September 2011 - Career Point

Volume - 7 Issue - 3September, 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 knows that setting goals will help you achieve more andadds excitement and meaning to life. But setting a goal is only thebeginning. We often fail to follow through and our goals turn intounfulfilled daydreams. To eliminate that pitafall, here is asystematic, approach that will help you turn your goals intorealities.• Decide what you want to achieve. Determine exactly what youwant. Be specific. Be sure your goal is measurable, so you can tellwhen you're making progress. Pick a target date for a achieving it.Be sure it is realistically achievable.• Ask yourself why it is important for your to achieve this goal.How you will benefit from reaching this goal ? Knowing why youwant something raises your level of motivation. The higher yourmotivation level, the more likely you are to act on your goal.• Consider what obstacles, problems or personal shortcomingsmight block your progress. List every one you can think of someobstacles will be real, others may be only imaginary. You mustconquer both.• Examine the obstacles one at a time, and think about how youmight solve each problem.• List the people or organizations who could help you achieve yourgoal. Decide specifically what you will ask them to do.• Consider what information you need that you don't have now.Where will you get it? What could you read? Who could you talkto? What seminars could you attend?• Write out a detailed action plan for achieving your goal. What arethe priorities involved. ? Which tasks must be done first? Whenwill different actions take place?Setting a goal is a good step, but it is only the beginning. It takesall seven steps to make sure you actually follow through, and by sodoing achieve your goal.I guarantee that you will succeed and will secure a good rank inyour exams if you make a habit of never to postpone your work.Forever presenting positive ideas to your success.Yours trulyPramod Maheshwari,B.Tech., IIT DelhiTo determine your priorities, examine your goalsEditorialXtraEdge for IIT-JEE 1 SEPTEMBER 2011

Volume-7 Issue-3September, 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• If you don’t notice when you win, you willonly notice when you lose.• It’s not bragging if you can do it.• Feel the power of yet. As in “I don't knowhow to do this yet.”• The difficult we do immediately. Theimpossible takes a bit longer.• Some look down the rapids and see therocks. Hunters look down the rapids andsee the flow around the rocks.• To know what you are doing is anadvantage. To look like you know whatyou are doing is essential.• First law of expertise: Never ask a barber ifyou need a haircut.• If you think you can, you are probablyright. If you think you can't, you arecertainly right.• Don't do modesty unless you have earnedit.INDEXCONTENTSRegulars ..........PAGENEWS ARTICLE 3• IIT Kharagpur presents Development Planfor Cuttach• IIT Mandi to rid the peasantry off agrariancrisis ShimlaIITian ON THE PATH OF SUCCESS 5Mr. Prassanna PathmanathanKNOW IIT-JEE 6Previous IIT-JEE QuestionStudy Time........DYNAMIC PHYSICS 138-Challenging Problems [Set # 5]Students’ ForumPhysics Fundamentals• Current Electricity• Circular Motion, Rotational MotionCATALYSE CHEMISTRY 33Key Concept• Aliphatic Hydrocarbon• Oxygen & Hydrogen FamilyUnderstanding : Inorganic ChemistryDICEY MATHS 42Mathematical ChallengesStudents’ ForumKey Concept• Probability• Binomial TheoremTest Time ..........XTRAEDGE TEST SERIES 54Class XII – IIT-JEE 2012 PaperClass XI – IIT-JEE 2013 PaperXtraEdge for IIT-JEE 2 SEPTEMBER 2011

challenges for both high school andcollege level students in the area ofenvironmentally responsible aviation.These challenges were open to bothUS citizens and foreign students. Thegoal of the competition was to submitideas and designs for aircraft or engineconcepts and technologies that wouldassist in meeting the project’s goalsfor more environmentally friendlyaviation by the year 2020.Two senior students of Aerospaceengineering department of IndianInstitute of Technology Kanpursecured first position in foreigndivision of NASA’s EnvironmentallyResponsible (Green) AviationCollege Student Challenge. Theirdesign “Vaayu-An effort towardsgreener aviation” proposed a novelengine concept which bagged the firstprize.This would probably be the first timethat teams from India swept all awardsin foreign division category for aNASA competition as two seniorsfrom Anna University in Chennaicame in second for their blended wingbody design. Third place went to fourundergraduates from SRM Universityin Kattankulathur for their airlinerconcept.Among high school category studentsfrom India and Romania took tophonors among the foreign entries.First place went to 11th grader NitishKulkarni, from the OakridgeInternational School Hyderabad,India. Twelfth grader teams fromTudor Vianu National High School ofComputer Science in Bucharest,Romania came in second and third.This certainly bodes well for thefuture of science and technology inIndiaRevisiting protein foldingProtein folding is at least a six decadeold problem, since the times ofPauling and Anfinsen. At IIT Delhi, acompletely new direction has beenprovided towards obtaining a solutionto this problem. Rigorous analyses ofseveral thousand crystal structures offolded proteins reveal a surprisinglysimple unifying principle of backboneorganization. We find that proteinfolding is a direct consequence of anarrow band of stoichiometricoccurrences of amino-acids in primarysequences, regardless of the size andthe fold of the protein. We call thisnarrow band of stoichiometricoccurrences as the "margin of life". Ourfindings present a compelling case for anewer view of protein folding whichtakes into account solvent mediated andamino acid shape and size assistedoptimization of the tertiary structure ofthe polypeptide chain to make afunctional protein that leads to survivalof living systems over evolutionarytimescales.Ion Beam Complex At IIT Kanpurfor Micro and Nanoscale Science,Engineering & TechnologyAn Ion Beam Complex for ScienceEngineering and Technology (IBC-SET), equipped with different types oflow energy ion beam facilities rangingfrom few keV to few MeV, is being setupat IIT Kanpur for carrying outresearch in the interdisciplinary areasleading to technology development andprototype device fabrication.Under this complex a state of artfocused ion beam system wasTandetron accelerator equipped withmicro beam facility was inaugurated.Prof. V. S. Ramamurthy, Prof. G. K.Mehta, Prof. R. M. Singru, Dr. S.Kailas and other eminent scientists,faculty members of our institute andother institutions graced the occasion.Prof. S. G. Dhande presided over thefunction.Following research and technologydevelopment areas will be pursued bythe researchers within and outside theinstitute using this facility:i. Microfabrication /Micromachining(MEMS/NEMS)ii. Ion Beam synthesis of nano phases /surface engg.iii. Surface/interface studies byRBS/ERDA/PIXE/Channelling.iv. Defect and damage studies inmaterials.v. Bio Materials; damage studiesand 3D mapping.vi. Process optimization and systemautomation.Mr. Narayana Murthy as chief guestinaugurated the Golden Jubileecelebrations. A documentary on IITKpremiered during the function.Detials of the events and projects forthis period are at the website.Inauguration of DST Unit onNanosciences & TechnologryForum Initiative on FabrionicsDr. T. Ramasami, Secretary,Department of Science &Technology, Government of India,inaugurated DST Unit onNanosciences & Launching of Indo-US Science & Technologry ForumInitiative of Fabrionics atNanosciences Laboratory Building,IIT Kanpur. Dr. Samir K.Brahmachari, Director General,Council of Scientific & IndustrialResearch, Government of India wasthe Guest of Honor.IIT Guwahati Celebrates"Techniche 2011" from 1 st - 4 thseptember 2011Indian Institute of Technology(IIT), Guwahati is celebrating theTechno- Management extravaganza-“Techniche 2011” from 1st -4thSeptember 2011.Charmed by the beauty of innovationand the underlying essence ofgrowth, IIT Guwahati yet againbrings the Techno Managementextravaganza, the quintessentialcelebration of innovation, foreveryone to enjoy. Techniche 2011offers perfect opportunity to stretchone’s imagination, flaunt skills andprove mettle in the toughest ofcompetitions and be inspired by theleaders of innovation. In the lecturesseries, some of the biggest names oftoday enlighten all with their wordsof wisdom for everyone to appreciateand reflect in awe..XtraEdge for IIT-JEE 4 SEPTEMBER 2011

Success StoryThis article contains storys/interviews of persons who succeed after graduation from different IITsMr. Prassanna PathmanathanIIT success storyThis alumnus of Informatics Institute of Technology (IIT)is a PR specialist, lecturer, supervisor, research scholar,trainer, entertainer and journalist.Pathmanathan, a 25-year-old past graduate of IIT’s BSc inInformation Systems with Business Management (firstclass) from the University of Westminster UK and the goldmedallist in his year by toping the batch, now lecturesother students studying for the same degree and guidesthem through their final year research/thesis projects as aproject supervisor as well.The experience and knowledge gained from this degree,Prassanna was able to move into other disciplines such asbusiness administration, marketing and human resourcemanagement as well. He’s a Chartered Marketer andcurrently studying for a degree in psychology andmastering in organisational psychology as well.His first venture into the business world was in the fashionindustry as a writer. In 2009 he was the youngest editor-inchiefof a publication when he held that position at Agora– South Asia’s first online magazine. What started out as ahobby turned him towards fashion journalism and he hashad the privilege of interviewing and featuring manyprominent personalities in the field of fashion, sports andentertainment and he has now moved into fashion stylingas well.He was invited as a member of the jury this year to judgeMr. Earth 2011 this October in Central America. He iscurrently working as the Head of Public relations atCameron Pale and Medina (Pvt) Ltd., an all-inclusiveCommunication Agency.Success story is just one example of many other similarstories related to Informatics Institute of Technology (IIT).In fact Prassanna stated: “As a project supervisor I haveguided a few students in their final year projects. Thetopics and solutions that these students have come up withon their own are amazing in their complexity and advancedthinking capabilities. All these students received an ‘A’and what’s more industry representatives who had theopportunity of going through these projects are nowmaking arrangements to purchase these projects andimplement them in a real-time environment. This is madepossible only through the practical and advanced educationsystem implemented by IIT in BSc (Hons) InformationSystem with Business Management degree.”Is also in the process of writing research papers togetherwith his students to present at international businessresearch conferences. As a result, he is presenting threeresearch papers with his students in the InternationalConference on Business and Technology in November2011 at FRI University, Dehradun, New Delhi, India. Thiskind of exposure is made available to the students so thatthey receive local as well as international experience intheir studies.passionately about his work at IIT, Prassanna said: “It is nosecret that every industry, no matter what business it isengaged in, ultimately needs the assistance of informationtechnology in order to take the venture into the future. Invarying degrees of importance, information technology hasbecome a part and parcel of our everyday life. This iswhere the Information Systems with BusinessManagement BSc (Hons) Degree that IIT offers comesinto play. The degree is engineered in such a uniquemanner so as to enable the students to gain a well roundedknowledge of necessary IT components as well as a soundbusiness acumen that will provide a stepping stone intowhatever industry they wish to move into.”“popular perception regarding IIT’s Information Systemswith Business Management degree is that it is a technicaldegree for extreme IT savvy people. However, this is notso. Instead this degree gives more knowledge on businessmanagement essentials together with a view point of ITharmonisation. The syllabus demands a completeimmersion in the studies. However what sets this degreeoption apart is the practical angle that is inherent in everyaspect of the programme.“third year of the four year course is the placement year.During this year, students are assisted in getting into acompany of their choice in order that they might developand acquire a practical understanding of the modules thatthey are learning about, and its application in the realworld. The students get immense experience in workingfor the multinational giants.“The internship has proven to be a resounding success asthe depth of experience a student gains at that crucialmoment in their studies is an essential step. It has alsoenabled the students to better apply themselves to thecompulsory final year project that needs to be completedwithin a timeframe of eight to nine months.” So think big.Think different,” were Prassanna’s final words ofencouragement to hopeful students.XtraEdge for IIT-JEE 5 SEPTEMBER 2011

KNOW IIT-JEEBy Previous Exam QuestionsPHYSICS1. A transverse harmonic disturbance is produced in astring. The maximum transverse velocity is 3 m/s andmaximum transverse acceleration is 90 m/s 2 . If thewave velocity is 20 m/s then find the waveform.[IIT-2005]Sol. The wave form of a transverse harmonic disturbancey = a sin (ωt ± kx ± φ)Given v max = aω = 3 m/s...(i)A max = aω 2 = 90 m/s 2 ....(ii)Velocity of wave v = 20 m/s ...(iii)Dividing (ii) by (i)aω2 90= ⇒ ω = 30 rad/s ...(iv)aω3Substituting the value of ω in (i) we geta = 303 = 0.1 m ...(v)Now2 π 2π 2π ω 30 3k = = = = = =λ v / v vvv 20 2From (iv), (v) and (vi) the wave form is⎡ 3 ⎤y = 0.1 sin ⎢30t ± x ± φ⎥ ⎣ 2 ⎦...(vi)2. A 5m long cylindrical steel wire with radius 2 × 10 –3m is suspended vertically from a rigid support andcarries a bob of mass 100 kg at the other end. If thebob gets snapped, calculate the change in temperatureof the wire ignoring radiation losses. (For the steelwire : Young's modulus = 2.1 × 10 11 Pa; Density= 7860 kg/m 3 ; Specific heat = 420 J/kg-K).[IIT-2001]Sol. When the mass of 100 kg is attached, the string isunder tension and hence in the deformed state.Therefore it has potential energy (U) which is givenby the formula.U = 21 × stress × stain × volume21 (Stress)= × 2 Y22× πr 2 l1 (Mg / πr)= × πr 2 1 M g ll = ...(i)2 Y22πrYThis energy is released in the form of heat, therebyraising the temperature of the wireQ = mc ∆T...(iii)22From (i) and (iii) Since U = Q Therefore2 21 M g l∴ mc∆T =22 πrY2 21 M g l∴ ∆T =22 πrYcmHerem = mass of string = density × volume of string= ρ × πr 2 l2 21 M g∴ ∆T =2 22 ( πr) Ycρ1 (100×10)= × 2−3211(3.14×2×10 ) × 2.1×10 × 420×7860= 0.00457ºC3. The x – y plane is the boundary between twotransparent media. Medium –1 with z ≥ 0 has arefractive index 2 and medium –2 with z ≤ 0 has arefractive index 3 . A ray of light in medium –1given by the vector A = 6 3 ^i + 8 3^j – 10 ^k isincident on the plane of separation. Find the unitvector in the direction of the refracted ray inmedium –2.[IIT-2003]Sol.Y6^3 i + 8^3 jZ^ ^6 3 i + 8 3 j8 3^jOM ' XOX–10 ^KMFig(1)M '6 3^iFigure 1 shows vector2→^ ^ ^A = 6 3 i+8 3 j–10k6 3i+8 3^jFig(2)Figure 2 shows vector A → = 6 3^i + 8 3^j – 10^kThe perpendicular to line MOM' is Z-Axis which hasa unit vector of ^k . Angle between vector IO →and→ZO can be found by dot product→IO . ZO →= (IO) (ZO) cos i(6(63)2^+ (83)^3 i+8 3 j–10 k).(– k)2^+ (–10)2^(–1)2= cos i⇒ i = 60Unit vector in the direction MOM' from figure (1) isXtraEdge for IIT-JEE 6 SEPTEMBER 2011

^^^ 6 3 i+8 3 jn =2 2 1/ 2[(6 3) + (8 3) ]^ 3 ^ 4 ^n = i+j8 5To find the angle of refraction, we use snell's law3 sin i sin 60º= =⇒ r = 45º2 sin r sin rFrom the triangle ORS^r = (sin r) ^n – ( cos r) ^k⎡3^ 4 ^⎤= (sin 45º) ⎢ i+j⎥ – (cos 45º) ^k⎣55 ⎦1 ^ ^= [3^+ i 4 j– 5k]5 24. Along horizontal wire AB, which is free to move in avertical plane and carries a steady current of 20A, isin equilibrium at a height of 0.01 m over anotherparallel long wire CD which is fixed in a horizontalplane and carries a steady current of 30A, as shownin figure. Show that when AB is slightly depressed, itexecutes simple harmonic motion. Find the period ofoscillations.[IIT-2003]ABCDSol. When AB is steady,Weight per unit length = Force per unit lengthµ 0 2I1 I2weight per unit length =...(i)4πrwhen the rod is depressed by a distance x, then theforce acting on the upper wire increases and behaveas a restoring forceAA'r = 0.01 mF magmgI 1 = 20AB'BC I 2 = 30A Dµ 0 2I1 I2µ 0 I1IRestoring force/length =–4πr – x 4πrµ 0 ⎡ 1 1⎤= 2I 1 I 24π⎢ – ⎥⎣ r – x r ⎦µ 0 ⎡ r – (r – x) ⎤⇒ Restoring force/length = 2I 1 I 2 ⎢ ⎥4π⎣ (r – x)r ⎦µ 0 2I1I2x=4πr(r – x)when x is small i.e., x

CHEMISTRY6. A hydrated metallic salt A, light green in colour,gives a white anhydrous residue B after being heatedgradually. B is soluble in water and its aqueoussolution reacts with NO to give a dark browncompound C. B on strong heating gives a brownresidue and a mixture of two gases E and F. Thegaseous mixture, when passed through acidifiedpermanganate, discharges the pink colour and whenpassed through acidified BaCl 2 solution, gives awhite precipitate. Identify A, B, C, D, E and F.[IIT-1988]Sol. The given observations are as follows.(i) Hydrated metallic salt ⎯ →(A)(ii) Aqueous solution of B(iii) Salt B(iv)⎯Strong ⎯⎯heatingGaseous mixture(E) + (F)→⎯ heat(B)white anhydrous residue⎯ NO ⎯→)dark brown compound(CBrown residue +(D)acidified KMnO 4BaCl 2 solutionTwo gases(E) + (F)Pink colour isdischargedWhite precipitateThe observation (ii) shows that B must be ferroussulphate since with NO, it gives dark browncompound according to the reaction[Fe(H 2 O) 6 ] 2+ 2+ NO → [Fe(H 2 O) 5(NO)]+ + H 2 Odark brownHence, the salt A must be FeSO 4 . 7H 2 OThe observation (iii) is2FeSO 4 → Fe 2 O 3 + SO 2 + SO 3(D)brown(E) + (F)The gaseous mixture of SO 2 and SO 3 explains theobservation (iv), namely,2MnO −4pink colour+ 5SO 2 + 2H 2 O → 2Mn2no colour2−4+ + 5SO + 4H +2H 2 O + SO 2 + SO 3 4H + + SO 2– 2–3 + SO 4Ba 2+ + SO 2– 3 → BaSO ; Ba 2+ + SO – 4 → BaSO3white ppt.Hence, the various compounds areA. FeSO 4 . 7H 2 O B. FeSO 4C. [Fe(H 2 O) 5 NO]SO 4 D. Fe 2 O 3E and F SO 2 and SO 34white ppt.7. An organic compound (A) C 8 H 6 , on treatment withdil. H 2 SO 4 containing HgSO 4 gives a compound (B),which can also be obtained from a reaction ofbenzene with an acid chloride in the presence ofanhydrous AlCl 3 . The compound (B) when treatedwith iodine in aq. KOH, yields (C) and a yellowcompound (D). Identify (A), (B), (C) and (D) withjustification. Show, how (B) is formed from (A).[IIT-1994]Sol. The given reactions may be formulated as follows :Dil HC 8 H 2SO 4AlCl 6(B)3C∆ 6 H 6 + Acid chloride(A) HgSO 4∆ I 2 + KOH(C) + (D)The reaction of compound (B) with I 2 in KOH isiodoform reaction. The compound (B) must have a–COCH 3 group so as to exhibit iodoform reaction.Since (B) is obtained from benzene by Friedal-Craftsreaction, it is an aromatic ketone (C 6 H 5 COCH 3 ). Thecompound (C) must be potassium salt of an acid.The compound (A) may be represented as C 6 H 5 C 2 H.Since it gives C 6 H 5 COCH 3 on treating with dil.H 2 SO 4 and HgSO 4 , it must contain a triple bond(–C ≡ CH) in the side chain. Here, the given reactionsmay be formulated as follows :OHC≡CHdil H 2SO 4HgSO 4; H 2O(A)(B)Hence. (A)(C)O– C – CH 3C = CH 2CH 3COClAlCl 3; – HClBenzene+ 3I 2 + 4KOHCOCH 3Acetophenone(B)∆–3KI;–3H 2OCOOK+ CHI3(D)(C)Potassium benzoateC≡CHCOCH 3(B)Phenyl acetyleneCOOKPotassium benzoate(D)AcetophenoneCHI 3IdoformXtraEdge for IIT-JEE 8 SEPTEMBER 2011

8. (a) Write the chemical reaction associated with the"brown ring test".(b) Draw the structures of [Co(NH 3 ) 6 ] 3+ , [Ni(CN) 4 ] 2–and [Ni(CO) 4 ]. Write the hybridization of atomicorbital of the transition metal in each case.(c) An aqueous blue coloured solution of a transitionmetal sulphate reacts with H 2 S in acidic medium togive a black precipitate A, which is insoluble inwarm aqueous solution of KOH. The blue solution ontreatment with KI in weakly acidic medium, turnsyellow and produces a white precipitate B. Identifythe transition metal ion. Write the chemical reactioninvolved in the formation of A and B. [IIT-2000]Sol. (a) NaNO 3 + H 2 SO 4 → NaHSO 4 + HNO 32HNO 3 + 6FeSO 4 + 3H 2 SO 4 →3Fe 2 (SO 4 ) 3 + 2NO + 4H 2 O[Fe(H 2 O) 6 ]SO 4 .H 2 O + NOFerrous Sulphate⎯→ [Fe(H 2 O) 5 NO] SO 4 + 2H 2 O(Brown ring)(b) In [Co(NH 3 ) 6 ] 3+ cobalt is present as Co 3+ and itscoordination number is six.Co 27 = 1s 1 , 2s 2 2p 6 , 3s 2 3p 6 3d 7 , 4s 2Co 3+ ion = 1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 6HenceCo 3+ ion inComplex ionH 3 NNH 33d 4s 4p3d 4s 4p3+CoorH 3 N NH 3NH 3NH 3d 2 sp 3 hybridizationH 3 NH 3 NNH 3Co 3+NH3NH3NH 3In [Ni(CN) 4 2– nickel is present as Ni 2+ ion and itscoordination numbers is fourNi 28 =1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 8 , 4s 2Ni 2+ ion = 1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 8Ni 2+ ion =Ni 2+ ion inComplex ion3d 4s 4p3d 4s 4pdsp 2 hybridizationHence structure of [Ni(CN) 4 ] 2– isN ≡ CN ≡ CNi 2+C ≡ NC ≡ NIn [Ni(CO) 4 , nickel is present as Ni atom i.e. itsoxidation number is zero and coordination number isfour.Ni in3d 4s 4pComplexIts structure is as follows :OCCONisp 3 hybridizationCOCO(c) The transition metal is Cu 2+ . The compound isCuSO 4 .5H 2 OCuSO 4 + H 2 S ⎯Acidic ⎯⎯⎯medium ⎯ → CuS ↓ + H 2 SO 4Black ppt2CuSO 4 + 4KI ⎯→ Cu 2 I 2 + I 2 + 2K 2 SO 4(B) whiteI 2 + I – ⎯→ I – 3 (yellow solution)9. The solubility product of Ag 2 C 2 O 4 at 25ºC is1.29 × 10 –11 mol 3 l –3 . A solution of K 2 C 2 O 4containing 0.1520 mole in 500 ml water is shaken at25ºC with excess of Ag 2 CO 3 till the followingequilibrium is reached :Ag 2 CO 3 + K 2 C 2 O 4 Ag 2 C 2 O 4 + K 2 CO 3At equilibrium the solution contains 0.0358 mole ofK 2 CO 3 . Assuming the degree of dissociation ofK 2 C 2 O 4 and K 2 CO 3 to be equal, calculate thesolubility product of Ag 2 CO 3 . [IIT-1991]Sol. Ag 2 CO 3 + K 2 C 2 O 4 → Ag 2 C 2 O 4 + K 2 CO 3Moles at start Excess 0.1520 0 0Moles after reaction0.1520 – 0.0358 0.0358 0.0358= 0.11622–Molar concentration of K 2 C 2 O 4 or C 2 O 4 left0.1162unreacted =0.5= 0.2324 moles l –1[K 2 CO 3 ] = [CO 3 2– ] at equilibrium == 0.07156 moles l –10.03580.5XtraEdge for IIT-JEE 9 SEPTEMBER 2011

Given that K sp for Ag 2 C 2 O 4 = 1.29 × 10 –11 mol 3 l –3 at25ºCSo, [Ag + ] 2 [C 2 O 2– 4 ] = 1.29 × 10 –11or [Ag + ] 2 × 0.2324 = 1.29 × 10 –11Hence [Ag+] 2 1.29= × 10 –110.2324Then K sp for−11Ag 2 CO 3 = [Ag + ] 2 [CO 2– 1.29×103 ] =× 0.07160.2324= 3.794 × 10 –12 mol 3 l –310. The molar volume of liquid benzene(density = 0.877 g ml –1 ) increases by a factor of 2750as it vaporizes at 20ºC and that of liquid toluene(density = 0.867 g ml –1 ) increases by a factor of 7720at 20ºC. A solution of benzene and toluene at 20ºChas a vapour pressure of 46.0 torr. Find the molefraction of benzene in vapour above the solution.[IIT-1996]Sol. Given that,Density of benzene = 0.877 g ml –1Molecular mass of benzene (C 6 H 6 )= 6 × 12 + 6 × 1 = 7878∴ Molar volume of benzene in liquid form = ml 0.877=780.8771× 1000L = 244.58 LAnd molar volume of benzene in vapour phse=780.8772750× L = 244.58 L1000Density of toluene = 0.867 g ml –1Molecular mass of toluene (C 6 H 5 CH 3 )= 6 × 12 + 5 × 1 + 1 × 12 + 3 × 1 = 92∴ Molar volume of toluene in liquid form92 92 1= ml = × L0.867 0. 867 1000And molar volume of toluene in vapour phase=920.8677720× L = 819.19 L1000Using the ideal gas equation,PV = nRTAt T = 20ºC = 293 KFor benzene, P =0P B =nRTV1 × 0.082×293=244.58= 0.098 atm= 74.48 torr (Q 1 atm = 760 torr)Similarly, for toluene,P =0P T =nRTV1 × 0.082×293== 0.029 atm819.19= 22.04 torr (Q 1 atm = 760 torr)According to Raoult's law,0P B = PBx B = 74.48 x B0P T = PTx T = 22.04 (1 – x B )0 0And P M = PBx B + PTx Tor 46.0 = 74.48 x B + 22.04 (1 – x B )Solving, x B = 0.457According to Dalton's law,'P B = P M x B (in vapour phase)or mole fraction of benzene in vapour form,'x B =PPBM74 .48×0.457== 0.7446.0MATHEMATICSt –te + e e t – e–t11. For any real t, x = , y = is a point22on the hyperbola x 2 – y 2 = 1. Find the area boundedby this hyperbola and the lines joining its centre tothe points corresponding to t 1 and – t 1 . [IIT-1982]Sol. We have to find the area of the region bounded bythe curve x 2 – y 2 = 1 and the lines joining the centrex = 0, y = 0 to the point (t 1 ) and (– t 1 )yP(t 1 )–1CA1 NRequired areat⎡1 – te + e 1 ⎤⎢2 ⎥= 2 ⎢area of ∆PCN – ⎥⎢ ∫y dx⎥⎢1⎥⎣⎦⎡ ⎛t1– t⎞⎛⎞ ⎤= 2 ⎢ ⎜+1 t1– tt1 e e⎟⎜e – e1⎟ ⎥⎢∫ 1dx– y . dt2⎥⎣ ⎝ 2 ⎠⎝2 ⎠ dt1 ⎦⎡ 2t1–2tt⎛ ⎞ ⎤= ⎢⎜ ⎟ ⎥⎢ ∫ 121 t – te – e e – e–dt8⎥⎣0 ⎝ 2 ⎠ ⎦xXtraEdge for IIT-JEE 10 SEPTEMBER 2011

==ee2t1 – 2t–1e42t1 – 2t–1e42t1 – 2t1 t 12t–2 t–∫( e + e – 2) dt2–120⎡e⎢⎢⎣22t–2te–2⎤– 2t⎥⎥⎦e – e11 2t=– {1 –2te – e1– 4t1}= t 14 4Hence, the required area bounded by this hyperbolaand the lines joining its centre is 't 1 '.12. A swimmer S is in the sea at a distance d km fromthe closest point A on a straight spere. The house ofthe swimmer is on the shore at a distance L km fromA. He can swim at a speed of u km/hr and walk at aspeed of v km/hr (v > u). At what point on the shoreshould be land so that he reaches his house in theshortest possible time?[IIT-1983]Sol. Let the house of the swimmer be at B.∴ d (AB) = L km.Let the swimmer land at C, on the shore and letd (AC) = x kmSd∴ d(SC) =2x + dt1A x C (L–x) B222x + d and d (CB) = (L – x)distancetime =speedTime from S to B = time from S to C + time from Cto B.∴ T =x2 +udHence, we take f (x) =2+L – xv1ux02 2 L x+ + – v v1 1.2x1⇒ f ' (x) = .+ 0 –u 2 22 x + d vFor either a maximum or minimum f,f '(x) = 0⇒ v 2 x 2 = u 2 (x 2 + d 2 )i.e., x 2 u d=2 2v – u∴ f ' (x) = 0 at x = ±But22x ≠– udv2 – uud2 – u 2 v2d.(v > u)∴ we consider x =f '' (x) =ud2 – u 2 v1 du 2 2 2x + d ( x + d∴ f has minimum at x =1– cos mx13. Let I m =∫ dx1– cos xπ022)> 0 for all x.ud2 – u 2 vUse mathematical induction to prove thatI m = mπ, m = 0, 1, 2... [IIT-1995]π1– cos mxSol. I m =∫ dx1– cos x0for m = 0,I 0π1– cos(0)=∫ 1– cos x dx0π1–1=∫ 1– cos x dx0π0I 0 =∫ 1– cos x dx = 00for m = 1,π1– cos xI 1 =∫ dx1– cos x0(given)π=∫1dx = π0Therefore, the result is true for m = 0 and m = 1Assume that the result is true for all m ≤ k, k > 0.Let now m = k + 1. We haveπ1 – cos( k + 1) xI k+1 =∫dx1 – cos x01 – cos (k + 1) x= 1 – {cos kx cos x – sin kx sinx}= 1 + cos kx cos x + sin kx sin x – 2 cos kx cos x(add and subtract cos kx cos x)= 1 + cos (k – 1) x – 2cos kx cos x= 2 – [1 – cos (k – 1) x] – 2cos kx cos x= 2 – [1 – cos (k –1)x ] – 2 cos kx cos x+ 2 cos kx – 2 cos x= 2(1 – cos kx) – [1 – cos (k – 1)x]+ 2 cos kx (1 – cos kx)Hence,π1 – cos( k + 1) xI k+1 =∫dx1 – cos x0XtraEdge for IIT-JEE 11 SEPTEMBER 2011

π(1 – cos kx)1 – cos( k – 1) x= 2∫dx –1 – cos x ∫dx1 – cos x0= 2I k – I k – 1 + [ sin kx] π 02π0π(1– cos x)(coskx)+ 2∫dx(1– cos x)k= 2I k – I k–1 + 0= 2 kπ – ( k – 1) π= (k + 1)πThis shows that the result is true for m = k + 1.By the principle of mathematical a induction theresult is true for all non-negative integers m.14. A bird flies in a circle on a horizontal plane. Anobserver stands at a point on the ground. Suppose60º and 30º are the maximum and the minimumangles of elevation of the bird and that they occurwhen the bird is at the points P and Q respectivelyon its path. Let θ be the angle of elevation of thebird when it is at a point on the arc of the circleexactly midway between P and Q. Find thenumerical value of tan 2 θ. (Assume that the observeris not inside the vertical projection of the path of theSol.bird.)0[IIT-1998]Let OM = d, PU = MT = UQ = TN = TS = rAlso, let PM = RS = QN = hIn ∆ POMOWe have60º30º dM⇒ h = 3 dIn ∆QONtan 30º =P r U rRQTrNrSPM htan 60º = = OM dhd + 2r⇒ d + 2r = 3 h = 3d⇒ d = rAlso, OS 2 = OT 2 + TS 2= (d + r) 2 + r 2 = 5d 2In ∆ ROS, we haveRS 3d3tan θ = = =OS 5d5⇒ tan 2 θ = 3/515.⎡a0 1⎤⎡a1 1⎤A =⎢ ⎥⎢1 c b⎥, B =⎢ ⎥⎢0 d c⎥,⎢⎣1 d b⎥⎦⎢⎣f g h⎥⎦⎡ f ⎤ ⎡2a ⎤U =⎢ ⎥ ⎢ ⎥⎢g⎥, V = ⎢ 0 ⎥⎢⎣h⎥⎢ ⎥⎦ 0⎣ ⎦If there is a vector matrix X,such that AX = U has infinitely many solutions, thenprove that BX = V cannot have a unique solution. Ifa + d ≠ 0. Then prove that BX = V has no solution.[IIT-2004]Sol. Since AX = U has infinitely many solutions⇒ |A| = 0a 0 11 c b = 01 d b⇒ a(bc – bd) + 1(d – c) = 0⇒ (d – c) (ab – 1) = 0∴ ab = 1 or d = ca 0 fAgain, |A 1 | = 1 c g = 0 ⇒ g = h1 d ha f 1|A 2 | = 1 g b = 0 ⇒ g = h1 h band |A 3 | =f 0 1∴ g = h, c = d and ab = 1Now, BX = Va 1 1|B| = 0 d c = 0f g h∴ghcdbb= 0 ⇒ g = h(Since C 2 and C 3 are equal) using (i)BX = V has no solution.|B 1 | =a2001dg1ch= 0(Since c = d and g = h) using (i)|B 2 | =a0f(Since c = d )Since, adf ≠ 0 ⇒ |B 2 | ≠ 0∴ |B| = 0 and |B 2 | ≠ 0;∴ BX = V has no solution.a2001ch= a 2 cf = a 2 df...(i)XtraEdge for IIT-JEE 12 SEPTEMBER 2011

Physics Challenging ProblemsSet # 5This 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. Two point monochromatic and coherent sources oflight of wavelength λDeach are placed as shown.SIf initial phase difference 1 S 2Obetween the sources is zerodand (D >> d) then –screen7λ(A) If d = , O will be minima2(B) If d = λ , only one maxima can be observed onscreen(C) If d = 4.8λ,the total 10 minima will be observedon screen5λ(D) If d = , the intensity at O will be minimum22. Resonance occurs in a series LCR circuit when thefrequency of applied emf is 1000 Hz. Then –(A) When frequency is 900Hz, then current throughthe voltage source will be ahead of emf of source(B) The impedance of circuit will be minimum at 100Hz(C) Only at resonance voltage across L and current inC differ in phase by 180º(D) If the value of C is doubled resonance occurs at2000Hz3. For the followingsituation which of thefollowing is/are correct –(A) Potential ofqconductor is4π ∈ 0 (d + R)q(B) Potential of conductor is4π ∈ 0 d(C) Potential of conductor cannot be determined asdistribution of induced charged is not known(D) Potential at B due to induced charges on conductoris−qR4π ∈ 0 (d + R)dPassage # (Q. No. 4 to Q. No. 6)A non conducting vessel containing n moles of a diatomicgas ions fitted with a conducting piston. The cross-sectionalarea, thickness and thermal conductivity of piston are A, l+qdBR OHollow neutralconductorand k respectively. The right side of piston is open toatmosphere attemperature T 0 .Heat is suppliedto the gas bymeans of anelectric heater at aconstant rates q.4. Temperature of gas as a function of time is (if initialtemperature is T 0 )ql−(2KAt/ 7nRl)(A) T = T0+ [1 − e ]kAql−(7nRl/ 2KAt)(B) T = T0+ [1 − e ]kAql−(2KAt/ 7nRl(C) T = T0− [1 − e )]kA(D) None of these5. Maximum temperature of gas –qlql(A) Tmax= T0+ (B) Tmax= T0+ [1 − e]kAkAkAql(C) Tmax = T0+ (D) T max =qlkA6. The ratio of the maximum volume to the minimumvolume is –qlkAT0ql kAT(A) 1+ (B) 1+ (C) (D) 0kAT 0 ql kAT 0 ql7. The electric field in a region is radially outward withmagnitude E = αr. Calculate the charge contained in asphere in 10 –10 C of radius R centred at origin. UseVα = 100 and R = 0.30m.2m8. Magnetic flux linked with a stationary loop ofresistance R varies with time during time period T asfollows φ = at(T− t)Then the amount of heat generated in the loop duringtime T is (assume inductance of coil is negligible)aT 32 2 2 32 3a T a T T(A) (B) (C) (D)3R3R Ra3 RXtraEdge for IIT-JEE 13 SEPTEMBER 2011

8Questions1. ConceptualOption [A,B,C,D] is correct2. ConceptualOption [A,C,D] is correctR3. (A) τ = KC = RC2SolutionSet # 4Physics Challenging Problemswere Published in August Issue4. Option [A,B,C] is correct5. Option [A] is correct6. Final current isRV that distributes in inverse ratio toinductance.Option [C] is correct7. Option [B] is correctI(C,D) tan φ = VItan φ = ωRCCRv1ωt=v / R18. Option [B] is correctTransitions emitting photons of energy more thanwork function W = 2.3 eV can result in aphotoelectric.h hhλ = = ⇒ λ min =p 2mK2mK maxK max = 13.6 – 0.85Option [A,C,D] is correct-: CHEMISTRY JOKE :-If you didn't get the joke, you probably didn't understand the science behind it. If this is the case, it's a chance for youto learn a little chemistry.Chemistry Joke :This is no joke but a call to *BAN* dihydrogen monoxide, otherwise know as the invisible, killer substance. JupiterScientific's science joke webpage is probably not the place to post this protest, but the JS staff feels very strongly aboutthis issue. For your information, dihydrogen monoxide (DHMO) is colorless, odorless, tasteless, and kills thousands ofpeople every year. Most of these deaths are caused by accidental inhalation of DHMO in its liquid form, but thedangers of dihydrogen monoxide do not end there. Prolonged exposure to its solid form causes tissue damage andcontact with its gaseous form causes burns. DHMO use is widespread. For those who have become dependent on it,DHMO withdrawal means death. DHMO can be an environmental hazard: it is a major component of acid rain,contributes to the "greenhouse effect", leads to the erosion of natural landscapes and hastens the corrosion of mostmetals. Being so prevalent (quantities are found in every stream, lake and reservoir), DHMO contamination is atepidemic proportions. Despite the dangers, DHMO is often used as an industrial solvent, as a fire retardant, in nuclearpower plants and (can you believe this) in certain food products. Companies dump waste dihydrogen monoxide intorivers and the ocean, and nothing can be done to stop them because this practice is still legal. STOP THE HORRORNOW! The American government and the United Nations have refused to ban the production, distribution or use ofthis chemical due to its "economic importance." The navy and certain other military organizations are highly dependenton DHMO for various purposes. Military facilities receive tons of it through a sophisticated underground distributionnetwork. It is also stored in large quantities for military emergencies. BUT IT'S NOT TOO LATE! You can help. Act*NOW* to prevent further contamination. Write your representatives. Start and sign petitions. Send e-mails. Informyour friends about the dangers. What you don't know *CAN* hurt you and every individual throughout the world.XtraEdge for IIT-JEE 14 SEPTEMBER 2011

PHYSICSStudents'ForumExpert’s Solution for Question asked by IIT-JEE Aspirants1. Consider a sphere of radius R 0 and mass M 0 . Inside itis a spherical cavity of radius b. The distancebetween the two centers is R(see figure). Calculatethe gravitational force acting on a mass m inside thecavity.]bRO´M 0OR 0Sol. The main principle which we use in the solution ofthe problem is the superposition principle, where weconsider the spherical cavity as a sphere of negativemass. We denote the position vector of a point insidethe cavity as r r and the position vector as measuredfrom the solid sphere's center O as r (see figure). Theforce acting upon m isrRF r = F r 0 – F r 1...(1)where F r 0 is the force of the solid sphere and F r1the force of the cavity (which is filled with negativemass).According to Newton's gravitation law,F r GM(r)m0 = ( − rˆ )...(2)2rwhere M(r) is the mass up to radius r (see figure).r´isNote that this is the only mass which affects mpositioned at r. The mass density is uniform, thusM0ρ = . Therefore,4π3R 0334πM(r) = r3 M0⎛ r ⎞= M 0 ⎜ ⎟3 4π3RR 00⎝ ⎠3Substituting M(r) in Eq. (2), we obtain :F r GM0mGM0mr0 = – rrˆ= – 33R 0 R 0Similarly,F r 1 =GM mr3R0 r0Now, F r = F r 0 – F r GM0mr r1 = (r)3 −R 0Since R r = r – r ,F r GM0m= R r ≡ –C R r3R 0where C is a constant. We have arrived at theinteresting result that the gravitational force insidethe cavity is constant and is directed towards thesphere's center. Notice that for the special caseR r = 0 the force vanishes, as expected.2. Two masses, m 1 and m 2 , are tied to the ends of aspring whose force constant is k, and whose naturallength is a. This system placed horizontally on aperfectly smooth table, as shown in fig. At t = 0, m 1is bumped and receives a linear momentum ofp r = p 0 xˆ , where p 0 is a constant.1m 2 k m 1rR 0M(r)a(i) Write the equations of motion for m 1 and m 2 .What is the velocity of the center of mass?XtraEdge for IIT-JEE 15 SEPTEMBER 2011

(ii) Prove that the harmonic oscillation equation ofthe system is : µ( & x&2 – & x&1) = – k(x 2 – x 1 )where µ =m1mm + m122(iii) What is the oscillation amplitude of (x 2 – x 1 ) ?Sol. The motion of the system of the two bodies can beconveniently described by using the center of massframe of reference. The centre of mass moves in astraight line with constant velocity, due to theconservation of linear momentum. In the centre ofmass frame the two bodies perform simple harmonicoscillations. Denoting the position of the masses m 1and m 2 by x 1 and x 2 , respectively, we can express thedistance between the masses as x 2 – x 1 . The change inthe length of the spring is then x = x 2 – x 1 – a.(i) The forces applied by the spring on the twosystems are :m 1 & x& 1 = kx ...(1)m 2 & x& 2 = – kx ...(2)The signs are used according to the position of themass relative to the spring. Multiplying Eq.(1) by m 2 ,Eq. (2) by m 1 , and subtracting Eq. (1) from Eq. (2)we havem 1 m 2 ( & x&2 – & x&1) = – k(m 1 + m 2 )x ...(3)Since x = x 2 – x 1 – a, we have x& = x& 2 – x&1. And& x& = & x&2 – & x&1. Therefore,m 1 m 2 & x& = –k(m 1 + m 2 )x ...(4)and the solution to this equation is :x(t) = A cos(ω 0 t + φ) ...(5)m1 where ω 0 =+ m2k . Notice that x does notm1m2denote the position of any of the masses. It denotesthe difference between the distance between themasses and the initial state, so that x = (x 2 – x 1 ) – a.The velocity of the centre of mass is given by :v r cm =p10xˆm + m2...(6)(ii) The equation specified in the problem is easilyderived from Eq. (3), which we found in the firstsection. The constant µ is called the "reduced mass"of the system, and is defined as1 1 1 ≡ + ...(7)µ m1m2or in a different form,m1m2µ =...(8)m1+ m2(iii) Taking energy into consideration, we have2 21 ⎛ p0⎞ 1 pE k = m12 ⎜m⎟0= ...(9)⎝ 1 ⎠ 2 m1where E k is the initial kinetic energy. The kineticenergy of the center of mass is :21 p0E k(cm) =...(10)2 m1+ m2and therefore, the total kinetic energy in the centre ofmass frame becomes :m2 2E´k(cm) = E k – E k(cm) =p02m1(m1+ m2)The kinetic energy is proportional to the square ofampitude, E´k = 21 kA2Therefore, A =m2p0km (m + m13. (i) A mass m with kinetic energy E k collides withanother mass M, initially at rest, and sticks to it at themoment of contact. What is the total kinetic energyimmediately after the collision ?(ii) The mass M is used as the weighting surface of aspring-scale whose spring is ideal (a mass lessspring). A body of known mass m is released from acertain height from where it falls to hit M. The twomasses M and m stick together at the moment theytouch, and move together from then on. The oscillationsthey perform reach to height a above the original levelof the scale, and depth b below it (see figure).(a) Find the constant of force of the spring.(b) Find the oscillation frequency.(c) What is the height above the initial level fromwhich the mass m was released ?mmMMambMBefore122)After1Sol. (i) The initial kinetic energy is given by E k = p 2 ,2mand the final value of the kinetic energy is given by1E´k = p´2.2(m + M)Using linear-momentum conservation (p´ = p), we have:mE´k = E km + MXtraEdge for IIT-JEE 16 SEPTEMBER 2011

(ii) (a) Since the point of equilibrium changes, weknow that a ≠ b (it moves down from the originallevel due to the extra mass m). Furthermore, usinga – (–b) = a + b = 2A, where A is the amplitude ofoscillations, and a – y = A, where y is the height ofthe new point of equilibrium relative to the originalone (y < 0), we find that the point of equilibrium isb − abelow the original level of the scale. Applying2the equilibrium of forces, we can write :⎛ b − a ⎞k|y| = k ⎜ ⎟ = mg⎝ 2 ⎠And, therefore, k =2mgb − a.(b) The oscillation frequency is v =where ω is defined as :ω2 π,k2mgω = =m + M (m + M)(b − a)(c) The potential energy in harmonic motion is1known to be kx 2 . Therefore, the law of2conservation of energy yields21E(t = 0) = E´k + ky 2 k ⎛ b − a ⎞= E´k + ⎜ ⎟2 2 ⎝ 2 ⎠2k ⎛ b + a ⎞= E´´ = ⎜ ⎟2 ⎝ 2 ⎠where E´´ denotes the total energy when the massstops; i.e., when the amplitude is maximal and thekinetic energy vanishes. Using the result of the firstsection (E k = mgh and the relation between E k andE´k) along with Eq. we obtain :22m k ⎛ b + a ⎞ k ⎛ b − a ⎞E´k = mgh = ⎜ ⎟ – ⎜ ⎟m + M 2 ⎝ 2 ⎠ 2 ⎝ 2 ⎠= 2k abPlugging in the value of k, we arrive at :m + M abh =m b − a4. A constant current I flows through a cable consistingof two thin co-axial metallic cylinders of radii R and2R. Calculate(i) energy stored in it per unit length and(ii) inductance per unit lengthSol. Due to flow of current through the cable, show infigure (A), magnetic field is established in the spacebetween cylinders. Energy is stored in this space dueto magnetic field.Fig.(A)Since, energy stored per unit volume in magneticfields of induction B is equal to B 2 /2µ 0 , therefore,magnetic induction in the space must be known. Butmagnetic field in the space is not uniform.Hence, consider a thin cylindrical coaxial shell inthe space. Let radius of the shell be x and radialthickness dx as shown in figure(B).RdxxI2RFig.(B)According to Ampere's circuital law, magneticinduction B is given by B 2πx = µ 0 Iµ 0or B =I2πx∴ Energy stored per unit length of the shellconsidered isdU =2B2µ (2πx dx) = µ 0I 2dx4πx0∴ Energy stored per unit length of the cable,U =∫ dU =x=2R∫x=R2µ 0Idx =4πxµ 0I 2log e 24πAns. (i)Since, energy stored in magnetic field is U = 21 LI2where L is self inductance, therefore, inductance perunit length of the cable is2U µL = =0 log2e 2 Ans. (ii)I 2π5. A non-conducting thin spherical shell of radius Rhas uniform surface charge density σ. The shellrotates about a diameter with constant angularvelocity ω. Calculate magnetic induction B at thecentre of the shell.Sol. When the shell rotates, current is induced due tomotion of charge. To calculate magnetic induction atcentre of the shell, rotating shell can be assumed tobe composed of thin circular current carrying rings.Such a ring can be assumed as follows:XtraEdge for IIT-JEE 17 SEPTEMBER 2011

θωConsider a radius of the shell inclined at angle 'θ'with the axis of rotation. This radius is rotated aboutthe axis keeping θ constant. Thus a circle is tracedas shown in figure.Its radius, r = R sin θDistance of its centre from centre of the shell,x = R cos θ,Now consider another radius inclined at angle(θ + dθ). It is also rotated in the same way andanother circle is traced. The portion between twocircles forms a circular ring.Area of this ring = 2πr R dθ = 2πR 2 sin θ dθCharge on this ring , = dθ = σ.2πr 2 sin θ dθSince, angular velocity of the shell is ω, therefore, itωcompletes revolutions per second.2πHence, current associated with the ring considered,ωi =2πdQ = σωR2 sin θ dθSince, centre of the shell is a point lying on the axisof a circular coil of radius r, carrying current i at adistance x from centre of the coil, therefore,magnetic induction at centre of the shell due to thiscoil isdB =2(rµ ir20+ x22 3 / 2= 21µ0 σω R sin 3 θ dθ)Hence. resultant magnetic induction at centre of theshellB =∫ dBπ1 3= µ0 σωR 2 ∫sin θ dθ= 32µ0 σωR Ans.0-: Niobium :-Brief description : the name niobium was adoptedofficially by IUPAC in 1950, but a few commercialproducers still like to refer to it as columbium.Niobium is a shiny, white, soft, and ductile metal, andtakes on a bluish tinge when exposed to air at roomtemperatures for a long time. The metal starts tooxidize in air at high temperatures, and when handledhot must be done so under a protective atmosphere soas to minimize oxide production.Table : basic information about and classifications ofniobium.• Name : Niobium• Symbol : Nb• Atomic number : 41• Atomic weight : 92.90638 (2)• Standard state : solid at 298 K• Group in periodic table : 5• Group name : (none)• Period in periodic table : 5• Block in periodic table : d-block• Colour : grey metallic• Classification : MetallicISOLATIONIsolation : isolation of niobium appears to becomplicated. Niobium minerals usually contain bothniobium and tantalum. Since they are so similarchemically, it is difficult to separate them. Niobiumcan be extracted from the ores by first fusing the orewith alkali, and then extracting the resultant mixtureinto hydrofluoric acid, HF. Current methodologyinvolves the separation of tantalum from these acidsolutions using a liquid-liquid extraction technique. Inthis process tantalum salts are extracted into theketone MIBK (methyl isobutyl ketone, 4-methylpentan-2-one). The niobium remains in the HFsolution. Acidification of the HF solution followed byfurther extraction in MIBK gives an organic solutioncontaining niobium.After conversion to the oxide, metallic niobium can bemade by reduction with sodium or carbon.Electrolysis of molten fluorides is also used.XtraEdge for IIT-JEE 18 SEPTEMBER 2011

XtraEdge for IIT-JEE 19 SEPTEMBER 2011

PHYSICS FUNDAMENTAL FOR IIT-JEECurrent ElectricityKEY CONCEPTS & PROBLEM SOLVING STRATEGYReview of Concepts :Electric current is the rate of transfer of chargethrough a certain surface.The direction of electric current is as that of flowof positive charge.If a charge ∆q cross an area in time ∆t, then theaverage current = ∆q/∆tIts unit is C/s or ampere.Electric current has direction as well asmagnitude but it is a scalar quantity.Electric current obeys simple law of algebra.i.e., I = I 1 + I 2I 1I 1αTypes of Current :Steady state current or constant current : Thistype of current is not function of time.Transient or variable current : This type of currentpassing through a surface depends upon time.∆qdqi.e., I = f(t) or I = lim ⇒∆ t → 0 ∆ t dtElectric charge passing a surface in timet = q =∫ tI dt0Average current I =∫∫t0t0IdtConvection Current : The electric due tomechanical transfer of charged particle is calledconvection current. Convection current in differentsituation.Case I : If a point charge is rotating with constantangular velocity ω.q 2πI = ; T = T ωdt⇒ I =I 2qω2πCase II : If a non-conducting ring having λ chargeper unit length is rotating with constant angularvelocity ω about an axis passing through centre ofring and perpendicular to the plane of ring.I = R λω∆S∆Ior J =∆ScosθIts unit A/m 2Electric current can be defined as flux of currentdensity vector.→ →i.e., i =∫ j . dSRelation between drift velocity and current→jdensity v d = – enHere, negative sign indicates that drifting of electrontakes place in the opposite direction of currentdensity.The average thermal velocity of electron is zero.Electric resistance : Electric resistance (R) isdefined as the opposition to the flow of electriccharge through the material.It is a microscopic quantity.Its symbol isIts unit is ohm.(a)(b)ρlR = Awhere, R = resistance,ρ = resistivity of the material,l = length of the conductor,A = area of cross sectionContinuity Equation :→ →dq∫j .dS = –c dtThe continuity equation is based on conservationprinciple of charge.Drift Velocity (v d ) : When a potential differenceis applied between ends of metallic conductor, anĴθXtraEdge for IIT-JEE 20 SEPTEMBER 2011

electric field is established inside the metallicconductor. Due to this, electron modify theirrandom motion and starts to drift slowly in theopposite direction of electric field. The averagevelocity of drifting possessed by electron isknown as drift velocity.→where, vd→v d =→⎛ eτ ⎞⎜ ⎟ E⎝ m ⎠= drift velocity, e = electron,τ = relaxation time, m = mass of electron→E = electric fieldVariation of Resistance with Temperature :Let a metallic conductor of length l and crosssectionalarea A.R t = R 0 (1 + αt)where,R t = resistance of conductor at temperature tºC,R 0 = resistance of conductor at 0ºC,α = temperature coefficient.iSome Important Points :(a) 'α' is proportionality constant known astemperature coefficient of resistance variation.(b) The value of α does not depend upon initial andfinal resistance of the conductor.(c) The value of α depends upon the unit which ischosen.(d) The value of α may by negative.Electric Conductance (G) :1It is reciprocal of resistance, G = RIts unit is per ohm.1Electric conductivity σ = ρOhm's law in vector form :E = ρ → imwhere, ρ =2 = receptivity of materialne τAccording to ohm's law, electric current passingthrough a conductor is proportional to thepotential difference between end of the conductori.e., V = IRIn case of ohm´s law, V-I graph is straight line.AVαOhm's Law fails in tube, crystal diodes, thyristorsetc.EMF and PD of a Cell : A device which supplieselectric energy is called a seat of emf. The seat ofemf is also called a cell.A battery is a device which manages a potentialdifference between its two terminals.e = EMF of the battery is the work done by theforce per unit charge.When the terminals of a cell are connected to anexternal resistance, the cell is said to be in closedcircuit.E.M.F. has no electrostatic origin.Internal Resistance of a Cell (r) : Internalresistance of a cell is the resistance of itselectrolyte.The internal resistance of cell :(a) Varies directly as concentration of the solution ofthe cell.(b) Varies directly as the separation betweenelectrodes i.e., length of solution betweenelectrodes.(c) Varies inversely as the area of immersedelectrodes.(d) is independent of the material of electrodes.Potential difference across the cell :Potential difference across the first cellV 1 = E 1 + Ir 1 (discharging of cell)iRE 1 r 1 E 2 r 2Potential difference across the second cellV 2 = E 2 – Ir 2 (charging of cells)Concept of Rise up and Drop up of voltage:(a) Ideal cellRise up+EI–EDrop upXtraEdge for IIT-JEE 21 SEPTEMBER 2011

(b) Real cellr,EiRise upr,EDrop upE – ir–E – ir(c) Electric resistanceRi R iDrop up–IR+IRiRise upWhen a battery being charged, the terminalvoltage is greater than its emf V = E + Ir.Kirchhoff's Law : Kirchhoff's law is able tosolve complicated circuit problems.(i) First Law : Incoming current = Outgoing currentI 1 + I 2 = I 3 + I 4 + I 5I 2I5I 1I 3I 4This law is based upon conservation principle ofcharge.(ii) Second Law : (Loop rule or voltage law.) Thislaw is based upon conservation principle ofenergy.Grouping of resistors :Case I : Resistors in seriesR MN = R eq = R 1 + R 2M R 1 R 2 NIn general,R eq = R 1 + R 2 + ... + R nCase II : Resistors in parallel1R MN=In general,1R MN=1R eqMR 1=1 1 +R 2R 1R 1R 2N1 1 1+ + ... +R 2 R nProblem solving strategy. : Power and Energy in circuitsStep 1 Identify the relevant concepts :The ideas of electric power input and output can beapplied to any electric circuit. In most cases you’llknow when these concepts are needed, because theproblem will ask you explicitly to consider power orenergy.Step 2 Set up the problem using the following steps :Make a drawing of the circuit.Identify the circuit elements, including sources ofemf and resistors.Determine the target variables. Typically theywill be the power input or output for each circuitelement, or the total amount of energy put into ortaken out of a circuit element in a given time.Step 3 Execute the solution as follows :A source of emf ε delivers power εI into a circuitwhen the current I runs through the source from –to +. The energy is converted from chemicalenergy in a battery, from mechanical energy in agenerator, or whatever. In this case the source hasa positive power output to the circuit or,equivalently, a negative power input to thesource.A source of emf power εI from a circuit – that is,it has a negative power output, or, equivalently, apositive power input–when currents passesthrough the source in the direction from + to –.This occurs in charging a storage battery, whenelectrical energy is converted back to chemicalenergy. In this case the source has a negativepower output to the circuit or, equivalently, apositive power input to the source.No matter what the direction of the currentthrough a resistor, It removes energy from acircuit at a rate given by VI = I 2 R = V 2 /R, whereV is the potential difference across the resistor.There is also a positive power input to the internalresistance r of a source, irrespective of thedirection of the current. The internal resistancealways removes energy from the circuit,converting it into heat at a rate I 2 r.You may need to calculated the total energydelivered to or extracted from a circuit element ina given amount of time. If integral is just theproduct of power and elapsed time.Step 4 Evaluate your answer : Check your results,including a check that energy is conserved. Thisconservation can be expressed in either of two forms:“net power input = net power output” or “thealgebraic sum of the power inputs to the circuitelements is zero.”Problem solving strategy : Series and ParallelStep 1 Identify the relevant concepts : Many resistornetworks are made up of resistors in series, inparallel, or a combination of the two. The keyconcept is such a network can be replaced by a singleequivalent resistor.XtraEdge for IIT-JEE 22 SEPTEMBER 2011

Travel around the loop in the designateddirection, adding potential differences as youcross them. Remember that a positive potentialand a negative potential difference corresponds toa decrease in potential. An emf is counted aspositive when you traverse it from (–) to (+), andnegative when you go from (+) to (–). An IR termis negative if you travel through the resistor in thesame direction as the assumed current andpositive if you pass it in the opposite direction.Figure. summarizes these sign conventions. Ineach part of the figure “travel” is the directionthat we imagine going around a loop while usingKirchhoff’s loop law, not necessary the directionof current.Equate the sum is Step 2 to zero.If necessary, choose another loop to get adifferent relation among the unknowns, andcontinue until you have as many independentequations as unknowns or until every circuitelement has been included in a at least one of thechosen loops.Solve the equations simultaneously to determinethe unknowns. This step involves algebra, notphysics, but it can be fairly complex. Be carefulwith algebraic manipulations; one sign error willprove fatal to the entire solution.You can use this same bookkeeping system tofind the potential V ab of any point a with respectto any other point b. Start at b and add thepotential changes you encounter in going from bto a, using the same sign rules as in Step 2. Thealgebraic sum of the these changes is v ab = V a –V b .Step 4 Evaluate your answer : Check all the step inyour algebra. A useful strategy is to consider a loopother than the ones you used to solve the problem; ifthe sum of potential drops around this loop is notzero, you made an error somewhere in yourcalculations. As always, ask yourself whether isanswer make sense.TravelεTravelε– + + ε+ – – ε+TravelR––IRTravelR+IIWhen using Kirchhoff’s rules, follow these signconventions as you travel around a circuit loop.–+IR1. In a circuit shown in fig.(i) find the current drawn from the accumulator.(ii) find the current through the 3 ohm resistor,(iii) What happens when 3 ohm resistor is removedfrom the circuit ?2VASolved Examples2Ω1ΩB3Ω2Ω4ΩSol. The equivalent Wheatstone's bridge network of thegiven circuit is shown in fig.BA2Ω1ΩD3ΩD4Ω2Ω2 VoltHere the points B and D are at the same potential asthe bridge is balanced. So the 3Ω resistance in BDarm is ineffective and can be omitted from the circuit.The resistance of ABC branch is 2Ω + 4Ω = 6Ω asAB and BC are in series. Similarly the resistance ofA D C branch is 1Ω + 2Ω = 3Ω.The two resistances, i.e., 6 ohm and 3 ohm are inparallel. The equivalent resistance R is given by1 1 1 1= + = ∴ R = 2ΩR 6 3 2(i) The current drawn from 2 volt accumulator isi = RE = 22 = I amp.(ii) The current through 3Ω resistor is zero.(iii) When the 3Ω resistor is removed from thecircuit, there will be no change.2. A battery of e.m.f. 5 volt and internal resistance 20Ωis connected with a resistance R 1 = 50 Ω and aresistance R 2 = 40Ω. A voltmeter of resistance 1000 Ωis used to measure the potential difference across R 1 .What percentage error is made in the reading ?CCXtraEdge for IIT-JEE 24 SEPTEMBER 2011

Sol. The circuit is shown in fig.1000ΩVR 1 = 50Ω20ΩR 2 = 50Ω5VWhen voltmeter is not connectedEcurrent in the circuit i =r + R 1 + R 25 5 1∴ i == = A20 + 50 + 40 110 22Potential difference across R 1 = i × R 11= × 50 = 2.27 volt.22When the voltmeter is connected across R 1 .In this case the galvanometer resistance is in parallelwith R 1 . Hence1000×50Equivalent resistance = = 47.62 ohm1000 + 50Current in the circuit55== A20 + 40 + 47.62 107.62Potential difference measured by voltmeter5= × 47.62 = 2.21 volt.107.622.27− 2.21Percentage error =× 100 = 2.6%2.273. In the circuit fig. a voltmeter reads 30 V when it isconnected across 400 ohm resistance. Calculate whatthe same voltmeter will read when it is connectedacross the 300 Ω resistance ?30 V300 ΩV400 Ω60 VSol. Potential difference across 400 ohm = 30 VPotential difference across 300 ohm= (60 – 30) = 30 VThis shows that the potential difference is equallyshared.Let R be the voltmeter resistance. The resistance 400and voltmeter resistance R are in parallel. Theirequivalent resistance R´ is given by1 1 1 400 + R 400R= + = orR´ R 400 400R 400 + RBut R´ should be equal to 300 ohm. Hence400R= 300 ∴ R = 1200 ohm400 + RThus, voltmeter resistance is 1200 ohm.When the voltmeter is connected across 300 ohm, theeffective resistance R" is given by1 1 1 1+ 4 5= + = =R" 1200 300 1200 12001200∴ R´´ = = 240 ohm.5Now the potential difference is shared between 240ohm and 400 ohm.Potential diff. across 240 ohm : Potential differenceacross 400 ohm= 240 : 400 = 3 : 5As total potential is 60 V, hence potential differenceacross 240 ohm, i.e., across resistance 300 ohm willbe3 × 60 = 22.5 V.84. In the circuit shown in fig. E, F, G and H are cells ofe.m.f. 2, 1, 3 and 1 volt and their internal resistancesare 2, 1, 3 and 1 ohm respectively. CalculateFA+–+E –2Ω–H+D –C+G(i) The potential difference between B and D and(ii) the potential difference across the terminals ofeach of the cells G and H.Sol. Fig. shows the current distribution.Applying Kirchhoff's first law at point D, we havei = i 1 + i 2 ...(1)Applying Kirchhoff's second law to mesh andADBA, we have2i + 1i + 2i 1 = 2 – 1 = 1or 3i + 2i 1 = 1 ...(2)2Ω 2VAB1Ω1ViDi 1i 22Ω3V3ΩBC1V1ΩXtraEdge for IIT-JEE 25 SEPTEMBER 2011

Applying Kirchhoff's second law to mesh DCBD, weget3i 2 – 1i 2 – 2i 1 = 3 – 1or 4i 2 – 2i 1 = 2 ...(3)Solving eqs. (1), (2) and (3), we get• In late 2001, Associated Press reported, "NASA1 6 5i 1 = amp., i2 = amp. and i = amp.might allow McDonald's to put its logo on the13 13 13international space station galley in exchange for(i) Potential difference between B and DMcDonald's promoting space exploration to⎛ 1 ⎞ 2= 2i 1 = 2 ⎜ ⎟ = volt.⎝13⎠ 13kids". Err...Mine's a Big Mac Please.5∴ R = × 6 = 5 ohm. levels.6(ii) Potential difference across G• A 10 pound sack of flour on the moon would6×3= E – i 2 R = 3 – = 1.61 Vbake six times as much bread as a sack weighing13Potential difference across H10 pounds on earth.⎛ − 6 ⎞= 1 – ⎜ ⎟ (1) = 1.46 V.• The Comets that pass close to the Sun originally⎝ 13 ⎠came from one of two places; either the Oort5. Twelve equal wires, each of resistance 6 ohm arejoined up to form a skeleton cube. A current enters atCloud or the Kuiper Belt. Approximately adozen 'new' Comets are discovered every year.one corner and leaves at the diagonally oppositeBecause they are so far from the Sun, the Cometscorner. Find the joint resistance between the corners.Sol. The skeleton ABCDEFGH, is shown in fig.in the Oort Cloud take over 1 million years toE i/6 Fmake a single revolution around the Sun.i/3A i/6B i/6 i/3• There are stars as much as 400,000 brighter thani/3i/3i/6the sun and others as much as 400,000 timefainter if they could all be seen at the samei/6H i/3 Gidistance.i/3• A pulsar is a small star made up of neutrons soD i/6 Cdensely packed together that if one the size of aThis skeleton consists of twelve wires. Let theresistance of each wire be r. Here the current i enterssilver dollar landed on Earth, it would weighat corner A and leaves at corner G. The current i atapproximately 100 million tons.corner A is divided into three equal parts (i/3)because the resistance of each wire is the same. At B, • An exploding supernova can outshine an entireD and E, the current i/3 is divided into two equalgalaxy of stars.parts each having magnitude i/6. At the corners C, Fand H, the currents again combine to give currents,• There are 17 bodies in the solar system whoseeach of magnitude i/3 along CG, FG and HGrespectively. At corner G, all these currents combineradius is greater than 1000 km.so that the current leaving at G is i.Let R be the equivalent resistance between thecorners A and G. Taking any one of the paths sayABCG, we have• Over 90 per cent of the Universe consists ofinvisible 'dark matter'.V AG = V AB + V BC + V CGi i iiR = r + r + r3 6 3• In 1719 Mars was closer to Earth than it wouldbe until the year 2003.5 • By 2100, in the absence of emissions controlor R = r 6 policies, carbon dioxide concentrations areAccording to given problem r = 6 ohmprojected to be 30-150% higher than today'sXtraEdge for IIT-JEE 26 SEPTEMBER 2011

PHYSICS FUNDAMENTAL FOR IIT-JEECircular Motion, Rotational MotionKEY CONCEPTS & PROBLEM SOLVING STRATEGYCircular Motion :When a particle moves on a circular path withuniform speed, its is said to execute a uniformcircular motion.Angular Velocity : It is the rate of change ofangular displacements of the body. If the radial linein the adjoining figure rotates through an angleθ(radian) in time t (seconds) then its angular velocity.O θω = tθ radian / secondIf it takes the radial line a time T to complete onerevolution, then2πω = Tand if n revolutions are made in 1s thenn = T1 and ω = 2πnThe angular acceleration of the particle is given by∆ωα =∆tLinear Velocity :Linear velocity = angular velocity × radiusv = ω × rlinear acceleration of particle (a) = a × rCentripetal Acceleration : When a particle moveswith uniform speed v in a circle of radius r it is actedupon by an acceleration v 2 /r in the direction of centre.It is called centripetal acceleration. The accelerationhas a fixed magnitude but its direction iscontinuously changing. It is always directed towardsthe centre of the circle.Centripetal Forces : If the particle of mass m moveswith uniform velocity v in circle of radius r, thenmv 2force acting on it towards the centre is . This isrcalled centripetal force. It has a fixed magnitude andis always directed towards the centre.Without centripetal force, a body can not move on acircular path. Earth gets this force from thegravitational attraction between earth and sun;electron moves in circular path due to electrostaticattraction between it and nucleus. A cyclic or carwhile taking turn, gets the centripetal force from thefriction between road and type. To create this force,the vehicle tilts itself towards the centre. If it makesangle θ with the vertical in tilted position then thanθ = v 2 /rg. where v is its velocity and r is the radius ofthe path. In order to avoid skidding (or slipping), theangle of tilt θ with vertical should be less than angleof friction λ. i.e. tan θ < tan λv 2or < µ (since coefficient of friction µ = tan λ)rgIn limiting condition v 2 = µ or v = µ .r. grgThis is the maximum safe speed at the turn.Since centripetal force is provided by the friction, itcan never be more than the maximum valueµR = (µmg) or frictional force.Motion in a vertical circle : When a body tied at oneend of a string is revolved in a vertical circle, it hasdifferent speed at different points of the circular path.Therefore, the centripetal force and tension in thestring change continuously. At the highest point A ofmotion.Av amgT arT a + mg =mv 2 arCT bBmgv bor T a =mv 2 ar– mgThis tension, at highest point will be zero, for aminimum velocity v c given bymvc2 0 = – mg or v c = grrXtraEdge for IIT-JEE 27 SEPTEMBER 2011

This minimum speed is called critical speed (v c ). Ifthe speed at A is less than this value, the particle willnot reach up to the highest point. To reach with thisspeed at A, the body should have speed at B given bythe conservation laws viz.Decrease in kinetic energy = increase in potentialenergy1 2 1 mvb – 2 mva = mg.2r2 2v 2 b = v 2 a + 4grfor critical speed v a = v c =gr∴ v 2 b gr + 4gr or v b = 5 grTherefore, the body should have speed at B at least5 gr , so that it can just move in vertical circle.Tension in string at B is given by.T b – mg =mv 2 bror T b = mg +m5vgrr= 6mgThis means that the string should be able to stand to atension, equal to six times the weight of the bodyotherwise the string will break.At any other point P making angle θ with the vertical,from the figure.AT – mg cos θ =CBθmv 2 prTormgv p⎛At point A, θ = 180º; T a = m ⎜⎝ rPQmg cos θ⎛T = m ⎜v 2 r⎝ rv 2 a⎞− g⎟⎠⎞+ g cosθ⎟⎠⎛ ⎞At point B, θ = 0º; T b = m ⎜v 2 b+ g⎟⎝ r ⎠Conical pendulum :A conical pendulum consists of a string AB (fig.)whose upper end is fixed at A and other and B is tiedwith a bob. When the bob is drawn aside and is givena horizontal push. Let it describe a horizontal circlewith constant angular speed ω in such a way that ABmakes a constant angle θ with the vertical. As thestring traces the surface of a cone, it is known asconic pendulum.Let l be the length of string AB. The forces acting onthe bob are (i) weight mg acting downwards,(ii) tension T along the sting (horizontal) componentis T sin θ and vertical component is T cos θ).T cos θ = mgThe horizontal component is equal to the centripetalforce i.e.,Rotational Motion :rAhOTT sin θT cosθBmgCentre of mass of a system of particles :The point at which the whole mass of the body maybe supposed to be concentrated is called the centre ofmass.Consider the case of a body of an arbitrary shape of nXY plane as shown in fig. Let the body consist ofnumber ofY(xP 1(x 1, y 2, y 2)1)(x, y ) P 2P 3(x 3, y 3)XOparticles P 1 , P 2 , P 3 , .... of masses m 1 , m 2 , m 3 , ..... andcoordinates (x 1 , y 1 ), (x 2 , y 2 ), (x 3 , y 3 ), ..... If ( x, y)bethe coordinates of centre of mass, thenm1x1+ m2x2 + m3x3+ .... Σmnxnx ==m + m + m + ..... Σm12m1y1+ m2y2+ m3y3+ ... Σmnynand y ==m1+ m2+ m3+ .... ΣmnWhen there is a continuous distribution of massinstead of being discrete, we treat an infinitesimalelement of the body of mass dm whose position is(x, y, z). In such a case, we replace summation byintegration in above equations. Now we have,x =y =∫∫∫∫x dm=dmy dm=dm∫∫x dmMy dmM3nXtraEdge for IIT-JEE 28 SEPTEMBER 2011

z =∫∫z dm=dm∫z dmMwhere M is the total mass.Motion of centre of mass :Consider two particles of masses m 1 and m 2 locatedat position vectors r 1 and r 2 respectively with respectto origin. Now the position vector r of the centre ofmass is given by(m 1 + m 2 )r = m 1 r 1 + m 2 r 2 ...(1)Thus, the product of the total mass of the system andposition vector of the centre of mass is equal to thesum of the products of the individual masses andtheir respective position vectors. Hencem1r1+ m2r2r =...(2)m1+ m2Now the velocity of centre of mass of the system isdrgiven by v = dtThe acceleration of the centre of mass is given bydva = = dtddt⎛⎜⎝ddt⎟⎠⎞=d2dtThe equation describing the motion of the centre ofmass may be written asdvf(total) = M dtWhen no external force acts on the system, thendv dv0 = M or = 0dt dt∴ v = constantTherefore, when no external force acts on the system,the centre of mass of an isolated system move withuniform velocity.Moment of inertia and radius of gyration :Moment of Inertia : The moment of inertia of abody about an axis is defined as the sum of theproducts of the masses of the particles constitutingthe body and the square of their respective distancefrom the axis.Radius of Gyration : If we consider that the wholemass of the body is concentrated at a distance K fromthe axis of rotation, then moment of inertia I can beexpressed asI = MK 2where M is the total mass of the body and K is theradius of gyration. Thus the quantity whose squarewhen multiplied by the total mass of the body givesthe moment of inertia of the body about that axis isknown as radius of gyration.x2Theorems on moment of inertia :Theorem of parallel axes : According to thistheorem, the moment of inertia I of a body about anyaxis is equal to its moment of inertia about a parallelaxis through centre of mass I G plus Ma 2 where M isthe mass of the body and a is the perpendiculardistance between the axes, i.e., I = I G + Ma 2Theorem of perpendicular axes : According to thistheorem, the moment of inertia I of the body about aperpendicular axis is equal to the sum of moment ofinertia of the body about two axes right angles toeach other in the plane of the body and intersecting ata point where the perpendicular axis passes, i.e.,I = I x + I yTable of moment of inertia :Body1. Thin uniform rodof length l2. Thin rectangularsheet of sides aand b.3. Thickrectangular barof length l,breadth b andthickness t.4. Uniform solidsphere of radiusR5. Circular ring ofradius R.AxisThrough itscentre andperpendicular toits lengthThrough itscentre andperpendicular toits planeThrough itsmidpoint andperpendicular toits lengthAbout a diameterThrough itscentre andperpendicular toits plane6. Disc of radius R. Through itscentre andperpendicular toits plane7. Solid cylinder oflength l andradius R.(i) Through itscentre andparallel to itslength(ii) Through itscentre andperpendicular toits length.Moment ofinertiaMl212⎛2 2⎞M ⎜a b⎟+ ⎝ 12 12 ⎠⎛2 2⎞M ⎜l b⎟+ ⎝ 12 12 ⎠2 MR25MR 21 MR221 MR22⎡ RM ⎢⎢⎣422l ⎤+ ⎥ 12 ⎥⎦Angular momentum of a rotating body :In case of rotating body about an axis, the sum of themomentum of the linear momentum of all theXtraEdge for IIT-JEE 29 SEPTEMBER 2011

particles about the axis of rotation is called angularmomentum about the axis.Q Also the angular momentum of rigid body aboutan axis is the product of moment of inertia and theangular velocity of the body about that axis.L = r × p = IωTranslational and rotational quantities :Translational Motion Rotational MotionDisplacement = s Angular displacement = θVelocity = vAcceleration = aInertia = mForce = FMomentum = mvPower = FvKinetic energy = 21 mv2Angular velocity = ωAngular acceleration = αMoment of inertia = ITorque = τAngular momentum = IωRotational power = τωRotational K.E. = 21 Iω2Kinematics equation of a rotating rigid body :The angular velocity of a rotating rigid body isdefined as the rate of change of angular displacement,i.e., → ω = ( d→ θ / dt)Similarly, the angular acceleration is defined as therate of change of angular velocity, i.e.,2→ d → ω d → θα = =2dt dtLet a body be rotating with constant angularacceleration → α with initial angular velocity → ω 0 . If θis the initial angular displacement, then its angularvelocity → ω and angular displacement θ at any time isgiven by the following equationsω = ω 0 + αt0 = ω 0 t + 21 αt2and ω 2 = ω 2 0 + 2 αθThese equations are similar to usual kinematicsequation of translatory motion.v = u + at, s = ut + 21 at2and v 2 = u 2 + 2asProblem Solving Strategy : Rotational Dynamics forRigid Bodies :Our strategy for solving problems in rotationaldynamics is very similar to the strategy for solvingproblems that in involve Newton’s second law.Step-1 : Identify the relevant concepts : The equationΣτ = Iα z is useful whenever torques act on a rigidbody–that is, whenever forces act on a rigid body insuch a way as to change the state of the body’srotation.In some cases you may be able to use an energyapproach instead. However, if the target variable is aforce, a torque, an acceleration, an angularacceleration, or an elapsed time, the approach usingΣτ = Iα 2 is almost always the most efficient one.Step-2 : Setup the problem using the following steps:Draw a sketch of the situation and select the bodyor bodies to be analyzed.For each body, draw a free-body diagramisolating the body and including all the forces(and only those forces) that act on the body,including its weight. Label unknown quantitieswith algebraic symbols. A new consideration isthat you must show the shape of the bodyaccurately, including all dimensions and anglesyou will need for torque calculations.Choose coordinate axes for each body andindicate a positive sense of rotation for eachrotating body. If there is a linear acceleration, it’susually simplest to pick a positive axis in itsdirection. If you know the sense of α z in advance,picking it as the positive sense of rotationsimplifies the calculations. When you represent aforce in terms of its components, cross out theoriginal force to avoid including it twice.Step-3 : Execute the solution as follows :For each body in the problem, decide whether itunder goes translational motion, rotationalmotion, or both. Depending on the behavior of thebody in question, apply ΣF = m a r , Στ z = Iα z , orboth to the body. Be careful to write separateequations of motion for each body.There may be geometrical relations between themotions of two or more bodies, as with a stringthat unwinds from a pulley while turning it or awheel that rolls without slipping. Express theserelations in algebraic form, usually as relationsbetween two linear accelerations or between alinear acceleration and an angular acceleration.Check that the number of equations matches thenumber of unknown quantities. Then solve theequations to find the target variable(s).Step-4 : Evaluate your answer : Check that thealgebraic signs of your results make sense. As anexample, suppose the problem is about a spool ofthread. If you are pulling thread off the spool, youranswers should not tell you that the spool is turningin the direction the results for special cases orintuitive expectations. Ask yourself : Does this resultmake sense ?”XtraEdge for IIT-JEE 30 SEPTEMBER 2011

Problem Solving Strategy: Equilibrium of a Rigid BodyStep-1 : Identify the relevant concepts : The first andsecond conditions for equilibrium are usefulwhenever there is a rigid body that is not rotating andnot accelerating in space.Step-2 : Set up the problem using the following steps:Draw a sketch of the physical situation, includingdimensions, and select the body in equilibrium tobe analyzed.Draw a free-body diagram showing the forcesacting on the selected body and no others. Do notinclude forces exerted by this body on otherbodies. Be careful to show correctly the point atwhich each force acts; this is crucial for correcttorque calculations. You can't represent a rigidbody as a point.Choose coordinate axes and specify a positivesense of rotation for torques. Represent forces interms of their components with respect to the axesyou have chosen; when you do this, cross out theoriginal force so that you don't included it twice.In choosing a point to compute torques, note thatif a force has a line of action that goes through aparticular point, the torque of the force withrespect to that point is zero. You can ofteneliminate unknown forces or components fromthe torque equation by a clever choice of point foryour calculation. The body doesn't actually haveto be pivoted about an axis through the chosenpoint.Step-3 : Execute the solution as follows :Write equations expressing the equilibriumconditions. Remember that ΣF x = 0, ΣF y = 0, andΣτ z = 0 are always separate equations; never addx-and y-components in a single equation. Alsoremember that when a force is represented in termof its components, you can compute the torque ofthat force by finding the torque of eachcomponent separately, each with its appropriatelever arm and sign, and adding the results. This isoften easier than determining the lever arm of theoriginal force.You always need as many equations as you haveunknowns. Depending on the number ofunknowns, you may need to compute torqueswith respect to two or more axes to obtain enoughequations. Often, there are several equally goodsets of force and torque equations for a particularproblem; there is usually no single "right"combination of equations. When you have asmany independent equations as unknowns, youcan solve the equations simultaneously.Step-4 : Evaluate your answer : A useful way tocheck your results is to rewrite the second conditionfor equilibrium, Στ z = 0, using a different choice oforigin. If you've done everything correctly, you'll getthe same answers using this new choice of origin asyou did with your original choice1. A particle a moves along a circle of radius R = 50 cmso that its radius vector r relative to point O (fig.)rotates with theA YAOrSolved ExamplesCROrθθR(a) (b)constant angular velocity ω = 0.40 rad/sec. Find themodulus of the velocity of the particle and modulusand direction of its total acceleration.Sol. Consider X and Y axes as shown in fig. Using sinelaw in triangle CAO, we getr R r R= or =sin( π − 2θ)sin θ 2sin θcosθsin θ∴ r = 2 R cos θNow r = r cos θ i + r sin θ j= 2 R cos 2 θ i + 2R cos θ sin θ jdr dθ dθNow, v = – 4R cos θ sin θ i + 2R cos 2θ jdtdtdt= – 2 R sin 2 θ ω i + 2 R cos 2θ ω j∴ |v| = 2 ω RFurtherdv dθ dθa = = 4 R cos 2 θ i – 4 R ω sin 2 θ jdtdtdt= –4 R ω 2 cos 2 θ i – 4R ω 2 sin 2θ j|a| = 4 Rω 22. A particle describes a horizontal circle on the smoothinner surface of a conical funnel as shown in fig. Ifthe height of the plane of the circle above the vertex9.8 mark cm, find the speed of the particle.Sol. The forces acting on the particle are shown in fig.They aremv 2 /rR sin αmgαR cos αRrh=9.8 cmα(i) weight m g acting vertically downwards.(ii) normal reaction R of smooth surface of the cone.(iii) reaction of the centripetal force (mv 2 /r) actingradially outward.Hence, R sin α = m g ...(1)and R cos α = (mv 2 /r) ...(2)CXXtraEdge for IIT-JEE 31 SEPTEMBER 2011

Dividing eq. (1) by eq. (2), we get⎛ g r ⎞tan α = ⎜ ⎟2⎝ v ⎠...(3)From figure, tan α = (r/h) ...(4)From eqs. (3) and (4), we getr g r =2h vor v = ( g h)∴ v = [9.8 × (9.8 × 10 –2 )] 1/2 = 9.98 m/s3. A particle of 10 kg mass is moving in a circle of 4mradius with a constant speed of 5m/sec. What is itsangular momentum about (i) the centre of circle (ii) apoint on the axis of the circle and 3 m distant from itscentre ? Which of these will always be in samedirection ?Sol. The situation is shown in fig.L 1L 2530 4(a) We know that L = r × mvL = m v r sin θHere m = 10 kg, r = 4 m, v = 5 m/secand θ = 90º∴ L = 10 × 5 × 4 × 1 = 200 kg-m 2 /sec.(b) In this case r = (42 2+ 3 ) = 5m.∴ L = 10 × 5 × 5 = 250 kg-m 2 /sec.From figure it is obvious that angular momentum infirst case always has same direction but in secondcase the direction changes.4. A symmetrical body is rotating about its axis ofsymmetry, its moment of inertia about the axis ofrotation being 1 kg-m 2 and its rate of rotation 2rev./sec. (a) what is its angular momentum ? (b) whatadditional work will have to be done to double itsrate of rotation ?Sol. (a) As the body is rotating about its axis ofsymmetry, the angular momentum vector coincideswith the axis of rotation.∴ Angular momentum L = Iω ...(1)Kinetic energy of rotation E = 21 Iω2or 2E = Iω 2∴ I 2 ω 2 = 2IE or Iω = ( 2IE)...(2)From eqs. (1) and (2), L = ( 2IE)...(3)ω = 2 rev/sec = 2 × 2π∴ E = 21 × 1 × (4π) 2 = 8π 2 jouleor 4π radian/sec.Now L = ( 2IE)= (2×1×8π)2= (16π) = 4π= 12.57 kg.m 2 /sec.(b) When the rate of rotation is doubled, i.e., 4rev/sec or 8π radians/sec, the kinetic energy ofrotation is given byE = 21 × 1 × (8π) 2 = 32π 2 jouleAdditional work required= Final K.E. of rotation – Initial K.E. of rotation= 32π 2 – 8π 2= 24 π 2 = 236.8 joule5. A thin horizontal uniform rod AB of mass m andlength l can rotate freely about a vertical axis passingthrough its end A. At a certain moment the end Bstarts experiencing a constant force F which is alwaysperpendicular to the original position of the stationaryrod and directed in the horizontal plane. Find theangular velocity of the rod as a function of itsrotation angle φ counted relative to the initialposition.Sol. The situation of the rod at an angle φ is shown in fig.Herer = i l cos φ + + j l sin φand F = j F(Force is always perpendicular to rod)Y FlθA→τ = r × F = (i l cos φ + j l sin φ) × (j F)= l F cos φ k| → τ | = l F cos φWe know that τ = 1 αHere I = 31 m l 2 (for rod) and α = ω (dω/dφ)∴ l F cos φ = 31 m l 2 . ω (dω/dφ)1or l F cos φ dφ = m l 2 . ω dω3Integrating within proper limits, we havel F∫ φ 1cos φ dφ= m l 32∫ ω ωdω01l F sin φ = m l 2 (ω 2 ⎡6Fsin φ⎤/2) ∴ ω = 3⎢ ⎥⎣ ml ⎦FB02XXtraEdge for IIT-JEE 32 SEPTEMBER 2011

OrganicChemistryFundamentalsALIPHATICHYDROCARBONAddition of hydrogen halides to Alkenes :Markovnikov’s RuleHydrogen halides (HI, HBr, HCl, and HF) add to thedouble bond of alkenes :C = C + HX ⎯→ – C – C –H HThese additions are sometimes carried out bydissolving the hydrogen halide in a solvent, such asacetic acid or CH 2 Cl 2 , or by bubbling the gaseoushydrogen halide directly into the alkene and using thealkene itself as the solvent. HF is prepared aspolyhydrogen fluoride in pyridine. The order ofreactivity of the hydrogen halides is HI > HBr > HCl> HF, and unless the alkene is highly substituted, HClreacts so slowly that the reaction is not one that isuseful as a preparative method. HBr adds readily, thereaction may follow an alternate course. However,adding silica gel or alumina to the mixture of thealkene and HCl or HBr in CH 2 Cl 2 increases the rateof addition dramatically and makes the reaction aneasy one to carry out.The addition of HX to an unsymmetrical alkenecould conceivably occur in two ways. In practice,however, one product usually predominates. Theaddition of HBr to propene, for example, couldconceivably lead to either 1-bromopropane or2-bromopropane. The main product, however is2-bromopropane :CH 2 = CHCH 3 + HBr ⎯→ CH 3 CHCH 3Br2-BromopropaneWhen 2-methylpropene reacts with HBr, the mainproduct is tert-butyl bromide, not isobutyl bromide :H 3 CH 3 CCH 3C = CH 2 + HBr ⎯→ CH 3 – C – CH 32-Methylpropene(isobutylene)Brtert-Butyl bromideConsideration of many examples like this led theRussian chemist Vladimir Markovnikov in 1870 toformulate what is now known as Markovnikov’srule. One way to state this rule is to say that in theaddition of HX to an alkene, the hydrogen atomadds to the carbon atom of the double bond thatalready has the greater number of hydrogenatoms. The addition of HBr to propene is anillustration :Carbon atomwith thegreaternumber ofhydrogen atomsCH 2 = CHCH 3 ⎯→ CH 2 – CHCH 3HBrH BrMarkovnikov additionproductReactions that illustrate Markovnikov’s rule are saidto be Markovnikov additions.A mechanism for addition of a hydrogen halide to analkene involves the following two steps :Step 1 :HC = Cslow+ H – X ⎯→+ –C – C – + XThe π electrons of the alkene form a bond with a protonfrom HX to form a carbocation and a halide ionStep 2 :HH–X + + C – C – fast⎯→ – C – C –The halide ion reacts with the carbocation bydonating an electron pair; the result is an alkyl halideModern Statement of Markovniov’s Rule :According to Modern statement of Markovnikov’srule, In the ionic addition of an unsymmetricalreagent to a double bond, the positive portion ofthe adding reagent attaches itself to a carbonatom of the double bond so as to yield the morestable carbocation as an intermediate. Because thisis the step that occurs first (before the addition of thenucleophilic portion of the adding reagent), it is thestep that determines the overall orientation of thereaction.Notice that this formulation of Markovnikov’s ruleallows us to predict the outcome of the addition of asuch as ICl. Because of the greater electro negativityof chlorine, the positive portion of this molecule isiodine. The addition of ICl to 2-methylpropene takesplace in the following way and produces 2-chloro-1-iodo-2-methylpropane :XXtraEdge for IIT-JEE 33 SEPTEMBER 2011

H 3 CH 3 Cδ+ δ–C = CH 2 + I – Cl ⎯→2-Methylpropene⎯→H 3 CC – CH 2 –IH 3 C +CH 3CH 3 – C – CH 2 – ICl2-Chloro-1-iodo-2-methylpropaneCl –An Exception to Markovnikov’s Rule :This rule exception concerns the addition of HBr toalkenes when the addition is carried out in thepresence of peroxides (i.e., compounds with thegeneral formula ROOR). When alkenes are treatedwith HBr in the presence of peroxides, an anti-Markovnikov addition occurs in the sense that thehydrogen atom becomes attached to the carbon atomwith the fewer hydrogen atoms. With propene, forexample, the addition takes place as follows :CH 3 CH = CH 2 + HBr ⎯⎯→CH 3 CH 2 CH 2 BrThis addition occurs by a radical mechanism, and notby the ionic mechanism. This anti-Markovnikovaddition occurs only when HBr is used in thepresence of peroxides and does not occursignificantly with HF, HCl, and HI even whenperoxides are present.Alcohols from Alkenes through Oxymercuration-Demercuration Markovnikov Addition :A useful laboratory procedure for synthesizingalcohols from alkenes that avoids rearrangement is atwo-step method called oxymercuration -demercuration.Alkenes react with mercuric acetate in a mixture oftetrahydrofurane (THF) and water to produce(hydroxyalkyl) mercury compounds. These(hydroxyalkyl) mercury compounds can be reducedto alcohols with sodium borohydride :Step 1 : OxymercurationC = C⎯ ROOR⎛⎜⎜O||⎞⎟⎟+ H 2 O + Hg ⎝OCCH 3 ⎠2– C – C –OHO Hg – OCCH 3Step 2 : Demercuration– C – C –OHO Hg – OCCH 3+ OH – + NaBH 4THFO+ CH 3 COH– C – C – + Hg + CH 3 CO –HO HIn the first step, oxymercuration, water and mercuricacetate add to the double bond; in the second step,demercuration, sodium borohydride reduces theacetoxymercury group and replaces it with hydrogen.(The acetate group is often abbreviated – OAc.)Both steps can be carried out in the same vessel, andboth reactions take place very rapidly at roomtemperature or below. The first step–oxymercuration–usually goes to completion within aperiod of 20s – 10 min. The second step –demercuration – normally requires less than an hour.The overall reaction gives alcohols in very highyields, usually greater than 90%.Oxymercuration–demercuration is also highlyregioselective. The net orientation of the addition ofthe elements of water, H – and –OH, is in accordancewith Markovnikov’s rule. The H– becomes attachedto the carbon atom of the double bond with thegreater number of hydrogen atoms :HRC = CHHH H(1) Hg(OAc) 2/THF–H 2OR– C – C – H(2) NaBH 4, OH –+HO – HThe following are specific examples :CH3(CH2)2CH= CH 21−PenteneHg(OAc) 2THF−H2O(15 s)O⎯ ⎯⎯⎯→HONaBH 4CH 3 (CH 2 ) 2 CH − CH 2 ⎯ ⎯⎯→−| |OH(1 h)OH HgOAcCH 3Hg(OAc) 2THF-H 2O(20 s)1-MethylcyclopenteneHCH 3 (CH 2 ) 2 CHCH 3+ Hg|OH2-Pentanol (93%)H 3 C OHHgOAcH 3 CNaBH 4+ HgOH –(6 min)1-MethylcyclopentanolRearrangements of the carbon skeleton seldom occurin oxymercuration - demercuration. TheHOHXtraEdge for IIT-JEE 34 SEPTEMBER 2011

oxymercuration - demercuration of 3, 3-dimethyl-1-butene is a striking example illustrating this feature.It is in direct contrast to the hydration of 3, 3-dimethyl-1-butene.CH3CH 3C| − CH = CH|CH33, 3-Dimethyl-1-butene2(1)Hg(OAc) 2 / THF − H2O⎯⎯⎯⎯⎯⎯⎯⎯→(2) NaBH4, OHCH3|CH3C — CHCH| |CH OH3,3-Dimethyl-2-butanol (94%)Analysis of the mixture of products by gaschromatography failed to reveal the presence of any2, 3-dimethyl-2-butanol. The acid-catalyzedhydration of 3, 3-dimethyl-1-butene, by contrast,gives 2, 3-dimethyl-2-butanol as the major product.A mechanism that accounts for the orientation ofaddition in the oxymercuration stage, and one thatalso explains the general lack of accompanyingrearrangements. Central to this mechanism is anelectrophilic attack by the mercury species, H + gOAc ,at the less substituted carbon of the double bond (i.e.,at the carbon atom that bears the greater number ofhydrogen atoms), and the formation of a bridgedintermediate.Hydroboration : Synthesis of AlkylboranesHydroboration of an alkene is the starting point for anumber of useful synthetic procedures, including theanti-Markovnikov syn hydration procedure.Hydroboration was discovered by Herbert C. Brown,and it can be represented in its simplest terms asfollows :C = C + H — B hydroboration — C — C —AlkeneBoron hydride33HBAlkylboraneHydroboration can be accomplished with diborane(B 2 H 6 ), which is a gaseous dimer of borane (BH 3 ), ormore conveniently with a reagent prepared bydissolving diborane in THF. When diborane isintroduced to THF, it reacts to form a Lewis acid–base complex of borane (the Lewis acid) and THF.The complex is represented as BH 3 : THF.B 2 H 6 + 2 ODiboraneTHF(tetrahydrofuran)H– +2H – B – OHBH 3 : THFSolutions containing the BH 3 : THF complex can beobtained commercially. Hydroboration reactions areusually carried out in ethers : either in diethyl ether(CH 3 CH 2 ) 2 O, or in some higher molecular weightether such as “diglyme” [(CH 3 OCH 2 CH 2 ) 2 O,diethylene glycol dimethyl ether]. Great care must beused in handling diborane and alkylboranes becausethey ignite spontaneously in air (with a green flame).The solution of BH 3 : THF must be used in an inertatmosphere (e.g., argon or nitrogen) and with care.Stereochemistry of Hydroboration :H B+HH — BHWecan see the results of a syn addition in our examplesinvolving the hydroboration of1-methylcyclopentene ring :Space Quick Facts1. There has only been one satellite destroyed by ameteor, it was the European Space Agency’sOlympus in 1993.2. The International Space Station orbits at 248miles above the Earth.3. The Earth orbits the Sun at 66,700mph.4. Venus spins in the opposite direction comparedto the Earth and most other planets. This meansthat the Sun rises in the West and sets in theEast.5. The Moon is moving away from the Earth atabout 34cm per year.6. The Sun, composed mostly of helium andhydrogen, has a surface temperature of 6000degrees Celsius.7. A manned rocket reaches the moon in less timethan it took a stagecoach to travel the length ofEngland.8. The nearest known black hole is 1,600 lightyears (10 quadrillion miles/16 quadrillionkilometers) away.XtraEdge for IIT-JEE 35 SEPTEMBER 2011

KEY CONCEPTInorganicChemistryFundamentalsOXYGEN &HYDROGEN FAMILYSinglet dioxygen can add to a diene molecule in theOxygen :H 2 O 2 + OCl – ⎯ EtOH ⎯ →O 2 ( 1 ∆ g ) + Η 2 Ο + Cl –Oxygen occurs as two non-metallic forms, dioxygen 1, 4 positions, rather like a Diels–Alder reaction. ItO 2 and ozone O 3 . Dioxygen O 2 is stable as a diatomic may add 1, 2 to an alkene which can be cleaved intomolecule, which accounts for it being a gas. The two carbonyl compounds.bonding in the O 2 molecules is not as simple as itCHO – O2 CHmight at first appear. If the molecule had two2 + singlet O 2covalent bonds, then all electron would be paired andthe molecule should be diamagnetic.CH – CHSinglet dioxygen may be involved in biologicalO + O → O O or O = Ooxidations.Ozone ODioxygen is paramaginetic and therefore contains3 is the triatomic allotrope of oxygen. It isunpaired electrons. The explanation of thisunstable, and decomposes to O 2 . The structure of O 3phenomenon was one of the early successes of theis angular, with an O – O – O bond angle of 116º48´.Both O – O bond lengths are 1.28 Å, which ismolecular orbital theory.intermediate between a single bond (1.48 Å in H 2 O 2 )Liquid dioxygen is pale blue in colour, and the solid and a double bond (1.21 Å in O 2 ). The older valenceis also blue. The colour arises from electronic bond representation as resonance hybrid now seldomtransitions which excite the ground state (a triplet used. The structure is described as the central O atomstate) to a singlet state. This transition is 'forbidden' using sp 2 hybrid orbitals to bond to the terminal Oin gaseous dioxygen. In liquid or solid dioxygen a atoms. The central atom has one lone pair, and thesingle photon may collide with two molecules terminal O atoms have two lone pairs. This leavessimultaneously and promote both to excited states,four electrons for π bonding. The p z atomic orbitalsabsorbing red – yellow – green light, so O 2 appearsfrom the three atoms form three delocalizedblue. The origin of the excited singlet states in O 2 liesmolecular orbitals covering all three atoms. One MOin the arrangement of electrons in the antibondingis bonding, one non-bonding, and one antibonding.π*2p y and π*2p z molecular orbitals, and is shownThe four π electron fill the bonding and non-bondingbelow.MOs and thus contribute one delocalized π bond tothe molecule in addition to the two σ bonds. Thus theSecond excited π*p y π*p zstate (electronsState Energy /kJ bond order is 1.5, and the π system is described as ahave oppositesinglet+four-electron three-centre bond.Σ g 157spinsTrioxides of Sulphur :First excitedThe only important trioxide in this group, SO 3 , isstate (electronssinglet∆ g 92obtained by reaction of sulfur dioxide with molecularpaired )oxygen, a reaction that is thermodynamically veryfavorable but extremely slow in the absence of aGround statecatalyst. Platinum sponge, V(electrons havetriplet–2 O 5 , and NO serve asΣ g 0catalysts under various conditions. Sulfur trioxideparallel spins)reacts vigorously with water to form sulfuric acid.Commercially, for practical reasons, SO 3 is absorbedSinglet O 2 is excited, and is much more reactive than in concentrated sulfuric acid, to give oleum, which isnormal ground state triplet dioxygen. Singlet then diluted. Sulfur trioxide is used as such fordioxygen can be generated photochemically by preparing sulfonated oils and alkyl arenesulfonateirradiating normal dioxygen in the presence of a detergents. It is also a powerful but generallysensitizer such as fluorescein, methylence blue or indiscriminate oxidizing agent; however, it willsome polycyclic hydrocarbons. Singlet dioxygen can selectively oxidize pentachlorotoluene and similaralso be made chemically :compounds to the alcohol.XtraEdge for IIT-JEE 36 SEPTEMBER 2011

The free molecule, in the gas phase, has a planar,triangular structure that may be considered to be aresonance hybrid involving pπ – pπ S – O bonding,with additional π bonding via overlap of filledoxygen pπ orbitals with empty sulfur dπ orbitals, toaccount for the very short S – O distance of 1.41 Å:O: :O: :O:SSSO O O O O OIn view of this affinity of S in SO 3 for electrons, it isnot surprising that SO 3 functions as a fairly strongLewis acid toward the bases that it does notpreferentially oxidize. Thus the trioxide givescrystalline complexes with pyridine, trimethylamine,or dioxane, which can be used, like SO 3 itself, assulfonating agents for organic compounds.The structure of solid SO 3 is complex. At least threewell-defined phase are known. γ-Sulfur trioxide,formed by condensation of vapors at – 80ºC orbelow, is an icelike solid containing cyclic trimerswith structure.OOO SS OOOSOOOA more stable, asbestos-like phase (β-SO 3 ) hasinfinite helical chains of linked SO 4 tetrahedra andthe most stable form, α-SO 3 , which also has anasbestos-like appearance, presumably has similarchains crosslinked into layers.O O O– S – O – S – O – S – O –O O OLiquid γ-SO 3 , which is a monomer-trimer mixture,can be stabilized by the addition of boric acid. In thepure state it is readily polymerized by traces of water.Compounds of Sulphur and Nitrogen :A number of ring and chain compounds containing Sand N exist. The elements N and S are diagonallyrelated in the periodic table, and have similar chargedensities. Their electronegativities are close (N 3.0, S2.5) so covalent bonding is expected. The compoundsformed have unusual structures which cannot beexplained by the usual bonding theories.Attempting to work out oxidation states is unhelpfulor misleading.The best known is tetrasulphur tetranitride S 4 N 4 , andthis is starting point for many other S – Ncompounds. S 4 N 4 may be made as follows :6SCl 2 + 16NH 3 → S 4 N 4 + 2S + 14NH 4 ClCCl6S 2 Cl 2 + 16NH 3 ⎯⎯ 4 → S 4 N 4 + 8S + 12NH 4 Cl6S 2 Cl 2 + 4NH 4 Cl → S 4 N 4 + 8S + 16HClNSNSS 4 N 4 is a solid, m.p. 178ºC, It is thermochromic, thatis it changes colour with temperature. At liquidnitorgen temperatures it is almost colourless, but atroom temperature it is orange-yellow, and at 100ºC itis red. It is stable in air, but may detonate with shock,grinding or sudden heating. The structure is aheterocylic ring. This is cradle shaped and differsstructurally from the S 8 ring, which is crown shaped.The X-ray structure shows that the average S – Nbond length is 1.62 Å. Since the sum of the covalentradii for S and N is 1.78 Å, the S – N bonds seem tohave some double bond character. The fact that thebonds are of equal length suggest that this isdelocalized. The S.....S distances at the top andbottom of the cradle are 2.58 Å. The van der waals(non-bonded) distance S ... S is 3.30 Å, and the singlebond distance S – S is 2.08 Å.This indicates weak S – S bonding, and S 4 N 4 is thus acage structure.Many different sizes of rings exist, for examplecyclo-S 2 N 2 , cyclo-S 4 N 2 , cyclo-S 4 N 3 Cl, cyclo-S 3 N 3 Cl 3 . In addition bicyclo compounds S 11 N 2 ,S 15 N 2 , S 16 N 2 ,S 17 N 2 and S 19 N 2 are known. The lastfour may be regarded as two heterocyclic S 7 N ring,with the N atoms joined through a chain of 1 – 5Satoms.S 4 N 4 is very slowly hydrolysed by water, but reactsrapidly with warm NaOH with the break-up of the ring:S 4 N 4 + 6NaOH + 3H 2 O → Na 2 S 2 O 3 + 2Na 2 SO 3 + 4NH 3If S 4 N 4 is treated with Ag 2 F in CCl 4 solution thenS 4 N 4 F 4 is formed. This has an eight-membered S – Nring, with the F atoms bonded to S. This results frombreaking the S – S bonds across the ring. Similarlythe formation of adducts such as S 4 N 4 .BF 3 orS 4 N 4 .SbF 5 (in which the extra group is bonded to N)breaks the S – S bonds and increases the mean S – Ndistance from 1.62 Å to 1.68 Å. This is presumablybecause the electron attracting power of BF 3 or SbF 5withdraws some of the π electron density.Reduction of S 4 N 4 with SNCl 2 in MeOH givestetrasulphur tetraimide S 4 (NH 4 ). Several imides canbe made by reacting S 4 N 4 with S, or S 2 Cl 2 with NH 3 .These imides are related to an S 8 ring in which one ormore S atoms have been replaced by imide NHgroups, for example in S 7 NH, S 6 (NH) 2 , S 5 (NH) 3 andS 4 (NH) 4 .NSSNXtraEdge for IIT-JEE 37 SEPTEMBER 2011

If S 4 N 4 is vaporized under reduced pressure andpassed through silver wool, then disulphur dinitrogenS 2 N 2 is formed.S 4 N 4 + 4Ag → S 2 N 2 + 2Ag 2 S + N 2S 2 N 2 is a crystalline solid, which is insoluble in waterbut soluble in many organic solvents. It explodeswith shock or heat. The structure is cyclic and thefour atoms are very nearly square planar.The most important reaction of S 2 N 2 is the slowpolymerization of the solid or vapour to formpolythiazyl (SN) x . This is a bronze coloured shinysolid that looks like a metal. It conducts electricityand conductivity increases as the temperaturedecreases, which is typical of a metal. It becomes asuperconductor at 0.26 K. The crystal structure showsthat the four-membered rings in S 2 N 2 have openedand polymerized into a long chain polymer. Theatoms have a zig-zag arrangement, and the chain isalmost flat. Conductivity is much greater along thechains than in other directions, and so the polymerbehaves as a one-dimensional metal. The resistivity isquite high at room temperature.Ortho and Para Hydrogen :The hydrogen molecule H 2 exists in two differentforms known as ortho and para hydrogen. Thenucleus of an atom has nuclear spin, in a similar wayto electrons having a spin. In the H 2 molecule, thetwo nuclei may be spinning in either the samedirection, or in opposite directions. This gives rise tospin isomerism, that is two different forms of H 2 mayexist. These are called ortho and para hydrogen. Spinisomerism is also found in other symmetricalmolecules whose nuclei have spin momenta, e.g. D 2 ,N 2 , F 2 , Cl 2 . There are considerable differencesbetween the physical properties (e.g. boiling points,specific heats and thermal conductivities) of the orthoand para forms, because of differences in theirinternal energy. There are also difference in the bandspectra of the ortho and para forms of H 2 .The para form has the lower energy, and at absolutezero the gas contains 100% of the para form. As thetemperature is raised, some of the para form changesinto the ortho form. At high temperatures the gascontains about 75% ortho hydrogen.Para hydrogen is usually prepared by passing amixture of the two forms of hydrogen through a tubepacked with charcoal cooled to liquid airtemperature. Para hydrogen prepared in this way canbe kept for weeks at room temperature in a glassvessel, because the ortho-para conversion is slow inthe absence of catalysts. Suitable catalysts includeactivated charcoal, atomic hydrogen, metals such asFe, Ni, Pt and W and paramagnetic substances or ions(which contain unpaired electrons) such as O 2 , NO,NO 2 , Co 2+ and Cr 2 O 3 .What do Aliens Look Like?Aliens are the extraterrestrial beings believed to exist.Some give accounts of having seen them visit ourworld. Then, what do aliens look like? Want to know?The read on…Aliens have always aroused the interestfor many. With new discoveries inastronomy, man has been able toexplore the extraterrestrial world andexamine the chances of the existenceof aliens.On one hand, the existence of extraterrestrial life isconsidered hypothetical while on the other hand,aliens have been sighted on a few occasions. Therehave been news about the aliens visiting Earth; therehave been some people claiming to have seen thealiens. The concept of ‘aliens’ remains alien!The sightings of aliens have brought aboutdescriptions of their appearance. What they look like,has been a question in the minds of one and all andnews have many a time answered it by givingaccounts of people witnessing aliens. We know offilms and television shows, which have depictedaliens as being humanoid in appearance.What do Aliens Look Like?Aliens are largely described as resembling humanbeings. Their height is approximately same as theaverage height of human beings. Like any normalhuman beings, aliens have a pair of eyes, a nose, amouth, a pair of arms and a pair of feet. There arecitations of aliens having wings or wheels instead offeet and other such abnormalities. It is believed thataliens have a rough lizard-like skin. Their skin colorsare believed to vary from gray, white, tan to gold,pink or red. Their skin is believed to glow in the dark.Their eyes are considered to resemble those ofhumans, lizards or insects. Some have documentedaliens as having webbed fingers while others believethat aliens have suction cups for fingertips or claws.Aliens have been documented as being variedly sizedand shaped. Some have documented them as 3 inchestall while others say that they are about 15 feet tall. Insome places aliens have been documented as beingshaped like balls of light, while in other places theyhave been shown as resembling robots or metalobjects. Some believe that aliens look like animals orlarge insects while some think of aliens as human-likefigures clothed in uniforms. Many believe that alienscan float through walls.XtraEdge for IIT-JEE 38 SEPTEMBER 2011

UNDERSTANDINGInorganic Chemistry1. For SO 2 (g) at 273 K and 1 atm pressure, the dielectricconstant (or relative permittivity) is 1.00993. Thismolecule has a permanent dipole moment of 1.63 D.Assuming that SO 2 behaves as an ideal gas, calculateper mol of (a) total, (b) orientation, (c) inducedpolarizations, and (d) distortion polarizability.Sol. We haveεε r = = 1.00993ε 0p = 1.63 D = 1.63(3.3356 × 10 –30 Cm)V m = 22414 cm 3 mol –1 at 1 atm and 273 K(a) Total polarization,εr−1MP total =εr+ 2 ρ1.00993 −1=× 22.414 cm 3 mol –11.00993 + 2= 73.95 cm 3 mol –1(b) Orientation polarization,⎛2N ⎞AP 0 = ⎜p⎟3ε0 ⎝ 3kT⎠⎪⎧23{6.023×10 mol= ⎨−122⎪⎩ 3(8.854×10 C N−1−1}m−2⎪⎧−302(1.63×3.3356×10 Cm) ⎪⎫⎨−23−1⎬⎪⎩ 3(1.38×10 J K )(273K) ⎪⎭= 59.31 × 10 –6 m 3 mol –1 = 59.31 cm 3 mol –1(c) Induced polarization,P ind = P total – P 0= 73.95 cm 3 mol –1 – 59.31 cm 3 mol –1= 14.64 cm 3 mol –1(d) Distortion polarizability,Pindα d =(1/ 3ε)N0A–614.64×10 m mol=−122 −1−223 −1{1/(3×8.854×10 C N m )}(6.023×10 mol )= 6.46 × 10 –40 C 2 N –1 m2. A green coloured compound (A) gave the followingreactions :(i) (A) dissolves in water to give a green solution.The solution on reaction with AgNO 3 gives a whiteppt. (B) which dissolves in NH 4 OH solution andreappears on addition of dil. HNO 3 . It on heating with3−1⎪⎫⎬⎪⎭K 2 Cr 2 O 7 and conc. H 2 SO 4 produced a red gas whichdissolves in NaOH to give yellow solution (C).Addition of lead acetate solution to (C) gives ayellow ppt. which is used as a paint.(ii) The hydroxide of cation of (A) in borax bead testgives brown colour in oxidising flame and greycolour in reducing flame.(iii) Aqueous solution of (A) gives a black ppt. onpassing H 2 S gas. The black ppt. dissolves inaquaregia and gives back (A).(iv) (A) on boiling with NaHCO 3 and Br 2 water givesa black ppt. (D)(v) (A) on treatment with KCN gives a light greenppt. (E) which dissolves in excess of KCN to give(F). (F) on heating with alkaline bromine water givesthe same black ppt. as (D).Identify compounds (A) to (F) and give balancedequations of the reactions.Sol. Reaction (i) indicates that (A) contains Cl – ionsbecause, it gives white ppt. soluble in NH 4 OH. It isagain confirmed because it gives chromyl chloridetest. The colour of oxidising and reducing flamesindicate that (A) also contains Ni 2+ ions. Hence, (A)is NiCl 2 . The different reactions are :(i) NiCl 2 + 2AgNO 3 → 2AgCl + Ni(NO 3 ) 2AgCl + 2NH 3 → [ Ag(NH3)2 ] ClAg(NH 3 ) 2 Cl + 2HNO 3 →SolubleAgCl ↓ + 2NH 4 NO 3whiteppt.(B)The equations of chromyl chloride tests are :NiCl 2 + Na 2 CO 3 → 2NaCl + NiCO 34NaCl + K 2 Cr 2 O 7 + 6H 2 SO 4 → 4NaHSO 4 + 2KHSO 4+ 3H 2 O + 2 CrO 2 Cl 2CrO 2 Cl 2 + 4NaOH →Na 2 CrO 4 + (CH 3 COO) 2 Pb →Na 2CrO 4Yellowsolution(C)Redgas+ 2NaCl + 2H 2 OPbCrO 4 + 2CH 3 COONaYellowppt.(ii) Na 2 B 4 O 7 . 10H 2 O∆Na 2 B 4 O 7 + 10H 2 ONa 2 B 4 O∆7 2 NaBO1442 + B4442O3NiO + B 2 O 3Ni(BO 2 ) 2 + C∆∆Transparent beadNi (BO 2 ) 2 [Oxidising flame]Nickelmetaborate(Brown)Ni + B 2 O 3 + COGrey[Reducing flame]XtraEdge for IIT-JEE 39 SEPTEMBER 2011

(iii) NiCl 2 + H 2 S → 2HCl + NiS ↓Black ppt.NiS + 2HCl + [O] → NiCl + H 2 S ↑(iv)(v)2(A)2(A)NiCl + 2NaHCO 3 → NiCO 3 + 2NaCl2NiCO 3 + 4NaOH + [O]2(A)+ CO 2 + H 2 O∆Ni ↓2 O 3Black ppt.(D)+ 2Na 2 CO 3 + H 2 ONiCl + 2KCN → Ni (CN) 2 + 2KClNi(CN) 2 + 2KCN → KGreen ppt.(E)2 [Ni(CN) 4](F)NaOH + Br 2 → NaOBr + HBr2K 2 [Ni(CN) 4 ] + 4NaOH + 9NaOBr∆Ni ↓ + 4KCNO + 9NaBr + 4NaCNO2O 3(D)3. A metal (A) gives the following observations :(i) It gives golden yellow flame.(ii) It is highly reactive and used in photoelectric cellsas well as used in the preparation of Lassaiganesolution.(iii) (A) on fusion with NaN 3 and NaNO 3 separately,yields an alkaline oxide (B) and an inert gas (C). Thegas (C) when mixed with H 2 in Haber's process givesanother gas (D). (D) turns red litmus blue and giveswhite dense fumes with HCl.(iv) Compound (B) react with water forming onalkaline solution (E). (E) is used for thesaponification of oils and fats to give glycerol and ahard soap.(v) (B) on heating at 670 K give (F) and (A). Thecompound (F) liberates H 2 O 2 on reaction with dil.mineral acids. It is an oxidising agent and oxidisesCr(OH) 3 to chromate, manganous salt to manganate,sulphides to sulphates.(vi) (B) reacts with liquid ammonia to give (G) and(E). (G) is used for the conversion of 1, 2dihaloalkanes into alkynes.What are (A) to (G)? Explain the reactions involved.Sol. (i) (A) appears to be Na as it gives the golden yellowflame. It is also used in the preparation of Lassaiganesolution which is sodium extract of organiccompounds.Na + C + N → NaCNNa + Cl → NaCl2Na + S → Na 2 S(ii) Compound (B) is Na 2 O and (C) is N 2 while (D) isNH 3 , as (D) is alkaline and turns red litmus blue andgives white fumes with HCl(C) + H 2 → NH 3N 2 + 3H 22 NH 3(D)NH 3 + HCl → NH 4 ClWhite fumes(iii) is prepared from Na as follows.2NaNO 3 + 10 Na → 6 Na 2 O + N3NaN 3 + NaNO 2 →2O(B)(B)2 Na 2 O +(B)2(C)N 2(C)(iv) Compound (E) is NaOH as it is used in thepreparation of soaps.Na + H 2 O → 2 NaOH(E)CH 2 OOCC 17 H 35CH 2 OH∆CHOOCC 17 H 35 + 3NaOH CH 2 OH + 3C 17 H 35 COONa(soap)CH 2 OOCC 17 H 35CH 2 OH(v) (F) is sodium peroxide as only peroxides givesH 2 O 2 on reaction with dil. acids.2 Na 2 O(B)⎯Na 2O 2 +(F)670 ⎯⎯K →∆(F)Na 2O 2 +2 Na(A)H 2 SO 4 → H 2O 2 + Na 2 SO 4dil.(F) gives the following oxidations :Cr(OH) 3 + 5OH – → CrO 2– 4 + 4H 2 O + 3e –Mn 2+ + 8OH – → MnO – 4 + 4H 2 O + 5e –S 2– + 8OH – → SO 2– 4 + 4H 2 O + 8e –The reduction equation of (F) isO 2– 2 + 2H 2 O + 2e – → 4OH –(vi) (G) is sodamide because it is used in thedehydrohalogenation reactions.Na + NH 3 (l) →2O(B)CH 3 – CH – CH 2 + 2NaNH 2BrBrNaNH +(G)2∆NaOH(E)CH 3 – C ≡ CHPropyne+ 2NaBr + 2NH 34. A white amorphous powder (A) when heated, gives acolourless gas (B), which turns lime water milkywhich dissolves on passing excess of gas (B) and theresidue (C) which is yellow while hot but white whencold. The residue (C) dissolves in dilute HCl and theresulting solution gives a white precipitate onaddition of potassium ferrocyanide solution. (A)dissolves in dil. HCl with the evolution of a gaswhich is identical in all respects to gas (B). Thesolution of (A) in dil. HCl gives a white ppt. (D) onaddition of excess of NH 4 OH and on passing H 2 Sgas. Another portion of this solution gives initially awhite ppt. (E) on addition of NaOH solution whichdissolves in excess of NaOH, the solution on passingagain H 2 S gas gives back the white ppt. of (E), thewhite ppt. on heating with dil. H 2 SO 4 give a gas usedin II and IV group analysis. What are (A) to (F) ?Give balanced chemical equations of the reaction.XtraEdge for IIT-JEE 40 SEPTEMBER 2011

Sol. All the above information can be formulated asfollows :(a) A ⎯⎯→∆ B + CWhitepowderColurless gasturning limewater milkyA residue yellowin hot and whitein coldK 4 Fe(CN )(b) C ⎯Dil ⎯. ⎯HCl → a solution ⎯⎯⎯⎯ 6⎯ → a white ppt.(c)dil.HClASolution + B(D)White ppt.NH 4 OH + H 2 SNaOH(E)White ppt.NaOHHdissolves ⎯→ 2S(F)From observations of section (a) one may concludethat the colourless gas is CO 2 because it turns limewater milky, due to formation of insoluble CaCO 3 .Ca (OH) 2 + CO → CaCO ↓ + H 2 OLime water2(B)3milkyCaCO 3 is soluble in excess of CO 2 due to formationof soluble calcium bicarbonate.CaCO 3 + CO 2 + H 2 O ⎯→ Ca (HCO 3 ) 2SolubleThe compound (C) is zinc oxide (ZnO) because it isyellow in hot and white in cold, hence the initialcompound (A) is zinc carbonate (ZnCO 3 ).From section (b), it is inferred that (C) is a salt ofZn (II) which dissolves in dil. HCl and white ppt.obtained after addition of K 4 Fe(CN) 6 is due to zincferrocyanide, a test of Zn ++ cation.The data of collection (C) proved that (A) is ZnCO 3because on treatment with dil. HCl it gives gas (B),i.e., CO 2 , while Zn (II) goes in solution, i.e., ZnCl 2 ,on passing H 2 S gas in presence of NH 4 OH, it gives awhite ppt. of ZnS(D). ZnS on heating with dil. H 2 SO 4evolves H 2 S, which is used for the precipitation ofsulphides of group II in acidic medium and of IVgroup in alkaline medium. ZnCl 2 , reacts with NaOHto give a ppt. of Zn(OH) 2 which dissolves in NaOH,as Zn(OH) 2 is amphoteric in nature. The solutionNa 2 ZnO 2 again gives ZnS on passing H 2 S gas into it.Different chemical equations concerned to the abovenumerical are given below :ZnO +(C)ZnCO 3 ⎯→ ZnO + CO(A)⎯ ∆2 HCl ⎯→dil.(C)2(B)ZnClSolution 2 + H 2 O2ZnCl 2 + K 4 Fe(CN) 6 → Zn2 [Fe(CN) 6]White ppt.↓ + 4KClZnS +H 2 SO 4dil.⎯∆ ⎯→ZnCl 2 + 2NaOH →Zn (OH) 2 + 2NaOH →Amphoteric(E)ZnSO 4 + H 2 S ↑Zn(OH) 2 ↓ + 2NaCl(E)Na 2 ZnO 2 + 2H 2OSolubleNa 2 ZnO 2 + H → ZnS ↓ + 2NaOH(F)2 SGas5. Compound (X) on reduction with LiAlH 4 gives ahydride (Y) containing 21.72% hydrogen along withother products. The compound (Y) reacts with airexplosively resulting in boron trioxide. Identify (X)and (Y). Give balanced reactions involved in theformation of (Y) and its reaction with air. Draw thestructure of (Y).Sol. Since B 2 O 3 is formed by reaction of (Y) with air, (Y)therefore should be B 2 H 6 in which % of hydrogen is21.72. The compound (X) on reduction with LiAlH 4gives B 2 H 6 . Thus it is boron trihalide. The reactionsare shown as:4 BX 3 + 3LiAlH 4 →(X)2 B 2 H 6 + 3LiX + 3AlX 3(Y)B 2H 6 + 3O 2 → B 2 O 3 + 3H 2 O + heat(Y)Structure of B 2 H 6 is as follows:H tH torH tBBH bH bH b1.33Å97ºBBH tH tH t121.5º(X = Cl or Br)H t 1.19Å H bH t1.77ÅThus, the diborane molecule has four two-centretwoelectron bonds (2c-2e – bonds) also called usualbonds and two three-centre-two-electron bonds (3c-2e – bonds) also called banana bonds. Hydrogenattached to usual and banana bonds are called H t(terminal H) and H b (bridged H) respectively.ZnCO +(A)3ZnCl 2 + H 2 S2 HCl ⎯⎯→∆ ZnCl 2 +dil.NH 4 OH⎯⎯⎯⎯ →White ppt.(D)CO 2 + H 2 O(B)ZnS + 2HClXtraEdge for IIT-JEE 41 SEPTEMBER 2011

`tà{xÅtà|vtÄ V{tÄÄxÇzxá5 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 conics through the intersection of tworectangular hyperbolas are also rectangularhyperbolas. If A, B, C & D be the four points ofintersection of these two rectangular hyperbolas, thenfind the orthocentre of the triangle ABC.2. Find the area of a right angle triangle if it is knownthat the radius of circle inscribed in the triangle is rand that of the circumscribed circle is R.3. Q is any point on the line x = a. If A is the point(a, 0) and QR, the bisector of the angle OQA, meetsOX in R, then prove that the locus of the foot of theperpendicular from R to OQ has the equation(x – 2a) (x 2 + y 2 ) + a 2 x = 04. Show that the equationz 4 + 2z 3 + 3z 2 + 4z + 5 = 0 with (z ∈ C) have nopurely real as well as purely imaginary root.5. Prove that∞⎛ a x ⎞ ln x∫f ⎜ + ⎟ dx = lna⎝ x a ⎠ x0∞∫0⎛ a x ⎞f ⎜ + ⎟⎝ x a ⎠dxx6. A straight line moves so that the product of theperpendiculars on it form two fixed points is aconstant. Prove that the locus of the foot of theperpendiculars from each of these points upon thestraight line is a circle, the same for each.7. Prove the identity :x∫0ezx−z2x x2dz = e4∫e−function f (x) =∫esolving it.x00zx−z2z / 4dz, deriving for the2dz a differential equation and8. Let α, β be the roots of a quadratic equation, suchα β athat αβ = 4 and + =α −1 β −1aof values of a for which α, β ∈ (1, 4)22− 7− 4Find the set9. Investigate the function f (x) = x 5/3 – 5x 2/3 for pointsof extremum and find the values of k such that theequation x 5/3 – 5x 2/3 = k has exactly one positive root.10. Let A = {1, 2, 3, ....., 100}. If X is a subset of Acontaining exactly 50 elements then show that∑ p min = 101 C 51 .p∈xDo you know• The largest meteorite crater in the world is inWinslow, Arizona. It is 4,150 feet across and 150feet deep.• The human eye blinks an average of 4,200,000times a year.• Skylab, the first American space station, fell to theearth in thousands of pieces in 1979. Thankfullymost over the ocean.• It takes approximately 12 hours for food to entirelydigest.• Human jaw muscles can generate a force of 200pounds (90.8 kilograms) on the molars.• The Skylab astronauts grew 1.5 - 2.25 inches (3.8 -5.7 centimeters) due to spinal lengthening andstraightening as a result of zero gravity.• An inch (2.5 centimeters) of rain water isequivalent to 15 inches (38.1 centimeters) of dry,powdery snow.• Tremendous erosion at the base of Niagara Falls(USA) undermines the shale cliffs and as a resultthe falls have receded approximately 7 miles overthe last 10,000 years.• 40 to 50 percent of body heat can be lost throughthe head (no hat) as a result of its extensivecirculatory network.XtraEdge for IIT-JEE 42 SEPTEMBER 2011

MATHEMATICAL CHALLENGESSOLUTION FOR AUGUST ISSUE (SET # 4)1. Let x 2 + y 2 = a 2 ....(1)and x 2 + y 2 + 2gx + c = 0 ....(2)are two circles they cut orthogonally.Hence c – a 2 = 0so from (2) x 2 + y 2 + 2gx + a 2 = 0 ...(3)Let P (a cos α, a sin α) be any point on 1st circle. It’spolar w.r.t. 2 nd circle isax cos α + ay sin α + g(x + a cos α) + a 2 = 0 ....(4)other end of diameter of 1 st circle through P isQ(– a cos α, – a sin α)This satisfies eqn.(4)Hence proved.2. Let A be at origin & position vectors of B, C, D are ,b r , c r & d r respectively. Perpendicular from B and Cto the faces ACD and ABD meet at H with positionvector h r ,ABbcCr r r rso BH ⊥ AC ⇒ ( h − b).c = 0 ⇒ h cr r r . = b . cr r rBH ⊥ AD ⇒ ( h − b).d = 0 ⇒ h r . dr= b r . drr r rCH ⊥ AB ⇒ ( h − c).b = 0 ⇒ h r . br r= c br.r r r r r rand CH ⊥ AD ⇒ ( h − c). d = 0 ⇒ h.d = c br.so b r . dr= c r dr r r r . ⇒ ( b − c ) . d = 0 ...(1)so BC ⊥ AD , proved.Now, let any point M (with position vector m ) beon BCD such that AM ⊥ plane BCD thenr rr r. m . ( c − b)= 0 = m . ( d − c)so m . b r = m . c r = m . d r ...(2)Now, let P be any point on AM with position vectort m such that DP is perpendicular to ABC, then (t m –d r ). b r = 0 = (t m – d r ). c r so t m . b r = b r . d r & t m . c r= d r . c r which are same equations using (1) and (2) inr rb . dthem. So such a scalar t can be obtained as t = rmb .dDπ3. I (u) =∫02 )l n ( 1 − 2ucos x + u dx ;I (– u) =∫ π l n ( 1 + 2ucos x + u dxUse∫ a00I (u) = I (– u)I (u) + I (– u)a2 )f ( x)dx =∫f ( a − x)dx0=∫ π l n ( 1 − 2ucos x + u ) (1 + 2ucos x + u ) dx0π=∫022l n[( 1 − u ) − 4ucos x]dx4 2 2 2=∫ π ln[1+ u + 2u− 4ucos x]dx024=∫ π ln [1 − 2ucos 2x+ u ] dx0Now let 2x = y222⇒ Ι (u) + I (– u) = [1 2ucos y u ]2 ∫ π l n − + dy221241= I (u 2 2) + [1 2 cos ]2 ∫ π l u y u2n − + dy0Now let y = 2π – t01 40∫1 24π1I (u) + I (– u) = I (u 2 ) + l n [1 − 2ucost+ u ] (– dt)2 2= 21 I (u 2 ) + 21 I (u 2 )2I (u) = I (u 2 ) as I(u) = I (– u)(or using f (2a – x) = f (x) Prop).so I (u) = 21 I(u 2 )π∫0similarly find 3ln [1 − 2ucos x + u ] dx1& show I (u) = I (– u) = I (u 2 ) = I(u )2 n22221 2 nXtraEdge for IIT-JEE 43 SEPTEMBER 2011

4. Let f (x) = (x – α) (x – β)so f (n) f (n + 1) = (n – α) (n –β) (n + 1 – α)(n + 1– β)= (n – α) (n + 1 – β) (n – β) (n + 1 – α)= [n (n + 1) – n(α + β) – α + αβ] [n(n + 1)– n (α + β) – β + αβ]= [n (n + 1) + na + b – α] [n (n + 1) + an + b – β]= (m – α) (m – β) ; let m = n(n + 1) + an + b= f (m)5. Let xy = c 2 be rectangular hyperbola.Let A (ct 1, c/t 1) and b (ct 2, c/t 2) be two fixed points onit. and P (ct, c/t) be any variable point.Line AP : x + y t 1t = c (t 1 + t)Line BP : x + y t 2t = c (t 2+ t)These lines intersect with x- axis at M (c(t 1+ t), 0)and N(c(t 2+ t), 0). Length MN = |c (t 1– t 2) | which isa constant.Similarly intercept on y-axis can be obtained as6.⎛ 1 1 ⎞c⎜ −⎟ . Hence proved.⎝ t1t2⎠y dx − x dy x+xy⇒x2y ( y dx − x dy)+2x y2dy − y( x − y)dy + xy dx − xy dx − y( x − y)2222dx= 0dx − xydy + xydyy⇒ . d(x/y) + xx ( x dy + y dx)− y ( ydx + x dy)− xy(dx − dy)= 02( x − y)⇒⇒d (x / y)x / yd ( x / y)+x / yx.d ( xy)− y d ( xy)− xy+2( x − y)d ( x / y)⎛ x y⇒ + dx / y⎟ ⎞⎜ = 0⎝ x − y ⎠x y⇒ ln (x/y) + = cx − y22d ( x − y)= 0( x − y)d(xy)− xy d(x − y)= 02( x − y)x y7. Let +2 2 = 1 be the ellipse anda by = m 1 (x – ae) and y = m 2 (x – ae) are two chordsthrough its focus (ae, 0). Any conic through theextremities of these chords can be defined as{y – m 1 (x – ae)} {y – m 2 (x – ae)} +⎛2 2⎞λ⎜x y+ −1⎟= 0 ...(1)2 2⎝ a b ⎠It it passes through origin, thenm 1 m 2 a 2 e 2 – λ = 0 ...(2)Solving (1) with x- axis⎛2⎞m 1m 2(x – a e) 2 + λ ⎜x⎟−12= 0⎝ a ⎠using (2) in it(x – ae)2 + e 2 (x 2 – a 2 ) = 0(1 + e 2 ) x 2 – 2ae x = 02aex = 0 & x =21 + eso other point on x-axis through which this conic⎛ 2ae⎞passes is ⎜ ,0⎟ which is a fixed point.2⎝1+ e ⎠Hence proved.8. Let x = c ∈ Rf ( c ± h)− f ( c)f ′ (c ±) = limh→0± h⎛ ⎛ h ⎞⎞f⎜c⎜ 1 ± ⎟ f cc⎟ − ( )⎝ ⎝ ⎠⎠= lim ; c ≠ 0h→0± hf (2c)f (1 + h / c)− f ( c)= lim2h→0± h⎛ h ⎞f (2c)f ⎜1± ⎟ − 2 f ( c)⎝ c ⎠= limh→0± 2 h⎛ h ⎞f (2c)f ⎜1± ⎟ − f (2c)f (1)⎝ c ⎠= lim ;h→0± 2 h(using x = 2c & y = 1 in⎛ xy ⎞f ⎜ ⎟ =⎝ 2 ⎠f ( x).f ( y))2⎛ h ⎞f ⎜1± ⎟ − f (1)⎝ c ⎠ f (2c)f '(1)= f (2c) lim =h→02h2c± . ccf (2c)f ' (1)=; as given f ′(1) = f (1)2cx f ( c)f ′ (c ±) =2cSo f (x) is differentiable for ∀ x ∈ R except x = 0f ( x)Now f ′(x) =xf '(x)1⇒ =f ( x)xso ln f(x) = ln x + ln c⇒ f (x) = cx⎛ ⎞Now as f ⎜xy f ( x)f ( y)⎟ =⎝ 2 ⎠ 2XtraEdge for IIT-JEE 44 SEPTEMBER 2011

let y = 1 in it2. f (x/2) = f (x) (f (1)c x2 =cx f(1) 2so f (1) = 1 as c ≠ 0so f (x) = x9. Let i lines are there, no two of which are parallel andno three of which are coincident. Introduction of(i + 1) th line will introduce (i + 1) new parts. Let P idenotes the number of parts in which plane is beingdivided by i lines, thenP i + 1 = P i + (i + 1)P i + 1 – P i= i + 1using i = 1, 2, 3, ......, n – 1P 2 – P = 2, 1P 1 – P 2 = 3MMP n – P n–1 = nAdd these equationP n – P i = 2 + 3 +.....+ nP n = P i + 2 + 3 + ....+ nn ( n +1)= 2 + 2 + 3+ .....+ n = 1 +21P n = (n 2 + n + 2)2Confidence Tips• Can you enjoy success if you don’t know whatyou wanted?• Can't is the worst of the 4-letter words.• Can you tell whether someone else lacks selfconfidence?• Can you act confident even when you are not?• Meal check: How many successes have you hadsince your last meal?• Meal check: What will you accomplish in thenext hour?• Meal check: Take the next thing you will do.How will you see that it is well done?• Meal check: How many times have you thought“I can’t” since you last ate?• You already have all to brain tools you need.You just need to find the tools that fit the job.10.1 sin x sin 3xsin 9x1 tan x + + + = tan 27x2 cos3xcos9xcos 27x2L.H.S. : Consider on1 sin x 1 sin x sin xtan x + = +2 cos3x2 cos x cos3xsin x cos 3x+ 2sin x cos x=2 cos x cos 3x2 sin x cos3x+ 2sin 2x=4 cos x cos3xsin 4x− sin 2x+ 2sin 2x=4cos x cos 3xsin 4x+ sin 2x=4cos x cos 3x2sin 3xcos x 1== tan 3x4 cos x cos 3x21 sin 3x1similarly tan 3x + = tan 9x2 cos9x21 sin 9x1and tan 9x + = tan 272 cos 27x2on adding all these we getsin 3xsin 3xsin 9x1+ + = (tan 27x – tan x)cos3xcos9xcos 27x2Proved.• You don't have to know everything. You justhave to know how to find out anything.• You will know more tomorrow. How will youuse that insight today?• If you want to be smart, find friends who aresmarter than you are.• List 5 ways to undermine your own selfconfidence.• Will what you seek be worth the work?• Pay attention to what you say about yourself.Would you say that about someone else?• Pay attention to what you say about what youcan do. Why do you believe it?• Those who say it can't be done should notinterrupt those who are doing it.• Don't should on yourself .• Some focus on what they can. Others focus onwhat they can't. What do you do you?• Tell yourself what you can't do. Hear a stopsign. Tell yourself what you can do. What doyou hear?XtraEdge for IIT-JEE 45 SEPTEMBER 2011

Students' ForumMATHSExpert’s Solution for Question asked by IIT-JEE Aspirants1. The sum of the digits of a seven-digit number is 59.Find the probability that this number is divisible by11.Sol. As 7 × 8 = 56 and the sum of seven digits is 59,clearly at least three of the digits must be 9.Obviously, the seven digits of the number will be asfollows;(a) 9,9,9,8,8,8,8 (b) 9,9,9,9,8,8,7(c) 9,9,9,9,9,7,7 (d) 9,9,9,9,9,8,6(e) 9,9,9,9,9,9,5∴ the total number of ways to form a seven-digitnumber whose sum of digits is 59==7 ! 7! +3!4! 4!2!7.6.53.2++7.6.527!5!2!++7.627 ! 7! +5! 6!+ 7.6 + 7= 210 ...(i)A number is divisible by 11 if the difference of thesum of the digits in odd places and that of the digitsin even places is divisible by 11.As the number is of seven digits we must have (forfavourable cases), sum of four digits in odd places–sum of three digits in even places= 0, 11, 22, 33, 44, 55.If the two sums are denoted by x and y respectivelythenx + y = 59 x + y = 59 x + y = 59x – y = 0 or x – y = 11 or x – y = 22(i) (ii) (iii)x + y = 59 x + y = 59 x + y = 59or x – y = 33 or x – y = 44 or x – y = 5 5(iv) (v) (vi)Clearly, (i), (iii) and (iv) do not give integral valuesof x and y.(ii) ⇒ x = 35, y = 24; (iv) ⇒ x = 46, y = 13;(vi) ⇒ x = 57, y = 2.Obviously, from (a), (b), (c), (d) and (e) we get, sumof three digits y cannot be 2 or 13.Hence, only favourable case takes place when thesum of four digits in the odd places = 35 and thesum of the three digits in even places = 24.∴ in the favourable numbers we will get,9,9,9,8 in odd places and 8,8,8 in even placesor 9,9,9,8 in odd places and 9,8,7 in even placesor 9,9,9,8 in odd places and 9,9,6 in even places∴ the number of numbers divisible by 114 ! 4 ! 4 ! 3!= + × 3 + ×3! 3!3!2!= 4 + 24 + 12 = 40 ...(ii)∴ from (i) and (ii)40 4the required probability = = . 210 212. The medians of a triangle ABC make angles α,β,γwith each other. Prove thatcot α + cot β + cot γ + cot A + cot B + cot C = 0.Sol. Here, G is the centroid of the ∆ABC and ∠BGC =α, etc.We know from geometry,AB 2 + AC 2 = 2(AD 2 + BD 2 ), etc.,and AG = 23 AD, BG = 32 BE, CG = 32 CF.AFEGαBCDNow, BC 2 + BA 2 = 2 (BE 2 + CE 2 )or a 2 + c 2 ⎪⎧2 2⎛ 3 ⎞ ⎛ b ⎞ ⎪⎫= 2 ⎨⎜BG ⎟ + ⎜ ⎟ ⎬⎪⎩ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎪⎭= 21 (9BG 2 + b 2 )∴ 9BG 2 = 2(a 2 + c 2 ) – b 2or BG 2 = 91 (2a 2 + 2c 2 – b 2 ).Similarly, CG 2 = 91 (2a 2 + 2b 2 – c 2 )XtraEdge for IIT-JEE 46 SEPTEMBER 2011

Now, cos α ==2 +BG CG – BC2BG .CG2 2 2 1 2(2a + 2c– b ) + (2a+ 2b992BG.CG1 2 2 2221 b + c – 5a1= . . sin α9 122. BG.CG.sin α221 1 ( b + c – 5a)sin α= . . 9 4 ar( ∆BGC)22222222– c) – a1 ( b + c – 5a)sin α= . 36 1ar( ∆ABC)3{Q ar(∆ABC) = 3 . ar(∆BGC)}2 2 2cosαb + c – 5a∴ =sin α 12∆2 2 2b + c – 5a∴ cot α =12∆2 2 2c + a – 5bSimilarly, cot β =,12∆2 2 2a + b – 5ccot γ =12∆2 2 2a + b + c∴ cot α + cot β + cot γ = –.4∆cos A cos B cosCAgain cot A + cot B + cot C = + +sin A sin B sin C=b2 +c2 – a2 2R.2bca+c2 +a2 – b2 2R.2caba2 + b2 – c2 2R+.2abc2R= (b 2 + c 2 – a 2 + c 2 + a 2 – b 2 + a 2 + b 2 – c 2 )2abcR= (a 2 + b 2 + c 2 )abc1= (a 2 + b 2 + c 2 ⎧ abc ⎫) ⎨QR = ⎬4∆⎩ 4 ∆ ⎭∴ cot α + cot β + cot γ + cot A + cot B + cot C2 2 2 2 2 2a + b + c a + b + c= –+= 0.4∆4∆3. A ray of light is coming along the line y = b, (b > 0)from the positive direction of the x-axis and strikes aconcave mirror whose intersection with the x-yplane is the parabola y 2 = 4ax, (a > 0). Find theequation of the reflected ray and show that it passesthrough the focus of the parabola.Sol. Let P be the point of incidenceThen P is the intersection of the line y = b and theparabola y 2 = 4ax.YQθPθTy 2 = 4axy = b⎛2⎞∴ P = ⎜b , b⎟⎝ 4a⎠∴ the equation of the tangent PT at P is⎛2⎞y . b = ⎜b2ax + ⎟⎝ 4 a ⎠2bor by = 2ax +...(i)22 a2'm' of (i) is . So, tan θ =bbaLet the slope of the reflected ray PQ be m.2am –∴ tan θ =b2a1+m.bor∴or m –∴ m –and m –∴2am –2a=bb 2a1+m.b2am –b 2a= ±2a1+m.bb2ab2ab2ab⎛2⎞= ± ⎜2a4a+ m.⎟2⎝ b b ⎠== –2ab2ab⎛2⎞⎜4am ⎟1–2=⎝ b ⎠+ m .– m .4ab24a2b24a2bX⎛2⎞and ⎜4am ⎟1 +2= 0.⎝ b ⎠XtraEdge for IIT-JEE 47 SEPTEMBER 2011

But m ≠ 0;⎛2⎞∴ ⎜4a⎟4am 1 – =2⎝ b ⎠ b4ab∴ m =2 2b – 4a∴ the equation of the reflected ray PQ is⎛24ab⎞y – b = ⎜bx – ⎟2 2b – 4a⎝ 4a⎠or (b 2 – 4a 2 )y – b(b 2 – 4a 2 ) = 4abx – b 3or (b 2 – 4a 2 )y – 4abx + 4a 2 b = 0.This will pass through the focus (a, 0) if(b 2 – 4a 2 )0 – 4ab . a + 4a 2 b = 0, which is true∴ the reflected ray passes through the focus.4. If a, b ∈ R be fixed positive numbers such thatf (a + x)= b + [b 3 + 1 – 3b 2 . f (x) + 3b {f (x)} 2 – {f (x)} 3 ] 1/3for all x ∈ R then prove that f (x) is a periodicfunction.Sol. Here, {f (a + x) – b} 3= b 3 + 1 – 3b 2 f (x) + 3b{f (x)} 2 – {f (x)} 3= 1 – [{f (x)} 3 – 3b . {f (x)} 2 + 3b 2 . f (x) – b 3 ]= 1 – {f (x) – b} 3∴ {f (a + x) – b} 3 + {f (x) – b} 3 = 1 ...(i)This is true for all x.Putting a + x for x in (i), we get,{f (2a + x) – b} 3 + {f (a + x) – b] 3 = 1 ...(ii)Subtracting (i) from (ii),{f (2a + x) – b} 3 – {f (x) – b} 3 = 0or {f (2a + x) – b} 3 = {f (x) – b} 3or f (2a + x) – b = f (x) – bor f (2a + x) = f (x)∴ f (x) is a periodic function5. Let PM be the perpendicular from the point P(1,2,3)to the x-y plane. If OP makes an angle θ with thepositive direction of the z-axis and OM makes anangle φ with the positive direction of the x-axis,where O is the origin, then find θ and φ.ZOφθ rP(1,2,3)Y∴ 1 = r sin θ . cos φ, 2 = r sin θ . sin φ, 3 = r cos θ.⇒ 1 2 + 2 2 = r 2 sin 2 θ . cos 2 φ + r 2 sin 2 θ . sin 2 φ= r 2 sin 2 θ(cos 2 φ + sin 2 φ) = r 2 sin 2 θ.∴ 5 = r 2 sin 2 θ. ∴ r sinθ = 5(clearly, θ and φ are acute).∴r sin θ=r cos θ⎛∴ θ = tan 1 ⎜⎝Also, 21 =535 ⎞⎟3⎠– .r sin θ.cosφr sin θ.sinφor tan θ =5 .3∴ cot φ = 21 . ∴ tan φ = 2 ∴ φ = tan –1 2.6. Take two sets of m and n parallel lines, lines of oneset cutting those of the other. Thus m × n smallrhombuses are formed. Two rhombuses are calledneighbouring rhombuses if they have a commonside. In each of these rhombuses a natural number isput such that the number written in a rhombus is theAM of the numbers written in all its neighbouringmembers. Show that all the numbers put in therhombuses are equal.Sol. There are two possibilities :(i) all the numbers are equal(ii) at least two numbers are differentIf all the numbers are equal, there is nothing toprove.If at least two numbers are different, there will bethe smallest number. Select the rhombus with thesmallest number p in it. As p is the AM of thenumbers in the neighbouring rhombuses, p must begreater than at least one of the numbers in theneighbouring rhombuses.a + b(Q AM of two numbers a, b is which is2bigger than one but smaller than the other; AM ofa + b + cthree numbers a, b, c is which cannot be3smaller than a, b as well as c, etc.). In that case p isnot the least among the numbers put in all therhombuses.Thus the possibility (ii) is absurd.Hence, all the numbers are equal.XMSol. We know that if P = (x, y, z) thenx = r sin θ . cos φ, y = r sin θ. sin φ, z = r cos θ.XtraEdge for IIT-JEE 48 SEPTEMBER 2011

MATHSPROBABILITYMathematics FundamentalsSome Definitions :Experiment : A operation which can produce somewell defined outcomes is known as an experiment.Random experiment : If in each trail of anexperiment conducted under identical conditions, theoutcome is not unique, then such an experiment iscalled a random experiment.Sample space : The set of all possible outcomes inan experiment is called a sample space. For example,in a throw of dice, the sample space is {1, 2, 3, 4, 5, 6}.Each element of a sample space is called a samplepoint.Event :An event is a subset of a sample space.Simple event : An event containing only a singlesample point is called an elementary or simple event.Events other than elementary are called composite orcompound or mixed events.For example, in a single toss of coin, the event ofgetting a head is a simple event.Here S = {H, T} and E = {H}In a simultaneous toss of two coins, the event ofgetting at least one head is a compound event.Here S = {HH, HT, TH, TT} and E = {HH, HT, TH}Equally likely events : The given events are said tobe equally likely, if none of them is expected to occurin preference to the other.Mutually exclusive events : If two or more eventshave no point in common, the events are said to bemutually exclusive. Thus E 1 and E 2 are mutuallyexclusive in E 1 ∩ E 2 = φ.The events which are not mutually exclusive areknown as compatible events.Exhaustive events : A set of events is said to betotally exhaustive (simply exhaustive), if no event outside this set occurs and at least one of these eventmust happen as a result of an experiment.Independent and dependent events : If there areevents in which the occurrence of one does notdepend upon the occurrence of the other, such eventsare known as independent events. On the other hand,if occurrence of one depend upon other, such eventsare known as dependent events.Probability :In a random experiment, let S be the sample spaceand E ⊆ S, then E is an event.The probability of occurrence of event E is defined asnumber of distinct elements in E n(E)P(E) ==number of distinct element in S n(S)number of outocomes favourable to occurrence of E=number of all possible outcomesNotations :Let A and B be two events, thenA ∪ B or A + B stands for the occurrence of atleast one of A and B.A ∩ B or AB stands for the simultaneousoccurrence of A and B.A´ ∩ B´ stands for the non-occurrence of both Aand B.A ⊆ B stands for "the occurrence of A impliesoccurrence of B".Random variable :A random variable is a real valued function whosedomain is the sample space of a random experiment.Bay’s rule :Let (H j ) be mutually exclusive events such thatP(H j ) > 0 for j = 1, 2, ..... n and S = U n H j . Let A bej=1an events with P(A) > 0, then for j = 1, 2, .... , n⎛ H ⎞P⎜j P(H j)P(A / H j)⎟=n⎝ A ⎠∑ P(H ) P(A / H )k=1kBinomial Distribution :If the probability of happening of an event in a singletrial of an experiment be p, then the probability ofhappening of that event r times in n trials will ben C r p r (1 – p) n – r .Some important results :Number of cases favourable to event A(A) P(A) =Total number of cases=n(A)n(S)kXtraEdge for IIT-JEE 49 SEPTEMBER 2011

P( A) ==Number of cases not favourable to event ATotal number of casesn(A)n(S)(B) Odd in favour and odds against an event : As aresult of an experiment if “a” of the outcomes arefavourable to an event E and b of the outcomes areagainst it, then we say that odds are a to b in favourof E or odds are b to a against E.Thus odds in favour of an event ENumber of favourable cases a==Number of unfavourable cases bSimilarly, odds against an event ENumber of unfavourable cases b==Number of favorablecases aNote :If odds in favour of an event are a : b, then theprobability of the occurrence of that event isaand the probability of non-occurrence ofa + bathat event isa + b.If odds against an event are a : b, then theprobability of the occurrence of that event isband the probability of non-occurrence ofa + bathat event isa + b.(C) P(A) + P( A ) = 10 ≤ P(A) ≤ 1P(φ) = 0P(S) = 1If S = {A 1 , A 2 , ..... A n }, thenP(A 1 ) + P(A 2 ) + .... + P(A n ) = 1If the probability of happening of an event in onetrial be p, then the probability of successivehappening of that event in r trials is p r .(D) If A and B are mutually exclusive events, thenP(A ∪ B) = P(A) + P(B) orP(A + B) = P(A) + P(B)If A and B are any two events, thenP(A ∪ B) = P(A) + P(B) – P(A ∩ B) orP(A + B) = P(A) + P(B) – P(AB)If A and B are two independent events, thenP(A ∩ B) = P(A) . P(B) orP(AB) = P(A) . P(B)If the probabilities of happening of n independentevents be p 1 , p 2 , ...... , p n respectively, then(i) Probability of happening none of them= (1 – p 1 ) (1 – p 2 ) ........ (1 – p n )(ii) Probability of happening at least one of them= 1 – (1 – p 1 ) (1 – p 2 ) ....... (1 – p n )(iii) Probability of happening of first event and nothappening of the remaining= p 1 (1 – p 2 ) (1 – p 3 ) ....... (1 – p n )If A and B are any two events, then⎛ B ⎞P(A ∩ B) = P(A) . P ⎜ ⎟ or⎝ A ⎠⎛ B ⎞P(AB) = P(A) . P ⎜ ⎟⎝ A ⎠⎛ B ⎞Where P ⎜ ⎟ is known as conditional probability⎝ A ⎠means probability of B when A has occured.Difference between mutually exclusiveness andindependence : Mutually exclusiveness is usedwhen the events are taken from the sameexperiment and independence is used when theevents are taken from the same experiments.(E) P(A A ) = 0P(AB) + P( AB ) = 1P( A B) = P(B) – P(AB)P(A B ) = P(A) – P(AB)P(A + B) = P(A B ) + P( A B) + P(AB)Some important remark about coins, dice and playingcards :Coins : A coin has a head side and a tail side. Ifan experiment consists of more than a coin, thencoins are considered to be distinct if not otherwisestated.Dice : A die (cubical) has six faces marked 1, 2,3, 4, 5, 6. We may have tetrahedral (having fourfaces 1, 2, 3, 4,) or pentagonal (having five faces1, 2, 3, 4, 5) die. As in the case of coins, If wehave more than one die, then all dice areconsidered to be distinct if not otherwise stated.Playing cards : A pack of playing cards usuallyhas 52 cards. There are 4 suits (Spade, Heart,Diamond and Club) each having 13 cards. Thereare two colours red (Heart and Diamond) andblack (Spade and Club) each having 26 cards.In thirteen cards of each suit, there are 3 face cards orcoart card namely king, queen and jack. So there arein all 12 face cards (4 kings, 4 queens and 4 jacks).Also there are 16 honour cards, 4 of each suit namelyace, king, queen and jack.XtraEdge for IIT-JEE 50 SEPTEMBER 2011

MATHSBINOMIAL THEOREMMathematics FundamentalsBinomial Theorem (For a positive Integral Index) :If n is a positive integer and x, a are two real orcomplex quantities, then(x + a) n = n C 0 x n + n C 1 x n – 1 a + n C 2 x n – 2 a 2 +... + n C r x n–r a r + .... + n C n – 1 x a n–1 + n C n a n ..(1)The coefficient n C 0 , n C 1 , ......, n C n are called binomialcoefficients.Properties of Binomial Expansion :There are (n + 1) terms in the expansion of(x + a) n , n being a positive integer.In any term of expansion (1), the sum of theexponents of x and a is always constant = n.The binomial coefficients of term equidistantfrom the beginning and the end are equal, i.e.n C r = n C n – r (0 ≤ r ≤ n).The general term of the expansion is (r + 1) th termusually denoted by T r + 1 = n C r x n – r a r (0 ≤ r ≤ n).The middle term in the expansion of (x + a) n(a) If n is even then there is just one middle term, i.e.th⎛ n ⎞⎜ +1⎟term.⎝ 2 ⎠(b) if n is odd, then there are two middle terms, i.e.thth⎛ n ⎞ ⎛ n + 3 ⎞⎜ +1⎟term and ⎜ ⎟ term.⎝ 2 ⎠ ⎝ 2 ⎠The greatest term in the expansion of (x + a) n ,x, a ∈ R and x, a > 0 can be obtained as below :∴orTr1rTr1Tr+ n − r + 1 a=r x+ ( n + 1) a − r(a + x)– 1 =Trx( a + x)⎧(n + 1)a ⎫= ⎨ − r⎬=rx ⎩ a + x ⎭( n + 1)awhere k =a + xNow, suppose that( n + 1)a(i) k = is an integer. We havea + xa + x |k – r|,rxT r + 1 > T r ⇔Tr1Along , T r + 1 = T r ⇔+> 1 ⇔ r < k (i.e. 1 ≤ r < k)TrTr1+= 1 ⇔ r = k,Ti.e. T k + 1 = T k > T k–1 > .... > T 3 > T 2 > T 1In this case there are two greatest terms T k andT k+1 .( n + 1)a(ii) k = is not an integer. Let [k] be thea + xgreatest integer in k. We haveT r+1 > T r ⇔Tr1Trr+⇔ r < k = [k] + (fraction)⇔ r ≤ [k]i.e. T 1 < T 2 < T 3 < ..... < T [k] – 1 < T [k] < T [k] + 1In this case there is exactly one greatest term viz.([k] + 1) th term.Term independent of x in the expansion of(x + a) n – Let T r + 1 be the term independent of x.Equate to zero the index of x and you will find thevalue of r.The number of term in the expansion of(x + y + z) n ( n + 1)( n + 2)is, where n is a positive2integer.Pascal TriangleIn(x + a) n when expanded the various coefficientswhich occur are n C 0 , n C 1 , n C 2 , .... The Pascal trianglegives the values of these coefficients for n = 0, 1, 2,3, 4, 5, ....n = 0 1n = 1 1 1n = 2 1 2 1n = 3 1 3 3 1n = 4 1 4 6 4 1n = 5 1 5 10 10 5 1n = 6 1 6 15 20 15 6 1n = 7 1 7 21 35 35 21 7 1n = 8 1 8 28 56 70 56 28 8 1XtraEdge for IIT-JEE 51 SEPTEMBER 2011

Rule : It is to be noted that the first and least terms ineach row is 1. The terms equidistant from thebeginning and end are equal. Any number in any rowis obtained by adding the two numbers in thepreceding row which are just at the left and just at theright of the given number, e.g. the number 21 in therow for n = 7 is the sum of 6 (left) and 15 (right)which occur in the preceding row for n = 6.Important Cases of Binomial Expansion :If we put x = 1 in (1), we get(1 + a) n = n C 0 + n C 1 a + n C 2 a 2 + ..................+ n C r a r + ........... n C n a n ...(2)If we put x = 1 and replace a by – a, we get(1 – a) n = n C 0 – n C 1 a + n C 2 a 2 – ..................+ (–1) r n C r a r + .... + (–1) n n C n a n ...(3)Adding and subtracting (2) and (3), and thendividing by 2, we get1 {(1 + a) n + (1 – a) n } = n C 0 + n C 2 a 22+ n C 4 a 4 + .... ...(4)1 {(1 + a) n – (1 – a) n } = n C 1 a + n C 3 a 32+ n C 5 a 5 + ...... ...(5)Properties of Binomial Coefficients :If we put a = 1 in (2) and (3), we get2 n = n C 0 + n C 2 + ...... + n C r + ....+ n C r + ..... n C n–1 + n C nand 0 = n C 0 – n C 1 + n C 2 – ...... + ...... + (–1) n n C n∴ n C 0 + n C 2 + .... = n C 1 + n C 3 + .... = 21 [2 n ± 0]= 2 n – 1 ...(6)Due to convenience usually written asC 0 + C 2 + C 4 + ... = C 1 + C 3 + C 5 + .... = 2 n – 1and C 0 + C 1 + C 2 + C 3 + ...... + C n = 2 nWhere n C r ≡ C r ==n!r!(n − r)!n ( n −1)(n − 2)...( n − r + 1)r !Some other properties to remember :C 1 + 2C 2 + 3C 3 + ... + nC n = n . 2 n – 1C 1 – 2C 2 + 3C 3 – .... = 0C 0 + 2C 1 + 3C 2 + ... + (n + 1)C n = (n + 2) 2 n –1C 0 C r + C 1 C r+1 + ... + C n – r C n =C 0 2 + C 1 2 + C 2 2 + .... + C n2=(2n)!( n − r)!.(n + r)!(2n)!( n!)2C 2 0 – C 2 1 + C 2 2 – C 2 3 + ...= ⎨ ⎧ 0, if n is oddn/2 n⎩ (–1) . C , if n is evenn / 2Binomial Theorem for Any Index :The binomial theorem for any index states that(1 + x) n nx n( n −1)= 1 + + x 21! 2!n( n −1)(n − 2)+x 3 +..... ...(7)3!Where n is any index (positive or negative)The general term in expansion (7) isn ( n −1)......(n − r + 1)T r + 1 =x rr!In this expansion there are infinitely many terms.This expansion is valid for |x| < 1 and first termunity.When x is small compared with 1, we see that theterms finally get smaller and smaller. If x is verysmall compared with 1, we take 1 as a firstapproximation to the value of (1 + x) n or 1 + nx asa second approximation.Replacing n by – n in the above expansion, we get(1 + x) –n n( n +1)= 1 – nx + x 2 n( n + 1)( n + 2)–x 32!3!+ ... + (–1) r n(n + 1)( n + 2)...( n + r −1)x r +r!...Replacing x by – x in this expansion, we get(1 – x) –n n( n +1)= 1 + nx + x 2 n( n + 1)( n + 2)+x 3 +2!3!n(n + 1)( n + 2)...( n + r −1)....... +x r + ....r!Important expansions for n = –1, –2 are :(1 + x) –1 = 1 – x + x 2 – x 3 + ...+ (–1) r x r + .... to ∞(1 – x) –1 = 1 + x + x 2 + x 3 + ..... + x r + .... to ∞(1 + x) –2 = 1 – 2x + 3x 2 – .... + (–1) r (r + 1)x r +….(1 – x) –2 = 1 + 2x + 3x 2 + ..... + (r + 1)x r + .... to ∞(1 + x) –3 = 1 – 3x + 6x 2 – 10x 3 + ...+ (–1) r ( r + 1)( r + 2)x r + .....2 !(1 – x) –3 = 1 + 3x + 6x 2 + 10x 3 + ......... +( r + 1)( r + 2)x r + ....2 !XtraEdge for IIT-JEE 52 SEPTEMBER 2011

XtraEdge for IIT-JEE 53 SEPTEMBER 2011

Based on New PatternIIT-JEE 2012XtraEdge Test Series # 5Time : 3 HoursSyllabus : Physics : Laws of motion, Friction, Work Power Energy, Gravitation, S.H.M., Laws of Conservations of Momentum,Rotational Motion (Rigid Body), Elasticity, Fluid Mechanics, Surface Tension, Viscosity, Refl. At Plane surface, Ref. at Curvedsurface, Refraction at Plane surface, Prism (Deviation & Dispersion), Refraction at Curved surface, Wave Nature of Light:Interference. Chemistry : Gaseous state, Chemical Energetics, Oxidation-Reduction, Equivalent Concept, Volumetric Analysis,Reaction Mechanism, Alkane, Alkene, Alkyne, Alcohol, Ether & Phenol, Practical Organic Chemistry, Aromatic Hydrocarbons,Halogen Derivatives, Carboxylic Acid & Its Derivatives, Nitrogen Compounds, Amines, Carbohydrates, Amino Acid, Protein &Polymers. Mathematics : Logarithm & Modulus Function, Quadratic Equation, Progressions, Binomial Theorem, Permutation &Combination, Complex Number, Indefinite Integration, Definite Integration, Area Under the Curve, Differential Equations.Instructions : [Each subject contain]Section – I :Section – II :Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will beawarded for correct answer and -1 mark for wrong answer.Question 9 to 12 are multiple choice question with multiple correct answer. +4 marks will be awarded forcorrect answer and -1 mark for wrong answer.Section – III : Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded forcorrect answer and -1 mark for wrong answerSection – III : Question 19 to 20 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 wronganswer.PHYSICSQuestions 1 to 8 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. A block of mass M is kept in gravity free space andtouches the two springs as shown in the figure.Initially springs are in its natural length. Now, theblock is shifted by small displacement 'x' from thegiven position in such a way it compresses a springand released. The time-period of oscillation of masswill be –KM4 Kl 0 2l 02. With what minimum speed should m be projectedfrom point C in presence of two fixed masses Meach at A and B as shown in the figure such thatmass m should escape the gravitational attraction ofA and B ?(A)(C)MA2GMRRC mR30ºR(B)GM2 (D)R2MB22V min2GMRGMR(A)(C)π23π2MKMK(B) 2π(D) πM5KM2K3. If the optic axis of convex and concave lenses areseparated by a distance 5 mm as shown in thefigure. Find, the co-ordinate of the final imageformed by the combination if parallel beam of lightis incident on lens, origin is at the optical center ofconvex lens.XtraEdge for IIT-JEE 54 SEPTEMBER 2011

f = 20 cm(0, 0)5mmf = –10Principle axisof concave lensPrinciple axisof convex lens30 cm(A) (25 cm, 0.5 cm) (B) (25 cm, 0.25 cm)(C) (25 cm, 0.5 cm) (D) (25 cm, –0.25 cm)4. A non viscous and incompressible liquid is flowingthrough a section of tube as shown. Area of crosssectionat A is 3 cm 2 whereas it is 1 cm 2 at B.Velocity of liquid at B is 6 m/s. Find the forceexerted by liquid on curved surface of conicalsection. Given the density of liquid is 10 3 kg/m 2 ,pressure at A is P A = 18 × 10 3 M/m 2 and pressure atB is P B .ABP A P B6 m/s1 cm 23 cm 2(A) 2.8 N(B) 2.4 N(C) 6.4 N(D) none of the above5. Let S 1 and S 2 be the two slits in Young's double slitexperiment. If central maxima is observed at P andangle S 1 P S 2 = θ, (θ is small) find the y-coordinatesof the 3 rd minima assuming the origin at the centralmaxima. (λ = wavelength of monochromatic lightused)(A) ±2λθ(B)5λ± 2 θ(C)5± λθ 2(D) ± 2 λθ6. The mirror of length 2l makes 10 revolutions perminute about the axis crossing its mid point O andperpendicular to the plane of the figure There is alight source in point A and an observer at point B ofthe circle of radius R drawn around centre OR(∠AOB = 90º). What is the proportion if the lobserver B first sees the light source when the angleof mirror ψ = 15º ?A7. A capillary tube is made of glass with the index ofrefraction 3, outer radius of the tube is 30 cm. Thetube is filled with a liquid with the index ofrefraction 2. What should be the minimum internalradius of the tube so that any ray that hits the tubewould enter the liquid -(A) 15 cm (B) 10 cm (C) 20 cm (D) 45 cm8. A ray of light when incident upon a prism suffers aminimum deviation of 39°. If the shaded halfportion of the prism is removed, then the same raywill -(A) suffer a deviation of 19.5°(B) suffer a deviation of 39°(C) not suffer any deviation(D) will be totally internally reflectedQuestions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.9. A smooth inclined plane has an angle of inclinationθ from the horizontal. From point A on the inclinedplane, body 1 is projected horizontally with speed uand from point B on the inclined plane, body 2 isreleased from rest. Distance AB is equal to l. If themagnitude of relative velocity of body 2 w.r.t. body1 and magnitude of relative acceleration of body 2w.r.t body 1 at time t are v and ‘a’, respectively,then which of the following is/are true ?xy2θuB1 Al(A) 2 (B)ψlO12lRB(C) 2 2 (D)212(A) v =u2(B) a = zero(C) a = g cos θ(D) v =− 2ug sin θcosθt+ g222cos2θt( gsin θ t − u) + (gsin θcosθt− gt)22XtraEdge for IIT-JEE 55 SEPTEMBER 2011

10. A particle is performing linear SHM. When itsspeed is v 1 , then its acceleration is a 1 . At anotherinstant, its speed is v 2 and acceleration is a 2 . Findthe amplitude and angular frequency of SHM.(A) ω =av2121− a− v2222(B) ω =( v1− v2)2 2 2 2(C) A = (a 2v1− a1v2)2 2 2(a − a )221( v2− v1) 2 2 2 2(D) A = (a 2v1− a1v2)2 2 2(a − a )22122av2221− a2122− v11. If a force field → E is defined as → k →E = − r (r > 1),n+1rthen -(A) the change in potential energy corresponding tothis force field can be written as,k ⎡ 1 1 ⎤U(r 2 ) – U(r 1 ) = − ⎢ − ⎥(n −1)n−1n−1⎢⎣r2 r1⎥⎦(B) the potential energy corresponding to this forcefield can be written as,U(r) = − kn−1(n −1)r, with U(∞) = 0(C) the force attracts the particle towards the→origin r = 0(D) the force repels the particle away from the→origin r = 012. Coefficient of friction between the two block is 0.3whereas the surface AB is smooth. (g = 10m/s 2 ) :2kgT 2A3kgT 110kg(A) Acceleration of 10 kg mass is 5.86 m/s 2(B) Acceleration of 10 kg mass is 7.55 m/s 2(C) Tension T 1 in the string is 17.7 N(D) Tension T 2 in the string is about 41.4 NThis section contains 2 paragraphs; each has 3multiple choice questions. (Questions 13 to 18) 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. + 4 marks will be given for each correctanswer and – 1 mark for each wrong answer.BPassage # 1 (Ques. 13 to 15)The following figure shows a horizontal pipe ofvarying cross-sectional area. The water flows in thepipe and flow remains laminar for whole the length.The pipe is completely smooth, i.e. no pressureenergy losses occur in the pipe.Here small tubes are connected at each crosssectionto measure the pressure at each crosssection.Neglect the capillary rise. Also, thedifference in heights of different cross-section aresmall, so we may assume that they are at the sameheight.(The flow in pipe is assumed incompressible, nonviscousand irrotational.)1 2 3A 1 = 0.4 m 2 A 2 = 0.1 m 2 A 3 =1 m 2v 2 = 5 m/sρ = 1000 kg/m 313. What is the mass flow rate in the pipe ?(A) 2 kg/s(B) 20 kg/s(C) 200 kg/s (D) 2000 kg/s14. What is the difference in kinetic energy per unitvolume between sections 2 and 3, i.e. K 2 – K 3 ?(A) 98 kJ(B) 198 kJ(C) 396 kJ(D) 1.98 ×10 5 kJ15. What is the difference in water levels of small tubesconnected at sections 2 and 3 ?(A) 9.8 m(B) 19.8 m(C) 20 m(D) 39.6 mPassage # 2 (Ques. 16 to 18)The diagram shows an equilateral prism. Themedium on one side of the prism is µ 1 . The4refractive index of the prism is µ = . The3diagram shows variation of magnitude of angle ofdeviation with respect to µ 1 . Consider the light rayto be normally incident on the first face.(Angle of deviation)ββ 1β 20 1 k 1 k 2µ 1XtraEdge for IIT-JEE 56 SEPTEMBER 2011

16. Value of k 2 is –8(A) (B)332(C)43(D)6320. Match the column :Column-I(A)Column-II(P) Converging17. Value of k 1 is –8(A) 2 (B) 3(C) 35(D) 34Rµ R18. Value of β 1 – β 2 is –(A) 60° (B) 30° (C) 20° (D) 90°(B)R µ(Q) Concave-convexThis section contains 2 questions (Questions 19, 20).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.+ 8 marks will be given for complete correct answer(i.e. +2 marks for each correct row) and No Negativemarks for wrong answer.19. A particle moving on the smooth horizontal tablestrikes the rod AB kept freely at an endA perpendicularly. Match the following –(A)(B)(C)(D)Column-IAngular momentumof system (particle +rod) about any pointisLinear momentum ofsystem (particle +rod) isAngular momentumof rod about its COMisKinetic energy ofsystem (particle +rod) except duringcollision isColumn-II(P) Conserved forelastic collision(Q) Not conserved forelastic collision(R) Conserved forperfect inelasticcollision(S) Not conserved forperfect inelasticcollision(T)Conserve in anytype of collision(C)(D)R2Rµµ2RR(R) Convex-concave(S) Diverging(T) NoneCHEMISTRYQuestions 1 to 8 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. Select the correct statement -(A) When attractive forces are dominant in a realgas U internal energy increases by increasingthe average molecular separation(B) Isothermal contraction decreases the internalenergy when repulsive forces betweenmolecules exist(C) For an ideal gas at constant T internal energy isfunction of volume only.(D) In case of attractive forces between particle, ifthe volume of the sample expands isothermallyits dU< 02. A piece of plumber’s solder (containing Pb & Snonly)weighing 3 gm was dissolved in dil HNO 3 and thentreated with dilute H 2 SO 4 — the ppt. of PbSO 4 isproduced , which after washing and drying weighed2.93 gm. The remaining solution was neutralized –a white ppt of stannic acid is produced , which onheating yields 1.27 gm of SnO 2 . Therefore mass % ofPb and Sn in the plumber’s solder isAt wt. Pb = 207 ; Sn = 118.7XtraEdge for IIT-JEE 57 SEPTEMBER 2011

(A) % Pb = 66.67 % , % Sn = 33.33 %(B) % Pb = 33.33 % , Sn = 66.67 %(C) % Pb = 50% , % Sn = 50 %(D) % Pb = 75 % , % Sn = 25 %3. The major product of following reaction is:OCF 3 CO 3 H7. 15 ml of gaseous hydrocarbon required forcomplete combustion 357 ml of air containing O 2 ,21 % by volume and the gaseous products occupied327 ml. If all volumes are measured at STP, themolecular formula of the hydrocarbon is -(A) C 3 H 6 (B) C 4 H 10(C) C 4 H 8 (D) C 3 H 8(A)OO(B)OO8. Which of the following is nylon 6, 6 -H(A)NNOH4OOO(C)(D) None of above(B)NHNH44. In the reaction MnO – 4 + SO 2– 3 + H + ⎯⎯→ SO 2– 4+ Mn 2+ + H 2 O(A) MnO– 4 and H + both are reduced(B) MnO – 4 is reduced and H + is oxidised(C) MnO – 4 is reduced and SO 2– 3 is oxidised(D) MnO– 4 is oxidised and SO 2– 3 is reduced(C)(D)OOCONHNHNH45.⎯ ⎯H 2 /Ni ⎯→A. A is -D DDD(A) CH 3 – (CH 2 ) 4 – CH 3 (B)HHDH(C)(D)HH HD6. Which of the following alcohols would you expectto form a corbocation most readily in H 2 SO 4 -OHOHQuestions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.9. The graph represent Boyle's law is/are -(A)1V1P(B) PVP(A)(B)OHOH(C) logP(D) V 1/PO(C)(D)logVXtraEdge for IIT-JEE 58 SEPTEMBER 2011

10. 0.5 M KI solution reacts with excess of H 2 SO 4 andKIO 3 solution according to the equation -6H + −+ 5 I + IO3− –→ 3I 2 + 3H 2 Owhich of the following statements is /are true -(A) 200 ml of KI solution reacts with 0.10 moleKIO 3(B) 0.5 litre of KI solution produces 0.15mole of I 2(C) Volume of 0.5M KIO 3 used to react completelywith 1 litre KI solution is 200 ml(D) 100 ml of the KI solution reacts with 0.06 MH 2 SO 4 solution11. When nitrobenzene is treated with Br 2 in presenceof FeBr 3 the major product formed is m-bromonitrobenzene. Statements which are relatedobtain the m-isomer are -(A) The electron density on meta carbon is morethan on ortho and para position.(B) The intermediate carbonium ion formed afterinitial attack of Br ⊕ attack the meta position isleast destabilized(C) Loss of aromaticity when Br + attcks at theortho and para positions and not at metaposition.(D) Easier loss of H + to regain aromaticity form themeta position than from ortho and parapositions. [A, B]12. Reaction (1)3mol alc.KOH∆ClCl⎯⎯⎯⎯⎯→Reaction (2)3mol alc.KOHClCl⎯⎯⎯⎯⎯→Reaction (3)3mol alc.KOHClCl[A][B]ClClClClClCl⎯⎯⎯⎯⎯→ [C]The correct statement is –(A) A and B are sameClClClClClCl(B) B and C are different(C) B and C are same (D) A and B are differentThis section contains 2 paragraphs; each has 3multiple choice questions. (Questions 13 to 18) 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. + 4 marks will be given for each correctanswer and – 1 mark for each wrong answer.Passage # 1 (Ques. 13 to 15)In the following reaction sequence, products I, Jand L are formed. K represents a reagent.Hex-3-ynal⎯⎯→K Me⎯1.NaBH4⎯2.PBr3⎯⎯ →IPd / BaSOO quinoline1.Mg / ether2.CO⎯⎯⎯ ⎯+→H3.H O42J32⎯ ⎯→ L13. The structure of the product I is-(A) Me(C) MeBrBr(B) Me(D) Me14. The structures of compounds J and K, respectively,are-(A) Me COOH and SOCl 2OH(B) Meand SOCl 2O(C) Meand SOCl 2COOH(D) Me COOH and CH 3SO 2 Cl15. The structure of product L is-(A) Me CHO(B) Me(C) Me(D) MeCHOCHOCHOPassage # 2 (Ques. 16 to 18)When (C–H) σ electrons are in conjugation to pibond, this conjugation is known ashyperconjugation.For any compound to show hyperconjugation(i) Compound should have one sp 2 -hybridisedcarbon(ii) α-carbon with respect to sp 2 should be sp 3BrBrXtraEdge for IIT-JEE 59 SEPTEMBER 2011

(iii) α-carbon should contain at least one hydrogenatomNo. of α-carbon ∝ stability of cation and alkene16. Which of the following cations ishyperconjugation destabilized ?(A) CH 3 – C Θ ΘH 2 (B) CH 3 –CH–CH 3CH 3(C) CH 3 –C ⊕⊕(D)CH 317. Which of the following alkyl benzene hasmaximum electron density ?CH 3CH 2 CH 3(A)(B)(C) H 3C–CH–CH 3CH 3(D) H 3C–C–CH 318. Which of the following orders is correct for magnitudeof +M power among these compounds ?O'X'(A) Z > Y > X(C) X > Y > ZO'Y'NCH 3'Z'(B) Y > X > Z(D) Z >X > YThis section contains 2 questions (Questions 19, 20).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.+ 8 marks will be given for complete correct answer(i.e. +2 marks for each correct row) and No Negativemarks for wrong answer.19. Match the column :Column-IColumn-II(A) Hydrogen gas (p) Compressibilityfactor ¹ 1(P = 200 atm AndT = 273 K)(B) Hydrogen gas (Q) attractive forces are(P ~ O and T = 273 K) dominant(C) CO 2 (P = 1atm (R) PV = nRTand T = 273 K)(D) Real gas with (S) P ( V – nb) = nRTvery large Molarvolume(T) P (V + nb) = nRT20. Given below are the reactions of oxidising agentswith excess KI and the liberated iodine is titratedagainst the standard reducing agent, sodiumthiosulphate-Column-IColumn-II(A) K 2 Cr 2 O 7 + KI + H 2 SO 4 (P) 20 milli moles of(20ml, 0.2 M)thiosulphate⎯→ Cr 2 (SO 4 ) 3 + K 2 SO 4+ H 2 O + I 2(B) CuSO 4 + KI(Q) 24 millimoles of(20 ml, 1.2M) thiosulphate→ Cu 2 I 2 + K 2 SO 4 + I 2(C) KMnO 4 + KI + H 2 SO 4 → (R) 20 milliequivalents(20 ml, 0.2 M)K 2 SO 4 + MnSO 4 + H 2 O + I 2(D) H 2 O 2 + KI + H 2 SO 4 → (S) 24 milliequivalents(20 ml, 0.2 M)K 2 SO 4 + H 2 O + I 2 (T) 36 milliMATHEMATICSQuestions 1 to 8 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 number of odd consecutive integers whose sumis 45 2 – 21 2 is :(A) 24 (B) 23 (C) 25 (D) NoneXtraEdge for IIT-JEE 60 SEPTEMBER 2011

2. The remainder when 5 99 is divided by 13 is :(A) 6 (B) 8 (C) 9 (D) 103.x∫π 2lim2x→π/2 x//(A)(C)∫(2(2π 4log e 2π2ln 2π–cost–1) dtt – π / 2) dt=(B)ln 22 π(D) None4. A shopkeeper sells three varieties of perfumes andhe has a large number of bottles of the same size ofeach variety in his stock. There are 5 places in arow in his showcase. The number of different waysof displaying the three varieties of perfumes in theshowcase is :(A) 6 (B) 50 (C) 150 (D) None5. Let f (a) denotes the area of the region bounded byy = a 2 x 2 + ax + 1, co-ordinate axes and the linex = 1. Then f (a) will be least at a =(A) – 3/4 (B) – 1/4 (C) – 1 (D) None(1 + x cos x)6.∫dx =2 2sin xx(1–x e )(A) log |xe sinx | – 21 log |1 – x 2 e 2sinx | + c(B) log |xe sinx | + 21 log |1 – x 2 e 2sinx | + c(C) log |xe sinx | – log |1 – x 2 e 2sinx | + c(D) None of these7. The solution of the equationdy + x(x + y) = x(x + y) 3 – 1 is ln|f (x)| = x 2 + c,dxwhere f (x) =( x + y + 1)( x – y + 1) ( x + y + 1)( x + y –1)(A)(B)32( x + y)( x + y)( x + y + 1)( x – y –1)(C)(D) None of these2( x + y)8. For a > 0, ≠ 1 the roots of the equationlog ax a + log x a 2 + loga 2 a 3 = 0 are given by :x(A) a –3/4 (B) a –4/3 , a 1/2(C) a –1/2 , a –4/3 (D) None of theseQuestions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.19.1Let I =∫ π/21+x0dx , then(A) I < π/4 (B) I > π/4(C) I < log e 2 (D) I < log e 210. Let f (x) =3x – 2+4x – 3has(A) two roots in (2, 4)(B) only one root in (2, 3)(C) only one root in (3, 4)(D) no root in (2, 4)+5x – 4. Then f (x) = 011. Let z 1 , z 2 , z 3 be three complex numbers such that|z 1 | = |z 2 | = |z 3 | = 1 and z 1 + z 2 + z 3 = 0. If z 1 , z 2 , z 3denote the vertices of a ∆ABC, then :(A) origin is orthocentre(B) arg(C)z– z3 2z3– z12 2 21 z2z3= argzzz + + = z 1 z 2 + z 2 z 3 + z 3 z 1(D) area of ∆ABC =3412. A point P(x, y) moves in such a way that [|x|] + [|y|]= 1, [.] = G.I.F. Area of the region representing allpossible positions of the point P is equal to :(A) 8 (B) 4 (C) 16 (D) NoneThis section contains 2 paragraphs; each has 3multiple choice questions. (Questions 13 to 18) 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. + 4 marks will be given for each correctanswer and – 1 mark for each wrong answer.Passage # 1 (Ques. 13 to 15)21If the length of an arc of a curve measured from afixed point A(x = a) to another point P(x, y) on thecurve be s, then we have the relationds ⎛ dy ⎞2= 1+ ⎜ ⎟dx ⎝ dx ⎠XtraEdge for IIT-JEE 61 SEPTEMBER 2011

⎛ dy ⎞∴ arc AP = s =∫ 21+⎜ ⎟ dx.⎝ dx ⎠xaP(x,y)A(x = a)Using the above information, answer the followingquestions :13. Length of the arc of the parabola y 2 = 4x cut off byits latus-rectum is :(A) 2[ 2 + log e ( 2– 1)](B) 2[ 2 + ln( 2+ 1)](C) [ 2 + ln( 2+1)](D) [ 2 + ln( 2–1)]14. Whole length of the astroid x 2/3 + y 2/3 = 4 is :(A) 12 (B) 48(C) 6(D) none of these15. Length of the curve y = log e (sec x) between x = 0to x = π/3 is :(A) ln(2 + 3 ) (B) ln(1 + 3 )(C) 21 ln 3 (D) ln(2 – 3 )Passage # 2 (Ques. 16 to 18)Multinomial theorem states that for n ∈ N anda 1 , a 2 ,...........,a p ∈ C, then(a 1 + a 2 + ...+ a p ) nn!= ∑n1n2n pa 1 a2..... a pn1 ! n2!....... n p !where n 1 , n 2 ,...,n p are non-negative integers suchthat n 1 + n 2 + .... + n p = nAnswer the following questions :16. If x 1 , x 2 ,.....x n are independent, then number ofterms in the expansion of (x 1 + x 2 + .... + x n ) 5 is :(A) n+1 C 5 (B) n+4 C 5 (C) n+5 C 4 (D) n+4 C 417. Coefficient of x 3 y 3 z in the expansion of (1 + x – y + z) 9is :(A) 2520 (B) 2800 (C) – 2800 (D) – 140018. Coefficient of a 3 b 4 c 5 in the expansion of (ab + bc +ca) 6 is :(A) 60 (B) 40 (C) 0 (D) NonesThis section contains 2 questions (Questions 19, 20).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.+ 8 marks will be given for complete correct answer(i.e. +2 marks for each correct row) and No Negativemarks for wrong answer.19. Match the column :Column-IColumn-II(A) The largest value of n such (P) 2that 2010 n divides 2010!6r12 C(B) N = ∑ (–1)r=0r (Q) 3(C) The number of positive (R) 5integer triplets (a,b,c)such that a + b + c ≤ 0(D) A set contains 6 elements. (S) 7The number of ordered pairs(A,B) such that A ∩ B = φ (T) 1120. Match the column :Column-IColumn-II(A) Area of the region defined (P) 0by the inequality1|y| + ≤ 2 1– | x | is p/q,then p – q =(B) Sum of the values of α for (Q) 5/6which the system of inequalitiesx 2 + 2x + α ≤ 0; x 2 – 4x – 6α ≤ 0has unique solution iscos nx(C) If I n =∫dx, thensin x(R) 12cos pxI 7 – I 5 = , where p =p(D) The number of 5 digit (S) 4/5numbers made up of thedigits 3, 4, 5, 6,7 withoutrepetitions which are multipleof 11 is ab. Then b – a = (T) 1/6XtraEdge for IIT-JEE 62 SEPTEMBER 2011

Based on New PatternIIT-JEE 2013XtraEdge Test Series # 5Time : 3 HoursSyllabus : Physics : Laws of motion, Friction, Work Power Energy, Gravitation, S.H.M., Laws of Conservations ofMomentum, Rotational Motion (Rigid Body), Elasticity, Fluid Mechanics, Surface Tension, Viscosity.Chemistry : Gaseous state, Chemical Energetic, Oxidation-Reduction, Equivalent Concept, Volumetric Analysis.Mathematics : Logarithm & Modulus Function, Quadratic Equation, Progressions, Binomial Theorem, Permutation &Combination, Complex Number.Instructions : [Each subject contain]Section – I : Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be awardedfor correct answer and -1 mark for wrong answer.Section – II : Question 9 to 12 are multiple choice question with multiple correct answer. +4 marks will be awarded for correctanswer and -1 mark for wrong answer.Section – III : Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded forcorrect answer and -1 mark for wrong answerSection – III : Question 19 to 20 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 wronganswer.PHYSICSQuestions 1 to 8 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. A child is standing at one end of a long trolleymoving with a speed v on a smooth horizontaltrack. If the child starts running towards the otherend of the trolley with a speed u, the centre of massof the system (trolley + child) will move with aspeed -(A) zero (B) (v + u) (C) (v – u) (D) v2. A particle strikes a horizontal frictionless floor witha speed u, at an angle θ with the vertical, andrebounds with a speed v, at an angle φ with thevertical. The coefficient of restitution between theparticle and the floor is e. The angle φ is equal to –uθ φ v3. Two blocks A and B of mass m and 2m areconnected by a massless spring of force constant k.They are placed on a smooth horizontal plane.Spring is stretched by an amount x and thenreleased. The relative velocity of the blocks whenthe spring comes to its natural length is –⎛(A) ⎜⎝(C)3k⎞⎟ x2m⎠2kxmA B⎛(B) ⎜⎝(D)2k3m3km2x⎞⎟ x⎠4. Three identical blocks A, B and C are placed onhorizontal frictionless surface. The blocks B and Care at rest but A is approaching towards B with aspeed 10 m/s. The coefficient of restitution for allcollisions is 0.5. The speed of the block C just aftercollision is –A B C(A) θ (B) tan –1 [e tan θ](C)− ⎡1⎤tan 1⎢ tan θ⎥⎣e⎦(D) (1 + e)θ(A) 5.6 m/s(C) 8 m/s(B) 6 m/s(D) 10 m/sXtraEdge for IIT-JEE 63 SEPTEMBER 2011

5. Two bodies of mass m 1 = 3 kg and m 2 = 2 kg movealong mutually perpendicular directions withvelocities 4 m/s and 3 m/s respectively as shown inthe figure. As a result of collision the bodies sticktogether. The amount of heat liberated ism 1 4 m/s3 m/sm 2(A) 10 J (B) 15 J (C) 20 J (D) 25 J6. A thin rod of mass m and length l is hinged at oneend point which is at a distance h (h < l) above thehorizontal surface. The rod is released from restfrom the horizontal position. If e is the co-efficientof restitution, the angular velocity of rod just aftercollision will be (h = 1m, l = 2m, e = 1) –lhinged(A)(C)35883g3g(B)h683g(D) none of these7. A body of cross-sectional area A, height H anddensity ρ S , is immersed to depth h in a liquid ofdensity (ρ L ) and tied to bottom with string. Thetension in string is –Area = AHString(A) ρgha(B) (ρ S – ρ L ) ghA(C) (ρ L – ρ S ) ghA (D) (ρ L h – ρ S H) gA8. If two wires of same length l and area of crosssection A with Young modulus Y and 2Y connectin series and one end is fixed on roof and other endwith mass m. Make simple harmonic motion, thenthe time period is -mlml(A) 2π (B) 2πYA3YA3ml(C) 2π (D) 2π2YAhml2YAQuestions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.9. A pendulum suspended from the roof of an elevatorat rest has a time period T 1 ; when the elevatormoves up with an acceleration a its time periodbecomes T 2 ; when the elevator moves down with anacceleration a; its time period becomes T 3 then –(A) T 3 > T 2 and T 1 (B) T 2 > T 3 > T 1(C) T 1 =T2T322 2T2+ T3(D) T 1 =2 2T 2 + T310. Two concentric spherical shells have masses m 1and m 2 and radii r 1 and r 2 . Then–m 2r 1m 1 m 3PrOr 2(A) Outer shell will have no contribution ingravitational field at point(B) Force on P is directed towards O(C) Force on P is(D) Force on P isGm m12rGm mr1211. A body of mass m was hauled up the hill withconstant speed v by a force F which at each pointwas directed along tangent to the path. If length ofbase is l and height of hill is h, then which of thefollowing are correct ?(A) Work done by gravity is – mgh(B) Work done by friction is – µmgl(C) Work done by gravity is path independent(D) Work done by friction is path independent12. The potential energy U (in joule) of a particle ofmass 1 kg moving in x-y plane obeys the law U =3x + 4y, where (x, y) are the co-ordinates of theparticle in metre. If the particle is at rest at (6, 4) attime t = 0, then -(A) the particle has constant acceleration(B) the particle has zero acceleration(C) the speed of particle when it crosses the y-axisis 10 m/s(D) coordinates of the particle at t = 1sec are (4.5 , 2)23XtraEdge for IIT-JEE 64 SEPTEMBER 2011

This section contains 2 paragraphs; each has 3multiple choice questions. (Questions 13 to 18) 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. + 4 marks will be given for each correctanswer and – 1 mark for each wrong answer.Passage # 1 (Ques. 13 to 15)ABCD is a vertical frame of mass M. Two stringsbetween block of mass m and frame are ideal andlies in vertical plane. The frame ABCD is placed onhorizontal smooth surface and a horizontal force Fis applied as shown. Points P and Q are in samevertical line and two strings are of equal lengthPm θQ13. Minimum force F required to make both string tightis -(A) Mg tanθ (B) Mg cot θ(C) (m + M)g tanθ (D) (m + M)g cot θ14. If the lower string is just tight then tension in theupper string is -mgmg(A)(B)sin θcosθ(C) mg sinθ(D) mg cosθ15. For what value of force F tension in upper string istwice than tension in lower string.(A) 3(m + M)g tanθ (B) 2(m + M)g tanθ(C) 4(m + M)g tanθ (D) None of thesePassage # 2 (Ques. 16 to 18)Two rods 1 and 2 are released from rest as shown infigure. Given : l 1 = 4l, m 1 = 2m, l 2 = 2l and m 2 = m.There is no friction between the two rods. If α bethe angular acceleration of rod 1 just after the rodsare released. Then :A12l16. What is the normal reaction between the two rods atthis instant ?4mlα(A) 16 3 mlα (B)3(C)32mlα33(D) 12FB3 mlα17. What is the horizontal force on rod 1 by hinge A atthis instant ?⎛ ⎞(A) ⎜32 −123 ⎛ ⎞⎟ mlα (B) ⎜16 − 2 3⎟ mlα⎝ 3 3⎠ ⎝ 3 ⎠(C) (14 + 2 3 ) mlα (D) 3 mlα18. What is initial angular acceleration of rod 2 interms of the given parameters in the question ?⎡ 2 3g ⎤ ⎡33g ⎤(A) ⎢ + 2 3α⎥(B) ⎢ − 3α⎥⎢⎣2l⎥⎦⎢⎣l ⎥⎦⎡63g ⎤ ⎡33g 8 ⎤(C) ⎢ + 5 3α⎥(D) ⎢ − α⎥⎢⎣8l⎥⎦⎢⎣8l3 ⎥⎦This section contains 2 questions (Questions 19, 20).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.+ 8 marks will be given for complete correct answer(i.e. +2 marks for each correct row) and No Negativemarks for wrong answer.19. A solid sphere is rotatingInsectabout an axis as shown infigure. An insect follows thedotted path on thecircumference of sphere asshown. Match the following:Column-IColumn-II(A) Moment of inertia (P) will remain cons.(B) Angular velocity (Q) will first increasethen decrease(C) Angular momentum (R) will first decreasethen increase(D) Rotational kinetic (S) will continuouslyenergydecrease(T) will continuouslyincreaseXtraEdge for IIT-JEE 65 SEPTEMBER 2011

20. The cubical container filled with water is givenacceleration → a = a 0 î + a 0 ĵ + a 0 kˆ , then : (neglectthe effect of gravity)EFyDCHGBA zColumn-IColumn-II(A) Pressure at point A is (P) less than pressureat point G(B) Pressure at point D is (Q) less than pressureat point F(C) Pressure at point E is (R) Greater thanpressure at point C(D) Pressure at point H is (S) Greater thanpressure at point B(T) NoneCHEMISTRYQuestions 1 to 8 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. For which of the following change ∆H ≠ ∆E ?(A) H 2 (g) + I 2 (g) → 2HI(g)(B) HCl(aq) + NaOH(aq) → NaCl(s) + H 2 O(l)(C) C(s) + O 2 (g) → CO 2 (g)(D) N 2 (g) + 3H 2 (g) → 2NH 3 (g)2. The value of ∆H° for the hypothetical reaction,1A 2 B(s)→ 2A(s)+ B2(g)is, + 7.3 kJ2The value of ∆U° for the reaction is -(A) > 7.3 kJ (B) < 7.3 kJ(C) = 7.3 kJ (D) unpredictable3. 2.8 L of N 2 gas at 300 K and 20 atm was allowed toexpand isothermally against the external pressure of1 atm. The value of ∆U of the process is -(a) 10 J (B) 0 (C) –10 J (D) 4.18 J4. What is the volume of O 2 (g) required at STP for theoxidation of 1L of SO 2 (g) at 298K and1 atm ? 2 SO 2 (g) + O 2 (g) ⎯→ 2 SO 3 (g)(A) equal to 0.5 L (B) greater than 0.5 L(C) lesser than 0.5 L (D) equal to 1Lx5. For a given one mole of ideal gas kept at 6.5 atm ina container of capacity 2.463 L. The Avogadroproportionality constant for the hypothesis is (seefigure)nV(A) 0.406 (B) 2.46(C) 22.4(D) none of these6. At low pressure if RT = 2 a. p , (a is vander Waal'sconstant) then the volume occupied by a real gas is-(A)(C)2RTPRT2P2P(B) RT(D)2RT7. According to classical concept, oxidation involves -(A) Addition of oxygen(B) Addition of electronegative radical(C) Removal of either hydrogen orsomeelectropositive radical(D) All of these8. According to modern concept, oxidation is -(A) Electronation(B) Deelectronation(C) Addition of oxygen(D) Addition of electronegative elementQuestions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.9. An adiabatic expansion is one in which -(A) all energy is transferred as heat(B) no energy is transferred as heat(C) the temperature of a gas decreases in areversible adiabatic expansion(D) du ≠ dwPXtraEdge for IIT-JEE 66 SEPTEMBER 2011

10. Select the correct statement -(A) Heat capacity of diatomic gas is higher thanthat of monoatomic gas(B) Standard molar enthalpy of formation of CO 2 isequal to standard molar enthalpy ofcombustion of carbon (graphite)(C) Thermodynamics, a process is called reversiblewhen surrounding and system change into eachother(D) Change in state is completely defined when theinitial and final states are specified11. Which of the following is/are correct regardingindicators ?(A) pH range of phenolphthalein is 8.3 – 10(B) pH range of methyl orange is 3.2 – 4.4(C) By the phenolphthalein equivalence point ofNa 2 CO 3 is indicated(D) For the suitable indicator neutral point ofindicator must coinside with the equivalencepoint of the titration12. In which of the following species the oxidationstate of Cr is equal to +6?(A) CrO 5 (B) K 2 CrO 4(C) CrO 3 (D) Cr 2 O 3This section contains 2 paragraphs; each has 3multiple choice questions. (Questions 13 to 18) 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. + 4 marks will be given for each correctanswer and – 1 mark for each wrong answer.Passage # 1 (Ques. 13 to 15)For the substance X(g), molar specific heat atconstant temperature, Cp, increases linearly withthe absolute temperature, T.Informations regarding X(s)Molar specific heat = 6.4 JK –1 mol –1The latent heat of fusion of X (s) = 40 KJ/mol.at the melting point.The melting point of X (s) = 100 KInformation regarding X(l)Molar specific heat = 4.5 JK –1 mol –1Boiling point of X (l) = 250 KThe latent heat of vapourisation of X(l) = 60 KJ/molat the boiling point.Information regarding X(g)For X(g) at 1 atm,when T = 200 K, Cp = 0.04 JK –1 mol –1when T = 300 K, Cp = 0.05 JK –1 mol –113. What is the molar enthalpy of sublimation of X(s) -(A) 100 kJ/mol (B) 20 kJ/mol(C) 50 kJ/mol (D) – 20 kJ/mol14. What is the molar enthalpy of vapourisation ofX(l) at 400 K ?(A) Slightly lesser than 60 kJ(B) Double the enthalpy of vapourisation at 250 K(C) Slightly higher than 60 kJ(D) Equal to 60 kJ because enthalpy is a statefunction15. In which of the following phase Transition changeof entropy is maximum ?(A) Heating of solid(B) Fusion of X (s)(C) Vapourisation of X(l)(D) Heating of X (g)Passage # 2 (Ques. 16 to 18)For a fixed mass of gas under isobaric (constantpressure) condition, the volume varies withcentigrade temperature as followsV = V 0 (1 + α t)V 0 represent the volume at 0ºC at constant pressure16. For every 1ºC rise in temperature volume of the gasas per charles law -1(A) increases times of its volume at final273temperature1(B) increases times of its volume at 0ºC273(C) increases 273 times of its volume at 0ºC1(D) decreases times of its volume at 0ºC27317. How does pressure vary with the centrigradetemperature (t) for the definite mass of gas underisochoric (constant volume) condition ?P 0 is the pressure of same mass of gas at 0ºC⎞(A) P ∝ t (B) P = P 0 ⎜⎛ 2731 + ⎟⎝ t ⎠⎛ t ⎞t(C) P = P 0 ⎜1 + ⎟ (D) P = P 0 ×⎝ 273 ⎠ 27318. Under the isochoric condition for the definite massof gas pressure varies with absolute temperature (T)as follows :PV 1V 2V 3TWhich of the following is/are correct ?(A) V 1 = V 2 = V 3 (B) V 1 > V 2 > V 3(C) V 1 < V 2 < V 3 (D) V 2 < V 1 < V 3XtraEdge for IIT-JEE 67 SEPTEMBER 2011

This section contains 2 questions (Questions 19, 20).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.+ 8 marks will be given for complete correct answer(i.e. +2 marks for each correct row) and No Negativemarks for wrong answer.19. Match the column :Column-IColumn-II(A) T c /P c(P) Z(B) T c /V c(Q) a/Rb(C) T B(R) 8b/R(D) T i /T B (S) 8a/81 Rb 2(T) a/5Rb20. Match the column :Column- I(A) Permanent hardness(B) Temporary hardness(C) React on heating(D) Not responsible forhardnessColumn-II(P) Na + (aq)(Q) Ca 2+ (aq)(R) Mg 2+ (aq)−(S) HCO 3 (aq)(T) Al 3+MATHEMATICSQuestions 1 to 8 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. Solution set of the equation 2 |x| – |2 x–1 –1| = 2 x–1 + 1is :(A) [1, ∞) (B) (– ∞, – 1] ∪ [1, ∞)(C) [1, ∞) ∪{–1} (D) None2. Let f (x) = ax 2 + bx + c, a, b, c ∈ R, a ≠ 0. Iff (1) + f (2) = 0, the quadratic equation f (x) = 0 has:(A) no real root (B) 1 and 2 as real roots(C) two equal roots (D) two distinct real roots3. If (2n + r)r, n ∈ N, r ∈ N is expressed as the sum ofk consecutive odd natural numbers, then k is equalto :(A) r (B) n (C) r + 1 (D) n + 14. Sum of common roots of the equations z 3 + 2z 2 + 2z+ 1 = 0 and z 100 + z 32 + 1 = 0 is equal to :(A) 0 (B) – 1 (C) 1 (D) Nonenn–r5. If x + y = 1, then ∑ r Crx y equals :r=0(A) nx (B) ny (C) n (D) None6. The tens digit of 1! + 2! + 3! + .... + 29! is :(A) 1 (B) 2 (C) 3 (D) 47. Number of solutions of the equation2(2x–x + 1)2n(2x–x + 1)2x–x25 + 9 = 34.15(A) 2 (B) 3 (C) 4 (D) None8. The equation |x – x 2 – 1| = |2x – 3 – x 2 | has :(A) infinite solution (B) one solution(C) two solutions (D) no solutionQuestions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.9. Solution set of inequation log 2 log 1/2 (2 x – 15/16) ≤ 2is :(A) (0, log(31/16)) (B) [1, 31/16)(C) [0, log(31/16)) (D) [0, log(31/16)]10. Let S n = C 0 –C r = n C r . Then :(A) S n =(C) S n =2n2 ( n!)(2n + 1)!C 1 C +23 52r(B) S n =2n .Sn–1 (D) S n =2n + 12– ......+ (– 1)n2n22 ( n!)(2n–1)!2n2n–1Sn–1Cn,2n + 111. In a triangle, the lengths of the two larger sides are10 and 9, respectively. If the angles are in A.P.,then the length of the third side can be :(A) 5 – 6 (B) 3 3 (C) 5 (D) 5 + 612. If α,β,γ are roots of x 3 + ax + 3x – 1 = 0 (α ≤ β ≤ γ)which are in H.P. Then :(A) one root must be 1(B) all roots must be equal(C) a must be a negative integer(D) α,β,γ also in A.P.XtraEdge for IIT-JEE 68 SEPTEMBER 2011

This section contains 2 paragraphs; each has 3multiple choice questions. (Questions 13 to 18) 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. + 4 marks will be given for each correctanswer and – 1 mark for each wrong answer.Passage # 1 (Ques. 13 to 15)Two consecutive numbers from 1,2,3,...n areremoved. A.M. of the remaining numbers is 105/4.Answer the following questions :13. n =(A) 7 (B) 8 (C) 50 (D) 5114. G.M. of removed numbers is :(A) 42 (B) 56 (C) 72 (D) none15. If roots of the equation x 2 + kx + 1 = 0 lies betweenremoved numbers, then set of values of k is :⎛ – 50 ⎞ ⎛ – 50 65 ⎞(A) ⎜ – ∞,⎟ (B) ⎜ ,– ⎟⎝ 7 ⎠ ⎝ 7 8 ⎠⎛ – 57 ⎞(C) ⎜ ,–2⎟ ⎝ 7 ⎠(D) no such k existsPassage # 2 (Ques. 16 to 18)Binomial expansion is a powerful tool to solvedivisibility problems, For example,(1 + x) n – 1 – nx = (1 + nx + n C 2 x 2 + ...) – 1 – nx= ( n C 2 + n C 3 x+...)x 2⇒ (1 + x) n – 1 – nx is divisible by x 2 for all n ≥ 2.Answer the following questions :16. The last three digits of 3 1000 must be :(A) 249 (B) 001 (C) 003 (D) 27917. Let f (n) = 5 2n+2 – 24n – 25, then f (n) is divisible forall n by (select the best option)(A) 12 (B) 24 (C) 576 (D) 24 318. For n ≥ 5, f (n) = 2 2n+1 – 9n 2 + 3n – 2 must bedivisible by(A) 36 (B) 48 (C) 24 (D) noneThis section contains 2 questions (Questions 19, 20).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.+ 8 marks will be given for complete correct answer(i.e. +2 marks for each correct row) and No Negativemarks for wrong answer.19. Match the column :Column-IColumn-II(A) If |z 1 – 6i| = 3, 0 ≤ arg z 2 ≤ π/6 (P) 3 3 – 3then minimum value of|z 1 – z 2 | is(B) If |z – (1 + i)| = r and (Q) 3π/6 ≤ arg z ≤ π/3, thenmaximum value of r is(C) The radius of the circle (R)3 –12z z – (2 – 3i)z – (2 + 3i) z + 4 is(D) If z satisfies the equations (S) 3 3⎛ z – 2 ⎞arg z = π/6 and arg ⎜ ⎟ = 2π/3,⎝ 2 ⎠then |z| is (T) 320. Consider all possible permutations of the letters ofthe word ENDEANOEL.Match the Column-I with the Column-IIColumn-IColumn-II(A) The number of permutations (P) 5!containing the word ENDEA is(B) The number of permutations (Q) 2 × 5!in which the letter E occurs inthe first and the last postions is(C) The number of permutations in (R) 7 × 5!which none of the letters D, L, Noccurs in the last five positions is(D) The number of permutations in (S) 21 × 5!which the letters A, E, O occuronly in odd positions is(T) 5 × 4!XtraEdge for IIT-JEE 69 SEPTEMBER 2011

XtraEdge Test SeriesANSWER KEYIIT- JEE 2012 (September issue)PHYSICSQues 1 2 3 4 5 6 7 8 9Ans C B B A B A A A A,BQues 10 11 12 13 14 15 16 17 18Ans B,C A,BC C,D D B B C B BColumnMatchQues 19 20Ans(A) → P,R,T, (B) → P,R,T(C) → Q,S (D) → P,S(A) → P, (B) → P, (C) → P,Q, (D) → R,SCHEMISTRYQues 1 2 3 4 5 6 7 8 9Ans A A A C B B D A A,B,CQues 10 11 12 13 14 15 16 17 18Ans B,C A,B A,C D A C D A AColumn Ques 19 20Match Ans (A) → P,S (B) → R (C) →P,Q (D) → P,S (A) →Q,S (B) →Q,S (C) →P,R, (D) →P,RMATHEMATICSQues 1 2 3 4 5 6 7 8 9Ans A B A C A A B C A,DQues 10 11 12 13 14 15 16 17 18Ans A,B,C A,C,D A B B A B C AColumn Ques 19 20Match(A) →P,Q,R, (B) →P,Q,S,TAns(C) →P,Q,R,S,T (D) →Q(A) → R, (B) → R, (C) → T, (D) → RIIT- JEE 2013 (September issue)PHYSICSQues 1 2 3 4 5 6 7 8 9Ans D C A A B D D C A,DQues 10 11 12 13 14 15 16 17 18Ans A,B,D A,C A,D C B A C A DColumn Ques 19 20Match Ans (A) → Q (B) → R (C) → P (D) → R (A) → P,Q (B) → R,S (C) → Q, (D) → SCHEMISTRYQues 1 2 3 4 5 6 7 8 9Ans D B B C B C D B B,CQues 10 11 12 13 14 15 16 17 18Ans A,B,D A,B A,B,C A A B B C CColumn Ques 19 20Match Ans (A) → R (B) → S (C) → Q (D) → P (A) →Q,R, (B) →Q,R,S (C) →S (D) →P,SMATHEMATICSQues 1 2 3 4 5 6 7 8 9Ans C D A B A A C B CQues 10 11 12 13 14 15 16 17 18Ans A,C A,D A,B,C,D C B D C C CColumn Ques 19 20Match Ans (A) → P (B) → R (C) → T (D) → Q (A) → P,T (B) → S, (C) → Q, (D) → QXtraEdge for IIT-JEE 70 SEPTEMBER 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 71 SEPTEMBER 2011

Subscription Offer for SchoolsXtraEdge for IIT-JEEIIT JEE becoming more competitive examination day by day.Regular change in pattern making it more challenging."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 72 SEPTEMBER 2011