x = – kx is naturally in radians, so φ must beas well.If you need to find the values of x, v x , and a x atvarious times, use Eqs.f =ω2π=12πkmdx, v x = = – ωA sin (ωt + φ)dt2dvand a x = x d x = = –ω 2 A cos (ωt + φ).2dt dtIf the initial position x 0 and initial velocity v 0x areboth given, you can determine the phase angle⎛ v ⎞and amplitude from Eqs. φ = arctan ⎜0x⎟⎝ ωx0 ⎠20x22 v(phase angle in SHM) and A = x 0 +ω(amplitude in SHM). If the body is given aninitial positive displacement x 0 but zero initialvelocity (v 0x = 0), then the amplitude is A = x 0and the phase angle is φ = 0. If it has an initialpositive velocity v 0x but no initial displacement(x 0 = 0), the amplitude is A = v 0x / ω and thephase angle is φ = –π/2.Step 4: Evaluate your answer : Check your results tomake sure that they're consistent. As an example,suppose you've used the initial position and velocityto find general expressions for x and v x at time t. Ifyou substitute t = 0 into these expressions, youshould get back the correct values of x 0 and v vx .Simple Harmonic Motion IIThe energy equation1 2 1E = mv x + kx 2 1= kA 2 = constant ...(i)2 2 2is a useful alternative relation between velocity andposition, especially when energy quantities are alsorequired. If the problem involves a relation amongposition, velocity, and acceleration without referenceto time, it is usually easier to use Eq.2d x ka x = = –2 x ...(ii)dt m(from Newton's second law) or eq. (i) (from energyconservation) than to use the general expressions forx, v x , and a x as functions of time [Eqs.x = A cos (ωt + φ) (displacement in SHM),dxv x = = –ωA sin (ωt + φ) (velocity in SHM) anddt2dva x = x d x= = – ω 2 A cos (ωt + φ) (acceleration2dt dtin SHM), respectively ]. Because the energy equationinvolves x 2 and v 2 x , it cannot tell you the sign of x orv x , you have to infer the sign from the situation. Forinstance, if the body is moving from the equilibriumposition towards the point of greatest positivedisplacement, then x is positive and v x is positive.XtraEdge for IIT-JEE 25 NOVEMBER 2011

Solved Examples1. A body of mass 1 kg is executing simple harmonicmotion which is given by x = 6.0 cos (100 t + π/4)cm. What is the (i) amplitude of displacement,(ii) frequency, (iii) initial phase, (iv) velocity,(v) acceleration, (vi) maximum kinetic energy ?Sol. The given equation of S.H.M. isx = 6.0 cos (100 t + π/4) cmComparing it with the standard equation of S.H.M.,x = a cos (ωt + φ), we have(i) amplitude a = 6.0 cm (ii) frequency ω = 100 /sec(iii) initial phase φ = π/4(iv) velocity v = ω (a2 − 2x ) = 100 2(36 − x )(v) acceleration = –ω 2 x = – (100) 2 x = – 10 4 x(vi) kinetic energy = 21 mv 2 = 21 mω 2 (a 2 – x 2 )When x = 0, the kinetic energy is maximum, i.e.,1(K.E.) max = mω 2 a 2 1= × 1 × 10 4 ⎛ 36 ⎞× ⎜ metre⎟ 2 2 ⎝100⎠= 18 joules2. A particle of mass 0.8 kg is executing simpleharmonic motion with an amplitude of 1.0 metre andperiodic time 11/7 sec. Calculate the velocity and thekinetic energy of the particle at the moment when itsdisplacement is 0.6 meter.2 −Sol. We know that, v = ω (a y )Further ω = 2π/T∴ v = 2 π(a2 − y2 ) =T22×3.14(11/ 7)[(1.0)2 −= 3.2 m/secKinetic energy at this displacement is given by(0.6)K = 21 mv 2 = 21 × 0.8 × (3.2) 2 = 4.1 joule3. A person normally weighing 60 kg stands on aplatform which oscillates up and down harmonicallyat a frequency 2.0 sec –1 and an amplitude 5.0 cm. If amachine on the platform gives the person's weightagainst time, deduce the maximum and minimumreading it will show, take g = 10 m/sec 2 .Sol. Acceleration of the platform a = ω 2 yMaximum accelerationa max = ω 2 A (A = Amplitude)∴ a max = (2πν) 2 A (ν = frequency)= 4(3.14) 2 (2) 2 × 0.05 = 7.88 m/sec 2m(g+ a max ) 60 (10 + 7.88)Maximum reading ==g10= 107.3 kgM(g− a max ) 60(10− 7.88)Minimum reading ==g10= 12.7 kg.2]4. A particle of mass m is located in a unidimensionalpotential field where the potential energy of theparticle depends on the coordinate x asU(x) = U 0 (1 – cos C x); U 0 and C are constants. Findthe period of small oscillations that the particleperforms about the equilibrium position.Sol. Given that U(x) = U 0 (1 – cos C x)dV(x)We know that F = ma = –dx1 ⎡ dU(x)⎤ 1∴ a = m⎢ − ⎥ = [– U0 C sin C]⎣ dx ⎦ mor a = –U 0 C[C x ] = –mU0C2mx (Q sin Cx ≈ Cx)Here acceleration is directly proportional to thenegative of displacement. So, the motion is S.H.M.Time period T is given by2πT = ω=(U2π0C2/ m)= 2π⎛⎜m⎝ U20 C5. Find the period of small oscillations in a verticalplane performed by a ball of mass m = 40 g fixed atthe middle of a horizontally stretched string l = 1.0 min length. The tension of the string is assumed to beconstant and equal to F = 10 N.Sol. The situation is showing in fig. The components of Tin upwards direction are T cos θ and T cos θ. Hencethe force acting on the ball = 2 T cos θl/2 l/2Tθxθ2Fx∴ma = –2 2( l / 4 + x )xQ T = F and cos θ =2 2( l / 4 + x )As x is small, x 2 can be neglected from thedenominator.2Fx ⎛ 4F ⎞∴ a = – = – ⎜ ⎟ xm( l / 2) ⎝ ml ⎠or a = – ω 2 x where ω 2 = (4 F/ml)Here acceleration is directly proportional to thenegative of displacement x. Hence the motion is S.H.M.2π2π⎛ ml ⎞T = == π ⎜ ⎟ ω (4F / ml)⎝ F ⎠Substituting the given values, we get⎛−2⎞T = 3.14 × ⎜(4×10 )(1.0)⎟ = 0.2 sec.⎝ 10 ⎠T⎞⎟⎠XtraEdge for IIT-JEE 26 NOVEMBER 2011

OrganicChemistryFundamentalsNITROGEN COMPOUNDBasicity of Amines :Amines are relatively weak bases. They are strongerbases than water but are far weaker bases thanhydroxide ions, alkoxide ions, and alkanide anions.A convenient way to compare the base strengths ofamines is to compare the acidity constants (or pK avalues) of their conjugate acids, the correspondingalkylaminium ions. The expression for this acidityconstant is as follows :R N + H 3 + H 2 O RNH 2 + H 3 O ++[RNH 2 ][H3O]K a =pK+a = – log K a[RNH3]The equilibrium for an amine that is relatively morebasic will lie more toward the left in the abovechemical equation than for an amine that is less basic.Consequently, the aminium ion from a more basicamine will have a smaller K a (larger pK a ) than theaminium ion of a less basic amine.When we compare aminium ion acidities in terms ofthis equilibrium, we see that most primaryalkylaminium ions (RNH + 3 ) are less acidic thanammonium ion (NH + 4 ). In other words, primaryalkylamines (RNH 2 ) are more basic than ammonia(NH 3 ):NH 3 CH 3 NH 2 CH 3 CH 2 NH 2 CH 3 CH 2 CH 2 NH 2Conjugate 9.26 10.64 10.75 10.67acid pK aWe can account for this on the basis of the electronreleasingability of an alkyl group. An alkyl groupreleases electrons, and it stabilizes the alkylaminiumion that results from the acid–base reaction bydispersing its positive charge. It stabilizes thealkylaminium ion to a greater extent than it stabilizesthe amine :R→– N – H + H – OHH–+R →– N – H + OHHHBy releasing electrons, R→– stabilizesthe alkylaminium ion throughdispersal of chargeThis explanation is supported by measurementsshowing that in the gas phase the basicities of thefollowing amines increases with increasing methylsubstitution :(CH 3 ) 3 N > (CH 3 ) 2 NH > CH 3 NH 2 > NH 3This is not the order of basicity of these amines inaqueous solution, however. In aqueous solution theorder is(CH 3 ) 2 NH > CH 3 NH 2 > (CH 3 ) 3 N > NH 3The reason for this apparent anomaly is now known.In aqueous solution the aminium ions formed fromsecondary and primary amines are stabilized bysalvation through hydrogen bonding much moreeffectively than are the aminium ions formed fromtertiary amines. The aminium ion formed from atertiary amine such as (CH 3 ) 3 NH + has only onehydrogen to use in hydrogen bonding to watermolecules, whereas the aminium ions from secondaryand primary amines have two and three hydrogens,respectively. Poorer solvation of the aminium ionformed from a tertiary amine more than counteractsthe electron-releasing effect of the three methylgroups and makes the tertiary amine less basic thanprimary and secondary amines in aqueous solution.The electron-releasing effect does, however, makethe tertiary amine more basic than ammonia.Basicity of ArylaminesAromatic amines (e.g., aniline and p- toluidine) aremuch weaker bases than the correspondingnonaromatic amine, cyclohexylamine :Cyclo-C 6 H 11 NH 2 C 6 H 5 NH 2 p-CH 3 C 6 H 4 NH 2Conjugate 10.64 4.58 5.08acid pK aWe can account for this effect, in part, on the basis ofresonance contributions to the overall hybrid of anarylamine. For aniline, the following contributors areimportant :NH 2NH 2+ +NH 2 NH 2–+NH 2–12345Structures 1 and 2 are the Kekule structures thatcontribute to any benzene derivative. Structures 3-5,however, delocalize the unshared electron pair of thenitrogen over the ortho and para positions of the ring.This delocalization of the electron pair makes it less–XtraEdge for IIT-JEE 27 NOVEMBER 2011

Enthalpyavailable to a proton, and delocalization of theelectron pair stabilizes aniline.When aniline accepts a proton it becomes ananilinium ion :C 6 H 5 N ⋅ ⋅H 2 + H 2 O C 6 H 5 N + H 3 + O –HAnilinium ionOnce the electron pair of the nitrogen atom acceptsthe proton, it is no longer available to participate inresonance, and hence we are only able to write tworesonance structures for the anilinium ion – the twoKekule structures:+ +NH 3 NH 3Structures corresponding to 3 – 5 are not possible forthe anilinium ion, and, consequently, althoughresonance does stabilize the anilinium ionconsiderably, resonance does not stabilize theanilinium ion to as great an extent as it does anilineitself. This greater stabilization of the reactant(aniline) when compare to that of the product(anilinium ion) means that ∆Hº for the reactionAniline + H 2 O → anilinium ion + OH –will be a larger positive quantity than that for thereactionCyclohexylamine + H 2 O→ cyclohexylaminium ion + OH –(See figure below) Aniline, as a result, is the weakerbase.Another important effect in explaining the lowerbasicity of aromatic amines is the electronwithdrawingeffect of a phenyl group. Because thecarbon atoms of a phenyl group are sp 2 hybridized,they are more electronegative (and therefore moreelectron withdrawing) than the sp 3 -hybridized carbonatoms of alkyl groups.+NH 3 + OH –+NH:NH Smaller resonance3 + OH –2 (1)stabilization of+ H anilinium ion2 O∆Hº 1∆Hº 2:NH 2 (2)Larger resonancestabilization of aniline + H 2 O∆Hº 2 > ∆Hº 1When primary, secondary and tertiary amines act asbases and react with acids, they form compoundscalled aminium salts. In an aminium salt thepositively charged nitrogen atom is attached to atleast one hydrogen atom :⋅ ⋅CH 3 CH 2 NH 2 + HCl⎯ ⎯⎯ → CHH2O(CH 3 CH 2 ) NH ⋅ ⋅+ HBr ⎯ ⎯⎯ →⋅ ⋅H2OH2(CH 3 CH 2 ) 3 N + HI ⎯ ⎯⎯ → (CHO+−3CH2N H3ClEthylaminium chloride(an aminium salt)+−3CH2 ) 2N H2BrDiethylaminium bromide(CH+−3CH2 ) 3N HITriethylaminium bromideWhen the central nitrogen atom of a compound ispositively charged but is not attached to a hydrogenatom, the compound is called a quaternaryammonium salt. For exampleCH 2 CH 3CH 3 CH 2 – N + – CH 2 CH 3 Br –CH 2 CH 3Tetraethylammonium bromide(a quaternary ammonium salt)Quaternary ammonium halides – because they do nothave an unshared electron pair on the nitrogen atom –cannot act as bases :+( CH−3 CH 2 ) 4 NBrTetra ethyl ammonium bromide(does not undergo reaction with acid)Quaternary ammonium hydroxides, however, arestrong bases. As solids, or in solution, they consistentirely of quaternary ammonium cations (R 4 N + ) andhydroxide ions (OH – ); they are, therefore, strongbases – as strong as sodium or potassium hydroxide.Quaternary ammonium hydroxides react with acids toform quaternary ammonium salts :(CH 3 ) 4+NOH – + HCl ⎯→ (CH 3 ) 4+NCl – + H 2 OBasicity of Aromatic Heterocyclic AminesIn aqueous solution, aromatic heterocyclic aminessuch as pyridine, pyrimidine, and pyrrole are muchweaker bases than nonaromatic amines or ammonia.In the gas phase, however, pyridine and pyrrole aremore basic than ammonia, indicating that solvationhas a very important effect on their relative basicities;NPyridinepK a = 5.23NNN HPyrimidine PyrrolepK a = 2.70 pK a = 0.40(Conjugate acid pK a)NQuinolinepK a = 4.5Aminium Salts and Quaternary Ammonium SaltsXtraEdge for IIT-JEE 28 NOVEMBER 2011

KEY CONCEPTInorganicChemistryFundamentalsNITROGEN FAMILYReaction of HNO 3 on Metals.(a) Metals lying below hydrogen in theelectrochemical series :Metals such as Na, K, Ca, Mg, Al, Zn, etc., lyingbelow hydrogen in the electrochemical seriesnormally displace hydrogen from dilute acids. Nitricacid also primarily behaves in the same manner. But,since it is a strong oxidising agent and hydrogen is areducing agent, secondary reactions take placeresulting in the reduction of nitric acid to give NO,N 2 O, N 2 or NH 3 , depending upon the nature of themetal, the temperature and the concentration of theacid. Thus, dilute nitric acid reacts with zinc in thecold giving N 2 O or N 2 according to the following eq.:4Zn + 10HNO 3 → 4Zn(NO 3 ) 2 + N 2 O + 5H 2 O5Zn + 12HNO 3 → 5Zn(NO 3 ) 2 + N 2 + 6H 2 OVery dilute nitric acid gives NH 3 which, of course, isneutralised by nitric acid to form NH 4 NO 3 .4Zn + 10HNO 3 → 4Zn(NO 3 ) 2 + 3H 2 O + NH 4 NO 3Similarly, iron and tin also give NH 4 NO 3 with dilutenitric acid. Lead gives nitric oxide with dilute nitricacid in cold. Magnesium and manganese givehydrogen. Concentrated nitric acid essentiallybehaves as an oxidising agent and metals likealuminium, iron, chromium, etc., are rendered'passive' probably due to surface oxidation.(b) Metals lying above hydrogen in theelectrochemical series. : Metals such as Cu, Bi, Hg,Ag, lying above hydrogen in the electrochemicalseries, do not liberate hydrogen from acids. In case ofthese metals, the action of nitric acid involves onlythe oxidation of the metals into the metallic oxideswhich dissolve in the acid to form nitratesaccompained by evolution of NO or NO 2 accordingas the acid is dilute or concentrated. For instance,concentrated acid attacks copper giving NO 2 .Cu + 4HNO 3 → Cu(NO 3 ) 2 + 2H 2 O + 2NO 2Dilute nitric acid gives NO.3Cu + 8HNO 3 → 3Cu(NO 3 ) 2 + 4H 2 O + 2NO(c) Noble metals : like Au, Pt, Rh and Ir are notattacked by nitric acid. Gold and platinum, however,are atacked by aqua regia (3 parts conc. HCl and 1part conc. HNO 3 ) which contains free chlorine.HNO 3 + 3HCl → 2H 2 O + 2Cl + NOClThis chlorine attacks gold and platinum forming solublechlorides which form complexes with HCl, e.g.,Au + 3Cl → AuCl 3AuCl 3 + HCl →HAuCl4Aurochloric acidHydroxylamine, NH 2 OH :It may be regarded as derived from ammonia by thereplacement of one H atom by an OH group.It is prepared by the reduction of nitrites with sulphurdioxide under carefully controlled conditions. Aconcentrated solution of sodium nitrite is mixed witha solution of sodium carbonate and sulphur dioxide ata temperature below 3ºC is passed till the solutionbecomes just acidic. The following reactions aresupposed to take place :Na 2 CO 3 + SO 2 + H 2 O → NaHSO 3 + NaHCO 3NaNO 2 + 3NaHSO 3 → HON (SO Na + Na 2 SO 3 + H 2 O3 ) 2Hydroxylaminesodium sulphonateThe sulphonate can be easily hydrolysed tohydroxylamine.HON(SO 3 Na) 2 ⎯H 2⎯→ ⎯O NH 2 OHAlternatively, it is prepared by the electrolyticreduction of nitric acid in 50% H 2 SO 4 usingamalgamated lead cathode.NO 2 – OH + 6H + + 6e – → NH 2 OH + 2H 2 OIt is a colourless solid melting at 33ºC. It is freelysoluble in water and lower alcohols. It is unstable anddecomposes violently even at 20ºC.3NH 2 OH → NH 3 + N 2 + 3H 2 OThe aqueous solution of hydroxylamine is less basicthan ammonia (K b = 1.8 × 10 –5 ). Thus,NH 2 OH(aq) + H 2 O NH 3 OH + + OH – ;K b = 6.6 × 10 –9Like H 2 O 2 , it acts as an oxidising as well as areducing agent depending upon circumstances.Nitrogen Trifluoride , NF 3 :It is conveniently prepared by fluorinating ammonia.4NH 3 + 3F 2 ⎯Cu ⎯catalyst ⎯⎯→NF 3 + 3NH 4 FIt can also be prepared by the electrolysis of NH 4 F.It is a colourless gas (m.p. –207ºC; b.p. –129ºC)which is quite stable thermodynamically.The gas acts as a fluorinating agent. It thus convertsCu into CuF.2NF 3 + 2Cu → N 2 F 4 + 2CuFAs, Sb and Bi also get fluorinated by interaction with NF 3 .XtraEdge for IIT-JEE 29 NOVEMBER 2011

NF 3 has a pyramidal structure with FNF angle = ~102º and dipole moment = 0.24 D, compared withHNH angle = ~ 107º and dipole moment = 1.48 D incase of NH 3 . The difference in the dipole moments ofNF 3 and NH 3 (both of which have pyramidalstructure though) is due to the fact that while thedipole moments due to N – F bonds in NF 3 are inopposite direction to the direction of the dipolemoment of the lone pair on N atom, the dipolemoments of N – H bonds in NH 3 are in the samedirection as the direction of the dipole moment of thelone pair on N atom, an illustrated below.••••NNF F H HFHBecause of its lower dipole moment, NF 3 is weakerligand than NH 3 .NF 3 is known to form complexes such as [NF 4 ] + andF 3 N→O. Thus,NF 3 + 2F 2 + SbF 200 º3 ⎯ ⎯⎯→ [NF + 4 ] [SbF 6 ] –2NF 3 + O 2High pressureElectric ⎯⎯⎯⎯⎯dischargelow temperature⎯ → 2F 3 N→ODinitrogen Difluoride, N 2 F 2 :Dinitrogen difluoride is best prepared by reactingNHF 2 with KF.KF + HNF low 2 ⎯ ⎯→ KF.HNF 2temperatureRoom temperature⎯⎯⎯⎯⎯⎯→N 2 F 2 + KHF 2The reaction yields both cis and trans isomers, viz.,FNNFFFBoth the isomeric forms are gases at roomtemperature, cis form boiling at – 106ºC and transform boiling at –112ºC. The cis form isthermodynamically more stable than the trans form.At above 70ºC, nearly 90% of N 2 F 2 is present in thecis form :Ntrans N 2 F 2> 70ºCcis N 2 F 2(~ 90%)If, however, the isomeric mixture is treated withAlCl 3 or the chlorides of bivalent Mn, Co, Ni and Fe,at low temperature, the major product is trans N 2 F 2 .The cis form reacts selectively with AsF 5 to form[N 2 F] + [AsF 6 ] – which when reacted with NaF – HF,regenerates the original compounds. The trans formdoes not react with AsF 5 .N + AsF 6 → [N2 F 2Mixture of cisand trans isomers+ −2F][AsF6]Formed bycis N2F2only[N 2 F] + [AsF 6 ] – ⎯NaF ⎯⎯−HF→Na + [AsF 6 ] – +N+N 2Ftrans 2N 2F 2(cis)Dinitrogen Tetrafluoride, N 2 F 4 :N 2 F 4 is prepared by reacting HNF 2 with NaClO.2HNF 2 ⎯⎯→N 2 F 4 + H 2 OHNF 2 , in turn, is obtained by first fluorinating ureaand then treating the fluorinated product withconcentrated sulphuric acid.N 2 F 4 exists both in the staggered and the gaucheconformations :N⎯ NaClOF F NFF F NFStaggered form (Side View) Gauche FormIt is a colourless gas (b.p. – 73ºC; m.p. –164ºC).It is a strong fluorinating agent. Thus,SiH 4 + N 2 F 4 → SiF 4 + N 2 + 2H 210 Li + N 2 F 4 → 4LiF + 2Li 3 NAsF 5 + N 2 F 4 → [N 2 F 3 ] + [AsF 6 ] –N 2 F 4 reacts with sulphur to give a number offluorinated sulphur compounds such as SF 4 andSF 5 .NF 3 .N 2 F 4 easily yields, at room temperature, the freeradical. NF 2 which reacts readily with a number ofcompounds. For example,Cl 22ClNF 2N 2 F 4 2(.NF 2 )–2NO2ON.NF 2Trisilylamine, N(SiH) 3 :Trisilylamine is prepared by reactingmonochlorosilane with ammonia.2SiH 3 Cl + 4NH 3 → N(SiH 3 ) 3 + 3NH 4 ClTrisilyamine is a trigonal planar compound with Norbitals in sp 2 hybrid state, unlike trimethyl ortriethylamine which is pyramidal and has N orbitalsin sp 3 hybrid state. There is considerable π overlapbetween the p orbital (containing the lone pair) of Natom and the vacant dπ orbitals of Si atoms. Thetrigonal planar structure of N(SiH 3 ) 3 is, thus,strengthened due to pπ–dπ bonding.Since the lone pairs of electrons of N atom areengaged in pπ-dπ bonding between N and Si, they areno longer available for donation to Lewis acids.Trisilylamine, therefore, behaves as much weakerbase compared to trimethylamine or triethylamine.Hence, trisilylamine does not form adducts with BH 3even at low temperature whereas trimethylamine ortriethylamine does so readily. Due to the samereason, N(SiH 3 ) 3 acts as a much weaker ligandcompared to N(CH 3 ) 3 and N(C 2 H 5 ) 3 .FFNXtraEdge for IIT-JEE 30 NOVEMBER 2011

UNDERSTANDINGPhysical Chemistry1. A container whose volume is V contains anequilibrium mixture that consists of 2 mol each ofPCl 5 , PCl 3 and Cl 2 (all as gases). The pressure is 3bar and temperature is T. A certain amount of Cl 2 (g)is now introduced, keeping the pressure andtemperature constant, until the equilibrium volume is2V. Calculate the amount of Cl 2 that was added and0the value of K p .Sol. At equilibrium, we havePCl52 molPCl +32 molCl22 molTotal amount = 6 mol2⎛ 2 ⎞⎜ p / pº ⎟(pPCl / pº )(pCl / pº )Thus Kº32p ==⎝ 6 ⎠(pPCl5/ pº ) ⎛ 2 ⎞⎜ p / pº ⎟⎝ 6 ⎠Substituting p = 3 bar, we get0K p = 1Let x be the amount of PCl 3 that combines when theamount y of chlorine is added keeping p and Tconstant. Thus, the amounts of PCl 3 , Cl 2 and PCl 5becomen(PCl 3 ) = 2 mol – xn(Cl 2 ) = y + 2 mol – xn(PCl 5 = 2 mol + xSince the final volume after the addition of Cl 2 istwice the initial volume, it follows that the totalamount of gases in 2V is 2 × 6 mol = 12 mol. Sincen(PCl 3 ) + n(PCl 5 ) is 4 mol, the total amount ofchlorine is 8 mol.Total amount = y + 6 mol – x = 12 molTheir partial pressures are2 mol − x 2 mol − xp PCl 3= p = × 3 bar12 mol 12 molp Cl 2=8 molp =12 mol8 mol× 3 bar = 2 bar12 mol2 mol + x 2 mol + xp PCl 5= p = × 3 bar12mol 12molSubstituting these in the expressionº K p =(pPCl / pº )(pCl3(pPCl52/ pº )/ pº )(where pº = 1 bar)we get º K p =⎛ 2 mol − x ⎞⎜⎟(2)⎝ 4 mol ⎠=⎛ 2 mol + x ⎞⎜⎟⎝ 4 mol ⎠or 4 – 2 (x/ mol) = 2 + (x/mol)or 3(x/mol) = 2 ⇒ x/mol = 2/3 = 0.67Therefore, the amount of Cl 2 addedy = 6 mol + x = 6.67 mol(2 mol − x)(2)= 1(2 mol + x)(as º K p = 1)2. A solution comprising 0.1 mol of naphthalene and0.9 mol of benzene is cooled until some solidbenzene freezes out. The solution is then decanted offfrom the solid, and warmed to 353 K, where itsvapour pressure is found to be 670 Torr. The freezingand normal boiling points of benzene are 278.5 K and353 K, respectively, and its enthalpy of fusion is10.67 kJ mol –1 . Calculate the temperature to whichthe solution was cooled originally and the amount ofbenzene that must have frozen out. Assumeconditions of ideal solution.Sol. The given data are :n 1 = 0.9 mol, n 2 = 0.1 mol,*T f = 278.5 K,*T b = 353 K,p* = 760 Torr, p = 670 Torr,∆ fus H 1,.m = 10.67 kJ mol –1From the relative lowering of vapour pressure,we obtain the amount fraction of the solute (i.e.naphthalene).p*− p 760 Torr – 670 Torrx 2 = = = 0.1185p* 760 TorrSince x 2 = n 2 /(n 1 + n 2 ), we getn 2 0.1moln 1 + n 2 = = = 0.844 molx 2 0.1185Since n 2 = 0.1 mol, we getn 1 = 0.844 mol – n 2= 0.844 mol – 0.1 mol = 0.744 molHence, the amount of benzene frozen out0.9 mol – 0.744 mol = 0.156 molThe freezing point depression constant of benzene isXtraEdge for IIT-JEE 31 NOVEMBER 2011

K f =M1RT∆ Hfus*2f1,m−1−1(0.078 kg mol )(8.314 J K mol=−1(10670 J mol )= 4.714 K kg mol –1Molality of the solution ism =nm21=n 2n M(0.1 mol)=(0.744 mol) (0.078 kg molFinally – ∆T f = K f m11−1)−1)(278.5K)= 1.723 mol kg –1= (4.714 K kg mol –1 )(1.723 mol kg –1 ) = 8.12 K3. What is the solubility of AgCl in 0.20 M NH 3 ?Given : K sp (AgCl) = 1.7 × 10 –10 M 2K 1 = [Ag(NH 3 ) + ] / [Ag + ] [NH 3 ] = 2.33 × 10 3 M –1 andK 2 = [Ag(NH 3 ) + 2 ]/[Ag(NH 3 ) + ][NH 3 ] = 7.14 × 10 3 M –1Sol. If x be the concentration of AgCl in the solution, then[Cl – ] = xFrom the K sp for AgCl, we deriveK−10 2[Ag + sp 1.7×10 M] = =−[Cl ] xIf we assume that the majority of the dissolved Ag +goes into solution as Ag(NH 3 ) + 2 then [Ag(NH 3 ) + 2 ] = xSince two molecules of NH 3 are required for everyAg(NH 3 ) + 2 ion formed, we have [NH 3 ] = 0.20 M – 2xTherefore,+[Ag ][NH3]K inst ==+[Ag(NH ) ]= 6.0 × 10 –8 M 2From which we derivex2322( 0.20M − 2x)2⎛⎜.7× 10−⎝ x2M ⎞⎟(0.20M − 2x)⎠x101 26.0×10=1.7×10−8−10MM22= 3.5 × 10 2which gives x = [Ag(NH 3 ) 2 + ] = 9.6 × 10 –3 M, whichis the solubility of AgCl in 0.20 M NH 34. The critical temperature and pressure for NO are 177K and 6.485 MPa, respectively, and for CCl 4 theseare 550 K and 4.56 MPa, respectively. Which gas (i)has smaller value for the van der Walls constant b;(ii) has smaller value of constant a; (iii) has largercritical volume; and (iv) is most nearly ideal inbehaviour at 300 K and 1.013 MPa.2Sol. We have T c (NO) = 177 K T c (CCl 4 ) = 550 Kp c (NO) = 6.485 MPa p c (CCl 4 ) = 4.56 MPa(i) SinceThus,pTcc=2a / 27b8a / 27Rb(177 K)(8.314 MPa cm Kb(NO) =(8)(6.485 MPa)= 28.36 cm 3 mol –1and=R therefore, b =8b3 –1 −1mol)3 −1−1T R550 K)(8.314 MPa cm K mol )b(CCl 4 ) =(8)(4.56MPa)= 125.35 cm 3 mol –1Hence b(NO) < b(CCl 4 )(ii) Since a = 27p c b 2thereforea(NO) = (27) (6.485 MPa) (28.36 cm 3 mol –1 ) 2= 140827 MPa cm 6 mol –2 ≡ 140.827 kPa dm 6 mol –2a(CCl 4 ) = (27) (4.56 MPa) (125.35 cm 3 mol –1 ) 2= 1934538 MPa cm 6 mol –2 ≡ 1934.538 KPa dm 6 mol –2Hence a(NO) < a(CCl 4 )(iii) Since V c = 3btherefore, V c (NO) = 3 × (28.36 cm 3 mol –1 )= 85.08 cm 3 mol –1V c (CCl 4 ) = 3 × (125.35 cm 3 mol –1 )= 376.05 cm 3 mol –1Hence V c (NO) < V c (CCl 4 )(iv) NO is more ideal in behaviour at 300 K and1.013 MPa, because its critical temperature is lessthan 300 K, whereas for CCl 4 the correspondingcritical temperature is greater than 300 K.5. Calculate ∆ r U, ∆ r H and ∆ r S for the process1 mole H 2 O (1,293 K, 101.325 kPa) →1 mol H 2 O (g, 523 K, 101.325 kPa)Given the following data :C p,m (1) = 75.312 J K –1 mol –1 ;C p,m (g) = 35.982 J K –1 mol –1∆ vap H at 373 K, 101.325 kPa = 40.668 kJ mol –1Sol. The changes in ∆ r U, ∆ r H and ∆ r S can be calculatedfollowing the reversible paths given below.Step I: 1 mole H 2 O(1,293 K, 101.325 kPa) →1 mole H 2 O(1,373 K, 101.325 kPa)q p = ∆ r H = C p,m (1) ∆T= (75.312 J K –1 mol –1 ) (80 K)= 6024.96 J mol –1c8pcXtraEdge for IIT-JEE 32 NOVEMBER 2011

XtraEdge for IIT-JEE 33 NOVEMBER 2011

CAREER POINTCorrespondence & Test Series Courses InformationCourses of IIT-JEECOURSE →INFORMATION↓Eligibility forAdmissionStudy MaterialPackageIIT-JEE 2012Class 12th or 12thPass StudentsAll India TestSeriesIIT-JEE 2012(At Center)Class 12th or 12thPass StudentsFor Class 12 th / 12 th Pass StudentsPostal All IndiaTest SeriesIIT-JEE 2012Class 12th or 12thPass StudentsMajor Test SeriesIIT-JEE 2012(At Center)Class 12th or 12thPass StudentsPostal Major TestSeriesIIT-JEE 2012(By Post)Class 12th or 12thPass StudentsCP Ranker'sPackageIIT-JEE 2012Class 12th or 12thPass StudentsStudyMaterialPackageIIT-JEE 2013Class 11thStudentsFor Class 11 th StudentsAll IndiaFoundationTest Series IIT-JEE 2013(At Center)Class 11thStudentsPostal AllIndia TestSeriesIIT-JEE 2013Class 11thStudentsMedium English or Hindi English English English English English English or Hindi English EnglishTest CenterPostalvisit our website orcontact usPostalvisit our website orcontact usPostal Postal Postalvisit our websiteor contact usIssue of ApplicationImmediate Immediate ImmediateImmediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate DispatchKitDispatch Dispatch DispatchDate of Dispatch/15 Jan 2012 15 Jan 2012Immediate 15 Aug 11 15 Aug 11Immediate Dispatch 15 Aug 11 onwards 15 Aug 11 onwardsImmediate DispatchFirst TestonwardsonwardsDispatch onwards onwardsCourse Fee 9000 3600 1500 1000 700 900 9800 4500 2500COURSE →INFORMATION↓Eligibility forAdmissionStudy MaterialPackageAIEEE 2012Class 12 th or 12 th PassStudentsAll India TestSeriesAIEEE 2012(At Center)Class 12 th or 12 thPass StudentsCourses of AIEEEPostal All IndiaTest SeriesAIEEE 2012Class 12 th or 12 thPass StudentsMajor Test SeriesAIEEE 2012(At Center)Class 12 th or 12 thPass StudentsPostal Major TestSeriesAIEEE 2012(By Post)Class 12 th or 12 thPass StudentsCP Ranker'sPackageAIEEE 2012Class 12 th or 12 thPass StudentsPostalFor Class 11 thStudentsStudyMaterialPackageAIEEE 2013Class 11 thStudentsMedium English or Hindi English or Hindi English or Hindi English or Hindi English or Hindi English or Hindi English or HindiTest CenterPostalvisit our website orcontact usFor Class 12 th / 12 th Pass StudentsPostalvisit our website orcontact usPostal Postal PostalIssue ofImmediateImmediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate DispatchApplication KitDispatchDate of Dispatch/15 Jan 2012 15 Jan 2012ImmediateImmediate Dispatch 15 Aug 11 onwards 15 Aug 11 onwardsImmediate DispatchFirst TestonwardsonwardsDispatchCourse Fee 9000 3600 1500 1000 700 900 9800Courses of Pre-MedicalFor Class 12 th / 12 th Pass StudentsFor Class 11 th StudentsCOURSE →INFORMATION↓Study MaterialPackageAIPMT 2012All India TestSeriesAIPMT 2012(At Center)Postal All IndiaTest SeriesAIPMT 2012Major Test SeriesAIPMT 2012(At Center)Postal Major TestSeriesAIPMT 2012(By Post)CP Ranker'sPackageAIPMT 2012StudyMaterialPackageAIPMT 2013All IndiaFoundationTest SeriesAIPMT 2013(At Center)Postal AllIndia TestSeriesAIPMT 2013Eligibility forAdmissionClass 12 th or 12 th PassStudentsClass 12 th or 12thPass StudentsClass 12 th or 12 thPass StudentsClass 12 th or 12 thPass StudentsClass 12 th or 12 thPass StudentsClass 12 th or 12 thPass StudentsClass 11 thStudentsClass 11 thStudentsMedium English or Hindi English or Hindi English or Hindi English or Hindi English English English or Hindi English EnglishClass 11 thStudentsTest CenterPostalvisit our website orcontact usPostalvisit our website orcontact usPostal Postal Postalvisit our websiteor contact usPostalIssue of ApplicationImmediate Immediate ImmediateImmediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate DispatchKitDispatch Dispatch DispatchDate of Dispatch/15 Jan 2012 15 Jan 2012Immediate 15 Aug 11 15 Aug 11Immediate Dispatch 15 Aug 11 onwards 15 Aug 11 onwardsImmediate DispatchFirst TestonwardsonwardsDispatch onwards onwardsCourse Fee 9000 3600 1500 1000 700 900 9800 4500 2500• For details about all correspondence & test Series Course Information, Please visit our website.: www.careerpointgroup.comXtraEdge for IIT-JEE 34 NOVEMBER 2011

XtraEdge for IIT-JEE 35 NOVEMBER 2011

XtraEdge for IIT-JEE 36 NOVEMBER 2011

∆ r S = C p,m lnTT21= (75.312 J K –1 mol –1 ⎛ 373K ⎞) × 2.303 × log⎜⎟⎝ 293K ⎠= 18.184 J K –1 mol –1∆ r U = ∆ r H – p∆ r V – ~ ∆ r HStep II: 1 mol H 2 O(1,373 K, 101.325 kPa)→ 1 mol H 2 O (g, 373K, 101.325 kPa)q p = ∆ vap H = 40.668 kJ mol –1–140668J mol∆ r S == 109.03 J K –1 mol –1373K∆ r U = ∆ r H – p∆ r V= 40668 J mol –1 – (101.325 kPa)⎛3 –1 373K ⎞⎜(22.414dmmol ) ×⎟⎝273K ⎠= 40668 J mol –1 – 3 103 J mol –1= 37565 J mol –1Step III: 1 mol H 2 O(g, 373 K, 101.325 kPa)→ 1 mol H 2 O(g, 523 K, 101.325 kPa)∆ r H = C p,m (g) ∆T= (35.982 J K –1 mol –1 ) (150 K) = 5397.3 J mol –1∆ r S = C p,m (g) lnTT21= (35.982 J K –1 mol –1 ⎛ 523K ⎞) × 2.303 × log⎜⎟⎝ 373K ⎠= (35982 J K –1 mol –1 ) × 2.303 × 0.1468= 12.164 J K –1 mol –1∆ r U = ∆ r H – R(∆T)= 5397.3 J mol –1 – (8.314 J K –1 mol –1 ) (150 K)= 5397.3 J mol –1 – 1247.1 J mol –1= 4 150.2 J mol –1Thus ∆U total = (6024.96 + 37565 + 4150.2) J mol –1= 47740.16 J mol –1∆H total = (6024.96 + 40668 + 5397.3) J mol –1= 52090.26 J mol –1∆S total = (18.184 + 109.03 + 12.164) J K –1 mol –1= 139.378 J K –1 mol –1Modern depiction of benzeneAs is standard for resonance diagrams, a doubleheadedarrow is used to indicate that the twostructures are not distinct entities, but merelyhypothetical possibilities. Neither is an accuraterepresentation of the actual compound, which is bestrepresented by a hybrid (average) of these structures,which can be seen at right. A C=C bond is shorterthan a C−C bond, but benzene is perfectlyhexagonal—all six carbon-carbon bonds have thesame length, intermediate between that of a singleand that of a double bond.A better representation is that of the circular π bond(Armstrong's inner cycle), in which the electrondensity is evenly distributed through a π-bond aboveand below the ring. This model more correctlyrepresents the location of electron density within thearomatic ring.The single bonds are formed with electrons in linebetween the carbon nuclei—these are called σ-bonds.Double bonds consist of a σ-bond and a π-bond. Theπ-bonds are formed from overlap of atomic p-orbitalsabove and below the plane of the ring. The followingdiagram shows the positions of these p-orbitals:Benzene electron orbitals :Since they are out of the plane of the atoms, theseorbitals can interact with each other freely, andbecome delocalised. This means that instead of beingtied to one atom of carbon, each electron is shared byall six in the ring. Thus, there are not enoughelectrons to form double bonds on all the carbonatoms, but the "extra" electrons strengthen all of thebonds on the ring equally. The resulting molecularorbital has π symmetry.Benzene orbital delocalisation :XtraEdge for IIT-JEE 37 NOVEMBER 2011

`tà{xÅtà|vtÄ V{tÄÄxÇzxá7 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, Kota1. Show that the lines 4x + y – 9 = 0, x – 2y + 3 = 0,5x – y – 6 = 0 make equal intercepts on any line ofgradient 2.2. ABC is a triangle with ∠A = 90°, AD is altitude.a acts along AB such that | a | =1/AB, b acts along1AC such that | b | = . Prove that a + b is aAC1vector along AD and | a + b | = . AD3. A circle passes through the origin O and cuts twolines x + y = 0 and x – y = 0 in P and Q respectively.If the straight line PQ always passes through a fixedpoint, find the locus of the centre of the circle.x x4. Let f (x) = a 1 tan x + a 2 tan + a2 3 tan +....+ an tan3x , where a1 , a 2 , a 3 ,...a n ∈ R and n ∈ N. If |f (x)| ≤n|tan x| for ∀ x ∈ ⎛ π π ⎞a⎜−, ⎟ , prove that⎝ 2 2 ⎠ ∑iii=1n≤ 1.5. Three digit numbers are formed. What is theprobability that the middle digit is largest.6. Prove that area of the region bounded by the curvey = log 2 (2 – x) and containing the points satisfyingthe inequality (x – |x|) 2 + (y – |y|) 2 ≤ 4 is⎛⎛ 2 ⎞⎜ π⎞⎜e e2 + − log ⎟⎟2 sq. units.⎜⎟⎝4 ⎝ 27 ⎠⎠7. r 1 , r 2 , r 3 be the radii of the circles drawn on thealtitudes respectively MD, ME and MF of thetriangles respectively ∆MBC, ∆MCA, ∆MAB, astheir diameters, where M be the circumcentreof the acute angled triangle ∆ABC. Prove that2a b +2r r2122c2+r 23≥ 144.8. Equilateral triangles are described externally on thesides BC, CA and AB of a given triangle ABC. Proveusing complex numbers that their centroids form anequilateral triangle9. Let a be a fixed real number satisfying 0 < a < π,a1−r cosusuch that T r =∫du21−2rcosu+ r−aProve that lim T r ,T 1 , lim T r form an A.P.r→1+r→1−10. Let a, b, c be real numbers such that the roots of thecubic equation x 3 + ax 2 + bx + c = 0 are all real.Prove that no one of these roots is greater than(2 a 2 − 3b– a)/3.Maths Funda1.62is the Golden Number, also called Phi.Golden Number property: ( + 1)/ = /1The fraction 1/998999 contains Fibonacci numbers,i.e.:1/998999=0.000001001002003005008013021034055089...Radii at 0° and approximately 222.49° divide a circlein the golden ratio: B/A = /1= ( 5 + 1)/2= 1 + (1 / (1 + (1 / (1 + (1 / (1 + ... ))))))= ( 4 + (4! - 4))/4= -2sin(666) ≈ F n+1 / F n (F = Fibonacci numbers)≈ 1.61803 39887 49894 84820 45868 34365 63811...The 3184th Fibonacci number is an apocalypsenumber (Apocalpyse numbers are numbers havingexactly 666 digits).XtraEdge for IIT-JEE 38 NOVEMBER 2011

MATHEMATICAL CHALLENGESSOLUTION FOR OCTOBER ISSUE (SET # 6)1. Utilize the formula : If a 1 + a 2 + ....... + a n = k(constant), then a 1 a 2 ..... a n has the greatest valuewhen a 1 = a 2 = ...... = a n = nk , where a1 , a 2 , ......, a nare all positive.(Using the concept of A.M. ≥ G.M.)Let E = (a – x) (b – y) (c – z) (ax + by + cz)Then abc E = (a 2 – ax)(b 2 – by)(c 2 – cz)(ax + by + cz)Now we have(a 2 – ax) + (b 2 – by) + (c 2 – cz) + (ax + by + cz)= a 2 + b 2 + c 2 (constant)a 2 – ax = b 2 – by = c 2 – cz = ax + by + cz2 2 2a + b + c=4∴ the greatest value of abc E =( a222+ b + c )256 abc2. xf ´(x) + f (x) = g(x) ...(1)xf ´(x) = g(x) – f (x) < 0; because f ´(x) < 0 & x > 0So g(x) < f (x)x g(x) < x f (x); as x > 0 ...(2)dNow from (1) (xf (x)) = g(x)dxxgso xf (x) =∫( x)dx use it is (2)0xgxg(x) 003. Let the eq n of chord bex + y = p ...(1)44. Let OC = → c|a| 2 = |b| 2 = |c| 2AB 2since = AC 1AaSo,O| a | θ 2=⎛ π ⎞ 1| a | ⎜ − θ⎟⎝ 2 ⎠2:1 Bwhere ∠AOB = θθ = π – 2θ ⇒ θ = π/3Hencea. b = |a| 2 π 1cos = |a|23 2b. c = |b| 2 πcos = |b| 6 23 2 =→→→bC3 |a| 2 & a. c = 02Let c = x a + y bSo a. c = x|a| 2 ⎛ y ⎞+ y a . b ⇒ ⎜ x + ⎟ |a| 2 = 0⎝ 2 ⎠y⇒ x = – 2andb. c = x a. b + y|b| 2⇒3 |a| 2 1= |a| 2 x + y|b| 22 2So, x + 2y = 3&yx = – 232135ºSo, x = – 2y + 2y = 3O(p, 0)⇒3y 2= 3 ⇒ y =23y 1Hence x = – = – 2 3In the limiting condition the line (1) will touch thecircle , Therefore p = 8,so as required |p| < 8So→c =→− a+32 →b3XtraEdge for IIT-JEE 39 NOVEMBER 2011

5. Sum will be odd if 1 out of 4 chosen numbers is oddand others are even or 1 is even & others are odd.P(O) =210 10C1.20C4C160 163P(E) = 1 – = 323 323Hence P(E) > P(O)3160= 3236. Let the point be P (x, y)so, 3x + 2y + 10 = 0 ...(1)since |PA – PB| is maximumhence P, A, B must be colinearx24y42111= 0⇒ – x – y + 6 = 0 ....(2)from (1) & (2)x = –22 & y = 28So, point P is (–22, 28)7. Let the point A be (x 1 , y 1 ) and the circle bex 2 + y 2 = a 2 Q 1P 1AP 2x − xLine AP 1 is 1 y − y= 1= rcosθsin θSolve it with circle.(x 1 + r cos θ) 2 + (y 1 + r sin θ) 2 = a 2Q 2x 1 2 + y 1 2 + 2rx 1 cos θ + 2ry 1 sin θ + r 2 = a 2so r 2 + (2x 1 cos θ + 2y 1 sin θ)r + x 1 2 + y 1 2 – a 2 = 0r 1 . r 2 = x 2 1 + y 2 1 – a 2 = ( S ) 21so, AP 1 . AQ 1 = ( S ) 21 sinceP 1 A . Q 1 A is independent on n, henceAP 1 . AQ 1 = AP 2 . AQ 2 = ........ = AP n . AQ n8. (AB) T = (BA) TB T A T = A T B Tso B T A T A = A T B T AB T = A T B T A (as AA T = I)AB T = AA T B T AAB T = B T A (again as AA T = I) Hence proved.bi − c9. iz =1+aiz + 1 bi − c + 1+a bi + a + (1 − c)==iz −1bi − c −1−a − ( c + 1) − ( a − ib)Now as given(a + ib) (a – ib) = 1 – c 2 = (1 – c) (1 + c)iz + 1=− iz + 1=2( a + b )bi + a +1+c21−c( c + 1) +a + ib( a + ib)[(1 + c)+ a − ib]2( c + 1) [ a + ib + 1−c]2⎛ a + ib=1 ⎟ ⎞ 1−iz⎜ . (using (1))⎝ c + ⎠ iz + 1iz + 1 a + ib =1−iz c + 122(Hence proved)...(i)10. Let n(n 2 – 1) = (n – 1) n (n + 1)Since n is odd so (n – 1) (n + 1) is the product of twoconsecutive even numbers, so it is divisible by 8.Since (n – 1) n (n + 1) is the product of 3 consecutiveintegers so it is divisible by 3 also Hence n(n 2 – 1) isdivisible by 24.Interesting Science Facts• The dinosaurs became extinct before the Rockiesor the Alps were formed.• Female black widow spiders eat their males aftermating.• When a flea jumps, the rate of acceleration is 20times that of the space shuttle during launch.• The earliest wine makers lived in Egypt around2300 BC.• If our Sun were just inch in diameter, the neareststar would be 445 miles away.• The Australian billy goat plum contains 100times more vitamin C than an orange.• Astronauts cannot belch - there is no gravity toseparate liquid from gas in their stomachs.• The air at the summit of Mount Everest, 29,029feet is only a third as thick as the air at sea level.• One million, million, million, million, millionthof a second after the Big Bang the Universe wasthe size of a …pea.• DNA was first discovered in 1869 by SwissFriedrich Mieschler.XtraEdge for IIT-JEE 40 NOVEMBER 2011

Students' ForumMATHSExpert’s Solution for Question asked by IIT-JEE Aspirants1. Let S(λ) be the area included between the parabolay = x 2 + 2x – 3 and the line y = λx + 1. Find the leastvalue of S(λ).Sol.αSolve y = x 2 + 2x – 3 & y = λx + 1.To get x 2 + (2 – λ) x – 4 = 0⇒ α + β = λ – 2, αβ = – 4S(λ) =∫ β [(λx + 1) – (x 2 + 2x – 3)] dxαS´(λ) =∫ β d [(λx + 1) – (x 2 + 2x – 3)]dxα dλ=∫ β β − αx dx =α 2( β − α)(β + α)S´(λ) =2( λ − 2)=22( λ − 2) + 16 = 0⇒ λ = 2Area is minimum for λ = 2min m 32Area = (find your self)32. Find the area enclosed by the curve,max {|x + y|, |x – y|} = 1Sol. Case I :If |x + y| ≥ |x – y|DCB2β2A⇒ |x + y| = 1⇒ locus of point (x, y) is two line segments AB andCD.Case IIIf |x + y| ≤ |x – y| ⇒ |x – y| = 1⇒ locus of point (x, y) is two line segments BD andAC.Then area bounded by the locus of (x, y) point is2 (unit) 2 , (because locus is a square of side one unit).3. If normals at the points P and Q of the parabolay 2 = 4ax meet at the point R of the parabola. Showthat the locus of centroid of the ∆ PQR is a ray. Findthe equation of the ray.Sol. Let P = (at 2 , 2at). Then Q is⎛⎤⎜ ⎛ ⎞ ⎛ ⎞a⎜2 2 2⎟ , 2a⎜⎟⎥⎜⎝ ⎝ t ⎠ ⎝ t ⎠⎥= ⎛ 4a 4a⎜ , ⎟ ⎞2⎦ ⎝ t t ⎠and R is (aT 2 2, 2aT). where T = – t – tcentroid of the ∆PQR⎡2 2 2at + 4a/ t + aT 2at+ 4a/ t + 2aT⎤= ⎢,⎥⎢⎣33 ⎥⎦y co-ordinates of the centroid4 a ⎛ 2 ⎞= 2at + + 2a ⎜−t − ⎟⎠ = 0t ⎝ tThus centrocid of the ∆PQR for any choice of P onthe parabola lies on the axis of the parabola.x-coordinates of the centroid⎡2a⎤2 4 ⎛ 2 ⎞= ⎢t+ + ⎜t+ ⎟ ⎥32⎢⎣t ⎝ t ⎠ ⎥⎦2a⎡ 2 4 ⎤ 2a= ⎢t+ + 232 ⎥ ≥ (4 + 2) = 4a⎣ t ⎦ 3Hence equation of the ray is given by y = 0, x > 4a4. In a class of 20 students, the probability that exactly xstudents pass the examination is directly proportionalto x 2 (0 ≤ x ≤ 20). Find out the probability that astudent selected at random has passed theexamination. If a selected students has been found topass the examination find out the probability thathe/she is only student to have passed theexamination.XtraEdge for IIT-JEE 41 NOVEMBER 2011

Sol. Let E x : event that exactly x out of 20 studentsand ∆pass the examination2 = 1 →| E B EA→ 1 → →× | = | a × b |22and A : event that a particular student passes→ →the examination| a × b |⇒ ∆⇒ P(E x ) = kx 2 2 =...(ii)(k is the proportionality constant)2Now, E 0 , E 2 , ....., E 20 are mutually exclusive andalso ∆ = 1 →| A C→ ( 1+ λ× B D | =1)(1+ λ2) → →| a × b |exhaustive events.22⇒ P(E 0 ) + P(E 1 ) + P(E 2 ) + ... + P(E 20 ) = 1→⇒ 0 + k(1) 2 + k(2) 2 + .... + k(20) 2 = 1| a →× b |∴ ∆ =( 1+λ1)(1+ λ 22)⎡ (20)(20 + 1)(40 + 1) ⎤⇒ k ⎢⎥ = 1⎣ 6 ⎦→ →| a× b |∆ =1+λ1+ λ2+ λ1λ212⇒ k = 2870λ1 + λwhere, 2≥202λ 1 λ 2Now, P(A) = ∑ P ( Ex ). P(A / E x )→ →x=0| a× b |∴ ∆ =1+2 λ1λ2 + λ1λ22020= ∑2 x k2kx . =3x=0 20 20∑xx = 0≥ 1 →| a × →b | +2( 1+λ1 λ2)221 ⎡ 20(20 + 1) ⎤ 63=20×2870⎢2⎥ = → →⎣ ⎦ 82| a × b | | →a× →b || λ= +1 λ 2 |P(E1).P(A / E1)22and P(E 1 /A) =P(A)∴ ∆ ≥ ∆ 2 + ∆ 1 {using (i) and (ii)1 1(1)2 .It is clear that equality holds if λ 1 = λ 2 and in this=2870 20 1case side AB and DC will become parallel.=63 441006. Let a 1 , a 2 , ......, a n be real constant, x be a real variable821 1and f (x) = cos(a 1 + x) + cos(a2 + x) + cos(a3 + x) +2 45. Let ABCD be any arbitrary plane quadrilateral in the1space having E as the point of intersection of its ...... + cos(an−1n + x). Given that f (x 1 ) = f (x 2 ) = 0,diagonals. If ∆ 1 and ∆ 2 be the areas of triangles DEC2and AEB, using vector method prove that prove that (x 2 – x 1 ) = mπ for integer m.Sol. f (x) may be written as,∆ ≥ ∆ 1 + ∆ 2 , where ∆ is the area of thenquadrilateral ABCD. Also discuss the case when the1f (x) =equality holds.∑ cos(ak + x)k−1k=12Sol. Let the position vector of the points A, B, C and andn1D with respect to E be → a , → b , – λ 1 . → → =a and –λ 2 b ; ∑ {cosak−1k . cos x – sin a k . sin x}where λ, λ 2 ∈ R +k=12⎛n⎞1 →Now, ∆ 1 = |→ λE C × E D | = 1λ2 → →| a × b |= cos x . ⎜cos a⎟⎜2∑k⎛n⎞– sin x ⎜sin a⎟k−1⎟2⎝ k=12⎜∑kk−1⎟⎠ ⎝ k=12 ⎠D(−λ2,b) C(−λ1,a)ncos a= A cos x – B sin x, where A = ∑kandk−1k=12E(0)nsin aB = ∑ksince f (xk−11 ) = f (x 2 ) = 0k=12A(a)B(b)⇒ A cos x 1 – B sin x 1 = 0 and A cos x 2 – B sin x 2 = 01⇒ ∆ 1 = |→a × →→ →A A| a× b |b || λ 1 λ2| =1 222λ λ ...(i) ⇒ tan x 1 = ⇒ tan x2 = B B⇒ tan x 1 = tan x 2 ⇒ (x 2 – x 1 ) = mπXtraEdge for IIT-JEE 42 NOVEMBER 2011

MATHSDIFFERENTIATIONMathematics FundamentalsDifferentiation and Applications of Derivatives :If y = f (x), thendy f ( x + h)− f ( x)1. = limdx h→0h2.3.⎛ dy ⎞⎜ ⎟⎝ dx ⎠⎛ dy ⎞⎜ ⎟⎝ dx ⎠x=ax=a==f ( x)− f ( a)limx − ax→af ( a + h)− f ( a)limhx→hIf u = f (x), v = φ(x), then1.d (k) = 0dx2.d du (ku) = kdx dx3.d du dv(u ± v) = ±dx dx dx4.d dv du(uv) = u + vdx dx dxdu ⎛ u ⎞5. ⎜ ⎟ =dx ⎝ v ⎠duvdx− uv2dvdxdy dy dx6. If x = f (t), y = φ (t), then = dx dt dtdy d7. If y = f[φ(x)], then = f´[φ(x)]. [φ(x)]dx dxdw dy8. If w = f (y), then = f ´(y) dx dxdy dy dx9. If y = f (x), z = φ(x), then = . dz dx dzdy dx dy10. . = 1 or =dx dy dx1.2.3.d (k) = 0dxd nx = nx n–1dxddx1 n = –1nxn+x1dx / dy4.ddx1x =2 x5.d e x = e xdx6.d a x = a x log adx7.d 1 log x =dx x8.ddx1 loga x = loga e x9.d sin x = cos xdxd10. dx cos x = – sin xd11. dx tan x = sec 2 xd12. cot x = – cosec 2 xdxd13. dx sec x = sec x tan xd14. dx cosec x = – cosec x cot xd15. sin –1 1x = dx21− xd16. cos –1 x = –dxd17. tan –1 x = dxd18. cot –1 x = –dxd19. sec –1 x = dx1+ xx11− x1d20. cosec –1 x = –dx211+ x122x 2 −1x1x 2 −1XtraEdge for IIT-JEE 43 NOVEMBER 2011

Suitable substitutions : The functions any also bereduced to simplar forms by the substitutions asfollows.1. If the function involve the term ( a − x ) , thenput x = a sin θ or x = a cos θ.2. If the function involve the term ( a + x ) , thenput x = a tan θ or x = a cot θ.3. If the function involve the term ( x − a ) , thenput x = a sec θ or x = a cosec θ.4. If the function involve the term222222a − x, then puta + xx = a cos θ or x = a cos 2θAll the above substitutions are also true, if a = 1Differentiation by taking logarithm :Differentiation of the functions of the following typesare obtained by taking logarithm.1. When the functions consists of the product andquotient of a number of functions.2. When a function of x is raised to a power which isitself a function of x.For example, let y = [f (x)] φ(x)Taking logarithm of both sides, log y = φ(x) log f (x)Differentiating both sides w.r.t 'x',1 dy f ´( x)= φ´(x) log f (x) + φ(x).y dxf ( x)= [f (x)] φ(x) log f(x).φ´(x) + φ(x) . [f (x) φ(x) – 1 .f´(x)dy= Differential of y treading f (x) as constantdx+ Differential of y treating φ(x) as constant.It is an important formula.Differentiation of implicit functions :1. If f(x, y) = 0 is a implicit function, thendy ∂f/ ∂x= –dx ∂f/ ∂yDiff.of f w.r.t. x keeping y constant= –Diff.of f w.r.t. y keeping x constantFor example, consider f (x, y) = x 2 + 3xy + y 2 = 0,thendy ∂f/ ∂x2x+ 3y= – = –dx ∂f/ ∂y3x+ 2y1. If y = f (x), then2.3.4.5.6.7.8.dy = y1 = f´(x),dxd 2 yndxddxddxddxddxddxddxddxnnnnnnnnnnnnnn= y n = f n (x)d2dx(ax + b) n = n ! a ny(ax + b) m = m(m – 1)e mx = m n e mxa mx = m n a mx (log a) nlog(ax + b) =2= y 2 = f´´(x), ......... (m – n + 1) a n (ax + b) m–n( −1)n−1na ( n −1)!( ax + b)sin (ax + b) = a n ⎛ nπ⎞sin ⎜ax+ b + ⎟⎝ 2 ⎠cos (ax + b) = a n ⎛ nπ⎞cos ⎜ax+ b + ⎟⎝ 2 ⎠Leibnitz's theorem : If u and v are any two functionsof x such that their desired differential coefficientsexist, then the n th differential coefficient of uv isgiven byD n (uv) = (D n u)v + n C 1 (D n–1 u)(Dv)+ n C 2 (D n–2 u)(D 2 v) +...... + u(D n v)Ability• We can accomplish almost anything win tin ourability if we but think we can.• He is the best sailor who can steer within fewestpoints of the wind, and exact a motive powerout of the greatest obstacles.• Our work is the presentation of our capabilities.• The wind and the waves are always on the sideof the ablest navigator.nXtraEdge for IIT-JEE 44 NOVEMBER 2011

MATHSSTRAIGHT LINE & CIRCLEMathematics FundamentalsDifferent standard form of the equation of a straightline :General form : Ax + By + C = 0where A, B, C are any real numbers not all zero.Gradient (Tangent) form : y = mx + cIt is the equation of a straight line which cuts off anintercept c on y-axis and makes an angle with thepositive direction (anticlockwise) of x-axis such thattan θ = m. The number m is called slope or thegradient of this line.Intercept form :x y + = 1a bIt is the equation of straight line which cuts offintercepts a and b on the axis of x and y respectively.Normal form (Perpendicular form) :x cos α + y sin α = pIt is the equation of a straight line on which thelength of the perpendicular from the origin is p and αis the angle which , this perpendicular makes with thepositive direction of x-axis.One point form :y – y 1 = m(x – x 1 )It is the equation of a straight line passing through agiven point (x 1 , y 1 ) and having slope m.Parametric equation :x − x 1 y − y= 1= rcosθsin θIt is the equation of a straight line passes through agiven point A(x 1 , y 1 ) and makes an angle θ withx-axis.Two points form :y2− y1y – y 1 = (x – x 1 )x2− x1It is the equation of a straight line passing throughy2− y1two given points (x 1 , y 1 ) and (x 2 , y 2 ), wherex2− x1is its slope.Point of intersection of two lines a 1 x + b 1 y + c 1 = 0and a 2 x + b 2 y + c 2 = 0 is given by⎛ b⎞⎜1 c2− b2c1a2c1− a1c2,⎟⎝ a1b2− a2b1a1b2− a2b1⎠Angle between two lines :The angle θ between two lines whose slopes are m 1and m 2 is given bym1− m2tan θ =1+m1m2If θ is angle between two lines then π – θ is also theangle between them.The equation of any straight line parallel to a givenline ax + by + c = 0 is ax + by + k = 0.The equation of any straight line perpendicular to agiven line, ax + by + c = 0 is bx – ay + k = 0.The equation of any straight line passing through thepoint of intersection of two given lines l 1 ≡ a 1 x + b 1 y+ c 1 = 0 and l 2 ≡ a 2 x + b 2 y + c 2 = 0 is l 1 + λl 2 = 0where λ is any real number, which can be determinedby given additional condition in the question.The length of perpendicular from a given point(x 1 , y 1 ) to a given line ax + by + c = 0 isax1+ by1+ c= p (say)2 2( a + b )In particular, the length of perpendicular from originc(0, 0) to the line ax + by + c = 0 is2 2a + bEquation of Bisectors :The equations of the bisectors of the angles betweenthe lines a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 area x + b y + c1a211 12+ b1= ±a x + b y + c2a222+ bDistance between parallel lines :Choose a convenient point on any of the lines(put x = 0 and find the value of y or put y = 0 and findthe value of x). Now the perpendicular distance fromthis point on the other line will give the requireddistance between the given parallel lines.Pair of straight lines :The equation ax 2 + 2hxy + by 2 = 0 represents a pair ofstraight lines passing through the origin.222XtraEdge for IIT-JEE 45 NOVEMBER 2011

Let the lines represented by ax 2 + 2hxy + by 2 = 0 bey – m 1 x = 0 and y – m 2 x = 0, then2ham 1 + m 2 = – and m 1 m 2 =bbGeneral equation of second degree in x, y isax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 ...(i)This equation represents two straight lines, if∆ = abc + 2fgh – af 2 – bg 2 – ch 2 = 0a h gor h b f = 0g f cand point of intersection of these lines is given by⎛ hf − bg − ⎞⎜⎟2⎝ −, hg afab h ab − h 2 ⎠The angle between the two straight lines representedby (i) is given by22 h − abtan θ = ±a + bIf ax 2 + 2hxy + by 2 + 2gx + 2f y + c = 0 represents apair of parallel straight lines, then the distancebetween them is given by22g − ac f − bc2 or 2a(a + b)b(a + b)Circle:Different forms of the equations of a circle :Centre radius form : the equation of a circle whosecentre is the point (h, k) and radius 'a' is(x – h) 2 + (y – k) 2 = a 2General equation of a circle : It is given byx 2 + y 2 + 2gx + 2f y + c = 0 ...(i)Equation (i) can also be written as2 + 2 | 2|x – (– g)| 2 + |y – (–f )| 2 = | g f − cwhich is in centre-radius form, so by comparing, weget the coordinates of centre (– g, – f ) and radius isg2 + f2 − c .Parametric Equations of a Circle :The parametric equations of a circle(x – h) 2 + (y – k) 2 = a 2 are x = h + a cos θ andy = k + a sin θ, where θ is a parameter.Lengths of intercepts on the coordinate axes made by22the circle (i) are 2 g − c and 2 f − cEquation of the circle on the line joining the pointsA(x 1 , y 1 ) and B(x 2 , y 2 ) as diameter is given by⎛ y − y ⎞⎜1⎛ y − y ⎞⎟⎝ x − x⎜2⎟ = 11 ⎠ ⎝ x − x2⎠If C 1 , C 2 are the centres and a 1 , a 2 are the radii of twocircles, then(i) The circles touch each other externally, ifC 1 C 2 = a 1 + a 2(ii) The circles touch each other internally, ifC 1 C 2 = |a 1 – a 2 |(iii) The circles intersects at two points, if|a 1 – a 2 | < C 1 C 2 < a 1 + a 2(iv) The circles neither intersect nor touch each other, ifC 1 C 2 > a 1 + a 2 or C 1 C 2 < |a 1 – a 2 |Equation of any circle through the point ofintersection of two given circles S 1 = 0 and S 2 = 0 isgiven by S 1 + λS 2 = 0 (λ ≠ –1) and λ can bedetermined by an additional condition.Equation of the tangent to the given circlex 2 + y 2 + 2gx + 2fy + c = 0 at any point (x 1 , y 1 ) on it,is xx 1 + yy 1 + g(x + x 1 ) + f (y + y 1 ) + c = 0The straight line y = mx + c touches the circle x 2 + y 2= a 2 , if c 2 = a 2 (1 + m 2 ) and the point of contact of thetangent y = mx ± a21+ m , is⎛⎜⎜⎝m ma,± a21+m 1+mLength of tangent drawn from the point (x 1 , y 1 ) to thecircle S = 0 is S 1 , whereS 1 = x 2 1 + y 2 1 + 2gx 1 + 2f y 1 + cThe equation of pair of tangents drawn from point(x 1 , y 1 ) to the circleS = 0 i.e. x 2 + y 2 + 2gx + 2f y + c = 0, is SS 1 = T 2 ,where T ≡ xx 1 + yy 1 + g(x + x 1 ) + f (y + y 1 ) + c and S 1 asmentioned above.Chord with a given Middle point :the equation of the chord of the circle S = 0 whosemid-point is (x 1 , y 1 ) is given by T = S 1 , where T andS 1 as defined a above.If θ be the angle at which two circles of radii r 1 and r 2intersect, then21222r + r − dcos θ =2r1r2where d is distance between their centres.Note: Two circles are said to be intersectorthogonally if the angle between their tangents attheir point of intersection is a right angle i.e.r 2 1 + r 2 2 = d 2 or2g 1 g 2 + 2f 1 f 2 = c 1 + c 2Radical axis : The equation of the radical axis of thetwo circle is S 1 – S 2 = 0 i.e.2x(g 1 – g 2 ) + 2y(f 1 – f 2 ) + c 1 – c 2 = 02⎞⎟⎟⎠XtraEdge for IIT-JEE 46 NOVEMBER 2011

Based on New PatternIIT-JEE 2012XtraEdge Test Series # 7Time : 3 HoursSyllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabusInstructions :Section – I :Question 1 to 10 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 11 to 15 are multiple choice question with multiple correct answer. +4 marks will be awarded forcorrect answer and -1 mark for wrong answer.Section – III : Question 16 to 21 are passage based single correct type questions. +3 marks will be awarded forcorrect answer and -1 mark for wrong answerSection – IV : Question 22 to 23 are Column Matching type questions. +8 marks will be awarded for the completecorrectly matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.PHYSICSQuestions 1 to 10 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. + 3 marks will be given for each correctanswer and – 1 mark for each wrong answer.1. Given are four arrangements of three fixed electriccharges. In each arrangement, a point labeled P isalso identified a test charge, +q, is placed at point P.All of the charges are the same magnitude Q. Butthey can be either +ve or –ve as shown. The chargesand point P all lie on a straight line. The distancebetween adjacent items, either between two chargesor between a charge and point P are all the same(I) P (II) P(III) P (IV) PCorrect order of choices in a decreasing order ofmagnitude force on P is -(A) II > I > III > IV (B) I > II > III > IV(C) II > I > IV > III (D) III > IV > I > II2. A segment of angle θ is cut from a half disc,symmetrically as shown. If the centre of mass ofthe remaining part is at a distance 'a' from O and thecentre of mass of the original half disc was atdistance d then it can be definitely said that :(A) a = d(B) a > d(C) a < d(D) all the above are possible depending on θ3. The moment of inertia of a hollow thick sphericalshell of mass M and its inner radius R 1 and outerradius R 2 about diameter is :5 55 52M( R2− R1) 2M( R2− R1)(A)(B)3 33 35( R − R )3( R − R )252324M( R −(C)5( R − R15R131 ))2523214M( R −(D)3( R − R5R131 )4. A cyclic process ABCA is shown in a V-T diagram.The corresponding PV diagram will beVB CAT(A)PPABCC(B)BVPPABB)CVθ(C)AV(D)ACVOXtraEdge for IIT-JEE 47 NOVEMBER 2011

5. A rectangular metal plate of dimension (a × b)having two holes of radii r 1 and r 2 (r 1 > r 2 ) and theirpositions are shown in the figure, now the plate isheated such that its temperature rises by ∆T. Thenseparation between the holes :da(A) decreases (B) increases(C) remain constant (D) can not say6. Assuming all surface to be smooth minimum valueof horizontal acceleration 'a' so that sphere lossescontact at P is –QαPα(A) g sinα (B) g tanα (C) g cotαa(D) g cosecα7. A uniform rod of mass M and length L lies flat on africtionless horizontal surface. Two forces F and 2Fare applied along the length of the rod as shown.The tension in the rod at the point P isLF(A)3F4(B) 3FP(C)5F4L/4(D)7F48. The K α wavelength of an element with atomicnumber z is λ z . The k α wavelength of an elementwith atomic number 3z is λ 3z . Then(A) λ z > 9λ 3z (B) λ z < 9λ 3z(C) λ z = 9λ 3z(D) Depending on z, λ z can be greater or smallerthen (9λ 3z ).9. The BE per nucleon of deutron ( 1 H 2 ) and heliumnucleus ( 2 He 4 ) is 1.1 MeV and 7 MeV respectively.If two deutron react to form a single heliumnucleus, then energy released is -(A) 23.6 MeV (B) 4.8 MeV(C) 25.8 MeV (D) None of these10. In a radioactive decay, let N represent the numberof residual active nuclei, D the number of daughternuclei, and R the rate of decay at any time t. Threecurves are shown in Fig. The correct ones are –Nt(1)Dt(2)RNt(3)2F(A) 1 and 3 (B) 2 and 3(C) 1 and 2(D) all threeQuestions 11 to 15 are multiple choice questions.Each question has four choices (A), (B), (C) and (D),out of which MULTIPLE (ONE OR MORE) iscorrect. Mark your response in OMR sheet againstthe question number of that question. + 4 marks willbe given for each correct answer and – 1 mark foreach wrong answer.11. Consider a hypothetical binding energy per nucleoncurve. Which of the following fission or fusion mayoccur ?(BE) n (MeV/n)1084 6250 100 150 200(A) X 250 → Y 160 + Z 90 (B) X 220 → Y 180 + Z 40(C) X 40 + Y 70 → Z 110 (D) X 40 + Y 120 → Z 16012. In series R-C circuit :RV = V 0 sin ωt(A) current leads the applied voltage byφ = tan –1 1ωRC(B) current lags the applied voltage by φ < 90º(C) V c < V0 0 , where V c 0and V 0 are the maximumvalues of the voltage across the capacitor andapplied voltage respectively(D) applied voltage, voltage across the resistor andcurrent are in phase.13. A sound wave of frequency 'f ' travels horizontallyto the right. It is reflected from a large verticalplane surface moving to the left with speed v 0 . Thespeed of sound in the medium is v. Choose thecorrect statement.(A) The number of waves striking the surface per⎛ v + v0⎞second is ⎜ ⎟ f⎝ v ⎠(B) The wavelength of the reflected wave is⎛ v + v ⎞⎜0⎟f⎝ v − v0⎠CXtraEdge for IIT-JEE 48 NOVEMBER 2011

(C) The frequency of the reflected wave is⎛ v + v ⎞⎜0⎟f⎝ v − v0⎠(D) The number of beats heard by a stationarylistener to the left of the reflecting surface is⎛ v ⎞⎜0⎟f⎝ v − v 0 ⎠14. Illustrated below is a uniform cubical block of massM and side a. Mark the correct statement(s)paDMAThis section contains 3 paragraphs; each has 2multiple choice questions. (Questions 16 to 21) Eachquestion has 4 choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. +3 marks will be given for each correctanswer and – 1 mark for each wrong answer.Passage # 1 (Ques. 16 - 17)Consider an electron moving in a circular orbit ofradius 'r' in an external uniform and steadymagnetic field B. Assume that Bohr's quantizationprinciple regarding the angular momentum is truein this case. Now answer the following question :16. If radius of n th orbit is r n and speed in this orbit is v nthen correct relationship between them(A) r n ∝ v n (B) r n ∝ v n2(C) r n 2 ∝ v n(D) r n ∝ 1/v nC(A) The moment of inertia about axis A, passingthrough the centre of mass is I A = 61 Ma2(B) The moment of inertia about axis B, whichbisect one of the cube faces is I B = 125 Ma2(C) The moment of inertia about axis C, along oneBof the cube edges is I C = 32 Ma2(D) The moment of inertia about axis D, whichbisects one of the horizontal cube faces is7 Ma21215. In the figure the block on the smooth table is setinto motion in a circular orbit of radius r round thecentre hole. The hanging mass is identical to themass on the table and remains in equilibrium,neglect friction. The string connecting the twoblocks is massless and intextensible. Select thecorrect answer.mr(A) the angular speed ω of the block in its circularmotion is g / r(B) kinetic energy as function of r is given bymgr/2(C) angular momentum about the hole is conservedeven if hanging block is pulled down(D) The block on table is in equillibriumm17. If total energy of e – in moving these orbit is sum ofKE and potential energy of interaction between themagnetic moment of orbital current and magneticfield B. Then total energy in n th orbit is(e = charge of electron, m = mass of e – , h = plank'sconstant)nheBnheB(A) E n =(B) E n =2πm4πm2nheB(C) E n = zero (D) E n =πmPassage # 2 (Ques. 18 - 19)When the strain is small (say < 0.01), the stress isproportional to the strain. This is the region whereHook's law is valid and where young's modulus isdefined. The point a on the curve represents theproportional limit up to which stress and strain areproportional. If the strain is increased a little bit, thestress is not proportional to the strain. However,the wire still remains elastic. This means, if thestretching force is removed, the wire acquires itsnatural length. This behaviour is shown up to apoint b on the curve known as the elastic limit onthe yield point. If the wire is stretched beyond theelastic limit, the strain increases much morerapidly. If the stretching force is removed, the wiredoes not come back to its natural length. Somepermanent increase in length takes places. In figure,we have shown this behaviour by the dashed linefrom C. The behaviour of the wire is now plastic.If the deformation is increased further, the wirebreaks at a point d known as fracture point. If largedeformation takes place between the elastic limitand the fracture point, the material is called ductile.If it breaks soon after the elastic limit is crossed, itis called brittle.XtraEdge for IIT-JEE 49 NOVEMBER 2011

stressc IIIplastic behavioura ba = proportional limitIIb = Elastic limitd = fracture pointI Elastic behaviourostrain0.3Permanent set18. A metal wire will retain its original shape whenload is removed in region.(A) oc (B) cd (C) ac (D) od19. Yield point and elastic limit point coincide for(A) ductile material (B) malleable material(C) brittle material (D) elastic materialPassage # 3 (Ques. 20 - 21)A uniform dense solid cylinder of mass m andradius R is released from rest on an inclined plane.After releasing from rest it starts performing purerolling (i.e. rolling without slipping). As there is noslipping the friction force acts is static in nature.Therefore the relative velocity between the pointsin contact is zero.We know that rolling is combined rotation andtranslation. During its downward journey along theincline the cylinder moves distance l along theincline. The angle ofinclination from horizontal isα and the coefficient offriction is given as µ. Theacceleration due to gravity isg downwards. Air resistance αis not present.20. The acceleration of centre of mass of cylinder is-(A) g sin α – µg cos α (B) g sin α(C)2g sin α3d(D) none of these21. (The final angular speed of cylinder is4 glsinα2 glsinα(A)(B)223 R3 R1 glsinα(C)(D) none of these23 RThis section contains 2 questions (Questions 22, 23).Each question contains statements given in two columnswhich have to be matched. Statements (A, B, C, D) inColumn I have to be matched with statements (P, Q, R,S, T) in Column II. The answers to these questions haveto be appropriately bubbled as illustrated in thefollowing example. If the correct matches are A-P, A-S,A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then thecorrectly bubbled 4 × 5 matrix should be as follows :P Q R S TA P Q R S TB P Q R S TC P Q R S TD P Q R S TMark your response in OMR sheet against thequestion number of that question in section-II.+ 8 marks will be given for complete correct answer(i.e. +2 marks for each correct row) and No Negativemarks for wrong answer.22. Assume that 2 bodies collide head on. The graph oftheir velocities with time are shown in column-Imatch them with appropriate situation in column II(A)(B)Column -Ivv(1) (2)(1)(2)v(C)(1)v(D) (1)tttColumn-IIm 1 m 2(P)(Q) m 1m 1 < m 2 , 0 < e v 2m 1 m 2m 1 > m 2 e = 1Column-II(P)(Q)XtraEdge for IIT-JEE 50 NOVEMBER 2011

(C) The charge is projectedperpendicular to E in a crossedE r and E rBq(D) The charge is projected at a nonzeroangle θ(< 90º) with themagnetic inductionqvBE(R)(S)3. Which of the following graphs are properlyrepresented ?(A)(B)RateRateTemperatureTemperatureFor normal reactionFor explosive reactionCHEMISTRY(T)(C)RateTemperatureFor all normal reactionQuestions 1 to 10 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. + 3 marks will be given for each correctanswer and – 1 mark for each wrong answer.1. The IUPAC name of compoundHO – C = O CH 3NH 2 – C ==== C ––– C – H is –NH 2 Cl(A) 2, 3 diamino-4-chloro-2-pentenoic acid(B) 4-chloro-3, 3-diamino-2-pentenoic acid(C) 3, 3–diamino-3-chloro-pentenoic acid(D) All of the above2. Identify the correct statements -H 3 CCH 3(A) The compoundfails to undergoCOOH Odecarboxylation(B) A Grignard reagent can be successfully madeBrfrom the following dibromideBr(C) Cyclopentan –1, 3– dione exists almost 100%in the enol form whereas diacetyl(CH 3 COCOCH 3 ) can exist in the keto form aswell as the enol form(D) Among the following resonance structure givenbelow, (ii) will be the major contributor to theresonance hybrid.Θ ..C–CH 3 ↔ C–CH 3(i)O : O ..– :(ii)(D)Rate45ºCTemperatureFor explosive reaction4. A hydrogen electrode is placed in a buffer solutionof sodium acetate and acetic acid in the ratio a : band other in the ratio b : a was taken. If electrodepotential values are found to be E 1 and E 2 thenwhich of the following is/are correct for the pK avalue of the acid.E1 − E2E1 + E 2(A)(B) –0.1180.118E 1E(C) × 0.118 (D)2 − E1E0. 11825. Identify the incorrect statement –(A) In solid state N 2 O 5 exist as NO 2 + and NO 3 – ions(B) Solid PCl 5 contains PCl 4 + and PCl 6 – ions(C) Solid PBr 5 has PBr 4 + and Br –(D) In N 4 S 4 all the bond angles are equal6. A bulb of constant volume is attached to amanometer tube open at other end as shown infigure. The manometer is filled with a liquid ofdensity (1/3 rd ) that of mercury. Initially h was 228cm.GasThrough a small hole in the bulb gas leakeddpassuming pressure decreases as = – kP. dthXtraEdge for IIT-JEE 51 NOVEMBER 2011

If value of h is 114 cm after 14 minutes. What is thevalue of k (in hour –1 ) ?[Use : ln(4/3) = 0.28 and density of Hg = 13.6g/mL](A) 0.6 (B) 1.2(C) 2.4(D) None of these7. The dipole moment of HCl is 1.03D, if H–Cl bonddistance is 1.26Å, what is the percentage of ioniccharacter in the H–Cl bond ?(A) 60% (B) 29% (C) 17% (D) 39%8. Arrange NH + 4 , H 2 O, H 3 O + , HF & OH – in increasingorder of acidic nature -(A) OH – < H 2 O < NH + 4 < HF < H 3 O +(B) H 3 O + > HF > H 2 O > NH + 4 > OH –(C) NH + 4 < HF < H 3 O + < H 2 O < OH –(D) H 3 O + < NH + 4 < HF < OH – < H 2 O9. A compression of an ideal gas is represented bycurve AB, which of the following is wrongB(v B )log PA(v A )log VVA(A) number of collision increases timesVB(B) number of moles in this process is constant(C) it is isothermal process(D) it is possible for ideal gas10. A compound containing only sodium, nitrogen andoxygen has 33.33% by weight of sodium. What isthe possible oxidation number of nitrogen in thecompound?(A) –3 (B) + 3 (C) –2 (D) + 5Questions 11 to 15 are multiple choice questions.Each question has four choices (A), (B), (C) and (D),out of which MULTIPLE (ONE OR MORE) iscorrect. Mark your response in OMR sheet againstthe question number of that question. + 4 marks willbe given for each correct answer and – 1 mark foreach wrong answer.11. Which of the following is/are correct regarding theperiodic classification of elements ?(A) The properties of elements are the periodicfunction of their atomic number(B) Non metals are lesser in number than metals(C) The first ionization energies of elements in aperiod do not increase with the increase inatomic numbers(D) For transition elements the d-subshells arefilled with electrons monotonically with theincrease in atomic number12. In the purification Zr and B, which of the followingis/are true ?(A) Zr + 2I 2 ⎯→ ZrI 4 (g)⎯passed over⎯ ⎯⎯⎯the whitehot Wthe pure Zr is deposited on W(B) 2B + 3I 2 ⎯→ 2BI 3 (g)⎯the pure B is deposited on W(C) Zr + 2I 2 ⎯→ ZrI 4 (s)(D) none of these⎯→passed over⎯ ⎯⎯⎯the whitehot Wmixed with W⎯ ⎯⎯⎯& then heated→→ZrI 4 is reduced to ZrI 213. Which of the following statements is correct ?(A) At 273ºC, the volume of a given mass of a gasat 0ºC and 1 atm. pressure will be twice itsvolume(B) At –136.5ºC, the volume of a given mass of agas at 0ºC and 1 atm. pressure will be half of itsvolume(C) The mass ratio of equal volumes of NH 3 andH 2 S under similar conditions of temperature andpressure is 1 : 2(D) The molar ratio of equal masses of CH 4 andSO 2 is 4 : 114. Dopamine of a drug used in the treatment ofparkinson's disease.NHHO CH 2 –CH 2COOHHODopaminWhich of the following statements about thiscompound are correct ?(A) It can exist in optically active forms.(B) One mole will react with three moles of sodiumhydroxide to form a salt(C) It can exist as a Zwitter ion in the aqueous solution(D) It gives nitroso compound on treatment withHNO 2 .15. In the given table types of H bonds and some Hbond energies are given and other H bond energiesare not given. You are to perdit the unknown H-bond energies.Types of H-bonds H-bond energies inKJ/mol(I) F – H …….. O –F – H …….. F 30(II) O – H …….. O –O – H …….. F 15(III) F – H …….. F – –(IV) N – H …….. N –Correct prediction are –(A) H-bond energy for (I) may be 20 kJ/mol(B) H-bond energy for (II) may be 25 kJ/mol(C) H-bond energy for (III) may be 113 kJ/mol(D) H-bond energy for (IV) may be 12 kJ/molXtraEdge for IIT-JEE 52 NOVEMBER 2011

This section contains 3 paragraphs; each has 2multiple choice questions. (Questions 16 to 21) Eachquestion has 4 choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. +3 marks will be given for each correctanswer and – 1 mark for each wrong answer.Passage # 1 (Ques. 16 - 17)Freezing point of a liquid is defined as thattemperature at which it is in equilibrium with itssolid phase.VapourPressureP = 1 atmsolventSolidsolution∆T fT T 0T 0 > TTemperaturePhase diagram for a pure solvent and solution fordepression in freezing point.16. The freezing point of the solvent is -∆H – ∆G∆H(A)(B)T∆S∆S∆G∆S(C)(D)∆S∆H17. Freezing point of the solution is smaller than thefreezing point of the solvent. Because -(A) ∆H of solution and ∆H of solvent are almostsame due to identical intermolecular forces(B) ∆S of the solution is larger than ∆S of solvent(C) ∆S of the solution is smaller than the ∆S ofsolvent(D) ∆H of the solution is much higher than the ∆Hof solvent but ∆S of solution is smaller than thatof the solventPassage # 2 (Ques. 18 - 19)Effect of temperature on the equilibrium process isanalysed by using the thermodynamics.From the thermodynamics relation∆G° = – 2.3 RT logK........(1) ∆G° = Standard freeenergy change∆G° =∆H° – T∆S°….(2) ∆H° = Standard heat ofthe reactionFrom (1) & (2)– 2.3 RT log K = ∆H° – T∆S° ; ∆S° : StandardEntropy change,∆H° ∆S°log K = + ........(3)2.3RT 2.3RClearly if a plot of log K vs 1/T is made then it is astraight line having slope– ∆H°∆S°= & y–intercept = .2.3R2.3RIf at a temperature T 1 equilibrium constant be K 1and at temperature T 2 equilibrium constant be K 2then, the above equation reduces to :– ∆H°∆S°⇒ log K 1 = + ........(4)2.3RT 1 2.3R– ∆H°∆S°⇒ log K 2 = + ........ (5)2.3RT 2 2.3RSubtracting (4) from (5) we get.K 2 ∆H° ⎛ 1 1 ⎞⇒ log =K12.3R⎜ –⎟⎝ T1 T 2 ⎠18. If standard heat of dissociation of PCl 5 is 230 Cal.then the slope of the graph of log K vs T1 is -(A) + 50 (B) – 50(C) 10(D) None of these19. For exothermic reaction of ∆S o < 0 then the sketchof log K vs T1 may be -log K(A)log K(C)1/T1/Tlog K(B)log K(D)1/T1/TPassage # 3 (Ques. 20 - 21)A pleasant smelling optically active compound,monoester 'F' has molecular weight 186. It doesn'treact with Br 2 in CCl 4 . Hydrolysis of 'F' gives twooptically active compounds 'G', which is soluble inNaOH and 'H'. H gives a positive iodoform test, buton warming with conc. H 2 SO 4 gives I with nodisastereomers. When the Ag + salt of 'G' is reactedwith Br 2 , racemic 'J' is formed. Optically active J isformed when 'H' is treated with tosyl chloride(TsCl), and then with NaBr.20. The pleasant smelling optically active compound, Fis -(A) (CH 3 ) 2 CH– CHCO– || O– CH – CH(CH 3 ) 2| |CH 3 CH 3O(B) (CH 3 ) 3 C–CH 2 C || –O– CH – CH(CH 3 ) 2|CH 3XtraEdge for IIT-JEE 53 NOVEMBER 2011

O(C) CH 3 CH 2 CH 2 CH – C || –O– CH – CH 2 CH 2 CH 3||CH3CH 3O||CH CH 2 – COCH2– CHCH 2 –CH 3(D) CH 3 CH 2 –|CH3|CH21. How would be the structure of F if I exists asdiastereomers ?O(A) (CH 3 ) 2 CH CHCO|| CHCH (CHCH | 3)23 CH| 3(B) (CH 3 ) 3 CCH 2 COO|| CHCH (CH3)2|CH3O(C) CH 3 CH 2 CH 2 CHCO||CH| CHCH22 33 CH| CH CH3(D) CH 3 CH 2 CHCH2C | COOCH || 2 CHCH 2H3CH| CH3This section contains 2 questions (Questions 22, 23).Each question contains statements given in two columnswhich have to be matched. Statements (A, B, C, D) inColumn I have to be matched with statements (P, Q, R,S, T) in Column II. The answers to these questions haveto be appropriately bubbled as illustrated in thefollowing example. If the correct matches are A-P, A-S,A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then thecorrectly bubbled 4 × 5 matrix should be as follows:P Q R S TA P Q R S TB P Q R S TC P Q R S TD P Q R S TMark your response in OMR sheet against thequestion number of that question in section-II. + 8marks will be given for complete correct answer (i.e.+2 marks for each correct row) and No Negativemarks for wrong answer.22. Match the Column :Column-IColumn-II(A) 5.4 g of Al(P) 0.5 N A electrons(B) 1.2 g of Mg 2+ (Q) 15.9994 amu(C) Exact atomic weight (R) 0.2 mole atomsof mixture of oxygen isotopes(D) 0.9 mL of H 2 O3(S) 0.05 moles(T) 3.1 × 10 23 electrons323. Match the Column :Column -IColumn-II(A) Two electron three centre (P) (BN) xbond(B) Four electron three centre (Q) B 2 H 6bond(C) sp 3 hybrid orbitals (R) AlCl 3(D) Inorganic graphite (S)(T)B 4 H 10HFMATHEMATICSQuestions 1 to 10 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. + 3 marks will be given for each correctanswer and – 1 mark for each wrong answer.1. The differential equation of all ellipse centred at theorigin is(A) y 2 + xy 1 2 – yy 1 = 0(B) xyy 2 + xy 1 2 – yy 1 = 0(C) yy 2 + xy 1 2 – xy 1 = 0(D) none of these2. The value of x for which the matrix⎡20 7⎤A =⎢ ⎥⎢0 1 0⎥is inverse of⎢⎣1 − 2 1⎥⎦⎡−x 14x7x⎤B =⎢⎥⎢0 1 0⎥is⎢⎣x − 4x− 2x⎥⎦1 1 1(A) (B) (C) 2 3 4(D) 513. If x = (7 + 4 3 ) 2n = [x] + f , then x(1– f ) if equal to(A) 1 (B) 2 (C) 3 (D) 44. The number of values of x ∈[0, nπ], n ∈ I thatsatisfy log |sinx| (1 + cos x) = 2 is(A) 0 (B) n (C) 2n (D) none5. Reflection of the line a z + a z = 0 in the real axis isa z(A) a z + az = 0 (B) = a z(C) (a + a ) (z + z ) = 0 (D) None of these6. The tangent to the curve x = a cos 2θcos θ,y = a cos 2θsin θ at the point corresponding toθ = π/6 is -(A) parallel to the x-axis (B) parallel to the y-axis(C) parallel to line y = x (D) none of theseXtraEdge for IIT-JEE 54 NOVEMBER 2011

7. The exponent of 7 in 100 C 50 is -(A) 0 (B) 2(C) 4(D) none of these⎧|x | for0< | x | ≤ 28. Let f (x) = ⎨, then at x = 0, f has -⎩ 1 for x = 0(A) a local maximum (B) no local maximum(C) a local minimum (D) no extremum9. If f (x) is a polynomial satisfyingf (x).f (1/x) = f (x) + f (1/x), and f (3) = 28, then f (4)is given by -(A) 63 (B) 65 (C) 67 (D) 6810. The integer n for which(cos x −1)(cosx − e )limnxx→0is a finite nonzero number is -(A) 1 (B) 2 (C) 3 (D) 4Questions 11 to 15 are multiple choice questions.Each question has four choices (A), (B), (C) and (D),out of which MULTIPLE (ONE OR MORE) iscorrect. Mark your response in OMR sheet againstthe question number of that question. + 4 marks willbe given for each correct answer and – 1 mark foreach wrong answer.11. If f (x) = 1 − 1 − x , then(A) f is continuous on [–1, 1](B) f is continuous at x = 0(C) f is not differentiable at x = 0(D) f is differentiable everywhere212. The line y = mx + c intersects the circle x 2 + y 2 = r 2at two real distinct points if(A) – r(C) – c21+ m < c ≤ 0 (B) 0 ≤ c < r21− m < r (D) r < c21+m21+m13. Let α, β be the roots of x 2 – 4x + A = 0 and γ, δ bethe roots of x 2 – 36x + B = 0. If α, β, γ, δ forms anincreasing G.P., then(A) B = 81 A (B) A = 3(C) B = 243 (D) A + B = 25114. Given an isosceles triangle with equal sides oflength b, base angle α < π/4, R, r the radii and O, Ithe centres of the circumcircle and incircle,respectively. Then(A) R = 21 b cosec α(B) ∆ = 2b 2 sin 2αx15. If∫ α dx A= + B (a ≠ 0). Then0 1 − cos α cos x sin αpossible values of A and B areπ π π(A) A = , B = 0 (B) A = , B = 2 4 4sin απ ππ(C) A = , B = (D) A = π, B =6 sin αsin αThis section contains 3 paragraphs; each has 2multiple choice questions. (Questions 16 to 21) Eachquestion has 4 choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. +3 marks will be given for each correctanswer and – 1 mark for each wrong answer.Passage # 1 (Ques. 16 - 17)We can derive reduction formulas for the integral ofnnnthe form∫sin x dx,cos x dx ,∫tan x dx,∫n∫cot x dx and other integrals of these form usingintegration by parts. In turn these reductionformulas can be used to compute x etc.5116. If∫sin x dx = – sin 4 x cos x + A sin 2 x cos x5– 158 cos x + C then A is equal to -(A) – 2/15 (B) – 3/5(C) – 4/15 (D) – 1/156117. If∫tan x dx = tan 5 x + A tan 3 x + tan x – x + C5then A is equal to -(A) 1/3 (B) 2/3(C) – 2/3 (D) – 1/3Passage # 2 (Ques. 18 - 19)Using differentiability and continuity of a functionf which satisfies certain functional equation, we candetermine in some cases the function explicity. E.g.If f satisfies f (x + y) = f (x) f (y) for all x, y ∈ R andf (x) ≠ 0 for any x ∈ R and f ′(0) = 1 thenf (x) = e x .⎛ x + y ⎞ 2 + f ( x)+ f ( y)18. If a function f satisfy f ⎜ ⎟ =⎝ 3 ⎠ 3for real x and y and f ′ (2) = 3, then f (x) is equal to -(A) – 121 x 3 + x 2 (B) 24 log (3x + 2)(C) r =bsin 2α2(1 + cosα)(D) OI =b cos(3α/ 2)2sin α cos( α / 2)(C) 3x + 2 (D) 43 x 2 + 2XtraEdge for IIT-JEE 55 NOVEMBER 2011

19. If f is a differentiable function on R and f ′(0) = 2f ( x)+ f ( y)satisfying f (x + y) =then f (π/8) is1−f ( x)f ( y)equal to -(A) 1/2 (B) 1 (C) 3/2 (D) tan π/8Passage # 3 (Ques. 20 - 21)Let k be the length of any edge of a regulartetrahedron. (A tetrahedron whose edges are allequal in length is called a regular tetrahedron.) Theangle between a line and a plane is equal to thecomplement of the angle between the line and thenormal to the plane whereas the angle between twoplanes is equal to the angle between the normals.Let O be the origin of reference and A, B and Cvertices with position vectors a, b and crespectively of the regular tetrahedron.20. The angle between any edge and a face notcontaining the edge is(A) cos –1 (1/2) (B) cos –1 (1/4)(C) cos –1 (1/ 3 ) (D) π/321. The value of [a b c] 2 is(A) k 2 (B) (1/2)k 2 (C) (1/3)k 2 (D) k 3This section contains 2 questions (Questions 22, 23).Each question contains statements given in two columnswhich have to be matched. Statements (A, B, C, D) inColumn I have to be matched with statements (P, Q, R,S, T) in Column II. The answers to these questions haveto be appropriately bubbled as illustrated in thefollowing example. If the correct matches are A-P, A-S,A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then thecorrectly bubbled 4 × 5 matrix should be as follows:P Q R S TABCDP Q R S TP Q R S TP Q R S TP Q R S TMark your response in OMR sheet against thequestion number of that question in section-II. + 8marks will be given for complete correct answer(i.e. +2 marks for each correct row) and No Negativemarks for wrong answer.22. Match the following-Column -IColumn -II(A) If the linesx – 2=1yy – 3=1z − 5x –1 – 4and = =λ 2 1intersect at (α, β, γ) then λ =z – 4λz – 5⎛ π –1⎛x + 1 ⎞(B) If⎟ ⎞lim 4x⎜ – tan ⎜ ⎟x→∞⎝ 4 ⎝ x + 2 ⎠⎠y 2 + 4y + 5 then y =1=(P) 0(Q) –1(C) If chord x + y + 1= 0 of parabolay 2 = ax subtends 90º at (0, 0)then a =(D) If → a = i ^+ j ^+ k ^, → a . → b = 1and → a × → b = j ^– k ^, then | → b | isequal to(R) –3(S) 1(T) 223. A is a set containing n elements. A subset P of A ischosen at random. The set A is reconstructed byreplacing the elements of the subset P. A subset Qof A is again chosen at random. The probabilitythat (where |x| = number of elements in X)Column-IColumn-II(A) P ∩ Q = φ(P) n(3 n–1 )/4 n(B) P ∩ Q is a singleton (Q) (3/4) n(C) P ∩ Q contains 2 (R) 2n C n /4 nelements(D) |P| = |Q| (S) 9n(n–1)/2(4 n )(T) NoneCOMPLEMENTARY COLOURSIf you arrange some colours in a circle, you get a"colour wheel". The diagram shows one possibleversion of this. An internet search will throw upmany different versions!Colours directly opposite each other on the colourwheel are said to be complementary colours. Blueand yellow are complementary colours; red andcyan are complementary; and so are green andmagenta.Mixing together two complementary colours of lightwill give you white light.What this all means is that if a particular colour isabsorbed from white light, what your eye detects bymixing up all the other wavelengths of light is itscomplementary colour. Copper(II) sulphate solutionis pale blue (cyan) because it absorbs light in the redregion of the spectrum. Cyan is the complementarycolour of red.The origin of colour in complex ionsTransition metal v other metal complex ionsXtraEdge for IIT-JEE 56 NOVEMBER 2011

Based on New PatternIIT-JEE 2013XtraEdge Test Series # 7Time : 3 HoursSyllabus : Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabusInstructions :Section – I :Question 1 to 10 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 11 to 15 are multiple choice question with multiple correct answer. +4 marks will be awarded forcorrect answer and -1 mark for wrong answer.Section – III : Question 16 to 21 are passage based single correct type questions. +3 marks will be awarded forcorrect answer and -1 mark for wrong answerSection – IV : Question 22 to 23 are Column Matching type questions. +8 marks will be awarded for the completecorrectly matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.PHYSICSQuestions 1 to 10 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. + 3 marks will be given for each correctanswer and – 1 mark for each wrong answer.1. When sound wave is refracted from air to water,which of the following will remain unchanged ?(A) wave number (B) wavelength(C) wave velocity (D) frequency2. In stationary wave –(A) All the particles of the medium vibrate inphase(B) All the nodes vibrate in phase(C) All the antinodes vibrate in phase(D) All the particles between consecutive nodesvibrate in phase3. A 20 N metal block is suspended by a spring balance.A beaker containing some water is placed on aweighing machine which reads 40 N. The springbalance is now lowered so that the block getsimmersed in water. The spring balance now reads16N. The reading of the weighing machine will be -(A) 36 N(B) 44 N(C) 60 N(D) None4. The springs shown in figure is unstretched whenJames bond starts pulling on the cord. The mass ofthe block is m. If he exerts a constant force F. Theamplitude of the motion of the block will be-(A)F2kkk(B) kFmSmooth(C)2FkF(const.)(D) None5. Ideal fluid flows along a flat tube of constant crosssection,located in a horizontal plane and bent asshown in figure (top view). The flow is steady. Thevelocities of fluid at point 1 and at point 2 are v 1and v 2 respectively then correct relation is –21(A) v 1 > v 2 (B) v 2 > v 1(C) v 1 = v 2(D) None of these6. A body is moving down a long inclined plane ofinclination θ. The coefficient of friction betweenthe body and the plane varies as µ = 0.5 x, where xis the distance moved down the plane. The bodywill have maximum velocity, when it has travelleda distance x given by -2(A) x = 2 tan θ (B) x =tan θ(C) x = 2 cot θ (D) x =2cot θXtraEdge for IIT-JEE 57 NOVEMBER 2011

7. A system consists of three particles, each of mass mand located at (1, 1), (2, 2) and (3, 3). The coordinatesof the centre of mass are -(A) (1, 1) (B) (2, 2)(C) (3, 3) (D) (6, 6)8. A thin wire of length l and mass m is bent in theform of a semicircle. Its moment of inertia about anaxis joining its free ends will be –(A) ml 2(C)mlπ22mOP(B) zero(D)ml2π9. What is the velocity v of a metallic ball of radius rfalling in a tank of liquid at the instant when itsacceleration is one half that of a freely fallingbody? (The densities of metal and of liquid areρ and σ respectively and the viscosity coefficient ofthe liquid is η) -(A)(C)r 2 g9ηr 2 g9η(ρ – 2σ) (B)(ρ – σ) (D)22r 2 g9η2r2 g9η(2ρ – σ)(ρ – σ)10. An anisotropic material has coefficient of linearthermal expansion α 1 and α 2 along x and yrespectively. Coefficient of superficial expansion ofits material will be equal to –(A) α 1 + α 2 (B) α 1 + 2α 2α1 + α(C) 3α 1 + 2α 2 (D) 22Questions 11 to 15 are multiple choice questions.Each question has four choices (A), (B), (C) and (D),out of which MULTIPLE (ONE OR MORE) iscorrect. Mark your response in OMR sheet againstthe question number of that question. + 4 marks willbe given for each correct answer and – 1 mark foreach wrong answer.11. A man is standing on a road and observes that therain is falling at angle 45º with the vertical. Theman starts running on the road with constantacceleration 0.5 m/s 2 . After a certain time from thestart of the motion, it appears to him that the rain isstill falling at angle 45º with the vertical, with speed2 2 m/s. Motion of the man is in the same verticalplane in which the rain is falling. Then which of thefollowing statement(s) are true -(A) It is not possible.(B) Speed of the rain relative to the ground is2 m/s(C) Speed of the man when he finds rain to befalling at angle 45º with the vertical, is 4 m/s(D) the man has travelled a distance 16 m on theroad by the time he again finds rain to befalling at angle 45º12. All the blocks shown in the figure are at rest. Thepulley is smooth and string is light. Coefficient offriction at all the contacts is 0.2. A frictional forceof 10N acts between A and B. The block A is aboutto slide on block B. The normal reaction andfrictional force exerted by the ground on the blockB is -A5kgC B(A) The normal reaction exerted by the ground onthe block B is 110 N.(B) The normal reaction exerted by the ground onthe block B is 50 N.(C) The frictional force exerted by the ground onthe block B is 20 N(D) The frictional force exerted by the ground onthe block B is zero.13. The value of mass m for which the 100 kg blockremain is static equilibrium is –(A) 35 kg(C) 83 kgµ = 0.310037º(B) 37 kg(D) 85 kg14. A particle is describing circular motion in ahorizontal plane in contact with the smooth insidesurface of a fixed right circular cone with its axisvertical and vertex down. The height of the plane ofmotion above the vertex is h and the semi verticalangle of the cone is α. The period of revolution ofthe particle –α(A) increases as h increases(B) decreases as h increases(C) increases as α increases(D) decreases as α increaseshXtraEdge for IIT-JEE 58 NOVEMBER 2011

15. A wire of density 9 × 10 3 kg / m 3 is stretchedbetween two clamps 1 m apart and is stretched toan extension of 4.9 × 10 –4 metre. Young's modulusof material is 9 × 10 10 N/m 2 then -(A) The lowest frequency of standing wave is 35 Hz(B) The frequency of 1st overtone is 70 Hz(C) The frequency of 1st overtone is 105 Hz(D) The stress in the wire is 4.41 × 10 7 N/m 2This section contains 3 paragraphs; each has 2multiple choice questions. (Questions 16 to 21) Eachquestion has 4 choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. +3 marks will be given for each correctanswer and – 1 mark for each wrong answer.Passage # 1 (Ques. 16 - 17)In the figure shown a uniform solid sphere isreleased on the top of a fixed inclined plane ofinclination 37º and height ‘h’. It rolls withoutsliding (take sin 37º = 3/5)37º16. The acceleration of the centre of the sphere is -(A)3g5(B)4g5(C)4g7(D)3g717. The speed of the point of contact of the sphere withthe inclined plane when the sphere reaches thebottom of the incline is -(A)2 gh(B)10gh7(C) Zero (D) 2 2ghPassage # 2 (Ques. 18 - 19)In a standing wave experiment, a 1.2 kg horizontalrope is fixed in place at its two ends (x = 0 andx = 2.0 m) and made to oscillate up and down in thefundamental mode at frequency 5.0 Hz. At t = 0, thepoint at x = 1.0 m has zero displacement and ismoving upward in the positive direction of y axiswith a transverse velocity 3.14 m/s18. Speed of the participating travelling wave on therope is -(A) 6 m/s (B) 15 m/s (C) 20 m/s (D) 24 m/s19. What is the correct expression of the standing waveequation -⎛ π ⎞(A) (0.1) sin ⎜ ⎟⎝ 2 ⎠(B) (0.1) sin (π) × sin (10π) t× sin ( 10π)t⎛ π ⎞(C) (0.05) sin ⎜ ⎟⎝ 2 × cos ( 10π)t⎠(D) (0.04) sin (π) × sin (10π) tPassage # 3 (Ques. 20 - 21)Transverse and longitudinal standing waves areeasily represented by sine waves. The distancebetween an adjacent node and antinode (N – A) is aquarter of the wavelength, 4λ .AN NNλ/4 Aλ/4λFundamental and first overtone for a pipe closed atone endLLAN A N A Nλ/4λ/4L/3λ λ L= L =44 3⎛ L ⎞Therefore λ 0 = 4L Therefore, λ 1 = 4 ⎜ ⎟⎝ 3 ⎠and ν 0 =V4Lovertones aremultiplies of ν 0 :⎡ v ⎤nµ 0 = n ⎢ ⎥⎣4L⎦This implies that theratio of natural frequencyis n = 1 : 3 : 5.........V ⎡ v ⎤& v 1 = = 3⎛ 4L ⎞⎢ ⎥⎣4L⎦⎜ ⎟⎝ 3 ⎠for the first overtonen = 3, the thirdharmonic.20. When an organ pipe is open at both ends, itresonates with a fundamental frequency of240 Hz. What is the fundamental frequency of thesame pipe if it is closed at one end-(A) 64 Hz(B) 120 Hz(C) 360 Hz(D) 480Hz21. A pipe resonates at 60 Hz, 100 Hz and 140 Hzconsecutive frequencies. How long is the pipe ?(A) 1.4 m(B) 2.8 m(C) 4.3 m(D) 8.5 mXtraEdge for IIT-JEE 59 NOVEMBER 2011

This section contains 2 questions (Questions 22, 23).Each question contains statements given in twocolumns which have to be matched. Statements (A, B,C, D) in Column I have to be matched withstatements (P, Q, R, S, T) in Column II. The answersto these questions have to be appropriately bubbledas illustrated in the following example. If the correctmatches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q andD-S, D-T then the correctly bubbled 4 × 5 matrixshould be as follows :ABCDP Q R S TP Q R S TP Q R S TP Q R S TP Q R S TMark your response in OMR sheet against thequestion number of that question in section-II. + 8marks will be given for complete correct answer (i.e.+2 marks for each correct row) and No Negativemarks for wrong answer.22. Let V and E denote the gravitational potential andgravitational field respectively at a point due tocertain uniform mass distribution described in fourdifferent situation of column – I, thenColumn -I(A) At centre of thinspherical shell(B) At centre of solidsphere(C) At the centre of athick hemispherical shell(D) At centre of linejoining two pointmasses of equalmagnitudeColumn -II(P) E = 0(Q) E ≠ 0(R) V ≠ 0(S)(T)V = 0None23. Two blocks A and B of mass m and 2m respectivelyare connected by a massless spring of springconstant K. This system lies over a smoothhorizontal surface. At t = 0 the block A has velocityu towards right as shown while the speed of blockB is zero, and the length of spring is equal to itsnatural length at that instant. In each situation ofcolumn-I, certain statements are given andcorresponding results are given in column-II,Match the statements in column-I to thecorresponding results in column-II :B K Am 2m usmooth horizontal surfaceColumn I(A) The velocity ofblock A(B) The velocity ofblock B(C) The kinetic energyof system oftwo blocks(D) The potentialenergy of springCHEMISTRYColumn II(P) Can never be zero(Q) may be zero atcertain instants oftime(R) is minimum atmaximumcompressionof spring(S) is maximum atmaximumextensionof spring(T) NoneQuestions 1 to 10 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. + 3 marks will be given for each correctanswer and – 1 mark for each wrong answer.1. At constant pressure P, A dissociates on heatingaccording to the equationA(g) B(g) + C(g)The equilibrium partial pressure of A at T K is 1/9P, the equilibrium K p at TK is8 64 16(A) P (B) P (C) P (D) 9 P9 9 92. Calculate the pH of 6.66 × 10 –3 M solution ofAl(OH) 3 . Its first dissociation is 100% where assecond dissociation is 50% and third dissociation isnegligible.(A) 2 (B) 12(C) 11 (D) 133. pH of the blood in the body is maintained by buffersolution of(A) glucose and salt concentration(B) protein and salt concentration(C) CO 33–and HCO 3–(D) Salt and carbonate ion4. IUPAC name of the following compound is :OHCH3(A) 2-methyl-3-cyclohexenol(B) 3-methyl-1-cyclohexen-4-ol(C) 4-hydroxy-3-methyl-1-cyclohexene(D) 2-hydroxy-1-methylcyclohexeneXtraEdge for IIT-JEE 60 NOVEMBER 2011

5. Which will form geometrical isomers ?Cl(A)(C)Cl(B) CH 3 CH = NOH(D) All6. The dissolution of Al(OH) 3 by a solution of NaOHresults in the formation of(A) [Al(H 2 O) 4 (OH) 2 ] + (B) [Al(H 2 O) 3 (OH) 3 ](C) [Al(H 2 O) 2 (OH) 4 ] – (D) [Al(H 2 O) 6 ](OH) 37. Helium-oxygen mixture is used by deep sea diversin preference to nitrogen-oxygen mixture because(A) helium is much less soluble in blood thannitrogen(B) nitrogen is highly soluble in water(C) helium is insoluble in water(D) nitrogen is less soluble in blood than helium8. SF 4 + BF 3 → (A). The compound 'A' is(A) [SF 5 ] – [BF 2 ] + (B)[SF 3 ] + [BF 4 ] –(C) SF 6 (D) S 2 F 412. 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 type13. The major product of reaction(A)(C)BrBrBrBrBr⎯⎯→2is –(B)(D) None of theseBrBr9. Which of the following reactions is a redoxreaction?(A) Cr 2 O 3 + 6HCl → 2CrCl 3 + 3H 2 O(B) CrO 3 + 2NaOH → Na 2 CrO 4 + H 2 O(C) 2CrO 4 2– + H + Cr 2 O 7 2– + OH –14.Br+ KOH (alc) —→(D) Cr 2 O 2– 7 + 6I – + 14H + 2Cr 3+ + 3I 2+ 7H 2 O10. The combustion reaction occurring in anautomobile is 2C 8 H 18 (s) + 5O 2 (g) → 16CO 2 (g) +18H 2 O(1). This reaction is accompanied with(A) ∆H = –ve, ∆S = + ve, ∆G = + ve(B) ∆H = + ve, ∆S = –ve, ∆G = + ve(C) ∆H = – ve, ∆S = +ve, ∆G = – ve(D) ∆Η = +ve, ∆S = +ve, ∆G = – veQuestions 11 to 15 are multiple choice questions.Each question has four choices (A), (B), (C) and (D),out of which MULTIPLE (ONE OR MORE) iscorrect. Mark your response in OMR sheet againstthe question number of that question. + 4 marks willbe given for each correct answer and – 1 mark foreach wrong answer.11. 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 → pyramidalBrWhich of the following can be formed.Br(A)(B)Br(C)Br(D)15. Reduction of But-2-yne with Na and liquid NH 3gives an alkene which upon catalytichydrogenation with D 2 / Pt gives an alkane. Thealkene and alkane formed respectively are -(A) cis-but-2-ene andrecemic-2, 3-dideuterobutane(B) trans-but-2-ene andmeso-2, 3-dideuterobutane(C) trans-but-2-ene andrecemic-2, 3-dideuterobutane(D) cis-but-2-ene andmeso-2, 3-dideuterobutaneBrXtraEdge for IIT-JEE 61 NOVEMBER 2011

This section contains 3 paragraphs; each has 2multiple choice questions. (Questions 16 to 21) Eachquestion has 4 choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. +3 marks will be given for each correctanswer and – 1 mark for each wrong answer.Passage # 1 (Ques. 16 - 17)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 .16. The possible value of s for electron is -(A) 1/2 (B) – 1/2(C) 0 (D) 117. 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 thesePassage # 2 (Ques. 18 - 19)The expression for the reaction quotient, Q, issimilar to that for equilibrium constant K. Thevalue of Q for the given composition of a reactionmixture helps us to know whether the reaction willmove forward or backward or remain inequilibrium. It also helps to predict the effect ofpressure on the direction of the gaseous reaction. Incertain reactions, addition of inert gas also favourseither the formation of reactants or products. Thevalue of equilibrium constant of a reaction changeswith change of temperature and the change is givenby van't Hoff equation, d ln K p /dT = ∆Hº/RT 2where enthalpy change, ∆Hº, is taken as constant inthe small temperature range.18. The reaction N 2 (g) + 3H 2 (g) 2NH 3 (g) is inequilibrium. Now the reaction mixture iscompressed to half the volume(A) More of ammonia will be formed(B) Ammonia will dissociate back into N 2 and H 2(C) There will be no effect on equilibrium(D)Equilibrium constant of the reaction willchange19. For the above reaction in equilibrium, helium gaswas added but the mixture was allowed to expandto keep the pressure constant. Then(A) More of ammonia will be formed(B) Ammonia will dissociate back into N 2 and H 2(C) There will be no effect on equilibrium(D) Equilibrium constant of the reaction will changePassage # 3 (Ques. 20 - 21)Lithium only forms monoxide when heated inoxygen. Sodium forms monoxide and peroxide inexcess of oxygen. Other alkali metals form superoxide with oxygen i.e., MO 2 . The abnormalbehaviour of lithium is due to small size. The largersize of nearer alkali metals also decides the role information of superoxides. The three ions related toeach other as follows :2O − 1/⎯ 2 O2Oxide ion⎯⎯→2O −2Peroxide ionO⎯⎯→22O −2SuperoxideAll the three ions abstract proton from water.20. Consider the following reaction :M + O 2 ⎯→ MO (M = alkali metal)2(super oxide)Select the correct statement :(A) M can not be Li and Na(B) M can not be Cs and Rb(C) M can not be Li and Rb(D) None of these21. Lithium does not form stable peroxide because :(A) of its small size(B) d-orbital are absent in it(C) it is highly reactive and form superoxide inplace of peroxide(D) covalent nature of peroxideThis section contains 2 questions (Questions 22, 23).Each question contains statements given in twocolumns which have to be matched. Statements (A, B,C, D) in Column I have to be matched withstatements (P, Q, R, S, T) in Column II. The answersto these questions have to be appropriately bubbledas illustrated in the following example. If the correctmatches are A-P, A-S, A-T;B-Q, B-R; C-P, C-Q and D-S, D-T then the correctlybubbled 4 × 5 matrix should be as follows :P Q R S TABCDP Q R S TP Q R S TP Q R S TP Q R S TMark your response in OMR sheet against thequestion number of that question in section-II. + 8marks will be given for complete correct answer (i.e.+2 marks for each correct row) and No Negativemarks for wrong answer.ionXtraEdge for IIT-JEE 62 NOVEMBER 2011

22. Match the Column :Column –IColumn II(A) pK b of X – (Ka of HX = 10 –6 ) (P) 6.9(B) pH of 10 –8 M HCl (Q) 8(C) pHof 10 –2 M acetic acid solution (R) 3.3(Ka = 1.6 × 10 –5 )(D) pH of a solution obtained by (S) 3.4mixing equal volumes of solutionwith pH 3 & 5.(t)23. Match the Column :Column IBr(A)(B)(C)(D)OOHBrOHCHONO 2Column II(p) Nucleopilic substitution(q) Elimination(r) Nucleophilic addition(s) Esterification withacetic anhydride(t) DehydrogenationMATHEMATICSQuestions 1 to 10 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. + 3 marks will be given for each correctanswer and – 1 mark for each wrong answer.1. In a certain test there are n questions. In this test 2 kstudents gave wrong answers to at least (n – k)questions, where k = 0, 1, 2, ...... , n. If the totalnumber of wrong answers is 4095, then value of n is(A) 11 (B) 12 (C) 13 (D) 152. The values of x between 0 and 2π which satisfy the2equation sin x 8 cos x = 1 are in A.P. Thecommon difference of the A.P. is(A) π/8 (B) π/4 (C) 3π/8 (D) 5π/83. If p 1 , p 2 , p 3 are respectively the perpendicular fromthe vertices of a triangle to the opposite sides, thencos A cos B cos C + + is equal top1p2p3(A) 1/r (B) 1/R (C) 1/∆ (D) None4. The positive integer n for which2 × 2 2 + 3 × 2 3 + 4 × 2 4 + .... + n × 2 n = 2 n + 10 is(A) 510 (B) 511(C) 512 (D) 5135. A vector c, directed along the internal bisector ofthe angle between the vectors a = 7i – 4j – 4k andb = –2i – j + 2k, with |c| = 5 6 , is -(A) 35 (i – 7j + 2k) (B) 35 (5i + 5j + 2k)(C) 35 (i + 7j + 2k) (D) 35 (–5i + 5j + 2k)6. The locus of the mid-point of the line segmentjoining the focus to a moving point on the parabolay 2 = 4ax is another parabola with directrix -(A) x = – a (B) x = – a/2(C) x = 0 (D) x = a/27. A flagstaff stands in the centre of a rectangular fieldwhose diagonal is 1200 m, and subtends angles 15ºand 45º at the mid points of the sides of the field.The height of the flagstaff is -(A) 200 m (B) 300 2 + 3 m(C) 300 2 − 3 m (D) 400 m8. If two vertices of a triangle are (–2, 3) and (5, – 1),orthocentre lies at the origin and centroid on theline x + y = 7, then the third vertex lies at(A) (7, 4) (B) (8, 14)(C) (12, 21) (D) None9. If tan x + tan(x + π/3) + tan (x + 2π/3) = 3, then -(A) tan x = 1 (B) tan 2x = 1(C) tan 3x = 1 (D) none of these10. The number of seven digit integers, with sum of thedigits equal to 10 and formed by using the digits 1,2 and 3 only, is(A) 55 (B) 66 (C) 77 (D) 88Questions 11 to 15 are multiple choice questions.Each question has four choices (A), (B), (C) and (D),out of which MULTIPLE (ONE OR MORE) iscorrect. Mark your response in OMR sheet againstthe question number of that question. + 4 marks willbe given for each correct answer and – 1 mark foreach wrong answer.11. The coordinates of a point on the linex −1 y + 1= = z at a distance 4 14 from the2 − 3point (1, –1, 0) are(A) (9, – 13, 4)(B) ( 8 14 + 1, –12 14 – 1, 4 14XtraEdge for IIT-JEE 63 NOVEMBER 2011

(C) (–7, 11, – 4)(D) (– 8 14 + 1, 12 14 – 1, –4 1412. If x 2 + 2hxy + y 2 = 0 represents the equations of thestraight lines through the origin which make anangle α with the straight line y + x = 0, then(A) sec 2α = h1+ h(B) cos α =2h(C) 2 sin α =(D) cot α =1+ hh1+hh −113. If z 1 , z 2 , z 3 , z 4 are the vertices of a square in thatorder, then(A) z 1 + z 3 = z 2 + z 4(B) |z 1 – z 2 | = |z 2 – z 3 | = |z 3 – z 4 | = |z 4 – z 1 |(C) |z 1 – z 3 | = |z 2 – z 4 |(D) (z 1 – z 3 )/(z 2 – z 4 ) is purely imaginary14. The equation of a tangent to the hyperbola3x 2 – y 2 = 3, parallel to the line y = 2x + 4 is -(A) y = 2x + 3(B) y = 2x + 1(C) y = 2x – 1(D) y = 2x + 215. If m is a positive integer, then [( 3 + 1) ] + 1,where [x] denotes greatest integer ≤ n, is divisibleby-(A) 2 m (B) 2 m+1(C) 2 m+2(D) 2 2mThis section contains 3 paragraphs; each has 2multiple choice questions. (Questions 16 to 21) Eachquestion has 4 choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. +3 marks will be given for each correctanswer and – 1 mark for each wrong answer.Passage # 1 (Ques. 16 - 17)f (x) = sin {cot –1 (x + 1)} – cos (tan –1 x)a = cos tan –1 sin cot –1 xb = cos (2 cos –1 x + sin –1 x)16. The value of x for which f (x) = 0 is(A) –1/2 (B) 0(C) 1/2 (D) 117. If a 2 = 26/51, then b 2 is equal to -(A) 1/25 (B) 24/25(C) 25/26 (D) 50/512mPassage # 2 (Ques. 18 - 19)To solve equation or inequality involvingexponential expression f(x) g(x) , we may uselogarithm or the identity x y y x= a log awhere a > 0, a ≠ 1.18. Solution set of the inequality 3 x (0.333 ....) x–3 ≤ (1/27) x is(A) [3/2, 5] (B) (– ∞, 3/2](C) (0, ∞)(D) None of these19. Solution set of the inequality2(25) x – 5(10 x ) + 2(4 x ) ≥ 0 is(A) (–1, ∞) (B) (0, ∞)(C) (2, ∞)(D) None of thesePassage # 3 (Ques. 20 - 21)2xC : x 2 + y 2 = 9, E :9+4= 1, L : y = 2x20. P is a point on the circle C, the perpendicular PL tothe major axis of the ellipse E meets the ellipse atMLM, then is equal toPL(A) 1/3 (B) 2/3(C) 1/2(D) none of these21. Equation of the diameter of the ellipse E conjugateto the diameter represented by L is(A) 9x + 2y = 0 (B) 2x + 9y = 0(C) 4x + 9y = 0 (D) 4x – 9y = 0This section contains 2 questions (Questions 22, 23).Each question contains statements given in twocolumns which have to be matched. Statements (A, B,C, D) in Column I have to be matched withstatements (P, Q, R, S, T) in Column II. The answersto these questions have to be appropriately bubbledas illustrated in the following example. If the correctmatches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q andD-S, D-T then the correctly bubbled 4 × 5 matrixshould be as follows :ABCD2yP Q R S TP Q R S TP Q R S TP Q R S TP Q R S TMark your response in OMR sheet against thequestion number of that question in section-II. + 8marks will be given for complete correct answer(i.e. +2 marks for each correct row) and No Negativemarks for wrong answer.XtraEdge for IIT-JEE 64 NOVEMBER 2011

22. Match the following -Column- I(A) If a, b, c are unequalpositive numbers and bis A.M. of a and c thenthe roots ofax 2 + 2bx + c = 0 are(B) If a ∈ R, then the rootsof the equationx 2 – (a +1) x – a 2 – 4 = 0 are(C) If a, b, c are unequalpositive numbers and bis H.M. of a and c thenthe roots ofax 2 + 2bx + c = 0 are(D) If |a ± b| < c and a = 0then the roots ofa 2 x 2 + (b 2 + a 2 – c 2 )x+ b 2 = 0 areColumn- II(P) of opposite signs(Q) rational numbers(R) real and unequal(S) imaginary(T) None23. Match the Column :Column-IColumn-II(A) Equation of the polar (P) 8x + 2y – 23 = 0of (–7, –9) withrespect to the circlex 2 + y 2 – 12x – 8y – 48 = 0(B) Equation of the (Q) 13x + 13y – 30 = 0common chord of thecircles x 2 + y 2+ 2x + 2y + 1 = 0 andx 2 + y 2 + 4x + 3y + 2 = 0(C) Equation of the (R) 2x + y + 1 = 0tangent at (–7, –9) tothe circle x 2 + y 2 + 12x+ 8y + 26 = 0(D) Equation of the radical (S) x + 5y + 52 = 0axis of the circles2x 2 + 2y 2 + 4x + 4y + 9 = 0and x 2 + y 2 + 6x+3y – 7 = 0(T) x + y – 1 = 0Regents PhysicsYou Should KnowNuclear Physics :• Alpha particles are the same as helium nucleiand have the symbol .• The atomic number is equal to the number ofprotons (2 for alpha)• Deuterium ( ) is an isotope of hydrogen( )• The number of nucleons is equal to protons +neutrons (4 for alpha)• Only charged particles can be accelerated in aparticle accelerator such as a cyclotron or VanDer Graaf generator.• Natural radiation is alpha ( ), beta ( )and gamma (high energy x-rays)• A loss of a beta particle results in an increase inatomic number.• All nuclei weigh less than their parts. This massdefect is converted into binding energy.(E=mc 2 )• Isotopes have different neutron numbers andatomic masses but the same number of protons(atomic numbers).• Geiger counters, photographic plates, cloud andbubble chambers are all used to detect orobserve radiation.• Rutherford discovered the positive nucleususing his famous gold-foil experiment.• Fusion requires that hydrogen be combined tomake helium.• Fission requires that a neutron causes uraniumto be split into middle size atoms and produceextra neutrons.• Radioactive half-lives can not be changed byheat or pressure.• One AMU of mass is equal to 931 meV ofenergy (E = mc 2 ).XtraEdge for IIT-JEE 65 NOVEMBER 2011

XtraEdge Test SeriesANSWER KEYIIT- JEE 2012 (November issue)PHYSICSQues 1 2 3 4 5 6 7 8 9 10 11Ans C C A A B B D A A B A,CQues 12 13 14 15 16 17 18 19 20 21Ans A,C,D A A,B,C A,B,C A A D C C AColumn Ques 22 23Match Ans A→ S; B→ R; C→ Q; D→ P A→ R; B→ P; C→ S; D→ QCHEMISTRYQues 1 2 3 4 5 6 7 8 9 10 11Ans A A B B D B C A A B A,B,DQues 12 13 14 15 16 17 18 19 20 21Ans A,B A,B,C,D A B,C B B B B A CColumn Ques 22 23Match Ans A→ R; B→ P,S,T; C→ Q; D→ P,T A→ Q,R; B→ S; C→ Q,R; D→ PMATHEMATICSQues 1 2 3 4 5 6 7 8 9 10 11Ans B D A A A A A D B C A,B,CQues 12 13 14 15 16 17 18 19 20 21Ans A,B A,B,C A,C,D A,B C D C B C BColumn Ques 22 23Match Ans A→ P,Q; B→ Q,R; C→ Q; D→ S A→ R; B→ P; C→ S; D→ RIIT- JEE 2013 (November issue)PHYSICSQues 1 2 3 4 5 6 7 8 9 10 11Ans D D B A B A B D A C C,DQues 12 13 14 15 16 17 18 19 20 21Ans A,D B,C A,C A,B D C C A B CColumn Ques 22 23Match Ans A→ P,R; B→ P,R; C→ Q,R; D→ P,R A→ P; B→ Q; C→ P,R; D→ Q,SCHEMISTRYQues 1 2 3 4 5 6 7 8 9 10 11Ans C B C A D C A B D C A,B,CQues 12 13 14 15 16 17 18 19 20 21Ans A,C,D A A,B,C,D C,D A B A B A AColumn Ques 22 23Match Ans A→ Q; B→ P; C→ S; D→ R A→ P,Q; B→ P,S,T; C→ R,S; D→ PMATHEMATICSQues 1 2 3 4 5 6 7 8 9 10 11Ans B B B B A C C D A C A,CQues 12 13 14 15 16 17 18 19 20 21Ans A,B,D A,B,C,D B,C A,B A B B D B BColumn Ques 22 23Match Ans A→ R; B→ P,R; C→ S; D→ R A→ Q; B→ R; C→ S; D→ PXtraEdge for IIT-JEE 66 NOVEMBER 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 67 NOVEMBER 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 68 NOVEMBER 2011