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Volume - 5 Issue - 10April, 2010 (Monthly Magazine)Editorial / Mailing Office :112-B, Shakti Nagar, Kota (Raj.) 324009Tel. : 0744-2500492, 2500692, 3040000e-mail : xtraedge@gmail.comEditor :Pramod Maheshwari[B.Tech. IIT-Delhi]Cover DesignGovind Saini, Om GocherLayout :Mohammed RafiqCirculation & AdvertisementAnkesh Jain, Praveen ChandnaPh (0744)- 3040007, 9001799502SubscriptionSudha Jaisingh Ph. 0744-2500492, 2500692© Strictly reserved with the publishers• No Portion of the magazine can bepublished/ reproduced without the writtenpermission of the publisher• All disputes are subject to the exclusivejurisdiction of the Kota Courts only.Every effort has been made to avoid errors oromission in this publication. In spite of this, errorsare possible. Any mistake, error or discrepancynoted may be brought to our notice which shall betaken care of in the forthcoming edition, hence anysuggestion is welcome. It is notified that neither thepublisher nor the author or seller will beresponsible for any damage or loss of action to anyone, of any kind, in any manner, there from.Unit Price Rs. 20/-Special Subscription Rates6 issues : Rs. 100 /- [One issue free ]12 issues : Rs. 200 /- [Two issues free]24 issues : Rs. 400 /- [Four issues free]Dear Students,To fly, we have to have resistanceEditorialMotivate YourselfOne of the greatest virtues of human beings is their ability to thinkand act accordingly. The emergence of the techno savvy human fromthe tree swinging ape has really been a long journey. This transitionhas taken a span of countless centuries and lots of thinking caps havebeen involved. Inquisitiveness and aspiration to come out withthe best have been the pillars for man's quest for development.Self-motivation is the sheer force, which pulled him apart anddistinguished him from his primitive ancestors.Many times, in our life, when we are reviving old memories we getinto a phase of nostalgia. We feel that we could have done betterthan what we had achieved. But thinking back won’t rewind thetireless worker called time. All we can do is promise ourselves thatwe will give our very best in the future. But do we really keep up toour mental commitments? I can guess that 90% answers are in thenegative. This is because of that creepy careless attitude which isslowly, but surely entering into our mind. We easily forget the painsof yesterday to relish the joys of today. This is the only time in ourlife, when we can control our fate, by controlling our mind. So it istime to pull up our socks and really motivate ourselves so that wecan give our best shot in the future. Self-motivation is the need of thehour. Only we can control and restrict ourselves. It’s up to us, howwe use our mental capabilities to the best of our abilities.Here are some Funda's for self-motivation. Don't just read themdigest each one of them and apply them and I bet it will make abetter YOU.• The ultimate motivator is defeat. Once you are defeated, youhave nowhere to go except the top.• Then only thing stopping you is yourself.• There is no guarantee that tomorrow will come. So do it today.• Intentions don't count, but action's do.• Don't let who you are, stunt what you want to be.• Success is the greatest motivator.• Your goals must be clear, but the guidelines must be flexible.Try to include these one liners in your scrapbook or on your favoriteposter. You will be sub-consciously tuned to achieve what you want.Also do keep in mind that nothing can control your destiny but you!With Best Wishes for Your Future.Yours trulyOwned & Published by Pramod Maheshwari,112, Shakti Nagar, Dadabari, Kota & Printedby Naval Maheshwari, Published & Printed at112, Shakti Nagar, Dadabari, Kota.Editor : Pramod MaheshwariPramod Maheshwari,B.Tech., IIT DelhiXtraEdge for IIT-JEE 1 APRIL 2010

XtraEdge for IIT-JEE 2 APRIL 2010

Volume-5 Issue-10April, 2010 (Monthly Magazine)NEXT MONTHS ATTRACTIONS Key Concepts & Problem Solving strategy for IIT-JEE. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics, Chemistry & Maths Much more IIT-JEE News. Xtra Edge Test Series for JEE – 2010 & 2011Success Tips for the Month• "The way to succeed is to double yourerror rate."• "Success is the ability to go from failure tofailure without losing your enthusiasm."• "Success is the maximum utilization of theability that you have."• We are all motivated by a keen desire forpraise, and the better a man is, the morehe is inspired to glory.• Along with success comes a reputation forwisdom.• They can, because they think they can.• Nothing can stop the man with the rightmental attitude from achieving his goal;nothing on earth can help the man withthe wrong mental attitude.• Keep steadily before you the fact that alltrue success depends at last upon yourself.INDEXCONTENTSRegulars ..........PAGENEWS ARTICLE 4• President buries 'Time Capsule' on IIT Kanpurcampus• IITs admission criteria set for an overhaulIITian ON THE PATH OF SUCCESS 6Mr. Krishnamurthy RengarajanKNOW IIT-JEE 7Previous IIT-JEE QuestionStudy Time........DYNAMIC PHYSICS 148-Challenging Problems [Set# 12]Students’ ForumPhysics FundamentalsCalorimetry, K.T.G., Heat TransferAtomic Structure, X-Ray & Radio ActivityCATALYST CHEMISTRY 30Key ConceptAromatic HydrocarbonSolubility ProductUnderstanding: Organic ChemistryDICEY MATHS 38Mathematical ChallengesStudents’ ForumKey ConceptCalculusAlgebraTest Time ..........XTRAEDGE TEST SERIES 52Mock Test IIT-JEE Paper-1 & Paper-2Mock Test AIEEEMock Test BIT SATSOLUTIONS 92XtraEdge for IIT-JEE 3 APRIL 2010

President buries 'TimeCapsule' on IIT KanpurcampusKanpur: President Pratibha Patilrecently buried a 'Time Capsule'on the Indian Institute ofTechnology, Kanpur (IIT-K)campus on the occasion of itsgolden jubilee celebrations andalso unveiled a nanosatellitedeveloped by the institute. Thecapsule, which is made of a specialmetal, contains pen-drives, chips,images and several other documentsrelated to the landmark achievementsof the IIT-K.Lauding the nanosatellite Jugnu'sdevelopment team, Patil said itprojects the complex nature oftasks that the students there wereequipped to handle.Congratulating IIT-K students andfaculty, Patil said that the institutehas come a long way in its 50years of its existence, and alsocalled upon the institute's studentsand faculty members to developsuch devices that can harnessenergy in efficient ways withminimal negative impact on theenvironment."It (IIT-K) had made an impact ontechnical education within thecountry, while its studentsthrough their innovations, haveplayed an important role in India,as well as around the world," thePresident said. Jugnu, developedby a team of 50 IIT students, willhelp in collection of informationrelated to floods, drought andother natural calamities.IITs admission criteria setfor an overhaulNew Delhi: The admission criteriafor admission to the premierengineering institutes of thecountry - the Indian Institutes ofTechnology (IITs), is all set for anoverhaul.According to the recommendationof a panel set up under IITKaharagpur's Director, Mr.Damodar Acharya, the JointEntrance Examination (JEE) forundergraduates and GraduateAptitude Test in Engineering(GATE) for postgraduates maysoon not be the only criteria fortheir admission. The committeeproposes to consider the Class XIImarks as well.Presently, the eligibility for astudent to sit for a JEE is 60% inclass XII after which a cut-off isdecided every year for admissioninto an IIT. Following theconsensus of the last meeting ofthe IIT Council to give moreweightage to the school leavingexamination, the committeeproposes to mull over the issue offactoring in class XII examinationresult in the cut-off for admissionto the IITs.The IIT Directors in a meetingwith the Human ResourceDevelopment (HRD) Minister,Kapil Sibal, in Manesar, alsoexpressed their discontent overthe existing pattern for GATE asthey felt that not enough studentswith research orientation werebeing picked up through this exam.Sibal said that each IIT shouldsubmit a proposal within a monthon one area of expertise in higherresearch. Although the admissioncriteria is becoming stringent withthe proposal of IIT-KharagpurDirector's recommendation, theHRD minister retreated with afirm 'No' to IIT-Kanpur Director'sproposal for a fee hike. Accordingto the fee-hike proposal, whichthe IIT-Kanpur's Director framed,it suggested to hike fees by eighttimes. At present, the B.Techstudents pay Rs.50,000 per year astheir fee which the committeeproposed to increase up to Rs.4lakh per annum over the period of10 years with a Rs.35,000 mark-upevery year."The proposal on fee hike shouldbe discussed as the governmentwas planning to set up a HigherEducation Funding Corporationwhich would address poorstudents," said a senior official.During the meeting, Sibal alsoasked the IITs to come upwith their plans for the futureendeavors in a specialized area inwhich they want to emerge asglobal giants by 2020 in fourweeks time.Nanotech is used to treatcancer - IIT-B & DocsIn Mumbai it could have beenIndia’s second nano success if itpasses the muster. Only this nanocreation is revealed in thehealthcare field and it was possiblebecause of partnership betweenoncologists and scientists of theIndian Institute of Technology-Bombay.The joint effort of the IIT -B anddoctors from Tata MemorialXtraEdge for IIT-JEE 4 APRIL 2010

Hospital in Parel and ApolloHospital in Hyderabad has thepossibility to transform treatmentof retinoblastoma—a rare cancerof the retina that mainly affectschildren under two years of age.They have developed a nanoparticlethat could conquer thechild killer.Shirin Thakur, Guntur-basedteenager was suffering fromrecurrent retinoblastoma sinceshe was two years old. Last week,she took the third injection of aspecial mixture into the tissuesaround her left eye whichwas made of nano-particles ofcarboplatin which is commonlyused to treat retinoblastoma.While she was standing with herdoctor in the Apollo hospital,Hyderabad said, “I have beensuffering from attacks ofretinoblastoma in my left eye sinceI was two. Even in the US, theytold me there is no hope but toremove my eye.’’ Now, she has“fuzzy’’ vision in the nearly blindeye. “My vision gets better everyday.’’PM was asked to fine-tuneJEE by Maths WizardPM Manmohan singh was givensuggestions on the improvementof IIT-JEE pattern by MathsWizard Anand Kumar. He pleadedthat poor aspirants should begiven at least three attempts forIIT-JEE exam, as they are usuallythe late starters.Every year Anand’s Super-30offers free accommodation to 30poor students and providescoaching to them free of cost inorder to help them to crack JEE.This noble initiative, has of latereported 100% success rate withall its 30 students making it toIIT’S, without seeking any financialsupport from government or nongovernmentsources.In his 15-minute interaction withthe PM in New Delhi, Anandenlightened him of his effortsusher in a new awakening in Biharby sending 182 poor students toIITs within the past seven years.He further informed the PM thathe has decided to increase thestudent intake to 60 from 30. Verysoon he would open schools fortalented poor children so as toprovide them the right momentumat the secondary school levelitself. He added that governmentshould run coaching programmeson the pattern of Super-30 fortalented students from thecountryside and it should not beconfined to IIT only.Audibly elated Anand told TOIover the phone that Pm pattedhim on his back and alsoinstructed PMO to look into hissuggestions.He said that many problems askedin JEE are of Maths Olympiad levelso students from villages find itdifficult to solve such problemseven if they have sound knowledgeof the subject at +2 level. Hepleaded that the exam should bedesigned in such a way thatconceptual or analytical problemsof the +2 level should be asked inJEE."IIT Rajasthan moves tostate varsity campusJaipur: The Indian Institute ofTechnology, Rajasthan (IIT-R) hasbeen temporarily shifted to theengineering faculty building of JaiNarain Vyas University accordingto a Memorandum of Understanding(MoU) signed in Jodhpur over theweekend. Students of IIT-R earlierhad to rely on IIT-Kanpur andother institutions to attendclasses.IIT-Rajasthan founder-director P.K. Kalra and JNV UniversityRegistrar Nirmala Meena, in thepresence of Chief Minister AshokGehlot who was on a visit toJodhpur, signed the MoU.The CM also inaugurated thework to extend the building.Gehlot pointed out in an officialpress release that theestablishment of IIT in Jodhpurwas recommended and headed bythe noted economist V.S. Vyasafter he studied the facilities andeducational standard of varioustowns in the State."The final decision to nod for anIIT in the city was accepted at thecentral level," added Gehlot. Hefurther announced that theconstruction of a new building inRajasthan would require sometime, and for the time being theclasses would be held in JNVUniversity while those in theextended portion of the buildingwould begin from May his year.Following special efforts made bythe state government Mr. Gehlotsaid that the union governmenthas also nodded to use the newlyconstructed A.S.K. Hostel in theuniversity for IIT students."The establishment of an IIT in thestate would definitely boost itshigher education and will prove tobe a milestone," said Gehlot.According to the MoU, itapproves the functioning of IIT-Rajasthan only on temporary basisfor two years in the engineeringfaculty building.State Technical Education MinisterMahendrajit Singh Malviya and JNVUniversity Vice-ChancellorNaveen Mathur signed the MoU aswitnesses. Among others, JodhpurMP Chandresh Kumari, Barmer-Jaisalmer MP Harish Chaudhary,Jodhpur Mayor RameshwarDadich, Deputy Mayor NiyazMohammed and Collector NaveenMahajan were present on theoccasion.XtraEdge for IIT-JEE 5 APRIL 2010

Success StoryThis article contains story of a person who is successful after graduation from different IIT'sMr. Krishnamurthy RengarajanIIT-B Gold MedallistKnowledge is indeed wealth. Who better exemplifies it thanKrishnamurthy Rengarajan,IIT-B gold medallist (B Techdual-degree course). Krishnamurthy's story is that of hardwork, sheer grit and determination. His undying passion forlearning and excellence has paid off. Coming from a lowermiddle class background, Krishnamurthy has made hisparents proud when he passed with flying colours.His father, who works as a typist at Bharatiya VidyaBhavan is overwhelmed by his son's achievement.Rengarajan, who hails from Tamil Nadu, came to Mumbai28 years ago and settled down in a distant Mumbai suburbof Dombivli. Though the family went through a lot ofhardships initially, he made sure that his children were welleducated."My son always wanted to join the IIT. When people askedhim what if you don't get through the entranceexaminations, he used to say, `there is no question of menot clearing the test," says his proud father. And, of course,he did top all the five years at IIT, a result of sheer hardwork and brilliance, says his mother, barely able to controlher excitement. "I am very happy for him," says RadhaRengarajan. Krishnamurthy did his schooling at the KidlandSchool in Dombivli and pre-degree from V G Vaze Collegeat Mulund. His favourite subject being mathematics it wasobvious that he would pursue a degree in engineering. Hewon the Rakesh Mathur award of Rs 1 lakh (Rs 100,000)during his third year and other scholarships throughout thefour years. Here's what Krishnamurthy had to say on his IITexperience.My IIT experienceThe five years I spent at IIT were the best in my life. I willcherish each and every moment here. I loved everythinghere: the professors are the best one can ever get, thefacilities to study and the extra-curricular activities areexcellent. I made best of friends and thoroughly enjoyedmy college life.I don't think I will ever get this experience anywhere else.On studiesBefore joining IIT, I used to study for 7 to 8 hours daily.After joining IIT, I used to spend about a couple of hours. Idid not go for anything coaching classes. I learnt throughBrilliant Tutorial correspondence course and my preparationbegan after I finished my 10th standard.Why IITIIT is one of the premier institutes in India. I always wantedto get good higher education, so I opted for IIT.My mantra for successThere is no short cut to success. One has to work veryhard, put in a lot of effort, should have a problem-solvingmentality and a right approach to every problem. Myparents always stood by me, their support has beeninvaluable and am overwhelmed.Advice to IIT aspirantsWork hard. You have to spend a lot of time preparing asexams are getting more and more competitive. You mustalso have problem-solving skills.InterestsSolving math puzzles, reading books. I used to play cricket,but now I don't get the time.Next moveMoney is the least important thing for Krishnamurthy. So nojobs for the time being. "I have been selected for thescholarship programme at Stanford University for a PhD inoperations research. I would like to research on optimisingcomputer networks and operation systems. Quality research isavailable abroad. After the PhD programme I would like tojoin any academia of good repute and continue my researchactivities. Among corporates, I admire Google. It is the onecompany that reflects perfection, hard work and efficiency."Will you come back to India?Of course, I will. The brain drain phenomenon is dying out.It's the time for reverse brain.XtraEdge for IIT-JEE 6 APRIL 2010

KNOW IIT-JEEBy Previous Exam QuestionsPHYSICS1. A spherical ball of mass m is kept at the highest pointin the space between two fixed, concentric spheres Aand B (see figure in solution). The smaller sphere Ahas a radius R and the space between the two sphereshas a width d. The ball has a diameter very slightly lessthan d. All surface are frictionless. The ball is given agentle push (towards the right see figure in solution)The angle made by the radius vector of the ball withthe upward vertical is denoted by θ [IIT-2002](a) Express the total normal reaction force exerted bythe sphere on the ball as a function of angle θ.(b) Let N A and N B denote the magnitudes of thenormal reaction forces on the ball exerted by thesphere A and B, respectively. Sketch the variations ofN A and N B as functions of cos θ in the range 0 ≤ θ ≤ πby drawing two separate graphs in your answer book,taking cos θ on the horizontal axes.Sol. The ball is moving in a circular motion. Thenecessary centripetal force is provided by(mg cos θ – N). ThereforeBARDCd/2θ θmgcosθmgN AVmgsinθmv 2mg cos θ – N A =…(i)⎛ d ⎞⎜R + ⎟⎝ 2 ⎠According to energy conservation1 mv 2 ⎛ d ⎞= mg ⎜R + ⎟ (1 – cos θ) …(ii)2 ⎝ 2 ⎠From (i) and (ii) N A = mg (3 cos θ – 2) …(iii)The above equation shows that as θ increases N Adecreases. At a particular value of θ, N A will becomezero and the ball will lose contact with sphere A. Thiscondition can be found by putting N A = 0 in eq. (iii)0 = mg (3 cos θ – 2)∴ θ = cos –1 ⎛ 2 ⎞⎜ ⎟⎝ 3 ⎠2mgN AmgThe graph between N A and cos θFrom eq (iii) when θ = 0, N A = mg.1cos θ –1When θ = cos –1 ⎛ 2 ⎞⎜ ⎟ ; NA = 0⎝ 3 ⎠The graph is a straight line as shown.When θ > cos –1 ⎛ 2 ⎞⎜ ⎟⎝ 3 ⎠mv 2N B – (– mg cos θ) =dR +2mv⇒ N B + mg cos θ =2⎛ d ⎞⎜R + ⎟⎝ 2 ⎠Using energy conservation1 2 ⎡⎛d ⎞ ⎛ d ⎞ ⎤mv = mg ⎢⎜R+ ⎟ − ⎜R+ ⎟ cos θ⎥ 2 ⎣⎝2 ⎠ ⎝ 2 ⎠ ⎦N B5mg2mgcos θ…(iv)mv 2 = 2 mg [1 – cos θ] …(v)⎛ d ⎞⎜R + ⎟⎝ 2 ⎠From (iv) and (v) we getN B + mg cos θ = 2 mg – 2mg cos θN B – mg (2 – 3 cos θ)When cos θ = 32 , NB = 0When cos θ = – 1, N B = 5 mgTherefore the graph is as shown.2. A cylindrical block of length 0.4 m and area of crosssection 0.04 m 2 is placed coaxially on a a thin metaldisc of mass 0.4 Kg and the same cross section. Theupper face of the cylinder is maintained at a constanttemperature of 400 K and the initial temperature ofthe disc is 300 K. If the thermal conductivity of thematerical of all cylinder is 10 watt/kg. K, how longwill it take for the temperature of the disc to increasesto 350 K? Assume, for purpose of calculation, thethermal conductivity of the disc to be very high andXtraEdge for IIT-JEE 7 APRIL 2010

v → →0 E →BBut because here, î = , ĵ = =v0E Bv0 × Band kˆ =v0B⎛ → ⎞⎛ ⎞→ ⎜ v0⎟ ⎛ qBt ⎞ ⎜→ E ⎟So, v 0 = ⎜ ⎟ v 0 cos ⎜ ⎟ +v0⎝ m⎜ ⎟ ⎠ E⎝ ⎠⎝ ⎠+v0 × B ⎛ qBt ⎞v 0 sin ⎜ ⎟v B ⎝ m ⎠0qE tm4. A long solenoid of radius 'a' and number of turns perunit length n is enclosed by cylindrical shell of radiusR, thickness d(d

CHEMISTRY6. From the following data, form the reaction betweenA and B.[IIT-1994][A] [B] Initial rate (mol L –1 s –1 )mol L –1 mol L –1 300 K 320 K2.5 ×10 –4 3.0 ×10 –5 5.0 ×10 –4 2.0 × 10 –35.0 × 10 –4 6.0 × 10 –4 4.0 × 10 –3 –1.0 × 10 –3 6.0 × 10 –5 1.6 × 10 –2 –Calculate(a) the order of reaction with respect to A and withrespect to B,(b) the rate constant at 300 K,(c) the energy of activation,(d) the pre exponential factor.Sol. Rate of reaction = k[A] l [B] mwhere l and m are the order of reaction with respectto A and B respectively. From the given data, weobtain following expressions :5.0 × 10 –4 = k[2.5 × 10 –4 ] l [3.0 × 10 –5 ] m ...(i)4.0 × 10 –3 = k[5.0 × 10 –4 ] l [6.0 × 10 –5 ] m ...(ii)1.6 × 10 –2 = k[1.0 × 10 –3 ] l [6.0 × 10 –5 ] m ..(iii)From eq. (ii) and eq. (iii), we get4.0×101.6×10−3−2⎛ 5.0×10= ⎜⎝ 1.0×10−4−3l⎞⎟⎠or 0.25 = (0.5) lor (0.5) 2 = (0.5) lor l = 2From eq. (i) and eq. (ii), we getor5.0×101.0×10−4−3⎛ 2.5 × 10= ⎜⎝ 5.0 × 10m1 1 ⎛ 1 ⎞= × ⎜ ⎟⎠8 4 ⎝ 2−4−42⎞ ⎛ 3.0 × 10⎟ ⎜⎠ ⎝ 6.0 × 10−5−5m1 ⎛ 1 ⎞or = ⎜ ⎟⎠2 ⎝ 2or m = 1(b) At T 1 = 300 K,−4Rate of reaction 5.0×10k 1 = =2 1−42[A] [B] [2.5×10 ] [3.0×10= 2.67 × 10 8 L 2 mol –2 s –1(c) At T 2 = 320 K,Rate of reactionk 2 =2 1[A] [B]−32.0×10=−42−5[2.5×10 ] [3.0×10 ]= 1.067 × 10 9 L 2 mol –2 s –1We know, 2.303 logkk21=E ⎛ ⎞a T2− T1⎜⎟R ⎝ T1T2 ⎠⎞⎟⎠m−5]91.067 × 10 ⎛ 320 − 300 ⎞or 2.303 log =8⎜ ⎟2.67 × 10 8.314 ⎝ 320×300 ⎠E aEor 2.303 × 0.6017 = a ⎛ 20 ⎞⎜ ⎟8.314 ⎝ 320×300 ⎠2 .303×0.6017×8.314×320×300or E a =20= 55.3 kJ mol –1(d) According to Arrhenius equation,/ RTk = Ae −E aor 2.303 log k = 2.303 log A – RTAt 300 K,2.303 log (2.67 × 10 8 355.3×10) = 2.303 log A –8.314×300or 2.303 × 8.4265 = 2.303 log A – 22.1719 .4062 + 22.17 41.5762or logA == = 18.05312.303 2.303A = Antilog 18.0531 = 1.13 × 10 18 s –17. 1 g of a mixture containing equal number of moles ofcarbonates of two alkali metals, required 44.4 ml of0.5 N HCl for complete reaction. The atomic weightof one metal is 7, find the atomic weight of othermetal. Also calculate the amount of sulphate formedon quantitative conversion of 1.0 g of the mixture intwo sulphates.[IIT-1972]Sol. Let, Mass of one alkali metal carbonate M 2 CO 3 = xgThen, mass of other alkali metal carbonate M 2´CO 3= (1 – x)gStep 1.Molecular mass 74Equivalent mass of M 2 CO 3 ==2 2= 37 (at mass of M = 7)E aMeq. of M 2 CO 3 = 37x ×1000xMoles of M 2 CO 3 = 74Step 2Molecular massEquivalent mass of M 2´CO 3 =22 m + 60=2( m = atomic mass of M´)2(1 − x)Meq. of M 2´CO 3 = × 10002m + 601−xMoles of M 2´CO 3 =2m + 60Step 3. Meq. of HCl = N HCl × V HCl = 0.5 × 44.4Step 4. According to the question,Moles of M 2 CO 3 = Moles of M 2´CO 3x 1−xor = ...(i)74 2m + 60XtraEdge for IIT-JEE 10 APRIL 2010

And Meq. of M 2 CO 3 + Meq . of M 2´CO 3= Meq. of HClorx 2(1 − x)× 1000 +37 2m + 60× 1000= 0.5 × 44.4 ...(ii)Solving eq. (i) and (ii), we getm = 23and x = 0.41∴ Mass of M 2 CO 3 = x = 0.41 gand Mass of M 2´CO 3 = 1 – x = 0.59 gStep 5.Molecular massEquivalent mass of M 2 SO 4 =2110= = 552WM Meq. of M 2 SO 4 = 2 SO 455× 1000But, Meq. of M 2 SO 4 = Meq. of M 2 CO 3∴WM 2 SO 40.41× 1000 = × 10005537or WM 2 SO4= 0.6095 gStep 6.Molecular massEquivalent mass of M 2´SO 4 =2142= = 712WMMeq. of M 2´SO 4 =2 ´ SO 471× 1000But, Meq. of M 2´SO 4 = Meq. of M 2´CO 3∴WM2 ´ SO 42× 0.59× 1000 =71106× 1000or W = 0.7904 gM 2 ´ SO 4∴ Total mass of sulphates = WM 2 SO+ W4 M2´SO4= 0.6095 + 0.7904= 1.3999 g8. 0.9 g of a solid organic compound (molecular mass90), containing carbon, hydrogen and oxygen, washeated with oxygen corresponding to a volume of224 ml at STP. After combustion the total volume ofthe gases was 560 ml at STP. On treatment withpotassium hydroxide, the volume decreased to 112ml. Determine the molecular formula of thecompound.[IIT-1972]Sol. Given that,Mass of solid organic compound = 0.9 gMolecular mass of organic compound = 90∴ No. of moles of organic compound available0. 9= = 0.01 90Volume of O 2 taken = 224 mlVolume of O 2 used = 224 – 112 = 112 ml22400 ml O 2 = 1 mol.112∴ 112 ml O 2 = = 0.005 mol22400112∴ At STP, no. of moles of O 2 used = 22400= 0.005 molVolume of CO 2 obtained = 560 – 112 = 448 ml448∴ At STP, no. of moles of CO 2 used = 22400= 0.02 mol0.01 mol organic compound yields = 0.02 mol CO 2 .∴ 1 mol organic compound yields= 2 mol CO 2 or 2 mol C∴ The molecular formula of organic compound isC 2 H y O z .The reaction is :C 2 H y O z + 21O2 → 2CO 2 + 2yH2 OEquating no. of oxygen atoms,z + 1 = 4 + 2yorz = 3 + 2yMolecular mass of C 2 H y O z = 2 × 12 + y × 1 + z × 16⎛ y ⎞Hence, 2 × 12 + y × 1 + ⎜3 + ⎟ × 16 = 90⎝ 2 ⎠or y = 2and z = 3 + 2y = 4∴ The molecular formula of organic compound isC 2 H 2 O 4 .9. (a) Write the intermediate steps for each of thefollowing reactions.H O(i) C 6 H 5 CHOHC ≡ CH ⎯ ⎯ +3→ C 6 H 5 CH=CHCHO(ii)H +OHO CH 3(b) Show the steps to carry out the followingtransformations :(i) Ethylbenzene → benzene(ii) Ethylbenzene → 2-phenylpropionic acid[IIT-1998]Sol. (a) (i)C 6 H 5 CH(OH)C ≡ CH⊕C 6 H 5 CH = C = CHOH –H +C 6 H 5 CH – C ≡ CHOH 2+⊕C 6 H 5 CH – C ≡ CH–H 2 OTautomerismC 6 H 5 CH = C = CHOH C 6 H 5 CH = CHCHOXtraEdge for IIT-JEE 11 APRIL 2010

(ii)OHH + CH 3OH⊕PorClPClClOCH 3OCH 3ClClCl(b) (i)C 2 H 5alk. KMnO 4Ethyl benzene(ii)CH 2 CH 3BrOHMg +Br 2/hvCOOHBenzoic acidCH 3 CHCOOH2-Phenylpropionic acidCHBrCH 3NaOH–H 2OMgEtherH 2O/H +H⊕COONaNaOH+CaOSodium benzoate+ Na 2 CaO 3BenzeneCH 3 CH – MgBrCO 2CH 3 CHCOOMgBr10. Interpret the non-linear shape of H 2 S molecule andnon-planar shape of PCl 3 using valence shell electronpair repulsion (VSEPR) theory. (Atomic numbers :H = 1, P = 15, S = 16, Cl = 17.) [IIT-1998]Sol. In H 2 S, no. of hybrid orbitals = 21 (6 + 2 – 0 + 0) = 4Hence here sulphur is sp 3 hybridised, so16S = 1s 2 , 2s 2 2p 6 2 2 1 1, 3 s 3px3py3pz14424343sp hybridisationSorH HH HDue to repulsion between lp - lp; the geometry ofH 2 S is distorted from tetrahedral to V-shape.In PCl 3 , no. of hybrid orbitals = 21 [5 + 3 – 0 + 0] = 4Hence, here P shows sp 3 -hybridisation15P = 1s 2 , 2s 2 2p 6 2 1 1 1, 3 s 3px3py3pz14424343sp hybridisationSThus due to repulsion between lp – bp, geometry isdistorted from tetrahedral to pyramidal.TMATHEMATICS11. 7 relatives of a man comprises 4 ladies and3 gentlemen; his wife has also 7 relatives; 3 of themare ladies and 4 gentlemen. In how many ways canthey invite a dinner party 3 ladies and 3 gentlemen sothat there are 3 of man’s relative and 3 of the wife’srelatives ?[IIT-1985]Sol. The possible cases are :Case I : A man invites 3 ladies and women invites 3gentlemen⇒ 4 C 3 . 4 C 3 = 16Case II : A man invites (2 ladies, 1 gentleman) andwomen invites (2 gentlemen, 1 lady)⇒ ( 4 C 2 . 3 C 1 ).( 3 C 1 . 4 C 2 ) = 324Case III : A man invites (1 lady, 2 gentlemen) andwomen invites (2 ladies, 1 gentleman)⇒ ( 4 C 1 . 3 C 2 ).( 3 C 2 . 4 C 1 ) = 144Case IV : A man invites (3 gentlemen) and womeninvites (3 ladies)⇒ 3 C 3 . 3 C 3 = 1∴ Total number of ways = 16 + 324 + 144 + 1 = 48512. Let n be a positive integer and(1 + x + x 2 ) n = a 0 + a 1 x + ..... + a 2n x 2nShow that a 2 0 – a 2 1 + ...... + a 2 2n = a n . [IIT-1994]Sol. (1 + x + x 2 ) n = a 0 + a 1 x + .... + a 2n x 2n ...(1)Replacing x by –1/x. we obtainn⎛ 1 1 ⎞ a⎜1− +2⎟ = a 0 – 1a +2a3 a2n– +...+ ..(2)⎝ x x ⎠ x2 32nx x xNow, a 2 0 – a 2 1 + a 2 2 – a 2 3 + ... + a 2 2n = coefficient ofthe term independent of x in[a 0 + a 1 x + a 2 x 2 + ... + a 2n x 2n ]⎡ a1a2a2n ⎤⎢a0− + −...+22n⎥⎣ x x x ⎦= coefficient of the term independent of x inn(1 + x + x 2 ) n ⎛ 1 1 ⎞⎜1−+2⎟⎝ x x ⎠nBut, R.H.S. = (1 + x + x 2 ) n ⎛ 1 1 ⎞⎜1−+2⎟⎝ x x ⎠2 n 2( 1+x + x ) (x − x +=2nx1)nXtraEdge for IIT-JEE 12 APRIL 2010

2 2 2 n[( x + 1) − x ]=2nx2 41+2x + x=2n2 4 n( 1+2x + x )=2n2( − x )xxThus, a 2 0 – a 2 1 + a 2 2 – a 2 23 + ... a 2n= coefficient of the term independent of x in12n (1 + x 2 + x 4 ) nx= coefficient of x 2n in (1 + x 2 + x 4 ) n= coefficient of t n in (1 + t + t 2 ) n = a n13. Solve for x the following equation :log (2x + 3) (6x 2 + 23x + 21)= 4 – log (3x + 7) (4x 2 + 12x + 9) [IIT-1987]Sol. log (2x + 3) (6x 2 + 23x + 21)= 4 – log (3x + 7) (4x 2 + 12x + 9)⇒ log (2x + 3) (2x + 3) . (3x + 7) = 4 – log (3x + 7) (2x + 3) 2⇒ 1 + log (2x + 3) )(3x + 7) = 4 – 2log (3x + 7) (2x + 3)Put log (2x + 3) (3x + 7) = y⇒ y + y2 – 3 = 0 ⇒ y 2 – 3y + 2 = 0⇒ (y – 1) (y – 2) = 0⇒ y = 1 or y = 2⇒ log (2x + 3) (3x + 7) = 1or log (2x + 3) (3x + 7) = 2⇒ 3x + 7 = 2x + 3 or (3x + 7) = (2x + 3) 2⇒ x = – 4 or 3x + 7 = 4x 2 + 12x + 94x 2 + 9x + 2 = 04x 2 + 8x + x + 2 = 0(4x + 1) (x + 2) = 0x = – 2, –1/4∴ x = – 2, –4, –1/4But, log exists only when, 6x 2 + 23x + 21 > 0,4x 2 + 12x + 9 > 0,2x + 3 > 0 and 3x + 7 > 0⇒ x > –3/2∴ x = –1/4 is the only solution.14. Let f[(x + y)/2] = {f(x) + f(y)} / 2 for all real x and y,if f´(0) exists and equals –1 and f(0) = 1, find f(2).[ΙΙΤ−1992]⎛ x + y ⎞ f (x) + f (y)Sol. f ⎜ ⎟ =⎝ 2 ⎠ 2∀ x, y ∈ R (given)Putting y = 0, we get⎛ x ⎞ f (x) + f (0)f⎜⎟ =⎝ 2 ⎠ 21= [1 + f(x)] [Q f(0) = 1]2⇒ 2f(x/2) = f(x) + 1⇒ f(x) = 2f(x/2) – 1 ∀ x, y ∈ R ...(1)Since f´(0) = –1, we getf (0 + h) − f (0)f (h) −1⇒ lim = – 1 ⇒ limh → 0 hh → 0 hnNow, let x ∈ R then applying formula ofdifferentiability.⎛ 2x + 2h ⎞f ⎜ ⎟ − f (x)f (x + h) − f (x)2f´(x) = lim = lim⎝ ⎠h → 0 hh→0hf (2x) + f (2h)− f (x)= lim2h→0h1 ⎧ ⎛ 2x ⎞ ⎛ 2h ⎞ ⎫⎨2f⎜ ⎟ −1+2f ⎜ ⎟ −1⎬− f (x)2 2 2= lim⎩ ⎝ ⎠ ⎝ ⎠ ⎭h → 0h[using equation (1)]1{2f (x) –1+2f (h) −1}− f (x)= lim 2h→0hf (h) −1= lim = –1h → 0 hTherefore f´(x) = – ∀ x ∈ R⇒∫f´(x)dx =∫−1 dx⇒ f(x) = – x + k where k is a constant.But f(0) = 1, therefore f(0) = – 0 + k⇒ 1 = k⇒ f(x) = 1 – x ∀ x ∈ R⇒ f(2) = – 115. If (a + bx)e y/x = x, then prove thatx 3 2d y ⎛ dy ⎞= ⎜ x − y⎟ [IIT-1983]2dx ⎝ dx ⎠Sol. (a + bx).e y/x = x ...(1)Differentiating both sides, we get⎧ dy ⎫⎪x− y ⎪(a + bx).e y/x .dx⎨ ⎬ + be y/x = 12⎪ x ⎪⎩⎭⎧ dy ⎫⎨x− y⎬dx⇒ x.⎩ ⎭+ be b/x = 1, (using (1))2xdy yor – + be y/x = 1,dx xAgain differentiation both sides,dy⎧ dy ⎫2 x − yd y–dx⎪x− y ⎪+ be y/x .dx22⎨ ⎬ = 0 ..(2)2dx x⎪ x ⎪⎩⎭dy2 x − yd yfrom (2), –dx ⎛ dy y ⎞22⎜ − ⎟ = 0dx x ⎝ dx x ⎠⇒ x 3 2d y ⎛ dy ⎞= ⎜ x − y⎟ 2dx ⎝ dx ⎠22XtraEdge for IIT-JEE 13 APRIL 2010

Physics Challenging ProblemsSet #12This 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 Sharma<strong>Solution</strong>s given in same issueDirector Academics, Jodhpur Branch1. Four infinite thin current carrying sheets are placed inYZ plane. The 2D view of the arrangement is as shownin fig. Direction of current has also been shown in thefigure. The linear current density. i.e. current per unitwidth in the four sheets are I, 2I, 3I and 4I, respectively.Y I II III IVB2µ 0 I(D)+µ 0 I–µ 0 Ia2a3a 4a 5aXaaaThe magnetic field as a function of x is bestrepresented by5µ 0 I(A)2µ 0 Iµ 0 I(B)3µ 0 Iµ 0 I–µ 0 I(C)+µ 0 IBBB4µ 0 I–µ 0 Iaaaa2a 3a 4a 5a2a 3a 4a 5a2a 3a4a 5aXXXX2. Match the columnColumn – I(A) a charge particle ismoving in uniformelectric and magneticfields in gravity freespace(B) a charge particle ismoving in uniformelectric, magneticand gravitationalfields(C) a charge particle ismoving in uniformmagnetic andgravitational fields(where electric fieldis zero)(D) A charge particle ismoving in onlyuniform electric fieldColumn – II(P) Velocity of theparticle may beconstant(Q) Path of the particlemay be straight line(R) Path of the particlemay be circular(S) Path of the particlemay be helical(T) None3. Magnetic flux in a circular coil of resistance 10Ωchanges with time as shown in fig. Cross indicates adirection perpendicular to paper inwards.Match the following :XtraEdge for IIT-JEE 14 APRIL 2010

10–10Column – Iφ(Magnetic flux)8 10 14 162 6t(s)× × × × × ×× × × × × ×× × × × × ×× × × × × ×× × × × × ×× × × × × ×Column – II(A) At 1s, induced current is (P) Clockwise(B) At 5s, induced current is (Q) Anticlockwise(C) At 9s, induced current is (R) Zero(D) At 15s, induced current is (S) 2A(T) None4. A conducting rod of length l is moved at constantvelocity v 0 on two parallel, conducting, smooth, fixedrails, which are placed in a uniform constant magneticfield B perpendicular to the plane of the rails asshown in figure. A resistance R is connected betweenthe two ends of the rail. Then which of the followingis/are correct ?R⊗ Bv 0⊗(A) The thermal power dissipated in the resistor isequal to the rate of work done by an externalperson pulling the rod(B) If applied external force is doubled, then a partof the external power increases the velocity ofthe rod(C) Lenz’s law is not satisfied if the rod isaccelerated by an external force(D) If resistance R is doubled, then power required tomaintain the constant velocity v 0 becomes half5. The x-z plane separates two media A & B ofrefractive indices µ 1 = 1.5 & µ 2 = 2. A ray of lighttravels from A to B. Its directions in the two media→are given by unit vectors µ 1 = a i + b j,µ 2 = c i + b j.Thena 4a 3(A) = (B) =c 3c 4(C)b 4b 3= (D) =d 3d 46. Two converging lenses of the same focal length f areseparated by distance 2f. The axis of the second lensis inclined at angle θ = 60ºwith respect to the axis ofthe first lens. A parallel paraxial beam of light isincident from left side of the lens. Then2f∧60º(A) Final image after all possible refraction willformed at optical centre of first lens(B) Final image after all possible refraction willformed at optical centre of second lens(C) Final image after all possible refraction willformed at distance f from second lens(D) Final image after all possible refraction willformed at distance f from first lens7. If C v for an ideal gas is given by C v = (3 + 2T)R,where T is absolute temperature of gas, then theequation of adiabatic process for this gas is(A) VT 2 = constant(B) VT 3 e -2T = C(C) VT 2 e 2T = constant(D) VT 3 e 2T = constant8. The pressure of one mole of ideal gas variesaccording to the law P = P 0 – αV 2 where P 0 & α arepositive constant constants. The highest temperaturethat gas may attain -(A)(C)2P0⎛ P0⎞⎜ ⎟3R ⎝ 3α⎠P0⎛ P0⎞⎜ ⎟R ⎝ 3α⎠1/ 21/ 2(B)(D)3P0⎛ P0⎞⎜ ⎟2R ⎝ 3α⎠P0 ⎛ P 0 ⎞⎜ ⎟R ⎝ α ⎠∧1/ 21/ 2→∧∧XtraEdge for IIT-JEE 15 APRIL 2010

8 Questions<strong>Solution</strong>Set # 11Physics Challenging Problemswere Published in March Issue1.[C] The given circuit as an R-L-C series circuit whenfrequency of the source varies the impedance of theR-L-C series circuit varies and correspondingly thecurrent in the circuit get variedImpedance variation and current variation are shownin figure.R = Zminimumf 1 f=fr f 2 fI =I0II 0f 1 f=fr f 2 f23[C] At frequency f 1 currentI 1 V 1 200 20± 0= . = . = = 102 2 R 2 10 2Watt less currentI µ = Isinφ1I µ = 10 2. = 10 Amp.2As φ = 45ºRbecause cos φ = R / Z = =R 22 Amp.V V 2004[A] At frequency current I = = = = 20 Amp.Zm R 10Potential difference across capacitor12∆f = f 2 -f 1∆f = f 2 -f 1V C = I C .X C = 1.X C = 20.X CAt frequency f 1 X C > X L Power factor – leadingnature of circuit is capacitanceAt frequency f 2 X L > X C Power factor – leadingnature of circuit is inductingAt frequency f 1 and frequency f 2 impedance Z=Because of the fact –V VAs I 0 = = ........(i)Zhm R⇒± =I 0 V=2 ZV / 2 V =2 Z2 . R⇒V V =R 2 Zf 1 < f < f 2and R 22.[A] ∆ f = f 2 − f1= B and width of R-L-C series circuit1= .R / L2πCharge on capacitor Q C = C.V C = C. 20X C1= C. (2v)ωC5.[C] Longest wavelengthλ6.[B] f20 20= =ω 2πf= (5π)= (5π)−1−120 1= =2π(50)5πCoulombcbv + vs350 × 0.8 × 5= =f 600v= fv − vmax =maxs350== 607Hz350 − 0.8 × 57.[A] 345.5/346.0 × 600 = 599 Hz8.[A]dydx= −1 dy d y 1& = +v dt2 2dx v2d2dty20.59mXtraEdge for IIT-JEE 16 APRIL 2010

XtraEdge for IIT-JEE 17 APRIL 2010

8 Questions<strong>Solution</strong>Set # 12Physics Challenging ProblemsPublished in this Issue1.[C] Magnetic field due to infinite current carryingµ Jsheet is given by B = 0, where J is linear current2density.II IIIµ 0 J2µ 0 J2µ 0 J2µ 0 J2(a) (b)Fig. (a) and (b) represent the direction of magneticfield due to current carrying sheets. For x < a,µ 0Jµ 0J(2J)µ 0 (3J) µ 0 (4J)Bresul tant = − − +2 2 2 2For a < x < 2a,µ 0Jµ 0 (2J) µ 0 (3J) µ 0 (4J)tant = − − + = −2 2 2 2For 2a < x < 3a,µ 0Jµ 0 (2J) µ 0 (3J) µ 0 (4J)Bresul tant = + − − = 02 2 2 2So, the required curve isBBresul µ 0Oa 2a 3a 4a 5aXJ(C) This situation is similar to part (i)(D) In a uniform electric field, path can be only3. A → Q B → R C → P D → Q(A) At t = 1s, flux is increasing in the inwarddirection, hence induced e.m.f. will be inanticlockwise direction.(B) At t = 5s, there is no change in flux, so inducede.m.f. is zero(C) At t = 9s, flux is increasing in upward directionhence induced e.m.f. will be in clockwise direction.(D) At t = 15s, flux is decreasing in upwarddirection, so induced e.m.f. will be inanticlockwise direction.4.[A, B, D]Rate of work done by external agent isde/dt = BIL.dx/dt = BILv and thermal powerdissipated in resistor = eI = (BvL) I clearly both areequal, hence (A).If applied external force is doubled, the rod willexperience a net force and hence acceleration. As aresult velocity increase, hence (B).Since, I = e/ROn doubling R, current and hence required powerbecome half.Since, P = BILvHence (D)2. A → P,Q,S B → P,Q,R,SC → P,Q,R,S D → Q(A) Velocity of the particle may be constant, if forcesof electric and magnetic fields balance eachother. Then, path of particle will be straight line.Also, path of particle may be helical if magneticand electric fields are in same direction. But pathof particle cannot be circular. Path can becircular if only magnetic field is present, or ifsome other forces is present which can cancel theeffect of electric field.(B) Here, all the possibilities are possible dependingupon the combinations of the three fields.5.[A]→∧^jxzn 1 sini = n 2 sinr→1.5(µ 1×j) = 2( µ 2×j)∧∧∧1.5(a i + b j) × j = 2[(c i + d j) × j]∧=∧1 .5a k 2c ka 20 4= =c 1.5 3∧∧∧∧XtraEdge for IIT-JEE 18 APRIL 2010

6.[A]I 2f–f sin60º f cos60º +fI 1xu = -f cos60ºf = +f1 1 1= −f v − f cos 60º1 1 2= +f v f1 2 1− =f f vv = -ff = cos 60ºxf= xcos 60ºx = 2f∴ final image will formed at optical centre of firstlens.7.[C] C v = (3 + 2T)RdQ = dU + PdVadiabatic process dQ = 00 = Rn (3 + 2T)dT + PdVnRT0 = Rn(3 + 2T)dT + dVVdV ⎛ 3 + 2T ⎞∫−= ∫⎜⎟dTV ⎝ T ⎠– log V = 3 logT + 2T + C– logV – logT 3 = 2T + Clog VT 3 = 2T + CVT 3 = e 2TVT 3 e -2T = C8.[A] P = P 0 − αVPV = RTRT2= P 0 − αVV3P0 V αVT = −R RdT = 0dV2P03αV− =R R0PV = 03 α2Now put V in T.• Saturn’s rings are made up of particles of ice, dustand rock. Some particles are as small as grains ofsand while others are much larger thanskyscrapers.• Jupiter is larger than 1,000 Earths.• The Great Red Spot on Jupiter is a hurricane-likestorm system that was first detected in the early1600’s.• Comet Hale-Bopp is putting out approximately250 tons of gas and dust per second. This isabout 50 times more than most comets produce.• The Sun looks 1600 times fainter from Pluto thanit does from the Earth.• There is a supermassive black hole right in themiddle of the Milky Way galaxy that is 4 milliontimes the mass of the Sun.• Halley’s Comet appears about every 76 years.• The orbits of most asteroids lie partially betweenthe orbits of Mars and Jupiter.• Asteroids and comets are believed to be ancientremnants of the formation of our Solar System(More than 4 billion years ago!).• Comets are bodies of ice, rock and organiccompounds that can be several miles in diameter.• The most dangerous asteroids, those capable ofcausing major regional or global disasters, usuallyimpact the Earth only once every 100,000 yearson average.• Some large asteroids even have their own moon.• Near-Earth asteriods have orbits that cross theEarth’s orbit. These could potentially impact theEarth.• There are over 20 million observable meteorsper day.• Only one or two meteorites per day reach thesurface of Earth.• The largest found meteorite was found in Hoba,Namibia. It weighed 60 tons.XtraEdge for IIT-JEE 19 APRIL 2010

PHYSICSSStudents'ForumExpert’s <strong>Solution</strong> for Question asked by IIT-JEE Aspirants1. A beam of length L, breadth b and thickness d whenloaded by a weight Mg in the middle, a depression eis produced in it. By measuring this depression e, thevalue of Young's modulus of the material of beamcan be calculated by using the expression3M g LY =34 b d eFollowing are the values of different physicalquantities obtained in one set of observations on thisexperiment :M = 1000 gms, L = 200 cm,b = 2.54 cm, d = 0.620 cm, e = 0.1764 cm.If M is measured by spring balance, L by metre scale,b by vernier calipers, d by screw gauge and e byspherometer, then what will be the maximumpossible percentage errors in Y ?3M g LSol. Given that Y =34 b d eTaking log on both sides of above equation, we getlog Y = log M + log g + 3 log L – log 4– log b – 3 log d – log eDifferentiating above equations, we have :∆Y ∆M ∆ L ∆ ∆ d ∆e= + 3 – – 3 –Y M L bbd eIn order to calculate maximum possible error, weshall convert negative sign into positive sign.∆Y ∆M ∆ L ∆ ∆ d ∆e∴ = + 3 + + 3 +Y M L bbd eNow, least counts of the different measuringinstruments used in the experiment are as under :Least count of spring balance = 5 gm i.e. ∆M = 5gmLeast count of metre scale = 0.1 cm i.e. ∆L = 0.1 cmLeast count of vernier callipers = 0.01 cmi.e. ∆b = 0.001 cmLeast count of screw gauge = 0.001 cmi.e. ∆d = 0.001 cmLeast count of spherometer = 0.005 cmi.e ∆e = 0.005 cm∆ Y 5 3× 0. 1 0. 01 3× 0.001 0. 005∴ = + + + +Y 1000 200 2.54 0.62 0.1764= 0.005 + 0.0015 + 0.00393 + 0.00484 + 0.02834= 0.0436 or 4.36%Hence the maximum possible percentage error is4.36%.2. A small glass ball is released from rest from the topof a smooth incline plane of constant base b. find theangle of inclination of the plane for minimum time ofmotion of the glass ball.AθCBbSol. Let the angle of inclination be θ. If, the glass ballreaches the bottom B of the inclined plane after atime, say t, the equation of motion along the plane isgiven asAB = (V A )t + 21 (g sin θ)t2AL = b sec θg sin θCθgbBSince the glass ball is released from rest V A = 0,hence AB = 21 (g sin θ)t2...(1)Putting AB = (BC) sec θ = b sec θ, in equation (1),we obtaint =2bg sin θcosθ=4bg sin 2θFor t to be minimum, sin 2θ is maximum∴ sin 2θ| max = 1ππor, 2θ = or, θ = 2 43. Two particles, both of mass m, attract each other withrthe force F(rr α) = – rˆr 2where α is a positive constant. At a certain moment(t = 0), the distance between the particles is R, andtheir velocities are⎧v1= v0xˆ⎨⎩v2= −2v0xˆXtraEdge for IIT-JEE 20 APRIL 2010

Assuming the two-particle system reaches aminimum of kinetic energy at a certain moment andat a certain finite distance between the particles (inthe laboratory frame), find the distance between theparticles at that moment and the value of thatminimal kinetic energy.Sol. For the sake of convenience, we will first solve theproblem in the frame of the centre of mass. Then, wewill transform the results into the laboratory frame.We will determine the velocity of the center of massby :v r mv0 2mvcm = − 0 1xˆ = – v0 xˆ ...(1)2m 2This velocity remains constant since there are noexternal forces acting on the whole system. In thevelocity of the centre of mass, the velocities of theparticles are :⎧ r r r 3⎪ u1= v1− vcm= v0xˆ⎨2...(2)r r r 3⎪u2 = v2− vcm= − v0xˆ⎩2By definition of a centre of mass frame, the totalmomentum of the particles is zero. The kineticenergy in this system at t = 0 is :K´ = 21 mu1 2 + 21 mu2 2 = 49 mv02...(3)Therefore, the total energy in the center of massframe at t = 0 isE´ = K´ + U´ = 49 mv0 2 – Rα...(4)where we define R ≡ r(t = 0). Note that since theforce is conservative, we have F r = – ∇ u. The scalarfunction is u = – α/r.The advantage of using the center of mass frame isevident when one inspects the moment of arrival at aminimal distance, t 0 . At that moment, in this system,the two particles stop and reverse their directions.The kinetic energy, therefore, vanishes at t 0 in thecenter of mass frame, or,'K´(t 0 ) = K min = 0 ...(5)αHence, E´(t 0 ) = – = E´ ...(6)R minPlugging in the value of E´, we find :4αRR max =...(7)24α − 9mv0RWe now transform the centre of mass frame to thelaboratory frame. Since R max is the relative distancebetween the two particles, it is unchanged by thetransformation. Recall that distance is an invariantquantity of the transformations of displacement and /or rotation. Therefore,α1K min – = E(0) = E = 2 1 mv0 + m(2v0 ) 2 α– 2 2 RR max....(8)The kinetic energy in the laboratory frame is,therefore :5K min = 2 α 4α − 9mv Rmv0 – +0α2 R 4αR⎛ 5 9 ⎞= ⎜ − ⎟ mv 2 1 20 = mv0 ...(9)⎝ 2 4 ⎠ 4Another way of finding the minimal kinetic energy isby using the following formula :1 2K = K´ + Mv cm...(10)2where K´ is the kinetic energy in the centre of massframe, K is the energy in the laboratory frame, and Mis the total mass of the system. In our case,12 1 2K = 0 + (2m)Mv cm = Mv 0 ...(11)24Note : Generally, in transforming from system S tosystem S´ with relative velocity V r , the kineticenergy is transformed as :p 2 2M ⎛ rrp ⎞K´ = K – + ⎜V− ⎟ ..(12)2M 2 ⎝ M ⎠where M is the total mass and p r is the totalmomentum in S. K´ is minimal in the center of massframe if we choose V r =K´ = K =2v r p rcm = . We obtain :Mp 2 ...(13)2M4. A hot body is being cooled in air according toNewton's law of cooling, the rate of fall oftemperature being K times the difference of itstemperature with respect to that of surroundings.Calculate the time after which the body will lose halfthe maximum heat it can lose. The time is to becounted from the instant t = 0.Sol. According to Newton's law of cooling, we havedθ = – K(θ – θ0 )dtwhere θ 0 is the temperature of the surrounding and θis the temperature of the body at time t. Supposeθ = θ 1 at time t = 0.Then,∫ θ dθ= –Kθ1 θ − θ ∫ t θ − θ0dt or, log = – Kt00 θ1− θ0or, θ – θ 0 = (θ 1 – θ 0 )e –Kt ...(1)The body continues to lose heat till its temperaturebecomes equal to that of the surroundings. The lossof heat in this entire period is dQ m = ms(θ 1 – θ 0 ).This is the maximum heat the body can lose. If thebody loses half this heat, the decreases in itstemperature will bedQ m θ1− θ= 02ms 2XtraEdge for IIT-JEE 21 APRIL 2010

If the body loses this heat in time t 1 , the temperatureat t 1 will beθ1 − θθ 1 – 0 θ1 + θ= 02 2Putting these values of time and temperature in (1) :θ1 + θ 0– θ 0 = (θ 1 – θ 0 ) e −Kt 12Ktor, e − 1 log 21= or t1 = 2 K5. In a certain region surrounding the origin of thecoordinates, → B = 5 × 10 –4 → k T and → E = kˆ V/m. Aproton enters the fields at the origin with an initialvelocity v → 0 = 2.5 × 10 5 î m/s. Describe the proton'smotion and give its position after three completerevolutions.Sol. The z-component of the force → F is a constantelectrical force. It produces a constant accelerationalong z-axis given as1Z = at 2 1 ⎛ eE ⎞= ⎜ ⎟ t22 2 ⎝ m ⎠E z ByThe other component of the force F is a magneticforce which provides the necessary centripetal forceto keep the proton in a circular path of radius r(say).The period of revolution of the proton2πrmv 20T = As F cp = = ev 0 Bv 0r∴ v 0 =eBrmHence, T =x2πmeBSince the particle (proton) moves in circular pathhaving a period of revolution T in x, y plane andmoves along z-axis with a constant accelerationa = eE/m, the path of the proton is a helix.After three revolutions, putting t = 3T, we obtain1 ⎛ eE ⎞z = ⎜ ⎟ (3T) 2 9 eET=2 ⎝ m ⎠ 2 mPutting T = 2πm/eB, we get218π Em 18×(22 / 7)z = =2−eB 1.610= 37m226× 5×1.66×10× (5×10−4)2−27Interesting Science Facts• The Universe contains over 100 billion galaxies.• Wounds infested with maggots heal quickly andwithout spread of gangrene or other infection.• More germs are transferred shaking hands thankissing.• The longest glacier in Antarctica, the Almbertglacier, is 250 miles long and 40 miles wide.• The fastest speed a falling raindrop can hit you is18mph.• A salmon-rich, low cholesterol diet means thatInuits rarely suffer from heart disease.• Inbreeding causes 3 out of every 10 Dalmationdogs to suffer from hearing disability.• The world’s smallest winged insect, the Tanzanianparasitic wasp, is smaller than the eye of ahousefly.• If the Sun were the size of a beach ball thenJupiter would be the size of a golf ball and theEarth would be as small as a pea.• It would take over an hour for a heavy object tosink 6.7 miles down to the deepest part of theocean.• There are more living organisms on the skin ofeach human than there are humans on thesurface of the earth.• The grey whale migrates 12,500 miles from theArtic to Mexico and back every year.• Quasars emit more energy than 100 giant galaxies.• Quasars are the most distant objects in theUniverse.• The Saturn V rocket which carried man to theMoon develops power equivalent to fifty 747jumbo jets.• Koalas sleep an average of 22 hours a day, twohours more than the sloth.• Light would take .13 seconds to travel around theEarth.• Neutron stars are so dense that a teaspoonfulwould weigh more than all the people on Earth.• One in every 2000 babies is born with a tooth.• Every hour the Universe expands by a billionmiles in all directions.• Somewhere in the flicker of a badly tuned TV setis the background radiation from the Big Bang.• The temperature in Antarctica plummets as lowas -35 degrees Celsius.• Space debris travels through space at over 18,000mph.XtraEdge for IIT-JEE 22 APRIL 2010

PHYSICS FUNDAMENTAL FOR IIT-JEECalorimetry, K.T.G., Heat transferKEY CONCEPTS & PROBLEM SOLVING STRATEGYCalorimetry :The specific heat capacity of a material is the amountof heat required to raise the temperature of 1 kg of itby 1 K. This leads to the relationQ = ms θwhere Q = heat supplied, m = mass, θ = rise intemperature.The relative specific heat capacity of a material is theratio of its specific heat capacity to the specific heatcapacity of water (4200 J kg –1 K –1 ).Heat capacity or thermal capacity of a body is theamount of heat required to raise its temperature by 1K. [Unit : J K –1 ]Thus heat capacity = Q/θ = msdθ 1 dQAlso = ×dt ms dti.e., the rate of heating (or cooling) of a body dependsinversely on its heat capacity.The water equivalent of a body is that mass of waterwhich has the same heat capacity as the body itself.[Unit : g or kg] This is given bym×sW =s wwhere m = mass of body, s = specific heat capacity ofthe body, s w = specific heat capacity of water.Principle of Calorimetry : The heat lost by onesystem = the heat gained by another system. Or, thenet heat lost or gainsed by an isolated system is zero.It system with masses m 1 , m 2 , ...., specific heatcapacities s 1 , s 2 , ...., and initial temperatures θ 1 , θ 2 , ....are mixed and attain an equilibrium temperature θthenθ =Σmsθ, for equal masses θ =Σms´ΣsθΣsNewton's law of cooling :The rate of loss of heat from a body in anenvironment of constant temperature is proportionalto the difference between its temperature and that ofthe surroundings.If θ = temperature of the surroundings thendθ– ms = C´(θ – θ0 )dtwhere C´ is a constant that depends on the nature andextent of the surface exposed. Simplifyingdθ C= –C(θ – θ0 ) where C = = constantdtmśKinetic theory of gases :The pressure of an ideal gas is given by p = 31 µnC2where µ = mass of each molecule, n = number ofmolecules per unit volume and C is the root squarespeed of molecules.p = 31 ρC2or pV = 31 mC2where ρ is the density of the gas and m = mass of thegas.Root Mean Square Speed of Molecules :This is defined as222C1 + C2+ C3+ ... + CNC =Nwhere N = total number of molecules. It can beobtained through these relationsC =3pρ=3RTMTotal Energy of an ideal gas (E) :This is equal to the sum of the kinetic energies of allthe molecules. It is assumed that the molecules do nothave any potential energy. This follows from theassumption that these molecules do not exert anyforce on each other.1E = mC 2 3 m 3= RT = pV2 2 M 2Thus, the energy per unit mass of gas = 21 C2The energy per unit volume = 23 pThe energy per mole = 23 pV = 23 RT2XtraEdge for IIT-JEE 23 APRIL 2010

Perfect gas equation :From the kinetic theory of gases the equation of anideal gas ispV = RT for a moleandpV = Mm RT for any mass mAvogadro number (N) and Boltzmann constant (k) :The number of entities in a mole of a substance iscalled the Avogadro number. Its value is6.023 × 10 23 mol –1 .The value of the universal gas constant per molecularis called Boltzmann constant (k). Its value is1.38 × 10 –23 J K –1 .Degrees of Freedom : Principle of equipartition ofenergy :The number of ways in which energy may be storedby a system is called its degrees of freedom.Principle of Equipartition of Energy : Thisprinciple states that the total energy of a gas inthermal equilibrium is divided equally among itsdegrees of freedom and that the energy per degree offreedom is kT/2 where T is the temperature of thegas. For a monoatomic atom the number of degreesof freedom is 3, for a diatomic atom it is 5, for apolyatomic atom it is 6.Hence the energy of a mole of a monoatomic gas is⎛ 1 ⎞ 3µ = N ⎜3 × kT ⎟ = RT⎝ 2 ⎠ 2Which is the same as that given by the kinetic theory.For a mole of diatomic gas µ⎛ 1 ⎞ 5= N⎜5 × kT ⎟ = RT⎝ 2 ⎠ 2For a mole of polyatomic gas µ⎛ 1 ⎞= N⎜6 × kT ⎟ = 3RT⎝ 2 ⎠When the irrational degrees of freedom are also takeninto account, the number of degrees of freedom= 6n – 6 for non-linear molecules= 6n – 5 for linear moleculeswhere n = number of atoms in a molecule.Kinetic Temperature :The kinetic temperature of a moving particle is thetemperature of an ideal gas in thermal equilibriumwhose rms velocity equals the velocity of the givenparticle.Maxwellian distribution of velocities :In a perfect gas all the molecules do not have thesame velocity, rather velocities are distributed amongthem. Maxwell enunciated a law of distribution ofvelocities among the molecules of a perfect gas.According to this law, the number of molecules withvelocities between c and c + dc per unit volume isdn = 4πna 3 2bce − c 2 dc wherem mb = and a =2kT 2πkTand the number of molecules with the velocity c perunit volume isn c = 4πna 3 2bce − c 2The plot of n c and c is shown in the figure. Thevelocity possessed by the maximum number ofmolecules is called the most probable velocityα = 2 kT / mThe mean velocityc =α c C rms8 kT / mπand v rms = 3 kT / mπConduction :The transfer of heat through solids occurs mainly byconduction, in which each particle passes on thermalenergy to the neighboring particle but does not movefrom its position. Very little conduction occurs inliquids and gases.θ 1 θ 2Qd AConsider a slab of area A and thickness d, whoseopposite faces are at temperature θ 1 and θ 2 (θ 1 > θ 2 ).Let Q heat be conducted through the slab in time t.⎛ θ1 − θ2⎞Then Q = λA ⎜ ⎟ t⎝ d ⎠where λ = thermal conductivity of the material.This has a fixed value for a particular material, beinglarge for good conductors (e.g., Cu, Ag) and low forinsulators (e.g., glass, wood).Heat Current : The quantity Q/t gives the heat flowper unit time, and is called the heat current.In the steady state, the heat current must be the sameacross every cross-section. This is a very usefulprinciple, and can be applied also to layers or slabs incontact.Q dθ dθ θ= – λA where the quantity =1 − θ 2t dxdx dcalled the temperature gradient.QisXtraEdge for IIT-JEE 24 APRIL 2010

Unit of λ : Different units are used,e.g., cal cm s ºC –1 , cal m –1 s –1 ºC –1 , jm´1 s –1 ºC –1 .Convection :It is a process by which heat is conveyed by theactual movement of particles. Particles closest to thesource receive heat by conduction through the wall ofthe vessel. They rise up-wards and are replaced bycolder particles from the sides. Thus, a circulation ofparticles is set up – hot particles constitute theupward current and cold particles, the side anddownward current.The transfer of heat by convection occurs only influids, and is the main mode of heat transfer in them.Most fluids are very poor conductors.Radiation :Thermal Radiation : Thermal radiations areelectromagnetic waves of long wavelengths.Black Body : Bodies which absorb the whole of theincident radiation and emit radiations of allwavelengths are called black bodies.It is difficult to realize a perfect black body inpractice. However, a cavity whose interior walls aredull black does behave like a black body.Absorption : Every surface absorbs a part or all ofthe radiation falling on it. The degree of absorptiondepends on the nature and colour of the surface. Dull,black surfaces are the best absorbers. Polished, whitesurfaces absorb the least. The coefficient ofabsorption for a surface isradiation absorbeda λ =radiation incidentThe suffix λ denotes the wavelength of the radiationbeing considered, Clearly, a λ = 1 for a black body, forall values of λ.Emission : Each surface emits radiation (radiates)continuously. The emissive power (e λ ) is defined asthe radiation emitted normally per second per unitsolid angle per unit area, in the wave-length range λand λ + dλ. Clearly, the emissive power of a blackbody (denoted by E λ ) is the maximum.Kirchhoff's Law : According to this law, for thesame conditions of temperature and wavelength, theratio e λ /a λ is the same for all surfaces and is equal toE λ . This simply means that good absorbers are goodemitters. Hence, a black body is the best emitter, anda polished white body, the poorest emitter.Prevost's Theory of Exchanges : All bodies emitradiations irrespective of their temperatures. Theyemit radiations to their environments and receiveradiations from their environments simultaneously. Inthe equilibrium state the exchange between a bodyand the environment of energy continues in equalamounts.Stefan-Boltzmann Law : If a black body at anabsolute temperature T be surrounded by anotherblack body at an absolute temperature T 0 , the rate ofloss of radiant energy per unit area isE = σ(T 4 – T 4 0 )where σ is a constant called Stefan constant and itsvalue is 5.6697 × 10 –8 W m –2 K –4The total energy radiated by a black body at anabsolute temperature T is given byE = σT 4 × surface area × timeNote : Remember that rate of generation of heat byelectricity is given by H = I 2 V 2R or or VI Js –1 or W.RSolved Examples1. An earthenware vessel loses 1 g of water per seconddue to evaporation. The water equivalent of thevessel is 0.5 kg and the vessel contains 9.5 kg ofwater. Find the time required for the water in thevessel to cool to 28ºC from 30ºC. Neglect radiationlosses. Latent heat of vaporization of water in thisrange of temperature is 540 cal g –1 .Sol. Here water at the surface is evaporated at the cost ofthe water in the vessel losing heat.Heat lost by the water in the vessel= (9.5 + 0.5) × 1000 × (30 – 20) = 10 5 calLet t be the required time in seconds.Heat gained by the water at the surface= (t × 10 –3 ) × 540 × 10 3(Q L = 540 cal g –1 = 540 × 10 3 cal kg –1 )∴ 10 5 = 540t or t = 185 s = 3 min 5s2. 15 gm of nitrogen is enclosed in a vessel attemperature T = 300 K. Find the amount of heatrequired to double the root mean square velocity ofthese molecules.Sol. The kinetic energy of each molecule with mass m isgiven by1 m2 3v rms = kT ...(1)2 2If we want to increase the r.m.s. speed to η times,then the temperature has to be raised to T´. Then,1 2 3 1mv rms = kT´ or mη2 2 3v rms = kT´2 2 2 2 ...(2)From eqs. (1) and (2), T´ = η 2 T ...(3)XtraEdge for IIT-JEE 25 APRIL 2010

The internal energy of n molecules at temperature Tis given bySimilarly,U = 25 nRTU´ = 25 nRT´∴ Change in internal energy ∆U = 25 nR[T´ – T]or ∆U = 25 nRT[η 2 – 1]= 25= 25⎛ m ⎞⎜ ⎟ RT[η 2 – 1]⎝ M ⎠⎛ 15 ⎞⎜ ⎟ (8.31) (300) [4 – 1] = 10 4 J⎝ 28 ⎠3. 10 gm of oxygen at a pressure 3 × 10 5 N/m 2 andtemperature 10ºC is heated at constant pressure andafter heating it occupies a volume of 10 litres (a) findthe amount of heat received by the gas and (b) theenergy of thermal motion of gas molecules beforeheating.Sol. (a) The states of the gas before and after heating arePV 1 = µM RT1 and PV 2 = µM RT2Solving these equations for T 2 , we have−35µV PT 2 = 2 32×(10×10 )(3×10 )= = 1156 K−33MR (10×10 )(8.31×10 )Now T 2 – T 1 = 1156 – 283 = 873 KThe amount of heat received by the gas is given by∆Q = µMCP (T 2 – T 1 )3× −(10 10 )29.08×10 × 873=32= 7.9 × 10 3 J(b) The energy of the gas before heatingM iE 1 = × × RT 1µ 2where i = number of degrees of freedom= 5 (for oxygen)−3−3(10×10 )5×(8.31×10 )(283)=2×32= 1.8 × 10 3 J4. A slab of stone of area 3600 sq cm and thickness 10cm is exposed on the lower surface of steam 100ºC.A block of ice at 0ºC rests on upper surface of theslab. In one hour 4800 gm of ice is melted. Calculatethe thermal conductivity of the stone.3Sol. The quantity of heat Q passing across the stone isgiven byKA(T1 − T2)tQ =dHere A = 3600 sq. cm = 0.36 m 2d = 10 cm = 0.10 m, (T 1 – T 2 ) = 100 – 0 = 100ºC andt = 1 hour = 3600 sec.K × 0.36×100×3600∴ Q =kilo-calories ...(1)0.10Now heat gained by the ice in one hour= mass of the ice × latent heat of ice= 4.8 × 80 kilo calories ...(2)From eqs. (1) and (2)K × 0.36×100×36004.8 × 80 =0.104.8×80×0.10or K =0.36×100×3600= 3 × 10 –4 kilo cal m –1 (ºC) –1 s –15. A flat bottomed metal tank of water is dragged alonga horizontal floor at the rate of 20m/sec. The tank isof mass 20 kg and contains 1000 kg of water and allthe heat produced in the dragging is conducted to thewater through the bottom plate of the tank. If thebottom plate has an effective area of conduction 1 m 2and the thickness 5 cm and the temperature of waterin the tank remains constant at 50ºC, calculate thetemperature of the bottom surface of the tank, giventhe coefficient of friction between the tank and thefloor is 0.343 and K for the material of the tank is 25cal m –1 s –1 K –1 .Sol. Frictional force = µ m g= 0.343 × (1000 + 20) × 9.81 = 3432 NThe rate of dragging, i.e., the distance travelled inone second = 20 m.∴ Work done per second= (3432 × 20) Nm/sec.This work done appears as heat at the bottom plate ofthe tank. Hence3432× 20H =cal/sec4.18KA(T1 − T2)But H =(Q t = 1 sec)d3432× 20 25×1×(T1 − T2)Now=4.18 0.053432×20×0.05∴ T 1 – T 2 == 32.844.18×25×1Temp. of bottom surface T 1 = 50 + 32.84= 82.84ºCXtraEdge for IIT-JEE 26 APRIL 2010

PHYSICS FUNDAMENTAL FOR IIT-JEEAtomic Structure, X-Ray & Radio ActivityKEY CONCEPTS & PROBLEM SOLVING STRATEGYAtomic Structure :According to Neil Bohr's hypothesis is the angularmomentum of an electron is quantised.n = ∞n = 7n = 6n = 5n = 4⎛mvr = n ⎜⎝2πr = nλh2π⎟⎠⎞or L = nh2πh ⎛ c ⎞ zv n = Zn = ⎜ ⎟ × ms–12πmr⎝137⎠ n⎛2r n = ⎜h2⎝ 4πmke2⎞⎟⎠n 2Z⎛ ⎞f n = ⎜ke 2 ⎟1 6.58×× = ⎝ hr ⎠ n n10n 2= 0.529 Å where k =Z15Hz14πε1 ke 2 Z − ke 2ke 2K.E. = ; P.E. = × Z; T.E. = – × Z2 rr2rn = 3n = 22−13.6ZT.E. = ev/atom where –13.62n= Ionisation energy+P.E.⇒ +T.E. = = – K.E.2Note : If dielectric medium is present then ε r has tobe taken into consideration.v 1 = = v =c λ⎡= RZ 2 1 1⎥ ⎥ ⎤⎢ −2 22⎢⎣n1 n ⎦K βK γK δL γL β4 2me z ⎡ 1 1⎥ ⎥ ⎤⎢ −2 3 2 28ε0h c ⎢⎣n1n 2 ⎦L αBalmer(Visible)p mv= = h hPaschen(I.R.)PfundBrackett (I.R.)(I.R.)Limiting line of Lyman seriesn = 1Lyman Series(U.V. rays)–0.85 eV–1.5 eV–3.4 eV–13.6 eV0The maximum number of electrons that can beaccommodated in an orbit is 2n 2 .X-rays :When fast moving electron strikes a hard metal,X-rays are produced. When the number of electronsstriking the target metal increases, the intensity of X-rays increases. When the accelerating voltage/kineticenergy of electron increases λ min decreases. X-rayshave the following properties :(a) Radiations of short wavelength (0.01 Å – 10Å);high pentrating power; having a speed of 3 × 10 8 m/sin vacuum.IntensityContinuous spectrum(Varies & depends onaccelerating voltage)λ minhc(b) λ min = = eVK βK αL γhcK. ECharacteristic spectrum(fixed for a target material)L βL α=λ12400 ÅV1(c) = R(Z – b)2 ⎡ 1 ⎤λ⎢1− ⎥⎣ n 2⎦b = 1 for k-line transfer of electron(d) Moseley law ν = a(z – b)R = R 0 A 1/3 where R 0 = 1.2 × 10 –15 mR = radius of nucleus of mass number A.* Nucleus density is of the order of 10 17 kg/m 3Isomers are nuclides which have identical atomicnumber and mass number but differ in their energystates.Nuclear binding energy ∆mc 2=NucleonAwhere ∆m = mass defect2[Zmp+ (A − Z)mn− M]c=AXtraEdge for IIT-JEE 27 APRIL 2010

* The binding energy per nucleon is small for smallnuclei.* For 2 < A < 20, there are well defined maximawhich indicate that these nuclei are more stable.* For 30 < A < 120 the average B.E./A is 8.5 MeV /nucleon with a peak value of 8.8 MeV for Iron.* For A > 120, there is a gradual decreases inB.E./nucleon.* More the B.E./A, more is the stability.Radioactivity :β particles are electrons emitted from the nucleus.(n → p + β)(a) N = N 0 e –λt(b)−dNdt⎛ 1 ⎞(c) N = N 0 ⎜ ⎟⎠⎝ 2⎛ 1 ⎞⇒ A = A 0 ⎜ ⎟⎠⎝ 2(d) T 1/2 =(e) τ = λ1dN= λN where = activity leveldtnn0.693λ(f) τ = 1.4 T 1/2(g) t =⎛ 1 ⎞ T 1/ 2= N 0 ⎜ ⎟⎠⎝ 22.303 Nlog 0 10 =λ Ntwhere A = activity level2.303 Alog 0 10λ A2.303 m= log 0λ m(h) If a radioactive element decays by simultaneous−dNemission of two particle then = λ 1 N + λ 2 NdtThe following parameters remain conserved during anuclear reaction(a) linear momentum(b) Angular momentum(c) Number of nucleons(d) Charge(e) The energy released in a nuclear reactionX + P → Y + Z + QQ = [m x + m p ) – (m y + m z )]c 2 = ∆m × c 2Q = ∆m × 931 MeV(f) In a nuclear fusion reaction small nuclei fuse togive big nuclei whereas in a nuclear fusion reaction abig nuclei breaks down.Thermal neutrons produce fission in fissile nuclei.Fast moving neutrons, when collide with atoms ofcomparable masses, transfer their kinetic energy tocolliding particle and slow down.According to Doppler's effect of lightPower, P = tE =η =out putIn putnhν =tnhcλt∆ λλSolved Examples=cv1. The energy of an excited hydrogen atom is –3.4 eV.Calculate the angular momentum of the electronaccording to Bohr theory.Sol. The energy of the electron in the nth orbit is2nE n = –13.6Here, – = –3.413.6eV2nor n = 2nh 2×6.63×10Angular momentum = =2π 2×3. 14= 2.11 × 10 –34 Js.−342. The wavelength of the first member of the Balmerseries in the hydrogen spectrum is 6563 Å. Calculatethe wavelength of the first member of the Lymanseries.Sol. For the first member of the Balmer series1 ⎡ 1 1 ⎤ 5R= Rλ⎢ − ⎥⎣ 2 23 2=⎦ 36For the first member of the Lyman series1λ ´= R ⎡ 1 1 ⎤⎢ − ⎥⎣122 2 ⎦=3R4Dividing Eq. (1) by Eq. (2)λ´ 5×4 5= =λ 36×3 27or λ´ = 275 λ = 275 × 6563 = 1215 Å...(1)...(2)XtraEdge for IIT-JEE 28 APRIL 2010

3. Hydrogen atom in its ground state is excited bymeans of a monochromatic radiation of wavelength970.6 Å. How many different wavelengths arepossible in the resulting emission spectrum ? Find thelongest wavelength amongst these.Sol. Energy the radiation quantum−34hc 6.6×10 × 3×10E = hv = = λ−10−19970.6×10 × 1.6×10= 12.75 eVEnergy of the excited sateE n = – 13.6 + 12.75 = – 0.85 eV13.6Now, we know that E n = –2nor n 2 13.6 −13.6= – = = 16E n − 0.85or n = 4The number of possible transition in going to theground state and hence the number of differentwavelengths in the spectrum will be six as shown inthe figure.n4321The longest wavelength corresponds to minimumenergy difference, i.e., for the transition 4 → 3.Now E 3 = –hcλmax13.623= E 4 – E 3= – 1.51 eV−346.6×10 × 3×10or λ max =−19(1.51−0.85) × 1.6×10= 18.75 × 10 –7 m = 18750 Å4. X-rays are produced in an X-ray tube by electronsaccelerated through a potential difference of 50.0 kV.An electron makes three collisions in the targetbefore coming to rest and loses half its kinetic energyin each of the first two collisions. Determine thewavelengths of the resulting photons. Neglect therecoil of the heavy target atoms.88Sol. Initial kinetic energy of the electron = 50.0 keVEnergy of the photon produced in the first collision,E 1 = 50.0 – 25.0 = 25.0 keVWavelength of this photon−34hc 6.6×10 × 3×10λ 1 = =−193E 1 1.6×10 × 12.5×10= 0.99 × 10 –10 m= 0.99 ÅKinetic energy of the electron after third collision = 0Energy of the photon produced in the third collision ,E 3 = 12.5 – 0 = 12.5 keVThis is same as E 2 . Therefore, wavelength of thisphoton, λ 3 = λ 2 = 0.99 Å5. In an experiment on two radioactive isotopes of anelements (which do not decay into each other), theirmass ratio at a given instant was found to be 3. Therapidly decaying isotopes has larger mass and anactivity of 1.0 µCi initially. The half lives of the twoisotopes are known to be 12 hours and 16 hours.What would be the activity of each isotope and theirmass ratio after two days ?Sol. We have, after two days, i.e., 48 hours,N 1 =40⎛1 ⎞N1⎜ ⎟⎝ 2 ⎠3=0⎛1 ⎞N 2 = N 2⎜⎟ =⎝ 2 ⎠Mass ratio =0N 1 /160N 2 /80N 1 =10N 2 N 20 0Now, A 1 = λ 1 N 1 = 1.0 µCiAfter two days,ButorN 8 3× 8 3. = = 16 162 20A 1 = λ 1 N 1 = λ 1 N 1 /16 =0A 2 = λ 2 N 2 = λ 2 N 2 /8λλ21=T 1 = 16T2λ 2 = 43λ1⎛ 3A 2 = ⎜⎝ 4λ 112 = 43⎞ ⎛ 1 0 ⎞ 1⎟ × ⎜ N 1 ⎟ ×⎠ ⎝ 3 ⎠ 81 0 1 0= λ1 N 1 = A 132 32= (1/32) µCi80A 1 /16 = (1/16)µCiXtraEdge for IIT-JEE 29 APRIL 2010

KEY CONCEPTOrganicChemistryFundamentalsAROMATICHYDROCARBONHalogenation of Benzene :The function of the Lewis acid can be seen in step 1.–H:Br – FeBr 3Br: Br:+ H – Br: + FeBr 3Benzene does not react with bromine or chlorineunless a Lewis acid is present in the mixture. (As aThe ferric bromide reacts with bromine to produce apositive bromine ion, Br + (and FeBr – 4 ). In step 2 thisconsequence, benzene does not decolorize a solution Br + ion attacks the benzene ring to produce anof bromine in carbon tetrachloride.) When Lewisacids are present, however, benzene reacts readilyarenium ion. Then, finally in step 3 a proton isremoved from the arenium ion by FeBr – 4 . This resultswith bromine or chlorine, and the reactions give in the formation of bromobenzene and hydrogenbromobenzene and chlorobenzene, respectively, in bromide the products of the reaction. At the samegood yields :time this step regenerates the catalyst, FeBr 3 .ClFeCl 3+ Cl 2+ HClNucleophilic Aromatic Substitution through an25ºCElimination – Addition Mechanism : BenzyneChlorobenzene (90%)Although aryl halides such as chlorobenzene andBrbromobenzene do not react with most nucleophilesFeBr 3+ Br 2 + HBrunder ordinary circumstances, they do react underheathighly forcing conditions. Chlorobenzene can beBromobenzene (75%)converted to phenol by heating it with aqueousThe Lewis acids most commonly used to effectsodium hydroxide in a pressurized reactor at 350ºC .chlorination and bromination reactions are FeCl 3 ,FeBr 3 , and AlCl 3 , all in the anhydrous form.ClONa OHA mechanism for the reaction :350ºCElectrophillic Aromatic Bromination :+ NaOHH 2OH 3O +Step 1Phenol+–Br – Br : + FeBr 3 ⎯→ :Br – Br – FeBr 3Bromobenzene reacts with the very powerful base,⎯→ :Br + –+ :Br – FeBr 3NH – 2 , in liquid ammonia :BrNH 2Bromine combines with FeBr 3 to form a complex that dissociates–to form a positive bromine ion and FeBr 4+ – -33ºC+ K :NHStep 22NH 3+ KBr+ slowAniline+ Br:These reactions take place through an elimination –HBr:HBr:+ H addition mechanism that involves the formation of anBr:interesting intermediate called benzyne (or+ +dehydrobenzene). We can illustrate this mechanismwith the reaction of bromobenzene and amide ion.Arenium ionThe positive bromine ion attacks benzene toform an arenium ionStep 3+A proton is removed from the arenium ionto become bromobenzeneIn the first step, the amide ion initiates an eliminationby abstracting one of the ortho protons because theyare the most acidic. The negative charge thatdevelops on the ortho carbon is stabilized by theinductive effect of the bromine. The anion then losesa bromide ion. This elimination produces the highlyunstable, and thus highly reactive, benzyne. Benzynethen reacts with any available nucleophile (in thiscase, an amide ion) by a two-step addition reaction toproduce aniline.XtraEdge for IIT-JEE 30 APRIL 2010

The Benzyne Elimination – Addition Mechanism :BrH–:NH 2(–NH 3)EliminationAdditionBr(–Br – )–Benzyne(or dehydrobenzene)–:NH 2NH 2–:NH 3NH 2 –+ :NH 2HEvidence for an elimination-addition mechanism :When 14 C-labeled (C*) chlorobenzene is treated withamide ion in liquid ammonia, the aniline that isproduced has the label equally divided between the 1and 2 positions.* NH 2* Cl K + –NH 2*– NH2(50%)NH 3*NH 2Elimination Addition (50%)When the ortho derivative 1 is treated with sodiumamide, the only organic product obtained ism-(trifluoromethyl) aniline :CF 31ClNaNH 2NH 3CF 3NH 2m-(Trifluoromethy)anilineThis result can also be explained by an elimination –addition mechanism. The first step produces thebenzyne 2 :CF 3CF 3ClNaNH 2+ Cl –NH 31 2This benzyne then adds an amide ion in the way thatproduces the more stable carbanion 3 rather than theless stable carbanion 4 :CF 3CF 3:NH 2NH 2–4Less stable carbanionCF 3–CF 32– :NH 3–NH 2NH 2 +:NH32More stable carbanion(The negative charge is closer to theelectronegative trifluoromethyl group)Carbanion 3 then accepts a proton from ammonia toform m-(trifluoromethyl) aniline.Carbanion 3 is more stable than 4 because the carbonatom bearing the negative charge is closer to thehighly electronegative trifluoromethyl group. Thetrifluoromethyl group stabilizes the negative chargethrough its inductive effect. (Resonance effects arenot important here because the sp 2 orbital thatcontains the electron pair does not overlap with the πorbitals of the aromatic system.)The Birch Reduction :Benzene can be reduced to 1, 4-cyclohexadiene bytreating it with an alkali metal (sodium, Lithium, orpotassium) in a mixture of liquid ammonia and analcohol.NaNH 3 , EtOHBenzene1, 4-cyclohexadieneA Mechanism for the Reaction :Brich Reduction :Naetc.– –BenzeneBenzene radical anionThe first electron transfer produces a delocalizedbenzene radical anion.HEtOHHetc.HHCyclohexadienyl radicalProtonation produces a cyclohexadienyl radical(also a delocalized species)NaH–Hetc. EtOHHHHHH–HCyclohexadienyl anion1,4-CyclohexadieneTransfer of another electron leads to the formation of a delocalizedcyclohexadienyl anion, and protonation of this produces the1,4-cyclohexadiene.XtraEdge for IIT-JEE 31 APRIL 2010

KEY CONCEPTPhysicalChemistryFundamentalsSOLUBILITYPRODUCTSolubility :The amount of a solute, dissolved in a given volumeof a solvent (water) in 100 mL or in 1L to form asaturated solution at a given temperature is termed asthe solubility of the solute.Solubility Product :Salts like AgI, AgCl, PbI 2 , BaSO 4 , PbSO 4 etc. areordinarily considered insoluble but they do possesssome solubility. These are sparingly soluble salts. Asaturated solution of sparingly soluble salt contains avery small amount of the dissolved salt. It is assumedthat whole of the dissolved salt is present in the formof ions, i.e., it is completely dissociated. Consider asparingly soluble salt like AgCl, the followingequilibrium occurs between the undissolved solid saltand the silver and chloride ions in the saturatedsolution.AgCl (s) Ag + (aq) + Cl – (aq)Applying the law of mass action to the ionicequilibrium,+ −[Ag ][Cl ]K =[AgCl(s)]or K[AgCl(s) = [Ag + ] [Cl – ]The concentration of solid AgCl in the solid state i.e.[AgCl(s)] is constant at a particular temperature, nomatter how much solid is present in contact with thesolution. It follows that[AgCl(s)] = K´ = constantHence, [Ag + ] [Cl – ] = KK´ = K sp (constant)K sp is termed as the solubility product. It is defined asthe product of the concentration of ions in a saturatedsolution of a salt at a given temperature. Consider, ingeneral, the salt of the type A x B y which is dissociatedas :A x B y x A y+ + y B x–Applying law of mass action,y+x x−y[A ] [B ]= K[A xBy]when the solution is saturated,[A x B y ] = K´ (constant)or [A y+ ] x [B x– ] y = K[A x B y ] = KK´ = K sp (constant)Thus, solubility product is defined as the product ofconcentrations of the ions raised to a power equal tothe number of times the ions occur in the equationrepresenting the dissociation of the salt at a giventemperature when the solution is saturated.Solubility product is not the ionic product under allconditions but only when the solution is saturated.Ionic product has a broad meaning since it isapplicable to all types of solutions, may be saturatedor unsaturated.Relationship between solubility and solubility product :The equilibrium for a saturated solution of anysparingly soluble salt be expressed as :A x B y x A y+ + y B x–Thus, solubility product, K sp = [A y+ ] x [B x– ] y .Let 's' mole per litre be the solubility of the salt, thenA x B y x A y+ + y B x–xs ysSo K SP = [xs] x [ys] y = x x .y y (s) x+ySince the solubility of a salt varies with temperature,the numerical value of K sp for a salt changes withtemperature; values usually recorded at 25ºC.Common Ion Effect :The suppression of the degree of dissociation of aweak acid or a weak base by the addition of a strongelectrolyte containing a common ion. The commonion effect play an important role in the qualitativeanalysis.Application of solubility product in qualitative analysis:Precipitation of sulphides of group II. Sulphides ofgroup II are precipitated by passing H 2 S gas throughthe solution of these cations in presence of dil HCl.H 2 S being a weak electrolyte ionizes only sligthtly,while HCl being a strong electrolyte is almostcompletely ionized.H 2 S 2H + + S 2– ; HCl ⎯→ H + + Cl –Thus, the common ion effect takes place. As a result,the degree of dissociation of H 2 S decreasessufficiently and the concentration of S 2– ions in thesolution becomes very small. But with this lowconcentration of second group and the sulphide ionsexceeds the very low solubility products of theircorresponding sulphides. Therefore, the cations ofgroup II get precipitated as their insoluble sulphides.On the other hand, the sulphides of the cations of theother groups (III, IV, V and Mg) are not precipitatedunder these conditions because their solubilityproducts are quite high.XtraEdge for IIT-JEE 32 APRIL 2010

Precipitation of the hydroxides of group III :Hydroxides of group III are precipitated by adding anexcess of solid NH 4 Cl to the solutions of thesecations followed by the addition of excess ofNH 4 OH. Being a weak electrolyte, NH 4 OH is onlyslightly ionised, whereas NH 4 Cl, being a strongelectrolyte, ionizes almost completely to give at largeconcentration of NH + 4 ions.NH 4 OH NH + 4 + OH – ; NH 4 Cl → NH + 4 + Cl –Due to the common ion effect, the degree ofdissociation of NH 4 OH gets suppressed and hence theconcentration of OH – ions in solution decreasesappreciably. But even with this low conc. of OH –ions, the ionic products of the cations of group III andOH – ions exceed the low values of the solubilityproducts of their corresponding metal hydroxides. Asa result, the cations of group III get precipitated astheir insoluble hydroxides.On the other hand, cations of groups IV, V and Mg,which require a large conc. of OH – ions due to theirhigh solubility products will not be precipitated.Precipitation of sulphides of group IV. Thesulphides of group IV are precipitated by passing H 2 Sthrough ammoniacal solution of these cations.Both H 2 S and NH 4 OH, being weak electrolytes,ionize only slightly as :H 2 S 2H + + S 2–+NH 4 OH NH 4 + OH –The H + ions and OH – ions combine to producepractically unionised molecules of waterH + + OH – → H 2 OAs a result, the above dissociation equilibriumreactions get shifted in the forward direction, so thatthe concentration of S 2– ions goes on increasing.Ultimately, the ionic product of the cations of groupIV and S 2– ions exceed the solubility products of theircorresponding metal sulphides and hence getprecipitated.Precipitation of carbonates of group V : Thecarbonates of group V are precipitated by adding(NH 4 ) 2 CO 3 solution to the solution of these cations inthe presence of NH 4 Cl and NH 4 OH. (NH 4 ) 2 CO 3 ,being a weak electrolyte ionises only slightly to givea small concentration of CO 2– 3 ions.(NH 4 ) 2 CO 3 2NH + 2–4 + CO 3On the other hand, NH 4 Cl being a strong electrolyte,ionises almost completely to give a large+concentration of NH 4 ions. Due to the common ioneffect, the dissociation of (NH 4 ) 2 CO 3 is suppressedand hence the concentration of CO 2–3 ions in thesolution decreases considerably. But even with thislow concentration of CO 2– 3 ions, the ionic products ofthese cations and CO 2– 3 ions exceed the low values ofthe solubility products of their corresponding metalcarbonates and thus get precipitated.However, under these conditions, Mg salts do not getprecipitated as MgCO 3 since its solubility product iscomparatively high and thus requires a highconcentration of CO 2–3 ions for precipitation. Thecarbonates of Na + , K + +and NH 4 ions are also notprecipitated because they are quite soluble.The necessity of adding NH 4 OH arises due to the factthat (NH 4 ) 2 CO 3 solution usually contains a largeamount of NH 4 HCO 3 . Thus, the cations of group Vwill form not only insoluble carbonates but solublebicarbonates as well. As a result, the precipitationwill not be complete. In order to convert NH 4 HCO 3to (NH 4 ) 2 CO 3 , NH 4 OH is always added.NH 4 HCO 3 + NH 4 OH → (NH 4 ) 2 CO 3 + H 2 OPreferential precipitation of Salts :A solution contains more than one ion capable offorming a precipitate with another ion which is addedto the solution. For example, in a solution containingCl – , Br – , and I – ions, if Ag + ions are added, then outof the three, the least soluble silver salt is precipitatedfirst. If the addition of Ag + ions is continued,eventually a stage is reached when the next lessersoluble salt starts precipitating along with the leastsoluble salt and so on. If the stoichiometry of theprecipitated salts is the same, then the salt with theminimum solubility product (and hence also theminimum solubility) will precipitate first followed bythe salt of next higher solubility product and so on.If the stoichiometry of the precipitated salts is not thesame, then, from the solubility product data alone, wecannot predict which ion will precipitate first. Take,for example, a solution containing Cl – and CrO 2– 4 .Both these ions form precipitates with Ag + . Throughthe solubility product product of AgCl is larger thanthat of Ag 2 CrO 4 , yet it is AgCl (lesser soluble) whichprecipitates first when Ag + ions are added to thesolution. In order to predict which ion (Cl – or CrO 2– 4 )precipitates first, we have to calculate theconcentration of Ag + ions needed to start theprecipitation through the solubility product data andthe given concentration of Cl – or CrO 2– 4 . Since squareroot is involved in the expression for computing Ag +for silver chromate, the quantity of Ag + needed tostart the precipitation of CrO 2– 4 is larger than that forCl – . Hence, as AgNO 3 is added to the solution, theminimum of the two concentrations of Ag + to startthe precipitation will be reached first and thus thecorresponding ion (Cl – in this case) will beprecipitated in preference to the other. During thecourse of precipitating, concentration of Cl –decreases and the corresponding concentration of Ag +to start the precipitation increases. Its concentrationeventually becomes equal to the value required forCrO 2– 4 . At this stage, practically the whole of Cl –ions have been precipitated. The addition of more ofAgNO 3 causes the precipitation of both the ionstogether.XtraEdge for IIT-JEE 33 APRIL 2010

UNDERSTANDINGOrganic Chemistry1. An organic compound (A), C 4 H 9 Cl, on reacting withaqueous KOH gives (B) and on reaction withalcoholic KOH gives (C) which is also formed onpassing vapours of (B) over heated copper. Thecompound (C) readily decolourise bromine water.Ozonolysis of (C) gives two compounds (D) and (E).Compound (D) reacts with NH 2 OH to give (F) andthe compound (E) reacts with NaOH to give analcohol (G) and sodium salt (H) of an acid. (D) canalso be prepared from propyne on treatment withwater in presence of Hg ++ and H 2 SO 4 . Identify (A) to(H) with proper reasoning.Sol.C 4 H 9 Cl(A)(Alkyl halide)Alc. KOH∆;– HClC 4 H 8(C)(Alkene)Aq. KOHCuC 4 H 9 OH∆; –KCl (B) ∆; –H 2 O(Alcohol)We know that p-alcohol on heating with Cu givesaldehyde while s-alcohol under similar conditionsgives ketone. Thus, (B) is a t-alcohol because it, onheating with Cu gives an alkene (C). Since at-alcohol is obtained by the hydrolysis of a t-alkylhalide, hence (A) is t-butyl chloride.Thus, (A) isClOH||CH3− C − CH3and (B), is CH3− C − CH3||CH3CH3The alkene (C) on ozonolysis gives (D) and (E),hence (C) is not symmetrical alkene. In thesecompound (E) gives Cannizaro's reaction withNaOH. So, (E) is an aldehyde which does not containα - H atom. Hence it is HCHO. Compound (D) canalso be prepared by the hydration of propyne in thepresence of acidic solution and Hg ++CH 3 – C ≡ CH + H 2 OHg ++H + CH 3 – C = CH 2OHCH 3 – C – CH 3O(D)Hence (D) is acetone and (E) is formaldehyde.Therefore, alkene (C) is 2-methyl propene.CH 3 – C = CH 2CH 3(C)(D) reacts with hydroxyl amine (NH 2 OH) to formoxime (F).CH 3–HC = O + H 2 NOH 2 O CH 3C = NOHCH 3CH 3(D)(F)Thus, (B) is CHReactions :ClCH 3 – C – CH 3CH 3(A)3OH|− C − CH|CH33and (A) is CHOHAq. KOHCH 3 – C – CH 3∆; –KClCH 3(B)CH 3 – C = CH 2 + H 2 OAlc. KOH/∆–KCl; –H 2 OCH 3 – C = CH 2(I) O 3(II) H 2O/ZnCH 3CH 3CH 3(C)C = O + H 2 NOH(D)CH 3CH 3CH 3∆–H 2O(C)3CH 3 – C = CH 2CH 3(C)Cl|− C − CH|CH3Cu/300ºC–H2OOC = O + H –C – H(D) (E)CH 3CH 3C = NOH(F)2HCHO + NaOH → CH 3 OH + HCOONa(E) (G) (H)OCH 3 – C ≡ CH + H 2 O Hg++ CH 3 – C – CH 3H +(D)3XtraEdge for IIT-JEE 34 APRIL 2010

2. An organic compound (A) C 7 H 15 Cl on treatment withalcoholic KOH gives a hydrocarbon (B) C 7 H 14 . (B)on treatment with O 3 and subsequent hydrolysis givesacetone and butyraldehyde. What are (A) and (B) ?Explain the reactions.Sol. In numerical, following data are given :C7HCl ⎯⎯⎯15(A)Alkyl halide∆Alc.KOH ⎯∆→C7H 14(B)AlkeneO⎯⎯→3C7H14O3Ozonide⎯H 2⎯O ⎯⎯/ Zn → CH 3 COCH 3 + CH 3 CH 2 CH 2 CHOThe alkene contains seven carbon atoms. Theposition of C = C double bond can be locatedas follows :HCH 3–2[O]C = O + O = C.CH 2 CH 2 CH 3CH 3CH 3 – C = CH.CH 2 CH 2 CH 3CH 3Thus, alkene (B) is 2-methyl hexene-2The ozonolysis reaction is as follows :CH 3 – C = CH.CH 2 CH 2 CH 3CH 3(B)O OCH 3 – C CH – CH 2 CH 2 CH 3OCH 3H 2O/Zn∆O 3CH 3 – C = O + O = C – CH 2 CH 2 CH 3CH 3 HSince alkene (B) is produced by the removal of onemol of HCl from alkyl halide (A) and thus (A) can beeither (I) or (II).Cl(I) CH 3 – C – CH 2 CH 2 CH 2 CH 3or(II)CH 3CH 3 – CH – CH – CH 2 CH 2 CH 3CH 3 ClThe dehydro halogenation reaction by(I) or (II) yeildsClKOHCH 3 – C – CH 2 CH 2 CH 2 CH 3alc.CH 3(I). CH 3 – C = CH.CH 2 CH 2 CH 3CH 3Main product (Saytzeff's rule)KOHCH 3 – CH – CH – CH 2 CH 2 CH 3Alc.CH 3 Cl(II) CH 3 – C = CH.CH 2 CH 2 CH 3Thus, both (I) and (II) give main product as 2-methylhexene-2, hence (A) is either (I) or (II).Cl(A) CH 3 – C – CH 2 CH 2 CH 2 CH 3or(B)CH 32-chloro-2-methyl hexaneCH 3 – CH – CH – CH 2 CH 2 CH 3CH 3Cl3-chloro-2-methyl hexaneCH 3 – C = CH.CH 2 CH 2 CH 3CH 32-methyl hexene-23. An unsaturated hydrocarbon (A), C 6 H 10 readily gives(B) on treatment with NaNH 2 in liquid NH 3 . When(B) is allowed to react with 1-chloro propane, acompound (C) is obtained. On partial hydrogenationin the presence of Lindlar catalyst (C) gives (D),C 9 H 18 . On ozonolysis (D) gives 2, 2-dimethylpropanal and butanal. Give structures of (A), (B), (C)and (D) with proper reasoning.Sol. The structure of compound (D) can be obtained byjoining the products of ozonolysis.CH 3–2[O]CH 3 – C – CH = O + O = CH.CH 2 CH 2 CH 3ButanalCH 32,2-dimethyl propanal CH 3CH 3 – C – CH = CH.CH 2 CH 2 CH 3CH 32,2-dimethyl heptene-3 (D)Ozonolysis equation of (D) is :CH 3(I) O 3CH 3 – C – CH = CHCH 2 CH 2 CH 3(II) H 2O/ZnCH 3(D)CH 3CH 3 – C – CHO + CH 3 CH 2 CH 2 CHOCH 3Alkene (D) is obtained by the partial hydrogenationof (C), thus (C) contains a – C≡C – triple bond at C 3 .CH 3CH 3 – C – C ≡ C – CH 2 CH 2 CH 3CH 3(C)CH 3H 2Lindlar catalystCH 3 – C – CH = CHCH 2 CH 2 CH 3CH 3Main product (Saytzeff's rule)CH 3(D)XtraEdge for IIT-JEE 35 APRIL 2010

The starting compound (A) reacts with NaNH 2 inpresence of liquid NH 3 . It means it contains one–C≡CH at the terminal carbon, and, therefore gives amono sodium derivative.C 6 H 10(A)NaNH 2⎯⎯NH3⎯ → C 4 H 9 –C ≡ C.Na(B)Compound (B) reacts with 1-chloro propane to givecompound (C) as follows :C 4 H 9 – C ≡ C – Na + Cl – CH 2 CH 2 CH 3(B)1-chloro propane∆–NaClC 4 H 9 – C≡C – CH 2 CH 2 CH 3(C)CH 3But, (C) is CH 3 – C – C ≡ C – CH 2 CH 2 CH 3CH 3Now, putting the value of C 4 H 9 as a t-butyl radical,we have :CH 3NaNHCH 3 – C – C≡C – H 2CH 3 – C – C≡CNa + NH 3Hence,(A)(B)(C)(D)CH 3(A)CH 3CH 3 – C – C≡CHCH 3(3,3-dimethyl butyne-1)CH 3CH 3 – C – C≡CNaCH 3CH 3CH 3 – C – C≡C – CH 2 CH 2 CH 3CH 3CH 3CH 3CH 3(B)CH 3 – C – CH = CHCH 2 CH 2 CH 3CH 34. A hydrocarbon (A) [C = 90.56%, V.D. = 53] wassubjected to vigrous oxidation to give a dibasic acid(B). 0.10 g of (B) required 24.10 ml of 0.05 N NaOHfor complete neutralization. When (B) was heatedstrongly with soda-lime it gave benzene. Identify (A)and (B) with proper reasoning and also give theirstructures.Sol. Determination of empirical formula of (A) :Element %Atomicwt.C 90.56 12H 9.44 1Relative no.of atoms90.56= 7.55129.441Simplest ratio7.55= 1 or 47.559.44= 1.25= 9.44 7. 55or 5The empirical formula of (A) = C 4 H 5Empirical formula weight = 48 + 5 = 53Molecular weight = V.D. × 2 = 53 × 2 = 106Molecular wt. 106Hence, n == = 2Empirical wt. 53Molecular formula = 2 × C 4 H 5 = C 8 H 10The given equation may be outlined as follows :COOHVigrous oxidationC 8 H 10C6[O]6 H 4(A)(B) COOH + 2H 2OMeq. of dicarboxylic acid = Meq. of NaOH0 .1×1,000= 24.1 × 0.05EEquivalent of acid = 83Molecular wt. = Basicity × Equivalent weight= 2 × 83 = 166Since (B) on heating with soda-lime gives benzene,the C 6 H 4 represents to benzene nucleus having twoside chains, thus (B) is a benzene dicarboxylic acid.There are three benzene dicarboxylic acids.COOH COOHCOOHCOOH COOHCOOHPhthalic acid Isophthalic acid Terphthalic acidAll the above three acids are obtained by theoxidation of respectively xylenes.CH 3 6[O] COOH+ 2H 2 OCH 3COOHo-xyleneCH 3CH 3m-xyleneCH 3CH 3p-xylene6[O]6[O]COOHCOOHCOOHCOOH+ 2H 2 O+ 2H 2 OXtraEdge for IIT-JEE 36 APRIL 2010

All the above three acids on heating with soda-limeyields only benzene.COOH COOH, ,COOHCOOHCOOHCOOHNaOH + CaO∆+ 2CO 2Of the three acids, one which on heating gives ananhydride, is o-isomer.COOHCOOH∆–H 2 OCOCOOne acid which on nitration gives a mono nitrocompound is p-dicarboxylic acid.COOHCOOHCOOHHNO3NO 2∆; H 2SO 4COOHOne acid which on nitration gives three mono nitrocompounds will be the m-isomer.COOHCOOHHNO 3H 2SO 4COOHNO 2COOHOCOOHNO 2COOHCOOHNO 2COOH5. Two moles of an anhydrous ester (A) are condensedin presence of sodium ethoxide to give a β-keto ester(B) and ethanol. On heating in an acidic solutioncompound (B) gives ethanol and a β-keto acid (C).(C) on decarboxylation gives (D) of molecularformula C 3 H 6 O. Compound (D) reacts with sodamideto give a sodium salt (E), which on heating with CH 3 Igives (F), C 4 H 8 O, which reacts with phenyl hydrazinebut not with Fehling reagent. (F) on heating with I 2and NaOH gives yellow precipitate of CHI 3 andsodium propionate. Compound (D) also givesiodoform, but sodium salt of acetic acid. The sodiumsalt of acetic acid on acidification gives acetic acidwhich on heating with C 2 H 5 OH in presence of conc.H 2 SO 4 gives the original ester (A). What are (A) to(F) ?Sol. (i) Acetic acid on heating with C 2 H 5 OH givesoriginal compound (A).CH 3 COOH + C 2 H 5 OH⎯ H ⎯ 2 SO ⎯4∆→CH3COOC2H5(A)+ H 2 O(ii) CH 3 COOC 2 H 5 (A) on heating with C 2 H 5 ONaundergoes Claisen condensation to give (B), which isaceto acetic ester.CH 3 CO OC 2 H 5 + H CH 2 COOC 2 H 5(A)C 2H 5ONaReflux+ C 2 H 5 OH + CH 3 COCH 2 COOC 2 H 5(B)(iii) (B) on heating in acidic solution gives (C) andethyl alcohol.CH + HOH ⎯→3COCH2COOC2H5(B)2(C)H⎯ +CH 3COCHCOOH + C 2 H 5 OH(iv) (C) on decarboxylation gives acetone (D).CH 3COCHCOOH2(C)∆⎯ ⎯→−CO 2CH 3COCH 3(D)(v) (D) reacts with NaNH 2 to form sodium salt (E),which on heating with CH 3 I gives butanone (F).CH 3COCH 3 + NaNH 2(D)∆⎯ ⎯→−NH 3CH3COCH2Na(E)CH 3⎯ I ⎯ ⎯–NaI → CH 3COCH2CH3(F)(vi) CH 3COCH2CH3+ 3I 2 + 4NaOH ⎯→(vii)(F)⎯ ∆CHI 3 + CH 3 CH 2 COONa + 3NaI + 3H 2 OCH 3COCH 3 + 3I 2 + 4NaOH ⎯⎯→∆(D)CHI 3 + CH 3 COONa + 3NaI + 3H 2 OCH 3 COONa ⎯ HCl ⎯→ CH 3 COOH + NaClThus, (A) CH 3 COOC 2 H 5(B) CH 3 COCH 2 COOC 2 H 5(C) CH 3 COCH 2 COOH(D) CH 3 COCH 3(E) CH 3 COCH 2 Na(F) CH 3 COCH 2 CH 3XtraEdge for IIT-JEE 37 APRIL 2010

`tà{xÅtà|vtÄ V{tÄÄxÇzxá12SetThis 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 Maheshwari<strong>Solution</strong>s published in this issueJoint Director Academics, <strong>Career</strong> <strong>Point</strong>, Kota1. Prove that, if n is a positive integer,a∫ − x0ne x dx =⎪⎧⎛⎞⎪⎫n ! ⎨ − − 2 na⎜a a1 e 1+a + + ... + ⎟⎬⎪⎩ ⎝ 2! n!⎠⎪⎭Also, deduce the value of∫ ∞e2. Let A ≡ (6, 5), B ≡ (2, –3) and C ≡ (–2, 1) be thevertices of a triangle. Find the point P in the interiorof the triangle such that ∆PBC is an equilateraltriangle.3. If S n = n C 1 + 2. n C 2 + 3. n C 3 + ....... + n n C n then findn∑n=1S . Also prove thatn0−xn C 1 . ( n C 2 ) 2 . ( n C 3 ) 3 .... ( n C n ) n ≤xndx⎛n2 ⎞⎜ ⎟n 1⎝ + ⎠4. Let z 1 , z 2 , z 3 be three distinct complex numberssatisfying|z 1 – 1| = |z 2 – 1| = |z 3 – 1|; Let A, B and C be thepoints represented in the argand plane correspondingto z 1 , z 2 and z 3 respectively. Prove that z 1 + z 2 + z 3 =3 if and only if ∆ABC is an equilateral triangle.5. Let A ≡ (r, 0) be a point on the circle x 2 + y 2 = r 2 andD be a given point inside the circle. If BC be anyarbitrary chord of the circle thorugh point D. Provethat the locus of the centroid of triangle ABC is acircle whose radius is less than r/3.n+1C2.6. A number is chosen at random from the set{1, 2, 3, …......, 2006}. What is the probability thatit has no prime factor in common with10 ! ?7. Two vertices of a triangle are a – î + 3ĵand 2 î + 5ĵand its orthocenter is at ( î + 2ĵ). Find the positionvector of third vertex.8. Show that an equilateral triangle is a triangle ofmaximum area for a given perimeter and a triangle ofminimum perimeter for a given area.9. The bottom of a tank with a capacity of 300 litres iscovered with a mixture of salt and some insolublesubstance. Assuming that the rate at which the saltdissolves is proportional to the difference betweenthe concentration at the given time and theconcentration of a saturated solution (1 kg of salt per3 litres of water) and that the given quantity of purewater disolves 1/3 kg of salt in 1 minute. Find thequantity of salt in solution at the expiration of onehour.10. An isosceles triangle with its base parallel to themajor axis of the ellipsex 2 +9y 2 = 1 is3circumscribed with all the three sides touchingthe ellipse. The least possible area of thetriangle is.XtraEdge for IIT-JEE 38 APRIL 2010

MATHEMATICAL CHALLENGESSOLUTION FOR MARCH ISSUE (SET # 11)1. Let z = x + iy so given1 ≤ x or 1 ≤ r cos θ , if z = ∑e iθ Now,1−x − iy (1 − x − iy)(1 + x − iy)=2 21+x + iy (1 + x) + y222(1 − x ) − yReal part≤ 0 as x ≥ 1 given and2 2(1 + x) + yimaginary part−(1− x)y − (1 + x)y −2y=≤ 02 22 2(1 + x) + y (1 + x) + yas y ≥ 0 given1−zso ∩ 0 is true.1+z2. As ∠ POQ = 90ºQso CP = (OC) . tan (90 – θ)= (OC) cot θ & CQ = (OC) tan θso CP. CQ = (OC) 2 = r 2when r is the radius of circle.3. f(0) = cf(1) = a + b+ c& f(–1) = a – b + cO90–θθsolving these, a = 21 [f(1) + f(–1) – 2f(0)]Cb = 21 [f(1) – f(–1)] & c = f(0)x (x + 1)so f(x) = f(1) + (1 – x 2 x(x−1)) f(0) + f(–1)222 | f(x) | ≤ |x| |x + 1| + 2 |1 – x 2 | + |x| |x – 1|; as|f(1)|, |f(0)|, |f(–1)| ≤ 1.2 |f(x)| ≤ |x| (x + 1) + 2(1 – x 2 ) + |x| (1 – x) asx ∈ [–1, 1]so 2 |f(x)| ≤ 2(|x| + 1 – x 2 5) ≤ 2 . 4so |f(x)| ≤ 45P4.=Now, as g(x) = x 2 ⎛ 1 ⎞f ⎜ ⎟⎝ x ⎠= 21 (1 + x) f(1) + (x 2 – 1) f(0) + 21 (1 – x) f(–1)so 2 | g(x) | ≤ |x + 1| + 2 |1 – x 2 | + |1 – x|2 | g(x) | ≤ x + 1 + 2(1 – x 2 ) + 1 – xas x ∈ [–1, 1]2 |g(x)| ≤ –2x 2 + 4 ≤ 4|g(x)| ≤ 21⎡⎤⎢2y1− y´⎥a ⎢ 2 2 ⎥⎣ 2 a − y ⎦=a +a12− y2.2−2y.y´a2− y2– y1 y´1⎡= y´a⎥ ⎥ ⎤⎢yy1−−⎢ 2 22 2 2 2⎣aa − y (a + a − y ) a − y y⎦2y´⎡y (a +⎢⎣y(a +a222ay(a +a2− y ) − y a − a(a +2a − y )22− ya − y22)a2a2− y− y22)a2− y= y´[y 2 a + y 2 a 2 − y2 – y 2 a – a 2 a 2 − y2 – a(a 2 – y 2 )]y (a +2y´ = –y´M =2a − y )a22ay− y−β− β22a − y = –y´(a 2 – y 2 ) (a +22222⎤⎥⎦a − y )y =−βx...(1)2 2a − βy = β ...(2)locus of intersection of these two lines isy =−yx2 2a − ya 2 – y 2 = x 2x 2 + y 2 = a 2XtraEdge for IIT-JEE 39 APRIL 2010

MATHEMATICAL CHALLENGESSOLUTION FOR THIS ISSUE (SET # 12)a −xna − x −1x n1. I n = ∫ e x dx0 = ( e − x x n ) a 0+ n ∫ e0 dx= – e –a a n + nI n – 1I n = –e –a a n + n[–e –a a n–1 + (n–1)I n–2 ]= –e –a [a n + na n–1 +n(n – 1)a n–2 + n(n – 1)(n – 2)a n–3+ ..... + n(n – 1)..... 2a] + n I 0=⎡⎪⎧n n−12 ⎤−xa −aa a a ⎪⎫n ⎢( − e )0− e ⎨ + + .... + + a⎬⎥⎢⎣⎪⎩ n n −12 ⎪⎭ ⎥⎦I n =⎡ ⎪⎧⎪⎫⎤⎢ − −2na a an 1 e ⎨1+ a + + ..... + ⎬⎥⎢⎣⎪⎩ 2 n ⎪⎭ ⎥⎦Now, I = ∫ ∞ −e x x n =2. Mid pt. M of BC = (0, –1)0C(–2, 1) 60ºM45ºLt I n = na→∞B(2, –3)−3−1slope of BC = = –12 + 2so slope of altitude of ∆PBC is = 1.length BC = 16 + 16 = 4 2Now altitude PM = 4 2 sin 60ºP= 4 2 .A(6, 5)3 = 2 62eq n . of PM line isx − 0 y + 1 = = r (as its slope is 1)1 12 2r rx = & y = – 12 2for req. pt. P take r = 2 6 .⎛⎞So pt. P ≡ ⎜2 6 2 6, −1⎟= ( 2 3, 2 3 − 1)⎝ 2 2 ⎠n3. S n = ∑r=0n1n=1r . n C r = n . 2 n–1so S = ∑S = ∑ nnn=1n.2−1S = 1 + 2.2 1 + 3.2 2 + 4.2 3 + .... + n . 2 n–12S = 2 + 2.2 2 + 3.2 3 + ..... + (n – 1). 2 n–1 + n.2 n(1 – 2)S = (1 + 2 + 2 2 + 2 3 + .... + 2 n–1 ) – n . 2 n2 n 1= 1 . – n . 2n2 −−1S = n . 2 n – 2 n + 1 = (n – 1) 2 n + 1Now A.M. ≥ G. M.n n nnC1+ 2. C2+ 3. C3+ .... + n. Cn≥1+2 + 3 + ..... + n⎧ n⎨ C1.⎩n 2 n 3 n n 1+2+... + n( C ) .( C ) ......( C )⎫23n(n+1)2n⎬⎭⎛ n−1⎞⎜ n.2 ⎟⎜ n(n + 1) ⎟ ≥ n C 1 ( n C 2 ) 2 ..... ( n C n ) n⎜ ⎟2⎝ ⎠⎛nso n C 1 . ( n C 2 ) 2 ..... ( n C n ) n 2 ⎞≤ ⎜ ⎟n 1⎝ + ⎠n+1C24. Let P be a point represented by 1.so as |z 1 – 1| = |z 2 – 1| = |z 3 – 1| so P is thez1 + z2+ z3circumcentre of ABC. Its centroid is3If the ∆ABC is equilateral then circumcentre = centroidsoz1 + z2+ z31 =3so z 1 + z 2 + z 3 = 3Now if z 1 + z 2 + z 3 = 3 then centroid of ∆ABC is 1which is point P and P is already the circumcentre of∆ABC. So now if they are same then ∆ABC isequilateral.5. let the centroid of ∆ABC be (h, k) then3h = r cos α + r cos β + r3hα + β α −β– 1 = 2 cos cosr2 2...(1)&3kr= sin α + sin βα −β α + β= 2 cos sin2 2...(2)1XtraEdge for IIT-JEE 41 APRIL 2010

α + β α + βline BC, x cos + y sin2 2Let point D be (a, b)thenB(β)a cosDα + β2+ b sinC(α)α + β2A(r, 0)= r cos= r cosα − β2α −β2....(3)α −βMultiply (3) by cos2α −β α + β α + β α − βa cos cos + b sin cos2 22 2= r cos 2 α − β2use (1) & (2)a ⎛ 3h ⎞ 3k⎜ −1⎟+ b = r cos 2 α − β...(4)2 ⎝ r ⎠ 2r2square & add (1) & (2)2 24 cos 2 α − β ⎛ 3h ⎞ 9k= ⎜ − 1⎟+...(5)22 ⎝ r ⎠ rfrom (4) & (5)2 21 ⎛ 3h ⎞ 9k a ⎛ 3h ⎞ b3k⎜ − 1⎟+ = ⎜ −1⎟ +24 ⎝ r ⎠ 4r 2r ⎝ r ⎠22rso req. locus is(3x – r) 2 + 9y 2 = 2a(3x – r) + 6.b.y9x 2 + 9y 2 – 6rx + r 2 = 6ax – 2ar + 6.b.yx 2 + y 2 2 2 r 2– (r + a)x – by + +3 3 9⎡ 1 ⎤⎢x− (r + a)3⎥⎣ ⎦=2(r + a)1 ⎤⎢⎡ + y − b3⎥⎣ ⎦222− r − 2ar + b92 +2=a2 +2ar9b92= 0a bIt is a circle and radius is3Since point D is interior of circle so a 2 + b 2 < r 2 ,so radius of this circle is less than r/3.6. Prime factors dividing 10 are 2, 3, 5, 7. As requiredthe number chosen should not be divisible by 2 or 3or 5 or 7. Define, the events asA : divisible by 2B : divisible by 3C : divisible by 5D : divisible by 72A ∪ B ∪ C ∪ D= A + B + C + D – A ∩ B – A ∩ C – A ∩ D– B ∩ C – B ∩ D – C ∩ D + A ∩ B ∩ C+ A ∩ C ∩ D + B ∩ C ∩ D + A ∩ B ∩ D– A ∩ B ∩ C ∩ DP(A ∪ B ∪ C ∪ D)1= [ 1003 + 668 + 401 + 286 – 334 – 2002006– 143 – 133 – 95 – 57 + 66 + 28 + 19 + 47 – 9]1547= 20067. Line BCBb = –i + 3jAaDHCc = 2i + 5j→ r = – î + 3ĵ+ t (3î + 2ĵ)...(1)any pt. D on it = (3t – 1) i + (3 + 2t) jAs HD ⊥ BC,so ((3t – 1 – 1) i + (3 + 2t – 2) j). (3i + 2j) = 03(3t – 2) + 2(2t + 1) = 0413 t – 4 = 0 ⇒ t = 13so point D = – 13i47 j+ 13Now line HD ⇒ → ⎛r = i + 2j + s´ ⎜−⎝= i + 2j + s(–14i + 21j)i1347 j ⎞+ − i − 2j⎟ 13 ⎠⇒ → r = i + 2j + λ(–2i + 3j)Any point A on it = (1 – 2λ)i + (2 + 3λ)jNow as AC ⊥ BHso [ (1 – 2λ – 2)i + (2 + 3λ – 5)j] . [2i – j] = 02 (–1 – 2λ) – (3λ – 3) = 01⇒ –7λ = 2 – 3 ⇒ λ = 75 i 17 jso pt. A = + 7 78. If A is the area of the triangle with sides a, b, and c,then A 2 = s(s – a) (s – b) (s – c);where 2s = a + b + cusing AM – GM inequality for s – a, s – b, s – c, wehave3A 2 ⎧(s− a) + (s − b) + (s − c) ⎫≤ s ⎨⎬⎩ 3 ⎭XtraEdge for IIT-JEE 42 APRIL 2010

A 2 ⎛ 3s − 2s ⎞ s≤ s ⎜ ⎟ =⎝ 3 ⎠ 3A ≤3s 23Let 2s = p,343p 2then A ≤12p 2A max = ,12 3As condition of equality holds iffs – a = s – b = s – c which happen if a = b = cso A max = ; for a = b = cNow again p ≥ 12 3Ap min. = 12 3A;and again equality holds if a = b = c.3at t = 60 min.⎡60⎛ 299 ⎞ ⎤x = 100 ⎢1 − ⎜ ⎟ ⎥ kg⎢⎣⎝ 300 ⎠ ⎥⎦10. Let P (3 cos θ, 3 sin θ)line BC : y = – 3 ; line AC :ABPC9. Let the amount of salt dissolved at any time t is x kg.xSo concentration is 300dx ⎛ 1 x ⎞ ⎛100− x ⎞so = k ⎜ − ⎟⎠ = k ⎜ ⎟dt ⎝ 3 300 ⎝ 300 ⎠dx100 − x= k dt300k– ln (100 – x) = t + C 300at t = 0, x = 0 so C = –ln 100kt100so = ln 100 – ln (100 – x) = ln300100 − xat t = 1 min., x = 1/3k 100so = ln3001100 −3k 100so = ln300 2993k 300so = ln300 299300so k = 300 ln 299300 100so ln . t = ln 299 100 − xso⎛ 300 ⎞⎜ ⎟⎠⎝ 299t100=100 − x⎛ 299 ⎞so 100 – x = 100 . ⎜ ⎟⎠⎝ 300⎡t⎛ 299 ⎞ ⎤so x = 100 ⎢1− ⎜ ⎟ ⎥⎢⎣⎝ 300 ⎠ ⎥⎦tx y cos θ + sin θ = 13 3⎛ 3(1 + sin θ)⎞pt. C ⎜ , − 3 ⎟⎝ cosθ⎠pt. A (0, 3 cosec θ )1 3(1 + sin θ)Area A = . 2 .. ( 3 cosec θ + 3 )2 cosθ223 3(1 + sin θ)6 3(1 + sin θ)==sin θcosθsin 2θ2dA 6 3(2(1 + sin θ)sin 2θcosθ − 2(1 + sin θ)cos 2θ)=2dθsin 2θ12 3(1 + sin θ)(sin 2θcosθ − cos 2θ − sin θcos 2θ)=2sin 2θ12 3(1 + sin θ)(sin(2θ − θ)− cos 2θ)=2sin 2θ212 3(1 + sin θ)(sinθ −1+2sin θ)=2sin 2θ12 3(1 + sin θ)(2sinθ −1)(sinθ + 1)=2sin 2θ212 3(1 + sin θ)(2sin θ −1)=2sin 2θπA min at θ = 6so A min =6⎛ 1 ⎞3⎜1+ ⎟⎝ 2 ⎠ 9= 12 . 4322= 27sq. units.= 0027 Ans.XtraEdge for IIT-JEE 43 APRIL 2010

MATHSStudents' ForumExpert’s <strong>Solution</strong> for Question asked by IIT-JEE Aspirants1. Let f : R → R and f(x) = g(x) + h(x), where g(x) is apolynomial and h(x) is a continuous anddifferentiable bounded function on both sides, thenf(x) is onto if g(x) of odd degree and f(x) is into ifg(x) is of even degree. Then check whether f(x) isone one, many one, onto or into.(i) f(x) = a 1 x + a 3 x 3 + a 5 x 5 + .... + a 2n + 1 x 2n + 1 – cot –1 xwhere 0 < a 1 < a 2 < a 3 ....... < a 2n + 144x(x + 1)(x + 1) + x + 2(ii) f(x) =2x + x + 1Sol. (i) f(x) = odd degree polynomial + bounded functioncot –1 x also f´(x) > 0⇒ y = f(x) will be one one and onto42(x + 1)(x + x + 1)(ii) f(x) == x 4 1+ 1 +22x + x + 1x + x + 1= even degree polynomial⎛ 4 ⎞+ bounded function ∈ ⎜0, ⎟⎝ 3 ⎠also f´(x) = 0 has at least one root real which is notrepeated since f´(x) is a polynomial of degree 7.⇒ f(x) = 0 has at least one point of extrema.⇒ many one & Into2. A rectangle ABCD of dimensions r and 2r is foldedalong the diagonal BD such that planes ABD andCBD are perpendicular to each other. Let the positionof the vertex A remains unchanged and C 0 is the newposition of C, then find the distance of C 0 from Aand shortest distance between the edges AB & C 0 D.(0, r) DN(2r, r)CA(0, 0)B (2r, 0)Sol. Let the rectangle ABCD lies on the plane xy. Afterfolding the rectangle along the BD co-ordinates ofpoints in 3-D are-A : (0, 0, 0), B : (2r, 0, 0), C : (2r, r, 0), D(0, r, 0)⎛ 2r r ⎞ ⎛ 2rand N : ⎜ , ,0⎟ and C 0⎝ 5 5⎜ ,⎠ ⎝ 5r52r ⎞, ⎟ ,5 ⎠Now AC 0 =85 r5and shortest distance =AC0.(AB×C0D)=| AB×C D |05 r3unit.3. Let f(x) is a polynomial one-one function such thatf(x).f(y) + 2 = f(x) + f(y) + f(xy) ∀ x, y ∈ R–{0},xf(1) ≠ 1, f´(1) = 3. Let g(x) = (f(x) + 3) – 4 ∫ xf (x) dx,then prove that g(x) is an identity for all givenx ∈ R –{0}.Sol. putting x = y = 1 in given condition we getf(1) 2 + 2 = f(1) + f(1) + f(1)⇒ {f(1)} 2 – 3{f(1)} + 2 = 0⇒ f(1) = 1 or 2 ⇒ f(1) = 2Now put y = 1/x,⎛ 1 ⎞⎛ 1 ⎞f(x) . f ⎜ ⎟ + 2 = f(x) + f ⎜ ⎟ + f(1)⎝ x ⎠ ⎝ x ⎠⇒ f(x) = 1 ± x nAccording to given conditions, f(x) = 1 + x 3xNow, g(x) = [1 + x 3 + 3] – 4 ∫ x(1 + x 3 ) dx = 0⇒ g(x) = 0 for ∀ x ∈ R –{0}4. Let three normals are drawn to the parabola y 2 = 4axat three points P, Q and R, from a fixed point A. Twocircles S 1 and S 2 are drawn on AP and AQ asdiameter. If slope of the common chord of the circlesS 1 and S 2 be m 1 and the slope of the tangent to theparabola at the point R be m 2 , then prove thatm 1 . m 2 = 2.Sol. Let A(h, k) be a fixed pointat 3 + (2a – h)t – k = 0{Q three normals are drawn from (h, k)}Let feet of normals P,Q and R are three points withparameters t 1 , t 2 and t 3 .Common chord of S 1 and S 2 = S 1 – S 2 = (t 1 + t 2 )x +2y – h(t 1 + t 2 ) – 2k = 02Tangent to the parabola at R = t 3 y = x + at 3⎛ t1 + t 2 ⎞ 1 1m 1 . m 2 = – ⎜ ⎟ . =⎝ 2 ⎠ t 3 2(Q t 1 + t 2 + t 3 = 0 for co-normal points).00XtraEdge for IIT-JEE 44 APRIL 2010

5. Let z 1 , z 2 and z 3 are unimodular complex numbersthen find the greatest value of |z 1 – z 2 | 2 + |z 2 – z 3 | 2 +|z 3 – z 1 | 2 .Sol. |z 1 – z 2 | 2 + |z 2 – z 3 | 2 + |z 3 – z 1 | 2= 2[|z 1 | 2 + |z 2 | 2 + |z 3 | 2 ] – [z 1 z 2 + z1z 2 + z1z 3+ z3z 1 + z 2 z 3 + z2z 3 ]= 6 – [z 1 z 2 + z 2 z 1 + z 1 z 3 + z 3 z 1 +z 2 z 3 + z 3 z 2 ]...(1)Now |z 1 + z 2 + z 3 | 2 ≥ 0⇒ z 1 z 2 + z 2 z 3 + z 2 z 1 + z 3 z 2 + z 1 z 3 + 1From (1) & (2)max m value of |z 1 – z 2 | 2 + |z 2 – z 3 | 2 + |z 3 – z 1 | 2= 6 – (–3) = 9z z 3 ≥ –3...(2)6. Consider following two infinite series in real θ and rC = 1 + r cos θ +22r cos 2θ r 3 cos3θ + + ....2! 3!r 2 sin 2θ r 3 sin 3θ S = r sin θ + + + .....2! 3!If a remains constant and r varies the prove thatdC dS(i) C + S = (C 2 + S 2 ) cos θdr dr⎛ dC ⎞(ii) ⎜ ⎟⎠⎝ drSol. We have2C + iS =C – iS =Now,C 2 + S 2 =⎛ dS ⎞+ ⎜ ⎟⎠⎝ dr2= C 2 + S 2iθree ...(1)−iθree ...(2)re i2e θ =ercosθ.e2i rsin θ= e 2r cos θ ...(3)Differentiating (1) w.r.t.r, we getdC dS + i = e iθ iθr.e. e ....(4)dr dr2⎛ dC ⎞ ⎛ dS ⎞iθiθr.e∴ ⎜ ⎟⎠ + ⎜ ⎟⎠ = e .e = e⎝ dr ⎝ dr= C 2 + S 2 (Form 3)multiply (2) and (4)⎛ dC dS ⎞⎜ + i ⎟ (C – iS) = e iθ iθr.e. e . e⎝ dr dr ⎠= e iθ i ir(e e )[eθ − θ+]2−iθr.e= e iθ 2r cosθ. eNow equating real parts in both sidesdC dSC + S = (C 2 + S 2 ) cosθdr drHence proved.2r cos θScience FactsSkin Deep StorageChip implants that keeptrack of personalinformation seem like anovelty but do they have amore useful future?These days, some people are following theirpets and getting tagged. Radio frequency identification(RFID) chips are the size of a grain of rice and can beloaded up with personal information like passwordsand implanted under the skin. Instead of having toremember a login code, an RFID reader can be set upto automatically detect it and grant you access to arange of things from your computer to your frontdoor. It seems like it could be useful to people withexceptionally poor memories, but right now thesechips are being snapped up by technology geeks likeAmal Graafstra. The 29-year-old businessman fromVancouver, Canada, is one of the first people to havean RFID implant and so far is happy with the results. "Ijust don't want to be without access to the things thatI need to get access to. In his chip, he has stored aunique identification number which can be used to loghim into various electronic devices. It didn't cost himan arm and a leg either: he got the whole set-up onthe internet for about $50 (£30), including the $2 costof the chip itself.The procedure to implant the chip is quitesimple and painless. Amal's chip was implanted underthe skin of his left hand while he was under a localanesthetic. It is possible to inject the chips using alarge enough needle, but in Amal's case the chip wasinserted by simply cutting through his skin with ascalpel. Other than complaining of sensitivity in thearea of the implant, Amal said that it doesn't hurt andhe expects that eventually the chip will be completelyunobtrusive.A hand implanted with anRFID chip and the chip reader.The chip is made of silicon and isdigitally encoded with information..A RFID reader, which is installed in a computer or anelectronic device like a reader by a front door, emits aradio signal of a particular frequency, just like radiostations each broadcast on their own frequency. Thechip acts passively when it is within 3 inches of thereader: the right incoming radio signal induces justenough energy in the antenna of the chip for a circuitin the chip to power up and produce a response. Thereader can then access the information on the chip andpass it on to the computer or device that requires it.XtraEdge for IIT-JEE 45 APRIL 2010

MATHSCALCULUSMathematics FundamentalsFunction, Limits, Continuity & Differentiability :If the domain of the function is in one quadrant thenthe trigonometrical functions are always one-one.If trigonometrical function changes its sign in twoconsecutive quadrants then it is one-one but if it doesnot change the sign then it is many one.In three consecutive quadrants tigonometricalfunctions are always many one.Any continuous function f(x), which has at least onelocal maximum, is many-one.Any polynomial function f : R → R is onto if degreeof f is odd and into if degree of f is even.An into function can be made onto by redefining thecordomain as the range of f is even.An into function can be made onto by redefining thecodomain as the range of the original function.1If f(x) is periodic with period T then is alsof (x)periodic with same period T.If f(x) is periodic with period T, f (x)is alsoperiodic with same period T.Period of x – [x] is 1. Period of algebraic functionsx , x 2 , x 3 + 5 etc. does't exist.If lim f (x)does not exist, then we can not removex→athis discontinuity. So this become a non-removablediscontinuity or essential discontinuity.If f is continuous at x = c and g is discontinuous at x= c, then(a) f + g and f – g are discontinuous(b) f.g may be continuousIf f and g are discontinuous at x = c, then f + g, f – gand fg may still be continuous.<strong>Point</strong> functions (domain and range consists one valueonly) is not a continuous function.If a function is differentiable at a point, then it iscontinuous also at that point.i.e., Differentiability ⇒ Continuity, but the converseneed not be true.If a function 'f' is not differentiable but is continuousat x = a, it geometrically implies a sharp corner orkink at x = a.If f(x) and g(x) both are not differentiable at x = athen the product function f(x).g(x) can still bedifferentiable at x = a.If f(x) is differentiable at x = a and g(x) is notdifferentiable at x = a then the sum function f(x) +g(x) is also not differentiable at x = a.If f(x) and g(x) both are not differentiable at x = a,then the sum function may be a differentiablefunction.Differentiation and Applications of Derivatives :dy d dis (y) in which is simply a symbol ofdx dx dxoperation and not 'd' divided by dx.If f´(x 0 ) = ∞, the function is said to have an infinitederivative at the point x 0 . In this case the line tangentto the curve of y = f(x) at the point x 0 isperpendicular to the x-axis.Of all rectangles of a given perimeter, the square hasthe largest area.All rectangles of a given area, the squares has theleast perimeter.A cone of maximum volume that can be inscribed in4 ra sphere of a given radius r, is of height . 3A right circular cylinder of maximum volume thatcan be inscribed in a square of radius r, is of height2 r .3If at any point P(x 1 , y 1 ) on the curve y = f(x), thetangent makes equal angle with the axes, then at theπ 3π dypoint P, ψ = or . Hence, at P tan ψ = = ±14 4dxIndefinite Integral :If F 1 (x) and F 2 (x) are two antiderivatives of afunction f(x) on an interval [a, b], then the differencebetween them is a constant.The signum function has an antiderivative on anyinterval which does not contain the point x = 0, anddoes not possess an anti=derivative on any intervalwhich contains the point.The antiderivative of every odd function is an evenfunction and vice-versa.XtraEdge for IIT-JEE 46 APRIL 2010

If I n =∫x n . e ax dx, then I n =nx eaIf I n =∫(log x) dx , then I n = x log x – xIf I n =∫1dx, thenlog xI n = log(logx) + logx +2(log x)2.(2!)ax+– anIn–13(log x)3.(3!)nIf I n =∫(log x) dx ; then I n = x(logx) n – n.I n–1+ ...Successive integration by parts can be performedwhen one of the functions is x n (n is positive integer)which will be successively differentiated and theother is either of the following sin ax, cos ax, e –ax , (x+a) m which will be successively integrated.Chain rule :∫u .v dx = uv 1 – u´v 2 + u"v 3 – u"'v 4 + ....+ (–1) n – 1 u n–1 v n + (–1) n ∫u n .v dxwhere u n stands for n th differential coefficient of uand v n stands for n th integral of v.∫axxe sin(bx + c)dx =cos(bx + c)] –(acos (bx + c)] + k∫ax2eax+ bxe cos(bx + c)dx =sin(bx + c)] –(asin (bx + c)] + k∫xeaxa=(loga)∫xeax2eax+ bsin(bx + c)dxxa=(loga)2+ b2cos(bx + c)dxx2+ ba cos x + bsin x∫dxccos x + d sin x222))22aaxe2ax+ b2[a sin(bx + c) – b[(a 2 – b 2 )sin (bx + c) – 2abx.e2ax+ b2[a cos(bx + c) – b[(a 2 – b 2 )cos (bx + c) – 2ab[(loga)sin(bx + c) – b cos(bx + c)] + k[(loga)cos(bx + c) + b sin(bx + c)] + kac + bd ad − bc= x + log |c cos x + d sinx| + k.2 2 2 2c + d c + dsinReduction formulae for I (n,m) =∫cosn−1nmxxdx is1I (n,m) =m − 1. sin x ( n −1)– .Im−1(n–2, m – 2 )cos x (m –1)Definite Integral and Area Under Curves :1 bThe number f(c) =− ∫f (x)dx is called the(b a) amean value of the function f(x) on the interval [a, b].If m and M are the smallest and greatest values of afunction f(x) on an interval [a, b], then m(b – a) ≤∫ baf (x)dx ≤ M(b – a).If f 2 (x) and g 2 (x) are integrable on [a, b], then∫ ba⎛f (x)g(x)dx ≤ ⎜⎝∫abf2⎞(x)dx ⎟⎠1/ 2⎛∫⎜⎝ab1/ 22 ⎞g (x)dx ⎟⎠Change of variables : If the function f(x) iscontinuous on [a, b] and the function x = φ(t) iscontinuously differentiable on the interval [t 1 , t 2 ] anda = φ(t 1 ), b = φ(t 2 ), then∫ bat2t1f (x)dx =∫f ( φ(t)φ´(t)dtLet a function f(x, α) be continuous for a ≤ x ≤ b andc ≤ α ≤ d. Then for any α ∈[c, d], if I(α) =∫abf (x, α)dx, then I´(α) =∫f ´(x, α)dx, whereI´(α) is the derivative of I(α) w.r.t. α and f´(x, α) isthe derivative of f(x, α) w.r.t α, keeping x constant.∫ ba∫bf ´(x)dx = (b – a)∫f [(b − a)t + a]dtf (x)dxf (x) + f (a + b − xa )10ab= 21 (b – a)The area of the region bounded by y 2 = 4ax, x 2 = 4by16abis sq. unit.3The area of the region bounded by y 2 = 4ax and y =28amx is sq. unit.33mThe area of the region bounded by y 2 = 4ax and its8a2latus-rectum is sq. unit.3The area of the region bounded by one arch of sin(ax)or cos (ax) and x-axis is 2/sq. unit.Area of the ellipse (x 2 /a 2 ) + (y 2 /b 2 ) = 1 is πab sq. unit.Area of region bounded by the curve y = sin x, x-axisand the line x = 0 and x = 2π is 4 sq. unit.XtraEdge for IIT-JEE 47 APRIL 2010

MATHSALGEBRAMathematics FundamentalsComplex Numbers :|z| ≥ |Re(z)| ≥ Re(z) and |z| ≥ |1m(z) | ≥ 1m(z)z| z |is always a unimodular complex number if z ≠ 0|Re(z) | + |1m(z) | ≤If2 |z|1z + = a, the greatest and least values of |z| arezrespectively|z 1 +2 21 z2a +a22 +z − | + |z 2 –4and2 21 z2− a +a22 +z − | = |z 1 + z 2 | + |z 1 – z 2 |If z 1 = z 2 ⇔ |z 1 | = |z 2 | or arg z 1 = arg z 2|z 1 + z 2 | = |z 1 – z 2 | ⇔ arg (z 1 ) – arg(z 2 ) = π/2.If |z 1 | ≤ 1, |z 2 | ≤ 1 then(i) |z 1 + z 2 | 2 ≤ (|z 1 | – |z 2 |) 2 + (arg (z 1 ) – arg (z 2 )) 2(ii) |z 1 + z 2 | 2 ≥ (|z 1 | + |z 2 |) 2 – (arg (z 1 ) – arg(z 2 )) 2|z 1 + z 2 | 2 = |z 1 | 2 + |z 2 | 2 + 2||z 2 | cos(θ 1 – θ 2 ).|z 1 – z 2 | 2 = |z 1 | 2 + |z 2 | 2 – 2|z 1 ||z 2 | cos(θ 1 – θ 2 ).If z 1 and z 2 are two complex numbers then|z 1 z 2 | = r 1 r 2 ; arg(z 1 z 2 ) = θ 1 + θ 2 and⎛ z ⎞arg⎜1⎟ = θ 1 – θ 2 and where |z 1 | = r 1 , |z 2 | = r 2 ,⎝ z2⎠arg(z 1 ) = θ 1 and arg(z 2 ) = θ 2 .The area of the triangle whose vertices are z, iz andz + iz is 21 |z| 2 .The area of the triangle with vertices z, wz and3z + wz is |z 2 |.4If z 1 , z 2 , z 3 be the vertices of an equilateral triangleand z 0 be the circumcentre, then2z 1 +2z 2 +2z 3 = 3 z 2 0 .If z 1 , z 2 , z 3 ....... z n be the vertices of a regularpolygon of n sides and z 0 be its centroid, then2z 1 +2z 2 + .......+2z n = n z 02 .If z 1 , z 2 , z 3 be the vertices of a triangle, then thetriangle is equilateral ifzz12=4rr12,or(z 1 – z 2 ) 2 + (z 2 – z 3 ) 2 + (z 3 – z 1 ) 2 = 02z 1 +2z 2 +2z 3 = z 1 z 2 + z 2 z 3 + z 3 z 11 1 1or + + = 0z 1 − z 2 z 2 − z 3 z 3 − z 1If z 1 , z 2 z 3 are the vertices of an isosceles triangle,2 2 2right angled at z 2 then z 1 + z 2 + z 3 = 2z 2 (z 1 + z 2 )If z 1 , z 2 , z 3 are the vertices of a right-angled isoscelestriangle, then (z 1 – z 2 ) 2 = 2(z 1 – z 3 )(z 3 – z 2 ).If z 1 , z 2 , z 3 be the affixes of the vertices A, B, Crespectively of a triangle ABC, then its orthocentre isa 3(secA)z1+ b(secB)z 2 + (csecC)za sec A + bsecB + csecCFor any a, b ∈ R(i)(ii)a + ib + a − ib = 2{ a + b + a}a + ib – a − ib = 2{ a + b − a}The sum and product of two complex numbers arereal simultaneously if and only if they are conjugateto each other.If ω and ω 2 are the complex cube roots of unity, then(i) (aω + bω 2 )(aω 2 + bω) = a 2 + b 2 – ab(ii) (a + b (aω + bω 2 )(aω 2 + b 2 ω) = a 3 + b 3(iii) (a + bω + cω 2 )(a + bω 2 + cω)= a 2 + b 2 + c 2 – ab – bc – ca(iv) (a + b + c) (a + bω + cω 2 ) (a + bω 2 + cω)= a 3 + b 3 + c 3 – 3abcIf three points z 1 , z 2 , z 3 connected by relationaz 1 + bz 2 + cz 3 = 0 where a + b + c = 0, then the threepoints are collinear.If three complex numbers are in A.P. then they lie ona straight line in the complex plane.Progression :If T k and T p of any A.P. are given, then formula forTn− T Tk p − Tkobtaining T n is = .n − k p − kIf pT p = qT q of an A.P., then T p + q = 0.If p th term of an A.P. is q and the q th term is p, thenT p+q = 0 and T n = p + q – n.If the p th term of an A.P. is 1/q and the q th term is 1/p,then its pq th term is 1.2222XtraEdge for IIT-JEE 48 APRIL 2010

The common difference of an A.P. is given byd = S 2 – 2S 1 where S 2 is the sum of first two termsand S 1 is the sum of first term or the first term.If sum of n terms S n is given then general termT n = S n – S n–1 , where S n–1 is sum of (n – 1) terms ofA.P.If for an A.P. sum of p term is q and sum of q termsis p, then sum of (p + q) terms is {–(p + q)}.If for an A.P., sum of p term is equal to sum of qterms, then sum of (p + q) terms is zero.If the p th term of an A.P. is 1/q and q th term is 1/p,then sum of pq terms is given by S pq = 21 (pq + 1).Sum of n A.M.'s between a and b is equal to n timesthe single A.M. between a and b.⎛ a + b ⎞i.e. A 1 + A 2 + A 3 + ...... + A n = n ⎜ ⎟ .⎝ 2 ⎠If T k and T p of any G.P. are given, then formula for1⎛ T n kn⎞ − ⎛ Tp⎞ p−kobtaining T n is⎜T⎟ = ⎜ ⎟⎝ k ⎠ T⎝ k ⎠Product of n G.M.'s between a and b is equal to n thpower of single geometric mean between a and bi.e. G 1 G 2 G 3 .... G n = ( ab ) n .The product of n geometric means between a and 1/ais 1.⎛ ⎞If n G.M.'s inserted between a and b then r = ⎜ ⎟⎝ a ⎠11b n+1Quadratic Equations and Inequations :An equation of degree n has n roots, real orimaginary.If f(α) = 0 and f´(α) = 0, then α is a repeated root ofthe quadratic equation f(x) = 0 and f(x) = a(x – α) 2 .In fact α = –b/2a.If α is repeated common root of two quadraticequations f(x) = 0 and φ(x) = 0, then α is also acommon root of the equations f´(x) = 0 and φ´(x) = 0.In the equation ax 2 + bx + = 0 [a, b, c ∈R], ifa + b + c = 0 then the roots are 1, c/a and if a – b + c= 0, then the roots are –1 and – c/a.If one root of the quadratic equation ax 2 + bx + c = 0is equal to the n th power of the other, thenn( ) 1ac n+ 1 + ( n 1c) n 1a + + b = 0.If one root is k times the other root of the quadratic2equation ax 2 (k + 1) b 2+ bx + c = 0, then = .k acIf an equation has only one change of sign, it has one+ve root and no more.Permutations and Combinations :n C 0 = n C n = 1, n C 1 = nn C r + n C r–1 = n+1 C rn C x = n C y ⇔ x = y or x + y = nn. n-1 C r–1 = (n – r + 1) n C r–1If n is even then the greatest value of n C r is n C n/2 .If n is odd then the greatest value of n C r isnC n − 1.2n C r = rn . n–1 C r–1 .nC 1n +or2Number of selection of zero or more things out of ndifferent things is, n C 0 + n C 1 + b C 2 + ... + n C n = 2 n .n C 0 + n C 2 + n C 4 + .... = n C 1 + n C 3 + n C 5 + .... = 2 n–1 .Gap method : Suppose 5 moles A, B, C, D, E arearranged in a row as × A × B × C × D × E ×. Therewill be six gaps between these five. Four in betweenand two at either end. Now if three females P, Q, Rare to be arranged so that no two are together weshall use gap method i.e., arrange them in betweenthese 6 gaps. Hence the answer will be 6 P 3 .Together : Suppose we have to arrange 5 persons ina row which can be done in 5! = 120 ways. But if twoparticular persons are to be together always, then wetie these two particular persons with a string. Thuswe have 5 – 2 + 1 (1 corresponding to these twotogether) = 3 + 1 = 4 units, which can be arranged in4! ways. Now we loosen the string and these twoparticular can be arranged in 2! ways. Thus totalarrangements = 24 × 2 = 48.If we are given n different digits (a , a 2 , a 3 ..... a n ) thensum of the digits in the unit place of all numbersformed without repetition is (n – 1)!(a 1 + a 2 + a 3 + ....+ a n ). Sum of the total numbers in this case can beobtatined by applying the formula (n – 1)!(a 1 + a 2 +a 3 + ..... + a n ). (1111 ......... n times).Binomial Theorem & Mathematical Induction :The number of terms in the expansion of (x + y) n are(n + 1).If the coefficients of p th , q th terms in the expansion of(1 – x) n are equal, then p + q = n + 2.For finding the greatest term in the expansion of(x + y) n . we rewrite the expansion in this formn(x + y) n = x n ⎡ y⎢ 1 ⎤ +x⎥ . Greatest term in (x + y) n = x n .⎣ ⎦n⎛ y ⎞Greatest term in ⎜1 + ⎟ .⎝ x ⎠There are infinite number of terms in the expansionof (1 +x) n , when n is a negative integer or a fraction.The number of term in the expansion of(x 1 + x 2 + .... + x 2 ) n = n+r-1 C r–1 .XtraEdge for IIT-JEE 49 APRIL 2010

Learning the MIghTy wayWhen you hear... you forget. When you see... you remember. When you do... you understand.– ConfuciusThis sounds very much the way we learntthings in our childhood. Our parents did not tell ushow to walk, but made us walk. They did not showus how to ride a bicycle, they made us ride it. Andthat’s how learning goes naturally: derivinginspiration by doing things, ‘Inspired Learning’.In this era of information overload,educational institutions hardly devote time andresources to make students follow this natural path oflearning. An exception to this is Manipal Institute ofTechnology (MIT), where the ‘been there, done that’spirit has helped students make a mark worldwide.While in other colleges, students were seeinghow a car is built, students at MIT designed andmanufactured a high performance car which made itsway into the ‘Formula Student 2009’ (Studentedition of Formula One), at Silverstone, UK. MITgot 65th position among teams from 126 universitiesof 23 countries. Several other feathers of innovationhave been added to MIT’s achievement hat withinthe last 2 years like: 1 st position in “Train Blazer-2008”; 1 st position in “GE Edison InnovationChallenge -2008”; 1 st and 2 nd position in “SchneiderElectric Innovation Challenge” 2008; 1 st prize for anInnovative proposal on “Solar Power embeddedStreet Lamp” and many more.The foundation of this hub of thinking andinnovation was laid 52 years ago, and since then isbeing strengthened by constructive effort and hardwork. The quality of technical education is evidentin the research activities going on in the campus: 6patents filed in the current year, 9 grants received forvarious activities and, several papers, books andawards added to its research portfolio. A majorlandmark is ‘Manipal University TechnologyBusiness Incubator’, which is being established witha huge funding from the Department of Science andTechnology, Government of India.While focusing on innovation, MIT has takencare of students by getting them absorbed into thebest industries, through quality industry projects andcollaborations with universities and companies.‘Practice School’ is one such concept, in sync withthe idea of ‘Inspired learning’. This was introducedat MIT in 2005, through which students are trained toeffectively link the theory learnt in classrooms withthe practice in the industry. This training helpsreduce the cost for a company, on in-house trainingof students who are their prospective employees. Dueto such efforts placement statistics have been movingup since 2005 reaching 95% in 2008. Even in therecession hit 2008-09, 56 companies turned upoffering jobs to 948 students, with the bestremuneration reaching 10.75 lakhs per annum. MITnow stands 7 th among the private Engineeringcolleges in India, and 22 nd overall.A student, Gaurav Sinha (4 th year, Electricaland Electronics) quotes “MIT has provided me withexcellent opportunities to develop to my fullpotential. The environment is conducive for ourgrowth and along with the state-of-art infrastructurefacilities available helps bring out the best in each ofus. The platform and exposure that I got at MIT hasgroomed my personality and prepared me to face anychallenge in future”. Already having produced strongpersonalities of industry like Rajeev Chandrasekhar(Chairman & CEO of Jupiter Capital; FICCIPresident; Founder & former-CEO of BPL mobile),Ravi Bapna (Executive Director of CITNE &Professor, Indian School of Business) , Amit Behl(Director, Intel India), etc. the excellentinfrastructure coupled with innovative learning andcourse content, holds promise for a much brighterfuture for the country.– MIT, ManipalXtraEdge for IIT-JEE 50 APRIL 2010

XtraEdge for IIT-JEE 51APRIL 2010

MOCK TEST FOR IIT-JEEPAPER - ITime : 3 Hours Total Marks : 240Instructions :• This question paper contains 60 questions in Chemistry (20), Mathematics (20) & Physics (20).• In section -I (8 Ques) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer.• In section -II (4 Ques) of each paper +4 marks will be given for correct answer –1 mark for wrong answer.• In section -III contains 2 groups of questions (2 × 3 = 6 Ques.) of each paper +4 marks will be given for eachcorrect answer & –1 mark for wrong answer.• In section -IV (2 Ques.) of each paper +8(2×4) marks will be given for correct answer & No Negative marking forwrong answer.CHEMISTRYCH 3CH 3sectionquestionwhich ONLYApropanedionefollowing(A)(C)An alkeneproductNH 2compound(A)(B)(C)SECTION – IStraight Objective Typecontains 8 multiple choice questions.choices (A), (B), (C) andONE is correct.bis-aldol dimerization of(C 6 H 5 COCOCH 3 ) gives?OC 6 H 5(B)C 6 H 5 C 6 H 5OO(D) C 6 H 5C 6 H 5O(A), C 16 H 16 on ozonolysisB(C 8 H 8 O). Compound (B) onH 2 SO 4 , ∆ gives N-methyl benzamide.'A', is –C = C HHCHC = CCH 3CH 2 –CH = CH–CH 2(D)Identifysequence:H 3 C–(A) H(B) H(C) H(D) HCH = CHproduct D3— CH 2 CH 2 OH⎯ 3⎯3⎯⎯⎯2 NH →CH3|C — CH 2 C ≡|CH3CH3N(CH| |C — CH 2 CHN(CH|CH3CH3O|– C — CH 2 C |||CH3CH3|C — CH 2 CH|CH3This Each 3. in the following reactionhas 4 (D), out ofCH1. 1-phenyl-1,2-|CK Cr2O7,H⎯⎯which of the|H2O,Heat→ + SOClA ⎯⎯→ 2 BCHI. LiAlH4,etherOC ⎯⎯⎯⎯⎯⎯→DC 6 H 5II. H2O3 C–NC 6 H O5C 6 H 53)22. gives only one3 C–3 ) 2reaction withOH, The3 C N(CH 3 ) 233CH 3 3 C–2 N(CH 3 ) 2XtraEdge for IIT-JEE 52APRIL 2010

4. Consider the following reactions :Reaction I :O OOH 3 COReaction II :OH 3 CCH CH2OH3CF 3⎯r1OO⎯⎯⎯3CF 3⎯r2→ H 3 CCH CH2OH⎯⎯⎯→O+F 3 COOHOOH 3 C OH+F 3 C OWhich of the following is a correct comparison of therate of reactions I and II ?(A) r I = r II(B) r 1 > r II(C) r 1 < r II(D) None5. A container containing HCl(g) is fitted with a funneland a long capillary tube as shown below. Capillaryis immersed in the water. When few drops of water isintroduced intothe container through the funnel then –AHCl(g)(A) there is a pressure drop in the container and liquidlevel in the capillary rises(B) there is a pressure drop in the container becauseHCl(g) effuse out more rapidly than the air effusein(C) HCl(g) undergoes spontaneous dissociation toH 2 (g) & Cl 2 (g), hence number of moles decreases,resulting pressure drop inside the container.Water level in the capillary rises(D) HCl(g) spontaneously mixed with water throughthe capillary therefore, water level in the capillaryremain same6. As per Boyle's law V ∝ 1/P at constant temperature,As per charles law V ∝ T at constant pressure.Therefore, by combining, one concluded that T ∝ 1/Phence, PT = constant(A) PT = constant is correct, because volume remainsame in both the laws(B) PT = constant is incorrect, because volumeremain same at the constant temperature and atthe constant pressure(C) PT = constant is correct, because volume atconstant temperaute and volume at constantpressure are not same(D) PT = constant is incorrect, because volume atconstant temperature and volume at constantpressure for the same amount of gas are different7. Which of the following statements is correct ?(A) (n – 1) d subshell has lower energy than nssubshell(B) (n – 1) d subshell has higher energy than nssubshell(C) (n + 1) d subshell has lower energy than nfsubshell(D) nf subshell has lower energy than (n + 2) ssubshell8. (A), (B) and (C) are elements in the third shortperiod. Oxide of (A) is ionic , that of (B) isamphoteric and of (C) a giant molecule. (A), (B) and(C) have atomic numbers in the order of -(A) (A) < (B) < (C) (B) (C) < (B) < (A)(C) (A) < (C) < (B) (D) (B) < (A) < (C)SECTION – IIMultiple Correct Answers TypeThis section contains 4 multiple correct answer(s) typequestions. Each question has 4 choices (A), (B), (C) and(D), out of which ONE OR MORE is/are correct.9.H+ Ph 3 C ⊕ BF – 4 →HH⊕BF 4 – + Ph 3 CHwhich of the following statements is/are correct -(A) the cation in reactant side is approximately10 11 times more stable than product side(B) cation in reaction side is non planer(C) it is acid base reaction(D) reaction must be exothermicXtraEdge for IIT-JEE 53APRIL 2010

10. Which of the following is/are correct ?(A) The efficiency of a solid catalyst dependsupon its surface area(B) Catalyst operates by providing alternate pathfor the reaction that involves lower activationenergy(C) Catalyst lowers the activation energy offorward reaction only without affecting theactivation energy of backward reaction(D) Catalyst does not affect the overall enthalpychange of the reaction11. For the three elements P, Q & R, ionizationenthalpy (IE) and electron gain enthalpies (∆ eg H)are given in the following table -Element IE in kJ/mol ∆eg H in kJ/molP 1680 –340Q 1100 –120R 500 –20(A) P is the highest electronegative elementamong P, Q and R(B) R is the least electronegative among P, Q & R(C) Electro negativity of P is approximately equal to 4(D) R may be chlorine12. Which of the following is/are correct ?⎛ dH ⎞(A) For the incompressible liquid ⎜ ⎟⎠⎝ dPapproximately equal to volume of liquid⎛ dH ⎞(B) For ideal gas ⎜ ⎟⎠ is equal to zero⎝ dP(C) For real gas if⎛ dH ⎞necessarily ⎜ ⎟⎠⎝ dP(D) None of theseTT⎛⎜⎝dE ⎞ ⎟⎠dVSECTION – IIIComprehension TypeTis equal to zeroTis= 0 then notThis section contains 2 groups of questions. Each grouphas 3 multiple choice questions based on a paragraph.Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out of which ONLY ONE is correct.Paragraph # 1 (Ques. 13 to 15)A pleasant smelling optically active compound,monoester 'F' has molecular weight 186. It doesn'treact with Br 2 in CCl 4 . Hydrolysis of 'F' gives twooptically active compounds 'G', which is soluble inNaOH and 'H'. H gives a positive iodoform test, buton warming with conc. H 2 SO 4 gives I with nodisastereomers. When the Ag + salt of 'G' is reactedwith Br 2 , racemic 'J' is formed. Optically active J isformed when 'H' is treated with tosyl chloride (TsCl),and then with NaBr.13. The pleasant smelling optically active compound, Fis -(A) (CH 3 ) 2 CH– CHCO– || O– CH – CH(CH 3 ) 2| |CH 3CH 3O(B) (CH 3 ) 3 C–CH 2 C || –O– CH – CH(CH 3 ) 2|CH 3O(C) CH 3 CH 2 CH 2 CH – C || –O– CH – CH 2 CH 2 CH 3||CH3CH 3O||CH CH 2 – COCH2– CHCH 2 –CH 3(D) CH 3 CH 2 –|CH3|CH14. How would be the structure of F if I exists asdiastereomers ?O(A) (CH 3 ) 2 CH CHCO|| CHCH (CHCH | 3)23 CH| 3(B) (CH 3 ) 3 CCH 2 COO|| CHCH (CH| 3)2CH3O(C) CH 3 CH 2 CH 2 CHCO||CH| CHCH 2 2 33 CH| CH CH3(D) CH 3 CH 2 CHCH2C | COOCH || 2 CHCH 2H3CH| CH333XtraEdge for IIT-JEE 54APRIL 2010

15. What would be the structure of F if H gives negativeidoform test ?O||(A) (CH 3 ) 2 CHCH2COCHCH(CH3)2|CH3CH| 3O||(B) CH 3 (CH 2 ) 3 CH 2 COCHCH CH(CH 3 ) 2|CH3O||(C) CH 3 CH 2 CH 2 CHCOCHCH2CH 2 CH 3| |CH3CH3O||(D) CH 3 CH 2 CHCH2COCH 2CHCH2CH 3||CH3CH3Paragraph # 2 (Ques. 16 to 18)To the 100 ml of 10 –2 (M) aqueous solution of HCl0.1 (M) HA (K a = 10 –2 ) is added in such a way sothat the final pH of the solution become 1.7Given log 2 = 0.316. What was the pH of 10 –2 (M) aqueous solution ofHCl ?(A) pH = 2 (B) pH < 2(C) pH > 2 (D) pH = 1.717. What volume of 0.1 (M) HA was required to add inaqueous HCl to reduced the final pH equal to 1.7 ?(A) 175 ml(B) 100 ml(C) 10 4 ml(D) 75 ml18. Which of the following solution is isohydric with0.1(M) aqueous solution of HA ?(A) 0.01(M) aqueous solution of HB (K a = 10 –2 )(B) 0.01(M) aqueous solution of HC (K a =2×10 –1 )(C) 0.01(M) aqueous solution of HNO 3(D) 1(M) aqueous solution of HD (K a = 10 –3 )SECTION – IVMatrix – Match TypeThis section contains 2 questions. Each questioncontains statements given in two columns, which haveto be matched. The statements in Column I are labeledA, B, C and D, while the statements in Column II arelabeled p, q, r, s and t. any given statement in Column Ican have correct matching with ONE OR MOREstatements (s) in column II. The appropriate bubbledcorresponding to the answers to these questions have tobe darkened as illustrated in the following example :If the correct matches are A – p, s and t; B – q and r;C – p and q; and D – s and t; then the correctdarkening of bubbles will look like the following.p q r s tABCD19. Match the following :Column-I(A)CH 3 CH 2 CH 2 NH 2pp(B) CH 3 CH 2 NHCH 3(C)CH 3 –N − CH 3|CH 3qqrrrssttColumn-II(p)Treatment of NaNO 2 , HClgives N-nitroso compound(q)Treatment of NaNO 2 , HClgives diazonium chloride(r) Treatment ofCH 3 I (excess)followed by AgOH;heat gives out alkene(D) NH 2(s) Treatment of HCl,heat gives dealkylation(t) Treatment with CHCl 3 /Alc.KOH gives isocyanide20. Match the following :Column-IColumn-II−OH ,Br2 (1 eq)(A) C 6 H 5 CH 2 CHO⎯⎯⎯⎯⎯→(p) Redox products(B) CH 3 CHO(C) CH 3 CH 2 –CH = OOH⎯⎯ − → (q) Enantiomeric(0−5ºC)products−OH ⎯⎯→(25ºC)Conc. OH→ −(D) CH 3 – CH – C – H|CH 3 O | |⎯⎯⎯⎯(r)K D⎯ αKαH⎯ → < 1(Primaryisotopic effect)(s) Diastereomericproducts(t) DisproportionationMATHEMATICSSECTION – IStraight Objective TypeThis section contains 8 multiple choice questions. Eachquestion has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.XtraEdge for IIT-JEE 55APRIL 2010

1. Let F denote the set of all onto functions from A ={a 1 , a 2 , a 3 , a 4 } to B = {x, y, z}. A function f is chosenat random from F. The probability that f –1 (x)consists of exactly two elements is(A) 2/3 (B) 1/3 (C) 1/6 (D) 02. A letter is known to have come from eitherTATANAGAR or CALCUTTA. On the envelope,just two consecutive letters TA are visible . Theprobability that the letter has come fromCALCUTTA is(A) 4/11 (B) 1/3 (C) 5/12 (D) None3. A person writes 4 letters and addresses 4 envelopes.If the letters are placed in the envelops at random, theprobability that not all letters are placed in correctenvelopes is(A) 1/24 (B) 11/24 (C) 5/8 (D) 23/244. The value of x for which the matrix⎡ 2 0 7⎤A =⎢ ⎥⎢0 1 0⎥is inverse of⎢⎣1 − 2 1⎥⎦⎡−xB =⎢⎢0⎢⎣x(A) 2114x1− 4x(B) 317x ⎤0⎥⎥− 2x⎥⎦is(C) 41(D) 515. The largest term in the expansion of (3 + 2x) 50 , wherex = 1/5, is(A) 5 th (B) 6 th (C) 8 th (D) 9 th6. If2z − = 2, then the greatest value of | z | isz(A) 1 + 2 (B) 2 + 2(C) 3 +1 (D) 5 + 116−17. The value of∫tan x −1 dx is(A)(C)116π + 3342 (B) π3– 2 3416π + 2 3 (D) π – 2 3338. The solution of(y + x + 5)dy = (y – x + 1) dx is(A) log ((y + 3) 2 + (x + 2) 2 ) + tan –1 y + 3x + 2= C(B) log ((y + 3) 2 + (x – 2) 2 ) + tan –1 y − 3x − 2= C(C) log ((y + 3) 2 + (x + 2) 2 ) + 2 tan –1 y + 3x + 2= C(D) log ((y + 3) 2 + (x + 2) 2 ) – 2 tan –1 y + 3x + 2= CSECTION – IIMultiple Correct Answers TypeThis section contains 4 multiple correct answer(s) typequestions. Each question has 4 choices (A), (B), (C) and(D), out of which ONE OR MORE is/are correct.9. If n > 1, then (1 + x) n – nx – 1 is divisible by -(A) x 2 (B) x 3(C) x 4 (D) x 51−sin 4A + 110. If y =, then one of the values of y is -1+sin 4A −1(A) – tan A (B) cot A⎛ π ⎞⎛ π ⎞(C) tan ⎜ + A ⎟ (D) – cot ⎜ + A ⎟⎝ 4 ⎠ ⎝ 4 ⎠11. Equation of a common tangent to the circlesx 2 + y 2 – 6x = 0 and x 2 + y 2 + 2x = 0 is -(A) x = 1 (B) x = 0(C) x + 3 y + 3 = 0 (D) x – 3 y + 3 = 02x t − 5t + 412. The points of extremum of∫are -0t2 + e(A) x = – 2 (B) x = 1(C) x = 0 (D) x = – 1SECTION – IIIComprehension TypeThis section contains 2 groups of questions. Each grouphas 3 multiple choice questions based on a paragraph.Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out of which ONLY ONE is correct.Paragraph # 1 (Ques. 13 to 15)Consider the region S 0 which is enclosed by the curve2y ≥ 1− x and max. {|x|, |y| } ≤ 4. If slope of afamily of lines is defined asm(t) = cos t where point (t, 2t + 0.4) lies inside theregion S 0 . Any member of this family of lines is calledL 1 = 0 if it passes through (π, max {t}) and L 2 = 0 if itpasses through the (π, min m {t}).13. Area of region of S 0 is -(A) 8 + π/2 sq. units (B) 8 – π/2 sq. units(C) 8 sq. units (D) 8 + π sq. units2XtraEdge for IIT-JEE 56APRIL 2010

14. If line lies inside the region S 0 , then -(A) t ∈ (0, 0.28) (B) t ∈ (0, 1)(C) t ∈ (0.28, 1) (D) t ∈ [.28, 1]15. L 1 = 0 having maximum slope, is -y −1y −1(A) = cos 1 (B) = cos (0.28)x − πx − πy − π(C) = cos (1) (D) None of thesex −1Paragraph # 2 (Ques. 16 to 18)Two circles S 1 = 0 and S 2 = 0 are touching to eachother externally at point T, with centreC 1 , C 2 and radii r 1 and r 2 respectively.If P and Q be the points of contact of a directcommon tangent to the two circles and PQ meets theline joining C 1 , C 2 in S. Tangent at common point Tis intersecting to the tangent PQ at R point and to otherdirect tangent at V point. Let S 1 = x 2 + y 2 – 6x = 0 andS 2 = x 2 + y 2 + 2x = 0.16. Angle between the two direct tangents is–(A) 90º (B) 30º(C) 60º(D) None of these17. Direct tangents are–(A) y = 3 x + 3 , y = – 3 x + 3x − x(B) y = – 3 , y = – 333x − x(C) y = + 3 , y = – 333(D) None of these18. A circle S = 0 of radius 1 units rolls on the outside ofthe circle S 2 = 0, touching it externally, locus of thecentre of this outer circle is –(A) Circle(B) Ellipse(C) Parabola (D) None of theseSECTION – IVMatrix – Match TypeThis section contains 2 questions. Each questioncontains statements given in two columns, which haveto be matched. The statements in Column I are labeledA, B, C and D, while the statements in Column II arelabeled p, q, r, s and t. any given statement in Column Ican have correct matching with ONE OR MOREstatements (s) in column II. The appropriate bubbledcorresponding to the answers to these questions have tobe darkened as illustrated in the following example :If the correct matches are A – p, s and t; B – q and r;C – p and q; and D – s and t; then the correctdarkening of bubbles will look like the following.ABCD19. Match the following-:Column- Ip q r s tpp(A) The sum of the series10∑r=020 C isr(B) The coefficient of x 53100100∑r=0rqqC (x − 3)rrr100−r(C) ( 10 C 0 ) 2 –( 10 C 1 ) 2 +…….…..–( 10 C 9 ) 2 + ( 10 C 10 ) 2 equals(D) The value of95 C 4 + ∑j=5100−j3120. Match the following -C isColumn -Ix − 2 y − 3 z − 4(A) If the lines = =1 1 λx −1 y − 4 z − 5and = =λ 2 1intersect at (α, β, γ) then λ =⎛ π −1⎛x + 1 ⎞(B) If lim 4xx→∞⎟ ⎞⎜ − tan ⎜ ⎟ =⎝ 4 ⎝ x + 2 ⎠⎠y 2 + 4y + 5 then y =2rssin(C) If chord x + y + 1= 0 of parabolay 2 = ax subtends 90º at (0, 0)then a =(D) If a = î + ĵ +kˆ , a . b = 1isand a × b = ĵ– kˆ , then | b | isequal tottColumn- II(p) – 10 C 5(q) 100 C 4(r) 2 19 + 21 20C 10(s) – 100 C 53(t) 100 C 47Column -II(p) 0(q) –1(r) –2(s) 1(t) –3XtraEdge for IIT-JEE 57APRIL 2010

PHYSICSSECTION – IStraight Objective TypeThis section contains 8 multiple choice questions. Eachquestion has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. ABCD is a smooth horizontal fixed plane on whichmass m 1 = 0.1 kg is moving in a circular path ofradius r = 1 m. It is connected by an ideal stringwhich is passing through a smooth hole and connectsof mass m 2 = 1/ 2 kg at the other end as shown. m 2also moves in a horizontal circle of same radius of 1m with a speed of 10 m/s. If g = 10 m/s 2 then thespeed of m 1 is-ABDm 1Cm 2(A) 10 m/s (B) 10 m/s(C)1 m/s (D) None of these102. A L shaped rod whose one rod is horizontal and otheris vertical is rotating about a vertical axis as shownwith angular speed ω. The sleeve shown in figure hasmass m and friction coefficient between rod andsleeve is µ. The minimum angular speed ω for whichsleeve cannot sleep on rod is –(A) ω =(C) ω =gµllµgωlm(B)sleeveω =µgl(D) None of these3. Two solid spherical balls of radius r 1 & r 2(r 2 < r 1 ), of density σ are tied up with a string andreleased in a viscous liquid of lesser density ρ andcoefficient of viscosity η, with the string just taut asshown. The terminal velocity of spheres is -2 r2 2 g(A) ( σ − ρ)9 η(C)293r 1r 2(r1 + r2) ( σ − ρ)g(D)r + r η123r1 g(B) ( σ − ρ)9 η2 2293(r1− r2) ( σ − ρ)gr − r η4. A block of mass m is attached to an ideal spring andsystem lies in vertical plane as shown. Initially thesupporting plane is placed so that spring remains inits natural length then the plane is moved very slowlydownwards. The graph showing variation of normalreaction applied by mass on supporting plane withdistance travelled by block is –N(A) mg(C)NmgM1Supporting planexxN(B) mg(D) None of these5. A massless container is filled with liquid of densityρ. It contains two holes as shown in figure. Containerrests on ground. Area of the two holes are A each.Container is filled with liquid upto height H. Then –H3H/4H/4(A)Torque produced by normal force betweencontainer & ground about center of gravity isρAgH 2into the plane of paper2(B) Torque produced by friction about center ofgravity is ρAgH 2 out of the plane of paper(C) Net torque produce by thrust force & frictionρAgH 2force about center of gravity is into the4plane of paper(D) Torque produced by normal force betweencontainer and ground about centre of gravity iszero32xXtraEdge for IIT-JEE 58APRIL 2010

6. A BTwo containers A & B contain ideal gases heliumand oxygen respectively. Volume of both containersare equal and pressure is also equal. Container A hastwice the number of molecules than container B thenif v A & v B represent the rms speed of gases incontainers A & B respectively, then -(A)(C)vvvvABAB= 2 (B)= 2 (D)vvvvABAB= 4= 87. A capacitor is composed of three parallel conductingplates. All three plates are of same area A. The first pairof plates are kept a distance d 1 apart and the spacebetween them is filled with a medium of a dielectric ε 1 .The corresponding data for the second pair are d 2 & ε 2respectively. What is the surface charge density on themiddle plate ?d 1 d 2(A) ε0ε 1 ε 2V 0V ⎡ ε1ε2⎤⎢ + ⎥⎣d1d(B) ⎡ ε1ε2− ε ⎢ + ⎥ ⎤0V2 ⎦ ⎣d1d2⎦⎡ ε1ε2⎤⎡ ε1ε2⎤(C) 2 ε0V⎢+ ⎥ (D) − 2 ε0V⎢+ ⎥⎣d1d2⎦ ⎣d1d2⎦8. 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 centreO (∠AOB = 90º)(A) 2 (B)l12OψAlRB(C) 2 2 (D)SECTION – IIMultiple Correct Answers TypeThis section contains 4 multiple correct answer(s) typequestions. Each question has 4 choices (A), (B), (C) and(D), out of which ONE OR MORE is/are correct.9. A body moves in a circular path of radius R withdeceleration so that at any moment of time itstangential and normal acceleration are equal inmagnitude. At the initial moment t = 0, the velocityof body is v 0 then the velocity of body at any timewill be –v0(A) v = at time t⎛ v t ⎞⎜1+0⎟⎝ R ⎠SR(B) v = v0e− after it has moved S meter(C) v = v 0 e –SR after it has moved S meter(D) None of these10. A cylinder block of length L = 1m is in twoimmiscible liquids. Part of block inside liquid(1) is1 1 m and in liquid (2) is m. Area of cross-section44of block is A. Densities of liquid (1) & (2) are ρ and2ρ respectively –ρ.Liquid (1)212What is the proportion lR if the observer B first seesthe light source when the angle of mirrorψ = 15º ?2ρ.Liquid (2)XtraEdge for IIT-JEE 59APRIL 2010

RP⊗BO ω 0Q19. A uniform solid cube is floating in a liquid as shown inthe figure, with part x inside the liquid. Some changes inparameters are mentioned in Column I. Assuming noother changes, match the following -x16. Magnitude of current as a function of time(A)(C)Bω l2RBω l8RWhere α =20 −αte20 −αte3B2 l 28RM(B)(D)Bω l16RBω l8R20 −2αte20 −2αte17. Total charge flow through resistance till rod PQ stoprotating .ω(A) 0 M ω(B) 0 M ω(C) 0 M ω(D) 0 M8B 3B 6B 9B18. Heat generated in the circuit by t = ∞(A)(C)Ml 2 ω2 024Ml 2 ω2 03(B)(D)Ml 2 ω2 08Ml 2 ω2 032SECTION – IVMatrix – Match TypeThis section contains 2 questions. Each questioncontains statements given in two columns, which haveto be matched. The statements in Column I are labeledA, B, C and D, while the statements in Column II arelabeled p, q, r, s and t. any given statement in Column Ican have correct matching with ONE OR MOREstatements (s) in column II. The appropriate bubbledcorresponding to the answers to these questions have tobe darkened as illustrated in the following example :If the correct matches are A – p, s and t; B – q and r;C – p and q; and D – s and t; then the correctdarkening of bubbles will look like the following.p q r s tABCDppqqrrrssttColumn IColumn II(A) If density of the liquid (p) Increasedecreases, x will(B) If height of the cube is (q) Decreasesincreased keeping base areaand density same, x will(C) If the whole system is (r) Remain sameaccelerated upward, thenx will(D) If the cube is replaced (s) May increaseby another cube of same or decreasesize but lesser density, x will(t) none20. A block of mass m = 1 kg is at rest with respect to arough wedge as shown in figure.mµθThe wedge starts moving up from rest with anacceleration of a = 2m/s 2 and the block remains at reswith respect to wedge then in 4 sec. of motion of wedgework done on block (assume angle of inclination ofwedge is θ = 30º and g = 10 m/s 2 ) –Column I(A) By gravity(B) By normal reaction(C) By friction(D) By all the forcesaColumn II(p) 144 J (in magnitude)(q) 32 J(r) 160 J(s) 48 J(t) noneXtraEdge for IIT-JEE 61APRIL 2010

MOCK TEST FOR IIT-JEEPAPER - IITime : 3 Hours Total Marks : 240Instructions :• This question paper contains 57 questions in Chemistry (19,) Mathematics (19) & Physics (19).• In section -I (4 Ques) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer.• In section -II (5 Ques) of each paper +4 marks will be given for correct answer & –1 mark for wrong answer• In section -III (2 Ques.) of each paper +8(2×4) marks will be given for correct answer. No Negative marking forwrong answer.• In section -IV (8 Ques.) of each paper +4 marks will be given for correct answer & –1 mark for wrong answer.CHEMISTRYSECTION – IStraight Objective TypeThis section contains 4 multiple choice questions. Eachquestion has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.The correct structures of 'A' and 'B' are,respectively–(A) CH2 =CH COOH ;CH=CHCOOHCOOH(B)CH 2 =C; CH=CH–COOH1.NOBr ProductNH 2The main product is –+ Enantiomer(A)Br(B)+ EnantiomerBrBr(C)+ EnantiomerNO(D)+ Enantiomer(C) CH=CHCOOH ; CH 2 =CH COOH(D) CH=CH–COOH ;C=CH 2COOH3. If H-He undergoes dissociation, which of thefollwoing product most expected to occur –2HHe products(A) He 2 + 2H (B) H 2 + 2He(C) D 2 + He 2 (D) 2H + 2He4. 100 ml solution (I) of buffer containing 0.1(M) HAand 0.2 (M) A – , is mixed with another solution (II) of100 ml containing 0.2(M) HA and 0.3(M) A – .Aftermixing what is the pH of resulting solution ?( GivenpK a of HA = 5)(A) 5 – log 5/3 (B) 5 + log 5/3(C) 5 + log 2/5 (D) 5 – log 5/22. There are two isomeric carboxylic acids– 'A' and 'B'C 9 H 8 O 2 . reacts with H 2 /Pd giving compounds,C 9 H 10 O 2 . 'A' gives a resolvable product and 'B' givesa non-resolvable product. Both isomers could byoxidised to PhCOOH.SECTION – IIMultiple Correct Answers TypeThis section contains 5 multiple correct answer(s) typequestions. Each question has 4 choices (A), (B), (C) and(D), out of which ONE OR MORE is/are correct.XtraEdge for IIT-JEE 62APRIL 2010

5. Dopamine of a drug used in the treatment ofparkinson's disease.NHHO CH 2 –CH 2COOHHODopaminWhich of the following statements about thiscompound are correct ?(A) It can exist in optically active forms.(B) One mole will react with three moles of sodiumhydroxide to form a salt(C) It can exist as a Zwitter ion in the aqueoussolution(D) It gives nitroso compound on treatment withHNO 2 .6. Which of the following method(s) would be usefulfor preparing ketones ?(A) Friedel-Crafts reaction of an acyl chloride withbenzene (AlCl 3 catalyst)(B) Reaction of methyllithium with the lithium salt ofcarboxylic acid, followed by hydrolysis(C) Reaction of R 2 CuLi with an acyl choride in etherat low temperature(D) Reaction of Grignard reagents with acyl chloridein ether followed by hydroylsis7. In the given table types of H bonds and some H bondenergies are given and other H bond energies are notgiven. You are to perdit the unknown H-bondenergies.Types of H-bonds H-bond energies in KJ/mol(I) F – H …….. O –F – H …….. F 30(II) O – H …….. O –O – H …….. F 15(III) F – H …….. F – –(IV) N – H …….. N –Correct prediction are –(A) H-bond energy for (I) may be 20 kJ/mol(B) H-bond energy for (II) may be 25 kJ/mol(C) H-bond energy for (III) may be 113 kJ/mol(D) H-bond energy for (IV) may be 12 kJ/mol8. During conductance measurement of an electrolytebased on the wheatstone bridge principle alternatingcurrent is used because direct current produces –(A) polymerisation(B) Ionisation(C) Electrolysis resulting in the chnge ofconcentration and in consequencae the resistance(D) Polarisation at the electrodes resulting in thechange of resistance9. Consider the cell :Ag | AgCl (s) | KCl (1M) | Hg 2 Cl 2 (s) | Hg (l) | Pt. Thecell potential :(A) increases on increasing concentration of Cl – ions(B) decreases on decreasing concentration of Cl – ions(C) is independent of concentration of Cl – ions(D) is independent of amounts of AgCl (s) andHg 2 Cl 2 (s)SECTION – IIIMatrix - Match TypeThis section contains 2 questions. Each questioncontains statements given in two columns, which haveto be matched. The statements in Column I are labeledA, B, C and D, while the statements in Column II arelabeled p, q, r, s and t. any given statement in Column Ican have correct matching with ONE OR MOREstatements (s) in column II. The appropriate bubbledcorresponding to the answers to these questions have tobe darkened as illustrated in the following example :If the correct matches are A – p, s and t; B – q and r;C – p and q; and D – s and t; then the correctdarkening of bubbles will look like the following.p q r s tABCDppqq10. Match the followingColumn IColumn II(A) HCOOH (p) Decarboxylationon heating(B) CH 3 COOH (q) Reation with Br 2(C)OHCOOH(r) Cu 2+ (alkaline)→Cu 2 O(D) PhCH 2 COOH11. Match the followingColumn I(A) XeO 4(B) XeF 4rrrsstt(s) Decarbonylation ordecarboxylation ontreatment withconc. H 2 SO 4(t) Reaction with I 2 +NaOHColumn II(p) Non polar(q) Having no lone pair ofelectron in the central atom(C) SO 3 (r) Planar(D) BF 3 (s) Having lone pair of(t)electrons in the central atomTetrahedralXtraEdge for IIT-JEE 63APRIL 2010

SECTION – IVInteger answer typeThis section contains 8 questions. The answer to each ofthe questions is a single-digit integer, ranging from 0 to9. The appropriate bubbles below the respectivequestion numbers in the ORS have to be darkened. Forexample, if the correct answers to question numbers X,Y, Z and W (say) are 6, 0, 9 and 2, respectively, thenthe correct darkening of bubbles will look like thefollowing :X Y Z W0123456789012345678912. 3 ampere current was passed through an aqueoussolution of an unknown salt of Pd for one hour,2.977g of Pd n+ was deposited at cathode. Find n.(At. wt. of Pd = 106.4)13. Consider the following reaction sequenceCa(OH)⎯⎯⎯∆⎯ 201234567890123456789NH NH2KOH / glycolO / H2O23O ⎯⎯⎯⎯ →W ⎯⎯⎯→X→ YC 5H COOOH6⎯⎯⎯⎯⎯→ZHow many carbons are present in the final product Z?14. For the reaction in the plant cells6CO 2 (g) + 12H 2 O (l) → C 6 H 12 O 6 (s) + 6O 2 (g)+ 6H 2 O(l) ∆ r Gº = 3000 kJ/mol3–ATP → ADP + PO 4 ∆ r Gº = – 30 kJ/molGlucose is stored in the plant cell as starch,(C 6 H 10 O 5 ) n . To produce 162 g of starch how manymoles of ATP are minimum required ? Give youranswer after divide actual answer by 100.15. The relative lowering of the vapour pressure of anaqueous solution containing a non-volatile solute is0.0125. What is the molality of the solution. Giveyour answer after multiplying actual answer by 10.16. A polyvalent metal weighing 0.1g and having atomicweight 51 reacted with dil. H 2 SO 4 to give 43.9 ml ofhydrogen at STP. The solution containing the metalin this lower oxidation state(X), was found to require58.8 mL of 0.1 N KMnO 4 for complete oxidation.What is the higher oxidation state (Y) of the metal ?17. The stopcock, connecting the two bulbs of volumes5 litres and 10 litres containing an ideal gas at 9 atmand 6 atm respectively, is opened. What is the finalpressure (in atm) in the two bulbs if the temperatureremained the same ?18. Sum of lone pairs present in XeOF 4 and XeO 3 is ….. .19. Total number of isomers are possible in [Co(en) 2 Cl 2 ]and [Co(C 2 O 4 ) 2 (NH 3 ) 2 ] – is.MATHEMATICSSECTION – IStraight Objective TypeThis section contains 4 multiple choice questions. Eachquestion has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.n41. Let ∑ r = f (n) , thenr=1∑ ( 2r −1)r=1is equal to(A) f(2n) – 16 f(n) (B) f (2n) – 7f (n)(C) f(2n – 1) – 8 f(n) (D) None of these⎛ α α ⎞2. If f r (α) = ⎜cos + isin ⎟ ×2 2⎝ r r ⎠thenn →nn⎛ 2α2α⎞ ⎛ α α ⎞⎜cos + isin ⎟ …. ⎜cos+ isin ⎟22⎝ r r ⎠ ⎝ r r ⎠lim f (π) equals∞(A) –1 (B) 1(C) – i(D) i3. If α, β, γ, δ are four complex numbers such that δγ isreal and αδ – βγ ≠ 0, then z =t ∈ R represents a(A) circle(C) ellipse4(B) parabola(D) straight lineα + βt,γ + δt4. The inflection points on the graph of functionxy =∫( t −1)(t− 2) 2 dt are0(A) x = –1 (B) x = 3/2(C) x = 4/3 (D) x =1SECTION – IIMultiple Correct Answers TypeThis section contains 5 multiple correct answer(s) typequestions. Each question has 4 choices (A), (B), (C) and(D), out of which ONE OR MORE is/are correct.XtraEdge for IIT-JEE 64APRIL 2010

5. If A and B are two invertible matrices of the same order,then adj (AB) is equal to -(A) adj (B) adj (A) (B) |B| |A| B –1 A –1(C) |B| |A| A –1 B –1 (D) |A| |B| (AB) –1⎡ cosα6. If A (α , β) =⎢⎢− sin α⎢⎣0sin αcosα00 ⎤0⎥⎥, then⎥⎦βe(A) A (α, β)′ = A (– α, β)(B) A (α, β) –1 = A (–α, –β)(C) Adj (A (α, β)) = e –β A (–α, –β)(D) A (α, β)′ = A (α, – β)7. For a positive integer n, if the expansion ofn⎛ 5 4 ⎞⎜ + x ⎟ has a term independent of x, then n can2⎝ x ⎠be(A) 18 (B) 21(C) 27 (D) 998. If k is odd then k C r is maximum for r equal to(A) 21 (k – 1) (B) 21 (k + 1)(C)k – 19. Let a n = ( 111...1) , then14243n times(A) a 912 is not prime(C) a 480 is not prime(D) k(B) a 951 is not prime(D) a 91 is not primeSECTION – IIIMatrix - Match TypeThis section contains 2 questions. Each questioncontains statements given in two columns, which haveto be matched. The statements in Column I are labeledA, B, C and D, while the statements in Column II arelabeled p, q, r, s and t. any given statement in Column Ican have correct matching with ONE OR MOREstatements (s) in column II. The appropriate bubbledcorresponding to the answers to these questions have tobe darkened as illustrated in the following example :If the correct matches are A – p, s and t; B – q and r;C – p and q; and D – s and t; then the correctdarkening of bubbles will look like the following.p q r s tABCDppqqrrrsstt10. Match the followingColumn-1Column-IIEquationGeneral <strong>Solution</strong>s(A) 2 sin θ – 3 = 0 (p) nπ + (–1) n π3(B) 2 sin 2θ + 3π(q) 2nπ – 3= 2 sin θ + 2 3 cos θπ(C) sin 2θ + cos 2θ + 4 sin θ (r) 2nπ + 3= 1 + 4 cos θ(D) cos 2 θ = 41(s) nπ – 4π(t) nπ + 4π11. Match the following :Column- I(A) If three unequal numbers a, b, care in A.P. and b – c,c – b, a are in G.P., then3 3 3a + b + cis equal to3abc(B) Let x be the arithmetic meanand y,z be two geometricmeans between any twopositive numbers, theny3 3+ zis equal toxyz(C) If a, b, c be three positivenumber which form threesuccessive terms of a G.P. andc> 4b –3a, then the commonratio of the G.P. can be equal to(D)⎪⎧lim tan ⎨n→∞⎪⎩⎪is equal ton∑r=1tan−1⎛ 1⎜⎝ 2r2⎞⎪⎫⎟ ⎬⎠ ⎭SECTION – IVInteger answer typeColumn- II(p) 1/3(q) 1(r) 2(s) 3(t) 1/2This section contains 8 questions. The answer to each ofthe questions is a single-digit integer, ranging from 0 to9. The appropriate bubbles below the respectivequestion numbers in the ORS have to be darkened. Forexample, if the correct answers to question numbers X,Y, Z and W (say) are 6, 0, 9 and 2, respectively, thenthe correct darkening of bubbles will look like thefollowing :XtraEdge for IIT-JEE 65APRIL 2010

X Y Z W0123456789012345678912. If z ≠ 0 and 2 + cos θ + i sin θ = 3/z, then find thevalue of 2 (z + z ) – |z| 2 .13. Find the sum of all the integral roots of(log 5 x) 2 + log 5x (5/x) = 1.14. Four different integers form an increasing A.P. suchthat one of them is the square of the remainingnumbers. Find the largest numbers.15. If n is a positive integer, andE = 2n + 1 C 1 + 2n + 1 C 2 + … + 2n + 1 C n – 2n + 1 C 2n + 1– 2n + 1 C 2n – … – 2n + 1 C n + 1 . Find |E|16. Find the number of values of t for which the systemof equations(a + 2t)x + by + by + cz = 0bx + (c + 2t)y + az = 0cx + ay + (b + 2t) z = 0has non-trivial solutions.17. A function f(x) is defined for x > 0 and satisfiesf(x 2 ) = x 3 for all x > 0. Then the value of f ′(4) is ___.18. If A is the area formed by the positive x-axis, and thenormal and tangent to the circle x 2 + y 2 = 4 at(1, 3 ) then A/ 3 is equal to _____.19. The area bounded by the curves x = y 2 andx = 3 – 2y 2 is ______.01234567890123456789PHYSICSSECTION – IStraight Objective TypeThis section contains 4 multiple choice questions. Eachquestion has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. A thin conducting plate is inserted in half way betweenthe plates of a parallel plates capacitor of capacitance C.Conducting plateWhat does the value of capacitance, if both the plateof capacitor is shortened ?(A) C(B) 2C(C) 3C(D) 4C2. White light is incident normally on a glass surface(n = 1.52) that is coated with a film ofmg F 2 (n = 1.38). For what minimum thickness of thefilm will yellow light of wavelength 550 nm (in air)be missing in the reflected light ?(A) 99.6 nm(C) 19.6 nmmgF 2glass(B) 49.8 nm(D) 10.6 nm3. A circuit element is placed in a closed box. At timet = 0, a constant current generator supplying a current of Iamp is connected across the box. Potential diff. across thebox varies according to graph as shown in the figure. Theelement in the box is -8Volts(V)23 t(sec)Time(A) a resistance of 2Ω (B) a battery of emf 6V(C) an inductance of 2H (D) a capacitance4. Consider a usual set-up of Young's double slitexperiment with slits of equal intensity as shown inthe figure. Take 'O' as origin and the Y axis asindicated. If average intensity between y 1 = λD/4dand y 2 = λD/4d equals n times the intensity ofmaximum, then n equal is (take average over phasedifference) -y(A)d1 ⎛ ⎞⎜1+ 2 ⎟2 ⎝ π ⎠S 1S 2DO⎛ ⎞(B) 2 ⎜1+ 2 ⎟⎝ π ⎠⎛ ⎞(C) ⎜ + 21 ⎛ ⎞1 ⎟ (D) ⎜1− 2 ⎟⎝ π ⎠ 2 ⎝ π ⎠SECTION – IIMultiple Correct Answers TypeThis section contains 5 multiple correct answer(s) typequestions. Each question has 4 choices (A), (B), (C) and(D), out of which ONE OR MORE is/are correct.XtraEdge for IIT-JEE 66APRIL 2010

5. A long round conductor of cross-sectional area A ismade of a material whose resistivity depends on theradial distance r from the axis of the conductor as ρα=r 2 , α is a constant. The total resistance per unitlength of the conductor is R and the electric fieldstrength in the conductor due to which a current Iflows in it is E.2πα4πα(A) R =(B) R =22AA2παI4παI(C) E =2A(D) E =2A6. Two infinite plates carry j ampere of current out ofthe age per unit width of the plate as shown. B P andB Q represent magnitude of field at points P and Qrespectively.(A) B P = 0(C) B Q = 0PQ(B) B P = µ 0 j(D) B Q = µ 0 j7. A bar magnet M is allowed to fall towards a fixedconducting ring C. If g is the acceleration due togravity, v is the velocity of the magnet at t = 2 s and sis the distance traveled by it in the same time then,M(A) v > 2g(C) s > 2g3gC(B) v < 2g(D) s < 2g8. In the network shown, the capacitor C is initiallyuncharged. The time constant of the circuit is τ andthe charge on C at time t after the switch S is closedis q.R 3R 1R 2SC9. Binding energy per nucleon vs mass number curvefor nuclei is shown in the figure. W, X, Y and Z arefour nuclei indicated on the curve. The process thatwould release energy is -Binding Energy/Nucleon in MeV8.58.07.55.00(A) Y → 2Z(C) W → 2YZYXW30 60 90 120Mass number of nuclei(B) W → X + Z(D) X → Y + ZSECTION – IIIMatrix - Match TypeThis section contains 2 questions. Each questioncontains statements given in two columns, which haveto be matched. The statements in Column I are labeledA, B, C and D, while the statements in Column II arelabeled p, q, r, s and t. any given statement in Column Ican have correct matching with ONE OR MOREstatements (s) in column II. The appropriate bubbledcorresponding to the answers to these questions have tobe darkened as illustrated in the following example :If the correct matches are A – p, s and t; B – q and r;C – p and q; and D – s and t; then the correctdarkening of bubbles will look like the following.p q r s tABCDppqq10. Capillary rise and shape of droplets on a plate due tosurface tension are shown in column II.Column IColumn II(A) Adhesive forces isrrrss(p)ttAB(A) τ = CR 1CVR 2 −t/ τ(C) q = (1 − e )R + R23⎛ R ⎞(B) τ = C ⎜2R3⎟R 1 +⎝ R 2 + R 3 ⎠CVR 1 −t/ τ(D) q = (1 − e )R + R12greater than cohesive forces(B) Cohesive forces is (q)greater than adhesive forcesABXtraEdge for IIT-JEE 67APRIL 2010

(C) Pressure at A > pressureat B(D) Pressure at B > Pressure(r) A mercury dropis pressed betweentwo parallel platesof glassB(s)at A(t) none11. Column I Column II(A) The coefficient of (p ) with decreasevolume expansion at in pressureconstant pressure is(B) Mean free path of (q) at all temperaturesmolecule increases(C) An ideal gas obeys (r) Same for all gasesBoyle’s and Charle’sLaw(D) A real gas behaves (s) At high temperatureas an ideal gas at lowpressure and(t) noneSECTION – IVInteger answer typeThis section contains 8 questions. The answer to each ofthe questions is a single-digit integer, ranging from 0 to9. The appropriate bubbles below the respectivequestion numbers in the ORS have to be darkened. Forexample, if the correct answers to question numbers X,Y, Z and W (say) are 6, 0, 9 and 2, respectively, thenthe correct darkening of bubbles will look like thefollowing :X Y Z W0123456789012345678912. A ball A is falling vertically downwards withvelocity v 1 . It strikes elastically with a wedge movinghorizontally with velocity v 2 as shown in figure. Findv 1the value of , so that the ball bounces back inv2vertically upward direction relative to the wedge.01234567890123456789BAAv 2v 130ºA13. Portion AB of the wedge shown in figure is roughand BC is smooth. A solid cylinder rolls withoutslipping from A to B. If AB = BC, then find the valueof ratio of translational kinetic energy to rotationalkinetic energy, when the cylinder reaches to point C.AD14. In what minimum time after its motion begins will aBparticle oscillating according to the law x = 7sin 21 πtmove from the position of equilibrium to the positionx = – 7/2 units ?15. A satellite is revolving in a circular equatorial orbit ofradius R = 2 × 10 4 km from east to west. Calculatethe interval (in hrs) after which it will appear at thesame equatorial town. Given that the radius of theearth = 6400 km and g (acceleration due to gravity)= 10 ms –2 .16. Three capillary tubes of same radius 1 cm but oflengths 1m, 2m and 3m are fitted horizontally to thebottom of a long cylinder containing a liquid atconstant pressure and flowing through these tubes.What is the length of a single tube which can replacethe three capillaries.17. Equations of a stationary and a traveling waves are asfollows : y 1 = a sin kx cos ωt and y 2 = a sin (ωt – kx)πThe phase difference between two points x 1 = and 3k3πx 2 = are φ1 and φ 2 respectively for the two waves.2kφ 1Then find the value of .φ218. A coil, a capacitor and an AC source of voltage 24 V(rms) are connected in series. By varying thefrequency of the source, a maximum rms current of6A is observed. If this coil is connected to a batteryof emf 12 V and internal resistance 4 Ω, then find thevalue of current flow through it.19. A potential difference of 10 3 V is applied across anX-ray tube. Calculate the value of ratio of thede-Broglie wavelength of the incident electrons to theshortest wavelength of X-rays produced.CXtraEdge for IIT-JEE 68APRIL 2010

MOCK TEST - AIEEE PATTERNSYLLABUS : Physics : Full syllabus Chemistry : Full syllabus Mathematics : Full syllabusTime : 3 Hours Total Marks : 432Instructions :• Part A – Physics (144 Marks) – Questions No. 1 to 2 and 9 to 30 consist FOUR (4) marks each and Question No.3 to 8 consist EIGHT (8) marks each for each correct response.Part B – Chemistry (144 Marks) – Questions No. 31 to 39 and 46 to 60 consist FOUR (4) marks each andQuestion No. 40 to 45 consist EIGHT (8) marks each for each correct response.Part C – Mathematics (144 Marks) – Questions No.61 to 82 and 89 to 90 consist FOUR (4) marks each andQuestion No. 83 to 88 consist EIGHT (8) marks each for each correct response• For each incorrect response, ¼ (one fourth) of the weightage marks allotted of the would be deducted.PHYSICS1. An P-N-P transistor circuit is arranged as shown. It isa –VPNPR L = 10 KV(A) common base amplifier circuit(B) common-emitter amplifier circuit(C) common-collector circuit(D) None2. A tuning fork and an air column in resonance tubewhose temperature is 51°C produces 4 beats in 1second when sounded together. When thetemperature of the air column decreases, the numberof beats per second decreases. When the temperatureremains 16°C, only 1 beat per second is produced.Then the frequency of the tuning fork is -(A) 55 Hz(B) 50 Hz(C) 68 Hz(D) none3. For wave propagation wrong statement is -(A) The wave intensity remains constant for a planewave(B) The wave intensity decreases as the inverse of thedistance from the source for a spherical wave(C) The wave intensity decreases as the inversesquare of the distance from the source forspherical wave(D) Total intensity of the spherical wave over thespherical surface centered at the source remainsconstant at all times4. Statement-1: Temperature of the body is loweredconsiderably if we put wet clothes.Statement -2: Specific heat of water is high.(A) Both statement -1 and statement -2 are trueand statement -2 is a correct explanation ofthe statement -1(B) Both statement -1 and statement -2 are truebut statement -2 is not a correct explanationof the statement -1(C) Both statement -1 and statement -2 arefalse(D) statement -1 is false but the statement -2 istrue.5. An observer starts moving with uniform acceleration atoward a stationary sound source emitting a whistle offrequency n. As the observer approaches source, theapparent frequency n' heard by the observer varieswith time as –(A)(C)n'n'timetime(B)(D)n'n'timetime6. The main scale of a spectrometer is divided into720 divisions in all. If the vernier scale consists of30 divisions, the least count of the instrument is(30 division of vernier scale coincide with29 division of main scale) -(A) 0.1°(B) 1''(C) 1'(D) 0.1''XtraEdge for IIT-JEE 69APRIL 2010

7. Experimental verification of Newton's law of coolingis valid for -(A) large temperature difference i.e.30°C to 85°C between hot liquid and surrounding(B) very large temperature difference i.e. 5°C to 95°Cbetween hot liquid and surrounding(C) small temperature difference i.e. 30°C to 35°Cbetween hot liquid and surrounding(D) any temperature difference8. While studying the dissipation of energy of a simplependulum by plotting a graph between square ofamplitude and time which of the following apparatusis not essential ?(A) Ticker timer (B) Meter scale(C) Vernier calliper (D) Stop watch9. An object is weighed on a balance whose pans arenot equal in masses when placed in the left pan, theobject appears to weigh 10.30g but when placed inthe right pan, it appears to weigh 12.62g. What is thecorrect mass of the object ?(A) 10.30 g (B) 12.62 g(C) 11.46 g (D) Can not find10. When jockey is put at two ends of the potentiometerwire, the galvanometer gives diflections in oppositedirections. It means that apparatus can -(A) not give a null point(B) give a null point(C) be faulty(D) be used after making some changes in the circuit11. A student performs an experiment to determine theYoung's modulus of a wire exactly 2cm long, by Searle'smethod. In a particular reading, the student measures theextension in the length of the wire to be 0.8 mm with anuncertainty of ± 0.05 mm at a load of exactly 1.0 kg.The student also measures the diameter of the wire tobe 0.4 mm with an uncertainty of ± 0.01 mm. Takeg = 9.8 m/s 2 (exact). The Young's modulus obtainedfrom the reading is -(A) (2.0 ± 0.3) × 10 11 N/m 2(B) (2.0 ± 0.2) × 10 11 N/m 2(C) (2.0 ± 0.1) × 10 11 N/m 2(D) (2.0 ± 0.05) × 10 11 N/m 212. A student measures the focal length of a convex lensby putting an object pin at a distance 'u' from the lensand measuring the distance 'v' of the image pin. Thegraph between 'u' and 'v' plotted by the student shouldlook like -(A)v(cm)Ou(cm)(B)u(cm)v(cm)O(C)v(cm)Ou(cm)(D)u(cm)v(cm)13. Two concave mirror each of focal length f. A pointsource is placed at a point midway between twomirror. The minimum value of d for which only oneimage of s is formed –(A) f(C) 3fsd(B) 2f(D) 4f14. In YDSE, if the intensity of central maxima is I O thenIthe y-coordinate of point where the intensity is O, 2d = 0.1 mm, D = 1m,λ = 5000A°(A) 1.5 mmd(C) 1.75 mmD(B) 2 mmPOy(D) 1.25 mm15. Two charges of –4µC and +4µC are placed at pointsA (1,0,4) and B (2, –1,5) located in an electric field→E = 0.20 î V/cm. The torque acting on the dipoleis-(A) 8 × 10 –5 N-m (B) 8/ 2 × 10 –5 N-m(C) 8 2 × 10 –5 N-m (D) 2 2 × 10 –5 N-m16. Three concentric spherical metallic shells A, B and Cof radii a, b and c (a < b < c) have surface chargedensities σ, – σ and σ respectively. If the shells Aand C are at same potential, then the correct relationbetween a, b and c is -(A) a + c = b (B) b + c = a(C) a – b = c (D) a + b = cOXtraEdge for IIT-JEE 70APRIL 2010

17. Five identical plates of equal area A are placedparallel to and at equal distance d from each other asshown in figure. The effective capacity of the systembetween the terminals A and B is -(A) 53(C) 35∈ o∈ oAdAd(B) 45(D) 54∈ o∈ oBAAdA18. Read the following statements carefully :Y : The resistivity of semiconductor decreases withincrease of temperature.Z : In a conducting solid, the rate of collisionbetween free electrons and ions increases withincrease of temperature.(A) Both Y and Z are true and Z is correctexplaination of Y(B) Both Y and Z are true but Z is not correctexplaination of Y(C) Y is true but Z is false(D) Y is false but Z is true19. A 1m long metallic wire is broken into two unequalparts A and B. The part A is uniformly extended intoanother wire C. The length of C is twice the length ofA and resistance of C is equal to that of B. The ratioof resistances of parts A and C is -(A) 4 (B) 41(C) 2 (D) 2120. A 600 cm long potentiometer wire is connected to acircuit as shown in figure. The resistance ofpotentiometer wire is 15r. The distance from pointA at which the jockey should touch the wire to getzero deflection in the galvanometer is –AE/2(A) 320 cm(C) 160 cmrEGJrdB(B) 230 cm(D) 460 cm21. A rectangular loop of metallic wire is of length a andbreadth b and carries a current i. The magnetic fieldat the centre of the loop is -(A)µ 0 i4π8µ 0 i 2(C)4π2a +2b(B)ab2a +2b(D)abµ 0 i 44πµ 0 i4π2a +2bab2a +2bab22. A short magnet produces a deflection of 30° when placedat certain distance in tanA position of magnetometer. Ifanother short magnet of double the length and thrice thepole strength is placed at the same distance in tanBposition of the magnetometer, the deflection producedwill be -(A) 60° (B) 30°(C) 45°(D) None of these23. A solenoid has 2000 turns wound over a length of 0.30 m.Its area of cross-section is 1.2 × 10 –3 m 2 . Around itscentral section a coil of 300 turns is wound. If aninitial current of 2A in the solenoid is reversed in0.25 sec, the emf induced in the coil is equal to -(A) 6 × 10 –4 Volt (B) 4.8 × 10 –2 Volt(C) 6 × 10 –2 Volt (D) 48 kV24. A 100 volt AC source of frequency 500 Hz isconnected to a L–C-R circuit with L = 8.1 mH,C = 12.5 µF and R = 10 Ω, all connected in series.The potential difference across the resistance is -(A) 100 V(B) 200 V(C) 300 V(D) 400 V25. Which one is correct ?(A) Resultant of two vectors of unequal magnitudecan be zero(B) Resultant of three non-coplanar vectors of equalmagnitude can be zero(C) Resultant of three coplanar vectors is always zero(D) Minimum number of non-coplanar vectors whoseresultant can be zero is four.26. A stone thrown with the velocity V 0 = 14 m/s at anangle 45° to the horizontal, dropped to the ground ata distance 'S' from the point where it was thrown.From what height should the stone be thrown inhorizontal direction with the same initial velocity sothat it fall at the same spot -(A) 14.2 m(B) 16.9 m(C) 10.0 m(D) 9.6 m27. A small body of mass 'm' is attached to one end of alight inelastic string of length l. The other end of thestring is fixed. The string is held initially taut andhorizontal and then body is released. The centripetalacceleration of the body and the tension in the stringwhen the string reaches vertical position will be -(A) g, mg(B) 2g, 3 mg(C) 3g, 2mg (D) 3g, 3 mgXtraEdge for IIT-JEE 71APRIL 2010

28. Assertion : A rocket moves forward by pushing thesurrounding air backwards.Reason : It derives the necessary thrust to moveforward, according to Newton's third law of motion.(A) Both Assertion and Reason are true and Reasonis a correct explanation of the Assertion(B) Both Assertion and Reason are true but Reason isnot a correct explanation of the Assertion(C) Both Assertion and Reason are false(D) Assertion is false but the Reason is true29. While slipping on rough spherical surface of radius'R', block A of mass 'm' comes with velocity 1 .4gRat bottom B. Work done in slipping the block from'B' to 'C' is –(A)mgR4(C) 1.3 mgRmAB(B) mgRC(D) 45 mgR30. A 2000 kg rocket in free space expels 0.5 kg of gasper second at exhaust velocity 400 ms –1 for 5seconds. What is the increase in speed of rocket inthis time -(A) 2000 ms –1 (B) 200 ms –1(C) 0.5 ms –1 (D) zeroCHEMISTRY31. Which of the following can act as a both Bronstedacid & Bronsted base -(A) Na 2 CO 3 (B) OH ––(C) HCO 3 (D) NH 332. In which compound the oxidation No. of Oxygenis + 21 -(A) OF 2 (B)O 2 F 2(C) O 2 [PtF 6 ] (D) KO 233. The favourable conditions for a spontaneousreactions are -(A) T ∆S > ∆H, ∆H = ⊕ , ∆S = ⊕(B) T ∆S > ∆H, ∆H = ⊕ , ∆S = Θ(C) T ∆S = ∆H, ∆H = Θ , ∆S = Θ(D) T ∆S = ∆H, ∆H = ⊕ , ∆S = ⊕34. The substance not likely to contains CaCO 3 is -(A) Dolomite (B) A marble statue(C) Calcined Gypsum (D) Sea shells35. On mixing 10ml of acetone with 50ml of CHCl 3 , thetotal volume of the solutions-(A) < 60ml(B) > 60ml(C) = 60ml(D) unpredictable36. On addition of He gas at constant volume to thereaction N 2 + 3H 2 2NH 3 at equilibrium-(A) The reaction stops(B) Forward reactions is favoured(C) Reaction remains unaffected(D) Backward reactions is favoured37. The half life of a reaction is 24 hours . If we startwith 10gm of reactant, How many grams of it willreaction after 96 hours, (I order reaction)(A) 0.625gm (B) 6.25gm(C) 1.25gm (D) 0.125gm38. The current is passed in Ag 2 SO 4 aqueous solution &1.6 gm O 2 is obtained. The amount of Ag depositedwill be- [Ag = 108gm](A) 107.8g (B) 1.6g (C) 0.8g (D) 21.6g39. In decinormal solution CH 3 COOH is ionized to theextent of 1.3% find the pH of solution.(A) 3.89 (B) 2.89 (C) 4.89 (D) 5.8940. A FCC element (atomic wt.= 60) has a cell edge of400pm. Its density is-(A) 6.23 g/cm 3 (B) 6.43 g/cm 3(C) 6.53 g/cm 3 (D) 6.63 g/cm 341. Which set of quantum No. is not possiblenl m s(A) 2 0 0 +1/2(B) 4 2 –3 –1/2(C) 3 2 –2 +1/2(D) 2 1 0 +1/242. Give simplest formula of compound whichcontaining 6gm C, 3.01×10 23 atom O and 2 mole Hatoms-(A) CH 2 O (B) CH 4 O (C) CHO (D) CH 3 O43. The IUPAC Name of(A) 1,2-dimethyl Cyclohexene(B) 2,3-dimethyl Cyclohexene(C) 1,2-dimethyl Cyclohex-2-ene(D) 5,6-dimethyl Cyclohex-1-ene44. Which of the following reagent can make distinctionbetween Pri. and Sec. amines ?(A) NH 3(B) NaNO 2 /HCl(C) HCl(D) AllXtraEdge for IIT-JEE 72APRIL 2010

45. Toluene reacts with Cl 2 in the presence of light togive -(A) Benzyl chloride (B) Benzoyl chloride(C) p-chlorotoluene (D) o- chlorotoluene46. Which compound is formed when excess of KCN isadded to an aqueous solution of copper sulphate(A) Cu (CN) 2 (B) K 2 [Ca(CN) 4 ](C) K [Cu(CN) 2 ] (D) K 3 [Cu(CN) 4 ]47. The Blue Print Process involves the use of-(A) Indigo dyes (B) Iron compound(C) Vat dyes (D) some other compounds48. The ionic radii of N 3– , O 2– , F – and Na + follow theorder-(A) N 3– > O 2– > F – > Na + (B) N 3– > Na + > O 2– > F –(C) Na + > O 2– > N 3– > F – (D) O 2– > F – > Na + > N 3–49. Reactionxy 2 xy + y(g) (g) (g)Initial pressure of xy 2 is 600 mm Hg & total pressureat equilibrium is 800 mm Hg. Kp of reaction is -(A) 50 (B) 100 (C) 166.6 (D) 40050. Cell: Zn|Zn +2 || Cu +2 | CuIf the correct reactions of Zn +2 & Cu +2 ions aredoubled, the emf of the cells:(A) doubled (B) halved(C) same(D) zero51. What is the pH of buffer solution containing 12gCH 3 COOH & 16.4g CH 3 COONa in 500ml ofsolution (Ka for CH 3 COOH = 1.8×10 –5 ).(A) 4.7447 (B) 4.4774(C) 4.4477(D) None52. How many moles of ferrous oxalate are completelyoxidized by 1 mole KMnO 4 in acidic medium-(A) 3/5 (B) 5/3(C) 1/5 (D) 553. In an irreversible process taking place at constant-T& P and which only pressure-volume work is beingdone, than (dG) and (dS), satisfy the criteria-(A) (dS) V, E > 0, (dG) T,P < 0(B) (dS) V, E = 0, (dG) T,P = 0(C) (dS) V, E = 0, (dG) T,P > 0(D) (dS) V, E < 0, (dG) T,P < 054. 3,3-dimethylbutan-2-ol, on reaction with Conc.H 2 SO 4 at 443K will give…… as major product-(A) 3,3-dimethyl but-1-ene(B) 2,3-dimethyl but-2-ene(C) 2,2-dimethyl but-2-ene(D) 2,2-dimethyl-1- butene55. Select the true statement about benzene fromamongst the following-(A) Because unsaturation benzene easily undergoesaddition reaction(B) There are two types of C-C bonds in benzenemolecule(C) There is a cyclic delocalisation of π es – inbenzene(D) Monosubstitution of benzene group gives threeisomeric substances56.(A)(C)OHZndustH CCl3⎯⎯⎯→B⎯⎯− ⎯⎯ →K4⎯ alk.KMnOAlCl3⎯⎯ ⎯⎯ →D. Identity ‘D’ ?CH 3CHO(B)(D)COOH57. In Lassaigne’s test, the organic compound isfirst fused with sodium metal. The sodiummetal is used because(A) The melting point of sodium metal is low(B) Sodium metal reacts with elements presentin organic compounds to form inorganiccompounds(C) All sodium salts are soluble in water(D) All the above58. Concentrated hydrochloric acid when kept in open airsometimes produces a cloud at white fumes theexplanation for it is that-(A) Oxygen in air reacts with the emitted HCl gas toform a cloud of chlorine gas(B) Strong affinity of HCl gas for moisture in airresults in forming of droplets of liquid solutionwhich appears like a cloudy smoke(C) Due to strong affinity for water concentratedhydrochloric acid pulls moisture of air towardsitself. This moisture forms droplets of water andhence the cloud.(D) Concentrated hydrochloric acid emits stronglysmelling HCl gas all the time59. Consider the following complex[Co(NH 3 ) 5 CO 3 ]ClO 4The coordination number, oxidation number, numberof d-electrons and number of unpaired electrons onthe metal respectively-(A) 6, 3, 6, 0 (B) 7, 2, 7, 1(C) 7, 1, 6, 4 (D) 6, 2, 7, 3XtraEdge for IIT-JEE 73APRIL 2010

60. <strong>Point</strong> out the incorrect statement about resonance(A) Resonance structure should have equal energy(B) In Resonance structure, the constituent atomsshould be in the same position(C) In Resonance structure there should not be samenumber of electron pairs(D) Resonance structure should differ only in thelocation of electrons around the constituentatomsMATHEMATICS61. If the angles of elevation of an aeroplane from twopoints 1 km apart be 60º and 30º, then the height ofthe aeroplane is -500(A) 500m(B) m3(C)2000 m (D) None362. Suppose a population A has 100 observations 101,102, ..........200 and another population B has 100observations 151, 152, ......... 250. If V A and V Brepresents the variances of the two populationVArespectively then is -V(A) 49(C) 32B(B) 94(D) 163. If vertices of a triangle are (1, 0), (2, b) & (c 2 , – 3)then centroid of triangle(A) can lie on y axis (B) always lie on x axis(C) lie on x axis if a + b = 3(D) lie on y axis if c =3 only64. The value of k in order thatf(x) = sin x – cos x – kx + b decreases for all realvalues is given by :(A) k < 1 (B) k ≥ 1(C) k > 2 (D) k < 2x65. x lim e 2 – cos x→ 0 2xis equal to -(A) 3/2 (B) 1/2(C) 2/3(D) None66. If ∆ =3aaa1233bbb123a + 2b − 3caa12+ 2b+ 2b121− 3c− 3c23ccc123bbb123, then(A) ∆(B) 2∆(C) – 3∆ (D) 0ccc123is equal to -67. In a ∆ABC, angle A is greater than angle B. If themeasures of angles A and B satisfy the equation 3sinx – 4 sin 3 x – k = 0, 0 < k < 1, then the measure ofangle C is -(A) 3π(B) 2π(C)2π3(D)5π668. If the p, q, r have truth values.F, F, T then the statement(p ↔ q) ∨ ~ r → (p ∧ r) will be(A) T (B) F (C) T, F (D) None69. The statement (p ∨ q) ↔ (q ∧ ~ p) is a(A) Tautology(B) Contradiction(C) Neither tautology nor contradiction(D) None of these70. If n th term of sequence12 ,271 ,131 20 51 , , .... is9 23 17then value of n is -(A) 20 (B) 10 (C) 5 (D) 1371. The ratio in which plane 2x – k = 0 divides the linejoining (–2, 4, 7) & (3, – 5, 8) is 9 : 1 then k equal to-(A) 4 (B) 5 (C) 6 (D) 772. If |a| = 2, |b| = 5 and |a × b| = 8 then |a – b| is equal to:(A) 12 (B) 15 (C) 17 (D) 573. The extremities of a line segment of length 6 move intwo fixed perpendicular lines. If locus of a point Pwhich divides this line segment in ratio 1 : 2 is anellipse then eccentricity of this ellipse is -(A) 21(B)12(C)32(D)74. The value of p such that the vertex of parabolay = x 2 + 2px + 13 is 4 units above x-axis & lies infirst quadrant is :(A) 3 (B) 4 (C) ± 3 (D) – 334XtraEdge for IIT-JEE 74APRIL 2010

75. If the lines represented by x 2 + 2λx + 2y 2 = 0 & linesrepresented by (1 + λ)x 2 – 8xy + y 2 = 0 are equallyinclined then λ equals :(A) – 2 (B) + 2(C) ± 2 (D) ± 484. In a class of 100 students there are 70 boys whoseaverage marks in a subject are 75.If the averagemarks of the complete class is 72, then what is theaverage of the girls.(A) 73 (B) 65(C) 68 (D) 7485. A letter is taken at random from the letters of word'STATISTICS' and a another letter is taken at randomfrom letters of word 'ASSISTANT'. The probabilitythat they are the same letter is -4(A) 1/45 (B) 13/90(C) 19/0 (D) 5/1886. If A = {x : x ∈ I ; – 2 ≤ x ≤ 2}2Β = { x : x ∈ I ; 0 ≤ x ≤ 3}(A) 2 (B) – 3C = {x : x ∈ N ; 1 ≤ x ≤ 2} and(C) – 5(D) NoneD = {(x, y) ∈ N × N; x + y = 8} then -(A) n(A ∪ (B ∪ C)) = 5 (B) n(D) = 6(C) n(B ∪ C) = 5 (D) None of these87. <strong>Solution</strong> of sec 2 dyy + 2x tan y = x 3 is -dx2x(A) tan y = ce − + (x 2 – 1)∫ 22–1xf (g(x)) f 'g(x). g'(x) dx where (B) tan y = ce − + (x 2 – 1)12xg(1) = g(2) is equal to -(C) tan y = ce − – (x 2 – 1)(A) 1 (B) 2(D) None of these(C) 0(D) None88. The equation of common tangent to the curves2 2x + sin xy 2 = 8x and xy = –1 is -sec 2 x dx and f(0) = 0 then2(A) 3y = 9x + 2 (B) y = 2x + 11 xf(1) =(C) 2y = x + 8 (D) y = x + 2(A) 1 – π/4 (B) π/4 – 1(C) tan1 – π/4 (D) None of these89. Let f be twice differentiable function such thatf "(x) = – f(x) and f '(x) = g(x)sec 1 xh(x) = (f(x)) 2 + (g(x)) 2 . If h(5) = 11is -x −[x]then h(10) is equal to(A) R(B) R – {(– 1, 1)I}(A) 22 (B) 11(C) R – I (D) R – [0, 1)(C) 0(D) None90. We are required to from different words with the help(A) D f = [– 1, 1] (B) R f = {–π/2, π/2}of letter of the word INTEGER. Let m 1 , be the(C) R f = {π/2} (D) None of thesenumber of words in which I and N are never togetherand m 2 be the number of words which begin withm1I and end with R. Thenm2is given by -76. locus of centre of a variable circletx 2 + ty 2 + 2(t 2 + 1)x – 2(t 2 – 1)y + t = 0 is a :(A) Straight line (B) Parabola(C) Ellipse(D) Hyperbola477. If∫f (x) dx = 4 and (3 f (x))− 1 ∫− dx = 7 then the2value of∫ 4f (x) dx is -78. Bisector of angle between lines 2x + y – 6 = 0 &4x – 2y + 7 = 0 which contains origin is -(A) acute angle bisector ; x = 5/8(B) acute angle bisector ; y = 19/4(C) obtuse angle bisector ; x = 5/8(D) obtuse angle bisector ; y = 19/479. The value of [ ]80. If f(x) =∫+81. The domain of the function f(x) =82. Let f(x) = sin –1 x + sec –1 x, then -83. If z 1 , z 2 , z 3 represents the vertices of an equilateraltriangle such that|z 1 | = |z 2 | = |z 3 | then -(A) z 1 + z 2 = z 3 (B) z 1 + z 2 + z 3 = 0(A) 30 (B) 1/30(C) 6 (D) 42XtraEdge for IIT-JEE 75APRIL 2010

MOCK TEST – BIT-SATTime : 3 Hours Total Marks : 450Instructions :• This question paper contains 150 questions in Physics (40) Chemistry (40), Mathematics (45), LogicalReasoning (10) & English (15). There is Negative Marking• Each question has four option & out of them, ONLY ONE is the correct answer. There is – ve marking.• +3 Marks for each correct & – 1 Mark for the incorrect answer.PHYSICS1. If the amplitude of a damped oscillator becomes halfin 2 minutes, the amplitude of oscillation w.r.t. initialone after 6 minutes is1 1 11(A) (B) (C) (D) 27 8 18 642. An infinite number of spring having force constantsas k, 2k, 4k, 8k ..... ∞ and respectively are connectedin series; then equivalent spring constant is(A) k (B) 2k (C) k/2 (D) ∞3. A point particle of mass 0.1 kg is executing SHM ofamplitude 0.1 m when the particle passes through themean position. Its kinetic energy is 8 × 10 –3 J. Theequation of motion of this particle when the initialphase of oscillation is 45º can be given by⎛ π ⎞⎛ π ⎞(A) 0.1 cos ⎜4 t + ⎟ (B) 0.1 sin ⎜4t + ⎟⎝ 4 ⎠ ⎝ 4 ⎠⎛ π ⎞⎛ π ⎞(C) 0.4 sin ⎜ t + ⎟ (D) 0.2 sin ⎜ + 2t ⎟⎝ 4 ⎠ ⎝ 2 ⎠4. A mass m is moving with constant velocity along aline parallel to x-axis away from the origin. It angularmomentum with respect to origin.(A) is zero(B) remains constant(C) goes on increasing (D) goes on decreasing5. A vessel containing oil (density = 0.8g/cm 3 ) overmercury (density = 13.6 g/cm 3 ) has a homogeneoussphere floating with half of its volume immersed inmercury and other half in oil. The density of materialof sphere in g/cm 3 is(A) 3.3 (B) 6.4 (C) 7.2 (D) 2.86. Two trains move towards each other with the samespeed, speed of sound is 340 ms –1 . If the pitch of thetone of the whistle of one is heard on the otherchanges by 9/8 times then the speed of each train isV V(A) 2 ms –1 (B) 2000 ms –1(C) 20 ms –1 (D) 200 ms –17. A sound level I differ by 4 dB from another sound ofintensity 10 nW cm –2 . The absolute value of intensityof sound level I in Wm –2 is(A) 2.5 × 10 –4 (B) 5.2 × 10 –4(C) 2.5 × 10 –2 (D) 5.2 × 10 –28. An ideal gas is taken through the cycle A → B → C→ A as shown. If the net heat supplied to the gas inthe cycle 5J, the work done by the gas in the processC → A2V(m 3 )1CP(N/m 2 ) 10(A) – 5 J (B) – 15 J (C) – 10 J (D) –20 J9. There are n electrons of charge e on a drop of oil ofdensity ρ. It is in equilibrium in an electric field E.Then radius of drop is1/ 21/ 2⎛ 2neE ⎞⎛ neE ⎞(A)⎜4 g⎟ (B)⎝ πρ⎜⎠g⎟⎝ ρ ⎠⎛ 3neE ⎞(C)⎜4 g⎟⎝ πρ ⎠1/ 3⎛ 2neE ⎞(D)⎜g⎟⎝ πρ ⎠BA1/ 310. Two identical cells of emf 1.5 V and internalresistance 1 Ω are in series. A third cell of similarparameters is connected in parallel to theXtraEdge for IIT-JEE 76APRIL 2010

combination. The terminal voltage of the cells A, B,C are1.5 V 1.5 VA1Ω B 1Ω1.5 V 1ΩC(A) 1, 1, 2 (B) 1.5, 1.5, 1.5(C) 1.5, 0, 0 (D) 2, 1, 111. A wire has resistance of R ohm at T kelvin. At whattemperature the resistance of wire be 2R ohm whentemperature coefficient of resistance is α per degreecentigrade.(273 − T) α + 1 (273 − T) α −1(A)(B)2α2α(273 − T) α −1(273 − T)2α + 1(C)(D)αα12. Two cells each of same emf but of internal resistancer 1 and r 2 are joined to form a series circuit through anexternal resistance R. Value of R in term of r 1 and r 2for which cell 1 has zero p.d. across it isE E1r 1 r22(A) R = r 1 – r 2 (B) R = r 1 + r 2r1r2r 1 + r2(C) = R (D)r + rr r= R12R1 213. A current i flows in the network shown. Resultingmangnetic induction at point p isA F aPaaa2aBµ(A) 04πia8 µ 0i(C) –2 πaE2aaa2µi(B) – 08πa2 µ i(D) 08 πa14. An alpha particle and a proton have same velocitywhen they enter a uniform magnetic field. The periodof rotation of proton will beDC(A) double of that of α particle(B) four times that of α particle(C) one half time that of α particle(D) same as that of α particle15. A coil of inductance 8.4 mH and resistance 6 Ω isconnected to a 12 V battery. The current in the coil is1A at approximate time(A) 500 s (B) 20 s (C) 35 ms (D) 1 ms16. A fish rising vertically up towards the surface withspeed 3ms –1 observe a bird diving vertically downtowards it with speed 9 m/s. The actual velocity ofbird is(A) 4.5 ms –1 (B) 5.4 ms –1(C) 3.0 ms –1 (D) 3.4 ms –117. A concave lens of glass, refractive index 1.5 has bothsurface of same radius of curvature R. on immersion ina medium of refractive index 1.75, it will behave as a(A) convergent lens of focal length 3.5 R(B) convergent lens of focal length 3R(C) divergent lens of focal length 3.5 R(D) divergent lens of focal length 3 R18. I is the intensity due to source of light at any point Pon the screen if light reaches the point P via twodifferent paths (a) direct (b) after reflection from aplane mirror then path difference between two pathsis 3λ/2, the intensity at P is(A) I (B) zero (C) 2I (D) 4I19. The surface of some material is radiated, in turn, bywaves of λ = 3.4 × 10 –7 m and λ = 5.4 × 10 –7 mrespectively. The ratio of stopping potential in twocases is 2 : 1, the work function is(A) 2.05 eV (B) 1.05 eV(C) 3.05 eV (D) None20. A X-ray tube has a working voltage of 40 × 10 3 V.The continuous spectrum limit of the emitted x-rays is(A) 0.17 Å (B) 0.13 Å (C) 0.13 Å (D) 0.31 Å21. The number of alpha and beta deca 88Ra 222experiences before turning into stable Pb 206 isotope is(A) 4, 2 (B) 2, 4 (C) 1, 3 (D) 6, 10yy´XtraEdge for IIT-JEE 77APRIL 2010

22. The displacement of interfering light waves arey 1 = 4 sin ωt and y 2 = 3 sin (ωt + π/2). The amplitudeof resultant wave is(A) 5 (B) 7 (C) 1 (D) 023. A beam of light of wavelength 600 nm from adistance source falls on a single slit 1 mm wide andresulting diffraction pattern is observed on a screen2m away. Distance between first dark fringe on eitherside of the central bright fringe.(A) 1.2 mm (B) 3.2 mm(C) 2.4 mm (D) 4.2 mm24. The intensity of light from one source is double ofthe other coherent source in a double slit experiment.The ratio of destructive to constructive interference inthe obtained pattern is(A) 34 (B) 1/34 (C) 17 (D) 1/1725. Two radioactive material of half life T are producedat different instants. Their activities area found to beA 1 and A 2 respectively when A 2 < A 1 . Their agedifference isA 2A1(A) 0.44 T log (B) 1.44 T logAAA(C) 4.44 T logA121A(D) 5.44 T logA26. Three concentric conducting spherical shell x, y and zhave radii a, b and c respectively such that c > b > a,their surface charge density are σ, –σ and σrespectively. Then potential V x is given byC22128. The potential energy of a particle varies with distanceA xx from a fixed origin as U = , where A and Bx2 + Bare dimensional constant then dimensional formulafor AB is(A) M L 7/2 T –2 (B) M L 11/2 T –2(C) M 2 L 9/2 T –2 (D) M L 13/2 T –329 A particle leaves the origin at t = 0 and moves in the+ve x axis direction. Velocity of the particle at any⎛ t ⎞instant is given by v = u ⎜1 − ⎟ . If u = 10 m/s and⎝ t´⎠t´ = 5 sec. Find the x coordinate of the particle at aninstant of 10s.vut´ = 5 sect(A) 0 (B) 10 m (C) 20 m (D) –10 m30. An aero-plane drops a parachutist. After covering adistance of 40 m, he opens the parachute and retardsat 2 ms –2 . If he reaches the ground with a speed of2ms –1 , he remains in the air for about(A)(C)σε0⎡a 2⎢⎢⎣c⎤− b + c⎥⎥⎦axyz(B)bσ [a – b + c]ε 0σ σ [a + b + c] (D) – [a + b – c]ε 0ε0(A) 16 s (B) 3 s (C) 13 s (D) 10 s31. A tank moves uniformly along x-axis. It fires a shotfrom origin at an angle of 30º with horizontal whilemoving along positive x-axis & the second shot isalso fired similarly that the tank moved alongnegative x-axis. If the respectively range of the shotsare 250 m and 200 m along x-axis, the velocity of thetank.27. A certain physical quantity is calculated from the30ºπformula (a 2 – b 2 )h, where h, a and b are all lengths.3The quantity being calculated is(A) velocity (B) length(C) area(D) volume(A) 9.4 m/s(C) 3.9 m/s(B) 4.9 m/s(D) 5.9 m/sXtraEdge for IIT-JEE 78APRIL 2010

32. A large number of particles are moving with samemagnitude of velocity v but having randomdirections. The average relative velocity between anytwo particles average over all the pairs is50 N(A) 4π v (B) 2π v (C) π3 v (D) π4 v33. A body is moving with uniform speed v on anhorizontal circle from A as shown in the fig. Changein the velocity in the first quarter revolution isOv 1AWE2 kg5 kg(A) 0, 2.5 ms –2 (B) 0, 0(C) 2.5 m/s 2 , 2.5 m/s 2 (D) 1 m/s 2 , 2.5 m/s 237. In the shown system m 1 > m 2 . Thread QR is holdingthe system. If this thread is cut, then just after cutting(A) v 2 north (B) 2 v south west(C) 2 v north-west (D) 2v westS34. A hollow vertical drum of radius r and height H has asmall particle in contact with smooth inner surface ofthe upper rim at point P. The particle is given ahorizontal speed u tangential to the rim. It leaves thelower rim at Q vertically below P. Taking n as aninteger for number of revolution we getP u(A) n =(C) n =2πrHu2π r2H / gHQ(B)(D) n =2πr2H / gu2πr2H / g35. A balloon is descending at a constant acceleration a.The mass of the balloon is M. When a mass m isreleased from the balloon is starts rising withacceleration a. Assuming that volume does notchange when the mass is released, what is the valueof m.(A)(C)2a⎛ a + g ⎞M (B) ⎜ ⎟ M(a + g)⎝ 2a ⎠2a(a + g)M(D)Maa + g36. Two blocks of masses 2 kg and 5 kg are at rest onground. The masses are connected by a string passingover a frictionless pulley which in under theinfluence of a constant upward force F = 50 N. Theaccelerations of 5 kg and 2 kg masses arem 2QR(A) Acceleration of mass m 1 is zero and that of m 2 isdirected upward(B) Acceleration of mass m 2 is zero and that of m 2 isdirected downward(C) Acceleration of both the block will be same(D) Acceleration of system is given by⎛ m ⎞⎜1 − m2⎟ kg, when k is the spring factor⎝ m1+ m2⎠38. A car of mass M accelerates starting from rest.1/ 2⎛ 2pt ⎞Velocity of the car is given by v = ⎜ ⎟ , where p⎝ M ⎠is the constant power supplied by the engine. Theposition of car as a function of time is given as⎛ 8p⎞(A) ⎜ ⎟⎠⎝ 9M⎛ 8p⎞(C) ⎜ ⎟⎠⎝ 9M1/ 2t 3/21/ 2m 1⎛ 9p⎞(B) ⎜ ⎟⎠⎝ 8Mt 2/3 ⎛ 9p ⎞(D) ⎜ ⎟⎝ 8M ⎠1/ 2t 3/239. Six identical uniform rods PQ, QR, RS and ST, TV,UP each weighing w are freely joined at their ends toform a hexagon. The rod PQ is fixed in a horizontalposition and middle points of PQ and ST areconnected by a vertical string. The tension in string isP QUT S(A) W (B) 3W (C) 2W (D) 4WRt 3XtraEdge for IIT-JEE 79APRIL 2010

40. A body of mass 2 kg is being dragged with a uniformvelocity of 2 ms –1 on a horizontal plane. Thecoefficient of friction between the body and thesurface is 0.2. Work in 5 sec. is(A) 39.2 J (B) 9.32 J (C) 23.9 J (D) 93.2 JCHEMISTRY1. 100 kg of iron ore (Fe 2 O 3 ) containing 20% impuritieson reduction with CO give iron equal to -(A) 112 kg(B) 80 kg(C) 100 kg(D) 56 kg2. Given : The mass of electron is 9.11 × 10 –31 kg,Planck’s constant is 6.626 × 10 –34 Js, the uncertaintyinvolved in the measurement of velocity within adistance of 0.1 Å is -(A) 5.79 × 10 7 ms –1 (B) 5.79 × 10 8 ms –1(C) 5.79 × 10 5 ms –1 (D) 5.79 × 10 6 ms –13. The van der Waal equation for 0.5 mol of real gas is -⎛ a ⎞ ⎛ V − b ⎞(A) ⎜P + ⎟ ⎜ ⎟ = RT2⎝ 4V ⎠ ⎝ 2 ⎠⎛(B) ⎜P+⎝a24V⎛ a ⎞(C) ⎜P + ⎟ =2⎝ 4V ⎠⎛ a ⎞(D) ⎜P + ⎟ =2⎝ 4V ⎠⎞⎟ (2V – b) = RT⎠RT2(V – 2b)2RT(2V − b)4. One mole of N 2 O 4 is enclosed in a 5L container. Atequilibrium, the container has 0.5 mol of N 2 O 4 . Theequilibrium constant for the decomposition of N 2 O 4[N 2 O 4 (g) 2NO 2 (g)] is-(A) 1 (B) 52(C) 3 (D) 515. Which one is the strongest Bronsted Lowry base outof the following -(A) ClO –−−−(B) ClO 2 (C) ClO 3 (D) ClO 46. The pH of a solution obtained by mixing 50 mL of0.4 M HCl with 50 ml of 0.2 N NaOH is -(A) – log 2 (B) – log 2 × 10 –1(C) 1.0 (D) 2.07. Oxidation number of sulphur in H 2 SO 5 is-(A) +2 (B) + 4 (C) +8 (D) +68. Equivalent mass of FeC 2 O 4 in the reactionFeC 2 O 4 → Fe 3+ + CO 2 is -(M.wt of FeC 2 O 4 = M)(A) M (B) M/2 (C) M/3 (D) 2M/39. The fraction of total volume occupied by the atoms ina simple cube is -π(A) 4(B)π2 8(C)2 6π(D) 6π10. In the diagram given below the value X is -Cu2+(A) 0.325 V(C) – 0.35 V+ 0.15V⎯⎯⎯→Cu++ 0.50V⎯⎯⎯→CuE° = X Volt(B) 0.65 V(D) – 0.65 V11. For a first order reaction, A → B, t 1/2 = 1 hr. Whatfraction of the initial conc. of A reacts in 4 hrs ?15 1 7 1(A) (B) (C) (D) 16 16 8 812. An azeotropic solution of two liquids has boilingpoint lower than that of either of them if it -(A) shows a –ve deviation from Raoult’s Law(B) shows a +ve deviation from Raoult’s Law(C) shows no deviation from Raoult’s Law(D) is saturated13. In multi-molecular colloidal solution atoms ormolecules are held together by -(A) Hydrogen bonding(B) Strong attraction forces(C) Van der Waal’s forces(D) Strong electrical forces14. Given : C + 2S → CS 2 ; ∆H° = + 117 kJC + O 2 → CO 2 ; ∆H° = –393 kJS + O 2 → SO 2 ; ∆H° = – 297 kJThe value of ∆H combustion of CS 2 in kJ mol –1 is(A) – 1104 (B) + 1104 (C) + 807 (D) – 80715. Aspirin is chemically -(A) Methyl salicylate(B) Ethyl salicylate(C) Acetyl salicylic acid(D) o-hydroxy benzoic acid16. Aniline when diazotized in cold and then treated withdimethyl aniline gives a colored product. It’sstructure would be -(A) (CH 3 ) 2 N N = N(B) (CH 3 ) 2 NNH(C) CH 3 NH N = N NHCH 3(D) CH 3 N = N NH 2XtraEdge for IIT-JEE 80APRIL 2010

32. Which one of the following is not an amphotericsubstance -−(A) HNO 3 (B) HCO 3 (C) H 2 O (D) NH 333. Which reaction cannot be used for the production ofhalogen acid –(A) 2KBr + H 2 SO 4 → K 2 SO 4 +2HBr(B) NaHSO 4 + NaCl → Na 2 SO 4 + HCl(C) NaCl + H 2 SO 4 → NaHSO 4 + HCl(D) CaF 2 + H 2 SO 4 → CaSO 4 + 2HF34. B(OH) 3 + NaOH NaBO 2 + Na[B (OH) 4 ] + H 2 OHow can this reaction is made to proceed in forwarddirection -(A) Addition of cis 1, 2-diol(B) Addition of borax(C) Addition of trans 1, 2-diol(D) Addition of Na 2 HPO 435. Sodium thiosulphate is prepared by -(A) Reducing Na 2 SO 4 solution with H 2 S(B) Boiling Na 2 SO 3 solution with S in alkalinemedium(C) Neutralising H 2 S 2 O 3 solution with NaOH(D) Boiling Na 2 SO 3 solution with S in acidic medium36. The critical temperature of water is higher than thatof O 2 because H 2 O molecule has -(A) Fewer electrons than oxygen(B) Two covalent bond(C) V-shape(D) Dipole moment37. Zone refining is a technique used primarily for whichone of the following process -(A) Alloying (B) Tempring(C) Sintering (D) Purification38. Which one of the following elements has the highestionization energy –(A) [Ne] 3s 2 3p 1 (B) [Ne] 3s 2 3p 2(C) [Ne] 3s 2 3p 3 (D) [Ar] 3d 10 4s 2 4p 239. The correct order of dipole moment is -(A) CH 4 < NF 3 < NH 3 < H 2 O(B) NF 3 < CH 4 < NH 3 < H 2 O(C) NH 3 < NF 3 < CH 4 < H 2 O(D) H 2 O < NH 3 < NF 3 < CH 440. If N x is the number of bonding orbitals of an atomand N y is the no. of the antibonding orbitals, then themolecule/atom will be stable if -(A) N x > N y (B) N x = N y(C) N x < N y (D) N x ≤ N yMATHEMATICS1. Consider the sequence (angles are measured inradians) sin log 10 2 , sin log 10 3 , sin log 10 4 ….then -(A) all the terms of this sequence are positive(B) all the terms of this sequence are negative(C) 1001 th term is negative(D) 10001 th term is negative2. The order relation between x, sin –1 x & tan –1 xx ∈(0 ,1) is -(A) tan –1 x < x < sin –1 x (B) sin –1 x < tan –1 x < x(C) x < sin –1 x < tan –1 x (D) None3. The smallest positive valve of x satisfying theequation log 2 cos x + log 2 (1 – tan x ) + log 2 (1 + tan x)– log 2 sin x = 1 is -(A) π/8 (B) π/6 (C) π/4 (D) π/64. A pole stands at a point A on the boundary of acircular park of radius r and subtends an angle α atanother point B on boundary. If arc AB subtends anangle α at the centre of the path, the height of thepole is -(A) r sin α/2 tan α (B) 2r sin α/2 tan α(C) 2r sin α/2 cot α (D) None of these5. The base of a triangle lies along the line x = a and isof length 2a. The area of the triangle is a 2 . If the thirdvertex lies on the line -(A) x = 0(B) x = – a(C) x = 2a, or x = 0 (D) x = 0 or x = – 2a6. If y = mx bisects an angle between the linesax 2 – 2hxy + by 2 m 2 –1= 0 then =mb – a b – b a + b(A) (B) (C) (D) Noneh h h7. If the circle x 2 + y 2 + 2gx + 2fy + c = 0 passesthrough all the four quadrant then -(A) g = – b (B) C > 0 (C) C < 0 (D) None8. The equation of the circle which has two normals(x–1) (y – 2) = 0 and a tangent 3x + 4y = 6 is(A) x 2 + y 2 – 2x – 4y + 4 = 0(B) x 2 + y 2 + 2x – 4y + 5 = 0(C) x 2 + y 2 = 5(D) (x –3) 2 + (y – 4) 2 = 59. Circles drawn on the diameter as focal distance ofany point lying on the parabola x 2 – 4x + 6y + 10 = 0will touch a fixed line whose equation is(A) y = 2 (B) y = –1(C) x + y = 2 (D) x – y = 2XtraEdge for IIT-JEE 82APRIL 2010

10. The foci of a hyperbola coincide with the foci of thex 2ellipse25y 2+ 9= 1. If eccentricity of thehyperbola is 2, then its equation is(A) x 2 – 3y 2 – 12 = 0 (B) 3x 2 – y 2 – 12 = 0(C) x 2 – y 2 – 4 = 0 (D) None of these11.→α and→β are two mutually perpendicular unit vectora → α + a → β + c( → α × → β ), → α + ( → α × → β ) and c → α + c → β+ b ( → α × → β ) are coplaner then c is(A) A.M. of a & b (B) G.M. of a & b(C) H.M. of a & b (D) None of these12. The point of contact of the spheresx 2 + y 2 + z 2 + 2x – 4y – 4z – 7 = 0x 2 + y 2 + z 2 + 2x – 4y – 16z + 65 = 0(A) (1, 2, 6) (B) (1, 2, –6)(C) (1, –2, 6) (D) (–1, 2, 6)13. If f(x) = 3 – 4{x 2 – 4x + 8} –1 then range of f(x) is(A) (–∞, 1) ∪ (3, ∞) (B) (2, 3)(C) [2, 3](D) None of these14. If x > 0 and g is a bounded function thennxf (x)e + g(x)limisn →∞ nxe + 1(A) 0 (B) f(x) (C) g(x) (D) None15. If a 1 = 1 and a n = n(1 + a n–1 ) ∀ n ≥ 2 than⎛ 1 ⎞⎛ 1 ⎞ ⎛ 1 ⎞lim ⎜1+⎟⎜1+⎟...⎜1+⎟ =n→∞⎝a1⎠⎝a 2 ⎠ ⎝ a n ⎠(A) 1 (B) e (C) 1/e (D) None16. Let f(x) = |2 sgn 2x| + 2 then f(x) has(A) removable discontinuity(B) infinte discontinuity(C) No discontinuity(D) essential discontinuity17. If f(x) = cosπ 3⎧⎨ [x] − x⎩2⎫⎬ , 1 < x < 2 and [.] = G.I.F.⎭⎛ ⎞then f´ ⎜π3 ⎟ is⎝ 2 ⎠(A) 0 (B) 3(π/2) 2/3(C) –3(π/2) 2/3 (D) None of these18. If ye x = cos x then, y 4 / y =(A) –1 (B) 2 (C) –4 (D) None19. Let f & g be differentiable function satisfying g´(a) = 2,g(a) = b and fog = I (Identity function), then f´(b) isequal to(A) 1/2 (B) 2 (C) 2/3 (D) None20. Tangents are drawn from origin to the curve y = sin xpoints of contact lie on the curve(A) x 2 + y 2 = x 2 y 2 (B) x 2 – y 2 = xy(C) x 2 – y 2 = x 2 y 2 (D) None of these21. Two positive numbers whose sum is 16 and sum ofwhose cubes is maximum are given by(A) 8, 8(B) no such number exist(C) 0, 16(D) None of these1 122. Let f(x) =2 , g(x) = on [a, b], 0 < a < b. Letx xf (b) − f (a) f´(c)= for same a < c < b then c isg(b) − g(a) g´(c)(A) A.M. of a & b(C) H.M. of a & b2(B) G. M. of a & b(D) None of these23.∫1+ cos x . sin 2x cos 2x dx =(A) 52 (1 + cos 2 x) 3/2 (3 – 2cos 2 x) 2 + c(B) 52 (1 + cos 2 x) 3/2 (3 – 2 cos 2 x) + c(C) 52 (1 + cos 2 x) 3/2 (3 + 2 cos 2 x) + c(D) None of these⎛ 2 424.∫ ⎟ ⎟ ⎞⎜ x x1− + .... dx⎜ 2 4⎝⎠(A) sin x (B) – sin x (C) cos x (D) None25.26.πx→2∫∫xπ / 22x2π / 4−cos t(2 −1)dtlim =(A)(log e 2(B)πt − π / 2)dtln22π2nlim 1 ⎞2/⎜1 ⎟n→∞2 ⎠(C)2ln2π(D) None224 / nn⎛ ⎛22 ⎞ ⎛2+ ⎜1⎟3 ⎞6/+ ⎜12 + ⎟⎝ nn2⎝ ⎠ n⎝ ⎠22n / n⎛2n ⎞.... ⎜1+ ⎟2n⎝ ⎠=(A) 4/e (B) 3/e (C) 2/e (D) NoneXtraEdge for IIT-JEE 83APRIL 2010

27. The area bounded by the curves y = 2x 2 | x |& y = xand x = 0 is equal to2 2 2(A) (B) 3 6(C)26(D) None28. Order and degree of the differential equationy " = (y´+ 3) 1/3 are respectively(A) 2, 2 (B) 2, 3 (C) 3, 2 (D) None29. If x 18 = y 21 = z 28 then 3, 3 log y x, 3 log z y, 7 log x z arein(A) A.P. (B) G.P. (C) H.P. (D) None30. If log 2 x + log 2 y ≥ 6 then least possible value of x + yis(A) 32 (B) 16 (C) 8 (D) None31. No. of real roots of the equationx 3 + x 2 + 10x + sin x = 0 is(A) 1 (B) 2 (C) 3 (D) ∞32. The roots of the equation ax 2 + bx + c = 0, a ∈R + aretwo consecutive odd positive integers then(A) |b| ≤ 4a (B) |b| ≥ 4a(C) |b| ≥ 2a(D) None of these33. The sum of the terms of an infinitely decreasing G.P.is equal to the greatest value of the functionf(x) = x 3 + 3x – 9 on the interval [–2, 3] and thedifference between the first two terms is f´(0) thensum of first terms is(A) 19 or – 37 (B) 19(C) –37(D) None of these34. If the complex number z 1 = a + i, z 2 = 1 + ib, z 3 = 0form an equilateral triangle (a, b are real numberbetween 0 & 1) then :(A) a = 3 – 1, b = 3 / 2(B) a = 2 – 3 , b = 2 − 31 3(C) a = , b = 2 4(D) None of thesen35. ∑(−1)r=0(A)21n −r n C r1⎛⎜1r⎝ 2(B)r r3 7 ⎞+ + + ..... ∞⎟is equal to2r 3r2 2⎠32n − 1(C) 22 1n −36. The coefficient of x 3 y 4 z in the expansion of(1 + x + y – z) 9 is(A) 2 . 9 C 7 . 7 C 4 (B) – 2 . 9 C 2 . 7 C 3(C) 9 C 7 . 7 C 4 (D) None of these(D) Nonee x37. If = B 0 + B 1 x + B 2 x 2 + .... then B n – B n–1 = ?1−x1 1 1(A) (B) (C) (D) Nonen n n −138. The number of point (x, y, z) in space whose eachcoordinate is a negative integer such that x + y + z +12 = 0 is(A) 55 (B) 110 (C) 75 (D) None39. Six boys and six girls sit along a line alternativelywith probability P 1 & along a circle (againalternatively) with probability P 2 then P 1 /P 2 is equalto(A) 1 (B) 1/5 (C) 6 (D) None40. If f(x) is a polynomial satisfyingf(x) = 21f (x)1⎛ 1 ⎞f ⎜ ⎟ − f (x)⎝ x ⎠and f(2) = 17⎛ 1 ⎞f ⎜ ⎟⎝ x ⎠then the value of f(5) is(A) 624 (B) –124 (C) 626 (D) 126⎡1x⎤41. If A = ⎢ ⎥ is idempotent then x =⎣02 ⎦(A) 0 (B) 2(C) no such x exist (D) None of these42. Let R be a relation on the set of integers given bya R b if a = 2 k b for some integer k then R is(A) an equivalence relation(B) reflexive and symmetric but not transitive(C) reflexive and transitive but not symmetric(D) symmetric and transitive but not reflxive43. Minimum value ofb + ca+c + ab+ve numbers a, b, c) is(A) 1 (B) 2 (C) 4 (D) 6+a + b , (for realc44. From mean value theorem f(b) – f(a) = (b – a) f´(x 1 );a < x 1 < b if f(x) = x1 then x1 =(A)ab (B)a + b2(C)2aba + b(D)b − aa + b45. If f(x) =∫cot 4 1x dx + cot 3 ⎛ π ⎞x – cot x and f ⎜ ⎟ 3 ⎝ 2 = π⎠ 2then f(x) is(A) π – x (B) x – π (C) x (D) NoneXtraEdge for IIT-JEE 84APRIL 2010

LOGICAL REASONING1. Fill in the blank spaces.11, 12, 17, 18, 23, 24, (?)(A) 12 (B) 29 (C) 30 (D) 352. Choose the best alternative.Dum-Dum : Calcutta : : Palam : ?(A) Kerala (B) Delhi (C) Madras (D) Bombay3. Pick the odd one out –(A) Wheat (B) Paddy (C) Towar(D) Mustard4. Which of the following figures (A), (B), (C) and (D)when folded along the lines, will produce the givenfigure (X) ?(C)(D)7. In the following question, find out which of theanswer figures (A), (B), (C) and (D) completes thefigure-matrix ??(A) (B) (C) (D)(X)(A) (B)(C)(D)8. The question that follow contain a set of three figuresX, Y and Z showing a sequence of a piece of paper.Fig. (Z) shows the manner in which the folded paperhas been cut. These three figures are followed by fouranswer figures from which you have to choose afigure which would most closely resemble theunfolded form of fig. (Z).5. In each of the following questions, choose the set offigures which follows the given rule.Rule : The series becomes complex as it proceeds :(A)X Y Z(B)(C)(D)(A)(C)(B)(D)6. In following question below, you are given a figure(X) followed by four figures (A), (B), (C) and (D)such that (X) is embedded in one of them. Trace outthe correct alternative.9. In the following question, complete the missingportion of the given pattern by selecting from thegiven alternatives (A), (B), (C) and (D).?(X)(X)(A)(B)(A) (B) (C) (D)10. In the following question, find out which of theanswer figures (A), (B), (C) and (D) complete thefigure - matrix ?XtraEdge for IIT-JEE 85APRIL 2010

(A) (B) (C) (D)ENGLISH1. Choose the one which best expresses the meaningfulconcept :The state's duty is to . . . . the safety of its Citizens.(A) assure (B) ensure (C) insure (D) accept2. Choose the one which best expresses the meaningfulconcept :The company went . . . . in the 1990's recession.(A) burst (B) bust (C) bursted (D) busted3. Choose the one which best expresses the meaningfulconcept :What can we . . . . from this evidence, Watson ?(A) deduce(B) deduct(C) reduce(D) conduce4. Choose the one which best expresses the meaningfulconcept in opposite meaning :Zenith :(A) Sky(B) Firmament(C) nadir(D) naive5. Pick up the correct Synonym for the following wordVoracious :(A) Hungry (B) Wild(C) Quick(D) Angry6. One who travels from place to place :(A) Journey man (B) Tramp(C) Itinerant (D) Mendicant7. Choose the one which best expresses the meaning ofthe given idiom/proverb :To fly off the candle :(A) To dislocate (B) To lose one's temper(C) To take off (D) To be indifferent8. Fill in the blanks with one of the options givenbelow:Gandhi Ji . . . . . . smoking in his youth.?(A) took to(C) took in(B) took for(D) took up9. Select the one which best expresses the samesentence in Passive or Active Voice.Get the box broken.(A) Get someone to break the box.(B) They have broken the box(C) Have the broken box(D) Break the box10. Choose the one which best expresses the correctanswer in the speech :He said, "How shabby you are looking!"(A) He asked how shabby I was looking(B) He exclaimed with disgust that I was lookingvery shabby(C) He exclaimed with sorrow that they werelooking much shabby(D) He told that I was looking much shabby11. Pick out the mis-spelt word –(A) Neigh (B) Rein (C) Neice (D) Neither12. Find out which part of the sentence has an error :I wonder / what he hasdone with the book /(a)(b)I lendhim /(c)NoError(d)(A) Wonder(B) What he has done with the book(C) I lend him(D) No Error13. Pick out the most appropriate pair to fill in theblanks in the same order, to make the sentencemeaningfully complete :She was . . . . . because all her plans had gone . . . . .(A) distraught, awry (B) Frustrated, Magnificently(C) Elated, wild (D) Dejected, splendidly14. Pick out the most effective word from the givenwords to fill in the blanks to make the sentencemeaningfully complete :Most of the issues discussed in the meeting weretrivial and only a few were :(A) Interesting (B) Practical(C) Complex (D) significant15. Pick out the most appropriate pair to fill in theblanks in the same order, to make the sentencemeaningfully complete :The . . . . . of glory lead but to the . . . . .(A) Paths, grave (B) Ways, happiness(C) Acts, Prosperity (D) Achievements, SufferingXtraEdge for IIT-JEE 86APRIL 2010

SOLUTION FOR MOCK TESTIIT-JEE (PAPER - I)1.[A]CHEMISTRYO||C 6 H 5 – CH 3 CC || OC||OCH 3C –C 6 H 5||O2.[B] A ⎯⎯⎯⎯→B3.[D]4.[C]⎯ OzonolysisOrientation foraldol condensationNH2OH,H2SO⎯⎯⎯⎯ ⎯⎯ →∆⎯4oximeformation followed by Beckmann rearrangement.C –NH–CH 3O ||N-methyl benzamideSo, B isA isandCCH | = O3CCH | = C3 CH |3K Cr2O7,H(CH 3 ) 3 CCH 2 CH 2 OH ⎯⎯⎯2 ⎯ → +H O, HeatSOCl ⎯⎯→⎯ 2(CH 3 ) 3 CC H 2 CH 2 N(CH 3 ) 2(D)2(CH 3 ) 3 CCH(A)2(CH 3 ) 3 CCH 2 COCl(B)(CH 3 ) 2 NHCOOHO(CH 3 ) 3 CCH2C || N(CH 3 ) 2(C)LiAlH4,ether← ⎯⎯⎯ ⎯⎯H OAttack of nucleophile is a rate determining stepO OF 3 CIOII2CH 35.[A]6.[D]7.[B]8.[A]I is more electron deficient and facilitates a fasterattack.As water introduces, water dissolves HCl(g) and apress drop is produced Liquid level in thecapillary rises.The energy of AOs depends on the(n + l) values n + l value of (n – 1) d = n + 1 ;n + l value of ns = n(n + l) value of (n + 1)d = n + 3 ; n + l value of nf= (n + 3)But due to lower value of principle quantum no.energy of nf < (n + 1) d∴ energy of (n + 2) s < nf.A, B and C are magnesium ,aluminium andsilicon. Magnesium form ionic oxide, MgO ;Aluminium forms amphoteric oxide ,Al 2 O 3 andsilicon forms a giant molecule SiO 2 .9.[ B,C,D] Due to resonance cyclohexatriene cation isaromatic which causes it's stability.10.[ A,B,D] Catalyst lower the activation energy offorward reaction & backward direction keepingthe same enthalpy of reaction.11.[ A,B,C] Greater the value of (IE – ∆ eg H) greater is theelectronegativity E.N. ofP = (1680 + 340)/(4.18) (125) ~ – 412.[A,B,C] H = E + PV⎛ dH ⎞ ⎛ dE ⎞⎜ ⎟⎠ = ⎜ ⎟⎠⎝ dP ⎝ dPT⎛ dH ⎞For liquid, ⎜ ⎟⎠⎝ dPTT⎛ dV ⎞+ P ⎜ ⎟⎠⎝ dP⎛ dE ⎞= ⎜ ⎟⎠⎝ dVTT+ V⎛ dV ⎞⎜ ⎟⎠⎝ dPT+⎛ dV ⎞P ⎜ ⎟⎠⎝ dP⎛ dV ⎞For incompressible liquid, ⎜ ⎟⎠ – ~ 0.⎝ dP⎛ dH ⎞∴ ⎜ ⎟⎠⎝ dPT~ – VTT+ VXtraEdge for IIT-JEE 87APRIL 2010

⎛ dH ⎞For ideal gas, ⎜ ⎟⎠⎝ dPT= 0⎛ dE ⎞For the real gas, if ⎜ ⎟⎠⎝ dV⎛ dH ⎞then ⎜ ⎟⎠⎝ dPTT= 0,⎛ dV ⎞= P ⎜ ⎟⎠⎝ dPT+ V ≠ 013.[A] F(Monoester) Molecular weight = 186No Br 2 reaction ⇒ SaturatedTwo oxygen ⇒ 2 × 16 = 32No. of CH 2 = (186 – 32)/14 = 11Hence, molecular formula of saturated monoesterF, is C 11 H 22 O 2HydrolysisGOptically active,soluble inNaOH ⇒ AcidWe have,Ag + Brsalt ⎯ → 2 ± JHunsdiecker reaction;radical intermediate,so racemic mixture(J contains onecarbon less than G)HOptically active,(Alcohols are notsoluble in NaOH)+ve iodoform test,it suggestsHO– CH – R On warning| with H 2 SO 4CH 3 (dehydration)I (no diastereomers meansno geometrical isomers)It suggests same alkyl groupon one of the doubly bondedcarbon atomsI. TsClH optically active ⇒ JII. NaBrIt suggests G contains one morecarbon atom than H.Hence, molecular formula of ester F, isC 5 H 11 – C || O–O–C 5 H 11OC 5 H 11 – C || –OH(G) C 5 H 11 OH(H)CH 3*|H CH 3⇒ HO – CH – CH – CH 3 ⎯H 2⎯SO 4⎯, → ∆ C=C (I)HCH 3 C CH 33No cis-trans isomersHTsCl⎯ →NaBrO||So, G is CH 3 – CH − CH − C − OH| | |CH3 CH3CH3⎯ CH 3 – CH – C H – BrCH | CH|*3 3(J)⎛O ⎞⎜* || ⎟⎜Br2CH3− CH − CH − C−OAg ⎯⎯→CH3− CH − CH − Br⎟⎜ | || | ⎟⎜ CH3CH3CH3CH ⎟3⎜⎟⎝( ± J)⎠O||F = CH 3 – CH − CH − C − O − CH − CH − CH3| || |CH CH CH CH3314.[C] I exists as diastereomers and H is optically active.So, H is HO– CH − CH 2 CH 2 –CH 3|CH 3H 2 SO 4∆HHC = C, andH 3 CCH 2 CH 315.[D]16.[A]G is CH 3 CH 2 CH 2 – CHCO−|| OH|CH 3Since H is optically active and gives negativeiodoform test, so H isHO–CH 2 – CH – CH 2 –CH 2 and|CH 3OG is CH 3 CH 2 CHCH2C || –OH|CH317.[D] Final pH = 1.7 or – log [H + ] = 1.7or log [H + ] = – 2 + 0.3 or [H + ] = 2 × 10 –2Let v ml of 0.1 (M) HA solution is mixed with100 ml of 10 –2 (M) HCl. In the mixed solution,[HA] =4v× 10− × 10 3 =(100 + v)−3330.1v(1003(M) and+ v)10 × 10 1[HCl]= =(100 + v) (100 + v)HA H + + A –0.1v(1 – α)(100 + v)∴ [H + ] =0.1v α(100 + v)1+(100 + v)0.1v α(100 + v)(0.1v α + 1)= 2 × 10 –2(100 + v)XtraEdge for IIT-JEE 88APRIL 2010

or 0.1 v α + 1 = 2 + 2 × 10 –2 × v(0.1vα + 1) 0.1vα×(100 + v) (1.00 + v) α –2K a == × 2 × 100.1v(1 − α)( 1−α)(100 + v)α∴ × 2 × 10 – 2= 10 –2 or( 1−α)or 2α = 1 – α or α = 1/3∴ 0.1 v × 31 + 1 = 2 + 2 × 10–2 × vα1− αv v 20vor – = 1 or = 130 50 30×50150or v = = 75 ml218.[D] For isohydric solution, Ka 1 C 1 = Ka 2 C 2= 21Column Matching19. [A] → r,s,t; [B] → p,r,s; [C] → s; [D] → q,tNaNO 2 / HClFor CH 3 CH 2 CH 2 NH 2 ⎯⎯⎯⎯→CH 3CH 3CH 3+CH3CH2CH2N2Diazonium ion issimply int ermediate;not the productCH 3 –CH 2 –CH 2 –OH+ O HN – CH 3 on heating doesn't give alkene.20. [A] → q,r; [B] → p,r; [C] → r,s; [D] → p,t−OH / Br 2C 6 H 5 CH 2 –CHO⎯⎯⎯→ C 6 H 5 – CH –CH = O|BrCH 3 CHOOH−⎯⎯⎯⎯→0 − 25º CCH 3 –OH|CH( ± )( ± )–CH 2 –CH = OCH 3 –CH 2 CH = O⎯ OH25ºC→ −CH 3 CH 2 –CH= C − CHO|CH 3Conc. OHCH 3 – CH − CH = O⎯⎯⎯⎯|CH 3OCH 3 – CH − C || –O – + CH 3 – CH − CH 2 OH||CH 3CH 3→ −MATHEMATICS1.[A] Let us first count the number of elements in F.Total number of functions from A to B is 3 4 = 81.The number of functions which do not containx(y) [z] in its range is 2 4 .∴ the number of functions which contain exactlytwo elements in the range is 3 . 2 4 = 48.The number of functions which contain exactlyone element in its range is 3.Thus, the number of onto functions from A to B is81 – 48 + 3 = 36[using principle of inclusion exclusion]n (F) = 36.Let f ∈ F. We now count the number of ways inwhich f –1 (x) consists of single element.We can choose preimage of x in 4 ways. Theremaining 3 elements can be mapped onto {y, z}is 2 3 – 2 = 6 ways.∴ f –1 (x) will consists of exactly one element in4 × 6 = 24 ways.Thus, the probability of the required event is24/36 = 2/32.[A]3.[D]4.[D]Let E 1 denote the event that the letter came fromTATANAGAR and E 2 the event that the lettercame from CALCUTTA. Let A denote the eventthat the two consecutive alphabets visible on theenvelope are TA. We have P(E 1 ) = 1/2, P(E 2 ) =1/2, P(A / E 1 ) = 2/8, P (A / E 2 ) = 1/7. Therefore,by Bayes' theorem we haveP(E 2)P(A / E2)P(E 2 / A) =P(E1)P(A / E1)+ P(E2) P(A / E2)4= 11Required probability = 1 – P (all the letters areput in correct envelops)The number of the ways of putting the letters inthe envelops = 4 P 4 = 4!The number of ways of putting letters in correctenvelops = 11 23∴ Required probability = 1 – = 24 24We have⎡5x0AB =⎢⎢0 1⎢⎣0 10x − 2⇒ x = 1/50 ⎤ ⎡10⎥⎥=⎢⎢05x⎥⎦⎢⎣00100⎤0⎥⎥1⎥⎦5.[B] Greatest term in the expansion of (x + y) n isk th ⎡(n+ 1)y ⎤term where k = ⎢ ⎥⎣ x + y ⎦XtraEdge for IIT-JEE 89APRIL 2010

In the present case⎡(50+ 1)(2x) ⎤ ⎡(51)(2 / 5) ⎤ ⎡102 ⎤k = ⎢ ⎥ =⎣ 3 + 2x⎢ ⎥ =⎦ ⎣ 3 + 2 / 5⎢ ⎥ = 6⎦ ⎣ 17 ⎦Thus, 6 th term is the largest term.6.[D] We have | z | =7.[D]4 4z − + ≤z z4z − +z4| z |4= 2 +| z |⇒ | z | 2 ≤ 2 | z | + 4 ⇒ (| z | – 1) 2 ≤ 5⇒ | z | – 1 ≤ 5 ⇒ | z | ≤ 5 + 1Also, for z = 5 + 14z −z= 2Therefore, the greatest value of | z | is 5 + 1.Integrating by parts, the given integral is equal to16 16x tan –1 x 1x − 1 −∫dx1 x 4 x x −16= π316= π316= π3161 dx–4 ∫x −1 131 4t(1 + t–4 ∫ t0– ( 3 + 3)= 31 12)dt ( x= 1 + t 2 )16 π – 2 38.[C] The intersection of y – x + 1 = 0 and y + x + 5 = 0is (– 2, –3). Put x = X – 2, y = Y – 3. The givendY Y − Xequation reduces to = . This is adX Y + Xhomogeneous equation, so putting Y = υX, we get2dυ υ +X = − dX υ + 11⎛ υ 1 ⎞ dX⇒ ⎜−− ⎟ dυ =2 2⎝ υ + 1 υ + 1 ⎠ X9.[A]⇒ – 21 log (υ 2 + 1) – tan –1 υ = log | X | + C⇒ log (Y 2 + X 2 ) + 2 tan –1 Y = CX⇒ log ((y + 3) 2 + (x + 2) 2 ) + 2 tan –1 y + 3x + 2= CWe have(1 + x) n – nx – 1= C 0 + C 1 x + C 2 x 2 + … + C n x n – nx – 1= x 2 [C 2 + C 3 x + … + C n x n–2 ][Q C 1 = n, C 0 = 1]Thus, (1 + x) n – nx – 1 is divisible by x 2 .(cos 2A − sin 2A) + 110.[A,B,C,D] y =2(cos 2A − sin 2A) −1± (cos 2A − sin 2A) + 1⇒ y =± (cos 2A − sin 2A) −1which gives us four values of y, say y 1 , y 2 , y 3 andy 4 . We havecos 2A − sin 2A + 1 (1 + cos 2A) − sin 2Ay 1 ==cos 2A − sin 2A −1(cos 2A −1)+ sin 2A22cos A − 2sin A cos A=2− 2sin A + 2sin A cos Acos A(cos A − sin A)== cot Asin A(cos A − sin A)−(cos 2A − sin 2A) + 1 (1 − cos 2A) + sin 2Ay 2 ==− (cos 2A + sin 2A) −1− (1 + cos 2A) − sin 2A22 sin A + 2 sin A cos A== – tan A2− 2 cos A − 2 sin A cos A(cos 2A − sin 2A) + 1 (1 + cos 2A) − sin 2Ay 3 ==− (cos 2A + sin 2A) −1− (1 + cos 2A) − sin 2A22cos A − 2sin A cos A cos A − sin A== –2− 2cos A − 2sin A cos A cos A + sin A1−tan A ⎛ π ⎞ ⎛ π ⎞= – = – tan ⎜ − A ⎟ = – cot ⎜ + A ⎟1+tan A ⎝ 4 ⎠ ⎝ 4 ⎠−(cos 2A − sin 2A) + 1 (1 − cos 2A) + sin 2Ay 4 ==(cos 2A + sin 2A) −1− (1 − cos 2A) + sin 2A22sin A + 2sin A cos A cos A + sin A==2− 2sin A + 2sin A cos A cos A − sin A1+tan A ⎛ π ⎞= = tan ⎜ + A ⎟ .1−tan A ⎝ 4 ⎠11.[B, C, D]Equations of the given circles can be written as(x – 3) 2 + y 2 = 3 2 (1)and (x + 1) 2 + y 2 = 1 2 (2)Equation of any tangent to circle (2) is(x + 1) cos θ + y sin θ = 1 (3)This will be a tangent to circle (1) also if(3 + 1)cosθ −1= ± 3 ⇒ 4 cos θ – 1 = ± 32 2cos θ + sin θThat is, cos θ = 1 or cos θ = – 21 . When cos θ = 1,we have sin θ = 0, and the equation of thecommon tangent (3) becomesx + 1 = 1 or x = 0 (4)When cos θ = –1/2, we have sin θ = ± 3 / 2 , andthe equations of the common tangents are1 3– (x + 1) ± y = 1 ⇒ x – 3 y + 3 = 0 (5)2 2and x + 3 y + 3 = 0 (6)2XtraEdge for IIT-JEE 90APRIL 2010

12.[A, B, C, D]13.[B]2x t − 5t + 4Let F(x) =∫dt0t2 + e⇒ F′(x) =2So from F′(x) = 0, we get x = 0 orHence x = 0, ±2, ±1.x = –4Shaded region is S 0 . Area of S 0= 4 × 2 – 21 π (1) 2 = 8 – π/214.[D] y ∈ [0, 4], x ∈ [–1, 1]m (t) = costlines y = 2x + 0.4 lies inside the region so⇒ t ∈ [0, 1]t 2 + (2t + 0.4) 2 – 1 ≥ 0 ⇒ t ∈ [0.28, 1]15.[B] (Slope) max. = (cos t) max = cos (0.28) and point is(π, 1)16.[C]18.[A]24 2x − 5x + 4.2xx2 + ex 2 5 ± 25 −165 ± 3== = 4, 12 2DCy = +4–1 1x = 4Ay = –4 B1 r1=2 r23= 13( −1)−1(3)− 6= = – 33 −12x + RTy −1= cos (0.28)x − πC SC Sx =17.[D] tangents = y = ±3tan30º = 32•C 2C 160º1(h + 1) 2 + k 2 = (1 + 2) 2 (circle)Column Matching19. [A] → r; [B] → p,r; [C] → s; [D] → r10(A) ∑ 20 C r = 20 C 0 + 20 C 1 +……+ 20 C 10r=0But, 20 C 0 + 20 C 1 +……+ 20 C 20 = 2 20(B) ∑r=Also, 20 C 20 = 1 = 20 C 0 , 20 C 19 = 20 C 1 , 20 C 18 = 20 C 2etc.∴ given sum = ( 20 C 0 + 20 C 1 +……+ 20 C 20 )– ( 20 C 11 +…..+ 20 C 20 )2 20 + 20 C 10 – ( 20 C 10 + 20 C 9 + ……+ 20 C 0 )∴ 2 ( 20 C 0 + 20 C 1 +…..+ 20 C 10 )= 2 20 + 20 C 101000100 C r (x –3) 100–r 2 r= ((x–3) +2) 100 = (x –1) 100 = (1 –x) 100100100∑r=0C r ( −x)= ∑=r100r0r( −1)( −1)r 100C x∴ Coeff. of x 53 = (–1) 53 100 C 53 = – 100 C 53(C) We have(1+ x) 10 = 10 C 0 + 10 C 1 x + 10 C 2 x 2 +……+ 10 C 10 x 10....(1)Also (1–x) 10 = 10 C 0 – 10 C 1 x + 10 C 2 x 2 +…….…..+ 10 C 10 x 10 ....(2)Multiplying, we get(1 –x 2 ) 10 = ( 10 C 0 + 10 C 1 x + 10 C 2 x 2 +……….+10 C 10 x 10 ) × ( 10 C 0 – 10 C 1 x +10 C 2 x 2 +……...+ 10 C 10 x 10)Equating the coefficients of x 10 , we get10 C 5 (–1) 5 = 10 C 10 0 C 10 – 10 C 10 1 C 9 + 10 C 10 2 C 8 +…….+ 10 C 10 10 C 0⇒ – 10 C 5 = ( 10 C 0 ) 2 – ( 10 C 1 ) 2 + ( 10 C 2 ) 2 +………+ ( 10 C 10 ) 2rrXtraEdge for IIT-JEE 91APRIL 2010

5100−j(D) 95 C 4 + ∑j=0C 3= 95 C 4 + 99 C 3 + 98 C 3 + 97 C 3 + 96 C 3 + 95 C 3= ( 95 C 4 + 95 C 3 ) + 96 C 3 + 97 C 3 + 98 C 3 + 99 C 3= ( 96 C 4 + 96 C 3 ) + 97 C 3 + 98 C 3 + 99 C 3= ( 97 C 4 + 97 C 3 ) + 98 C 3 + 99 C 3= ( 98 C 4 + 98 C 3 ) + 99 C 3= 99 C 4 + 99 C 3= 100 C 420. [A] → p,q; [B] → q,t; [C] → q; [D] → s(A) Given lines intersect if2 −13 − 4 4 − 51 1 λ = 0λ 2 1⇒ λ = 0, – 1⎛ x + 1 ⎞⎜ 1−(B) lim 4x ⎜−tan 1 x + 2x→∞⎜ x + 11+⎟ ⎟⎟⎟ =2 = y 2 + 4y + 5⎝ x + 2 ⎠⇒ y = –1, – 3(C) y 2 – ax (– x – y) = 0⇒ for perpendicular lines a + b = 0⇒ 1 + a = 0 ⇒ a = – 1(D) ( a × b ) × a = ( ĵ– kˆ ) × a⇒ ( a . a ) b – ( a . b ) a = ( ĵ– kˆ ) × aon solving,we get b = îPHYSICS1.[B] T = m 1 r 1 ω 2 2 1 also T = m 2 g + ω2r22.[A]∴ m 1 r 1 ω 1 2 = m 2or, 0.1 × ω 1 2 =222 2 2g + ( ω2r2)1 2 2( 10) + (10)20.1 ω 2 1 = 1ω 1 = 10 rad/s v 1 = r 1 ω 1 = 10 m/sAs f = µN = mgor, µmlω 2 = mg ⇒ ω =gµl5.[D]Net force acting on container due to liquidcoming out from the holes is given by⎡ 3H H ⎤F = ρA ⎢2 g × − 2g × ⎥ = ρgAH towads left⎣ 4 4 ⎦∴ F = f = ρgAH towards right.Now,6.[C] T A =7.[A]8.[A]τ F = ρgAH × 2H into the plane of paper.τ f = ρgAH × 2H out of plane of papers∴ τ F = τ f hence τ N = 0PA VAand T B =n RAPBVBn RGiven, P A = P B , V A = V B and n A = 2n B∴ T A =Now,VVABT B2ABEquivalent circuit1 1'22'Middle plateABT MB= × = 2T M3 3'11' 33'ε 1C 1 V 0V 0ε 2 C 2 V 02, 2'⎡ε1ε0Aε2ε0A⎤Total charge on 2 & 2' plate = ⎢ + ⎥V⎣ d1d2⎦⎡ εσ = ε 0 V ⎢⎣dC1130º C 15º45º15Oε+dA22⎥⎦⎤30B3.[C]4.[C]At terminal velocity net force is zero.6πη(r 1 + r 2 ) V T + 34 π (r1 3 + r 2 3 ) ρg = 34 π (r1 3 + r 2 3 ) σgAs supporting plane is lowered slowly∴ N = mg – kxA I30ºl120ºOSine ruleRBXtraEdge for IIT-JEE 92APRIL 2010

R l=sin 45º sin 30ºR sin 45º = = 2l sin 30º9.[A,B] As a T = a Nvdv − v 2∴ = which can alsods Rdv v 2be written as = – dt RIntegrating the above equations answer isobtained.10.[A,C] (A) ρ B LAg = ρ × 4L Ag + 2ρ × 4L Ag11.[B,D]12.[A,B,C]L 3L(B) F B = ρ × Ag + 2ρ × Ag4 47 FF B = ρLAg; a = B − mg4 m8V3Ω13Va•b6V← eq. ckt.1011Using Kirchoff's Law Solve the circuit.CVV–CVIf battery is disconnected and plate are pulledapart, then charge will remain constantQE =2A ∈ × 2 = Q0 Α ∈0∴ E remain same (A) is correct work is doneagainst attractive force+ −Felc. Fext←⎯⎯⎯ ⎯⎯⎯→by Fext. (B) is correct.U = 21 CV2V = constant [as battery is connected]∈ AC = 0dcas d increaseC decrease ∴ U decreaseoption (C) is correct.13.[C] As springs are in parallelFnet(k1+ k 2 )x∴ a = =mass (m + m )14.[A] Frictional force on m 2 will act in direction ofdisplacement ifk 2 x > m 2 a15.[A] as k 2 A – µm 2 g = m 2 a max⎡ k1+ k 2∴ k 2 A – µm 2 g = m 2 ⎢⎣ m1+ mSolve to get answer. ]16.[C] Induced emf across OP =17.[B](i) current =1Bωl28R2l / 22⎤⎥ A⎦21 ⎛ ⎞Bω⎜l ⎟ =2 ⎝ 2 ⎠Torque on the rod = 2∫Bi x dx =2204Bil24Bωl28… (i)…(ii)Ml dωB ωl∴ × = − [substituting τ = I α]12 dt 32Rω2 2 tdω3B l–∫=ω 8RM ∫dtω00Solving this eq. & eq. (i)2Bω i = 0 le –αt8R∞θ =∫idt018.[A] Heat generated =1I ω2Column Matching19. [A] → p; [B] → p; [C] → r; [D] → qAs cube is floating ρ s ALg = ρ L Axg⎛ ρ ⎞∴ x =⎜ s⎟ L⎝ ρL⎠20. [A] → r; [B] → p; [C] → s; [D] → q20S = 21 × 2 × 16 = 16 m| Wg | = mg S =W N = m(g + a) cos 2 θ. SW f = m(g + a) sin 2 θ . SXtraEdge for IIT-JEE 93APRIL 2010

XtraEdge for IIT-JEE 94APRIL 2010

1.[C]2.[B]3.[B]SOLUTION FOR MOCK TESTPAPER - IIIIT-JEE (PAPER - II)CH 2 =C'A''B'CHEMISTRYNH 2BrNOBrCOOHH 2 /PdPh+ EnantiomerNH 2CH 3 –CHBr –CH=CH–COOH H 2 /PdCH 2 –CH 2 –COOHOptically inacitve+COOH(resolvable)PhDistribution of electrons in the MO's in He 2 isσ 1s 2 σ∗ 1s 2 . He 2 is unstableDistribution of electrons in the MOS in H 2 is σ 1s2. H 2 is stable.Grignard reagent reacting with acyl halide usuallygives 3º alcohol.7.[B,C,D] The order of H-bond energiesF – H …….. F – > F – H …….. O > F – H …….. F >O – H …….. O > O – H …….. F > N – H …….. N8.[C,D]9.[C,D]RHE reaction : Hg 2 Cl 2 (s) + 2e → 2Hg(l) + 2Cl –LHE reaction : 2Ag(s) + 2Cl – → 2AgCl(s) + 2eNet reaction : Hg 2 Cl 2 (s) + 2Ag(s)⎯→ 2Hg (l) + 2AgCl(s)In case of same concentration of Cl – ions in thetwo half cells, E cell is independent on theconcentration of Cl – . Other substances are eitherpure solids or liquids, which have unit activitiesirrespective of their amounts.Column Matching :10. [A] → p, r, s ; [B] → q;[C] → p, q,,s ;[D] → q11. [A] → p, q,t ; [B] → p, r, s;[C] → p, q, r;[D] → p, q, r4.[B] After mixing total moles of A –= 100 × 0.2 × 10 –3 + 100 × 0.3 × 10 –3= 100 × 10 –3 × 0.5 molesAfter mixing total moles of HA= 100 × 0.1 × 10 –3 + 100 × 0.2 × 10 –3= 100 × 0.3 × 10 –3 moles5After mixing resulting pH = 5 + log 3NH 2HO CH 2 –CH*5.[A,B,C]HOCOOHIt contains 2 phenolic hydrogens and a carboxylicacidic hydrogens+NHCH 2 –HC 2HOCOO –HO Zwitter ion6.[A,B,C]Numerical Response type questions :12. [4]W Q 2.977 3 × 1×60×60= or =E 96500 106.4 / n 96500∴ n = 413. [8]O3/ H2O⎯⎯ ⎯⎯ →(Ozonolysis)HOHONH NH22O ⎯⎯ ⎯⎯→OOKOH / glycol(Wolff − Kishnerreduction)(X)(W)XtraEdge for IIT-JEE 95APRIL 2010

⎯ Ca(OH)⎯⎯⎯2∆→ OC 6 H COOOH⎯⎯⎯5(Baeyer−Villigeroxidation)(Y)O⎯⎯ →O(Z)14. [1]Molecular weight of starch = 162n∴ moles of C 6 H 12 O 6 to be produced = 1 mol3000∴ moles of ATP required = = 100 mol3015. [7]0P − P0 = x1 = mole fraction of solutePx 1 = 0.0125⎛ 1 ⎞ 1000⎜ −1⎟ =⎝ x1⎠ m×18∴ m = 0.7016. [5]Eq. of metal = Eq. of hydrogen0.1 43.9= ⇒ X = 251/ X 11200Now eq. of metal = eq. of KMnO 40.151/ Y − 2= 0 .1×58.81000∴ Y = 517. [7]For the first bulb p 1 v 1 = n 1 RT& second bulb p 2 v 2 = n 2 RTor p 1 v 1 + p 2 v 2 = (n 1 + n 2 ) RT …(1)Suppose equilibrium pressure at each bulb = p atmThen p(v 1 + v 2 ) = (n 1 + n 2 )RT …(2)From eq. (1) & (2)p 1 v 1 + p 2 v 2 = p(v 1 + v 2 )9 × 5 + 6 × 10 = p × 15p = 105/15 = 718. [2]••OF FXeXeOOFO••FXeO 3XeOF 419. [6]Each has two geometrical isomers and two opticalisomers (shown by cis-isomer).1. [A] We haveMATHEMATICSn2n4 4∑ 2r −1)= ∑r−∑r= 1r= 1 r=1n( (2r) = f(2n) – 16f(n)2. [D] Using De Moivre's theorem2i /f r (α) = e α r22i / re α i / r…. e α2(iα/ r )(1+2+...... + r)= e2(iα/ r )[r(r+1)/ 2] (iα/ 2)(1+1/ r)= e= e∴ lim f n ( π)= lim iπ/2(1+1/ n)en→∞n→∞= e iπ/2 ⎛ π ⎞ ⎛ π ⎞= cos ⎜ ⎟ + i sin ⎜ ⎟ = i⎝ 2 ⎠ ⎝ 2 ⎠α + βt3. [D] z = ⇒ ( γ + δt) z = α + βtγ + δt⇒ (δz – β)t = α – γzα − γz⇒ t =[Q αδ – βγ ≠ 0 ]δz− β4. [C]α − γz α − γzAs t is real, =δz− β δz− β⇒ (α – γz)( δ z – β ) = ( α –4γ z )(δz – β)⇒ ( γ δ – γ δ )z z +(γ β – α δ)z + (α δ – β γ ) z= (α β – α β) ...(1)Since δγ is real, δγ =δγ or γ δ – δ γ = 0Therefore (1) can be written as a z + a z = c ...(2)where a = i(α δ – β γ ) and c = i( α β – α β )Note that a ≠ 0 for if a = 0 thenα γ γ γaδ – β γ = 0 ⇒ = = [Q is real]β δ δ δ⇒ αδ – βγ = 0,which is against hypothesis.Also, note that c = i( α β – α β ) is a purely realnumber.α + βtThus, z = represents a straight line.γ + δtdy = (x – 1)(x – 2) 2 sodx2d y= (x – 2) × (3x – 4).2dxd yThe points of inflection are given by = 02dxSo x = 2, x = 4/3 are points of inflection.2XtraEdge for IIT-JEE 96APRIL 2010

5.[A,B,D] We have adj A = |A| A –1adj (AB) = |AB| (AB) –1 = |A| |B| (AB) –1= |A| |B| (B –1 A –1 )= ( |B| B –1 ) (|A| A –1 )= (adj B) (adj A)6.[A,B,C] We have⎡cosα − sin α 0 ⎤A(α, β)′ =⎢⎥⎢sin α cosα0⎥β⎢⎣0 0 e ⎥⎦⎡ cos( −α)sin( −α)0 ⎤=⎢⎥⎢− sin( −α)cos( −α)0⎥β⎢⎣0 0 e ⎥⎦= A (–α, β)Also, A (α, β) A( – α, –β)⎡ cosαsin α 0 ⎤ ⎡cosα − sin α=⎢⎥⎢− sin α cosα0⎢⎥ ⎢sin α cosαβ⎢⎣0 0 e ⎥⎦⎢⎣0 0⇒ A(α, β) –1 = A(–α, –β)Next, adj A(α, β) = |A (α, β)| A (α, β) –1= e β A(– α, – β).0 ⎤0⎥⎥= I⎥⎦−βe7.[A,B,C,D]n⎛ 5 4 ⎞Let (r + 1) th term of ⎜ + x ⎟ be2⎝ x ⎠independent of x. We haven−rT r + 1 = n ⎛ 5 ⎞C r ⎜ ⎟ (x 4 ) r2⎝ x ⎠= n C r 5 n – r 6r – 2nxFor this term to be independent of x, 6r – 2n = 0or n= 3r. As each of 18, 21, 27 and 99 is divisibleby 3, each of this can be a possible value of n.8.[A,B]Let k = 2n + 1, then 2n + 1 C r is maximum whenr = n. Also 2n+1 C n = 2n + 1 C n+1Thus, k C r is maximum whenr = 21 (k – 1) or r = 21 (k + 1).9.[A,B,C,D]As a 912 , a 951 and a 480 are divisible by 3, none ofthem is prime. For a 91 , we have1 1a 91 = (99....9) = (109 1424391 – 1) 991 times= 91 [(10 7 ) 13 – 1]⎡7 13(10 ) −1⎤⎡10 7 −1⎤= ⎢7⎥ ⎢ ⎥⎢⎣10 −1⎥⎦⎢⎣10 −1⎥⎦= [(10 7 ) 12 + (10 7 ) 11 + ….+ 10 7 + 1]× [10 6 + 10 5 + ….. + 10 + 1]= a 91 is not prime.Column Matching :10. [A] → p, r; [B] → p,q, r; [C] → t ; [D] → p, q, r3 πsin θ = = sin2 3⇒ θ = nπ + (–1) n π π = 2nπ +3 34 sin θ cos θ – 2 sinθ – 2 3 cos θ + 3 = 0⇒ (2 sin θ – 3 ) (2 cos θ – 1) = 03⇒ sin θ =2⇒ θ = nπ + (–1) n π 1 , cosθ =3 2π⇒ θ = 2nπ ± 3(C) sin 2θ + cos 2θ + 4 sin θ = 1 + 4 cos θ⇒ 2 sin θ cos θ + 1 – 2 sin 2 θ + 4 sin θ = 1+ 4cos θ⇒ 2 sin θ (cos θ– sin θ) – 4(cosθ– sinθ) = 0⇒ (2 sin θ –4) (cos θ – sin θ) = 0⇒ sin θ = 2 or sin θ = cos θ⇒ tan θ = 1 ⇒ θ = nπ + π/4, n ∈ I(D) cos 2 θ = 1/4= cos 2 π π⇒ θ = 2nπ ±3311. [A] → r ; [B] → r ; [C] → p ; [D] → q(A)We have b – a = c – b and (c – b) 2 = a(b – a)⇒ (b – a) 2 = a(b – a) ⇒ b = 2a and so c = 3a.Thus a : b : c = 1 : 2 : 3a + b(B) If the numbers are a and b, then x = and213b = ar 3 ⎛ b ⎞⇒ r = ⎜ ⎟⎠⎝ aNow,y3 3 3 3 3 6+ z a r + a r=2xyz x (ar)(ar )a (1 + r ) a + b= = = 2x a + b2(C) c > 4b –3a ⇒ ar 2 + > 4ar –3a⇒ r 2 – 4r + 3 > 0 ⇒ r < 1 or r > 3, But the termsare positive so r ∈ (0, 1) ∪ (3, ∞)(D) tan –1 ⎛ 1 ⎞⎜ ⎟ = tan –1 ⎛ 2 ⎞⎜ ⎟2⎝ 2r ⎠2⎝ 4r ⎠= tan –1 (2r + 1) − (2r −1)1+(2r + 1)(2r −1)= tan –1 (2r +1) –tan –1 (2r–1)∴n∑r=1−1⎛1 ⎞tan ⎜ ⎟ (2n+1) – tan –1 (1)2⎝ 2r ⎠= tan –1 (2n + 1) – 4π3XtraEdge for IIT-JEE 97APRIL 2010

⎪⎧∴ tan ⎨⎪⎩∴n∑r=1lim tann→∞⎪⎪⎫−1⎛1 ⎞ (2n + 1) −1tan ⎜ ⎟2 ⎬ ==⎝ 2r ⎠⎪⎭1+(2n + 1).1⎪⎧⎨⎪⎩n∑r=1nn + 1⎪⎫−1⎛1 ⎞tan ⎜ ⎟2 ⎬ =⎝ 2r ⎠lim nn→∞⎭n + 1= 1Numerical Response type questions :12. [3]We have92| z |= (2 + cos θ) 2 + sin 2 θ = 5 + 4 cos θ (1)and3 3 + = 4 + 2 cos θz z (2)Eliminating θ from (1) and (2), we get9 ⎛ 1 1 ⎞2 – 6 ⎜ + ⎟⎠ = – 3| z | ⎝ z z⇒ 3 = 2( z + z) – |z| 213. [6]Clearly x > 0 and x ≠ 1/5log55 − log5xlog 5x (5/x) =log55 + log5xPutting log 5 x = t, then equation becomest 2 1− t+ = 11+t⇔ t 3 + t 2 – 2t = 0⇔ t(t – 1) (t + 2) = 0⇔ t = 0, 1, – 2So integral roots are 1 and 5.14. [2]Let the A.P. be a – d, a, a + d, a + 2d. Note that aand d must be integers. Also as this is an increasinga + 2d is the largest. We havea + 2d = (a – d) 2 + a 2 + (a + d) 2= 3a 2 + 2d 2⇒ 3a 2 – a + 2d 2 – 2d = 0As a is real,1 – 8(d 2 – d) ≥ 0⇒ d 2 – d – 81 ≤ 0⇒2⎛ 1 ⎞ 3⎜d − ⎟ ≤⎝ 2 ⎠ 81 3 1 3⇒ – ≤ d ≤ +2 2 2 2 2 2As d is an integer, d = 0, 1But d ≠ 0, therefore, d = 1.Thus 3a 2 – a = 0 ⇒ a = 0 or a = 1/3.As a is an integer, a = 0.Hence, required number is 2.15. [1] We haveE = 2n + 1 C 1 + 2n + 1 C 2 + … + 2n + 1 C n – 2n + 1 C 0– 2n + 1 C 1 – … – 2n + 1 C n[using n C r = n C n – r ]= – 2n + 1 C 0 = – 1.∴ |E| = 116. [3] The given system of equations will have a nontrivialsolution ifa + 2t b c∆ = b c + 2t a = 0cab + 2tClearly ∆ is a cubic polynomial in t and has 3 roots.17. [3] None that it is not given that f is a differentiablefunction We havef (4 + h) − f (4)f ′(4) = limh→0h2f( 4 + h ) )= limh→0(4 + h)= limh→0h⎡8⎢1+= lim⎣h→018. [2]dy2x + 2y = 0 dx3/ 2h3 h2 4− f (22)− 8 8[(1 + h / 4)= limh→0h3/ 2−1]⎤ ⎡3⎤+ ...... −1⎥8⎢h + .....⎦ 8 ⎥= lim⎣ ⎦hh→0hdy x⇒ = –dx y⇒dydx(1,3)1= –3Therefore, the equation of the tangent at (1,3 ) isy – 3 = (–1/ 3 ) (x – 1)and the point of intersection of this tangent with thex-axis is (4, 0). The equation of the normal at(1, 3 ) is y – 3 = 3 (x – 1), and the point ofintersection of this normal with the x-axis is (0, 0).Hence the required area is1 . 4 3 = 2 3219. [4] The two curves represent parabolas withvertices at (0, 0) and (3, 0). They intersect at (1, 1)and (1, –1), so the required area isarea of OPMQO = 2 (area of OPMO).XtraEdge for IIT-JEE 98APRIL 2010

(1, 1)7λ 5.5× 10t min = =P4n 4×1. 38= 99.6 nm⎛ R ⎞⇒ E = I ⎜ total⎟ (By Ohm's Law)⎝ l ⎠t n mgF 22παI⇒ E =2A3.[D]dqI = ; dt q = it + a;q V = COM(3, 0)it + aV =C∴ V is proportional to timeQ4.[A] Phase difference corresponding to y 1 = –π/2 andthat for y 2 = + π/2∴ Average intensity between y 1 and y 2Fig.π / 212⎛ φ ⎞ ( π + 2)⎛ 13 ⎞= 2 ⎜3 − x=+ ⎟∫x dx0 ∫dxπ ∫I max cos ⎜ ⎟dφ= I max⎝ 2 ⎠2π1⎝2−π / 2⎠1 ⎛ ⎞⎛ 13Hence required ratio =⎞⎜1+ 2 ⎟= 2⎜ 2 3/ 2 1 2 3/ 2x − . (3 − x)⎟2 ⎝ π ⎠⎜ 33 ⎟⎝2M.C.Q. Type questions :01 ⎠5.[A,C]⎡ 1 2 2 4= 2 ⎛3/0 . 2⎞⎤⎛ ⎞a⎢ −⎥ = 2⎜+ ⎟ = 4.⎢⎣3⎜ −2 3⎟dr⎝⎠⎥⎦⎝ 3 3 ⎠rPHYSICSlConsider a cylindrical element of radius r,1.[D]thickness dr. If dR is the resistance of this1element then1′ρ(r)ld/22dR =2πr dr2′3Total resistance of the cylinder is given bya3′1 1 2πC =lR total∫ = dR αl∫r3 dr011 2π⎛ ⎞⇒= ⎜a 4⎟1′R3 C2total αl⎝ 4 ⎠l2′1 2αl3′⇒=4( ∈ A) 2= 2 × 0 × ∈ AR total πa= 4 . 0= 4CddR⇒total 2πα=l ( πa2 ) 2⎡ 2m+ 1⎤λ2.[A] 2t = ⎢2⎥2πα⎣ ⎦ n⇒ R =2AVSince E = (in magnitude)l6.[A,D] Consider only one plate as shown in figure.XtraEdge for IIT-JEE 99APRIL 2010

7.[B,D]8.[B,C]9.[C]The field on both sides of plate is shown in figure.Applying Ampere's Circuital Law to the contourC, we getL2BL = µ 0 (jL)µ jB = 02BSuperimposing, the field due to two plates we getat P both fields cancel each other and at Q, theyadded to give B 0 = µ 0 jv < g(t)⇒ v < 2g {due to Lenz's Law}Also, s < 21 gt2⇒ s < 2g {due to Lenz's Law}R 0 RR 3SR 2R 1C1CVR 2R3⇒ R 0 =R + R2⎛⇒ τ = C(R 1 + R 0 ) = C ⎜R⎝⇒ q = q 0 (1 – e –t/τ )⇒ q = CV 0 (1 – e –t/τ )CVR 2⇒ q = (1 – e –t/τ )R + R233V 0BCVR 2and V 0 =R + R123R 2R3+R + RIf it was AY → 2ZReactant : R = 60 × 8.5 = 510 MeVProduct : P = 2 × 30 × 5 = 300 MeV∆E = – 210 MeVENDOTHERMICIf it was BW → X + ZR = 120 × 7.5 = 900 MeVP = 90 × 8 + 30 × 5 = 870 MeV∆E = – 30 MeVENDOTHERMICIf it was CW → 2Y23⎞⎟⎠R = 120 × 7.5 = 900 MeVP = 2 × 60 × 8.5 = 1020 MeV∆E = 120 MeVEXOTHERMICIf it was DX → Y + ZR = 90 × 8.0 = 720 MeVP = 60 × 8.5 + 30 × 5.0 = 660 MeV∆E = – 60 MeVENDOTHERMICColumn Matching :10. [A] → p ; [B] → q, r, s; [C] → p,s ; [D] → q,rWhen cohesive forces are greater then adhesiveforces shape of meniscus is concave from liquidside and pressure is greater in concave side due tosurface tension.11. [A] → r ; [B] → p, s; [C] → q ; [D] → s⎛ dV ⎞ dT(A) ⎜ ⎟ = ⇒⎝ V ⎠PTkT(B) λ =2πd2 ρ1 ⎡dV⎤dT⎢V⎥⎣ ⎦P1=T(C) Ideal gas law are valid at all temperatures.(D) conceptualNumerical Response type questions :12. [1]13. [5]Mv12α v' 12v 1 α30ºN–vO230º→In the figure v 12 = velocity of ball w.r.t. wedge→before coolision, and v' 12 = velocity of ball w.r.t.wedge after collision, which must be in vergicallyupward direction as shown.→ →In elastic collision v 12 and v' 12 will make equalangle (say α) with the normal to the plane.We can show that α = 30º∴ ∠MON = 30ºv 11Now = tan 30º = ≈ 1v32hABhCXtraEdge for IIT-JEE 100APRIL 2010

K TFor a cylinder = 2 in case ofK Rpure rolling upto B :andK T = 32 mghK R = 31 mghAfter B : rotational kinetic energy is constanthowever translational kinetic energy increases.At C : K T = 32 mgh + mgh = 35 mghwhile K R = 31 mgh∴K T = 5KR14. [2]According to the formula, the position of theparticle is at the centre of the path. It goes by 7 unitsto the right and then by another 7 units back to theinitial position and then it goes to the negative sideby 7/2 units for the first time.So the minimum time is 2 × 4T plus the additionaltime (t) to cover 7/2 units on the negative side.π 2π πHere ω = or = or T = 4s2 T 24∴ t min = 2 × + t = 2 + t 47 π π π 1Now = 7sin t ⇒ t = sin 30º =sin ⇒t = s2 2 2 6 31∴ t min = 2 + = 2.33s ≈ 2315. [6] Let ω be the actual angular velocity of thesatellite from east to west and ω e be the angularspeed of the earth (west to east).Then ω relative = ω – (–w e ) = ω + ω e ⇒ ω = ω rel – w eBy the dynamics of circular motion2GMm2 = mω 2 R or ω 2 gR e=3 (Q GM 2 =gRe )RR⇒ ω =2e3gRR2∴ ω rel =122e3gRR+ ωe10×6.4 × 10⇒ ω rel =+ 7.27 × 10 –53 212 × 102π(Q ω e = = 7.27 × 10 –5 )86400⇒ ω rel =22.6×10 –5 +7.27×10 –5 = 30×10 –5 rad s –12π 2π∴ τ = = = 2.09×10 4 s = 5hr 48 min.−530×10w rel16. [1]444πPrπPrπPrV 1 = , V 2 = , V 3 =8ηl18ηl28ηl3πPr 4and V =8ηlNow V = V 1 + V 2 + V 3Substituting the values, we getl1l2l3l =l1l2 + l1l3+ l2l31×2×3 6== m1×2 + 1×3 + 2×3 1117. [1]π3πAt x 1 = and x 2 =3k2kSin k x 1 or sin k x 2 is not zero.Therefore, neither of x 1 or x 2 is a node⎛ 3 1 ⎞ π 7π∆x = x 2 – x 1 = ⎜ − ⎟ =⎝ 2 3 ⎠ k 6kSince2π π > ∆x >k kλλ > ∆x > 2Therefore, φ 1 = π7πand φ 2 = k.∆x = 6∴φ 1 6 =φ 72≈ 1⎛ 2π⎞⎜k= ⎟⎝ k ⎠18. [1]With an AC source, current in the circuit ismaximum whenZ = Z min = R (the resistance of coil)∴ R = 624 = 4ΩWhen connected with an dc source of emf 12V.12i =(r internal resistance of source)r + R12= = 1.5 A4 + 419. [1] For incident electronh h hλ 1 = = =p 2Km2mVehcand shortest wavelength of X-rays is λ 2 = Ve∴λ 1 1 =λ c2⎛ V ⎞⎛⎜ ⎟⎜⎝ 2 ⎠⎝emλ 1Substituting the values, we get = 1λ⎞⎟⎠2XtraEdge for IIT-JEE 101APRIL 2010

SOLUTION FOR MOCK TESTPAPER - II AIEEEPHYSICS12.[B]13.[B]1.[C]2.[B] v ∝ n ∝ T because λ = constantN + 4 324 18= =N + 1 289 1717N + 68 = 18N + 1850 = N3.[B]4.[A]5.[B] u = 0a = constantv 0 = u + at = at⎛ v + v0⎞ a n tn' = ⎜ ⎟ n = n +⎝ v ⎠ v6.[C] LC = 1 MSD – 1VSD⎛ 29 ⎞ 1LC = ⎜1 − ⎟ MSD = MSD⎝ 30 ⎠ 30720 MSD = 360°360°1 MSD = = 1/2° 7201 1° 1° 60 'So LC = × = = = 1'30 2 60 607.[C]8.[D]Ticker timer is a better device than a stop watch.9.[C]M MM =1 + 2 10 .30+12.62=22= 11.46 g10.[B]11.[B]∆ Y 2∆ D ∆ l= +Y D l∆ YY⎛ 0.01⎞⎛ 0.05 ⎞= 2 ⎜ ⎟ + ⎜ ⎟⎝ 0.4 ⎠ ⎝ 0.8 ⎠= 2× 0.025 + 0.0625∆ Y= 0.05 + 0.0625 = 0.1125Y∆Y = 2× 10 11 × 0. 1125 = 0.225 × 10 11So (2 ± 0.2) × 10 11 N/m 2fdIt is only possible if object and image coincide.14.[D] I = I 0 cos 2 φ215.[C]FI 0= I 0 cos 2 φ2 2φ 1cos = 2 22πφ = 90° = ∆xλπ 2 = ∆x2 λπλ y d λ∆x = ⇒ = 4 D 4−7λ D 5×10 × 1y = =4d−44×1×10y = 1.25 mmf= 45 × 10–3→ τ =→ p ×→E = q (2→ a ) ×→EHere 2 → a = (2 – 1) î + (– 1–0) ĵ + (5 – 4) kˆ= î – ĵ + kˆ→E = 0.20 î V/cm = 20 î V/m∴16.[D]→ τ = (4 × 10 –6 ) [( î – ĵ kˆ ) × 20 î]( kˆ + ĵ)Magnitude of torque τ = 8CBAc+ σ– σ+ σba+ = 8 × 10 –52 × 10 –5 N-mXtraEdge for IIT-JEE 102APRIL 2010

∴17.[C]Potential of shell A is,1 ⎛ 2 2 2V A =4π∈⎟ ⎟ ⎞⎜ 4 πaσ − 4πb σ + 4πcσ⎜o ⎝a b c⎠σ= (a – b + c)∈ 0Potential of shell C is,1 ⎛ 22 2V C =4 π∈⎟ ⎟ ⎞⎜4πaσ − 4πbσ + 4πcσ⎜0 ⎝c c c⎠σ ⎛ 2 2 ⎞= ⎜ a b− + c⎟∈ 0⎜ ⎟⎝c c⎠As V A = V Cσ σ ⎛ 2 2 ⎞(a – b + c) = ⎜ a − b+ c⎟∈ 0∈ ⎜⎟0 ⎝c c⎠a − b a + bor a + b = ccor a – b = ( )( )≡A543211 23 2 5 43 4∈ AThe capacity of each capacitor is, C 0 = 0d5 5 ∈From fig. it is clear that C eq = C0 = o A3 3 d18.[B] Resistivity of conductors increases withincrease in temperature because rate ofcollisions between free electrones and ionsincreases with increase of temperature.However, the resistivity of semiconductorsdecreases with increase in temperature becausemore and more covalent bonds are broken athigher temperatures.19.[B] let L A and L B be length of parts A and BRThenA L=A[as cross-section is same]R B LBNow L c = 2 L A and (volume) c = (volume) Pi.e.L c × A c = 2 L A × A c = L A × A Awhere A c = A A are cross-sectional area of part Cand A.∴ A c = A A /2BAB∴R AR C/ρ LAA=A L=A×ρ Lc/A L c CL=A A×A/ 2 1=2LAAA4ACAA20.[A] Current flowing through potential wire is –EI =15r + r= E16rPotential drop across potential wire is,15EV = I × 15 r =1615EPotential gradient, K =1660021.[A]E E 15E∴ = Kl or = × l2 2 16×60016×600or l = = 320 cm.15×2AD i Cφ 1OPaφ 2φ 1φ 2Q bµ 0 iB AB = B CD = (sin φ4π( b / 2)1 + sin φ 1 )µ=0 4ia.4πb 2 2a + bB BC = B DA =Bµ 0 4 i b. .4πa 2 2a + b∴ B = B AB + B BC + B CD + B DAµ=0 4i ⎡ab.4π2 2⎢ + +a + b ⎣ba=µ4π0 .2 +28i a bab22.[A] In tan A position,µ 0 2 M= B4π3 H tan 30° =dabb ⎤+a⎥⎦B H…….(1)3Magnetic moment of second magnet,M' = (3m)(2 × 2l) = 6MIn tan B position,µ 0 6 M= B4π3 H tanθ …… (2)ddividing eq. (2) by (1) we getXtraEdge for IIT-JEE 103APRIL 2010

6 tan θ = or tan θ = 3 or θ = 60°2 1/ 323.[B] Magnetic field of solenoid, B 1 =µ 0N1i1lMagnet flux of coil, φ 2 = N 2 B 1 A 2 = N 2⎛ µ 0 N1i1⎞⎜ ⎟ A 2⎝ l ⎠φ2µ 0 N1N 2A2As φ 2 = M i 1 , so M = =i ldi∴ induced emf, |e| = M 1dtµ 0 N1N2A2dior |e| =× 1l dt– 7−34π×10 × 2000×300 × 1.2×10=0.30= 4.8 × 10 –2 Volt224.[A] Z = R + ( X − X ) 225.[D]26.[C]27.[B]28.[D]LC1×40.25Here X L = 2πfL = 2 × 3.14 × 500 × (8.1 × 10 –3 )= 25.4 Ω11and X C = =2πfC−62×3.14×500×12.5×10= 25.4 Ω2 25.4 25. 4∴ Z = ( ) ( ) 210 + − = 10 ΩENow i rms = rms 100= = 10 AZ 10∴ V R = i rms × R = 10 × 10= 100 Vu 2 sin 2 θ 2gh = ug2(14) sin 29.8( 45)=Loss in P.E. = gain in K.E.mg r = 21 mv 2 ⇒ v 2 = 2g ra c =v 2r= 2gT – mg cos θ =2(h)14 ⇒ h = 10 m9.8m v2r⇒ T = 3 mg29.[C] Loss in P.E. =gain in K.E. + work done against frictionmg R = 21 m (1.4 gR) + WfW f = 0.3 mgR30.[C]Now,W B → C = mg R + 0.3 mgR= 1.3 mgRdm dv × u = Mdt dt0.5 × 400 = 2000 ×∆ v5⇒ ∆v = 0.5 ms –1CHEMISTRY31.[C] HCO 3 – can donate a proton to CO 3 2– and it canaccept a proton to form H 2 CO 3 .32.[C] O F 2 ⇒ O = +2O 2 F 2 ⇒ O = +1O 2 [PtF 6 ] ⇒ O 2 + + [PtF 6 ] –1O 2 + ⇒ 2x = +1x = +1/233.[A] ∆G = ∆H –T ∆Sif ∆H = ⊕ & ∆S = ⊕& T ∆S > ∆H than∆G = Θ & process is spontaneous.34.[C] Calcined Gypsum is calcium sulphate35.[A] Inter particles forces between CH 3 COCH 3 &CHCl 3 are strong H-bonding. Thus solutionshows negative deviation. ∆V mixing = Negative.36.[C] Addition of inert gas at constant volume doesnot cause any effect on the equilibrium.37.[A] For I order Reaction t 1/2 is constant10gm⎤⎥ t1/2 = 24↓ ⎦5g⎤⎥ t1/2 = 24↓ ⎦= 96 hours.2.5g ⎤⎥↓ ⎥ t1/2 = 241.25g⎥⎦38.[D]↓ ⎤⎥ t0.625g⎦1/ 2= 24Eq. wt. of Ag wt. of Ag=Eq. wt. of O wt. of108 W =8 1.6WAg = 21.6 gm,2 O 2XtraEdge for IIT-JEE 104APRIL 2010

39.[B] <strong>Solution</strong> is decinormal, that is N/10 & x factoris 1, so conc. = 0.1M[H + ] = c.α= 0.1 ×1.3/100 = 13×10 –4pH = –log [13×10 –4 ] = 2.8940.[A] The no. of atoms in fcc lattice = 4a = 400 pm = 4×10 –10 m= 4×10 –8 cmn × M 4×60d = =323–8 3No × a 6×10 × (4×10 )d = 6.23 g/cm 3 .41.[B] For any Value of l possible values of m arem = –l to + ll = 2, m = –2, –1, 0 +1, +2So option is (B)42.[B] C : O : H6gm : 3.01 × 10 23 atoms : 2 moleRatio ½ : ½ : 2of mole1 : 1 : 4COH 4 or CH 4 O43.[B]44.[B]45. [A]312NaNO 2 /HCl gives HNO 2 which gives differentproducts with Pri. and Sec. amines.CH 3+ Cl 2⎯⎯→h νCH 2 Cl +HCl46.[D] 2 CuSO 4 + 2 KCN ⎯→ Cu(CN) 2 + K 2 SO 42Cu(CN) 2⎯⎯→ 2CuCN ↓ + (CN) 2 ↑3 KCN + CuCN ⎯⎯→ K 3 [Cu(CN) 4 ]47. [B] Blue print process occurs with the help of IronCompound.48.[A] Effective nuclear charge increases thereforeionic radius follow the order.49.[B] xy 2 xy + yt = 0 : 600 0 0teq m : 600-P P PNow : 600 – P + P + P = 800P = 200 mm Hg(Pxy)(Py) 200×200Kp = =(Pxy ) 4002= 100mm50.[C] Cell reaction : Zn + Cu +2 ⎯→ Zn +2 + CuCell emf:Hg0.059 ⎛E cell = E° cell – log ⎜Znn⎝ CuSo doubling the conc. of ions.E cell remains same.+ 2+ 251.[A] pK a = –log K a = –log (1.8×10 –5 ) = 4.744712 mol[CH 3 COOH] = = 0.460×0.5 L16.4 mol[CH 3 COONa] = = 0.482× 0.5 L⎡ salt ⎤Now, pH = pK a + log ⎢ ⎥⎣acid⎦⎞⎟⎠⎡0.4⎤= 4.7447 + log ⎢ ⎥⎣0.4= 4.7447 ⎦Increase = 152.[B] FeC 2 O 4 ⎯→ Fe +3 + CO 2+2 +3 +3 +4Increase = 1Total Increase in O.N. = 3So valence factor of FeC 2 O 4 = 3KMnO 4 ⎯→Mn +2+7 +2 v.f. (KMnO 4 ) = 5gm E KMnO 4 = gm E of FeC 2 O 4Mole × v.f. = Mole × v.f.1× 5 = x × 3 ⇒ x = 5/353.[A] ∆G = ∆H – T ∆S∆G = Θ , ∆G < 0 (Spontaneous process)∆G = ∆H – T ∆S= ∆E + P ∆V – T ∆S(∆G) E,V = 0 + 0 – T ∆S(∆S) E,V = ⊕ ⇒ ∆G = Θ (Spontaneous process)54.[B]CH| 3H3C – C – CH – CH|3OH| CH3CH 3 |H++⎯⎯→CH3– C – C H – CH–H O32CH| 32°Carbocation1, 2-Methyl shift+ +H C – C C– CH –H3H C – C – CH – CHCH| =CH| 3 ←⎯⎯3CH| CH| 33 33 33°Carbocation55.[C] Factual Q.(Morestable)XtraEdge for IIT-JEE 105APRIL 2010

56.[D]OH Zndust ⎯⎯→H3C–Cl⎯⎯⎯→CH 3AlCl3alk. KMnO 4⎯ ⎯⎯ → COOH64.[C] f(x) = sin x – cos x – kx + bf '(x) = cos x + sin x – kf '(x) = 2 sin(π/4 + x) – kif f(x) is decreases for all x ∴ f '(x) is –vei. e. k > max. of 2 sin(π/4 + x)i. e. k > 257.[D]58.[A] 4 HCl + O 2 ⎯⎯→ 2 H 2 O + 2 Cl 2Chlorine is in the form of cloud.59.[A] Coordination no. = 6Oxidation no. = 3no. of d electron = 6no. of Unpairedd electron = 060.[C] Resonance structure should have same numberof electron pairs.61.[A]MATHEMATICS30º 60ºAOLet the height of the tower is h1000 = h cot 30º – h cot 60º1000⇒ h =cot 30º −cot60ºh = 500 3 m62.[D] since variance is independent of change inorigin. Hence variance of observations 101,102, ...... 200 is same as variance of 151, 152,.....250.∴ V A = V BV A⇒ = 1VB63.[C] centroid of triangle is given by⎛⎞⎜c 2 + 3 a + b − 3, ⎟⎝ 3 3 ⎠if centroid lie on y axis ⇒ abscissae = 0c 2 + 3 = 0 ⇒ no real c existif centroid lie on x axis ⇒ ordinate = 0⇒ a + b – 3 = 0⇒ a + b = 3Ph65.[A] x lim → 0= x lim → 0⎡⎢⎢⎣⎛⎜⎜⎝xe 2− cos x ⎤⎥2x ⎥⎦−11−cos x⎟ ⎟ ⎞+22x x⎠xe 221 + x lim 2sin x / 2 1 3→ 0= 1 +2=x2 266.[A] Second determinant has been obtainedfrom the first by the operationC 1 → C 1 + 2C 2 – 3C 3 . so its value remainsunchanged67.[C] given 3sin x – 4sin 3 x – k = 0⇒ 3sin x – 4sin 3 x = k⇒ sin3x = k.........(i)angle A and B satisfy the equation (i)∴ sin 3A = sin3B = k ⇒ sin3A = sin3BBut A > B ⇒ A BNow, sin3A = sin(π – 3B)3A = π – 3Bπ 2πA + B = ⇒ C = 3 368.[B]69.[C]given statement(p q) ∨ ~ r → (p ∧ r)(F ↔ F) ∨ F → (F ∧ T)T ∨ F → FT → F = FpTTFFqTFTFp ∨ qTTTF~ pFFTTq∧~ pFFTF(p ∨ q) ↔ q∧~ pHence neither tautology nor contradiction70.[D] Given sequence can be written as5 20 10 20 , , , , ..............2 13 9 2320 20 20 20or , , , ..........8 13 18 23FFTTXtraEdge for IIT-JEE 106APRIL 2010

71.[B]which is a H.P.n th term of corresponding A.P.8 5 5 n + 3a n = + (n – 1) = 20 20 20n th term of H.P. =1 20 =5n + 3a n20 5also =5n + 3 17175n + 3 = 20 × = 68 5n =65 = 135Let any point P divides line joiningA(–2, 4, 7) & (3, –5, 8) in ratio λ : 1⎛ − 2 + 3λ4 − 5λ8λ + 7 ⎞then P ⎜ , , ⎟⎝ λ + 1 λ + 1 λ −1⎠if P lies on plane 2x – k = 0k⇒ x = 2k −2+ 3λ=2 λ + 1given λ = 9⎛ 25 ⎞k = ⎜ ⎟ × 2 ⇒ k = 5⎝ 10 ⎠72.[C] |a × b| = |a| |b| sin θ8 4sin θ = =5.2 5⇒ cos θ = 53|a – b| 2 = |a| 2 + |b| 2 – 2a.b= 4 + 25 – 2|a| |b| cos θ= 29 – 2.2.5 53 = 29 – 12AP 1if = PB 2⎛ a 2b ⎞co-ordinates of point P given by ⎜ , ⎟⎝ 3 3 ⎠locus of point P, a 2 + b 2 = (3x) 2 ⎛ 3y⎞+ ⎜ ⎟⎠⎝ 2or22x y+ = 14 16b > a then eccentricity e =2a1− =2b74.[D] parabola y = x 2 + 2px + p 2 + 13 – p 2(y – (13 – p 2 )) = (x + p) 2vertex is given by (– p, 13 – p 2 )is 4 units above x-axis⇒ 13 – p 2 = 4 ⇒ p = ± 3also lies in I st quadrant ⇒ p < 0⇒ p = – 375.[C] If lines x 2 + 2λx + 2y 2 = 0& (1 + λ) x 2 – 8xy + y 2 = 0are equally inclined⇒ Their bisectors eq n must be same2⇒ eq n x − y1−2&222x − y=(1 + λ)−1⇒ x 2 – y 2 =−4⇒ – λ =λ⇒ λ 2 = 4⇒ λ = ± 2= λxyxy− 4are samexy & x 2 – y 2 =− λxy− 4 / λ232= 36are same73.[C]|a – b| = 17A(0, b)176.[D] given circle in standard form isx 2 + y 2 ⎛ 1⎞⎛ 1⎞+ 2 ⎜ t + ⎟ x – 2 ⎜ t − ⎟ y + 1 = 0⎝ t ⎠ ⎝ t ⎠⎛ 1 1⎞centre is given by ⎜−t − , t − ⎟⎝ t t ⎠given length AB = 6or a 2 + b 2 = 36P2B(a, 0)⎛ 1⎞1or h = – ⎜ t + ⎟ & k = t –⎝ t ⎠ t22h 2 – k 2 ⎛ 1⎞⎛ 1⎞= ⎜ t + ⎟ – ⎜ t − ⎟ = 4⎝ t ⎠ ⎝ t ⎠locus x 2 – y 2 = 4 which is a hyperbolaXtraEdge for IIT-JEE 107APRIL 2010

77.[C]4∫−1f (x) dx = 42⇒∫f (x)−12⇒∫f (x)−14dx +∫24dx = 4 –∫2f (x) dx = 4f (x) dx ...........(i)4Q∫(3 – f(x) ) dx = 7 (given)24 4⇒∫3 dx –∫f (x) dx = 72 24⇒ 3[x] 4 2 = 7 +∫f (x) dx24⇒ 3 × 2 = 7 +∫2.....(ii)f (x)4dx ⇒∫2put this value from (ii) in (i), we get2∫−1f (x) dx = 4 – (–1) = 5−1⇒∫f (x) dx = – 52f (x) dx = – 178.[A] lines – 2x – y + 6 = 0 & 4x – 2y + 7 = 0(make c 1 & c 2 +ve)now a 1 a 2 + b 1 b 2 = – 8 + 2 = – 6 < 0bisector by +ve is acute and contains origin eq n−2 x − y + 6 ( 4x − 2y + 7)given by= +52 5– 4x – 2y + 12 = 4x – 2y + 78x = 5∫79.[C] [ f g(x) ]21−1f 'g(x). g'(x) dxput f g(x) = t ⇒ f 'g(x) g'(x) dx = dt∴ required integral2[ log f g(x) ]1log (f g(2)) – log (f g(1)) = 0{Q g(1) = g(2)}2x + sin x80.[C] f(x) =∫sec 2 x dx21+x⎛22=∫ ⎟ ⎞⎜x + 1−cos xsec 2 x dx2⎝ 1+x ⎠281.[B]⎛2=∫ ⎟ ⎞⎜cos x1−sec 2 x dx2⎝ 1+x ⎠⎛ 2 1 ⎞=∫⎜secx − ⎟ dx2⎝ 1+x ⎠= tan x – tan –1 x + c∴ f(0) = 0 ⇒ tan0 –tan –1 0 + c = 0⇒ c = 0∴ f(x) = tan x – tan –1 x∴ f(1) = tan1 – tan –1 1= tan1 – π/4we know x –[x] = {x}∴ Domain of {x} is R & {x} ∈ [0, 1)but in f(x), {x} is in denominator & it shouldnot be equal to zero∴ {x} 0 ⇒ x Iand domain of sec –1 x is R –(–1, 1)∴ domain of f(x) isR – (–1, 1) – I82.[C] Q sin –1 x is defined for |x| ≤ 1 andsec –1 x is defined for |x| ≥ 1 thereforeboth defined for |x| = 1 ⇒ x = {1,–1}∴ D f = {–1, 1}further f(–1) = sin –1 (–1) + sec –1 (–1)= – π/2 + π = π/2and f(1) = sin –1 (1) + sec –1 (1) = π/2 + 0 = π/2Hence R f = {π/2}83.[B]Let A(z 1 ), B(z 2 ) and C(z 3 ) be the vertices of thetriangle then|z 1 | = |z 2 | = |z 3 |⇒ |OA| = |OB| = |OC|O being the origin⇒ O is circumcentre of the triangle, Also, thetriangle is equilateral, therefore circumcentrecoincide with the centroid⇒ origin is centroidz1 + z2+ z3⇒= 03⇒ z 1 + z 2 + z 3 = 084.[B] Total students n = 100Average marks x = 72Total boys n 1 = 70average marks of boys x 1 = 75Total girls n 2 = 30n1 x1+ n1x2now x =n∴ x 2 =x 2 =nx− n1x1n 2100 × 72 − 70×75= 6530XtraEdge for IIT-JEE 108APRIL 2010

85.[C] Letters of word 'STATISTICS' are1A, 2I, 1C, 3S, 3T total = 10Letters of word 'ASSISTANT' are2A, 1I, 1N, 3S, 2T total =9common letters are A, I, S & T1 2 2probability of choosing A is = × = 10 9 90probability of choosing I = 102 × 91 = 902probability of choosing S = 103 × 93 = 909probability of choosing T = 103 × 92 = 9062 2 9 6 19total = + + + = 90 90 90 90 9086.[D] A = { – 2, –1, 0, 1, 2} ⇒ n(A) = 5B = {0, 1, 2, 3} ⇒ n(B) = 4C = {1, 2} ⇒ n(C) = 2D = {(1, 7),(2, 6),(3, 5),(4, 4),(5, 3)(6, 2),(7, 1),}87.[B]⇒ n(D) = 7(A ∪ B ∪ C) = {– 2, – 1, 0, 1, 2, 3}⇒ n(A ∪ B ∪ C ) = 6n(D) = 7n(B ∪ C) = 4this is of the formdyf '(y) + f(y). f(x) = Q(x)dxput tan y = z∴ sec 2 dy dzy = dx dxdz∴ + 2xz = x3dx∴ e∫ pdx= ∫ 2xdxe =∴ sol n is z= 21= 21∫xe 22xe. x∫e t . tdt2=∫2x dx2xexe 2. x3dx= 21 e t (t – 1) + c2x 1 2x∴ tan y e = e (x 2 – 1) + c288.[D] The given curves are y 2 = 8x ...(1)and xy = – 1 ...(2)any tangent to (1) is2y = mx + m...(3)⎧Q tan gent to y2= 4ax⎪ a⎨is y = mx +⎪ m⎪∴4a= 8 ⇒ a = 2⎩we shall find m so that it touches (2)∴ from (2) & (3)⎛ 2 ⎞x⎜mx+ ⎟ = −1⇒ m 2 x 2 + 2x + m = 0 ...(4)⎝ m ⎠(3) touches (2) if quadratic (4) has equal roots∴ D = 0⇒ 4 – 4m 3 = 0 ⇒ m 3 = 1 ⇒ m = 1∴ required common tangent is y = x + 289.[B] h(x) = (f(x)) 2 + (g(x)) 2∴ h'(x) = 2f(x)f '(x) + 2g(x) g'(x)= 2f(x)g(x) + 2g(x)g'(x) {Q f '(x) = g(x)}Also f '(x) = g(x)∴ f"(x) = g'(x) ⇒ – f(x) = g'(x)∴ h'(x) = 2f(x) g(x) – 2g(x)f(x)h'(x) = 0∴ h(x) = constant for all xbut h(5) = 11 ∴ h(x) = 1 for all x∴ h(10) = 1190.[A] for m 1 : Let I & N are assumed single letter inIN order Now total no. of letter are IN T E G ER = 6 letterthese 6 letter can be arranged in row6= × 2 = 720 ways2∴ Letter of words INTEGER can be arranged in7a row = = 2520 ways2Now no. of ways in which IN are not together ism 1 = 2520 – 720 = 1800Now for m 2 : I - - - - - R → rest five can arrange5= = 60 ways22x⇒ tan y = c e − 1+ (x 2 – 1) 2∴m 1 =60m21800 = 30XtraEdge for IIT-JEE 109APRIL 2010

SOLUTION FOR MOCK TESTPAPER - II BIT-SAT1.[B]2.[C]PHYSICSIn case of damped vibrations, amplitudedecreases exponentially with time∴ A = A 0 e –bt Aor = e–btororHere1 = e b×2 and2A ´ = (e –2b ) 3 =A 0k e= k1A 0A ´A 0⎛ 1 ⎞⎜ ⎟⎠⎝ 2= e–b×b3= 811 1 1 = + + + ....... ∞k 21k 4k⎛11 1 ⎞⎜ + + + ...... ∞⎟⎝12 4 ⎠1 ⎛ 1 ⎞ 2= ⎜ ⎟ =k ⎝1−1/2 ⎠ k3.[ B] KE max = 21 kA2= 21 Mω 2 A 28 × 10 –3 = 21 × 1 × ω 2 × (0.1) 2i.e. k e = 2kω = 4 rad/secy = A sin(ωt + φ) or y = 0.1 sin (4t + π/4)4. [B] Here → → →→L = m ( r × v)= m v y(– k )(Where y is the vertical distance of particle fromx axis)Here m,v and y all are fixed so → L . remainsconstant.5. [B] Using, weight of floating body = weight ofliquid displaced.⎛ V ⎞ Vwe get V ρ g = ⎜ ⎟ (13.6) g – (0.8g)⎝ 2 ⎠ 2(buoyant forces of mercury and oil act inopposite direction)13.6− 0.8Then, ρ = = 6.426. [C] v´ =v + vv − v9 340 + v´i.e. = 8 340 − v´i.e. v´ = 20 ms –10sv We get νν´ =v − v0v − v∴ (v 0 = v s = v´)I2I27. [A] I log = dB; i.e. 4 = 10 log−94I1(10×10 × 10 )Then I 2 = 2.5 × 10 –4 Wm –28. [A] Here work done = pdv and area under the curvegiven work done∴ 10 + W CA = 5 or W CA = –5 J9. [C] For equilibriumF = qE = mg1/ 34or neE = πr 3 ⎛ 3neE ⎞ρg or r =3 ⎜4 g⎟⎝ πρ ⎠1.5 + 1.5 −1.510.. [A] i == 0.5 A1+1+1As the current has to from A to C to B,for kirchhoff's law,V A = 0.5 × 1 + 1.5 = 1V (Q v = E – ir)V B = 0.5 × 1 + 1.5 = 1VV C = 0.5 × 1 – (–1.5) = 2V11. [D] R = R 1 + (273 – T) α ...(i)or 2R = R 0 [1 + T´α]...(ii)1 1(273- T) αDeciding (i) by (ii) = 2 1+T´ αor 1 + T´α = 2 + (273 – T)2α(273- T)2α + 1or T´ =α12. [A] Current, I =2ER + r 1 + r 22Er1P.O. across cell 1 = Ir 1 =R + r1+ r2For zero p.d. the fall of potential should beequal to in emf.2Er1E =i.e. R = r 1 – r 2R + r + r12sXtraEdge for IIT-JEE 110APRIL 2010

13. [B] <strong>Point</strong> P lies on the arms CD and AF so inclusionat P due to them is zero.Magnetic induction at p due to currents in ABand BC is given byµB 1 = B 2 = 0 isin 45º4π(2a)µ= 0 i(Q distance of p from AB or BC is8 2πa2a)similarly due to DE and EFµB 3 = B 4 = 0 iµsin 45º = 0 i⊗4πa4 2πaNet induction = 2(B 1 – B 3 )µ= 2 0 i8 2π a– µ 0 i4 2π a= – 2µ0 i8πa14. [C] v = rω = r × T2πor T =NowT p = km pq α4m= k2q2π rv2πm =qB=kmqm α = 4m p and q α = 2qvmαand T α = kqPP= 2k⇒ T α = 2T p or15. [D] i = i 0 (1 – e –t/τ )v ⎛ −i = ⎜ t1 − eR ⎜⎝L / Rm Pq ααT p = 21Tα⎞ ⎛⎟ 12 ⎜=⎟ ⎜1−e6⎠⎝−t(Q i = 1A given)⇒ t = 0.97 × 10 –3 s , i.e. t ≈ 1ms−38.4×106⎞⎟⎟ = 1⎠16. [A] Optical distance between fish and the bird isDifferentiating w.r.t.ds dy µdy= +dt dt´dt4 dyi.e. 9 = 3 + 3 dtdx 3or = 6 × = 4.5 ms–1dt 417. [A]1 ⎛ µ g =f ⎟ ⎞ ⎛⎜−1⎝ µ⎜m ⎠ ⎝1R−1 R 21 ⎞⎟⎠1 ⎛ 1.5 ⎞ ⎛ 1 1 ⎞ 1= ⎜ −1 ⎟⎠ × ⎜ − ⎟ =f ⎝1.75⎝ − R R ⎠ 3.5Ri.e. f = 3.5 R.In the medium it behaves as a convergent lens.18. [D] Reflection of light from plane mirror givesadditional path difference of λ/2 between twowaves.3λ π∴ Total path difference = + = 2λ2 2which satisfy the condition of maxima.Resultant intensity ∝ (A 2 + A 2 ) [Q I ∝ Α 2 ]4Α 2 = 4Ι19. [B] Hereλλ21( λ( λ00− λ1)2=− λ ) 15.4 ( λ0− 3.5×10or3.7 ( λ0− 5.4×10or λ 0 = 11.8 × 10 –7 mBut ω =eV02−7−7))= 12−34hcλ = (6.6×10 )(3×10 )= 1.05−7−19(11.8×10 )(1.6 × 10 )20. [D] Let whole the energy of electrons be convertedin x-rays. eV = hvor eV = λhc−34hc (6.6×10 )(3×10 )or λ = = eV−193(1.0 × 10 )(40×10 )i.e. λ = 3.1 × 10 –11 m or λ = 0.31 Åβ⎯ −21. [A] Here 88 Ra 222 ⎯⎯→α 86R 218 ⎯→ 87Fr 218β⎯ −⎯⎯→α 85Al 214 ⎯→ 80Rn 214⎯⎯→α 82Pb 2064α deceys and 2β decays.22. [A] A 1 = 4, A 2 = 3 and θ = 2π = 90º∴ Resultant amplitude,2 2A1 2 + 1 2Α = + A 2A A cos90º==23. [C] Using d sin θ = nλsin θ = θ = Dλ∴224 + 3 = 25 = 5 unitdy = nλD−7nD or y = λd88⎯⎯→α 84PO 2102 21 A2A +1×2×(6×10 )i.e.= 1.2 × 10 –3 = 1.2 mm−31×10Distance between first minima on either side ofcentrar maxima ∆y = 2y = 2.4 mm24. [B] For constructive inteferenceI max = ( I ) 21 + I 2 = ( 2 I + I ) 2XtraEdge for IIT-JEE 111APRIL 2010

For destuctive interferenceI min = ( I ) 21 − I 2 = ( 2I− I ) 2ThenIIminmax=((2I −2I +25. [B] Using decay equationA 2 = A 1 e –λtor e –λt =time t =(4πa)26. [B] V x = ka= 4π⎛⎜Q k =⎝I)I)A 2Aor λt = log1A1A 222⎛ 2 1⎞= ⎜−⎟1= 1 2⎝ + ⎠ 34log A 1 / A 2⎛ A ⎞= 1.44T log eλ⎜1⎟ ⎝ A 2 ⎠2 σ14πε2 σk(4πb)–b14πε σ= (a – b + c)ε 027. [D] Let the given quantity be x 1 then,0⎞⎟⎠x = 3π (a 2 – b 2 )h = 3π (a 2 h – b 2 h)= 3π a 2 h – 3π b 2 h022 σk(4πc)+bσ(a – b + c)Each term has the dimension of x 1 then[x] = [a 2 h] = [L 2 L] = [L 3 ] and also[x] = [b 2 h] = [L 2 L] = [L 3 ] quantity is volume.A x U(x2 + B)28. [B] U = or A =x2 + BxHere dimensions of x 2 and B should be same.i.e. [B] = [x 2 ] = [L 2 ]⎡2 −2ML T ⎤Also [A] = ⎢1/ 2⎥ [L 2 ] = [ML 7/2 T –2 ]⎢⎣L ⎥⎦Then [AB] = [ML 7/2 T –2 ] [L 2 ] = [ML 11/2 T –2⎛ t ⎞ dx ⎞29. [A] v = u ⎜1 − ⎟ or = u ⎜⎛ t1 − ⎟⎠⎝ t´⎠ dt ⎝ t ´⎛2⎞integrating, x = u ⎜tt − ⎟ + C⎝ 2t´⎠at t = 0, n = 0 and c = 0⎛2⎞∴ x = u ⎜t⎟⎛ t ⎞t − = 10t ⎜ t − ⎟⎝ 2t´⎠ ⎝ 10´⎠Putting t = 10⎛ 10 ⎞x = 10 × 10 ⎜1 − ⎟ = 0⎝ 10 ⎠30. [A] using h = 21 gt 2 , we get t 1 =2hglet t 1 be the time taken from instants of jumpingto the opening of parachute, then2× 40t 1 = = 2.86 sec9.8His velocity at this point is given byv 2 1 = 2gh 1 = 2 × 9.8 × 40= 784 or v 1 = 28 ms –1for the remaining journey,v = v 1 + at 2v − u 2 − 28or t 2 = = = 13seca − 2∴ total time = t 1 + t 2 = 2.86 + 13= 15.86 ≅ 16 s31. [B] Let u be the velocity of projectile w.r.t. tanksvelocity v thenu x = u cos 30 + v; u y = u sin 30º2u sin 30ºand T =g2u sin 30ºRange, R 1 = u x T = (u cos 30º + v)gfor y axisu x´ = u cos 30º – v and u y´ = u sin 30º2u sin 30ºT =gRange e, R 2 = Tu´x2u sin 30º= (u cos 30º – v)g4u2Then R 1 + R 2 = (sin 30º cos 30º)g4uR 1 – R 2 = v sin 30ºgEliminating u we getv 2 ==g4 tan 30104 tan 30⇒ 4.9 ms –1(R1− R 2 )(R + R )1222(250 − 200)= 24 m 2 s –2(250 + 200)32. [D] Let α be the angle between velocities of pair ofparticles then relative velocity is given by2 2αv r = v + v − 2v×v×cosα= 2 v sin 22π2v(sin α / 2) 4average v r =∫dα =2πv π0dα∫0XtraEdge for IIT-JEE 112APRIL 2010

33. [B] For quarter revolution∆ V → = V → 2 – V→ 1angle between V → 1 andNWv 2a∴ ∆v =Also∴ ∆ → v =v 1–v 1S2 22 v1v + =tan –1 v = 45ºv34. [D] For vertical motionH = 21 gt 2 or t =→2V is 90ºE22 v south west2 H / g2v + v = 2 VFor horizontal motion, distance covered is givenby2πrn = utor 2πrn = uor u =42πr2 H / g2H / g35. [A] On descending(Mg – f) – Ma = 0(where f is the upthrust due to buoyancy)On ascending,f – (M – m)g – (M – m)a = 0⎛m =⎜⎝⎟ ⎞M+ g ⎠2aa36. [A] The masses will be lifted if the tension of thestring is more than the gravitational pull ofmass.50 N20 N 50 N50 – 2T = 0 or T = 25NSo, 5 kg weight cannot be lifted but 2 kg weightwill be lifted25 – 20 = 2a or a = 25 = 2.5 ms–237. [A] On cutting of string QR, the resultant force onm 1 remains zero because its weight mg isbalanced by the tension is the spring but onblock m 2 a resultant upward Force (m 1 – m 2 )g isden sped. The block m 1 will have no resultantacceleration where as m 2 does have an upward( m1− m2)gacceleration given bym⎛ 2pt ⎞38. [A] Here v = ⎜ ⎟⎝ M ⎠or1/ 21/ 2ds ⎛ 2pt ⎞ = ⎜ ⎟dt ⎝ M ⎠⎛ 2pt ⎞or ds = ⎜ ⎟⎝ M ⎠1/ 2dt⎛ 2p⎞integrating s = ⎜ ⎟⎠⎝ M1/ 2at t = 0, S = 0, so c = 01/ 2⎛ 8p⎞S = ⎜ ⎟⎠ t 3/2⎝ 9M22 t 3/2 + C339. [B] Let a small displacement be given to the systemin vertical plane of frame such that ST remainshorizontal then let vertical displacement ofcentres of rods up and QR be y then verticaldisplacment of centres of VT and RS will be 3yand that of TS will be 4y. Equating total verticalwork to zero we getWWPWTWQWSW(w + w)δy + (w + w)3 δy + w(4δy) – T(4δy) = 0or 2w + 6w + 4w = 4T or T = 3wWRyyyy40.[A] Normal reaction R = mg = 2 × 9.8 NFrictional force,F = µR = 0.2 × 2 × 9.8 = 3.92 NDistance traveled2 × 5 = 10 m∴ Work done = f × s = 3.92 × 10= 39.2 JXtraEdge for IIT-JEE 113APRIL 2010

CHEMISTRY1.[D] Fe 2 O 3 + 3CO → 2Fe + 3CO 2160 g 112 g100×80Pure Fe 2 O 3 in ore = = 80 kg10Iron produced from 80 kg of Fe 2 O 3112= × 80 = 56 kg1602.[D] ∆v =h4πm(∆x)–346.626×10=–31 −104(3.14)(9.11×10 )(10 × 0.1)= 5.79 × 10 6 ms –13.[B] Given n = 0.5⎛2⎞Then ⎜n a⎟P +2(V – nb) = nRT⎝ V ⎠2(0.5) a⇒ [P + ] [V – 0.5b] = 0.5 RT2V⎛ a ⎞⇒ ⎜P+2⎟ (2V – b) = RT⎝ 4V ⎠4.[B] N 2 O 4 (g) 2NO 2 (g)n 0 : 1 0n e : 1 – 0.5 2 × 0.5= 0.5 10.5 1C e : M55= 0.1 M = 0.2 M5.[A]6.[C]7.[D][NO2]then K C = =[N O ]2422(0.2)=(0.1)0.04= 0.40.1HClO < HClO2< HClO3< HClO4→AcidicstrengthWeakest acid has strongest conjugate base.HCl : N A × V A = (0.4 × 1) × 50 = 20 meq.NaOH : N B × V B = (0.2) × 50 = 10 meq.Q N A V A > N B V B∴ [H + NAVA− NBVB20 −10] ==V + V 100AB= 0.1 M = 10 –1 M, ∴ pH = 1(–2)(+1)O(–1) (+1)H – O – O – S (x) – O – H(–1)O(–2)(–2)2(+1) + 2 (– 1) + x + 3(–2) = 0⇒ x = + 68.[C]9.[D](+2)(+3)3+ (+4)FeC 2 O 4 → Fe 3+ + CO 2–1e–2e∴ Valence factor (FeC 2 O 4 ) = 1 + 2 = 3M∴ E( FeC 2 O 4 ) =3The fractional of total volume occupied in simplecube =volumeof particlesvolumeof cube=4 ⎛ a ⎞π⎜⎟3 ⎝ 2 ⎠3a10.[A] Cu 2+ + e – → Cu + 0; E 1 = 0.15 V0∴ ∆ G 1 = –1 × 0.15 × F = – 0.15 FCu + + e – 0→ Cu ; E 2 = 0.5 V0∴ ∆ G 2 = – 1 × 0.5 × F = – 0.5 FCu 2+ + 2e – → Cu ;0E 3 = ?00∴ ∆ G 3 = – 2 × E 3 × F = – 2F0 0 0Also, ∆ G 3 = ∆ G 1 + ∆ G 2= – 0.15 F + (– 0.5 F)= – 0.65 FNow – 2E0 3 F = – 0.65 For0E 3 =0.652= 0.325 F11.[A] 4hrs = 4 half lives1t 1⎯⎯→1/ 2t⎯⎯→1/ 2212.[B]14t⎯⎯→1/ 20E 3183= 6πt⎯⎯→1/ 241 ⎛ 1 ⎞fraction left after 4 half lives = or ⎜ ⎟⎠ 16 ⎝ 21 15fraction reacted in 4 half lives = 1 – = 16 16A solution showing +ve deviation has highervapour pressure and lower boiling point.13.[C] In multi molecular solutions the different layershold each other through van der Waal's forces.14.[A] CS 2 + 3O 2 → CO 2 + 2SO 2 ; ∆ r H = ?∆ r H° = Σ∆ – Σ∆ 00f H (P)f H (R)= [2(– 297) + (– 393)] – (117)= – 1104 kJ mol –115.[C] Aspirin chemically acetyl salicylic acidOCOCH 3COOH116XtraEdge for IIT-JEE 114APRIL 2010

16.[A]NH 2⎯ 0−5°C+ NaNO 2 + 2HCl ⎯⎯→N 2 Cl+ 2H 2 O + NaCl(CH 3 ) 2 N H + Cl – N = N ⎯ −HCl⎯ →26.[A]27.[B]OC NCH 3CH 3N,N-dimethyl cyclopropane carboxamide.NaCl + H 2 SO 4 → NaHSO 4 + HClK 2 Cr 2 O 7 + 2H 2 SO 4 → 2KHSO 4 + 2CrO 3 + H 2 OCrO 3 + 2HCl → CrO 2 Cl 2orange red vapour17.[A](CH 3 ) 2 N N = NOR – C – O – C 2 H 5 + CH 3 MgBr ⎯→:O – OC 2 H 5⎯⎯⎯⎯→R – C−−C 2 H 5 OR CH 3CH 3 MgX⎯⎯⎯⎯ →OCH 328.[C]29.[B]enenCoClCl+Cis-d-isomer−2 MnO 4 + 16H + +8H 2 OClClCoenenCis-l-isomer−C2O 4 ⎯→2Mn 2+ +2CO 2 ++XMgOCH 3+ HOHOH / H⎯⎯ ⎯⎯ →CH 330.[C] Lanthanoid contraction takes place.RCH 3RCH 331.[D] In nitrogen d orbital is absent.18.[C] It is cannizzaro reactionCHO⎯KOH ⎯⎯→ ∆COO – +ClClCH 2 OH19.[C] Phenol is less acidic than acetic acid andp-nitrophenol.OHO – C 2 H 5Anhy.20.[A] + C 2 H 5 I ⎯⎯⎯→C2H5OH40% H21.[C] CH ≡ CH2SO4CH 2 = CH – OH1% HgSO4⎯ Keto ⎯⎯⎯⎯⎯⎯⎯−enoltautomerism→ CH 3 – C – HOAcetaldehyde22.[C] The rate of nitration is greater inhexadeuterobenzene23.[D] Halogenation on alkene occurs by electrophilicaddition.24.[A] Twisted boat is chiral as it does not have planeof symmetry.Cl32.[A] HNO 3 is acidic in nature.33.[A] 2KBr + H 2 SO 4 → K 2 SO 4 + 2HBr34.[A] Due to formation of chelate compound it act asstrong acid and proceed in forward direction.CH – O O – CHBCH – O O – CH +H+35.[B] Na 2 SO 3 + S ⎯ NaOH ⎯⎯→Na 2 S 2 O 3Sod. thiosulphate36.[D] Critical temperature of water is more than O 2due to its dipole moment (Dipole moment ofwater= 1.84 D, Dipole moment of O 2 = 0D.)37.[D] By the process of zone refining, semiconductorslike Si, Ge and Ga are purified.38.[C] Half filled orbitals are more stable incomparison of partial filled.39.[A] The dipole moment ofCH 4 = 0NF 3 = 0.2 DNH 3 = 1.47 DH 2 O = 1.85 D25.[C]C – CH 3OAcetophenone has highest dipole moment.40.[A] Molecule existence is possible in such casewhen no. of bonding electron is greater thanantibonding.XtraEdge for IIT-JEE 115APRIL 2010

1.[D]MATHEMATICSClearly all term can Neither be positive nornegativeT 1001 = sin log 10 1000 = sin 3 > 0T 10001 = sin log 10 10000 = sin 4 > 0∴ (0 < 2 < π & π < 4 < 2π)2. [A] sin x < x < tan x in (0, 1) ⇒ tan –1 x < x < sin –1 xAltiter :f(x) = sin –1 x – xf '(x) =1– 1 > 0 ∀ x ∈ (0, 1)21– x∴ f(x) is increasing function∴ x > 0 ⇒ f(x) > f(0) = 0⇒ sin –1 x > xSimilarly g(x) = x – tan –1 x is increasing fu nand x > 0 ⇒ x – tan –1 x > 03.[A] log 2 cos x + log 2 (1 – tan x) + log 2 (1 + tan x)– log 2 sinx = 1⇒ log 2 (1 – tan 2 x) – log 2 tan x = 1(cos x > 0, sin x > 0, – 1 < tan x < 1)⇒ (1 – tan 2 1x) × = 2tan xtan 2 x + 2 tan x – 1 = 0tan x = –1 ± 2tan x = 2 – 1 (Q 0 < tan x < 1) ⇒ x = π/8∴ h = 0 or h = 2a∴ locus of P is x = 0 or x = 2a6. [B] Equation of angle bisector of the pair of straight2 2x − y xylines is =a − b − h⇒ hx 2 + (a – b)xy – hy 2 = 07.[C]2y ⎛ y ⎞⇒ h + (a – b) – h ⎜ ⎟⎠ = 0x ⎝ xNow, y = mx is one of the bisector∴ hm 2 – (a – b)m – h = 0h(m 2 – 1) = (a – b)m⇒m 2 − 1 a − b =m hx 2 + y 2 + 2gx + 2fy + c = 0 passes through allthe four quadrants⇒ origin in an interior point⇒ c < 08. [A] two normal are x – 1 = 0 and y – 2 = 0, theirpoint of intersection (1, 2) is the centre & radiusof circle perpendicular distance from centre(1, 2) on tangent 3x + 4y = 63 + 4.2 − 6== 19 + 16∴ equation of circle is(x – 1) 2 + (y – 2) 2 = 1 2⇒ x 2 + y 2 – 2x – 4y + 4 = 04. [B]5.[C]BrαAB = 2r sin α/2αh = AB tan α = 2r sin 2α tan αLet vertex P be (h, k), then perpendiculardistance of P from the base x = a is |h – a|∴ Since length of the base is 2a, we have1 × 2a|h – a| = a22⇒ |h – a| = a (a ≠ 0)So h – a = – a or h – a = arPA9.[B] x 2 – 4x + 6y + 10 = 0⇒ (x – 2) 2 = –6(y + 1)tangent to the vertex is y + 1 = 0circle drawn on focal distance as diameteralways touch the tangent at the vertex i.e. the liney + 1 = 0.x 210. [B] Given ellipse is + 25y 2 = 192 −a 2 = 25, b 2 a b 4= 9, e = =a 5⇒ ae = 4∴ Foci of ellipse are (± ae, 0) = (± 4, 0)For hyperbola e = 2⇒ 2a = 4 ⇒ a = 2Also b 2 = a 2 (e 2 – 1)⇒ b 2 = 4 × 3 = 12∴ equation of hyperbola4x 2 – 12y 2 = 1⇒ 3x 2 – y 2 – 12 = 02XtraEdge for IIT-JEE 116APRIL 2010

11.[B]→ →∴ α & β are two mutually perpendicularunit vector.→ →∴ α × β is a unit vector perpendicular toboth → α & → β . So we can consider → α , → β , → α ×→β as î , ĵ & kˆ .Given vector are coplanar soa a c1 0 1 = 0c c b⇒ a(– c) + a(c – b) + c 2 = 0 ⇒ c 2 = ab12. [D] S 1 = x 2 + y 2 + z 2 + 2x – 4y – 4z – 7 = 0centre C 1 = (–1, 2, 2) and radius r 1 = 4S 2 = x 2 + y 2 + z 2 + 2x – 4y – 16z + 65 = 0centre C 2 = (–1, 2, 8) radius r 2 = 2C 1 C 2 = 6; r 1 + r 2 = 6∴ sphere S 1 & S 2 touches externaly∴ point of contact divides C 1 C 2 in the ratio 4 : 2∴ point of contact = (–1, 2, 6)413.[D] y = f(x) = 3 –2x − 4x + 8⇒ (3 – y)x 2 – 4(3 – y)x + 20 – 8y = 0x ∈ R ∴ D ≥ 0⇒ 16(3 – y) 2 – 4(3 – y) (20 – 8y) ≥ 0, y ≠ 3⇒ –y 2 + 5y – 6 ≥ 0; y ≠ 3⇒ (y – 2) (y – 3) ≤ 0 ⇒ 2 ≤ y < 314. [B] x > 0, g(x) is boundednxf (x)e + g(x)∴ limn→∞nx1+enxf (x) + g(x) / elim= f(x)n→∞nx1+1 e[g(x) is bounded ⇒15.[B] a 1 = 1, a n = n(1 + a n–1 )a ng(x)nxefinite⇒ = 0] infinte⇒ 1 + a n–1 =n⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1∴ limn→∞⎜1+ ⎟⎝ a⎜1 + ⎟ ....1 ⎠ ⎝ a⎜1+2 ⎠ ⎝ a nlim ⎛ 1+an→∞⎟ ⎞⎜1⎛ 1+a ⎞⎝ a⎜2⎛ 1+a⎟ ....1 ⎠ ⎝ a⎜2 ⎠ ⎝ a n= lim a 2 a. 3 a........ n + 1 1 .n→∞2 3 n + 1 a1a2...a na n+1 (1 + a n )(1 + n)= lim = limn →∞ n + 1 n →∞ n + 1⎞⎟⎠n⎞⎟⎠⎛ ⎞= ⎜ 1 a ⎛nlim + ⎟ = ⎜ 1 1 a n−1lim + +n→∞⎜n n ⎟ n ∞⎜n n −1n −⎝ ⎠ ⎝⎪⎧1 1 1 1 a ⎪⎫1= lim ⎨ + + .... + + ⎬n→∞n n −12 1 1⎪⎩⎪⎭⎪⎧1 1 1 ⎪⎫= lim ⎨1+ + + .... ⎬ {Q a 1 = 1}n→∞1 2 n⎪⎩⎪⎭= e⎧−2⎪16.[A] 2 sgn 2x = ⎨ 0⎪⎩ 2;;;→ 1x < 0x = 0x > 0⎧4x ≠ 0f(x) = |2 sgn 2x| + 2 = ⎨⎩2x = 0∴ By defining f(0) = 4, f(x) will becomecontinuous function at x = 0 as thenf(0 – 0) = f(0 + 0) = 4Hence at x = 0 f(x) has removable discontinuity17.[A] Q 1 < x < 2 ⇒ [x] = 1π 3⎛ ⎞∴ f(x) = cos ⎜ − x ⎟ = sin x 3⎝ 2 ⎠f ´(x) = cos x 3 . 3x 22/ 3⎛ ⎞⇒ f ´ ⎜π⎟π ⎛ π3 = 3 cos . ⎝ 2 ⎠ 2 2 ⎟ ⎞⎜ = 0⎝ ⎠18.[C] ye x = cos x ...(1)ye x + y 1 e x = – sin x ...(2)again differentiatingye x + y 1 e x + y 1 e x + y 2 e x = – cos xye x + 2y 1 e x + y 2 e x = – ye x (from (1))⇒ 2y + 2y 1 + y 2 = 0 ...(3)again differentiating2y 1 + 2y 2 + y 3 = 0again differentiating2y 2 + 2y 3 + y 4 = 0 ...(4)from (1) (2) & (3)4y + y 4 = 0y∴ 4 = – 4y19.[A] fog = I ⇒ fog(x) = I(x) = x∴ fg(x) = x⇒ f´g(x) × g´(x) = 1⇒ f´g(a) × g´(a) = 1⇒ f´(b) × 2 = 1⇒ f´(b) = 21⎞⎟⎟⎠XtraEdge for IIT-JEE 117APRIL 2010

20.[C]Let P(h, k) be one of point of contact thenk = sin h ...(1)equation of tangent is y – k = cos h (x – h)which parries through origin∴ k = h cos h ...(2)from (1) & (2)2k+ k 2 = sin 2 h + cos 2 h = 12h⇒ k 2 + k 2 h 2 = h 2 ⇒ x 2 – y 2 = x 2 y 2∴ locus of (h, k) is x 2 – y 2 = x 2 y 224.[A] 1 –4 2= (1 + cos 2 x) 3/2 ⎡⎤⎢− (1 + cos x) + 2⎥ ⎣ 5⎦= 52 (1 + cos 2 x) 3/2 (3 – 2cos 2 x) + cx 2 +3x 44– .......... = cos x∴ I =∫cos x dx = sin x+ c21.[B] x + y = 16, x, y > 0s = x 3 + y 3 = x 3 + (16 – x) 3ds = 3x 2 – 3(16 – x) 2dxds = 0dx⇒ x 2 = 256 – 32x + x 2 ⇒ x = 822.[C]2d s= 6x – 6(16 – x) = 6[2x – 16]2dxat x = 823d s d s= 0 and = 12 ≠ 023dx dxHence there in no minimum exist.1 1 2− −2 2 3b a=c1 1 1− −2b a c1 1 2⇒ + = b a c= c22ab⇒ c =a + c∴ c is H.M. of a & b23.[B] I =∫1+ cos x sin 2x cos 2x dxput 1 + cos 2 x = t 2⇒ –2sin x cos x dx = 2tdt⇒ – sin 2x dx = 2tdt22∴ I = –∫t .(2t dt).(2cos x − 1)= –∫t.2t.(2t − 3) dt2225.[A]πx→226.[A] y =27.[B]∫x−cos t(2 −1)dtπ / 2⎛ 0 ⎞lim ⎜ ⎟2x⎛ π ⎞⎝ 0 ⎠∫⎜ t − ⎟ dt2π / 4⎝2 ⎠−cos x2 −1lim=πx→⎛ π ⎞2 ⎜ x − ⎟.2x⎝ 2 ⎠⎛ 1 ⎞lim ⎜1+2 ⎟⎝ n ⎠n→∞log y =nn→∞r=12limπx→2−cos xln2.sin x=4x− π2462 2 2 2 2n ⎛ 2 ⎞ n ⎛ 3 ⎞ n⎜1+ ⎟ ⎜1⎟2 +2⎜⎝n⎟⎠⎛ n.... ⎜1+⎝ n⎛2∑ ⎟ ⎞⎜rlim log1 + .2⎝ n ⎠n⎛= ∑⎜rlim log→∞1 +nr=1 ⎝ n1=∫2 x log(1 + x2 ) dx02=∫1⇒ y = 4/e22⎟ ⎞⎠log t dt = ( t log t − t) 2 1= 2 log 2 – 1 = log e4required area122⎞⎟⎠⎜⎝2n2n2r2n2 r 1. . n nn⎟⎠y = 2x 2ln2π⎡ 2t= –2 ⎢⎢⎣553t− 33⎤⎥⎥⎦+ c01/√2= – 54 (1 + cos 2 x) 5/2 + 2(1 + cos 2 x) 3/2 + c–1XtraEdge for IIT-JEE 118APRIL 2010

1/ 21 –2∫2 x 2 dx⇒⇒01 2 – 3 2] 1 /[ x 02 31 2 1 – .2 3 2 2=322=2 2628.[B] y ´´ = (y´ + 3) 1/3 ⇒ (y´´) 3 = (y´ + 3) 23⎛2d y ⎞⇒ ⎜ ⎟⎛ dy ⎞ dy–2dx⎜ ⎟⎠ – 6 – 9 = 0⎝ ⎠ ⎝ dx dx∴ order is 2 & degree = 329.[A] x 18 = y 21 = z 28⇒ 18 log x = 21 log y = 28 log z⇒ log y x = 7/6, log z y = 4/3, log x z = 9/14Now, 3, 3 log y x, 3 log z 7, 7 log x 27 4 9= 3, , 3 × , 7 × 2 3 147 9= 3, , 4,2 2which are in A.P.30.[B] log 2 x + log 2 y ≥ 6Here x > 0, y > 0∴ log 2 xy ≥ 6 ⇒ xy ≥ 2 6⇒ xy ≥ 64Now, A.M ≥ G.M.x + y∴ ≥ (xy) 1/22⇒ x + y ≥ 2(64) 1/2x + y ≥ 2 × 8⇒ x + y ≥ 16∴ (x + y) min = 1631.[A] f(x) = x 3 + x 2 + 10x + sinxf´(x) = 3x 2 + 2x + 10 + cos x32.[B]2⎛ 1 ⎞= 3 ⎜ x + ⎟ +⎝ 3 ⎠⇒ f(x) is strictly increasingAlso x → ∞ ⇒ f(x) → ∞,x → – ∞ ⇒ f(x) → – ∞∴ f(x) has only one real root.229 + cos x > 0 ∀ x3Let roots be (2k – 1) & (2k + 1) k ∈ Nthe Sum of roots : 4k = – abQ a ∈ R + , b < 0 as k ≥ 1We have – b = 4ak ⇒ – b ≥ 4a⇒ |b| ≥ 4a {b < 0 ∴ |b| = –b}33.[B] Let G.P. be a + ar + ar 2 ......G.P is infinite so – 1 < r < 1G.P. is decreasing⇒ r > 0 so 0 < r < 1 and therefore a > 0f´(x) = 3x 2 + 3 > 0⇒ f(x) is strictly increasing function∴ f(x) max on [–2, 3] is f(3) = 27 & f´(0) = 3a∴ = 27 & a – ar = 31−r2 4⇒ r = or Q r < 13 334.[B]∴ r = 32& if r = 32 ; a = 9∴ Sum of first three terms = 9 + 6 +4 = 19If z 1 , z 2 & z 3 vertex of equilateral triangle then2z 1 +2z 2 +2z 3 = z 1 z 2 + z 2 z 3 + z 3 z 1∴ (a + i) 2 + (1 + ib) 2 + 0 = (a + i) (1 + ib) + 0 +0⇒ a 2 – 1 + 2ia + 1 – b 2 + 2ib = a + iba + i – b⇒ a 2 – b 2 + i(2a + 2b) = a – b + i(ab + 1)on comparinga 2 – b 2 = a – b and 2a + 2b = ab + 1⇒ (a – b) (a + b – 1) = 0 & 2a + 2b = ab + 1⇒ a = b or a + b = 1 ....(1)⇒ 2a + 2b = ab + 1 .... (2)Now from (1) take a = b put in (2)2a + 2a = a 2 + 1⇒ a 2 – 4a + 1 = 0 ⇒ a = 2 ± 3Q a < 1 ⇒ a = 2 – 3Q a = 2 – 3 & a = b = 2 – 3It we take a + b = 1 & put in (2) then it becomesab = 0 which not possible because a & b liesbetween 0 and 1n ⎡⎤35.[A] S = ⎢⎪⎧r r⎛ ⎞ ⎛ ⎞ ⎪⎫∑ − r 1 3 7( 1) ⎨ + ⎜ ⎟ + ⎜ ⎟ + .... ⎬⎥n Crr=⎢⎥⎣⎪⎩2 ⎝ 4 ⎠ ⎝ 8r 0⎠ ⎪⎭ ⎦nr 1 n= ∑ ( 1) . Cr rr=02nr ⎛ 3 ⎞ n− + ∑( − 1) . ⎜ ⎟ . Cr⎝ 4 ⎠r=0nrr ⎛ 7 ⎞ n+ ( 1) . ⎜ ⎟ . Cr⎠∑ −r=0⎝ 8nnn⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 7 ⎞= ⎜1 − ⎟ + ⎜1 − ⎟ + ⎜1 − ⎟ + .....⎝ 2 ⎠ ⎝ 4 ⎠ ⎝ 8 ⎠=1 1 +nn222+13n2+ ..... =21n −r1+ ....XtraEdge for IIT-JEE 119APRIL 2010

36.[B] The general term in the expansion of (x 1 + x 2 +n...x n ) n p1p2pgiven ...... x1x2.... xmm ,p p ...37.[B]1 2 pnp 1 + p 2 + p 3 .... + p m = nNow in (1 + x + y – z) 9 , coefficient of x 3 y 4 z= coeft of u 0 x 3 .y 4 z 1 in (u + x + y – z) 99=× (–1) 1 = –2 . 9 C 2 . 7 C 30 3 4 1e x= B 0 + B 1 x + B 2 x 2 + .....1−x⇒ e x = B 0 – B 0 x + B 1 x – B 1 x 2 + B 2 x 2 – B 2 x 2 +....⇒ e x = B 0 + (B 1 – B 0 )x + (B 2 – B 1 )x 2 + .....⇒ 1 + x +x 2 +2x 3 +3x 4 + ....4= B 0 + (B 1 – B 0 )x + (B 2 – B 1 )x 2 + ...∴ B n – B n –1 is coeff. of x nOn comparing coeff. of x n =38.[A] x + y + z + 12 = 0x, y, z are negative integersLet x = – a, y = –b, z = – c,a, b, c are +ve integer then required number ofpoints (x, y, z)= Number of positive integral solution ofa + b + c = 12= 12–1 C 3–1 = 11 C 2 = 5539.[A] p 1 =pp1240.[C] 2f(x) =2 6 612p 2 =2 6 6= × 11 = 112 5 6f (x)1f (1/ x)511f (1/ x) − f (x)⎛ 1 ⎞ ⎛ 1 ⎞= f(x) . ⎜ ⎟ – f ⎜ ⎟ + f(x)⎝ x ⎠ ⎝ x ⎠⎛ 1 ⎞ ⎛ 1 ⎞⇒ f(x) + f ⎜ ⎟ = f(x) . f ⎜ ⎟⎝ x ⎠ ⎝ x ⎠⇒ f(x) = 1 ± x nf(2) = 17⇒ 1 ± 2 n = 17 ⇒ ± 2 4 = 16∴ +ve sign will be take⇒ 2 n = 16 ⇒ n = 4Now, ∴ f(x) = 1 + x 4⇒ f(5) = 5 4 + 1 = 62661n41.[C] A is idenpotent ⇒ A 2 = AA 2 ⎡1x⎤⎡1x⎤⎡13x⎤= ⎢ ⎥⎣02⎢ ⎥ =⎦ ⎣02⎢ ⎥ ≠ A⎦ ⎣04 ⎦∴ not possible for any x42.[A] for any a ∈z ⇒ a = 2 0 a⇒ a R a ∀ a ∈ z∴R is reflexivea R b ⇒ a = 2 k b, k ∈ z ⇒ b = a.2 –k , – k ∈ z⇒ b R a∴ R is symmetrickLet a R b, b R c ⇒ a = 2 1kb , b = 2 2 cka = 21 k2 2 kc = 21+k2c , k 1 + k 2 ∈ z⇒ a R c∴ R is transitiveHence R is an equivalence Relation.43.[D] Q A.m ≥ G.m∴1 ⎛ a b ⎞⎜ + ⎟ ≥2 ⎝ b a ⎠b a.a b= 1 ⇒ ba + ab ≥ 2Similarly cb + bc ≥ 2 & ca + ac ≥ 2Adding we geta b b c a c + + + + + ≥ 6b a c b c ab + c c + a a + b⇒ + +a b c∴ minimum value is 6.1 144.[A] f(x) = ⇒ f ´(x) = – x2xf (b) − f (a)∴= f ´(x 1 )b − a⇒≥ 61 1 ⎛– = (b – a)b a ⎟ ⎞⎜1 − 21;⎝ x ⎠a < x 1 < b⇒ x 1 2 = ab ⇒ x 1 =45.[C]∫cot 4 x dx =∫cot 2 x (cosec 2 x –1)dxab=∫cot 2 x cosec 2 x dx –∫(cosec x − 1)dx= – 31 cot 3 x + cotx + x + c∴f(x) = – 31 cot 3 x + cotx + x + c + 31 cot 3 x – cotx= x + c2XtraEdge for IIT-JEE 120APRIL 2010

8.[A]9.[A]10.[B]To take off :irrelevantTo be indifferent :Irrelevanttook to : (to be accustomed to/to be addicted to)Correct because it suits to the sentence whenGandhi Ji was addicted to smoking.took for : (to be mistaken while recognising)irrelevanttook in : (to deceive someone)irrelevanttook up : (to adopt)irrelevantGet someone to break the box.Correct answer because :• The given sentence is in Passive Voice whichrequires its answer in Active Voice.• The given verb is Causatives in Imperative.• This option is Causative Active Voice inimperative form.They have broken the box.Incorrect Answer : Because• The given verb is not causative.• The given sentence is not imperative.Have the broken box.Incorrect : Because• The verb is not causatives.• 'Broken' has been used as an adjective in thisoption.Break the boxIncorrect because :• The subject 'you' (implied) wan't break it butwill get someone to break it.He asked how shabby I was looking.(Incorrect option) because :The required answer (type of sentence) iswrong.He exclaimed with disgust that I was lookingvery shabby.(Correct answer because)• This option is exclamatory.• Past Indefinite Tense has been used.• The mood of the speaker is correct.He exclaimed with sorrow that they werelooking much shabby.(Incorrect option because) :• Mood of exclamatory sentence is wrong.• 'much' will be replaced with 'very'• 'They' won't be used as a singular subject isrequired.He told that I was looking much shabby.Incorrect answer because –• Type of sentence is assertive whereas therequired type is 'exclamatory'.• 'much' is to be replaced with 'very'.11.[C] Neigh : Correct spelling as it means 'the cry ofhorse'.Reign : Correct spelling as it means 'thecontrolling chord of an animal.'Niece : Incorrect spelling as the correct one is'niece'. (Opposite of 'Nephew').Neither : Correct spelling. It is a conjunction tobe used with 'nor' for one of two options.12.[C] I wonder :No error in it.What he has done with the book.No errorI lend him(Erroneous) because there is an error of'Tenses'. The word 'lend' is to be 'lent'.No error :There is an error.13.[A] Distraught, awry :Correct answer : 'Distraught' means to 'getupset' and 'awry' means in 'disorder'.Frustrated, Magnificently :Both are opposite. One is positive and the otherone is negative. Therefore, no meaningfulsentence.Elated, Wild :No co-ordination, therefore incorrect answer.Dejected, splendidly :No co-ordination, therefore incorrect answer.14.[D] Interesting :" . . . and only a few were . . . . " phrase showsthat something opposite is required here. Thegiven option is not opposite to 'trivial'. Hence,Irrelevant option.Practical :Like aforesaid logic, this option also isirrelevant.Complex :'Complex' can't be opposite to 'trivial'.Therefore, can't be relevant.Significant :This is the relevant option making the sentencequite meaningful with two contradictory words,i.e. trivial and 'Significant connected with thephrase ". . . . . and only a few were . . . . . "15.[A] Paths, grave :It is a meaningful pair of words to make thesentence idiomatically correct.Ways, happiness :Not a meaningful pair. Hence, irrelevant option.Acts, prosperity :Not a meaningful pair. Hence, irrelevant option.Achievements, Suffering :Not a meaningful pair. Hence, irrelevant option.Hence,inappropriateXtraEdge for IIT-JEE 122APRIL 2010

XtraEdge for IIT-JEE 123APRIL 2010

CAREER POINTIn associationwithLaunches Online Test Series forBIT-SAT[ For Admission to BITS Pilani, Hyderabad, Goa ]Please fill the following form to enrollApplication FormStudent Name : __________________________________________________Father’s Name : __________________________________________________Current Institute/Batch : __________________________________________________Paste yourRecentPassport sizePhotographCurrent Class : __________________________________________________Date of Birth : __________________________________________________Address (Local): __________________________________________________________________________________________________________________________________________________City : ___________________________________________ Pin : _________________________________________Email : ________________________________________________________________________________________Are you already a user of a2zExam.com? Yes/No (Mention User Name, if known_________________________________Send the above filled form toBIT SATC/o <strong>Career</strong> <strong>Point</strong>, CP Tower, Road No.1, IPIA, Kota Rajasthan – 324005Phone : 0744-3040000, Fax : 0744 -3040050XtraEdge for IIT-JEE 124APRIL 2010