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WORRY IS A MISUSE OF IMAGINATION. Volume - 7 Issue - 6 December, 2011 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) Tel. : 0744-2500492, 2500692, 3040000 e-mail : xtraedge@gmail.com Editor : Editorial Dear Students, Pramod Maheshwari [B.Tech. IIT-Delhi] • It’s the question you dreamed about when you were ten years old. • It’s the question your parents nagged you about during high school. Cover Design • It’s the question that stresses most of us out more and more the older we Satyanarayan Saini get. “What do you want to be when you grow up?” Layout There are people who are studying political science but hate politics, nursing majors who hate biology, and accounting majors who hate math. Rajaram Gocher Obviously, a lot of people are confused about what exactly it is that they want to spend their life doing. Think about it. If you work for 10 hours each Circulation & Advertisement day, you’re going to end up spending over 50% of your awake life at work. Praveen Chandna Personally, I think it’s important that we spend that 50% wisely. But how Ph 0744-3040000, 9672977502 can you make sure that you do? Here are some cool tips for how to decide what you really want to be when you grow up. Subscription Relax and Keep an Open Mind: Contrary to popular belief, you don’t have to “choose a career” and stick with it for the rest of your life. You Himanshu Shukla Ph. 0744-3040000 never have to sign a contract that says, “I agree to force myself to do this © Strictly reserved with the publishers for the rest of my life”. It’s your life. You’re free to do whatever you want and the possibilities are endless. So relax, dream big, and keep an • No Portion of the magazine can be open mind. published/ reproduced without the Notice Your Passions: Every one of us is born with an innate desire to do written permission of the publisher something purposeful with our lives. We long to do something that we’re • All disputes are subject to the passionate about; something that will make a meaningful impact on the exclusive jurisdiction of the Kota world. Courts only. Figure Out How to Use Your Passions for a Larger Purpose: You notice that this is one of your passions, so you decide to become a personal Every effort has been made to avoid errors or trainer. Making a positive impact on the world will not only ensure that omission in this publication. Inr spite of this, you are successful financially, it will also make you feel wonderful. It’s a errors are possible. Any mistake, error or proven principle: The more you give to the world, the more the world discrepancy noted may be brought to our will give you in return. notice which shall be taken care of in the Figure Out How You Can Benefit Once you’ve figured out what your forthcoming edition, hence any suggestion is passions are and how you can use those passions to add value to the welcome. It is notified that neither the world &to yourself , publisher nor the author or seller will be It’s time to take the last step: figure out how you can make great success responsible for any damage or loss of action to doing it. My most important piece of advice about this last step is to any one, of any kind, in any manner, there from. remember just that: it’s the last part of the decision process. I feel sorry for people who choose an occupation based on the average income for Unit Price ` 20/- that field. No amount of money can compensate for a life wasted at a job Special Subscription Rates that makes you miserable. However, that’s not to say that the money isn’t important. Money is important, and I’m a firm believer in the concept 6 issues : ` 100 /- [One issue free ] that no matter what it is that you love doing, there’s at least one way to 12 issues : ` 200 /- [Two issues free] make extraordinary money doing it. So be creative! 24 issues : ` 400 /- [Four issues free] Simply discover your passions, figure out how to use your passions to make an impact on the world & to yourself. Owned & Published by Pramod Maheshwari, 112, Shakti Nagar, Yours truly Dadabari, Kota & Printed by Naval Maheshwari, Published & Printed at 112, Shakti Nagar, Dadabari, Kota. Editor : Pramod Maheshwari Pramod Maheshwari, B.Tech., IIT Delhi XtraEdge for IIT-JEE 1 DECEMBER 2011 Volume-7 Issue-6 December, 2011 (Monthly Magazine) CONTENTS INDEX PAGE NEXT MONTHS ATTRACTIONS Much more IIT-JEE News. Regulars .......... Know IIT-JEE With 15 Best Questions of IIT-JEE NEWS ARTICLE 3 Challenging Problems in Physics,, Chemistry & Maths IIT-B'S Plan : Job support to Entrepreneurial Spirit IIT Students' Research Work to Be Available Online Key Concepts & Problem Solving strategy for IIT-JEE. Xtra Edge Test Series for JEE- 2011 & 2012 IITian ON THE PATH OF SUCCESS 6 Mr. Amitabha Ghosh CBSE Mock Test Paper KNOW IIT-JEE 7 Previous IIT-JEE Question Study Time........ DYNAMIC PHYSICS 15 8-Challenging Problems [Set# 8] S Students’ Forum Success Tips for the Months Physics Fundamentals Ray Optics Fluid Mechanics & Properties of Matter • "The way to succeed is to double your error rate." CATALYSE CHEMISTRY 34 • "Success is the ability to go from failure to failure without losing your enthusiasm." Key Concept Carboxylic Acid • "Success is the maximum utilization of the Chemical Kinetics ability that you have." Understanding : Inorganic Chemistry • We are all motivated by a keen desire for praise, and the better a man is, the more he DICEY MATHS 45 is inspired to glory. Mathematical Challenges • Along with success comes a reputation for Students’ Forum wisdom. Key Concept Monotonicity, • They can, because they think they can. Maxima & Minima Function • Nothing can stop the man with the right mental attitude from achieving his goal; Test Time .......... nothing on earth can help the man with the wrong mental attitude. XTRAEDGE TEST SERIES 58 • Keep steadily before you the fact that all Class XII – IIT-JEE 2012 Paper true success depends at last upon yourself. Class XI – IIT-JEE 2013 Paper Mock Test CBSE Pattern Paper -1 [Class # XII] XtraEdge for IIT-JEE 2 DECEMBER 2011 IIT-B'S Plan : Job support to IIT Students' Research Work to witnessed tremendous competition Entrepreneurial Spirit Be Available Online between the energy of the present generation and that of the alumni. The IIT-Bombay has hit upon an idea Students of Indian Institutes of event was full of activities comprising that could boost the spirit of Technology (IITs) will soon share of talks, alumni -students matches and entrepreneurship among its students. their research details for all institutes Inter-Sports Arm Wrestling and Tug of Its placement cell is weighing the via a common website to be launched war competition which was followed option of helping students whose in November. by dinner and group discussions in start-ups have not fired to be placed The two day long 'Inter-IIT Gymkhana SAC Lawns. in jobs after two years of Summit 2011' being held at IIT- experimenting with their ideas. The post match fundae sessions (be it Gandhinagar saw students of various related to the life post IITs or the As part of the scheme, students keen consortium discuss issues common to prospects and benefits of sports in on their own start-ups will be assigned all. The students included everyday life) were undoubtedly the mentors after graduating. These representatives from 10 IITs including highlight of the complete event. The experts-either people who have Madras, Mandi, Rajasthan, Mandi, commitment and dedication shown successfully floated their own Hyderabad, with IIT Kanpur and IIT by the Old Machine Guns from IITB companies or those with enough Bombay participating in online left the students sports fraternity exposure to new businesses-will sessions. speechless. evaluate their ideas to see if there's Among the things discussed, students any potential. Once the ideas are particularly discussed methods to share Alumni showed the perfect blend of approved, students can float their own research works of B.Tech, M.Tech and attitude, experience and sporting companies. After two years, if a start-up Phd students. Students are seen to spirit with some exceptional fails to take off, the student-entrepreneur repeat researches, being unaware of performances in all the sports. can participate in the regular placement past researchers. Sharing of past Results of this year’s meet like last process and get a job. researches will help avoid such year were pretty much dominated by repetitions. the alumni, winning in almost all the Ravi Sinha, professor in charge of sports. placements, said the idea, which is at a Sport's Alumni Student Meet nascent stage, can give students the IITians' Nano Satellite Launched The 2nd official annual Alumni assurance to float their ideas without IIT-Kanpur has made the 3kg Sports Day was full of tremendous hesitation. "Very few start-up ideas on satellite 'Jugnu'. It will generate competition between the energy of the campus turn out to be successful real time data on droughts, floods, the present generation and that of the ventures. Often, many good ideas are vegetation and forestation. alumni. The event was full of not commercially-viable. So, students activities comprising of talks, alumni Jugnu, the Nano satellite designed by are apprehensive about floating their -students matches and Inter-Sports IIT Kanpur students has been ideas," added Sinha. Arm Wrestling and Tug of war launched via PSLV C-18 rocket from The office of Society for Innovation competition which was followed by Shrihari Kota, Andhra Pradesh 11:00 and Entrepreneurship ( SINE), which dinner and group discussions in SAC hrs today, reports TNN. This is the promotes entrepreneurship on the Lawns. country's first Nano satellite and campus by promoting business weighs just 3 Kgs incorporated with “Good, better, best. Never let it rest. incubation, and the Entrepreneurship PSLV C-18. Until your good is better and your cell (E-cell) is working with the better is best. ” this quote perfectly fit The mission of developing Jugnu placement cell to check the feasibility the enthralling display of experience, was done under the guidance of Prof. of the project. "We have been working wit and never say die attitude of the N.S vyas. this is extremely a on the initiative for a couple of alumni at this year’s Alumni Sports complex process and this was a hard months. If it works out, student- Day. exercise. The Nano satellite has been entrepreneurs will be reasonably assured of getting a job through the Brainchild of the basketball fraternity of designed ethnically. institute's placement office," said IIT Bombay, this was the 2nd official "We have done our best. Jugnu has Sinha. annual Alumni Sports Day which undergone laborious tests like thermo XtraEdge for IIT-JEE 3 DECEMBER 2011 vacuum tests, vibration tests and arbitrary or unreasonable or may be made at a meeting between electronic magnetic interference test. discriminatory,” Justice Raveendran senior HRD ministry officials and Students have worked very hard in said. the IIT directors on May 5, the developing Jugnu. They have worked The apex court passed the judgment sources said. round-the-clock. We are hopeful of its while dismissing an appeal filed by The Council alone is empowered to successful launch," said Vyas. aspirant Sanchit Bansal, son of an IIT appoint IIT directors and take any The payload of the satellite has Kharagpur professor, who appeared in policy decisions binding across the uniquely designed camera for infra red IIT-JEE 2006 as a general category top engineering schools -- including remote sensing, a GPS receiver and an candidate. their fees, reforms, administrative inertial measurement unit. structure and any amendments to the Sanchit had secured 75 marks in Shatanu Agarwal, the team head of mathematics, 104 in physics and 52 in aw governing the Institutes. Jugnu said "About 25 minutes after chemistry, aggregating to 231. “The IIT Council -the highest the launch, Jugnu would will click the decision making body of the IITs -at The board had fixed the cut-off marks first image of its launch vehicle and its for admission at 37 for math, 48 for present depends on the HRD antennas would be deployed. It will physics and 55 for chemistry and the ministry to even invite members for then stabilize and start transmitting meetings or to prepare the Council's aggregate cut-off at 154. images to the ground station." agenda. All this will change,“ a As Sanchit did not secure the minimum source said. After 115 minutes of the take off cut-off in chemistry, he failed to qualify Jugnu was visible at the ground even though his aggregate was higher The IITs have decided to approach D station. Udaya Kumar, assistant professor at than required. "Jugnu's beacon (blinking signal at all IIT Guwahati and the designer of the He then challenged the procedure on new rupee symbol, to design a logo times all over the earth) will get the ground that candidates with for the IIT Council. switched on after 30 minutes of aggregates lower than his were separation from the launch vehicle. The move is a component of HRD selected. Five seconds after its separation, minister Kapil Sibal's larger plan to Jugnu will click its image," said Rejecting his plea, the court said: “For increase the functional autonomy of Shantanu. a layman, the above procedure may the IITs. The IIT Council -- appear to be highly cumbersome and consisting of all IIT directors and It took two years to develop this Nano complicated. chairman of boards, other eminent satellite and as people working on it But the object of the aforesaid academic administrators and used non space grade, commercial off the shelf (COTS) to make the research procedure for arriving at the cut-off scientists -- at present does not even marks is to select candidates well have an office of its own. It will now on Jugnu affordable. Jugnu has minimum number of redundant equipped in all the three subjects, with have an office - a location has been systems when compared to other reference to their merit, weighed identified in South Delhi's against the average merit of all the Chittaranjan Park. conventional satellites. candidates who appeared in the But this may only be the first step SC refuses to adjudicate on IIT- examination.” towards greater autonomy for the JEE selections IITs, sources indicated. “Once the IIT Council free from HRD NEW DELHI: The Supreme Court has independent IIT Council is capable ministry clutches refused to interfere with the ranking of handling matters, there is a and selection procedure adopted for NEW DELH: The apex decision possibility that we will empower it the IIT-JEE saying there was no making body of the Indian Institutes of with far greater powers and withdraw arbitrariness or ulterior motives in Technology (IITs) has broken free of from many administrative aspects of fixing the methodology, says a PTI the Human Resource Development the IIT governance system,“ a source report. (HRD) Ministry in a move that could said. be the first step towards allowing the (Courtesy : Hindustan Times) A bench of Justices R.V. Raveendran IITs to govern themselves. and A.K. Patnaik said courts would No permanent foreign faculty interfere with the procedure only if Empowered by independent staff and for IITs there was proven malafide, caprice or with a identity of its own, the IIT arbitrariness, which it said was Council will now no longer need the NEW DELHI: In a major setback to lacking in the present system adopted HRD ministry to take its administrative the Indian Institutes of Technology decisions under the move, government (IITs) plan, the Ministry for External by the Joint Admission Board, which conducts the exams. sources told HT. Affairs (MEA) has rejected a proposal to liberalise visa norms to “The fact that the procedure was The final decision on the plan -- aimed allow foreign teachers to take up complicated would not make it at creating an IIT Council Secretariat -- permanent posts at the IITs. XtraEdge for IIT-JEE 4 DECEMBER 2011 The MEA has refused to change the fold increase in fee for undergraduate “This will ensure timely selection of rules under which each foreign faculty programme of the IITs. professors,” the Council noted. member at the IITs needs to re-obtain It also decided that the appointment The Kakodkar committee, set up by the a work visa every five years, top government in October 2009 to study of directors should be through government and IIT sources have the roadmap for the autonomy and advertisements so that a wider base confirmed to HT. was created. future of the IITs, had recommended Human resource development minister that the fee be raised from Rs 50,000 to “It was decided that in principle Kapil Sibal had on September 11, Rs 2 to Rs 2.5 lakh per annum. approval may be granted for setting 2010 announced the plan to allow the As the committee report came for up an institute in Mauritius with the IITs to fill up to 10% of their discussion at the 42nd meeting of the help of the IITs,” an official said. permanent teaching posts with foreign IIT Council Sibal rejected the fee hike At the meeting, a presentation was faculty. proposal saying “such a hike would made on adopting cyber security as The proposal -first reported by HT on prove a deterrent to a large number of part of the curriculum for the IITs. September 2, 2010 -was approved by IIT aspirants,” a ministry official So it was decided that a committee the IIT Council -the highest decision said. The Council asked the committee be set up to develop a roadmap for making body of the IITs -and is aimed to rework the fee structure taking into the future and give a report in next at reducing a massive faculty crunch account the aspirations of all sections. three months. plaguing the IITs. During the meeting, Sibal announced “The committee would involve all But the MEA's refusal to allow setting up 50 research parks at a cost of educational institutions as well as foreign faculty to join with visas of Rs 200 crore during the 12th Five Year government departments,” the HRD longer duration than five years means Plan period. Ministry official said. that the IITs will not be able to offer Under the programme, industry will The meeting also could not discuss permanent posts to foreign faculty. undertake research on various subjects reform in the Joint Entrance “We will need to continue to offer with the support of experts from the Examination and curriculum as T contractual appointments something IITs. Ramaswamy, secretary, Department we wanted to, and quite frankly, need The research parks have been proposed of Science & Technology, was not to change,“ an IIT Director said. to be set up on public-private- present. Ramaswamy had prepared a partnership (PPP) model. One such report on the two issues. Each IIT is facing a faculty crunch research park has already come up in "The Indian Statistics Institute has between 15 and 40% with a total of Chennai. arrived at a formula for equalising over 1,000 faculty posts vacant across the premier engineering schools. The The meeting took note of the fact that marks in all boards. If one board Institutes have over the past year credit-based practices were being gives 90 percent as highest marks, however received a number of followed by the IITs to promote and the other gives 75 percent, the applications from foreign faculty, students from one semester to the next, marks will be equalized on the basis including Persons of Indian Origin and agreed that academic bodies of the of a formula," Sibal said. (PIOs) keen to teach at the IITs. The IITs should consider acquisition of Producing more research scho- IITs are arguing that permanent posts credits as a criteria for students and lars was one of the key issues taken would help them lure the best of granting of degrees to bring uniformity. up during the meeting, with the foreign teachers. The issue came up following submission council deciding to enhance the All foreign teachers are at present of Dhande committee report on uniform capacity of IITs to produce 10,000 and homogeneous criteria for promoting Ph.D. graduates annually from required to teach as visiting or ad-hoc faculty. students in the IITs. around 1,000 presently and increase (Courtesy : Hindustan Times) Kanpur IIT director Dhande, who faculty strength from around 4,000 headed two committees, presented presently to 16,000 by 2020. IANS Sibal rejects steep fee hike for reports on a “uniform criteria for IIT students promoting students from one semester Science Fact:- NEW DELHI: There will be no steep to the next in the IITs and on the fee hike for the students of the Indian “requirement of infrastructure for • The brain uses over a quarter Institutes of Technology, according to research”. Both reports have been of the oxygen used by the a decision taken by the IIT Council on accepted. Each IIT at present has its human body. January 21. own criteria for promotion. The Council decided that a panel for • Your heart beats around Chairing the IIT Council meet here, 100000 times a day, Human Resource Development visitor’s nominee for a particular department would be created which all 36500000 times a year and Minister Kapil Sibal rejected the Anil over a billion times if you Kakodkar committee proposal for five- IITs could use for faculty selection. live beyond 30. XtraEdge for IIT-JEE 5 DECEMBER 2011 Success Story This article contains storie/interviews of persons who succeed after graduation from different IITs Dr. Amitabha Ghosh • Post graduation in applied geology from IIT Kharagpur, • Working at NASA Dr Amitabha Ghosh was the only Asian on NASA's How India can we develop science and technology Mars Pathfinder mission. At present, he is a member of the sector : Mars Odyssey Mission and the Mars Exploration Rover It should be treated as a business. There should be more Mission. private participation. We must have an external review to During the Mars Pathfinder Mission, he conducted evaluate the system and make changes as science and chemical analysis of rocks and soil on the landing site. The technology can take the country forward. simple and unassuming 34-year-old planetary geologist We must check brain drain. About 80,000 students migrate has won several accolades, which include the NASA Mars to the US for further studies, and settle there. They find the Pathfinder Achievement Award in 1997 and the NASA facilities much better abroad. We need to reverse brain Mars Exploration Rover Achievement Award in 2004. drain by enhancing and upgrading institutes in India. The journey from India to NASA. The state of space research in India : It has been an intriguing experience. I was keen on I don't want to make controversial statements. All I can say geologic research data interpretation and solar system is India is not at the frontier of space research. We have formation. During my geological research days in India, I made commendable progress but there is a long way to go. had slept in railway stations while traveling to various We can do much better. I would be glad to be of help in places. any way. Investment in research is investment in After my post graduation in applied geology from IIT imagination. It is a matter of national pride and internal Kharagpur, I wrote a letter to a professor at NASA recognition. We need to allocate more funds to enhance expressing a desire to work at the space agency. research and development work. I made certain suggestions; in fact, it was a critical letter. We need good educational institutes like IITs and IIMs, In India, you can never imagine criticising your professor. but IITians don't rule the world. You must remember that My suggestions were approved, while I got an opportunity Microsoft co-founder (Bill Gates does not have a college to work at NASA. degree. I think one requires luck and to put in sincere effort to Youngsters must look around for role models and see what it is that they are doing right. Individuals must make use of achieve one's goals. Being in the right place at the right time is also important. their inherent strengths to succeed. In Mumbai for the Pravasi Bharatiya Divas, he spoke How can India become a leading global player : about his work at NASA and his vision for India. Globalisation will reap huge and long-term benefits and India must make the best use of the opportunities. At the PBD The Vision for India : seminar, I found people presenting grandiose plans. Instead, I feel there India has a great future. We have world-class we should look at the realities and immediate solutions. companies. Today, companies like Infosys can be compared with world leaders like Oracle. Like the The private sector has to be actively involved in the development of the country and the government has to Information Technology revolution, we can have a science or space revolution. We have the potential to bring about respond to the needs of the people. Fifteen years ago, we revolutions in other sectors as well. didn't have an Infosys, today we have many global companies. XtraEdge for IIT-JEE 6 DECEMBER 2011 KNOW IIT-JEE By Previous Exam Questions PHYSICS 2. A cubical box of side 1 meter contains Helium gas (atomic weight 4) at a pressure of 100 N/m2. 1. A non-viscous liquid of constant density 1000 kg/m3 During an observation time of 1 second, an atom flows in a streamline motion along a tube of variable travelling with the root-mean-square speed parallel cross section. The tube is kept inclined in the vertical to one of the edges of the cube, was found to make plane as shown in figure. The area of cross section of 500 hits with a particular wall, without any the tube two points P and Q at heights of 2 metres 25 and 5 metres are respectively 4 × 10–3 m2 and 8 × 10–3 collision with other atoms. Take R = J/mol-K m2. The velocity of the liquid at point P is 1 m/s. Find 3 the work doen per unit volume by the pressure and and k = 1.38 × 10–23 J/K [IIT-2002] the gravity forces at the fluid flows from point P to Q. (a) Evaluate the temperature of the gas. [IIT-1997] (b) Evaluate the average kinetic energy per atom. (c) Evaluate the total mass of helium gas in the Q box. Sol. The distance travelled by an atom of helium in 1 sec (times between two successive collision) P 500 5m is 2m. Therefore root mean square speed 2m Sol. Given that ρ = 1000 Kg/m3, h1 = 2 m, h2 = 5 m A1 = 4 × 10–3 m2 , A2 = 8 × 10–3 m2 . v1 = 1 m/s 1m Equation of continuity dis tan ce 2 A1 = 4 × 10–3 m2, A2 = 8 × 10–3 m2 , v1 = 1 m/s Crms = = = 1000 m/s time 1 / 500 Equation of continuity A1v1 3RT A1v1 = A1v2, ∴ v2 = = 0.5 m/s (a) But Crms = A2 M According to Bernouilli's theorem, 3 × 25 / 3 × T ⇒ 1000 = ⇒ T = 160 K 1 4 × 10 – 3 (p1 – p2) = ρg (h2 – h1) – ρ (v22 – v12) 2 (b) Average kinetic energy of an atom of a where (p1 – p2) = work done/vol. [by the pressure] 3 monoatomic gas = kT ρg (h2 – h1) = work done/vol. [by gravity forces.] 2 Now, work done/vol by gravity forces 3 ∴ Eav = kT = 3.312 × 10–12 Joules = ρg (h2 – h1) = 103 × 9.8 × 3 = 29.4 × 103 J/m3 2 and m 1 1 1 3 (c) From gas equation PV = RT ρ(v22 – v12) = × 103 – 1 = – × 103 M 2 2 4 8 ⇒ m = 0.3012 gm J/m3 = – 0.375 × 103 J/m3 ∴ Work done/vol. by pressure = 29.4 × 103 – 0.375 × 103 J/m3 = 29.025 × 103 m3. XtraEdge for IIT-JEE 7 DECEMBER 2011 3. A thin equiconvex lens of glass of refractive index 1 2 µ = 3/2 and of focal length 0.3 m in air is sealed into ⇒ ⇒ R = 0.3 m = (1.5 – 1) 0.3 R an opening at one end of a tank filled with water (µ = 4/3). On the opposite side of the lens, a mirror is Substituting the value of R in equation (iii) we get placed inside the tank on the tank wall perpendicular 4 3 – –1+ 2× to the lens axis, as shown in figure. The separation 1 4/3 3 2 – + = between the lens and the mirror is 0.8 m. A small (0.9) v 0.3 object is placed outside the tank in front of the lens at a distance of 0.9 m from the lens along its axis. Find –4 – 3 + 9 the position (relative to the lens) of the image of the 1 4 3 2 + = = object formed by the system. [IIT-1997] 0.9 3v 0.3 0.9 0.9 m 0.8 m ⇒ v = 0.2 m As shown in the figure this image will form as I" behind the mirror. But the ray will get reflected from the mirror in between and the final image formed will be I. Since CI" = 0.2 m and CM = 0.8 m Sol. ∴ MI" = 0.2 m v' ⇒ MI = 0.4 m [Q MI = MI"] ∴ CI = 0.4 m v 4. A 5 m long cylindrical steel wire with radius 2 × 10–3 m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If O C I M I'' I' the bob gets snapped, calculate the change in temperature of the wire ignoring radiation losses. u v (For the steel wire : Young's modulus = 2.1 × 1011 Pa ; Density = 7860 kg/m3 ; Specific heat = 420 The first refraction of the ray coming form the object J/kg-K). [IIT-2001] is suffered on the left side of convex lens. For this we Sol. When the mass of 100 Kg is attached, the string is can apply the equation under tension and hence in the deformed state. µg µg – µa Therefore it has potential energy (U) which is µ – a + = ...(i) given by the formula. u V' R 1 The image formed by this can be treated as a U= × stress × strain × volume virtual object and the refracting surface now is the 2 right of the convex lens. 1 (Stress) 2 = × × πr2l µg µ µw – µg 2 Y – + w = ...(ii) v' v –R 1 (Mg / πr 2 ) 2 1 M 2g 2l Adding (i) and (ii) = × πr2l = ...(i) 2 Y 2 πr 2 Y µa µg µg µw µg – µa – + – + = + This energy is released in the form of heat, thereby u v' v' v R raising the temperature of the wire µw – µg Q = mc∆T –R ...(ii) µa µ –µ w – µ a + 2µ g From (i) and (iii) Since U = Q Therefore ⇒ – + w = u v R 1 M 2g 2l ∴ mc∆T = But according to lens formula 2 πr 2 Y 1 f = ( µ – 1) R a g 1 – 1 R2 ∴ ∆T = 1 M 2g 2l 1 2 πr 2 Ycm ⇒ 1 f = ( µ – 1) R1 a g – 1 R2 Here m = mass of string = density × volume of string 1 = ρ × πr2l XtraEdge for IIT-JEE 8 DECEMBER 2011 1 M 2g 2 c c ∴ ∆T = ⇒ r1 = ⇒ r2 = ⇒ r 1 = r2 2 (πr 2 ) 2 Ycp 2 2 ⇒ R2 gives the most accurate value 1 (100 × 10) 2 = × 2 –3 2 (3.14 × 2 × 10 ) × 2.1× 1011 × 420 × 7860 = 0.00457º C CHEMISTRY 5. An unknown resistance X is to be determined using resistance R1, R2 or R3. Their corresponding 6. From the following data, form the reaction between null points are A, B and C. Find which of the A and B. [IIT-1994] above will give the most accurate reading and why? [IIT-2005] [A] [B] Initial rate (mol L–1s–1) mol L–1 mol L–1 300 K 320 K 2.5 ×10–4 3.0 ×10–5 5.0 ×10 –4 2.0 × 10–3 5.0 × 10–4 6.0 × 10–5 4.0 × 10–3 – 1.0 × 10–3 6.0 × 10–5 1.6 × 10 –2 – X R R = R1 or R2 or R3 Calculate (a) the order of reaction with respect to A and with G respect to B, A B C (b) the rate constant at 300 K, (c) the energy of activation, Sol. All Null point, the wheat stone bridge will be (d) the pre exponential factor. balanced Sol. Rate of reaction = k[A]l [B]m X R ∴ = where l and m are the order of reaction with respect r1 r2 to A and B respectively. From the given data, we r1 obtain following expressions : ⇒X=R r2 5.0 × 10–4 = k[2.5 × 10–4]l [3.0 × 10–5]m ..(i) where R is a constant r1 and r2 are variable. The maximum fraction error is 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 get l X R 4.0 × 10 −3 5.0 × 10 −4 = 1.0 × 10 −3 G 1.6 × 10 − 2 r1 r2 M N or 0.25 = (0.5)l A B C or (0.5)2 = (0.5) l R=R1 R=R2 R=R3 or l=2 ∆X ∆r ∆r From eq. (i) and eq. (ii), we get = 1 + 2 X r1 r2 2 m 5.0 × 10 −4 2.5 × 10 −4 3.0 × 10 −5 = 5.0 × 10 − 4 6.0 × 10 −5 Here ∆r1 = ∆r2 = y (say) then 4.0 × 10 −3 ∆X For to be minimum r1 × r2 should be max 1 1 1 m X or = × 8 4 2 [Q r1 + r2 = c (Constt.)] m Let E = r1 × r2 1 1 or = ⇒ E = r1 × (r1 – c) 2 2 dE or m=1 ∴ = (r1 – c) + r1 = 0 dr1 XtraEdge for IIT-JEE 9 DECEMBER 2011 (b) At T1 = 300 K, Hence, difference in energy between first and Rate of reaction second Bohr orbit for a H-atom is given by, k1 = ∆E = E n i – E n f = E2 – E1 [A]2 [B]1 5.0 × 10 −4 21.76 × 10 −19 21.76 × 10 −19 = =– + [2.5 × 10 − 4 ]2 [3.0 × 10 −5 ] 22 12 = 2.67 × 108 L2 mol–2 s–1 1 1 = – 21.76 × 10–19 2 − 2 = 16.32 ×10–19J 2 1 –8 (c) At T2 = 320 K, (b) For λ = 3.0 × 10 m Rate of reaction hc 6.626 × 10 −34 × 3 × 108 k2 = ∆E = = [A]2 [B]1 λ 3.0 × 10 −8 2.0 × 10 −3 = 6.626 × 10–18 J ....(i) = We know that, for H-like atoms, [2.5 × 10 − 4 ]2 [3.0 × 10 −5 ] En for H-like atom = En for H-atom × Z2 = 1.067 × 109 L2 mol–2 s–1 ∴ ∆E for H-like atom = Z2 × ∆E for H-atom k2 E T −T 1 1 We know, 2.303 log = a 2 1 = –Z2 × 21.76 × 10–19 2 − 2 k1 R T1T2 2 1 1.067 × 109 E a 320 − 300 = 16.32 × 10–19 Z2 ...(ii) or 2.303 log = From eq. (i) and (ii), 2.67 × 108 8.314 320 × 300 16.32 × 10–19 Z2 = 6.626 × 10–18 Ea 20 or Z=2 or 2.303 × 0.6017 = 8.314 320 × 300 Thus, hydrogen atom like species for Z = 2 is He+. 2.303 × 0.6017 × 8.314 × 320 × 300 8. An organic compound A, C8H4O3, in dry benzene or Ea = 20 in the presence of anhydrous AlCl3 gives compound = 55.3 kJ mol –1 B. The compound B on treatment with PCl5 (d) According to Arrhenius equation, followed by reaction with H2/Pd(BaSO4) gives compound C, which on reaction with hydrazine k = Ae − E a / RT gives a cyclised compound D(C14H10N2). Identify Ea A, B, C and D. Explain the formation of D from C. or 2.303 log k = 2.303 log A – RT [IIT-2000] At 300 K, Sol. The given reactions are as follows. 55.3 × 10 3 O O 2.303 log (2.67 × 108) = 2.303 log A – 8.314 × 300 AlCl3 O + or 2.303 × 8.4265 = 2.303 log A – 22.17 OH 19.4062 + 22.17 41.5762 O O or logA = = = 18.0531 2.303 2.303 C6H5 C6H5 A = Antilog 18.0531 = 1.13 × 1018 s–1 C O H2NNH2 PCl5 N 7. Estimate the difference in energy between 1st and O H2/Pd (BaSO4) C 2nd Bohr orbit for a H atom. At what minimum N atomic no., a transition from n=2 to n = 1 energy H level would result in the emission of X-rays with The formation of D from C may be explained as λ = 3.0×10–8 m. Which hydrogen atom like species follows. does this atomic no. corresponds to ? [IIT-1993] C6H5 O– O– Sol. (a) For H atom, C6H5 C6H5 + Z=1 O NH2 NH2 N–H O + ni = 2 NH2 NH2 N–H nf = 1 O– OH 21.76 × 10 −19 C6H5 En = – 2 J n N N XtraEdge for IIT-JEE 10 DECEMBER 2011 9. (a) Write the chemical reaction associated with the N≡C C≡N "brown ring test". 2+ (b) Draw the structures of [Co(NH3)6]3+, Ni [Ni(CN)4]2– and [Ni(CO)4]. Write the N≡C C≡N hybridization of atomic orbital of the transition metal in each case. (c) An aqueous blue coloured solution of a In [Ni(CO)4, nickel is present as Ni atom i.e. its transition metal sulphate reacts with H2S in oxidation number is zero and coordination number acidic medium to give a black precipitate A, is four. which is insoluble in warm aqueous solution of 3d 4s 4p KOH. The blue solution on treatment with KI Ni in in weakly acidic medium, turns yellow and Complex produces a white precipitate B. Identify the transition metal ion. Write the chemical sp3 hybridization reaction involved in the formation of A and B. Its structure is as follows : [IIT-2000] CO Sol. (a) NaNO3 + H2SO4 → NaHSO4 + HNO3 2HNO3 + 6FeSO4 + 3H2SO4 → 3Fe2(SO4)3 + 2NO + 4H2O Ni [Fe(H2O)6]SO4.H2O + NO OC CO Ferrous Sulphate → [Fe(H2O)5NO] SO4 + 2H2O (Brown ring) CO (b) In [Co(NH3)6]3+ cobalt is present as Co3+ and its (c) The transition metal is Cu2+. The compound is coordination number is six. CuSO4.5H2O Co27 = 1s1, 2s22p6, 3s23p63d7, 4s2 CuSO4 + H2S Acidic medium→ CuS ↓ + H2SO4 Co3+ion = 1s2, 2s22p6, 3s23p63d6 Black ppt 3d 4s 4p 2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4 Hence (B) white – – 3d 4s 4p I2 + I → I3 (yellow solution) Co3+ion in Complex ion 10. An organic compound A, C6H10O, on reaction with 2 3 CH3MgBr followed by acid treatment gives d sp hybridization compound B. The compound B on ozonolysis gives NH3 3+ NH3 compound C, which in presence of a base gives H3N NH3 H3N NH3 1-acetyl cyclopentene D. The compound B on reaction with HBr gives compound E. Write the Co or structures of A, B, C and E. Show how D is formed Co3+ from C. [IIT-2000] Sol. The given reactions are as follows. H3N NH3 NH3 H3N NH3 O OMgBr CH3 CH3 NH3 2– 2+ CH3 Br In [Ni(CN)4 nickel is present as Ni ion and its CH3MgBr H+ HBr coordination numbers is four –H2O Ni28 =1s2, 2s22p6, 3s23p63d8, 4s2 (A) (B) (E) Ni2+ ion = 1s2, 2s22p6, 3s23p63d8 3d 4s 4p 2+ Ni ion = COCH3 COCH3 CH3 O Base O O 3d 4s 4p Ni2+ion in (D) (C) Complex ion The conversion of C into D may involve the dsp2 hybridization following mechanism. Hence structure of [Ni(CN)4]2– is XtraEdge for IIT-JEE 11 DECEMBER 2011 COCH3 COCH3 COCH3 12. Solve for x the following equation [IIT-1987] 2 log(2x +3)(6x + 23x + 21) CH2 O HC O HC O– B+ BH+ = 4 – log(3x + 7) (4x2 + 12x + 9) –BH+ –B Sol. log(2x + 3) (6x2 + 23x + 21) (C) = 4 – log(3x + 7) (4x2 + 12x + 9) COCH3 ⇒ log(2x + 3) (2x + 3).(3x + 7) COCH3 COCH3 – = 4 – log(3x + 7) (2x + 3)2 OH +B OH –OH– ⇒ 1 + log(2x + 3) (3x + 7) –BH+ = 4 – 2 log(3x + 7) (2x + 3) (D) Put log(2x + 3) (3x + 7) = y 2 ⇒ MATHEMATICS y+ y –3=0 ⇒ y2 – 3y + 2 = 0 ⇒ (y – 1) (y – 2) = 0 11. Let the three digit numbers A28, 3B9, and 62C, where A, B and C are integers between 0 and 9, be ⇒ y = 1 or y=2 divisible by a fixed integer K. Show that the ⇒ log(2x +3) (3x + 7) = 1 determinant or log(2x + 3) (3x + 7) = 2 A 3 6 ⇒ 3x + 7 = 2x + 3 or (3x + 7) = (2x + 3)2 8 9 C is divisible by K. [IIT-1990] ⇒ x = – 4 or 3x + 7 = 4x2 + 12x + 9 2 B 2 4x2 + 9x + 2 = 0 4x2 + 8x + x + 2 = 0 Sol. We know, (4x + 1)(x + 2) = 0 A28 = A × 100 + 2 × 10 + 8 x = – 2, – 1/4 3B9 = 3 × 100 + B × 10 + 9 and 62C = 6 × 100 + 2 × 10 + C ∴ x = – 2, – 4, – 1/4 ...(i) 2 Since; A28, 3B9 and 62C are divisible by K, But, log exists only when, 6x + 23x + 21 > 0, therefore there exist positive integers m1, m2 and 4x2 + 12x + 9 > 0, m3 such that, 2x + 3 > 0 and 3x + 7 > 0 100 × A + 10 × 2 + 8 = m1K ⇒ x > – 3/2 ...(ii) 100 × 3 + 10 × B + 9 = m2K ∴ x = – 1/4 is the only solution. and, 100 × 6 + 10 × 2 + C = m3K ...(i) A 3 6 13. Find the equation of the normal to the curve ∴ ∆= 8 9 C y = (1 + x)y + sin–1 (sin2x) at x = 0. [IIT-1993] Sol. y = (1 + x)y + sin–1 (sin2x) (given) 2 B 2 Let y = u + v, where u = (1 + x)y, v = sin–1 (sin2 x). Applying R2 → 100R1 + 10R3 + R2 Differentiationg A 3 dy du dv = + ...(i) ⇒ ∆ = 100 A + 2 × 10 100 × 3 + 10 × B + 9 dx dx dx 2 B Now, u = (1 + x)y 6 take logarithm of both sides loge u = loge(1 + x)y 100 × 6 + 10 × 2 + C ⇒ loge u = y loge(1 + x) 2 1 du y dy A 3 6 ⇒ = + . {loge(1 + x)} u dx 1+ x dx = A28 3B9 62C , Using (i) du y dy 2 B 2 ⇒ =u + log e (1 + x) dx 1 + x dx A 3 6 A 3 6 du y dy = m1 K m2 K m3 K = K m1 m2 m3 ⇒ = (1 + x)y + log e (1 + x) ...(ii) dx 1 + x dx 2 B 2 2 B 2 Again, v = sin–1sin2 x ∴ ∆ = mK, Hence determinant is divisible by K. ⇒ sin v = sin2x XtraEdge for IIT-JEE 12 DECEMBER 2011 dv π/3 ⇒ dx cos v dx = 2 sin x cos x I = 2π ∫ 2 – cos( x + π / 3) 0 dv 1 ⇒ = [2sin x cos x] 2π / 3 dt π dx cos v 2 sin x cos x 2 sin x cos x = 2π π/3 ∫ 2 – cos t , where x + 3 =t ⇒ = = ...(iii) 1 – sin v 2 4 1 – sin x t 2π / 3 sec 2 dt Put these values in equation (i) = 2π ∫ 2 π / 3 1 + 3 tan 2 t dy y dy 2 sin x cos x 2 = (1 + x)y + log e (1 + x) + dx 1 + x dx 1 – sin 4 x 3 dy y (1 + x) y –1 + 2 sin x cos x / 1 – sin 4 x = 2π ∫ 2du 1 + 3u 2 = 4π 3 . { 3 tan –1 3u } 3 1 ⇒ = 1/ 3 3 dx 1 – (1 + x) y ln (1 + x) 4π 4π 1 At x = 0, = (tan–1 3 – tan–1 1) = tan–1 3 3 2 y = (1 + 0)y + sin–1 sin (0) = 1 π/3 π + 4 x3 4π 1 ⇒ dy 1(1 + 0)1–1 + 2 sin 0. cos 0 / (1 – sin 4 0) = ∴ ∫ π – π / 3 2 – cos | x | + dx = 3 tan –1 . 2 dx 1 – (1 + 0)1 ln (1 + 0) 3 dy ⇒ =1 dx 15. The position vectors of the vertices A, B and C of a Again the slope of the normal is tetrahedron ABCD are i + j + k, i and 3i, respectively. 1 The altitude from vertex D to the opposite face ABC m=– =–1 meets the median line through A of the triangle ABC dy / dx at a point E. If the length of the side AD is 4 and the Thus, the required equation of the normal is 2 2 y – 1 = (–1) (x – 0) volume of the tetrahedron is , find the position 2 i.e., y + x – 1 = 0. vector of the point E for all its possible positions. [IIT-1996] π/3 π + 4x3 i + 3i 14. Evaluate ∫ π – π / 3 2 – cos | x | + dx. [IIT-2004] Sol. F is mid-point of BC i.e., F = 2 = 2i and 3 AE ⊥ DE (given) A(i+j+k) Sol. D π/3 π/3 πdx x 3dx λ Let, I= ∫ π – π / 3 2 – cos | x | + + 4 ∫ π – π / 3 2 – cos | x | + E 3 3 1 a 0, f (– x) = – f ( x) a –a ∫ (Using f ( x)dx = ∫ 2 f ( x)dx, f (– x) = f ( x) B(i) F(2i) C(3i) 0 Let E divides AF in λ : 1. Then position vector of π/3 E is given by πdx ∴ I= 2 ∫ π +0 2 λi + 1(i + j + k ) = 2λ + 1 i+ 1 j+ 1 k 0 2 – cos | x | + λ +1 λ +1 λ +1 3 λ +1 Now, volume of the tetrahedron π/3 3 1 x dx as ∫ π – π / 3 2 – cos | x | + is odd = 3 (area of the base) (height) 3 2 2 1 ⇒ = (area of the ∆ ABC) (DE) 3 3 XtraEdge for IIT-JEE 13 DECEMBER 2011 1 → → But area of the ∆ ABC = | BC × BA | WHAT ARE EARTHQUAKES? 2 1 = |2i × (j + k)| = |i × j + i × k| 2 = |k – j| = 2 2 2 1 Therefore, = ( 2) (DE) 3 3 ⇒ DE = 2 Since ∆ ADE is a right angle triangle, AD2 = AE2 + DE2 ⇒ (4)2 = AE2 + (2)2 Earthquakes like hurricanes are not only super ⇒ AE2 = 12 destructive forces but continue to remain a mystery → 2λ + 1 1 1 But AE = i+ j+ k – (i + j + k) in terms of how to predict and anticipate them. To λ +1 λ +1 λ +1 understand the level of destruction associated with λ λ λ earthquakes you really need to look at some = i– j– k λ +1 λ +1 λ +1 examples of the past. → 1 3λ2 If we go back to the 27th July 1976 in Tangshan, ⇒ | AE | 2 = [λ2 + λ2 + λ2] = (λ + 1) 2 (λ + 1) 2 China, a huge earthquake racked up an official 3λ2 death toll of 255,000 people. In addition to this an Therefore, 12 = estimated 690,000 were also injured, whole (λ + 1) 2 families, industries and areas were wiped out in the ⇒ 4(λ + 1)2 = λ2 blink of a second. The scale of destruction is hard to ⇒ 4λ2 + 4 + 8λ = λ2 imagine but earthquakes of all scales continue to ⇒ 3λ2 + 8λ + 4 = 0 happen all the time. ⇒ 3λ2 + 6λ + 2λ + 4 = 0 ⇒ 3λ(λ + 2) + 2 (λ + 2) = 0 So what exactly are they ? Well the earths outer ⇒ (3λ + 2) (λ + 2) = 0 layer is made up of a thin crust divided into a ⇒ λ = – 2/3, λ = – 2 number of plates. The edges of these plates are Therefore, when λ = – 2/3, position vector of E is referred to as boundaries and it’s at these given by boundaries that the plates collide, slide and rub 2λ + 1 1 1 against each other. Over time when the pressure at i+ j+ k λ +1 λ +1 λ +1 the plate edges gets too much, something has to give which results in the sudden and often violent 2.(–2 / 3) + 1 1 1 = i+ j+ k tremblings we know as earthquakes. – 2 / 3 +1 – 2 / 3 +1 – 2 / 3 +1 –4 / 3 + 1 1 1 The strength of an earthquake is measured using a = i+ j+ k – 2+3 – 2+3 – 2+3 machine called a seismograph. It records the 3 3 3 trembling of the ground and scientists are able to –4 + 3 measure the exact power of the quake via a scale = 3 i+ 1 j+ 1 k known as the richter scale. The numbers range from 1/ 3 1/ 3 1/ 3 1-10 with 1 being a minor earthquake (happen = – i + 3j + 3k multiple times per day and in most case we don’t and when λ = – 2, even feel them) and 7-10 being the stronger quakes Position vector of E is given by, (happen around once every 10-20 years). There’s a 2 × (–2) + 1 1 1 –4 + 1 lot to learn about earthquakes so hopefully we’ll i+ j+ k= i–j–k – 2 +1 – 2 +1 – 2 +1 –1 release some more cool facts in the coming months. = 3i – j – k Therefore, – i + 3j + 3k and +3i – j – k are the answer. XtraEdge for IIT-JEE 14 DECEMBER 2011 Physics Challenging Problems Set # 8 This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants. By : Dev Sharma Solutions will be published in next issue Director Academics, Jodhpur Branch Passage # 1 (Que. 1 to 3) 4. When barium is irradiated by a light of λ = 4000Å all 1. In the circuit shown C1 = 6µF, C 2 = 3µF and battery the photoelectrons emitted are bent in a circle of radius 50cm by a magnetic field of 5.26 × 10-6 T. Then E = 20V. The switch S1 is first closed. It is then opened and afterwards S2 is closed them the final (A) the kinetic energy of fastest photoelectrons is 0.6eV charge on C2 is (B) work function of the metal is 2.5eV C2 (C) the maximum velocity of photoelectrons is 0.46 × 106 m/s C1 S2 (D) the stopping potential for photoelectric effect is 0.6V 5. Suppose the potential energy between on electron S1 ke 2 and a proton at a distance r is given by . E = 20 V 3r 3 (A) 120µC (B) 80µC Application of Bohr’s theory to hydrogen atom in this case shows that (C) 40µC (D) 20µC (A) Energy in nth orbit is proportional to n6 2. A current in a coil of self-inductance 2H is (B) Energy is proportional to m-3 (m = mass of electron) increasing as i = 2 sin( t 2 ) . The amount of energy (C) Energy in nth orbit is proportional to n-2 spent during the period when current changes from (D) Energy is proportional to m3 (m = mass of electron) 0 to 2A is (A) 1J (B) 2J (C) 3J (D) 4J Passage # (Q. No. 6 to Q. No. 8) 3. A particle of charge q and mass m is projected with An ideal gas is taken round a cyclic process ABCA a velocity v forwards a circular region having as shown. If the internal energy of the gas at point A uniform magnetic field B perpendicular and into the is assumed zero while at B it is 50J. The heat plane of paper from point P as shown. R is the absorbed by the gas in the process BC is 90J. radius and O is the centre of the circular region. If the line OP makes an angle θ with the direction of 30 C v then the value of v so that particle passes through O is 20 P(N/m2) A 10 B × × × × × D E × O× × O 1 2 3 × R × × q v × θ× × V(m)3 m P × × 6. Heat energy absorbed by the gas in process AB is qBR qBR (A) 50J (B) 70J (C)30J (D) None (A) (B) m sin θ 2m sin θ 7. Heat energy rejected by gas in process CA is 2qBR 3qBR (A) 180J (B) 140J (C) 220J (D) None (C) (D) m sin θ 2m sin θ 8. The net work done by gas in the complete cycle ABCA is (A) 20J (B) zero (C) 40J (D) -20J XtraEdge for IIT-JEE 15 DECEMBER 2011 8 Solution Set # 7 Physics Challenging Problems Qu e s tio ns we r e Pub lis he d in Nov emb er I ss ue 3R R B 0 cos r = A 0 (1 + cos φ + cos 2φ) 1. I AB = I and I DC = I 4R 4R B 0 sin r = (A 0 (sin φ + sin 2φ)) Option [C] is correct 2 2 ∴ B 0 = A 0 (1 + 4 cos φ + 4 cos 2 φ) 2. L.∆I = area under v – t curve ∴ I θ = I 0 (1 + 4 cos φ + 4 cos 2 φ) 1 ∴ (I θ ) max = 9I 0 2[i − 0] = × 10 x 2 2 Iθ 1 1 Option [A,C] is correct also = = (1 + 4 cos φ + 4 cos 2 φ) (I θ ) max 2 9 3. By conservation of energy ⇒ cos φ = 0.56 2 gR − GMm 1 1 GMm × m + = mV + 2 2 R 2 8. A → P; B → R ; C → Q; D → S 2 R R+ 4 5 5 GM v net = v0 = by conservation of angular momentum 4 4 r0 gR R −GMm 1 5 GM −3GMm m. .R sin θ = mV R + ∴ total energy = − + m× . = 2 4 r0 2 4 r0 8r0 Option [A] is correct when particle is at maximum or minimum distance r, then energy conservation 2q 2 1 1 1 1 1 GMm 3GMm 4. U= − + − + + ..... mv 2 − =− …….(1) 4π ∈0 d d 2d 3d 4d 2 r 8r0 Option [C] is correct Angular momentum conservation mvr 5 Passage Based Question: = m. v 0 .r0 …….(2) 4 5. Option [A] is correct 6. Option [D] is correct 7. Option [A] is correct Science Jokes Let A and w be the amplitude and angular frequency of A chemistry professor couldn't resist interjecting a the wave incident on the three slit grating. Let φ be little philosophy into a class lecture. He interrupted the phase difference between the diffracted waves his discussion on balancing chemical equations, from S1 and S2 and φ between those from S2 and S2. If saying, "Remember, if you're not part of the θ is angle of diffraction then using. solution, you're part of the precipitate!" 2πd sin θ 1. 1 couldn't resist interjecting a little philosophy φ= λ into a class lecture. He interrupted his A 1 = A 0 sin ωt discussion on balancing chemical equations, saying, "Remember, if you're not part of the A 2 = A 0 sin(ωt + φ) solution, you're part of the precipitate!". A 3 = A 0 sin(ωt + 2φ) 2. Q. What is volume of a person who lost all his A = A 1 + A 2 + A 3 = A 0 [sin ωt (1 + cos φ + cos 2φ) memory ? A. 1/3 πr2h + cos ωt (sin φ + sin 2 φ)] Because he keeps on saying, “main CONE A = B 0 sin(ωt + r ) hu!" where XtraEdge for IIT-JEE 16 DECEMBER 2011 XtraEdge for IIT-JEE 17 DECEMBER 2011 Students Forum Expert’s Solution for Question asked by IIT-JEE Aspirants PHYSICS 1. Oxygen is used as working substance in an engine Heat supplied to gas during the process, working on the cycle shown in Figure 7 Q23 = nCP(T3 – T2) = nRT0 2 2 3 Work done by gas during the process, W23 = nR(T3 – T2) = nRT0 P 4 Now considering adiabatic process 3 → 4, 1 V3 = 2V0 , T3 = 2T0 V4 = 5V0 , T4 = ? V γ–1 Processes 1-2, 2-3, 3-4 and 4-1 are isothermal, Using T.V = constant isobaric, adiabatic and isochoric, respectively. If (2T0) (2V0)γ – 1 = T4(5V0)γ – 1 ratio of maximum to minimum volume of oxygen or T4 = 2 (0.4)0.4 T0 during the cycle is 5 and that of maximum to Work done by the gas during the process, minimum absolute temperature is 2, assuming P V – P4 V4 nR (T3 – T4 ) oxygen to be an ideal gas, calculate efficiency of the W24 = 3 3 = engine. γ –1 γ –1 Given, (0.4)0.4 = 0.693 and loge 5 1.6094 = 5 nRT0 (1 – (0.4)0.4) T0 Sol. Volume of gas is minimum at state 2 during the During isochoric process 4 → 1, no work is done by cycle. Let it be V0. Then maximum volume of gas the gas and heat is rejected from the gas. during the cycle will be equal to 5V0 which is at Hence, W41 = 0 and Q41 is negative states 4 and 1. Therefore, V4 = V1 = 5V0. ∴ Net work done by the gas during the cycle, Temperature during the cycle is maximum at the end W = W12 + W23 + W34 + W41 = of isobaric process 2 → 3 i.e. state 3 and minimum nRT0 {6 – loge 5 – 5 × (0.4)0.4} at the end of isochoric cooling process 4 → 1 i.e. state 1. Let minimum absolute temperature be T0. Heat supplied to the gas during heating process, Then T1 = T0 and T3 = 2T0. 7 QS = Q23 = ( 6 – loge 5 – 5 × 0.40.4) = 0.2642 Since gas is Oxygen which is di-atomic, therefore, 2 5 7 7 W 2 Cv = R, Cp = R and γ = or η = = ( 6 – loge 5 – 5 × 0.40.4) = 0.2642 2 2 5 Qs 7 Since, process 1 → 2 is isothermal, therefore, or η = 26.42 % Ans. temperature during the process remains constant. Hence temperature T2 is also equal to T0. 2. Calculate speed of sound in a mixture containing n1 Considering n mole of the gas, mole of He, n2 mole of H2 and n3 mole of CO2 at Work done by the gas during isothermal process temperature T 1 → 2, (i) when n1 = 1, n2 = 2, n3 = 3 and T = 434 K, V (ii) when n1 = 3, n2 = 2, n3 = 1 and T = 415 K W12 = nRT1.log 2 = – nRT0 loge 5 V1 Gas constant, R = 8.3 J (mole)–1 K–1 But for isothermal process Q = W, therefore, Sol. Sound waves in gases are longitudinal waves and Q12 = – nRT0 loge 5 speed of longitudinal waves in a gas is given by Now considering isobaric process 2 → 3 γRT v= . V3 T M = 3 =2 or V3 = 2V0 V2 T2 Hence, to calculate speed of sound waves in the mixture, its mean molar mass M and adiabatic exponent γ must be known. XtraEdge for IIT-JEE 18 DECEMBER 2011 Molar mass of He is M1 = 4 gm/mole and its molar Sol. First, incident waves are received by the wall and 3 then reflected back. These reflected waves are specific heat at constant volume is C v1 = R. received by the source. 2 5 First, consider incident waves received by the wall. These for H2 are M2 = 2 gm/mole and C v 2 = R Since, source and wall (observer) both are 2 respectively and for CO2, M3 = 44 gm/mole and approaching towards each other, therefore, C v3 = 3R respectively. frequency received by the wall is given by, v + v0 v+u Mean molar mass of the mixture, n 1 = n0 v – v = n0 v – u = 272 Hz M m + m 2 n 2 + M 3n 3 s M= 1 1 ...(i) n1 + n 2 + n 3 Now waves are reflected by the wall. For reflected waves wall works as a sonic source of frequency and Mean molar specific heat at constant volume, n1 = 272 Hz, which is approaching towards the n1C v1 + n 2 C v 2 + n 3C v3 receiver R with velocity u. Receiver is also Cv = ...(ii) n1 + n 2 + n 3 approaching the wall (source) with the same speed. (i) substituting n1 = 1, n2 = 2, n3 = 3 in equation (i) Therefore, frequency n2 of reflected waves received and (ii) by the receiver is given by 140 140 v + v0 v+u M= gm/mole or × 10–3 kg/mole n 2 = n1 v – v = n1 v – u = 289 Hz Ans. 6 6 s 31 Since, receiver is approaching the wall with velocity Cv = R 12 u to receive reflected waves, therefore, velocity of 43 wave propagation relative to the receiver is But Cp = Cv + R, therefore, Cp = R v' = (v + u) = 340 ms–1. 12 C 43 Let wavelength of reflected waves by λ'. ∴ Adiabatic exponent, γ = P = 20 Cv 31 Using, v' = n2λ' or λ' = m Ans. 17 γRT ∴ Speed of longitudinal waves = 4. Both ends of a solid cylindrical glass rod of diameter M d = 10 cm are made hemispherical. When a luminous = 214140 ms–1 object is placed on axis of the rod at a distance or 462.75 ms–1 Ans. a = 20 cm from one end, its real image is obtained at a (ii) Substituting n1 = 3, n2 = 2, n3 = 1 in equation (i) distance b = 40 cm from the other end. If the refractive and (ii) index of glass µ = 1.5. Calculate length l of the rod. M = 10 gm/mole or 10 × 10–3 kg/mole Sol. Since image is real, therefore, it is formed by 25 37 convergence of transmitted rays. If object is placed Cv = R and Cp = Cv + R = R 12 12 on left of left end of the rod, image will be on right Cp 37 of the right end of rod. γ= = Cv 25 Since diameter of the rod is d and ends are made hemispherical, radius of curvature of each end is γRT equal to d/2. Speed of longitudinal waves M First considering refraction at left end, = 83 74 ms–1 or 714 ms–1 Ans. d u = – a, µ1 = 1, µ2 = µ = 1.5, R = + , v = v1 (let) 3. A sonic source of natural frequency n0 = 256 Hz and 2 a receiver are moving along the same line with speed l u = 10 ms–1 towards a reflecting surface. Their line of (µ) motion is normal to the surface and the surface is also approaching towards them with the same speed as O I1 I shown in figure. If speed of sound in air is v = 330 ms–1, calculate frequency and wavelength of reflected v1 waves received by the receiver. µ2 µ µ – µ1 u u Using formula, – 1 = 2 v u R S v1 = + 30 cm R It means that image I1 formed after refraction at left u end is on right of left end at a distance of 30 cm. XtraEdge for IIT-JEE 19 DECEMBER 2011 This image works as an object for refraction at the dφ 1 other end. e=– = π ωa2 B sin (ωt) dt 2 Then, for refraction at right end, since, resistance of the circuit is negligible, µ1 = 1.5, µ2 = 1, v = + b = + 40 cm, therefore, Potential difference across capacitor is d equal to induced emf in the circuit. R=– = – 5 cm, u = ? 2 ∴ Charge on the capacitor at time t is q = C.e. µ µ µ – µ1 1 Substituting in 2 – 1 = 2 = πωa2 CB sin (ωt) v u R 2 u = – 20 cm dq 1 But current I = = π ω2 a2 CB cos (ωt) Ans. Negative sign indicates that object for right end is on dt 2 its left or image I1 is on left of right end of the rod Due to flow of current, semicircle experience a and at a distance of 20 cm. moment. Therefore, power is required to keep the ∴ Length of the rod = v1 + 20 cm = 50 cm Ans. semi circle rotating with constant angular velocity. In fact, power required to rotate the semicircle is 5. A copper rod is bent into a semi-circle of radius a and equal to electrical power generated in the circuit. at ends straight parts are bent along diameter of the semi-circle and are passed through fixed, smooth and 1 ∴ Power required, P = e.I = π2ω3a4 CB2 sin (2ωt) conducting rings O and O' as shown in figure. A 8 capacitor having capacitance C is connected to the Ans. rings. The system is located in a uniform magnetic field of induction B such that axis of rotation OO' is perpendicular to the field direction. At initial moment of time (t = 0), plane of semi-circle was normal to the field direction and the semi-circle is set in rotation with Dimensional Formulae of Some constant angular velocity ω. Neglecting resistance and Physical Quantities inductance of the circuit, calculate current I flowing through the circuit at time t and instantaneous power Dimensional Physical Quantity required to rotate the semi-circle. Formulae Work (W) [ML2T–2] O O' Stress [ML–1T–2] a Torque (τ) [ML2T–2] Moment of Inertia (I) [ML2] ×B Coefficient of viscosity (η) [ML–1T–1] C Gravitational constant (G) [M–1L3T–2] Sol. When the copper rod is rotated, flux linked with the Specific heat (S) [L2T–2θ–1] circuit varies with time. Therefore, an emf is Coeficient of thermal conductivity (K) [MLT–3θ–1] induced in the circuit. Universal gas constant (R) [ML2T–2θ–1] At time t, plane of semi-circle makes angle ωt with Potential (V) [ML2T–3A–1] the plane of rectangular part of the circuit. Hence, component of the magnetic induction normal to Intensity of electric field (E) [MLT–3A–1] plane of semi circle is equal to B.cos ωt. Permittivity of free space (ε0) [M–1L–3T4A2] ∴ Flux linked with semicircular part is Specific resistance (ρ) [ML3T–3A2] 1 Magnetic Induction (B) [MT–2A–1] φ1 = πa2.B cos ωt 2 Planck's constant (h) [ML2T–1] Let area of rectangular part of the circuit be A. Boltzmann's constant (k) [ML2T–2θ–1] Then flux linked with this part is φ2 = BA ∴ Total flux linked with the circuit is φ = φ1 + φ2 Entropy (S) [ML2T–2θ–1] 1 Decay constant (λ) [T–1] or φ = πa2 B.cos(ωt) + B.A 2 Bohr magnetic (µB) [L2A] ∴ Induced emf in the circuit, Thermmionic current density (J) [AL–2] XtraEdge for IIT-JEE 20 DECEMBER 2011 P HYSICS F UNDAMENTAL F OR IIT-J EE Ray Optics KEY CONCEPTS & PROBLEM SOLVING STRATEGY Reflection : Key Concepts : n αα (a) Due to reflection, none of frequency, wavelength P P' and speed of light change. α Real n α Virual (b) Law of reflection : Object Object Incident ray, reflected ray and normal on incident point are coplanar. For solving the problem, the reference frame is The angle of incidence is equal to angle of chosen in which optical instrument (mirror, lens, etc.) reflection is in rest. The formation of image and size of image is Incident n Reflected n independent of size of mirror. Ray Ray θ θ θ θ Tangent Visual region and intensity of image depend on size at point P of mirror. P Plane surface Convex surface P P' n n θ θ αα α Convex A α surface Tangent at point P Some important points : In case of plane mirror For real object, image is virtual. If the plane mirror is rotated through an angle θ, the For virtual object, image is real. reflected ray and image is rotated through an angle 2θ The converging point of incident beam behaves as a in the same sense. object. If mirror is cut into a number of pieces, then the focal If incident beam on optical instrument (mirror, lens length does not change. etc) is converging in nature, object is virtual. The minimum height of mirror required to see the full If incident beam on optical instrument is diverging in image of a man of height h is h/2. nature, the object is real. Rest The converging point of reflected or refracted beam from an optical instrument behaves as image. v If reflected beam or refracted beam from an optical Object v Image instrument is converging in nature, image is real. n vsinθ P P P v Rest P vsinθ Real n θ Virtual Real Virual Object Object Object Object Object vcosθ vcosθ Image If reflected beam or refracted beam from an optical instrument is diverging in nature, image is virtual. vm 2vm–v Object v Image XtraEdge for IIT-JEE 21 DECEMBER 2011 y y vm 2vm x' x x' x Object Image In rest y' y' 1 1 1 The mirror formula is + = v u f Object vm v 2vm+v Also, R = 2f Image These formulae are only applicable for paraxial rays. (c) Number of images formed by combination of All distances are measured from optical centre. It two plane mirrors : The images formed by means optical centre is taken as origin. combination of two plane mirror are lying on a The sign conventions are only applicable in given circle whose centre is at the meeting points of values. mirrors. Also, object is lying on that circle. The transverse magnification is 360º image size −v Here, n = β= = θ object size u where θ = angle between mirrors. 1. If object and image both are real, β is negative. 360º 2. If object and image both are virtual, β is negative. If is even number, the number of images is θ 3. If object is real but image is virtual, β is positive. n – 1. 4. If object is virtual but image is real, β is positive. 360º If is odd number and object is placed on 5. Image of star; moon or distant object is formed at θ focus of mirror. bisector of angle between mirror, then number of images is n – 1. If y = the distance of sun or moon from earth. D = diameter of moon or sun's disc 360º If is odd and object is not situated on f = focal length of the mirror θ bisector of angle between mirrors, then the d = diameter of the image number of images is equal to n. θ = the angle subtended by sun or moon's disc (d) Law of reflection in vector form : D d Then tan θ = θ = = ˆ Let e1 = unit vector along incident ray. y f ˆ e 2 = unit vector along reflected ray Here, θ is in radian. ˆ n = unit vector along normal on point of Incidence Sun D F Then e 2 = e1 − 2(e1.n ) n ˆ ˆ ˆ ˆ ˆ θ θ n d nˆ ˆ e1 ˆ e2 Problem solving strategy : Image formation by mirrors (e) Spherical mirrors : Step 1: Identify the relevant concepts : There are two different and complementary ways to solve It easy to solve the problems in geometrical optics problems involving image formation by mirrors. One by the help of co-ordinate sign convention. approach uses equations, while the other involves y y y drawing a principle-ray diagram. A successful problem solution uses both approaches. x' x x' x x' x Step 2: Set up the problem : Determine the target variables. The three key quantities are the focal y' y' y' length, object distance, and image distance; typically XtraEdge for IIT-JEE 22 DECEMBER 2011 you'll be given two of these and will have to Note that the same sign rules (given in section) determine the third. work for all four cases in this chapter : reflection Step 3: Execute the solution as follows : and refraction from plane and spherical surfaces. The principal-ray diagram is to geometric optics Step 4: Evaluate your answer : You've already what the free-body diagram is to mechanics. In checked your results by using both diagrams and any problem involving image formation by a equations. But it always helps to take a look back mirror, always draw a principal-ray diagram first and ask yourself. "Do these results make sense ?". if you have enough information. (The same advice should be followed when dealing with Refraction : lenses in the following sections.) Laws of Refraction : It is usually best to orient your diagrams consistently with the incoming rays traveling The incident ray, the refracted ray and normal on from left to right. Don't draw a lot of other rays at incidence point are coplanar. random ; stick with the principal rays, the ones you know something about. Use a ruler and µ1 sin θ1 = µ2 sin θ2 = ... = constant. measure distance carefully ! A freehand sketch will not give good results. θ1 If your principal rays don't converge at a real µ1 image point, you may have to extend them straight backward to locate a virtual image point, µ2 as figure (b). We recommend drawing the extensions with broken lines. Another useful aid θ2 is to color-code the different principal rays, as is done in figure(a) & (b). Snell's law in vector form : Q I ˆ n 4 2 3 C P' F v P 2 ˆ e1 4 Q' µ1 3 µ2 1 ˆ e2 (a) ˆ Let, e1 = unit vector along incident ray Q 1 1 3 ˆ e 2 = unit vector along refracted. Q' 2 4 2 v ˆ n = unit vector along normal on incidence point. P P' F C 4 ˆ ˆ ˆ ˆ Then µ1( e1 × n ) = µ2( e 2 × n ) (b) Some important points : 1 1 1 (a) The value of absolute refractive index µ is always Check your results using Eq. + = and the s s' f greater or equal to one. y' s' magnification equation m = = − . The (b) The value of refractive index depends upon y s material of medium, colour of light and results you find using this equation must be consistent with your principal-ray diagram; if not, temperature of medium. double-check both your calculation and your (c) When temperature increases, refractive index diagram. decreases. Pay careful attention to signs on object and image distances, radii or curvature, and object and image (d) Optical path is defined as product of geometrical heights. A negative sign on any of these path and refractive index. quantities always has significance. Use the equations and the sign rules carefully and i.e., optical path = µx consistently, and they will tell you the truth ! XtraEdge for IIT-JEE 23 DECEMBER 2011 ti t t Σ = 1 + 2 + ... (e) For a given time, optical path remains constant. µi µ1 µ2 i.e., µ1x1 = µ2x2 = ... constant ti The apparent depth due to a number of media is Σ dx dx µi ∴ µ1 1 = µ2 2 dt dt The lateral shifting due to a slab is d = t sec r sin(i – r). ∴ µ1c1 = µ2c2 (where c1 and c2 are speed Critical angle : When a ray passes from denser of light in respective mediums) medium (µ2) to rarer medium (µ1), then for 90º angle µ2 c ∴ = 1 of refraction, the corresponding angle of incidence is µ1 c2 critical angle. 1 i.e., µ∝ µ1 c Mathematically, sin c = µ2 (f) The frequency of light does not depend upon medium. (i) When angle of incidence is lesser than critical ∴ c1 = fλ1, c2 = fλ2 angle, refraction takes place. The corresponding µ1 c λ deviation is ∴ = 2 = 2 µ2 c1 λ1 µ δ = sin–1 2 sin i – i µ for i < c 1 1 ∴ µ∝ λ (ii) When angle of incidence is greater than critical When observer is rarer medium and object is in angle, total internal reflection takes place. The denser medium : corresponding deviation is real depth Then µ= δ = π – 2i when i < c apparent depth The δ – i graph is : When object is in rarer and observer is in denser (i) Critical angle depends upon colour of light, medium : material of medium, and temperature of medium. apparent position µ= (ii) Critical angle does not depend upon angle of real position incidence 1 The shift of object due to slab is x = t 1 – µ (a) This formula is only applicable when observer is in rarer medium. δ (b) The object shiftiness does not depend upon the c i π/2 position of object. (c) Object shiftiness takes place in the direction of Refractive surface formula, incidence ray. µ2 µ µ − µ1 – 1 = 2 The equivalent refractive index of a combination of a v u r Σt i Here, v = image distance, number of slabs for normal incidence is µ = t Σ i u = object distance, µi r = radius of curvature of spherical surface. Here, Σti = t1 + t2 + ... (a) For plane surface , r = ∞ (b) Transverse magnification, XtraEdge for IIT-JEE 24 DECEMBER 2011 Im age size µv (b) This formula is only applicable when medium on m= = 1 object size µ2u both sides of lens are same. (c) Refractive surface formula is only applicable for (c) Intensity is proportional to square of paraxial ray. aperture. Lens : (d) When lens is placed in a medium whose refractive Lens formula : index is greater than that of lens. i.e., µ1 > µ2. 1 1 1 Then converging lens behaves as diverging lens – = and vice versa. v u f (a) Lens formula is only applicable for thin lens. (e) When medium on both sides of lens are not same. (b) r = 2f formula is not applicable for lens. Then both focal lengths are not same to each other. image size v (c) m = = (f) If a lens is cut along the diameter, focal length object size u does not change. (d) Magnification formula is only applicable when object is perpendicular to optical axis. (g) If lens is cut by a vertical, it converts into two lenses of different focal lengths. (e) lens formula and the magnification formula is only applicable when medium on both sides of 1 1 1 i.e., = + lenses are same. f f1 f2 (h) If a lens is made of a number of layers of different (f) refractive index number of images of an object by the lens is equal to number of different media. f(+ve) f(–ve) (i) The minimum distance between real object and (i) (ii) real image in is 4f. (j) The equivalent focal length of co-axial combination of two lenses is given by f(–ve) f(+ve) 1 1 1 d = + – F f1 f 2 f1f 2 (iii) (iv) (k) If a number of lenses are in contact, then 1 1 1 = + + ...... F f1 f 2 f(–ve) f(+ve) (v) (vi) 1 (l) (i) Power of thin lens, P = F (g) Thin lens formula is applicable for converging as well diverging lens. Thin lens maker's formula : 1 (ii) Power of mirror is P = – F 1 µ − µ1 1 1 = 2 µ − r r (m) If a lens silvered at one surface, then the system f 1 1 2 behaves as an equivalent mirror, whose power µ1 µ1 P = 2PL + Pm µ2 Here, PL = Power of lens µ − µ1 1 1 = 2 µ − r r (a) Thin lens formula is only applicable for paraxial 1 1 2 ray. XtraEdge for IIT-JEE 25 DECEMBER 2011 1 1 1 1 Pm = Power of silvered surface = – Now = + Fm f1 u1 v1 1 1 1 Here, Fm = r2/2, where r2 = radius of silvered surface. or = + 20 25 v1 P = – 1/F v1 = 100 cm. Here, F = focal length of equivalent mirror. As v1 is positive, hence the image is real. In the absence of convex mirror, the rays after reflection from concave mirror would have formed a real image I1 at distance 100 cm from the mirror. Due to the Solved Examples presence of convex mirror, the rays are reflected and appear to come from I2. 1. Rays of light strike a horizontal plane mirror at an (ii) For convex mirror, angle of 45º. At what angle should a second plane In this case, I1 acts as virtual object and I2 is the mirror be placed in order that the reflected ray finally virtual image. be reflected horizontally from the second mirror. The distance of the virtual object from the convex Sol. The situation is shown in figure mirror is 100 – 50 = 50 cm. Hence u2 = –50 cm. C G As focal length of convex mirror is negative and A D hence f2 = –30/2 = –15 cm. Here we shall calculate S θ θ the value of v2. Using the mirror formula, we have 45º N 1 1 1 − = − + 45º 15 50 v 2 P Q B or v2 = –21.42 cm The ray AB strikes the first plane mirror PQ at an As v2 is negative, image is virtual. So image is angle of 45º. Now, we suppose that the second formed behind the convex mirror at a distance of mirror SG is arranged such that the ray BC after 21.43 cm. reflection from this mirror is horizontal. 3. There is a small air bubble in side a glass sphere (n = From the figure we see that emergent ray CD is parallel to PQ and BC is a line intersecting these 1.5) of radius 10 cm. The bubble is 4 cm below the parallel lines. surface and is viewed normally from the outside So, ∠DCE = ∠CBQ = 180º ∠DCN + ∠NCB + ∠CBQ = 180º (Fig.). Find the apparent depth of the air bubble. θ + θ + 45º = 180º ∴ θ = 67.5º P A n2 = 1 As ∠NCS = 90º, therefore the second mirror should I be inclined to the horizontal at an angle 22.5º. O 2. An object is placed exactly midway between a C concave mirror of radius of curvature 40 cm and a n1 = 1.5 convex mirror of radius of curvature 30 cm. The mirrors face each other and are 50 cm apart. Determine the nature and position of the image formed by the successive reflections, first at the Sol. The observer sees the image formed due to refraction concave mirror and then at the convex mirror. at the spherical surface when the light from the Sol. The image formation is shown in figure. bubble goes from the glass to air. Here u = – 4.0 m, R = – 10 cm, n1 = 1.5 and n2 = 1 50cm We have [(n2/v) – (n1/u) = (n2 – n1)/R I2 P2 C F or (1/v) – (1.5/ –4.0 cm) = (1 – 1.5)/ (– 10 cm) P1 25cm or (1/v) = (0.5/10 cm) – (1.5/4.0 cm) r = 40 cm or v = – 3.0 cm r = 30 cm I1 Thus, the bubble will appear 3.0 cm below the surface. (i) For concave mirror, u1 = 25 cm, f1 = 20 cm and v1 = ? XtraEdge for IIT-JEE 26 DECEMBER 2011 4. A convex lens focuses a distance object on a screen placed 10 cm away from it. A glass plate (n = 1.5) of DEEPEST LAKE IN THE WORLD ? thickness 1.5 is inserted between the lens and the screen. Where should the object be placed so that its image is again focused on the screen ? Sol. The situation when the glass plate is inserted between the lens and the screen, is shown in fig. The lens forms the image of object O at point I1 but the glass plate intercepts the rays and forms the final image at I on the screen. The shift in the position of image after insertion of glass plate Screen Lake Baikal (Baikal) in Siberia, Russia is the deepest O I1 I lake in the world measuring 1620m deep at its deepest point. This makes it not only deep but also the oldest lake in the world estimated to be around 25 10 cm million years old. At over 636 kilometers long and 80 kilometers wide this fresh water lake holds over 20 1 1 percent of all the fresh water in the world and is I1I = t 1 − = (1.5 cm) 1 − = 0.5 cm. n 1.5 second in size only to the Caspian Sea (the caspian is called a sea but is technically a lake). Thus, the lens forms the image at a distance of 9.5 cm from itself. Using To put things into perspective the lake is so big that if all the rivers in the world flowed into its basin it 1 1 1 1 1 1 1 1 would take almost 1 year to fill. We all know Siberia – = , we get = – = – v u f u v f 9.5 10 isn’t the warmest of places so you can imagine what or u = – 190 cm. a phenomenal site it is when in the winter months the lake freezes over holding ice up to 115 meters thick. i.e. the object should be placed at a distance of Now that’s a lot of ice! 190 cm. from the lens. 5. A candle is placed at a distance of 3 ft from the wall. WHICH IS THE HIGHEST Where must a convex lens of focal length 8 inches be WATERFALL IN THE WORLD ? placed so that a real image is formed on the wall ? Sol. According to formula for refraction though a lens 36 – v v f = 8" d = 3 ft = 36" 1 1 1 1 1 1 The highest waterfall in the world is the Angel Falls – = or – = v u f v − (36 − v) 8 in Venezuela. At a towering height of 979m did you know that each drop of water takes 14 seconds to fall 1 1 1 36 − v + v 1 or + = or = from the top to the bottom. The water flows from the v 36 − v 8 v(36 − v) 8 top of a “Tepui” which is a flat topped mountain with or, v2 – 36 v + 8 × 36 = 0 vertical sides. or v = 12" or 24" = 1 ft or 2 ft. The waterfall which despite being known to the local indians for thousands of years was originally called ∴ u = 24" or 12" = 2 ft or 1 ft the “Churun Meru” but for some reason they were Hence, lens should be placed at either 1 ft or 2 ft renamed by an American bush pilot called Jimmy away from the wall. Angel, who noticed them in 1935 whilst flying over the area looking for gold. XtraEdge for IIT-JEE 27 DECEMBER 2011 P HYSICS F UNDAMENTAL F OR IIT-J EE Fluid Mechanics & Properties of Matter KEY CONCEPTS & PROBLEM SOLVING STRATEGY Fluid Mechanics : Fluid dynamics : Fluid statics : 1 2 p Bernoulli's Theorem : v + gh + = a constant Pressure at a point inside a Liquid : p = p0 + ρgh 2 ρ where p0 is the atmospheric pressure, ρ is the density for a streamline flow of a fluid (liquid or gas). of the liquid and h is the depth of the point below the Here, v is the velocity of the fluid, h is its height free surface. above some horizontal level, p is the pressure and ρ p0 is the density. p1 h v1 p2 p h1 ρ v2 h2 v2 > v1 p2 < p1 Pressure is a Scalar : The unit of pressure may be atmosphere or cm of mercury. These are derived According to this principle, the greater the velocity, units. The absolute unit of pressure is Nm–2. Normal the lower is the pressure in a fluid flow. atmospheric pressure, i.e, 76 cm of mercury, is It would be useful to remember that in liquid flow, approximately equal to 105 Nm–2. the volume of liquid flowing past any point per Thrust : Thrust = pressure × area. Thrust has the unit second is the same for every point. Therefore, when of force. the cross-section of the tube decreases, the velocity Laws of liquid pressure increases. (a) A liquid at rest exerts pressure equally in all Note : Density = relative density directions. or specific gravity × 1000 kg m–3. (b) Pressure at two points on the same horizontal line Surface tension and surface energy : in a liquid at rest is the same. Surface Tension : The property due to which a (c) Pressure exerted at a point in a confined liquid at liquid surface tends to contract and occupy the rest is transmitted equally in all directions and minimum area is called the surface tension of the acts normally on the wall of the containing vessel. liquid. It is caused by forces of attraction between the This is called Pascal's law. A hydraulic press molecules of the liquid. A molecule on the free works on this principle of transmission of surface of a liquid experiences a net resultant force pressure. which tends to draw it into the liquid. Surface tension The principle of floating bodies (law of flotation) is is actually a manifestation of the forces experienced that W = W´, that is, weight of body = weight of by the surface molecules. displaced liquid or buoyant force. The weight of the If an imaginary line is drawn on a liquid surface then displaced liquid is also called buoyancy or upthrust. the force acting per unit length of this line is defined Hydrometers work on the principle of floating as the surface tension. Its unit is, therefore, newton / bodies. This principle may also be applied to gases metre. This force acts along the liquid surface. For (e.g., a balloon). curved surfaces, the force is tangent to the liquid Liquids and gases are together called fluids. The surface at every point. important difference between them is that liquids Surface Energy : A liquid surface possesses cannot be compressed, while gases can be potential energy due to surface tension. This energy compressed. Hence, the density of a liquid is the per unit area of the surface is called the surface same everywhere and does not depend on its energy of the liquid. Its units is joule per square pressure. In the case of a gas, however, the density is metre. The surface energy of a liquid has the same proportional to the pressure. numerical values as the surface tension. The surface XtraEdge for IIT-JEE 28 DECEMBER 2011 tension of a liquid depends on temperature. It The upward force by which a liquid surface is pulled decreases with rise in temperature. up in a capillary tube is 2πrTcos θ, and the downward Excess of Pressure : Inside a soap bubble or a gas force due to the gravitational pull on the mass of bubble inside a liquid, there must be pressure in liquid in the tube is (πr2h + v)ρg, where v is the excess of the outside pressure to balance the tendency volume above the liquid meniscus. If θ = 0º, the of the liquid surface to contract due to surface meniscus is hemispherical in shape. Then v = tension. difference between the volume of the cylinder of radius r and height r and the volume of the 1 1 p(excess of pressure) = T + in general hemisphere of radius r r r 1 2 2 3 1 3 = πr3 – πr = πr where T is surface tension of the liquid, and r1 and r2 3 3 are the principal radii of curvature of the bubble in two mutually perpendicular directions. When θ ≠ 0, we cannot calculate v which is generally very small and so it may be neglected. For For a spherical soap bubble, r1 = r2 = r and there are equilibrium two free surfaces of the liquid. (πr2h + v) ρg = 2πrT cos θ 4T ∴ p= When a glass capillary tube is dipper in mercury, the r meniscus is convex, since the angle of contact is For a gas bubble inside a liquid, r1 = r2 = r and there obtuse. The surface tension forces now acquire a is only one surface. downward component, and the level of mercury 2T inside the tube the falls below the level outside it. the ∴ p= relation 2T cos θ = hρgr may be used to obtain the r fall in the mercury level. For a cylindrical surface r1 = r and r2 = ∞ and there Problem Solving Strategy are two surfaces. Bernoulli's Equations : 2T ∴ p= Bernoulli's equation is derived from the work-energy r theorem, so it is not surprising that much of the Angle of Contact : The angle made by the surface of problem-solving strategy suggested in W.E.P. also a liquid with the solid surface inside of a liquid at the applicable here. point of contact is called the angle of contact. It is at Step 1: Identify the relevant concepts : First ensure this angle that the surface tension acts on the wall of that the fluid flow is steady and that fluid is the container. incompressible and has no internal friction. This case The angle of contact θ depends on the natures of the is an idealization, but it hold up surprisingly well for liquid and solid in contact. If the liquid wets the solid fluids flowing through sufficiently large pipes and for (e.g., water and glass), the angle of contact is zero. In flows within bulk fluids (e.g., air flowing around an most cases, θ is acute (figure i). In the special case of airplane or water flowing around a fish). mercury on glass, θ is obtuse (figure ii). Step 2: Set up the problem using the following steps Always begin by identifying clearly the points 1 θ θ and 2 referred to in Bernoulli's equation. Define your coordinate system, particular the fig. (i) fig. (ii) level at which y = 0. Rise of Liquid in a Capillary Tube : In a thin Make lists of the unknown and known quantities (capacity) tube, the free surface of the liquid becomes 1 1 in Eq. p1 + ρgy1 + ρv12 = p2 + ρgy2 + ρv22 curved. The forces of surface tension at the edges of 2 2 the liquid surface then acquire a vertical component. (Bernoulli's equation) T θ meniscus θ T The variables are p1, p2, v1, v2, y1 and y2, and the constants are ρ and g. Decide which unknowns are your target variables. θ θ Step 3: Execute the solutions as follows : Write h Bernoulli's equation and solve for the unknowns. In some problems you will need to use the continuity equation, Eq. A1v1 = A2v2 (continuity equation, incompressible fluid), to get a relation between the r two speeds in terms of cross-sectional areas of pipes XtraEdge for IIT-JEE 29 DECEMBER 2011 or containers. Or perhaps you will know both speeds F and need to determine one of the areas. You may also dV need to use Eq. = Av (volume flow rate) to find dt F F the volume flow rate. Step 4: Evaluate your answer : As always, verify that the results make physical sense. Double-check that you have used consistent units. In SI units, pressure is F V in pascals, density in kilograms per cubic meter, and (c) Shear strain = φ speed in meters per second. Also note that the pressures must be either all absolute pressure or all gauge pressures. φ Properties of matter : Shear strain Key Concepts : Stress : Stress-strain graph : The restoring force setup inside the body per unit From graph, it is obvious that in elastic limit, stress is area is known as stress. proportional to strain. This is known as Hooke's law. Restoring forces : If the magnitude of applied ∴ Stress ∝ Strain deforming force at equilibrium = F ∴ Stress = E .strain F stress Then, Stress = ∴ E= A strain In SI system, unit of stress is N/m2. where E is proportionality dimensional constant Difference between pressure and stress : known as coefficient of elasticity. (a) Pressure is scalar but stress is tensor quantity. Plastic (b) Pressure always acts normal to the surface, but region Breaking stress may be normal or tangential. strength B C (c) Pressure is compressive in nature but stress may Elastic A Stress be compressive or tensile. limit Strain : change in dimension Strain = original dimension O Strain ∆L Types of coefficient of elasticity : (a) Longitudinal strain = L logitudinal stress L (a) Young's modulus = Y = longitudinal strain F F F FL Longitudinal strain is in the direction of ∴ Y= = deforming force but lateral strain is in ∆L A∆L A perpendicular direction of deforming force. L Poisson ratio : lateral strain ∆d/D σ= = L longitudinal strain ∆L/L Here ∆d = change in diameter. ∆V ∆L (b) Volumetric strain = V F volumetric stress (b) Bulk modulus = B = volumetric strain Compressibility = 1/B XtraEdge for IIT-JEE 30 DECEMBER 2011 F shear stress Surface tension : (c) Modulus of rigidity = η = = Aφ shear strain F T= (d) For isothermal process, B = P. L F Here L = length of imaginary line drawn at the surface of liquid. and F = force acting on one side of φ line (shown in figure) φ (a) Surface tension does not depend upon surface F area. (e) For adiabatic process, B = γP (b) When temperature increases, surface tension decreases. Adiabatic bulk modulus (f) =γ (c) At critical temperature surface tension is zero. Isothermal bulk modulus (g) Esolid > Eliquid > Egas F (h) Young's modulus Y and modulus of rigidity η exist only for solids. (i) Bulk modulus B exist for solid, liquid and gas. L (j) When temperature increases, coefficient of F elasticity (Y, B, η) decreases. 1 3 9 Rise or fall of a liquid in a capillary tube : (k) + = B η Y 2T cos θ (l) Y = 2(1 + σ)η h= rρg (m) Poisson's ratio σ is unitless and dimensionless. Here θ = angle of contact. 1 r = radius of capillary tube Theoretically, –1 < σ < 2 ρ = density of liquid 1 For a given liquid and solid at a given place, Practically, 0<σ< 2 hr = constant (n) Thermal stress = Yα∆θ Surface energy : (o) Elastic energy stored, Surface energy density is defined as work done 1 1 1 against surface tension per unit area. It is numerically U= × load × extension = Fx = kx2 equal to surface tension. 2 2 2 = stress × strain × volume W = work = surface tension × area For twisting motion, (a) For a drop of radius R, W = 4πR2T 1 (b) For a soap bubble, W = 8πR2T U= × torque × angular twist 2 Excess pressure : 1 1 2T = τ × θ = cθ2 (a) For drop, P = 2 2 R Elastic energy density, 4T (b) For soap bubble, P = 1 1 R u= × stress × strain J/m3 = Y × strain2J/m3 2 2 Viscosity : Thermal stress = Yα∆θ and Thermal strain = α∆θ (a) Newton's law of viscous force : Work done in stretching a wire : dv F = – ηA 1 dy (a) W = F∆L 2 dv where = velocity gradient 1 dy (b) Work done per unit volume = × stress × strain 2 A = area of liquid layer (c) Breaking weight = breaking stress × area η = coefficient of viscosity The unit of coefficient of viscosity in CGS is poise. XtraEdge for IIT-JEE 31 DECEMBER 2011 (b) SI unit of coefficient of viscosity Sol. Draw a horizontal line through the mercury-glycerine = poiseuille = 10 poise. surface. This is a horizontal line in the same liquid at rest namely, mercury. Therefore, pressure at the (c) In the case of liquid, viscosity increases with points A and B must be the same. density. (d) In the case of gas, viscosity decreases with density. h (e) In the case of liquid, when temperature increases, 10 cm viscosity decreases. (1 – h) (f) In the case of gas, when temperature increases, viscosity increases. A B Poiseuille's equation : Pπr 4 Pressure at B V= 8ηL = p0 + 0.1 × (1.3 × 1000) × g where V = the volume of liquid flowing per second Pressure at A through a capillary tube of length L and radius r = p0 + h × 800 × g + (0.1 – h) × 13.6 × 1000g ∴ p0 + 0.1 × 1300 × g η = coefficient of viscosity = p0 + 800gh + 1360g – 13600 × g × h and P = pressure difference between ends of the tube ⇒ 130 = 800h + 1360 – 13600h Stoke's law : 1230 ⇒ h= = 0.096 m = 9.6 cm The viscous force acting on a spherical body moving 12800 with constant velocity v in a viscous liquid is F = 6πηrv 2. A liquid flows out of a broad vessel through a narrow where r = radius of spherical body vertical pipe. How are the pressure and the velocity of the liquid in the pipe distributed when the height Determination of η : of the liquid level in the vessel is H from the lower 2r 2 (ρ − σ)g end of the length of the pipe is h ? η= Sol. Let us consider three points 1, 2, 3 in the flow of 9v water. The positions of the points are as shown in the where r = radius of spherical body moving with figure. constant velocity v in a viscous liquid of coefficient Applying Bernoulli's theorem to points 1, 2 and 3 of viscosity η and density ρ and σ = density of spherical body •1 Critical velocity (v0) : kη x v0 = H ρr •2 h where k = Reynold's number for narrow tube, k ≈ 1000. (a) For stream line motion, flow velocity v < v0. •3 2 (b) For turbulant motion, flow velocity v > v0. p0 1 2 p 1 + v 1 + gH = + v 2 + g (h – x) 2 ρ 2 ρ 2 p 1 2 = 0 + v3 + 0 ρ 2 Solved Examples By continuity equation v 1A1 = A2v 2 = A2v 3 1. A vertical U-tube of uniform cross-section contains Since A1 >> A2,v1 is negligible and v2 = v3 = n (say). mercury in both arms. A glycerine (relative density p0 p 1 1.3) column of length 10 cm is introduced into one of ∴ + gH = 2 + v2 + g (h – x) ρ ρ 2 the arms. Oil of density 800 kg m–3 is poured into the other arm until the upper surface of the oil and p0 1 2 = + v glycerine are at the same horizontal level. Find the ρ 2 length of the oil column. Density of mercury is ∴ v = 2gH (i) 13.6 × 103 kg m–3. XtraEdge for IIT-JEE 32 DECEMBER 2011 p0 p = 4.99 × 10–3 cm and + gH = 2 + gH + g (h – x) ρ ρ Also, elastic limit for copper = 1.5 × 109 dynes/cm2 ⇒ p0 + p2 + ρg (h – x) If d' is the minimum diameter, then maximum stress ⇒ p2 = p0 – ρg (h – x) (ii) F 4F on the wire = = Thus pressure varies with distance from the upper πd '2 / 4 πd' 2 end of the pipe according to equation (ii) and velocity is a constant and is given by (i). 4F Hence, = 1.5 × 109 πd' 2 3. Calculate the difference in water levels in two 4F 4 × 5.0 × 1000 × 980 communicating tubes of diameter d = 1 mm and or d'2 = 9 = d = 1.5 mm. Surface tension of water = 0.07 Nm–1 π × 1.5 × 10 3.142 × 1.5 × 109 –4 and angle of contact between glass and water = 0º. = 41.58 × 10 2T cos θ d' = 0.0645 cm. Sol. Pressure at A = p0 – r2 5. A uniform horizontal rigid bar of 100 kg in supported (Q pressure inside a curved surface is greater than horizontally by three equal vertical wires A, B and C that outside) each of initial length one meter and cross-section 2T cos θ 1 mm2. B is a copper wire passing through the centre Pressure at B = p0 – r1 of the bar; A and C are steel wires and are arranged 1 1 symmetrically one on each side of B YCu = 1.5 × 1012 ∴ pressure difference = 2T cos θ − r r dynes / cm2, Ys = 2 × 1012 dynes/cm2. Calculate the 1 2 tension in each wire and extension. Sol. The situation is shown in figure. Because the rod is B A horizontally supported, hence extensions in all the wires must be equal i.e., strains in all the wires are equal as initial lengths are also equal. Stress As Y = Strain Let this pressure difference correspond to h units of C A B the liquid. S Cu S 1 1 Then 2T cos θ − = ρgh r r 1 2 100 Kg 2T cos θ 1 1 FCu / A ⇒ h= − ρg r1 r2 Hence, YCu = Strain … (1) 2 × 0.07 1 1 Fs / A ∴ h= −3 − −3 = 4.76 mm and Ys = … (2) 1000 × 9.8 1× 10 1.5 × 10 Strain YCu FCu 1.5 3 4. A mass of 5 kg is suspended from a copper wire of 5 ∴ = = = or 4FCu = 3FS ...(3) mm diameter and 2 m in length. What is the YS FS 2 4 extension produced in the wire ? What should be the According to figure, we can write minimum diameter of the wire so that its elastic limit 2FS + FCu = 100 g or 2 × (4/3) FCu + FCu = 100 g is not exceed ? Elastic limit for copper = 1.5 × 109 dynes/cm2. Y for copper = 1.1 × 1012 dynes/cm2. or [(8/3) + 1] FCu = 100 g Sol. Given that Y = 1.1 × 1012 dynes/cm2, ∴ FCu = (3/11) × 100g L = 2m = 200 cm, d = 5 mm = 0.5 cm = (3/11) × 100 Kgwt = 27.28 Kgwt or r = d/2 = 0.25 cm, F = 5.0 × 1000 × 980 dynes. and FS = (4/3) FCu = (4/3) × (3/11) × 100g FL = (400/11)g = 36.36 Kgwt Y= Extension in each wire, πr 2l FL 5.0 × 1000 × 980 × 200 FCu L 27280 × 980 × 100 or l = = l= = = 0.178 cm πr 2 Y 3.142 × (0.25) 2 × 1.1× 10 2 AYCu 10 − 2 × 1.5 × 1012 XtraEdge for IIT-JEE 33 DECEMBER 2011 KEY CONCEPT Organic Chemistry CARBOXYLIC ACID Fundamentals Acidity of carboxylic acids. (1.27Å) which is nearly intermediate between C O Fatty acids are weak acids as compared to inorganic and C—O bond length values. This proves resonance acids. The acidic character of fatty acids decreases in carboxylate anion. with increase in molecular weight. Formic acid is the O – strongest of all fatty acids. O The acidic character of carboxylic acids is due to H C H C Na+ resonance in the acidic group which imparts electron OH O deficiency (positive charge) on the oxygen atom of Formic acid Sodium formate the hydroxyl group (structure II). It is important to note that although carboxylic acids O O– and alcohols both contain –OH group, the latter are + R C O H not acidic in nature. It is due to the absence of R C O H resonance (factor responsible for acidic character of I II –COOH) in both the alcohols as well as in their Non-equivalent structures (Resonance less important) corresponding ions (alkoxide ions). R—O—H R—O– + H+ O– Alcohol Alkoxide ion R C O+H + (No resonance) (No resonance) Relative acidic character of carboxylic acids with The positive charge (electron deficiency) on oxygen common species not having —COOH group. atom causes a displacement of electron pair of the O—H bond towards the oxygen atom with the result RCOOH > Ar—OH > HOH > ROH > the hydrogen atom of the O—H group is eliminated HC CH > NH3 > RH as proton and a carboxylate ion is formed. Effect of Substituents on acidity. Once the carboxylate ion is formed, it is stabilised by The carboxylic acids are acidic in nature because of means of resonance. stabilisation (i.e., dispersal of negative charge) of O O– carboxylate ion. So any factor which can enhance the R C R C dispersal of negative charge of the carboxylate ion O– O will increase the acidity of the carboxylic acid and Resonating forms of carboxylate ion (Equivalent structures) vice versa. Thus electron-withdrawing substitutents (Resonance more important) (like halogens, —NO2, —C6H5, etc.) would disperse the negative charge and hence stabilise the – O carboxylate ion and thus increase acidity of the R C parent acid. On the other hand, electron-releasing O substituents would increase the negative charge, destabilise the carboxylate ion and thus decrease Resonance hybrid of carboxylate ion acidity of the parent acid. Due to equivalent resonating structures, resonance in – carboxylate anion is more important than in the O parent carboxylic acid. Hence carboxylate anion is X C more stabilised than the acid itself and hence the O equilibrium of the ionisation of carboxylic acids to The substituent X withdraws electrons, disperses negative the right hand side. charge, stabilises the ion and hence increases acidity RCOOH RCOO– + H+ – O The existence of resonance in carboxylate ion is Y C supported by bond lengths. For example, in formic O acid, there is one C=O double bond (1.23 Å) and one C—O single bond (1.36Å), while in sodium formate The substituent Y releases electrons, intensifies negative both of the carbon-oxygen bond lengths are identical charge, destabilises the ion and hence decreases acidity XtraEdge for IIT-JEE 34 DECEMBER 2011 Now, since alkyl groups are electron-releasing, their Comparison of nucleophilic substitution (e.g., presence in the molecule will decrease the acidity. In hydrolysis) in acid derivatives. Let us first study the general, greater the length of the alkyl chain, lower mechanism of such reaction. shall be the acidity of the acid. Thus, formic acid O O (HCOOH), having no alkyl group, is about 10 times (i) Addition step stronger than acetic acid (CH3COOH) which in turn R C Z + Nu R C Nu is stronger than propanoic acid (CH3CH2COOH) and so on. Similarly, following order is observed in Z chloro acids. O Cl Cl (ii) Elimination step R C Nu + Z Cl C CO2H > Cl C CO2H (where Z= —Cl, —OCOR, —OR, —NH2 and Nu= A nucleophile) Cl H Nucleophilic substitution in acid derivatives pKa 0.70 1.48 O O OH H H Nu H+ R C R' R C Nu R C Nu > Cl C CO2H > H C CO2H R' R' H H (where R' = H or alkyl group) pKa 2.86 4.76 Nucleophilic addition on aldehydes and ketones Decreasing order of aliphatic acids The (i) step is similar to that of nucleophilic addition (i) O2NCH2COOH > FCH2COOH > ClCH2COOH in aldehydes and ketones and favoured by the > BrCH2COOH presence of electron withdrawing group (which would stabilise the intermediate by developing (ii) HCOOH > CH3COOH > (CH3)2 CHCOOH negative charge) and hindered by electron-releasing > (CH3)3CCOOH group. The (ii) step (elimination of the leaving group (iii) CH3CH2CCl2COOH > CH3CHCl.CHCl.COOH Z) depends upon the ability of Z to accommodate > ClCH2CHClCH2COOH electron pair, i.e., on the basicity of the leaving group. Weaker bases are good leaving groups, (iv) F3CCOOH > Cl3CCOOH > Br3CCOOH hence weaker a base, the more easily it is removed. Benzoic acid is somewhat stronger than simple Among the four leaving groups (Cl–, –OCOR, –OR, aliphatic acids. Here the carboxylate group is and –NH2) of the four acid derivatives, Cl– being the attached to a more electronegative carbon (sp2 weakest base is eliminated most readily. The relative hybridised) than in aliphatic acids (sp3 hybridised). order of the basic nature of the four groups is HCOOH > C6H5COOH > CH3COOH. – NH2 > –OR > –O.COR > Cl– Nucleophilic substitution at acyl carbon : Hence acid chlorides are most reactive and acid It is important to note that nucleophilic substitution amides are the least reactive towards nucleophilic (e.g., hydrolysis, reaction with NH3, C2H5OH, etc.) in acyl substitution. Thus, the relative reactivity of acid acid derivatives (acid chlorides, anhydrides, esters derivatives (acyl compounds) towards nucleophilic and amides) takes place at acyl carbon atom substitution reactions is (difference from nucleophilic substitution in alkyl ROCl > RCO.O.COR > RCOOR > RCONH2 halides where substitution takes place at alkyl carbon Acid Acid Esters Acid atom). Nucleophilic substitution in acyl halides is chlorides anhydrides amides faster than in alkyl halides. This is due to the OH– being stronger base than Cl–, carboxylic acids presence of > CO group in acid chlorides which (RCOOH) undergo nucleophilic substitution facilitate the release of halogen as halide ion. (esterfication) less readily than acid chlorides. δ– O δ– δ+ δ– R C Cl R Cl δ+ Alkyl chloride Acid chloride XtraEdge for IIT-JEE 35 DECEMBER 2011 KEY CONCEPT Physical Chemistry CHEMICAL KINETICS Fundamentals The temperature dependence of reaction rates : behaviour is a signal that the reaction has a complex The rate constants of most reactions increase as the mechanism. temperature is raised. Many reactions in solution fall The temperature dependence of some reactions is somewhere in the range spanned by the hydrolysis of non-Arrhenius, in the sense that a straight line is not methyl ethanoate (where the rate constant at 35ºC is obtained when ln k is plotted against 1/T. However, 1.82 times that at 25ºC) and the hydrolysis of sucrose it is still possible to define an activation energy at any (where the factor is 4.13). temperature as (a) The Arrhenius parameters : d ln k Ea = RT2 .......(ii) It is found experimentally for many reactions that a dT plot of ln k against 1/T gives a straight line. This This definition reduces to the earlier one (as the slope behaviour is normally expressed mathematically by of a straight line) for a temperature-independent introducing two parameters, one representing the activation energy. However, the definition in eqn.(ii) intercept and the other the slope of the straight line, is more general than eqn.(i), because it allows Ea to and writing the Arrhenius equaion. be obtained from the slope (at the temperature of Ea interest) of a plot of ln k against 1/T even if the ln k = ln A – ......(i) Arrhenius plot is not a straight line. Non-Arrhenius RT behaviour is sometimes a sign that quantum The parameter A, which corresponds to the intercept mechanical tunnelling is playing a significant role in of the line at 1/T = 0(at infinite temperature, shown in the reaction. figure), is called the pre-exponential factor or the (b) The interpretation of the parameters : 'frequency factor'. The parameter Ea, which is We shall regard the Arrhenius parameters as purely obtained from the slope of the line (–Ea/R), is called empirical quantities that enable us to discuss the the activation energy. Collectively the two quantities variation of rate constants with temperature; are called the Arrhenius parameters. however, it is useful to have an interpretation in mind ln A and write eqn.(i) as k = Ae − E a / RT .......(iii) Slope = –Ea/R To interpret Ea we consider how the molecular potential energy changes in the course of a chemical ln k reaction that begins with a collision between molecules of A and molecules of B(shown in figure). 1/T Ea Potential energy A plot of ln k against 1/T is a straight line when the reaction follows the behaviour described by Reactants the Arrhenius equation. The slope gives –Ea/R and the intercept at 1/T = 0 gives ln A. The fact that Ea is given by the slope of the plot of ln k against 1/T means that, the higher the activation energy, the stronger the temperature dependence of Products the rate constant (that is, the steeper the slope). A high activation energy signifies that the rate constant Progress of reaction depends strongly on temperature. If a reaction has A potential energy profile for an exothermic zero activation energy, its rate is independent of reaction. The height of the barrier between temperature. In some cases the activation energy is the reactants and products is the activation negative, which indicates that the rate decreases as energy of the reaction the temperature is raised. We shall see that such XtraEdge for IIT-JEE 36 DECEMBER 2011 XtraEdge for IIT-JEE 37 DECEMBER 2011 CAREER POINT Correspondence & Test Series Courses Information Courses of IIT-JEE For Class 12th/ 12th Pass Students For Class 11th Students COURSE → All India All India Test Postal Major Test Study Postal All Study Material Postal All India Major Test Series CP Ranker's Foundation INFORMATION Series Series Material India Test Package Test Series IIT-JEE 2012 Package Test Series IIT- IIT-JEE 2012 IIT-JEE 2012 Package Series ↓ IIT-JEE 2012 IIT-JEE 2012 (At Center) IIT-JEE 2012 JEE 2013 (At Center) (By Post) IIT-JEE 2013 IIT-JEE 2013 (At Center) Eligibility for Class 12th or 12th Class 12th or 12th Class 12th or 12th Class 12th or 12th Class 12th or 12th Class 12th or 12th Class 11th Class 11th Class 11th Admission Pass Students Pass Students Pass Students Pass Students Pass Students Pass Students Students Students Students Medium English or Hindi English English English English English English or Hindi English English visit our website or visit our website or visit our website Test Center Postal Postal Postal Postal Postal Postal contact us contact us or contact us Issue of Application Immediate Immediate Immediate Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Kit Dispatch Dispatch Dispatch Date of Dispatch/ 15 Jan 2012 15 Jan 2012 Immediate 15 Aug 11 15 Aug 11 Immediate Dispatch 15 Aug 11 onwards 15 Aug 11 onwards Immediate Dispatch First Test onwards onwards Dispatch onwards onwards Course Fee 9000 3600 1500 1000 700 900 9800 4500 2500 Courses of AIEEE For Class 11th For Class 12th/ 12th Pass Students COURSE → Students All India Test Postal Major Test Study Study Material Postal All India Major Test Series CP Ranker's INFORMATION Series Series Material Package Test Series AIEEE 2012 Package ↓ AIEEE 2012 AIEEE 2012 Package AIEEE 2012 AIEEE 2012 (At Center) AIEEE 2012 (At Center) (By Post) AIEEE 2013 Eligibility for Class 12th or 12th Pass Class 12th or 12th Class 12th or 12th Class 12th or 12th Class 12th or 12th Class 12th or 12th Class 11th Admission Students Pass Students Pass Students Pass Students Pass Students Pass Students Students Medium English or Hindi English or Hindi English or Hindi English or Hindi English or Hindi English or Hindi English or Hindi visit our website or visit our website or Test Center Postal Postal Postal Postal Postal contact us contact us Issue of Immediate Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Application Kit Dispatch Date of Dispatch/ 15 Jan 2012 15 Jan 2012 Immediate Immediate Dispatch 15 Aug 11 onwards 15 Aug 11 onwards Immediate Dispatch First Test onwards onwards Dispatch Course Fee 9000 3600 1500 1000 700 900 9800 Courses of Pre-Medical For Class 12th/ 12th Pass Students For Class 11th Students COURSE → All India All India Test Postal Major Test Study Postal All Study Material Postal All India Major Test Series CP Ranker's Foundation INFORMATION Series Series Material India Test Package Test Series AIPMT 2012 Package Test Series AIPMT 2012 AIPMT 2012 Package Series ↓ AIPMT 2012 AIPMT 2012 (At Center) AIPMT 2012 AIPMT 2013 (At Center) (By Post) AIPMT 2013 AIPMT 2013 (At Center) Eligibility for Class 12th or 12th Pass Class 12th or 12th Class 12th or 12th Class 12th or 12th Class 12th or 12th Class 12th or 12th Class 11th Class 11th Class 11th Admission Students Pass Students Pass Students Pass Students Pass Students Pass Students Students Students Students Medium English or Hindi English or Hindi English or Hindi English or Hindi English English English or Hindi English English visit our website or visit our website or visit our website Test Center Postal Postal Postal Postal Postal Postal contact us contact us or contact us Issue of Application Immediate Immediate Immediate Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Kit Dispatch Dispatch Dispatch Date of Dispatch/ 15 Jan 2012 15 Jan 2012 Immediate 15 Aug 11 15 Aug 11 Immediate Dispatch 15 Aug 11 onwards 15 Aug 11 onwards Immediate Dispatch First Test onwards onwards Dispatch onwards onwards Course Fee 9000 3600 1500 1000 700 900 9800 4500 2500 • For details about all correspondence & test Series Course Information, Please visit our website.: www.careerpointgroup.com XtraEdge for IIT-JEE 38 DECEMBER 2011 XtraEdge for IIT-JEE 39 DECEMBER 2011 XtraEdge for IIT-JEE 40 DECEMBER 2011 As the reaction event proceeds, A and B come into ratio of the two rates, and therefore of the two rate contact, distort, and begin to exchange or discard constants : atoms. The reaction coordinate is the collection of [P2 ] k motions, such as changes in interatomic distances and = 2 bond angles, that are directly involved in the [P1 ] k1 formation of products from reactants. (The reaction This ratio represents the kinetic control over the coordinate is essentially a geometrical concept and proportions of products, and is a common feature of quite distinct from the extent of reaction.) The the reactions encountered in organic chemistry where potential energy rises to a maximum and the cluster reactants are chosen that facilitate pathways of atoms that corresponds to the region close to the favouring the formation of a desired product. If a maximum is called the activated complex. After the reaction is allowed to reach equilibrium, then the maximum, the potential energy falls as the atoms proportion of products is determined by rearrange in the cluster and reaches a value thermodynamic rather than kinetic considerations, characteristic of the products. The climax of the and the ratio of concentration is controlled by reaction is at the peak of the potential energy, which considerations of the standard Gibbs energies of all corresponds to the activation energy Ea. Here two the reactants and products. reactant molecules have come to such a degree of The kinetic isotope effect closeness and distortion that a small further distortion will send them in the direction of products. The postulation of a plausible mechanism requires This crucial configuration is called the transition careful analysis of many experiments designed to state of the reaction. Although some molecules determine the fate of atoms during the formation of entering the transition state might revert to reactants, products. Observation of the kinetic isotope effect, a if they pass through this configuration then it is decrease in the rate of a chemical reaction upon inevitable that products will emerge from the replacement of one atom in a reactant by a heavier encounter. isotope, facilitates the identification of bond-breaking events in the rate-determining step. A primary We also conclude from the preceding discussion that, kinetic isotope effect is observed when the rate- for a reaction involving the collision of two determining step requires the scission of a bond molecules, the activation energy is the minimum involving the isotope. A secondary isotope effect is kinetic energy that reactants must have in order the reduction in reaction rate even though the bond to form products. For example, in a gas-phase involving the isotope is not broken to form product. reaction there are numerous collisions each second, In both cases, the effect arises from the change in but only a tiny proportion are sufficiently energetic to activation energy that accompanies the replacement lead to reaction. The fraction of collisions with a of an atom by a heavier isotope on account of kinetic energy in excess of an energy Ea is given by changes in the zero-point vibrational energies. the Boltzmann distribution as e − E a / RT . Hence, we First, we consider the origin of the primary kinetic can interpret the exponential factor in eqn(iii) as the isotope effect in a reaction in which the rate- fraction of collision that have enough kinetic energy determining step is the scission of a C–H bond. The to lead to reaction. reaction coordinate corresponds to the stretching of The pre-exponential factor is a measure of the rate at the C–H bond and the potential energy profile is which collisions occur irrespective of their energy. shown in figure. On deuteration, the dominant Hence, the product of A and the exponential factor, change is the reduction of the zero-point energy of e − E a / RT , gives the rate of successful collisions. the bond (because the deuterium atom is heavier). The whole reaction profile is not lowered, however, Kinetic and thermodynamic control of reactions : because the relevant vibration in the activated In some cases reactants can give rise to a variety of complex has a very low force constant, so there is products, as in nitrations of mono-substituted little zero-point energy associated with the reaction benzene, when various proportions of the ortho-, coordinate in either isotopomeric form of the meta-, and para- substituted products are obtained, activated complex. depending on the directing power of the original substituent. Suppose two products, P1 and P2, are produced by the following competing reactions : Ea(C–H) Potential energy C–H A + B → P1 Rate of formation of P1 = k1[A][B] C–D Ea(C–D) A + B → P2 Rate of formation of P2 = k2[A][B] The relative proportion in which the two products have been produced at a given state of the reaction (before it has reached equilibrium) is given by the Reaction coordinate XtraEdge for IIT-JEE 41 DECEMBER 2011 UNDERSTANDING Inorganic Chemistry 1. An inorganic halide (A) gives the following Bi + NaSnO3 + H2O + 3NaCl reactions. (G) Black ppt. (i) The cation of (A) on raction with H2S in HCl 2Bi + 6HCl ∆ → 2BiCl3 + 3H2 medium, gives a black ppt. of (B). (A) neither gives ppt. with HCl nor blue colour with (G) (A) K4Fe(CN)6. Hence, (ii) (B) on heating with dil.HCl gives back (A) is BiCl3, compound(A) and a gas (C) which gives a black (B) is Bi2S3, ppt. with lead acetate solution. (C) is H2S, (iii) The anion of (A) gives chromyl chloride test. (D) is Bi(NO3)2, (iv) (B) dissolves in hot dil. HNO3 to give a solution, (E) is Bi(OH)3, (F) is BiOCl and (G) is Bi (D). (D) gives ring test. (v) When NH4OH solution is added to (D), a white 2. A colourless solid (A) on heating gives a white solid precipitate (E) is formed. (E) dissolves in (B) and a colourless gas (C). (B) gives off reddish- minimum amount of dil. HCl to give a solution of brown fumes on treating with H2SO4. On treating (A). Aqueous solution of (A) on addition of water with NH4Cl, (B) gives a colourless gas (D) and a gives a whitish turbidity (F). residue (E). The compound (A) on heating with (vi) Aqueous solution of (A) on warming with (NH4)2SO4 gives a colourless gas (F) and white alkaline sodium stannite gives a black precipitate residue (G). Both (E) and (G) impart bright yellow of a metal (G) and sodium stannate. The metal colour to Bunsen flame. The gas (C) forms white (G) dissolves in hydrochloride acid to give powder with strongly heated Mg metal which on solution of (A). hydrolysis produces Mg(OH)2. The gas (D) on Identify (A) to (G) and give balanced chemical heating with Ca gives a compound which on equations of reactions. hydrolysis produces NH3. Identify compounds (A) to Sol. Observation of (i) indicates that cation (A) is Bi3+ (G) giving chemical equations involved. because it does not give ppt. with HCl nor blue Sol. The given information is as follows : colour with K4Fe(CN)6, hence it is neither Pb2+ nor (i) A Heat → B + C Cu2+. Since anion of (A) gives chromyl chloride test, Colourless Solid Colourless hence it contains Cl– ions. Thus, (A) is BiCl3. Its Solid gas different reactions are given below : (i) 2BiCl3 + 3H2S → Bi2S3 + 6HCl (ii) B + H2SO4 ∆ Reddish brown gas → (A) (B) (iii) B + NH4Cl ∆ → D + E (ii) Bi2S3 + 6HCl → 3H2S + 2 BiCl3 Colourless gas (B) (C) (A) (iv) A + (NH4)2SO4 ∆→ F + G ∆ (iii) Bi2S3 + 8HNO3 → 2Bi(NO3)3 + 2NO olourless gas White (B) (D) + 3S + 4H2O Residue (iv) Bi(N O3)3 + 3NH4OH → (v) E and G imparts yellow colour to the flame. (D) Bi(OH)3 ↓ + 3NH4NO3 (vi) C + Mg Heat → White powder (E) White ppt. ∆ O H→ Mg(OH)2 2 Bi(OH)3 + 3HCl → BiCl3 + 3H2O H O (vii) D + Ca Heat → Compound → NH3 2 Dil. (A) Information of (v) indicates that (E) and (G) and also BiCl3 + H2O → BiOCl + 2HCl (A) are the salts of sodium because Na+ ions give (A) (F) yellow coloured flame. Observations of (ii) indicate Bismuth oxychloride that the anion associated with Na+ in (A) may be (White turbidity) NO3–. Thus, the compound (A) is NaNO3. (v) BiCl3 + 2NaOH +Na2[SnO2] → The reactions involved are as follows : (A) XtraEdge for IIT-JEE 42 DECEMBER 2011 (i) 2NaNO3 ∆ → 2NaNO2 + O2 ↑ H2S + 2HNO3 → 2NO2 + 2H2O + S (D) (A) (B) (C) (C) White turbidity (ii) 2NaNO2 + H2SO4 → Na2SO4 + 2HNO2 CuSO4 + H2S → CuS ↓ + H2SO4 (B) Dil. (E) (B) (F) 3HNO2 → HNO3 + H2O + 2NO↑ Black ppt. 2NO + O2 → 2NO2 ↑ 3CuS + 8HNO3 → Reddish brown Dil. 3Cu(NO3)2 + 2NO + 3S + 4H2O Fumes Cu++ + 4NH3 → [Cu(NH3)4]2+ (iii) NaNO2 + NH4Cl → NaCl + N2 ↑ + 2H2O (G) Blue colour (B) (E) (D) [Cu(NH3)4]2+ + 4CH3COOH → Cu2+ + 4CH3COONH4 (iv) 2NaNO3 + (NH4)2SO4 ∆ → Na2SO4 + 2NH3 Cu2+ + [Fe(CN)6]4– → Cu2[Fe(CN)6] (A) (G) (F) (H) 2HNO3 Chocolate colour (v) O2 + 2Mg ∆ H O → 2MgO → Mg(OH)2 2 CuSO4(aq) + BaCl2(aq) → BaSO4 ↓ + CuCl2 (C) (E) White ppt. (vi) N2 + 3Ca ∆→ Ca3N2 Insuluble in HNO3 (D) Hence, Ca3N2 + 6H2O → 3Ca(OH)2 + 2NH3 ↑ (A) is FeS, (B) is H2S, (C) is HNO3, (D) is S, Hence, (E) is CuSO4, (F) is CuS, (G) is [Cu(NH3)4]SO4 and (H) is Cu2[Fe(CN)6] (A) is NaNO3, (B) is NaNO2, 4. (i) An inorganic compound (A) is formed on passing (C) is O2, a gas (B) through a concentrated liquor containing (D) is N2, Na2S and Na2SO3. (E) is NaCl, (ii) On adding (A) into a dilute solution of AgNO3, a (F) is NH3 and (G) is Na2SO4. white ppt. appears which quickly changes into 3. A black coloured compound (A) on reaction with dil. black coloured compound (C). H2SO4 gives a gas (B) which on passing in a solution (iii) On adding two or three drops of FeCl3 into of an acid (C) gives a white turbidity (D). Gas (B) excess of solution of (A), a violet coloured when passed through an acidified solution of a compound (D) is formed. This colour disappears compound (E), gives ppt.(F) which is soluble in quickly. dilute nitric acid. After boiling this solution an excess (iv) On adding a solution of (A) into the solution of of NH4OH is added, a blue coloured compound (G) is CuCl2, a white ppt. is first formed which dissolves produced. To this solution, on addition of CH3COOH on adding excess of (A) forming a compound (E). and aqueous K4[Fe(CN)6], a chocolate ppt. (H) is Identify (A) to (E) and give chemical equations for produced. On addition of an aqueous solution of the reactions at steps (i) to (iv) BaCl2 to aqueous solution of (E), a white ppt. Sol. (i) The compound (A) appears to be Na2S2O3 from its insoluble in HNO3 is obtained. Identify compounds method of preparation given in the problem. (A) to (H). Na2S + Na2SO3 + I2 → 2NaI + Na2S2O3 Sol. From the data on compounds (G) and (H), it may be (B) (A) inferred that the compound (E) contains cupric ions (Cu2+), i.e., (E) is a salt of copper. Since the addition or Na2SO3 + 3Na2S + 3SO2 → 3Na2S2O3 of BaCl2 to (E) gives a white ppt. insoluble in HNO3, (B) (A) it may be said that the anion in the salt is sulphate ion (ii) White ppt. of Ag2S2O3 is formed which is (SO42–). Hence, (E) is CuSO4. hydrolysed to black Ag2S The gas (B) which is obtained by adding dil. H2SO4 Na2S2O3 + 2AgNO3 → 2NaNO3 + Ag2S2O3 ↓ to a black coloured compound (A), may be H2S since White ppt it can cause precipitation of Cu2+ ions in acidic Ag2S2O3 + H2O → Ag2S + H2SO4 medium. The black coloured compound (A) may be (C) ferrous sulphide (iron pyrite). (iii) A violet ferric salt is formed. Hence, the given observation may be explained from 3Na2S2O3 + 2FeCl3 → Fe2(S2O3)3 + 6NaCl the following equations. (D)(violet) Fe S + H2 SO4 → FeSO4 + H2S (iv) 2CuCl2 + 2Na2S2O3 → 2CuCl + Na2S4O6 + 2NaCl (A) Dil. (B) White ppt. XtraEdge for IIT-JEE 43 DECEMBER 2011 2CuCl + Na2S2O3 → Cu2S2O3 + 2NaCl 3Cu2S2O3 + 2Na2S2O3 → Na4[Cu6(S2O3)5] (E) SCIENCE TIPS or 6CuCl + 5Na2S2O3 → Na4[Cu6(S2O3)5] + 6NaCl (E) • What is the expression for growing current, in LR Hence, (A) is Na2S2O3, − t R circuit ? I = I0 1 − e L (B) is I2 or SO2, (C) is Ag2S, (D) is Fe2(S2O3)3 and • What is the range of infrared spectrum ? (E) is Na4[Cu6(S2O3)5]. This covers wavelengths from 10–3 m down to 7.8 × 10–7 m 5. A white amorphous powder (A) on heating yields a • What is the nature of graph between electric field colourless, non-combustible gas (B) and solid (C). and potential energy (U) ? The latter compound assumes a yellow colour on The nature of the graph heating and changes to white on cooling. 'C' dissolves will be parabola having in dilute acid and the resulting solution gives a white symmetry about U-axis precipitate on adding K4Fe(CN)6 solution. 'A' dissolves in dilute HCl with the evolution of gas, • Why no beats can be heard if the frequencies of which is identical in all respects with 'B'. The gas 'B' the two interfering waves differ by more than ten ? turns lime water milky, but the milkiness disappears this is due to persistence with the continuous passage of gas. The solution of of hearing 'A', as obtained above, gives a white precipitate (D) • Why heating systems based on steam are more on the addition of excess of NH4OH and and passing efficient than those based on circulation of hot H2S. Another portion of the solution gives initially a water ? This is because steam white precipitate (E) on the addition of sodium has more heat than water hydroxide solution, which dissolves on futher a the same temperature addition of the base. Identify the compounds A, B, D and E. [IIT-1979] • Can the specific heat of a gas be infinity ? Yes ∆ Sol. (i) ZnCO3 → ZnO + CO2 • What is the liquid ascent formula for a capillary ? (A) (B) 2T cos θ r h= – (ii) ZnO + 2HCl → H2O + ZnCl2 γpg 3 (C) (soluble) where h is the height through (iii) 2ZnCl2 + K4[Fe(CN)6] which a liquid of density ρ and → 4KCl + Zn2[Fe(CN)6]↓ surface tension T rises in a (white ppt) capillary tube of radius r (iv) ZnCO3 + HCl → CO2 + ZnCl2 • What is the expression for total time of flight (T) (A) (soluble) 2u sin θ (v) CO2 + Ca(OH)2 → CaCO3 + H2O for oblique projection ? T= g (B) (Milky) (vi) CaCO3 + CO2 + H2O → Ca(HCO3)2 • The space charge limited current iP in the diode value is given by iP = k Vp3/2 (soluble) (vii) ZnCl2 + H2S → 2HCl + ZnS↓ 4 NH OH • What is an ideal gas ? An ideal gas is one in (white) which intermolecular forces are absent (viii) ZnCl2 + 2NaOH → 2NaCl + Zn(OH)2↓ (white) • Can a rough sea be calmed by pouring oil on its (ix) Zn(OH)2 + 2NaOH → Na2ZnO2 + H2 surface ? Yes sod. ziniate • What is the expression for fringe width (β) in (soluble) Young's double slit experiment? β=Dλ/d where D is the distance between the source and screen and d is distance between two slits XtraEdge for IIT-JEE 44 DECEMBER 2011 Set `tà{xÅtà|vtÄ V{tÄÄxÇzxá 8 This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants. By : Shailendra Maheshwari Solutions will be p ublished in next issue Joint Director Academics, Career Point, Kota 1. Show that the six planes through the middle point of 1 each edge of a tetrahedron perpendicular to the opposite edge meet in a point. 8. ∫ If n ≥ 2 and In = (1 − x 2 ) n cos mx dx, then show that −1 m2In = 2n(2n – 1) In–1 – 4n(n – 1) In–2. 2. Prove that if the graph of the function y = f (x), defined throughout the number scale, is symmetrical 9. Find the sum to infinite terms of the series about two lines x = a and x = b, (a < b), then this 3 5 7 9 11 function is a periodic one. + + + + + ........ ∞ 4 36 144 400 900 3. Show that an equilateral triangle is a triangle of 10. ABC is a triangle inscribed in a circle. Two of its maximum area for a given perimeter and a triangle of sides are parallel to two given straight lines. Show minimum perimeter for a given area. that the locus of foot of the perpendicular from the centre of the circle on to the third side is also a circle, 4. Let az2 + bz + c be a polynomial with complex concentric to the given circle. coefficients such that a and b are non zero. Prove that the zeros of this polynomial lie in the region |z|≤ b + c MEMORABLE POINTS a b • The vector relation between linear velocity and 5. An isosceles triangle with its base parallel to the → → → x2 y2 angular velocity is v =ω× r major axis of the ellipse + = 1 is a2 b2 • In the case of uniform circular motion the angle between circumscribed with all the three sides touching the → → → ω and r is always 90º(hence | v | = ωr ellipse. Find the least possible area of the triangle. • The relation between Faraday constant F, Avogadro 6. If one of the straight lines given by the equation number N and the electronic charge e is F = Ne ax2 + 2hxy + by2 = 0 coincides with one of those given by a′x2 + 2h′xy + b′y2 = 0 and the other lines • Depolariser used in Lechlanche cell is represented by them be perpendicular, show that Manganese dioxide ha´b´ h´ab´ = • The absorption or evolution of heat at a junction of b´− a´ b−a two dissimilar metals when a current is passed is 7. Prove that known as Peltier effect n m n m + 1 n m + 2 • The part of the human ear where sound is transduced + 0 n 1 n + 2 n + ......... is the Cochlea .... to (n + 1) terms • Similar trait resulting from similar selection pressure n m n m n m acting on similar gene pool is termed = + 2 + 22 + ........ 0 0 1 1 2 2 Parallel evolution ..... to (n + 1) terms • Group of related species with the potential, directly or indirectly, of forming fertile hybrids with one another is called Coenospecies XtraEdge for IIT-JEE 45 DECEMBER 2011 MATHEMATICAL CHALLENGES SOLUTION FOR NOVEMBER ISSUE (SET # 7) 3. The line PQ always passes through (α, β) so it is 1. Let the line be y = 2x + c y –β = m(x – α) 9 − c 9 + 2c Let the circle be x2 + y2 – 2hx – 2ky = 0 Point A , 6 3 Joint equation of OP and OQ. 2c − 3 + c − 6 ( y − mx) Point B , x2 + y2 – 2 (hx + ky) =0 −3 −3 β − mα c + 6 5c + 12 P Point C , 3 3 1 2c − 3 c + 6 O mid point of B & C is . + , 2 −3 3 (h,k) 1 − c + 6 5c + 12 9 − c 2c + 9 +3 + 2 3 = 6 , 3 Q which is point A, so AB and AC are equal. 2k 2 h − mk 2hn 2 β − mα y – 2 β − mα xy + 1 + β − mα x = 0 1 − A 2 2 It must represent y – x = 0 2. b h − mk a so = 0 ⇒ m = h/k ...(1) β − mα C 2k 2hm B D and 1– = –1 – β − mα β − mα 1 AB 1 AC ⇒ β – mα – 2k = –β + mα – 2hm a +b = . + . AB AB AC AC ⇒ –β + mα + k – hm = 0 1 1 ⇒ –β + k + h/k(α – h) = 0 (using (1) in it) = AB + AC AB2 AC 2 ⇒ k2 – βy + αh – h2 = 0 so required locus is 1 1 x2 – y2 – αx + βy = 0 = (AD + DB) + (AD + DC) AB 2 AC 2 π π 4. As |f (x)| ≤ |tan x| for ∀ x ∈ − , 1 1 DB DC 2 2 = + AD + + AB 2 2 AC BD.DC CD.CB so f (0) = 0 so |f (x) – f (0)| ≤ |tan x| 1 1 DB DC 1 divides both sides by |x| = 2 + 2 AD + + BD CD BC AB AC f ( x ) − f ( 0) tan x ⇒ ≤ 1 1 x x = AD . + BD.DC CD.CB f ( x ) − f ( 0) tan x ⇒ lim ≤ lim AD 1 1 x →0 x x →0 x = . + CD BD CD ⇒ |f (0)| ≤ 1 1 1 1 AD DC + BD ⇒ a1 + a 2 + a 3 + ..... + a n ≤ 1 = . 2 3 n BC BD.CD n ai = AD BD.CD = AD 2 = AD 1 . AD AD ⇒ ∑i ≤1 AD i =1 1 so it is vector along AB with magnitude . 5. Let the number is xyz, here x < y and z < y. AD Let y = n, then x can be filled in (n – 1) ways. 1 (i.e. from 1 to (n – 1)) and z can be filled in n ways |a +b | = AD (i.e. from 0 to (n – 1)) XtraEdge for IIT-JEE 46 DECEMBER 2011 here 2 ≤ n ≤ 9 3 (tan A . tan B . tan C)1/3 so total no. of 3 digit numbers with largest middle and for a triangle tan A + tan B + tan C digit = tan A . tan B . tan C 9 9 9 so (tan A . tan B . tan C)2/3 ≥ 3 = ∑ n=2 n(n − 1) = ∑n=2 n2 – ∑n n=2 ⇒ tan A . tan B . tan C ≥ 3 3 ⇒ tan2A + tan2B + tan2C 9.10.19 9.10 = – ≥ 3(tan A. tan B tan C)2/3 ≥ 3.3 6 2 a2 b2 c2 = 285 – 45 = 240 so from (1), 2 + 2 + 2 ≥ 144. 240 r1 r2 r3 required probability = 9 × 10 × 10 8. Z1, Z2, Z3 are centroids of equilateral triangles ACX, 8 ABY and BCZ respectively. = 30 Z − Z A iπ/6 4 Z1 – ZA = (ZC – ZA) 1 e = ZC − ZA 15 ZA x 6. The region bounded by the curve y = log2(2 – x) and the inequality (x – |x|)2 + (y – |y|)2 ≤ 4 is required area is A Z1 y Z2 (–1,log23) B C ZB ZC Z3 (–1,0) (1,0) z (3/2,–1) (0,–1) 1 3 i Z1 – ZA = (ZC – ZA) + ...(1) 1 0 3 2 2 1 ∫ log ∫ (2 − 2 y = 2 (2 − x) dx + ) dy + π similarly, 4 −1 −1 1 3 i 2 1 π Z2 – ZA = (ZB – ZA) − ...(2) = log 2 3 − + 2 log 2 3 + 2 – + 3 2 2 ln 2 2ln2 4 1 i e2 e π So, Z1 – Z2 = (ZC – ZB) + (ZC + ZB – 2ZA) = – log2 +2+ sq units 2 2 3 27 4 ...(3) A 1 similarly Z2 – Z3 = (ZA – ZC) 2 F E i 7. + (ZA + ZC – 2ZB) ..(4) M 2 3 To prove ∆xyz as equilateral triangle, we prove that B C (Z3 – Z2)eiπ/3 = Z1 – Z2 D 1 So, (Z3 – Z2)eiπ/3 = ( (ZC – ZA) 2 ∠BMC = 2∠BAC = 2∠BMD 1 i 3 BD BC BC a – (ZA + ZC – 2ZB)) + i so tan A = = = = 2 3 2 2 MD 2MD 4r1 4r1 1 i a2 = (ZC – ZB) + (ZC + ZB – 2ZA) so = tan2A 2 2 3 r12 = Z1 – Z2 a2 b2 c2 so + + r12 r22 r32 2 = 16 (tan A + tan2B + tan2C) ...(1) Now as tan A + tan B + tan C ≥ XtraEdge for IIT-JEE 47 DECEMBER 2011 a 1 − r cos u 9. Tr = 2 ∫ 1 − 2r cos u + r 2 du. ...(1) FRACTIONAL DISTILLATION a 0 OF AIR 1 − 2r cos u + r 2 − r 2 + 1 = ∫ 0 1 − 2r cos u + r 2 du a 1− r2 = ∫ 1 + 1 − 2r cos u + r 2 0 du = a + (1 – r2) × a sec 2 u / 2 ∫ (1 + r 0 2 )(1 + tan 2 u / 2) − 2r (1 − tan 2 u / 2) a sec 2 u / 2 = a + (1 – r2) ∫ (1 + r ) 0 2 tan 2 u / 2 + (1 − r ) 2 a 1− r2 sec 2 u / 2 du =a+ (1 + r ) 2 ∫ (1 − r ) 2 Did you know that the air we breathe isn’t just 0 tan 2 u / 2 + oxygen, infact it’s made up of a number of different (1 + r ) 2 gases such as nitrogen, oxygen, carbon dioxide, Let tan u/2 = t argon, neon and many others. Each of these tan a / 2 1− r2 2 dt gases carry useful properties so separating them so, Tr = a + (1 + r ) 2 ∫ 1− r 2 from the air around us is extremely beneficial. 0 t2 + 1+ r The process is called fractional distillation and tan a / 2 2(1 − r ) 1 + r −1 1 + r 2 consists of two steps, the first relies on cooling the =a+ tan t 1− r air to a very low temperature (i.e. converting it into (1 + r ) 2 1 − r 0 a liquid), the second involves heating it up thus 2(1 + r )(r − 1) π Now lim Tr = a – = a–π allowing each gas within the mixture to evaporate at r →1 + (1 + r )(r − 1) 2 its own boiling point. The key to success here is that 2(1 − r )(r + 1) π every element within air has its own unique boiling and lim Tr = a + = a+π r →1+ (1 + r )(r − 1) 2 temperature. As long as we know these boiling a temperatures we know when to collect each gas. and (from (1)) T1 = ∫ du = a 0 So what are the real world benefits of separating Hence lim Tr, T1, lim Tr form an A.P. with and extracting these gases? Well liquid oxygen is r →1+ r →1− used to power rockets, oxygen gas is used in common difference π. breathing apparatus, nitrogen is used to make 10. Let α, β, γ be the three real roots of the equation fertilizers, the nitric acid component of nitrogen is without loss of generality, it can be assumed that used in explosives. α ≤ β ≤ γ. so x2 + ax2 + bx + c = (x – γ) (x2 + (a + γ) x + (γ2 + aγ + b)) The other gases all have their own uses too, for where – γ (γ2 + aγ + b) = c, as γ is the root of given example argon is used to fill up the empty space in equation, so x2 + (a + γ) x + (γ2 + aγ + b) = 0 must most light bulbs (thanks to its unreactive nature). have two roots i.e. α and β. So its discriminant is non Carbon dioxide is used in fire extinguishers and is negative, thus great for putting out fires in burning liquids and (γ + a)2 – 4(γ2 + aγ + b) ≥ 0 ⇒ 3γ2 + 2aγ – a2 + 4b ≤ 0 electrical fires. There really are too many uses to list − a + 2 a 2 − 3b so γ ≤ but suffice it to say that fractional distillation is an 3 extremely useful process for humans the world so greatest root is also less than or equal to over. − a + 2 a 2 − 3b . 3 XtraEdge for IIT-JEE 48 DECEMBER 2011 Students' Forum Expert’s Solution for Question asked by IIT-JEE Aspirants MATHS 1. A man parks his car among n cars standing in a row, The line of intersection OC of the planes OAC and his car not being parked at an end. On his return he OBC, i.e., y + z = 0 & x + y = 0, is given by x = – y = z, finds that exactly m of the n cars are still there. What x–0 y–0 z–0 is the probability that both the cars parked on two i.e., = = . ...(ii) 1 –1 1 sides of his car, have left? The line of intersection OB of the planes OBC and Sol. Clearly, his car is at one of the crosses (×). OAB, i.e., x + y = 0 and z + x = 0, is given by |× × × . . . × × ×| – x = y = z, The number of the ways in which the remaining x–0 y–0 z–0 m–1 cars can take their places (excluding the car of i.e., = = . ...(iii) the man) = n–1Cm–1 {Q there are n – 1 places for the –1 1 1 m – 1 cars}. The number of ways in which the The line BC is the intersection of the planes remaining m – 1 cars can take places keeping the two O (0,0,0) places on two sides of his car vacant = n–3Cm–1 1,1,–1 P ∴ the required probability A n –3 n( E ) C m –1 C = = n –1 n( S ) C m –1 (0,0,a) (n – 3) ! (m – 1) ! (n – m) ! = × (m – 1)!(n – m – 2) ! (n – 1) ! Q 1,–1,0 (n – m)(n – m – 1) B = . (n – 1)(n – 2) x + y = 0 and x + y + z = a ⇒ y = – x, z = a 2. Show that the shortest distance between any two x y z–a opposite edges of the tetrahedron formed by the ⇒ = = . ...(iv) 1 –1 0 planes y + z = 0, z + x = 0, x + y = 0 and x + y + z = a 2a Similarly, CA has the equations is . y + z = 0, x + y + z = a 6 x–a y z Sol. Clearly, the planes y + z = 0, z + x = 0 and x + y = 0 ⇒ = = ...(v) pas through the origin O(0, 0, 0,). 0 1 –1 O AB has the equation z + x = 0, x + y + z = a x y–a z ⇒ = = –1 0 1 C Let PQ be the shortest distance between OA and BC, A and its line has the direction cosines l, m, n. Then using perpendicularity, l .1 + m. 1 + n . (–1) = 0 B and l . 1 + m (–1) + n . 0 = 0 The line of intersection OA of the planes OAC and OAB, ⇒ l + m – n = 0 and l – m = 0 i.e., y + z = 0 and z + x = 0, l m n l2 + m2 + n2 1 ⇒ = = = = . is given by x = y = – z, i.e., 1 1 2 2 2 2 6 1 +1 + 2 x–0 y–0 z–0 = = . ...(i) 1 1 –1 XtraEdge for IIT-JEE 49 DECEMBER 2011 1 1 2 ( y '– x' ) 2 1 y '–x' ∴ l= ,m= ,n= . + + + =1 6 6 6 2 4 2 ∴ the shortest distance PQ 1 1 1 = projection of OC on the line PQ or {(x' + y')2 + (y' – x')2} + + (y' – x' – x' – y') 2 2 2 = |(x2 – x1) l + (y2 – y1)m + (z2 – z1)n| 3 3 + 3(y'2 – x'2) – + . 2x' = 1 1 1 2 2a 2 2 = ( 0 – 0) + (0 – 0) + ( a – 0) = 6 6 6 6 or x'2 + y'2 – 2 x' + 3y'2 – 3x'2 + 3 2 x' = 2 Similarly, other shortest distances between opposite edges are or 4y'2 – 2x'2 + 2 2 x' = 2 1 2 2 2a or x'2 – 2y'2 – 2 x' + 1 = 0 (0 – 0) + ( a – 0) + (0 – 0) ' i.e., ' 6 6 6 6 ∴ the changed equation of the locus is (obtained by dropping dashes) 2 1 1 2a and (a – 0) + (0 – 0) + (0 – 0) ' i.e., . x2 – 2y2 – 2 x + 1 = 0. 6 6 6 6 Hence, the problem. 4. A wheel with diameter AB touches the horizontal ground at A. The rod BC is fixed at B, ABC being 3. Find the changed equation of the locus x2 + 6xy + y2 = 1 vertical. A man from a point P on the ground at a when the lines x + y = 0 and x – y + 1 = 0 are taken as distance d from A, finds that the angle of elevation of the new x and y axes respectively. C is α. The wheel turns about the fixed centre O of Sol. Here the lines x + y = 0 and x – y + 1 = 0 are the wheel such that C turns away from the man and perpendicular to each other. its angle of elevation is β when it is about to disappear. Find PC when C is about to disappear. x – y +1 So, take x' = ...(i) Sol. When C is about to disappear, let it be at C' and then 12 + (–1) 2 PC' touches the wheel at Q. C' C x– y and y' = ...(ii) 12 + 12 From (i), x – y + 1 = 2 x' ...(iii) B Q From (ii), x + y = 2 y' ...(iv) B' (iii) + (iv) ⇒ 2x + 1 = 2 ( x'+ y ' ) Ο A' x'+ y ' 1 β/2 β α ∴ x= – 2 2 Α d P From the question, we have to find PC'. (iv) – (iii) ⇒ 2y – 1 = 2 (y' – x') Clearly, PQ = PA = d and OC' = OC = AC – r, y '–x' 1 r being the radius of the wheel. ∴ y= + 2 2 AC In the ∆CAP, tan α = ; ∴ AC = d tan α ∴ Putting these in the equation of the locus, we get d x'+ y ' 1 2 x'+ y ' 1 2 y '–x' 1 ∴ OC + r = d tan α; ∴ OC' = d tan α – r – + 6 – + 2 2 2 2 2 2 ∴ C'Q = OC' 2 –OQ 2 = (d tan α – r ) 2 – r 2 2 y '– x' 1 ∴ PC' = PQ + C'Q = d + (d tan α – r ) 2 – r 2 ...(i) + + =1 2 2 Also, from the ∆OAP, ( x'+ y ' ) 2 1 x'+ y ' β r β or + – tan = ; ∴ r = d tan . 2 4 2 2 d 2 β y ' 2 – x' 2 1 1 ∴ PC' = d + d 2 tan 2 α – 2d tan α . d tan + 6 – + ( x'+ y '– y '+ x' ) 2 2 4 2 2 XtraEdge for IIT-JEE 50 DECEMBER 2011 {from (i)} 1 Putting 1 + = z, β x4 = d 1 + tan 2 α – 2 tan α. tan 2 –4 dx 1 5 dx = dz, i.e., 5 =– dz x x 4 β = d 1 + tan α. 1 – 2 cot α . tan . 1 –1 1 –3 / 4 2 ∴ I= ∫z 3/ 4 . 4 dz = – 4 z ∫dz 2 1 n c 5. If n ∈ N and ck = Ck find the value ofn k =1 k . k . c k –1 ∑ 3 =– 1 . 4 1 +c z4 n 4 ck Ck n! (k – 1) !(n – k + 1) ! Sol. = n = . 1 1 ck –1 C k –1 k !( n – k ) ! n! 1 4 = – z4 + c = – 1 + 4 + c n – k +1 x = k 1 n 2 n 2 (1 + x 4 ) 4 ck n – k +1 =– + c. ∴ ∑ k . c k =1 3 = k –1 ∑ k . k =1 3 k x (b) Here we have second and third roots of x. n = ∑ k (n – k + 1) k =1 2 The LCM of 2 and 3 = 6 So, Put x = z6; then dx = 6z5 dz n = ∑ k{(n + 1) k =1 2 – 2(n + 1)k + k 2 } ∴ I= ∫z 3 z3 + z2 . 6z5dz n n n z6 = (n + 1)2 ∑ k =1 k – 2(n + 1) ∑ k =1 k2 + ∑ k =1 k3 =6 ∫ z +1 dz n(n + 1)(2n + 1) ( z 6 – 1) + 1 = (n + 1) . n(n + 1) 2 2 – 2(n + 1) . 6 =6 ∫ z +1 dz n 2 (n + 1) 2 ( z 3 + 1)( z 3 – 1) dz + 4 = 6 ∫ z +1 dz + 6 ∫ z +1 = n(n + 1) 2 2 2 .(n + 1) – (2n + 1) + 3 n 2 ∫ = 6 ( z 2 – z + 1)( z 3 – 1) dz +6log(1 + z) = 6 ∫ (z 5 – z 4 + z 3 – z 2 + z – 1) dz + 6log (1 + z) n(n + 1) 2 6(n + 1) – 4(2n + 1) + 3n = . 2 6 z6 z5 z 4 z3 z 2 1 =6 – + – + – z + 6 log (1+z)+ c = n(n + 1) 2 . (n + 2) 6 5 4 3 2 12 6 5/ 6 3 dx =x– x + x 2 / 3 – 2x1/2 + 3x1/3 – 6x1/6 6. Evaluate (a) ∫x 2 (1 + x 4 ) 3 / 4 5 2 + 6 log (1 + x1/6) + c x (b) ∫ x +3 x dx dx Sol. (a) I = ∫ 1 3/ 4 x 2 .x 3 4 + 1 x 1 dx = ∫ 1 3/ 4 . x5 . 1 + 4 x XtraEdge for IIT-JEE 51 DECEMBER 2011 MATHS MONOTONICITY, MAXIMA & MINIMA Mathematics Fundamentals Monotonic Functions : or negative according as f (x) is monotonic increasing A function f (x) defined in a domain D is said to be or decreasing at x = a. (i) Monotonic increasing : So at x = a, function f (x) is x < x 2 ⇒ f ( x1 ) ≤ f ( x 2 ) monotonic increasing ⇔ f ´(a) > 0 ⇔ 1 ∀ x1, x2 ∈ D x1 > x 2 ⇒ f ( x1 ) ≥ f ( x 2 ) monotonic decreasing ⇔ f ´(a) < 0 y (ii) In an interval : In [a, b], f (x) is y monotonic increasing ⇔ f '(x) ≥ 0 monotonic decreasing ⇔ f '(x) ≤ 0 ∀ x ∈ (a, b) constant ⇔ f '(x) = 0 x x Note : O O (i) In above results f ´(x) should not be zero for all x < x 2 ⇒ f ( x1 ) > f ( x 2 ) / values of x, otherwise f (x) will be a constant i.e., ⇔ 1 ∀ x1, x2 ∈ D x1 > x 2 ⇒ f ( x1 ) < f ( x 2 ) / function. (ii) Monotonic decreasing : (ii) If in [a, b], f ´(x) < 0 at least for one value of x and f ´(x) > 0 for at least one value of x, then f (x) x < x 2 ⇒ f ( x1 ) ≥ f ( x 2 ) will not be monotonic in [a, b]. ⇔ 1 ∀ x1, x2 ∈ D x1 > x 2 ⇒ f ( x1 ) ≤ f ( x 2 ) Examples of monotonic function : y y If a functions is monotonic increasing (decreasing ) at every point of its domain, then it is said to be monotonic increasing (decreasing) function. In the following table we have example of some monotonic/not monotonic functions x x O O Monotonic Monotonic Not increasing decreasing monotonic x < x 2 ⇒ f ( x1 ) < f ( x 2 ) / i.e., ⇔ 1 ∀ x1, x2 ∈ D x3 1/x, x > 0 x2 x1 > x 2 ⇒ f ( x1 ) > f ( x 2 ) / x|x| 1 – 2x |x| A function is said to be monotonic function in a ex e –x ex + e–x domain if it is either monotonic increasing or monotonic decreasing in that domain. log x log2x sin x Note : If x1 < x2 ⇒ f (x1) < f (x2) ∀ x1, x2 ∈ D, then sin h x cosec h x, x > 0 cos h x f (x) is called strictly increasing in domain D and [x] cot h x, x > 0 sec h x similarly decreasing in D. Method of testing monotonicity : Properties of monotonic functions : (i) At a point : A function f (x) is said to be If f (x) is strictly increasing in some interval, then in monotonic increasing (decreasing) at a point x = a of that interval, f – 1 exists and that is also strictly its domain if it is monotonic increasing (decreasing) increasing function. in the interval (a – h, a + h) where h is a small If f (x) is continuous in [a, b] and differentiable in positive number. Hence we may observer that if f (x) (a, b), then is monotonic increasing at x = a then at this point f ´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f (x) is monotonic increasing tangent to its graph will make an acute angle with in [a, b] x-axis where as if the function is monotonic decreasing there then tangent will make an obtuse f ´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f (x) is monotonic angle with x-axis. Consequently f ´(a) will be positive decreasing in [a, b] XtraEdge for IIT-JEE 52 DECEMBER 2011 If both f (x) and g (x) are increasing (or decreasing) in Least value = min. {f (a), f (b), f (c)} [a, b] and gof is defined in [a, b], then gof is where x = c is a point such that f´(c) = 0. increasing. If a continuous function has only one maximum If f (x) and g (x) are two monotonic functions in [a, b] (minimum) point, then at this point function has such that one is increasing and other is decreasing its greatest (least) value. then gof, it is defined, is decreasing function. Monotonic functions do not have extreme points. Maximum and Minimum Points : Conditions for maxima and minima of a function The value of a function f (x) is said to be maximum at Necessary condition : A point I = a is an extreme x = a if there exists a small positive number δ such point of a function f (x) if f ´(a) = 0, provided f ´(a) that f (a) > f (x) exists. Thus if f ´(a) exists, then y x = a is an extreme point ⇒ f ´(a) = 0 or f ´(a) ≠ 0 ⇒ x = a is not an extreme point But its converse is not true i.e. f ´(a) = 0 ⇒ x = a is an extreme point. / For example if f (x) = x3, then f ´(0) = 0 but x = 0 is x O (a) (b) (c) not an extreme point. Also then the point x = a is called a maximum point Sufficient condition : For a given function f (x), a for the function f (x). point x = a is Similarly the value of f (x) is said to be minimum at x = b a maximum point if f ´(a) = 0 and f´´(a) < 0 if there exists a small positive number δ such that a minimum point if f´(I) = 0 and f ´´(a) > 0 f (b) < f (x) ∀ x ∈ (b – δ, b + δ) not an extreme point if f ´(a) = 0 = f ´´(a) and Also then the point x = b is called a minimum point f ´´´(a) ≠ 0. for f (x) Note : If f ´(a) = 0, f ´´(a) = 0, f ´´´(a) = 0 then the Hence we find that : sign of f(4)(a) will determine the maximum or minimum point as above. (i) x = a is a maximum point of f (x) Working Method : f ( a ) – f (a + h) > 0 ⇔ Find f ´(x) and f ´´(x). f ( a ) – f (a – h) > 0 Solve f ´(x) = 0. Let its roots be a, b, c, ... (ii) x = b is a minimum point of f(x) Determine the sign of f ´´(x) at x = a, b, c, .... and f (b) – f (b + h) > 0 decide the nature of the point as mentioned above. ⇔ Properties of maxima and minima : f (b) – f (b – h) > 0 If f (x) is continuous function, then (iii) x = c is neither a maximum point nor a minimum point Between two equal values of f (x), there lie atleast one maxima or minima. f (c ) – f (c + h ) Maxima and minima occur alternately. For example ⇔ and have opposite signs. if x = –1, 2, 5 are extreme points of a continuous f (c ) – f (c – h ) function and if x = –1 is a maximum point then x = 2 will be a minimum point and x = 5 will be a Where h is a very small positive number. maximum point. Note : When x passes a maximum point, the sign of dy/dx The maximum and minimum points are also changes from + ve to – ve, where as when x passes known as extreme points. through a minimum point, the sign of f ´(x) changes A function may have more than one maximum from –ve to + ve. and minimum points. If there is no change in the sign of dy/dx on two sides A maximum value of a function f (x) in an of a point, then such a point is not an extreme point. interval [a, b] is not necessarily its greatest value If f (x) is maximum (minimum) at a point x = a, then in that interval. Similarly a minimum value may 1/f (x), [f (x) ≠ 0] will be minimum (maximum) at that not be the least value of the function. A minimum point. value may be greater than some maximum value If f (x) is maximum (minimum) at a point x = a, then for for a function. any λ ∈ R, λ + f (x), log f (x) and for any k > 0, k f (x), The greatest and least values of a function f (x) in an interval [a, b] may be determined as follows : [f (x)]k are also maximum (minimum) at that point. Greatest value = max. {f (a), f (b), f (c)} XtraEdge for IIT-JEE 53 DECEMBER 2011 MATHS FUNCTION Mathematics Fundamentals Definition of a Function : f + g, f – g, fg, f /g, fog Let A and B be two sets and f be a rule under which and they are defined as follows : every element of A is associated to a unique element (f + g) (x) = f (x) + g(x) of B. Then such a rule f is called a function from A to B and symbolically it is expressed as (f – g) (x) = f (x) – g(x) f :A→B (f g) (x) = f (x) f (g) or A f → B (f /g) (x) = f (x)/g(x) (g(x) ≠ 0) Function as a Set of Ordered Pairs (fog) (x) = f [g(x)] Every function f : A → B can be considered as a set Formulae for domain of functions : of ordered pairs in which first element is an element Df ± g = Df ∩ Dg of A and second is the image of the first element. Thus Dfg = Df ∩ Dg f = {a, f (a) /a ∈ A, f (a) ∈ B}. Df/g = Df ∩ Dg ∩ {x |g(x) ≠ 0} Domain, Codomain and Range of a Function : Dgof = {x ∈ Df | f(x) ∈ Dg} If f : A → B is a function, then A is called domain of D = Df ∩ {x |f (x) ≥ 0} f f and B is called codomain of f. Also the set of all images of elements of A is called the range of f and it Classification of Functions is expressed by f (A). Thus 1. Algebraic and Transcendental Functions : f (A) = {f (a) |a ∈ A} Algebraic functions : If the rule of the function obviously f (A) ⊂ B. consists of sum, difference, product, power or roots of a variable, then it is called an algebraic Note : Generally we denote domain of a function f by function. Df and its range by Rf. Transcendental Functions : Those functions Equal Functions : which are not algebraic are named as Two functions f and g are said to be equal functions transcendental or non algebraic functions. if 2. Even and Odd Functions : domain of f = domain of g Even functions : If by replacing x by – x in f (x) there in no change in the rule then f (x) is called codomain of f = codomain of g an even function. Thus f (x) = g(x) ∀ x. f (x) is even ⇔ f (– x) = f (x) Algebra of Functions : Odd function : If by replacing x by – x in f (x) If f and g are two functions then their sum, there is only change of sign of f (x) then f (x) is difference, product, quotient and composite are called an odd function. Thus denoted by f (x) is odd ⇔ f (– x) = – f (x) XtraEdge for IIT-JEE 54 DECEMBER 2011 3. Explicit and Implicit Functions : Function Period Explicit function : A function is said to be sin x, cos x, sec x, cosec x, 2π explicit if its rule is directly expressed (or can be expressed( in terms of the independent variable. tan x, cot x π Such a function is generally written as sinnx, cosn x, secn x, cosecn x 2π if n is odd y = f (x), x = g(y) etc. π if n is even Implicit function : A function is said to be tann x, cotnx π∀n∈N implicit if its rule cannot be expressed directly in terms of the independent variable. Symbolically |sin x|, |cos x|, |sec x|, |cosec x| π we write such a function as |tan x|, |cot x|, π f (x, y) = 0, φ(x, y) = 0 etc. |sin x| + |cos x|, sin4x + cos4x π 4. Continuous and Discontinuous Functions : 2 |sec x| + |cosec x| Continuous functions : A functions is said to be |tan x| + |cot x| π continuous if its graph is continuous i.e. there is 2 no gap or break or jump in the graph. Discontinuous Functions : A function is said to x – [x] 1 be discontinuous if it has a gap or break in its Period of f (x) = T ⇒ period of f (ax + b)= T/|a| graph atleast at one point. Thus a function which is not continuous is named as discontinuous. Period of f1(x) = T1, period of f2(x) = T2 5. Increasing and Decreasing Functions : ⇒ period of a f1(x) + bf2(x) ≤ LCM {T1, T2} Increasing Functions : A function f (x) is said to Kinds of Functions : be increasing function if for any x1, x2 of its domain One-one/ Many one Functions : x1 < x2 ⇒ f (x1) ≤ f (x2) A function f : A → B is said to be one-one if different elements of A have their different or x1 > x2 ⇒ f (x1) ≥ f (x2) images in B. Decreasing Functions : A function f (x) is said to Thus be decreasing function if for any x1, x2 of its domain a≠b ⇒ f (a) ≠ f (b) f is one-one ⇔ or x1 < x2 ⇒ f (x1) ≥ f(x2) f (a) = f (b) ⇒ a=b or x1 > x2 ⇒ f (x1) ≤ f (x2) A function which is not one-one is called many Periodic Functions : one. Thus if f is many one then atleast two A functions f (x) is called a periodic function if there different elements have same f -image. exists a positive real number T such that Onto/Into Functions : A function f : A → B is f (x + T) = f (x) ∀ x said to be onto if range of f = codomain of f Also then the least value of T is called the period of Thus f is onto ⇔ f (A) = B the function f (x). Hence f : A → B is onto if every element of B Period of f (x) = T (co-domain) has its f –preimage in A (domain). ⇒ Period of f (nx + a) = T/n A function which is not onto is named as into Periods of some functions : function. Thus f : A → B is into if f (A) ≠ B. i.e., XtraEdge for IIT-JEE 55 DECEMBER 2011 if there exists atleast one element in codomain of f Domain and Range of some standard functions : which has no preimage in domain. Function Domain Range Note : Polynomial R R Total number of functions : If A and B are finite function sets containing m and n elements respectively, Identity R R then function x total number of functions which can be defined Constant R {c} from A to B = nm. function c Reciprocal R0 R0 total number of one-one functions from A to B function 1/x n P if m≤n x2, |x| R R+ ∪ {0} = m 0 if m>n x3, x |x| R R Signum R {–1, 0, 1} total number of onto functions from A to B function (if m ≥ n) = total number of different n groups of x + |x| R R+ ∪ {0} m elements. x – |x| R R– ∪ {0} Composite of Functions : [x] R Z Let f : A → B and g : B → C be two functions, then x – [x] R [0, 1) the composite of the functions f and g denoted by gof, x [0, ∞) [0, ∞) is a function from A to C given by gof : A → C, (gof) (x) = g[f (x)]. ax R R+ Properties of Composite Function : log x R+ R The following properties of composite functions can sin x R [–1, 1] easily be established. cos x R [–1, 7] Composite of functions is not commutative i.e., tan x R – {± π/2, ± 3π/2, ...} R cot x R – {0, ± π. ± 2π, ..... R fog ≠ gof sec x R – (± π/2, ± 3π/2, ..... R – (–1, 1) Composite of functions is associative i.e. cosec x R – {0, ±π, ± 2π, ......} R –(–1, 1) (fog)oh = fo(goh) sinh x R R Composite of two bijections is also a bijection. cosh x R [1, ∞) Inverse Function : tanh x R (–1, 1) If f : A → B is one-one onto, then the inverse of f i.e., coth x R0 R –[1, –1] f –1 is a function from B to A under which every b ∈ B sech x R (0, 1] is associated to that a ∈ A for which f (a) = b. cosech x R0 R0 –1 –1 Thus f : B → A, sin x [–1, 1] [–π/2, π/2] f –1(b) = a ⇔ f (a) = b. cos–1x [–1, 1] [0, π] –1 tan x R (–π/2, π/2} –1 cot x R (0, π) –1 sec x R –(–1, 1) [0, π] – {π/2} –1 cosec x R – (–1, 1) (– π/2, π/2] – {0} XtraEdge for IIT-JEE 56 DECEMBER 2011 XtraEdge for IIT-JEE 57 DECEMBER 2011 Based on New Pattern a IIT-JEE 2012 XtraEdge Test Series # 8 Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 6 are passage based questions. +4 marks will be awarded for correct answer and –1 mark for wrong answer. • Question 7 to 9 are Reason and Assertion type question with one is correct answer. +4 marks and –1 mark for wrong answer. • Question 10 to 15 are Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answer and –1 mark for wrong answer. • Question 16 to 18 are Numerical Response Question (four digit Ans. type) +6 marks will be awarded for correct answer and –1 mark for wrong answer. (C) If the speed of water is 4 m/sec then direction is which PHYSICS 3 he should row his boat is cos–1 w.r.t. bank 4 This section contains 2 paragraphs; each has 3 multiple (D) If the speed of water is 4 m/sec then direction in choice questions. (Questions 1 to 6) Each question has 4 3 choices (A), (B), (C) and (D) out of which ONE OR which he should row his boat is sin–1 w.r.t bank MORE THAN ONE may be correct. Mark your 4 response in OMR sheet against the question number of 2. The velocity of boatman is still water is 3m/sec and that question. + 4 marks will be given for each correct river is flowing at 2 m/sec. To cross the river in answer and – 1 mark for each wrong answer. minimum time. Passage # 1 (Ques. 1 to 3) (A) The boatman should move at an angle → 2 There is a river which is flowing at the rate v r . If a θ = sin–1 w.r.t bank 3 → boatman starts to row his boat at the speed v br with (B) The boatman should move at an angle respect to river then the velocity of boatman with 2 θ = cos–1 w.r.t bank → → → respect to ground can be given by, v b = v br + v r . 3 Now a boat can travel at a speed of 3m/s in still (C) the batman should move at an angle 90º w.r.t bank water. A boatman has to across the river to reach the (D) None of these other side. Now give the answer of following 3. Two boats A & B move away from a buoy anchored question at the middle of river along the mutually perpendicular 1 straight lines : the boat A along the 1. The boatman wants to cross the river in such a way river & boat B across the river. Having moved off an that he should cover the shortest possible distance. equal distance from the buoy the boats returned. (A) If the speed of water is 2m/sec then direction in which Then the ratio of times of boats tA/tB. If the velocity 2 of each boat w.r.t flow of water ‘x’ times greater than he should row his boat is cos–1 w.r.t. bank the stream velocity tA/tB is - 3 x x (B) If the speed of water is 2m/sec then direction in which (A) (B) 2 x −1 x2 +1 2 he should row his boat is sin–1 w.r.t bank 1 1 3 (C) (D) 2 x −1 x2 +1 XtraEdge for IIT-JEE 58 DECEMBER 2011 Passage # 2 (Ques. 4 to 6) 9. Assertion (A) : Work done by the static friction is If two masses A & B are drawn in their attached always zero. 0.2t 2 Reason (R) : when the body is stationary, there is no cables with a = m / sec where t is in second. → → 3 displacement. Hence w = F ⋅ S = zero This section contains 6 questions (Q.10 to 15). +4 marks will be given for each correct answer and –1 mark for each wrong answer. The answer to each of the D questions is a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the C respective question numbers in the OMR have to be darkened. For example, if the correct answers to E question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, A h B respectively, then the correct darkening of bubbles will 4. The speed of block ‘E’ when it reaches a height of look like the following : h = 4 m starting from rest is X Y Z W (A) 4 m/sec (B) 2 m/sec 0 0 0 0 (C) 1 m/sec (D) None of these 1 1 1 1 2 2 2 2 5. The approx speed of pulley D is 3 3 3 3 (A) – 2.5 m/sec (B) – 3.5 m/sec 4 4 4 4 (C) – 4.5 m/sec (D) – 5.5 m/sec 5 5 5 5 6 6 6 6 6. If vA = KvD then K, is 7 7 7 7 (A) 2 (B) 3 (C) 4 (D) 5 8 8 8 8 9 9 9 9 This section contains 3 questions numbered 7 to 9, (Reason and Assertion type question). Each question contains Assertion and Reason. Each question has 4 10. A particle starts oscillating simple harmonically from choices (A), (B), (C) and (D) out of which ONLY ONE its equilibrium position then the KE of the particle is is correct. Mark your response in OMR sheet against ‘n’ times the P.E. of particle at the time T/12, find the the question number of that question. + 4 marks will be value of n (T : time period) given for each correct answer and – 1 mark for each 11. A bird flies for 4 sec with a velocity of wrong answer. |t – 2| m/sec in a straight line where t is in second the The following questions given below consist of an distance covered by the bird is ….. (in m) "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer. 12. An object of mass 0.2 kg executes SHM along the x- (A) If both (A) and (R) are true, and (R) is the axis with a frequency 25/π Hz. At the position correct explanation of (A). x = 0.04 m, the object has K.E., 0.5 J and P.E., (B) If both (A) and (R) are true but (R) is not the 0.4 J. The amplitude of oscillation in cm will be …... correct explanation of (A). (PE is zero at mean position) (C) If (A) is true but (R) is false. 13. Difference between nth & (n + 1)th Bohr's radius of (D) If (A) is false but (R) is true. H atom is equal to its (n – 1)th Bohr's radius. The value of n is …….. 7. Assertion (A) : A lighter and a heavier bodies moving with same momentum and experiencing 14. In the arrangement shown in the figure m1 = 1kg m2 same retarding force have equal stopping times. = 2 kg. Pulley are mass less & strings are light for Reason (R) : For a given force and momentum, what value of M, the mass m1 moves with constant stopping time is independent of mass. velocity (in kg) 8. Assertion (A) : The shortest wavelength of x-rays M emitted from x-ray tube is independent of voltage applied to tube. Reason (R) : wavelength of characteristic spectrum depends upon the atomic number of target. m1 m2 XtraEdge for IIT-JEE 59 DECEMBER 2011 15. A block of mass 1 kg is attached to one end of a electron easily. The oxidation state of any element spring of force constant k = 20 N/m. The other end of can never be in fraction. If oxidation number of any the spring is attached to a fixed rigid support. This element comes out be in fraction, it is average spring block system is made to oscillate on a rough oxidation number of that element which is present in horizontal surface with µ = 0.04. The initial different oxidation states. displacement of block from the (Eq.) mean position 1. The oxidation state of Fe in Fe3O4 is - is a = 30 cm. How many times the block will pass (A) 2 and 3 (B) 8/3 from the mean position before coming to rest ? (C) 2 (D) 3 (g = 10 m/sec2) 1 This section contains Numerical response type questions N 3 (Q. 16 to 18). +6 marks will be given for each correct 2. N–H, In this compound HN3 (hydrazoic acid), answer and –1 mark for each wrong answer. Answers to N 2 this Section are to be given in the form of nearest integer- oxidation state of N1, N2 and N3 are - in four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write (A) 0, 0, 3 (B) 0, 0, –1 0092; 2.1 write 0002) (C) 1, 1, –3 (D) –3, –3, –3 235 16. If 92 U reactor takes 30 days to consume 4 kg of fuel 3. Equivalent weight of chlorine molecule in the and each fission gives 185 MeV of unstable energy equation then find the output power [× 107 W] 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O 17. A bullet in fired at a plank of wood with a speed of (A) 42.6 (B) 35.5 200 m/sec. After passing through the plank, its speed (C) 59.1 (D) 71 is reduced to 150 m/sec. Another bullet of same mass & size but moving with a speed of Passage # 2 (Ques. 4 to 6) 100 m/sec is fired at the same plank. What would be Secondary and tertiary alcohols always give E1 the speed of bullet after passing through the plank ? reaction in dehydration. Primary alcohols whose β- Assume that the resistance offered by plank is same carbon is 3º or 4º also give E1 reaction. However, the for both the bullets. primary alcohols whose β-carbon is 1º or 2º give E2 18. If the end of the cord A is pulled down with 2m/sec reaction. Dehydrating agents like conc. H2SO4, Al2O3 then the velocity of block will be : anhydrous ZnCl2 are used. ( × 10–1 m/sec) The reactivity of alcohols for elimination reaction lies in following sequence : Tertiary alcohol > secondary alcohol > primary alcohol Electron attracting groups present in alcohols increase the reactivity for dehydration. Greater is the A –I effect of the group present in alcohol, more will be its reactivity. Both E1 and E2 mechanism give the product according to Saytzeff's rule, i.e., major product is the most substituted alkene. 2 m/sec OH B CH3 – CH – CH – CH3 conc. H2SO4 Above 413K CHEMISTRY CH3 CH3 – C = CH – CH3 + CH3 – CH – CH = CH2 This section contains 2 paragraphs; each has 3 multiple CH CH3 CH3 choice questions. (Questions 1 to 6) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR Major product Minor product MORE THAN ONE may be correct. Mark your response in OMR sheet against the question number of 4. Arrange the reactivity of given four alcohols in that question. + 4 marks will be given for each correct decreasing order for dehydration. answer and – 1 mark for each wrong answer. OH OH NO2 Passage # 1 (Ques. 1 to 3) Redox reactions are those in which oxidation and NO2 reduction take place simultaneously. Oxidising agent (a) (b) can gain electron whereas reducing agent can lose XtraEdge for IIT-JEE 60 DECEMBER 2011 OH OH 7. Assertion (A) : The value of van der Waals constant 'a' is larger for ammonia than for nitrogen. Reason (R) : Hydrogen bonding is present in ammonia. NO2 NO2 (c) (d) 8. Assertion (A) : 3-hydroxy - butan-2-one on (A) a > b > c > d (B) d > c > b > a treatment with [Ag(NH3)2]⊕ cause precipitation of (C) c > b > d > a (D) b > c > a > d silver. CH3 Reason (R) : [Ag(NH3)2] ⊕ oxidises 3-hydroxy butan-2-one to butan-2-3-dione 5. In the reaction, CH3 – C – CH2OH conc. H2SO4 9. Assertion (A) : HBr adds to 1,4-pentadiene at a CH3 faster rate than to 1,3-pentadiene the product obtained will be : Reason (R) : 1,4-pentadiene is less stable than (A) CH2 = C – CH2 – CH3 1,3-pentadiene. CH3 This section contains 6 questions (Q.10 to 15). (B) CH3 – C = CH – CH3 +4 marks will be given for each correct answer and –1 mark for each wrong answer. The answer to each of the CH3 questions is a SINGLE-DIGIT INTEGER, ranging CH3 from 0 to 9. The appropriate bubbles below the (C) CH3 – CH – CH = CH2 respective question numbers in the OMR have to be darkened. For example, if the correct answers to (D) all of these question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, 6. Which of the following dehydration product is/are respectively, then the correct darkening of bubbles will correct ? look like the following : CH2OH X Y Z W conc.H2SO4 0 0 0 0 (A) ∆ 1 1 1 1 CH3 CH3 CH3 2 2 2 2 (B) CH3 – C — CH – CH3 conc. H2SO4 CH3 – C – CH = CH2 3 3 3 3 ∆ CH3 CH3 4 4 4 4 conc. H2SO4 5 5 5 5 (C) CH3CH2CH2CH2OH ∆ 6 6 6 6 CH3 – CH = CH – CH3 7 7 7 7 CH3 CH3 8 8 8 8 CH3 conc.H2SO4 (D) 9 9 9 9 OH ∆ CH3 10. Equal volumes of 0.02 M AgNO3 & 0.02 M HCN This section contains 3 questions numbered 7 to 9, were mixed. If the [Ag+] at equilibrium was 10–n. (Reason and Assertion type question). Each question Find n. Given Ka(HCN) = 4 × 10–10, Ksp(AgCN) contains Assertion and Reason. Each question has 4 = 4 × 10–16. choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against 11. Haemoglobin contains 0.25% iron by weight. the question number of that question. + 4 marks will be The molecular weight of haemoglobin is 89600. given for each correct answer and – 1 mark for each Calculate the number of iron atoms per molecule of wrong answer. haemoglobin. The following questions given below consist of an 12. Two liquids A and B form an ideal solution at "Assertion" (A) and "Reason" (R) Type questions. Use temperature T. When the total vapour pressure above the following Key to choose the appropriate answer. the solution is 600 torr, the amount of A in the (A) If both (A) and (R) are true, and (R) is the vapour phase is 0.35 and in the liquid phase is 0.70. correct explanation of (A). What is the vapour pressure of pure A ? Express your (B) If both (A) and (R) are true but (R) is not the answer after divide actual answer by 100. correct explanation of (A). (C) If (A) is true but (R) is false. 13. The value of x in the complex Hx[Co(CO)4] is (D) If (A) is false but (R) is true. XtraEdge for IIT-JEE 61 DECEMBER 2011 14. Calculate the emf of the cell B, one or more than one choice is (are) correct out of Cd|Cd2+ (0.10M)1|| H+(0.20M)| Pt, H2(0.5 atm) 4 choices and the student is awarded 3 marks if he [Given: EºCd2+/Cd = – 0.403 V, (she) ticks only the correct choice and all the correct 2.303RT choices. There is no negative marking. = 0.0591] 1. In how many ways can a student answer to any F Round off your answer after multiplying actual question of section B answer by 10. (A) 11 (B) 5C2 + 5C4 (C) 24 – 1 (D) 15 15. Calculate enthalpy change (in calories) adiabatic 2. If a student attempts 3 particular questions-one from compression of one mole of an ideal monoatomic gas section A and two from section B, the probability against constant external pressure of 2 atm starting that he will get marks in only two questions is from initial pressure of 1 atm and initial temperature (assuming all ways to answer a question to be of 300 K. (R = 2 cal/mol degree) Give your answer equally likely) after divide actual answer by 100. 31 23 23 (A) (B) (C) (D) None 900 900 484 This section contains Numerical response type questions (Q. 16 to 18). +6 marks will be given for each correct 3. The probability that a student gets 10 marks if he answer and –1 mark for each wrong answer. Answers to attempts only 4 questions is this Section are to be given in the form of nearest integer- 3 3 in four digits. Please follow as per example : (i.e. for 1 1 4 1 answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write (A) (B) 5 15 3 11 0092; 2.1 write 0002) 3 3 16. A current of 4 A flows in a coil when connected to a 1 1 1 (C) 16 (D) 12V dc source. If the same coil is connected to a 60 4 15 12V, 50 rad/s ac source a current of 2.4 A flows in the circuit. Also find the power developed in the Passage # 2 (Ques. 4 to 6) circuit if a 2500 µF capacitor is connected in series a a with the coil. ∫ ∫ f ( x) dx = ( f ( x) + f (− x)) dx 17. A capacitor of capacity 2 µF is charged to a potential −a 0 difference of 12V. It is then connected across an a inductor of inductance 0.6 mH. What is the current in the circuit at a time when the potential difference (i) If f(x) is odd, then ∫ f ( x) dx = 0 −a across the capacitor is 6.0 V ? a a 18. A ball of mass 100 g is projected vertically upwards (ii) If f(x) is even, then ∫ −a ∫ f ( x) dx = 2 f ( x) dx 0 from the ground with a velocity of 49 m/s. At the same lime another identical ball is dropped from a This is one of the important property for the height of 98 metre to fall freely along the same path integrable function f(x) 1 as that followed by the first ball. After some time the ∫ (x + x | cos x |)(sin −1 x 2 ) dx is equal to : 3 two balls collide and stick together and finally fall to 4. ground. Find the time of flight of the masses. −1 (g = 9.8 m/s2) (A) 1 (B) – 1 (C) 0 (D) 2 1/ 4 MATHEMATICS 5. If ∫ 16 x 2 + 16 x + 41 + 40 x 2 + x + 1 + −1 / 4 This section contains 2 paragraphs; each has 3 multiple 1 choice questions. (Questions 1 to 6) Each question has 4 16 x 2 + 16 x + 41 − 40 x 2 + x + 1 cosx dx = k sin , choices (A), (B), (C) and (D) out of which ONE OR 4 MORE THAN ONE may be correct. Mark your then the value of k is : response in OMR sheet against the question number of (A) 10 (B) 0 (C) 20 (D) 30 that question. + 4 marks will be given for each correct π/ 4 ∫ ( 1 + sin 2 x + ) answer and – 1 mark for each wrong answer. 6. 1 − sin 2 x dx is equal to Passage # 1 (Ques. 1 to 3) −π / 4 An objective test contains two sections : A and B each consisting of 10 questions. In section A, only 4 (A) 2 (B) 2 2 (C) 4 2 (D) one choice out of 4 choices is correct and student is 2 awarded 1 mark for every correct answer. In section XtraEdge for IIT-JEE 62 DECEMBER 2011 This section contains 3 questions numbered 7 to 9, (Reason 10. The position vectors of two points A and C are and Assertion type question). Each question contains ˆ j ˆ ˆ ˆ 9i − ˆ + 7k and 7i − 2 ˆ + 7 k respectively. The point j Assertion and Reason. Each question has 4 choices (A), (B), of intersection of the lines containing vectors (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that ˆ j ˆ ˆ j ˆ AB = 4i − ˆ + 3k and CD = 2i − ˆ + 2k is P. If vector question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. PQ is perpendicular to AB and CD and PQ = 15 The following questions given below consist of an units, then possible position vectors of Q are "Assertion" (A) and "Reason" (R) Type questions. Use ˆ ˆ ˆ ˆ x1i + x2 ˆ + x3 k and y1i + y 2 ˆ + y3 k . Find the value j j the following Key to choose the appropriate answer. 3 (A) If both (A) and (R) are true, and (R) is the correct explanation of (A). of ∑ (x + y ) . i =1 i i (B) If both (A) and (R) are true but (R) is not the x −1 y −3 z−4 correct explanation of (A). 11. Let image of the line = = in the 3 5 2 (C) If (A) is true but (R) is false. plane 2x – y + z + 3 = 0 be L. A plane (D) If (A) is false but (R) is true. 7x + By + Cz + D = 0 is such that it contains the line L 7. Assertion (A) : Let f (x) be an even function which and perpendicular to the plane 2x – y + z + 3 = 0 then x find the value of (B + C + D)/10. is periodic, then g (x) = ∫ f (t ) dt is also periodic. 12. A circle touches the hypotenuse of a right angled a triangle at its middle point and passes through the Reason (R) : If α(x) is a differentiable and periodic middle point of shorter side. If 3 unit and 4 unit be function, then α′(x) is also periodic. the length of the sides and 'r' be the radius of the circle, then find the value of '3r'. 8. Assertion (A) : The locus represented by xy + yz = 0 is a pair of perpendicular planes. 13. The remainder when 2740 is divided by 12 Reason (R) : If a1x + b1y + c1z + d1 = 0 and 14. Let f (x) = x3 – x2 – 3x – 1 and h(x) = f (x)/g(x) a2x + b2y + c2z + d2 = 0 are perpendicular then where h is a function such that a1a2 + b1b2 + c1c2 = 0 (a) it is continuous every where except when x = – 1 9. Assertion (A) : Locus of center of a variable circle (b) lim h(x) = ∞ and (c) lim h(x) = 1/2 x →∞ x → –1 touching two circles (x – 1)2 + (y – 2)2 = 25 and (x – 2)2 + (y – 1)2 =16 is an ellipse. FInd lim (4h(x) + f (x) + 2g(x)). x →0 Reason (R) : If a circle S2= 0 lies completely inside the circle S1= 0 then locus of center of a variable circle 15. In a triangle ABC, if sin A cos B = 1/4 and S = 0 which touches both the circles is an ellipse. 3 tan A = tan B, then cot2A = This section contains 6 questions (Q.10 to 15). This section contains Numerical response type questions +4 marks will be given for each correct answer and –1 (Q. 16 to 18). +6 marks will be given for each correct mark for each wrong answer. The answer to each of the answer and –1 mark for each wrong answer. Answers to questions is a SINGLE-DIGIT INTEGER, ranging this Section are to be given in the form of nearest integer- in four digits. Please follow as per example : (i.e. for from 0 to 9. The appropriate bubbles below the answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write respective question numbers in the OMR have to be 0092; 2.1 write 0002) darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, 16. The vertices B and C of a triangle ABC lie on the respectively, then the correct darkening of bubbles will lines 3y = 4x and y = 0 respectively and the side BC look like the following : passes through the point (2/3, 2/3). If ABOC is a X Y Z W rhombus, O being the origin. If co-ordinates of vertex 0 0 0 0 A is (α, β), then find the value of 5(α + β). 1 1 1 1 17. Number of different words that can be formed using 2 2 2 2 all the letters of the word "DEEPMALA", I two 3 3 3 3 vowels are together and the other two are also together but separated from the first two. 4 4 4 4 5 5 5 5 18. If a complex number z satisfies the conditions 6 6 6 6 4i 1+ i 2 – 3i z – 4i 7 7 7 7 = 1 and z = x + – 1 + i – 2i 3 + 4i , z + 4i 8 8 8 8 – 2 – 3i – 3 + 4i 0 9 9 9 9 then x = XtraEdge for IIT-JEE 63 DECEMBER 2011 XtraEdge for IIT-JEE 64 DECEMBER 2011 Based on New Pattern IIT-JEE 2013 XtraEdge Test Series # 8 Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 6 are passage based questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 7 to 9 are Reason and Assertion type question with one is correct answer. +4 marks and –1 mark for wrong answer. • Question 10 to 15 are Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answer and –1 mark for wrong answer. • Question 16 to 18 are Numerical Response Question (four digit Ans. type) +6 marks will be awarded for correct answer and –1 mark for wrong answer. 3. If some of heat is allowed to escape to surrounding PHYSICS (temperature of surrounding is 20ºC) then this amount of steam (mentioned in question 22) is This section contains 2 paragraphs; each has 3 multiple increase the temperature to - choice questions. (Questions 1 to 6) Each question has 4 (A) greater than 80ºC (B) less than 80ºC choices (A), (B), (C) and (D) out of which ONE OR (C) equal to 80ºC (D) can't say anything MORE THAN ONE may be correct. Mark your response in OMR sheet against the question number of Passage # 2 (Ques. 4 to 6) that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. A body of mass 1 kg moving along x-axis has velocity 4 m/s at x = 0. The acceleration and Passage # 1 (Ques. 1 to 3) potential energy of body varies as shown in In espresso coffee machines steam is passed into milk diagrams. at room temperature for a brief time interval. Some of the steam condenses and the temperature or milk a(m/s2) U(J) rises. Since the time for which the steam is passed is 120 brief, one can ignore the heat lost to the environment 2 and assume that the usual assumption of calorimetry : Heat lost = Heat gain is valid. 4 8 4 8 x(m) x(m) –120 1. Steam at 100ºC is passed into milk to heat it. The amount of heat required to heat 150 g of milk from 4. Work done by conservative forces when body moves room temperature (20ºC) to 80ºC is (specific heat of from x = 0m to x = 8m is – capacity of milk = 4.0 kJ kg–1 K–1 specific latent heat (A) 0 J (B) 120 J (C) 240 J (D) – 240 J of steam = 2.2 MJ kg–1 , specific heat capacity of water = 4.2 × 103 J kgK–1) 5. Work done by external forces when body moves (A) 3.6 × 104 J (B) 3.6 × 103 J from x = 0m to x = 8m is – 2 (C) 3.6 × 10 J (D) None of these (A) 120 J (B) – 120 J (C) – 112 J (D) None of these 2. How many grams of steam condensed into water in 6. The change in kinetic energy when body moves from above question - x = 0m to x = 8m is – (A) 1.57 g (B) 15.7 g (A) 256 J (B) 240 J (C) 157 g (D) None of these (C) 128 J (D) 120 J XtraEdge for IIT-JEE 65 DECEMBER 2011 This section contains 3 questions numbered 7 to 9, 10. If y = 4x2 – 4x + 7. Find the minimum value of 'y'. (Reason and Assertion type question). Each question contains Assertion and Reason. Each question has 4 11. A block of mass 1 kg starts slipping on a circular choices (A), (B), (C) and (D) out of which ONLY ONE track of radius 2m and it is observed that when θ = is correct. Mark your response in OMR sheet against 60º its speed is 4 m/s as shown in figure. Assuming the question number of that question. + 4 marks will be size of block to be O r = 2m given for each correct answer and – 1 mark for each negligible and coefficient of friction between block and θ wrong answer. The following questions given below consist of an track is 0.5 frictional force "Assertion" (A) and "Reason" (R) Type questions. Use (in N) on block when v the following Key to choose the appropriate answer. θ = 60º is (g = 10 m/s2) – (A) If both (A) and (R) are true, and (R) is the correct explanation of (A). 12. A projectile is fired with speed 'u' at angle 60º with (B) If both (A) and (R) are true but (R) is not the horizontal. Velocity of projectile when it makes an correct explanation of (A). angle 120º with initial direction of velocity is u' then (C) If (A) is true but (R) is false. ratio b/w u : u' is . (D) If (A) is false but (R) is true. 13. A block of mass 'm' is hanged vertically by a wire. Potential energy stored in wire u1. Potential energy 7. Assertion (A) : In the flow-tube as the cross-section stored in wire is u2. When mass hanged is doubled. area decreases the flow velocity increases. Ratio u2 : u1 is. Reason (R) : In ideal fluid flow the total energy per unit mass remains constant. 14. An electrically heating coil is placed in calorimeter containing 360 gm of H2O at 10ºC. The coil 8. Assertion (A) : When a spring is elongated work consumes energy at the rate of 90 W. The water done by spring is negative but when it compressed equivalent of calorimeter and the coil is 40 g. The work done by spring is positive. temperature of water after 10 minutes will be n/8.5 Reason (R) : Work done by spring is path then find the value of 'n' - independent. 15. Variation of pressure at certain point is space is given 9. Assertion (A) : If in some case work done by a force by : is path independent then it must be conservative. P = P0 is 2πt cos 212 πt sin (220πt – π/2) Reason (R) : Work done by conservative forces in a The Beat Frequency is - round trip must be zero. This section contains Numerical response type questions (Q. 16 to 18). +6 marks will be given for each correct This section contains 6 questions (Q.10 to 15). answer and –1 mark for each wrong answer. Answers to +4 marks will be given for each correct answer and –1 this Section are to be given in the form of nearest integer- mark for each wrong answer. The answer to each of the in four digits. Please follow as per example : (i.e. for questions is a SINGLE-DIGIT INTEGER, ranging answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write from 0 to 9. The appropriate bubbles below the 0092; 2.1 write 0002) respective question numbers in the OMR have to be 16. A conical container of radius R darkened. For example, if the correct answers to R = 1m and height H = 5m is question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, filled completely with liquid. respectively, then the correct darkening of bubbles will There is a hole at the bottom of look like the following : H container of area π × 10–3 m2 (see X Y Z W figure). Time taken to empty the conical container (in sec) is…… 0 0 0 0 ………….. Take g = 10 m/s2 – 1 1 1 1 2 2 2 2 17. A pendulum of length l is 3 3 3 3 given a horizontal velocity 4 4 4 4 kgl at the lowest point of 5 5 5 5 vertical circular path as O 6 6 6 6 shown. In the subsequent 7 7 7 motion the string gets slag at l 7 8 8 8 8 a certain point and the pendulum bob strikes the v = kgl 9 9 9 9 point of suspensión then the value of k is – XtraEdge for IIT-JEE 66 DECEMBER 2011 18. A small body is released from point A of the smooth 2. Instead of developing a catalyst, why didn’t Haber parabolic path y = x2. Where y is vertical axis and x increase the rate of the reaction by raising the is horizontal axis at ground as shown. The body temperature ? leaves the surface from point B. If g = 10 m/s2 then (A) The Haber process is exothermic, so raising the the total horizontal distance travelled by body before temperature would have lowered the rate of the it hits ground is – reaction y (B) The Haber process is exothermic, so raising the temperature would have reduced the yield of the A reaction (C) Higher temperatures would have caused an B increased in pressure lowering the yield of the reaction –2m O +1m x (D) Higher temperatures might have decomposed the hydrogen 3. Which of the following graphs could also accurately CHEMISTRY reflect the establishment of equilibrium between nitrogen, hydrogen, and ammonia ? This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 1 to 6) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR H2 Conc MORE THAN ONE may be correct. Mark your (A) NH3 response in OMR sheet against the question number of N2 that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. Time Passage # 1 (Ques. 1 to 3) The Haber Process shown below : N2 Conc N2(g) + 3H2(g) 2 NH3(g) (B) NH3 In 1912 Fritz Haber developed the Haber process for H2 making ammonia from nitrogen and hydrogen. His development was crucial for the German war effort Time of world War Ι, providing the Germans with ample H2 Conc fixed nitrogen for the manufacture of explosives. The Haber process takes place at 500° C and 200 (C) NH3 atm. It is an exothermic reaction .The graph below N2 shows the change in concentrations of reactants and products as the reaction progresses. Time N2 Conc Concentration (D) NH3 A H2 B Time C Passage # 2 (Ques. 4 to 6) Time Entropy is measure of degree of randomness. Entropy is directly proportional to temperature. Every system 1. Even at the high temperatures, the conversion of tries to acquire maximum state of randomness or nitrogen and hydrogen to ammonia was slow. In disorder. Entropy is measure of unavailable energy. order to make the process industrially efficient Fritz Unavailable energy = Entropy × Temperature Haber used a metal oxide catalyst. Which of the The ratio of entropy of vapourisation and boiling following did not accomplished by use of the metal point of substance remains almost constant. oxide catalyst ? (A) The rate of production of ammonia increased 4. Which of the following process have ∆S = – ve ? (B) The energy of activation was raised (A) Adsorption (C) The equilibrium shifted to the right (B) Dissolution of NH4Cl in water (D) activation energy equals to zero (C) H2 → 2H (D) 2NaHCO3(s) → Na2CO3 + CO2 + H2O XtraEdge for IIT-JEE 67 DECEMBER 2011 5. Observe the graph and identify the incorrect This section contains 6 questions (Q.10 to 15). statement(s) +4 marks will be given for each correct answer and –1 mark for each wrong answer. The answer to each of the Β ∆Svap questions is a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the Entropy respective question numbers in the OMR have to be Α ∆Sfusion darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : T1 T2 X Y Z W Temperature 0 0 0 0 (A) T1 is melting point, T2 is boiling point 1 1 1 1 (B) T1 is boiling point, T2 is melting point 2 2 2 2 (C) ∆Sfusion is more than ∆Svap 3 3 3 3 (D) T2 is lower than T1 4 4 4 4 5 5 5 5 6. The law of Thermodynamics invented by Nernst, 6 6 6 6 which helps to determine absolute entropy is 7 7 7 7 (A) Zeroth law (B) 1st law nd 8 8 8 8 (C) 2 law (D) 3rd law 9 9 9 9 This section contains 3 questions numbered 7 to 9, (Reason 10. For the Ca atom calculate total No. of e– which have and Assertion type question). Each question contains m = –1 Assertion and Reason. Each question has 4 choices (A), (B), 11. The molecular formula of a non-stoichiometric tin (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that oxide containing Sn (II) and Sn (IV) ions is Sn4.44O8. question. + 4 marks will be given for each correct answer Therefore, the molar ratio of Sn (II) to Sn (IV) is and – 1 mark for each wrong answer. approximately The following questions given below consist of an 12. How much volume (in mL) 0.001 M HCl should we "Assertion" (A) and "Reason" (R) Type questions. Use add to 10 cm3 of 0.001 M NaOH to change its pH by the following Key to choose the appropriate answer. one unit ? (A) If both (A) and (R) are true, and (R) is the correct explanation of (A). 13. What is the sum of total electron pairs (b.p. + l.p.) present in XeF6 molecule ? (B) If both (A) and (R) are true but (R) is not the correct explanation of (A). 14. The number of geometrical isomers of (C) If (A) is true but (R) is false. CH3CH=CH–CH=CH–CH=CHCl is. (D) If (A) is false but (R) is true. 15. No. of π bond in the compound H2CSF4 is. 7. Assertion (A) : At zero degree Kelvin the volume This section contains Numerical response type questions occupied by a gas is negligible. (Q. 16 to 18). +6 marks will be given for each correct Reason (R) : All molecular motion ceases at 0 K. answer and –1 mark for each wrong answer. Answers to this Section are to be given in the form of nearest integer- 8. Assertion (A) : Compressibility factor for hydrogen in four digits. Please follow as per example : (i.e. for varies with pressure with positive slope at all answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write pressures. 0092; 2.1 write 0002) Reason (R) : Even at low pressures, repulsive forces 16. The ion An+ is oxidised to AO3– by MnO4–, changing dominate hydrogen gas. to Mn2+ in acidic solution. Given that 2.68 × 10–3 mol of An+ requires 1.61 × 10–3 mol of MnO4–. What is 9. Assertion (A) : Enthalpy of graphite is lower than the value of n ? that of diamond. Reason (R) : Entropy of graphite is greater than that 17. Standard heat of formation of HgO(s) at 298 K and at of diamond. constant pressure is –90.8 kJ mol–1. Excess of HgO(s) absorbs 41.84 kJ of heat. Calculate the mass of Hg (in g) that can be obtained at constant volume and at 298 K. (Hg = 200.6g mol–1) XtraEdge for IIT-JEE 68 DECEMBER 2011 18. A near ultraviolet photon of 300 nm is absorbed by a 6. If BL2 + CM2 + AN2 = x and CL2 + AM2 + BN2 = y gas and then re-emitted as two photons. One photon then is red with wavelength 760 nm. What would be the (A) x + y = 0 wavelength (in nm) of the second photon ? (B) x – y = 0 (C) x + y = 24R2 – 2(a2 + b2 + c2) (D) none of these MATHEMATICS This section contains 3 questions numbered 7 to 9, This section contains 2 paragraphs; each has 3 multiple (Reason and Assertion type question). Each question choice questions. (Questions 1 to 6) Each question has 4 contains Assertion and Reason. Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR choices (A), (B), (C) and (D) out of which ONLY ONE MORE THAN ONE may be correct. Mark your is correct. Mark your response in OMR sheet against response in OMR sheet against the question number of the question number of that question. + 4 marks will be that question. + 4 marks will be given for each correct given for each correct answer and – 1 mark for each answer and – 1 mark for each wrong answer. wrong answer. The following questions given below consist of an Passage # 1 (Ques. 1 to 3) "Assertion" (A) and "Reason" (R) Type questions. Use A circle C1 of radius 2 units rolls on the outerside of the following Key to choose the appropriate answer. the circle C2 : x2 + y2 + 4x = 0, touching it externally. (A) If both (A) and (R) are true, and (R) is the 1. If the line joining the centres of C1 and C2 makes an correct explanation of (A). angle of 60º with the x-axis, equation of a common (B) If both (A) and (R) are true but (R) is not the tangent to them is - correct explanation of (A). (A) x + 3y–2=0 (C) If (A) is true but (R) is false. (B) 3x–y+4+2 3 =0 (D) If (A) is false but (R) is true. (C) 3x–y–4+2 3 =0 7. Assertion (A) : If a, b > 0 and a3 + b3 = a – b, then a2 + b2 < 1 (D) 3x–y–4–2 3 =0 1 Reason (R) : If a, b > 0, then ab < (a + b) 2 2. Area of the quadrilateral formed by a pair of tangents from a point on C3 to the circle C2 with a pair of radii at the points of contact of the tangents is - 8. Assertion (A) : The line bx – ay = 0 will not meet the x2 y2 (A) 2 3 sq. units (B) 4 3 sq. units hyperbola 2 – 2 = 1 (a > b > 0) a b (C) 3 sq. units (D) 3 3 sq. units Reason (R) : The line y = mx + c does not meet the x2 y2 3. If the line joining the centres of C1 and C2 is hyperbola 2 – 2 = 1 if c2 = a2 m2 – b2. perpendicular to the x-axis; equation of the chord of a b contact of the tangents drawn from the centre of C2 to the circle C1 is - 9. Assertion (A) : The solution set of the inequality (A) y – 2 = 0 (B) y + 2 = 0 x2 + x (C) y – 3 = 0 (D) y + 3 = 0 log 0.7 log 6 < 0 is (– 4, – 3) ∪ (8, ∞). x+4 Passage # 2 (Ques. 4 to 6) Reason (R) : For x > 0, loga x is an increasing function AL, BM and CN are diameter of the circumcircle of a if a > 1 and a decreasing function if 0 < a < 1. triangle ABC. ∆1, ∆2, ∆3, and ∆ are the areas of the triangles BLC, CMA, ANB and ABC respectively, This section contains 6 questions (Q.10 to 15). +4 marks will be given for each correct answer and –1 4. ∆1 is equal to mark for each wrong answer. The answer to each of the (A) 2R2 sin A cos B cos C questions is a SINGLE-DIGIT INTEGER, ranging (B) 2R2 sin A sin B cos C from 0 to 9. The appropriate bubbles below the (C) 2R2 cos A cos B sin C respective question numbers in the OMR have to be (D) 2R2 sin A sin B sin C darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, 5. ∆1 + ∆2 + ∆3 is equal to respectively, then the correct darkening of bubbles will (A) 2∆ (B) 3∆ look like the following : (C) ∆ (D) none of these XtraEdge for IIT-JEE 69 DECEMBER 2011 X Y Z W 0 1 0 1 0 1 0 1 Elements Named for Places 2 2 2 2 This is an alphabetical list of element toponyms or 3 3 3 3 elements named for places or regions. Ytterby in 4 4 4 4 Sweden has given its name to four elements: Erbium, 5 5 5 5 Terbium, Ytterbium and Yttrium. 6 6 6 6 7 7 7 7 • Americium : America, the Americas 8 8 8 8 • Berkelium : University of California at Berkeley 9 9 9 9 • Californium : State of California and University 10. If two of the lines represented by of California at Berkeley x4 + x3 y + cx2 y2 – xy3 + y4 = 0 bisect the angle between the other two, then the value • Copper : probably named for Cyprus of |c| is • Darmstadtium : Darmstadt, Germany 11. For n > 3, a, b ∈ R, let • Dubnium : Dubna, Russia n n(n – 1)...(n – r + 1) S(n, a, b)= ∑ r =0 (–1) r r! (a – r) (b – r) • Erbium : Ytterby, a town in Sweden • Europium : Europe 1 1 Find S 5, , 2 7 • Francium : France 12. Find the degree of the remainder when x2007 – 1 is • Gallium : Gallia, Latin for France. Also named divided by (x2 + 1) (x2 + x + 1). for Lecoq de Boisbaudran, the element's discoverer (Lecoq in Latin is gallus) 13. Let l be the length of the interval satisfying the inequality • Germanium : Germany 1 log6(x + 2) (x + 4) + log1/6 (x + 2) < log 6 (7). • Hafnium : Hafnia, Latin for Copenhagen 2 Find the value of l. • Hassium : Hesse, Germany 14. The number of pairs (x, y) satisfying the equation • Holmium : Holmia, Latin for Stockholm sin x + sin y = sin (x + y) and |x| + |y| = 1 is. • Lutetium : Lutecia, ancient name for Paris 15. If a = ( 0, 1, – 1) and c = (1, 1, 1) are given vectors, the |b|2 where b satisfies a × b + c = 0 and a . b = 3 is. • Magnesium : Magnesia prefecture in Thessaly, Greece This section contains Numerical response type questions (Q. 16 to 18). +6 marks will be given for each correct • Polonium : Poland answer and –1 mark for each wrong answer. Answers to this Section are to be given in the form of nearest integer- • Rhenium : Rhenus, Latin for Rhine, a German in four digits. Please follow as per example : (i.e. for province answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002) • Ruthenium : Ruthenia, Latin for Russia 16. The number of natural numbers which are smaller • Scandium : Scandia, Latin for Scandinavia than 2.108 and which can be written by means of the digits 1 and 2 is ............. . • Strontium : Strontian, a town in Scotland 1 • Terbium : Ytterby, Sweden 17. If z = ( 3 – i), find the smallest value of positive 2 • Thulium : Thule, a mythical island in the far integer n for which (z89 + i97)94 = zn. north (Scandinavia?) 18. If α + β = γ and tan γ = 22, a is the arithmetic and b is • Ytterbium : Ytterby, Sweden the geometric mean respectively between tan α and a3 • Yttrium : Ytterby, Sweden tan β, then the value of is equal to - (1 – b 2 ) 3 XtraEdge for IIT-JEE 70 DECEMBER 2011 MOCK TEST PAPER-1 CBSE BOARD PATTERN CLASS # XII SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS Solutions will be p ublished in next issue General Instructions : Physics & Chemistry • Time given for each subject paper is 3 hrs and Max. marks 70 for each. • All questions are compulsory. • Marks for each question are indicated against it. • Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each. • Question numbers 9 to 18 are short-answer questions, and carry 2 marks each. • Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each. • Question numbers 28 to 30 are long-answer questions and carry 5 marks each. • Use of calculators is not permitted. General Instructions : Mathematics • Time given to solve this subject paper is 3 hrs and Max. marks 100. • All questions are compulsory. • The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each. Section B comprises of 12 questions of four marks each. Section C comprises of 7 questions of six marks each. • All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. • There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2 question of six marks each. You have to attempt only one of the alternatives in all such questions. • Use of calculators is not permitted. PHYSICS 10. Explain how the interference fringe width in young's double slits experiment will change if 1. What is the phase difference between two points on a (i) separation between the two slits is decreased wave front (ii) wavelength of light increased. 2. Why stars seems to be twinkling ? 11. In an atom, two electrons moves around the nucleus 3. Name the characteristics of electromagnetic wave that in a circular orbit of radius R and 4R. Calculate the (i) increases ratio of speeds of two electrons. (ii) remains constant 12. Which type of semiconductor is better out of in the electromagnetic spectrum as one moves from p-type and n-type ? infrared to ultraviolet region. 4. What is total energy of an electron revolving in first 13. Relate input frequency and output frequency of a half orbit of Hydrogen atom . wave rectifier and a full wave rectifier 5. What are digital and analog signals ? 14. Define the term modulation index for an AM wave. 6. Name any two basic properties of electric charge. What would be the modulation index for an AM wave for which the maximum amplitude is a while 7. Out of the two bulbs marked 25 W and 100 W which the minimum amplitude is 'b' ? one has higher resistance ? 8. What is displacement current ?. 15. Distinguish between point to point and broadcast communication modes. Give one example of each 9. A thin glass prism has a minimum deviation δm in air. State with reason, how the angle of minimum 16. Why does the conductivity of a semi conductor deviation will change if the prism is immersed in a change with the rise in temperature ? liquid of refractive index greater than 1 XtraEdge for IIT-JEE 71 DECEMBER 2011 17. An electric dipole is held in uniform electric field. Why objective lens of an astronomical telescope is of (i) Show that no translatory force acts on it. large size ? what defects can be remoned by using (ii) Derive an expression for the torque acting on it. reflecting telescope in place of refracting telescope ? IF length of an astronomical telescope is 100 cm and 18. Define angle of dip at a given place. What is the its magnifying power is 19 then calculate focal value of angle of dip on the equator ? lengths of objective and eye-piece lens. 19. When a Uranium nucleus (U238) originally at rest decays by emitting alpha particle having speed u. 29. Explain with the help of a labelled diagram, the Find the recoil speed of residual nucleus. principle, construction and working of a transformer. OR 20. If focal lengths of objective and eye-piece lens of a An a.c. generator consists of a coil of 50 turns and compound microscope is 2cm and 3cm respectively area 2.5 m2 rotating at an angular speed of 60 rad s–1 and distance between both the lenses is 15cm then in a uniform magnetic field B = 0.30 T between two calculate distance of object from objective lens if fixed pole pieces. The resistance of the circuit final image forms at infinity. including that of the coil is 500 Ω. (i) Find the maximum current drawn from the generator. (ii) What 21. If a gas is at temp T, then derive an expression for will be the orientation of the coil w.r.t. the magnetic debroglie wavelength of its molecule of mass m. field to have (a) maximum (b) zero magnetic flux. (iii) Would the generator work if the coil were stationary 22. For a common emitter transistor amplifier current and instead the poles were rotated with same speed as gain is 72. Calculate the base current for which above. emitter current is 8.9mA. 30. Calculate the electric field intensity for following points 23. Give one use of each of the following due to a uniformly charged non-conducting sphere. (i) microwaves Represent the results by graph – (ii) infra-red waves (A) at any point outside the sphere (iii) ultraviolet radiation (B) at any point on the surface of sphere (iv) gamma rays. (C) at any point inside the sphere. OR 24. State the principle of potentiometer with the help of circuit diagram, describe a method to find the What is a Capacitor ? Explain its principles. Derive internal resistance of a primary cell. the relations for equivalent capacity of series and r parallel combinations of capacitors. 25. A proton is shot into the magnetic field B = 0.8ˆ T j with a velocity (2 × 106 ˆ + 3 × 106 ˆ) ms–1. Calculate i j CHEMISTRY the radius and pitch of the helix path followed by proton. 1. Write the IUPAC names of the following : Br 26. Derive an expression for the torque on a rectangular coil of area A, carrying a current I placed in a magnetic field B. The angle between the direction of OH B and vector perpendicular to plane of coil is θ. COOC2H5 27. Using Kirchhoff's laws in the given network, calculate the values of I1, I2 and I3. 2. How would you convert methylamine into ethylamine ? A B C 3. What are essential amino acids ? Give two example. 5Ω I1 3Ω I2 2Ω I3 4. What is meant by the term peptization ? 12V 6V 5. Identify the reaction order if the unit of rate constant F E D is sec–1. 28. With the help of ray diagram, explain the 6. In an alloy of gold and cadmium if gold crystallizes phenomenon of total interval reflection. Obtain the in cubic structure occupying the corners only and relation between critical angle and refractive index of cadmium fits into edge centre voids, what is the the medium. formula of the alloy ? OR XtraEdge for IIT-JEE 72 DECEMBER 2011 7. Which oxide of nitrogen has oxidation number of N 19. Name the reagents which are used in the following same as that in nitric acid. conversions : (i) A primary alcohol to an aldehyde 8. What is froth floatation for which ores it is used ? (ii) Butan-2-one to butan-2-ol (iii) Phenol to 2,4,6-tribromopheno 9. Explain as to why haloarenes are much less reactive than haloalkanes towards nucleophilic substitution 20. Give the structures of following polymers. reactions. (A) Perlon (B) Orlon (C) Neoprene 10. Assign a reason for each of the following statements : 21. What are tranquilizers ? Explain with example (i) Alkylamines are stronger bases than arylamines tranquilizers are neurologically active drugs which are (ii) Acetonitrile is preferred as solvent for carrying used to reduce strain or anxiety. out several organic reactions. 22. Name any three fat soluble vitamins & their 11. Write one chemical equation each to exemplify the deficiency disease. following reactions : (i) Carbylamine reaction 23. (a) State the products of electrolysis obtained on the (ii) Hofmann bromamide reaction cathode and the anode in the following cases : [2 + 1] 12. Identify A and B in the following : (i) A dilute solution of H2SO4 with platinum CH2Br electrodes. (ii) An aqueous solution of AgNO3 with silver CN– LiAlH4 (i) A B electrodes (b) Write the cell formulation and calculate the (ii) R2CO NH3 A Ni/H2 B standard cell potential of the galvanic cell in which the following reaction takes place : 13. Write any two feature which distinguish Fe2+ (aq) + Ag+ (aq) → Fe3+ (aq) + Ag (s) physisorption from Chemisorption. Calculate ∆rGº for the above reaction. [Given : E o + / Ag = + 0.80 V ; Ag 14. The decomposition of a compound is found to follow a first order rate law. If it takes 15 minutes for 20% of E o 3+ / Fe 2 + = + 0.77 V ; 1 F = 96500 C mol–1] Fe the original material to react, calculate- 24. (a) Explain each of the following with a suitable (i) Specific rate constant. example : (ii) the time at which 10% of the original material (i) Paramagnetism (ii) Frenkel defect in crystals remains unreacted. (b) An element occurs in bcc structure with cell edge 300 pm. The density of the element is 15. Prove that the time required for the completion of 5.2 g cm–3. How many atoms of the element does 200 3/4th of the reaction of first order is twice the g of the element contain ? time required for the completion of half of the reaction. 25. (a) Describe the preparation of KMnO4 from pyrolusite ore. 16. When a certain conductivity cell was filled with 0.1 M (b) Among ionic species, Sc+3, Ce+4 and Eu+2, which KCl, it has a resistance of 85 Ω at 25ºC. When the one is good oxidizing agent ? same cell was filled with an aqueous solution of 0.052 (Atomic numbers : Sc = 21, Ce = 58, Eu = 63 ) M unknown electrolyte the resistance was 96 Ω. 26. (a) Assign a reason for each of the following Calculate the molar conductivity of the unknown statements : electrolyte at this concentration. (Specific (i) Di-methyl amine is stronger base than conductivity of 0.1 M KCl = 129 × 10–2 ohm–1cm–1). tri-methyl amine in aq. solution. (ii) Explain why ? Benzamide is more basic in 17. Give methods of preparation of XeO3 and XeOF4. comparision to acetamide. 18. Explain giving reason : (b) How would you convert aryl amine into (i) Copper (I) is diamagnetic whereas copper (II) is cynobenzene. paramagnetic (ii) K2PtCl6 is a well known compound whereas the corresponding Ni compound does not exist. XtraEdge for IIT-JEE 73 DECEMBER 2011 27. (a) Why chelated complexes are more stable than unchelated complexes ? MATHEMATICS (b) Write IUPAC names of : (i) K3 [Al(C2O4)3] (ii) [Mn (H2O)6] S Section A 28. You are provided with four reagents : LiAlH4, I2/NaOH, NaHSO3 and Schiff's reagent 1. Find order and degree of diff. equation 2 (a) Write which two reagents can be used to dy 1 distinguish between the compounds in each of the + =2 dx dy following pairs : dx [3+2=5] (i) CH3CHO and CH3COCH3 ∫ tan −1 (ii) CH3CHO and C6H5CHO 2. Evaluate : x dx . (iii) C6H5COCH3 and C6H5COC6H5 (b) Account for the following : (i) The order of reactivity of halogen acids with ether 3. Find unit vector in the direction of vector → is HI > HBr > HCl. ˆ a = 2i + 3 ˆ +k . ˆ j (ii) The pKa value of chloroacetic acid is lower than the pKa value of acetic acid. → → → → Or 4. ˆ j ˆ ˆ If a = i + ˆ – 3 k and b = ˆ + 2 k . Find |2 b × a | j (a) An organic compound contains 69.77% carbon, 11.63% hydrogen and the rest is oxygen. The molecular mass of the compound is 86. It does 5. Find the angle between the two planes not reduce Tollens' reagent but forms an addition 3x – 6y – 2z = 7 and 2x + 2y – 2z = 5 compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous 6. Construct a matrix of order 3 × 3, whose element aij oxidation it gives ethanoic and propanoic acids. is given by rule aij = (i + j)2. Deduce the possible structure of the organic compound. 7. Find a matrix X such that A + 2B + X = 0, where (b) State reasons for the following : 2 − 1 − 1 1 A= ;B= . (i) Monochloroethanoic acid has a higher pKa 3 5 0 2 value than dichloroethanoic acid. (ii) Ethanoic acid is a weaker acid than benzoic acid. sin 10º − cos 10º 8. Find value of . 29. (a) Density of 0.8 M aqueous solution of H2SO4 is sin 80º cos 80º 1.06 g mL–1. Calculate the concentration of solution in (i) mol fraction (ii) molality (molar 9. Let f = {(1, 3), (2, 1), (2, 1), (3, 2)} and mass of H2SO4 = 98 g mol–1). g = {(1, 2), (2, 3), (3, 1)} then find gof (1). (b) Heptane and octane form ideal solution. At 373 K, the vapour pressures of the two liquid 10. Find derivative of sin–1 (2 x 1 − x 2 ) w.r.to x. components are 105.2 kPa and 46.8 kPa, respectively. What will be the vapour pressure, in bar, of a mixture of 25.0 g of heptane and 35.0 g of octane ? Section B 30. (a) Explain the following giving reasons : 11. A and B are mutually exclusive events of an (i) H3PO3 is diprotic. experiment. If P (Not A) = 0.65, P (A ∪ B) = 0.65 (ii) Nitrogen does not form pentahalides and P (B) = p. Find the value of p. (iii) SF6 is well known but SH6 is not known (b) Complete the equations : x 1 + sin x cos x (i) XeF6 + PF5 → 12. Evaluate ∫e cos 2 x dx. (ii) AlN + H2O → cos θ 13. Evaluate ∫ (2 + sin θ)(3 + sin θ) dθ OR XtraEdge for IIT-JEE 74 DECEMBER 2011 3x + 1 Evaluate : ∫ 2x 2 − 2x + 3 dx . Section C 23. The bag A contains 5 red and 3 green balls and bag B 14. Find the diff. equation of the family of curves contains 3 red and 5 green balls. One ball is drawn y = a sin (bx + c), a and c being parameter. from bag A and two from bag B. Find the Probability OR that of the three balls drawn two are red and one is green. dy Solve the differential equation : x − y − 2x 3 = 0 . dx 24. Find area between x2 + y2 = 4 and line x = 3 y in → → → ˆ j ˆ ˆ j ˆ ˆ j ˆ 15. If a = i – ˆ + 2 k , b =2 i + ˆ – 3 k , c = i +2 ˆ – k , first quadrant OR → → → → → → → → → verify that a × ( b × c ) = ( a . c ) b –( a . b ) c . Find the area of the region [(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9] 16. Find the vector equation of the plane passing through → 25. Find the image of the point (1, 6, 3) in the line ˆ ˆ the intersection of the planes r .(2 i –7 ˆ + 4 k ) =3 j x y −1 z − 2 → = = . ˆ ˆ and r .(3 i – 5 ˆ + 4 k ) + 11=0 and passing through j 1 2 3 the point (–2, 1, 3). 26. A dealer wishes to purchase a number of fans and OR radios. He has only Rs 5,760 to invest and has a Find the distance of the point (2, 3, 4) from the plane space for at most 20 items. A fan costs him Rs. 360 3x + 2y + 2z + 5 = 0, measured parallel to the line and a radio Rs. 240. His expectation is that he can x+3 y−2 z sell a fan at a profit of Rs. 22 and a radio at a profit of = = Rs. 18. Assuming that he can sell all the items he 3 6 2 buys, how should he invest his money for maximum profit ? Translate the problem as LPP and solve it 17. Using properties of determinats, prove that graphically. 1+ a 1 1 1 1 1 27. Evaluate the following integrals as limit of sums 1 1+ b 1 = abc 1 + + + a b c 2 1 1+ c ∫ (x 1 2 + 3) dx . OR 0 4 − 5 − 11 1 − 1 0 If A = 1 − 3 1 find A–1 28. Given that A = 2 3 4 and 2 3 − 7 0 1 2 2 2 − 4 18. If f (x) = cos x, g(x) = x2, then show that B = − 4 2 − 4 find A.B. Use this to solve the fog (x) ≠ gof (x) 2 −1 5 19. Show that curve xy = a2 and x2 + y2 = 2a2 touch each following system of equations other. x–y=3 2x + 3y + 4z = 17 20. Find the derivative of xex from first principle. y + 2z = 7 OR 21. If cos y = x cos (a + y), then prove that Using matrix method solve the following system of linear equations dy cos 2 (a + y ) = . 2x – 3y + 5z = 11 dx sin a 3x + 2y – 4z = –5 22. Determine f (0) so that function f (x) is defined by x + y – 2z = –3 (4 x − 1) 3 29. If the sum of the lengths of the hypotenuse and a side f (x) = becomes continuous at x = 0. x x2 of right angled triangle is given, show that the area of sin log1 + 4 3 triangle is maximum when the angle between them is π/3. XtraEdge for IIT-JEE 75 DECEMBER 2011 XtraEdge Test Series ANSWER KEY IIT- JEE 2012 (December issue) PHYSICS Ques 1 2 3 4 5 6 7 8 9 Ans A,C C A D A A A D D Ques 10 11 12 13 14 15 16 17 18 Ans 3 4 6 4 8 7 0012 0049 0005 C HE M ISTR Y Ques 1 2 3 4 5 6 7 8 9 Ans A B A C A,B A,B,C A A D Ques 10 11 12 13 14 15 16 17 18 Ans 4 4 3 1 4 6 0108 0001 0041 MATHEMATICS Ques 1 2 3 4 5 6 7 8 9 Ans B,C,D A C,D C C B,D D C D Ques 10 11 12 13 14 15 16 17 18 Ans 6 3 5 9 6 3 0012 1440 0000 IIT- JEE 2013 (December issue) PHYSICS Ques 1 2 3 4 5 6 7 8 9 Ans A A B C B D B D D Ques 10 11 12 13 14 15 16 17 18 Ans 6 7 1 4 5 4 0200 0004 0008 C HE M ISTR Y Ques 1 2 3 4 5 6 7 8 9 Ans B,C,D B A A B,C,D D C A B Ques 10 11 12 13 14 15 16 17 18 Ans 4 1 1 6 1 4 0002 0094 0496 MATHEMATICS Ques 1 2 3 4 5 6 7 8 9 Ans A,B,C B C,D A C B,C B C A Ques 10 11 12 13 14 15 16 17 18 Ans 6 0 3 5 6 6 0766 0010 1331 XtraEdge for IIT-JEE 76 DECEMBER 2011