Nov 2009

If you can't make a mistake, you can't make anything.

Volume - 5 Issue - 5 November, 2009 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009

Editorial

Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected] Editor :

Dear Students,

Pramod Maheshwari

It's the question you dreamed about when you were ten years old. It's the question our parents nagged you about during high school. It's the question that stresses most of us out more and more the older we get. "What do you want to be when you grow up?" There are people who are studying political science but hate politics, nursing majors who hate biology, and accounting majors who hate math. Obviously, a lot of people are confused about what exactly it is that they want to spend their life doing. Think about it. if you work for 10 hours each day, you're going to end up spending over 50% of your awake life at work. Personally, I think it's important that we spend that 50% of your awake life at work. Personally, I think it's important that we spend that 50% wisely. But how can you make sure that you do? Here are some cool tips for how to decide that you really want to be when you grow up. • Relax and Keep an Open Mind: Contrary to popular belief, you don't have to "choose a career" and stick with it for the rest of your life. You never have to sign a contract that says, "I agree to force myself to do this for the rest of my life" You're free to do whatever you want and the possibilities are endless. So relax, dream big, and keep an open mind. • Notice Your Passions: Every one of us is born with an innate desire to do something purposeful with our lives. We long to do something that we're passionate about; something that will make a meaningful impact on the world. • Figure Out How to Use Your Passions for a Larger Purpose: You notice that this is one of your passions, so you decide to become a personal trainer. Making a positive impact on the world will not only ensure that you are successful financially, it will also make you feel wonderful. It's proven principle: The more you give to the world, the more the world will give you in return. • Figure Our How You Can Benefit: Once you've figured out what your passions are and how you can use those passions to add value to the world & to yourself, it's time to take the last step: figure out how you can make great success doing it. my most important piece of advice about this last step is to remember just that: It's the last part of the decision process. I feel sorry for people who choose an occupation based on the average income for that field. No amount of money can compensate for a life wasted at a job that makes you miserable. However, that's not to say that the money isn't important. Money is important, and I'm a firm believer in the concept that no matter what it is that you love doing, there's at least one way to make extraordinary money doing it. So be creative! No matter how successful you become, how great your life is, or how beautiful you happen to be... there will still be times when you simply feel like you're an ugly mess. But when those times come, remember that all you need to get yourself back on track is a positive outlook, a dash of self confidence, and the willingness to make yourself feel better as soon as you know how. Simply discover your passions, figure out how to use your passions to make an impact on the world & to yourself. Presenting forever positive ideas to your success.

[B.Tech. IIT-Delhi] Analyst & Correspondent Mr. Ajay Jain Cover Design & Layout Mohammed Rafiq Om Gocher, Govind Saini 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.

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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

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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NOVEMBER 2009

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Volume-5 Issue-5 November, 2009 (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

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NEWS ARTICLE

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IITian ON THE PATH OF SUCCESS

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IIT-K signs MoU with US university White House names IIT-ian Arun Majumdar as America's Green Czar

Much more IIT-JEE News. Xtra Edge Test Series for JEE-2010 & 2011

Mr. S. Janakiraman

KNOW IIT-JEE

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Previous IIT-JEE Question

Study Time........ S Success Tips for the Months • If you can't make a mistake, you can't make anything. • Sometimes a big step is safer; you can't cross a ditch in small jumps • Self-confidence grows not from what you can do, but what you know you can do.

DYNAMIC PHYSICS 8-Challenging Problems [Set# 7] Students’ Forum Physics Fundamentals Electromagnetic Induction & A.C. Simple Harmonic Motion

CATALYSE CHEMISTRY

• You never need to feel fear if you don't want to do anything. • You got to know when to hold ‘em and know when to fold ‘em…

DICEY MATHS

• Defeat is advance payment for victory.

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Mathematical Challenges Students’ Forum Key Concept Differentiation Straight Line & Circle

• An ounce of success is worth a pound of positive thinking. • To understand motivation, know the power of the Hunter.

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Key Concept Nitrogen Compounds Nitrogen Family Understanding : Physical Chemistry

• Children focus on what they can’t do. Adults focus on what they can do. • The secret of confidence is to know your resources.

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Test Time .......... XTRAEDGE TEST SERIES

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Class XII – IIT-JEE 2010 Paper Class XII – IIT-JEE 2011 Paper

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NOVEMBER 2009

IIT reviews expel rule The IITs are considering scrapping a weeding process they practice to expel weak students mid-course, stung by a string of legal challenges and allegations of caste discrimination. Officials across the IITs have held two rounds of discussions on a proposal to replace expulsion of weak students with performance checks, administrators, including two directors Around 10 students are dismissed from each IIT on an average every year for failing to earn a minimum number of credits required at midcourse stages, the administrators said. An end to the system would mean that students, once admitted, would not be dismissed. The move follows caste allegations against the IITs — a majority of students expelled during their courses for poor performance belong to Scheduled Castes or Scheduled Tribes. It also comes at a time when the Supreme Court has held that the IITs cannot “throw out” SC/ST students on the basis of poor performance. In cases when students are asked to leave, they are given the option to quit the BTech course, and instead opt for a less reputed diploma. Hardly any student opts for it, officials said. The students argue that as they have cleared the IIT entrance test, their ability to pursue the BTech course cannot be challenged.

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IIT-K signs MoU with US university KANPUR: Indian Institute of Technology-Kanpur signed an MoU with the University of Texas, San Antonio, US, here on Saturday. Director, IIT-K, SG Dhande represented the institute and CM Agarwal (an alumnus of IIT-K of the 1982 batch), dean, College of Engineering, represented the University of Texas. They hoped that the two educational institutes will be benefited from the agreement. "The areas identified for collaborative research include fields like bio-material, where the focus will be on the research collaboration, faculty and student exchange programme. The cooperative research programme in other areas of mutual interest will be conceived later," Dhande said while talking to TOI. Agarwal said that Indo-US center for bio-material has been established which will concentrate on the development of material which will be useful in treating patients suffering from knee problems like arthritis. "The long-term goal is that the implants done in a human body should end as knee replacement in India is a costly affair and costs between Rs 30,000 and Rs 2 lakh. The best part would be that the human body itself reaches a stage where it can regenerate bone. This is called tissue engineering and doctors, scientists and biologists are working together for this cause," Agarwal said. Next month, a delegation will visit IIT-K to hold talks with Dhande, Agarwal added. 4

White House names IIT-ian Arun Majumdar as America's Green Czar WASHINGTON: There was more than a hint of irony in the Obama White House on Friday naming Arun Majumdar, a product of the best engineering schools in India and US, as the first Director of the US Department of Energy's Advanced Research Projects Agency-Energy (ARPA-E), an agency tasked with reducing America's reliance on foreign energy supplies, cutting greenhouse gas emissions, and improving energy efficiency. That an (Indian) immigrant engineer-scientist should head the premier agency at a time Washington is hectoring the world, principally India and China, to cut emissions, amid a growing trade and job protectionism, says something about the United States – and Majumdar was quick to articulate it. "It is a rare privilege and an honor when the President asks you to serve the nation in such a capacity," said Majumdar of his nomination, which, while needing to be confirmed by the Senate, sent ripples across the country's scientific-academic community. "I came to this country as an immigrant and am deeply appreciative and indebted to this nation for opening the doors and welcoming me with open arms. I have received so much. This is my way of stepping up and paying back." Not that the IIT-Mumbai graduate has forsaken his roots – in fact, his roots may well have been responsible for his nomination.

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Ever since he joined the University of California (UC), Berkeley faculty in 1997, where he holds the Almy and Agnes Maynard Chair Professorship in the College of Engineering and heads the Environmental Energy Technologies Division, Majumdar has cemented the Lab's role as a world-renowned leader in energy efficiency research in close collaboration with India and China – a feat the White House has been quick to recognize and reward. Among the lab's partnerships is the Berkeley-India Joint Leadership on Energy and the Environment announced last year, which brings together researchers from Berkeley Lab and UC Berkeley, and other US and Indian universities and institutions, with a goal to reduce greenhouse gas emissions while maintaining sustained economic growth in both nations. Another partnership between the Lab and China's Tsinghua University is to promote the shared development and implementation of building energy efficiency, a move intended to reduce energy consumption and greenhouse gas emissions in the US and China. In fact, Majumdar's mentor in academia was Professor Chang-Lin Tien, a legendary Chinese Don who went to become the Chancellor of UC Berkeley in 1990, the first Asian to head a major university in the United States. Majumdar's India-China connections is what appears to have driven the Obama White House to choose him for the new job, considering the two countries are thought to be pivotal in the upcoming energy debate. "He has had a highly distinguished research career in the science and engineering of energy conversion, transport, and storage ranging from molecular and nanoscale level to large energy systems," the

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White House said in its announcement. "At Berkeley Labs and UC Berkeley, he helped shape several strategic initiatives in the areas of energy efficiency, renewable energy as well as energy storage." For more than a decade, Majumdar, who is also the founding chair of the American Society of Mechanical Engineers' Nanotechnology Institute, has been the country's leading materials scientist, making spectacular advances in energy conservation. He was recently credited with developing a way to use silicon nanowires to capture and use the energy lost as heat during the production of electricity. The futuristic technology could someday be used to convert the large amounts of waste heat into useful electricity. A graduate (mechanical engineering) of IIT-Mumbai, Majumdar came to the US in 1985 and received a Ph.D. in mechanical engineering from the University of California, Berkeley in 1989. His nomination continues the steady march of Indian geeks and academics in the higher echelons of administration. The Obama administration picked IndianAmericans for the post of White House Chief Information Officer (Vivek Kundra) and Chief Technology Officer (Aneesh Chopra). Majumdar will effectively be the Chief Energy Officer.

IIT-B looks to solar power To help facilitate cost-effective solar thermal power generation, IIT Bombay plans to develop a megawatt-scale solar thermal power facility, which is being sponsored by the ministry of new and renewable energy. "The idea is to help create costeffective solar power. There is a huge gap between the demand and supply of electricity and one option worth developing is solar power," said a faculty member of 5

IIT Bombay. The plan to build the plant was proposed by IIT-B last year and it will come up at the Solar Energy Centre in Gurgaon. It will be connected to a grid and supply around a megawatt to the national grid. The test and simulation facility will be set up by a consortium involving different Indian industries and IIT-B. "While the US and Europe have already built such consortia, it will be a first in India. This facility is expected to help in developing inexpensive solar power plants in the future," he said. Even as the test facility will enable assessment of new technologies, components, and systems for solar thermal power, the simulation can be used to scale up designs and optimise use of solar power. "Besides developing indigenous capability, it is expected to provide the experience in concentrated solar power, which has the potential to provide a sustainable energy solution for India's power system," said the faculty member. The project, which will last for five years, is expected to start in another two years.

Signature drive launched for IIT status to ISM RANCHI: A signature campaign seeking IIT status for the Indian School of Mines (ISM), Dhanbad, was launched , On the first day of the three-day campaign, students managed to collect 12,500 signatures in favour of their demand. According to the campaign activists, a total of 1,25,000 signatures have been obtained so far in favour of the demand from across the state. Out of this, about 75,000 signatures were obtained from Dhanbad alone. The campaign was simultaneously launched at Dhanbad, Jamshedpur and Bokaro on Monday last to NOVEMBER 2009

garner the support of the people of Jharkhand towards the cause. "A delegation of ISM students met Congress MP Sachin Pilot on Friday and he too extended support to the cause," said a campaign activist. Earlier, a delegation had called on Union minister of state for human resources development D Purandeshwari in this regard. The minister had promised to look into the matter.

Cong tussle home for IIT

holds

up

New Delhi: A tussle between Congress leaders is depriving IIT Rajasthan of a permanent home two years after its conception, replacing an earlier battle the party waged over the institute with the BJP when it ruled the state. Coaching hub Kota, proposed by the former BJP state government as the venue for the IIT but dismissed by the UPA at the Centre, has now found powerful supporters within the Congress. The reason behind their demand: they had promised Kota an IIT in the Lok Sabha elections. The problem: others in the Congress had campaigned on bringing the same IIT to Jodhpur, home of current chief minister Ashok Gehlot. The Congress won in both Kota and Jodhpur seats in the Lok Sabha polls and is well ensconced in Rajasthan, where it is also in power. IIT Rajasthan, launched last year, however, continues to reside as a tenant on the IIT Kanpur campus. “It is a strange situation where we don’t know how to proceed. So we are likely to just sit on any decision for the time being,” a top government official said. Senior Congress leader Digvijay Singh has, in his capacity as party

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general secretary, written to human resource development minister Kapil Sibal, requesting that the new IIT be set up in Kota.

IIT-K students the benefits of technology

Digvijay was a Congress observer for Rajasthan during the Lok Sabha polls and his letter represents concerns of several party MPs and MLAs from the southern parts of the state, sources said.

KANPUR: "The MEMS (MicroElectro Mechanical System) technology is widely used in the cellphones, which are being used by a large percentage of population nowadays. The major use of the MEMS technology is in the field of IT followed by the consumer electronics sector," said Dr V K Aatre, former scientific adviser to the defence minister, who was on his visit to the IIT-K, here on Thursday. Dr Aatre further informed that All India Institute of Medical Sciences (AIIMS) and IIT-Bombay are together working on a device called cardiac monitor (it's a temperature measuring device) which will provide the temperature of the arteries. This cardiac monitor would based on the MEMS technology.

Kota MP Ijyaraj Singh, too, has met junior HRD minister D. Purandeswari, requesting that the IIT be set up in his constituency. But accepting Kota as the venue for the new IIT will not prove easy for the HRD ministry, sources said, because of a clear position it had earlier taken against the town’s eligibility. The Rajasthan IIT is one of eight promised by the UPA under the Eleventh Five Year Plan and was announced in 2007. Vasundhara Raje Scindia, who was then chief minister, proposed Kota, the nearest big town to her family fief in Jhalawar, as the venue. But a central team sent to examine prospective sites advised against Kota, arguing that it was poorly connected and would not attract top teachers, students and industry. Scindia accused then HRD minister Arjun Singh of “playing politics” over the IIT’s location. Last December, after the Congress wrested Rajasthan from the BJP in the Assembly polls, new chief minister Gehlot appointed a team to propose afresh a venue for the new institute. The Gehlot-appointed panel recommended Jodhpur, the chief minister’s hometown. In the Lok Sabha polls that followed a few months later, the Congress campaign for party candidate Chandresh Kumari in Jodhpur promised the IIT to the city. 6

learn MEMS

"AIIMS and IIT-Bombay will now be going for the animal trials of the cardiac monitor before going for a human trial which is a genuine procedure," added Dr Aatre. He also added that MEMS based technology is seeing its use in the fields of automobile, electronics, bio-medical, defence etc. Dr Vikram Kumar, former director of NPL (it is one of the labs of CSIR) and professor, department of Physics at IIT-Delhi said, "We carry MEMS-based gadgets with us on a daily basis but we are hardly aware of it. The mobile phones and I-Pods are the best examples of the same. The MEMS technology is also used in Plasma TVs and other household consumer electronics." Dr Kumar further said that the pressure sensors based on MEMS technology is widely used in the automobile industry. "At present the pressure sensors are being imported from the other companies of the world. The NOVEMBER 2009

pressure sensors have varied applications, especially in the fields like automobile, aerospace, acceleration etc", he added. "The pressure sensors are of various kinds and even used in a rocket," he said. Dr N S Vyas, professor and HOD of the department of Mechanical Engineering, IIT-Kanpur emphasised on the fact that MEMS technology has several potential applications in the railways and automotive sector.

IITs strategise for more PhD scholars Efforts include joint MTech and PhD degrees and streamlining policies so that thesis papers are cleared within two months With research becoming a clear focus area at all Indian Institutes of Technology (IITs) and with the 20-30 per cent growth in sponsored research, the premier technology institutes are now targeting a 10-30 per cent increase in PhD scholars. Globally, China produces the maximum number of research scholars per year. It is widely recognised that there will be substantially more PhD engineers and scientists in China in 2010 than in the United States, as China produces three times the number of engineers per year. Smalley, a nobel prize-winning scientist from Rice University, recently concluded that by 2010, 90 per cent of all PhD physical scientists and engineers in the world will be Asians living in Asia, and among Asian PhD engineers and scientists, most will be produced by China. India, therefore, is in a hurry to catch up. IIT-Kharagpur (IIT-KGP), for instance, awarded 212 PhDs this year, of which nearly 70 per cent had studied engineering. Last year, the institute had awarded 167 PhDs.

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“We want at least 30 per cent of our students to be research scholars, double of what it is right now. We are making several enticements for that, like joint MTech and PhD degrees and streamlining policies so that thesis papers are cleared within two months instead of one year which is usually the norm,” said Damodar Acharya, director of IITKGP. The institute has also introduced joint degree programmes with other reputed universities in India and abroad. A student admitted to such joint degree programmes has to spend upto two years in the partnering university and would have a joint guide. Through this programme, the institute aims at producing high quality faculty who will have exposure to at least two different environments. The institute from its own fund supports written airfare up to two visits of the students to the partnering university. The local expenses of the student are taken care of by the partnering university. At IIT-Bombay, 140 PhDs graduated in 2007, 200 in 2008 and around 175 in 2009. "We are incubating our PhD students using their intellectual properties. This should encourage students and make them feel more secure about their research findings," said Rangan Banerjee, dean of research and development at IIT Bombay. At IIT Madras, from 2006 to 2009 there has been a 50 per cent increase in PhD intake. Currently the institute has around 1100 PhD scholars, informed Job Kurian, dean of sponsored research at IIT Madras. IIT Madras aims to have a 1:1 ratio between research scholars and undergraduates, from 1:5 ratio currently, said Kurian. IIT-Delhi has seen a 23 per cent increase in the number of PhD degrees given out this year. A total number of 181 PhD degrees was awarded as compared to 147 last year. "This is a phenomenal 7

achievement and is contrary to belief that we are very poor on research output," said M Balakrishnan, dean of postgraduate studies at IIT-Delhi. IIT Bhubaneswar, one of the newest IITs in India, is encouraging faculty to join the institute with their own research scholars. M Chakraborty, director of IIT Bhubaneswar, said that the institute is also making provision for upto Rs 5 lakh research grant to a faculty. This would help them to invest in necessary infrastructure they require to carry out their research, like softwares, hardwares, books and journals, etc. Student researchers get a grant of Rs 15,000 per month. International exposure for faculty and student researchers and presenting their research papers at international conferences is another priority area for IIT Bhubaneswar. IIT Gandhinagar (IIT-G), another new IIT, has also started focusing on establishing the institute as a preferred destination for research students by initiating quality research activities on the campus. The institute, which was established in 2008, has just received its second batch of undergraduate students, but is already working on lines of creating a centre for research. Human Resource Development Minister Kapil Sibal had recently said the country's premier Indian Institutes of Technology (IITs) must focus on quality research and act as a catalyst to boost technical education in India. "This is not only necessary for the economic growth of the country but also for the IITs to make the transition as creator of knowledge. Without a large base of well educated undergraduates in the country it is difficult to imagine any significant growth in research output from these institutions," Sibal said. NOVEMBER 2009

Success Story This article contains story of a person who get succeed after graduation from different IIT's

Mr. S. Janakiraman B.E., M.E.(IIT – Chennai) President & Group CEO – Product Engineering Services, Mindtree Ltd.

Janakiraman (Jani) heads Product Engineering Services (PES) of MindTree as the President and Group CEO of

areas for customers like Alcatel-Lucent, AOL, Apple, Cisco, Microsoft, Real Networks, Symantec, Texas

the business unit. PES comprises of R&D Services (RDS) and Outsourced Product Development Services (OPD). PES under Jani covers whole spectrum of product and technology development services covering Semiconductor, Embedded System, Middleware and Application level Products for the Hi-Tech Product and

Instruments, Toshiba, UTC, Vendavo and Volvo. Jani has rich 28 years of experience in building R&D and Product Engineering Services organizations through setting up multiple dedicated development centers for the Engineering units for semiconductor, system and application product vendors. The services include IP

Independent Software Vendor (ISV) organizations. In addition the Research Units under PES build ready to use Intellectual Properties (IPs) and re-usable Technology Building Blocks (TBBs) covering segments like short range wireless like Bluetooth and UWB, Video surveillance platforms including management and analytics, virtualization and cloud computing.

licensing, Architecture-Design-Development, Independent Testing, Packaging, and Technical Support.

Starting from scratch, Jani has built the Product Engineering Services organization of MindTree through organic and inorganic means to over 3000 technology professionals executing projects in leading technology

of Commerce & Industries (IJCCI), Karnataka and a member of India Semiconductor Association (ISA) Executive Council.

Jani holds Bachelor's degree in Electronics and Communications from the National Institute of Technology (NIT), Trichy, India, and Master's degree in Electronics from the Indian Institute of Technology (IIT), Chennai, India. He is the President of Indo Japan Chamber

"Time is a circus, always packing up and moving away." XtraEdge for IIT-JEE

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NOVEMBER 2009

KNOW IIT-JEE By Previous Exam Questions

PHYSICS 1.

∴ 1 molecule will have mass

One gram mole of oxygen at 27º and one atmospheric pressure is enclosed in a vessel. [IIT-1983] (i) Assuming the molecules to be moving with vrms , Find the number of collisions per second which the molecules make with one square metre area of the vessel wall. (ii) The vessel is next thermally insulated and moved with a constant speed v0. It is then suddenly stopped. The process results in a rise of the temperature of the gas by 1ºC. Calculate the speed v0.

F Sol. (i) We know that P = A ∴ F = P × A = 105 × 1 = 105 N But

F=

∴ n=

= nC V ∆T= 1 × Cv × 1 = Cv From (i) and (ii)

...(ii)

1 Mv 20 = C V 2 Cp – C V = R

... (i)

∆p ∆t mv

Now, Cp C R ⇒ – V = CV CV CV ∴ CV =

... (iii)

⇒ γ–1=

R γ −1

R CV

... (iv)

From (iii) and (iv) 1 R Mv 20 = 2 γ −1 ∴ v0 =

2R = M (γ − 1)

2 × 8.314 32 × (1.41 − 1) 100

γ = 1.41 for O2 (diatomic gas) ⇒ v0 = 35.6 m/s

105 2mv

2.

3RT 3 × 8.314 × 300 = = 483.4 m/s M 32 / 1000

Hot oil is circulated through an insulated container with a wooden lid at the top whose conductivity K = 0.149 J/(m-ºC-sec), thickness t = 5 mm, emissivity = 0.6. Temperature of the top of the lid is maintained at Tl = 127ºC. If the ambient temperature Ta = 27ºC. [IIT-2003]

According to mole concept 6.023 × 1023 molecules will have mass 32 g

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105 × 6.023 × 10 23 = 1.97 × 1027 2 × 32 × 483.4

1 Mv 20 ... (i) 2 Where v0 is the velocity with which the vessel was moving. The heat gained by 1 gm mole of molecules at constant volume

Root mean square velocity v=

g

K.E. of 1 gm mole of oxygen =

∴ ∆p = F × ∆t = F × 1 = 105 [From (i)] ...(ii) Let m be the mass of one molecule and v be the r.m.s. velocity. Now Momentum change per second m2 (∆p) = n × 2mv ... (iii) Where n is the number of collisions per second per square metre area From (ii) and (iii) n × 2mv = 10 5 n=

6.023 ×10 23

(ii) The kinetic energy of motion of molecules will be converted into heat energy.

mv

∴

32

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NOVEMBER 2009

Tl = 127º

T0

− π R 2ρ g 3g =– x 4 2R πR 3ρs 3 F 4 [Q acceleration a = and m = π R3 ρs ] m 3

∴ a=

Ta = 27ºC Hot Oil

Calculate : (a) rate of heat loss per unit area due to radiation from the lid. 17 × 10–8) 3 Sol. (a) The rate of heat loss per unit area per second due to radiation is given by Stefan's-Boltzmann law

4.

(b) temperature of the oil. (Given σ =

3g 2R [Comparing with a = – ω 2 x]

∴

ω=

or

v=

1 2π

Two identical prisms of refractive index 3 are kept as shown in the figure. A light ray strikes the first prism at face AB. Find, [IIT-2005] B D

E = εσ(T4 – T04 )

60º

Sol.

595 × l 595 × 5 ×10 −3 +T= + 400 k 0.149 = 419.83 K

A sphere of radius R is half submerged in liquid of density ρ. if the sphere is slightly pushed down and released, find the frequency of oscillation. [IIT-2004] At equilibrium net force is zero ∴ Fmg = Fbuouncy or ρm ×

sin i = 3 × sin 30º =

3 2

⇒ i = 60º B

4 3 2 π r × g = ρ × π r3 g 3 3

P

ρ ⇒ ρm = 2 Let the sphere is slightly displaced downward by x.

i

60º r M Q

N

60º C A (b) When the prism DCE is rotated about C in anticlockwise direction, as shown in the figure, then the final emergent ray SR becomes parallel to the incident ray TM. Thus, the angle of deviation becomes zero.

∴ Fres = – π R2 xρg [Q Volume of submerged portion of sphere increases by πR2 x, hence buouncy increases by π R2 xρg]

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60º

C A E (a) the angle of incidence, so that the emergent ray from the first prism has minimum deviation. (b) through what angle the prism DCE should be rotated about C so that the final emergent ray also has minimum deviation. Sol. (a) For minimum deviation of emergent ray from the first prism MN is parallel to AC ∴ ∠ BMN = 90º ⇒ ∠ r = 30º Applying Snell's law at M sin i µ= sin r sin i = µ sin r

KA(Toil − T ) = 595 × A where A is the area of l the top of lid

3.

60º

60º

17 = 0.6× × 10–8 [(400)4 – (300)4] 3 = 595 Watt/m2 (b) Let Toil be the temperature of the oil. Then rate of heat flow through conduction = Rate of heat flow through radiation.

⇒ Toil =

3g 2R

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NOVEMBER 2009

5.

A neutron of kinetic energy 65eV collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90º with respect of its original direction. [IIT-1993] (i) Find the allowed values of the energy of the neutron and that of the atom after the collision. (ii) If the atom get de-excited subsequently by emitting radiation, find the frequencies of the emitted radiation. [Given: mass of He atom –4×(mass of neutron), Ionization energy of H atom = 13.6eV]

Case (3) ∆E3 = – 3.4 – (–54.4eV) = 51.1 eV ⇒ K1 + K2 = 14 eV Solving with (3), we get K2 = 15.8 eV; K1 = – 1.8 eV But K.E. can never be negative therefore case (3) is not possible. Therefore the allowed values of kinetic energies are only that of case (1) and case (2) and electron can jump upto n = 3 only. n=4 –3.4eV n=3 –6.04eV

Sol. m

K2 θ

m 4m

4m

x

–54.4eV

K1

2 Km = 2(4 m)K1 cos θ

n3 → n1 or n3 → n2 and n2 → n1 The frequencies will be E3 − E 2 E − E1 E − E1 ν2= 3 ν3= 2 h h h 15 15 = 1.82×10 Hz = 11.67×10 Hz = 9.84×10 15 Hz ν1 =

...(i)

Now applying conservation of linear momentum in Y-direction piy = pfy

CHEMISTRY

0 = 2 K 2 m – 2(4 m)K 1 sin θ ⇒

2 K 2 m = 2(4 m)K 1 sin θ

...(ii) Estimate the difference in energy between 1st and 2nd Bohr orbit for a H atom. At what minimum atomic no., a transition from n=2 to n = 1 energy level would result in the emission of X-rays with λ = 3.0×10–8 m. Which hydrogen atom like species does this atomic no. corresponds to ? [IIT-1993] Sol. (a) For H atom, Z=1 ni = 2 nf = 1 6.

Squaring and adding (i) and (ii) 2Km + 2K2m = 2(4m)K1 + 2(4m)K1 K1 + K2 = 4K1 ⇒ K = 4K1 – K2 ⇒ 4K1 – K2 = 65 ...(iii) When collision takes place, the electron gains energy and jumps to higher orbit. Applying energy conservation K = K1 + K2 + ∆E ⇒ 65 = K1 + K2 + ∆E ...(iv) + Possible value of ∆E For He Case (1) ∆E1 = – 13.6 – (54.4eV) = 40.8 eV ⇒ K1 + K2 = 24.2 eV from (4) Solving with (3), we get K2 = 6.36 eV; K1 = 17.84 eV Case (2) ∆E2 = – 6.04 – (–54.4 eV) = 48.36 eV ⇒ K1 + K2 = 16.64 eV from (4) Solving with (3), we get K2 = 0.312 eV; K1 = 16.328 eV

XtraEdge for IIT-JEE

n=1

For He+

(ii) Thus when electron jumps back there are three possibilities

Applying conservation of linear momentum in horizontal direction (Initial Momentum) x = (Final Momentum) x pix = p fx ⇒

n=2

–13.6eV

y

En = –

21.76 ×10 −19

J n2 Hence, difference in energy between first and second Bohr orbit for a H-atom is given by, ∆E = E n i – E n f = E2 – E1 =–

21.76 ×10 −19 22

+

21.76 ×10 −19 12

1 1 = – 21.76 × 10–19  2 − 2  = 16.32 × 10–19J 1  2 12

NOVEMBER 2009

(b) For λ = 3.0 × 10–8 m −34

Also, 1 mol of O2 yields = 1 mol of N2O4 ∴ Number of moles of N2O4 formed = 0.00320 mol N2O4 condenses on cooling, ∴ 0.350 L (0.1 + 0.250) contains only 0.00429 mol of NO At T = 220 K, Pressure of the gas,

8

hc 6.626 × 10 × 3 × 10 = λ 3.0 ×10 −8 = 6.626 × 10–18 J ....(i) We know that, for H-like atoms, En for H-like atom = En for H-atom × Z2 ∴ ∆E for H-like atom = Z2 × ∆E for H-atom ∆E =

1 1 = –Z2 × 21.76 × 10–19  2 − 2  1  2 = 16.32 × 10–19 Z2 ...(ii) From eq. (i) and (ii), 16.32 × 10–19 Z2 = 6.626 × 10–18 or Z=2 Thus, hydrogen atom like species for Z = 2 is He+.

P= 8.

An organic compound C xH2yOy was burnt with twice the amount of oxygen needed for complete combustion to CO2 and H2O. The hot gases when cooled to 0 ºC and 1 atm pressure, measure 2.24 L. The water collected during cooling weighed 0.9 g. The vapour pressure of pure water at 20ºC is 17.5 mm of Hg and is lowered by 0.104 mm when 50 g of the organic compound are dissolved in 1000 g of water. Give the molecular formula of the organic compound. [IIT-1983] Sol. According to the question, an organic compound C xH2yOy was burnt with twice the amount of oxygen. Hence, C xH2yOy + 2x O2 → xCO2 + yH2O + xO2 Volume of gases after combustion = 2.24 L (given) Volume of gases left after combustion = xCO2 + xO2 ∴ x + x = 2.24 or x = 1.12 L 22.4 L CO2 = 1 mol CO2

7.

At room temperature, the following reactions proceed nearly to completion : 2NO + O2 → 2NO2 → N2O4 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 the gases to behave ideally) [IIT-1992] Sol. According to the gas equation, PV = nRT

or

n=

PV RT

1.12 = 0.05 mol CO2 22.4 and 18 g H2O = 1 mol H2O 0.9 ∴ 0.9 g H2O = = 0.05 mol H2O 18 Thus, the empirical formula of the organic compound is CH2O. Empirical formula mass = 12 + 2 + 16 = 30 Vapour pressure of the pure liquid, ∴

At room temperature, For NO, P = 1.053 atm, V = 250 ml = 0.250 L 1.053 × 0.250 ∴ Number of moles of NO = 0.0821× 300 = 0.01069 mol For O2, P = 0.789 atm, V = 100 ml = 0.1L 0.789 × 0.1 ∴ Number of moles of O2 = 0.0821 × 300 = 0.00320 mol According to the given reaction, 2NO + O2 → 2NO2 → N2O4 Composition of gas after completion of reaction, Number of moles of O2 = 0 1 mol of O2 react with = 2 mol of NO ∴ 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

XtraEdge for IIT-JEE

nRT 0.00429 × 0.0821× 220 = = 0.221 atm V 0.350

1.12 L CO2 =

PA0 = 17.5 mm of Hg Lowering in vapour pressure PA0 –PA =0.104mm of Hg Mass of organic compound = 50 g Mass of water = 1000 g 50 / M 50 1000 + M 18 where M is the molecular mass of the organic compound, the molecular mass of water being 18. ∴ Mole fraction of organic compound =

13

NOVEMBER 2009

We know, PA0 − PA

The reactions are : CH3CH2CH2COOCH2CH3

= Mole fraction of organic compound

PA0

Ethyl butanoate (A)

0.104 50 / M = 50 1000 17.5 + M 18

∴

Solving,

CH3CH2CH2CH2OH + CH 3CH2OH Reduction

CH3CH(OH)CH 2CHO

M = 150.5 ≈ 150

CH3CHO Ethanal (D) [O]

CH3COOH Ethanoic acid (F)

= 5(CH2O) = C5H10O5

10. A white substance (A) reacts with dil. H2SO4 to produce a colourless gas (B) and a colourless solution (C). The reaction between (B) and acidified K2Cr2O7 solution produces a green solution and a slightly coloured precipitate (D). The substance (D) burns in air to produce a gas (E) which reacts with (B) to yield (D) and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous NH3 or NaOH to (C) produces first a precipitate, which dissolves in the excess of the respective reagent to produce a clear solution in each case. Identify (A), (B), (C), (D) and (E). Write the equations of the reactions involved. [IIT-2001] Sol. (A) is ZnS,

Compound (A) (C 6H12O2) on reduction with LiAlH4 yielded two compounds (B) and (C). The compound (B) on oxidation gave (D) which on treatment with alkali (aqueous) and subsequent heating furnished (E). The later on catalytic hydrogenation gave (C). The compound (D) was oxidised further to give (F) which was found to be monobasic acid (m. wt. 60.0). [IIT – 1990]

Sol. Clue 1. Compound (F) is a monobasic acid molecular mass 60. ∴

aq.NaOH

Aldol

∴ Molecular formula or organic compound

Deduce structures of (A) to (E).

[O]

Crotonaldehyde

Molecular mass 150 n= = =5 Empirical formula mass 30

9.

Ethanol (B)

Butanol (C)

CH3CH = CHCHO

17.5 1000 × M =1+ 0.104 18 × 50

or

LiAlH4

CnH2n+1COOH = 60

or n × C + (2n + 1) × H + C + 2 × O + H = 60 or 12n + 2n + 1 + 12 + 2 × 16 + 1 = 60

ZnS + H2SO4 → ZnSO 4 + H 2S

60 − 46 or n = =1 14

( A)

(C )

(B )

3H 2S + K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3

∴ (F) is CH3COOH.

( B)

+ 7H2O +

Clue 2. (D) on oxidation gives (F), therefore (D) is CH3CHO.

3S

White grey (D )

S + O 2 → SO 2

Clue 3. (B) on oxidation gives (D), therefore (B) is

(D )

CH3CH2OH.

Air

( E)

SO 2 + 2H 2S →

Clue 4. (D) undergoes aldol condensation and on

( E)

heating gives (E), therefore (E) is CH3CH = CHCHO.

(B )

2H 2 O Colourless liquid(C )

+ 3S (D)

5H2O + CuSO 4 → CuSO 4 .5H 2 O

Clue 5. (E) on reduction gives (C), therefore (C) is

White

Blue

CH3CH2CH2CH2OH.

ZnSO4 + 2NaOH → Zn(OH)2 + Na 2SO4

Clue 6. (A) having formula C6H12O2 on reduction

Zn(OH)2 + 2NaOH → Na2ZnO2 (soluble) + 2H2O Also in excess of NH4OH it forms soluble complex [Zn(NH3)4](OH)2.

yields (B) and (C). ∴ (A) is CH3CH2CH2COOCH2CH3.

XtraEdge for IIT-JEE

14

NOVEMBER 2009

MATHEMATICS

P(C) = probability that C will hit A = P(E) = probability that A will be hit

11. Find the centre and radius of the circle formed by all the points represented by z = x + iy satisfying the relation

⇒ P(E) = 1 – P( B ). P( C )

z−α =k(k ≠ 1), where α and β are constant z −β

=1–

⇒

2

| z −α | = k2 | z − β |2

⇒

(z – α)( z – α ) = k2(z – β)( z – β )

+ (|α|2 – k2|β|2) = 0 (α − k 2 β) (1 − k 2 )

(α − β k 2 )

z–

(1 − k 2 ) 2

+

2

| α | −k | β |

z

dy du dv = + dx dx dx

2

(1 − k 2 )

=0

...(i)

whose centre is (–a) and radius = | a | 2 − b ∴ centre for (i) α − k 2β 1− k2

and radius

 α − k 2 β  α − k 2 β  α α − k 2 β β  − =  2  2  1− k2  1 − k  1 − k  radius =

⇒

loge u = y loge (1 + x)

⇒

1 du y dy = + . {loge (1 + x)} u dx 1 + x dx

⇒

du dy  y  =(1+x)y  + loge (1 + x ) .....(2) dx 1 + x dx  v = sin–1 sin2 x

Again,

k (α − β) 1− k 2

⇒

sin v = sin2 x

⇒

cos v

⇒

dv = 2. sin x cos x dx dv 1 = [2 sin x cos x] dx cos v

⇒

12. A is targeting to B, B and C are targeting to A. Probability of hitting the target by A, B and C are 2 1 1 , and respectively. If A is hit, then find the 3 2 3 probability that B hits the target and C does not. [IIT-2003] Sol. Here,

=

2 sin x cos x 2

1 − sin v

=

2 sin x cos x 1 − sin 4 x

.....(3)

Put these values in equation (1) dy dy  y  2 sin x cos x = (1 + x)y  + loge (1 + x ) + dx 1 + x dx   1 − sin 4 x

P(A) = probability that A will hit B =

2 3

⇒

P(B) = probability that B will hit A =

1 2

At

XtraEdge for IIT-JEE

.....(1)

Now, u = (1 + x) take logarithm of both sides loge u = loge (1 + x)y

On comparing with equation of circle, |z|2 + a z + α z + b = 0

=

1 2 . P(B ).P(C) 2 3 1 = = 2 P (E ) 2 3

13. Find the equation of the normal to the curve y = (1 + x)y + sin–1 (sin2 x) at x = 0 [IIT-1993] y –1 2 Sol. y = (1 + x) + sin (sin x) (given) y Let y = u + v, where u = (1 + x) , v = sin–1 (sin2 x). Differentiating

or |z|2 (1 – k2) – (α – k2β) z – ( α – β k2) z

|z|2 –

P(B ∩ C / E)

⇒

|z|2 – α z – α z + |α|2 = k2(|z|2 – β z – β z+ |β|2)

⇒

1 2 2 . = 2 3 3

Probability if A is hit by B and not by C.

complex numbers given by α = α1 + iα2, β = β 1 + iβ2. [IIT-2004] Sol. As we know; |z|2 = z. z ⇒

1 3

15

dy y (1 + x ) y −1 + 2 sin x cos x / 1 − sin 4 x = dx 1 − (1 + x ) y ln(1 + x ) x=0 y = (1 + 0)y + sin–1 sin (0) = 1 NOVEMBER 2009

⇒

dy 1(1 + 0)1−1 + 2 sin 0. cos 0 / (1 − sin 4 0) = dx 1 − (1 + 0)1 ln(1 + 0)

Also,

dy =1 dx Again the slope of the normal is

⇒

m=–

1 =–1 dy / dx

Thus, the required equation of the normal is y – 1 = (– 1) (x – 0) i.e., y + x – 1 = 0. 14. Determine the equation of the curve passing through the origin in the from y = f(x), which satisfies the dy differential equation = sin (10x + 6y) [IIT-1996] dx dy Sol. = sin (10x + 6y) dx Let 10x + 6y = t (given) .....(1) dy  dt  ⇒ 10 + 6 =  dx  dx  dy 1  dt  =  −10  dx 6  dx  Now the given differential equation becomes 1  dt  sin t =  −10  6  dx 

I1 =

6sin t =

dt – 10 dx

∫

=

∫ (1+ u

=

2 5

∫u

=

2 5

∫u

=

2 5

∫

=

2 5

∫

=

2 5  u +3/5 . tan–1   5 4  4/5 

=

1  5u + 3  tan–1   2  4 

Let

I1 =

dt

∫ 3 sin t + 5 = x + c

⇒ ⇒ ⇒

.....(2)

+ (6 / 5)u + 1

2

du 6 9 9 + u+ − +1 5 25 25 du 2

3  16 u +  + 5 25  du 2

3  4 u +  +  5  5 

2

t   5 tan + 3   2 ⇒ tan–1   = 4x + 4c 4    

dt

∫ 3 sin t + 5

1 [5 tan (5x + 3y) + 3] = tan (4x + 4c) 4 ⇒ 5tan (5x + 3y) + 3 = 4 tan (4x + 4c) When x = 0, y = 0 we get 5 tan 0 + 3 = 4 tan (4c) ⇒

1 sec2 t/2 dt = du 2 2du dt = sec 2 t / 2 2du dt = 1 + tan 2 t / 2 2du dt = 1+ u 2

XtraEdge for IIT-JEE

du 2

1 I1 = x + c 2 t   5 tan + 3   1 –1 2 ⇒ tan   = x+c 4 4    

Put tan t/2 = u ⇒

)(5u 2 + 6 u + 5)

Now

∫

1 2

2

1  5 tan t / 2 + 3  tan–1   2 4   Putting this in (2)

dt = 6 sin t + 10 dx dt ⇒ = dx apply variable separable 6 sin t + 10 Integrating both the sides, we get dt = dx 6 sin t + 10 ⇒

(1 + tan 2 t / 2)dt t t (6 tan + 5 + 5 tan 2 ) 2 2 2(1 + u 2 )du

=

⇒

∫

dt 2 tan t / 2   +5  1 + tan 2 t / 2 

=

⇒

⇒

dt

∫ 3 sin t + 5 = ∫ 3

⇒ ⇒ 16

3 = tan 4c 4 3 4c = tan–1 4 NOVEMBER 2009

Then, 5 tan (5x + 3y) + 3 = 4 tan (4x + tan–1 3/4)

Therefore, coordinates of A are (3cosθ.3sinθ) Now, the joint equation of the tangents at A is given by T2 = SS 1

4 3 tan (4x + tan–1 3/4) – 5 5

⇒

tan (5x + 3y) =

⇒

3 4 5x + 3y = tan–1  {tan(4 x + tan −1 3 / 4} −  5 5 

⇒

3 4 3y = tan–1  {tan(4 x + tan −1 3 / 4} −  –5x 5 5

⇒

y=

2   h2 k2   hx ky   x 2 y 2 i.e.,  + − 1 =  + − 1  + − 1 .....(5) 3 3 3  6   6   6  In equation (5)  h 2 1  h 2 k 2 coefficient of x2 = –  + − 1 36 6  6 3 

1 –1  4 3  5x tan  {tan(4 x + tan −1 3 / 4} −  – 3 5 3 5

15. A tangent to the ellipse x2 + 4y2 = 4 meets the ellipse x2 + 2y2 = 6 at P and Q. Prove that the tangents at P and Q of the ellipse x2 + 2y2 = 6 arc at right angles. [IIT-1997] Sol. x2 + 4y2 = 4 (given)

3

P –2 O – 6

2

6

x

2

x + 2y = 6 ⇒

2

 k 2 1  h 2 k 2 –  + − 1 9 3  6 3 

–1

SCIENCE TIPS

– 3 2

1 k2 – 6 18

k2 h2 k2 1 h2 1 – – + =– + 9 18 9 3 18 3 Again coefficient of x2 + coefficient of y2 1 2 1 1 =– (h + k2) + + 18 6 3 1 1 =– (9cos2 θ + 9sin2 θ) + 18 2 9 1 =– + 18 2 1 1 =– + =0 2 2 which shows that two lines represent by (5) are at right angles to each other.

Q

1

=

=

.....(2)

90º

h2 h2 k2 1 – – + 36 36 18 6

coefficient of y2 =

x2 y2 ⇒ + =1 .....(1) 4 1 Equation of any tangent to the ellipse on (1) can be written as x cos θ + ysinθ = 1 2 Equation of second ellipse y A

=

(given) 2

x y + =1 6 3

• An electron is moving along X-axis in a magnetic field acting along Y-axis. What is the direction of magnetic force acting on it. ® Along Z-axis

.....(3)

Suppose the tangents at P and Q meet in A(h, k).

• What is the equation of a plane progressive simple harmonic wave traverlling in + x direction?

h x ky + =1 .....(4) 6 3 But (4) and (2) represent the same straight line, so comparing (4) and (2)

® y = a sin

h/6 k/3 1 = = cos θ / 2 sin θ 1

• What type of magnetic material is used in making permanent magnets? ® Ferromagnetic

⇒ h = 3cos θ and k = 3sinθ Equation of the chord of contact of the tangents through A(h, k) is

XtraEdge for IIT-JEE

2π  t x (vt – x) = a sin 2π  −  λ T λ

• A wire kept along north-south is allowed to fall freely. Will an induced emf be set up? ® No • Which of A.C. or D.C. is blocked by a capacitor? ® D.C.

17

NOVEMBER 2009

Physics Challenging Problems

Set # 7

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

Sol ut i ons w il l be publ i s he d i n ne xt is s ue Passage # 1 (Ques. 1 to 3)

5.

Upper Branch V1

V1

V1 ...............up to infinite.

a

b Lower Branch

1. 2.

3. 4.

V

Resistance of volt meters V1, V2, V3 ...... are R R R R R R, , , , , ...... respectively 3 5 ln e 2 4 ln e (2 ) 16 ln e (2 ) then Find the resistance of voltmeter V such that the current in upper branch is same as in lower branch. If the reading of voltmeter is V1 is X and the sum of reading of all the voltmeters in upper branch except voltmeter X is Y then. Is X = Y or not. Write the reason to support your answer. If the resistance of the voltmeter V is R then write the relation between the reading of voltmeter V1 and V. A rod is rotating about axis YY′ as shown below the linear charge density at distance x is λ(x) = 3x and it is rotating with angular speed ω about axis YY′ then

Part-A ω ω

Part-B ω

ω

Q

Q

σ

σ

C-1

C-2

C-3

C-4

(A) Equivalent current of C-1 and C-3 is same irrespective to value of σ (B) Equivalent current of C-1 and C-2 is same (C) Equivalent current of C-3 and C-4 is different (D) Equivalent current of Part - A and Part-B can be same dependent on value of σ Passage # 2 (Ques. 6 to 8) A multirange voltmeter is shown below. The galvanometer is having the resistance of it's coil as 10Ω and the maximum potential difference that can be applied across the galvanometer is 50mv then

Y x. A

Part-A and Part-B, a pair of closed and open cone is shown. In Part - A the charge on both the cones is same and in part - B surface charge density on both the cones is same All the cones are rotating with angular speed ω as shown in figure then

B

a. b.

CT

3ω 2 2 (b – a ) 4π (B) If length of the rod varies keeping (a + b) as 4π constant and angular speed ω = then 3 equivalent current i is directly proportional to length of rod (C) Charge on rod is 3(b2 – a2) (D) Charge on rod is 3/2(b2 – a2) (A) Equivalent current i =

XtraEdge for IIT-JEE

G R1

Y'

R3

R2 a 5V

b

c

CT is the common terminal 6. 7. 8.

18

If the range between CT and a is 5volt then the value of resistance R1. If the range between CT and b is just double as the range between CT and a then the value of resistance R2. If the value of R3 is 3000Ω then what will be the range between CT and c. NOVEMBER 2009

XtraEdge for IIT-JEE

19

NOVEMBER 2009

1.

8

Solution

Set # 6

Physics Challenging Problems Que s t i ons we r e Publ is he d i n Oc t obe r I s s ue

As shown in graph, the relation of U v/s PV is

4.

linear So,

f = b = C 0 + C 1t2, 2

As C V = b R and

U

df = 4C1t dt

f = 2C 0 + 2C 1t2 and

df / dt v/s t graph is a straight line with slope 4C1. (0, a)

As v and are B mutually perpendicular so path will be circular but due to presence of resistive

PV

U = (tan φ).PV + a

medium speed decreases and radius of circular

as (tan φ) = b

So, U = b. PV + a

path decreases.

Using ideal gas equation PV = nRT

So, path is spiral of decreasing radius

U = b (nRT) + a

Option (D) is correct [For Ans. 6, 7, 8]

Differentiate it, dU = nbRdT

The equivalent circuit diagram is

As

dU = n C V dT

As 'a' is grounded

So

C V = bR =

f R 2

⇒

f =b 2

a As 'a' grounded

Degrees of freedom of the gas, f = 2b as b = 3

Degrees of freedom are 6 so it is tri-atomic

3.

2 2 ⇒γ=1+ f 2b

So,

γ = 1 + b–1

d As 'd' is grounded

Y

x R=10Ω

as f = 2b 6.

Current through 'R' is from Y to x i=

As C V = bR it is not dependent on 'a' so if a varies there is no change in the value of C V. CV

120 = 12 Amp. 10

7.

Potential difference across a and c is 60 volt.

8.

Charge on deutron is e so energy of deutron is 120eV.

a

XtraEdge for IIT-JEE

e2 = vB(2e)=2vBI = 60 volt

b

non- linear gas. γ=1+

c e1 = 2v(B)1 = 2v BI = 60V

So f = 6

2.

→

→

5.

φ

20

NOVEMBER 2009

Students' Forum Expert’s Solution for Question asked by IIT-JEE Aspirants

PHYSICS

A stone is projected with velocity v0 at an angle θ0 from the horizontal. Find the angular velocity of the stone relative to the point of projection, while it is at its maximum height. Sol. Method 1 : The position vector of the particle at any arbitrary instant when it is at point P is given by : r(P) = xi + yj, r = y tanθi + yi Differentiating w.r.t. time we get, dr dy dy = tanθ i + y sec2θ θ& i + j dt dt dt At the highest point of trajectory. 1.

2.

dy v 2 sin 2 θ 0 = 0 and y = H = 0 dt 2g So, at the sought point, y ur

θ O

y x

a

H

m

(a) Find the time when the bead will leave the rod if the co-efficient of friction between bead and the rod is µk. (Neglect the weight of the bead) (b) Do the part (a) of this problem without neglecting the weight of the bead. Sol. Let us work in the frame of the rod. (a) In this case in the absence of friction, under the action of the inertial force – ma the bead will move up along the rod, hence the kinetic friction will act along the rod down (fig.) From F x = ma x ma cos α – µN = ma rel (1) From, Fy = may N = ma sin α (2) Using (2) in (1), we get arel = a(cos α – µ sin α) x y N

uθ x

R

v0 cos θ0 i = H sec θ . θ& i v cos θ 0 So, θ& = 0 cos2θ H  2g cos θ 0   cos2θ or, θ& =   v sin 2 θ   0 0  Method 2 : At an arbitrary instant, when the particle is at point P tan θ = x/y Differentiating Eqn. (1) w.r.t. time we get dy dx x −y 2 & dt dt sec θ θ = y2 2

a

m

fr α From kinematic equation in the frame of the rod. 1 1 ∆x = v0xt + a xt2 or, l = 0 + arel t2 2 2 2l 2l t= = a rel a (cos α − µ sin α)

dx   dy −y x  θ& =  dt 2 dt  cos2θ   y     dx dy At position P, = v0 cos θ0, = 0, dt dt x = R/2 and y = H

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α l

v

P

v 20 sin 2θ 0 v 2 sin 2 θ 0 and H = 0 g 2g Substituting the above values in Eqn.(2)  2g cos θ 0   cos2θ θ& =   v sin 2 θ   0 0  A bead of mass m is fitted onto a rough rod of length of 2l and can move along it only. At the initial moment the bead is in the middle of the rod. The rod moves translationally in space with the constant acceleration a in a direction forming an angle α with the rod (fig.) where R =

21

NOVEMBER 2009

(b) In the frame of rod, the bead is under the action of three forces N, –ma and mg, except kinetic friction. Case (i): If a cos α > g sin α, then the kinetic friction having the value of µkN will act to resist the upward motion of the bead along the rod. (Fig.) From, Fy = may (for the bead) N – mg cos α – ma sin α = 0 N = m a (sin α + g cos α) (1) and from F x = cosα – µN – mg sinα = ma rel (2) x y

=

2l g (sin α − µ cos α) − a (µ cos α + µ sin α)

[In this part of problem a cosα = g sinα is not relevant] 3.

Determine the period of oscillations of mercury of mass m poured into a bent tube whose arms form the angles θ1 and θ 2 with the vertical respectively (Fig. a). The cross-sectional area of the tube is s. neglect the viscosity of the mercury. θ1

θ1 θ2

N

x x

ma

a

fr mg α Using (1) in (2), we get arel = a cosα – µ(a sinα + g cosα) – g sinα or, arel = a(cosα + µsinα) – g(µ cosα + sinα) (3) From kinematic equation, (in the frame of rod) 1 ∆x = v0xt + a xt2 2 1 or, l = 0 + arel t2 2 2l 2l t= = a rel a (cos α − µ sin α ) − g(µ cos α + sin α)

•

1 • m( x )2 = Constant 2 sρg 2 1 • or, x (cosθ2 +cosθ 2) + m( x )2= Constant 2 2 Differentiating w.r.t. time, we get • •• sρg (cos θ1 + cos θ 2 ) 1 • 2x x + m 2 x x = 0 2 2 •• sρg (cos θ1 + cos θ 2 ) or x=– x m Thus, the sought time period, m T = 2π sρg (cos θ1 + cos θ 2 ) Method 2: If the mercury rises in the left arm by x, obviously it must fall by the same length in the right arm. At this position the total pressure difference in the two arms will be ρg x cos θ 1 + ρg x cos θ2 = ρg x (cos θ1 + cos θ2) This will give rise to a restoring force – ρg x (cos θ1 + cos θ2) s This must equal mass times acceleration which can be obtained from work-energy principle. +

fr ma α

mg

x' or, a'rel =g(sinα – µ(g cosα) – a(cosα + sinα)) (6) From the kinematic equation, 1 ∆x' = v 0x' t + a x' t2 2 1 or, l = 0 + a'rel t2 2 2l So, t' = a ' rel

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••

Sol. Method 1: ( x = dx/dt, x = d2x/dt2) If at an arbitrary moment during the oscillations, the mercury rises in the left arm by x it must fall by the same length in the other arm. As the viscosity of the mercury is negligible, the mechanical energy of oscillations of the given system is conserved, i.e. U(x) + T(x, x) = Constant. Taking the P.E. of oscillations zero at the equilibrium position and using the mathematical trick of negative mass, the conservation of M.E. of oscillations gives: x cos θ1  x cos θ 2  (s x ρ) g – sx (−ρ)g  2 2  

Case (ii): When g sinα > a cosα , the bead will move down along the rod, so the kinetic friction will act upward along the rod. (Fig.) From, Fy' = may' N = m (g cosα + a sinα) (4) and from F x' = ma x' mg sinα – µN – mg cosα = m arel (5) Using (4) in (5), we get a'rel = g sinα – µ(g cosα + a sinα) – a cosα N

θ2

The K.E. of the mercury 1 in the tube is clearly : m(x)2 2 22

NOVEMBER 2009

F = q (v × B)

••

So mass times acceleration must be : m x Hence,

 µ i  F = q  v x i + v y j ×  0 ( −k )    2π x   µ qi = 0 (–vy i + vxj) (A) 2π x µ q i vy µ qi vy Fx = – 0 so, ax = – 0 2π mx 2π mx v x dv x µ 0 q i v y = dx 2π mx 2 2 v x + v y = v2

{

or,

••

m x + ρ g x(cos θ1 + cos θ 2) s = 0 which is the Eq. of S.H.M. with time period

or,

m T = 2π ρ g s(cos θ1 + cos θ 2 )

so,

4.

A sphere and a cube of the same material and same total surface area are placed in the same space turn by turn, after heating them to the same temperature. Compare their initial rate of cooling in the enclosure. Sol. Rate of emission of energy  dT  σ T4 S = msphere c  −   dt sphere Also So,

or, But

But and

Vcube = a3 =

S3 / 2 63/ 2

v

∫

x

dvy =

0

dx

∫x

x0

2πm x v = ln µ0 qi x0 2 π mv/µ0 qi Hence, x = x0 e Note : Instead of F x we may write Fy from Eqn. (A) and then proceed in similar fashion. or,

3/2

, because, S=4πr2

( 4π −1/ 2 × 6 3 / 2 ) = 1.38 3

MEMORABLE POINTS

5.

A positive point charge q of mass m, kept at a distance x0 (in the same plane) from a fixed very long straight current i is projected normally away from it with speed v. Find the maximum separation between the wire and the particle. Sol. We know that a moving charge in magnetic field experiences a side way force given by the formula F = q (v × B) at a certain instant of time. As the magnetic field is not uniform, the particle does not follow the circular path but the speed (v) of the particle is constant. Here the magnetic field set up by the straight current is directed normally into the page (i.e., along the negative z-axis) and the initial velocity of the particle is along x-axis and further the force F is always in the x-y plane, so the motion of the particle is confined in the xy plane. The force at time t (Fig.) after starting from point P is. y

• The vector relation between linear velocity and →

→

→

® v =ω× r

angular velocity is

• In the case of uniform circular motion the angle between →

→

ω and r is always

→

® 90º(hence | v | = ωr

• The relation between Faraday constant F, Avogadro number N and the electronic charge e is ® F = Ne • Depolariser used in Lechlanche cell is ® Manganese dioxide • The absorption or evolution of heat at a junction of two dissimilar metals when a current is passed is known as ® Peltier effect • The part of the human ear where sound is transduced is the ® Cochlea • Similar trait resulting from similar selection pressure acting on similar gene pool is termed

i O

2πm µ0 qi

or,

, because, S = 6a2

Hence the required ratio =

(1)

so, 2v xdv x = 2vydvy = 0 or, v xdv x = – vydvy (2) From Eqns. (1) and (2) dv y 2πm = dx µ0 qi x

 dT  σ T4 S = mcube c  −   dt cube ( −dT / dt )sphere msphere Vsphere = = ( −dT / dt )cube mcube Vcube 4πr 3 4π  S  Vsphere= =   3 3  4π 

}

® Parallel evolution x0.

• Group of related species with the potential, directly or indirectly, of forming fertile hybrids with one another is called ® Coenospecies

x x.

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23

NOVEMBER 2009

P HYSICS F U NDAMENTAL F OR IIT-J EE

Electromagnetic Induction & A.C. KEY CONCEPTS & PROBLEM SOLVING STRATEGY Electromagnetic Induction (E.M.I.) Faraday's law states that the induced emf in a closed loop equals the negative of time rate of change of magnetic flux through the loop. This relation is valid whether the flux change is caused by a changing magnetic field, motion of the loop, or both. ε=–

When an emf is induced by a changing magnetic flux through a stationary conductor, there is an induced r electric field E of non-electrostatic origin. This field is non conservative and cannot be associated with a potential.

dΦ B dt

∫

r r dΦ B E.d l = – dt

A

B

G

φ

E I

ε

r

E

Lenz's law states that an induced current or emf always tends to oppose or cancel out the change that caused it. Lenz's law can be derived from Faraday's law, and is often easier to use.

B

E When a bulk piece of conducting material, such as a metal, is in a changing magnetic field or moves through a field, currents called eddy currents are induced in the volume of the material.

Change in B B (increasing)

B0

I

ε

I´

I0 Binduced If a conductor moves in a magnetic field, a motional emf is induced. ε = vBL r (conductor with length L moves rin uniform B field, r r L and v both perpendicular to B and to each other) r r r ε = (v × B ).d l r (all or part of a closed loop moves in a B field) ×B × × a × × × + a × × × × × × F=qvB × × × × × × L q v × × × × × ×

B´

A time-varying electric field generates a displacement current iD, which acts as a source of magnetic field in exactly the same way as conduction current.

∫

×

×

×

×

×

×

×

×

×

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–

iD = ε

Alternating Current (A.C.) An alternator or ac source produces an emf varies sinusoidally with time. A sinusoidal voltage or current can be represented of the by a phasor, a vector that rotates counterclockwise with constant angular velocity ω equals to the angular frequency of the sinusoidal quantity. Its projection on the horizontal axis at any instant represents the instantaneous value of the quantity.

× × × F = qE × × × ×

×

dΦ E (displacement current) dt

×

24

NOVEMBER 2009

V = IZ

ω I

Z = R 2 + ( X L − X C ) 2 = R 2 + [ωL − (1 / ωC )]2 tan φ =

ωt O i=I cos ωt For a sinusoidal current, the rectified average and rms (root-mean-square) currents are proportional to the current amplitude I. Similarly, the rms value of a sinusoidal voltage is proportional to the voltage amplitude V. 2 Irav = I = 0.637 I π I V Irms = ; Vrms = 2 2 In general, the instantaneous voltage between two points in an ac circuit is not in phase with the instantaneous current passing through points. The quantity φ is called the phase angle of the voltage relative to the current. i = I cos ωt v = V cos(ωt + φ)

ωL − 1 / ωC R V L = IXL

V = IZ I φ

V L – VC

VR = IR O

Pav = ½ VIcosφ

φ ω

O The voltage across a resistor R is in phase with the current. The voltage across an inductor L leads the current by 90º (φ = + 90º), while the voltage across a capacitor C lags the current by 90º(φ = –90º). The voltage amplitude across each type of device is proportional to the current amplitude I. An inductor has inductive reactance XL = ωL and a capacitor has capacitive reactance XC = 1/ωC. VR = IR; VL = IXL; VC = IXC Resistor connected to Inductor connected to ac source ac source

a

R

b a L Capacitor connected to ac source

i

an angular frequency ω 0 = 1/ LC called the resonance angular frequency. This phenomenon is called resonance. At resonance the voltage and current are in phase, and the impedance Z is equal to the resistance R. I(A) 0.5 0.4 0.3 0.2 0.1 0

b

200 Ω 500 Ω 2000 Ω

ω (rad/s) 1000 2000 A transformer is used to transform the voltage and current levels in an ac circuit. In an ideal transformer with no energy losses, if the primary winding has N1 turns and the secondary winding has N2 turns, the amplitudes (or rms values) of the two voltages are related by Eq. The amplitudes (or rms values) of the primary and secondary voltages and currents are related by Eq. V2 N = 2 ; V1I1 = V2I2 V1 N1

q –q i

a C b In a general ac circuit, the voltage and current amplitude are related by the circuit impedance Z. In an L-R-C series circuit, the values of L, R, C, and the angular frequency ω determine the impedance and the phase angle φ of the voltage relative to the current.

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t

p

In an L-R-C series circuit, the current becomes maximum and the impedance becomes minimum at

i

i

?

p

V cosφ ωt

i

V C = IXC

The average power input Pav to an ac circuit depend on the voltage and current amplitudes (or, equivalently, their rms values) and the phase angle φ of the voltage relative to the current. The quantity cos φ is called the power factor. 1 Pav = VI cos φ = Vrms Irms cos φ 2 v, i, p

I V φ

ωt

25

NOVEMBER 2009

Problem Solving Strategy (P.S.S.) : Faraday' Law Step 1: Identify the relevant concepts: Faraday's law applies when there is a changing magnetic flux. To use the law, make sure you can identify an area through which there is a flux of magnetic field. This will usually be the area enclosed by a loop, usually made of a conducting material. As always, identify the target variable(s). Step 2: Set up the problems using the following steps Faraday's law relates the induced emf to the rate of change of magnetic flux. To calculate this rate of change, you first have to understand what is making the flux change. Is the conductor moving? Is it changing orientation? Is the magnetic field changing? Remember that it's not the flux itself that counts, but its rate of change. r r Choose a direction for the area vector A or dA . The direction must always be perpendicular to the plane of the area. Note that you always have two choice of direction. For instance, if the plane of r the area is horizontal, A could point straight up or straight down. It's like choosing which direction is the positive one in a problem involving motion in a straight line; it doesn't matter which direction you choose, just so you use it consistently throughout the problem. Step 3: Execute the solution as follows : Calculate the magnetic flux using Eq. r r r φB = B . A = BA cos φ if B is uniform over the r r area of the loop or eq. φ B = B . dA = B dA cos φ

∫

when an inductor is included in a circuit, all the voltages, currents, and capacitor charges are in general functions of time, not constants as they have been in most of our previous circuit analysis. But Kirchhoff's rules, which we studied in section, are still valid. When the voltages and currents vary with time, Kirchhoff's rules hold at each instant of time. Step 2: Set up the problem using the following steps Draw a large circuit diagram and label all quantities known and unknown. Apply the junction rule immediately at any junction. Determine which quantities are the target variables. Step 3: Execute the solution as follows : Apply Kirchhoff's loop rule to each loop in the circuit. As in all circuit analysis, getting the correct sign for each potential difference is essential. To get the correct sign for the potential difference between the terminals of an inductor, remember Lenz's law and the sign rule described in section di in conjunction with eq. ε = –L (self-induced dt emf) and fig. In Kirchhoff's loop rule, when we go through an inductor in the same direction as the assumed current, we encounter a voltage drop equal to L di/dt, so the corresponding term in the loop equation is –L di/dt. When we go through an inductor in the opposite direction from the assumed current, the potential difference is reversed and the term to use in the loop equation is + L di/dt. a

∫

if it is not uniform, being mindful of the direction you chose for the area vector. Calculate the induced emf using Eq. ε = –

dφ B dt

dφ B . dt If your conductor has N turns in a coil, do not forget multiply by N. Remember the sign rule for the positive direction of emf and use it consistently. If the circuit resistance is known, you can calculate the magnitude of the induced current I using ε = IR. Step 4: Evaluate your answer : Check your results for the proper units, and double-check that you have properly implemented the sign rules for calculating magnetic flux and induced emf. P.S.S. :: Inductors in Circuits : Step 1: Identify the relevant concepts : An inductor is just another circuit element, like a source of emf, a resistor, or a capacitor. One key difference is that

b

(Faraday's law of induction) or ε = –N

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L

i

di dt Inductor with current i following from a to b: If di/dt > 0 : potential drops from a to b If di/dt < 0: potential increases from a to b If i is constant (di/dt = 0): no potential difference As always, solve for the target variables. Step 4: Evaluate your answer : Check whether your answer is consistent with the way that inductors behave. If the current through an inductor is changing, your result should indicate that the potential difference across the inductor opposes the change. If not, you probably used an incorrect sign somewhere in your calculation. Vab = L

26

NOVEMBER 2009

P.S.S. :: Alternating –Current Circuits : Step 1: Identify the relevant concepts: All of the concepts that we used to analyze direct-current circuits also apply to alternating current circuits. However, we must be careful to distinguish between the amplitudes of alternating currents and voltages and their instantaneous values. We must also keep in mind the distinctions between resistance (for resistors), reactance (for inductors or capacitors), and impedance (for composite circuits). Step 2: Set up the problem using the following steps Draw a diagram of the circuit and label all known and unknown quantities. Determine the target variables. Step 3: Execute the solution as follows : In ac circuit problem it is nearly always easiest to work with angular frequency ω. If you are given the ordinary frequency f, expressed in Hz, convert it using the relation ω = 2πf. Keep in mind a few basic facts about phase relationships. For a resistor, voltage and current are always in phase, and the two corresponding phasor in a diagram always have the same direction. For an inductor, the voltage always leads the current by 90º (i.e., φ = + 90º), and the voltage phasor is always turned 90º counterclockwise from the current phasor. For a capacitor, the voltage always lags the current by 90º (i.e., φ = –90º), and the voltage phasor is always turned 90º clockwise from the current phasor. Remember that with ac circuits, all voltages and currents are sinusoidal functions of time instead of being constant, but Kirchhoff's rules hold nonetheless at each instant. Thus, in a series circuit, the instantaneous current is the same in all circuit elements; in a parallel circuit, the instantaneous potential difference is the same across all circuit elements. Inductive reactance, capacitive reactance, and impedance are analogous to resistance; each represents the ratio of voltage amplitude V to current amplitude I in a circuit element or combination of elements. Keep in mind, however, that phase relations play an essential role. The effect of resistance and reactance have to be combined by vector addition of the corresponding voltage phasors, as in fig(i) & (ii). When you have several circuit elements in series, for example, you can't just add all the numerical values of resistance and reactance to get the impedance; that would ignore the phase relations.

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V = IZ

VL = IXL

VR = IR

I VL – VC

φ

ωt VR = IR

O

VC = IXC

Phasor diagram for the case XL> XC

φ

V = IZ I

VL = IXL ωt O

VL–VC VC=IXC Phasor diagram for the case XL< X C

Fig. (i) Fig. (ii) Evaluate your answer : When working with a series L-R-C circuit, you can check your results by comparing the values of the inductive reactance XL and the capacitive reactance XC. If XL > XC, then the voltage amplitude across the inductor is greater than that across the capacitor and the phase angle φ is positive (between 0 and 90º). If XL < XC , then the voltage amplitude across the inductor is less than that across the capacitor and the phase angle φ is negative between (0 and –90º).

Solved Examples 1.

A coil of 160 turns of cross-sectional area 250 cm2 rotates at an angular velocity of 300 rad/sec. about an axis parallel to the plane of the coil in a uniform magnetic field of 0.6 weber/metre2. What is the maximum e.m.f. induced in the coil. If the coil is connected to a resistance of 2 ohm, what is the maximum torque that has to be delivered to maintain its motion ?

Sol. We know that, e max = NABω = 160 × 0.6 × (250 × 10–4) × 300 = 720 volt. e max 720 = = 360 amp R 2 τ(torque) = NiBA sin θ Now imax =

τmax = NiBA = 160 × 360 × 0.6 × (250 × 10–4) = 864 newton metre This torque opposes the rotation of the coil (Lenz's Law). Hence to maintain the rotation of the coil, an equal torque must be applied in opposite direction. So the torque required is = 864 newton metre. 2.

27

A closed coil having 50 turns, area 300 cm2, is rotated from a position where it plane makes an angle of 45º with a magnetic field of flux density 2.0 weber/metre2 to a position perpendicular to the field in a time 0.1 sec. What is the average e.m.f. induced in the coil ? NOVEMBER 2009

Sol. The flux linked initially with each turn of the coil is

(a) the potential difference across R, L and C (b) the impedance of the circuit (c) the voltage of A.C. supply (d) phase angle Sol. (a) Potential difference across resistance VR = iR = 5 × 16 = 80 volt Potential difference across inductance

Φ = B.A = BA cos θ = BA cos 45º Substituting the values, we get  weber  Φ = 2.0   × (300 × 104 metre–2)×(0.7071) 2  metre  = 4.24 × 10–2 weber The final flux linked with each turn of the coil

VL = i × (ωL) = 5 × 24 = 120 volt Potential difference across condenser

Φ´ = BA cos 0º = BA = 2.0 × (300 × 10–4) = 6.0 × 10–2 weber

VC = i × (1/ωC) = 5 × 12 = 60 volt

Change in flux = Φ´ – Φ = (6.0 × 10–2) – 4.24 × 10–2 = 1.76 × 10–2 Weber This change is carried out in 0.1 sec. The magnitude of the e.m.f. induced in the coil is given by

(b)

=

3.

1.76 ×10 −5 = 8.8 volt. 0.1

 ωL − (1 / ωC )  φ = tan–1   R  

A vertical copper disc of diameter 20 cm makes 10 revolution per second about a horizontal axis passing through it centre. A uniform magnetic field 10–2 weber/m2 acts perpendicular to the plane of the disc. Calculate the potential difference between its centre and rim in volt.

 24 − 12  = tan–1    16  = tan–1(0.75) = 36º46´ 5.

A 100 volt A.C. source of frequency 500 hertz is connected to a L-C-R circuit with L = 8.1 millinery, C = 12.5 microfarad and R = 10 ohm, all connected in series. Find the potential difference across the resistance. Sol. The impedance of L-C-R circuit is given by

Sol. The magnetic flux Φ linked with the disc is given by Φ = BA The induced e.m.f. (potential difference) between rim and centre ∴ e=– where ∴

dΦ d dA = – (BA) = B (numerically) dt dt dt

where XL = ωL = 2πfL = 2 × 3.14 × 500 × (8.1 × 10–3) = 25.4 ohm

dA = πr2 × number of revolutions per second dt

and XC =

= 3.14 × (0.1)2 × 10

=

= 0.314 ∴

2

2

e = (10 weber/m ) × (0.314 m /sec) ∴

A resistance R and inductance L and a capacitor C all are connected in series with an A.C. supply. The resistance of R is 16 ohm and for a given frequency, the inductive reactance of L is 24 ohm and capacitive reactance of C is 12 ohm, If the current in the circuit is 5 amp., find

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1 1 = ωC 2πfC 1 2 × 3.14 × 500 × (12.5 × 10 − 6 )

= 25.4 ohm

= 3.14 × 10–3 volt. 4.

[R 2 + (X L − X C ) 2 ]

Z=

dA is the area swept out by the disc in unit time. dt

–2

[(16) 2 + (24 − 12) 2 ] = 20 ohm

(c) The voltage of A.C. supply is given by E = iZ = 5 × 20 = 100 volt (d) Phase angle

d (Φ´−Φ ) e=N dt = 50 ×

Z =

2  1   2  R +  ωL −   ωC    

Z =

∴ ir.m.s. =

[(10) 2 + (25.4 − 25.4) 2 ] = 10 ohm E r.m.s 100 volt = = 10 amp. Z 10 ohm

Potential difference across resistance VR = ir.m.s. × R = 10 amp × 10 ohm = 100 volt.

28

NOVEMBER 2009

P HYSICS F U NDAMENTAL F OR IIT-J EE

Simple Harmonic Motion KEY CONCEPTS & PROBLEM SOLVING STRATEGY Periodic motion is motion that repeats itself in a definite cycle. It occurs whenever a body has a stable equilibrium position and a restoring force that acts when it is displaced from equilibrium. Period T is the time for one cycle. Frequency f is the number of cycles per unit time. Angular frequency ω is 2π times the frequency. 1 1 2π f = or T = ; ω = 2πf = T f T y y y a n n n a x x x mg F O O mg O F mg

In SHM, the displacement, velocity, and acceleration are sinusoidal functions of time. The angular

If the net force is a restoring force F that is directly proportional to the displacement x, the motion is called simple harmonic motion (SHM). In many cases this condition is satisfied if the displacement from equilibrium is small. F k F x = –kx; ax = x = – x m m

Energy is conserved in SHM. The total energy can be expressed in terms of the force constant k and amplitude A. 1 1 1 E = mv x2 + kx2 = kA2 = constant 2 2 2 Energy E = K+U U

frequency is ω = k / m ; the amplitude A and phase angle φ are determined by the initial position and velocity of the body. x = A cos(ωt + φ) x A O –A

Restoring force Fx x 0

Displacement x 0

K –A

Fx < 0

The circle of reference construction uses a rotating vector called a phasor, having a length equal to the amplitude of the motion. Its projection on the horizontal axis represents the actual motion of a body in simple harmonic motion. y

O

t

O

k 1 k and f = I 2π I A simple pendulum consists of a point mass m at the end of a massless string of length L. Its motion is approximately simple harmonic for sufficiently small amplitude; the angular frequency, frequency, and period depend only on g and L, not on the mass or amplitude. ω=

Q

A

2T

x A In angular simple harmonic motion, the frequency and angular frequency are related to the moment of inertia I and the torsion constant k.

x>0

Displacement

T

P

x x= A cos θ

θ T

The angular frequency, frequency, and period in SHM do not depend on the amplitude, but only on the mass m and force constant k. ω=

k ω 1 ; f= = m 2π 2π

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L

x

mg sinθ

k 1 m ; T= = 2π m f k

m θ mg cosθ

mg

29

NOVEMBER 2009

ω=

g ; L

f=

ω 1 = 2π 2π

If you need to find the values of x, v x, and a x at various times, use Eqs.

g L

2π 1 L = = 2π ω f g

T=

f=

mgd ; I

T = 2π O

d sinθ mg sinθ

and a x =

I mgd

(phase angle in SHM) and A =

x 02 +

v 20 x

ω2 (amplitude in SHM). If the body is given an initial positive displacement x0 but zero initial velocity (v0x = 0), then the amplitude is A = x0 and the phase angle is φ = 0. If it has an initial positive velocity v0x but no initial displacement (x0 = 0), the amplitude is A = v0x / ω and the phase angle is φ = –π/2. Step 4: Evaluate your answer : Check your results to make sure that they're consistent. As an example, suppose you've used the initial position and velocity to find general expressions for x and v x at time t. If you substitute t = 0 into these expressions, you should get back the correct values of x0 and vv x. Simple Harmonic Motion II The energy equation

z

θ d cg mg cosθ

mg Problem Solving Strategy : Simple Harmonic Motion I : Step 1: Identify the relevant concepts : An oscillating system under goes simple harmonic motion (SHM) only if the restoring force is directly proportional to the displacement. Be certain that this is the case for the problem at hand before attempting to use any of the results of this section. As always, identify the target variables. Step 2: Set up the problem using the following steps Identify the known and unknown quantities, and determine which are the target variables. It's useful to distinguish between two kinds of quantities. Basic properties of the system include the mass m and force constant k. (In some problems, m, k, or both can be determined from other information.) They also include quantities derived from m and k, such as the period T, frequency f, and angular frequency ω. Properties of the motion describe how the system behaves when it is set into motion in a particular way. They include the amplitude A, maximum velocity vmax, and phase angle φ, as well as the values of x,v x, and a x at the particular time. If necessary, define an x-axis as. Step 3: Execute the solution as follows :

1 1 1 mv 2x + kx2 = kA2 = constant ...(i) 2 2 2 is a useful alternative relation between velocity and position, especially when energy quantities are also required. If the problem involves a relation among position, velocity, and acceleration without reference to time, it is usually easier to use Eq. E=

d2x

k x ...(ii) m dt (from Newton's second law) or eq. (i) (from energy conservation) than to use the general expressions for x, v x, and a x as functions of time [Eqs. ax =

2

=–

x = A cos (ωt + φ) (displacement in SHM), vx =

dx = –ωA sin (ωt + φ) (velocity in SHM) and dt

dv x d2x = = – ω 2A cos (ωt + φ) (acceleration 2 dt dt in SHM), respectively ]. Because the energy equation involves x2 and v x2, it cannot tell you the sign of x or v x, you have to infer the sign from the situation. For instance, if the body is moving from the equilibrium position towards the point of greatest positive displacement, then x is positive and v x is positive.

Use the equations T = 1/f and ω = 2πf = 2π/T to solve for the target variables. If you need to calculate the phase angle, be certain to express it in radians. The quantity ωt in Eq. F x = – kx is naturally in radians, so φ must be as well.

XtraEdge for IIT-JEE

k dx , vx = = – ωA sin (ωt + φ) m dt

dv x d2x = 2 = –ω 2A cos (ωt + φ). dt dt If the initial position x0 and initial velocity v0x are both given, you can determine the phase angle v  and amplitude from Eqs. φ = arctan  0 x   ωx 0 

A physical pendulum is a body suspended from an axis of rotation a distance d from its center of gravity. If the moment of inertia about the axis of rotation is I, the angular frequency and period for small-amplitude oscillations are independent of amplitude. ω=

ω 1 = 2π 2π

ax =

30

NOVEMBER 2009

4.

A particle of mass m is located in a unidimensional potential field where the potential energy of the particle depends on the coordinate x as U(x) = U0(1 – cos C x); U0 and C are constants. Find the period of small oscillations that the particle performs about the equilibrium position. Sol. Given that U(x) = U0(1 – cos C x) dV( x ) We know that F = ma = – dx 1  dU( x )  1 ∴ a= − = [– U0 C sin C]   m  dx  m

Solved Examples 1.

A body of mass 1 kg is executing simple harmonic motion which is given by x = 6.0 cos (100 t + π/4) cm. What is the (i) amplitude of displacement, (ii) frequency, (iii) initial phase, (iv) velocity, (v) acceleration, (vi) maximum kinetic energy ? Sol. The given equation of S.H.M. is x = 6.0 cos (100 t + π/4) cm Comparing it with the standard equation of S.H.M., x = a cos (ωt + φ), we have (i) amplitude a = 6.0 cm (ii) frequency ω = 100 /sec (iii) initial phase φ = π/4

U 0C U C2 [C x] = – 0 x (Q sin Cx ≈ Cx) m m Here acceleration is directly proportional to the negative of displacement. So, the motion is S.H.M. Time period T is given by or a = –

(iv) velocity v = ω (a 2 − x 2 ) = 100 (36 − x 2 ) (v) acceleration = –ω 2 x = – (100)2x = – 10 4 x 1 1 (vi) kinetic energy = mv2 = mω 2(a2 – x2) 2 2 When x = 0, the kinetic energy is maximum, i.e., 1 1  36  (K.E.)max = mω 2a2 = × 1 × 104 ×  metre  2 2 100   = 18 joules 2.

T=

 m = 2π   U C2 (U 0 C 2 / m)  0 2π

   

5.

Find the period of small oscillations in a vertical plane performed by a ball of mass m = 40 g fixed at the middle of a horizontally stretched string l = 1.0 m in length. The tension of the string is assumed to be constant and equal to F = 10 N. Sol. The situation is showing in fig. The components of T in upwards direction are T cos θ and T cos θ. Hence the force acting on the ball = 2 T cos θ

A particle of mass 0.8 kg is executing simple harmonic motion with an amplitude of 1.0 metre and periodic time 11/7 sec. Calculate the velocity and the kinetic energy of the particle at the moment when its displacement is 0.6 meter.

Sol. We know that, v = ω (a 2 − y 2 )

l/2

Further ω = 2π/T 2π 2× 3.14 ∴ v= (a 2 − y 2 ) = [(1.0) 2 − (0.6) 2 ] T (11 / 7) = 3.2 m/sec Kinetic energy at this displacement is given by 1 1 K = mv 2 = × 0.8 × (3.2)2 = 4.1 joule 2 2 3. A person normally weighing 60 kg stands on a platform which oscillates up and down harmonically at a frequency 2.0 sec–1 and an amplitude 5.0 cm. If a machine on the platform gives the person's weight against time, deduce the maximum and minimum reading it will show, take g = 10 m/sec2. Sol. Acceleration of the platform a = ω 2y Maximum acceleration amax = ω 2A (A = Amplitude) ∴ amax = (2πν)2A (ν = frequency) = 4(3.14)2 (2)2 × 0.05 = 7.88 m/sec2 m(g + a max ) 60(10 + 7.88) Maximum reading = = g 10 = 107.3 kg M(g − a max ) 60(10 − 7.88) Minimum reading = = g 10 = 12.7 kg.

XtraEdge for IIT-JEE

2π = ω

l/2 x θ

T

∴ma = –

θ

T

2Fx (l 2 / 4 + x 2 )

Q T = F and cos θ =

x 2

(l / 4 + x 2 ) As x is small, x2 can be neglected from the denominator. 2Fx  4F  ∴ a=– =–  x m(l / 2)  ml  or a = – ω 2x where ω 2 = (4 F/ml) Here acceleration is directly proportional to the negative of displacement x. Hence the motion is S.H.M. 2π  ml  =π   (4F / ml)  F  Substituting the given values, we get T=

2π = ω

T = 3.14 ×

31

 ( 4 ×10 − 2 )(1.0)    = 0.2 sec.   10  

NOVEMBER 2009

KEY CONCEPT

Nitrogen Compounds

Preparation of Amines : Through Nucleophilic Substitution Reactions Alkylation of Ammonia Salts of primary amines can be prepared from ammonia and alkyl halides by nucleophilic substitution reactions. Subsequent treatment of the resulting aminium salts with a base gives primary amines : •• NH3 + R — X

+

R — NH3 X–

OH–

Step 1

O C ••

N—H

C

KOH

C

C

O

O

RNH2

C ••

N—R

C

+

+

•• •• NH2NH2 ethanol reflux (several steps)

O

••

O

O

+ •• CH3CH2NH 2 + NH4

N

H +

•• CH 3CH2NH2+CH 3CH 2 —Br

N –

(CH 3CH 2)2NH2+Br , etc.

– •• S 2

•• + – •

• (− X )

•

Na / alcohol

Phthalazine-1,4dione

••

LiAlH4

Alkyl azide

Then the alkyl azide can be reduced to a primary amine with sodium and alcohol or with lithium aluminum hydride. A word of caution: Alkyl azides are explosive, and low-molecular-weight alkyl azides should not be isolated but should be kept in solution. Sodium azide is used in automotive airbags. The Gabriel Synthesis: Potassium phthalimide (see the following reaction) can also be used to prepare primary amines by a method known as the Gabriel synthesis. This synthesis also avoids the complications of multiple alkylations that occur when alkyl halides are treated with ammonia:

XtraEdge for IIT-JEE

••

+ R—NH2 H Primary amine

(several steps)

C—NHNH2 ••

C—N—R O H

Phthalimide is quite acidic (pKa = 9); it can be converted to potassium phthalimide by potassium hydroxide (step 1). The phthalimide anion is a strong nucleophile and (in step 2) it reacts with an alkyl halide by an S N2 mechanism to give an N-alkylphthalimide. At this point, the N-alkylphthalimide can be hydrolyzed with aqueous acid or base, but the hydrolysis is often difficult. It is often more convenient to treat the N-alkylphthalimide with hydrazine (NH2NH2) in refluxing ethanol (step 3) to give a primary amine and phthalazine-1, 4-dione. Syntheses of amines using the Gabriel synthesis are, as we might expect, restricted to the use of methyl, primary, and secondary alkyl halides. The use of tertiary halides leads almost exclusively to eliminations.

N • R—X + •N=N=N   → R–N=N=N   → R NH2 •  − •  or

Azide ion (A good nucleophile)

H

O

Alkylation of Azide ion and Reduction: A much better method for preparing a primary amine from an alkyl halide is first to convert the alkyl halide to an alkyl azide (R—N3) by a nucleophilic substitution reaction: •

R—X (–KX)

N-Alkylphthalimide

CH 3CH 2 — N — H + NH3

+

K

Step 3

O

CH3CH2 — NH3 + Br–

H

••–

+

– N

Phthalimide

This method is of very limited synthetic application because multiple alkylations occur. A Mechanism for the Alkylation of NH3 •• NH3 + CH 3CH 2 — Br

Step 2

O •• ••

Organic Chemistry Fundamentals

Reaction with Nitrous Acid Amines of different classes react with nitrous acid to yield different products. Nitrous acid being an unstable acid is prepared in situ by the reaction of sodium nitrite and dilute hydrochloric acid. 32

NOVEMBER 2009

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NOVEMBER 2009

Primary aromatic amines react with nitrous acid at low temperature (273-278 K) to give aromatic diazonium salts. This reaction is known as diazotisation. +

NH2

N2 Cl 273 − 278 K

+NaNO2 +2HCl    →

Stork Enamine Reactions Aldehydes and ketones react with secondary amines to form compounds called enamines. The general reaction for enamine formation can be written as follows:

–

••O••

+ NaCl+2H2O Benzene diazonium chloride

+

NaNO2 / HCl

Aldehyde or ketone

••

••

H

H

H

Piperidine

Morpholine

Cyclohexanone, for example, reacts with pyrrolidine in the following way: ••

O

H

N–NO+H2O

p-TsOH, –H2O N-(1-Cyclohexenyl)pyrrolidine (an enamine)

Enamines are good nucleophiles. Examination of the resonance structures that follow show that we should expect enamines to have both a nucleophilic nitrogen and a nucleophilic carbon.

N-Nitrosodiethylamine

The nitrosoamines on warming with a little phenol and concentrated sulphuric acid produce red solution which turns blue on treatment with sodium hydroxide. This colour change provides an excellent test for secondary amines and is known as Liebermann's nitroso reaction. Tertiary aliphatic amines on reaction with nitrous acid form nitrites while tertiary aromatic amines undergo electrophilic substitution at the ring.

••

N

Contribution to the hybrid made by this structure confers nucleophilicity on nitrogen

Triethylammonium nitrite HNO

Ring substitution

N,N-Dimethylaniline

XtraEdge for IIT-JEE

+

N ••–

(C2H5) 3N + HNO2  → [(C2H5)3NH]+NO2– 2 N(CH3)2 → ON

N

N

CH3

(C2H5) 2NH + HNO2  → (C2H5)2N—NO + H2O

Triethyl amine

••

N

Pyrrolidine

N-Nitroso-N-methylaniline

Diethyl amine

Enamine

N

N

Ethene

N-Methylaniline

+H2O R

••

H2 O

Secondary aliphatic and aromatic amines react with nitrous acid to produce nitrosoamines that are insoluble in the aqueous solution and separate out as a yellow oily layer.

→ NH + HNO2 

2ºAmine

••

••

C2H5OH + CH2 = CH2 + N2 + H2O

( NaNO2 +HCl)

C=C

O

(Unstable)

CH3

N

R

The secondary amines most commonly used to prepare enamines are cyclic amines such as pyrrolidine, piperidine and morpholine:

C2H5NH ] → 2  → [C2H 5 N 2 C l

Ethanol

H

R

H

Primary aliphatic amines also react with nitrous acid to form diazonium salt, however the aliphatic diazonium salts being unstable, decompose to yield mixture of alcohols and alkenes, and nitrogen gas is evolved.

C–C–N

R

R

••

C + HN–R

C

Aniline

OH ••

Contribution to the hybrid made by this structure confers nucleophilicity on carbon and decreases nucleophilicity of nitrogen

The nucleophilicity of the carbon of enamines makes them particularly useful reagents in organic synthesis because they can be acylated, alkylatd, and used in Michael additions. Enamines can be used as synthetic equivalents of aldehyde or ketone enolates

N(CH3) 2

p-Nitroso-N,N-dimethylaniline

37

NOVEMBER 2009

because the alkene carbon of an enamine reacts the +

same way as does the α-carbon of an aldehyde or ketone enolate, and after hydrolysis, the products are the same. When an enamine reacts with an acyl halide or an acid anhydride, the product is the C-acylated compound. The iminium ion that forms hydrolyzes when water is added, and the overall reaction

N (a)

(a)

N-Alkylated product

••

N

provides a synthesis of β-diketones:

heat

+ R – CH2 – X (b)

(b)

CH2R + X–

+

N CH2R + X–

R=CH2=CH– or C6H5–

C-Alkylated product

O

••

N

+

O–

N

+CH3C–Cl

+

H2O

O

N

Cl

C

O

C

CH2R

CH3

CH3

+ N

Iminium salt

+ Cl

–

H Enamine alkylations are SN2 reactions; therefore, when we choose our alkylating agents, we are usually restricted to the use of methyl, primary, allylic, and benzylic halides. α-Halo esters can also be used as the alkylating agents, and this reaction provides a convenient synthesis of γ-keto esters :

O

O

C

H2 O

CH3 +

+

N

H

H

+ Cl–

2-Acetylcyclohexanone (a β-diketone)

Although N–acylation may occur in this synthesis, the N-acyl product is unstable and can act as an acylating agent itself: –

O

••

N

+CH3C

+

N

+Br—CH2COC2H5

–

Cl +

O

••

N

N

Cl O

+

O

N

CH2COC2H5 + Br –

heat

••

N O

CCH3 +

CH2COC2H5

H2 O

N-Acylated C-Acylated iminium salt enamine

Enamine

O

Enamine

A γ-keto ester (75%)

As a consequence, the yields of C-acylated products

Enamines can also be used in Michael additions. An example is the following :

are generally high. Enamines can be alkylated as well as acylated. Although alkylation may lead to the formation of a

••

considerable amount of N-alkylated product, heating

+CH2 =CHCN

the N-alkylated product often converts it to a C-alkyl compound.

This

rearrangement

is

C2H5OH

N CH2CH2CN

reflux

particularly

favoured when the alkyl halide is an allylic halide,

O

benzylic halide, or α-haloacetic ester:

XtraEdge for IIT-JEE

+

N

CH2CH2CN

38

NOVEMBER 2009

KEY CONCEPT

Inorganic Chemistry Fundamentals

Nitrogen Family

Reaction of HNO3 on Metals. (a) Metals lying below hydrogen in the electrochemical series : Metals such as Na, K, Ca, Mg, Al, Zn, etc., lying below hydrogen in the electrochemical series normally displace hydrogen from dilute acids. Nitric acid also primarily behaves in the same manner. But, since it is a strong oxidising agent and hydrogen is a reducing agent, secondary reactions take place resulting in the reduction of nitric acid to give NO, N2O, N2 or NH3, depending upon the nature of the metal, the temperature and the concentration of the acid. Thus, dilute nitric acid reacts with zinc in the cold giving N2O or N2 according to the following eq.: 4Zn + 10HNO3 → 4Zn(NO3)2 + N2O + 5H2O 5Zn + 12HNO3 → 5Zn(NO3)2 + N2 + 6H2O Very dilute nitric acid gives NH3 which, of course, is neutralised by nitric acid to form NH4NO3. 4Zn + 10HNO3 → 4Zn(NO3)2 + 3H2O + NH4NO3 Similarly, iron and tin also give NH4NO3 with dilute nitric acid. Lead gives nitric oxide with dilute nitric acid in cold. Magnesium and manganese give hydrogen. Concentrated nitric acid essentially behaves as an oxidising agent and metals like aluminium, iron, chromium, etc., are rendered 'passive' probably due to surface oxidation. (b) Metals lying above hydrogen in the electrochemical series. : Metals such as Cu, Bi, Hg, Ag, lying above hydrogen in the electrochemical series, do not liberate hydrogen from acids. In case of these metals, the action of nitric acid involves only the oxidation of the metals into the metallic oxides which dissolve in the acid to form nitrates accompained by evolution of NO or NO2 according as the acid is dilute or concentrated. For instance, concentrated acid attacks copper giving NO2. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 Dilute nitric acid gives NO. 3Cu + 8HNO3 → 3Cu(NO3)2 + 4H2O + 2NO (c) Noble metals : like Au, Pt, Rh and Ir are not attacked by nitric acid. Gold and platinum, however, are atacked by aqua regia (3 parts conc. HCl and 1 part conc. HNO3) which contains free chlorine. HNO3 + 3HCl → 2H2O + 2Cl + NOCl This chlorine attacks gold and platinum forming soluble chlorides which form complexes with HCl, e.g., Au + 3Cl → AuCl3

XtraEdge for IIT-JEE

AuCl3 + HCl →

HAuCl4 Aurochloric acid

Hydroxylamine, NH2OH : It may be regarded as derived from ammonia by the replacement of one H atom by an OH group. It is prepared by the reduction of nitrites with sulphur dioxide under carefully controlled conditions. A concentrated solution of sodium nitrite is mixed with a solution of sodium carbonate and sulphur dioxide at a temperature below 3ºC is passed till the solution becomes just acidic. The following reactions are supposed to take place : Na2CO3 + SO2 + H2O → NaHSO3 + NaHCO3 NaNO2 + 3NaHSO3 → HON(SO3 Na ) 2 + Na2SO3 + H2O Hydroxylamine sodium sulphonate

The sulphonate hydroxylamine.

can be easily hydrolysed

to

O 2 HON(SO3Na)2 H → NH2OH Alternatively, it is prepared by the electrolytic reduction of nitric acid in 50% H2SO4 using amalgamated lead cathode. NO2 – OH + 6H+ + 6e– → NH2OH + 2H2O It is a colourless solid melting at 33ºC. It is freely soluble in water and lower alcohols. It is unstable and decomposes violently even at 20ºC. 3NH2OH → NH3 + N2 + 3H2O The aqueous solution of hydroxylamine is less basic than ammonia (Kb = 1.8 × 10–5). Thus, NH2OH(aq) + H2O NH3OH+ + OH–; Kb = 6.6 × 10–9 Like H2O2, it acts as an oxidising as well as a reducing agent depending upon circumstances. Nitrogen Trifluoride , NF3 : It is conveniently prepared by fluorinating ammonia.

4NH3 + 3F2 Cu catalyst  → NF3 + 3NH4F It can also be prepared by the electrolysis of NH4F. It is a colourless gas (m.p. –207ºC; b.p. –129ºC) which is quite stable thermodynamically. The gas acts as a fluorinating agent. It thus converts Cu into CuF. 2NF3 + 2Cu → N2F 4 + 2CuF As, Sb and Bi also get fluorinated by interaction with NF3.

39

NOVEMBER 2009

NF3 has a pyramidal structure with FNF angle = ~ 102º and dipole moment = 0.24 D, compared with HNH angle = ~ 107º and dipole moment = 1.48 D in case of NH3. The difference in the dipole moments of NF3 and NH3 (both of which have pyramidal structure though) is due to the fact that while the dipole moments due to N – F bonds in NF3 are in opposite direction to the direction of the dipole moment of the lone pair on N atom, the dipole moments of N – H bonds in NH3 are in the same direction as the direction of the dipole moment of the lone pair on N atom, an illustrated below. •• ••

Dinitrogen Tetrafluoride, N2F4 : N2F 4 is prepared by reacting HNF 2 with NaClO. 2HNF 2 NaClO → N2F4 + H2O HNF 2, in turn, is obtained by first fluorinating urea and then treating the fluorinated product with concentrated sulphuric acid. N2F 4 exists both in the staggered and the gauche conformations : F F

F N

N

F H H F H Because of its lower dipole moment, NF 3 is weaker ligand than NH3. NF3 is known to form complexes such as [NF 4]+ and F3N→O. Thus,

Staggered form

2F3N→ O

KF + HNF2 temperatur low →  e KF.HNF2 re Room temperatu   → N2F 2 + KHF2 The reaction yields both cis and trans isomers, viz.,

F N

F

N

N

> 70ºC

cis N 2F2 (~ 90%)

If, however, the isomeric mixture is treated with AlCl3 or the chlorides of bivalent Mn, Co, Ni and Fe, at low temperature, the major product is trans N2F 2. The cis form reacts selectively with AsF5 to form [N2F]+[AsF6]– which when reacted with NaF – HF, regenerates the original compounds. The trans form does not react with AsF5. N 2 F2 + AsF6 → [N 2 F]+ [ AsF6 ]− + N 2 F2 Mixture of cis and trans isomers

Formed by cis N 2 F2 only

trans

−HF [N2F]+[AsF6]– NaF  → Na+[AsF 6]– + N 2 F2 ( cis)

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Gauche Form

Trisilylamine, N(SiH)3 : Trisilylamine is prepared by reacting monochlorosilane with ammonia. 2SiH3Cl + 4NH3 → N(SiH3)3 + 3NH4Cl Trisilyamine is a trigonal planar compound with N orbitals in sp2 hybrid state, unlike trimethyl or triethylamine which is pyramidal and has N orbitals in sp3 hybrid state. There is considerable π overlap between the p orbital (containing the lone pair) of N atom and the vacant d π orbitals of Si atoms. The trigonal planar structure of N(SiH3)3 is, thus, strengthened due to p π–d π bonding. Since the lone pairs of electrons of N atom are engaged in p π-d π bonding between N and Si, they are no longer available for donation to Lewis acids. Trisilylamine, therefore, behaves as much weaker base compared to trimethylamine or triethylamine. Hence, trisilylamine does not form adducts with BH3 even at low temperature whereas trimethylamine or triethylamine does so readily. Due to the same reason, N(SiH3) 3 acts as a much weaker ligand compared to N(CH3)3 and N(C 2H5)3.

N

F Both the isomeric forms are gases at room temperature, cis form boiling at – 106ºC and trans form boiling at –112ºC. The cis form is thermodynamically more stable than the trans form. At above 70ºC, nearly 90% of N2F2 is present in the

cis form : trans N2F2

F

(Side View)

F

10 Li + N2F 4 → 4LiF + 2Li3N AsF5 + N2F 4 → [N2F 3]+ [AsF6]– N2F 4 reacts with sulphur to give a number of fluorinated sulphur compounds such as SF4 and SF5.NF 3. N2F 4 easily yields, at room temperature, the free radical. NF2 which reacts readily with a number of compounds. For example, Cl2 2ClNF2 N2F4 2(.NF2)– 2NO 2ON.NF2

Dinitrogen Difluoride, N2F2 : Dinitrogen difluoride is best prepared by reacting NHF 2 with KF.

F

N

SiH4 + N2F4 → SiF 4 + N2 + 2H2

High pressure

discharge O2 Electric   → low temperature

N

It is a colourless gas (b.p. – 73ºC; m.p. –164ºC). It is a strong fluorinating agent. Thus,

NF3 + 2F 2 + SbF 3 200  º→ [NF4+] [SbF6]– 2NF3 +

F F

N

F

F

N

40

NOVEMBER 2009

UNDERSTANDING

Physical Chemistry

2.

At 353 K, the vapour pressures of pure ethylene bromide and propylene bromide are 22.93 and 16.93 kN m–2, respectively, and these compounds form a nearly ideal solution. 3 mol of ethylene bromide and 2 mol of propylene bromide are equilibrated at 353 K and a total pressure of 20.4 kN m–2 (a) What is the composition of the liquid phase? (b) What amount of each compound is present in the vapour phase ? Sol. The given data are p*A = p*ethylene = 22.93 kN m–2

1.

A solution comprising 0.1 mol of naphthalene and 0.9 mol of benzene is cooled until some solid benzene freezes out. The solution is then decanted off from the solid, and warmed to 353 K, where its vapour pressure is found to be 670 Torr. The freezing and normal boiling points of benzene are 278.5 K and 353 K, respectively, and its enthalpy of fusion is 10.67 kJ mol–1. Calculate the temperature to which the solution was cooled originally and the amount of benzene that must have frozen out. Assume conditions of ideal solution. Sol. The given data are : n1 = 0.9 mol, n2 = 0.1 mol, * Tf = 278.5 K, Tb* = 353 K, p* = 760 Torr, p = 670 Torr, ∆fus H1,.m = 10.67 kJ mol–1 From the relative lowering of vapour pressure, we obtain the amount fraction of the solute (i.e. naphthalene). p * −p 760 Torr – 670 Torr x2 = = = 0.1185 p* 760 Torr Since x2 = n2/(n1 + n2), we get n 0.1 mol n1 + n2 = 2 = = 0.844 mol x2 0.1185 Since n2 = 0.1 mol, we get n1 = 0.844 mol – n2 = 0.844 mol – 0.1 mol = 0.744 mol Hence, the amount of benzene frozen out 0.9 mol – 0.744 mol = 0.156 mol The freezing point depression constant of benzene is Kf = =

bromide

p*B

=

(0.078 kg mol −1 )(8.314 J K −1mol −1 )( 278.5 K) 2 (10670 J mol −1 )

−1

= 1.723 mol kg–1

(0.744 mol) (0.078 kg mol ) Finally – ∆Tf = Kfm = (4.714 K kg mol–1)(1.723 mol kg–1) = 8.12 K

XtraEdge for IIT-JEE

0.578 × 22.93 kNm −2

= 0.6497 20.4 kN m − 2 Let nA and nB be the amounts of vaporized ethylene bromide and propylene bromide, respectively, when p = 20.4 kN m–2. Hence we have nA yA = = 0.6497 nA + nB 3 mol − n A xA = (3 mol − n A ) + (2mol − n B ) 3 mol − n A 3 mol − n A = = 5mol − ( n A + n B ) 5mol − n A / 0.6497 = 0.578 nA   or (3 mol – nA) = 0.578  5mol −  0.6497   Therefore 3 mol − 0.578 × 5mol nA = = 0.9967 mol 1 − 0.578 / 0.6497

M1RTf*2 ∆fus H1,m

(0.1 mol)

bromide

nA = 3 mol nB = 2 mol p = 20.4 kN m–2 (a) We have p = pA + pB = xA p*A + xB p*B = p*B + ( p*A – p*B )xA Therefore p − p* xA = * B* pA − pB Substituting the given data, we have 20.4 kNm −2 − 16.93kNm −2 xA = = 0.578 22.93kNm −2 − 16.93kNm −2 xB = 1 – 0.578 = 0.422 (b) Now p x p* yA = A = A A p p

= 4.714 K kg mol–1 Molality of the solution is n n2 m= 2 = m1 n1M1 =

= p*propylene = 16.93 kN m–2

41

NOVEMBER 2009

Since

Substituting p = 3 bar, we get K 0p = 1

nA = 0.6497, we therefore, have nA + nB

Let x be the amount of PCl3 that combines when the amount y of chlorine is added keeping p and T constant. Thus, the amounts of PCl3, Cl2 and PCl5 become n(PCl3) = 2 mol – x n(Cl2) = y + 2 mol – x n(PCl5 = 2 mol + x Since the final volume after the addition of Cl2 is twice the initial volume, it follows that the total amount of gases in 2V is 2 × 6 mol = 12 mol. Since n(PCl3) + n(PCl5) is 4 mol, the total amount of chlorine is 8 mol. Total amount = y + 6 mol – x = 12 mol Their partial pressures are 2 mol − x 2 mol − x p PCl3 = p= × 3 bar 12 mol 12 mol

nA – nA 0.6497  1  (0.9967 mol)(0.3503) − 1 = = nA  (0.6497)  0.6497  = 0.5374 mol

nB =

3.

For the following cell Pb |PbCl2(s)| PbCl2(soln.)|AgCl(s)| Ag the potential at 298 K is 0.490 V and the variation of emf with temperature is given by E = a – (1.86 × 10–4V K–1)(T – 25 K) Write the equation for the cell reaction and calculate ∆G, ∆H and ∆S for the reaction at 298 K. Sol. For the given cell, we have Electrode Reduction reaction Right 2AgCl(s) + 2e– = 2Ag(s) + 2Cl–(aq) ..(1) Left PbCl2(s) + 2e– = Pb(s) + 2Cl–(aq) ...(2) Subtracting Eq. (2) from Eq. (1), we get 2AgCl(s) + Pb(s) = 2Ag(s) + PbCl2(s) Now since E = a – (1.86 × 10–4V K–1)(T – 25 K), therefore  ∂E    = – 1.86 × 10–4 V K–1  ∂T  p

p Cl2 =

2 mol + x 2 mol + x p= × 3 bar 12 mol 12 mol Substituting these in the expression ( pPCl 3 / pº )( pCl 2 / pº ) K ºp = (where pº = 1 bar) (pPCl 5 / p º ) p PCl5 =

Hence ∆G = –nFE = –2(96500 C mol–1) (0.490 V) = – 94570 J mol–1   ∂E   ∆H = –nF E − T     ∂T  p   = –2(96500 C mol–1) [(0.490 V) – (298 K) (–1.86 × 10–4 V K–1)] = – 105267.6 J mol–1  ∂E  ∆S = nF   = 2(96500 C mol–1)(–1.86 ×10–4 V K–1)  ∂T  p

 2 mol − x   (2) 4 mol  ( 2 mol − x )( 2)  we get K ºp = = =1  2 mol + x  (2 mol + x )    4 mol  (as K ºp = 1) or 4 – 2 (x/ mol) = 2 + (x/mol) or 3(x/mol) = 2 ⇒ x/mol = 2/3 = 0.67 Therefore, the amount of Cl2 added y = 6 mol + x = 6.67 mol

= – 35.9 J K–1 mol–1 4.

5.

Calculate the resonance energy of benzene compared with one Kekule structure. Given the following data ∆fHº(methane, g) = –74.85 kJ mol–1 ∆fHº(ethane, g) = –84.68 kJ mol–1 ∆fHº(ethylene, g) = 52.3 kJ mol–1 ∆fHº(benzene, g) = 82.93 kJ mol–1 Enthalpy of sublimation of carbon (graphite) = 718.39 kJ mol–1 Dissociation enthalpy of H2 = 435.89 kJ mol–1 Sol. In order to calculate the resonance energy of benzene, we need to compute ∆fHº from the bond enthalpy data. For this, we need C – C, C = C and C – H bond enthalpies. These can be calculated as follows : (i) Bond enthalpy of C – H from ∆fHº(methane) : We have CH4(g) → C(graphite) + 2H2(g); ∆rHº = + 74.85 kJ mol–1

A container whose volume is V contains an equilibrium mixture that consists of 2 mol each of PCl5, PCl3 and Cl2 (all as gases). The pressure is 3 bar and temperature is T. A certain amount of Cl2(g) is now introduced, keeping the pressure and temperature constant, until the equilibrium volume is 2V. Calculate the amount of Cl2 that was added and the value of K 0p .

Sol. At equilibrium, we have PCl 5 PCl 3 + Cl 2 2 mol

2 mol

2 mol

Total amount = 6 mol 2

Thus

2   p / pº  ( pPCl / p º )( pCl / p º ) 6   3 2 K ºp = = 2  (pPCl 5 / p º )  p / pº  6 

XtraEdge for IIT-JEE

8 mol 8 mol p= × 3 bar = 2 bar 12 mol 12 mol

42

NOVEMBER 2009

C(graphite) → C(g); ∆rHº = + 718.39 kJ mol–1 2H2(g) → 4H(g); ∆rHº = 2×435.89 kJ mol–1 Adding, we get, CH4 (g) → C(g) + 4H(g) ∆rH1º = 1665.02 kJ mol–1 Now ∆rH1º = 4 ∈C – H, therefore

TRUE OR FALSE 1. 2.

−1

1665.02 kJ mol = 416.255 kJ mol–1 4 (ii) Bond enthalpy of C – C from ∆fHº(ethane) : We have C2H6(g) → 2C(graphite) + 3H2(g) ∆rHº = 84.68 kJ mol–1 2C(graphite) → 2C(g) ∆rHº = 2 × 718.39 kJ mol–1 3H2(g) → 6H(g); ∆rHº = 3 × 435.89 kJ mol–1 Adding, we get C2H6(g) → 2C(g) + 6H(g) ∆rH2º = 2829.13 kJ mol–1 Now ∆rH2º = ∈C–C + 6 ∈C–H Thus ∈C–C = [2829.13 – 6 × 416.255] kJ mol–1 = 331.60 kJ mol–1 (iii) Bond enthalpy of C = C from ∆fHº(ethylene) : We have C2H4(g) → 2C(graphite) + 2H2(g); ∆rHº = –52.3 kJ mol–1 2C(graphite) → 2C(g) ∆rHº = 2 × 718.39 kJ mol–1 2H2(g) → 4H(g) ∆rHº = 2 × 435.89 kJ mol–1 Adding, we get C2H4(g) → 2C(g) + 4H(g) ∆rH2º = 2256.26 kJ mol–1 Now ∆rH3º = ∈C=C + 4∈C–H Therefore ∈C=C = (2256.26 – 4 × 416.255) kJ mol–1 = 591.24 kJ mol–1 (iv) ∆fHº (benzene) from the bond enthalpy data. We have 6C(g) + 6H(g) → C6H6(g) ∆rHº = –(3∈C–C + 3∈C=C + 6∈C–H) = –5266.05 kJ mol–1 6C(graphite) → 6C(g) ∆rHº = 6 × 718.39 kJ mol–1 3H2(g) → 6H(g); ∆rHº = 3 × 435.89 kJ mol–1 Adding, we get 6C(graphite) + 3H2(g) → C6H6(g) ∆fHº = + 315.96 kJ mol–1 (v) Resonance energy of benzene (g) : Actual value of ∆fHº = 82.93 kJ mol–1 Calculated value of ∆fHº = 351.96 kJ mol–1 Thus, benzene become more stable by + 269.03 kJ mol–1. Therefore, its resonance energy is 269.03 kJ mol–1.

∈C–H =

XtraEdge for IIT-JEE

3. 4. 5.

6. 7. 8.

When a concentrated solution of a solute is diluted by adding more of solvent, the number of moles of solute remains unchanged. Acetate ion is stronger conjugated base than the sulphate ion. Mass of 60% HCl (by mass), required to completely react with 0.2 mol of zinc is 14.6 gm. Ferric hydroxide precipitate when agitated with dilute ferric chloride solution, gives negatively charged colloidal solution. Vapour pressures of ethanol and benzene at 293K are 43.9 mm and 74 mm respectively indicating stronger intermolecular forces in benzene as compared to ethanol. A mixture of molten zinc and lead is a heterogeneous system. The energy required to increase the surface area of a liquid by unit amount is called the surface tension of liquid. In aqueous solution cuprous ions disproportionate into cupric ions and metallic copper.

Sol. 1. [True] The number of moles of solute remains unchanged on dilution but molarity changes because No. of moles Molarity (M) = Volumes of solution(L ) 2. [True] Since acetic acid is a weak acid as compared to sulphuric acid and hence the conjugate base of a weak acid will be stronger than that of strong acid. 3. [False] Zn + 2 HCl → ZnCl2 + H2 Pure HCl required = 14.6 gm 100 ×14.6 60% Cl required will be = = 24.33 gm 60 4. [False] Positively charged sol is formed as Fe(OH)3 + FeCl3 → [Fe(OH)3]Fe 3+ : 3 Cl– Positively charged sol

5.

6. 7.

8.

43

[False] Higher the vapour pressure, weaker are the intermolecular forces and hence intermolecular forces in ethanol will be stronger than those in benzene. [True] [True] 2 Cu+ Cu2+ + Cu Since oxidation and reduction of Cu+ ion takes place simultaneously and hence it is known as the disproportion reaction. [False] Thomson through his experiment determined the charge to mass ratio of an electron and the value of 3/m is equal to 1.76 × 108 coulomb/gm. Hence one gm of electrons have charge 1.76 × 108 C. 1.60 × 10–19 coulomb is the charge on one electron. NOVEMBER 2009

Set

7

`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 Sol ut i ons w il l be publ i s he d i n ne xt is s ue Joint Director Academics, Career Point, Kota 1.

2.

3.

4.

Show that the lines 4x + y – 9 = 0, x – 2y + 3 = 0, 5x – y – 6 = 0 make equal intercepts on any line of gradient 2.

using complex numbers that their centroids form an equilateral triangle 9.

ABC is a triangle with ∠A = 90°, AD is altitude. a acts along AB such that | a | =1/AB, b acts along 1 AC such that | b | = . Prove that a + b is a AC 1 vector along AD and | a + b | = AD

a

such that Tr =

for ∀ x ∈  − π , π  , prove that  2 2

∑ i =1

r→1+

Three digit numbers are formed. What is the probability that the middle digit is largest.

6.

Prove that area of the region bounded by the curve y = log2 (2 – x) and containing the points satisfying the inequality (x – |x|)2 + (y – |y|)2 ≤ 4 is   2   2 + π − log  e e   sq. units. 2   4  27   

7.

r1, r2, r3 be the radii of the circles drawn on the altitudes respectively MD, ME and MF of the triangles respectively ∆MBC, ∆MCA, ∆MAB, as their diameters, where M be the circumcentre of the acute angled triangle ∆ABC. Prove that a 2 b2 c 2 + 2 + 2 ≥ 144. r12 r2 r3

8.

Equilateral triangles are described externally on the sides BC, CA and AB of a given triangle ABC. Prove

XtraEdge for IIT-JEE

r→1−

10. Let a, b, c be real numbers such that the roots of the cubic equation x3 + ax2 + bx + c = 0 are all real. Prove that no one of these roots is greater than (2 a 2 − 3b – a)/3.

P R E F IXE S Prefix

Symbol Sub multiplies deci d centi c milli m micro µ nano n pico p femto f atto a zepto z yocto y Multiples deca da hecto h kilo k mega (or million) M giga (or billion) G tera (or trillion) T peta P exa E zetta Z Yotta Y

ai ≤ 1. i

5.

1 − r cos u

∫ 1 − 2r cos u + r 2 du −a

Prove that lim Tr ,T1, lim Tr for an A.P.

A circle passes through the origin O and cuts two lines x + y = 0 and x – y = 0 in P and Q respectively. If the straight line PQ always passes through a fixed point, find the locus of the centre of the circle. x x x Let f(x) = a1 tan x + a2 tan + a3 tan +....+ an tan , 2 3 n where a1, a2, a3,...an ∈ R and n ∈ N. If |f(x)| ≤ |tan x| n

Let a be a fixed real number satisfying 0 < a < π,

44

Powers of 10 10 –1 10 –2 10 –3 10 –6 10 –9 10–12 10–15 10–18 10–21 10–24 101 102 103 106 109 10 12 10 15 10 18 10 21 10 24

NOVEMBER 2009

MATHEMATICAL CHALLENGES SOLUTION FOR OCTOBER ISSUE (SET # 6)

1.

Utilize the formula : If a1 + a 2 + ....... + an = k (constant), then a1a2 ..... an has the greatest value when a 1 = a 2 = ...... = an =

In the limiting condition the line (1) will touch the circle , Therefore p = 8, so as required |p| < 8

k , where a1, a2, ......, an are n

all positive.

4.

|a|2 = |b|2 = |c|2

(Using the concept of A.M. ≥ G.M.) Let E = (a – x) (b – y) (c – z) (ax + by + cz) Then abc E = (a2 – ax)(b2 – by)(c2 – cz)(ax + by + cz) Now we have (a2 – ax) + (b2 – by) + (c2 – cz) + (ax + by + cz) = a2 + b2 + c 2 (constant) a2 – ax = b2 – by = c 2 – cz = ax + by + cz =

AB 2 = AC 1

so xf(x) = xg(x) <

So,

A b O

(a 2 + b 2 + c 2 ) 4 256 abc where ∠AOB = θ

d (xf(x)) = g(x) dx

θ = π – 2θ

⇒ θ = π/3

Hence a.b = |a|2 cos

π 1 = |a|2 3 2

b.c = |b|2 cos

x

∫0 g( x) dx use it is (2)

x

π 3 2 3 2 = |b| = |a| & a.c = 0 6 2 2 →

→

c = xa + yb

y  So a.c = x|a|2 + y a.b ⇒  x +  |a|2 = 0 2 

Let the eqn of chord be x+ y= p

y 2

⇒ x=–

...(1)

and

⇒

135º

O (p, 0)

XtraEdge for IIT-JEE

→

Let

for x > 0

32

B C

a

∫0 g( x) dx ;

| a |θ 2 = π  1 | a |  − θ 2 

2:1

xf´(x) + f(x) = g(x) ...(1) xf´(x) = g(x) – f(x) < 0; because f´(x) < 0 & x > 0 So g(x) < f(x) x g(x) < x f(x); as x > 0 ...(2) Now from (1)

3.

since

a 2 + b 2 + c2 4

∴ the greatest value of abc E = 2.

→

Let OC = c

45

b.c = x a.b + y|b|2 3 2 1 2 |a| = |a| x + y|b|2 2 2

So,

x + 2y =

&

x=–

3

y 2

NOVEMBER 2009

y + 2y = 2

So,

x=–

⇒

3y = 2

3 ⇒ y=

so r2 + (2x1 cos θ + 2y1 sin θ)r + x12 + y12 – a2 = 0 2

r1 . r2 = x12 + y12 – a 2 =

3

y 1 =– 2 3

Hence x = –

→

−a

→

So

x12 + y12 + 2rx1 cos θ + 2ry1 sin θ + r2 = a2

3

c =

3

+

2 3

so, AP1 . AQ1 =

2

1

(AB)T = (BA)T

and others are even or 1 is even & others are odd.

so BTATA = ATBTA

2 10C1. 10C 3 20

C4

=

since

AP1 . AQ1 = AP 2 . AQ2 = ........ = APn . AQn

b

BTAT = AT BT

P(E) = 1 –

BT = ATBT A (as AAT = 1)

160 323

ABT = AATBT A ABT = BT A (again as AAT = 1) Hence proved.

160 163 = 323 323

Hence P(E) > P(O) 6.

(S)

Sum will be odd if 1 out of 4 chosen numbers is odd

P(O) =

2

1

P1A . Q1A is independent on n, hence →

8. 5.

(S)

9.

Let the point be P (x, y) so,

3x + 2y + 10 = 0

since

|PA – PB| is maximum

iz =

bi − c 1+ a

iz + 1 bi − c + 1 + a bi + a + (1 − c) = = iz − 1 bi − c − 1 − a − (c + 1) − (a − ib)

...(1)

...(i)

Now as given

hence P, A, B must be colinear

(a + ib) (a – ib) = 1 – c2 = (1 – c) (1 + c)

x y 1 2 4 1 =0 4 2 1

iz + 1 = − iz + 1

⇒ –x– y+6 =0 from (1) & (2) x = –22 & y = 28 So, point P is (–22, 28)

(a 2 + b 2 ) 1+ c 1− c2 (c + 1) + a + ib

bi + a +

....(2) =

(a + ib) 2 [(1 + c) + a − ib] (c + 1) 2 [a + ib + 1 − c] 2

7.

 a + ib  1 − iz = (using (1))  .  c + 1  iz + 1

Let the point A be (x1, y1) and the circle be x2 + y2 = a2 P1 A

iz + 1 a + ib = 1 − iz c +1

Q1

P2

(Hence proved)

Q2

10. Let n(n2 – 1) = (n – 1) n (n + 1) Since n is odd so (n – 1) (n + 1) is the product of two

Line AP 1 is

x − x1 y − y1 = =r cos θ sin θ

consecutive even numbers, so it is divisible by 8. Since (n – 1) n (n + 1) is the product of 3 consecutive

Solve it with circle.

integers so it is divisible by 3 also Hence n(n2 – 1) is

(x1 + r cos θ)2 + (y1 + r sin θ) 2 = a2

divisible by 24.

XtraEdge for IIT-JEE

46

NOVEMBER 2009

Students' Forum Expert’s Solution for Question asked by IIT-JEE Aspirants

MATHS 1.

Let a variable chord from (–1, 0) point to the circle (x – 2)2 + y2 = 1, makes a intercept of length 'l' on the circle and length of perpendicular from centre of the circle to chord is 'p'. find the range of 'λ ' such that l2 + 3λ p2 + 5 = 0. Sol. We have OB 2 = OD2 + BD2 B D A p (–1, 0)

3.

Let f(x) be a function which satisfy the equation f(xy) = f(x) + f(y) for all x > 0, y > 0 such that f ´(1) = 2. Find the area of region bounded by the curves y = f(x), y = |x3 – 6x2 + 11x – 6| and x = 0 Sol. Take x = y = 1 ⇒ f(1) = 0

 1 1 Now f  x.  = f(x) + f    x x 1 ⇒ f   = – f(x) x

0

x 1 f   = f(x) + f   = f(x) – f(y) y y f ( x + h ) − f (x ) f ´(x) = lim h →0 h  h f 1 +  − f (1) 1  x+h  x = lim f   = lim h →0 h  h  h →0 h .x x f ´(1) 2 = = x x ⇒ f(x) = 2 log |x| + c ⇒ c = 0 {when x = 1; as f(1) = 0} ⇒ f(x) = 2logx ∴ Required area

l2 4 − l2 ⇒ p2 = 4 4 we have been given, l2 + 3λ p2 + 5 = 0 3λ  4 − l 2  l2 + +5=0 4  4  12λ + 20 l2 = 3λ − 4 clearly 0 ≤ l2 < 4 4(λ + 5 / 3) ⇒ 0≤ 1 1 9–a ≥ 3 ⇒ a ≤ – 2

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1

=

3(1 + cos θ)x + 5 is increasing. Sol. We have, az2 + z + 1 = 0 ...(i) 2 ⇒ az + z + 1 = 0 {Taking conjugate on both sides} ⇒a z 2+ z +1 =0 ...(ii) Eliminating z from eq. (i) and (ii) by cross multiplication rule, ( a – a)2 + 2(a + a ) = 0 47

NOVEMBER 2009

2

Sol. Let, f(x) = (x – α1) (x – α2) ..... (x – αn) where α1, α2 ........, αn are the roots of f(x) = 0 since |f(0)| = f(1) ∴ α1 . α2 ...... αn = (1 – α1) (1 – α2) ...... (1 – αn) ⇒ (α1 . α2 ....... αn)2 ⇒ α1(1 – α1) α2(1 – α2) ...... αn(1 – αn) ⇒ ( Π αi)2 = II αi(1 – αi) . {i = 1, 2, ..... n}

2

a−a a+a ⇒   +  =0  2   2  2

2

 a −a  a+a ⇒ –  +  =0 2 i    2  ⇒ – sin2θ + cos θ = 0 ⇒ cos θ = sin2θ ...(iii) Now, f(x) = x3 – 3x2 + 3(1 + cos θ)x + 5 f ´(x) = 3x2 – 6x + 3 (1 + cos θ) ∴ Discriminate (D) = 36 – 36(1 + cos θ) = – 36 cos θ = – 36 sin2θ < 0 ⇒ f(x) is increasing ∀ x ∈ R

 α + (1 − α i )  Now, ( Π αi)2 = Π αi(1 – αi) ≤ Π   2   1 = 2n 2 Since GM ≤ AM 1 ⇒ ( Π αi) ≤ n 2

5.

Let a1, a2, ......, an be real constant, x be a real variable 1 1 and f(x) = cos(a1 + x) + cos(a2 + x) + cos(a3 + x) 2 4 1 +...... + n −1 cos(an + x). Given that f(x1) = f(x2) = 0, 2 prove that (x2 – x1) = mπ for integer m. Sol. f(x) may be written as, n

f(x) =

1

∑ k −1 k =1 2

k

∑

  n sin a k  – sin x  k −1     k =1 2

∑ n

= A cos x – B sin x, where A = n

∑ k =1

• •

{cosak. cos x – sin ak . sin x}

 n cos a k = cos x .  k−1   k=1 2

B=

INTERESTING SCIENCE FACTS

cos(a + x) ∑ k −1 k =1 2 n

=

1

∑ k =1

   

cos a k 2

k −1

• •

and

•

sin a k

2 k −1 since f(x1) = f(x2) = 0 ⇒ A cos x1 – B sin x1 = 0 and A cos x2 – B sin x2 = 0 A ⇒ tan x1 = B A ⇒ tan x2 = B ⇒ tan x1 = tan x2 ⇒ (x2 – x1) = mπ 6.

2

• • • • • •

Let a0, a1, .... an – 1 be real numbers where n ≥ 1 and het f(x) = xn + an – 1 xn –1 + ..... + a0 be such that : |f(0)| = f(1) and each root of f(x) = 0 is real and lies between 0 and 1. Prove that the product of the roots 1 does not exceed n . 2

XtraEdge for IIT-JEE

• • • 48

The speed of light is 186,000 miles per second. It takes 8 minutes 17 seconds for light to travel from the Sun’s surface to the Earth. In October 1999 the 6 billionth person was born. 10 percent of all human beings ever born are alive at this very moment. The Earth spins at 1,000 mph but it travels through space at an incredible 67,000 mph. Every year over one million earthquakes shake the Earth. The largest ever hailstone weighed over 1 kg and fell in Bangladesh in 1986. Every second around 100 lightning bolts strike the Earth. Every year lightning kills 1000 people. In October 1999 an Iceberg the size of London broke free from the Antarctic ice shelf. If you could drive your car straight up you would arrive in space in just over an hour. All the hydrogen atoms in our bodies were created 12 billion years ago in the Big Bang. The Earth is 4.56 billion years old…the same age as the Moon and the Sun. Alfred Nobel invented dynamite in 1866. NOVEMBER 2009

MATH

DIFFERENTIATION Mathematics Fundamentals

Differentiation and Applications of Derivatives : If y = f(x), then

4.

d dx

5.

d x e = ex dx

f ( x ) − f (a )  dy  2.   = lim x → a dx x −a   x =a

6.

d x a = a x log a dx

f (a + h ) − f (a )  dy  3.   = lim x → h h  dx  x =a

7.

d 1 log x = dx x

8.

d 1 logax = logae dx x

9.

d sin x = cos x dx

10.

d cos x = – sin x dx

11.

d tan x = sec2x dx

12.

d cot x = – cosec2x dx

13.

d sec x = sec x tan x dx

14.

d cosec x = – cosec x cot x dx

15.

d sin–1x = dx

16.

d 1 cos–1x = – dx 1− x2

17.

d 1 tan–1x = dx 1+ x 2

1.

dy f ( x + h ) − f (x ) = lim h →0 dx h

If u = f(x), v = φ(x), then 1. 2. 3. 4.

d (k) = 0 dx d du (ku) = k dx dx d du dv (u ± v) = ± dx dx dx d dv du (uv) = u +v dx dx dx

du  u  5.  = dx  v 

v

du dv −u dx dx v2

6. If x = f(t), y = φ (t), then 7. If y = f[φ(x)], then 8. If w = f(y), then

dy dy = dt dx

dx dt

dy d = f´[φ(x)]. [φ(x)] dx dx

dw dy = f´(y) dx dx

9. If y = f(x), z = φ(x), then

dy dy dx = . dz dx dz

x =

1 2 x

1 1− x2

10.

dy dx dy 1 . = 1 or = dx dy dx dx / dy

18.

1.

d (k) = 0 dx

d 1 cot–1x = – dx 1+ x 2

19.

2.

d n x = nxn–1 dx

d 1 sec–1x = dx x x 2 −1

20.

3.

d 1 n = – n +1 n dx x x

d 1 cosec–1x = – dx x x 2 −1

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49

NOVEMBER 2009

Suitable substitutions : The functions any also be reduced to simplar forms by the substitutions as follows. 1. If the function involve the term

d2y dx n

(a 2 − x 2 ) , then

2.

put x = a sin θ or x = a cos θ. 2. If the function involve the term

2

3.

2

(a + x ) , then

put x = a tan θ or x = a cot θ. 3. If the function involve the term

dx n dn dx n

(ax + b)n = n ! an (ax + b) m = m(m – 1) .... (m – n + 1) an(ax + b) m–n

(x 2 − a 2 ) , then

4.

put x = a sec θ or x = a cosec θ. 4. If the function involve the term

dn

= yn = fn(x)

a−x , then put a+x

5.

x = a cos θ or x = a cos 2θ All the above substitutions are also true, if a = 1 Differentiation by taking logarithm : Differentiation of the functions of the following types are obtained by taking logarithm. 1. When the functions consists of the product and quotient of a number of functions. 2. When a function of x is raised to a power which is itself a function of x.

6. 7.

dn dx n dn dx

n

dn dx n

emx = mne mx amx = mna mx (log a)n log(ax + b) =

( −1) n −1 a n (n − 1) ! (ax + b ) n

dn

nπ   sin (ax + b) = an sin  ax + b +  2  dx  n

dn

nπ   cos (ax + b) = an cos  ax + b +  2  dx  Leibnitz's theorem : If u and v are any two functions of x such that their desired differential coefficients exist, then the nth differential coefficient of uv is given by Dn(uv) = (Dnu)v + nC1(Dn–1u)(Dv) + nC2(Dn–2u)(D2v) +...... + u(Dnv) 8.

For example, let y = [f(x)] φ(x) Taking logarithm of both sides, log y = φ(x) log f(x) Differentiating both sides w.r.t 'x', 1 dy f ´(x ) = φ´(x) log f(x) + φ(x). y dx f ( x)

n

= [f(x)] φ(x) logf(x).φ´(x) + φ(x) . [f(x) φ(x) – 1.f´(x)

Do you know

dy = Differential of y treading f(x) as constant + dx Differential of y treating φ(x) as constant. It is an important formula. Differentiation of implicit functions : 1. If f(x, y) = 0 is a implicit function, then

• •

dy ∂f / ∂x =– dx ∂f / ∂y

•

Diff . of f w.r.t. x keeping y constant =– Diff. of f w.r.t. y keeping x constant For example, consider f(x, y) = x2 + 3xy + y2 = 0, then

•

dy 2x + 3y ∂f / ∂x =– =– dx ∂f / ∂y 3x + 2 y

•

1. If y = f(x), then

•

dy d2 y = y1 = f´(x), = y2 = f´´(x), ..... dx dx 2

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50

Did you know that there are 206 bones in the adult human body and there are 300 in children (as they grow some of the bones fuse together). Flea's can jump 130 times higher than their own height. In human terms this is equal to a 6ft. person jumping 780 ft. into the air. The most dangerous animal in the world is the common housefly. Because of their habits of visiting animal waste, they transmit more diseases than any other animal. Snakes are true carnivorous because they eat nothing but other animals. They do not eat any type of plant material. The world's largest amphibian is the giant salamander. It can grow up to 5 ft. in length. The smallest bone in the human body is the stapes or stirrup bone located in the middle ear. It is approximately .11 inches (.28 cm) long.

NOVEMBER 2009

MATH

STRAIGHT LINE & CIRCLE Mathematics Fundamentals

Different standard form of the equation of a straight line : General form : Ax + By + C = 0 where A, B, C are any real numbers not all zero. Gradient (Tangent) form : y = mx + c It is the equation of a straight line which cuts off an intercept c on y-axis and makes an angle with the positive direction (anticlockwise) of x-axis such that tan θ = m. The number m is called slope or the gradient of this line. Intercept form :

 b1c 2 − b 2 c1 a 2 c1 − a 1c 2    ,  a1 b 2 − a 2 b1 a1 b 2 − a 2 b1  Angle between two lines : The angle θ between two lines whose slopes are m1 and m2 is given by tan θ =

If θ is angle between two lines then π – θ is also the angle between them. The equation of any straight line parallel to a given line ax + by + c = 0 is ax + by + k = 0. The equation of any straight line perpendicular to a given line, ax + by + c = 0 is bx – ay + k = 0. The equation of any straight line passing through the point of intersection of two given lines l1 ≡ a1x + b1y + c1 = 0 and l2 ≡ a2x + b2y + c 2 = 0 is l1 + λ l 2 = 0 where λ is any real number, which can be determined by given additional condition in the question. The length of perpendicular from a given point (x1, y1) to a given line ax + by + c = 0 is

x y + =1 a b It is the equation of straight line which cuts off intercepts a and b on the axis of x and y respectively. Normal form (Perpendicular form) : x cos α + y sin α = p It is the equation of a straight line on which the length of the perpendicular from the origin is p and α is the angle which , this perpendicular makes with the positive direction of x-axis. One point form : y – y1 = m(x – x1) It is the equation of a straight line passing through a given point (x1, y1) and having slope m. Parametric equation :

ax 1 + by1 + c (a 2 + b 2 )

= p (say)

In particular, the length of perpendicular from origin c (0, 0) to the line ax + by + c = 0 is a 2 + b2

x − x1 y − y1 = =r cos θ sin θ It is the equation of a straight line passes through a given point A(x1, y1) and makes an angle θ with xaxis. Two points form :

Equation of Bisectors : The equations of the bisectors of the angles between the lines a1x + b1y + c1 = 0 and a 2x + b2y + c2 = 0 are a 1 x + b1 y + c1 a 12 + b12

y − y1 y – y1 = 2 (x – x1) x 2 − x1

=±

a 2x + b2 y + c 2 a 22 + b 22

Distance between parallel lines : Choose a convenient point on any of the lines (put x = 0 and find the value of y or put y = 0 and find the value of x). Now the perpendicular distance from this point on the other line will give the required distance between the given parallel lines. Pair of straight lines : The equation ax2 + 2hxy + by2 = 0 represents a pair of straight lines passing through the origin.

It is the equation of a straight line passing through y − y1 two given points (x1, y1) and (x2, y2), where 2 x 2 − x1 is its slope. Point of intersection of two lines a1x + b 1y + c1 = 0 and a2x + b2y + c 2 = 0 is given by

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m1 − m 2 1 + m1m 2

51

NOVEMBER 2009

Let the lines represented by ax2 + 2hxy + by2 = 0 be y – m1x = 0 and y – m2x = 0, then 2h a m1 + m2 = – and m1m2 = b b General equation of second degree in x, y is ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ...(i) This equation represents two straight lines, if ∆ = abc + 2fgh – af2 – bg2 – ch2 = 0 a or h g

h b f

If C 1, C2 are the centres and a1, a2 are the radii of two circles, then (i) The circles touch each other externally, if C1C 2 = a1 + a2 (ii) The circles touch each other internally, if C1C 2 = |a1 – a 2| (iii) The circles intersects at two points, if |a1 – a 2| < C 1C 2 < a1 + a2 (iv) The circles neither intersect nor touch each other, if C1C 2 > a1 + a2 or C1C2 < |a1 – a2| Equation of any circle through the point of intersection of two given circles S1 = 0 and S 2 = 0 is given by S 1 + λ S2 = 0 (λ ≠ –1) and λ can be determined by an additional condition. Equation of the tangent to the given circle x2 + y2 + 2gx + 2fy + c = 0 at any point (x1, y1) on it, is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0 The straight line y = mx + c touches the circle x2 + y2 = a2, if c2 = a2(1 + m2) and the point of contact of the

g f =0 c

and point of intersection of these lines is given by  hf − bg hg − af  ,    ab − h 2 ab − h 2  The angle between the two straight lines represented by (i) is given by 2 h 2 − ab a +b If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of parallel straight lines, then the distance between them is given by tan θ = ±

2

 m ma ± a  tangent y = mx ± a 1 + m 2 , is  ,   2 1 + m2   1+ m Length of tangent drawn from the point (x1, y1) to the

g 2 − ac f 2 − bc or 2 a (a + b) b (a + b)

circle S = 0 is 2

S1 = x1 + + 2gx1 + 2fy1 + c The equation of pair of tangents drawn from point (x1, y1) to the circle S = 0 i.e. x2 + y2 + 2gx + 2fy + c = 0, is SS1 = T2, where T ≡ xx1 + yy1 + g(x + x1) + f(y + y1) + c and S1 as

Circle: Different forms of the equations of a circle : Centre radius form : the equation of a circle whose centre is the point (h, k) and radius 'a' is (x – h)2 + (y – k)2 = a 2 General equation of a circle : It is given by x2 + y2 + 2gx + 2fy + c = 0 ...(i) Equation (i) can also be written as

mentioned above.

Chord with a given Middle point : the equation of the chord of the circle S = 0 whose mid-point is (x1, y1) is given by T = S1, where T and S1 as defined a above.

|x – (–g)|2 + |y – (–f)|2 = | g 2 + f 2 − c |2

If θ be the angle at which two circles of radii r1 and r2 intersect, then

which is in centre-radius form, so by comparing, we get the coordinates of centre (–g, – f) and radius is g2 + f 2 − c .

cos θ =

Parametric Equations of a Circle : The parametric equations of a circle (x – h)2 + (y – k)2 = a 2 are x = h + a cos θ and y = k + a sin θ, where θ is a parameter. Lengths of intercepts on the coordinate axes made by 2

Note — Two circles are said to be intersect orthogonally if the angle between their tangents at their point of intersection is a right angle i.e. r12 + r22 = d2 or

2

2g1g2 + 2f1f2 = c1 + c2

Equation of the circle on the line joining the points A(x1, y1) and B(x2, y2) as diameter is given by

XtraEdge for IIT-JEE

r12 + r22 − d 2 2 r1r2

where d is distance between their centres.

the circle (i) are 2 g − c and 2 f − c

 y − y1   y − y 2     x − x1   x − x 2

S1 , where

y12

Radical axis : The equation of the radical axis of the two circle is S1 – S2 = 0 i.e.

  = 1 

2x(g1 – g2) + 2y(f1 – f2) + c1 – c2 = 0

52

NOVEMBER 2009

Based on New Pattern

a

IIT-JEE 2010 XtraEdge Test Series # 7

Time : 3 Hours Syllabus :

Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus

Instructions :

Section - I • Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 9 to 12 are multiple choice questions with multiple correct answer. +4 marks and -1 mark for wrong answer. • Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer. Section - II • Question 19 to 20 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer and No Negative marks for wrong answer. However, +2 marks will be given for a correctly

marked answer in any row. 4.

PHYSICS Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

2.

It require 1 mJ of work to move identical positive charges +q from infinity so that they are separated by a distance a. How much work is required to move three identical positive charges +q from infinity so that they are arranged at the vertices of an equilateral triangle with edge length a ? (A) 2 mJ (B) 3 mJ (C) 4 mJ (D) 9 mJ Two capacitor C1 & C2, charged with q 1 & q 2 are connected in series with an uncharged capacitor C, as shown in figure. As the switch S is closed C +q C1 – 1

Average torque on a projectile of mass m, initial speed u and angle of projection θ between initial and final positions P and Q as shown in figure about the point of projection is y

u P

(A)

5.

q2 +C1 –

x

(B) mu2 cosθ

(C) mu2 sinθ

(D)

mu 2 cos θ 2

Portion AB of the wedge shown in figure is rough and BC is smooth. A solid cylinder rolls without slipping from A to B. If AB = BC, then ratio of translational kinetic energy to rotational kinetic energy, when the cylinder reaches point C is A B

What is the radius of the imaginary concentric sphere that divides the electrostatic field of a metal sphere of a radius 20 cm & a charge of 8 µC in two regions of identical energy ? (A) 30 cm (B) 40 cm (C) 60 cm (D) 80 cm

XtraEdge for IIT-JEE

Q

mu 2 sin 2θ 2

S (A) C gets charged in any condition (B) C gets charged only when q 1C 2 > q 2 C1 (C) C gets charged only when q 1C 2 < q 2C1 (D) C gets charged when q1C2 ≠ q 2C 1

3.

θ

C

D

(A) 3/5 (C) 7/5

53

(B) 5 (D) 8/3

NOVEMBER 2009

6.

7.

8.

Two bodies of masses m1 and m2 are initially at rest placed infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity when they are r distance apart is (A)

2G (m1 + m2 ) r

(B)

(C)

G(m1 + m 2 ) r

(D)

(C) The magnitude of acceleration at a distance 3 units from the fixed point is 27 unit (D) The motion of the particle is periodic but not simple harmonic

2 G m1 m 2

Find the quantum number 'n' corresponding to the exciting state of He+ ion. If on transition to the ground state that ion emits two photons in succession with wavelength 1026.7Å and 304 Å. (assume R = 1.096 × 107/m). (A) 4 (B) 6 (C) 2 (D) 1

Four identical bulbs A, B, C, D are connected in a circuit as shown in figure. Now whenever any bulb fails, then it cannot conduct current through it. ThenB A

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 13 to 15) One particle of mass 1 kg is moving along x-axis with velocity 3 m/s and another particle of mass 2 kg is moving along y-axis with 6 m/s. At t = 0, 1 kg mass is at (3m, 0) and 2 kg at (0, 9m). Here x – y plane is horizontal plane 13. The centre of mass of two particle is moving in a straight line (at t = 0) (A) y = x + 2 (B) y = 4x + 2 (C) y = 2x – 4 (D) y = 2x + 4

D

(A) Brightness of bulb C is highest (B) If C fails, brightness of bulb D increases (C) If C fails, brightness of all bulbs remain same (D) If A fails, B will not glow The speed v of a particle moving along a straight line, when it is at a distance x from a fixed point on the line is v 2 = 144 – 9x2. Select the correct alternative(s) : (A) The motion of the particle is SHM with time 2π period T = unit 3 (B) The maximum displacement of the particle from the fixed point is 4 unit

XtraEdge for IIT-JEE

A uniform rod AB of length l is free to rotate about a horizontal axis passing through A. The rod is released from rest from horizontal position. If the rod gets broken at mid-point when it becomes vertical then just after breaking of rod A B

(A) angular velocity of upper part starts to decrease while that of lower part remains constant (B) angular velocity of upper part starts to decrease which that of lower part starts to increase (C) angular velocity of both the parts is identical (D) angular velocity of lower parts becomes zero

Ideal Battery

10.

12.

G m1 m 2 (m1 + m2 )r

Which of the following statements are correct for an X-ray tube (A) on increasing potential difference between filament and target, photon flux of X-Rays increases (B) on increasing potential difference between filament and target, frequency of X-Ray increases (C) on increasing filament current, cut off wavelength increases (D) on increasing filament current, intensity of X-Rays decreases

C

An electron orbiting in a circular orbit around the nucleus of an atom: (A) has a magnetic dipole moment (B) exerts an electric force on the nucleus equal to that on it by the nucleus (C) does produces a magnetic induction at the nucleus (D) has a net energy inversely proportional to its distance from the nucleus

(m1 + m 2 )r

Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 9.

11.

14. If both particle have same value of coefficient of friction µ = 0.2. The centre of mass will stop after time (A) 1.5 sec (B) 4.5 sec (C) 3 sec (D) 2 sec 15. The coordinates of centre of mass when it will stop (A) (2m, 14.25 m) (B) (2.25 m, 10m) (C) (3.75 m, 9m) (D) (1.75 m, 12 m) 54

NOVEMBER 2009

Passage : II (No. 16 to 18)

[if R = 1 m and data in column II are rounded off and g = 10 m/s 2]

A particle with charge +7.60 nC is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved 8.00 cm, the additional force has done 6.50 × 10–5 J of work and the particle has 4.35 × 10–5 J of kinetic energy.

D

O

16. What work was done by the electric force ? (A) + 2.15 × 10–5 J (B) – 2.15 × 10–5 J –5 (C) – 4.30 × 10 J (D) + 4.30 × 10–5 J

A

Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

Which of the following is correct order of the reactivity towards electrophillic substitution ? O

O

C–NH–CH2 –CH3

(i)

NH–C–CH2–CH3

(ii) O

NH–CH2–C–CH3

Column- II (P) Monoenergetic particles (Q) Poly energetic particles are emitted (R) Angular momentum is conserved (S) Can take place inside and outside nucleus (T) none

(iii)

(A) i > ii > iii > iv (C) iii > iv > ii > i

20. A small body of mass m = 2 kg is thrown with speed u from point A along a smooth circular track as shown. The body after moving through the points B, C and D comes back at hits point A. Length AB is x. When x = 3R then u = v0 and normal reaction at point C is N1. The minimum value of x = x0 and in this case normal reaction at point C is N2 then,

XtraEdge for IIT-JEE

Column-II [In SI units] (P) 85 (Q) 8 (R) 2 (S) 60 (T) none

CHEMISTRY

This section contains 2 questions (Questions 19 to 20). 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, T) 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, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T

(D) Electron capture

B

x

Column-I (A) v0 (B) x0 (C) N1 (D) N2

18. What is the magnitude of the electric field ? (A) 3.54 × 102 V/m (B) 3.54 × 104 V/m 6 (C) 3.54 × 10 V/m (D) 3.54 × 103 V/m

(C) Positron emission

C

u

17. What is the potential of the starting point with respect to the end point ? (A) + 2.83 kV (B) + 5.66 kV (C) – 2.83 kV (D) – 5.66 kV

19. Column- I (A) Alpha Decay (B) Beta Decay

R= 1m

55

O CH2–NH–C–CH3

(iv)

(B) iii > ii > iv > i (D) None

2.

Decomposition of A follows first order kinetics by the following equation. 4A(g) → B(g) + 2C(g) If initially, total pressure was 800 mm of Hg and after 10 minutes it is found to be 650 mm of Hg. What is half-life of A? (Assume only A is present initially) (A) 10 minutes (B) 5 minutes (C) 7.5 minutes (D) 15 minutes

3.

An electron has velocity x ms–1. For a proton to have the same de-Broglie wavelength, the velocity will be approximately – 1840 x (A) (B) (C) 1840x (D) x x 1840 NOVEMBER 2009

Select the most ionic and most covalent compounds respectively from the following. CrO5, Mn2O7, PbO, P4O10, SnO2 (A)CrO5, Mn2O7 (B) PbO, Mn2O7 (C)CrO5, P4O10 (D) SnO2 ,CrO5

5.

2 gm of complex [Cr(H2O)5Cl]Cl2.H2O was passed through a cation exchanger to produce HCl. The acid liberated was diluted to 0.5 litre. The normality of acid solution will be : [Molar mass of complex = 266.5] 1 2 8 4 (A) (B) (C) (D) 266.5 266.5 266.5 266.5

6.

To effect the conversion

CH 3 N(CH 3 ) 2 | | (B) H3C– C — CH2 CHN (CH 3 ) 2 | CH 3 CH 3 O | || (C) H3C – C — CH2 C N(CH3)2 | CH 3 CH 3 | (D) H3C– C — CH2CH2N(CH3)2 | CH 3 Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

O

O

CH(CH2)5 CH3

CH(CH2) 4COCH3

O

O

9.

7.

There are two isomeric carboxylic acids– 'A' and 'B' C9H8O2. reacts with H2/Pd giving compounds, C9H10O2. 'A' gives a resolvable product and 'B' gives a non-resolvable product. Both isomers could by oxidised to PhCOOH. The correct structures of 'A' and 'B' are, respectively– (A) CH 2=CH

COOH ;

(C) (D)

8.

COOH ;

CH=CH–COOH ;

COOH

C=CH2 COOH

Identify product D in the following reaction sequence: CH 3 | SOCl2 K2 Cr2 O7 , H+ H3C– C — CH2CH2OHH   → A → B O , Heat 2 | CH 3 ( CH ) NH

Z

11. A 100 ml mixture of CO and CO2 is passed through tube containing red hot charcoal. The volume now becomes 160 ml. The volumes are measured under the same condition of temperature and pressure. Amongst the following, select the correct statements(A) Mole percent of CO2 in the mixture is 60 (B) Mole fraction of CO in the mixture is 0.40 (C) The mixture contains 40 ml of CO2 (D) The mixture contains 40 ml of CO

I. LiAlH , ether

32→ C 4 → D II. H2 O

CH 3 | (A) H3C– C — CH2C ≡ N | CH 3

XtraEdge for IIT-JEE

PB

10. At high temperature the compound S4N4 decomposes completely into N2 and sulphur vapour. If all measurement are made under same T and P, it is found that for each volume of S 4N4 decomposed, 2.5 volume of gaseous products are formed. Which statements are true(A) Molecular formula of sulphur is S8 (B) Volume of N2 obtained is 2 times that of volume of S4N4 taken (C) Volume of sulphur obtained is equal to the volume of N2 obtained (D) Volume of sulphur obtained is half of the volume of S4N4 consumed

CH=CH–COOH

CH=CHCOOH ; CH 2=CH

PA

S 1.0 0.0 (A) vapour pressure of A = SZ (B) vapour pressure of B = ZY (C) vapour pressure of B = SY (D) vapour pressure of solution at X = SZ + SY

CH=CHCOOH

(B) CH2=C

Consider the following vapour pressure composition graph. Hence X PBº PS PAº Y vapour pressure

Which of the following reagents is best suitable ? (A) Zn-Hg, concentrated HCl (B) LiAlH4 (C) NH2NH2,KOH,DMSO (D) NaBH4

vapour pressure

4.

56

NOVEMBER 2009

12. The major product of reaction

15. The resolvable organic compound, G is – CH3 CH2CH3 Br CH3 H Br (A) CH3 (B) Br Br H CH 3 CH2CH3 CH 3 CH 3

Br

2 → is Br

Br

(A)

(B) Br

CH3 (C) H

Br Br

Br

(C)

(D) None of these

Passage : I (No. 13 to 15) C (Resolvable) HBr,Peroxide

HBr,R2O 2 (Excess)

D(C6H12) O3;Zn,H2O

G Resolvable

F nonresolvable

18. How much excess reagent remains (A) 0.59 g Al (B) 5.9 kg Cr2O3 (C) 0.59 g Cr2O3 (D) None of these This section contains 2 questions (Questions 19 to 20). 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, T) 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, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T

O || CH 3–C–CH3

13. Organic compound 'A' is – CH2Br

(A)

Br

(B) CH2Br

(C)

Br

(D)

14. The resolvable organic compound 'C' is – CH2Br Br (A) (B)

19.

Br

Br

Br

(C) Br

Br

XtraEdge for IIT-JEE

(D)

CH 3

17. Which reagent is limiting reagent (A) Al (B) Cr2O3 (C) Both (A) and (B) (D) None of these

HBr B (Non-resolvable) Zn,Heat

Br H

16. How much metallic chromium can be made (A) 9.6 kg Cr (B) 9.6 g Cr (C) 0.96 kg Cr (D) None of these

A(C6H 11Br) Decolourise Br2 water and connot be resolved

a single possible product E

CH H (D) Br

Passage : II (No. 16 to 18) Aluminium react metallic oxide to release high amount of energy so it is used in metallurgy process named Aluminothermite process. Reduction of Cr2O3 by Aluminium can be given by following reaction. 2Al + Cr2O3 → Al2O3 + 2Cr If 5 kg aluminium and 20 kg Cr2O3 react with each other to form aluminium oxide then give answer following questions.

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

alc. KOH

CH 2Br H CH3 CH2Br

Br

57

Column-I Molecules (A) PH3 (B) H2O (C) PF5 (D) IF7

(P) (Q) (R) (S) (T)

Column-II Bond Angle ≈ 90º or = 90º 100º < B.A. < 109º28' 120º 72º 180º NOVEMBER 2009

20.

Column-I (A) 2A + B → C + 3D Rate = k (B) 2A + 2B → C + D Rate = k CA (C) A + 2B → 3C + 4D Rate = k CACB k 1C A (D) 2A + 2B → 3C + 3D Rate = 1+ k 2C A Column-II (P) Unit of rate constant possesses concentration unit (Q) Rate constant for the reaction of both the reactants are equal (R) Rate of consumption of at least one of the reactant is equal to rate of production of atleast one of the products (S) If both reactants are taken in stoichiometric ratio, half life for both reactants are equal (T) Both rate constant and order are not defined

6.

7.

MATHEMATICS Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

2.

8.

4.

Given the function f(x) = 1/(1–x), the number points of discontinuity of the composite function y = f 3n(x), where f n(x) = fof ... of (n times) are (n ∈ N) (A) 0, 1 (B) 2n (C) 3n (D) 2n + 1

5.

The tangent to the curve x = a cos 2θ cos θ,

9.

a + b tan 2 x

dx (a > b), then I equals -

(A)

 a + b tan 2 x sin–1   b −a b −a 

 + C  

(B)

 a + b tan 2 x cos–1   b −a b −a 

 + C  

(C)

 a + b tan 2 x tan–1   b −a b −a 

 + C  

(D)

 b−a tan–1  2  b −a  a + b tan x

 +C  

If I =

1

1

1

1

sin 2θ

π/4

∫0

2

sin θ + cos 4 θ (B) π/ 3

dθ , then I equals -

(C) π/2 3

(D) π/3 3

x y – , then a b (A) a – b is a factor of u (B) a + b is a factor of u (C) a + ib is a factor of u (D) a – ib is a factor of u Suppose (x + iy)1/5 = a + ib and u =

10. Consider the system of linear equations in x, y and z : (sin 3θ) x – y + z = 0 (cos 2θ) x + 4y + 3z = 0 2x + 7y + 7z = 0 The values of θ for which the system of equations has a non-trivial solution are (A) {nπ:n ∈ I} (B) {mπ + (–1) m π/6 : m ∈ I} (C) {nπ + (–1)m π/3 : m ∈ I} (D) none of these 11. Let [x] denote the greatest integer less than or equal to x. If f(x) = [x sin π x], then f(x) is (A) continuous at x = 0 (B) continuous in (–1, 0) (C) differentiable at x = 1 (D) differentiable in (–1, 1)

y = a cos 2θ sin θ at the point corresponding to θ = π/6 is (A) parallel to the x-axis (B) parallel to the y-axis (C) parallel to line y = x (D) none of these

XtraEdge for IIT-JEE

tan x

∫

Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

 cos α sin α 0    If A(α, β) = − sin α cos α 0  , then A(α, β)–1 is  0 0 eβ  equal to (A) A(–α, β) (B) A(–α, –β) (C) A(α, –β) (D) A(α, β) If f(x) is a polynomial satisfying f(x).f(1/x) = f(x) + f(1/x), and f(3) = 28, then f(4) is given by (A) 63 (B) 65 (C) 67 (D) 68

If I =

(A) π/2

If coefficients of x20 in (1 – x + x2)20 and in (1 + x – x2) 20 are respectively a and b, then (A) a = b (B) a > b (C) a < b (D) a + b = 0

3.

| x | for 0 2 (D) R must be 2 ^

^

5.

A projectile thrown with initial velocity (a i +b j ) and its range is twice the maximum height attained by it then a (A) b = (B) b = a (C) b = 2a (D) b = 4a 2

6.

Two identical heavy spheres of equal mass are placed on smooth cup of radius 3r where r is radius of each sphere as shown. Then the ratio of reaction force between cup and any sphere to reaction force between two sphere is – 0

∆T

Temperatrue change

(A)

→

→

2

2.

→

also if | R | = R, then (A) R < 0 (C) 0 ≤ R≤ 2

3

1

→

3r 3r r r

11R 6 7R (D) 6 (B)

(A) 1 60

(B) 2

(C) 3

(D) none NOVEMBER 2009

7.

The distance of centres of mass of two square plates system a shown from point O. If masses of plates are 2m and m is (their edges are 'a' and '2a' respectively)

a

(A) 1.6 3 cm (C) 4.8 cm 11.

A projectile is thrown from point P on horizontal ground at angle θ with horizontal then (A) the projectile moves always from point P for any values of θ (B) the projectile moves always from point P for some values of θ (C) for some value of θ projectile first moves always from point P then comes closer to point P for some time interval (D) none of these

12.

As shown in figure pulley is ideal and strings are massless. If mass m of hanging block is the minimum mass to set the equilibrium of system then – (g = 10 m/s2)

2m O m

(A) 8.

a 2

(B) a

(C)

2a 3a 2

(D)

2a 3

A particle is moving in a circular path and its acceleration vector is making an angle of 30° with the velocity vector, then the ratio of centripetal acceleration to its tangential acceleration is – (A)

1 2

(B)

3 2

(C)

1 3

(D)

(B) 1.6 cm (D) none of these

3 θ= 37º

Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 9.

θ= 37º

Figure shows cyclic process. From c to b 40 J is transferred as heat from b to a, 130 J is transferred as heat, and work done is 80 J from a to c, 400 J is transferred as heat then – P c

20 kg µ = 0.5 θ = 37º

(A) m = 2.5 kg (B) m = 5 kg (C) force applied by 20 kg block on inclined plane is 179 N (D) force applied by 20 kg block on inclined plane is 223 N

b

a

V (A) work done in process a to c is 310 J (B) Net work done in cycle is 230 (C) Net change in internal energy in cycle is 130 J (D) Thermal efficiency is 57.5%

10.

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Two identical ideal springs of spring constant 1000 N/m as connected by an ideal pulley as shown and system is arranged in vertical plane. At equilibrium θ is 60º and masses m1 and m2 are 2kg and 3kg respectively. Then elongation in each spring when θ is 60º is – θ

Passage : I (No. 13 to 15) One mole of monoatomic gas is taken through above cyclic process. TA = 300 K Process AB is defined as PT = constant. P

θ

3P 0 P0

B

C A

T 13. Work done in process AB is (A) 400 R (B) – 400 R (C) 200 R (D) – 300 R

m1 m2

XtraEdge for IIT-JEE

m

61

NOVEMBER 2009

14. Change in internal energy in process CA (A) 900 R (B) 300 R (C) 1200 R (D) zero

19. In the equation, y = A sin the following : Column-I (A) Frequency of wave (B) Wavelength of wave (C) Phase difference between two points 1/4a distance apart (D) Phase difference of a point after a time interval of 1/8b

15. Heat transferred in the process BC is (A) 1000 R (B) 500 R (C) 2000 R (D) 1500 R Passage : II (No. 16 to 18) An external force F is applied at an angle θ with the horizontal as shown on the block of mass 'm'. The coefficient of friction between block and wall is µ. 16. The minimum value of force f required to keep the block at rest is –

θ

m

mg µ cos θ

(C)

mg sin θ − µ cos θ

mg sin θ − µ cos θ

A

rough (µ)

B

(A) Adhesive forces is (B) (D)

(D)

(P)

greater than cohesive forces

mg sin θ + µ cos θ

A

mg µ tan θ

(B) Cohesive forces is

(Q) B

greater than adhesive forces (C) Pressure at A > pressure at B

mg µ tan θ

(R) A mercury drop is pressed between two parallel plates of glass B

18. The value of force F for which friction force between block and wall is zero mg mg mg (A) mg (B) (C) (D) sin θ cos θ tan θ

(D) Pressure at B > Pressure at A

(S)

B

A

A

(T) none

CHEMISTRY

This section contains 2 questions (Questions 19 to 20). 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, T) 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, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T

XtraEdge for IIT-JEE

(S) π/2

20. Capillary rise and shape of droplets on a plate due to surface tension are shown in column II. Column-I Column-II

17. The maximum value of force F up to which block remains at rest is mg mg (A) (B) µ cos θ sin θ + µ cos θ (C)

Column-II (P) a (Q) b (R) π

(T) none

F

(A)

2 π (ax + bt + π/4) match

Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. Kinetic energy and potential energy of an electron in an orbit and the centrifugal force experienced by it are respectively Ze 2 Ze 2 mv2 ,– , 2r r r 2 2 Ze Ze mv2 (C) ,– , r 2r r (A)

62

Ze 2 Ze 2 mv2 , , 2r r r 2 2 Ze Ze mv2 (D) – , , r 2r r (B) –

NOVEMBER 2009

2. 3 L of a gas mixture consisting of propane and butane on complete combustion produced 10 dm3 (cubic decimeter) CO2 under identical conditions. The volume of propane in the mixture is (A) 2L (B) 1L (C) 1.5 L (D) 0.5 L

9.

Br

3. Chloropicrin CCl3NO2 can be made cheaply for use as an insectiside by a process which utilizes the reaction CH3NO2 + 3Cl2 ––→ CCl3NO2 + 3HCl How much nitromethane, CH3NO2 is needed to form 300 g of chloropicrin (A) 55 g (B) 111g (C) 222 g (D) None of these

(B) XeF4

(C) SF4

(D) I3−

Br

Which of the following can be formed. Br

(A)

+

(A) ⊕

+

N O–

– +

(B) O

(D)

+

+

N

Br

Which of the following methods yield saturated hydrocarbons -

–

O

O–

6. 2 mole each of SO3, CO, SO2 and CO2 is taken in one litre vessel. If Kc for SO3 + CO SO2 + CO2 is 1/9 then (A) total no. of moles at equilibrium are less than 8 (B) n(SO3) + n(CO2) = 4 (C) [n(SO2) / n(CO)] < 1 (D) Both (B) & (C)

(i ) BH

3 (A) RCH = CH2   →

(ii )CH3 COOH CH N

(B) R–CH=CH2 2 2 → ∆

Na / Ether

(C) Br–CH2(CH2)3–CH2–Br → ∆

7. Which of the following statements is correct for a solution saturated with AgCl and AgBr if their solubilities in moles per litre in separate solutions are x and y respectively ? (A) [Ag+] = x + y (B) [Ag+] = [Br–] + [Cl–] (D) [Br–] = y (D) [Cl–] > x

(D)

COONa

NaOH / CaO

→ ∆

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

8. The entropy change accompanying the heating of one mole of Helium gas, assuming ideal behaviour from a temperature of 300 K to a temperature of 1000 K at constant pressure. (A) 25.17 J K–1 mol–1 (B) 20 kJ K–1 mol–1 (C) 2.517 J K–1 mol–1 (D) 0.2517 J K–1 mol–1

Passage : I (No. 13 to 15) Photoelectric effect is the phenomenon in which the surface of alkali metals like potassium and cesium emit electrons when a beam of light with high frequency is made to fall on them. The ejected electrons are called photoelectrons Energy of photon = Work function + Maximum kinetic energy of ejected electrons. 1 ⇒ mv2max = h(ν – ν0) 2

Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

XtraEdge for IIT-JEE

Br

12.

O

N

(D)

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 recemic-2, 3-dideuterobutane (B) trans-but-2-ene and meso-2, 3-dideuterobutane (C) trans-but-2-ene and recemic-2, 3-dideuterobutane (D) cis-but-2-ene and meso-2, 3-dideuterobutane

O– +

(C) Br

O

O– –

(C)

+

N

(B)

11.

5. The least stable resonance structure is –

+ KOH (alc) —→

10.

4. In which of the following, the maximum number of lone pairs is present on the central atom ? (A) ClO −3

Pick up the correct statement(s) (A) Pb4+ salts are better oxidising agents (B) As5+ salts are oxidising agents (C) Tl3+ salts are oxidising agents (D) Ga+ salts are reducing agents

63

NOVEMBER 2009

13.

Lithium does not show photoelectric effect due to (A) small size and high charge density (B) high ionization energy (C) low ionization energy (D) None

14.

The K.E. of photoelectrons depend on (A) Wavelength of light (B) Frequency of light (C) Intensity of light (D) None of these

15.

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800Å. The threshold frequency (ν0) is(A) 2.92 × 10–19 s–1 (B) 4.41 × 1014 s–1 19 –1 (C) 7.18 × 10 s (D) 5.84 × 105 s–1

(C) over all pressures (D) over a specific temperature and specific pressure 18.

This section contains 2 questions (Questions 19 to 20). 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, T) 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, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T

Passage : II (No. 16 to 18) Real gases deviate from ideal behaviour because of the following two faulty assumptions of kinetic theory: (i) The actual volume occupied by molecules is negligible as compared to the total volume of the gas. (ii) The forces of attraction and repulsion between molecules of the gas are negligible The extent of deviation of a real gas from ideal behaviour is expressed in terms of compressibility factor (z). Hence, suitable corrections were applied to the ideal gas equation so that it can also explain the behaviour of real gases. The equation obtained by applying the two corrections to the usual gas equations is known as van der Waal equation 1. Volume correction Corrected (ideal) volume = (V – b) where b is the effective volume of the molecules. 2. Pressure correction (intermolecular attraction correction) Corrected (ideal pressure = P + p a However, p = 2 V ∴ The van der Waal's equation becomes 2    P + an  (V – nb) = nRT  V 2   On the basis of the above work-up answer the following questions : 16.

17.

19.

20.

0.5 value of compressibility factor (z) indicates that the gas : (A) shows positive deviation from the ideal gas (B) negative deviation from the ideal gas behaviour (C) either of the two (D) the factor is insufficient van der Waal equation is obeyed by the real gases: (A) over a wide range of temperature and pressure (B) over all temperatures

XtraEdge for IIT-JEE

At low pressure, the van der Waal's equation is a a (A) PV = RT+ (B) PV = RT – V V a a (C) PV = RT + 2 (D) PV = RT – 2 V V

Column-I (A) No. of ion in 1 mole K4[Fe(CN)6] (B) No. of atoms in Ca3(PO4) 2 in 0.2 mole of this compound (C) No. of electron in 0.5 mole H2O (D) No. of protons in 0.1 mole CH4 (NA = 6.0 × 1023)

Column-II (P) NA

Column-I (A) Ratio of energy of electron in 3rd orbit of H-atom and 4th orbit of Li2+ ion (B) Ratio of de-Broglie wavelengths of electron in 2nd orbit of H-atom to 3rd orbit of He+ ion (C) Ratio of 3rd and 4th separation energies for H-atom (D) Ratio of frequencies of revolution of electrons in 2nd orbit of H-atom and 3rd orbit of He+ ion

Column-II (P) 4 : 3

(Q) 5NA (R) 2.6 NA (S) 1.56 × 1024 (T) 0.3 × 1025

(Q) 25 : 16

(R) 27 : 32 (S) 4 : 9

(T) 2 : 3 64

NOVEMBER 2009

MATHEMATICS

7.

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola y2 = 4ax is another parabola with directrix (A) x = – a (B) x = – a/2 (C) x = 0 (D) x = a/2

8.

If PQ is a double ordinate of the hyperbola

Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

If z1, z2, z3 are three distinct complex numbers and a, b, c are three positive real numbers such that

x2

a2 b2 c2 + + is z 2 − z 3 z 3 − z1 z1 − z 2 (A) 0 (C) w2 2.

y2

= 1 such that OPQ is an equilateral a 2 b2 triangle, O being the centre of the hyperbola. Then the eccentricity e of the hyperbola, satisfies -

a b c = = , then value of | z 2 − z 3 | | z 3 − z1 | | z1 − z 2 |

(B) w (D) none of these

–

(A) 1 < e < 2 / 3

(B) e = 2 / 3

(C) e = 3 / 2

(D) e > 2 / 3

Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

If a ∈ R, and the equation (a – 2) (x – [x])2 + 2(x – [x]) + a2 = 0 (1) (where [x] denotes the greatest integer ≤ x) has no integral solution and has exactly one solution in (2, 3), then a lies in the interval (A) (–1, 2) (B) (0, 1) (C) (–1, 0) (D) (2, 3)

9.

Let α be a repeated root of p(x) = x3 + 3ax2 + 3bx + c = 0, then (A) α is a root of x2 + 2ax + b = 0

n

3.

Value of

(C k )(C k −1 ) is ∑ k =1

(A) 2nCn 2n

(C) Cn+2 4.

5.

6.

(B) α =

1 2n+2 ( Cn+1) – 2nCn 2 (D) none of these (B)

(C) α =

(B) 5 3 /3cm

(C) 2 3 /3cm

(D)

ab − c a2 −b

10. Let N denote the number of ways in which n boys can be arranged in a line so that 3 particular boys are separated. Then (A) 3!|N (B) (n – 2)!|N n–2 (C) C3|N (D) (n – 3)2 (n – 4)2|N

3 cm

x 1  The value of cos–1 x + cos–1  + 3 − 3x 2  2 2  (1/2 ≤ x ≤ 1) is equal to (A) π/6 (B) π/3 (C) π (D) 0

11. The equation 3 sin2 x+10 cos x – 6 = 0 is satisfied if(A) x = nπ + cos–1 (1/3) (B) x = nπ – cos–1 (1/3) (C) x = 2nπ + cos–1 (1/3)

Two rods of lengths a and b slide along the x-axis and y-axis respectively in such a manner that their ends are concyclic. The locus of the centre of the circle passing through the end points is (A) 4(x2 + y2) = a2 + b 2 (B) x2 + y2 = a 2 + b2 (C) 4(x2 – y2) = a 2 – b2 (D) x2 – y2 = a2 – b 2

XtraEdge for IIT-JEE

2(a 2 − b)

(D) α is a root of ax2 + 2bx + c = 0

In a triangle with one angle 2 π/3, the lengths of the sides form an A.P. If the length of the greatest side is 7 cm, the radius of the circumcircle of the triangle is (A) 7 3 /3cm

c − ab

(D) x = 2nπ – cos–1 (1/3) (n ∈ I) 12. The Cartesian equation of the curve whose parametric equation is x = 2t – 3 and y = 4t2 – 1 is given by (A) (x+3)2 – y – 1 = 0 (B) x2 + 6x – y + 8 = 0 (C) (y+1)2 + x + 3 = 0 (D) y2 + 6x – 2y + 4 = 0

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This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

This section contains 2 questions (Questions 19 to 20). 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, T) 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, A-T, B-Q, B-R, C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A P Q R S T B P Q R S T C P Q R S T D P Q R S T

Passage : I (No. 13 to 15) A and B are two points on the boundary of a circular field of radius R and centre O. ∠AOB = θ. A circle with centre A and radius AB meets the circular field again at C and the line AO produced at E.L., M are points on the boundary of the field lying between C and A, A and B, respectively. 13. AB is equal to (A) R sin θ (C) R cos θ

(B) 2R sin (θ/2) (D) 2R cos (θ/2)

14. Area of the segment AMB is equal to (A) (1/2) R2 θ (B) (1/2)R2 sin θ (C) (1/2) R2 (θ – sin θ) (D) none of these

19. Let a1, a2, a 3, ..... be a geometric progression such that 1 1 and log10(an) = n m for two fixed positive integer m and n, with m < n, then Column-I Column-II (A) a2m+n (P) 101/m–1/n (B) amn (Q) 10 (C) am+n (R) 102/n + 1/m (D) an–m (S) 101/n + 1/m log10(am) =

15. If the area AMBECL is 1/nth of the field, then sin θ + ( π – θ) cos θ is equal to (A) nπ (C) (n – 1)π

n −1 π n (D) (n + 1)π (B)

Passage : II (No. 16 to 18) A(x1, y1), B(x2, y2), C(x3, y3) are the vertices of a triangle ABC. lx + my + n = 0 is an equation of the line L.

20. Column-I Column-II (A) Equation of the polar (P) 8x + 2y – 23 = 0 of (–7, –9) with respect to the circle x2+y2 –12x–8y–48 = 0 (B) Equation of the (Q) 13x + 13y – 30 = 0 common chord of the circles x2 + y2 + 2x + 2y + 1 = 0 and x2+y2 +4x + 3y + 2 = 0 (C) Equation of the (R) 2x + y + 1 = 0 tangent at (–7, –9) to the circle x2 + y2 + 12x + 8y + 26 = 0 (D) Equation of the radical (S) x + 5y + 52 = 0 axis of the circles 2x2 + 2y2 + 4x + 4y + 9 = 0 and x2 + y2 + 6x+3y – 7 = 0

16. If L intersects the sides BC, CA and AB of the triangle ABC at P, Q, R respectively then BP CQ AR × × is equal to PC QA RB (A) – 1 (B) – 1/2 (C) 1/2 (D) 1 17. If the centroid of the triangle ABC is at the origin and algebraic sum of the lengths of the perpendiculars from the vertices of the triangle ABC on the line L is equal to 1 then sum of the squares of the intercepts made by L on the coordinate axes is equal to (A) 0 (B) 4 (C) 9 (D) 16 18. If P divides BC in the ratio 2 : 1 and Q divides CA in the ratio 1 : 3 then R divides AB in the ratio (A) 2 : 3 internally (B) 2 : 3 externally (C) 3 : 2 internally (D) 3 : 2 externally

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XtraEdge Test Series ANSWER KEY IIT- JEE 2010 (November issue) PHYSICS Ques Ans Ques Ans 19 20

1 B 11 A, B , C , D A → P,R A→ Q

2 D 12 A, C

3 B 13 B B → Q,R,S B→R

4 A 14 C

5 B 15 D C → Q,R C→P

6 A 16 B

7 B 17 C D → P,R D→S

8 B 18 B

9 A, C , D

10 A, B , C

7 B 17 A D → S,P D → S,T

8 D 18 B

9 A, C , D

10 A, B , D

7 C 17 D D→R D→P

8 D 18 D

9 A, B , C , D

10 A, B

7 B 17 C D→T D → Q,R

8 C 18 B

9 A, D

10 A

9 A, C , D

10 A, B

C H EM I STR Y Ques Ans Ques Ans 19 20

1 B 11 A, B , D A→ P A → P,R,S

2 B 12 A

3 B 13 B B→Q B → Q,S

4 B 14 B

5 C 15 C C → P,R C → P,S

6 C 16 A

MATHEMATICS Ques Ans Ques Ans 19 20

1 B 11 A, B , D A→ R A→ S

2 B 12 B,D

3 B 13 C B→P B→R

4 A 14 B

5 A 15 D C→S C→Q

6 D 16 C

IIT- JEE 2011 (November issue) PHYSICS Ques Ans Ques Ans 19 20

1 A 11 B,C A→ Q A→ P

2 B 12 A, C

3 D 13 B B→T B → Q,R,S

4 C 14 A

5 C 15 C C→S C → P,S

6 B 16 B

C H EM I STR Y Ques Ans Ques Ans 19 20

1 A 11 C A → Q,T A→ R

2 A 12 A, B , C , D

3 B 13 B B → R,S B→P

4 D 14 B

5 A 15 B C→Q C→Q

6 D 16 B

7 B 17 A D→P D→R

8 A 18 B

MATHEMATICS Ques Ans Ques Ans 19 20

1 A 11 C,D A→ R A→ Q

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2 C 12 A, B

3 B 13 B B→Q B→R

4 A 14 C

5 B 15 B C→S C→S

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6 C 16 A

7 C 17 C D→P D→P

8 D 18 D

9 A, B , D

10 A, B , C , D

NOVEMBER 2009

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