Survey

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Survey

Document related concepts

Work (physics) wikipedia , lookup

Nordström's theory of gravitation wikipedia , lookup

Hydrogen atom wikipedia , lookup

Partial differential equation wikipedia , lookup

Derivation of the Navier–Stokes equations wikipedia , lookup

Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup

Transcript

Teachers open the door. You enter by yourself. Volume - 5 Issue - 7 January, 2010 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009 Editorial Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected] Editor : Pramod Maheshwari [B.Tech. IIT-Delhi] Cover Design & Layout Govind Saini, Om Gocher, Mohammed Rafiq Circulation & Advertisement Ankesh Jain, Praveen Chandna Ph (0744)- 3040007, 9001799502 Subscription Sudha Jaisingh Ph. 0744-2500492, 2500692 © Strictly reserved with the publishers • No Portion of the magazine can be published/ reproduced without the written permission of the publisher • All disputes are subject to the exclusive jurisdiction of the Kota Courts only. Every effort has been made to avoid errors or omission in this publication. In spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from. Unit Price Rs. 20/- Special Subscription Rates 6 issues : Rs. 100 /- [One issue free ] 12 issues : Rs. 200 /- [Two issues free] 24 issues : Rs. 400 /- [Four issues free] Owned & Published by Pramod Maheshwari, 112, Shakti Nagar, Dadabari, Kota & Printed by Naval Maheshwari, Published & Printed at 112, Shakti Nagar, Dadabari, Kota. Editor : Pramod Maheshwari XtraEdge for IIT-JEE Dear Students, Are You an Optimist or a Pessimist ? I have been giving some thought lately to optimism and pessimism. Basically, these are attitudes — attitudes that shape and formulate our entire existence. I mean, have you ever met a happy pessimist? Of course not. In short, our optimism or pessimism is this: The way we interpret the past. The way we experience and view the present. The way we imagine the future Have you given much thought about how your attitude, whether you are an optimist or a pessimist, affects you business, organization or school? Have you thought about how it affects you personally? And what about the team you are a part of? What is optimism? It is the belief that things in our past were good for us, even if that means they were hard and taught us lessons. It is also the belief that things will be better in the future. Here are some contrasts between optimism and pessimism and how they affect us: Optimism breathes life into you each day. Pessimism drains you. Optimism helps you to take needed risks. Pessimism plays it safe and never accomplishes much. Optimism improves those around you. Pessimism drags them down. Optimism inspires people to great heights. Pessimism deflates people to new lows. There is only one way that optimism and pessimism are the same, and that is that they are both self-fulfilling. If you are an optimist, you will generally find that good things happen to you. And if you are a pessimist, you will find yourself in the not-so-good situations more often than not. So can a person just become an optimist? Yes! We can choose to look at the world any way we want to. We can choose to look at the world and think the worst, or we can tell ourselves the good things about each situation. As you find yourself looking at your enterprise, begin to view it through the eyes of an optimist, and you will reap the rewards listed above, and so will the people around you. There are tremendous benefits to being an optimist, as stated above. But there are some pessimists out there who will say, “But that isn’t realistic.” I say, “Who cares?” If things go awry, at least I have spent my time beforehand enjoying life and not worrying about it. And, being an optimist, I would view the “negative” situation as an opportunity to grow and learn. So I can even look forward to my failures because they will be steppingstones and learning tools to be applied to my future success. Have you ever met a successful pessimist? Become an optimist, and see your world change before your eyes! Have a blessed day! Let's make the uncommon knowledge common Yours truly Pramod Maheshwari, B.Tech., IIT Delhi 1 JANUARY 2010 XtraEdge for IIT-JEE 2 JANUARY 2010 Volume-5 Issue-7 January, 2010 (Monthly Magazine) NEXT MONTHS ATTRACTIONS CONTENTS INDEX Regulars .......... Key Concepts & Problem Solving strategy for IIT-JEE. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics, Chemistry & Maths Much more IIT-JEE News. Xtra Edge Test Series for JEE-2010 & 2011 Mock Test CBSE Pattern Cllass XII PAGE NEWS ARTICLE 4 IITian ON THE PATH OF SUCCESS 8 KNOW IIT-JEE 10 IIT boys draw power & water from sewage Alumni of IIT-Madras to come together on December 26 Dr. Alok Aggarwal Previous IIT-JEE Question Study Time........ DYNAMIC PHYSICS Success Tips for the Month • It is more important to know where you are going than to get there quickly. • The secret of success is constancy to purpose. • 8-Challenging Problems [Set# 8] Students’ Forum Physics Fundamentals Refraction at plane & curved surfaces Properties of Matter CATALYST CHEMISTRY of all success. We cannot discover new oceans unless we DICEY MATHS have the courage to lose sight of the Mathematical Challenges Students’ Forum Key Concept Differential Equations Trigonomatrical Rations shore. • One person with a belief is equal to 99 who have only interests • • Keep your eyes on the stars and your feet on the ground. XtraEdge for IIT-JEE 48 Test Time .......... A thousand mile journey begins with one step. Start today. 33 Key Concept Carbohydrates Salt Analysis Understanding : Organic Chemistry Action without planning is the cause of all failure. Action with planning is the cause • 17 XTRAEDGE TEST SERIES 59 Class XII – IIT-JEE 2010 Paper Class XII – IIT-JEE 2011 Paper Mock Test CBSE Pattern Paper-2 [Class # XII] Mock Test CBSE Pattern Paper-1 (Solution)[Class # XII] 3 JANUARY 2010 IIT boys draw power & water from sewage When the campus placements Anand (energy engineering) and a cash prize of Rs 3 lakh. "LOCUS Mohan Yama (PhD student of is a green tech development, biotechnology). which Renowned is sustainable both biotech faculty member Debabrata economically and environmentally happen in 2011 at IIT Kharagpur, Das enthusiastically joined them in and serves as an ideal integration these five students will give it a the pursuit. to address the key issues of miss. They would have by then started their own company. They have developed a bio cell' (battery) that can not only treat sewage water but also generate electricity a that could offer a onestop solution to the water and energy crises. Their bio-product has won them The idea is simple. The specially designed bio cell (LOCUS) will be set up in the form of a plant, through which the sewage water of a housing complex would be flowed in. The genius of this invention which is awaiting patent lies in the design of the cell that will automatically grow millions of rave accolades from the ministry anaerobic bacteria that multiply of science and technology and a through respiration. The bacteria cash award to carry their research clean up the sewage water and in forward. the process generate free harnessed, these It was while working on microbes electrons. that can be used as purifying electrons agents that Manoj Mandelia, a electricity. fourth year student of biotech "We worked on this concept for engineering, stumbled upon the nearly a year before we readied idea that if a bio cell can be the cell and applied to the ministry developed to treat sewage water If can generated to enter its annual business plan for use, it would solve one of the competition that focuses on bio biggest problems of the present technology times. Mandelia, who is pursuing sustainable development," Mandelia an integrated M Tech programme products for said. at the institute, started looking for Twenty teams, mostly corporate like minded boys for his project. houses dealing with bio products, He participated soon found Prateek Jain (agriculture and food engineering department), (electrical Shobhit engineering), XtraEdge for IIT-JEE Singhal in this premier competition. The IIT-Kgp team managed to come second and won wastewater treatment and energy gap," reads the award citation. The cell, at this stage, can clean up 50,000 litres of sewage water, about the amount generated by 100 flats in a day. The water produced this way compared with can that be supplied provided by a civic body, the students say. "The purified water has been tested and has been certified to be clean and fit for household use. It is, however, not fit for drinking," Mandelia explained. The IIT-Kgp team has even produced electricity with the bio cell. "A township of 100,000 people needs about 2.3 megawatts of electricity a day. It will be years before we reach that stage. But we have already been able to generate electricity. By next year, we aim to generate 350 units, enough to meet 50% of the demand of a 100-flat complex. When we say this we are not taking airconditioners into consideration," said Prateek. Pulkit 4 JANUARY 2010 Alumni of IIT-Madras to for the first year, the initiative visited come being executed by the department College here on Friday. Sources of management studies is designed said they discussed the make-shift to act as a catalyst to provide requirements and took stock of initial momentum. the together on December 26 The "first ever" gathering of the alumni of IIT-Madras (IITM) will be held here on December 26. According to a release here, more than 32,000 students have graduated from the institute in the past 50 years and most were expected to attend the reunion, co-organised by the Office of Alumni Affairs (OAA) and IIT-M Alumni Association. Speaking at the launch of the initiative here on Friday, L S Ganesh, professor-in-charge of Cell for Technology, Innovation, Development and Entrepreneurship (C-TIDES), Support said IIT-Madras is perhaps the first institution to start a formal programme in entrepreneurship in the 1980s. The day-long event would see the Emphasizing the importance of screening developing of two films, one small and medium highlighting alumni contributions enterprises, IIT-Madras director M to the campus and the other their S Ananth said they generated 10 socially-relevant work, especially times the employment per unit of in the rural sector. investment. "We need to the MBM facilities Engineering available in the college. The team will submit the report to the Director (IIT-R), who is scheduled to visit the city in the first week of December to consolidate the ground for shifting. MBM dean Arvind Roy said that the team was here to oversee the proposed shifting early next year. The team had a thorough visit of the entire college campus buildings, classrooms, laboratories and other rooms housing different departments and collected a map of the college. Roy added that their main focus, decentralise these aspects to get however, innovations started with centres facilities, for which they also such as this. Also, innovators need visited the AIIMS site, where the to be aware of their privileges and students are initially proposed to Alumni from different parts of the what they are entitled to," he said. be shifted, in case the building world would make presentations According to professor Thillai of the IIT-R is not ready by then. on various issues, the release said Rajan, It is evident that there has been an adding an "IIT Madras Heritage department Quiz" would also be held. studies, detailed workshops would IIT-R will be shifted to the said IIT-M launches seed fund be for prospective college here from next year. A incubates before issuing a call for Central team, headed by Union proposals. additional secretary (HRD) Ashok IIT-M director Prof MS Ananth is scheduled to deliver the inaugural address, the release said. of $ 0.8 mn to help budding entrepreneurs Students, researchers and faculty assistant of conducted professor, management IIT-R shifting to Jodhpur was on understanding Thakur, the hostel in-principle recently visited that the college and the hostels which are members from IIT-Madras now gradually taking shape have the chance to turn innovative JODHPUR: ideas into sustainable businesses referred to as one step further arrangements there. with the Micro, Small and Medium towards shifting the IIT-R from The visit of this team of these sites Enterprises Kanpur team again has further strengthened administrative that fact that the plans of shifting initiative. (MSME) Intended incubation to provide In to what Jodhpur, comprising can a complete be and said to have expressed satisfaction with the seed-stage funding of up to $ 0.8 in-charge (Kanpur) are gradually taking shape. The million (Rs 3.5 crore) per venture Niraj Gupta and two others, team will explore and identify the XtraEdge for IIT-JEE of IIT-R 5 JANUARY 2010 necessary additions and alterations total number of PD seats in 2009 Signal engineering is the that will have to be made into the across all IITs was 251, only 138 existing structure of the college to candidates could qualify at the JEE, backbone of Rlys: SCR accommodate IIT-R and prepare a despite report to be submitted to the candidates taking the exam as director. compared to 2008. IITs worried as reserved "Despite the 50% relaxation for seats remain unfilled PD candidates (from the last Even as the process for applying to the Joint Entrance Examination (JEE) 2010 of the Indian Institutes of Technology (IITs) has begun, 12% more disabled general candidate), only 44 could its productivity to meet the increase in demand for rail traffic, Sudesh Kumar, member (Electrical) Railway Board has said. celerations of the Indian Railways unfilled," said Kumar. Institute of Signal Engineering and scores of PD candidates were with physical disabilities (PD), got relaxed by another 50% to enable wasted. The reason is that not candidates to qualify for the enough disabled candidates qualify, preparatory course, which was and started for disabled candidates in be enhance IITs. This left around 207 seats 137 seats meant for candidates cannot Railway Speaking at the 52nd Annual Day To make up for the shortfall, seats Indian be given admission across all the figures from JEE 2009 reveal that these Technology changes have helped Telecommunications (IRISET) here today, he said signalling has evolved from an ordinary means of communication to start a train and later emerged as a symbol of trust and safety on railway tracks over the years. converted into general seats. 2009. In case of OBCs, 51 seats "Unless the rules are changed and across the IITs in 2009 were Participating in the event, the new the IITs are allowed to transfer converted into general seats due South the vacant PD seats to general or to unavailability of OBC students General Manager M S Jayanth said non-PD seats within the category, even IRISET over a period of time has we will not be able to stop this relaxation. Again, over 1,000 seats loss," said Anil Kumar, JEE 2010 for SCs and STs which remained chairman, IIT Bombay. vacant were transferred to the As per a judgment passed by the chief commissioner for persons after giving full 10% preparatory course and were filled after lowering the bar by 50%. IIT Guwahati director Gautam to treat disabled candidates on a Barua said that a resolution was par adopted at the last joint admission scheduled caste tribe (ST) (SC)/scheduled board meeting that the PD candidates and give them similar commissioner relaxation from 2009, including approached admission conversion of PD seats to general for preparatory courses. should for Railway (SCR) evolved as a quality institute. ``Signal engineering happens to be the backbone of railways, since this is vital for reliability and safety of rails,’’ Jayanth said. with disabilities, IITs were directed with Central be possible category in future. "However, the According to him, advancement made in telecommunication has palyed vital role in the information management of Indian Railways, especially in passenger reservation system, fright and other services. IRISET Director V Balaram said signalling is very specific to each issue is complex since there are category [general category, SC, ST certain PD seats within SC and ST railways. and and these cannot be converted," IRISET is the only institute in said Barua. railways, which trains both officers Currently, other 3% seats backward in classes (OBC)] are reserved for disabled students in each IIT. While the XtraEdge for IIT-JEE and supervisory staff. ``During the 6 JANUARY 2010 last 52 years, about 10,036 officers It helps him determine the amount site. But with the Australian and 44,906 signal and telecom of anaesthesia, and can also be newspaper for which we made an supervisiors have been trained,’’ tweaked app, we not only created an app Balaram said. requirements. This app is now but V G K Murti, former Dean of IIT used by around 15,000 doctors required to their Chennai, said in the earlier days internationally. In the future, we operation so that railways used to be the most have plans to make it possible for delivered faster to the user. When preferred place of employment for doctors we approach a client we don't most IIT students. He delivered profiles," Avinash says. his speech targeting young trainees of IRISET, by passing on knowledge inputs. The function was attended by many senior officials of the SCR. for to other medical exchange patient applications for an Australian newspaper and a top computer manufacturer, among others. Making apps for the enterprise Bangalore: One of the most discussed issues when it came to developing applications for mobile phones was the lack of any reasonable returns on apps made for popular platforms like the iPhone. A lot of developers said they hardly made any money from developing apps for these high-end instead of selling it to the public via an appstore is a different ballgame altogether. The app store was an instant success as it allowed a single developer to roll out an app and be able to sell it without worrying about marketing. All the developer has to worry about, is the competition phones. Turns out they have got --similar apps available. But while their strategy wrong. Endeavour, a making apps for enterprises, it's city-based company, sees good the business adaptability that counts. in making apps for enterprises. T functionality and the changes backend data was developers, but instead, tell them we can come up with a mobile strategy for their company. These enterprise apps will never make it to the app store, but are a company," he adds. The company has also developed a few fun apps like upcoming app Ambience which allows users mix and match sound effects like that of a waterfall, the sea, birds, etc to create a relaxing atmosphere. "As of now, most of the enterprises still use Blackberries.But we're seeing a slow shift towards touch screen devices like the iPhone," he says. Avinash says they have not made any apps for Indian companies till "The cost of an app can range now, as very few Indians use high- from end smartphones or other mobile- $10,000 to $3,50,000 he founders, all IIT graduates, depending on the depth and the started the company in 2002. functionality required. We don't According to Avinash Misra, one just create an app, we see to it of the founders, their first app was that targeted at doctors. "We made it operational in co-ordination with a doctor environment as well. It's very easy based in the US using which he to make a news application by could make a profile of a patient. using the RSS feed available on the XtraEdge for IIT-JEE the distributed to all the employees in Enterprises need mobile strategy made introduce ourselves as mobile app The company has also developed mobile also their back-end for 7 is such made an enabled devices. "But in 18 months, I'm sure India will have one of the largest markets in terms of mobile usage. Indians just might skip the desktop internet revolution and switch over to the mobile internet altogether," Avinash adds. JANUARY 2010 Success Story This article contains story of a person who get succeed after graduation from different IIT's Dr. Alok Aggarwal Electrical Engineering from IIT Delhi in 1980 Ph.D. in Electrical Engineering and Computer Science, Hopkins University (1984) Director : IBM Research Division worldwide Emerging Business Opportunities for IBM Research Division worldwide. Dr. Alok Aggarwal received his Bachelor’s Degree in Electrical Engineering from IIT Delhi in 1980. He obtained his Ph.D. in Electrical Engineering and Computer Science from Johns Hopkins University in 1984. Dr. Alok Aggarwal has published 86 Research papers and he has also been granted 8 patents from the US Patents and Trademark Office. Along with his colleagues at Evalueserve, in 2003, he has pioneered the concept of “Knowledge Process Outsourcing (KPO)” and wrote the first article in this regard. Dr. Aggarwal has served as a Chairperson of the IEEE Computer Society's Technical Committee on Mathematical Foundations of Computing and has been on the editorial boards of SIAM Journal of Computing, Algorithmica, and Journal of Symbolic Computation. During 1998-2000, Dr. Aggarwal was a member of Executive Committee on Information Technology of the Confederation of the Indian Industry (CII) and also of the Telecom Committee of Federation of Indian Chamber of Commerce and Industry (FICCI). He is currently a Chartered Member of The Indus Entrepreneur (TiE) organization. Dr. Alok Aggarwal is the Founder and Chairman of Evalueserve - a company that was started in December 2000 and that provides various kinds of research and analytics services to clients in North America, Europe and Asia Pacific from its five research centers in DelhiGurgaon, India; Shanghai, China; Cluj, Romania; SantiagoValparaiso, Chile; and New York, USA. Dr. Alok Aggarwal joined IBM Research Division in Yorktown Heights New York in 1984. During the fall of 1987 and 1989, he was on sabbatical from IBM and taught two courses (in two terms) at the Massachusetts Institute of Technology (MIT) and also supervised two Ph.D. students. During 1991 and 1996, along with other colleagues from IBM, he created and sold a "Supply Chain Management Solution" for paper mills, steel mills and other related industries. In July 1997, Dr. Aggarwal "Founded" the IBM India Research Laboratory that he setup inside the Indian Institute of Technology Delhi. Dr. Aggarwal started this Laboratory from "ground zero" and by July 2000, he had built it into a 60-member team (with 30 PhDs and 30 Masters in Electrical Engineering, Computer Science, and in Business Administration). In August 2000, Dr. Aggarwal became the Director of In honouring Dr. Alok Aggarwal, IIT Delhi recognizes the outstanding contributions made by him as an Entrepreneur and Researcher. Through his achievements, Dr. Alok Aggarwal has brought glory to the name of the Institute. Anything you can hunt, I can hunt better. XtraEdge for IIT-JEE 8 JANUARY 2010 XtraEdge for IIT-JEE 9 JANUARY 2010 KNOW IIT-JEE By Previous Exam Questions The direction of B is given by Right hand palm rule no. 1. PHYSICS 1. A charge +Q is fixed at the origin of the co-ordinate system while a small electric dipole of dipole moment A P → O P pointing away from the charge along the x-axis is set free from a point far away from the origin. (a) Calculate the K.E. of the dipole when it reaches to a point (d, 0). (b) Calculate the force on the charge +Q at this [IIT-2003] moment. Sol. (a) Potential energy of the dipole-charge system U i = 0 (Since the charge is far away) E= d A → 2pQ 4πε 0 d 3 1 pQ 4πε 0 d 2 f r → → → B p = B PA + B PB + B PC î where → B PA = magnetic field at P due to A → B PB = magnetic field at P due to B Three infinitely long thin wires, each carrying current i in the same direction, are in the x-y plane of a gravity free space. The central wire is along the yaxis while the other two are along x = ± d. (i) Find the locus of the points for which the [IIT-1997] magnetic field B is zero. (ii) If the central wire is displaced along the Zdirection by a small amount and released, show that it will execute simple harmonic motion. If the linear mass density of the wires is λ, find the frequency of oscillation. Sol. (i) We know that magnetic field due to an infinitely long current carrying wire at distance r is given by → B PC = magnetic field at P due to C. Bp = For µ 0 2I 1 1 1 (– k̂ ) + + 4π d + x x d − x → BP = 0 On solving we get x = ± d 3 . (ii) The force per unit length between two parallel current carrying wires is given by µ 0 2I1I 2 = f(say) 4π r and is attractive if currents are in the same direction. So when the wire B is displaced along z-axis by a small distance z, the restoring force per unit length µ 0 2I 4π r XtraEdge for IIT-JEE X Z θ θ Z Hence in case of three identical wires resultant field can be zero only if the point P is between the two wires otherwise field B due to all the wires will be in the same direction and so resultant B cannot be zero. Hence, if point P is at a distance x from the central wire as shown in fig. then, 2. B= r d B 1 2p î 4πε 0 d 3 F =QE = B f Now, force on charge Q is given by → d d (b) Electric field at origin due to dipole → X Z 1 p Uf = – Q × 4πε 0 d 2 ∴ K.E. = |Uf – Ui | = C B F/l on the wire B due to wires A and C will be 10 JANUARY 2010 The magnetic flux linked with the solenoid z as cos θ = r µ 2I I F z = 2f cos θ = 2 0 1 2 × l 4π r r or F µ0 4I 2 = z [as I1 = I2 = I and r2 = d2 + z2] l 4 π (d 2 + z 2 ) or µ 2I F = 0 z [as d>>z and F is opposite to z] l 4π d → 2 dφ = π µ 0 n a2 i0 ω cos ωt dt The same rate of change of flux is linked with the cylindrical shell. By the principle of electromagnetic induction, the induced emf produced in the cylindrical shell is ...(1) Since F ∞ –z the motion is simple harmonic. Comparing eq. (1) with the standard equation of S.H.M. which is F = – mω2z i.e., F m = – ω2z = – λω2z, we get l l λω2 = ⇒ 3. µ 0 4I 2 ⇒ 4π d 2 I 2πn = d µ0 ⇒ πλ ω= ×××× µ0 I 2 πd 2 λ 1 n= 2πd I TOP VIEWS µ0 πλ dφ = – πµ0 n a2 i0 ω cos ωt … (i) dt The resistance offered by the cylindrical shell to the flow of induced current I will be e=– A long solenoid of radius a and number of turns per unit length n is enclosed by cylindrical shell of radius R, thickness d(d <<R) and length L. A variable current i = i0 sin ωt flows through the coil. If the resistivity of the material of cylindrical shell is ρ, find [IIT- 2005] the induced current in the shell. R a Here, 2πR Ld The induced current I will be ∴ R=ρ I= L I= ⇒ I= Sol. The magnetic field in the solenoid is given by B = µ0 ni R a … (ii) [ πµ 0 na 2 i 0 ω cos ωt ] × Ld |e| = R ρ × 2πR πµ 0 na 2 Ld i 0 ω cos ωt 2πRρ µ 0 na 2 Ld i 0 ω cos ωt 2ρR 4. An object is moving with velocity 0.01 m/s towards a convex lens of focal length 0.3 m. Find the magnitude of rate of separation of image from the lens when the object is at a distance of 0.4 m from the lens. Also calculate the magnitude of the rate of change of the [IIT - 2004] lateral magnification. Sol. Using lens formula d L 1 1 1 = = v − 0.4 0.3 B = µ 0 n i0 sin ωt ⇒ [Q i = i0 sin ωt given] XtraEdge for IIT-JEE l A l = 2 π R, A= L×d R=ρ d ⇒ → φ= B.A = B A cos 90º = (µ 0 n i0 sin ωt) ( πa 2) ∴ The rate of change of magnetic flux through the solenoid 11 v = 1.2 m JANUARY 2010 Now we have A B 1 1 1 – = , differentiating w.r.t. t v u f 1 dv 1 du + =0 v 2 dt u 2 dt we have – given du = 0.01 m/s dt ⇒ dv (120) × 0.01 = 0.09 m/s = dt (0.4) 2 d = 1 cm (b) Charge on plate B at t = 10 sec Qb = 33.7 × 10–12 – 5 × 107 × 1.6 × 10–19 = 25.7 × 10 –12 C also Q a = 8 × 10–12C 2 So, rate of separation of the image (w.r.t. the lens) = 0.09 m/s v m= u Now, udv vdu − dm ⇒ = dt 2 dt dt u (0.4)(0.09) − (1.2)(0.01) (0.4) 2 E= = = – 0.35 CHEMISTRY Two metallic plates A and B, each of area 5 × 10–4 m2, are placed parallel to each other at a separation of 1 cm. Plate B carries a positive charge of 33.7×10–12C. A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0 so that 10 16 photons fall on it per square meter per second. Assume that one photoelectron is emitted for every 106 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remain constant at the [IIT-2002] value 2eV. Determine 6. At room temperature, the following reactions proceed nearly to completion : 2NO + O2 → 2NO2 → N2O 4 The dimer, N2O4, solidified at 262 K. A 250 ml flask and a 100 ml flask are separated by a stopcock. At 300 K, the nitric oxide in the larger flask exerts a pressure of 1.053 atm and the smaller one contains oxygen at 0.789 atm. The gases are mixed by opening the stopcock and after the end of the reaction the flasks are cooled to 200 K. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at 220 K. (Assume [IIT-1992] the gases to behave ideally) Sol. According to the gas equation, (a) the number of photoelectrons emitted to t = 10 s, (b) the magnitude of the electric field between the plates A and B at t = 10 s, and (c) the kinetic energy of the most energetic photoelectron emitted at t = 10 s when it reaches plate B. PV = nRT Neglect the time taken by the photoelectron to reach plate B. Takes ε0 = × 10–12 C2/N-m2. or Sol. (a) Number of electrons falling on the metal plate 16 10 6 XtraEdge for IIT-JEE PV RT For NO, P = 1.053 atm, V = 250 ml = 0.250 L ∴ Number of photoelectrons emitted from metal plate A upto 10 second is (5 × 10 −4 ) × 1016 n= At room temperature, –4 A = 10 × (5 × 10 ) ne = 17.7 ×10 −12 = 2000 N/C 5 × 10 − 4 × 8.85 ×10 −12 (c) K.E. of most energetic particles = (hν – φ) + e(Ed) = 23 eV [(hν – φ) is energy of photo electrons due to light. e(Ed) is the energy of photoelectrons due to work done on photoelectrons between the plates]. So magnitude of the rate of change of lateral magnification = 0.35. 5. σB σ 1 – A = (Q B – QA) 2ε 0 2ε 0 2Aε 0 ∴ Number of moles of NO = 1.053 × 0.250 0.0821× 300 = 0.01069 mol For O2, P = 0.789 atm, V = 100 ml = 0.1L × 10 = 5 × 10 7 12 JANUARY 2010 ∴ Number of moles of O 2 = 0.789 × 0.1 0.0821× 300 = 0.00320 mol According to the given reaction, γ=1+ ∴ 0.00320 mol of O2 react with = 2 × 0.00320 = 0.0064 mol of NO Number of moles of NO left = 0.01069 – 0.0064 = 0.00429 mol Also, 1 mol of O2 yields = 1 mol of N2O4 f= 2 f 2 =5 0.4 P1V1γ = P 2V2γ P1 = P V1 = V V2 = 5.66 V P P P [using eq.(1)] = = γ 1.4 11 .32 (5.66) (5.66) Hence, work done by the gas during adiabatic expansion P V −P V = 1 1 2 2 = γ –1 An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V, while its temperature falls to T/2. (a) How many degrees of freedom do the gas molecules have ? (b) Obtain the work done by the gas during the expansion as a function of the initial pressure P and volume V. [IIT-1990] Sol. (a) According to adiabatic gas equation, PV − P × 5.66 V 11.32 1.4 – 1 PV 2 = PV = 1.25 PV 0.4 2 × 0.4 PV − = 8. (a) A white solid is either Na2O or Na2O2. A piece of red litmus paper turns white when it is dipped into a freshly made aqueous solution of the white solid. (i) Identify the substance and explain with balanced equation. (ii) Explain what would happen to the red litmus if the white solid were the other compound. (b) A, B and C are three complexes of chromium (III) with the empirical formula H12O 6Cl3Cr. All the three complexes have water and chloride ion as ligands. Complex A does not react with concentrated H2SO4, whereas complexes B and C lose, 6.75% and 13.5% of their original mass, respectively, an treatment with conc. H2SO4. Identity A, B and C. [IIT-1999] Sol. (a) The substance is Na2O2 TVγ–1 = constant T1V1γ–1 = T2V2γ–1 T1 = T ; T2 = T/2 V1 = V V2 = 5.66 V T × (5.66V)γ–1 2 T × (5.66)γ–1 × V γ–1 2 or (5.66)γ–1 = 2 Taking log, or or P2 = 7. = 2 = 1.4 – 1 = 0.4 f 1.4 = 1 + Hence, PVγ = P2 × (5.66V)γ = P2 ×(5.66)γ × Vγ nRT 0.00429 × 0.0821× 220 = = 0.221 atm V 0.350 Hence, TVγ–1 = or or Here, ∴ 0.350 L (0.1 + 0.250) contains only 0.00429 mol of NO At T = 220 K, Pressure of the gas, and 2 f (b) According to adiabatic gas equation, ∴ Number of moles of N 2O4 formed = 0.00320 mol N2O4 condenses on cooling, or Here, log 2 0.3010 = = 0.4 log 5.66 0.7528 or γ = 1.4 If f, be the number of degrees of freedom, then 2NO + O2 → 2NO2 → N2O4 Composition of gas after completion of reaction, Number of moles of O 2 = 0 1 mol of O2 react with = 2 mol of NO P= γ–1= or ...(1) (i) Na2O2 + 2H2O → 2NaOH (strong base) (γ – 1)log 5.66 = log 2 + H2O2 (Weak acid) H2O2 + red litmus → White H2O2 → H2O + [O] XtraEdge for IIT-JEE 13 JANUARY 2010 Nascent oxygen bleaches the red litumus. 4s 4p 3d 4s 4p 3+ Co ion in Complex ion d2sp3 hybridization H3N NH3 NH3 3+ 18 × 100 = 6.75% 266.5 C = [Cr(H2O)4Cl]Cl2.2H2O Conc. H2SO 4 removes its 2H2O which are outside of the coordination sphere. H3N H3N or Co % loss = NH3 NH3 NH3 NH3 Co3+ H3N NH3 NH3 In [Ni(CN)42– nickel is present as Ni2+ ion and its coordination numbers is four Ni28 =1s2, 2s22p 6, 3s23p63d8, 4s2 Ni2+ ion = 1s2, 2s22p 6, 3s23p63d8 3d 4s 4p 2+ Ni ion = 18 × 100 = 13.5 % 266.5 Hence complexes A = [Cr(H2O)6]Cl3 B = [Cr(H2O)5Cl]Cl2.H2O C = [Cr(H2O)4Cl2]Cl.2H2O % loss = 2 × 3d 9. (a) Write the chemical reaction associated with the "brown ring test". (b) Draw the structures of [Co(NH3)6]3+, [Ni(CN)4]2– and [Ni(CO)4]. Write the hybridization of atomic orbital of the transition metal in each case. (c) An aqueous blue coloured solution of a transition metal sulphate reacts with H2S in acidic medium to give a black precipitate A, which is insoluble in warm aqueous solution of KOH. The blue solution on treatment with KI in weakly acidic medium, turns yellow and produces a white precipitate B. Identify the transition metal ion. Write the chemical reaction involved in the formation of A and B. [IIT-2000] Sol. (a) NaNO3 + H2SO 4 → NaHSO4 + HNO3 2HNO 3 + 6FeSO4 + 3H2SO4 → 3Fe2(SO4)3 + 2NO + 4H2O [Fe(H2O)6]SO4.H2O + NO Ferrous Sulphate → [Fe(H2O)5NO] SO4 + 2H2O (Brown ring) 3+ (b) In [Co(NH3)6] cobalt is present as Co3+ and its coordination number is six. Co27 = 1s1, 2s22p6, 3s23p 63d7, 4s2 Co3+ion = 1s2, 2s22p 6, 3s23p63d6 XtraEdge for IIT-JEE 3d Hence (ii) Na2O + H2O → 2NaOH NaOH solution turns colour of red litmus paper into blue due to stronger alkaline nature. (b) A = [Cr(H2O)6]Cl3. It has no reaction with conc. H2SO 4 as its all water molecular are present in coordination sphere. B = [Cr(H2O)5Cl]Cl2.H2O Conc. H2SO4 removes its one mol of H2O as it is outside the coordination sphere. Molecular Weight of complex = 266.5 4s 4p 2+ Ni ion in Complex ion dsp2 hybridization Hence structure of [Ni(CN)4]2– is C≡N N≡C Ni2+ C≡N N≡C In [Ni(CO)4, nickel is present as Ni atom i.e. its oxidation number is zero and coordination number is four. 3d 4s 4p Ni in Complex sp3 hybridization Its structure is as follows : CO Ni OC CO CO 14 JANUARY 2010 (c) The transition metal is Cu2+. The compound is CuSO4.5H2O CH3CHCH3 OH 2-propanol CuSO4 + H2S Acidic medium → CuS ↓ + H2SO4 Black ppt and hence the structure of (A) should be 2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4 CH3CHCH3 (B) white NH2 Propan-2-amine I2 + I– → I3– (yellow solution) 10. A basic volatile, nitrogen compound gave a foul smelling gas when treated with CHCl3 and alcoholic KOH. A 0.295 g sample of the substance dissolved in aqueous HCl and treated with NaNO 2 solution at 0ºC liberated a colourless; odourless gas whose volume corresponds to 112 ml at STP. After the evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and iodine gave yellow precipitate. Identify the original substance. Assume it contains one N-atom [IIT-1993] per molecule. MATHEMATICS 11. Find the values of a and b so that the function x + a 2 sin x, 0 ≤ x ≤ π/ 4 f(x) = 2x cot x + b, π/4 ≤ x ≤ π/2 a cos 2 x − b sin x , π / 2 < x ≤ π is continuous for 0 ≤ x ≤ π Sol. As, f(x) is continuous for 0 ≤ x ≤ π π π ∴ R.H.L. at x = = L.H.L. at x = 4 4 Sol. Clue 1. Nitrogen compound gave foul smelling gas when treated with CHCl3 and alc. KOH (carbylamine reaction), thus it is a primary amine. π π π π ⇒ 2. cot + b = + a 2 . sin 4 4 4 4 Clue 2. This compound when treated with HCl + NaNO2 solution (nitrous acid test) at 0ºC liberates colourless and odourless gas. ⇒ π π +b= +a 2 4 ⇒ a–b= HCl + NaNO 2 CnH2n+1NH2 → ROH + N 2 ↑ Alcohol [IIT-1989] π 4 ....(i) π π also, R.H.L at x = = L.H.L at x = 2 2 Nitrogen At STP, 112 ml of N2 is evolved from = 0.295 g CnH2n+1NH2 2π π π π ⇒ a cos − b sin = 2. . cot + b 2 2 2 2 ∴ 22400 ml of N2 is evolved from ⇒ –a–b =b ⇒ a + 2b = 0 = 0.295 × 22400 = 59 g CnH2n+1NH2 112 ∴ CnH2n+1NH2 = 59 or n × C + (2n + 1) × H + N + 2 × H = 59 12. Find or 12n + 2n + 1 + 14 + 2 × 1 = 59 or n = 42 =3 14 dy at x = –1, when dx (sin y) Thus the molecular formula of nitrogen compound is C3H7NH2. π sin x 2 + π sin x 2 + Sol. Here, Clue 3. Alcohol obtained gives iodoform test positive, thus it is a secondary alcohol and its structure should be XtraEdge for IIT-JEE ...(ii) 3π −3π and b = From (i) and (ii), a = 2 4 3 sec–1(2x) + 2x tan ln (x + 2) = 0 2 [IIT-1991] 3 sec–1(2x) + 2x tan (log (x + 2)) = 0 2 Differentiating both sides, we get (sin y) 15 JANUARY 2010 π sin x (sin y) 2 14. If exp {(sin2x + sin4x + sin6x + ...... ∞). ln 2} satisfies the equation x2 – 9x + 8 = 0, find the value of π cos x ,0<x< . [IIT-1991] cos x + sin x 2 Sol. exp {(sin2x + sin4x + sin6x + ...... ∞) loge2 π π . log(sin y) . cos x . 2 2 π sin x −1 dy π + sin x (sin y) 2 . cos y . dx 2 2 2 x . sec 2 (log( x + 2)) 3 + + . ( x + 2) 2 ( 2 | x |) 4 x 2 − 2 x + 2 log 2 . tan (log(x + 2)) = 0 3 , we get putting, x = −1, y = − π dy dx −1, − 3 π − 3 π = 2 3 1− π 2 = 3 sin(2A + B) = sin(C – A) = –sin(B + 2C) = e log e 2 ⇒ ⇒ 2 tan x satisfy x 2 – 9x + 8 = 0 x = 1, 8 ∴ ⇒ 2 tan x = 1 and 2 tan x = 8 tan2x = 0 and tan2x = 3 ⇒ π x = nπ and tan2x = tan 3 and x = nπ ± 1 2 ⇒ 2 2 2 x= 2 π 3 π 2 π π ∈ 0, 3 2 1 cos x 1 2 ∴ = × = cos x + sin x 1+ 3 1 3 + 2 2 Sol. Given that in ∆ABC, A, B and C are in A.P. A + C = 2B A + B + C = 180º ⇒ tan 2 x Neglecting x = nπ as 0 < x < If A, B and C are in Arithmetic Progression, determine the values of A, B and C. [IIT-1990] = B = 60º Also given that, ∴ sin (2A + B) = sin (C – A) = – sin (B + 2C) = 1/2 ...(1) 3 −1 3 −1 3 −1 2 cos x = cos x + sin x 3 −1 2 15. Find the value of : 1 ⇒ sin (2A + 60º) = sin (C – A) = – sin (60º + 2C) = 2 cos (2 cos–1 x + sin–1x) at x = ⇒ 2A + 60º = 30º, 150º and –π/2 ≤ sin–1x ≤ π/2 Sol. cos{2cos–1x + sin–1x} {neglecting 30º, as not possible} ⇒ ⇒ π π2 − 3 13. ABC is a triangle such that also ⇒ sin 2 x . log e 2 2 e 1−sin x 2A + 60º = 150º 1 , where 0 ≤ cos–1x ≤ π 5 [IIT-1981] again from (1), sin (60º + 2c) = –1/2 π π = cos cos −1 x + , as cos–1x + sin–1x = 2 2 –1 = – sin(cos x ) ⇒ 60º + 2C = 210º, 330º = – sin(sin–1 ⇒ C = 75º or 135º 1 = – sin sin −1 1 − 2 5 ⇒ A = 45º Also from (1) sin (C – A) = ½ C – A = 30º, 150º, 195º 2 6 2 6 = – sin sin −1 = 5 5 for A = 45º, C = 75º and C = 195º (not possible) ∴ 1− x 2 ) C = 75º Hence, A = 45º, B = 60º, C = 75º XtraEdge for IIT-JEE 16 JANUARY 2010 Physics Challenging Problems Set # 9 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 Director Academics, Jodhpur Branch So lu tio n s wi ll b e p ub lish ed in n ex t is su e Q.1 An imaginary closed loop is shown with current carrying conductors then 4A c d mv then the particle will hit the upper qB plate at P 1 if it is negatively charged (C) If d ≥ mv then it will never hit upper plate and qB it's motion will be clockwise for positively charged particle as seen in plane mirror (D) If d ≥ b a 3A Closed Loop 1A (A) Line integral of magnetic field over the closed loop abcda is zero (B) Surface integral of the magnetic field over the closed loop abcda is non zero (C) Line integral of magnetic field and surface integral of magnetic field over the closed loop abcda both are non zero (D) Surface integral of magnetic field over the closed loop is zero but the line integral of 1 magnetic field over the closed loop is 2 C .ε 0 Here C - speed of light in air / free space / vacuum ε0 - Absolute permittivity of air / free space / vacuum Q.3 The principal axis of the given concave mirror is along X-axis. The details about the mirror are also shown. In the shaded portion there is the coexistence of uniform magnetic field and electric field the details are written below → E = E0 î , Plate-2 d Plate-1 × × × P × × × × × × P1 × × × E 0 : B0 = 19.6 : 1 A charged particle is projected from the point → (30 cm, 0, 0) with the velocity of v = v 0 ˆj , v0 is a positive constant. If the particle moves undeviated then the particle is Y-axis → × × × B × × × Plane Mirror Concave mirror (A) (B) (C) (D) (A) If the particle is positively charged it's motion as seen in plane mirror is anticlockwise and on circular path mv (B) If d ≤ then the particle will hit the upper qB plate at P 2 if it is negatively charged XtraEdge for IIT-JEE C F 10cm 20cm Z-axis Charged particle B = B 0 k̂ E0 and B0 are positive constants Q.2 A charged particle is entering through the tiny hole, in the given magnetic field between the plates P2 → 17 30cm 40cm X-axis F = Focus C= centre of curvature Positively charged Negatively charged May be positive of negative Particle can not pass undeviated through the pair of transverse magnetic field and electric field JANUARY 2010 Q.4 In previous ques.(Q.3) if charged particle is projected from point (30cm, 0, 0) with the velocity Particle-A → of v = v 0 ˆj and it goes un deviated then (here v0 is positive constant) (A) Particle should be negatively charged and v0 = 19.6 m/s (B) Particle should be positively charged and Particle-B (A) Proton Electron (B) Deutron Electron (C) Monoionized helium atom Electron (D) Doubly ionized Electron v0 = 19.6 m/s helium atom (C) Particle is negatively charged and it will reach up to the maximum height of 10m parallel to Y-axis if gravity is taken into account Q.7 A circular current carrying coil is placed in uniform magnetic field as shown in left column (D) Particle is negatively charged and when viewed through the concave mirror it is going parallel to negative Y-axis maximum 20m below Column-I i (A) Q.5 An R-L series circuit is shown in figure Column-II → r o B The R-L circuit is in discharging mode and current i = 18 amp. then o r (B) B A 3R (C) L↑i 6R R2 R1 L × × × i i (P) Magnetic force τm = 0 → B (Q) Magnetic torque τm = 0 × × × B→ ×o r × × × × × (R) Expansion of coil i R1 = R2 = R = 1Ω L = 1/3 Henry r o (D) A, B = Terminals of resistance R 1 (S) Compression of coil G = Ground terminal In column II, quantities are in SI units (A) (B) (C) (D) Column-I Total energy dissipated in resistances Time constant for discharging mode Potential drop across R1 initially Potential drop across R2 initially Q.8 Match the followings Column-I Column-II Column-II (P) 0.25 (A) (Q) 12 O × × × × × × × × × × × × × × × × × × × × × × × × (x, 0, 0) × × × × × × s (P) C2µ0 q net Gauss law (C = speed of light in Air/Free space/Vacuum) (R) 108 (S) 18 Q.6 Two charged particles having same de-Broglie wavelengths enters in given transverse magnetic field as shown below y-axis → B × × × × × × → → ∫ E. d → → (B) ∫ B. d (C) ∫ B. d s → → e (Q) Zero (R) Magnetic monopole is impossible (D) × × × × × × → → ∫ E. d e (S) Induced emf Faraday's law of electromagnetic induction (T) x-axis 1 C 2 .ε 0 .i net Ampere's circuital law z-axis XtraEdge for IIT-JEE 18 JANUARY 2010 1. 8 Solution Physics Challenging Problems Q u es tio n s wer e Pu b lish ed in D ecem b er I ss u e As information given in: 1st Bright Fringe occurs in front of a slit so, x As 2d. n = nλ for nth Bright Fringe D x d d2 = λb ⇒ λb = 2. So, 2d. 1 = 1λb ⇒ 2d D D D For the missing wavelength, destructive interference should occur in front of slit λ missing d So, 2d. = (2n – 1). D 2 2d 2 1 ⇒ λmissing = 2. . D ( 2n − 1) = If R eq = 1 = (a , b ) Slit s1 = 5. 6. vab = i. R eq = 1 (1) = 1 volt ε1 ε1 ε ε R ε – =0 ⇒ 1 = 2 ⇒ 1 = 1 R1 R1 R1 R 2 R 2 ε2 ε1 R1 R C ε 2 R2 Option (A) is correct D For n = 1 λmissing = 2λb ≡ = more than λb 2 n = 2 λmissing = . λb ≡ less than λb 3 (λmissing) max. = 2λb , Option (C) is correct 2. As(lmissing) max. = 2λb Option (A) is correct 3. If Slit width are not equal then I1 ≠ I2 and a1 ≠ a2 So Imin. = Intensity of Dark fringe = (a1 ~ a2)2 ≠ 0 So, dark fringe will be of blue colour Option (B) is correct 7. R1R2/R1+R2 R XtraEdge for IIT-JEE ⇒ C R2 C R2 Option (D) is correct 8. R To calculate time constant Replace voltage source by short circuit mean by zero resistance and then find Req with C and time constant τ = Req.C R R τ = 1 2 + R C R1 + R 2 R1 The equivalent circuit is R (a , b ) (a , b ) If vab= 0, then irrespective to the value of capacitance C energy stored will be zero. ε / R + (− ε 2 / R 2 ) =0 vab = 1 1 1 / R1 + 1 / R 2 So, 2 R eq R eq = 1Ω So, x1 R R 5R = = 2R + 2 2 2 Ω, 5 R= (a , b ) 1st B/F R 2 Ω 5 Option (B) is correct Slit =s a So R = As, O 4. 5R 2 Option (A) is correct 2 2 2d 2 = . . λb 2n − 1 D 2n − 1 Screen d Set # 8 b 19 Maximum current through the resistance Imax v (ε / R ) + ( −ε 2 / R 2 ) / 1 / R 1 + 1 / R 2 ε eq = = ab = 1 1 R R R Option (C) is correct. JANUARY 2010 Students' Forum Expert’s Solution for Question asked by IIT-JEE Aspirants PHYSICS 1. A sphere of mass 50 g is attached to one end of a steel wire, 0.315 mm diameter and one metre long. In order to form a conical pendulum, the other end is attached to a vertical shaft which is set rotating about its axis. Calculate the number of revolutions necessary to extend the wire by 1 mm. Young's modulus of elasticity of steel = 2 × 1012 dynes/cm 2 and g = 980 cm/sec2. T sin θ = or, 1.559 × 106 × 1 = 50 × v 2 100.1 or 1.559 ×10 6 × 100.1 50 v = 1766 cm/sec ∴ Period of revolution = ∴ Sol. Let T be the tension in the wire, when the extension is 1 mm. According to definition, mv 2 r Now, v2 = 2πr 2π × 100.1 = v 1766 = 0.3561 s So, Frequency of revolution = θ θ T cos θ θ = 2.808/sec L = 1m T 2. T sin θ mg Y= A wave travels out in all directions from a point source. Justify the expression y = (a0/r) sin K(r – vt), at a distance r from the source. Find the speed, periodicity and intensity of the wave. Sol. If P be the power of the source then intensity Tensile stress T/A T L = × = Tensile strain l/L A l I= L = × 2 l πr T ∴ T= 2 × 1012 × π × (0.01575) 2 × 0.1 Yπr 2 l = L 100 = 1.559 × 10 6 dynes ∴ sin θ = or cos θ = 4πr 2 1 r2 or, I∝ But I ∝ a2, so a∝ a 1 or a = 0 r r a0 sin K(r – vt) r comparing this equation with y = a sin (Kr – ωt) y= mg 50 × 980 = T 1.559 ×10 6 Now, (1 − cos 2 θ) = 0.9998 = 1.00 (nearly) We have, ω = Kv or ∴ Radius of the circle described = r = (L + l) sin θ K= = 100.1 cm XtraEdge for IIT-JEE P Where a0 is constant. The equation in standard form is, y = a sin K (r – vt) Therefore, above equation is written as : When the sphere is revolving, it is acted upon by two forces namely the tension T along the wire and its weight mg acting vertically downwards. Resolving T into vertical and horizontal components, we get T cos θ = mg 1 1 = period 0.3561 20 2π λ or n= Kv 2π λ= 2π K and JANUARY 2010 4. Kv 2π Speed c = nλ = × = v 2π K 1 2π = n Kv Thus, intensity is given by Also, or 3. T= I= 1 2 2 1 a2 ρa ω c = ρ 20 . K2v2 . v 2 2 r I= 1 ρ a 20 K 2 v 3 2 r2 Two coherent light sources emit light of wavelength 550 nm which produce an interference pattern on a screen. The sources are 2.2 mm apart and 2.2 m from the screen. Determine whether the interference at the point O is constructive or destructive. Calculate the fringe width. S1 D l A conducting bar of mass m, length l is pushed with a speed v0 on a smooth horizontal conducting rail containing an inductance L. If the applied magnetic field has inward field of induction B, find the maximum distance covered by the bar before it stops. S2 2d O Sol. The path difference at O is given by ∆ = S2O – S1O From figure, S2O = [l 2 + (2d)2]1/2 m L B ⊗ v0 l Sol. If the bar slides a distance dx, the flux linkage – dφ = Bldx The induced e.m.f. = or, dx dI =–L dt dt LdI = Bl dx or, LI = Bl x or, Bl x I= L ∴ – Bl ∴ B2 l 2 v dv = – v0 ML or, ∫ ∴ v s = ( mL ) 0 Bl XtraEdge for IIT-JEE n= β= ∆ 1.1× 10 −6 =2 = λ 5.5 × 10 −7 ∫ λD 5.5 ×10 −7 × 2.2 = 2d 2.2 × 10 −3 = 5.5 × 10–4 m = 0.55 mm dv B2 l 2 =– x dx L 0 (2.2 ×10 −3 ) 2 = 1.1 × 10–6 m 2 × 2.2 Fringe width, 5. Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find (a) the energy of the photons causing photoelectric emission Bl F = IlB = – x l B L mv 1 2 d 2 (2 d ) 2 ∆ = l 1 + − 1 = 2l 2 l The difference will be constructive if path difference is an integral multiple of wavelength i.e., n = 1, 2, 3, ……. This induced current interacts with the applied magnetic field of induction and imparts a restoring (magnetic) force or, Now, = dI dt 1 2d 2 = l 1 + 2 l 2d 2 S2O = l 1 + l dφ dx = – Bl dt dt Since the induced e.m.f. across the inductor = – L 1/ 2 ∴ B2 l 2 s 2 x dx = – 0 2mL s (b) the quantum numbers of the two levels involved in the emission of these photons 21 JANUARY 2010 (c) the change in the angular momentum of the hydrogen atom in the above transition, and p Ph = (d) the recoil speed of the emitting atoms assuming it to be at rest before the transition. hν 2.55 × 1.6 × 10 −19 J = 1.36 × 10–27 Kg.m/s = c 3 ×10 8 m / s According to the law of conservation of momentum, the recoil momentum of a hydrogen atom will be equal and opposite to the momentum of the emitted photon. (Ionization potential of hydrogen is 13.6 volt and the mass of the hydrogen atom is 1.67 × 10 –27 Kg, 1 eV = 1.6 × 10 –19 J) → → → ( p Ph + p A = 0 or Sol. (a) According to Einstein’s photo-electric equation, the maximum kinetic energy Ek of the emitted electrons is given by → p A = – p Ph ) Hence, the recoil speed of the atoms is : → → | Momentum | | p A | | p Ph | = = V= mass mA mA EKmax = hν – W, Where hν is the energy of photons causing the photoelectric emission and W is the work-function of the emitting surface. = 1.36 × 10 −27 kg − m / s 1.67 × 10 − 27 kg = 0.814 m/s Given that, EKmax = 0.73 eV and W = 1.82 eV ∴ hν = E Kmax + W = 0.73 eV + 1.82 eV = 2.55 eV SCIENCE TIPS (b) These photons (where energy is 2.55 eV) are emitted by hydrogen atoms. As • A porcelain funnel used for filtration by suction is known as ® Bucher Funnel • What is diazomethane ? (I.E.)H = 13.6 eV, hence E1H = – (I.E.)H = – 13.6 eV The energy of higher levels is given by E aH = 13.6 = – 3.4 eV 4 • Reforming of a gasoline fraction to increase branching in presence of AlCl3 is known as ® Isomerization 13.6 – = – 1.5 eV 9 • A condenser consisting of glass tube surrounded by another glass tube through which cooling water ® Liebig condenser flows is known as 13.6 = – 0.85 eV 16 The energy of the emitted photon is 2.55 eV (c) – • A drying chamber, containing chemicals such as concentrated sulphuric acid or silica gel is known as ® Desiccator n2 Hence, E H 2 =– E 3H = + ® [CH 2 = N = N or CH 2 N 2 ] E1H and EH 4 =– Now H EH 4 – E 2 = – 0.85 – (– 3.4) = 2.55 eV • For wattles current what should be the value of the ® Zero power factor of the circuit ? Thus, the quantum numbers of two levels involved in the emission of photon of energy 2.55 eV are 4 and 2. • For which colour is the critical angle of light, pasing ® Violet from glass to air, minimum ? The electron transition causing the emission of photon of energy 2.55 eV is from n = 4 level to n = 2 level. Now, according to Bohr’s 2 nd postulate, the angular momentum of electron in the hydrogen atom is (n h/2π). Thus, the change in angular momentum in above transition is • Give an example of application of mutual induction in any device. ® Transformer • What is the correct sequence of the semiconductors silicon, tellunium and germanium in the increasing order of their energy gap ? ® Tellurium, germanium, silicon • Which ammeter is used to measure alternating current ? ® Hot wire ammeter 4h 2h h – = ∆L = 2π 2π π • What quantity has the ampere-second as its unit ? (d) The momentum of the photon emitted from the hydrogen atom XtraEdge for IIT-JEE ® Quantity of electricity 22 JANUARY 2010 XtraEdge for IIT-JEE 23 JANUARY 2010 P HYSICS F UNDAMENTAL F OR IIT-J EE Refraction at plane & curved surface KEY CONCEPTS & PROBLEM SOLVING STRATEGY Laws of Refraction : (e) For a given time, optical path remains constant. The incident ray, the refracted ray and normal on incidence point are coplanar. i.e., µ1x1 = µ2x2 = ... constant µ 1 sin θ1 = µ 2 sin θ2 = ... = constant. θ1 µ1 ∴ µ1c1 = µ2c2 ∴ µ2 c = 1 µ1 c2 i.e., µ∝ µ1 µ2 θ2 Snell's law in vector form : (where c1 and c2 are speed of light in respective mediums) 1 c (f) The frequency of light does not depend upon medium. n̂ ê1 dx1 dx = µ2 2 dt dt ∴ µ1 ê 2 ∴ c1 = fλ 1, ∴ µ1 c λ = 2 = 2 µ2 c1 λ1 ∴ µ∝ µ2 c2 = fλ 2 1 λ When observer is rarer medium and object is in denser medium : Let, ê1 = unit vector along incident ray ê 2 = unit vector along refracted. Then n̂ = unit vector along normal on incidence point. µ= real depth apparent depth Then µ1( ê1 × n̂ ) = µ 2( ê 2 × n̂ ) Air Observer Some important points : (a) The value of absolute refractive index µ is always greater or equal to one. (b) The value of refractive index depends upon material of medium, colour of light and temperature of medium. Denser medium (µ) P Object (c) When temperature increases, refractive index decreases. When object is in rarer and observer is in denser medium : (d) Optical path is defined as product of geometrical path and refractive index. µ= i.e., optical path = µx XtraEdge for IIT-JEE Apparent depth P Real depth 24 apparent position real position JANUARY 2010 1 The shift of object due to slab is x = t 1 – µ 90º t µ P Q P´ Mathematically, Object shiftness c Denser µ2 sin c = µ1 µ2 Rarer medium (µ1) (a) This formula is only applicable when observer is in rarer medium. r (b) The object shiftiness does not depend upon the position of object. (c) Object shiftiness takes place in the direction of incidence ray. number of slabs for normal incidence is µ = µ1 t1 µ2 t2 i c i<c i=c i i Denser medium (µ2) (i) When angle of incidence is lesser than critical The equivalent refractive index of a combination of a angle, refraction takes place. The corresponding deviation is Σt i t Σ i µi µ δ = sin–1 2 sin i – i µ 1 for i < c (ii) When angle of incidence is greater than critical angle, total internal reflection takes place. The corresponding deviation is δ = π – 2i Σti = t1 + t2 + ... Here, Rarer µ1 when i < c The δ – i graph is : t t t Σ i = 1 + 2 + ... µi µ1 µ2 (i) Critical angle depends upon colour of light, material of medium, and temperature of medium. The apparent depth due to a number of media is Σ ti µi (ii) Critical angle does not depend upon angle of incidence The lateral shifting due to a slab is d = t sec r sin(i – r). i µ δ t r i d c π/2 Refractive surface formula, µ2 µ µ − µ1 – 1 = 2 v u r Critical angle : When a ray passes from denser medium (µ2) to rarer medium (µ 1), then for 90º angle of refraction, the corresponding angle of incidence is critical angle. Here, v = image distance, u = object distance, r = radius of curvature of spherical surface. XtraEdge for IIT-JEE 25 JANUARY 2010 (a) For plane surface , r = ∞ (a) Thin lens formula is only applicable for paraxial ray. (b) Transverse magnification, m= (b) This formula is only applicable when medium on both sides of lens are same. Im age size µv = 1 object size µ2u (c) Intensity aperture. (c) Refractive surface formula is only applicable for paraxial ray. is proportional to square of (d) When lens is placed in a medium whose refractive Lens : index is greater than that of lens. i.e., µ 1 > µ2. Then converging lens behaves as diverging lens and vice versa. Lens formula : 1 1 1 – = v u f (e) When medium on both sides of lens are not same. Then both focal lengths are not same to each other. (a) Lens formula is only applicable for thin lens. (b) r = 2f formula is not applicable for lens. (c) m = image size v = object size u (f) If a lens is cut along the diameter, focal length 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 lenses are same. i.e., 1 1 1 = + f1 f2 f (f) + f(+ve) f(–ve) (i) (ii) f1 f f1 (h) If a lens is made of a number of layers of different refractive index (shown in figure) f(+ve) f(–ve) (iii) µ1 µ2 µ3 (iv) +++ +++ µ4 µ5 f(–ve) f(+ve) µ6 (v) (vi) Then number of images of an object by the lens is equal to number of different media. (g) Thin lens formula is applicable for converging as well diverging lens. Thin lens maker's formula : µ − µ1 1 = 2 f µ1 (i) The minimum distance between real object and real image in is 4f. 1 1 − r1 r2 µ1 (j) The µ1 focal length of co-axial 1 1 1 d = + – F f1 f 2 f1f 2 µ2 XtraEdge for IIT-JEE equivalent combination of two lenses is given by 26 JANUARY 2010 f1 f2 d<f1 In 1 st case, d<f2 o2 o1 In 2 nd case, d (k) If a number of lenses are in contact, then 1 1 1 = + + ...... F f1 f 2 (l) (i) Power of thin lens, P = 1 F (ii) Power of mirror is P = – 1 F 2. u − a q +1 = f q (Q m = p) or, or, p +1 a q +1 – = p f q or, a pq + q − pq − p q − p = = f pq pq ∴ f= u a q +1 – = f f q or, = a p +1 q +1 – – f p q apq q−p A convex refracting surface of radius of curvature 30 cm separates two media of refractive indices n1 = 4/3 and n2 = 3/2 respectively. Find the position of image formed by refraction of an object placed at a (m) If a lens silvered at one surface, then the system behaves as an equivalent mirror, whose power P = 2P L + Pm distance of (i) 280 cm and (ii) 80 cm, from the surface. Here, PL = Power of lens µ − µ1 = 2 µ1 u p +1 = f p 1 1 − r1 r2 Pm = Power of silvered surface = – Sol. (i) Given that n1 = 4/3, n2 = 3/2, |u| = 280 cm, |R| = 30cm 1 Fm For refraction through a spherical surface : n 2 n 1 (n 2 − n 1 ) – = v u R Here, Fm = r2/2, where r2 = radius of silvered surface. Here, u = – 280 cm, R = + 30 cm. Hence 3 4 [(4 / 3) − (3 / 2)] – = 2 v − 3 × 280 +3 P = – 1/F Here, F = focal length of equivalent mirror. Solved Examples 1. object magnified p times. The magnification becomes q when the lens is moved nearer to the object by a distance a. Calculate the focal length of the lens. XtraEdge for IIT-JEE or ∴ 3 1 1 1 = – = 2 v 180 210 1260 or, v = (3/2) × 1260 = 1890 cm = 18.9 m (ii) In this case, u = – 80 cm, R = + 30 cm Again from the formula for refraction through a surface, (3/2v) – [4/–3 × 80)] = [{(4/3) – (3/2)} / + 30] or (3/2v) + (1/60) = (1/180) or 3/2v = [(1/180) – (1/60)] Sol. The magnification (m) produced by a lens in terms of u and f i : given by f u−f 3 1 1 + = 2 v 210 180 As v is positive, hence the image is real and is formed in second medium at a distance of 18.9 m from the refracting surface. A thin converging lens forms the image of a certain m= or u m +1 = f m or 27 v = (3/2) × (– 90) = – 135 cm JANUARY 2010 Screen As v is negative, hence the image is virtual and is formed in the first medium of refractive index 4/3 at a distance of 135 cm from the pole. 3. O I1 I There is a small air bubble in side a glass sphere (n = 10 cm 1.5) of radius 10 cm. The bubble is 4 cm below the surface and is viewed normally from the outside 1 1 I1I = t 1 − = (1.5 cm) 1 − = 0.5 cm. n 1.5 (Fig.). Find the apparent depth of the air bubble. P I A Thus, the lens forms the image at a distance of 9.5 cm n2 = 1 from itself. Using O C 1 1 1 – = , we get v u f n1 = 1.5 or 1 1 1 1 1 = – = – u v f 9.5 10 u = – 190 cm. i.e. the object should be placed at a distance of Sol. The observer sees the image formed due to refraction 190 cm. from the lens. at the spherical surface when the light from the bubble goes from the glass to air. Here u = – 4.0 m, We have 5. R = – 10 cm, n1 = 1.5 and n2 = 1 Where must a convex lens of focal length 8 inches be [(n2/v) – (n1/u) = (n2 – n1)/R or (1/v) – (1.5/ –4.0 cm) = (1 – 1.5)/ (– 10 cm) or (1/v) = (0.5/10 cm) – (1.5/4.0 cm) or v = – 3.0 cm A candle is placed at a distance of 3 ft from the wall. placed so that a real image is formed on the wall ? Sol. According to formula for refraction though a lens 36 – v v Thus, the bubble will appear 3.0 cm below the surface. f = 8" d = 3 ft = 36" 4. A convex lens focuses a distance object on a screen 1 1 1 – = v u f placed 10 cm away from it. A glass plate (n = 1.5) of thickness 1.5 is inserted between the lens and the screen. Where should the object be placed so that its or image is again focused on the screen ? 1 1 1 + = v 36 − v 8 or 1 1 1 – = v − (36 − v) 8 or 36 − v + v 1 = v(36 − v) 8 or, v2 – 36 v + 8 × 36 = 0 Sol. The situation when the glass plate is inserted between the lens and the screen, is shown in fig. The lens or v = 12" or 24" = 1 ft or 2 ft. forms the image of object O at point I1 but the glass ∴ or 12" = 2 ft or 1 ft plate intercepts the rays and forms the final image at I Hence, lens should be placed at either 1 ft or 2 ft on the screen. The shift in the position of image after away from the wall. insertion of glass plate XtraEdge for IIT-JEE u = 24" 28 JANUARY 2010 P HYSICS F UNDAMENTAL F OR IIT-J EE Properties of Matter KEY CONCEPTS & PROBLEM SOLVING STRATEGY Key Concepts : Stress : The restoring force setup inside the body per unit area is known as stress. Restoring forces : If the magnitude of applied deforming force at equilibrium = F φ Shear strain Stress-strain graph : From graph, it is obvious that in elastic limit, stress is proportional to strain. This is known as Hooke's law. ∴ Stress ∝ Strain F A In SI system, unit of stress is N/m2. Difference between pressure and stress : (a) Pressure is scalar but stress is tensor quantity. (b) Pressure always acts normal to the surface, but stress may be normal or tangential. (c) Pressure is compressive in nature but stress may be compressive or tensile. Strain : Then, ∴ Stress = E .strain ∴ E= change in dimension original dimension (a) Longitudinal strain = stress strain where E is proportionality dimensional constant known as coefficient of elasticity. Plastic region Breaking B C strength Stress Strain = Stress = ∆L L Elastic limit A L F F O Longitudinal strain is in the direction of deforming force but lateral strain is in perpendicular direction of deforming force. Poisson ratio : σ= Types of coefficient of elasticity : (a) Young's modulus = Y = lateral strain ∆d/D = longitudinal strain ∆L/L ∴ Here ∆d = change in diameter. (b) Volumetric strain = Strain Y= logitudinal stress longitudinal strain F FL = ∆L A∆L A L ∆V V L F ∆L F F F (c) Shear strain = φ XtraEdge for IIT-JEE F (b) Bulk modulus = B = V volumetric stress volumetric strain Compressibility = 1/B 29 JANUARY 2010 (c) Modulus of rigidity = η = Surface tension : F shear stress = Aφ shear strain F L Here L = length of imaginary line drawn at the surface of liquid. and F = force acting on one side of line (shown in figure) (a) Surface tension does not depend upon surface area. (b) When temperature increases, surface tension decreases. (c) At critical temperature surface tension is zero. T= (d) For isothermal process, B = P. F φ φ F (e) For adiabatic process, B = γP (f) Adiabatic bulk modulus =γ Isothermal bulk modulus (g) Esolid > E liquid > E gas 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. (j) When temperature increases, coefficient of elasticity (Y, B, η) decreases. (k) L F 1 3 9 + = B η Y Rise or fall of a liquid in a capillary tube : h= (l) Y = 2(1 + σ)η (m) Poisson's ratio σ is unitless and dimensionless. Theoretically, Here 1 –1 < σ < 2 1 1 1 × load × extension = Fx = kx2 2 2 2 = stress × strain × volume For twisting motion, U= (a) For a drop of radius R, W = 4πR2T (b) For a soap bubble, W = 8πR2T Excess pressure : 1 U= × torque × angular twist 2 1 1 τ × θ = cθ2 2 2 Elastic energy density, (a) For drop, P = = 2T R (b) For soap bubble, P = 1 1 × stress × strain J/m3 = Y × strain2J/m3 2 2 Thermal stress = Yα∆θ and Thermal strain = α∆θ Work done in stretching a wire : u= 4T R Viscosity : (a) Newton's law of viscous force : F = – ηA 1 F∆L 2 where 1 × stress × strain 2 (c) Breaking weight = breaking stress × area (b) Work done per unit volume = XtraEdge for IIT-JEE θ = angle of contact. r = radius of capillary tube ρ = density of liquid For a given liquid and solid at a given place, hr = constant Surface energy : Surface energy density is defined as work done against surface tension per unit area. It is numerically equal to surface tension. W = work = surface tension × area 1 Practically, 0<σ< 2 (n) Thermal stress = Yα∆θ (o) Elastic energy stored, (a) W = 2T cos θ rρg dv dy dv = velocity gradient dy A = area of liquid layer η = coefficient of viscosity The unit of coefficient of viscosity in CGS is poise. 30 JANUARY 2010 or r = d/2 = 0.25 cm, F = 5.0 × 1000 × 980 dynes. (b) SI unit of coefficient of viscosity = poiseuille = 10 poise. (c) In the case of liquid, viscosity increases with density. (d) In the case of gas, viscosity decreases with density. (e) In the case of liquid, when temperature increases, viscosity decreases. (f) In the case of gas, when temperature increases, viscosity increases. Poiseuille's equation : V= Y= or l = Pπr 4 8ηL FL = πr 2 Y Hence, or d'2 = 5.0 ×1000 × 980 × 200 3.142 × (0.25) 2 × 1.1× 10 2 4F πd ' 2 = 1.5 × 10 9 4F 2. F = 6πηrv where r = radius of spherical body Determination of η : 2r 2 (ρ − σ)g 9v where r = radius of spherical body moving with constant velocity v in a viscous liquid of coefficient of viscosity η and density ρ 9 A uniform horizontal rigid bar of 100 kg in supported horizontally by three equal vertical wires A, B and C each of initial length one meter and cross-section I mm2. B is a copper wire passing through the centre of the bar; A and C are steel wires and are arranged symmetrically one on each side of B YCu = 1.5 × 1012 dynes / cm2, Y s = 2 × 10 12 dynes/cm2. Calculate the tension in each wire and extension. kη ρr As Y = where k = Reynold's number for narrow tube, k ≈ 1000. (a) For stream line motion, flow velocity v < v0. (b) For turbulant motion, flow velocity v > v0. Stress Strain A S Solved Examples B Cu C S 100 Kg A mass of 5 kg is suspended from a copper wire of 5 mm diameter and 2 m in length. What is the extension produced in the wire ? What should be the minimum diameter of the wire so that its elastic limit is not exceed ? Elastic limit for copper = 1.5 × 10 9 dynes/cm2. Y for copper = 1.1 × 1012 dynes/cm2. FCu / A Strain Hence, YCu = and Ys = ∴ YCu FCu 1.5 3 = = = YS FS 2 4 … (1) Fs / A Strain … (2) or 4FCu = 3FS ...(3) According to figure, we can write 2FS + FCu = 100 g or 2 × (4/3) FCu + FCu = 100 g Sol. Given that Y = 1.1 × 10 12 dynes/cm2, L = 2m = 200 cm, d = 5 mm = 0.5 cm XtraEdge for IIT-JEE 4 × 5.0 × 1000 × 980 Sol. The situation is shown in figure. Because the rod is 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. and σ = density of spherical body Critical velocity (v 0) : 1. = π × 1.5 × 10 3.142 × 1.5 × 10 9 –4 = 41.58 × 10 d' = 0.0645 cm. η = coefficient of viscosity and P = pressure difference between ends of the tube Stoke's law : The viscous force acting on a spherical body moving with constant velocity v in a viscous liquid is v0 = πr 2 l = 4.99 × 10–3 cm Also, elastic limit for copper = 1.5 × 10 9 dynes/cm2 If d' is the minimum diameter, then maximum stress F 4F = on the wire = πd '2 / 4 πd' 2 where V = the volume of liquid flowing per second through a capillary tube of length L and radius r η= FL or 31 [(8/3) + 1] FCu = 100 g JANUARY 2010 ∴ 4. FCu = (3/11) × 100g = (3/11) × 100 Kgwt = 27.28 Kgwt FS = (4/3) FCu = (4/3) × (3/11) × 100g and = (400/11)g = 36.36 Kgwt Extension in each wire, l= 3. FCu L 27280 × 980 × 100 = 0.178 cm = AYCu 10 − 2 × 1.5 × 1012 A ring is cut from a platinum tube 8.5 cm internal and 8.7 cm external diameter. It is supported horizontally from a pan of a balance so that it comes in contact with the water in a glass vessel. What is the surface tension of water if an extra 3.97 gm weight is required to pull it away from water (g = 980 cm/sec2). Sol. The ring is in contact with water along is inner and outer circumference. So when pulled out the total force on it due to surface tension will be A copper rod of length L and radius r is suspended from the ceiling by one of its ends. Find: (a) the elongation of the rod due to its own weight when ρ and Y are density and Young's modulus of the copper respectively, (b) the elastic potential energy stored in the rod due to its own weight. F T Sol. (a) Consider any length x of the rod from the fixed end. Weight of lower portion of rod will exert stretching force on the upper portion. ∴ T= where m is the mass per unit length of the rod (m = πr 2ρ) 5. [Q Strain = (Stress/Y)] Hence, increase in length in elementary length dx at πr 2 Y dx = L 0 ( L − x )mg h= πr 2 Y L2 πr 2ρgL2 ρgL2 = = 2Y πr 2 Y 2 2πr 2 Y mg (b) Energy density at x = = ∫ 1 × Strain × Stress 2 ∴ h= hr = 1 (L − x ) 2 ρ 2 g 2 × πr2dx 2 Y ∴Total potential energy stored in the rod of length L, dx = XtraEdge for IIT-JEE ∫ L 0 (L − x ) 2 dx = … (1) 2 × (7.0 × 10 −2 N / m ) ( 2.5 ×10 − 3 ) × (1× 10 3 Kg / m 3 ) × (10 N / Kg ) = 5.6 mm According to equation (1), for the same liquid, we have 1 (L − x ) 2 ρ 2 g 2 = . 2 Y ∴ Energy stored in the volume of length element 1 πr 2ρ 2 g 2 2 Y 2T cos θ rρg where T is surface tension, ρ is density and θ is angle of contact of water-glass which can be assumed zero. For the first tube, r = 2.5 mm = 2.5 × 10 –3 m 1 (L − x ) mg ( L − x )mg × × 2 πr 2 πr 2 Y U= Two long capillary tubes of diameter 5.0 mm and 4.0 mm are held vertically inside water one by one. How much high the water will rise in each tube ? (g = 10 m/s 2, surface tension of water = 7.0 × 10–2 N/m). Sol. Height of water column in a capillary tube of radius r is given by dx ∴ Total increase in length = mg 3.97 × 980 = 2π(r1 + r2 ) 2 × 3.14 × (8.5 + 8.7) = 72.13 dyne/cm. ∴ Strain at x = (L – x)mg/πr2Y (L − x ) mg T F = T (2πr1 + 2πr2) ∴ Stress over the portion at a distance x from fixed end = (L – x) mg/πr2 x= F 2T cos θ = Constant ρg If a liquid rises to a height h1 in a capillary tube of radius r 1 and to a height of h2 in a capillary tube of radius r 2, then 1 πr 2ρ 2 L3g 2 6 Y h1r1 = h2r2 32 or h2 = h1r1 5.6 × 2.5 = = 7.0 mm r2 2.0 JANUARY 2010 KEY CONCEPT Organic Chemistry Fundamentals CARBOHYDRATES By 1895 it had become clear that the picture of D(+)-glucose as a pentahydroxy aldehyde had to be modified. Among the facts that had still to be accounted for were the following: (a) D-(+)-Glucose fails to undergo certain reactions typical of aldehydes. Although it is readily oxidized, it gives a negative Schiff test and does not form a bisulfite addition product. (b) D-(+)-Glucose exists in two isomeric forms which undergo mutarotation. When crystals of ordinary D-(+)-glucose of m.p. 146ºC are dissolved in water, the specific rotation gradually drops from an initial + 112º to + 52.7º. On the other hand, when crystals of D-(+)-glucose of m.p. 150ºC (obtained by crystallization at temperatures above 98ºC) are dissolved in water, the specific rotation gradually rises from an initial + 19º to + 52.7º. The form with the higher positive rotation is called α-D-(+)-glucose and that with lower rotation β-D-(+)-glucose. The change in rotation of each of these to the equilibrium value is called mutarotation. (c) D-(+)-Glucose forms two isomeric methyl Dglucosides. Aldehydes react with alcohols in the presence of anhydrous HCl to form acetals. If the alcohol is, say methanol, the acetal contains two methyl groups : H H H Definition and Classification : are polyhydroxy aldehydes, Carbohydrates polyhydroxy ketones, or compounds that can be hydrolyzed to them. A carbohydrate that cannot be hydrolyzed to simpler compounds is called a monosaccharide. A carbohydrate that can be hydrolyzed to two monosaccharide molecules is called a disaccharide. A carbohydrate that can be hydrolyzed to many monosaccharide molecules is called a polysaccharide. A monosaccharide may be further classified. If it contains an aldehyde group, it is known as an aldose; if it contains a keto group, it is known as a ketose. Depending upon the number of carbon atoms. It contains, a monosaccharide is known as a triose, tetrose, pentose, hexose, and so on. An aldohexose, for example, is a six-carbon monosaccharide containing an aldehyde group; a ketopentose is a five-carbon monosaccharide containing a keto group. Most naturally occurring monosaccharides are pentoses or hexoses. Carbohydrates that reduce Fehling’s (or Benedict’s) or Tollens’ reagent are known as reducing sugars. All monosaccharides, whether aldose or ketose, are known as reducing sugars. Most disaccharides are reducing sugars; sucrose (common table sugar) is a notable exception, for it is a non-reducing sugar. (+)-Glucose : an aldohexose : Because it is the unit of which starch, cellulose, and glycogen are made up, and because of its special role in biological processes, (+)-glucose is by far the most abundant monosaccharide- there are probably more (+)-glucose units in nature than any other organic group–and by far the most important monosaccharide. Cyclic structure of D-(+)-glucose. Formation of glucosides : D-(+)-glucose is a pentahydroxy aldehyde. D-(+)glucose had been definitely proved to have structure. CHO H OH H HO OH H OH H –C=O –C–OCH3 OH CH3 OH,H+ –C–OCH3 OCH3 Aldehyde Hemiacetal Acetal When D-(+)-glucose is treated with methanol and HCl, the product, methyl D-glucoside, contains only one –CH3 group; yet it has properties resembling those of a full acetal. It does not spontaneously revert to aldehyde and alcohol on contact with water, but requires hydrolysis by aqueous acids. Furthermore, not just one but two of these monomethyl derivatives of D-(+)-glucose are known, one with m.p. 165ºC and specific rotation + 158º, and the other with m.p. 107 ºC and specific rotation –33º. The isomer of higher positive rotation is called methyl α-D-glucoside, and the other is called methyl β-D-glucoside. These glucosides do not undergo mutarotation, and do not reduce Tollens’ or Fehling’s reagent. CH2 OH D-(+)-Glucose XtraEdge for IIT-JEE CH3 OH,H+ 33 JANUARY 2010 We shall study four disaccharides : (+)-maltose (malt sugar), (+)-cellobiose, (+)-lactose (milk sugar), and (+)-sucrose (cane or beet sugar). D-(+)-Glucose has the cyclic structure represented crudely by IIa and IIIa, more accurately by IIb and IIIb. 1 2 3 4 5 H H HO H H 6 OH OH O H OH 6 CH2OH 4 H OH H 3 HO H (+)-Maltose : O H 5 2 H (+)-Maltose can be obtained, among other products, by partial hydrolysis of starch in aqueous acid. (+)Maltose is also formed in one stage of the fermentation of starch to ethyl alcohol; here hydrolysis is catalyzed by the enzyme diastase, which is present in malt (sprouted barley). 1 OH OH CH2OH IIa IIb α-D-(+)-Glucose (m.p. 146 ºC, [α] = +112º) 1 HO 2 H 3 HO 4 H 5 H 6 H OH O H OH 6 CH2OH 4 H OH H HO 5 3 H CH2OH IIIa Let us look at some of the facts from which the structure of (+)-maltose has been deduced. O OH 2 H (+)-Maltose has the molecular formula C12H22O 11. It reduces Tollens’ and Fehling’s reagents and hence is a reducing sugar. It reacts with phenylhydrazine to yield an osazone, C12H20O9(=NNHC6H5)2. It is oxidized by bromide water to a monocarboxylic acid, (C11H21O10)COOH, maltobionic acid. (+)-Maltose exists in alpha ([α] = + 168º) and beta ([α] = + 112º) forms which undergo mutarotation in solution (equilibrium [α] = + 136º). 1 H OH IIIb β-D-(+)-Glucose (m.p. 150 ºC, [α] = +19º) D-(+)-Glucose is the hemiacetal corresponding to reaction between the aldehyde group and the C-5 hydroxyl group of the open-chain structure. It has a cyclic structure simply because aldehyde and alcohol are part of the same molecule. (+)-Cellobiose : When cellulose (cotton fibers) is treated for several days with sulfuric acid and acetic anhydride, a combination of acetylation and hydrolysis takes place; there is obtained the octaacetate of (+)cellobiose. Alkaline hydrolysis of the octaacetate yields (+)-cellobiose itself. There are two isomeric forms of D-(+)-glucose because this cyclic structure has one more chiral centre than Fisher’s original open-chain structure. αD-(+)-Glucose and β-D-(+)-glucose are diastereomers, differing in configuration about C-1. Such a pair of distereomers are called anomers. Like (+)-maltose, (+)-cellobiose has the molecular formula C12H22O 11, is a reducing sugar, forms an osazone, exists in alpha and beta forms that undergo mutarotation, and can be hydrolyzed to two molecules of D-(+)-glucose. The sequence of oxidation, methylation, and hydrolysis (as described for (+)-maltose) shows that (+)-cellobiose contains two pyranose rings and glucoside linkage to an –OH group on C–4. As hemiacetals, α-and β-D-(+)- glucose are readily hydrolyzed by water. In aqueous solution either anomer is converted –via the open-chain form–into an equilibrium mixture containing both cyclic isomers. This mutarotation results from the ready opening and closing of the hemiacetal ring. The typical aldehyde reactions of D-(+)-glucose – osazone formation, and perhaps reduction of Tollens’ and Fehling’s reagents– are presumably due to a small amount of open-chain compound, which is replenished as fast as it is consumed. The concentration of this open-chain structure, however, is too low (less than 0.5%) for certain easily reversible aldehyde reactions like bisulfite addition and the Schiff test. (+)-Cellobiose differs from (+)-maltose in one respect : it is hydrolyzed by the enzyme emulsin (from bitter almonds), not by maltase. Since emulsin is known to hydrolyze only β-glucoside linkages. (+)-Lactose : (+)-Lactose makes up about 5% of human milk and of cow’s milk. It is obtained commercially as a byproduct of cheese manufacture, being found in the whey, the aqueous solution that remains after the milk proteins have been coagulated. Milk sours when lactose is converted into lactic acid (sour, like all acids) by bacterial action (e.g., by Lactobacillus bulgaricus). Disaccharides : Disaccharides are carbohydrates that are made up of two monosaccharide units. On hydrolysis a molecule of disaccharide yields two molecules of monosaccharide. XtraEdge for IIT-JEE 34 JANUARY 2010 (+)-Lactose has the molecular formula C12H22O11, is a reducing sugar, forms an osazone, and exists in alpha and beta forms which undergo mutarotation. Acidic hydrolysis or treatment with emulsin (which splits β linkages only) converts (+)-lactose into equal amounts of D-(+)-glucose and D-(+)-galactose. (+)Lactose is evidently a β-glycoside formed by the union of a molecule of D-(+)glucose and a molecule of D-(+)-galactose. influence of enzymes, the components of starch are hydrolyzed progressively to dextrin (a mixture of low-molecular-weight polysaccharides), (+)-maltose, and finally D-(+)-glucose. (A mixture of all these is found in corn sirup, for example.) Both amylose and amylopectin are made up of D-(+)-glucose units, but differ in molecular size and shape. Cellulose : Cellulose is the chief component of wood and plant fibers; cotton, for instance, is nearly pure cellulose. It is insoluble in water and tasteless; it is a nonreducing carbohydrate. These properties, in part at least, are due to its extremely high molecular weight. (+)-Sucrose : (+)-Sucrose is our common table sugar, obtained from sugar cane and sugar beets. Of organic chemicals, it is the one produced in the largest amount in pure form. Cellulose has the formula (C6H10O5)n. Complete hydrolysis by acid yields D-(+)-glucose as the only monosaccharide. Hydrolysis of completely methylated cellulose gives a high yield of 2, 3, 6-triO-methyl-D-glucose. Like starch, therefore, cellulose is made up of chains of D-glucose units, each unit joined by a glycoside linkage to C–4 of the next. (+)-Sucrose has the molecular formula C12H22O11. It does not reduce Tollen’s or Fehling’s reagent. It is a non-reducing sugar, and in this respect it differs from the other disaccharides we have studied. Moreover, (+)-sucrose does not form an osazone, does not exist in anomeric forms, and does not show mutarotation in solution. All these facts indicate that (+)-sucrose does not contain a “free”aldehyde or ketone group. Cellulose differs from starch, however, in the configuration of the glycoside linkage. Upon treatment with acetic anhydride and sulfuric acid, cellulose yields octa-O-acetylcellobiose. (+)-Sucrose is made up of a D-glucose unit and a Dfructose unit; since there is no “free” carbonyl group, if must be both a D-glucoside and a D-fructoside. Reactions of cellulose : Polysaccharides : Like any alcohol, cellulose form esters. Treatment with a mixture of nitric and sulfuric acid converts cellulose into cellulose nitrate. The properties and uses of the product depend upon the extent of nitration. Polysaccharides are compounds made up of manyhundreds or even thousands-monosaccharide units per molecule. Polysaccharides are naturally occurring polymers, which can be considered as derived from aldoses or ketoses by polymerization with loss of water. A polysaccharide derived from hexoses, for example, has the general formula (C6H10O5)n. In the presence of acetic anhydride, acetic acid, and a little sulfuric acid, cellulose is converted into the triacetate. Partial hydrolysis removes some of the acetate groups, degrades the chains to smaller fragments (of 200–300 units each), and yields the vastly important commercial cellulose acetate (roughly a diacetate). Cellulose acetate is less flammable than cellulose nitrate and has replaced the nitrate in many of its applications, in safety-type photographic film, for example. When a solution of cellulose acetate in acetone is forced through the fine holes of a spinnerette, the solvent evaporates and leaves solid filaments. Threads from these filaments make up the material known as acetate rayon. The most important polysaccharides are cellulose and starch. Both are produced in plants from carbon dioxide and water by the process of photosynthesis. Starch : Starch occurs as granules whose size and shape are characteristic of the plant from which the starch is obtained. When intact, starch granules are insoluble in cold water; if the outer membrane has been broken by grinding, the granules swell in cold water and form a gel. Industrially, cellulose is alkylated to ethers by action of alkyl chlorides (cheaper than sulfates) in the presence of alkali. Considerable degradation of the long chains is unavoidable in these reactions. Methyl, ethyl, and benzyl ethers of cellulose are important in the production of textiles, films, and various plastic objects. In general, starch contains about 20% of a watersoluble fraction called amylose, and 80% of a waterinsoluble fraction called amylopectin. These two fractions appear to correspond to different carbohydrates of high molecular weight and formula (C6H10O5)n. Upon treatment with acid or under the XtraEdge for IIT-JEE 35 JANUARY 2010 KEY CONCEPT Inorganic Chemistry Fundamentals SALT ANALYSIS 2. Identification of acidic radicals For the identification of the acidic radicals, the following scheme is followed. Group I : The radicals which are analysed by dilute H2SO 4 or dilute HCl. These are (i) carbonate (ii) sulphite, (iii) sulphide, (iv) nitrite, and (v) acetate Group II : The radicals which are analysed by concentrated H2SO4 . These are (i) chloride, (ii) bromide, (iii) iodide (iv) nitrate, and (v) oxalate Group III : The radicals which are not analysed by dilute and concentrated H2SO4. These are (i) sulphate, (ii) Phosphate, (iii) borate, and (iv) fluoride. Group I : Add dilute HCl or H2SO4 to a small amount of substance and warm gently, observe. 1. Carbonate or CO32– : The carbonates are decomposed with the effervescence of carbon dioxide gas. Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2 ↑ When this gas is passed through lime water, it turns milky with the formation of calcium carbonate. Ca(OH)2 + CO2 → CaCO3 + H2O Lime water K2Cr2O7 + H2SO4 + 3SO2 → K2SO4 + Cr2 (SO4)3 + H2O The sulphite also gives white precipitate with BaCl2, Soluble in dil. HCl 3. Na2SO3 + BaCl2 → 2 NaCl + BaSO 3 ↓ Sulphide, S–2: The sulphide salts form H2S which smells like rotten eggs. Na2S + H2SO4 → Na2SO4 + H2S ↑ On exposure to this gas, the lead acetate paper turns black due to the formation of lead sulphide. Pb(CH3COO)2 + H2S → PbS ↓ + 2CH3COOH black ppt. The sulphides also turn sodium nitroprusside solution violet (use sodium carbonate extract for this test). White ppt. If the CO2, gas is passed in excess, the milky solution becomes colourless due to the formation of soluble calcium bicarbonate. CaCO3 + H2O + CO 2 → Ca(HCO3)2 White ppt. Sulphite : The sulphites give out sulphur dioxide gas having suffocating smell of burning sulphur. CaSO3 + H2SO4 → CaSO4 + H2O + SO2 ↑ When acidified potassium dichromate paper is exposed to the gas, it attains green colour due to the formation of chromic sulphate. Na2S + Na2[FeNO(CN)5] → Na4 [Fe(NOS) (CN)5] Sulphides of lead, calcium, nickel, cobalt, antimony and stannic are not decomposed with dilute H2SO4. Conc. HCl should be used for their test.However brisk evolution of H2S takes place even by use of dilute H2SO 4 if a pinch of zinc dust is added. Soluble Note : Carbonates of bismuth and barium are not easily decomposed by dilute H2SO 4. Dilute HCl should be used. Sulphur dioxide evolved from sulphites also turns lime water milky. Ca(OH)2 + SO2 → CaSO3 + H2O Zn + H2SO4 → ZnSO4 + 2H HgS + 2H →Hg + H2S ↑ 4. White ppt. However SO 2 can be identified by its pungent odour of burning sulphur. PbCO3 reacts with HCl or H2SO4 to give in the initial stage some effervescence but the reaction slows down due to formation of a protective insoluble layer of PbCl2 or PbSO4 on the surface of remaining salt or mixture. – Nitrite, NO2 : The nitrites yield a colourless nitric oxide gas which in contact with oxygen of the air becomes brown due to the formation of nitrogen dioxide. 2KNO2 + H2SO4 →`K2SO4 + 2HNO2 Nitrous acid 3HNO2 → H2O + 2NO + HNO3 2 NO + O2 → 2NO 2 ↑ brown coloured gas XtraEdge for IIT-JEE 36 JANUARY 2010 On passing the gas through dilute FeSO4 solution, brown coloured complex salt is formed. 1. FeSO4.7H2O + NO → [Fe(H2O)5NO].SO4 + 2H2O NaCl + H2SO4 → NaHSO4 + HCl ↑ The gas evolved forms white fumes of ammonium chloride with NH4OH. Brown coloured (panta aquo nitroso ferrous sulphate) When a mixture of iodide and nitrite is treated with dilute H2SO4, the iodides are decomposed giving violet vapours of iodine, which turns starch iodide paper blue. NH4OH + HCl White fumes 2KI + H2SO4 → K2SO 4 + 2HI + AgNO3 + HCl → AgCl ↓ + HNO3 Yellowish : green chlorine gas with suffocating odour is evolved on addition of MnO2 to the above reaction mixture. 2NO Violet vapours I2 + Starch → Blue colour 5. → NH4Cl + H 2O The gas evolved or solution of chloride salt forms a curdy precipitate of silver chloride with silver nitrate solution. 2NaNO2 + H2SO4 → Na2SO4 + 2HNO2 2HNO 2+ 2HI → 2H2O + I2 Chloride Cl–: Colourless pungent fumes of hydrogen chloride are evolved. NaCl + H2SO4 Acetate : Acetates decompose to give acetic acid vapours having characteristic smell of vinegar. –→ NaHSO4 + HCl MnO2 + 4HCl –→ MnCl2 + 2H2O + Cl2 Note : The curdy precipitate of AgCl dissolves in ammonium hydroxide forming a complex salt. 2CH3COONa + H2SO4 → 2CH3COOH + Na2SO4 All acetates are soluble in water and their aqueous solution on addition to neutral FeCl3 solution develops a blood red colour due to the formation of ferric acetate. Ag(NH3)2Cl + 2H2O AgCl + 2NH4OH → The solution having the silver complex on acidifying with dilute nitric acid gives again a white precipitate of silver chloride. Ag(NH3)2Cl + 2HNO3 → AgCl + 2NH4NO3 Chromyl chloride Test : When solid chloride is heated with conc. H2SO4 in presence of K2Cr2O7, deep red vapours of chromyl chloride are evolved. FeCl3 + 3CH3COONa → (CH3COO)3Fe + 3NaCl Blood Red colour Acetates are also decomposed with oxalic acid and give off acetic acid. NaCl + H2SO4 → NaHSO4 + HCl 2CH3COONa + H2C2O4 → Na2C2O4 + 2CH3COOH K2Cr2O7 + 2H2SO4 → 2KHSO4 + 2CrO3 + H2O CrO3 + 2HCl → CrO2Cl2 + H2O Note : Chromyl chloride The ferric chloride solution supplied in the laboratory is always acidic due to hydrolysis. It is made neutral by the addition of dilute solution of NH4OH drop by drop with constant stirring till the precipitate formed does not dissolve. The filtrate is called neutral ferric chloride solution. Before testing acetate in the aqueous solution by FeCl3, it must be made sure that the solution does not contain (i) CO3–2, (ii) SO 3–2 (iii) PO4–3, (iv) I– Since these also combine with Fe+3. Therefore , the test of acetate should be performed by neutral ferric chloride solution only after the removal of these ions by AgNO 3 solution. These vapours on passing through NaOH solution, form the yellow solution due to the formation of sodium chromate. CrO2Cl2 + 4NaOH → Na2CrO4 +2NaCl+ 2H2O Yellow colour The yellow solution neutralised with acetic acid gives a yellow precipitate of lead chromate with lead acetate. Na2CrO4 + Pb(CH3COO)2 –→ PbCrO4 + 2CH3COONa Yellow ppt. Note : This test is not given by the chloride of mercuric, tin, silver, lead and antimony. The chromyl chloride test is always to be performed in a dry test tube otherwise the Group II: Add concentrated H2SO4 to a small amount of the salt or mixture and warm gently, observe. XtraEdge for IIT-JEE 37 JANUARY 2010 chromyl chloride vapours will be hydrolysed in the test tube. 4. CrO2Cl2 + 2H2O →H2CrO4 + 2HCl Bromides and iodides do not give this test. 2. NaNO3 + H2SO4 → NaHSO4 + HNO3 4 HNO 3 → 2H2O + 4 NO2 + O 2 These fumes intensify when copper turnings are added. Bromide, Br– : Reddish- brown fumes of bromine are formed. NaBr + H2SO4 → NaHSO4 + HBr Cu + 4HNO3 → Cu(NO3)2 + 2NO 2 + 2H2O Ring Test : An aqueous solution of salt is mixed with freshly prepared FeSO4 solution and conc. H2SO4 is poured in test tube from sides, a brown ring is formed on account of the formation of a complex at the junction of two liquids. 2HBr + H2SO4 →Br2 + 2H2O + SO2 More reddish brown fumes of bromine are evolved when MnO2 is added. 2NaBr + MnO2 + 3H2SO4 → 2NaHSO 4 + MnSO4 + 2H2O + Br2 The aqueous solution of bromide or sodium carbonate extract gives pale yellow precipitate of silver bromide which partly dissolves in excess of NH4OH forming a soluble complex. NaBr + AgNO3 → AgBr ↓ + NaNO3 + H2SO4 → NaHSO4 + HNO3 6 FeSO4 + 2HNO3 + 3H2SO4 → 3Fe2 (SO4)3 + 4H2O + 2NO [Fe(H2O)6]SO4. H2O + NO → NaNO 3 Ferrous sulphate Pale yellow ppt. 3. [Fe(H2O)5 NO]SO 4 + 2H2O AgBr +2NH4OH → Ag(NH3)2Br + 2H2O Iodide, I– : Violet vapours of iodine are evolved. Brown ring The nitrates can also be tested by boiling nitrate with Zn or Al in presence of concentrated NaOH solution when ammonia is evolved which can be detected by the characteristics odour. 2KI + H2SO4 → 2KHSO4 + 2HI 2 HI + H2SO4 → I2 + SO2 + 2H2O Violet vapours with starch produce blue colour. Zn + 2NaOH → Na2ZnO 2 + 2H Al + NaOH + H2O → NaAlO 2 + 3H I2 + Starch → Blue colour More violet vapours are evolved when MnO2 is added. 2KI + MnO2 + 3H2SO 4 –→ 2KHSO4 + MnSO 4 + 2H2O + I2 5. Aqueous solution of the iodide or sodium carbonate extract gives yellow precipitate of AgI with silver nitrate solution which does not dissolve in NH4OH. NaNO3 + 8H → NaOH + 2H2O + NH3 Note : Ring test is not reliable in presence of nitrite, bromide and iodide. Oxalate, C2O4–2 : A mixture of CO and CO2 is given off. The CO burns with blue flame. Na2C2O4 + H2SO4 → Na2SO4 + H2C2O4 H2C2O4 + [H2SO4] → CO + CO2 + H2O + [H2SO4] NaI + AgNO3 → AgI + NaNO3 2CO + O2 → 2CO 2 Yellow ppt. A solution of oxalates give the white precipitate with CaCl2 solution. This precipitate get dissolved in dil. H2SO4 and decolourises KMnO4 (acidified) solution. Note : Sodium carbonate extract of bromide and iodide on addition of CHCl3 and chlorine water gives brown or violet layer to CHCl3 respectively. Na2C2O4 + CaCl2 → CaC2O4 ↓ + 2NaCl 2NaBr + Cl2 → 2NaCl + Br2 ; CaC2O4 + H2SO 4 → CaSO4 + H2C2O4 Br2 + CHCl3 → Brown 5H2C2O4 + 2KMnO4 + 3H2SO4 → 2 MnSO4 + K2SO4+ 8 H2O + 10CO2 2NaI + Cl2 → 2NaCl + I2 ; I2 + CHCl3 → Violet Excess of chlorine water should be avoided as the layer may become colour less due to conversion of Br2 into HBrO and I2 into HIO 3. Group III : 1. Sulphate ,SO4–2 : Add conc. HNO3 to a small amount of substance or take sodium carbonate extract and then add BaCl2 solution. A white precipitate of BaSO 4 insoluble in conc. acid is obtained. Br2 + 2H2O + Cl2 → 2HBrO + 2HCl I2 + 5Cl2 + 6H2O → 2HIO3 + 10 HCl XtraEdge for IIT-JEE Nitrate, NO3– : Light brown fumes of nitrogen dioxide are evolved. 38 JANUARY 2010 Na2SO 4 + BaCl2 → 2NaCl + BaSO4 3SiF4 + 4H2O → H4SiO4 + 2H2SiF6 White ppt. 2. Silicic acid (white) Note : Silver and lead if present, may be precipitated as silver chloride and lead chloride by the addition of barium chloride. To avoid it, barium nitrate may be used in place of barium chloride. Borate : To a small quantity of the substance (salt or mixture), add few ml. of ethyl alcohol and conc. H2SO4. Stir the contents with a glass rod. Heat the test tube and bring the mouth of the test tube near the flame. The formation of green edged flame indicates the presence of borate. 2Na3BO3 + 3H2SO4 → 3Na2SO 4 + 2H3BO3 Note : The test should be performed in perfectly dry test tube otherwise waxy white deposit will not be formed on the rod. HgCl2 and NH4Cl also give white deposits under these conditions, but these are crystalline in nature. Sodium carbonate extract : One part of the given substance is mixed with about 3 parts of sodium carbonate and nearly 10 to 15 ml. of distilled water. The contents are then heated for 10-15 minutes and filtered. The filtrate is known as sodium carbonate extract or soda extract and this contains soluble sodium salts due to exchange of partners in between sodium carbonate and salts. CaCl2 + Na2CO3 → CaCO3 + 2NaCl H3BO3 + 3C2H5OH → (C2H5)3BO3 + 3H2O Ethyl borate 3. Phosphate : Add conc. HNO 3 to a small amount of substance or take sodium carbonate extract, heat and then add ammonium molybdate. A canary yellow precipitate of ammonium phospho molybdate is formed. Ca3(PO4)2 + 6HNO 3 → 3Ca (NO 3)2 + 2H3PO4 Insoluble (soluble) PbSO4 + Na2CO3 → PbCO3 + Na2SO4 Insoluble H3PO 4 + 12 (NH4)2 MoO4 + 21 HNO 3 → (NH4)3 PO4. 12 MoO 3 + 21 NH4NO3 + 12 H2O Sodium sulphate (Soluble) BaCl2 + Na2CO3 → BaCO3 + 2 NaCl Canary yellow ppt. 4. Sodium chloride Insoluble Sodium chloride Note : Arsenic also yields a yellow precipitate of (NH4)3. AsO4.12 MoO3 (Ammonium arseno molybdate).Thus in presence of As, phosphate is tested in the filtrate of second group. The precipitate of ammonium phosphomolybdate dissolves in excess of phosphate. Thus, the reagent (ammonium molybdate) should always be added in excess. HCl interferes in this test. Hence, before the test of phosphate is to be performed, the solution should be boiled to remove HCl. Reducing agents such as sulphites, sulphides, etc., interfere as they reduce Mo+6 to molybdenum blue (Mo3O8.xH2O). The solution, therefore, turns blue. In such cases, the solutions should be boiled with HNO 3 so as to oxidise them before the addition of ammonium molybdate. Fluoride : Take small amount of the substance in dry test tube and add an equal amount of sand and conc. H2SO 4.Heat the contents and place a glass rod moistened with water over the mouth of the test tube. A gelatinous waxy white deposit on the rod is formed. 2NaF + H2SO4 → Na2SO4 + H2F2 (Soluble) The carbonates of the cations of the mixtures are mostly insoluble in water and are obtained in the residue. On the other hand, sodium salts of the anions (acidic radicals) of the mixture being soluble in water are obtained in the filtrate. The sodium carbonate extract is basic in nature and before it is used for the analysis of a particular acidic radical, it is first neutralised by the addition of small quantity of an appropriate acid. The acid is added to the extract till the effervescence cease to evolve. Advantages of preparing sodium carbonate extractThe preparation of sodium carbonate extract affords a convenient method for bringing the anions of the mixture into solution which were otherwise insoluble with cation of salt. It removes the basic radicals (usually coloured) which interferes in the usual tests of some of the acidic radicals. The residue can be used for the tests of basic radicals of I to VI groups. Such a solution does not involve the problem of removing interfering radicals like oxalate, fluoride, borate and phosphate. SiO2 + 2H2F2 → SiF4 + 2H2O XtraEdge for IIT-JEE 39 JANUARY 2010 UNDERSTANDING Organic Chemistry 1. A hydrocarbon (A) [C = 90.56%, V.D. = 53] was subjected to vigrous oxidation to give a dibasic acid (B). 0.10 g of (B) required 24.10 ml of 0.05 N NaOH for complete neutralization. When (B) was heated strongly with soda-lime it gave benzene. Identify (A) and (B) with proper reasoning and also give their structures. Sol. Determination of empirical formula of (A) : Element % C 90.56 H 9.44 CH3 Phthalic acid Isophthalic acid CH3 COOH + 2H2O CH3 COOH m-xylene COOH CH3 6[O] + 2H2O CH3 COOH All the above three acids on heating with soda-lime yields only benzene. COOH COOH , , COOH COOH COOH NaOH + CaO ∆ + 2CO2 COOH Of the three acids, one which on heating gives an anhydride, is o-isomer. COOH CO ∆ O CO COOH –H2O One acid which on nitration gives a mono nitro compounds is p-dicarboxylic acid. COOH COOH NO2 HNO3 ∆; H2SO4 COOH COOH One acid which on nitration gives three mono nitro compounds will be the m-isomer. COOH COOH COOH HNO3 NO2 COOH , COOH , COOH H2SO4 NO2 COOH COOH NO2 Terphthalic acid 2. All the above three acids are obtained by the oxidation of respectively xylenes. XtraEdge for IIT-JEE 6[O] CH3 The empirical formula of (A) = C4H5 Empirical formula weight = 48 + 5 = 53 Molecular weight = V.D. × 2 = 53 × 2 = 106 Molecular wt. 106 Hence, n = = =2 Empirical wt. 53 Molecular formula = 2 × C4H5 = C8H10 The given equation may be outlined as follows : HOOC oxidation C 8 H 10 Vigrous → C6H4+ 2H2O 6[O ] HOOC (A) (B) Meq. of dicarboxylic acid = Meq. of NaOH 0.1× 1000 = 24.1 × 0.05 E Equivalent of acid = 83 Molecular wt. = Basicity × Equivalent weight = 2 × 83 = 166 Since (B) on heating with soda-lime gives benzene, the C6H4 represents to benzene nucleus having two side chains, thus (B) is a benzene dicarboxylic acid. There are three benzene dicarboxylic acids. COOH COOH COOH COOH COOH + 2H2O COOH o-xylene Atomic Relative no. Simplest ratio wt. of atoms 12 90.56 7.55 = 7.55 = 1 or 4 12 7.55 1 9.44 9.44 = 9.44 = 1.25 1 7.55 or 5 COOH 6[O] 40 COOH An organic compound (A) contains 69.42% C, 5.78% H and 11.57% N. Its vapour density is 60.5. It evolves NH3 when boiled with KOH. On heating with P2O5, it gives a compound (B) having C = 81.55%, H = 4.85% and N= 13.59%. On reduction JANUARY 2010 with Na + C2H5OH (B) gives a base, which reacts with HNO 2 giving off N2 and yielding an alcohol (C). The alcohol can be oxidised to benzoic acid. Explain the above reactions and assign structural formulae to (A), (B) and (C) Sol. (i) Calculation of empirical formula of (A) : Element C At. wt. 12 % 69.42 H 1 5.78 N 14 11.57 O 16 13.23 Relative no. of atoms 69.42 = 5.785 12 5.78 = 5.78 1 11.57 = 0.826 14 13.23 = 1.827 16 The formula of benzoic acid indicates that the compound (A) is an aromatic amide. Hence, the reactions are : COOK CONH2 KOH Boil Benzamide (A) Simplest ratio 5.785 =7 0.826 5.78 =7 0.826 0.826 =1 0.826 0.827 =1 0.826 CONH2 C At. wt. 12 % 81.55 H 1 4.85 N 14 13.59 Relative no. of atoms 81.55 = 6.80 12 4.85 = 4.85 1 13.59 = 0.97 14 + 4[H] XtraEdge for IIT-JEE N2 + (C) CH2OH HNO2 + N2 + H2O Benzyl alcohol (C) COOH CH2OH 2[O] – H2O (C) 3. Simplest ratio 6.80 =7 0.97 4.85 =5 0.97 0.97 =1 0.97 A (C 6 H12 ) HCl → Benzoic acid B+C (C 6 H13Cl) KOH B alc . → D isomer of A D Ozonolysis → E (it gives negative test with Fehling solution but responds to iodoform test) A Ozonolysis → F + G Both gives positive Tollen's test but do not give iodoform test. . NaOH F + G Conc → HCOONa + primary alcohol Identify to A to G Sol. A (C 6 H12 ) HCl → B+C (C 6 H13Cl) B Ozonolysis → E (it gives negative test with Fehling solution but responds to iodoform test) A Ozonolysis → F + G Both gives positive Tollen's test but do not give iodoform test. ( B) (B) Benzyl amine (Base) CH2NH2 . NaOH F + G Conc → HCOONa + primary alcohol Both F and G are aldehydes because they give positive Tollen's test and do not give iodoform test. These aldehydes give Cross Cannizzaro's reaction. So they do not have α-hydrogen atoms. In cross Cannizzaro's reaction HCOONa is formed along with p-alcohols. So in these an aldehyde is HCHO and The above reaction confirms that (A) is an amide, and the remaining reaction are : COOH HNO2 Na +C2H5OH (B) C 7 H 7 ON → C 7 H 5 N + H 2 O [H] CH2NH2 C≡N P2 O5 C7H5N + H2O Benzonitrile (B) (A) Hence, empirical formula of (B) = C7H5N (v) Determination of structural formulae : (a) Since compound (A) on heating with KOH gives NH3, a characteristic test of amide, hence the compound (A) is an amide (–CONH2). (b) Since compound (B) is obtained by heating (A) with P2O5, a dehydrating agent. (A) C≡N P2O5 ∆ Hence, empirical formula of (A) = C7H7NO Empirical formula wt. = 84 + 7 + 14 + 16 = 121 (ii) Calculation of molecular weight of (A) : Molecular weight = 2 × V.D. = 2 × 60.5 = 121 (iii) Determination of molecular formula of (A): Molecular wt. 121 = =1 n= Empirical wt. 121 Hence, molecular formula = empirical formula i.e., C7H7NO (iv) Calculate of empirical formula of (B) : Element + NH3 ↑ [O] Alcohol 41 JANUARY 2010 Hence, another is (CH3) 3C.CHO. F and G are obtained by ozonolysis of A. Therefore compound 'A' is CH2 = CH – C(CH3)3. Structure of compound 'A' is CH3 CH3 – C – CH = CH2 CH3 Compound 'A' on reaction with HCl gives comp. B and C which have molecular formula C 6H13Cl. Thus, CH3 CH3 CH3 Compound 'A' = CH3 – C – CH = CH2 (C6H12) CH3 CH3 Compound 'B' = CH3 – C – CH – CH3 CH3 Cl CH3 HCl CH3 – C – CH = CH2 → CH3 – C — CH – CH3 CH3 Compound 'C' = CH3 – C – CH2 – CH2Cl CH3 Cl Comp. 'B' + CH3 CH3 Compound 'D' = CH3 – C – CH2 – CH2Cl CH3 – Boil (–Cl ) Compound 'F' = + CH3 – C — CH – CH3 CH3 Cl Compound 'G' = CH3 – C – CH2OH CH3 CH3 – C – CH + CH3 H+ CH3 CH3 CH3 C=C CH3 4. CH3 CH3 Compound 'D' Compound 'D' on ozonolysis to give compound 'E' CH3 Ozonolysis CH3 C=C 2CH3 – C – CH3 CH3 CH3 O Sol. Compound 'E' 'F' (ii ) H 2 O ( B) 2H H – C ≡ C – H − → C3H5 – C ≡ C – C3H5 + C 6 H10 the C3H5 – corresponds to cyclopropyl (∆ ) radical, hence compund (A) is CH2 CH2 CH – C≡C – CH CH2 CH2 Ozonolysis CH3 – C — CHO + CH2O + conc. NaOH → 'F' Comp. 'G' i ) O3 A(C8H10) ( → C 4 H 6 O 2 C=C or a – C ≡ C – bond. should have either If it was alkene its formula should be C8H16 (CnH2n), and if it was alkyne it should have the formula C8H14; it means it is neither a simple alkenen or simple alkyne. However it is definite that the compound has an unsaturated group, it appears that it is a cyclosubstituted ethyne. Compound G and F gives crossed Cannizzaro's reaction with conc. NaOH solution. CH3 CH3 A hydrocarbon (A) of the formula C8H10, on ozonolysis gives compound (B), C4H6O2, only. The compound (B) can also be obtained from the alkyl bromide, (C) (C3H4Br) upon treatment with Mg in dry ether, followed by treatment with CO2 and acidification. Identify (A), (B) and (C) and also equations for the reactions. Since compound (A) adds one mol of O3, hence it Compound 'E' has methyl ketonic groups (–COCH3) so it gives positive iodoform test and does not give the test with Fehling solution due to absence of –CHO group. Compound 'A' on ozonolysis to give compounds F and G as follows : Ozonolysis (CH3)3CCH = CH2 → (CH3)3C – CHO + CH2O Comp. 'G' CH3 H–C–H CH3 Sec. carbonium ion CH3 CH3 Compound 'E' = CH3 – C – CH3 O Compound 'B' gives 'D' on dehydrohalogenation with alc. KOH. CH3 CH3 alc. KOH CH3 C=C O Comp. 'C' CH3 – C — CH – CH3 CH3 1,2-dicyclopropyl ethyne CH3 The ozonolysis of above compound would give two moles of cyclopropane carboxylic acid (C4H 6O2). HCOONa + CH3 – C – CH2OH CH3 XtraEdge for IIT-JEE 42 JANUARY 2010 CH2 CH2 CH2 CH – C≡C – CH CH2 (A) O CH2 CH2 CH2 CH2 CH2 CH – C — C – CH (A) O CH – C – C – CH O 2 H2O CH2 Warm O O compound (E) gives Cannizaro's reaction with NaOH. So, (E) is an aldehyde which does not contain α–H atom. Hence it is HCHO. Compound (D) can also be prepared by the hydration of propyne in the presence of acidic solution and Hg++. ++ CH3 – C ≡ CH + H2O Hg → CH 3 − C = CH 2 + H | OH → CH 3 − C − CH 3 || O (i) O3 CH2 CH2 CH2 CH2 + H2O2 (D) Hence (D) is acetone and (E) is formaldehyde. Therefore, alkene (C) is 2-methyl propene. (CH3)2–C=CH2 (D) reacts with hydroxyl amine (NH2OH) to form oxime (F). CH3 –H2O CH3 C = O + H2 NOH C = NOH CH3 CH3 CH – COOH (B) Compound (B) is prepared from cyclopropyl bromide as follows : CH2 CH2 CH – Br Mg ether O CH2 CH . MgBr CH2 C=O ∆ (D) Cyclopropyl magnesium bromide CH2 CH2 CH .COOMgBr HOH dil. HCl; –MgBrOH CH2 CH2 CH–COOH Addition compound 5. Sol. An organic compound (A), C4H9Cl, on reacting with aqueous KOH gives (B) and on reaction with alcoholic KOH gives (C) which is also formed on passing vapours of (B) over heated copper. The compound (C) readily decolourises bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH2OH to gives (F) and the compound (E) reacts with NaOH to give an alcohol (G) and sodium salt (H) of an acid. (D) can also be prepared from propyne on treatment with water in presence of Hg++ and H 2SO4. Identify (A) to (H) with proper reasoning. C4H9Cl Alc. KOH ∆; –KCl (A) (Alkyl halide) Aq.KOH ∆; –KCl (A ) ( B) Cu / 300 º C → CH 3 − C = CH 2 + H 2 O − H 2O | CH 3 ( C) Alc.KOH / ∆ → CH 3 − C = CH 2 − KCl; − H 2O | CH 3 ( C) C4H8 O (C) (Alkene) Cu C4H9OH ∆; –H2O (B) (Alcohol) CH3 – C = CH2 CH3 (I) O3 CH3 C=O+H–C–H (II) H2O/Zn CH3 (E) (D) (C) We know that p-alcohol on heating with Cu gives aldehyde while s-alcohol under similar conditions gives ketone. Thus, (B) is a t-alcohol because it, on heating with Cu gives an alkene (C). Since a talcohol is obtained by the hydrolysis of a t-alkyl halide, hence (A) is t-butyl chloride. Cl OH | | (A) = CH 3 − C − CH 3 and (B) = CH 3 − C − CH 3 | | CH 3 CH 3 The alkene (C) on ozonolysis gives (D) and (E), hence (C) is not symmetrical alkene. In these XtraEdge for IIT-JEE (F) OH Cl | | (B) = CH 3 − C − CH 3 and (A) = CH 3 − C − CH 3 | | CH 3 CH 3 Reactions : OH Cl | | CH 3 − C − CH 3 CH 3 − C − CH 3 Aq.KOH → | | ∆;– KCl CH 3 CH 3 CH3 CH3 C = O + H2NOH (D) CH3 ∆ –H2O CH3 C = NOH (F) 2HCHO + NaOH → CH 3OH + HCOONa (E) (G) ( H) O Hg + + CH3 – C ≡ CH + H2O → CH3 – C – CH3 + H 43 (D) JANUARY 2010 Set 9 `tà{xÅtà|vtÄ V{tÄÄxÇzxá 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 So lu tio n s w ill b e p ub lis h ed in n ex t iss u e Joint Director Academics, Career Point, Kota 1. ( x − b ) ( x − c) ( x − b ) (a − c ) Given that φ (x) = Show that the dip θ (angle with horizontal) of the oil f (a) + bed which is assumed to be a plane is given by tan θ . a < c < b and f ′′(x) exists at all points in (a, b) . lengths of the sides CA and AB respectively and A is a< that there exists a number c 2 + b 2 − 2 yz cos A where b and c are the bc the angle between CA and AB. µ, µ < b , such that f (a ) f ( b) + (a − b ) (a − c ) ( b − c) ( b − a ) 1 f ( c) + f ′′(µ). = (c − a ) (c − b ) 2 ∫ cos 8x − cos 7 x 1 + 2 cos 5x 6. Evaluate : 7. Let f (x) be an even function such that f ′ (x) is continuous, find y for which d2y dx 2 x = ∫ f (t) dt −x An unbiased die is tossed until it lands the same way 8. up twice running. Find the probability that it requires Prove the inequality (aα + bα)1/α < (aβ + bβ)1/β, for a > 0, b > 0 & α > β > 0. r tosses. 3. z2 sin A = Prove 2. y2 ( x − c) ( x − a ) ( x − a ) ( x − b) f (b)+ f (c) - f (x) where ( b − c) ( b − a ) (c − a ) (c − b ) 9. Given the base of a triangle and the sum of its sides A circle of radius 1 rolls (without sliding) along prove that the locus of the centre of its incircle is an the x-axis so that its centre is of the form (t, 1) ellipse. with t increasing. A certain point P touches the x-axis at the origin as the circle rolls. As the circle 4. Let f (x) = ax2 + bx + c & g (x) = cx2 + bx + a, such rolls further, the point P passes through the point that | f (0) | ≤ 1, | f (1) | ≤ 1 and |f (-1) | ≤ 1 , prove that | (x, 1/2). Find x, when it passes through (x, 1/2) f (x) | ≤ 5/4 and | g (x) | ≤ 2. 5. first time. In order to find the dip of an oil bed below the surface of the ground, vertical borings are made from 10. Find all positive integers n for which n −1 + the angular points, A, B, C of a triangle ABC which n + 1 is rational. is in horizontal plane. The depth of the bed at these points are found to be x, x + y and x + z respectively. XtraEdge for IIT-JEE 44 JANUARY 2010 MATHEMATICAL CHALLENGES SOLUTION FOR NOVEMBER ISSUE (SET # 8) 1. 3. If A is the area of the triangle with sides a, b and c, then A2 = s (s − a) (s − b) (s − c) ; A a where 2s = a + b + c. using AM - GM inequality for s − a, s − b, s − c, we have dd Bb 3 (s − a ) + (s − b) + (s − c) A2 ≤ s 3 C c Plane through mid pt of AB, ⊥ to CD is r r r r r ( r – 1/2 ( a + b )).( c – d ) = 0; Let centroid r r r r a +b+c+d = 0 at origin 4 r r r r r ( r + 1/2 ( c + d )). ( c – d ) = 0 r r r r | r + c |2 = | r + d |2 Let 2s = p , then A ≤ r r it is the locus of pt. equidistance from – c & – d s − a = s − b = s − c which happen if a = b = c. 3 s4 3s − 2s A2 ≤ s = 3 3 3 similarly. r r r r r r r r | r + c |2 = | r + d |2 = | r + a |2 = | r + b |2 r r r r so the pt. is equidistant from – a , – b , – c , – d r r r r (i.e. circumcentre of tetrahedron a , – b , – c , – d ) 2. ................(1) f (b + x ) = f (b − x) ................(2) 12 3 so Amax = p min = 12 3 s2 3 3 p2 12 3 , As condition of equality holds iff ; for a = b = c 12 3 A 12 3 A , and again equality holds if a = b = c. b 2 − 4ac ≤ | b | 4. & 2ac ≤ | b | 1 + 2 b As it is defined for x ∈ R. Let x = b − a − t in (1) use (2) in it so that − f (b + t) = f (2a − 2b + b + t) so the function is periodic & its possible period = may be |2a − 2b| = 2b - 2a (as b > a). 45 1− 4ac b 2 ≤|b| 1+ 4ac b2 b 2 − 4ac ≤ | b | + so f (b − t) = f (2a − b + t) XtraEdge for IIT-JEE p2 Now again p ≥ As the function is symmetrical about x = a & x = b lines so f (a + x) = f (a − x) p2 Amax = ⇒ A≤ 2ac b b b 2 − 4ac b b c ≤ + + ± 2a 2a b 2a 2a b c + a b JANUARY 2010 Hence the solutions of az2 + bz + c = 0 satisfy condition | z | ≤ + 5. = b c + . a b sin 2 θ cos 2 θ [– cos2θ + sin2θ + sin3θ + cos 2θ sinθ] for max / min . P (a cos θ , b sin θ) Equation of AC ⇒ ab (1 + sin θ) x y cos θ + sin θ = 1 a b dA =0 dθ sinθ (cos2θ + sin2θ) + sin2θ − cos2θ = 0 sinθ + sin2θ − (1 − sin2θ) = 0 A ⇒ 2sin2θ + sinθ – 1 = 0 (2 sinθ − 1) (sin θ + 1) = 0 as sin θ ≠ −1 sin θ = 1/2 ; θ = π/6 P(θ) C B D Point A : (0 , b cosec θ) ⇒ Equation of BC Point C = x= when θ > π/6 ; dA >0 dθ when θ < π/6 ; dA <0 dθ y = −b x cos θ − sin θ = 1 a min. area. (1 + sin θ) a cos θ Amin = a (1 + sin θ) ,−b Point C cos θ Area A = = =3 6. ab (1 + sin θ) 2 sin θ cos θ A = ab (1 + sin θ) 2 sin θ cos θ 1 3 4. . 2 2 3 ab sq. units. ax2 + 2hxy + by2 = 0; (y – M1x) (y – M2x) = 0 2h a & M1M2 = b b M2y2 + (1 – M1M2)xy – M1x2 = 0 Compare it with a´x 2 + 2h´xy + b´y2 = 0 −M1 M2 1 − M1M 2 = = b´ 2h´ a´ 2 sin θ cos θ sin 2 θ cos 2 θ ab × 9 (y – M1x)(M2y + x) = 0 2 ab (1 + sin θ) = Now as given the second pair must be given by 2 (1 + sin θ) cos 2 θ sin θ − (1 + sin θ) 2 (cos 2 θ − sin 2 θ) = 1/ 2 . 3 / 2 where M1 + M2 = – dA = dθ ab . ab (1 + sin θ) 2 sin θ cos θ ab . (1 + 1 / 2) 2 1 AD . BC = AD . DC 2 b a (1 + sin θ) = b+ . sin θ cos θ = π ; is the pt of min. 6 so θ = so [2cos2θ sinθ − (1 + sinθ) cos2θ + (1 + sinθ) sin3θ)] XtraEdge for IIT-JEE = 46 −M1 M2 M + M2 −2h = 1 =– = b´ a´ b´−a´ b( b´−a´) 1 − M1M 2 1− a / b = 2h´ 2h´ JANUARY 2010 M2 = – 2hb´ (b − a )a´ & M1 = – b( b´−a´) 2h´b 9. a 2hb´ (b − a )a´ a so . = b b( b´−a´) 2h´b b Since M1M2 = = ha´b´ h´ab = b´−a´ b−a Thus LHS = coeff. of xn in [ nC0(1 + x)m + nC1(1 + x)m+1 + 7. n m+n .... + Cn(1 + x) tn = ] 2n + 1 n . (n + 1) 2 2 1 n 2 Sn = 1− Sn = 1− − 1 ( n + 1) 2 1 1 1 1 1 + 2 – 2 + 2 − 2 ................. 22 2 3 3 4 1 ( n + 1) 2 Required sum = Lim S n = 1. n m n n n →∞ n = coeff. of x in (1 + x) [ C0 + C1(1 + x) + ..... + C0 (1 + x)n] 10. Let the given circle be x2 + y2 = r2 & parametric = coeff. of xn in (1 + x)m(2 + x)n = coeff of xn in (1 + x)m n ∑ n angles of A, B, C are respectively θ1 , θ2 & θ3. Let the slopes of the given two lines are m1 & m2. Sides AB & BC are parallel to these lines. C r x n − r .2 r r =0 n m n A(θ1) m = C0 . C0 + C1 . C1 . 2 + nC 2 . mC 2 . 22 + .... + nCnmC n . 2n 1 8. ∫ (1 − x In = 2 n cos mx dx ) −1 ∫ Equation of AB; 2n =0+ m − x (1 − x 2 ( ) n −1 x cos 1 1 cos mx 1 + n −1 m ∫ (2 (n − 1)x −1 C(θ3) B(θ2) 1 1 sin mx 2n + = (1 − x 2 ) n x (1 − x 2 ) n −1 sin mx dx m −1 m −1 2 (1 − x 2 ) n −2 + (1 − x 2 ) n −1 cos mx dx θ1 + θ 2 θ + θ2 θ − θ2 + y sin 1 = r cos 1 2 2 2 so m1 = – cot θ1 + θ 2 = θ1 + θ2 = α 2 θ2 + θ3 =θ2 + θ3 = β 2 Here α, β are constants as m1 & m2 are constants. similarly : m2 = − cot = = = (2 n − 1) 2n m2 2n m 2 1 ∫ (1− x 2 n −2 ) [(−2n + 2) x 2 ] + 1 − x 2 cosmx dx −1 Now equation of AC ; 1 ∫ (1− x 2 n −2 ) [(−2n + 1) x + 1] cos mx dx θ +θ θ − θ3 θ + θ3 x cos 1 + y sin 1 3 = r cos 1 2 2 2 2 −1 θ +θ θ + θ3 x cos 1 + y sin 1 3 = rk 2 2 2n m2 1 ∫ (1 − x 1 2 n −1 ) cos mx dx − (2 n − 2) −1 ∫ (1 − x 2 ) n −2 −1 α−β (i . e. constant) 2 so foot of the perpendicular from centre of given θ + θ3 θ +θ circle on AC r k cos 1 , r k 1 3 is 2 2 which lies on x2 + y2 = (rk)2. where k = cos cos mx dx m2In = (2n (2n − 1) In–1 − 4n (n − 1) In–2. Hence proved. XtraEdge for IIT-JEE 47 JANUARY 2010 Students' Forum Expert’s Solution for Question asked by IIT-JEE Aspirants MATHS 1. x + x 5 + x 6 y 4 + y5 + y 6 G is 4 , 3 3 There are two die A and B both having six faces. Die A has 3 faces marked with 1, 2 faces marked with 2 and 1 face marked with 3. Die B has 1 face marked with 1, 2 faces marked with 2 and 3 faces marked with 3. Both dice are thrown randomly once. If E is the event of getting sum of the numbers appearing on top faces equal to x and let P(E) be the probability of event E, then α − α1 β − β1 G is , 3 3 The point dividing OG. in the ratio 3 : 1 is α β , ≡ (2,1) ⇒ h + k = 3 4 4 3. (i) find x, when P(E) is maxm. (ii) find x, when P(E) is minm conditions f(x + y) = Sol. X can be 2, 3, 4, 5, 6. the set of values of x where f(x) is differentiable and can occur are the coefficients of x2, x3, x4, x5, x6 is (3x + 2x2 + x3) (x + 2x2 + 3x3) 3 4 5 also find the value of lim [f(x)]x. x →∞ 6 Sol. First put x = 0, y = 0 ⇒ f(0) = 0 = 3x + 8x + 14x + 8x + 3x This shows that sum that occurs most often is 4, and sum that occurs minimum times is 2 or 6. 2. Now, 6 ∑x i =1 6 i = 8 and ∑y i f ( h ) − f (0) 1 − {f ( x )}2 = lim x →0 h − 0 − 1 + f ( x ).f (h ) = 4. i =1 The line segment joining orthocentre of a ∆ made by = 1 – {f2(x)} any three points and the contrioid of the ∆ made by other three points passes through a fixed points (h, k), then find h + k. 6 Sol. Let ∑ integrating we get 6 x i = α and i =1 ∑ ⇒ f(x) = yi = β i =1 1 1 + f ( x ) ln =x+c 2 1 − f ( x) ex − e−x e x + e−x clearly f(x) is differentiable for all x ∈R. let 0 be orthocentre of ∆ made by (x1, y1), (x2, y2) and (x3, y3) e x − e−x lim [f ( x )] x = lim x x →∞ x →∞ e + e − x ⇒ 0 is (x1 + x2 + x3, y1 + y2 + y3) = (αi , β 1) similarly let G be the centroid of the ∆ made by other three points. XtraEdge for IIT-JEE f (x + h) − f (x) x →0 h f´(x) = lim f (x) + f (h) − f ( x) 1 + f ( x ).f (h ) = lim x →0 h Six points (xi, yi), i = 1, 2, 3, .... 6 are taken on the circle x2 + y2 = 4 such that f ( x ) + f ( y) ∀ x, y and 1 + f ( x ).f ( y) f´(0) = 1. Also – 1 < f(x) < 1 for all x ∈R, then find The no of ways in which sum 2, 3, 4, 5, 6. 2 Suppose a function f(x) satisfies the following = e 48 ex −e− x lim x →∞ e x + e − x x x=1 JANUARY 2010 XtraEdge for IIT-JEE 49 JANUARY 2010 CAREER POINT’s Correspondence & Test Series Courses Information Courses of IIT-JEE For Class 12th / 12th Pass Students CP Ranker's Package IIT-JEE 2010 Study Material Package IIT-JEE 2011 All India Foundation Test Series IIT JEE 2011 (AT CENTER) Postal All India Test Series IIT-JEE 2011 Class 12th or 12th pass students Class 12th or 12th pass students Class 11th students Class 11th students Class 11th students English English English English OR Hindi English English Postal Postal Postal Visit our website Postal Immediate Dispatch Dispatch 17-Jan-10 Onwards Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch 25-Oct-09 Immediate Dispatch Rs. 9,500/including S.Tax Rs. 4,500/including S.Tax Rs. 2,500/including S.Tax COURSE INFORMATION All India Test Series IIT JEE 2010 (AT CENTER) Postal All India Test Series IIT-JEE 2010 Major Test Series IIT JEE 2010 (AT CENTER) Eligibiltiy for Admission Class 12th or 12th pass students Class 12th or 12th pass students Class 12th or 12th pass students Medium English OR Hindi English English Test Center Postal Visit our website Postal Issue of Application Kit Immediate Dispatch Immediate Dispatch Immediate Dispatch Date of Dispatch/ First Test Immediate Dispatch 25-Oct-09 Immediate Dispatch Course Fee Rs.8,500/including S.Tax Rs. 3,500/including S.Tax For Class 11th Students Postal Major Test Series IIT JEE 2010 (By POST) Class 12th or 12th pass students Study Material Package IIT-JEE 2010 Visit our website Immediate Dispatch 17-Jan-10 Rs. 1,500/Rs. 1,550/Rs. 700/Rs. 850/including S.Tax including S.Tax including S.Tax including S.Tax Courses of AIEEE For Class 11th Students For Class 12th / 12th Pass Students COURSE INFORMATION Study Material Package AIEEE 2010 Eligibiltiy for Admission Class 12th or 12th pass students All India Test Series AIEEE 2010 (AT CENTER) Class 12th or 12th pass students Postal All India Test Series AIEEE 2010 Major Test Series AIEEE 2010 (AT CENTER) Class 12th or 12th pass students Class 12th or 12th pass students Postal Major Test Series AIEEE 2010 (BY POST) Class 12th or 12th pass students CP Ranker's Package IIT-JEE 2010 Study Material Package AIEEE 2011 Class 12th or 12th pass students Class 11th students English OR Hindi Medium English OR Hindi English/Hindi English/Hindi English/Hindi English/Hindi English Test Center Postal Visit our website Postal Visit our website Postal Postal Postal Issue of Application Kit Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Date of Dispatch/ First Test Immediate Dispatch 25-Oct-09 Immediate Dispatch 17-Jan-10 Immediate Dispatch Immediate Dispatch Course Fee Rs. 7,500/including S. Tax Rs. 4,000/including S. Tax Rs. 1450/including S. Tax Rs. 1000/including S. Tax Immediate Dispatch Dispatch 17-Jan-10 Onwards Rs. 550/including S. Tax Rs. 850/including S.Tax Rs. 8,500/including S.Tax Courses of Pre-Medical For Class 11th Students For Class 12th / 12th Pass Students COURSE INFORMATION Study Material Package Pre-Medical 2010 All India Test Series Pre-Medical 2010 (AT CENTER) Postal Test Series Pre-Medical 2010 Major Test Series Pre-Medical 2010 (AT CENTER) Postal Major Test CP Ranker's Package Series Pre-Medical IIT-JEE 2010 2010 (BY POST) Study Material Package Pre-Medical 2011 Eligibiltiy for Admission Class 12th or 12th pass students Class 12th or 12th pass students Class 12th or 12th pass students Class 12th or 12th pass students Class 12th or 12th Class 12th or 12th pass students pass students Class 11th students Medium English OR Hindi English/ Hindi English/ Hindi English/ Hindi English English English OR Hindi Test Center Postal Visit our wesite for list of test centers Postal Visit our wesite for list of test centers Postal Postal Postal Issue of Application Kit Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch Date of Dispatch Commencement / First Test Immediate Dispatch 25-Oct-09 Immediate Dispatch 17-Jan-10 Dispatch 17-Jan-10 Onwards Immediate Dispatch Immediate Dispatch Course Fee Rs.7,200/including S. Tax Rs. 3,600/including S. Tax Rs. 1400/including S. Tax Rs. 900/including S.Tax Rs. 8,000/including S.Tax Rs. 1100/- including Rs. 600/- including S. Tax S. Tax ◆For details about all Correspondence & Test Series Course Information, please visit our website.: www.careerpointgroup.com XtraEdge for IIT-JEE 50 JANUARY 2010 XtraEdge for IIT-JEE 51 JANUARY 2010 XtraEdge for IIT-JEE 52 JANUARY 2010 4. Let z1, z2, z3 be the complex nos. represent the given that vertices of the ∆ABC which is circumscribed by the circle |z| = 1. Altitude from A meets the side BC at D a=– (ii) the complex no. of point E. 1 1 ,b= 4 2 2x − x 2 4 f(x) = A(z1) 1 since 4x + 4y – 5 = 0 passes through A 1, and 4 O D 1 B ,1 so area bounded is OAB = 2OAC 4 C(z3) = 2[ar(OCP) + ar(CAQP) – ar(OAQ)] E(z4) 1 5 5 1 5 1 3 =2 × × + + − 2 8 8 2 8 4 8 Sol. (i) we know that the image of orthocentre about any side of the ∆ lies on the circum circle of ∆. Point ...(2) from (1) and (2) (i) the complex no. of point P. B (z2) ...(1) 0 = 2a + b and circum circle at E. Let P be the image of E about BC, then find P 1 =a+b 4 P = z1 + z2 + z3 = (ii) Let O (origin) be the circum centre of ∆. ∫ 1 2x − x 2 0 4 dx 37 (unit)2 96 ∠BOE = π – 2B and ∠AOC = 2B 6. z1 = ei 2B z3 ...(1) z4 = ei(π–2B) z2 ...(ii) i for i = 0, 1, 2 ..... n. If n is odd than find i +1 the value of P(n + 1). P(i) = Sol. Let Q(x) = (x + 1) P(x) – x z z z1 z 4 . = –1 ⇒ z4 = – 3 2 z3 z 2 z1 5. clearly Q(x) is polynomial of degree n + 1. Also Q(i) = (i + 1) If A be the area bounded by y = f(x), y = f–1(x) and the line 4x + 4y – 5 = 0 where f(x) is a polynomial of 2 nd degree passing through the origin and having maximum value of y = f (x) Q(x) = kx(x – 1) (x – 2) ...... (x – n) where k is a constant. 1 at x = 1, then find A. 4 Now Q(–1) = k(–1)(–2)(–3) ...... (–1 – n) 1 = (–1)n + 1 k(n + 1) ! y=x ⇒ k= 1 ( n + 1) ! Thus P(x) = A y=f(x) O i – i = 0 for i = 1, 2, 3 .....n i +1 Thus we can assume –1 BC Let P(x) be a polynomial of degree n such that P Q x=1 1 x ( x − 1)(x − 2)....( x − n ) + x , x +1 ( x + 1) ! where n is odd ∴ P(n + 1) = 1 Sol. Let f(x) = ax 2 + bx XtraEdge for IIT-JEE (Q n is odd) 53 JANUARY 2010 MATH DIFFERENTIAL EQUATIONS Mathematics Fundamentals Differential Equation : Formation of Differential Equation : An equation involving independent variable x, dependent variable y and the differential coefficients (1) Write down the given equation. (2) Differentiate it successively with respect to x that number of times equal to the arbitrary constants. dy d 2 y , , .... is called differential equation. dx dx 2 (3) And hence on eliminating arbitrary constants results a differential equation which involves x, y, Examples : (1) dy =1+x+y dx (2) dy + xy = cot x dx dy d 2 y , ..... dx dx 2 Solution of Differential Equation : A solution of a differential equation is any function which when put into the equation changes it into an identity. 3 d4 y dy + 4y = 5 cos 3x (3) 4 – 4 dx dx d2y (4) x2 2 + dx dy 1+ dx General and particular solution : The solution which contains a number of arbitrary constant equal to the order of the equation is called general solution by giving particular values to the constants are called particular solutions. 2 =0 Order of a Differential Equation : Several Types of Differential Equations and their Solution : The order of a differential equation is the order of the highest derivative occurring in the differential equation. For example, the order of above differential equations are 1, 1, 4 and 2 respectively. (1) Solution of differential equation dy = f(x) is y = dx Degree of a Differential Equation : (2) Solution of differential equation The degree of the differential equation is the degree of the highest derivative when differential coefficients are free from radical and fraction. For example, the degree of above differential equations are 1, 1, 3 and 2 respectively. dy = f(x) g(y) is dx dy = f(ax + by + c) by dx dy 1 dv putting ax + by + c = v and = −a dx b dx A differential equation in which the dependent variable and its differential coefficients occurs only in the first degree and are not multiplied together is called a linear differential equation. The general and nth order differential equation is given below : dn y dx n + a1(x) d n −1y dx n −1 + .... + an – 1 dy ∫ g( y) = ∫ f ( x) dx + c (3) Solution of diff. equation Linear and Non-linear Differential Equation : a0(x) ∫ f (x)dx + c dv = dx a + bf ( v) dy dx Thus solution is by integrating dv ∫ a + bf (v) = ∫ dx + an(x)y + φ(x) = 0 Those equations which are not linear are called nonlinear differential equations. XtraEdge for IIT-JEE 54 JANUARY 2010 (4) To solve the homogeneous differential equation f ( x, y) dy , substitute y = vx and so = g ( x, y) dx Solve a b obviously = = 2 a´ b´ dy dv =v+x . dx dx Put x – 3y = v dv Thus v + x = f(v) dx ⇒ ⇒1–3 dx dv = x f ( v) − v Therefore solution is dy 2x − 6 y + 7 2( x − 3y) + 7 = = dx x − 3y + 4 x − 3y + 4 ∫ dx = x ∫ Solution of the linear differential equation : dv +c f ( v) − v dy + Py = Q, where P and Q are either constants or dx functions of x, is Equation reducible to homogeneous form : A differential equation of the form ye ∫ dy a x + b1y + c1 = 1 , dx a 2x + b2 y + c2 x = X + h, y = Y + k, so that dY dy = dX dx P dx ∫ Qe ∫ = Where e ∫ a b where 1 ≠ 1 , can be reduced to homogeneous a2 b2 form by adopting the following procedure : Put dy dv = (Now proceed yourself) dx dx P dx P dx dx + c is called the integrating factor. Equations reducible to linear form : Bernoulli's equation : A differential equation of dy the form + Py = Qyn, where P and Q are dx functions of x alone is called Bernoulli's equation. Dividing by yn, we get y–n dy + y–(n – 1). P = Q dx a X + b1Y + (a 1h + b1k + c1 ) dY = 1 dX a 2 X + b 2 Y + (a 2 h + b 2 k + c 2 ) Putting y–(n – 1) = Y, so that (1 − n ) dy dY = , dx y n dx Now choose h and k such that a1h + b1k + c1 = 0 and a2h + b2k + c2 = 0. Then for these values of h and k, the equation becomes we get The equation then transformed to dY + (1 – n)P. Y = (1 – n)Q dx dY a X + b1Y = 1 dX a 2X + b2Y which is a linear differential equation. This is a homogeneous equation which can be solved by putting Y = vX and then Y and X should be replaced by y – k and x – h. dy + P. f(y) = Q . g(y), where P and Q are dx functions of x alone, we divide the equation by g(y) and get If the given equations is of the form Special case : 1 dy f ( y) + P. =Q g( y) dx g ( y) ax + by + c dy a b = = = m (say), i.e. and dx a´x + b´y + c´ a´ b´ when coefficient of x and y in numerator and denominator are proportional, then the above equation cannot be solved by the discussed before because the values of h and k given by the equations will be indeterminate. If f ( y) = v and solve. g ( y) Solution of the differential equation : Now substitute d2y = f(x) is obtained by integrating it with respect dx 2 to x twice. In order to solve such equations, we proceed as explained in the following example. XtraEdge for IIT-JEE 55 JANUARY 2010 MATH TRIGONOMETRICAL RATIOS Mathematics Fundamentals Some Important Definitions and Formulae : Trigo. Function sin x Measurement of angles : The angles are measured in degrees, grades or in radius which are defined as follows: Degree : A right angle is divided into 90 equal parts and each part is called a degree. Thus a right angle is equal to 90 degrees. One degree is denoted by 1º. A degree is divided into sixty equal parts is called a minute. One minute is denoted by 1´. A minute is divided into sixty equal parts and each parts is called a second. One second is denoted by 1´´. Thus, 1 right angle = 90º (Read as 90 degrees) 1º = 60´ (Read as 60 minutes) 1´ = 60´´ (Read as 60 seconds). Grades : A right angle is divided into 100 equal parts and each part is called a grade. Thus a right angle is equal to 100 grades. One grade is denoted by 1g. A grade is divided into 100 equal parts and each part is called a minute and is denoted by 1´. A minute is divided into 100 equal parts and each part is called a second and is denoted by 1" Thus, 1 right angled = 100g (Read as 100 grades) 1 g = 100´ (Read as 100 minutes) 1´ = 100´´ (Read as 100 seconds) Radians : A radian is the angle subtended at the centre of a circle by an arc equal in length to the radius of the circle. Domain and Range of a Trigono. Function : If f : X → Y is a function, defined on the set X, then the domain of the function f, written as Domf is the set of all independent variables x, for which the image f(x) is well defined element of Y, called the co-domain of f. Range R, the set of all the real number –1 ≤ sin x ≤ 1 cos x R – 1 ≤ cos x ≤ 1 tan x π R – ( 2n + 1) , n ∈ I 2 R cosec x R – {n π, n ∈ I} R – {x : –1 < x < 1} sec x π R – ( 2n + 1) , n ∈ I 2 R – {x : –1 < x < 1} cot x R – {n π, n ∈ I} R Relation between Trigonometrically Ratios and identities: sin θ cos θ ; cot θ = cos θ sin θ sin A cosec A = tan A cot A = cos A sec A = 1 tan θ = sin2θ + cos2θ = 1 or sin2θ = 1 – cos2θ or cos2θ = 1 – sin2θ 1 + tan2θ = sec2θ or sec2θ – tan2θ = 1 or sec2θ – 1 = tan2θ 1 + cot2θ = cosec2θ or cosec2θ – cot2θ = 1 or cosec2θ – 1 = cot2θ Since sin 2A + cos2A = 1, hence each of sin A and cos A is numerically less than or equal to unity. i.e. | sin A| ≤ 1 and | cos A | ≤ 1 or –1 ≤ sin A ≤ 1 and – 1 ≤ cos A ≤ 1 Note : The modulus of real number x is defined as |x| = x if x ≥ 0 and |x| = – x if x < 0. Since sec A and cosec A are respectively reciprocals of cos A and sin A, therefore the values of sec A and cosec A are always numerically greater than or equal to unity i.e. Range of f : X → Y is the set of all images f(x) which belongs to Y, i.e., sec A ≥ 1 or sec A ≤ – 1 and cosec A ≥ 1 or cosec A ≤ – 1 In other words, we never have –1 < cosec A < 1 and –1 < sec A < 1. Range f = {f(x) ∈ Y : x ∈ X} ⊆ Y The domain and range of trigonmetrical functions are tabulated as follows : XtraEdge for IIT-JEE Domain 56 JANUARY 2010 Trigonometrical Ratios for Various Angles : Formulae Involving Double, Triple and Half Angles : θ 0 π 6 π 4 π 3 π 2 π 3π 2 2π sin θ 0 1 2 1 3 2 1 0 –1 0 cos θ 1 3 2 1 1 2 0 –1 0 1 tan θ 0 3 ∞ 0 ∞ 0 2 2 1 1 3 2 tan θ 1 + tan 2 θ cos 2θ = cos2 θ – sin 2 θ = 2 cos2θ – 1 sin 2θ = 2 sin θ cos θ = = 1 – 2 sin2θ = Trigonometrical Ratios for Related Angles : θ –θ π±θ π ±θ 2 1 − tan 2 θ 1 + tan 2 θ sin θ =± 2 θ 1 − cos θ ; cos =± 2 2 tan θ =± 2 1 − cos θ 1 + cos θ tan 2θ = 2 tan θ 1 − tan 2 θ sin 3θ = 3 sin θ – 4 sin3θ 2π ± θ 3π ±θ 2 sin – sin θ cos θ m sin θ – cos θ ± sin θ cos cos θ m sin θ – cos θ ± sin θ cos θ tan – tan θ m cot θ ± tan θ m cot θ ± tan θ 1 (3 sin θ – sin 3θ) 4 cos 3θ = 4 cos3θ – 3 cos θ cot – cot θ m tan θ ± cot θ m tan θ ± cot θ or cos3θ = or sin3θ = Addition and Subtraction Formulae : sin (A ± B) = sin A cos B ± cos A sin B tan 3θ = cos (A ± B) = cos A cos B m sin A sin B tan (A ± B) = 1 (3 cos θ + cos 3θ) 4 3 tan θ − tan 3 θ 1 − 3 tan 2 θ π θ ≠ nπ + 6 Trigonometrical Ratios for Some Special Angles : tan A ± tan B 1 m tan A tan B θ cot A cot B m 1 cot B ± cot A sin (A + B) sin (A – B) = sin2A – sin2B = cos2B – cos2A cos (A + B) cos (A – B) = cos2A – sin 2B = cos2B – sin2A Formulae for Changing the Sum or Difference into Product : cot (A ± B) = sin θ cos θ tan θ 7 1º 2 4− 2 − 6 22 3 −1 2− 2 2 2 2 2 2 3 +1 4+ 2 + 6 2 2 ( 3– 2) ( 2 –1) 2 2 2+ 2 2 2– 3 2 –1 C+D C−D cos 2 2 θ sin C – sin D = 2 cos C+D C−D sin 2 2 sin θ 5 −1 4 10 − 2 5 4 cos θ 10 + 2 5 4 5 +1 4 tan θ 25 − 10 5 5 5−2 5 C+D C−D cos 2 2 C+D D−C sin 2 2 Formulae for Changing the Product into Sum or Difference : 2 sin A cos B = sin (A + B) + sin (A – B) 2 cos A sin B = sin (A + B) – sin (A – B) 2 cos A cos B = cos (A + B) + cos (A – B) 2 sin A sin B = cos (A – B) – cos(A + B) cos C – cos D = 2 sin XtraEdge for IIT-JEE 18º 1º 2 15º sin C + sin D = 2 sin cos C + cos D = 2 cos 1 + cos θ 2 36º Important Points to Remember : Maximum and minimum values of a sin x + b cos x are + respectively. 57 a 2 + b2 , – a 2 + b2 JANUARY 2010 sin2x + cosec2x ≥ 2 for every real x. 2 cos x + sec x ≥ 2 for every real x. tan2x + cot2x ≥ 2 for every real x If x = sec θ + tan θ, then A B C A B C + cos2 – cos2 = 2cos cos sin 2 2 2 2 2 2 5. If x + y + z = π/2, then sin2x + sin2y + sin2z = 1 – 2 sin x sin y sin z cos2x + cos2y + cos2z = 2 + 2 sin x sin y sin z sin 2x + sin 2y + sin 2z = 4 cos x cos y cos z cos2 1 = sec θ – tan θ x 1 = cosec θ – cot θ x cos θ . cos 2θ . cos 4θ . cos 8θ If x = cosec θ + cot θ, then 6. If A + B + C = π, then tan A + tan B + tan C = tan A tan B tan C cot B cot C + cot C cot A + cot A cot B = 1 sin 2 n θ 2 n sin θ .... cos 2n–1θ = 1 sin 3θ 4 sin θ sin (60º – θ) sin (60º + θ) = tan 1 cos 3θ 4 tan θ tan (60º – θ) tan (60º + θ) = tan 3θ cos θ cos (60º – θ) cos (60º + θ) = A B C cos cos 2 2 2 sin A + sin B – sin C = 4 sin A B C sin sin 2 2 2 cos A+ cos B+ cos C = 1 + 4 sin A B C sin sin 2 2 2 cos A+ cos B – cos C = –1 + 4cos A B C cos sin 2 2 2 A B C A B C + cot + cot = cot cot cot 2 2 2 2 2 2 7. (a) For any angles A, B, C we have sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C cos (A + B + C) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C tan(A + B + C) = sin(A + B + C) = sin π = 0 and cos (A + B + C) = cos π = –1 then (a) gives sin A sin B sin C = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C and (a) gives 1 + cos A cos B cos C = cos A sin B sin C + sin A cos B sin C + sin A sin B cos C Method of Componendo and Dividendo : 3. If A + B + C = π, then sin2A + sin 2B – sin2C = 2 sin A sin B cos C cos2A + cos2B + cos2C = 1– 2 cos A cos B cos C sin2A + sin 2B + sin2C = 2 + 2 cos A cos B cos C cos2A + cos2B – cos2C = 1 – 2 sin A sin B cos C p a = , then by componendo and dividendo we q b can write If p−q a −b q−p b−a = = or p+q a +b q+p b+a 4. If A + B + C = π, then A B C A B C +sin2 + sin2 = 1– 2 sin sin sin 2 2 2 2 2 2 cos2 A B C A B C +cos2 +cos2 =2 + 2 sin sin sin 2 2 2 2 2 2 XtraEdge for IIT-JEE tan A + tan B + tan C – tan A tan B tan C 1 − tan A tan B − tan B tan C − tan C tan A (b) If A,B, C are the angles of a triangle, then cos A cos B cos C + + =2 sin B sin C sin C sin A sin A sin B sin2 B C C A A B tan + tan tan + tan tan = 1 2 2 2 2 2 2 cot Conditional Identities : 1. If A + B + C = 180º, then sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C sin 2A + sin 2B – sin 2C = 4 cos A cos B sin C sin (B + C – A) + sin (C + A – B) + sin (A+B – C) = 4 sin A sin B sin C cos 2A + cos 2B + cos 2C = –1 – 4 cos A cos B cos C cos 2A+ cos 2B – cos 2C = 1 – 4 sin A sinB cosC 2. If A + B + C = 180º, then sin A + sin B + sin C = 4 cos A B C A B C +sin2 –sin2 = 1 – 2cos cos cos 2 2 2 2 2 2 sin2 2 or 58 p+q a+b q+p b+a = or = p−q a−b q−p b−a JANUARY 2010 Based on New Pattern IIT-JEE 2010 XtraEdge Test Series # 9 Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 10 to 14 are multiple choice questions with multiple (one or more than one) correct answer. +4 marks and -1 mark for wrong answer. • Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer. Section - II • Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly matched answer and No Negative marks for wrong answer. However, +1 marks will be given for a correctly marked answer in any row. (A) PHYSICS (C) Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. a sin θ + b cos θ If x = , then – 1. a+b (A) The dimension of a and x are same (B) The dimension of b and x are same (C) Both (A) and (B) (D) x is dimensionless 2. ABCD is a smooth horizontal fixed plane on which mass m1 = 0.1 kg is moving in a circular path of radius 1 m. It is connected by an ideal string which is passing through a smooth hole and 1 kg at the other end as connects mass m2 = 2 shown. m2 also moves in a horizontal circle of same radius of 1 m with a speed of 10 m/s. If g = 10 m/s2, then the speed of m1 is – A B m1 D C m2 XtraEdge for IIT-JEE 59 10 m/s 1 m/s 10 (B) 10 m/s (D) None of these 3. If x grams of steam at 100ºC becomes water at 100ºC which converts y grams of ice at 0ºC into water at 100ºC, then the ratio x/y will be – 1 27 (B) (A) 3 4 4 (C) 3 (D) 27 4. A gas is at pressure P and temperature T. Coefficient of volume expansion of one mole of gas at constant pressure is – 1 1 (D) T2 (B) T (C) 2 (A) T T 5. A source of light is placed at double focal length from a convergent lens. The focal length of the lens is f = 30 cm. At what distance from the lens should a flat mirror be placed so that ray reflected from the mirror are parallel after passing through the lens for the second time ? (A) Beyond 2 F (B) Between lens and F (C) Between F and 2F near 2F (D) Between F and 2F equidistant from F and 2F JANUARY 2010 6. Consider a toroid of circular cross-section of radius b, major radius R much greater than minor radius b, (see diagram) find the total energy stored in magnetic field of toroid – Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 10. Path of three projectiles are shown. If T1, T2 and T3 are time of flights and ignoring air resistances y 1 2 3 x 7. (A) B2 π 2 b 2 R 2µ 0 (B) B2 π 2 b 2 R 4µ 0 (C) B2 π 2 b 2 R 8µ 0 (D) B2 π 2 b 2 R µ0 (A) T1 > T3 T +T (C) T2 = 1 3 2 11. A small charged ball is hovering in the state of equilibrium at a height h over a large horizontal uniformly charged dielectric plate. What would be the instantaneous acceleration of the ball if a disc of radius r = 0.001 h is removed from the plate directly underneath the ball – g r (A) 2 h 2 g r 4 h 2 (C) 8. R 1.5R 2R g h (B) 2 r 2 g h 4 r 2 (D) (A) K = 8J (C) e = 0.5 a = 13. A solenoid is connected to a source of constant emf for a long time. A soft iron piece is inserted into it. Then – (A) self inductance of the solenoid gets increased (B) flux linked with the solenoid increases hence steady state current gets decrease (C) energy stored in the solenoid gets increased (D) magnetic moment of the solenoid increased → → 3 î + ĵ and | b | = 10 units while θ = 23º, → → then the value of R = | a + b | is nearly – y → b → a θ O (A) 12 x (B) 13 XtraEdge for IIT-JEE (B) K = 16J (D) e = 0 A constant voltage is applied between two ends of a uniform conducting wire. If both the length and radius of the wire are doubled – (A) the heat produced in the wire will be doubled (B) the electric field across the wire will be doubled (C) the heat produced will remain unchanged (D) the electric field across the wire will become half For the vectors a and b shown in figure, → t 12. (D) 10 30/sec → Two blocks of masses 2 kg each are moving in opposite direction with equal speed collides at t = 5 sec. The magnitude of relative velocity (v) is plotted against time 't'. The loss in kinetic energy is K and coefficient of restitution in e, then – v t = 5 sec electron and positron incident photon must have minimum frequency of the order of – (B) 10 21/sec (A) 10 18/sec 9. (D) T1 = T2 = T3 4 m/s For pair production i.e. for the production of (C) 10 25/sec (B) T1 < T3 (C) 14 (D) 15 60 JANUARY 2010 14. In radioactivity decay according to law N = N0e–πt which of the following is/are true ? (A) Probability that a nucleus will decay is 1 – e–λt (B) Probability that a nucleus will decay four half lives is 15/16 (C) Fraction nuclei that will remain after two half lives is zero (D) Fraction of nuclei that will remain after two half-lives is 1/4 Scientists can observe the spectral lines of atoms that are dominant in far-away galaxies. Due to the speed at which these galaxies are travelling, these lines are shifted, but their pattern remains the same. This allows researchers to use the spectral pattern to determine which atoms they are seeing. Table - 1 below shows spectroscopic measurements made by researchers trying to determine the atomic makeup of a particular faraway galaxy. Light energy is not measured directly, but rather is determined from measuring the frequency of light, which is proportional to the energy. Table – 1 This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 15 to 17) ε 10 Ω Frequencies Measured 868440 880570 879910 856390 V 6 10×106Ω Al foil Voltmeter is ideal and two aluminum foil is given at certain separation. In this set up upper aluminum foil jumps to the lower Al foil at the potential difference between the plates (or foils) of 500 V. 15. When emf ε of the battery is 400 volt and foil has not jumped, approximate reading of the voltmeter is (A) 500 V (B) 400 V (C) 0 V (D) 250 V 16. What is the reading of voltmeter just after the foil has jumped and connected the two plates (A) 500 V (B) less than 500 V (C) more than 500 V (D) 600 V 17. What is the emf of the battery just after the foil has jumped and connected the two plates (A) 500 V (B) 600 V (C) 700 V (D) 800 V 18. For each of three hypothetical atoms (Atom 1, Atom 2 and Atom 3), Figure 1 depicts the (A) number of electrons and the amount of energy the atom contains (B) distance an electron travels from one part of the atom to another (C) energy released by the atom as an electron as it moves from one energy state to another (D) frequency with which the atom’s electrons move from one energy state to another 19. Based on the spectroscopic measurements shown in Table - 1, which of the atoms in Figure (i) (Atom 1, Atom 2, or Atom 3) is most similar to the one the scientists were observing, and why ? (A) Atom 2, because it contains four different energy levels (B) Atom 3, because it contains four different energy levels (C) Atom 1, because the frequencies listed in Table 1 indicate a high level of atomic activity (D) Atom 3, because there is a comparatively small difference between exactly two of the four frequencies listed in Table 1 20. The laws of atomic physics prohibit electron movements between certain energy states. In atomic physics, these prohibitions are called “forbidden transitions.” Based on Figure (i), which of the following is most accurate ? (A) Atom 2 has the same number of forbidden transitions as Atom 1 Passage : II (No. 18 to 20) Figure (i), below depicts three hypothetical atoms. Energy levels are represented as horizontal segments. The distance between the segments is representative of the energy difference between the various levels. All possible transitions between energy levels are indicated by arrows. Atom # 1 Atom # 2 Atom # 3 Fig (i) XtraEdge for IIT-JEE 61 JANUARY 2010 (B) Atom 2 has more forbidden transitions than Atom 3 (C) Atom 3 has the same number of forbidden transitions as Atom 1 (D) Atom 1 has fewer forbidden transitions than Atom 2 CHEMISTRY Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. This section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows : P Q R S A P Q R S B P Q R S 1. Brown ppt. (A) dissolve in HNO3 gives (B) which gives white ppt. (C) with NH4OH. (C) on reaction with HCl gives solution (D) which gives white turbidity on addition of water. What is (D) ? (A) BiCl3 (B) Bi(OH)3 (C) BiOCl (D) Bi(NO3)3 2. Histidine (A) has pKa values as indicated O CH2 6.04 N C P Q R S D P Q R S 21. CH C OH + NH NH3 9.17 1.82 (A) What will be its form at pH = 4 ? Column I Column II (A) Pair production (P) Few MeV (B) Inverse photoelectric (Q) 20 KeV effect (C) De-excitation of Be+3 (R) 54 eV atom from second excited state (S) 0.1 eV (D) Kα – X-ray photons of molybdenum Z = 42 O CH2–CHCOH (A) HN + ⊕ NH3 NH O – CH2CHC O (B) HN + + NH NH3 O CH2CHC (C) N 22. + NH – O NH3 O µ1 µ2 60° 30° (A) µ1 µ1 µ2 µ2 θc (B) Column I (Α) θc 1 (Β) sin–1 3 (C) Refractive index of 1 with respect to 2 (D) Total internal reflection XtraEdge for IIT-JEE – CH2CHC O (D) N α β (C) α > θc The correct order of M–C π bond and strength in given metal carbonyl is(A) [Fe(CO)4]2– > [Co(CO)4]– > [Ni(CO)4] (B) [Ni(CO)4] > [Co(CO)4]– > [Fe(CO)4]2– (C) [Fe(CO)4]2– > [Ni(CO)4] > [Co(CO)4]– (D) [Ni(CO)4] > [Co(CO)4]– = [Fe(CO)4]2– 4. Compound 'A' (molecular formula C 3H 8O) is treated with acidified potassium dichromate to form a product B (molecular formula C3H6O). 'B' form a shining silver mirror on warming with ammonical silver nitrate 'B' when treated with an aqueous solution of H2NCONHNH2.HCl and sodium acetate gives a product 'C'. Identify the structure of C. (Q) Critical angle (S) α = β 62 NH2 3. Column II (P) 45° 1 (R) 3 NH JANUARY 2010 (A) CH3C NCONHNH2 + CH3 (B) CH3CH2CH N NHCONH2 (C) CH3 C NNHCONH2 9. The product 'A' is (A) Me CH3 (D) CH3CH2CH = NCONHNH2 5. (C) Ph P + n a (V – nb) = nRT 2 V 2 Q Q (D) (C) T 6. T A solution containing NaOH and Na2CO3 was titrated against HCl using phenolphthalein as an indicator. The tire value of HCl solution was found to be x ml. At the end point, methyl orange was added and the titration continued. A further y ml of HCl solution was required to get the end point with methyl orange. The volume of HCl solution used with Na2CO 3 during the whole process is (A) 2x (B) 2y (C) x (D) y – x 7. For crystallisation of a solid from the aqueous solution, if the values of ∆H and ∆S are –x J mol–1 and – y J K–1 mol–1 respectively, which of the following relationships is correct (A) x = T × y (B) x > T × y (C) x < T × y (D) None of these 8. Reduction of but-2-yne with Na and liquid NH3 gives an alkene which upon catalytic hydrogenation with D2/Pt gives an alkane. The alkene and alkane formed respectively are (A) cis but-2-ene and racemic-2, 3-dideuterobutane (B) trans but-2-ene and meso 2, 3-dideuterobutane (C) trans but-2-ene and racemic 2, 3-dideuterobutane (D) cis but-2-ene and meso 2, 3-dideuterobutane XtraEdge for IIT-JEE Ph OH H Me (D) H Ph 10. The following complexes are given (1) trans – [Co(NH3)4Cl2]+ (2) cis – [Co(NH3)2 (en)2] 3+ (3) trans – [Co(NH3)2(en)2]3+ (4) NiCl42– (5) TiF62– (6) CoF63– Choose the correct code : (A) (1), (2) are optically active, (3) is optically inactive (B) (2) is optically active, (1), (3) are optically inactive (C) (4), (6) are coloured and (5) is colourless (D) (4) is coloured and (5), (6) are colourless (B) Q (B) Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. T T OH OH ab with temperature is Plot of quantity Q = a+b (A) Ph O Vander Waal's equation for a real gas is Q 1. AlCl3 A 2. H+/H2O O 11. Cellulose is made of glucose units joined together by β-1, 4-glycosidic linkages. These molecules are held by(A) ionic bonds (B) hydrogen bonds and van der Waal's forces (C) weak van der Waal's forces only (D) All of these 12. Which of the following statements is/are correct. ? (A) The monoatomic gas He has lower entropy than the triatomic gas CO2, which has lower entropy than gaseous benzene (B) For single atoms, the absolute entropy increases as the number of electrons and the protons increases (C) Among the solid elements the absolute entropy generally increases as the atomic number increases (D) All metallic solids have entropies below 85 J. mole–1K–1 13. In the Libermman test for phenols, the blue or green colour produced is due to the formation of - 63 (A) OH OH (B) O = = NOH JANUARY 2010 (C) O = =N– –+ – ONa (D) O = =N– – OH 15. Aldehydes with H at α-carbon do not undergo this type of reaction O R 16. Identify the products in the following reactions ? O D C (i) O (A) D C (B) H H O (C) H (ii) R C R C (D) D C R C O CH3OH H O CH2DOH D C O– CH2DOH D C O– CD3OH O CCl3 H C C H (ii) O H C H Cl (iv) (iii) The reaction is not given by(A) (i), (ii) (B) (ii), (iii) (C) (iii), (i) (D) only (iii) – O OH + R C H Passage : II (No. 18 to 20) Equilibrium constant are given (in atm) for the following reaction at 0ºC : SrCl2. 6H2O(s) SrCl2.2H2O(s) + 4H2O(g) : Kp = 5 × 10–12 Na2HPO4.12H2O(s) NaSO4.7H2O(s) +5H2O(g) : CH2O O C O– CH2DOH C CH2 H R O– CH2DOH C O – O O OH + R C O– CH2DOH C O – (i) H O C C H H OH (iii) R C (iv) 17. Consider the following aldehydes : O O O H + R O– CH2D–OH D O OH O (iii) O O OH H + OH – C (ii) O R C O + R CH2OH All non-enolisable aldehydes undergo such type of ractions in strongly basic medium. These type of reactions are disproportionation reaction in which one molecule gets oxidised and other reduced. – O O (i) R O + OH → acid salt(iii) + alcohol(iv) C 2D Passage : I (No. 15 to 17) O O H+R D + OH → acid salt(i) + alcohol(ii) C 2H This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. C H OH (no reaction) C because (A) aldehyde is enolised in basic condition (B) bond energy of C–H increase (C) steric hinderance increase (D) all the above 14. Which of the following statements are correct for SN2 reaction. (A) Increasing the polarity of solvent causes a large increase in the rate of SN2 attack by NH3 on alkyl halide (B) Increasing the polarity of solvent causes a large decrease in the rate of SN2 attack by OH– ion on trimethyl sulphonium ions (C) Increasing the polarity of solvent causes a small decrease in the rate of S N2 attack by trimethyl amine on trimethyl sulphonium ion (D) Increasing the polarity of solvent causes a large increase in the rate of S N2 attack by OH– ion on trimethyl sulphonium ion R CH2 O + R CH2OH Kp = 2.43 × 10–13 XtraEdge for IIT-JEE 64 JANUARY 2010 Na2SO 4.10H2O(s) Na2SO4 (s) + 10H2O(g) : MATHEMATICS Kp = 1.024 × 10 –27 The vapour pressure of water at 0ºC is 4.56 torr. 18. 19. Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Which is the most effective drying agent at 0ºC ? (A) SrCl2.2H2O (B) Na2HPO4.7H2O (C) Na2SO4 (D) all equal 1. At what relative humidities will Na2SO4.10H2O lose water of hydration when exposed to air at 0ºC? 20. (A) above 33.33% (B) below 33.33% (C) above 66.66% (D) below 66.66% z1 and z2 lie on a circle with centre at the origin. The point of intersection z3 of the tangents at z1 and z2 is given by (A) 1 ( z1 + z 2 ) 2 (C) 1 2 At what humidities will Na2SO4 absorb moisture when exposed to air at 0ºC ? (A) above 33.33% (C) above 66.66% 2. (B) below 33.33% (D) below 66.66% 3. 2z1z 2 z1 + z 2 (D) z1 + z 2 z1z 2 The set of all x satisfying the equation x log 3 x This section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows : P Q R S A P Q R S B P Q R S 1 1 + z1 z 2 (B) 2 + (log 3 x ) 2 −10 = 1/x2 is (A) {1, 9} (B) {1, 9, 1/81} (C) {1, 4, 1/81} (D) {9, 1/81} Value of S=1×2×3×4+2×3×4×5 + … + n (n + 1) (n + 2) (n + 3) is - C P Q R S D P Q R S (A) 1 n (n + 1) (n + 2) (n + 3) (n + 4) 5 (B) 1 n+3 ( C5) 5! (C) 1 n+4 ( C4) 5 (D) none of these −1 21. Column –I (A) NH3 → NO3– (B) Fe2S3 → 2FeSO4 + SO2 (C) KMnO4 in acidic medium (D) CuS → CuSO4 Column-II (P) M/20 (Q) M/5 (R) M/8 (S) M 22. Column-I (A) CuCl2.2H2O Column-II (P) Colourless and diamagnetic (Q) Green coloured and paramagnetic (R) Calamine (S) Black in colour, basic in nature (B) Cu2Cl2 (C) CuO (D) ZnCO 3 XtraEdge for IIT-JEE 4. tan θ − tan θ 1 1 a − b = If , θ − θ tan 1 tan 1 b a then (A) a = b = 1 (B) a = cos 2θ, b = sin 2θ (C) a = sin 2θ, b = cos 2θ (D) a = 1, b = sin 2θ 5. A natural number x is chosen at random from the first one hundred natural numbers. The probability ( x − 20)( x − 40) < 0 is that x − 30 (A) 1/50 (B) 3/50 (C) 3/25 65 (D) 7/25 JANUARY 2010 6. Two flagstaffs stand on a horizontal plane. A and B are two points on the line joining their feet and between them. The angles of elevation of the tops of the flagstaffs as seen from A are 30º and 60º and as seen from B are 60º and 45º. If AB is 30 m, then distance between the flagstaffs in metres is - 12. (A) 30 + 15 13. (A) K = –1/4 (C) M = – 2 (D) 60 + 15 3 (C) 60 – 15 3 7. M cot x + C then - (B) 45 + 15 3 3 If g(x) is a polynomial satisfying g(x) g(y) = g(x) + g(y) + g(xy) – 2 for all real x and y and g(2) = 5 then lim g(x) is - 14. 8. 9. 1+ x 2 −1 , then x (A) y′ (1) = 1 (B) y′ (1) = 1/4 (D) y′ (1) does not exist x, y, z ∈ {1, 2, …, n, n + 1} is - (C) 12 + 22 + … + n2 1 n (n + 1) (2n + 1) 6 (D) 2(n+2C3) – n+1C2 1 1 If f(x) = cos 2 x cos 2 x cosec 2 x then 1 π/4 (A) cos 2 x cot 2 x 17. Let m, n ∈ N and p(x) = 1 + x + … + xm. p(x) will divide p(xn) if (A) hcf (m, n) = 1 (B) hcf (m + 1, n) = 1 (C) hcf (m, n + 1) = 1 (D) hcf (m+1, n+1) = 1 1 ∫ f ( x)dx = 16 (3π + 8) −π / 4 (B) f ′ (π/2) = 0 (C) Maximum value of f(x) is 1 (D) Minimum value of f(x) is 0 XtraEdge for IIT-JEE (D) none of these 16. Sum of the rational roots of 133x − 78 x5 = 133 − 78x is (A) 2/9 (B) 9/2 (C) 13/6 (D) 6/13 sec 2 x 11. x3/2 15. If p(x) is a reciprocal polynomial of odd degree, then one of the roots of p(x) = 0 is (A) – 1 (B) 1 (C) 0 (D) (n + 1)/2 The number of ways of choosing triplets (x, y, z) such that z ≥ max {x, y} and (B) x3/2 3 a A polynomial p(x) = a0 xn + a1 xn–1 + … + an–1 x + an, a0 ≠ 0 is said to be a reciprocal equation if ai = an – i for 0 ≤ i ≤ [n/2] where [x] denote the greatest integer ≤ x. Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. C3 3 a −2 Passage : I (No. 15 to 17) (B) 2 (D) none of these n+2 2 (B) y + C = This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. The degree of the differential equation satisfying (A) n+1C3 + (B) a multiple of π/2 (D) a multiple of π The orthogonal trajectories of the system of curves (C) y + C = 1 − x 2 + 1 + y 2 = a(x – y) is - 10. 0 (A) 9a(y + C)2 = 4x3 If y = tan –1 (A) 1 (C) 3 π/2 dx is 1 + tan x (A) a multiple of π/4 (C) equal to π/4 ∫ 2 (B) 25 (D) none of these (C) y′ (1) = 0 (B) L = 2 (D) none of these dy = a/x are dx x →3 (A) 9 (C) 10 ∫ If I = sec 2 x cosec4 x dx = K cot3 x + L tan x + 66 JANUARY 2010 Passage : II (No. 18 to 20) This section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows : P Q R S A P Q R S B P Q R S A straight line is called an asymptote to the curve y = f(x)( if the distance from the variable point M of the curve to the straight line approaches zero as the point M recedes to infinity along some branch of the curve. We have three kinds of asymptotes; vertical, horizontal and inclined. There are three types of asymptotes vertical, horizontal and inclined. Vertical asymptotes If at least one of lim f(x) or x →a + C P Q R S D P Q R S lim f(x) is equal to infinity then x = A is a vertical x →a − asymptote. If lim f(x) = A then y = A is a x →± a 21. f ( x) =k2, lim x →± a x x →± a [f(x) – kx] = b 2 then y = k 2x + b 2 is an inclined asymptote. horizontal asymptote. If limits lim 18. Let y = Column-II (A) locus of point of intersection (P) of the lines x = at2, y = 2at (B) locus of the point of (Q) intersection of the perpendicular tangents to the circle x 2 + y2 = a2 (C) locus of the point of (R) intersection of the lines x cos θ = y cot θ = a (D) The locus of the mid (S) points of the chords of the circle x2 + y2 – 2ax = 0 passing through the origin 3x + 3x be curve 1 and y = xe1/x be curve 2 x −1 then (A) curve 1 has no horizontal asymptote and curve 2 has no vertical asymptote (B) y = 3x + 3 and y = x + 2 inclined asymptotes to curve 1 curve 2 (C) y = 3x + 3 and y = x + 1 are inclined asymptotes to curve 1 and curve 2 22. (D) y = x + 1 and y = 3x + 3 are inclined asymptotes to curve 1 and curve 2 19. Let y = Column-I Column-I (A) f(x) = 5x then x −3 x2 + y2 = 2a2 y2 = 4ax x2 + y2 = ax x2 – y2 = a2 Column-II ( x + 3) 2 x 2 +1 , (P) 0 ≤ f(x) ≤ 3 –∞<x<∞ (B) R = {(x, y): x, y ∈ R, (A) There are no vertical asymptotes 2 (Q) 3 ≤ f(x) ≤ 9 2 x + y ≤ 25} (B) There are no horizontal asymptotes R′ = {(x, y): x, y ∈ R, (C) There are no inclined asymptotes y ≥ 4x2/9} (D) x = 3 and y = 5 are the only asymptotes and let (x, f(x)) = R ∩ R′ 2 20. Let y = x + 1 sin 1/x (C) f(x) = (A) There are no horizontal asymptotes 9 2 − cos 3x (S) 0 ≤ f(x) ≤ 5 (D) f(x) = (B) There is only one horizontal asymptote (C) There are only two horizontal asymptotes 3 2 sin (R) 0 ≤ f(x) ≤ 10 ( π 2 / 16) − x 2 (D) There is one vertical asymptote XtraEdge for IIT-JEE 67 JANUARY 2010 Based on New Pattern IIT-JEE 2011 XtraEdge Test Series # 9 Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 10 to 14 are multiple choice questions with multiple (one or more than one) correct answer. +4 marks and -1 mark for wrong answer. • Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer. Section - II • Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly matched answer and No Negative marks for wrong answer. However, +1 marks will be given for a correctly marked answer in any row. y PHYSICS Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. L Two blocks A and B, have equal masses. They are x placed next to each other on a horizontal frictionless fixed table. Compare the velocities of L the blocks as each of them reaches the opposite end of the table – A 4F B F (A) vA = 2 vB (C) vA = 8 vB 2. A B XtraEdge for IIT-JEE 68 L L (C) , 3 3 L L (D) , 3 6 A body of mass 1 kg starts moving from rest at t = 0, in a circular path of radius 8 m. Its kinetic energy varies as a function of time as : KE = 2t2 Joules, where t is in seconds. Then (A) Tangential acceleration = 4 m/s2 (B) Power of all forces at t = 2 sec is 8 watt (C) First round is completed in 2 sec (D) Tangential force at t = 2 sec is 4 newton 4. With what minimum velocity should block be projected from left end A towards end B such that it reaches the other end B of conveyer belt moving with constant velocity v. Friction coefficient between block and belt is µ . Centre of mass of two thin uniform rods of same of co-ordinates – 2L L , (B) 3 2 3. (B) vA = 4 vB (D) vA = 16 vB length but made up of different materials and kept as shown, can be, if the meeting point is the origin L L (A) , 2 2 JANUARY 2010 A v0 m touching (event 1) or make an elastic head-on collision (event 2) (A) we can never make out which event has occurred (B) we can not make out which event has occurred only if v 1 = v 2 (C) we can always make out which event has occurred (D) we can make out which event has occurred only if v1 = v2 B µ v L 5. (A) µgL (B) 2µgL (C) 3µgL (D) 2 µgL 9. A block of mass m is attached to a pulley disc of equal mass m, radius r by means of a slack string as shown. The pulley is hinged about its centre on a horizontal table and the block is projected with an initial velocity of 15 m/s. Its velocity when the string becomes taut will be – The escape velocity from the earth is about 11 km/s. The escape velocity from a planet having twice the radius and the same mean density as the earth is (A) 22 km/s (B) 11 km/s (C) 5.5 km/s (D) 15.5 km/s Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. (A) 5 m/s (C) 7.5 m/s (B) 6 m/s (D) 10 m/s 6. A beaker containing water is placed on the platform of a spring balance. The balance reads 1.5 kg. A stone of mass 0.5 kg and density 500 kg/m3 is completely immersed in water without touching the walls of beaker. Now the balance reading will be (A) 2 kg (B) 1 kg (C) 2.5 kg (D) 3 kg 7. A uniform rod of mass M1 is hinged at its upper end. A particle of mass M2 moving horizontally strikes the rod at its mid point elastically. If the particle comes to rest after collision, the value of M1 is – M2 10. The potential energy of a particle of mass 0.1 kg, moving along x-axis is given by U = 5x (x – 4) J, where x is in meters. It can be concluded that (A) The particle is acted upon by a constant force (B) The speed of the particle is maximum at x=2m (C) The speed of the particle is maximum at x=0m (D) The period of oscillation of the particle is π sec 5 11. A particle of mass m is at rest in a train moving with constant velocity with respect to ground. Now the particle is accelerated by a constant force F0 acting in the direction of motion of train for time t0. A girl in the train and a boy on the ground measure the work done by this force. Which of the following are incorrect ? (A) Both will measure the same work (B) Boy will measure higher value than the girl (C) Girl will measure higher value than the boy (D) Data are insufficient for the measurement of work done by the force F0 12. In figure, two blocks M1 and m2 are tied together with an inextensible and light string. The mass M1 is placed on a rough horizontal surface with coefficient of friction µ and the mass m2 is hanging vertically against a smooth vertical wall. The pulley is frictionless – v M2 (A) 8. 3 4 M1 4 (B) 3 (C) 2 3 (D) 3 2 Two identical spheres move in opposite directions with speeds v1 and v2 and pass behind an opaque screen, where they may either cross without XtraEdge for IIT-JEE 69 JANUARY 2010 B M1 1.5 m Rough (µ) A Smooth C m2 1m 4m D Choose the correct statement (s) related to the tension T in the string (A) When m2 < µM1, T = m2g 15. (B) When m2 < µM1, T = M1g (C) When m2 > µM1, µM1g < T < m2g (D) When m2 > µM1 , m2g < T < µM1g 13. 14. Overall changes in volume and radii of a uniform cylindrical steel wire are 0.2% and 0.002 % respectively when subjected to some suitable force. If Young’s modulus of elasticity of steel is Y = 2.0 × 1011 N/m2, then (A) Longitudinal tensile stress acting on the wire is 4.08 × 108 N/m2 (B) Longitudinal tensile stress acting on the wire is 3.92 × 108 N/m2 (C) Longitudinal strain is 0.204 % (D) Longitudinal strain is 0.196 % (A) 4 5 m/s (B) (C) 5 2 m/s (D) 10 m/s 130 m/s 16. Gauge pressure at the highest point B is (A) – 52 kPa (B) – 44 kPa (C) – 20 kPa (D) – 12 kPa 17. Gauge pressure at point C is (A) – 32 kPa (B) 8 kPa (C) 20 kPa (D) 0 Passage : II (No. 18 to 20) Two pulse are traveling in opposite direction with speed 1 m/s. Figure shows the shape of pulse at t = 0. y A particle falls freely near the surface of the earth. Consider a fixed point O (not vertically below the particle) on the ground (A) Angular momentum of the particle about O is increasing (B) Torque of the gravitational force on the particle about O is decreasing (C) The moment of inertia of the particle about O is decreasing (D) The angular velocity of the particle about O is increasing 1 m/s 10 mm 5mm x – 5mm 2 4 6 8 10 12 14 16 Distance (in cm) This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 15 to 17) A siphon tube is discharging a liquid of specific gravity 0.8 from a reservoir as shown in figure. (Take g = 10 m/s2). XtraEdge for IIT-JEE Velocity of the liquid coming out of siphon at D is - 70 18. Speed of particle at x = 2 cm and t = 0 is (A) 1 m/s (B) 0.75 m/s (C) 0.5 m/s (D) 0.25 m/s 19. Displacement of particle at x = 8 cm and t = 6 sec is (A) 10 mm (B) 5 mm (C) – 5 mm (D) zero 20. Speed of particle at x = 8 cm and t = 6 sec (A) zero (B) 0.125 m/s (C) 0.25 m/s (D) None of these JANUARY 2010 This section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows : P Q R S A P Q R S B P Q R S Column – I (A) (B) → v1 (P) = î + ĵ , Acceleration of block → Normal force by 22. Net force on the wedge = – î – ˆj log P A(vA) log V (Q) >θ (A) number of collision increases (R) θ 2. A compound containing only sodium, nitrogen and oxygen has 33.33% by weight of sodium. What is the possible oxidation number of nitrogen in the compound? (A) –3 (B) + 3 (C) –2 (D) + 5 3. How many moles of nitrogen is produced by the oxidation of one mole of hydrazine by 2/3 mole bromate ion 1 2 (A) (B) 1 (C) 1.5 (D) 3 3 4. For the reaction (S) 2 y BaSO4 (s) 3 BaSO4 (aq) The equilibrium moles of BaSO4(aq) were 0.2 moles. The equilibrium constant of the above x 4 XtraEdge for IIT-JEE VA times VB (B) number of moles in this process is constant (C) it is isothermal process (D) it is possible for ideal gas A rigid cylinder is kept on a smooth horizontal surface as shown. If column I indicates velocities of various points (3-centre of cylinder, 2-top point, 4-bottom point, 1-on the level of 3 at the rim) on it shown. Choose correct state of motion from column – II. 1 (S) Not possible A compression of an ideal gas is represented by curve AB, which of the following is wrong B(vB) θ block on wedge (D) (R) Rolling without slipping to right Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. A relative to wedge (C) slipping to left = – î A relative to ground (B) (Q) Rolling without CHEMISTRY 1. Acceleration of the block centre → θ (A) (P) Pure rotation about = 2 î (D) v 4 = 0, A Column-II = î + ĵ , → In the situation shown, all surfaces are frictionless and triangular wedge is free to move. In column II the direction of certain vectors are shown. Match the direction of quantities in column I with possible vector in column II. Column-I Column – II (C) v 2 = î , v 3 = 0 C P Q R S D P Q R S 21. → v1 → v2 → v1 → v3 reaction is 71 JANUARY 2010 (A) 0.2 (C) 2 × 10 –4 mol L–1 5. (B) 0.2 mol L–1 (D) Data insufficient 12. The order of Keq values for the following keto-enol equilibrium constants is k CH3 CHO 1 CH2 CH OH O O k2 CH3 C CH2 C CH3 OH O Which of the following correctly explain the nature of boric acid in aqueous medium 2O (A) H3BO3 H → H3O + + H2BO3– 2 H 2O (B) H3BO3 → 2H3O + + HBO 32– CH3 O 3– H 2O (C) H3BO3 3 → 3H3O+ + BO 3 CH3 – 2O (D) H3BO3 H → B(OH)4 + H+ 6. 7. (B) Square planar (D) See-Saw CH3 C (B) K2 > K3 > K1 (D) K1 > K3 > K2 the pure Zr is deposited on W passed over (B) 2B + 3I2 → 2BI3(g) the white → hot W the pure B is deposited on W mixed with W (C) Zr + 2I2 → ZrI4 (s) & then → heated ZrI4 is reduced to ZrI2 The correct order of increasing boiling point is (A) NH3 > HF > H2O (B) H2O > HF > NH3 (D) none of these (D) HF > H2O > NH3 14. Which of the following statements is correct ? (A) At 273ºC, the volume of a given mass of a gas at 0ºC and 1 atm. pressure will be twice its volume (B) At –136.5ºC, the volume of a given mass of a gas at 0ºC and 1 atm. pressure will be half of its volume (C) The mass ratio of equal volumes of NH3 and H2S under similar conditions of temperature and pressure is 1 : 2 (D) The molar ratio of equal masses of CH4 and SO2 is 4 : 1 Oxidation states of carbon and nitrogen in KCN are, respectively (A) – 3, + 2 (B) + 2, – 3 (C) + 1, – 2 (D) zero each Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 10. Which of the following samples of reducing agents is/are chemically equivalent to 25 mL of 0.2 N KMnO4, to be reduced to Mn2+ + H2O ? (A) 25 mL of 0.2 M FeSO4 to be oxidized to Fe3+ (B) 50 mL of 0.1 MH3AsO3 to be oxidized to H3AsO 4 (C) 25 mL of 0.2 M H2O2 to be oxidized to H+ and O2 (D) 25 mL of 0.1 M SnCl2 to be oxidized to Sn4+ This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 15 to 17) Entropy is measure of degree of randomness. Entropy is directly proportional to temperature. Every system tries to acquire maximum state of randomness or disorder. Entropy is measure of unavailable energy. Unavailable energy = Entropy × Temperature The ratio of entropy of vapourisation and boiling point of substance remains almost constant. 11. A sample of water has a hardness expressed as 77.5 ppm Ca2+. This sample is passed through an ion exchange column and the Ca2+ is replaced by H+. Select correct statement(s) (A) pH of the water after it has been so treated is 2.4 (B) Every Ca2+ ion is replaced by one H+ ion (C) Every Ca2+ ion is replaced by two H+ ions (D) pH of the solution remains unchanged XtraEdge for IIT-JEE CH3 passed over (C) NH4+ < HF < H3O+ < H2O < OH– (D) H3O+ < NH4+ < HF < OH– < H2O 9. CH2 C (A) Zr + 2I2 → ZrI4(g) the white → hot W Arrange NH4+, H2O, H3O +, HF & OH– in increasing (C) NH3 > H2O > HF k3 CH 13. In the purification Zr and B, which of the following is/are true ? order of acidic nature (A) OH– < H2O < NH4+ < HF < H3O+ (B) H3O + > HF > H2O > NH4+ > OH– 8. CH3 (A) K1 > K2 > K3 (C) K2 > K1 > K3 The shape of TeCl4 is (A) Linear (C) Tetrahedral C C OH 72 JANUARY 2010 15. Which of the following process have ∆S = – ve ? (A) Adsorption (B) Dissolution of NH4Cl in water (C) H2 → 2H (D) 2NaHCO3(s) → Na2CO3 + CO2 + H2O 16. Observe the statement(s) graph and identify the ⇒ log 18. correct Entropy Β ∆Svap 19. Α ∆Sfusion 1 1 – T1 T2 If standard heat of dissociation of PCl5 is 230 Cal. 1 then the slope of the graph of log K vs is T (A) + 50 (B) – 50 (C) 10 T1 T2 Temperature (A) T 1 is melting point, T2 is boiling point (B) T 1 is boiling point, T2 is melting point (C) ∆ Sfusion is more than ∆ S vap (D) T 2 is lower than T1 (D) None of these For exothermic reaction of ∆S o < 0 then the 1 sketch of log K vs may be T log K log K (A) (B) 1/T 1/T log K log K (C) (D) 1/T 17. The law of Thermodynamics invented by Nernst, which helps to determine absolute entropy is (A) Zeroth law (B) 1 st law nd (D) 3 rd law (C) 2 law 20. Passage : II (No. 18 to 20) Effect of temperature on the equilibrium process is analysed by using the thermodynamics. From the thermodynamics relation ∆G° = – 2.3 RT logK........(1) ∆ G° = Standard free energy change ∆G° = ∆H° – T ∆S°….(2) ∆H° = Standard heat of the reaction From (1) & (2) – 2.3 RT log K = ∆H° – T ∆S° ; ∆S° : Standard Entropy change, ∆H ° ∆S° + ........(3) log K = 2.3RT 2.3R Clearly if a plot of log K vs 1/T is made then it is – ∆H° & a straight line having slope = 2.3R ∆S° . y–intercept = 2.3R If at a temperature T1 equilibrium constant be K1 and at temperature T2 equilibrium constant be K2 then, the above equation reduces to : – ∆H ° ∆S° ⇒ log K1 = + ........(4) 2.3RT1 2.3R 1/T If for a particular reversible reaction if Kc = 57 at 355°C and Kc = 69 at 450°C then (A) ∆H < 0 (B) ∆H > 0 (C) ∆H = 0 (D) ∆H sign can't be determined This section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows : P Q R S A P Q R S B P Q R S C P Q R S D P Q R S 21. Match the temperature (in column I) with its value (in column II) Column –I Column II (A) Critical temperature (P) a/Rb (B) Boyle temperature (Q) 2a/Rb (C) Inversion temperature (R) T/Tc (D) Reduced temperature (S) 8a/27Rb ∆S° – ∆H ° + ........ (5) 2.3RT2 2.3R Subtracting (4) from (5) we get. ⇒ log K2 = XtraEdge for IIT-JEE ∆H ° K2 = K1 2.3R 73 JANUARY 2010 22. Match the half-reaction (in column I) with equivalent mass (molar mass = M) (in column II) Column –I Column II (A) Cr2O72– → Cr3+ (P) M (B) C2O42– → CO2 (Q) M/2 (C) MnO4– → MnO2 (R) M/6 (D) HC2O 4– → C2O42– (S) M/3 (A) 3(x – 21) = 3y + 92 = 3z – 32 MATHEMATICS 8. Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. 2. 3. 4. 5. 6. (B) 3 (C) 2 (D) 1 If log3 x + log3 y = 2 + log3 2 and log3 (x + y) = 2 then - 9. (C) x − 21 y − (92 / 3) z + (32 / 3) = = 1/ 3 1/ 3 1/ 3 (D) x − 2 y + 3 z −1 = = 1/ 3 1/ 3 1/ 3 A vector c, directed along the internal bisector of the angle between the vectors a = 7i – 4j – 4k and (A) 5 (i – 7j + 2k) 3 (B) 5 (5i + 5j + 2k) 3 (C) 5 (i + 7j + 2k) 3 (D) 5 (–5i + 5j + 2k) 3 If a and b are two unit vectors such that a + 2b and (A) x = 1, y = 8 (B) x = 8, y = 1 5a – 4b are perpendicular to each other then the (C) x = 3, y = 6 (D) x = 9, y = 3 angle between a and b is (A) 45º The exponent of 7 in 100C50 is (A) 0 (B) 2 (C) 4 (D) none of these –1 (C) cos (1/3) The equation cos 2x + a sin x = 2a – 7 possesses a solution if (A) a < 2 (B) 2 ≤ a ≤ 6 (C) a > 6 (D) a is any integer 10. In a triangle ABC, 2a2 + 4b2 + c2 = 4ab + 2ac, then numerical value of cos B is equal to (A) 0 (B) 1/8 (C) 3/8 (D) 7/8 11. (D) cos–1 (2/7) Suppose a, b, c are positive integers and f(x) = ax2 – bx + c = 0 has two distinct roots in (0, 1), then (A) a ≥ 5 (B) b ≥ 5 (C) abc ≥ 25 (D) abc ≥ 250 The coefficient of xk (0 ≤ k ≤ n) in the expansion of E = 1 + (1 + x) + (1 + x)2 + … + (1 + x)n is - The tangent at the point P(x1, y1) to the parabola y2 = 4ax meets the parabola y2 = 4a (x + b) at Q and R, the coordinates of the mid-point of QR are - (A) n+1Ck+1 (C) (A) (x1 – a, y1 + b) 12. (C) (x1 + b, y1 + a) (D) (x1 – b, y1 – b) n+1 Cn–k (B) nCk (D) nCn–k–1 The equation of a tangent to the hyperbola 3x2 – y2 = 3, parallel to the line y = 2x + 4 is (A) y = 2x + 3 (B) y = 2x + 1 Equation of the line of shortest distance between the (C) y = 2x – 1 x y z x − 2 y −1 z + 2 lines = is = = = and −5 2 −3 1 3 2 (D) y = 2x + 2 XtraEdge for IIT-JEE (B) 60º Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. (B) (x1, y1) 7. x − (62 / 3) y − 31 z + (31 / 3) = = 1/ 3 1/ 3 1/ 3 b = –2i – j + 2k, with |c| = 5 6 , is - If iz3 + z2 – z + i = 0, then |z| equals (A) 4 (B) 74 JANUARY 2010 13. The plane passing through the point (–2, –2, 2) and containing the line joining the points (1, 1, 1) and (1, –1, 2) makes intercepts of lengths a, b, c respectively on the axes of x, y and z respectively, then - Passage : II (No. 18 to 20) (A) a = 3b (B) b = 2c (C) a + b + c = 12 (D) a + 2b + 2c = 0 18. The value of x for which f(x) = 0 is (A) –1/2 (B) 0 (C) 1/2 f(x) = sin {cot–1 (x + 1)} – cos (tan–1 x) a = cos tan–1 sin cot–1 x b = cos (2 cos–1 x + sin–1 x) (D) 1 2 14. 19. If f(x) = 0 then a is equal to (A) 1/2 (B) 2/3 (C) 5/9 (D) 9/5 Let a = 4i + 3j and b be two vectors perpendicular to each other in xy-plane. The vectors c in the same plane having projections 1 and 2 along a and c are (A) – 2 11 i+ j 3 2 (B) 2 i – j (C) – 2 11 i+ j 5 5 (D) 20. If a2 = 26/51, then b 2 is equal to (A) 1/25 (B) 24/25 (C) 25/26 (D) 50/51 2 11 i+ j 3 2 This section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows : P Q R S A P Q R S B P Q R S C P Q R S This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 15 to 17) For k, n ∈ N, we define B(k, n) = 1.2.3 … k + 2.3.4 …(k + 1) + … + n(n + 1) … (n + k – 1), So (n) = n and S k(n) = 1k + 2k + … + nk To obtain value of B(k, n), we rewrite B(k, n) as follows: B(k, n) = k! kk + kk+1 + kk+ 2 + ... n +kk −1 [( ) ( ) ( ) ( = k! ( ) )] D P n+k k +1 21. n (n + 1).....(n + k ) = k +1 (A) B(2, n) 1 (B) B(2, n) 2 1 (C) B(2, n) 6 (D) none of these (B) 17. (A) B(3, n) (B) B(3, n) – 2B(2, n) (C) B(3, n) – 2B(1, n) (D) B(3, n) + 2B(1, n) z + 3 + 4i (D) arg z + 5 − 2i 22. ( ) S (n) + ( ) S (n) + … + ( ) S (n) + ( ) S (n) equals k k–1 k +1 k (A) (n + 1)k (C) nk – (n – 1)k XtraEdge for IIT-JEE 1 k +1 k +1 z −1 =2 z +1 (C) z z – (1 + i)z – (1 – i) z + 7 = 0 16. S 3(n) + 3S 2(n) equals - k +1 2 Centre of circle Column-I (A) |z – 2|2 + |z – 4i|2 = 20 15. S 3(n) + S1(n) equals - k +1 1 Q R S 0 (B) (n + 1)k – 1 (D) (n + 1)k – (n – 1)k (Q) 5/3 + 0i (R) – 4 – i (S) 1 + 2i cos α + cos β = a, sin α + sin β = b Column-I Column-II (A) cos (α + β) (P) 2ab/(a2 + b2) (B) sin (α + β) (C) cos (α – β) (Q) b/a (R) (a2 – b2)/(a2 + b 2) (D) tan 75 Column-II (P) 1 – i α+β 2 (S) (a2 – b2 – 2)/2 JANUARY 2010 MOCK TEST PAPER-2 CBSE BOARD PATTERN CLASS # XII SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS S OLU TIONS WI LL BE PU BLI SH ED IN N EXT I S SU E 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 1. 2. 3. Write the name of the quantity whose SI Unit is Amp m–1. d → → ∫ B . ds → Here B = magnetic field strength. → ds = small area vector over a closed surface. 4. 5. If an electron is having the energy of 10 eV then find its De-Broglie wavelength. 6. Why the 'CORE' of the transformer is laminated. 7. Draw the symbol of photodiode. 8. Write the name of maxwell's fourth equation for electro-magnetic waves. 9. G/D is known as ground current detector. Find the reading of it for the given circuit diagram +3v +3v +3v If a thin foil of metal, parallel to capacitor plates get introduced between the two plates of air capacitor then write the effect on capacitance C of air capacitor. XtraEdge for IIT-JEE C = ε0A/d b a Name the material for which magnetic susceptibility is negative. Write the value of A A 1Ω G/D 76 JANUARY 2010 10. α particle, β particle and Deutron are placed on the vertices of an equilateral triangle of side 'a' find the electric potential energy of the given system. Also find the work done to place these particles from equilateral triangle of side 'a' to the equilateral triangle of side '2a'. α particle β particle 15. Find the value of potential difference across tertninals A and B. C C 5V 11. Draw the truth table of the following gates A Y (i) B A B X R R R R 10V A (ii) B 16. If potential difference across terminals X and B is 10V then find the potential difference across A and B. R R R R Deutron a C A 17. If the potential difference across the resistance R = 10 Ω is 100 volt then (i) find the current passing through R–L–C series circuit. (ii) Write the power factor of circuit. Y B R 12. In which case the bulb will glow and why ? (i) If V A = 0 volt (ii) If V A = 3 volt Here V A is the potential of A and B is the bulb +5V L ~ V= 100√2 sin 314t volt B 18. Write Einstein's eqn for photoelectric effect. A photon of energy 5eV falls on the surface a metal whose work function is 2eV. Calculate the value of stopping potential for metal. RC A 19. Appratus of YDSE is shown in figure, the interference fringes are observed on the screen. The relation between x and y is given below y → f(t) : y = xt for 0 < t ≤ 4 sec. 13. Calculate the no. of electric field lines emitted by the stationary proton. 14. An equilateral triangle current carrying coil of side 'b' is placed near the intinite current carrying conductor which is stationary then in which direction the triangular loop will move (towards the conductor or away from the conductor) and why ? x y S2 Find the intensity ratio of maxima to minima on the screen at (i) t = 1s (ii) t = 4s a XtraEdge for IIT-JEE S1 Screen i2 i1 C 77 JANUARY 2010 20. A convex lens is made of two different materials which are having the absolute refractive indices as µ 1 and µ 2, this is placed in a media of ARI µ3 as shown below, then Incident rays Medium - 1 Lens µ1 Medium - 2 µ3 µ3 µ2 Surrounding 24. (i) Find the terminal voltages of Batteries BT-1 and BT-2. (ii) Why terminal voltage of battery BT-1 is more than it's emf and in case of battery BT-2 the terminal voltage is less than it's emf. BT-1 Medium - 3 10V Incident rays (i) Which is the most dense media (ii) Arrange µ1, µ 2 and µ3 in increasing order i Medium -1 air 26. State inconsistency in Ampere's circuital law. What is meant by displacement current? Prove that displacement current is equals to conduction current. Refracted ray ip, polarizing angle for air water 27. Find the electric flux passing through the cube for the given arrangement of charges. (i) Calculate the value of ip if wµ a = 3/4 (ii) Find the value of φ, φ is the angle between reflected ray and refracted ray. (iii)Out of reflected and refracted ray which of the following is plane polarized. (iv)Write the relation between polarizing angle and critical angle for the two media interface. +1C –4C +3C +5C a V0 23. Draw the wave shape of the output signal Y. The wave shapes of inputs A and B are shown. A 1 B t 0 XtraEdge for IIT-JEE –8C 29. Explain the principle, construction and working of Vande Graff Generator. A B 1 +7C 28. What type of feedback is used in transistor Amplifier. Draw the circuit diagram of transistor oscillator and explain it's working. What type of feedback is used in transistor oscillator. Write the expression for frequency of signals generated by transistor oscillator. 10.3V 1Ω 0 –2C -6C 22. Calculate the output voltage V 0 2Ω 7Ω 2Ω 25. Explain the followings (i) Why a metallic spring get shrinked when current is flown through it (ii) Why N-P-N transistor is preferred over P-N-P transistor in electronic industry Medium -2 water Ge. Diode 20V RL = 2Ω Reflected ray ip 1Ω Load Resistance 21. A monochromatic light Ray is incident as shown in figure then Normal incident ray BT-2 30. Explain construction and working of Cyclotron. Why cyclotron can not be used to accelerate the electrons. Y t 78 JANUARY 2010 16. Write the chemical equations for all the steps involved in the rusting of iron. CHEMISTRY 1. What are the co-ordination no. of Ca+2 & F– ions in CaF2 structure ? 2. Name of ionisation isomer of [Cr(H2O)5Br]SO4. 3. Write IUPAC name of following (a) O2N – C6H4 – OCH3(p) 17. Give the reactions : (a) When phenol is treated with excess of Br2 water. (b) Diethyl ether heated with conc. HI. 18. From Fehling's solution, Schiff's reagent, Tollen's reagent & Grignard reagent, which reagent react with both aldehyde & ketone. CH3 CH3 H (b) CH2 = C — C — C — C — OH OH CH3 Br 19. Why 2-pentanone give iodoform test but 3-pentanone not ? CH3 20. Give the order of basicity in the following compound. O 4. Why aromatic ketones are not react with NaHSO3 ? 5. Why sulphanilic are amphoteric in nature ? 6. Give monomer of Glyptal. N 7. Explain terms Antacids. H (I) 8. Why carbohydrates are optically active ? 9. N (II) H (III) N H (IV) 21. 0.2 molal acid HX is 20% ionised in solution, Kf = 1.86 K molality–1. Calculate the freezing point of the solution. + In a compound AX, the radius of A ion is 95 pm and that of X– is 181 pm. Predict the crystal structure of AX and write the co-ordination number of each of the ions. 22. How long a current of 3-ampere has to be passed through a solution of AgNO3 to coat a metal surface of 80 cm2 with a 0.005 mm thick layer. Density of silver is 10.5 g/cm3. 10. MgO has structure of NaCl and TlCl has the structure of CsCl. What are the co-ordination number of ions in MgO & TlCl ? 23. Explain that the rate of physisorption increases with decrease in temperature. 11. What is meant by mole fraction of solute & solvent? 12. A solution containing one mole per litre of each Cu(NO3)2, AgNO3, Hg2(NO3)2 is being electrolyzed by using inert electrodes, the value of standard electrode potential in volts (reduction potential) are 2Hg/Hg2++ = + 0.79 V Ag/Ag+ = 0.80 V, +2 Cu/Cu = 0.34 V, Mg/Mg++ = – 2.37 V with increasing voltage, give the sequence of deposition of metal on cathode ? 24. Explain that boric acid is monobasic & weak lewis acid. 25. Give the reason : (a) VOCl2 & CuCl2 give same colour in aqueous solution. (b) CuSO4 decolourise on addition of KCN. 13. Give the structure of dichromate dianion. 26. Complete the reaction. CHO CHO 14. Which compound is form when excess of KCN is added to aqueous solution of CuSO4. (i) NaOH/100ºC (ii) H+/H2O 15. Give the number & structure of possible enantiomeric pairs that can be produced during monochlorination of 2-methyl butane. XtraEdge for IIT-JEE N CHO 79 CHO JANUARY 2010 27. Aspartame, an artificial sweetner, is a peptide having following structure. CH2 – C6H5 y y x x H2N – CH – CONH – CH – COOCH3 (a) CH2 – COOH (b) (i) Identify the four functional group. (ii) Write the zwitter ionic structure. (iii) Write the structure of the amino acids obtained from the hydrochloride of aspartame. 2. 28. The rate constant for the first order decomposition of a certain reaction is described by the equation 3. A matrix A of order 3 × 3 has determinant 5. What is the value of |3A| ? 4. For what value of x, the following matrix is singular? 2π 2π cos–1 cos + sin–1 sin ? 3 3 1.25 × 10 4 . T (i) What is the energy of activation for this reaction. (ii) At what temperature will its half-life period be 256 minutes. log (k) = 14.34 – 29. 5 − x x + 1 2 4 The Haber process can be represented by following NH3 + H2O B CaCO3 → CaO + CO2 A 5. Find the point on the curve y = x2 – 2x + 3, where the tangent is parallel to x-axis. 6. What is the angle between vectors a & b with NaCl H2O What is the principal value of → magnitude → → 3 and 2 respectively ? Given a . b = 3. NaHCO3 + D C + H2O + CO2 7. Cartesian equations of a line AB are. 2x − 1 4 − y z +1 = = 2 7 2 NH3 + H2O + E Identify A, B, C, D & E. Write the direction ratios of a line parallel to AB. 30. An alkene 'A' on ozonolysis yields acetone and an aldehyde, the aldehyde is easily oxidised to an acid (B). When (B) is treated with Br2 in presence of phosphorus it yields a compound (C) which on hydrolysis give a hydroxy acid (D). This acid can also be obtained from acetone by the reaction with HCN followed by hydrolysis. Identify compound A, B, C, & D. 3 log x (x ) dx 4 Write a value of 9. Write the position vector of a point dividing the line → → b externally in the ratio 1 : 4, → → where a = 2î + 3 ĵ + 4k̂ and b = − î + ĵ + k̂ 2 1 4 10. If A = and B = 4 1 5 Section A 3 − 1 2 2 . 1 3 Write the order of AB and BA. Which one of the following graphs represent the function of x ? Why ? XtraEdge for IIT-JEE ∫e 8. segment joining A and B with position vectors a & MATHEMATICS 1. → 80 JANUARY 2010 1 most. Its semi vertical angles is tan–1 . Water is 2 poured into it at a constant rate of 5 cubic meter per minute. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 10m. Section B 11. Show that the function f : R → R defined by 2x − 1 , x ∈ R is one-one function. Also find f(x) = 3 the inverse of the function f. OR Examine which of the following is a binary operation a+b , a, b ∈ N (i) a * b = 2 a+b , a, b ∈ Q (ii) a * b = 2 for binary operation check the commutative and associative property. 18. Evaluate sum ∫ 2 1 the following integral as limit of 2 (3x − 1) dx 19. Evaluate ∫ π/ 2 0 log sin x dx 20. Find the vector equation of the line parallel to the line x −1 3 − y z +1 = = and passing through (3, 0, –4). 5 2 4 Also find the distance between these two lines. 12. Prove that 63 5 3 tan–1 = sin–1 + cos–1 16 13 5 → → 21. In a regular hexagon ABCDEF, if AB = → → → → → a and → → BC = b , then express CD , DE , EF , FA , AC , → 13. Using elementary transformations, find the inverse of 2 − 6 1 − 2 a 2 + ac b 2 + bc c 2 + bc − ac a 2 + ab b 2 + ab c 2 + ac = (ab + bc + ca)3 − ab 14. Find all the points of discontinuity of the function f defined by x + 2, x ≤1 f(x) = x − 2, 1 < x < 2 0, x≥2 15. If x pyq = (x + y)p+q, prove that → → → 22. A football match may be either won, drawn or lost by the host country's team. So there are three ways of forecasting the result of any one match, one correct and two incorrect. Find the probability of forecasting at least three correct results for four matches. OR A candidate has to reach the examination centre in time. Probability of him going by bus or scooter or by 3 1 3 , , respectively. other means of transport are 10 10 5 1 1 and The probability that he will be late is 4 3 respectively, if he travels by bus or scooter. But he reaches in time if the uses any mode of transport. He reached late at the centre. Find the probability that he travelled by bus. OR Using properties of determinants, prove that − bc → AD , AE and CE in terms of a and b . dy y = dx x OR Find 1 + x 2 + 1− x 2 dy , if y = tan–1 dx 1 + x 2 − 1− x 2 16. Evaluate ( x 2 + 1)( x 2 + 4) ∫ (x 2 + 3)( x 2 − 5) , 0 < |x| < 1 Section C 23. Find the matrix P satisfying the matrix equation 2 1 − 3 2 1 2 3 2 P 5 − 3 = 2 −1 dx 24. Find all the local maximum values and local minimum values of the function π π f(x) = sin 2x – x, – < x < 2 2 17. A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lower XtraEdge for IIT-JEE 81 JANUARY 2010 29. A catering agency has two kitchens to prepare food at two places A and B. From these places 'Mid-day Meal' is to be supplied to three different schools situated at P, Q, R. The monthly requirements of the schools are respectively 40, 40 and 50 food packets. A packet contains lunch for 1000 students. Preparing capacity of kitchens A and B are 60 and 70 packets per month respectively. The transportation cost per packet from the kitchens to schools is given below : OR A given quantity of metal is to be cast into a solid half circular cylinder (i.e., with rectangular base and semicircular ends). Show that in order that the total surface area may be minimum, the ratio of the length of the cylinder to the diameter of its circular ends is π : (π + 2). 25. Sketch the graph of | x − 2 | +2, x ≤ 2 f(x) = 2 x>2 x − 2, Evaluate ∫ 4 0 Transportation cost per packet (in rupees) To f ( x ) dx. What does the value of this integral represent on the graph ? 26. Solve the following differential equation dy – xy = x2, dx given y = 2 when x = 0 (1 – x2) A B P 5 4 Q 4 2 R 3 5 How many packets from each kitchen should be transported to school so that the cost of transportation is minimum ? Also find the minimum cost. 27. Find the foot of the perpendicular from P(1, 2, 3) on the line x −6 y−7 z−7 = = 3 2 −2 MEMORABLE POINTS Also obtain the equation of the plane containing the line and the point (1, 2, 3) 1. 2. 3. 28. Let X denote the number of colleges where you will apply after your result and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that 4. 5. if x = 0 or 1 kx P(X = x) = 2kx , k is +ve constant if x = 2 k(5 – x) if x = 3 or 4 6. 7. 8. (a) Find the value of k. (b) What is the probability that you will get admission in exactly two colleges ? (c) Find the mean and variance of the probability distribution. 9. OR The bags A and B contain 4 white 3 black balls and 2 white and 2 black balls respectively. From bag A two balls are transferred to bag B. Find the probability of drawing 10. 11. 12. (a) 2 white balls from bag B ? 13. 14. (b) 2 black balls from bag B ? (c) 1 white & 1 black ball from bag B ? XtraEdge for IIT-JEE From 82 MECHANICS Weight (force of gravity) decreases as you move away from the earth by distance squared. Mass and inertia are the same thing. Constant velocity and zero velocity means the net force is zero and acceleration is zero. Weight (in newtons) is mass x acceleration (w=mg). Mass is not weight! Velocity, displacement [s], momentum, force and acceleration are vectors. Speed, distance [d], time, and energy (joules) are scalar quantities. The slope of the velocity-time graph is acceleration. At zero (0) degrees two vectors have a resultant equal to their sum. At 180 degrees two vectors have a resultant equal to their difference. From the difference to the sum is the total range of possible resultants. Centripetal force and centripetal acceleration vectors are toward the center of the circlewhile the velocity vector is tangent to the circle. An unbalanced force (object not in equilibrium) must produce acceleration. The slope of the distance-tine graph is velocity. The equilibrant force is equal in magnitude but opposite in direction to the resultant vector. Momentum is conserved in all collision systems. Magnitude is a term use to state how large a vector quantity is. JANUARY 2010 XtraEdge for IIT-JEE 83 JANUARY 2010 MOCK TEST PAPER SOLUTION FOR PAPER – 1 PUBLISHED IN DECEMBER ISSUE 13. From Einsten's equation Kmax = hv – hv0 eV0 = hv – hv0 hν 0 φ hν hν V0 = – = – e e e e φ = work function. From the graph. metal-2 has high value of φ ∴ threshold wavelength of metal-1 is high. (ii) θ1 = θ2 because slope of the graph is constant for all metals. PHYSICS 1. Relation is not valid because charge is independent of motion of particle. 2. No change because focal length of mirror is independent of refractive index. 3. Galium Aresnide 4. Yes, it get observed but the fringes are of diffraction type. These are not the interference fringes. 5. Zener diode is used as a voltage regulator to obtain constant voltage output. 6. Davison – Germer experiment. 7. n=0 8. Because for alloys temperature coefficient of resistivity is nearly constant for wide range of temperature. 9. 14. Ep ( R.B T ∫ 0 i = 1 + 3 2 sin (314 t + 30º) G = 1, i0 = 3 2 from the above formulae 17. (i) Lenz's law : The direction of magnetic induction in a circuit is such that so as to oppose the cause of change in magnetic flux. (ii) On increasing current i inwards magnetic field increases. Direction of induced current is such that to produce outwards magnetic field i.e. anticlockwise direction. 1/ 2 = (1 + 9)1/2 = 10 12. At P1 drift speed of the electron is maximum for a 1 conductor vd ∝ (I = neAVd) A At P3 current density is minimum. XtraEdge for IIT-JEE ) K2 16. (i) Algebric sum of currents meeting at a point is equal to zero. (ii) Current in 2 Ω resistor is 3A ∴ Potential difference is = 2 × 3 = 6V 1 2 i (f ) dt T ( 15. (i) Angle of Dip: It is the angle which the direction of resultant intensity of earth's magnetic field subtends with horizontal line in magnetic meridian at the given place. B (ii) BH = B cos φ ⇒ tan φ = v BH Bv = B sin φ Bv = BH ⇒ tan φ = 1 If φ = 45º 10. (i) NPN transistor (ii) Yes, the transistor is properly biased because emitter-base junction is in forward bias and the collector base junction is in reverse bais i2 irms G 2 + 0 2 Rh G (i) V-m = electric flux (ii) C-m = Dipole moment 11. irms = K1 ) 18. (i) The order of colours in secondary rainbow is opposite to that of primary rainbow (ii) This is due to total internal reflection. 84 JANUARY 2010 19. (i) Potential difference on 1µF and 3µF is same 1 ∴ from energy = CV2 2 E∝C E2 1 = E1 3 (ii) Parallel combination of 1 µF and 3 µF is 4 µF 12 3µF 4 30V Q = CV = 90µC The refractive index is related to the angle ip called the polarising angle by a relation known as Brewster's law µ = tan ip. Plane polarised light ip µ 30V (ii) sin θc = d 90 30 = = = 7.5 Volt c 12 4 (iii) Energy supplied by battery E = QV = 9 ×10–6 × 30 = 2.7 × 10–3 J P.d. on 12µF ⇒ V = θc = sin–1(cot ip) (iii) It is not correct. 23. Need for modulation (i) Frequency of signal : The audio frequency signal (20 Hz to 20 KHz) cannot the transmitted without distortion over long distance due to less energy carried by low frequency audio waves. (ii) Number of channels: Audio frequencies are concentrated in the range 20 Hz to 20 kHz. This range is so narrow that there will be overlapping of signals. In order to separated the various signals it is necessary to convert all of them to different portions of the electromagnetic spectrum 20. (i) Magnetic field in the solenoid is along the axis → → ∴ angle between v and B is 0º → → 1 1 = µ tan i p ∴ F = q( v × B ) = 0 (ii) When current flows through the spring current in different coils of spring flows in same direction. Therefore due to magnetic force spring gets compressed. 21. (i) 10Ω 5Ω EC EC x Carrier Mo dulating signal (100 – x) 1 x = 2 100 – x 2x = 100 – x, x = 33.33 cm A let P be the null point (ii) A.M. wave Y B A 0 0 1 1 B 0 1 0 1 24. To convert a galvanometer into an ammeter of range I a small resistance S is connected in parallel with the galvanometer so that the current passing through the galvanometer G becomes equal to its null scale deflection value Ig. S Y 1 0 1 1 (I – Ig) 22. (i) When unpolarised light is incident on a transparent surface of refractive index µ at a certain angle ip such that the refracted light ray and the reflected light ray are perpendicular to each other, the reflected light is plane polarised as shown here. XtraEdge for IIT-JEE I 85 G Ig I JANUARY 2010 28. Working : During the positive half cycle of the input signal, the forward bias across the emitterbase junction will be increased while during the negative half cycle of the signal, the forward bias across emitter-base junction is decreased. Hence more electrons flow from the emitter to the collector via the base during positive half cycle. The increased collector current will produce a large voltage drop across the load resistance RL. However during the negative half cycle of the collector, current decreases resulting in the decreased output voltage. Hence an amplified output is obtained across the load. IgG I − Ig 25. (i) Equivalent ckt is R/2 R/2 R R R R R 4R/3 4R/3 R Ans. 8R/7 (ii) Material will be constantan because for alloys value of α varies very slightly with temperature as compared to metals. 27. (i) BE O VCE high frequency AC source D2 2a D1 path of accelerated proton +q P net = P 2 + P 2 + 2P 2 cos 60º = (ii) Electric potential energy –2q 3P= +q +q 2a kq (−2q ) kq (−2q ) kq 2 + + 2a 2a 2a = − 2kq 2 − 2kq 2 kq 2 – + 2a 2a 2a = − 3kq 2 2a B positive charged beam Principle : Cyclotron device is based on the fact that heavy positive ions can be accelerated to the high energies with a comparatively smaller alternating potential difference by making them to cross the field again and again using strong magnetic field. Here the magnetic field used in cyclotron maintain the charged particles in circular paths while the electric field imparts them energy periodically. Construction : Cyclotron consists of two D shaped hollow metallic enclosures D1 and D 2 called dees. These dees have their diometric edge parallel to one other and are separated by a small gap. These dees are connected to the terminals of a high frequency alternating potential difference. This potential difference creates an electric field of high frequency in the gap between the dees. The whole apparatus is placed between N-S poles of a powerful electro magnet which produces strong magnetic field. 3 q(2a) 2a 2a XtraEdge for IIT-JEE IC 29. Cyclotron 2a –2q can be assumed as two –q charges placed at the point p = q(2a) U = IC IE t Pnet +q RL ~ –q –q Pt IB Signal 26. (i) Angle between electric field line and equipotential surface is 90º. (ii) Electric field directed in the direction of decreasing potential. So electric potential is maximum at point a and proton will have maximum potential energy at point a . (iii) Electric field is maximum at point c. Thus proton will have maximum force at c. 2a IC Vi Output IgG = (I – Ig) × S ⇒ S = 86 JANUARY 2010 t Cyclotron can not be used to accelerate electrons because it is used to accelerate heavy ions. 4. When two different molecules participate in the polymerization process it is called copolymerization. 5. A metal which is more electropositive than iron such as Al, Zn, Mg can be used in cathodic protection of iron against rusting. 6. Lower value of bond dissociation energy of F2 is due to the strong repulsion between the non bonding electrons of F atoms in the small sized F 2 molecule. Also there is no multiple bonding due to absence of d-orbitals. 7. Hydrolysis of sucrose produces change in optical nature form dextro rotatory to laevorotatory, the process is called inversion of sugar. → 30. Ampere's circuital law : The circulation of B along a closed loop of any arbitary shap is µ0 times the algebric sum of current embraced by the loop. Determination of magnetic field for a solenoid : l b c P a d b c B C6H12O6 + C6H12O6 Glucose Fructose Turns of Solenoid Axis a H+ C12H22O11 ((cane sugar) + H2O ) → Invert sugar d dl Let p be the point where B is to be determined. From Ampere's law → → ∫ B . dl = µ i ∫ 0 net [ ∫ b→ → B . dl = a and ∫ ∫ ∫ d→ → c c → B . dl a→ → d 9. Mechanism of the formation of diethyl ether from ethanol : The formation of ether is a nucleophilic bimolecular reaction (SN2) involving the attack of alcohol molecule on a protonated alcohol, as indicated below B . dl → B . dl = 0; B ⊥ dl (i) → H CH3 – CH2 – O – H + H+ → CH3 – CH2 – +O – H Ethanol B . dl = 0, Q B = 0] a ∫ B . dl = B dl cos 0 = BI ⇒ µ0(nil) H + → (ii) CH3 CH2 – O + CH3 – CH2 – O | H H d ⇒ B = µ0ni CHEMISTRY 1. Three types of lattice imperfections are possible (a) Schottky defect (b) Frenkel defect (c) Interstitial defects c→ → b ∫ B . dl + d→ → a→ → d d→ → c ∫ B . dl + a ∫ + b→ → 8. 5 4 3 2 1 CH3 – C = CH – C – CH3 || | O CH3 4-methyl-pent-3-en-2-one 2. CH3 – CH2 – CH = CH2 + HCl → CH3 – CH2 – CH – CH3 | Cl 3. Enzyme streptokinase can dissolve blood clots. So it is useful in medicines for checking heart attacks caused by blood clotting. XtraEdge for IIT-JEE + CH3 CH2 – O – CH2 CH3 + H2O | H (iii) CH3 CH2 – +O – CH2 CH3 → | H CH3 CH2 – O – CH2 CH3 + H+ Diethyl ether (Ethoxyethane) 87 JANUARY 2010 10. The order of basicity in gaseous phase is (i) (CH3)3 N > (CH3)2 NH > CH3NH2 > NH3 due to +I effect of alkyl group, there is more density at N at tertiary amine. (ii) The order of basicity in aqueous state is (CH3)2 NH > CH3 NH2 > (CH3) 3 N > NH3 The inductive effect, solvation effect, Hbonding and steric hinderance of the alkyl group decides the basic strength of alkyl amines in the aqueous state. 13. (i) Treat the compound with Lucas reagent (conc. HCl + anhy. ZnCl2) 2-propanol gives turbidity in 5 min whereas ethanol gives no turbidity at room temperature ZnCl 2 → No reaction CH3CH2OH + HCl ZnCl 2→ CH3CHCH3 + HCl | OH CH3 – CH – CH3 + H2O | Cl Turbidity appears in 5 min (ii) Acetaldehyde reduces Tollen’s reagent to silver mirror but acetone does not. 11. (i) Monomer of Teflon is tetrafluro ethylene. Teflon is a addition polymer. (ii) Monomer of Bakelite is formaldehyde and phenol. Bakelite is a condensation polymer. (iii)Monomer of natural rubber is isoprene (2methyl-1, 3-butadiene). Natural rubber is a addition polymer. CH3CHO + 2 [Ag (NH3)2]+ + OH– –→ CH3COO + 2H2O + 2Ag ↓ + 4NH3 Silver mirror ONa Tollen’s reagent CH3COCH3 → No action 4 − 7 atm + CO2 → 12. (i) 400 K 14. Sod. phenoxide OH Multimolecular Colloids The particles of this type of colloids are aggregates of atoms or molecules with diameter less than 1 nm. Examples : Sol of sulphur conists of colloidal particles which are aggregate of 58 molecules. The atoms of molecules are held together with Van der Waal’s forces + COONa H → H O 2 OH COOH sod. salicylate salicylic acid OH CHCl , aq NaOH 3 → 340 K (ii) Phenol ONa CHCl2 Examples : proteins etc. Starch, Covalent bonds are present in one chain and different chains have the force like H-bonds, dipole-dipole interaction and salt bridge etc. 15. Pyrophosphoric acid is prepared by the removal of H2O from two molecules of orthophosphoric acid (having tetrahedral shape). Hence two tetrahedra are attached through an oxygen. OH OH ONa CHO H3O + → NaOH → Macromolecular Colloids The particles of this type of colloids are themselves large molecules of colloidal dimension. OH O P O P O CHO O Salicylaldehyde XtraEdge for IIT-JEE 88 O JANUARY 2010 OR Formation of Ni (CO)4 OH OH | | O=P–O–P=O | | OH OH ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 0.059 [Cu 2+ ] log 2 [Cu ] 19. (i) Markownikoff’s rule : According to this rule, when addition across an unsymmetrical double bond takes place, the positive part of the addendum goes to the carbon atom with the larger number of hydrogen atoms. 0.059 0.1 0.059 1 = 0.34 + log log 2 1 2 10 0.059 [Q log 10 = 1] = 0.34 + × (−1) 2 = 0.34 – 0.0295 = 0.3105 V When the concentration of Cu2+ ions is decreased, the emf for copper electrode decreases. = 0.34 + Absence of → CH3 – CH = CH2 + HBr Peroxide propene Br | CH3 – CH – CH3 2-Bromopropane (ii) Hofmann Bromide Reaction : When amide is treated with bromide in on alkaline solution, an amide yields an amine containing one carbon less than the starting amide. O || R – C – NH2 + Br2 + 4 KOH Amide RNH2 + K2 CO3 + 2KBr + 2H2O Amine For example : O || CH3 CH – C – NH2 + Br2 + 4 KOH Propanamide CH3 CH2 NH2 + K2 CO3 + 2KBr + 2H2O Ethylamine 17. Adsorption isobar for physical adsorption shows that the extent of adsorption decreases with the increase in temperature. The adsorption isobar of chemical adsorption shows that the extent of adsorption first increases and then decreases with the increase in temperature. The initial unexpected increase in the extent of adsorption with temperature is due to the fact that the heat supplied acts as activation energy required for chemical adsorption which is much more than that of physical adsorption. Extent of adsorption Physical adsorption isobar Extent of adsorption Temperature Chemical adsorption isobar 20. (a) (i) CH2 = CH2 + H2SO4 –→ CH3CH2HSO4 conc. Ethene Ethyl Sulphuric hydrogen acid Temperature H O 18. In the complex [Ni (CO)4], the oxidation state of Ni is ‘0’. Its electronic configurations is [Ar] 3d8 4s2. A.O. of Ni (28) 3d 4s 4p ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ 2→ CH3CH2OH + H2SO4 Boil Ethanol (ii) OCOCH3 OH + CH3 COCl sp hybridised obritals of Ni Phenol ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ sp (iii) 3 hybridisation 89 CH3 H Pyridine Acetyl chloride 3 XtraEdge for IIT-JEE × × × Four electrons pairs from four CO molecules The resulting complex has tetrahedral shape and is diamagnetic due to absence of unpaired electrons. 16. Cu2+ (aq) + 2e– → Cu (s) E Cu 2+ / Cu = E°Cu 2+ / Cu + × Phenyl ethanoate + CH MgI CH3 Dry ether H C = O 3→ Ethanal + HCl C OMgI CH3 JANUARY 2010 H + , H 2O CH3 → − Mg(OH) I H C (b) On dilution, the degree of ionization of the weak electrolyte increases. This increases the molar conductance of the solution sharply. OH CH3 2-Propanol 24. In [Co (NH3)6]3+, the oxidation state of Co is + 3. Co (Z = 27) atom is ground state 21. (i) A is CH2 = CH2 B is CH2 – CH2 | | Br Br 3d ↑↓ ↑↓ ↑↓ (Benzene diazonium chloride) ↑↓ ↑↓ NH3+ HSO4– (Anilium hydrogen ↑ ↑ sp3 d2 hybridisation ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ × × × × × Due to sp 3 d2 hybridisation, the complex ion has octahedral shape. Due to the presence of four unpaired electrons, the complex ion is paramagnetic. NH2 (Sulphanilic acid) 26. (i) A deep red sol of ferric hydroxide is obtained by the hydrolysis of ferric chloride. The sol particles are propositively charged because of preferential adsorption of Fe3+ ions. (ii) Adsorption is an endothermic process. So the rate of physical adsorption decreases with the rise in temperature in accordance with Le Chatelier’s principle. (iii)River water contains charged colloidal particles of sand, clay, etc. As river water comes in contact with saline sea water, the electrolytes of sea water coagulate the suspended colloidal particles which settle down at the point of contact resulting in the rise of river bed. So water adopts a different course and a delta is formed in due course of time. SO3H 22. (a) Electronic configuration of Ti in [Ti (H2O)6]3+ is Ti3+ (d1) Two vacant d orbitals are available for octahedral hybridization with 4s and 4p orbitals. (b) The colour of the complex is purple. The colour of complex is due to the jumping of electron from lower level to higher level. When an electron from a lower energy of orbital is excited to higher energy of level, the energy of excitation corresponds to the frequency of light absorbed. This frequency lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed. 27. Solid catalysts are used in a number of gaseous reactions. Such catalytic reactions called heterogeneous reactions. Examples of heterogeneous catalysis are (i) Manufacture of ammonia from N2 and H2 by Haber’s process in the presence of catalyst. 23. (a) According to electrochemical theory during the formation of just the impure iron surface behaves like a small electrochemical cell in the presence of water containing dissolved oxygen or carbon dioxide. In such alls, pure iron acts as anode and impure surface acts as cathode. Moisture containing dissolved oxygen or CO 2 in the electrolytic solution. Hence rusting is an electrochemical phenomenon. XtraEdge for IIT-JEE ↑ Six pairs of electrons from six NH3 molecules sulphate) B is ↑ 4d ↑↓ Formation of [Co(NH3)6]3+ ion (Benzonitrile) (iii) A is ↑ Hybridisation CN B is ↑ Co3+ ion + N2Cl– (ii) A is 4p 4s ↑ Fe(s) N2(g) + 3H2(g) → 2NH3(g) (ii) V2O5 catalyst is used in the manufacture of H2SO4 by contact process 90 JANUARY 2010 2Na2 CrO4 + H2 SO4 –→ Na2 Cr2 O7 + Na2 SO4 + H2O Na2 SO4 is separated by fractional crystallisation. Sod. dichromate is converted into potassium dichromate by heating with KCl. Na2 Cr2 O7 + 2KCl –→ K2 Cr2 O7 + 2NaCl Potassium dichromate being less soluble is obtained by fractional crystallisation. (i) Cr2 O72– + 14 H+ + 6I– –→ 2Cr3+ + 7H2 O + 3I2 (ii) Cr 2 O72– + 4Fe2+ + 14 H+ – → 2Cr3+ + 6Fe3+ + 7H2 O Uses the potassium dichromate : (i) In volumetric analysis for the estimation of Fe2+ and I– ions. (ii) In chrome tanning in leather industry. V O (s) 2SO 2(g) + O2 2 5 → 2SO3(g) Solid catalyst helps in the following ways : (a) Simultaneous adsorption of reactants increases the concentration at the surface of the catalyst which increases the reaction rate. (b) Adsorption of reactant molecules makes the attack of other molecules on it easier. (c) Some adsorbed molecules dissociate into atoms which are very reactive. (d) Heat of adsorption released provides activation energy for the reaction. 28. (a) A is aldehyde or Ketone. A gives Tollen’s test hence it is an aldehyde CH3 – CH = CH – CH2 – CHO CH2 CH3 (b) 30. (i) At elevated temperatures, sulphur vapours exists, as S2 molecules which are paramagnetic like O2. (ii) This is due to reluctance of silicon to form pπ – pπ multiple bonds because of large size of silicon atom. Hence, silicon exists only in diamond structure. (iii) Xe has relatively lower ionization energy among inert gases and thus the outermost shell electrons of Xe are excited to d-subshell and thereby showing unpaired electronic structure. (iv) Nitrogen shows a little tendency for catenation, due to weakness of N – N single bond whereas phosphorus shows a clear tendency for catenation due to its unexpectedly high bond energy. KMn O / OH − 4 → (i) Ethylbenzene COOH Benzoic acid CaO + NaOH → Benzene Dil NaOH (ii) 2CH3CHO → Acetaldehyde (Aldol condensati on ) 4 3 2 1 NaBH 4 → CH 3 − CHOH − CH 2 − CHO (Reduction ) 4 3 2 1 CH3 – CHOH – CH2 CH2OH Butane-1, 3-diol CH3 C = O + H2 –→ CH3 CHOH CH3 (iii) CH3 Acetone MATHEMATICS Section – A 1. Conc. H SO 4 2 → CH3 CH = CH2 Dehydration Propene 2. 29. Preparation of K2 Cr2 O7 : Chromite ore is fused with molten NaOH in the presence of air to get sodium dichromate. 4 Fe Cr2 O4 + 16 NaOH + 7O2 – → Chromite ore 8 Na2 Cr O4 + 2Fe2 O3 + 8H2 O Sod. chromate The fused mass is dissolved in water. The filtrate is treated with dil H2 SO4 XtraEdge for IIT-JEE Let (a, b) ∈ R ∴ (b, a) ∈ R Hence R is symmetric. [Q (1, 2) ∈ R ] [Q (2, 1) ∈ R ] x2 1 x2 . tan −1 x – dx 2 x2 +1 2 (Integrating by parts) ∫ x. tan −1 x dx = = 1 2 1 x tan −1 x − 2 2 ∫ ∫ x 2 + 1−1 dx x 2 +1 1 1 1 1− = x 2 tan −1 x − dx 2 2 x 2 + 1 ∫ = 91 1 2 1 x tan −1 x − ( x − tan −1 x ) + C. 2 2 JANUARY 2010 3. The given curves are x2 + y2 = 2ax Here, a is arbitrary constant Diff. (1) w.r.t. x dy 2x + 2y = 2a dx Substituting for 2a in (1), we get 4 3 5 ∴ | A | = O ⇒ 3 − 2 7 10 − 1 x ⇒ 4 (–2x + 7) –3 (3x – 70) + 5 (–3 + 20) = 0 ⇒ x = 19 ….(1) dy x2 + y2 = x 2x + 2 y dx 8. [Q tan( π – θ) = – 9. 7. → → → → → → → a + b+ c → 3π 4 θ= 3π 4 → → a .b → 3 = → 3.2 = 3 2 π 3 10. Let the equal angle = x ∴ l = cos α ; m = cos α and n = cos α ∴ l2 + m 2 + n 2 = 1 cos2 α + cos2 α + cos2 α = 1 ⇒ 3cos2α = 1 1 ⇒ cos α = ± Hence direction cosines 3 1 3 1 3 ± − 2 − 2 1 + − 4 0 3 − 1 − 1 1 3 , ± 1 3 , ± 1 3 Section – B 11. We have f(x) = 2x, g (y) = 3y + 4 and h (z) = sin z L.H.S. = ho (gof) (x) = h ( g ( f ( x )) = h ( g (2x)) = h (3 (2x) +4) = h (6x + 4) = sin (6x + 4) R.H.S. = (hog) of (x) = (hog) (f (x)) = (hog) 2x = h (g (2x) = h (3 (2x) +4) = h (6x + 4) = sin (6x + 4) ∴ L.H.S. = R.H.S. {Order of matrix n=3} 1− x 12. Let y = cot–1 1+ x 1− x Put : =u 1+ x 4 3 5 Let A = 3 − 2 7 10 − 1 x Q the matrix A is singular XtraEdge for IIT-JEE 2 3 3 and | b | = 2 ; a . b = 3 | a || b | Q | adj A | = | A | n – 1 64 = | A | n – 1 64 = | A |3 – 1 | A |2 = 64 |A|=8 2(î + ĵ + k̂ ) 3 → Here | a | = cos θ = 3 1 − 2 and B = A= 0 2 0 2A – B + C = 0 ⇒ C = –2A + B 3 1 − 2 ⇒ C = –2 + 0 2 0 = î + ĵ + k̂ → → 3π 4 − 6 ⇒ C= 0 − 8 C= 0 6. → = ∴ the principle value of tan–1 (–1) is 5. → | a + b+ c | tanθ] x= → ∴ unit vector = Let x = tan–1 (–1) tan x = –1 π tan x = – tan 4 π tan x = tan π − 4 tan x = tan → a + b + c = î + ĵ + ĵ + k̂ + k̂ + î = 2 ( î + ĵ + k̂ ) a + b + c = 2 1+ 1 + 1 = 2 3 dy + x2 – y2 = 0 ⇒ 2xy dx which is the reqd. diff. eq. 4. → 92 ...(1) JANUARY 2010 ∴ y = cot–1 u Diff. w.r.t. u −1 dy = = du 1 + u 2 = ∴ The reqd. sol. of eq. (1) is ∫ 2x .x dx + c = ∫ 2 x dx + c = x + c 1− x 1+ 1+ x − (1 + x ) 2 2 (1 + x ) + (1 − x ) 2 2 = 2 (1 + x ) 2(1 + x 2 ) ...(2) π/4 ∫ sin 2 x sin 3 x dx 14. …(3) 0 From (2) and (3), dy dy du = ⋅ dx du dx 13. The given diff. eq. is (1 + e2x) dy + ex (1 + y2) dx = 0 dy 1+ y 2 + e 1 + e 2x ∫ 0 sin 5x π / 4 sin x − 5 0 = 1 2 5π π 1 sin 4 − 5 sin 4 − 0 1 π 1 π 1 1 1 3 sin + sin = 1 + = 2 4 5 4 2 5 2 5 2 OR (where t = ex) π/4 Sol. Let I = ∫ log(1 + tan x )dx …..(1) 0 a By the formula, ∫ a ∫ f ( x )dx = f (a − x )dx 0 π/4 = 0 π 1 − tan x 2 ∫ log 1 + tan 4 − x dx 0 π/4 = ∫ log 1 + 1 + tan x dx 0 π/ 4 = ∫ log 1 + tan x dx 0 π/ 4 ...(1) = ∫ [log 2 − log(1 + tan x)]dx 0 π/ 4 = log 2 ∫ π/4 dx − 0 = log2 [x ] 1 −1 ∫ (cos x − cos 5x) 1 2 = ∴ Pdx ∫ − dx I.F. = e ∫ = e x = e − log x = e log x 0 π/4 = dx = 0 Integrating, we get dt tan–1 y + =c 1+ t2 ⇒ tan–1 y + tan –1 t = c ⇒ tan–1 y + tan –1 ex = c when x = 0, y =1 ⇒ tan–1 + tan–1 e° =c π π π ⇒ c= + = 4 4 2 π tan–1 y + tan–1 ex = . 2 This is the reqd. sol of (1) OR Sol. The given diff. eq. is dy x − y − 2x 3 = 0. dx dy 1 − y = 2x 2 ⇒ dx x This is a linear diff. eq. dy On comparing by, + Py = Q dx 1 Here, P = – , Q = 2x2 x ∫ (2 sin 3x sin 2x)dx 1 2 = ...(1) π/4 1 2 = −2 (1 + x ) 2 1 ⋅ = =– . 2 2(1 + x ) (1 + x ) 1+ x2 ⇒ y = x3 + cx. ⇒ 2 Diff. (1) w.r.t. x du (1 + x ).(−1) − (1 − x ).1 2 = = 2 dx (1 + x ) (1 + x ) 2 x −1 2 y . x–1 = −1 2I = ∫ log(1 + tan x )dx 0 π/4 0 − I [By (1)] π π log2 − 0 ⇒ I = log 2 8 4 = x −1 XtraEdge for IIT-JEE 93 JANUARY 2010 15 Let I = ∫ 2x 3x + 1 2 17 dx − 2x + 3 3 d 5 (2x2 – 2x + 3) + Here : 3x + 1 = . 4 dx 2 3 5 ( 4x − 2) + 2 dx ∴ I= 4 2 2x − 2x + 3 3 4x − 2 5 dx dx + = 4 2x 2 − 2 x + 3 2 2x 2 − 2 x + 3 3 5 = I1 + I 2 + c …(1) 4 2 4x − 2 dx where I1 = 2 2 x − 2x + 3 Put : 2x2 – 2x + 3 = t ⇒ (4x – 2)dx = dt dt ∴ I2 = = log | t | t ....(3) = log | 2x2 – 2x + 3 | dx 1 dx = and I 2 = 2 3 2 2 2x 2 − x + 1 5 − + x 2 2 2 x →π − x →π − lim f ( x ) = lim x 3 + 3 = 3 ∫ x →0 x →0 Hence f(x) continuous at x = 0. π x −1 −1 x + 1 18. tan–1 = + tan x−2 x+2 4 ∫ x −1 x + 1 + π tan −1 x − 2 x + 2 = x 1 x 1 − + 4 1 − . x − 2 x + 2 1 x− 1 1 −1 2 = ⋅ tan 2 ( 5 / 2) 5/2 ∫x dx 2 +a 2 = 2x − 1 tan −1 5 5 From (1), (2) and (3), we get x2 + x − 2 + x2 − x − 2 …(3) 2 x − 4 − x +1 d2 y dx At θ = 2 = d θ dθ cot . dθ 2 dx =– 1 θ 1 cosec2 . 2 a (1 − cos θ) 2 = − 1 θ cos ec 4 . 4a 2 19. 4 − 5 − 11 A = 1 − 3 1 2 3 − 7 | A | = 4 (21 – 3) + 5 (–7 – 2) –11 ( 3 + 6) = 72 – 45 – 99 = – 72 9 9 18 Adj. A = − 68 − 6 − 22 − 38 − 15 − 7 2 1 π d y 1 π ; = − cos ec 4 = – . 4a 2 dx 2 a 4 XtraEdge for IIT-JEE π Q 4 = 1 =1 2x − 4 =1 −3 2 2x – 4 = –3 2x2 = –3 + 4 = 1 2x2 = 1 1 x2 = 2 1 ⇒x=± 2 Given : x = a (θ – sin θ); y = a (1 – cosθ) dx dy = a (1– cos θ); = a sin θ ⇒ dθ dθ θ dy dy / dθ a sin θ ⇒ = cot . = = dx dx / dθ a (1 − cos θ) 2 Now 2 2 2x − 1 3 5 + c log | 2 x 2 − 2x + 3 | + tan −1 4 2 5 I= 16 ( x − 1)( x + 2) + ( x + 1)( x − 2) π ( x − 2)( x + 2) = tan 4 ( x − 2)( x + 2) − ( x − 1)(x + 1) ( x − 2)( x + 2) 1 x tan −1 a a 1 = x →0 Also f(0) = 1 ∴ lim f ( x ) = f (0) ∫ By the formula, x → π+ K · π + 1= cos π πK+1 =–1 π K = –2 −2 K = π OR Sol. Since the function is defined at x = 0 ∫ ∫ x →π + lim Kx + 1 = lim cos x ∫ ∫ Q f(x) is continuous at x = π ∴ lim f ( x ) = lim f ( x ) = f (π) 94 T JANUARY 2010 θ 2 → → θ | a + b | = 2 cos 2 → → θ 1 cos = | a + b | 2 2 → 18 − 68 − 38 ∴ adj A = 9 − 6 − 15 9 − 22 − 7 19 − 68 − 38 adjA − 1 = 9 − 6 − 15 A 72 9 − 22 − 7 OR y z a + x Sol. ∆ = x a+y z x y a + z ∴ A –1 = 2 = 4C3 3 operate C2 → C2 + C 1 o o a ∆ = x a + y + x z x y+ x a + z ∆ = a [(a + y + x) (a + z) – z (y + x)] = a [a2 + az + ( y + x) a + (y + x ) z – z (y + x)] = a2 ( a + x + y + z) Proved. → → → 3 4 [Q P (r) = nCr qn– r Pr] 4 3 3 1 · 40 10 4 = = 3 1 3 1 1 1 3 + · + · + ·0 40 30 10 4 10 3 5 3 3 120 9 = 40 = × = . 9 + 4 40 13 13 120 →→ Section – C → = 1 + 1 + 2 | a | | b | cos θ π/2 23. Let = 2 + 2 (1) (1) cos θ = 2 (1 + cosθ) θ = 2 . 2 cos2 2 XtraEdge for IIT-JEE 1 3 2 1 1 · +1 · 1 · 3 27 81 8 1 9 1 = + = = 81 81 81 9 OR Sol. Let A, B and C be the events of candidate going by bus, scooter and other means of transport. Let E be the event of getting late. 3 1 3 P (A) = , P (B) = , P (C) = 10 10 5 1 1 P (E/A) = , P(E/B) = , P(E/C) = 0 4 3 P (that he traveled by bus) = P (A/E) P( A ) P( E / A ) = P( A) P( E / A) + P( B) P( E / B) + P (C)P (E / C) | a + b |2 = | a |2 + | b | 2 +2 a b → 0 4−4 =4· 21. Q | a | = | b | = 1 (Given) → 3 1 4 2 + C4 3 3 1 ( 2 + 1) 2 + (3 + 3) 2 + ( 4 − 2) 2 = 49 = 7 → 4 −3 2 1 2 1 = 4C1 + 4 C 0 3 3 3 3 20. The given plane is 3x + 2y + 2z + 5 = 0 ...(i) line through P (2, 3, 4) and parallel to the line : x+3 y−2 z = = is 3 6 2 x −2 y−3 z −4 = = = k (say) ….(ii) 3 6 2 Any point on it is Q (3k +2, 6k + 3, 2k + 4) Let it lie on (i) ∴ 3(3k + 2) +2 (6k +3) +2 (2k + 4) +5 = 0 ⇒ 25 k + 25 = 0 ⇒ k=–1 ∴ Q (–1, –3, 2) ∴ The required distance = PQ → Proved. 22. p = P (correct forcasting) = 1/3 q = P (two incorrect forecasting) = 2/3 n=4 Let r be the number of correct forecast P (at least three correct results) = P ( r = 3) + P (r = 4) operate R1 → R1 – R2 o a − a ∆ = x a + y z y a + z x = → | a + b |2 = 4 cos2 I= ∫ 0 cos x dx (1 + sin x )(2 + sin x ) ⇒ cos x dx = dt π Also x = 0 ⇒ t = 0 and x = ⇒t=1 2 Put : sin x = t 95 JANUARY 2010 1 ∴ I= ∫ 0 dt = (1 + t )(2 + t ) 1 4 3 16 ∴ Reqd. Area = Area of circle – + π 3 3 1 1 1 + t − 2 + t dt 0 ∫ (Resolving into partial fractions) = [log | 1 + t | – log | 2 + t = 16 π – |] 10 = (log 2 – log 3) – (log 1 – log 2) = 2 log 2 – log 3 = log 2 2 – log 3 4 = log . 3 2 32 4 3 4 π− = (8π − 3 ) sq. units. 3 3 3 OR = Sol. 2 24. x +y = 16 y2 = 6x. (1) and (2) intersect, where …(1) ...(2) x2 4 3 16 − π 3 3 y2 + =1 a2 b2 x y + =1 a b ….(1) …..(2) Y 2 Y B (0, b) 2 X' 1 A (2, 2 3 ) 1 X' x=4 x=2 x=0 O D C Y' We shall find the shaded area (Area of the smaller region) a ∫ a (Q x ≠ –8) 2 0 b = a a ∫ 0 1− x − b1 − dx a a x2 2 a x a − x dx − b 1 − dx a 0 2 ∫ 2 a 16 − x 2 dx = b a2 a2 −1 sin 1 − b a − a 2 2a = ab π ab 1 . − = ab(π − 2)sq.units 2 2 2 4 25. Let S, V, r and h be the surface area, volume, radius of the base and height of the given cylinder respectively. Then S = 2 πrh + πr2 (Given) [Q Cylinder is open at the top] 2 6 2.2 3 1 = 2 .2 2 + {0 + 8 sin −1 1} − + 8 sin −1 2 2 3 16 3 π π + 16. − (4 3 + 16. ) 3 2 6 ⇒ 4 3 16 = + π sq. units. 3 3 XtraEdge for IIT-JEE − y 2 )dx a b x a2 − x2 a2 x x2 = + sin −1 − b x − a 2 2 a 2a 0 0 4 2 2 3/ 2 x 16 − x 2 16 −1 x = 2 6. x + + sin 3 2 2 4 0 2 = ∫ b = ∫ ∫ 1 0 4 2 = 2 y 2 dx + y1dx 0 2 4 ∫ (y = Q A (2, 2 3 ) and B (2, –2 3 ) Also C (4, 0). Area OBCAO = 2 (Area ODA + Area DCA) 2 = 2 6 x dx + 0 A (a, 0) x=0 Y' ∫ X x=a X B (2, –2 3 ) x2 + 6x – 16 = 0 ⇒ (x + 8) (x – 2) = 0 ⇒ x=2 O h= S − πr 2 2πr V = πr2h = πr 2 96 S − πr 2 1 = [Sr − πr 3 ] 2 2πr JANUARY 2010 Diff. w.r.t r x 4 − 5 1 4 y = 1 2 0 − 2 0 10 z 2 5 3 2 2 dV 1 d V = [S − 3πr 2 ] and 2 = −3πr 2 dr dr For max. or min., dV = 0 dr ⇒ S – 3πr2 = 0 S 3π ⇒ r= ∴ x= For this value of r, d 2V dr 2 ∴ V is maximum and h= 16 − 0 + 2 18 1 1 = 8 + 0 − 4 = 4 10 10 8 + 0 + 6 14 <0 3πr 2 − πr 2 =r 2πr 27. Let A and B the events of getting letter from Tata Nagar and Calcutta 1 1 ∴ P(A) = , P (B) = 2 2 Let E be the event of visibility of letter TA 2 7 2.2 1 = = P (E | A) = 7.6.5.4.3.2.1 1260 2 [Using (1) and (2) ] 26. Sol. Part-I 1 − 1 1 Given matrix A = 2 1 − 3 1 1 1 1 −1 [Q Total no. of events in Tata Nagar = 1 |A|= 2 1 −3 1 1 1 TA = 1 (1 + 3) + 1 ( 2 + 3) +1 ( 2 – 1 ) =4+5+1 = 10 ∴ A–1 exists C11 = (–1)2 (1 + 3) = 4 C12 = (–1)3 (2 + 3) = –5 C13 = (–1)4 (2 – 1) = 1 C21 = (–1)3 (–1 –1) = 2 C22 = (–1)4 (1 – 1) = 0 C23 = (–1)5 (1 + 1) = –2 C31 = (–1)4 (3 – 1) = 2 C32 = (–1)5 (–3 –2) = 5 C33 = (–1)6 (1 + 2) = 3 2 2 4 1 1 –1 A = adj A = − 5 0 5 A 10 1 − 2 3 P (E | B) = 4 − 5 1 1 ∴ A = 2 0 − 2 10 2 5 3 XtraEdge for IIT-JEE TA 1 7 2 = NAGAR 7 2 as has only 2 A's ] 2 1 = 7.6.5.4.3.2.1 2520 [Q Total no. of events in Calcutta = 7 2 as CALCUT TA has only 2 C's ] (i) P (that letter has come from Tata Nagar) = (A | E) = P (A ).P( E | A) P ( A ) P ( E | A ) + P ( B) P ( E | B) 1 1 1 · 1260 2 1260 = = 2 +1 1 1 1 1 · + · 2520 2 1260 2 2520 1 2520 2 × = 1260 3 3 (ii) P (that letters has come from Calcutta) = P (B | E) = 1 – P (A | E) 2 1 =1– = 3 3 OR Sol. White balls = 4 Red balls = 6 Total balls = 4 + 6 = 10 Let X be the number of drawing 3 white balls = Part II The given equation can be written as AX = B ⇒ X = A–1 B 2 1 1 x 4 Where A = − 1 1 1 x = y B = 0 1 − 3 1 z 2 –1 18 9 4 2 14 7 = , y= = , z= = . 10 5 10 5 10 5 [Using part I] 97 JANUARY 2010 ∴ X = 0, 1, 2, 3 4 P (X = 0) = 5λ = 2 – 12 5λ = –10 ⇒ λ = –2 Putting the value of λ, we get. The coordinate of P (–2 + 3, –4 + 4, –4 + 5) = P (1, 0 1) The required distance = |AP| 6 C 0 · C3 10 = C3 1 . 5 6.5.4 3.2.1 15 . = = 3.2.1 10.9.8 90 30 4 P (x = 1) = 4 P (x = 2) C1·6 C2 10 C3 C 2 · 6 C1 10 C3 =4· = 6.5 3.2.1 45 15 . = = 2.1 10.9.8 90 30 4.3 6.3.2.1 27 9 = = 2.1 10.9.8 90 30 4 P (X = 3) = (3 − 1) 2 + ( 4 − 0) 2 + (5 − 1) 2 = 4 + 16 + 16 = 36 = 6 unit. 6 C3 · C 0 10 C3 = 29. Given problem can be tabulated as 3.2.1 3 1 4.3.2 .1. = = 3.2.1 10.9.8 90 30 Therefore, required probability distribution is X 0 1 2 3 5 15 9 1 P (X) 30 30 30 30 Wheat Rice 28. Equation of the given line is 2x = y = z x y z = = 1 2 2 Equation of line passes through A (3, 4, 5) and parallel to line ( 1 ) is x −3 y− 4 z −5 = = = λ(say) 1 2 2 ∴ Any point on the line (2) is P (λ + 3, 2λ + 4, 2λ + 5) Since the point P lies on the plane x+ y+ z= 2 ∴ (λ + 3) + (2λ + 4) + (2λ + 5) = 2 λ + 3 + 2λ + 4 + 2λ + 5 = 2 Proteins 0.01 0.05 Min50g Carbohydrates 0.025 0.5 Min200 g Cost Rs. 4kg = 0.4 P/g Rs. 6 kg = 0.6 P/g Let the quantity of wheat = x gms and the quantity of rice = y gms Min cost z = 0.4x + 0.6y subject to constraints 0.1x + 0.05y ≥ 50 0.25x + 0.5y ≥ 200 x, y ≥ 0 Table for 0.1x + 0.05y = 50 Calculation for mean and variance X P(x) XP(x) X2P(x) 5 0 0 0 30 15 15 15 1 30 30 30 9 18 36 2 30 30 30 1 3 9 3 30 30 30 36 6 60 Total 1 = =2 30 5 30 6 Mean µ = Σ X P(x) = = 1.2 5 Variance = Σ X2 P (x) – [ Σ X P (X)]2 = 2 – (1.2)2 = 2 – 1.44 = 0.56 XtraEdge for IIT-JEE = x y 0 500 1000 0 Table for 0.25x + 0.5y = 200 x y 0 400 800 0 X 1000 800 (0, 500) 600 400 (1000, 0) 200 …..(1) 200 400 600 800 1000 the point The corner points of feasible region is (0, 500) and (1000, 0) Now evaluate z at the corner points …..(2) Corner point (0, 500) (1000, 0) ….(3) Z = 0.4x + 0.6y Z = 0 + 300.0 = 300 ← Min Z = 400.0 + 0 = 400 Minimum cost = 300 paise = Rs. 3. when x = 0 gms, y = 500 gms. 98 JANUARY 2010 XtraEdge Test Series ANSWER KEY IIT- JEE 2010 (January issue) PHYSICS Ques An s Ques An s 21 22 1 D 11 A,D A→P A→Q 2 B 12 A,D 3 A 13 A ,C , D B→Q B→Q 4 A 14 A ,B , D 5 D 15 C C→R C→R 6 D 16 B 7 A 17 A D→Q D → Q,S 8 B 18 C 9 A 19 D 10 C,D 20 B C HEM I ST RY Ques An s Ques An s 21 22 1 A 11 B A→R A→Q 2 B 12 A ,B ,C ,D 3 A 13 C B→P B→P 4 B 14 A,B,C 5 A 15 A C→Q C→S 6 B 16 D 7 B 17 D D→R D→R 8 C 18 A 9 A 19 B 10 B ,D 20 A MATHEMATICS Ques An s Ques An s 21 22 1 B 11 A,B,C,D A→Q A→R 2 B 12 A ,C 3 A 13 A ,C B→P B→S 4 B 14 A,B,C 5 D 15 A C→S C→Q 6 D 16 C 7 C 17 B D→R D→P 8 B 18 C 9 A 19 D 7 A 17 A D→S D → P,Q,R 8 A 18 D 9 A 19 D 10 A ,B ,C ,D 20 C IIT- JEE 2011 (January issue) PHYSICS Ques An s Ques An s 21 22 1 A 11 A,C A→Q A→S 2 D 12 A,C 3 B 13 A,C B→P B→S 4 B 14 A ,C , D 5 D 15 D C→R C→P 6 C 16 A 10 B,D 20 B C HEM I ST RY Ques An s Ques An s 21 22 1 A 11 A,C A→S A→R 2 B 12 B 3 B 13 A,B B→P B → P,S 4 D 14 A , B ,C ,D 5 D 15 A C→Q C→Q 6 D 16 A 7 A 17 D D→R D→Q 8 B 18 B 9 B 19 B 10 A ,D 20 B MATHEMATICS Ques An s Ques An s 21 22 1 D 11 A,C A→S A→R XtraEdge for IIT-JEE 2 C 12 B,C 3 A 13 A ,B , C B→Q B→P 4 B 14 B,C 5 D 15 A C→P C→S 99 6 B 16 C 7 A 17 B D→R D→Q 8 A 18 A 9 B 19 C 10 A ,B ,C 20 B JANUARY 2010 XtraEdge for IIT-JEE 100 JANUARY 2010